Logarithm real analytic on $(0,\infty)$

Question

I am working on a Tao Analysis II question. I have to prove that $log$ the inverse function of $exp$ is real analytic on $(0,\infty)$. I have already proven that $$ \forall x \in(-1,1): ln(1-x) = - \sum_{n=1}^\infty \frac{x^n}{n} $$ and that $$ \forall x \in (0,2): ln(x)= \sum_{n=1}^\infty \frac{(-1)^{n+1}}n (x-1)^n $$ Does this help ? Further i may not make use of complex numbers.

Answer 1

Suppose that $f'=g$ on an interval $(a,b)$, and $g$ is real analytic. For each $c\in(a,b)$, there is an $r>0$ and a sequence $(a_n)_n$ such that for all $x\in(c-r,c+r)$, $g(x)=\sum\limits_{n=0}^\infty a_n(x-c)^n$. The power series $h(x)=\sum\limits_{n=0}^\infty \frac{a_n}{n+1}(x-c)^{n+1}$ also converges for $x\in(c-r,c+r)$, and $h'=g$ in this interval (this requires justification). Therefore $f(x)=f(c)+h(x)=f(c)+\sum\limits_{n=0}^\infty \frac{a_n}{n+1}(x-c)^{n+1}$ for all $x\in(c-r,c+r)$ (e.g., this follows from the Mean Value Theorem). This shows that $f$ is real analytic.

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$g(x)=1/x$ is real analytic on $(0,\infty)$, because for each $c\neq 0$, $g(x)=\sum\limits_{n=0}^\infty\frac{(-1)^n}{c^{n+1}}(x-c)^n$ when $|x-c|<|c|$, by the geometric series identity. (More generally, you may have seen that a quotient of real analytic functions is real analytic away from zeros of the denominator.)

Answer 2

Since I re-worked this example and was supposed to give the formal Power-Series $\sum_n c_n(x-a)^n$ which converges to $\log(x)$ if $|x-a| < R$ for some $R > 0$ I give my answer here:

We know that $x \in (0,2)$ implies

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

Let $z \in (a-a,a+a) = (0,2a)$ i.e. $R = a$. Then we have that $z = xa$ for some $x \in (0,2)$. Thus $$ \log(z) = \log(xa) = \log(a) + \log(x) = \log(a) + \sum_{n=1}^\infty \frac {{-1}^{n+1}}{na^n}(ax-a)^n $$ which equals $$ \log(z) = \log(a) + \sum_{n=1}^\infty \frac {{-1}^{n+1}}{na^n}(z-a)^n $$

Setting $c_0 := \log(a)$ and for $n>0$ $c_n:= \frac {{-1}^{n+1}}{na^n}$ we have $$ \forall a \in (0,+\infty) \forall z \in (a-R,a+R): \log(z) = \sum_{n=0}^\infty c_n(z-a)^n $$ with $R > 0$ which proves that $\log$ is real analytic on $(0,+\infty)$.

Answer 3

We first establish that $\ln$ is real analytic at $x=1$ with radius of convergence $1$. We start from the geometric series $\forall x\in(-1,1), \sum_{n=0}^\infty x^n = \frac{1}{1-x}$

Integrate over $[0,x]$ on both sides, we obtain

$\forall x \in (-1,1), -\sum_{n=1}^\infty \frac{x^n }{ n} = \ln(1-x)$

This is equivalent to

$\forall x \in (0,2), \sum_{n=1}^\infty (-1)^{n+1}\frac{(x-1)^n }{ n} = \ln(x)$.

We then proceed to show that $\ln$ is real analytic over $\mathbf R$.

Let $b\in \mathbf R^+$ and $x \in (0,2)$.

Now $\ln (bx) = \ln b + \ln x = \ln b + \sum_{n=0}^\infty c_n (x-1)^n = \ln b + \sum_{n=0}^\infty c_n b^{-n} (bx-b)^n$

Define $d_n := \ln b + c_0$ for $n=0$ and $d_n := c_n b^{-n}$ for $n \geq 1$,

Note that $\ln(bx) = \sum_{n=0}^\infty d_n (bx - b)^n$. By definition, $\ln$ is real analytic at $b$ with radius of convergence $R := \frac{1}{\limsup_{n\to\infty} |d_n|^{1/n}} = \frac{1}{\limsup_{n\to\infty} |c_n b^{-n}|^{1/n}} = b$.

Since $b$ is arbitrary, we conclude that $\ln$ is real analytic.