## Area of a circle – proof by exhuastion

Once $\pi$ has been defined as the ratio circumference to diameter, the area of a circle must be $\pi r^{2}$.

A proof relies on an infinite, limiting process which paves the way to some calculus-like ideas.

A circle is cut up into 6 sectors which are then rearranged into a near rectangle,

What doesn’t look too close to a rectangle at 6 sectors looks better at 26:

As the number of sectors becomes very big the shape becomes indistinguishable from a rectangle and the argument is complete.

I think that Archimedes used such arguments, or, proof by exhaustion, in some of his solid geometry work.

## Integration by Substitution

Current UK exam textbooks pass over proofs and mathematical discussions in a hurry to show the ‘how to’ of exam questions.

Integration by substitution is a little more than just backwards chain-rule and deserves a fuller treatment.

Try this,

Let,

$y=\displaystyle \int \textnormal{f}(x) \ \textnormal{d}x$

then,

$\dfrac{\textnormal{d}y}{\textnormal{d}x}=\textnormal{f}(x)$

Suppose that there exists a function g, of another variable $u$, such that $x=\textnormal{g}(u)$ and let, $\textnormal{f}(x)=\textnormal{f}(\textnormal{g}(u))=\textnormal{F}(u)$. So that,

$\dfrac{\textnormal{d}y}{\textnormal{d}x}=\textnormal{F}(u)$

Now, by the chain rule,

$\dfrac{\textnormal{d}y}{\textnormal{d}u}=\dfrac{\textnormal{d}y}{\textnormal{d}x}\times \dfrac{\textnormal{d}x}{\textnormal{d}u}=\textnormal{F}(u)\dfrac{\textnormal{d}x}{\textnormal{d}u}$

Hence,

$y=\displaystyle \int \dfrac{\textnormal{d}y}{\textnormal{d}u} \ \textnormal{d}u=\displaystyle \int \textnormal{F}(u)\dfrac{\textnormal{d}x}{\textnormal{d}u} \ \textnormal{d}u$

i.e.

$y=\displaystyle \int \textnormal{f}(x) \ \textnormal{d}x=\displaystyle \int \textnormal{F}(u)\dfrac{\textnormal{d}x}{\textnormal{d}u} \ \textnormal{d}u$

## Differentiation From First Principles

The gradient of a smooth curve, $\textnormal{f}(x)$, at a point $x$ is the gradient of the tangent to the curve at the point $x$. Point $P$ is on the curve and $Q$ is a neighbouring point whose $x$ value is displaced a small quantity, $\delta x$.

The idea behind differentiation is that as $\delta x$ becomes very small, the gradient of $PQ$ tends towards the gradient of the curve. In the limit as $\delta x$ becomes infinitesimally close to zero, the gradient $PQ$ becomes the gradient of the curve.

We write:

$\textnormal{gradient f}(x)=\dfrac{\textnormal{d}y}{\textnormal{d}x}=\lim_{\delta x \rightarrow 0}\left(\dfrac{\delta y}{\delta x}\right)=\lim_{\delta x \rightarrow 0}\left(\dfrac{\textnormal{f}(x+\delta x)-\textnormal{f}(x)}{\delta x}\right)$

there is a fair bit of analytic work missing (higher education) to make these ideas sound.

We also write:

$\dfrac{\textnormal{d}y}{\textnormal{d}x}=\textnormal{f}'(x).$

STANDARD RESULTS

Standard results can be proved for different functions.

If $\textnormal{f}(x)=x^{n}$ then

If $\textnormal{f}(x)=\sin x$, then we need to consider the small angle approximation that is if $\delta x$ radians is very small (infinitesimal), then $\delta x\approx\sin \delta x$ and $\cos \delta x \approx 1$, and compound trigonometry from which follows,

The differentiation process described above is linear and extends to more complicated functions. That is to say that if, $y=a\textnormal{f}(x)+b\textnormal{g}(x)$ where $a,b \in \mathbb{R}$,
$\dfrac{\textnormal{d}y}{\textnormal{d}x}=a\textnormal{f}'(x)+b\textnormal{g}'(x)$

## Parallelograms generalise Pythagoras to any triangle

There’s always space on the inter-web for another proof of Pythagoras’s Theorem.  Here’s one that uses the following equal areas property of parallelograms.

This kind of area chopping and shape translation is a feature of Euclidean geometry and our senses support it’s veracity at the order of size of the classroom.

The squares on the sides of a right-angle triangle set up a system of parallel lines which can then be used to demonstrate the Theorem using the above equal areas property.

The thread does not stop here though.  Taking the parallel line structure which makes this work we get a generalisation of Pythagoras to non-right-angled triangles with the area of the parallelogram on the longest side being the sum of the areas of those constructed on the other two sides.