Mathematical Immaturity

1.17 Exercises

• In the real linear space $C(0, 2\pi)$ with inner product $(f, g) = \int_0^{2\pi} f(x) g(x) \, dx,$ let $f(x) = x.$ In the subspace spanned by $u_1(x) = 1,$ $u_2(x) = \cos x,$ $u_3(x) = \sin x,$ find the trigonometric polynomial nearest to $f.$
• Let $S$ be the subspace of $C(0, 2\pi)$ spanned by $\{u_1, u_2, u_3\}.$ Then, we can use the Gram-Schmidt process to construct an orthogonal basis for $S:$ \begin{align*} y_1 &= u_1 = 1 \\ y_2 &= u_2 - \frac{(u_2, y_1)}{(y_1, y_1)}y_1 \\ &= \cos x - \frac{\int_0^{2\pi}\cos x\,dx}{\int_0^{2\pi}1\,dx} \\ &= \cos x \\ y_3 &= u_3 - \frac{(u_3, y_2)}{(y_2, y_2)}y_2 - \frac{(u_3, y_1)}{(y_1, y_1)}y_1 \\ &= \sin x - \frac{\int_0^{2\pi}\sin x \cos x\,dx}{\int_0^{2\pi}\cos^2 x\,dx}\cos x - \frac{\int_0^{2\pi}\sin x\,dx}{\int_0^{2\pi}1\,dx} \\ &= \sin x - \frac{\frac{1}{2}\int_0^{2\pi}\sin 2x \,dx}{\int_0^{2\pi}\cos^2 x\,dx}\cos x \\ &= \sin x \end{align*} As we can see, the set $\{u_1, u_2, u_3\}$ is an orthogonal basis for $S.$ To construct an orthonormal basis, we divide each element by its norm: \begin{align*} e_1 &= \frac{1}{[\int_0^{2\pi}1\,dx]^{1/2}} \\ &=\frac{1}{\sqrt{2\pi}} \\ e_2 &= \frac{\cos x}{\left[\int_0^{2\pi}\cos^2 x\,dx\right]^{1/2}} \\ e_3 &= \frac{\sin x}{\left[\int_0^{2\pi}\sin^2 x\,dx\right]^{1/2}} \end{align*} Using integration by parts, with $u = \cos x$ and $dv = \cos x \,dx$ we have: \begin{align*} \int_0^{2\pi}\cos^2 x\,dx &= \left[\sin x \cos x \right]_0^{2\pi} + \int_0^{2\pi}\sin^2 x\, dx \\ &= \int_0^{2\pi}\left[1 - \cos^2 x\right]\, dx \end{align*} Rearranging variables, we can see that $\int_0^{2\pi}\cos^2 x\,dx = \frac{1}{2}\int_0^{2\pi}1\,dx = \pi,$ giving us: \begin{align*} \\ e_2 &= \frac{\cos x}{\sqrt{\pi}} \end{align*} To normalize $y_3 = \sin x,$ we perform another integration by parts on $\int_0^{2\pi}\sin^2 x\,dx$ with $u = \sin x$ and $dv = \sin x\,dx:$ \begin{align*} \int_0^{2\pi}\sin^2 x\,dx &= \left[-\sin x \cos x \right]_0^{2\pi} + \int_0^{2\pi}\cos^2 x\,dx \end{align*} But from our earlier calculations, we know that the right-hand side is equal to $\pi,$ giving us: \begin{align*} e_3 &= \frac{\sin x}{\sqrt{\pi}} \end{align*} Now, to find the trigonometric polynomial nearest to $f(x) = x,$ we find the projection of $f$ on the subspace $S:$ \begin{align*} g &= (f, e_1)e_1 + (f, e_2)e_2 + (f, e_3)e_3 \\ &= \frac{1}{2\pi}\int_0^{2\pi}x\,dx \\ &+ \frac{\cos x}{\pi}\int_0^{2\pi}x\cos x\,dx \\ &+ \frac{\sin x}{\pi}\int_0^{2\pi}x\sin x\,dx \end{align*} The first integral can be evaluated directly to give $\frac{1}{2\pi}\int_0^{2\pi}x\,dx = \frac{4\pi^2}{4\pi} = \pi.$ The remaining two integrals we will evaluate using integration by parts.

  For $\int_0^{2\pi}x\cos x\,dx,$ let $u = x$ and $dv = \cos x,dx.$ We then have: \begin{align*} \int_0^{2\pi}x\cos x\,dx &= \left[x \sin x \right]_0^{2\pi} - \int_0^{2\pi} \sin x\,dx \\ &= 0 + \cos x \biggr|_0^{2\pi} \\ &= 0 \end{align*} For $\int_0^{2\pi}x\sin x\,dx,$ let $u = x$ and $dv = \sin x,dx.$ We then have: \begin{align*} \int_0^{2\pi}x\sin x\,dx &= -x \cos x \biggr|_0^{2\pi} + \int_0^{2\pi} \cos x\,dx \\ &= -2\pi + \sin x \biggr|_0^{2\pi} \\ &= -2\pi \end{align*} Putting these values back into our expression for $g(x),$ we have: \begin{align*} g(x) &= (f, e_1)e_1 + (f, e_2)e_2 + (f, e_3)e_3 \\ &= \frac{1}{2\pi}\int_0^{2\pi}x\,dx \\ &+ \frac{\cos x}{\pi}\int_0^{2\pi}x\cos x\,dx \\ &+ \frac{\sin x}{\pi}\int_0^{2\pi}x\sin x\,dx \\ &= \pi - 2\sin x \quad \blacksquare \end{align*}