Chapter 10.3, Problem 37E

To determine

(a)

To prove:

The integrals ${\displaystyle \underset{-\pi }{\overset{\pi }{\int }}{F}_N{}^2\left(x\right)dx}=\pi \left(\frac{{\alpha }_0{}^2}{2}+{\alpha }_1{}^2+g+{\alpha }_N{}^2+{\beta }_1{}^2+g+{\beta }_N{}^2\right)$ and ${\displaystyle \underset{-\pi }{\overset{\pi }{\int }}f\left(x\right)F_N\left(x\right)dx}=\pi \left(\frac{{\alpha }_0{a}_0}{2}+{\alpha }_1{a}_1+\dots +{\alpha }_N{a}_N+{\beta }_1{b}_1+g+{\beta }_N{b}_N\right)$.

To determine

(b)

To prove:

The integral $E*={\displaystyle \underset{-\pi }{\overset{\pi }{\int }}{f}^2\left(x\right)dx}-\pi \left(\frac{a_0{}^2}{2}+a_1{}^2+\dots +a_N{}^2+b_1{}^2+\dots +b_N{}^2\right)$.

To determine

(c)

To prove:

The equation $E-E*=\pi \left\{\frac{{\left({\alpha }_0-a_0\right)}^2}{2}+{\left({\alpha }_1-a_1\right)}^2+\dots +{\left({\alpha }_N-a_N\right)}^2+{\left({\beta }_1-b_1\right)}^2+\dots +{\left({\beta }_N-b_N\right)}^2\right\}$.