Mathematical Immaturity

1.13 Exercises

• In the real linear space $C(-1, 1)$, let $(f, g) = \int_{-1}^1 f(t) g(t) \, dt$. Consider the three functions $u_1, u_2, u_3$ given by $$u_1(t) = 1, \quad u_2(t) = t, \quad u_3(t) = 1 + t.$$ Prove that two of them are orthogonal, two make an angle $\pi/3$ with each other, and two make an angle $\pi/6$ with each other.
• Recall from Section 11:
  $1. \quad$For an element $x$ in a Euclidean space $V$ with inner product $(x, x),$ the nonnegative number $\|x\| = (x, x)^{1/2}$ is called the norm of $x.$
  
  $2.\quad$ In a real Euclidean space $V,$ the angle between two nonzero elements $x$ and $y$ is defined to be that number $\theta$ in the interval $0 \leq \theta \leq \pi$ which satisfies the equation \begin{align*} \cos \theta &= \frac{(x, y)}{\|x\|\|y\|}. \end{align*}
• 
  $\quad 1. \quad u_1$ and $u_2$ are orthogonal.
  $\quad 2. \quad u_1$ and $u_3$ form angle $\pi/6.$
  $\quad 3. \quad u_2$ and $u_3$ form angle $\pi/3.$
  
  Proof.$\quad (u_1, u_2)$ is given by \begin{align*} (u_1, u_2) &= \int_{-1}^{1}u_1(t)u_2(t)\,dt \\ &= \int_{-1}^{1}t\,dt \\ &= \frac{t^2}{2}\Biggr|_{-1}^{1} \\ &= 0. \end{align*} So by definition, $u_1$ and $u_2$ are orthogonal. $\quad \blacksquare$
  
  To find the elements that make an angle of $\pi/3$ and $\pi/6$ with one-another, we first note that as a result of the Cauchy-Schwarz inequality, the angles $\pi/3$ and $\pi/6$ are the only values in the interval $[0, \pi]$ whose cosines are $1/2$ and $\sqrt{3}/2,$ respectively.
  
  To calculate the cosine of the angle between $u_1$ and $u_3,$ we first find the inner product of the two elements: \begin{align*} (u_1, u_3) &= \int_{-1}^{1}1 + t\,dt \\ &= t + \frac{t^2}{2}\,\Biggr|_{-1}^1 \\ &= 2 \end{align*} Then, we calculate the norms $\|u_1\|$ and $\|u_3\|:$ \begin{align*} \|u_1\| &= (u_1, u_1)^{1/2} \\ &=\left(\int_{-1}^{1}\,dt\right)^{1/2} \\ &= \left(t\,|_{-1}^1\right)^{1/2} \\ &= \sqrt{2} \\ \\ \|u_3\| &= (u_3, u_3)^{1/2} \\ &=\left(\int_{-1}^{1}1 + 2t + t^2\,dt\right)^{1/2} \\ &= \left(t + t^2 + \frac{t^3}{3}\,\biggr|_{-1}^1\right)^{1/2} \\ &= \left(2 + \frac{2}{3}\right)^{1/2} \\ &= \frac{2\sqrt{2}}{\sqrt{3}} \end{align*} Combining these values, we get the cosine of the angle $\theta$ between $u_1$ and $u_3:$ \begin{align*} \cos \theta &= \frac{(u_1, u_3)}{\|u_1\|\|u_3\|} \\ &= \frac{2}{4/\sqrt{3}} \\ &= \frac{\sqrt{3}}{2} \end{align*} As noted previously, $\pi/6$ is the only angle in the interval $[0, \pi]$ with cosine $\sqrt{3}/2.$ Thus, we can conclude that $u_1$ and $u_3$ make an angle of $\pi/6. \quad \blacksquare$
  
  To calculate the cosine of the angle between $u_2$ and $u_3,$ we first find the inner product of the two elements: \begin{align*} (u_2, u_3) &= \int_{-1}^{1}t + t^2\,dt \\ &= \frac{t^2}{2} + \frac{t^3}{3}\,\Biggr|_{-1}^1 \\ &= \frac{2}{3} \end{align*} Then, we calculate the norms $\|u_2\|$ and $\|u_3\|:$ \begin{align*} \|u_2\| &= (u_2, u_2)^{1/2} \\ &=\left(\int_{-1}^{1}t^2\,dt\right)^{1/2} \\ &= \left(\frac{t^3}{3}\,\biggr|_{-1}^1\right)^{1/2} \\ &= \frac{\sqrt{2}}{\sqrt{3}} \\ \\ \|u_3\| &= (u_3, u_3)^{1/2} \\ &=\left(\int_{-1}^{1}1 + 2t + t^2\,dt\right)^{1/2} \\ &= \left(t + t^2 + \frac{t^3}{3}\,\biggr|_{-1}^1\right)^{1/2} \\ &= \left(2 + \frac{2}{3}\right)^{1/2} \\ &= \frac{2\sqrt{2}}{\sqrt{3}} \end{align*} Combining these values, we get the cosine of the angle $\theta$ between $u_2$ and $u_3:$ \begin{align*} \cos \theta &= \frac{(u_2, u_3)}{\|u_1\|\|u_3\|} \\ &= \frac{(2/3)}{(4/3)} \\ &= \frac{1}{2} \end{align*} And as noted previously, $\pi/3$ is the only angle in the interval $[0, \pi]$ with cosine $1/2.$ As such, we can conclude that $u_2$ and $u_3$ make an angle of $\pi/3.\quad \blacksquare$
  
  This completes the proof.