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I am learning about real analytic functions on my own right now. I've been having trouble with one of the exercises, and it isn't much help that most of the resources online for analytic functions are for Complex Analysis. I am talking about real analytic functions.

For reference, here is a definition that I have been using:

A real function $f(x)$ is analytic at $x_{0}$ if there is a $r > 0$:

$$f(x) = \sum_{n = 0}^{\infty} a_{n}(x - x_{0})^{n}, |x - x_{0}| < r$$

i.e. there is some power series which converges to the function.

Using this definition, I want to solve the following exercise problem:

Find a function $f$ analytic at $x_{0} = 0$ so that $f\left(\frac{1}{n}\right) = \frac{n}{n + 1}$, $n = 1, 2, \ldots$.

Show that such a function cannot be analytic on $(-2, 0)$.

So, working backwards, I found out that $\frac{1}{1 + x}$ satisfies the property $f(\frac{1}{n}) = \frac{n}{n + 1}$. I'm really not so sure how to prove the analytic properties though. I think that now I need to show $\frac{1}{1 + x}$ is analytic, and then I need to prove the second part of the claim, which is that such a function cannot be analytic on $(-2, 0)$.

I have an example in my book which shows $1 + x + x^{2} + x^{3} + \cdots$ is analytic. Here's how they do it:

A prototypical example is the geometric series $$1 + x + x^{2} + \cdots = \lim_{n\to\infty} 1 + x + x^{2} + \ldots x^{n} = \lim_{n\to\infty}\frac{1 - x^{n + 1}}{1 - x}$$

for which it is well known equals $f(x) = \frac{1}{1 - x}$ for $|x| < 1$. To verify that the function is analytic, we need to expand about any point $x_{0} \neq 1$:

$$ \begin{align*} \frac{1}{1-x} = \frac{1}{1 - x_{0} - (x - x_{0})} = \frac{1}{1- x_{0}}\left(1 - \frac{x - x_{0}}{1 - x_{0}}\right)^{-1} \end{align*} $$

I tried outlining this example, but I couldn't make any progress. I would really appreciate some sort of help.

Vincent
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  • Your function is $f(z)=\frac1{1+z}$ for $z_0=0$. – xpaul Nov 12 '18 at 16:30
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    It seems the answer is already inside your own post! You know how to show that $g(x) = 1/(1-x)$ is analytic at $0$, now you only need to realize that $f(x) = 1/(1+x)$ can be related to this function $g$ by $f(x) = g(-x)$ – Vincent Nov 12 '18 at 16:33
  • @Vincent I don't see how it helps. I had also made that observation but couldn't make any progress. I also don't know how to show such a function cannot be analytic on (-2, 0). –  Nov 12 '18 at 16:38
  • @joseph What can you use? Can you use the Identity Theorem for analytic functions? – Batominovski Nov 12 '18 at 17:30
  • I cannot use it. I did not learn it in here. –  Nov 12 '18 at 17:31
  • Maybe I am missing something obvious, but I'd say that the power series expansion $f(x) = \sum_k (-x)^k$ coming from the geometric series proves that $f$ is analytic on whatever the domain of convergence of this series is. – Vincent Nov 13 '18 at 12:44
  • As for the second question, why is it not analytic on $(-2, 0)$, I guess it suffices to find one point $x_0$ in that interval where it is not analytic. My money would be on $x_0 = -1$. – Vincent Nov 13 '18 at 12:46
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    @Vincent I think it's because it's not defined at $x_{0} = -1$ –  Nov 13 '18 at 16:58
  • I don't think it's supposed to be as difficult as the current solution posted. It is classified as an "easy" problem in the book. –  Nov 13 '18 at 17:00
  • Hey @Vincent. In an earlier comment, you said I showed $1/(1 - x)$ is analytic at $0$, but I thought the proof showed that the function showed analyticity anywhere but for $0$ (because the function isn't defined at $0$). I think an easy way to solve this problem is to use the fact that $1/(1 - x)$ is analytic at $x = 1$ (I don't know how to show it, I'm not sure if my original post already shows it) and then the fact that composition of analytic functions is analytic with $f(x) = -x$ and be done. –  Nov 14 '18 at 02:35
  • Almost there... $1/(1-x)$ IS defined at $x = 0$: substituting $x = 0$ in the expression yields $1/1$ which is a perfectly well defined number (better known as simply $1$). On the other hand, it is not defined at $x = 1$. – Vincent Nov 14 '18 at 08:41
  • The main point is: the equality $1/(1-x) = \sum_{n=0}^\infty x^n$ shows that the left hand side is analytic whenever the right hand side converges, so in particular at $x = 0$. This is the definition of analyticity at $0$: you can think of the $x^n$ on the right hand side as $(x - 0)^n$, which is the special case $x_0 = 0$ of the expression $(x - x_0)^n$ appearing in the definition of analyticity you quote. Now that you know that $1/(1-x)$ is analytic, you can indeed use composition of functions to show that $1/(1+x = 1/(1 - (-x))$ is analytic as well. – Vincent Nov 14 '18 at 08:45

1 Answers1

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This is a solution without the Identity Theorem. It is a bit lengthy, but I have no other tricks. Combinatorial identities I use in my proof can be proven using the technique from here.

Note that $$f(x)=\dfrac{1}{1+x}\tag{*}$$ for all $x$ such that $\dfrac{1}{x}$ is a positive integer. We also see that $$f(0)=\lim_{n\to\infty}\,f\left(\dfrac1n\right)=1\,.$$ This means (*) is true when $x=0$ too.

We can try to find $f^{(k)}(0)$ for $k=1,2,3,\ldots$ by noting that $$f^{(k)}(0)=\lim_{h\to 0^+}\,\frac{1}{h^k}\,\sum_{r=0}^k\,\binom{k}{r}\,(-1)^r\,f\big((k-r)h\big)\,.$$ Taking $h:=\dfrac{1}{m\cdot k!}$ for some positive integer $m$, we have $$f^{(k)}(0)=\lim_{m\to \infty}\,(m\cdot k!)^k\,\sum_{r=0}^k\,\binom{k}{r}\,(-1)^r\,f\left(\frac{k-r}{m\cdot k!}\right)\,.$$ Since $\dfrac{m\cdot k!}{k-r}$ is a positive integer for all $r=0,1,2,\ldots,k-1$, and $f(0)=1$, we get $$f^{(k)}(0)=\lim_{m\to \infty}\,(m\cdot k!)^k\,\sum_{r=0}^k\,\binom{k}{r}\,(-1)^r\,\left(\frac{1}{1+\frac{k-r}{m\cdot k!}}\right)\,.$$ Because $\sum\limits_{r=0}^k\,\binom{k}{r}\,(-1)^r\,(k-r)^t=0$ for $t=0,1,2,\ldots,k-1$, we get $$f^{(k)}(0)=\lim_{m\to \infty}\,(m\cdot k!)^k\,\sum_{r=0}^k\,\binom{k}{r}\,(-1)^r\,\left(\frac{1}{1+\frac{k-r}{m\cdot k!}}-\sum_{t=0}^{k-1}\,(-1)^t\,\left(\frac{k-r}{m\cdot k!}\right)^t\right)\,.$$ Using Taylor's Theorem, we have $$\frac{1}{1+\frac{k-r}{m\cdot k!}}-\sum_{t=0}^{k-1}\,(-1)^t\,\left(\frac{k-r}{m\cdot k!}\right)^t=(-1)^k\,\left(\frac{k-r}{m\cdot k!}\right)^k+\mathcal{O}\left(\frac{1}{m^{k+1}}\right)\,.$$ That is, $$f^{(k)}(0)=\lim_{m\to\infty}\,\left((-1)^k\,(m\cdot k!)^k\,\sum_{r=0}^k\,\binom{k}{r}\,(-1)^r\,\left(\frac{k-r}{m\cdot k!}\right)^k+\mathcal{O}\left(\frac{1}{m}\right)\right)\,.$$ Ergo, $$f^{(k)}(0)=(-1)^k\,\sum_{r=0}^k\,\binom{k}{r}\,(-1)^r\,\left(k-r\right)^k=(-1)^k\,k!\,.$$ Therefore, $$f(x)=\sum_{k=0}^\infty\,\frac{f^{(k)}(0)}{k!}\,x^k=\sum_{k=0}^\infty\,(-1)^k\,x^k=\frac{1}{1+x}$$ for all $x\in (-1,+1)$ because the radius of convergence is $1$. This proves the existence and the uniqueness of $f$.

To show that there is no analytic function with the given property that is defined on $(-2,0)$, you need to show that such a function $f$ cannot be defined at $-1$. One way to do this is noting that $x=-1$ is a pole of $\dfrac{1}{1+x}$. Therefore, $x=-1$ is a natural boundary of $f(x)$.

Batominovski
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