# The sum of series involving cosine

• thanksie037
In summary, the student is trying to find the sum of the series 1+cos(x)+cos2x+...+cos3x which converges to 1. They are also trying to find the sum of the series e^{ix}=cos x+i sin x.
thanksie037

## Homework Statement

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

## The Attempt at a Solution

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.

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 ?

thanksie037 said:
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?

thanksie037 said:
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 don't know how to deal with the factorial

thanksie037 said:
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?

thanksie037 said:
and how do i even find the sum of this series. i only know how to find a sum of a geometric series...i don't 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?

thanksie037 said:
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...

gabbagabbahey said:
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.

Last edited:
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.

## What is the formula for finding the sum of a series involving cosine?

The formula for finding the sum of a series involving cosine is S = a + a cos(x) + a cos(2x) + a cos(3x) + ... + a cos(nx), where a is the amplitude and x is the angle in radians.

## How do you determine if a series involving cosine converges or diverges?

A series involving cosine converges if the limit of the absolute value of the terms as n approaches infinity is equal to 0. If the limit is any other value, the series diverges.

## Can the sum of a series involving cosine be negative?

Yes, the sum of a series involving cosine can be negative. This depends on the values of a and x in the formula. If a is negative, the sum will also be negative.

## Are there any special cases when finding the sum of a series involving cosine?

Yes, there are a few special cases when finding the sum of a series involving cosine. One is when x is equal to 0, in which case the sum will simply be (n+1)a. Another is when a is equal to 0, in which case the sum will be 0 regardless of the value of x.

## Can the sum of a series involving cosine be simplified?

Yes, the sum of a series involving cosine can be simplified in some cases. For example, if the series involves cos(nx), it can be simplified using the double angle formula for cosine. Additionally, if the series involves cos(x), it can be simplified using the power series expansion for cosine.

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