I claim no originality to the ideas below

## Triangle inequalities

I like to write the triangle inequalities this way. While the absolute sign on the right hand side of the triangle inequality is unnecessary, it adds symmetry to the inequalities.

\begin{align*} |a+b| &\leq \big| |a| + |b| \big| \\ |a-b| &\geq \big| |a| - |b| \big| \end{align*}

In particular / as a mnemonic,

$\bigg| \frac{a + b}{a - b} \bigg| \leq \bigg| \frac{|a|+|b|}{|a|-|b|}\bigg|$

## Some more inequalities

Given a function $$f(x)+c$$ for some constant $$c$$, we may sometimes have a bound of the form $$f(x/2) \leq f(x)+c \leq f(2x)$$ for large $$x$$. For example, $$x^2 + 1 \leq (2x)^2$$ or $$(x/2)^2 \leq x^2 - 1$$ as $$\vert{x}\vert\to \infty$$.

For example for $$x > 1$$,

$e^{x/2} \leq e^x - 1 \leq e^{2x}$

So $$\frac{1}{e^x - 1}$$ is integrable on the interval $$[1, \infty)$$ as $$\frac{1}{e^{x/2}} = e^{-x/2}$$ is integrable on $$[1, \infty)$$

## Vector equalities

Particularly useful for dealing with limits in fluid mechanics / electromagnetism.

\begin{align*} \frac{1}{|u|} + \frac{1}{|v|} &= \frac{|u|+|v|}{|u||v|} \\ \frac{1}{|u|} - \frac{1}{|v|} &= \frac{|v|^2 - |u|^2}{|u||v|(|u|+|v|)} \end{align*}

## Trigonometric functions

One way I like to visualise the definitions is by considering

\begin{align*} e^z &= 1 + z + \frac{z^2}{2} + \frac{z^3}{6} + \frac{z^4}{24} + \dots \\ \cos z &= 1 \phantom{+ z }- \frac{z^2}{2} \phantom{+ \frac{z^3}{6} }+ \frac{z^4}{24} + \dots \\ \sin z &= \phantom{1 + }z \phantom{+ \frac{z^2}{2} }- \frac{z^3}{6} + \phantom{\frac{z^4}{24} }+ \dots\\ \cosh z &= 1 \phantom{+ z }+ \frac{z^2}{2} \phantom{+ \frac{z^3}{6} }+ \frac{z^4}{24} + \dots \\ \sinh z &= \phantom{1 + }z \phantom{+ \frac{z^2}{2} }+ \frac{z^3}{6} + \phantom{\frac{z^4}{24} }+ \dots\\ \end{align*}

The above aids getting an intuitive picture of the various identities for these functions, such as $$\sinh z = -i\sin(iz), \cosh z = \cos(iz), \frac{d}{dz} \cosh z = \sinh z$$ etc

## Integration by parts

Starting from chain rule we have

\begin{align*} (uv)' &= uv' + u'v \\ [uv]^b_a &= \int^b_a uv' + u'v dx \end{align*}

Using this form may reduce likelihood of careless mistakes. As a classic example,

\begin{align*} [x \ln x]^c_0 &= \int^c_0 \ln x + x \frac{1}{x} dx \\ c \ln c - c &= \int^c_0 \ln x \; dx \\ \text{So } \int \ln x \; dx &= x \ln x - x + C \end{align*}

## Polar and Cartesian dynamical systems

We can compute that

\begin{align*} \dot x &= \tfrac{d}{dt}(r \cos \theta) = \dot r \cos \theta - r \dot \theta \sin \theta = -y\dot \theta + x \dot r / r, \\ \dot y &= \tfrac{d}{dr}(r \sin \theta) = \dot r \sin \theta + r \dot \theta \cos \theta = x \dot \theta + y \dot r / r. \end{align*}

Geometrically there are two velocity fields at play. One is $$\dot \theta(-y, x)$$ which is perpendicular to $$(x,y)$$ and has magnitude $$r \dot \theta$$. This aligns with what we know about uniform circular motion. Finally we also have $$\dot r / r (x,y)$$ which is parallel to $$(x,y)$$ and has magnitude $$\dot r$$. This is the radial velocity.

## Graph transformations

Thinking about transformations of the graph $$y=f(x)$$ is generally a bit annoying, especially deducing transformations involving the x-axis. Sometimes I find it easier to think as follows.

We can treat the graph $$y=f(x)$$ as the set $$S = \{ (t, f(t)) \; \forall t \in \mathbb{R}\}$$. This lets us tap into intuition from coodrinate geometry / linear algebra. We can replace $$(t,f(t))$$ by the following to perform the various transformations.

\begin{align*} (t, f(t)+c) &\text{ Translating upwards by c units} \\ (t+c, f(t))&\text{ Translating rightwards by c units} \\ (t, cf(t)) &\text{ Scaling along y-axis by c} \\ (ct, f(t)) &\text{ Scaling along x-axis by c} \\ \end{align*}

It’s only when we try to write $$y(t)$$ in terms of $$x(t)$$ do we get

\begin{align*} (t, f(t)+c) &\text{ Translating upwards by c units} \\ (t, f(t-c))&\text{ Translating rightwards by c units} \\ (t, cf(t)) &\text{ Scaling along y-axis by c} \\ (t, f(t/c)) &\text{ Scaling along x-axis by c} \\ \end{align*}

We substituted $$t$$ with $$t-c$$ for the second row. This substitution is bijective in that every point from the original graph $$S$$ is going to be associated to a unique point in the new graph.

Similarly, we substituted $$t$$ with $$ct$$ for the fourth row. This substitution is bijective as long as $$c\neq 0$$.

Credit to Ben Zhang for inspiring me on the symmetries in graph transformations.