How do you use a power series to find the exact value of the sum of the series $\frac{π}{4} - \frac{{(\frac{π}{4})}^{3}}{3!} + \frac{{(\frac{π}{4})}^{5}}{5!} - \frac{{(\frac{π}{4})}^{7}}{7!} + \dots$ $pi/4-(pi/4)^3/(3!)+(pi/4)^5/(5!)-(pi/4)^7/(7!) + …$ ?

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How do you use a power series to find the exact value of the sum of the series $\frac{π}{4} - \frac{{(\frac{π}{4})}^{3}}{3!} + \frac{{(\frac{π}{4})}^{5}}{5!} - \frac{{(\frac{π}{4})}^{7}}{7!} + \dots$ $pi/4-(pi/4)^3/(3!)+(pi/4)^5/(5!)-(pi/4)^7/(7!) + …$ ?

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Answer 1 · 2014-09-25T21:56:56

Since

$x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \dots = sin x$ $x-x^3/{3!}+x^5/{5!}-x^7/{7!}+cdots=sinx$ ,

$\frac{π}{4} - \frac{{(\frac{π}{4})}^{3}}{3!} + \frac{{(\frac{π}{4})}^{5}}{5!} - \frac{{(\frac{π}{4})}^{7}}{7!} + \dots = sin (\frac{π}{4}) = \frac{1}{\sqrt{2}}$ $pi/4-(pi/4)^3/{3!}+(pi/4)^5/{5!}-(pi/4)^7/{7!}+cdots=sin(pi/4)=1/sqrt{2}$

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How do you use a power series to find the exact value of the sum of the series $\frac{π}{4} - \frac{{(\frac{π}{4})}^{3}}{3!} + \frac{{(\frac{π}{4})}^{5}}{5!} - \frac{{(\frac{π}{4})}^{7}}{7!} + \dots$ $pi/4-(pi/4)^3/(3!)+(pi/4)^5/(5!)-(pi/4)^7/(7!) + …$ ?

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How do you use a power series to find the exact value of the sum of the series $\frac{π}{4} - \frac{{(\frac{π}{4})}^{3}}{3!} + \frac{{(\frac{π}{4})}^{5}}{5!} - \frac{{(\frac{π}{4})}^{7}}{7!} + \dots$ $pi/4-(pi/4)^3/(3!)+(pi/4)^5/(5!)-(pi/4)^7/(7!) + …$ ?