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Given the function $$f (x) = x\bigg(x − {1\over4}\bigg)\bigg(x − {1\over2}\bigg)$$

How can I interpolate $f(x)$ with $p(x) = a_0T_0(x) + a_1T_1(x) + a_2T_2(x) + a_3T_3(x)$ to show that

$$a_0 = -{3\over8},\ \ a_1 = {7\over8},\ \ a_2 = −{3\over8},\ \ a_3 = {1\over4}$$

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It may help if you define what each $T$ is. –  Daryl Nov 13 '12 at 7:05
    
$T_k(x)$ are the Chebyshev polynomials, plus they cancel out when you do the integral –  StealzHelium Nov 13 '12 at 7:18

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First, note that, the Chebyshev polynomials have the property

$$\int_{-1}^1 T_n(x)T_m(x)\,\frac{dx}{\sqrt{1-x^2}}= \begin{cases} 0 &: n\ne m \\ \pi &: n=m=0\\ \pi/2 &: n=m\ne 0 \end{cases}\rightarrow (*)\,.$$

The first four polynomials are $ T_0(x)=1, T_1(x)=x, T_2=2x^2-1, T_3=4 x^3-3x.$

Now, we have

$$ f(x) \sim a_0T_0(x) + a_1T_1(x) + a_2T_2(x) + a_3T_3(x) \rightarrow (1). $$

To find $a_i, i=0..3$, we exploit the above property. We start by finding $a_0$. Multiply both sides of $(1)$ by $T_0$ and integrate from $-1$ to $1$ w.r.t. the weight function $\frac{1}{\sqrt{1-x^2}}$ gives

$$ \int_{-1}^{1} \frac{f(x)T_0(x)}{\sqrt{1-x^2}}dx = a_0\int_{-1}^{1} \frac{{T_0}^2(x)}{\sqrt{1-x^2}}dx + a_1\int_{-1}^{1} \frac{{T_0}(x)T_1(x)}{\sqrt{1-x^2}}dx+\dots. $$

$$ \implies -\frac{3}{8}\pi = a_0 \pi + a_1.0 + a_2.0+a_3.0 \implies a_0=-\frac{3}{8} .$$

Notice that, we used the orthogonal property $(*)$ to evaluate the integrals on the right hand side in the above equation.

To find $a_1$, we multiply both sides of equation $(1)$ by $T_1$ and integrate from $-1$ to $1$ w.r.t. the weight function $\frac{1}{\sqrt{1-x^2}}$

$$ \int_{-1}^{1} \frac{f(x)T_1(x)}{\sqrt{1-x^2}}dx = a_0\int_{-1}^{1} \frac{{T_0}(x)T_1(x)}{\sqrt{1-x^2}}dx + a_1\int_{-1}^{1} \frac{{T_1}(x)T_1(x)}{\sqrt{1-x^2}}dx+\dots. $$

$$ \implies \frac{7}{16}\pi = a_0.0 + a_1\frac{\pi}{2}+a_2.0+a_3.0 \implies a_1 = \frac{7}{8}. $$

You can find the rest of $a_2$ and $a_3$ the same way.

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