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Questions on advanced topics - beyond those in typical introductory courses: higher degree algebraic number and function fields, Diophantine equations, geometry of numbers / lattices, quadratic forms, discontinuous groups and automorphic forms, Diophantine approximation, transcendental numbers, elliptic curves and arithmetic algebraic geometry, exponential and character sums, Zeta and L-functions, multiplicative and additive number theory, etc.

0 votes
1 answer
121 views

How to find derivatives such as $\frac{\partial }{\partial p} \int_0^1 \frac{1-x^p}{1-x}dx$?

I've been doing some work in analytic number theory, and I ended up deriving this expression:$$\frac{\partial }{\partial p} \int_0^1 \frac{1-x^p}{1-x}dx$$ This looks insanely difficult. I plugged it i …
Alexandra's user avatar
  • 499
2 votes
0 answers
78 views

Cases where transcendental numbers can add up to a rational number? [closed]

Other than sums like $π + (1 - π)$, obviously. Can two transcendental numbers add up to a rational number? Or how about an infinite series of them?
Alexandra's user avatar
  • 499
3 votes
0 answers
112 views

How to tell if an infinite series sum will be rational or irrational?

I plugged in the following series into a calculator: $$\sum_{n=1}^\infty \ln(1+\frac{1}{n^2})$$ and got a result of approximately $1.29686$. That's nice and all, but I want to know: is this result irr …
Alexandra's user avatar
  • 499
0 votes
0 answers
45 views

Analytic continuation of product operator? [duplicate]

Everybody knows that $\Gamma(z) = (z-1)!$. Seeing as factorials can be analytically continued to all complex numbers (with a few exceptions), can a similar thing be said for product operators such as …
Alexandra's user avatar
  • 499
0 votes
1 answer
97 views

Can the infinite product of the powers of a transcendental number ever be transcendental?

I plugged the following product into a calculator: $$\prod_{n=1}^\infty e^{\frac{1}{n^2}}$$ and got a result of roughly 5.1806683. I would like to say that this result is transcendental, as its only …
Alexandra's user avatar
  • 499