-
Tomorrow we will discuss Problem 17 in Section 4.7 which is based on Problem 38 in Section 4.3 and Problem 34 in Section 4.5. In these problems you are asked to consider the following trigonometric functions
\begin{alignat*}{2}
\mathbf{a}_0 &= 1, & \qquad \mathbf{b}_0 &= 1, \\
\mathbf{a}_1 &= \cos t, & \qquad \mathbf{b}_1 &= \cos t, \\
\mathbf{a}_2 &= (\cos t)^2, & \qquad \mathbf{b}_2 &= \cos(2t), \\
\mathbf{a}_3 &= (\cos t)^3, & \qquad \mathbf{b}_3 &= \cos(3t), \\
\mathbf{a}_4 &= (\cos t)^4, & \qquad \mathbf{b}_4 &= \cos(4t), \\
\mathbf{a}_5 &= (\cos t)^5, & \qquad \mathbf{b}_5 &= \cos(5t), \\
\mathbf{a}_6 &= (\cos t)^6, & \qquad \mathbf{b}_6 &= \cos(6t). \\
\end{alignat*}
Thus, the functions $\mathbf{a}_0,\ldots,\mathbf{a}_6,$ are powers of the cosine function and the functions $\mathbf{b}_0,\ldots,\mathbf{b}_6,$ are multiple angle cosine functions. In fact, there is no particular reason to limit our considerations to these seven functions. We could consider any positive integer $n$, and consider
\begin{alignat*}{2}
\mathbf{a}_k &= (\cos t)^k, & \qquad \mathbf{b}_k &= \cos(kt), \\
\end{alignat*}
for all $k \in \{0,1,\ldots,n\}.$
-
The first remarkable fact about these functions is that for an arbitrary positive integer $n$ we have
\[
\mathbf{b}_n \in \operatorname{span} \bigl\{ \mathbf{a}_0, \mathbf{a}_1,\ldots,\mathbf{a}_n \bigr\}.
\]
To prove this claim we need trigonometric identities presented in the textbook in Problem 34 in Section 4.5. I will prove those identities here. The proof is based on the addition formulas for the cosine function. The method I present here is called Chebyshev method. Let $n$ be a positive integer greater than $1.$ Then we have
\begin{align*}
\cos(nt) = \cos\bigl((n-1)t + t \bigr) & = \cos\bigl((n-1)t \bigr) \cos t - \sin\bigl((n-1)t \bigr) \sin t \\
\cos\bigl((n-2)t \bigr) = \cos\bigl((n-1)t - t \bigr) & = \cos\bigl((n-1)t \bigr) \cos t + \sin\bigl((n-1)t \bigr) \sin t \\
\end{align*}
Adding the last two identities yields
\[
\cos(nt) + \cos\bigl((n-2)t \bigr) = 2 \cos\bigl((n-1)t \bigr) \cos t
\]
and, more importantly, we have
\begin{equation*} \tag{rr}
\cos(nt) = 2 \cos\bigl((n-1)t \bigr) \cos t - \cos\bigl((n-2)t \bigr).
\end{equation*}
This formula gives $\cos(nt)$ in terms of two previous multiple angle cosines $\cos\bigl((n-1)t \bigr)$ and $\cos\bigl((n-2)t \bigr).$ Such a formula is called a recursion or a recurrence relation. For $n=2$ we have \[
\cos(2t) = 2 \cos(1t) \cos t - \cos(0t) = 2(\cos t)^2 - 1.
\]
Or, in our notation
\[
\mathbf{b}_2 = - \mathbf{a}_0 + 2 \mathbf{a}_2.
\]
That is $\mathbf{b}_2 \in \operatorname{span} \bigl\{\mathbf{a}_0, \mathbf{a}_2\bigr\}.$ Next we apply the recursion (rr) to $n=3$ and we get (using already established formula for $\cos(2t)$)
\[
\cos(3t) = 2 \cos(2t) \cos t - \cos t = 2\bigl( 2(\cos t)^2 - 1 \bigr) \cos t - \cos t = 4 (\cos t)^3 - 3 \cos t.
\]
That is
\[
\mathbf{b}_3 = - 3 \mathbf{a}_1 + 4 \mathbf{a}_3,
\]
implying that $\mathbf{b}_3 \in \operatorname{span} \bigl\{\mathbf{a}_1, \mathbf{a}_3\bigr\}.$ Continuing to $n=4$ we get
\begin{align*}
\cos(4t) & = 2 \cos(3t) \cos t - \cos(2t) \\
& = 2 \bigl( 4 (\cos t)^3 - 3 \cos t \bigr) \cos t - \bigl(2(\cos t)^2 - 1 \bigr) \\
& = 8 (\cos t)^4 - 8 (\cos t)^2 + 1.
\end{align*}
That is
\[
\mathbf{b}_4 = \mathbf{a}_0 - 8 \mathbf{a}_2 + 8 \mathbf{a}_4,
\]
implying that $\mathbf{b}_4 \in \operatorname{span} \bigl\{\mathbf{a}_0, \mathbf{a}_2, \mathbf{a}_4 \bigr\}.$ Continuing to $n=5$ we get
\begin{align*}
\cos(5t) & = 2 \cos(4t) \cos t - \cos(3t) \\
& = 2 \bigl( 8 (\cos t)^4 - 8 (\cos t)^2 + 1 \bigr) \cos t - \bigl( 4 (\cos t)^3 - 3 \cos t \bigr) \\
& = 16 (\cos t)^5 - 20 (\cos t)^3 + 5 \cos t.
\end{align*}
That is
\[
\mathbf{b}_5 = 5 \mathbf{a}_1 - 20 \mathbf{a}_3 + 16 \mathbf{a}_5,
\]
implying that $\mathbf{b}_5 \in \operatorname{span} \bigl\{\mathbf{a}_1, \mathbf{a}_3, \mathbf{a}_5 \bigr\}.$ Continuing to $n=6$ we get
\begin{align*}
\cos(6t) & = 2 \cos(5t) \cos t - \cos(4t) \\
& = 2 \bigl( 16 (\cos t)^5 - 20 (\cos t)^3 + 5 \cos t \bigr) \cos t - \bigl( 8 (\cos t)^4 - 8 (\cos t)^2 + 1 \bigr) \\
& = 32 (\cos t)^6 - 48 (\cos t)^4 + 18 (\cos t)^2 - 1.
\end{align*}
That is
\[
\mathbf{b}_6 = - \mathbf{a}_0 + 18 \mathbf{a}_2 - 48 \mathbf{a}_4 + 32 \mathbf{a}_6,
\]
implying that $\mathbf{b}_6 \in \operatorname{span} \bigl\{\mathbf{a}_0, \mathbf{a}_2, \mathbf{a}_4, \mathbf{a}_6 \bigr\}.$
In general, for a positive integer $n$ we have \begin{equation*} \cos(n t) = 2^{n-1} n \sum_{k=0}^{\lfloor n/2 \rfloor} \frac{(-1)^k}{4^k (n-k)} \binom{n-k}{k} (\cos t)^{n-2k}, \end{equation*} Here $\lfloor \cdot \rfloor$ denotes the floor function and $\displaystyle \binom{m}{k}$ denotes a binomial coefficient, that is \[ \binom{m}{k} = \frac{m!}{k! (m-k)!}. \] The above formula for $\cos(nt)$ I derived from Viète's formulas for sine and cosine of multiple angles. I think that Viète's formulas are not difficult to prove by Mathematical induction.
For an even $n = 2m$ a slight manipulation of the above formula for $\cos(nt)$ expressed in terms of the functions $\mathbf{a}_k$ gives \begin{equation*} \mathbf{b}_{2m} = (-1)^{m} m \sum_{k=0}^{m} \frac{(-4)^k}{m+k} \binom{m+k}{m-k} \mathbf{a}_{2k}. \end{equation*} Thus $\mathbf{b}_{2m} \in \operatorname{span} \bigl\{\mathbf{a}_0, \mathbf{a}_2, \cdots, \mathbf{a}_{2m}\bigr\}.$ Similarly for an odd $n=2m+1$ we have \begin{equation*} \mathbf{b}_{2m+1} = (-1)^{m} (2m+1) \sum_{k=0}^{m} \frac{(-4)^k}{m+1+k} \binom{m+1+k}{m-k} \mathbf{a}_{2k+1}. \end{equation*} Thus $\mathbf{b}_{2m+1} \in \operatorname{span} \bigl\{\mathbf{a}_1, \mathbf{a}_3, \ldots, \mathbf{a}_{2m+1}\bigr\}.$ -
Another remarkable fact about the functions $\mathbf{a}_0, \mathbf{a}_1, \ldots, \mathbf{a}_{n}$ is that they are linearly independent. You are asked to prove this with $n=6$ in Problem 38 in Section 4.3.
Here is a proof.
We will prove that the only linear combination of the funcions $\mathbf{a}_0, \mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3,\mathbf{a}_4, \mathbf{a}_5, \mathbf{a}_{6}$ which results in the zero function is the trivial linear combination. Assume that $\alpha_0, \alpha_1, \ldots, \alpha_6 \in \mathbb R$ are such that \begin{equation*} \alpha_0 \mathbf{a}_0 + \alpha_1 \mathbf{a}_1+\alpha_2 \mathbf{a}_2 + \alpha_3 \mathbf{a}_3+\alpha_4\mathbf{a}_4+\alpha_5 \mathbf{a}_5+\alpha_6 \mathbf{a}_{6} = \mathbf 0. \end{equation*} The preceding equality means that for all $t \in \mathbb R$ we have \begin{equation*} \alpha_0 + \alpha_1 (\cos t) +\alpha_2 (\cos t)^2 + \alpha_3 (\cos t)^3 +\alpha_4 (\cos t)^4 +\alpha_5 (\cos t)^5+\alpha_6 (\cos t)^6 = 0. \end{equation*} Now we introduce the substitution $s = \cos t.$ Since the range of $\cos t$ is the interval $[-1,1]$, the preceding equality means that for all $s \in [-1,1]$ we have \begin{equation*} \alpha_0 + \alpha_1 s +\alpha_2 s^2 + \alpha_3 s^3 +\alpha_4 s^4 +\alpha_5 s^5+\alpha_6 s^6 = 0. \end{equation*} Differentiating both sides of the preceding equality sith respect to $s$ yields that for all $s \in (-1,1)$ we have \begin{align*} \alpha_1 + 2 \alpha_2 s + 3 \alpha_3 s^2 + 4 \alpha_4 s^3 + 5 \alpha_5 s^4 + 6 \alpha_6 s^5 & = 0 \\ 2 \alpha_2 + 6 \alpha_3 s + 12 \alpha_4 s^2 + 20 \alpha_5 s^3 + 30 \alpha_6 s^4 & = 0 \\ 6 \alpha_3 + 24 \alpha_4 s + 60 \alpha_5 s^2 + 120 \alpha_6 s^3 & = 0 \\ 24 \alpha_4 + 120 \alpha_5 s + 360 \alpha_6 s^2 & = 0 \\ 120 \alpha_5 + 720 \alpha_6 s & = 0 \\ 720 \alpha_6 & = 0 \\ \end{align*} Now, substituting $s = 0$ in the last 7 equalities yields \begin{align*} \alpha_0 & = 0 \\ \alpha_1 & = 0 \\ 2 \alpha_2 & = 0 \\ 6 \alpha_3 & = 0 \\ 24 \alpha_4 & = 0 \\ 120 \alpha_5 & = 0 \\ 720 \alpha_6 & = 0 \\ \end{align*} Hence, $\alpha_0 = \alpha_1 = \alpha_2 = \alpha_3 = \alpha_4 =\alpha_5 = \alpha_6 = 0.$ This proves that the functions $\mathbf{a}_0, \mathbf{a}_1, \ldots, \mathbf{a}_{n}$ are linearly independent.
The method that we used for 7 functions $\mathbf{a}_0, \mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3,\mathbf{a}_4, \mathbf{a}_5, \mathbf{a}_{6}$ can be used for $n$ functions $\mathbf{a}_k$ with $k \in \{0,1,\ldots,n\}$ for every positive integer $n.$ Assume that \[ \sum_{j=0}^n \alpha_j \mathbf{a}_j = \mathbf 0. \] That is, assume that for every $t \in \mathbb R$ we have \[ \sum_{j=0}^n \alpha_j (\cos t)^j = 0. \] Substitution $s = \cos t$ yields that for every $s \in [-1,1]$ we have \[ \sum_{j=0}^n \alpha_j s^j = 0. \] Let $k \in \{0,1,\ldots,n\}$ be arbitrary. Taking the $k$-th derivative with respect to $s$ of both sides of the preceding equality yields that for all $s \in (-1,1)$ we have \[ \sum_{j=k}^n \frac{j!}{(j-k)!} \alpha_j s^{j-k} = 0. \] Now, substituting $s = 0$ in the last equality gives \[ k! \alpha_k = 0. \] Since $k! \neq 0$, we obtain $\alpha_k = 0.$ Since $k \in \{0,1,\ldots,n\}$ was arbitrary we have proved that $\alpha_k = 0$ for all $k \in \{0,1,\ldots,n\}.$ Thus, we have proved the implication \[ \sum_{j=0}^n \alpha_j \mathbf{a}_j = \mathbf 0 \quad \Rightarrow \quad \alpha_k = 0 \ \ \text{for all} \ \ k \in \{0,1,\ldots,n\}. \] That is, we proved that the vectors $\mathbf{a}_0, \mathbf{a}_1, \ldots, \mathbf{a}_{n}$ are linearly independent. - Since the vectors $\mathbf{a}_0, \mathbf{a}_1, \ldots, \mathbf{a}_{n}$ are linearly independent, the set \[ \mathcal B_n = \bigl\{ \mathbf{a}_0, \mathbf{a}_1, \ldots, \mathbf{a}_{n} \bigr\} \] is a basis of the space \[ \mathcal M_n = \operatorname{span} \bigl\{ \mathbf{a}_0, \mathbf{a}_1, \ldots, \mathbf{a}_{n} \bigr\}. \] Here $n$ is an arbitrary positive integer.
- Now recall that we established the following facts: The set \[ \mathcal B_6 = \bigl\{ \mathbf{a}_0, \mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3, \mathbf{a}_4, \mathbf{a}_5, \mathbf{a}_{6} \bigr\} \] is a basis of the space \[ \mathcal M_6 = \operatorname{Span} \bigl\{ \mathbf{a}_0, \mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3, \mathbf{a}_4, \mathbf{a}_5, \mathbf{a}_{6} \bigr\}. \] The following identities hold: \begin{align*} \mathbf{b}_0 & = \mathbf{a}_0 \\ \mathbf{b}_1 & = \mathbf{a}_1 \\ \mathbf{b}_2 & = - \mathbf{a}_0 + 2 \mathbf{a}_2 \\ \mathbf{b}_3 & = - 3 \mathbf{a}_1 + 4 \mathbf x_y3 \\ \mathbf{b}_4 & = \mathbf{a}_0 - 8 \mathbf{a}_2 + 8 \mathbf x_yy4 \\ \mathbf{b}_5 & = 5 \mathbf{a}_1 - 20 \mathbf{a}_3 + 16 \mathbf{a}_5 \\ \mathbf{b}_6 & = - \mathbf{a}_0 + 18 \mathbf x_y2 - 48 \mathbf{a}_4 + 32 \mathbf{a}_6 \\ \end{align*} In particular \[ \operatorname{Span} \bigl\{ \mathbf{b}_0, \mathbf{b}_1, \mathbf{b}_2, \mathbf{b}_3, \mathbf{b}_4, \mathbf{b}_5, \mathbf{b}_{6} \bigr\} \subseteq \operatorname{Span} \bigl\{ \mathbf{a}_0, \mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3, \mathbf x_y4, \mathbf{a}_5, \mathbf{a}_{6} \bigr\} = \mathcal M_6. \] Since $\mathcal B_6$ is a basis of $\mathcal M_6$, the formulas for $\mathbf y$-s in terms of $\mathbf x$-s can be expressed as coordinate vectors as follows: \[ \bigl[ \mathbf{b}_0\!\bigr]_{\mathcal B_6}\!\!=\!\!\left[\!\begin{array}{r} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\!\right]\!\!, \bigl[ \mathbf{b}_1\!\bigr]_{\mathcal B_6}\!\!=\!\!\left[\!\begin{array}{r} 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\!\right]\!\!, \bigl[ \mathbf{b}_2\!\bigr]_{\mathcal B_6}\!\!=\!\!\left[\!\!\begin{array}{r} -1 \\ 0 \\ 2 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\!\right]\!\!, \bigl[ \mathbf{b}_3\!\bigr]_{\mathcal B_6}\!\!=\!\!\left[\!\!\begin{array}{r} 0 \\ -3 \\ 0 \\ 4 \\ 0 \\ 0 \\ 0 \end{array}\!\right]\!\!, \bigl[ \mathbf{b}_4\!\bigr]_{\mathcal B_6}\!\!=\!\!\left[\!\!\begin{array}{r} 1 \\ 0 \\ -8 \\ 0 \\ 8 \\ 0 \\ 0 \end{array}\!\right]\!\!, \bigl[ \mathbf{b}_5\!\bigr]_{\mathcal B_6}\!\!=\!\!\left[\!\!\begin{array}{r} 0 \\ 5 \\ 0 \\ -20 \\ 0 \\ 16 \\ 0 \end{array}\!\right]\!\!, \bigl[ \mathbf{b}_6\!\bigr]_{\mathcal B_6}\!\!=\!\!\left[\!\!\begin{array}{r} -1 \\ 0 \\ 18 \\ 0 \\ -48 \\ 0 \\ 32 \end{array}\!\right]\!\!. \]
-
The coordinate vectors in the previous item answer the question (a) in Problem 34 in Section 4.5. To answer the question (b) in this problem we need to prove that the functions $\mathbf{b}_0, \mathbf{b}_1, \mathbf{b}_2, \mathbf{b}_3, \mathbf{b}_4, \mathbf{b}_5, \mathbf{b}_{6}$ are linearly independent.
Here is a proof that the functions $\mathbf{b}_0, \mathbf{b}_1, \mathbf{b}_2, \mathbf{b}_3, \mathbf{b}_4, \mathbf{b}_5, \mathbf{b}_{6}$ are linearly independent.
Let $\beta_0, \beta_1, \beta_2, \beta_3, \beta_4, \beta_5, \beta_6 \in \mathbb R$ and assume that \[ \beta_0 \mathbf{b}_0 + \beta_1 \mathbf{b}_1+ \beta_2 \mathbf{b}_2+ \beta_3 \mathbf{b}_3+ \beta_4 \mathbf{b}_4+ \beta_5 \mathbf{b}_5+ \beta_6 \mathbf{b}_{6} = \mathbf 0. \] Now apply the coordinate mapping $[\cdot]_{\mathcal B_6} : \mathcal M_6 \to \mathbb R^7$ defined by $\mathbf v \mapsto [\mathbf v]_{\mathcal B_6}$ where $\mathbf v$ is an arbitrary function in $\mathcal M_6.$ \[ \bigl[\beta_0 \mathbf{b}_0 + \beta_1 \mathbf{b}_1+ \beta_2 \mathbf{b}_2+ \beta_3 \mathbf{b}_3+ \beta_4 \mathbf{b}_4+ \beta_5 \mathbf{b}_5+ \beta_6 \mathbf{b}_{6}\bigr]_{\mathcal B_6} = \bigl[\mathbf 0\bigr]_{\mathcal B_6}. \] Next, recall that the coordinate mapping is linear to conclude that \[ \beta_0 \bigl[\mathbf{b}_0\bigr]_{\mathcal B_6} + \beta_1 \bigl[\mathbf{b}_1\bigr]_{\mathcal B_6}+ \beta_2 \bigl[\mathbf{b}_2\bigr]_{\mathcal B_6} + \beta_3 \bigl[\mathbf{b}_3\bigr]_{\mathcal B_6}+ \beta_4\bigl[\mathbf{b}_4\bigr]_{\mathcal B_6}+ \beta_5 \bigl[\mathbf{b}_5\bigr]_{\mathcal B_6}+ \beta_6 \bigl[\mathbf{b}_{6}\bigr]_{\mathcal B_6} = \bigl[\mathbf 0\bigr]_{\mathcal B_6}. \] Now we rewrite the previous equality using the coordinate vectors from the previous item: \[ \beta_0 \left[\!\begin{array}{r} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\!\right]\!+\beta_1\!\left[\!\begin{array}{r} 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\!\right]\!+\beta_2\!\left[\!\!\begin{array}{r} -1 \\ 0 \\ 2 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\!\right]\!+\beta_3\!\left[\!\!\begin{array}{r} 0 \\ -3 \\ 0 \\ 4 \\ 0 \\ 0 \\ 0 \end{array}\!\right]\!+\beta_4\!\left[\!\!\begin{array}{r} 1 \\ 0 \\ -8 \\ 0 \\ 8 \\ 0 \\ 0 \end{array}\!\right]\!+\beta_5\!\left[\!\!\begin{array}{r} 0 \\ 5 \\ 0 \\ -20 \\ 0 \\ 16 \\ 0 \end{array}\!\right]\!+\beta_6\!\left[\!\!\begin{array}{r} -1 \\ 0 \\ 18 \\ 0 \\ -48 \\ 0 \\ 32 \end{array}\!\right] = \left[\!\begin{array}{r} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\!\right]. \] The preceding vector equation is equivalent to the following matrix equation \[ \left[\!\begin{array}{rrrrrrr} 1 & 0 & -1 & 0 & 1 & 0 & -1 \\ 0 & 1 & 0 & -3 & 0 & 5 & 0 \\ 0 & 0 & 2 & 0 & -8 & 0 & 18 \\ 0 & 0 & 0 & 4 & 0 & -20 & 0 \\ 0 & 0 & 0 & 0 & 8 & 0 & -48 \\ 0 & 0 & 0 & 0 & 0 & 16 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 32 \end{array}\!\right] \left[\!\begin{array}{r} \beta_0 \\ \beta_1 \\ \beta_2 \\ \beta_3 \\ \beta_4 \\ \beta_5 \\ \beta_6 \end{array}\!\right] = \left[\!\begin{array}{r} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\!\right]. \] Since the matrix in the last equation is upper triangular with the nonzero entries on the diagonal, the above equation has only the trivial solution. That is \[ \beta_0 = \beta_1 = \beta_2 = \beta_3 = \beta_4 = \beta_5 = \beta_6 = 0. \] Thus, we have proved the implication \[ \sum_{j=0}^6 \beta_j \mathbf{b}_j = \mathbf 0 \quad \Rightarrow \quad \beta_k = 0 \ \text{for all} \ k \in\{0,1,\ldots,n\}. \] Consequently, the functions $\mathbf{b}_0, \mathbf{b}_1, \mathbf{b}_2, \mathbf{b}_3, \mathbf{b}_4, \mathbf{b}_5, \mathbf{b}_{6}$ are linearly independent.
Since $\mathbf{b}_0, \mathbf{b}_1, \mathbf{b}_2, \mathbf{b}_3, \mathbf{b}_4, \mathbf{b}_5, \mathbf{b}_{6} \in \mathcal M_6$, $\dim \mathcal M_6 = 7$ and $\mathbf{b}_0, \mathbf{b}_1, \mathbf{b}_2, \mathbf{b}_3, \mathbf{b}_4, \mathbf{b}_5, \mathbf{b}_{6}$ are linearly independent, by Theorem 12 in Section 4.5 (The Basis Theorem) the set \[ \mathcal C_6 = \{\mathbf{b}_0, \mathbf{b}_1, \mathbf{b}_2, \mathbf{b}_3, \mathbf{b}_4, \mathbf{b}_5, \mathbf{b}_{6}\} \] is a basis for $\mathcal M_6.$ - As before, there is nothing special about the number $6$ in the previous item. The conclusion holds for an arbitrary positive integer $n$: the functions $\mathbf{b}_k$ with $k \in \{0,1,\ldots,n\}$ are linearly independent. The set of functions \[ \mathcal C_n = \{\mathbf{b}_0, \mathbf{b}_1, \ldots, \mathbf{b}_{n}\} \] is a basis of the space $\mathcal M_n.$ The proofs are very similar to the special case of $n=6.$
- By Theorem 15 in Section 4.7 and the previous calculations we have \[ \underset{\mathcal{B}_6\leftarrow\mathcal{C}_6}{P} = \left[\!\begin{array}{rrrrrrr} 1 & 0 & -1 & 0 & 1 & 0 & -1 \\ 0 & 1 & 0 & -3 & 0 & 5 & 0 \\ 0 & 0 & 2 & 0 & -8 & 0 & 18 \\ 0 & 0 & 0 & 4 & 0 & -20 & 0 \\ 0 & 0 & 0 & 0 & 8 & 0 & -48 \\ 0 & 0 & 0 & 0 & 0 & 16 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 32 \end{array}\!\right] \] Since $\underset{{\mathcal C}_6 \leftarrow {\mathcal B}_6}{P} = \Bigl(\underset{{\mathcal B}_6 \leftarrow {\mathcal C}_6}{P}\Bigr)^{-1},$ we have \[ \underset{{\mathcal C}_6 \leftarrow {\mathcal B}_6}{P} = \left[ \begin{array}{ccccccc} 1 & 0 & \frac{1}{2} & 0 & \frac{3}{8} & 0 & \frac{5}{16} \\ 0 & 1 & 0 & \frac{3}{4} & 0 & \frac{5}{8} & 0 \\ 0 & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & \frac{15}{32} \\ 0 & 0 & 0 & \frac{1}{4} & 0 & \frac{5}{16} & 0 \\ 0 & 0 & 0 & 0 & \frac{1}{8} & 0 & \frac{3}{16} \\ 0 & 0 & 0 & 0 & 0 & \frac{1}{16} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{32} \end{array} \right] \] Again, by Theorem 15 in Section 4.7 we have \begin{align*} \mathbf{a}_0 & = \mathbf{b}_0 \\ \mathbf{a}_1 & = \mathbf{b}_1 \\ \mathbf{a}_2 & = \frac{1}{2} \bigl( \mathbf{b}_0 + \mathbf{b}_2 \bigr) \\ \mathbf{a}_3 & = \frac{1}{4} \bigl( 3 \mathbf{b}_1 + \mathbf{b}_3 \bigr) \\ \mathbf{a}_4 & = \frac{1}{8} \bigl( 3 \mathbf{b}_0 + 4 \mathbf{b}_2 + \mathbf{b}_4 \bigr) \\ \mathbf{a}_5 & = \frac{1}{16} \bigl( 10 \mathbf{b}_1 + 5 \mathbf{b}_3 + \mathbf{b}_5 \bigr) \\ \mathbf{a}_6 & = \frac{1}{32} \bigl( 10 \mathbf{b}_0 + 15 \mathbf{b}_2 +6 \mathbf{b}_4 + \mathbf{b}_6 \bigr) \end{align*} This solves Problem 17 in Section 4.7.
- It is nice to rewrite the last formulas of the previous item in terms of the specific funcions which are involved: \begin{align*} 1 & = 1 \\ \cos t & = \cos t \\ (\cos t)^2 & = \frac{1}{2} \bigl( 1 + \cos(2t) \bigr) \\ (\cos t)^3 & = \frac{1}{4} \bigl( 3 \cos(t) + \cos(3t) \bigr) \\ (\cos t)^4 & = \frac{1}{8} \bigl( 3 + 4 \cos(2t) + \cos(4t) \bigr) \\ (\cos t)^5 & = \frac{1}{16} \bigl( 10 \cos(t)+ 5 \cos(3t) + \cos(5t)\bigr) \\ (\cos t)^6 & = \frac{1}{32} \bigl( 10 + 15 \cos(2t) +6 \cos(4t) + \cos(6t)\bigr) \end{align*} It is truly remarkable that in Chapter 6 we will develop a completely different way of deriving formulas like these. See Section 6.7, specifically An Inner Product for $C[a,b]$ on page 433 and Section 6.8, specifically Fourier Series on page 440. It turns out that in the sense of Section 6.7 the functions $\cos(kt)$ are mutually orthogonal. Using the method from Section 6.8 we will be able to derive the following formulas \begin{align*} (\cos t)^{2m} & = \frac{1}{2^{2m}} \binom{2m}{m} + \frac{1}{2^{2m-1}} \sum_{k=1}^{m} \binom{2m}{m-k} \cos(2kt) \\ (\cos t)^{2m+1} & =\frac{1}{2^{2m}} \sum_{k=0}^{m} \binom{2m+1}{m-k} \cos((2k+1)t) \end{align*} for all nonnegative integers $m.$
-
And finally, it would be nice to plot these fourteen functions:
And all seven multiple angle cosine functions on one graph:
${\mathbf y}_0 = 1$
${\mathbf x}_0 = 1$
${\mathbf y}_1 = \cos t$
${\mathbf x}_1 = \cos t$
${\mathbf y}_2 = \cos(2t)$
${\mathbf x}_2 = (\cos t)^2$
${\mathbf y}_3 = \cos(3t)$
${\mathbf x}_3 = (\cos t)^3$
${\mathbf y}_4 = \cos(4t)$
${\mathbf x}_4 = (\cos t)^4$
${\mathbf y}_5 = \cos(5t)$
${\mathbf x}_5 = (\cos t)^5$
${\mathbf y}_6 = \cos(6t)$
${\mathbf x}_6 = (\cos t)^6$
Finally it is nice to illustrate how gradually partial approximations of $(\cos t)^6$ by multiple cosines gradually approach this function.
$(\cos t)^6 \approx \frac{5}{16}$
$(\cos t)^6 \approx \frac{5}{16}+\frac{15}{32} \cos(2t)$
$(\cos t)^6 \approx \frac{5}{16}+\frac{15}{32} \cos(2t)+\frac{3}{16} \cos(4t)$
$(\cos t)^6 = \frac{5}{16}+\frac{15}{32} \cos(2t)+\frac{3}{16} \cos(4t)+\frac{1}{32} \cos(6t)$
-
The first remarkable fact about these functions is that for an arbitrary positive integer $n$ we have
\[
\mathbf{b}_n \in \operatorname{span} \bigl\{ \mathbf{a}_0, \mathbf{a}_1,\ldots,\mathbf{a}_n \bigr\}.
\]
To prove this claim we need trigonometric identities presented in the textbook in Problem 34 in Section 4.5. I will prove those identities here. The proof is based on the addition formulas for the cosine function. The method I present here is called Chebyshev method. Let $n$ be a positive integer greater than $1.$ Then we have
\begin{align*}
\cos(nt) = \cos\bigl((n-1)t + t \bigr) & = \cos\bigl((n-1)t \bigr) \cos t - \sin\bigl((n-1)t \bigr) \sin t \\
\cos\bigl((n-2)t \bigr) = \cos\bigl((n-1)t - t \bigr) & = \cos\bigl((n-1)t \bigr) \cos t + \sin\bigl((n-1)t \bigr) \sin t \\
\end{align*}
Adding the last two identities yields
\[
\cos(nt) + \cos\bigl((n-2)t \bigr) = 2 \cos\bigl((n-1)t \bigr) \cos t
\]
and, more importantly, we have
\begin{equation*} \tag{rr}
\cos(nt) = 2 \cos\bigl((n-1)t \bigr) \cos t - \cos\bigl((n-2)t \bigr).
\end{equation*}
This formula gives $\cos(nt)$ in terms of two previous multiple angle cosines $\cos\bigl((n-1)t \bigr)$ and $\cos\bigl((n-2)t \bigr).$ Such a formula is called a recursion or a recurrence relation. For $n=2$ we have \[
\cos(2t) = 2 \cos(1t) \cos t - \cos(0t) = 2(\cos t)^2 - 1.
\]
Or, in our notation
\[
\mathbf{b}_2 = - \mathbf{a}_0 + 2 \mathbf{a}_2.
\]
That is $\mathbf{b}_2 \in \operatorname{span} \bigl\{\mathbf{a}_0, \mathbf{a}_2\bigr\}.$ Next we apply the recursion (rr) to $n=3$ and we get (using already established formula for $\cos(2t)$)
\[
\cos(3t) = 2 \cos(2t) \cos t - \cos t = 2\bigl( 2(\cos t)^2 - 1 \bigr) \cos t - \cos t = 4 (\cos t)^3 - 3 \cos t.
\]
That is
\[
\mathbf{b}_3 = - 3 \mathbf{a}_1 + 4 \mathbf{a}_3,
\]
implying that $\mathbf{b}_3 \in \operatorname{span} \bigl\{\mathbf{a}_1, \mathbf{a}_3\bigr\}.$ Continuing to $n=4$ we get
\begin{align*}
\cos(4t) & = 2 \cos(3t) \cos t - \cos(2t) \\
& = 2 \bigl( 4 (\cos t)^3 - 3 \cos t \bigr) \cos t - \bigl(2(\cos t)^2 - 1 \bigr) \\
& = 8 (\cos t)^4 - 8 (\cos t)^2 + 1.
\end{align*}
That is
\[
\mathbf{b}_4 = \mathbf{a}_0 - 8 \mathbf{a}_2 + 8 \mathbf{a}_4,
\]
implying that $\mathbf{b}_4 \in \operatorname{span} \bigl\{\mathbf{a}_0, \mathbf{a}_2, \mathbf{a}_4 \bigr\}.$ Continuing to $n=5$ we get
\begin{align*}
\cos(5t) & = 2 \cos(4t) \cos t - \cos(3t) \\
& = 2 \bigl( 8 (\cos t)^4 - 8 (\cos t)^2 + 1 \bigr) \cos t - \bigl( 4 (\cos t)^3 - 3 \cos t \bigr) \\
& = 16 (\cos t)^5 - 20 (\cos t)^3 + 5 \cos t.
\end{align*}
That is
\[
\mathbf{b}_5 = 5 \mathbf{a}_1 - 20 \mathbf{a}_3 + 16 \mathbf{a}_5,
\]
implying that $\mathbf{b}_5 \in \operatorname{span} \bigl\{\mathbf{a}_1, \mathbf{a}_3, \mathbf{a}_5 \bigr\}.$ Continuing to $n=6$ we get
\begin{align*}
\cos(6t) & = 2 \cos(5t) \cos t - \cos(4t) \\
& = 2 \bigl( 16 (\cos t)^5 - 20 (\cos t)^3 + 5 \cos t \bigr) \cos t - \bigl( 8 (\cos t)^4 - 8 (\cos t)^2 + 1 \bigr) \\
& = 32 (\cos t)^6 - 48 (\cos t)^4 + 18 (\cos t)^2 - 1.
\end{align*}
That is
\[
\mathbf{b}_6 = - \mathbf{a}_0 + 18 \mathbf{a}_2 - 48 \mathbf{a}_4 + 32 \mathbf{a}_6,
\]
implying that $\mathbf{b}_6 \in \operatorname{span} \bigl\{\mathbf{a}_0, \mathbf{a}_2, \mathbf{a}_4, \mathbf{a}_6 \bigr\}.$