O limita dificila

[Bogdan Cebere]

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

[Cezar Lupu]

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.

[Bogdan Cebere]

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.

[Cezar Lupu]

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