8
votes
4answers
213 views

Real-analytic $f(z)=f(\sqrt z) + f(-\sqrt z)$?

Are there nonconstant real-analytic functions $f(z)$ such that $$ f(z)=f(\sqrt z) + f(-\sqrt z)$$ is satisfied near the real line ? Also can such functions be entire ? And/Or can they be periodic ...
1
vote
1answer
189 views

Solve the functional equation $\,\,f(2x)=2x f^\prime(x)$

If $f(x)$ is a real analytic functions on $\mathbb R$, and $$2xf'(x)=f(2x),$$ then find $f(x)$. My idea: express $f$ as: $$f(x)=\sum_{n=0}^{\infty}a_{n}x^n.$$ Thank you
27
votes
2answers
645 views

How prove there is no continuous functions $f:[0,1]\to \mathbb R$, such that $f(x)+f(x^2)=x$.

Prove that there is no continuous functions $f:[0,1]\to R$, such that $$ f(x)+f(x^2)=x. $$ My try. Assume that there is a continuous function with this property. Thus, for any $n\ge 1$ and all ...
3
votes
1answer
240 views

How to identify this power series as $k\sin(k/x)$?

In this question, a functional equation is solved for functions with a power series. We find a recursive formula: (copied from the answer by user achille hui) \begin{align} ( 2^1 - 3 ) a_2 &= 0\\ ...
1
vote
0answers
52 views

A functional power series equation

I am interested in solving the following functional equation: $$(1-w-zw)F(z,w)+zw^nF(z,w^2)=z-zw\,.$$ Here $n\geq2$ is a fixed integer and $F(z,w)$ is a power series in two variables with complex ...
8
votes
1answer
618 views

Equation about generating functions and subfactorial

Suppose $G_n(w)$ is a formal power series (really a probability generating function, see the following explanation) of variable $w$, try to solve out $G_n(w)$ for all $n\ge0$ from the ...
8
votes
3answers
599 views

How to calculate $f(x)$ in $f(f(x)) = e^x$?

How would I calculate the power series of $f(x)$ if $f(f(x)) = e^x$? Is there a faster-converging method than power series for fractional iteration/functional square roots?
0
votes
2answers
131 views

How to solve the following system?

I need to find the function c(k), knowing that $$\sum_{k=0}^{\infty} \frac{c(k)}{k!}=1$$ $$\sum_{k=0}^{\infty} \frac{c(2k)}{(2k)!}=0$$ $$\sum_{k=0}^{\infty} \frac{c(2k+1)}{(2k+1)!}=1$$ ...