User:LebesgueStieltjes/proofs

Proof of Pithagoras's Theorem using differential equation
The statement of the theorem is well-konwn: Given a right triangle with legs $$a,b$$ and hypotenuse $$c$$, the following equality holds

$$\LARGE a^2+b^2=c^2$$

As seen on the figure, if $$da$$ is sufficiently small, the triangle $$AA'D$$ can be considered as a right-angled triangle, similar to the original triangle with the following correspondence:

$$\LARGE\frac{a}{c}=\frac{dc}{da}$$

This leads to the integral formula as follows:

$$\Large \int a\text{ } \mathrm{d}a = \int c\text{ } \mathrm{d}c $$

The solution gives us the $$a^2 + C = c^2$$ equation. The value of the constant $$C$$ is determined by the initial value problem, because if $$a=0$$, then $$c=b$$. This leads us to the desired result $$a^2+b^2=c^2$$

Can computers answer any question?
One might ask: if I can describe a problem or ask a question (yes/no for the sake of simplicity) in a well-defined language, is there always an algorithm which answers it correctly? Formally, given an alphabet, a language $$\cal L$$ is a set of words, each word consisting of letters of that alphabet. We say, that a language is decidable if for an arbitrary word there is an algorithm, that decides whether the word is in $$\cal L$$ or not. Is there such an algorithm for every language? Assume, that every algorithm is a word of the given alphabet - for example if the alphabet is binary, we know that every algorithm can be translated into ones and zeros. An algorithm works as follows: it gets a word as input and answers with yes or no or runs forever - think of endless loops. Now fix a special language $$\cal D$$. $$d \in \cal D$$ If and only if $$d$$ is a code of an algorithm and if we give $$d$$ as an input for that algorithm - remember, $$d$$ is also a word - then the answer is no. Now assume, that the algorithm, that decides if $$d \in \cal D$$ for every word, exists. By the existence of that algorithm we also mean, that it is also a word - denoted by $$W$$ -, as every algorithm is. If $$W \in \cal D$$, then the answer for $$W$$ is yes, but it also means, that the algorithm $$W$$ answers no for the input $$W$$. Likewise, if the answer is no, i.e. $$W \notin \cal D$$, then $$W$$ should be accepted by itself which leads to contradiction in both cases, so we found a question with no algorithm deciding it.

Explicit formula for the Fibonacci numbers
The Fibonacci-series is defined by a recursion: $$F_n = F_{n-1}+F_{n-2}$$ and $$F_0=0$$, $$F_1 = 1$$. Of course, we can easily calculate the first few element, but there is an exact formula, that calculates $$F_n$$ without knowing the previous ones.

First of all, we calculate the generator function: $$f(x) = \sum\limits_{k=0}^{\infty}F_nx^n$$. Using the definition of $$F_n$$ we obtain, that $$f(x) = x + \sum\limits_{k=2}^{\infty}F_{n-1}x^n+F_{n-2}x^n = x + x\sum\limits_{k=0}^{\infty}F_n x^n + x^2\sum\limits_{k=0}^{\infty}F_n x^n = x + xf(x) + x^2f(x)$$, thus $$f(x) = \frac{x}{1-x-x^2}$$. Factoring the denominator we get $$\LARGE f(x)=\frac{1}{\sqrt{5}}\left(\frac{1}{1-\frac{2x}{\sqrt{5} - 1}}-\frac{1}{1-\frac{2x}{\sqrt{5}+1}}\right)$$

Thus the generator function is the sum of two geometric series: $$f(x) = \frac{1}{\sqrt{5}}\left(\sum\limits_{k=0}^{\infty}\left(\frac{2x}{\sqrt{5}-1}\right)^{n}-\left(\frac{-2x}{\sqrt{5}+1}\right)^n\right)=\frac{1}{\sqrt{5}}\left(\sum\limits_{k=0}^{\infty}x^n\left(\left(\frac{2}{\sqrt{5}-1}\right)^{n}-\left(\frac{-2}{\sqrt{5}+1}\right)^n\right)\right)=\sum\limits_{k=0}^{\infty}F_nx^n$$

Comparing the coefficient of $$x^n$$ leads to $$F_n=\frac{1}{\sqrt{5}}\left(\left(\frac{2}{\sqrt{5}-1}\right)^{n}-\left(\frac{-2}{\sqrt{5}+1}\right)^n\right)$$