# 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

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