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Let there be the following bilinear form: $\int_0^1f(x)g(x)x\,dx$, which acts on the polynomials with degree $\leq2$. I needed to prove it's an inner product and then find an orthonormal basis. I needed to use Gram-Schmidt proccess.

So, when I make the vectors I find to be of length one, what's the inner product I use? Lets say some vector for basis if $h$, then the normal of the vectors is $\sqrt{\int_0^1h(x)h(x)x\,dx}$, or is it the 'standard' inner product $\sqrt{\int_0^1h(x)h(x)\,dx}$? In other words, when a basis is orthnormal, it is orthogonal and of length one with accordance to some specific inner product and not necessarily others?

Thanks in advance!

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1 Answer 1

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The definite integral is your Inner Product. Now, in order to build all polynomials that have degree less or equal than 2 you need a basis like $x^0$, $x$ and $x^2$.

$$B_{x^0}^N=1,\text{ as }\int\limits_0^1~1\cdot1=1$$

$$B_{x^1}=x-\frac{1}{2}~~=~~x-\frac{\langle 1,x\rangle}{\langle 1,1\rangle}\cdot1$$

Now, the norm of $x-\frac{1}{2}$ is $\sqrt{\frac{1}{12}}$, given by the integral you posted.

$$B_{x^1}^N=2\sqrt{3}·x - \sqrt{3}$$

I'll let you guess the vector that spans the $x^2$ part.

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  • $\begingroup$ My inner products is $\int_0^1f(x)g(x)xdx$. So I normalize vectors using that inner product? Notice it's -not- the standard inner products, and that the polynomials are also multiplied by $x$. $\endgroup$
    – Jim
    Feb 1, 2014 at 6:05

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