# Simplifying AlgebraWatch

#1
Hi,

can you help me simplify

please I'm having a bit of trouble
0
11 years ago
#2
Multiply the bottom with 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)
0
11 years ago
#3
There's a property with

z^2 + z^-2

I'll go find my text book. hang on
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#4
(Original post by nota bene)
Multiply the bottom with 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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11 years ago
#5
(Original post by Goldenratio)
Hi,

can you help me simplify

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?
0
#6
it's from

where I have substituted
I hope that helps
0
11 years ago
#7
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!).
0
#8
Not easy, for sure! 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

so here is what it says it should look like

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

Spoiler:
Show
t-substitution
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11 years ago
#10
Yeah just about to suggest that.
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#11
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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11 years ago
#12
There are three identities

Where

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.
0
#13
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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11 years ago
#14
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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11 years ago
#15
(If you were to do it using you're original substitution, it's supposed to be

at which point you can multiply top and bottom by and the factorise the denominator to get the desired result, I think. )
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#16
(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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11 years ago
#17
(Original post by ukgea)
(If you were to do it using you're original substitution, it's supposed to be

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

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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#18
Thank you guys! that was really helpful!
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11 years ago
#19
(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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