# After Queneau’s Elements of Style II

Mathematics (informal)

There exist four men, A, B, C, and I. I observes an altercation between A and B, followed by A sitting down away from B. I conjectures that A was dismissed by B after A had accused B of deliberately jostling A on a crowded bus. Later I observes C and A. I conjectures that C is advising A on a matter of dress.

# After Queneau’s Elements of Style I

Mathematics (formal)

There exist a and b
b contacts a
a berates b
b retorts
b wins
a flees
There also exists c

# The difficulties of number and the number of difficulties

This morning in a well known DIY store I heard two members of staff discussing the metre-cubed measure of volume. One said ‘what is a metre cubed?’ and the conversation then went:

‘Do you know what a square meter is?’

‘Yes’

‘Well that’s, like, with 2 and a metre-cubed is, like, with 3.’

‘Oh I see’

If this is a hard concept then, in the words of a well known fictional army reservist, ‘we’re doomed, doomed I say.’

# Differentiating under the integral sign with variable limit

The formula

$\frac{d}{dt}\int_0^t \phi(t,s) u(s)\, ds = \phi(t,t)u(t) + \int_0^t \phi_t(t,s) u(s)\, ds$

is not new but it doesn’t seem so easy to track down a simple derivation. A simple (formal) derivation of this runs as follows. First define,

$F(t,\xi) := \int_0^t f(\xi, s)\, ds = \digamma(\xi, t) - \digamma(\xi,0)$

where $\digamma$ is a primitive such that $\digamma_s(\cdot,s) = f(\cdot,s)$. Then

$\frac{dF}{dt}= \frac{\partial F}{\partial t} + \frac{\partial F}{\partial\xi}\frac{d\xi}{d t}$

But on substituting

$\frac{\partial F}{\partial t}(t,\xi) = \frac{\partial\digamma}{\partial t}(\xi,t) = f(\xi,t)$

and

$\frac{\partial F}{\partial\xi}(t,\xi) = \int_0^t f_\xi(\xi,s)\, ds$

this becomes

$\frac{dF}{dt}= f(\xi,t) + \frac{d\xi}{d t} \int_0^t f_\xi(\xi,s)\, ds.$

That’s it really. As a corollary take $\xi = t$ and $f(\xi,s) = \phi(t,s)u(s)$ and then we see that

$\frac{d}{dt}\int_0^t \phi(t,s)u(s)\, ds = \phi(t,t)u(t) + \int_0^t(\phi(t,s)u(s))_t\, ds$

which is what we set out to demonstrate.

# The Legendre Transform

This topic doesn’t get anywhere near the attention it deserves. I like it because it leads to a useful family of inequalities. Given a strictly convex function $f\colon\mathbb{R}\to\mathbb{R}$ (i.e. $f'' > 0$ and $f'$ is monotone) we define its Legendre transform as

$\displaystyle g(y) = \max_{x\in\mathbb{R}}\left\{xy - f(x)\right\}$.

The maximum can be calculated by differentiating and requiring that $\frac{d}{dx}(xy-f(x))=0$. Hence we can write $y = y(x) := f'(x)$ and use the fact that this function is invertible to write $x = x(y)$. With these we can check that this was indeed a maximisation because $\frac{d^2}{dx^2}(xy-f(x))=-f''(x) < 0$ by assumption.

Now, with $y = y(x) = y(x(y))$ as above we get the Legendre transform:

$g(y) = y x(y) - f(x(y))$.

Omitting the display of the functional dependency we get the symmetric relationship $g(y) + f(x) = xy$ and this suggests (correctly) that the Legendre transform is its own inverse.

For more information you can check out the wikipedia page at http://en.wikipedia.org/wiki/Legendre_transform or the interesting article ‘Making Sense of the Legendre Transform’ available on the arxiv at http://arxiv.org/abs/0806.1147.

To finish we’ll use the Legendre transform to prove the following useful inequality.

Proposition. If $p,q\in(1,\infty)$ satisfy $\frac{1}{p} + \frac{1}{q} = 1$ then $xy \le \frac{x^p}{p} + \frac{y^q}{q}$ for all real $x,y > 0$.

Proof. Define $f(x) = x^p/p$ and then $y(x) = x^{p-1}$ gives $x(y) = y^{q-1}$ since it is readily shown that $(p-1)(q-1)=1$. Therefore $g(y) = y x(y) - f(x(y)) = y^q - (y^{q-1})^p/p = y^q(1-1/p) = y^q/q$ since $(q-1)p = q$. Therefore, by definition, $g(y) = \max_x\{xy - f(x)\}\ge xy-f(x)$ implies that $x^p/p + y^q/q \ge xy\ \forall x,y > 0$. $\square$

# A simpler demo of Euler’s formula

Following on from yesterday I found the following at:
http://www.proofwiki.org/wiki/Euler’s_Formula

Differentiate $\displaystyle\frac{\cos\theta + i\sin\theta}{e^{i\theta}}$ to get zero. Then put $\theta=0$ to show that the constant value of $\displaystyle\frac{\cos\theta + i\sin\theta}{e^{i\theta}}$ must be $1$.

That’s it!

# A simple derivation of Euler’s formula

The usual way of proving Euler’s formula, $\exp(i\theta) = \cos\theta + i\sin\theta$, is to use the Maclaurin series expansions of $\exp(x)$, $\sin(x)$ and $\cos(x)$. But suppose you want to introduce this formula to a class that has a good grasp of complex arithmetic and calculus, but hasn’t yet met Mclaurin’s series. The following is a non-rigorous demonstration of this result using only basic tools, and could form an extended homework.

Three basic results first need to be established. The first two were taken from the wikipedia page on Euler’s formula (1 August 2013).

First the student should show that

$\displaystyle \int\frac{dx}{1+ax} = \frac{1}{a}\ln(1+ax)$

(the arbitrary constant isn’t important here). Second, that

$\displaystyle\frac{2}{1+x^2} = \frac{1}{1-ix} + \frac{1}{1+ix}$

noting that this can be shown by direct manipulation and does not require a knowledge of partial fractions. Thirdly, by differentiating both sides for example, that

$\displaystyle \int\frac{dx}{1+x^2} = \tan^{-1}(x)$

Then, step-by-step, show that:

1. $\displaystyle 2\tan^{-1}(x) = \int\frac{dx}{1-ix} + \int\frac{dx}{1+ix}$.
2. $\displaystyle 2i\tan^{-1}(x) = \ln\left(\frac{1+ix}{1-ix}\right)$.
3. $\displaystyle \frac{1+ix}{1-ix} = \frac{1-x^2 + 2ix}{1+x^2}$.
4. (by putting $x=\tan(\alpha)$) $\displaystyle \frac{1+ix}{1-ix} = \cos(2\alpha)+i\sin(2\alpha)$.
5. $\displaystyle 2i\tan^{-1}(x) = 2i\alpha$.
6. $2i\alpha = \ln(\cos(2\alpha)+i\sin(2\alpha))$.

Writing $\theta=2\alpha$ and taking antilogs then results in Euler’s formula:

$e^{i\theta} = \cos(\theta) + i\sin(\theta)$.

If you can find a simpler `easy’ demonstration of this result I would appreciate hearing about it.