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Let $f: \mathbb{R} \setminus \{1\} \to \mathbb{R}$ defined by $f(x) : = 1 /(1-x)$. Show that this function is real analytic on all of $\mathbb{R} \setminus \{1\}$.

Real analyic functions: Let $E$ be a subset of $\mathbb{R}$, and let $f: E \to \mathbb{R}$ be a function. If $a$ is an interior point of $E$, we say that $f$ is real analytic at $a$ if there exists an open interval $(a-r, a+r)$ in $E$ for some $r>0$ such that there exists a power series $\sum_{n=0}^\infty c_n(x-a)^n$ centered at $a$ which has a radius of convergence greater than or equal to $r$, and which converges to $f$ on $(a-r, a+r)$.

The author show that $f$ is real analytic at $2$ because we have a power series $\sum_{n=0}^\infty (-1)^{n+1} (x-2)^n$ which converges to $\frac{-1}{1-(-(x-2))} = \frac{1}{1-x} = f(x)$ on the interval $(1, 3)$.

Thus, in order for $f$ to be real analytic on all of $\mathbb{R} \setminus \{1\}$, I need to find $c_n(a)$ such that $\sum_{n=0}^\infty c_n(a)(x-a)^n = \frac1{1-x}$ for every $a \in \mathbb{R}\setminus \{1\}$. How can I find such $c_n(a)$?

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  • $\begingroup$ en.wikipedia.org/wiki/… ? $\endgroup$
    – Sera Gunn
    Commented Mar 28, 2020 at 0:59
  • $\begingroup$ @TrevorGunn In my book, a real analytic function can be represented as a taylor expansion. So, I think that it circular if I define $c_n(a) = f^{(n)}(a)/n!$. $\endgroup$
    – shk910
    Commented Mar 28, 2020 at 1:11
  • $\begingroup$ Why is it circular? You definite it that way and then prove the series converges. $\endgroup$
    – Sera Gunn
    Commented Mar 28, 2020 at 1:12

3 Answers 3

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The Taylor series of an analytic function is

$$ \sum_{n = 0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n. $$

Here $f'(x) = (1 - x)^{-2}$ and $f''(x) = 2(1 - x)^{-3}$ and $f'''(x) = 3!(1 - x)^{-4}$ and so on. By induction, we have $f^{(n)}(x) = \frac{n!}{(1 - x)^{n + 1}}$. This gives us a Taylor series of

$$ \sum_{n = 0}^{\infty} \frac{1}{(1 - a)^{n + 1}} (x - a)^n. $$

This is a geometric series so you should be easily able to confirm it converges to $\frac{1}{1 - x}$ and that the radius of convergence is positive (namely $R = |1 - a|$).

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  • $\begingroup$ $f(x) = \sum_{n = 0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n$ if $f$ is real analytic. But, we don't know yet that $f$ is real analytic, but can we use this argument? (I might be wrong, and I am just confused.) $\endgroup$
    – shk910
    Commented Mar 28, 2020 at 1:21
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    $\begingroup$ @shk910 It's not that I'm saying $f(x) = \sum_{n = 0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n$ it's that if $f(x)$ were to equal a power series, that's the power series it must be equal to. So we start with that series and then show that it converges to $f(x)$ in some neighbourhood of $a$. $\endgroup$
    – Sera Gunn
    Commented Mar 28, 2020 at 1:24
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$\forall a \in \mathbb{R} \setminus \{1\},$ let $d = 1 / (1 - a).$

Then $$f(x) = 1 / (1 - x) = \frac{d}{1-d(x-a)}$$

$$ = \sum_{n = 0}^{\infty} d(d(x - a))^n = \sum_{n = 0}^{\infty} d^{n + 1}(x - a)^n, \;\;\;\;\forall x \in (a - 1/d, a + 1/d),$$ by the root test.

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If $f(x)$ is real analytic at $x=a$ then $f$ possesses derivatives of all orders at $x=a$. For the representation $f(x)=\sum_{n=0}^\infty c_n (x-a)^n$ the coefficient $c_n$ is essentially $f^n(a)/n!$ i.e. the Taylor's coefficient.

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