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.

160530AreaOfParallelogram

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.

160529PythagProof

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.

160530PythagGeneralised

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.

160522Berkeley

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.

 

 

 

 

Yachting and similar triangles

A yacht has two parallel masts which are both perpendicular to a level deck.  One mast is 12 m high, the other is 8 m.  Stays are rigged from the head if each mast to the foot of the other.  Find the perpendicular height of the crossing point above the deck.

SailBoat

Finding the height of the crossing point is a good similar triangles challenge for bright maths pupils.  In fact that the distance between the two masts is not given introduces two variables and makes this a subtle challenge.

160508SimilarSailboat

I find the independence of ‘h’ with the distance between the masts intriguing.

 

The linear combination of a sine and a cosine is itself a sine wave

A linear combination of two functions, f(x) and g(x) is a sum involving constant multiples of the functions.  That is,

a f(x)+b g(x)

where a, b \in \mathbb{R}.

So, in the case of \sin x and \cos x, we would have,

a\sin(x)+b\cos(x).

It is a slightly surprising fact that the linear combination of two sine waves is itself a sine wave.  The set of sine waves is closed under linear combination.

The in the featured animation, a\sin(x) is green and dotted and b\cos(x) is red and dotted.  The resulting linear combination is the continuous blue line.  The value a is set to 2.5, whereas the value b is animated.

This principle occurs in A level maths, Core 3, and is responsible for many long and complex questions.

Exdexcel Formula Sheet lays Pooh Traps for Further Mathematics Pupils

FP3 integration standard forms could lead unwary pupils and teachers into a Pooh Trap. The following issue arises each year as the finer points of this course get sharpened.

When consulting the Edexcel formula sheet to find the integral,

\displaystyle \int \dfrac{1}{\sqrt{a^{2}+x^{2}}} \textnormal{d}x

two anti-derivatives are given,

\textnormal{arsinh}\left(\dfrac{x}{a}\right)\ \ \ \textnormal{and}\ \ \ \ln\{x+\sqrt{x^{2}+a^{2}}\}

Some might be tempted to infer that these functions are equal, this is the Pooh Trap, because whilst they both differentiate to \dfrac{1}{\sqrt{a^{2}+x^{2}}} they are not equal.

Play with exponential functions and quadratic equations give us the logarithmic form of \textnormal{arsinh}\ x which is also represented in formula books.

\textnormal{arsinh}\ x=\ln\{x+\sqrt{x^{2}+1}\}

Then,
\textnormal{arsinh} \left(\dfrac{x}{a}\right)=\ln\left\{\dfrac{x}{a}+\sqrt{\left(\dfrac{x}{a}\right)^{2}+1}\right\}=\\\ln\left\{\dfrac{x+\sqrt{x^{2}+a^{2}}}{a}\right\}=\ln\left\{x+\sqrt{x^{2}+a^{2}}\right\}-\ln a
The two anti-derivatives differ by the constant \ln a.

In a definite integration, this constant is added and then taken away making no difference as to which anti-derivative the student uses. In a particular solution to a differential equation, the evaluation of the integral using boundary condition would lead to two different constant values.

The worse case is that a student takes the integration standard from and reads it as the logarithmic form of the inverse of \sinh x . I have seen this happen.