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# The Zeta Function: Finding a general formula Part 1

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.

Note by Bogdan Simeonov
2 years, 9 months ago

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In 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? · 1 year, 9 months ago

Oh, sorry about that.Also it's weird that people still check out this post, it's like a year old :D · 1 year, 9 months ago

Really interesting. Now how can we necessarily states that $$\frac{\sin x}{x}=\left(1-\frac{x}\pi\right)\left(1+\frac{x}\pi\right)\left(1-\frac{x}{2\pi}\right)\left(1+\frac{x}{2\pi}\right)\dots$$? · 2 years, 9 months ago