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I have a question asking to find an orthonormal basis of $p_2$ with respect to the inner product =2 X integral from 0 to 1 p(x)q(x)dx. What do I do with the 2 in front of the integral? When I solve for v2 do I use the integral from 0 to 1 p(x)q(x)dx without multiplying by 2? Do I only multiply when solving for the norm.

In other words, using the basis $/{1,x,x^2/},$

would my $v_2$ be $x-(1/2) $ or $x-1.$

William Chang
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user123204
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  • If my v2 is x-1, then would my orthonormal v2 be xsqrt3-sqrt3? – user123204 Mar 13 '14 at 00:25
  • They force you to use $\sqrt2$, that's it. Anyway, the critical part is very hard to read, what is that letter X, for example? And, please use $ sign to enclose mathematical formulas, \int_0^1, etc.. – Berci Mar 13 '14 at 00:26
  • sorry X means times – user123204 Mar 13 '14 at 00:34
  • http://math.stackexchange.com/questions/710038/finding-an-orthonormal-basis-for-the-space-p-2-with-respect-to-a-given-inner-p/710062#710062 This is where i posted last. why is TZakrevskiy's vector 2 x-1/2 and not x-1 since shouldn't you be multiplying it by 2 – user123204 Mar 13 '14 at 00:35

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The multiplier $2$ is not effective when you look for orthogonal vectors, it effects only the norm.

For the constant polynomial $1$, we get $\|1\|^2=2\cdot\int_0^1 1\cdot 1=2$, so $\|1\|=\sqrt2$, meaning that $v_1$ should rather be the constant $\displaystyle\frac1{\sqrt2}\,$.

For $v_2$, we have $\ 2\int_0^1 1\cdot x\ =\ 1 $, so $2\int_0^1 1\cdot(x-1/2)=0$.
[But we would get the same without the $2$ in front, because $\int_0^1x$ is the half of $\int_0^11$.]

Then again, we have to norm it and finally set $$v_2:=\frac{x-1/2}{\|x-1/2\|}\,.$$

Berci
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