The Zeta Function: Finding a general formula Part 1

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

Now let's get started.Remember the Taylor series for sinxsin x?

sinx=xx33!+x55!...\sin x=x- \frac{x^3}{3!}+\frac{x^5}{5!}-...

Now let's divide both sides by x.

sinxx=1x23!+x45!...\frac{\sin x}{x}=1- \frac{x^2}{3!}+\frac{x^4}{5!}-...

The zeros of the sine function occur when x=kπx=k\cdot\pi, where k is an integer. So sinxx=(1xπ)(1+xπ)(1x2π)(1+x2π)...=(1x2π2)(1x222π2)...\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=πsx=\pi\cdot s

Then sinπ.sπ.s=(1s2)(1s222)...=k=1(1s2k2)\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!

lnsinπ.sπ.s=ln(k=1(1s2k2))=k=1ln(1s2k2)\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

lnsinπ.s=ln(π.s)+k=1ln(1s2k2)\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

π.cosπ.ssinπ.s=1s+k=11(1s2k2).2sk2\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 ss and we get

π.scosπ.ssinπ.s=1+k=11(1s2k2).2s2k2\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
7 years, 3 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?

Isaac Thomas - 6 years, 3 months ago

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

Bogdan Simeonov - 6 years, 3 months ago

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Really interesting. Now how can we necessarily states that sinxx=(1xπ)(1+xπ)(1x2π)(1+x2π)\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 - 7 years, 3 months ago

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

Bogdan Simeonov - 7 years, 3 months ago

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