Using Power series to find a Maclaurin series

intervade

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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

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

f(x)=cos2x\displaystyle f(x)=cos^2x

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

cosx1x22!+x44!x66!+...+(1)nx2n(2n)!\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..

cos2x=1+cos(2x)2\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 ...

xcosxxx32!+x54!x76!+...\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!
 
intervade said:
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

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

f(x)=cos2x\displaystyle f(x)=cos^2x

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

cosx1x22!+x44!x66!+...+(1)nx2n(2n)!\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..

cos2x=1+cos(2x)2\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 ...

xcosxxx32!+x54!x76!+...\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!

cos(2x)1(2x)22!+(2x)44!(2x)66!+...+(1)n(2x)2n(2n)!\displaystyle cos(2x)\approx 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \frac{(2x)^6}{6!} + ... +\frac{(-1)^n(2x)^{2n}}{(2n)!}
 
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