The sum of series involving cosine

1. Feb 1, 2009

thanksie037

1. The problem statement, all variables and given/known data

Find the sum of the series s(x) = 1 +cos(x)+ (cos2x)/2!+(cos3x)/3!...

2. Relevant equations

3. The attempt at a solution

2. Feb 1, 2009

gabbagabbahey

Hi thanksie, what have you tried? What methods have you learned? Are you allowed to compare it to other known series?

3. Feb 1, 2009

thanksie037

Hi, Yes we can compare it to known series. I've tried a comparison with the harmonic series and the expansion for e^x looks promising. I'm just not sure how to get started.

4. Feb 1, 2009

gabbagabbahey

Comparing it to the series for e^x sounds good to me.....try writing $\cos(nx)$ in terms of complex exponentials....what does that give you?

5. Feb 1, 2009

thanksie037

would x= cosnx? is there a property for cosine where (cos x)^n=cos nx ?

6. Feb 1, 2009

gabbagabbahey

Not quite...try writing $\cos(nx)$ in terms of complex exponentials....what do you get?

7. Feb 1, 2009

thanksie037

Do you mean:

x^ni=cos nx + sin nx?

8. Feb 1, 2009

gabbagabbahey

Huh?! Why on Earth would you think that were true?

Have you not seen the formulas $$e^{i\theta}=\cos\theta+i\sin\theta$$ and $$\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}$$ before?

9. Feb 1, 2009

thanksie037

sorry i was using that first one. i was confused as to what you meant with cos nx.

using the second one makes a lot more sense. are you saying i should:

cos n(theta) = (e^ni(theta) +e^-ni(theta))/2

10. Feb 1, 2009

thanksie037

and how do i even find the sum of this series. i only know how to find a sum of a geometric series...i dont know how to deal with the factorial

11. Feb 1, 2009

gabbagabbahey

Yes, so now you have

$$1+\cos(x)+\frac{\cos(2x)}{2!}+\ldots=\sum_{n=0}^{\infty} \frac{\cos(nx)}{n!}=\frac{1}{2}\sum_{n=0}^{\infty} \frac{e^{i n x}+e^{-i n x}}{n!}$$

correct?

Try breaking it into two separate sums:

$$\frac{1}{2}\sum_{n=0}^{\infty} \frac{e^{i n x}+e^{-i n x}}{n!}=\frac{1}{2}\sum_{n=0}^{\infty} \frac{e^{i n x}}{n!}+\frac{1}{2}\sum_{n=0}^{\infty} \frac{e^{-i n x}}{n!}$$

Now use the fact that $$e^{n\theta}=\left(e^{\theta}\right)^n$$ and compare each sum to the series for e^u.

12. Feb 1, 2009

thanksie037

That makes sense. Thanks for all your help.

One more unrelated quick question:
do the series n/(n+1) and n^2/(n^2+1)converge or diverge?

13. Feb 1, 2009

gabbagabbahey

Well, try applying some of the convergence tests that you've learned....

14. Feb 1, 2009

Gregg

I looked at this and became interested but got stuck.

$e^{ix} = \cos x + i \sin x$

$e^{-ix} = \cos x - i \sin x$

$\cos x = \frac{e^{ix} + e^{-ix}}{2}$

$\displaystyle\sum_{n=0}^\infty \frac{\cos (nx)}{n!} = \displaystyle\sum_{n=0}^\infty \frac{e^{inx} + e^{-inx}}{2n!}$

$\displaystyle\sum_{n=0}^\infty \frac{e^{inx} + e^{-inx}}{2n!} = \displaystyle\sum_{n=0}^\infty \frac{e^{inx}}{2n!} + \displaystyle\sum_{n=0}^\infty \frac{e^{-inx}}{2n!}$

$\displaystyle\sum_{n=0}^\infty \frac{e^{inx} + e^{-inx}}{2n!} = \frac{1}{2} \left[ \displaystyle\sum_{n=0}^\infty \frac{e^{inx}}{n!} + \displaystyle\sum_{n=0}^\infty \frac{e^{-inx}}{n!} \right]$

$e^x = \displaystyle\sum_{n=0}^\infty \frac{x^n}{n!}$

$\displaystyle\sum_{n=0}^\infty \frac{e^{inx}}{n!} = e^{e^{ix}}$

$\displaystyle\sum_{n=0}^\infty \frac{e^{-inx}}{n!} = e^{e^{-ix}}$

$\displaystyle\sum_{n=0}^\infty \frac{e^{inx} + e^{-inx}}{2n!} = \frac{1}{2} \left[e^{e^{-ix}} + e^{e^{ix}} \right]$

Edit: I think this is OK actually.

Last edited: Feb 1, 2009
15. Feb 1, 2009

thanksie037

yeah I was going to say that's what I got... lots of thanks to gab, I would have put my work in here but I don't know how to use this forum very well.