2 Sep 2020
In mathematics, Euler’s identity is the equality \[e^{i \pi} + 1 = 0.\]
Euler’s identity is a special case of Euler’s formula from complex analysis, which states that for any real number \(x\), \[e^{ix} = \cos x + i \sin x.\]
\[(x+y)^n = \sum_{k=0}^n {n \choose k} x^{n - k} y^k.\]
\[e^x = \lim_{n \to \infty} \left( 1+ \frac{x}{n} \right)^n.\]
\[\left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)\]
\[P(A \mid B) = \frac{P(B \mid A) \, P(A)}{P(B)}.\]
Theorem 1 (Euler’s summation formula). If \(f\) has a continuous derivative \(f'\) on the interval \([y, x]\), where \(0 < y < x\), then
\[\begin{aligned} \sum_{y < n \le x} f(n) = & \int_y^x f(t) dt + \int_y^x (t - [t]) f'(t) dt \notag \\ & + f(x)([x] - x) - f(y)([y] - y). \label{theorem} \end{aligned}\]
Proof. Let \(m = [y]\), \(k = [x]\). For integers \(n\) and \(n - 1\) in \([y, x]\) we have \[\begin{aligned} \int_{n-1}^n [t] f'(t) dt & = \int_{n-1}^n f'(t) dt \\ & = (n - 1) \bigl( f(n) - f(n - 1) \bigr) \\ & = \bigl( n f(n) - (n - 1) f(n - 1) \bigr) - f(n). \end{aligned}\] Summing from \(n = m + 1\) to \(n = k\) we find \[\begin{aligned} \int_{m}^k [t] f'(t) dt & = \sum_{n = m + 1}^k \bigl( n f(n) - (n - 1) f(n - 1) \bigr) - \sum_{y < n \le x} f(n) \\ & = k f(k) - m f(m) - \sum_{y < n \le x} f(n). \end{aligned}\] Hence, \[\begin{aligned} \sum_{y < n \le x} f(n) & = - \int_{m}^k [t] f'(t) dt + k f(k) - m f(m) \notag \\ & = - \int_{y}^x [t] f'(t) dt + k f(x) - m f(y). \label{summation} \end{aligned}\] Integration by parts gives us \[\int_y^x f(t) dt = x f(x) - y f(y) - \int_y^x t f'(t) dt.\] When this is combined with \(\eqref{summation}\) we obtain \(\eqref{theorem}\). ◻
Here is an example of "hello, world" program written in
the C programming language:
#include <stdio.h>
int main()
{
printf("hello, world\n");
return 0;
}