Mathematical Immaturity

1.17 Exercises

• Let $V$ be the linear space of all real functions $f$ continuous on $[0, +\infty)$ and such that the integral $\int_0^\infty e^{-t} f^2(t) \, dt$ converges. Define $(f, g) = \int_0^\infty e^{-t} f(t) g(t) \, dt,$ and let $y_0, y_1, y_2, \dots,$ be the set obtained by applying the Gram-Schmidt process to $x_0, x_1, x_2, \dots,$ where $x_n(t) = t^n$ for $n \geq 0.$ Prove that $y_0(t) = 1,$ $y_1(t) = t - 1,$ $y_2(t) = t^2 - 4t + 2,$ $y_3(t) = t^3 - 9t^2 + 18t - 6.$
• We recall the Gram-Schmidt process from Section 1.14: \begin{align*} (1.16) \qquad y_1 = x_1, \quad y_{r+1} &= x_{r+1} - \sum_{i=1}^r \frac{(x_{r+1}, y_i)}{(y_i, y_i)}y_i \quad \text{for} \quad r = 1, 2, \dots, k-1 \end{align*} If we begin the index at $k = 0,$ we get: \begin{align*} y_0 &= x_0 \\ y_1 &= x_1 - \frac{(x_1, y_0)}{(y_0, y_0)}y_0 \\ y_2 &= x_2 - \frac{(x_2, y_1)}{(y_1, y_1)}y_1 - \frac{(x_2, y_0)}{(y_0, y_0)}y_0 \\ y_3 &= x_3 - \frac{(x_3, y_2)}{(y_2, y_2)}y_2 - \frac{(x_3, y_1)}{(y_1, y_1)}y_1 - \frac{(x_3, y_0)}{(y_0, y_0)}y_0 \end{align*}

  Referring to Section 1.13, Exercise 11, we recall that if $x_n = t^n$ for $n = 0, 1, 2, \dots,$ we have \begin{align*} (x_m, x_n) &= \int_0^{\infty}e^{-t}t^{m + n}\,dt \\ &= (m + n)! \end{align*} As such, we get the following values for $y_0, y_1, y_2, y_3:$ \begin{align*} y_0 &= x_0 = 1 \quad \blacksquare \end{align*}
  \begin{align*} y_1 &= x_1 - \frac{(x_1, y_0)}{(y_0, y_0)}y_0 \\ &= t - \frac{\int_0^{\infty}te^{-t}\,dt}{\int_0^{\infty}e^{-t}\,dt}(1) \\ &= t - \frac{(1+0)!}{(0 + 0)!}(1) \\ &= t - 1 \quad \blacksquare \end{align*}
  \begin{align*} y_2 &= x_2 - \frac{(x_2, y_1)}{(y_1, y_1)}y_1 - \frac{(x_2, y_0)}{(y_0, y_0)}y_0 \\ &= t^2 - \frac{\int_0^{\infty}e^{-t}t^2(t - 1)\,dt}{\int_0^{\infty}e^{-t}(t-1)^2\,dt}(t-1) - \frac{\int_0^{\infty}e^{-t}t^2\,dt}{\int_0^{\infty}e^{-t}\,dt} \\ &= t^2 - \frac{\int_0^{\infty}e^{-t}(t^3 - t^2)\,dt}{\int_0^{\infty}e^{-t}(t^2- 2t + 1)\,dt}(t-1) - \frac{\int_0^{\infty}e^{-t}t^2\,dt}{\int_0^{\infty}e^{-t}\,dt} \\ &= t^2 - \frac{(3! - 2!)}{2! - 2(1!) + 1!}(t - 1) - \frac{2!}{0!} \\ &= t^2 - 4t + 2 \quad \blacksquare \end{align*}
  \begin{align*} y_3 &= x_3 - \frac{(x_3, y_2)}{(y_2, y_2)}y_2 - \frac{(x_3, y_1)}{(y_1, y_1)}y_1 - \frac{(x_3, y_0)}{(y_0, y_0)}y_0 \\ &= t^3 - \frac{\int_0^{\infty}e^{-t}t^3(t^2 - 4t + 2)\,dt}{\int_0^{\infty}e^{-t}(t^2 - 4t + 2)^2\,dt}(t^2 - 4t + 2) \\ &- \frac{\int_0^{\infty}e^{-t}t^3(t-1)\,dt}{\int_0^{\infty}e^{-t}(t^2- 2t + 1)\,dt}(t-1) \\ &- \frac{\int_0^{\infty}e^{-t}t^3\,dt}{\int_0^{\infty}e^{-t}\,dt} \\ &= t^3 - \frac{5! - 4(4!) + 2(3!)}{4! - 8(3!) + 20(2!) - 12}(t^2 - 4t + 2) \\ &- \frac{4! - 3!}{2! - 1}(t - 1) \\ &- \frac{3!}{1} \\ &= t^3 - 9(t^2 - 4t + 2) - 18(t - 1) - 6 \\ &= t^3 - 9t^2 + 18t - 6 \quad \blacksquare \end{align*}