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>> No.7943550 [View]
File: 155 KB, 600x752, emma-stone-paddle-17dec13-03.jpg [View same] [iqdb] [saucenao] [google]
7943550

Say you have to solve the differential equation f’(x)=f(x) with f(0)=1.
You naturally make the ansatz

[math] f(x) = 1 + c_1 \, x + c_2 \, x^2 + c_3 \, x^3 + \dots [/math],

[math] f’(x) = c_1 + 2\,c_2\, x + 3\,c_3\, x^2 + \dots[/math].

Comparing coefficients, this implies that the solution to f’(x)=f(x) must have, for example [math] 3 \, c_3 = c_2 [/math]. In fact all coefficients are determined this way, by the recursive relation

[math] \dfrac { c_{n+1} } {c_n} = \dfrac {1} {n+1} [/math]

With the polynomial q(n) := n+1, this means
[math] c_n = \frac{1} { \prod_{k=1}^n q(k) } c_0 = \dfrac {1} {n!} [/math]
and hence
[math] f(x) = \sum_{n=0}^\infty \dfrac {1} {n!} x^n [/math].

Such an approach to solve a differential equation will often look like this. A whole lot of function have series coefficients [math]c_n[/math], such that

[math] \dfrac { c_{n+1} } {c_n} = \dfrac {p(n)} {q(n)} [/math]

where p and q are some polynomials. Any (arbitrary product of) polynomials in n can be written as a product of terms [math] (a_i-n) [/math]. They are the solutions to differential equations with recursive character:

[math] \dfrac {d}{dz} D_b f(z) = D_a f(z) [/math]

with

[math] D_b := \prod_{n=1}^{q} \left ( z \dfrac {d} {dz} + b_n-1 \right) [/math],

[math] D_a := \prod_{n=1}^{p} \left( z \dfrac {d} {dz} + a_n \right) [/math].

is solved by

[math] {}_p F_q [a_1,…,a_p; b_1,…,b_q] (z) := \sum_{n=0}^\infty c_n z^n [/math]

with

[math] c_n = \prod_{m=0}^{n-1} \dfrac{1} { (1+m) } \dfrac { \prod_{k=1}^p (a_k+m) }{ \prod_{j=1}^q(b_j+m) } [/math]

E.g. a1 = a2 = b1 = 1 characterized some simple differential equation that characterized the log, as

[math] \dfrac { \prod_{m=0}^{n-1} (a_1+m) } { \prod_{m=0}^{n-1} (b_1+m) } = \dfrac {n!} { (n+1)! } = \dfrac {1} {n+1} [/math]

and then

[math] {}_2 F_1 [1, 1; 2](z) = \sum_{n=0}^\infty \dfrac {1} {n+1} z^n = \dfrac {1} {(-z)} \sum_{k=1}^\infty \dfrac{ (-1)^{k+1} } {k} (-z)^k = - \dfrac {1} {z} \log(1-z) [/math]

>> No.7747114 [View]
File: 155 KB, 600x752, emma-stone-paddle-17dec13-03.jpg [View same] [iqdb] [saucenao] [google]
7747114

>>7747084

Bernhard Riemann was a master of fourier analysis, a more applied child of complex analysis, and from there he reached out into different mathematical subjects. For example, Riemann surfaces are two-dimensional real surfaces and give us a good image of what complex valued functions ought to look like. And Riemann geometry now famously provides the tools for general relativity.
About 150 years ago, Riemann must have been bored as he wrote a short paper on number theory, the only number theory paper he wrote in his life. It’s not really a number theory paper in the sense people did number theory before. It’s about complex analysis, and more like a conglomerate of what the fuck moment. „Oh btw. guys, have you tried this?“
The function in (for now) a real variable s > 1 defined by
[math] \zeta(s) := \sum_{n=1}^\infty n^{-s} [/math]
was looked at by Euler and owes it’s relevance to [math] (y · z)^x = y^x · z^x [/math]. The sum is over all numbers and those are multiplicatively build up from primes and you should really read it as

[math] 1 + 2^{-s} + 3^{-s} + 2^{-s} \, 2^{-s} + 5^{-s} + 2^{-s} \, 3^{-s} + 7^{-s} + 2^{-s} \, 2^{-s} \, 2^{-s} + 3^{-s} \, 3^{-s} + \dots[/math]

The Euler-Product is a compact product representation of that sum,

[math]\prod_{p} \frac {1} {1-p^{-s}} = \frac {1} { (1-2^{-s}) (1-3^{-s}) (1-5^{-s}) (1-7^{-s})\dots} [/math]

It’s a product over all primes.
By the rules of the logarithm, a log of a product is a sum of logs, and that’s the starting point for Riemann.
He first considers the function in s for complex numbers - this works for all s except at s=1. What comes next is the transition of something inherently discrete to his own playground of function theory. And it’s a little mad.

The object of interest is a function that counts primes up to some real number x.

>> No.7706735 [View]
File: 155 KB, 600x752, emma-stone-paddle-17dec13-03.jpg [View same] [iqdb] [saucenao] [google]
7706735

>He thinks I don't try to acknowledge the water

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