An inner product problem

Given two polynomials $f$ and $g$, define their inner product by
$$\langle f,g\rangle =\int _0^{\pi }f(x)g(x)\sin (x)dx$$

1. Prove that for any two polynomials $f$ and $g$, $\langle f,g\rangle =\langle g,f\rangle$.

2. Prove that for any three polynomials $f$, $g$, and $h$ and any two real number $a$ and $b$, $\langle a f+b g,h\rangle =a\langle f,h\rangle +b\langle g,h\rangle$.

3. If I use this inner product to apply the Gramm-Schmitt orthogonalization process to the set $\left\{1,x,x^2\right\}$, I obtain the following three polynomials.
   \begin{eqnarray*}
   p_0(x) & = & \frac{1}{\sqrt{2}} \\
   p_1(x) & = & \sqrt{\frac{2}{\pi ^2-8}} x-\frac{\pi }{\sqrt{2 \left(\pi ^2-8\right)}} \\
   p_2(x) & = & \frac{x^2}{\sqrt{40-4 \pi ^2}}-\frac{\pi x}{\sqrt{40-4 \pi ^2}}+\frac{2}{\sqrt{40-4 \pi ^2}}
   \end{eqnarray*}
   Suppose I now wish to express $f(x)=x(\pi -x)$ as a linear combination of these polynomials. Thus:
   $$f(x) = a_0p_0(x)+a_1p_1(x)+a_2p_2(x).$$
   Write down the integral that you would use to compute $a_2$. You do not have to evaluate the integral!
   
   
   
   
   
   
   
   

Problem 1: Let $f, g$ be polynomials. By definition, the inner product is: $$\langle f,g\rangle = \int _0^{\pi }f(x)g(x)\sin (x)dx$$

From here, the following steps are pretty obvious:
$$\int _0^{\pi }f(x)g(x)\sin (x)dx = \int _0^{\pi }g(x)f(x)\sin (x)dx = \langle g,f\rangle$$
Thus $\langle f,g\rangle= \langle g,f\rangle$.

Problem 2: This proof is somewhat similar to what we did in problem 1. This time, we let $f,g, h$ be polynomials and let $a,b$ be real numbers. Using our definition of the inner product, we get
$$\langle af + bg,h\rangle = \int _0^{\pi }(af(x)+bg(x))h(x)\sin (x)dx$$
$$= \int _0^{\pi }af(x)h(x)\sin (x)+bg(x)h(x)\sin (x)dx$$
$$= \int _0^{\pi }af(x)h(x)\sin (x)dx+ \int _0^{\pi }bg(x)h(x)\sin (x)dx$$
$$= a\int _0^{\pi }f(x)h(x)\sin (x)dx+ b\int _0^{\pi }g(x)h(x)\sin (x)dx$$
$$=a\langle f,h\rangle + b\langle g,h\rangle$$
Thus $\langle af + bg,h\rangle = a\langle f,h\rangle + b\langle g,h\rangle$.

Problem 3: This time around we need to recall a fact that Professor McClure talked about back when we went over the inner product, that $C_n = \langle V,V_n\rangle$, as this is important in finding the integral for this problem. $V_n$ is a basis, and since we're looking for $a_2$, in this case it's $p_2(x)$. In addition, our $C_n$ this time is $a_2$, and in this case, $V = f(x) = x(\pi - x)$.

With all of this mentioned, we can now find our integral representation of $a_2$:
$$a_2 = \langle x(\pi - x), \frac{x^2 - \pi x + 2}{\sqrt{40-4\pi^2}} \rangle = \int_0^\pi x(\pi - x) \frac{x^2 - \pi x + 2}{\sqrt{40-4\pi^2}}\sin(x)dx$$

You lost me on problem 3. Can you explain why $C_n$ is $a_2$, $V_n$ is $p_2(x)$, and V is $f(x)$?

In order to compute the Fourier series coefficients $a_n$ where $g(x) = \sum_n a_n f_n$, we can always find the coefficients by

$$
a_n = \langle f_n, \; g(x) \rangle.
$$

We found this formula by exploiting the fact that the functions $f_n$ we use (in this case polynomial functions) must be orthogonal. Since we want to find $a_2$ for $g(x) = x(\pi-x)$, this means we should compute

$$a_2 = \langle f_2, \; x(\pi-x)\rangle$$

using the definition of the inner product we were provided with. Mark told us we don't need to actually compute this inner product fully, just write out the integral and stop.