James Arathoon
• Member for 7 years
• Last seen more than a month ago
• Ceredigion, United Kingdom
Stats
3,659
reputation
195
reached
1
2
questions
Communities
View all

Control and Automation Engineer

Main Mathematical Interests:

Integration, Infinite Series and Elementary Number Theory.

One particular interest is finding new Infinite Series for Apéry's Constant and other Zeta Constants e.g. this especially strange one I found for Apéry's Constant last year (2016),

$$\zeta(3)=$$ $$\frac{1}{2}\frac{8}{7} \left(\frac{\pi}{2}\right)^3 \left(\frac{\pi}{4}\right)^3 \left( b_{1} \lvert B_{4}\rvert + b_{2} \lvert B_{6} \rvert \left(\frac{\pi}{4}\right)^2 + b_{3} \lvert B_{8} \rvert \left(\frac{\pi}{4}\right)^4+ \; ... \right)$$

where $$b_k=\left( \frac{2^{2k+2}\left( 2^{2k+2}-1\right)\left( 2^{2k+1}-\left( k+2\right)\right)}{\left( 2k+2\right)! \; \left( k+1\right)\left( k+2\right)} \right)$$ and where $$\lvert B_{2k+2}\rvert =\left(-1 \right)^k \; B_{2k+2}$$ are unsigned Bernoulli Numbers.

Other simpler slowly converging series I have found for $$\lambda(3)$$ are: $$\lambda(3)=\frac{\pi^2}{4}\sum_{k=1}^{\infty} \frac{\beta(2k-1)}{(2k-1)(2k+1)}$$ and $$\lambda(3)=\frac{\pi^2}{4}\sum_{k=1}^{\infty} \frac{\lambda(2k)}{(k+1)(2k+1)}$$ where $$\lambda(n)=\sum_{k=1}^{\infty} \frac{1}{(2k-1)^n}$$ and $$\beta(n)=\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{(2k-1)^n}$$

The analogous series for $$\lambda(5)$$ are $$\lambda(5)=\frac{\pi^4}{3\times2^4}\sum_{k=1}^{\infty} \frac{\beta(2k-1)(2k+4)}{(2k-1)(2k+1)(2k+3)}$$ and $$\lambda(5)=\frac{\pi^4}{4!}\sum_{k=1}^{\infty} \frac{\lambda(2k)(2k+7)}{(2k+1)(2k+3)(2k+4)}$$

Also interested in learning more about the relation between Maths and Physics.

This user doesn’t have any gold badges yet.
7
6
Top tags
0
Score
1
Posts
33
Posts %
0
Score
1
Posts
33
Posts %
0
Score
1
Posts
33
Posts %
-25
Score
1
Posts
33
Posts %
-25
Score
1
Posts
33
Posts %
-25
Score
1
Posts
33
Posts %
Top posts
-25
Jan 30, 2022
Top network posts
View all network posts