How do you use a power series to find the exact value of the sum of the series $1 - \frac{{(\frac{π}{4})}^{2}}{2!} + \frac{{(\frac{π}{4})}^{4}}{4!} - \frac{{(\frac{π}{4})}^{6}}{6!} + \dots$ $1-(pi/4)^2/(2!)+(pi/4)^4/(4!) -(pi/4)^6/(6!) + …$ ?

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How do you use a power series to find the exact value of the sum of the series $1 - \frac{{(\frac{π}{4})}^{2}}{2!} + \frac{{(\frac{π}{4})}^{4}}{4!} - \frac{{(\frac{π}{4})}^{6}}{6!} + \dots$ $1-(pi/4)^2/(2!)+(pi/4)^4/(4!) -(pi/4)^6/(6!) + …$ ?

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Answer 1 · 2014-09-12T11:43:06

Since the Maclaurin series
$1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + \dots = cos x$ $1-x^2/{2!}+x^4/{4!}-x^6/{6!}+cdots=cosx$ ,
$1 - \frac{{(\frac{π}{4})}^{2}}{2!} + \frac{{(\frac{π}{4})}^{4}}{4!} - \frac{{(\frac{π}{4})}^{6}}{6!} + \dots = cos (\frac{π}{4}) = \frac{1}{\sqrt{2}}$ $1-(pi/4)^2/{2!}+(pi/4)^4/{4!}-(pi/4)^6/{6!}+cdots=cos(pi/4)=1/sqrt{2}$ .

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How do you use a power series to find the exact value of the sum of the series $1 - \frac{{(\frac{π}{4})}^{2}}{2!} + \frac{{(\frac{π}{4})}^{4}}{4!} - \frac{{(\frac{π}{4})}^{6}}{6!} + \dots$ $1-(pi/4)^2/(2!)+(pi/4)^4/(4!) -(pi/4)^6/(6!) + …$ ?

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