The sum of series involving cosine

[thanksie037]

Homework Statement

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

Homework Equations

The Attempt at a Solution

 

[gabbagabbahey]

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

 

[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.

 

[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?

 

[thanksie037]

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

 

[gabbagabbahey]

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?

 

[thanksie037]

Do you mean:

x^ni=cos nx + sin nx?

 

[gabbagabbahey]

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?

 

[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

 

[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 don't know how to deal with the factorial

 

[gabbagabbahey]

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 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.

 

[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?

 

[gabbagabbahey]

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...

 

[Gregg]

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.

 

[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.