Suppose we have a function give by a power series \begin{align} f(x)= \sum_{n=0}^\infty (-1)^n a_n x^{2n}, \text{ for } x\in \mathbb{R} \end{align} where $a_n>0$.

My question is when does the following ratio \begin{align} \frac{f(bx)}{f(x)}= \frac{\sum_{n=0}^\infty (-1)^n a_n (bx)^{2n}}{\sum_{n=0}^\infty (-1)^n a_n x^{2n}}, \text{ where } b>1 \end{align}

no longer continuous at some $x$. Specifically, can we find conditions on coefficients $a_n$ and constant $b$ that would give necessary condition for continuity.

My Approach . My idea was to express $g(x)=\frac{f(ax)}{f(x)}$ as a power series and show that the radius of convergense is finite.

So, if

\begin{align} g(x)=\frac{f(ax)}{f(x)}=\frac{\sum_{n=0}^\infty (-1)^n a_n (bx)^{2n}}{\sum_{n=0}^\infty (-1)^n a_n x^{2n}}= \sum_{i=0}^\infty d_i x^{2n} \end{align}

Then we have that \begin{align} \sum_{n=0}^\infty (-1)^n a_n (bx)^{2n} = \sum_{n=0}^\infty (-1)^n a_n x^{2n} \sum_{i=0}^\infty d_i x^{2n} = \sum_{n=0}^\infty \left( \sum_{i=0}^n (-1)^i a_i d_{n-i} \right) x^{2n} \end{align} Comparing coefficients we have that \begin{align} (-1)^n a_n b^{2n}=\sum_{i=0}^n (-1)^i a_i d_{n-i}. \end{align} Ideal one would like to solve for $d_i$'s from here. Unfortuanatly, I am don't know how to do it an easy way, since we have to do some kind of discrete deconvolution.

My next step was to use ratio test on $d_n$'s and find radius of convergence, that is \begin{align} R=\lim_{n \to \infty} \left|\frac{d_n}{d_{n+1}}\right| \end{align} Perhaps there is a techniques that can solve for $R$ without explicitly solving for every $d_n$.


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