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Principles Of Mathematical Analysis-Walter Rudin

  1. Jan 6, 2014 #1
    I am reading Principles Of Mathematical Analysis by Walter Rudin.
    Chapter 8, Equation 28 needs some clarification.

    We are taking a lim as h->0 on a series. Value of series is lim as n->INF on the squence of partial
    sums. So first we take lim as n->INF on a sequence of partial sums and then we are taking
    lim as h->0. But can we interchange the order of limits. If so result is immediate.

    But how to prove that we can change the order of limits? Is it that we have to use Thearem 7.11?

    Please help me.
     
  2. jcsd
  3. Jan 6, 2014 #2

    jbunniii

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    You are correct. You need to use theorem 7.11. Rudin likes to leave gaps like this for the reader to fill in, which can be annoying sometimes. Here is a detailed explanation. I assume you are OK with the first equality:
    $$\lim_{h \rightarrow 0} \frac{E(z+h) - E(z)}{h} = E(z) \lim_{h \rightarrow 0} \frac{E(h) - 1}{h}$$
    As Rudin indicates, this follows directly from ##E(z+h) = E(z)E(h)## and doesn't require interchanging any limits. So consider this limit
    $$\lim_{h \rightarrow 0} \frac{E(h) - 1}{h}$$
    By definition, we have
    $$E(h) = \sum_{n=0}^{\infty}\frac{h^n}{n!}$$
    and the ratio test shows that this series converges absolutely for all ##h##. Therefore
    $$\begin{align}
    \frac{E(h) - 1}{h} &= \frac{1}{h} \left(\sum_{n=0}^{\infty} \frac{h^n}{n!} - 1\right) \\
    &= \frac{1}{h}\sum_{n=1}^{\infty} \frac{h^n}{n!}\\
    &= \sum_{n=0}^{\infty} \frac{h^n}{(n+1)!} \\
    &= 1 + \sum_{n=1}^{\infty} \frac{h^n}{(n+1)!}
    \end{align}$$
    Let
    $$S(h) = \sum_{n=1}^{\infty} \frac{h^n}{(n+1)!}$$
    We're interested in what happens as ##h \rightarrow 0##, so we may restrict our attention to a neighborhood of zero, say ##B_{\epsilon}(0)##. Since we are dealing with a power series, the convergence is uniform on ##B_{\epsilon}(0)##, by theorem 8.1. [Technically you need to reprove theorem 8.1 for complex power series, but the proof is exactly the same.] Thus ##S## is continuous on ##B_{\epsilon}(0)##, by theorem 7.11. Therefore
    $$\lim_{h \rightarrow 0}S(h) = S(0)$$ and the result follows.

    [edit]: Actually you can dispense with theorem 7.11, because his theorem 8.1 includes the result that ##S## is continuous within the radius of convergence.
     
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