## Calculus, 10th Edition (Anton)

$\pi x - \frac {\pi^2 x^2} {2!} + \frac {\pi^4 x^4} {4!} - \frac {\pi ^6 x^6 } {6!} + ... = \sum\limits_{n=0}^{\infty} (-1)^n$ $\frac{(x \pi)^{2n}} {2n!}$
Maclauren Series is $\sum\limits_{n=0}^{\infty} \frac{f^n(0)x^n} {n!} = f(0) + f'(0)x + \frac{(f''(0)x^2)}{2!} + ..\\$ Given function, $f(x)=cos\pi x$ Differentiate w.r.t x $f'(x)=-\pi sin\pi x$ $f''(x)=-\pi^2 cos\pi x$ $f'''(x)=\pi^3 sin\pi x$ $f''''(x)=\pi^4 cos\pi x$ Now $f(0)=\pi$ , $f'(0)x =0$ , $f''(0)-\pi^2$, $f'''(0)=0$, $f''''(x)=\pi^4$ Hence the Maclaurin Series for the function is $\pi x - \frac {\pi^2 x^2} {2!} + \frac {\pi^4 x^4} {4!} - \frac {\pi ^6 x^6 } {6!} + ... = \sum\limits_{n=0}^{\infty} (-1)^n$ $\frac{(x \pi)^{2n}} {2n!}$