Mathematical Immaturity

1.13 Exercises

• In the linear space of all real polynomials, define $(f, g) = \int_0^\infty e^{-t} f(t) g(t) \, dt.$
  $\text{(a)}\quad$ Prove that this improper integral converges absolutely for all polynomials $f$ and $g.$
  $\text{(b)}\quad$ If $x_n(t) = t^n$ for $n = 0, 1, 2, \ldots,$ prove that $(x_n, x_m) = (m + n)!$.
  $\text{(c)}\quad$ Compute $(f, g)$ when $f(t) = (t + 1)^2$ and $g(t) = t^2 + 1.$
  $\text{(d)}\quad$ Find all linear polynomials $g(t) = a + bt$ orthogonal to $f(t) = 1 + t.$
• 
  $\text{(a)}\quad$ For any choice of real polynomials $f(t)$ and $g(t),$ $f(t)g(t)$ is a real polynomial of degree $\leq n$ for some integer $n \geq 0.$ Accordingly, we can rewrite $f(t)g(t)$ as \begin{align*} f(t)g(t) &= \sum_{k=0}^n a_kt^k \end{align*} By extension of the triangle inequality, we have \begin{align*} |f(t)g(t)| &= \left|\sum_{k=0}^n a_kt^k\right| \\ &\leq \sum_{k=0}^n |a_kt^k| \end{align*} And because $e^{-t} \gt 0$ for all real $t,$ we have $|e^{-t}| = e^{-t}.$ Hence, \begin{align*} |e^{-t}f(t)g(t)| &= e^{-t}\left|\sum_{k=0}^n a_kt^k\right| \\ &\leq e^{-t} \sum_{k=0}^n |a_kt^k| \\ \\ &\text{and} \\ \\ \int_0^{\infty}|e^{-t}f(t)g(t)| &\leq \int_0^{\infty} e^{-t} \sum_{k=0}^n |a_kt^k| \end{align*} Integrating by parts with $u = \sum_{k=0}^n |a_kt^k|$ and $dv = e^{-t}\,dt,$ we get \begin{align*} \int_0^{\infty} e^{-t} \sum_{k=0}^n |a_kt^k| &= uv\, \biggr|_0^{\infty} - \int_0^{\infty}v\,du \\ &= \left[-e^{-t} \sum_{k=0}^n |a_kt^k|\right]_0^{\infty} + \int_0^{\infty} e^{-t} \sum_{k=1}^n k\left|a_kt^{k-1}\right| \\ &= |a_0| + \int_0^{\infty} e^{-t} \sum_{k=1}^n k\left|a_kt^{k-1}\right| \end{align*} Performing integration by parts on the rightmost term with $u = \sum_{k=1}^n k\left| a_kt^{k-1}\right|$ and $dv = e^{-t}\,dt,$ we get \begin{align*} \int_0^{\infty} e^{-t} \sum_{k=0}^n |a_kt^k| &= |a_0| + \int_0^{\infty} e^{-t} \sum_{k=1}^n k\left|a_k t^{k-1}\right| \\ &= \left|a_0\right|(0!) + \left|a_1 \right|(1!) + \int_0^{\infty} e^{-t} \sum_{k=2}^n k(k-1)\left|a_kt^{k-2}\right| \end{align*} Continuing this process for $k = 2, \dots, n,$ we find that \begin{align*} \int_0^{\infty} e^{-t} \sum_{k=0}^n |a_kt^k| &= \sum_{k=0}^n \left|a_k\right|k! \end{align*} But this means that the improper integral $\int_0^{\infty}|e^{-t}f(t)g(t)|\,dt$ is bounded above by $\sum_{k=0}^n \left|a_k\right|k!,$ hence the improper integral $\int_0^{\infty}e^{-t}f(t)g(t)\,dt$ converges absolutely for any real polynomials $f(t)$ and $g(t).\quad \blacksquare$
  
  $\text{(b)}\quad$ The inner product $(x_n, x_m)$ is given by: \begin{align*} (x_n, x_m) &= \int_0^{\infty}e^{-t}x_n(t)x_m(t)\,dt \\ &= \int_0^{\infty}e^{-t}t^{m + n}\,dt \end{align*} As in part (a), we can integrate by parts with $u = t^{m + n}$ and $dv = e^{-t}\,dt$ to get: \begin{align*} \int_0^{\infty} e^{-t} t^{m + n}\,dt &= uv\, \biggr|_0^{\infty} - \int_0^{\infty}v\,du \\ &= \left[-e^{-t}t^{m + n}\right]_0^{\infty} + \int_0^{\infty}(m + n)e^{-t}t^{(m + n - 1)}\,dt \\ &= (m + n)\int_0^{\infty}e^{-t}t^{(m + n - 1)}\,dt \end{align*} Integrating by parts once again with $u = t^{m + n - 1}$ and $dv = e^{-t}\,dt,$ we get: \begin{align*} (m + n)\int_0^{\infty}e^{-t}t^{(m + n - 1)}\,dt &=(m + n) \left[uv\, \biggr|_0^{\infty} - \int_0^{\infty}v\,du\right] \\ &= (m + n)\left[-e^{-t}t^{m + n - 1}\right]_0^{\infty} + (m + n)(m + n - 1)\int_0^{\infty}e^{-t}t^{(m + n - 2)}\,dt \\ &= (m + n)(m + n - 1)\int_0^{\infty}e^{-t}t^{(m + n - 2)}\,dt \end{align*} If we repeat this process an additional $(m + n - 2)$ times, we reach the following integral for $(x_n, x_m):$ \begin{align*} (x_n, x_m) &= (m + n)!\int_0^{\infty}e^{-t}\,dt \\ &= (m + n)!\left[-e^{-t}\right]_{0}^{\infty} \\ &= (m + n)!(0 + 1) \\ &= (m + n)! \quad \blacksquare \end{align*}
  
  $\text{(c)}\quad$ When $f(t) = (1 + t)^2 = 1 + 2t + t^2$ and $g(t) = 1 + t^2,$ we have $f(t)g(t) = 1 + 2t + 2t^2 + 2t^3 + t^4.$ And because the coefficients of $f(t)g(t)$ are all positive, we can use the solution from part (a) directly to get: \begin{align*} (f, g) &= \int_0^{\infty}e^{-t}f(t)g(t) \\ &= \sum_{k=0}^n a_k k! \\ &= 1(0!) + 2(1!) + 2(2!) + 2(3!) + 1(4!) \\ &= 43 \quad \blacksquare \end{align*}
  
  $\text{(d)}\quad$ As we have shown in parts (a) through (c), the inner product $(f, g)$ given by \begin{align*} (f, g) &= \int_0^{\infty} e^{-t}f(t)g(t)\,dt \end{align*} can be written as the following sum: \begin{align*} (f, g) &= \sum_{k=0}^{n}a_k k! \end{align*} If $f(t) = 1 + t$ and $g(t) = a + bt,$ we have the product $f(t)g(t) = a + (a+b)t + bt^2.$ For $g$ to be orthogonal to $f,$ their inner product must be zero. In other words, we wish to find real scalars $a$ and $b$ that satisfy: \begin{align*} a(0!) + (a + b)(1!) + b(2!) &= 2a + 3b \\ &= 0. \end{align*} This relation is satisfied when $b = -2a/3$ for any real $a$ As such, for any real $a,$ $$g(t) = \left(1 - \frac{2}{3}t\right)a$$ is orthogonal to $f(t) = 1 + t.\quad\blacksquare$