\(\TeX\) Samples
Binomial theorem
$$ (x+y)^n = \sum_{k=0}^n {n \choose k} x^{n - k} y^k $$
Exponential function
$$ e^x = \lim_{n \to \infty} \left( 1+ \frac{x}{n} \right)^n $$
Cauchy–Schwarz inequality
$$ \left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right) $$
Bayes' theorem
$$ P(A \mid B) = \frac{P(B \mid A) \, P(A)}{P(B)} $$
Euler's summation formula
Theorem. Euler's summation formula. If $f$ has a continuous derivative $f'$ on the interval $[y, x]$, where $0 < y < x$, then \begin{align} \label{theorem} \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). \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} \label{summation} \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). \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}.
