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## Homework Statement

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

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- #1

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

- #2

gabbagabbahey

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- #4

gabbagabbahey

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- #5

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

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gabbagabbahey

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

- #7

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Do you mean:

x^ni=cos nx + sin nx?

x^ni=cos nx + sin nx?

- #8

gabbagabbahey

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

x^ni=cos nx + sin nx?

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

- #9

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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

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- #11

gabbagabbahey

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Yes, so now you have

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

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

[tex]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!}[/tex]

correct?

Try breaking it into two separate sums:

[tex]\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!}[/tex]

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

- #12

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One more unrelated quick question:

do the series n/(n+1) and n^2/(n^2+1)converge or diverge?

- #13

gabbagabbahey

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

- #14

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I looked at this and became interested but got stuck.Huh?! Why on Earth would you think that were true?

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

[itex] e^{ix} = \cos x + i \sin x [/itex]

[itex] e^{-ix} = \cos x - i \sin x [/itex]

[itex] \cos x = \frac{e^{ix} + e^{-ix}}{2} [/itex]

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

[itex]\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!}[/itex]

[itex]\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][/itex]

[itex] e^x = \displaystyle\sum_{n=0}^\infty \frac{x^n}{n!}[/itex]

[itex] \displaystyle\sum_{n=0}^\infty \frac{e^{inx}}{n!} = e^{e^{ix}} [/itex]

[itex] \displaystyle\sum_{n=0}^\infty \frac{e^{-inx}}{n!} = e^{e^{-ix}} [/itex]

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

Edit: I think this is OK actually.

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