# The sum of series involving cosine

## Homework Statement

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

## The Attempt at a Solution

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

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.

gabbagabbahey
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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?

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

gabbagabbahey
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would x= cosnx? is there a property for cosine where (cos x)^n=cos nx ?
Not quite...try writing $\cos(nx)$ in terms of complex exponentials....what do you get?

Do you mean:

x^ni=cos nx + sin nx?

gabbagabbahey
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Do you mean:

x^ni=cos nx + sin nx?
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?

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

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

gabbagabbahey
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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
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?

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
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.

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?

gabbagabbahey
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Gold Member
do the series n/(n+1) and n^2/(n^2+1)converge or diverge?
Well, try applying some of the convergence tests that you've learned....

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?
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.

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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.