# Textbook Proposition on Product of Real Analytic Functions

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.

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 $$B_{N-j}=g(x)-R_{N-j}$$.

• Amazing. Except that I believe that (3) should start with $\sum_{j=0}^N$ rather than $\sum_{m=0}^N$
– clay
Dec 5, 2018 at 22:02
• @clay: You're welcome. Typo corrected. Thanks. Dec 5, 2018 at 22:11

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 $$N$$

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

• Awesome! Thank you so much!
– clay
Dec 5, 2018 at 22:03