All Questions
Tagged with products real-analysis
7 questions
0
votes
1
answer
266
views
On a type of sequence of integrals inspired in the Borwein integral and an integral due to Furdui
This post is inspired in the Borwein integral, and in a problem proposed by Ovidiu Furdui in Crux Mathematicorum, that is the Problem 3707, in page 151.
I've considered integrals of the form $$\int_0^...
5
votes
2
answers
441
views
Intuitive explanation of regularized products
I've come across some regularized product during study of zeta regularization .
We can prove various results like :
$ \infty != \prod_{k=1}^\infty k = \sqrt{2\pi} $
I also know the proof using $\...
user145059
-2
votes
1
answer
214
views
About infinite products and Euler Gamma functions [closed]
I am interested in knowing how to calculate infinite products like (or reading any reference about it):
$$\prod_{j=1}^{\infty}\left( 1-\left( \frac{x}{a+j\pi} \right) ^2 \right)$$
Inserting it into ...
10
votes
2
answers
1k
views
Why is $\sup f_- (n) \inf f_+ (m) = \frac{5}{4} $?
This question is an old question from mathstackexchange.
Let $f_- (n) = \Pi_{i=0}^n ( \sin(i) - \frac{5}{4}) $
And let
$ f_+(m) = \Pi_{i=0}^m ( \sin(i) + \frac{5}{4} ) $
It appears that we have
$$\sup ...
- 769
2
votes
0
answers
1k
views
Is there an infinite product like this for $\cos x$?
There are infinite products of iterated square roots for $\log x$ and $\arccos x$ as functions of $x$. For example
$$\log x = \frac{x - 1}{\sqrt{x}\sqrt{\frac{1}{2} + \frac{1}{2}\left ( \frac{1 + x}{...
2
votes
3
answers
260
views
Nonzero solutions of an infinite product
Let $-\frac{1}{2}\le a \le\frac{1}{2}$ and $b\in[0,\infty)$.
Definitions: $$f_k(a;b):=\frac{(2k+\frac{1}{2}+a)^2+b}{(2k+\frac{1}{2}-a)^2+b}(\frac{k}{k+1})^{2a},$$
$$f(a;b):=\prod\limits_{k=1}^\infty ...
- 293
3
votes
0
answers
511
views
Two (strictly related) proofs by induction of inequalities
This is a question I originally asked on MSE, receiving no answer, even with a bounty (which expired) on it. Therefore I am crosslinking in order to prevent duplication of effort: see here for the ...
