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I am currently working through some of the content in Murty and Rath's Transcendental Numbers, and in their section entitled "Some Applications of Baker's Theorem" they present the following excercise:

Suppose that the sum $$ F(z;x) = \sum_{n=1}^{\infty}\frac{z^n}{n+x} $$ converges. If $z$ is algebraic and $x$ is rational, show that the sum is either zero or transcendental.

Presumably I can prove this using Baker's Theorem, but I am honestly at a loss on how to approach this, and any of the other questions from that section for that matter. This isn't homework, so I'd really like a hand-holding walkthrough on how I'm suppose to tackle a problem like this.

Edit: Added new tag for visibility.

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  • $\begingroup$ The main idea is to represent the sum as an integral and prove that the integral is a linear combination of $1$ and logarithms of algebraic numbers. $\endgroup$ – Sungjin Kim May 26 '17 at 22:42
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I think the problems in the book has some missing assumptions, but their hints are great. For correct statements and the proofs, see this paper: http://www.sciencedirect.com/science/article/pii/S001935770180001X

The problem here with some extra assumptions, can be solved by the same idea.

Suppose that $z$ is algebraic, $0<|z|\leq 1$ and $x\in \mathbb{Q}$ with $0< x\leq 1$. We write $x=\frac ba$ with $a, b$ integers, $(a,b)=1$, $0< b\leq a$. Then if $f(z)$ converges, then $$f(z)=\sum_{n=1}^{\infty} \frac{z^{an+b}}{an+b}$$ is in $\mathbb{Q}(z)$ or it is transcendental.

Then the derivative $f'(z)$ when $|z|<1$. (The result is extended to the boundary $|z|=1$ can be done by Abel's theorem provided that the series $f(z)$ converges) is $$ f'(z)= \sum_{n=1}^{\infty} z^{an+b-1} = \frac{z^{b-1+a}}{1-z^a}=-z^{b-1}+\frac{z^{b-1}}{1-z^a}. $$ Thus, by integrating it after partial fraction: $$\frac{t^{b-1+a}}{1-t^a} = -t^{b-1}+ \frac{A_1}{1-\zeta_1 t} + \cdots + \frac{A_a}{1-\zeta_a t}$$ with $\zeta_1=\zeta=\exp(2\pi i /a)$, $\zeta_2, \ldots, \zeta_a$ are distinct roots of $1-t^a=0$, $A_i\in\mathbb{Q}(\zeta)$, we obtain that $$ f(z) =\int_0^z \frac{t^{b-1+a}}{1-t^a}dt = -\frac{z^b}b + B_1 \log(1-\zeta_1 z)+ \cdots + B_a \log(1-\zeta_a z) $$ for some $B_i\in \mathbb{Q}(\zeta)$.

Therefore, $f(z)+\frac{z^b}b$ is a nontrivial $\overline{\mathbb{Q}}$-linear combination of logarithm, which is by Baker's theorem, $0$ or transcendental. (Corollary 20.1 in Murty & Rath)

Hence, $f(z) = \frac{z^b}b$ or it is transcendental, the result follows.

However, I do not know how to prove that the sum is $0$ or transcendental.

Now, we can drop the condition $0<x<1$.

Suppose that $z$ is algebraic, $0<|z|\leq 1$ and $x\in \mathbb{Q}$. We write $x=\frac ba$ with $a, b$ integers, $(a,b)=1$. Then if $f(z)$ converges, then $$f(z)=\sum_{n=1}^{\infty} \frac{z^{an+b}}{an+b}$$ is in $\mathbb{Q}(z)$ or it is transcendental.

For any given $x\in\mathbb{Q}$, there is $k\in\mathbb{Z}$ such that $0< x+k \leq 1$. Then apply the first result.

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