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## #1 2008-12-05 07:38:25

Daniel123
Member
Registered: 2007-05-23
Posts: 663

### Chebyshev polynomials

.

I've done the first part using induction. The polynomials I chose were:

However, the last part is a bit tricky.

I can't see how to write this in the form

though. I know that
and

Can anyone help? (Small hints preferred) Thanks.

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## #2 2008-12-06 06:09:40

Daniel123
Member
Registered: 2007-05-23
Posts: 663

No one?

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## #3 2008-12-06 07:14:05

Kurre
Member
Registered: 2006-07-18
Posts: 280

### Re: Chebyshev polynomials

Cos2x+Cos3x=0
Cos2x=-Cos3x
Cos2x=Cos(pi-3x)
1)2x=3x-pi->x=pi->cosx=-1
2)2x=pi-3x->x=pi/5
looks like something is wrong with your equation...

Last edited by Kurre (2008-12-06 07:14:54)

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## #4 2008-12-06 08:17:51

Daniel123
Member
Registered: 2007-05-23
Posts: 663

### Re: Chebyshev polynomials

Kurre wrote:

Cos2x+Cos3x=0
Cos2x=-Cos3x
Cos2x=Cos(pi-3x)
1)2x=3x-pi->x=pi->cosx=-1
2)2x=pi-3x->x=pi/5
looks like something is wrong with your equation...

You're right.

I'm actually rather stupid.

The factor should be (x+1), in which case x = -1 does work, as we can let x = cosθ (which we have to do anyway). Also,

Thanks

The point of the question was obviously to work in the trigonometric form, as the value of phi s.t

isn't obvious.

Last edited by Daniel123 (2008-12-06 08:20:13)

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