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I have the Legendre polynomials base $\{p_0 = 1, p_1 = x, p_2 = x^2 - \frac{1}{3}\}$ which is known to be orthogonal on $[-1,1]$ with respect to $\int_{-1}^1f(x)g(x)\text{ }dx$ and I want a base $\{\phi_1, \phi_2,\phi_3\}$, orthogonal on $[a,b]$ with respect to $\int_a^bf(x)g(x)\text{ }dx$, given that $$\langle p_0,p_0\rangle=2, \langle p_1,p_1\rangle=\frac{2}{3}\text{ and } \langle p_2,p_2\rangle=\frac{8}{45}$$ for inner product on $[-1,1]$

I know I could straighly use Gram-Schmidt orthogonalization process, but I want to get $\phi_1, \phi_2, \phi_3$ with the info given above to minimize calculations (as I would do in a test).

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    $\begingroup$ Try a linear change of variables. $\endgroup$ – Angina Seng May 31 '17 at 4:49
  • $\begingroup$ I yet don't know how to use the given info, can you give more information? $\endgroup$ – AnalyticHarmony May 31 '17 at 4:54
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Define $$\phi_1(x) = 1$$ $$ \phi_2(x)=\frac{2(x-a)}{b-a}-1$$ $$ \phi_3(x)=\left(\frac{2(x-a)}{b-a}-1\right)^2-\frac{1}{3}$$

Which we obtained by the substitution $t\to \frac{2(t-a)}{b-a}-1$ which maps $[a,b] \to [-1,1]$

Check that this preserves the inner product.

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  • $\begingroup$ The functions defined this way are not orthogonal on [a,b]. For example: for $[2,3]$ we have $\phi_0 = 1$ and $\phi_1 = x-2$ and $\int_2^3 \phi_0 \phi_1 dx = \int_2^3 x-2 dx = \frac{1}{2}$. $\endgroup$ – AnalyticHarmony May 31 '17 at 6:21
  • $\begingroup$ Yes, but you can normalize.(Find constants $c_0,c_1,c_2$ to divide these functions) $\endgroup$ – Guy May 31 '17 at 6:26
  • $\begingroup$ I can't understand, I want them orthogonal, not with norm 1 $\endgroup$ – AnalyticHarmony May 31 '17 at 6:28
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    $\begingroup$ Sorry, I meant to map to $[-1,1]$. editing $\endgroup$ – Guy May 31 '17 at 6:28
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    $\begingroup$ @FryanBury check now $\endgroup$ – Guy May 31 '17 at 6:31

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