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This problem appears in the book:

Linear Algebra and its applications - David C. Lay - Fourth Edition

It appears in: Chapter 4 (Vector Spaces), Section 4.7 (Change of Basis), Exercise 18

$(4.7), \mathbf{[18]:}$

(Calculus Required) Recall from calculus that integrals such as

$$\tag{10}\int\left(5\cos^3 t-6\cos^4 t+5\cos^5 t-12\cos^6 t\right)\,\mathrm{d}t$$ are tedious to compute. (The usual method is to apply integration by parts repeatedly and use the half-angle formula.) Use the matrix $P$ or $P^{-1}$ from Exercise 17 to transform $(10)$; then compute the integral.

$(4.7), \mathbf{[17]:}$

Let $\mathcal{B}=\left\{\mathbf{x}_0,...,\mathbf{x}_6\right\}$ and $\mathcal{C}=\left\{\mathbf{y}_0,...,\mathbf{y}_6\right\}$, where $\mathbf{x}_k$ is the function $\cos^k t$ and $\mathbf{y}_k$ is the function $\cos kt$. Exercise 34 in Section 4.5 showed that both $\mathcal{B}$ and $\mathcal{C}$ are bases for the vector space $H=\text{Span}\left\{\mathbf{x}_0,...,\mathbf{x}_6\right\}$.

a. Set $P=\begin{bmatrix} {[\mathbf{y}_0]}_\mathcal{B} \quad \dots \quad {[\mathbf{y}_6]}_\mathcal{B}\end{bmatrix}$, and calculate $P^{-1}$.

b. Explain why the columns of $P^{-1}$ are the $\mathcal{C}$-coordinate vectors of $\mathbf{x}_0,...,\mathbf{x}_6$. Then use these coordinate vectors to write trigonometric identities that express powers of $\cos t$ in terms of the functions in $\mathcal{C}$.

$(4.5), \mathbf{[34]:}$

Let $\mathcal{B}=\left\{1, \cos t, \cos^2 t, \dots, \cos^6 t\right\}$ and $\mathcal{C}=\left\{1, \cos t, \cos 2t, \dots, \cos 6t\right\}$. Assume the following trigonometric identities (see Exercise 37 in Section 4.1).

$\cos 2t=-1+2\cos^2 t$

$\cos 3t=-3\cos t+4\cos^3 t$

$\cos 4t=1-8\cos^2 t+8\cos^4 t$

$\cos 5t=5\cos t-20\cos^3 t+16\cos^5 t$

$\cos 6t=-1+18\cos^2 t-48\cos^4 t+32\cos^6 t$

Let $H$ be the subspace of functions spanned by the functions in $\mathcal{B}$. Then $\mathcal{B}$ is a basis for $H$, by Exercise 38 in Section 4.3.

a. Write the $\mathcal{B}$-coordinate vectors of the vectors in $\mathcal{C}$, and use them to show that $\mathcal{C}$ is linearly independent set in $H$.

b. Explain why $\mathcal{C}$ is a basis for $H$.

$(4.1), \mathbf{[37]:}$

The vector space $H=\text{Span}\left\{1,\cos^2 t,\cos^4 t,\cos^6 t\right\}$ contains at least two interesting functions that will be used in a later exercise:

$\mathbf{f}(t)=1-8\cos^2 t+8\cos^4 t$

$\mathbf{g}(t)=-1+18\cos^2 t+48\cos^4 t+32\cos^6 t$

Study the graph of $\mathbf{f}$ for $0\le t\le 2\pi$, and guess a simple formula for $\mathbf{f}(t)$. Verify your conjecture by graphing the difference between $1+\mathbf{f}(t)$ and your formula for $(t)$. (Hopefully, you will see the constant function $1$.) Repeat for $\mathbf{g}$.

$(4.3), \mathbf{[38]:}$

Show that $\left\{1,\cos^2 t,\cos^4 t,\cos^6 t\right\}$ is linearly independent set of functions defined on $\mathbb{R}$. Use the method of Exercise 37. (This result will be needed in Exercise 34 in Section 4.5.)

$(4.3), \mathbf{[37]:}$

Show that $\left\{t,\sin t,\cos 2t,\sin t\cos t\right\}$ is a linearly independent set of functions defined on $\mathbb{R}$. Start by assuming that

$\tag{5} c_1\cdot t + c_2\cdot\sin t + c_3\cdot\cos 2t + c_4\cdot\sin t\cos t = 0$

Equation $(5)$ must hold for all real $t$, so choose several specific values of $t$ (say, $t=0,0.1,0.2$) until you get a system of enough equations to determine that all $c_j$ must be zero.

My Attempt:

My attempt to compute the integral is the classic way:

Using the results:

$$\cos^2 t = \frac{1}{2}(1+\cos 2t)$$

$$\cos a \cos b = \frac{1}{2}(\cos(a+b)+\cos(a-b))$$

$$\int\cos at \,\mathrm{d}t=\frac{1}{a}\sin at+C$$

$$\dots\text{some tedious steps (not hard)}\dots$$

$$5\cos^3(t)-6\cos^4(t)+5\cos^5(t)-12\cos^6(t)=-6 + \frac{55}{8}\cos t - \frac{69}{8}\cos 2t + \frac{45}{16}\cos 3t - 3\cos 4t + \frac{5}{16}\cos 5t - \frac{3}{8}\cos 6t$$

Therefore,

$$\int\left(5\cos^3(t)-6\cos^4(t)+5\cos^5(t)-12\cos^6(t)\right)\,\mathrm{d}t$$

$$=\int\left(-6 + \frac{55}{8}\cos t - \frac{69}{8}\cos 2t + \frac{45}{16}\cos 3t - 3\cos 4t + \frac{5}{16}\cos 5t - \frac{3}{8}\cos 6t\right)\,\mathrm{d}t$$

$$=-6t + \frac{55}{8}\sin t - \frac{69}{16}\sin 2t + \frac{15}{16}\sin 3t - \frac{3}{4}\sin 4t + \frac{1}{16}\sin 5t - \frac{1}{16}\sin 6t + C$$

I am not sure how can we make use of the previous exercises in order to make the integral simpler. Using those exercises, I believe, will increase the difficulty.

Your help on how to integrate this, using the above exercises, would be appreciated. Thanks.

ultralegend5385
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Hussain-Alqatari
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2 Answers2

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The given integrand has the following vector in the basis $\mathcal{B}$: $$\begin{pmatrix}0\\0\\0\\5\\-6\\5\\-12\end{pmatrix}$$ To convert it into the basis $\mathcal{C}$, use the matrix $P$ or $P^{-1}$ (which one should you use?). Once you have the coordinates of it in the basis $\mathcal{C}$, you now have it as an expression $$c_0+c_1\cos t+c_2\cos 2t+\cdots+c_6\cos6t$$ Then, integrating is easy.

Hope this helps. :>

ultralegend5385
  • 10,494
3

Transform the coefficients $\displaystyle{\large{x}}_{k}$ to $\displaystyle{\large{y}}_{k}$ using the matrix $\displaystyle{P}^{-1}$,
Where $\displaystyle{\sum_{k=0}^{6}{\large{x}}_{k}\cos^{k}{t}}={\sum_{k=0}^{6}{\large{y}}_{k}\cos{k\,t}}$ . $${\left[{\,{\large{x}}_{k}\,}\right]={\,P\,}{\large{.}}\left[{\,{\large{y}}_{k}\,}\right]}\implies{\left[{\,{\large{y}}_{k}\,}\right]={\,P^{-1}\,}{\large{.}}\left[{\,{\large{x}}_{k}\,}\right]}$$ Sence, from exercise ${(4.5){\small-}\mathbf{[34]}}$, calc $P\,$: $$P=\begin{bmatrix} 1 & 0 & -1 & 0 & 1 & 0 & -1 \\ 0 & 1 & 0 & -3 & 0 & 5 & 0 \\ 0 & 0 & 2 & 0 & -8 & 0 & 18 \\ 0 & 0 & 0 & 4 & 0 & -20 & 0 \\ 0 & 0 & 0 & 0 & 8 & 0 & -48 \\ 0 & 0 & 0 & 0 & 0 & 16 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 32 \end{bmatrix}$$ Hence, from exercise ${(4.7){\small-}\mathbf{[17]}}$, calc $P^{-1}\,$: $$\begin{bmatrix} {\large{y}}_0 \\ {\large{y}}_1 \\ {\large{y}}_2 \\ {\large{y}}_3 \\ {\large{y}}_4 \\ {\large{y}}_5 \\ {\large{y}}_6 \end{bmatrix}=\underbrace{\frac{1}{32}\begin{bmatrix} 32 & 0 & 16 & 0 & 12 & 0 & 10 \\ 0 & 32 & 0 & 24 & 0 & 20 & 0 \\ 0 & 0 & 16 & 0 & 16 & 0 & 15 \\ 0 & 0 & 0 & 8 & 0 & 10 & 0 \\ 0 & 0 & 0 & 0 & 4 & 0 & 6 \\ 0 & 0 & 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}}_{\color{blue}{P^{-1}}}{\large{.}}\begin{bmatrix} 0 \\ 0 \\ 0 \\ 5 \\-6 \\5 \\-12 \end{bmatrix}=\frac{1}{16}\begin{bmatrix} -96 \\ 110 \\ -138 \\ 45 \\ -48 \\ 5 \\ -6 \end{bmatrix}$$ Thus: $$ \begin{align} {I=}&{\,\int{\left(0\cos^0t+0\cos^1t+0\cos^2t+5\cos^3t-6\cos^4t+5\cos^5t-12\cos^6t\right)dt}} \\ =&{\,\frac{1}{16}\int{\left(-96\cos{0t}+110\cos{1t}-138\cos{2t}+45\cos{3t}-48\cos{4t}+5\cos{5t}-6\cos{6t}\right)dt}} \\ =&{\,\frac{1}{16}\left(-96\,t+110\,\frac{\sin{1t}}{1}-138\,\frac{\sin{2t}}{2}+45\,\frac{\sin{3t}}{3}-48\,\frac{\sin{4t}}{4}+5\,\frac{\sin{5t}}{5}-6\,\frac{\sin{6t}}{6}\right)+C} \end{align} $$

Hazem Orabi
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