# Compilation of miscellaneous ideas

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 \([1, \infty)\)

## 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,

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

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