Inner product space - minimization.

The question is : If the vector space C[1,1] of continuous real valued functions on the interval [1,1] is equipped with the inner product defined by (f,g)=1−1 intf(x)g(x)dx

Find the linear polynomial g(t) nearest to f(t) = e^t?

So I understand the solution will be given by (u1,e^t).u1/||u1|| + (u2,e^t).u2/||U2||

But I am having trouble understanding what u1 and u2 should be. I understand they must be othorgonal and basis for a subspace S ∈ C[-1,1].

However Im not too sure what dimension this basis should be of, and not 100% sure what is meant by the vector space C[-1,1].

(The solution uses 1 and t as u1 and u2....)

Many thanks in advance for any assistance.

You clearly haven't organised your thoughts very well. I will try to summarise for you.

- The vector space C[-1,1] is defined in the first line of your post. It is the space of continuous real valued functions on the interval [-1,1]. An element in this space is therefore a continuous function f: [-1,1] --> R and scalar multiplication is given by multiplication on the image of your function i.e.

$\lambda \cdot f$ is the function given by $x \mapsto \lambda f(x)$.

- The question is asking you to find the linear polynomial that is nearest to e^t. By linear polynomial, I am assuming they mean a function of the form g(t) = at+b.

- The distance or metric in an inner product space is the one induced by the norm induced by the inner product unless otherwise stated. In more detail, in this instance, the distance between two functions f and g is given by
$\sqrt{(f-g,f-g)} = \sqrt{\int_{[-1,1]}(f-g)(t)(f-g)(t)\mathrm{d}t}$.

- You therefore need to substitute f(t)=e^t and g(t)=at+b and simply find the values of a and b which minimise the resulting distance.

Why a function of the form g(t)=at + b? How do we know the polynomial is of degree 1-dimension 2 ? Is this from f:[-1,1]---> R? Im also still not sure what this means? What is the interval representing?

How do you then go about solving for a and b ?

Thanks alot.

Why a function of the form g(t)=at + b? How do we know the polynomial is of degree 1-dimension 2 ? Is this from f:[-1,1]---> R? Im also still not sure what this means? What is the interval representing?

How do you then go about solving for a and b ?

Thanks alot.

The question is : If the vector space C[1,1] of continuous real valued functions on the interval [1,1] is equipped with the inner product defined by (f,g)=1−1 intf(x)g(x)dx

Find the linear polynomial g(t) nearest to f(t) = e^t?

So I understand the solution will be given by (u1,e^t).u1/||u1|| + (u2,e^t).u2/||U2||

But I am having trouble understanding what u1 and u2 should be. I understand they must be othorgonal and basis for a subspace S ∈ C[-1,1].

However Im not too sure what dimension this basis should be of, and not 100% sure what is meant by the vector space C[-1,1].

(The solution uses 1 and t as u1 and u2....)

Many thanks in advance for any assistance.

As Davos says, you are told that the function is linear.

Yes, you are looking for a function g:[-1,1] ---> R since this is the space you are working in.

A function f:[-1,1] ---> R is something that associates to each real number from -1 to 1 a real number. The interval just represents all numbers between (and including) those two values... there isn't much more to say.

Have you actually calculate the distance between e^t and an arbitrary linear function yet? Actually calculate it and then worry about how you can find the coefficients making the distance as small as possible. If you do the calculation - you will probably see how. In any case, come back and asked when you tried yourself if you get stuck - I am not completely answering your homework for you.

1)Okay so using pythagorous to calculate the distance
I have:
||(f-g),(f-g)||=(||f^2||-||g^2||)^1/2
Evaluating ||f||^2 and ||g||^2 independently and using the RHS gives the distance as
((1/2(e^2-e^-2)-2a(^2)/3-2b^2))^1/2
- So I assume to minimise this a^(2)/3=-b^2, but this looks different than the solution

2) So the interval specified has no influence on the answer?

You have calculated the distance wrong and misapplied pythagoras which you don't even need here because it is just as simple to do it directly (as suggested already) just try it.

What do you mean the interval has no influence? The interval is used in the definition of the space. I think your problem is that you don't understand what a function is or what an integral is. A function consists of a domain and a codomain and a an association of one element of the latter to each of the former. The inner product space that this question is about is defined to be the set of all continuous functions with a specific domain i.e. [-1,1] and codomain R.

The space you are considering is the space of functions on that interval. When you work out the distance, you integrate over that interval. Look, if you are really interested, rephrase the question by replacing the interval [-1,1] with some other interval and check the answer for yourself.