## Perspectives on a Complete Graph   ## 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 $a,b \in \mathbb{R}$, $\dfrac{\textnormal{d}y}{\textnormal{d}x}=a\textnormal{f}'(x)+b\textnormal{g}'(x)$

## The Fundamental Theorem of Calculus Integration is introduced as the reversal of differentiation i.e. in solving a differential equation, $\dfrac{\textnormal{d}y}{\textnormal{d}x}=\textnormal{g}(x)$. The link between integration and area is often passed over and is the subject of the Fundamental Theorem of Calculus. [The following discussion can be adapted for a decreasing function or, piece-wise, a function which successively increases or decreases.]

Consider and area function, $A(x)$, defined by the area under $\textnormal{f}(x)$ between $a$ and and a general point, $x$. If a small increment, $\delta x$, is applied to $x$ giving a small element, $\delta A$ of area. Now, $\textnormal{f}(x)\delta x \leqslant \delta A \leqslant \textnormal{f}(x+\delta x)\delta x$

dividing though by $\delta x$, gives, $\textnormal{f}(x) \leqslant \dfrac{\delta A}{\delta x} \leqslant \textnormal{f}(x+\delta x),$

a limit sandwich where, as $\delta x \rightarrow 0$, $\dfrac{\textnormal{d}A}{\textnormal{d}x}=\textnormal{f}(x)$

The curve function, $\textnormal{f}(x)$ is the derivative of the area function; hence the area function is the anti-derivative of the curve function and, $\displaystyle\int \textnormal{f}(x) \textnormal{d}x=A(x).$

## Compound Trigonometry – proof without words  $\sin(x+y)=\sin x \cos y+\cos x\sin y$ $\cos(x+y)=\cos x \cos y-\sin x\sin y$

## Carr’s Synopsis and Sixth Term Entrance Papers

Students in search of extension work in maths, particularly as far as Sixth Term Entrance Papers (STEP) are concerned, should peruse Carr’s Synopsis of results in Elementary Pure Mathematics.

This book is famous for being the volume used by the autodidact Ramanujan to teach himself mathematics.  The book is merely a list of results and student how works through them, proving each one, will develop a strong ability in the subject.  This life line into higher mathematics could not be produced in the same form in today’s educational and publishing context but, perhaps, Carr deserves a place as one of the subject’s great educators.

Higher mathematics is about generalisation and abstraction.  Because of this, school mathematics students can find university text books unreadable.  In such a context how can harder, higher level questions be set using the frame work of sixth form mathematics? The answer to this question is to wind the clock back to the 1880s and see how the subject was constructed then.  Carr does this for us; many useful stages in school mathematics are listed here with extensions and generalisation written in generally sixth form recognisable form.

Take question 1 from this year’s (2016) STEP I exam.  Pupils will only have to be familiar with the first six results to tackle this quickly. The question play is all about the divisibility of, $x^{2n+1}+1$

which result 6, above, tells us is $(x+1)(x^{2n}-x^{2n-1}+\dots+1)$.

Question game play leads us to factorise large numbers (without use of calculator): $\dfrac{300^{3}+1}{301}=89911\times 90091$

and $\dfrac{7^{49}+1}{7^{7}+1}=[(7^{7}+1)^{3}-7^{4}(7^{14}+7^{7}+1)][(7^{7}+1)^{3}+7^{4}(7^{14}+7^{7}+1)]$

in a difference of two squares (result 1) construction.

Full solution is attached.

step-i-2016-q1

## Transformations of Sine

Functional notation and transformations is always tricky to teach and understand.  GCSE students will meet this in Year 11.   In general, transformations applied after the function are more easily understood: $y=f(x)+a$  or $y=af(x)$

Something very un-intuitive happens when the transformation is applied to the argument of the function: $y=f(x-a)$  or $y=f(ax)$

with things stretching when they look like they should be compressing and other things moving the wrong way.

Mathematics is not always obvious, if it were we wouldn’t need it.

Try this for transformations of Sine.

## 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. ## The Ghosts of Departed Quantities and the difficulties of teaching and learning caculus

Bishop Berkeley writes this attack on the apparent supernatural reasoning involved in calculus.  The infidel was probably Halley (of comet fame) or Newton.

If pupils find the subject difficult to understand at school, and teachers find it difficult to teach, then the reason may be articulated in this book by the great man. Quotes include:

“Now to conceive a Quantity infinitely small, that is, infinitely less than any sensible or imaginable Quantity, or any the least finite Magnitude, is, I confess, above my Capacity. But to conceive a Part of such infinitely small Quantity, that shall be still infinitely less than it, and consequently though multiply’d infinitely shall never equal the minutest finite Quantity, is, I suspect, an infinite Difficulty to any Man whatsoever”

and.

“They are neither finite Quantities nor Quantities infinitely small, nor yet nothing. May we not call them the Ghosts of departed Quantities?”

Some sympathy for the thesis is gained by Berkeley’s examination of tangent reasoning:

“Therefore the two errors being equal and contrary destroy each other; the first error of defect being corrected by a second error of excess. ……. If you had committed only one error, you would not have come at a true Solution of the Problem. But by virtue of a twofold mistake you arrive, though not at Science, yet at Truth. For Science it cannot be called, when you proceed blindfold, and arrive at the Truth not knowing how or by what means.”

The student of sixth form level mathematics who is eager to see how this is resolved must continue their path to mathematical enlightenment by studying the $\epsilon-\delta$ Analysis of Cauchy, Riemann and Weierstrass.

## Euler and the Properties of the Second Derivative

The summer examination season sees pupils searching for maximums and minimums on their text papers.

This long-standing pursuit was initiated by the likes of Newton and Leibniz in their calculus.

It can be all too difficult to think about for some:

“our modern Analysts are not content to consider only the Differences of finite Quantities: they also consider the Differences of those Differences, and the Differences of the Differences of the first Differences. And so on ad infinitum.”

Bishop Berkeley, The Analyst, 1734

The example above is one in which Euler demonstrates the geometrical significance of the first and second derivatives.

Note that the Point of Inflection is where $f''(x)=\dfrac{d^{2}y}{dx^{2}}=0$ and is a change from upward to downward convex curvature, or vice versa.  There is no need for $f'(x)=\dfrac{dy}{dx}=0$ too.