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2600	https://math.stackexchange.com/questions/449927/limit-of-recursive-sequence-a-n1-fraca-n1-a-n	calculus - Limit of recursive sequence $a_{n+1} = \frac{a_n}{1- {a_n}}$ - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Limit of recursive sequence a n+1=a n 1−{a n}a n+1=a n 1−{a n} Ask Question Asked 12 years, 2 months ago Modified5 years, 11 months ago Viewed 2k times This question shows research effort; it is useful and clear 34 Save this question. Show activity on this post. Consider the following sequence: let a 0>0 a 0>0 be rational. Define a n+1=a n 1−{a n},a n+1=a n 1−{a n}, where {a n}{a n} is the fractional part of a n a n (i.e. {a n}=a n−⌊a n⌋{a n}=a n−⌊a n⌋). Show that a n a n converges, and find its limit. We can show it converges as follows: suppose a n=p n/q n=k n+r n/q n a n=p n/q n=k n+r n/q n, where p n=k n q n+r n p n=k n q n+r n, 0≤r n<q n 0≤r n<q n. Then a n+1=p n/q n 1−r n/q n=p n q n−r n,a n+1=p n/q n 1−r n/q n=p n q n−r n, so the denominator will keep decreasing until it is a divisor of p 0 p 0 (maybe 1). Also, note we may take p n=p 0 p n=p 0 for all n n. Further, the limit will be ≤p 0 gcf(p 0,q 0)≤p 0 gcf⁡(p 0,q 0), because if f∣p 0 f∣p 0 and f∣q n f∣q n, then f∣(p 0−k n q n)=r n f∣(p 0−k n q n)=r n, so f∣q n−r n=q n+1 f∣q n−r n=q n+1. But the limit may be strictly smaller; for instance, a 0=30/7 a 0=30/7 converges right away to 6. Can we say anything else about the limit of a sequence starting with a 0 a 0? This was a problem on a qualifier, so I suspect there is more to the answer, but maybe not. calculus sequences-and-series elementary-number-theory recurrence-relations Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Aug 22, 2013 at 17:59 Eric AuldEric Auld asked Jul 23, 2013 at 1:21 Eric AuldEric Auld 29.1k 12 12 gold badges 86 86 silver badges 224 224 bronze badges 5 1 Let f(p,q)f(p,q) be the limit of the sequence when a 1=p/q a 1=p/q; then I believe that for fixed p p, the function g p(q)=f(p,q)g p(q)=f(p,q) is periodic with period p p, while for fixed q q, the function h q(p)=f(p,q)h q(p)=f(p,q) is periodic with period lcm[1,2,…,q][1,2,…,q].Greg Martin –Greg Martin 2013-07-23 08:25:43 +00:00 Commented Jul 23, 2013 at 8:25 Greg's belief is correct: for fixed p p, if q=p k+r q=p k+r with k≥0 k≥0 and 0<r<p 0<r<p then the a n a n sequence begins p/q,p/(q−p),p/(q−2 p),...,p/(q−k p)=p/r p/q,p/(q−p),p/(q−2 p),...,p/(q−k p)=p/r, so indeed the limit is determined by p p and r r. If q q is fixed, and p p and P P are congruent to one another mod lcm(1,2,...,q)(1,2,...,q), then p p and P P are congruent to one another modulo any integer between 1 1 and q q, so they're congruent mod q n q n for all n n. Thus the sequences of q n q n's and r n r n's for a 1=p/q a 1=p/q are identical to the corresponding sequences for a 1=P/q a 1=P/q.Michael Zieve –Michael Zieve 2013-08-20 18:55:41 +00:00 Commented Aug 20, 2013 at 18:55 Greg's functions g p g p seems to be surjective as a function on {1,…,p}{1,…,p} into the divisors of p p.Uwe Stroinski –Uwe Stroinski 2013-08-22 11:55:37 +00:00 Commented Aug 22, 2013 at 11:55 So where is the question here?Norbert –Norbert 2013-08-22 16:33:59 +00:00 Commented Aug 22, 2013 at 16:33 1 @Norbert. The second sentence from the end looks like a question to me (though a bit open-ended, I'll admit).Rick Decker –Rick Decker 2013-08-22 16:51:41 +00:00 Commented Aug 22, 2013 at 16:51 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 5 Save this answer. +250 This answer has been awarded bounties worth 250 reputation by Community Show activity on this post. Wanted to record some observations here... If we consider the sequence p=a 1 q+r 1 p=a 1 q+r 1 p=a 2(q−r 1)+r 2 p=a 2(q−r 1)+r 2 p=a 3(q−r 1−r 2)+r 3 p=a 3(q−r 1−r 2)+r 3 etc. and try to solve for the remainders in the form r i=c i p−d i q r i=c i p−d i q, there is a nice recursive relation: First, c 1=1 c 1=1 and d 1=a 1 d 1=a 1, and in general, c j=1+a j(c 1+…+c j−1)c j=1+a j(c 1+…+c j−1) and d j=(1+a 1)(1+a 2)⋯(1+a j−1)a j d j=(1+a 1)(1+a 2)⋯(1+a j−1)a j We can also write c j(1+a j)=c 1+…+c j c j(1+a j)=c 1+…+c j so that c j=1+a j a j−1(c j−1(1+a j−1)−1)c j=1+a j a j−1(c j−1(1+a j−1)−1). This can be expanded further to obtain the form c j=1+a j[1+(1+a j−1)[1+(1+a j−2)[1+…[1+(1+a 2)]]…]c j=1+a j[1+(1+a j−1)[1+(1+a j−2)[1+…[1+(1+a 2)]]…], or even c j=1+a j+a j(1+a j−1)+a j(1+a j−1)(1+a j−2)+…+a j(1+a j−1)(1+a j−2)⋯(1+a 2)c j=1+a j+a j(1+a j−1)+a j(1+a j−1)(1+a j−2)+…+a j(1+a j−1)(1+a j−2)⋯(1+a 2) Note that the a i a i are strictly increasing in the sequence. Suppose the procedure terminates at the n n-th step (when r n=0 r n=0). The limit is then p/a n p/a n, and 0=r n=c n(p/q)−d n 0=r n=c n(p/q)−d n or that p/q=d n/c n p/q=d n/c n. I still don't have a closed form expression, but for instance: For rationals p/q p/q for which the sequence terminates in the first step, 0=r 1=c 1(p/q)−d 1=p/q−a 1 0=r 1=c 1(p/q)−d 1=p/q−a 1, so that q q is a divisor of p p, and the limit is p/q p/q. For termination at the second step, 0=r 2=c 2(p/q)−d 2=(1+a 2)(p/q)−(1+a 1)a 2 0=r 2=c 2(p/q)−d 2=(1+a 2)(p/q)−(1+a 1)a 2, or that p q=(1+a 1)a 2 1+a 2 p q=(1+a 1)a 2 1+a 2. If there exists a 1<a 2 a 1<a 2 satisfying this equality, then the sequence terminates in the second step. One such criteria is if d d is a divisor of p p, q=d+1 q=d+1 and p/d−1<d p/d−1<d, then the sequence terminates in the second step to p/d=(1+a 1)p/d=(1+a 1). For examples: 30/7=(1+4)6/(1+6)30/7=(1+4)6/(1+6). Also, 30/4=120/16=(1+7)15/(1+15)30/4=120/16=(1+7)15/(1+15). Third step termination: p q=(1+a 1)(1+a 2)a 3 1+a 3(1+(1+a 2))p q=(1+a 1)(1+a 2)a 3 1+a 3(1+(1+a 2)), limit is (1+a 1)(1+a 2)(1+a 1)(1+a 2), and etc. Also, it appears that if you plug in a n=d a n=d in any formula, you can generate for which p p the process converges to d d by substituting any a 1<a 2<…<a n−1<d a 1<a 2<…<a n−1<d. Maybe there's more special structure... Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Aug 26, 2013 at 19:02 answered Aug 24, 2013 at 22:19 EvanEvan 3,911 15 15 silver badges 16 16 bronze badges Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Let n∈N n∈N, q i q i for 0≤i≤n 0≤i≤n be a non-decreasing sequence of positive integers and a 0=(∑j=0 n 1 q 0⋯q j)−1 a i+1=a i 1−{a i}a 0=(∑j=0 n 1 q 0⋯q j)−1 a i+1=a i 1−{a i} Then a i=q n a i=q n for every i≥n i≥n. This follows by proving, by induction, that for every i≤n i≤n we have a i=(∑j=i n 1 q i⋯q j)−1(1)(1)a i=(∑j=i n 1 q i⋯q j)−1 This is clearly true for i=0 i=0. Assuming (1)(1), then clearly a i≤q i a i≤q i. On the other hand, j≥i j≥i implies q j≥q i>1 q j≥q i>1 (the case q i=1 q i=1 is trivial), hence 1 a i≤∑j=i n 1 q j+1 i<1 q i−1 1 a i≤∑j=i n 1 q i j+1<1 q i−1 from which q i−1<a i≤q i q i−1<a i≤q i that's ⌈a i⌉=q i⌈a i⌉=q i. Consequently, a i+1=a i 1−{a i}=a i q i−a i=(∑j=i+1 n 1 q i+1⋯q j)−1 a i+1=a i 1−{a i}=a i q i−a i=(∑j=i+1 n 1 q i+1⋯q j)−1 hence the assertion follows by induction on i i. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Oct 31, 2019 at 21:57 answered Oct 31, 2019 at 11:49 Fabio LucchiniFabio Lucchini 16.6k 1 1 gold badge 32 32 silver badges 43 43 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Your proof is correct if a0 is rational number. It will not converge for irrational numbers like e or pi. Assuming rationality, I don't think there is a closed form solution of converging number (until and unless you embed loop and conditional kind of behavior in closed form). There is equivalence class of converging point given q0 where solution will converge: the equivalence class are factors (I mean all factors not only prime factors) of p0 including itself. Therefore, if p0 is prime it will converge to itself. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Aug 22, 2013 at 16:49 vinash85vinash85 161 4 4 bronze badges 3 It is stated at the beginning of the question that a 1 a 1 is rational (I guess this was changed to a 0 a 0 later).Jonas Meyer –Jonas Meyer 2013-08-22 17:44:45 +00:00 Commented Aug 22, 2013 at 17:44 @JonasMeyer Thanks for pointing out the mistake in the question...changed the a 1 a 1 to a 0 a 0.Eric Auld –Eric Auld 2013-08-22 18:00:17 +00:00 Commented Aug 22, 2013 at 18:00 how to show this is not convergent for irrational starting point? i have shown this seq is increasing so it must not be bounded above.CHOUDHARY bhim sen –CHOUDHARY bhim sen 2019-02-25 12:54:40 +00:00 Commented Feb 25, 2019 at 12:54 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. 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2601	https://pubmed.ncbi.nlm.nih.gov/7925671/	Parvovirus B19 outbreak on an adult ward - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Parvovirus B19 outbreak on an adult ward C Seng1,P Watkins,D Morse,S P Barrett,M Zambon,N Andrews,M Atkins,S Hall,Y K Lau,B J Cohen Affiliations Expand Affiliation 1 PHLS Communicable Disease Surveillance Centre, London. PMID: 7925671 PMCID: PMC2271527 DOI: 10.1017/s0950268800051773 Item in Clipboard Parvovirus B19 outbreak on an adult ward C Seng et al. Epidemiol Infect.1994 Oct. Show details Display options Display options Format Epidemiol Infect Actions Search in PubMed Search in NLM Catalog Add to Search . 1994 Oct;113(2):345-53. doi: 10.1017/s0950268800051773. Authors C Seng1,P Watkins,D Morse,S P Barrett,M Zambon,N Andrews,M Atkins,S Hall,Y K Lau,B J Cohen Affiliation 1 PHLS Communicable Disease Surveillance Centre, London. PMID: 7925671 PMCID: PMC2271527 DOI: 10.1017/s0950268800051773 Item in Clipboard Full text links Cite Display options Display options Format Abstract In November and December 1992, an outbreak of parvovirus B19 infection occurred among patients and staff on an adult mixed surgical ward at a large hospital in London. Three patients and 15 staff members were serologically confirmed as acute cases. The attack rate among susceptible members of staff was 47%. In those infected, arthralgia (80%) and rash (67%) were the most common symptoms. Of six susceptible in-patients on the ward, three became infected. One of the in-patients who had carcinoma of the mouth was viraemic for more than 10 days with marrow suppression resulting in the postponement of chemotherapy until intravenous immunoglobulin was given and he was no longer viraemic. Control measures taken included closure of the ward to new admissions, transfer of only immune staff to the ward, and restriction of the ward nursing staff to working only on that ward. Although no specific exposure was conclusively identified as a risk factor, there was a suggestion of an increased risk of acquiring parvovirus B19 infection among those staff who did not adopt strict hand washing procedures after each physical contact with a patient (RR = 2.33; P = 0.07). Knowledge of parvovirus B19 among interviewed health care workers was poor: only 42% reported knowing about parvovirus B19 and only 38% could name a patient category at risk of a severe outcome following infection. This is the first report of a nosocomial outbreak affecting an adult ward and of possible transmission of parvovirus B19 infection from staff to in-patients. Hospital control of infection teams should include parvovirus B19 in their outbreak containment plans. PubMed Disclaimer Similar articles Human parvovirus B19 nosocomial outbreak in healthcare personnel in a paediatric ward at a national tertiary referral centre in Thailand.Sungkate S, Phongsamart W, Rungmaitree S, Lapphra K, Wittawatmongkol O, Pumsuwan V, Wiruchkul N, Assanasen S, Rongrungruang Y, Onlamoon N, Horthongkham N, Lermankul W, Kongstan N, Chokephaibulkit K.Sungkate S, et al.J Hosp Infect. 2017 Jun;96(2):163-167. doi: 10.1016/j.jhin.2017.03.014. Epub 2017 Mar 18.J Hosp Infect. 2017.PMID: 28412176 Parvovirus B19 outbreak in a children's ward.Pillay D, Patou G, Hurt S, Kibbler CC, Griffiths PD.Pillay D, et al.Lancet. 1992 Jan 11;339(8785):107-9. doi: 10.1016/0140-6736(92)91009-w.Lancet. 1992.PMID: 1345828 Parvovirus B19 infection in hospital workers: community or hospital acquisition?Dowell SF, Török TJ, Thorp JA, Hedrick J, Erdman DD, Zaki SR, Hinkle CJ, Bayer WL, Anderson LJ.Dowell SF, et al.J Infect Dis. 1995 Oct;172(4):1076-9. doi: 10.1093/infdis/172.4.1076.J Infect Dis. 1995.PMID: 7561182 Are Daycare Workers at a Higher Risk of Parvovirus B19 Infection? A Systematic Review and Meta-Analysis.Romero Starke K, Kofahl M, Freiberg A, Schubert M, Groß ML, Schmauder S, Hegewald J, Kämpf D, Stranzinger J, Nienhaus A, Seidler A.Romero Starke K, et al.Int J Environ Res Public Health. 2019 Apr 17;16(8):1392. doi: 10.3390/ijerph16081392.Int J Environ Res Public Health. 2019.PMID: 30999694 Free PMC article. Parvovirus B19 in pregnancy.Goff M.Goff M.J Midwifery Womens Health. 2005 Nov-Dec;50(6):536-8. doi: 10.1016/j.jmwh.2005.06.008.J Midwifery Womens Health. 2005.PMID: 16260369 Review. See all similar articles Cited by Structural and functional studies of the main replication protein NS1 of human parvovirus B19.Zhang Y, Fan B, Gao Y, Yang J, Zhang W, Su S, Li L, Li H, Luo Z, Tang G, Wang C, Zhang X, Liu H, Gan J.Zhang Y, et al.Nucleic Acids Res. 2025 Jun 20;53(12):gkaf562. doi: 10.1093/nar/gkaf562.Nucleic Acids Res. 2025.PMID: 40568941 Free PMC article. Structures and implications of the nuclease domain of human parvovirus B19 NS1 protein.Zhang Y, Shao Z, Gao Y, Fan B, Yang J, Chen X, Zhao X, Shao Q, Zhang W, Cao C, Liu H, Gan J.Zhang Y, et al.Comput Struct Biotechnol J. 2022 Aug 27;20:4645-4655. doi: 10.1016/j.csbj.2022.08.047. eCollection 2022.Comput Struct Biotechnol J. 2022.PMID: 36090819 Free PMC article. First report on severe septic shock associated with human Parvovirus B19 infection after cardiac surgery.Xiang C, Wu X, Wei Y, Li T, Tang X, Wang Y, Zhang X, Huang X, Wang Y.Xiang C, et al.Front Cell Infect Microbiol. 2023 Apr 5;13:1064760. doi: 10.3389/fcimb.2023.1064760. eCollection 2023.Front Cell Infect Microbiol. 2023.PMID: 37091672 Free PMC article. Use of ward closure to control outbreaks among hospitalized patients in acute care settings: a systematic review.Wong H, Eso K, Ip A, Jones J, Kwon Y, Powelson S, de Grood J, Geransar R, Santana M, Joffe AM, Taylor G, Missaghi B, Pearce C, Ghali WA, Conly J.Wong H, et al.Syst Rev. 2015 Nov 7;4:152. doi: 10.1186/s13643-015-0131-2.Syst Rev. 2015.PMID: 26546048 Free PMC article. Adult acute respiratory distress syndrome due to human parvovirus B19 infection after cardiac surgery: a case report.Ma M, Ma X, Jia M, Hou X, Wang H.Ma M, et al.BMC Infect Dis. 2022 Mar 7;22(1):231. doi: 10.1186/s12879-022-07213-9.BMC Infect Dis. 2022.PMID: 35255838 Free PMC article. See all "Cited by" articles References J Infect Dis. 1980 Jan;141(1):98-102 - PubMed Infect Control Hosp Epidemiol. 1992 Jun;13(6):343-8 - PubMed J Pediatr. 1981 Jul;99(1):100-3 - PubMed Lancet. 1981 Sep 19;2(8247):595-7 - PubMed J Hyg (Lond). 1983 Aug;91(1):113-30 - PubMed Show all 21 references MeSH terms Adult Actions Search in PubMed Search in MeSH Add to Search Aged Actions Search in PubMed Search in MeSH Add to Search Aged, 80 and over Actions Search in PubMed Search in MeSH Add to Search Cross Infection / diagnosis Actions Search in PubMed Search in MeSH Add to Search Cross Infection / epidemiology Actions Search in PubMed Search in MeSH Add to Search Cross Infection / transmission Actions Search in PubMed Search in MeSH Add to Search Disease Outbreaks Actions Search in PubMed Search in MeSH Add to Search Erythema Infectiosum / diagnosis Actions Search in PubMed Search in MeSH Add to Search Erythema Infectiosum / epidemiology Actions Search in PubMed Search in MeSH Add to Search Erythema Infectiosum / transmission Actions Search in PubMed Search in MeSH Add to Search Female Actions Search in PubMed Search in MeSH Add to Search Hand Disinfection Actions Search in PubMed Search in MeSH Add to Search Humans Actions Search in PubMed Search in MeSH Add to Search Infectious Disease Transmission, Professional-to-Patient Actions Search in PubMed Search in MeSH Add to Search London / epidemiology Actions Search in PubMed Search in MeSH Add to Search Male Actions Search in PubMed Search in MeSH Add to Search Middle Aged Actions Search in PubMed Search in MeSH Add to Search Parvovirus B19, Human Actions Search in PubMed Search in MeSH Add to Search Personnel, Hospital Actions Search in PubMed Search in MeSH Add to Search Risk Factors Actions Search in PubMed Search in MeSH Add to Search Related information MedGen LinkOut - more resources Full Text Sources PubMed Central Medical MedlinePlus Health Information Full text links[x] Free PMC article [x] Cite Copy Download .nbib.nbib Format: Send To Clipboard Email Save My Bibliography Collections Citation Manager [x] NCBI Literature Resources MeSHPMCBookshelfDisclaimer The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). 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2602	https://www.convertunits.com/from/cubic+angstrom/to/meters%5E3	Convert cubic angstrom to meters^3 - Conversion of Measurement Units Convert cubic angstrom to cubic metre Please enable Javascript to use the unit converter. Note you can turn off most ads here: \| \| \| --- \| \| \| cubic angstrom \| \| \| meters^3 \| \| \| More information from the unit converter How many cubic angstrom in 1 meters^3? The answer is 1.0E+30. We assume you are converting between cubic angstrom and cubic metre. You can view more details on each measurement unit: cubic angstrom or meters^3 The SI derived unit for volume is the cubic meter. 1 cubic angstrom is equal to 1.0E-30 cubic meter. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between cubic angstroms and cubic meters. Type in your own numbers in the form to convert the units! Want other units? You can do the reverse unit conversion from meters^3 to cubic angstrom, or enter any two units below: Enter two units to convert \| \| \| --- \| \| From: \| \| \| To: \| \| \| \| \| Common volume conversions Definition: Cubic angstrom A cubic angstrom is defined as the volume of a cube with edges one angstrom in length. The definition of a angstrom is as follows: An angstrom or ångström (Å) is a non-SI unit of length equal to 10-10 metres, 0.1 nanometres or 100 picometres. Definition: Cubic meter The cubic metre (symbol m³) is the SI derived unit of volume. It is the volume of a cube with edges one metre in length. Older equivalents were the stere and the kilolitre. Metric conversions and more ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 70 kg, 150 lbs, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!
2603	https://math.stackexchange.com/questions/580545/suppose-a-nb-n-converges-does-a-nb-n-converges-also	calculus - Suppose $a_n+b_n$ converges. Does $a_nb_n$ converges also? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Suppose a n+b n a n+b n converges. Does a n∗b n a n∗b n converges also? Ask Question Asked 11 years, 10 months ago Modified11 years, 10 months ago Viewed 635 times This question shows research effort; it is useful and clear 5 Save this question. Show activity on this post. a n,b n a n,b n - sequences Suppose a n+b n a n+b n converges. Does a n b n a n b n converge also? I tried thinking if I can learn something about a n a n and b n b n by the assumption a n b n a n b n converges. I also tried to develop this equation \|a n b n−L\|<ϵ\|a n b n−L\|<ϵ assuming it is converging. I didn't get any bright conclusions. Will be glad help. calculus limits Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Nov 25, 2013 at 19:24 lily 4,511 4 4 gold badges 19 19 silver badges 26 26 bronze badges asked Nov 25, 2013 at 13:27 captain dragoncaptain dragon 769 1 1 gold badge 5 5 silver badges 16 16 bronze badges 1 What does the first "sentence" mean? Certainly those are sequences, not sets.Marc van Leeuwen –Marc van Leeuwen 2013-11-25 13:51:41 +00:00 Commented Nov 25, 2013 at 13:51 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 21 Save this answer. Show activity on this post. a n=n,b n=−n a n=n,b n=−n a n+b n=0 a n+b n=0 but then... a n b n=−n 2→−∞a n b n=−n 2→−∞ Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Nov 25, 2013 at 13:39 Clayton 25.1k 7 7 gold badges 60 60 silver badges 116 116 bronze badges answered Nov 25, 2013 at 13:32 user87543 user87543 15 1 When we are talking about convergence, I think it needs to converges to a real number. A sequences that tends to infinity does not converges Giiovanna –Giiovanna 2013-11-25 13:34:44 +00:00 Commented Nov 25, 2013 at 13:34 1 Yes, indeed. This only shows an example when the product does not converges. But it is really easy to show one that the product does converges. Just take 2 constant sequences. The sum is a constant and so does the product. Then, as we can see, the product can or not converge Giiovanna –Giiovanna 2013-11-25 13:39:13 +00:00 Commented Nov 25, 2013 at 13:39 3 Could you please let me know what are you expecting when you say "general way"user87543 –user87543 2013-11-25 14:04:31 +00:00 Commented Nov 25, 2013 at 14:04 3 @Giiovanna: The question is if a n+b n a n+b n, does a n b n a n b n also converge? Certainly, anyone can come up with cases where both converge. However, to negate a false implication, one needs to satisfy the hypotheses (in this case, a n+b n a n+b n converges), yet show that the conclusion is not necessarily true (in this case, a n b n a n b n does not converge). Finding a pair of sequences for which a n b n a n b n converges does not apply to the question.robjohn –robjohn♦ 2013-11-25 15:34:23 +00:00 Commented Nov 25, 2013 at 15:34 1 @Giiovanna If you were asked to prove that 2x is not equal to x^2 how would you do that? Then what would you say to the person that says well it's true for x=2!Cruncher –Cruncher 2013-11-25 18:01:03 +00:00 Commented Nov 25, 2013 at 18:01 \|Show 10 more comments This answer is useful 15 Save this answer. Show activity on this post. As shown in Praphulla Koushik's answer a n=n,b n=−n a n=n,b n=−n cancellation is a problem. However, even if you restrict a n,b n≥0 a n,b n≥0 then the answer is no. For example, if a n=2+(−1)n,b n=3−(−1)n a n=2+(−1)n,b n=3−(−1)n then a n+b n=5 a n+b n=5 yet a n b n a n b n oscillates between 4 4 and 6 6. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 25, 2013 at 13:47 robjohn♦robjohn 355k 38 38 gold badges 499 499 silver badges 892 892 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions calculus limits See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 2Limit of a sequence: a n≤b n≤c n a n≤b n≤c n 1Prove that 0≤a n,b n≤1 0≤a n,b n≤1 converge to 1 1 given lim n→∞a n b n=1 lim n→∞a n b n=1 1∑a n∑a n converges but a n=b n−c n a n=b n−c n for suitable (b n),(c n)(b n),(c n) is impossible 0lim n→∞a n b n=∞lim n→∞a n b n=∞, ∑a n∑a n converges, does ∑b n∑b n converge? 3(a n)∞n=1(a n)n=1∞&(b n)∞n=1(b n)n=1∞ are seq st (a n)∞n=1(a n)n=1∞&[(a n)∞n=1+(b n)∞n=1][(a n)n=1∞+(b n)n=1∞] con. Prove (b n)(b n) con 7Does there exists a sequence b n b n, s.t. lim b n=0 lim b n=0 and for every divergent series ∑a n∑a n, The series ∑a n b n∑a n b n also diverges? 0Suppose a n b n a n b n converges to a limit S. Show that ∑N n=0 a n∑N n=0 b n∑n=0 N a n∑n=0 N b n converges to that same limit S 2If two sequences {a n}{a n} and {b n}{b n} be such that a n>b n a n>b n then show that lim a n≥lim b n lim a n≥lim b n Hot Network Questions Xubuntu 24.04 - Libreoffice Numbers Interpreted in Smallest Valid Base Direct train from Rotterdam to Lille Europe How exactly are random assignments of cases to US Federal Judges implemented? Who ensures randomness? Are there laws regulating how it should be done? In the U.S., can patients receive treatment at a hospital without being logged? "Unexpected"-type comic story. Aboard a space ark/colony ship. Everyone's a vampire/werewolf Should I let a player go because of their inability to handle setbacks? Is it ok to place components "inside" the PCB How do you emphasize the verb "to be" with do/does? Why does LaTeX convert inline Python code (range(N-2)) into -NoValue-? 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2604	https://cs.brown.edu/courses/csci1951-w/lec/lec%203%20notes.pdf	CSCI 1951-W Sublinear Algorithms for Big Data Fall 2020 Lecture 3: Concentration Inequalities and Mean Estimation Lecturer: Jasper Lee Scribe: Ross Briden 1 Overview In this lecture, we’ll quickly recap how an optimization algorithm that fails with constant probability can be adapted into a high probability one. Then, we’ll introduce two concentra-tion inequalities – Hoeﬀding’s inequality and Bernstein’s inequality – for analyzing the sample complexity of the sample mean for estimating the mean of an underlying distri-bution. We also state (without proof) a sample complexity lower bound that demonstrates the sample mean of independent random variables, in non-asymptotic regimes, is a poor estimator for their true mean, and introduce a more robust estimator known as the Median of Means algorithm. 2 Recap from Last Class Suppose we have an optimization problem speciﬁed by an objective function f : S →R subject to a set of constraints {C1, . . . , Ck}, where S is any set and Ci ⊆S for all constraints. Deﬁnition 3.1 We say that y ∈S is a feasible solution to our optimization problem if y ∈Tk i=1 Ci. The objective of the optimization problem is to ﬁnd a feasible solution x that either maxi-mize or minimize f(x). WLOG suppose we want to maximize f. Now, suppose there exists an algorithm D that, with probability ≥2 3, returns an answer z so that f(z) ≥OPT−ϵ with a promise that z ∈Tk i=1 Ci, where OPT is the global maximum of f. Notice that the promise ensures that the algorithm returns a value that is feasible if it succeeds. Now, we can design a new algorithm D′ that uses D as a subroutine to solve the opti-mization problem with high probability. In particular, given δ ∈(0, 1), D′ should run D for n = Θ(log( 1 δ)) iterations and return the largest feasible value returned by D. To see why D′ is correct, ﬁrst notice that the promise ensures that every time D succeeds, the solution returned by D is feasible. Therefore, the maximum of feasible solutions is also feasible and will also be greater than OPT−ϵ by deﬁnition. Hence, D′ returns a valid solution if after n runs of D, at least one valid solution is returned by D. It then follows that the probability D′ fails is (1 −2 3)n = 1 3n . Therefore, 1 3n ≤δ ⇐ ⇒−n log(3) ≤log(δ) ⇐ ⇒n ≥log 1 δ / log(3) So, if D′ runs D for Θ( 1 δ) iterations, D′ will return a correct solution with probability at least 1 −δ, as desired. 1 This example, along with the examples we saw last lecture, demonstrates that many con-stant probability algorithms can be converted to high probability algorithms relatively eas-ily. So, before designing a high probability algorithm, you should always consider whether a constant probability algorithm would be suﬃcient for your problem. Moreover, we can generalize to the following this heuristic: Heuristic 3.2 When designing high probability algorithms, the sample / query complexity q(δ) of your algorithm should be at least as good as q(δ) = O(q( 1 3) log( 1 δ)). Otherwise, a constant probability algorithm combined with a boosting technique (e.g. ﬁnding the median or max of solutions returned by a constant probability algorithm) will improve the δ dependence to the multiplicative log( 1 δ) factor. Moreover, note that the log( 1 δ) term is not tight for all problems; we can sometimes do better than this! 3 Concentration Inequalities for Mean Estimation First, we’ll derive two concentration inequalities – Hoeﬀding’s inequality and Bernstein’s inequality – and use them to evaluate the sample complexity of the sample mean when estimating the true mean of a collection of i.i.d. random variables. Remark 3.3 For the theorems below, we’ll be assuming that all random variables are one dimensional. Lemma 3.4 (Hoeﬀding’s Lemma) Let X be a real-valued random variable so that P(X ∈[a, b]) = 1 for some a < b and E[X] = 0. Then, E[etX] ≤e t2(b−a)2 8 . Remark 3.5 The proof of Hoeﬀding’s lemma is complex and involves some calculus, so it was not be covered in class. Nevertheless, the intuition is to notice x →ext is convex and thus Jensen’s inequality can be applied. Theorem 3.6 (Hoeﬀding’s Inequality) Let {Xi}n i=1 be independent random variables so that P(Xi ∈[a, b]) = 1 for some a < b, and let ϵ > 0. Then: 1. P(Xn −E[Xn] ≥ϵ) ≤e −2nϵ2 (b−a)2 2. P(Xn −E[Xn] ≤−ϵ) ≤e −2nϵ2 (b−a)2 3. P(\|Xn −E[Xn]\| ≥ϵ) ≤2e −2nϵ2 (b−a)2 Proof. We will ﬁrst show (1) and use it to easily prove (2) and (3). Now, P(Xm −E[Xn] ≥ϵ) ≤inf t>0 E[et(Xn−E[Xn])] etϵ = inf t>0 E[e t n (Pn i=1 Xi−E[Xi])] etϵ = inf t>0 Qn i=1 E[e t n (Xi−E[Xi])] etϵ 2 by Lemma (2.8) and since X1, . . . , Xn are independent. By applying Hoeﬀding’s lemma to this inequality, we have that inf t>0 Qn i=1 E[e t n (Xi−E[Xi])] etϵ ≤inf t>0 Qn i=1 e (t2/n2)(b−a)2 8 etϵ = inf t>0 e (nt2/n2)(b−a)2 8 etϵ = inf t>0 e (t2/n)(b−a)2 8 −tϵ Recall that ex is an increasing function, so the inﬁmum of e (t2/n)(b−a)2 8 −tϵ will be determined by the minimum value of (t2/n)(b−a)2 8 −tϵ. Furthermore, this is a quadratic function, which implies that the minimum of (t2/n)(b−a)2 8 −tϵ will be −y 2x where x = (b−a)2 8n and y = ϵ. There-fore, the minimum occurs at t = 4nϵ (b−a)2 . Moreover, since t > 0, we can bound the inﬁmum by t = 4nϵ (b−a)2 . Now, it follows that inf t>0 e (t2/n)(b−a)2 8 −tϵ = e −2nϵ2 (b−a)2 Hence, inequality (1) follows. By negating Xn and translating, we get inequality (2). Finally, by combining inequalities (1) and (2), we have that P(\|Xn −E[Xn]\| ≥ϵ) = P(Xn −E[Xn] ≥ϵ) + P(Xn −E[Xn] ≤−ϵ) ≤2e −2nϵ2 (b−a)2 which proves inequality (3). Corollary 3.7 Suppose that {Xi}n i=1 are i.i.d. random variables so that P(Xi ∈[a, b]) = 1, with a < b, holds for all Xi. Then, given ϵ > 0, the sample complexity of the sample mean required to estimate E[X] to an additive error ϵ while only failing with probability at most δ is n = O( σ2 ϵ2 + M ϵ ) log( 1 δ). Proof. Let ϵ > 0. Now, the probability of failure for this problem is P(\|Xn −E[Xn]\| ≥ϵ). Furthermore, by Hoeﬀding’s inequality, we have that P(\|Xn −E[Xn]\| ≥ϵ) ≤2e −2nϵ2 (b−a)2 . Hence, it follows that 2e −2nϵ2 (b−a)2 ≤δ ⇐ ⇒n ≥(b −a)2 2ϵ2 log 2 δ Thus, we should select n = O (b−a)2 ϵ2 log 1 δ to estimate the mean of {Xi}n i=1 to an additive error ϵ. Is this a good sample complexity? In other words, can we exploit any more information about {Xi}n i=1 to reduce the sample complexity of this problem? Yes, in many cases we can further reduce the sample complexity. To see why this is intuitively true, suppose you have a collection of i.i.d. random variables X1, . . . , Xn that are drawn from a distribution that has high probability mass in the interval [0, 1] and zero mass outside of this interval, 3 Figure 1: How outliers can hurt sample complexity when using Hoeﬀding’s inequality. In this case, the interval [a, b] must be large enough to contain the points located in the green probability mass, making [a, b] quite large and thus increasing the sample complexity. In this case, M represents the distance from the mean of the distribution to the center of the outlier probability mass. with the exception of one small interval [α, β] with non-zero probability mass and 1 ≪α. Then, since P(Xi ∈[α, β]) > 0 for all Xi, b is forced to be greater than or equal to β (See Figure 1 for a visual example of this). However, if the mass in [α, β] is suﬃciently small, then n samples will not see anything outside [0, 1], making Hoeﬀding’s inequality very loose. The variance is a much more robust way to measure the ”width” of a distribution than the interval covering all its probability mass. Hence, the robustness of variance motivates our next inequality: Bernstein’s Inequal-ity, which uses variance information to achieve a tighter bound. Theorem 3.8 (Bernstein’s Inequality) Let {Xi}n i=1 be independent random variables so that P(\|Xi−E[Xi]\| ≤M) = 1 holds for all Xi for some M ≥0. Also, let σ2 = 1 n Pn i=1 Var[Xi]. Then, P(Xn −E[Xn]) ≤exp −nϵ2 2σ2 + 2Mϵ 3 Remark 3.9 The proof of this is quite complicated and was not covered in lecture. Corollary 3.10 Suppose that {Xi}n i=1 are i.i.d. random variables so that P(\|Xi −E[Xi]\| ≤ M) = 1 holds for all Xi for some M ≥0. Then, given ϵ > 0, the sample complexity of the sample complexity required to estimate E[X] to an additive error ϵ while only failing with probability at most δ is n = O( σ2 ϵ2 + M ϵ ) log( 1 δ). Proof. Fix ϵ > 0. Then, the probability of failure is P(Xn −E[Xn] ≥ϵ). By Bernstein’s inequality, we have that P(Xn −E[Xn] ≥ϵ) ≤exp −nϵ2 2σ2+ 2Mϵ 3 . Therefore, it follows that exp −nϵ2 2σ2 + 2Mϵ 3 ≤δ ⇐ ⇒ nϵ2 2σ2 + 2Mϵ 3 ≥log 1 δ ⇐ ⇒n ≥ 2σ2 ϵ2 + 2 3 M ϵ log 1 δ Therefore, n = O ( σ2 ϵ2 + M ϵ ) log( 1 δ) , as desired. 4 So, making essentially the same assumption that the random variables are bounded in an interval of width O(M) = O(b −a), we have two possible sample complexities for the sample mean to choose from: n = O (b−a)2 ϵ2 log 1 δ (derived from Hoeﬀding) and n = O ( σ2 ϵ2 + M ϵ ) log( 1 δ) (derived from Bernstein) when estimating the mean of independent random variables. It’s therefore natural to ask if and when the second sample complexity is better than the ﬁrst. The fundamental insight is that when the standard deviation σ of the random variables is signiﬁcantly smaller than M, in other words σ ≪M, the sample complexity provided by Bernstein’s inequality is tighter since the bound provided by Hoeﬀding’s inequality grows quadratically with M. This is intuitive because factoring in information about variance will allow us to better approximate the underlying behavior of the random variables, requiring fewer samples to approximate their mean. In the regime where σ ≈M, the two sample complexities are asymptotically the same, so Bernstein’s inequality will provide minimal beneﬁt over Hoeﬀding’s inequality. Nevertheless, Bernstein’s inequality provides a sample complexity at least as good as that given by Hoeﬀding’s inequality because σ2 ≤O(M2) always holds. 4 Comparison with CLT; Median of Means Method In statistics, the sample mean of a collection of independent random variables is often used to estimate their true mean. Moreover, the central limit theorem states that the sample mean Xn of a collection of i.i.d. random variables X1, . . . , Xn exhibits Gaussian-like behavior 1 as n →∞. Yet, as we’ll see, in the non-asymptotic regime, when estimating E[Xn] with high probability, Xn does not behave necessarily like a Gaussian – no matter what concentration inequality you use to bound the sample mean. The upside, however, is that in the constant probability regime (i.e. when the probability of error is ﬁxed), Xn does provide Gaussian-like guarantees on sample complexity! Using this, we will introduce the Median of Means algorithm which estimates E[Xn] with high probability while also giving Gaussian-like performance. To start our analysis, recall that given i.i.d. Gaussian random variables X1, . . . , Xn ∼ N(µ, σ2), we have that P(\|Xn −E[Xn]\| ≥ϵ) ≈exp −nϵ2 2σ2 Therefore, exp −nϵ2 2σ2 ≤δ ⇐ ⇒n ≥2σ2 ϵ2 log 1 δ So, the sample complexity of the sample mean is n ≈2σ2 ϵ2 log 1 δ when X1, . . . , Xn are Gaussian. Given this, we can show that the sample complexity derived from Bernstein’s inequality is not Gaussian in all regimes: Proposition 3.11 n = O(( σ2 ϵ2 + M ϵ ) log( 1 δ)) is worse than n = O( σ2 ϵ2 log 1 δ ) and thus the sample complexity of the sample mean given by Bernstein’s inequality does not give Gaussian-behavior in all regimes. Proof. There are three distinct cases: 1A similar statement holds for independent but not identical random variables, requiring additional constraints on the moments of these random variables. See Lyapunov’s CLT for more information. 5 1. Case 1: ϵ ≫σ2 M . Since σ2 ϵ2 has an inverse square dependence on ϵ, we have that M ϵ ≫σ2 ϵ2 . Therefore, O ( σ2 ϵ2 + M ϵ ) log( 1 δ) provides a worse sample complexity than O σ2 ϵ2 log 1 δ in this case. Note that this is the case that concretely shows that Bernstein’s inequality does not give a Gaussian-like sample complexity for the sample mean. 2. Case 2: ϵ = Θ( σ2 M ). Then, M = Θ( σ2 ϵ ). So then follows that M ≈ σ2 ϵ . So by substituting this value of M into O σ2 ϵ2 + M ϵ , we have that O σ2 ϵ2 + M ϵ ≈O σ2 ϵ2 Therefore, Bernstein’s inequality gives roughly a Gaussian sample complexity for the sample mean in this case. 3. Case 3: ϵ ≪σ2 M . Since σ2 ϵ2 has an inverse square dependence on ϵ, we have that M ϵ ≪σ2 ϵ2 . Therefore, since σ2 ϵ2 dominates M ϵ , Bernstein’s inequality gives the same sample complexity for the sample mean as the Gaussian case. Since Bernstein’s inequality does not give Gaussian-like performance in some regimes, it’s pertinent to consider whether any concentration inequality can guarantee a Gaussian-like sample complexity for the sample mean. Catoni 2012 provides a lower bound on the sample complexity for the sample mean, which answers this question negatively: Theorem 3.12 (Catoni 2012)2 Given a collection X1, . . . , Xn of independent random variables, the sample mean Xn needs Ω( σ2 ϵ2δ) samples, assuming that the second moment of Xn is ﬁnite. Given this result, we know that there exists a bad distribution D so that the sample mean requires at least Ω( σ2 ϵ2δ) samples to estimate the mean of D to additive error ϵ, with proba-bility at least 1 −δ. Hence, in the high probability regime (when δ is not ﬁxed), the sample complexity of the sample mean varies inverse linearly in δ, which blows up when δ is made small. Conversely, the Gaussian sample complexity for the sample mean grows at a much slower rate of log( 1 δ). Therefore, the sample mean is a poor estimator of the true mean of a collection of independent random variables because it does not have Gaussian-like perfor-mance, as claimed by the CLT, for some distributions. However, by examining the lower bound Ω( σ2 ϵ2δ) closely, you’ll notice that if we ﬁx δ as a constant, the lower bound has the same form as the sample complexity of the Gaussian case! Namely, the Gaussian sample complexity for the sample mean 2 σ2 ϵ2 log( 1 δ) diﬀers from σ2 ϵ2δ only by a constant factor when δ is a ﬁxed constant. So, in the constant probability regime, the sample mean could be a Gaussian-like estimator. In fact, we can easily demon-strate that the sample mean has Gaussian sample complexity in this regime. If we assume {X1, . . . , Xn} is a collection of i.i.d. random variables with variance σ2, then P(\|Xn −E[Xn]\| ≥ϵ) ≤σ2 ϵ2n 2 6 Figure 2: Concentration of the Sample Mean by Chebyshev’s inequality. Given this bound on Xn, it follows that the sample complexity is σ2 ϵ2n ≤δ ⇐ ⇒n ≥σ2 ϵ2δ So, if we ﬁx δ as a constant, then the sample complexity of the sample mean becomes n = O( σ2 ϵ2 ), which is Gaussian. Therefore, in the constant probability regime, the sample mean is a good estimator of the true mean of X1, . . . , Xn. The above observations can alternatively be summarized by Figure 2, which shows a “bump” denoting the distribution of the sample mean normalized to variance 1, namely √n σ Xn. We can divide the bump into two regimes: the “body” which is within a constant number of standard deviations from the mean, and the “tail” which is the rest of the distribution. Thus, by Chebyshev’s inequality, the body is always Gaussian-like, in a big-O sense. The tail, on the other hand, can be heavy (having a lot of mass, decaying very slowly) and badly-behaved, from Catoni’s lower bound. Moreover, since the sample mean provides a good constant probability estimate for the true mean, we can combine it with the median trick from last lecture to design a high probability algorithm known as Median of Means: Algorithm 1: Median of Means input : X1, . . . , Xn samples where n = O( σ2 ϵ2 log( 1 δ)) output: A mean estimate µ so that \|µ −E[Xn]\| ≥ϵ with probability less than δ steps: 1. Divide samples in to m = Θ(log(1 δ)) groups 2. Compute sample mean Si for each group i, where each group is of size O( σ2 ϵ2 ) 3. Output median of {Si}m i=1 In the median of means algorithm, the sample mean of each group Si is within ϵ of the true mean with probability ≥2 3. This is done by selecting a group size of 3σ2 ϵ2 = O( σ2 ϵ2 ) in Step 2 of the algorithm. Thus, using the median technique from last lecture, the success probability of the entire algorithm is at least 1 −δ. 7 Remark 3.13 In conclusion, the median of means algorithm should be preferred over the sample mean for estimating the mean of a collection of independent random variables with high probability, as it provides Gaussian-like performance that is unattainable with the sample mean in general. 5 Conclusions • Constant probability algorithms are incredibly useful and can be used to build eﬃcient high probability algorithms; moreover, when designing high probability algorithms, a constant probability counterpart should serve as a baseline for performance. • While concentration inequalities are very useful, they are only an analysis technique and shouldn’t be blindly applied. • Don’t forget about Chebyshev’s inequality, which can be powerful (and tight) when used correctly with other tools. 6 Additional Content Theorem 3.14 (McDiarmid’s Inequality) Let {Xi}n i=1 be independent random vari-ables, where each Xi : Ω− →Si for some set Si. Consider f : Qn i=1 Si − →R, and suppose that f is Ci-lipschitz in the i-th coordinate of the input for all i. Then, for any ϵ > 0, P(f(x1, . . . , xn) −E[f] ≥ϵ) ≤exp −2ϵ2 Pn i=1 C2 i Remark 3.15 By Ci-lipschitz, we mean that given a Ci ∈R≥0 and any two points x = (x1, . . . , xi, . . . , xn), x′ = (x1, . . . , x′ i, . . . , xn) ∈Qn i=1 Si that diﬀer by only their i-th corrdinate, we have that \|f(x) −f(x′)\| ≤Ci Remark 3.16 Notice the connection between McDiarmid’s inequality and Hoeﬀding’s in-equality. Simply set f(x1, . . . , xn) = 1 n Pn i=1 xi and you’ll get back Hoeﬀding’s inequality. 8
2605	https://eceseniordesign2022spring.ece.gatech.edu/sd22p35/trpbalkew.pdf	Directivity and Gain of Antennas and Applications of Parabolic Dish Antennas in Satellite Tracking and Communication I. Introduction The purpose of this document is to provide a basic technical review on the current state of antenna technology centered around satellite tracking applications. In particular, the directivity and gain of parabolic dish antennas are emphasized and analyzed. In the context of ground-based satellite tracking, these two parameters are critical in being able to predict various design requirements and specifications for a given tracking system, including but not limited to: input signal variability tolerances, tracking model error tolerance, system response time, and others. II. Fundamentals of Antenna Directivity and Gain The directivity of an antenna can be defined as its ability to radiate in a particular direction relative to an isotropic source, which is a theoretical antenna that emits electromagnetic radiation with the same intensity in all directions . The latter does not exist in practice but is widely used as the standard measuring device for quantifying the directivity of practical antennas. Antenna gain is directly related to directivity, in that the gain of an antenna is equal to its directivity up to a constant. This constant is the product of the dielectric and conduction losses associated with a given antenna. Furthermore, it is important to note that a physical antenna is incapable of converting 100% of the energy it receives from a power source into “information” (in the form of radiation) that can be accurately interpreted by a receiver. The quantity used to describe power losses of this kind is known as the reflection efficiency, which is used to compute what is known as realized gain . This in unison with the fact that gain is a function of received electromagnetic power (unlike directivity from a single source, which emphasizes the directionality of transmitted radiation) demonstrates that the gain parameter takes into account realistic non-idealities associated with satellite tracking and communication. III. Industry Usage and Applications to Project A variety of antennas are currently used in industry for satellite tracking. This document shall focus on a specific type known as parabolic dish antennas, since the primary system of interest later on will be a parabolic antenna. Parabolic antennas can be found in numerous aspects of day-to-day life, such as TV satellite dishes, cellphone communications, cellular data, and so on. At the precipice of these applications resides a fundamental minimum directivity and gain associated with these parabolic dish antennas. According to , high-performance parabolic antennas maintain a high efficiency due to the geometry of the device. In particular, the component of the antenna responsible for receiving and “guiding” the data to the desired destination (otherwise known as the reflector) is able to operate at higher efficiencies due to the symmetric parabolic profile of the antenna . Recent work done by Nagasaka, et al. have demonstrated gain values (aperture efficiency) as high as 34.0 dB at 12 GHz and 38 dB at 21 GHz. Also, work done by Jablon et al. demonstrated antenna gains as high as 40.5 dB at approximately 13 GHz using radar altimeter antennas, devices that are stated to be well-approximated, performance-wise, by wide dish parabolic antennas . This along with constructive wave interference that occurs prior to transmission (EM waves become in phase) leads to very high directivity and high gain associated with these antennas. In addition, as outlined in , parabolic antennas exist in numerous variations. One type, known as a Cassegrain antenna, is particularly efficient as a transmitting device due to the unique hyperboloid concave reflector of the structure. Lastly, it is important to note that both the directivity and the gain of a given parabolic antenna increases as the size (usually look at diameter) of the parabolic profile increases. This is primarily due to similar effects of the antenna’s physical structure that was outlined earlier. Due to the geometry-based increases in directivity and gain as well as the drop in radiated beamwidth, parabolic antennas are ideally designed for very high frequency applications in which high-frequency gain losses are prevalent. It is fundamental for antennas utilized in satellite tracking to possess and maintain high directivities and gains in order for a designed auto-tracking system to produce models capable of meeting design specifications. For example, formulating a system that incorporates both RF design and Machine Learning algorithms to keep a grounded satellite dish locked onto a low-Earth orbit satellite would first require an antenna with a high directivity and gain. This would allow for less non-idealities to factor into the RF design process, thereby decreasing the required complexity and robustness (with respect to handling input data) of the implemented ML algorithm. IV. Collaboration with VIASAT This project is to be done in collaboration with an industry leader in antenna design, analysis, and synthesis – VIASAT. As such, the important antenna characteristics and parameters outlined in this review shall serve as an initial foundation for discussions with VIASAT moving forward. V. Key Takeaways To conclude, the numerous advantages of parabolic dish antennas in relation to satellite tracking and communication and the work collaboration with VIASAT place this research effort at the forefront of improving performance and efficiency in this area across the board. The unique characteristics and advantages of parabolic dish antennas provide a research-driven advantage within our area of interest. VI. References “Antenna gain,” Antenna Gain - an overview \| ScienceDirect Topics. [Online]. Available: [Accessed: 08-Oct-2021]. E. Notes, “Parabolic reflector antenna: Dish antenna,” Electronics Notes. [Online]. Available: [Accessed: 08-Oct-2021]. M. Nagasaka, S. Nakazawa and S. Tanaka, "Study on 12/21-GHz Dual-circularly Polarized Receiving Antenna for Satellite Broadcasting," 2018 IEEE International Symposium on Antennas and Propagation & USNC/URSI National Radio Science Meeting, 2018, pp. 339-340, doi: 10.1109/APUSNCURSINRSM.2018.8609229 C. Balanis, Antenna theory: Analysis and design. Hoboken, NJ: Wiley-Interscience, 2016. Allan R. Jablon and Robert K. Stillwell, “Spacecraft Reflector Antenna Development: Challenges and Novel Solutions.” journal article Antenna theory - parabolic reflector. [Online]. Available: [Accessed: 08-Oct-2021].
2606	https://ozdic.com/collocation/negotiate	negotiate - OZDIC - English collocation examples, usage and definition OZDIC x Discover more Idiomatic expressions guide English learning community Online English language tutors English literature anthologies Writing improvement workshop Collocation dictionary subscription IELTS preparation course Dictionary mobile app Natural English phrases Professional editing services negotiateverb 1 try to reach an agreement ADV. carefullya carefully negotiated series of concessions\| successfully \| freely \| individually, separatelyRents are individually negotiated between landlord and tenant.\| jointly \| directlynegotiating directly with the rebels\| secretly \| constantly, continuallyThe parameters of the job are being continually negotiated. VERB + NEGOTIATE be able to \| be prepared to, be willing toAre the employers really willing to negotiate?\| attempt to, seek to, try to \| manage to \| help (to) PREP. betweento negotiate between the two sides\| forWe are negotiating for the release of the prisoners.\| onThey have refused to negotiate on this issue.\| on behalf ofnegotiating on behalf of Britain\| withI managed to negotiate successfully with the authorities. 2 successfully get over/past sth ADV. easily, safelyHe safely negotiated the slippery stepping stones. VERB + NEGOTIATE be difficult toThe flight of steps was quite difficult to negotiate with a heavy suitcase. Discover more English Dictionaries Online PTE Academic Resources find English tutor online English Grammar Book Improve English Fluency Vocabulary Builder App English Collocation Dictionary English Writing Software academic writing help services buy advanced English dictionary ↺ handlegainthingtimefashionissue ⇄ ∅
2607	https://www.scribd.com/document/467618286/Ladder-problems-m2	Ladder Problems m2 \| PDF \| Ladder \| Center Of Mass Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0 ratings 0% found this document useful (0 votes) 939 views 2 pages Ladder Problems m2 Ladder problems use moments and force resolution to analyze situations involving ladders resting against walls and on surfaces. 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2608	https://books.google.com/books/about/Bailey_Scott_s_Diagnostic_Microbiology.html?id=l8UZEAAAQBAJ	Sign in Try the new Google Books Check out the new look and enjoy easier access to your favorite features Try it now No thanks Try the new Google Books Try the new Google Books My library Help Advanced Book Search EBOOK FROM $48.19 Get this book in print Elsevier Health Sciences Amazon.com Barnes&Noble.com Books-A-Million IndieBound Find in a library All sellers » \| \| \| Bailey & Scott's Diagnostic Microbiology Patricia M. Tille Elsevier Health Sciences, Feb 4, 2021 - Medical - 1184 pages Textbook and Academic Authors Association (TAA) Textbook Excellence Award Winner, 2024Selected for Doody's Core Titles® 2024 in Laboratory TechnologyPerfect your lab skills with the essential text for diagnostic microbiology! Bailey & Scott's Diagnostic Microbiology, 15th Edition Is known as the #1 bench reference for practicing microbiologists and as the preeminent text for students in clinical laboratory science programs. With hundreds of full-color illustrations and step-by-step methods for procedures, this text provides a solid, basic understanding of diagnostic microbiology and also covers more advanced techniques such as matrix-assisted laser desorption time-of-flight mass spectrometry. Written by noted CLS educator Dr. Patricia Tille, Diagnostic Microbiology has everything you need to get accurate lab test results in class and in clinical practice. - More than 800 high-quality, full-color illustrations help you visualize concepts. - Expanded sections on parasitology, mycology, and virology allow you to use just one book, eliminating the need to purchase other microbiology textbooks for these topics. - Hands-on procedures show exactly what takes place in the lab, including step-by-step methods, photos, and expected results. - Case studies allow you to apply your knowledge to diagnostic scenarios and to develop critical thinking skills. - Genera and Species boxes provide handy, at-a-glance summaries at the beginning of each organism chapter. - Learning objectives at the beginning of each chapter provide measurable outcomes to achieve by completing the chapter material. - A glossary defines terms at the back of the book and on the Evolve companion website. - New! Updated content includes infectious disease trends and new illustrations such as culture plate images of real specimens, complex gram stains, lactophenol cotton blue microscopy, and more. - NEW COVID-19 information has been added. - UPDATED topics include the Human Microbiome Project, expanded MALDI-TOF applications and molecular diagnostics in conjunction with traditional microbiology, additional streps, and significant news in mycology. - EXPANDED glossary defines terms on the Evolve companion website. Preview this book » \| Selected pages Page 63 Title Page Table of Contents Index References Contents \| \| \| --- \| \| I Basic Medical Microbiology \| 1 \| \| \| \| II General Principles in Clinical Microbiology \| 44 \| \| \| \| III Bacteriology \| 208 \| \| \| \| IV Parasitology \| 601 \| \| \| \| V Mycology \| 791 \| \| \| \| \| \| --- \| \| VI Virology \| 884 \| \| \| \| VII Diagnosis by Organ System \| 953 \| \| \| \| VIII Clinical Laboratory Management \| 1085 \| \| \| \| Index \| 1126 \| \| \| \| Copyright \| \| \| Other editions - View all \| \| \| Bailey & Scott's Diagnostic MicrobiologyPatricia TilleNo preview available - 2021 \| Common terms and phrases acid-fast addition aerobic amplification anaerobic antibiotic antibody antigen antimicrobial antimicrobial agents assays associated aureus bacilli bacteremia bacteria beta-lactam biochemical blood culture broth Brucella Campylobacter catalase cause cephalosporins Chapter characteristics chocolate agar Clin Microbiol clinical microbiology clinical specimens CLSI cocci coli colonies containing detection diagnosis differentiate disk endocarditis Enterobacterales enterococci enzyme factors fermentation fluid fluorescent gene genera genus glucose Gram stain gram-negative gram-positive grow growth Haemophilus host human hybridization identification immune incubation infections inhibit inoculated isolated laboratory Legionella MacConkey agar MALDI-TOF medium membrane methods microbial microorganisms molecular morphology Mycobacterium Mycoplasma negative Neisseria normal microbiota nucleic acid organisms pathogens patients plate pneumoniae positive probe procedures produce protein Pseudomonas reaction resistance respiratory rods sample sequence serum sheep blood agar skin species Spectrum of Disease Staphylococcus sterile Streptococcus subsp swab Table target therapy tion tissue toxin tract tube tuberculosis vancomycin virulence About the author (2021) Department Vice Chair, Clinical and Health Information Sciences, Graduate Program Director/Professor, Medical Laboratory Science, University of Cincinnati, Cincinnati, Ohio; Chair, Microbiology Advisory Committee, International Federation of Biomedical Laboratory Science; Editor in Chief, International Journal of Biomedical Laboratory Science, International Federation of Biomedical Laboratory Science; President, American Society of Clinical Laboratory Science. Bibliographic information \| \| \| --- \| \| Title \| Bailey & Scott's Diagnostic Microbiology \| \| Author \| Patricia M. Tille \| \| Edition \| 15 \| \| Publisher \| Elsevier Health Sciences, 2021 \| \| ISBN \| 0323681069, 9780323681063 \| \| Length \| 1184 pages \| \| Subjects \| › Allied Health Services › Medical Technology Medical / Allied Health Services / Medical Technology \| \| \| \| \| Export Citation \| BiBTeX EndNote RefMan \| About Google Books - Privacy Policy - Terms of Service - Information for Publishers - Report an issue - Help - Google Home
2609	https://nvlpubs.nist.gov/nistpubs/Legacy/TN/nbstechnicalnote667.pdf	• W ^ -fc NBS TECHNICAL NOTE 667 U.S. DEPARTMENT OF COMMERCE / National Bureau of Standards QC /OO .6/5753 />o. 6)67 F>t. I / 97 5 Upper-Bound Errors in Far-Field Antenna Parameters Determined From Planar Near-Field Measurements PART 1: Analysis NATIONAL BUREAU OF STANDARDS The National Bureau of Standards 1 was established by an act of Congress March 3, 1901. The Bureau's overall goal is to strengthen and advance the Nation's science and technology and facilitate their effective application for public benefit. To this end, the Bureau conducts research and provides: (1) a basis for the Nation's physical measurement system, (2) scientific and technological services for industry and government, (3) a technical basis for equity in trade, and (4) technical services to promote public safety. The Bureau consists of the Institute for Basic Standards, the Institute for Materials Research, the Institute for Applied Technology, the Institute for Computer Sciences and Technology, and the Office for Information Programs. THE INSTITUTE FOR BASIC STANDARDS provides the central basis within the United States of a complete and consistent system of physical measurement; coordinates that system with measurement systems of other nations; and furnishes essential services leading to accurate and uniform physical measurements throughout the Nation's scientific community, industry, and commerce. The Institute consists of the Office of Measurement Services, the Office of Radiation Measurement and the following Center and divisions: Applied Mathematics — Electricity — Mechanics — Heat — Optical Physics — Center for Radiation Research: Nuclear Sciences; Applied Radiation — Laboratory Astrophysics 2 — Cryogenics" — Electromagnetics- — Time and Frequency 1 '. THE INSTITUTE FOR MATERIALS RESEARCH conducts materials research leading to improved methods of measurement, standards, and data on the properties of well-characterized materials needed by industry, commerce, educational institutions, and Government; provides advisory and research services to other Government agencies; and develops, produces, and distributes standard reference materials. The Institute consists of the Office of Standard Reference Materials, the Office of Air and Water Measurement, and the following divisions: Analytical Chemistry — Polymers — Metallurgy — Inorganic Materials — Reactor Radiation — Physical Chemistry. THE INSTITUTE FOR APPLIED TECHNOLOGY provides technical services to promote the use of available technology and to facilitate technological innovation in industry and Government; cooperates with public and private organizations leading to the development of technological standards (including mandatory safety standards), codes and methods of test; and provides technical advice and services to Government agencies upon request. The Insti-tute consists of the following divisions and Centers: Standards Application and Analysis — Electronic Technology — Center for Consumer Product Technology: Product Systems Analysis; Product Engineering — Center for Building Technology: Structures, Materials, and Life Safety; Building Environment; Technical Evalua-tion and Application — Center for Fire Research: Fire Science; Fire Safety Engineering. THE INSTITUTE FOR COMPUTER SCIENCES AND TECHNOLOGY conducts research and provides technical services designed to aid Government agencies in improving cost effec-tiveness in the conduct of their programs through the selection, acquisition, and effective utilization of automatic data processing equipment; and serves as the principal focus within the executive branch for the development of Federal standards for automatic data processing equipment, techniques, and computer languages. The Institute consists of the following divisions: Computer Services — Systems and Software — Computer Systems Engineering — Informa-tion Technology. THE OFFICE FOR INFORMATION PROGRAMS promotes optimum dissemination and accessibility of scientific information generated within NBS and other agencies of the Federal Government; promotes the development of the National Standard Reference Data System and a system of information analysis centers dealing with the broader aspects of the National Measurement System; provides appropriate services to ensure that the NBS staff has optimum accessibility to the scientific information of the world. The Office consists of the following organizational units: Office of Standard Reference Data — Office of Information Activities — Office of Technical Publications — Library — Office of International Relations — Office of International Standards. 1 Headquarters and Laboratories at Gaithersburg. Nfaryland. unless otherwise noted: mailing address Washington, DC. 20234. - Located at Boulder, Colorado 80302. Upper-Bound Errors in Far-Field Antenna Parameters Determined From Planar Near-Field Measurements PART 1: Analysis A.D. Yaghjian Electromagnetics Division Institute for Basic Standards National Bureau of Standards Boulder, CO 80302 Sponsored by Air Force Avionics Laboratory Air Force Wright Aeronautical Laboratories Air Force Systems Command Wright-Patterson Air Force Base, Ohio 45433 ;\ OT BTANOAfiDf tTBRART DEC 3, 1 1975 Qc/co , US7S3 ftot 6 6 7 p-f I U.S. DEPARTMENT OF COMMERCE, Rogers C. B. Morton, Secretary James A. Baker, III, Under Secretary Dr. Betsy Ancker-Johnson, Assistant Secretary for Science and Technology NATIONAL BUREAU OF STANDARDS, Ernest Ambler, Acting Director Issued October 1975 NATIONAL BUREAU OF STANDARDS TECHNICAL NOTE 667 Nat. Bur. Stand. (U.S.), Tech Note 667, 120 pages (Oct. 1975) CODEN: NBTNAE For sale by the Superintendent of Documents, U.S. Government Printing Office , Washington, DC. 20402 (Order by SD Catalog No. C13. 46:667) $1.85 CONTENTS Page List of Figures and Tables iv I. Introduction 1 CI. Relationship of Errors in Gain Function, Sidelobe Level, Polarization Ratio, and Beamwidth to the Far-Field Error in Electric Field 4 II. Error Analysis 8 A. Finite Scan Errors 8 1. Mathematical Formulation of the Problem 9 2. The Fields of Electrically Large Aperture Antennas 13 3. Evaluation of n(r) 18 B. Position and Instrumentation Errors 30 1. Position Errors 33 2. Instrumentation Errors 48 C. Multiple Reflections 58 IV. Summary 69 Appendix A --Near-Fields of a Circular Antenna with Uniform Aperture Distribution 96 Appendix B --Position and Instrumentation Errors for the Null Depth of Difference Patterns 105 References 111 Acknowledgment 113 in LIST OF FIGURES AND TABLES Page Figure 1. Main beam of a hypothetical test antenna 82 Figure 2. Schematic of scanning geometry 82 Figure 3. Schematic of aperture antenna 83 Figure 4. Definition of a , 9 , D 83 to m ' m ' m Figure 5. Schematic of aperture and scan areas 84 Figure 6. Circular reflector antenna 84 Figure 7a. Near-field centerline data (constrained lens) (z = 25 cm) 85 Figure 7b. Near-field centerline data (reflector antenna) (z = 43.18 cm) 85 Figure 8a. Change in gain vs. decreasing scan length (constrained lens) 86 Figure 8b. Change in gain vs. decreasing scan length (reflector antenna) 86 Figure 9a. Constrained lens sum port near-field log ampli-tude, f = 9.2 GHz, z = 25.0 cm, no radome 87 Figure 9b. Near-field phase, constrained lens sum port, f = 9.2 GHz, z = 25 cm 87 Figure 9c. Constrained lens antenna and probe 87 Figure 9d. Near-Field Centerline Data, z = 25 cm 88 Figure 10. Position errors in on-axis gain 89 Figure 11. Comparison with Rodrigue et al . for random phase errors 89 Figure 12. Comparison with Newell for quadratic phase errors 90 Figure 13. Amplitude errors in on-axis gain 90 Figure 14. On-axis error in far-field from multiple reflections 91 Figure 15. Far-field errors from multiple reflections (a/A = 12) 91 IV LIST OF FIGURES AND TABLES (continued) Page Figure Al . Definition of t and a 102 Figure A2. Dotted line shows region in which eq . (A12a) holds 102 Figure A3. Near-field amplitude of circular antenna 103 Figure A4 . Near-field phase of circular antenna (a/A = 12)-104 Table 1. Finite Scan Errors from Centerline Data of a Typical X-Band Antenna 92 Table 2. XY-and Z-Position Errors for a Typical X-and K-Band Antenna 9 3 Table 3. Instrumentation Errors in Measuring Amplitude 94 Table 4. Total Far-Field Errors for a Typical X-and K-Band Antenna 9 5 v UPPER-BOUND ERRORS IN FAR- FIELD ANTENNA PARAMETERS DETERMINED FROM PLANAR NEAR- FIELD MEASUREMENTS Part I : Analysis A.D. Yaghj ian ABSTRACT General expressions are derived for estimating the errors in the sum or difference far-field pattern of electrically large aperture antennas which are measured by the planar near-field scanning tech-nique. Upper bounds are determined for the far-field errors pro-duced by 1) the nonzero fields outside the finite scan area, 2) the inaccuracies in the positioning of the probe, 3) the distortion and nonlinearit ies of the instrumentation which measures the amplitude and phase of the probe output, and 4) the multiple reflections. Computational errors, uncertainties in the receiving characteristics of the probe, and errors involved with measuring the input power to the test antenna are briefly discussed. Key words: Antennas; error analysis; far-field pattern; near-field measurements; planar scanning; planewave spectrum. vr I . Introduction The planar near-field scanning method has been used to measure with high accuracy the electromagnetic fields of microwave antennas. High accuracy is possible primarily because so few restrictive approximations are involved in the formulation and application of the near-field techniques. Moreover, the far-field errors associated with each approximation can be estimated because the approximations themselves can be expressed in convenient mathematical form. This report essentially derives and evaluates some of these mathematical expressions under the given laboratory conditions to determine quan-titatively the accuracy with which the far-field of antennas can be measured by the planar near-field scanning method. The error analysis was undertaken for two main reasons. One, it was desired to find general upper-bound expressions for the limits of accuracy in computing far-fields from planar near-field measurements without resorting to direct far-field comparisons. It has long been the feeling of those involved with the near-field measurement techniques at the NBS that these techniques were often more accurate than measurements taken on conventional "far- field" ranges with a standard antenna. Thus comparisons with patterns measured on conventional far-field ranges would not give a re-liable evaluation of the near-field techniques which do not have to cope with proximity corrections, ground reflections, or the cali-bration of standard far-field antennas. Two, the upper-bound ex-pressions could be used to stipulate design criteria for the con-struction of new near-field scanning facilities. This meant that the upper-bounds for the accuracy in a given far-field parameter should be expressed in terms of measured near-field data and/or the computed far-field, the frequency and dimensions of the antenna-probe system, the variation in the positioning of the scanner, and the precision of the instrumentation which measures the probe output. The design engineer could then compute from the upper-bound ex-pressions the near-field tolerances required to insure a given far-field accuracy for the range in size and frequency of the antennas he was considering. There have been two major computer studies performed in the past to estimate the errors involved with planar near-field scanning measurements [11,16]. Rodrigue, Joy and Burns have introduced position and instrumentation errors into a hypothetical near-field distribution in order to compute their effects on the far-field. Newell and Crawford [11,17] performed a similar analysis with mea-sured near- field data which included an estimate of errors involved with truncating the scan plane. (Jensen has also made a numerical investigation into the accuracy of near-field measurement techniques but for nonplanar scanning only.) A major drawback of the computer studies, as well as direct far-field comparisons, is that they apply to particular antennas and their results do not necessarily represent upper-bounds to errors which will hold for large general classes of antennas. It should be emphasized, however, that the computer studies are an extremely useful aid in giving direction to the general analysis and checking its results. This report does not attempt to estimate the accuracy of the extrapolation technique for measuring the gain of antennas . An error analysis involving the extrapolation technique has been performed recently by Kanda . If we assume that the antennas under consideration are linear, of finite extent, and operating in a single mode at fixed frequency and amplitude, and that Maxwell's equations for free space describe the region in which the antenna is situated, the only approxima-tions involved in formulating and applying the planar scanning method are : 1) The fields outside the finite scan area are zero. 2) The scanner is aligned and positioned with infinite precision . 3) The instrumentation introduces no distortion and measures the amplitude and phase of the probe output with perfect accuracy . 4) Multiple reflections between the test antenna and "probe" antenna are zero . 5) Computation errors in "deconvolut ing" the measured near-field data to get the far-field are nil. Errors caused by uncertainties in the receiving characteristics of the probe, and errors involved with measuring the input power to the test antenna are discussed in Section IV, which, incidently, contains a summary of the major results and conclusions of this report. The far- field errors introduced by the computations (approxima-tion 5) are so much smaller than the combined errors in the far-field caused by a finite scan area, positioning, instrumentation, and multiple reflections (approximations 1-4) that they are of little consequence. With the help of the sampling theorem, Fast Fourier Transform, and computers accurate to many places, the necessary deconvolution of the measured data can be performed with insignificant error. The amplitude of the near-field multiple reflections (approxi-mation 4) can be estimated in practice by changing the distance be-tween the probe and test antennas. Any periodic variations in the amplitude of the received signal repeating about every A/2 (A = wavelength) would be caused primarily by the multiple reflections. In Section III.C the effects on the far-field of multiple reflections are discussed and estimated analytically. Although they cannot be eliminated completely, multiple reflections can be reduced by using efficient absorber material, by decreasing the size of the probe, or by increasing the distance between the probe and test antenna. Also, it is likely that the effect of multiple reflections on the far-field can be reduced by averaging the far-field patterns ob-tained by scanning on a number of near- field planes which are separated by a small fraction of a wavelength. Far-field errors caused by the errors in the scanner position-ing and instrumentation (approximations 2 and 3) are determined from a common set of equations. These equations are derived and eval-uated in Section III.B for both systematic and random near-field Without the advantage of the FFT, typically the computer would add 20,000 terms to calculate the variable which yields the far-field. Even if we make the absurd hypothesis that every round-off error adds in the same direction, a computer of 10 -place accuracy adding 20,000 numbers would retain more than 5-place accuracy (.001%) for the sum. Of course, in practice the FFT algorithm vastly reduces the number of necessary computations. There is also the approximation associated with applying the samp-ling theorem, which assumes that the output of the probe is the Fourier transform of a band-limited function. For a separation dis-tance between test antenna and probe of more than a few wavelengths, this can be shown to be an extremely good approximation which intro-duces negligible errors into the far-field. errors. "Position errors" may be reduced by scanning along both vertical and horizontal lines and appropriately averaging the two sets of measurements. "Instrumentation errors" can be reduced if the distortion and nonlinearit ies of the receivers can be determined and included as part of the computer program that deconvolutes the near-field data. The first portion of the error analysis (Section III. A) is devoted to determining the maximum far- field errors introduced by neglecting the fields outside the scan area. As part of the analy-sis, asymptotic expressions are derived for the near-field in front of large aperture radiators . All errors are assumed small enough so that the individual con-tribution to the far-field error from each source of near-field error can be estimated independently and then combined to give the total far-field error. We shall find that all the individual errors in the far- field combine linearly except for the on-axis sum pattern position and instrumentation phase errors which combine quadratic-ally (see footnote 12). 1 1 . Relationship of Errors in Gain Function, Sidelobe Level, Polarization Ratio, and Beamwidth to the Far-Field Error in Electric Field Let E (r ) denote the electric field (to within the limits of error) of an antenna radiating into free-space and ± AE(r) the limits of error involved with the measurement of E(r) (e time dependence has been suppressed). The magnitude of the fractional error n(r) in the electric field amplitude can be defined for small errors as , n(r) = E±AE AE(r) \|E(F) (1) (Henceforth, the < sign will be omitted when eq . (1) or similar ex-pressions are used.) The Hermitian amplitude \|E\| = /E # E (the asterisk denotes the complex conjugate) is related to the power S radiated per unit area in the far-field or any other locally plane-wave field in free-space by S = /e /y \| E \| 2 From the values of n(r) as r approaches infinity and the approximate amplitude of the far-field \|E(r)\| , all other errors in far-field quantities can be found. Four such quantities of interest are the gain func-tion, the sidelobe level, the polarization ratio, and the half-power beamwidth. Of course, these four are not always the only far-field quantities of interest, but they will be used as typical examples to demonstrate how the error in any far- field quantity can be re-lated easily to n(r) and \|E(r)\| . (In eqs . (2)-(5) below, where the error in these four quantities is derived, the subscript M r+°°" is understood, but not shown explicitly.) The error n^ in the gain function G(r), which is proportional to the square of the far-field amplitude, may be written immediately as (for small \|AE~\|/\|e\|, so that \|AE\| 2 terms can be neglected) E ± AE El 2 1 I AE , G = ± 2 U±Ba. g (2a) or in decibels = ±2n(r)G(r), dB _ -, n -. n G = 10 log E ± AE 2 ^ = 20 log(l±n(r)) (2b) For small errors, n << 1 , and (2c) The sidelobe level, when meaningful, is usually stated in deci bels and is defined as the ratio of the maximum far-field intensity of the largest sidelobe to the maximum far-field intensity of the main beam For SL = 'side The error r\ in sidelobe level SL is s 'side ± AE side side E ± AE max max E \| 2 max ' VIEmax and small r\ (r) , r, s = ±2[n(r s ) + n(r Q )]SL, (3a) with n(r ) and n(r ) denoting the values of n at the sidelobe maxi mum and main beam maximum respectively. In most cases n(r ) will be much larger than n (r ) so that r\ = ±2ri(r )SL. (3b) In decibels dB n = 20 log side AE s ide E ± AE max max log 'side max or for small n , and n (r ) >> n (r ), dB 7 n(r s ) (3c) i.e., as we might expect, simply the error in the gain of the side-lobe itself. The polarization ratio (Pol) of the electric field is defined as the ratio of the minor to major axis of its polarization ellipse . For a given amplitude of error field \| AE \| the maximum error in the polarization ratio at a point in the far-field occurs when the field AE has the same polarization ratio as E but with major and minor axes interchanged and the direction of rotation of the electric-field vector reversed. (Also the electric field vectors of AE and E must be colinear along the major-minor axes.) Under these conditions, the maximum error (APol) in polarization ratio is realized and can be expressed for small errors in the form (4a) APol = ±[1 + (Pol) l ] n(r) . The upper-bound error in polarization is never greater than 2n (cir-cularly polarized) and for antennas with small cross polarization the upper-bound error is approximately n . It is an interesting result that if the maximum error in polari-zation ratio were realized, the rotational shift in the polarization ellipse caused by the error field would be a minimum, i.e., zero. Fortunately, as part of the derivation of eq. (4a) the maximum possible rotational shift \b in the polarization ellipse for a r r max r r given \| A E \| can also be determined. Specifically, 4>max 1 + (Pol) 2 ' (PoI) 2 J radians (4b) For a far-field of approximately circular polarization (Pol -1) the shift in polarization ellipse introduced by the error field may This result, the proof of which is straightforward but rather lengthy, was obtained by Flemming Jensen and brought to my attention by Flemming Larsen both of the Technical University of Denmark, Lyngby, Denmark. become large --as one might expect. Finally, it is mentioned that footnote 11 should be consulted when applying eqs . (4) to instrumen-tation amplitude errors. The half-power beamwidth is defined for a plane containing the line along the maximum intensity of the beam. It is simply "the angle between the two directions in which the radiation intensity is one half the maximum value of the beam" . If we know E(r) and AE(r) in the far- field we can determine the maximum errors in measur-ing the beamwidth. In figure 1 and \|E ± AE 2 E AE (neglecting \| AE \| 2 terms, as usual) are plotted for the main beam of an arbitrary test antenna. The beamwidth is 0,+6 2 . From figure 1, we see that the error in 9-. is A0 1 = ± 2\|AE(9 1 )\| \|E(6 1 ) AE(6 1 )\| \|E(8 1 ) tan a E(9 1 ) d\|E(9 1 ) ±n(9 1 )\|E(9 1 ) dlECe^l radians . ~dQ d9 A corresponding expression holds for the error in A9~. The total fractional error n R in beamwidth may be written (5a) (5b) or, for symmetrical beams (9=9,= 9~), simply n B = ±A9 9 . ±ri(9) E(9) fi d\|E(9)\| d9 For large circular or square apertures of uniform amplitude and phase distribution, \|F(0)\|/ e^-L equal 1, and n™becomes simply, phase distribution, \|E(0)\|/ 9 L-t4 j [ can be shown to approximately ! B :n(8) (5c) As expected, eqs. (2)- (5) verify that a knowledge of \|E(r)\| and n(r) as r -» °° (as mentioned above, the "r > °°" is suppressed in eqs. (2) -(5)) is all that is required to determine the errors in the gain function, sidelobe level, polarization ratio, and beamwidth. If desired, errors in other far-field parameters can equally well be expressed in terms of \|E(r)\| and n (r) . The far-field \|F\| may be computed from the measured near-field data or estimated analyti-cally. Thus, the problem of finding the far-field errors reduces to the problem of evaluating n (r ) of eq. (1) in the far-field. Here we are neglecting the error caused by the change in the_maximum value of the beam. If this error is significant compared to AE(9,) it should be added to it . III. Error Analysis A. Finite Scan Errors The purpose of the present section is to estimate the maximum errors in the far-field introduced by scanning in the near-field over a plane of finite area. "Finite" is the key word. In prin-ciple, the planar scan method requires that the output of the probe be recorded over an infinite plane in front of the test antenna. In practice, of course, only a finite area of the plane is scanned and the fields outside that finite area are set equal to zero. The scan area is usually, but not necessarily, rectangular with the boundaries commonly chosen where the output of the probe antenna is down 30 or 40 dB or more from its maximum. For many microwave an-tennas such a scan area turns out to be about two or three times the aperture area of the antenna. The present analysis will be re-stricted to electrically large (average width/wavelength > 10 will do) aperture-type antennas, usually but not necessarily operating at microwave frequencies. The phase will be assumed fairly uniform across the aperture with the amplitude of the field either uniform or reaching a maximum near the center of the aperture and convexly 2 tapered toward the edge to reduce the sidelobe radiation . Near-field, centerline, amplitude and phase data for the sum pattern of a typical antenna measured at NBS can be seen in figure 9d . Re-flectors, large horns, and broadside arrays are probably the most common examples of the test antennas under consideration. The re-sults of this finite scan part of the error analysis apply to antennas with their boresight direction steered at an arbitrary angle with respect to the scan plane. Initially we are assuming the antenna is operating in a sum mode or pattern. For finite scan errors it will be shown that difference patterns need not be considered separately since they are formed by the superposition of two sum patterns with wavefronts slightly skewed to create a "null" in the boresight direction. For position and instrumentation errors, however, Section III.B along with Appendix B shows that sum and difference patterns must be con-sidered separately, at least when determining far-field errors near the boresight direction. The approach which is used to estimate the finite scan errors involves finding an upper-bound to the appropriate integral of the fields outside the finite scan area. Physical optics and the geo-metrical theory of diffraction are used to show that the fields out-side the scan area are determined chiefly by edge diffracted fields of the antenna. For each antenna these edge diffracted fields are different. However, they all can be expressed in a general form (eq. (19)) which allows upper-bound expressions (eqs. (32) and (36)) to be found by evaluating the integral outside the scan area in terms of the probe output on the edge of the scan area. Outside the "solid angle" formed by the edge of the aperture and edge of the scan area, the evaluation of the integral can be done by the method as stationary phase to show that in this region the far-fields com-puted from the near-field data cannot be relied upon with any con-fidence. Well within the solid angle the integral evaluation can be performed through integration by parts to yield an upper-bound expression for finite scan errors from both centerline data scans (eq. (32)) and full scans (eq. (36) or (32)). 1 . Mathematical Formulation of the Problem Suppose we want to determine the radiation pattern of a given antenna of aperture area A bounded by the curve C, as in figure 2. The task is accomplished experimentally by recording the output of an arbitrary but known probe antenna (for two orientations, in general) as the probe scans in front of the radiating test antenna on a plane of area A' bounded by C 1 . The z-axis will always be chosen perpendicular to the scan plane with origin in the antenna aperture. The scan plane, however, may not always be chosen parallel to the aperture plane. For beams steered off -axis, larger scan areas may be required if the scan plane does not lie perpendicular (approx-imately) to the boresight direction. After taking a double Fourier transform of the probe output in each orientation, the radiating characteristics (S, (K) ) of the test antenna are found simply by solving simultaneously the two resulting linear equations. In particular, if the probe were a perfect dipole, the output of the probe would be proportional to the electric field at the dipole in the direction of the dipole moment. The expression for S, (K) (defined with respect to the reference plane z = 0, and the transverse part of the propagation vector denoted by K) then becomes S l0 (K) = —7— e iYd / E (P,d)e" iK ' P dP, (6) o f - ioot j , . (e time dependence) where E (P,d) is the electric field, i.e., the output of the dipole probe in two mutually perpendicular orientations, transverse to the z-axis at the point (P,d) in the scan plane A'. (The equation of the A' plane is z = d.) The amplitude of the input mode to the test antenna is designated by a , and the variable y is defined by 1, 7 TT Y = (k 2 -K 2 )'2 (k = 00/c = —r) , where the radical is chosen to keep y positive real or imaginary. ' For the sake of simplifying the theory we will assume a perfect electric dipole as the probe. The dipole gives information about the electric field only at a point. All physically realizable probes respond to a weighted average of the field near the probe. Thus it is expected that the errors in the computed far-field introduced by omitting a part of the infinite scan plane are as great or greater for a perfect dipole than for any other probe antenna, and that the following conclusions and resulting upper bound expressions (32) and (36) hold for arbitrary probes. Also, at this point in the error 3aThe sampling theorem shows that to obtain the far-field pattern the double integral in eq . (6) can essentially be replaced by a double summation over points in A' separated by about A/2 or less. Thus, in practice, data need be taken only at a finite number of discrete points for eq . (6) to be evaluated. However, for most of the error analysis_we prefer (somewhat arbitrarily) the integral representation of S. to the summation. r lo Strictly speaking, the Fourier transform in eq. (6) and eqs . (7), (10), and (11) below may not converge^ to a unique value as the scan area approaches infinity because E+-(P,d) has a 1/P dependence in a lossless medium as P->°°, and thus gives rise to a rapidly oscillating part in the transform as P-». Usually this oscillatory term can be ignored with impunity because it vanishes upon integration when taking the inverse transform. However, it does determine the limit-ing value of the finite scan errors (see footnote 9). 10 analysis any uncertainty in the receiving characteristics (S') of the measuring probe are ignored. For eq. (6) to represent S, (K) exactly, the probe scan and subsequent integration would have to be performed over the infinite plane. Thus the error (AS, ) produced in S, by a scan of finite r lo^ r lo J area may be written formally as AS-= —^— e" iYd / E (P,d)e~ iK ' P dP, (7) 10 4^ 2 a A' t o c where A 1 is the infinite area outside or complementary to A' . The far-fields can be found from S, (K) by taking the inverse double Fourier transform of eq . (6) and evaluating the resulting expression by the method of stationary phase for double integrals. So doing yields ikr E.(r) = -27ia ik cosB S, (kR/r) (8) ^ v j lo v J r X" -> co and from eq. (8) ikr AE^(r) = -2iTa ik cosGAS, (kR/r) . (9) t K J o lo v r r -> oo (cos9 = z/r, z>0) Substitution of S, and AS, from eqs. (61 and (7) into eqs. (8) and lO lO ivy v. j -i v j (9), respectively, produces expressions for the far -electric-field and its error in terms of the near-field data: E (7) = - ik C056 e ik(r - d C0S ^/ E t (P,d)e" lFR ' P dP (10) Z 27rr A' r X" --°o AEJ7) = -i\^-e ik(r " d COs9) / E t (P,d)e" lFR ' P dP. (11) c X -> oo (For a dipole probe, E (P,d) and the output of the probe are identi-cal.) Equations (10) and (11) express mathematically the well-known 11 "Fourier optics" result that the far-field amplitude is proportional to the spatial Fourier transform of the near-field times cos0. Division of eq . (10) by eq. (11) results in the fractional error r\ in the transverse part of the far-electric -field n t (r) AE t (r) E t (r)\| -i-R-P / E.(P,d)e r dP A' z -i-R-P \| E (P,d)e r dP A' Z (12a) or simply cos6 n t (r) = -i- R-P { E. (P,d)e r dP A' z __c Ar \|1\ (r) \| (12b) Of course, the fractional error r\ in the total far-electric field is not necessarily n f except on the z-axis. However we can find n in terms of n f by a simple argument. In the far-field \|E (r) differs from \|E(r)\| at most by a factor cosG, i.e. \|E(r) \|cos6 _< \|E (r) \| < \|E(r) Similarly AElcosG < AE t \| < AE (12c) (12d) Consequently ri is greater than n t by at most a factor Q , and from eqs . (12) we can express the maximum possible n in the far-field as n(r) = AE / E.(P,d) A' Z -i- R-P e r dP Xr\|E(r)\| (13) •]"->• 00 For the denominator of eq . (13) we can use the far-field estimated analytically or computed from the measured near-field data. Thus, the problem of finding n reduces to that of estimating E (P,d) on A^, i.e., on the area outside the finite scan plane. 12 2 . The Fields of Electrically Large Aperture Antennas If the behavior of the electric field E (P,d) were known in the area A, the integral of eq. (13), r I = / E.(P,d")e dP, (14) A t L Ac could be evaluated and the far-field errors could be found immediately from eqs . (l)-(5). Even asymptotic methods cannot be applied to eq . (14) until the behavior of E (P,d) is determined. Fortunately, the electric field outside the aperture of electrically large aperture antennas can be determined analytically from the electric field dis-tribution across the aperture. In fact, it will be shown shortly that just the electric field at the edge of the aperture distribution is required. As a first step in finding the electric field E (P,d) to use in eq. (14), consider the aperture antenna drawn schematically in figure 3. (The boundary or rim of the antenna is assumed to lie in a plane with e here chosen perpendicular to the plane. These restrictions are relaxed later.) It is well-known that the electric field everywhere to the right of the infinite plane A^ can be expressed in terms of an integral of the electric field over A^, EOO = - ^ / [e z xE t (R')] x VG(F,R») dR' , (15) 00 where lklr-R' I G = SL-! ' \|r-R»\| and E. (R' ) is the transverse electric field on the plane A^ emanat-ing from sources to the left of A^ (e.g., electric fields from a feed located to the right of A^ would not be included in E.(R'), except, of course, indirectly as reflected fields from the antenna). For an aperture antenna, whether it be a reflector, large horn, or broadside array, the components of the electric field E. (R T ) are 13 4 slowly varying in phase and amplitude across the aperture, except possibly right near the edge. The amplitude often tapers toward the edge of the aperture and drops abruptly (in a distance less than a wavelength) to zero or near zero beyond the edge. Thus the limits of integration in eq. (15) reduces to the aperture area A or at most a couple of wavelengths beyond A. Also, since eq. (14) requires only the transverse component of E (r) , we can ignore the z-components of eq. (15) and write E t (r) = ' 17 \|y / E t (R')G(F,R') d R' . (16) For electrically large apertures, eq . (16) can be evaluated asymp-totically. Before doing this a couple of remarks are in order. The first has to do with neglecting the fields outside the aperture area A when in reality there may be scattering from the edge of the antenna aperture. In those cases it may seem unreasonable to neglect, initially, the fields beyond the aperture area A as part of the procedure to calculate the fields beyond the scan area A' . The "canonical" problems that can be solved exactly, such as scatter-ing by an infinite wedge or elliptical disk, indicate, however, that at short wavelengths the scattered fields are caused predominantly by high intensity fields within a wavelength or so of the edge, and indeed the fields more than this distance beyond the edge may be neglected . Of course, this assumption has been confirmed by experiment and constitutes the basic postulate of the geometrical theory of diffraction . In terms of the radiation from large aperture antennas, it means simply that the fields everywhere to the right of the aperture plane depend solely upon the fields within and near the edge of the aperture, even when appreciable scattering from the edges is present. The second remark concerns antennas that have part of their feeding mechanism mounted to the right of the aperture plane--as in the case of reflector antennas. As mentioned above, the direct radiation from the feed must not be included as part of E.(R') for eq . (16) to be correct. However, scattering from feed mounts 4For electrically steered arrays the planes of "uniform" phase may be skewed with respect to the aperture plane. This situation is considered at the end of the section. 14 (including the feed itself) of the fields reflected by the antenna are not taken into account by eq . (16). Although the mount scattered fields can be ignored if they are small compared to the fields scattered from the aperture edge, there is no reason to believe this will always be the case even in the region A'. Fortunately, the results of the geometrical theory of diffraction (GTD) can also be used to determine the behavior of the fields diffracted from the mounting in front of the aperture as well as from any sharp edges at the boundary of the aperture . Moreover, we shall find that the final expression (eqs. (32) and (36)) for the far-field error does not require an explicit estimation of either the mount or edge diffracted fields. Of course, for many antennas such as horns and arrays there are no obstructions in front of the aperture. For the moment we shall ignore the problems of the exact nature of the edge diffraction and diffraction from feed mounts and return to evaluate eq. (16) for a smoothly tapered amplitude within A and zero outside A. For apertures which are many wavelengths across, the double integral (16) can be expanded in an asymptotic series. The first three terms in the series for integrals like eq. (16) have been derived by Van Kampen . Keller, Lewis and Seckler apply Van Kampen' s results to eq. (16) specifically. The final expression has been confirmed by the present author using an approach different from Van Kampen' s. In the shadow region, i.e., the entire half space z > excluding the cylindrical volume formed by the projection of the aperture area A along the z-axis, the electric field in eq. (16) is approximated by / E\ (F) = I -t v J L 2tt m am a +D sin6 mm W H cos9 m E. m tm e sin0m m 4 (17) The variables in eq. (17) are defined with the help of figure 4. The distance from the point r to the edge of an aperture with a smooth boundary has at least one relative maximum and one relative minimum. The subscript m simply refers to the mth relative extremum point on the edge of the aperture. D is the distance from r to the ° r m mth relative extremum, a is the radius of curvature of the edge (in 7 m b K 15 the plane of the aperture) at the mth relative extremum, 9 is the _ m angle between D and the z-direction, and E is the value of the transverse electric field at the edge of the aperture excluding the fields diffracted from the edge. The radius of curvature a is taken b m positive if the distance D is a relative minimum and negative if a am 1 1/2 m relative maximum. The radical a +D sin0 mm m is taken positive if real and negative if imaginary. Of course, all subscripted quanti-ties are, in general, functions of position r. Actually, in order for expression (17) to remain valid, must be greater than a few A/£ (I e /A) , and D no closer than a couple of wavelengths X from the edge. This latter restriction says simply that eq. (17) does not describe the reactive fields. The former restriction can be understood physically by dividing the aperture into Fresnel zones for the direction . For greater ave mm 6 than a few X/l there are enough Fresnel zones (even for the far-field) to assure that the main contribution to the fields in the shadow region are from the fields near the edge of the aperture rather than from the fields well within the aperture. The expression (17) does not necessarily account correctly for scattering from edges that may be present on the boundary or rim of the antenna. The necessary modification of eq . (17), for example, when scattering from sharp conducting edges occurs can be extracted from Kellers GTD . Kouyoumj ian [9a] has written Keller's results in a form similar to eq. (17). The GTD expression differs from eq. (17) only in that the factor cos0 E. is replaced by v J J m tm r J l' 1 g (0 ) + e}-g. (0 ), (18) tm 5 II v nr tm 6_L V nr ' K J —II —J-where El and E^ are the transverse components of the incident tm tm r electric field parallel and perpendicular to the edge of the aper-ture. The factors g and g can be found from either reference [6a] or [9a] , but it is not necessary to know them explicitly for our purposes. Moreover, scattering from other than sharp conducting edges can also be handled by changing appropriately the factors & and gj_ in eq. (18) . 5Note that in a similar manner it can be argued that the far-fields for less than a couple X/£max (&max = maximum width of aperture) are determined mainly by the near- fields well within the boundary of the aperture. 16 Thus it is seen from eqs . (17) and (18) that regardless of the nature of the scattering from the edges, the fields in the shadow zone appear to emanate from points along the edge of the aperture. For our purposes it proves convenient to write simply the one equa-tion valid for the scattered field in the shadow zone ikDm E t (r) = I F (r)e m . (19) m F is defined by comparing eq. (19) with eqs. (17) and (18). The essential property of F (r ) , which allows the asymptotic evaluation ikD — of eq. (14), is that unlike e m it varies slowly with r. In fact, for the error analysis it turns out that this is the only property of F (r) that is required. Never is it necessary to evaluate F. (r) explicitly, although for reasons of general interest it has been evaluated in Appendix A for the circular aperture of uniform dis-tribution. Results from Appendix A are also used in Section III.B.l. Before carrying out the integration of eq . (14) it should be mentioned that eqs. (17) and (18) are not valid for large D if the radius of curvature a approaches infinity. For example, the ex-pressions would be modified if the aperture were rectangular. How- _ ikDm ever, the modifications occur in F (r) but not e . Thus, eq. (19) remains valid for all shaped apertures even when part of the edge is a straight line or has infinite radius of curvature. Also if the edge of the aperture has points where the radius of curvature is much smaller than a wavelength (e.g., the corners of a rectangular aperture) , these corners and tips contribute to the field. For large apertures at least it can be shown [5-9] that their contribution is usually much smaller (higher order in X/ I ) than the edge fields of eq. (19), and thus can usually be neglected for the purpose of evaluating eq. (14). However, even if they can-not be neglected, the fields from corners and tips can be expressed in the same form as eq. (19). Electronically steered aperture antennas (broadside phased arrays) are also covered by eq. (19) with the appropriate modifica-tions of F (see, for example, reference which deals with an arbitrary phase of the field across the aperture) . When the axis of the main beam is steered away from the perpendicular to the aper-17 ture, the region of validity of eq. (19) changes to the entire half space in the direction of the new axis, excluding the projection of the aperture along that axis, i.e., eq . (19) is still valid in the shadow zone regardless of what direction the beam is steered. Eq. (19) also applies to antennas operating in a difference pattern where the on-axis field drops to a sharp minimum. As in the case of the sum pattern, the fields in the shadow zone are determined by whatever value of electric field impinges upon the edge of the aperture. And again the variation in electric field separates into ikDm _ a rapidly oscillating part e and a slowly varying part F . Eq . (17) (or (17) modified by (18)) represents the first term in an asymptotic expansion of the electric field. The higher order terms are assumed so much smaller than the first that they are neg-lected. However, if the electric field at the edge of the aperture becomes too small, the second term in the asymptotic expansion must be included. Even then, for reasonably smooth aperture distribu-tions, this second term also has the form of eq . (19) but with F depending on the derivative of the electric field at the edge rather than the field itself. This "slope diffraction coefficient" has been derived for the GTD by Hwang and Kouyoumj ian [9b]. In brief, eq . (19) describes the electric field in the shadow zone of nearly all large aperture antennas including electronically steered arrays, antennas excited in a difference pattern, and an-tennas with aperture distributions that taper to zero at the edge. 3 . Evaluation of n(r) The evaluation of the integral in eq. (14) and subsequently n(r) in eq. (13) is accomplished by first substituting the electric field from eq. (19) into eq. (14), 2tt °° ik[D -p sinBcos (4>-<> )] ; (V d) e m P tm L ' J r -' 'P I = I J "J F t fP,d)e m P ' J p dp deb. (20) m The vectors P and r have been written in cylindrical (p,(b ) and spherical coordinates (r,0,-<f) )] occur at 3D ^ = p sine sin (cf) -cf)) (21a) ^p p 3D 3p^ = sine cos(cj)-cj) ) . (21b) 3D Consider the derivative ^ , which can be interpreted physically by referring to figure 5. Let the vector P be the perpendicular pro-jection of the line D such that J m Then p = /p 2 -Ji 2 . m r m 3D 3D 3p p-£ m m ± in it m _ m _m ' m 3p 3P 3Pm 3P Pm C0SYm , m x x m where y is the angle between P and D . Ordinarily, the scan areas 'm & mm J ' are appreciably larger than the aperture area so that 9£m Pm ~ i and (21b) may be written cosy = sine cos(cf)-(() ). (22) 3Dm Similarly, it can be argued that -^— is much less than p for scan 3c\|)p areas appreciably larger than the aperture area. Thus for 6 not too small eq. (21a) implies the critical point must be near $=<{>. Now cosy has a minimum value greater than zero because y never reaches 'm b m 90°. The minimum value of cosy occurs at the maximum value of 'm 19 ym = Ym v or on tne boundary of the scan area near for 9 < 90 -v fd>1 . For > 90 -y critical points exist at max no cf) -<J) P 90 m m and eq. (20) can be evaluated immediately by the method of station-ary phase for double integrals . Such a procedure yields after some rearrangement F. AD ik(d cos6+x sin9) f-v tm m v m J I = I -r^r^ e m cos6 (23) Actually, eq. (23) was derived using the approximation D -/(p±x ) +d -d X and — ^—— << 1, where x and d are defined in figure 5. The above p dcp m to approximations simplify the mathematics but do not alter by a great amount the amplitude of the final expression (nor the following conclusions) . The implications of eq . (23) prove to be quite significant be-cause it shows that for 9 > 90 -v () "the magnitude of the inte-max v J b gral I is of the same order as the amplitude of the electric field itself with each term multiplied by A D /cos9. By referring back to eqs . (12) and (13) we see that this result implies the following: In the re gion outside of the envel ope , which is formed by the ra ys running from the edge of the aperture throu gh the bounda ry of the scan area , the fractional error n(r) is on the or der of unity. Thus , outside the envelope the fax fields COmputed from a planar :5can in the near- field cannot be relied upon with any con fidence . Moreover, diffraction from feed mounts, if present, does not affect the above conclusion, because the radiation scattered by the feed mounts grazes the boundary of the scan area at a wider angle than the radiation from the edge of the aperture. The conclusion says, essentially, that the planar scan tech-nique does not give information about the fields outside the solid 20 angle formed by the edge of the aperture antenna and the boundary of the scan area. (We use the term "solid angle" loosely since it will not be a solid angle in the strict mathematical sense unless the extension of its sides meet at a single point.) As an example consider a circular reflector antenna of radius "a" scanned in the near-field on a larger concentric circular area of radius a/2. Suppose the scan area were a distance d = a in front of the aper-ture. Then, as shown in figure 6, the above results reveal imme-diately that the data from the near-field scan would not contain reliable information about the fields outside the angle 9 = 22.5°. b m Newell and Crawford reached the same conclusion from ex-perimental data taken on scan planes at different distances in front of the same microwave antenna. It appears from the above analysis that their conclusion is a general result which holds for all elec-trically large aperture radiators. It should be emphasized that the above results were derived for the sum and difference pattern of electrically large aperture an-tennas and for a scan area that extends well beyond the main near-field beam region. Also the main near-field beam has been assumed to be characterized by planes of fairly uniform phase. The above conclusion would not necessarily apply, for instance, to broadbeam horns with dimensions on the order of a wavelength or less, to beams steered nearly to the edge of the scan area and scan areas just covering the main near-field beam, to apertures on a finite ground plane, to defocussed antennas, or to electrically large aperture antennas with a diverging or converging lens placed within or directly in front of the aperture. Fortunately, special situations and classes of antennas such as these can often be analyzed sep-arately within the framework of the preceding analysis and results. The special cases mentioned above are discussed in the following paragraph. Specifically, an analysis similar to the preceding shows that the encircled conclusion applies to the latter three classes of antennas (the defocussed antennas and antennas with a ground plane or lens) , provided the scan area extends well beyond the edge of the antenna, the ground plane or lens, and provided the edge of the ground 21 plane or lens is used as the base perimeter for the solid angle when the edge of the ground plane or lens extends appreciably beyond the edge of the aperture and significant scattering occurs at these edges. Also, it can be shown that the encircled con-clusion applies to broadbeam horns if the center of the horn is taken as the base of the solid angle instead of the perimeter of the horn. (For scan areas much larger than the aperture of the horn, there is little difference in size between these two solid angles.) Similarly, for antennas with their mainbeam steered close to the edge of the scan area it may be more accurate to choose a point nearer the center of the aperture rather than the edge to determine the side of the solid angle near the direction of the main beam. Again it makes little difference for large scan areas. In general, when the scan area is close to the boundary of the main beam, the base of the solid angle outside of which the computed far-field pattern is unreliable tends to shift from the perimeter of the aperture (ground plane or lens) toward the center. Often the in-crease in solid angle is slight, however. In brief, the above encircled conclusion (in its stated or slightly modified forms just explained) applies to a very large variety of antennas, including electrically large aperture antennas operating in a sum or difference pattern (with or without beam steer-ing, defocussing, a finite ground plane, or modifying lens) and broadbeam horns. Next we want to evaluate I of eq. (20) for points within the solid angle formed by the edges of the antenna and scan area. Within this solid angle the integrand contains no critical points of the first kind. Consequently, the p integration can be done by parts to yield ik[D'-p'sin9 cos (-(J> ) ] , 2tt F (p' ,<j) )p' e m p t-wtIJ tm p <v (24) m o [cosy'-sinG cos(d)-cb )] F 3D' where again cosy' has replaced -^—^ , and the primes refer to points on the boundary of the scan area. It should be noted that eq. (24) represents the first term in an asymptotic expansion of eq. (20), 22 (24) remains valid for y 1 right up to 90°. and when cosy 1 gets too small eq. (24) no longer represents a good approximation to eq. (20). We can get an idea of the largest per-missible y' by realizing that the result (24), if valid, must be much smaller in amplitude than the amplitude of the integrand of eq. (20) multiplied by the change in distance p as D changes from D' to D' + A/2. A little mathematics shows that this condition is always m y~Y 7 satisfied if cosy' is greater than about V-p-, where d' is the per-pendicular distance from the edge of the aperture to the scan plane. However, when used to find an upper bound expression for \|l\| (see eq. (28) below) , eq Eq . (24) has been derived under the condition that the fields emanate or at least appear to emanate from points on the edge of the antenna aperture. If the fields are also scattered from the feed mounts, eq. (24) must include these fields as well associated with these mount scattered fields will always be equal to or greater than that of the edge diffracted fields--since the fields scattered by feed mounts make larger angles with the plane of the scan area at its boundary than the edge diffracted fields. Thus, the integration by parts remains valid (as an upper bound ex-pression within the solid angle formed by the edges of the aperture and scan area), and eq . (24) holds even when there exists appreciable radiation scattered from feed mounts in front of the antenna. For angles near the z-axis (sin0 << cosy ) eq . (24) becomes The cosy 1 'm I = \ 2tt 7T1 J F + e tm ikD' m cosy' 'm J p' e ikp' sin9 cos ((f) -(j) ) d (25) Upon taking the amplitude of eq. (25) we find , 2tt ih ! 2tt F. e tm ikD' m cosym p' d(f) . P (26) For scan areas whose boundary is well outside the edge of the aper-ture, cosy' -1. For scan areas with boundaries somewhat close to ' 'm the edge of the aperture, the term (in the summation over m) which corresponds to the minimum D' (maximum y') predominates (assuming f m v. i m ; f to 23 fairly uniform illumination around the edge of the aperture) , so that I m F + e tm ikD' m cosy' 'm cosy m 'max ^ Ft ikD' m m In either case, we see from eq . (19) that iyp',y tm ikD' m cosy' m cosymax where F (p',4> ) is the electric field at the boundary of the scan area Eq . (26) can now be written it-i . A 2tt cosymax 2tt J o E t (p', )\|p' d(j) , (27) or A Lmax 2 cosy max (28) max with \|E c denoting the maximum amplitude of the transverse elec-trie field found on the boundary C of the scan area, and L is the maximum width of the scan area. If the output of the measuring probe at the limits of the scan area is down at least X dB from its maximum output, then eq. (28) may be rewritten A Lmax 10 X_ 20 2 cosy 'to' (29) max where E represents the highest amplitude of the transverse electric tO y g field found on the scan area. ' Equation (29) combines with eqs . (13) and (14) to yield an upper bound expression for the error n in 'For an arbitrary probe (i.e., not necessarily a dipole probe) E-to represents the highest output amplitude of the probe on the scan area and X the largest output amplitude of the probe at the edge of the scan area measured in dB down. ^The maximum electric field on the scan area in the near-field (z << A/A) of an electrically large aperture is very nearly equal to the maximum electric field on the aperture itself. 24 the far-field produced by neglecting the fields outside the finite scan area: __X , T max ., n 2 „-f=\ nCF)<-A-L 10 _gW 9 (30a) E t 2 cosy If ^— da I 1 max ' i E . ' A to o where g(r) is the ratio of the amplitude of the maximum far-electric field to the far-electric-field in the given direction r. In other words, it is the inverse of the normalized far-field pattern. Use has been made of eq. (10) which shows that for sum patterns Aril" I -1/ E\ da I , (30b) o where \|E I is the amplitude of the maximum far-field, and the o r -"00 integration is performed in the near-field over the area A , that part of the beam which has nearly uniform phase. Since we are in the near- field, A - A cos0 , where A is the aperture area. (The factor cos0 accounts for the reduction in effective aperture area o r for beams which are steered off-axis electronically through an angle . ) 6 o Because \|E. I/E < 1, where E is the highest amplitude of elec-1 t ' o — ' o 6 r trie field on the scan area, and E. -E cos0 , we can write too o = a cos0 /A = a/A, (31) J E^- da A c to o where the factor a is greater than its minimum value of 1.0 (for apertures of uniform amplitude and phase) but less than 5 for most tapered aperture distributions found in practice. For example, the tapered distributions found in Table IX of have a maximum a of 4.0. Newell has found that the factor we have called a has not been greater than about 2 A/A for any microwave antenna he has measured. Since the effective area A (see for a definition of effective area) is less than the aperture area A and greater than . 5A for most aperture antennas , the experience of Newell 25 also indicates that a is less than 4 or 5 for nearly all electrically large aperture antennas found in practice. Equation (31) combines with eq. (30a) to give the final expres-sion for n(r) nO) x , T max t A 2 r— -, a A L 10 g(r) 2 A cosymax (32) max max where A = area of the antenna aperture. A = wavelength. = maximum width of the scan area. = maximum acute angle between the plane of the scan area and any line connecting the edges of the aperture and scan area. = the largest amplitude of the probe output at the edge of the scan area, measured in dB down from the maximum amplitude of probe output in the scan plane . = a "taper" factor--equal to a minimum of 1.0 (for apertures of uniform amplitude and phase) and less than 5 for most tapered distributions found in practice. (See eq. (31) for the pre-cise definition of a.) g(r) = ratio of the amplitude of the maximum far-electric-field to the far-electric-field at the given direction r, i.e., the inverse of the normalized far-field pattern. (g(r) = 1 for the center of the main beam, or beams if a difference pattern.) (If desired the errors in the gain function sidelobe level, polariza-tion ratio, and beamwidth may be calculated from eqs . (2) -(5) once n and the far- field pattern is known . ) Equation (32) has been derived for antennas operating in a sum pattern. But since a difference pattern can be divided into two, approximately equal, sum patterns with wavefronts slightly skewed, eq. (32) holds for difference patterns as well. (For a difference pattern one should still use the taper factor a of the constituent sum patterns . ) In summary, eq. (32) applies to either sum or difference pat-terns of all electrically large aperture antennas (including antennas 26 with their boresight direction steered away from the axis perpendicu-lar to the scan area) within a solid angle (sine << cosy ) about the axis perpendicular to the scan area. Again, it has been assumed that the scan area extends well beyond a main near-field beam which is characterized by planes of fairly uniform phase. It can be shown that eq. (32) and eq. (36) below apply even to defocussed antennas, apertures in a finite ground plane, and to antennas with a diverging or converging lens, provided the angle y is chosen in accord with max the second paragraph preceding that of eq . (24). The condition (sine << cosy ) can be made more specific by returning to eqs. v max ^ J t>n (24) and (25). For all practical purposes, eq . (25) follows from eq. (24) if (33) sine < ~- cosy = T sinf 2 ' max 2 max (6 = 90 -y ) v max 'max For example, if y were 45°, condition (33) becomes 6 < 20°, which is a large enough angle to include many side lobes of most microwave antennas (assuming their boresight direction at = 0) . Roughly speaking, eq. (32) represents a valid upper bound within the region < i 2 max As an upper bound, eq. (32) remains valid for y right up to max 90°. However, from the discussion immediately following eq. (24), it is unlikely that eq. (32) would remain small enough to be very useful when (34) where d refers to the minimum perpendicular distance from the mm r r edge of the aperture to the scan plane. The error n given by eq . (32) can be compared with the results of the empirical error analysis performed by Newell and Crawford They took "centerline" data on a near-field scan plane 25 cm in front of a circular, fixed-beam "constrained lens" array, which was 80 cm in diameter (see figure 9). The centerline was 213 cm long. Assuming a rectangularly separable field pattern, they first used the 213 cm centerline data to compute the far-field pattern. Successively, more and more of the 213 cm distance was deleted and 27 the corresponding far-field pattern computed. In that way they could get an idea of the errors involved in scanning on a near-field plane of finite area. The envelope of their near-field centerline amplitude is repro-duced in figure 7a. The on-axis gain change computed by Newell and Crawford by deleting distances from the scan line is reproduced by the dashed line in figure 8a. The solid lines in figure 8a represent the maximum envelope of on-axis gain change calculated from eqs . (32) and (2c) of the present report. The values of X and Lmax (see eq. (32)) were taken from figure 7a. The value of a was estimated from eq. (31) and figure 7a to be about 3. The remaining parameters needed to calculate n from eq. (32) are contained in reference : -if d Y „ = tan t r max T max — a U z J J A = 3.26 cm A = ira 2 a = 40 cm d = 25 cm g(r) 1. Figure 8a confirms the result that the fractional error r\ of eq. (32) represents a reasonable upper bound. In the region max L /2a > 1.7 (y < 42°) the upper-bound error from eq . (32) is max no more than double the error estimated by computer "deconvolut ion" of the near-field data. The upper bound error grows inordinately mo y large, however, for L ' /2a much less than about 1.2 (y > 75°), max as eq. (34) predicts. Figure 8b shows the same comparison as that made in figure 8a but for a 46 cm (18 in) reflector antenna operating at 60 GHz (A = .5 cm). (The envelope of the amplitude for a centerline scan of this antenna is shown in figure 7b. The value of a is about 2.) The dashed line in figure 8b represents the on-axis gain change which Newell computed by deleting distances from the centerline scan length taken 43.18 cm in front of the aperture. The agreement be-tween his computations and the upper bound solid curve calculated from eq. (32) is even closer than for that of the constrained lens 28 H antenna (figure 8a). The upper bound error is no more than double ) which is very the computed error for L /2a > 1.05 (y ov < « max close to the value of y < 84° predicted for the range of useful-1 max r & ness by eq. (34) . It appears from this somewhat limited experimental evidence with centerline data that the simple formula (32) provide s a useful upper bound error at least for scan areas with cosy > 'max It should be noted, however, that centerline data do not accountror changes in phase of the field around the perimeter of the scan area, and thus would predict larger finite scan errors in most cases than the com-plete 2-dimensional scan data. An upper bound expression of smaller magnitude generally than eq. (32) that takes the phase changes into account can be derived by returning to eq. (24). Under the condition of eq. (33), eq. (24) becomes — 2tt cosymax 2tt / E t (p',-<}> ) p 'p' d<\|> (35) Again E (p',(J> ) refers to the transverse electric field, i.e., output of the dipole probe, at the boundary of the scan area. Substitution of eq. (35) into eq. (13) yields the fractional error n (r) in the far-field for 1 max (90 Y ) 'max ; ' nO) ; 2tt -ikp'sinQ cos(-cf) ) J E t (p',cf> )e P p» d 27rr I E (r) I cosy i ^ j i r->oo 'max (36) The far-field pattern r\|E(r)\| in eq. (36) can be approximated AE to r ^°° analytically by — or found by deconvoluting the measured near-aAg(r) _ field data. The field E t (p',(j> ) at the boundary of the scan area z p can be taken from the measured near-field data. Thus, in practice both the numerator and denominator of eq. (36) can be determined straightforwardly. Although eq. (36) involves more computations than eq. (32) even for the on-axis value for which 6=0, it could be com-puted by a simple routine added to the program which deconvolutes the near-field data, since for arbitrary probes E (p',cj) ) can be replaced by the output of the probe (for two orientations, in 29 general) en the perimeter of the scan area. In cases where the antenna pattern is assumed separable in xy coordinates and only centerline data is taken, eq. (36) cannot be applied but eq . (32) still can. 9 In Table 1 are listed the errors calculated from eq. (32) for some of the far-field parameters of a typical X-band antenna operat-ing in a sum and difference pattern. The finite scan errors are proportional to wavelength, so changing the wavelength while holding the other antenna dimension the same merely changes the values in Table 2 proportionately. Of course, such an isolated change is rather unrealistic. B . Position and Instrumentation Errors To determine the radiating fields of an unknown antenna by scanning on a near-field plane with a given probe, the output b ' (P) of the probe must be recorded throughout the scan area A' (see foot-note 3a). In principle, the data points should lie in a plane and the position of the probe should be recorded exactly as it scans from point to point. And, ideally, the instrumentation used to measure the phase and amplitude of b' should do so with perfect accuracy . Obviously, in practice, neither the position of the probe nor the phase and amplitude of the probe output b' can be measured exactly. Regardless of how small the uncertainties in the measure-ment of b'(P) they will introduce errors into the calculated far-field. It is the purpose of this section to derive general expres-sions which estimate the magnitude of the errors in the far-field produced by the inaccuracies in measuring the position and output of the probe in the near-field. Specifically, we want to evaluate AE in the far-field so that the fractional error n(r) of eq. (1) can be determined. (In Section y It is interesting to note that ri in both eqs . (32) and (36) does not approach zero but simply an insignificantly small number as the scan boundary approaches infinity. This limiting value of r\ represents the contribution of the oscillatory part of the Fourier transform men-tioned in footnote 3b. Consequently, once the edges of the scan area reach the region where the fields behave as 1/p', there may be no advantage to scanning on a larger area at least if only the pattern near the boresight direction is required. It is shown in reference that eqs. (32) and (36) slightly modified apply to broadbeam horn antennas as well as electrically large aperture antennas. For these broadbeam antennas the limiting value of n can be significant. ilk II the errors in gain function, sidelobe level, polarization rati, and beamwidth are derived in terms of n(r) and the far-field pat-tern.) In order to simplify the theoretical analysis, the errors will be evaluated as if the probe were a perfect electric dipole. The justification for choosing a perfect dipole in the analysis is similar to that stated in Section III. A. The dipole measures the electric field components at a point. All physically realizable probes respond to a weighted average of the fields near the probe. Thus any small error in position would be expected to change the output of a perfect dipole by as much or more than any other probe. As for instrumentation errors, they remain essentially independent of the particular measurement probe. Also, as in the previous sec-tions, any uncertainty in probe receiving characteristics (S^-, ) will be ignored for this part of the error analysis. Under the above conditions the far-field error in AE may be found with the help of eq. (10): A E .(F) = - ik C0S6 e ik(r " d C0SG) / AE t (P,d)e~ lFR ' P dP. (37) Z 2ttt A' r Y -> oo AE (P,d) is the difference between the actual electric field at the point (P,d) and the measured output of the hypothetical dipole probe (for two orientations in general) at the point (P,d). For errors near rn O -y" the z-axis (0 < 2A/L , see footnote 5) eq . (37) becomes iic iicrr <n --"i^R-P AE(r) = -i£-e 1Kir " aj J AE (P,d) e r dP, (38) 2i\r A r - °° o where A designates that part of the scan area over which the major variations in phase are relatively small. (For the position and instrumentation error analysis the scan plane is assumed to lie nearly parallel to the near-field planes of "uniform" phase, i.e., perpendicular to the center of the main beam of sum patterns or to the null axis of difference patterns.) For scan planes in the very near-field, A is approximately equal to A cosG , i.e., the projected area of the antenna aperture. (Only electrically large aperture 31 antennas are being considered.) 6 is the angle between the perpen-O /\ dicular to the aperture and the perpendicular (e ) to the scan plane. 0=0 for beams which are not steered off-axis electronically. The scan area outside A can be neglected for this part of the o error analysis because the rapidly changing phase in the region out-side A contributes little when integrated to find the far-field and the errors in the far-field near the z-axis. It follows from foot-note 5, or more rigorously from an asymptotic analysis like the one performed in Appendix A for a circular aperture of uniform distribu-tion, that eq. (38) can be used as an upper bound expression for AE in the region given approximately by < 2A/L where L is the maximum breadth of the partial scan area A . This region is large enough to include the first sidelobe maximum of many electrically large aperture radiators. For example, a circular aperture of uniform distribution and radius a -L /2 has its first sidelobe maximum at o an angle -1.7X/L radians. Physically this 2A/Lmax condition says that for most electrically large aperture antennas, the part of the near-field (A ) over which the phase is fairly uniform strongly influences the far-field within the angle 2A/Lmax . Beyond this angle the edge diffracted fields dominate the far-fields to a greater and greater extent until in the far sidelobes the fields are determined essentially by the edge dif-fracted fields alone. To find either the position or instrumentation errors, the integral -i^R-P T = J AE.(P,d) e r dP (39) A z o in eq. (38) must be evaluated. Of course, the integration can be performed only after AE (P,d) is found. In Section 1 below, AE t , T , and thus n (r ) are evaluated for position errors, and in Section 2 for instrumentation errors. The approach that is taken is quite straightforward. For posi-tion errors AE is expanded in a Taylor series about (P,d) assuming the deviation in the position of the scanner is small compared to a wavelength. The Taylor series and the integral (39) into which it 32 is substituted divides naturally into a longitudinal or z-position part and a transverse or xy-position part. The upper-bound for each integral and thus for the z-and xy-position errors are then determined as a function of the measured near-field data and the computed far-field pattern. For instrumentation errors, AFT. is expressed in terms of the amplitude and phase errors introduced by the nonlinearities in the receivers which measure these quantities as the probe traverses the scan area. Mathematically, the integrals involving the receiver phase errors are handled in the same way as z-position errors, and thus the final upper-bound expressions have the same form. The integrals involving the amplitude errors have their upper-bound determined by characterizing the receiver nonlinearity in measuring amplitude in dB of error per dB down from the maximum amplitude of the probe output on the scan area. 1. Position Errors Consider the scanner which moves the probe throughout the near-field scan area. Typically the scanner covers the area by travers-ing a grid or raster of lines while the probe output is recorded at given points along each line. Ideally, all the scan lines lie per-fectly straight and parallel in a single plane, and the position of the data points along each line is recorded exactly. In reality, of course, none of these idealizations hold, basically because the scan lines will never be perfectly straight and the data must always be recorded over an interval rather than at a point. Regardless of the reason for the position errors, they all effect the near-field data simply by positioning the probe at points (P+AP, d+Az) rather than (P,d) . In other words, the difference AE (P,d) in eq. (39) can be written for position errors as AE t (P,d) = E t (P+AP,d+Az) -E t (P,d), (40) where, as usual, E (P~,z) is the electric field at the point (P,z) in the near-field. In general, AP~ and Az, which will be referred to as displacement errors, are functions of the transverse position P. 33 Before continuing with the analysis it should be pointed out that displacement errors caused by a small initial translation of the entire scanner with respect to the test antenna will not cause a change in the computed far-field amplitude, as eqs . (6) and (8) indicate. Also, an initial rotation of the entire scanner through a small angle will have no effect on the far-field pattern other than to rotate the entire pattern through that same angle. Since the displacement errors must be much smaller in magnitude than a wavelength, the right hand side of eq . (40) can be expanded in a three dimensional Taylor series. By letting Ar = AP+Aze and keeping only the first two terms in the series, eq. (40) becomes AE t (P,d) = Ar-VE t (P,d) + j Ar • [Ar «VVE (P,d) ] (41) where VVE^ = (WE )e + (WE )e t v x^ x v y y and VE\ = (VE )e + (VE )e . t v x J x y y Substitution of eq. (41) into eq. (39) yields a useful expression for I , 1 -i-R-P = / Ar- [VE. +yAr-WE. ] e r dP. A z L z o (42) Again it is emphasized that Ar is, in general, a function of P, the transverse coordinates over which eq . (42) is integrated. Also, as we shall see shortly, it is necessary to retain the second term in the integrand of eq. (42) when on-axis errors for sum patterns are considered. As a check on eq. (42) let R = (on-axis) and Ar be constant over A so that Ar can be taken outside the integral. The o terms containing the transverse part of the gradient operator con-vert to line integrals around the boundary of A . These line inte-grals must equal zero because in effect eq. (42) assumes negligible fields outside A . Only the "z" derivatives, which can also be taken outside the integral, are left, and eq. (42) may be written 34 ^B Az \|-+ Az 2 \|-r 8z 3z / E dP. A z o With the aid of eq. (10) , eq. (43) converts to (43) I = -(Az+ikAz 2 )2^re lk ( r ~ d ) E^ (r) o J t K J X" -> oo (44) Equations (38), (39), (42) and (44) combine to show that E(r)+AE(r) T-X30 1 + C^Az) 8 E(r) f -»-00 i.e., the error in the far field amplitude calculated from eq. (42) I A -p I is of higher order than (-U-<-) 2 , i.e., it is negligible when Ar is constant --a result which must hold if eq. (42) is a valid expres-sion for I , because, as mentioned above, a translation of the entire scanner has a negligible effect on the far-field pattern. Equation (42) also checks in a similar way for R f 0, but the proof is more involved. In order to evaluate T of eq. (42) exactly, both Ar and ET (F) in the near- field would have to be known. Fortunately, an upper bound approximation can be found for T without detailed information about Ar or E (r) . Consider one component, say E , of the integrand of eq . (42). For an electrically large aperture antenna, we can express E within .A. the area A as o E (P,z) (1+AA (P,z))E (P) e i(kz + cJ)ox +Act> x (P,z)) (45) where d> is a real arbitrary phase constant, and AA , Ad> and E r ox J r x' r x ox are real functions of the indicated near-field coordinates. The functions AA (P,z) and A<J) (P) typically oscillate over a distance equal to or greater than a wavelength, each with a magnitude usually much less than one. E (P) is the smoothly tapered amplitude func-tion. In other words, eq. (45) simply states that the near-field across the area A of an electrically large aperture antenna has a phase equal essentially to kz and a smoothly tapered amplitude except for small variations which oscillate over distances equal to or 35 greater than a wavelength. (Recall that for the position and instru-mentation error analysis we are assuming that the perpendicular to the scan area is approximately aligned with the boresight direction of the antenna.) In Appendix A eq. (45) is verified for a circular aperture of radius "a" and uniform aperture distribution (E = E ) . In that 1 ^—r— ox case AA and A<\|> are on the order of — /A/a or less. Rusch and Potter report similar results for the circular aperture . Equa-tion (45) has been verified experimentally by the substantial near-field scan and extrapolation data taken at the NBS . The measured amplitude and phase on a near- field plane of a typical microwave antenna are plotted in figures 9a,b,d. This particular scan was taken 25 cm in front of the circular "constrained lens array" (see figure 9c) of radius 40 cm operating at 9.2 GHz (A = 3.26 cm). This antenna is the same one described in the paragraph following eq. (34) and whose centerline amplitude envelope is plotted in figure 7a. When eq. (45) is substituted into the integrand of eq . (42) and all terms higher than either second order in \|Ar\|/A or first order in \|Ar\|//A~ are discarded, the following expression for the x component of I remains I -/ ox [Ar-V\|E x 2ttAz X ox + i '2ttAz' A I j . i. -i-R-P r where E i4>. dP, (46) x E e x or from eq . (45) \|E I = fl + AA )E 1 X ' v x ; ox d> = kd + + A . yx Y ox r x It has been assumed that the transverse (P) and longitudinal (z) displacement errors are of the same order of magnitude. Eq. (46) can i i(kd + (j)ox ) be simplified further by writing e as e (cosAcfj^+i sinA<\|> x ) , to put I in the form r ox 36 i(kd+ + i sin A<\|> )e r dP . (47) (6 = 2-ttAz/A) First consider on-axis errors for antennas operating in a sum mode, that is ^--<< 1 or 6 less than about A/ (10 Lmax ) . Then the exponential in eq. (47) can be approximated by unity. The arbitrary phase constant i> can be chosen so that Ad> varies about zero r ox Tx throughout the area A . Furthermore, since the variations in Ad> te o T x are small, oscillate many times across the area A , and remain com-pletely independent of the variations in the displacement errors Ar of the scanner, to a high probability, c\|> can be chosen such that the integration in eq. (47) which is multiplied by sinAcj) can be neglected compared to the maximum possible value of the cosAcf> inte-gration. In addition, since E cosAc}> remains positive throughout A for a sum pattern, the reference plane (z=d) for 6 can be chosen to make J 6E cosAcJ) dP = 0. (48) ; ox Yx • J o Thus, under the above conditions the iS term of eq . (47) is elimi-nated entirely and eq. (47) simplifies to i(kd + ) n I = e ox j ( Ap . v E _ i 6 2 E ^ cos A(J) dp> ox i v t ox 2 ox ; x o (A = A +e \|-) • t Z dZ J (49) E in the first term of eq . (49) has replaced \|E I =(1+AA ) E of ox n v J r ' x 1 v x^ ox eq. (47) because the many oscillations of AA would also (to a high probability) eliminate upon integration all but higher order con-tributions from the V(AA ) term. v x' It is interesting that eq. (48) expresses the same condition chosen by Ruze in his well-known work on "antenna tolerance theory," and under the above assumptions the z-position errors always lower the on-axis gain value of the sum pattern. Of course, in his work 6 represented a small "arbitrary 37 phase error or aberration" (what we call A ) rather than the dis-placement errors of a near- field scanner. It would be desirable that the reference planes of Ax and Ay (AP = Axe +Aye ) be chosen such that / AP«V E cos A<\|> dP" became zero also. Unfortunately, such a A Z 0X X 9E choice is impossible (in general) because neither derivative, ox 8E 9x or ox , stay the same sign throughout A . From eqs . (38) and (39) it is seen that the fractional error n of eq. (1) can be written as n(r) Ar \|E(r) ox + \|I \| 2 1 oy X->oo (50) The amplitude of I is found from eq . (49) to be \|I \| = \|/ (AP-V.E - \ 6 2 E )cos Ac}) dP 1 OX ' A t OX 2 OX ; X o < AP X6 / \|V^E I dP + \ 6 2 max i ' t ox ' 2 max i ox A / E_ dP, Ao (51) where AP and max 2tt max are the maximum magnitudes of the transverse (P) and longitudinal (z) displacement errors, respectively, intro-duced by the near- field scanner. The last integral in eq. (51) can be related to the maximum far-field \|E I , i.e. from eq . (10) or (30b) xo r--°o 7 / E dP = ArlE I ^ ox ' xo ' r^-°° o (52) The remaining integral, J A V^E dP, t ox ' ' (53) is a bit more troublesome. However, it can also be estimated by expressing the integral in polar coordinates (p,) as follows: / \|V.E I dP = f A t °X A o o 3E ox 3P -, 8E 1 ox ^ 2 P 94) pdpdcj) (54) 38 If we assume that the amplitude of the aperture distribution tapers with much greater slope in the radial direction (p) than in the azi-muthal direction () , the second term under the radical sign of eq . (54) can be neglected, leaving 9E _ f ox Ao 3P pdpdcj) . (55a) 9E The negative sign is present in eq . (55a) because ox 3p is negative for amplitude distributions which taper toward the boundary of A . (We are assuming p = at the maximum of E .) Integration of eq. O -A. (55a) by parts with respect to p yields 8E •J A ox 3p where E and L xo o o max pdpdcj) = J E dpdc}> A ox o tt_ r T max 2-C 1j xo o (55b) denote the maximum amplitude of E and the maxi-ox mum breadth of the partial scan area A , respectively. With the aid of eq. (55b), eq . (54) becomes / \|V F \| dP < £ Lmax E { ' t ox 1 — 2 o xo o (56) Since the area A is in the very near- field of the aperture antenna, the maximum amplitude E on A is approximately equal to the maxi-mum amplitude on the aperture itself. maximum far-field by (see eqs . (30b) and (31)) Thus E is related to the xo aAr Exo ' r-xo A and eq. (56) becomes (57) J \|V^E I dP < { ' t ox 1 — o , T max i ~ i TTaArL E o ' xo ' r-»°° 2 A (58) Substitution of eqs. (52) and (58) into eq. (51) yields r . „ . max TraAP n L I I < ArlE I OX ' — ' xo ' ]f->OQ TTaAP 1/ max o 2 A + i 6 2 2 max (59) 39 A corresponding inequality holds for ll I with IE re-r ? n 7 ' oy ' ' yo ' r->°° placing \|E \| ^oo . At first sight, one may raise the objection that the reference plane for <5 cannot necessarily be chosen so that both eq. (48) and the corresponding equation with E are satisfied simultaneously. It should be noted, however, that the two equations need not be satisfied simultaneously. It is only necessary that one reference plane can be found that allows eq . (48) to hold, and that a second reference plane can be found that allows the corresponding equation with E to hold. If the two reference planes are a slight distance apart, the relative phase of the x and y components of the far-electric-field will be in slight error but not the Hermitian amplitude of the far-electric-field, i.e., the far-field pattern will remain unchanged even though the far-field polarization ratio will be shifted slightly. Eq. (59) and the corresponding equation for \|l \| combine with eq. (50) to give an expression for the maximum position error n in the far-field: n(r) An T max TraAP L -. max o 1 x-2 + -7T 2 A 2 maX A J 10Lmax o (sum patterns) (60a) Equation (60a) was derived assuming the antenna was operating in a sum pattern and does not apply to difference patterns since eq. (48) may not be satisfiable for difference patterns. However, in Appendix B it is shown that an error expression similar to eq . (60a) may be derived near the boresight direction (null axis) of difference patterns as well. Specifically, n(r) An T max TraAP L max o 2 A + 4AF6 max g(r), 6 < J 10Lmax where AF and g(r) are defined below after eq. (61) Finally for the angular region > A/10 Lmax o (difference pattern) (60b) but still less than 2A/L (see eqs . (37) and (38)), we can derive an upper bound expression similar to eq. (60a). Specifically, the magnitude of I Qx in eq. (47) may be written 40 J<\| \|Ar-V\|E \|\| dP + \ j \|6E \| dP. A X A 0X o o ox The term second order in <5 has been ignored since now the first order term in 6 is the major contributor to the z-displacement error. When we carry the analysis through in a way similar to that described between eqs . (51) to (59), the following expression for the frac-tional error for both sum and difference patterns results : n(r) < An T max ^ TTO.AP L max o o g(r) 2 A » max 1QLmax 2A max o (60c) Although eq. (60c) was derived for 9 < 2A/L , it remains valid as an upper-bound expression all the way to = tt/2. This result follows from the fact, which will not be proven here, that the con-tribution from the integration in eq. (37) outside the area A is negligible compared to the upper-bound contribution obtained from inside A and expressed by eq. (60c). (It is interesting to note that the maximum possible errors in even the far sidelobes are determined by the displacement errors across A , whereas the field itself in the far sidelobe region is determined by the near-field outside A .) o ' Thus, if we combine eqs. (60a), (60b), and (60c); insert the approximations, L A 2ttAPmax max o Imax A (£max ) 2 into eqs. (60); and let max A far-field hemisphere emerges an upper-bound expression for r\ valid over the entire n(r) < A + n max max l z 2£ 6 2 max g(r) 2 ^z =( 8AF 5 max max (sum patterns) (difference patterns) (sum and difference patterns) 10£max 10£max <e< (61) 41 where X = wavelength „max = maximum width of the antenna aperture A™a Y = 2lTAP rr1 o V / A » where AP is the maximum amplitude of the trans-IlLdA IlLdA IIldA verse displacement errors within the partial scan area A . o (A is that part of the scan area over which the phase is fairly uniform. For near-field scans parallel to the aper-ture A -A, the aperture area.) 6 m o V = Z^^z /\ , where Az m „ v is the maximum amplitude of the IIldA IlLdA ITldX longitudinal displacement errors within the partial scan area A . o AF = fractional difference between the amplitude of the two main far-field lobes of the difference pattern (see Appendix B) a = a "taper" factor --equal to a minimum of 1.0 (for apertures of uniform amplitude and phase) and less than 5 for most tapered distributions found in practice. (See eq . (31) for the precise definition of a; for a difference pattern one should still use the taper factor of the constituent sum patterns . ) g(r) = ratio of the amplitude of the maximum far-electric-field to the far- electric- field at the given direction r, i.e., the inverse of the normalized far-field pattern. (g(r) = 1 for the center of the main beam, or beams if a difference pattern.) (If desired, the errors in the gain function, sidelobe level, polari-zation ratio, and beamwidth may be calculated from eqs . (2) -(5) once n and the far-field pattern are known.) The expression (61) represents an upper bound to the far-field error n caused by inaccuracies in the position of the near-field scanner. It applies to both sum and difference patterns in the entire forward hemisphere of all electrically large aperture antennas. Equation (61) was derived under the assumption that the scan plane is parallel or nearly parallel to the near-field planes of "uniform" phase, i.e., the plane perpendicular to the electrical boresight direction. In addition, eq. (61) holds for arbitrary (random as well as systematic) errors in the positioning of the scanner, since only the 42 maximum magnitude of displacement errors A and 6 are required to r max max l to evaluate eq. (61). However, it proves useful to derive an ex-pression similar to eq. (61) which separates the effects of the systematic and random errors in position. To do this and also clarify what is meant by systematic and random errors, consider the motion of the scanner as it takes mea-surements along lines in the near-field plane. As the scanner moves along each line it will deviate from a perfectly straight -line by a gently varying curve that will contain at most a few oscillations from one end of the scan line to the other. These curves, which may also change gently from scan line to scan line, represent the sys-tematic errors in position of the scanner. For example, a slight warp or deformation of the scanner frame would create a systematic error . Superimposed upon the systematic deviations from the straight-line would be position errors which changed randomly (within limits) from measurement point to measurement point. These random errors in position have zero or nearly zero mean and could result, for example, from vibrations of the entire scanner or from a slight play in the drive mechanisms. The maximum magnitude of the random errors are often smaller than that of the systematic errors. However, it is possible that the scanner is aligned so precisely, that essentially all but the random errors are eliminated. Return to eq. (47) and separate AP and 6 into systematic and random displacement errors, i.e., AP = AP S + AP rn (63a) 5 = 5 S + 6 rn . ( 63b ) —rn All the integrals in eq. (47) which contain linear terms in AP or tji rn rn <5 can be dropped. Since AP "" and 6 change randomly from measure-ment point to measurement point (over distances less than A/2) , the integrals actually summations --see footnote 3a) containing these linear terms will be extremely small compared to the largest possible errors produced by the remaining integrals (summations). Thus, sub-stituting eqs . (63) into eq. (47) and proceeding as we did before with eq. (47), yields the desired expression for n separated into 43 systematic and random errors, n(r) aX . s 21max max s ,rn g(r) 2 (6 ) 2 +(6 in ) 2 (sum patterns) max' v max' r ' i S ' rn =< 8AF 6 : max max (difference patterns) (sum and difference patterns) 6 < X 1(Umax A — <e< 10JImax (64) All symbols have the same definitions found after eq. (61). The superscripts "s" and "rn" refer to "systematic" and "random" errors respectively Note from eq. (64) that to the given order of approximation the longitudinal random errors (6 ) , but not the transverse random errors (A ), cause an error in the on-axis far-field of sum pat-r n terns (provided, of course, that A is of the same order of magni-rn tude or less than 6 ). In addition, eqs . (64) and (61) above show that the maximum possible transverse (xy) position errors do not depend upon wavelength for a given g(r) since A behaves as 1/X. max Also note that in expressions (64) and (61) above, the z-position ma v error is not continuous across the angle 6 = X/lOi . There lies no contradiction in this fact since the expressions remain valid as inequalities. The jump between the two regions merely indicates that the upper-bound was determined by a different method in each region. Obviously expressions (61) and (64) do not represent a rn o -y-least upper- bound throughout the region 9 > X/KU . This region will be discussed further at the end of the section. One can get an idea of the magnitude of the position errors by plotting the on-axis gain from eqs. (64) and (2c) for the fixed-beam constrained lens array described in the paragraph following eq. (34) and shown in figure 9. In that case, A/£max = .04 g(r) = 1, 44 and with a equal to 3, eq. (64) combines with eq. (2c) to give s nf < ± 8.7 A" + J-(^ S ) max 2 K max ; 2^ max ; (65) I£ we assume 6 written rn max = and let 6=6' o max = A" max eq. (65) may be dB < ±i 7 6 o (.03 + f « ). (66) dB particular antenna, 6 must be less than about r ' o The solid curve in figure 10 depicts the maximum value of r\ r in u eq. (66) versus 6 , i.e. when the random errors are negligible. The o dB s s dashed line represents the maximum error r\ n when the A = 6 = r G max max rn and 6=6 , i.e., when the systematic errors are negligible, o max' ' J & & To insure that the on-axis gain is less than .01 dB for this 027. That is, each component of systematic displacement error of the scanner must be less than .027 A/2tt, or less than about .004A. For this constrained lens antenna A = 3.26 cm, so the errors in each component of dis-placement must be less than about .14 mm for better than .01 dB accuracy in the on-axis far-field. For random errors only (the dashed line) the z -displacement errors must be less than about .26 mm for .01 dB accuracy in the on-axis far-field for this particular wavelength. As a matter of fact, when the systematic errors are negligible, eq. (64) shows that dB the "random" error in gain x\ n for the sum-pattern main-beam of an & G,rn r arbitrary (electrically large aperture) antenna can be written simply as dB < + 8^ f6 rn . 2 G,rn — 2 ^ max-' (6 rn = 2TiAz rn /A) • max max } (67) dB rn rn For x\ n < ± .01, 6 must be less than about .05 radians or Az G,rn — ' max max less than .008 of a wavelength (3 degrees). That is, the random errors in z-position should be no greater than about ± .01A to insure a .01 dB accuracy in the gain of the main beam. Equation (67) re-veals that the random error in the on-axis far-field gain of sum patterns increases as the square of the random error in the Actually, as mentioned above, the z-position or phase errors always lower (under the stated assumptions) the on-axis gain of sum patterns, but the + sign will be included throughout to keep the same form to all the error expressions. 45 z-position of the near-field scanner. For example, a position error of ± .025A -10° (@ 3.125 x .008A) leads to a maximum r^ B of about 2 G,rn ± . 1 dB (@ (3.125) x .01) for the center of the main beam. To date, little experimental data are available with which to compare the results of eqs . (61), (64) or (67). However, a compari-son can be made between eq . (67) and the computations performed by Rodgrigue, Joy, and Burns . They introduced errors into a hypothetical near-field distribution in order to compute the effects of the errors on the far-field. The results of their computations for the effect of random phase errors (or equivalently z-position errors) on on-axis gain are plotted in figure 3-14 or A-21 of , and are reproduced here by the dashed line in figure 11 below. The solid line is n^ plotted from eq . (67). One can see from figure G,rn r n v J & 11 that the two curves are in close agreement. The solid line lies slightly above the computed dashed line as indeed it should if eq (67) represents an upper bound. Figure 12, which will be explained in greater detail below under instrumentation errors, shows a comparison between the effects on the on-axis gain of a sum pattern from systematic (quadratic in this case) phase errors introduced into actual near-field data by Newell [17,21] and the corresponding upper-bound results calculated from eq. (64). Again agreement is close, with the upper-bound curve lying just above the actual curve for small deviations in phase Next, let's calculate the effect of systematic z-displacement errors on the depth of the null for an antenna operaing in a dif-ference pattern. From eq . (64) tv B < ± 35 AF 6 s g(r) , 'G,s — max & • J ' where g ( r ) is no longer equal to unity because r is in the direction of the boresight null. Typically, AF is about .01 and the depth of the null is 25 dB down from the main beams of the difference pat-tern. Thus g(7) ~-10 5/4 -18, and n„ < ±b . 3 o l G,s — max 46 rn for the null depth. For example, if Az = .01 A. (which would cor-max respond to an accuracy of about .27 dB for the main beams) eq. (68) insures that the null depth would be accurate to within .40 dB . This surprisingly high insensitivity of the null depth to z-displacement errors (or, equivalently , phase errors) has been ob-served by Newell upon introducing phase errors into the measured near-field data of a number of antennas operating in the difference mode. To understand the reasons for this high accuracy in null depth, one must refer to the derivation in Appendix B. There appears to be no simple way to explain this result heuristically . It is also found in Appendix B that the maximum shift 6 , . r . ri^ shift in the direction of the far-field null of a difference pattern caused by z-position errors is given by the simple expression which is not a function of wavelength: 4Az'max 2X shift Imax tt£max max radians . (68) In general, the shift in the null caused by all other sources of error are negligible compared to this shift caused by the z-position error or, equivalently, the phase errors. Table 2 lists the upper-bound position errors in a number of far-field parameters for a typical X-band and K-band antenna. The values in the table were calculated from expressions (61) and (68) . As eqs . (61) and (64) show, the maximum possible transverse (xy) position errors do not depend upon frequency for a given g(r). Table 2 also shows quite dramatically that the z-position or phase errors everywhere except in the boresight direction can be extremely large compared to all other representative sources of error --com-pare Tables 1-4. (This is also true of the shift in the difference pattern null.) Especially note that the error in a -25 dB sidelobe can be several dB for phase errors (2ttAz /A) of just a few degrees max Whether or not these maximum possible far-field errors are actually experienced in practice depend strongly upon the shape of the near-field z-position or phase error throughout the scan area. For example, we shall find in the next section that receiver phase dis-tortion usually has a functional dependence which introduces 47 negligible errors into the sidelobe fields. It is important to know exactly what effect various distributions of near-field phase errors have on the far-field in order to avoid experimentally, if possible, the distributions which produce large far-field errors in the directions of interest. Such a detailed study of the depen-dence of the far-field errors on the functional form of the near-field z-position or phase errors will not be included as part of this report but will be contained in a forthcoming report by Newell 2. Instrumentation Errors The amplitude and phase of the probe output are measured at discrete points as the probe moves back and forth across the near-field scan area. The receivers are capable of sampling and recording the amplitude and phase to within a certain accuracy only. The errors in the near-field data, caused by the inaccuracies in the receivers or instrumentation used to measure the probe output, pro-duce errors in the computed far-field. This section estimates the far-field errors under given limits of accuracy of the instrumenta-tion which measures the amplitude and phase of the probe output. It is emphasized that the errors produced by the imperfect positioning of the scanner were determined in the previous section and are not considered as part of the instrumentation error analysis of this section. Also the instrumentation errors associated with convert-ing analogue to digital information is assumed negligible. Under the conditions explained in Section III.B, the far-field errors can be found in the region e < x/(icumax ) by evaluating the integral (39) I = / AE.(P,d) dP. ° A z o Here, AE (P,d) represents the difference between the measured (E t ) and "actual" (E ) output of the probe at the point (P,d), i.e. AE (P,d) = E™eas (P,d) -E t (P,d). (69) 48 If we look at just the x -component first, the integral I and eq. (69) combine to give I ox = / (E^eaS -Ex ) dP. (70) o By writing and Ex = \|E \|-e X (71a) i(cf> +ACJ) 1 ) Emeas = M E i + AA ± ]e x x (71b) X L ' X ' X J ' K J where AA and Acf> are the errors in amplitude and phase respectively -A. -A. introduced by the measuring instrumentation, eq. (70) becomes I« v = / [Cl'E l+AA^e X -\|E \|]e x dP. (72) ox i L v ' x ' x' ' X ' J • J o I I In general, both AA and Ac)) are functions of the transverse coordi-.A. -A. -p ' nates ¥. For small errors A<J) << 1, so that yC x « 1 + iAcf) 1 -x— Y x 2 and I ox = / [AA + i(\|E x \| + AA^)A^ - -^(A^Me X dP. (73) o All terms higher than second order have been neglected in eq . (73). icj> x If a sum pattern is assumed and the exponential e is written as in eq. (47) , i x i(kd+<\|> ) e = e (cos A + i sinA(j) ), -A. A with the arbitrary phase constant chosen as in eq. (47), then to OX a high probability, the integration in eq. (73) which is multiplied by the oscillating quantity sinA can be neglected compared to the -A. 49 maximum possible value of the cosA x = 0, (74) o which again expresses essentially the same condition as that chosen by Ruze in his work on antenna tolerance theory. Under the above conditions, the imaginary term within the brackets of eq. (73) is eliminated and the magnitude of I reduces to to ox T l E I T \|I 0X I = 1/ [AAx - ^^(A4> x ) 2 ]cosAcJ) x dP\| (75) 1{ Kl d¥+ iKmax^ 2 I l E xl d¥ -o ' o Acf> ov is the maximum value (in absolute value) of the instrumentation xmax -. n phase error within the partial scan area A . Since it can be shown that (see eq . (10), (30b) or (52)) / \|E I dP = ArlE I (76) o where \|E I is the magnitude of the maximum x-component of far-xo r^"°° field, I I I may be rewritten ' ' ox ' ; Because the errors in measuring phase are usually greatest at points in the scan area where the amplitude is least, it is desirable to choose the partial scan area A as small as possible when estimating Acj) 1 . The far-field within <2A/£max is hardly effected by the T xmax ' ' J near-field outside that part of the scan area where the amplitude of electric field is equal to the edge taper down from the maximum amplitude. Thus for the sake of designating maximum errors in phase measurement on a near-field scan plane, the area A need be chosen no larger than this effective "edge taper" scan area. 50 I I < J I AA 1 t dP + tCAcf) 1 ) 2 Ar\|E \| ox ' — i ' x 1 2^ Y xmax ; ' xo o p->oo (77) It is usually possible to express the errors in amplitude in dB per dB change of amplitude. That is, the receivers are assumed to read the correct (zero error) amplitude of the probe output at its maximum value point on the area A . At all other points the ampli-tude lies a certain number of dB down from its maximum value. Typically the dB errors in measuring the amplitude are linearly related to this number of dB down from the maximum amplitude, and thus the errors can be expressed in dB per dB down. Even if the actual amplitude error curve is not linear, for the sake of the upper bound expression (77), it can be replaced by a straight line (linear curve) which is equal to or greater than the actual error curve . Specifically, if N, R designates the amplitude error in the number of dB per dB down, and A JT) the amplitude in dB down from the j Old maximum amplitude, AA can be expressed as (for small errors) AA I "x x NdB W 8 - 7 (78) By definition AdB 20 log xo xo IE X (79) where E denotes the maximum of E on the scan area xo ' x ' tion of eq. (79) into eq. (78) and the result into eq. Substitu-(77) yields ox < N I dB / (E -IE A xo ) dP + =-(Acj) ) 2 Ar\|E I J 2 V Y xmax ; ' xo ' r-^°° (80a) With the help of eqs . (57) and (76), eq . (80a) becomes 'aA I < ox ' — NdB A -1 + ^(Acf) 1 ) 2 ^ Txmax ; Ar E xo ' r--°° (80b) By combining eq. (80b) and the corresponding equation for \|l \| with o eq. (50) , assuming E _.-£, and approximating —r- by 1, the upper-bound expression for the fractional far-field error n caused by the 51 instrumentation is obtained 11 n(r) < [Nj B (a-l) ^(A<ax ) 2 ] 9 <^max (sum patterns) . (81) Equation (81), which is analagous to the position error equation max (60a), holds only for sum patterns within an angle A/(10£ ) of the boresight direction. For difference patterns as well, and for > A/(10£ ), the following upper bound expression analagous to the position error equation (61) applies: n(r) < [2N^ B (3 : ) + n] ^^ (A(J> ) where dB 8AF A(J) A<(> max r = < max (a-D 2/g(r) (a-l)/2 (sum patterns) (difference patterns) (sum and difference patterns) (sum patterns) e (difference patterns) A 10£max 10£ A max <e< 10£max A (sum and difference patterns) aALmax 101 10A max <e< 2A max 3A <e< 1 V. Imax (82) = wavelength. = the maximum instrumentation errors involved in measuring the amplitude of the probe output --N,„ is expressed in error per dB amplitude down from the maximum amplitude on the scan area. (For the present purposes, the amplitude error is designated as zero at the maximum amplitude.) Hwhen the maximum value of the probe output on the scan area for the x-orientation is very different from that of the y-orientation (E / E ) , it can be shown that eq . (81) remains valid as an upper yo xo n v bound for the errors in magnitude of the far-field. But for errors T in polarization ratio an extra term, N , R \| x \| /8 . 7, must be added to eq. (81), where xp is the difference between the maximum probe out-puts measured in dB for the two orientations. 52 Ad) = the maximum instrumentation errors (expressed in radians) Ymax r J involved in measuring the phase of the probe output on the effective scan area A (see footnotes 10 and 13) . o J -fractional difference between the amplitude of the two main far-field lobes of the difference pattern (see Appendix B) . 'taper" factor --equal to a minimum of 1.0 (for apertures AF a = a g(r) max max of uniform amplitude and phase) and less than 5 for most tapered distributions found in practice. (See eq . (31) for the precise definition of a; for a difference pattern one should still use the taper factor of the constituent sum patterns . ) = ratio of the amplitude of the maximum far-electric-field to the far- electric-field at the given direction r, i.e., the inverse of the normalized far-field pattern. (g(r) = 1 for the center of the main beam, or beams if a difference pattern.) = maximum width of the antenna aperture. = maximum width of scan area. = area of antenna aperture. The derivation of r] is identical to that done for z-position errors z I X 2 X in Section III.B.l. The derivation of 3 for <0< is io£max £max ax accomplished by the same procedure used above for 9 < X/(10£m ). In the far sidelobe region, 9 > 10A/&max the far-field errors be-come approximately equal to the corresponding errors in the near-field amplitude, and after using eqs. (30b), (31), and (79) 3 in this region can be written as shown in eq. (82) . Between 9 equal to may rn o "Y" 2X/£ and lOX/il the value of 3 can be estimated by connecting rnn v rn o "V" a straight line from its value at 2A/£ to its value at 10A/JI 3 near the boresight direction < X/(10£ ) of difference patterns is derived in Appendix B. (Note that the error factor 3 is generally much smaller near the boresight direction or null axis of difference patterns than near the center of the main beam of sum patterns. This result occurs, as Appendix B shows, because the instrumentation dis-torts the amplitude on the "positive" and "negative" sides of a dif-ference pattern by approximately the same amount.) Appendix B also shows that we can write an upper-bound expression for the null shift of difference patterns caused by the instrumentation errors: 53 2Ad) X Ymax , . u . £ . < radians. shift — n max (83) Equation (83) is identical to eq . (68) with the instrumentation phase error replacing the z-position error. Equation (82) represents an upper bound to the far-field errors produced by the instrumentation which measures the amplitude and phase of the probe output. It applies to either sum or difference patterns in the forward hemisphere of all electrically large aper-ture antennas. As with eq. (61), eq . (82) was derived for scan planes which are parallel or nearly parallel to the plane perpendicular to the electrical boresight direction. (If desired, the errors in the gain function, sidelobe level, polarization ratio, and beamwidth may be calculated from eqs . (2) -(5) once n and the far-field pattern are known . ) A comparison of eq. (82) with eq . (61) shows that the instrumen-tation phase error Ad) in eq . (82) has taken the place of 6 eq in max l A max (61). This result acts as a check on eqs. (61) and (82) because 6 simply represents a phase error caused by a z-displacement max -i -j -p error in the position of the scanner. Also, Ad) can be separated r max r into a random and systematic part to get a result analagous to eq (64). However, the random phase errors introduced by the instrumen-tation are usually much smaller than the systematic phase errors, and thus can usually be neglected. Figure 11 reveals that the phase (Acf> ) part of eq. (82) for max the on-axis gain of a sum pattern is in good agreement (as an upper bound) with the computer error analysis performed by Rodrigue, Joy and Burns on a hypothetical near-field distribution with random phase errors. Newell [17,21] has introduced phase errors which are quadratic with respect to the xy coordinates, into the actual near-field data of the 60 GHz, 46 cm (18 inch) reflector antenna whose 12 Note that the "phase error terms" in eqs. (61) and (82) for sum patterns and < A/(10£max ) depend on the square of 8 and A6 JIlci.A. IIld-A. respectively. Thus, these two squared terms should not be added directly when estimating the total error in the far-field. Instead, 6 should be added to Acf) 1 before the square is taken. max max 54 near-field amplitude envelope is shown in figure 7b. The resulting JD errors in r\ r for on-axis gain computed by Newell are shown with the dashed line in figure 12. The solid line represents the maximum phase error plotted from eqs . (82) and (2c). Again it appears that the expression (82) represents a useful upper bound estimate for the instrumentation phase errors in on-axis gain, especially when the measured phase is accurate to within a few degrees across the effective scan area A (as is usually the case in practice). The accuracy with which the receiver can measure the phase of the probe output is related to the amplitude of the probe output. Specifically, the smaller the amplitude the larger the phase errors usually become (see footnote 10). For example, a typical receiver used at the NBS near-field range measures phase to within ± .001 radians (± .05°) at the maximum amplitude on the near-field scan plane, and ± .01 radians (.5°) at an amplitude 20 dB down from the maximum. Thus, for an edge taper 20 dB down, AcJ) =.01, and eq . (82) max shows that the error in the main beam of a sum pattern caused by the errors in measuring the phase of the probe output are negligible (r^ = Z^L (-01) 2 < ± .001 dB) . The same is true for the null depth and shift of difference patterns. Of course, the errors in sidelobe level could be affected to a greater extent by the phase errors, depending on the shape and distribution of the phase errors across the scan area. However, since receiver phase errors usually increase monotonically with decreasing amplitude, it can be shown as a consequence that the upper-bound off-axis or sidelobe phase errors (A<j) . /2 g(r)) given in eq. (82) represents a much larger error in far-field than would usually occur in practice. Thus , in general , receiver phase errors have a relatively small effect over the entire far- field of sum or difference patterns . In fact, compared to the maximum possible effect that typical z-position errors can have on the off-axis far- fields, instrumentation phase errors can be ignored completely in the off-axis region. Even for aperture antennas with edge tapers greater than 20 dB down, it is unlikely that the near-fields outside these -20 dB points (or even the -15 dB points) have a significant effect on the maximum pos-sible far- field errors. In other words, regardless of how large the edge taper, the effective scan area A need not extend beyond about the -15 or -20 dB points. 55 For receivers which measure phase with high accuracy, the phase part of the instrumentation error can be neglected and only the amplitude error remains in eq. (82), i.e. n(r) < N^Cg 1 ) g(F). Amplitude errors for < 2\/ I do not depend directly on frequency or the size of the antenna, only on the taper factor a of the near-field beam, the receiver inaccuracy N, R , and the inverse of the normalized far-field pattern g(r). (For null depth of a difference pattern the amplitude errors are extremely small and do not even in-volve g(r). These results are proven in Appendix B.) Also, in the far sidelobe region, 6 > 10A/& , the instrumentation amplitude errors are relatively small, usually less than a few tenths of a dB for N, R less than a few thousands of a dB per dB . The far-field error in the on-axis (g(r) = 1) gain for sum patterns is found from the above equation and (2c) to be n£ B < ± 8.7 N B (a-l). (84) Note that for a=l (uniform amplitude distribution) the far-field error, caused by the instrumentation errors in measuring near-field amplitude, equals zero--as it should since the receivers measure essentially at a constant amplitude across the effective scan area. The maximum error in on-axis gain (eq. (84)), which is linear with respect to N, R , is plotted with the solid lines in figure 13 for different values of a. Rodrigue et al . have also computed linear amplitude errors for their hypothetical near-field distribu-tion, which has an a exactly equal to 3.0. Their results (see figures 3-5 or A-5 of ) are reproduced by the dotted line in figure 13. The errors computed by Rodrigue et al . and the maximum errors predicted by the analytically derived expression (84) are in good agreement. The solid line for a=3 lies above the dotted line --as it must if eq. (84) represents a valid upper bound expression for the errors To insure an on-axis gain error less than ± .01 dB , the error in measuring the near-field amplitude should be kept less than about ± .001 dB per dB down. Unfortunately, the accuracy of most receiver 56 systems is of the order of ± .01 dB per dB down (rather than ± .001) which, according to figure 13, can lead to errors in the on-axis gain of about ± .1 dB . This value of far-field error can be larger than the errors from all other sources combined. Thus, if high accuracy is desired, special effort should be devoted to designing a receiver system which can measure the amplitude of the probe output to better than ± .001 dB per dB down. Alternatively, the amplitude calibration curve for the receivers could be determined to within ± .001 dB per dB down and the errors in amplitude compensated for by including the calibration curve as part of the computer program that deconvolutes the near-field data. This latter correction procedure has been adopted by A.C. Newell et al. at the National Bureau of Standards. The difference between the on-axis gain computed by Newell [17,21] with and without the ampli tude calibration curve is shown by the dashed line in figure 13 for the 46 cm (18 in) reflector antenna operating at 60 GHz. (Actually, Newell found a .112 dB on-axis gain difference for an amplitude cali-bration curve that deviated by at most .02 dB per dB down over the effective scan area. The dashed line in figure 13 assumes linearity and simply connects the origin to the point .112 dB at .02 dB per dB . ) The value of a for this antenna was estimated at 2 from the measured near- field data shown in figure 7b. Again it is seen from figure 13 that the computed errors for this particular antenna correspond quite well with the maximum possible errors predicted by the general expression (84) for a = 2. In brief, the computations of both Rodrigue et al . and Newell indicate that eq. (82) yields reasonable values for the maximum far-field errors expected from instrumentation errors in measuring the amplitude and phase of the probe output. Table 3 lists some representative far-field amplitude errors calculated from eq. (82) for an antenna with taper factor a equal to 3, and a receiver nonlinearity in measuring amplitude (N, R ) equal to .002 dB per dB . Note the extremely small effect that amplitude errors have on the null depth of difference patterns. As Appendix B shows, this small effect on the null depth is due to the fact that the receiver which measures the amplitude of the probe output distorts the opposite sides of the near-field difference pattern by approxi-mately the same amount. 57 C. Multiple Reflections Consider a probe which scans on a plane in the near-field of a radiating test antenna. The radiation that the probe receives can be described by an infinite series of rapidly decreasing terms, with the first term equal to the unperturbed field of the test antenna. This unperturbed field scatters from the probe, reflects from the test antenna and other nearby objects, and returns to the probe to give the second term in the series. The return radiation again scatters from the probe, reflects, and returns to the probe to yield the third term. The process repeats ad infinitum. In order to determine the far-field of the test antenna by "deconvoluting" the measured near-field data (without knowing the detailed scattering properties of the probe or test antenna) , the multiple reflections must be neglected. That is, the second and higher order terms in the infinite series just described are assumed negligible when applying the planar near-field scanning techniques Multiple reflections can be reduced by decreasing the size of the probe antenna, by increasing the distance between the probe and test antenna, and by appropriately covering the scanning range with efficient absorber material. These measures will not, however, eliminate the multiple reflections entirely. It is the purpose of this section to estimate the effects of the multiple reflections on the far-field which is computed from the near-field scan data under the assumption of zero multiple reflections As we shall see shortly, it is a fairly simple matter to esti-mate maximum and minimum values expected for the far-field errors produced by the multiple reflections. However, it is impossible to derive an accurate estimate of the far-field errors analytically without knowing the phase of the multiply reflected fields throughout the scan area. Thus, the only reliable way to get an accurate esti-mate of the far-field errors caused by multiple reflections is through measurement. Specifically, a number of near-field scans could be 14in principle, the effect of multiple reflections could be eliminated at microwave frequencies by the use of gated sinewaves instead of CW. Unfortunately, the speed of electromagnetic propagation (c) is so great that the necessary gating times are too short for the present-day elec-tronics to handle. It is possible, however, in the analagous measure-ment of electroacoustic transducers to eliminate the problem of multiple reflections by the use of gated sinewaves because the speed of sound is much smaller than that of light . 58 taken on parallel planes separated by about 1/4 wavelength and the far-field computed from each scan. Any differences observed in the far-fields computed from the separate near-field scans would indicate the extent of the effect of multiple reflections (assuming the scan area is large enough so that changes in finite scan errors are neg-ligible) . In this way the far-field errors can be determined straightforwardly and accurately. Of course, the main disadvantage of this "straightforward method" of determining errors lies in the time and effort it takes to record data and compute far-fields from several near-field scan planes for every antenna that has to be mea-sured. Because of this disadvantage it may prove worthwhile, par-ticularly when the multiple reflections are very small, to derive the following very approximate, yet general, upper and lower bound expressions for the far-field errors caused by the multiple reflec-tions. In addition to the upper and lower limits of errors, the far-field errors will be derived for multiply reflected fields which satisfy a certain class of hypothetical near-field distributions. Consider a probe antenna scanning on a plane in the near-field of an electrically large, aperture antenna. Assume once again, for the sake of simplifying the mathematics, that the probe behaves as an electric dipole, i.e. its output in one orientation is propor-tional to E and in a second orientation proportional to E . Then x_ y the error (AE) in the computed far-electric- field can be expressed with the aid of eqs . (37) and (12d) , AE < "A e ik(r-dcos6) j Agnr (F^ e " 1 rR " P dF< (85a) ;f->-co rn "p The superscript "mr" on AE denotes that part of the transverse near-electric-field caused by multiple reflections, and A' refers to the scan area. The determination of (AE) requires the evaluation of the integral -i-R-P I = / AE™r (P,d)e r dP, (85b) A' Z 59 or if the x- component is concentrated on first I = / AE I ' 1I (P,d)e r dP. (85c) x The amplitude and phase of AEmr can be written explicitly as • ^nir ,pinr . .mr r x AE = AA e x x to recast eq. (85c) in the form i(cJ)mr -^R.P) I = / AA™r e x r dP. (86) x A , x The maximum value of the integral in eq . (86) occurs when <pmT = —R»P, i.e. I < / AA™r dP. (87) x A , x If polarization is not changed drastically upon reflection from the mr probe and test antenna, AA will be roughly proportional to the magnitude of the x-component of electric field at the probe. That is, Mmr a £ mr \|E , } X X ' X ' ' v mr where e is an average proportionality constant between the ampli-tude of the multiply reflected x-component of electric field and the amplitude of the total x-component of electric field as the probe traverses the scan area. Substitution of eqs . (88) and (52) into eq. (87) yields I < emr A r IE I . (89a) x — x ' xo ' r->°° v In the same manner, the expression for I is found to be I < e mr A r IE I . (89b) y — y l yo ' r->°° An upper bound expression for the fractional error n emerges when t eqs. (89) and (85) are combined with eq . (1): n(F) < e mr g(F), (90) 60 mr where e is the average proportionality constant between the ampli-tude of the multiply reflected electric field (probe output) and the amplitude of the total electric field (probe output) as the probe traverses the scan area. The value of e can be estimated experi-mentally by changing the distance between the probe and test antenna by a few wavelengths at various locations within the scan area. Periodic variations in the amplitude of the probe output which re-peated about every A/2 would be caused primarily by the multiple re-flections. As usual, g(r) is the ratio of the amplitude of the maxi-mum far-electric-field to the far-electric -field at the given direction r. Equation (90) represents an upper bound expression for the errors in the far-field caused by multiple reflections. It applies to the entire far-field of electrically large aperture antennas. However, for nearly all antenna-probe interactions it will give much too large an estimate of the far-field errors because it was derived under the unrealistic assumption that the multiply reflected fields possessed just the right phase across the scan area to maximize the far-field errors. In reality, as the probe traverses the scan area it will usually experience large variations in the phase of the multiply reflected fields throughout the scan area that will greatly reduce their effect on the far field. We emphasize the word "usually" because there exist some antennas (like the constrained lens, fixed-beam array show.n in figure 9) which present a rather flat reflective surface to the radiation scattered from the probe, and thus a rather constant effective path length and phase for the multiply reflected fields as the probe scans on a plane parallel to the aperture. In that case the maximum error could be experienced at the center of the main beam. The multiple reflections would have a minimum effect on the far-field when their phase varied radically over the scan area, or rn y V more precisely, when cj) - — R«P in eq. (86) has no critical points and varies rapidly with position P on the scan area. Then eq. (86) written in polar coordinates (p, cj) ) can be integrated by parts with respect to p to give X x s 271/ X 3f/dp' P e P P' dV (91) 2tt AA™r (p',cj) ) ikf(p\ct ) o 61 where p' refers to the distance from the origin within the scan area to the point on the boundary of the scan area at the angle $ . The function f(p', ), which is assumed to possess no stationary points p with respect to p, is defined by kf = mr x - y R ' p -(92) As mentioned above, the smallest values of I occur, in general, when the phase function f varies rapidly with p and cf> . Experience at the NBS indicates that it is unlikely that the average phase variations in multiple reflections which occur in practice ever exceed 360° per wavelength across the scan area. Thus it is also unlikely that \| I \| will be smaller than its value when f is chosen as a function which changes an average of about 360° per wavelength of motion in any direction across the scan area. For example, one such function is f = p(l + cos 2 9 ) . (93) Substitution of the function (93) into eq . (91) shows that the magni-tude of I may be expressed as x 7 ^ >-£ 2, AA^p', ) / -x— p' e ikp' (l+cos 2 9^) d(J) l+cos^e p (94) Equation (94) is written as an inequality to emphasize that for nearly all antennas \|l \| would be larger than the right side of eq (94). In order to get an idea of the value of eq . (94), assume that << p' so that the points of stationary phase of p ' (l + cos 2 ^ ) d(J) p with respect to /A L ve AAave (95a) with L equal to the average of the width of the scan area, and AAave equal to the average amplitude of the x-component of the multi-ply reflected electric field at the four points <> = 0, tt/2, tt and 3tt/2 62 on the boundary o£ the scan area. If we approximate AA by e \|E I , where \|E \| refers to the average amplitude of the x-XX X component of electric field on the boundary of the scan area, eq. (95a) becomes x A /, , ave mrip i ave > t~ vX L e El — 2tt x ' x ' (95b) Since II = /\|I : + \| I \| 2 , eq. (95b) combines with the corresponding x y equation for \| I \| to give y I > -^ /A L 1 — 2tt ave A yC (96) The average amplitude \|E \| of the transverse electric field may be expressed in terms of the maximum transverse electric field (E^ ) on r to the scan plane, -Xave E t \| ave = 10 20 'to (97) .ave with X denoting the number of dB down from that maximum (see footnote 7) . Equations (30b) and (31) can be utilized in conjunction with eqs . (97) and (96) to give the final expression for the minimum value of Hi , Xave I I > /A L ave 10 20 mr e a A 2 r I E I (98) From eqs. (98), (85) and (1) we find the minimum value expected for the fractional error n in the far-field caused by multiple reflections Xave n(r) > f A 1 ave 3/2 10 20 mr ,— . a £ SO) (99) (The aperture area A has been taken as . 5(L ) 2 .) The inequality (99) represents a lower bound expression for n in the sense that the actual far-field errors caused by multiple reflections would, to a high probability, be greater than but could lie reasonably close to 63 the value of the right side of eq. (99). Of course, there remains the possibility that for some points in the far-field the effects of the multiple reflections will cancel to such a degree that n would actually be less than the right side of eq . (99). These exceptional points must be ignored if eq . (99) is accepted as a valid lower bound Between eqs . (90) and (99) we have approximate upper and lower limits to the value of n, Xave A 1 ^L ave 3/2 a 10 20 mr ,—. . ,—, . mr e g(r) n(r) e g(r) (100) where mr X a ave ave = wavelength. = the average ratio of the amplitude of the multiply reflected probe output to the amplitude of the total probe output as the probe traverses the scan area. (Its value can be estimated experimentally by changing the distance between the probe and test antenna at various locations within the scan area, and calculating one-half the fractional peak to peak height of the variations in amplitude that repeat about every X/ . ) = average width of the scan area. = the average amplitude of the probe output at the boundary of the scan area measured in dB down from the maximum amplitude of probe output in the scan plane. = a "taper" factor --equal to a minimum of 1.0 (for apertures of uniform amplitude and phase) and less than 5 for most tapered distributions found in practice (see eq. (31) for the precise definition of a; for a difference pattern one should still use the taper factor of the constituent sum patterns) g(r) = ratio of the amplitude of the maximum far-electric-field to the far-electric-field at the given direction r, i.e., the inverse of the normalized far-field pattern. (g(r) = 1 for the center of the main beam, or beams if a difference pattern.) 64 (If desired, upper and lower limits for the errors in the gain func-tion, sidelobe level, polarization ratio, and beamwidth may be found by combining eq. (100) with eqs . (2) -(5) and the far-field pattern.) The value of the factor „ave x 1 3/2 a 10 20 T avej TT for a typical microwave antenna and scan area is on the order of .001; in which case eq. (100) becomes .001 emr g(r) < n(r) < emr g(r). (101) It is clear from eq. (101) that the errors in the far-field produced by multiply reflected fields of a given relative amplitude, i.e. a given e , can span an extremely wide range of values depend-ing on the variation in phase of the multiply reflected fields across the scan area. (For example, if the multiple reflections are down 40 dB (e = .01), eqs. (101) and (2c) show that the multiple re-flection error in the on-axis gain of the main beam lies between about ± .0001 and ± .1 dB . ) Essentially, the right side of eq. (101) gives the far-field errors when the effective phase of the multiply reflected fields is uniform, and the left side when the phase varies an average of 360° every wavelength across the scan area. Because of the extremely large range in the possible value of n, it appears unlikely that a precise value of the far-field errors produced by the multiple reflections can be obtained by any method other than direct measurement. As was mentioned at the beginning of this sec-tion, data could be taken on a number of parallel scan planes separa-ted by about A./4, and the far-fields computed for each plane. Any differences noted in the computed far-fields for the separate scan planes would be caused primarily by the multiple reflections. In addition, it appears likely that the effect of multiple reflections could be reduced appreciably by averaging the far-fields obtained from a number of different scan planes separated by a small fraction of a wavelength over a distance of one wavelength. Finally, we shall evaluate the far-field errors for multiply reflected fields which are described by a class of hypothetical 65 near-field distributions. In particular, assume that the fields are mr linearly polarized and that cb in eq . (86) has the functional dependence , cb = 2tt 1 X 2 J (102) where A, and A~ are arbitrary real constants. Also, assume that the test antenna has a circular aperture of radius "a" and uniform ampli-mr tude distribution. Then for scans in the very near-field, AA in mr x eq. (86) can be approximated by a constant (AE ) inside the aperture area and zero outside. Under these conditions, the integral (85b) may be written in polar coordinates as ,mr ztt a 1 I^ 3 [3—cos 8 +-r—sincb -sin0cos(cb -cb)] A L A I = AE" 11 / / e J J o o p A pdpdcb (103) The ) = B cos(cb -b) A, p A ? p p p (104) where and B = /(A/A^sinecoscb) 2 + (A/ \^- sinBsincb) (105a) b = tan t— -sinG sincb A -, sine cosi (105b) Using the integral representation of the Bessel function, 2tt J (z) = J^ / e iz cosip 2tt dijj eq. (103) reduces to ,mr I = 2ttAE ]111 / J O J O A o 2TrpB p dp (106) 66 and eq. (106) becomes I = AaAEmr 1 2-rraB A (107) By combining eqs . (107), (85), and (1), the fractional error n in the far-field takes the form aAE mr n(r) = J 1 27raB A Br \|E(r) (108) ;£->-oo For a circular aperture with uniform amplitude distribution E , e< (10) can be integrated to yield the far-field, E(r) I = E -(cos 2 9 + sin 2 cos 2 (j)) J 2Tra A sin6 cose (109) The definition of n(r) loses significance when its value becomes larger than 1, i.e. when the far-field amplitude \|E(r)\| becomes less than the amplitude of the far-field error. To avoid this 9-TT Q situation, J-, (—r— sine) in eq . (109) can be approximated by its envelope , ffiz -\|J]_(z) I ~ J x (z) = ' 7TZ \ z 2 ) z < 2 z > 2 (110) Substitution of eq. (110) into eq . (109) and the result into eq. (108) transforms the expression for n into mr . „ e sine n(r) = (2^aB^ 1 X 1 B(cos 2 e + sin 2 e cos 2 4)) J 2-rra sin6 (111) r mr A -cmr /X2 ^ (e = AE /E ) o ' o J If A, and X ? (see eq . (102)) are chosen such that A, = A ? = A , and (J) is set equal to zero, B can be written from eq . (105a) as A ^ o sine + sin 67 and r\ becomes mr -a e smQ n(6) = j 2Tra 4-o t— sine o +sin^0 X X X X -sinf +sin 2 6 J^ 2fTa X (112) sin9 We can compare this result with the previous maximum and "minimum" values estimated for r\ in eq . (100) when g(r) = a = 1, and X =0. For X = °° (uniform phase for the multiply reflected fields) eq. (112) should equal the right side of eq . (100). For X = X eq. (112) should be of the same order of magnitude as the left side of eq . (100). Indeed, when X eq. (112) becomes emr (the right side of eq. (100)), and when A = X it reduces to approxi' mately .06 (A/a) 3/2 (100) since A = na 2 was taken as .5(L ) in eq which is nearly equal to the left side of eq. (100). This agreement between eqs . (112) and (110) supports the validity of both expressions Equation (112) reveals that the on-axis far-field error n(o) can be expressed in the especially simple form, n(o) = emr J X (113) Figure 14 shows the on-axis error n(o) plotted against the variable /2i\a/X . Note that for X < 2a, n(o) is given approximately by .06 3/2 cos 8.9 X mr mr and never gets larger than about .1 z" x . That is, if the phase of the multiply reflected fields changes 360° or more across the diameter of the aperture area, the error in the on-axis far-electric- field is mr mr less than one-tenth e , where e is the ratio of the amplitude of the multiply reflected electric field (probe output) to the total electric field (probe output) as the probe traverses the effective scan area. Only when the phase of the multiply reflected fields is uniform (X = °°) across the scan area does the on-axis far-field o , mr error equal e In figure 15, n(6) is plotted for different values of a/X Q from eq. (112) for a/A = 12. When X is greater than a couple of aperture AC diameters, the envelope of the errors from multiple reflections are mr on the order of e throughout the far-field. As X becomes less & o than a couple of diameters, the near-axis errors grow much smaller mr than e but the envelope of errors in the far sidelobes remains at mr about e . Finally, when X gets as small as the free-space wave-mr length A, the far-field errors are much smaller than e all the way to 6 = 30° in the far-field. IV. Summary The far-field characteristics of a radiating test antenna can be determined throughout the forward hemisphere by scanning on a near-field plane of the test antenna with a probe antenna of arbi-trary but known receiving characteristics. The amplitude and phase of the probe output are recorded on the near-field scan plane and the far-field pattern is computed by "deconvolving" the near-field data. The accuracy of the computed far-field depends upon the size of the scan area, the accuracy with which the scanner positions the probe, the accuracy with which the instrumentation measures the amplitude and phase of the probe output, the extent of the multiple reflections, the computation errors involved with deconvoluting the near-field data, and, of course, the accuracy with which the probe is calibrated and the input power to the test antenna is measured. Essentially, this report derives upper bound expressions for the errors in the far-field pattern produced by these sources of error in the near-field measurements. The upper-bound expressions are written in a form that can be used to stipulate design criteria for the construction of near-field scanning facilities. In particular, the limits of accuracy in a given far-field parameter are expressed in terms of the measured near-field data and/or the computed far-field, the frequency and dimensions of the antenna-probe system, the systematic and random variation in the positioning of the scanner, and the precision of the instrumentation which measures the probe output. In order to simplify the mathematics, the probe was usually assumed to be an electric dipole, although the resulting upper bound expressions hold for arbitrary probes. 69 The analysis and resulting upper-bound expressions are not re-stricted to a particular antenna as previous computer studies [11,16] and direct far-field comparisons have been, but apply generally to electrically large aperture antennas which can be operating in either a sum or difference pattern. The results for position and instrumen-tation errors apply only to nonscanning antennas in the sense that the analysis for these two sources of error assumed that the scan plane lay perpendicular or nearly perpendicular to the boresight direction of the antenna pattern. Position and instrumentation errors for beams which are steered away from the perpendicular to the scan plane will be included in a subsequent report by Newell Except for the position and instrumentation errors, however, the re-sults of the present report, and in particular the error expressions pertaining to the truncation of the scan plane, apply to arbitrarily steered antennas. Broadbeam antennas where the wavelength is on the same order of magnitude as the dimensions of the aperture are not examined in this report. But it is shown in a report by Crawford et al . that many of the conclusions and upper bound expressions derived here apply directly or in slightly modified form to broadbeam antennas It is emphasized that the upper-bound expressions derived in this report determine the limits of accuracy of the far-field com-puted from the planar near-field scanning technique without resorting to comparisons with direct far-field measurements-. This is probably the foremost purpose of the report along with the report acting as an aid in deciding design criteria and tolerances for the construc-tion of new near-field scanning facilities. It has been the feeling of those involved with near-field measurement techniques at the NBS that often the near-field techniques determine the far-field more accurately than conventional far-field measurements with a standard antenna. Thus comparison with measurements made on conventional 'far- field" ranges would not be a reliable method, even if it were feasible, for estimating the accuracy of the near-field techniques, which do not have the problems of proximity corrections, ground re-flections, or the need of a standard far-field antenna. A brief summary of the major conclusions and results of the report follows: 70 Errors in computation can be ignored immediately. A simple exercise in Section I showed that their effect on the far-field is extremely small compared to the effects of the other sources of error. Far-field errors from the approximation involved in applying the sampling theorem are also negligible (see footnote 1) . In Section II it was demonstrated that errors in various far-field parameters could be expressed conveniently in terms of the fractional far-electric-field error (n = I AE I / I E I ) and the approxi-vi I I'll p-^-oo ' r tr mate far-field pattern. In particular, the errors in gain function, sidelobe level, polarization ratio, and beamwidth were expressed in terms of n (see eqs. (l)-(5)). In Section III. A the far-field errors associated with neglecting the near-fields outside the finite scan area were investigated. First it was shown analytically that no reliable information about the far-field pattern could be obtained by the planar near-field scan method outside the "solid angle" formed by the edge of the antenna aperture and the boundary of the scan area. Well within this solid angle (0 < ~-, see eq. (33) and following), reasonable upper l m cix bound expressions for the finite scan errors were found that could be applied to center-line data (eq. (32)) as well as full-scan data (eq. (36) or (32)). As part of the finite scan analysis, asymptotic expres sions for the near-fields in front of a circular antenna of uniform aperture distribution were derived in Appendix A and plotted in figures A3 and A4 . A comparison was made between the empirical analysis of finite scan errors performed by Newell and Crawford with centerline data and the maximum errors calculated from the upper bound expression (32). Agreement is quite reasonable, as figures 8a and 8b indicate. Table 1 shows the finite scan error for various far-field parameters of a typical x - band antenna. The finite scan errors are proportional to wavelength, so changing the wavelength while holding the other antenna dimensions the same merely changes the values in Table 1 proportionately. Of course, such an isolated change is rather unrealistic. Deviations in the position of the probe from its assumed posi-tion in the scan area will produce errors in the near-field data which show up as errors in the computed far-field pattern. Section 71 III.B.l derives an upper bound expression throughout the forward hemisphere for the far-field errors produced by both systematic and random errors in the positioning of the probe (see eqs . (61), (64), (68) and figures 10, 11, and 12). (Of course, a uniform displacement of the scanner does not alter the far-field pattern, and a uniform rotation of the scanner simply rotates the entire far-field pattern through the same angle.) The contribution from the inaccuracies in the position of the scanner divides naturally into a transverse or xy (parallel to the scan area) and longitudinal or z (perpendicular to the scan area) part. For sum patterns, the on-axis far-field errors were found pro-portional to the xy-displacement errors of the scanner but propor-tional to the square of the z-displacement errors (normalized to wavelength) . This latter result is analogous to that obtained by Ruze in his classical work on "antenna tolerance theory" for a small arbitrary phase error or aberration in the surface of an an-tenna. This result does not imply that the z-displacement errors generally have a much smaller effect on the main beam than the xy errors, because the multiplying factor is, in general, much larger for the z term. It does imply, however, that random and systematic displacement errors in the longitudinal or z direction weigh equally in their contribution to errors near the center of the main beam of sum far-field patterns. Unlike the z errors, random errors in the transverse or xy-displacement of the scanner have a negligible effect throughout the far-field compared to systematic transverse errors of the same order of magnitude, and the xy-position errors do not depend on wavelength. The xy-position errors also differ from the z-position errors in that the same xy error expression applies throughout the far-field hemisphere whereas the z errors are given by one expression close to the main beam, i.e., near the boresight direction, and another expression for the errors off -axis. Specifically, the maxi-mum possible off-axis (or sidelobe) z-position errors depend linearly upon the systematic z-displacement errors of the scanner and are much larger than the on-axis errors, which are proportional to the square of the z-displacement errors. In fact, as Tables 2 and 4 indicate, these off-axis errors caused by displacements in the 72 z-position of the scanner can be much larger than the total off-axis errors from all other sources combined. The present report does not examine in detail the relationship between the distribution of z-displacement errors and the off -axis or sidelobe errors, but such an analysis will be included as part of the subsequent report by Newell . For difference patterns, the effect of the near-field displace-ment errors on the far-field were given by the same upper-bound expressions derived for sum patterns, except for the effect of z-displacement errors on the null depth. This error in null depth for difference patterns was shown in Appendix B to be independent of the value of the null depth itself and proportional to the z -displacement errors, but, in general, proportional by a very small proportionality constant. In fact, for most antennas and reasonably accurate scan-ning systems, the z-displacement errors do not affect the null depth by more than a few tenths of a dB . This rather surprising result, which is rather difficult to explain without going through the mathe-matics of Appendix B, has been confirmed by an empirical error analysis performed by Newell on the measured near-field data of a number of antennas operating in the difference mode. Although the z-displacement errors do not have a strong influence on the depth of the null of difference patterns, eq . (68), which was also derived in Appendix B, shows that they do have a strong influence on the direc-tion of this null. In fact, the effect of all other sources of errors combined on the null direction of difference patterns is generally negligible compared to the effect of z-position errors, although instrumentation phase errors can sometimes shift the null an appreciable amount as well. Table 2 shows the xy-and z-position error in various far-field parameters for a typical X-band and K-band antenna. Again, note the strong influence that z-position or phase errors can have on the off-axis far-field parameters (sidelobe level, beamwidth, mainlobes of difference patterns) and on the null shift of the difference pattern. Note also that the maximum possible null shift caused by z-position errors is independent of frequency. 73 The far-field errors caused by the inaccuracies of the receivers in measuring the phase and amplitude of the probe output are esti-mated in Section III.B.2 (see eqs. (82), (83) and figures 11-13). (It should be mentioned that instrumentation errors associated with converting analog to digital information is assumed negligible.) As would be expected, the instrumentation errors in measuring phase contribute to the far-field errors in exactly the same way as longi-tudinal or z-position errors of the scanner. However, since the errors which a typical receiver introduces into the phase are small and increase monotonically with decreasing amplitude of the probe output, calculations show that their effect is often negligible throughout the far-field. Although typical instrumentation errors in measuring phase are small and often introduce insignificant errors into the far-field, typical instrumentation errors in measuring amplitude can have a pronounced effect on the far-field (see figure 13), except for the far sidelobe region and in the null depth of difference patterns. In general, if high accuracy is desired, the receiving system which measures the amplitude of the probe output should be calibrated and the calibration curves included as part of the computer program which deconvolutes the near-field data to get the far-field. It is significant, however, that the instrumentation errors in measuring near-field amplitude have a relatively small affect on the null depth of difference patterns, as eq . (82) and Table 3 demonstrate. The reason for this is that the receiver distorts the opposite lobes of the near-field amplitude by approximately the same amount (see Appendix B) . Table 3 also displays the amplitude error in various other far-field parameters for a typical microwave antenna. The in-strumentation amplitude errors for 8 < 2X/£ do not depend directly upon frequency or the size of the test antenna, only upon the taper factor a of the near-field amplitude, the receiver inaccuracy N, R , and, except for the null depth of difference patterns, the inverse of the normalized far-field pattern g(r). It should also be pointed out that the instrumentation amplitude errors in the far sidelobe region, > 10X/ imax , are relatively small, usually less than a few tenths of a dB for N, R less than a few thousands of a dB per dB 74 Whenever comparisons were possible, the expressions for both position and instrumentation errors (eqs. (61), (64), (68), (82), and (83)) agreed well as an upper-bound with the results of the empirical error analysis of Newell et al . [11,17,21] at the National Bureau of Standards, and with the error analysis performed by Rodrigue et al. at the Georgia Institute of Technology with a hypothetical near-field distribution (see figures 11-13). In Section III.C upper and "lower" bound expressions were derived for the far-field errors caused by multiple reflections (see eq. (100)). The upper and lower bounds were extremely far apart because the phase of the multiply reflected fields has a strong in-fluence on the far-field errors. Since the phase of the multiply reflected fields for a given test and probe antenna interaction would, in general, be difficult to measure or estimate, it was con-cluded that the only reliable way to get an accurate estimate of the far-field errors from multiple reflections would be to use the fol-lowing straightforward but tedious procedure. Take several near-field scans on parallel planes separated by about 1/4 wavelength or less. Any deviations in the far-field patterns computed from the data on the separate scan planes would be caused primarily by the multiple reflections (assuming the scan area is large enough so that changes in finite scan errors are negligible). If necessary, the effect of multiple reflections could probably be reduced appreciably by averaging the amplitude of the far-fields obtained from these different scan planes separated by a small fraction of a wavelength and covering a total change in separation distance of one wavelength In addition to the upper and lower limits of error, the far-field errors were derived for multiply reflected fields which satisfy a certain class of hypothetical near-field distributions (see eq. (112) and figures 14 and 15) . The major analytical results of this error analysis study can be combined into one long upper-bound expression for the fractional far-electric- field error n(6,(\|0. (6 and specify the angular spherical coordinates of the far-field pattern.) 75 n(e,cj)) , T max, n aAL 10 20 aAA' max A sine n probe max 11max 2N dB (3l) + 2 mr 2 input null shift of 2A shift difference pattern — max 7T£ 6 S +A X max Tmax radians (114a) (114b) (6 max a<kLj 2 + max (6 rn ) v max ; 8AF(6 S +A0 1 ) v max Y max ; r B 1 -< max (o-l) 2/g(e,<fr) (a-l)/2 , . max aAL 3A (sum patterns) (difference patterns) (sum and difference patterns) (sum patterns) (difference patterns) (sum and difference patterns) 10Amax 10Amax 10£max A 10£ 10A max <e< 2A ,max £max <e< I Equations (114) represent essentially an amalgamation of eqs . (32), (64), (68), (82), (83) and (100), under the extreme condition that the various near-field errors combine in such a way as to create the maximum possible far-field errors. If desired, the tighter upper-bound, eq . (36), for the finite scan error could be used for the first term in eq . (114a) instead of eq. (32). Recall from Section III. A. 3 that the finite scan error term represents a valid upper-bound halfway or more within the "solid angle" formed by the edges of the aperture and the boundary of the scan area (9 < _ 9 = ~-(90-y ) ; and r ' v "• 2 max 2 K 'max' outside this solid angle region the planar near-field scanning tech-nique cannot be relied upon with any confidence to yield accurate far-fields. The instrumentation amplitude factor 3 is not given explicitly by eqs. (114) in the region between 9 equal to 2A/£ and 10A/£ . However 3 can be estimated in this region by connect-ing a straight line from its value at 2X/1 to its value at 10A/£ The detailed derivation of each of the terms in eq. (114), except 76 n , and n. . , can be found in the part of the main text from probe input r which each particular term came. The extra terms n , and n. r probe input will be explained below. The definition of the various parameters in eqs . (114) can also be found from the preceeding main text: X = wavelength. A = area of the antenna aperture. TH £L X I = maximum width of the antenna aperture. L = maximum width of the scan area. y = maximum acute angle between the plane of the scan area 'max & r and any line connecting the edges of the aperture and scan area (0 = 90-y ) . ^ max 'max' X = the largest amplitude of the probe output at the edge of the scan area, measured in dB down from the maximum amplitude of probe output in the scan plane. a = a "taper" factor- -equal to a minimum of 1.0 (for apertures of uniform amplitude and phase) and less than 5 for most tapered distributions found in practice. (See eq. (31) for the precise definition of a; for a difference pattern one should still use the taper factor of the constituent sum patterns . ) A = 2ttAP /A, where AP is the maximum amplitude of the max max ' max r transverse (xy) displacement errors within the effective scan area A . (A is that part of the scan area over o^o ^ which the phase is fairly uniform. For near-field scans parallel to the aperture A -A.) 8 = 2ttAz /A, where Az is the maximum amplitude of the max max ' max r longitudinal (z) displacement errors within the effective scan area A . I ° Acb = the maximum instrumentation errors (expressed in radians) Y max y r j involved in measuring the phase of the probe output on the effective scan area A (see footnotes 10 and 13) . o AF = fractional difference between the amplitude of the two main far-field lobes of the difference pattern (see Appendix B) . Njt, = the maximum instrumentation errors involved in measuring the amplitude of the probe output --N, R is expressed in dB error per dB amplitude down from the maximum amplitude 77 on the scan area. (The amplitude error is designated as zero at the maximum amplitude; see footnote 11.) mr z = the average ratio of the amplitude of the multiply re-flected probe output to the amplitude of the total probe output as the probe traverses the scan area. (Its value can be estimated experimentally by changing the distance between the probe and test antenna at var-ious locations within the scan area, and calculating one-half the fractional peak to peak height of the varia-tions in amplitude which repeat about every A/2.) g(9,) = 1 for the center of the main beam, or beams if a difference pattern.) The superscripts "s" and "rn" refer to the "systematic" and "random" parts of the displacement errors respectively. The instrumentation phase and amplitude errors (A\ , N, n ) are assumed systematic in I max ai5 i max nature. The phase error Ac}) does not show up in n for 9 > A/(10£ ) in cix. z because, as explained in Section III.B.2, the shape of the near-field instrumentation phase error is such that it has negligible effect in this off-axis region compared to the maximum effect of typical sys-s tematic z-position errors 6 . Random position errors have non-r max v negligible effect only near the boresight direction of sum patterns. mr The multiple reflection ratio e is enclosed in a box in eq. (114a) to emphasize that for most antennas it represents an unreal-istically large upper bound. The error n , (9,cb) simply represents the uncertainty in the probe v ,rJ r j r receiving characteristics of the probe in the direction corresponding to (9,(J)). For example, if the receiving characteristic S'-. of the probe (see reference ) was known to an accuracy of 1% for the direction (9,6), then n , would be .01 for that direction. k. r j y 'probe The error n . . arises in normalizing the amplitude of the input r probe output to the input power or amplitude \|a \| of the input mode to the test antenna. Such a measurement is necessary whenever abso-lute values of the gain function are required. Probably the simplest and most accurate method of performing this normalization is to con-nect the input waveguide of the test antenna directly to the output waveguide of the probe through a variable attenuator. If, as ex-plained in Section III.B.2, the receiver which measures the amplitude of the probe output is specified arbitrarily to have zero error when the probe output is at its maximum amplitude on the scan area, then the normalization can be accomplished by measuring the attenuation needed to reduce the amplitude of the direct input from the trans-mitter to the level of the maximum amplitude of the probe output on the scan area. Of course, "mismatch factors" of any consequence must also be measured. The quantity n. . merely denotes the com-n 7 input J bined fractional error of the variable attenuator and of the devices used to measure the necessary mismatch factors. By using a high precision attenuator, the fractional error n. . can usually be r ' input ' kept below a few thousandths. Table 4 shows the total maximum possible error in a number of far- field parameters for a typical X-band and K-band antenna and a reasonably accurate scanning facility. The table was computed from eqs. (114) for the representative values of the near-and far-field parameters listed above the table. It is emphasized that the values shown in Table 4 are the maximum possible upper-bounds to the far-field errors computed under the extreme condition that each of the sources of near-field error produce its maximum possible change in the far- field and then all these maximum changes in the far-field add in phase. The z-position errors have been separated and under-lined when they represent the dominant contribution to the upper-bound errors, because whether or not these maximum possible z-position errors actually occur depends strongly on the far-field direction of interest and on the shape of the deviation in z-position throughout the scan area. As mentioned above, a detailed analysis of this de-pendence will be included as part of a subsequent report by Newell . 79 The error Cn be + ^ ut ) in the receiving characteristic of the probe and in normalizing to the input power of the test antenna was chosen as . 1 dB in Table 4. For the error in on-axis gain of the sum patterns, Table 4 shows that this contribution of .1 dB is about as large as all the other errors combined. The same is true for the gain of the mainlobes of the difference pattern if the z-position part of the errors is ignored. Thus, for the situation described by Table 4, greater accuracy in the calibration of the probe and in the measuring of the input power to the test antenna could be a first step in significantly reducing the errors in the direction of maximum gain. It is interesting to compare the maximum equivalent reflected signal allowable in a conventional "far-field" range or anech-oic chamber to get accuracies comparable to those shown in Table 4 for the near-field scanning technique. It is a simple matter to show that the maximum equivalent reflected signal (ERS) measured in dB down from the direct signal is related to the values of ri, R by n dB ERS = 20 log -X—~ (assuming small ri,J . For example, near the boresight direction the ERS would have to be 33 dB down for the sum pattern and 15 dB down for the difference pattern to give the corresponding errors shown in Table 4. Of course, there would also be errors on conventional ranges due to proximity effects, uncertainties in the calibration of the standard antenna, and instrumentation errors. In conclusion, it can be seen from Tables 1-4 that for a reason-ably accurate scanning system no one source of error dominates over all the others in the boresight direction of both sum and difference patterns. However, in the far-field region away from the boresight direction, but well within the solid angle region formed by the edge of the aperture and boundary of the scan area, the deviation in z-position of the scanner can, in principle, cause far-field errors which are much larger than the combined errors of all other sources. In practice, however, these maximum possible errors seldom occur. Moreover, it will be shown in the subsequent report by Newell 80 that the scanner can be designed and utilized to keep the off-axis, z-position errors far below the upper-bound values given by eqs . (114) and shown in Tables 2 and 4. Of course, beyond the solid angle region formed by the edge of the aperture and boundary of the scan area, it was shown in Section III. A that the far-field computed from the near-field data cannot be relied upon with any confidence. Ef-2 E AE Figure 1. Main beam of a hypothetical test antenna, Aperture of Test Antenna »-z Figure 2. Schematic of scanning geometry 82 Test Antenna Figure 3. Schematic of aperture antenna, Figure 4. Definition of a , 6 , D . 6 m m m 83 P.Pm Dm /j m point / rm r^-] t "Xem R ^\ 9 >o'm »-0' j rl 0' % Figure 5. Schematic of aperture and scan areas. (Although A and A' are drawn parallel, it is not a necessary requirement of the theory.) e=tan~'(-/2-l) Reflector Figure 6. Circular reflector antenna. 84 -5 -10 GO -a S " l5 <u -o B -20 "5. <f -25 <-o 2-30 o a> >-35 UJ -40 -45 -80 -60 -40 -20 20 40 60 80 Probe X- position in cm Figure 7a. Near-field centerline data (constrained lens) (z = 25 cm) . -50 -40 -30 -20 -10 10 20 Probe X-position in cm 50 Figure 7b. Near-field centerline data (reflector antenna) (z = 43.18 cm). .15 CD T3 C .05 Q> o> c D t_ o c o o X o c " .05 o 15 Upper bound \ \ Newell & Crawford 2.6 24 2.2 2.0 1.4 1.0 (L /2a) -Scan Length /Aperture Diameter Figure 8a. Change in gain vs. decreasing scan length (constrained lens) CD c o .c o c "o CO c o .25 20 10 .05 .05 10 15 .20 .25 \ A\ s Newel Crav i a ford Upper boun d / \ i \ \ \N i s 2.05 1.85 1.65 1.45 1.25 1.05 .85 .65 45 .25 .05 (Lmox/2a)- Scan Length /Aperture Diameter Figure 8b. Change in gain vs. decreasing scan length (reflector antenna) 86 Figure 9b Figure 9a. Constrained lens sum port near-field log amplitude, £=9.2 GHz, z-25.0 cm, no radome. Near-field phase, constrained lens sum port, f = 9.2 GHz, z = 25 cm. Figure 9c. Constrained lens antenna and probe 87 288 252 216 d) CD w m <v IHO o c a; 144 in o si a. IOR 100 -80 -60 -40 -20 20 40 60 80 100 Probe X-Position in cm Figure 9d. Near-Field Centerline Data, z = 25 cm 88 ±1.4 ±1.2 ±1.0 ±0.8 ±0.6 ±0.4 ±0.2 Figure 10. Position errors in on-axis gain. ±.5i— ±10 ±20 Figure 11. Comparison with Rodrigue et al. for random phase errors. 89 ±.25 ±20 CO "3 ±15 o I c m P- ±10 ±.05 Upper bound Newell mm ——— — ±1 ±2 ±3 ±4 , ±5 ±6 ±7 ±8 Degrees (±A<£ max ) Figure 12. Comparison with Newell for quadratic phase errors. ±07 ±06 ±.05 ±04 CD ±.03 ±.02 ±.01 a=3.0 y a=2.5 /Rod / rigueetaL (a=3) >X / a =2.0 X / V Newell (a=2) .^ / a=l.5 Jr " X ±.001 ±.002 ±.003 ±.004 ±.005 ±.006 ±007 ±008 ±009 N^B (dB per dB down) Figure 13. Amplitude errors in on-axis gain. 90 Figure 14. On-axis error in far-field from multiple reflections. a/X =I.O I.Oc env elope \ -K — 2>r<«h r 0° 6° 12° 18° 24° 30° 6 a/X =8.0 Figure 15. 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CD • O r< • v— v— • rH v_ v— Jh f +1 +1 + 1 + 1 + 1 + 1 + i PU rH 03 o 4-> -e-O E ,— » H u ,— , oo LO to , N vO • o ' v t~-(XI (XI . vO I— 1 • + + CD • rO VXD • + O0 1— 1 en CD + 3" II 00 rH • CD CXI CD LO CM r-i <-< • '—" v— • -— 1 p ^ + 1 + 1 +1 + 1 + 1 +1 + 1 o •H 4-> 03 i-> 5-, CU +-> d /• v CU / N O (O E CQ •H CU 03 ,-— N T3) +-> CU fH PQ v— 03 S-. 03 X N , s tie Dh — rH •H PQ CU ,— s CU rH T3 "d PQ t3 fi > o 03 v— v— Xi rH •H CU e-r— 1 v— CU 03 I— 1 o .d +-> •H bJD x P +-> <-H (0 PU CU M P. •H CU I to x> T3 to CU H X) 5-. •H o •H •H T3 CO o 03 X rH £ X I— 1 PL, 03 CU E 03 rH r— i c i xs 03 i rH r— 1 •H c •H CU d 3 3 rt o en PQ o Z z "'• CU u d d c !-H ca Jh E cu f-H CU P +-> CU +-> CO +-> Mh -m 03 Mh oj a, •H a \ 03 c c cu M d < d 03 CQ d 03 X o3 U • H p-X E-03 H O M-t (O !h O H LU T3 PL, i 03 •M O CU rH 03 95 Appendix A Near-Fields of a Circular Antenna with Uniform Aperture Distribution The purpose of this Appendix is to derive expressions for the near-field of a circular antenna of uniform aperture distribution. Specifically, the amplitude and phase are constant within the aper-ture and zero outside. The antenna is electrically large in the sense that the diameter of the aperture is many wavelengths across. The transverse electric field to the right of the aperture is given exactly by eq. (16). If we assume that the aperture fields are linearly polarized in the x-direction, eq . (16) becomes E (r) = x v J 2tt 9 z -E . 2tt a iklr-R' O 8 r r Q l a 7 i J . _ _ , R'dR'dtJ)' (Al) o o \| r-R ' for the circular aperture of radius a and uniform amplitude E . The element of area has been written in terms of the polar coordinates (R',cf)') of the vector R'. In addition it can be proven from eq. (15) that if the aperture fields have longitudinal components which are small compared to the transverse components, then the longitudinal components remain relatively small in the near-field (z << a 2 /A) , provided a/A >> 1. The change of variable t = R'-R, where R is the transverse part of r, converts eq. (Al) to E (r) x v J -2iE o 3z ikz 7T ik/t 2 (a) + z 2 / e da o Ra (A2b) The variables t , t, and t and the limit of integration a, are o 1 2 l defined in figures Ala and Alb. Apparently, eqs . (A2) were first derived by Schoch For large k, the integrals in eqs. (A2) can be evaluated by the method of stationary phase. The integrand of eq. (A2a) has two stationary points, one at a = and one at a = tt. The two terms in 96 the integrand of eq. (A2b) each have a single stationary point at a = 0. After carrying through the details of the method of station-ary phase, eqs . (A2) become E (r) = E [e lkz +AI] Ra, where AI is given by AI = ^ M 2tt R i [k/z 2 +(R+a) 2 -£] i [k/z 2 + (R-a) 2 +J] (R+a)y^ 2 +(R+a) 2 (R-a) /z 2 + (R-a) 2 (A3a) (A3b) (A3c) Equation (A3b) is identical to eq . (17) of the main text when the following substitutions are made: D 1 = /z 2 +(R+a) D 2 = /z 2 +(R-a) 2 a. cosO-, = z/(R+a) cos6~ = z/(R-a) Strictly speaking, eqs. (A3) are valid only as k = 2tt/A approaches infinity. For finite k they represent first term approximations to an infinite asymptotic series, and the first term approximations are valid only in certain regions of the near-field. These regions of validity can be estimated by returning to eqs. (A2) . First concen-trate on eq. (A2a) . In eq. (A2a) t ranges from (a-R) to (a+R) . Thus if the rate of change of t is somewhat uniform with a on the & o interval a = to it, the method of stationary phase will yield a good approximation when /(R+ a) 2 + z /(a-R) 2 +z 2 >> 1. (A4) Unfortunately t does not change uniformly with a from a = to it, but it does so in the separate intervals to y and y to it. Since t 2 = a 2 -R 2 at a = y, the condition (A4) must be replaced by the two conditions , 97 /(R+a) 2 + z 2 A 2 R 2 + z 2 > 3 (A5a) , , , , /a 2 -R 2 +z 2 -/(a-R) 2 +z 2 > 3. (A5b) (Actually the right side of eqs . (A5) should be >> 1 but experience with the method of stationary phase indicates it remains a good approximation down to where the quantities in brackets in eqs. (A5) are just a few wavelengths -- specif ically 3A is chosen in eqs. (A5).) Manipulation of eqs. (A5) shows that they are satisfied for large a/A if 3A < R < [a- (12A/a)A] (A6a) z < ^y /R 2 -(3A) 2 . (A6b) That is, eq. (A3a) approximates eq . (A2a) in the region defined by eqs. (A6) . A similar analysis with eq . (A2b) shows that it is approximated by eq. (A3b) in the region defined by R > [a+(12A/a) A] (A7a) a z < (R-a) 3A' (A7b) The regions of validity defined by eqs. (A6) and (A7) are shown in figure A2 for an a/A equal to 12. Each region is labeled by the equation which approximates the field in that region. In addition to the regions defined by eqs. (A6) and (A7) two other regions--one near the z-axis, R < 7a R < the larger of 2a z or 3(^) 2/3 A (A8a) (A8b) and one near the edge of the aperture are shown in figure A2 z > 2 r a z > Tra a-R R-a 1 < 4 a (A9a) (A9b) (A9c) 98 In the axis region defined by eqs . (A8) , the exponent in the integrand of eq. (A2a) can be approximated by the first two terms of its expansion in R/a, and integrated to give E (r) = E x v J o ikz ze i k /z 2 + a : J /z 2 ka R ^2+a 2 ^ (A10) z"+a' Similarly, in the edge region defined by eqs. (A9) , the eq. (A2b) can be integrated approximately to give E (r) = J x v J 2 ikz ze ik/z 2 + 2a 2 J /z 2 + 2; ka : Vz 2 + 2a 2^ (All) In the near-field in front of the aperture (z less than about Yy a 2 /A) , eqs. (A3a) and (A10) reveal that the electric field can be approximated by a single equation, with E (r) = E [e lkz + AI ' ] , x v J o L J ' (A12a) f i[k/z 2 +(R+a) 2 -J] ( AI za /2 y { (R+a)/z 2 +(R+a) 2 i[k/z 2 +(R-a) 2 +\|] ka R •/z^+CR+a) 2 (R-a)/z 2 +(R-a) 2 ka R Vz 2 + (R-a) and J defined as o J (z) = o v J (1 - i z 2 ) z < 1.46 z > 1.46. (A12b) (A12c) Essentially, eqs. (A12) represent the total electric field for the linearly polarized circular aperture in the region R < a-A (A13a) 1 a 2 z < 12 A (A13b) 99 The amplitude \|E \| and phase 6 of E v may be found from eqs. CA121 , 1— [Q! cos(k/z 2 + (R+a) 2 -kz-\|) -Q 2 cos (k/z 2 + (R-a) 2 -kz +}1 a E (A14a) kz za /2 coskz Q 1 sin(k/z 2 + (R+a) 2 -^) -Q sin(k/z 2 + (R-a) 2 - J) sinkz Q 1 cos(k/z 2 +(R+a) 2 -\|) -Q 2 cos (k/z 2 + (R-a) 2 +j) with (A14b) J' ka /z 2 +(R+a) 2 R (A14c) (R+a)/z 2 + (R+a) 2 Q? = T e ka p «j _ K u /z 2 +(R-a) 2 (A14d) (R-a)/z 2 +(R-a) 2 Equations (A14) hold (to a first order approximation) everywhere in the near-field region defined by eqs. (A13) . Except within a wavelength or so of the z-axis, eqs. (A14) reveal that in this region 1 X the maximum amplitude fluctuations are about ± — / — , and the phase varies slightly from kz. The amplitude of the electric field given by eq. (A14a) is plotted in figure A3 for a/A = 12 and 15, and for increments of z/a from to y~ a/A. The amplitude curves show large fluctuations within a wavelength of the points of maxima and minima along the z-axis. These z-axis maxima and minima, which are a well-known phenomenon for the circular aperture of uniform distribution, are less pronounced for noncircular apertures, or for apertures with a tapered distribution (see reference , Section 1-F). Figure A3 also shows that the amplitude variations across the aperture repeat about every wavelength in the very near-field (z < a/4) , but spread out as the distance from the aperture gets larger. This behavior has been observed for tapered distributions as well, by the many near- field measurements on microwave antennas performed at the National Bureau of Standards 100 The phase in the near-field of a circular antenna with a/A = 12 is plotted in figure A4 . Except near the zeros in on-axis amplitude, the phase across the beam is uniform to within a few degrees of oscillation which repeat about every wavelength in the very near-field. 101 (a) b) Figure Al . Definition of t and a. Aperture Figure A2. Dotted line shows region in which eq . (A12a) holds 102 A/\ f\s\P , r r (z/a = .75) \j .2 .4 .6 R/a w\A/ S. S". /\ / \J V ) --. vv (z/a = .5) ' \/ c) .; > .4 .( R/a 3 .8 (z/a = .25) .2 .4 .6 R/a \|E X I p rw\ A A /\ z' n ^V (z/a = 1.25) 2 4 6 R/a (z/a 1.0) r^sA?\P\/ 2 .4 6 R/a IE, A/\ f\~riA>k / 1/v^ (z/a= .75) vy .2 .4 .6 R/a IE J \i'>^V/I A fl A /I A n U/Q--5) ^/v .2 .4 .6 R/a (z/a = .25) .2 4 6 R/a (a) a/A = 12 ^ a/A = 1S Figure A3. Near-field amplitude of circular antenna. 103 77 . .2 .4 .6 .8 R/a (z/a=.5) 77 . -7T 2 .4 R/a (z/a = .25) 6 .8 IT +. 77" .2 .4 6 .8 R/a (z/a=I.O) 77 . -77 .2 .4 .6 .8 R/a (z/a=75) Figure A4 . Near-field phase of circular antenna (a/A = 12). 104 Appendix B Position and Instrumentation Errors for the Null Depth of Difference Patterns 1 . Position Errors The errors in null depth of a difference pattern caused by transverse displacement errors (AP) of the near-field scanner can be derived in the same way and are given by the same upper bound ex-pression as that of the transverse errors for sum patterns (eq. (60a)). However, the sum pattern derivation of Section III.B.l for longitudinal displacement errors cannot be applied to finding errors in the null depth of difference patterns. The main reasons for this are twofold. The first and most obvious is that the equation for difference patterns analogous to eq . (48) may not be satisfiable be-cause the near- field of a difference pattern changes phase by 180° across the aperture. Secondly, as the analysis below shows, the greatest effect of a longitudinal displacement error is a slight shift in the position of the null rather than a change in the depth of the null. If the longitudinal position errors were zero and the perpen-dicular (e ) to the scan plane were parallel to the null axis, the far-field in the direction of the "null" is given in terms of the near-field by eq. (10): null m 1 e ik(r-d) j p- d dp -(B1) Ao where, as usual, A refers essentially to that part of the scan area that covers the antenna aperture, and just the x-component of electric-field will be considered first. Now if the amplitude of E (P,d) on opposite halves of the aper-x ture (and thus A ) were equal and the phase difference exactly 180 the field in the direction of the null axis would actually be zero. In reality the amplitudes are not exactly equal and the phase dif-ference, even on the average, is not exactly 180°. Specifically, we can write E (P,d) as 2v 105 Ex (P,d) i( +kd) (A +AA1 e (over one half, A ,) A. A 01 A e x i( +kd+Aip ) ^ r ox r x' (over second half, A 2 ) (B2) with AA /A and Axp both << 1. For simplicity has been chosen XX X ox constant, although it can be shown that the results do not change significantly if cf> is allowed to vary slightly across A . Substi-0X iAip ° tution of eq. (B2) into (Bl) and approximating e x by (1 + iAip ) yields ,null i(kr+c\|) ) r y Y ox^ i Xr A AA dP x ol A / iAx Aij, x dP o2 (B3) for the field in the null direction. Experimentally, it has been found at the NBS that for many if not all antennas operating in a difference mode the first integral in (B3) predominates , but for the sake of the error analysis we must retain both integrals. When longitudinal displacement errors (Az) are introduced, eq (10) shows that eq . (Bl) must be replaced by .null 1 x Ar i(kr+(J) ) ik(Az-sin9e D »P) ox / E fP,d) e K dP, A X o (B4) where E represents the null field computed from the actual near-field data containing errors in the z-position of the scanner. 6 can no longer be set equal to zero because the error Az may shift the angular position of the null axis. After substituting eq. (B2) into (B4) , expanding the exponential in a power series, and discarding error terms higher than second order, eq . (B4) becomes, 106 .null' Jx i(kr+cf> ) r ox Ar / AA dP -/ iA Aip dP A^i A , X X ol o2 (B5) + / (±A +AA + ) A X X o ik(Az-6e D 'P) + k 2 Az0e p P (kAz) 2 _ Ck9eR -P) , 2 2 'R dP R k / Ax A^ x (Az-9e R -P) dP A o2 with AA in A , x ol AA in Ao2 and the + and -sign before A being used in the areas A , and A ~ 6 x to ol o2 respectively. The first order error term in eq . (B5) can be made zero by choosing 0=6 such that to s / (±Ax+AAx ) Az dP = s / (±Ax +AAx )(e R -P) dP o o (B6) corresponds essentially to the shift in the direction of the null caused by longitudinal position errors. If we take the weighted i — i III £LX average value of [£P\| at about I /4, eq. (B6) implies that for most antennas Imax < Azmax ' ,max where I denotes the maximum width of the antenna aperture defining 6 = 2ttAz /X, eq. (B7) becomes approximately, max max (B7) By (B8) With the choice of given by eq. (B6) the amplitude, ,null ' r,null i x E"" \| , of the error field in the null direction may be written by subtracting eq. (B3) from eq. (B5) , 107 pnull ' pnull x x <h k 2 / ± A A x Az0 e n P s R (Az) Ce s e R .P) 2 dP k / Ax AiP x (Az-6 s eR .P) dP o2 (B9) The first term in the first integral of eq . (B9) can be made zero by merely shifting the reference plane from which Az is measured. The third term in the first integral is identically zero because it is an odd function over the scan area A o c null ' ^null i . k x x ' — Ar Thus eq . (B9) reduces to k / A A^j, (Az-6 \|P\|) dP + / ± A (Az) A , X X S Z A X o2 o (BIO) The second integral in eq. (BIO) is also negligible if we assume the rms value of (Az) is approximately equal on the "positive" and "nega-tive" sides of A . Finally, since the maximum value of is 26 A/tt£, J s max ma) eq. (BIO) becomes cnull' c null x x ave k 6 A max Ar Imax / A Aij, \|P\| dP A X X o2 x a i a ve o Alp max x Ar / A dP, A , X o2 (Bll) with AiJj denoting the average phase difference from it radians of the probe output between the positive and negative sides of the partial scan area which is perpendicular to the null axis . The factor t— / A dP is approximately equal to the maximum Ko2 field in the mainbeams of the difference pattern. And since an ex-pression analagous to eq. (Bll) holds for the y-component of the field, we can write the fractional error ri(r) near the null of a difference pattern as ^ 1 6 max A^ave ^ r ) ' (B12a) where, as in the main text, g(r) is the ratio of the amplitude of the maximum far-electric-field to the far-electric-field at the given direction r. Here r is essentially the direction of the null axis and thus g(r) the ratio of maximum far-field to null axis far-field. Now eq. (B12) is a very simple expression. However, the average phase difference A^ of the probe output may be a difficult 108 quantity to estimate accurately from the near-field phase data. For-tunately, it can be shown that Aip is related to the relative size of the two mainlobes in the far-field of the difference pattern. Specifically, a straightforward but rather lengthy manipulation (which will not be shown here) of the near-field of a difference pat-tern reveals that Ail» can be approximated by r ave ^ r J Ail; -4AF y ave where AF is the fractional difference between the amplitude of the two main far-field lobes of the difference pattern. For example, if the amplitude of one mainlobe is 10 on some linear scale and the amplitude of the other mainlobe is 10.1 then AF would equal (10.1-10)/10 or .01. Equation (B12a) can now be written in the alternative form ^ i 46max AF ^' (B12b) 2 . Instrumentation Errors The inaccuracies in measuring near-field phase have the same effect on the null-depth as the longitudinal position errors. That is, A , the maximum instrumentation error in measuring phase, simply replaces 6 in eqs . (B8) and (B12) . IT13.X The inaccuracies in measuring amplitude affect the null depth differently, however, than the transverse position errors because the nonlinearities in the instrumentation distort the amplitude on each side of the difference pattern in nearly the same way and thus their effect on the null depth is much smaller than might first be expected. Specifically, errors in measuring amplitude show up in the null depth through distortion of only the AA part (see eq. (B2)) of the near field. Carrying through an analysis similar to that per-formed above for z-position errors yields the following expression for the maximum change AE. in null electric field caused by ampli-tude errors: \|AEnull\| 1 d\| ; \|AA\|) dp £ dB max ol ? (B13) Aol Xr 109 where AA is the maximum of AA across A , and N JT. is the maximum max ol dB instrumentation error in measuring the amplitude of the probe output (see Section III.B.2). If we approximate \|E. \|, the x-component AAmaxA j of which is given in eq. (B3) , by ^ , then the fractional z Ar error in null depth caused by instrumentation errors in measuring amplitude can be written (B14) Note that the instrumentation amplitude error in null depth does not depend on the null depth itself. In addition, it can be shown that, unlike phase errors, the instrumentation amplitude errors have negligible effect on the angular position of the null direction. 110 References IEEE Standard Definitions of Terms for Antennas , IEEE Transac-tions on Antennas and Propagation, AP- 22 , 1 (January 1974). DeSize, L.K. and J.F. Ramsay, in Microwave Scanning Antennas , Ed. R.C. Hansen, Academic Press--New York, London (1964), Vol. 1, Ch. 2-II.B. Kerns, D.M., "Correction of Near-Field Antenna Measurements Made with an Arbitrary But Known Measuring Antenna," Electronics Letters, 6, 11, pp. 346-347 (28th May 1970). See e.g. Jackson, J.D., Classical Electrodynamics , John Wiley § Sons Inc., New York, London (1962), Section 9.6. a Braunbek, W. , "Zur Beugung an der Kreisscheibe , " Zeitschrift Fur Physik, 122, pp. 405-415 (1950). b Ufimtsev, P. la., "Approximate Computation of the Diffraction of Plane Electromagnetic Waves at Certain Metal Bodies," Soviet Physics--Technical Physics, 2, 8, pp. 1708-1718 (August 1957). a Keller, J.B., "Diffraction by an Aperture," Journal of Applied Physics, 2^, 4, pp. 426-444 (April 1957). b Keller, J.B., "Geometrical Theory of Diffraction," Journal of the Optical Society of America, 5_2, 2, pp. 116-130 (February 1962) . Van Kampen, N.G., "An Asymptotic Treatment of Diffraction Problems," Physica, 14, 9, pp. 575-589 (January 1949). Keller, J.B., R.M. Lewis, and B.D. Seckler, "Diffraction by an Aperture II," Journal of Applied Physics, 2_8, 5, pp. 570-579 (May 1957) . a Kouyoumjian, R.G., and P.H. Pathak, "A Uniform Geometrical Theory of Diffraction for an Edge in a Perfectly Conducting Surface," Proceedings of the IEEE, 62, 11, pp. 1448-1461 (November 1974) . b Hwang, Y.M. and R. G. Kouyoumjian, "A Dyadic Coefficient for an Electromagnetic Wave Which is Rapidly-Varying at an Edge," URSI 1974 Annual Meeting, Boulder, Colorado. See e.g., Born M. , and E. Wolf, Principles of Optics , Pergamon Press, Oxford, New York (1970), 4th Ed., Appendix III. Ill Newell, A.C. and M.L. Crawford, "Planar Near-Field Measurements on High Performance Array Antennas," NBSIR 74-380, National Bureau of Standards, Boulder, Colorado (July 1974). Kraus , J.D., Antennas , McGraw-Hill, Inc., New York, Toronto, London (1950), Chs . 3 and 12. Hansen, R.C., (Ch. 1, Sect. 2-1 in ref ): A /A = G/G in Table IX, p. 66. Rusch, W.V.T. and P.D. Potter, Analysis of Reflector Antennas , Academic Press, New York -London (1970), Section 2.52. Ruze, John, "Antenna Tolerance Theory --A Review," Proceedings IEEE, 54, 4, pp. 633-640 (April 1966). Rodrigue, G.P., E.B. Joy, and C.P. Burns, An Investigation of the Accuracy of Far-Field Radiation Patterns Determined from Near-Field Measurements , Report --Georgia Institute of Technology, Atlanta, Georgia (August 1973). Newell, A.C, private communication, National Bureau of Stand-ards, Boulder, Colorado. Kerns, D.M., "Scattering-Matrix Description and Near-Field Mea-surements of Electroacoustic Transducers," The Journal of the Acoustical Society of America, 5_7, 2, pp. 497-507 (February 1975) Schoch, V.A. , "Betrachtungen liber das Schallfeld einer Kolbenmembran," Akustische Zeitschrift, 6, pp. 318-326 (1941). Crawford, M.L., A.C. Newell, J.W. Greene, and A.D. Yaghjian, "Experimental Design Study for a Near-Field Broadbeam Antenna Pattern Calibrator," AFAL-TR- 75-179 , Air Force Avionics Lab., Wright-Patterson Air Force Base, Ohio 45433 (to be published). Newell, A.C, "Planar Near-Field Measurement Techniques on High Performance Arrays --Part II," Air Force Technical Report, Air Force Avionics Laboratory -- Wright Patterson Air Force Base, Ohio (to be published) Newell, A.C, R.C Baird, and P.F. Wacker, "Accurate Measurement of Antenna Gain and Polarization at Reduced Distances by an Extrapolation Technique," IEEE Transactions on Antenna and Propagation, AP-21 , 4, pp. 418-431 (July 1973). 112 Kanda, M., "Accuracy Considerations in the Measurement of the Power Gain of a Large Microwave Antenna," IEEE Transactions on Antennas and Propagation, AP-23 , 3, pp. 407-411 (May 1975). Appel -Hansen, J., "Reflectivity Level of Radio Anechoic Chambers," IEEE Transactions on Antennas and Propagation, AP-21 , pp. 490-498 (July 1973) . Jensen, Frank, "Electromagnetic near- field-far-field correla-tions," Ph.D. Dissertation, Technical University of Denmark, Lyngby, Denmark (July 1970). ACKNOWLEDGMENT The author is grateful to Allen C. Newell for his invaluable consultation throughout the preparation of this report. Much appreciation is extended to John .W. Greene and Douglas P. Kremer for the information they provided about the NBS near-field scanning facilities. Mark T. Ma and Carl F. Stubenrauch of the Department of Commerce, and Kenneth Grimm of the Air Force Avionics Laboratory contributed many helpful suggestions in reviewing the report. Many thanks also go to Janet R. Jasa for carefully and patiently typing the manuscript . 113 U.S. DEPT. OF COMM. BIBLIOGRAPHIC DATA SHEET 1. PUBLK A 1 ION OK HI- PORT NO. NBS Tech Mote 667 2. Ciov't Accession No. 4. TITLE AND SUB' Upper-Bound Errors in Far-Field Antenna Parameters Determined from Planar Near-Field Measurements, Part I : Analysis 3. Recipient's Accession No. 5. Publication Date October 1975 6. Performing Organization Cod< 276.05 7. AU rHOR(S) Arthur D Y a g h j i a n 8. Performing Organ. Report No. TN-667 9. PERFORMING ORGANIZATION NAME AND ADDRESS NATIONAL BUREAU OF STANDARDS, DEPARTMENT OF COMMERCE WASHINGTON, D.C. 20234 Bou 1 der Labs 10. Projccr/Task/Work Unit No. 2765276 11. Contract Grant No. 12. Sponsoring Organization Name and < omplete Address (Street, City, State, ZIP) Air Force Avionics Laboratory Air Force Wright Aeronautical Laboratories Air Force Systems Command Wright-Patterson Air Force Base, Ohio 45433 13. Type of Report & Period C overed 7/73 -7/74 14. Sponsoring Agency ( od 15, SUPPLEMENTARY NOTES 16. ABSTRACT (A 200-word or less factual summary of most significant inforniation. If document includes a significant bibliography or literature survey, mention it here.) General expressions are derived for estimating the errors in the sum or difference far-field pattern of electrically large aperture antennas which are measured by the planar near-field scanning tech-nique. Upper bounds are determined for the far-field errors produced by 1) the nonzero fields outside the finite scan area, 2) the inac-curacies in the positioning of the probe, 3) the distortion and non-linearities of the instrumentation which measures the amplitude and phase of the probe output, and 4) the multiple reflections. Computa-tional errors, uncertainties in the receiving characteristics of the probe, and errors involved with measuring the input power to the test antenna are briefly discussed. 17. KEY WORDS (six to twelve entries; alphabetical order; capitalize only the first letter of the first key word unless a proper name; separated by semicolons) Antennas; error analysis; far^field pattern; near-field measurements; planar scanning; plane wave spectrum. 18. AVAILABILITY [jfc Unlimited _ For Official Distribution. Do Not Release to NTIS PX 1 Order From Sup. of Doc, U.S. Government Pointing Dlfice Washington, D.C. 20 402, SO Cat. No. CM. . 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2610	https://www.khanacademy.org/science/hs-chemistry/x2613d8165d88df5e:solutions-acids-and-bases/x2613d8165d88df5e:molarity/e/understand-molarity	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails.
2611	https://math.stackexchange.com/questions/4578882/given-the-vectors-a-and-b-such-that-ab-and-2a-b-are-perpendicular-and-a-b-and-4	linear algebra - Given the vectors a and b such that a+b and 2a-b are perpendicular and a-b and 4a+b are perpendicular, find the angle between a and b - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Given the vectors a and b such that a+b and 2a-b are perpendicular and a-b and 4a+b are perpendicular, find the angle between a and b Ask Question Asked 2 years, 10 months ago Modified2 years, 10 months ago Viewed 566 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. (a+b)⋅(2 a−b)=0(a+b)⋅(2 a−b)=0 (a−b)⋅(4 a+b)=0(a−b)⋅(4 a+b)=0 2 a⋅a+a⋅b−b⋅b=0 2 a⋅a+a⋅b−b⋅b=0 4 a⋅a−3 a⋅b−b⋅b=0 4 a⋅a−3 a⋅b−b⋅b=0 b⋅b=2 a⋅a+a⋅b b⋅b=2 a⋅a+a⋅b then I found a⋅a=2 a⋅b a⋅a=2 a⋅b b⋅b=5 a⋅b b⋅b=5 a⋅b Then I dont know how to continue to find an angle. Can someone help. Appreciate that linear-algebra vectors inner-products angle Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Nov 17, 2022 at 17:56 MeltedStatementRecognizing 729 1 1 gold badge 4 4 silver badges 17 17 bronze badges asked Nov 17, 2022 at 16:27 yoloyolo 1 1 1 bronze badge 3 For some basic information about writing mathematics at this site see, e.g., here, here, here and here.Another User –Another User 2022-11-17 16:28:45 +00:00 Commented Nov 17, 2022 at 16:28 Does the OP realize that this is an exercise about inner products of vectors?M. Wind –M. Wind 2022-11-17 16:57:03 +00:00 Commented Nov 17, 2022 at 16:57 Hints: 1) The cosine of the angle equals the dot product divided by the product of the magnitudes. 2) The dot product of v v with itself equals the square of the magnitude of v v. Then just substitute and simplify to get the desired cosine.Ned –Ned 2022-11-17 23:52:11 +00:00 Commented Nov 17, 2022 at 23:52 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. 0=(a⃗+b⃗)⋅(2 a⃗−b⃗)=2 a⃗⋅a⃗−b⃗⋅b⃗+a⃗⋅b⃗=2 a 2−b 2+a b cos(θ)0=(a→+b→)⋅(2 a→−b→)=2 a→⋅a→−b→⋅b→+a→⋅b→=2 a 2−b 2+a b cos⁡(θ) 0=(a⃗−b⃗)⋅(4 a⃗+b⃗)=4 a⃗⋅a⃗−b⃗⋅b⃗−3 a⃗⋅b⃗=4 a 2−b 2−3 a b cos(θ)0=(a→−b→)⋅(4 a→+b→)=4 a→⋅a→−b→⋅b→−3 a→⋅b→=4 a 2−b 2−3 a b cos⁡(θ) Multiply the first equation by 3 3, then add the second equation. It follows that 10 a 2=4 b 2 10 a 2=4 b 2, hence b=1 2 10−−√a b=1 2 10 a. Substitute this result into one of the two equations and you get cos(θ)=1 10 10−−√cos⁡(θ)=1 10 10. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Nov 18, 2022 at 4:37 M. WindM. Wind 4,305 1 1 gold badge 16 16 silver badges 21 21 bronze badges 0 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions linear-algebra vectors inner-products angle See similar questions with these tags. 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Sound Waves and Music » Reflection, Refraction, and Diffraction Sound Waves and Music - Lesson 3 Behavior of Sound Waves Reflection, Refraction, and Diffraction Interference and Beats The Doppler Effect and Shock Waves Boundary Behavior Reflection, Refraction, and Diffraction Getting your Trinity Audio player ready... Hold down the T key for 3 seconds to activate the audio accessibility mode, at which point you can click the K key to pause and resume audio. Useful for the Check Your Understanding and See Answers. Like any wave, a sound wave doesn't just stop when it reaches the end of the medium or when it encounters an obstacle in its path. Rather, a sound wave will undergo certain behaviors when it encounters the end of the medium or an obstacle. Possible behaviors include reflection off the obstacle, diffraction around the obstacle, and transmission (accompanied by refraction) into the obstacle or new medium. In this part of Lesson 3, we will investigate behaviors that have already been discussed in a previous unit and apply them towards the reflection, diffraction, and refraction of sound waves. Reflection and Transmission of Sound When a wave reaches the boundary between one medium another medium, a portion of the wave undergoes reflection and a portion of the wave undergoes transmission across the boundary. As discussed in the previous part of Lesson 3, the amount of reflection is dependent upon the dissimilarity of the two media. For this reason, acoustically minded builders of auditoriums and concert halls avoid the use of hard, smooth materials in the construction of their inside halls. A hard material such as concrete is as dissimilar as can be to the air through which the sound moves; subsequently, most of the sound wave is reflected by the walls and little is absorbed. Walls and ceilings of concert halls are made softer materials such as fiberglass and acoustic tiles. These materials are more similar to air than concrete and thus have a greater ability to absorb sound. This gives the room more pleasing acoustic properties. Reflection of sound waves off of surfaces can lead to one of two phenomena - an echo or a reverberation. A reverberation often occurs in a small room with height, width, and length dimensions of approximately 17 meters or less. Why the magical 17 meters? The effect of a particular sound wave upon the brain endures for more than a tiny fraction of a second; the human brain keeps a sound in memory for up to 0.1 seconds. If a reflected sound wave reaches the ear within 0.1 seconds of the initial sound, then it seems to the person that the sound is prolonged. The reception of multiple reflections off of walls and ceilings within 0.1 seconds of each other causes reverberations - the prolonging of a sound. Since sound waves travel at about 340 m/s at room temperature, it will take approximately 0.1 s for a sound to travel the length of a 17 meter room and back, thus causing a reverberation (recall from Lesson 2, t = d/v = (34 m)/(340 m/s) = 0.1 s). This is why reverberations are common in rooms with dimensions of approximately 17 meters or less. Perhaps you have observed reverberations when talking in an empty room, when honking the horn while driving through a highway tunnel or underpass, or when singing in the shower. In auditoriums and concert halls, reverberations occasionally occur and lead to the displeasing garbling of a sound. But reflection of sound waves in auditoriums and concert halls do not always lead to displeasing results, especially if the reflections are designed right. Smooth walls have a tendency to direct sound waves in a specific direction. Subsequently the use of smooth walls in an auditorium will cause spectators to receive a large amount of sound from one location along the wall; there would be only one possible path by which sound waves could travel from the speakers to the listener. The auditorium would not seem to be as lively and full of sound. Rough walls tend to diffuse sound, reflecting it in a variety of directions. This allows a spectator to perceive sounds from every part of the room, making it seem lively and full. For this reason, auditorium and concert hall designers prefer construction materials that are rough rather than smooth. Reflection of sound waves also leads to echoes. Echoes are different than reverberations. Echoes occur when a reflected sound wave reaches the ear more than 0.1 seconds after the original sound wave was heard. If the elapsed time between the arrivals of the two sound waves is more than 0.1 seconds, then the sensation of the first sound will have died out. In this case, the arrival of the second sound wave will be perceived as a second sound rather than the prolonging of the first sound. There will be an echo instead of a reverberation. Reflection of sound waves off of surfaces is also affected by the shape of the surface. As mentioned of water waves in Unit 10, flat or plane surfaces reflect sound waves in such a way that the angle at which the wave approaches the surface equals the angle at which the wave leaves the surface. This principle will be extended to the reflective behavior of light waves off of plane surfaces in great detail in Unit 13 of The Physics Classroom. Reflection of sound waves off of curved surfaces leads to a more interesting phenomenon. Curved surfaces with a parabolic shape have the habit of focusing sound waves to a point. Sound waves reflecting off of parabolic surfaces concentrate all their energy to a single point in space; at that point, the sound is amplified. Perhaps you have seen a museum exhibit that utilizes a parabolic-shaped disk to collect a large amount of sound and focus it at a focal point. If you place your ear at the focal point, you can hear even the faintest whisper of a friend standing across the room. Parabolic-shaped satellite disks use this same principle of reflection to gather large amounts of electromagnetic waves and focus it at a point (where the receptor is located). Scientists have recently discovered some evidence that seems to reveal that a bull moose utilizes his antlers as a satellite disk to gather and focus sound. Finally, scientists have long believed that owls are equipped with spherical facial disks that can be maneuvered in order to gather and reflect sound towards their ears. The reflective behavior of light waves off curved surfaces will be studies in great detail in Unit 13 of The Physics Classroom Tutorial. Diffraction of Sound Waves Diffraction involves a change in direction of waves as they pass through an opening or around a barrier in their path. The diffraction of water waves was discussed in Unit 10 of The Physics Classroom Tutorial. In that unit, we saw that water waves have the ability to travel around corners, around obstacles and through openings. The amount of diffraction (the sharpness of the bending) increases with increasing wavelength and decreases with decreasing wavelength. In fact, when the wavelength of the wave is smaller than the obstacle or opening, no noticeable diffraction occurs. Diffraction of sound waves is commonly observed; we notice sound diffracting around corners or through door openings, allowing us to hear others who are speaking to us from adjacent rooms. Many forest-dwelling birds take advantage of the diffractive ability of long-wavelength sound waves. Owls for instance are able to communicate across long distances due to the fact that their long-wavelength hoots are able to diffract around forest trees and carry farther than the short-wavelength tweets of songbirds. Low-pitched (long wavelength) sounds always carry further than high-pitched (short wavelength) sounds. Scientists have recently learned that elephants emit infrasonic waves of very low frequency to communicate over long distances to each other. Elephants typically migrate in large herds that may sometimes become separated from each other by distances of several miles. Researchers who have observed elephant migrations from the air and have been both impressed and puzzled by the ability of elephants at the beginning and the end of these herds to make extremely synchronized movements. The matriarch at the front of the herd might make a turn to the right, which is immediately followed by elephants at the end of the herd making the same turn to the right. These synchronized movements occur despite the fact that the elephants' vision of each other is blocked by dense vegetation. Only recently have they learned that the synchronized movements are preceded by infrasonic communication. While low wavelength sound waves are unable to diffract around the dense vegetation, the high wavelength sounds produced by the elephants have sufficient diffractive ability to communicate long distances. Bats use high frequency (low wavelength) ultrasonic waves in order to enhance their ability to hunt. The typical prey of a bat is the moth - an object not much larger than a couple of centimeters. Bats use ultrasonic echolocation methods to detect the presence of bats in the air. But why ultrasound? The answer lies in the physics of diffraction. As the wavelength of a wave becomes smaller than the obstacle that it encounters, the wave is no longer able to diffract around the obstacle, instead the wave reflects off the obstacle. Bats use ultrasonic waves with wavelengths smaller than the dimensions of their prey. These sound waves will encounter the prey, and instead of diffracting around the prey, will reflect off the prey and allow the bat to hunt by means of echolocation. The wavelength of a 50 000 Hz sound wave in air (speed of approximately 340 m/s) can be calculated as follows wavelength = speed/frequency wavelength = (340 m/s)/(50 000 Hz) wavelength = 0.0068 m The wavelength of the 50 000 Hz sound wave (typical for a bat) is approximately 0.7 centimeters, smaller than the dimensions of a typical moth. Refraction of Sound Waves Refraction of waves involves a change in the direction of waves as they pass from one medium to another. Refraction, or bending of the path of the waves, is accompanied by a change in speed and wavelength of the waves. So if the media (or its properties) are changed, the speed of the wave is changed. Thus, waves passing from one medium to another will undergo refraction. Refraction of sound waves is most evident in situations in which the sound wave passes through a medium with gradually varying properties. For example, sound waves are known to refract when traveling over water. Even though the sound wave is not exactly changing media, it is traveling through a medium with varying properties; thus, the wave will encounter refraction and change its direction. Since water has a moderating effect upon the temperature of air, the air directly above the water tends to be cooler than the air far above the water. Sound waves travel slower in cooler air than they do in warmer air. For this reason, the portion of the wavefront directly above the water is slowed down, while the portion of the wavefronts far above the water speeds ahead. Subsequently, the direction of the wave changes, refracting downwards towards the water. This is depicted in the diagram at the right. Refraction of other waves such as light waves will be discussed in more detail in a later unit of The Physics Classroom Tutorial. Jump To Next Lesson: Natural Frequency Tired of Ads? 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2613	https://stats.stackexchange.com/questions/467233/combinatory-probability-group-composed-by-different-balls-of-different-colors	combinatorics - Combinatory probability, group composed by different balls of different colors - Cross Validated Join Cross Validated By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR == Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… 1. 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You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Combinatory probability, group composed by different balls of different colors Ask Question Asked 5 years, 3 months ago Modified5 years, 3 months ago Viewed 642 times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. Let's say you have a group of M M balls of different colors in a box. For example, 20 balls are red, 15 are blue, 10 are green, 5 are grey, 5 are yellow and 5 violet, for a total of M=60 M=60 balls. You pick 1⩽n⩽M 1⩽n⩽M of them without replacement. The order of the colors does not count, so for instance, if n=2 n=2 and you pick red then gray, it is the same as picking gray then red. How do I calculate the probability of all the possible outcomes for n n elements? Is there a general formula for this problem? In particular, if M M and n n are large then the number of possible combinations is large, so how can I find the most probable combinations? probability combinatorics Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications edited May 19, 2020 at 22:47 Ben 141k 7 7 gold badges 277 277 silver badges 636 636 bronze badges asked May 18, 2020 at 20:06 FabrizioFabrizio 123 5 5 bronze badges 2 1 "Probability of all the possible outcomes" seems either too inclusive or too vague. Specifically, if 4 balls drawn without replacement, then probability of getting exactly 2 red, 1 green and 1 yellow is (20 2)(10 1)(5 1)(25 0)(60 4).(20 2)(10 1)(5 1)(25 0)(60 4). –BruceET Commented May 18, 2020 at 21:22 1 See Wikipedia on 'multinomial distribution' for more. –BruceET Commented May 18, 2020 at 21:30 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. Yes, there is a general formula. Consider an urn containing M M balls, where M 1 M 1 balls have the color c 1 c 1, M 2 M 2 balls have the color c 2 c 2,..., M r M r balls have the color c r c r, and M 1+⋯+M r=M M 1+⋯+M r=M. If you draw a sample of size n library(extraDistr) K <- 10 # sample size x <- subset(expand.grid(red=0:20, blue=0:15, green=0:10, gray=0:5, yellow=0:5, violet=0:5), red+blue+green+gray+yellow+violet==K) dim(x) 2625 6 head(x) red blue green gray yellow violet 11 10 0 0 0 0 0 31 9 1 0 0 0 0 51 8 2 0 0 0 0 71 7 3 0 0 0 0 91 6 4 0 0 0 0 111 5 5 0 0 0 0 tail(x) red blue green gray yellow violet 739201 0 0 0 2 3 5 753986 1 0 0 0 4 5 754006 0 1 0 0 4 5 754321 0 0 1 0 4 5 757681 0 0 0 1 4 5 776161 0 0 0 0 5 5 p <- dmvhyper(x, n=c(20,15,10,5,5,5), k=K) max(p) 0.008930581 x[which.max(p),] red blue green gray yellow violet 159646 3 2 2 1 1 1 dmvhyper(x[which.max(p),], n=c(20,15,10,5,5,5), k=K) 0.008930581 choose(20,3)choose(15,2)choose(10,2)555/choose(60,K) 0.008930581 Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited May 20, 2020 at 13:30 answered May 19, 2020 at 22:05 Image 8: Sergio's user avatar SergioSergio 6,151 2 2 gold badges 16 16 silver badges 29 29 bronze badges 2 \ Usually, comments are not for this but your answer is absolutely beautiful and impressive. I was able to get a result but yours is amazing and efficient. –Fabrizio Commented May 20, 2020 at 13:32 \ You are welcome. Happy to be useful! –Sergio Commented May 20, 2020 at 13:35 Add a comment\| . In the following case, I extracted 12 elements and I did a statistics using 10 million extractions. library(parallel) library(plyr) name of the elements of the set acolors=c("B","G", "O", "R", "Y", "Gr") generating a real set with a certain composition alist=c(rep("B",20), rep("G", 15), rep("O", 10), rep("R",5), rep("Y",5), rep("Gr",5) ) number of extraction to simulate pulls=10000000 parallel version of the extraction all_res=mclapply(1:pulls, function(x, alist, acolors){ ares=NULL asamp=list(table(sample(alist, 12))) for(ac in acolors){ if(is.na(asamp[ac])){ ares=c(ares,0) } else{ ares=c(ares,asamp[ac]) } } return(ares) }, alist=alist, acolors=acolors, mc.cores=8) tdata store the result of each extraction. Each column has a given name corresponding to "acolors" a line can look like 4,2,3,1,1,1, that mean 4 from color B, 2 from color G and so on... tdata=as.data.frame(do.call(rbind, all_res)) colnames(tdata)=acolors now we do the statistics of the result, we count how many times a given line is duplicated stat_res=as.data.frame(ddply(tdata,.(B, G, O, R, Y, Gr),nrow)) we sort the data frame from the most probable to the least probable stat_res=stat_res[order(stat_res$V1, decreasing = TRUE ),] we calculate the frequency of each line stat_res$frequency=stat_res$V1/sum(stat_res$V1) ``` With the results is then possible to use the binomial formula given by @BruceET and calculate the probability for the most frequent results. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered May 19, 2020 at 20:15 FabrizioFabrizio 123 5 5 bronze badges 0 Add a comment\| Your Answer Thanks for contributing an answer to Cross Validated! Please be sure to _answer the question_. Provide details and share your research! But _avoid_ … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. MathJax reference. To learn more, see our tips on writing great answers. Draft saved Draft discarded Sign up or log in Sign up using Google Sign up using Email and Password Submit Post as a guest Name Email Required, but never shown Post Your Answer Discard By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. 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2614	https://www.ck12.org/flexi/precalculus/sums-of-geometric-series/	Sums of Geometric Series Arithmetic Series Counting with Permutations and Combinations Concept Summary: \| \| \| A geometric series is a sum of numbers whose consecutive terms form a geometric sequence, and it can be finite or infinite. The sum of a finite geometric series can be calculated using the formula: where is the first term, is the common ratio, and is the number of terms. An infinite geometric series can be either convergent or divergent, depending on the value of the common ratio If the infinite geometric series is divergent, and its sum does not go to a specific number. If \|r\| < 1, the infinite geometric series is convergent, and its sum can be calculated using the formula: A partial sum of an infinite sum is the sum of all the terms up to a certain point. \| Represent geometrically the following numbers on the number line : Represent geometrically the following numbers on the number line : Represent geometrically the following numbers on the number line : Is the sequence geometric? What is the sum of the geometric series? What is the sum of the infinite geometric series? What is the sum of an infinite geometric series? Which geometric series converges? Which geometric series represents 0.4444 as a fraction? How do you find the sum of an infinite geometric series? How do you find the sum of a geometric series? Evaluate: Infinite Geometric Series Determine whether the infinite geometric series is convergent or divergent. If it is convergent, find its sum. Infinite Geometric Series Determine whether the infinite geometric series is convergent or divergent. If it is convergent, find its sum. Infinite Geometric Series Determine whether the infinite geometric series is convergent or divergent. If it is convergent, find its sum. Infinite Geometric Series Determine whether the infinite geometric series is convergent or divergent. If it is convergent, find its sum. Directions: Consider the following function: f(x) = (1-x)^{-2}. Find the Radius of Convergence for this series. If necessary, write INF for ∞. R=1 Directions: Consider the following function: f(x) = (1-x)^{-2}. Write out the Maclaurin series for f(x) (in sigma notation). f(x) = ∑_{n=0}^{∞} ____ Find a power series solution of the differential equation given below. Determine the radius of convergence of the resulting series, and use the series given below to identify the series in terms of familiar elementary functions. (10x-1)y' + 10y = 0 Determine the maximum value of the sum S, given by , over all sequences of nonnegative real numbers satisfying . Evaluate: When written out in full, the number has 4041 digits. What is the sum of the digits of this 4041-digit number? Find the nth term of the sequence 1/1, 1/2, 1/3,... The common ratio of a geometric progression is 2, and the first term is 3. What is the fourth term of the sequence? Evaluate: $$\sum_{n=1}^{6} 4(3)^{n-1}$$ Evaluate: S = Find the missing term in the geometric sequence: 6, [?], Consider a geometric series 4, 12, 36, ..... What is the common ratio of the sequence? Find the 7th term of this geometric sequence: -1, 2, -4, 8, ... Find the 6th term of the Geometric progression 3, 9, 27. Evaluate . How to tell if a given series is geometric? How do you find a finite geometric series? How to find the common ratio in a geometric series? How do you find r in a geometric series? How can geometric series in sigma notation be evaluated? How do you find the common ratio in a geometric series? How can you tell if a series is geometric? How do you evaluate Geometric Series in Sigma Notation? What is the process for evaluating a geometric series? How can an infinite geometric series be evaluated? How do you evaluate a geometric series? How to tell if a series is geometric? What is a geometric series? What is the difference of two squares theorem? What is a common ratio? What is the method for finding the common ratio? What is the difference between two squares? What is an infinite geometric series? How to solve Fourier series problems? How to find partial sum of infinite series? How do you solve the difference of two squares? How do you factor using the difference of two squares? How can I find the common ratio? How can Fourier Series problems be solved? Define an infinite geometric series. CK-12 Foundation is a non-profit organization that provides free educational materials and resources. 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2615	https://www.immunopathol.com/Article/ipp-41724	Association between metabolic syndrome and risk of endometrial cancer; a systematic review and meta-analysis Home Journal Information For Reviewers Human and Animal Rights Open Access Publication Ethics and Malpractice Statement Publisher Guidelines Informed Consent Peer Review Process Conflicts of Interest DISCLAIMER Editorial Workflow Peer Review Process Publishing schedule/Archiving Contact Us Create Account Login Please don't use dangerous characters Search eISSN: 2423-8015 Read This Article PDF Article History Submitted: 29 Jul 2024 Accepted: 17 Nov 2024 ePublished: 25 Nov 2024 Google Scholar Profile Articles by Abbasi S Articles by Rezaei J Articles by Bazi F Articles by Mousavi MA Articles by Rassouli S Articles by Zamanpour Z Articles by Hosseini E Articles by Noori M Articles by Hamidi Madani Z PubMed Profile Articles by Abbasi S Articles by Rezaei J Articles by Bazi F Articles by Mousavi MA Articles by Rassouli S Articles by Zamanpour Z Articles by Hosseini E Articles by Noori M Articles by Hamidi Madani Z Share This Article! Export Citation EndNote) (Enw Format - Win & Mac) BibTeX) (Bib Format - Win & Mac) Bookends) (Ris Format - Mac only) EasyBib) (Ris Format - Win & Mac) Medlars) (Txt Format - Win & Mac) Mendeley Web Mendeley (Ris Format - Win & Mac) Papers) (Ris Format - Win & Mac) ProCite) (Ris Format - Win & Mac) Reference Manager) (Ris Format - Win only) Refworks) (Refworks Format - Win & Mac) Zotero) (Ris Format - Firefox Plugin) Cited By Google Scholar Cited by CrossRef (1) Cited by Scopus (1) Select Language▼ Immunopathol Persa. 2025;11(2): e41724. doi:10.34172/ipp.2025.41724 Scopus ID:86000186538 Citations Citation Indexes: 1 Captures Readers: 2 see details Abstract View: 1618 PDF Download: 816 1 CITATION 1 Total citation 1 Recent citation n/a Field Citation Ratio n/a Relative Citation Ratio Meta-analysis Association between metabolic syndrome and risk of endometrial cancer; a systematic review and meta-analysis Sheida Abbasi 1, Jalal Rezaei 2, Fariba Bazi 3, Moloud Alsadat Mousavi 4, Sadaf Rassouli 5, Zeinab Zamanpour 6, Elmira Hosseini 7, Marziyeh Noori 8, Zahra Hamidi Madani 9 1 Department of Gynecology and Obstetrics, School of Medicine, Tehran University of Medical Sciences, Tehran, Iran 2 Department of Critical Care Nursing, School of Nursing and Midwifery, Tehran University of Medical Sciences, Tehran, Iran 3 Department of Reproductive Health and Midwifery, Faculty of Medical Sciences, Tarbiat Modares University, Tehran, Iran 4 Department of Gynecology and Obstetrics, Imam Hossein Hospital, School of Medicine, Shahid Beheshti University of Medical Sciences, Tehran, Iran 5 Department of Gynecology and Obstetrics, Imam Khomeini Hospital, School of Medicine, Sari University of Medical Sciences, Sari, Iran 6 Department of Gynecology and Obstetrics, School of Medicine, Jundishapur University of Medical Sciences, Ahvaz, Iran 7 Department of Gynecology and Obstetrics, School of Medicine, Urmia University of Medical Sciences, Urmia, Iran 8 Department of Gynecology and Obstetrics, Shahid Akbarabadi Hospital, School of Medicine, Iran University of Medical Sciences, Tehran, Iran 9 Reproductive Health Research Center, Department of Gynecology and Obstetrics, School of Medicine, Guilan University of Medical Sciences, Rasht, Iran Corresponding Author: Zahra Hamidi Madani, Email: Tannaz.hamidi@yahoo.com Abstract Introduction: Endometrial cancer is one of the most prevalent female malignancies, with various factors, including metabolic syndrome, contributing to its incidence. Thus, this study aims to evaluate the association between metabolic syndrome and the risk of endometrial carcinoma. Materials and Methods:In this systematic review and meta-analysis, two independent authors searched electronic databases, including Cochrane, PubMed, ProQuest, Web of Science, and the Google Scholar search engine up to May 16, 2024. Data analysis was conducted using STATA 14 software with a significance level of P< 0.05 for all tests. Results: A pooled analysis of 12 observational studies found that metabolic syndrome elevated the risk of endometrial carcinoma by 37% overall (OR: 1.37, 95% CI: 1.33, 1.42), 35% in cohort studies (OR: 1.35, 95% CI: 1.29, 1.42), and 40% in case-control studies (OR: 1.40, 95% CI: 1.33, 1.48). However, hypertension increased the risk of endometrial carcinoma by 25% (OR: 1.25, 95% CI: 1.18, 1.33), fasting hyperglycemia by 25% (OR: 1.25, 95% CI: 1.15, 1.37), hypertriglyceridemia by 17% (OR: 1.17, 95% CI: 1.13, 1.21), low high density lipoprotein (HDL) by 20% (OR:1.20, 95% CI: 1.12, 1.28), increased waist circumference by 59% (OR:1.59, 95% CI: 1.43, 1.77), pre-menopausal period by 67% (OR:1.67, 95% CI: 1.38, 2.02), and post-menopausal period by 61% (OR: 1.61, 95% CI: 1.17, 2.21). Likewise, obesity almost doubled the risk of endometrial carcinoma (OR: 2.13, 95% CI: 1.65, 2.75). Conclusion: Metabolic syndrome increases the risk of endometrial carcinoma, with obesity being the most dangerous risk factor for endometrial cancer. Thus, managing metabolic disorders in women can be an important step toward reducing the incidence of endometrial cancer. Registration: This study has been compiled based on the PRISMA checklist, and its protocol was registered on the PROSPERO (ID: CRD42024551509) and Research Registry (UIN: reviewregistry1838). Keywords:Metabolic syndrome, Endometrial neoplasms, Endometrial carcinoma, Reaven syndrome X, Insulin resistance syndrome X, Endometrium cancer Citation: Abbasi Sh, Rezaei J, Bazi F, Alsadat Mousavi M, Rassouli S, Zamanpour Z, Hosseini E, Noori M, Hamidi Madani Z. Association between metabolic syndrome and risk of endometrial cancer; a systematic review and meta-analysis. Immunopathol Persa. 2025;11(2):e41724. DOI:10.34172/ipp.2025.41724. × Association between metabolic syndrome and risk of endometrial cancer; a systematic review and meta-analysis Immunopathol Persa. 2025;11(2): e41724. Metrics Statistics Comments Citations Citation Indexes: 1 Captures Readers: 2 see details Citations Scopus - Citation Indexes: 1 Captures Mendeley - Readers: 2 - see details First Name Please don't use dangerous characters Last Name Please don't use dangerous characters Email Address Please enter a email address Comments Please don't use dangerous characters Security code The entered code is invalid. 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2616	https://www.jendodon.com/article/S0099-2399(05)60951-X/fulltext	Pulpal Pain Diagnosis—A Review - Journal of Endodontics Skip to Main ContentSkip to Main Menu Login to your account Email/Username Your email address is a required field. E.g., j.smith@mail.com Password Show Your password is a required field. Forgot password? [x] Remember me Don’t have an account? Create a Free Account If you don't remember your password, you can reset it by entering your email address and clicking the Reset Password button. 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Ok Clinical ArticlesVolume 26, Issue 3p175-179 March 2000 Download Full Issue Download started Ok Pulpal Pain Diagnosis—A Review I.B.Bender, DDS I.B.Bender, DDS Affiliations Dr. Bender is Chairman Emeritus, Department of Dentistry, Albert Einstein Medical Center, and Professor Emeritus, School of Dental Medicine, University of Pennsylvania, Philadelphia, PA Search for articles by this author Affiliations & Notes Article Info Dr. Bender is Chairman Emeritus, Department of Dentistry, Albert Einstein Medical Center, and Professor Emeritus, School of Dental Medicine, University of Pennsylvania, Philadelphia, PA Footnotes: This study was supported by a grant from the I. B. Bender Research Endowment Fund, Albert Einstein Medical Center (Philadelphia, PA). DOI: 10.1097/00004770-200003000-00012 External LinkAlso available on ScienceDirect External Link Copyright: © 2000 The American Association of Endodontists. Get Access Outline Outline Abstract References Article metrics Related Articles Share Share Share on Email X Facebook LinkedIn Sina Weibo Add to Mendeley bluesky Add to my reading list More More Get Access Cite Share Share Share on Email X Facebook LinkedIn Sina Weibo Add to Mendeley Bluesky Add to my reading list Set Alert Get Rights Reprints Download Full Issue Download started Ok Previous articleNext article Show Outline Hide Outline Abstract References Article metrics Related Articles Abstract Evidence gathered from our studies and the work of others appears to support the presence of two distinct nerve pain pathways in the dental pulp, represented by fast conducting A-delta and slow conducting C-fibers. Each of these types of fibers has different pain characteristics: A-delta fibers evoke a rapid, sharp, lancinating pain reaction, and C-fibers cause a slow, dull, crawling pain. Pain response thresholds vary in different regions of the tooth, and thermal, osmotic, ionic, and electric stimuli involve different mechanisms to provoke nerve excitation of the dental pulp. Evidence also points to the fact that the incidence of pain increases as the histopathosis worsens. On interrogation, patients who manifest severe or referred pain almost always give a previous history of pain in the tooth with the ache. Eighty percent of patients who give a previous history of pain manifest histopathologic evidence of chronic partial pulpitis with partial necrosis, the untreatable category, for which endodontics or extraction is indicated. The other 20% exhibit histopathosis of the pulp with slight inflammation to chronic partial pulpitis without necrosis, a treatable category. Clinically, one can determine the degree of pulp histopathosis by asking the patient about a previous history of pain in the involved tooth. This history of previous pain adds another dimension in diagnosis for the clinician as to whether the painful pulpitis is reversible. This information also aids in referred pain localization. Get full text access Log in, subscribe or purchase for full access. Get Access References 1. Narhi, M ∙ Virtanen, A ∙ Kuhta, J ... Electrical stimulation of teeth with a pulp tester in the cat Scand J Dent Res. 1979; 87:32-38 PubMed Google Scholar 2. Byers, MR ∙ Dong, WK Autoradiographic location of sensory nerve endings in dentin of monkey teeth Anat Rec. 1983; 205:441-454 Crossref Scopus (69) PubMed Google Scholar 3. Bender, IB ∙ Landau, MA ∙ Fonseca, S ... The optimum placement site of the electrode in electric pulp testing of the 12 anterior teeth J Am Dent Assoc. 1989; 118:305-310 Abstract Full Text (PDF) Scopus (73) PubMed Google Scholar 4. Trowbridge, HO ∙ Franks, M ∙ Korostoff, E ... Sensory response to thermal stimulation in human teeth J Endodon. 1980; 6:405-412 Abstract Full Text (PDF) Scopus (94) PubMed Google Scholar 5. Fuss, Z ∙ Trowbridge, HO ∙ Bender, IB ... Assessment of reliability of electrical and thermal pulp testing agents J Endodon. 1986; 12:301-305 Abstract Full Text (PDF) Scopus (147) PubMed Google Scholar 6. Rickoff, B ∙ Trowbridge, HO ∙ Baker, J ... Effects of thermal vitality on human dental pulps J Endodon. 1988; 14:482-485 Abstract Full Text (PDF) Scopus (36) PubMed Google Scholar 7. Kim, S Neurovascular interactions in the dental pulp in health and inflammation J Endodon. 1990; 16:48-53 Abstract Full Text (PDF) Scopus (115) PubMed Google Scholar 8. Guyton, AC Basic neuroscience WB Saunders Co., Philadelphia, 1991; 129 Google Scholar 9. Bender, IB Pulp Biology Conference: a discussion J Endodon. 1978; 4:37-52 Abstract Full Text (PDF) Scopus (10) PubMed Google Scholar 10. Pashley, D ∙ Michelich, V ∙ Kehl, T Dentin permeability: effects of smear layers removal J Prosthet Dent. 1981; 46:531-537 Full Text (PDF) Scopus (251) PubMed Google Scholar 11. Meyers, FH ∙ Jawetz, E ∙ Goldfine, R Review of medical pharmacology Lange Medical Publishers, Los Altos, 1978; 469 Google Scholar 12. Albright, F ∙ Reifenstein, EC The parathyroid glands and metabolic bone disease Williams & Wilkins, Baltimore, 1948; 65 Google Scholar 13. Olgart, LM ∙ Haegerstam, G ∙ Edwall, L The effect of extracellular calcium on thermal excitability of the sensory units in the tooth of the cat Acta Physiol Scand. 1974; 91:116-122 Crossref Scopus (20) PubMed Google Scholar 14. Guyton, AC Basic neuroscience WB Saunders, Philadelphia, 1991; 131-132 Google Scholar 15. Markowtiz, K ∙ Kim, S The role of selected cations in the desensitization of intradental nerves Proc Finn Dent Soc. 1992; 88:39-54 Google Scholar 16. Trowbridge, H ∙ Edwall, L ∙ Panopoulos, P Effect of zinc oxide-eugenol and calcium hydroxide on intradental nerve activity J Endodon. 1982; 8:403-406 Full Text (PDF) Scopus (45) PubMed Google Scholar 17. Hahn, CL ∙ Falkler, WA ∙ Siegle, M Study of T and B cells in pulpal pathosis J Endodon. 1989; 15:20-26 Abstract Full Text (PDF) Scopus (62) PubMed Google Scholar 18. Olgart, LM The role of local factors in dentin and pulp in intradental pain mechanisms J Dent Res. 1985; 64:572-578 PubMed Google Scholar 19. Van Hassel, HJ ∙ Harrington, GW Localization of pulpal sensations Oral Surg. 1969; 28:153-160 Abstract Full Text (PDF) Scopus (18) Google Scholar 20. Byers, M ∙ Maeda, T Periodontal innervation: regional specializations, ultrastructure, cytochemistry and tissue interactions Acta Med Dent Helv. 1997; 2:116-133 Google Scholar Figures (7)Figure Viewer Show all figures Hide figures Article metrics Related Articles View full text Open in viewer Hide CaptionDownloadSee figure in Article Toggle Thumbstrip Download Hi-res image Download .PPT Go to Go to Show all references Expand All Collapse Expand Table Authors Info & Affiliations Home Access for Developing Countries Articles and Issues Articles in Press Current Issue List of Issues Supplements JOE Publication Awards For Authors/Reviewers About Open Access Author information Copyright Permissions Guidelines for Publishing Papers in JOE Language Editing Reviewer Information Researcher Academy Submit Your Manuscript Understanding the Publishing Process Journal Info About the Journal About Open Access Activate Online Access Contact Us Editorial Board Information for Advertisers New Content Alerts Pricing Information Subscribe AAE About the AAE AAE Website Online CE for JOE Submit Your Manuscript Follow Us AAE Connection Twitter Facebook Instagram The content on this site is intended for healthcare professionals. 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2618	https://www.frontiersin.org/journals/global-womens-health/articles/10.3389/fgwh.2021.669826/full	Your new experience awaits. Try the new design now and help us make it even better MINI REVIEW article Front. Glob. Women’s Health, 28 October 2021 Sec. Women's Mental Health Volume 2 - 2021 \| This article is part of the Research TopicPerinatal Mood Symptoms and Postpartum Maternal Functioning: Describing the Evidence Related to Effective and Ineffective InterventionsView all 8 articles Dysphoric Milk Ejection Reflex: The Psychoneurobiology of the Breastfeeding Experience Reem Deif1Emily Michelle Burch1Jihan Azar1Nouran Yonis1Macy Abou Gabal1Nabila El Kramani2Duaa DakhlAllah1 1Institute of Global Health and Human Ecology, School of Sciences and Engineering, The American University in Cairo, Cairo, Egypt 2Department of Biology, School of Sciences and Engineering, The American University in Cairo, Cairo, Egypt Breastfeeding, given its biochemical and physiological basis, is known for its many benefits for both the lactating mother and the infant. Among the many challenges new breastfeeding mothers experience is the feeling of aversion in response to their newborn's suckling which has been termed dysphoric milk-ejection reflex (D-MER). Characterized by intense feelings of dysphoria which may eventually interfere with the mother's ability to breastfeed regularly, evidence suggests both the neurobiological and psychological basis of D-MER in an attempt to explain its complexity. Biologically, breastfeeding is expressed by the intracerebral release of oxytocin, an increased expression of oxytocin receptors in specific brain regions, increased mesocorticolimbic reward region activation, the secretion of prolactin and possibly the inhibition of dopamine. Hence, different theories explain D-MER in terms of disrupted neurotransmitter and hormonal activity. Breastfeeding has also proven to influence mood and stress reactivity in nursing mothers with a potential link with postpartum depression. Psychological theories attempt to explain D-MER from a sociopsychosexual lense shedding light on the significance of mother-infant attachment, the sexualization of the female body and the motherhood experience as a developmental stage in a woman's lifespan. The aim of this review is to provide a literature update of D-MER incorporating both neurobiological and psychological theories calling for raising awareness about the complexity of breastfeeding and for the need for mother-centered interventions for the management of D-MER and other postpartum-specific conditions. Introduction Breastfeeding, given its biochemical and physiological basis, is known for its many benefits for both the lactating mother and the infant. Despite the recommendations of the WHO and the American Academy of Pediatrics (1) for exclusive breastfeeding till the age of 6 months, percentages of breastfed infants show a decrease over the first months of life (2) and statistics suggest that only 41% of infants between the ages of 0–6 are exclusively breastfed (3). Such rates may reflect the fact that many mothers may initiate breastfeeding, yet stop at a certain point. Research has examined medical and physical challenges in breastfeeding that might predict early cessation, yet little attention has been given to the emotional experiences that might affect the course of breastfeeding. Among the many challenges new breastfeeding mothers experience is the feeling of aversion in response to their newborn's suckling and their need to appropriately position the newborn, nurse and be patient. Clinically, this is referred to as the Dysphoric Milk Ejection Reflex (D-MER) which is characterized by dysphoria starting shortly before ejection of milk and progressing for several minutes. It is likely to recur with each milk ejection response, or in certain cases, only the initial milk ejection response of each feeding session (4). Such sensations create a hollow or churning sensation in the pit of the stomach (5, 6). Symptoms may diminish by 3 months or may persist during the course of breastfeeding (7). Such symptoms have been described from a narrative perspective in different ways; as overwhelming, uncontrollable feelings that have strong influence on the mother, as a sense of obligation toward the baby, as an abnormal experience based on the assumption that a nursing mother should enjoy breastfeeding and finally, as giving a sense of achievement when done. Such feelings may put a mother under psychological pressure and impact her sense of self which may have implications on her relationship with her infant (8). As an underresearched area, the first and only study examining the epidemiology of D-MER was published in 2019 suggesting a prevalence of 9.1% (9). Symptoms of D-MER may interfere with the mother's ability to maintain an appropriate breastfeeding schedule. For example, women may breastfeed less, and wean earlier (6). Research suggests a prevalence of 20% of postpartum depression among new mothers (10); but whether or not lactation might be a contributing factor is questionable. Although emotional and physical symptoms, such as nausea, associated with D-MER distinguish it from other breast feeding manifestations and postpartum depression (7), symptoms of postpartum depression may contribute to or follow premature weaning in breastfeeding mothers (11), and may overlap with symptoms of other affective problems including dysphoric milk ejection reflex (12). Given the complex and possibly mutual relationship between breastfeeding and maternal depression, it is likely that breastfeeding complications may influence maternal affect and mood. For instance, Brown et al. (13) observed that breastfeeding cessation in mothers is associated with high depression scores among mothers who quit breastfeeding due to physical difficulties and discomfort while breastfeeding. At 8 weeks postpartum, another research examined breastfeeding difficulties and maternal mood and reported that breastfeeding difficulties were correlated with worse maternal mood, or other comorbid physical problems (14). However, although breastfeeding is linked with maternal mood and postpartum depression, it is hard to ascertain whether the effects are triggered by breastfeeding or maternal mood given the complicated relationship between both. Research findings on the link between breastfeeding and postpartum mental health is still inconsistent. For example, breastfeeding is less prevalent among mothers with depression, it might help prevent postpartum depression (15), but it is still related to postpartum depression in some cases (16). The function of oxytocin is significantly involved, not only in breastfeeding, but also in postnatal depression (17) by reducing neuroendocrine stress signaling and anxiety-related and depressive symptoms (18). Looking into the history of D-MER, it was first identified by Alia Macrina Heise in 2007. In 2008, Heise created the web domain, www.d-mer.org, to educate the public about this condition and provide support for mothers (19) and in 2010, the first D-MER case study was issued. Two other case analyses were released in 2011 and 2018 (5, 19). Unfortunately, there is not enough evidence confirming D-MER as a clinical phenomenon given the limited case cases and anecdotal data (4). In this regard, the aim of this paper is to provide a literature update on D-MER from a biopsychosocial perspective. A literature search has been performed in PubMed, ScienceDirect, MedLine, and APA PsycArticles, and given the limited literature available on this subject, research needs are also highlighted toward the end of this review. Breastfeeding From a Hormonal/Endocrine Perspective Lactation or lactogenesis is established in two phases. Stage one is known as secretory differentiation and it is followed by stage two or secretory activation (20). The first stage of lactogenesis begins around 15–20 weeks of gestation and is primarily driven by hormones (21). Secretory differentiation can be observed as the breast develops the capacity to synthesize colostrum, breastmilk, and other milk products like lactose, casein, a-lactalbumin, and lactoferrin (20). The mammary gland is an exocrine gland that produces breast milk in humans. This gland faces a number of developmental changes when preparing for lactation. Gland maturation and alveologenesis are among two of these changes. The formation of alveoli, milk secreting cells, is primarily driven by a hormone called progesterone and this process occurs early on in pregnancy (22). Lactation in alveolar cells is induced by a hormone called prolactin. Mammary glands are sufficiently developed around 20 weeks of gestation by way of both prolactin and progesterone. Following the development of the mammary glands, the accumulation of colostrum begins. Colostrum production is usually initiated about halfway through pregnancy and continues through the third trimester (23). Breast size and volume usually increase as a result of these factors. The amount of change that occurs can vary significantly between mothers. It is important to note that the size of the breast or the increase in volume do not determine the amount of milk that will be synthesized postpartum (20). Secretory activation begins anywhere from 24 to 72 h after birth and is triggered by the birth of the infant and the delivery of the placenta (20). Women feel increased breast fullness in this second stage. After birth, progesterone levels decrease rapidly and prolactin and oxytocin levels increase (23). Prolactin is able to induce the secretory activity of alveolar cells and promote the release of milk into alveoli and smaller ducts (21). Prolactin responds to nipple stimulation, infant suckling, and expression of breastmilk. Oxytocin triggers the let down reflex or the release of milk when the nipple areola is stimulated (23). Although prolactin and oxytocin are the primary hormones in this process, Insulin, cortisol, and thyroxine are also involved (21). The first h following the birth of a child is often referred to as the “Magical hour.” This first h can be a strong determinant of breastfeeding success. The reason this hour is so important is because the composition of this first feed will be colostrum since secretory activation, or the arrival of breast milk has not occurred yet. Colostrum is very rich in antibodies and high in nutritional value that is sufficient to meet a newborns' nutritional needs immediately following delivery (21). The “Magical Hour” is more than the introduction of colostrum and breastfeeding alone. It provides a vital time for skin-to-skin contact between the mother and her newborn. It allows the newborn to bond and to learn to utilize their olfactory sense to properly latch onto the mothers breast. Srirama et al. showed that early lactation stimulates the development of more prolactin receptors. The increase in prolactin receptors may enhance the potential for future breast milk production (21). Hormones are key elements in both secretory differentiation and secretory activation regardless of whether breastfeeding is initiated. Following these two stages, ongoing milk production is also controlled by hormones, however unlike the first two stages it is determined by the removal of milk from the breasts (21). Ongoing milk production is an issue of supply and demand. When the milk has been secreted out of the breast, hormonal signals are sent to the hypothalamus to initiate more milk production. Lactation will continue as long as milk is being secreted and removed from the breast. The breast capacity to store milk can range anywhere from 80 to 600 ml. In breasts that have lower storage capacity, milk synthesis will be a more rapid process when compared to breasts with a larger storage capacity (23). This is to ensure that there is sufficient milk to meet the infant's needs. It is recommended that infants are fed on demand based on their hunger cues. It is estimated that a newborn will feed at least 10–12 times a day (21). Breast milk provides the appropriate nutrients to an infant based on their developmental stage. Breast milk also provides proteins, growth factors, antimicrobial peptides and proteins to support an infant's immune system and gastrointestinal system (24). Exclusive breast feeding is recommended for the first 6 months of life (25). Introducing formula early on can adversely affect breast milk production and can lead to an increased risk of breastfeeding cessation. Formula should be introduced only if there is a medical need indicating that supplementation is required. This will optimize milk supply and increase the chances of exclusive breastfeeding until it is time for the introduction of complementary foods (21). The Milk Ejection Reflex (MER) is an essential element in milk production. MER helps to both establish and maintain milk production. MER may also be referred to as the “let down reflex.” This is a neuro-endocrine reflex that is brought on by the stimulation of the nipple and areola, the circular area around the nipple (21). The negative pressure related to the infants sucking results in the 4th intercostal nerve sending signals to the hypothalamus to release oxytocin. Oxytocin is released from the posterior pituitary gland. The release of oxytocin can be enhanced by a variety of factors, such as, hearing the baby cry, thinking about the baby or preparing to breastfeed. The release of oxytocin can also be inhibited by fear, pain, embarrassment or anxiety of the mother (21). The Psychology of Breastfeeding On the healthy end of the spectrum, breastfeeding is believed to foster maternal responsiveness and healthy mother-child attachment (26, 27). For instance, breastfeeding mothers hold their babies more, are more receptive, and spend more time nursing in mutual gaze than bottle-feeding mothers (28). On the neural level, brain mapping shows higher brain activity in different limbic brain regions in breastfeeding mothers when responding to the infant's cries (29). Breastfeeding has also been shown to influence mood and stress reactivity in different ways. For example, in comparison to formula-feeding mothers, breastfeeding mothers may show less agitation, depressive disposition, stress (30), better modulation of cardiac vagal tone, lowered blood pressure, and decreased heart rate, and an overall relaxed and non-anxious physiological state (31). Additionally, breastfeeding has shown to be associated with a diminished reaction to cortisol when dealing with social tension (32) and longer and higher quality sleep cycles (33). In terms of social cognition, breastfeeding may positively affect a mother's reaction to others as suggested by research showing an association between extended breastfeeding durations and positive responses to happy facial expressions and less responsiveness to angry ones (34). From a neuropsychological perspective, it can be confidently claimed that oxytocin does not only play a role in the process of milk production, but also in the emotional attachment between the mother and the newborn during lactation. This also confirms the link between breastfeeding and low cortisol levels in order to promote low-stress interaction between the mother and the infant (35). In order to understand the psychopathology of D-MER, motherhood should first be conceptualized from a psychological perspective as an independent developmental stage in a woman's lifespan. Breastfeeding in itself can be regarded as a part of the emerging motherhood identity. This also means that some lactating mothers may not have a personal meaning of breastfeeding and may, therefore, discontinue breastfeeding or introduce formula at an early age. Such identity problems may also affect maternal mental health and may result in less pleasurable experiences of breastfeeding (36). Different psychological factors may come into play when a mother makes the decision of whether or not to breastfeed (37). One systematic review highlighted the effect of self-efficacy, postpartum mental health, social support, intention and attitudes toward breastfeeding as psychological predictors of exclusive breastfeeding duration (38). Other concerns which might interfere with the mother's decision to breastfeed include the mother's feelings of incompetence, anxiety about the baby's nutrition and/or embarrassment about her bodily exposure (39), maternal sense of autonomy (40), and the mother's perception of the social context needs and the needs of others (41). Exposure to intimate partner violence has also shown to increase the risk of breastfeeding avoidance by double (42). The sexualization of a woman's breasts besides the need to expose them to breastfeed, depending on the level of privacy provided, may create a sense of confusion for the nursing mother. That is why embarrassment and high levels of modesty may interfere with a mother's willingness to breastfeed (43). Supporting such a hypothesis, more recent research has examined self-objectification and “reproductive shame” associated with different reproductive roles including breastfeeding (44). On the other hand, women who prefer not to breastfeed may have less positive attitudes toward sex as suggested by Sears et al. (45). One study aiming to explore the mother's narratives of their negative embodied emotional sensations while breastfeeding using interpretative phenomenological analysis suggests the influence of breastfeeding on initiating conflicting thoughts and emotions which in itself can impact the mother's self image and the way they relate to their newborns. Three themes were identified; breastfeeding as an unexpected trigger of intense embodied emotional sensations incongruent with view of self; fulfilling maternal expectation and maintaining closeness with the child; and making sense of embodied emotional sensations essential to acceptance and coping. Put together, such findings call for raising awareness about the complexity of the breastfeeding experience and for the need for mother-centered interventions (6). The Neurobiology of Breastfeeding Different hypotheses of mechanisms of action underlying dysphoria have been proposed, but dopamine has been the most probable candidate (19). The secretion of prolactin, which also happens during lactation, is dependent on the inhibition of dopamine (5). Pseudoephedrine, which stops milk production, was seen to eliminate D-MER without acting on oxytocin, but rather by reducing levels of prolactin. Whether this occurs by increasing dopamine levels is unknown (19). Dopamine is thought to be more abrupt in rising and falling during breastfeeding, therefore it might correlate more with D-MER (19). Transient reduction in dopamine in the hypophyseal portal veins has been observed resulting in elevated prolactin levels (19). Breastfeeding is generally associated with the upregulation of endogenous oxytocin levels. Research also shows that mothers' genetic variance in oxytocin gene influences the pace at which cortisol declines throughout breastfeeding sessions. Similar reductions in infants were also observed (28). Research has shown that intranasal administration of oxytocin was shown to reduce anxiety in non-breastfeeding contexts (46, 47). Such findings might be implicated in relation to reducing aversion in cases of D-MER by regulating levels of oxytocin. During breastfeeding, there is an increase in intracerebral release of oxytocin and the expression of oxytocin receptors in specific brain locations. Nipple stimulation is also believed to stimulate oxytocin production. Higher circulating oxytocin levels correlate with increased mesocorticolimbic reward region activation (48). Hence, reduction in oxytocin might be implicated in the dysphoric symptoms in D-MER. Dopamine was also found to stimulate the release of oxytocin in rats while lowering it in others. However, the neurotransmitter release pattern is difficult to determine since the brain is not as easily penetrated as the blood (19). Glutamate might also play a role in the release of oxytocin. The effect of glutamate on dopamine is still rather unknown. Emotions, ranging from desire to disgust, triggered by dopamine are mainly due to the effect of glutamate on different locations in the brain. To strengthen the hypothesis that dopamine is responsible for the dysphoria caused during breastfeeding, is the fact that dopamine reuptake inhibitors such as bupropion, which increases the availability of dopamine, resulted in the elimination of D-MER symptoms in one woman. Herbals such as Rhodiola rosea or golden root acts as monoamine oxidase inhibitors which inhibits the breakdown of dopamine and other neurotransmitters, and may improve symptoms of D-MER in nursing mothers (19). Discussion and Implications Due to poor public awareness of D-MER and the scarcity of evidence-based literature, many mothers may mistake D-MER for postpartum depression especially given its atypical symptomatic manifestations, and lactation practitioners and health care providers may also barely recognize D-MER (49). Another challenge in the management of D-MER is that mental health professionals may lack knowledge about lactation or training in lactation management (12). This makes it necessary to educate mothers because educated mothers are usually better at handling postpartum situations if they are prepared in advance. Given the variety of emotions experienced by breastfeeding mothers with D-MER, the fact that symptoms may persist up until weaning and the potential risk of suicidality, professional intervention should always be recommended and tailored based on the mother's needs and the severity of symptoms (19). Generally, research shows the positive effects of breastfeeding support on increasing the overall breastfeeding duration (50) and the effectiveness of relaxation techniques on improving maternal and newborn outcomes (51). Other psychosocial-based breastfeeding promotion programs exist, yet the extent to which such programs are evidence-based and effective is unclear (52). Although not specific to D-MER, such findings may give hope to the introduction of more specialized interventions for depending on the severity of D-MER symptoms, risk factors or comorbid postpartum mental health problems. Preliminary recommendations highlight the need to alleviate the stress that is triggered during breastfeeding through creating a safe environment for the lactating mother (53). Other forms of psychological management which have been suggested include lifestyle adjustments (54) as mothers may still perceive themselves as “bad mothers” blaming themselves for experiencing such emotions. Conclusion Dysphoric milk ejection reflex is a common, yet not a commonly researched, condition among breastfeeding mothers that is often confused with other postpartum conditions. Emotional and physical symptoms may help distinguish D-MER from other disorders and should be considered for the effective management of this condition. Preliminary research shows the involvement of different hormonal, neurobiological and psychological mechanisms in the pathology of D-MER paving the way for future research to attempt to understand overlaps between different mechanisms and to suggest evidence-based prevention and treatment protocols. The authors are aware of the limitations of this review given the lack of experimental and epidemiological research in the area of D-MER. A limited number of theoretical research was available through different scholarly databases and all were included in this review. The foundations have been presented to raise awareness about the complexity of breastfeeding, as a physiological function, health behavior and emotional experience, and to encourage more evidence-based research in this area. Author Contributions RD has put the outline for this review and contributed to the sections of the psychological aspects of D-MER with NE. EB focused on the part of breastfeeding from a hormonal/endocrine perspective. JA, MA, and NY worked on the neurobiology of breastfeeding and D-MER specifically. DD worked on revising the outline of the paper and editing the sections accordingly, adding recommendations, and proofreading the paper. All authors contributed to the article and approved the submitted version. Conflict of Interest The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest. Publisher's Note All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article, or claim that may be made by its manufacturer, is not guaranteed or endorsed by the publisher. References American Academy of Pediatrics. Breastfeeding and the use of human milk. Pediatrics. (2012) 129:e827–41. doi: 10.1542/peds.2011-3552 PubMed Abstract \| CrossRef Full Text \| Google Scholar Centers for Disease Control and Prevention (2020). Breastfeeding. (accessed October 5, 2021). Google Scholar World Health Organization. WHO Breastfeeding Website. (2020). Available online at: (accessed October 5, 2021). Google Scholar Heise AM. Before the Letdown: Dysphoric Milk Ejection Reflex and the Breastfeeding Mother. New York, NY: Independently Published. (2017). Ureño TL, Buchheit TL, Hopkinson SG, Berry-Cabán CS. 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Available online at: (accessed October 5, 2021). Google Scholar Keywords: lactation, breastfeeding, postpartum, maternal, dysphoric milk-ejection reflex Citation: Deif R, Burch EM, Azar J, Yonis N, Abou Gabal M, El Kramani N and DakhlAllah D (2021) Dysphoric Milk Ejection Reflex: The Psychoneurobiology of the Breastfeeding Experience. Front. Glob. Womens Health 2:669826. doi: 10.3389/fgwh.2021.669826 Received: 19 February 2021; Accepted: 11 October 2021; Published: 29 October 2021. Edited by: Pamela A. Geller, Drexel University, United States Reviewed by: Yolanda Contreras-García, University of Concepcion, Chile Ghada Hassan, Ain Shams University, Egypt Copyright © 2021 Deif, Burch, Azar, Yonis, Abou Gabal, El Kramani and DakhlAllah. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms. Correspondence: Reem Deif, reem.deif@aucegypt.edu Disclaimer: All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article or claim that may be made by its manufacturer is not guaranteed or endorsed by the publisher.
2619	https://openstax.org/books/introductory-statistics-2e/pages/4-4-geometric-distribution	Skip to ContentGo to accessibility pageKeyboard shortcuts menu Introductory Statistics 2e 4.4 Geometric Distribution Introductory Statistics 2e4.4 Geometric Distribution Search for key terms or text. There are four main characteristics of a geometric experiment. A trial is repeated until a success occurs. Think of this as one or more Bernoulli trials with all failures except the last one, which is a success. In other words, you keep repeating what you are doing until the first success. Then you stop. For example, you throw a dart at a bullseye until you hit the bullseye. The first time you hit the bullseye is a "success," so you stop throwing the dart. It might take six tries until you hit the bullseye. You can think of the trials as failure, failure, failure, failure, failure, success, STOP. In theory, the number of trials could go on forever. The repeated trials are independent of each other. The probability, p, of a success and the probability, q, of a failure is the same for each trial. and For example, the probability of rolling a three when you throw one fair die is . This is true no matter how many times you roll the die. Suppose you want to know the probability of getting the first three on the fifth roll. On rolls one through four, you do not get a face with a three. The probability for each of the rolls is the probability of a failure. The probability of getting a first three on the fifth roll is The random variable X represents the number of the trial in which the first success occurs. That is, X = the number of independent trials until the first success. The following are additional attributes of the geometric distribution: The random variable is discrete. The random variable may be defined in two ways depending upon the analyst’s interest. In the example above for throwing a die, the question was “What is the probability that the first success will be on the fifth throw?” Alternatively, the question could be asked as “What is the probability that it takes four failures before a success?” We will see that each way of asking the question will alter the form of the geometric probability density function slightly and will change the mean and standard deviation of the geometric pdf. Implicit in the random variable is that the probability of a success is constant and therefore so is the probability of a failure. For flipping a coin this is obvious, but in experiments that require skill, such as hitting a baseball or throwing a dart, one might consider that learning during the experiment would alter the probability of a success. The geometric distribution cannot capture “learning” thus the historical probability of success is assumed to be constant. The geometric distribution is “memoryless.” There are very few probability density functions that are what is known as “memoryless,” and the Geometric distribution is the only one with a discrete random variable that is memoryless. As an example, historically Major League Baseball player Jones has a record of hitting the ball for at least an advance to first base with a probability of 0.20. Jones has not had a hit in his last 10 times at bat. What is the probability that Jones will get a hit in his third time at bat? The answer ignores his 10 previous failures. All events prior to the events in current time are irrelevant and thus are considered “memoryless.” Formally: where k = number of previous failures Jones’s probability of a hit begins anew each time he comes to bat. This feature of the geometric distribution results in a curious result: Drawing parts from a manufacturing process to test for parts that are defective, the geometric distribution begins with a clean slate each time the tests begin with no consideration of previous test results. More on this when we get to the exponential probability density function. Example 4.17 You play a game of chance that you can either win or lose (there are no other possibilities) until you lose. Your probability of losing is p = 0.57. What is the probability that it takes five games until you lose? Let X = the number of games you play until you lose (includes the losing game). Then X takes on the values 1, 2, 3, ... (could go on indefinitely). The probability question is P(x = 5). Try It 4.17 You throw darts at a board until you hit the center area. Your probability of hitting the center area is p = 0.17. You want to find the probability that it takes eight throws until you hit the center. What values does X take on? Example 4.18 Problem A safety engineer feels that 35% of all industrial accidents in the plant are caused by failure of employees to follow instructions. They decide to look at the accident reports (selected randomly and replaced in the pile after reading) until they find one that shows an accident caused by failure of employees to follow instructions. On average, how many reports would the safety engineer expect to look at until they find a report showing an accident caused by employee failure to follow instructions? What is the probability that the safety engineer will have to examine at least three reports until they find a report showing an accident caused by employee failure to follow instructions? Let X = the number of accidents the safety engineer must examine until they find a report showing an accident caused by employee failure to follow instructions. X takes on the values 1, 2, 3, .... The first question asks you to find the expected value or the mean. The second question asks you to find P(x ≥ 3). ("At least" translates to a "greater than or equal to" symbol). Solution On average, how many reports would the safety engineer expect to review until they find a report showing an accident caused by something other than failure to follow instructions? This question asks you to find the expected value or the mean. This is the average number of failures from something other than following instructions. The formula for the mean of this geometric distribution is:Note that the mean does not need to be a whole number although the random variable is discrete and must be a counting number What is the probability that the safety engineer will have to examine at least three reports until they find a report showing an accident caused by employee failure to follow instructions?This question is answered by first defining the random variable. Let X = the number of accidents the safety engineer must examine until they find a report showing an accident caused by employee failure to follow instructions. X can take on the values 1, 2, 3, 4, … ∞. Unlike the binomial distribution with a fixed number of trials, the geometric distribution may have an infinite number of failed trials before a success occurs.This second question asks you to find . ("At least" translates to a "greater than or equal to" symbol.)In a binomial distribution, we saw the solution to questions of “more than” or “less than” would be to calculate each probability individually and add from zero to three. If the probability of interest is “greater than or equal to” (≥) the individual probabilities are added from zero to three and then subtracted from one.Because we cannot add the full range of the geometric distribution random variable because it goes through infinity, an alternative solution is developed. An alternative geometric distribution pair of formulas provides solutions for questions asking for probabilities “more than” and “less than.”If the question is:What is the probability it takes MORE THAN n events to get first success?What is the probability it takes LESS THAN n event for success? Try It 4.18 An instructor feels that 15% of students get below a C on their final exam. They decide to look at final exams (selected randomly and replaced in the pile after reading) until they find one that shows a grade below a C. We want to know the probability that the instructor will have to examine at least ten exams until they find one with a grade below a C. What is the probability question stated mathematically? Example 4.19 Suppose that you are looking for a student at your college who lives within five miles of you. You know that 55% of the 25,000 students do live within five miles of you. You randomly contact students from the college until one says they live within five miles of you. What is the probability that you need to contact four people? This is a geometric problem because you may have a number of failures before you have the one success you desire. Also, the probability of a success stays the same each time you ask a student if they live within five miles of you. There is no definite number of trials (number of times you ask a student). Problem a. Let X = the number of ____________ you must ask ____________ one says yes. b. What values does X take on? c. What are p and q? d. The probability question is P(_______). Solution a. Let X = the number of students you must ask until one says yes. b. 1, 2, 3, …, (total number of students) c. p = 0.55; q = 0.45 d. P(x = 4) Try It 4.19 You need to find a store that carries a special printer ink. You know that of the stores that carry printer ink, 10% of them carry the special ink. You randomly call each store until one has the ink you need. What are p and q? Notation for the Geometric: G = Geometric Probability Distribution Function X ~ G(p) Read this as "X is a random variable with a geometric distribution." The parameter is p; p = the probability of a success for each trial. CASE I: Random Variable X Is Event of First Success In this case we ask, “What is the probability that we will have some number x of events of interest to us of failures before a success?” The geometric pdf tells us the probability that the first occurrence of success requires x number of failure independent trials, each with probability . If the probability of success on each trial is p, then the probability that the xth trial (out of x trials) is the first success is: for Like the binomial distribution, the geometric distribution has the parameters of the mean and standard deviation. The expected value of X, the mean for Case I, is This tells us how many failed trials to expect until we get the first success. This count includes in the count of trials the trial that results in success. The above form of the geometric distribution is used for modeling the number of trials until the first success. The number of trials includes the one that is a success: x = all trials including the one that is a success. This can be seen in the form of the formula. If X = number of trials including the success, then we must multiply the probability of failure, , times the number of failures, that is . The standard deviation of Case I of the geometric distribution is: CASE II: Random Variable X Is Number of Failures BEFORE a Success By contrast to Case I, the following form of the geometric distribution used for modeling number of failures until the first success is: for In this case the trial that is the success is not counted as a trial in the formula: x = number of failures. The expected value, the mean, of this distribution is . This tells us how many failures to expect before we have a success. In either case, the sequence of probabilities is a geometric sequence. In Case II, the standard deviation parameter is: Figure 4.4 The y-axis in Figure 4.4 contains the probability of x, and the x-axis is the random number components tested. For example, at the probability it will be found to be defective is 0.0196. With two components tested, the probability the second component is defective is graphed at a probability of 0.0196 at on the x axis. For the probability that the third component is defective we find (The first two are the same because of rounding in the computations.) Notice on Figure 4.4 that the probabilities decline by the same step down with each change in the value of x. This increment is called the common ratio. This exists for the geometric probability distribution uniquely. The common ratio, called r, can be calculated by dividing any value by the previous value, e.g., For this set of data, therefore, the common ratio is 0.98. The common ratio then multiplied by any other probability value will provide the next probability value in the sequence. For example, the probability that the sixth component tested is a failure is 0.018078. Check this using the formula from Case I. Now we have the . Knowing this and the common ratio we can calculate the by simple multiplication: , the same value we found earlier by using the geometric probability distribution. This common ratio increment is the same ratio between every number and is called a geometric progression and thus the name for this probability density function. Once the common ratio is calculated, any one desires to know can be easily found. The number of components that you would expect to test until you find the first defective component is the mean, for this case of defective components. The formula for the mean of the geometric distribution for the random variable defined as number of failures before first success is See Example 4.20 for an example where the geometric random variable is defined as number of trials until first success. The expected value of this formula for the geometric distribution will be different from this version of the distribution. Case II also has a variance, but this is changed from the Case I formula. This formula for the variance is: for The formula for the variance is The standard deviation is Example 4.20 Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the first defect is caused by the seventh component tested. How many components do you expect to test until one is found to be defective? Let X = the number of computer components tested until the first defect is found. X takes on the values 1, 2, 3, ... where p = 0.02. Find P(x = 7). P(x = 7) = 0.0177. Using the TI-83, 83+, 84, 84+ Calculator To find the probability that x = 7, Enter 2nd, DISTR Scroll down and select geometpdf( Press ENTER Enter 0.02, 7); press ENTER to see the result: P(x = 7) = 0.0177 To find the probability that x ≤ 7, follow the same instructions EXCEPT select E:geometcdf(as the distribution function. The probability that the seventh component is the first defect is 0.0177. The formula for the variance is σ2 = = = 2,450 The standard deviation is σ = = = 49.5 Try It 4.20 The probability of a defective steel rod is 0.01. Steel rods are selected at random. Find the probability that the first defect occurs on the ninth steel rod. Use the TI-83+ or TI-84 calculator to find the answer. Example 4.21 Problem The lifetime risk of developing cancer is about one in 67 (1.5%). Let X = the number of people you ask until one says they have cancer. Then X is a discrete random variable with a geometric distribution: or What is the probability of that you ask ten people before one says they have cancer? What is the probability that you must ask 20 people? Find the (i) mean and (ii) standard deviation of X. Solution Try It 4.21 The literacy rate for a nation measures the proportion of people age 15 and over who can read and write. The literacy rate for women in Afghanistan is 12%. Let X = the number of Afghani women you ask until one says that she is literate. What is the probability distribution of X? What is the probability that you ask five women before one says she is literate? What is the probability that you must ask ten women? Find the (i) mean and (ii) standard deviation of X. PreviousNext Order a print copy Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. 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2620	https://www.chemguide.co.uk/basicorg/acidbase/acids.html	\| \| \| THE ACIDITY OF ORGANIC ACIDS This page explains the acidity of simple organic acids and looks at the factors which affect their relative strengths. Why are organic acids acidic? Organic acids as weak acids For the purposes of this topic, we are going to take the definition of an acid as "a substance which donates hydrogen ions (protons) to other things". We are going to get a measure of this by looking at how easily the acids release hydrogen ions to water molecules when they are in solution in water. An acid in solution sets up this equilibrium: \| \| Note:We are writing the acid as AH rather than HA, because, in all the cases we shall be looking at, the hydrogen we are interested in is at the right-hand end of a molecule. \| \| A hydroxonium ion is formed together with the anion (negative ion) from the acid. This equilibrium is sometimes simplified by leaving out the water to emphasise the ionisation of the acid. If you write it like this, you must include the state symbols - "(aq)". Writing H+(aq) implies that the hydrogen ion is attached to a water molecule as H3O+. Hydrogen ions are always attached to something during chemical reactions. The organic acids are weak in the sense that this ionisation is very incomplete. At any one time, most of the acid will be present in the solution as un-ionised molecules. For example, in the case of dilute ethanoic acid, the solution contains about 99% of ethanoic acid molecules - at any instant, only about 1% have actually ionised. The position of equilibrium therefore lies well to the left. Comparing the strengths of weak acids The strengths of weak acids are measured on the pKa scale. The smaller the number on this scale, the stronger the acid is. Three of the compounds we shall be looking at, together with their pKa values are: Remember - the smaller the number the stronger the acid. Comparing the other two to ethanoic acid, you will see that phenol is very much weaker with a pKa of 10.00, and ethanol is so weak with a pKa of about 16 that it hardly counts as acidic at all! Why are these acids acidic? In each case, the same bond gets broken - the bond between the hydrogen and oxygen in an -OH group. Writing the rest of the molecule as "X": \| \| Note:If you aren't sure about coordinate covalent (dative covalent) bonding, you might like to follow this link. It isn't, however, particularly important to the rest of the current page. Use the BACK button on your browser to return to this page later. \| \| So . . . if the same bond is being broken in each case, why do these three compounds have such widely different acid strengths? Differences in acid strengths between carboxylic acids, phenols and alcohols The factors to consider Two of the factors which influence the ionisation of an acid are: the strength of the bond being broken, the stability of the ions being formed. In these cases, you seem to be breaking the same oxygen-hydrogen bond each time, and so you might expect the strengths to be similar. \| \| Note:You've got to be a bit careful about this. The bonds won't be identically strong, because what's around them in the molecule isn't the same in each case. \| \| The most important factor in determining the relative acid strengths of these molecules is the nature of the ions formed. You always get a hydroxonium ion - so that's constant - but the nature of the anion (the negative ion) varies markedly from case to case. Ethanoic acid Ethanoic acid has the structure: The acidic hydrogen is the one attached to the oxygen. When ethanoic acid ionises it forms the ethanoate ion, CH3COO-. You might reasonably suppose that the structure of the ethanoate ion was as below, but measurements of bond lengths show that the two carbon-oxygen bonds are identical and somewhere in length between a single and a double bond. To understand why this is, you have to look in some detail at the bonding in the ethanoate ion. \| \| Warning!If you don't already understand about the bonding in the carbon-oxygen double bond, you would be well advised to skip this next bit - all the way down to the simplified structure of the ethanoate ion towards the end of it. It goes beyond anything that you are likely to want for UK A level purposes. If you do choose to follow this link, it will probably take you to several other pages before you are ready to come back here again. Use the BACK button (or HISTORY file or GO menu) on your browser to return to this page. \| \| Like any other double bond, a carbon-oxygen double bond is made up of two different parts. One electron pair is found on the line between the two nuclei - this is known as a sigma bond. The other electron pair is found above and below the plane of the molecule in a pi bond. Pi bonds are made by sideways overlap between p orbitals on the carbon and the oxygen. In an ethanoate ion, one of the lone pairs on the negative oxygen ends up almost parallel to these p orbitals, and overlaps with them. This leads to a delocalised pi system over the whole of the -COO- group, rather like that in benzene. All the oxygen lone pairs have been left out of this diagram to avoid confusion. Because the oxygens are more electronegative than the carbon, the delocalised system is heavily distorted so that the electrons spend much more time in the region of the oxygen atoms. So where is the negative charge in all this? It has been spread around over the whole of the -COO- group, but with the greatest chance of finding it in the region of the two oxygen atoms. Ethanoate ions can be drawn simply as: The dotted line represents the delocalisation. The negative charge is written centrally on that end of the molecule to show that it isn't localised on one of the oxygen atoms. The more you can spread charge around, the more stable an ion becomes. In this case, if you delocalise the negative charge over several atoms, it is going to be much less attractive to hydrogen ions - and so you are less likely to re-form the ethanoic acid. Phenol Phenols have an -OH group attached directly to a benzene ring. Phenol itself is the simplest of these with nothing else attached to the ring apart from the -OH group. When the hydrogen-oxygen bond in phenol breaks, you get a phenoxide ion, C6H5O-. \| \| Warning!You need to understand about the bonding in benzene in order to make sense of this next bit. If your syllabus says that you need to know about the acidity of phenol, then you will have to understand the next few paragraphs - which in turn means that you will have to understand about benzene. If it doesn't mention phenol, skip it! If you follow this link, you may have to explore several other pages before you are ready to come back here again. Use the BACK button (or HISTORY file or GO menu) on your browser to return to this page. \| \| Delocalisation also occurs in this ion. This time, one of the lone pairs on the oxygen atom overlaps with the delocalised electrons on the benzene ring. This overlap leads to a delocalisation which extends from the ring out over the oxygen atom. As a result, the negative charge is no longer entirely localised on the oxygen, but is spread out around the whole ion. Why then is phenol a much weaker acid than ethanoic acid? Think about the ethanoate ion again. If there wasn't any delocalisation, the charge would all be on one of the oxygen atoms, like this: But the delocalisation spreads this charge over the whole of the COO group. Because oxygen is more electronegative than carbon, you can think of most of the charge being shared between the two oxygens (shown by the heavy red shading in this diagram). If there wasn't any delocalisation, one of the oxygens would have a full charge which would be very attractive towards hydrogen ions. With delocalisation, that charge is spread over two oxygen atoms, and neither will be as attractive to a hydrogen ion as if one of the oxygens carried the whole charge. That means that the ethanoate ion won't take up a hydrogen ion as easily as it would if there wasn't any delocalisation. Because some of it stays ionised, the formation of the hydrogen ions means that it is acidic. In the phenoxide ion, the single oxygen atom is still the most electronegative thing present, and the delocalised system will be heavily distorted towards it. That still leaves the oxygen atom with most of its negative charge. What delocalisation there is makes the phenoxide ion more stable than it would otherwise be, and so phenol is acidic to an extent. However, the delocalisation hasn't shared the charge around very effectively. There is still lots of negative charge around the oxygen to which hydrogen ions will be attracted - and so the phenol will readily re-form. Phenol is therefore only very weakly acidic. Ethanol Ethanol, CH3CH2OH, is so weakly acidic that you would hardly count it as acidic at all. If the hydrogen-oxygen bond breaks to release a hydrogen ion, an ethoxide ion is formed: This has nothing at all going for it. There is no way of delocalising the negative charge, which remains firmly on the oxygen atom. That intense negative charge will be highly attractive towards hydrogen ions, and so the ethanol will instantly re-form. Since ethanol is very poor at losing hydrogen ions, it is hardly acidic at all. Variations in acid strengths between different carboxylic acids You might think that all carboxylic acids would have the same strength because each depends on the delocalisation of the negative charge around the -COO- group to make the anion more stable, and so more reluctant to re-combine with a hydrogen ion. In fact, the carboxylic acids have widely different acidities. One obvious difference is between methanoic acid, HCOOH, and the other simple carboxylic acids: \| \| \| pKa \| \| HCOOH \| 3.75 \| \| CH3COOH \| 4.76 \| \| CH3CH2COOH \| 4.87 \| \| CH3CH2CH2COOH \| 4.82 \| Remember that the higher the value for pKa, the weaker the acid is. Why is ethanoic acid weaker than methanoic acid? It again depends on the stability of the anions formed - on how much it is possible to delocalise the negative charge. The less the charge is delocalised, the less stable the ion, and the weaker the acid. The methanoate ion (from methanoic acid) is: The only difference between this and the ethanoate ion is the presence of the CH3 group in the ethanoate. But that's important! Alkyl groups have a tendency to "push" electrons away from themselves. That means that there will be a small amount of extra negative charge built up on the -COO- group. Any build-up of charge will make the ion less stable, and more attractive to hydrogen ions. Ethanoic acid is therefore weaker than methanoic acid, because it will re-form more easily from its ions. The other alkyl groups have "electron-pushing" effects very similar to the methyl group, and so the strengths of propanoic acid and butanoic acid are very similar to ethanoic acid. \| \| Note:If you want more information about the inductive effect of alkyl groups, you could read about carbocations (carbonium ions) in the mechanism section of this site. Use the BACK button on your browser to return to this page if you choose to follow this link. \| \| The acids can be strengthened by pulling charge away from the -COO- end. You can do this by attaching electronegative atoms like chlorine to the chain. As the next table shows, the more chlorines you can attach the better: \| \| \| pKa \| \| CH3COOH \| 4.76 \| \| CH2ClCOOH \| 2.86 \| \| CHCl2COOH \| 1.29 \| \| CCl3COOH \| 0.65 \| Trichloroethanoic acid is quite a strong acid. Attaching different halogens also makes a difference. Fluorine is the most electronegative and so you would expect it to be most successful at pulling charge away from the -COO- end and so strengthening the acid. \| \| \| pKa \| \| CH2FCOOH \| 2.66 \| \| CH2ClCOOH \| 2.86 \| \| CH2BrCOOH \| 2.90 \| \| CH2ICOOH \| 3.17 \| The effect is there, but isn't as great as you might expect. Finally, notice that the effect falls off quite quickly as the attached halogen gets further away from the -COO- end. Here is what happens if you move a chlorine atom along the chain in butanoic acid. \| \| \| pKa \| \| CH3CH2CH2COOH \| 4.82 \| \| CH3CH2CHClCOOH \| 2.84 \| \| CH3CHClCH2COOH \| 4.06 \| \| CH2ClCH2CH2COOH \| 4.52 \| The chlorine is effective at withdrawing charge when it is next-door to the -COO- group, and much less so as it gets even one carbon further away. \| \| \| Questions to test your understanding If this is the first set of questions you have done, please read the introductory page before you start. You will need to use the BACK BUTTON on your browser to come back here afterwards. questions on organic acids \| Where would you like to go now? To the acids and bases menu. . . To menu of basic organic chemistry. . . To Main Menu . . . \| \| \| You might also be interested in: The properties and reactions of: carboxylic acids . . . amino acids . . . phenol . . . alcohols . . . \| © Jim Clark 2000 (last modified December 2012) \|
2621	https://www.bytelearn.com/math-topics/exponent-rules	Sign inSign up Resources TestimonialsPlans Math Topics Exponent Rules Exponent Rules Introduction What Are the Different Exponent Rules? Zero Power Rule Product of Powers Rule Quotient of Powers Rule Power of a Product Rule Power of a Quotient Rule Power of a Power Rule Negative Exponent Rule Practice Problems Frequently Asked Questions Introduction Exponent rules are also called as exponent laws. These are fundamental laws used to simplify expressions involving exponents. These rules enable us to carry out arithmetic operations like addition, subtraction, multiplication, and division more efficiently when dealing with expressions containing exponents. By applying these rules, we can simplify complex expressions with fractional, decimal, and root exponents, making mathematical computations quicker and easier. Let's understand the various rules of exponents. What Are the Different Exponent Rules? There are some rules that help in solving problems related to exponents. Zero Power Rule If a non-zero number is raised to 0, its value equals 1. Therefore, (a^0 = 1) for any non-zero number (a).Example: Evaluate 7^0Explanation: Apply the zero power rule, a^0=1So, 7^0=1 Product of Powers Rule If two numbers having the same base are multiplied, then we add the exponents of the two numbers.Consider two powers with the same base (a), represented as (a^m) and (a^n), where (m) and (n) are exponents. When we multiply (a^m) by (a^n), it's like multiplying (a) to itself (m) times and then multiplying (a) to itself (n) more times:[a^m \times a^n = \overbrace{a \times a \times \ldots \times a}^{m \text{ times}} \times \overbrace{a \times a \times \ldots \times a}^{n \text{ times}}]Since we're combining the factors, we end up with (a) multiplied by itself a total of (m + n) times:[a^m \times a^n = \underbrace{a \times a \times \ldots \times a}_{(m + n) \text{ times}} = a^{m + n}]Example: Simplify 2^5\cdot 2^3Solution: Use the product of powers rule, a^m\cdot a^n=a^{m+n}So, 2^5\cdot2^3=2^{5+3}=2^8 Quotient of Powers Rule If two numbers having the same base are divided, then we subtract the exponents of the two numbers.Let's consider two powers with the same base (a), represented as (a^m) and (a^n), where (m) and (n) are exponents.When we divide (a^m) by (a^n), it's akin to canceling out (a) from each (a^m) (n) times:[a^m ÷ a^n = \frac{\overbrace{a \times a \times \ldots \times a}^{m \text{ times}}} {\underbrace{a \times a \times \ldots \times a}_{n \text{ times}}}]As each (a) in the numerator cancels out with an (a) in the denominator, we're left with (a) multiplied by itself (m - n) times:[a^m ÷ a^n = \underbrace{a \times a \times \ldots \times a}_{(m - n) \text{ times}} = a^{m - n}]Example: Simplify \frac{5^7}{5^3}Solution: Apply the quotient of power rule, \frac{a^m}{a^n}=a^{m-n}So, \frac{5^7}{5^3}=5^{7-3}=5^4 Power of a Product Rule When raising a product to a power, we can distribute the exponent over the different factors.When we have a product (ab) raised to a power (n), represented as ((ab)^n), it means we're multiplying (ab) by itself (n) times.\begin{align\}(ab)^n &= \underbrace{ab \times ab \times \ldots \times ab}\_{n \text{ times}} \\&= \underbrace{a \times a \times \ldots \times a}\_{n \text{ times}} \times \underbrace{b \times b \times \ldots \times b}\_{n \text{ times}} \\&= a^n \times b^n\end{align\}Example: Simplify \left(2\cdot6\right)^3Solution: Power of a product rule, \left(a\cdot b\right)^n=a^n\cdot b^n``\left(2\cdot6\right)^3=2^3\cdot6^3 Power of a Quotient Rule When raising a quotient to a power, we can distribute the exponent over the numerator and the denominator.When we raise this fraction to a power (n), it means multiplying (\frac{a}{b}) by itself (n) times:\begin{align\}\left(\frac{a}{b}\right)^n &= \underbrace{\frac{a}{b} \times \frac{a}{b} \times \ldots \times \frac{a}{b}}\_{n \text{ times}} \\&= \frac{\underbrace{a \times a \times \ldots \times a}\_{n \text{ times}}}{\underbrace{b \times b \times \ldots \times b}\_{n \text{ times}}} \\&= \frac{a^n}{b^n}\end{align\}Example: Simplify \left(\frac{6}{5}\right)^7Solution: Use power of a quotient rule, \left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}So, \left(\frac{6}{5}\right)^7=\frac{6^7}{5^7} Power of a Power Rule If a number with an exponent is raised to another exponent, then we multiply the exponents.Think of a number (x) raised to a power (m). It means multiplying (x) by itself (m) times. Now, if we raise that expression (x^m) to another power (n), it's like multiplying (x^m) by itself (n) times. Since each (x^m) means (x) multiplied by itself (m) times, doing it (n) times is like multiplying (x) by itself (m \times n) times altogether.[(x^m)^n = \underbrace{x^m \times x^m \times \ldots \times x^m}_{n \text{ times}}=x^{m\times n}]Example: Simplify \left(3^4\right)^2Solution: Apply the power of a power rule, \left(a^m\right)^n=a^{m\cdot n}So, \left(3^4\right)^2=3^{4\cdot2}=3^8 Negative Exponent Rule If a number has a negative number as an exponent, the exponent can be converted into a positive number by taking its reciprocal.Imagine (x) as a fraction with a numerator of 1. When raised to the power of (n), it's like multiplying the numerator (which is 1) by itself (n) times. So, (x^{-n}) becomes (\frac{1}{x^n}), where (x) is the base and (n) is the exponent.So, (x^{-n}=\frac{1}{x^n})Example: Write 2^{-5} using positive exponent.Solution: Apply the negative exponent rule, a^{-m}=\frac{1}{a^m}So, 2^{-5}=\frac{1}{2^5} Practice Problems Q1: Simplify the expression (3^4 \times 3^2). 6561 729 2187 72 Answer: b Q2: Simplify (5^7 \div 5^3). 625 25 7/3 125 Answer:a Q3: Simplify the expression ((2^3)^4). 128 256 24 4096 Answer:d Q4: Determine the value of ((4 \times 5)^2). 40 400 2048 100 Answer: b Frequently Asked Questions Q1. Why are the laws of exponents important? Answer: Understanding the laws of exponents is like having a superpower in math! These rules make it easier for us to solve problems involving numbers with powers. They're important because they help us simplify math problems and find solutions faster. Q2. Are there any special cases involving exponents? Answer:Yes, when we raise any nonzero number to the power of zero, it always equals 1. And if we raise any number to the power of one, it stays the same. Q3. How can I practice applying the exponent laws? Answer:You can practice by solving lots of different problems involving exponents. Try simplifying expressions, figuring out the values of different powers, and solving equations with exponents. Related Topics Algebra - Order Of Operations Algebra - Distributive Property X And Y Axes Geometry - Scalene Triangle Common Multiple Geometry - Quadrant
2622	https://en.wikipedia.org/wiki?title=Talk:Binomial_theorem	Talk:Binomial theorem - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk [x] Toggle the table of contents Contents move to sidebar hide (Top) 1 Multiple issues2 comments 2 Original Research in substitution of of e^ax and e^bx in general Leibniz formula section of generalizations4 comments 3 Applications1 comment 4 History section49 comments 5 Perhaps a dedicated history article would be helpful1 comment Talk:Binomial theorem [x] Add languages Page contents not supported in other languages. Article Talk [x] English Read Edit Add topic View history [x] Tools Tools move to sidebar hide Actions Read Edit Add topic View history General What links here Related changes Upload file Permanent link Page information Get shortened URL Download QR code Expand all Print/export Download as PDF Printable version In other projects From Wikipedia, the free encyclopedia hide This level-5 vital article is rated C-class on Wikipedia's content assessment scale. It is of interest to the following WikiProjects: showMathematicsMid‑priority Mathematics portal This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.Mathematics Wikipedia:WikiProject Mathematics Template:WikiProject Mathematics mathematics MidThis article has been rated as Mid-priority on the project's priority scale. Archives 1 This page has archives. Sections older than 1095 days may be auto-archived by Lowercase sigmabot III if there are more than 5. Multiple issues [edit] NOTE: The version of the page being discussed Currently only a history is provided with enough citations and the rest is poorly sourced. Would be nice of someone adds a couple of links out there. DAVRONOV A.A.✉⚑08:27, 17 June 2019 (UTC)[reply] That's only one issue :p. Abstractly you are certainly right; concretely, are there any particular places you think are suspect (either, might be wrong, or with inappropriate weight, or original research)? Because "please make the article better" alone is not very constructive. --JBL (talk) 14:28, 17 June 2019 (UTC)[reply] Original Research in substitution of of e^ax and e^bx in general Leibniz formula section of generalizations [edit] Hi! The substitution of e^ax and e^bx into the general Leibniz formula seems to have no sources and thus seems to be original research. Although I understand that this substitution is correct mathematically, but there are no published sources having the same idea. Please take a moment to review my edit and verify my changes. The two changes I made are: Added [original research?] tag to the e^ax, e^bx substitution claim Added [verification needed] to the source "Calculus in One and Several Variables" by Robert Seeley since the book does not contain the Leibniz formula at all. The "Calculus for One Variable" can be found here, available for digital borrowing. This is single-variable calculus part of the combined book. Therefore, the Leibniz formula is expected to be in here. However, after extensive searching, no record of the Leibniz formula was found here. I understand that a similar issue was addressed under the "Multiple issues" section, but I wanted to keep a concrete discussion just for this change. If a citation is found, please feel free to enter a citation. Otherwise I will be forced to remove the content according to the Wikipedia: Original research policy. Thank you. --DhPhoenix (talk) 17:40, 20 June 2020 (UTC)[reply] This isn't OR; I'm having trouble finding a good ref for it, but a quick google search finds a lot of message-board type expositions of this fact. While a source would be ideal, the ease with which one finds non-RS discussion + WP:CALC probably lets this one stand. As for the Leibniz rule, there's another source at the main article that you could probably use if you think it's better. You could always just cite Abramowitz & Stegun too, or probably hundreds of other Calculus textbooks. –Deacon Vorbis(carbon•videos) 18:54, 20 June 2020 (UTC)[reply]Ref added for the binomial theorem as a corollary of the Leibniz rule. –Deacon Vorbis(carbon•videos) 19:38, 20 June 2020 (UTC)[reply]Yes, I'm sorry: this is not original research. Thank you for the source! And yes, I find the main article Leibniz rule citation better: I'll correct that.--DhPhoenix (talk) 05:47, 21 June 2020 (UTC)[reply] Applications [edit] The section on the infinite series for e is alright up until "This indicates that e can be written as a series:", at which point rigor and accuracy is lost. See Rudin, Priniciples of Mathematical Analysis, Section 3.3.1, for a proof. The mistake in the page as it exists is that it is not sufficient to say that the in the limit kth term is 1/k! implies that the entire series of 1/k! terms is equal to the original series. —Preceding unsigned comment added by 192.80.55.86 (talk) 18:20, 29 November 2021 (UTC)[reply] History section [edit] @Jacobolus: Hi, I just came to see your recent edits. In my humble opinion, sources like Amulya Kumar Bag, Biggs ... cannot challenge the views of prominent experts of the history of mathematics like Roshdi Rashed, Robertson, O'Connor as per WP:WEIGHT. According to these sources, Al-Karaji discovered the binomial theorem. please let me know if you think I'm missing something. Best.---Wikaviani(talk)(contribs)21:18, 2 December 2024 (UTC)[reply] You can read the historical Indian sources or their translations directly for yourself. It is not at all controversial to claim that what is described is binomial coefficients / "Pascal's triangle". I haven't looked at Rashed's book (paper?) but this sounds like a miscommunication.It is entirely plausible though that these topics were developed independently in India, the Islamic world, and China. Who counts as "discoverer" or "inventor" seems like a pointless argument; we should just try to describe precisely what various historical scholars did, and let readers draw their own conclusions. –jacobolus(t)21:46, 2 December 2024 (UTC)[reply]I have checked your sources, honestly, nothing very impressive, some obscure mathematicians of Indian origin, mainly. Also, the historical section should give a fair representation of the history of this theorem, the discovery of it might be pointless for you, but not for our readers. The page you're looking for is 63 in Rashed's work. Here, another source repeating what Rashed says:"THE BINOMIAL THEOREM: A WIDESPREAD CONCEPT IN MEDIEVAL ISLAMIC MATHEMATICS"(PDF). core.ac.uk. p.401. Retrieved 2019-01-08..---Wikaviani(talk)(contribs)22:00, 2 December 2024 (UTC)[reply]The authors involved here are world class experts in the history of Indian mathematics. I find your dismissive tone pretty insulting to be honest, and elevating one or another religious or ethnic group for point-scoring purposes by erasing the contributions of other groups is one the things I find personally most unpleasant and unfortunate about discussions of math history both off and especially on Wikipedia. Mathematics is a great human achievement which should be celebrated rather than fought over. Rashed does good work, and is well worth citing as an authority on developments among medieval Islamic mathematicians, but it's also not like he's omniscient. –jacobolus(t)23:44, 2 December 2024 (UTC)[reply]It's about discussing the quality of the sources, not about making it personal, and that's a legit concern of mines. So according to you, "Amulya Kumar Bag" is a world class expert? Wow, so how many influencial books has this guy published? with what publisher? how many awards has he? how about his academic career? The source number 3 that you use repeatedly in the section about Indian maths is a 60 years old paper with barely any information about its publisher."Mathematics is a great human achievement": agreed, so what?---Wikaviani(talk)(contribs)10:27, 3 December 2024 (UTC)[reply]I'm going to stop engaging with this now, because I find your sarcasm very unpleasant. I recommend you adopt a more agreeable attitude before engaging in Wikipedia talk page discussion. Don't delete paragraphs of material from this page without consensus. –jacobolus(t)16:23, 3 December 2024 (UTC)[reply]So convenient, you refuse to achieve a consensus here, ok then, I'll revert back to the status quo version. This is not "sarcasm" but legit concerns about the quality of the sources you cite as I told you already above. Also, even Amyula kumar Bag says that some sources say that the so-called Pascal triangle found by Indian mathematicians has nothing to do with it. Best.---Wikaviani(talk)(contribs)18:26, 3 December 2024 (UTC)[reply]@JayBeeEll Do you want to take this one on? I'm not in the mood to deal with persistent rudeness today. @Wikaviani, please read Wikipedia:Civility, Wikipedia:What Wikipedia is not §Wikipedia is not a battleground and Wikipedia:Here to build an encyclopedia before making further edits or comments. –jacobolus(t)18:38, 3 December 2024 (UTC)[reply]Well, I'm quite baffled to see an editor who has been editing here for 20 years blatantly throwing baseless accusations of rudeness while faced with an editor who tries to discuss the quality of the sources ... How about complying with WP:CONSENSUS? WP:UNDUE? WP:BURDEN? WP:RS? WP:ASPERSIONS? and so on? You have made many edits to the article without any consensus and while some of them are cosmetic, others aren't. @JayBeeEll: I would appreciate your input about the reliability of Amyula kumar Bag for this topic. Thanks. Best.---Wikaviani(talk)(contribs)22:05, 3 December 2024 (UTC)[reply] "Tries to discuss" would be a more convincing summary if you had skipped the unjustified content removal, edit warring, repeated insults, sarcasm, ~~air quotes~~scare quotes, and unexplained purity tests about which historians are "true" enough.From a content perspective, please just be straight forward about your goals. My speculation is that you don't think the history of combinations (n choose k) and Pascal's triangle should be discussed in a page about the binomial theorem. You can try to build consensus for this position but you should explain yourself clearly and give some clear rationale for your position, instead of just blanking content and revert warring about it.The sources are clearly fine: Bag's paper in the Indian Journal of History of Science was written by a professional historian in a respectable peer reviewed journal, has been widely cited since, and unquestionably meets WP:RS. Bibhutibhushan Datta, Radha Charan Gupta, Kripa Shankar Shukla were all celebrated and decorated mathematical historians. I'm sure if you put in some effort could find yet further secondary and tertiary sources discussing this topic, including by European/American authors if you don't like Indian historians. –jacobolus(t)23:50, 3 December 2024 (UTC)[reply] "would be a more convincing summary if you had skipped the unjustified content removal, edit warring, repeated insults, sarcasm, air quotes, and unexplained purity tests about which historians are "true" enough.": Uh? are you aware of WP:PA? you must know that baseless accusations qualify as personal attacks? where are my so-called insults? please clarify or this will end up at ANI. "The sources are clearly fine: Bag's paper in the Indian Journal of History of Science was written by a professional historian in a respectable peer reviewed journal, has been widely cited since, and unquestionably meets WP:RS. Bibhutibhushan Datta, Radha Charan Gupta, Kripa Shankar Shukla were all celebrated and decorated mathematical historians. I'm sure if you put in some effort could find yet further secondary and tertiary sources discussing this topic, including by European/American authors if you don't like Indian historians": I never said I don't like Indian historians, I just said that Bag has not the expertise to challenge prominent historians of maths like Robertson, Rashed etc baseless accusations of yours, again ... Interestingly, you have not included Bag in your blue links lists above by the way ... Also, please read WP:ONUS, I don't need to achieve consensus since you are the one who make repeated changes to the article, not me, thus you need to achieve consensus, not me. I will step out this article for now as you are not in the mood to have a constructive discussion. I will wait for more input from other editors. ---Wikaviani(talk)(contribs)08:55, 4 December 2024 (UTC)[reply]We got into this because you blanked a whole relevant and RS-sourced paragraph, based on the complaint that "none of the cited authors is a prominent expert of this topic", because Jean-Claude Martzloff[fr] is a mathematical historian primarily focused on China rather than India and Norman L. Biggs is a professional mathematician (though in this case writing a peer-reviewed paper in a top mathematical history journal).That blanking was inappropriate, so was quickly reverted by @JayBeeEll, but you edit-warred to re-blank the paragraph, this time saying "sorry, but I insist" because you don't consider Martzloff or Biggs to be a "true historian of maths". In my opinion these sarcastic comments about both Martzloff and Biggs were insulting and out of line.But whatever, fair enough. Since you didn't like these quasi-tertiary sources, I expanded the section adding closer secondary sources by subject experts. I don't know much about Bag – he is a professional mathematical historian specializing on the history of Indian mathematics who wrote a good number of widely cited papers in the 60s–70s which are still being cited today, and remains active. The other cited authors included Bibhutibhushan Datta and Radha Charan Gupta. In an edit summary you sarcastically and insultingly called Bag's paper a "'source'" with gratuitous ~~air quotes~~scare quotes, and then in your comment here you insultingly summarized all of these sources as "nothing very impressive, some obscure mathematicians of Indian origin". Neither Datta nor Gupta is obscure: these are two of the most famous, celebrated, and prolific historians of Indian mathematics.I was at that point quite unhappy with your repeated rude comments, but what put you over the line was "So according to you, 'Amulya Kumar Bag' is a world class expert? Wow, so how many influencial books has this guy published? with what publisher? how many awards has he? how about his academic career? The source number 3 that you use repeatedly in the section about Indian maths is a 60 years old paper with barely any information about its publisher." – Here you put more sarcastic ~~air~~ scare quotes, plus this time a string of sarcastic and insulting rhetorical questions. (To answer them though: Yes Amulya Kumar Bag is a world-class expert on the history of Indian mathematics and science. He doesn't have a Wikipedia article about him yet (feel free to write one), but here's a Google Scholar page. According to a quick web search he's a fellow of the Indian Academy of Sciences and was the editor of the Indian Journal of History of Science from 2002–2018. In particular his 1979 book Mathematics in ancient and medieval India has been very widely cited. I'm not sure which awards he has won, I'll leave you to research that. Both Datta and Gupta won multiple awards for their mathematical history work (e.g. Gupta won the Kenneth O. May Prize in 2009; Rashed won the same prize in 2017), and Datta's book History of Hindu Mathematics (with Singh) is one of the most influential ever written about the subject. As for that "its publisher", we are talking about the Indian Journal of History of Science published by the Indian National Science Academy.) I found your tone here to be completely unacceptable, insulting both to me personally and to these professional scholars. Since you are now making threats, I demand that you retract your insults.After that, we have more sarcastic and exaggerated language from you which I found to be insulting: "So convenient," "baffled [...] blatantly", "Uh?" " Interestingly,". Please cut that out now.––jacobolus(t)17:38, 4 December 2024 (UTC)[reply]I guess you and me don't have the same definition of the word insult. Let me be clear about that, if my comments were insults, go ahead and report me, otherwise, stop casting aspersion by labelling as "insults" legit concerns of another editor about that 60 years old source. Also, for your information, a mathematician is not a historian of maths, Biggs and Bag may be respectable mathematicians, but they are not expert in the field of history of maths. You seem to consider that since a publisher is reliable, that's enough to make the source reliable, this is wrong and you probably know that. Let me remind you what is a reliable source here, for Wikipedia (WP:SOURCEDEF):"When editors talk about sources that are being cited on Wikipedia, they might be referring to any one of these three concepts: The piece of work itself (the article, book) The creator of the work (the writer, journalist) The publisher of the work (for example, Random House or Cambridge University Press) Any of the three can affect reliability. Reliable sources may be published materials with a reliable publication process, authors who are regarded as authoritative in relation to the subject, or both. These qualifications should be demonstrable to other people."This is cristal clear, being published by a good publisher is not enough to make a source reliable, thus, while the Indian journal of history of science might be a great publisher, the author (Bag) is all but a world class expert and quite outdated since the sources that come after him, like Rashed or Robertson don't share his views (WP:AGEMATTERS). Besides, even if Bag was a world class expert like you claim, his views should not be given such an undue weight since they are not supported by the mainstream of reliable sources that are cited just after, especially when that source makes such an extraordinary claim about about the discovery of something equivalent to the triangle of Pascal in the 10th century. By the way, you keep editing the article while there is an ongoing dispute here, which is all but a constructive behaviour.---Wikaviani(talk)(contribs)08:59, 5 December 2024 (UTC)[reply]I don't intend to "report" you, which seems like a waste of time and energy. Let's assume (per WP:AGF) that your repeated inaccurate, sarcastic, condescending (and in my opinion quite insulting both to me personally and to the scholars cited here) comments were an honest mistake. I am merely asking you to please stop now. Specifically: I expect no more ~~air~~ scare quotes; no more sarcasm or condescension (along the lines of "so convenient", "interestingly", or most recently "for your information"); no more comments like "nothing very impressive"; no more sarcastic rhetorical questions. If you can cut the attitude and engage respectfully, we have no problem.Amulya Bag is a professional historian of mathematics and science who was for an extended time the editor of a respected journal of mathematical history."extraordinary claim" Well let's just look at the translated 10th century text directly, shall we: "After drawing a square on the top, two squares are drawn below (side by side) so that half of each is extended on either side. Below it three squares, below it (again) four squares are drawn and the process is repeated till the desired pyramid is attained. In the (topmost) first square the symbol for one is to be marked.. Then in each of the two squares of the second line figure one is to be placed. Then in the third line figure one is to be placed on each of the two extreme squares. In the middle square (of the third line) the sum of the figures in the two squares immediately above is to be placed; this is the meaning of the term pūrṇa. In the fourth line one is to be placed in each of the two extreme squares. In each of the two middle squares, the sum of the figures in the two squares immediately above, that is, three, is placed. Subsequent squares are filled in this way.What is described here is precisely Pascal's triangle – indeed, even in a more "modern" form than Pascal himself used."You seem to consider that ..." – You are putting words in my mouth, but you are also setting up a standard that has nothing to do with WP:RS. Scholarly papers about mathematical history written in reputable mathematical history journals clearly meet the wikipedia reliable source standard.I think you are somewhat mixing up what these various sources are claiming (and in particular the claims we are repeating here); to be specific, I think you are interpreting claims made about the history of combinations (which are numerically the same as binomial coefficients) as claims about the binomial theorem per se. Binomial coefficients, as numbers, occur in multiple contexts in mathematics. One of the context where they appear is in combinatorics, in counting the number of subsets of size ⁠k{\displaystyle k}⁠ of a set of size ⁠n{\displaystyle n}⁠. The paragraph about Indian examples is mainly about this combinatorial occurrence of these numbers. A distinct context is the algebra of polynomials, where these numbers appear as the coefficients of a binomial multiplied by itself some number of times.There's really no question that there are multiple Indian sources ranging over a wide time period discussing binomial coefficients as numbers, especially in a combinatorial context. The Bhagavati Sutra describes combinations up through ⁠n=4{\displaystyle n=4}⁠. (Here's O'Connor & Robertson on this since you consider them an acceptable source: "Permutations and combinations are used in the Sthananga Sutra. In the Bhagabati Sutra rules are given ... in the cases where n = 2, 3, and 4. The author then says that one can compute the numbers in the same way for larger n."). Pingala's description of making verses with various metres is in my understanding somewhat cryptic, but the elaboration given by his 10th century commentator describes exactly Pascal's triangle with the standard method of generation. (O'Connor and Robertson's summary: "In a commentary on this third century work in the tenth century, Pascal's triangle appears in order to give the coefficients of the binomial expansion." [O'Connor and Robertson make a mistake here: the commentary in question is of Pingala, not of the Bhagabati Sutra – I think they probably accidentally mixed up nearby sections of whatever secondary source they were working from]) The fraction-multiplication rule for finding n choose k is very clear and explicit in multiple sources from the 8th–9th century (here's a page by Ian G Pearce on O'Connor and Robertson's website discussing Mahavira's version).Where there's some amount of ambiguity is the question of what counts as the "binomial theorem" per se. Amulya Bag makes the claim that Pingala's rules for verses should be considered as binomial expansion, because they involve forming metres as arrangements of two types of syllables, so for instance with length three we get all of the syllable patterns (aaa, baa, aba, aab, bba, bab, bba, bbb), which can be grouped by how many a's and b's they have and then counted, yielding the counts (1, 3, 3, 1), the 3rd row of Pascal's triangle. For Bag, this is a form of binomial expansion. However, there's an argument that could be made that these aren't binomials in the sense of sums of variable numerical quantities ⁠x+y{\displaystyle x+y}⁠ to be expontiated like (x+y)3={\displaystyle (x+y)^{3}={}!!}x x x+y x x+x y x+x x y+{\displaystyle xxx+yxx+xyx+xxy+{}!!!}y y x+y x y+x y y+y y y={\displaystyle yyx+yxy+xyy+yyy={}!!}x 3+3 x 2 y+3 x y 2+y 3{\displaystyle x^{3}+3x^{2}y+3xy^{2}+y^{3}}. (Partly for this reason, I didn't repeat Bag's more opinionated claim in the article, but only his clearly factual claims. I think discussion in detail of this point is too in the weeds, and should be relegated to a more detailed history of binomial coefficients article if it is to be discussed at all.)Per Rashed, the oldest known description of the the binomial theorem per se, i.e. taking powers of a binomial sum of numbers, at least for exponents greater than 3, can be found in al-Samawʾal's book from the 12th century, credited by al-Samawʾal to al-Karajī.These different claims are not contradictory, and we don't need to reject the claim that Indians were working on combinatorics or expressed Pascal's triangle in order to accept Rashed's claim that the earliest version of something pretty close to the binomial theorem as we think of it today is al-Samawʾal/al-Karajī. Frankly we don't even need to reject Bag's claim that Pingala's verses are a kind of binomial expansion to also accept Rashed's claim. These claims are just not in any kind of conflict.As for your concern about sources though: These sources are clearly "reliable" by Wikipedia standards because they are written by reputable mathematical historians in reputable peer-reviewed history journals or published in scholarly books, have been widely cited by other scholars, and are based directly on primary historical sources whose translations we can easily read directly and understand. We don't need to rely on anyone's subjective interpretation: the texts are right there in black and white. –jacobolus(t)19:10, 5 December 2024 (UTC)[reply]This wall of text is mainly your own interpretation of this topic, what I see on my end, is an editor who is not capable to provide serious sources for the claim "The Chandaḥśāstra by the Indian lyricist Piṅgala (3rd or 2nd century BC) somewhat crypically describes a method of arranging two types of syllables to form metres of various lengths and counting them; as interpreted and elaborated by Piṅgala's 10th-century commentator Halāyudha his "method of pyramidal expansion" (meru-prastāra) for counting metres is Pascal's triangle.".Bag is not an expert historian of maths and is a bit outdated while Jayant shah's field of expertise is "computer vision" (Jayant shah is source number 12). I will remove this sentence but leave in the rest of your work since it is quite well-sourced. Best.---Wikaviani(talk)(contribs)08:27, 9 December 2024 (UTC)[reply]Bag is not an expert historian of maths – This is a falsehood which you now know to be false because we have been over this several times. Bag is a professional historian of mathematics who spent his career in the field and was the editor of a respected history of mathematics journal.editor who is not capable to provide serious sources – I demand that you retract your insults and desist from further personal attacks against me or other people.If you remove this perfectly fine sentence you will be reverted. Your personal understanding of sourcing policy does not accord with WP:RS and your behavior and comments here continue to well outside Wikipedia policy and norms. –jacobolus(t)15:54, 9 December 2024 (UTC)[reply]"If you remove this perfectly fine sentence you will be reverted. Your personal understanding of sourcing policy does not accord with WP:RS and your behavior and comments here continue to well outside Wikipedia policy and norms." Said the guy who already refused to comply with WP:RS, WP:CON, WP:ASPERSIONS, WP:NPA, WP:WEIGHT and so on ...What I see is a list of sources (mainly with no precise page numbers) who are either unreliable or outdated. Alsdorf Ludwig 90 years old source, outdated Bag: outdated and unreliable Shah Jayant: unreliable (field: computer vision, not historian of maths) Bose: unreliable (he was a physicist) Edwards A. W. F. unreliable (British statistician and geneticists) Fowler: historian of Greek maths Ranjan Roy: mathematician, not historian Divakaran P. P.: Mathematician Not a single historian of maths specialiced in the history of maths in Asia and no precise page numbers to allow me to verify what the sources really say. Please stop edit warring and self revert until a consensus is found here.---Wikaviani(talk)(contribs)11:21, 10 December 2024 (UTC)[reply]Not a single historian of maths specialiced in the history of maths in Asia – this is outright false. Amulya Kumar Bag is and Samarendra Nath Sen was a career expert on the history of South Asian mathematics and science. Sen's obituary was titled "The Doyen of Research in the History of Science in India".You've replaced the ordinary WP:RS standard with a new "Wikaviani must personally vouch for every source" standard. You were upset that sources were too old, so I added sources from a wide range of dates, but you're not happy with newer sources either.Your pick-and-choose criterion lets you reject any and all sources you want: this one is out because it is by a mathematician, the next one is out because it is by a historian but their specialty is the wrong country, the third one is out because it is by a historian specializing in the right topic but it was published in a book edited by a physicist, etc.And now you made a formal complaint at Wikipedia:Administrators' noticeboard/Incidents §Jacobolus and WP:ASPERSION about my asking you repeatedly to stop using sarcastic insulting language.My interpretation is that you pre-determined what outcome you want, and now you're going to make up whatever excuse you need or (mis)use whatever tool you can to obtain that outcome, irrespective of truth or community norms. –jacobolus(t)15:58, 10 December 2024 (UTC)[reply]My goal is to find a compromise, looking at your last comment, you sound like you own this article.---Wikaviani(talk)(contribs)16:52, 10 December 2024 (UTC)[reply]I threw more sources in. We can keep piling more sources on, but I really don't see the point. –jacobolus(t)18:11, 9 December 2024 (UTC)[reply]I added that the indian method was equivalent to Pascal's triangle. This is in accordance with the sources cited at Pascal's triangle, namely: Encyclopaedia of the History of Science, Technology, and Medicine in Non-Western Cultures (Helaine Selin): "Other, lost works of al-Karaji's are known to have dealt with inderterminate algebra, arithmetic, inheritance algebra, and the construction of buildings. Another contained the first known explanation of the arithmetical (Pascal's) triangle; the passage in question survived through al-Sama'wal's Bahir (twelfth century) which heavily drew from the Badi." The Development of Arabic Mathematics Between Arithmetic and Algebra (Roshdi Rashed): "The first formulation of the binomial and the table of binomial coefficients, to our knowledge, is to be found in a text by al-Karaji, cited by al-Sama'wal in al-Bahir." From Alexandria, Through Baghdad (Nathan Sidoli and Glen Van Brummelen): "However, the use of binomial coefficients by Islamic mathematicians of the eleventh century, in a context which had deep roots in Islamic mathematics, suggests strongly the table was a local discovery - most probably of al-Karaji." Hope this compromise is ok for you. Best.---Wikaviani(talk)(contribs)22:10, 10 December 2024 (UTC)[reply]I don't really see the difference, but sure.Rashed is making a point about a "formulation of the binomial" which doesn't say anything one way or the other about earlier examples of Pascal's triangle interpreted in other fashions.Len Berggren (1985), the paper you are quoting, which was reprinted in the book edited by Sidoli & Van Brummelen, says: "some had suggested that the table was a Chinese import. However, the use of binomial coefficients by Islamic mathematicians of the eleventh century, in a context which had deep roots in Islamic mathematics, suggests strongly the table was a local discovery—most probably of al-Karajī." This is only making the point that binomial coefficients as they appear in Islamic work were an independent discovery, not based on earlier work elsewhere, and the evidence for this is that the context related to topics/methods which had "deep roots" in other algebra work from the Islamic world – this is not the same as a claim that nobody else ever explored the same numbers in any other context. It's clear that the same table of numbers has been independently invented and reinvented repeatedly in a wide variety of cultures by a wide range of authors (in India, China, Persia, Italy, etc.).Jacques Sesiano's encyclopedia entry about al-Karajī (in the encylopedia edited by Selin) says that the "first known explanation" of the arithmetical triangle is from a lost work by al-Karajī. I don't know what Sesiano means by an explanation.(Aside: note Berggren, while being a fine mathematical historian, was professionally employed by a math department and as far as I can tell primarily taught pure math courses. All of your criticisms above about mathematicians not being historians would apply equally here (which is to say, are just as inapplicable in either case). Selin, the editor of the encyclopedia you mentioned, was a librarian. Many people have a career split between fields and manage to do good professional work in an area outside their job title.)None of Sesiano, Berggren, or Rashed is an expert in ancient Indian mathematics, so it's not surprising that they wouldn't go out of their way to discuss it in survey sources focusing on Islamic contributions. –jacobolus(t)02:53, 11 December 2024 (UTC)[reply]Brummelen, Rashed, Berggren and Selin are all historians of science and reliable for this topic. Your above quote of Rashed is incomplete, he says "The first formulation of the binomial and the table of binomial coefficients, to our knowledge, is to be found in a text by al-Karaji, cited by al-Sama'wal in al-Bahir." meaning that the binomial coefficients and Pascal's triangle were first found in a work by al-Karaji. I'm going to remove Pascal's triangle from this article and move it to the relevant article. All views should be fairly represented with sentences like some sources claim that Pascal's triangle first appeared in indian works (cite sources) while others claim that it first appeared in al-Karaji's work (cite sources). Since I am not conviced by the reliability of many of the sources you mentioned, I will exclude those (like Shah jayant, Edwards etc) who, in my opinion, obviously have no expertise for this specific topic, feel free to take them at WP:RSN in order to include them. However, keep in mind WP:CITEOVERKILL since many sources are already in the article. Best.---Wikaviani(talk)(contribs)08:03, 12 December 2024 (UTC)[reply]I'm going to remove Pascal's triangl I don't support this at all. The history of these topics is closely intertwined. What we could do however is summarize the whole history section in a more compressed way, and elaborate on it at a dedicated article called something like History of binomial coefficients, as I suggested below.My take is that you are backing away from your supposed "compromise" toward your previous maximalist (and in my opinion inappropriate and unjustified) content removal of before.obviously have no expertise – this is, once again, nonsensical and extremely insulting. Please stop. –jacobolus(t)13:02, 12 December 2024 (UTC)[reply]I kept the sentence about Pascal's triangle (I already added it to the article about the triangle) but reworded it in order to represent the different views of the sources about this claim, I strongly advise you to desist from reverting again with no legit reason. Also, you should stop from making baseless comments about me insulting the sources while I only say that they are not expert for this specific topic (i.e. the history of sciences), as it is starting to become disruptive. your mass revert of all my work is all but constructive, you could just add back the content about Pascal's triangle without undoing all my work.---Wikaviani(talk)(contribs)14:17, 12 December 2024 (UTC)[reply] In my opinion it is awkward, not according with expected encyclopedic tone/style, and unhelpful to throw a bunch of "this is contradicted" caveats to the end of every sentence of the paragraph about combinatorics in India which mention the topic of al-Samawʾal, since the latter topic is described clearly in the immediately following paragraph, which the reader is expected to immediately encounter. The resulting prose reads like a confused author with multiple personalities is having an argument with himself and the reader is being involuntarily dragged along as a spectator. The source provided (Rashed's book) does not address combinatorics in India at all, and I think you are unjustifiably putting words in Rashed's mouth by implying his work says something which it very clearly does not. If I were Rashed I would be unhappy to have my work mischaracterized in this way. –jacobolus(t)20:29, 12 December 2024 (UTC)[reply]Aside: I started a discussion at WikiProject Mathematics asking for some help resolving this dispute from uninvolved editors. –jacobolus(t)20:32, 12 December 2024 (UTC)[reply]Ping also @Slawekb, who somewhat expanded this section in July 2015 and may be interested to weigh in / may have other recommended sources. –jacobolus(t)01:23, 4 December 2024 (UTC)[reply] Apologies @Jacobolus and Wikaviani: I am aware of this discussion but I am travelling and so I can't log in and don't have the time to digest and engage properly. --JBL writing from 158.144.178.11 (talk) 15:06, 12 December 2024 (UTC)[reply] No need for apologies as far as I'm concerned, enjoy your trip JBL!---Wikaviani(talk)(contribs)15:12, 12 December 2024 (UTC)[reply]No worries, there's really no rush here. –jacobolus(t)17:43, 12 December 2024 (UTC)[reply]Wikaviani's edits such as this one, use sources claiming al-Karaji to have produced the first copy of Pascal's triangle as the basis for adding text to the article that casts doubt on unrelated claims of the appearance of binomial coefficients in earlier Indian mathematics. My feeling is that this is a serious violation of both WP:SYN and MOS:DOUBT. Those sources cannot be used to back up statements that Indian mathematics did not do these things, because they are not about Indian mathematics at all. They cannot be used to back up statements that the Indian accomplishments contradict scholarly consensus, because they do not make that comparison. To back up such statements we need sources explicitly comparing the Indian and Persian contributions, and we do not have such sources. We cannot point to the difference between what the sources say and say ourselves that it is a contradiction; that is WP:SYN. To be blunt, Wikaviani: please stop making these badly-sourced claims of contradiction unless you can back it up with reliable sources that make the same claims.Taking a step back, part of the issue may be that the binomial theorem, binomial coefficients, multiplicative method of computing binomial coefficients, additive method of computing binomial coefficients, Pascal's triangle, and combinations from sets, are six different but closely related things. Even when we have a statement saying that the first appearance of Pascal's triangle was in one place, and even if we take that statement at face value as accurate, it implies nothing about the first appearance of any of the other aspects of the subject. —David Eppstein (talk) 03:22, 13 December 2024 (UTC)[reply]Well, don't get me wrong, I have much respect for you as an admin here, but with all due respect, I'm quite speechless when i read you. First, that 200 years old source you posted at Wiki project maths and now your above post, saying that so many prominent historians of sciences are wrong when they deal with the history of maths while we, editors, know better than them. I hope I'm missing something but I don't know what exactly. Best.---Wikaviani(talk)(contribs)15:34, 13 December 2024 (UTC)[reply]Ok, so you are digging in. But you are wrong when you use a source about the Persians and the binomial theorem to make inferences about the Indians and binomial coefficients. Inferences are what our sources do. You should not be doing them here. Stop. —David Eppstein (talk) 17:41, 13 December 2024 (UTC)[reply]Ok, let's say that the sources I provided are not reliable (even if I'm not convinced by that) but how about the sources like Jayant Shah, Bose, Edwards etc? They seem to be good in their fields (i.e. maths, physics etc), but they are not really historians and they are about Indian maths, not Islamic maths.---Wikaviani(talk)(contribs)08:07, 16 December 2024 (UTC)[reply]Roshdi Rashed's 1972 paper is an excellent source about the content and relevance of the works of al-Samawʾal and al-Karajī, which is why it has been widely cited and its content has been incorporated into later surveys of Islamic mathematics and the history of algebra and combinatorics. But it, and sources drawing on it, aren't relevant about topics they never discussed.Likewise, papers by Bag and Shah are excellent sources about the works of Piṅgala and later Indian combinatorialists, but would not be relevant to cite about topics they did not mention. Aside: Bose was the chief editor of a book (not an author), and S. N. Sen, the author of that broad survey (also another editor for the whole book), was a celebrated historian of Indian mathematics. Edwards's book has been cited hundreds of times (including by historians) because it is one of the best high level surveys about Pascal's triangle. –jacobolus(t)12:53, 16 December 2024 (UTC)[reply]If all involved editors agree with that, then it's all good for me. Howrver, while Rashed's 1972 work does not discuss Indian maths, sources like Bag, Sen or Edwards don't discuss about Islamic sciences either. Best.---Wikaviani(talk)(contribs)19:41, 16 December 2024 (UTC)[reply]Do you have a point to this, or is this just culture warring? Your comment comes across as a total non-sequitur, because we are not using any of those people in the sources to the paragraph about Islamic mathematics. —David Eppstein (talk) 19:51, 16 December 2024 (UTC)[reply]No, David, this is not "culture warring", I'm just talking about what you said above:"Those sources cannot be used to back up statements that Indian mathematics did not do these things, because they are not about Indian mathematics at all. They cannot be used to back up statements that the Indian accomplishments contradict scholarly consensus, because they do not make that comparison.". I just say that if the Islamic sources don't make a comparative study with India, the Indian ones don't do a comparative study with Islamic sciences either.---Wikaviani(talk)(contribs)20:41, 16 December 2024 (UTC)[reply]That's precisely the point. So the Wikipedia article shouldn't speculate about relaions between the two or set up a direct comparison. The standard thing Wikipedia articles do in this kind of situation is to mention both topics, separately and independently. For Wikipedia to misattribute Wiki editors' speculation to scholars is a serious problem, and especially so when the speculation is an (implicit or explicit) criticism of sources that the cited source never mentioned. –jacobolus(t)21:16, 16 December 2024 (UTC)[reply]Leaving aside what your goal is, you keep saying things that are wrong. Edwards's book discusses both of these topics. The part discussing developments in India is pages 27–33, while the part mentioning al-Karajī is on page 52:"[...] In India, Brahmegupta (A.D. 628) gave ⁠(a+b)3{\displaystyle (a+b)^{3}}⁠ in his Arithmetic . ¶ It is to Persia, however, that the European thread can be traced back. The Al-bahir of Al-Samawal (died about 1180) is reported as containing a calculation of the coefficients, resulting in the Binomial Triangle, which had been discovered by Al-Karaji some time soon after 1007. It is possible that Al-Karaji was inspired to make his discovery by hearing of Brahmegupta's result for the cube of a binomial, for it is believed that Brahmegupta's work had been brought to Baghdad in the eighth century, and Al-Karaji, who worked in Baghdad, drew much else from Hindu sources . ¶ Al-Kashi , who died in Samarkand [...]" I don't have any idea whether or not Al-Karaji was inspired by Brahmagupta, and I'm not sure such speculation is worth mentioning in this article, whose history section should remain relatively concise in my opinion. We probably should mention Jamshīd al-Kāshī and Naṣīr al-Dīn al-Ṭūsī, however. –jacobolus(t)20:21, 16 December 2024 (UTC)[reply]Perhaps,but if we are talking about A. W. F. Edwards, I see "Anthony William Fairbank Edwards, FRS (born 1935) is a British statistician, geneticist and evolutionary biologist. Edwards is regarded as one of Britain's most distinguished geneticists, and as one of the most influential mathematical geneticists in the history." Again, this scholar has no expertise in the field of history of sciences, thus, I'm not convinced by what he can say in this field.---Wikaviani(talk)(contribs)20:53, 16 December 2024 (UTC)[reply]You keep saying "X, therefore not Y", first in your "Persia, therefore not India" edits and now in your "mathematician, therefore not historian of mathematics". It is fallacious, it is ad hominem, and it is unconvincing. Focus on the work, not on its author. Its reviews in e.g. MR0930876, Zbl0641.01004, and Zbl1032.01013 treat it seriously and positively; there is no reason to consider it as unreliable. And if you read our article on Edwards more carefully than just skimming its lead and looking for gotchas, you would see that "He has also written extensively on the history of genetics and statistics" (which apparently he considers binomial coefficients to fall under, since he also published about their history earlier than his book in International Statistical Review, focusing on the significance of Pascal's work on his triangle in the later history of probability and statistics). —David Eppstein (talk) 21:07, 16 December 2024 (UTC)[reply]You are putting words in my mouth David, I never said such things like "Persia therefore not India" or "mathematician therefore not historian of mathematics", rather the opposite. Again, if all involved editors agree with the reliability of the sources proposed by Jacobolus, then it's all good for me.---Wikaviani(talk)(contribs)21:58, 16 December 2024 (UTC)[reply]@David Eppstein: There are 8 sources in number 11, this is not WP:OVERKILL while the 4 sources are? If the formatting was your concerns, you should have formatted it in a way that is ok for you. Anyway, I think we're done here.---Wikaviani(talk)(contribs)22:32, 17 December 2024 (UTC)[reply]Let me be more clear since you apparently failed to understand my edit summary. There were three reasons why your edit adding a reference was bad. 1. It duplicated a reference that was already in the article. If the footnote was kept, it should have re-used the existing footnote using named references, instead of making another footnote with another copy of the reference. 2. It was very badly formatted. But it should not have been reformatted because of the duplication issue. 3. You added it as a footnote to a sentence that already had three footnotes, each reliable and with a detailed supporting quote. We do not need a fourth footnote on that one sentence. Putting another footnote there does not improve the article in any way. Because of the third problem, I undid your addition rather than reformatting your new footnote or reusing the old one. —David Eppstein (talk) 22:53, 17 December 2024 (UTC)[reply]I think the issue is the number of consecutive footnotes rather than the number of sources, but the previous claim really also does not need so many sources. The reason I kept adding more was because you kept objecting to sources one after another (for what I found to be inappropriate and mysterious reasons), but if we're all okay with the paragraph about Indian combinatorics we can thin out some of those. In case we want them later, I'm taking out: Alsdorf, Ludwig (1991) . "The Pratyayas: Indian Contribution to Combinatorics"(PDF). Indian Journal of History of Science. 26 (1): 17–61. Translated by S. R. Sarma from "π Die Pratyayas. Ein Beitrag zur indischen Mathematik". Zeitschrift fur Indologie und Iranistik. 9: 97–157. 1933. Sen, Samarendra Nath (1971). "Mathematics". In Bose, D. M. (ed.). A Concise History Of Science In India. Indian National Science Academy. Ch. 3, pp.136–212, esp. "Permutations, Combinations and Pascal Triangle", pp.156–157. Fowler, David H. (1996). "The Binomial Coefficient Function". The American Mathematical Monthly. 103 (1): 1–17, esp. §4 "A Historical Note", pp. 10–17. doi:10.2307/2975209. JSTOR2975209. Alsdorf was one of the earliest sources about this topic. Fowler's paper might be worth citing elsewhere in this article, either for the historical note included or for his more mathematical content. –jacobolus(t)00:18, 18 December 2024 (UTC)[reply]Re "one of the earliest sources about this topic": I mentioned this also on WT:WPM, but see Burrow, Reuben (1790), "A Proof that the Hindoos had the Binomial Theorem", Asiatick Researches: 487–497 I don't think we should cite it here but it may be worth keeping if we ever split out a history of binomial coefficients article. —David Eppstein (talk) 00:27, 18 December 2024 (UTC)[reply]Yeah, this one is focused on an (anonymous?) historical source with similar content to that found in Śrīdhara or Mahāvīra which we currently mention here, notably the formula ⁠(n k)=n 1×n−1 2×⋯×n−k+1 n−k{\displaystyle {\tbinom {n}{k}}={\tfrac {n}{1}}\times {\tfrac {n-1}{2}}\times \cdots \times {\tfrac {n-k+1}{n-k}}}⁠. Though that historical source does also mention ⁠∑k=1 n(n k)=2 n−1{\displaystyle \textstyle \sum _{k=1}^{n}{\tbinom {n}{k}}=2^{n}-1}⁠, which could also be worth mentioning in a dedicated history article. By "one of the earliest sources" I meant specifically about the counting of poetic metres. But you are right that Burrow's paper is quite a bit older than Alsdorf. –jacobolus(t)00:53, 18 December 2024 (UTC)[reply] Perhaps a dedicated history article would be helpful [edit] It might be useful to have an article History of binomial coefficients to consolidate Binomial coefficient §History and notation, Pascal's triangle §History, and Binomial theorem §History. That would leave more room to describe specific historical versions and their context in more detail and list more historical examples without unduly burdening these articles which otherwise are in a hurry to get to the mathematical content. This article could then make do with a more concise summary of 2–3 paragraphs. I don't have the motivation to write such a thing in the near future, but if someone feels inspired I'd be supportive of that effort. –jacobolus(t)17:46, 3 December 2024 (UTC)[reply] Retrieved from " Categories: C-Class level-5 vital articles Wikipedia level-5 vital articles in Mathematics C-Class vital articles in Mathematics C-Class mathematics articles Mid-priority mathematics articles Hidden category: All Wikipedia vital articles This page was last edited on 9 August 2025, at 15:52(UTC). Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Privacy policy About Wikipedia Disclaimers Contact Wikipedia Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search [x] Toggle the table of contents Talk:Binomial theorem Add topic
2623	https://www.doubtnut.com/qna/644852647	Simplify : 1+2i1−2i−1−2i1+2i More from this Exercise The correct Answer is:N/a To simplify the expression (1+2i)/(1−2i)−(1−2i)/(1+2i), we can follow these steps: Step 1: Write the expression We start with the expression: 1+2i1−2i−1−2i1+2i Step 2: Find a common denominator The common denominator for the two fractions is (1−2i)(1+2i). So we rewrite the expression: (1+2i)(1+2i)−(1−2i)(1−2i)(1−2i)(1+2i) Step 3: Expand the numerators Now we expand the numerators: 1. For (1+2i)(1+2i): (1+2i)2=12+2⋅1⋅2i+(2i)2=1+4i−4=−3+4i 2. For (1−2i)(1−2i): (1−2i)2=12−2⋅1⋅2i+(−2i)2=1−4i−4=−3−4i Step 4: Combine the results Now we substitute back into the expression: (−3+4i)−(−3−4i)(1−2i)(1+2i) This simplifies to: −3+4i+3+4i(1−2i)(1+2i)=8i(1−2i)(1+2i) Step 5: Simplify the denominator Now we simplify the denominator: (1−2i)(1+2i)=12−(2i)2=1−(−4)=1+4=5 Step 6: Final expression Thus, we have: 8i5 This can be written as: 0+85i Final Answer The simplified expression is: 85i --- To simplify the expression (1+2i)/(1−2i)−(1−2i)/(1+2i), we can follow these steps: Step 1: Write the expression We start with the expression: 1+2i1−2i−1−2i1+2i Step 2: Find a common denominator The common denominator for the two fractions is (1−2i)(1+2i). So we rewrite the expression: (1+2i)(1+2i)−(1−2i)(1−2i)(1−2i)(1+2i) Step 3: Expand the numerators Now we expand the numerators: 1. For (1+2i)(1+2i): (1+2i)2=12+2⋅1⋅2i+(2i)2=1+4i−4=−3+4i 2. For (1−2i)(1−2i): (1−2i)2=12−2⋅1⋅2i+(−2i)2=1−4i−4=−3−4i Step 4: Combine the results Now we substitute back into the expression: (−3+4i)−(−3−4i)(1−2i)(1+2i) This simplifies to: −3+4i+3+4i(1−2i)(1+2i)=8i(1−2i)(1+2i) Step 5: Simplify the denominator Now we simplify the denominator: (1−2i)(1+2i)=12−(2i)2=1−(−4)=1+4=5 Step 6: Final expression Thus, we have: 8i5 This can be written as: 0+85i Final Answer The simplified expression is: 85i Topper's Solved these Questions Explore 9 Videos Explore 29 Videos Explore 20 Videos Explore 8 Videos Similar Questions Simplify: (14+2i)(7+12i) Simplify: (1+i1−i)200 Simplify 1−iroot32+2i Simplify: (−12−√32i)2 Simplify: 1i−1i2+1i3−1i4 The modulus and amplitude of 1+2i1−(1−i)2 are Simplify: 1i+1i2+1i3+1i4 Change the following complex numbers into polar form 1+2i1−(1−i)2 Simplify the following: (−i)(2i)(−18i)3 Express the following complex number in the polar form: (i) 1+7i(2−i)2 (ii) 1+3i1−2i NAGEEN PRAKASHAN ENGLISH-COMPLEX NUMBERS AND QUADRATIC EQUATION -MISCELLANEOUS EXERCISE Simplify : (1+2i)/(1-2i)-(1-2i)/(1+2i) Evaluate : [i^(18)+(1/i)^(25)]^3 For any two complex numbers z1and z2, prove that R e(z1z2)=R ez1R e z2... Reduce (1/(1-4i)-2/(1+i))((3-4i)/(5+i))to the standard form. If under root of (a+i b)/(c+i d)=x+i y , Prove (a^2+b^2)/(c^2+d^2)=(x... Convert the following in the polar form : (i) (1+7i)/((2-i)^2) (ii) (... Solve the equation : 3x^2-4x+(20)/3=0 Solve the equation :x^2-2x+3/2=0 Solve the equation :27 x^2-10 x+1=0 Solve the following quadratic: 21 x^2-28 x+10=0 If z1=2-i ,z2=1+i ,find \|(z1+z2+1)/(z1-z2+i)\| If a + i b =((x+i)^2)/(2x^2+1),prove that a^2+b^2=((x^2+1)^2)/((2x^2+... If z1=2-i ,\ +2=-2+i , find : R e((z1z2)/(z1)) Find the modulus and argument of the complex number (1+2i)/(1-3i). Find the real numbers x and y if (x-i y)(3+5i)is the conjugate of -6-... Find the modulus of (1+i)/(1-i)-(1-i)/(1+i) If (x+i y)^3=u+i v ,then show that u/x+v/y=4(x^2-y^2). If alphaand betaare different complex numbers with \|beta\|=1,then fin... Find the number of non-zero integral solution of the equation \|1-i\|^x=... If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that (a^2 +... If ((1+i)/(1-i))^m=1, then find the least positive integral value of m... Exams Free Textbook Solutions Free Ncert Solutions English Medium Free Ncert Solutions Hindi Medium Boards Resources Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc NCERT solutions for CBSE and other state boards is a key requirement for students. Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. Get PDF and video solutions of IIT-JEE Mains & Advanced previous year papers, NEET previous year papers, NCERT books for classes 6 to 12, CBSE, Pathfinder Publications, RD Sharma, RS Aggarwal, Manohar Ray, Cengage books for boards and competitive exams. Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation Contact Us
2624	https://www.sfu.ca/~lockhart/richard/450/17_3/lectures/decision_theory_for_HT/web.pdf	Hypothesis Testing and Decision Theory • Decision set is D = {0, 1} • Loss is L(d, θ) = 1(make an error) • Or more generally L(0, θ) = ℓ11(θ ∈Θ1) L(1, θ) = ℓ21(θ ∈Θ0) for two constants ℓ1, ℓ2 > 0. • Make decision space convex: decision is a probability measure on D. • Any such measure can be speciﬁed by δ = P(reject) so D = [0, 1]. • The loss function of δ ∈[0, 1] is L(δ, θ) = (1−δ)ℓ11(θ ∈Θ1)+δℓ01(θ ∈Θ0) . 1 Simple hypotheses • Prior is π0 > 0 and π1 > 0 with π0+π1 = 1. • Procedure: map from sample space to D – a test function. • Risk function of procedure φ(X) is a pair of numbers: Rφ(θ0) = E0(L(δ, θ0)) and Rφ(θ1) = E1(L(δ, θ1)) • We ﬁnd Rφ(θ0) = ℓ0E0(φ(X)) = ℓ0α and Rφ(θ1) = ℓ1E1(1 −φ(X)) = ℓ1β • Bayes risk of φ is π0ℓ0α + π1ℓ1β 2 • Minimized by likelihood ratio test: φ(X) = 1(f1(X)/f0(X) > π0ℓ0/(π1ℓ1)) • These tests are Bayes and admissible. • The risk is constant if βℓ1 = αℓ0; you can use this to ﬁnd the minimax test in this context. 3
2625	https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/mit8_01scs22_chapter15.pdf	Chapter 15 Collision Theory 15.1 Introduction........................................................................................................... 1 15.2 Reference Frames and Relative Velocities.......................................................... 1 15.2.1 Relative Velocities .......................................................................................... 3 15.2.2 Center-of-mass Reference Frame................................................................. 4 15.2.3 Kinetic Energy in the Center-of-Mass Reference Frame........................... 5 15.2.4 Change of Kinetic Energy and Relatively Inertial Reference Frames...... 5 15.3 Characterizing Collisions..................................................................................... 7 15.4 One-Dimensional Collisions Between Two Objects........................................... 7 15.4.1 One Dimensional Elastic Collision in Laboratory Reference Frame........ 7 15.4.2 One-Dimensional Collision Between Two Objects – Center-of-Mass Reference Frame ..................................................................................................... 11 15.5 Worked Examples............................................................................................... 12 Example 15.1 Elastic One-Dimensional Collision Between Two Objects.......... 12 Example 15.2 The Dissipation of Kinetic Energy in a Completely Inelastic Collision Between Two Objects ............................................................................. 13 Example 15.3 Bouncing Superballs ....................................................................... 14 15.6 Two Dimensional Elastic Collisions .................................................................. 17 15.6.1 Two-dimensional Elastic Collision in Laboratory Reference Frame...... 17 Example 15.5 Elastic Two-dimensional collision of identical particles.............. 20 ................................................................................................................................... 21 Example 15.6 Two-dimensional elastic collision between particles of equal mass Example 15.7 Two dimensional collision between particles of unequal mass ... 22 15.7 Two-Dimensional Collisions in Center-of-Mass Reference Frame ................ 24 15.7.1 Two-Dimensional Collision in Center-of-Mass Reference Frame........... 24 15.7.2 Scattering in the Center-of-Mass Reference Frame ................................. 25 Example 15.8 Scattering in the Lab and CM Frames ......................................... 26 Chapter 15 Collision Theory Despite my resistance to hyperbole, the LHC [Large Hadron Collider] belongs to a world that can only be described with superlatives. It is not merely large: the LHC is the biggest machine ever built. It is not merely cold: the 1.9 kelvin (1.9 degrees Celsius above absolute zero) temperature necessary for the LHC’s supercomputing magnets to operate is the coldest extended region that we know of in the universe—even colder than outer space. The magnetic field is not merely big: the superconducting dipole magnets generating a magnetic field more than 100,000 times stronger than the Earth’s are the strongest magnets in industrial production ever made. And the extremes don’t end there. The vacuum inside the proton-containing tubes, a 10 trillionth of an atmosphere, is the most complete vacuum over the largest region ever produced. The energy of the collisions are the highest ever generated on Earth, allowing us to study the interactions that occurred in the early universe the furthest back in time.1 Lisa Randall 15.1 Introduction When discussing conservation of momentum, we considered examples in which two objects collide and stick together, and either there are no external forces acting in some direction (or the collision was nearly instantaneous) so the component of the momentum of the system along that direction is constant. We shall now study collisions between objects in more detail. In particular we shall consider cases in which the objects do not stick together. The momentum along a certain direction may still be constant but the mechanical energy of the system may change. We will begin our analysis by considering two-particle collision. We introduce the concept of the relative velocity between two particles and show that it is independent of the choice of reference frame. We then show that the change in kinetic energy only depends on the change of the square of the relative velocity and therefore is also independent of the choice of reference frame. We will then study one- and two-dimensional collisions with zero change in potential energy. In particular we will characterize the types of collisions by the change in kinetic energy and analyze the possible outcomes of the collisions. 15.2 Reference Frames and Relative Velocities  We shall recall our definition of relative inertial reference frames. Let R be the vector from the origin of frame S to the origin of reference frame S′ . Denote the 1 Randall, Lisa, Knocking on Heaven's Door: How Physics and Scientific Thinking Illuminate the Universe and the Modern World, Ecco, 2011. 15-1  position vector of the jth particle with respect to the origin of reference frame S by rj jth and similarly, denote the position vector of the particle with respect to the origin of reference frame S′ by r ′ j (Figure 15.1). S rj jth particle rj S R Figure 15.1 Position vector of jth particle in two reference frames. The position vectors are related by  rj =  r′ j +  R . (15.2.1) The relative velocity (call this the boost velocity) between the two reference frames is given by   dR V = . (15.2.2) dt Assume the boost velocity between the two reference frames is constant. Then, the relative acceleration between the two reference frames is zero,    dV A = = 0 . (15.2.3) dt When Eq. (15.2.3) is satisfied, the reference frames S and S′ are called relatively inertial reference frames. jth Suppose the particle in Figure 15.1 is moving; then observers in different reference frames will measure different velocities. Denote the velocity of jth particle in  frame S by v j = dr  j / dt , and the velocity of the same particle in frame S′ by  v′ j = dr ′ j / dt . Taking derivative, the velocities of the particles in two different reference frames are related according to    v j = v′ j + V . (15.2.4) 15-2 15.2.1 Relative Velocities Consider two particles of masses m1 and m2 interacting via some force (Figure 15.2). Figure 15.2 Two interacting particles Choose a coordinate system (Figure 15.3) in which the position vector of body 1 is given   by r 1 and the position vector of body 2 is given by r2 . The relative position of body 1    with respect to body 2 is given by r 1 2 , = r 1 − r2 .    Figure 15.3 Coordinate system for two bodies.  During the course of the interaction, body 1 is displaced by dr 1 and body 2 is displaced  by dr2 , so the relative displacement of the two bodies during the interaction is given by    dr 1 2 , = dr 1 − dr2 . The relative velocity between the particles is     dr 1 2 , dr 1 dr2   v1 2 , = = − = v1 − v2 . (15.2.5) dt dt dt We shall now show that the relative velocity between the two particles is independent of the choice of reference frame providing that the reference frames are relatively inertial.  The relative velocity v12 ′ in reference frame S′ can be determined from using Eq. (15.2.4) to express Eq. (15.2.5) in terms of the velocities in the reference frame S′ , v1 ′ v1, 2 ′   = v1 v1, 2 − v2 = (v  1 ′ + V) − (v ′ 2 +  V) = −  v′ 2 = (15.2.6) 15-3 and is equal to the relative velocity in frame S . For a two-particle interaction, the relative velocity between the two vectors is independent of the choice of relatively inertial reference frames. 15.2.2 Center-of-mass Reference Frame Let  r cm be the vector from the origin of frame S to the center-of-mass of the system of particles, a point that we will choose as the origin of reference frame S cm , called the center-of-mass reference frame. Denote the position vector of the jth particle with respect to origin of reference frame S by  rj and similarly, denote the position vector of the jth particle with respect to origin of reference frame S cm by ′ rj (Figure 15.4). S cm r cm rj jth particle rj S Figure 15.4 Position vector of jth particle in the center-of-mass reference frame. th j The position vector of the particle in the center-of-mass frame is then given by    −  r′ j = rj r . (15.2.7) cm The velocity of the jth particle in the center-of-mass reference frame is then given by   −  v′ j = v j v . (15.2.8) cm There are many collision problems in which the center-of-mass reference frame is the most convenient reference frame to analyze the collision. Consider a system consisting of two particles, which we shall refer to as particle 1 and particle 2. We can use Eq. (15.2.8) to determine the velocities of particles 1 and 2 in the center-of-mass, µ v1 ′ −   v cm = v1 −   m 1v1 + m2v2 m2 = m 1 + m2 m 1 + m2 ( v1, − v2 ) = m 1  v1, 2 .  (15.2.9) = v1 15-4 where  v12 =  v1 − v2 is the relative velocity of particle 1 with respect to particle 2 . A similar result holds for particle 2 : v′ 2  −   v cm = v2 − m 1v1 + m2v2 m 1 = −   m 1 + m2 m 1 + m2 ( v1 − v2 ) = − µ m2  v1, 2 . (15.2.10)  v2 = The momentum of the system the center-of-mass reference frame is zero as we expect,      m 1v1 ′ + m2v′ 2 = µ − µ = 0 . (15.2.11) v12 v12 15.2.3 Kinetic Energy in the Center-of-Mass Reference Frame The kinetic energy in the center of mass reference frame is given by 1  1  Kcm = m 1v1 ′ ⋅ v  1 ′ + m2v′ 2 ⋅ v ′ 2 . (15.2.12) 2 2 We now use Eqs. (15.2.9) and (15.2.10) to rewrite the kinetic energy in terms of the  relative velocity v  12 ′ = v1 ′ − v ′ 2 , ⎛ ⎞⎛ ⎞ 1 ⎛ ⎞⎛ ⎠ ⎟⋅ − µ ⎞ 1  v  v  v  v µ m 1 µ m 1 − µ K ⎠ ⎟ ⋅ ⎝ ⎜ ⎠ ⎟ + m 1 m2 = ⎝ ⎜ ⎝ ⎜ ⎝ ⎜ ⎠ ⎟ 1, 2 1, 2 1, 2 1, 2 2 2 cm m2 m2 . (15.2.13) ⎛ 1 1 ⎞ 1 1  v  v 2 2 ⋅ ⎝ ⎜ + m 1 m2 ⎠ ⎟ = µ µv1, 2 = 1, 2 1, 2 2 2 where we used the fact that we defined the reduced mass by 1 1 1 ≡ + . (15.2.14) µ m 1 m2 15.2.4 Change of Kinetic Energy and Relatively Inertial Reference Frames The kinetic energy of the two particles in reference frame S is given by KS = 1 m 1v1 2 + 1 m2v2 2 . (15.2.15) 2 2 We can take the scalar product of Eq. (15.2.8) to rewrite Eq. (15.2.15) as 15-5 1     1     KS = m 1(v1 ′ + v )⋅(v1 ′ + v ) + m2(v′ 2 + v )⋅(v′ 2 + v ) cm cm cm cm 2 2 . (15.2.16) 1 1 1    = 2+ 2+ )v 2 + (m 1 )⋅ v m 1v1 ′ m2v2 ′ (m 1 + m2 cm v1 ′ + m2v′ 2 cm 2 2 2 The last term is zero due to the fact that the momentum of the system in the center of mass reference frame is zero (Eq. (15.2.11)). Therefore Eq. (15.2.16) becomes 1 1 1 2 KS = m 1v1 ′ 2+ m2v2 ′ 2+ (m 1 + m2)v . (15.2.17) 2 2 2 cm The first two terms correspond to the kinetic energy in the center of mass frame, thus the kinetic energies in the two reference frames are related by KS = K + 1 (m 1 + m2)v 2 . (15.2.18) cm cm 2 We now use Eq. (15.2.13) to rewrite Eq. (15.2.18) as 1 1 2 KS = 2 + (m 1 + m2)v (15.2.19) µv1, 2 cm 2 2 Even though kinetic energy is a reference frame dependent quantity, because the second term in Eq. (15.2.19) is a constant, the change in kinetic energy in either reference frame is equal to 1 2 2 ΔK = µ(( ) − ( ) ) . (15.2.20) v1, 2 v1, 2 2 f i This generalizes to any two relatively inertial reference frames because the relative velocity is a reference frame independent quantity, the change in kinetic energy is independent of the choice of relatively inertial reference frames. We showed in Appendix 13A that when two particles of masses m1 and m2 interact, the work done by the interaction force is equal to 1 2 2 W = µ(( ) − ( ) ) . (15.2.21) v1, 2 v1, 2 2 f i Hence we explicitly verified that for our two-particle system W = ΔKsys . (15.2.22) 15-6 15.3 Characterizing Collisions In a collision, the ratio of the magnitudes of the initial and final relative velocities is called the coefficient of restitution and denoted by the symbol e , vB e = . (15.3.1) vA If the magnitude of the relative velocity does not change during a collision, e = 1, then the change in kinetic energy is zero, (Eq. (15.2.21)). Collisions in which there is no change in kinetic energy are called elastic collisions, ΔK = 0, elastic collision . (15.3.2) If the magnitude of the final relative velocity is less than the magnitude of the initial relative velocity, e < 1, then the change in kinetic energy is negative. Collisions in which the kinetic energy decreases are called inelastic collisions, ΔK < 0, inelastic collision . (15.3.3) If the two objects stick together after the collision, then the relative final velocity is zero, e = 0 . Such collisions are called totally inelastic. The change in kinetic energy can be found from Eq. (15.2.21), 1 2 1 m 1m2 2 ΔK = − µ vA = − vA , totally inelastic collision . (15.3.4) 2 2 m 1 + m2 If the magnitude of the final relative velocity is greater than the magnitude of the initial relative velocity, e > 1, then the change in kinetic energy is positive. Collisions in which the kinetic energy increases are called superelastic collisions, ΔK > 0, superelastic collision . (15.3.5) 15.4 One-Dimensional Collisions Between Two Objects 15.4.1 One Dimensional Elastic Collision in Laboratory Reference Frame Consider a one-dimensional elastic collision between two objects moving in the x direction. One object, with mass m1 and initial x -component of the velocity v1x,i , collides with an object of mass m2 and initial x -component of the velocity v2x,i . The scalar components v1x,i and v1x,i can be positive, negative or zero. No forces other than the interaction force between the objects act during the collision. After the collision, the 15-7 final x -component of the velocities are v1x, f and v2 x, f . We call this reference frame the “laboratory reference frame”. Figure 15.5 One-dimensional elastic collision, laboratory reference frame For the collision depicted in Figure 15.5, v1x,i > 0 , v2x,i < 0, v1x, f < 0 , and v2 x, f > 0. Because there are no external forces in the x -direction, momentum is constant in the x direction. Equating the momentum components before and after the collision gives the relation m 1v1x, i + m2v2 x, i = m 1v1x, f + m2v2x, f . (15.4.1) Because the collision is elastic, kinetic energy is constant. Equating the kinetic energy before and after the collision gives the relation 1 1 1 1 2 2 2 2 m 1v1x,i + m2v2x,i = m 1v1x, f + m2v2 x, f (15.4.2) 2 2 2 2 Rewrite these Eqs. (15.4.1) and (15.4.2) as m 1(v1x,i − v1x, f ) = m2(v2 x, f − v2x,i ) (15.4.3) 2 2 2 2 m 1(v1x,i − v1x, f ) = m2(v2 x, f − v2x,i ) . (15.4.4) Eq. (15.4.4) can be written as m 1(v1x,i − v1x, f )(v1x,i + v1x, f ) = m2(v2 x, f − v2x,i )(v2 x, f + v2x,i ) . (15.4.5) Divide Eq. (15.4.4) by Eq. (15.4.3), yielding v1x,i + v1x, f = v2 x,i + v2x, f . (15.4.6) Eq. (15.4.6) may be rewritten as v1x,i − v2 x,i = v2x, f − v1x, f . (15.4.7) Recall that the relative velocity between the two objects is defined to be  vrel ≡ v ≡ v − v2 . (15.4.8) 1,2 1 15-8 where we used the superscript “rel” to remind ourselves that the velocity is a relative velocity (and to simplify our notation). Thus vx rel ,i = v1x,i − v2 x,i is the initial x -component of the relative velocity, and vx rel , f = v1x, f − v2 x, f is the final x -component of the relative velocity. Therefore Eq. (15.4.7) states that during the interaction the initial relative velocity is equal to the negative of the final relative velocity rel rel vi = −v  f , (1− dimensional energy-momentum prinicple) . (15.4.9) Consequently the initial and final relative speeds are equal. We shall call this relationship between the relative initial and final velocities the one-dimensional energy-momentum principle because we have combined these two principles to realize this result. The energy-momentum principle is independent of the masses of the colliding particles. Although we derived this result explicitly, we have already shown that the change in kinetic energy for a two-particle interaction (Eq. (15.2.20)), in our simplified notation is given by 1 rel ) f rel )i 2) ΔK = µ((v 2 − (v (15.4.10) 2 Therefore for an elastic collision where ΔK = 0 , the square of the relative speed remains constant rel )2 f rel )2 i (v = (v . (15.4.11) For a one-dimensional collision, the magnitude of the relative speed remains constant but the direction changes by 180 . We can now solve for the final x -component of the velocities, v1x, f and v2 x, f , as follows. Eq. (15.4.7) may be rewritten as v2 x, f = v1x, f + v1x,i − v2x,i . (15.4.12) Now substitute Eq. (15.4.12) into Eq. (15.4.1) yielding m 1v1x,i + m2v2x,i = m 1v1x, f + m2(v1x, f + v1x,i − v2 x,i ) . (15.4.13) Solving Eq. (15.4.13) for v1x, f involves some algebra and yields m 1 − m2 2 m2 v1x, f = v1x,i + v2x,i . (15.4.14) m 1 + m2 m 1 + m2 15-9 To find v2 x, f , rewrite Eq. (15.4.7) as v1x, f = v2x, f − v1x,i + v2x,i . (15.4.15) Now substitute Eq. (15.4.15) into Eq. (15.4.1) yielding m 1v1x,i + m2v2 x,i = m 1(v2x, f − v1x,i + v2x,i )v1x, f + m2v2x, f . (15.4.16) We can solve Eq. (15.4.16) for v2 x, f and determine that m2 − m 1 2 m 1 v2 x, f = v2x,i + v1x,i . (15.4.17) m2 + m 1 m2 + m 1 Consider what happens in the limits m 1 >> m2 in Eq. (15.4.14). Then v1x, f → v1x,i + 2 m2v2 x,i ; (15.4.18) m 1 the more massive object’s velocity component is only slightly changed by an amount proportional to the less massive object’s x -component of momentum. Similarly, the less massive object’s final velocity approaches v2 x, f →−v2x,i + 2v1x,i = v1x,i + v1x,i − v2 x,i . (15.4.19) We can rewrite this as − v1x,i = v1x,i − v2x,i = vrel . (15.4.20) v2 x, f x,i i.e. the less massive object “rebounds” with the same speed relative to the more massive object which barely changed its speed. If the objects are identical, or have the same mass, Eqs. (15.4.14) and (15.4.17) become v1x, f = v2x,i , v2x, f = v1x,i ; (15.4.21) the objects have exchanged x -components of velocities, and unless we could somehow distinguish the objects, we might not be able to tell if there was a collision at all. 15-10 15.4.2 One-Dimensional Collision Between Two Objects – Center-of-Mass Reference Frame We analyzed the one-dimensional elastic collision (Figure 15.5) in Section 15.4.1 in the laboratory reference frame. Now let’s view the collision from the center-of-mass (CM) frame. The x -component of velocity of the center-of-mass is m 1 v1x,i + m2 v2x,i vx,cm = . (15.4.22) m 1 + m2 With respect to the center-of-mass, the x -components of the velocities of the objects are m2 v1 ′ x,i = v1x,i − v = (v1x,i − v2x,i ) x,cm m 1 + m2 (15.4.23) m 1 v2 ′ x,i = v2x,i − v = (v2x,i − v1x,i ) . x,cm m 1 + m2 In the CM frame the momentum of the system is zero before the collision and hence the momentum of the system is zero after the collision. For an elastic collision, the only way for both momentum and kinetic energy to be the same before and after the collision is either the objects have the same velocity (a miss) or to reverse the direction of the velocities as shown in Figure 15.6. Figure 15.6 One-dimensional elastic collision in center-of-mass reference frame In the CM frame, the final x -components of the velocities are m2 v1 ′ x, f = −v1 ′ x,i = (v2x,i − v1x,i ) m 1 + m2 (15.4.24) m 1 v2 ′ x, f = −v2 ′ x,i = (v2 x,i − v1x,i ) . m 1 + m2 The final x -components of the velocities in the “laboratory frame” are then given by 15-11 v1x, f = v1 ′ x, f + v x,cm m2 m 1 v1x,i + m2 v2x,i = (v2x,i − v1x,i ) + (15.4.25) m 1 + m2 m 1 + m2 m 1 − m2 2 m2 = v1x,i + v2x,i m 1 + m2 m 1 + m2 as in Eq. (15.4.14) and a similar calculation reproduces Eq. (15.4.17). 15.5 Worked Examples Example 15.1 Elastic One-Dimensional Collision Between Two Objects 1 2 ˆ i v1,i = v1,x,i ˆ i v2,i = 0 initial state m2 = 2m 1 ˆ i ˆ i final state v1, f = v1,x, f v2, f = v2,x, f ˆ i m2 = 2m 1 1 2 Figure 15.7 Elastic collision between two non-identical carts Consider the elastic collision of two carts along a track; the incident cart 1 has mass m 1 and moves with initial speed v1,i . The target cart has mass m2 = 2 m 1 and is initially at rest, = 0, (Figure 15.7). Immediately after the collision, the incident cart has final v2,i speed v1, f and the target cart has final speed v2, f . Calculate the final x -component of the velocities of the carts as a function of the initial speed v1,i . Solution The momentum flow diagram for the objects before (initial state) and after (final state) the collision are shown in Figure 15.7. We can immediately use our results above with m2 = 2 m 1 and v2,i = 0 . The final x -component of velocity of cart 1 is given by Eq. (15.4.14), where we use v1x,i = v1,i 15-12 v1x, f 1 = − v1,i . 3 (15.5.1) The final x -component of velocity of cart 2 is given by Eq. (15.4.17) v2 x, f = 2 3 v1,i . (15.5.2) Example 15.2 The Dissipation of Kinetic Energy in a Completely Inelastic Collision Between Two Objects ˆ i = 0 v1,i v2,i initial state 1 2 ˆ i final state v f 1 2 Figure 15.7b Inelastic collision between two non-identical carts An incident cart of mass m1 and initial speed v1, i collides completely inelastically with a cart of mass m2 that is initially at rest (Figure 15.7b). There are no external forces acting on the objects in the direction of the collision. Find ΔK / Kinitial = (Kfinal − Kinitial ) / Kinitial . Solution: In the absence of any net force on the system consisting of the two carts, the momentum after the collision will be the same as before the collision. After the collision the carts will move in the direction of the initial velocity of the incident cart with a common speed v f found from applying the momentum condition m1v1, i = (m1 + m2)vf ⇒ m1 (15.5.3) = vf v1, i . m1 + m2 The initial relative speed is vi rel = v1, i . The final relative velocity is zero because the carts stick together so using Eq. (15.3.4), the change in kinetic energy is ΔK = − 2 1 µ(vi rel )2 = − 1 m1m2 v1, 2 i . (15.5.4) 2 m1 + m2 15-13 The ratio of the change in kinetic energy to the initial kinetic energy is then m2 = − . ΔK / Kinitial m1 + m2 (15.5.5) As a check, we can calculate the change in kinetic energy via 1 1 2 2 − ΔK = (K f − Ki ) = (m1 + m2 )vf 2 v1, i 2 2 1 ⎛ m1 ⎞ 1 2 2 = (m1 + m2 ) − v1, i 2 v1, i 2 ⎝ ⎜ ⎠ ⎟ m1 + m2 ⎛ ⎞ m1 ⎛ 1 1 2 ⎞ m1m2 2 = −1 . ⎠ ⎟ = − ⎝ ⎜ 2 m1v1, i v1, i ⎝ ⎜ ⎠ ⎟ m1 + m2 2 m1 + m2 (15.5.6) in agreement with Eq. (15.5.4). Example 15.3 Bouncing Superballs 1 2 g M2 >> M1 Figure 15.8b Two superballs dropping Consider two balls that are dropped from a height hi above the ground, one on top of the other (Figure 15.8). Ball 1 is on top and has mass M1, and ball 2 is underneath and has mass M2 with M2 >> M1. Assume that there is no loss of kinetic energy during all collisions. Ball 2 first collides with the ground and rebounds. Then, as ball 2`starts to move upward, it collides with the ball 1 which is still moving downwards (figure below left). How high will ball 1 rebound in the air? Hint: consider this collision as seen by an observer moving upward with the same speed as the ball 2 has after it collides with ground. What speed does ball 1 have in this reference frame after it collides with the ball 2? 15-14 Solution The system consists of the two balls and the earth. There are five special states for this motion shown in the figure below. part a) Initial State: the balls are released from rest at a height hi above the ground. State A: the balls just reach the ground with speed va = = 0 ⇒ΔK = −ΔU . Thus (1 / 2)mv2 − 0 = −mgΔh = mghi ΔEmech a 2ghi . This follows from ⇒ v a = 2ghi . State B: immediately before the collision of the balls. Ball 2 has collided with the ground and reversed direction with the same speed, va , but ball 1 is still moving downward with speed va . State C: immediately after the collision of the balls. Because we are assuming that m2 >> m 1, ball 2 does not change its speed as a result of the collision so it is still moving upward with speed va . As a result of the collision, ball 1 moves upward with speed vb . Final State: ball 1 reaches a maximum height hf = vb 2 / 2g above the ground. This again follows from ΔK = −ΔU ⇒ 0 − (1/ 2)mv2 = −mgΔh = −mgh ⇒ h = v2 / 2g . b f f b Choice of Reference Frame: As indicated in the hint above, this collision is best analyzed from the reference frame of an observer moving upward with speed va , the speed of ball 2 just after it rebounded with 15-15 the ground. In this frame immediately, before the collision, ball 1 is moving downward with a speed vb ′ that is twice the speed seen by an observer at rest on the ground (lab reference frame). va ′ = 2va (15.5.7) The mass of ball 2 is much larger than the mass of ball 1, m2 >> m 1. This enables us to consider the collision (between States B and C) to be equivalent to ball 1 bouncing off a hard wall, while ball 2 experiences virtually no recoil. Hence ball 2 remains at rest in the reference frame moving upwards with speed va with respect to observer at rest on ground. Before the collision, ball 1 has speed va ′ = 2va . Since there is no loss of kinetic energy during the collision, the result of the collision is that ball 1 changes direction but maintains the same speed, = 2v . (15.5.8) vb ′ a However, according to an observer at rest on the ground, after the collision ball 1 is moving upwards with speed = 2v + v = 3v . (15.5.9) vb a a a While rebounding, the mechanical energy of the smaller superball is constant (we consider the smaller superball and the Earth as a system) hence between State C and the Final State, ΔK + ΔU = 0 . (15.5.10) The change in kinetic energy is 1 )2 ΔK = − m 1(3va . (15.5.11) 2 The change in potential energy is ΔU = m 1 g hf . (15.5.12) So the condition that mechanical energy is constant (Equation (15.5.10)) is now − 1 m 1(3v1a )2 + m 1 g hf = 0 . (15.5.13) 2 We can rewrite Equation (15.5.13) as 1 )2 g hf = 9 ( v . (15.5.14) m 1 m 1 a 2 15-16 Recall that we can also use the fact that the mechanical energy doesn’t change between the Initial State and State A yielding an equation similar to Eq. (15.5.14), m 1 g hi = 1 m 1( v )2 . (15.5.15) 2 a Now substitute the expression for the kinetic energy in Eq. (15.5.15) into Eq. (15.5.14) yielding m 1 g hf = 9 m 1 g hi . (15.5.16) Thus ball 1 reaches a maximum height hf = 9 hi . (15.5.17) 15.6 Two Dimensional Elastic Collisions 15.6.1 Two-dimensional Elastic Collision in Laboratory Reference Frame Consider the elastic collision between two particles in which we neglect any external forces on the system consisting of the two particles. Particle 1 of mass m 1 is initially  moving with velocity and collides elastically with a particle 2 of mass that is v1, i m2 initially at rest. We shall refer to the reference frame in which one particle is at rest, ‘the target’, as the laboratory reference frame. After the collision particle 1 moves with   velocity v1, f and particle 2 moves with velocity v2, f , (Figure 15.9). The angles θ1, f and θ2, f that the particles make with the positive forward direction of particle 1 are Figure 15.9 Two-dimensional collision in laboratory reference frame called the laboratory scattering angles. 1, f 2, f 1 1 2 2 v1, i v1, f v2, f  Generally the initial velocity of particle 1 is known and we would like to determine v1, i   the final velocities and v2, f , which requires finding the magnitudes and directions v1, f 15-17 of each of these vectors, v1, f , v2, f , θ1, f , and θ2, f . These quantities are related by the two equations describing the constancy of momentum, and the one equation describing constancy of the kinetic energy. Therefore there is one degree of freedom that we must specify in order to determine the outcome of the collision. In what follows we shall express our results for v1, f , v2, f , and θ2, f in terms of v1, i and θ1, f . The components of the total momentum  psys i = m 1  v1,i + m2  v2,i in the initial state are given by psys x,i = m 1v1,i (15.6.1) sys py,i = 0. The components of the momentum  psys f = m 1  v1, f + m2  v f 2, in the final state are given by sys p v1, f cosθ1, f v2, f cosθ2, f x, f = m 1 + m2 (15.6.2) psys sinθ1, f sinθ2, f . y, f = m 1 v1, f − m2 v2, f There are no any external forces acting on the system, so each component of the total momentum remains constant during the collision, sys sys = (15.6.3) px,i px, f sys sys p = p . (15.6.4) y,i y, f Eqs. (15.6.3) and (15.6.4) become m 1 = m 1 v1, f cosθ1, f + m2 v2, f cosθ2, f , (15.6.5) v1,i 0 = m 1 v1, f sinθ1, f − m2 v2, f sinθ2, f . (15.6.6) The collision is elastic and therefore the system kinetic energy of is constant sys sys Ki = K f . (15.6.7) Using the given information, Eq. (15.6.7) becomes 1 1 1 2 2 2 m 1 = m 1 + m2 . (15.6.8) v1,i v1, f v2, f 2 2 2 Rewrite the expressions in Eqs. (15.6.5) and (15.6.6) as m2v2, f cosθ2, f = m 1 − v1, f cosθ1, f ), (15.6.9) (v1,i 15-18 sinθ2, f . m2v2, f = m 1v1, f sinθ1, f (15.6.10) Square each of the expressions in Eqs. (15.6.9) and (15.6.10), add them together and use the identity cos2 θ + sin2 θ = 1 yielding 2 m 1 2 2 2 = − 2v1,iv1, f cosθ1, f ) . (15.6.11) v2, f 2 (v1,i + v1, f m2 Substituting Eq. (15.6.11) into Eq. (15.6.8) yields 2 1 2 1 2 1 m 1 2 2 = + − 2v1,i v1, f cosθ1, f ) . (15.6.12) m 1v1,i m 1v1, f (v1,i + v1, f 2 2 2 m2 Eq. (15.6.12) simplifies to ⎛ ⎞ ⎛ ⎞ m 1 2 m 1 m 1 2 0 = 1+ ⎠ ⎟ v1, f − 2v1,i v1, f cosθ1, f − 1− ⎠ ⎟ v1,i , (15.6.13) ⎝ ⎜ m2 m2 ⎝ ⎜ m2 Let α = m 1 / m2 then Eq. (15.6.13) can be written as 0 = (1+ α )v1, 2 f − 2αv1,i v1, f cosθ1, f − (1−α )v1, 2 i , (15.6.14) The solution to this quadratic equation is given by 1/2 ± α 2 2 2 αv1,i cosθ1, f ( v1,i cos2 θ1, f + (1−α 2)v1,i ) = . (15.6.15) v1, f (1+ α ) Divide Eq. (15.6.10) by Eq. (15.6.9): sinθ2, f sinθ1, f v2, f v1, f = . (15.6.16) v2, f cosθ2, f − v1, f cosθ1, f v1,i Eq. (15.6.16) simplifies to sinθ1, f v1, f tanθ2, f = . (15.6.17) − v1, f cosθ1, f v1,i The relationship between the scattering angles in Eq. (15.6.17) is independent of the masses of the colliding particles. Thus the scattering angle for particle 2 is 15-19 ⎛ sinθ1, f ⎞ v1, f = tan−1 ⎜ ⎟ (15.6.18) θ2, f ⎝ − v1, f cosθ1, f ⎠ v1,i We can now use Eq. (15.6.10) to find an expression for the final velocity of particle 2 sinθ1, f αv1, f = . (15.6.19) v2, f sinθ2, f Example 15.5 Elastic Two-dimensional collision of identical particles 1 1 2 2 v1, i v1, f v2, f 1, f 2, f ˆ i ˆ j Figure 15.10 Momentum flow diagram for two-dimensional elastic collision = 30 Object 1 with mass is initially moving with speed m a 1 v1,i = 3.0m ⋅s−1 and collides elastically with object 2 that has the same mass, m2 = m 1, and is initially at rest. After the collision, object 1 moves with an unknown speed v1, f at an angle θ1, f with respect to its initial direction of motion and object 2 moves with an unknown speed v2, f , at an unknown angle θ2, f (as shown in the Figure 15.10). Find the final speeds of each of the objects and the angle θ2, f . Solution: Because the masses are equal, α = 1. We are given that v1,i = 3.0 m ⋅s−1 and θ1, f = 30o . Hence Eq. (15.6.14) reduces to v1, f = v1,i cosθ1, f = (3.0 m ⋅s−1)cos30 = 2.6 m ⋅s−1 . (15.6.20) Substituting Eq. (15.6.20) in Eq. (15.6.17) yields 15-20 ⎛ sinθ1, f ⎞ v1, f = tan−1 θ2, f ⎜ ⎟ ⎝ cosθ1, f ⎠ v1,i − v1, f (2.6 m ⋅s−1)sin(30 ) = tan−1 ⎛ ⎞ (15.6.21) θ2, f ⎝ ⎜ 3.0 m ⋅s−1 − (2.6 m ⋅s−1)cos(30 )⎠ ⎟ = 60 . The above results for v1, f and θ2,f may be substituted into either of the expressions in Eq. (15.6.9), or Eq. (15.6.11), to find v2, f = 1.5m ⋅ s−1 . Eq. (15.6.11) also has the solution v2, f = 0 , which would correspond to the incident particle missing the target completely. Before going on, the fact that θ1, f +θ2, f = 90 , that is, the objects move away from the collision point at right angles, is not a coincidence. A vector derivation is presented in Example 15.6. We can see this result algebraically from the above result. Substituting Eq. (15.6.20) v1, f = v1,i cosθ1, f in Eq. (15.6.17) yields cosθ1, f sinθ1, f tanθ2, f = 2 = cotθ1, f = tan(90 −θ1, f ) ; (15.6.22) 1− cosθ1, f showing that θ1, f +θ2, f = 90 , the angles θ1, f and θ2, f are complements. Example 15.6 Two-dimensional elastic collision between particles of equal mass Show that the equal mass particles emerge from a two-dimensional elastic collision at right angles by making explicit use of the fact that momentum is a vector quantity. 1 1 2 2 v1, i v1, f v2, f 1, f 2, f ˆ i ˆ j Figure 15.11 Elastic scattering of identical particles 15-21 Solution: Choose a reference frame in which particle 2 is initially at rest (Figure 15.11). There are no external forces acting on the two objects during the collision (the collision forces are all internal), therefore momentum is constant   sys sys pi = p f , (15.6.23) which becomes m 1  v  v  v (15.6.24) + m 1 = m 1 . 1, i 1, f 2, f Eq. (15.6.24) simplifies to  v =  v  v (15.6.25) + . 1,i 1, f 2, f Recall the vector identity that the square of the speed is given by the dot product  v ⋅ v = v2 . With this identity in mind, we take the dot product of each side of Eq. (15.6.25) with itself,  v ⋅ v = ( v  v  v f 1,  v f 1,  v )⋅( ) + + 1,i 1,i 1, f 2, f 2, f (15.6.26) =  v  v  v  v ⋅ v ⋅ + 2 ⋅ + 2, f . 1, f 1, f 2, f 2, f This becomes 2 2 2 + 2 ⋅  . (15.6.27) v1,i = v1, f v1, f v2, f + v2, f Recall that kinetic energy is the same before and after an elastic collision, and the masses of the two objects are equal, so constancy of energy, (Eq. (15.4.2)) simplifies to 2 2 2 . (15.6.28) v1,i = v1, f + v2, f Comparing Eq. (15.6.27) to Eq. (15.6.28), we see that   v1, f ⋅ v2, f = 0 . (15.6.29) The dot product of two nonzero vectors is zero when the two vectors are at right angles to each other justifying our claim that the collision particles emerge at right angles to each other. Example 15.7 Two dimensional collision between particles of unequal mass Particle 1 of mass m1 , initially moving in the positive x -direction (to the right in the figure below) with speed v1,i , collides with particle 2 of mass m = m /3, which is 2 1 initially moving in the opposite direction (Figure 15.12) with an unknown speed v2,i . Assume that the total external force acting on the particles is zero. Do not assume the collision is elastic. After the collision, particle 1 moves with speed v1, f / 2 in the = v1,i negative y -direction. After the collision, particle 2 moves with an unknown speed v2, f , 15-22 at an angle θ2, f = 45o with respect to the positive x -direction. (i) Determine the initial speed v2,i of particle 2 and the final speed v2, f of particle 2 in terms of v1,i . (ii) Is the collision elastic? ˆ i ˆ j before after m 1 = m 1 2 1 2 = 45 m2 = m / 3 v1,i v1, f = v1,i / 2 v2, f v2,i Figure 15.12 Two-dimensional collision between particles of unequal mass Solution: We choose as our system the two particles. We are given that v1, f / 2 . We = v1,i apply the two momentum conditions, m 1 − (m 1 / 3)v2,i = (m 1 / 3) v2, f ( 2 / 2) (15.6.30) v1,i 0 = m 1 − (m 1 / 3) v2, f ( 2 / 2) . (15.6.31) v1, f Solve Eq. (15.6.31) for v2, f = (15.6.32) v2, f = 3 2v1, f 3 2 v1,i 2 Substitute Eq. (15.6.32) into Eq. (15.6.30) and solve for v2,i =(3/ 2)v1,i . (15.6.33) v2,i The initial kinetic energy is then 1 2 1 2 7 2 = + (m 1 / 3)v2,i = . (15.6.34) Ki m 1v1,i m 1v1,i 2 2 8 The final kinetic energy is 1 1 1 3 7 2 2 2 2 2 K f = m 1 + m2 = m 1 + m 1 = m 1 . (15.6.35) v1, f v2, f v1,i v1,i v1,i 2 2 8 4 8 Comparing our results, we see that kinetic energy is constant so the collision is elastic. 15-23 15.7 Two-Dimensional Collisions in Center-of-Mass Reference Frame 15.7.1 Two-Dimensional Collision in Center-of-Mass Reference Frame Consider the elastic collision between two particles in the laboratory reference frame (Figure 15.9). Particle 1 of mass m 1  is initially moving with velocity v i 1, and collides elastically with a particle 2 of mass m2 that is initially at rest. After the collision the particle 1 moves with velocity v  1, f and particle 2 moves with velocity v  2, f . In section 15.7.1 we determined how to find v1, f , v2, f , and θ2, f in terms of v1, i and θ2, f . We shall now analyze the collision in the center-of-mass reference frame, which is boosted form the laboratory frame by the velocity of center-of-mass given by  v i 1, m 1 cm m 1 + m2  v = . (15.6.36) Because we assumed that there are no external forces acting on the system, the center-of mass velocity remains constant during the interaction. 1 1 2 2 v2, f v1, f v1,i v2,i cm Figure 15.13 Two-dimensional elastic collision in center-of-mass reference frame Recall the velocities of particles 1 and 2 in the center-of-mass frame are given by (Eq.,(15.2.9) and (15.2.10)). In the center-of-mass reference frame the velocities of the two incoming particles are in opposite directions, as are the velocities of the two outgoing particles after the collision (Figure 15.13). The angle Θ cm between the incoming and outgoing velocities is called the center-of-mass scattering angle. 15-24 15.7.2 Scattering in the Center-of-Mass Reference Frame  Consider a collision between particle 1 of mass and velocity and particle 2 of m1 v1,i mass m2 at rest in the laboratory frame. Particle 1 is scattered elastically through a scattering angle Θ in the center-of-mass frame. The center-of-mass velocity is given by   m 1v1,i v cm = . (15.6.37) m 1 + m2 In the center-of-mass frame, the momentum of the system of two particles is zero      0 = m 1 + m2 = m 1 + m2 . (15.6.38) v1, ′ i v′ 2,i v1, ′ f v′ 2, f Therefore  m2  (15.6.39) v1, ′ i = − v′ 2,i . m 1  m2  = − (15.6.40) v1, ′ f v′ 2, f m 1 The energy condition in the center-of-mass frame is 1 1 2 1 2 1 2 2 + = + . (15.6.41) m 1v1, ′ i m2v2, ′ i m 1v1, ′ f m2v2, ′ f 2 2 2 2 Substituting Eqs. (15.6.39) and (15.6.40) into Eq. (15.6.41) yields = (15.6.42) v1, ′ i v1, ′ f . (we are only considering magnitudes). Therefore = . (15.6.43) v2, ′ i v2, ′ f Because the magnitude of the velocity of a particle in the center-of-mass reference frame is proportional to the relative velocity of the two particles, Eqs. (15.6.42) and (15.6.43) imply that the magnitude of the relative velocity also does not change , (15.6.44) = v  1, 2, ′ i v  1, 2, ′ f verifying our earlier result that for an elastic collision the relative speed remains the same, (Eq. (15.2.20)). However the direction of the relative velocity is rotated by the center-of-mass scattering angle Θ cm . This generalizes the energy-momentum principle to two dimensions. Recall that the relative velocity is independent of the reference frame, 15-25   −  −  = (15.6.45) v1, i v2, i v1, ′ i v′ 2, i   In the laboratory reference frame = 0 , hence the initial relative velocity is v2,i    = = , and the velocities in the center-of-mass frame of the particles are then v1, 2, ′ i v1, 2, i v1, i  µ  = (15.6.46) v1, ′ i v1, i m 1  = − µ  . (15.6.47) v′ 2, i v1, i m2 Therefore the magnitudes of the final velocities in the center-of-mass frame are µ µ µ = = = = (15.6.48) v1, ′ f v1, ′ i v1, 2, ′ i v1, 2, i v1, i . m 1 m 1 m 1 µ µ µ = = = = (15.6.49) v2, ′ f v2, ′ i v1, 2, ′ i v1, 2, i v1, i . m2 m2 m2 Example 15.8 Scattering in the Lab and CM Frames  Particle 1 of mass and velocity by a particle of mass m2 at rest in the laboratory m1 v1,i frame is scattered elastically through a scattering angle Θ in the center of mass frame, (Figure 15.14). Find (i) the scattering angle of the incoming particle in the laboratory frame, (ii) the magnitude of the final velocity of the incoming particle in the laboratory reference frame, and (iii) the fractional loss of kinetic energy of the incoming particle. v1, f ˆ j v1, f ˆ i 1 1 1, 1 v i 1, f v1,i cm 2 1 2 v2,i 2 2, f cm 2 v2, f v2, f Figure 15.14 Scattering in the laboratory and center-of-mass reference frames 15-26 Solution: i) In order to determine the center-of-mass scattering angle we use the transformation law for velocities   −  = v . (15.6.50) v1, ′ f v1, f cm In Figure 15.15 we show the collision in the center-of-mass frame along with the laboratory frame final velocities and scattering angles. 1 v1, f v1,i 1, f v cm v1, f v2,i cm 2 v2, f 2, f cm ˆ i ˆ j v cm 1 2 v2, f Figure 15.15 Final velocities of colliding particles Vector decomposition of Eq. (15.6.50) yields v1, f cosθ1, i = v1, ′ f cosΘ − v , (15.6.51) cm cm v1, f sinθ1, i = v1, ′ f sinΘ cm . (15.6.52) where we choose as our directions the horizontal and vertical Divide Eq. (15.6.52) by (15.6.51) yields sinθ1, i sinΘ v1, f v1, ′ f cm tanθ1, i = = (15.6.53) v1, f cosθ1, i v1, ′ f cosΘ − v cm cm Because = , we can rewrite Eq. (15.6.53) as v1, ′ i v1, ′ f v1, ′ i sinΘ cm tanθ1, i = (15.6.54) v1, ′ i cosΘ − v cm cm We now substitute Eqs. (15.6.48) and v / (m 1 ) into Eq. (15.6.54) yielding cm = m 1v1, i + m2 15-27 m2 sinΘ cm tanθ1, i = . (15.6.55) cosΘ cm − m 1 / m2 Thus in the laboratory frame particle 1 scatters by an angle ⎛ ⎞ m2 cm θ1, i = tan−1 ⎝ ⎜ sinΘ ⎠ ⎟ . (15.6.56) cosΘ cm − m 1 / m2 ii) We can calculate the square of the final velocity in the laboratory frame  v1, f ⋅ v1, f = ( v1, ′ f +  v cm ) ⋅( v1, ′ f +  v cm ) . (15.6.57) which becomes 2 2 2 v1, f = v1, ′ f 2 + 2v  1, ′ f ⋅ v  cm + vcm = v1, ′ f 2 + 2v1, ′ f vcm cosΘ cm + vcm . (15.6.58) We use the fact that = = (µ / m 1 = (µ / m 1 = (m2 / m 1 + m2 to rewrite v1, ′ f v1, ′ i )v1,2, i )v1, i )v1, i Eq. (15.6.58) as v1, f 2 = ⎝ ⎜ ⎛ m2 ⎠ ⎟ ⎞ 2 v1, i 2 + 2 m2m 1 )2 v1, i cosΘ cm + m 1 2 )2 v1, i 2 . (15.6.59) m 1 + m2 (m 1 + m2 (m 1 + m2 Thus 1/2 2 (m2 2 + 2m2m 1cosΘ + m 1 ) = cm . (15.6.60) v1, f v1, i m 1 + m2 (iii) The fractional change in the kinetic energy of particle 1 in the laboratory frame is given by 2 2 K1, f − K1, i v1, f 2 − v1, i m2 2 + 2m2m 1cosΘ cm + m 1 2m2m 1(cosΘ cm −1) = = −1 = .(15.6.61) 2 )2 )2 K1, i v1, i (m 1 + m2 (m 1 + m2 We can also determine the scattering angle Θ cm in the center-of-mass reference frame from the scattering angle θ1, i of particle 1 in the laboratory. We now rewrite the momentum relations as v1, f cosθ1, i + v = v1, ′ f cosΘ , (15.6.62) cm cm v1, f sinθ1, i = v1, ′ f sinΘ cm . (15.6.63) In a similar fashion to the above argument, we have that 15-28 sinθ1, f v1, f tanΘ cm = . (15.6.64) v1, f cosθ1, f + v cm Recall from our analysis of the collision in the laboratory frame that if we specify one of the four parameters v1, f , v2, f , θ1, f , or v1, f , then we can solve for the other three in terms of the initial parameters and With that caveat, we can use Eq. (15.6.64) to v1, i v2, i . determine Θ . cm 15-29 MIT OpenCourseWare 8.01 Classical Mechanics Spring 2022 For information about citing these materials or our Terms of Use, visit:
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Previous Next Ch. 08 The Law of Contract SECTION 1 GENERAL APPLICATION A. Singapore contract law largely based on English contract law 8.1.1 Contract law in Singapore is largely based on the common law of contract in England. Hence, the rules developed in the Singapore courts do bear a very close resemblance to those developed under English common law. Indeed, where there is no Singapore authority specifically on point, it will usually be assumed that the position will, in the first instance, be no different from that in England. Unlike its neighbours Malaysia and Brunei, following Independence in 1965, Singapore’s Parliament made no attempt to codify Singapore’s law of contract. Accordingly, much of the law of contract in Singapore remains in the form of judge-made rules. B. Certain aspects of Singapore contract law based on specific statutes 8.1.2 In some circumstances, these judge-made rules have been modified by specific statutes. Many of these statutes are English in origin. To begin with, 13 English commercial statutes have been incorporated as part of the Statutes of the Republic of Singapore by virtue of s 4 of the Application of English Law Act (Cap 7A, 1993 Rev Ed). These are listed in Part II of the First Schedule of this Act. Other statutes, eg the Contracts (Rights of Third Parties) Act (Cap 53B, 2002 Rev Ed), are modelled upon English statutes. There are also other areas where statutory development based on non-English models has taken place, eg the Consumer Protection (Fair Trading) Act (Cap 52A, 2004 Rev Ed) (which was largely drawn from fair trading legislation enacted in Alberta and Sasketchewan). SECTION 2 OFFER AND ACCEPTANCE A. Contract is essentially agreement between two or more parties 8.2.1 A contract is essentially an agreement between two or more parties, the terms of which affect their respective rights and obligations which are enforceable at law. Whether the parties have reached agreement, or a meeting of the minds, is objectively ascertained from the facts. The concepts of offer and acceptance provide in many, albeit not all, cases the starting point for analysing whether agreement has been reached. B. Offer 1. Definition of offer 8.2.2 An offer is a promise, or other expression of willingness, by the ‘offeror’ to be bound on certain specified terms upon the unqualified acceptance of these terms by the person to whom the offer is made (the ‘offeree’). Provided the other formation elements (ie consideration and intention to create legal relations) are present, the acceptance of an offer results in a valid contract. 2. Offer must be made with objectively ascertained intention to be bound 8.2.3 Whether any particular statement amounts to an offer depends on the intention with which it is made. An offer must be made with the intention to be bound. On the other hand, if a person is merely soliciting offers or requesting for information, without any intention to be bound, at best, he or she would be making an invitation to treat. Under the objective test, a person may be said to have made an offer if his or her statement (or conduct) induces a reasonable person to believe that the person making the offer intends to be bound by the acceptance of the alleged offer, even if that person in fact had no such intention. 3. Offer may be terminated by various ways of withdrawal prior to acceptance 8.2.4 An offer may be terminated by withdrawal at any time prior to its acceptance, provided there is communication, of the withdrawal to the offeree, whether by the offeror or through some reliable source. Rejection of an offer, which includes the making of a counter-offer or a variation of the original terms, terminates the offer. In the absence of an express stipulation as to time, an offer will lapse after a reasonable time. What this amounts to depends on the particular facts of the case. Death of the offeror, if known to the offeree, would render the offer incapable of being accepted by the offeree. Even in the absence of such knowledge, death of either party terminates any offer which has a personal element. C. Acceptance 1. Definition of acceptance 8.2.5 An offer is accepted by the unconditional and unqualified assent to its terms by the offeree. This assent may be expressed through words or conduct, but cannot be inferred from mere silence save in very exceptional circumstances. 2. Acceptance must generally be communicated to offeror 8.2.6 As a general rule, acceptance must be communicated to the offeror, although a limited exception exists where the acceptance is sent by post and this method of communication is either expressly or impliedly authorised. This exception, known as the ‘postal acceptance rule’, provides that acceptance takes place at the point when the letter of acceptance is posted, whether or not it was in fact received by the offeror. D. Certainty and Completeness 1. Essential terms must be sufficiently certain for contract to be enforceable 8.2.7 Before the agreement may be enforced as a contract, its terms must be sufficiently certain. At the least, the essential terms of the agreement should be specified. Beyond this, the courts may resolve apparent vagueness or uncertainty by reference to the acts of the parties, a previous course of dealing between the parties, trade practice or to a standard of reasonableness. On occasion, statutory provision of contractual details may fill the gaps. For more on implication of terms, see Paragraphs 8.5.5 to 8.5.8 below. 2. Agreement must be complete for contract to be enforceable 8.2.8 An incomplete agreement also cannot amount to an enforceable contract. Agreements made ‘subject to contract’ may be considered incomplete if the intention of the parties, as determined from the facts, was not to be legally bound until the execution of a formal document or until further agreement is reached. E. Electronic Transactions Act 8.2.9 The Electronic Transactions Act (Cap 88, 1999 Rev Ed) (‘ETA’) clarifies that, except with respect to the requirement of writing or signatures in wills, negotiable instruments, indentures, declarations of trust or powers of attorney, contracts involving immovable property and documents of title (s 4(1)), electronic records may be used in expressing an offer or acceptance of an offer in contract formation (s 11). A declaration of intent between contracting parties may also be made in the form of an electronic record (s 12). The ETA also clarifies when an electronic record may be attributed to a particular person (s 13) and how the time and place of despatch and receipt of an electronic record are to be determined (s 15). SECTION 3 CONSIDERATION A. Definition of consideration 8.3.1 A promise contained in an agreement is not enforceable unless it is supported by consideration or it is recorded in a written document executed as a deed. Consideration is something of value (as defined by the law), requested for by the party making the promise (the ‘promisor’) and provided by the party who receives it (the ‘promisee’), in exchange for the promise that the promisee is seeking to enforce. Thus, it could consist of either some benefit received by the promisor, or some detriment to the promisee. This benefit/detriment may consist of a counter promise or a completed act. B. Reciprocity: Causal relation between consideration and promise 8.3.2 The idea of reciprocity that underlies the requirement for consideration means that there has to be some causal relation between the consideration and the promise itself. Thus, consideration cannot consist of something that was already done before the promise was made. However, whether this is so is not determined solely by the chronology of events. An act performed the making of a firm promise may still be good consideration if it was done on the understanding that it would form part of a binding exchange. C. Consideration need not be sufficient 8.3.3 Whether the consideration provided is sufficient is a question of law, and the court is not, as a general rule, concerned with whether the value of the consideration is commensurate with the value of the promise. The performance of, or the promise to perform, an existing public duty imposed on the promisee does not, without more, constitute sufficient consideration in law to support the promisor’s promise. The performance of an existing obligation that is owed contractually to the promisor is capable of being sufficient consideration, if such performance confers a real and practical benefit on the promisor. If the promisee performs or promises to perform an existing contractual obligation that is owed to a third party, the promisee will have furnished sufficient consideration at law to support a promise given in exchange. D. Promissory estoppel 1. Promissory estoppel renders promise unsupported by consideration binding 8.3.4 Where the doctrine of promissory estoppel applies, a promise may be binding notwithstanding that it is not supported by consideration. 2. When promissory estoppel applies 8.3.5 This doctrine applies where a party to a contract makes an unequivocal promise, whether by words or conduct, that he or she will not insist on his or her strict legal rights under the contract, and the other party acts, and thereby alters his or her position, in reliance on the promise. The party making the promise cannot seek to enforce those rights if it would be inequitable to do so, although such rights may be reasserted upon the promisor giving reasonable notice. The doctrine prevents the enforcement of existing rights, but does not create new causes of action. SECTION 4 INTENTION TO CREATE LEGAL RELATIONS A. Contractual intention necessary for enforceable contract 8.4.1 In the absence of contractual intention, an agreement, even if supported by consideration, cannot be enforced. Whether the parties to an agreement intended to create legally binding relations between them is a question determined by an objective assessment of the relevant facts. B. Commercial arrangements: Presumption of intention to be legally bound 8.4.2 In the case of agreements in a commercial context, the courts will generally presume that the parties intended to be legally bound. However, the presumption can be displaced where the parties expressly declare the contrary intention. This is often done through the use of honour clauses, letters of intent, memoranda of understanding and other similar devices, although the ultimate conclusion would depend, not on the label attached to the document, but on an objective assessment of the language used and on all the attendant facts. C. Social arrangements: No presumption of intention to be legally bound 8.4.3 The parties in domestic or social arrangements are generally presumed not to intend legal consequences. SECTION 5 TERMS OF THE CONTRACT A. Express terms 1. Ascertainment of terms: Distinction between term or representation 8.5.1 The rights and obligations of contracting parties are determined by first, ascertaining the terms of the contract, and secondly, interpreting those terms. In ascertaining the terms of a contract, it is sometimes necessary, especially where the contract has not been reduced to writing, to decide whether a particular statement is a contractual term or a mere representation. Whether a statement is contractual or not depends on the intention of the parties, objectively ascertained, and is a question of fact. In ascertaining the parties’ intention, the courts take into account a number of factors including the stage of the transaction at which the statement was made, the importance which the representee attached to the statement and the relative knowledge or skill of the parties vis-à-vis the subject matter of the statement. 2. Interpretation of terms: Objective test to determine meaning 8.5.2 Once the terms of a contract have been determined, the court applies an objective test in construing or interpreting the meaning of these terms. What is significant in this determination therefore is not the sense attributed by either party to the words used, but how a reasonable person would understand those terms. In this regard, Singapore courts have consistently emphasised the importance of the factual matrix within which the contract was made, as this would assist in determining how a reasonable man would have understood the language of the document. 3. Parol evidence rule determines if statement forms part of written contract 8.5.3 Where the parties have reduced their agreement into writing, whether a particular statement (oral or written) forms part of the actual contract depends on the application of the parol evidence rule. In Singapore, this common law rule and its main exceptions are codified in s 93 and s 94 of the Evidence Act (Cap 97, 1997 Rev Ed). Section 93 provides that where ‘the terms of a contract…have been reduced …to the form of a document…, no evidence shall be given in proof of the terms of such contract …except the document itself’. Thus, no evidence of any oral agreement or statement may be admitted in evidence to contradict, vary, add to, or subtract from the terms of the written contract. However, secondary evidence is admissible if it falls within one of the exceptions to this general rule found in the proviso to s 94. Some controversy remains as to whether s 94 is an exhaustive statement of all exceptions to the rule, or whether other common law exceptions not explicitly covered in s 94 continue to be applicable. 8.5.4 It should, however, be noted that the scope of s 93 and s 94 has been circumscribed by Parliament in certain circumstances. See, for example, s 17, Protection (Fair Trading) Act (Cap 52A, 2009 Rev Ed). B. Implied terms 8.5.5 In addition to those expressly agreed terms, the court may sometimes imply terms into the contract. 1. Implied term must not contradict express term 8.5.6 Generally, any term to be implied must not contradict any express term of the contract. 2. Term implied in fact only if necessary 8.5.7 Where a term is implied to fill a gap in the contract so as to give effect to the presumed intention of the parties, the term is implied in fact and depends on a consideration of the language of the contract as well as the surrounding circumstances. A term will be implied only if it is so necessary that both parties must have intended its inclusion in the contract. The fact that it would be reasonable to include the term is not sufficient for the implication, as the courts will not re-write the contract for the parties. 3. Terms also implied if required statutorily or on public policy considerations 8.5.8 Terms may also be implied because this is required statutorily, or on public policy considerations. The terms implied by the Sale of Goods Act (Cap 393, 1994 Rev Ed) (eg s 12(1) – that the seller of goods has a right to sell the goods) provide examples of the former type of implied terms. As for the latter, whilst there has been no specific authority on the point, it is not inconceivable that Singapore courts, like their English counterparts, may imply ‘default’ terms into specific classes of contracts to give effect to policies that define the contractual relationships that arise out of those contracts. C. Classification of terms 1. Terms classified into conditions, warranties or intermediate terms 8.5.9 The terms of a contract may be classified into conditions, warranties or intermediate (or innominate) terms. Proper classification is important as it determines whether the contract may be discharged or terminated for breach [as to which see Paragraphs 8.8.11 to 8.8.12 below]. 2. Contracting parties may expressly stipulate classification of term 8.5.10 The parties may expressly stipulate in the contract how a particular term is to be classed. This is not, however, conclusive unless the parties are found to have intended the technical meaning of the classifying words used. 3. Court determines objectively classification of term in absence of express stipulation 8.5.11 In the absence of express stipulation, the courts will look objectively at the language of the contract to determine how, in light of the surrounding circumstances, the parties intended a particular term to be classed. There are also instances where statutes may stipulate whether certain kinds of terms are to be treated as conditions or warranties, in the absence of any specific designation by the contracting parties. D. Exception clauses 1. Definition of exception clauses 8.5.12 Exception clauses that seek to exclude or limit a contracting party’s liability are commonly, but not exclusively, found in standard form agreements. The law in Singapore relating to such clauses is essentially based on English law. The English Unfair Contract Terms Act 1977, which either invalidates an exception clause or limits the efficacy of such terms by imposing a requirement of reasonableness, has been re-enacted in Singapore as the Unfair Contract Terms Act (as Cap 396, 1994 Rev Ed). 2. Exception clause must be incorporated into contract 8.5.13 Whether an exception clause will have its intended effect depends on a number of factors. The threshold requirement is that the clause must have been incorporated into the contract. There are generally three ways in which such incorporation may occur. Where a party has signed a contract which contains an exception clause, the signatory is bound by the clause, even if he or she had not read or was unaware of the clause. An exception clause may also be incorporated, in the absence of a signed contract, if the party seeking to rely on the clause took reasonably sufficient steps to draw the other party’s attention to the existence of the clause. The determination of this issue is heavily dependent on the facts of the particular case. Finally, exception clauses may be incorporated because there has been a consistent and regular course of dealing between the parties on terms that incorporate the exception clause. Even if no steps were taken to incorporate the clause in a particular contract between such parties, it may have been validly incorporated by the parties’ prior course of dealing. 3. Exception clause must on proper construction apply to situation concerned 8.5.14 The next consideration is one of construction (or interpretation). This is necessary to determine if the liability, which the relevant party is seeking to exclude or restrict, falls within the proper scope of the clause. Here, the courts adopt the contra proferentum rule of construction, and will construe exception clauses strictly against parties seeking to rely on them. Nevertheless, the Singapore courts appear to construe clauses which seek to limit liability more liberally than those which seek to completely exclude liability. 4. Unfair Contract Terms Act 8.5.15 Finally, the limits placed by the Unfair Contracts Terms Act (Cap 396, 1994 Rev Ed) (the ‘UCTA’) on the operation and efficacy of exceptions clauses must be considered. It should be noted that the UCTA generally applies only to terms that affect liability for breach of obligations that arise in the course of a business or from the occupation of business premises. It also gives protection to persons who are dealing as consumers. Under the UCTA, exception clauses are either rendered wholly ineffective, or are ineffective unless shown to satisfy the requirement of reasonableness. Whether the exception clause is reasonable is a question that is highly fact-dependent. In considering this issue, the courts have generally given consideration to a number of factors, including those listed in the UCTA itself. These factors include the relative bargaining positions of the parties, and whether there was an inducement to agree to the clause. Terms that attempt to exclude or restrict a party’s liability for death or personal injury resulting from that party’s negligence are rendered wholly ineffective by the UCTA, while terms that seek to exclude or restrict liability for negligence resulting in loss or damage other than death or personal injury, and those that attempt to exclude or restrict contractual liability, are subject to the requirement of reasonableness. The reasonableness of the exception clause is evaluated as at the time at which the contract was made. The actual consequences of the breach are therefore, in theory at least, immaterial. SECTION 6 CAPACITY TO CONTRACT Minors 8.6.1 For purposes of contracting, a minor is a person under the age of 18. The validity of contracts entered into by minors is governed by the common law, as modified by the Minors’ Contracts Act (Cap 389, 1994 Rev Ed). Contracts with Minors 8.6.2 As a general rule, contracts are not enforceable against minors. However, where a minor has been supplied with necessaries (ie goods or services suitable for the maintenance of the station in life of the minor concerned: see also s 3(3), Sale of Goods Act (Cap 393, 1999 Rev Ed)),he must pay for them. Contracts of service which are, on the whole, for the minor’s benefit are also valid against him. Further, the minor is bound by certain types of contracts (ie contracts concerning land or shares in companies, partnership contracts and marriage settlements) unless he repudiates the contract before attaining majority or within a reasonable time thereafter. Minors’ Contracts Act 8.6.3 Under s 2 of the Minors’ Contracts Act, a guarantee given in respect of a minor’s contract, which may not be enforceable against the minor, is nevertheless enforceable against the guarantor. Section 3(1) of the Minors’ Contracts Act empowers the court to order restitution against the minor if it is just and equitable to do so. Mental Incapacity and Drunkards 8.6.4 A contract entered into by a person of unsound mind is valid, unless it can be shown that that person was incapable of understanding what he or she was doing and the other party knew or ought reasonably to have known of the disability. In this case, the contract may be avoided at the option of the mentally unsound person (assisted by a court-sanctioned representative where necessary). The same principle applies in the case of inebriated persons. Under s 3(2) of the Sale of Goods Act, persons incapacitated mentally or by drunkenness are required to pay a reasonable price for necessaries supplied. Corporations 8.6.5 Subject to any written law and to any limits contained in its constitution, a company has full capacity to undertake any business, do any act or enter into any transaction (s 23 – Companies Act, Cap 50, 1994 Rev Ed). Where there are restrictions placed on the capacity of a company and the company acts beyond its capacity, s 25 of the Companies Act validates such ultra vires transactions if they would otherwise be valid and binding. Contracts purportedly entered into by a company prior to its incorporation may be ratified and adopted by the company after its formation (s 41 – Companies Act). 8.6.6 A limited liability partnership is also a body corporate under Singapore law – see Limited Liability Partnerships Act 2005 (Act No 5 of 2005). It may, in its own name: sue and be sued in its own name; acquire, own, hold and develop property; hold a common seal; and may do and suffer such other acts and things as any body corporate may lawfully do and suffer – see s 5(1). Section 5(2) also extends s 41 of the Companies Act to apply to a limited liability partnership. SECTION 7 PRIVITY OF CONTRACT Third party Enforcement of Contractual Rights Generally not Permitted 8.7.1 As a general proposition, only persons who are party (ie ‘privy’) to a contract may enforce rights or obligations arising from that contract. This is sometimes referred to as the ‘privity rule.’ 8.7.2 A third party who is not privy to a contract is generally not allowed to bring any legal action in his or her own name for breach of contract against a contracting party who fails to perform his or her contractual obligations, even if such failure of performance has caused the third party to suffer a loss. When is Someone Party or Privy to a Contract? 8.7.3There is no clear definition as to when a person is/is not privy to a contract. Generally, a party who is an offeror or offeree will be privy to the contract. However, it seems that merely being mentioned in the contract is not enough. 8.7.4 It is, nevertheless, possible to have a multilateral contract where there are multiple offerees (one or more of whom accept the offer on behalf of the others) or where there are multiple offerors (one or more of whom make the offer on behalf of the others). In either case, each offeree or offeror is a joint party to the contract and the privity rule will not apply to them. Non-statutory Exceptions to the Privity Rule 8.7.5 The privity rule is not absolute. It is subject to many exceptions. Apart from the possibility of a multilateral or multi-party contract (mentioned above), some other exceptions can be found in the law relating to: (a) agency; (b) trusts; or (c) land (in relation to covenants which ‘run’ with the land or lease). For an in depth discussion of these other legal techniques to circumvent the privity rule, please see Chapters 15 and 18. Statutory Exceptions to the Privity Rule 8.7.6 There are also statutory exceptions. Most of these are only applicable to specific and narrowly defined cases. Two examples of such statutes include: (a) the Bills of Exchange Act (Cap 23, 1985 Rev Ed) [see Chapter 22 on Banking Law]; and (b) the Bills of Lading Act (Cap 384, 1994 Rev Ed) [see Chapter 25 on Shipping Law]. Of more general application, the Singapore Parliament enacted the Contracts (Rights of Third Parties) Act (Cap 53B, 2002 Rev Ed) in 2001. Contracts (Rights of Third Parties) Act 8.7.7 Section 1 provides that the Contracts (Rights of Third Parties) Act has no retrospective effect – it cannot apply to any contract formed before 1 January 2002. Section 1 also provides that the Act does not apply to any contracts which were formed on or after 1 January 2002, but before 1 July 2002, unless the contracting parties expressly provided in their contract for it to do so. Contracts formed on or after 1 July 2002 are always subject to the Act. 8.7.8 Where the Act applies, it gives a third party a statutory right to enforce a term of a contract against a party who is in breach of his or her obligations under the contract (the ‘promisor’), even though even though the third party is a volunteer who has not provided any contractual consideration – see s 2(5). 8.7.9 This may occur if either: (a) the contract expressly provides that the third party may enforce a term of the contract in his or her own right – s 2(1) (a) ; or (b) the contract, ‘purports to confer a benefit on the third party’ – s 2(1) (b). However, s 2(1) (b) is qualified: a third party will not be granted the direct statutory right of suit in the absence of an express provision permitting him or her to do so, ‘if, on a proper construction of the contract, it appears that the parties did not intend the term to be enforceable by the third party.’ – s 2(2). 8.7.10 This statutory right of enforcement is not just limited to cases where the promisor is under an obligation to act to confer a positive benefit on the third party. ‘Negative’ benefits, such as the benefit of a term excluding or limiting the third party’s legal liabilities to the promisor, may also be enforced –s 2(5). 8.7.11 The third party’s statutory right of enforcement against the promisor is qualified in a number of ways. First, the third party’s statutory right of recovery may be qualified by a defence or set-off which the promisor would have been able to assert vis-à-vis the other party to the contract (the ‘promisee’) – s 4. Second, any sum to be recovered by the third party pursuant to the Act may be reduced to take into account sums recovered by the promisee from the promisor in respect of the promisor’s breach – s 6. 8.7.12 Once third party rights are created under the Act, certain restrictions are imposed on the ability of the parties to the contract to vary or rescind their contract if this would extinguish or alter the third party’s rights under the Act – s 3. 8.7.13 Though wider in its scope than many of the other legal techniques for circumventing privity, the Act is not of universal application. Section 7 of the Act sets out a number of situations where the Act does not apply. Excluded cases include: (a) contracts on a bill of exchange, promissory note or other negotiable instrument; (b) the statutory contract binding a company and its members under s 39 of the Companies Act (Cap 50, 2006 Rev Ed) ; (c) limited liability partnership agreements as defined under the Limited Liability Partnerships Act (Cap 163A, 2006 Rev Ed) ; (d) third party enforcement of any term of an employment contract against an employee; and (e) third party enforcement of any term (apart from any exclusion or limitation of liability for the benefit of the third party) in a contract for carriage of goods by sea, or a contract for the carriage of goods or cargo by rail, road or air, if such contract is subject to certain international transport conventions. SECTION 8 DISCHARGE OF CONTRACT Discharge by Performance 8.8.1 If all the contractual obligations as defined by the terms of the contract are fully performed, the contract is brought to an end or ‘discharged’ by performance. In theory, such performance must be precise. However, trivial defects in performance may be ignored as being negligible or ‘de minimis.’ In addition, where full performance is only possible with the cooperation of the other party (as is almost invariably the case with obligations of payment or delivery), tender of performance in circumstances where the other party refuses to accept it is generally deemed to be equivalent to full performance so as to discharge the contract. Non- or Defective Performance 8.8.2 In the event that a contractual obligation is not performed or is performed defectively in a non-trivial fashion, Singapore law provides for a variety of legal responses and remedies, depending on the nature of the failure of performance. Lawful Excuses for Breach of Contract 8.8.3 If the failure of performance is not subject to any lawful excuse, the contract is said to be ‘breached.’ In this context, ‘lawful excuses’ may take the following forms. Discharge by Agreement 8.8.4 First, just as parties are free to agree to bind themselves to a contract, they are free to negotiate with each other to release themselves from the obligations of that contract. Such agreement may well have been built into the original contract, for example, where parties agree that their original agreement be terminable by giving notice of termination, or upon lapse of a specified period of time. Alternatively, contracting parties may release themselves from the obligations of the original contract by entering into a subsequent contract of release. Where each contractual party is still subject to contractual obligations which have yet to be performed, the mutual release of their outstanding obligations is generally effective under Singapore law without the need for any further formalities or any other consideration. However, where the party who is owed the obligation in question does not have any outstanding obligations under the original contract, the party seeking to be released from that obligation will have to provide some form of valuable consideration in exchange for the release. In the alternative, the release must be executed under seal to be effective. 8.8.5 Secondly, it may be that the obligation which has not been performed is conditional upon the prior occurrence of certain specified events: these may be external events, or some contractually specified counter-performance by the other party to the contract. 8.8.6 Thirdly, the parties may contractually provide for non-performance following from certain events to be excused so as not to amount to a breach, for example, in the form of a ‘force majeure’ clause. At the very least, such a clause will hold all parties innocent of liability for non-performance following the specified force majeure event. More detailed force majeure clauses may also make provision for issues such as the return and refund of advance payments, reimbursements for expenses incurred in preparation of the performance of the contract, and so forth. Such provisions will generally be given effect by Singapore law. Discharge by Frustration 8.8.7 Fourthly, where the reason for the failure of performance lies in events beyond the control of the contracting parties and which neither party could have reasonably foreseen, the contract is said to be ‘frustrated’. In such cases, there are statutory rules which set out the extent to which advance payments made before the frustrating event intervened may be refunded and work done in preparation of the performance of the contract in advance of the frustrating event may be reimbursed – see Frustrated Contracts Act (Cap 115, 1985 Rev Ed) s 2(2) and s 2(4) respectively. Section 2(3) of the Frustrated Contracts Act also empowers the Singapore courts to make valuations of any non-money benefits which may have been conferred by one contracting party on another, prior to the frustrating event, and to order the recipient of those benefits to pay for such value received. Effects of a Breach of Contract 8.8.8 In the absence of a lawful excuse, a breach of contract has two significant effects. Contract Damages 8.8.9 First, if the breach of contract by one contracting party (the ‘party-in-breach’) causes loss to the other (the ‘aggrieved party’), the party-in-breach may be ordered by the courts to compensate the aggrieved party in money damages for those losses, in lieu of the primary obligations left unperformed under the contract. However, contractual damages (which are compensatory and not punitive in nature), is not the only judicial remedy available. Other types of remedies may be available in lieu, or sometimes, in addition to damages, depending on the nature of the obligation which has been breached. [See Section 13 below]. Right to Elect to Discharge for Breach 8.8.10 Second, the breach may give the aggrieved party the right to bring the contract to an end, ie to discharge the contract for breach. In this connection, it is useful to distinguish actual breaches of contract (wherein the breach occurs at the actual time of performance as specified by the contract) from anticipatory breaches of contract (wherein the breach is said to occur in advance of the contractually stipulated time of performance). Actual Breach Giving Rise to Right of Discharge 8.8.11 In the case of an actual breach of contract, the aggrieved party may elect to discharge the contract for breach if the contractual term which has been breached is: (a) a ‘condition’; or (b) an ‘innominate term,’ the breach of which deprives the aggrieved party of substantially the whole of the benefit of the contract. In such a case, the aggrieved party may choose to discharge the contract for breach. 8.8.12 The aggrieved party has no such power of election if the contractual term which has actually been breached is: (a) a ‘warranty’; or (b) an ‘innominate term,’ the breach of which does not deprive the aggrieved party of substantially the whole of the benefit of the contract. In such a case, the contract will persist despite the breach (unless the contract is brought to an end by some other event). 8.8.13 For details as to how a contract term may be categorised as a ‘condition,’ a ‘warranty’ or an ‘innominate term,’ see Paragraphs 8.5.9 to 8.5.10 above. Discharge by Actual Breach 8.8.14 If the aggrieved party is entitled to discharge the contract and elects to do so, the contract is brought to an end prospectively. That is, the contract ceases to bind the parties to the contract from the time the election is effectively communicated to the other contracting parties. Such communication may take the form of words, acts, or even (in exceptional cases) silence. Prior to that time, such an election may be withdrawn. Following an effective discharge, the parties are released from all outstanding contractual obligations. Affirmation of Contract Following an Actual Breach 8.8.15 The aggrieved party may choose, however, not to discharge the contract. Instead, the aggrieved party may choose to affirm the contract, thereby giving the party-in-breach another opportunity to rectify the non-performance or defective performance. If so, the entire contract is kept alive and the aggrieved party loses the right to have the contract discharged (although the right to sue the party-in-breach and recover money damages for any losses incurred as a result of the delay in procuring full performance may well be retained, unless the aggrieved party also elects to waive his or her right to compensatory money damages). Anticipatory Repudiatory Breach 8.8.16 A breach of contract may also occur anticipatorily (in advance of the time of actual performance). If this breach is also repudiatory (where the evidence demonstrates that one party intends not to be bound by the terms of the contract, nor to honour his or her contractual obligations as and when they fall due), the aggrieved party has the right to choose whether to discharge or to affirm the contract. ‘Repudiatory’ intentions will be more readily proved where there are clear and express communications by the purported party-in-breach to such effect. However, they can also be inferred from actions or steps taken by the purported party-in-breach which render it impossible for his or her obligations to be performed when they become due. Effect of Discharge by Anticipatory Repudiatory Breach 8.8.17 Significantly, a party aggrieved by an anticipatory repudiatory breach may exercise his or her right to discharge the contract immediately without waiting until the time of actual performance. If the aggrieved party elects to discharge the contract, the contract is immediately and prospectively brought to an end. The aggrieved party is then entitled to sue the party-in-breach for damages as compensation for any loss suffered by the aggrieved party as a result of the non-performance of the contract. Effect of Affirmation Following an Anticipatory Repudiatory Breach 8.8.18 On the other hand, the aggrieved party may elect to affirm the contract. If so, the contract continues to bind all parties to the contract and the anticipatory breach is ignored. Consequently, once the aggrieved party affirms the contract, there can be no liability for money damages for that anticipatory breach since it is treated as if the breach never occurred. Limits on Right of Election to Affirm Contact 8.8.19 Although the aggrieved party’s right of election to discharge/affirm a contract following an actual/anticipatory breach is largely unqualified, the English case of White & Carter (Councils) Ltd v McGregor AC 413 suggests that this right is limited under English law. However, it is arguable that the limitation is less strict in Singapore. In MP-Bilt Pte Ltd v Oey Widarto 3 SLR 592, the Singapore High Court adopted the limitations set out in White & Carter v McGregor that the aggrieved party may only elect to affirm a contract (despite the other contracting party’s breach) if the aggrieved party was reasonably able to perform his or her part of the contract without the need for any cooperation from the party-in-breach and if the aggrieved party had a legitimate interest in doing so. However, the High Court stated that these limitations would not apply when the aggrieved party ‘is under a legal obligation or practical compulsion to complete performance of the contract in question and other contracts he has entered into on the basis of the contract in question.’ – at p 607. Consequently, it appears that an aggrieved party’s freedom to elect to affirm a contract may be less strongly curtailed in Singapore as compared with the case in England. SECTION 9 MISTAKE Introduction 8.9.1 If one or both parties enter into a contract under a misapprehension of its basis, or of an important aspect of the transaction, the contract may either be completely void, or voidable. In the latter case, the contract is valid until it is rescinded (or set aside) by the mistaken party. This distinction is critical for determining third party rights – see Paragraph 8.9.12 below. Whether a mistake has the effect of rendering a contract void or voidable depends on the manner in which the mistake arises. Mutual Mistake 8.9.2 If A contracts with B believing that he is purchasing X but B is in fact intending to sell Y to A, there is no contract between A and B because they have failed to reach any agreement on the subject matter of the contract. Mistakes of this nature are commonly referred to as ‘mutual mistakes’. A transaction entered into under a mutual mistake (relating to a fundamental aspect of the contract) is void. Common Mistake 8.9.3 A ‘common mistake’ arises when an agreement is reached on the basis of a mistaken assumption or belief shared by both parties. This occurs, for instance, when A contracts to sell a consignment of goods to B but unknown to both parties, the goods had been destroyed before the contract was formed. In this situation, owing to the destruction or non-existence of the subject matter, the contract may justifiably be regarded as invalid and void even though it is otherwise properly formed. 8.9.4 The more problematic situation arises when the common mistake relates to a less fundamental matter, such as the quality of a subject matter of the contract (as opposed to its existence). Here, the law has to strike an appropriate balance between doing justice to the party disadvantaged by the mistake and protecting the counter party’s legitimate expectation that the contractual bargain would be upheld. The common law and principles of equity respond to this problem in different ways (on the distinction between common law and equitable rules, see [Chapters 1 and 18 – Singapore Legal System and Trusts]). Common Mistake at Common Law 8.9.5 At common law, precedence is given to upholding bargains. Thus, a common mistake as to quality would not, in general, render a contract void unless the mistake has the effect of rendering the subject matter of the contract essentially and radically different from what the parties believed it to be. The ambit of the common law doctrine is therefore extremely narrow, having little application outside cases involving non-existent or destroyed subject matter. Common Mistake in Equity 8.9.6 Equity, in comparison, permits a more liberal approach: even if a mistake is not sufficiently fundamental to render a contract void at common law, it may still be set aside provided that the mistake is sufficiently serious. 8.9.7 Distinguishing between the different degrees of ‘fundamental’ mistakes that are operative at common law and in equity is a difficult task. Nevertheless, the Singapore Court of Appeal’s recent observations appear to favour the retention of this two-prong approach (Chwee Kin Keong v Digilandmall.com Pte Ltd 1 SLR 502). This may be contrasted with the position in England, where the more flexible equitable rule appears to have been abolished (Great Peace Shipping Ltd v Tsavliris Salvage (International) Ltd QB 679). Unilateral Mistake 8.9.8 A contract may also be affected by a ‘unilateral mistake’, that is when only one party is acting under a mistake. For purposes of discussion, it is convenient to distinguish between the following two cases: (a) where the mistake relates to the identity of a contracting party, and (b) those where the mistake relates to a term of the contract. Unilateral Mistake as to Identity 8.9.9 It is useful to note, for a start, that unilateral mistakes as to identity typically involve cases where one party’s consent to an agreement is procured by deception. If A agrees to sell his car to B (who has deceived A into believing that B is C), the contract is affected by A’s unilateral mistake as to B’s true identity provided that it is clear that B’s identity is material, ie an important factor which induced the contract. As between A and B, it is not essential to determine whether such a mistake renders the contract void or voidable, since A, the mistaken party, would have the right to set aside the contract in either case. However, the distinction becomes critical if B has sold the car to T (an innocent third party who acquires the car without notice of B’s deception) before A discovers the fraud. If the mistake has the effect of rendering the contract between A and B void, A will be able to recover the car from T because B, not having acquired any property right in the car, has nothing to sell to T. In the converse situation where the contract between A and B is merely voidable, B would have acquired property rights in the car, which he could subsequently transfer to T. A is therefore unable to recover against T in this instance. 8.9.10 Disputes involving mistakes as to identity are invariably ‘hard’ cases because they often require the court to prefer one of two innocent parties. Nevertheless, it may be observed that the general approach in these cases requires examination of the facts to ascertain whether there is in fact an agreement between the mistaken party and the (fraudulent) counter party. Thus, if A intends to sell his car only to C, then no agreement is reached between A and B when B attempts to purchase the car by pretending to be C. Such intention may, for instance, be inferred from the fact that A’s offer is expressly addressed to C, or where there is a written contract purportedly made between A and C (although fraudulently signed by B on C’s behalf). However, where A and B transact face-to-face, there is a presumption that they intend to deal with the physical person present, in which case A is presumed to have intended to contract with B, the fraudster. Such a presumption may, however, be rebutted by clear evidence to the contrary. Unilateral Mistake as to a Term 8.9.11 Unilateral mistakes may also arise in relation to the terms of a contract. If A enters into a contract under a misapprehension as to a particular important term (other than the identity of the other party, B), and the mistake is known to B, such a mistake may render the contract void at common law. The Singapore Court of Appeal has recently clarified (in Chwee Kin Keong v Digilandmall.com Pte Ltd 1 SLR 502) that this common law doctrine is confined to cases where the non-mistaken party, B, has actual knowledge of A’s mistake. In addition, if a case does not fall within the ambit of the common law doctrine (because, for instance, it has not been established that B has actual knowledge of A’s mistake), the court may nevertheless exercise its equitable power to set the contract aside if B is guilty of unconscionable conduct. This may arise where B suspects that A is labouring under a mistake but consciously omits to disabuse A of his error. Documents Mistakenly Signed 8.9.12 Generally, a person of full age and understanding who has signed a written contract is bound by it even if he or she has not read it. Exceptionally, a signatory to a contract may be able to set it aside if it is fundamentally or radically different from what the signatory believed it to be, as may occur if the signatory’s understanding is limited by some innate incapacity, or when he or she has been tricked into signing it. This defence cannot, however, be invoked by a person who has been negligent in signing the document. Documents Mistakenly Recorded 8.9.13 If a written contract does not, by reason of a mistake, accurately record the agreement between the parties, the court may rectify the contract so as to give effect to the parties’ true intention. Originally, the remedy of rectification was only available in cases where the mistake is shared by both parties, but was subsequently extended to situations where only one party is mistaken, and such mistake is known to the other party. SECTION 10 MISREPRESENTATION The Nature of the Representation 8.10.1 A contract which is induced by a misrepresentation may be set aside, and may give rise to an action for damages. A misrepresentation occurs when one party to a contract makes a false statement of fact to the other contracting party which induces the latter to enter into the contract. To be operative, the false representation must relate to a past or present fact. It follows that a vague or exaggerated statement that is in the nature of a ‘puff’ does not suffice. Generally, a statement of a party’s intention or opinion is also not a sufficient ground for relief. However, if the representor does not honestly hold such intention or opinion, there is a misrepresentation of fact as to the representor’s state of mind. A statement of opinion may also be actionable if it is made by a person who professes to have special skill or knowledge in the matter stated. Statements of law appear still to be excluded from the ambit of operative representations, although the correctness of this position must now be doubted in light of the abolition of this distinction in the context of mistakes (see [Chapter 19 on Restitution – Mistaken Payments]). 8.10.2 A representation may be express, or it may be inferred from the representor’s conduct. On its own, silence or non-disclosure does not usually constitute a representation. There are, however, exceptions to this general rule. If a party makes a positive but incomplete disclosure, the omitted disclosure may amount to a misrepresentation if it has the effect of distorting the truth of the information disclosed. Similarly, a failure to correct an earlier ( and continuing) representation that was true at the time it was made but which has subsequently become incorrect is actionable. A failure to disclose material facts whilst negotiating contracts uberrimae fidei, such as insurance contracts, would also give rise to an action for misrepresentation. 8.10.3 Generally, a misrepresentation must also be material, in the sense that it relates to a matter which would influence a reasonable person’s decision whether to enter into the contract. If a representation is ambiguous and may be interpreted in two (or more) ways, of which one is true and the other false, it is not a misrepresentation unless the representor has intended it to be understood in the sense that is false. The Falsehood Must Have Induced the Contract 8.10.4 Misrepresentation is a ground for relief only where it has induced a contract. Clearly, if a person is unaware of the representation, or knows that it is untrue, or does not believe it to be true, he or she cannot reasonably have relied on, or be induced by, the representation to enter into the contract. Reliance may also be negated if the representee has independently verified the truth of the representation, although the failure to verify (when the opportunity to do so is available) is not in itself a bar to relief. If the misrepresentation has in fact induced the representee to enter into the contract, it does not matter that it is not the sole inducing factor. The persons who may rely on a representation are not confined to those directly addressed by the representor, but include any person whom the representor intends to reach and influence, even if such a person learns of the representation indirectly from a third party. The Right to Rescind 8.10.5 Once it is established that a contract has been induced by a misrepresentation (whether innocent, negligent or fraudulent), the party induced may elect to rescind (ie set it aside) or affirm it. The effect of rescission is to release the parties from their contractual obligations, and to restore the parties to their respective positions prior to the making of the contract. The right to rescind will, however, be lost if: (a) the induced party has affirmed the contract; (b) innocent third parties have acquired (for value) rights in the subject matter of the contract; (c) it is no longer possible to restore the parties to their respective prior positions; and (d) (except in the case of fraud) an inordinate period of time has lapsed. It should also be noted that the court may, pursuant to s 2(2) of the Misrepresentation Act (Cap 390, 1994 Rev Ed), award damages in substitution for the right to rescind. Damages for Fraudulent Misrepresentation 8.10.6 Whether damages may be awarded for misrepresentation depends on whether the misrepresentation is fraudulent, negligent or innocent. At common law, damages may be awarded for fraudulent misrepresentations. A fraudulent misrepresentation is a false representation that is made: (1) knowingly, (2) without belief in its truth, or (3) recklessly, careless whether it be true or false. In such a case, the representor would have committed the tort of deceit and the representee is permitted to recover for all losses incurred as a consequence of the fraudulent misrepresentation, even for losses which might not have been reasonably foreseeable. Common Law Damages for Negligent Misrepresentation 8.10.7 Where an operative misrepresentation results from negligence, the party who has relied on it may obtain damages by commencing an action in the tort of negligence. This requires proof that there is a ‘special relationship’ between the parties which places the representor under a duty to take reasonable care in furnishing information or advice to the representee, and that the representor has failed to do so. A more extensive survey of the legal principles relating to this branch of the law is contained in [See Chapter 20 on Tort – Negligence]. Recovery in such a case would, however, be restricted to losses which are reasonably foreseeable. Statutory Damages for Negligent Misrepresentation 8.10.8 Alternatively, a party who has entered a contract in reliance on a negligent misrepresentation may claim damages under 2(1) of the Misrepresentation Act (Cap 390, 1994 Rev Ed). In fact, where the issue arises as between contracting parties, this statutory action is generally the preferred route for recovering damages as its requirements are less onerous than those of the common law (tortious) action outlined in Paragraph 8.10.7 above. Under s 2(1), the claimant only has to establish that he or she has contracted in reliance on the other party’s misrepresentation, whereupon the latter has the onus of proving that he or she was not negligent in that he or she had reasonable ground for believing in the truth of the statement. In contrast, the claimant in a tortious action bears the burden of proof of all elements of the action, including the existence of a special relationship between the parties, as well as the other party’s negligence. The language of the provision suggests that the measure of damages under s 2(1) should be the same as that for fraudulent misrepresentations, which is more liberal than the measure which applies in contract cases [see Paragraph 8.13.10 below] or in cases based on the tort of negligence [see Paragraph 8.10.7 above]. As a matter of principle, however, the contract measure appears to be the more appropriate option. Innocent Misrepresentations 8.10.9 Misrepresentations may also be made innocently. In such a case, the claimant is not entitled to damages at common law, but where the claimant still has the right to rescind ( and it appears beneficial to do so), the claimant may persuade the court to exercise its discretion under s 2(2) of the Misrepresentation Act to award damages in lieu of rescission. If the court is not so persuaded and the contract is rescinded, the claimant may be compensated for expenses incurred in performing the contract in the form of an ‘indemnity’. Misrepresentations and Terms 8.10.10 Misrepresentations are usually pre-contractual statements made to induce a person to contract with the representor. A pre-contractual statement which has induced a contract may also have been incorporated as a term of the contract. If so, the person who made the statement would now also be in breach of the contract if the statement turns out to be false. In such an event, damages for breach of contract may be claimed, and s 1 of the Misrepresentation Act makes it clear that the representee may still rescind the contract for misrepresentation. For the test for distinguishing between terms and representations, see Paragraph 8.5.1. Excluding Liability For Misrepresentation 8.10.11 Parties to a contract may agree to contractual terms which exclude or limit their liability for misrepresentation, but s 3 of the Misrepresentation Act requires such a term to satisfy the test of reasonableness set out in s 11(1) of the Unfair Contract Terms Act (Cap 396, 1994 Rev Ed). This test has been discussed in Paragraph 8.5.15 above. SECTION 11 DURESS, UNDUE INFLUENCE & UNCONSCIONABILITY Duress 8.11.1 If A enters into a contract with B as a result of B’s coercion (often taking the form of threats of unlawful acts), the contract may be set aside by A on the ground of duress. The types of unlawful or improper pressure that may have this effect include actual or threatened harm to a person, a person’s goods or his or her economic interests. 8.11.2 The recognition that economic duress can suffice as a ground for avoiding a contract is a relatively recent development, justified by the concern to prevent a party with strong bargaining power from exploiting the weaker position of the counter party. However, it is not the case that economic duress arises whenever a contract is entered into between parties of unequal bargaining strength. The law recognises that a measure of commercial pressure is inherent in every transaction between such parties, and inequality in bargaining power is a well-accepted ( and perhaps necessary) facet of modern commercial life. A plea of economic duress will therefore only succeed in the exceptional case, where a party has used his or her superior bargaining position a way that is illegitimate. 8.11.3 That said, the line between illegitimate pressure and mere commercial ( and legitimate) pressure is extremely fine, and where it falls is often dependent on the particular facts of the case. In general, the reasonableness of the parties’ respective conduct appears to be an important consideration. For instance, a party who threatens to breach a contract with another if the latter does not agree to its request for increased payments is not exerting illegitimate pressure if, owing to acute financial conditions, that is the only course available to him. However, where the dominant party makes the same demand for no reason other than an opportunistic desire to exploit the counter party’s vulnerability for financial gain, such conduct is less likely to be viewed favourably. Undue Influence 8.11.4 The doctrine of undue influence guards against the victimisation of persons by those who exercise dominance or influence over them. The pressure so exerted is generally less direct and acute than that which occurs in cases involving duress. Traditionally, cases involving undue influence fall into two main categories. Actual Undue Influence 8.11.5 Under the first category, a contract may be set aside if one utilises one’s dominant position over another to procure the latter’s consent to the contract. The victim has the burden of proving that the guilty party so dominates the victim’s will as to substantially undermine the victim’s independence of mind. It is, however, unnecessary to establish that such dominance arises out of a special relationship between the parties, nor that the resulting transaction is manifestly unfair to the victim. Presumed Undue Influence 8.11.6 The second category is concerned with situations involving relationships of trust and confidence between the parties. In such cases, if the nature of the transaction is such that it cannot be reasonably accounted for even in the light of the nature of that particular relationship, a presumption that one party has acted under the undue influence of another arises. The effect of the presumption is to shift the burden to the defendant to prove that no undue influence has been exercised. There are two classes of such relationships. In the first class, the presumption of influence arises automatically, as a matter of law, from the proof of the existence of certain relationships that are characterized by strong elements of confidence and influence. These include parent-child, guardian-ward, trustee-beneficiary, doctor-patient, lawyer-client, director-company, and religious adviser-disciple relationships. In the second class, although the parties’ relationship does not fall into the first-mentioned group, the presumption may nevertheless arise if the claimant is able to establish that he or she has in fact reposed trust and confidence on the other party. It is, however, unsettled as to whether the claimant would also have to establish that the transaction is one which is manifestly disadvantageous. Rebutting the Presumption 8.11.7 The presumption may be rebutted by showing that the dominant party did not abuse his or her position and that the subservient party understood what he or she was doing and was in a position to exercise a free judgment based on full information. Generally, it would suffice to demonstrate that the subservient party had the opportunity to receive independent legal advice prior to making the contract. Third Parties and the Doctrine of ‘Infection' 8.11.8 If A improperly influences B to contract with C (usually for the benefit of A), B may seek to set aside the contract on the ground of undue influence if it can be shown either (a) that A was acting as the agent of C; or that (b) C had either actual or constructive notice of A’s misconduct. If the transaction is one which is, on its face, disadvantageous to B, and C knows of reasons why B could have reposed trust and confidence in A (where, for instance, B is A’s wife), then C would be fixed with constructive notice of the improper influence, unless C has taken reasonable steps to ensure that B’s consent was in fact obtained independently. This will entail, at the very least, explaining the transaction to B in a private meeting, and advising her to seek independent legal advice. Effects of Duress and Undue Influence 8.11.9 Contracts that are procured by duress, undue influence or unconscionable conduct are voidable. In each case, the improper conduct must be a significant or decisive cause of the victim’s consent. This right to rescind may, however, be lost in certain circumstances (see Paragraph 8.10.5 above). Unconscionable Bargains 8.11.10 Apart from instances involving duress or undue influence, equity may also relieve parties from ‘unconscionable bargains’. Such bargains typically involve the exploitation of one party’s weakness, though the mere fact that the parties are of unequal bargaining power does not suffice. The exact ambit of this jurisdiction is unclear, but it has traditionally been applied narrowly to cases involving expectant heirs and improvident transactions. SECTION 12 ILLEGALITY AND PUBLIC POLICY Statutory Illegality 8.12.1 A contract may be said to be ‘illegal’ in a number of different contexts. For example, there may be a statutory prohibition as to the formation of contracts which would entail carrying out certain socially undesirable activities. 8.12.2 In such cases, the statute may clearly provide that the ‘illegal’ contract is void. That is to say, it is to be treated in law as if it had never been formed. If the statutory wording is clear, there is no need to go any further to ascertain the intention of the legislature as to the status of the contract. 8.12.3 Difficulties arise, however, where the statutory wording is unclear, particularly where the statute in question does not clearly specify whether its object is to prohibit the formation of the contract, or the performance of the obligations under that contract. The true parliamentary intention underlying the statutory prohibition will have to be ascertained. In the former case, the contract is void. Illegality at Common Law 8.12.4 At common law, certain strands of public policy prohibit the formation of certain types of contract. 8.12.5 Such contracts are completely void and examples include: (a) contracts prejudicial to the administration of justice – these include contracts to stifle prosecution, or contracts savouring of maintenance (where one person supports another in bringing or resisting an action – as by paying the costs of it – which is permissible only if the party providing the support has a legitimate and genuine interest in the result of the action and the circumstances are such as reasonably to warrant such support) or champerty (which is a species of maintenance where the maintainer seeks to make a profit out of another man's action – by taking the proceeds of it, or part of them, for himself or herself) ; (b) contracts to deceive public authorities; (c) contracts to oust the jurisdiction of courts (although contracts or agreements to arbitrate, or agreements to confer exclusive jurisdiction over a dispute in favour of a foreign court are not caught by this prohibition) ; (d) contracts to commit a crime, tort or fraud; (e) contracts prejudicial to public safety; and (f) contracts promoting sexual immorality. Effect of Statutory Illegality or Illegality at Common Law 8.12.6 Where a contract is rendered void by statute or common law, the general starting point is to treat the contract as if it had never existed. Any outstanding or unperformed obligations under that contract are extinguished. In other words, in so far as enforcement of such outstanding obligations would have required reliance on the illegal contract, no judicial enforcement is possible. Judicial enforcement may still be available, notwithstanding the illegality, if it is possible to do so without referring to the illegal contract, ie by relying on an independent and separate cause of action. 8.12.7 Conversely, the question arises whether any recovery may be had for benefits which have been conferred under an illegal contract. On one view, such benefits will have been conferred without any basis. It may well be that, in some cases, some form of recovery pursuant to the law of unjust enrichment is possible. This is very likely to be allowed in instances where one party repents of the illegal contract and withdraws from it before the illegal purpose of the contract is fulfilled. If such repentance is genuine, voluntary and timely, before any part of the illegal purpose has been carried out, restitutionary recovery pursuant to the principles of unjust enrichment is likely to be allowed [see Chapter 19 on Unjust Enrichment]. Contracts in Restraint of Trade 8.12.8 A contract which is wholly in restraint of trade is contrary to public policy and is illegal at common law. Such a contract is, as a general rule, void. However, it is accepted that, in some contexts, some degree of restraint of trade may be necessary to protect legitimate interests. 8.12.9 Thus, a ‘reasonable’ restraint of trade clause which seeks to protect: (a) the interests of the parties concerned; (b) and the interests of the public will not be void. The assessment of the reasonableness of the clause must be made from both these perspectives. 8.12.10 This determination will vary from case to case, but significant factors will include the geographic scope as well as the length of time for which the restraint of trade is to apply. The wider and longer the restraint, the more difficult it will be to prove that the restraint is reasonable. Severance of the Illegal Terms 8.12.11 Sometimes, the illegality taints only part of a contract, eg, attempts to restrain competition from ex-employees. Such restraints of trade are often incorporated as a covenant or term in an otherwise unobjectionable employment or service contract. 8.12.12 If the restraint of trade covenant is found to be unreasonable, and hence void, the ‘illegal’ covenant can be severed from the rest of the contract, thereby maintaining the contract’s validity. Such severance, however, is only possible if the severed covenant does not form the whole or the main consideration for the contract. If the severed covenant does form the whole or the main consideration for the contract, no severance can take place and the entire contract is void. 8.12.13 Severance may also take effect in a more limited form within the confines of a particular covenant or term. This more limited form of severance is akin to taking a ‘blue-pencil’ to strike out those words which would render the covenant ‘unreasonable.’ In doing so, however, the court will not go so far as to re-write the contractual bargain which had been reached by the contracting parties. SECTION 13 JUDICIAL REMEDIES FOR BREACH OF CONTRACT Judicial Remedies Contrasted with Self-help Remedies 8.13.1Following a breach of a condition of a contract, or where the breach causes one party to be deprived of substantially the whole of the benefit of the contract, the aggrieved party may elect to bring the contract to an end. When this happens, both the aggrieved party and the party-in-breach will be released from any outstanding obligations under the contract. This is said to be a ‘self-help’ remedy because the release is effected without the need for any court approval or intervention. 8.13.2 Where the aggrieved party has suffered financial losses as a result of the breach, or where release of the party-in-breach from outstanding obligations will cause financial loss, discharge of contract alone may not be an adequate remedy. Recourse to other judicial remedies may be needed. Types of Judicial Remedies 8.13.3 In relation to contract law, the following types of judicial remedy are commonly sought: (a) the common law remedy of damages; (b) the common law remedy of an action for a fixed sum; (c) the equitable remedy of specific performance; and (d) the equitable remedy of injunction. It is important to draw the distinction between the common law and the equitable remedies because, while the former are available as of right, the latter are discretionary. Availability of Judicial Remedies – Time bars, Limitation Periods and Laches 8.13.4 Urgency should be the order of the day when seeking judicial remedies as access to judicial remedies may be barred by lapse of time. 8.13.5 Generally speaking, no action may be brought for a breach of contract after 6 years have lapsed from the time when the contract was breached – s 6 of the Limitation Act (Cap 163, 1996 Rev Ed). This bars access to the court insofar as the remedies of damages or an action for a fixed sum are concerned. [See Chapter 2 on Court Procedure for a fuller discussion]. 8.13.6 In relation to the equitable remedies of specific performance and injunction, the equitable doctrine of laches applies. Shortly put, applicants who delay applying for equitable relief from the courts may be turned away if the delay is inordinate and inexcusable, such that it would be inequitable to grant such relief. Indeed, an application for an order for specific performance might be denied if the application is not made as soon as the nature of the case might permit. Damages – Compensation for Pecuniary Loss 8.13.7 Contractual damages are awarded to an aggrieved party in the form of a sum of money, in compensation for any pecuniary losses which have been incurred as a result of the breach of contract. Compensation Only 8.13.8 In general, damages are compensatory in nature. It remains an open question whether, in the appropriate case, damages might be awarded for breach of contract on any other basis. Liquidated Compared with Unliquidated Damages 8.13.9 In some cases, compensation for losses resulting from breach may have been pre-agreed by the contracting parties as a term of the contract. If the agreed sum is a genuine pre-estimate of the loss which could be suffered as a result of a breach of the contract, the court will order that sum to be paid in compensation as liquidated damages. However, if the sum is intended to be a penalty aimed at ‘punishing’ the party-in-breach, the court will strike down the ‘penalty’ clause and award unliquidated damages instead to compensate the aggrieved party. Quantification and Measure of Unliquidated Damages 8.13.10 The court will usually quantify unliquidated damages so as to place the aggrieved party, as far as money can do so, in the position he or she would have been had the contract been performed fully. Therefore, if the aggrieved party would have expected to make a profit by resale of goods which had been purchased from the party-in-breach, but where such profit is not earned because of non-delivery and breach, the aggrieved party’s ‘expectation loss’ in the form of the loss of profit may be recovered. Alternatively, where the aggrieved party has to incur additional costs, over and above what was expected under the contract by reason of having to pay for a replacement supply of goods or services following the failure by the party-in-breach to perform his or her contractual obligations, those additional expenses may be recovered by the aggrieved party in compensation as a form of expectation loss. As a further alternative, an aggrieved party may choose to quantify his or her damages on the basis of expenses which were incurred in reliance on the other party performing his or her contractual obligations, instead of on an expectation basis (unless it is demonstrated that the aggrieved party had made a bad bargain and the reliance expenditure would have exceeded any expected gain). Time of Quantification 8.13.11 In most instances, unliquidated damages will be assessed as at the time of the breach although, in appropriate cases, the court may take into account events occurring after the breach. Restrictions on Recovery of Unliquidated Damages 8.13.12 It is not the case, however, that unliquidated damages are available for all losses. Recovery is subject to certain restrictions. Non-pecuniary Loss 8.13.13 First, non-pecuniary losses (ie for hurt feelings, disappointment, mental distress, and so forth), are generally not compensable except in certain limited circumstances – for example, where the contractual obligation itself related to non-pecuniary matters, as in the case of a contract for a package holiday. Remoteness of Loss 8.13.14 Second, losses which are too remote are not compensable. Losses which arise in the usual course of things as a result of the breach are not too remote, and are compensable. Losses which are out of the ordinary and which would not ordinarily have been in the contemplation of either party to the contract at the time of its formation are usually too remote, and therefore not recoverable. However, losses which are out of the ordinary may be found to be not too remote ( and therefore recoverable), if it can be shown that the special circumstances of the innocent party which led to his sustaining those losses were known to the party-in-breach, at the time of contract formation. Mitigation of Loss 8.13.15 Third, losses which the aggrieved party has avoided by taking measures to mitigate them, are not compensable since there is nothing to compensate. That said, losses which the aggrieved party could have taken reasonable steps to avoid, but did not, are also not compensable. This is to encourage mitigation of losses, that is, steps by the aggrieved party to reduce his or her losses. The duty is to take all reasonable steps to minimise one’s loss. If, in taking objectively reasonable steps to mitigate, the aggrieved party incurs greater loss than if no steps been taken at all, such increased losses will still be recoverable from the party-in-breach. Action for a Fixed Sum 8.13.16 Damages, whether liquidated or unliquidated, are not the only remedy at common law. Where the contractual breach relates solely to an obligation to pay a fixed sum of money, damages are not usually available as a remedy. Instead of damages, the court will order that the fixed sum, due and owing, be paid. 8.13.17 In such cases, generally, there will be no damages for the delay in payment, apart from any court ordered interest on the judgment sum, or any contractual interest (if the contract expressly provides for the payment of interest on any delayed payment of the sum owed). Specific Performance 8.13.18 Sometimes, damages will not be an adequate remedy for a breach of contract. This may be the case where the breach involves delivery of property which is unique (such as a piece of land). In such instances, the aggrieved party may make an application for the court to make an order of specific performance – ie an order to the party-in-breach (or threatening to be in breach) to perform in accordance with the terms of his or her contractual promise. 8.13.19 Specific performance is, however, not available as against the Singapore Government in any civil proceedings to which the state is a party – see s 27(1) (a) of the Government Proceedings Act (Cap 121, 1985 Rev Ed). Nor shall the court in any civil proceedings grant an injunction against an officer of the Government if the effect of making such an order would have been to give relief against the Government which could not have been obtained in proceedings against the Government – see s 27(2) of the Government Proceedings Act (Cap 121, 1985 Rev Ed). Limits on Availability of Specific Performance 8.13.20 Specific performance is a discretionary remedy. It may be withheld if, in all the circumstances of the case, it would be inequitable to make such an order. As has been mentioned above, substantial delay in applying for such relief may be enough to cause the court to withhold such relief. Relief may also be withheld if the applicant does not come to court with ‘clean hands’. The order for specific performance may also be made on terms, so as to balance the interests of the parties to the dispute. 8.13.21 Specific performance might also be refused in a number of other instances, most notably where: (a) the proposed order would require constant supervision by the court; (b) the court is not able to specify the terms of the order which is to be complied with; (c) the proposed order would require the performance of something which is impossible to achieve; and (d) the order relates to a contract of personal service because such an order could amount to judicial compulsion of involuntary servitude. Injunction 8.13.22 Not all contractual obligations are susceptible to orders of specific performance. Sometimes, the contractual obligation in question is a negative one, where the party-in-breach fails to honour his or her promise not to do something. In such circumstances, an application for a prohibitory injunction may be made by the aggrieved party. 8.13.23 In the absence of factors such as those mentioned above in Paragraph 8.13.20, prohibitory injunctions are likely to be granted unless: (a) the remedy would be inequitable or oppressive; or (b) the balance of convenience does not favour making such an order. 8.13.24 If the breach of the negative obligation lies wholly in the past, the aggrieved party may seek a mandatory injunction instead. Such an order requires the party-in-breach to reverse the effects of the breach so as to restore the aggrieved party to the position he or she would have been, had the negative obligation not been breached. 8.13.25 The discretion whether to issue a mandatory injunction is also generally subject to the ‘balance of convenience’ test. 8.13.26 In general, injunctions will also be refused in relation to contracts of personal service – where the practical effect of the proposed injunction would be to compel the performance of a contract for personal service for which no order of specific performance would have been made in the first place. Updated as at 30 April 2015 By: Lee Pey Woan Associate Professor, School of Law Singapore Management University Pearlie Koh Associate Professor, School of Law Singapore Management University Tham Chee Ho Associate Professor, School of Law Singapore Management University Ch. 08A Remedies for Breach of Contract Print 600498 Categories Categories 39RSSEditor's PickExpand/Collapse 100RSSHeadlinesExpand/Collapse 92RSSCommentariesExpand/Collapse 1RSSCivil Litigation UpdatesExpand/Collapse 28RSSLegislationExpand/Collapse 43RSSJudgmentsExpand/Collapse 14RSSBusinessExpand/Collapse 7RSSInternationalExpand/Collapse 10446RSSJudgmentsExpand/Collapse 1766RSSCourt of AppealExpand/Collapse 8390RSSHigh CourtExpand/Collapse 123RSSIntellectual Property Office of SingaporeExpand/Collapse 154RSSPersonal Data Protection CommissionExpand/Collapse 13RSSTax BoardsExpand/Collapse 192RSSLegislationExpand/Collapse 3RSSBills IntroducedExpand/Collapse 189RSSSubsidiary LegislationExpand/Collapse 23RSSNotices and DirectionsExpand/Collapse 3RSSSupreme Court Practice DirectionsExpand/Collapse 2RSSState Courts Practice DirectionsExpand/Collapse 4RSSFamily Justice Courts Practice DirectionsExpand/Collapse 2RSSSupreme Court Registrar's CircularsExpand/Collapse 3RSSState Courts Registrar's CircularsExpand/Collapse 2RSSFamily Justice Courts Registrar's CircularsExpand/Collapse 7RSSGeneral NoticesExpand/Collapse 109RSSContinuing Legal EducationExpand/Collapse 4RSSAbout Singapore LawExpand/Collapse 7RSSOverviewExpand/Collapse 25RSSCommercial LawExpand/Collapse 8RSSSingapore Legal SystemExpand/Collapse 3RSSCivil Practice & ADRExpand/Collapse 4RSSCivil Litigation UpdateExpand/Collapse 5RSSVC Investment Model AgreementsExpand/Collapse 9RSSVenture Capital Investment Model Agreements 2.0Expand/Collapse 3RSSVCC Model ConstitutionsExpand/Collapse 1RSSSingapore Shipping LawExpand/Collapse 1RSSCoalExpand/Collapse Advertise Frequently Asked Questions Contact Us Tags Administrative & Constitutional LawAssociations & SocietiesBanking & Financial ServicesBusiness & CommerceCivil Law & ProcedureCommodities & EnergyCompany LawCompetitionConstruction & InfrastructureCriminal LawData ProtectionDispute ResolutionEmployment LawEnvironmentEquity & TrustsFamily LawFundsGovernment & PoliticsHealth Care & Life SciencesInsolvencyInsuranceIntellectual PropertyInternational Law & TradeProfessional Practice & EducationPropertyProperty LawSecurities & FuturesShippingStatutory InterpretationTaxTMTTortTransportationWills & Probate Archive «)September 2025») Sun Mon Tue Wed Thu Fri Sat 31123456 78910111213 14151617181920 21222324 25 26 27 28 29 30 1 2 3 4 5 6 7 8 9 10 11 Download the app or follow us Terms Of UsePrivacy StatementCopyright 2025 by Singapore Academy of Law Back To Top
2627	https://www.vocabulary.com/dictionary/retrospect	Retrospect - Definition, Meaning & Synonyms \| Vocabulary.com SKIP TO CONTENT Log inSign up Dictionary Vocabulary Lists VocabTrainer™ Word Finder All Dictionary Vocabulary Lists Random Word Featured Dictionary Vocabulary Lists More results Any words Any words 2-letter words 3-letter words 4-letter words 5-letter words 6-letter words 7-letter words 8-letter words 9-letter words 10-letter words 11-letter words 12-letter words 13-letter words 14-letter words 15-letter words Starting with Starting with Ending with Including Containing in order Advanced Search Find Word Part of speech- [x] noun - [x] verb - [x] adjective - [x] adverb Syllable range Between and Restrict to Synonym Synonym Antonym of Definition contains Rhymes with Example of Ends with Parts of Type of Find Word Random Word retrospect Add to listShare /ˌrɛtrəˈspɛkt/ /ˈrɛtrəʊspɛkt/ IPA guide Other forms: retrospects; retrospecting; retrospected In retrospect — that is, in looking back and contemplating the past — we sometimes find ourselves wishing that we had done some things differently. Though this word most commonly appears as a noun in the phrase "in retrospect," it can also be used as a verb. The prefix retro- means “back," and spect is a component of the words inspect, spectator, spectacles, and perspective, among others, which all have to do with looking or seeing. So it makes sense that retrospect means to look back in time, or to remember. Definitions of retrospect noun contemplation of things past “in retrospect” see more see lesstype of:contemplation, musing, reflection, reflexion, rumination, thoughtfulnessa calm, lengthy, intent consideration verb look back upon (a period of time, sequence of events); remember synonyms:look back, review see more see lesstype of:remember, think backrecapture the past; indulge in memories Cite this entry Style: MLA MLA APA Chicago "Retrospect."Vocabulary.com Dictionary, Vocabulary.com, Accessed 25 Jul. 2025. Copy citation Test your knowledge of retrospect and thousands of other words. Think you know retrospect? Answer a question to start your personalized learning plan. ASSESSMENT : 100 POINTS retrospect means : standard or typical examplethe scientific study of languagecontemplation of things pastthe ability to read and write Not sure? Get a hint:50/50Word in the WildDefinition Examples from books and articles All sources All sources Fiction Arts / culture News Business Sports Science / Med Technology loading examples... It was, to her, funny only in retrospect. Americanah by Chimamanda Ngozi Adichie In retrospect, it is odd how little power the dead farmer exercised over an imagination as morbid and hysterical as my own. The Secret History by Donna Tartt In retrospect, Fenn told me, “One of George Washington’s most brilliant moves was to inoculate the army against smallpox during the Valley Forge winter of ’78.” 1491 by Charles C. Mann Though, in retrospect, he actually kind of does. Simon vs. the Homo Sapiens Agenda by Becky Albertalli < prev \| next > DISCLAIMER: These example sentences appear in various news sources and books to reflect the usage of the word ‘retrospect'. Views expressed in the examples do not represent the opinion of Vocabulary.com or its editors. Send us feedback Word Family retrospectretrospectsretrospectingretrospectedretrospectiveretrospection the "retrospect" family Vocabulary lists containing retrospect Common Senses: Spec, Spect, Spic ("Look") The words on this list all derive from the Latin verb specere, meaning "to look." Here are links to the complete set of Common Senses lists: Hearing: Phon / Aud / Son Sight: Vid, Vis / Spec, Spect, Spic / Op, Ops, Opt Touch: Path / Sent, Sens / Tact, Tang Wonder R.J. Palacio In this novel, a boy with facial anomalies tries to navigate the sometimes treacherous world of public school. Fantastic Beasts and Where to Find Them J.K. Rowling Set in the Harry Potter universe, this book purports to be an encyclopedia of magical creatures written by the wizard Newt Scamander. MORE VOCABULARY LISTS 2 million people are mastering new words.Master a word Sign up now (it’s free!) Whether you’re a teacher or a learner, Vocabulary.com can put you or your class on the path to systematic vocabulary improvement. 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2628	https://eaglepubs.erau.edu/introductiontoaerospaceflightvehicles/chapter/lifting-line-theory/	Skip to content 36 Lifting Line Theory Introduction When a freestream flow approaches a finite wing, characterized by a definitive span from tip to tip, the downstream flow develops a trailing wake system containing swirling flows known as wingtip vortices, as shown in the figure below. These vortices induce a vertical downwash velocity over the wing’s surface, particularly near the wing tips. This downwash reduces the effective angle of attack of the wing, diminishes the lift, and increases drag, which is called induced drag. A wing’s aerodynamic performance, including its lift and drag characteristics, is significantly influenced by its geometric design. Factors such as planform shape (chord distribution), span and aspect ratio, and spanwise twist are all essential in determining the wing’s aerodynamic efficiency. The wing design process must carefully balance these elements to optimize lift generation and minimize induced drag, among other factors, thereby achieving the wing’s best aerodynamic performance for the intended application. Lifting line theory is another cornerstone of classical aerodynamics, explaining and predicting the aerodynamic behavior of finite wings at low speeds, i.e., without considering compressibility effects. Unlike the two-dimensional airfoil theory, which assumes infinite span and aspect ratio, the lifting line theory accounts for the three-dimensional effects of finite wings, including the formation of trailing vortices and the creation of induced drag. By modeling the wing as a bound vortex line with a spanwise variation in circulation, along with the associated effects of the trailed vortex wake, this theory provides a sound mathematical framework for determining the spanwise lift distribution, induced drag, and overall aerodynamic efficiency of a finite wing. Learning Objectives Understand how trailing vortices generate downwash and affect the angle of attack of a wing. Mathematically predict how and why the lift varies along the span of a finite wing and its impact on total lift and induced drag. Appreciate why an elliptical lift distribution on a wing minimizes its induced drag. Determine the total lift and induced drag of a finite wing of arbitrary shape. Apply the lifting line solutions to non-elliptical solutions, such as those for a flying wing. Review the extension of lifting line theory to lifting surface theory. History Lifting line theory, initially developed by Ludwig Prandtl and his students in the early 1900s, revolutionized the understanding of finite wing aerodynamics, marking a significant milestone in aeronautical developments. Before Prandtl’s work, aerodynamic theories were primarily based on two-dimensional airfoil analysis, which assumed a wing of infinite span and ignored the effects of the wing tip vortices and the three-dimensional impact caused by the wake. The work of Frederick Lanchester on the roll-up of wingtip vortices, as illustrated in a figure from his book shown below, was influential in establishing the importance of adequately modeling finite wing aerodynamics. Furthermore, as aviation advanced, particularly during WWI, including the transition from biplanes to higher-performance monoplanes by the 1920s, the predictive limitations of existing aerodynamic theories quickly became evident. Designers needed a way to better predict and optimize the performance of monoplane wings with finite span, particularly in terms of their lift and drag, as well as other factors such as aeroelastic characteristics and stalling behavior. Prandtl’s lifting line theory, which was published in 1904 in the proceedings of the Third International Congress of Mathematicians in Heidelberg, Germany. This paper introduced the fundamental ideas of the lifting-line theory.") addressed this need by modeling a finite wing as a bound vortex line with a spanwise circulation distribution. Prandtl’s advance over Lanchester’s work was in the theoretical modeling of the wing, with its wake represented as a sheet of vortices trailing behind it. This wake produced a downwash flow, decreased the effective angle of attack along the entire wing, and created induced drag, a drag component directly associated with lift production. Prandtl also demonstrated that an elliptical lift distribution over the wing span minimized the induced drag, setting a practical goal for efficient wing design. Prandtl’s original published works also show an “ideal” elliptical wing planform, which may have influenced the design of the Supermarine Spitfire in the early 1930s. In the 1920s, Hermann Glauert further refined and extended Prandtl’s work, making it more accessible and applicable to practical engineering problem-solving. The use of straight vortex lines, as shown in the figure below from his book, allowed the induced velocity over the wing to be calculated using the Biot-Savart law, which formed the cornerstone of the lifting line theory. He then clarified the fundamental importance of the elliptical lift distribution and introduced systematic methods for analyzing wings of arbitrary planform and twist using a Fourier series, much like the approach employed in the thin airfoil theory. Glauert’s detailed methods for calculating induced drag helped bridge the gap between theoretical predictions and practical wing designs, making lifting line theory one of the first valuable approaches for analyzing three-dimensional wings. Additionally, the theory was written in a form that later allowed it to be adapted to swept wings and non-planar configurations and extended to lifting surface methods that model the chordwise distribution of circulation. Though more advanced techniques, such as computational fluid dynamics (CFD), are now widely used, the lifting line theory remains a cornerstone of aerodynamic modeling. Its mathematical simplicity and elegance make it valuable for preliminary wing design and essential to aerodynamic education. Helmholtz’s Vortex Theorems Helmholtz’s vortex theorems describe the fundamental properties of vortex motion in an ideal fluid. First formulated by Hermann von Helmholtz in 1858, they remain central to understanding vorticity dynamics in fluid mechanics and aerodynamics. These theorems are derived from the Euler equations and express how vortex lines behave in an ideal flow. First Helmholtz Theorem (Conservation of Circulation): The circulation around a material loop moving with the fluid remains constant over time, i.e., , where is the circulation. This theorem implies that vortices cannot spontaneously appear or vanish in an inviscid fluid. Circulation can only be introduced through boundaries or viscous effects, which are excluded in the ideal model. This result is consistent with Kelvin’s Circulation Theorem, which forms the foundation for Helmholtz’s first theorem. While Kelvin’s theorem asserts the conservation of circulation around any material loop in an ideal fluid, Helmholtz’s first theorem applies this principle specifically to vortex tubes, emphasizing that their circulation (vortex strength) remains constant along their length as the flow convects them. Second Helmholtz Theorem (Conservation of Vortex Lines): Vortex lines move with the fluid. If a vortex line passes through a fluid element at some initial time, it will continue to pass through that fluid element at all later times. This implies that vortex lines are “frozen” into the fluid, i.e., the topology of the vortex structure is conserved as it is convected with the flow. Third Helmholtz Theorem (Strength of Vortex Tubes): The strength of a vortex tube (the circulation around any closed loop within the tube) is constant along its length and does not vary with time. This theorem follows from the divergence-free nature of the vorticity field in an ideal flow, i.e., , which implies that vortex tubes cannot begin or end in the fluid interior, i.e., they must either form closed loops or terminate at boundaries. Helmholtz’s theorems are foundational in the study of vortex dynamics, including the formation and evolution of wingtip vortices and the inviscid modeling of aerodynamic flows. They are essential in developing the Biot-Savart law, the lifting-line theory, and other potential flow models that underpin much of classical aerodynamics. While real fluids are viscous and may violate these theorems under certain conditions, Helmholtz’s laws provide an essential idealized framework for understanding the behavior of vortex flows. Theory of Vortex Lines The line vortex is a fundamental singularity in fluid dynamics used to model velocity fields induced by a circulation, . The ideas of circulation and the effects associated with vortices have been previously introduced. A vortex line is a curve of any shape and tangent to the vortex, as shown in the figure below. Notice that a line vortex in three dimensions is analogous to a point vortex in two dimensions, but instead of being a singular point, it extends along a curve. The circulation around any closed curve is equal to the sum of the strengths of the line vortices intersecting the surface bounded by the curve. Consequently, a line vortex cannot terminate within a fluid. It must form a closed loop, end at a solid boundary, or extend to infinity; the principles are formally embodied in Helmholtz’s theorems. The Biot-Savart law describes the velocity field induced by such a circulation distribution, . Regarding the situation shown in the figure above, the increment in the downwash, , from the vortex element can be mathematically expressed as (1) where is the circulation or vortex strength, is a differential element of the line vortex, is the distance of the point P from the element, and is the angle between the direction of the element and the straight line joining the element to the point P. Notice that the element of a line vortex cannot exist by itself and only forms the basis for integrating the effects along a line vortex of finite length. Straight Line Vortices Straight-line vortices are used in the development of the finite-wing theory. Consider the evaluation of the induced velocity of a straight-line vortex of finite length AB, as shown in the figure below. If PN has a perpendicular length , then the induced velocity at P from the element at point Q is (2) To show this result, the element may be geometrically expressed as (3) where the angle is as shown in the figure above, so that (4) The total induced velocity is then obtained by integration along the length of the vortex line using (5) which gives (6) Doubly-Infinite Straight Line Vortex If, as a special case, the line is of doubly infinite length, as shown in the figure below, then it will be apparent that , so the general result reduces to (7) which is consistent with the result of a two-dimensional point vortex. Singly-Infinite Straight Line Vortex For a singly infinite vortex, which starts at point N and then extends to infinity only in one direction, as shown in the figure below, then and , and the downwash is given by (8) This latter result for the downwash from a semi-infinite line vortex is extensively used in developing the lifting line theory, in which several line vortices are combined appropriately to represent the physics of the flow produced by a finite wing. Lifting Line Theory The following exposition of the lifting line theory and the development of the monoplane wing equation follows the work of Gluert. In addressing the problem of a wing of finite span in three-dimensional flow, the following assumptions were made: The wing’s chord is relatively minor compared to its span, resulting in a high aspect ratio. The wing’s spanwise axis can be considered a straight line perpendicular to the freestream flow, i.e., the wing is unswept. The wing planform is symmetrical laterally about its centerline, i.e., the left and right wing panels mirror each other. Apart from these restrictions, the planform (chord), pitch angle (including any twist angle), and zero-lift angle of the airfoil section may vary arbitrarily across the wing’s span. Bound & Trailed Vortices If a wing generates lift, there must be a flow circulation around each airfoil section comprising the wing, effectively creating a line vortex or a set of line vortices along the wing’s span. These line vortices, which are attached to the wing, as shown in the figure below, are referred to as the bound vortices and create lift in accordance with the Kutta-Joukowsky theorem. Physically, these bound vortices are formed by the integrated effect of the vorticity in the boundary layer or equivalent vortex sheet surrounding the airfoil’s surface, as in the manner used to develop the thin airfoil theory. According to the general theory of vortex flows and Helmholtz’s theorems, these bound vortices cannot terminate at the tips of the wing but must extend off the wing tips into the fluid as free line vortices. These are referred to as the trailing or trailed vortices, as illustrated in the figure below. This entire vortex system, which extends to infinity far behind the wing, is completed by a transverse vortex parallel to the span. However, this vortex has no consequence in the steady state as its induced effects are asymptotically zero. For practical purposes, the trailing vortices can be considered to align with the freestream flow and extend downstream to infinity as a singly infinite line vortex. Horseshoe Vortex System Glauert describes the simplest type of vortex system as occurring when the circulation around each airfoil section has a constant value across the wing’s span. The bound vortex system can then be represented as a single “lumped” line vortex with strength , which, to be consistent with the thin airfoil theory, can be positioned at the aerodynamic center along the 1/4-chord of the wing, as shown in the figure below. The trailing vortices will consist of two line vortices, each with the same strength, originating from the tips of the wing and extending downstream in the direction of the freestream flow. This representation is called a “horseshoe vortex system.” In practice, the vortices trailing from the wing tips will generally not be straight lines because of variations in the downwash at different distances behind the wing. Experiments show that as they develop downstream, the trailing vortices tend to descend below the wing and contract inward. However, as Glauert explains, they can be approximated as straight lines parallel to the direction of motion for most practical purposes. This assumption leads to the simplified representation of the wing and its wake as a horseshoe vortex system, which forms the fundamental basis of the lifting line theory. Nested Horseshoe System In reality, the vortex system of an aerofoil is more complex because the circulation is not uniform across the span. Instead, it typically reaches its maximum value at the center and tapers to zero at the tips. This distribution can be represented by superimposing multiple simple “horseshoe vortex systems,” as shown in the figure below. This setup yields a more accurate representation of the vortex system over the wing and in its wake. This model then comprises interlaced bound vortices, all colocated at the aerodynamic center on the 1/4-chord axis, and a sheet of trailing vortices extending from the wing’s trailing edge. Glauert used this representation of the vortex system as the fundamental premise in developing the mathematics of the lifting line model. Wake Rollup As Glauert explains, the origin of the trailing vortex wake system can also be understood from the perspective of the pressure difference between the upper and lower surfaces of the wing. Suppose the lift distribution across the span of a wing reaches a maximum at its centerline. In that case, there will be a significant decrease in pressure above the wing and an increase in pressure below it, as shown in the figure below. These pressure differences decrease towards the tips of the wing because of the influence of the trailing wing tip vortices. As the streamlines pass above the wing and flow inward toward the center, while those below the wing flow outward, they leave the trailing edge and form a discontinuity surface, as shown in the figure above. The trailing vortices from the wing then represent the vorticity of this discontinuity surface. The sheet of trailing vortices rolls up into a pair of concentrated vortices downstream, forming the characteristic pair of vortices behind a wing, as observed in practice using flow visualization methods. Nearer to the wing, the influence of the trailing vortex system can be modeled by assuming that the individual trailing vortices extend downstream as straight lines. For the wake farther from the wing, which tends to contract somewhat, it is perhaps more accurate to assume a horseshoe vortex system with a span a little shorter than the wing span. Both representations are sufficiently representative of the actual flow physics for modeling purposes. Calculating the Induced Velocity In general, the circulation will vary across the span of a wing, being symmetrical about the centerline and decreasing to zero at the tips. Consistent with the nested horseshoe vortex model, between the points and of the span of the wing, the circulation can be assumed to decrease by an amount (9) Therefore, a trailed wake filament of this strength originates from the element of the span, as shown in the figure below. Therefore, in the aggregate, a sheet of trailing vortices will extend behind the entire wingspan because of the continuous change in along the wing, i.e., . The induced downwash at any point of the span will then be obtained as the sum of the effects of all the trailing vortices of this sheet. For example, consider some reference point on the wing, say . Using Eq. 8, the downwash at this point from all of the trailing line vortices comprising the sheet will be (10) This latter result will then apply at every section, , along the span of the wing, and different values of will be obtained depending on the distribution, , and the proximity to the wing tips. It will be apparent from the previous discussion that the circulation gradient, i.e., , is much higher at the wing tips, so the trailed wake will have a greater circulation here. This behavior is consistent with Lanchester’s interpretation of the wake generated behind a wing. The question now is how this downwash distribution can be calculated, as it affects the lift and drag at each wing section, , and the overall wing lift and drag. Sectional Effects of the Induced Velocity The induced velocity from the trailed wake will affect the aerodynamic angle of attack at each section across the wing. Adding the downwash flow vector to the freestream flow vector produces a resultant flow velocity (or local relative wind) that turns through an angle , called the induced angle of attack, as shown in the figure below. It will be apparent that the resultant flow now approaches the wing at a different angle, creating a reduced “effective” angle of attack, . From the geometry, as shown above, the induced angle, , is given by (11) which will, in general, be different at each station on the wing, i.e., and . For small angles, which is typical for a wing, it is sufficient to write that (12) Therefore, the effective (and lower) angle of attack of the wing section is now (13) This outcome means that the corresponding lift per unit span will be reduced from its two-dimensional value, i.e., the lift obtained without the effects of the downwash flow. Notice that because will typically be much smaller than , then the resultant velocity, , can be assumed to be equal to , i.e., (14) It is particularly significant in this situation to note that the lift vector at each section assumes a new orientation and is tilted slightly rearward from its original direction in two-dimensional flow, thereby reducing the vertical component of the lift for a given angle of attack. The consequence is that there is now a component of the lift that acts in the downstream direction, which explains the origin of the induced drag, , i.e., (15) Development of the Monoplane Wing Equation With an understanding of the aerodynamic elements that comprise the problem of modeling a finite wing, the development of the fundamental equation of lifting line theory can be established, known as the monoplane wing equation. Connection Between Circulation and Downwash As previously shown, if is the circulation produced around any section of the wing, the downwash from the trailing wake system at a point of the span is determined using (16) A representative wing section, therefore, experiences the lift force corresponding to the two-dimensional conditions at the effective angle of attack, i.e., (17) If the angles of attack and are measured from the angle of zero lift of the airfoil section,, then (18) Therefore, the lift coefficient of the wing section will be (19) where is the slope of the curve of lift coefficient against the angle of attack for the wing section in two-dimensional flow. Furthermore, the circulation around the wing section can be connected to the lift coefficient, , using (20) This means that (21) Therefore, it is possible to determine the circulation and the downwash for any wing in terms of the chord and angle of attack of the wing sections, which may vary across its span. When the circulation and the downwash of any monoplane wing have been determined, the lift and induced drag are obtained by evaluating the integrals (22) Use of the sectional lift curve slope Strictly, the lift curve slope, , for which Glauert assigns the symbol , depends on the shape(s) of the wing section(s). However, two-dimensional thin airfoil theory has shown that per radian angle of attack. It is known from experiments that is approximately equal to for all practical wing sections at low Mach numbers, and hence any normal variations in may be neglected without any appreciable loss of accuracy. Nevertheless, because a wing section may differ from the theoretical value , it is best to retain the value of as a variable, and the theoretical value of is generally only used only in theoretical solutions. Method of Solution Glauert used a transformation to replace the coordinate , measured to the left wing tip along the span of the wing from its center, by the angle , as defined by (23) where is the semi-span. It will be apparent that as varies from to , then varies from to across the span from the left tip to the right. The circulation, , which is a function of , can then be expressed as the Fourier series (24) where the values of the coefficients must be determined by connecting the distribution of “bound” to the wing with produced by the trailed wake in a consistent manner. Notice that the series chosen for the circulation satisfies the condition that the circulation decreases to zero at the wing tips. As shown in the figure below, if the wing is flying in level flight, then the spanwise loading will be symmetrical about its centerline, and only odd integral values of will occur in the Fourier series. It will be apparent that the primary mode is the or mode, corresponding to an elliptical circulation distribution over the wing. Because , then (25) Simplifying gives (26) and so (27) which is elliptical. The higher symmetric modes, i.e., , for which and are shown in the figure below, represent a symmetric modification to the primary loading, i.e., a deviation from the elliptically distributed form. The downwash at the point or of the wing now becomes (28) because (29) Therefore, at the general point of the wing, the downwash is (30) Monoplane Wing Equation The equation connecting the circulation and the downwash velocity now becomes (31) which gives (32) and finally (33) Equation 33 is called the fundamental equation of lifting line theory or the monoplane wing equation and can be used to determine the coefficients values for any monoplane wing. It is generally solved numerically. The equation must be satisfied at all points of the wing, but because the wing is symmetrical about its mid-point in level flight, it is usually sufficient to consider values of between and . In this case, only the symmetric modes, i.e., , must be evaluated. The dimensionless Glauert parameter , which is proportional to the chord , and the angle of attack , will generally depend on wing twist, e.g., if the wing has wash-out. Lift and Induced Drag The lift on the wing is given by (34) Switching to the spanwise angle coordinate, , where and , the lift becomes (35) Substituting the general series expression for circulation gives (36) and so (37) Using the orthogonality property of sine functions, i.e., (38) then the only term that contributes to the integral is , so (39) The coefficient, , which is the elliptical loading mode, is related to the lift coefficient, , using (40) The induced drag is determined using (41) Switching to the coordinate gives (42) Recall that the downwash velocity is given by (43) Substituting and gives (44) Expanding the summation gives (45) Again, the orthogonality of functions ensures that only terms where are retained, so that (46) It is convenient to write (47) so the induced drag, , is (48) In terms of drag coefficient, then (49) The total drag coefficient includes both profile drag and induced drag. If the profile drag coefficient varies across the span, the profile drag coefficient for the wing is (50) Adding the induced drag gives (51) Asymmetric Loading Modes The even modes in the lifting line theory represent an asymmetric distribution of lift. These modes can arise from aileron deflections and angular velocity of the wing in roll, yaw, sideslip, or other asymmetries along the wing’s span. This situation then leads to lift asymmetry, influencing rolling moments, possible yawing moments, and induced drag. For example, ailerons are control surfaces located near the wingtips, which are deflected asymmetrically to induce a rolling moment about the aircraft’s longitudinal axis. The figure below shows that their operation produces asymmetric modes in the lifting-line theory, particularly and other higher even-numbered modes. The total circulation distribution can be expressed as (52) where represents the symmetric lift distribution from the freestream velocity, and captures the antisymmetric lift contribution from aileron deflection, as shown in the figure below. The primary contribution from the ailerons is (53) The coefficient depends on the aileron deflection angle , which modifies the local angle of attack near the wingtips, i.e., (54) where is the aspect ratio. The constant of proportionality depends on the actual wing’s geometry and aerodynamic characteristics. For small deflection angles, is approximately linear in . The resulting rolling moment, , about the longitudinal axis, is directly related to this antisymmetric lift distribution, i.e., (55) where is the lift difference from aileron deflection at spanwise location . This rolling moment scales with , i.e., (56) From a design perspective, proper sizing and placement of ailerons are crucial to ensure sufficient rolling authority without excessive induced drag or adverse yaw effects. To this end, the lifting line theory can provide valuable insight into the aileron designs that may be required. Worked Example #1 – Further understanding of circulation modes An untwisted elliptical wing planform is installed symmetrically in a wind tunnel with its centerline along the tunnel axis. If the air in the wind tunnel has a freestream axial velocity and also has a small angular velocity about the tunnel axis, show that there will be an extra asymmetric distribution of circulation along the wing given by Determine in terms of and the wing parameters. Show solution/hide solution. From the lifting-line theory, the spanwise circulation distribution can be expressed as a Fourier sine series, i.e., where is the semi-span of the wing, is the spanwise angular coordinate related to the spanwise position by , and are the Fourier coefficients determined by the flow and wing properties. For an elliptic wing, the baseline lift distribution from the freestream flow, , corresponds to the term and perturbations introduced by that will manifest as higher harmonics, starting with .The angular velocity induces a spanwise velocity component at a location , given by This spanwise velocity modifies the local effective angle of attack , and the change in angle of attack is Therefore, the angular velocity introduces a spanwise perturbation in , which varies linearly with . The local circulation (y) or is proportional to the effective angle of attack , so the perturbation in circulation from will be Substituting gives For an elliptical wing, the spanwise variation contributes to the term in the series expansion. This means that where is the second harmonic, specifically where is the lift curve slope (per radian). Substituting , then Elliptic Spanwise Loading The lift and induced drag of a wing are given by (57) For a wing of a given size, the coefficient has a fixed value independent of the wing’s shape. The induced drag will be minimized when all higher-order coefficients () in the series for circulation are zero, i.e., . In this case, the distribution of circulation across the span of the wing simplifies to (58) so the circulation is elliptically loaded. Putting into Eq. 33 gives (59) and so (60) The corresponding induced drag coefficient will be (61) Notice that the result of the downwash becomes (62) and the equivalent angle of attack is (63) along the span. This result implies that the local lift coefficient, , is constant across the span, as shown in the figure below. To produce a constant and an elliptic circulation distribution, the chord length must vary elliptically along the span. This result can be obtained by using (64) where it will be apparent that (65) so that a solution is (66) where is the chord length at the wing root. This special case of elliptic loading is significant for several reasons: It results in the minimum possible induced drag for a given total lift, i.e., . Most conventional wing designs closely approximate an elliptic load distribution, making it a valid first-order approximation. The results derived from the assumption of elliptic loading represent the optimal scale for minimizing induced drag while remaining applicable to practical wing designs. Elliptic loading (or almost elliptic) can also be achieved with wings of non-elliptic planforms by suitably varying the twist angle along the span. Effect of Aspect Ratio The results developed for the lift and drag coefficients of an elliptic wing can be used to calculate the effect of a change in aspect ratio. If the aspect ratio is reduced from to , the changes in the drag coefficient at a given value of the lift coefficient is (67) It will be apparent that wings with a higher aspect ratio have aerodynamic advantages, all other factors being equal, e.g., wing area and planform shape. While this formula for the drag coefficient applies only to wings with an elliptical loading, the lift distribution curves for rectangular and most airplane wings do not differ significantly from the elliptic form. Therefore, the formula may be used more generally to calculate the effect of a modest aspect ratio change. Effect on Lift Curve Slope The assumptions are: 1. The two-dimensional lift curve slope for the airfoil is , which assumes no three-dimensional effects. 2. The lift curve slope for the finite wing is , which includes the effects of finite aspect ratio (). The lifting line theory introduces an induced angle of attack caused by the downwash from the wing’s trailing vortices. The net angle of attack, , is related to the effective angle of attack, , and the induced angle, , by (68) The total lift coefficient is then (69) where the two-dimensional lift curve slope has been replaced with using Gluert’s notation. The induced angle of attack is related to the lift coefficient and aspect ratio (70) Substituting this back into the lift equation gives (71) and rearranging gives (72) Dividing through by gives (73) Therefore, for wings with a high aspect ratio or a small , the lift curve slope approaches the two-dimensional value, corrected by an aspect ratio term to account for three-dimensional effects. Worked Example #2 – Aerodynamic characteristics of an elliptical wing An elliptical wing has the following characteristics: Span, = 10 m. Planform area, = 8 m. Angle of attack, = 5. Zero-lift angle of attack, = -0.5. 2-dimensional lift curve slope, = = 2 per radian. Compute: 1. The wing’s lift coefficient, . 2. The corresponding induced drag coefficient, . Show solution/hide solution. The aspect ratio of the wing is The lift curve slope of the wing adjusted for 3-dimensional effects is The lift coefficient is For an elliptical wing, the induced drag coefficient is Numerical Solution to the Monoplane Wing Equation The governing equation for the monoplane wing is (74) where: are the coefficients of the circulation expansion. is referred to as Glauert’s non-dimensional parameter. is the geometric angle of attack, or more correctly, pitch angle. is the chord length (may vary with and so ). is the semi-span of the wing. To solve numerically, the domain is divided into discrete points, i.e., (75) At each discrete point , the governing equation becomes (76) This equation can be rewritten in matrix form as (77) where is the vector of unknown coefficients; is the right-hand side vector containing the the wing properties; is the influence coefficient matrix. The components of the matrix equation are (78) and (79) The matrix equation is expanded as (80) where and The unknown coefficients are found by solving (81) Efficient numerical methods, such as LU decomposition, are typically used for inverting when is large. Once the coefficients are determined, the following aerodynamic quantities can be computed: The spanwise circulation distribution, , i.e., (82) 2. The lift distribution, , and the lift coefficient, . The total lift coefficient is directly related to , i.e., (83) 3. The induced drag coefficient, which is computed using (84) Worked Example #3 – Numerical solution of the monoplane wing equation A rectangular wing has the following geometric and operating conditions: Span, = 10 m. Chord, = 1.0 m. Angle of attack, = 12. Zero-lift angle of attack, = -0.5. Lift curve slope, = 2 per radian. Free-stream velocity, = 50 m/s. Air density, = 1.225 kg/m. Set up a numerical solution of the monoplane wing equation to determine: 1. The spanwise circulation distribution, . 2. The lift coefficient, . 3. The induced drag coefficient, . Use five spanwise modes to approximate your answer. Show solution/hide solution. The semi-span is so the non-dimensional parameter is The geometric angle of attack (allowing for the zero-lift angle) is Dscretize into into equally spaced angular points, i.e., Therefore, The governing equation with odd Fourier coefficients is This is written in matrix form as where as well as The right-hand side vector is The influence coefficient matrix is Solving gives The spanwise circulation distribution is Substituting the coefficients gives Therefore, the lift coefficient is The corresponding induced drag coefficient is and substituting the values of the coefficients gives Wing Shape & Loading Distributions The wing’s planform, twist, and interference effects strongly influence the spanwise lift distribution and overall wing performance. The planform shape, including the wing’s aspect ratio, taper ratio, and sweep angle, determines the baseline distribution of lift along the span and affects induced drag and aerodynamic efficiency. Whether geometric or aerodynamic, wing twists allow designers to tailor the angle of attack distribution, optimizing lift and reducing induced drag under specific flight conditions. Interference effects, such as those arising from wing-body junctions, nacelles, or wingtip devices, further modify the flow field around the wing, impacting both the spanwise lift distribution and the overall drag characteristics. These factors must be carefully considered in wing design to strike a balance between aerodynamic efficiency, structural feasibility, and operational requirements. Planform Effects The theoretically best aerodynamic efficiency () and lowest induced drag are obtained with a wing planform that is elliptical in planform shape with no twist. This wing shape gives an elliptical spanwise aerodynamic loading (i.e., lift per unit span) and uniform downwash over the wing, which, as previously mentioned, is theoretically the minimum induced drag condition, and so . However, this value is unobtainable in any practical wing design. Values of for a plain wing can range from about 0.01 to 0.1, as shown in the figure below; anything more than 0.1 would usually be classified as poorly designed. However, even a rectangular planform wing will have a reasonably good value of (probably between 0.05 and 0.1) if it has a decent aspect ratio, i.e., , and also uses some wing twist or wash-out to control the spanwise lift distribution. The shape of the wing (i.e., its planform) will affect the distribution of lift and the formation of the trailing vortex system, hence the magnitude of the induced drag on the wing. The idea is to use variations in wing chord along the span and perhaps wing twist and airfoil sections to approximate the ideal elliptical spanwise loading closely and minimize the value of . The actual effective aspect ratio of the wing can also be improved somewhat by paying attention to the shape of the wing tips, which can lower the values of if they are suitably shaped, e.g., with more rounded contours or perhaps with the addition of a winglet. Incorporating Wing Twist In the lifting line theory, wing twist affects the circulation distribution by altering the local geometric angle of attack along the span, which modifies the local lift coefficient. The effective angle of attack at any spanwise location is now given by (85) where is the local geometric angle of attack, including the effect of twist, and the induced angle of attack is (86) Wing twist, denoted as , modifies the geometric angle of attack by (87) where is the root angle of attack. Substituting this, the effective angle of attack becomes (88) Notice that in the case of no twist, i.e., (), the geometric angle of attack is constant, and the circulation distribution depends only on the wing shape and induced effects. In the case of wash-out, i.e., decreases toward the tip, this reduces lift at the wingtips, redistributes lift inboard, as shown in the figure below, and also decreases induced drag. If the wing has wash-in, i.e., increases toward the tip (which is unusual), it increases lift at the wingtips, leading to higher induced drag. Twist allows designers to optimize the lift distribution and induced drag for specific flight conditions. Interference Effects The photograph below illustrates the nature of the spanwise loading over a wing from the process of natural condensation in the low-pressure zones, which is nominally elliptically distributed, even on this relatively low aspect ratio wing. The highest lift (highest pressure difference) is at midspan, and the lowest lift (lowest pressure difference) is at the wing tips. The problem is, however, that even minor deviations from the ideal elliptical form can result in higher induced drag from the wing. Such variations can occur because of interference effects on the wing from the fuselage, engines, undercarriage, external stores, and other components. Examples of the effects of spanwise interference are shown in the figure below, which are for the same approximate value of total wing lift. Notice that the fuselage can produce significant deviations from the ideal elliptical form, perhaps increasing the value of to over 1.15. However, local effects along the wing, such as those caused by the aerodynamic interference effects produced by an engine, tend to have a minor impact on the value of . What is aerodynamic twist? Aerodynamic twist refers to the variation in the airfoil’s camber or thickness distribution along the wing span, which alters the local zero-lift angle of attack, , such as using a different airfoil section. Unlike geometric twist, which physically pitches the wing sections, aerodynamic twist modifies the airfoil shape to tailor the lift characteristics. By modifying , the effective angle of attack and, consequently, the spanwise lift distribution can be modified. In the context of lifting line theory, aerodynamic twist directly impacts the governing equation by introducing a spanwise variation in . This changes the effective angle of attack, which determines the local lift coefficient and the distribution of circulation. The primary advantage of aerodynamic twist is its ability to achieve specific spanwise loading patterns, such as near-elliptic lift distributions, without changing the wing’s physical orientation. This minimizes induced drag, optimizes load redistribution, and enhances performance for non-elliptic planforms. Additionally, aerodynamic twist improves stall behavior by reducing lift near the tips, ensuring the wing root stalls first. It also provides flexibility in wing design, as it does not require a physical twist of the wing sections like a geometric twist. By incorporating aerodynamic twist into lifting line theory, designers can achieve a better balance between aerodynamic efficiency, structural feasibility, and flight performance. Wing Sweep The sweep of a wing has a significant effect on the aerodynamic characteristics of a wing, especially at higher subsonic and transonic speeds. The sweep angle, , alters the effective flow direction, affecting the airfoil sections and thereby reducing the component of freestream velocity normal to the leading edge, which modifies the aerodynamic response. Sweep can be included within the framework of lifting-line theory by modifying the freestream velocity components and the associated expressions for lift and circulation. For a wing with sweep angle , the local airfoil sections respond primarily to the component of freestream velocity normal to the 1/4-chord line. The component of the freestream velocity normal to the local quarter-chord is (89) Therefore, the effective angle of attack at each section is reduced to (90) where is the geometric angle of attack and is the induced angle of attack at spanwise location . The sectional lift per unit span is then given by the Kutta-Joukowsky theorem, as before, but in this case using (91) The Biot-Savart law still governs the induced velocity (downwash), but must be paired with the modified form of lift. To solve for the spanwise circulation , the lifting-line integral equation is modified to include the sweep factor, i.e., (92) This formulation preserves the structure of the original theory but incorporates the effect of sweep via the terms, which appear in both the effective angle of attack and the circulation-induced downwash. It can be shown that the effect of a constant wing sweep angle on the lift curve slope from the lifting-line theory gives (93) where and is the wing’s aspect ratio. At low Mach numbers, where 1, this reduces to the incompressible result. Sweep also reduces the effective aspect ratio of the wing to (94) This reduction also increases the induced drag coefficient, i.e., (95) where is the span efficiency factor. Winglets Winglets reduce the induced drag of finite wings by modifying the location and structure of the trailed vortex system generated as a consequence of lift generation. Although winglets increase the wing’s wetted surface area and give a small increment to overall skin friction and profile drag, they typically produce a net reduction in total airplane drag for no increase in wingspan. This tradeoff explains the widespread adoption of winglets on modern commercial airliners as an effective strategy for reducing drag and saving fuel. In the following analysis, the wing and its wake are modeled using a classical lifting-line horseshoe vortex system, modified to include winglets by geometrically displacing the trailing vortex legs, as illustrated in the figure below. The induced velocity is evaluated at the mid-span of the wing. While this effective span winglet theory does not represent the spanwise variation in downwash, it provides a valid approximation of the total induced drag using the principles of the Biot-Savart law. It highlights the aerodynamic benefit of winglets through their influence on the geometry of the trailing vortex system. The winglet itself is assumed to generate no lift or drag; its effect is modeled purely as a geometric shift in the starting position of the tip vortex, effectively moving it farther from the lifting wing. As such, this model isolates the influence of vortex positioning on the downwash field (albeit a mild violation of Helmholtz’s laws) without accounting for the winglet’s detailed aerodynamic behavior or the tip vortex’s roll-up. The wing tip vortices are still modeled as semi-infinite straight filaments extending downstream, so the induced vertical velocity field can be calculated using the Biot-Savart law. When the wing has regular tips (no winglets), the trailing vortices originate in the plane of the wing at , the effects of any dihedral angle being neglected. The induced vertical velocity at mid-span is (96) where is the semi-span, representing the lateral displacement from the mid-span at to the origin of the tip vortices. Now, consider the case when the winglets relocate the vortex tips along a length at a cant angle from the vertical. The trailed vortex filaments, i.e., the wing tip vortices, now start from . Notice that this approach assumes that the spatial separation of the trailing vortex system governs induced drag. This formulation is consistent with Prandtl-Jones minimum induced drag theory, in which the induced drag is inversely proportional to the distance between the trailing vortices. Using the Biot-Savart law, the vertical velocity induced at mid-span by one vortex leg is (97) where (98) Including both tip vortices, the induced velocity becomes (99) which gives the induced angle of attack as (100) The lift generated by the wing is (101) As a consequence, the induced drag is (102) and so compared to the baseline wing without the winglet, then (103) where . In coefficient form, an effective aspect ratio can also be defined based on the inverse of the downwash ratio (to reflect a correct reduction of induced drag), i.e., (104) where is the baseline aspect ratio of the wing without the winglet. Therefore, the induced drag coefficient becomes (105) Special cases include: No winglet: (106) 2. Vertical winglet: (107) 3. Flat winglet (full span extension): (108) Minimizing induced drag from the use of a winglet corresponds to maximizing , which depends on the effective location of the trailed vortex system. Define the geometric function (109) Expanding and simplifying this gives (110) This function is minimized for a given winglet length when is minimized, i.e., at (111) which corresponds to a downward-pointing winglet. However, such a configuration is impractical because of ground clearance, among other limitations. Therefore, to determine the optimum cant angle under the current assumptions, define the performance metric (112) which measures induced drag reduction per unit increase in span. The induced drag reduction is (113) and the effective span added by the winglet is (114) Substituting yields in ratio form, i.e., (115) Maximizing corresponds to solving (116) which leads to a transcendental equation in that must be solved numerically. The results summarized in the figure below indicate that the reduction in induced drag increases with the cant angle and winglet length but diminishes beyond a cant angle of approximately . A commonly used approach in airplane performance analysis to estimate the effect of a winglet is to use a linear correction to the aspect ratio of the form (117) where is the corrected or “effective” aspect ratio, is the height of the winglet, and is a coefficient that depends on the winglet geometry. This expression is widely used in preliminary design because of its simplicity. This model, given by Eq. 117, can be reconciled with the theoretical effective span winglet theory by performing a Taylor expansion of the more general result. Recall that the effective aspect ratio based on the trailed wingtip vortex geometry is (118) where . Expanding this expression for small gives (119) Comparing this expansion with Eq. 117, the linear winglet correction coefficient is identified as (120) This result shows that the linear model is simply a first-order approximation to the vortex-induced effective span theory. For a typical cant angle of , this yields , which aligns with values often cited in the literature for a classic winglet. Therefore, while Eq. 117 provides a convenient approximation; the full geometric expression is more accurate, especially for larger winglet lengths or varying cant angles. In both cases, however, increasing the spatial separation of the trailed vortices, whether through increased span and/or vertical offset, results in reduced induced drag on the wing, which is consistent with the proven aerodynamic effects of winglets. Worked Example #4 – Aerodynamic benefit of a typical winglet Consider a wing with semi-span = 30 ft and a winglet of height = 2 ft, giving . If the winglet has a cant angle of , estimate the improvement in effective aspect ratio using both the vortex-induced effective span theory and its linear approximation. Then, assuming the wing has a baseline aspect ratio of , a wing reference area = 124 m, cruise lift coefficient = 0.5, and fuel flow rate of 2.6 kg/s at cruise, determine the reduction in induced drag using the winglet and the corresponding fuel savings over a 3-hour flight. Assume the induced drag accounts for 40% of the total drag. Show solution/hide solution. Using the vortex-induced effective span model, then (121) Putting in the values gives (122) For comparison, using the linear approximation gives (123) The vortex-induced effective span model predicts an effective aspect ratio of . The ratio of induced drag coefficients is (124) This corresponds to an 8.6% reduction in induced drag. If the induced drag is 40% of the total drag, then the total drag reduction is (125) The corresponding reduction in fuel flow rate is (126) Therefore, over a 3-hour flight (10,800 s), the total fuel saved is (127) Notice that a modest 2-foot winglet at a 25 cant angle yields a 9.4% increase in effective aspect ratio (vortex-induced effective span mode), leading to an 8.6% reduction in induced drag and a 3.4% reduction in total drag. Over a typical 3-hour flight, this translates to nearly one metric ton of fuel saved, highlighting the aerodynamic and operational value of even modest winglet designs. Biplane Wing Theory While there are few biplane types of airplanes flying today, other than in general aviation, their aerodynamics are interesting and instructive because of the dual-lift system and mutual aerodynamic interference between the two wings. A simple biplane consists of two wings, typically aligned almost vertically one above the other, and spaced apart by a vertical distance called the gap, , as shown in the figure below. Each wing can be assumed to have a span and a chord , although the wings may have different spans and chords. The wings may be offset horizontally by a geometric parameter called stagger, . Another geometric parameter, the decalage angle, refers to the angular offset between the incidences, or pitch angles, of a biplane’s upper and lower wings. If the upper wing has an incidence angle and the lower wing , then the decalage angle, , is defined as the difference (128) History There are subtle differences between biplanes that have been built over time. One of the most iconic biplanes of WWI, the Sopwith Camel, employed moderate positive stagger and slight positive decalage, improving forward visibility and enhancing longitudinal trim. Thanks to its relatively short wingspan, the Camel was known for its agility and tight turning radius. The Royal Aircraft Factory SE5a, as shown in the photograph below, featured a more pronounced stagger and relatively large gap. This configuration reduced interference drag and gave it more airspeed in level flight. The Curtiss JN-4 “Jenny” was used primarily as a trainer during WWI. Its small wing gap and relatively low stagger led to greater aerodynamic interference. However, its docile handling characteristics were ideal for training, though its low performance reflected the compromise made in aerodynamic efficiency. The Fokker D.VII was more unusual because it used minimal stagger but a large gap. This design significantly reduced the interference drag, and it was one of the few biplanes that was nearly as efficient as a monoplane. A modern example of a biplane is the Pitts Special, a compact, high-performance aerobatic aircraft. Unlike early biplanes, the Pitts Special prioritizes extreme maneuverability and high agility. It features positive wing stagger and decalage, improving pilot visibility and trim stability during upright flight. The aircraft features low-aspect-ratio wings and symmetric airfoils, which facilitate rapid roll rates in both upright and inverted attitudes. In such designs, the induced drag is not minimized; instead, the aerodynamic layout is optimized for maneuverability, agility, and light control response under high-loading maneuvers. The Pitts Special exemplifies how biplane configurations persist in niche roles where their unique design features offer specific performance advantages. Other biplanes, such as the Christen (Aviat) Eagle and the Waco YMF-5, further demonstrate the continued relevance of the biplane layout in specialized applications. The Christen Eagle, also designed for aerobatics, features similar structural compactness and maneuvering performance as the Pitts. The Waco YMF-5, in contrast, is a modern reproduction of a 1930s-era luxury sport biplane, emphasizing nostalgic appeal and stable touring characteristics rather than aerobatic performance. These examples highlight the biplane configuration’s versatility in recreational and competitive aviation, despite the overwhelming dominance of monoplanes in mainstream civil and military aviation. Aerodynamics of Biplanes In a biplane configuration, each wing generates a trailing vortex system, producing a downwash field in the wake. This downwash alters the local flow field encountered by the other wing, effectively reducing its angle of attack. Consequently, the lift produced by each wing is lower than it would be if the wings operated independently in isolation. This mutual aerodynamic interference is a defining feature of biplane aerodynamics; it becomes more pronounced as the vertical gap between the wings decreases or as their stagger and decalage angles are reduced. The interference reduces the total lift for a given geometric angle of attack. It increases the induced drag because the downwash fields reinforce each other, contributing to a greater wake deflection. Consequently, the aerodynamic efficiency of the biplane is significantly lower than that of an equivalent monoplane with the same total area and span. The extent of this degradation can be quantified through theoretical models, such as Prandtl’s biplane interference formula or Glauert’s extensions of lifting-line theory, which incorporate correction factors for gap, stagger, and wing loading distribution. Understanding these interference effects is crucial for evaluating the performance of historical biplane designs and assessing the trade-offs between structural compactness and aerodynamic efficiency. Classic Theory Ludwig Prandtl was the first to address the aerodynamics of biplanes, deriving a formula for the induced drag, i.e., (129) where and refer to the lift and span of the upper wing, and and refer to the lift and span of the lower wing, respectively. The term is a biplane interference factor. While not all biplanes have the same wingspan, if and , then (130) The ratio of biplane to monoplane induced drag is then (131) In the worst case, where = 1, the biplane has four times the induced drag of an equivalent monoplane, i.e., (132) More typically, if to , then (133) This latter result highlights why biplanes, while structurally efficient and compact, suffer aerodynamically from mutual interference between the two wings. The theory shows that they have a lift-to-drag ratio less than half that of a monoplane. The same result can be determined in terms of aerodynamic coefficients, perhaps a more general and familiar approach, i.e., (134) where and are the lift coefficients of the upper and lower wings, respectively, and is the aspect ratio of each wing (assumed equal). If the wings are identical and equally loaded, so that , then (135) The induced drag coefficient of a monoplane with the same lift coefficient and aspect ratio is (136) Therefore, the ratio of induced drag for the biplane to the monoplane becomes (137) General Theory Other early aerodynamicists, such as Hermann Glauert and Max Munk, developed models to further quantify the biplane interference effect using extensions of lifting-line theory. For a monoplane wing, the three-dimensional lift-curve slope can be shown to be (138) where is the two-dimensional value (often referred to as , and accounts for the deviation of the spanwise loading from the elliptical. For a biplane with no stagger or decalage, an additional interference term appears as (139) The value of is referred to as an aerodynamic interference factor, so a penalty that increases as the gap, , decreases. For the monoplane, the induced drag coefficient is (140) The lift coefficient is (141) and so substituting gives (142) where the aspect ratio is . For the biplane, with symmetric lift sharing and span for each wing and total lift , the induced drag is (143) and the corresponding drag coefficient is (144) Using , then (145) Assuming , then (146) This latter result shows that the induced drag coefficient for a biplane is increased by a factor of four relative to a monoplane with the same aspect ratio, assuming symmetric lift sharing, equal lift coefficient, and . Munk’s Stagger Theorem Munk’s Stagger Theorem is a classical result from early biplane theory, addressing the induced drag characteristics of multiple lifting surfaces arranged with longitudinal stagger. The theorem was formulated by Max Munk in the 1920s and remains a foundational concept in theoretical aerodynamics for multi-wing configurations. It states: “For two or more coplanar lifting surfaces with the same total span and the same total circulation, the induced drag of the system is independent of the stagger between the wings, provided that the wake remains planar and aligned with the freestream.” As Munk states, this result holds under several, perhaps idealized, assumptions. The flow must be steady, incompressible, inviscid, and irrotational. All lifting surfaces are assumed to lie in the same horizontal plane, with no vertical separation between them. The total span and total bound circulation of the system are held constant. The wake shed by the lifting surfaces is assumed to remain flat and aligned with the freestream direction, and the flow is modeled using classical lifting-line theory. In lifting-line theory, the induced drag is expressed as (147) where is the bound circulation at the spanwise location , and is the downwash induced by the trailing vortex system. When the total span and circulation are held fixed, introducing a fore-aft offset between the wings, known as stagger, modifies the local circulation distribution and downwash but leaves the total integral unchanged. Because induced drag depends only on this integral, it remains invariant under idealized changes in stagger. The local induced angle of attack on each wing is affected by both its own trailing vortices and those of the other wing. The net induced angle of attack at spanwise location is (148) where arises from the downwash induced by the other wing. For a vortex filament at vertical gap and horizontal stagger , the mutual downwash is approximated by (149) As stagger increases, this induced velocity becomes more asymmetric between the wings, enhancing the induced angle of attack and, therefore, increasing induced drag. In terms of a Fourier-mode representation of the spanwise circulation, the induced drag is given by (150) where are the coefficients of the loading modes, and is a stagger-dependent interference term. Because increases with stagger, it follows that . In the non-staggered case (), the trailing vortices from the upper and lower wings are vertically aligned, and their downwash fields partially cancel each other. In contrast, stagger skews the wake and increases mutual interference, thereby enhancing the overall downwash and, consequently, the induced drag. Munk’s theorem implies that one cannot simply reduce induced drag on a biplane by staggering its wings. Although stagger changes local aerodynamic interference, the total induced drag remains the same if the total circulation and span remain the same. Consequently, modifications such as introducing vertical separation (gap), adjusting twist, or redistributing circulation spanwise are required to reduce induced drag in multi-wing configurations in a meaningful manner. It is essential to recognize the limitations of the theorem, as it applies only within the framework of lifting-line theory. In real flows, effects such as mutual wake distortion introduce deviations from the idealized assumptions. Three-dimensional wake behavior at high angles of attack can also invalidate the assumption of a flat, undisturbed wake. Nonetheless, Munk’s Stagger Theorem offers a valuable theoretical insight into the behavior of coplanar lifting surfaces and remains a benchmark in the aerodynamic analysis of biplanes and multiplanes. Decalage Decalage refers to the angular difference in incidence or geometric pitch between the upper and lower wings of a biplane. This geometric offset directly affects the lift distribution, typically increasing the lift on the wing with greater incidence. However, decalage also introduces asymmetric aerodynamic interference. The downwash produced by one wing modifies the effective angle of attack on the other in a non-uniform manner, depending on the gap, stagger, and relative circulation strengths. Consequently, the mutual influence between the wings becomes more complex, and the induced drag may increase. The interference factor , which accounts for the degradation in aerodynamic efficiency because of wing-on-wing interactions, must be carefully re-evaluated when decalage is present. In addition to altering lift sharing, decalage significantly impacts the pitching moment and longitudinal trim, and must be chosen with consideration of both aerodynamic performance and stability requirements. Design Trades with Biplanes Despite Munk’s theorem, actual biplane configurations often incorporate stagger and decalage for practical reasons that extend beyond minimizing induced drag. Stagger can improve the pilot’s overall visibility by moving the upper wing forward. It also allows for shorter and lighter interplane struts, which reduce structural weight and drag. Adjustments in stagger and decalage can also help shift the center of pressure to achieve desirable trim and stability characteristics. In addition, physical constraints can influence the choice of wing placement. Positive decalage, in which the upper wing is set at a higher geometric angle of incidence than the lower wing, is commonly used to achieve longitudinal trim or pitch stability. This configuration generates more lift from the upper wing than from the lower wing. As a result, the aerodynamic center shifts, and the aircraft gains more favorable moment characteristics for trimmed, stable flight. Aircraft with short wings and low aspect ratio experience reduced roll damping compared to longer-span designs. Roll damping arises from the differential lift generated across the wing span during a rolling rate motion, with one wing experiencing an increased angle of attack and the other a decreased one. This effect is pronounced on high-aspect-ratio wings, which generate significant aerodynamic damping in roll, e.g., a sailplane. However, the roll damping is much smaller with two short wings rather than one longer and higher aspect ratio wing of the same wing area. This characteristic is advantageous in aerobatic aircraft, such as the Pitts Special, where rapid roll response and minimal resistance to roll acceleration are essential for precise maneuvering. The two wings give the needed lift, and the low roll damping contributes directly to the aircraft’s high agility. Worked Example #5 – Biplane aerodynamic performance A biplane has two identical wings, each with span = 6 m and chord = 1.2 m. The total mass of the airplane is = 1,000 kg. The aircraft flies at MSL ISA conditions at an airspeed of = 80 mph. Assume: 1. Two-dimensional lift curve slope: = 2. 2. Span efficiency factor for monoplane: = 0.05. 3. Interference correction factor for biplane: = 0.30. 4. Effective span efficiency factor for the biplane: = 0.2. Determine the lift coefficient based on the total reference area . Compute the lift curve slope of separate monoplane wings, , and for the biplane, . Determine the induced drag coefficient for the separate monoplanes, , and for the biplane . Comment on the aerodynamic efficiency of this biplane compared to a monoplane of the same span and total area. Show solution/hide solution. The area of one wing is (151) so the aspect ratio is (152) The total reference area for the biplane is (153) The total lift required equals the airplane’s weight, i.e., (154) The airspeed is , so the dynamic pressure is (155) The lift coefficient is then (156) 2. The lift curve slope for a monoplane wing with span efficiency is (157) Substituting the values gives (158) The biplane’s lift curve slope, accounting for interference, is (159) Substituting the values gives (160) 3. The induced drag coefficient for the monoplane is (161) The induced drag coefficient for the biplane is (162) 4. Compared to two separate monoplane wings, the biplane has a lower lift curve slope ( vs. ) and a significantly higher induced drag coefficient ( vs. ). The aerodynamic penalty comes from mutual downwash interference between the wings, quantified by and . Although compact, the biplane is substantially less efficient in cruise than an equivalent monoplane. Note on Triplane Configurations A triplane, a relatively rare type of airplane, features an additional lifting surface positioned above or below the existing biplane configuration. As shown in the photograph below, the Sopwith Triplane was a British single-seat fighter aircraft designed and manufactured by the Sopwith Aviation Company during WW1. It has the distinction of being the first military triplane to see operational service. The motivation for a triplane is to increase the total wing area without increasing the span, which benefits both structural stiffness and maneuverability. However, interference effects become more severe with the third wing. In this case, each wing contributes to the downwash seen by the others, and the mutual interference becomes more complex. Theoretical treatments typically model the triplane as three lifting lines, where each line experiences downwash induced by the other two. The induced drag of a triplane can be estimated by extending lifting line theory to three wings. If each of the three wings carries an equal share of the aircraft’s weight, i.e., = and the span is , then the induced drag is approximated as (163) where is an effective span efficiency factor for the triplane configuration. The factor of 3 arises because the three lifting surfaces contribute cumulatively to the total induced drag and experience mutual aerodynamic interference. In general, (164) indicating a progressive loss in aerodynamic efficiency compared to a biplane and monoplane. In coefficient form, using a total reference area , the induced drag coefficient becomes (165) This latter result shows that while the triplane has the same functional form for as a monoplane, the span efficiency factor is significantly higher because of the added interference and downwash effects from the multiple wings. The net result is a higher induced drag penalty for the same lift coefficient and aspect ratio. The Fokker “Dr.I” Triplane, the most famous triplane, used short spans and closely spaced wings. It had an impressive roll rate and climb capability, but suffered from high drag, which limited its speed and maneuverability. Its triplane configuration reflected a specific design solution focused on turning agility in dogfights. Triplanes had a brief period of popularity during WWI but were quickly abandoned thereafter as monoplane structures improved and interference drag became better understood. Flying Wings Flying wings, also known as tailless aircraft, are a distinctive and relatively unusual class of aircraft in which the fuselage and empennage, as well as all the flight control surfaces, are fully integrated into the wing structure, thereby eliminating the need for a traditional tail. This design offers several advantages, including reduced drag, a lower structural empty weight fraction, and an increased payload fraction. The Northrop Grumman B-2 Spirit stealth bomber is a prime example of a modern flying wing design, which cruises just below the speed of sound. Its smooth, blended aerodynamic surfaces minimize drag and also reduce its radar cross-section, making it highly effective in evading enemy radar. However, the tailless flying wing configuration also introduces unique technical challenges that must be addressed to ensure acceptable flight performance, particularly in terms of its directional stability and control. Aerodynamics of Flying Wings Flying wing designs benefit from reduced parasitic drag compared to conventional airplanes of similar size and weight. Their continuous wing surfaces and carefully shaped aerodynamic profiles eliminate many of the typical sources of drag and flow interference found on traditional configurations. However, while conventional wing design has long aimed for an elliptical spanwise lift distribution to minimize induced drag, flying wings require a slightly different approach. The absence of a fuselage and tail surfaces means that the spanwise lift distribution must be tailored not only for good aerodynamic efficiency but to ensure adequate stability and control. In contrast to the elliptical spanwise lift distribution, which is ideal for minimizing induced drag on conventional wings, flying wings typically employ a bell-shaped spanwise loading to give the wing sufficient directional stability and control, as illustrated in the schematic below. This distribution concentrates more lift toward the inboard sections of the wing, reducing structural bending moments and enabling a lighter wing design. However, this comes at the cost of increased induced drag. One way to offset this penalty is to use a slightly longer wing with a higher aspect ratio, within the constraints of an allowable bending moment and wing weight, thereby regaining aerodynamic efficiency while maintaining overall structural efficiency. But creating such a bell-shaped spanwise loading, which is achieved through suitable variations in wing chord and twist, serves another important purpose in that it establishes upwash regions over the outer wing panels. This upwash generates forward-acting lift vectors in those regions, effectively producing negative induced drag. It is this forward-acting lift force at the wingtips that provides the flying wing with essential directional (yaw) stability and control. In addition, it leads to proverse yaw in response to aileron or flaperon inputs. a feature critical for flying wings, which lack a vertical tail and rudder. Bell-Shaped Loading The aerodynamic consequences of the bell-shaped spanwise loading require further exposition. A starting point is the extended spanwise loading solutions discussed by Prandtl in 1933, which stemmed from Munk’s original formulation of the lifting-line theory, who concluded that if the wing span or aspect ratio is not a design constraint, other forms of the spanwise lift distribution may help optimize the aerodynamic and structural efficiency of a wing. However, he makes no mention of any application to tailless airplanes or flying wings. Prandtl introduced a more general spanwise circulation loading function given by (166) where = 1/2 gives the classic elliptical loading and = 3/2 gives a “bell-shaped” loading. The bell-shaped loading distribution has a higher peak near the wing root compared to the elliptical loading. It then quickly decreases towards the wingtip, where the lift over the outer part of the wing is much lower, thereby giving it the characteristic “bell shape,” as shown in the figure below. Therefore, the two idealised spanwise circulation loadings of interest for further comparative analysis are the classical elliptical form (as a reference) with = 1/2, i.e., (167) and the bell-shaped form with = 3/2, i.e., (168) where and are the respective loadings (circulation values) at mid-span. A standard symmetric wing of semi-span is assumed. Downwash There is a non-uniform downwash over the wing with the bell-shaped spanwise loading (compared to the uniform downwash obtained with the elliptical loading), with an upwash being produced over the outer part of the wing. For the well-known elliptical span loading, the downwash, , is constant across the span, i.e., (169) For an untwisted elliptical planform with uniform downwash, it also results in a constant induced angle of attack and a constant local lift coefficient across the span, representing the minimum induced drag configuration for the wing. The downwash corresponding to the bell-shaped loading can be obtained from the lifting-line method discussed previously, which expresses the circulation as (170) where the coefficients can be determined from the prescribed loading. The bell spanwise loading distribution is (171) Using the standard transformation with the lifting line theory that where , then . Furthermore, using the trigonometric identity that , then only the first and third harmonics appear in the loading distribution, i.e., (172) Matching coefficients gives (173) For any general circulation distribution, then the downwash is (174) and substituting for the bell-shaped loading gives (175) Using the trigonometric identity that , then , and so (176) Because , then (177) This result means that the downwash with the bell-shaped spanwise loading is a maximum at the wing root and becomes an upwash at the tip outboard of = = 0.707, as shown in the figure below. Wing Lift The lift on a wing, according to the Kutta-Joukowsky theorem, is given by (178) The standard lifting line transformation that can be used, where . In this case, the total wing lift is (179) Substituting for the elliptical loading gives (180) In terms of the coordinate, the bell-shaped spanwise circulation distribution is given by (181) Substituting this distribution into the lift equation gives (182) If the elliptically loaded and bell-loaded wings produce the same lift, then , so their peak values of circulation are related by (183) Induced Drag Distribution Because of the low geometric angles and upwash generated by the bell-shaped spanwise loading over the tip regions, they produce a forward-acting lift vector, resulting in negative induced drag. The downwash has been previously derived in Eq. 177, so the form of the local induced drag coefficient follows directly. For the elliptical loading, then (184) where , and for the bell loading (185) where is the local wing chord. As shown in the figure below, for which a rectangular planform with = constant is assumed, the induced drag becomes negative outboard of = 0.707. Notice that when using Eq. 183, then (186) It is the forward-acting lift vector at the wing tips, as shown for the sectional level in the figure below, that proves necessary to give stability and control to a flying wing. However, the net drag on the wing is still positive for a given span and aspect ratio, the bell-shaped form of wing loading producing higher overall induced drag compared to a wing with elliptical loading, i.e., the spanwise loading factor, , will be greater than zero, as will be shown later. Lift Coefficient Distribution For a rectangular planform where the chord and section lift-curve slope are constant, the local section lift coefficient is directly proportional to the bound circulation. Consequently, the spanwise lift-coefficient distributions corresponding to the two loadings are (187) Normalizing by the mid-span value gives the purely geometric forms as where . Therefore, the bell-shaped loading will generally have the lower values at the wing tips than the elliptical loading for the same wing planform. This distribution gives a flying wing good stall qualities because the inboard sections stall first, allowing the wing to maintain its directional stability while the flaperons remain effective. Wing Twist Distribution It is necessary to use a planform shape and nonuniform wing twist to produce the bell-shaped form of spanwise loading, with significant nose-down twist being needed over the outer span. The general form of the nonlinear twist needed to create the bell-shaped loading is where and depend on the details of the wing planform as well as the airfoil section as it affects the zero-lift angle of attack. The approach to finding the needed twist on the wing to achieve the bell-shaped loading is straightforward. With the bell circulation distribution, then where . If a rectangular wing of unit chord is assumed, then . The section lift coefficient is and in terms of section angles, then where is the geometric angle of attack or pitch measured relative to the reference or datum, . Therefore, the needed twist distribution can be expressed as Using = 2/rad for the sectional lift curve slope, then For the bell-shaped spanwise loading, then , so the needed twist is where the lift coefficient, , can be interpreted as the design lift coefficient for the wing, such as in cruise flight. This equation represents the unique geometric twist that creates both the bell spanwise loading and its unique downwash distribution for a wing of constant chord, as illustrated in the figure below. If the wing has taper and/or camber, which is typical, the lifting-line problem and the needed spanwise twist can be solved numerically; however, the fundamental approach remains the same. Wing Bending Moments In his 1933 paper, Prandtl minimized the integral of the square of the bending moment (an energy proxy) while maintaining a constant total lift. The key argument in the use of other than an elliptical wing loading is to minimize wing bending and hence reduce wing weight, although at the expense of some loss of aerodynamic efficiency. Prandtl remarked in his paper that one may “regard the integrated bending moment as a measure of spar weight.” A reasonable assumption, therefore, is that wing weight is proportional to the characteristic wing bending moment. Later, R.T. Jones made similar arguments with similar outcomes. The wing root bending moment is obtained by integrating the aerodynamic loading from the tips to the root, i.e., using For the elliptical loading, where , then For the bell-shaped loading, where , then To produce the same total lift, the peak circulation of the bell-shaped loading must be , as shown in the previous figure. Substituting these relationships gives so the ratio of the bending moments is This result indicates that the bell-shaped loading yields a lower overall spanwise bending moment and a 20% reduction in the root bending moment compared to the elliptical loading, for the same total wing lift, as illustrated in the figure below. This outcome has significant implications for the design of a wing with a bell-shaped spanwise loading because it can be lighter than if it had the elliptical loading with some increase in induced drag for the same span. Total Induced Drag The induced drag for a wing with the bell-shaped spanwise loading distribution can now be determined. For the elliptical loading, where , then For the bell-shaped loading, where , the induced drag is To ensure equal total lift, the peak circulation values must be scaled using . Substituting gives Therefore, the ratio of induced drag values is Because, in general, the induced drag coefficient is and because = 0 for the elliptic loading, then it can be concluded that = 1.219. This result shows that the bell-shaped spanwise loading produces 21.9% more induced drag than the elliptical distribution, keeping in mind that this is for the same total lift and wing span. This is an expected outcome based on the significant spanwise deviations of the bell-shaped loading from the elliptical ideal. Design Application The primary reason for using a bell-shaped spanwise loading is that for flying wings or other tailless airplane configurations, the upwash at the wing tips helps give yaw stability and eliminates adverse yaw effects. Ideally, this goal should be achieved without either aerodynamic or structural weight penalty. Therefore, a final step is to determine a new wing span with a bell-shaped spanwise loading that yields the same structural root bending moment as the elliptical loading, while maintaining the same total wing lift, wing area, and without as much of an induced drag penalty. From Eq. 197, it is apparent that the wing root bending moment scales with the square of the semi-span, i.e., , or equivalently . From the previous results, it was shown that the root bending moment for the bell-shaped loading is 80% of that for the elliptical loading, i.e., . Therefore, the span of a bell-loaded wing that produces the same root bending moment as an elliptically loaded wing can be increased by which corresponds to an 11.8% increase in wing span. Consequently, the wing aspect ratios are related by if the wing area is held constant. The aspect ratio scales with the square of the span, so for the bell-shaped loading, then and for an 11.8% increase in span, then Therefore, the aspect ratio increases by 25% with the wing area being held constant. In this case, the induced drag coefficients, which are both for the bell-shaped loading, will be related using Eq. 206 by This result shows that by adopting a bell-shaped spanwise loading and not allowing the structural root bending moment to increase over that obtained with the elliptical loading, the wing span can be increased by 11.8% (for the same wing area), resulting in a 25% increase in aspect ratio and a 2.5% reduction in induced drag. It will be apparent that, in general, then as shown in the plot below. Such is the powerful advantage of the aspect ratio as a means of offsetting less-than-ideal spanwise loading distributions on a wing. This is an interesting and potentially important result in airplane design, not limited to flying wings. When geometric span alone constrains the wing design, the elliptical spanwise loading remains optimal. However, the main reason for using an alternative spanwise loading, in this case, the bell-shaped loading, is the upwash it produces at the wing tips, which inclines the lift vectors forward rather than aft, as previously described. This condition produces directional stability on a tailless flying wing and also gives desirable proverse yaw effects with the application of flaperons or ailerons. This characteristic is crucial for the success of flying wings. It is the primary advantage of designing the wing to give a bell-shaped spanwise loading, despite the penalty of a higher induced drag factor, , of 0.219. But, if the wing span is not a design constraint, then the induced drag penalty from the use of a bell-shaped spanwise loading, or any other non-ideal loading for that matter, can be offset by using a wing of modestly higher span and aspect ratio. The consideration of increased wing weight, however, still remains an issue. It has been demonstrated that the root bending moment is proportional to the product of lift and span. However, a 12% increase in span incurs no net wing weight penalty with the bell-shaped loading and reduces the induced drag as a result of the higher aspect ratio. Further increasing the wing span, say by 20%, increases the bending moment by the same fraction for the same wing lift. The weight of a wing grows roughly with the cube of its span, an outcome of the square-cube law. Taking also the lower inboard bending moment achieved with the bell loading (i.e., ), a first-order wing weight estimate is Even after benefiting from the bell loading’s lower inboard bending moment, when the span grows too much, the cubic term in Eq. 213 produces a 40% increase in wing weight. Therefore, when both drag and weight are considered, the optimum wing design typically involves using only enough extra wing span to make the structural cost aerodynamically worthwhile. Introduction to Lifting Surface Theory The progression from lifting line theory to lifting surface theory and eventually to incorporating thickness represents a series of refinements in aerodynamic modeling using vortex methods. Each step increases the model’s fidelity by including additional aspects of three-dimensional flow around finite wings. Lifting line theory is the simplest of these models, in which the wing is represented as a single line of bound circulation located at the quarter-chord along the span. While straightforward and analytically elegant, lifting line theory does not account for chordwise loadings or the presence of wing thickness and camber, limiting its applicability to more complex wing geometries, including those with sweep or non-planar geometries. Lifting surface theory, also known as the vortex lattice or vortex panel method, extends this framework by treating the wing as a surface. The circulation is now distributed across both the span and the chord, denoted by , allowing for a more realistic spatial representation of aerodynamic loading. As shown in the figure below, this approach utilizes distributed vortex sheets in the form of quadrilateral panels to model both spanwise and chordwise variations in lift. The induced downwash at a control point at is obtained from the net effect of all vortex elements using where is an influence coefficient matrix associated with the unit-strength vortex element at . Each entry is a scalar representing the vertical velocity induced at the control point by a unit-strength vortex element at , and the matrix is assembled from these scalar entries. The actual strengths of the vortex elements are then solved for by satisfying the flow tangency (the no-penetration condition) at control points located at the centroids of each panel. This condition is expressed as where is the net velocity at the surface and is the local unit normal vector. The net velocity is composed of the freestream and induced components, i.e., Substituting into the boundary condition gives In many practical implementations, such as the classical vortex lattice method, the surface is assumed to lie approximately in the – plane, and the unit normal vector is taken to point in the direction. Therefore, the boundary condition reduces to a scalar equation involving the vertical component of velocity, i.e., For small angles, the boundary condition is simplified to where is the local induced angle of attack resulting from the downwash. This relation is based on the small-angle approximation and the linear relationship Therefore, the equations to be solved are Solving this set of linear simultaneous equations yields the vortex strengths over the lifting surface, from which the spanwise and chordwise loading distributions can then be determined. Solving the linear system typically involves assembling the influence matrix and applying boundary conditions to determine using matrix inversion or iterative solution methods. Lifting surface theory significantly improves the fidelity of aerodynamic predictions, particularly for wings with moderate or low aspect ratios, but it continues to assume a thin airfoil shape. However, as also illustrated in the figure above, the lifting surface model can be extended to include wing thickness and camber. Wing thickness introduces an additional layer of aerodynamic complexity because a wing with finite thickness alters the chordwise and spanwise pressure distributions, and hence the lift distribution and induced drag. Alternatively, source distributions can be added to the thin vortex sheet representation to model thickness effects. In this case, the total potential is then expressed as Here, represents the bound vortex strength, as before, and denotes the distributed source strength over the surface . The quantity is the distance from the field point to the source point, defined as Such lifting surface models, with thickness and camber, form the foundation of many modern aerodynamic “panel methods” for analyzing arbitrary aircraft configurations. While classical lifting surface theory models the wing as a distribution of bound vortices, modern panel methods incorporate vortex and source distributions to model thickness and camber effects. Summary & Closure The lifting line theory provides a foundational framework for analyzing the aerodynamic behavior of finite wings. By modeling the wing as a spanwise distribution of bound circulation and accounting for the induced downwash generated by the trailing vortex system, the theory allows for estimating essential aerodynamic quantities such as the lift distribution and induced drag Although the theory relies on simplifying assumptions, including high aspect ratio, small angles of attack, and inviscid, incompressible flow, it represents the essential physics governing three-dimensional wing aerodynamics with reasonable accuracy. It remains a core analytical tool in aircraft design because of its simplicity, analytical tractability, and predictive capability. For multiple lifting surfaces, such as biplanes, the theory must be extended to account for aerodynamic interference between the wings. Biplane theory models each wing as a lifting line and includes the mutual downwash that one wing induces on the other. This coupling modifies the local angle of attack and alters the spanwise circulation distribution, resulting in increased induced drag compared to an isolated wing. The aerodynamic penalty is commonly expressed through an interference factor that quantifies the degradation in performance relative to a monoplane of equivalent span and total lift. Parameters such as gap, stagger, decalage, and chord ratio significantly influence the interference and, therefore, affect the overall efficiency of the wing configuration. To better represent three-dimensional effects and more general wing geometries, the lifting line theory can be extended to a lifting surface theory, which treats the wing as a two-dimensional surface rather than a line. This approach discretizes the wing into a grid of panels or vortex elements, allowing for the simultaneous resolution of both spanwise and chordwise variations in vorticity. Lifting surface methods, such as vortex lattice models, account for non-uniform chord, sweep, taper, and airfoil thickness because they enforce the flow tangency condition across the entire wing surface. These models are intermediate between simple analytical theories and full computational fluid dynamics (CFD) simulations. 5-Question Self-Assessment Quickquiz For Further Thought or Discussion How might the assumptions used in lifting line theory (inviscid, incompressible flow) affect the quality of predictions of actual wings? Why does the elliptical circulation distribution minimize induced drag, and what are the practical challenges in achieving it for non-elliptic wing planforms? How does increasing a wing’s aspect ratio improve aerodynamic efficiency but introduce structural and design challenges? How do wings with rectangular or linearly tapered shapes deviate from elliptic loading, and how does this impact induced drag and overall wing performance? Why is the lifting line theory inapplicable in supersonic flows, and how are these issues addressed in modern aerodynamics? How might winglets be modeled within the framework of the lifting line theory? What types of modifications, if any, may be required? How does wing twist affect wing performance and allow for loading optimization under specific flight conditions? What are some examples of aircraft that achieve near-elliptic wing loading? How is the lifting line theory or its derivatives applied in other contexts, such as rotor blades, wind turbines, or hydrofoils? Other Useful Online Resources To understand more about the aerodynamics of finite wings and wing theory, explore some of these online resources:” Finite Wings – Lifting Line Theory” from the Postcard Professor.”Prandtl’s Lifting Line Theory” by Professor Sid G. “Lifting Line Theory” by Professor Lakshmi Sankar at GaTech. Lecture notes from MIT OpenCourseWare introduces Prandtl’s lifting line theory, including geometry and fundamental definitions. A comprehensive overview of lifting line theory, including its principles, derivation, and applications. An explanation of Prandtl’s lifting line theory, focusing on three-dimensional potential flow, provided by the University of Cambridge. Lecture notes from Stanford University discussing finite wing theory and related details, including lifting line theory. An educational website offering resources on various aerodynamics topics, including lifting line theory. A GitHub repository containing Python code implementing Prandtl’s lifting line theory for aerodynamic analysis. Lecture notes from MIT OpenCourseWare focusing on force calculations within lifting line theory. A tutorial example providing practical insights into applying lifting line theory. Lecture notes offering a detailed explanation of lifting line theory concepts. A collection of MATLAB scripts and functions implementing lifting line theory for aerodynamic analysis. The addition of a winglet will cause the tip vortex to form at the tip of the winglet. ↵ Prandtl's original publication on the lifting-line theory was "Über Tragflügeltheorie" ("On Wing Theory"), which was published in 1904 in the proceedings of the Third International Congress of Mathematicians in Heidelberg, Germany. This paper introduced the fundamental ideas of the lifting-line theory. ↵ Hermann Glauert was a British aerodynamicist and a pioneer in theoretical aerodynamics. He worked at the Royal Aircraft Establishment (RAE) and is best known for his contributions to thin airfoil theory and lifting line theory. His influential book, "The Elements of Aerofoil and Airscrew Theory," remains a classic. ↵ Glauert systematically presented this theory in his 1926 book The Elements of Aerofoil and Airscrew Theory. ↵ This transformation is analogous to the one used in the thin airfoil theory. ↵ There is also a biplane wing theory, which is considered later in this chapter. ↵ R. T. Jones, "The Spanwise Distribution of Lift for Minimum Induced Drag of Wings Having a Given Lift and a Given Bending Moment," NACA TN-2249, 1950. ↵ Proverse yaw occurs because the increased lift and reduced drag on the down-going aileron facilitated by the upwash eliminate the adverse yaw effects that would typically occur with a conventional airplane. ↵ Prandtl, L., "Über die Tragfähigkeit der Flügel (On the Lift of Wings)," Zeitschrift für Flugtechnik und Motorluftschiffahrt, 1933. See also: Hunsaker, D. F., and Phillips, W., "Ludwig Prandtl’s 1933 Paper Concerning Wings for Minimum Induced Drag, Translation and Commentary," AIAA 2020-0644. AIAA Scitech 2020 Forum. January 2020. ↵ Jones, R. T., "The Spanwise Distribution of Lift for Minimum Induced Drag of Wings Having a Given Lift and a Given Bending Moment," NACA Technical Note 2249, Dec. 1950. ↵
2629	https://medium.com/mathematica-stories/why-is-2017-an-interesting-number-dc3cfc135853	Sitemap Open in app Sign in Sign in ## Mathematica Stories · Follow publication Dedicated to the exposition of mathematics in and as narrative. Follow publication Why is 2017 an interesting number? David Radcliffe 5 min readDec 31, 2016 Some species of cicada emerge once every 17 years. It is believed that this long life cycle evolved in response to predators. Since 17 is a prime number, a predator with a shorter life cycle cannot reliably prey on the cicadas. At the dawn of the new year 2017, let us reflect that 2017 is also a prime number. Perhaps we can take a lesson from the cicadas, and resolve this year to break out of our old destructive cycles. In this article, I will discuss some of the interesting mathematical properties of the number 2017. 2017 is prime The first property to note is that 2017 is a prime number. This means that 2017 is not divisible by any number except for 1 and itself. (At this point, I am obligated to mention that 1 is not considered to be prime.) If a number N >1 is not prime, then one of its factors is less than or equal to the square root of N. So if we wish to prove that N is prime, it suffices to show that N is not divisible by any prime number up to the square root of N. For 2017, it suffices to check for divisibility by 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, and 43. We can stop at 43, because the next prime 47 is greater than the square root of 2017. Here is a short Python script to check if a number is prime. Keep in mind that this is not an efficient method for large numbers, but it works fine for small numbers like 2017. Curiously, 2017π and 2017e both yield prime numbers when rounded to the nearest integer. 2017π ≈ 6337 and 2017e ≈ 5483. 2017 is a sum of two squares Since 2017 is a prime number of the form 4k+1, it can be written uniquely as the sum of two squares, by Fermat’s Theorem on sums of two squares. In this case we have 2017 = 44² +9². We can factor 2017 in the Gaussian integers as (44+9i)(44–9i). A Gaussian integer is a complex number of the form a + bi, where a and b are integers, and i is the imaginary unit. This shows that the property of being prime depends on the number system, not just the number. 2017 is prime in the ordinary integers, but it is not prime in the Gaussian integers. Since 9 is a square, we can write 2017 as the sum of a square and a fourth power: 2017 = 44² + 3⁴. Primes of this form are known as Friedlander-Iwaniec primes, because Friedlander and Iwaniec proved in 1997 that there are infinitely many primes of this form. We can also write 2017 as the sum of three cubes: 2017 = 11³+7³+7³. Heath-Brown proved in 2001 that there are infinitely many primes of the form x³+2y³. 2017 is a lazy caterer number This picture shows a pancake that has been divided into 7 pieces by 3 straight cuts. How many pieces can be made using N straight cuts? This question can be answered by counting the number of additional pieces created by each cut. In the beginning, there is only 1 piece. The first cut adds 1 piece, the second cut adds 2 pieces, the third cut adds 3 pieces, and so on. So after 3 cuts, there are 1 + 1 + 2 + 3 = 7 pieces. Get David Radcliffe’s stories in your inbox Join Medium for free to get updates from this writer. But what if we wanted 2017 pieces? It turns out that we can cut a pancake into 2017 pieces with exactly 63 cuts: 1 + 1 + 2 + 3 + 4 + 5 + … + 63 = 2017 The numbers 1, 2, 4, 7, 11 representing the number of pieces after N cuts is known as the lazy caterer’s sequence. 2017 is the 63rd lazy caterer’s number. An odd-number spiral Arrange the odd numbers in a square spiral, as shown below. If you start at the 1 in the center and move down 16 cells then you will land on the number 2017. Neat, huh? (How this is related to the lazy caterer’s sequence?) Here is another question related to arranging numbers in a grid. Suppose that we want to arrange the numbers 1–16 in a 4×4 grid so that the numbers are increasing along each row, each column, and both diagonals. Would you believe that there are 2017 ways to accomplish this? [Source] Pattern-avoiding permutations There are 5040 ways to arrange the numbers 1–7. To see this, note that there are 7 choices for the first number, 6 choices for the second number, 5 choices for the third number, and so on. Multiplying the number of choices at each stage, we get 7×6×5×4×3×2×1 = 5040. Mathematicians use the word permutation to refer to arrangement of the numbers 1 through N. A permutation is said to contain the pattern 123 if it contains three consecutive digits in ascending order. For example, the permutation 351462 contains the pattern 123 because 1 < 4 < 6, but 351642 does not contain this pattern. Mathematicians are very interested in counting permutations that avoid various patterns. It turns out that there are 2017 permutations of 1–7 that avoid the pattern 123. Some additional properties of 2017 2017 is also prime when interpreted in base 8. (2017+1)/2 and (2017+2)/3 are prime. There are 2017 solutions to \|w-x\| = \|x-y\| in integers between 0 and 36. 2017 is a palindrome in base 31 (232) and base 32 (1V1), and it is the smallest number that is a palindrome in both bases. 2017 = (10×9×8×7×6) / (5+4+3×2) + 1. [Source] There are 2017 horizontally convex polyominoes with 8 cells. [Source] 2017 remains prime when 7 is inserted anywhere (27017, 20717, 20177). The sum of all odd primes up to 2017 is also prime. The set {1, 2, …, 25} has 2017 three-element subsets with no common factor. There are 2017 seventh powers which have 27 digits. Mathematics Math ## Published in Mathematica Stories 210 followers ·Last published Jan 7, 2017 Dedicated to the exposition of mathematics in and as narrative. ## Written by David Radcliffe 365 followers ·260 following I am an independent software developer and former college math instructor. Still trying to find the answers to life's persistent questions. Responses (1) Write a response What are your thoughts? Cliff Raymond Jan 17, 2017 ``` Interesting stuff, but I’m curious why you used the term “odd prime” in number 8 of the additional properties. There’s no such thing as an even prime, is there? ``` More from David Radcliffe and Mathematica Stories David Radcliffe ## Maintaining State in Erlang One of the distinguishing features of Erlang is that variables do not vary. A variable can only be assigned once during a function… Aug 25, 2018 3 In Mathematica Stories by Tiago Matos ## Faster Sieve of Eratosthenes About a decade ago I was participating in a lot of programming challenges and one of those challenges was posted on a programming forum… Dec 20, 2016 124 In Mathematica Stories by Colin Howard ## We Need More Films like “Hidden Figures” The film brings to light three Black female mathematicians as the unacknowledged reason NASA won the Space Race. Jan 7, 2017 22 In Mathematica Stories by αλεx π ## Some of 2016 books I’ve read several 2016 book reviews (one of them by Alvaro Videla) and thought I have read several books that have influenced my life… Jan 1, 2017 21 See all from David Radcliffe See all from Mathematica Stories Recommended from Medium In Cantor’s Paradise by Kasper Müller ## Numbers the Way God Intended: The Surreal Numbers Are we using the wrong number system? Aug 4 464 11 Saurav Mandal ## Do Hard Things if You Want an Easy Life The one skill that changes everything Jun 13 6.8K 225 In ThinkArt by Nnamdi Samuel ## Never Use a Calculator for Square Roots Again Square Roots Made Ridiculously Fast — 5 Seconds Aug 7 765 18 Keith McNulty ## Possibly The Most Useful Beast In The History of All Mathematics The logarithm was an invention which unlocked incredible untapped computational potential Jun 30 787 11 In by Worthy Things Jun 28 754 11 In Science Spectrum by Anshul Sanghi Aug 5 273 5 See more recommendations Text to speech
2630	https://gamedev.stackexchange.com/questions/72528/how-can-i-project-a-3d-point-onto-a-3d-line	Skip to main content How can I project a 3D point onto a 3D line? Ask Question Asked Modified 3 years ago Viewed 47k times This question shows research effort; it is useful and clear 29 Save this question. Show activity on this post. Let's say I have a line defined by two points, A and B, both in the form (x, y, z). These points represent a line in 3D space. I also have a point P, defined in the same format, that isn't on the line. How would I calculate the projection of that point on to the line? I'm aware of how to do this in 2D but 3D seems to have bugger all resources on it. mathematics geometry Share CC BY-SA 3.0 Follow this question to receive notifications edited Mar 27, 2014 at 15:19 user1430 asked Mar 27, 2014 at 11:36 EndOfTheZonersEndOfTheZoners 29111 gold badge33 silver badges33 bronze badges 0 Add a comment \| 4 Answers 4 Reset to default This answer is useful 44 Save this answer. Show activity on this post. You simply need to project vector AP onto vector AB, then add the resulting vector to point A. Here is one way to compute it: ``` A + dot(AP,AB) / dot(AB,AB) AB ``` This formula will work in 2D and in 3D. In fact it works in all dimensions. Share CC BY-SA 3.0 Follow this answer to receive notifications answered Mar 27, 2014 at 11:51 sam hocevarsam hocevar 24k22 gold badges6565 silver badges9595 bronze badges 5 thank you mr Sam - how did you derive the above formula? – BenKoshy Commented May 4, 2017 at 5:31 3 I did not derive it, it is a well known formula you can find in many handbooks. – sam hocevar Commented May 4, 2017 at 12:13 Is there a sample on how to write that in a programming language like C++? – Vinicius Rocha Commented Jul 5, 2019 at 14:47 1 @ViniciusdeMeloRocha dot would be a.xb.x+a.yb.y+a.zb.z ... everything else is as straightforward as per-coordinate operation between vectors. – Ocelot Commented Aug 12, 2019 at 1:26 2 Sort of obvious, but maybe worth mentioning that the scalar dot(AP,AB) / dot(AB,AB) represents a parameterization along the AB vector. So if the points actually define a line segment, you can quickly determine if the projection lies within the segment (or maybe you need to clamp the projection to one of the endpoints) by comparing the scalar to range [0,1]. – mcmcc Commented Jun 10, 2021 at 15:41 Add a comment \| This answer is useful 7 Save this answer. Show activity on this post. Here's a fast and easy way to do it in python: ``` from numpy import def ClosestPointOnLine(a, b, p): ap = p-a ab = b-a result = a + dot(ap,ab)/dot(ab,ab) ab return result ``` Use floats; If your vectors contain integers the division will be an integer division, and the results will be incorrect. Share CC BY-SA 3.0 Follow this answer to receive notifications edited Mar 14, 2018 at 18:05 CommunityBot 1 answered May 9, 2016 at 23:01 caleb c.caleb c. 17111 silver badge11 bronze badge 2 1 /dot(ab,ab) is redundant – Waldo Bronchart Commented Jul 4, 2017 at 23:20 4 Answering to Waldo, /dot(ab,ab) is ONLY redundant if ab has norm 1, so that dot(ab,ab) = 1. So unless you are 100% sure that is the case and you are looking for super hyper performance (assuming this function would have to be run 1M times / second), don't remove it... – Ginés Hidalgo Commented Mar 27, 2020 at 15:56 Add a comment \| This answer is useful 2 Save this answer. Show activity on this post. Just for an explanation about the formula from Sam Hocevar: If A,B are on the line, then the vector u⃗ =AB→/∥∥∥AB→∥∥∥ is the unit vector for this line (don't forget that ∥∥∥AB→∥∥∥=AB→⋅AB→−−−−−−−√ or respectively AB→⋅AB→=∥∥∥AB→∥∥∥2). The line equation follows: k⋅u⃗ +A for any k scalar. Then when projecting P on this line, the projected point I is defined as perpendicular to the line's unit vector, that is: IP→⋅u⃗ =0 and I=u⃗ ⋅∥∥∥AI→∥∥∥+A Since projecting P on the line is the dot product of AP→ and AB→, we can compute: ∥∥∥AI→∥∥∥=AP→⋅AB→/∥∥∥AB→∥∥∥ since AP→=AI→+IP→ and IP→⋅u⃗ =IP→⋅AB→=0, then: AP→⋅AB→=AI→⋅AB→=AI→⋅u⃗ ∗∥∥∥AB→∥∥∥=∥∥∥AI→∥∥∥∗∥∥∥AB→∥∥∥ That means that only the colinear component of AP→ to AB→ is participating to the scalar result Which finally gives: IxIyIz=AxAyAz+⎛⎝⎜Px−AxPy−AyPz−Az⎞⎠⎟⋅⎛⎝⎜Bx−AxBy−AyBz−Az⎞⎠⎟⋅⎛⎝⎜uxuyuz⎞⎠⎟/∥∥∥AB→∥∥∥ This is the formula from Sam above. Share CC BY-SA 4.0 Follow this answer to receive notifications edited Oct 5, 2021 at 12:56 answered Oct 1, 2021 at 9:30 xryl669xryl669 12122 bronze badges Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. on this one (i cannot cooment) ``` ap = p-a ab = b-a result = a + dot(ap,ab)/dot(ab,ab) ab return result ``` If you first make the "versor" of ab you do not need dot(ab,ab) right? or you just do the normal somehow you can rewrite as: ``` ap = p-a ab = normal(b-a) result = a + dot(ap,ab) ab return result ``` (I learned it with versors (normalized vectors) so was kind of weird. Share CC BY-SA 4.0 Follow this answer to receive notifications edited Jul 29, 2022 at 23:07 answered Jul 29, 2022 at 23:02 Pablo SalasPablo Salas 122 bronze badges 2 This looks more like a question, not an answer. Try posting it as a question instead if you'd like feedback on whether you can do away with dot(ab,ab) this way. (Hint: consider how would you compute the versor without computing this dot product under the hood, or an equivalent/more expensive operation) – DMGregory ♦ Commented Jul 30, 2022 at 0:56 This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review – liggiorgio Commented Jul 30, 2022 at 10:14 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions mathematics geometry See similar questions with these tags. 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2631	https://lifeinfreshwater.net/freshwater-bryozoans/	Freshwater bryozoans or moss animals - LIFE IN FRESHWATER LIFE IN FRESHWATER Macro photography of aquatic insects and other freshwater invertebrates Freshwater bryozoans or moss animals Freshwater bryozoans or moss animals (phylum Bryozoa, class Phylactolaemata) Feeding: Moss animals use a crown of tentacles to filter algae, detritus and microscopic organisms from the water. Habitat:Bryozoans occur in both still and running waters. Their presence indicates good water quality. Movement:Most bryozoans are sessile and immobile, but some colonies are able to slowly glide on the substrate. Size:Individual zooids are about 0.5 mm long. Life cycle:The life cycle includes both sexual and asexual reproduction. Sexually produced “larvae” undergo metamorphosis into adults, which grow the new colonies by budding clones of themselves. Moss animals also reproduce asexually by creating stratoblasts (capable of surviving freezing or desiccation). Stratoblasts are produced by adult zooids in the fall, after which the colony dies. Introduction:The basic body plan of bryozoans superficially resembles marine corals or freshwater hydra. In fact, the similarity is only apparent. They are unrelated and each one belongs to different phyla. What seems to be an individual animal is actually a colony of zooids (not polyps as in corals and hydra). The bryozoan zooid is a complex animal with cell layers, tissues and organs. They are capable of independent feeding, digestion and reproduction. Zooids share certain tissues and fluids which unify the colony. Therefore it is impossible to distinguish precisely, where one zooid ends and the adjacent one begins. Bryozoans use a food-gathering structure (lophophore) bearing the crown of ciliated tentacles to filter the water and trap algae, small particles of detritus, diatoms and other microscopic organisms. Mouth is situated at the base of tentacles. Food is processed in a stomach, intestine, and passes out of the body through anus. Digestive tract is U-shaped, with anus located below the ring of tentacles. All members of the colony are genetically identical clones, produced by asexual reproduction. The colony grows and expands by budding new zooids from parental tissues. If a piece of bryozoan colony breaks off, the part (with at least one living zooid) drifts in the current until it encounters a solid object, to which the zooid may adhere. If conditions permit, zooid will continue to grow by creating buds and establishes a new colony. Fission is similar process of reproduction, which occurs in gelatinous colonies. The body wall is constricted by muscular contractions and divided parts of the colony move slowly in different directions. Another method of asexual reproduction is forming small encapsulated structures called statoblasts. Dormant statoblasts remain viable for long periods and can survive variable conditions of freshwater environments. These seed-like masses of cells can be transported across the land by animals, wind, or currents. When they reach appropriate environment (or suitable conditions return into the habitat), stratoblasts develop into zooids and start to form new colonies. Bryozoans are also capable of sexual reproduction. All members of the colony are hermaphrodites producing both eggs and sperm. Cristatella mucedo: Plumatella fruticosa: Plumatella repens:
2632	https://www.khanacademy.org/math/precalculus-tx/xecf67014514b60db:trigonometric-functions/xecf67014514b60db:reference-angles-and-standard-position/v/identifying-reference-angles-of-positive-angles-in-standard-position	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Top Voted
2633	https://math.stackexchange.com/questions/1212283/what-makes-the-cauchy-schwarz-inequality-so-important	Skip to main content What makes the Cauchy-Schwarz inequality so important? Ask Question Asked Modified 10 years, 1 month ago Viewed 7k times This question shows research effort; it is useful and clear 18 Save this question. Show activity on this post. The Cauchy-Schwarz inequality is (a⋅b)2≤\|a\|2\|b\|2. Why is this considered such an important inequality: to quote my textbook it's "one of the most important inequalities in all of mathematics". But why? Doesn't it just immediately follow from the definition of the dot product: a⋅b=\|a\|\|b\|cos(θ)? And even if you define the dot product differently, like maybe a⋅b=a1b1+a2b2+..., it still doesn't seem all THAT important to me. So what makes this particular inequality so important/ interesting? real-analysis linear-algebra analysis Share CC BY-SA 3.0 Follow this question to receive notifications edited Mar 30, 2015 at 3:04 Cameron L. Williams 30.5k55 gold badges6262 silver badges113113 bronze badges asked Mar 30, 2015 at 1:49 user227406user227406 18111 silver badge44 bronze badges 10 1 For one, it essentially helps to generalize the triangle inequality that is so often used in analysis. Also, in dealing with statistics, it can be used to show that the linear correlation between two random variables is between −1 and 1. – Brent Commented Mar 30, 2015 at 2:11 1 Inequalities (including this one) are important in analysis, as they help bound things. – Sujaan Kunalan Commented Mar 30, 2015 at 2:26 OK. I guess I'll just have to wait until I get to real analysis to see where this inequality comes in handy. – user227406 Commented Mar 30, 2015 at 2:29 2 The notion of the dot product (or generally, any inner product) is more general than any reasonable notion of an "angle" that you might have. The fact that a⋅b=\|a\|\|b\|cosθ is not a definition. The definition is a1b1+a2b2+⋯, and the cosine formula follows (in 3 dimensions or fewer) from the law of cosines. – Ben Grossmann Commented Mar 30, 2015 at 2:54 There are some pretty good answers given here. – Ben Grossmann Commented Mar 30, 2015 at 2:58 \| Show 5 more comments 4 Answers 4 Reset to default This answer is useful 7 Save this answer. Show activity on this post. The triangle inequality is an application of the Cauchy-Schwarz inequality. The triangle inequality is very important, especially since it is a condition for metric spaces. It is also very useful in probability theory with regards to the variance of Y where Y is a random variable. The Cauchy-Schwarz inequality also is important because it connects the notion of an inner product with the notion of length. Share CC BY-SA 3.0 Follow this answer to receive notifications edited Jul 25, 2015 at 11:03 Daniel Fischer 212k1919 gold badges306306 silver badges436436 bronze badges answered May 3, 2015 at 0:31 Mike El JacksonMike El Jackson 59511 gold badge55 silver badges1616 bronze badges Add a comment \| This answer is useful 3 Save this answer. Show activity on this post. The Cauchy-Schwarz inequality holds for much wider range of settings than just the two- or three-dimensional Euclidean space R2 or R3. In fact, it holds for all kind of spaces, where an inner product (an abstract concept) is defined. Thus it can be applied to bound things in wide number of settigns. Share CC BY-SA 3.0 Follow this answer to receive notifications answered Mar 30, 2015 at 9:08 dawdaw 55k22 gold badges4545 silver badges8686 bronze badges 1 My thesis advisor used to say that the Cauchy-Schwarz inequality and integration by parts is all you need to do analysis. – Julián Aguirre Commented Mar 30, 2015 at 9:10 Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. Those questions are rather hard to answer. An answer "The property A is very important, because it can be used to prove properties B, C and D that are very important" does not clarify much but rise more questions why the latter are important. A definition of an "important result", that is the easiest to grasp by a non-professional, may come from a statistical estimation: if we count all known results (theorems etc) in analysis to be N, and let M be the number of those that use a particular result, e.g. the Cauchy-Schwarz inequality, in their simplest proof (i.e. it would be much harder to prove without it), then MN may serve as a measure of importance. This kind of statistics is, of course, never calculated explicitly, but rather understood intuitively by an individuum who has a background in a certain subject, and others have to trust him until they accumulate their own statistical data. Those who say that the Cauchy-Schwarz inequality is important may refer to that the ratio MN for this result is relatively high compared to the average. However, one can argue that the inequality x2≥0, ∀x∈R is much more important in this sense. An alternative (qualitative instead of quantitative) way to measure importance is perhaps to judge the influence of a particular notion and possibility to generalize it to other, more general constructions, that helps to study those. In this sense, the CS inequality is as important as the notion of a norm which was generalized from the plane and the 3D space to Rn first, and then to an abstract vector space, giving rise to Hilbert spaces, Banach spaces and a huge and very successful area of mathematics called functional analysis. Share CC BY-SA 3.0 Follow this answer to receive notifications edited Jul 25, 2015 at 12:23 answered Jul 25, 2015 at 12:18 A.Γ.A.Γ. 30.4k44 gold badges5050 silver badges8383 bronze badges Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. Well, with a⋅b=a1b1+a2b2+... it is just Cauchy's inequality. Then there is one with integrals, Buniakovskii's inequality. Finally one with abstract inner products, Schwarz' inequality. All of them are important. For example, to find many applications, consult the classic book Inequalities by Hardy, Littlewood & Polya. Share CC BY-SA 3.0 Follow this answer to receive notifications answered May 3, 2015 at 0:40 GEdgarGEdgar 117k99 gold badges128128 silver badges274274 bronze badges Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions real-analysis linear-algebra analysis See similar questions with these tags. 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2634	https://www.goldbooknews.com/mob/articles/detail/6902e566-e9a2-40fd-beae-1993d39a8763	如何理解事件的独立性？首页悬赏任务试题库试题试卷文章库古文文章素材库图片视频音频复合碎片学科控件电子书课件店铺多学科编辑器 AI商城统计 App 帮助登录发布任务0登录注册登录如何理解事件的独立性？文章二维码店铺名称：齐鲁化工发布时间：2024-09-03 13:57:30 作者：产品内测作者二维码质量指数： 5.0 校对指数： 5.0 分享到： #### 微信扫一扫：分享微信里点“发现”，扫一下二维码便可将本文分享至朋友圈。推荐用途：如何理解事件的独立性？同学们在学习概率这一章内容时会遇到一个概念一一随机事件的独立性。书中对其的解释是：直观地，如果两个随机事件A，B是否发生互相不影响，就认为它们是独立的，这时它们同时发生的概率等于它们各自发生的概率的乘积，即成立在这里，用“A，B同时发生的概率等于它们各自发生的概率的乘积”来定义事件A与事件B(相互)独立在数学上是没有任何问题的。但是，该如何理解“A，B是否发生互相不影响”这句话呢？今天就和大家来聊聊随机事件的独立性。 1.错误的理解在最初学习事件独立性这个概念时，同学们往往会和互斥混淆，互斥指的是两个随机事件A，B没有共同的基本事件，也即这两个事件不可能同时发生从集合观点看，即两个子集不相交，即满足用韦恩图表示如下(其中Ω为样本空间)：同学们在此处往往会对此产生某种错误的“联想”：从而将独立性误解为两事件对应的集合不相交，这实际上是将独立性与互斥两个概念混淆了。我们可以举出一个简单的例子来说明两者的区别。比如，生一个孩子，现有两事件A，B： A：孩子是男的 B：孩子是女的那么显然，此时事件A，B就是互斥的，这两个事件不可能同时发生。再比如，生两个孩子，也有两事件A，B： A：第一个孩子是男的 B：第二个孩子是女的那么此时事件A，就是独立的，因为不难理解：第一个孩子的性别和第二个孩子的性别，这两者肯定是互不影响的，而且性别为男或女的概率都为。现在我们来研究一下，其含义是“生两个孩子，第一个孩子是男的，第二个孩子是女的”的概率很直观，其实就是：这里，实际上用到了独立性的定义进行计算在韦恩图中可以有如下表示：实际上，事件A还包含了另一种情况(左)：第一个孩子和第二个孩子都是男的。同样地，事件B也包含了另一种情况(右)：第一个孩子和第二个孩子都是女的。 (当然，还有第四种情况：第一个孩子是女的，第二个孩子是男的这种情况也在样本空间Ω中，但它在A，B两个圈之外) 我们发现，虽然在这里事件A，B互相独立，但是两者是有交集的，这是区别于互斥事件最重要的一点。 2.条件概率那么，为什么会产生这种现象呢？要充分理解这件事，我们先要学习一个概念——条件概率。现在考虑这样一个问题：袋中有2个白球，2个黑球，现在从袋中不放回地连续拿两次球，求两次都拿到白球的概率。这个问题需要注意的是“不放回”这个条件，也就是说，第一次抽取一定会影响第二次抽取。我们对此进行简单的分类讨论便可得知：依据题意，我们希望发生的是路径①。那么，路径①的概率该如何计算呢？我们记事件A，B为： A：第一次抽到白球 B：第二次抽到白球显然，第一次抽到白球的概率是接下来，我们希望计算的是当第一次抽到白球后，第二次也能抽到白球的概率，也即“A已经发生的条件下B发生的概率”，我们称之为条件概率，用来表示由于A发生后，袋中剩余是“黑、黑、白”，所以那么，路径①(两次均抽到白球)的概率即为其他三种情况的概率也可以用相同的方法计算出来，实际上，总的四种情况的概率之和为1。 3.如何理解事件的独立性？了解条件概率后，我们就可以理解什么是事件的独立性了。上述例子的事件A，B显然不是相互独立的，因为A的发生对B造成了影响并且由前述讨论，的概率需用条件概率计算这让我们联想到两事件相互独立所满足的公式对照这两个式子，我们能得到一个结论：如果A，B两事件不独立，就一定有其实在白球黑球这个例子中，显然是不等于的，其根本原因就在于这个问题一开始就限定了“不放回”的游戏规则。如果说把游戏规则改为“放回”，则情况便截然不同了。由于放回这个机制的存在，每次抽球的概率都是恒定的，永远为也就是说，在放回的条件下，A，B两事件互不影响，它们便是相互独立的。换句话说，如果A，B两事件相互独立，就一定有实际上，这便是相互独立事件的本质。从韦恩图来理解，指的是B在样本空间Ω所占的比例。而指的则是B在A中所占的比例。 A，B两事件相互独立便意味着，这两个比例是一致的，形象地说，对于B来说，A就是Ω的“缩影。现在你可以理解什么是事件的独立性了吗？声明：发布此样例仅用于展示云章多学科编辑器在理工科内容数字化编辑加工的内容样例效果，样例中若有来源标注错误或侵犯了您的合法权益，请作者及时与本网联系，我们将及时更正、删除，谢谢。分享到： #### 微信扫一扫：分享微信里点“发现”，扫一下二维码便可将本文分享至朋友圈。公司名称：北京云章科技有限公司联系电话：010-82609567 82609568联系邮箱：mail@pzcp.com公司地址：北京市海淀区苏州街18号院-4楼6层601室云章科技公众号 Copyright©2016-2025云章科技goldbooknews.com 版权所有京ICP备15007995号-2经营许可证编号：京B2-20214000京公网安备 11010802030858号本网站正在研发中，所有数据皆为测试数据，如果侵犯了您的合法权益，请作者及时与本网联系，我们将及时更正、删除，谢谢。交易详情素材交易协议登录登录
2635	https://arxiv.org/pdf/2005.06308	arXiv:2005.06308v2 [math.CO] 7 Aug 2020 Hamiltonicity of a coprime graph M. H. Bani Mostafa A. a Ebrahim Ghorbani a,b,∗ a Department of Mathematics, K. N. Toosi University of Technology, P. O. Box 16765-3381, Tehran, Iran b University of Hamburg, Department of Mathematics, Bundesstrae 55 (Geomatikum), 20146 Hamburg, Germany August 10, 2020 Abstract The k-coprime graph of order n is the graph with vertex set {k, k + 1 , . . . , k + n − 1} in which two vertices are adjacent if and only if they are coprime. We characterize Hamiltonian k-coprime graphs. As a particular case, two conjectures by Tout, Dabboucy, Howalla (1982) and by Schroeder (2019) on prime labeling of 2-regular graphs follow. A prime labeling of a graph with n vertices is a labeling of its vertices with distinct integers from {1, 2, . . . , n } in such a way that the labels of any two adjacent vertices are relatively prime. Keywords: Prime labeling, Coprime graph, 2-regular graph AMS Mathematics Subject Classification (2010): 05C78, 11B75 1 Introduction Let G be a simple graph with n vertices. A prime labeling of G is a labeling of its vertices with distinct integers from {1, 2, . . . , n } in such a way that the labels of any two adjacent vertices are relatively prime. We say that G is prime if it has a prime labeling. The coprime graph of integers (see [16, Section 7.4]) has the set of all integers as vertex set where two vertices are adjacent if and only if they are relatively prime. So, for an n-vertex graph, being prime is equivalent to being a subgraph of the induced subgraph on {1, 2, . . . , n } by the coprime graph of integers. Many properties of the coprime graph of integers including investigating its subgraphs were ∗Corresponding author E-mail Addresses :m.bani.mostafa@ut.ac.ir, ebrahim.ghorbani@uni-hamburg.de 1studied by Ahlswede and Khachatrian [1, 2, 3], Erd˝ os [6, 7, 8], Erd˝ os, S´ ark¨ ozy, and Szemer´ edi [10, 11], Szab´ o and T´ oth , Erd˝ os and S´ ark¨ ozy [9, 12], and S´ ark¨ ozy . For a survey on the known results on this subject see . The notion of prime labeling originated with Entringer and was introduced in . Entringer around 1980 conjectured that all trees are prime. Haxell, Pikhurko, and Taraz proved that there is an integer n0 such all trees with at least n0 vertices are prime. Besides that, several classes of graphs have been shown to be prime, see for more details. The prime labeling of r-regular graphs have been studied so far for r ≤ 3. For r = 3, i.e., for cubic graphs, Schroeder confirmed a conjecture of Schluchter and Wilson and classified prime cubic graphs: a cubic graph G is prime if and only if G is bipartite and G 6 = K3,3. This result, in particular provides an additional proof that the ladder graph (the Cartesian product Pn × P2) is prime for all n ≥ 1, as conjectured by Varkey (see [13, 22]) and first proved by Ghorbani and Kamali . Classification of prime 2-regular graphs has remaind open to date. A 2-regular graph G must be a disjoint union of cycles: G = Cn1 ∪ · · · ∪ Cnm , where each ni is at least 3. The following conjecture was first given by Tout, Dabboucy, and Howalla: Conjecture 1 () . Let G = Cn1 ∪ · · · ∪ Cnm be a 2-regular graph. Then G is prime if and only if at most one ni is odd. Note that if a graph G with n vertices is prime, the vertices with even labels form an independent set. Thus its independence number satisfies α(G) ≥ ⌊ n/ 2⌋. For cycles, we have α(Cℓ) = ⌊ℓ/ 2⌋. Consequently, if in G = Cn1 ∪ · · · ∪ Cnm more than one ni is odd, then α(G) < ⌊n/ 2⌋. Hence, the necessity of the condition that “at most one ni is odd” is obvious. Some partial cases of Conjecture 1 have been settled in the literature which are reported below: • m ≤ 4 (); • (i) m ≤ 7 provided that all n1, . . . , n m are even; (ii) for arbitrary m when n1 = · · · = nm is a sufficiently large even integer (); • (i) if each ni is even; (ii) if nm is odd and gcd( nm − 1, n ) = 1; (iii) if nm is odd and nm can be written as 2 r + ps, for some r ≥ 1 and odd prime p which is relatively prime to 2 r − 1(). Let n1, . . . , n m−1 be even and nm be odd. By the above result of , Cn1 ∪ · · · ∪ Cnm−1 is prime. So to prove Conjecture 1 it suffices to show that Cnm has a prime labeling with labels k, . . . , k +nm −1 with k = n1 +· · · +nm−1 +1. Note that here both k and nm are odd. Therefore, as observed in , Conjecture 1 follows from the following conjecture: Conjecture 2 () . If n, k are odd integers and n ≥ 3, then Cn has a prime labeling using the labels k, k + 1 , . . . , k + n − 1. 2Motivated by Conjecture 2, we define the k-coprime graph of order n, denoted CPG( k, n )as the graph with vertex set {k, k + 1 , . . . , k + n − 1} in which two vertices a, b are adjacent if and only if they are coprime, i.e. gcd( a, b ) = 1. Here n can be any positive integer and k any integer. If 0 happen to be a vertex of our graph, it has at most two neighbors, namely −1 and 1, because for every nonzero integer a, gcd( a, 0) = \|a\|.As the main result of this paper, we characterize Hamiltonian k-coprime graphs as follows. Theorem 3. Let k and n ≥ 3 be integers. Then CPG( k, n ) is Hamiltonian if and only if either (i) both n and k are odd, or (ii) n is even and each of k and k + n − 1 is not divisible by some odd prime less than n. Obviously, from Part (i) of Theorem 3, Conjecture 2 and consequently Conjecture 1 follow. 2 Proofs In this section, we present the proof of Theorem 3 which is organized as follows: in Theorems 6 and 12 we shall prove that if n, k satisfy the conditions (i) or (ii) of Theorem 3, respectively, then CPG( k, n ) is Hamiltonian; in Theorem 7, we show that if n, k do not satisfy (i) and (ii), then CPG( k, n ) is not Hamiltonian. We start with the following useful lemma. Lemma 4. Let G be a graph with a Hamiltonian path v1, v 2, . . . , v n. If there is a sequence of indices 1 < i 1 < · · · < i k < n such that G contains the edges v1vi1+1 , v i1 vi2+1 , . . . , v ik−1 vik +1 , v ik vn, (1) then G is Hamiltonian. Proof. Consider the induced subgraph of G by the edges of the path v1, v 2, . . . , v n together with the edges given in (1). If we remove the edges vi1 vi1+1 , v i2 vi2+1 , . . . , v ik vik +1 from this subgraph, what is left is a cycle with n edges, and so we are done. In what follows, we frequently use the fact that for any distinct nonzero integer a, b ,gcd( a, b ) = gcd( a, b − a). (2) Lemma 5 () . Any odd integer greater than 1 and less than 149 can be written as 2r + ps where r ≥ 1, s ≥ 0, and p is an odd prime with p ∤ 2r − 1. In the next theorem, we prove that if n and k satisfy the condition (i) of Theorem 3, then CPG( k, n ) is Hamiltonian. 3Theorem 6. Let k and n ≥ 3 be odd integers. Then CPG( k, n ) is Hamiltonian. Proof. Let G = CPG( k, n ) and k′ := k + n − 1. The graph G contains the Hamiltonian path k, . . . , k ′. We define the sequence a0, a 1, . . . as follows. We set a0 = k. Assume that ai−1 is already defined, we choose ai in such a way that ai−1 < a i ≤ k′ − 1, gcd( ai + 1 , a i−1) = 1 , and ai is odd . We assume that m is the largest index for which am can be defined. If gcd( am, k ′) = 1, then we have the edges {a0, a 1 + 1 }, {a1, a 2 + 1 }, . . . , {am−1, a m + 1 }, {am, k ′} in G and thus we are done by Lemma 4. Hence we assume that gcd( am, k ′) > 1. In what follows, for simplicity we write a for am. First suppose that k′ − a ≤ 32. Since k′ − a + 1 is an odd integer ≥ 3, by Lemma 5, k′ − a + 1 = 2 r + ps for some r ≥ 1, s ≥ 0, and an odd prime p with p ∤ 2r − 1. If s = 0, then k′ = a + 2 r, and a being odd implies that gcd( a, k ′) = 1 which is not the case. Hence s ≥ 1. Note that we have either gcd( a, a + ps) = 1 or gcd( k′, a + 2 r − 1) = 1, since otherwise we have gcd( a, a + ps) > 1 and so gcd( a, p s) > 1 which implies that p \| a. Also we have 1 < gcd( k′, a +2 r −1) \| k′ −(a+2 r −1) = ps which in turn implies that p \| k′. It turns out that p divides k′ − a = 2 r + ps − 1 and so p \| 2r − 1, a contradiction. Now, if gcd( a, a + ps) = 1, then am+1 can be defined as am+1 = a + ps − 1which is not possible by our choice of m. Therefore, gcd( k′, a + 2 r − 1) = 1. It follows that we have the edges {k, a 1 + 1 }, {a1, a 2 + 1 }, . . . , {am−1, a + 1 }, {a, a + 2 r}, {a + 2 r − 1, k ′} in G and again we are done by Lemma 4. Next, suppose that k′ − a ≥ 33. Note that for every odd prime p < k ′ − a, we have p \| a (since if there is an odd prime p < k ′ − a with p ∤ a, then am+1 can be defined as am+1 = a + p − 1, a contradiction). As a is odd, by (2) it is seen that all the integers a + 2 , a + 4 , a + 8 , a + 16 , a + 32 (3) are coprime to a and so they are coprime to every prime p < k ′ − a. It follows that any integer b of the list (3) is coprime to all positive integer less than k′ − a. Now, if a + 1 ≤ c ≤ k′ and c 6 = b, then 1 ≤ \| b − c\| < k ′ − a. So, in view of (2), gcd( b, c ) = gcd( b, \|b − c\|) = 1. Therefore, b is coprime to all the integers in {a + 1 , a + 2 , . . . , k ′} \ { b}. Similarly, using (2), we see that a + 1 is coprime to all odd integers in the range a, . . . , k ′. It turns out that G contains a path P on the vertices a, a + 1 , k ′, k ′ − 1, k ′ − 2, . . . , a + 33 , a + 2 , a + 3 , a + 32 , a + 31 , . . . , a + 4 . 4If a = k, the path P together with the edge {a, a + 4 } give rise to a Hamiltonian cycle of G.Otherwise, since gcd( a − 1, a ) = 1 and 5 \| a, we have 5 ∤ a − 1 and thus gcd( a − 1, a + 4) = 1. This shows that P together with the path a + 4 , a − 1, a − 2, . . . , k + 1 , k give a Hamiltonian path of G. Since we have the edges {k, a 1 + 1 }, {a1, a 2 + 1 }, . . . , {am−1, a m} in G, in view of Lemma 4, it follows that G is Hamiltonian. By ϑ(n), we denote the product of all odd primes less than n. This function has a role in Hamiltonicity of CPG( k, n ). Theorem 7. In the following cases, CPG( k, n ) is not Hamiltonian: (i) n odd and k even; (ii) n even and either ϑ(n) \| k or ϑ(n) \| k + n − 1.Proof. (i) If n is odd and k even, then CPG( k, n ) has an independent set of size ( n + 1) /2consisting of even vertices. Note if a graph with n vertices has an independent set of size larger than n/ 2, then it cannot be Hamiltonian. So CPG( k, n ) is not Hamiltonian in this case. (ii) Let n be even. Then G = CPG( k, n ) has an independent set of size n/ 2 consisting of even vertices. It follows that if G is Hamiltonian, then in any Hamiltonian cycle of G, every edge should join two vertices with opposite parities. In particular, any vertex of G should have at least two neighbors with opposite parity. First assume that ϑ(n) \| k. We show that in this case, the vertex k has only one neighbor with opposite parity, namely k + 1, which in turn implies that G is not Hamiltonian. To see this, let ℓ be a neighbor of k with opposite parity and k + 1 < ℓ ≤ n + k − 1. So ℓ − k is an odd integer with 3 ≤ ℓ − k ≤ n − 1. Hence, there is some odd prime p such that p \| ℓ − k. As p < n , we have p \| ϑ(n) and thus p \| k. It follows that p \| ℓ, too. So k and ℓ cannot be adjacent, a contradiction. In the case ϑ(n) \| k + n − 1, with a similar proof as given above, we see that the vertex k + n − 1 has only one neighbor with opposite parity. Thus, G cannot be Hamiltonian. Here, we give another property of ϑ(n) in connection with Hamiltonicity of CPG( k, n ). Lemma 8. Let n ≥ 4 be even. Then, v1, . . . , v n, v 1 is a Hamiltonian cycle of CPG( k, n ) if and only if v1 + ϑ(n), . . . , v n + ϑ(n), v 1 + ϑ(n) is a Hamiltonian cycle of CPG( k + ϑ(n), n ).Proof. Since n is even, as it is observed in the proof Theorem 7, in any Hamiltonian cycle of a coprime graph of order n, the ends of every edge have opposite parities. Let a, b be two integers with opposite parities and k ≤ a < b ≤ k + n − 1. We claim that gcd( a, b ) = 1 if and only if gcd( a+ ϑ(n), b + ϑ(n)) = 1, from which the result follows. To see this, suppose that gcd( a, b ) = 1 and an odd prime p divides gcd( a + ϑ(n), b + ϑ(n)). Then p \| (b + ϑ(n) − a − ϑ(n)) = b − a.5Since b − a < n , p \| ϑ(n) and thus p should divide both a, b and so p = 1, a contradiction. Thus gcd( a + ϑ(n), b + ϑ(n)) = 1. The other direction is similar. We need further properties of ϑ(n). Lemma 9. For every integer n ≥ 6, ϑ(n) ≥ 2n + 1 .Proof. For n = 6 , 7, the inequality holds: ϑ(6) = ϑ(7) = 15. We first verify by induction that ϑ(2 i) > 2i+2 for i ≥ 3. For i = 3, we have ϑ(8) = 3 · 5 · 7 > 32. By the Bertrand’s postulate, there is a prime p with 2 i < p < 2i+1 . Therefore, by induction we have ϑ(2 i+1 ) ≥ p · ϑ(2 i) > p · 2i+2 > 2i+3 , for i ≥ 3. Now, for any integer n ≥ 8, choose i ≥ 3 such that 2 i ≤ n < 2i+1 . Then ϑ(n) ≥ ϑ(2 i) > 2i+2 > 2n + 1 . We use the standard notation π(n) to denote the number of primes less than n. Lemma 10. Let n ≥ 12 be even, and t be the number of prime factors of n−1. Then π(n) ≥ t+4 .Proof. We proceed by induction on t. If t = 1, we are done as π(n) ≥ π(12) = 5 . If t = 2, we have n ≥ 16 and thus π(n) ≥ π(16) = 6. If t = 3, then π(n) ≥ π(n − 1) ≥ π(2 · 3 · 5) > 7. Let p1, p 2, . . . be the sequence of primes. Then π(p1 · · · pt) ≥ π(p1 · · · pt−1) + 1 because there is a prime between p1 · · · pt−1 and p1 · · · pt (as a consequence of the Bertrand’s postulate). Now, for t ≥ 4, if n − 1 has t prime factors, then by the induction hypothesis, π(n) ≥ π(n − 1) ≥ π(p1 · · · pt) ≥ π(p1 · · · pt−1) + 1 ≥ t + 4 . Lemma 11. Let k and n ≥ 5 be odd integers. Then in the graph CPG( k, n ), at least one of k or k + n − 1 has at least two neighbors with opposite parity. Proof. The vertices k and k′ := k + n − 1 have the neighbors k + 1 and k′ − 1, respectively. So it is enough to show that either of k or k′ have some other neighbor with opposite parity. Since the first and the last vertex of G := CPG( k, n ) are odd, G has t even and t + 1 odd vertices, for some t ≥ 2. By Theorem 6, G has a Hamiltonian cycle C. Since even vertices form an independent set of G, 2 t edges of C are between even vertices and odd vertices. Hence C has a unique edge e whose ends are both odd. If either of k or k′ is not on e, we are done. Therefore, suppose that e = {k, k ′}. So, by (2), gcd( k, n − 1) = 1. If n − 1 has an odd factor p, then k + p is adjacent with k and we are done. Otherwise, n − 1 is a power of 2. Therefore, we have either 3 ∤ k or 3 ∤ k′ (since otherwise 3 \| (k′ − k) = n − 1, a contradiction). Hence, either k is adjacent with k + 3 or k′ is adjacent with k′ − 3, as desired. 6Finally, we prove that if n, k satisfy the condition (ii) of Theorem 3, then CPG( k, n ) is Hamiltonian. This together with Theorems 6 and 7 complete the proof of Theorem 3. Theorem 12. Let n ≥ 4 be even. If ϑ(n) ∤ k and ϑ(n) ∤ k+n−1, then CPG( k, n ) is Hamiltonian. Proof. Let k′ := k + n − 1 and G := CPG( k, n ). Since ϑ(n) ∤ k, there is an odd prime p < n such that p ∤ k. It follows that k + p is adjacent to k in G. Let ¯ p be the smallest such prime and ℓ := k + ¯ p. Indeed, ℓ is the smallest neighbor of k with opposite parity other than k + 1. Similarly, we can define ¯ p′ and ℓ′ := k′ − ¯p′ as the largest neighbor of k′ with opposite parity other than k′ − 1. We may assume that ¯p′ ≤ ¯p or equivalently k′ − ℓ′ ≤ ℓ − k, (4) since otherwise we can consider CPG( −k′, n ) instead which is isomorphic to G. Furthermore, we may assume that k′ is odd, otherwise, by Lemma 8, we can consider the graph CPG( k + ϑ(n), n )in which the last vertex, i.e. k′ + ϑ(n) is odd. By induction, we prove the stronger statement that G = CPG( k, n ) contains a Hamiltonian cycle including the edges {k, k + 1 } and {k′ − 1, k ′}. We call such a Hamiltonian cycle special .At first, we need to prove the statement for n ≤ 10. n = 4: Since 3 = ϑ(4) ∤ k, we have the special Hamiltonian cycle k, k + 1 , k + 2 , k + 3 , k in G. n = 6: We have 3 · 5 = ϑ(6) ∤ k and ϑ(6) ∤ k′ = k + 5 . If 5 ∤ k, then we have the special Hamiltonian cycle k, . . . , k + 5 , k in G. If 5 \| k, then necessarily 3 ∤ k, 3 ∤ k′ and thus k, k + 1 , k + 2 , k + 5 , k + 4 , k + 3 , k is a special Hamiltonian cycle. n = 8: We have 3 · 5 · 7 = ϑ(8) ∤ k and ϑ(8) ∤ k′ = k + 7. If 7 ∤ k, then k, . . . , k + 7 , k is a special Hamiltonian cycle of G. So let 7 \| k. Note that 3 cannot divide both k, k ′ as k′ − k = 7. By (4), we may assume that 3 ∤ k′. If further 5 ∤ k, then we have the special Hamiltonian cycle k, k + 1 , k + 2 , k + 3 , k + 4 , k + 7 , k + 6 , k + 5 , k. If 5 \| k, then we have have necessarily 3 ∤ k, 5 ∤ k′ and thus k, k + 3 , k + 4 , k + 5 , k + 6 , k + 7 , k + 2 , k + 1 , k is a special Hamiltonian cycle . n = 10: We have 3 · 5 · 7 = ϑ(10) ∤ k and ϑ(10) ∤ k′ = k + 9. If 3 ∤ k, then we have the special Hamiltonian cycle k, . . . , k + 9 , k in G. So we assume that 3 \| k (and so 3 \| k′). Also, 5 cannot divide both k, k ′ and again by (4), we can assume that 5 ∤ k′. If further 5 ∤ k, then we have the special Hamiltonian cycle k, k +1 , k +2 , k +3 , k +4 , k +9 , k +8 , k +7 , k +6 , k +5 , k .If 5 \| k, since 3 \| k, too, then necessarily 7 ∤ k. If further 7 ∤ k + 2, then we have the special Hamiltonian cycle k, k + 1 , k + 2 , k + 9 , k + 8 , k + 3 , k + 4 , k + 5 , k + 6 , k + 7 , k . If 7 \| k + 2, since we already have 15 \| k, then k ≡ 75 (mod 105). By Lemma 8, we only need to consider 1 ≤ k ≤ ϑ(10) = 105, so we may assume that k = 75 in which case 75 , 76 , 81 , 80 , 77 , 78 , 83 , 84 , 79 , 82 , 75 is a special Hamiltonian cycle. 7In what follows, we assume that n ≥ 12. We consider two cases. Case 1. ℓ′ > ℓ .Let n′ = ℓ′ − k + 2 which is even as k ≡ ℓ′ (mod 2). First, assume that ϑ(n′) ∤ ℓ′ + 1. Since k has a neighbor with opposite parity between k + 1 and ℓ′ + 1, we have ϑ(n′) ∤ k (if ϑ(n′) \| k, then gcd( k, k + r) > 1 for every odd r, 1 < r < n ′). Hence CPG( k, n ′) satisfies the induction hypothesis and so it has a special Hamiltonian cycle. In particular, CPG( k, n ′) has a Hamiltonian path P between ℓ′ and ℓ′ + 1, where P includes the edge {k, k + 1 }. Now, if we let P ′ be the path ℓ′ + 1 , ℓ ′ + 2 , . . . , k ′ − 1, k ′, ℓ ′, then P ∪ P ′ gives rise to a Hamiltonian cycle of G including both the edge {k, k + 1 } and {k′ − 1, k ′}.Next, assume that ϑ(n′) \| ℓ′ + 1. Consider ¯ p′ = k′ − ℓ′ which is an odd prime. We claim that ¯p′ = 3. For a contradiction, assume that ¯ p′ ≥ 5. Consider the graph G′′ := CPG( ℓ′ + 1 , ¯p′). In G′′ the smallest vertex ℓ′ + 1 is odd and the largest vertex k′ has only one neighbor with opposite parity, namely k′ − 1. Thus, by Lemma 11, ℓ′ + 1 has at least one neighbor with opposite parity other than ℓ′ + 2. Suppose ℓ′ + 1 + r is this neighbor, i.e. gcd( ℓ′ + 1 , ℓ ′ + 1 + r) = 1. Here r should be an odd integer ≥ 3 and further by (4), r < k ′ − ℓ′ ≤ ℓ − k < ℓ ′ − k < n ′. We may assume that r is a prime since otherwise r can be replaced by any of its prime factors. Thus r \| ϑ(n′) and so r \| ℓ′ + 1. This yields r \| ℓ′ + 1 + r. By this contradiction, the claim follows, that is ¯ p′ = 3. Therefore, ℓ′ + 1 = k′ − 2. So ϑ(n′) \| k′ − 2. It turns out that for any prime p < n ′, we have p ∤ k′. Given that k′ − (ℓ − 1) = ℓ′ − ℓ + 4 ≤ ℓ′ − (k + 3) + 4 < n ′, it follows that gcd( ℓ − 1, k ′) = 1, and thus we have the following special Hamiltonian cycle in G: k, k + 1 , . . . , ℓ − 1, k ′, k ′ − 1, . . . , ℓ, k. Case 2. ℓ′ < ℓ (having opposite parities, ℓ = ℓ′ is not possible). We claim that if k, k ′ have neighbors a, a ′, respectively, with a 6 ≡ a′ (mod 2) and a − a′ ≥ 5, then G has a special Hamiltonian cycle. To see this, let m := a − a′ + 1 which is an even integer ≥ 6. We observe that ϑ(m) does not divide either both of a + 1 , a ′ + 1, or both of a − 1, a ′ − 1 (if this does not hold, ϑ(m) should divide at least one of the integers 2 , m − 3, m − 1, m + 1 which is a contradiction in view of Lemma 9). If ϑ(m) ∤ a + 1 , a ′ + 1, then from the induction hypothesis, it follows that the graph CPG( a′ + 1 , m ) contains a Hamiltonian path P between a and a + 1. This together with the path a + 1 , a + 2 , . . . , k ′ − 1, k ′, a ′, a ′ − 1, . . . , k + 1 , k, a, give rise to a special Hamiltonian cycle in G. If ϑ(m) ∤ a − 1, a ′ − 1, then from the induction hypothesis, it follows that the graph CPG( a′ − 1, m ) contains a Hamiltonian path P between a′ 8and a′ − 1. This together with the path a′ − 1, a ′ − 2, . . . , k, a, a + 1 , . . . , k ′ − 1, k ′, a ′ give rise to a special Hamiltonian cycle in G and thus the claim follows. Now, if ℓ′ = ℓ − 1, then G contains the following special Hamiltonian cycle: k, ℓ, ℓ + 1 , . . . , k ′ − 1, k ′, ℓ ′, ℓ ′ − 1, . . . , k + 1 , k. So we assume that ℓ − ℓ′ ≥ 3. Let t be the number of prime factors of n − 1. Since n ≥ 12, by Lemma 10, π(n) ≥ t + 4. Hence the is a prime q < n such that q ∤ n − 1 and q 6 ∈ { 2, ¯p, ¯p′}. As k′ − k = n − 1, it follows that either q ∤ k or q ∤ k′. If q ∤ k, then a = k + q is a neighbor of k with opposite parity other than ℓ = k + ¯ p. By the definition of ℓ, we must have a ≥ ℓ + 2. Therefore, a − ℓ′ ≥ 5 and we are done in view of the above claim. If q ∤ k′, we are done in a similar manner. Acknowledgements The second author carried this work during a Humboldt Research Fellowship at the University of Hamburg. He thanks the Alexander von Humboldt-Stiftung for financial support. References R. Ahlswede and L.H. Khachatrian, On extremal sets without coprimes, Acta Arith. 66 (1994), 89–99. R. Ahlswede and L.H. 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2636	https://openstax.org/books/calculus-volume-1/pages/3-5-derivatives-of-trigonometric-functions	Skip to Content Go to accessibility page Keyboard shortcuts menu Log in Calculus Volume 1 3.5 Derivatives of Trigonometric Functions Calculus Volume 1 3.5 Derivatives of Trigonometric Functions Search for key terms or text. Learning Objectives 3.5.1 Find the derivatives of the sine and cosine function. 3.5.2 Find the derivatives of the standard trigonometric functions. 3.5.3 Calculate the higher-order derivatives of the sine and cosine. One of the most important types of motion in physics is simple harmonic motion, which is associated with such systems as an object with mass oscillating on a spring. Simple harmonic motion can be described by using either sine or cosine functions. In this section we expand our knowledge of derivative formulas to include derivatives of these and other trigonometric functions. We begin with the derivatives of the sine and cosine functions and then use them to obtain formulas for the derivatives of the remaining four trigonometric functions. Being able to calculate the derivatives of the sine and cosine functions will enable us to find the velocity and acceleration of simple harmonic motion. Derivatives of the Sine and Cosine Functions We begin our exploration of the derivative for the sine function by using the formula to make a reasonable guess at its derivative. Recall that for a function f(x),f(x), f′(x)=limh→0f(x+h)−f(x)h. f′(x)=limh→0f(x+h)−f(x)h. Consequently, for values of hh very close to 0, f′(x)≈f(x+h)−f(x)h.f′(x)≈f(x+h)−f(x)h. We see that by using h=0.01,h=0.01, ddx(sinx)≈sin(x+0.01)−sinx0.01 ddx(sinx)≈sin(x+0.01)−sinx0.01 By setting D(x)=sin(x+0.01)−sinx0.01D(x)=sin(x+0.01)−sinx0.01 and using a graphing utility, we can get a graph of an approximation to the derivative of sinxsinx (Figure 3.25). Figure 3.25 The graph of the function D(x)D(x) looks a lot like a cosine curve. Upon inspection, the graph of D(x)D(x) appears to be very close to the graph of the cosine function. Indeed, we will show that ddx(sinx)=cosx. ddx(sinx)=cosx. If we were to follow the same steps to approximate the derivative of the cosine function, we would find that ddx(cosx)=−sinx. ddx(cosx)=−sinx. Theorem 3.8 The Derivatives of sin x and cos x The derivative of the sine function is the cosine and the derivative of the cosine function is the negative sine. ddx(sinx)=cosx ddx(sinx)=cosx (3.11) ddx(cosx)=−sinx ddx(cosx)=−sinx (3.12) Proof Because the proofs for ddx(sinx)=cosxddx(sinx)=cosx and ddx(cosx)=−sinxddx(cosx)=−sinx use similar techniques, we provide only the proof for ddx(sinx)=cosx.ddx(sinx)=cosx. Before beginning, recall two important trigonometric limits we learned in Introduction to Limits: limh→0sinhh=1andlimh→0cosh−1h=0. limh→0sinhh=1andlimh→0cosh−1h=0. The graphs of y=(sinh)hy=(sinh)h and y=(cosh−1)hy=(cosh−1)h are shown in Figure 3.26. Figure 3.26 These graphs show two important limits needed to establish the derivative formulas for the sine and cosine functions. We also recall the following trigonometric identity for the sine of the sum of two angles: sin(x+h)=sinxcosh+cosxsinh. sin(x+h)=sinxcosh+cosxsinh. Now that we have gathered all the necessary equations and identities, we proceed with the proof. ddxsinx=limh→0sin(x+h)−sinxhApply the definitionof the derivative.=limh→0sinxcosh+cosxsinh−sinxhUse trig identity for the sine of the sum of two angles.=limh→0(sinxcosh−sinxh+cosxsinhh)Regroup.=limh→0(sinx(cosh−1h)+cosx(sinhh))Factor outsinxandcosx.=sinx·0+cosx·1Apply trig limit formulas.=cosxSimplify. ddxsinx=limh→0sin(x+h)−sinxh=limh→0sinxcosh+cosxsinh−sinxh=limh→0(sinxcosh−sinxh+cosxsinhh)=limh→0(sinx(cosh−1h)+cosx(sinhh))=sinx⋅0+cosx⋅1=cosxApply the definitionof the derivative.Use trig identity for the sine of the sum of two angles.Regroup.Factor outsinxandcosx.Apply trig limit formulas.Simplify. □ Figure 3.27 shows the relationship between the graph of f(x)=sinxf(x)=sinx and its derivative f′(x)=cosx.f′(x)=cosx. Notice that at the points where f(x)=sinxf(x)=sinx has a horizontal tangent, its derivative f′(x)=cosxf′(x)=cosx takes on the value zero. We also see that where f(x)=sinxf(x)=sinx is increasing, f′(x)=cosx>0f′(x)=cosx>0 and where f(x)=sinxf(x)=sinx is decreasing, f′(x)=cosx<0.f′(x)=cosx<0. Figure 3.27 Where f(x)f(x) has a maximum or a minimum, f′(x)=0f'(x)=0 that is, f′(x)=0f'(x)=0 where f(x)f(x) has a horizontal tangent. These points are noted with dots on the graphs. Example 3.39 Differentiating a Function Containing sin x Find the derivative of f(x)=5x3sinx.f(x)=5x3sinx. Solution Using the product rule, we have f′(x)=ddx(5x3)·sinx+ddx(sinx)·5x3=15x2·sinx+cosx·5x3. f'(x)=ddx(5x3)⋅sinx+ddx(sinx)⋅5x3=15x2⋅sinx+cosx⋅5x3. After simplifying, we obtain f′(x)=15x2sinx+5x3cosx. f′(x)=15x2sinx+5x3cosx. Checkpoint 3.25 Find the derivative of f(x)=sinxcosx.f(x)=sinxcosx. Example 3.40 Finding the Derivative of a Function Containing cos x Find the derivative of g(x)=cosx4x2.g(x)=cosx4x2. Solution By applying the quotient rule, we have g′(x)=(−sinx)4x2−8x(cosx)(4x2)2. g′(x)=(−sinx)4x2−8x(cosx)(4x2)2. Simplifying, we obtain g′(x)=−4x2sinx−8xcosx16x4=−xsinx−2cosx4x3. g′(x)=−4x2sinx−8xcosx16x4=−xsinx−2cosx4x3. Checkpoint 3.26 Find the derivative of f(x)=xcosx.f(x)=xcosx. Example 3.41 An Application to Velocity A particle moves along a coordinate axis in such a way that its position at time t is given by s(t)=2sint−t for 0≤t≤2π. At what times is the particle at rest? Solution To determine when the particle is at rest, set s′(t)=v(t)=0. Begin by finding s′(t). We obtain s′(t)=2cost−1, so we must solve 2cost−1=0for0≤t≤2π. The solutions to this equation are t=π3 and t=5π3. Thus the particle is at rest at times t=π3 and t=5π3. Checkpoint 3.27 A particle moves along a coordinate axis. Its position at time t is given by s(t)=√3t+2cost for 0≤t≤2π. At what times is the particle at rest? Derivatives of Other Trigonometric Functions Since the remaining four trigonometric functions may be expressed as quotients involving sine, cosine, or both, we can use the quotient rule to find formulas for their derivatives. Example 3.42 The Derivative of the Tangent Function Find the derivative of f(x)=tanx. Solution Start by expressing tanx as the quotient of sinx and cosx: f(x)=tanx=sinxcosx. Now apply the quotient rule to obtain f′(x)=cosxcosx−(−sinx)sinx(cosx)2. Simplifying, we obtain f′(x)=cos2x+sin2xcos2x. Recognizing that cos2x+sin2x=1, by the Pythagorean theorem, we now have f′(x)=1cos2x. Finally, use the identity secx=1cosx to obtain f′(x)=sec2x. Checkpoint 3.28 Find the derivative of f(x)=cotx. The derivatives of the remaining trigonometric functions may be obtained by using similar techniques. We provide these formulas in the following theorem. Theorem 3.9 Derivatives of tanx,cotx,secx, and cscx The derivatives of the remaining trigonometric functions are as follows: ddx(tanx)=sec2x (3.13) ddx(cotx)=−csc2x (3.14) ddx(secx)=secxtanx (3.15) ddx(cscx)=−cscxcotx. (3.16) Example 3.43 Finding the Equation of a Tangent Line Find the equation of a line tangent to the graph of f(x)=cotx at x=π4. Solution To find the equation of the tangent line, we need a point and a slope at that point. To find the point, compute f(π4)=cotπ4=1. Thus the tangent line passes through the point (π4,1). Next, find the slope by finding the derivative of f(x)=cotx and evaluating it at π4: f′(x)=−csc2xandf′(π4)=−csc2(π4)=−2. Using the point-slope equation of the line, we obtain y−1=−2(x−π4) or equivalently, y=−2x+1+π2. Example 3.44 Finding the Derivative of Trigonometric Functions Find the derivative of f(x)=cscx+xtanx. Solution To find this derivative, we must use both the sum rule and the product rule. Using the sum rule, we find f′(x)=ddx(cscx)+ddx(xtanx). In the first term, ddx(cscx)=−cscxcotx, and by applying the product rule to the second term we obtain ddx(xtanx)=(1)(tanx)+(sec2x)(x). Therefore, we have f′(x)=−cscxcotx+tanx+xsec2x. Checkpoint 3.29 Find the derivative of f(x)=2tanx−3cotx. Checkpoint 3.30 Find the slope of the line tangent to the graph of f(x)=tanx at x=π6. Higher-Order Derivatives The higher-order derivatives of sinx and cosx follow a repeating pattern. By following the pattern, we can find any higher-order derivative of sinx and cosx. Example 3.45 Finding Higher-Order Derivatives of y=sinx Find the first four derivatives of y=sinx. Solution Each step in the chain is straightforward: y=sinxdydx=cosxd2ydx2=−sinxd3ydx3=−cosxd4ydx4=sinx. Analysis Once we recognize the pattern of derivatives, we can find any higher-order derivative by determining the step in the pattern to which it corresponds. For example, every fourth derivative of sin x equals sin x, so d4dx4(sinx)=d8dx8(sinx)=d12dx12(sinx)=…=d4ndx4n(sinx)=sinxd5dx5(sinx)=d9dx9(sinx)=d13dx13(sinx)=…=d4n+1dx4n+1(sinx)=cosx. Checkpoint 3.31 For y=cosx, find d4ydx4. Example 3.46 Using the Pattern for Higher-Order Derivatives of y=sinx Find d74dx74(sinx). Solution We can see right away that for the 74th derivative of sinx,74=4(18)+2, so d74dx74(sinx)=d72+2dx72+2(sinx)=d2dx2(sinx)=−sinx. Checkpoint 3.32 For y=sinx, find d59dx59(sinx). Example 3.47 An Application to Acceleration A particle moves along a coordinate axis in such a way that its position at time t is given by s(t)=2−sint. Find v(π/4) and a(π/4). Compare these values and decide whether the particle is speeding up or slowing down. Solution First find v(t)=s′(t): v(t)=s′(t)=−cost. Thus, v(π4)=−1√2. Next, find a(t)=v′(t). Thus, a(t)=v′(t)=sint and we have a(π4)=1√2. Since v(π4)=−1√2<0 and a(π4)=1√2>0, we see that velocity and acceleration are acting in opposite directions; that is, the object is being accelerated in the direction opposite to the direction in which it is travelling. Consequently, the particle is slowing down. Checkpoint 3.33 A block attached to a spring is moving vertically. Its position at time t is given by s(t)=2sint. Find v(5π6) and a(5π6). Compare these values and decide whether the block is speeding up or slowing down. Section 3.5 Exercises For the following exercises, find dydx for the given functions. 175. y=x2−secx+1 y=3cscx+5x 177. y=x2cotx y=x−x3sinx 179. y=secxx y=sinxtanx 181. y=(x+cosx)(1−sinx) y=tanx1−secx 183. y=1−cotx1+cotx y=cosx(1+cscx) For the following exercises, find the equation of the tangent line to each of the given functions at the indicated values of x. Then use a calculator to graph both the function and the tangent line to ensure the equation for the tangent line is correct. 185. [T] f(x)=−sinx,x=0 [T] f(x)=cscx,x=π2 187. [T] f(x)=1+cosx,x=3π2 [T] f(x)=secx,x=π4 189. [T] f(x)=x2−tanx,x=0 [T] f(x)=5cotx,x=π4 For the following exercises, find d2ydx2 for the given functions. 191. y=xsinx−cosx y=sinxcosx 193. y=x−12sinx y=1x+tanx 195. y=2cscx y=sec2x 197. Find all x values on the graph of f(x)=−3sinxcosx where the tangent line is horizontal. Find all x values on the graph of f(x)=x−2cosx for 0<x<2π where the tangent line has slope 2. 199. Let f(x)=cotx. Determine the points on the graph of f for 0<x<2π where the tangent line(s) is (are) parallel to the line y=−2x. [T] A mass on a spring bounces up and down in simple harmonic motion, modeled by the function s(t)=−6cost where s is measured in inches and t is measured in seconds. Find the rate at which the spring is oscillating at t=5 s. 201. Let the position of a swinging pendulum in simple harmonic motion be given by s(t)=acost+bsint where a and b are constants, t measures time in seconds, and s measures position in centimeters. If the position is 0 cm and the velocity is 3 cm/s when t=0, find the values of a and b. After a diver jumps off a diving board, the edge of the board oscillates with position given by s(t)=−5cost cm at t seconds after the jump. Sketch one period of the position function for t≥0. Find the velocity function. Sketch one period of the velocity function for t≥0. Determine the times when the velocity is 0 over one period. Find the acceleration function. Sketch one period of the acceleration function for t≥0. 203. The number of hamburgers sold at a fast-food restaurant in Pasadena, California, is given by y=10+5sinx where y is the number of hamburgers sold and x represents the number of hours after the restaurant opened at 11 a.m. until 11 p.m., when the store closes. Find y′ and determine the intervals where the number of burgers being sold is increasing. [T] The amount of rainfall per month in Phoenix, Arizona, can be approximated by y(t)=0.5+0.3cost, where t is months since January. Find y′ and use a calculator to determine the intervals where the amount of rain falling is decreasing. For the following exercises, use the quotient rule to derive the given equations. ddx(cotx)=−csc2x ddx(secx)=secxtanx ddx(cscx)=−cscxcotx Use the definition of derivative and the identity cos(x+h)=cosxcosh−sinxsinh to prove that d(cosx)dx=−sinx. For the following exercises, find the requested higher-order derivative for the given functions. 209. d3ydx3 of y=3cosx d2ydx2 of y=3sinx+x2cosx 211. d4ydx4 of y=5cosx d2ydx2 of y=secx+cotx 213. d3ydx3 of y=x10−secx Previous Next Order a print copy Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution-NonCommercial-ShareAlike License and you must attribute OpenStax. 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Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Monte Carlo Simulation of Irregular Shape Ask Question Asked 12 years ago Modified12 years ago Viewed 2k times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. need a working example of how to do this, i have a bmp file showing the shape of a lake, the bmp's size is the rectangular area and known. i need to take this picture and estimate the lake size. so far, i have a script that generates a giant matrix of each pixel, telling me whether or not it's in the lake - but this isn't monte carlo! i need to generate random points and compare them against the shape somehow, this is where i'm getting stuck. i don't understand how to compare here, i don't have an equation for the shape or lines, i only have exact point information - either it is or isn't in the lake. so i guess i have the exact area already, but i need to find a way to compare random points against this. ```matlab function Yes = Point_In_Lake(x,y,image_pixel) [pHeight,pWidth]=size(image_pixel); %pHeight = Height in pixel %pWidth = Width in pixel width = 1000; %width is the actual width of the lake height = 500; %height is the actual height of the lake %converting x_value to pixel_value in image_pixel point_x_pixel = xpWidth/width; xl = floor(point_x_pixel)+1; xu = min(ceil(point_x_pixel)+1,pWidth); %converting y_value to pixel_value in image_pixel point_y_pixel = ypHeight/height; yl = floor(point_y_pixel)+1; yu = min(ceil(point_y_pixel)+1,pHeight); %Finally, perform the check whether the point is in the lake if (image_pixel(yl,xl)~=0)&&(image_pixel(yl,xu)~=0)&&(image_pixel(yu,xl)~=0)&&(image_pixel(yu,xu)~=0) Yes=0; else Yes=1; end ``` matlab math simulation montecarlo Share Share a link to this question Copy linkCC BY-SA 3.0 Improve this question Follow Follow this question to receive notifications asked Sep 17, 2013 at 2:47 afrotaintafrotaint 296 1 1 gold badge 3 3 silver badges 13 13 bronze badges 4 I don't quite get where your problem is. I would assume the whole purpose of the Monte Carlo method is not having to evaluate all pixels but only a randomly choosen part of it. If you are able to check if a specific pixel is inside the lake or not you got everything you need. Just generate some uniformly choosen points, evaluate whether they are in the lake or not and extrapolate from there - or am I missing something?Gigo –Gigo 2013-09-17 02:55:10 +00:00 Commented Sep 17, 2013 at 2:55 The problem lies within what you stated "evaluate whether they are in the lake or not" - this is what i need an example of.afrotaint –afrotaint 2013-09-17 03:00:13 +00:00 Commented Sep 17, 2013 at 3:00 2 How do you determine whether a given pixel is in lake or not when you iterate through all the pixels? That logic would be the same regardless of how you arrived at the pixel. In Monte Carlo you just generate the pixel locations at random. Count what proportion of your random pixels are lake pixels, divide by the total number you generated. That proportion times the image size is an estimate of the size of the lake.pjs –pjs 2013-09-17 13:40:58 +00:00 Commented Sep 17, 2013 at 13:40 If your goal is to determine lake size without evaluating too many pixels, monte carlo is probably not a very efficient way. You would for example like to spread out your sampling points instead of selecting them randomly (and thus perhaps sampling many from a small area)Dennis Jaheruddin –Dennis Jaheruddin 2013-09-19 08:07:53 +00:00 Commented Sep 19, 2013 at 8:07 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 0 Save this answer. Show activity on this post. Here was the solution: ```matlab binaryMap = image_pixel for i = 1:numel(image_pixel) xrand = randperm(size(image_pixel,1),1); yrand = randperm(size(image_pixel,2),1); Yes(i) = Point_In_Lake(xrand,yrand,binaryMap); end PercentLake = length(find(Yes==1))/length(Yes); LakeArea = (PercentLake 500000)/43560; ``` Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications answered Oct 1, 2013 at 12:41 afrotaintafrotaint 296 1 1 gold badge 3 3 silver badges 13 13 bronze badges Comments Add a comment Your Answer Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. To learn more, see our tips on writing great answers. 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2638	https://ocw.mit.edu/courses/18-034-honors-differential-equations-spring-2009/resources/mit18_034s09_lec03/	MIT OpenCourseWare 18.034 Honors Differential Equations Spring 2009 For information about citing these materials or our Terms of Use, visit: � � � � � � � LECTURE 3. FIRST-ORDER LINEAR EQUATIONS First-order linear differential equations. We will give a systematic method of solving first-order differential equations (of normal form) (3.1) y� + p(x)y = f (x) on a given interval I, where p, f are continuous functions. First, the homogeneous equation (3.2) y� + p(x)y = 0 is solved by quadrature . Let P (x) = p(x)dx be an indefinite integral of p(x). Then, d (eP (x)y) = eP (x)y� + p(x)eP (x)y = eP (x)(y� + p(x)y) = 0 dx if and only if y solves (3.2), since eP (x) = 0 . Theorem 3.1. Let p(x) be continuous on an interval I and P (x) = p(x)dx be an anti-derivative of p(x). Then, R φ(x) = ce −P (x) = ce − p(x)dx is a solution of (3.2) for any constant c. Conversely, all solutions of (3.2) are of this form. Exercise. Show that a solution y(x) of (3.2), where p(x) is continuous, is: either y(x) = 0 for all x or y(x) = 0 � for all x. Next, we treat the inhomogeneous equation (3.3) y� + p(x)y = f (x) by variation of parameters . Let P (x) = p(x)dx as before. Then, d (eP (x)y) = eP (x)(y� + p(x)y) = eP (x)f (x). dx x Hence, ye P (x) = y0 + eP (s)f (s)ds for some y0 and for some x0. x0 Theorem 3.2. Let p(x) be as in Theorem 3.1 and let f (x) be continuous on an interval x0 ∈ I. Then, the general solution of (3.3) is given by x y(x) = y0e−P (x) + e−P (x) eP (s)f (s)ds. x0 Moreover, y(x0) = y0 if and only if P (x) = x x 0 p(s)ds . Example 3.3. Consider the differential equation (3.4) y� + y = x + 3 . By trying y(x) = ax + b one easily finds a solution y = x + 2 of (3.4). If y = φ(x) is another solution of (3.4), then z = y − (x + 2) must satisfy the corresponding homogeneous equation z� + z = 0 . By Theorem 3.2, then, z = ce −x. Therefor, the general solution of (3.4) is y = ce −x + x + 2 . 1 Exercise. (Bernoulli equations) 1. If p and q are continuous functions of x and n =� 1 is constant, show that the Bernoulli equation y� + p(x)y = q(x)y n , y > 0 m can be reduced to a linear equation upon substitution y = u for a suitable constant m. The condition y > 0 ensures that the function u = y1/m is meaningful. When n = 1 the original equation is already linear. Solve the differential equation y� + y = xy 3 , y > 0. The logarithmic spiral. Suppose a curve r = f (θ) in polar coordinates cuts the radius at a con stant angle, say ψ, as shown in the figure below. Figure 3.1. The logarithmic spiral. The game is to find the equation of this curve. If ψ = 0 the curve is a ray extending from the origin to ∞ and it cannot be represented in the form r = f (θ). If ψ = π/ 2 or −π/ 2 the curve is a circle centered at he origin. If (3.5) −π/ 2 < ψ < π/ 2, ψ = 0 �, 1 we write tan ψ = . The differential equation is k rdθ tan ψ = . dr (See Figure 3.2.) 1 Using tan ψ = we obatin k rdθ 1 dr (3.6) = , = kr, dr k dθ and therefore r = ce kθ .A curve of this kind is called a logarithmic spiral . The steps in the discussion above are reversible, and hence a curve r = f (θ) cuts the radius at a constant angle ψ satisfying (3.5) if, and only if, it is a logarithmic spiral. Since k is arbitrary, this gives a geometric interpretation of all processes obeying the exponential law of growth. A logarithmic spiral looks like a snail shell (Figure 3.1), and this is not a coincidence. A snail shell has the characteristic that it grows only at one end, and the part of the shell already laid down does not change. The growth is specified by its rate in two perpendicular directions - radial Lecture 3 218.034 Spring 2009 Figure 3.2. and transverse. Let us suppose that both rates are proportional to the size at time t. The precise meaning of ”size” will not matter, and we use the weight W as a quantity that is easily measured. If arc length in the radial direction is s1 and in the transverse direction is s2, then (3.7) ds 1 = k1W and ds 2 = k2W, dt dt where k1, k 2 are constants. The arc length in polar coordinates thus satisfies ds 2 = dr 2 + ( rdθ )2 . Setting dθ = 0 we get ds 1 = dr and setting dr = 0 we get ds 2 = rdθ . Dividing the second equation in (3.7) by the first, we arrive at rdθ k2 = . dr k1 This is the same equation as in (3.6) with k = k1/k 2.
2639	https://alatius.com/ls/	Lewis & Short Please consult the preface at Wikisource for advice on how to read and use this dictionary. See also the Wikipedia article for information on its origin and relation to other dictionaries. The Latin-English dictionary by Lewis & Short was originally digitized by the Perseus Project. This is simply an alternative interface to that digital version. There are many others as well on the net: The original version at Perseus Digital Library Pollux: Archimedes Project Dictionary Access Numen - The Latin Lexicon ΛΟΓΕΙΟΝ glossa [a latin dictionary] philolog.us/ I made this page primarily for my own use, not being content with the alternatives. The main raison d’être for yet another version is that I have here tried to normalize and restore the original layout of the printed dictionary, thus reverting or at least disguising many of the (unfortunately frequent) errors in the original digitized document. This is most obvious in the way definitions are listed; I have also tried to fix some of the garbled Greek. In my opinion, this means that this version gives the most faithful and clear presentation of the dictionary currently available. To see what I’m talking about, search for, e.g., “accedo” and compare the results from the different sites. Nevertheless, all users should be aware that there are still a lot of proofreading errors in all digital versions of this dictionary. I urge anyone, whenever in doubt, to consult the printed book. A scanned copy can be found (and read online) at the Internet Archive. It is also worth mentioning that this dictionary, while being frequently refered to, really has been superseded by the Oxford Latin Dictionary, which, however, is not as easily available online.
2640	https://nata.com.au/files/2021/07/Guidance-on-Significant-Figures-for-Laboratory-Reports2021.pdf	Guidance on Significant Figures for Laboratory Reports The information below is provided as a guide for the reporting of significant figures for laboratories that undertake analytical procedures in the area of Life Sciences. In particular, the analytical procedures of interest are chemical and biological analyses, but it is applicable to the reporting of most analytical results. The primary reason that this guidance has been produced is because there is an observed level of difference in the way that laboratories report significant figures. This can create an issue for laboratory clients as they typically compare the analytical results they receive to a standard or guideline value, and whether a result is deemed compliant, or not, to the relevant standard or guideline value can have a range of downstream effects for the client and various regulatory agencies that they may report to. By detailing a recommended approach to the reporting of significant figures, NATA is seeking to reduce one potential area of variation when an analytical result is compared to a standard or guideline value. What are significant figures? The significant figures (which are also known as the significant digits or precision) of a number written in positional notation are those digits that carry meaningful contributions to its measurement resolution. This includes all digits except leading and trailing zeros.” Measurement resolution is an important phrase here, as the resolution is indicative of the accuracy of the analytical procedure, and can also be used to elude measurement uncertainty or the level of confidence of the analysis at a particular level. For example, a reported value of 12345 mg/kg would indicate a relatively high degree of accuracy in the area of environmental testing, and could be interpreted (possibly by the untrained client) as meaning that the analytical method that was used to derive this result is capable of differentiating between a concentration of 12345 mg/kg and 12344mg/kg, or 12346 mg/kg. In terms of percentage error, this relates to a 0.0081% difference in the results. As most Environmental Chemists will be well aware, such a low level of analytical error is highly unlikely and, therefore, the advice is to use less significant figures to better reflect the level of confidence in the significant figures presented (for example 12300 mg/kg). Whilst the piece of analytical equipment used to derive the result of 12345 mg/kg may have presented this as the result, some caution and due diligence needs to be applied to the result to ensure that it accurately reflects the limit of reporting, detection or quantification for the analytical method that was used to derive the result, as well as any limitations with respect to the instrument that was used. Reference documents There is some guidance in the literature on the reporting of significant figures; for example, the American Public Health Association’s “Standard Methods for the Examination of Water and Wastewater” (widely known as APHA) states: “To avoid ambiguity in reporting results or in presenting directions for a procedure, it is customary to use “significant figures.” All digits in a reported result are expected to be known definitely, except for the last digit, which may be in doubt. Such a number is said to contain only significant figures. If more than a single doubtful digit is reported, the extra digit or digits are not significant. This is an important distinction. Extra digits should be carried in calculation….” “Report only such figures as are justified by the accuracy of the work.” Therefore, APHA gives an indication of how to report significant figures with the final digit being one that may be in doubt. American Standard Testing Methods (ASTM) E29-13, section 7.4, provides more clearly defined mechanisms on how many significant figures should be reported. For example, precision based on validation data and/or ongoing QC data (e.g. control charts) could be used to extract standard deviations (sd) at various levels; e.g. at or near the reporting limit (if applicable) and at a more accurate level in the reporting range. For example from ASTM E29.13 section 7.4:- Option 1:- The aim is that the rounding interval should not be greater than 0.5 x sd, but not less than 0.05 x sd, the example below may illustrate this better:- Test result = 123.456, sd = 7.52, hence if the rounding interval is 1, then [0.5 x 7.52 = 3.76] > 1 > [0.05 x 7.52 = 0.376] is acceptable. Hence one can report the result as 123 (three significant figures) using this rule, as the rounding interval is 1. For a rounding interval of 0.1 (where you would report the result as 123.4 i.e. four significant figures), you have 0.05 x 7.52 = 0.376 which is >0.1 and hence is not aligned to this ruling. Option 2:- There is another option where you look at the sd, if the first digit of the sd is >2 you would report to the significant figures of that digit, but if the sd is <2 then you’d be able to report a further significant figure e.g. Test result = 123.456, sd = 2.52, can potentially report 123, as the 1st digit of the sd is >2. Test result = 123.456, sd = 1.52, can potentially report 123.4, as the 1st digit of the sd is <2. An example of where too many significant figures are reported is in a recent trace level proficiency program for PFAS in a biota sample, a matrix that is challenging for such analytes given the spike level or entrained level. A laboratory reported the significant figures as below: Lab Indent. Result (µg/Kg) Uncertainty (µg/Kg) Surrogate Recovery % XYZ 4.8485 0.8425 216.35 Therefore, based on Option 1, to report a result of 4.8485µg/Kg, the sd for relevant validation data (e.g. replicate fortified matrices near this concentration level) and/or ongoing QC data would need to be <0.02. Expressed as a percentage, 0.02/4.8485 = 0.412% relative standard deviation (%RSD). Such precision for trace level PFAS analysis in a validation is highly unlikely, and in routine QC data even less likely. Hence, the use a lower number of significant figures is justified and should have been used. Typically, the Uncertainty reported should have less significant figures than the reported Result, as there is a significant error associated with Uncertainty determinations. In the opinions of the authors, in this specific case, the result should be rounded to two significant figures i.e. 4.8 µg/Kg to reflect the resolution of the analytical procedure in this circumstance. There are always some professional and defendable judgements that can be applied, but a conservative approach is recommended in order to avoid misleading the client and the level of confidence in the reported values! Another document that provides advice on significant figures is the Australian Drinking Water Guidelines (ADWG), in its section on guidance on the rounding of results. This section states that the: The vast majority of numerical guideline values in the ADWG are rounded to a single significant figure. Consistent with standard rounding convention, mid-way values are rounded up. For example, 1.5 is rounded to 2 and 25 is rounded to 30. Trailing zeros in numbers where there is no decimal point should not be taken as significant (e.g., nitrate, 50 mg/L). Practically all of the health-based guideline values were established using data and assumptions with a precision of one significant figure (e.g., volume of water consumed by an adult = 2 L/day). Furthermore, the vast majority include the incorporation of safety factors, which are applied at the precision of ‘order of magnitude’ (e.g., 10 for interspecies extrapolation and 10 for intra-species variation). Quoting more significant figures misrepresents the degree of calculated precision and may lead to unfounded concern when guidelines are exceeded at the second or third significant figure. It is noted that exceptions to this may be necessary for some chemicals. These will be considered on a case-by-case basis and the reasons for the deviation from the convention of rounding to a single significant figure will be explained in the [related] fact sheet. It is noted that aesthetic guidelines are generally based on direct information on palatability to consumers, including appearance, taste and odour, and so do not need to be rounded. Recommended approach to significant figures In the first instance, the number of significant figures that are quoted in an analytical result should reflect the accuracy of the analytical technique used to derive the result. If in doubt as to how many significant figures should quoted in an analytical result, apply the significant rules described in ASTM E29-13. In consultation with the client, or relevant regulatory agency, and where it is justified to do so, if the analytical result is going to be compared to a standard or guideline value in order to determine compliance with the standard or guideline value, the analytical result should contain the same number of significant figures as standard or guideline value, or no more than one additional significant figure.
2641	http://www.hsc.edu.kw/student/materials/Physics/website/hyperphysics%20modified/hbase/electric/elefie.html	\| \| \| \| --- \| Electric Field Electric field is defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force it would exert on a positive test charge. The electric field is radially outward from a positive charge and radially in toward a negative point charge. \| \| \| Using Gauss' law for electric field calculation \| \| IndexElectric field conceptsElectromagnetic force \| \| \| \| \| --- \| \| HyperPhysics Electricity and Magnetism \| R Nave \| \| Go Back \| \| \| \| \| \| \| \| --- --- --- \| \| Electric Field of Point Charge \| \| \| --- \| \| The electric field of a point charge can be obtained from Coulomb's law: The electric field from any number of point charges can be obtained from a vector sum of the individual fields. A positive number is taken to be an outward field; the field of a negative charge is toward it. \| \| This electric field expression can also be obtained by applying Gauss' law. \| \| \| --- \| \| Other electric field geometries \| Multiple point charges \| \| IndexElectric field concepts \| \| \| \| \| --- \| \| HyperPhysics Electricity and Magnetism \| R Nave \| \| Go Back \| \| \| \| \| --- \| Electric and Magnetic Constants In the equations describing electric and magnetic fields and their propagation, three constants are normally used. One is the speed of light c, and the other two are the electric permittivity of free space and the magnetic permeability of free space, . The magnetic permeability of free space is taken to have the exact value \| \| \| See also relative permeability \| and then the electric permittivity takes the value given by the relationship This gives a value In the presence of polarizable or magnetic media, the effective constants will have different values. In the case of a polarizable medium, called a dielectric, the comparison is stated as a relative permittivity or a dielectric constant. In the case of magnetic media, the relative permeability may be stated. \| IndexElectric field concepts \| \| \| \| \| --- \| \| HyperPhysics Electricity and Magnetism \| R Nave \| \| Go Back \|
2642	https://spiral.ece.cmu.edu/pub-spiral/pubfile/paper_27.pdf	IN SEARCH OF THE OPTIMAL WALSH-HADAMARD TRANSFORM Jeremy Johnson Mathematics and Computer Science Drexel University Philadelphia, PA 19104 jjohnson@mcs.drexel.edu Markus P¨ uschel Electrical and Computer Engineering Carnegie Mellon University Pittsburgh, PA 15213 pueschel@ece.cmu.edu ABSTRACT This paper describes an approach to implementing and optimizing fast signal transforms. Algorithms for com-puting signal transforms are expressed by symbolic ex-pressions, which can be automatically generated and translated into programs. Optimizing an implemen-tation involves searching for the fastest program ob-tained from one of the possible expressions. In this pa-per we apply this methodology to the implementation of the Walsh-Hadamard transform. An environment, accessible from MATLAB, is provided for generating and timing WHT algorithms. These tools are used to search for the fastest WHT algorithm. The fastest algorithm found is substantially faster than standard approaches to implementing the WHT. The work re-ported in this paper is part of the SPIRAL project (see an ongoing project whose goal is to automate the implementation and op-timization of signal processing algorithms. 1. INTRODUCTION In this paper we present an approach to implementing and optimizing fast signal transforms in general and the Walsh-Hadamard transform (WHT) in particular. We have chosen the WHT because it simple yet impor-tant (for applications of the WHT to signal processing and coding theory see an respectively). Fast al-gorithms for computing the WHT are similar to the fast Fourier transform (FFT) and its variants . The only diﬀerence is that there are no twiddle factors and bit-reversal is not necessary. By removing the extra complexity of the twiddle factors and bit-reversal we can concentrate on divide and conquer strategies and iterative versus recursive algorithms. Our approach to implementing the WHT is to cre-ate a ﬂexible software architecture that can be con-ﬁgured to implement many diﬀerent algorithms. Inter-nally algorithmic choices are represented in a tree struc-ture similar to the plan data structure of the FFTW package . Externally algorithmic choices are described by a simple grammar which can be parsed to create dif-ferent algorithms that can be executed and timed. This is similar to the approach advocated in the design of TPL, a language for designing and implementing FFT algorithms, and its predecessors [6, 3, 1]. Our package is written in C; however, the external interface allows the user to explore algorithmic choices without writing C code. A MATLAB interface which allows the user to interact and experiment with our package is provided. After reviewing the WHT in Section 2 and describ-ing our package in Section 3, we summarize empirical data illustrating performance and optimization of the WHT. In our approach, optimizing the WHT becomes a search problem over the space of special class of trees called partition trees. While the space of trees is too large to search exhaustively, we can use dynamic pro-gramming and other techniques to prune the search. Our data shows that the performance varies dramati-cally from algorithm to algorithm and from machine to machine. The search process is similar to what is done by FFTW; however, we allow a larger search space and study it more systematically. 2. THE WALSH-HADAMARD TRANSFORM The Walsh-Hadamard transform of a signal x, of size N = 2n, is the matrix-vector product WHTN·x, where WHTN = n ⊗ i=1 DFT2 = n z }\| { DFT2⊗· · · ⊗DFT2 . The matrix DFT2 = [ 1 1 1 −1 ] is the 2-point DFT matrix, and ⊗denotes the tensor or Kronecker product. The tensor product of two matrices is obtained by replacing each entry of the ﬁrst matrix by that element multiplied by the second matrix. Thus, for example, WHT4 = [ 1 1 1 −1 ] ⊗ [ 1 1 1 −1 ] =     1 1 1 1 1 −1 1 −1 1 1 −1 −1 1 −1 −1 1    . Algorithms for computing the WHT can be derived using properties of the tensor product [9, 6]. A re-cursive algorithm for the WHT is obtained from the factorization WHT2n = (WHT2⊗I2n−1)(I2⊗WHT2n−1) (1) This equation corresponds to the divide and conquer step in a recursive FFT. An iterative algorithm for com-puting the WHT is obtained from the factorization WHT2n = n ∏ i=1 (I2i−1⊗WHT2⊗I2n−i), (2) which corresponds to an iterative FFT. More generally, let n = n1 + · · · + nt, then WHT2n = t ∏ i=1 (I2n1+···+ni−1⊗WHT2ni ⊗I2ni+1+···+nt ) (3) This equation encompasses both the iterative and re-cursive algorithm and provides a mechanism for explor-ing diﬀerent breakdown strategies and combinations of recursion and iteration. Alternative algorithms are ob-tained through diﬀerent sequences of the application of Equation 3. Each algorithm obtained this way can be represented by a tree, called a partition tree. The root of the partition tree corresponding to an algorithm for computing WHTN, where N = 2n is labeled with n. Each application of Equation 3 corresponds to an expansion of a node into children whose sum equals the node. Figure 1 shows the trees for a recursive and iterative algorithm for computing WHT16. In this paper we explore all WHT implementations corresponding to all possible partition trees. The to-tal number of partition trees of size n is given by the recurrence Tn = 1 + ∑ n1+···+nk=n Tn1 · · · Tnk (4) Table 1 lists the ﬁrst few values of Tn. The generating function, T(z), for Tn satisﬁes the functional equation T(z) = z/(1 −z) + T(z)2/(1 −T(z)), (5) 4 @ P P P 1 1 1 1 1 1 @ 1 2 @ 1 3 @ 4 Figure 1: Partition Trees for Iterative and Recursive WHT Algorithms n 1 2 3 4 5 6 7 8 Tn 1 2 6 24 112 568 3032 16768 Table 1: Number of Partition Trees for WHT2n and consequently Tn = Θ(αn/n3/2), where α = 4 + √ 8 ≈6.828427120. Even if we restrict to binary par-tition trees, the number of trees is Θ(5n/n3/2). Hence it is impossible to exhaustively search all possible trees for the optimal algorithm. 3. DESIGN OF THE WHT PACKAGE The design of our WHT package is based on the design of an FFT package from the thesis of Sebastion Egner , and is similar in spirit to the design of FFTW . Algorithm alternatives are represented syntactically using a grammar for describing the breakdown strategy. W(n) ::= small[n] \| split[W(n1),...,W(nt)] # n=n1+...+nt The nonterminal symbol W(n) gets expanded into a string, called a WHT expression, corresponding to an algorithm for computing WHT2n. Algorithms are built up from the symbol small[n], which corresponds to a sequence of unrolled straight-line code for computing WHT2n. The string split[W(n1),...,W(nt)] corre-sponds to an application of Equation 3. For example, the strings split[small,small,small,small] split[small,split[small, split[small,small]]] corresponds to the iterative and recursive algorithms for computing WHT16 depicted in Figure 1. Let N = N1 · · · Nt, where Ni = 2ni, and let xM b,s denote the vector (x(b), x(b + s), . . . , x(b + (M −1)s)). Then evaluation of x = WHTN ·x using Equation 3 is performed using R = N; S = 1; for i = 1, . . . , t R = R/Ni; for j = 0, . . . , R −1 for k = 0, . . . , S −1 xNi jNiS+k,S = WHTN · xNi jNiS+k,S; S = S ∗Ni; This scheme assumes that the algorithm works in-place and is able to accept stride parameters. Several code generators are provided for producing code for small[n]. The resulting functions, similar to the codelets in FFTW, are straight-line code sequences without the overhead of loop control or recursion. The elimination of the control overhead makes them more eﬃcient for small transform sizes than general purpose code; however, the size of the instruction cache even-tually causes straight-line code to become slower. A parser is provided for reading WHT expressions and translating them into a data structure called a WHT tree, which is a partition tree with additional information and is related to an FFTW plan. Each algorithm described by a WHT tree can be used to compute the WHT using an apply function. Timing and veriﬁcation programs are provided, which take as input a WHT expression. The timing pro-gram can be accessed through MATLAB. MATLAB programs to generate WHT expressions and analyze the resulting algorithms are available. 4. THE SEARCH FOR THE OPTIMAL WHT ALGORITHM This section summarizes our performance experiments and outlines our search for the optimal WHT algo-rithm. The key observations are: (1) There is a wide distribution of computing times for diﬀerent algorithms. (2) The optimal algorithm depends on the computer and compiler. (3) Substantial performance improve-ment can be obtained by choosing an appropriate al-gorithm. (4) By providing an environment to generate algorithmic choices and intelligently search for the one with optimal performance it is possible to take advan-tage of the previous 3 observations. The WHT package is written in C and is available for download at . Computer experiments were per-formed on a 233 MHz Pentium II running Linux and a 400 MHz UltraSPARC II running Solaris. On the Pentium gcc version 2.7.2.3 with -O6 was used, and on the UltraSPARC version 5.0 of Sun’s C compiler with -fast -xO5 was used. All algorithms operated in-place on double precision real data. In our ﬁrst experiment, we compared the iterative algorithm from Equation 2 with the recursive algorithm 0 2 4 6 8 10 12 14 16 18 20 0.6 0.8 1 1.2 1.4 1.6 1.8 2 rec/it Figure 2: Ratio of Recursive/Iterative on Pentium from Equation 1. From Figure 2 we see that initially the iterative algorithm is faster, but eventually the recursive algorithm becomes faster. The switch over point occurs when the input no longer ﬁts in L2 cache. The recursive algorithm utilizes cache better, however, the overhead for recursion seems to be signiﬁcant for small sizes. Further improvement can be obtained by using larger unrolled code sequences. We generated straight-line code for small transform sizes and observed improve-ment over the iterative algorithm up to size 28 on both platforms. Code generators based on a recursive algo-rithm were superior (due to better register utilization) optimization ﬂags The amount of improvement, up to a factor of 7.4, depended on the compiler. We observed that Sun’s C compiler performed signiﬁcantly better than gcc. The ﬁrst two experiments suggest a recursive design that switches to an iterative algorithm inside the cache boundary which in turn is built from straight-line code for transforms up to size 256. However, performance can be highly unpredictable and consequently the best thing to do is systematically search all possible imple-mentations to determine the optimal combination of straight-line code, iteration, and recursion. Since it is infeasible to do an exhaustive search we used dynamic programming to search for an optimal algorithm. We assumed that the optimal algorithm for a given transform size is independent of the context in which it is being called. Under this assumption, we can search for the optimal algorithm for a given size by considering all possible applications of Equation 3 with the previously determined best algorithms used 0 500 1000 1500 2000 2500 1 1.5 2 2.5 3 3.5 relative runtime of splits of 12 runtime relative to best Figure 3: Distribution of all Split Times on Pentium for recursive evaluations. The number of cases for size n is equal to 2n−1, the number of ordered partitions of n. Figure 3 shows the times compared to the best time for all possible splits for n = 12. The same exper-iment run on the Sun produces a ﬁgure with diﬀerent characteristics. When n > 12 examing all possible splits becomes too expensive. Thus we restricted dynamic program-ming to binary splits (there are only n(n −1)/2 binary splits). Up to size 12 there was no penalty for this re-striction; however, we expect additional improvements when considering non-binary trees for larger sizes. The optimal binary split on the Sun for n = 20 was split[split[split[small,small], small],small] and the resulting computing time was 1.4875e-01 sec-onds, while the optimal split on the Pentium was split[small,split[small,split[small, split[small,split[small,split[small, small]]]]]] with computing time equal to 6.4500e-01. Compared to the iterative algorithm the improvement was a factor of 10.4 on the Sun. Observe that for the Sun the tree has a leftmost expansion, while the tree for the Pentium has a rightmost expansion. On both machines small[n] was optimal up to size n = 7 (small was not optimal despite being faster than the iterative algorithm). Despite this, the large values of small were not utilized when building larger transforms. On the Pentium all leftmost factors in the optimal formulas were equal to 2 when the size of the input no longer ﬁt in L1 cache. When comparing the Pentium to the Sun we see that the ratio of the run-times of the optimal formulas is much larger than the ratios in CPU speed. Finally we remark that the dynamic programming assumption is not always true. We observed that the runtimes of small[n] depend upon the stride at which they are applied. Nevertheless, so far we have been un-able to ﬁnd substantially better algorithms than those determined by dynamic programming. Additional ex-perimental results and details are available from . 5. REFERENCES L. Auslander, J. R. Johnson, and R. W. John-son. Automatic implementation of FFT al-gorithms. Technical report, Department of Mathematics and Computer Science, Drexel University, Philadelphia, PA, June 1996. K.G. Beauchamp. Applications of Walsh and re-lated functions. Academic Press, 1984. D. L. Dai, S. K. S. Gupta, S. D. Kaushik, J. H. Lu, R. V. Singh, C.-H. Huang, P. Sadayappan, and R. W. Johnson. EXTENT: A portable program-ming environment for designing and implementing high-performance block recursive algorithms. In Su-percomputing 1994, pages 49–58, 1994. S. Egner. Zur Algorithmischen Zerlegungstheorie Linearer Transformationen mit Symmetrie. PhD thesis, Univ. Karlsruhe, Informatik, 1997. Matteo Frigo and Steven G. Johnson. FFTW: An adaptive software architecture for the FFT. In ICASSP ’98, volume 3, pages 1381–1384, 1998. J. R. Johnson, R. W. Johnson, D. Rodriguez, and R. Tolimieri. A methodology for designing, mod-ifying, and implementing Fourier transform algo-rithms on various architectures. Circuits, Systems, and Signal Processing, 9(4):449–500, 1990. J. R. Johnson and Markus P¨ uschel. WHT: An adaptable library for comput-ing the Walsh-Hadamard transform, 1999. F.J. MacWilliams and N.J. Sloane. The theory of error-correcting codes. North-Holl. Publ. Co., 1992. C. Van Loan. Computational Frameworks for the Fast Fourier Transform, volume 10 of Frontiers in Applied Mathematics. Society for Industrial and Applied Mathematics, Philadelphia, 1992.
2643	https://www.statology.org/confidence-interval-difference-in-proportions/	Confidence Interval for the Difference in Proportions by Zach Bobbitt Posted on A confidence interval (C.I.) for a difference in proportionsis a range of values that is likely to contain the true difference between two population proportions with a certain level of confidence. This tutorial explains the following: The motivation for creating this confidence interval. The formula to create this confidence interval. An example of how to calculate this confidence interval. How to interpret this confidence interval. C.I. for the Difference in Proportions: Motivation Often researchers are interested in estimating the difference between two population proportions. To estimate this difference, they’ll go out and gather a random sample from each population and calculate the proportion for each sample. Then, they can compare the difference between the two proportions. However, they can’t know for sure if the difference in the sample proportons matches the true difference in the population proportions which is why they may create a confidence interval for the difference between the two proportions. This provides a range of values that is likely to contain the true difference between the population proportions. For example, suppose we want to estimate the difference in the proportion of residents who support a certain law in county A compared to the proportion who support the law in county B. Since there are thousands of residents in each county, it would take too long and be too costly to go around and survey every individual resident in each county. Instead, we might take a simple random sample of residents from each county and use the proportion in favor of the law in each sample to estimate the true difference in proportions between the two counties: Since our samples are random, the difference in proportions between the two samples is not guaranteed to exactly match the difference in proportions between the two populations. So, to capture this uncertainty we can create a confidence interval that contains a range of values that are likely to contain the true difference in proportions between the two populations. C.I. for the Difference in Proportions: Formula We use the following formula to calculate a confidence interval for a difference between two population proportions: Confidence interval = (p1–p2) +/- z√(p1(1-p1)/n1+ p2(1-p2)/n2) where: p1, p2: sample 1 proportion, sample 2 proportion z: the z-critical value based on the confidence level n1, n2: sample 1 size, sample 2 size The z-value that you will use is dependent on the confidence level that you choose. The following table shows the z-value that corresponds to popular confidence level choices: \| Confidence Level \| z-value \| --- \| \| 0.90 \| 1.645 \| \| 0.95 \| 1.96 \| \| 0.99 \| 2.58 \| Notice that higher confidence levels correspond to larger z-values, which leads to wider confidence intervals. This means that, for example, a 95% confidence interval will be wider than a 90% confidence interval for the same set of data. C.I. for the Difference in Proportions:Example Suppose we want to estimate the difference in the proportion of residents who support a certain law in county A compared to the proportion who support the law in county B. Here is the summary data for each sample: Sample 1: n1 = 100 p1 = 0.62 (i.e. 62 out of 100 residents support the law) Sample 2: n2 = 100 p2 = 0.46 (i.e. 46 our of 100 residents support the law) Here is how to find various confidence intervals for the difference in population proportions: 90% Confidence Interval: (.62-.46) +/- 1.645√(.62(1-.62)/100 + .46(1-.46)/100) =[.0456, .2744] 95% Confidence Interval: (.62-.46) +/- 1.96√(.62(1-.62)/100 + .46(1-.46)/100) = [.0236, .2964] 99% Confidence Interval: (.62-.46) +/- 2.58√(.62(1-.62)/100 + .46(1-.46)/100) =[-0.0192, 0.3392] Note:You can also find these confidence intervals by using the Confidence Interval for the Difference in Proportions Calculator. C.I. for the Difference in Proportions: Interpretation The way we would interpret a confidence interval is as follows: There is a 95% chance that the confidence interval of [.0236, .2964] contains the true difference in the proportion of residents who favor the law between the two counties. Since this interval does not contain the value “0” it means that it’s highly likely that there is a true difference in the proportion of residents who support this law in County A compared to county B. Zach Bobbitt Hey there. My name is Zach Bobbitt. I have a Masters of Science degree in Applied Statistics and I’ve worked on machine learning algorithms for professional businesses in both healthcare and retail. I’m passionate about statistics, machine learning, and data visualization and I created Statology to be a resource for both students and teachers alike. My goal with this site is to help you learn statistics through using simple terms, plenty of real-world examples, and helpful illustrations. One Reply to “Confidence Interval for the Difference in Proportions” Hi Zach, Thanks for your article!I was curious to know whether one could the same test method if one gets 2 proportions, but just one sample. Typically imagine you have a sample n=200 of a large country population.100 support law A120 support law B (it is possible to support both laws).Could you still use the same logic? Reply Leave a Reply Cancel reply
2644	https://byjus.com/physics/unit-of-speed/	Speed is defined as the rate of change of distance with time. It has the dimension of distance by time. Thus, the SI unit of speed is given as the combination of the basic unit of distance and the basic unit of Time. Thus, the SI unit of speed is metre per second. In everyday life, kilometre per hour or in countries like US and UK miles per hour are used as the unit of speed. \| \| \| Table of Contents SI Unit of Speed Other Speed Units Frequently Asked Questions – FAQs \| SI Unit of Speed The SI unit of speed can be derived from the formula of velocity. Basically, velocity is the vector equivalent of speed. Mathematically, velocity is given as the ratio of displacement to the time taken. d is the displacement measured using the SI unit of distance; that is a metre. t is the time interval measured using the SI unit of time; that is seconds. Here, we can see that length and time are base quantities, thus we can say that speed can be measured with the SI unit of length (metres) over the SI unit of time (seconds). Or in other words, to derive an SI unit of velocity or speed we will simply substitute corresponding units in the formula of velocity. Hence, the SI Unit of speed is metre per second or m/s or m.s-1. Dimensional Formula is M0.L1.T-1 Other Speed Units However, to mention a few, there are other units of speed such as kilometres per hour, feet per second, mach, miles per hour, rpm, and knots. Related articles: \| \| \| Unit Of Velocity \| \| Unit Of Distance \| \| Value Of C \| Stay tuned with BYJU’S for more such interesting articles. Also, register to “BYJU’S-The Learning App” for loads of interactive, engaging physics-related videos and unlimited academic assistance. Frequently Asked Questions – FAQs Q1 What is speed? Speed is defined as the rate in change of distance with time. It has the dimension of distance by time. Thus, the SI unit of speed is given as the combination of the basic unit of distance and the basic unit of Time. Q2 What is uniform speed? When an object covers equal distance in equal time intervals it is said to be in uniform speed. Q3 What is instantaneous speed? When an object is moving with variable speed, then the speed of that object at any instant of time is known as instantaneous speed. Q4 Define average speed? Average speed is defined as the ratio of total distance travelled by an object to the total time taken by the object. Q5 Define velocity? Velocity can be defined as the rate of change of the object’s position with respect to a frame of reference and time. Test your Knowledge on unit of speed Q5 Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin! Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz Congrats! Visit BYJU’S for all Physics related queries and study materials Your result is as below 0 out of 0 arewrong 0 out of 0 are correct 0 out of 0 are Unattempted Login To View Results Did not receive OTP? Request OTP on Login To View Results Comments Leave a Comment Cancel reply Niraj kumar August 23, 2022 at 7:34 pm If there is rate of change then it is necessary to mention with time Reply Deepa August 24, 2022 at 6:10 pm The rate of change can be measured with respect to quantities/variables other than time. For example, the rate at which the area of a circle changes with respect to its radius. Reply
2645	https://nvlpubs.nist.gov/nistpubs/bulletin/02/nbsbulletinv2n2p161_A2b.pdf	CALCULATION OF THE SELF-INDUCTANCE OF SINGLE-LAYER COILS. By Edward B. Rosa. 1. THE FORMULA. A well-constructed standard of self-inductance may be measured with considerable precision in terms of a resistance. If its value be computed from its dimensions, the two results should agree, provided the resistance is known in absolute measure. This affords a method of determining resistance in absolute measure, the success of which depends, of course, on how accurately the self-inductance can be computed, as well as upon the accuracy of the measurements. KirchhofT's method of determining resistance in absolute measure consists essentially in finding by experiment the mutual inductance of a pair of coils in terms of a certain resistance, the value of the mutual inductance having been calculated from the dimensions of the coils and their distance apart. Placing these two values equal to each other, the value of the given resistance becomes known in absolute measure. The mutual inductance of such a pair of coils can be determined experimentally by two measurements of self-inductance, in one of which the current flows in the same direction in the two coils and in the other it flows in opposite directions. If L and L' are the two values of self-inductance determined experi-mentally, L x and L2 the self-inductances of the two coils separately, and M their mutual inductance, then L' = L X-2M+L 2 n, L~ L '' 161 l62 Bulletin of the Bureau ofStandards. [ Vol. 2, He. 2. Such a pair of coils may have a large radius and relatively small rectangular cross section, such as used by Rowland and Glazebrook, or they may be coaxial solenoids of equal length, or one may be long and the other quite short, the short one being either within or without the long one. All these cases may be calculated quite accu-rately when the dimensions are accurately known. The simplest method, however, is to use a single coil of relatively large dimensions, wound with a single layer of wire, (i) calculating its self-inductance from its dimensions and (2) measuring its self-inductance directly in terms of a resistance. The dimensions of such a coil may be measured with great precision, supposing it wound on an accurately ground marble cylinder. In an article in the last number of this Bulletin Professor Coffin gave a description of such a standard of inductance, and calculated its value by two different formulae, one being an absolute • ; 1 i 2 i 3 ' 4 ' 5 ' 6 ' 7 ' 8 formula using elliptic integrals, first given by Lorenz, 2 and the second a formula de-rived by Coffin in the form of a converging series, being an extension of Rayleigh's formula for a cylindrical winding of a single layer. These two formulae give results agreeing to within 1 part in 50,000, an agreement which is highly satisfactory. As the measurements of the coil may readily be made of sufficient accuracy, the only question remaining is whether the two closely agreeing formulae are as accurate when applied to the coil in ques-tion as they appear to be. In deriving the formulae the current is supposed to be uniformly distributed over the surface of the cylinder, whereas the coil is actually wound with 661 turns of round wire, having a diameter of 0.0634 cm and insulated by a covering 0.0030 cm thick. Instead of the single-current sheet ab first assumed in deriving the formulae, the current sheet is subsequently supposed to be divided into n sections having altogether a self-inductance ;z 2 times as great as the single cuirent sheet. This can be realized if we assume a winding of n turns of a flat strip, l\n wide (where / is the length of the wind-Fig. 1. 1 No. 4, March, 1906. Wied. Annalen, 7, p. 170; 1879. £osa.] Self-Inductance ofSingle-Layer Coils. 163 ing), and no thickness, wound uniformly over the cylinder, at a distance of — cm from the surface of the cylinder. If the edges of the strip come together but do not make electrical contact, such a winding would be equivalent to a uniform current sheet and the formulae for the latter would apply if n 2 is inserted as a factor in Z, n being the number of turns. In the case of the coil constructed for the Bureau of Standards, when ;z = 66i, the wire is so small and the number of turns so large it was assumed by Coffin that the actual winding is substantially equivalent to the current sheet to which the formulae strictly apply. 2. CASE OF SHORT COIL. Let us first examine the case of a short coil having a radius of 25 cm and a length of 1 cm, wound with 10 turns of wire, the bare wire being 0.08 cm and the covered wire 0.10 cm in diameter. Coffin's formula for a short single-layer coil of radius a and length b reduces in this case to -^mmmmmmmm L = \izn % AXIS OF CYLINDER Fig. 2. the terms neglected amounting to less than one part in a hundred million of L. This is the formula given by Rayleigh 3 in 1881. Substituting # = 25, £=1, tz=io, we have for the self-inductance Z= 4TO Xiooj(i+^^)l0ge 2oo+I28 ^ 625 -o. 5 or, L = /.7raX 479.8595 cm. This value assumes the current flowing in a sheet, as though the coil were wound with 10 turns of very thin tape 1 mm wide, the radius of the winding being 25 cm. 3 Proc. Roy. Soc, 32, 1881, and Collected Papers. 164 Bulletin of the Bureau of Standards. \ Vol. 2, No. 2. The summation formula for the self-inductance of a single-layer coil of n turns is as follows: L= nL1+2(n-i)M 12+2(n-2)M 13+2(n- 3)M li+ . . . + 2M ln (2) where Lx is the self-inductance of a single turn, M 1% is the mutual inductance of any two adjacent turns, M 13 is the mutual inductance of the first and third or any two turns separated by one, and M ln is the mutual inductance of the first and last turns. For a coil of 10 turns this becomes £=ioA+i8^ 12+ 16^3+14^' + 2^110 ( 3 ) If we calculate Lx and the nine Ms, we can compute the self-induc-tance of the actual coil wound with wire of any particular size, the only assumption being that the current is uniformly distributed over the cross section of the wire. Max Wien's formula 4 for the self-inductance of a single circular turn of wire of radius a and radius of cross section p is (4) z=H( i+6hT-i - 75-°o83? Substituting # = 25, ^ = .04, we have A = 4^j( I +^2)loge5ooo- I .75--oo83 (-^?} = 47m X 6.76720 cm. Maxwell's formula for the mutual inductance of two coaxial cir-cles near each other reduces to the following when the two radii are equal, a being the common radius and b the distance between the circles. M =47r<zJ(i + -4 8tf T Hence, — \ira X 5.600924 cm. 2.0-2000 t b[ 16 a 1 (5) 1 .01 16 625 4 \Vied. Annalen, 53, p. 934; 1894. Rosa.} Self-Inductance of Single-Layer Coils. 165 Giving b the successive values 0.2, 0.3, 0.4, etc., the other values of the Ms are derived. The following values of the 10 terms of the summation formula (3) are thus obtained: 10 L x = 67.6720 iSM 12= 100.8166 i6M JM = 78.5254 14^14= 63-0344 12 Af 1B= 50.5787 ioM 16 = 39.9189 lis is less than the value SM 17 6Mn \M X , 2M 110 = 30-4779 = 21.9346 = 14.0898 = 6.8099 Sum = 473.8582 = L /[nra L Dr 4-ira [t will by formula (1) by-be noticed that the found 6.0013, which is about 1.25 per cent, values of the mutual inductances are independent of the size of the wire with which the coil is wound, but that the self-inductance is not. Thus, if the wire were only half a millimeter in diameter, 10 Lx would be 79.3034. In the latter case the total would be 5.6139 more by the summation formula than by the current sheet formula ; that is, it would be more than 1 per cent greater instead of being 1.25 a'? «b per cent smaller, as when the wire is }^^^^f3^£%0. 0.08 cm in diameter. It is evident that the summation formula, which takes account of the actual size and position of the wires on the coil, gives the true values and that the current sheet formula, which is very exact for a current sheet or for a winding of thin strip which is equiva-lent to a current sheet, can not be applied without modification to a winding of round wires. Recognizing the fact that the two cases are not identical, Coffin applies a correction by reducing the length of the solenoid, supposing the equivalent current sheet to extend only to the centers of gravity of the outer semicircles of the end wires ; that is, from a to b, Fig. 3. In this case the length of the winding would be shortened by 0.046 cm, making it 0.954 cm in-stead of 1.0 cm. Using this value of b in formula (1), we find the self-inductance of the coil to be Z = 4^x I oo\|(i + ( -^4)!\ l0ge ^? + ^954r_ o . 5l (6) \|\ 20,000/ &e .954 80,000 J j — 47rax 484.567 1 66 Bulletin of the Bureau of Standards. [Vol. 2, jvo. 2. This is 4.707 (about 1 per cent) more than the former value, which is already too great. Thus, the correction makes the matter worse. It is evident that a series of round wires can not be regarded as equivalent to a current sheet, no matter how fine the wires are ; for the magnetic field is very intense close to the wires, and the smaller the wire the stronger is the field and the greater is the self-induc-tance. For the case of mutual inductance between two coils at a distance from each other, the size of the wires is, of course, imma-terial (if not too great), and hence single-layer coils, so far as mutual inductance is concerned, are equivalent to current sheets. 3. SELF-INDUCTANCE OF SINGLE TURNS OF THIN STRIP. In order properly to compare the results given by the summation formula with the formula for the current sheet, we may apply the former to the case of a winding of very thin tape, to which the lat-ter applies strictly. Let the tape be 1 mm wide, of infinitesimal thickness, and let 10 turns be applied to the cylinder of radius 25 cm. The length b is then 1 cm, and the value found above by formula (1) is accurate, namely, ^=479-8595-To find 10 4, the first term of the summation formula, we use equation (5), replacing M by L and b by R, the geometric mean distance of the strip from itself ; for the self-inductance of a con-ductor is equal to the mutual inductance of two filaments having a distance apart equal to the geometric mean distance of the con-ductor from itself. The g. m. d. R of a flat strip of negligible thick-ness is 0.223130 of its breadth. 5 Hence 17 , 3 ^ 2\i Sa 1 R 21 , x =47™ log ^-2 L R 2 neglecting terms in — , which here amount to less than one part in a million. 5 Maxwell II, \ 692. Rosa.] Self-Inductance of Single-Layer Coils. 167 Here a — 25 cm and R = .0223130 cm. Substituting these numer-ical values, we find = 71.0000. \ira y 4. THE GEOMETRIC MEAN DISTANCE OF ONE STRIP FROM ANOTHER. In order to calculate the mutual inductance of the several strips upon one another it is necessary to know their geometric mean dis-tance from one another. Let Fig-. 4 rep-t , . ^ T ? C dx D resent the section of the first and third \|° a"" ""« V & turns of the winding, having a width a. ' "^ Find first the g. m. d. of a point P dis-tant b from the end taken as origin, from the line CD. Then a\o^R— I log (x—b) dx — (x—b) log (x—b)— (x—b)\ = ($a— &) log ($a— b) — (2a— b) log (2^— £)— a (8) where i? is the g. m. d. of the point P from the line CD. To find the g. m. d. of all points in AB from the line CD we integrate again, this time along the line AB, first changing b in (8) to x. Thus, «2 2 logR 2 = I (3a and 4«, and o and 0", respectively, and we should get log R9= — log 4«— 9 log 30+ ^ log 20-^ (10) 1 68 Bulletin of the Bureau ofStandards. [Voi. 2,No.2. The general expression is log Rn — ' log(;z-f-i)#— n 2 \og na--^-'- log(;z— i)a— $ (n) where na is the distance from the center of one element to the center of the other, a being the breadth of each element. Making n in the general expression equal to o, i, 2, 3, 4, etc., successively, we may find the values of the geometric mean distances of the first strip from the other nine. The above formula is, however, not well adapted to numerical cal-culation when n is large, as the logarithm of R is the difference between large positive and negative terms, and unless the latter are calculated with extreme accuracy there is likely to be an appreciable error in the differences. The expression for log R may, however, be transformed into a series well adapted to numerical calculation for all values of n greater than unity. Expanding the coefficients of equation (11) and recombining the terms we have, putting a equal to unity, log #n =J log 2-i)-7 ^g »+» l°gj^~+~ log O 2- 1)-! =^ log ( I -i 5 )+ «log («±i)-«log(^)+ log ,-2 (I2) Expanding the first three terms on the right of equation (12), 101 3 7Z 2 +l/l III \ log Rn = log n — °— — ( -\|,4- —j-f—.4-—.+ I + a(s+^+^+ 7?+ ) = log »-l-(l+^+&+$?+£?+ ) ~(+^ 2+ 4^ + 6V+ 8P + ) / 2 2 2 2 \ ^iogjea=iog«-(^+^+^+^+gg^+ )(i 3 ) Rosa. Self-Inductance ofSingle-Layer Coils. The coefficients of the denominators of the series are formed as follows: I i 2 I a+ 4""3 = ia I I 2 I 4+ 6~5 = 6o 112 I / 1 68 etc. The law of formation is evident, and any number of values can readily be calculated. The series is, however, very converge?it for all values of 11 greater than I. Thus when n= 2, five terms are sufficient to give an accurate value of R2 . These terms are as follows: log R 2 = log 2 — (.0208333 -f- .001041 7 + .0000930+ .0000108 + .0000015) whence ^3=1.95653. When n equals 3 or 4, four terms suffice; when n is 5 or 6, three terms suffice, and for larger values of n two or even one term is suf-ficient. Thus log ^5 =log 5 -(.0033333+ .0000267+ .0000004) log Rs — log 8— (.001 302 1 + .0000041) log Ru = log 14— (.00042 5 2 + .0000004) log R2i = log 24— (.000 1 44 7+ .0000000) The following values of the geometric mean distances (calling a unity) for the case under consideration were thus found: ^0 = 0.22313 ^ = 0.89252 ^=i;95653 ^3=2.97171 ^ = 3-9789o ^ = 4.98323 i?6 = 5.98610 R~ = 6.98806 #8 = 7-9957 ^ = 8.99076 5. CALCULATION OF THE MUTUAL INDUCTANCES FOR THE ASSUMED WINDING OF FLAT STRIP. . We can now calculate the last nine terms of formula (3), using formula (5), but putting the above values of R successively in place of the values of #, which in the case of round wires were 0.1, 0.2, 0.3 .... cm. 170 Bulletin of the Bureau ofStandards. [ Vol. 2, No. 2. Thus, ^= 4-{(I+f 6^gfi) log-200 089252 I (.089252) l6 625 = 47r0X5.7I463 , i8J/ 12 . and ^=102.8633. /.ira (^4) Proceeding in this way we find the values for the ten terms of equation (3). These values are given in column 2 of Table I. The values previously found for these terms for the winding of round wire are given in column 3. The differences between the corre-sponding terms are given in column 4, the sum of these differences being 6.0014, or 1.25 per cent, which represents the error of the current sheet formula when applied to this particular winding of round wire. TABLE I. For Strip For Round "Wires Differences 10 L t 71.0090 67.6720 3.3370 18MI2 102.8634 100.8166 2.0468 16MI3 78.8771 78.5254 .3517 14MI4 63.1671 63.0344 .1327 12 M, s 50.6421 50.5787 .0634 10M l6 39.9525 39.9189 .0336 8M I7 30.4965 30.4779 .0186 6M l8 21.9449 21.9346 .0103 4MI9 14.0950 14.0898 .0052 2MII0 6.8120 6.8099 .0021 Total 479.8596 473.8582 6.0014 Thus, we have the value of for flat strips, by the two formulae, as follows: 47T# By the current sheet formula (1), 479.8595 By the summation formula (4), 479.8596 The difference between these results, amounting to less than one part in a million, is inappreciable. This discrepancy is of quite a Rosa.] Self-Inductance of Single-Layer Coils. 171 different order of magnitude from the difference found between the current sheet formula and the summation formula for the case of round wires amounting to over 1.25 per cent. It is thus seen that this method of deducing the self-inductance of a coil by the method of summation is a practicable one, and in the case of a coil wound with flat strip it leads to correct results. There is no reason why it should not be equally exact in the case of round wires covered by insulation. We may therefore be sure that the value 473.8582 for the coil of ten turns of round wires is an accurate value for that case. 6. COIL OF LARGE NUMBER OF TURNS. When the coil has a large number of turns of wire, it becomes impracticable to use the summation formula, because of the large number of terms to calculate. Thus, in the inductance standard of the Bureau of Standards having 661 turns there would be 661 terms to calculate. But we may calculate the differences shown in Table I, between the self and mutual inductances for round wires and for flat strip, and apply their sum as a correction to the value found by the current sheet formula, and so obtain the desired value with a mod-erate amount of labor ; for the differences become very small after the distance becomes appreciable. Thus, at 1 cm in the above case the difference for a single pair of wires was only 1 per cent of its value at 1 mm. In calculating these differences in the case of the large coil, we should of course stop as soon as the difference becomes inappreciable. The standard of inductance of the Bureau of Stand-ards has three sections, which may be used singly or in combinations. Thus, there are six different cases. The number of turns and length of each coil is given in Table II, together with the value of the inductances as calculated by the current sheet formula. The mean radius of the coils is 27.0862 cm. The wire is round and has a diameter of 0.0634 cm bare, 0.0694 cm covered. As already indicated, the above values of the inductances, calculated by formulae which are correct for a winding which is equivalent to a current sheet, are too great for a winding of round wires, in which the thickness of the insulation is small. We have found Coffin's correction to be wrong, as it makes the corrected value larger in-stead of smaller than the value for a current sheet. We shall now 172 Bulletin of the Bureaic ofStandards, _voi. 2,No. 2 . TABLE II. Coil No. of Turns Length Inductance by Current-Sheet Formulae 1 221 15.3347 Cm 0.0361941 henry 2 251 17.3565 " .0441703 " 3 189 13.1945 " .0282220 " 1+2 472 32.6912 " .112722 " 2+3 440 30.5510 " .101810 " 1+2+3 661 45.8857 " .179615 " proceed to calculate the correction that must be applied to the above values to give the true values for this particular winding. This correction consists of two parts and may be written as follows: Ls—L = JLt+4M where Ls is the value of the inductance calculated from the current sheet formula, L is the true value of the inductance, JLX is the correction depending on the self-inductance of the n single turns of the coil, and z/Af is the correction depending on the mutual induc-tances of each of the n turns on the (n—i) other turns. The self-inductance of any coil of n turns may therefore evidently be written L=Le JLx -zlM the corrections <dLx and 4M being subtracted from the value of the self-inductance given by the current sheet formula. The total self-inductance of any coil of n turns may also be written where 2Z X is the sum of the self-inductances of all the n turns taken separately, and ^M is the sum of the mutual inductances of each of the n turns on the (;/— 1) other turns. The first term of the correc-tion, z/Z x , is thus the difference between the value of 2ZX for a wind-ing of flat strip which would exactly represent the current sheet and its value for the actual winding of round insulated wire; while the second term is the difference between 1M for the winding of strip and the actual winding of wire. Rosa.} Self-Inductance of Single-Layer Coils. 173 7. TO CALCULATE THE CORRECTION z/L. Maxwell's formula for the self-inductance of a single circular turn of round wire, which is practically equivalent to Wien's, is derived from the formula for the mutual inductance of two parallel coaxial circles, by replacing b the distance apart of their planes by R the geometric mean distance of the section, which in this case is a circle. Thus L= 47ra[ i + ~? 3 # 16 a Jlog_- 2_^-\| (I7 ) The geometric mean distance R for the circular section of straight wire is pe~% = . 778801/3 where p is the radius of the section. This is not quite exact where the conductor is a circle, but is sufficiently exact when a is large and p relatively small. Wien's formula is derived directly by integrating the expression for the mutual induc-tance of two parallel circles (of infinitesimal section) twice over the area of the circular cross section of the conductor. By comparing the results of the two formulae we may get an idea of the magnitude of the error arising from using the g. m. d. of the circle as the same as that for a rectilinear conductor. If ^ = 25 cm and £ = .05 cm, Z, the self-inductance of one turn of wire, is 654.40537^ cm by Wien's formula and 654.40533^ cm by Maxwell's. The difference is inap-preciable. We need not hesitate, therefore, to use Maxwell's formula in the present case, where p is less than 0.04 cm. The self-inductance Lx of a round wire of radius p is given by formula (17), where R will be written Rx and will have a value 0.778801/3. Similarly, the self-inductance Z2 of a single turn of flat strip of width d wound on a circle of radius a will be given by the same equation, except that R (here written R2) will have a value 0.223130^. The difference between the two will be Z8-Z 1 = 47ralog— 1 , (18) neglecting small quantities, which here amount to less than one part in a million. Where the bare wire has a diameter 0.0634, p = .OT ) iy. The strip has a width D equal to the diameter of the covered wire, = 0.0694. -J 174 Bulletin of the Bureau ofStandards. [vol. 2, no. 2. Hence J^1 = .yyS8oi p and ^ = .223130 D •'-#= 1-594-Thus the excess of the self-inductance of the n turns taken sepa-rately of a coil wound with flat strip of width 0.0694 cm over the self-inductance of the same number of turns of round wire of 0.0634 cm diameter (0.0694 covered) is or, ^JL1 = 47ran loge 1.594. (19) In the standard of inductance under consideration a = 27.0862 cm, and the number of turns of wire in each of the six sections is given in Table II. Substituting these values in equation (19), we have the following values of the corrections JL\ Coil I ^A= 35073 cm 2 ^A= 39833 " 3 JL,= 29993 " 1 + 2 JL,= 74904 " 2+ 3 JLh= 69827 " 1 + 2 + 3 JL,= 104900 " 8. TO CALCULATE THE CORRECTION z/M. The corrections JM are found in a similar manner. The mutual inductance of two parallel circles of round wire is given by equation (5), where b is the distance apart of the centers of the wires, which is also the geometric mean distance, supposing the radius a is large and b relatively small. In the case of the two strips, forming part of a current sheet, we have found the g. m. d. to be less than the distance apart of their centers. The mutual inductance of two such strips will therefore always be greater than that of two round wires, the distance apart of their centers being supposed the same in each case. The mutual inductance in the case of the wires will be, ^=H( I+ A'?) log J-'-rei] < 2°) Rosa.] Self-Inductance ofSingle-Layer Coils. 175 and for the strip, putting kb for the geometric mean distance of the strips, where k is always less than unity, The difference is ^ s-^i= 4™{iog l+^ x^+ 3^ ^ S~ 3 log ¥)) (22) In the above expression for M 2—M t) k does not differ from unity appreciably except where b is very small ; that is, where the strips are very near together. Thus the second part of the expression is 1 b 2 . negligible in all cases, for when the coefficient —? — 2 is more than 0.000001 (its value for b=i mm) k is so nearly unity that the quantity within the parentheses is very small. Thus, for b=i cm, the coefficient is 0.0001, £= 0.999575 and the quantity in the parentheses is about 0.001, so that the term amounts to 0.0000001 and can be neglected. The correction BM for any pair of wires is thus SM=47ralogj (23) Since the value of k depends upon the distance apart of the two turns of wire or strip under consideration, it is evident that there will be as many different terms as there are different distances. Thus hM^ — ^ira log — — /.'Trahl BM 2 = 47ra log -r — A^nrah^ etc. (24) /e2 where hM x is the correction for a pair of adjacent wires, the distance of their centers being Z>, the diameter of the covered wire, and Dkx their g. m. d. ; &M 2 is the correction for a pair of wires distant 2Z>, 2Dk2 being their g. m. d., etc. If there are n turns of wire on the coil we shall have 29572—06 2 176 Bulletin of the Bureau of Standards. Woi. 2, No. 2. JM=2(n—i)hM x+ 2(n—2)hM 2 --2{n—? ))hMz+ = 87ra [(n-i)81 +(n-2)82+(n- 3)8s+ ] (25) In a coil of n turns there would be (n— 1) terms in this equation, but inasmuch as k rapidly approaches unity when the distance is increased, the correction terms decrease rapidly in value, so that only a limited number of terms need be calculated. In Table III the consecutive values of the geometric mean dis-tances are given up to R9 and then every fifth value is given up to TABLE III. Geometric Mean Distances and Corrections Depending upon Them. Geometric Mean Distances R -4" 1 k •-lofcj R z 0.89252 0.89252 1.12042 0.11371 R2 1.95653 .978265 1.02223 .02198 R3 2.97171 .99057 1.00952 .00948 R4 3.97890 .99473 1.00529 .00528 Rs 4.98323 .99665 1.00336 .00336 R6 5.98610 .99768 1.00233 .00233 R7 6.98806 .99829 1.00171 .00171 R8 7.98957 .99869 1.00131 .00131 R9 8.99076 .99897 1.00103 .00103 Ru 13.99405 .999575 1.000425 .000425 R19 18.99531 .999753 1.000247 .000247 RaA 23.99653 .999855 1.000145 .000145 R29 28.99724 .999905 1.000095 .000095 R3A 33.99754 .999928 1.000072 .000072 R39 38.99785 .999945 1.000055 .000055 RAA 43.99811 .999957 1.000043 .000043 R49 48.99828 .999965 1.000035 .000035 R99 98.99920 .999992 1.000008 .000008 7?49 , corresponding to the first and fiftieth turns of wire, respectively. The second column gives the coefficients k obtained by dividing each geometric mean distance by the corresponding distance between centers of the wire, indicated by its subscript; thus, R% is divided Rosa.] Self-Inductance of Single-Layer Coils. 177 by 2, etc. The third column gives the values of -r and the fourth column gives the corrections S, the natural logarithms of — . These fc 120 100 90 70 60 50 30 Fig. 5. Curves showing relative values of the correction factors 82 , 8 2 , etc. The ordinate at the point 1 is 5 lt at 2 is <52 , etc. The ordinates of curve B are 100 times greater than those of A for the same abscissae. The corrections above 10 are very small, but have been included up to 50. 2 3 4 5 6 7 20 30 values are plotted in Fig. 5. The curve shows the values of 8 up to ^ = 45. This curve shows how rapidly the quantity 8 falls off i 7 8 Bulletin of the Bureau of Standards. [ Vol. 2, No. 2. as the distance increases. For the first and fiftieth wires it is only three ten-thousandths of its value for the first pair of wires, whereas the correction for the first and one-hundredth wire is less than one ten-thousandth of the first pair. In Tables IV and V the products (n— i)^, (n— 2)S2 , etc., are given TABLE IV. Corrections to Mutual Inductance. B. S. Standard Coil. Sections i, 2, 3. T „<r Coil 1, n = 22i Coil 2, n=25i Coil 3, n = i8g L°g k n-i Product n-i Product n-i Product S 1= . 11371 220 25.016 250 28.428 188 21.377 5 2= . 02198 219 4.814 249 5.473 187 4.110 5 3= . 00948 218 2.066 248 2.351 186 1.763 5 4=.00528 217 1.146 247 1.304 185 0.977 5 5= .00336 216 0.726 246 0.827 184 0.618 56=.00233 215 0.501 245 0.571 183 0.426 5 7= . 00171 214 0.366 244 0.417 182 0.311 5 8= .00131 213 0.279 243 0.318 181 0.237 59= .00103 212 0.218 242 0.249 180 0.185 2510 u =.00321 209 0.671 239 0.767 177 0.568 2515 19—00163 204 0.332 234 0.381 172 0.280 S520 24 =. 00096 199 0.191 229 0.220 167 0.143 ^"25-100 0.493 0.576 0.405 z/M. Zira = 36.819 41.882 31.400 87rft=680.75, \^/M = 25,064 cm 28,511 cm 21,375 cm for the six sections of the N. B. S. standard of inductance, as indi-cated in equation (25). The sum of the terms and Sira times the sum are given for each coil, the latter being the correction JM for mutual inductance. Each term is given separately up to 89 , after that the sum of five consecutive terms is given, as 2S10 14 , and finally the correction for all the turns from 25 to 100 is given in one. It will be noticed that the first four terms amount to about 90 per cent of the whole correction, and that the sum of the terms after S20 amounts to only a little more than 1 per cent of the whole correc-tion, or 1 part in 100,000 of the whole inductance. Rosa.] Self-Inductance ofSingle-Layer Coils. 179 The geometric mean distances given are those between parallel straight wires, and not parallel circles. But the quantity k is the ratio between the g. m. ds. of flat strips and round wires, and this TABLE V. Corrections to Mutual Inductance. S. Standard Coil. Sections 1+2, 2+3, 1+2+3. Coils 1 + 2, n=472 Coils 2+3, n=44o Coils 1+2+3, n=66i LogJL n-i (n-2)etc. Product n-i (n-2)etc. Product n-i (n-2)etc. Product 51=.11371 471 53.557 439 49.919 660 75.049 5 2= .02198 470 10.331 438 9.627 659 14.485 5 3= . 00948 469 4.446 437 4.143 658 6.238 <5 4 =.00528 468 2.471 436 2.301 657 3.469 <55 =.00336 467 1.569 435 1.462 656 2.204 <5 6= .00233 466 1.086 434 1.011 655 1.526 5 7=.00l7l 465 0.795 433 0.740 654 1.118 88=. 00131 464 0.608 432 0.566 653 0.855 89=. 00103 463 0.477 431 0.444 652 0.672 2S10_U=.00321 460 1.477 428 1.374 649 2.083 251519=.00163 455 0.742 423 0.689 644 1.050 2d. 20_2i=.00096 450 0.432 418 0.401 639 0.614 -^25-100 1.180 1.095 1.700 87ta ~ 79.171 73.770 111.063 87ra=680.75> Z/M = 53,896 cm 50,219 cm 75,606 cm can not be appreciably different for the case of parallel circles at moderate distances from its value for straight conductors. For the difference in the g. m. ds. is very small, and the difference in their ratios would be a small quantity of the second order. Hence the corrections calculated by (25) and given in Tables IV and V must be very exact. In Table VI is given a summary of the values of the two correc-tions 4L X and JM for the six sections of the standard coil, together with the values of the inductances given by the current sheet for-mula and the final corrected values. It will be noticed that the i8o Bulletin of the Bureau of Standards. [ Vol. 2, No. 2. whole correction varies in amount from i part in 600 for the smaller section to 1 part in 1,000 for the whole coil. This is a very large quantity in a standard when we remember the extreme TABLE VI. Summary of Corrections 4LX and 4M for the Six Sections of the N. B. S. Standard of Inductance, and the Corrected Values of the Inductances Coil No. 1 n=22i Coil No. 2 11 = 251 Coil No. 3 n = i8g Correction for self-in-ductance 4'L 1 35073 cm 39833 cm 29993 cm Correction for mutual inductance JM 25064 " 28511 " 21375 " Total correction JL 60137 " 68344 " 51368 " Total correction JL 0.0000601 henry 0.0000683 henry 0.0000514 henry Inductance by current sheet formula L s 0.0361941 " 0.0441703 " 0.0282220 " Corrected inductance L 0.0361340 " 0.0441020 " 0.0281706 " Coils 1+2 n= 472 Coils 2+3 n=440 Coils 1+2+3 n=66i Correction for self-in-ductance J x 74904 cm 69827 cm 104900 cm Correction for mutual inductance JM 53896 " 50219 " 75606 " Total correction JL 128800 " 120046 " 180506 " Total correction JL 0.000129 henry 0.000120 henry 0.000180 henry Inductance by current sheet formula Ls 0.112722 " 0.101810 " 0.179615 " Corrected inductance L 0.112593 " 0.101690 " 0.179435 " precision with which the measurements of the dimensions were made and the high sensibility obtainable in measuring self-inductance. Rosa.\ Self-Inductance of Single-Layer Coils. 181 9. THE CORRECTION TABLES. It is possible to put these two correction terms into such form that they may be quickly applied to any single layer coil, by the aid of tables of constants. Since 0.77880/0 = 0.3894^, where d is the diam-eter of the bare wire, equation (19) may be written ^A = 4^log e^g4j = \^an log e ( 1-7452-^ J (26) From equation (25) we have n-50 AM=%waSHn log \|j n — 1 ^ ' The sum of these two terms may be written AL=4L x --4M=$nran_A + B~\ (27) where A stands for log e I 1.7452— ) and B stands for - 2( n log — ) \ D) n \ k), the summation being carried from n= (n— 1) to n— 1 for coils of less than 50 turns and up to («— 50) for coils of more than 50 turns. The values of the constants A are given in Table VII with the ratios — as arguments. D is the width of the current sheet corre-sponding to one turn of wire. If the wire is wound so that the con-secutive turns are in contact, D is also the diameter of the insulated wire. The mean length of the coil divided by the number of turns gives the value of D to be used. The mean diameter of the bare wire is d. The two corrections AL X and AM are to be subtracted from Ls . When the ratio — is less than about 0.57, A is negative and hence AL X is negative, and it is added; AM\s always positive. The values of the constant B are given in Table VIII with n the total number of turns in the coil as argument. By means of these two tables and equation (27) it is easy to find the correction for any particular case, as the following illustrations will show. 182 Bulletin of the Bureau of Standards. [Voi. 2,No.2. TABLE VII. Values of Correction Term A , Depending on the Ratio — of the Diameters of Bare and Covered Wire on the Single Layer Coil. d A Ai d A Al D D 1.00 0.5568 100 .70 0.2001 144 .99 .5468 101 .69 .1857 146 .98 .5367 103 .68 .1711 148 .97 .5264 104 .67 .1563 150 .96 .5160 105 .66 .1413 152 .95 .5055 106 .65 .1261 155 .94 .4949 107 .64 .1106 157 .93 .4842 108 .63 .0949 160 .92 .4734 109 .62 .0789 163 .91 .4625 110 .61 .0626 166 .90 .4515 112 .60 .0460 168 .89 .4403 113 .59 .0292 171 .88 .4290 114 .58 .0121 174 .87 .4176 116 .57 — .0053 177 .86 .4060 117 .56 — .0230 180 .85 .3943 118 .55 — .0410 184 .84 .3825 120 .54 — .0594 187 .83 .3705 121 .53 — .0781 190 .82 .3584 123 .52 ' — .0971 194 .81 .3461 124 .51 — .1165 198 .80 .3337 .3211 126 .50 — .1363 .79 127 .50 — .1363 .78 .3084 1053 129 .45 — .2416 .77 .2955 1178 131 .40 — .3594 .76 .2824 1335 133 .35 — .4928 .75 .2691 1542 134 .30 — .6471 .74 .2557 1823 136 .25 — .8294 .73 .2421 2232 138 .20 —1.0526 .72 .2283 2877 140 .15 — 1.3403 .71 .2143 4054 142 .10 — 1.7457 .70 .2001 Rosa.} Self-Inductance of Single-Layer Coils. TABLE VIII. Values of the Correction Term B, Depending on the Number of Turns of Wire on the Single-Layer Coil. 183 Number of Turns B Number of Turns B 1 0.0000 50 0.3186 2 .1137 60 .3216 3 .1663 70 .3239 4 .1973 80 .3257 5 .2180 90 .3270 6 .2329 100 .3280 7 .2443 125 .3298 8 .2532 150 .3311 9 .2604 175 .3321 10 .2664 200 .3328 15 .2857 300 .3343 20 .2974 400 .3351 25 .3042 500 .3356 30 .3083 600 .3359 35 .3119 700 .3361 40 .3148 800 .3363 45 .3169 900 .3364 50 .3186 1000 .3365 10. EXAMPLES ILLUSTRATING THE USE OF THE CORRECTION TABLES. Example 1.—Coil of 10 turns, radius 25 cm, length 1 cm, diameter of insulated wire 0.1 cm = Z), diameter of bare wire 0.08 cm — d; thus — = 0.8. From Table VII, ^ = 0.3337 " VIII, .#= 0.2664 ^-f-^ = 0.6001 n(A + B) =6.001 .-. z/Z, —\iraY. 6.001. This value of z/Z, is to be subtracted from Ls to obtain the true value L of the self-inductance. 184 Bulletin of the Bureau of Standards. [Vol. 2, jvo. 2. This is the value found already (p. —), where the correction was determined by calculating L by the summation formula (2) and Ls by the current sheet formula (1). Example 2.—Coil of 50 turns, radius a — 20 cm, length £ = 5 cm, D — 0.1 cm, ^=0.075 cm; thus — = 0.75. By formula (1) [/ 2<5 Y 160 , 2S = 4™x 2 5ooj(i + ^-2 )loge 32+ L-. 5o Ls = 47raX 2500 X 2.972945 = 47r<2X 7432.36 From Table VII, ^ = 0.2691 " VIII, .# = 0.3186 .4 +^ = 0.5877 ; (.4 + ^) = 29.39 .-. L = Ls-JL = 47ra (7432.36-29.39) 4^=251.3274 .-. L= 1860570 cm = 1.86057 millihenrys. The correction to Ls here amounts to 0.4 per cent. Example j.—As an extreme case to test the method we may cal-culate the self-inductance of a single turn of wire. Let us take the particular case already calculated by Wien's and Maxwell's formulae, (4) and (7), page— . The radius a = 25 cm, the diameter of the bare wire = 1 mm. We may now assume that the wire is covered and that the diameter D is 2 mm. Then — = 0.5. In using Ray-leigh's current sheet formula we take the length of the equivalent current sheet as equal to D. We thus have = 4H( I +^fc5) log^+^6 2 -5 " a5 = 4H( I+^^) 6 - 9O7755+2^0^- a5 = 47TtfX 6.4O777. Rosa.] Self-Inductance of Single-Layer Coils. 185 From Tables VII and VIII A = —0.1363 and B = o. Thus, since «=i, z/Z = 47rax (— 0.1363), and being negative is added to L& . Hence Z=47T« (6.4O777 + O.I 363) = 47T«X 6.544O7 = 654.407W. This is practically identical with the values given by the other formulae (p. —), the slight difference being due to the fact that the correction term A is carried only to four places of decimals. If we had taken the bare wire of diameter 0.1 cm as equivalent to a current sheet o. 1 cm long in the above formulae for Zs , we should have obtained a different value for Zs , but in that case — would be D unity and A would be +-5568. The resulting value of L would, however, be the same as before. Example 4.—Take the first section of the B. S. standard. Here n — 221, ^=.0634, Z)=.o6o4. Hence — = .9135. From Table VII ^ = 0.4663 " VIII ,9= 0.3332 A +£ = o.j995 n (A + B) = 176.690 47m = 340.375 .. JL—/\iran (A--E) — 60,141 cm = .0000601 henry. This is practically identical with the correction calculated directly for this section (Table VI). Taking the whole coil, for which ;z = 66i, we have ^ = 0.4663 ^= 0.3360 ^ +^ = 0.8023 n (^+^ = 530.32 4^ = 340.375 JL — ^iran (A--B)= 180,508 cm = .0001805 henry which is the same value found previously for JL. 1 86 Bulletin of the Bureau ofStandards. [Vol. 2, no. 2. These examples are sufficient to illustrate the use and the accuracy of the correction Tables VII and VIII. By their use an accurate value of the self-inductance of a single layer coil of any number of turns can be calculated, if the proper current sheet for-mula is employed. Rayleigh's formula (1) already used is the most convenient one for short coils ; that is, for coils whose length is small compared with the radius. Coffin's formula is an extension of Rayleigh's, and may be used where the length is too great to tf b 6 b s omit terms in — 4 , — 6 , and -g (b being the length and a the radius). Lorenz's formula is an absolute one, and may be used for coils of any length, being more exact than Coffin's for coils whose length is as great as the diameter, but agreeing with Coffin's very exactly for all lengths up to b—a. For convenience of reference I here give these three formulae. 1. Rayleigh's formula: z.= 4„,,-{ 1o8 5-; +^,„4«+ ;)} <a8, 2. Coffin's formula: T a (\ 8a 1 . b 2 L Sa . i\ 1 PL 8a i\ L8= 4,an'{log --+(log T+-)-— -(log y--) 131072 a\ s b 120/ 4,194,304 « 8\ B b 420/J v y; 3. Lorenz's formula : L 47r;^( f I -J% Y(V?-V)E+d& F-W\ (30) In the above formulae a = radius of the coil and b — length of the coil, the length being the mean over-all length including the insulation on the first and last wires. In formula (3), d is the diagonal of the coil = ^/4^ 2+^ 2 and E and F are the complete elliptical integrals of the first and second kinds, respectively, to modulus k, where A_2#_ 2a k —~d~ I z\ip The subscript s? is attached to L in each case as Rosa.] Self-Inditctance ofSingle-Layer Coils. 187 a reminder that each is a current sheet formula, and the value of the inductance must in every case be corrected by formula (27) and Tables VII and VIII in order to give the true inductance of a wind-ing of round wires. A winding of square or rectangular wire would also require correction, the only winding for which the formulae are correct being a winding of strip of infinitesimal thickness in which the edges of the strip come together without making electrical con-tact, and so fulfilling the current sheet conditions assumed in deriv-ing the formula. In a subsequent paper I shall discuss the case of coils having more than one layer. oc ^a a /
2646	https://www.quora.com/Justify-the-sum-1+2+3+-+n-n%C3%97-n+1-2-How-does-the-formula-take-its-form-as-it%E2%80%99s-written	Justify the sum 1+2+3+…+n= n×(n+1) /2. How does the formula take it's form as it’s written? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Summation Identity Proofs (mathematics) Arithmetic Sequences Formulas in Maths Series (mathematics) Mathematical Equations Arithmetic Summation of Series 5 Justify the sum 1+2+3+…+n= n×(n+1) /2. How does the formula take it's form as it’s written? All related (46) Sort Recommended Carl Bryan B.S. in Mathematics, Carnegie Mellon University (Graduated 1972) · Upvoted by Horst H. von Brand , PhD Computer Science & Mathematics, Louisiana State University (1987) · Author has 3.3K answers and 4.7M answer views ·7y Originally Answered: How do you prove 1+2+3+…+n terms=n(n+1) /2? · I have always liked this way. Write the series twice, with the second time having the terms in reverse order. 1 + 2 + 3 + … + (n-2) + (n-1) + n n + (n-1) + (n-2) + … + 3 + 2 +1 Now add the two series together term by term. (n+1) + (n-1+2) + (n-2+3) + … + (3+n-2) + (2+n-1) + (1+n) = (n+1) + (n+1) + (n+1) + … + (n+1) + (n +1) + (n+1) You have added (n+1) a total of n times, so the sum is n(n+1). You added the series twice, so adding the series once will give you n(n+1)/2 Upvote · 999 135 9 3 9 3 Related questions More answers below How do you prove 1+2+3+…+n terms=n(n+1) /2? What is a formula for the sum S n=1(1)(2)(3)+1(2)(3)(4)+1(3)(4)(5)+⋯+1 n(n+1)(n+2))S n=1(1)(2)(3)+1(2)(3)(4)+1(3)(4)(5)+⋯+1 n(n+1)(n+2)) ? Using the formula S=n (n + 1) /2, what is the sum of 1 + 2 + 3 + … + 575? Given 1+2+3+…+n=n(n+1) /2, what is the sum of 105 entries? How do I prove that (2 n+2 n−1)(2 n+1−2 n)=3/2?(2 n+2 n−1)(2 n+1−2 n)=3/2? Doug Dillon Ph.D. Mathematics · Author has 12.4K answers and 11.4M answer views ·6y Originally Answered: How do you prove 1+2+3+…+n terms=n(n+1) /2? · Upvote · 99 12 9 1 Sherwin Dsouza B.Tech in Computer Engineering, Vishwakarma Institute of Technology, Pune (Graduated 2021) · Author has 94 answers and 325.2K answer views ·7y Let's take S n=1+2+3+…+(n−1)+n S n=1+2+3+…+(n−1)+n Taking the reverse(as addition follows commutative law), S n=n+(n−1)+(n−2)+…+2+1 S n=n+(n−1)+(n−2)+…+2+1 Adding both the sums, 1 s t 1 s t term with 1 s t 1 s t term, 2 n d 2 n d term with 2 n d 2 n d term and so on, 2 S n=(n+1)+(n+1)+(n+1)+…+(n+1)2 S n=(n+1)+(n+1)+(n+1)+…+(n+1) Here there are exactly n such terms because we're adding term by term so there will be ’n’ (n+1) terms. 2 S n=n(n+1)2 S n=n(n+1) S n=n(n+1)2 S n=n(n+1)2 There's another method to get the sum which involves using telescoping series, which is used as a general method to calculate sums like sum of n² and sum of n³, but that method is too long, this is the simplest one. Upvote · 9 1 Barry Zennia Wanderer at Earth (planet) (1993–present) · Author has 181 answers and 295.4K answer views ·7y Suppose, n is odd.. Say 1,2,3,4,5… Then you can add always first term with last term, second term and last but term and all of them yield the same result. The result is n+1. Since you are pairing up you will have (n-1)/2 such pairings and the middle term is left out which is (n+1)/2 In above example, the pairs are (1,5), (2,4) & 3. Thus, the total sum is (n-1)(n+1) + (n+1)/2 Taking out (n+1) common, we have n(n+1)/2. Suppose n is even… Here too we have the same situation but there is middle term. There are exactly n/2 pairs each of which sum up to (n+1). Hence total sum is n(n+1)/2 Thus, whether n is Continue Reading Suppose, n is odd.. Say 1,2,3,4,5… Then you can add always first term with last term, second term and last but term and all of them yield the same result. The result is n+1. Since you are pairing up you will have (n-1)/2 such pairings and the middle term is left out which is (n+1)/2 In above example, the pairs are (1,5), (2,4) & 3. Thus, the total sum is (n-1)(n+1) + (n+1)/2 Taking out (n+1) common, we have n(n+1)/2. Suppose n is even… Here too we have the same situation but there is middle term. There are exactly n/2 pairs each of which sum up to (n+1). Hence total sum is n(n+1)/2 Thus, whether n is even or odd, the sum is n(n+1)/2. Upvote · Related questions More answers below Is there is any formula for: 1! +2! +3! +4! +⋯+n! And 1! 2+2! 2+3! 2+⋯+n! 2? What is the sum of 1+1/2+1.3…+n/n+1? When is 2 n−1+1≡0(mod n)2 n−1+1≡0(mod n)? What's the difference between these formula n (n+1) /2 or, n (1+n) /2? What is the sum of the 2^10th, (2 power 10th) and is this correct formula to solve this question X.n (n+1) /2? Nick Shales Combinatorics enthusiast · Author has 419 answers and 1.5M answer views ·Updated 6y Originally Answered: How do you prove 1+2+3+…+n terms=n(n+1) /2? · Here is a lovely combinatorial “proof without words”, but with some words. :) There are ∑n k=1 k∑k=1 n k green balls in total, as evidenced by the summation on the right hand edge of the figure which adds green balls layer by layer. On the other hand we may uniquely specify any one of these green balls with the intersection of the right and left slanted lines (in that order) as shown. Said lines must emanate from similarly slanted arrows beneath the bottom layer of green balls, these arrows may have their 2 ordered positions chosen from the n+1 n+1 positions shown in (n+1 2)=1 2(n+1)n(n+1 2)=1 2(n+1)n Continue Reading Here is a lovely combinatorial “proof without words”, but with some words. :) There are ∑n k=1 k∑k=1 n k green balls in total, as evidenced by the summation on the right hand edge of the figure which adds green balls layer by layer. On the other hand we may uniquely specify any one of these green balls with the intersection of the right and left slanted lines (in that order) as shown. Said lines must emanate from similarly slanted arrows beneath the bottom layer of green balls, these arrows may have their 2 ordered positions chosen from the n+1 n+1 positions shown in (n+1 2)=1 2(n+1)n(n+1 2)=1 2(n+1)n ways. The number of choices of 2 positions for the ordered arrows is therefore exactly the number of green balls, hence: n∑k=1 k=(n+1 2)=1 2(n+1)n.■◼∑k=1 n k=(n+1 2)=1 2(n+1)n. Upvote · 99 27 Bhupendra Charan BS-MS from Indian Institute of Science Education and Research, Pune (IISER-P) (Graduated 2018) · Author has 913 answers and 7.4M answer views ·7y Consider the sum: 1+2+3+⋯+n 1+2+3+⋯+n ⋆⋆ This is an Arithmetic progression having the first term as 1 1 and common difference 1 1 ⋆⋆ We know that sum of an A.P. is given by n 2[2 a+(n−1)d]n 2[2 a+(n−1)d] where, n=n= no. of terms a=a= first term d=d= common difference In this case, we have, a=1 a=1 d=1 d=1 On substituting Continue Reading Consider the sum: 1+2+3+⋯+n 1+2+3+⋯+n ⋆⋆ This is an Arithmetic progression having the first term as 1 1 and common difference 1 1 ⋆⋆ We know that sum of an A.P. is given by n 2[2 a+(n−1)d]n 2[2 a+(n−1)d] where, n=n= no. of terms a=a= first term d=d= common difference In this case, we have, a=1 a=1 d=1 d=1 On substituting these values, we get, n 2[2⋅1+(n−1)⋅1]n 2[2⋅1+(n−1)⋅1] ⟹n 2[2+n−1]⟹n 2[2+n−1] ⟹n 2[1+n]⟹n 2[1+n] ⟹n(n+1)2⟹n(n+1)2 Hence Justified. ¨⌣⌣¨ Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes Upvote · 9 2 Mohammad Afzaal Butt B.Sc in Mathematics&Physics, Islamia College Gujranwala (Graduated 1977) · Author has 24.6K answers and 22.9M answer views ·4y Originally Answered: How can I prove that 1+2+3+4+5+6+7+8+9+…+n=n(n+1) /2 is true? · Let S=1+2+3+⋯+(n−2)+(n−1)+n Let S=1+2+3+⋯+(n−2)+(n−1)+n ⟹S=n+(n−1)+(n−2)+⋯+3+2+1⟹S=n+(n−1)+(n−2)+⋯+3+2+1 ⟹2S=(n+1)+(n+1)+(n+1)+⋯+(n+1)+(n+1)+(n+1)⟹2S=(n+1)+(n+1)+(n+1)+⋯+(n+1)+(n+1)+(n+1) ⟹2S=n(n+1)⟹2S=n(n+1) ⟹S=1 2 n(n+1)⟹S=1 2 n(n+1) ⟹1+2+3+⋯+(n−2)+(n−1)+n=1 2 n(n+1)⟹1+2+3+⋯+(n−2)+(n−1)+n=1 2 n(n+1) Upvote · 9 1 Naveen Mukesh Mukesh B.E in Electronics&Instrumentation Engineering, Madras Institute of Technology, Tamil Nadu, India (Graduated 2021) ·7y The first n natural numbers are:1,2,3,……..n. Their sum is given by 1+2+3+……………(n-2)+(n-1)+n If we add the first and last number we get (n+1) Similarly if we add second number and the second number from last,third number and third number from last we get the same (n+1). So from n natural numbers we get (n/2) pairs of (n+1). So to get the total sum we multiply the sum(ie (n+1)) with total number of terms Therefore sum=(n/2)(n+1)=n(n+1)/2. Hence the proof. Upvote · Soham Joshi Author has 155 answers and 109.8K answer views ·4y Originally Answered: How can I prove that 1+2+3+4+5+6+7+8+9+…+n=n(n+1) /2 is true? · S=1+2+3+…+n S=1+2+3+…+n 0=n−n 0=n−n 1=n−(n−1)1=n−(n−1) 2=n−(n−2)2=n−(n−2) .. .. n=n−0 n=n−0 ⟹S=(n+n+n+..+n)−S⟹S=(n+n+n+..+n)−S 2 S=n(n+1)2 S=n(n+1) S=n(n+1)2 S=n(n+1)2 Upvote · 9 1 Joyneel Bepari Aspiring to compete in the official IMO... · Author has 745 answers and 220.8K answer views ·1y Related What is the proof for the formula 1+2+3+…+n=n(n+1)/2 1+2+3+…+n=n(n+1)/2? Many people will definitely insist on using mathematical induction, which is cool, but there are cooler ways, so I’ll be showing you two other methods. Gauss’ Proof Sequence Definition GAUSS’ PROOF Let s=1+2+3+⋯+n s=1+2+3+⋯+n. We can form two equations: s=1+2+3+⋯+n(Eq. 1)(Eq. 1)s=1+2+3+⋯+n s=n+(n−1)+(n−2)+⋯+1(Eq. 2)(Eq. 2)s=n+(n−1)+(n−2)+⋯+1 Adding Eq.1+2 Eq.1+2 by summing up each corresponding term in both sums, we get: 2 s=(n+1)+(n+1)+⋯+(n+1)n times=n(n+1)2 s=(n+1)+(n+1)+⋯+(n+1)⏟n times=n(n+1) SEQUENCE DEFINITION Let’s define [math]s_n = 1 + 2 +[/math] Continue Reading Many people will definitely insist on using mathematical induction, which is cool, but there are cooler ways, so I’ll be showing you two other methods. Gauss’ Proof Sequence Definition GAUSS’ PROOF Let [math]s = 1 + 2 + 3 + \dots + n[/math]. We can form two equations: [math]s = 1 + 2 + 3 + \dots + n\tag{Eq. 1}[/math] [math]s = n + (n - 1) + (n - 2) + \dots + 1\tag{Eq. 2}[/math] Adding [math]\text{Eq. }1 + 2[/math] by summing up each corresponding term in both sums, we get: [math]2s = \underbrace{(n + 1) + (n + 1) + \dots + (n + 1)}_{n\text{ times}} = n(n + 1)\tag{}[/math] [math]\boxed{\therefore s = \dfrac{n(n + 1)}{2}}\tag{}[/math] SEQUENCE DEFINITION Let’s define [math]s_n = 1 + 2 + 3 + \dots + n[/math] to be a sequence for all [math]n \in \mathbb{W}[/math]. Listing out some terms of this sequence, we get: [math]s_1 = 1\tag{}[/math] [math]s_2 = 3\tag{}[/math] [math]s_3 = 6\tag{}[/math] [math]s_4 = 10\tag{}[/math] [math]\vdots\tag{}[/math] This is very much a quadratic sequence: [math]1, 3, 6, 10, 15, 21, ...\tag{}[/math] The first difference goes by the AP: [math]2, 3, 4, 5, ...[/math], and the second difference is a constant of [math]1[/math]. Since [math]s_n = an^2 + bn + c[/math] (as it is a quadratic sequence), we know the value of [math]a[/math]. It is [math]a = 1/2[/math]. Now, list each sequence down. [math]\begin{align}s_n &\in {1, 3, 6, 10, ...} \tag{Seq. 1}\ \dfrac{n^2}{2} &\in \left{\dfrac{1}{2}, 2, \dfrac{9}{2}, 8, ...\right}\tag{Seq. 2}\end{align}[/math] Subtracting [math]\text{Seq. 2}[/math] from [math]\text{Seq. 1}[/math], we get: [math]bn + c \in \left{\dfrac{1}{2}, 1, \dfrac{3}{2}, ...\right}\tag{}[/math] So, [math]b = 1/2[/math] and [math]c = 0[/math]. [math]\boxed{\therefore s_n = \dfrac{n^2}{2} + \dfrac{n}{2} = \dfrac{n(n + 1)}{2}}\tag{}[/math] Hope this was helpful! Upvote · 9 6 9 3 Karl Smith-Petersen Former Program Manager MS C/C++ Comp./linker, (1994–1998) · Author has 162 answers and 142.9K answer views ·5y Originally Answered: How do you prove 1+2+3+…+n terms=n(n+1) /2? · I look at it this way: if n is even: this makes n elements, an even number each pair moving from outside in adds to (n + 1) there are n/ 2 pairs, each adding to (n + 1) the sum can be calculated by: n/2 (n + 1) = n(n + 1)/2 if n is odd: add zero at the beginning to make an even count of elements without effecting the sum this makes (n + 1 ) elements, an even number each pair moving from outside in adds to n there are (n + 1)/2 pairs, each adding to n the sum can be calculated by: (n + 1)/2 n = n(n + 1)/2 Even or odd, the formula for the sum of integers 1 through n is n(n + 1)/2 QED for the set of all p Continue Reading I look at it this way: if n is even: this makes n elements, an even number each pair moving from outside in adds to (n + 1) there are n/ 2 pairs, each adding to (n + 1) the sum can be calculated by: n/2 (n + 1) = n(n + 1)/2 if n is odd: add zero at the beginning to make an even count of elements without effecting the sum this makes (n + 1 ) elements, an even number each pair moving from outside in adds to n there are (n + 1)/2 pairs, each adding to n the sum can be calculated by: (n + 1)/2 n = n(n + 1)/2 Even or odd, the formula for the sum of integers 1 through n is n(n + 1)/2 QED for the set of all positive even and odd integers: all positive integers. Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes Upvote · 9 3 Jesus Solorzano 4y Originally Answered: How can I prove that 1+2+3+4+5+6+7+8+9+…+n=n(n+1) /2 is true? · I know two ways. One is by induction and the other is by accommodating the sum. 1.Let [math]S=1+2+3+\cdots+n[/math]. Then, [math]S=n+(n-1)+(n-2)+\cdots+1[/math]. Adding member to member, we have to, [math]2S=(n+1)+(n-1+2)+(n-2+3)+\cdots +(n+1)=(n+1)+(n+1)+\cdots +(n+1)[/math]. the point is that you are adding the term math[/math], n-times. So, [math]2S=n(n+1)\Rightarrow S=\dfrac{n(n+1)}{2}.[/math]. By induction. For case [math]n=1[/math] is easy. Suppose that is true for [math]n[/math], and and let's show that it is true for math[/math]. In fact, [math]\begin{align}1+2+3+\cdots +n+(n+1)&=(1+2+\cdots +n)+(n+1)\ &=\dfrac{n(n+1)}{2}+(n+1)\ &=\dfrac{n(n+1)+2(n+1)}{2}\ &=\dfrac{(n+1)(n+2)}{[/math] Continue Reading I know two ways. One is by induction and the other is by accommodating the sum. 1.Let [math]S=1+2+3+\cdots+n[/math]. Then, [math]S=n+(n-1)+(n-2)+\cdots+1[/math]. Adding member to member, we have to, [math]2S=(n+1)+(n-1+2)+(n-2+3)+\cdots +(n+1)=(n+1)+(n+1)+\cdots +(n+1)[/math]. the point is that you are adding the term math[/math], n-times. So, [math]2S=n(n+1)\Rightarrow S=\dfrac{n(n+1)}{2}.[/math]. By induction. For case [math]n=1[/math] is easy. Suppose that is true for [math]n[/math], and and let's show that it is true for math[/math]. In fact, [math]\begin{align}1+2+3+\cdots +n+(n+1)&=(1+2+\cdots +n)+(n+1)\ &=\dfrac{n(n+1)}{2}+(n+1)\ &=\dfrac{n(n+1)+2(n+1)}{2}\ &=\dfrac{(n+1)(n+2)}{2}.\end{align}[/math] This shows that it is true for [math]n + 1[/math]. Which ends the proof. Upvote · 9 2 Jonathan Davis Blah blah blah ·4y Originally Answered: How do you prove 1+2+3+…+n terms=n(n+1) /2? · I will give two proofs, one is geometric, and the other is algebraic(the first way I solved it): 1.Geometric(easiest method) Let the series be represented as blocks instead, then you would get a “blocky” triangle. You can estimate it with [math]\frac{1}{2}n^2[/math] To get the pesky error term, then it is easy to deduce that it must amount to [math]\frac{1}{2}n[/math] , thus the answer is [math]\frac{n(n+1)}{2}[/math] Algebraic(Harder) Instead begin with the series of squares: [math]S(n)=\sum_{i=0}^n i^2 = \sum_{i=0}^n (n-i)^2 = \sum_{i=0}^{n-1} (n^2 - 2ni + i^2)[/math] Using this then you can get: [math]S=n^3 - 2n \sum_{i=0}^{n-1} i + S(n-1)[/math] ,or [math]n^2(n-1) = 2n [/math] Continue Reading I will give two proofs, one is geometric, and the other is algebraic(the first way I solved it): 1.Geometric(easiest method) Let the series be represented as blocks instead, then you would get a “blocky” triangle. You can estimate it with [math]\frac{1}{2}n^2[/math] To get the pesky error term, then it is easy to deduce that it must amount to [math]\frac{1}{2}n[/math] , thus the answer is [math]\frac{n(n+1)}{2}[/math] Algebraic(Harder) Instead begin with the series of squares: [math]S(n)=\sum_{i=0}^n i^2 = \sum_{i=0}^n (n-i)^2 = \sum_{i=0}^{n-1} (n^2 - 2ni + i^2)[/math] Using this then you can get: [math]S=n^3 - 2n \sum_{i=0}^{n-1} i + S(n-1)[/math] ,or [math]n^2(n-1) = 2n \sum_{i=0}^{n-1} i[/math] , hence giving [math]\frac{n(n-1)}{2} = \sum_{i=0}^{n-1} i[/math] Upvote · 9 2 Related questions How do you prove 1+2+3+…+n terms=n(n+1) /2? What is a formula for the sum S n=1(1)(2)(3)+1(2)(3)(4)+1(3)(4)(5)+⋯+1 n(n+1)(n+2))S n=1(1)(2)(3)+1(2)(3)(4)+1(3)(4)(5)+⋯+1 n(n+1)(n+2)) ? Using the formula S=n (n + 1) /2, what is the sum of 1 + 2 + 3 + … + 575? Given 1+2+3+…+n=n(n+1) /2, what is the sum of 105 entries? How do I prove that (2 n+2 n−1)(2 n+1−2 n)=3/2?(2 n+2 n−1)(2 n+1−2 n)=3/2? Is there is any formula for: 1! +2! +3! +4! +⋯+n! And 1! 2+2! 2+3! 2+⋯+n! 2? What is the sum of 1+1/2+1.3…+n/n+1? When is 2 n−1+1≡0(mod n)2 n−1+1≡0(mod n)? What's the difference between these formula n (n+1) /2 or, n (1+n) /2? What is the sum of the 2^10th, (2 power 10th) and is this correct formula to solve this question X.n (n+1) /2? Given a number N, what is the formula for getting the sum of 1+2+…+N? How do you derive (n−1)+(n−2)+(n−3)+...+2+1=n(n−1)2(n−1)+(n−2)+(n−3)+...+2+1=n(n−1)2? Why is 1+2+3+∙∙∙+n = 1/2 (n²+n)? How can I prove that (n r)=∑(n+1−r)n r−2=0(∑(n+1−r)−n r−2 n r−3=0(∑(n+1−r)−(n r−3+n r−2)n r−4=0(...(∑(n+1−r)−∑r−2 k=1 n k n 0=0(n 0))))))(n r)=∑n r−2=0(n+1−r)(∑n r−3=0(n+1−r)−n r−2(∑n r−4=0(n+1−r)−(n r−3+n r−2)(...(∑n 0=0(n+1−r)−∑k=1 r−2 n k(n 0)))))) is true? What is the formula of the sum 1.2 + 2.3 +…+n(n-1)? Related questions How do you prove 1+2+3+…+n terms=n(n+1) /2? What is a formula for the sum S n=1(1)(2)(3)+1(2)(3)(4)+1(3)(4)(5)+⋯+1 n(n+1)(n+2))S n=1(1)(2)(3)+1(2)(3)(4)+1(3)(4)(5)+⋯+1 n(n+1)(n+2)) ? Using the formula S=n (n + 1) /2, what is the sum of 1 + 2 + 3 + … + 575? Given 1+2+3+…+n=n(n+1) /2, what is the sum of 105 entries? How do I prove that (2 n+2 n−1)(2 n+1−2 n)=3/2?(2 n+2 n−1)(2 n+1−2 n)=3/2? Is there is any formula for: 1! +2! +3! +4! +⋯+n! And 1! 2+2! 2+3! 2+⋯+n! 2? What is the sum of 1+1/2+1.3…+n/n+1? When is 2 n−1+1≡0(mod n)2 n−1+1≡0(mod n)? What's the difference between these formula n (n+1) /2 or, n (1+n) /2? What is the sum of the 2^10th, (2 power 10th) and is this correct formula to solve this question X.n (n+1) /2? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
2647	https://www.cuemath.com/ncert-solutions/if-cos-alpha-plus-beta-equalto-0-then-sin-alpha-minus-beta-can-be-reduced-to-a-cosbeta-b-cos-2beta-c-sinalpha-d-sin-2alpha/	If cos (α + β) = 0, then sin (α - β) can be reduced to a. cos β b. cos 2β c. sin α d. sin 2α Solution: Given, cos(α + β) = 0 We have to find the value of sin(α - β) From the trigonometric ratios, Given, cos(α + β) = 0 (α + β) = cos-1(0) Cos is zero at 90° So, (α + β) = 90° Now, α = 90° - β Substitute the value of α, sin(α - β) = sin (90° - β - β) = sin(90° - 2β ) Using the trigonometric ratio of complementary angles, sin (90° - A) = cos A So, sin (90° - 2β ) = cos 2β Therefore, sin (α - β) = cos 2β ✦ Try This: If sin (α + β) = 1, then cos (α - β) can be reduced to Given, sin(α + β) = 1 We have to find the value of cos(α - β) From the trigonometric ratios of angles, (α + β) = sin-1(1) (α + β) = 90° So, α = 90° - β Substituting the value of α, Now, cos (α - β) = cos (90° - β - β) = cos (90° - 2β) Using the trigonometric ratio of complementary angles, cos(90° - A) = sin A So, cos (90° - 2β) = sin 2β Therefore, cos (α - β) can be reduced to sin 2β. ☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 8 NCERT Exemplar Class 10 Maths Exercise 8.1 Problem 5 If cos (α + β) = 0, then sin (α - β) can be reduced to a. cos β, b. cos 2β, c. sin α, d. sin 2α Summary: If cos (α + β) = 0, then sin (α - β) can be reduced to cos 2β ☛ Related Questions:
2648	https://www.education.com/common-core/CCSS.MATH.CONTENT.5.OA.B/	Choose an Account to Log In Notifications 5.OA.B Worksheets, Workbooks, Lesson Plans, and Games CCSS.MATH.CONTENT.5.OA.B These worksheets, workbooks, and lesson plans can help students practice this Common Core State Standards skill. Worksheets Lesson Plans Workbooks Games No games found for this common core code. Exercises No exercises found for this common core code. Add to collection Create new collection New Collection New Collection Sign up to start collecting! Bookmark this to easily find it later. Then send your curated collection to your children, or put together your own custom lesson plan. Educational Tools Support Disable Cookies Warning - you are about to disable cookies. If you decide to create an account with us in the future, you will need to enable cookies before doing so. Connect About IXL Comprehensive K-12 personalized learning Rosetta Stone Immersive learning for 25 languages Wyzant Trusted tutors for 300 subjects Vocabulary.com Adaptive learning for English vocabulary ABCya Fun educational games for kids SpanishDictionary.com Spanish-English dictionary, translator, and learning Emmersion Fast and accurate language certification TPT Marketplace for millions of educator-created resources Copyright © 2025 Education.com, Inc, a division of IXL Learning • All Rights Reserved.
2649	https://www.degruyterbrill.com/document/doi/10.1515/1569392042572177/html?lang=en&srsltid=AfmBOooli_DHefE6u58Wat-phfkmPgRn24tAwLUO7q1BgOAUGFW1BFO2	On the number of solutions of the equation (x1 + . . . + xn)m = ax1 . . . xn in a finite field Skip to main content Manage Access Not authenticated Institutional AccessHow does access work? 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Not Authenticated through an institution. Add institutional access User Account Log inRegister a new account Cart 0 item items 0 items Language ENDE Currency EUR €GBP £USD $ You are not authenticated through an institution. Should you have institutional access?Here's how to get it We are currently experiencing platform issues that may prevent some users from accessing certain products. We're working to resolve this as quickly as possible. Your purchase has been completed. Your documents are now available to view. HomeOn the number of solutions of the equation (x1 + . . . + xn)m = ax1 . . . xn in a finite field Article Licensed Unlicensed Requires Authentication Cite this On the number of solutions of the equation (x 1 + . . . + x n)m = ax 1 . . . x n in a finite field Yu. N. BaulinaYu. N. Baulina Search for this author in: De Gruyter De Gruyter Brill\| Google Scholar Published/Copyright:October 1, 2004 or Purchase Article 30,00 € Purchase Article £23.00 Purchase Article $42.00 Published by Become an author with De Gruyter Brill Author InformationExplore this Subject From the journalVolume 14 Issue 5 MLA MLA APA Harvard Chicago Vancouver MLA APA Harvard Chicago Vancouver Baulina, Yu. N.. "On the number of solutions of the equation (x 1 + . . . + x n)m = ax 1 . . . x n in a finite field" Discrete Mathematics and Applications, vol. 14, no. 5, 2004, pp. 501-508. Baulina, Y. (2004). On the number of solutions of the equation (x 1 + . . . + x n)m = ax 1 . . . x n in a finite field. Discrete Mathematics and Applications, 14(5), 501-508. Baulina, Y. (2004) On the number of solutions of the equation (x 1 + . . . + x n)m = ax 1 . . . x n in a finite field. Discrete Mathematics and Applications, Vol. 14 (Issue 5), pp. 501-508. Baulina, Yu. N.. "On the number of solutions of the equation (x 1 + . . . + x n)m = ax 1 . . . x n in a finite field" Discrete Mathematics and Applications 14, no. 5 (2004): 501-508. Baulina Y. On the number of solutions of the equation (x 1 + . . . + x n)m = ax 1 . . . x n in a finite field. Discrete Mathematics and Applications. 2004;14(5): 501-508. Copy Copied to clipboard BibTeXEndNoteRIS Article We consider the equation (x 1 + . . . + x n)m = ax 1 . . . x n, where a is a nonzero element of the finite field Fq, n ≥ 2, and m is a positive integer. Explicit formulas for the number of solutions of this equation in under the condition d ∈ {1, 2, 3, 6}, where d = gcd(m – n, q – 1), are found. Moreover, we obtain formulas for the number of solutions for arbitrary d> 2 if there exists positive integer l such that d \| (p l + 1), where p is the characteristic of Fq . 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2650	https://www.omnicalculator.com/math/vector-projection	Last updated: Vector Projection Calculator This vector projection calculator finds the orthogonal projection of one vector onto the other. We start with two vectors, a and b, which are not on the same line. Imagine a light source above the vectors. Now, think of the vector projection of a onto b as the shadow that vector a projects on the direction of vector b. This is why this type of vector is known as an orthogonal projection vector (marked as proj in the image): Another, more technical, way of intuiting the idea of vector projection is to ask a question: how much of the vector a goes in the same direction of vector b? Let us now move on to the details and discuss the vector projection formula. What is the vector projection formula? Here is the orthogonal projection formula you can use to find the projection of a vector a onto the vector b: proj = (a·b / b·b) × b The formula utilizes the vector dot product, a·b, also called the scalar product. You can visit the dot product calculator to find out more about this vector operation. But where did this vector projection formula come from? In the image above, there is a hidden vector. This is the vector orthogonal to vector b, sometimes also called the rejection vector (denoted by ort in the image): How to derive the vector projection formula? Here is a step-by-step procedure on how to get to the vector projection formula: Decompose the vector a into a sum of the projection and the rejection vectors: a = proj + ort Express proj as proj = a - ort Also, since proj is parallel to b, we can write it as proj = C × b, where C is some unknown factor we want to determine Now we have C × b = a - ort Calculate the dot product of both sides of this equation with the vector b: C × b·b = a·b - ort·b Since the vector ort is orthogonal to the vector b, their dot product is zero, and we have C × b·b = a·b Finally, we have C = a·b / b·b. When you insert this expression for C into proj = C × b, you get the formula: proj = (a·b / b·b) × b! Since this formula uses the dot product, which can be defined for vectors of any integer dimension, this formula covers vectors of any dimensionality. Its practical applications are for 2-D and 3-D vectors, which is why our calculator is designed for vectors with two or three components. Also, please be aware that this formula is sometimes called the orthogonal projection formula. If we followed this terminology, we'd have to call our calculator the orthogonal projection calculator. There isn't a big difference. Either way, it'll work the same :)! How to use the vector projection calculator Let a and b be a = 2i -3j + 5k and b = 3i + 6j - 4k. From this point on, we will write these 3-D vectors in their component form a = [2, -3, 5] and b = [3, 6, -4]. Use the vector projection calculator to find out the projection of vector a onto vector b. The result we get is: proj = [-1.5738, -3.1475, 2.0984] If you want to calculate the projection by hand, use the vector projection formula p = (a·b / b·b) × b and follow this step-by-step procedure: Calculate the dot product of vectors a and b: a·b = 2×3 + (-3)×6 + 5×(-4) = -32 Calculate the dot product of vector b with itself: b·b = 3×3 + 6×6 + (-4)×(-4) = 61 Insert these two dot products into the vector projection formula to get proj = (-32/61) × [3, 6, -4] = [-96/61, -192/61, 128/61]. This result, when expressed decimally, equals the result obtained by this calculator. Maybe you wondered what happens when a is orthogonal to b; what does the projection proj look like? In this case, the shadow of vector a that is perpendicular to b should be non-existent, so we should get the null vector. This is reflected in the vector projection formula: if a and b are orthogonal, their dot product is zero: a·b = 0. Now we have: proj = (0 / b·b) × b = 0 × b = 0. Explore this special case is shown in the following example: a = [2, 6, -3] and b = [6, 4, 12]. You shouldn't be surprised when [0, 0, 0] pops out since this is the correct result! Vector projection in physics – playing with forces One of the most straightforward, and maybe the most interesting, applications of the vector projection formula comes from the physics of forces and their decomposition. Imagine a cart rolling down the slope of the hill. What is the force one should use to counterbalance the force of the cart? Of course, this counterforce is dependent on the weight of the cart: the more massive the cart, the harder it will be to stop. Also, the slope of the hill itself plays a role - the steeper the hill is, the more force you will need to stop the cart. Here is an image of the situation: Here α is the angle of the hill's slope relative to the ground, and F the force of gravity between the cart and the Earth. Notice that vector F is perpendicular to the ground, not to the hill's slope itself. Note that you can find the slope using math, e.g., in the slope calculator. To find the force needed to counterbalance gravity's action on the cart, we should calculate the projection of the force vector F along the direction of the hill. Let's say that F = 400 N (which corresponds to the cart with a mass of around m = 40 kg) and that the hill's slope is α = 45°. The question is, how do you find a vector along the direction of the hill's slope, the one to project the force vector F on? We can use any vector that has the same direction as the hill's slope, so the most convenient one will be the unit vector u = [cos 45°, sin 45°], marked as the blue vector on the above image. Check the unit vector calculator to find more information about this kind of object. Also, for the force vector F we take F = [0, -400]. The negative sign here means that force F is directed downwards. Use the vector projection calculator and choose to work with vectors in two dimensions, since we're dealing with a two-dimensional problem. So, the vector a will be equal to the force vector F; a = [0, -400], and the vector b is obtained from the vector u, using the trigonometric functions calculator to find the values for sin 45° and cos 45°: b = [0.70710678, 0.70710678]. Working with these numbers, we get the projection vector: proj = [-200, -200]. Since we want to know the magnitude of the counterbalancing force, we can use the vector magnitude calculator to find the magnitude of proj. It is equal to 282.84 N. If you play a bit with the hill's slope, i.e., with the angle α, you will see that the less narrow the hill, the smaller force you will need, while in the case of a larger α – a steeper hill – you will need a larger required reactive force. Besides physics and the problem of finding a component of force along a given direction, vector projection is widespread in one of the most fashionable real-world applications of science, namely data science. What are the applications of vector projections in data science? Vector projection is used in the important statistical problem of linear regression, as well as in principal component analysis, one of the most used methods in data science. Here, one is confronted with the task of representing many, possibly high-dimensional, data points within a space of lower dimensionality in such a way that there is as minimal information loss as possible. FAQs How do I calculate the projection of vector a on b? To determine the projection of a vector a onto the vector b: Compute the dot product a·b. Compute the dot product b·b. This is the same as the squared norm of b. Divide the result of Step 1 by that of Step 2. That is, a·b / b·b. The final result is the coefficient from Step 3 times the vector b onto which we project: proj = (a·b / b·b) × b. What is the length of vector projection? You can find the length of the projection of a vector a onto the vector b using the formula a·b / \|b\|, where a·b is the dot product and \|b\| is the length of the vector b (the one onto which we project). What is the projection of [3,4] onto [1,1]? The answer is [3.5, 3.5]. To derive this result, we apply the formula proj = (a·b / b·b) b with a = [3,4] and b = [1,1]. Indeed, we have a·b = 7 and b·b = 2, so the answer follows easily. Dimensions Vector a Vector b Projection of a onto b Did we solve your problem today? Check out 46 similar coordinate geometry calculators 📈 Average rate of change Bilinear interpolation Catenary curve Share Calculator Vector Projection Calculator Share Calculator Vector Projection Calculator
2651	https://www.droracle.ai/articles/189430/can-you-explain-the-pathophysiology	What is the pathophysiology of metabolic alkalosis with vomiting? Select Language▼ What is the pathophysiology of metabolic alkalosis with vomiting? Medical Advisory BoardAll articles are reviewed for accuracy by our Medical Advisory Board Educational purpose only • Exercise caution as content is pending human review Article Review Status Submitted Under Review Approved Last updated: July 1, 2025 • View editorial policy From the Guidelines Metabolic alkalosis from vomiting occurs primarily due to the loss of gastric acid and subsequent compensatory mechanisms, leading to a decrease in hydrogen ions in the body and a shift in pH balance toward alkalinity. When a person vomits, hydrochloric acid from the stomach is lost, initiating this process. The kidneys then attempt to maintain the new acid-base balance through several mechanisms, including increased bicarbonate reabsorption in the renal tubules to preserve elevated bicarbonate levels, and activation of the renin-angiotensin-aldosterone system due to hypovolemia from fluid loss, which increases sodium reabsorption and potassium excretion 1. This potassium depletion, defined as K<3.5 mEq/L, further exacerbates the alkalosis as hydrogen ions move into cells to replace the lost potassium. Additionally, chloride depletion from vomiting maintains the alkalosis, as the kidneys require adequate chloride to excrete bicarbonate. Key factors in the pathophysiology of metabolic alkalosis with vomiting include: Loss of gastric acid (hydrochloric acid) from the stomach Decrease in hydrogen ions in the body Increase in bicarbonate reabsorption by the kidneys Activation of the renin-angiotensin-aldosterone system Potassium depletion (hypokalemia) due to increased potassium excretion Chloride depletion from vomiting, which hinders bicarbonate excretion by the kidneys Laboratory tests, as mentioned in the evaluation of patients with bowel obstruction and potential vomiting 2, are crucial for identifying electrolyte imbalances, elevated urea nitrogen, and metabolic alkalosis that may occur as a consequence of vomiting and dehydration. Addressing the underlying cause of vomiting, correcting volume depletion with normal saline, and replenishing electrolytes, particularly potassium and chloride, are essential in managing metabolic alkalosis. In severe cases, administration of acetazolamide may be considered to increase bicarbonate excretion, though this is rarely needed once volume and electrolytes are restored. From the Research Pathophysiology of Metabolic Alkalosis with Vomiting The pathophysiology of metabolic alkalosis with vomiting involves several key factors: Generation of alkalosis: This can occur due to excessive hydrogen ion loss by the gastrointestinal tract, such as through vomiting 3, 4, 5, 6. Maintenance of alkalosis: This is due to the kidney's inability to excrete excess bicarbonate, which can be caused by factors such as hypovolemia, chloride depletion, hypokalemia, hyperaldosteronism, and renal failure 3, 4, 5, 6. Role of the kidney: The kidney plays a crucial role in maintaining acid-base balance through bicarbonate reclamation and generation 4, 5, 6. Factors affecting bicarbonate reabsorption: These include effective arterial blood volume, glomerular filtration rate, chloride, and potassium 4, 5. Clinical States Associated with Metabolic Alkalosis Metabolic alkalosis can be associated with various clinical states, including: Vomiting: This is a common cause of metabolic alkalosis, particularly when it leads to chloride depletion 3, 4, 5, 6, 7. Mineralocorticoid excess: This can lead to increased bicarbonate reabsorption and contribute to metabolic alkalosis 4, 5, 6. Licorice ingestion: This can cause mineralocorticoid excess and lead to metabolic alkalosis 4, 5, 6. Diuretic administration: Certain diuretics can lead to hypokalemia and hypochloremia, contributing to metabolic alkalosis 4, 5, 6. Treatment of Metabolic Alkalosis Treatment of metabolic alkalosis involves: Correcting existing depletions: This includes replenishing potassium, chloride, and other electrolytes as needed 3, 4, 5, 6. Preventing further losses: This may involve addressing the underlying cause of vomiting or other factors contributing to metabolic alkalosis 3, 4, 5, 6. Administering carbonic anhydrase inhibitors or acid infusion: These may be used in severe cases of metabolic alkalosis 4, 5, 6. References 1 Guideline Guideline Directed Topic Overview Dr.Oracle Medical Advisory Board & Editors, 2025 2 Guideline Guideline Directed Topic Overview Dr.Oracle Medical Advisory Board & Editors, 2025 3 Research The patient with metabolic alkalosis. Acta clinica Belgica, 2019 4 Research Metabolic alkalosis. Journal of nephrology, 2006 5 Research Metabolic alkalosis. Respiratory care, 2001 6 Research Metabolic Alkalosis Pathogenesis, Diagnosis, and Treatment: Core Curriculum 2022. American journal of kidney diseases : the official journal of the National Kidney Foundation, 2022 7 Research Severe Multifactorial Metabolic Alkalosis in the Emergency Department: A Case Report. The Journal of emergency medicine, 2024 Related Questions What is the meaning of a pH of 7.48, Carbon Dioxide (CO2) level of 47, and Bicarbonate (HCO3) level of 34.7, indicating alkalosis with elevated bicarbonate levels?What are the common causes of metabolic alkalosis?What causes metabolic alkalosis?What are the causes of metabolic alkalosis?What causes metabolic alkalosis?What is the management of a pruriginous rash in a patient receiving Polymyxin B, Aztreonam, and Ceftazidime (Avibactam)?When should hypoglycemia be treated in cardiac arrest?What is the efficacy of rituximab (Rituxan) in treating vasculitis, specifically granulomatosis with polyangiitis (GPA) or microscopic polyangiitis (MPA), as shown in the ADVOCATE trial?What are the safe antidepressants for pregnant women?Can I start a patient with a hemoglobin A1c (HbA1c) level of 16.1% on 20 units of basal insulin if they are already on metformin (Metformin) 500 mg extended release (ER)?What is the treatment for orthostatic hypotension in an 87-year-old patient? 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2652	https://www.cuemath.com/calculators/cubic-feet-calculator/	Cubic Feet Calculator Cubic Feet Calculator is an online tool that calculates the volume of a cuboid and performs unit conversions to represent the volume in cubic feet. The volume of a cuboid can be computed by multiplying the length, width, and height. What is a Cubic Feet Calculator? Cubic feet calculator is used to compute the volume of a cuboid and express it in cubic feet. A cuboid is a three-dimensional solid shape bound by 6 quadrilateral faces. To use the cubic feet calculator, enter the values in the input boxes and choose the unit from the drop-down list. Cubic Feet Calculator How to Use the Cubic Feet Calculator? Follow the steps listed below to find the volume of a cuboid and convert its units by using the online cubic feet calculator. How Does Cubic Feet Calculator Work? The volume of most 3-dimensional objects can be calculated by multiplying the base area by its height. Suppose we have a cuboid with length (l), width (b), and height (h). The formula for the volume is given as: Volume of a cuboid = l × b × h. Here, the base area (rectangular in shape) is given by l × b. The following conversion factors are used to convert the respective units to cubic feet. To convert units from one to another we will need to use the unitary method. Depending upon the type of conversion we will either have to multiply or divide the volume by the given conversion factors. Book a Free Trial Class Solved Examples on Cubic Feet Example 1: Find the volume of a box of dimensions 4 ft × 4 ft × 5 ft and verify it using the online cubic feet calculator. Solution: Volume of the box = 4 × 4 × 5 = 80 ft3. Example 2: Find the volume of a box of dimensions 3 m × 5 m × 6 m in ft3 and verify it using the online cubic feet calculator. Solution: Volume of the box = 3 m × 5 m × 6 m = 90m3. We have to multiply the volume by a conversion factor of 35.315 to convert it from m3 to ft3. This volume of the box = 90 × 35.315 = 3178.35 ft3 Now, try to find the volume in the given units by using the online cubic feet calculator: ☛ Math Calculators: \| \| \| --- \| \| Completing the Square Calculator \| Normal Distribution Calculator \| \| Percentage off Calculator \| Volume of a Cylinder Calculator \| \| Trigonometry Calculator \| Permutation Calculator \| Completing the Square Calculator Normal Distribution Calculator Percentage off Calculator Volume of a Cylinder Calculator Trigonometry Calculator Permutation Calculator
2653	https://www.youtube.com/watch?v=268JDhHU-tE	A Rational Functional Equation Solved in Two Ways SyberMath 155000 subscribers 606 likes Description 18824 views Posted: 19 Mar 2022 Join this channel to get access to perks:→ My merch → Follow me → Subscribe → Suggest → If you need to post a picture of your solution or idea: ChallengingMathProblems #FunctionalEquations PLAYLISTS 🎵 : Number Theory Problems: Challenging Math Problems: Trigonometry Problems: Diophantine Equations and Systems: Calculus: 73 comments Transcript: Introduction hello everyone in this video we're going to be evaluating a function at two different points we are given f of the quantity ax minus b divided by bx minus a and that equals x squared minus 3x plus 1 and we're going to evaluate or find f of 1 plus f of negative 1. now i'm going to be presenting two methods and the first method is just going to be a little bit more painful just to introduce the idea i mean sometimes the one of the methods could look like an overkill but the idea is to present different approaches so let's First Method start with the first method and here of course we do need some conditions like bx minus a should not be equal to zero you know so on and so forth so certain things need to be satisfied for a and b in order for in order for this function to be well defined my first method involves uh substitution and i want to get f of x f of something so i'm going to set this whole thing the you know that rational expression i'm going to set it equal to z i use a different variable you can use a y as well but f of x equals y is sometimes a standard way of writing that's why i'm avoiding y here that's y alright let's go ahead and cross multiply here we get a x minus b equals b x z minus a z here my goal is to solve for x and be able to write x in terms of z so that i can substitute that on both sides obviously on the left hand side if you substitute something for x whatever you find from here it's going to give us z inside the parentheses no doubt about it right but on the right hand side is more important so let's go ahead and solve for x let's bring the x z here and put the b on the right hand side and then go ahead and factor out x a minus b z equals b minus a z and then divide both sides by a minus b z to find x remember our goal was to solve for x in terms of z now one thing that i would like you to do here is to negate both the numerator and the denominator because you want to have the main variable first kind of like you write it in standard form sort of you know like how we write mx plus b for linear functions but we we don't want to write it as b plus mx even though it is correct it doesn't look standard so let's go ahead and write it as a z minus b divided by bz minus a you can obtain this by multiplying both the top and the bottom by negative one which is equivalent to multiplying by one great so x is equal to that now what i can do is i can replace this x with that so i know that x equals something so i can go ahead and replace x with that so that's going to give me the following f of now remember this whole thing was called z right so we call this z therefore it's just going to be f of z on the left hand side which is cool i don't have to worry about it i mean if you want to verify that it's going to be easy you can go ahead and plug it in but that's not needed so now x replaced with a z minus b divided by bz minus a of course we have x squared so we have to square that and then what else do we have minus three x plus one minus three times x which is this plus one awesome so that gives us pretty much f of z and you can go ahead and simplify this like square it make a common denominator so and so forth but that's not needed that's kind of unnecessary let's go ahead and find out f of one and f of negative one from here because we have an expression for f of z now at this point if you replace z with x you're gonna get f of x but that doesn't matter again and that could be confusing for some people even though it shouldn't be because uh we already called x something but that's not the same x these are just dummy variables you use them and throw them away and then use them again and every time you use it it has a different value it may have a different value okay great anyways let's just go ahead and find f of one this is replace z with one that's going to give you f of one equals a minus b divided by b minus a quantity squared minus 3 times a minus b divided by b minus a plus 1. now notice that a minus b divided by b minus a is negative 1 because those are opposites so it's going to give you negative 1 squared which is 1. this is going to give you negative 1. so it's going to be plus 3 plus 1. so f of 1 is going to equal 5. awesome let's go ahead and find f of negative 1 by replacing z with negative one if you replace z with negative one in this expression you get uh a times negative one so it's gonna be negative a minus b divided by at the bottom you're gonna get uh negative b minus a squared minus three times the same thing pretty much negative b minus a plus one now notice that this time you're not getting a negative one but you're getting a positive one because this is positive one obviously if um a plus b is zero this is not gonna work so you don't want that to happen so on and so forth but anyways this is gonna be negative one now i mean positive one so what am i talking about uh positive one squared is one one minus three plus 1 and from here f of negative 1 is going to be negative 1. so we got the two values we're supposed to add them f of 1 plus f of negative 1 is just going to be 4 in this case and that brings us to the Second Method end of the first method not to the end of the video yet because we're going to do the second method now let's go and talk about the second method now these two methods are commonly used with functions and obviously the second method is shorter so we have this equality and we're supposed to find f of one so instead of trying to solve for f of something like a single variable why don't we just try to set what is inside the parenthesis equal to 1. so we can directly do this so here's how it works i want f of 1. so let's go ahead and set this equal to 1 and find the x value that satisfies it so ax minus b divided by bx minus a i want this to equal 1. from here cross multiplying gives us x minus b equals b x minus a now our goal is to solve for x so that we can replace x with something on the right hand side so let's put the ax bx together and then on the right hand side we're going to get have b minus a and then if you take out an x here x times a minus b equals b minus a now b minus a you can write as negative 1 times a minus b so you can kind of write it like this and then provided that a does not equal b of course you can divide both sides by a minus b and get the value of x from here so x is going to be negative one for uh for the parentheses uh inside the parenthesis to be positive one you need to use negative one so in other words this means f of one equals negative one squared minus 3 times negative 1 plus 1 that is equal to 1 plus 3 plus 1 which is 5. so f of 1 is going to be 5 and let's go ahead and find f of negative 1 the same way so we have this equation again and we're trying to find f of negative 1 to set this equal to negative 1 means this whole thing is equal to negative 1. let's set up an equation for that ax minus b divided by bx minus a is equal to negative 1 cross multiply ax minus b is equal to negative bx plus a and then from here we get ax plus bx equals a plus b and clearly from here x is going to be positive 1 as long as a plus b does not equal 0. so if x is equal to 1 i can go ahead and substitute that here and here to find f of negative 1 and f of negative 1 is going to be 1 squared minus 3 times 1 plus 1 and that is equal to negative 1 as before and we got the two values f of negative 1 is negative 1 and f of 1 is 5. therefore their sum is going to be 4 as before so f of 1 plus f of negative 1 is going to be 5 plus negative 1 and that is equal to 4. and this brings us to the end of this video thank you for watching i hope you enjoyed it please let me know don't forget to comment like and subscribe i'll see you tomorrow with another video until then be safe take care and bye bye
2654	https://www.englishclub.com/vocabulary/food-dining.php	Dining Vocabulary \| Learn English 🔍e-BooksLEARNTEACH Games 🧩Quizzes 🤔e⌁BooksGrammarVocabPronunciationNEWJoin us!ForumsListeningSpeakingReadingWritingArticles7 SecretsGuestEasyEnglisheQuizTEFL.net EnglishClub : Learn English : Vocabulary : Topic : Food : Dining Vocabulary Dining Vocabulary with word definitions, example sentences and quiz Photo: Dining table set up outdoors. Breakfast, lunch and dinner In most parts of the world people have three meals a day. The first is breakfast, then lunch in the middle of the day and dinner in the evening. Most people have breakfast at home, and some take a home-made lunch to work or school. Most people also have dinner at home, and for many modern families dinnertime is the only chance we have to get together and talk to one another. At family dinners we don't have to follow the rules of etiquette for formal dining, but if you go to expensive restaurants or travel for business it's useful to know these rules. You might also need to know them if you're hosting a dinner party or having special guests for dinner. Table settings and dining etiquette Houses and large apartments often have a dining room with a dining table that can seat many people. Dining rooms are sometimes used for everyday meals, but often they're only used for big family gatherings, dinner parties or formal meals. If you're hosting a meal like this, you'll probably use your best set of dinnerware and cutlery. Before your guests arrive you'll need to set the table. Begin by covering the table with a tablecloth and then place a centrepiece such as candle sticks or a flower arrangement in the middle of the table. Then arrange place settings around the table, making sure there's a place for everyone, including yourself. Place settings for formal dinners usually include a place mat, or an empty space, in the middle with a dinner fork and a smaller salad fork to the left, a table knife and soup spoon to the right, and a dessert fork and spoon across the top. Each place setting also has a bread plate and butter knife on the left and a drinking glass or tumbler and a wine glass on the right. Sometimes these glasses are on coasters, but only if place mats are used instead of a tablecloth. The place settings for dinner in most hotel dining rooms and fancy restaurants are also like this. Salad or soup bowls, dinner plates and dessert bowls are often brought to the table during the meal, and if so they shouldn't be on the table before the meal begins. There should, however, be a folded napkin at each guest's place, and if you're serving steak or fish you can add a steak knife or fish knife as well. Salt and pepper shakers and jugs full of drinking water should also be on the table. If you're serving tea or coffee after the meal, bring cups and saucers as well as a sugar bowl, teaspoons and a small jug of milk or cream to the table. bowl butter knife candlesticks centrepiece coasters cup and saucer dessert fork and spoon dinner fork dinner plate dinnerware set fish knife napkin or serviette placemat place setting salt and pepper shakers setting the table soup spoon steak knife sugar bowl table knife tablecloth tumbler water jug wine glass bowl (noun): a round dish for soup, salad or a serving of dessert - Where are the soup bowls? bread plate (noun): a small plate for buttering bread rolls - We need a bread plate for each guest. butter knife (noun): a knife with a blunt, rounded end for spreading butter - There should be a butter knife with each guest's bread plate. candlestick (noun): a holder, usually tall and thin, for one or more candles - If it's a romantic dinner for two, you'll need candlesticks. centrepiece (US spelling "centerpiece") (noun): a display placed in the middle of a dining table - Did you arrange the flower centrepiece yourself? coaster (noun): a small mat or flat object under a bottle or glass that protects the table - Put out some coasters or there'll be round stains on the table. cup and saucer (noun): a small plate and matching cup for tea or coffee - Do dinnerware sets usually include cups and saucers? dessert fork and spoon (noun): a three-pronged fork with an oval spoon for eating desserts from a bowl or plate - Can we have an extra dessert fork and spoon, please? dining room (noun): the room in a house or hotel where meals are eaten - Do we really need such a big dining room? dining table (also "dinner table") (noun): a table at which several people can sit together to eat - How many people can you fit around your dining table? dinner fork (noun): an eating utensil with four prongs used during a main course - Why do Americans hold a dinner fork in the right hand? dinner party (noun): a dinner in someone's home to which guests are invited - Can you come to our dinner party on Saturday night? dinner plate (noun): a flat dish for eating a meal's main course - You haven't broken another dinner plate, have you? dinnertime (noun): the time of day usually spent eating dinner - I only see my kids at dinnertime these days. dinnerware (also "dishware or table service") (noun): a set of matching plates, bowls, cups, saucers etc. for several people - Where did you get this beautiful dinnerware? etiquette (noun): rules for behaving correctly in social situations - I had to learn all the etiquette before my first business trip to Europe. fish knife (noun): a knife with a wide blade for eating fish - Can you lend us your fish knives for the dinner party? formal (adjective): requiring official or social rules be followed - If it's a formal occasion, wear a suit and tie. napkin (also UK "serviette") (noun): a cloth or paper towel for wiping your mouth and hands while eating - Can you bring some more napkins, please? place mat (noun): a flat piece of cloth, plastic or wood at the centre of each place setting on a dining table - There's a lovely set of eight place mats in that shop. place setting (noun): dishes, glasses and cutlery arranged in place for one person - Do you know how to make place settings for a formal meal? set the table (verb): to arrange place settings for everyone on a dining table - Can you set the table? Dinner's nearly ready. shaker (noun): a container for salt, pepper, chili powder, etc. with small holes from which the contents are shaken out - I love those salt and pepper shakers shaped like dogs. soup spoon (noun): a round or oval spoon for eating soup - Do you put soup spoons on the left or the right when you set the table? steak knife (noun): a knife with a serrated blade for diners to cut steak at the table - The waiter forgot to bring us steak knives. tablecloth (noun): a large cloth for covering a dining table - Have we got a clean tablecloth? This one's got stains on it. table knife (noun): the knife used for eating a main course - Table knives are bigger than butter knives, aren't they? tumbler (noun): a drinking glass with straight sides and no stem or handle - The tumblers are in a cupboard in the kitchen. water jug (also US "pitcher") (noun): a large container with a handle from which drinking water is poured - The water jug's nearly empty. wine glass (noun): a glass with a long stem and a wide base for drinking wine - I poured water in my wine glass and someone laughed. I didn't know the etiquette. Dining Vocabulary Quiz Food-related Topics and Vocabulary Types of Food—Vocabulary Cooking Vocabulary Kitchens and Kitchenware Vocabulary Dining Vocabulary Restaurant Vocabulary Food and Health Vocabulary Contributor: Matt Errey. Matt is the author of several books including 1000 Phrasal Verbs in Context and Common English Idioms for learners, and Matt's ESL Games and Quizzes for teachers. 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2655	https://www.masterclass.com/articles/led-vs-lead	Skip To Main Content MasterClass logo \| Articles Get At Work Log In Get All Community and Government Wellness Design & Style Arts & Entertainment Writing Sports & Gaming Science & Tech Music Food Home & Lifestyle Business Writing ‘Lead’ vs. ‘Led’: Differences Between ‘Lead’ and ‘Led’ Written by MasterClass Last updated: Jul 19, 2021 • 2 min read The words ”lead” and “led” are often used interchangeably, but there are differences between the two. Learn when and how to use “lead” and “led.” Learn From the Best Teaches Screenwriting Teaches Creating Outside the Lines Teaches Filmmaking Teaches Documentary Filmmaking Teaches Writing Teaches Writing for Television Teaches Dramatic Writing Teaches Writing Teaches Investigative Journalism Teaches Writing Teaches Writing for Young Audiences Teaches Creative Writing Teaches Writing Thrillers Teaches Reading and Writing Poetry Teaches Mystery and Thriller Writing Teaches the Art of the Short Story Teaches Storytelling and Humor Teaches Fiction and Storytelling Teaches Storytelling and Writing Teaches Writing for Social Change Teaches Fiction, Memory, and Imagination Teaches Fantasy and Science Fiction Writing Teaches Poetic Thinking Teaches Writing and Performing Poetry Tell a Great Story Icons and Their Influences Scripting Your Own Success Teaches Screenwriting Teaches Creating Outside the Lines Teaches Filmmaking Teaches Documentary Filmmaking Teaches Writing Teaches Writing for Television Teaches Dramatic Writing Teaches Writing Teaches Investigative Journalism Teaches Writing Teaches Writing for Young Audiences Teaches Creative Writing Teaches Writing Thrillers Teaches Reading and Writing Poetry Teaches Mystery and Thriller Writing Teaches the Art of the Short Story Teaches Storytelling and Humor Teaches Fiction and Storytelling Teaches Storytelling and Writing Teaches Writing for Social Change Teaches Fiction, Memory, and Imagination Teaches Fantasy and Science Fiction Writing Teaches Poetic Thinking Teaches Writing and Performing Poetry Tell a Great Story Icons and Their Influences Scripting Your Own Success Get What Does ‘Lead’ Mean? The word “lead” has a few definitions. As a verb, lead can mean to guide, show, or usher (such as “She will lead the way”) or to cause and bring forth (like “Job layoffs can lead to depression”). As a noun, “lead” can refer to the metallic element or mean the foremost position, such as “He was the lead in the film.” Lead will take on different pronunciations; as a verb it has a long e, sounding more like “leed” and rhyming with “seed” or “need.” That pronunciation is often the same in the noun forms except for when referring to the element and graphite, in which case it has a short e sound, rhyming with “bed” or “wed.” Meet One of Your New Instructors Get Get How to Use ‘Lead’ in a Sentence Use “lead” for instances where someone is in charge or taking charge. “A good team captain leads by example.” Lead is used as a verb in this sentence and is synonymous with “guide.” “Exposure to even low levels of lead can have long-lasting damage.” This sentence uses lead as a noun and refers to the metallic element that is harmful for humans to consume. “I will take the lead if the manager falls ill.” Lead is again used as a noun here; the subject will step into the chief position if the current leader is unable to perform their duties. What Does ‘Led’ Mean? “Led” is the past tense form of the verb “lead.” For example, “Last night, he led the horses to the stable.” “Led” is also used in infinitive phrases, such as “He wants to be led.” You can use “led” in the passive voice, for the present or future tenses. When capitalized, “LED” is an acronym standing for “light-emitting diode,” a type of light that glows when voltage is applied. How to Use ‘Led’ in a Sentence Understanding that “led” is often the past tense of “lead” helps clarify when you should use it: “When we were on vacation, the tour guide led us through the forest.” This sentence is in the past tense, making “led” the appropriate verb form. “A dog may be led with a leash if it is too easily excited.” Led here is used as the passive present tense of the verb. “After Marsha retires, the podcast team will be led by a new CEO.” In this example sentence, “led” takes on the passive future tense. ‘Lead’ vs. ‘Led’: What’s the Difference? “Lead” and “led” can be homophones, making them easily confused words in the English language. Despite their phonetic similarities, “led” and “lead” have different meanings. In general, “lead” is the present tense version of the verb that means to spearhead, guide, or take the first position, while “led” is the passive, infinitive, and past tense version of the verb. How to Choose Between ‘Lead’ and ‘Led’ The word “present” is longer than the word “past,” and the word “lead” is longer than “led.” That may be one helpful way to distinguish “lead” and “led,” which are commonly confused because when “lead” refers to the type of metal, the two are homophones. Simply put, the verb “lead” is in the present tense, and “led” is its past tense form. Want to Learn More About Writing? Become a better writer with the MasterClass Annual Membership. Gain access to exclusive video lessons taught by the world’s best, including David Baldacci, David Sedaris, Amy Tan, Roxane Gay, Walter Mosley, Margaret Atwood, Dan Brown, and more. Thanks for signing up! We’ll let you know about new instructors, classes, and promotions. Featured MasterClass Instructor David Baldacci In his MasterClass, bestselling thriller author David Baldacci teaches you how he fuses mystery and suspense to create pulse-pounding action. 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2656	https://math.stackexchange.com/questions/4064602/how-to-get-an-original-function-from-the-limit-definition-of-a-derivative	calculus - How to get an original function from the limit definition of a derivative? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How to get an original function from the limit definition of a derivative? Ask Question Asked 4 years, 6 months ago Modified3 years, 4 months ago Viewed 4k times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. Say I have lim h→0 e h−1 h lim h→0 e h−1 h If d d x(e x)\|x=0=lim h→0 e 0+h−e 0 h,d d x(e x)\|x=0=lim h→0 e 0+h−e 0 h, how would I “back engineer” the derivative limit definition to satisfy the expression meaning expression-I equals expression-II. But we’re only given expression-II the limit expression so how do we find the “mystery” expression-I from expression-II. I basically want to remove the guess work required to satisfy the two expressions, and if there’s even a general “algorithm” to follow? calculus limits derivatives algorithms reverse-math Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited May 29, 2022 at 16:20 Rohit Singh 1,141 2 2 gold badges 7 7 silver badges 16 16 bronze badges asked Mar 16, 2021 at 20:55 Canadian BaconCanadian Bacon 23 1 1 silver badge 5 5 bronze badges 13 I'm sorry I don't quite understand what you are asking. Do you want to integrate e x e x?vitamin d –vitamin d 2021-03-16 20:58:00 +00:00 Commented Mar 16, 2021 at 20:58 I haven’t learned about integration yet so I’m sorry if the question doesn’t make any sense but I basically have this original function that’s being solved for its derivative so d/dx f(x) that I have no idea what it is and I want to find what it is(f(x)). I am only given the equivalent but in limit definition form. So I want to find that original function being used to find that limit we are given. I hope that helps sorry if it doesn’t lmk if there anything I can clarify, thank you!Canadian Bacon –Canadian Bacon 2021-03-16 21:08:35 +00:00 Commented Mar 16, 2021 at 21:08 Do you have d/dx f(x)=e x e x?vitamin d –vitamin d 2021-03-16 21:11:27 +00:00 Commented Mar 16, 2021 at 21:11 Ahh I see thank you so much and I don’t have d/dx f(x) only the lim h->0. But this solves my problem appreciate all the help!Canadian Bacon –Canadian Bacon 2021-03-16 21:13:56 +00:00 Commented Mar 16, 2021 at 21:13 I think your comment got edited but to confirm I would need integration to “discover” the original expression?Canadian Bacon –Canadian Bacon 2021-03-16 21:24:00 +00:00 Commented Mar 16, 2021 at 21:24 \|Show 8 more comments 2 Answers 2 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. If I'm understanding your question correctly, it boils down to (1) identifying that a given limit is the derivative of a function f(x)f(x) at some point x=a x=a, namely f′(a)=lim h→0 f(a+h)−f(h)h f′(a)=lim h→0 f(a+h)−f(h)h and (2) precisely picking out what the function f(x)f(x) is supposed to be. In the given example, lim h→0 e h−1 h=lim h→0 e 0+h−e 0 h=lim h→0 f(0+h)−f(0)h⟹f(x)=e x and a=0 lim h→0 e h−1 h=lim h→0 e 0+h−e 0 h=lim h→0 f(0+h)−f(0)h⟹f(x)=e x and a=0 and hence this limit is exactly the derivative of e x e x at x=0 x=0. If you know that (e x)′=e x(e x)′=e x, then the limit is simply e e. So the strategy is to recast the given limit into an expression involving the forward difference f(a+h)−f(a)f(a+h)−f(a). Another example: lim h→0(1+h)1 3−1 h=lim h→0(1+h)1 3−1 1 3 h⟹f(x)=x 1 3 and a=1 lim h→0(1+h)1 3−1 h=lim h→0(1+h)1 3−1 1 3 h⟹f(x)=x 1 3 and a=1 Then this limit would be 1 3 1 3, or the derivative (x 1 3)′=1 3 x 2 3(x 1 3)′=1 3 x 2 3 at x=1 x=1. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Mar 16, 2021 at 22:05 user170231user170231 26.6k 3 3 gold badges 48 48 silver badges 90 90 bronze badges Add a comment\| This answer is useful 3 Save this answer. Show activity on this post. The derivative of a function f f at a point a a is defined as f′(a)=lim h→0 f(a+h)−f(a)h.f′(a)=lim h→0 f(a+h)−f(a)h. Setting f(x)=e x f(x)=e x and a=0 a=0 this yields d d x e x∣0=lim h→0 e 0+h−e 0 h=lim h→0 e h−1 h.d d x e x∣0=lim h→0 e 0+h−e 0 h=lim h→0 e h−1 h. This would be the solution to your problem. If I understood you correctly you want to have a step by step derivation, from the limit to the derivative. I don't think that's the aim of the book, it's about observing that these two expressions are equal and not about calculations. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Mar 16, 2021 at 22:06 vitamin dvitamin d 5,913 8 8 gold badges 16 16 silver badges 38 38 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. 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2657	https://sites.math.washington.edu/~putnam/HandOutTrig.pdf	Pythagorean identities sin2 x + cos2 x = 1 1 + tan2 x = sec2 x 2. Sum-Difference formulas sin(x ± y) = sin x cos y ± sin y cos x cos(x ± y) = cos x cos y ∓sin x sin y tan(x ± y) = tan x±tan y 1∓tan x tan y cot(x ± y) = cot x cot y∓1 cot x±cot y arctan(α) ± arctan(β) = arctan α±β 1∓αβ 3. Double Angle formulas sin 2x = 2 sin x cos x cos 2x = cos2 x −sin2 x = 2 cos2 x −1 = 1 −2 sin2 x tan 2x = 2 tan x 1−tan2 x 4. Half Angle formulas sin2 x = 1−cos(2x) 2 cos2 x = 1+cos(2x) 2 tan2 x = 1−cos(2x) 1+cos(2x) 5. Sum to product formulas sin x+sin y = 2 sin( x+y 2 ) cos( x−y 2 ) sin x−sin y = 2 cos( x+y 2 ) sin( x−y 2 ) cos x+cos y = 2 cos( x+y 2 ) cos( x−y 2 ) cos x−cos y = −2 sin( x+y 2 ) sin( x−y 2 ) 6. Product to sums formulas sin x sin y = 1 2(cos(x −y) −cos(x + y)) cos x cos y = 1 2(cos(x −y) + cos(x + y)) sin x cos y = 1 2(sin(x + y) + sin(x −y)) 7. Another useful trick A sin x+B cos x = √ A2 + B2 sin(x+α), where α = arccos A √ A2+B2 1
2658	https://condor.depaul.edu/glancast/444class/docs/lectures/lecOct02.html	Use of NFA's for Closure Properties of Regular Languages The set of regular languages is closed under the following operations: Complement Intersection Union Concatenation Star (Kleene Closure) L1 = { w ∊ {0,1} \| w contains the substring "1101" } L2 = { 0n1m \| n >= 0 and m >= 0} L3 = { aa, ab, ba, bb } L4 = { a, aa, aaa} ∅ (the empty set; not the empty string) What is the complement of L1? What is L3 ∩ L4? What is L3 o L4 (concatenation)? What is L3 What is ∅ How does one prove that regular languages are closed under each of these operations? NFAs are useful to show regular languages are closed under the last three operations (union, concatenation, star). Union An NFA to recognize L1 ∪ L2 s1 is the old start state of L1. s2 is the old start state of L2. f1, f2 are old final states of L1. f is the old final state of L2. Concatenation An NFA to recognize L1 o L2. s1 is the old start state of L1. s2 is the old start state of L2. f1, f2 are old final states of L1. f is the old final state of L2. Star (Kleene Closure) An NFA to recognize L s1 is the old start state of L. f1, f2 are old final states of L. Equivalences NFAs DFAs regular expressions regular grammars Regular Expressions (1.3) Regular expressions over an alphabet ∑ represents sets of strings in the alphabet; that is, a formal language over the alphabet. So a regular expression describes a formal language. In the notation for regular expressions, an element a ∈ ∑ is also used to represent the regular expression that denotes the set of strings consisting of 'a' alone: { a }. Formal Definition The formal definition of a regular expression over an alphabet ∑ is (page 64): R is a regular expression if R is a for some a in the alphabet ∑, ε, ∅, (R1 ∪ R2), where R1 and R2 are regular expressions, (R1 o R2), where R1 and R2 are regular expressions, or (R1), where R1 is a regular expression ∅ represents the empty language ε represents the language { ε } a ∈ ∑ represents the language { a } Operations and Examples The operations on regular expressions are union concatenation star (closure) ∑ = {0,1} Union: 0 ∪ 1 or 0 \| 1 represents the set {0} ∪ {1} = {0,1} Concatenation: (0 ∪ 1)1 represents the set {01, 11} Star: (01\|11) represents the set {ε, 01, 11, 0111, 1101, ... } ∅ represents the set { ε } (0 ∪ 1)1101(0 ∪ 1) What language does this describe? Theorem A language is regular if and only if some regular expression describes it. : Proof requires two parts. : First Part: If a language is regular, then it is described by some regular expression. : Second Part: If a language is described by some regular expression, then it is a regular language. GNFAs A Generalized Nondeterministic Finite Automaton is similar to an NFA but the transition function takes a state and a regular expression in the alphabet instead of a state and an alphabet element. A GNFA for strings in {1,0} that contain the substring "1101". The idea is that in state q0 the transition to state q1 can be taken if the next input matches the regular expression 1101. The formal definition is given (on page 73) by: A generalized nondeterministic finite automaton is a 5-tuple, (Q,∑, δ, qstart, qaccept), where Q is the finite set of states, ∑ is the input alphabet δ : (Q - {qaccept} x (Q - { qstart }) → R is the transition function, where R is the set of all regular expressions over ∑ qstart is the start state qaccept is the accept state Note that there is only one accept state. However, this is no real restriction for a nondetrministic automaton. (Why?) On the other hand, the transition function is defined on a different arguments than is the case for an ordinary NFA. GNFA Transition Function Example The transition function, δ, for the example GNFA: is \| \| q1 \| q2 \| q3 \| --- --- \| \| q0 \| ε \| ⌀ \| ⌀ \| \| q1 \| (0 \| 1) \| 1101 \| ⌀ \| \| q2 \| ⌀ \| (0 \| 1) \| ε \| Language of a GNFA The definition of the language of a GNFA is technically different than that of an NFA because the transition function is defined differently. However, the idea is really similar, but extended to allow regular expressions on the transitions. The formal definition is given by (page 73): A GNFA accepts a string w in ∑ if w = w1w2...wk, where each wi is in ∑ and a sequence of states q0, q1, ..., qk exists such that q0 is the start state qk is the accept state for each i, wi ∊ L(Ri) where Ri = δ(qi-1, qi); that is, where Ri is the expression on the arrow from qi-1 to qi. Example String Accepted by GNFA The string 0011011 is accepted by If 0011011 is partitioned into blocks w1=ε w2=0 w3=0 w4=1101 w5=1 w6=ε; that is, 0011011 = ε 0 0 1101 1 ε then \| \| \| \| \| --- --- \| \| \| State \| Input \| Next \| \| w1=ε ∊ L(δ(q0, q1)) \| q0 \| w1 \| q1 \| \| w2=0 ∊ L(δ(q1, q1)) \| q1 \| w2 \| q1 \| \| w3=0 ∊ L(δ(q1, q1)) \| q1 \| w3 \| q1 \| \| w4=1101 ∊ L(δ(q1, q2)) \| q1 \| w4 \| q2 \| \| w5=1 ∊ L(δ(q2, q2)) \| q2 \| w5 \| q2 \| \| w6=ε ∊ L(δ(q2, q3)) \| q2 \| w6 \| q3 \| For example, w1=ε ∊ L(δ(q0, q1)) means for state q0 and input w1, next state is q1 Converting a DFA to an GNFA A DFA can be converted to an equivalent GNFA by Adding a new start state with an ε transition to the old start state. Adding a new accepting state and adding ε transitions from all old accepting states to the new one. Definining the transition function δ' for the GNFA in terms of the transition function δ for the DFA by δ'(qi, qj) = a if and only if δ(qi, a) = qj Example conversion of DFA to GNFA Steps 1 and 2 require adding a new start state and a new accepting state. Step 3 require no changes in the diagram except labeling the transitions out of the new start state and into the new accepting state. Eliminating a state in an GNFA Any state except the start state and the accept state of a GNFA can be eliminated to obtain a new equivalent GNFA with one fewer states. For example, to eliminate state q1 in replace each existing path q to q1 to q' by a path from q to q' Label the new path with a regular expression that describes the strings that would cause a transition from state q to q1 to q' There are 3 arrows coming into state q1 from other states: q0, q2, and q4. If state q1 is removed, paths must be replaced by new transitions: q0 to q2 q2 to q2 q4 to q2 Determining the Regular Expression Labels Eliminating q1, what regular expressions are needed? GNFA Regular Expression Labels Suppose qa is to be removed and the path being considered goes from qi to qa to qj, transition from qi to qa is labeled by the regular expression Ria transition from qa to qj is labeled by the regular expression Raj transition from qi to qj is labeled by the regular expression Rij (Note this may be the empty regular expression, ࣼ) Then the path from qi to qj should be labeled ``` RiaRaRaj ∪ Rij ``` Result Original DFA (converted to a GNFA): After eliminating state q1: Now the Theorem Again A language is regular if and only if some regular expression describes it. : Proof requires two parts. : First Part: If a language is regular, then it is described by some regular expression. : Second Part: If a language is described by some regular expression, then it is a regular language. DFA to Regular Expression To show the first part, if we are given an DFA, we need to show that there is a regular expression that describes exactly the language of the DFA. The construction of the regular expression begins with the DFA and each step will eliminate one state of the DFA until the only state(s) remaining are the start state and a final state. Example Exercise 1.21 Convert this DFA to a regular expression that describes the same language. Regular Expresssion to NFA To show that for any regular expression there is an NFA that recognizes the same language described by the regular expression, the proof describes a procedure for constructing the NFA from the regular expression. Parse the regular expression into its parts based on the regular expression operator precedence and parentheses used if any to determine the operands of each operator. star is highest, then concatenation, and union is lowest precedence. For each of the operators use the construction described in showing the closure properties of regular languages to construct an NFA for each operator and its operands. See Lemma 1.55 in the text. Example regular expression ``` a(ba)b ∪ bab ``` Construct the NFA for (ba): concatenation of b and a Construct the NFA for (ba): star of (1) Construct the NFA for a(ba): concatenation of a and (2) Construct the NFA for a(ba)b: concatenation of (3) and b Construct the NFA for ba: concatenation of b and a Construct the NFA for bab: concatenation of 5 and b Construct the NFA for a(ba)b ∪ bab: union of (4) and (6) More Text Exercises 1.19 1.20 Is this language L over the alphabet {a, b} regular? L = {anbm \| n >= 0, m >= 0 and n ≠ m } Pumping Lemma Pumping Lemma for Regular Languages: If A is a regular language, then there is a number p such that if s is any string in A of length at least p, then s may be divided into three pieces, s = xyz with the properties: for each i >= 0, xyiz ∈ A, \|y\| > 0, and \|xy\| <= p Strategies Strategies for deciding whether a formal language is regular. It may be easier to show whether the complement is regular. Use the pumping lemma to show the language is not regular. Example 1: { anbn \| n >= 0 } is not regular. Example 2: {anbm \| n >= 0, m >= 0 and n ≠ m } is not regular.
2659	https://www.gdandtbasics.com/perpendicularity/	← Back to All Symbols by GD&T Basics on December 14, 2014. Special Note: Perpendicularity in Geometric Dimensioning and Tolerancing can mean two very different things depending on which reference feature is called out. The normal form or Surface Perpendicularity is a tolerance that controls Perpendicularity between two 90° surfaces, or features. Surface Perpendicularity is controlled with two parallel planes acting as its tolerance zone. Axis Perpendicularity is a tolerance that controls how perpendicular a specific axis needs to be to a datum. Axis Perpendicularity is controlled by a cylinder around a theoretical perfectly parallel axis. Pay close attention if a hole or pin is referenced since axis perpendicularity is commonly called out on these features. GD&T Symbol: Relative to Datum: Yes MMC or LMC applicable: Yes GD&T Drawing Callout: Surface Perpendicularity: Axis Perpendicularity: Description: Surface: Perpendicularity is a fairly common symbol that requires the referenced surface or line to be perpendicular or 90° from a datum surface or line. Perpendicularity can reference a 2D line, but more commonly it describes the orientation of one surface plane perpendicular to another datum plane. The tolerance of the perpendicularity callout indirectly controls the 90° angle between the parts by controlling the location where the surfaces have to lie. See the tolerance zone below for more details. Note: Perpendicularity does not control the angle of the referenced feature –the tolerance is in distance units. (mm/in) Axis: Axis control can also be called out for Perpendicularity and is one of the more common forms of axes call outs. When it is referenced for a circular feature, the feature control frame will contain the diameter (Ø) symbol. Axis Perpendicularity can be applied to a positive feature (pin/boss) or to a negative feature (a hole). When Perpendicularity is referenced for axial control of a feature, the symbol now specifies a cylindrical boundary where the axis of the referenced feature must lie. This cylindrical boundary is formed by taking a line that is directly perpendicular to the datum feature. When this version of Perpendicularity is called out it is to be used with maximum material condition to enable easy gauging of the part. See example 2 below for how these particular parts are gauged. GD&T Tolerance Zone: Surface: Two parallel planes or lines which are oriented perpendicular to the datum feature or surface. The planes are held perpendicular to the datum, but only ensure that the entire feature falls into the tolerance zone. Remember: Perpendicularity does not directly control the angle of the referenced surface; it controls the envelope (like flatness) where the surface needs to be. A cylinder surrounding a referenced theoretical axis which is directly perpendicular to the datum feature. The tolerance zone is the diameter of this symbol in which the central axis of the measured feature must lie. Gauging / Measurement: Surface: Perpendicularity is measured using a height gauge, similar to flatness, however, the gauge (or part) is locked to a 90° datum to measure how perpendicular the surface is. The entire surface has to be measured if it is a flat feature. Axis: To ensure that a part or feature is axially perpendicular, Maximum material condition is most often called out on axis perpendicularity to allow easy measurement with a gauge. This allows it to be designed for either a negative (hole) or positive (pin) feature and can take into account a bonus tolerance. Gauge size for an internal feature (like a hole): Gauge Ø (pin gauge)= Min Ø of hole (MMC) – Perpendicularity Tolerance Gauge size for an external feature (like a pin): Gauge Ø (hole gauge) = Max Ø of pin (MMC) + Perpendicularity Tolerance See Example #2 below for a good example Axis Perpendicularity using MMC. Note on Bonus Tolerance: When a functional gauge is used for Perpendicularity, any difference the actual feature size is from the maximum material condition would be a bonus tolerance. The goal of a maximum material condition callout is to ensure that when the part is in its worst tolerances, the orientation and size of the hole/pin will always assemble together. This means that if you make a pin smaller, you make more bonus tolerance for yourself. This bonus can be added to the GD&T tolerance and would widen the perpendicularity tolerance. Bonus Tolerance = Difference between MMC & Actual condition (See Example 2 Below) Confused yet? No worries! For more detail, see the sections on Maximum Material Condition. Relation to Other GD&T Symbols: Surface: Perpendicularity is a specific form of Angularity at 90°. All of the orientation symbols (Angularity, Parallelism, and Perpendicularity) all call out the particular feature envelope referenced to a datum. The Perpendicular Symbol is also closely related to flatness when referenced/measured surface is a surface plane. When you call out Perpendicularity, flatness is implied (you are measuring a surface variation between two parallel planes = Flatness) Perpendicularity is always measured with respect to a datum, where flatness is not. Axis: Perpendicularity is closely related to all the other orientation GD&T symbols when called on an axis. The tolerance zone now refers to the uniformity and cylindrical envelope of a central axis. Perpendicularity and Parallelism can be called out on holes and cylindrical pins, often with MMC added. When Used: Surface: Whenever two surfaces needing a constant 90° angle, Perpendicularity is effective. Flange bearings and critical square edges usually reference it. Perpendicularity is also commonly called out on the corners of cylinders where the flat bottom must be perpendicular with the curved sides. Axis: Perpendicularity is very commonly called out on the center axis of a hole. Almost always, your hole needs to be perpendicular to the surface it is drilled into. When this is the case, it is called out alongside MMC to ensure that if a pin or bolt needs to be inserted into this hole, the part can enter the whole perpendicular at and always fit in. See example 2 for this explanation. Surface Perpendicularity Example: The edge of a stopping block for a rail must form a 90° to ensure proper mating contact takes place. The base of the block is will be our datum and the face where the stopping block makes contact is our referenced surface. To ensure that this face is always perpendicular and flat to make good contact, you would need to both tightly control the angle and the dimensional width of the part. Ensuring perpendicular/flat surfaces without GD&T symbol With perpendicularity, you can open up the width dimension and control the face’s angle containing the part very tightly. Your tolerance zone remains the same, but your part is now easier to control and fabricate. Controlling the perpendicularity with GD&T symbol. Axis Perpendicularity Example with MMC: If you have a critical hole feature that needs to remain parallel to the surface that is formed into, perpendicularity can be called out to ensure that the hole is straight. In this example, a bolt hole is specified to remain perpendicular to its surface. Perpendicularity on a hole under MMC Without an MMC callout, you would need to control just the center axis of the hole and measure it to ensure it is at 90° to the bottom surface. However, when MMC is called out on the print, you are controlling both the size and the orientation of the hole. You now can check both tolerances using a functional gauge with the following dimensions: Formula for a perpendicularity functional gauge: Gauge Ø (pin gauge) = Min hole Ø – Perpendicularity Tolerance Gauge Ø = 9.9 – 0.2 = 9.7 Hole Ø + Hole Perpendicularity > 9.7 (Pin Ø) to be in spec. Due to the Max Material Condition callout, if you have a hole that is larger than the MMC of 9.9 you will have bonus tolerance that can be added on to your perpendicularity. (According to print Hole Ø cannot be above 10.1 though) In the example below – the hole is at the least material condition (largest hole size) with the hole at the LMC, your bonus tolerance that can be added to the perpendicularity is calculated as follows: Bonus Tolerance = Actual Part Size – Max Material Condition Bonus Tolerance = 10.1 – 9.9 = 0.2 Adding this bonus tolerance to your perpendicularity means your “gauged” perpendicularity tolerance can go up to 0.4 when the part is at its largest diameter. Final Notes: Very Common: Perpendicularity is very common in its surface and axis form. You will see this commonly on many mechanical engineering drawings. Features of Size: Perpendicularity will most likely have an MMC or LMC callout if gauge control is used in a production environment. It allows both size and orientation to be measured quickly on the line, as opposed to having to measure perpendicularity with a CMM. Axis Controls: Straightness, Axis Angularity, Axis Parallelism, True Position, and Axis Perpendicularity can all be called out to control a center axis. Usually, when this is a case in a production environment, MMC is also called out so that a functional gauge can be used. However, the only callouts with this case that you would see commonly are perpendicularity and straightness. Be The Go-To Engineer at Your Company Learn GD&T at your own pace and apply it with confidence in the real world. Get GD&T Training All Symbols Share On Categories GD&T Symbol Rules and Examples Orientation Tags GDandT Symbol orientation perpendicularity 67 replies on “Perpendicularity” Do we have to consider the perpendicularity defined for a bolt hole while doing worst case stackup analysis?? Does perpendicularity still apply when the datum is a curved surface, not a flat surface? Can an orientation control have three datum references? I am under the impression that it cannot for example, if I am orienting a hole on a cube (length, width and depth is not important for the sake of the example) and datum features A primary, B secondary, C tertiary are all nominally perpendicular to each other. (Simple coordinate system here…) I came across a drawing where engineering specified this and I am under the impression that they meant position relative to ABC. I just feel it’s a super odd way of orientating a hole with respect to 3 datum feature references. I mean, I guess I can control something to be oriented with respect to three datum features but why not just call out position? I’d like to note that the drawing has a general note, CAD model is master all dimensions not shown on the drawing are to be within .040 profile relative to ABC. The perpendicularity tolerance is .010. Does this mean the hole is to be located within the profile of .040 with respect to ABC but orientation has to be refined within .010 relative to ABC? (Just an odd ball… hope I’m not overthinking it) You are correct! orientation really could only have 2 datums to fully control u,v,w in rotation. An exception would be if something is related to a sphere (no rotation controlled) but most geometric datums, (cylinders, planes, tabs/slots) are going to control 2 orientations with and primary datum and secondary datum Can an (axis) perpendicular control frame have a zero tolerance?ex: perp Ø 0.00m A (-A- is a datum plane w/ cylinder perpendicular to it – Ø53.400 +0.00/-0.005 -B-) Kim – Yes, I suspect that what you are seeing is referred to as ‘zero tolerance at MMC’. If there is no MMC next to the tolerance in the feature control frame then there is a problem. The MMC condition invokes the concept of ‘bonus’ tolerance, wherein, as the feature departs from MMC towards LMC you gain additional tolerance equal to the amount of departure from MMC. Thus, in your example, if your cylindrical feature was produced at 53.400 the axis of the cylinder would have to be theoretically perfect to datum A. Now, if you feature was produced at 53.395, your tolerance zone would be .005. Reference Section 6.4.4 in the ASME Y14.5 – 2009 standard for more information. Alternatively, our GD&T Advanced course has a lot of good content to help on this and many other topics if you’re interested. Hope this helps! Cheers,Matt 5. I have a pin 1.500″ long and 0.500″ in diameter. There is a call out for perpendicularity of 0.003 at the end of the part to the datum A, which is the outside cylinder of the part. I measure this by placing the end of the pin with the call out on a flat surface and recording the Min and MAX reading at the other end of part, 1.500″ from the call out end. Is this a direct reading, or must I divide by 3 and the diameter is 1/3 of the length? 6. I am trying to tolerance the axis of a tapered hole in a disc. The tapered hole mounts on a rotating tapered shaft so the size of the tapered hole will have no effect on how the disc tracks relative to the rotating shaft. The goal is to have the tapered hole perpendicular to the face of the disc so there is no wobble of the disc as it rotates. I guess the tolerance I’m looking for is the amount of wobble, ie. the perpendicularity of the tapered hole to the face of the disc. Thank you for your time,Dave 7. I have an old drawing (1950’s) which is relating a flange to a bore with the note “THIS SURFACE MUST BE FLAT AND SQUARE WITH BORE WITHIN .0003 PER INCH” The flatness I understand, but how do you apply a linear requirement (.0003 per inch) to perpendicularity? JLK – Ha! I also have some experience with trying to interpret drawings that date back to before men landed on the moon. It’s always a bit of detective work to try and figure out what they meant. Take this with a grain of salt, but I suspect that they are trying to control the perpendicularity of the surface to the bore within a +/- zone of .0003 and vice versa. I would say that the axis of the bore must lay within that +/- .0003 zone. This is a great example of why the GD&T system was developed. How many different ways can that note be interpreted? With a universal symbology the is clear with no room for interpretation. I’ll also go on to say that even today in my industry mis-application and flat out wrong GD&T is used all the time. With all drawings (especially old drawings) you’ve got to keep your eyes open to the bigger picture as to what the draftsman’s intent was for the drawing. You can often deduce the end result even with incorrect and improperly applied controls. Hope this helps. Cheers,Matt 8. Hi, I have a CMM soft dilemma, Lets say we have a block 100x100x100mm with a thru hole in center of 20 mm diameter. The frontal face of the block is datum A and the hole is datum B.If I callout perpendicularity Datum A / Datum B it gives a value (ex: 0.1 ) and if I callout perpendicularity Datum B / Datum A it gives a value 0.02.My question is why there is different results if I switch the callout on the same datums ? Best regards, 3D – Student – The answer is in your question essentially. Due to imperfections in the manufacturing process nothing is truly perfect. There is always some roughness, some angle etc. to various faces. Those datums are indicating in which order they are to contact the datum simulator: first A, then B, then C (if that’s how your datum structure is called out). Depending on the particular imperfections in the part you’ll get a different result from each setup. Note that this is one of the fundamental reasons behind GD&T as a whole. You want consistent, repeatable results. You are telling the inspector with your feature control frame that the primary datum is to contact the datum feature simulator first, then the secondary datum and lastly the tertiary. This way everyone is inspecting the part the same way and the results will be the same. It prevents rejection of good parts and acceptance of bad parts. I hope this helps clear things up a bit. Cheers,Matt 9. Hi, would the workpiece thickness influence the perpendicularity of holes (I am trying to find some references or online sources which could explain that). in my opinion, if the workpiece is thinner then it is more likely to have increase bending in the workpice during drilling which could influence the perpendicularity of the machined hole. Regards Khaled Yes, that is correct, when the material will have less structural control over the piece as well. Think a piece of sheet metal with a bolt in it. The perpendicularity of the hole does not matter too much since the bolt would easily bend the sheet metal if it was at an off angle. One solution to this is functional testing or Finite Element Analysis, which could tell you the strain that the hole will have with a fastener. Location matters too since a hole that is very offset would need to have its fastener installed at an angle. Perpendicularity is usually best when the depth exceeds the diameter of the fastener as a rule of thumb. I have a question about determining what value of perpendicularity tolerance to put on a drawing. I have a part with a blind hole for an M5 thread-forming screw. I can determine the size and position tolerance on the hole based on thread-engagement and lining up the hole with the mating part, but I’m not sure what considerations inform the perp tolerance. Considering that the screw is about 12mm long, it will naturally align to the blind hole into which it is threading. If the hole is at a cocked angle, it could put asymmetrical compressive force on the mating part under the screw-head. I imagine that determining the perp tolerance properly would probably necessitate calculations that involve the compressive yield stress of the mating part, the size of the screw head, and other factors, but I’m hoping there might be a simpler “rule of thumb” I can reference? 11. Himy query is on our drg. hole perpendicularity is given in value over a length. i.e 0.03/100 . Is it correct to mention perpendicularity of hole ? i think as per your guidance it should be diameter not length. 12. i have a cylinder product, which L is approx. 57 mm. Customer requires us to measure perpendicularity of inner diameter by using one of the end of cylinder as the datum. We do not have a CMM now. I have no idea on how to measure for this condition. 13. Hi,I have a long bolt. The axis of the bolt is referenced as Datum A. A small surface is of shank is defined with 0.15mm Perpendicularity. Is this a correct definition of perpendicularity?This datum A is now added with datum modifier. How the perpendicularity will be affected due to this datum modifier? 14. Matt great info!!Quick question. Can perp of a hole to a primary datum be measured by position? Ie if I have 2 dimensions one from secondary and one from tertiary datums, but the axis is off 90 degree. Would this not show one of the dimensions out of position? Or does the perpendicularity have to be measured separately and then geometric tol calculated? Nick (aka Spacechicken) – Actually, perpendicularity is controlled via position which is why you will frequently see a composite control with a refinement of the perpendicularity. The way the datum structure works is that you are locating a theoretically exact cylinder at basic dimensions to your secondary and tertiary datums and perfectly orthogonal to your primary datum. Within this cylinder the axis of the hole may shift left/right/up/down or tilt as long as it remains entirely within the cylinder throughout the length of the part. It is in this manner that the perpendicularity of the feature is being restricted. When you perform the inspection of this requirement through the use of either a functional gage or by using a CMM you are also inspecting the perpendicularity. As an example, if your part was drilled perfectly for location and size but at a 25 degree angle the axis of this hole would be out of bounds of the tolerance zone cylinder I had described earlier and the part would be rejected. We’ve got a great introduction to the basics of GD&T, I’d encourage you to enroll and learn more. Let us know how else we can help. Cheers,Matt 15. This has been a great source of GD&T knowledge for an unexperienced quality engineer like me. I have a question regarding your axis perpendicularity examples: Does having 0.2 bonus tolerance mean that I should now use a functional gauge Ø = 9.5? Since Min. hole Ø – Perpendicular Tolerance is now equal to 9.9 – 0.4. This question extends to all bonus tolerance applications, really. Should the bonus tolerance be considered in choosing the appropriate functional gauge? Aaron – All functional gages are based off of the virtual condition. Also, remember that in order to use fixed functional gaging you have to have either the MMC or LMC modifier called out in the feature control frame. Here’s a quick rundown on what the virtual condition is for both MMC/LMC for internal and external features. MMCExternal Feature: Largest size + tolerance (from feature control frame)Internal Feature: Smallest size – tolerance (from feature control frame) LMCExternal Feature: Smallest size – tolerance (from feature control frame)Internal Feature: Largest size + tolerance (from feature control frame) Hope this helps. Cheers,Matt 16. Thank you for your website and comments. I find them very helpful.My question is this, does a perpendicularity callout (with dia symbol) for a hole feature provide control for both orientation AND location? If so, would that be the case when it just happens that you want to control perpendicularity and location with the same cylindrical tolerance zone? Wouldn’t you accomplish the same thing if a position callout was used? Walter – So this is actually one of the most common misunderstandings of the perpendicularity control. It does NOT control the position of a feature of size. It only controls the orientation. The simplest way to explain this is that perpendicularity and parallelism are both very specific cases of angularity. Would you be able to support the statement that either parallelism or angularity locate features of size? Same story for perpendicularity. It is for this reason that you often see perpendicularity as a refinement of the position tolerance for features of size. The only situation I can think of offhand where you have only a perpendicuarlity control applied to a feature of size would be on datum. An example of this would be if you were specifying a perpendicularity requirement for the inner or outer diameter (as a datum) of a washer to the flat primary surface. You would then use position to locate the other diameter of the washer back to datum A and B. I hope this helps. Let us know if you have any further questions, we’re happy to help. Cheers,Matt 17. thanks a lot Matt, that really helped. does that mean that a perfect hole would have near 0 perpendicularity or ideally zero. I will make sure to add any updates here if I get any new information regarding the matter. Cheers Khaled 18. Hi All., I got some data obtained by a CMM machine for machined holes, one of the measured parameters is called PERP. I assumed it is hole perpendicularity. but I am not sure what is the unit for the data. is it in mm since the other data is in mm or is it in degrees.sorry I am not familiar with perpendicularity before and I am trying to make a study from collected results. sorry for asking such a questions but if for example for a hole the perpendicularity is 10 microns, what can this tell me about the hole?many thanks Khaled – Perpendicularity as it applies to holes describes the orientation of a hole axis relative to a specified datum. The tolerance zone is represented by a cylinder located at the basic location. The axis of the controlled hole must then lie entirely within this tolerance zone cylinder in order to be considered ‘in spec’. So when the CMM is telling you the perpendicularity value of the hole in mm I believe it is telling you the actual as manufactured condition. To describe it a little bit better it is telling you what the smallest diameter the axis of your hole would fit in. Unless you have programmed it otherwise you’ll have to then compare the as measured value against the drawing requirement. Use a little bit of caution and common sense here as I’m not all that familiar with CMM and inspection equipment. I hope this helps. If you find out otherwise let us know and contribute to the community knowledge base. Cheers,Matt 19. We have a six sided block, all angles 90 degrees. Perpendicularity of a surface relative to an adjacent surface is described as 0.00174 RAD (followed by the first surface reference letter, then the 2nd surface reference letter) Does RAD describe the maximum variation in Radians? (so .00174 RAD is .0997 degrees?) If not, I cannot determine how RAD is used in this case. Any help is appreciated! Thank youDillon Dillon – I can’t really think what else RAD would stand for other than radians. Taking the radian value and multiplying by 180/pi will give you degrees. What’s confusing here is that in GD&T all of the orientation controls (angularity, parallelism and perpendicularity) all state tolerances in terms of width between 2 parallel planes NOT degrees. Can you provide any more info? Is there a chance there is a setting in your software that isn’t what it should be? I hope this helps. Sorry I wasn’t able to provide conclusive direction here. Cheers,Matt 20. Hi I really Enjoy to read our GD&T Lesson,but i have doubt How you will Define Geometrical Tolerance value? for examPerpendicular Tolerance value is 0.03 Ramesh – A geometric tolerance zone is best described as: the distance between two parallel planes or lines, a sphere (very rare) cylinder or circle of diameter equal to the specified tolerance value. I’ve listed out the geometric controls and the types of tolerance zones they can use. Flatness of a Surface – Two parallel planesFlatness of a Feature of Size (FOS) – Two parallel planesStraightness of a Surface – Two parallel linesStraightness of a FOS – A cylinderCircularity – Two coaxial circlesCylindricity – Two coaxial cylindersPerpendicularity of a Surface – Two parallel planesPerpendicularity of a FOS – Two parallel planes or a cylinder if the dia symbol precedes the toleranceAngularity of a Surface – Two parallel planesAngularity of a FOS – Two parallel planes or a cylinder if the dia symbol precedes the toleranceParallelism of a Surface – Two parallel planesParallelism of a FOS – Two parallel planes or a cylinder if the dia symbol precedes the tolerancePosition – Two parallel planes or a cylinder if the dia symbol precedes the toleranceConcentricity – A cylinderSymmetry – Two parallel planesCircular Runout – Two coaxial circlesTotal Runout – Two coaxial circles when applied to a diameter and two parallel planes when applied to a surface that is 90 deg to your datum axis I hope this helps, let us know if you have any further questions. Cheers,Matt 21. not remedy what i mean is common solution 22. sir, What are commom problem why we encountered no good perpendiculary with our cnc lathe process and what are the remedy 23. Hello, I’ve the question on the perpendicular. Normally we measure the perpendicular of a part is placing the datum to granite table, let say the measure result is 0.1. However, when the part is to be mount on the machine (with tighten screw on the datum side), the perpendicular value is run out till 0.2 due to the flatness of the datum is not good. In this case, is it i need to specify flatness on the datum as well? Thanks. Loh – I’m not positive I understand your question. You do seem to have a firm grasp on how things work though. In general it is a good practice to place a flatness tolerance on planar datums to make them more stable and reliable on the inspection table. I believe this should solve your issue. I would also encourage you to look at the mating surface as well if the perpendicularity during assembly is critical. I hope this helps. Cheers,Matt 24. Can perpendicularity have 2 datum? thanks Shrenik – Absolutely, perpendicularity is allowed to use multiple datums. Think of a coordinate system with X, Y, and Z axes. With these three axes you define three separate planes, that in a right coordinate system are all mutually orthogonal. Any single one of these planes can tilt while still being perpendicular to the other. It is in this sense that you can restrict or limit the angularity and perpendicularity to multiple datums. I hope this helps. Cheers,Matt 25. I have a cylindrical part and I wish to specify the perpendicularity of the end faces. Intuitively, I want to use the axis of the part as the datum and specify perpendicularity of the faces. This seems to be the opposite of the description of normal use. Would I be in error if I followed my intuitive approach? John – There are two ways you can accomplish what you want. The first of which is to reference the diameter of the cylinder as a datum and then specify a perpendicularity control on the normal face. The second and more roundabout way of doing the same thing would be to use a total runout control on the normal face. The net result is the same, you’re just not used to thinking of surfaces in this way. I hope this helps! Cheers,Matt 26. Hello, very good information from this site and well explained, I have a doubt, I have a drawing that requires to measure perpendicularity of a hole with cylindrical tolerance, but the reference datum is a cylinder (axis), the piece to be measured does not have any planes since it’s completely cylindrical (tube). Is it possible to have a evaluation of perpendicularity on a cylindrical zone between 2 axis? Oscar – Absolutely. The axis of one feature is based on the orientation of the other. I’m assuming that the perpendicular hole is for a pin/setscrew/bolt etc. The main axis (the datum) moves within it’s tolerance zone and has a direct affect on the lower level axis. It is theoretically located at 90 from the primary datum and is free to move within the confines of its own tolerance zone. Hope this helps. Cheers, Matt 27. We have a plane being dimensioned perpendicular to a cylinder but they also have a flatness callout on the same surface, i.e. Perp. .0005 and Flatness .0005. Based on the example above, these appear to be overlapping callouts? My initial thought was the surface must be flat within .0005 and the plane must be perpendicular within .0005, not all the individual points of the surface. Mark – Not necessarily and are you sure it’s not the other way around. What is the datum, the plane or the cylinder? A perpendicularity control requires the use of a datum and it’s fairly normal to throw a flatness control onto a datum to improve stability on the inspection table. Now, in the event that it’s the plane that is being double controlled as you have it. It would appear that these are conflicting requirements, where in reality the flatness would be serving as a refinement of the perpendicularity. You are correct in that you are viewing the perpendicularity control as an oriented flatness. However, what you are forgetting is that flatness does not use any datum and is only holding the flatness of the surface relative to itself. So it’s entirely possible that you can live with your part have a surface perpendicular to within .04, but then, within that zone be flat relative to itself (the surface) to .2. I hope this helps. Cheers,Matt 28. Hello, If a perpendicularity specification is called out for a hole but with no symbol of diameter, and besides, the only datum is a line (not a plane), how can we proceed? How can we interpret this? Is it an axis perpendicularity? The drawing is ruled by ASME Y14.5. Thanks a lot Jorge – If a perpendicularity specification is applied to a circular feature of size it must include a diameter symbol before the tolerance. This is in accordance with both ASME Y14.5 specs in common use (1994 and 2009). Not to throw stones but I believe the drawing to be in error. Further, having a single line (edge?) as a datum isn’t correct either. I hope you mean axis, this is really the only way this control works. My recommendation to you would be to go back to the designer/engineer and ask for clarification and gently point out areas you are struggling with. Now, as for what I think the engineer/designer is trying to achieve? I think they are trying to control the perpendicularity of the circular feature basically relative to a cylindrical tolerance zone. The thought process isn’t necessarily wrong, just the implementation. I hope this helps, let me know. Cheers,Matt 29. What if i do not have access to CMM. I need to measure the perpendicularity of a hole. What is the old school method of doing this? The easiest way to check perpendicularity of a feature without a CMM is through the use of a functional gage, where the pin of the gage is set to the virtual condition of the hole and perpendicular to the relevant datum reference frame of the gage. This won’t tell you the amount of perpendicularity error, but it will provide either a pass/fail check. This method is often used in high quantity production runs as it requires no interpretation, if it fits on the gage the part meets requirements, if it doesn’t, it fails. This assumes that you have the MMC modifier called out in the feature control frame next to the allowable tolerance. There are a few other methods you can use to actually determine the amount of perpendicularity error, but they become somewhat complicated. 30. if there is no call out or symbol for perpendicular between a hole and surface how is a tolerance determened Typically you ca add a general loose tolerance for perpendicularity, and Profile that apply to the entire drawing. If nothing is called out though it should be added. For an ISO drawing, there are standards that can apply a general tolerance to the drawing, but there is no such standard for the ASME y14.5 What about implied 90drg. angle hole is .500-501 dia. how is tolerance determined Implied 90 NEEDS a tolerance of perpendicularity either in the notes, title block or on the dimension. If this is missing then it is an untoleranced dimension and can essentially be whatever you want. If the datum is the hole, then the surface needs to be within two parallel planes on the surface (of whatever tolerance (x) that is called) if the datum is the surface, then your hole’s axis needs to be within a perfect cylinder of diameter (x) perpendicular to the surface. Without a tolerance though this is wide open and ambiguous. We have a Dwg callout of perpendicularity of Threaded collar for a bolt to be inserted ref to a Datum surface that the collar is welded onto – .50mm to datum A . There is no other indications on the dwg for bonus tolerance or any other symbol – There is a great deal of opinion as to the correct use of this symbol Vs Angle . We want the threads inside the collar to be straight to datum surface – Please advise if you can – Thank you . The theoretical definition would be that there is an axis that you want perpendicular to a datum surface. What you are controlling with this is a cylindrical tolerance zone that is perfectly perpendicular to datum A. it can translate anywhere along datum A but not tilt. The axis of your threaded bolt must fall within this cylindrical tolerance zone that can translate along the datum surface. For Perpendicularity, the axis is derived from the Unrelated Actual Mating Envelope which is another way of saying the smallest perfect cylinder that can fit around the thread. Here is a picture for more detail: A common misconception when people hear “perpendicularity” is they think of an angle. Angle really has nothing to do about it. You are taking an imperfect part and seeing if it fits a perfectly perpendicular size condition. Hope this helps! 32. Supposing you have a cylindrical feature extending off of a flat surface that you want to put a perpendicularity spec on. If I am understanding the explanation correctly, the axis perpendicularity spec defines the diameter of a cylinder that the axis of the feature must fit entirely within. Therefore, am I correct in understanding that for the same perpendicularity spec (say, 0.030 like in the given examples), the allowable angle of tilt for the cylindrical feature is also dependent on its length? i.e. a short cylinder could be tilted at a larger angle than a tall cylinder, since the shorter cylinder’s axis can be further off of perpendicular without falling outside of the specified position? Yes that is correct. A smaller work-piece would allow a larger angle of offset. However, if they were properly located in the same initial spot, the top of the cylinder would not go past a certain limit. forgot to say Datum A is a Plane I have this drawing that has a callout of perpendicularity of a hole to Datum A only, but the Ø symbol is not there (tolerance type), so is it possible to inspectthis hole with the planar tolerance type zone?. The software i am using says is good but the hole is tilted about 20 degrees and the perp. tol. is very small. if i switch to Ø tol type zone then it rejects it as it should. cut this be the wrong callout for this feature?thanks. Perpendicularity of an axial feature (to anything) should have a Ø symbol. Now this could be a typo, but according to the standard it is required. The tilt of the angle is only indirectly influencing it. It needs to fall within a cylindrical tolerance zone. I have two grips of 200 mm each and both touched in uper portion and lower portion . Total tolerance limit is .24mm. In between each jaw V slot of 100 mm is cut which has measuring error between .1mm to .25 mm.. pls inform what will be my tolerance limit Along with the datum A, if another datum B is there. Then the surface should also be perpendicular to datum B also.Which means that the surface should be Perpendicular to both A and B datum….?? Yes – you need to consider restraining all the degrees rotation of the surface plane. If your function requires the orientation requirement to both datums to fully constrain the rotation. How would you measure the length of a cylinder’s perpendicularity to the center line of the cylinder? I have access to a CMM but when I try to use the tolerance buttons for it, it won’t allow that action. TO measure axis perpendicularity, you must have a datum and then establish your axis to this. The axis is established by the software by taking measurement points of the cylinder (the more the better). I have not learned CMM programming so I cannot help you too much with the specifics. Hi,I really love your website and your content which is very much easier to apprehend. In the surface perpendicularity content mentioned on this page, there shouldn’t be any diametrical tolerance zone as we are controlling the surface instead of an axis. But in the diagram Ø symbol is followed by 0.2 tolerance. Correct me if I am wrong as I am a novice in this field. Thanks & Regards,Chandy You are 100% correct – this is a typo. Good catch on that! the Ø symbol only applies when there is a cylindrical tolerance zone. Perpendicularity is almost always shown using holes as an example. What about something like a protrusion from a surface that needs to be perpendicular would the same concepts apply. I’ve got a heating element part that has two legs that go thru thin sheet metal and those need to remain perpendicular to the main heating element surface such that when the nuts are installed on the threaded fitting on the legs it does not cause the main heating element surface to come out of a certain flatness range. I’m unsure what GD&T call out would accomplish this. Perpendicularity can indeed be used on more than just a cylinder. A cylinder is probably the most common application since you are usually defining some sort of datum feature (hole or pin) making it perpendicular to a surface. However you can use it to control, say, the perpendicularity of a flange or fin based on its central plane. Whenever the feature you are controlling is a box-like protrusion, you would be controlling its central plane. I will think of a simple example to show this and add it on here soon. Thanks for the feedback! Comments are closed. ← Newer Older →
2660	https://fiveable.me/ap-calc/unit-1/removing-discontinuities/study-guide/InY5nKkX7B44HNb3T0fD	Removing Discontinuities - AP Calc Study Guide \| Fiveable \| Fiveable ap study content toolsprintables upgrade ♾️AP Calculus AB/BC Unit 1 Review 1.13 Removing Discontinuities All Study Guides AP Calculus AB/BC Unit 1 – Limits and Continuity Topic: 1.13 ♾️AP Calculus AB/BC Unit 1 Review 1.13 Removing Discontinuities Written by the Fiveable Content Team • Last updated September 2025 Verified for the 2026 exam Verified for the 2026 exam•Written by the Fiveable Content Team • Last updated September 2025 print study guide copy citation APA ♾️AP Calculus AB/BC Unit & Topic Study Guides AP Calculus AB/BC Exams Unit 1 – Limits and Continuity Unit 1 Overview: Limits and Continuity 1.1 Introducing Calculus: Can Change Occur at An Instant? 1.2 Defining Limits and Using Limit Notation 1.3 Estimating Limit Values from Graphs 1.4 Estimating Limit Values from Tables 1.5 Determining Limits Using Algebraic Properties of Limits 1.6 Determining Limits Using Algebraic Manipulation 1.7 Selecting Procedures for Determining Limits 1.8 Determining Limits Using the Squeeze Theorem 1.9 Connecting Multiple Representations of Limits 1.10 Exploring Types of Discontinuities 1.11 Defining Continuity at a Point 1.12 Confirming Continuity over an Interval 1.13 Removing Discontinuities 1.14 Connecting Infinite Limits and Vertical Asymptotes 1.15 Connecting Limits at Infinity and Horizontal Asymptotes 1.16 Working with the Intermediate Value Theorem (IVT Calc) Unit 2 – Fundamentals of Differentiation Unit 3 – Composite, Implicit, and Inverse Functions Unit 4 – Contextual Applications of Differentiation Unit 5 – Analytical Applications of Differentiation Unit 6 – Integration and Accumulation of Change Unit 7 – Differential Equations Unit 8 – Applications of Integration Unit 9 – Parametric Equations, Polar Coordinates, and Vector–Valued Functions (BC Only) Unit 10 – Infinite Sequences and Series (BC Only) Frequently Asked Questions Previous Exam Prep Study Tools Exam Skills AP Cram Sessions 2021 Live Cram Sessions 2020 practice questions print guide report error 1.13 Removing Discontinuities In this guide, we’ll be diving into the fascinating world of discontinuities. Ever looked at a graph and noticed it suddenly jumps or has holes? These are discontinuities, and sometimes, we can remove them. Let's explore how we can smooth out these mathematical speed bumps and make our functions continuous. more resources to help you study practice questionscheatsheetscore calculator ⭕What are Discontinuities? Before we patch things up, let's first understand what we're dealing with. Discontinuities occur where a function "breaks." There are three types: removable, jump, and infinite. Today, we are going to focus on removable discontinuities. 💡 Jump back to 1.12 if you need a refresher on confirming continuity over an interval! An image depicting the three types (removable, jump, and infinite) of discontinuities. Image courtesy of LibreTexts Mathematics/02%3A_Limits/2.04%3A_Continuity) 📍Removable Discontinuities Removable discontinuities are also sometimes called “holes” in a graph. They exist when a function is undefined at a point, but its limit exists. We can fill this hole to make the graph continuous. Here’s an example: A graph with a removable discontinuity. Image courtesy of LibreTexts Mathematics/02%3A_Limits/2.04%3A_Continuity) We can see in this graph, the limit as x x x approaches a a a exists, it’s just undefined at that point. But, if we color in the circle (in other words, remove the discontinuity), the graph will be continuous over the interval in the image. 🌀Filling the Gap To remove a discontinuity, we redefine the function's value at that point to equal the limit of the function as x approaches that point. For example: f(x)=x−1 x 2−1 for x≠1 there’s a hole at x=1 f(x) = \frac{x - 1}{x^2 - 1} \quad \text{for} \quad x \neq 1 \text{ there's a hole at}\quad x = 1 f(x)=x 2−1 x−1for x=1 there’s a hole at x=1 To remove this discontinuity, we can factor the denominator and cancel like terms, like so: x−1 x 2−1=x−1(x−1)(x+1)=1 x+1\frac{x-1}{x^2-1}=\frac{x-1}{(x-1)(x+1)}=\frac{1}{x+1}x 2−1 x−1=(x−1)(x+1)x−1=x+1 1 This function no longer has a discontinuity at x=1 x=1 x=1, instead, it is defined as 1 2\frac{1}{2}2 1. ✏️ Practice Filling the Gap Consider the function g(x)g(x)g(x) defined as follows: g(x)=(x 2−4 x+3)(x−1)for x≠1 g(x) = \frac{(x^2 - 4x + 3)} {(x - 1)} \text{ for x} \neq 1 g(x)=(x−1)(x 2−4 x+3)for x=1 g(x)=k for x=1 g(x) = k \text{ for x = 1}g(x)=k for x=1 where k k k is a constant. Determine the value of k k k that would make g(x)g(x)g(x) continuous at x=1 x = 1 x=1. Here's the solution! ⬇️ First, we need to factor our numerator. This will allow us to see whether we can cancel out the denominator, which is what is causing our discontinuity. x 2−4 x+3=(x−3)(x−1)x^2-4x+3=(x-3)(x-1)x 2−4 x+3=(x−3)(x−1) We see that there is an x−1 x-1 x−1 term in our numerator, so we can cancel it with the x−1 x-1 x−1 in the denominator! (x−3)(x−1)(x−1)=x−3\frac{(x-3)(x-1)} {(x - 1)}=x-3(x−1)(x−3)(x−1)=x−3 For x=1 x=1 x=1, g(x)=k=1−3=−2 g(x)=k=1-3=\boxed{-2}g(x)=k=1−3=−2 📈 Piecewise Functions For piecewise functions, we can ensure continuity by checking the right and left limit of the function, and assuring that the value is the same as the one defined at the point. ✔️Ensuring Continuity Consider f(x)f(x)f(x) defined by two pieces of a function on either side of x=a x = a x=a: On the left side, f(x)f(x)f(x) approaches L L L as x x x approaches a a a. On the right side, f(x)f(x)f(x) approaches M M M as x x x approaches a a a. For f(x)f(x)f(x) to be continuous at x=a x = a x=a, we need L=M=f(a)L = M = f(a)L=M=f(a). For example, this image depicts the continuous, piecewise function f(x)={x 2−3,x≤2 x−1+2,x>2}f(x) = \begin{Bmatrix}x^2-3, & x\leq2 \ x-1 + 2, & x> 2\\end{Bmatrix}f(x)={x 2−3,x−1+2,x≤2 x>2} Continuous piecewise function defined by x 2−3 x^2-3 x 2−3 for x≤2 x≤2 x≤2 and x−1 x-1 x−1 for x>2 x>2 x>2. Image courtesy of mathcoachblog We can see the the limit of f(x)f(x)f(x) as it approaches the point a=2 a=2 a=2 from the left is equal to 1 1 1, and similarly, from the right, it is also 1 1 1. Finally, f(x)f(x)f(x) is defined at a=2 a=2 a=2 as 1 1 1. Thus, our function is continuous. ✏️Practice Ensuring Continuity Consider the function, f(x)={x 2+5 x+4 a(x+4),x≠2 a,x=2}f(x) = \begin{Bmatrix}\frac{x^2+5x+4}{a(x+4)}, & x\neq2 \ a, & x=2\\end{Bmatrix}f(x)={a(x+4)x 2+5 x+4,a,x=2 x=2} What must a a a be set to for the function to be continuous? To solve this problem, we will use the first part of the piecewise function to solve for a a a at x=2 x=2 x=2. First, let’s factor and cancel some terms: x 2+5 x+4 a(x+4)=(x+1)(x+4)a(x+4)=x+1 a\frac{x^2+5x+4}{a(x+4)}=\frac{(x+1)(x+4)}{a(x+4)}=\frac{x+1}{a}a(x+4)x 2+5 x+4=a(x+4)(x+1)(x+4)=a x+1 Now, let’s plug in x=2 x=2 x=2: x+1 a=2+1 a=3 a\frac{x+1}{a}=\frac{2+1}{a}=\frac{3}{a}a x+1=a 2+1=a 3 Finally, we set this equal to $a$ and solve: 3 a=a→3=a 2→3=a\frac{3}{a}=a\rightarrow3=a^2\rightarrow \boxed{\sqrt{3}=a}a 3=a→3=a 2→3=a 📷Visualizing Continuity Graphs are an excellent way to see continuity (or the lack thereof). Use graphing tools to visualize the function and identify discontinuities. A continuous graph can be drawn without lifting your pencil! ✏️Practice Visualizing Continuity Graph the function f(x)=x 2−9 x+3 f(x)=\frac{x^2-9}{x+3}f(x)=x+3 x 2−9 over the interval [−5,5][-5,5][−5,5]. Is it continuous? Can you make it continuous? The graph of this function looks like so: Graph of the function described above. Image courtesy of Emery You can see that there is a discontinuity at x=−3 x=-3 x=−3. But, we can remove it by factoring! x 2−9 x+3=(x+3)(x−3)x+3=x−3\frac{x^2-9}{x+3}=\frac{(x+3)(x-3)}{x+3}=x-3 x+3 x 2−9=x+3(x+3)(x−3)=x−3 Plugging in x=−3 x=-3 x=−3, we find that f(x)=−6 f(x)=-6 f(x)=−6 at x=−3 x=-3 x=−3, making our function continuous. ⭐ Closing 📚 AP Calc is practice-driven! Attempt more problems, especially from past AP exams, to strengthen your understanding. Always check for continuity and practice "patching up" those functions. Great work! 👏 Frequently Asked Questions How do I remove a discontinuity from a function? A removable discontinuity (a “hole”) can be fixed only when the limit exists at that point. Quick steps: 1. Identify the point a where f is undefined or piecewise changes. 2. Compute the two one-sided limits: lim_{x→a^-} f(x) and lim_{x→a^+} f(x). If they’re equal, call that common value L. (AP expects you check one-sided limits for piecewise boundaries—EK LIM-2.C.1 and EK LIM-2.C.2.) 3. If the limit L exists, redefine f(a) = L to remove the discontinuity. If the one-sided limits differ or limit is infinite, it’s not removable. 4. For algebraic rational functions, factor and cancel the common (x−a) factor, then evaluate the simplified expression at a to find L. Example: f(x) = (x^2−1)/(x−1). Factor to (x+1), so lim_{x→1} f(x)=2. Redefine f(1)=2 to remove the hole. For piecewise functions, solve for parameters so left-hand expression at the boundary equals the right-hand value and the function value there. More practice: see the Topic 1.13 study guide ( and unit review ( What's the difference between removable and non-removable discontinuities? A removable discontinuity (a “hole”) is a point where the two one-sided limits agree (so the overall limit exists) but the function’s value is either undefined or not equal to that limit. You can “remove” it by redefining the function at that x to equal the limit—common with rational functions after factor cancellation. This is exactly EK LIM-2.C.1: if the limit exists at the discontinuity, you can redefine the value to make the function continuous. A non-removable discontinuity is one you can’t fix just by redefining the function at that single x. Examples: a vertical asymptote (limit is infinite) or a jump/step where left- and right-hand limits are different (so the limit does not exist). For piecewise functions, continuity at a boundary requires left value = right value = function value (EK LIM-2.C.2); if that fails and the one-sided limits differ, it’s non-removable. For extra practice and examples, see the Topic 1.13 study guide ( and lots of problems ( When can I actually fix a discontinuity and when can't I? You can “fix” a discontinuity only when the limit exists at that x—meaning the left-hand and right-hand limits are equal. If lim f(x) as x→a exists but f(a) is missing or different, you can redefine f(a) to equal that limit and remove the hole (a removable discontinuity; see EK LIM-2.C.1). Typical algebraic case: a factor cancels in a rational function so you get a hole you can fill by taking the simplified value. You can’t fix it when one-sided limits differ (a jump) or when the function blows up (vertical asymptote/infinite limit)—there’s no single finite value to assign that makes the function continuous. For piecewise functions, continuity at a boundary needs left limit = right limit = function value at the boundary (EK LIM-2.C.2). For more examples and AP-style practice, check the Topic 1.13 study guide ( and unit review ( Practice problems are at ( How do I solve for the parameter that makes a piecewise function continuous? To make a piecewise function continuous at a boundary x = a, you need the left-hand limit, right-hand limit, and the function’s value at a to all match (CED EK LIM-2.C.2). Practically: 1. Compute lim(x→a−) f(x) using the expression for x a. 3. If the function’s value f(a) is given as a parameter, include it; otherwise you may be allowed to redefine f(a). Set the two one-sided limits equal and solve for the parameter. If those limits match a simple value L, then set f(a)=L (this “removes” a removable discontinuity per EK LIM-2.C.1). If one-sided limits don’t agree, no single parameter can make it continuous. Quick tip: with rational pieces, simplify (factor/cancel) before taking limits. For more examples and practice aligned to the AP CED, see the Topic 1.13 study guide ( the Unit 1 overview ( and lots of practice problems ( I don't understand how to make a function continuous at a point - can someone explain step by step? Short version: continuity at a point means limit exists there and the function’s value equals that limit. Step-by-step: 1. Identify the point a. Check one-sided limits: lim x→a− f(x) and lim x→a+ f(x). If they’re not equal, the limit doesn’t exist → not removable. (CED: one-sided limits, limit exists.) 2. If one-sided limits match, compute L = lim x→a f(x). If f(a) is undefined or f(a) ≠ L, you can remove the discontinuity by redefining f(a) = L (removable hole / removable discontinuity). 3. For algebraic functions (like rationals), simplify: factor and cancel common (x−a) terms, then plug in a to find L. 4. For piecewise functions, set the left-expression and right-expression equal at the boundary and solve for any parameter(s) so both sides (and f(a) if given) equal the same value (CED: continuity at a boundary). 5. Always mention one-sided limits on the AP—they assess LIM-2.C and piecewise continuity. For worked notes and examples see the Topic 1.13 study guide ( For more review and practice, check the Unit 1 overview ( and practice problems ( What's the formula for removing discontinuities in rational functions? If a rational function has a removable discontinuity (a “hole”) at x = a, the fix is: redefine the function value at a to equal the limit as x → a. Practically: 1. Factor numerator and denominator and cancel any common factor (x − a). 2. Evaluate the simplified expression at x = a to get L = lim_{x→a} f(x). 3. Redefine f(a) = L. That removes the hole if the one-sided limits agree (i.e., the limit exists). In symbols: if f(x) = p(x)/q(x) and p(x) = (x−a)·r(x), q(x) = (x−a)·s(x) with s(a) ≠ 0, then for x ≠ a f(x) = r(x)/s(x) and lim_{x→a} f(x) = r(a)/s(a). Define f(a) := r(a)/s(a) to make f continuous at a (EK LIM-2.C.1, factor cancellation). For piecewise functions, set the boundary value equal to the common one-sided limit (EK LIM-2.C.2). For more examples and practice, see the Topic 1.13 study guide ( and Unit 1 overview ( How do I know if a discontinuity is removable just by looking at the graph? Look for a hole (an open circle) in the curve where the left- and right-hand behavior match. If the graph approaches the same y-value from both sides but either there's an open dot (no value) or a filled dot at a different y, that point is a removable discontinuity—you can redefine f(a) to equal the common limit. If the left and right limits are different (the two sides go to different y-values) the discontinuity is nonremovable. For piecewise functions, check the boundary value: the two side-expressions and f(a) must agree for continuity (EK LIM-2.C.2). In AP language: a discontinuity is removable exactly when the limit exists at that x (EK LIM-2.C.1). For practice and quick examples, see the Topic 1.13 study guide ( and try problems at ( When do I use limits to fix discontinuities vs just plugging in values? Use limits whenever the function’s value at the point is undefined or doesn’t match the behavior of nearby x-values. Key CED ideas: - If f(a) is defined and lim_{x→a} f(x) exists and equals f(a), the point is continuous (EK LIM-2.C.1). If f(a) ≠ lim, you can “remove” the hole by redefining f(a) to equal the limit. - If you get an indeterminate form like 0/0, algebraically simplify (factor/cancel) and then plug the simplified expression into x = a to find the limit—that limit is the value you’d assign to remove a removable discontinuity. - If one-sided limits differ (jump) or either side is infinite, the two-sided limit doesn’t exist so you can’t fix continuity by one value. - For piecewise functions, require left-hand limit = right-hand limit = function value at the boundary (EK LIM-2.C.2). Quick checklist: Is f(a) defined? Is lim_{x→a} f(x) finite? Do one-sided limits match? If yes, plug that limit in; if no, you must use limits to analyze and you can’t always “fix” the discontinuity. More on examples and practice in the Topic 1.13 study guide ( For broad review: ( For extra practice problems: ( How do I make a piecewise function continuous at the boundary points? Focus on each boundary point c. For a piecewise function you need three things at x = c: 1) left-hand limit exists: lim x→c− f(x), 2) right-hand limit exists: lim x→c+ f(x), 3) those two one-sided limits are equal and equal to the function value f(c). So if your pieces are g(x) for x c (and maybe a value A at x=c), require lim x→c− g(x) = lim x→c+ h(x) = A. If A is not given or you can choose a parameter (say k), set g(c) = h(c) (after simplifying) and solve for k, then define f(c) to be that common limit. Typical algebra: factor and cancel to find limits for rational expressions (removable hole). Quick checklist for the AP: always check one-sided limits at boundaries (CED LIM-2.C / EK LIM-2.C.1,2). If you want worked examples and practice, see the Topic 1.13 study guide ( and try problems at What does it mean that the limit exists at a discontinuity but the function value doesn't? If the limit exists at a discontinuity but the function value doesn’t, it means the function’s behavior approaching that x-value settles to a single number L (left-hand and right-hand limits equal L), but f(a) is either undefined or equals some other number. Graphically that’s a removable discontinuity (a “hole”)—the curve approaches a point but there’s no dot (or the dot is in the wrong place). By EK LIM-2.C.1 you can “remove” the discontinuity by redefining f(a) = L so the function becomes continuous at a. Example: f(x) = (x^2 − 1)/(x − 1) simplifies to x + 1 for x ≠ 1, so lim x→1 f(x) = 2, but f(1) is undefined. Redefine f(1)=2 to remove the hole. For AP-style piecewise problems, remember EK LIM-2.C.2: both one-sided expressions and the function value must match at the boundary (see the Topic 1.13 study guide: For extra practice try the AP practice question bank ( I'm confused about redefining function values - how do I know what to redefine it as? You redefine the value at a point so the function equals the limit as x approaches that point—but only when that two-sided limit exists (EK LIM-2.C.1). Practically: - Check the limit. If lim x→a f(x) exists, define f(a) = that limit to “fill the hole.” - For piecewise functions, make left- and right-hand expressions match at the boundary and set f(boundary) equal to that common value (EK LIM-2.C.2). Common algebra steps: simplify the expression (factor and cancel removable factors), then plug in x = a. Example: f(x) = (x^2 − 1)/(x − 1) is undefined at 1, but simplifies to x + 1, so lim x→1 f(x) = 2—redefine f(1)=2. On the AP exam you’ll be asked to solve for parameter values or boundary values that make a function continuous; show the one-sided limits when relevant and state the value you redefine to (the limit). For more practice and worked examples, see the Topic 1.13 study guide ( and the Unit 1 overview ( For extra problems, try the practice bank ( How do I solve problems where I need to find the value of k that makes the function continuous? You want k so the function has no removable hole at x = a—so make f(a) equal the limit as x → a. Steps: 1. Check the two one-sided limits (if piecewise) or the two-sided limit. For continuity you need lim x→a− f(x) = lim x→a+ f(x) = L and f(a) = L (EK LIM-2.C.1 & LIM-2.C.2). 2. If f is algebraic/rational with a factor (x − a) causing 0/0, factor and cancel to find the simplified expression g(x). Then L = g(a). Set k = L. 3. If the function is piecewise, evaluate each side at the boundary and solve for k so left value = right value = k. 4. If one-sided limits differ or limit doesn’t exist, no value of k can make it continuous. Example outline: f(x) = { (x^2−1)/(x−1) for x≠1; k for x=1 } → simplify to x+1, so L = 2, choose k = 2. For guided examples and AP-aligned practice, see the Topic 1.13 study guide ( and more practice problems ( What's the step-by-step process for checking if a piecewise function is continuous everywhere? Step-by-step check for continuity of a piecewise function: 1. Identify pieces and their domains. List the boundary points where the formula changes (and any endpoints or domain holes). 2. For every x inside a piece (not on a boundary), check the defining expression is continuous there (polynomials, exponentials, trig, rational without zero denom.—use algebra). If each piece’s rule is continuous on its open interval, those interior points are fine. 3. At each boundary point a between pieces: - Compute the left-hand limit L = lim_{x→a^-} f(x). - Compute the right-hand limit R = lim_{x→a^+} f(x). - Evaluate f(a) if it's defined. - For continuity at a you need L = R = f(a). If L = R but f(a) ≠ that common value (or f(a) is undefined), it’s a removable discontinuity—redefine f(a) = L to make it continuous (EK LIM-2.C.1/2). 4. Check domain endpoints similarly using one-sided limits. 5. If any one-sided limits differ, the point is a jump/nonremovable discontinuity. On the AP exam, be explicit: show the one-sided limits and compare with f(a) (CED LIM-2.C language). For practice and worked examples, see the Topic 1.13 study guide ( and more problems ( Why can't I remove jump discontinuities but I can remove holes? A removable discontinuity (a hole) happens when the two one-sided limits are equal—the limit exists—but the function’s value at that x is missing or different. Because EK LIM-2.C.1 says you can redefine f(a) to equal the limit, you “remove” the hole and make the function continuous there. A jump discontinuity happens when the left-hand and right-hand limits are different (the one-sided limits don’t match). The two-sided limit does not exist, so there’s no single number you can assign to f(a) that matches both sides. That’s why a jump can’t be removed by redefining the function at a single point. For piecewise boundaries (EK LIM-2.C.2) you need the left expression, right expression, and the value at the point all equal to be continuous—if the left and right values disagree, continuity is impossible at that point. For more practice and examples (including graphs and algebraic cancellation for holes), check the Topic 1.13 study guide ( and try problems on Fiveable’s AP Calc practice page ( How do I use the three conditions for continuity to remove discontinuities? The three conditions for continuity at a point a are: (1) f(a) is defined, (2) the limit lim_{x→a} f(x) exists (i.e., left-hand and right-hand limits are equal), and (3) f(a) = lim_{x→a} f(x). To remove a removable discontinuity, use them in this order: 1. Check one-sided limits. Compute lim_{x→a^-} f(x) and lim_{x→a^+} f(x). If they aren’t equal the discontinuity isn’t removable. 2. If they are equal, that common value L = lim_{x→a} f(x) exists. For algebraic functions, simplify (factor/cancel) to find L. 3. Redefine or choose the parameter so f(a) = L. For piecewise functions, set the expression values from each side equal at the boundary and solve for the parameter; then set the function value at a to that common value. Example: f(x) = (x^2−1)/(x−1) for x≠1. Simplify to x+1, so limit as x→1 is 2. Define f(1)=2 to remove the hole. This is exactly what the CED expects for LIM-2.C tasks. 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2661	https://www.math.ucla.edu/~njhu/notes/na/interp/divdiff.pdf	Divided differences Nicholas Hu Let be a set of distinct abscissae (also called "nodes") and let denote the real vector space of polynomials of degree at most with real coefficients. If is a real-valued function defined on , then there exists a polynomial that interpolates at the abscissae in the sense that for each : namely, , where . Furthermore, it is the only such polynomial, for if also interpolates at the same abscissae, then has distinct roots and must therefore be the zero polynomial. The polynomials constitute a basis of known as the Lagrange basis. Other bases of include the monomial basis , given by , and the Newton basis , given by . Divided differences are quantities that can be used (among other things) to compute the coefficients of in the Newton basis. We define the divided difference as the coefficient of in the monomial basis representation of (which for simplicity we will refer to as the "leading coefficient" of despite the fact that it may be zero). Below we present several properties of divided differences. Coefficients of polynomial interpolant in Newton basis Proof. Write . For each , the polynomial interpolates at since for all . Clearly, its leading coefficient is , so by definition, . ∎ Notably, divided differences obey a recurrence relation that allows for their recursive computation. Recurrence relation Proof. Suppose that . If interpolates at and interpolates at , then the polynomial given by interpolates at . For the base case, observe that the constant polynomial interpolates at . ∎ Interestingly, we observe that is a convex combination of and for . The formula above also yields a recursive algorithm for evaluating known as Neville's algorithm. Namely, if interpolates at so that , then , where . An explicit formula for divided differences can also be deduced by inspecting the Lagrange basis representation of . Linearity If , then Proof. If and interpolate and , respectively, at , then interpolates at . ∎ Symmetry If is a permutation of , then Proof. In view of the definition above, this is immediate since . ∎ Factor property If and , then Proof. If interpolates at , then the polynomial given by interpolates at . ∎ In fact, the recurrence relation (excluding the base case) can be derived solely from linearity, symmetry, and the factor property: Thus, these three properties along with the property determine the values of all divided differences. The identity suggests a relationship between divided differences and derivatives: if, say, , , and exists on , then the mean value theorem amounts to the assertion that for some . This generalizes readily to divided differences and derivatives of higher order. Mean value theorem Let and . If and exists on , then for some . Proof. By symmetry, we may assume that . If interpolates at , then has distinct zeroes in . By (repeated applications of) Rolle's theorem, has a zero . ∎ Polynomial interpolation error Let and . If and exists on , then for all there exists a for which Proof. If , the conclusion is trivial; otherwise, the polynomial interpolates at and the conclusion follows from the mean value theorem. ∎ In the same vein, as if is differentiable at , the geometric interpretation being that the slope of the secant line through and tends to that of the tangent line through . This observation along with the identity allows us to recover the product rule for derivatives. More generally, we have the following identity for divided differences. Product rule Proof. If interpolates at , then since agrees with on . By linearity and the factor property, we have By Taylor's theorem, if is times differentiable at , where denotes the forward difference operator with step (that is, ). As a result, and from the above we can derive the generalized product rule for derivatives, .
2662	https://www.heldermann-verlag.de/jca/jca10/jca0334.pdf	Journal of Convex Analysis Volume 10 (2003), No. 1, 265–273 Convex Bodies of Optimal Shape G. Carlier Universit´ e Paris IX Dauphine, Ceremade, Place de Lattre de Tassigny, 75775 Paris Cedex 16 carlier@ceremade.dauphine.fr T. Lachand-Robert Universit´ e Pierre et Marie Curie, Laboratoire d’Analyse Num´ erique, 75252 Paris Cedex 05 www.ann.jussieu.fr\~lachand lachand@ann.jussieu.fr Received May 4, 2001 Given a continuous function f : Sn−1 →R, we consider the minimization of the functional R ∂A f(νA)dHn−1 with respect to A ⊂Rn, included in a class of convex bodies deﬁned by surface or shape conditions. This corresponds to non-parametric formulations of older problems, including Newton’s problem of the body of minimal resistance, following an approach due to G. Buttazzo and P. Guasoni . We establish existence and uniqueness results and some characterizations of the minimizers. 1. Introduction One of the oldest problem in the Calculus of Variations is to ﬁnd a body of minimal resistance, under various assumptions. It was ﬁrst stated by I. Newton in his Principia, and has been widely studied since. In this sort of problem, it is necessary to impose global constraints on the admissible set of bodies, such as convexity . Usually the considered bodies are taken as epigraphs of convex functions on a given domain Ω⊂R2. Then the problem is: inf u∈C∩K Z Ω g(∇u(x)) dx (1) where g(p) = gN(p) = 1/(1 + \|p\|2) in the original formulation, and C = {u : Ω→ R, u convex } and K expresses an additional condition, for instance 0 ≤u ≤M, a height constraint, or R q 1 + \|∇u(x)\|2 = k, a surface area constraint. The latter case is considered in , with g(p) = j( q 1 + \|p\|2), j convex, and u ∈BV (Ω). The authors prove there that minimizers exist, that functions of the forms c dist(x, ∂Ω) are minimizers, and they are the only convex ones vanishing on the boundary. They also consider a more elaborate model with frictional eﬀects caused by the front and lateral surfaces, that is, minimize Z Ω dx 1 + \|∇u(x)\|2 + c1 Z Ω q 1 + \|∇u(x)\|2 dx + c2 Z ∂Ω u(x) dHn−1(x). (2) Those parametric formulations actually consider bodies with a ﬁxed face Ω× {0} even though there is no requirement of this sort in the original problem. In the authors ISSN 0944-6532 / $ 2.50 c ⃝Heldermann Verlag 266 G. Carlier, T. Lachand-Robert / Convex Bodies of Optimal Shape consider a generalization of this problem in a non-parametric form: inf A∈A Z ∂A f(x, νA(x)) dHn−1(x) where A is a class of n-convex bodies of Rn with volume or obstacle constraints. Here νA(x) is the unit outward normal vector at x ∈∂A. They prove various existence results and suggest to take Ωas an unknown in the problem. Their problem corresponds (in a way explained hereafter) to (1) with f(x, ν) = f(ν) = −νng(−ν−1 n ν′) (where ν = (ν′, νn) ∈Sn−1, ν′ = (ν1, . . . , νn−1)) and A = n (x′, xn) ∈Rn, 0 ≥xn ≥u(x′) o , νA(x) =      (∇u(x′), −1) q 1 + \|∇u(x′)\|2 for x ∈graph u = {(x′, u(x′))}, en = (0, . . . , 0, 1) for x ∈Ω× {0}. Since in these cases f does not depend on x, we study in this paper the autonomous case with prescribed surface area: inf A∈Ai Z ∂A f(νA(x)) dHnA−1(x) (3) where i = 1 or i = 2 and A1 := {A convex ⊂Rn, HnA−1(∂A) = 1} A2 := n A ∈A1, A ⊂{xn ≤0}, HnA−1(Π[xn=0]A) = α = 1 −HnA−1(∂A ∩{xn < 0}) o Here Π[xn=0] designates the projection on the hyperplane {xn = 0}. Also for any convex set A ⊂Rn, nA is the dimension of the aﬃne space spanned by A. Note that A ∈A2 implies that there exists a convex set ΩA ⊂Rn−1 ≡{xn = 0}, with LnA−1(ΩA) = α (the (nA −1)-dimensional Lebesgue measure), and a convex function uA : ΩA →R−such that A = {x = (x′, xn) ∈ΩA × R ; 0 ≥xn ≥uA(x′)}; moreover the condition α = 1 −HnA−1(∂A ∩{xn < 0}) implies: Z ΩA q 1 + \|∇uA\|2 dLnA−1(x′) ≤1 −α. Note also that we must have α ∈(0, 1/2), otherwise A2 is empty. In this formulation, only the area of ΩA is given, not its shape. Note that in the deﬁnitions of A1, the convexity constraint is a normalization of the considered sets: see Remark 2.1 below. G. Carlier, T. Lachand-Robert / Convex Bodies of Optimal Shape 267 Concerning problem (3), our main concerns are existence and uniqueness of a solution; existence of solution of full dimension (nA = n); existence of symetrical solutions for special values of f. Our main result reads as follows: Theorem 1.1. For any f ∈C0(Sn−1), both problems (3) (i = 1, 2) have a solution. Moreover, there exists a Gδ dense subset X ⊂C0(Sn−1) such that this solution is unique (up to translations) and is a n-simplex. We will actually prove a more detailed result: see Theorems 2.2 and 2.3 for detailed statements. 2. Equivalent formulation We recall that any convex compact set A can be associated with a measure µA ∈ M+(Sn−1), such that ∀ϕ ∈C0(Sn−1), Z Sn−1 ϕ(y) dµA(y) = Z ∂A ϕ(νA(x)) dHnA−1(x). (4) In particular we have R Sn−1 dµA(y) = HnA−1(∂A) and Z Sn−1 y dµA(y) = 0. (5) For example, if A is a polyhedron, then µA = P αiδνi, where νi is the unit normal vector to the face i, and αi its (relative) surface area. Conversely, given any measure µ ∈M+(Sn−1), µ ̸= 0, satisfying (5), there exists a unique convex compact set A (up to translations), with dimension nA = dim supp µ ∈{1, . . . , n}, such that µ = µA. This follows from Alexandrov’s Theorem [1, Theorem 19.2]. This allows us to express problem (3) as a linear programming problem on a subclass of M+(Sn−1): inf µ∈Mi Z Sn−1 f(y) dµ(y) (6) where i = 1 or i = 2 and M1 := º µ ∈M+(Sn−1); Z Sn−1 dµ(y) = 1, Z Sn−1 y dµ(y) = 0 » (7) M2 := n µ ∈M1; µ ¬ ¬ Sn−1 + = α δen o . (8) Here Sn−1 + = Sn−1 ∩{xn > 0}, δen is the Dirac mass at en := (0, . . . , 0, 1). Remark 2.1. From Alexandrov’s Theorem we could drop the convexity constraint in the deﬁnition of Ai. Indeed, to any suﬃciently regular compact set B corresponds a convex set A such that µA = µB; this implies that the inﬁmum of (6) is the same for the wider class with no convexity constraint. 268 G. Carlier, T. Lachand-Robert / Convex Bodies of Optimal Shape Theorem 2.2. A convex compact set A ∈Ai (i = 1 or i = 2) is a solution of (3) if and only if the corresponding µA is a solution of (6). For any f ∈C0(Sn−1), both problems have a solution. Theorem 2.3. There exists a Gδ dense subset X ⊂C0(Sn−1) such that f ∈X implies problem (3) has a unique solution, up to translations, and this solution is a n-simplex. Proof of Theorem 2.2. In the case i = 1, the Theorem follows directly from (4) and Alexandrov’s Theorem recalled above. If i = 2, we ﬁrst observe that A ∈A2 implies that the projection of A on Pn := {xn = 0} is equal to A ∩Pn. For otherwise, we would have HnA−1(A ∩Pn) < HnA−1(Π A) = α (where Π = Π[xn=0]); this contradicts the other condition on the surface area of A and A ∩{xn < 0}. Now note that if x = (x′, xn) ∈∂A∩{xn < 0} then since A is convex and Π(x) = (0, xn) ∈ A we have ⟨νA(x), Π(x) −x⟩≤0, so that ⟨νA(x), en⟩≤0. (9) Let ϕ ∈C0(Sn−1) such that ϕ has support in Sn−1 + ; using (9), we have: Z Sn−1 ϕ(y)dµA(y) = Z Π(A) ϕ(en)dHn−1(x) = ⟨αδen, ϕ⟩ so that µA ∈M2. Conversely assume that µ ∈M2 and let us prove that there exists A ∈A2 such that µ = µA. Let A ∈A1 be such that µA = µ. Up to a translation of direction en, we may assume 0 ∈∂A and A ⊂{xn ≤0}. Let us prove that (up to HnA−1 negligible set of ∂A) one has {νA = en} = ∂A ∩{xn = 0}. First, the inclusion {νA = en} ⊃∂A ∩{xn = 0} is straightforward. Assume now that x ∈∂A and νA(x) = en, by convexity we get A ⊂{y ∈Rn : yn ≤xn}. Since 0 ∈A we get xn = 0, hence the desired result. Now, using µA ∈M2 we get: HnA−1(∂A ∩{xn = 0}) = µA({en}) = α, HnA−1(∂A ∩{xn < 0}) = 1 −α. (10) To show that A ∈A2 it remains to prove that Π(A) = ∂A ∩{xn = 0} and obviously it is enough to prove that Π(A) ⊂A. With (10) we obtain that the set B := ¨ x ∈∂A ∩{xn ≤0} : νA(x) exists and ⟨νA(x), en⟩≤0 © is of full HnA−1 measure in ∂A ∩{xn ≤0}. Since A is a convex compact set, we have: A = {x ∈Rn: xn ≤0, and ⟨νA(z), x −z⟩≤0, for all z ∈B} Now let x = (x′, xn) ∈A with xn < 0, y := Π x = (x′, 0) ∈Π(A). For any z ∈B, we have ⟨νA(z), y −z⟩= ⟨νA(z), x −z⟩−xn ⟨νA(z), en⟩≤0 G. Carlier, T. Lachand-Robert / Convex Bodies of Optimal Shape 269 so that y ∈A. This proves that A ∈A2. Existence of solutions for (3) immediately follows since (6) consists in minimizing a con-tinuous linear functional over a convex weakly ∗compact subset of M(Sn−1). 3. An example of uniqueness Proposition 3.1. Assume that f achieves a strict minimum at exactly n + 1 points x0, . . . , xn ∈Sn−1, such that ∀ξ ∈Sn−1, x0 · ξ, . . . , xn · ξ are not all nonnegative. Then (3) has a unique minimizer, which is a n-simplex. We will prove in the following that this simple case is actually the generic one for unique-ness, up to the addition of an aﬃne function to f. The assumption on x0, . . . , xn means that they do not belong to the same closed half-hypersphere. This condition is equivalent to the fact that 0 is an interior point of Conv(x0, . . . , xn), from Hahn-Banach Theorem. Proof. Since 0 is an interior point of Conv(x0, . . . , xn), the barycentric coordinates α0, . . . , αn of 0 in (x0, . . . , xn) are positive, and are the only solutions of                  n X k=0 αkxk i = 0, i = 1, . . . , n n X k=0 αk = 1, αk ≥0, k = 0, . . . , n. Hence µ := Pn k=0 αkδxk belongs to M1, and is clearly the unique minimizer of the prob-lem (6). The corresponding convex set is a n-simplex whose unit normal vectors are the xk, with faces surface areas equal to αk. 4. Optimality conditions and consequences Lemma 4.1. Let µ ∈M1. Then µ is a minimizer of (6) with i = 1 if and only if there exists (λ0, . . . , λn) ∈Rn+1 such that f(y) = λ0 + n X i=1 λiyi µ a.e. in Sn−1, (11) f(y) ≥λ0 + n X i=1 λiyi ∀y ∈Sn−1. (12) Lemma 4.2. Let µ ∈M2. Then µ is a minimizer of (6) with i = 2 if and only if there exists (λ0, . . . , λn−1) ∈Rn such that f(y) = λ0 + n−1 X i=1 λiyi + λn(α + (1 −α)yn) µ a.e. in Sn−1 − . (13) f(y) ≥λ0 + n−1 X i=1 λiyi + λn(α + (1 −α)yn) ∀y ∈Sn−1 − . (14) 270 G. Carlier, T. Lachand-Robert / Convex Bodies of Optimal Shape Here Sn−1 − = Sn−1 ∩{xn ≤0}. In both lemmas, note that λ0 is the optimal value of the functional. Proof. We prove the last lemma ﬁrst. Let µ be a solution, and λ0 := R f dµ, the optimal value of the functional. Deﬁne g := f −λ0; then µ is a minimizer of µ 7→ R g dµ in M2, with a minimal value of zero. In other terms, g is non-negative on M2, that is g ∈(R+M2)+, the positive polar set of the cone R+M2. Since R+M2 is the intersection of M+(Sn−1) and a ﬁnite number of hyperplanes of M(Sn−1): R+M2 = n µ ∈M+(Sn−1); Z Sn−1 − (α + (1 −α)yn) dµ = 0, Z Sn−1 yi dµ = 0, i = 1, . . . , n −1 o we get, in M(Sn−1)′: (R+M2)+ = clos ¢ M+(Sn−1)+ + Span{y1, . . . , yn−1, w} £ with w(y) := (α + (1 −α)yn) 1Sn−1 − (y). Since C0 + is dense in M+(Sn−1)+, there exists sequences (Λk) ⊂C0 +, (ak) ∈R, (λk i ) ⊂R (i = 1, . . . , n −1) such that g = lim k gk with gk := Λk + akw + n−1 X i=1 λk i yi. Since α ∈(0, 1/2), w changes sign in Sn−1 − ; more precisely, w > 0 for yn ∈(β, 0], where β := α/(α −1), and w < 0 for yn ∈[−1, β). Then since Z yn∈(β,0) Λk + ak Z yn∈(β,0) w = Z yn∈(β,0) gk converges to R yn∈(β,0) g, it is bounded. Hence ak is bounded from above (recall that Λk ≥ 0). Similarly we obtain a lower bound for ak by integrating with respect to yn ∈[−1, β). Up to subsequences, we can assume that ak →a as k →∞. Then, if µ ∈N := {µ ∈M(Sn−1); R w dµ = 0, R yi dµ = 0 ∀i < n}, we have Z Sn−1 − g dµ = lim Z Sn−1 − gk dµ = Z Sn−1 − Λk dµ. Hence (Λk) is bounded in N ′ and therefore admits a convergent subsequence in the weak- topology, with limit Λ. Since µ ∈N and R g dµ = 0, we get R Λ dµ = 0. Since Λ dµ ≥0, that yields Λ = 0, µ a.e. Now gk converges to g, ak is bounded and R Λk is bounded; so (λk i ) is bounded, and we can assume, extracting subsequences, that λk i →λi, i = 1, . . . , n−1. Therefore g−P λiyi is equal to Λ + aw in M′. This implies that Λ is a continuous function, and g − X λiyi = aw = aα + a(1 −α)yn µ a.e. in Sn−1 − . Then (14) follows from the nonnegativity of Λ. This ends the proof of the second lemma. G. Carlier, T. Lachand-Robert / Convex Bodies of Optimal Shape 271 The proof of the ﬁrst lemma is similar, except that i = 1, . . . , n, and the term λn plays a role similar to the other λi. Let us give a few simple consequences of these lemmas. Proposition 4.3. Assume that f is continuous, and not aﬃne. Then any optimal set in problem (3) is singular, in the sense that νA(∂A) has a complement with nonempty interior in Sn−1. Note that this indicates that the set of minimizers is not generic: for a smooth (C1) n-dimensional convex set A, we have νA(∂A) = Sn−1. From Lemma 4.1 there exists an aﬃne function θ such that f = θ in the support of µA. By assumption {f ̸= θ} has nonempty interior. Proposition 4.4. Let f be continuous, i = 1 or 2. There exists µ ∈Mi, solution of (6), such that the support of µ contains the support of any other solution of (6). In particular, in Lemmas 4.1, 4.2, the coeﬃcients (λi)i can be chosen independently from the minimizer. Proof. Let M be the minimal set of the functional, which is a convex closed subset of M(Sn−1). Let (µk) ⊂M be a dense sequence in M, and µ := P k αkµk where αk > 0, P αk = 1. Then µ ∈M, the support of µ is the closure of the union of the supports of µk, and it contains the support of any element of M; for if not, that would contradict the density of (µk) in M. Corollary 4.5. Let Σ be the support of the solution µ deﬁned in the last proposition. Then the set of solutions of (6) is exactly M(Σ) ∩Mi. In particular, the solution of (6) is unique if and only if Card Σ = 1 + dim Span Σ ≤n + 1. Note that Σ is not known explicitely (it comes from the set of solutions). Consequently, 0 lies in the relative interior of co Σ by construction. The additional condition in the corollary means that 0 has a unique barycentric decomposition 0 = P P∈Σ αPP, with αP > 0. On the other hand, Σ = (f −θ)−1(0), where θ(y) = λ0 + P i λiyi, since adding a measure with support in the latter set does not change the value of the functional. Note that, if all minimizers of (3) have empty interior in Rn, then Σ = (f −θ)−1(0) is included in an hyperplane of Rn. We are now in position to prove our second main theorem. Proof of Theorem 2.3. Generic uniqueness is a consequence of Mazur’s Theorem (see , see also [5, Theorem 1.20]) which states that both (concave Lipschitz) functionals f 7→inf µ∈Mi ⟨f, µ⟩, i = 1, 2 are GÝ ateaux-diﬀerentiable on a Gδ dense subset of C0(Sn−1) so that the corresponding minimizing measure is unique. Corollary 4.5 shows that, in case of uniqueness, the corresponding body is a k-simplex, with k ≤n. On the other hand, the case k < n occurs only if (f −θ)−1(0), the minimal set 272 G. Carlier, T. Lachand-Robert / Convex Bodies of Optimal Shape of f−θ, lies in a hyperplane. The opposite of this property is generic. Indeed if f ∈C0(S1) has only one minimum, then there exists r1, r2 ∈Q such that max[r1,r2] f ≤minS1[r1,r2] f; for each r1, r2 the set of such f is closed and has empty interior. 5. The cylindrical case We now specialize to the case where i = 2 and f, in problem (3), depends only on yn. This is particularly interesting in view of Newton’s problem of the body of minimal resistance, where f(y) = −y3 n. In the following we write f(y) = g(yn). Theorem 5.1. Assume f(y) = g(yn) for y ∈Sn−1 − , with g strictly convex on [−1, 0]. The problem (3) with i = 2 and f(y) = g(yn) admits a unique solution A among convex sets invariant by SOn−2, the group of rotations of Sn−2 = Sn−1 ∩{xn = 0}. Moreover, this solution is a cone, that is, up to translations, ΩA is a disc with center 0, and uA(x) = k1 −k2 dist(x, 0). As explained in the introduction, a similar result was already established in the parametric case in . Note that the proof here is completely diﬀerent. Proof. Let µ be a solution. Let us deﬁne µ as follows: for all ϕ ∈C0(Sn−1), ⟨µ, ϕ⟩= Z SOn−2 ⟨µ, ϕ ◦R⟩dR where dR is the Haar probability measure on SOn−2. Then µ is rotationnaly invariant, and R g(yn) dµ = R g(yn) dµ: therefore µ is is a mini-mizer. From(13), there exists a constant c such that g(yn)−cyn is an aﬃne function of y1, . . . , yn−1, µ a.e. This implies λi = 0, i = 1, . . . , n −1 since µ is rotationnaly invariant. Now g(yn) −cyn = λ0, µ a.e and from (14) g(yn) −cyn ≥λ0 in [−1, 0]. Therefore the support of µ in Sn−1 − reduces to the minimal set of g(yn) −cyn, which is a sphere with dimension n −2 since g is strictly convex. The corresponding convex body is a cone as indicated in the statement of the Theorem. Remark 5.2. The proof of the theorem can be easily extended to the case where f is invariant under the action of an arbitrary subgroup G ⊂On. Then µ is also invariant under the action of the group. For instance, if f ◦σ = f for some symmetry σ, then µ ◦σ = µ and σ(A) = A for the corresponding body. 6. Frictional eﬀects In , the authors suggest to take into account frictional eﬀects on the front and lateral sides of the body, minimizing the functional in (2) among all function u ≤0, u ∈W 1,∞ loc (Ω). In a non-parametric formulation, this can be expressed as the minimizing problem: inf A∈A3 Z ∂A f(νA(x)) dHnA−1(x) + + c1HnA−1({x; νA(x) · en > 0}) + c2HnA−1({x; νA(x) · en = 0}). G. Carlier, T. Lachand-Robert / Convex Bodies of Optimal Shape 273 where A3 is an appropriate set of bodies. Here again we can normalize the minimizers by considering convex sets. Assuming that A is the intersection of the graph of a convex function u with {xn ≤0} is equivalent to requiring that µA ¬ ¬ Sn−1 + = αδen for some α > 0 as before. Then the compact set A is a minimizer if and only if µA minimizes F3(µ) := Z Sn−1 f(y) dy + c1µ({yn < 0}) + c2µ({yn = 0}). among all µ in M3 := ¨ µ ∈M+(Sn−1); R Sn−1 y dµ(y) = 0; µA ¬ ¬ Sn−1 + = αδen © . Clearly, all the results in previous sections remain true for this functional. In particular, for f(y) = −y3 n, c1 > 0, c2 > 0, the case considered in , there exists a unique cylindrical minimizer which can be found as follows: write µ(y) = αδen +η(−yn), with η ∈M+([0, 1]) and minimizes H(η) := Z 1 0 t3 dη(t) + c1η([0, 1]) + c2η({0}) under the constraint R 1 0 t dη(t) = α. This obviously implies η(0) = 0 for the minimizer. Moreover there exists λ ∈R such that t3 + c1 = λt, η a.e. in [0, 1], and t 7→t3 −λt + c1 is minimal on the support of η. Then H(η) = (1−α)λ+c1η([0, 1]), which implies that λ must be the smallest number such that t3 −λt + c1 has a nonnegative root. The corresponding root a satisﬁes a3 −λa + c1 = 0 and 3a2 −λ = 0, that is λ = 3a2 and a = (c1/2)1/3, if c1 ≤2. If c1 ≥2, then a = 1. Hence the optimal body is ﬂat if c1 ≥2. If c1 < 2, it is a cone with slope p given by 1/ p 1 + p2 = νn = a = (c1/2)1/3 i.e. p = ((2/c1)2/3 −1)1/2. This result was obtained in using diﬀerent arguments. References I. J. Bakelman: Convex Analysis and Nonlinear Geometric Elliptic Equations, Springer-Verlag (1994). G. Buttazzo, V. Ferone, B. Kawohl: Minimum problems over sets of concave functions and related questions, Math. Nachrichten 173 (1993) 71–89. G. Buttazzo, P. Guasoni: Shape optimization problems over classes of convex domains, J. Convex Anal. 4(2) (1997) 343–351. V. Ferone, B. Kawohl: Bodies of minimal resistance under prescribed surface area, Z. Angew. Math. Mech. 79(4) (1999) 277–280. R. R. Phelps: Convex Functions, Monotone Operators and Diﬀerentiability, Lect. Notes Math. 1364, 2nd ed., Springer-Verlag (1993). S. Mazur: ¨ Uber konvexe Mengen in linearen normierten R¨ aumen, Studia Math. 4 (1933) 70–84.
2663	https://artofproblemsolving.com/wiki/index.php/Jensen%27s_Inequality?srsltid=AfmBOopMu-rGvGc9IqcTXiexml9VIcq-bI-tJmnKqg8eVYSn5PjYPtBw	Art of Problem Solving Jensen's Inequality - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Jensen's Inequality Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Jensen's Inequality Jensen's Inequality is an inequality discovered by Danish mathematician Johan Jensen in 1906. Contents [hide] 1 Inequality 2 Proof 3 Example 4 Problems 4.1 Introductory 4.1.1 Problem 1 4.1.2 Problem 2 4.2 Intermediate 4.3 Olympiad Inequality Let be a convex function of one real variable. Let and let satisfy . Then If is a concave function, we have: Proof We only prove the case where is concave. The proof for the other case is similar. Let . As is concave, its derivative is monotonically decreasing. We consider two cases. If , then If , then By the fundamental theorem of calculus, we have Evaluating the integrals, each of the last two inequalities implies the same result: so this is true for all . Then we have which is exactly what we want! Hooray!😀😀😀 Example One of the simplest examples of Jensen's inequality is the quadratic mean - arithmetic mean inequality. Taking , which is convex (because and ), and , we obtain Similarly, arithmetic mean-geometric mean inequality (AM-GM) can be obtained from Jensen's inequality by considering . In fact, the power mean inequality, a generalization of AM-GM, follows from Jensen's inequality. Problems Introductory Problem 1 Prove AM-GM using Jensen's Inequality Problem 2 Prove the weighted AM-GM inequality. (It states that when ) Intermediate Prove that for any , we have . Show that in any triangle we have Olympiad Let be positive real numbers. Prove that (Source) Retrieved from " Categories: Algebra Inequalities Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
2664	https://miamioh.ecampus.com/harrisons-principles-internal-medicine/bk/9780071356589?srsltid=AfmBOoqdKMX05RvqNLOkeYNa-sfMi8tHmGEcGdp4xWQZ6j-RzDjN9AwF	Harrison's Principles of Internal Medicine Companion Handbook Skip Navigation MIAMI UNIVERSITY OFFICIAL BOOKSTORE Login/Sign Up Home Miami Alumni Miami Athletics Shopping Cart (0) Shop MENU Shop Textbooks Campus Locations Login/Sign Up Back\| Campus Locations Home Home Miami Alumni Miami Alumni Miami Athletics Miami Athletics Back\| Shop Clothing Clothing Accessories Accessories Gifts Gifts Graduation Graduation Supplies Supplies Back\| Clothing Kids Men Sweatshirts Women View All Back\| Accessories For You For Your Car For Your Home For Your Pet For Your Tech View All Back\| Gifts Artwork Cooking Essentials Games Gift Wraps Holiday Home Decor Mascot Office Decor Outdoor/Recreation View All Back\| Graduation Graduation Gear Graduation Gifts View All Back\| Supplies Art Supplies For Your Office Medical Supplies Office Supplies School Supplies View All Kids Back\| Kids Bibs Bottoms Dresses Headwear Hoodies Matching Sets Onesies Shirts Sweatshirts Men Back\| Men Bottoms Footwear Hoodies Jerseys Mens Apparel Outerwear Polos Shirts Sweatshirts T-Shirts Sweatshirts Back\| Sweatshirts Women Back\| Women Bottoms Dresses Headwear Hoodies Outerwear Pants Shirts Sweatshirts Undergarments View All Clothing > Miami Merger Myaamia Heritage Collection Alumni Collection Miami Regionals Custom Items Miami Cradle of Coaches collection For You Back\| For You Backpacks Bags Buttons Drinkware Fan Gear Flags Hair Accessories Headwear ID Holders Jewelry Keychains Knitwear Lanyards Lapel Pins Pennants Socks Ties Umbrellas Wallets For Your Car Back\| For Your Car Decals License Plates For Your Home Back\| For Your Home Banners Blankets Cooking Essentials Decals Drinkware Flags Frames Home Decor Magnets Pillows Signs Stickers For Your Pet Back\| For Your Pet Beds Bowls Charms Clothes Toys For Your Tech Back\| For Your Tech Computer Accessories Phone Accessories View All Accessories > Artwork Back\| Artwork Ornaments Wall Art Cooking Essentials Back\| Cooking Essentials Food Games Back\| Games Balls Puzzles Gift Wraps Back\| Gift Wraps Gift Bags Holiday Back\| Holiday Ornaments Stocking Home Decor Back\| Home Decor Candles Mascot Back\| Mascot Plushies Office Decor Back\| Office Decor Desk Accessories Outdoor/Recreation Back\| Outdoor/Recreation Tailgate View All Gifts > Graduation Gear Back\| Graduation Gear Caps and Gowns Hoods Stoles Study Abroad Sashes Tassel Tassels Graduation Gifts Back\| Graduation Gifts Diploma Frames Yard Signs View All Graduation > Art Supplies Back\| Art Supplies Art Supply Products For Your Office Back\| For Your Office Desk Accessories Pens Medical Supplies Back\| Medical Supplies Kits Office Supplies Back\| Office Supplies Padfolios School Supplies Back\| School Supplies Folders Notebooks Planners View All Supplies > Erin Condren Textbooks Search Shopping Cart (0) Write a Review Harrison's Principles of Internal Medicine Companion Handbook byFauci, Anthony S., Md.; Braunwald, Eugene, M.D.; Isselbacher, Kurt J., M.D.; Wilson, Jean D., M.D.; Martin, Joseph B., M.D.; Kasper, Dennis L., M.D.; Hauser, Stephen L., M.D.; Longo, Dan L., M.D. Edition: 14th ISBN13: 9780071356589 ISBN10: 0071356584 Format: Paperback Pub. 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Summary -- The ultimate on-the-spot reference now lies flat for ease-of-use; it puts Harrison's authoritative, concise guidelines on pathophysiology, clinical signs, diagnosis, and disorder management at users' fingertips -- Includes many of the new features found in the latest edition of the Harrison's textbook -- 60 new chapters, a state-of-the-art look at AIDS/HIV, expanded therapy and oncology coverage, new treatment algorithms, and more -- Each chapter is linked to the corresponding chapter in the Harrison's text to make additional reading a snap Table of Contents Abbreviated Contents v Contributors xv Preface xix SECTION 1 IMPORTANT SIGNS AND SYMPTOMS Pain and Its Management 1(5) Chest Pain 6(5) Abdominal Pain 11(4) Headache and Facial Pain 15(7) Low Back Pain 22(10) Fever and Hyperthemia 32(4) Skin Rash 36(1) Pain or Swelling of Joints 37(5) Syncope and Seizures 42(3) Dizziness and Vertigo 45(2) Visual Loss and Disorders of Ocular Motility 47(5) Paralysis and Movement Disorders 52(5) Alterations in Consciousness (Confusion, Stupor, and Coma)57(4) Dyspnea 61(2) Cough and Hemoptysis 63(6) Cyanosis 69(2) Edema 71(5) Nausea and Vomiting 76(2) Weight Loss and Weight Gain 78(3) Diarrhea, Constipation, and Malabsorption 81(8) Gastrointestinal Bleeding 89(5) Jaundice and Evaluation of Liver Function 94(6) Ascites 100(5) Overt Manifestations of Renal Disease 105(6) Electrolytes/Acid-Base Balance 111(16) Lymphadenopathy and Splenomegaly 127(4) SECTION 2 MEDICAL EMERGENCIES Cardiovascular Collapse and Sudden Death 131(5) Shock 136(6) Acute Pulmonary Edema 142(3) Emergency Room Evaluation of Stroke 145(4) Increased Intracranial Pressure and Head Trauma 149(2) Status Epilepticus 151(3) Poisoning and Its Management 154(19) Diabetic Ketoacidosis and Hyperosmotic Coma 173(3) Hypoglycemia 176(4) Oncologic Emergencies 180(4) Drug Overdose 184(6) Drowning and Near-Drowning 190(3) Anaphylaxis and Transfusion Reaction 193(3) Bites, Envenomations, Stings, and Ectoparasite Infestations 196(14) Hypothermia 210(3) SECTION 3 DERMATOLOGY General Examination of the Skin 213(6) Common Skin Conditions 219(8) SECTION 4 NUTRITION Nutritional Requirements and Assessment 227(5) Anorexia Nervosa and Bulimia 232(4) Obesity 236(4) Diet Therapy 240(7) SECTION 5 HEMATOLOGY AND ONCOLOGY Examination of Blood Smears and Bone Marrow 247(3) Red Blood Cell Disorders 250(8) Leukocytosis and Leukopenia 258(4) Bleeding and Thrombotic Disorders 262(6) Transfusion and Pheresis Therapy 268(3) Cancer Chemotherapy 271(4) Myeloid Leukemias, Myelodysplasia, and Myeloproliferative Syndromes 275(9) Lymphoid Malignancies 284(14) Skin Cancer 298(3) Head and Neck Cancer 301(2) Lung Cancer 303(4) Breast Cancer 307(4) Tumors of the Gastrointestinal Tract 311(12) Genitourinary Tract Cancer 323(4) Gynecologic Cancer 327(4) Prostate Hyperplasia and Carcinoma 331(4) Cancer of Unknown Primary Site 335(4) Sarcomas of Bone and Soft Tissues 339(3) Paraneoplastic Syndromes 342(3) Neurologic Manifestations of Systemic Neoplasia 345(6) SECTION 6 INFECTIOUS DISEASES Diagnosis 351(11) Antimicrobial Therapy 362(10) Immunization and Advice to Travelers 372(9) Sepsis and Septic Shock 381(6) Infective Endocarditis 387(9) Intrabdominal Infections 396(3) Infectious Diarrheas 399(6) Sexually Transmitted Diseases 405(16) Infections of Skin, Soft Tissues, Joints, and Bones 421(8) Infections in the Immunocompromised Host 429(6) HIV Infection and AIDS 435(12) Hospital-Acquired Infections 447(3) Pneumococcal Infections 450(4) Staphylococcal Infections 454(10) Streptococcal and Enterococcal Infections and Diphtheria 464(9) Meningococcal and Listerial Infections 473(4) Infections Caused by Haemophilus, Bordetella, Moraxella, and HACEK Group Organisms 477(4) Diseases Caused by Gram-Negative Enteric Bacteria, Pseudomonas, and Legionella 481(9) Diseases Caused by Other Gram-Negative Bacteria 490(5) Anaerobic Infections 495(8) Nocardiosis and Actinomycosis 503(4) Tuberculosis and Other Mycobacterial Infections 507(13) Lyme Disease and Other Spirochetal Infections 520(7) Rickettsial Infections 527(9) Mycoplasma Infections 536(2) Chlamydial Infections 538(3) Herpesvirus Infections 541(8) Cytomegalovirus and Epstein-Barr Virus Infections 549(5) Influenza and Other Viral Respiratory Diseases 554(7) Rubeola, Rubella, Mumps, and Parvovirus Infections 561(7) Enteroviral Infections 568(3) Insect-and Animal-Borne Viral Infections 571(12) Fungal Infections 583(10) Pneumocystis carinii Infection 593(2) Protozoal Infections 595(17) Helminthic Infections 612(9) SECTION 7 CARDIOVASCULAR DISEASES Physical Examination of the Heart 621(6) Electrocardiography and Echocardiography 627(8) Preoperative Evaluation of Cardiovascular Disease 635(3) Arrhythmias 638(11) Congestive Heart Failure and Cor Pulmonale 649(6) Congenital Heart Disease in the Adult 655(3) Valvular Heart Disease 658(7) Cardiomyopathies and Myocarditis 665(5) Pericardial Disease 670(4) Acute Myocardial Infarction 674(14) Chronic Coronary Artery Disease 688(8) Hypertension 696(10) Diseases of the Aorta 706(4) Peripheral Vascular Disease 710(5) SECTION 8 RESPIRATORY DISEASES Respiratory Function and Diagnosis 715(7) Asthma and Hypersensitivity Pneumonitis 722(5) Environmental Lung Diseases 727(3) Chronic Bronchitis, Emphysema, and Acute Respiratory Failure 730(5) Upper and Lower Respiratory Tract Infections 735(10) Pulmonary Thromboembolism and Primary Pulmonary Hypertension 745(5) Interstitial Lung Disease (ILD)750(3) Diseases of the Pleura, Mediastinum, and Diaphragm 753(6) Disorders of Ventilation, including Sleep Apnea 759(5) Pulmonary Insufficiency and Acute Respiratory Distress Syndrome (ARDS)764(3) SECTION 9 RENAL DISEASES Approach to Patient with Renal Disease 767(5) Acute Renal Failure (ARF)772(5) Chronic Renal Failure (CRF) and Uremia 777(4) Dialysis and Transplantation 781(5) Glomerular Diseases 786(9) Renal Tubular Disease 795(5) Urinary Tract Infections 800(3) Renovascular Disease 803(4) Nephrolithiasis 807(3) Urinary Tract Obstruction 810(3) SECTION 10 GASTROINTESTINAL DISEASES Esophageal Diseases 813(5) Peptic Ulcer and Related Disorders 818(8) Inflammatory Bowel Diseases 826(5) Colonic and Anorectal Diseases 831(4) Cholelithiasis, Cholecystitis, and Cholangitis 835(5) Pancreatitis 840(5) Acute Hepatitis 845(7) Chronic Hepatitis 852(4) Cirrhosis and Alcoholic Liver Disease 856(5) Portal Hypertension 861(6) SECTION 11 ALLERGY, CLINICAL IMMUNOLOGY, AND RHEUMATOLOGY Diseases of Immediate-Type Hypersensitivity 867(4) Primary Immunodeficiency Diseases 871(5) SLE, RA, and Other Connective Tissue Diseases 876(10) Vasculitis 886(5) Sarcoidosis 891(2) Ankylosing Spondylitis 893(3) Degenerative Joint Disease 896(3) Gout, Pseudogout, and Related Diseases 899(4) Psoriatic Arthritis 903(2) Reactive Arthritis and Reiter's Syndrome 905(2) Other Arthritides 907(4) Amyloidosis 911(2) SECTION 12 ENDOCRINOLOGY AND METABOLISM Disorders of the Anterior Pituitary and Hypothalamus 913(10) Disorders of the Posterior Pituitary 923(5) Diseases of the Thyroid 928(8) Diseases of the Adrenal Cortex 936(7) Diabetes Mellitus 943(4) Disorders of the Testes and Prostate 947(7) Disorders of the Ovary and Female Genital Tract 954(5) Hyper-and Hypocalcemic Disorders 959(10) Metabolic Bone Disease 969(3) Disorders of Lipid Metabolism 972(8) Inherited Metabolic Diseases 980(5) SECTION 13 NEUROLOGY The Neurologic Examination 985(7) Diagnostic Methods in Neurology 992(6) Seizures and Epilepsy 998(8) Cerebrovascular Diseases 1006(8) Neoplastic Diseases of the Central Nervous System 1014(3) Bacterial Infections of the Central Nervous System 1017(6) Aseptic Meningitis, Viral Encephalitis, and Prion Diseases 1023(9) Multiple Sclerosis (MS)1032(6) Alzhemier's Disease and Other Dementias 1038(6) Parkinson's Disease 1044(3) Ataxix Disorders 1047(3) Motor Neuron Disease, including ALS 1050(3) Aphasia 1053(4) Cranial Nerve Disorders 1057(7) Metabolic Encephalopathy 1064(3) Disorders of the Autonomic Nervous System 1067(6) Disorders of Sleep and Circadian Rhythms 1073(5) Spinal Cord Diseases 1078(6) Peripheral Neuropathies including Guillain-Barre Syndrome 1084(5) Myopathies and Muscular Dystrophies 1089(7) Myasthenia Gravis (MG)1096(3) SECTION 14 PSYCHIATRIC DISORDERS AND PSYCHOACTIVE SUBSTANCE USE Approach to the Patient with Psychiatric Symptoms 1099(8) Psychiatric Medications 1107(12) Alcoholism 1119(3) Narcotic Abuse 1122(3) SECTION 15 ADVERSE DRUG REACTIONS Adverse Drug Reactions 1125(14) SECTION 16 LABORATORY VALUES Laboratory Values 1139(8) Index 1147 We are currently experiencing difficulties. 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2665	https://kconrad.math.uconn.edu/blurbs/ugradnumthy/pythagtriple.pdf	PYTHAGOREAN TRIPLES KEITH CONRAD 1. Introduction A Pythagorean triple is a triple of positive integers (a, b, c) where a2 + b2 = c2. Examples include (3, 4, 5), (5, 12, 13), and (8, 15, 17). Below is an ancient Babylonian tablet listing 15 Pythagorean triples. It is called Plimpton 322 (George Arthur Plimpton donated it to Columbia University). For more information about it, see . Plimpton 322 Some Pythagorean triples are scalar multiples of other triples: (6, 8, 10) is twice (3, 4, 5). We call a triple (a, b, c) primitive when the three integers have no common factor. For any triple (a, b, c), if d is the greatest common divisor of all three terms then (a/d, b/d, c/d) is a primitive triple and the original triple is a scalar multiple of this, so ﬁnding all Pythagorean triples is basically the same as ﬁnding all primitive Pythagorean triples. Our goal is to describe the primitive Pythagorean triples. We will be using diﬀerent characterizations of primitive triples, as described in the fol-lowing lemma. Lemma 1.1. For a Pythagorean triple (a, b, c), the following properties are equivalent: (1) a, b, and c have no common factor, i.e., the triple is primitive, (2) a, b, and c are pairwise relatively prime, (3) two of a, b, and c are relatively prime. Proof. To show (1) implies (2), we prove the contrapositive. If (2) fails then two of a, b, and c have a common prime factor. Suppose a prime p divides a and b. Then c2 = a2 + b2 is divisible by p, and since p \| c2 also p \| c, so p is a common factor of a, b, and c. A similar argument works using a common prime factor of a and c or of b and c. Thus (1) fails. It is easy to see that (2) implies (3) and (3) implies (1). □ 1 2 KEITH CONRAD Theorem 1.2. If (a, b, c) is a primitive Pythagorean triple then one of a or b is even and the other is odd. Taking b to be even, (1.1) a = k2 −ℓ2, b = 2kℓ, c = k2 + ℓ2 for integers k and ℓwith k > ℓ> 0, (k, ℓ) = 1, and k ̸≡ℓmod 2. Conversely, for such integers k and ℓthe above formulas yield a primitive Pythagorean triple. k ℓ a b c 2 1 3 4 5 3 2 5 12 13 4 1 15 8 17 4 3 7 24 25 Table 1. Examples of Primitive Triples Table 1 lists examples of k and ℓfor some primitive triples. Notice the triple (8, 15, 17) occurs in the table as (15, 8, 17) due to the convention in Theorem 1.2 that the middle term of the triple is even. Since k ̸≡ℓmod 2, one of k and ℓis even and the other is odd. The ﬁrst two rows in Table 1 shows either k or ℓcan be even; it depends on the triple (a, b, c). How it depends on the triple will be seen at the end of Section 3. Let’s ﬁrst check that the formula in Theorem 1.2 always yields primitive Pythagorean triples. For all k and ℓin Z, the formula (k2 −ℓ2)2 + (2kℓ)2 = (k2 + ℓ2)2 is true, so (k2 −ℓ2, 2kℓ, k2 + ℓ2) is a Pythagorean triple when k > ℓ> 0.1 When (k, ℓ) = 1 and k ̸≡ℓmod 2, let’s check (k2 −ℓ2, 2kℓ, k2 + ℓ2) is a primitive triple. This follows (by Lemma 1.1) from showing k2 −ℓ2 and k2 + ℓ2 are relatively prime. If d is a (positive) common divisor of k2 −ℓ2 and k2 + ℓ2 then d divides their sum and diﬀerence, which are 2k2 and 2ℓ2. Since k2 + ℓ2 and k2 −ℓ2 are odd, d is odd, so d \| k2 and d \| ℓ2. Since k and ℓ are relatively prime, so are k2 and ℓ2, so d = 1. If we relax any of the conditions on k and ℓin Theorem 1.2 we won’t get a primitive Pythagorean triple. For instance, taking k = 3 and ℓ= 1 produces the triple (8, 6, 10). The condition k ̸≡ℓmod 2 is violated here. In the next two sections we will show in two ways, one algebraic and the other geometric, that every primitive Pythagorean triple (a, b, c) arises in the form given by Theorem 1.2: using unique factorization in Z and using intersections of lines and circles. Then we will put the parametric formula in Theorem 1.2 to work and see some generalizations. 2. Proof of Theorem 1.2 by algebra To show that one of a and b is odd and the other is even, suppose a and b are both odd. Then a2 ≡b2 ≡1 mod 4, so c2 = a2 + b2 ≡2 mod 4. However, 2 mod 4 is not a 1Euclid, in Lemma 1 to Prop. 29 in Book X of his Elements, says a geometric formula equivalent to (a, b, c) = (mn, (m2 −n2)/2, (m2 + n2)/2), where m ≡n mod 2, produces Pythagorean triples. This is (1.1) with m = k + ℓand n = k −ℓ. See for a conversion of Euclid’s geometric Lemma 1 to an algebraic formula. Euclid did not claim his formula gives all primitive triples. Who showed it does? See PYTHAGOREAN TRIPLES 3 square. Thus a or b is even. If both a and b are even the triple (a, b, c) is not primitive, a contradiction. That shows one of a or b is odd and the other is even, so c2 = a2 + b2 is odd. Taking b to be the even choice among a and b (swap their order if necessary), rewrite a2 + b2 = c2 as b2 = c2 −a2 = (c + a)(c −a). Both a and c are odd, so c + a and c −a are even. Dividing by 4, (2.1) b 2 2 = c + a 2 c −a 2 . The two integers on the right, (c+a)/2 and (c−a)/2, are relatively prime: if d is a common divisor then it divides their sum and diﬀerence, which are c and a, and those are relatively prime, so d = 1. Since c > a > 0, both factors on the right side of (2.1) are positive. Now we’ll use the following lemma, which is a consequence of unique factorization in Z. Lemma 2.1. If xy = z2 where x, y, z ∈Z+ and (x, y) = 1, then x and y are both squares. Proof. The result is clear when x or y is 1, so let x, y > 1. Write them both in terms of their prime factorizations: x = pe1 1 · · · pek k , y = qf1 1 · · · qfℓ ℓ for distinct primes p1, . . . , pk and q1, . . . , qℓand positive exponents ei and fj. No pi and qj are equal since (x, y) = 1. We have xy = pe1 1 · · · pek k qf1 1 · · · qfℓ ℓ= z2. In z2, each prime factor of z has even multiplicity, so the ei’s and fj’s are even by unique factorization, say ei = 2e′ i and fj = 2f′ j. Then x = (pe′ 1 1 · · · p e′ k k )2, y = (qf′ 1 1 · · · q f′ ℓ ℓ)2, so x and y are both squares. □ Applying the lemma to x = (c + a)/2,y = (c −a)/2, and z = b/2 in (2.1), (2.2) c + a 2 = k2, c −a 2 = ℓ2 for some k and ℓin Z+. We noted already that (c + a)/2 and (c −a)/2 are relatively prime, so k and ℓare relatively prime. Adding and subtracting the equations in (2.2), c = k2 + ℓ2 , a = k2 −ℓ2 , so b 2 2 = k2ℓ2 = ⇒b2 = 4k2ℓ2 = ⇒b = 2kℓ, the last step holding because b, k, and ℓare positive. It remains to show k ̸≡ℓmod 2. If k ≡ℓmod 2, then k and ℓare both even or both odd. Since k and ℓare relatively prime, they are not both even. If they were both odd then k2 + ℓ2, 2kℓ, and k2 −ℓ2 would all be even, contradicting primitivity of (a, b, c). Thus k ̸≡ℓmod 2. 4 KEITH CONRAD 3. Proof of Theorem 1.2 by geometry Pythagorean triples are connected to points on the unit circle: if a2 + b2 = c2 then (a/c)2 + (b/c)2 = 1. So we get a rational point (a/c, b/c) on the unit circle x2 + y2 = 1. For a primitive Pythagorean triple (a, b, c), the ﬁrst paragraph of the previous proof shows we can take a odd and b even. Having done that, draw the line through (−1, 0) and (a/c, b/c) as in the ﬁgure below. Its slope is m = b/c 1 + a/c = b a + c and the line passes through (−1, 0), so the equation of the line is y = m(x + 1). (−1, 0) (a/c, b/c) x y Substituting the equation for the line into the equation for the unit circle gives us an equation whose roots are the x-coordinates of the two points on both the line and the circle: 1 = x2 + y2 = x2 + (m(x + 1))2 = (1 + m2)x2 + 2m2x + m2, so (3.1) (1 + m2)x2 + 2m2x + m2 −1 = 0. Since (−1, 0) and (a/c, b/c) lie on both the line and the circle, the two roots of (3.1) are −1 and a/c. The sum of the roots is −2m2/(1 + m2), so −1 + a c = −2m2 1 + m2 . Thus a c = 1 − 2m2 1 + m2 = 1 −m2 1 + m2 . Since (a/c, b/c) is on the line y = m(x + 1), b c = m a c + 1 = m 1 −m2 1 + m2 + 1 = 2m 1 + m2 . Since m is the slope of a line connecting (−1, 0) to a point on the unit circle in the ﬁrst quadrant, 0 < m < 1. Write m = b/(a + c) in reduced form as m = ℓ/k with relatively prime k and ℓin Z+. Since m < 1, k > ℓ. Formulas for the reduced fractions a/c and b/c in terms of m are (3.2) a c = 1 −(ℓ/k)2 1 + (ℓ/k)2 = k2 −ℓ2 k2 + ℓ2 PYTHAGOREAN TRIPLES 5 and (3.3) b c = 2(ℓ/k)2 1 + (ℓ/k)2 = 2kℓ k2 + ℓ2 . By deﬁnition, k and ℓare relatively prime. Let’s check that k ̸≡ℓmod 2. If k ≡ℓmod 2 then k and ℓare both even or both odd, and they are not both even since they are relatively prime. So k and ℓare odd. Then the fraction 2kℓ/(k2 + ℓ2) in (3.3) has an even numerator and denominator, so it simpliﬁes further: 2kℓ k2 + ℓ2 = kℓ (k2 + ℓ2)/2. The numerator on the right is odd, so the numerator of the reduced form of this fraction (divide numerator and denominator by their greatest common factor) is odd. The reduced form is b/c by (3.3), so b is odd. But b is even by hypothesis, so we have a contradiction. Thus k and ℓare not both odd, so k ̸≡ℓmod 2. From the end of Section 1, the three numbers k2−ℓ2, 2kℓ, and k2+ℓ2 form a Pythagorean triple and the ﬁrst and third numbers are relatively prime, so the triple is primitive by Lemma 1.1. Therefore the fractions on the right in (3.2) and (3.3) are in reduced form. The fractions a/c and b/c are also in reduced form since the triple (a, b, c) is primitive, so the (positive) numerators and denominators match: a = k2 −ℓ2, b = 2kℓ, c = k2 + ℓ2. This concludes the geometric proof of Theorem 1.2. Remark 3.1. While this derivation was motivated by geometry, the calculations were pure algebra and we could formulate them without any reference to a picture. That is, if we have numbers x and y where x2 + y2 = 1 and x ̸= −1, we can (without motivation) deﬁne the number m by declaring it to ﬁt the condition y = m(x + 1), namely deﬁne m = y/(x + 1). Substituting m(x + 1) for y in x2 + y2 = 1 tells us x2 + m2(x + 1)2 = 1, and expanding the square gives 0 = (1 + m2)x2 + 2m2x + (m2 −1) = (1 + m2)(x + 1) x + m2 −1 1 + m2 , so from x ̸= −1 we have x = (1 −m2)/(1 + m2) and y = m(x + 1) = 2m/(1 + m2). One nice consequence of our geometric derivation of the formula for primitive Pythagorean triples (a, b, c) is that it tells us which of k or ℓwill be even. The formula m = b a + c = ℓ k says that ℓ/k is the reduced form fraction for b/(a + c). We know either ℓor k is even. We will have k even when a+c is divisible by a higher power of 2 than b is, and ℓis even when b is divisible by a higher power of 2 than a+c is. For instance, in the triple (3, 4, 5), a+c = 8 and b = 4, so k is even. (Here k = 2 and ℓ= 1.) The triple (5, 12, 13) has a + c = 18 = 2 · 9 and b = 12 = 22 · 3, so ℓis even. (Here k = 3 and ℓ= 2.) 6 KEITH CONRAD 4. Applications Using the parametric formula for primitive Pythagorean triples, we can address questions concerning relations among the sides of a primitive right triangle. The famous triples (3, 4, 5), and (5, 12, 13), have consecutive terms. What are all the Pythagorean triples (a, b, c) with a pair of consecutive terms (either a and b, or b and c)? Such a triple is primitive since consecutive integers are relatively prime. First suppose a and b (the two legs) are consecutive. That a −b = ±1 means (k2 −ℓ2) −2kℓ= ±1. This can be rewritten as (k −ℓ)2 −2ℓ2 = ±1, where k −ℓis positive and odd and ℓis positive. Conversely, if we have a solution to x2 −2y2 = ±1 in Z+, then necessarily x is odd and (x, y) = 1. Let k = x + y and ℓ= y, so k > ℓ> 0, (k, ℓ) = 1, and k ̸≡ℓmod 2. Thus (k2 −ℓ2, 2kℓ, k2 + ℓ2) is a primitive triple, so ﬁnding Pythagorean triples whose legs diﬀer by 1 is the same as ﬁnding positive integer solutions to the Pell equation x2 −2y2 = ±1. Some examples are in Table 2. x y k ℓ a b c 1 1 2 1 3 4 5 3 2 5 2 21 20 29 7 5 12 5 119 120 169 17 12 29 12 697 696 985 41 29 70 29 4059 4060 5741 Table 2. Consecutive Legs Even if two legs in a primitive triple don’t diﬀer by 1, the formula for their diﬀerence is still (k −ℓ)2 −2ℓ2, so the possible diﬀerences between legs in a primitive triple are precisely the odd values of x2 −2y2 for positive integers x and y. Not every odd number can arise in this way, e.g., the equation x2 −2y2 = 5 has no integral solution (why?), so no primitive Pythagorean triple has its legs diﬀering by 5. When a leg and hypotenuse diﬀer by 1, the story is much simpler. The hypotenuse is odd, so it must diﬀer by 1 from the even leg. The diﬀerence is k2 + ℓ2 −2kℓ= (k −ℓ)2, which is an odd square. This is 1 only if k = ℓ+ 1 (recall k > ℓby convention), leading to the triple (2ℓ+ 1, 2ℓ2 + 2ℓ, 2ℓ2 + 2ℓ+ 1). The ﬁrst four examples are in Table 3. ℓ 2ℓ+ 1 2ℓ2 + 2ℓ 2ℓ2 + 2ℓ+ 1 1 3 4 5 2 5 12 13 3 7 24 25 4 9 40 41 Table 3. Consecutive Leg and Hypotenuse PYTHAGOREAN TRIPLES 7 Now we look at a connection between Pythagorean triples and pairs of reducible quadratic polynomials, taken from . Here “reducible” means integral coeﬃcients. While x2 + 4x + 3 = (x + 1)(x + 3) and x2 + 4x −3 is irreducible, we have x2 + 5x + 6 = (x + 2)(x + 3) and x2 + 5x −6 = (x −1)(x + 6). Are there more examples like the second one, where integral polynomials x2 + mx + n and x2 + mx −n both factor? Here m and n are nonzero. If we ask what happens when the sign of m changes, the answer is not interesting. Indeed, if x2 +mx+n = (x−r1)(x−r2) then x2 −mx+n = (x+r1)(x+r2), so either both of these factor (with integral coeﬃcients) or both don’t. Therefore in our question on x2 + mx ± n, we may assume m > 0. Since the issue is about sign changes on n, we may take n > 0 also. By the quadratic formula, the roots of x2 + mx ± n are −m ± √ m2 ± 4n 2 , which are integers exactly when m2 ± 4n = □. (Note m2 ± 4n ≡m mod 2, so the numerator is even when the discriminant is a perfect square.) So we can factor x2 + mx + n and x2 + mx −n if and only if m2 −4n = d2, m2 + 4n = e2, d and e ∈Z. Then d2 + e2 = 2m2, so d ≡e mod 2. Solving, m2 = d2 + e2 2 = e + d 2 2 + e −d 2 2 . Thus we have a Pythagorean triple (without a speciﬁed even term) e −d 2 , e + d 2 , m , e −d 2 < e + d 2 < m. As an exercise, show this Pythagorean triple is primitive if and only if (m, n) = 1. There is a one-to-one correspondence Pythagorean triples (a, b, c) with a < b < c ← →reducible x2 + mx ± n, given by (a, b, c) 7→x2 + cx ± ab 2 , x2 + mx ± n 7→ e −d 2 , e + d 2 , m , with m2 −4n = d2 and m2 + 4n = e2. The table below shows some corresponding Pythagorean triples and reducible x2 + mx ± n. In particular, the example x2 + 5x ± 6 corresponds to the (3, 4, 5) triple and thus is the simplest example. a b c m n x2 + mx + n x2 + mx −n 3 4 5 5 6 (x + 2)(x + 3) (x −1)(x + 6) 5 12 13 13 30 (x + 3)(x + 10) (x −2)(x + 15) 8 15 17 17 60 (x + 5)(x + 12) (x −3)(x + 20) 20 21 29 29 210 (x + 14)(x + 15) (x −6)(x + 35) Using the formula for primitive Pythagorean triples, we can now write a formula for all the reducible pairs of polynomials x2 + mx ± n where (m, n) = 1: x2 + (k2 + ℓ2)x ± kℓ(k2 −ℓ2), where k > ℓ> 0, (k, ℓ) = 1, and k ̸≡ℓmod 2. As an exercise, work out the factorization of this polynomial explicitly in terms of k and ℓ. 8 KEITH CONRAD 5. Generalizations Our two derivations of the parametric formula for primitive Pythagorean triples, one algebraic and the other geometric, are each worthwhile: they let us extend Theorem 1.2 in diﬀerent directions.2 The algebraic proof of Theorem 1.2 carries over to Pythagorean triples of polynomials in R[T]. These are polynomials f(T), g(T), and h(T) in R[T] that satisfy f(T)2 + g(T)2 = h(T)2. One example of such a triple is (T 2 −1, 2T, T 2 + 1). Can we describe all polynomial Pythagorean triples? There are two features of the Pythagorean triples of integers that play no role when we look at Pythagorean triples of polynomials: (1) The special attention to positive solutions of a2 + b2 = c2 is ignored. We allow any polynomial solutions to f2 + g2 = h2 with the only restriction being that f, g, and h are all nonzero. (2) The number 2 is an invertible constant in R[T], so the even/odd aspect that occurs for integral Pythagorean triples simply drops out of consideration. Being divisible by 2 is not important in R[T] since everything is divisible by 2: f(T) = 2(f(T)/2) in R[T]. We will call a Pythagorean triple of polynomials (f, g, h) primitive if the terms in it are pairwise relatively prime. This is the same as any two of the polynomials being relatively prime, and is also the same as (f, g, h) not being a non-constant multiple of another triple (polynomial analogue of Lemma 1.1). Theorem 5.1. The primitive Pythagorean triples (f(T), g(T), h(T)) in R[T] are given by the formulas f(T) = c(k(T)2 −ℓ(T)2), g(T) = ±2ck(T)ℓ(T), h = ±c(k(T)2 + ℓ(T)2) where c ∈R× and k(T) and ℓ(T) are relatively prime in R[T]. Proof. This is left as an exercise in adapting the techniques in the algebraic proof of Theorem 1.2. Note that if two polynomials are relatively prime and multiply to a squared polynomial, the two polynomials have to be squares up to constant multiple. This follows from unique factorization in R[T]. □ The geometric proof of Theorem 1.2 gives us a non-trigonometric parametrization of the points on the unit circle: for each point (x0, y0) on the circle x2 +y2 = 1 other than (−1, 0), has the form (5.1) x0 = 1 −m2 1 + m2 , y0 = 2m 1 + m2 where m = y0/(x0 + 1). The formulas in (5.1) for x0 and y0 arise by looking at the intersection points of the circle x2 + y2 = 1 and the line y = m(x + 1), where the linear 2Other ways of getting the parametric formula use more advanced ideas: a proof based on factoring in the Gaussian integers is in Theorem 8.3 in pdf and a proof based on Hilbert’s Theorem 90 from Galois theory is at the end of Section 4 of https:// kconrad.math.uconn.edu/blurbs/galoistheory/linearchar.pdf. PYTHAGOREAN TRIPLES 9 equation describes the lines through (−1, 0) other than a vertical line. The correspondences (5.2) m ⇝ 1 −m2 1 + m2 , 2m 1 + m2 , (x, y) ⇝m = y 1 + x, between real numbers m and points on the unit circle (x, y) other than (−1, 0) are inverses of each other, as the reader can check. Moreover, the formulas have arithmetic content: m is rational if and only if x0 and y0 are both rational since the formulas in (5.2) both have rational output if there is rational input. So (5.1) is well-suited to describe the rational points on x2 + y2 = 1. See Table 4. Compare this to the trigonometric parametrization (cos θ, sin θ), where nearly all rational points have messy corresponding angles θ. m x y 1/2 3/5 4/5 1/3 4/5 3/5 2/5 21/29 20/29 7/9 16/65 63/65 Table 4. Rational points from slopes on x2 + y2 = 1 Comparing the rational parametrization of the unit circle in (5.1) and the trigonometric parametrization (cos θ, sin θ), we can write (with some assistance from (5.2)) (5.3) cos θ = 1 −m2 1 + m2 , sin θ = 2m 1 + m2 , m = sin θ 1 + cos θ. A point on the unit circle at angle θ relative to the origin and the positive x-axis makes an angle θ/2 relative to the point (−1, 0) and the x-axis, as shown in the ﬁgure below. The slope of a line equals the tangent of the angle the line makes with the x-axis, so m = tan(θ/2) . (−1, 0) (cos θ, sin θ) θ θ/2 x y The formulas in (5.3) have a connection to calculus. In terms of the substitution u = tan(θ/2), which is motivated by the formula for the slope m above, we can write cos θ = 1 −u2 1 + u2 , sin θ = 2u 1 + u2 , and du = sec2 θ 2 dθ 2 = 1 + tan2 θ 2 dθ 2 = (1 + u2) dθ 2 = ⇒dθ = 2 1 + u2 du. 10 KEITH CONRAD With these formulas, integrals of rational functions of cos θ and sin θ can be turned into inte-grals of rational functions of u. This method of integration is called a tan(θ/2)-substitution or a Weierstrass substitution.3 It is based on having two diﬀerent parametrizations of the unit circle, which allow us to express an integral involving one of the parameterizations (with trig functions) as an integral involving the other parametrization (with rational functions). What we just did with rational points on the unit circle is not true just for that curve. If we take any smooth conic in the plane (a smooth curve of degree 2: an ellipse, hyperbola, or parabola) that is deﬁned by an equation having rational coeﬃcients and we know one rational point on the conic, then slopes of lines through that point let us parametrize all the other rational points on the conic. Here is an example of this idea using another circle. Theorem 5.2. The rational points on the circle x2 + y2 = 2 other than (−1, 1) can be described by the formula 1 + 2m −m2 m2 + 1 , m2 + 2m −1 m2 + 1 where m ∈Q. A rational point (x, y) other than (−1, ±1) arises this way using m = (y + 1)/(x + 1), and (−1, −1) arises this way using m = −1. Proof. The point (−1, −1) lies on the circle. A non-vertical line through (−1, −1) has the form y = m(x + 1) −1. Any line through (−1, −1) intersects the circle in a second point, except for the tangent line y = −x −2 (where m = −1). x y (−1, −1) (−1, 1) To ﬁnd the second point of intersection, substitute the equation of the line into the equation x2 + y2 = 2 and solve for the two roots of the resulting quadratic in x: one root is 1, and we can ﬁnd the other root by the same method as in the geometric proof of Theorem 1.2. This will give the x-coordinate formula above, and substituting this into y = m(x + 1) −1 gives the y-coordinate. The only rational point on x2 + y2 = 2 that we don’t ﬁnd by this method is (−1, 1), which is the second intersection point of the circle with the vertical line through (−1, −1). This vertical line is not described by an equation of the form y = m(x + 1) −1. The parameter value that gives the point (−1, −1) itself is m = −1, which is the slope of the tangent line to x2 + y2 = 2 at (−1, −1) (touching the point (−1, −1) twice, in a sense). □ Remark 5.3. Since x2 +y2 = 2 if and only if ((x−y)/2)2 +((x+y)/2)2 = 1, the geometric proof of Theorem 1.2 yields (x −y)/2 = (1 −m2)/(1 + m2) and (x + y)/2 = 2m/(1 + m2). Adding and subtracting these equations recovers the formulas for x and y in Theorem 5.2. 3See substitution. PYTHAGOREAN TRIPLES 11 Armed with the description of the rational points on x2 + y2 = 2 in Theorem 5.2, we can get a description of the primitive integral solutions of a2 + b2 = 2c2, where primitive means a, b, and c have no common factor (equivalently, this means any two of a, b, and c are relatively prime, i.e., an analogue of Lemma 1.1 holds). In a primitive triple (a, b, c), a and b both must be odd, so 2c2 ≡2 mod 8, hence c is odd as well. For Pythagorean triples, the middle term was chosen to be the even one. The analogue of that here is that we take c > 0 and pick the signs on a and b so that a ̸≡b mod 4. With these conventions, we have a = k2 + 2kℓ−ℓ2, b = ℓ2 + 2kℓ−k2, c = k2 + ℓ2 where (k, ℓ) = 1, k ̸≡ℓmod 2, and k > 0. Conversely, any triple deﬁned by these formulas is a primitive solution to a2 + b2 = 2c2 with b ̸≡a mod 4. k ℓ a b c 1 2 1 7 5 1 −2 −7 −1 5 1 4 −7 23 17 4 1 23 −7 29 2 5 −1 41 29 Table 5. Primitive Solutions to a2 + b2 = 2c2 Integers satisfying a2 + b2 = 2c2 are not the sides of a right triangle, but they have an arithmetic interpretation: since c2 = 1 2(a2 + b2), c2 is the average of a2 and b2. In other words, a2, c2, b2 is an arithmetic progression of squares (taking a2 < b2). From Table 5, squaring the entries in the a, b, and c columns yield three such progressions: 1, 25, 49 (common diﬀerence 24), 49, 289, 529 (common diﬀerence 240), and 1, 841, 1681 (common diﬀerence 840). There are inﬁnitely many 3-term arithmetic progressions of squares in Z+, and we can ﬁnd all of them from the parametrization of the rational points on x2 + y2 = 2 in Theorem 5.2. What about 4-term arithmetic progressions of squares? That is a more complicated story. See References W. Casselman, J. L. Poet and D. L. Vestal, Jr., “Curious Consequences of a Miscopied Quadratic,” College Math. J. 36 (2005), 273–277.
2666	https://accesspharmacy.mhmedical.com/content.aspx?sectionid=39910772&bookid=449	Chapter 6. Biotransformation of Xenobiotics \| Casarett & Doull's Essentials of Toxicology, 2e \| AccessPharmacy \| McGraw Hill Medical Skip to Main Content MCGRAW HILL ACCESS MCGRAW HILL ACCESS McGraw Hill Medical Home Explore More SitesAccessAnesthesiology AccessAPN Accessartmed AccessCardiology AccessDermatologyDxRx AccessEmergency Medicine AccessHemOnc AccessMedicina AccessMedicine AccessNeurology AccessObGyn AccessPediatrics AccessPharmacy AccessPhysiotherapy AccessSurgery AccessWorldMed Case Files Collection Clinical Sports Medicine Collection F.A. Davis AT Collection F.A. 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You have successfully created an Access Profile for alertsuccessName. Features of Access include: Remote Access Favorites Save figures into PowerPoint Download tables as PDFs Go to My DashboardClose HomeBooksCasarett & Doull's Essentials of Toxicology, 2e Previous Chapter \| Next Chapter Chapter 6. Biotransformation of Xenobiotics Add to Favorites Andrew Parkinson; Brian W. Ogilvie Sections Download Chapter PDF Share Email Twitter Facebook Linkedin Reddit Get CitationCitation Disclaimer: These citations have been automatically generated based on the information we have and it may not be 100% accurate. Please consult the latest official manual style if you have any questions regarding the format accuracy. AMA Citation Parkinson A, Ogilvie BW. Parkinson A, & Ogilvie B.W. Parkinson, Andrew, and Brian W. Ogilvie.Chapter 6. Biotransformation of Xenobiotics. In: Klaassen CD, Watkins JB, III. Klaassen C.D., & Watkins J.B., III(Eds.),Eds. Curtis D. Klaassen, and John B. Watkins, III.eds._Casarett & Doull's Essentials of Toxicology, 2e_. The McGraw-Hill Companies; 2010. Accessed September 28, 2025. APA Citation Parkinson A, Ogilvie BW. Parkinson A, & Ogilvie B.W. Parkinson, Andrew, and Brian W. Ogilvie. (2010). Chapter 6. biotransformation of xenobiotics. Klaassen CD, Watkins JB, III. Klaassen C.D., & Watkins J.B., III(Eds.),Eds. Curtis D. Klaassen, and John B. Watkins, III._Casarett & Doull's Essentials of Toxicology, 2e_. The McGraw-Hill Companies. MLA Citation Parkinson A, Ogilvie BW. Parkinson A, & Ogilvie B.W. Parkinson, Andrew, and Brian W. Ogilvie. "Chapter 6. Biotransformation of Xenobiotics." _Casarett & Doull's Essentials of Toxicology, 2e_ Klaassen CD, Watkins JB, III. Klaassen C.D., & Watkins J.B., III(Eds.),Eds. Curtis D. Klaassen, and John B. Watkins, III. The McGraw-Hill Companies, 2010, Download citation file: RIS (Zotero) EndNote BibTex Medlars ProCite RefWorks Reference Manager Mendeley © Copyright Tools Search Book Annotate Clip Autosuggest Results Sections View Full Chapter Figures Tables Videos Annotate Full Chapter Figures Tables Videos Supplementary Content Add to Favorites ++ Biotransformation is the metabolic conversion of endogenous and xenobiotic chemicals to more water-soluble compounds. Xenobiotic biotransformation is accomplished by a limited number of enzymes with broad substrate specificities. Phase I reactions involve hydrolysis, reduction, and oxidation. These reactions expose or introduce a functional group (—OH, —NH 2, —SH, or —COOH), and usually result in only a small increase in hydrophilicity. Phase II biotransformation reactions include glucuronidation, sulfonation (more commonly called sulfation), acetylation, methylation, and conjugation with glutathione (mercapturic acid synthesis), which usually result in increased hydrophilicity and elimination. Add to Favorites ++ Biotransformation is the metabolic conversion of endogenous and xenobiotic chemicals to more water-soluble compounds. Generally, the physical properties of a xenobiotic are changed from those favoring absorption (lipophilicity) to those favoring excretion in urine or feces (hydrophilicity). An exception to this general rule is the elimination of volatile compounds by exhalation. ++ Chemical modification of a xenobiotic by biotransformation may alter its biological effects. Some drugs undergo biotransformation to active metabolites that exert their pharmacodynamic or toxic effect. In most cases, however, biotransformation terminates the pharmacologic effects of a drug and lessens the toxicity of xenobiotics. Enzymes catalyzing biotransformation reactions often determine the intensity and duration of action of drugs and play a key role in chemical toxicity and chemical tumorigenesis. Add to Favorites +++ Basic Properties of Xenobiotic Biotransforming Enzymes ++ Xenobiotic biotransformation is accomplished by a limited number of enzymes with broad substrate specificities. The synthesis of some of these enzymes is triggered by the xenobiotic (by the process of enzyme induction), but in most cases the enzymes are expressed constitutively (i.e., they are synthesized in the absence of a discernible external stimulus). Although the synthesis of steroid hormones is catalyzed by cytochrome P450 enzymes in steroidogenic tissues, this family of enzymes in the liver converts steroid hormones into water-soluble metabolites to be excreted. ++ The structure (i.e., amino acid sequence) of a given biotransforming enzyme may differ among individuals, which can give rise to differences in rates of xenobiotic biotransformation. The study of the causes, prevalence, and impact of heritable differences in xenobiotic biotransforming enzymes is known as pharmacogenetics. +++ Biotransformation versus Metabolism ++ The terms biotransformation and metabolism are often used synonymously, particularly when applied to drugs. The term metabolism is often used to describe the total fate of a xenobiotic, which includes absorption, distribution, biotransformation, and elimination. However, metabolism is commonly used to mean biotransformation, which is understandable from the standpoint that the products of xenobiotic biotransformation are known as metabolites. Furthermore, individuals with a genetic enzyme deficiency resulting in impaired xenobiotic biotransformation are described as poor metabolizers rather than poor biotransformers. +++ Stereochemical Aspects of Biotransformation ++ Stereochemical properties influence the interaction between a xenobiotic and its biotransforming enzyme. Many xenobiotics, especially drugs, contain one or more chiral centers and can exist in two mirror-image stereoisomers or enantiomers. The biotransformation ... Your Access profile is currently affiliated with [InstitutionA] and is in the process of switching affiliations to [InstitutionB]. Please select how you would like to proceed. 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2667	https://faculty.cs.niu.edu/~hutchins/csci241/eval.htm	# Using a Stack to Evaluate an Expression We often deal with arithmetic expressions written in what is called infix notation: Operand1 op Operand2 We have rules to indicate which operations take precedence over others, and we often use parentheses to override those rules. It is also quite possible to write arithmetic expressions using postfix notation: Operand1 Operand2 op With postfix notation, it is possible to use a stack to find the overall value of an infix expression by first converting it to postfix notation. Example: Suppose we have this infix expression Q: 5 ( 6 + 2 ) - 12 / 4 The equivalent postfix expression P is: 5 6 2 + 12 4 / - This discussion assumes all our operations are binary operations (2 arguments each). Notice that we also sometimes use unary operations such as ++ or -- or the unary + and -. We are not including the possibility of array elements in this discussion. (The subscript can be an expression which would have to be evaluated.) One way to think of an expression is as a list or sequence of items, each of which is a left parenthesis, right parenthesis, argument, or operator. An argument can be a constant or the name of a variable. Presumably it would be necessary at some point to replace each variable with its value. There are two algorithms involved. One converts an infix expression to postfix form, and the other evaluates a postfix expression. Each uses a stack. Transform an infix expression to postfix notation Suppose Q is an arithmetic expression in infix notation. We will create an equivalent postfix expression P by adding items to on the right of P. The new expression P will not contain any parentheses. We will use a stack in which each item may be a left parenthesis or the symbol for an operation. Start with an empty stack. We scan Q from left to right. While (we have not reached the end of Q) If (an operand is found) Add it to P End-If If (a left parenthesis is found) Push it onto the stack End-If If (a right parenthesis is found) While (the stack is not empty AND the top item is not a left parenthesis) Pop the stack and add the popped value to P End-While Pop the left parenthesis from the stack and discard it End-If If (an operator is found) If (the stack is empty or if the top element is a left parenthesis) Push the operator onto the stack Else While (the stack is not empty AND the top of the stack is not a left parenthesis AND precedence of the operator <= precedence of the top of the stack) Pop the stack and add the top value to P End-While Push the latest operator onto the stack End-If End-If End-While While (the stack is not empty) Pop the stack and add the popped value to P End-While Notes: At the end, if there is still a left parenthesis at the top of the stack, or if we find a right parenthesis when the stack is empty, then Q contained unbalanced parentheses and is in error. Evaluate a postfix expression Suppose P is an arithmetic expression in postfix notation. We will evaluate it using a stack to hold the operands. Start with an empty stack. We scan P from left to right. While (we have not reached the end of P) If an operand is found push it onto the stack End-If If an operator is found Pop the stack and call the value A Pop the stack and call the value B Evaluate B op A using the operator just found. Push the resulting value onto the stack End-If End-While Pop the stack (this is the final value) Notes: At the end, there should be only one element left on the stack. This assumes the postfix expression is valid. How can this be implemented? Work like this is usually done by an assembler, compiler or interpreter. A programmer uses an expression in her or her code, and evaluating it is someone else's problem. Suppose it is our problem (maybe we are writing an interpreter). The interpreter is reading a line at a time from a file as a string, such as A = ((B + C) / 3 - 47 % E) (F + 8) The string needs to be parsed--that is, we need to break it up into substrings, each of which is one meaningful part. These substrings are often called tokens. The tokens are separated by spaces, in many cases, but also a token ends if we find a left or right parenthesis or the symbol for an operator. Thus for instance, in the above example, we have "E)", and this consists of two tokens "E" and ")". Bear in mind that the symbol for an operator can be more than one character. We then have a list of tokens, perhaps in an array or a linked list. Somewhere we will have an Evaluate function which takes such a list as an argument and returns a numeric value.
2668	https://www.health.state.mn.us/diseases/cytomegalovirus/childcareeduc.html	Information for Child Care and Education Professionals - MN Dept. of Health Skip to main content MENU Main navigation Home Data, Statistics, and Legislation Diseases and Conditions Health Care Facilities, Providers, and Insurance Healthy Communities, Environment, and Workplaces Individual and Family Health About Us News and Announcements Translated Materials MENU Search MDH site Main navigation mobile Data, Statistics, and Legislation Diseases and Conditions Health Care Facilities, Providers, and Insurance Healthy Communities, Environment, and Workplaces Individual and Family Health About Us News and Announcements Translated Materials Breadcrumb Home Diseases and Conditions print Topic Menu Cytomegalovirus (CMV) CMV Home About CMV and Congenital CMV CMV Prevention For Families and Caregivers For Health Professionals For Childcare and Education Professionals Related Programs Newborn Screening Children and Youth with Special Health Needs Early Hearing Detection and Intervention (EHDI) Cytomegalovirus (CMV) CMV Home About CMV and Congenital CMV CMV Prevention For Families and Caregivers For Health Professionals For Childcare and Education Professionals Related Programs Newborn Screening Children and Youth with Special Health Needs Early Hearing Detection and Intervention (EHDI) Contact Info Children and Youth with Special Health Needs 651-201-3650 800-728-5420 (toll-free) health.cyshn@state.mn.us Contact Info Children and Youth with Special Health Needs 651-201-3650 800-728-5420 (toll-free) health.cyshn@state.mn.us Information for Child Care and Education Professionals Professionals who work in child care and educational settings are at increased risk of contact with young children with cytomegalovirus (CMV). CMV infection is a special concern for workers who are or may become pregnant. Learning about CMV and how exposures occur is an important step in reducing your risk of getting CMV. Teaching and supporting children and families impacted by the virus are also important roles for child care and educational professionals. How is CMV transmitted in child care and education settings? CMV is passed from person to person through contact with bodily fluids such as urine or saliva. Babies and young children are a common source of CMV because they often get CMV from other kids. The virus can stay in a child's saliva and urine and be passed to others for months after the infection. Up to 70% of healthy children aged one to three years in child care settings may shed, or release, CMV in their saliva and urine even if they don’t seem sick. CMV can be spread through close contact, such as diaper changing, kissing, sharing food, and other activities where a person could have contact with the urine or saliva of a child with CMV. People who work closely with young children in places like child care centers, schools, or who provide family child care within their home have a higher risk of getting a CMV infection than those who don’t work in such settings. How can I reduce my risk of getting CMV? Wash your hands often with soap and water for 15 to 20 seconds, especially after: Changing a diaper or helping a young child to use the toilet, Wiping a young child's nose or mouth, and Handling children's toys or touching a surface that may have a child's saliva or urine on it. Do not put things in your mouth that have just been in a child’s mouth, such as a pacifier or toothbrush. Do not share food, drinks, utensils, and straws with a young child. Kiss a child on the forehead instead of the lips to avoid saliva. Properly disinfect toys, changing tables, and other surfaces that may have a child’s urine or saliva on them. Discuss CMV with your health care provider if you are pregnant or thinking about becoming pregnant. How can I help educate parents about preventing CMV and congenital CMV? Contact with the saliva or urine of young children is a major cause of CMV infection in pregnant people, but only about 10% of women have ever heard of CMV. Child care and education professionals should help teach parents of babies and young children about CMV and ways that they can reduce their risk of getting CMV. How can I best support parents of children identified with congenital CMV? Finding out that a child has congenital CMV may cause anxiety or stress for some parents. One of the best ways to show support is to educate yourself about congenital CMV and the possible long-term impacts and follow-up that affected children and their families might experience. Also, remember that CMV is a common childhood virus and that children known to have congenital CMV should not be treated differently from other children in your care. Should children with CMV stay home from child care or school? Children known to have congenital CMV or who get CMV after birth should not be treated differently from other children. They should not be expected to stay home. CMV is common in babies and young children, and children who are known to have CMV do not transmit the virus more often than any other child. Resources CMV and Congenital CMV Fact Sheet for Child Care Providers (PDF) Hmong (Lus Hmoob) Somali (Af Soomaali) Spanish (Español) CMV and Congenital CMV Fact Sheet for School Nurses (PDF) Are you pregnant or thinking about becoming pregnant? Learn about CMV (PDF) Hmong (Lus Hmoob) Somali (Af Soomaali) Spanish (Español) Karen (S’gaw Karen) Minnesota Low Incidence Projects: Congenital Cytomegalovirus Educational webinars and podcasts Webinar: All Hands On Deck! Supporting the Development of Children with Congenital CMV Webinar: Potential Impacts of Congenital CMV in Early Childhood Special Education in Minnesota Podcast: CMV and Congenital CMV Part 1 Podcast: CMV and Congenital CMV Part 2 Tags cytomegalovirus Last Updated: 01/02/2025 Get email updates Enter Email Address Submit Sign up for GovDelivery notifications Privacy Policy Equal Opportunity Translated Materials Feedback Form About MDH Minnesota.gov Facebook Twitter Linked In Instagram Youtube
2669	https://www.dummies.com/article/academics-the-arts/science/chemistry/rules-for-assigning-oxidation-numbers-to-elements-194221/	Meet Earth’s mightiest heroes! Avengers For Dummies is here! Order your copy today. Book & Article Categories Collections Custom Solutions Dummies AI Main Menu Book & Article Categories Technology Academics & The Arts Home, Auto, & Hobbies Body, Mind, & Spirit Business, Careers, & Money Dummies AI Main Menu Book & Article Categories Technology Academics & The Arts Home, Auto, & Hobbies Body, Mind, & Spirit Business, Careers, & Money Dummies AI Main Menu Collections Explore all collections BYOB (Be Your Own Boss) Be a Rad Dad Career Shifting Contemplating the Cosmos For Those Seeking Peace of Mind For the Aspiring Aficionado For the Budding Cannabis Enthusiast For the College Bound For the Exam-Season Crammer For the Game Day Prepper Dummies AI Home Academics & The Arts Articles Science Articles Chemistry Articles Rules for Assigning Oxidation Numbers to Elements By No items found. Updated 2021-07-16 15:13:29 From the book No items found. Share Download E-Book Personal Finance For Dummies Explore Book Chemistry All-in-One For Dummies (+ Chapter Quizzes Online) Explore Book Buy Now Buy on AmazonBuy on Wiley Subscribe on Perlego Download E-Book Personal Finance For Dummies Explore Book Chemistry All-in-One For Dummies (+ Chapter Quizzes Online) Explore Book Buy Now Buy on AmazonBuy on Wiley Subscribe on Perlego Oxidation numbers are bookkeeping numbers. They allow chemists to do things such as balance redox (reduction/oxidation) equations. Oxidation numbers are positive or negative numbers, but don’t confuse them with positive or negative charges on ions or valences. Oxidation numbers are assigned to elements using these rules: Rule 1: The oxidation number of an element in its free (uncombined) state is zero — for example, Al(s) or Zn(s). This is also true for elements found in nature as diatomic (two-atom) elements and for sulfur, found as: Rule 2: The oxidation number of a monatomic (one-atom) ion is the same as the charge on the ion, for example: Rule 3: The sum of all oxidation numbers in a neutral compound is zero. The sum of all oxidation numbers in a polyatomic (many-atom) ion is equal to the charge on the ion. This rule often allows chemists to calculate the oxidation number of an atom that may have multiple oxidation states, if the other atoms in the ion have known oxidation numbers. Rule 4: The oxidation number of an alkali metal (IA family) in a compound is +1; the oxidation number of an alkaline earth metal (IIA family) in a compound is +2. Rule 5: The oxidation number of oxygen in a compound is usually –2. If, however, the oxygen is in a class of compounds called peroxides (for example, hydrogen peroxide), then the oxygen has an oxidation number of –1. If the oxygen is bonded to fluorine, the number is +1. Rule 6: The oxidation state of hydrogen in a compound is usually +1. If the hydrogen is part of a binary metal hydride (compound of hydrogen and some metal), then the oxidation state of hydrogen is –1. Rule 7: The oxidation number of fluorine is always –1. Chlorine, bromine, and iodine usually have an oxidation number of –1, unless they’re in combination with an oxygen or fluorine. These rules give you another way to define oxidation and reduction — in terms of oxidation numbers. For example, consider this reaction, which shows oxidation by the loss of electrons: Notice that the zinc metal (the reactant) has an oxidation number of zero (rule 1), and the zinc cation (the product) has an oxidation number of +2 (rule 2). In general, you can say that a substance is oxidized when there’s an increase in its oxidation number. Reduction works the same way. Consider this reaction: The copper is going from an oxidation number of +2 to zero. A substance is reduced if there’s a decrease in its oxidation number. About This Article This article is from the book: No items found. About the book author: No items found. This article can be found in the category: Chemistry No items found. Get a Subscription
2670	http://ch302.cm.utexas.edu/physEQ/solutions/selector.php?name=henrys-law	Mixtures of liquids and gases (Henry's Law) When gases dissolve into liquids we have a slightly different situation from the dissolution of a solid or a liquid into another liquid. The biggest issue is the concentration of the "pure" solute. In the case of the solid or a liquid the pure substance is the pure substance. The "molecules" are all next to each other and the concentration is the same at all conditions. For a gas, the "pure" substance has a concentration that is extremely dependent on the conditions. Thus the most important factor for a gas dissolving into a liquid is the concentration of the gas above the liquid. In chemistry we express concentrations of gases as partial pressure. The solubility of gas depends on the partial pressure of the gas above the solvent. The solubility of a gas in the liquid is quantified using Henry’s Law. Henry’s Law states that the mole fraction of gas dissolved in the liquid (like the concentration) is directly proportional to the pressure of the gas over the liquid. [P_{\rm gas} = K X_{\rm gas}] The constant K depends on the IMFs between the solute and the solvent and thus it depends on both the chemical structure of the gas and the liquid. Xgas is the mole fraction which is the (moles of gas)/(total number of moles of the mixture). Henry's Law is sometimes given with different units (but it is exactly the same otherwise). You may find it written as [C_{\rm gas} = k_{\rm H} \; P_{\rm gas} ] Here the Henry's Law constant has units of M/atm. It might also be written as above, but with molarity instead of mole fraction. You might also find it given as the 2nd equation but with mole fraction instead of molarity. Or any number of other ways! Watch the units. (and note that mole fraction has no units). The consequences of Henry's Law are fairly straight forward. Double the pressure. Double the concentration (mole fraction). Gases dissolved in liquids seems a bit of a random topic, but in fact Henry’s Law shows up in every day life. Have you ever wondered how they get all the CO2 into carbonated beverages? They apply a big pressure of CO2 above the liquid and the CO2 dissolves. When the pressure goes away (the pssst when you open the can), the amount of CO2 dissolved decreases and bubbles appear. Also, if you are a scuba diver, you know you do not come up from great depths too quickly. This is because when your body is at higher pressure, N2 dissolves into your blood and tissue. Should you come up to the surface too quickly the pressure drops and, just like the soda can, bubbles form. N2 bubbles in your body can be very painful. Worse, bubbles in your brain tissue and spinal cord are very bad for your health. Finally, Henry’s law is important for fish (and marine life in general). Fish, if you didn’t know, manage to get O2 out of the water to live. As the earth’s atmosphere has a partial pressure of oxygen of around 0.18 atm there is always some O2 dissolved in water (rivers, lakes, oceans, the glass of water you are drinking,…) However, as we will discuss later the temperature dependence of solubility depends on the enthalpy. The fact that dissolution of gases is exothermic means that their solubility decreases with increasing temperature. Thus rising ocean temperatures mean lower dissolved O2 levels. This is turn leads to tough times for marine life. Note: This is a very easy (if a bit boring) experiment to do at home. Put a pan of water on the stove and heat it up. What happens? Way before the solution begins to boil, little bubbles form on the bottom of the pan. Why? This is the dissolved air coming out of the water. As you raise the temperature the solubility drops and the gas comes out. (Note: The bubbles tend to form in the same place. This is because the surface tension of water makes the bubble formation slow. So the bubbles will form at little defects or scratches in the bottom of your pan.) Alternatively (and more amusing) open a very cold can of soda and a very warm can of soda and notice the difference in the solubility of the CO2 in the two situations. More details Other differences for gases dissolving into liquids. For a gas dissolving in a liquid the entropy of solution is negative, (\Delta S _{\rm solution} < 0). The solute is in the gas phase which is the highest entropy state. So even though we are making a mixture, the entropy will be going down. A solution will form only if (\Delta G _{\rm solution} < 0). However, now the entropy term is causing the free energy to increase. The only way that we can a have a decrease in the free energy is if the solution process is exothermic. This is in fact the gas for gas dissolution. The reason is a key difference for the enthlapy of gases compared to other solutes. For gases we don’t have to overcome any intermolecular forces in pulling the solute molecules away from each other. They are in the gas phase and already apart. Thus (\Delta H _{\rm solution} = \Delta H_{\rm solvation}) since the lattice-energy term is zero (because it is a gas with no IMFs). Since the solvation term is negative, we find that (\Delta H_{\rm solution} < 0). A solution will form to a small extent thanks to the enthalpy term being exothermic. Henry's Law © 2013 mccord/vandenbout/labrake
2671	https://www.uv.es/~borrasj/ingenieria_web/temas/tema_1/lecturas_comp/p938.pdf	Research: Science and Education 938 Journal of Chemical Education • Vol. 80 No. 8 August 2003 • JChemEd.chem.wisc.edu The ionization energy of an atom depends on its atomic number and electronic configuration. Ionization energies tend to decrease on descending groups in the s and p blocks (with exceptions) and group 3 in the d block of the periodic table. Successive ionization energies increase with increasing charge on the cation. This paper describes some less familiar aspects of ionization energies of atoms and atomic ions. Apparently irregular first and second ionization energies of transition metals and rare earth metals are explained in terms of the electronic configurations of the ground states. A semiquan-titative treatment of pairing, exchange, and orbital energies accounts for discontinuities at half-filled p, d, and f electron shells and the resulting zigzag patterns. We begin with a reminder of the difference between ion-ization potential and ionization energy. Ionization potential is the electric potential (measured in volts) required to sepa-rate an electron from the orbital system in free space with the kinetic energy remaining unchanged. Ionization energy is the work done in removing the electron at zero tempera-ture and is measured conveniently in electronvolts, where 1 eV = 1.6022 × 1019 J. The molar ionization energy, or change in molar internal energy, is NA eV = 96.485 kJ mol1 where NA is the Avogadro constant. Ionization wavenumbers (reciprocal wavelengths) are derived from series limits of atomic spectral lines. The en-ergy per cm1 is 1.2398 × 104 eV or 1.9864 × 1023 J. The molar energy per cm1 is 11.962 J mol1. Sources of Data Three volumes containing wavenumbers and atomic en-ergy levels (1) preceded the critical survey by Moore of ion-ization limits from ground state to ground state for atoms and atomic ions. Ionization energies derived from optical and mass spectroscopy and calculations ranging from crude ap-proximations to complex equations based on quantum me-chanical theory are accompanied by assessments of reliability and a bibliography (2). Martin, Zalubus, and Hagan reviewed energy levels and ionization limits for rare earth elements (3). Handbook of Chemistry and Physics (4) contains authoritative data from these and later sources. For example, an experi-mental value for the second ionization energy of cesium, 23.157 eV (5), supersedes 25.1 eV (2). Periodicity Third ionization energies of atoms from lithium (Z = 3) to hafnium (Z = 72) are plotted against atomic number, Z, in Figure 1. Values are from reference (4) except those for Cs (Z = 55) and Ba (Z = 56) from the Journal of the Optical Society of America (5, 6). The first peak occurs at Be2+ (Z = 4), which has the electronic configuration 1s2. Peaks corre-sponding to filled s, p, d, f, and half-filled d and f orbitals illustrate the shell model of the atom (7). Figure 1 provides a more compelling demonstration of periodicity than plots of first ionization energy (8) where transition metals and rare earth metals do not show zigzag patterns. For M2+ ions, 3d orbitals have lower energies than 4s orbitals (9), 4d orbitals have lower energies than 5s orbitals, and 5p orbitals have lower energies than 4f orbitals. s Electrons Figure 2 shows how first ionization energies decrease from hydrogen to cesium and from helium to barium. Straight lines joining ionization energies from five pairs of group 1 and 2 atoms have intercepts of approximately 2.6 Ionization Energies of Atoms and Atomic Ions Peter F. Lang and Barry C. Smith School of Biological and Chemical Sciences, Birkbeck College (University of London), Malet Street, London WC1E 7HX, England; smithbc@talk21.com Figure 1. Third ionization energies from lithium to hafnium. 0 20 40 60 80 100 120 140 160 30 20 10 50 40 60 70 Atomic Number Ionization Energy / eV Research: Science and Education JChemEd.chem.wisc.edu • Vol. 80 No. 8 August 2003 • Journal of Chemical Education 939 eV. The intercept for lithium and beryllium is 1.46 eV but there is no reason to believe that the Moore values are incor-rect. First ionization energies of francium and radium, not shown in Figure 2, are greater than those of cesium and barium respectively as a result of poor shielding by 4f elec-trons. First ionization energies of atoms from hydrogen to be-ryllium are plotted in Figure 3. The ionization energy of he-lium is greater than that of hydrogen but less than four times as great (10) because the electrons provide some screening for each other as mutual repulsion pushes them away from the nucleus. The outer electron of lithium occupies a new shell screened by two electrons and the ionization energy is lower than that of hydrogen or helium. Similarly, the first ionization energy of beryllium is higher than that of lithium but much lower than that of helium. Second ionization en-ergies from helium to boron are higher and follow a similar pattern. Other points correspond to third, fourth, and fifth ionization energies of the respective atoms. Ionization energies of atomic hydrogen and one-electron atomic ions, at the left of Figure 3, are approximately pro-portional to the electron–nucleus attraction, Z 2. They are re-produced with reasonable accuracy by the following expression, where RM is the appropriate Rydberg constant and α is the Sommerfeld fine structure constant (11): M 2 2 1 1 4 R Z Z Z + − ( ) α        Screening (electron–electron repulsion) reduces electron– nucleus attractions in helium and two-electron atomic ions but ionization energies are not functions of simple squares, (Z − S)2, where S is a screening constant (12). The correct expression takes account of relaxation by the remaining elec-tron (13): − + Z Z 2 5 4 5 16 The square roots of the first ionization energies plotted against atomic number for six isoelectronic series are shown in Figure 4. The one-electron plot falls close to a straight line through the origin. Differences between square roots of suc-cessive ionization energies for the other series confirm increas-ing curvature from left to right. Gradients for 2s series approach one half and gradients for 3s series approach one third of the gradient for the one-electron series. Their ion-ization energies are based on quadratic expressions, where n is the principal quantum number of the electron, and b and c are constants characteristic of the series: − + Z n bZ c 2 p Electrons First ionization energies of atoms from the first three periods of groups 13 to 18 (2p, 3p, and 4p electron series) form zigzag patterns in Figure 5. First and second ionization energies of atoms from the next two periods (5p and 6p se-ries) appear in Figure 6. Gallium has a slightly higher first ionization energy than aluminum because of relatively poor Figure 2. First ionization energies of group 1 and group 2 atoms. 0 5 10 15 20 25 30 0 1 2 Number of s Electrons 1s 2s 3s 4s 5s 6s Ionization Energy / eV Figure 3. Energies to remove 1s or 2s electrons from atoms or ions. Roman numerals denote the ionization number. 0 50 100 150 200 250 300 350 400 H He Li Be B C N O Atom V IV III II I Ionization Energy / eV Figure 4. Square roots of energies to remove 1s, 2s, or 3s electrons. 0 5 10 15 20 25 H He Li Be B C N O F Ne Na Mg Al Si P S Cl Atom 1s 2s 3s Ionization Energy / eV Research: Science and Education 940 Journal of Chemical Education • Vol. 80 No. 8 August 2003 • JChemEd.chem.wisc.edu Figure 8. First ionization energies from boron to neon. -5 0 5 10 15 20 25 B C N O F Ne Atom ionization energy modified ionization energy pairing energy exchange energy Ionization Energy / eV shielding by 3d electrons. Thallium and lead have higher first ionization energies than indium and tin, respectively, because of poor shielding by 4f electrons. First ionization energies vary in the order B > Al < Ga > In < Tl and C > Si > Ge > Sn < Pb but decrease with increasing atomic number down groups 15 to 18. Lead and bismuth have greater second ion-ization energies than tin and antimony respectively. The first ionization energy of bismuth appears to be anomalous. The increase from thallium to lead is followed by a decrease to bismuth rather than the expected increase to approximately 8 eV (14). It has been claimed that spin– orbit coupling by the Russell–Saunders scheme would lower the ground state of Bi+ by 0.8 eV and that the lower ioniza-tion energy is correct (15). Condon and Shortley identified differences between theoretical and experimental values for a number of atoms (16) but spin–orbit coupling effects of this magnitude were not observed. First ionization energies at the bottom of Figure 7 in-crease from boron to nitrogen, decrease to oxygen, and in-crease to neon. Second ionization energies increase from carbon to oxygen, decrease to fluorine, and increase to so-dium. Third, fourth, and fifth ionization energies of the re-spective atoms show similar patterns. Discontinuities at half-filled p orbitals are convention-ally attributed to repulsion between electrons of opposite spin occupying the same orbital in the second half of a subperiod. Figure 7. Energies to remove 2p electrons. Roman numerals de-note the ionization number. 0 20 40 60 80 100 120 140 160 180 B C N O F Ne Na Mg Al Si Atom II III IV V I Ionization Energy / eV Figure 5. Energies to remove first p electron. 0 5 10 15 20 25 13 14 15 16 17 18 Group 2p 3p 4p Ionization Energy / eV Figure 6. Energies to remove first and second p electrons. Roman numerals denote the ionization number. 0 5 10 15 20 25 13 14 15 16 17 18 19 Group 5p 6p II I Ionization Energy / eV Research: Science and Education JChemEd.chem.wisc.edu • Vol. 80 No. 8 August 2003 • Journal of Chemical Education 941 Less attention has been paid to quantum mechanical exchange interactions, where more energy is required to ionize an elec-tron in a group with parallel spins as a result of increased electron–nuclear attractions (17). The semiquantitative ap-proach developed here considers double occupancy and ex-change interactions as discussed by Blake (18) and takes account of the work of Johnson (19) and Cann (20). Table 1 summarizes numbers of pairing interactions (pairs of electrons occupying the same orbital), p0 and p1 (subcripts denote ionization numbers); numbers of exchange interactions between pairs of electrons having parallel spins, e0 and e1; and changes on ionization, ∆p and ∆e, for boron to neon and their singly charged cations. We assume that in-dividual pairing energies, P, and exchange energies, Eex, are constant across the subperiod. Figure 8 shows how ioniza-tion energies from boron to neon are lowered by changes in pairing energy, P∆p, and raised by changes in exchange en-ergy, Eex∆e, where P = 2Eex = 1.778 eV. Modified ioniza-tion energies, adjusted for pairing and exchange interactions, lie on a curve that increases smoothly from boron to neon. Modified second ionization energies produce a curve that increases from carbon through oxygen and fluorine to so-dium, where P = 2Eex = 2.354 eV. Similar curves can be derived for other p electron series, where P = 2Eex = 1.038 eV for first ionization energies from aluminum to argon and P = 2Eex = 0.784 eV for first ion-ization energies from gallium to krypton. Transition Metals First ionization energies of calcium, strontium, barium, and transition metals are plotted against group number in Figure 9. Apparent irregularities across the periods (21) are caused by different ground-state electronic configurations (1) that are summarized in Table 2. Hafnium and other post-lanthanum atoms have higher first ionization energies than Figure 9. First ionization energies of alkaline earth and transition metals. 5 6 7 8 9 10 11 Ca to Zn Sr to Cd Ba to Hg 4 2 3 Ionization Energy / eV Group 4 5 6 7 8 9 10 11 12 s n o i t a r u g i f n o C c i n o r t c e l E e t a t S d n u o r G . 2 e l b a T s n o I d n a s m o t A l a t e M -n o i t i s n a r T f o Z m o t A n M M+ M2+ M3+ s 4 d 3 s 4 d 3 s 4 d 3 s 4 d 3 0 2 a C 0 2 0 1 0 0 0 1 2 c S 1 2 1 1 1 0 1 0 0 2 2 i T 2 2 2 1 2 0 2 0 1 3 2 V 3 2 3 0 4 0 3 0 2 4 2 r C 4 1 5 0 5 0 4 0 3 5 2 n M 5 2 5 1 5 0 5 0 4 6 2 e F 6 2 6 1 6 0 6 0 5 7 2 o C 7 2 7 0 8 0 7 0 6 8 2 i N 8 2 8 0 9 0 8 0 7 9 2 u C 9 1 0 1 0 0 1 0 9 0 8 0 3 n Z 0 1 2 0 1 1 0 1 0 0 1 0 9 s 5 d 4 s 5 d 4 s 5 d 4 s 5 d 4 8 3 r S 0 2 0 1 0 0 0 9 3 Y 1 2 1 2 0 1 0 0 0 0 4 r Z 2 2 2 1 2 0 2 0 1 1 4 b N 3 1 4 0 4 0 3 0 2 2 4 o M 4 1 5 0 5 0 4 0 3 3 4 c T 5 2 5 1 5 0 5 0 4 4 4 u R 6 1 7 0 7 0 6 0 5 5 4 h R 7 1 8 0 8 0 7 0 6 6 4 d P 8 0 0 1 0 9 0 8 0 7 7 4 g A 9 1 0 1 0 0 1 0 9 0 8 8 4 d C 0 1 2 0 1 1 0 1 0 0 1 0 9 s 6 d 5 s 6 d 5 s 6 d 5 s 6 d 5 6 5 a B 0 2 0 1 0 0 0 7 5 a L 1 2 1 0 2 0 1 0 0 2 7 f H 2 2 2 2 1 0 2 0 1 3 7 a T 3 2 3 0 4 4 7 W 4 2 4 0 5 5 7 e R 5 2 5 1 5 6 7 s O 6 2 6 1 6 7 7 r I 7 2 7 0 8 8 7 t P 8 1 9 0 9 0 8 0 7 9 7 u A 9 1 0 1 0 0 1 0 9 0 8 0 8 g H 0 1 2 0 1 1 0 1 0 0 1 0 9 s n o i t c a r e t n I e g n a h c x E d n a g n i r i a P . 1 e l b a T n o e N o t n o r o B c i m o t A f o n o i t a z i n o I n o m o t A p0 e0 n o I p1 e1 ∆p ∆e B 0 0 B+ 0 0 0 0 C 0 1 C+ 0 0 0 1 N 0 3 N+ 0 1 0 2 O 1 3 O+ 0 3 1 0 F 2 4 F+ 1 3 1 1 e N 3 6 e N + 2 4 1 2 Research: Science and Education 942 Journal of Chemical Education • Vol. 80 No. 8 August 2003 • JChemEd.chem.wisc.edu corresponding atoms in the first and second periods, presum-ably because of poor shielding by 4f electrons. Observed ion-ization energies from ground state to ground state (4) will appear as filled circles in Figures 10–12. Other symbols rep-resent ionization energies derived from wavenumbers of ex-cited states (1). First Transition Period Most atoms in the first transition period have outer elec-tronic configurations M(3dn4s2), where n represents the po-sition of the atom in the d block. First and second ionization energies are examined in Figure 10. Calcium, scandium, ti-tanium, manganese, iron, and zinc lose both s electrons in processes M(3dn4s2) → M+(3dn4s1) → M2+(3dn4s0). Their ionization energies, with energies derived from wavenumbers of excited states for other atoms (1), fall on curves IA and IIA. Chromium (n = 4) and copper (n = 9) lose one s elec-tron in the process M(3dn+14s1) → M+(3dn+14s0) and their first ionization energies form part of curve IB, which is slightly lower than curve IA. Vanadium, cobalt, and nickel lose two s electrons and gain one d electron in processes M(3dn4s2) → M+(3dn+14s0). Their first ionization energies form part of zigzag pattern IC, which has a minimum at chro-mium and a maximum at manganese. This is the reverse of zigzag pattern IIB formed by second ionization energies of atoms that lose one d electron in the process M+(3dn+14s0) → M2+(3dn4s0) and where chromium is higher than manga-nese. Zinc (n = 10) can not contain more than ten 3d elec-trons and does not form part of curve IB or zigzag patterns IC or IIB. Third ionization energies from Figure 1 are plotted on a larger scale in Figure 11. The systematic loss of one d elec-tron in the process M2+(3dn4s0) → M3+(3dn−14s0) gives a zig-zag pattern that excludes calcium (Z = 20, n = 0), increases s n o i t c a r e t n I e g n a h c x E d n a g n i r i a P . 3 e l b a T d o i r e P n o i t i s n a r T t s r i F e h t f o s n o I r o f s l o b m y S m r e T d n a n o I p2 e2 m r e T n o I p3 e3 m r e T ∆p ∆e c S 2+ 0 0 2D c S 3+ 0 0 1S 0 0 i T 2+ 0 1 3F i T 3+ 0 0 2D 0 1 V2+ 0 3 4F V3+ 0 1 3F 0 2 r C 2+ 0 6 5D r C 3+ 0 3 4F 0 3 n M 2+ 0 0 1 6S n M 3+ 0 6 5D 0 4 e F 2+ 1 0 1 5D e F 3+ 0 0 1 6S 1 0 o C 2+ 2 1 1 4F o C 3+ 1 0 1 5D 1 1 i N 2+ 3 3 1 3F i N 3+ 2 1 1 4F 1 2 u C 2+ 4 6 1 2D u C 3+ 3 3 1 3F 1 3 n Z 2+ 5 0 2 1S n Z 3+ 4 6 1 2D 1 4 from scandium to manganese, decreases to iron, and increases to zinc. As with p electrons, discontinuities are convention-ally attributed to repulsion between paired electrons of op-posite spin in the second half of the subperiod. The energy of removal of d electrons was discussed by Catalan et al. in 1954 (22). Our semiquantitative approach involves pairing and exchange energies as before and incorporates orbital en-ergies (19, 20, 23). Table 3 summarizes numbers of pairing interactions, p2 and p3; exchange interactions, e2 and e3; changes on ioniza-tion, ∆p and ∆e; and term symbols for doubly and triply charged ions. Figure 11 shows the third ionization energies modified by pairing and exchange energies, P∆p and Eex∆e, so that scandium, vanadium, manganese, iron, nickel, and zinc form a smooth, almost-linear curve with titanium and Figure 11. Third ionization energies of scandium to zinc. Ionization Energy / eV 24 26 28 30 32 34 36 38 40 42 Sc Ti V Cr Mn Fe Co Ni Cu Zn Atom third ionization energy adjusted for exchange and pairing energy adjusted for exchange, pairing, and orbital energies Figure 10. First and second ionization energies of calcium to zinc. Roman numerals denote the ionization number. See text for expla-nation of the curves A–C. 0 5 10 15 20 25 Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Atom Ionization Energy / eV observed A B C II I Research: Science and Education JChemEd.chem.wisc.edu • Vol. 80 No. 8 August 2003 • Journal of Chemical Education 943 cobalt above the curve and chromium and copper below. These irregularities are the result of changes in orbital en-ergy, Eorb, since F → D transitions are higher and D → F transitions are lower in energy. Application to titanium, chro-mium, cobalt, and copper gives a nearly linear curve that in-creases from scandium to zinc, where P = 4Eex = ±4Eorb = 2.32 eV. Orbital energy changes were not relevant for p elec-trons and cannot be identified for D → S and S → D transi-tions involving scandium, manganese, iron, and zinc, the first and last members of each arm of the zigzag pattern. Fourth ionization energies increase from titanium to iron, decrease to cobalt, and increase to gallium. Similar modification by pairing, exchange, and orbital energies gives an almost linear curve increasing from titanium to gallium, where P = 4Eex = ±4Eorb = 2.80 eV. Second Transition Period First and second ionization energies of the second tran-sition period are examined in Figure 12. Strontium, zirco-nium, technetium, and cadmium lose both s electrons in processes M(4dn5s2) → M+(4dn5s1) → M2+(4dn5s0). Their ionization energies, with energies derived from wavenumbers of excited states for other atoms (1), fall on curves IA and IIA. Niobium, molybdenum, ruthenium, rhodium, and sil-ver lose one s and then one d electron in the processes M(4dn+15s1) → M+(4dn+15s0) → M2+(4dn5s0). Their ioniza-tion energies form part of curve IB and of zigzag pattern IIB. Yttrium follows the pathway M(4d15s2) → M+(4d05s2) → M2+(4d05s1). The first ionization energy falls near curve IA and the second on IIB. Palladium loses two d electrons in the processes M(4d105s0) → M+(4d95s0) → M2+(4d85s0). The first ionization energy falls near curve IA and the second forms part of curve IB. Cadmium (n = 10) does not form part of curve IB or zigzag pattern IIB. Third ionization energies involve loss of one d electron in M2+(4dn5s0) → M3+(4dn−15s0) transitions. The zigzag pat-tern in Figure 1, excluding strontium (Z = 38, n = 0), in-creases from yttrium to technetium, decreases to ruthenium, and increases to cadmium. Modification by pairing, exchange, and orbital energies gives a smooth curve increasing from yt-trium through technetium and ruthenium to cadmium, where P = −4Eex = ±4Eorb = 1.46 eV. Third Transition Period Barium and lanthanum precede the lanthanides and have lower first ionization energies than other atoms in the third transition period. Hafnium loses one d electron and then loses two s electrons and gains one d electron in the processes M(5d26s2) → M+(5d16s2) → M2+(5d26s0). Tantalum and tungsten lose two s electrons and gain one d electron in the process M(5dn6s2) → M+(5dn+16s0); rhenium, osmium, iri-dium, and mercury lose one s electron in the process M(5dn6s2) → M+(5dn6s1); and platinum and gold lose one s electron and then one d electron in the processes M(4dn+16s1) → M+(5dn+16s0) → M2+(5dn6s0). Mercury loses two s elec-trons in the processes M(5d106s2) → M+(5d106s1) → M2+(5d106s0). Rare Earth Metals Table 4 summarizes outer electronic configurations of atoms and ions from the lanthanum and actinium series (2– 4). Eleven pairs of atoms have similar configurations. First ionization energies are plotted in Figure 13 against number of post-lanthanum or post-actinium electrons, n, which rep-resents the position of the atom in the f block. Most actinides have higher ionization energies than corresponding lan-thanides. Figure 12. First and second ionization energies of strontium to cad-mium. Roman numerals denote the ionization number. See text for explanation of the curves A and B. Ionization Energy / eV 0 5 10 15 20 25 Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd Atom observed A B II I Figure 13. First ionization energies of lanthanum and actinium series. 5.5 5.0 6.5 7.0 6.0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 “Ideal” Number of f Electrons in Series lanthanum series actinium series Ionization Energy / eV Research: Science and Education 944 Journal of Chemical Education • Vol. 80 No. 8 August 2003 • JChemEd.chem.wisc.edu s e i r e S m u n a h t n a L e h t f o s n o I r o f s l o b m y S m r e T d n a s n o i t c a r e t n I e g n a h c x E d n a g n i r i a P . 5 e l b a T n o I p2 e2 m r e T n o I p3 e3 m r e T ∆p ∆e a L + 2 0 0 2F a L + 3 0 0 1S 0 0 e C + 2 0 1 3H e C + 3 0 0 2F 0 1 r P + 2 0 3 4I r P + 3 0 1 3H 0 2 d N + 2 0 6 5I d N + 3 0 3 4I 0 3 m P + 2 0 0 1 6H m P + 3 0 6 5I 0 4 m S + 2 0 5 1 7F m S + 3 0 0 1 6H 0 5 u E + 2 0 1 2 8S u E + 3 0 5 1 7F 0 6 d G + 2 1 1 2 7F d G + 3 0 1 2 8S 1 0 b T + 2 2 2 2 6H b T + 3 1 1 2 7F 1 1 y D + 2 3 4 2 5I y D + 3 2 2 2 6H 1 2 o H + 2 4 7 2 4I o H + 3 3 4 2 5I 1 3 r E + 2 5 1 3 3H r E + 3 4 7 2 4I 1 4 m T + 2 6 6 3 2F m T + 3 5 1 3 3H 1 5 b Y + 2 7 2 4 1S b Y + 3 6 6 3 2F 1 6 u L + 2 2S u L + 3 1S s e i r e S m u i n i t c A d n a m u n a h t n a L e h t r o f s n o i t a r u g i f n o C c i n o r t c e l E e t a t S d n u o r G . 4 e l b a T n Z m o t A M M+ M + 2 M + 3 f 4 d 5 s 6 f 4 d 5 s 6 f 4 d 5 s 6 f 4 d 5 s 6 0 7 5 a L 0 1 2 0 2 0 0 1 0 0 0 0 1 8 5 e C 1 1 2 1 2 0 2 0 0 1 0 0 2 9 5 r P 3 0 2 3 0 1 3 0 0 2 0 0 3 0 6 d N 4 0 2 4 0 1 4 0 0 3 0 0 4 1 6 m P 5 0 2 5 0 1 5 0 0 4 0 0 5 2 6 m S 6 0 2 6 0 1 6 0 0 5 0 0 6 3 6 u E 7 0 2 7 0 1 7 0 0 6 0 0 7 4 6 d G 7 1 2 7 1 1 7 1 0 7 0 0 8 5 6 b T 9 0 2 9 0 1 9 0 0 8 0 0 9 6 6 y D 0 1 0 2 0 1 0 1 0 1 0 0 9 0 0 0 1 7 6 o H 1 1 0 2 1 1 0 1 1 1 0 0 0 1 0 0 1 1 8 6 r E 2 1 0 2 2 1 0 1 2 1 0 0 1 1 0 0 2 1 9 6 m T 3 1 0 2 3 1 0 1 3 1 0 0 2 1 0 0 3 1 0 7 b Y 4 1 0 2 4 1 0 1 4 1 0 0 3 1 0 0 4 1 1 7 u L 4 1 0 2 4 1 0 2 4 1 0 1 4 1 0 0 f 5 d 6 s 7 f 5 d 6 s 7 0 9 8 c A 0 1 2 0 0 2 1 0 9 h T 0 2 2 0 2 1 2 1 9 a P 2 1 2 3 2 9 U 3 1 2 4 3 9 p N 4 1 2 5 4 9 u P 6 0 2 6 5 9 m A 7 0 2 7 6 9 m C 7 1 2 8 7 9 k B 9 0 2 9 8 9 f C 0 1 0 2 0 1 9 9 s E 1 1 0 2 1 1 0 0 1 m F 2 1 0 2 2 1 1 0 1 d M 3 1 0 2 3 1 2 0 1 o N 4 1 0 2 4 1 3 0 1 r L 4 1 1 2 Research: Science and Education JChemEd.chem.wisc.edu • Vol. 80 No. 8 August 2003 • Journal of Chemical Education 945 Lanthanum Series First and second ionization energies of the lanthanum series are examined in Figure 14. Filled circles represent NBS (National Bureau of Standards) values (3, 4). Other symbols represent ionization energies derived from wavenumbers of excited states for neighboring atoms (3). Most atoms lose both s electrons during processes M(4f n+15d06s2) → M+(4f n+15d06s1) → M2+(4f n+15d06s0). Their ionization en-ergies fall on curves IA and IIA with derived energies for lan-thanum, cerium, and gadolinium but not lutetium (n = 14). Gadolinium loses both s electrons during the processes M(4f 75d16s2) → M+(4f 75d16s1) → M2+(4f 75d16s0) and ion-ization energies fall on curves IB and IIB with energies de-rived from ref 3 for most atoms from terbium to lutetium. The A and B curves increase across the series and are approxi-mately parallel with differences that might depend on shield-ing by 6s or 4f electrons. Lutetium has filled 4f orbitals and loses one d electron followed by one s electron in M(4f 145d16s2) → M+(4f145d06s2) → M2+(4f145d06s1) transitions. These ion-ization energies fall on curves IC and IIC (approximately par-allel to IIA and IIB) with energies derived from wavenumbers (3) for erbium, thulium, and ytterbium. Lanthanum and ce-rium lose two s electrons and gain one d electron in M(4f n5d16s2) → M+(4f n5d26s0) transitions. Their ionization energies fall on curve ID with derived energies for praseody-mium and neodymium. Lanthanum loses secondly one d electron and falls on curve IID with derived energies for ce-rium, praseodymium, and neodymium. Cerium loses sec-ondly two d electrons while gaining one f electron and forms part of reverse zigzag pattern IIE with derived energies for lanthanum, praseodymium, and neodymium. Third ionization energies in Figure 15 refer to loss of one f electron in M2+(4f n+15d06s0) → M3+(4f n5d06s0) tran-sitions. They increase from lanthanum to praseodymium, flat-ten to promethium, increase to europium, and decrease to gadolinium with similar behavior from gadolinium to ytter-bium giving a distorted zigzag pattern. These energies de-rived for excited states of lanthanum and gadolinium (3) differ slightly from ground-state ionization energies in Figure 1. Table 5 summarizes numbers of pairing interactions (pairs of electrons occupying the same orbital), p2 and p3; numbers of exchange interactions between pairs of electrons having parallel spins, e2 and e3; changes on ionization, ∆p and ∆e; and term symbols for ions of the lanthanide series. Corrections for exchange and pairing energies remove the principal discontinuity but as with d electrons, orbital ener-gies must be considered (19, 23). Corrections involving H → F transitions, Eorb1, and I → H transitions, Eorb2, and cor-responding negative transitions produce a surprisingly smooth curve, where P = 6Eex = ±4Eorb1 = ±3Eorb2 = 2.625 eV. Esti-mated experimental errors for third ionization energies in-clude ±0.4 eV for Pm and ±0.3 eV for Nd, Sm, and Dy (3). Fourth ionization energies of the lanthanides show a similar distorted zigzag pattern. Fourteen atoms from cerium through lutetium but not lanthanum (n = 0), lose one f elec-tron in M3+(4f n5d06s0) → M4+(4f n-15d06s0) transitions. Cor-rections for pairing, exchange, and orbital energies produce a reasonably smooth curve when P = 6Eex = ±4Eorb1 = ±3Eorb2 = 2.6 eV, despite estimated experimental errors of ±0.7 eV (Sm and Gd), ±0.6 eV (Pm, Eu, and Ho), and ±0.4 eV (Nd, Dy, Er, and Tm). Actinium Series Electronic configurations of lanthanum and actinium are similar for atoms but different for ions, and actinium has the lower ionization energy. An irregular pattern from thorium to neptunium and americium and curium in Figure 13 sug-gests that cations have different configurations. Ionization energies of plutonium and berkelium to nobelium fall on a smooth curve in the second half of the series that is approxi-mately parallel to the curve from europium to ytterbium and suggests loss of one s electron in M(5f n+16d07s2) → M+(5f n+16d07s1) transitions. We are unable to find a reliable ionization energy for lawrencium. Figure 15. Third ionization energies of lanthanum series. 18 19 20 21 22 23 24 25 26 27 28 La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Atom Ionization Energy / eV third ionization energy adjusted for exchange and pairing energy adjusted for exchange, pairing, and orbital energies Figure 14. First and second ionization energies of lanthanum series. 4 6 8 10 12 14 16 La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Atom observed A B C D E I II Ionization Energy / eV Research: Science and Education 946 Journal of Chemical Education • Vol. 80 No. 8 August 2003 • JChemEd.chem.wisc.edu Conclusion This paper reviews the ionization energies of atoms and atomic ions in the s, p, d, and f blocks of the periodic table. The apparently irregular behavior of the first and second ion-ization energies across the transition-metal and rare earth periods is explained in terms of electronic configurations of the atoms and ions. A semiquantitative treatment of pairing energies, P , exchange energies, Eex, and orbital energies, Eorb, explains discontinuities at half-filled p, d, and f electron shells. Some of the great advances of the last century in the understanding of atomic spectra and ionization energies are summarized in the Appendix at the suggestion of a reviewer. Acknowledgments We thank C. D. Flint and P . J. Heard for helpful dis-cussions. Literature Cited 1. Moore, C. E. Atomic Energy Levels; NBS Circular 467; U.S. Department of Commerce: Washington DC; (a) 1949; Vol. 1. (b) 1952; Vol. 2. (c) 1958; Vol. 3. 2. Moore, C. E. Ionization Potentials and Ionization Limits De-rived from the Analyses of Optical Spectra; NSRDS-NBS 34; U.S. Department of Commerce: Washington DC, 1970. 3. Martin, W. C.; Zalubus, R.; Hagan, L. Atomic Energy Levels– The Rare Earth Elements; NSRDS-NBS 60; U.S. Department of Commerce: Washington DC, 1978. 4. Handbook of Chemistry and Physics, 81st ed.; Lide, D. R., Ed.; CRC: Boca Raton, FL, 2000. 5. Reader, J. J. Opt. Soc. Am. 1975, 65, 638. 6. Reader, J.; Epstein, G. L. J. Opt. Soc. Am. 1976, 66, 590. 7. Gillespie, R. J.; Moog, R. S.; Spencer, J. N. J. Chem. Educ. 1998, 75, 539–540. 8. Ahrens, L. H. Ionization Potentials; Pergamon: Oxford, En-gland, 1983. 9. Pilar, F. L. J. Chem. Educ. 1978, 55, 2–6. 10. Rioux, F.; DeKock, R. L. J. Chem. Educ. 1998, 75, 537–539. 11. Lang, P . F.; Smith, B. C. Inorg. Nucl. Chem. Letters 1981, 17, 27–29. 12. Agmon, N. J. J. Chem. Educ. 1988, 65, 42–44. 13. Ali, M. E. S.; Lang, P . F.; Smith B. C. J. Chem. Soc., Faraday 2 1984, 80, 1089–1091. 14. Born, M. (revised Blin-Stoyle, R. J.; Radcliffe, J. M.) Atomic Physics, 8th ed.; Blackie: London, 1969; Chapter 6. 15. Smith, D. W. J. Chem. Educ. 1975, 52, 576–577. 16. Condon, E. U.; Shortley, G. H. Theory of Atomic Spectra; Cam-bridge University: Cambridge, 1955; Chapter 7. 17. Atkins, P . J. Quanta: A Handbook of Concepts, 2nd ed.; Ox-ford University: Oxford, England, 1980; p 113. 18. Blake, A. B. J. Chem. Educ. 1981, 58, 393–398. 19. Johnson, D. A. Some Thermodynamic Aspects of Inorganic Chemistry, 2nd ed.; Cambridge University: Cambridge, England, 1982; Chapter 6. 20. Cann, P . J. Chem. Educ. 2000, 77, 1056–1061. 21. Lang, P . F.; Smith B. C. Educ. Chem. 1986, 23, 50–53. 22. Catalan, M. A.; Rohrlich, R.; Shenstone, A. G. Proc. Roy. Soc. 1954, 221A, 421–437. 23. Bills, J. L. J. Chem. Educ. 1998, 75, 589–593. 24. Series, G. W. Spectrum of Atomic Hydrogen; Oxford University Press: Oxford, England, 1957; Chapter 3. 25. Hertzberg, G. Atomic Spectra and Atomic Structure, 2nd ed.; Dover: New York, 1944; Chapter 1. 26. Sanders, J. H. The Fundamental Atomic Constants; Oxford Uni-versity Press: Oxford, England, 1961; Chapter 2. 27. Born, M. (revised Blin-Stoyle, R. J., Radcliffe, J. M.) Atomic Physics, 8th ed.; Blackie: London, 1969; Chapter 4. 28. Sommerfeld, A. Atomic Structure and Spectral Lines; Methuen: London, 1934; Chapter 2. 29. Heisenberg, W. Physical Principals of the Quantum Theory; Do-ver: New York, 1930. 30. Atkins, P . W. Molecular Quantum Mechanics; Oxford Univer-sity Press: Oxford, England, 1970; Chapter 8. 31. Bethe H. A.; Salpeter, E. E. Quantum Mechanics of One and Two Electron Atoms; Plenum: New York, 1977. Appendix In the late 19th century, it was shown that atomic spec-tra consist of discrete lines rather than continuous emission or absorption. Paschen, Bracket, Pfund, and others made no-table contributions (24), and Balmer identified four lines as members of a converging series (25). Rydberg discovered the constant that bears his name (26). At the beginning of the 20th century, Bohr proposed that electrons in an atom oc-cupy discrete energy states and that radiation is emitted or absorbed on transition from one discrete (quantum) state to another (27). The energy change, ∆E, for hydrogen or a hydrogen-like ion is given by the following expression, where R is the Rydberg constant, Z is the atomic number, and n1 and n0 are integers: ∆ = RZ E 2 1 1 12 n 02 n − Agreement between theory and experiment was well within the limits of error in the measurement of atomic constants at that time. Sommerfeld and others developed general laws of atomic spectroscopy (28). Further advances came from the development of wave mechanics and matrix mechanics (29). The Schrodinger equa-tion can be solved exactly only for one-electron ions. Approxi-mate methods were used to determine energy levels and ionisation energies in other systems. In 1930, Slater devised an approximate method for estimating the extent to which electrons are shielded from the nucleus (30). He assumed that each electron is situated in the field of an effective nuclear charge, Z, and that an effective quantum number, n, could be assigned to each electron. The energy required to ionize an electron, I, could be estimated as: 2 I R Z N = Slater’s rules are based on simple assumptions about shielding and effective nuclear charge. They are unable to ac-count for the zigzag patterns of ionization energies illustrated in Figure 1. More sophisticated equations allow ionisation en-ergies in multielectron atoms to be calculated (31).
2672	https://www.reddit.com/r/askscience/comments/21rdia/can_someone_please_explain_to_me_substrate_level/	Reddit - The heart of the internet Skip to main content Open menu Open navigationGo to Reddit Home r/askscience A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to askscience r/askscience r/askscience Ask a science question, get a science answer. Members Online • [deleted] Can someone please explain to me substrate level phosphorylation in the Krebs cycle? Biology At the moment I know it's the formation of ATP directly via the conversion of one substrate to another, but I don't really know what that means...I think it's that one reaction is causing another to happen but I really don't know. Thankyou! Read more Archived post. New comments cannot be posted and votes cannot be cast. Share Related Answers Section Related Answers Definition of substrate level phosphorylation Latest breakthroughs in quantum physics Impact of climate change on ocean currents Mechanics behind black hole formation Advances in renewable energy technologies New to Reddit? Create your account and connect with a world of communities. Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community 0 0 Top Posts Reddit reReddit: Top posts of March 30, 2014 Reddit reReddit: Top posts of March 2014 Reddit reReddit: Top posts of 2014 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
2673	https://www.gauthmath.com/solution/1707130316300294/Prove-that-xp-q-xq-r-xr-p-1	Solved: Prove that : x^(p-q) x^(q-r) x^(r-p)=1 [Math] Drag Image or Click Here to upload Command+to paste Upgrade Sign in Homework Homework Assignment Solver Assignment Calculator Calculator Resources Resources Blog Blog App App Gauth Unlimited answers Gauth AI Pro Start Free Trial Homework Helper Study Resources Math Questions Question Prove that : x^(p-q) x^(q-r) x^(r-p)=1 Expert Verified Solution 100%(491 rated) Answer Explanation Helpful Not Helpful Explain Simplify this solution Related Prove that. 2 xa-b xb-a xc-a=1 b xp-q xq-r xr-p=1 c frac xa+bxb+c2frac xb+cxc+a2frac xc+axa+b2=1 d frac xaxba+bfrac xbxcb+cfrac xcxac+a=1 100% (2 rated) Simplify a2.a3 f c a16-a1 a 21' b d x+yx+y3 c 2x+3y5n2x+3y5n 22.21.24 CHA 1 g 3a2 xy7 ab2 1 h 1 3/2 1 k a/b 2 x+y/x-y Less o m 523 a23 [a+b] n 6. Simplify each of the following by removing negative index and radieg 1 1 1 1 a a-23 b x-7-4 c x-2y2z2 d frac x-2y3x-2y-7 c x- 2/3 +y- 1/2 0 1 1 7. Prove that: a xa-b xb-c xc-a=1 b xp-q xq-r xr-p=1 c frac xa+bxb-c2frac xb+cxb-a2frac xa+cxa+b2=1 d frac xaxba+bfrac xbxcb+cfrac xcxcc+a=1 W . c ax / a-3x-3 a3 / a-2y-2 a2 / a-4z-4=1 8. I f x=-1,y=2 and z=3 , calculate the following c Z1 a y2 b y2 square d xn c 3x 2x 3x f xy2 A N S W E R S J a 34 b 34 c a4 2. a 2 2 2 a a a b 6 6 6 6 c x x y y=y a 6075 b l c 1000 100% (1 rated) There is a triangular field with boundaries 17 m, 28 m, and 24 m. It wants to fence its boundaries by 5 times by wire. Find the total ler a. if xneq 0 then write the rule of xm / xn. b.prove that xp-q xq-r xr-p=1 100% (3 rated) Write the formula to find the total surface area of the cuboid. 1 4 b. Find the total surface area of the box. 1 c. Is there any difference in the total surface area when it is opened or closed at the top? justify. 13 cm d. There is a triangular field with boundaries 17 m, 28 m, and 24 m. It's land owne wants to fence its boundaries by 5 times by wire. Find the total length of wire. 6. a. if xneq 0 then write the rule of xm / xn. b.prove that xp-q xq-r xr-p=1 100% (2 rated) How much amou 5. In the figure a cuboid with length breadth and height are 6cm, 3cm, and 4 cm. a. Write the formula to find the total surface area of the cuboid. 1 b. Find the total surface area of the box. 1 c. Is there any difference in the total surface area when it is opened or closed at the top? justify. 1 d. There is a triangular field with boundaries 17 m, 28 m, and 24 m. It's land owner wants to fence its boundaries by 5 times by wire. Find the total length of wire. ₹2 6. a. if xneq 0 then write the rule of xm / xn. 1 b.prove that xp-q xq-r xr-p=1 100% (2 rated) Algebra Let's write the numbers in the box in order. State whether the find the correct number in the box. mathematical sentence is 'true' or 'not true' for each number. Then a 2+square =5 Write 1, 2 and 3 respectively in the box. b square +4=9 Write 1, 2, 3, 4 and 5 respectively in the box. c 12-square =8 Write 1, 2, 3 and 4 respectively in the box. d square -3=7 Write 1 to 10 respectively in the box. e 5 square =40 Write 1 to 8 respectively in the box. f square 9=54 Write 1 to 6 respectively in the box. g 45 / square =9 Write 1 to 5 respectively in the box. h square / 4=2 Write 1 to 8 respectively in the box. 100% (1 rated) 60+ Happy Get Well Soon Quotes for Crush Writing Examples Essential Guide to Hofstra University Personal Statement Length and Requirements Writing Examples What is a Personal Essay How to Impress College Admission Officers Blog Grammar Rules to Follow When Writing an Essay: Your Guide to Polished Work That Persuades Readers Blog Gauth it, Ace it! contact@gauthmath.com Company About UsExpertsWriting Examples Legal Honor CodePrivacy PolicyTerms of Service Download App
2674	https://www.youtube.com/watch?v=zhFCq-ZnriI	Find the Parametric Equations for a Line Segment Given an Orientation Mathispower4u 327000 subscribers 179 likes Description 94718 views Posted: 18 Dec 2013 This video explains who to determine the parametric equations of a line segment given the orientation. Site: 6 comments Transcript: Suppose parametric equations for the line segment between the point 8 comma 5 and the point 2 comma 1 have the form x= a + bt and y = c + dt. If the parametric curve or in this case segment starts at the point 8a 5 when t = z which would be this point here and ends at the point 2 comma 1 when t = 1 which would be this point here. We want to find a, b, c, and d. Before we do this though, notice how as t increases, the segment will be traced in this direction here, which is called the orientation of the curve, or in this case, the orientation of the segment. Because we have to be at the point 8, 5 when t equals z. This means x of 0 must equal the x coordinate 8 and y of 0 must equal the y-coordinate 5. And when t equals 1, we must be at this point, which means x of 1 must equal 2 and y of 1 must equal 1. Let's begin by determining x of zero and y of zero. x of 0 would be equal to a plus b t. But we know t is zero. So we'd have b 0. Again, must be equal to the x coordinate when t is zero, which is equal to 8. Notice how this tells us that a must be equal to 8. Now let's find y of zero, which we know must equal 5. y of 0 must equal c plus d t, which is zero, and this must equal 5. So notice how this tells us that c equals 5. So now that we know a= 8 and c = 5, let's find x of 1 and y of 1, where x of 1 must equal 2 and y of 1 must equal 1. So x of 1 would be equal to a + b t. But t is 1. So we have b 1. Again, this must equal 2. But we also know a= 8. So this gives us the equation 8 + b = 2. Subtracting 8 on both sides, notice how b = -6. And now we'll find y of 1, which we know must equal 1. y of 1 = c + d 1, which must equal 1. And we know C is 5. So 5 + D = 1. Subtracting five on both sides, D = -4. So now we have the values of A, B, C, and D. Where our parametric equations would be x or if we want x of t equals a + bt, which would be 8 minus 6 t. and y or y of t would be equal to c + dt or 5 - 4t. I do want to point out one more thing before we go though. Notice how the equations for x and t are linear, meaning they do fit the form y mx plus b, where in this form m is the slope, which tells us the change of y with respect to x. But in this case, B would be the change of X with respect to T and D would be the change of Y with respect to T. So we could have found B, which again would be the change of X divided by the change of T by finding the change of the x coordinates and divide by the change of t. Notice how the change of x would be 2 - 8 and the change of t would be 1 - 0 which equals -6 which again is the value of b that we found. We could also have found d by determining the change of y divided by the change of t where the change of y would be 1 - 5 and the change of t would be 1 minus 0 giving us -4 for d which again we already found but I think it is important to recognize that the values of b and d do represent the change of x with respect to t and the change of y with respect to P in our parametric equations. I hope you found this helpful.
2675	https://docs.lib.purdue.edu/cgi/viewcontent.cgi?article=2354&context=cstech	Published Time: Tue, 13 Oct 2020 19:27:30 GMT Purdue University Purdue University Purdue e-Pubs Purdue e-Pubs Department of Computer Science Technical Reports Department of Computer Science 1997 Matching for Run-Length Encoded Strings Matching for Run-Length Encoded Strings Alberto Apostolico Gad M. Landau Steven Skiena Report Number: 97-018 Apostolico, Alberto; Landau, Gad M.; and Skiena, Steven, "Matching for Run-Length Encoded Strings" (1997). Department of Computer Science Technical Reports. Paper 1355. This document has been made available through Purdue e-Pubs, a service of the Purdue University Libraries. Please contact epubs@purdue.edu for additional information. MATCHING FOR RUN-LENGTH ENCODED STRINGS Alber(o Apostolico Gad. M. Landau Steven Skiena CSD-TR 97-018 March 1997 Matching for Run-Length Encoded Strings Alberto Apostolico 1 Motivation Gad M. Landau i Feb. 15, 1997 Steven Skiena j Measuring the similarity between two strings, through such standard measures as Hamming distance, edit distance, and longest common subsequence, i.s one of the fundamental problems in pattern matching. Tn this paper, we consider the problem of finding the longest common subsequence of two strings. The standard dynamic progr<tTTlming algorithm computes the longest common subsequence of strings X and Y in O(IXI . IYI) time. Here, we develop significantly faster algorithms for a special class of strings which emerge frequently in pattern matching problems. 1\ string is run-length encollcr! if it is described as an ordered sequence of pairs, each consisting of an alpha.bet symbol (J and an integer counting the number of consecutive occurrences of (J. For example, the string aaaabbbbcccabbbbcc can be encoded as a4b4c3alb4c2. Such a run-length encoded string can be significantly shorter than the expanded string representation. Indeed, run-length coding serves as a popular image compression technique, since some classes of images, e.g., binary images in facsimile transmission, typically contain large patches of identically-valued pixels. The need to approximately match run-length encoded strings emerged during develop-ment of an OCR system in concert with Data Capture Systems Inc. , which has been designed to achieve a low substitution error-rate via fixed-font character recognition. The ·Department of Computer Sciences, Purdue University, Computer Sciences Building, 'West Lafayette, IN 47907, USA and Dipartimento di Elettronica e Informatica, Universita di Padova, Padova, Italy. axalllcs. purdue. edu. Work supported in part by NSF Grant CCR-9201 078, by NATO Grant CRG 900293, by the National Research Council of Italy, and by the ESPRIT lIT Basic Research Programme of the EC under contract No. 9072 (Project GEPPCOM). IDepartment of Computer Science, Polytechnic University, 6 MetroTech Center, Brooklyn, NY 11201, USA, and Department of Mathematics and Computer Science, IIaifa University, Haifa 31905, Israel. landaullIpoly. edu. Work supported in part by NSF grant CCR-9305873. ~ Dept. of Computer Science, State University of New York, Stony Brook, NY, 11791-1\1\00 skienal!lcs.sunysb.edu. This work is partially supported by ONR award 400xI16yipOl and NSF Grant CCR-9625669. 1ith row or column of pixels in a given query character image will define a hina,}"y string con-tainiTlg a small number of white-black transitions. By comparing this run-length encoded string against the ith row or column of each of the character image-models, we can identify similar characters. Since a typical row/column contains approximately 50 pixels but only 3-4 white-black transitions, a time savings of roughly two orders of magnitude would follow by matching in time proportional to t.he product of the run lengths, instead of t.he full string lengths. This problem of Tn<Ltching of run-length encoded strings is a TlcLLural generalization of the Ol"iginal sLring matching problem. Indeed, any matching algorithm which tal\:es time proportional to the product of the run lengths on encoded strings would have the samc worst-case complexity as standard matching algorithms while exploiting any runs which happen to exist. Our problem is a simplified version of the previously studied Set LCS and the Set-Set LCS problems [6, 9]. In this paper, we present the first algorithm for finding the longest common subsequence of strings X and Y which runs in time polynomial in the si~c of the compressed strings. Our final algorithm HIllS in O(kllog(kl)) time, where k and 1 are the compressed lengths of striTlgs X and Y, and is a substantial improvement on the previously best algorithm of BUTlke and Csirik , which runs in Q(l!YI + klXI) time. Our algorithm is elegant but non-trivial, and suitable for implementation. 2 Previous Work Throughout this paper, we usc the following notation. Let XIX",! ... Xl denote the run length encoding of string X, where Xi is a maximal run of identical characters and IX;! denotes the length of this run. The length of string X, denoted lXI, represents the total number of characters in X, so IXI = L~=l IX;]. Let Xi denote the unique character comprising run Xi. Similarly Yi y; ... Y k denoLes the run length encoding of string Y. A string W is said to be a subsequence of X if W can be obtained from X by deleting one ot' more symbols. The Longest Common Subsequence (LeS) problem for input strings X and Y consists of finding a longest string Hf which is a subsequence of both X and Y. String editing and LCS problems have been extcnsively studied, resulting in a copious literature for which we refer, e.g., to . When the size of the alphabet :E is unbounded, an D(IXllog IX)) lower bound for com-puting LCS applies, due to Hirschberg . The best known lower bound for bounded :E is linear. Aho, Hirschberg and Ullman showed that, for unbounded alphabets, any algo-rithm using only "equal-unequal" comparisons must take D(lXI 2 ) time in the worst case. The asymptotica]]y fastest general solution rests on the corresponding solution by Masek and Paterson [71 to the string editing, and hence takes time O(!XI'log log IXI/log IXI). In practice, the following 0(IXI x IYI) dynamic programming algorithm from Hirschberg is used. The algot'ithm starts with a matrix L[O ... [Yj, D... IXI] fi]]ed with zeroes, and then transforms L in such a, way that L[i,j] (1 .::; i .::; JYI, 1 .::; j .::; IXI) contains the length of an 2LCS between XIXZ ... Xi and VIYz ... Yh as follows. far i = 1 ta WI da far j = 1 ta IXI da if Xi # Yj then L[i,j] elseL[i,j] = L[i-l,j-1J + 1Max {L[i,j -1]'L[i -1,j)] The correctness of this paradigm follows from the following relations: T,[i -1,j] T,[i,j - lJ L[i-l,j-l] < L[i,j] < L[i - l,j] + I; < L[i,j] < L[i,j -1] + 1; < L[i,j] < L[i - l,j - I] +1. 3 Longest Common Subsequence - initial algorithm In this section, we present an algorithm for computing the longest common subsequence of run length encoded strings X = X 1X 2 ... XI and Y = Y1Y; ... Y;' in O(kl(k + I)) time. This algorithm maintains an I x k matrix M of blocks, such that 111[i,j] contains the value of an optimal solution between prefixes X(i) = X 1 X 2 ••• Xi and y(j) = yj y; ... l'i. The correctness of our algorithm follows because Ail contains all the essential information of the standard IXI x IYI alignment matrix L associated with the uncompressed strings. A .. __ .---A =- __ -= --= ,, ,= ciifiiiil =.;-~= :-==;.- ------~~---_ ... A ~:-~---.--.-•. g;g===:.~f-: :.:-:'~ ----.. __ :=: =.- - --- :~:.=--->-?-~~. j Figure 1: Light and dark blocks dcIined by strings X and Y. Figure 1 illustrates l.his matrix of blocks for input strings X = a 6 b3 as b3 and Y = a 3 b6 c1 a 4 • We say that block (i,j) is dark if the corresponding characters match, i.e. Xj = Vj. Block (i,j) is light if Xi f Vj. Any common subsequence defines a monotonically non-decreasing path fmm (0,0) to (IXI,IYI). Each rightward step on this path denotes the deletion of acharacter from X, and each downward step a deletion from Y. The matched characters in the common subsequence correspond to diagonal down-right steps across lVI, hence the LCS maximizes the total number of such diagonal steps through the dark blocks of J1tI. 3Any such path can exit a dark block in one of three ways - at the lower right corner, along the bottom side, or along the right side. The longest common subsequence of Figure 1(shown as the solid line), happens to enter and exit each dark block only through its corners. An optimal path with this additional constraint is computed easily in O(kl) by dynamic programming. However, paths which exit dark blocks through sides are more complicated to account for, since the llllTnber of possible exit points on either side of a block can dominate the number of blocks OIl very long runs. \"'le now consider t.wo special classes of paths across lVI. vVe define a corner path as one which enters dark blocks only at the upper-left corner and exits only through the lower-right corner. Vve say that a path beginning at the upper-left corner of dark block (i,j) is f01'ced if it exits through a side of (i,j), and proceeds to the next dark block by a straight horizontal or vertical "leap", according to the case. As illustrated by t.he dotted line in Figure 1, there is precisely one forced path beginning from the upper lefthand corner of any dark block. A subpath Pi ... Pi of path P is a contiguous chain of edges from P. Subpaths of forced and corner paths can be composed to define an interesting class of paths through iV!: Lemnla 1 l'he1'e is always a longest common subsequence vll of X and Y such that 111 18 defined by a path composed of subpaths of forced and corner paths. Proof: Consider any path through .M which defines the longest common subsequence of X and Y. \Ve now describe a sequence of transformations which reduce it to a path of the prescribed shape. First, consider any maximal subpath passing only through light blocks. Such a subpath consists only of l·ight.wa.rd and downward moves, for it contributes no matched charaders to the longest. common subsequence. Therefore, wit.hout loss of generality, all of the rightward moves can be collected t.o appear before any of the downward moves. Figure 2: Converting an arbitrary subpath into a forced subpath. Second, consider ilny maximal subpath through dark block (oi,j). This path cannot contain both a rightward and a downward move, since by replacing these with a diagonal 4move we increase the length of the putative longest common subsequence. Therefore, without loss of generality, all of the diagonal moves can be collected to appear before any of the verticalfhoriwntal moves. Finally, we consider Lhe dark blocks in Lhe order they are enconntered on the paLh from (0,0) to (IXI,IYI). Consider the first dark block which is either (1) not enLered through its uppcr-Iefthand corner or (2) is not exited through its lower-righthand corner. Case (1) cannot occur in a longest common subsequence, since the subsequence will be lengthened by entering in the upper-lefthand corner. Ca.sc (2) describes the start of .. t forced subpath, unless dark blocks are noL completely traversed. The reduction of Figure 2 converts this subpath into a forced subpath, thus giving the claimed result. I Theorenl 2 A longest common subsequence oJ Tun len!]th encoded strings X = X t X 2 ..• XI and Y = Yi Y2 ••• yj. can be computed in O(kl(k + 1)) time. Proof: Lemma 1 guara,ntees that a lOTlgcst common subsequence of X <md y' can always be obtained by concatenation of subpaths of forced and comer paths. The following algorithm exhaustively constructs all such subpaths via dynamic programming: LCSI(X, Y) l1t[[i,j] = 0, 1::; i ::; I, 1 $ j $ k for i = 1 to k for j = I 1.0 I if (color(i,j) == "lighl") t.hen M[i,j] = max(M[i -I,j], M[i,j - I]) else begin ( dark block ) d = min(IX;j, Wil) M[i,j] = max(M[i - I,j - I] +d, M[i,j], 111[i - I,j]' 111[i,j - 1]) ForcedPat.h\Jpdate(i, j, M) end The procedure ForcedPathUpdate explicitly traces out Lhe forced path originating at block (i,j), proceeding vertically if IX,. I > llil a.nd horizontally if IXiI < II'jl, until the next dark block (say (i',j)) is encountered. On exiting each dark block (i',j) along this forced path, Lhe block value is updated where !H'-[i',j] = max(M[i l ,j],.111[i,j] + dl), where ell is the diagonal length of the forced path through (il,j). This process continues unLil the forced path exits the cornel' of a block, or the end of one of Lhe strings is encountered. This ForcedPathUpdate operation can be computed in O(k + I) time for any block (i.,j). Each light block requires constant time to update, while each dark block takes O(k + I). The total time complexity follows since there are O(kl) dark blocks_ I 54 Longest Common Subsequence - a faster algorithm In this section we present an algori thm that computes the LCS of the run length encoded strings in O( kllog( kl)) time. In the previous algorithm, each iteration (i,j) was compnted in 0(1) if colol'(i,j) is "light.". When color(i,j) is dark the iteration computed Jl1[i,j] in 0(1) time before per-forming a ForcedPathlTpclate operation in O(k + I) time. In this scdioTl, we show how to replace this ForccdPathUpdate by a much morc efficient operation. The ForccdPathUpdate operation starts from (1:,j) and updates all JH[i',lls encountered 011 the way toward thE' lower right corner. Eventually, each dark J!I[i',J'] is updated hy all forced paths that cross its block. In this improved algorithm, the ForcedPaLh UpdaLe is eliminated. vVhile compnting l11[i,j]' only two [arced piLths from previous iterations will be considered, and their relevant values will be computed upon request. Figure 3: Two forced paths that match the character A. Lemma 3 All chamcfers which a1'C mnlched on any gi1Jen forced path will be identical. Also, two forced paths which p'/'Oceed on matches oj lhe srune charadeI' will nevel' (;ro-5S each other (Fig S). Consider a forced path that starts in an upper lefL comer of a dark block (i,j) that matches the character 0'. Its initial value v is 111[i - 1,j - 1], This path moves down and La the right in light blocks and diagonally on dark blocks that match 0' s. By Lemma 3, this path will not cross blocks that match characters other then CY. A record is kept for each forced path, including the following information: (a) (i,j) - starting location of the path; (b) the letter of the mi1.teh; and (c) its initial value v. Define '1'OPi(o:) to be the number of occurrences o[ the letter 0: in the uncompressed version of Xl", Xj, and LEFT'(o) to be the number of occurrences of the letter 0: in Y1 ••• 1-';'. For example. when string Y = aaaabbbbcccabbbbcc is encoded as a"'b-ld Ja l b4 c2 , 6LEP1'5(b) is 8. LEFT'-(O') will be defined only when Yi = 0' or }';-+1 = 0', and TOpi(a) defined only when X j = 0' or Xi+! = 0'. Consider a forced path which stmts at (i,j) and matches 0: with an initial value v. When this path crosses column j' > j its value will be Vi = V + TOpi'(a) - TOpi-l(a) (Fig 3). Moreover, it crosses column j' al row i"', where i- is the minimum row such that Similarly, when this ]l i, its valuc will be v' = v + LEFTi'(a) -LE1"'1'i-1(0:), and it crosses row i ' on column j- such that Lemm.a 4 Consider (l Jorced path which slaTls nl (i,j) nnrl mnlchcs 0' with an initial value v. Given a column j' (1'010 if), the value of each !on;ed path that (TOSSeS this column (row) can be computed in O( 1) time following O(k + l) time prepmcessing. Proof: By a performing a prefix-slITJI computation, the functions Lb'FT and TO P CiUJ be precomputed in O(k + I) time, such that TOpi(a) or LEF'l'i(a) can retrieved in consLant lime. The appropriate values can be computed using the formulas above. I As described in Section 3, l\l[i,j] is the maximurn of M[i - 1,j], l1I[i,j - 1] and the forced paths that cross its block, including the one that starts on iLs upper left corner. The set of forced p;.tLhs can he divided into two groups. The first group contains all paths Lhat cross column j above row i, while the second group cont.ains <tIl p<tths that cross row i left La column j. Our goal is to find the path with the highcst score in each group, so that M[i,j] can be computed in 0(1) time. Bclow, we discuss only how to find the highest in the rirst group, considering forced paths that match the character 0'; the second group and olher characters can be handled in the same way. Since two forced paths that match the same character never intersect, the forced paths of character 0: define a top-down order. vVe define the order of a path starting from 111[i,j] as ORDER(o.; i,j) = TO pi-lea) - LEFT i - 1 (a). The paths intersect any column j' <tccording to the value of ORDER. However, the values of Lhe forced paths at column j' do noL increase monotonically in their crossing order, bccause certain paths may hegin wilh lower ioi tial values, and they maintain the following property: Lemnla 5 Consider two f01'ced paths with values 'V~ and v~ when they CI'OSS column /, and V~' and v~ when they cross column iff. These paths mainlctin the equality: v~ - v~ = v:' - v~ (Fig 3). Therefore, if a forced path Pl intersect column i' lower than another forced path P2 and ils value on j' is smaller than the value of P2 on j', then path Ih can be deleLed. Om goal is to ma,intain, in order: only the paths which have higher values than the paths above them. A balanced binary search tree can be built with the records of the forced paths mat.ching 0', 7with the key of each pal.h defined by its ORDER function. This tree will be pruned so as to insure that for any given column j', the values of the paths in the nodes increase during an in-order traversal. vVe will maintain two balance binary sea.rch trees for each IcLter (x, one maintaining the order of paths crossiTlg columns, the other maintaining the order of paths crossing rows. These same two trees will be used in dealing with all dark blocks that match (X. For each such block lV1[i, j], we insert, separately, to both trees a new forced path that st.trts from the upper left corner of 111[i,j]. Then we get the highest scores crossing the lower a,nd right sides of JVJ[i, j], one from each tree. 'Vhen computing a dark block 1\11 [i, j] the following operations are performed: • Step 1. Insed a Tlew forced path. • Step TI. Find the highest score (0) of the forced paths on column j, ahove row i. • Step ITI. Find the highest score (R) of the forced paths on row i, left to column j. • Step TV. 111(i,jl = max(111[i - l,j],111[i,j - 11, C, R). Step [ - Insertion of a new path. (a) Compute ORDE R(a; i,j) := l'OPi-' (a) - LEFl'i-'(a). (b) Compute v:= 111[i -1,j -1]. (c) Insert the ncw path into the tree. (d) Compute the valuE' of the path that is stored in the leaf on the left. If its value is greater th<tn v delete the TlCW path. (e) Compute the value of the path that is stored in the leaf on the right. If its value is smaller than v delete Ute old path. Continue till you reach a path with a greater value. Step II - Finding the highest score of the forced paths on column j, above row i. (a) Compute 0:= TOpi(a) - LEFT'(a). (b) Find the locatioTl of 0 in the tree. (c) Compute the value 0, of the path that is stored in the leaf all the left. Step III is computed ill <tn analogous way to SLcp II. Theorem 6 A longest common subsequence of 1'1Ln length encod(;d strings X = X I X 2 .•• XI and Y = Y 1 Y 2 .. _Y k can be computed in O( kllog( k + I)) t.ime. Proof: The correctness of this procedure follows from the fact that all the relevant forced paths from the algorithm of Theorem 2 are evaluated in the dynamic programming phase of the current algorithm. The time complexity may be analyzed as follows. Precomputing the variables LEFT and 'FOP as in Lemma 4 takes O(k + I) time. Each of the 2· E balanced 8binary search trees has at most kl nodes, so any insertion, deletion or membership operation takes O(log( kl)) time. We perform Steps I to IV for each of the Jd blocks. Step I takes O(log( kl) + (number of deletions) log( kl)) time. Since each deleted block mnst previously have been inserted, the total number of deletions is O(kl). Steps 11 and lIT are computed in O(1og(kl)) while Step IV requires 0(1) time. Therefore, O((kl) log(kl)) time suffices to compute the longest common subsequence of X and Y. 5 Open Problems vVhat can be said abont more general versions of string matching, in particular edit distance computations? References A.V. Aho, D.S. Hirschberg and J.D. Ullman. Bounds on the complexity of the longest common subsequence problem. J. Assoc. Com]lul. l}[ach., 23 pp. 1-12 (HJ76). A. AposLolico. String editing and longest common subsequences. Ilandbook oj FOTlned Languages (G. Rozenbc1'9 and 11. Salomaa, Eds.), Vol II, pp. 361-398, Springer-Verlag (1996). H. Uunke, and J. Csirik. An improved algorithm for computing the edit distance of run-length coded strings. JPL, 54, 1'1'. 93-96 (1995). (4] D.S. Hirschberg. An infonnation theoretic lower bound for the longest common subse-quence problem. TPTJ 7,1, pp. 40-41 (1978). [5J D.S. Hirschberg. !\ linear space algorithm for computing maximal common subse-quences. Com11wniccttions of the CA Ci1l1 18, 6, pp. 311-3/1a (1975). D.S. Hirschberg, fllld L.L. Larmore. The set-set LCS problem. AI.qol'ifhmica, 4, PI'. 503-510 (1989). [7J vV.J. Masek, andlVI. S. Paterson. A faster algorithm compnting string edit distances. J. Comput. Sysfe/ll Sci., 20, pp. 18-31 (19S0). C. Sazaklis, E. Arkin, J. Mitchell, and S. Skiena. Geometric decision trees for optical character recognitioll. submitted for publication, 1996. B. Wang, G. Chen. and K. Parlc On the set LCS and set-set LCS prohlems. J. of AIgo,.ilhms, 14, Pl'. '166-177 (1993). 9
2676	https://bmcmedresmethodol.biomedcentral.com/articles/10.1186/s12874-025-02611-4	BMC Medical Research Methodology Over- and under-estimation of vaccine effectiveness Download PDF Download PDF Research Open access Published: Over- and under-estimation of vaccine effectiveness Hilla De-Leon1,2 & Dvir Aran1,3 BMC Medical Research Methodology volume 25, Article number: 163 (2025) Cite this article 556 Accesses 1 Altmetric Metrics details Abstract Background The effectiveness of SARS-CoV-2 vaccines against infection has been a subject of debate, with varying results reported in different studies, ranging from 60–95% vaccine effectiveness (VE). This range is striking when comparing two studies conducted in Israel at the same time, as one study reported VE of 90–95%, while the other study reported only ~ 80%. We argue that this variability is due to inadequate accounting for indirect protection provided by vaccines, which can block further transmission of the virus. Materials and methods We developed a novel analytic heterogenous infection model and extended our agent-based model of disease spread to allow for heterogenous interactions between vaccinated and unvaccinated across close-contacts and regions. We applied these models on real-world regional data from Israel from early 2021 to estimate VE using two common study designs: population-based and secondary infections. Results Our results show that the estimated VE of a vaccine with efficacy of 85% can range from 70–95% depending on the interactions between vaccinated and unvaccinated individuals. Since different study designs capture different levels of interactions, we suggest that this interference explains the variability across studies. Finally, we propose a methodology for more accurate estimation without knowledge of interactions. Discussions and conclusions Our study highlights the importance of considering indirect protection when estimating vaccine effectiveness, explains how different study designs may report biased estimations, and propose a method to overcome this bias. We hope that our models will lead to more accurate understanding of the impact of vaccinations and inform public health policy. Clinical trial number Not applicable. Peer Review reports Introduction Hundreds of millions of individuals have been infected with the SARS-CoV-2 virus, and millions of them died because of the infection since it was first detected in China at the end of 2019. In response, several companies and research institutions developed vaccines to help suppress the pandemic, and less than a year after identifying of the virus, several clinical trials reported results of proven vaccine efficacy [1, 2]. The randomized clinical trials of the Pfizer/BioNTech and Moderna vaccines showed vaccine efficacy of around 95% against SARS-CoV-2 infection in the first months after the second dose; however, the studies had relatively low numbers of infected individuals, and this level of efficacy can only be regarded as a proxy to the real-world vaccine effectiveness (VE) across divergent populations, demographics, and clinical features. By the end of 2020, many countries launched mass vaccination campaigns. The data collected through these mass vaccination campaigns was used to evaluate the effectiveness of the vaccines in a wide range of populations 3,4,5,6,7,8,9,100 . PMID: 34355689; PMCID: PMC8343550."),[11:2100452. . PMID: 34142647; PMCID: PMC8212595."),12:396."),13:1431–3.")]. Individual-level study designs for VE evaluation can be split into two main approaches: [1:2603–15.")] A population-based approach (PB), where the number of vaccinated and unvaccinated infections are counted after correcting for possible confounders [3:1412–23.")]; [2:403–16.")] A secondary infections approach (SI), which allows to overcome a major limitation of the first approach of unknown exposure. In this approach, infections are counted after a known exposure, such as secondary household infections [4:e734–40.")]. Interestingly, when examining the VE estimations from studies that use each approach, there are major differences. In PB studies, VE was usually estimated at 90%−95% [1:2603–15."),2:403–16."),3:1412–23."), 5], while in SI studies, much lower VE estimations are usually reported (60%−80%) [4:e734–40."), 8:5456–60."), 10:2100640. . PMID: 34355689; PMCID: PMC8343550.")]. Moreover, observational studies in nursing home residents reported varying VE levels ranging from 53 to 92% against SARS-CoV-2 infection [6 Variant — National Healthcare Safety Network, Mar 1–Aug 1, 2021. MMWR Morb Mortal Wkly Rep. 2021;70(34):1163–6."), 11:2100452. . PMID: 34142647; PMCID: PMC8212595."), 12:396.")]. The difference between the two effectiveness levels is emphasized when we compare the 90%−95% effectiveness of Dagan et al. [3:1412–23.")] with the 80% effectiveness of Gazit et al. [4:e734–40.")]. Both studies evaluated the effectiveness of the same vaccine in a similar population in Israel at the same time, and only the study design differed. Multiple factors likely contribute to the observed differences in vaccine effectiveness estimates across studies. Household studies often detect milder or asymptomatic infections that may be missed in population-based studies focusing on medically attended cases, potentially leading to lower VE estimates in household settings. Additionally, vaccine-induced immunity wanes over time, which can result in differing VE estimates depending on the time horizon of the study [14, 15]. The circulation of different SARS-CoV-2 variants during different study periods can also affect VE estimates, as vaccines may offer varying levels of protection against different variants . Furthermore, population differences in age structure, comorbidity prevalence, and prior infection history across studies could influence the observed effectiveness . However, these explanations cannot fully account for differences that persist when studies are conducted in similar populations during the same time period with the same circulating variants, as in the case of Dagan et al. and Gazit et al. We hypothesize that the main reason for this specific gap can be explained by the proportion of interactions of vaccinated individuals with other vaccinated individuals that diverges between the two settings. This factor—the interaction patterns between vaccinated and unvaccinated individuals—has received less attention in the literature but may play an important role in explaining these persistent differences. VE estimates can be influenced not only by the direct protection vaccines provide to individuals but also by indirect effects that emerge at the population level. These indirect effects occur when vaccinated individuals reduce onward transmission, effectively shielding others from infection [17, 18]. The magnitude of such effects and how they impact VE estimates may vary substantially depending on interaction patterns between vaccinated and unvaccinated individuals. Our hypothesis is based on one feature of the vaccines that has been broadly neglected in current VE studies. While the vaccines were designed mainly to reduce symptomatic disease, severe disease, and death, studies from 2021 have shown that they also provide efficient transmission-blocking. A study from Sweden found that the number of immune members in a family was negatively correlated with the likelihood of incidence of infection of non-immune family members . Similarly, a study from Israel found reduced infection rates in communities as vaccination rates increased . Previous studies before COVID-19 have shown how interference, the potential outcomes of one individual are affected by the treatment assignment of other individuals, can affect VE estimations . It seems that an underlying assumption made by the COVID-19 VE studies is that there is no interference, which is an unfounded assumption. Previous studies presented causal inference frameworks with interference to deal with the effects of indirect protection on VE estimations [21,22,23,24,25]. The main novelty of this work stems from our attempt to use a numerical model that incorporates real-world data on vaccination and infection rates to demonstrate how different study designs may lead to divergent estimations of VE. Here, we developed both an analytical approach and a numerical approach which utilizes our Monte Carlo Agent-based Model (MAM) to examine the influence of interference between vaccinated and unvaccinated individuals on data of COVID-19 vaccinations and infections in Israel . Our analyses show that different assumptions may lead to significant under- or over-estimations and might explain the differences in estimations between the different study design approaches. We suggest that this divergence is inversely related to the rate of mixing between vaccinated and unvaccinated individuals, in line with previous studies on interference [22, 23, 25]. Since it is not simple to estimate the level of mixing, which diverges across time and communities, we developed a simple approach that can overcome this issue and provide accurate estimations of VE regardless of mixing levels. Methods The main aim of this paper is to quantify VE estimations in terms of interference between vaccinated and unvaccinated individuals. Interference occurs when the treatment received by other individuals influences the effects of an exposure or intervention on an individual. Because interference can affect the relationship between exposure and outcome, it can be harder to determine causality. In terms of this work, we define interference as the effect on the risk of infection for a vaccinated individual due to exposure to an unvaccinated individual. Our model focuses specifically on VE against SARS-CoV-2 infection as the primary outcome. This focus allows for direct comparison with the studies by Dagan et al. and Gazit et al., which both reported effectiveness against infection. We acknowledge that VE can vary substantially depending on the outcome of interest (infection, symptomatic disease, severe disease, hospitalization, or death), with effectiveness typically being higher against more severe outcomes [1, 2]. The interaction patterns we describe would likely affect estimates of effectiveness against all outcomes, though the magnitude of bias might differ. For instance, if vaccinated individuals maintain some protection against severe outcomes even when breakthrough infections occur, the bias in VE estimates for severe outcomes may be less pronounced than for infection. We also note that our analytical model presents a simplified approach to illustrate the core mechanism of bias due to non-random mixing. Real-world VE studies typically employ various adjustment methods to address potential confounding factors, such as matching, stratification, or regression adjustment for demographic and clinical characteristics. Our model does not explicitly incorporate these adjustment strategies, as our focus is on demonstrating how interaction patterns—which are rarely measured or adjusted for in practice—can influence VE estimates even when other confounders are properly addressed. The basic mechanism of bias we describe would persist in adjusted analyses if the underlying interaction patterns differ systematically between study designs or populations. All the data and code for modelling and figures presented in this manuscript are available on: Analytical infection model First, we developed a simple heterogenous analytical infection model to find a quantifiable relationship between the amount of interference in a network and the effectiveness of vaccines (see Supplementary Methods; Table S1-2 for more information). Within the model, we define a network of ({N}_{all}={N}_{in}+{N}_{out}) vaccinated and unvaccinated nodes. Each node is strongly connected to all the nodes in its clique (({N}_{in}-1)) with a value of 1, and weakly connected to all ({N}_{out}) other nodes with a value of (\delta), which is the ratio between the number of infections that originate from the inner clique and the other infections. Using the model we can define the observed vaccine effectiveness, (\widehat{VE}), as a function of the average fraction of edges with unvaccinated nodes to the vaccinated nodes (\langle I\rangle) (see Supplementary Methods): $$\widehat{VE} =VE\times \frac{{N}_{in}{\prime}\cdot \left(1+\delta \right)+VE\cdot \delta {\prime}\cdot \nu \cdot {N}^{all}+VE\cdot \langle I\rangle \cdot {N}_{in}{\prime}\cdot \left(1-\delta {\prime}\right) }{{N}_{in}{\prime}\cdot \left(1+\delta \right)+VE\cdot \delta {\prime}\cdot \left(\nu \cdot {N}_{all}-1\right)+VE\cdot \left[1+\langle I\rangle \left(1-\frac{1}{\nu }\right)\right] \cdot {N}_{in}{\prime}\cdot \left(1-\delta {\prime}\right)}$$ (1) With (\nu) as the fraction of unvaccinated people in the total population, ({\delta }{\prime}=\delta \frac{{N}_{in}-1}{{N}_{all}-{N}_{in}}) and ({N}_{in}{\prime}={N}_{in}-1). Equation 1 quantifies how the observed vaccine effectiveness ((\widehat{VE})) relates to the true VE as a function of the interaction patterns between vaccinated and unvaccinated individuals. The key parameter in this equation, (\langle I\rangle), represents the average fraction of interactions that vaccinated individuals have with unvaccinated individuals. Intuitively, the equation captures how indirect protection modifies observed effectiveness: when (\langle I\rangle) is low (vaccinated individuals mostly interact with other vaccinated individuals), the observed effectiveness is higher than the true effectiveness because vaccinated individuals benefit from additional indirect protection; when (\langle I\rangle) is high (vaccinated individuals mostly interact with unvaccinated individuals), the observed effectiveness is lower than the true effectiveness because vaccinated individuals face higher exposure while simultaneously providing protection to the unvaccinated. The parameter (\delta) in the equation represents the ratio between infections occurring inside a person's close contact network versus outside of it, with (\delta) = 1 indicating that approximately half of all infections occur through close contacts (such as within households). The parameter (\nu) represents the fraction of unvaccinated individuals in the population, which influences the baseline probability of interaction with unvaccinated individuals. The terms ({N}_{in}{\prime}) and (\delta {\prime}) are derived constants related to network structure that help quantify how interaction patterns affect transmission dynamics. To summarize, this analytical model is a relatively simple model that simulates infections in a population with non-uniform vaccination rates and different transmission risk levels across various types of interactions. It distinguishes between high-risk interactions (such as within households) and lower-risk interactions (such as casual community contacts) through parameters ({N}_{in}) and (\delta). The model considers the risk of infection in vaccinated households and allows us to quantify the population (\widehat{VE}), corrected by the interference, as opposed to the individual-level VE. MAM – a flexible Monte-Carlo agent based model To model the infections occurred in the real-world we use MAM, a flexible Monte-Carlo Agent based Model for modelling COVID-19 spread that accurately predicts infection dynamics in Israel . In this work, we expand this model to include complex interaction networks, such as differing between household interactions, which represent around half of infections in Israel , and other interactions. The spatial feature of MAM allows it to be used to predict both the local and global spread of COVID-19 by simultaneously modeling spatial interactions between families (high interaction), neighborhoods (medium interaction), and cities (low interaction). By applying MAM to public data from Israel, we could model the actual spread of COVID-19 during Israel’s first vaccination campaign for 9.2 million particles across 1,578 separate communities, where each community has its own vaccination rate. Data description and characteristics Real-world data from Israel was collected during the early phase of its vaccination campaign from December 2020 to March 2021. We obtained publicly available data from the Israeli Ministry of Health that included daily COVID-19 vaccination rates and confirmed cases for 1,578 distinct statistical areas (communities) across Israel . Each statistical area represents approximately 5,000 individuals, providing high-resolution data on the heterogeneity of vaccination uptake. The data reveals substantial variation in vaccination rates across communities, with some communities quickly reaching high vaccination coverage while others lagged significantly. This heterogeneity in vaccination rates across communities created natural variation in interaction patterns between vaccinated and unvaccinated individuals, which our model leverages to understand the impact on vaccine effectiveness estimates. The vaccination campaign in Israel initially prioritized older individuals and gradually expanded to younger age groups. By February 2021, a significant portion of adults over 60 had been vaccinated, while vaccination coverage in younger age groups remained lower. This age-stratified vaccination pattern, combined with demographic variations across communities, further contributed to heterogeneous mixing patterns between vaccinated and unvaccinated individuals. Results Most transmissions of the virus leading to infection are in close interactions. These interactions are often cliques, where all individuals in the clique have close interaction with each other (i.e., households). Heterogenous vaccination rates between those cliques bias the estimations of VE. Consider this simple example: two nursing homes, each with 60 people that all interact with each other. Assuming that in one home 50 people are vaccinated with a vaccine with 80% efficacy and the rest are unvaccinated, while in the other it is the opposite (10 vaccinated). Both homes had an outbreak, in the first house one vaccinated and one unvaccinated were infected, which is what we expect with 80% protection. In the second home 25 unvaccinated and one vaccinated were infected, again translates to 80% protection. However, if we calculate (\widehat{VE}) for both homes together (120 people in total), we substantially overestimate (\widehat{VE}) at 92.3%, which is 2.6-fold risk reduction. Similar analysis can be performed with secondary infection models to show any desired level of observed (\widehat{VE}). We present here an analytical framework to model this observation. Analytical infection model For illustration purposes, we present three possible mixtures described using three networks. In all networks, there is a similar amount of vaccinated and unvaccinated nodes, and there are precisely four close connections for each node (({N}_{in}=5), ({N}_{out}=20)) (Fig. 1A-C). In each network all the nodes are numbered, such that for the ({i}^{th}) node we indicate with vector ({A}^{i}) its vaccination status and the number of connections it has with the unvaccinated nodes in the network (see Supplementary Methods). The only difference between the three networks is the dispersal of the vaccinated and unvaccinated nodes: in network (A), there is complete segregation between vaccinated and unvaccinated nodes ((\langle I\rangle =0)). This network might simulate nursing homes that were vaccinated early, while most of the rest of the population was not yet vaccinated. In network (B), for each vaccinated node, half of the edges are to vaccinated nodes and half to unvaccinated nodes ((\langle I\rangle =0).5). In network (C), each vaccinated node is connected to one vaccinated node and three unvaccinated nodes ((\langle I\rangle =0).75). Network (C) might simulate households where two members are vaccinated (e.g., parents) and three are not, e.g., children. The ({A}^{i}) values for each node in the networks are provided in the Supplementary Methods (Table S3). For each of the networks we count the average fraction of edges with unvaccinated nodes to the vaccinated nodes,(\langle I\rangle), and calculate (\widehat{VE}\left(\langle I\rangle ,\delta \right)) for (VE) = 0.85 and (\nu =0.6) (proportion of the population that are unvaccinated), and (\delta =1) (ratio between inner clique and other infections, thus 1 represents that half of the infections are household infections). Using the equations described in the analytical infection model, we find that the risk of infection in (B) for vaccinated nodes is 0.072 and 0.52 for unvaccinated nodes, which translates to(\widehat{VE}=0.86), similar to(VE). However, in (A) and (C) we see a diversion from(VE): in (A) risk for vaccinated is 0.046 and 0.63 for unvaccinated, which translates to an overestimation of almost double reduced risk ((\widehat{VE}=0.93)), and in (C) these numbers are 0.1 and 0.4, respectively, which translate to an underestimation of VE ((\widehat{VE}=0.75)). Generally, we observed that for (\langle I\rangle \text{around}) 0.6 (VE) is similar to (\widehat{VE}) (Fig. 1D and Eq (S-18) in the Supplementary Methods), which notably corresponds to the proportion of unvaccinated individuals in our model population ((\nu =0.6)). This relationship is not coincidental—our analytical model demonstrates that unbiased VE estimation occurs when ⟨I⟩ ≈ ν, representing a scenario where vaccinated individuals interact with unvaccinated individuals at a rate proportional to their prevalence in the population. Figure 1D shows (\widehat{VE}\left(\langle I\rangle ,\delta \right)) for (0\le \langle I\rangle \le 1) and (0.2\ge \delta \ge 2) Lower (\langle I\rangle) levels produce overestimations of (\widehat{VE}) up to 95%, which occurs when vaccinated individuals interact primarily with other vaccinated individuals, thereby receiving additional indirect protection. Conversely, higher (\langle I\rangle) levels produce underestimations towards 70%, as vaccinated individuals interact disproportionately with unvaccinated individuals, increasing their exposure risk while simultaneously providing indirect protection to the unvaccinated population, which all stem from the same VE value ((85\%)). Numerical analysis: using MAM to estimate vaccine effectiveness in Israel We next attempted to model the infections occurred in real-world and show how calculations of VE can be skewed based on different assumptions regarding (\langle I\rangle). Using MAM, we simulated the spread of COVID-19 infections in Israel from the beginning of the vaccination campaign on December 20, 2020. The particles were numbered and represented by the vector ({A}^{i}), similarly to the analytical model. However, since this is a time-dependent model, and the model predicts the spread of infection while the population is actively being vaccinated, we use ({A}^{i}\left(t\right)) where the vaccination rate is determined by the actual vaccination rates across 1,578 communities in Israel (Figure S1). We created ten different mixing scenarios between vaccinated and unvaccinated individuals in each community. For each scenario, we created a different array of ${A}^{i}\left(t\right)$ according to the real vaccination rates, which differ from each other in their ${I}^{i}(t)$, which represents the level of close interactions of a vaccinated individual $i$ with unvaccinated individuals as a function of time $t$ (Supplementary Methods). We simulated different scenarios of infections throughout February 2021. In each date we calculated the observed vaccine effectiveness(, \widehat{VE}), using both the population-based approach (PB) and the secondary infection approach (SI). In the PB-simulated analysis, we performed a simple adjustment to eliminate the heterogeneity in vaccination rates among communities by matching the number of vaccinated and unvaccinated in each community, which can be seen as 1:1 matching. In the SI-simulated analysis we only considered infections within the close-contact circuit, which is similar to a situation of (\delta =0) in the analytical model (Supplementary Methods). Thus, low (\langle I\rangle) represents a scenario where most close-contacts circuits are partially vaccinated (i.e., a family with only one parent vaccinated), and high (\langle I\rangle) represents a scenario where most close-contact circuits are either fully vaccinated or not vaccinated at all. In all analyses, we observed clear negative correlations between (\widehat{VE}) and(\langle I\rangle), and (\widehat{VE}) ranged between ~ 70% and ~ 95% (Fig. 2A). We can see that the input (VE) was only obtained for (\langle I\rangle) at levels around 0.5–0.6 for both the PB and SI simulations. However, the PB-based Dagan et al. study probably had an (\langle I\rangle) of ~ 0.25, which is in line with the lockdown that was imposed at this time, while the SI-based Gazit et al. study probably had an (\langle I\rangle) of ~ 0.75, which is in line with the limited availability of vaccinations to adults only at this time. As we can see, even careful matching would be insufficient to retrieve the actual(VE). This is due to the additional circuits we added in the modelling that aim to mimic household contacts. This creates heterogeneity of (\langle I\rangle) at the close contacts-level, not just at the community-level. To overcome this issue, we derived an additional analysis that considers the level of mixing. Since it is not simple to adjust for different (\langle I\rangle) levels in real-world data for PB analysis, we performed the analysis by including only close-contact circuits with over 50% vaccinated individuals. This analysis showed that for PB, it is possible to achieve relatively accurate estimations of (VE) with this relatively simple approach, regardless of (\langle I\rangle) and time (Fig. 2A and Figure S2). Of note, in SI analysis, both (\langle I\rangle) and (\nu) can be theoretically obtained for each family (and (\delta =0)), therefore, (\widehat{VE}) can be adjusted more easily in real-world data (full analysis for (\widehat{VE}) for PB and SI is given in the Supplementary Methods). Finally, we analyzed how (\widehat{VE}) changed over time in the different (\langle I\rangle) scenarios (Fig. 2B). Interestingly, even for (\langle I\rangle \approx 0.5), where the average (\widehat{VE}) in February was ~ 85% (PB method; Fig. 2A), we observed underestimation in early February and overestimation in late February. In general, we noticed that as vaccination rates increased with time, (\widehat{VE}) estimations increased accordingly, in line with early assessments of VE in Israel compared to later assessments. This temporal relationship between the distribution of (\langle I\rangle) and the vaccination rate adds another layer to the complexity of VE estimation and can explain divergent results across studies that were performed in different time points. Discussion and conclusions We demonstrated here that VE estimations are sensitive to the dispersion of vaccinations across the population. We found that the level of mixing between vaccinated and unvaccinated individuals can alter the observed VE. High level of mixing creates indirect protection to unvaccinated individuals and on the other hand increases the risk of infection for vaccinated people, since they frequently interact with unvaccinated people, which in turn leads to underestimation of VE, while low level of mixing leads to additional protection to the already protected individuals, and in turn to overestimation of VE. Regardless of what is the real mixing between vaccinated and unvaccinated, which we assume is far from random, it is clear from our analyses that the observed VE can significantly differ from the real individual-level protection offered by the vaccine. An important finding from our analysis is that the relationship between the interaction patterns and VE estimation bias is directly tied to the proportion of unvaccinated individuals in the population. Specifically, we demonstrate that when vaccinated individuals interact with unvaccinated individuals at a rate precisely matching their prevalence in the population, the estimated VE closely matches the true VE. This can be understood intuitively: under such random mixing conditions, neither group receives disproportionate indirect protection or exposure risk. If population-level vaccination coverage changes—for instance, as vaccination campaigns progress—the threshold for unbiased effectiveness estimates would shift accordingly. This relationship highlights the dynamic nature of estimation bias in real-world settings where vaccination rates continually evolve, and emphasizes the importance of considering interaction patterns when interpreting VE estimates from different time periods or populations with varying vaccination coverage. This study is subject to several limitations. First, this work did not consider that part of the population is protected by previous infection (i.e., recovered). However, the infections data from Israel indicates that by November 2020, a relatively low proportion of the population (no more than 5%) had been infected and recovered. It may affect the assumptions and calculations in this work, but this oversight should result in only a minor error, since we assume that some of the recovered population is also vaccinated. Second, it is not simple to estimate the level of mixing in the population. However, we note that during the time of first estimations of VE coming from Israel , Israel was mostly under a lockdown that was imposed between January 8, 2021, and February 9, 2021. Under the lockdown there were limited interactions overall, which enable us to assume relatively low interaction within the population in Israel and especially low number of interaction between the vaccinated elderly population and the rest of the population which were vaccinated later. It is therefore safe to assume that (\langle I\rangle) at the time of the study was trending towards 0, or at least much lower than later on. Based on this, we argue that the high VE reported 95% for adults over 70 years old is substantially overestimated. Furthermore, the lower VE reported for the 40–69 years old group in that study (90%) is a result of the higher mixing rate of that group. Later on, Israel imposed “green passports”, which limit entrance of unvaccinated individuals to crowded places. This is another measure that reduces vaccined-unvaccined interactions and may again skew towards overestimation of VE. On the other hand, we argue that study designs of secondary household infections underestimate VE. Gazit et al. used this approach to estimate VE in Israel and found VE to be 80% . In households, especially in a country with relatively big families as in Israel, most interactions of vaccinated individuals are with unvaccinated children that were not vaccinated at the time of the study. This translates to high (\langle I\rangle) levels, and as we show this leads to significant underestimation. We note that some of the lower VE in household infections can be attributed to the prolonged exposure, and in theory, the vaccination is less effective in such scenario. We are aware that the number of family members in Israel is not uniform, however, since the average number of children in Israel is around 3 , assuming five people per household is a reasonable assumption. Our findings align with the literature on indirect vaccination effects and can be understood within established frameworks of direct, indirect, overall, and total vaccine effects [21, 30]. What we describe as 'overestimation' when vaccinated individuals primarily cluster together corresponds to measuring overall effectiveness rather than direct effectiveness alone. In these scenarios, vaccinated individuals benefit from both direct protection and indirect protection from their vaccinated social environment. Conversely, 'underestimation' occurs when vaccinated individuals frequently interact with unvaccinated individuals, effectively serving as 'shields' for the unvaccinated while simultaneously facing increased exposure risk themselves. Our study is not the first to caution about VE estimations confounded by indirect protection. Previous studies developed a causal inference framework with interference to deal with this issue [19,20,21,22,23]. These frameworks specifically address how the potential outcomes of one individual are affected by the treatment assignment of other individuals. Interestingly, although this issue is well known, in our literature review of VE studies for COVID-19, we did not identify any study that takes this issue into account. Recent work by Lin et al. further quantifies these vaccination effects in SIR models, demonstrating how indirect effects vary with both basic reproduction number and vaccination coverage. While multiple factors contribute to varying effectiveness estimates across studies—including detection of different infection severities, vaccine waning, and variant escape—we argue that interaction patterns represent an important factor that has received less attention in applied vaccine studies. What distinguishes our work is our demonstration that different study designs inherently capture different mixing patterns, leading to systematic differences in effectiveness estimates even when measuring the same outcome in the same population at the same time. We believe our analytical model provides a novel and relatively simple framework for dealing with vaccine interference, and, as we show here, can be implemented in real-world settings. The model is structured to differentiate between different types of interactions (close and remote), thereby better representing real-world infection networks. While our analytical model demonstrates how interaction patterns can bias VE estimates, we recognize that real-world studies employ various methodological approaches to address confounding factors. Population-based studies typically use matching or statistical adjustment for age, sex, comorbidities, geographic location, and timing of vaccination to ensure that vaccinated and unvaccinated groups are comparable . Similarly, household studies may adjust for household size, age of household members, and other factors that could influence transmission dynamics . However, these standard adjustment methods typically do not account for the interaction patterns we describe. Even with perfect matching on demographic and clinical characteristics, if vaccinated individuals systematically differ from unvaccinated individuals in their interaction networks—which is likely, given clustering of vaccination status within households and communities—VE estimates may still be biased. The underlying mechanism of bias due to non-random mixing would persist in adjusted analyses if mixing patterns are not directly measured and accounted for. Our simplified analytical model isolates this effect, enabling researchers to understand and potentially correct for it in future studies. In conclusion, we argue here that observed VE estimations of 90%−95% and 80% stems from a vaccine that probably provides individual-level protection of around 85%. We suggest that by adjusting to the fraction of unvaccinated individuals in a household, in addition to demographic and clinical features, it is possible to reduce the bias when estimating VE. We believe that the implications of this study extend beyond COVID-19 to vaccination programs broadly, including routine childhood immunizations and seasonal influenza campaigns. By revealing how social interaction patterns systematically bias effectiveness estimates, these findings contribute to more accurate evaluation of protection across all vaccines where family or community vaccination decisions tend to cluster together. These insights may help public health officials better interpret effectiveness data, design more targeted interventions, and communicate more precisely about vaccine benefits at both individual and population levels. We hope that future studies will consider the possible bias we highlight in this study and attempt to correct for it, as the rate of effectiveness of the vaccines have high impact on strategies for controlling the spread of infectious diseases and optimizing public health resources. Data availability No datasets were generated or analysed during the current study. References Polack FP, Thomas SJ, Kitchin N, Absalon J, Gurtman A, Lockhart S, et al. Safety and efficacy of the BNT162b2 mRNA Covid-19 Vaccine. N Engl J Med. 2020;383(27):2603–15. Article CAS PubMed Google Scholar 2. Baden LR, el Sahly HM, Essink B, Kotloff K, Frey S, Novak R, et al. Efficacy and safety of the mRNA-1273 SARS-CoV-2 Vaccine. N Engl J Med. 2021;384(5):403–16. Article CAS PubMed Google Scholar 3. 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First indication of the effect of COVID-19 vaccinations on the course of the outbreak in Israel. MedRxiv. 2021. 37. Caspi G, Dayan A, Eshal Y, Liverant-Taub S, Twig G, Shalit U, et al. Socioeconomic disparities and COVID-19 vaccination acceptance: a nationwide ecologic study. Clin Microbiol Infect. 2021;27(10). Article CAS PubMed PubMed Central Google Scholar Download references Acknowledgements DA is supported by the Azrieli Faculty Fellowship and is a Deloro Fellow. We thank Yaniv Erlich, Barak Rave and Michael Geller for fruitful discussions. Funding No specific funding. Author information Authors and Affiliations Faculty of Biology, Technion-Israel Institute of Technology, Technion-Israel Institute of Technology, 21 Sderot Rose, Haifa, 3095230, Israel Hilla De-Leon & Dvir Aran 2. Current address: Public Health Services, Ministry of Health, Jerusalem, Israel Hilla De-Leon 3. The Taub Faculty of Computer Science, Technion-Israel Institute of Technology, Haifa, Israel Dvir Aran Authors Hilla De-Leon View author publications Search author on:PubMed Google Scholar 2. Dvir Aran View author publications Search author on:PubMed Google Scholar Contributions H.D.L. and D.A. conceived and designed the study. H.D.L conducted all the calculations needed for this study. Both H.D.L and D.A wrote the manuscript. D.A. supervised the study and edited the manuscript. Both authors reviewed and approved the final manuscript. Corresponding author Correspondence to Dvir Aran. Ethics declarations Ethics approval and consent to participate Not applicable. Consent for publication Not applicable. Competing interests The authors declare no competing interests. Additional information Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Supplementary Information 12874_2025_2611_MOESM1_ESM.pdf Supplementary Material 1. [31,32,33,34,35,36,37]. Rights and permissions Open Access This article is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, which permits any non-commercial use, sharing, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if you modified the licensed material. You do not have permission under this licence to share adapted material derived from this article or parts of it. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit Reprints and permissions About this article Cite this article De-Leon, H., Aran, D. Over- and under-estimation of vaccine effectiveness. BMC Med Res Methodol 25, 163 (2025). Download citation Received: Accepted: Published: DOI: Share this article Anyone you share the following link with will be able to read this content: Sorry, a shareable link is not currently available for this article. Provided by the Springer Nature SharedIt content-sharing initiative Keywords Vaccine effectiveness COVID-19 Interference Population-based studies Agent-based modeling Indirect protection BMC Medical Research Methodology ISSN: 1471-2288 Contact us General enquiries: journalsubmissions@springernature.com
2677	https://pmc.ncbi.nlm.nih.gov/articles/PMC6844175/	Secondary and Tertiary Hyperparathyroidism in Chronic Kidney Disease: An Endocrine and Renal Perspective - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide New Try this search in PMC Beta Search View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice Indian J Endocrinol Metab . 2019 Jul-Aug;23(4):391–399. doi: 10.4103/ijem.IJEM_292_19 Search in PMC Search in PubMed View in NLM Catalog Add to search Secondary and Tertiary Hyperparathyroidism in Chronic Kidney Disease: An Endocrine and Renal Perspective Manju Chandran Manju Chandran 1 Department of Endocrinology, Osteoporosis and Bone Metabolism Unit, Singapore General Hospital, Singapore Find articles by Manju Chandran 1,✉, Jiunn Wong Jiunn Wong 1 Department of Renal Medicine, Singapore General Hospital, Singapore Find articles by Jiunn Wong 1 Author information Copyright and License information 1 Department of Endocrinology, Osteoporosis and Bone Metabolism Unit, Singapore General Hospital, Singapore 1 Department of Renal Medicine, Singapore General Hospital, Singapore ✉ Address for correspondence: Dr. Manju Chandran, Department of Endocrinology, Osteoporosis and Bone Metabolism Unit, Singapore General Hospital, 20 College Road, Academia, 169856, Singapore. E-mail: manju.chandran@singhealth.com.sg Copyright: © 2019 Indian Journal of Endocrinology and Metabolism This is an open access journal, and articles are distributed under the terms of the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 License, which allows others to remix, tweak, and build upon the work non-commercially, as long as appropriate credit is given and the new creations are licensed under the identical terms. PMC Copyright notice PMCID: PMC6844175 PMID: 31741895 Abstract Secondary Hyperparathyroidism (SHP) seen as a frequent complication in Chronic Kidney Disease (CKD) has many pathogenetic peculiarities that are still incompletely defined and understood. During the long course of chronic renal failure, SHP can also transform sometimes into the hypercalcemic state characterized by quasi-autonomous production of Parathyroid Hormone from the parathyroid glands: a disorder that is termed Tertiary Hyperparathyroidism. The clinical consequences of SHP in CKD are protean, encompassing bone and mineral abnormalities but as recently identified, also several metabolic and cardiovascular problems, the most important of which is vascular calcification. There have been several advances in the therapeutic armamentarium available for the treatment of SHP, though clear demonstration of a benefit regarding major clinical outcomes with any of the new agents is still lacking. This narrative review summarizes the current understanding about this disorder and highlights some of the recent research on the subject. Keywords: CKD, FGF-23, parathyroid hyperplasia, renal failure, secondary hyperparathyroidism, tertiary hyperparathyroidism I NTRODUCTION The parathyroid gland(s) play a pivotal role in bone mineral homeostasis through its secretion of parathyroid hormone (PTH). PTH increases calcium efflux from the bone, increases tubular reabsorption of calcium and phosphate excretion in the kidneys, and by stimulating the renal production of 1,25 dihydroxy vitamin D [(1,25 (OH)2 D], increases gastrointestinal absorption of calcium. It is important to distinguish between a primary disorder of the parathyroid glands in which there is dysregulated and excessive production of PTH (as in the case of primary hyperparathyroidism, PHPT), and situations in which the parathyroid glands respond secondarily to a stimulus such as malabsorption or renal failure and reacts by increasing PTH secretion. These latter forms of hyperparathyroidism are collectively known as secondary hyperparathyroidism (SHP) [Table 1]. Table 1. Causes of secondary hyperparathyroidism Chronic Kidney Disease Decreased Calcium Intake Decreased Absorption of Calcium Vitamin D deficiency Bariatric Surgery Celiac disease Pancreatic diseases with fat malabsorption Renal Calcium losses -Idiopathic hypercalciuria -Loop Diuretics Secondary to Phosphate Replacement therapy in conditions such as X-linked Hypophosphatemia, Autosomal Dominant Hypophosphatemia, Tumour Induced Osteomalacia etc. Open in a new tab Tertiary hyperparathyroidism refers to the hypercalcemic state in which, after longstanding SHP, the stimulated parathyroid glands assume a quasiautonomous role akin to that seen in PHPT. The differentiation of tertiary from PHPT is usually made possible since in the former; a clearly identifiable longstanding disorder such as malabsorption or renal failure is present predating the onset of hypercalcemia [Table 2]. Table 2. Biochemical differentiation between primary, secondary and tertiary hyperparathyroidism \| Biochemical Parameter \| Primary hyperparathyroidism \| Secondary hyperparathyroidism \| Tertiary hyperparathyroidism \| :---: :---: \| \| Calcium \| ↑ \| ↓ \| ↑ \| \| phosphate \| ↓ \| ↑ \| ↑ \| \| iPTH \| ↑ \| ↑ \| ↑ \| Open in a new tab N.B. SHP in patients with normal renal function (unlike as in those with CKD) is usually associated with low levels of phosphate given the inhibitory effect of PTH on Sodium-Phosphate co-transporters in the renal tubules Though SHP and its eventual progression to tertiary hyperparathyroidism has many causes as outlined in Table 1, this review will focus on the pathogenic, clinical and therapeutic aspects of these conditions in the setting of chronic kidney disease (CKD). The SHP associated with CKD is characterized by a complicated, multifaceted and as, yet incompletely understood pathophysiology. It is estimated that 30%-50% of stage 5 CKD patients have iPTH levels of >300 pg/ml. As the kidneys fail, gross derangements in fluid and solute clearance occur. An initial adaptive response, it becomes maladaptive over time and leads to the clinical syndrome termed as Chronic Kidney Disease-Metabolic Bone disorder (CKD-MBD), defined as a systemic disorder of mineral and bone metabolism, manifested by either one or a combination of the following: Abnormalities of calcium, phosphorus, PTH or vitamin D metabolism. Abnormalities in bone turnover, mineralization, volume, linear growth or strength Vascular or other soft tissue calcification. N ORMAL PHYSIOLOGY The secretion of PTH is closely regulated by extracellular ionized calcium through the calcium sensing receptor (CaSR) on parathyroid cells. PTH synthesis and secretion are also influenced by 1,25 (OH)2 D produced in renal proximal tubular cells by the conversion of 25(OH)D under the influence of the enzyme 1-alpha hydroxylase. 1,25 (OH)2 D binds to the vitamin D receptor (VDR) in parathyroid tissue and inhibits PTH mRNA synthesis. Inorganic Phosphate (Pi) may also act as an important regulator of PTH although the exact sensing mechanism through which it does this still remains to be elucidated. In addition, the relatively recently identified and characterized fibroblast growth factor 23 (FGF-23), an osteocyte and osteoblast derived phosphaturic hormone has been shown to decrease PTH synthesis and secretion by acting on the parathyroid glands through its receptor Klotho-FGF.[7,8] P ATHOPHYSIOLOGY It was long considered that the failure of the kidney to excrete serum phosphate with resultant hyperphosphatemia is the key driver of SHP in CKD patients. The formation of calciumphosphate salts with a reduction in serumionized calcium, the inhibitory effect of Pi on the enzyme CYP27B1 (25(OH) D1alpha hydroxylase) that is involved in the conversion of 25hydroxy D to 1,25(OH)2 D in the proximal renal tubular cell and the decreased viable renal mass.[12,13] in chronic renal insufficiency with resultant lesser 1alpha hydroxylase activity, all are believed to result in hypocalcemia and initiate the cascade of events that leads to dysregulation of parathyroid hormone in SHP. However, these postulations do not explain the clinical observation that serum 1,25(OH)2 D begins to decline even in early kidney disease before overt hyperphosphatemia develops and that hyperparathyroidism develops early in chronic renal failure at a time when plasma calcium and phosphorous are within normal limits. The identification and characterization of FGF-23 in the last decade has provided important clues towards understanding the early phases in the pathogenesis of SHP.[14,15] This 22.5 kDa protein, encoded by the FGF-23 gene located on chromosome 12 is secreted mainly by osteocytes and osteoblasts. Its synthesis and release is mainly stimulated by 1,25(OH)2 D and also by Pi, PTH and calcium by as yet incompletely defined mechanisms though it is thought that PTH induces FGF-23 transcription through activation of the orphan receptor Nurr1 and through activation of PKA and Wnt signaling in bones thereby constituting a bone-parathyroid-endocrine loop.[16,17,18] FGF23 together with its obligate coreceptor, the membrane bound αKlotho function to induce phosphaturia through downregulation of sodiumphosphate cotransporters. The FGF-23 αKlotho complex also inhibits 1,25(OH)2 D synthesis in the kidney by inhibiting 1alpha hydroxylase and, by stimulating 24hydroxylase, the catabolism of active vitamin sterols. This leads to hypocalcemia and stimulation of the parathyroid gland [Figure 1]. A soluble and circulating form of αKlotho produced mainly by the kidney may also have additional autonomous (i.e., independent of FGF-23) phosphaturic and anti-calciuric effects. A progressive renal reduction in production of both membrane-bound and circulating αklotho, increasing levels of FGF-23 secondary to its reduced renal clearance and due to Pi retention, and resistance to the phosphaturic effect of FGF-23 due to deficiency of αKlotho characterize CKD progression. Figure 1. Open in a new tab Schematic representation of the current understanding regarding the pathophysiology of SHP As FGF-23 level increases, a “trade-off” occurs between maintaining normophosphatemia versus 1,25 hydroxy vitamin D deficiency, with the latter progressing relentlessly and causing SHP. Phosphate level will start to rise only when the adaptive compensation by FGF-23 becomes inadequate. At this stage, hyperphosphatemia, continued decreased 1,25(OH)2 D and hypocalcemia all contribute to increasing PTH mRNA levels and PTH synthesis. Initial diffuse parathyroid cell hyperplasia and ultimately nodular hyperplasia results. A separate CaSR independent mechanism for hypocalcemia to stimulate PTH secretion may also be through microRNA (miRNA) dysregulation within the parathyroid glands. Though conflicting data also exists, FGF-23 may also act directly on the parathyroid gland to suppress PTH secretion through the Klotho–FGFR1 complex. In patients with advanced SHP, the parathyroid expression of the Klotho–FGFR1 complex is downregulated. This likely contributes to the resistance to the inhibitory effect of FGF-23 on PTH secretion in progressive and advanced SHP. As parathyroid hyperplasia progresses, both CaSR and VDR on the parathyroid glands become downregulated and reduced expression of these has been observed in the most severe forms of SHP.[26,27,28,29,30] The size of the parathyroid glands progressively increases as SHP worsens and gland size is positively correlated with serum PTH levels [Figure 2]. The cellular etiology of tertiary hyperparathyroidism is unknown, but it is postulated to be due to a monoclonal expansion of parathyroid cells in which the set point of the CaSRs has been altered such that semi-autonomous secretion of PTH occurs despite high serum calcium levels. Monoclonal chief cell growth results in the formation of nodules. Nodular glands have less VDRs and CaSRs[26,27,28,29,30] compared to diffusely hyperplastic glands and this exacerbates parathyroid gland resistance to calcitriol and calcium. Figure 2. Open in a new tab The stages in the evolution of secondary and tertiary hyperparathyroidism C LINICAL F EATURES Skeletal manifestations PTH binds to the PTH/PTHrP receptor on osteoblasts and thus by indirectly stimulating osteoclastic activity leads to a high turnover bone disease. Fragility fractures have been reported to be 2-4 times more frequent in patients with SHP when compared with age- and gender-matched normal populations. This increased risk is associated with an increased risk of mortality and an association between PTH levels and fracture risk has been observed, with intact PTH levels above 900 pg/ml shown to be independently associated with an increased risk of incident fractures in the DOPPS study. It has to be remembered however that the bone fragility in CKD may have several causes other than SHP; such as metabolic acidosis, anemia, hypogonadism, inflammation, beta 2 microglobulin associated amyloidosis, vitamin D deficiency, bone formation inhibition secondary to Wnt inhibition in osteocytes, etc. to name a few. Extra-skeletal manifestations Elevated PTH levels may be associated with an increased sympathetic drive, and endothelial stress and SHP may play a causal role in the development of vascular calcifications, ischemic cardiovascular events and cardiac failure. In contrast to the intimal calcification seen in aging individuals with normal renal function, what is seen in patients with CKD is calcification of the medial layer. Elevated PTH levels have been found to be independently associated with anemia, which is a hallmark of CKD and severe SHP is associated with a resistance to erythropoietin therapy in CKD. It should be borne in mind that, no clear causal relationship between SHP and these extra-skeletal manifestations has been established, neither has it been shown that correction of SHP can result in a complete remission of these clinical conditions. C LINICAL E VALUATION The diagnoses of secondary and tertiary hyperparathyroidism are purely biochemical and therefore, accurate measurement of PTH is essential. PTH is a hormone of 84 amino acids. Over the last few decades, three generations of PTH assays have been developed that measure different parts of the molecule [Figure 3]. The first radioimmunoassay for PTH was developed in 1963 by generating a single polyclonal antibody against epitopes in the carboxy terminal (C terminal) end of the PTH molecule. The first-generation assays had poor specificity as the antibodies used mainly targeted the non-bioactive portion of the PTH molecule; the C-terminal which is retained in CKD. The second-generation assays that were developed to overcome this problem use two sets of antibodies: (a) capture antibodies against epitopes located within the C-terminal and (b) detection antibodies directed to amino acid sequences 12-20 within the amino terminal end. This assay is currently the most widely used and is called intact PTH assay (iPTH) as it is assumed that it captures intact PTH 1-84. However, the detection antibodies have been found to cross-react with PTH 7-84 fragments that also tend to accumulate in patients with CKD.[37,38] It has also become apparent that high concentrations of 7-84 PTH and some other C-terminal PTH fragments may oppose the biochemical and bone-metabolic effects of 1-84 PTH; aggravating the potential undesirable clinical consequences of overestimating 1-84 PTH concentrations in renal failure patients,[39,40] that is, the physician might mistakenly assume the erroneously high PTH reading as the correct value and may institute further PTH lowering therapies with disastrous consequences. To overcome these shortcomings, third-generation PTH assays such as the whole PTH assay and the Bio-Intact PTH assay have been developed. Though the capture antibody used in the third-generation assay is the same as that used in the second-generation assay, the detection antibody used is directed towards the first four amino acids. These assays do not thus recognize 7-84 PTH and are therefore considered more specific to 1-84 PTH than second-generation assays. Two automated third-generation PTH assays are now available. However, there is little evidence to show that they provide any better clinical information than the second-generation assays with regard to the diagnosis of CKD-MBD and therefore have not been adopted for use in current guidelines for management of SHP. In general, PTH levels measured with second-generation assays are higher than those obtained with third-generation ones. The ratio of whole (biointact) to intact PTH levels has been noted to be between 0.6-0.7 in dialysis patients, though exceptions to this rule have been reported in patients with severe SHP with a new molecular form of PTH with an intact N-terminus[45,46] that can be detected by third-generation PTH assays but not by second-generation ones identified in these patients. Thus in these patients, PTH levels measured with third-generation assays are paradoxically higher than those with second-generation ones. Existing clinical data suggest that an over-production of N-PTH may be associated with rapid progression of SHP and that N-PTH has significant bio-activity.[47,48] Figure 3. Open in a new tab Schematic representation of the three generations of PTH assays KDIGO guidelines earlier recommended target iPTH levels of 2-9 times of the upper limit of normal for the given assay to noninvasively monitor bone status in dialysis patients. iPTH values above this target suggest high bone turnover bone disease with specificity of 86%, and values below the target values suggest low bone turnover with sensitivity of 66%. However, the latest update to the CKD-MBD guideline published by the KDIGO in 2017 suggest that in patients with CKD G3a–G5D, treatment for CKD-MBD should be based on serial assessments of phosphate, calcium, and PTH levels, considered together and not absolute values of any of these parameters. T REATMENT OPTIONS A paradigm shift has occurred in the approach to the treatment of secondary and tertiary hyperparathyroidism with the understanding that the alterations in calcium and phosphate metabolism in CKD do not only cause renal osteodystrophy and bone abnormalities but also are linked to increased risk of cardiovascular disease and all-cause mortality potentially mediated through vascular calcification. The complex pathophysiology of secondary and tertiary hyperparathyroidism makes it necessary that their treatment should be multi-pronged. The three main targets are thus phosphate, 1,25 vitamin D and PTH. Controlling Phosphate Levels The management of hyperphosphatemia has formed the cornerstone of therapy for SHP for decades. Dietary phosphate restriction and treatment with oral phosphate binders can decrease PTH levels up to stages 3 and 4 CKD. The reduction of protein intake (particularly protein of animal origin) has been the basis of dietary prescriptions in CKD. However, such a diet is difficult to maintain, and such dietary restrictions should be counter-balanced by the awareness that they may be associated with an increased risk of malnutrition in CKD patients. Nevertheless, all attempts should be made to have a diet that contains more vegetal than animal proteins and to avoid processed foods. Most often, as CKD progresses and hyperphosphatemia ensues, dietary phosphate restriction alone is not helpful and phosphate binding medications are needed. An ever-increasing number of phosphate binders have been developed over the last few decades. They can be broadly grouped into calcium based and non–calcium-based agents. All these agents are more or less equally effective in the control of phosphate levels and SHP. Calcium-based phosphate binders such as calcium carbonate and calcium acetate are widely prescribed and very effectively lower Pi levels. However, their benefits must be weighed against possible adverse effects of hypercalcemia. KDOQI guidelines recommend that the total dose of elemental calcium provided through calcium-based phosphate binders should not exceed 1500 mg daily. One of the most commonly used non–calcium-based binders is Sevalamer; a phosphate binding resin. A meta-analysis of randomized controlled trials comparing calcium-based phosphate binders to non–calcium-based ones surmised that the use of calcium-based binders was associated with higher all-cause mortality compared to Sevalamer. However, due to lack of placebo-controlled trials, the debate on whether calcium-based Pi binders are associated with higher risk for vascular calcifications and consequently for cardiovascular mortality continues. Another non–calcium-based phosphate binder; Lanthanum carbonate has also been shown to control hyperphosphatemia. However, Lanthanum can be systemically absorbed and may accumulate in liver and bone. This limits its use as a first-line phosphate binder. Enteral phosphate binders may lead to up-regulation of the intestinal sodium-phosphate NPT2b transporter thereby increasing active enteral phosphate absorption, off-setting some of their beneficial effects. Extended release Niacin, a NPT2b inhibitor has shown phosphate and FGF-23 lowering effects in adult CKD patients. However, the recent CKD Optimal Management with Binders and Nicotinamide (COMBINE) study designed to assess whether the addition of NPT2b blockade to Lanthanum improves phosphate and FGF-23 levels showed disappointing results with no significant differences in any of the three active treatment groups (Nicotinamide plus Lanthanum Carbonate, Nicotinamide plus Placebo, or Lanthanum Carbonate plus Placebo) compared with the double placebo group though the findings could have been affected by the high non-compliance rates observed amongst the double-treatment group. Tenapanor is a new agent that inhibits intestinal absorption of phosphate through the inhibition of intestinal sodium/hydrogen exchanger isoform 3 has recently been shown to effectively reduce phosphate in patients who are on maintenance dialysis. It opens up a potential new therapeutic option in controlling of serum phosphate in patients with CKD. Phosphate is also removed during dialysis. Hence it is vital that dialysis dose is adequate to optimize phosphate control. This can be achieved by adjusting dialysis time as well as blood flow settings during dialysis. It has been shown that patients who are on long/frequent dialysis have much better control of phosphate than their counterparts who are on conventional dialysis. Vitamin D analogues Treatment with Vitamin D Receptor Activators (VDRAs) has long been a very important therapeutic strategy in the management of SHP. Calcitriol, the first synthetic VDRA to be developed decreases serum PTH levels[57,58] in CKD and also has been shown to reduce bone turnover and to thus ameliorate osteitis fibrosa in dialysis patients though its effect on risk of fractures in CKD has not been adequately studied. The inhibitory effect on PTH synthesis is mediated through binding of calcitriol to its specific receptor (VDR) and subsequent regulation of gene transcription and inhibition of PTH mRNA synthesis. This is important to know because in advanced SHP, with nodular hyperplasia of the parathyroid glands, there is decreased expression of CaSR and VDRs in the parathyroid gland[29,30], and in such a situation, VDRAs are not as effective in suppressing PTH secretion. The clinical utility of Calcitriol is limited by its potential to cause hypecalcemia. The newer selective VDRAs such as paricalcitol (19-nor-1,25-dihydroxyvitamin D2) and maxacalcitol (22-oxa-1,25-dihydroxyvitamin D3) may be preferable in this regard because they have more modest effects on serum calcium levels though it has to be noted that these agents can also cause hypercalcemia. Although it has long been known that vitamin D deficiency is common in CKD patients, the common belief has been that the need for its correction is not as stringent in this clinical setting provided that active vitamin D is administered. However, the administration of native vitamin D may have other pleuripotent benefits and it has also been demonstrated that use of native vitamin D esters are effective in lowering PTH levels at least in the early stages of SHP. Calcimimetics The introduction of calcimimetics, agents that allosterically activate the calcium sensing receptor, has significantly mitigated the need for high doses of activated vitamin D and the risk of hypercalcemia in SHP. These agents “mimic” calcium and increase the sensitivity of calcium sensing receptors on the parathyroid gland. Currently, the only oral calcimimetic approved by the FDA is Cinacalcet. Cinacalcet effectively reduces PTH levels and serum calcium levels in patients with SHP. Notably, cinacalcet is effective even in patients with marked parathyroid hyperplasia thus making it an acceptable alternative to parathyroidectomy for treatment of severe SHP. Cinacalcet may cause gastrointestinal adverse effects such as nausea and vomiting. The introduction of a new intravenous calcimimetic, Etelcalcetide offers a therapeutic alternative to oral cinacalcet. Etelcalcetide has a longer half-life than cinacalcet and can be administered intravenously every other day at the end of dialysis treatment thus overcoming the problem of compliance with a daily oral regimen. Etelcalcetide has been shown to markedly decrease PTH levels in patients on hemodialysis with moderate to severe SHP and may be superior to Cinacalcet in this regard though further studies are needed to assess clinical outcomes as well as long-term efficacy and safety of this agent. The effects of Cinacalcet on bone turnover and bone histology in patients with CKD and evidence of high-turnover bone disease have been studied in the Bone Histomorphometry Assessment for dialysis patients with Secondary Hyperparathyroidism of End-Stage Renal Disease (BONAFIDE) study. This study demonstrated that long-term treatment with cinacalcet lowers biochemical markers of high bone-turnover and improves bone histology in this setting. In the Evaluation of Cinacalcet Hydrochloride therapy to Lower Cardiovascular Events (EVOLVE) trial, a randomized controlled trial to assess the effects of cinacalcet on clinical outcomes, though no significant effect of cinacalcet in the primary intention to treat analysis was seen, a significant reduction in the risk of fracture in the cinacalcet group was found when differences in baseline characteristics, multiple fractures and/or events prompting discontinuation of study drug were taken into account. The results of the EVOLVE trial also suggested a beneficial effect of Cinacalcet with regard to reduction in the risk of death or cardiovascular outcomes though it has to be noted that this again was not in the primary unadjusted intention to treat analysis but in the log-censoring analysis. Parathyroidectomy Despite the availability of newer vitamin D analogues and calcimimetics, parathyroidectomy continues to be a necessity in certain patient groups. It is estimated that parathyroidectomy is required in about 15% of patients after 10 years and in 38% of patients after 20 years of ongoing dialysis therapy. Successful surgical treatment results in a dramatic reduction in PTH levels and improvement of clinical symptoms such as bone pain and itching. Parathyroidectomy is also associated with better patient survival[70,71,72] and reduced risk of fractures in patients with severe SHP. A description of the surgical techniques for parathyroidectomy is beyond the scope of this article. The choice of surgical technique viz subtotal parathyroidectomy versus total parathyroidectomy with auto transplantation ultimately depends on operator experience and expertise and must be individualized to the patient. However, parathyroidectomy is not without its risk. The most commonly seen is the phenomenon of Hungry Bone Syndrome characterized by severe hypocalcemia post parathyroidectomy. The abrupt withdrawal of very high and sustained levels of PTH following parathyroidectomy, turns off osteoclast activity and bone resorption in the remodeling space. However, osteoblast activity and new bone formation continues, which leads to the influx of calcium, phosphate and magnesium into bone resulting in their abrupt drops in the serum. This condition remains poorly defined and the prevalence of this condition has been reported to range from 8% to 87% following parathyroidectomy for SHP. The other concern is the occurrence of adynamic bone disease and hypoparathyroidism post parathyroidectomy. Hypoparathyroidism typically is reported following surgery for PHPT and there is no study that reports the incidence or prevalence of this condition in patients who undergo parathyroidectomy for SHP. Low turnover bone disease, however, has been reported to occur post parathyroidectomy and has been associated with worsening of vascular calcification in hemodialysis patients.[76,77,78] Chemical ablation of parathyroid gland Percutaneous fine needle ethanol injection of parathyroid gland was first reported in 1985 in 12 patients with SHP. In 2003, the Japanese Society of Parathyroid Intervention published its guideline for selective percutaneous ethanol injection therapy of the parathyroid glands in chronic dialysis patients in which they recommended that enlarged parathyroid glands with nodular hyperplasia could be “selectively” destroyed by ethanol injection, and other glands with diffuse hyperplasia could be then managed by medical therapy. Percutaneous injection using the Vitamin D analogue-Calcitriol instead of alcohol has also been described. The rationale behind this approach is to introduce high level of vitamin D around the parathyroid gland without the systemic complications that could potentially be caused by its systemic administration. These local approaches could be considered in patients who refuse or are not candidates for surgery although long-term control of SHP is unlikely to be obtained. P ERSISTENT HYPERPARATHYROIDISM AFTER KIDNEY TRANSPLANTATION Persistent hyperparathyroidism even after renal transplantation is most likely to be present in patients with advanced SHP with nodular hyperplasia of the parathyroid glands before transplant. It is the most common cause of hypercalcemia in renal transplant patients and may result in poor graft outcomes and progression of vascular calcification. Surgical parathyroidectomy should be considered in kidney transplant patients with persistent hyperparathyroidism especially when it is associated with severe hypercalcemia. Cinacalcet appears to be a promising therapeutic option for patients with persistent hypercalcemia post transplantation at least as a bridging agent before parathyroidectomy.[84,85,86] C ONCLUSION Our knowledge of the pathophysiology of secondary and tertiary hyperparathyroidism has vastly improved during the past few years. They may be caused by various conditions, however, that associated with CKD has been the one most studied and yet remains incompletely defined. The clinical consequences of these disorders of the parathyroid gland are not limited to the musculo-skeletal system but are multifold and systemic. It is however difficult to define clearly whether there is a causal relationship between the elevated levels of PTH seen in these disorders and the protean clinical manifestations or whether they simply are associations in a complex clinical setting. 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[DOI] [PMC free article] [PubMed] [Google Scholar] Articles from Indian Journal of Endocrinology and Metabolism are provided here courtesy of Wolters Kluwer -- Medknow Publications ACTIONS View on publisher site PDF (917.5 KB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract INTRODUCTION NORMAL PHYSIOLOGY PATHOPHYSIOLOGY CLINICAL FEATURES CLINICAL EVALUATION TREATMENT OPTIONS PERSISTENT HYPERPARATHYROIDISM AFTER KIDNEY TRANSPLANTATION CONCLUSION REFERENCES Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
2678	https://bshook-92488.medium.com/pulcher-is-one-of-the-latin-words-for-beautiful-handsome-and-pulchritudo-its-noun-form-from-bb1d3f560db5	pulcher is one of the Latin words for "beautiful/handsome," and pulchritudo its noun form, from which pulchritude. It's one of the English words I'd tell my Latin students that they should recognize… - Brian S. Hook - Medium Sitemap Open in app Sign up Sign in Write Search Sign up Sign in I've always found it odd, this word - it's such an unattractive-sounding word, for meaning beauty. 7 2 Joe Guay - Dispatches From the Guay Life! Brian S. Hook Follow Jul 21, 2025 9 1 Listen Share pulcher is one of the Latin words for "beautiful/handsome," and pulchritudo its noun form, from which pulchritude. It's one of the English words I'd tell my Latin students that they should recognize in English but never use, except in cases like Patricia's, who deploys it perfectly. Get Brian S. Hook’s stories in your inbox Join Medium for free to get updates from this writer. Subscribe Subscribe It's exactly as you say, Joe: its sound is contrary to what it expresses. A professor in college made the same observation about the German word for butterfly--so airy in other languages, papillon and farfalla--but Schmetterling auf Deutsch. I'll never forget that! 9 9 1 Follow Written by Brian S. Hook ------------------------ 3K followers ·235 following Dad, classicist, mountain dweller, erstwhile triathlete, wannabe woodworker, follower of Socrates and Jesus (two famous non-writers), writing to avoid raveling Follow Responses (1) Write a response What are your thoughts? Cancel Respond Joe Guay - Dispatches From the Guay Life! Jul 21 Ah, thanks, Brian. Happy to know I have a good ear but also intrigued to know the origin! -- Reply More from Brian S. Hook In Age of Empathy by Brian S. Hook How I Lost My Job As A Tenured Professor ---------------------------------------- ### Where Does An Expert In A Dying Field Go Next? Jul 26, 2024 376 In Catharsis Chronicles by Brian S. Hook A Taxonomy of Tears ------------------- ### A poem Jan 8 12 In ILLUMINATION by Brian S. Hook Stop, Traveler, And Read ------------------------ ### Why So Many Latin Tombstones Address the Passersby Jan 10, 2024 17 In Beyond the Scoreboard by Brian S. Hook The Problem of Court Storming ----------------------------- ### ‘This gotta change’ and my proposal might do just that Feb 26, 2024 10 See all from Brian S. Hook Recommended from Medium In An Injustice! by J. Henderson 10 Charlie Kirk Quotes, Ranked from Simply Bad to Utterly Horrible ------------------------------------------------------------------ ### Everyone has the right to free speech, but what happens when it turns into radicalism and extremism? Sep 20 370 In Long. Sweet. Valuable. by Ossai Chinedum I’ll Instantly Know You Used Chat Gpt If I See This --------------------------------------------------- ### Trust me you’re not as slick as you think May 16 1441 Will Lockett The AI Bubble Is About To Burst, But The Next Bubble Is Already Growing ----------------------------------------------------------------------- ### Techbros are preparing their latest bandwagon. Sep 14 346 In Bitchy by Maria Cassano Studies Show That Predators Target Women Based on One Thing ----------------------------------------------------------- ### Attackers all chose similar victims — and it wasn’t what they were wearing. 5d ago 296 In How To Profit AI by Mohamed Bakry Your ChatGPT History Just Went Public on Google. Here’s What I Did in 10 Mins to Fix It. ---------------------------------------------------------------------------------------- ### Safety/Privacy Check Prompt Template Is Included 5d ago 326 In Fourth Wave by Toni Crowe I Came Across My First 2025 “Whites Only” Sign ---------------------------------------------- ### “Separate but Equal” is alright with me Sep 18 315 See more recommendations Help Status About Careers Press Blog Privacy Rules Terms Text to speech
2679	https://www.youtube.com/watch?v=T_Bpt0zl88E	9.4 Geometry - Compositions of Isometries Math is FUNdamental 1370 subscribers 14 likes Description 3361 views Posted: 3 Mar 2015 In this Geometry lesson you will learn about Compositions of Isometries. Examples are provided. 4 comments Transcript: Hank laughs welcome to math is fundamental today we're going to be talking about compositions of isometry x' so this is going to be where we take different types of transformations and stick them together compose them together so when isometry or we've called them rigid motions is a transformation that preserves distance or length so in the transformation the shape does not does not change size or shape it stays the same size of shape translations reflections rotations are all isometry x' or rigid motions we're gonna call them mysamma trees now all right a composition of isometry says let me combine the transformations one after the other so an example for the notation really important that we understand our notation of a composition of mysamma trees so here's an example reflection of triangle PQ are across the x-axis after it's translated down three units so this is written as here we've got a reflection across y equals zero or x axis and then an open circle that little open circle means it's composed with the other transformation which in this case is a translation zero horizontally and negative three vertically so three down and here's our triangle PQR really really important the order is backwards when you see the notation of a composition of isometry so you actually have to do the second I saw on a tree first and then do the first so here we would do the translation first and then reflect it okay so let's look at an example we're gonna graph the coordinates of triangle d EF and then I do the two reflections so here we have negative 1 negative 4 so negative 1 1 2 3 negative 4 so here is D all right so we have E at negative 2 negative 5 and f is at negative 4 negative 3 all right so here is triangle d EF now when we're reflecting we did a video on reflections a couple videos ago so you can go back and review reflection across y equals negative 2 we always want to draw our line of reflection so here's y equals negative 2 and I'm going to reflect it across this line so D is going to go straight to the line and all the way on the other side so here's D prime B prime 1 2 3 1 2 3 so we count how far it is from our line of reflection and then count the same distance on the other line so f prime is 1 2 the line so 1 after the line so here is our first reflection over that line y equals negative 2 now I have a couple different let's color the next one green alright so now I have the line y equals 2 so y equals positive 2 is going to be here again we want to always draw our lines of reflections it always helps alright and then reflect that second one D prime a prime F prime over that line so here we've got D double prime double prime and 1 2 3 - 3 f double-prime so here is our resulting figure and look at the resulting figure compared to the preimage the very first de f that we started with so here our theorem states that a composition of reflections across two parallel lines is actually just a translation so we can write this composition of reflections if we want to we can say okay so we have two reflections now again we have to write the second reflection first so we did y equals to first and then we do a little open circle and then the first one was y equals negative 2 so there's our composition of reflections then we have triangle de f and then triangle d double prime Y double prime F double prime so there is our composition of reflections but we could have also written this as a translation so our translation well to get from one to the next I translated this let's see this ended up all the way at one two three four so we went from negative 4 to positive 4 vertically we didn't translate anything horizontally so 0 positive 8 it traveled 8 up of triangle de f equals triangle G double prime e double prime F double prime alright let's look at our second example here so we're going to start with D EF so let's graph that so negative 1 negative 4 is d e is negative 2 negative 5 and X is negative 4 negative 3 so same triangle that we started with last time all right and then let's look at what he wants us to do this time so we've got reflect over y equals negative - so that's the same thing we had to do last time we draw our reflection y equals negative 2 and I go 1 2 1 2 so there's my D Prime and E prime 1 2 3 1 2 3 and s Prime and here is our first reflected figure now the second reflection again I'm gonna change the color here and make it green just so it's a different color for us ah they want us to reflect it over there line x equals 0 now this x equals 0 is actually just the y axis so this time we have our two lines are not parallel that we're reflecting so let's reflect this second figure over that new line so here's D double Prime e double Prime and 1 2 3 4 - the lines so 1 2 3 4 1 2 3 4 on the other side of the line f double Prime alright and there is our resulting figure now if we look at our first figure in our second figure this from D to D prime is let's see let's see if I can draw this if I draw a line from D the D Prime it's a line that goes straight through that intersection point there and if we draw all of our other lines it's going to give us the same result it's going to be a line that's straight through and if we remember back to our other videos our last couple videos this is actually a rotation of 180 degrees so composition of reflections across two intersecting lines is a rotation now it's not always a rotation of 180 degrees it happens to be a rotation of the two angles tween the lines so in this case my lines were perpendicular so if I take a ninety degree and then another ninety degree between our two figures that's going to give us 180 degrees so it's like adding those angles together so we could write the composition of reflections as a rotation the rotation in this case would be a plus C rotation is lowercase R and we have our angle which is 180 degrees and our center of rotation in this case is the point 0 negative 2 0 negative 2 so we've got kind of a more complicated there and then triangle de f equals triangle D double prime e double Prime at double Prime all right so a little bit more complicated there but as you can see our two reflections ended up as a rotation in this case alright so the last piece of our lesson is what we call a glide reflection so any composition of isometry x' can be represented by either a reflection a translation a rotation or a glide reflection and if we think about the word the wide reflection a glide reflection is a composition of a translation or a glide and the reflection across a line that has to be parallel to the direction of the translation so there's kind of the kick that we have to make sure is true that our reflection is across the line parallel to our direction of translation so let's look at this last example here we have the coordinates of triangle ABC and then we want to do our glide reflections so a is going to be at negative 1 negative 3 B is negative 4 negative 1 and C is negative 6 negative 4 all right so here is our preimage ABC now remember when we do a composition of transformations or of yeah transformations we want to do the second one first so in this case we have a reflection and a translation but we want to do the translation first and they want us to go ten horizontally and nothing vertically so I'm gonna go ten in the horizontal direction so I'm going to let's just take a from negative one all the way to positive nine so there's a prime P goes from negative 4 to positive 6 so I'm just adding 10 horizontally and C goes from negative 6 to positive 4 C Prime alright so here is my translation oops and I went horizontally so my direction of translation was horizontal which means in order for this to be glad inflection I need to reflect it over horizontal line now the x axis it's telling me to reflected across the x-axis is a horizontal line so there's my x axis which is parallel to my direction of translation so we are good there let's reflect it over so we've got V double Prime 1 2 3 1 2 3 a double prime and 1 2 3 4 1 2 3 4 C double Prime and there we go thank you so much for watching that concludes our lesson for today and remember math is fundamental
2680	https://math.answers.com/other-math/How_can_you_determine_if_the_given_ordered_pair_is_a_solution_to_the_system_of_equations	How can you determine if the given ordered pair is a solution to the system of equations? - Answers Create 0 Log in Subjects>Math>Other Math How can you determine if the given ordered pair is a solution to the system of equations? Anonymous ∙ 13 y ago Updated: 4/28/2022 Plug your ordered pair into both of your equations to see if you get they work. Wiki User ∙ 13 y ago Copy Add Your Answer What else can I help you with? Search Continue Learning about Other Math ### Determine if the ordered pair y3x 5 yx 9 211 isa solution to the system of equations? Tell whether the ordered pair (5, -5) is a solution of the system ### What is an ordered pair that makes all equations in a system true? That would be the "solution" to the set of equations. ### What is the solution to a linear equation? The solution to a system on linear equations in nunknown variables are ordered n-tuples such that their values satisfy each of the equations in the system. There need not be a solution or there can be more than one solutions. ### How can you determine whether a point is a solution to system of equations? You substitute the coordinates of the point in the equation. If the result is true then the point is a solution and if it is false it is not a solution. ### What is the solution of a system of linear equations in two variables? The solution of a system of linear equations is a pair of values that make both of the equations true. Related Questions Trending Questions What is the next number in the sequence 2 5 28 257 3126?Imagine you have 25 beads you have to make a 3 digit number on an abacus you must use all the beads how many 3 digit numbers can you make?How do you Write 79938 in expanded form?When graphing inequalities when do you shand upward and when do you shade downward?What is half of 69?Key code for modal of ohnor AAA?What is the area of a square that is 16cm?What is 16 x 69?What is 200 divided by 2.2?What number is ten less than one hundred and one?How can you compare fractions with denominator of 100 and 10?Which describes the two parts of a measurement?How do you write four billion four hundred ninety-eight million two hundred fifty-two thousand nine hundred in standard form?What number logically comes next to the sequence 4 6 9 5 4 2 3 9?What is greater 4.24 or 4.42?How many face are on a triangular pyramid?How many UK pennies are in a meter?What is 453 divided by 1000?How many hrs I s between 815 to 530?How many times can 65 go into 26? Resources LeaderboardAll TagsUnanswered Top Categories AlgebraChemistryBiologyWorld HistoryEnglish Language ArtsPsychologyComputer ScienceEconomics Product Community GuidelinesHonor CodeFlashcard MakerStudy GuidesMath SolverFAQ Company About UsContact UsTerms of ServicePrivacy PolicyDisclaimerCookie PolicyIP Issues Copyright ©2025 Answers.com. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Answers.
2681	https://www.youtube.com/watch?v=7QZs__0c2P8	Mathematical Number Pattern 5 \|\| Number 12345679 \|\| RHS contain 1's, 2's, 3's etc. only \|\| Math1089 Math1089 1670 subscribers 23 likes Description 1380 views Posted: 17 Sep 2022 This is the fifth video under the heading Number pattern. The pattern is given below: 12345679 × 09 = 111111111 12345679 × 18 = 222222222 12345679 × 27 = 333333333 12345679 × 36 = 444444444 12345679 × 45 = 555555555 12345679 × 54 = 666666666 12345679 × 63 = 777777777 12345679 × 72 = 888888888 12345679 × 81 = 999999999 Thank you for liking the previous videos. LHS contains the number 12345679, whereas the RHS contains 1's, 2's, 3's etc. only. The link for other four number pattern videos are given below. Number Pattern 1 Number Pattern 2 Number Pattern 3 Number Pattern 4 Many more videos can be found in the channel Math1089. Please browse. Many interesting write ups can be found in the blog Math1089. Please browse. Please share the works in your network, if you find them useful. To follow us through social media, use the following links. Blog: Facebook: Instagram: Linkedin: Twitter: Telegram: math, #maths, #mathematics, #algebra, #geometry, #calculus, #trigonometry, #numbers, #derivatives, #symbol, #mathematical, #arithmetic, #stem, #gcse, #CBSE, #cbseboard, #ICSE, #gcsemaths, #math1089, #mathteacher, #mathstudent, #mathtricks, #mathisfun, #mathematicians, #mathfacts, #mathnerd, #mathskills, #mathproblems, #zerotohero 4 comments Transcript: hello everyone welcome to the YouTube channel map1089 definitely this is a mathematics YouTube channel and the name also consists the number 1089 so far in this YouTube channel we have uploaded the properties of few important numbers their various representations using the digits 0 to 9 or maybe one to nine or something similar so far we have considered various number patterns or pyramidal structure patterns of numbers today's video is also about number pattern and the number pattern is here in this number pattern we can see one number and that number is one two three four five six seven nine and every row starts with this particular number the second number in any row is multiple of 9. for example in the first row it is 9 second row 18 which is equal to 9 into 2 third row 27 9 into 3 4th row 36 9 into 4 then 9 into 5 9 into 6 9 into 7 9 into 8 9 into 9. the right hand side value of the first row consisting of all ones so it's a sequence of one how many ones are there suddenly we can see there are nine ones nine ones similarly nine two nine threes nine fours nine five nine six nine seven nine eight and nine nine are there in the last row now this pattern entirely depends on the first row now one thing is sure if we multiply the two numbers of the left hand side like one two three four five six seven nine and nine clearly we will get the right side number which is a number consisting of all ones in fact nine ones but here we will show the same thing in a different way let us apply the distributive property we can write 0 9 0 9 as 10 minus 1 so 10 minus 1 is equal to 9 now if we apply distributive property then one two three five six seven nine into ten will give us the number one two three four five six seven nine ten zero so the zero of ten of the number 10 will come at the end and then multiplying by 1 that is one two three four five six seven nine will come as a subtraction now we will apply a particular method that we will write both the numbers one below the other we so that we can see the value of the subtraction is equal to on one's in fact nine ones which is the right hand side or as expected now it is easy to find the value of the second row or the third row or the fourth row or like this because the second row is a multiple of the first row how come let's see the second row looks like one two three four five six seven nine into 18. now 18 means nine into two nine into two if we take two numbers together which two number one two three four five six seven nine and nine so one two three four five six seven into nine if we multiply them how much we will get from row number one we will get all ones in fact all nine ones and if we multiply that particular number by two surely we will get all those in fact nine twos and that is the right hand side so in the same way we can find the values of other rows also thank you for watching hope you enjoyed the video please subscribe to our Channel and don't forget to share this video thank you
2682	https://www.accessdata.fda.gov/drugsatfda_docs/label/2011/005010s050lbl.pdf	CII Demerol® (meperidine hydrochloride, USP) WARNING: May be habit forming DESCRIPTION Meperidine hydrochloride, is a white crystalline substance with a melting point of 186° C to 189° C. It is readily soluble in water and has a neutral reaction and a slightly bitter taste. The solution is not decomposed by a short period of boiling. The tablets contain 50 mg or 100 mg of DEMEROL brand of meperidine hydrochloride. Inactive Ingredients: Calcium Sulfate, Dibasic Calcium Phosphate, Starch, Stearic Acid, Talc. Chemically, DEMEROL is 4-Piperidinecarboxylic acid, 1-methyl-4-phenyl-, ethyl ester, hydrochloride and has the following structure: structural CLINICAL PHARMACOLOGY Meperidine hydrochloride is a narcotic analgesic with multiple actions qualitatively similar to those of morphine; the most prominent of these involve the central nervous system and organs composed of smooth muscle. The principal actions of therapeutic value are analgesia and sedation. There is some evidence which suggests that meperidine may produce less smooth muscle spasm, constipation, and depression of the cough reflex than equianalgesic doses of morphine. INDICATIONS AND USAGE DEMEROL is indicated for the relief of moderate to severe pain. CONTRAINDICATIONS DEMEROL is contraindicated in patients with hypersensitivity to meperidine. Meperidine is contraindicated in patients who are receiving monoamine oxidase (MAO) inhibitors or those who have recently received such agents. Therapeutic doses of meperidine have occasionally precipitated unpredictable, severe, and occasionally fatal reactions in patients who have received such agents within 14 days. The mechanism of these reactions is unclear, but may be related to a preexisting hyperphenylalaninemia. Some have been characterized by coma, severe respiratory depression, cyanosis, and hypotension, and have resembled the syndrome of 1 Reference ID: 2931372 This label may not be the latest approved by FDA. For current labeling information, please visit acute narcotic overdose. In other reactions the predominant manifestations have been hyper excitability, convulsions, tachycardia, hyperpyrexia, and hypertension. Although it is not known that other narcotics are free of the risk of such reactions, virtually all of the reported reactions have occurred with meperidine. If a narcotic is needed in such patients, a sensitivity test should be performed in which repeated, small, incremental doses of morphine are administered over the course of several hours while the patient’s condition and vital signs are under careful observation. (Intravenous hydrocortisone or prednisolone have been used to treat severe reactions, with the addition of intravenous chlorpromazine in those cases exhibiting hypertension and hyperpyrexia. The usefulness and safety of narcotic antagonists in the treatment of these reactions is unknown.) WARNINGS Meperidine should not be used for treatment of chronic pain. Meperidine should only be used in the treatment of acute episodes of moderate to severe pain Prolonged meperidine use may increase the risk of toxicity (e.g., seizures) from the accumulation of the meperidine metabolite, normeperidine. DEMEROL is an opioid agonist and a Schedule II controlled substance with an abuse liability similar to morphine. DEMEROL can be abused in a manner similar to other opioid agonists, legal or illicit. This should be considered when prescribing or dispensing DEMEROL in situations where the physician or pharmacist is concerned about an increased risk of misuse, abuse, or diversion. Misuse, Abuse, and Diversion of Opioids Meperidine is an opioid agonist of the morphine-type. Such drugs are sought by drug abusers and people with addiction disorders and are subject to criminal diversion. Meperidine can be abused in a manner similar to other opioid agonists, legal or illicit. This should be considered when prescribing or dispensing DEMEROL in situations where the physician or pharmacist is concerned about an increased risk of misuse, abuse, or diversion. DEMEROL has been reported as being abused by crushing, chewing, snorting, or injecting the dissolved product. These practices will result in the uncontrolled delivery of the opioid and pose a significant risk to the abuser that could result in overdose or death (see WARNINGS and DRUG ABUSE AND ADDICTION). Concerns about abuse, addiction, and diversion should not prevent the proper management of pain. Healthcare professionals should contact their State Professional Licensing Board or State Controlled Substances Authority for information on how to prevent and detect abuse or diversion of this product. Interactions with Alcohol and Drugs of Abuse Meperidine may be expected to have additive effects when used in conjunction with alcohol, other opioids, or illicit drugs that cause central nervous system depression. 2 Reference ID: 2931372 This label may not be the latest approved by FDA. For current labeling information, please visit Head Injury and Increased Intracranial Pressure: The respiratory depressant effects of meperidine and its capacity to elevate cerebrospinal fluid pressure may be markedly exaggerated in the presence of head injury, other intracranial lesions, or a preexisting increase in intracranial pressure. Furthermore, narcotics produce adverse reactions which may obscure the clinical course of patients with head injuries. In such patients, meperidine must be used with extreme caution and only if its use is deemed essential. Asthma and Other Respiratory Conditions: Meperidine should be used with extreme caution in patients having an acute asthmatic attack, patients with chronic obstructive pulmonary disease or cor pulmonale, patients having a substantially decreased respiratory reserve, and patients with preexisting respiratory depression, hypoxia, or hypercapnia. In such patients, even usual therapeutic doses of narcotics may decrease respiratory drive while simultaneously increasing airway resistance to the point of apnea. Hypotensive Effect: The administration of meperidine may result in severe hypotension in the postoperative patient or any individual whose ability to maintain blood pressure has been compromised by a depleted blood volume or the administration of drugs such as the phenothiazines or certain anesthetics. Usage in Ambulatory Patients: Meperidine may impair the mental and/or physical abilities required for the performance of potentially hazardous tasks such as driving a car or operating machinery. The patient should be cautioned accordingly. Meperidine, like other narcotics, may produce orthostatic hypotension in ambulatory patients. Usage in Pregnancy: Meperidine should not be used in pregnant women prior to the labor period, unless in the judgment of the physician the potential benefits outweigh the possible risks, because safe use in pregnancy prior to labor has not been established relative to possible adverse effects on fetal development. Labor and Delivery: Meperidine crosses the placental barrier and can produce depression of respiration and psychophysiologic functions in the newborn. Resuscitation may be required (See OVERDOSAGE). Therefore meperidine is not recommended during labor. Nursing Mothers: Meperidine appears in the milk of nursing mothers receiving the drug. Due to the potential for serious adverse reactions in nursing infants, a decision should be made whether to discontinue nursing or to discontinue the drug, taking into account the potential benefits of the drug to the nursing woman. PRECAUTIONS General Opioid analgesics can have a narrow therapeutic index in certain patient populations, particularly when combined with CNS depressant drugs. The use of these products should be reserved for cases where the benefits of opioid analgesia outweigh the known risks of respiratory depression, altered mental state, and postural hypotension. Use of DEMEROL may be associated with increased potential risks and should be used with caution in the following conditions: sickle cell anemia, pheochromocytoma, acute alcoholism; 3 Reference ID: 2931372 This label may not be the latest approved by FDA. For current labeling information, please visit adrenocortical insufficiency (e.g., Addison’s disease); CNS depression or coma; delirium tremens; debilitated patients; kyphoscoliosis associated with respiratory depression; myxedema or hypothyroidism; prostatic hypertrophy or urethral stricture; severe impairment of hepatic, pulmonary, or renal function; and toxic psychosis . The administration of meperidine may obscure the diagnosis or clinical course in patients with acute abdominal conditions. All opioids may induce or aggravate seizures in some clinical settings. Interactions with other CNS Depressants DEMEROL should be used with caution and consideration should be given to starting with a reduced dosage in patients who are concurrently receiving other central nervous system depressants including sedatives or hypnotics, general anesthetics, phenothiazines, other tranquilizers, and alcohol. Drug-drug interactions may result in respiratory depression, hypotension, profound sedation, or coma if these drugs are taken in combination with the usual doses of DEMEROL. Interactions with Mixed Agonist/Antagonist Opioid Analgesics Agonist/antagonist analgesics (i.e., pentazocine, nalbuphine, butorphanol, and buprenorphine) should be administered with caution to a patient who has received or is receiving a course of therapy with a pure opioid agonist analgesic such as meperidine. In this situation, mixed agonist/antagonist analgesics may reduce the analgesic effect of meperidine and/or may precipitate withdrawal symptoms in these patients. Supraventricular Tachycardias: Meperidine should be used with caution in patients with atrial flutter and other supraventricular tachycardias because of a possible vagolytic action which may produce a significant increase in the ventricular response rate. Convulsions: Meperidine may aggravate preexisting convulsions in patients with convulsive disorders. If dosage is escalated substantially above recommended levels because of tolerance development, convulsions may occur in individuals without a history of convulsive disorders. Acute Abdominal Conditions: The administration of meperidine or other narcotics may obscure the diagnosis or clinical course in patients with acute abdominal conditions. Tolerance and Physical Dependence Tolerance is the need for increasing doses of opioids to maintain a defined effect such as analgesia (in the absence of disease progression or other external factors). Physical dependence is manifested by withdrawal symptoms after abrupt discontinuation of a drug or upon administration of an antagonist. Physical dependence and tolerance are not unusual during chronic opioid therapy. The opioid abstinence or withdrawal syndrome is characterized by some or all of the following: restlessness, lacrimation, rhinorrhea, yawning, perspiration, chills, myalgia, mydriasis. Other symptoms also may develop, including: irritability, anxiety, backache, joint pain, weakness, abdominal cramps, insomnia, nausea, anorexia, vomiting, diarrhea, or increased blood pressure, respiratory rate, or heart rate. 4 Reference ID: 2931372 This label may not be the latest approved by FDA. For current labeling information, please visit In general, opioids used regularly should not be abruptly discontinued. Use in Drug and Alcohol Addiction DEMEROL is an opioid with no approved use in the management of addictive disorders. Its proper usage in individuals with drug or alcohol dependence, either active or in remission, is for the management of pain requiring opioid analgesia. DEMEROL should be used with caution in patients with alcoholism and other drug dependencies due to the increased frequency of narcotic tolerance, dependence, and the risk of addiction observed in these patient populations. Abuse of DEMEROL in combination with other CNS depressant drugs can result in serious risk to the patient. Information for Patients/Caregivers If clinically advisable, patients receiving DEMEROL (meperidine hydrochloride) tablets or their caregivers should be given the following information by the physician, nurse, pharmacist, or caregiver: 1. Patients should be aware that DEMEROL tablets contain meperidine, which is a morphine-like substance. 2. Patients should be advised to report pain and adverse experiences occurring during therapy. Individualization of dosage is essential to make optimal use of this medication. 3. Patients should be advised not to adjust the dose of DEMEROL without consulting the prescribing professional. 4. Patients should be advised that DEMEROL may impair mental and/or physical ability required for the performance of potentially hazardous tasks (e.g., driving, operating heavy machinery). 5. Patients should not combine DEMEROL with alcohol or other central nervous system depressants (sleep aids, tranquilizers) except by the orders of the prescribing physician, because dangerous additive effects may occur, resulting in serious injury or death. 6. Women of childbearing potential who become, or are planning to become pregnant should be advised to consult their physician regarding the effects of analgesics and other drug use during pregnancy on themselves and their unborn child. 7. Patients should be advised that DEMEROL is a potential drug of abuse. They should protect it from theft, and it should never be given to anyone other than the individual for whom it was prescribed. 8. Patients should be advised that if they have been receiving treatment with DEMEROL for more than a few weeks and cessation of therapy is indicated, it may be appropriate to taper the DEMEROL dose, rather than abruptly discontinue it, due to the risk of precipitating withdrawal symptoms. Their physician can provide a dose schedule to accomplish a gradual discontinuation of the medication. 9. Patients should be instructed to keep DEMEROL in a secure place out of the reach of children. When DEMEROL is no longer needed, the unused tablets should be destroyed by flushing down the toilet. 5 Reference ID: 2931372 This label may not be the latest approved by FDA. For current labeling information, please visit Drug Interactions: Also see WARNINGS. Acyclovir: Plasma concentrations of meperidine and its metabolite, normeperidine, may be increased by acyclovir, thus caution should be used with concomitant administration. Cimetidine: Cimetidine reduced the clearance and volume of distribution of meperidine and also the formation of the metabolite, normeperidine, in healthy subjects and thus, caution should be used with concomitant administration. Phenytoin: The hepatic metabolism of meperidine may be enhanced by Phenytoin. Concomitant administration resulted in reduced half-life and bioavailability with increased clearance of meperidine in healthy subjects, however, blood concentrations of normeperidine were increased. Ritonavir: Plasma concentrations of the active metabolite normeperidine may be increased by ritonavir, thus concomitant administration should be avoided. Opioid analgesics, including DEMEROL, may enhance the neuromuscular blocking action of skeletal muscle relaxants and produce an increased degree of respiratory depression. Special Risk Patients: Meperidine should be given with caution and the initial dose should be reduced in certain patients such as the elderly or debilitated, and those with severe impairment of hepatic or renal function, Sickle Cell Anemia, hypothyroidism, Addison’s disease, Pheochromocytoma and prostatic hypertrophy or urethral stricture. In patients with pheochromocytoma, meperidine has been reported to provoke hypertension. Usage in Hepatically Impaired Patients: Accumulation of meperidine and/or its active metabolite, normeperidine, can occur in patients with hepatic impairment. Meperidine should therefore be used with caution in patients with hepatic impairment. Usage in Renally Impaired Patients: Accumulation of meperidine and/or its active metabolite, normeperidine, can also occur in patients with renal impairment. Meperidine should therefore be used with caution in patients with renal impairment. Carcinogenesis, mutagenesis, impairment of fertility: Studies to assess the carcinogenic or mutagenic potential of meperidine have not been conducted. Studies to determine the effect of meperidine on fertility have not been conducted. Pregnancy: Teratogenic effects. Pregnancy Category C: Animal reproduction studies have not been conducted with meperidine. It is also not known whether DEMEROL can cause fetal harm when administered to a pregnant woman or can affect reproduction capacity. DEMEROL should be given to a pregnant woman only if clearly needed. Labor and Delivery: See WARNINGS. Nursing Mothers: See WARNINGS. Pediatric Use: The safety and effectiveness of meperidine in pediatric patients has not been established. Literature reports indicate that meperidine has a slower elimination rate in neonates and young infants compared to older children and adults. Neonates and young infants may also 6 Reference ID: 2931372 This label may not be the latest approved by FDA. For current labeling information, please visit be more susceptible to the effects, especially the respiratory depressant effects. If meperidine use is contemplated in neonates or young infants, any potential benefits of the drug need to be weighed against the relative risk to the patient. Geriatric Use: Clinical studies of DEMEROL during product development did not include sufficient numbers of subjects aged 65 and over to evaluate age-related differences in safety or efficacy. Literature reports indicate that geriatric patients have a slower elimination rate compared to young patients and they may be more susceptible to the effects of meperidine. A reduction in the total daily dose of meperidine may be required in elderly patients, and the potential benefits of the drug weighed against the relative risk to a geriatric patient. ADVERSE REACTIONS The major hazards of meperidine, as with other narcotic analgesics, are respiratory depression and, to a lesser degree, circulatory depression; respiratory arrest, shock, and cardiac arrest have occurred. The most frequently observed adverse reactions include lightheadedness, dizziness, sedation, nausea, vomiting, and sweating. These effects seem to be more prominent in ambulatory patients and in those who are not experiencing severe pain. In such individuals, lower doses are advisable. Some adverse reactions in ambulatory patients may be alleviated if the patient lies down. Other adverse reactions include: Nervous System: Euphoria, dysphoria, weakness, headache, agitation, tremor, uncoordinated muscle movements (e.g. muscle twitches, myoclonus), severe convulsions, transient hallucinations and disorientation, visual disturbances. Gastrointestinal: Dry mouth, constipation, biliary tract spasm. Cardiovascular: Flushing of the face, tachycardia, bradycardia, palpitation, hypotension (see WARNINGS), syncope. Genitourinary: Urinary retention. Allergic: Pruritus, urticaria, other skin rashes, wheal and flare over the vein with intravenous injection. Hypersensitivity reactions, anaphylaxis, shock. Histamine release leading to hypotension and/or tachycardia, flushing, sweating, and pruritus. DRUG ABUSE AND ADDICTION DEMEROL contains meperidine, a mu-agonist opioid with an abuse liability similar to morphine and is a Schedule II controlled substance. Meperidine, like morphine and other opioids used in analgesia, can be abused and is subject to criminal diversion. Drug addiction is characterized by compulsive use, use for non-medical purposes, and continued use despite harm or risk of harm. Drug addiction is a treatable disease, utilizing a multi disciplinary approach, but relapse is common. 7 Reference ID: 2931372 This label may not be the latest approved by FDA. For current labeling information, please visit “Drug seeking” behavior is very common in addicts and drug abusers. Drug-seeking tactics include emergency calls or visits near the end of office hours, refusal to undergo appropriate examination, testing or referral, repeated “loss” of prescriptions, tampering with prescriptions and reluctance to provide prior medical records or contact information for other treating physician(s). “Doctor shopping” to obtain additional prescriptions is common among drug abusers and people suffering from untreated addiction. Abuse and addiction are separate and distinct from physical dependence and tolerance. Physicians should be aware that addiction may not be accompanied by concurrent tolerance and symptoms of physical dependence in all addicts. In addition, abuse of opioids can occur in the absence of true addiction and is characterized by misuse for non-medical purposes, often in combination with other psychoactive substances. DEMEROL, like other opioids, has been diverted for non-medical use. Careful record-keeping of prescribing information, including quantity, frequency, and renewal requests is strongly advised. Abuse of DEMEROL poses a risk of overdose and death. This risk is increased with concurrent abuse of DEMEROL with alcohol and other substances. Due to the presence of talc as one of the excipients in tablets, parenteral abuse of crushed tablets can be expected to result in local tissue necrosis, infection, pulmonary granulomas, and increased risk of endocarditis and valvular heart disease. In addition, parenteral drug abuse is commonly associated with transmission of infectious diseases such as hepatitis and HIV. Proper assessment of the patient, proper prescribing practices, periodic re-evaluation of therapy, and proper dispensing and storage are appropriate measures that help to limit abuse of opioid drugs. OVERDOSAGE Symptoms: Serious overdosage with meperidine is characterized by respiratory depression (a decrease in respiratory rate and/or tidal volume, Cheyne-Stokes respiration, cyanosis), extreme somnolence progressing to stupor or coma, skeletal muscle flaccidity, cold and clammy skin, and sometimes bradycardia and hypotension. In severe overdosage, particularly by the intravenous route, apnea, circulatory collapse, cardiac arrest, and death may occur. Treatment: Primary attention should be given to the reestablishment of adequate respiratory exchange through provision of a patent airway and institution of assisted or controlled ventilation. The narcotic antagonist, naloxone hydrochloride, is a specific antidote against respiratory depression which may result from overdosage or unusual sensitivity to narcotics, including meperidine. Therefore, an appropriate dose of this antagonist should be administered, preferably by the intravenous route, simultaneously with efforts at respiratory resuscitation. An antagonist should not be administered in the absence of clinically significant respiratory or cardiovascular depression. Oxygen, intravenous fluids, vasopressors, and other supportive measures should be employed as indicated. In cases of overdosage with DEMEROL tablets, the stomach should be evacuated by emesis or gastric lavage. 8 Reference ID: 2931372 This label may not be the latest approved by FDA. For current labeling information, please visit NOTE: In an individual physically dependent on narcotics, the administration of the usual dose of a narcotic antagonist will precipitate an acute withdrawal syndrome. The severity of this syndrome will depend on the degree of physical dependence and the dose of antagonist administered. The use of narcotic antagonists in such individuals should be avoided if possible. If a narcotic antagonist must be used to treat serious respiratory depression in the physically dependent patient, the antagonist should be administered with extreme care and only one-fifth to one-tenth the usual initial dose administered. DOSAGE AND ADMINISTRATION For Relief of Pain Dosage should be adjusted according to the severity of the pain and the response of the patient. Meperidine is less effective orally than on parenteral administration. The dose of DEMEROL should be proportionately reduced (usually by 25 to 50 percent) when administered concomitantly with phenothiazines and many other tranquilizers since they potentiate the action of DEMEROL. Adults: The usual dosage is 50 mg to 150 mg orally, every 3 or 4 hours as necessary. Pediatric Patients: The usual dosage is 1.1 mg/kg to 1.8 mg/kg orally, up to the adult dose, every 3 or 4 hours as necessary. SAFETY AND HANDLING DEMEROL (meperidine HCl) tablets contain meperidine hydrochloride which is a controlled substance. Like morphine, meperidine is controlled under Schedule II of the Controlled Substances Act. Meperidine, like all opioids, is liable to diversion and misuse and should be handled accordingly. Patients and their families should be instructed to flush DEMEROL tablets that are no longer needed. DEMEROL has been targeted for theft and diversion by criminals. Healthcare professionals should contact their State Professional Licensing Board or State Controlled Substance Authority for information on how to prevent and detect abuse or diversion of this product. HOW SUPPLIED For Oral Use Tablets are white, round and convex. The 50 mg tablet has a stylized “W” on one side and “M” score “35” on the other side. The 100 mg tablet has a stylized “W” on one side and “D” score “37” on the other side. Tablets of 50 mg, bottles of 100 (NDC 0024-0335-04) and 100 mg, bottles of 100 (NDC 0024 0337-04). Store at 25° C (77° F); excursions permitted to 15° - 30° C (59° - 86° F) [See USP Controlled Room Temperature]. Rx only 9 Reference ID: 2931372 This label may not be the latest approved by FDA. For current labeling information, please visit Revised September 2010 Manufactured for: sanofi-aventis U.S. LLC Bridgewater, NJ 08807 ©2010 sanofi-aventis U.S. LLC Reference ID: 2931372 10 This label may not be the latest approved by FDA. For current labeling information, please visit
2683	https://www.youtube.com/watch?v=fcooyOIvtc4	Basic Trigonometry \| AOPS Introduction To Geometry Problem 18.18 \| Ronchen Math Ronchen's Studio 179 subscribers 5 likes Description 298 views Posted: 6 Apr 2022 In this video, we are going to learn some basic trigonometry and a very useful theory to calculate the area of a triangle. Thanks for watching! Trigonometry #AOPS #Geometry Transcript:
2684	https://www.gauthmath.com/solution/1743092868241413	The number 2.2 10^(-7) is written in scientific notation. [Math] Drag Image or Click Here to upload Command+to paste Upgrade Sign in Homework Homework Assignment Solver Assignment Calculator Calculator Resources Resources Blog Blog App App Gauth Unlimited answers Gauth AI Pro Start Free Trial Homework Helper Study Resources Math Arithmetic Questions Question The number 2.2 10^(-7) is written in scientific notation. Gauth AI Solution 94%(283 rated) Answer The answer is The number $$2.2\times 10^{-7}$$2.2×1 0−7 is written in scientific notation. Explanation This question asks to confirm if the given number is written in scientific notation. Understand the definition of scientific notation. Scientific notation is a way of expressing numbers that are too large or too small to be conveniently written in decimal form. It is commonly used by scientists, mathematicians, and engineers, in part because it can simplify certain arithmetic operations. A number is written in scientific notation when it is expressed in the form $$a \times 10^b$$a×1 0 b, where $$a$$a is a number greater than or equal to 1 and less than 10 (i.e., $$1 \le \|a\| < 10$$1≤∣a∣<10), and $$b$$b is an integer. Analyze the given number. The given number is $$2.2 \times 10^{-7}$$2.2×1 0−7 Check if the number conforms to the definition of scientific notation. In the given number, $$a = 2.2$$a=2.2 and $$b = -7$$b=−7 We need to check if $$1 \le \|a\| < 10$$1≤∣a∣<10. Since $$a = 2.2$$a=2.2, we have $$\|a\| = \|2.2\| = 2.2$$∣a∣=∣2.2∣=2.2 The condition $$1 \le 2.2 < 10$$1≤2.2<10 is true. We also need to check if $$b$$b is an integer. Since $$b = -7$$b=−7, and $$-7$$−7 is an integer, this condition is also true. Conclude whether the number is written in scientific notation. Since both conditions for scientific notation are met, the number $$2.2 \times 10^{-7}$$2.2×1 0−7 is indeed written in scientific notation. Explain Simplify this solution Gauth AI Pro Back-to-School 3 Day Free Trial Limited offer! Enjoy unlimited answers for free. Join Gauth PLUS for $0 Related Perform the indicated operation. Write the answer in SCIENTIFIC notation. 5.45 10-24.09 10-6 2.2 10-7 2.2 1012 2.23 10-7 2.23 1012 100% (4 rated) Simplify the expression frac 2.2 10-7 4021.7 106 1.3 106 using scientific notation and express your answer in scientific notation. Round your answer to the nearest thousandth 5 1018 2.2 10-7=square Use scientific notation. Use the multiplication symbol in the math palette as nee 100% (1 rated) 5 1017 2.2 10-7=square Use scientific notation. Use the multiplication symbol in the math palette as needed. 100% (1 rated) My IAL Learing Eighth grade D.3 Compare numbers written in scientific notation RHT Which sign makes the statement true? 2.2 10-7 0.000000022 1015 < = Submit 98% (23 rated) Write this number as a whole number or a decimal. 2.2 10-7=square 100% (3 rated) 6.4 108 divided by 2.2 10-7= _ 100% (3 rated) How may different arrangements are there of the letters in The number of possible arrangements is MISSISSIPPI？ 100% (2 rated) Which sign makes the statement true? 0.00000022 2.2 10-7 >< = 100% (5 rated) Solve the following inequality algebraically. 5x-5/x+2 ≤ 4 What is the solution? -2,13 Type your answer in interval notation. Simplify your answer. Use integers or fractions for any numbers in the expression. 100% (4 rated) Gauth it, Ace it! contact@gauthmath.com Company About UsExpertsWriting Examples Legal Honor CodePrivacy PolicyTerms of Service Download App
2685	https://math.stackexchange.com/questions/1598035/how-to-calculate-the-drop-as-one-moves-in-a-straight-line-from-the-surface-of-th	geometry - How to calculate the drop as one moves in a straight line from the surface of the earth - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How to calculate the drop as one moves in a straight line from the surface of the earth Ask Question Asked 9 years, 9 months ago Modified9 years, 9 months ago Viewed 3k times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. I am trying to come up with a formula that will calculate the drop as one moves in a straight line from earth. In other words, if one travels in a straight line from the surface of the earth, to what distance will the surface fall away as a function of the distance traveled? It will have to involve the formula for calculating the length of a chord of a circle. Put in geometric terms, the question would be: "What is the distance from a line tangent to a circle to the surface of the circle, the distance being perpendicular to the tangent?" The formula for calculating the length of a chord is l = 2 r 2–d 2−−−−−√r 2–d 2 where r is the radius of the circle and d is the perpendicular distance of the chord from the center of the circle. The radius of the earth is 3,959 miles. How would one come up with the formula that would do this? geometry mathematical-astronomy Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Jan 3, 2016 at 8:04 PaulPaul asked Jan 3, 2016 at 7:41 PaulPaul 123 1 1 silver badge 4 4 bronze badges 8 When you say perpendicular distance, is this distance perpendicular to the surface, or to the tangent? For any positive distance, those two are not exactly the same.Brian Tung –Brian Tung 2016-01-03 07:44:11 +00:00 Commented Jan 3, 2016 at 7:44 I mean the distance perpendicular to the tangent in the case of the tangent, and perpendicular to the chord in the case of the chord.Paul –Paul 2016-01-03 07:45:49 +00:00 Commented Jan 3, 2016 at 7:45 The question has been updated accordingly, thanks.Paul –Paul 2016-01-03 08:08:34 +00:00 Commented Jan 3, 2016 at 8:08 A 'straight' line doesn't exists on the surface of the earth, is it ok to consider a geodesic?mrprottolo –mrprottolo 2016-01-03 08:36:52 +00:00 Commented Jan 3, 2016 at 8:36 I don't understand the question.Paul –Paul 2016-01-03 08:56:17 +00:00 Commented Jan 3, 2016 at 8:56 \|Show 3 more comments 2 Answers 2 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. It is not totally clear what you are asking but looking at this: then a=r 2+d 2−−−−−−√−r a=r 2+d 2−r b=r−r 2−d 2−−−−−−√b=r−r 2−d 2 though b b is only real when d≤r d≤r. When d d is much smaller than r r then both a a and b b are about d 2 2 r d 2 2 r. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 3, 2016 at 10:25 HenryHenry 171k 10 10 gold badges 139 139 silver badges 297 297 bronze badges 5 Ok, thanks for this. So, would you know how to reformulate this if one is given a radius, which in this case for the earth would be 3,959 miles? In other words, after substituting 3,959 for r, how could the formula be simplified?Paul –Paul 2016-01-03 22:08:08 +00:00 Commented Jan 3, 2016 at 22:08 I have seen a formula b = 8in. d ^2, where d is the distance in miles. This seems to check out with my results on a graphing calculator, etc., but I am wondering how it is derived.Paul –Paul 2016-01-03 22:24:49 +00:00 Commented Jan 3, 2016 at 22:24 d 2 2 r d 2 2 r would give about 0.0001263 d 2 0.0001263 d 2 miles if d d is measured in miles. Since there are 63360 63360 inches in a mile, you might think of that as about 8.002 d 2 8.002 d 2 inches if d d is measured in miles.Henry –Henry 2016-01-03 23:55:16 +00:00 Commented Jan 3, 2016 at 23:55 Forgive my ignorance, how does it happen that if d is much smaller the r you get the above formula, and what are the steps to get to it?Paul –Paul 2016-01-04 01:53:31 +00:00 Commented Jan 4, 2016 at 1:53 1 If d d is much less than r r then r−r 2−d 2−−−−−−√≈r−r 2−d 2+d 4 4 r 2−−−−−−−−−−−√=r−(r 2−d 2 2 r)2−−−−−−−−−√=d 2 2 r r−r 2−d 2≈r−r 2−d 2+d 4 4 r 2=r−(r 2−d 2 2 r)2=d 2 2 r. Similarly r 2+d 2−−−−−−√−r≈r 2+d 2+d 4 4 r 2−−−−−−−−−−−√−r=(r 2+d 2 2 r)2−−−−−−−−−√−r=d 2 2 r r 2+d 2−r≈r 2+d 2+d 4 4 r 2−r=(r 2+d 2 2 r)2−r=d 2 2 r Henry –Henry 2016-01-04 08:17:14 +00:00 Commented Jan 4, 2016 at 8:17 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. If I understood what you are asking, you want a formula to find x x, given R R the radius of the Earth and d d the distance traveled. First notice that x=R−R cos θ x=R−R cos⁡θ and since θ=d/R θ=d/R the formula is x=R(1−cos d R).x=R(1−cos⁡d R). For example, if you travel d=200 k m d=200 k m : x=∼3 k m.x=∼3 k m. Notice that this does not mean that every time you travel 200 k m 200 k m you 'fall' down of 3 k m 3 k m, this is only true when you travel along a maximal cricle (called geodesic). In the other cases the falling distance will be smaller. This is because you can't talk about straight line on the surface of earth, since a straight line lies on a plane and there is no plane contained in the surface of a sphere. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jan 3, 2016 at 10:36 answered Jan 3, 2016 at 10:31 mrprottolomrprottolo 3,812 1 1 gold badge 15 15 silver badges 22 22 bronze badges 2 For d d much less than R R you too have x≈d 2 2 R x≈d 2 2 R, as with my results.Henry –Henry 2016-01-04 08:18:51 +00:00 Commented Jan 4, 2016 at 8:18 Interesting and thanks for this, although what I was trying to ask was what Henry answered, i.e. not the drop for the distance traveled over the surface of the earth, but the drop for the distance traveled if one travels through space in a straight line from earth.Paul –Paul 2016-01-06 01:39:38 +00:00 Commented Jan 6, 2016 at 1:39 Add a comment\| You must log in to answer this question. 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2686	https://www.youtube.com/watch?v=5g7ktrauxJA	Problem 3 on IC engine, volumetric efficiency, indicated thermal efficiency, mechanical efficiency Edutech Guru Engineering Learning-By Priyanka ma'am 13400 subscribers 14 likes Description 869 views Posted: 24 Sep 2024 Find the speed at which a four cylinder engine using gas as a fuel can develop a brake Power of 50 kW working under following conditions: air-gas ratio 9:1; calorific value of fuel=34 MJ/m^3 compression ratio 10:1 volumetric efficiency 70% indicated thermal efficiency 35% mechanical efficiency 80% total volume of the engine 2 litres Please refer my following Playlists , Links are given: 1. Theory of Machines or Kinematics of Machines play list Fluid mechanics and Hydraulic machines playlist Refrigeration and air conditioning playlist Renewable Energy engineering or Non-convensional energy sources play list Dynamics of Machinery play list Engineering graphics or Engineering drawing play list Engineering Mechanics Playlist Power plant engineering play list Engineering Thermodynamics play list Heat and mass transfer play list Operation research play list Strength of material playlist Control Engineering/ Control systems engineering playlist 5 comments Transcript: given question find the speed at which four cylinder engine using gas as a fuel develop a brake power of 50 kilow working under following conditions air to gas ratio 9 is to1 calorific value of the field 34 m per M Cube compression ratio 10 is to 1 volumetric efficiency 70% indicated thermal effic efficiency 35% and mechanical efficiency 80% total volume of the engine is 2 L let us understand given data this is 4 cylinder engine that is number of cylinders is equal to 4 brake power BP is equal to 50 kilowatt air to gas ratio here gas is used as a fuel so we can say that a a by F ratio is 9 is 1 calorific value of Fu so calorific value is given 34 m per M Cube but the standard unit is K per M Cube so how to convert it we will write here as a 34 into 103 K per M Cube then compression ratio R is = 10 is 1 volume volumetric efficiency n v 70% indicated thermal efficiency n it 35% mechanical efficiency n m 80% and the total volume of Engine 2 L so we will first calculate the total volume per cylinder now the unit is given in lit so we have to convert it into CM Cub so how to convert it so how to convert the liter so 1 L is = 10 3 cm Cub so this is the relation now we have to calculate this total volume per cylinder so there number of cylinders four so we have to divide it by four so 2 L that is 2 into 10 3 cm Cub / 4 is equal 500 cm Cub so CM Cube we will write here as a cc then compress ratio is given now we get this total volume now we have to find out the swept volume so how to find out the swept volume so with the help of compression ratio so compression ratio R is equal to total volume divided by clearance volume so when we put the value R is equal to 10 is to 1 that is 10 by 1 is equal to this 500 / VC so VC is equal to 50 cm CU now we know that this total volume VT is equal to vs + VC so we will get the value of vs when we put the value of VT VT is 500 and VC is 50 so swept volume is equal to 450 CM Cub so this is vs and that is required for the further calculation now we will move to the next volume volumetric efficiency now this volumetric efficiency is given 70% so what is this volumetric efficiency that is the volume of air taken in per cycle that is V air that is volume of air taken in per cycle divided by swept volume so this is the ratio of volume volume of air taken in divided by swept volume is known as volumetric efficiency so when we put the value so volumetric efficiency 70% that is 70 by 100 that is 0.7 is equal to V air ided vs that is 450 so V A that is the volume of air taken in per cycle is equal to 315 CM Cub so this is the value of volume of air taken in so how to calculate the volume of gas taken in so for that air to fuel ratio is given so in this case fuel is known as gas that is fuel used as a gas so air to fuel ratio is given so we can say that a by f is equal to volume of air taken in per cycle divided by volume of gas taken in per cycle now from this we can calculate the volume of gas taken in per cycle so volume of air taken in 315 divided by volume of gas taken in as it is is equal to a by f a a by f is 9 is 1 so volume of gas taken in per cycle when we calculate we will get the answer 35 CM Cub when we multiply volume of gas taken in with calorific value of fuel then we will get the en supplied per cylinder so this energy supplied per cylinder e is equal to this volume of gas taken in per cycle that is 35 CM Cub multiplied by calorific value so it's a unit is K per M Cube 34 into 10 3 K per M Cube and this is in cm Cube so we will convert it into M Cube so how to convert it so 1 cm that is equal to 10 - 2 m so we will write here 35 into 10 - 2 m and we have to take its Cube that is bracket ra to 3 so 3 2 that is 6 so we will write here 35 into 10 - 6 34 into 10 3 so it's a unit is M Cube so this meter cube and this denominator me Cub is getting cancelled and we will get the answer 1.19 K so this is the energy supplied per cylinder now we have to find out this speed in so we will take the another formula to find out energy supplied per cylinder with the help of this indicated thermal efficiency so indicated thermal efficiency n it is equal to indicated power divided by energy supplied per cylinder per second so here one extra unit that is per second now how to find out this IP because it is not mentioned in the question so n mechanical that is the mechanical efficiency n m isal to BP / IP so instead of Ip we will write BP ided by n m so BP is given that is 50 kilowatt then n m is 80% that is 0.8 and here per cylinder that means we have to divide it by number of cylinders so here four so how to write it IP is equal to BP that is 50 ided 0.8 so instead of Ip we will write 50 / 0.8 then divided by daa I because we have to calculate this so this term for this term we have to take this in the denominator Nita indicated thermal efficiency that is 35% that is 0.35 and we have to also divide with number of cylinders that is four so when we solve this then we will get energy supplied per cylinder per second is equal to 4 34.6 K now we have to compare these two terms energy supplied per cylinder and this is the equation number one and energy supplied per cylinder per second that is 44.6 K so we will say this is the equation number two now we have to convert this term into energy supplied per cylinder so how to convert this we have to remove REM this per second term we will write the formula as energy supplied per cylinder per second divided by number of Power Stroke per second is equal to energy supplied per cylinder so if we observe if we divide it with number of power strokes per second then we will get energy supplied per cylinder so energy supplied per cylinder per second is 44.6 K so we have to write this now how to calculate the number of Power Stroke per second so this speed n is in RPM but we have to take number of Power Stroke per second so we have to divide this n with 60 and we have to take the number of Power Stroke for this four stroke engine that is n by 2 so we will take n / by 2 multiplied by 60 so when we simplify this then we will get here this 44.6 into 120 divided by n and that is equal to 53 56.8 / n now if we observe this is the value of energy supplied per cylinder now if we compare the equation 1 and equation three then the left hand side is energy supplied per cylinder is same so we can write here 53 56 6.8 / n is = 1.19 therefore n is = 4,500 RPM
2687	https://settingthestandard.us/elapsed-time-using-a-number-line/	Elapsed Time Using a Number Line - Setting the Standard Skip to content HOME BLOG ABOUT SUBSCRIBE CONTACT SHOP Menu HOME BLOG ABOUT SUBSCRIBE CONTACT SHOP March 6, 2022 Elapsed Time Using a Number Line Elapsed time is one of those skills that stumps so many of my students! What makes it tough for students is they want to simply subtract to find their answer. To help avoid this error, I like to teach elapsed time using a number line. Using visual models such as number lines work well because they are simple, to the point, and students can accurately solve problems using them. Elapsed Time on a Number Line I like to start my lesson by introducing the new vocabulary terms related to the skill. I add vocabulary posters like these to my math word wall so students can refer to them as needed during our unit (you can read more about why math vocabulary is important here). Elapsed Time Vocabulary Posters I think it’s always best to start with some fairly simple problems. We start off learning about elapsed time using analog clocks. To do this, students note the start and end times. Then, they add that information to the number line. Next, they start at the start time and count forward using increments of 1’s, 5’s, or 10’s until they reach the end time. Students can then count the increments of time to calculate the elapsed time. In this case, the elapsed time is 37 minutes. I like starting off with analog clocks because if students need extra support, they can draw the number line wrapping around the start time clock and transfer it to the number line. If you are looking for some guided practice for your class, you can grab this free elapsed time worksheet here. If your students need to review how to tell time to the minute, you can read more about that here. Elapsed Time Worksheet Elapsed Time Word Problems After students get the concept of elapsed time, we move on to elapsed time word problems. For example, Lily starts her homework at 5:03. She finishes her homework at 5:39. How many minutes does Lily spend on her homework? Students start by plotting the start and end times on a number line. Then, students begin with the start time and count forward in increments of 1’s, 5’s, or 10’s until they reach the end time. Students count the increments of time to calculate the elapsed time. In this case, the elapsed time is 36 minutes- Lily spent 36 minutes working on her homework. Elapsed Time Word Problems on a Number Line When your students are ready to practice their new skills on their own, I like using these Elapsed Time Boom Cards for whole class or independent practice. Your students will love the self-correcting practice, and you will easily be able to monitor who still needs extra support with this skill. Elapsed Time Word Problems I hope this strategy brings your students success with elapsed time! Leave a Reply Cancel reply You must be logged in to post a comment. Hi, I'm Kelie! I love to create fun, engaging, and meaningful educational resources to help save you time and make teaching easier. My standards-based digital and printable activities are geared towards 2nd, 3rd, & 4th-grade students. If you’re searching for high-quality and budget-friendly resources for your classroom, you’ve come to the right place! Let's Connect! InstagramPinterest Subscribe to My Email List Interested in Fun Freebies and Tips? Subscribe to get our latest content by email. Subscribe We won't send you spam. Unsubscribe at any time. Built with Kit Setting the Standard offers educational resources and teaching tips for the middle grades. Subscribe to get my latest teaching tips and freebies. You’ll receive my Multiplication Fluency freebie with the link below. 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2688	https://derivative-calculator.org/proofs/cosh/	Derivative of cosh(x) - Proof and Explanation \| Derivative-Calculator.org Derivative Calculator Differentiation Rules Proofs en English Español Français Italiano Português 日本語 Derivative Calculator Differentiation Rules Proofs Proofs of Common Derivatives Derivative of cosh(x) - Proof and Explanation Derivative of cosh(x) - Proof and Explanation Proof The hyperbolic cosine function is defined as: cosh(x)=e x+e−x 2 To find the derivative, we differentiate using the sum rule: d d x cosh(x)=d d x(e x+e−x 2) Differentiating each term: 1 2(d d x e x+d d x e−x)=1 2(e x−e−x) This simplifies to: sinh(x) Thus, the derivative of cosh(x) is: d d x cosh(x)=sinh(x) Explanation The hyperbolic cosine function, cosh(x), is defined as e x+e−x 2. This formula combines the exponential functions e x and e−x. To differentiate cosh(x), we use basic differentiation rules. The function can be broken down into 1 2(e x+e−x). Here, 1 2 is a constant factor that we can factor out during differentiation. We then apply the differentiation rule to each part of the expression. The derivative of e x is simply e x, and the derivative of e−x is −e−x due to the chain rule. Putting these results together, the derivative becomes 1 2(e x−e−x). This expression is the definition of sinh(x), the hyperbolic sine function. Therefore, the derivative of cosh(x) is sinh(x). Q.E.D. © 2023 – 2024 Derivative-Calculator.org The information provided by the derivative calculator on this website is for educational purposes only, and no liability or responsibility is assumed for the accuracy or correctness of the calculations. Please always double-check your results!
2689	https://www.youtube.com/watch?v=S_5YDo6Co9c	Prove that x 0 = 0 for all real numbers x (ILIEKMATHPHYSICS) ILIEKMATHPHYSICS 8060 subscribers 130 likes Description 3029 views Posted: 6 Jan 2025 This video is part of the “Real Analysis” series I am making. Thanks and enjoy the video! Real Analysis Playlist: Proof of the cancellation law of addition video: Here is our description of the real number system using the ten axioms. This approach is similar to the one given in Apostol's Calculus book, Volume I, Second Edition. We are given a set R, whose elements are called real numbers. We equip R with two binary operations, addition (+) and multiplication (•), so that for every pair of real numbers a and b, a + b and a • b are real numbers. In short, we may write a • b as ab. Axiom 1 (Commutative Laws). For all a,b in R, a + b = b + a and ab = ba. Axiom 2 (Associative Laws). For all a,b,c in R, (a + b) + c = a + (b + c) and (ab)c = a(bc). Axiom 3 (Distributive Law). For all a,b,c in R, a(b + c) = ab + ac. Axiom 4 (Existence of Zero). There exists an element 0 in R such that for all x in R, x + 0 = x. Axiom 5 (Existence of Negatives). For all x in R, there exists an element -x in R such that x + (-x) = 0. Axiom 6 (Existence of One). There exists an element 1 in R distinct from 0 such that for all x in R, 1 • x = x. Axiom 7 (Existence of Reciprocals). For all nonzero x in R, there exists an element x^-1 in R such that x • x^-1 = 1. Definition. For each a,b in R, we define a - b = a + (-b). Definition. For each a,b in R where b ≠ 0, we define a/b = a • b^-1. We equip R with a subset R+, whose elements we call positive real numbers, which satisfy the following axioms: Axiom 8 (Closure of R+ under + and •). For all x,y in R+, x + y and xy belong to R+. Axiom 9 (Trichotomy). For all x in R, exactly one of the following is true: x belongs to R+, x = 0, or -x belongs to R+. Definition. Let a,b in R. We say “a ﹥ b” means a - b belongs to R+. We say “a ﹤ b” means b ﹥ a. We say “a ≥ b” means a ﹥ b or a = b. We say “a ≤ b” means b ≥ a. Definition. We define R- = {x in R \| -x in R+}. The elements of this set we call negative real numbers. Definition. Let S be a subset of real numbers and b is an element of S. We say b is a smallest element of S if for all x in S, x ≥ b. Using properties of order, it is easy to check that b is unique. Definition. Let S be a subset of real numbers and w is a real number. (1) We say w is an upper bound of S if for all x in S, x ≤ w. (2) We say w is a lower bound of S if for all x in S, x ≥ w. (3) We say S is bounded above if S has an upper bound. We say S is bounded below if S has a lower bound. We say S is bounded if S is both bounded above and bounded below. Axiom 10 (Completeness of R). Every nonempty subset S of real numbers that is bounded above has a least upper bound. That is, the set of all upper bounds of S has a smallest element. 11 comments Transcript: hello in this video we are going to prove the following theorem for all real numbers x x 0 is equal zero now in this series we are using a list of 10 axum for the real number system and I'll leave that list of axioms in the description of the video below now in this video we are going to be using the following axioms Axiom 3 is just the distributive law Axiom 4 tells us that there exists a real number that we call zero that has the property that x + 0al x for all real numbers X and we're also going to use a property that we proved about the real number system and that property was the cancellation law of addition we have proven for all real numbers a b and c if a plus Bal a plus c then B is equal to see okay so now let's get into proving this theorem to start the proof since we're trying to prove a statement about every real number let's give ourselves an arbitrary real number I'll call it X from here we want to show that x 0 is equal to Z and applying Axiom 3 and four the observation is as follows first of all by axium 4 we know that x 0 + 0 is equal to x 0 applying axm 4 again we can replace 0 with 0 + 0 and then applying axm 3 we can distribute X across this parentheses so we have x 0 + 0 = x 0 + x 0 and now as you can see we're in in the position to apply the cancellation law of addition applying the cancellation law of addition we can cancel out the X time zeros so we're left with 0 = x 0 or in other words x 0 = 0 so we have shown that x 0 is equal 0 for an arbitrary real number X since X was arbitrary this means we have shown for all real numbers x x 0 is equal 0 and that is exactly what we wanted to prove so this completes the proof and so yeah that's pretty much it for this video
2690	https://brainly.com/question/39785020	[FREE] Which nitrogenous waste requires hardly any water for its excretion? A. Ammonia B. Urea C. Uric Acid D. - brainly.com 7 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +87,8k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +28,4k Ace exams faster, with practice that adapts to you Practice Worksheets +6,2k Guided help for every grade, topic or textbook Complete See more / Chemistry Textbook & Expert-Verified Textbook & Expert-Verified Which nitrogenous waste requires hardly any water for its excretion? A. Ammonia B. Urea C. Uric Acid D. Creatinine 1 See answer Explain with Learning Companion NEW Asked by spierboy8327 • 10/09/2023 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 2321851 people 2M 0.0 0 Upload your school material for a more relevant answer Final answer: 3) Uric acid is the nitrogenous waste that requires hardly any water for excretion. It is produced primarily by birds, reptiles and terrestrial arthropods, and is less toxic than ammonia. The conversion of ammonia to uric acid is believed to have evolved as a method of conserving water in arid environments. Explanation: The nitrogenous waste that requires hardly any water for its excretion is uric acid. Nitrogenous wastes are produced by the breakdown of proteins and nucleic acids. In animals, these wastes are typically in the form of ammonia, urea, or uric acid. Ammonia, although toxic, is soluble in water and is therefore easier for aquatic animals to excrete. Mammals, including humans, convert the toxic ammonia into urea, which is easier to handle by the body. However, it is the process of converting ammonia into uric acid that makes it the most efficient in terms of water conservation. Birds, reptiles, and most terrestrial arthropods convert toxic ammonia into uric acid or guanine, which are considerably less toxic than ammonia, and helps conserve more water as they are insoluble in water and are excreted as a white paste or powder. This form of excretion is thought to have evolved in response to arid conditions as a means of conserving water. Learn more about Nitrogenous Waste here: brainly.com/question/34651541 SPJ11 Answered by rahulkkushwaha123 •14.5K answers•2.3M people helped Thanks 0 0.0 (0 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 2321851 people 2M 0.0 0 Biology - John W. Kimball Anatomy and Physiology - J. Gordon Betts, Kelly A. Young, James A. Wise, Eddie Johnson, Brandon Poe, Dean H. Kruse, Oksana Korol, Jody E. Johnson, Mark Womble, Peter Desaix Biology for AP® Courses - Julianne Zedalis, John Eggebrecht Upload your school material for a more relevant answer The nitrogenous waste requiring hardly any water for excretion is uric acid. This form is produced mainly by birds, reptiles, and terrestrial arthropods to conserve water efficiently. It is excreted as a semi-solid paste, which is advantageous in dry conditions. Explanation The nitrogenous waste that requires hardly any water for its excretion is C. Uric Acid. Explanation: Nitrogenous wastes are the byproducts of metabolism, specifically from the breakdown of proteins and nucleic acids. In animals, these wastes are typically in the form of ammonia, urea, or uric acid, and the method of excretion varies by species based on their habitat and water conservation needs. Ammonia: This is the most toxic form of nitrogenous waste. It is highly soluble in water and is efficiently excreted by aquatic animals because they live in a watery environment. For these animals, excreting ammonia does not require significant water conservation. Urea: Mankind and other mammals primarily excrete urea, which is less toxic than ammonia. The liver converts ammonia into urea via the urea cycle, which requires some water for excretion but is still more efficient than excreting ammonia directly. Uric Acid: This form of nitrogenous waste is much less toxic and is largely insoluble in water. As a result, animals that excrete uric acid—such as birds, reptiles, and terrestrial arthropods—can do so in a paste-like form. This adaptation allows them to conserve water extremely effectively, making it especially useful in arid environments. In conclusion, while urea is a common form of nitrogenous waste in many mammals, uric acid is the waste that conserves water to the greatest extent. This adaptation is crucial for animals living in environments where water is scarce, allowing them to survive with minimal water loss. Examples & Evidence Examples of animals that excrete uric acid include birds like sparrows and reptiles like lizards. These organisms produce uric acid from the breakdown of nitrogenous compounds, allowing them to thrive in dry habitats without losing much water. Scientific studies and biological texts validate that uric acid is less toxic, less soluble in water, and often excreted as a paste, making it the most efficient nitrogenous waste product in terms of water conservation. Thanks 0 0.0 (0 votes) Advertisement spierboy8327 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Chemistry solutions and answers Community Answer Which of the following nitrogenous wastes has the lowest concentration (mg/dL) in normal urine?a. creatinineb. uric acid c. ammoniad. urea Community Answer 4.2 19 A drink that contains 4 1/2 ounces of a proof liquor… approximately how many drinks does this beverage contain? Community Answer 5.0 7 Chemical contamination is more likely to occur under which of the following situations? When cleaning products are not stored properly When dishes are sanitized with a chlorine solution When raw poultry is stored above a ready-to-eat food When vegetables are prepared on a cutting board that has not been sanitized Community Answer 4.3 189 1. Holding 100mL of water (ebkare)__2. Measuring 27 mL of liquid(daudgtear ldnreiyc)____3. Measuring exactly 43mL of an acid (rtube)____4. Massing out120 g of sodium chloride (acbnela)____5. Suspending glassware over the Bunsen burner (rwei zeagu)____6. Used to pour liquids into containers with small openings or to hold filter paper (unfenl)____7. Mixing a small amount of chemicals together (lewl letpa)____8. Heating contents in a test tube (estt ubet smalcp)____9. Holding many test tubes filled with chemicals (estt ubet karc) ____10. Used to clean the inside of test tubes or graduated cylinders (iwer srbuh)____11. Keeping liquid contents in a beaker from splattering (tahcw sgasl)____12. A narrow-mouthed container used to transport, heat or store substances, often used when a stopper is required (ymerereel kslaf)____13. Heating contents in the lab (nuesnb bneurr)____14. Transport a hot beaker (gntos)____15. Protects the eyes from flying objects or chemical splashes(ggloges)____16. Used to grind chemicals to powder (tmraor nda stlepe) __ Community Answer Food waste, like a feather or a bone, fall into food, causing contamination. Physical Chemical Pest Cross-conta Community Answer 8 If the temperature of a reversible reaction in dynamic equilibrium increases, how will the equilibrium change? A. It will shift towards the products. B. It will shift towards the endothermic reaction. C. It will not change. D. It will shift towards the exothermic reaction. Community Answer 4.8 52 Which statements are TRUE about energy and matter in stars? Select the three correct answers. Al energy is converted into matter in stars Only matter is conserved within stars. Only energy is conserved within stars. Some matter is converted into energy within stars. Energy and matter are both conserved in stars Energy in stars causes the fusion of light elements Community Answer 4.5 153 The pH of a solution is 2.0. Which statement is correct? Useful formulas include StartBracket upper H subscript 3 upper O superscript plus EndBracket equals 10 superscript negative p H., StartBracket upper O upper H superscript minus EndBracket equals 10 superscript negative p O H., p H plus P O H equals 14., and StartBracket upper H subscript 3 upper O superscript plus EndBracket StartBracket upper O upper H superscript minus EndBracket equals 10 to the negative 14 power. New questions in Chemistry Consider the chemical equations shown here. C H 4(g)+2 O 2(g)→C O 2(g)+2 H 2O(g)Δ H 1=−802 k J 2 H 2O(g)→2 H 2O(l)Δ H 2=−88 k J Which equation shows how to calculate Δ H r x n for the equation below? C H 4(g)+2 O 2(g)→C O 2(g)+2 H 2O(l) What is Δ H r x n for the overall reaction? Consider the chemical equations shown here. 2 H 2(g)+O 2(g)→2 H 2O(g)Δ H 1=−483.6 k J+2=−241.8 k J/m o l 3 O 2(g)→2 O 3(g)Δ H 2=284.6 k J+2=142.3 k J/m o l What is the overall enthalpy of reaction for the equation shown below? 3 H 2(g)+O 3(g)→3 H 2O(g)□ kJ Consider the chemical equations shown here. P 4(s)+3 O 2(g)→P 4O 6(s)Δ H 1=−1,640.1 k J P 4O 10(s)→P 4(s)+5 O 2(g)Δ H 2=2,940.1 k J What is the overall enthalpy of reaction for the equation shown below? Round the answer to the nearest whole number. P 4O 6(s)+2 O 2(g)→P 4O 10(s) Prepare a "Coal and Petroleum" fact file. What is a double bond? A. A covalent bond between 2 atoms where each atom contributes 2 electrons. B. A covalent bond between 2 atoms where each atom contributes 4 electrons. C. An ionic bond between 2 atoms where 2 electrons have been transferred. D. A covalent bond between 2 atoms where each atom contributes 1 electron. 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2691	https://www.themathdoctors.org/generalizing-and-summing-the-fibonacci-sequence/	Typesetting math: 100% Skip to content Generalizing and Summing the Fibonacci Sequence Continuing our look at the Fibonacci sequence, we’ll extend the idea to “generalized Fibonacci sequences” (with different starting numbers), and see that the ratio of consecutive terms is the same in general as in the usual special case. Then we’ll look at the sum of terms of both the special and general sequence, turning it into a series. This turns out to be as simple as you could imagine! Generalized Fibonacci sequences Recall that the Fibonacci sequence is defined by specifying the first two terms as F1=1 and F2=1, together with the recursion formula Fn+1=Fn+Fn−1. We have seen how to use this definition in various kinds of proofs, and also how to find an explicit formula for the nth term, and that the ratio between successive terms approaches the golden ratio, ϕ, in the limit. Similar questions can be raised about any sequence that has the same recursion as the Fibonacci sequence, but starts with a different pair of numbers. That is the subject of a question from 2002: ``` Generalised 'Fibonacci' Series and Phi I have searched but I can't find an answer to this. I have shown with a spreadsheet that a Fibonacci-style series that starts with any two numbers at all, and adds successive items, produces a ratio of successive items that converges to phi in about the same number of terms as for the 1 1 2 3 5 etc. basic Fibonacci series. I can't find reference to this - maybe I don't know how to look - surely this is well known? Is it - and is it provable? ``` None of these series converges to ϕ in a finite number of terms, but presumably Stuart means that it converges to about the same precision in about the same number of steps. Here is an example of a spreadsheet showing this (which is very easy to make): The first column is Fibonacci proper, starting with 1, 1; these are the numbers we call the Fibonacci numbers. The second starts with 1, 5 (which we’ll be seeing later), and the third, just to be very different, starts with 3, 100. The ratios in each case rapidly converge toward 1.618034…, which is ϕ. Doctor Floor answered: ``` Hi, Stuart, Thanks for your question. To prove your conjecture we will delve into formulas of generalized Fibonacci sequences (sequences satisfying X(n) = X(n-1) + X(n-2) ). Let's first investigate two very special members of the family of generalized Fibonacci sequences, those of the form X(1)=g=g^1, X(2)=g^2, X(3)=g^3, ..., X(n)=g^n [g nonzero] ``` In other words, is it possible for a sequence with this recursion to also be a geometric sequence? That’s not the first question I’d think to ask about generalized Fibonacci sequences; but it turns out to be interesting: ``` Since X(n)=X(n-1)+X(n-2), we must have: g^n = g^(n-1) + g(n-2) g^n - g^(n-1) - g(n-2) = 0 g^(n-2) (g^2 - g - 1) = 0 g^2 - g - 1 = 0 [we can divide by g^(n-2) because g is nonzero] So g must be one of the roots of g^2 - g - 1 = 0, so it must be one of the following numbers: Phi = (1+SQRT(5))/2 (the Golden Ratio) 1-Phi = (1-SQRT(5))/2 ``` Only these two numbers can work as the common ratio. We could actually start the sequence with any number we like (including 1), so there are many such geometric “Fibonacci” sequences (which would be multiples of these); but taking the sequences as An=ϕn and Bn=(1−ϕ)n will keep the formulas simple. ``` So we have two very special generalized Fibonacci sequences: Phi^1, Phi^2, Phi^3, ..., Phi^n, ... (1-Phi)^1, (1-Phi)^2, ..., (1-Phi)^n, ... I leave it for you to check that, when you have two generalized Fibonacci sequences A(n) and B(n), and two numbers a and b, then aA(n) + bB(n) is a generalized Fibonacci sequence. ``` Let’s do that: If An and Bn satisfy the recursions An=An−1+An−2 and Bn=Bn−1+Bn−2, and Cn=aAn+bBn, then Cn−1+Cn−2=(aAn−1+bBn−1)+(aAn−2+bBn−2)=a(An−1+An−2)+b(Bn−1+Bn−2)=aAn+bBn=Cn. So any linear combination of any two generalized Fibonacci sequences is also a generalized Fibonacci sequence. ``` In fact, all generalized Fibonacci sequences can be calculated in this way from Phi^n and (1-Phi)^n. This can be seen from the fact that any two initial terms can be created by some a and b from two (independent) pairs of initial terms from A(n) and B(n), and thus also from Phi^n and (1-Phi)^n. So each generalized Fibonacci sequence can be written as X(n) = aPhi^n + b(1-Phi)^n. ``` For example, suppose (as in my second column above) that we want the first two terms of our sequence to be X1=1 and X2=5. Then we want to find a and b such that aϕ+b(1−ϕ)=1aϕ2+b(1−ϕ)2=5. These equations simplify (using the fact that ϕ2=ϕ+1) to (a−b)ϕ+b=1(a−b)ϕ+(a+2b)=5. Subtracting to eliminate the first term, we find that a+b=4. (We could have obtained the same result trivially by using X0=4!) Replacing b with 4−a in the first equation, we get (2a−4)ϕ+(4−a)=1(2ϕ−1)a+(4−4ϕ)=1a=4ϕ−32ϕ−1=25–√−15–√b=4−25–√−15–√=25–√+15–√. So our sequence is Xn=25–√−15–√ϕn+25–√+15–√(1−ϕ)n. This can be done for any initial pair of terms. ``` Now your statement follows for all sequences for which a<>0, when we realize that \|1-Phi\|<1, so that (1-Phi)^n --> 0 for n --> infinity. So we see that there are exceptions: sequences b(1-Phi)^n. ``` This is the same reasoning we used before for the limit of the Fibonacci sequence proper, at the end of The Golden Ratio and Fibonacci. And if it turns out that a=0, then the limit is zero rather than ϕ. We’ll be seeing more on the generalized Fibonacci sequence below! Summing the Fibonacci series It is common to mistakenly use the term “series” (which refers to a sum) in place of the word “sequence” (which is what the Fibonacci numbers are, an ordered list). We even see that in the original title for the question above, which was taken from the question. But this time we’re really going to talk about the Fibonacci series: What do you get when you add up the Fibonacci numbers? This question is from 2000: ``` Sum of Fibonacci Series Is there any formula for the sum of the first n numbers in the Fibonacci sequence? I've been looking for one without success. I did find a quartic equation that worked for the first few (five or six), but it got progressively more inaccurate. I have also tried plotting them against the respective terms of the Fibonacci sequence, which almost gave a straight line, but it was not quite close enough. ``` Edward has been trying to guess at a formula; I tried fitting the sequence to a quartic polynomial using Excel, which did fairly well for the first 15 terms as in my table above, but not exact: If I plotted more Fibonacci numbers with the same polynomial, they would rapidly diverge – because the Fibonacci sequence is closely related to exponential functions (geometric sequence) as we’ve already seen, and therefore grows faster than any polynomial. Doctor Mitteldorf answered first, with a hint and a challenge: ``` Dear Edward, Believe it or not, the Fibonacci series is the sum of two geometric series, each one of which separately is not even an integer. Given the formula, it's not too hard to prove that it works (use proof by induction). But where does the formula come from? The formula involves the square root of 5, which I write as sqrt. The nth Fibonacci is: ax^n + (1-a)y^n where a = (3+sqrt)/(5+sqrt) x = (1+sqrt)/2 y = (1-sqrt)/2 ``` We’ve already seen such a formula both “discovered” and proved; this version is different because, for some reason, he chose to start with 1 and 2 rather than 1 and 1. (You might like to try obtaining the formula using the method above for the generalized sequence!) So my challenges to you are: first, to prove that the formula works, and second, to derive the equation you want for the sum of the first n terms, using the formula for the sum of a geometric series. We don’t need to prove the formula; and I’ll leave you to find the formula for the sum, based either on Doctor Mitteldorf’s version or Binet’s. As a reminder, here is the formula for the sum of a finite geometric series with a0=a and common ratio r: ∑k=0nark=a+ar+ar2+ar3+⋯+arn=a(rn+1−1)r−1. For example, ∑k=033⋅2k=3+6+12+24=3(24−1)2−1=45. Then Doctor Floor, writing at the same time, gave a fuller answer: ``` Hi, Edward, Thanks for writing. We know that the nth Fibonacci number F(n) = (PHI^n - (1 - PHI)^n) / sqrt where PHI = (1+sqrt)/2 = 'Golden ratio' ``` This, of course, is the usual Binet formula for the sequence starting with 1, 1, which is the difference of two geometric series. ``` I will use the value of F(0) in my sum of the first n Fibonacci numbers. That doesn't matter because F(0) = 0. So we can derive the following: SUM{i = 0 to n} F(n) = SUM{i = 0 to n} (PHI^i - (1 - PHI)^i) / sqrt5 = (1/sqrt5){[SUM{i = 0 to n} PHI^i] - [SUM{j = 0 to n} (1-PHI)^j]} ``` Here we will be applying the formula I stated above, with a=1 for each series, and with common ratios ϕ and (1−ϕ) respectively: ∑k=0nϕk=ϕn+1−1ϕ−1 ∑k=0n(1−ϕ)k=(1−ϕ)n+1−1(1−ϕ)−1=(1−ϕ)n+1−1−ϕ ``` Using the formula for a geometric series we know: SUM{i = 0 to n} PHI^i = (PHI^(n+1)-1)/(PHI - 1) and SUM{j = 0 to n} (1-PHI)^j = ((1-PHI)^(n+1) - 1)/-PHI so we conclude: (1/sqrt){[SUM{i = 0 to n} PHI^i] - [SUM{j = 0 to n} (1-PHI)^j]} ( PHI^(n+1) - 1 (1-PHI)^(n+1) - 1 ) 1 = ( ------------- + ----------------- ) ------- ( PHI-1 PHI ) sqrt PHI^(n+2) - PHI - (1-PHI)^(n+2) - PHI + 1 = ----------------------------------------- [use PHI(PHI-1) = 1] sqrt5PHI(PHI-1) PHI^(n+2) - (1-PHI)^(n+2) 1 - 2PHI = ------------------------- + --------- sqrt sqrt = F(n+2) - 1 ``` After a bit of fraction work and recognizing the Binet formula coming back to us, the result is amazingly simple – not a polynomial in n, but one less than a subsequent term in the sequence. ``` We see that there comes a very simple formula {F(0) +} F(1) + F(2) + ... + F(n) = F(n+2) - 1 _^^^^^^^^_ not needed Now that we found the formula, it seems that you can prove it more easily by mathematical induction. Try it! ``` Series sum by induction Let’s do it! We want to prove that for any positive integer n, the sum of the first n terms of the Fibonacci sequence is Fn+2−1. That is, ∑i=1nFi=Fn+2−1 Base case: We’ll use weak recursion (not strong as we have often used with Fibonacci), and only need one case: ∑i=11Fi=F1=1F1+2−1=F3−1=2−1=1 Inductive hypothesis: Suppose that the sum of the first k terms is Fk+2−1: ∑i=1kFi=Fk+2−1 We want to show that the sum of the first (k+1) terms is F(k+1)+2−1=Fk+3−1: ∑i=1k+1Fi=Fk+3−1 Induction: The sum of the first (k+1) terms is the sum of the first k terms plus the (k+1)st: ∑i=1k+1Fi=∑i=1kFi+Fk+1=(Fk+2−1)+Fk+1=(Fk+2+Fk+1)−1=Fk+3−1 by the recursion. That was easy! Summing the generalized Fibonacci series Our final question (from 2003) combines these ideas, summing a generalized series. ``` Sum of Sequence: a, b, a+b, a+2b, 2a+3b... Is there a formula to find the sum of the first n terms of the sequence a, b, a+b, a+2b, 2a+3b... ? If I know a and b, and how many terms, is there a way to find the sum? Question: Find the sum of the first 30 terms of the sequence: 1, 5, 6, 11, 17, 28... if the 30th term is 2888956 and the 31st term is 4674429. Thanks. ``` Do you recognize that, though the name wasn’t used, this is about the generalized Fibonacci sequence, starting with any two numbers a and b? Will it be as simple as for the standard sequence? Doctor Jacques answered: ``` Hi Daniel, Although your sequence is not the Fibonacci sequence, it looks a lot like it and the formula is the same: a[n+2] = a[n+1] + a[n] We can also write it as: a[n] = a[n+2] - a[n+1] Let us write some terms, starting at the end: a = a - a a = a - a a = a - a ...... a = a - a a = a - a We notice that, on the right side, the first term of each equation (starting at the second) also appears with a minus sign in the previous equation. This means that, if we add together all the equations, a lot of terms will cancel each other. ``` This is a less formal way to do the same thing we did in the recursion earlier. Just imagine crossing off both a’s, then both a’s, and so on down to the a’s. What’s left? ``` On the left-hand side, we will have a + ... + a - this is almost what we are looking for. On the right-hand side: The first (a) will remain there All the terms from a down to a will cancel out The last term (-a) will remain. To summarize, after all simplifications, we get: a + ... + a = a - a As we know a, a, and, of course, a, you should be able to complete the calculation. ``` Turning this into an explicit formula for the sum in the general case (since n in our example is 30), A1+A2+⋯+An−1=An+1−A2 so that, add An to both sides, A1+A2+⋯+An−1+An=An+An+1−A2=An+2−A2 This is exactly equivalent to the standard case above. For the particular numbers we were given, A1=1, A2=5, and A32=A30+A31 =2,888,956+4,674,429 =7,563,385, so the sum is A32−A2=7,563,385−5 =7,563,380. While we’re here, let’s check whether the numbers we were given are correct. Remember the specific sequence we worked out earlier? That was for X1=1 and X2=5, and we found that Xn=25–√−15–√ϕn+25–√+15–√(1−ϕ)n. Therefore, X30=25–√−15–√ϕ30+25–√+15–√(1−ϕ)30=2888955.9999987+0.0000013=2888956m which is what we were told it was. The author of the problem either used the formula or, more likely, used a spreadsheet to make the sequence. They didn’t just toss out a fake number figuring no one would ever check. 1 thought on “Generalizing and Summing the Fibonacci Sequence” Pingback: Fibonacci Word Problems I: Basic – The Math Doctors Leave a Comment Cancel Reply
2692	https://stats.stackexchange.com/questions/24033/to-use-discrete-fourier-transform-to-invert-a-covariance-matrix	r - To use Discrete Fourier Transform to invert a covariance matrix - Cross Validated Join Cross Validated By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more To use Discrete Fourier Transform to invert a covariance matrix Ask Question Asked 13 years, 7 months ago Modified13 years, 7 months ago Viewed 3k times This question shows research effort; it is useful and clear 8 Save this question. Show activity on this post. I am working on a problem that its difficult part is to invert a covariance matrix (in R). I could not use usual approches like SVD and Chol. Then, I decided to use a Discrete Fourier Transform (DFT) approach. But I couldn't understand how to apply the method, especially in R. So, your comments and possible examples in R, would be appreciated. many thanks in advance. r matrix-inverse Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Improve this question Follow Follow this question to receive notifications edited Mar 3, 2012 at 4:29 mpiktas 35.7k 6 6 gold badges 90 90 silver badges 146 146 bronze badges asked Mar 3, 2012 at 2:24 hbaghishanihbaghishani 681 6 6 silver badges 12 12 bronze badges 2 Could you please explain why you cannot use the "usual approaches"?whuber –whuber♦ 2012-03-03 03:26:16 +00:00 Commented Mar 3, 2012 at 3:26 First, using, for example, choleski decomposition lead to an error regarding singularity. Second, the structure of the matrix is circulant and using DFT would be recommended.hbaghishani –hbaghishani 2012-03-03 07:52:43 +00:00 Commented Mar 3, 2012 at 7:52 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 6 Save this answer. Show activity on this post. A circulant is a matrix whose first column is a vector x x and its subsequent columns are obtained by rotation of one element to the right. Here is R code to produce any circulant from its first column x: r rotate <- function(x,k) {c(tail(x,-k), head(x,k))} circulant <- function(x) { n=length(x) apply(matrix(0:(n-1),1,n), 2, function(k) rotate(x,n-k)) } # Returns the circulant matrix of which x is the first column For example, ```r circulant(c(2,3,5,7)) [,1] [,2] [,3] [,4] [1,] 2 7 5 3 [2,] 3 2 7 5 [3,] 5 3 2 7 [4,] 7 5 3 2 ``` It is inverted by changing to an eigenbasis. The diagonal elements are the entries of the Fourier Transform of x, so we invert them individually and change back to the original basis: r reciprocal <- function(x) {i <- which(x!=0); x[i] <- 1/x[i]; x} inverse.circulant <- function(x) { n <- length(x) # x is the first column of the circulant i <- (0:(n-1)) %o% (1:n) # Powers of exp(2 Pi I/n) in the eigenbasis q q <- matrix(exp(complex(real=-log(n)/2, imaginary=2pii / n)), n, n) w <- reciprocal(fft(x)) # Reciprocals of nonzero eigenvalues Re(t(q) %% diag(w) %% Conj(q))# Convert back to the original basis } # Returns a generalized inverse to circulant(x) For example, we demonstrate this works by multiplying its output by the original circulant and checking that the identity matrix is obtained (up to negligible floating point error): ```r a <- c(2,3,5,7) zapsmall(inverse.circulant(a) %% circulant(a)) [,1] [,2] [,3] [,4] [1,] 1 0 0 0 [2,] 0 1 0 0 [3,] 0 0 1 0 [4,] 0 0 0 1 ``` Be aware that ill-conditioning will still plague this approach due to floating point roundoff in fft. That is why I have implemented a reciprocal function: it refuses to compute 1/x 1/x when x=0 x=0. As such, inverse.circulant computes a generalized inverse, exactly as in MASS::ginv: ```r The following determines a nonsingular but ill-conditioned circulant: (a <- c(1, -200000/200001, -2500000/500001, 5000000/1000003)) 1.000000 -0.999995 -4.999990 4.999985 1 / rcond(circulant(a)) # HUGE condition number! 5.404306e+16 library(MASS) inverse.circulant(a) - ginv(circulant(a)) [,1] [,2] [,3] [,4] [1,] 6.938894e-18 -2.081668e-17 -4.163336e-17 2.775558e-17 [2,] 1.387779e-17 -6.938894e-18 -3.469447e-18 1.387779e-17 [3,] 1.387779e-17 4.163336e-17 -6.938894e-18 1.040834e-17 [4,] 3.469447e-17 -2.775558e-17 -1.387779e-17 -2.081668e-17 ``` Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Mar 5, 2012 at 21:13 answered Mar 5, 2012 at 17:09 whuber♦whuber 342k 66 66 gold badges 823 823 silver badges 1.4k 1.4k bronze badges 6 Thank you so much for your clear answer. However, when I ran your code to test it works properly, for checking identity matrix, I got this error: "Error in t(q) %% w : non-conformable arguments"hbaghishani –hbaghishani 2012-03-05 19:27:45 +00:00 Commented Mar 5, 2012 at 19:27 Sorry; I made a typo when I introduced the reciprocal function as an afterthought. In so doing I left out the call to diag. I fixed this, copied the text of the edited message into R, and ran the examples again as a test.whuber –whuber♦ 2012-03-05 20:07:17 +00:00 Commented Mar 5, 2012 at 20:07 awesome! I didn’t know anything of all that.Elvis –Elvis 2012-03-05 20:24:11 +00:00 Commented Mar 5, 2012 at 20:24 Elvis: as the Wikipedia article helpfully points out, circulants can be viewed as elements of the left regular representation of the group algebra of Z/n Z. Everything follows from that via the theory of group representations. Having an FFT available doesn't really speed things up asymptotically: at best, it halves the computation time.whuber –whuber♦ 2012-03-05 20:42:01 +00:00 Commented Mar 5, 2012 at 20:42 Thank you so much whuber. The code has still an error. For checking it, the line "a <- circulant(c(2,3,5,7))" could be "a <- c(2,3,5,7)". Best.hbaghishani –hbaghishani 2012-03-05 21:08:32 +00:00 Commented Mar 5, 2012 at 21:08 \|Show 1 more comment This answer is useful 1 Save this answer. Show activity on this post. Have you tried a correction- by adding a small ϵ perturbation to the diagonal of the matrix you are trying to invert. This is a standard processing routine used to defer the singularity issue while inverting a covariance matrix or a hessian. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Mar 4, 2012 at 13:49 answered Mar 3, 2012 at 13:40 hearsehearse 2,555 1 1 gold badge 21 21 silver badges 34 34 bronze badges 2 Yes, but the result remains the same.hbaghishani –hbaghishani 2012-03-03 14:58:41 +00:00 Commented Mar 3, 2012 at 14:58 1 Can you tell us the (reciprocal) condition number of your matrix? In R, you can do rcond(A).Zen –Zen 2012-03-03 19:20:37 +00:00 Commented Mar 3, 2012 at 19:20 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Is the Wikipedia article on circulant matrices clear? This is something that in a time series context is discussed, for instance, in the first pages of Hannan's Time Series book. The eigenvalues of a circulant matrix are given by the Fourier transform of what (again in a time series context) would be the autocovariances. So to invert the matrix you have to take the reciprocals of the eigenvalues and pre- and post-multiply by the matrix whose columns are the eigenvectors. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Mar 5, 2012 at 0:10 whuber♦ 342k 66 66 gold badges 823 823 silver badges 1.4k 1.4k bronze badges answered Mar 3, 2012 at 21:18 F. TusellF. Tusell 8,778 27 27 silver badges 38 38 bronze badges Add a comment\| Your Answer Thanks for contributing an answer to Cross Validated! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. MathJax reference. To learn more, see our tips on writing great answers. 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2693	https://www.youtube.com/watch?v=gLZr0W-oCC8	Math 6- Constructing and Interpreting a Pie Graph Teacher Jhaniz 5090 subscribers 129 likes Description 16961 views Posted: 7 Jun 2021 Module No. 4 : Week 4 : Fourth Quarter Math 6 Pie Graph Constructing and Interpreting a Pie Graph 7 comments Transcript: Intro [Music] [Music] good day everyone welcome to our math class for this week's lesson we will focus on constructing and interpreting a pie graph [Music] with our learning target constructs a pie graph based on a given set of data and interpreted Learning Target to start let's have a review on how to draw a fraction draw the fraction given use the circle number one one fourth number two five sixth [Music] and number three two fifths What is a Pie Graph what we have drawn is similar to what we are going to discuss this time about a pie graph now what is a pie graph a pie graph or a circle graph is a good way of showing how a whole or one hundred percent is divided into fractional parts it presents data in a circular chart which is divided into sections called sectors each sector represents a part or a percentage of a whole to construct a pie graph we need to follow these steps step 1 arrange the data or information step change each data into percent step 3 compute for the number of degrees for each percent remember a circle has 360 degrees step 4 use compass or protractor to make a circle then plot the data step 5 label the parts and make a title or label for the pie graph Example here's the example maria's homework is to make a pie graph of her daily allowance she divides her 100 pesos into four parts 30 pesos per food 20 pesos for transportation 20 passes for school materials and 10 pesos for her savings here's the solution first we'll do the step one arrange the data by using a table listed here are the expenses amount then later on you will find this percent and degrees step two is Percent to change data into present how to solve for percent here is the solution for food we have 30 pesos divide that amount by the total amount 80 pesos then multiply by 100 we get 37.5 for the transpo divide amount 20 pesos by the total amount 80 pesos then multiply by 100 we get 25 for the school materials divide the amount 20 pesos by the total amount 80 pesos then multiply by 100 we get 25 for the savings divide the amount 10 pesos by the total amount 80 pesos then multiply by 100 we get 12.5 a total of 100 percent Degrees for the step 3 compute for the number of degrees for each percent now how to solve for degrees here's the solution for food multiply the percent 37.5 by 360 degrees we cannot just multiply 37.5 percent by 360 degrees so first we will change the percent to decimal by dividing it by 100 or moving two decimal places to the left like this one and two 37.5 will become [Music] 375 thousands and that's the time that we will multiply the decimal by 360 degrees and we get 135 degrees for food next for transportation we will do the same step or process we will change first the percent to decimal 25 becomes 25 hundredths then multiply it by 360 degrees we get 90 degrees for school materials 25 change it to decimal form we have 25 hundredths and multiply by 360 degrees we get 90 degrees for the savings 12.5 percent change it to decimal form that becomes 125 thousands times 360 degrees we get 45 degrees a total of 360 degrees for step four we will use a compass or a protractor to make a circle then plot the data we'll do the step five label the parts and make a title or label for the pie graph savings 12.5 transportation 25 school materials and are 25 and food for 37.5 a point to remember class is when you construct or you make a pie graph you start from the least to greatest just like this we start from 12.5 25 under 25 to 37.5 this is maria's daily expenses Summary why do we need to use a circle graph or a pie graph yes pie graph or a circle graph makes the presentation of data more easily understandable for your activity do exercises and let's apply complete the table the table shows the favorite games of 20 children next is using the table in net supply make a pie graph good job kids [Music] that's all for today and see you for the next topic goodbye thank you for watching don't forget to like share and subscribe
2694	https://teachy.ai/en/summaries/high-school/10th-grade/physics-en/kinematics-relative-velocity-or-traditional-summary-0cc8d	Log In Summary of Kinematics: Relative Velocity Lara from Teachy Subject Physics Physics Source Teachy Original Teachy Original Topic Kinematics: Relative Velocity Kinematics: Relative Velocity Kinematics: Relative Velocity \| Traditional Summary Contextualization Relative velocity is a fundamental concept in physics, especially in the study of kinematics. It refers to the speed of one object in relation to another and is crucial for understanding how different moving objects behave in relation to each other. In our daily lives, we are constantly observing and calculating relative speed, even if unconsciously. For example, when driving on a road, we may notice whether we are overtaking another car or if it is overtaking us, based on the difference in speed between the vehicles. Understanding relative velocity is essential in various areas, such as aviation and navigation. Airplane pilots and ship captains need to calculate relative speed concerning the ground and water to ensure safe and efficient navigation. Moreover, in sports like motorsport, relative velocity is a determining factor for overtaking strategies and safety. Therefore, understanding this concept not only enhances our theoretical knowledge but also has significant practical applications in our everyday lives. Definition of Relative Velocity Relative velocity is defined as the speed of one object in relation to another. This concept is essential for understanding how different moving objects are perceived relative to each other. For instance, when two cars are moving in the same direction but at different speeds, relative velocity helps determine which car is faster and how much faster it is compared to the other. In mathematical terms, relative velocity can be calculated by subtracting the speed of one object from the speed of the other when both are moving in the same direction and sense. This is expressed by the formula: V_rel = V_object1 - V_object2. When the objects move in opposite directions, the formula changes to: V_rel = V_object1 + V_object2. These calculations are fundamental for solving practical problems in kinematics. Understanding relative velocity not only facilitates the analysis of motion in physics but also has significant practical applications. In traffic, for example, the relative speed between two vehicles can influence road safety since perceiving relative speed helps drivers make informed decisions about overtaking and maintaining a safe distance. Relative velocity is the speed of one object in relation to another. For objects in the same direction and sense: V_rel = V_object1 - V_object2. For objects in opposite directions: V_rel = V_object1 + V_object2. Relative Velocity in Movements in the Same Direction and Sense When two objects are moving in the same direction and sense, the relative velocity is determined by the difference between the speeds of the two objects. This means that if one object is moving faster than the other, the relative velocity will be positive, indicating that one object is distancing itself from the other. For example, if two cars are moving in the same direction, with car A at 30 m/s and car B at 20 m/s, the relative velocity of car A concerning car B will be 10 m/s (30 m/s - 20 m/s). This means that from car B's perspective, car A is moving away at a speed of 10 m/s. This concept is fundamental for understanding how different speeds influence the perception of motion in a common scenario, such as on a road. The ability to calculate relative velocity allows drivers to make safe decisions about overtaking and maintaining an adequate distance from other vehicles. Relative velocity is the difference between the speeds of two objects in the same direction and sense. Example: Car A at 30 m/s and Car B at 20 m/s results in V_rel = 10 m/s. Important for safety decisions in traffic. Relative Velocity in Movements in Opposite Directions When two objects are moving in opposite directions, relative velocity is determined by the sum of the speeds of the two objects. This occurs because the objects are moving in opposite directions, which increases the relative speed between them. For example, if two airplanes are moving towards each other, with airplane X at 200 m/s and airplane Y at 250 m/s, the relative velocity between them will be 450 m/s (200 m/s + 250 m/s). This means that from either airplane's perspective, the other is approaching at a combined speed of 450 m/s. Understanding relative velocity in opposing movements is crucial for safety in various situations, such as in aviation and navigation, where precise speed perception is vital to avoid collisions and ensure safe navigation. Relative velocity is the sum of the speeds of two objects moving in opposite directions. Example: Airplane X at 200 m/s and Airplane Y at 250 m/s results in V_rel = 450 m/s. Crucial for safety in aviation and navigation. Importance of Relative Velocity Relative velocity is a concept that goes beyond theory and has practical applications in various fields. In road safety, for instance, understanding relative velocity can help prevent accidents. Drivers who understand how to calculate relative velocity are capable of making safer decisions about when to overtake another vehicle or how to maintain a safe distance. In air navigation, pilots need to calculate the relative speed of the aircraft concerning the ground and wind to ensure safe and precise navigation. This calculation is essential for adjusting the route and speed of the aircraft, avoiding possible collisions and ensuring flight efficiency. In sports like motorsport, relative velocity is crucial for racing strategy. Drivers who can calculate the relative speed of competitors can plan overtakes and other maneuvers more accurately, improving their performance and safety on the track. Relative velocity has practical applications in road safety, air navigation, and sports. Helps drivers make safe decisions while overtaking and maintaining distance. Essential for pilots to adjust routes and speeds in flights. To Remember Relative Velocity: Speed of one object in relation to another. Kinematics: Branch of physics that studies the motion of bodies. Motion: Change in position of an object over time. Direction and Sense: Orientation and route of the motion of an object. Conclusion The lesson on relative velocity addressed the definition and methods for calculating the relative velocity between two moving objects. We discussed how to calculate relative velocity in movements in the same direction and sense, as well as in opposite directions, using specific formulas. Practical examples, such as analyzing two cars on the road or airplanes approaching each other, helped illustrate these concepts clearly and applied them to everyday life. Understanding relative velocity is essential for various practical areas, such as road safety, air navigation, and racing sports. Knowing how to calculate relative velocity allows for informed and safe decisions, whether when overtaking another vehicle or adjusting an aircraft's route. Therefore, this knowledge not only improves students' theoretical understanding but also has a direct impact on real-life situations. We encourage students to continue exploring the topic of relative velocity and its applications in different contexts. Mastering this concept is fundamental for a deeper understanding of kinematics and can open doors to advanced studies in physics and related areas. Continuous practice and resolving additional problems will further strengthen this understanding. Study Tips Review the practical examples discussed in class and try solving similar problems with different values. This will help consolidate the concept of relative velocity. Read supplementary materials and watch educational videos on kinematics and relative velocity. Visual resources can facilitate the understanding of abstract concepts. Practice solving relative velocity problems in different contexts, such as traffic, aviation, and sports. Practical application reinforces theoretical learning. Want access to more summaries? 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2695	https://artofproblemsolving.com/wiki/index.php/Extreme_principle?srsltid=AfmBOoogMdHC7vykWFC2YiQFPgRjuMcbE_OazOj2GbU3keXJ9SJt9PpC	Art of Problem Solving Extreme principle - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Extreme principle Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Extreme principle The extreme principle (or extremal principle) is a problem-solving technique that involves looking at objects with extreme properties, such as the largest or smallest element. For example, consider the following problem: Example: Imagine an infinite chessboard that contains a positive integer in each square. If the value in each square is equal to the average of its four neighbors to the north, south, west, and east, prove the values in all the squares are equal. Solution: Consider the square containing the minimal value. Then its four neighbors must all have this minimal value. Similarly, their neighbors must also have this minimal value, and so on ad infinitum. Thus, every square in the chessboard has the same value. Problems Suppose you are given a finite set of coins in the plane, all with different diameters. Show that one of the coins is tangent to at most five of the others. (Solution) A palindrome is a number or word that is the same when read forward and backward, for example, "176671" and "civic." Can the number obtained by writing the numbers from 1 to in order (for some ) be a palindrome? (Russia, 1996) Place the integers (without duplication) in any order onto an chessboard, with one integer per square. Show that there exist two adjacent entries whose difference is at least . (Adjacent means horizontally or vertically or diagonally adjacent.) There are points in the plane, not all collinear. Prove that there exists a line passing through exactly points. There are blue points and red points in the plane (no 3 are collinear). Prove that there are non-intersecting line segments, each connecting a red point to a blue point (each point is on 1 segment). There is a set of points in the plane with the property that any triangle with vertices in has area at most 1. Prove that there exists a triangle with area 4 containing all the points in . (Korea, 1995) We are given an array of real numbers. An operation consists of changing the sign (positive to negative, and vice versa) of all the entries in any row or column. Prove that we can perform a number of such operations such that in the resulting array, the sum of the entries in each row and in each column is nonnegative. Retrieved from " Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
2696	https://www.youtube.com/watch?v=nx-Y2L2psH4	2.2 Part 4: Greater Integer Function (step function); Graph Using Desmos and TI84+ Lemon Math 1450 subscribers 8 likes Description 607 views Posted: 28 Aug 2022 Transcript: greatest integer function sometimes we call this a step function greatest integer function is not widely used it's not super popular but it is good to know what it is so there are two ways to write it the textbook that i read they used a double square bracket so f of x right here f of x is equal to u we have x and then for the greatest integer the symbol that the textbook i use they use a double square bracket so this double square bracket one on the left side one on the right side and then i i also look up online some sometimes people use one square bracket sometimes people use two square bracket uh depends on what textbook you use i think just um just stick with the preference of the book if your textbook use two then use two if your textbook use one then then use one okay so in this video i show you two ways to sketch the graph of the greatest integer function the first one that i show is using decimal so this one using decimals d-e-s-m-o-s is a free graphing calculator app you can download this to your phone or you can use your computer just look it up on google you can do it on on your um on your computer as well so the way that what you have to type is you have to type here do you see this you have to type y is equals to floor x i don't know why they named that floor but it's programming stuff so floor x and then they also show you f of x equals to f they show you the table value so when x is equal to negative five y is equal to negative 5. when x is equals to negative 2 y is equal to negative 2. they might be saying okay so i just looked through the table it looks like what whatever x equals to y just equals to that okay so let's take a look at the graph so take a look at the graph it looks like a staircase right steps so step function it looks like a step so let's take a look at uh let's let's pick a piece to to take a look let's pick this piece all right so we have uh from zero right so this is a zero and then this is a one right so when x is between zero and one including zero but not including one y is equals to zero when x is between one and two including one dot including two y is just equals to one and then this piece when x is equal between two and three and then y is just equals to three so how is that going to work out the greatest integer function right so i gave you some examples over here so let's take a look when f of 2 is just equals to 2 so we we have this point when x is equal to 2 y is equal to 2 but if you plug in something like that 2.1 2.579 so you are putting numbers right here do you see that right right here i am talking about this piece so if you're plugging some not real decimals like 2.1 2.5 2.009 so you have numbers right here the corresponding y value is all equal to 2. so as long as you don't reach to 3 the corresponding y value is equal to 2 do you see that 2.1 2.579 and then two point a bunch of nine so all of the corresponding y values are equals to two but once you reach to three then it will just equals to 3 so if you go back to the graph once you wish to 3 you will just get here when x is equal to 3 y is equal to 3 because the greatest in integer so think think about this you are standing at 2 so let's say uh you are standing up right now okay so you are standing up straight and then you are your foot is stepping on the two just picture this in your mind you are looking you are standing up or you can do it do do it right now stand up look forward okay and then look down just imagine that your shoes is stepping on the two okay so x is equal to two so between two and three imagine that in some distance like ahead of you there is there there is a three so as long as you don't step on the three as long as you are walking between your current steps and without touching the three so you are stepping forward a little bit and then you just keeps trying different places so you mix one step two step three step just don't touch the three as long as you are not stepping on the three y is still equal to two another way to think about this is as long as you do not step on three so every time you take a little step forward 2.1 2.01 2.9090 you will just bounce back to a two again you are taking a step two point five one one you step forward and that evil just bounce you back to a two all right just like playing a video game it will just bounce you back to a two but once you step at a three you are at three like you are you are at the next step once you step at three but if you pick something like x equals to three point zero zero one you take us you take an extremely small step forward it will bounce you back to a three so that means for a step function so if you picture this as a staircase as long as this is a two right so the entire step is x equals to 2 right so this in here the entire step oh sorry not x equals to 2 is y equals to 2 right so as long as your foot is not on the next step so as long as your foot is not touching the next step you are still at you all right so just like walking up on a staircase as long as you are not touching the next step you are still at two okay so let's take a look at um the the other screenshot like right over here so this one i show you how to graph that using a ti 84 plus graphing calculator so the place that you will have to go is first you will have to plus the y equal key do you see that y equal key and then make sure there is nothing just go up and up up and down make sure there is no other function type in and then you will have to press the math key so you press the math key and then you go to num and um just press the right arrow go to num and then select int integer so once you select int click enter and then type x x is right next to the alpha key the green key right next to alpha and then you close the interval heat graph uh the downside of using a graphing calculator is they don't show you um they don't show you the open circle in case you might be wondering hey how come there is an open circle in in every step like at the tail of every step so here is do you still remember the vertical line test have used have you heard about the vertical line test every function must satisfy the one-to-one property if i draw a vertical line like that the vertical line should have one intersection only so i have one intersection right over here if i use a closed circle then i will have two intersection that means this is not a function so therefore you will have to close this you have to end this with an open circle if you end up with a solid circle then the one to one property is not satisfied but the graphing calculator doesn't show you this so that's why for this function i prefer you to graph that or look at that study that using decimals it's free anyway all right so that is the end of this video if you like it give me a thumbs up subscribe truly appreciate your subscription and i will meet you all in the next video take care
2697	https://mathoverflow.net/questions/88220/special-arithmetic-progressions-involving-perfect-squares	diophantine equations - Special arithmetic progressions involving perfect squares - MathOverflow Join MathOverflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community MathOverflow helpchat MathOverflow Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Special arithmetic progressions involving perfect squares Ask Question Asked 13 years, 7 months ago Modified4 months ago Viewed 3k times This question shows research effort; it is useful and clear 15 Save this question. Show activity on this post. Prove that there are infinitely many positive integers a a, b b, c c that are consecutive terms of an arithmetic progression and also satisfy the condition that a b+1 a b+1, b c+1 b c+1, c a+1 c a+1 are all perfect squares. I believe this can be done using Pell's equation. What is interesting however is that the following result for four numbers apparently holds: Claim. There are no positive integers a a, b b, c c, d d that are consecutive terms of an arithmetic progression and also satisfy the condition that a b+1 a b+1, a c+1 a c+1, a d+1 a d+1, b c+1 b c+1, b d+1 b d+1, c d+1 c d+1 are all perfect squares. I am curious to see if there is any (decent) solution. Thanks. diophantine-equations discrete-mathematics nt.number-theory arithmetic-progression Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Improve this question Follow Follow this question to receive notifications edited Jan 11, 2016 at 6:46 Cosmin PohoataCosmin Pohoata asked Feb 11, 2012 at 19:02 Cosmin PohoataCosmin Pohoata 645 6 6 silver badges 22 22 bronze badges 6 //Apologies if the tags are not appropriate; didn't know where elementary nt fits better (if it fits at all on this website)Cosmin Pohoata –Cosmin Pohoata 2012-02-11 19:05:50 +00:00 Commented Feb 11, 2012 at 19:05 What do you mean by "apparently holds"? Have you checked it computationally, or can you derive it from the three-term progression case?Seva –Seva 2012-02-11 19:12:22 +00:00 Commented Feb 11, 2012 at 19:12 I checked it computationally, sorry for not mentioning.Cosmin Pohoata –Cosmin Pohoata 2012-02-11 19:24:55 +00:00 Commented Feb 11, 2012 at 19:24 2 I added the "diophantine-equations" tag. See also my comment on duje's answer below relating to the case of triples in AP.Michael Stoll –Michael Stoll 2012-02-22 19:31:29 +00:00 Commented Feb 22, 2012 at 19:31 math.stackexchange.com/questions/1607501/…individ –individ 2016-01-11 12:03:33 +00:00 Commented Jan 11, 2016 at 12:03 \|Show 1 more comment 8 Answers 8 Sorted by: Reset to default This answer is useful 19 Save this answer. Show activity on this post. Starting from the equations in my previous answer, we get, by multiplying them in pairs, (x−y)x(x+y)(x+2 y)+(x−y)x+(x+y)(x+2 y)+1=(z 1 z 6)2,(x−y)x(x+y)(x+2 y)+(x−y)x+(x+y)(x+2 y)+1=(z 1 z 6)2, (x−y)x(x+y)(x+2 y)+(x−y)(x+y)+x(x+2 y)+1=(z 2 z 5)2,(x−y)x(x+y)(x+2 y)+(x−y)(x+y)+x(x+2 y)+1=(z 2 z 5)2, (x−y)x(x+y)(x+2 y)+(x−y)(x+2 y)+x(x+y)+1=(z 3 z 4)2.(x−y)x(x+y)(x+2 y)+(x−y)(x+2 y)+x(x+y)+1=(z 3 z 4)2. Write u=z 1 z 6 u=z 1 z 6, v=z 2 z 5 v=z 2 z 5, w=z 3 z 4 w=z 3 z 4 and take differences to obtain 3 y 2=u 2−v 2 and y 2=v 2−w 2.3 y 2=u 2−v 2 and y 2=v 2−w 2. The variety C C in P 3 P 3 described by these two equations is a smooth curve of genus 1 whose Jacobian elliptic curve is 24a1 in the Cremona database; this elliptic curve has rank zero and a torsion group of order 8. This implies that C C has exactly 8 rational points; up to signs they are given by (u:v:w:y)=(1:1:1:0)(u:v:w:y)=(1:1:1:0) and (2:1:0:1)(2:1:0:1). So y=0 y=0 or w=0 w=0. In the first case, we do not have an honest AP (y y is the difference). In the second case, we get the contradiction a b c d+a d+b c+1=0 a b c d+a d+b c+1=0 (a,b,c,d a,b,c,d are supposed to be positive). So unless I have made a mistake somewhere, this proves that there are no such APs of length 4. Addition: We can apply this to rational points on the surface. The case y=0 y=0 gives a bunch of conics of the form x 2+1=z 2 1,z 2=±z 1,…,z 6=±z 1;x 2+1=z 1 2,z 2=±z 1,…,z 6=±z 1; the case w=0 w=0 leads to a d=−1 a d=−1 or b c=−1 b c=−1. The second of these gives a d+1<0 a d+1<0, and the first gives a c+1=(a 2+1)/3 a c+1=(a 2+1)/3, which cannot be a square. This shows that all the rational points are on the conics mentioned above; in particular, (weak) Bombieri-Lang holds for this surface. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Feb 12, 2012 at 10:34 answered Feb 12, 2012 at 9:48 Michael StollMichael Stoll 11.4k 2 2 gold badges 46 46 silver badges 60 60 bronze badges 5 2 @Michael: Very nice! Let X X denote the surface given by your earlier answer. Then if I understand correctly, you've given a dominant rational map X→C X→C, where C C has genus 1 and #C(\mathbb{Q})=8#C(\mathbb{Q})=8. It seems a little surprising that a surface of general type should map onto a genus 1 curve, although I guess the fact that X X is an intersection of quadrics gives it a chance. But did you have some a priori reason to suspect that X X admitted such a covering?Joe Silverman –Joe Silverman 2012-02-12 12:52:38 +00:00 Commented Feb 12, 2012 at 12:52 1 @Joe: I didn't have some good a priori reason to suspect that X X would map to a genus 1 curve, but the observation that the analogous problem for five terms admits such a map (to the "four squares in AP" curve) provided some motivation to try a little bit longer.Michael Stoll –Michael Stoll 2012-02-12 13:13:41 +00:00 Commented Feb 12, 2012 at 13:13 Bravo! The identities look a bit less mysterious if the arithmetic progression is written symmetrically as (x−3 y,x−y,x+y,x+3 y)(x−3 y,x−y,x+y,x+3 y), when the products (z i z 7−i)2(z i z 7−i)2 are invariant under y↔−y y↔−y (Then u 2−v 2 u 2−v 2 and v 2−w 2 v 2−w 2 are 12 y 2 12 y 2 and 4 y 2 4 y 2, which comes to the same curves over Q Q.)Noam D. Elkies –Noam D. Elkies 2012-02-12 19:50:19 +00:00 Commented Feb 12, 2012 at 19:50 Can magma (or other software) detect the nontrivial H 1 H 1 of this surface which makes this analysis possible?Noam D. Elkies –Noam D. Elkies 2012-02-13 01:16:13 +00:00 Commented Feb 13, 2012 at 1:16 Hm. I guess you would want the $H^1$ of its desingularization, which might be a fairly complicated object. As damiano pointed out to me by email, the fact that there is a map to an elliptic curve must be related to the singularities being sufficiently bad (otherwise the surface would be simply connected).Michael Stoll –Michael Stoll 2012-02-15 11:10:23 +00:00 Commented Feb 15, 2012 at 11:10 Add a comment\| This answer is useful 9 Save this answer. Show activity on this post. Already for three-term progressions it's somewhat surprising that there are infinitely many solutions, because the usual probabilistic guess for the expected number of solutions leads to a convergent sum: a random number of size M M is a square with probability about M−1/2 M−1/2, so we're summing something like 1/(a b c)1/(a b c) over all three-term progressions (a,b,c)(a,b,c), etc. To be sure such a guess cannot account for non-random patterns arising from polynomial identities, but it does suggest that past a certain point such identities will be the only source of solutions. Now a mindless exhaustive search over progressions (x,x+y,x+2 y)(x,x+y,x+2 y) with 0<x,y<10 4 0<x,y<10 4 finds only the first six examples (1,7),+(4,26),+(15,97),+(56,362),+(209,1351),+(780,5042)(1,7),+(4,26),+(15,97),+(56,362),+(209,1351),+(780,5042) of an infinite family associated with the solutions (2,1)(2,1), (7,4)(7,4), (26,15)(26,15), (97,56)(97,56), (362,209)(362,209), (1351,780)(1351,780), etc. of the Pell equation x 2−3 y 2=1 x 2−3 y 2=1. If it can be proved that these are the only solutions then it will immediately follow that there are no four-term arithmetic progressions with the same property. But that seems like a very hard problem. Here's the gp code; with a bound of 10 4 10 4 it takes only a few minutes. One can surely do better with a more intelligent search procedure (e.g. start by finding all solutions of a b+1=r 2 a b+1=r 2 by factoring r 2−1 r 2−1). H = 10^4 progsq(x,y,n) = sum(i=0,n-2,sum(j=i+1,n-1,issquare((x+iy)(x+jy)+1))) for(x=1,H,for(y=1,H,if(progsq(x,y,3)==3,print([x,y])))) Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Feb 11, 2012 at 19:39 Noam D. ElkiesNoam D. Elkies 81.2k 15 15 gold badges 288 288 silver badges 381 381 bronze badges 5 Isn't there a similar problem involving 1,3,8,120, that says something like there is no fifth member such that the product of any two is one less than a square? It might be also that there are finitely many quadruplets, with the nontrivial ones being not arithmetic progressions. Gerhard "Ask Me About Obscure Puzzles" Paseman, 2012.02.11 Gerhard Paseman –Gerhard Paseman 2012-02-11 20:37:26 +00:00 Commented Feb 11, 2012 at 20:37 1 I was reminded of this 1,3,8,120 puzzle too, because Martin Gardner wrote about it in one of his columns decades ago and reported that the nonexistence of a fifth positive integer was proved only with difficulty (including several 1000+ digit computations, back when that was impressive). But there are infinitely many such integer quadruplets. A few Google searches turned up vixra.org/pdf/0907.0024v1.pdf with citations going as far as Euler and Diophantus! But naturally none of these are four-term arithmetic progressions...Noam D. Elkies –Noam D. Elkies 2012-02-11 21:43:28 +00:00 Commented Feb 11, 2012 at 21:43 Thanks for the reply! This is essentially what I did to claim that I "checked it manually": I verified the correspondence between the solutions of the problem in question and x 2−3 y 2=1 x 2−3 y 2=1 up to 10 9 10 9 and then just assummed this to be true. I was hoping however to find some slick way of doing the four number case without refering to the three case :).Cosmin Pohoata –Cosmin Pohoata 2012-02-12 05:10:57 +00:00 Commented Feb 12, 2012 at 5:10 You're welcome. You saw that by now M.Stoll has completely solved the 4-term problem, even over the rationals. For the 3-term problem, you write that you "checked manually" up to 10 9 10 9; how did you do that? I used the factorization of t 2−1 t 2−1 in b c+1=t 2 b c+1=t 2 to search up to t=10 8 t=10 8 in a few hours, still finding only the Pell solutions up to (a,b,c)=(7865521,58709048,109552575)(a,b,c)=(7865521,58709048,109552575); so t=10 9 t=10 9 (if that's what you meant) is feasible too, though it would take a while (or a number of CPU's running in parallel) to finish − and it's certainly not "manual" in the usual sense of "by hand"...Noam D. Elkies –Noam D. Elkies 2012-02-12 19:56:43 +00:00 Commented Feb 12, 2012 at 19:56 progsq3(x,y)=issquare(x(x+y)+1) && issquare(x(x+2y)+1) && issquare((x+y)(x+2y)+1) then do(lim)=lim\=1; for(r=3,sqrtint(2lim^2+1), my(t=r^2-1); fordiv(t,x, if(x>lim,break); my(y=t/x-x); if(y<1, next); if(progsq3(x,y),print([x,y])))) lets you run do(1e4) in less than a second and do(1e6) in a few minutes.Charles –Charles 2025-05-13 16:35:32 +00:00 Commented May 13 at 16:35 Add a comment\| This answer is useful 7 Save this answer. Show activity on this post. According to Magma, the projective closure of the variety associated to the problem (given by the equations x(x−y)+1=z 2 1,(x+y)(x−y)+1=z 2 2,(x+2 y)(x−y)+1=z 2 3,x(x−y)+1=z 1 2,(x+y)(x−y)+1=z 2 2,(x+2 y)(x−y)+1=z 3 2, (x+y)x+1=z 2 4,(x+2 y)x+1=z 2 5,(x+2 y)(x+y)+1=z 2 6)(x+y)x+1=z 4 2,(x+2 y)x+1=z 5 2,(x+2 y)(x+y)+1=z 6 2) is an irreducible surface in P 8 P 8 with 34 isolated singularities. Since it is a complete intersection of six quadrics, it should be of general type (and it has trivial rational points with x=y=0 x=y=0 and slightly less trivial ones with y=0 y=0, so reduction methods will not work), which makes it very likely that this is a hard question. Added later: You may want to look at Question 73346 for an explanation by Noam Elkies of the reasoning behind this. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Apr 13, 2017 at 12:58 CommunityBot 1 2 2 silver badges 3 3 bronze badges answered Feb 11, 2012 at 21:17 Michael StollMichael Stoll 11.4k 2 2 gold badges 46 46 silver badges 60 60 bronze badges 4 This suggests that indeed the rational points are likewise fully accounted for by finitely many curves. The question asked only for integer points, which could conceivably be more tractable, though I don't see how.Noam D. Elkies –Noam D. Elkies 2012-02-11 21:39:02 +00:00 Commented Feb 11, 2012 at 21:39 @Noam: The problem with three terms should give a K3 Surface. Are there any K3-related methods that might prove that (up to signs) all the integral points are on the curve that comes from the Pell equation?Michael Stoll –Michael Stoll 2012-02-11 22:01:40 +00:00 Commented Feb 11, 2012 at 22:01 1 There are no solutions with five terms a,b,c,d,e a,b,c,d,e in arithmetic progression, since then a b+1 a b+1, a c+1 a c+1, a d+1 a d+1, a e+1 a e+1 would have to be four squares in arithmetic progression. Of course, this doesn't really help...Michael Stoll –Michael Stoll 2012-02-11 22:23:36 +00:00 Commented Feb 11, 2012 at 22:23 @Michael: It might be an interesting K3 surface but I don't see how to use this to prove anything about its integral points (for which it's of "log-general type").Noam D. Elkies –Noam D. Elkies 2012-02-12 20:39:51 +00:00 Commented Feb 12, 2012 at 20:39 Add a comment\| This answer is useful 4 Save this answer. Show activity on this post. Dujella has written many papers on Diophantine m-tuples, check out his webpage. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Feb 11, 2012 at 22:00 remoteremote 41 1 1 bronze badge Add a comment\| This answer is useful 4 Save this answer. Show activity on this post. Michael Stoll has given a nice answer, but here is a 17th century argument. Let a a, b b, c c, d d be an arithmetic progression with common difference Δ≠0 Δ≠0. Suppose that 1+a b=z 2 1,1+a c=z 2 2,1+a d=z 2 3,1+a b=z 1 2,1+a c=z 2 2,1+a d=z 3 2, 1+b c=z 2 4,1+b d=z 2 5,1+c d=z 2 6.1+b c=z 4 2,1+b d=z 5 2,1+c d=z 6 2. Consider the following quantities: A=2 z 2 2 z 2 5−z 2 1 z 2 6,A=2 z 2 2 z 5 2−z 1 2 z 6 2, B=z 2 z 3 z 4 z 5−Δ⋅z 1 z 6,B=z 2 z 3 z 4 z 5−Δ⋅z 1 z 6, C=z 1 z 3 z 4 z 6−2 Δ⋅z 2 z 5,C=z 1 z 3 z 4 z 6−2 Δ⋅z 2 z 5, D=z 1 z 2 z 5 z 6−3 Δ⋅z 3 z 4.D=z 1 z 2 z 5 z 6−3 Δ⋅z 3 z 4. Then one may easily compute that A 2,B 2,C 2,D 2 A 2,B 2,C 2,D 2 is an arithmetic progression. In particular, if the z i z i are rational, then we have constructed an arithmetic progression of four squares, contradicting a theorem of Fermat. To be careful, we also have to check that this arithmetic progression of squares has non-trivial difference. A little algebra shows this can only happen (assuming Δ≠0 Δ≠0) if one of the following equations holds: (1+a d)(1+b c)=0,(1+a d)(1+b c)=2 Δ 2.(1+a d)(1+b c)=0,(1+a d)(1+b c)=2 Δ 2. The first of these equations leads to no rational solutions (as in Michael's answer), the second says that a square (z 2 3 z 2 4 z 3 2 z 4 2) is equal to twice a (non-zero) square, which is also impossible. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Jan 15, 2016 at 5:11 Shadow of FelipeShadow of Felipe 646 4 4 silver badges 5 5 bronze badges 1 To relate this to Michael's answer, Fermat's four square problem is well known to equivalent to computing X 0(24)(Q)X 0(24)(Q). To see the equivalence, if 1 1, 1+t 1+t, 1+2 t 1+2 t, and 1+3 t 1+3 t are squares, then one has a rational point on s 2=(1+t)(1+2 t)(1+3 t)s 2=(1+t)(1+2 t)(1+3 t) which is X 0(24)X 0(24). The converse is slightly more subtle; one has to take points in the corresponding coset of the 2 2-Selmer group.Shadow of Felipe –Shadow of Felipe 2016-01-15 05:12:04 +00:00 Commented Jan 15, 2016 at 5:12 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. In Diophantine quadruple a<b<c<d a<b<c<d, it holds d≥4 b c d≥4 b c (see e.g. Lemma 14 in ). Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Feb 12, 2012 at 11:01 dujeduje 635 9 9 silver badges 17 17 bronze badges 1 1 To complete this: From $d \ge 4bc$, it follows easily that a,b,c,d a,b,c,d cannot be an AP. Note also the following: Acdording to Lemma 13 in Dujella's paper linked to in his answer above, if a,b,c a,b,c is a Diophantine triple with a<b<c a<b<c, then $c = a + b + 2 \sqrt{ab+1}$ or $c \ge 4ab$. If a,b,c a,b,c form an AP, then the second possibility cannot occur, and the first (together with the AP condition) then implies that b b is even and $y^2 - 3(b/2)^2 = 1$, where y=b−a=c−b y=b−a=c−b. So it is indeed the case that all such triples come from this Pell equation. (This gives another proof.)Michael Stoll –Michael Stoll 2012-02-22 19:21:24 +00:00 Commented Feb 22, 2012 at 19:21 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. I think it is better to solve a more formal task. We write the system. ⎧⎩⎨⎪⎪a b+T=x 2 a c+T=y 2 b c+T=z 2{a b+T=x 2 a c+T=y 2 b c+T=z 2 We need to find solutions a,b,c a,b,c - that was an arithmetic progression. This will help the solution of the equation Pell. p 2−3 s 2=T p 2−3 s 2=T Knowing any solution of the equation Pell (p 0;s 0)(p 0;s 0) you can always find the following formula. p 1=2 p 0+3 s 0 p 1=2 p 0+3 s 0 s 1=p 0+2 s 0 s 1=p 0+2 s 0 Having any decision - can immediately write down the solution of this system. a=2 s−p a=2 s−p b=2 s b=2 s c=2 s+p c=2 s+p x=s−p x=s−p y=s y=s z=s+p z=s+p Interesting here is that the x,y,z x,y,z looks like an arithmetic progression. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Jan 12, 2016 at 13:33 individindivid 526 1 1 gold badge 4 4 silver badges 10 10 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. My friend Huang showed me a clever solution to the second question: If a>b>c>d a>b>c>d are consecutive terms of an arithmetic progression , then we have (a b+1)(a c+1)=a 2(b c+1)+a(b+c−a)+1=a 2(b c+1)+a d+1>a 2(b c+1)(a b+1)(a c+1)=a 2(b c+1)+a(b+c−a)+1=a 2(b c+1)+a d+1>a 2(b c+1) where (a b+1)(a c+1)(a b+1)(a c+1) and a 2(b c+1)a 2(b c+1) are both perfect squares. Therefore, a d=(a b+1)(a c+1)−a 2(b c+1)−1≥2 a b c+1−−−−−√a d=(a b+1)(a c+1)−a 2(b c+1)−1≥2 a b c+1 Thus 4 b c+4≤d 2<b c 4 b c+4≤d 2<b c , a contradiction. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered May 13 at 15:01 Tong LinglingTong Lingling 838 2 2 silver badges 16 16 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. 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2698	https://python-forum.io/thread-3868.html	Calculating time difference between times and splitting the HH and MM View Active Threads View Today's Posts Home Forums View New Posts View Today's Posts My Discussions Unanswered Posts Unread Posts Active Threads Mark all forums read Staff List Member List Help Calendar Search Statistics Interpreter IRC Calculating time difference between times and splitting the HH and MM Python Forum Python Coding General Coding Help Thread Rating: 0 Vote(s) - 0 Average 1 2 3 4 5 Thread Modes Calculating time difference between times and splitting the HH and MM Bass Not Blown Up Yet Posts: 58 Threads: 10 Joined: Mar 2017 Reputation: 3 #1 Jul-02-2017, 08:30 PM (This post was last modified: Jul-02-2017, 08:30 PM by Bass.) Hi, Can you help with a time comparrison quesion in Python 3.6? I want to calculate the hours and minutes between two time and then display the difference,, by "splitting" the hours and minutes into a text string along the lines of "The market opens in HH hours and MM minutes" The solution seems to start off with putting both times into the same format, so I can subtract one from the other. This will handle the minutes element correctly. The difficulty is being able to "split" the result into Hours and Minutes, to display the difference into the required text string "The market opens in HH hours and MM minutes" with time.split but this fails with a str error. If I extract the hour elements and then the minute elements from both times and then compare them, this causes problems with carrying over the minutes. I have tried to manipulate the output "HH:MM", but have failed. Any help would be appreciated. Thank you Bass 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28import``datetime as datetime local_time``=``datetime.datetime.now() # The Sydney Opening time is 22H00M # This time is put into a valid time string (via .replace) # in order to subtract the number of hours and minutes until the # market opens in this line of code: countdown = sydney_open - local_time # This calculates the correct difference in hours/minutes. But prevents me # from splitting the result into hours and then minutes. # One alternative could be to multiply the hours by 60 and adding the # minutes element, for both times (current and opening time), then # calculating the hours by dividing by 60 and then deriving the # minutes to provide a result in hours and minutes. But this is # converluted - so looking for a smarter method. sydney_open``=``datetime.datetime.strptime(``'22:00:00'``,``"%H:%M:%S"``) sydney_open``=``local_time.replace(hour``=``sydney_open.time().hour, minute``=``0``, second``=``0``, microsecond``=``0``) countdown``=``sydney_open``-``local_time print``(``"local_time = "``, local_time) print``(``"sydney_open = "``, sydney_open) print``(``"Countdown...."``, countdown) # I want to display the countdown as "The Market opens in HH hours and MM minutes # Currently the output is: Sydney opens in 2:04:34.950393 # The preferred output is "Sydney opens in 2 hours and 4 minutes" "The good thing about standards is that you have so many to choose from" Andy S. Tanenbaum Reply Find Reply Larz60+ aetate et sapientia Posts: 12,089 Threads: 490 Joined: Sep 2016 Reputation: 452 #2 Jul-02-2017, 09:49 PM take a look at: Reply Find Reply Bass Not Blown Up Yet Posts: 58 Threads: 10 Joined: Mar 2017 Reputation: 3 #3 Jul-02-2017, 09:54 PM Thanks Larz60, I did look at this and the other two links that you (I think it was you?) posted before, but didn't spot a solution. BUT I have been working on this for most of the day so probably missed it. I'll look at it with fresh eyes! Appreciate your help. Bass "The good thing about standards is that you have so many to choose from" Andy S. Tanenbaum Reply Find Reply ichabod801 Bunny Rabbit Posts: 4,216 Threads: 97 Joined: Sep 2016 Reputation: 273 #4 Jul-02-2017, 10:04 PM It seems to be calculating correctly. If the only problem is getting the hours and minutes into a string, you can calculate that from the seconds attribute of the timedelta object: 1 2 3hours``=``countdown.seconds``/``/``3600 minutes``=``(countdown.seconds``/``/``60``)``%``60 print``(``'The market opens in {} hours and {} minutes.'``.``format``(hours, minutes)) Craig "Ichabod" O'Brien - xenomind.com I wish you happiness. Recommended Tutorials: BBCode, functions, classes, text adventures Reply WebsiteFind Reply Bass Not Blown Up Yet Posts: 58 Threads: 10 Joined: Mar 2017 Reputation: 3 #5 Jul-02-2017, 10:29 PM Thanks Guys, It's now working. Appreciate your help. Bass "The good thing about standards is that you have so many to choose from" Andy S. Tanenbaum Reply Find Reply snippsat Posts: 7,369 Threads: 123 Joined: Sep 2016 Reputation: 508 #6 Jul-02-2017, 11:04 PM (This post was last modified: Jul-02-2017, 11:04 PM by snippsat.) pendulum is what i think is best date/times library for Python. All python standard library datetime work in pendulum. So as a example can use sydney_open made in datetime and subtract if from pendulum.now(). 1 2 3 4 5 6 7 8 9>>>``import``pendulum >>> now``=``pendulum.now() >>> now <Pendulum [``2017``-``07``-``03T00``:``56``:``43.582239``+``02``:``00``]> >>> countdown``=``sydney_open``-``now >>>``print``(``f``"Sydney opens in {countdown.hours} hours and {countdown.minutes} minutes"``) Sydney opens``in``21``hours``and``3``minutes Reply Find Reply DeaD_EyE So-and-so of the Yard Posts: 2,161 Threads: 11 Joined: May 2017 Reputation: 234 #7 Jul-02-2017, 11:10 PM (This post was last modified: Jul-02-2017, 11:10 PM by DeaD_EyE.) > (Jul-02-2017, 10:04 PM)ichabod801 Wrote: It seems to be calculating correctly. If the only problem is getting the hours and minutes into a string, you can calculate that from the seconds attribute of the timedelta object: > > > > > > 1 > > > > 2 > > > > 3hours``=``countdown.seconds``/``/``3600 > > > > minutes``=``(countdown.seconds``/``/``60``)``%``60 > > > > print``(``'The market opens in {} hours and {} minutes.'``.``format``(hours, minutes)) Alternate way. 1 2 3 4 5 6 7 8 9from``collections``import``namedtuple def``to_hms(seconds): st_time``=``namedtuple(``'time'``,``'hours minutes seconds'``)`minutes, seconds=divmod(seconds,60)` ```hours, minutes=divmod(minutes,60)```return``st_time(hours, minutes, seconds) countdown``=``to_hms(``3660``) print``(``f``"Sydney opens in {countdown.hours} hours and {countdown.minutes} minutes"``) Output:Sydney opens in 1 hours and 1 minutesBtw: namedtuple is underused. But the interesting part is divmod. Almost dead, but too lazy to die: All humans together. We don't need politicians! Reply WebsiteFind Reply Bass Not Blown Up Yet Posts: 58 Threads: 10 Joined: Mar 2017 Reputation: 3 #8 Jul-03-2017, 01:26 PM (This post was last modified: Jul-03-2017, 01:27 PM by Bass.) Thanks for all the additional help. This is all useful advice for my routine. Sydney opening on a Sunday evening is significant as it opens the global currency market (which is closed Friday evening until Sunday evening). The market becomes more active/volatile as each trading day open across the world opens/closes, and therefore I can use the coutdown throught the day to provide alerts for say New York (which is perhaps the most volatile, followed by London). Pendulum looks like a useful module, especially as it is designed, using both the standard tools available for the datetime modules as well as additional features. So hopefully is/will become a sort of "one stop shop" for date and time manipulation. Thankssnippsat Let me know if anyone else is using Python with FX, happy to share solutions for API's and FX analysis. Thank you for all your help, much appreciated. Bass PS Starting to try and use tags etc. but hey learning is all about making mistakes, like highlighting whole paragraphs! "The good thing about standards is that you have so many to choose from" Andy S. 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Search Resources Search Advanced Search Filter By Education Standards Education Standards: Level Learning Domain Alignment Tag Grade Learning Domain Alignment Tag Grade Learning Domain Alignment Tag Grade Learning Domain Alignment Tag Grade Learning Domain Alignment Tag Grade Learning Domain Alignment Tag Grade Learning Domain Alignment Tag Grade Learning Domain Alignment Tag Grade Learning Domain Alignment Tag Grade Science Domain Topic Practice Alignment Tag Grade Learning Domain Alignment Tag Grade Learning Domain Alignment Tag Grade Learning Domain Alignment Tag Grade Learning Domain Alignment Tag Learning Domain Alignment Tag Level Learning Domain Alignment Tag Level Learning Domain Alignment Tag Grade Science Domain Topic Alignment Tag Grade Learning Domain Topic Alignment Tag Grade Learning Domain Alignment Tag Grade Learning Domain Alignment Tag Grade Learning Domain Alignment Tag Grade Science Domain Topic Practice Cross-Cutting Concept Disciplinary Core Idea Alignment Tag Grade Category Alignment Tag Grade Learning Domain Alignment Tag Grade Learning Domain Alignment Tag Grade Learning Domain Alignment Tag Grade Learning Domain Alignment Tag Grade Learning Domain Alignment Tag Subject Area [x] Applied Science [x] Architecture and Design [x] Engineering [x] Health, Medicine and Nursing [x] Education [x] Mathematics [x] Algebra [x] Geometry [x] Measurement and Data [x] Ratios and Proportions [x] Trigonometry [x] Physical Science [x] Physics Education Level [x] High School [x] Middle School [x] Adult Education [x] Career / Technical [x] Community College / Lower Division Material Type [x] Activity/Lab [x] Lesson Plan [x] Assessment [x] Lesson [x] Homework/Assignment [x] Module [x] Lecture [x] Reading [x] Teaching/Learning Strategy [x] Lecture Notes [x] Student Guide [x] Textbook License Types [x] Unrestricted Use [x] Creative Commons BY [x] Conditional Remix & Share Permitted [x] Creative Commons BY-NC [x] Creative Commons BY-NC-SA [x] Only Sharing Permitted [x] Creative Commons BY-NC-ND [x] Read the Fine Print [x] Educational Use Permitted Content Source [x] Open Author 2.0 [x] Content Provider Resources Primary User [x] Teacher [x] Student [x] Other [x] Parent Media Format [x] Text/HTML [x] Downloadable docs [x] Video [x] Graphics/Photos [x] Audio [x] Interactive [x] Other Educational Use [x] Curriculum/Instruction [x] Assessment [x] Informal Education Language [x] English [x] Spanish [x] Arabic Providers Open filtersClose filters 40 Results [x] Save Please log in to save materials. Log in Per page Sort By View Selected filters: pythagorean-theorem [x] Unrestricted Use CC BY Tangent Lines and the Radius of a Circle Rating 0.0 stars This task presents a foundational result in geometry, presented with deliberately sparse … This task presents a foundational result in geometry, presented with deliberately sparse guidance in order to allow a wide variety of approaches. Teachers should of course feel free to provide additional scaffolding to encourage solutions or thinking in one particular direction. We include three solutions which fall into two general approaches, one based on reference to previously-derived results (e.g., the Pythagorean Theorem), and another conducted in terms of the geometry of rigid transformations. MoreLess Subject:GeometryMathematicsMaterial Type:Activity/LabProvider:Illustrative MathematicsProvider Set:Illustrative MathematicsAuthor:Illustrative Mathematics Date Added:11/13/2012 MoreLess [x] Unrestricted Use CC BY Pythagorean Theorem Student-Made PowerPoint Template Rating 0.0 stars This is a short PowerPoint presentation template that I have created for … This is a short PowerPoint presentation template that I have created for my students to revise and make functional. After spending a day working with the Pythagorean Theorem, I want my students to work in groups to revise this presentation so that it could be used to introduce the Pythagorean Theorem in a classroom setting. MoreLess Subject:MathematicsMaterial Type:LessonDate Added:03/20/2015 MoreLess [x] Pythagorean Theorem in the Movies Rating 0.0 stars Reviewing the Pythagorean Theorem. Did you know the Scarecrow states the Pythagorean … Reviewing the Pythagorean Theorem. Did you know the Scarecrow states the Pythagorean Theorem incorrectly in The Wizard of Oz? At the end is a clip of Homer Simpson also stating the Pythagorean Theorem incorrectly MoreLess Subject:GeometryMathematicsMaterial Type:LectureProvider:Jeff Clark's Math in the MoviesAuthor:Jeff Clark Date Added:11/23/2016 MoreLess [x] Conditional Remix & Share Permitted CC BY-NC-SA CTE Architecture: Access Ramp Rating 0.0 stars This task was developed by high school and postsecondary mathematics and design/pre-construction … This task was developed by high school and postsecondary mathematics and design/pre-construction educators, and validated by content experts in the Common Core State Standards in mathematics and the National Career Clusters Knowledge & Skills Statements. It was developed with the purpose of demonstrating how the Common Core and CTE Knowledge & Skills Statements can be integrated into classroom learning - and to provide classroom teachers with a truly authentic task for either mathematics or CTE courses. MoreLess Subject:Applied ScienceArchitecture and DesignGeometryMathematicsTrigonometryMaterial Type:Activity/LabAssessmentHomework/AssignmentLesson PlanProvider:National Association of State Directors of Career Technical Education ConsortiumProvider Set:Career Technical EducationDate Added:03/05/2012 MoreLess [x] Unrestricted Use CC BY Pythagorean Theorem Lesson Plan Rating 0.0 stars A visual/physical inroduction to the origins and use of the Pythagorean Theorem … A visual/physical inroduction to the origins and use of the Pythagorean Theorem . MoreLess Subject:GeometryMaterial Type:Activity/LabHomework/AssignmentAuthor: Michael Wixted Date Added:05/06/2025 MoreLess [x] Conditional Remix & Share Permitted CC BY-NC-SA Módulo 2 de grado 8: El concepto de congruencia Rating 0.0 stars (Nota: Esta es una traducción de un recurso educativo abierto creado por … (Nota: Esta es una traducción de un recurso educativo abierto creado por el Departamento de Educación del Estado de Nueva York (NYSED) como parte del proyecto "EngageNY" en 2013. Aunque el recurso real fue traducido por personas, la siguiente descripción se tradujo del inglés original usando Google Translate para ayudar a los usuarios potenciales a decidir si se adapta a sus necesidades y puede contener errores gramaticales o lingüísticos. La descripción original en inglés también se proporciona a continuación.) En este módulo, los estudiantes aprenden sobre traducciones, reflexiones y rotaciones en el avión y, lo que es más importante, cómo usarlas para definir con precisión el concepto de congruencia. A lo largo del tema A, sobre las definiciones y propiedades de los movimientos rígidos básicos, los estudiantes verifican experimentalmente sus propiedades básicas y, cuando son factibles, profundicen su comprensión de estas propiedades utilizando el razonamiento. Todas las lecciones del tema B demuestran a los estudiantes la capacidad de secuenciar varias combinaciones de movimientos rígidos mientras mantienen las propiedades básicas de los movimientos rígidos individuales. Los estudiantes aprenden que la congruencia es solo una secuencia de movimientos rígidos básicos en el Tema C, y el Tema D comienza el aprendizaje del Teorema Pitagórico. Encuentre el resto de los recursos matemáticos de Engageny en English Description: In this module, students learn about translations, reflections, and rotations in the plane and, more importantly, how to use them to precisely define the concept of congruence. Throughout Topic A, on the definitions and properties of the basic rigid motions, students verify experimentally their basic properties and, when feasible, deepen their understanding of these properties using reasoning. All the lessons of Topic B demonstrate to students the ability to sequence various combinations of rigid motions while maintaining the basic properties of individual rigid motions. Students learn that congruence is just a sequence of basic rigid motions in Topic C, and Topic D begins the learning of Pythagorean Theorem. Find the rest of the EngageNY Mathematics resources at MoreLess Subject:GeometryMathematicsMaterial Type:ModuleProvider:New York State Education DepartmentProvider Set:EngageNYDate Added:09/21/2013 MoreLess [x] Conditional Remix & Share Permitted CC BY-NC Distance Formula (Dan Meyer 3-act-play) (Real) Rating 0.0 stars The 3 act math is a perfect way to incorporate the See-Think-Wonder … The 3 act math is a perfect way to incorporate the See-Think-Wonder activity. Dan Meyer is the originator of the 3 act math plays and the Pearson Envisions Math textbook incorporates one every Topic/Chapter. I am writing this lesson plan with the idea of using the Taco Cart 3 act math play.The link to the 3-act-math play is below.Taco Cart MoreLess Subject:MathematicsMaterial Type:Lesson PlanAuthor: Jill Oliver Date Added:11/05/2018 MoreLess [x] Conditional Remix & Share Permitted CC BY-NC The Pythagorean Formula Mini-Flexbook Rating 0.0 stars This Flexbook is community contributed through ck12.org. It covers three lessons on … This Flexbook is community contributed through ck12.org. It covers three lessons on the Pythagorean Theorem. 1) Introduction and Determining if the Triangle is a Right Triangle, 2) Finding the Hypotenuse, and 3) Finding a leg. It includes step by step instructions, application problems, and answers (at the end of each lesson). Ck12.org material is downloadable, editable, and accessible offline and online. MoreLess Subject:AlgebraGeometryMathematicsMaterial Type:Lecture NotesLessonReadingProvider:CK-12 FoundationProvider Set:CK-12 FlexBookAuthor:Shana Friend Date Added:05/25/2018 MoreLess [x] Conditional Remix & Share Permitted CC BY-NC-SA 8.GM.B.7 Unpacked Rating 0.0 stars Use the Pythagorean Theorem to determine unknown side lengths in right triangles … Use the Pythagorean Theorem to determine unknown side lengths in right triangles in problems in two- and three-dimensional contexts MoreLess Subject:MathematicsMaterial Type:Activity/LabAuthor:Liberty Public Schools Date Added:04/13/2021 MoreLess [x] Conditional Remix & Share Permitted CC BY-NC-SA Pythagoras and the Juice Seller Rating 0.0 stars This video lesson presents a real world problem that can be solved … This video lesson presents a real world problem that can be solved by using the Pythagorean theorem. The problem faces a juice seller daily. He has equilateral barrels with equal heights and he always tries to empty the juice of two barrels into a third barrel that has a volume equal to the sum of the volumes of the two barrels. This juice seller wants to find a simple way to help him select the right barrel without wasting time, and without any calculations - since he is ignorant of Mathematics. The prerequisite for this lesson includes knowledge of the following: the Pythagorean theorem; calculation of a triangles area knowing the angle between its two sides; cosine rule; calculation of a circle's area; and calculation of the areas and volumes of solids with regular bases. MoreLess Subject:GeometryMathematicsMeasurement and DataMaterial Type:LectureProvider:MITProvider Set:MIT BlossomsAuthor:Ghada Sulaiman Abdullah Marmash Date Added:06/02/2012 MoreLess [x] Only Sharing Permitted CC BY-NC-ND Proofs of the Pythagorean Theorem Rating 0.0 stars This lesson unit is intended to help teachers assess how well students … This lesson unit is intended to help teachers assess how well students are able to produce and evaluate geometrical proofs. In particular, this unit is intended to help you identify and assist students who have difficulties in: interpreting diagrams; identifying mathematical knowledge relevant to an argument; linking visual and algebraic representations; and producing and evaluating mathematical arguments. MoreLess Subject:MathematicsMaterial Type:AssessmentLesson PlanProvider:Shell Center for Mathematical EducationProvider Set:Mathematics Assessment Project (MAP)Date Added:04/26/2013 MoreLess [x] Conditional Remix & Share Permitted CC BY-NC-SA Grade 8 Module 2: The Concept of Congruence Rating 0.0 stars In this module, students learn about translations, reflections, and rotations in the … In this module, students learn about translations, reflections, and rotations in the plane and, more importantly, how to use them to precisely define the concept of congruence. Throughout Topic A, on the definitions and properties of the basic rigid motions, students verify experimentally their basic properties and, when feasible, deepen their understanding of these properties using reasoning. All the lessons of Topic B demonstrate to students the ability to sequence various combinations of rigid motions while maintaining the basic properties of individual rigid motions. Students learn that congruence is just a sequence of basic rigid motions in Topic C, and Topic D begins the learning of Pythagorean Theorem. Find the rest of the EngageNY Mathematics resources at MoreLess Subject:GeometryMathematicsMaterial Type:ModuleProvider:New York State Education DepartmentProvider Set:EngageNYDate Added:09/21/2013 MoreLess [x] Read the Fine Print Educational Use Let’s Build an Aqueduct! Rating 0.0 stars Students explore in detail how the Romans built aqueducts using arches—and the … Students explore in detail how the Romans built aqueducts using arches—and the geometry involved in doing so. Building on what they learned in the associated lesson about how innovative Roman arches enabled the creation of magnificent structures such as aqueducts, students use trigonometry to complete worksheet problem calculations to determine semicircular arch construction details using trapezoidal-shaped and cube-shaped blocks. Then student groups use hot glue and half-inch wooden cube blocks to build model aqueducts, doing all the calculations to design and build the arches necessary to support a water-carrying channel over a three-foot span. They calculate the slope of the small-sized aqueduct based on what was typical for Roman aqueducts at the time, aiming to construct the ideal slope over a specified distance in order to achieve a water flow that is not spilling over or stagnant. They test their model aqueducts with water and then reflect on their performance. MoreLess Subject:GeometryMathematicsMaterial Type:Activity/LabProvider:TeachEngineeringAuthor:Lauchlin Blue Malinda Zarske Nathan Coyle Date Added:02/07/2017 MoreLess [x] Conditional Remix & Share Permitted CC BY-NC-SA Geometry - TI Activities (Teacher's Edition) Rating 0.0 stars CK-12's Texas Instruments Geometry Teachers Edition Flexbook is a useful collection of … CK-12's Texas Instruments Geometry Teachers Edition Flexbook is a useful collection of exercises intended to enrich a student's understanding of basic geometric principles. MoreLess Subject:GeometryMathematicsMaterial Type:TextbookProvider:CK-12 FoundationProvider Set:CK-12 FlexBookAuthor:Jordan, Lori Date Added:12/15/2010 MoreLess [x] Unrestricted Use CC BY Applying the Pythagorean Theorem in a mathematical context Rating 0.0 stars This task requires students to apply the Pythagorean Theorem. Subject:GeometryMathematicsMaterial Type:Activity/LabProvider:Illustrative MathematicsProvider Set:Illustrative MathematicsAuthor:Illustrative Mathematics Date Added:05/01/2012 MoreLess [x] Conditional Remix & Share Permitted CC BY-NC Curriculum Guide: The Pythagorean Theorem Rating 0.0 stars The purpose of this curriculum guide is to provide OER lesson material … The purpose of this curriculum guide is to provide OER lesson material and support activities for Pythagorean Theorem instruction. It is geared towards GED® requirements. Both printable and online options are provided. MoreLess Subject:MathematicsMaterial Type:Activity/LabAssessmentLessonReadingTeaching/Learning StrategyDate Added:05/25/2018 MoreLess [x] Only Sharing Permitted CC BY-NC-ND Equations of Circles 1 Rating 0.0 stars This lesson unit is intended to help teachers assess how well students … This lesson unit is intended to help teachers assess how well students are able to: use the Pythagorean theorem to derive the equation of a circle; and translate between the geometric features of circles and their equations. MoreLess Subject:GeometryMathematicsMaterial Type:AssessmentLesson PlanProvider:Shell Center for Mathematical EducationProvider Set:Mathematics Assessment Project (MAP)Date Added:04/26/2013 MoreLess [x] Only Sharing Permitted CC BY-NC-ND The Pythagorean Theorem: Square Areas Rating 0.0 stars This lesson unit is intended to help teachers assess how well students … This lesson unit is intended to help teachers assess how well students are able to: use the area of right triangles to deduce the areas of other shapes; use dissection methods for finding areas; organize an investigation systematically and collect data; deduce a generalizable method for finding lengths and areas (The Pythagorean Theorem.) MoreLess Subject:GeometryMathematicsMaterial Type:AssessmentLesson PlanProvider:Shell Center for Mathematical EducationProvider Set:Mathematics Assessment Project (MAP)Date Added:04/26/2013 MoreLess [x] Conditional Remix & Share Permitted CC BY-NC-SA Módulo 7 de grado 8: Introducción a los números irracionales usando geometría Rating 0.0 stars (Nota: Esta es una traducción de un recurso educativo abierto creado por … (Nota: Esta es una traducción de un recurso educativo abierto creado por el Departamento de Educación del Estado de Nueva York (NYSED) como parte del proyecto "EngageNY" en 2013. Aunque el recurso real fue traducido por personas, la siguiente descripción se tradujo del inglés original usando Google Translate para ayudar a los usuarios potenciales a decidir si se adapta a sus necesidades y puede contener errores gramaticales o lingüísticos. La descripción original en inglés también se proporciona a continuación.) El módulo 7 comienza con el trabajo relacionado con el teorema de Pitágoras y los triángulos rectos. Antes de que se presenten las lecciones de este módulo a los estudiantes, es importante que las lecciones en los módulos 2 y 3 sean relacionadas con el teorema de Pitágoras se imparten (M2: Lecciones 15 y 16, M3: Lecciones 13 y 14). En los módulos 2 y 3, los estudiantes usaron el teorema de Pitágoras para determinar la longitud desconocida de un triángulo derecho. En los casos en que la longitud lateral era un entero, los estudiantes calcularon la longitud. Cuando la longitud lateral no era un entero, los estudiantes dejaron la respuesta en forma de x2 = c, donde C no era un número cuadrado perfecto. Esas soluciones se revisan y son la motivación para aprender sobre las raíces cuadradas y los números irracionales en general. Encuentre el resto de los recursos matemáticos de Engageny en English Description: Module 7 begins with work related to the Pythagorean Theorem and right triangles. Before the lessons of this module are presented to students, it is important that the lessons in Modules 2 and 3 related to the Pythagorean Theorem are taught (M2: Lessons 15 and 16, M3: Lessons 13 and 14). In Modules 2 and 3, students used the Pythagorean Theorem to determine the unknown length of a right triangle. In cases where the side length was an integer, students computed the length. When the side length was not an integer, students left the answer in the form of x2=c, where c was not a perfect square number. Those solutions are revisited and are the motivation for learning about square roots and irrational numbers in general. Find the rest of the EngageNY Mathematics resources at MoreLess Subject:GeometryMathematicsMaterial Type:ModuleProvider:New York State Education DepartmentProvider Set:EngageNYDate Added:02/02/2014 MoreLess [x] Read the Fine Print Educational Use History and Geometry of Roman Aqueducts Rating 0.0 stars Students see that geometric shapes can be found in all sorts of … Students see that geometric shapes can be found in all sorts of structures as they explore the history of the Roman Empire with a focus on how engineers 2000 years ago laid the groundwork for many structures seen today. Through a short online video, brief lecture material and their own online research directed by worksheet questions, students discover how the Romans invented a structure known today as the Roman arch that enabled them to build architecture never before seen by humankind, including the amazing aqueducts. Students calculate the slope and its total drop and angle over its entire distance for an example aqueduct. Completing this lesson prepares students for the associated activity in which teams build and test model aqueducts that meet specific constraints. This lesson serves as an introduction to many other geometry—and engineering-related lessons—including statics and trusses, scale modeling, and trigonometry. MoreLess Subject:GeometryMathematicsMaterial Type:LessonProvider:TeachEngineeringAuthor:Lauchlin Blue Malinda Zarske Nathan Coyle Date Added:02/07/2017 MoreLess Load more Discover Resources Collections Providers Community All Hubs All Groups Create Open Author Submit a Resource Our Services About Hubs About OER Commons OER 101 Help Center My Account My Items My Groups My Hubs Subscribe to OER Newsletter Subscribe Connect with OER CommonsFacebook, Opens in new windowTwitter, Opens in new window Donate to ISKME Powered By Privacy PolicyTerms of ServiceCollection PolicyDMCA © 2007 - 2025, OER Commons A project created by ISKME. Except where otherwise noted, content on this site is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 License. ` No restrictions on your remixing, redistributing, or making derivative works. Give credit to the author, as required. Your remixing, redistributing, or making derivatives works comes with some restrictions, including how it is shared. Your redistributing comes with some restrictions. Do not remix or make derivative works. Most restrictive license type. Prohibits most uses, sharing, and any changes. Copyrighted materials, available under Fair Use and the TEACH Act for US-based educators, or other custom arrangements. Go to the resource provider to see their individual restrictions. × Sign in / Register Your email or username: Did you mean ? Password: Forgot password?- [x] Show password or Sign In through your Institution Create an account Register or Sign In through your Institution

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