Textbook Proposition on Product of Real Analytic Functions

Question

Let \begin{align*} \sum\limits_{j=0}^\infty a_j (x-c)^j && \sum\limits_{j=0}^\infty b_j (x-c)^j \\ \end{align*}

be two power series with intervals of convergence $\mathcal{C}_1$ and $\mathcal{C}_2$ centered on at $c$. Let $f_1(x)$ be the function defined by the first series on $\mathcal{C}_1$ and $f_2(x)$ the function defined by the second series on $\mathcal{C}_2$. Then, on their common domain $\mathcal{C} = \mathcal{C}_1 \cap \mathcal{C}_2$, it holds that \begin{align*} f(x) \cdot g(x) &= \sum\limits_{m=0}^\infty \sum\limits_{j+k=m} (a_j \cdot b_k) (x-c)^m \\ \end{align*}

Proof: Let \begin{align*} A_N &= \sum\limits_{j=0}^N a_j (x-c)^j & B_N &= \sum\limits_{j=0}^N b_j (x-c)^j \\ \end{align*}

be respectively, the Nth partial sums of the power series that define $f$ and $g$. Let \begin{align*} D_N &= \sum\limits_{m=0}^N \sum\limits_{j+k=m} (a_j \cdot b_k) (x-c)^m & R_N &= \sum\limits_{j=N+1}^\infty b_j (x-c)^j \\ \end{align*}

We have: \begin{align*} D_N &= a_0 B_N + a_1 (x-c) B_{N-1} + \cdots + a_N (x-c)^N B_0 \\ &= a_0 (g(x) - R_N) + a_1 (x-c) (g(x) - R_{N-1}) + \cdots + a_N (x-c)^N (g(x) - R_0) \\ \end{align*} [snip]

I've transcribed the above from my Real Analysis textbook and stopped at the part I'm having trouble understanding. How do you get from the first definition of $D_N$ to the second one? They seem like different equations. I've tried to do sequence manipulations to get from one definition to the other, but I've not had success.

Answer 1

We start with the first representation of $D_n$ and obtain the second one.

We obtain \begin{align*} \color{blue}{D_n}&\color{blue}{=\sum_{m=0}^N\left(\sum_{{j+k=m}\atop{j,k\geq 0}} a_jb_k\right)(x-c)^m}\\ &=\sum_{m=0}^N\left(\sum_{j=0}^m a_jb_{m-j}\right)(x-c)^m\tag{1}\\ &=\sum_{0\leq j\leq m\leq N} a_jb_{m-j}(x-c)^m\tag{2}\\ &=\sum_{j=0}^N\sum_{m=j}^N a_jb_{m-j}(x-c)^m\tag{3}\\ &=\sum_{j=0}^Na_j\sum_{m=0}^{N-j}b_m(x-c)^{m+j}\tag{4}\\ &=\sum_{j=0}^N a_j(x-c)^j\sum_{m=0}^{N-j}b_m(x-c)^m\tag{5}\\ &\,\,\color{blue}{=\sum_{j=0}^N a_j(x-c)^jB_{N-j}}\tag{6} \end{align*} and we finally got the second representation of $D_n$.

Comment:

• In (1) we eliminate $k$ by substituting $k\to m-j$.

• In (2) we write the index region somewhat more conveniently.

• In (3) we exchange the order of summation.

• In (4) we shift the index $m$ to start with $m=0$.

• In (5) we factor out the terms which are not dependent on $m$.

We also obtain for $0\leq j\leq N$ \begin{align*} \color{blue}{g(x)}&=\sum_{k=0}^\infty b_k(x-c)^k\\ &=\sum_{k=0}^{N-j} b_k(x-c)^k+\sum_{k=N-j+1}^\infty b_k(x-c)^k\\ &\,\,\color{blue}{=B_{N-j}+R_{N-j}} \end{align*} and the last line of OP's post follows from (6) by replacing $B_{N-j}=g(x)-R_{N-j}$.

Answer 2

You can actually prove this by induction

Works for $N = 1$

\begin{eqnarray} D_1 &=& a_0b_0 + (a_0b_1 + a_1 b_0)(x - c) \\ &=& a_0 [b_0 + b_1(x - c)] + a_1[b_0(x - c)] \\ &=& a_0 B_1 + a_1 B_0 \end{eqnarray}

Assume that it works for $N - 1$

$$ \sum_{m = 0}^{N - 1}\sum_{j + k = m}a_j b_k(x - c)^m = a_0 B_{N - 1} + a_1(x - c) B_{N - 2} + \cdots + a_{N-1}(x - c)^{N-1}B_0 \tag{1} $$

Now let's prove it for $N$

\begin{eqnarray} \sum_{m = 0}^{N}\sum_{j + k = m}a_j b_k(x - c)^m &=& \sum_{m = 0}^{N - 1} \sum_{j + k = m}a_j b_k(x - c)^m + \sum_{j + k = N}a_j b_k(x - c)^N \\ &\stackrel{(1)}{=}& a_0 B_{N - 1} + a_1(x - c) B_{N - 2} + \cdots + a_{N-1}(x - c)^{N-1}B_0 \\ && + [a_0 b_N + a_1 b_{N - 1} + \cdots + a_N b_0](x - c)^N \\ &=& a_0 \left[ B_{N - 1} + b_N(x - c)^N \right] + a_1(x - c) \left[ B_{N - 2} + b_{N - 1}(x - c)^{N - 1} \right] \\ && + \cdots + a_{N}(x - c)^{N}[b_0] \\ &=&a_0 B_N + a_1(x - c) B_{N - 1} + \cdots + a_{N}(x - c)^{N}B_0 \tag{2} \end{eqnarray}