James Arathoon's user avatar
James Arathoon's user avatar
James Arathoon's user avatar
James Arathoon
  • Member for 6 years, 3 months
  • Last seen this week
  • Ceredigion, United Kingdom

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.

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