Need to find a base to a certain space

[Tomer]

Homework Statement

I'm probably incredibly rusty, but I can't seem to solve this.

I've been given the space V = C[-[itex]\pi,\pi[/itex]] (continous functions on the closed segment) with the next linear transformation:

T(f(x)) = g(x) = [itex]\int_{-\pi}^{\pi}[1+cos(x-t)]f(t)dt[/itex]

I ought to prove that T(V), the range of V, has a finite dimension, and find an appropriate basis.

Homework Equations

Can't really think of any.

The Attempt at a Solution

I'm really simply blocked. How do I start getting the formation of functions in this space? Intuitively it seems like the addition of a certain number with a cos function, but I can't see any way to directly prove it, nor do I really have an obvious basis.

I'm sorry if it's dumb :-)

Thanks a lot,
Tomer.

 

[tiny-tim]

Hi Tomer!

Have you tried starting with a basis for V ?

 

[Tomer]

It looks like V's dimension is infinite... I recall that functions on this segment can be described as infinite sums of trigonometric functions...
:-)

 

[tiny-tim]

ok, start there

 

[Tomer]

All I'm supposed to know up to now, if I'm not mistaken, is that the projection of a function f(x) on a subspace of V spanned by {1,sinx,sin2x,...sinnx,cosx,cos2x,...cosnx}, with the dimension of 2n+1, is the best approximation to f(x) in this subspace... (the "distance" between the two is the minimal)
When I said "I recall", I meant that I remember it from previous studies (Fourier series). But this is not what I'm rehearsing right now - I'm rehearsing some linear algebra, and I therefore I don't think I'm meant to show f(x) as an infinite series or something like that.

Any chance I'll get a thicker hint? :-)

And thanks for replying!

 

[tiny-tim]

All I'm supposed to know up to now, if I'm not mistaken, is that the projection of a function f(x) on a subspace of V spanned by {1,sinx,sin2x,...sinnx,cosx,cos2x,...cosnx}, with the dimension of 2n+1, is the best approximation to f(x) in this subspace... (the "distance" between the two is the minimal)

That's correct …

{sin(nx)} and {cosnx} and {1} are a basis for V (they span V and they're linearly independent) …

so find T of each of them …

my guess is that you'll get only one or two independent results. ​

 

[Tomer]

Ok, I know what you mean, but the thing is, the actual fact that I can write: f(x) = (sum of trigonometric functions) is based on material which I'm pretty sure I'm not supposed to use here.
It is mentioned nowhere in the book that any continuous function in a closed segment can be represented as this sum. They only talk about approximations to finite subspaces, and mention that in the finite subspace spanned by {1,sinx,sin2x...sinnx,cosx,cos2x...cosnx} (finite!) the projection of f(x) on it would be the best approximation to f(x)...

I don't see how I can take it formally onwards. If you tell me I have to use the fact that any such function can be represented as an infinite sum, then I can solve it of course in no time.

Thanks and sorry for bugging :-)

 

[tiny-tim]

Hi Tomer!

(sorry for not replying earlier )

Try it either for cos(nt) and sin(nt), or just for tⁿ, to see what results you get …

you should find that the dimension (ie the number of independent solutions) is very low, and once you realize what the solutions are, you should be able to see a quick and simple way of proving it.

What solutions do you get? ​

 

[Tomer]

The solutions I get are {1,sinx,cosx}. But I can't seem to be able to prove it without expanding f(x) as this infinite series which, as I've mentioned, I don't think I can. But maybe I'm wrong and I can? :)

 

[tiny-tim]

The solutions I get are {1,sinx,cosx}.

That's what I got too.

ok, ignore the {1} for the moment …

what familiar equation has the general solution C₁cosx + C₂sinx ? ​

 

[Tomer]

f''(x) + f(x) = 0 ? :-)

 

[tiny-tim]

(just got up :zzz: …)

Yup!

T(f(x)) = g(x) = [itex]\int_{-\pi}^{\pi}[1+cos(x-t)]f(t)dt[/itex]

ok, so what is g''(x) ?

(for a given f)

 

[Tomer]

I'd never manage to see it myself :-)
so for a given f, g(x) must be a solution of the equation: g''(x) - g(x) = C. (C is the integral of f from -pi to pi, and is a constant).
Therefore g has the form: g(x) = C1sinx + C2cosx + C3.
Which of course holds as a proof... correct? :-)

Thanks a lot!

 

[tiny-tim]

Therefore g has the form: g(x) = C1sinx + C2cosx + C3.
Which of course holds as a proof... correct? :-)

Completely!

The moral of all this is:

in future similar questions, try differentiating wrt to x to get a simple equation

and more generally always try a few simple cases to see if you can see a pattern! ​

 

[Tomer]

Thanks a lot :-) I'll remember that!