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{align} \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{align}
Proof. Let \( m = [y] \), \( k = [x] \). For integers \( n \) and \( n - 1 \) in \( [y, x] \) we have \begin{align*} \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{align*}
Summing from \( n = m + 1 \) to \( n = k \) we find \begin{align*} \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{align*}
Hence, \begin{align} \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{align}
Integration by parts gives us \begin{equation*} \int _y^x f(t) dt = x f(x) - y f(y) - \int _y^x t f'(t) dt. \end{equation*} 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;
}