Ben Challenor
Posted on Monday, 12 July, 2004 - 06:19 pm:
I'm doing some complex number proofs for P5 and I have two problems left in the proof. For it to be complete I have to show that:

(1)
LaTeX Image

and:

(2)
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Mathematica tells me that (1) is:

LaTeX Image

and (2):

LaTeX Image

But I don't know how to reach these intermediate steps. Any ideas?

Thanks.
Vicky Neale
Posted on Monday, 12 July, 2004 - 06:57 pm:
I haven't really thought about these yet, but what was the original question? Is it possible that there's a method that doesn't involve proving these?

Vicky
Vicky Neale
Posted on Monday, 12 July, 2004 - 07:18 pm:
Ok, it does actually fall out. Are you happy about using de Moivre's theorem to get an expression for sin(5q) as a polynomial in sin(q)? I've put it in white here, so that you can check:

sin(5q)=16sin5 q- 20sin3 q+5sinq

Now letting q = 2p/5, we get a polynomial in sin(2p/5 (which is what we want) equal to 0 (because sin(5q) is cunningly zero). Hopefully you can see how to solve this (it reduces to a quadratic very easily), and you can get cos(2p/5) from this, and the other angles.

If you want this explaining any more, let us know!

Vicky

Ben Challenor
Posted on Monday, 12 July, 2004 - 10:56 pm:
OK, thanks, I've just worked it through to the end and that makes perfect sense.

Although I'm not sure of the mathematical validity of how I 'solved' the quadratic.

You have:

sin(5q)=16sin5q-20sin3q +5sinq

And setting q = 2p/5 you get:

16sin5(2p/5)-20sin3(2p/5)+5sin(2p/5) = 0

Therefore:

sin(2p/5)[16sin4(2p/5)-20sin2(2p/5)+5] = 0

Now, must you say that either

sin(2p/5)=0

or

16sin4(2p/5)-20sin2(2p/5)+5=0

Or can you simply divide both sides by sin(2p/5) as we 'know' already that sin(2p/5) ¹ 0?

I guess I'm confused between constants and variables.

Vicky Neale
Posted on Tuesday, 13 July, 2004 - 09:11 am:
sin(2 pi/5) is a constant, so why not call it s? Then you have a polynomial that looks a lot less scary. I think it's reasonable to 'know' that s is not 0. Now, as you have said, you have something eminently soluble. You're going to solve a quadratic in s2 , so you know that you need the positive root there (s2 > 0). You will have to decide whether you want the positive or negative square root to actually get s; I'd think (at least for P5) that it's reasonable to look at the graph for sin and deduce that s had better be positive. Does that help?

Vicky
Ben Challenor
Posted on Tuesday, 13 July, 2004 - 05:33 pm:
Those are the decisions that I made, I just wondered if they could be justified. Thanks for your help.