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I am a little stuck on the following problem:

By using the Gram-Schmidt Orthogonalization, find an orthonormal basis for the subspace of $L^2[0,1]$ spanned by $1,x, x^2, x^3$.

OK, so I have defined:

$$e_1 = 1$$

I would then assume that we proceed as follows:

$$e_2 = x - \frac{\langle x,1 \rangle}{\langle 1,1 \rangle} \cdot 1$$

And we have:

$$\langle x,1 \rangle = \int_{0}^{1}x dx = \frac{1}{2}$$

And:

$$\langle 1,1 \rangle = 1$$

So:

$$e_2 = x - \frac{1}{2}$$

But already at this point, my answer is wrong. According to my book, the correct answer here should be:

$$e_2 = \sqrt{12}(x - \frac{1}{2})$$

And this obviously makes all my subsequent answers also incorrect.

So what is it I'm doing wrong here? I simply don't see where the $\sqrt{12}$ term comes from. Any help would be greatly appreciated!

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    Hint: What is $\Vert x - \frac{1}{2} \Vert$?2012-09-24
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    It is required so that $\Vert e_2 \Vert = 1$2012-09-24
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    Ah, I think I get it. We have: $\| x - \frac{1}{2} \| = \sqrt{\int_{0}^{1} (x - \frac{1}{2})^2 dx} = \frac{1}{\sqrt{12}}$. And from there the answer follows. Isn't this correct?2012-09-24
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    Yes it does, @Kristian. The Gram-Schmidt process requires to normalize every vector we get *after* you've done what you did to get $\,e_2\,$...that's all you were missing.2012-09-24
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    Great! Thanks a lot for the help, everyone!2012-09-24
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    Can I please just ask a brief follow-up question (instead of starting a new thread). I see that on most websites, the norm in the $L^2$-space is defined as $\|f(x)\| = \int_{a}^{b}f(x)^{2} dx$. However, in order to get the desired answer above, I have to take the square root of this. So wouldn't my procedure above be incorrect?2012-09-24

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