Consider two real-valued real analytic functions $f$ and $g$. I want to prove that there exists a greatest common divisor $d$, which is a real analytic function. By greatest common divisor, I mean the following: Common divisor: There exist real analytic functions $q_1, q_2$ such that $f = dq_1, g = dq_2$, and Greateast: If […]

Let $f$ be a holomorphic function on the punctured unit disk. If the imaginary part of $f$ is bounded, is it true that $f$ has a removable singularity at 0? I see that $|e^{-if}|=e^{Im\;f}$ so $e^{-if}$ is a bounded holomorphic function on the punctured unit disk and it follows that $e^{-if}$ has a removable singularity […]

I want to verify the proof of this result and get some start ideas to overcome the different steps of this proof. Lemma: Let $g$ be a real analytic function. Then we have the equivalence $((a)∧(b))⇔(c)$, where the statements $(a),(b)$ and $(c)$ are given by: (a) $g$ has infinitely many real zeros and the set […]

This question already has an answer here: Show that if $f(z)$ is a continuous function on a domain $D$ such that $f(z)^N$ is analytic for some integer $N$, then $f(z)$ is analytic on $D$. 1 answer

One could prove the following theorem in the smooth setting: Theorem Let $(M,m)$ be a $d$ dimensional $C^\infty$ manifold with smooth volume $m$. Let $\{F_i\}_{i=1}^k$ and $\{G_i\}_{i=1}^k$ be two systems of disjoint open subsets, satisfying $m(F_i)=m(G_i)$. Assume $\bar{F}_i$ and $\bar{G}_i$ are diffeomorphic to the $d$ dimensional closed ball in $\mathbb{R}^m$ for $i=1,\ldots k$. Then, given […]

This problem is taken from Section VIII.4 of Theodore Gamelin’s Complex Analysis: Let $f(z)$ be an analytic function on the open unit disk $\mathbb{D}=\{|z|<1\}$. Suppose there is an annulus $U = \{r<|z|<1\}$ such that the restriction of $f(z)$ to $U$ is one-to-one. Show that $f(z)$ is one-to-one on $\mathbb{D}$. Any hints?

Does there exist an infinitely differentiable function $f:U\to\mathbb{R}$, where $U$ is open subset of $\mathbb{R}$, such that the Taylor series of $f$ at $x=x_0\in U$ has radius of convergence $R>0$ $f$ equals its Taylor series only on the subinterval $(x_0-r,x_0+r)$, where $\color{red}{0<}r<R$ The customary examples of smooth real functions that fail to be analytic, e.g. […]

I am reviewing Calc $2$ material and I came across a problem which asked me to explain why $x^\pi$ does not have a Taylor Series expansion around $x=0$. To me it seems that it would have an expansion but it would just be $0$, so maybe it’s not a suitable expansion. It doesn’t have any […]

As evident $f_n=\frac{1}{n!}\frac{d^n}{dx^n}g(x)(at x=0)$.If I use Cauchy’s integral formula to find the $nth$ derivative,then I’m stuck,because there also the derivative crops up while finding the residue.

I have a question, how can we prove that a function, here specificaly the function $\frac{1}{1+x^2}$ is analytic? I know we must show that for any $x_0$ in $\mathbb R$, the series $\sum_{k=0}^{\infty}\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k$ has a convergent radius greater than zero, but how to show that? I appreciate your solutions.

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