How did Rudin change the order of the double sum $\sum_{n=0}^\infty c_n\sum_{m=0}^n\binom nma^{n-m}(x-a)^m$?

Question

I see many people change the order of sum but I don't understand how they did that.

Is there is a way to change the order of the sum, $\sum\limits_{k=a}^n\sum\limits_{j=b}^m X_{j,k}$ and $\sum\limits_{k=a}^n\sum\limits_{j=b}^k X_{j,k}$?

An example of this is Theorem 8.4 from Principles of Mathematical Analysis.

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Theorem 8.4: Suppose $$f(x)=\sum_{n=0}^\infty c_n x^n$$ the series converging in $|x|<R$. If $-R<a<R$, then f can be expanded in a power series about the point $x=a$ which converges in $|x-a|<R-|a|$, and $$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x-a)^n\ \ \ (|x-a|<R-|a|)$$

Proof we have $$f(x)=\sum_{n=0}^\infty c_n[(x-a)+a]^n$$ $$=\sum_{n=0}^\infty c_n\sum_{m=0}^n\binom nma^{n-m}(x-a)^m$$ $$\color{red}{=\sum_{m=0}^{\infty}\Big{[}\sum_{n=m}^\infty\binom nmc_na^{n-m}\Big{]}(x-a)^m}$$

Answer 1

You should think about this as analogous to changing the order of integration in a double integral. Sketch the region $0\le y<\infty$, $0\le x\le y$. It is an infinite triangle, if you like, between the line $x=y$ and the $y$-axis. You can equally well write it as $0\le x<\infty$, $x\le y<\infty$.

The same thing works with the region defined with integer coordinates. $\{0\le n<\infty, 0\le m\le n\}$ is precisely the same set as $\{0\le m<\infty, m\le n<\infty\}$.

Answer 2

If you're looking for the intuition as to how to find the alternative representation of the first two sums, then you're looking for a double counting argument: Make a table where the cell in the i-th row and j-th column is $X_{i,j}$. Now you get the first representation by summing up the table elements row after row, and the second representation by summing up the table elements column after column.

If you're looking for proofs, then there are many ways. One can e.g. try to turn the above into a rigorous proof or reduce the problem to set notation.

For your second example, take a look at this question, which is a slightly simpler variant.

If you in the same vein try to reduce your first example to set notation, you will find that both $\sum\limits_{k=a}^n\sum\limits_{j=b}^m X_{j,k}$ and $\sum_{j=b}^m \sum_{k=a}^n X_{k,j}$ reduce to the sum over the elements of the set $\{(j,k)\in\mathbb N^2\mid b\le j\le m \land k\le a \le n\}$.

Answer 3

It is enough to prove that $$ \sum_{k=0}^n \sum_{m=0}^k f(k,m)=\sum_{m=0}^n \sum_{k=m}^n f(k,m) $$ to show that firstly we assume $f(k,m)=0 \text{ for } m>k$ and using the fact $$ \sum_{k=0}^a f(k)=\sum_{k=0}^b f(k)-\sum_{k=a+1}^{b} f(k) $$ So $$ \sum_{m=0}^k f(k,m)=\sum_{m=0}^n f(k,m)-\sum_{m=k+1}^n f(k,m)=\sum_{m=0}^n f(k,m)$$ then $$ \sum_{k=0}^n \sum_{m=0}^k f(k,m)=\sum_{k=0}^n \sum_{m=0}^n f(k,m)=\sum_{m=0}^n \sum_{k=0}^n f(k,m) $$ now we have $$ \sum_{k=0}^n f(k,m)=\sum_{k=m}^n f(k,m)-\sum_{k=0}^{m-1} f(k,m)=\sum_{k=m}^n f(k,m) $$ finally we get (with vanished the function $f(n,m) \text{ for } m>n$ ) $$ \sum_{k=0}^n \sum_{m=0}^k f(k,m)=\sum_{m=0}^n \sum_{k=m}^n f(k,m) $$

Answer 4

Just think that

\begin{align*}\sum_{k=a}^n \sum_{j=b}^m X_{k,j}&=[X_{a,b}+X_{a,b+1}+...+X_{a,m}]+...+[X_{n,b}+X_{n,b+1}+...+X_{n,m}]\\&=\sum_{j=b}^m \sum_{k=a}^n X_{k,j}\end{align*}