Ytukyg

Let
begin{align*}
sumlimits_{j=0}^infty a_j (x-c)^j && sumlimits_{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) &= sumlimits_{m=0}^infty sumlimits_{j+k=m} (a_j cdot b_k) (x-c)^m \
end{align*}

Proof: Let
begin{align*}
A_N &= sumlimits_{j=0}^N a_j (x-c)^j & B_N &= sumlimits_{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 &= sumlimits_{m=0}^N sumlimits_{j+k=m} (a_j cdot b_k) (x-c)^m &
R_N &= sumlimits_{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}^Nleft(sum_{{j+k=m}atop{j,kgeq 0}} a_jb_kright)(x-c)^m}\
&=sum_{m=0}^Nleft(sum_{j=0}^m a_jb_{m-j}right)(x-c)^mtag{1}\
&=sum_{0leq jleq mleq N} a_jb_{m-j}(x-c)^mtag{2}\
&=sum_{j=0}^Nsum_{m=j}^N a_jb_{m-j}(x-c)^mtag{3}\
&=sum_{j=0}^Na_jsum_{m=0}^{N-j}b_m(x-c)^{m+j}tag{4}\
&=sum_{j=0}^N a_j(x-c)^jsum_{m=0}^{N-j}b_m(x-c)^mtag{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 $kto m-j$.




• 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 $0leq jleq N$
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 $B_{N-j}=g(x)-R_{N-j}$.

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}