Show that $1 /(1-x)$ is real analytic

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 $$c_n(a)$$?

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

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 $$R = |1 - a|$$).

• $f(x) = \sum_{n = 0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n$ 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.) Commented Mar 28, 2020 at 1:21
• @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 $a$. Commented Mar 28, 2020 at 1:24

$$\forall a \in \mathbb{R} \setminus \{1\},$$ 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.

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.