# Getting a base of polynomials on [a,b] from a base of Legendre polynomials (orthogonal on [-1,1])

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).

• Try a linear change of variables. – Angina Seng May 31 '17 at 4:49
• I yet don't know how to use the given info, can you give more information? – AnalyticHarmony May 31 '17 at 4:54

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 $[a,b] \to [-1,1]$
• 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 $\int_2^3 \phi_0 \phi_1 dx = \int_2^3 x-2 dx = \frac{1}{2}$. – AnalyticHarmony May 31 '17 at 6:21
• Yes, but you can normalize.(Find constants $c_0,c_1,c_2$ to divide these functions) – Guy May 31 '17 at 6:26
• Sorry, I meant to map to $[-1,1]$. editing – Guy May 31 '17 at 6:28