Scalar product problem

**arbolis** · June 6th 2008, 05:03 PM

I feel sorry to ask help for this problem, since I believe it's a simple one. But I go nowhere when I think alone.
The problem states :
Let $\langle \cdot ,\cdot\rangle$ be one of the following scalar product in the space of polynomials with real coefficients of degree $\leq n$ .
a) $\langle f,g \rangle=\sum_{j=0}^n f(x_j)g(x_j)$ , with $x_i$ different from $x_j$ when $i$ different from $j$ .
b) $\langle f,g \rangle=\int_0^1 f(x)g(x)dx$ .
Prove that { $1,x,x^2,...,x^n$ } is not an orthogonal set if $n\geq 2$ .
--------------------------
First : I don't know how to start. Second : in the item a), I don't understand why there is the condition of the "i", if it isn't even in the equation. Maybe an error of text, so it should be $f(x_i)g(x_j)$ ? Third : About what I must prove. It says, an orthogonal set, wouldn't it be an orthogonal basis? But even having said that, I still don't understand at all what they ask me to do. Please help me!

**TheEmptySet** · June 6th 2008, 09:46 PM

Originally Posted by arbolis

I feel sorry to ask help for this problem, since I believe it's a simple one. But I go nowhere when I think alone.
The problem states :
Let $\langle \cdot ,\cdot\rangle$ be one of the following scalar product in the space of polynomials with real coefficients of degree $\leq n$ .
a) $\langle f,g \rangle=\sum_{j=0}^n f(x_j)g(x_j)$ , with $x_i$ different from $x_j$ when $i$ different from $j$ .
b) $\langle f,g \rangle=\int_0^1 f(x)g(x)dx$ .
Prove that { $1,x,x^2,...,x^n$ } is not an orthogonal set if $n\geq 2$ .
--------------------------
First : I don't know how to start. Second : in the item a), I don't understand why there is the condition of the "i", if it isn't even in the equation. Maybe an error of text, so it should be $f(x_i)g(x_j)$ ? Third : About what I must prove. It says, an orthogonal set, wouldn't it be an orthogonal basis? But even having said that, I still don't understand at all what they ask me to do. Please help me!

Two Vectors are orthogonal if there inner product is zero

For b) I think your inner product should be

$<f,g>=\int_{-1}^{1}f(x)g(x)dx$

Then the vectors $1 \\\ x$ are orthogonal

$<1,x>=\int_{-1}^{1}1\cdot x dx=\frac{1}{2}x^2 \bigg|_{-1}^{1}=\frac{1}{2}(1^2-(-1)^2)=0$

but if you try

$<br /> <1,x^2>=\frac{2}{3} \ne 0$ so they are not orthogonal.

I hope this helps.

**Isomorphism** · June 6th 2008, 09:58 PM

Originally Posted by TheEmptySet

For b) I think your inner product should be

$<f,g>=\int_{-1}^{1}f(x)g(x)dx$
.

Why do you think so?

**Moo** · June 7th 2008, 02:44 AM

Originally Posted by Isomorphism

Why do you think so?

It wouldn't be orthogonal for any n, whereas the question asks for n>2 :/

**arbolis** · June 7th 2008, 05:40 AM

I hope this helps.

Yes it does. But I insist, I think it's $\int_0^1 f(x)g(x)dx$ . If the lower limit of the integral would be $-1$ , then I must show that the product of 2 different functions is an odd function. It wouldn't be always true, but as you shown, if $n=2$ then it doesn't work. So it's not even worth looking for further n.
I think I'm starting to understand the problem. So please answer me to :

it says, an orthogonal set, wouldn't it be an orthogonal basis?

Now for the a). If $j=0$ , I get $f(1)g(1)$ . How can they want me to say it's equal to 0? There's no way to suppose this!

**TheEmptySet** · June 7th 2008, 07:53 AM

The set $\{ 1,x,x^2,...x^n\}$ does not form a orthogonal basis for the innerproduct space $P_n$ with respect to the innerproduct $<f,g>=\int_{-1}^{1}f(x)g(x)dx$

Look for the Orthogonality property here

Legendre polynomials - Wikipedia, the free encyclopedia

and for the Gram-Schmidt process here

Gramâ€“Schmidt process - Wikipedia, the free encyclopedia

If you use the Gram-Schmidt process on the above set you will obtain an orthogonal set with respect to the inner product mentioned it turns out to be the Legendre polynomials.

**arbolis** · June 7th 2008, 09:56 AM

Thanks a lot TheEmptySet. It looks very interesting to me even if quite hard to understand.
And well... I give up for the a). I'll ask help to a friend on Monday, maybe he understood this.

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