Simplifying Algebra Watch

Goldenratio
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#1
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#1
Hi,

can you help me simplify

\displaystyle\frac{\frac{1}{2}(z  ^3+z^{-3})}{(5-\frac{1}{2}(z+z^{-1})*iz)}

please I'm having a bit of trouble
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nota bene
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#2
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Multiply the bottom with 5+\frac{1}{2}(z+z^{-1}\times iz to rationalise the denominator?

Haven't done it, but looks like a way to go (if z isn't complex itself, which it probably is)
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sugar-pie
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There's a property with

z^2 + z^-2

I'll go find my text book. hang on
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Goldenratio
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(Original post by nota bene)
Multiply the bottom with 5+\frac{1}{2}(z+z^{-1}\times iz to rationalise the denominator?

Haven't done it, but looks like a way to go (if z isn't complex itself, which it probably is)
You think? lets have a go
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DFranklin
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(Original post by Goldenratio)
Hi,

can you help me simplify

\displaystyle\frac{\frac{1}{2}(z  ^3+z^{-3})}{(5-\frac{1}{2}(z+z^{-1})*iz)}

please I'm having a bit of trouble
Without some more context, it's hard to get any further I think - in other words, I'm not sure you can get simpler than the expression you have. Where has this expression come from? What exactly is z?
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Goldenratio
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it's from \int_0 ^{2\pi} \displaystyle\frac{cos3\theta d\theta}{5-4cos\theta}

where I have substituted z=e^{i\theta}
I hope that helps
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DFranklin
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OK, good to know, because my temptation was going to be to try to simplify towards your original expression!

What makes you think this substitution is the right way of solving the integral?

(It might be, but it doesn't look easy, that's for sure!).
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Goldenratio
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#8
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Not easy, for sure!:p: To be honest it's the only substitution i've seen so far. I cant really think what else would be good. any ideas?

However, I do have the finished product up my sleeve, but it doesn't help me know how it was simplified. I've got about two sheets of paper full of workings and they all end up wrong (according to this result). suffice to say it asks me to prove the integral is equal to \frac{\pi}{12}

so here is what it says it should look like

\displaystyle\frac{-1}{2i}\displaystyle\oint_C\displ  aystyle\frac{z^6+1}{z^3(2z-1)(z-2)}dz

where c is the contour. this is not what it looks like for me
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Dystopia
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#9
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I did it using a far simpler method.

Spoiler:
Show
t-substitution
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insparato
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#10
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Yeah just about to suggest that.
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Goldenratio
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#11
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t-substitution seems interesting. can someone post their approach please?

ps i sorted the simplification. but this other method looks nice!
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insparato
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#12
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There are three identities

 cosx = \frac{1-t^2}{1+t^2}

 sinx = \frac{2t}{1+t^2}

 tanx = \frac{2t}{1-t^2}

Where  t = tan\frac{x}{2}

So let t = tan x/2, and follow through on the integration by substitution, making sure you substitute the trig functions for the identities above.
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Goldenratio
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#13
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Thank you! looks very useful! a silly question though.

if i have cos3x how does this relate to the identities? I started to use cos(2x+x) and simplifying and substituting... it looked horrid! :P
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insparato
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I just scribbled down some working, i think by my working it will come out but its abit of algebraic manipulation starting from cos3x = 4cos^3x - 3cosx.
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ukgea
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#15
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(If you were to do it using you're original substitution, it's supposed to be

\displaystyle \frac{\cos\theta d\theta}{5-4\cos\theta} = \frac{\frac{1}{2}\left(z^3+z^{-3}\right)}{5-2\left(z+z^{-1}\right)}\frac{dz}{iz}

at which point you can multiply top and bottom by z^3 and the factorise the denominator to get the desired result, I think. )
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Goldenratio
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#16
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(Original post by insparato)
I just scribbled down some working, i think by my working it will come out but its abit of algebraic manipulation starting from cos3x = 4cos^3x - 3cosx.
thank you! I always forget !
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DFranklin
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#17
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(Original post by ukgea)
(If you were to do it using you're original substitution, it's supposed to be

\displaystyle \frac{\cos\theta d\theta}{5-4\cos\theta} = \frac{\frac{1}{2}\left(z^3+z^{-3}\right)}{5-2\left(z+z^{-1}\right)}\frac{dz}{iz}

at which point you can multiply top and bottom by z^3 and the factorise the denominator to get the desired result, I think. )
I think so:

\displaystyle \frac{\frac{1}{2}\left(z^3+z^{-3}\right)}{5-2\left(z+z^{-1}\right)}\frac{dz}{iz} = \frac{z^6+1}{2iz^4(5-2(z+z^{-1}))}

\displaystyle = \frac{z^6+1}{2iz^3(5z-2z^2-2)} = \frac{-1}{2i} \frac{z^6+1}{z^3(2z^2-5z+2)} = \frac{-1}{2i} \frac{z^6+1}{z^3(z-2)(2z-1)}

But I'm not sure I'd call that particularly "simpler", although it's going to be easy to find the residues in that form...
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Goldenratio
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#18
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Thank you guys! that was really helpful!
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Dystopia
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#19
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(Original post by Goldenratio)
Thank you! looks very useful! a silly question though.

if i have cos3x how does this relate to the identities? I started to use cos(2x+x) and simplifying and substituting... it looked horrid! :P
I'm not sure if you decided to do this, but I think it's easier if you use polynomial division before using the substitution. You end up with a quadratic in cos x and a fraction containing only one cos x term which is made trivial by the substitution.
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