# General formulas and bounds¶

This section collects some results from real and complex analysis that are useful when deriving error bounds. Beware of typos.

## Error propagation¶

We want to bound the error when $$f(x+a)$$ is approximated by $$f(x)$$. Specifically, the goal is to bound $$f(x + a) - f(x)$$ in terms of $$r$$ for the set of values $$a$$ with $$|a| \le r$$. Most bounds will be monotone increasing with $$|a|$$ (assuming that $$x$$ is fixed), so for brevity we simply express the bounds in terms of $$|a|$$.

Theorem (generic first-order bound):

$|f(x+a) - f(x)| \le \min(2 C_0, C_1 |a|)$

where

$C_0 = \sup_{|t| \le |a|} |f(x+t)|, \quad C_1 = \sup_{|t| \le |a|} |f'(x+t)|.$

The statement is valid with either $$a, t \in \mathbb{R}$$ or $$a, t \in \mathbb{C}$$.

Theorem (product): For $$x, y \in \mathbb{C}$$ and $$a, b \in \mathbb{C}$$,

$\left| (x+a)(y+b) - x y \right| \le |xb| + |ya| + |ab|.$

Theorem (quotient): For $$x, y \in \mathbb{C}$$ and $$a, b \in \mathbb{C}$$ with $$|b| < |y|$$,

$\left| \frac{x}{y} - \frac{x+a}{y+b} \right| \le \frac{|xb|+|ya|}{|y| (|y|-|b|)}.$

Theorem (square root): For $$x, a \in \mathbb{R}$$ with $$0 \le |a| \le x$$,

$\left| \sqrt{x+a} - \sqrt{x} \right| \le \sqrt{x} \left(1 - \sqrt{1-\frac{|a|}{x}}\right) \le \frac{\sqrt{x}}{2} \left(\frac{|a|}{x} + \frac{|a|^2}{x^2}\right)$

where the first inequality is an equality if $$a \le 0$$. (When $$x = a = 0$$, the limiting value is 0.)

Theorem (reciprocal square root): For $$x, a \in \mathbb{R}$$ with $$0 \le |a| < x$$,

$\left| \frac{1}{\sqrt{x+a}} - \frac{1}{\sqrt{x}} \right| \le \frac{|a|}{2 (x-|a|)^{3/2}}.$

Theorem (k-th root): For $$k > 1$$ and $$x, a \in \mathbb{R}$$ with $$0 \le |a| \le x$$,

$\left| (x+a)^{1/k} - x^{1/k} \right| \le x^{1/k} \min\left(1, \frac{1}{k} \, \log\left(1+\frac{|a|}{x-|a|}\right)\right).$

Proof: The error is largest when $$a = -r$$ is negative, and

\begin{align}\begin{aligned}x^{1/k} - (x-r)^{1/k} &= x^{1/k} [1 - (1-r/x)^{1/k}]\\&= x^{1/k} [1 - \exp(\log(1-r/x)/k)] \le x^{1/k} \min(1, -\log(1-r/x)/k)\\&= x^{1/k} \min(1, \log(1+r/(x-r))/k).\end{aligned}\end{align}

Theorem (sine, cosine): For $$x, a \in \mathbb{R}$$, $$|\sin(x+a) - \sin(x)| \le \min(2, |a|)$$.

Theorem (logarithm): For $$x, a \in \mathbb{R}$$ with $$0 \le |a| < x$$,

$|\log(x+a) - \log(x)| \le \log\left(1 + \frac{|a|}{x-|a|}\right),$

with equality if $$a \le 0$$.

Theorem (exponential): For $$x, a \in \mathbb{R}$$, $$|e^{x+a} - e^x| = e^x (e^a-1) \le e^x (e^{|a|}-1)$$, with equality if $$a \ge 0$$.

Theorem (inverse tangent): For $$x, a \in \mathbb{R}$$,

$|\operatorname{atan}(x+a) - \operatorname{atan}(x)| \le \min(\pi, C_1 |a|).$

where

$C_1 = \sup_{|t| \le |a|} \frac{1}{1 + (x+t)^2}.$

If $$|a| < |x|$$, then $$C_1 = (1 + (|x| - |a|)^2)^{-1}$$ gives a monotone bound.

An exact bound: if $$|a| < |x|$$ or $$|x(x+a)| < 1$$, then

$|\operatorname{atan}(x+a) - \operatorname{atan}(x)| = \operatorname{atan}\left(\frac{|a|}{1 + x(x+a)}\right).$

In the last formula, a case distinction has to be made depending on the signs of x and a.

## Sums and series¶

Theorem (geometric bound): If $$|c_k| \le C$$ and $$|z| \le D < 1$$, then

$\left| \sum_{k=N}^{\infty} c_k z^k \right| \le \frac{C D^N}{1 - D}.$

Theorem (integral bound): If $$f(x)$$ is nonnegative and monotone decreasing, then

$\int_N^{\infty} f(x) \le \sum_{k=N}^{\infty} f(k) \le f(N) + \int_N^{\infty} f(x) dx.$

## Complex analytic functions¶

Theorem (Cauchy’s integral formula): If $$f(z) = \sum_{k=0}^{\infty} c_k z^k$$ is analytic (on an open subset of $$\mathbb{C}$$ containing the disk $$D = \{ z : |z| \le R \}$$ in its interior, where $$R > 0$$), then

$c_k = \frac{1}{2\pi i} \int_{|z|=R} \frac{f(z)}{z^{k+1}}\, dz.$

Corollary (derivative bound):

$|c_k| \le \frac{C}{R^k}, \quad C = \max_{|z|=R} |f(z)|.$

Corollary (Taylor series tail): If $$0 \le r < R$$ and $$|z| \le r$$, then

$\left|\sum_{k=N}^{\infty} c_k z^k\right| \le \frac{C D^N}{1-D}, \quad D = \left|\frac{r}{R}\right|.$

## Euler-Maclaurin formula¶

Theorem (Euler-Maclaurin): If $$f(t)$$ is $$2M$$-times differentiable, then

$\sum_{k=L}^U f(k) = S + I + T + R$
$S = \sum_{k=L}^{N-1} f(k), \quad I = \int_N^U f(t) dt,$
$T = \frac{1}{2} \left( f(N) + f(U) \right) + \sum_{k=1}^M \frac{B_{2k}}{(2k)!} \left(f^{(2k-1)}(U) - f^{(2k-1)}(N)\right),$
$R = -\int_N^U \frac{B_{2M}(t - \lfloor t \rfloor)}{(2M)!} f^{(2M)}(t) dt.$

Lemma (Bernoulli polynomials): $$|B_n(t - \lfloor t \rfloor)| \le 4 n! / (2 \pi)^n$$.

Theorem (remainder bound):

$|R| \le \frac{4}{(2\pi)^{2M}} \int_N^U \left| f^{(2M)}(t) \right| dt.$

Theorem (parameter derivatives): If $$f(t) = f(t,x) = \sum_{k=0}^{\infty} a_k(t) x^k$$ and $$R = R(x) = \sum_{k=0}^{\infty} c_k x^k$$ are analytic functions of $$x$$, then

$|c_k| \le \frac{4}{(2\pi)^{2M}} \int_N^U |a_k^{(2M)}(t)| dt.$