
#56
Jul2311, 07:04 AM

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C_{1} is the coefficient of cosx, so it's 0.




#58
Jul2311, 07:47 AM

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As an "appendix", here's the alternative result using complex numbers and Euler's formula
you won't understand it yet, but anyone else following this might be interested sin^{3}xsin3xThis works for sin^{k}xsinkx for any value of k sin^{k}xsinkx = (1/2^{k}) ∑ ^{k}C_{m} cos2mx (so what's sin^{k}xcoskx, and is there a quicker way of doing it?) 



#59
Jul2311, 07:53 AM

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#60
Jul2311, 08:18 AM

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As another aside, this series is called the Fourier cosine series expansion.
Fourier was a mathematician who found that any function can be expressed as a cosine series of this type, and that the coefficients are unique. Many mathematical programmatic tools have this built in, such as Mathematica. It's also available in WolframAlpha: http://www.wolframalpha.com/input/?i...s+sin^3x+sin3x although I found that you won't be able to get the full expansion here, because WolframAlpha will time out before you'll get there. :( However, I'm afraid it will be quite a while before you get to learn this, and maybe you will never learn this, depending on what you're going to study exactly. The application of Fourier analysis is typically in spectral analysis, signal processing, image processing and electronical engineering, where it's used for frequency related analysis and operations. 



#61
Jul2311, 10:36 AM

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Hi, tinytim, is not there a sign error in your derivation?
Just to show off my way: [tex]sinx=\frac{e^{ix}e^{ix}}{2i}[/tex], [tex]cosx=\frac{e^{ix}+e^{ix}}{i}[/tex], [tex] sin(x)^3 sin(3x)=(\frac {e^{ix}e^{ix}}{2i})^3 \frac{e^{3ix}e^{3ix}}{2i}=[/tex] [tex]=\frac {e^{3ix}3 e^{ix}+3 e^{ix}e^{3ix}}{8i}\frac{e^{3ix}e^{3ix}}{2i}=[/tex] [tex]1/16(e^{6ix}3 e^{4ix}+3 e^{2ix}11+3 e^{2ix}3 e^{4ix}+e^{6ix})=[/tex] [tex]1/8(cos(6x)3cos(4x)+3cos(2x)1)[/tex] ehild 



#62
Jul2311, 10:53 AM

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@tinytim: Nice improvement on the calculation btw. It's way less work!
First, I think that PranavArora is very well able to follow this one and do it himself. It only requires a basic grasp on complex numbers, and beyond that it's just exponentiation rules, which PranavArora's has already practiced quite a bit. And it also shows that in my humble opinion calculating with sines, cosines and angles is kind of obsolete. I found that calculating with imaginary epowers (or vectors in geometric problems) is much easier. I'm interested in examples where you would really want to use sines, cosines and angles to do your calculations.... Does any of you have any? 



#63
Jul2311, 11:18 AM

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ehild 



#64
Jul2311, 11:30 AM

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What I'm looking for is actually calculating with it. Usually I convert an angle as soon as I can to a vector, then calculate with it, and possibly convert it back into an angle to present a result. To me it feels a bit like converting inches to metric before calculating, and converting back to present a result. And even here... If you measure an angle, you use a tangent to convert distance to height. But what you actually measure, is a distance on a measuring device that is marked as an angle, but it might as well be marked as a distance at a specific short distance, or as a tangent ratio. Multiply by the ratio in distance and you get the height without ever using an angle or the tangent function. 



#65
Jul2311, 12:14 PM

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hi ehild!and it's certainly shorter for k = 3 (but not so good for large k) 



#66
Jul2311, 12:15 PM

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Please try to use euler's forumla and u will realise that it is made of some even m (2,4,6).
So n=6 



#67
Jul2311, 01:43 PM

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For k is odd we have: [tex]\begin{eqnarray} \sin^kx \sin kx &=& \left(\frac {e^{ix}  e^{ix}} {2i} \right)^k \left(\frac {e^{ikx}  e^{ikx}} {2i} \right) \\ &=& \frac 1 {(2i)^{k+1}} \left(\sum_{m=0}^k \binom k m (1)^m e^{i(k2m)x} \right) (e^{ikx}  e^{ikx}) \\ &=& \frac 1 {(2i)^{k+1}} \sum_{m=0}^k \binom k m (1)^m (e^{i(2k2m)x} + e^{i(2k2m)x}) \\ &=& \frac 2 {(2i)^{k+1}} \sum_{m=0}^k \binom k m (1)^m \cos(2k2m)x \\ &=& (1)^{\frac {k+1} 2} \frac 1 {2^k} \left(\binom k 0 \cos 2kx  \binom k 1 \cos(2k2)x + \binom k 2 \cos(2k4)x  ... + \binom k {k1} \cos 2x  1 \right) \end{eqnarray}[/tex] As you can see, I've also corrected the sign error. 



#68
Jul2311, 03:05 PM

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mmm
let's compare length
sin^{k}xsinkx = sin^{k}x Im((cosx + isinx)^{k}) = Im((cosxsinx + isin^{2}x)^{k}) = (1/2^{k}) Im((±sin2x + i(1  cos2x))^{k}) = (1/2^{k}) Im(i^{k}(1  cos2x ± isin2x)^{k}) (I lost an i^{2} here originally ) = (1/2^{k}) Im(i^{k}(1  e^{2ix})^{k}) = (1/2^{k}) Im(i^{k} ∑ ^{k}C_{m} (1^{m}) e^{2imx}) = (1/2^{k})(1)^{(k1)/2} ∑ ^{k}C_{m} (1^{m}) cos2mx for k oddooh, yes, yours is 2 lines shorter (and for k even, there's a signchange somewhere) though we can then use ^{k}C_{m} = ^{k}C_{km} to go straight to [tex]\frac 2 {(2i)^{k+1}} \sum_{m=0}^k \binom k m (1)^{km} \cos2mx[/tex] 



#69
Jul2311, 03:18 PM

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I just wanted to show this, but you both beat me while I watched Poirot on TV
ehild 



#70
Jul2311, 03:23 PM

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ehild 



#71
Jul2311, 08:08 PM

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Hi!!
I had read the complex numbers chapter from my textbook and i think i now have a basic idea of them. So may i know how ehild has got this relationship: [tex]sinx=\frac{e^{ix}e^{ix}}{2i}[/tex] 


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