Mathematical Immaturity

1.17 Exercises

• In the linear space $V$ of Exercise 5, let $f(x) = e^{-x}$ and find the linear polynomial that is nearest to $f.$
• Recall from Exercise 5 that if $V$ is the Euclidean space of all real functions $f$ continuous on $[0, +\infty)$ such that the integral $\int_0^{\infty}e^{-t}f^2(t)\,dt$ converges, with inner product $(f, g) = \int_0^{\infty}e^{-t}f(t)g(t)\,dt,$ then, if $S$ is the subspace of $V$ containing all linear polynomials, $\{1, t - 1\}$ is an orthogonal set that spans $S.$

  $\quad$ Dividing each element by its norm, we construct an orthonormal basis for $S:$ \begin{align*} e_1 &= \frac{1}{\left[\int_0^{\infty}e^{-t}\,dt\right]^{1/2}} \end{align*} Referring to Section 1.13, Exercise 11, we know that $\int_0^{\infty}e^{-t}t^{m + n}\,dt = (m + n)!,$ which makes $\int_0^{\infty}e^{-t}\,dt = 1$ and $e_1 = 1.$ Taking the inner product $(t - 1, t - 1),$ we get: \begin{align*} \int_0^{\infty}e^{-t}(t - 1)^2\,dt &= \int_0^{\infty}t^2e^{-t}\,dt - 2\int_0^{\infty}te^{-t}\,dt + \int_0^{\infty}e^{-t}\,dt \\ &= 2! - 2(1!) + 0! \\ &= 1 \end{align*} Making $e_2 = (t - 1)/1 = t - 1.$ Then, from Section 15, we know that the element $g$ of $S$ nearest to $f(t) = e^{-t}$ is the projection of $f$ on $S:$ \begin{align*} g &= (f, e_1)e_1 + (f, e_2)e_2 \\ &= \int_0^{\infty}e^{-2t}\,dt + (t - 1)\int_0^{\infty}e^{-2t}(t - 1)\,dt \\ &= \frac{1}{2} + (t - 1)\left[\frac{1}{2}\int_0^{\infty} e^{-2t}\,dt-\frac{1}{2}e^{-2t}(t - 1)\biggr|_0^{\infty}\right] \\ &= \frac{1}{2} - \frac{1}{4}(t - 1) \\ &= \frac{3}{4} - \frac{1}{4}t \quad \blacksquare \end{align*}