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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.

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You know how to show that $\log$ is differentiable with derivative $\frac1{x}$? You know how to show that $\frac1{x}$ is real analytic? You know how to show that an antiderivative of a real analytic function is real analytic? – Jonas Meyer Nov 26 '12 at 14:59
1) Yes. 2) Should be possible. 3) No :D But i will see if i can figure out your last point. – Epsilon Nov 26 '12 at 15:03
But this should be possible by some other method, since Tao has not mentioned your last statement earlier in the text. – Epsilon Nov 26 '12 at 15:11
André: True, it is possible by other methods, but have you seen term by term differentiation of power series? – Jonas Meyer Nov 26 '12 at 15:14
I know the following: Given some real analytic function i know that the k-th derivative of that function is in terms of an real analytic function. – Epsilon Nov 26 '12 at 15:18
up vote 1 down vote accepted

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.

$g(x)=1/x$ 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.)

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Thank you very much :) – Epsilon Nov 27 '12 at 9:20
One question. You say by the mean value theorem. But then i wonder if this works, too: We assume $f' = g$ on $(a,b)$. Then $\frac{f(x)-f(c)}{x-c} = f'(z) = g(z)$ for some $z \in (c,x) \subset (c-r,c+r)$ if $x > c$ so that $f(x) = (x-c)g(z)+f(c)$ and thus is $f$ real analytic because $g$ is real analytic and $x-c$ is a constant. Is this true ? – Epsilon Nov 27 '12 at 10:29
@André: I don't have time to read or think about your comment, so I'll just briefly state that MVT implies that if $k'=0$ on an interval, then $k$ is constant, and as a consequence, if $f'=h'$ on an interval, then $f$ and $h$ differ by a constant (consider $k=f-h$...). – Jonas Meyer Nov 27 '12 at 16:51

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 $(0,+\infty)$.

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