# Maths hacks

## Handy inequalities

If $$a \geq 0$$ and $$b \geq 0$$ are nonnegative real numbers and if $$p>1$$ and $$q>1$$ are real numbers such that $$\frac{1}{p}+\frac{1}{q}=1$$, then $a b \leq \frac{a^p}{p}+\frac{b^q}{q} .$ Equality holds if and only if $$a^p=b^q$$. $$a=b=2$$ arises a lot in statistics.

My favourite generalisation: If $$0 \leq p_i \leq 1$$ with $$\sum_i p_i=1$$ then $\prod_i a_i^{p_i} \leq \sum_i p_i a_i$ Equality holds if and only if all the $$a_i$$s with non-zero $$p_i$$s are equal.

## exp, sinh, cos etc

Avoid having to integrate by parts twice

Suppose $$f(x)$$ and $$g(x)$$ are functions that are each proportional to their second derivative. These include exponential, circular, and hyperbolic functions. Then the integral of $$f(x) g(x)$$ can be computed in closed form with a moderate amount of work. There's a formula that can compute all these related integrals in one fell swoop. Suppose $f^{\prime \prime}(x)=h f(x)$ and $g^{\prime \prime}(x)=k g(x)$ for constants $$h$$ and $$k$$. …Then $\int f(x) g(x) d x=\frac{1}{h-k}\left(f^{\prime}(x) g(x)-f(x) g^{\prime}(x)\right)+C \text {. }$

## Maclaurin integration

consider the integral $\int e^{x^2} d x$ The most common approach to evaluating this integral is to expand it as a power series and integrate term-by-term, which yields $C+x+\frac{x^3}{3}+\frac{x^5}{10}+\frac{x^7}{42}+\frac{x^9}{216}+\cdots$ as the antiderivative, with $$C$$ as the constant of integration. Maclaurin Integration is an alternative solution by series (although not a power series, since it involves the function itself in the solution) that eliminates the nuisance of calculation completely.

…the formula is simply: $\int f(x) d x=-\sum_{u=0}^{\infty}\left(\frac{d^u}{d x^u}(x f(x)) \sum_{n=0}^{\infty} \frac{(1-x)^{n+u+1}}{\prod_{v=1}^{u+1}(n+v)}\right)+C,$ where $$C$$ is the constant of integration.…

This formula is valid only if: 1. $$f(x)$$ is defined on the domain $$(0,2)$$, 2. $$f(x)$$ is continuous on $$(0,2)$$, and 3. $$x f(x)$$ has derivatives of all orders on $$(0,2)$$.

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