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Let $V = \mathcal{C}[-1, 1]$ be the real vector space of real-valued continuous functions defined on the closed interval $[-1, 1]$. $V$ is an inner product space with the inner product $\langle f, g\rangle = \int_{-1}^{1} f(x)g(x) dx$.

Find the least square approximation to $p = x^{1/3}$ in $W = \textrm{span} \left\{q_o = 1, q_1=x, q_2 = x^2 - \frac{1}{3} \right \}$.

I thought about using $\textrm{proj}_W p$, but I ran into the trouble of evaluating $\int_{-1}^{1} x^{1/3} dx$.

Any ideas?

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    What trouble? An antiderivative is $(3/4)x^{4/3}$.2012-03-20
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    The domain of $x^{1/3}$ is x > 0 only.2012-03-20
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    It is a matter of definition. Certainly $(-1/2)^{\pi}$ makes no sense. But often $x^{p/q}$ is thought of as defined even for negative $x$, if $p$ and $q$ are integers and $q$ is odd.2012-03-20
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    http://www.wolframalpha.com/input/?i=Integrate[x^%281%2F3%29%2C[x%2C-1%2C1}]2012-03-20
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    I would not pay much attention to what Wolfram Alpha says. When it is good, it is very very good, and the price is right. However, $\dots$.2012-03-20
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    I think you are implying that the odd integer with the symmetric bounds will imply it is 0, but it doesn't seem that way to me because (-1)^r where r is rational is not defined.2012-03-20
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    Well, as I wrote, it is a matter of definition. But the question, as partly quoted, seems to treat $x^{1/3}$ as a function continuous on $[-1,1]$. So the problem setter's definition seems in this case to coincide with mine.2012-03-20
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    So I am not embarrassed to ask, how do I get a real number solution then?2012-03-21
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    No problem, the integral you were worried about is $0$.2012-03-21

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