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.
Logarithm real analytic on $(0,\infty)$
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0I 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. – 2012-11-26
2 Answers
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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0@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 $k=f-h$...). – 2012-11-27
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 $(0,+\infty)$.