Question

MacLaurin series for 15xCos((1/11)x^2)

Answer 1

\[\cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}\], can you take it from here?

Answer 2

Im kinda new to this so no :/, do you just square it and multiply by 1/11?? Thanks for responding.

Answer 3

Before you do anything, replace \(x\) by \(\frac{x^2}{11}\). That will take care of the argument of Cosine and make it what you need it to be. After that, treat the above statement as what it is, an equation.

Answer 4

how does it remove the argument of cosine?

Answer 5

What @jlvm was getting at:
\[\cos x=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}~~\Rightarrow~~\cos\frac{x^2}{11}=\sum_{n=0}^{\infty}\frac{(-1)^n\left(\frac{x^2}{11}\right)^{2n}}{(2n)!}\]
There's still a matter of the \(15x\) in front, but that's easy:
\[15x\cos\frac{x^2}{11}=15x\sum_{n=0}^{\infty}\frac{(-1)^n\left(\frac{x^2}{11}\right)^{2n}}{(2n)!}\]
Typically, the \(x\) would be written so that it's absorbed into the series.
\[x(x^2)^{2n}=x(x^{4n})=x^{4n+1}\]