I will be writing about the zeta function and finding a general formula for \(s=2n\).All you need is a bit of knowledge in Calculus and Trigonometry.

Now let's get started.Remember the Taylor series for \(sin x\)?

\(\sin x=x- \frac{x^3}{3!}+\frac{x^5}{5!}-...\)

Now let's divide both sides by x.

\(\frac{\sin x}{x}=1- \frac{x^2}{3!}+\frac{x^4}{5!}-...\)

The zeros of the sine function occur when \(x=k\cdot\pi\), where k is an integer. So \(\frac{\sin x}{x}=(1-\frac{x}{\pi})(1+\frac{x}{\pi})(1-\frac{x}{2\pi})(1+\frac{x}{2\pi})...=(1-\frac{x^2}{\pi^2})(1-\frac{x^2}{2^2\pi^2})...\)

Now let's substitute \(x=\pi\cdot s\)

Then \(\frac{\sin \pi.s}{\pi.s}=(1-s^2)(1-\frac{s^2}{2^2})...=\displaystyle\prod_{k=1}^{\infty} (1-\frac{s^2}{k^2})\)

Nobody likes to deal with infinite products, but if we take the logarithm of a product, it becomes a sum!

\(\ln\frac{\sin \pi.s}{\pi.s}=\ln (\displaystyle\prod_{k=1}^{\infty} (1-\frac{s^2}{k^2})) =\sum_{k=1}^{\infty} \ln(1-\frac{s^2}{k^2}) \)

That makes

\(\ln\sin \pi.s=\ln (\pi.s)+\displaystyle\sum_{k=1}^{\infty} \ln(1-\frac{s^2}{k^2}) \)

Now if we differentiate both sides with respect to s (I will not go through the derivation of this, just bash with the chain rule a bit), we will get

\(\pi.\frac{\cos \pi.s}{\sin \pi.s}=\frac{1}{s}+\displaystyle\sum_{k=1}^{\infty} \frac{1}{(1-\frac{s^2}{k^2})}.-\frac{2s}{k^2} \)

Now we multiply both sides by \(s\) and we get

\(\pi.s\frac{\cos \pi.s}{\sin \pi.s}=1+\displaystyle\sum_{k=1}^{\infty} \frac{1}{(1-\frac{s^2}{k^2})}.-\frac{2s^2}{k^2} \).

To be continued in Part 2.

## Comments

Sort by:

TopNewestIn the line before you talk about differentiation, shouldnt you take the natural log of the product instead of making it a product of natural logs? – Isaac Thomas · 2 years, 1 month ago

Log in to reply

– Bogdan Simeonov · 2 years, 1 month ago

Oh, sorry about that.Also it's weird that people still check out this post, it's like a year old :DLog in to reply

Really interesting. Now how can we necessarily states that \(\frac{\sin x}{x}=\left(1-\frac{x}\pi\right)\left(1+\frac{x}\pi\right)\left(1-\frac{x}{2\pi}\right)\left(1+\frac{x}{2\pi}\right)\dots\)? – Cody Johnson · 3 years, 1 month ago

Log in to reply

– Bogdan Simeonov · 3 years, 1 month ago

Well, I cannot exactly prove it.It's like a continuation of the Fundamental Theorem of Algebra, but for infinite polynomials.Log in to reply