How do you find the Maclaurin series of $f (x) = cos (x^{2})$ $f(x)=cos(x^2)$ ?

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How do you find the Maclaurin series of $f (x) = cos (x^{2})$ $f(x)=cos(x^2)$ ?

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Answer 1 · 2014-09-12T21:09:27

We have the Maclaurin series
$cos x = \infty \sum n = 0 {(- 1)}^{n} \frac{x^{2 n}}{(2 n)!}$ $cosx=sum_{n=0}^infty(-1)^n{x^{2n}}/{(2n)!}$

by replacing $x$ $x$ by $x^{2}$ $x^2$ ,
$cos (x^{2}) = \infty \sum n = 0 {(- 1)}^{n} \frac{x^{4 n}}{(2 n)!}$ $cos(x^2)=sum_{n=0}^infty(-1)^n{x^{4n}}/{(2n)!}$

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How do you find the Maclaurin series of $f (x) = cos (x^{2})$ $f(x)=cos(x^2)$ ?

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How do you find the Maclaurin series of $f (x) = cos (x^{2})$ $f(x)=cos(x^2)$ ?