## The Chain Rule and the Theory

**I FOUND THIS ACCOUNT OF THE CHAIN RULE FROM SAM SHAH**

** https://samjshah.com/2018/02/08/there-might-be-light-at-the-end-of-the-chain-rule-tunnel-maybe/**

**The partial text is as follows:**

Find two points close to each other, like (x,g(f(x)) and (x+0.001,g(f(x+0.001)).

Find the slope between those two points: {g(f(x+0.001)-g(f(x))}/{(x+0.001)-x}.

There we go. An approximation for the derivative! (We can use limits to write the exact expression for the derivative if we want.)

But that doesn’t help us understand that {d/dx}[g(f(x)]=g'(f(x))f'(x) on any level. They seem disconnected!

But I’m on my way there. I’m following things in this way: x >>> f >>> g

Check out this thing I whipped up after school today. The diagram on top does x >>> f and the diagram on the bottom does f >>> g. The diagram on the right does both. It shows how two initial inputs (in this case, 3 and 3.001) change as they go through the functions f and g.

At the very bottom, you see the heart of this.

It has {δg}/{δf} times {δf}/{δx}={δg}/{δx}

(END OF SAM SHAH’S BIT)

**Proof of the chain rule.**

** We can be more exact and use the derivatives, and show that the formula is true for Y = G(F(X)) with F(X) = X ^{ N} and G(F) = F^{ M}**

**The direct solution for Y’ is the derivative of X**

^{ MN}, that is MNX^{ MN – 1}**The formula solution is MF**

^{ M – 1}times NX^{ N – 1}, giving MNX^{ N(M-1)}times X^{ N-1}**and finally MNX**

^{ MN – 1}**SOME DEFINITIONS:**

** A function is a process which converts an input into a corresponding output.**

** In symbols, input –> f –> output**

**Examples:**

** The input is transformed (converted) into the output,**

** usually by a formula or an expression using the values of the input:**

** output = 2(input) + 5**

** or you can write this as output = 2 times input + 5**

**The input and the output are both expressions, which can be**

** a: a number, or**

** b: a single variable, x or y or z … , or**

** c: a more complicated expression**

**The commonest form for the input-output relationship for a function f**

** is, as an example, f(x) = 3x + 4, where x is an input**

** and the corresponding output is 3x + 4**

** f is the label of the function,**

** and f(x) is the expression whose value is 3x + 4**

** f(x) = 3x + 4 is then an equation**

** The equation can be seen for example as f(8) = 3 x 8 + 4,**

** or f(y) =3y + 4 using y as the input,**

** or f(z ^{2} + 5z + 7) = 3(z^{2} + 5z + 7) + 4 using an expression.**

**In its most simple formulation the input is not present and the equation is simply f = 3y +4, where the input**

** is identified as the ‘y’.**

** Now if an equation has a single variable on the left and an expression on the right then**

** a) it can be interpreted as a function (functional form) with f(y) = 3y + 4, and**

** b) the expression on the right can be substituted for the variable on the left.**

**Example**

** Let g(A) = A + 2 be a function g with output A + 2**

** Then it can be identified with the equation g = A + 2**

** Let g have the input x ^{2} – 4x + 3**

**Then the output is (x + 2)**

^{2}– 4(x + 2)x + 3**which is x**

^{2}– 1 (surprise, surprise)Filed under Uncategorized

## The Chain Rule and the theory.

#### I found this account of the chain rule from Sam Shah

https://samjshah.com/2018/02/08/there-might-be-light-at-the-end-of-the-chain-rule-tunnel-maybe/

**The partial text is as follows:**

Find two points close to each other, like (x,g(f(x)) and (x+0.001,g(f(x+0.001)).

Find the slope between those two points: {g(f(x+0.001)-g(f(x))}/{(x+0.001)-x}.

There we go. An approximation for the derivative! (We can use limits to write the exact expression for the derivative if we want.)

But that doesn’t help us understand that {d/dx}[g(f(x)]=g'(f(x))f'(x) on any level. They seem disconnected!

But I’m on my way there. I’m following things in this way: x >>> f >>> g

Check out this thing I whipped up after school today. The diagram on top does x \rightarrow f and the diagram on the bottom does f \rightarrow g. The diagram on the right does both. It shows how two initial inputs (in this case, 3 and 3.001) change as they go through the functions f and g.

At the very bottom, you see the heart of this.

It has {δg}/{δf} times {δf}/{δx}={δg}/{δx}

#### (end of Sam Shah’s bit)

We can be more exact and use the derivatives, and show that the formula is true for y = g(f(x)) with f(x) = x^{ n} and g(f) = f^{ m}

The direct solution for y’ is the derivative of x^{ mn}, that is mnx^{ mn – 1}

The formula solution is mf^{ m – 1} times nx^{ n – 1}, giving mnx^{ n(m-1)} times x^{ n-1}

and finally mnx^{ mn – 1}

#### SOME DEFINITIONS:

A function is a process which converts an input into a corresponding output.

In symbols, input –> f –> output

#### Examples:

The input is transformed (converted) into the output,

usually by a formula or an expression using the values of the input:

output = 2(input) + 5

or you can write this as output = 2 x input + 5

#### The input and the output are both expressions, which can be

a: a number, or

b: a single variable, x or y or z … , or

c: a more complicated expression

#### The commonest form for the input-output relationship for a function f

is, as an example, f(x) = 3x + 4, where x is an input

and the corresponding output is 3x + 4

#### f is the label of the function,

and f(x) is the expression whose value is 3x + 4

f(x) = 3x + 4 is then an equation

The equation can be seen for example as f(8) = 3 x 8 + 4,

or f(y) =3y + 4 using y as the input,

or f(z^{2} + 5z + 7) = 3(z^{2} + 5z + 7) + 4 using an expression.

#### In its most simple formulation the input is not present and the equation is simply f = 3y +4, where the input

is identified as the ‘y’.

**Now if an equation has a single variable on the left and an expression on the right then**

** a) it can be interpreted as a function (functional form) with f(y) = 3y + 4, and**

** b) the expression on the right can be substituted for the variable on the left.**

** Example**

** Let g(A) = A + 2 be a function g with output A + 2**

** Then it can be identified with the equation g = A + 2**

** Let g have the input x ^{2} – 4x + 3**

**Then the output is (x + 2)**

^{2}– 4(x + 2)x + 3**which is x**

^{2}– 1 (surprise, surprise)Filed under Uncategorized

## Sorry we keep lying to you…

I like “Add a zero”, but lying in maths is wonderfully stupid. Read on !!!!!

It’s well documented that many school students and adults alike are less than fond of mathematics. It tends to be a theme I discuss on here. I’ve singled out over-emphasis on speed of processes, misguided attempts at trying to convince students it’s entirely relevant to their daily lives, and of course, not explaining things, as factors. These are all ways in which we, the teachers, are sometimes subconsciously influencing things – but it’s certainly not all our fault though. We live in a country where, typically, it’s a badge of honour to be crap at maths and still miraculously live a normal life. Our bloated curriculum and imbalanced subject hierarchy don’t exactly help either. I was reading a maths book for trainee teachers the other day and I came across a familiarly painful explanation of the column method of subtraction, stating that you ‘cannot subtract three from…

View original post 409 more words

Filed under Uncategorized

## Parabola, it’s scarily simple…

No distances, no circles, and you can easily derive an equation.

Just a right angled triangle.

First, the definition of a parabola from the focus and directrix.

Pick a line, the directrix, and a point (B) not on that line (the focus):

Find the line at right angles, passing through a point (C) on that line.

Now find the line from B to C, and the midpoint of BC, which will be D.

Find the line at right angles to BC from D, and the intersection of this line and the vertical line, E, is a point on the parabola.

As point C is moved the parabola is traced out.

The picture is completed with the line BE. Check it!

Filed under bisecting, conic sections, conics, construction, definitions, Uncategorized

## Inversion in a circle

Diagram, then text:

The circle has radius 1 and centre at the origin.

The line is x = a

Now 1/a is the inverse of a, so a * 1/a = 1, and is fixed.

The line z from (1/a,0) to (x,y) is orthogonal to the radial line, so r/z = Y/a and the two triangles are similar

and r/(1/a) = a/R

Hence rR = a * (1/a) = 1, and both conclusions are true.

The point (x,y) follows a circular path, and rR = 1

Filed under Uncategorized

## https://fee.org/articles/some-refreshing-honesty-about-the-purpose-of-mass-schooling/

This is a reblog.

Do read it.

It’s quite short!

Filed under Uncategorized

## Multitasking: Good for computers, bad for people.

See if you can read this to the end without answering the phone, noticing a notification, etcetera:

Filed under computer, confusion, Uncategorized

## Area models for completing the square, dynamic approach.

An area model, or a dot array model (same thing really) is one way of illustrating the algebraic completion of a square.

I have used dots as they are easier to create.

The quadratic is viewed initially as the “standard form”, and then rebuilt dynamically line by line into the “square plus a bit over” form, as shown in the following sequence:

The odd valued coefficient of x in the original expression can appear as a row and a column of half-dots, or half squares in the area model form.

Filed under algebra, arithmetic, completing the square, Uncategorized

## Geometry and Numbers – the theory

## Multiplication, the theory – by Thales’ theorem

The diagram can be simplified by using an acute triangle.

Thales’ theorem

Proof of Thales theorem :

If a line is drawn parallel to one side of a triangle and it intersects the other two sides at two distinct points then it divides the two sides in the same ratio.

Given : In ∆ABC , DE || BC and intersects AB in D and AC in E.

Prove that : AD / DB = AE / EC

Construction : Join BC,CD and draw EF ┴ BA and DG ┴ CA.

Statements Reasons

1) EF ┴ BA 1) Construction

2) EF is the height of ∆ADE and ∆DBE 2) Definition of perpendicular

3)Area(ADE) = (AD.EF)/2 3)Area = (Base .height)/2

4)Area(DBE) =(DB.EF)/2 4) Area = (Base .height)/2

5)(Area(ADE))/(Area(DBE)) = AD/DB 5) Divide (3) by (4)

6) (Area(ADE))/(Area(DEC)) = AE/EC 6) Divide (3) by Area(DEC)

7) ∆DBE ~∆DEC 7) Both the ∆s are on the same base and

between the same || lines.

8) Area(∆DBE)=area(∆DEC) 8) So the two triangles have equal areas

9) AD/DB =AE/EC 9) From (5) and (6) and (7)

Not only this but also AD/AB = DE/BC

I borrowed this from http://www.ask-math.com/basic-proportionality-theorem.html

Some adjustments, but the Thales theorem is well done. I liked it.

Filed under arithmetic, construction, definitions, education, geometrical, geometry, geostruct, Uncategorized