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  1. Mathematics
    1.1 The Riemann Hypothosis and its relation to the distribution of prime nubmers
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The Riemann Hypothosis and its relation to the distribution of prime nubmers

QuantaMagazine | NumberPhile | 3Blue1Brown

todo: 1. visualize user input zeta function down below 2. visualize points moving from input to output 3. visualize analytic continuation 4. visualize similarity of zeta zero distribution to eulers prime log counting The Zeta Function $$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}$$ Riemann expanded it to the complex plane. So s can be a complex number like $$z = a + ib$$ This coordsys shows what happens when you put in some complex numbers into the zeta function. The thicker the line gets the higher the upper limit of the infinite sums was set.

So the beginning of the first light line is the value for the zeta function's first summand.

Each animation frame represents the zeta function's output with one additional summand.

Each sprial represent the zeta function with another input. The real part is represented by the red share and the imaginary part by the blue share.

For example the sprial in the bottom right is $$s = 2 + i$$

To calculate a number x's complex exponent we need eulers formula and three general rules concerning logarithms and exponents.

  1. \(x = e^{ln(x)}\)
  2. \(log_{b}(m^n) = n * log_{b}(m)\)
  3. \(x^{a+b} = x^a * x^b\)
  4. \(e^{ix} = cos(x) + isin(x)\) Euler's Formula
$$\large{x^{a+ib} = e^{ln(x^{a+ib})}}$$ $$\large{x^{a+ib} = e^{(a+ib) * ln(x)}}$$ $$\large{x^{a+ib} = e^{aln(x) + ibln(x)}}$$ $$\large{x^{a+ib} = e^{aln(x)} * e^{ibln(x)}}$$ $$\large{x^{a+ib} = e^{ln(x^a)} * e^{ibln(x)}}$$ $$\large{x^{a+ib} = x^a * e^{ibln(x)}}$$ $$\large{x^{a+ib} = x^a * (cos(bln(x)) + isin(bln(x)))}$$ $$\large{x^{a+ib} = x^acos(bln(x)) + x^aisin(bln(x)))}$$


Bernhard Riemann