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Demonstrati ca \( \frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+.... \)

Foarte frumos .

Problema postata de Gordon C. mai sus nu este altceva decat celebra serie Leibniz-Gregory .

Daca consideram seria geometrica urmatoare:

\( 1-x^{2}+x^{4}-x^{6}+\ldots=\frac{1}{1+x^{2}} \)

si integram, vom obtine ca

\( \arctan(x)=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}+\ldots \).
Luand \( x=1 \), obtinem exact ce vrem.

P.S. Exista si o alta solutie pentru seria \( \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2n-1}=\frac{\pi}{4} \) bazata pe considerarea unui sir de integrale.

Demonstrati ca \( \frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+.... \)

Solutie

Lema
Daca n este par, atunci ctg\( \frac{\pi}{4n} \)-ctg\( \frac{3\pi}{4n} \)+ctg\( \frac{5\pi}{4n} \)-\( \dots \)+ctg\( \frac{(2n-3)\pi}{4n} \)-ctg\( \frac{(2n-1)\pi}{4n} \)=n.
Demonstratie
Se poate demonstra, dezvoltand \( (cos x+i {\dot}sin x)^n \) cu ajutorul formulei lui Moivre si a binomului lui Newton, ca
\( x^n-C_n^1{ x^{n-1}}-C_n^2{ x^{n-2}}+C_n^3{ x^{n-3}}+C_n^4{ x^{n-4}}-{\dots}= \) \( (x-ctg\frac{\pi}{4n})(x+ctg\frac{3\pi}{4n})(x-ctg\frac{5\pi}{4n}){\dots}(x-ctg\frac{(2n-3)\pi}{4n})(x+ctg\frac{(2n-1)\pi}{4n}) \)
iar din fomula lui Viete pentru coeficientul lui \( x^{n-1} \) obtinem relatia din enuntul lemei.

Revenind la problema noastra avem:
\( ctg {\alpha}-ctg(\beta)=\frac{sin({\beta}- {\alpha})}{sin{\alpha}{\dot} sin{\beta}}=sin({\beta}-{\alpha})cosec({\alpha})cosec({\beta}) \)
deci din identitatea din lema vom avea:
\( sin(\frac{\pi}{2n})[cosec(\frac{\pi}{4n})cosec({\frac{3\pi}{4n}})+cosec(\frac{5\pi}{4n})cosec({\frac{7\pi}{4n}})+{\dots+}cosec(\frac{(2n-3)\pi}{4n})cosec({\frac{(2n-1)\pi}{4n}})]=n \)
sau\( cosec(\frac{\pi}{4n})cosec({\frac{3\pi}{4n}})+cosec(\frac{5\pi}{4n})cosec({\frac{7\pi}{4n}})+{\dots+}cosec(\frac{(2n-3)\pi}{4n})cosec({\frac{(2n-1)\pi}{4n}})={\frac{n}{sin{\frac{\pi}{2n}}}} \)

Pe de alta parte
\( ctg{\alpha}-ctg{\beta}=tg({\alpha}-{\beta})(1+ctg{\alpha}ctg{\beta}) \)
Rezulta din aceeasi aceeasi identitate ca
\( tg(\frac{\pi}{2n})[ctg(\frac{\pi}{4n})ctg(\frac{3\pi}{4n})+ctg(\frac{5\pi}{4n})ctg(\frac{7\pi}{4n})+{\dots}+ctg(\frac{(2n-3)\pi}{4n})ctg(\frac{(2n-1)\pi}{4n})+\frac{n}{2}]=n \)
sau
\( ctg(\frac{\pi}{4n})ctg(\frac{3\pi}{4n})+ctg(\frac{5\pi}{4n})ctg(\frac{7\pi}{4n})+{\dots}+ctg(\frac{(2n-3)\pi}{4n})ctg(\frac{(2n-1)\pi}{4n})=\frac{n}{tg{\frac{\pi}{2n}}}-\frac{n}{2} \).
Din \( sinx<x<tgx \) avem \( cosecx>\frac{1}{x}>ctgx \), deci din relatiile obtinute anterior avem:
\( \frac{n}{sin{\frac{\pi}{2n}}}>{\frac{4n}{\pi}}{\dot}{\frac{4n}{3\pi}}+{\frac{4n}{5\pi}}{\dot}{\frac{4n}{7\pi}}+{\dots}+{\frac{4n}{(2n-3)\pi}}{\dot}{\frac{4n}{(2n-1)\pi}}>{{{\frac{n}{tg({\frac{\pi}{2n}})}}-\frac{n}{2} \)
sau
\( {\frac{\pi^2}{8n}}{\frac{n}{sin{\frac{\pi}{2n}}}}>{\frac{2}{1* 3}}+{{\frac{2}{5*7}}}+{\dots}+{\frac{2}{(2n-3)*(2n-1)}}>{\frac{\pi^2}{8n}}{{\frac{n}{tg({\frac{\pi}{2n})}-\frac{\pi^2}{16n} \)
sau
\( {\frac{\pi}{4}}{\frac{{\frac{\pi}{2n}}}{sin{\frac{\pi}{2n}}}}>1-{\frac{1}{3}}+{\frac{1}{5}}-{\dots}+{\frac{1}{2n-3}}-{\frac{1}{2n-1}}>{\frac{\pi}{4}}({{{\frac{{\frac{\pi}{2n}}}{tg({\frac{\pi}{2n})}}-\frac{\pi}{4n}) \)
Trecand la limita inegalitatea, obtinem concluzia.

P.S. Aceasta solutie este prezentata si in cartea "Probleme neelementare tratate elementar" de A.M.Iaglom si I.M. Iaglom.

Misto solutia postata de Gordon C. Habar nu aveam de ea pana acuma , desi o cam intuiam in mare. Dupa cum spuneam, exista si o solutie care foloseste calculul integral, dupa cum urmeaza:

Consideram sirul \( (I_{n})_{n\geq 1} \) definit prin
\( I_{n}=\int_{0}^{\frac{\pi}{4}}\tan^{n}xdx \) pentru \( n\geq 1 \).
Facand schimbarea de variabila \( \tan x=t \), integrala noastra devine
\( I_{n}=\int_0^1\frac{t^{n}}{t+1}dt \). Este cat se poate de evident ca \( \lim_{n\to\infty}I_{n}=0 \). Mai mult, avem egalitatea
\( I_{n}+I_{n+2}=\int_{0}^{\frac{\pi}{4}}tan^{n}x\sec^{2}xdx=\int_0^1t^{n}dt=\frac{1}{n+1} \). Construim sirul \( (J_{n})_{n\geq 1} \) definit astfel \( J_{n}=(-1)^{n-1}I_{2n} \). Se observa cu ochiul liber ca \( J_{n}-J_{n-1}=\frac{(-1)^{n-1}}{2n-1} \).
Mai mult, avem ca \( J_{1}=I_{2}=1-\frac{\pi}{4} \). Astfel, prin recurenta, va rezulta ca

\( J_{n}=\sum_{k=1}^{n}\frac{(-1)^{k-1}}{2k-1}-\frac{\pi}{4} \).

Concluzia se impune