Using Power series to find a Maclaurin series

[intervade]

Ok, so I am suppose to use a power series function that we were given in our book to help me find a Maclaurin series for the following functions

\(\displaystyle f(x)=xcos(2x)\) and

\(\displaystyle f(x)=cos^2x\)

Now, I assume I am going to use the following power series (which was given to me)..

\(\displaystyle cosx\approx 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + ... +\frac{(-1)^nx^{2n}}{(2n)!}\)

Now I have a good idea of how I am going to do cos^2(x) .. using trig identities..

\(\displaystyle cos^2x = \frac{1 + cos(2x)}{2}\)

But.. Im not really sure how to go about doing the xcos(2x) .. would that be basically ...

\(\displaystyle xcosx\approx x - \frac{x^3}{2!} + \frac{x^5}{4!} - \frac{x^7}{6!} + ...\)

Then multiplied by a 2^n ?

Just sort of checking my work here , help would be much appreciated!

 

[Deleted member 4993]

Ok, so I am suppose to use a power series function that we were given in our book to help me find a Maclaurin series for the following functions

\(\displaystyle f(x)=xcos(2x)\) and

\(\displaystyle f(x)=cos^2x\)

Now, I assume I am going to use the following power series (which was given to me)..

\(\displaystyle cosx\approx 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + ... +\frac{(-1)^nx^{2n}}{(2n)!}\)

Now I have a good idea of how I am going to do cos^2(x) .. using trig identities..

\(\displaystyle cos^2x = \frac{1 + cos(2x)}{2}\)

But.. Im not really sure how to go about doing the xcos(2x) .. would that be basically ...

\(\displaystyle xcosx\approx x - \frac{x^3}{2!} + \frac{x^5}{4!} - \frac{x^7}{6!} + ...\)

Then multiplied by a 2^n ?

Just sort of checking my work here , help would be much appreciated!

\(\displaystyle cos(2x)\approx 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \frac{(2x)^6}{6!} + ... +\frac{(-1)^n(2x)^{2n}}{(2n)!}\)