# Prove that $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+R_8(x)$ where $|R_8(x)|\leq \frac{x^8}{8!}$

Prove that $$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+R_8(x)$$

where $|R_8(x)|\leq \frac{x^8}{8!}$

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Which formulas for the error term in a Taylor expansion do you know? –  Hagen von Eitzen Mar 7 '13 at 19:22
$R_n(x)=\frac{f^{(n)}(y)}{n!} x^n$ –  Kasper Mar 7 '13 at 19:25
Well, when $f(x)=\cos(x)$ what is $f^{(8)}(x)$? –  Thomas Andrews Mar 7 '13 at 19:27
$f^{(8)}(x)=\cos x$ so I get $|R_n(x)|=|\frac{\cos y}{8!}x^8|\leq \frac{x^8}{8!}$ –  Kasper Mar 7 '13 at 19:30
@Kasper exactly what you wanted to have –  Dominic Michaelis Mar 7 '13 at 19:31

The formula you have is the Taylor polynomial of order 7, use the legendre error term and that $|\cos(x)|\leq 1$ for all $x$.

The second derivative of $\cos(x)$ is $-\cos(x)$ so the eight derivative is $\cos(x)$.

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Hint: $f(x)=\sum_{i=0}^n\frac{f^ {(i)}(x)x^i}{i! }$
$\text if$ $$|f^{n+1}(x)|\le M_{n+1}$$then $$|R(x)|\le \frac{M_{n+1}|x|^n}{(n+1)!}$$

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Sorry but now i don't unterstand your first line anymore. Your second display formula is wrong (you surely meant $|x|^n$ instead of $|x|n$ and the stuff is pretty much the same as mine –  Dominic Michaelis Mar 7 '13 at 19:40
@Dominic Michaelis:R(x) is error term of general form of finite teylor sequence –  Maisam Hedyelloo Mar 7 '13 at 19:46
yeah i know but you say $f(x)=\sum \dots$ which is wrong as you don't mention the error term –  Dominic Michaelis Mar 7 '13 at 19:48
@Dominic Michaelis :thanks i modify it –  Maisam Hedyelloo Mar 7 '13 at 19:50