uni maths - sum of series...need a genius to help! Watch

hohoho11
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#1
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Show that the sum of the series 1 + cos x/1! + cos 2x/2! + cos 3x/3!......
is (e^cos x)cos x(sin x), where x is real. Hence show the integral of (e^cos x)cos x(sin x), between pi and 0 = pi.

No idea how to do this at all! Any help would be grately appreciated!
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apd35
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look it as e^(cos x) = sumr (cos x)^r/r!

and work from there would seem logical to me. Haven't worked it out, but I'd guess you are just a few trig identities away after that. The integration is easy, integrate term by term and you get each term being zero.
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hohoho11
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i understand what your saying, but i just dont understand how to start it...
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hohoho11
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i understand what your saying, but i just dont understand how to start it...
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dvs
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Do you perhaps mean e^(cos(x)) cos(sin(x)) instead?
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dvs
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Well, assuming that you do, consider
\sum_{n=0}^{\infty} [(e^(ix))^n]/n! = e^(e^(ix))
= e^(cos(x) + i sin(x))
= e^(cos(x)) e^(i sin(x))
= e^(cos(x)) (cos(sin(x)) + i sin(sin(x)))

Take the real part of that equality, and you get
\sum_{n=0}^{\infty} (cos(nx))/n! = e^(cos(x)) cos(sin(x))

I'll let you figure out the next part on your own.
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hohoho11
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Yes that is what i meant sorry

Ok i idnt really understand what you wrote, but so far i've done...

Let C = 1 + cos x/1! + cos 2x/2!...

Let S = a + sin x/1! + sin 2x/2!...

Then C + iS = 1 + i + [cos x + i sin x]/1! + [cos 2x + i sin 2x]/2!...

Using e^(i theta) = cos x + i sin x

C + iS = 1 + i + e^(i theta)/1! + e^(2i theta)/2! + e^(3i theta)/3!

However i cant see any relationship between these terms so i have no idea where to go from here!
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evariste
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(Original post by hohoho11)
Yes that is what i meant sorry

Ok i idnt really understand what you wrote, but so far i've done...

Let C = 1 + cos x/1! + cos 2x/2!...

Let S = a + sin x/1! + sin 2x/2!...

Then C + iS = 1 + i + [cos x + i sin x]/1! + [cos 2x + i sin 2x]/2!...

Using e^(i theta) = cos x + i sin x

C + iS = 1 + i + e^(i theta)/1! + e^(2i theta)/2! + e^(3i theta)/3!

However i cant see any relationship between these terms so i have no idea where to go from here!
C + iS = 1 + i + e^(i theta)/1! + e^(2i theta)/2! + e^(3i theta)/3!
=1+e^(itheta)+[e^(itheta)]^2/2!+[e^(itheta)]^3/3!+....
now e^x=1+x+x^2/2!+x^3/3!+...
so your expression is
C+iS=e^{e^(itheta)}+i
then refer to dvs post to find the real part of e^{e^(itheta)}
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