Coded Data Proofs: Mean & Standard Deviation, y=x/k

Coded Data Poofs (2):

Say y=x/k and that: x={p, q}, y={p/k, q/k}.

This would mean that:

$\frac { \Sigma y }{ n } =\frac { \frac { p }{ k } +\frac { q }{ k } }{ n } \\ \\ =\frac { \frac { 1 }{ k } \left( p+q \right) }{ n } \\ \\ =\frac { 1 }{ k } \cdot \frac { \left( p+q \right) }{ n } \\ \\ =\frac { 1 }{ k } \cdot \frac { \Sigma x }{ n } \\$

It would also mean that:

$\frac { \Sigma { y }^{ 2 } }{ n } =\frac { { \left( \frac { p }{ k } \right) }^{ 2 }+{ \left( \frac { q }{ k } \right) }^{ 2 } }{ n } \\ \\ =\frac { \frac { { p }^{ 2 } }{ { k }^{ 2 } } +\frac { { q }^{ 2 } }{ { k }^{ 2 } } }{ n } \\ \\ =\frac { \frac { 1 }{ { k }^{ 2 } } \left( { p }^{ 2 }+{ q }^{ 2 } \right) }{ n } \\ \\ =\frac { 1 }{ { k }^{ 2 } } \cdot \frac { \left( { p }^{ 2 }+{ q }^{ 2 } \right) }{ n } \\ \\ =\frac { 1 }{ { k }^{ 2 } } \cdot \frac { \Sigma { x }^{ 2 } }{ n } \\$

And if the above is true:

${ \sigma }_{ y }=\sqrt { \frac { \Sigma { y }^{ 2 } }{ n } -{ \left( \frac { \Sigma y }{ n } \right) }^{ 2 } } \\ \\ =\sqrt { \frac { 1 }{ { k }^{ 2 } } \cdot \frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { 1 }{ k } \cdot \frac { \Sigma x }{ n } \right) }^{ 2 } } \\ \\ =\sqrt { \frac { 1 }{ { k }^{ 2 } } \cdot \frac { \Sigma { x }^{ 2 } }{ n } -\frac { 1 }{ { k }^{ 2 } } \cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 } } \\ \\ =\sqrt { \frac { 1 }{ { k }^{ 2 } } \left( \frac { \Sigma { x }^{ 2 } }{ n } -{ \left\{ \frac { \Sigma x }{ n } \right\} }^{ 2 } \right) } \\ \\ =\sqrt { \frac { 1 }{ { k }^{ 2 } } } \cdot \sqrt { \frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { \Sigma x }{ n } \right) }^{ 2 } } \\ \\ =\frac { 1 }{ k } \cdot { \sigma }_{ x }\\$

Coded Data Proofs, Mean & Standard Deviation: y=kx

Coded Data Proofs (1):

Say y=kx, and also that: x={p, q}, y={kp, kq}.

This would mean that:

$\frac { \Sigma y }{ n } =\frac { kp+kq }{ n } \\ \\ =\frac { k\left( p+q \right) }{ n } \\ \\ =k\cdot \frac { \Sigma x }{ n } \\$

Therefore, when y=kx:

$\frac { \Sigma y }{ n } =k\cdot \frac { \Sigma x }{ n } \\$

What we’d also be able to conclude is that:

$\frac { \Sigma { y }^{ 2 } }{ n } =\frac { { \left( kp \right) }^{ 2 }+{ \left( kq \right) }^{ 2 } }{ n } \\ \\ =\frac { { k }^{ 2 }{ p }^{ 2 }+{ k }^{ 2 }{ q }^{ 2 } }{ n } \\ \\ =\frac { { k }^{ 2 }\left( { p }^{ 2 }+{ q }^{ 2 } \right) }{ n } \\ \\ ={ k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } \\$

When considering the above, we can deduce that:

${ \sigma }_{ y }=\sqrt { \frac { \Sigma { y }^{ 2 } }{ n } -{ \left( \frac { \Sigma y }{ n } \right) }^{ 2 } } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } -{ \left( k\cdot \frac { \Sigma x }{ n } \right) }^{ 2 } } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } -{ k }^{ 2 }\cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 } } \\ \\ =\sqrt { { k }^{ 2 }\left( \frac { \Sigma { x }^{ 2 } }{ n } -{ \left\{ \frac { \Sigma x }{ n } \right\} }^{ 2 } \right) } \\ \\ =\sqrt { { k }^{ 2 } } \cdot \sqrt { \frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { \Sigma x }{ n } \right) }^{ 2 } } \\ \\ =k\cdot { \sigma }_{ x }\\$

Therefore, when y=kx:

${ \sigma }_{ y }=k\cdot { \sigma }_{ x }\\$

Further Pure Maths: Complex Number Proof (1)

In this post, I’ll be proving that: $\left| { z }_{ 1 }\cdot { z }_{ 2 } \right| =\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right|$

First of all, let’s say that:

${ z }_{ 1 }=x+iy$

Whereby, $\left\{ x\in R,\quad y\in R \right\}$.

And also that:

${ z }_{ 2 }=p+iq$

Whereby, $\left\{ p\in R,\quad q\in R \right\}$.

If this is the case, this means that:

${ z }_{ 1 }\cdot { z }_{ 2 }=\left( x+iy \right) \left( p+iq \right) \\ \\ =px+iqx+ipy+{ i }^{ 2 }qy\\ \\ =px-qy+i\left( qx+py \right)$

Therefore:

$LHS\\ \\ =\left| { z }_{ 1 }\cdot { z }_{ 2 } \right| \\ \\ =\sqrt { { \left( px-qy \right) }^{ 2 }+{ \left( qx+py \right) }^{ 2 } } \\ \\ =\sqrt { \left( px-qy \right) \left( px-qy \right) +\left( qx+py \right) \left( qx+py \right) } \\ \\ =\sqrt { { p }^{ 2 }{ x }^{ 2 }-2pqxy+{ q }^{ 2 }{ y }^{ 2 }+\left\{ { q }^{ 2 }{ x }^{ 2 }+2pqxy+{ p }^{ 2 }{ y }^{ 2 } \right\} } \\ \\ =\sqrt { { p }^{ 2 }{ x }^{ 2 }+{ q }^{ 2 }{ y }^{ 2 }+{ q }^{ 2 }{ x }^{ 2 }+{ p }^{ 2 }{ y }^{ 2 } } \\ \\ =\sqrt { \left( { x }^{ 2 }+{ y }^{ 2 } \right) \left( { p }^{ 2 }+{ q }^{ 2 } \right) } \\ \\ =\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \cdot \sqrt { { p }^{ 2 }+{ q }^{ 2 } } \\ \\ =\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| \\ \\ =RHS$

Hence we’ve proven that:

$\left| { z }_{ 1 }\cdot { z }_{ 2 } \right| =\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right|$

How to derive the formula for the area of an equilateral triangle

In this post I’ll be showing you how to derive the formula for the area of an equilateral triangle – in easy steps. In order to understand this derivation properly, you need to be familiar with Pythagoras’ theorem and also a few algebraic rules. What you’ll also need is a ruler, pair of compasses, a pencil and a sheet of paper.

Ready? Let me begin…

Step 1: Put a point on a blank sheet of paper and name it A.

Step 2: Put the needle of your compass on the point A and draw a circle around it.

Step 3: Add a point B to this circle, on its edge.

Step 4: Put the needle of your compass on the point B and your pencil on the point A.

Step 5: Draw another circle with a radius the length AB.

Step 6: Now add a few extra points to your drawing. Call these points C and D.

Step 7: Connect the points A, B and C forming a triangle.

Step 8: Draw a line going through the points C and D.

Step 9: Where the line going through C and D intersects the triangle, place the point E.

Step 10: Now look at your latest work very carefully… What you will notice is that the lengths AB, AC and BC are all equal to one another. This is because both the circles you drew – are exactly the same size. They each have radiuses equal in proportion. In simple terms, AB=AC=BC.

What you have to do now is name these lengths (r) for radius. Here’s the thing though, because the line going through C and D splits the triangle (equilateral, as each of its sides has the same length) down its middle, the length AE is equal to 1/2 x r, and similarly the length BE is equal to 1/2 x r. Together, the length AE + BE = AB = r.

Step 11: Remember that I said that the line going through C and D splits the triangle down its middle. Also, notice that this exact line is perpendicular to the length AB. Now, because of this, at the point E, you’ve got two right angles. Name these two right angles big R.

[Knowing that these two angles are equal to 90 degrees is vital – because you’ll be able to use Pythagoras’ theorem to find the length CE.]

Step 12: Find the length CE using Pythagoras’ theorem, Adjacent² + Opposite² = Hypotenuse². You will need this length to find the area of the equilateral triangle you’ve produced.

*Algebraic skills will be required from this point…

${ AE }^{ 2 }+{ CE }^{ 2 }={ AC }^{ 2 }\\ \\ \Rightarrow \quad { \left( \frac { 1 }{ 2 } r \right) }^{ 2 }+{ CE }^{ 2 }={ r }^{ 2 }\\ \\ \Rightarrow \quad { CE }^{ 2 }={ r }^{ 2 }-{ \left( \frac { 1 }{ 2 } r \right) }^{ 2 }\\ \\ \Rightarrow \quad { CE }^{ 2 }=\frac { 4r^{ 2 } }{ 4 } -\frac { { r }^{ 2 } }{ 4 } \\ \\ \Rightarrow \quad { CE }^{ 2 }=\frac { 3{ r }^{ 2 } }{ 4 } \\ \\ \Rightarrow \quad CE=\sqrt { \frac { 3{ r }^{ 2 } }{ 4 } } \\ \\ \therefore \quad CE=\frac { r\sqrt { 3 } }{ 2 }$

Step 13: Derive the formula for the area (A) of the equilateral triangle. Remember that the area of a right angled triangle is L x W x 1/2.

$A=\frac { 1 }{ 2 } r\cdot \frac { r\sqrt { 3 } }{ 2 } \cdot \frac { 1 }{ 2 } +\frac { 1 }{ 2 } r\cdot \frac { r\sqrt { 3 } }{ 2 } \cdot \frac { 1 }{ 2 } \\ \\ =\frac { 1 }{ 8 } { r }^{ 2 }\sqrt { 3 } +\frac { 1 }{ 8 } { r }^{ 2 }\sqrt { 3 } \\ \\ =2\cdot \frac { 1 }{ 8 } { r }^{ 2 }\sqrt { 3 } \\ \\ =\frac { 1 }{ 4 } { r }^{ 2 }\sqrt { 3 }$

Presto!!! Keep in mind that you can transform the variable (r) into any variable you wish. This variable (r) is the length of each side of the equilateral triangle you were working with. The formula you’ve derived can be used to find the area of any equilateral triangle.

Proof: Thales’ Theorem

In this post I’ll be demonstrating how you can prove that Thales’ Theorem is true. To follow the steps in this post (11 in total), what you will require is a ruler, pair of compasses and a pencil.

Step 11: Prove that the angle at point D is equal to 90 degrees.

Thales’ Theorem is as follows:

Because AC is the diameter of the circle you drew, the angle at the point D (α+β) must be equal to 90 degrees. In more specific and general terms, if you have the points A, C and D lying on a circle – and the line AC is in fact the diameter of this circle – then the angle at point D (α+β) must be a right angle.

Proof (which must be derived using the diagram you’ve created):

All angles within a triangle (in 2 space) must add up to 180 degrees.

Mathematically, this means that:

$\alpha +\alpha +\beta +\beta =180\\ \\ \Rightarrow \quad 2\alpha +2\beta =180\\ \\ \Rightarrow \quad 2\left( \alpha +\beta \right) =180\\ \\ \Rightarrow \quad \frac { 2\left( \alpha +\beta \right) }{ 2 } =\frac { 180 }{ 2 } \\ \\ \therefore \quad \alpha +\beta =90$

And as a result, Thales’ theorem must be true. The angle α+β is the angle at point D.

How to prove that the two angles below the apex of an isosceles triangle are equivalent

*You will need a pair of compasses, a ruler, pen and pencil to formulate this proof.

How would you go about proving that an isosceles triangle has two angles (below its apex) equal to one another?

Well, first of all – let’s start off by drawing a circle…

Now… We can tell that the circle we’ve just drawn has a centre (point at the centre). Next, what we have to do is add a couple of points to the edge of this circle. Like this…

Let’s name all these points A, B and C…

Now, let’s connect these points together with a few lines – to create an isosceles triangle ABC…

Ok… So far, so good… What you will need to do now is – place the needle of your compass on the point C and your pencil on the point B, like this…

Now spin your compass – and create an arc…

Next, get the needle of your compass and place it on the point B and put your pencil on the point C…

Draw another arc, like this…

Where the two arcs you’ve just drawn intersect, create a point… Call this point D…

Now, draw a line going through the points A and D. Call this line L. Line L will be perpendicular to the line BC…

Where the line L intersects the line BC, create a point E…

Now, it turns out, within the isosceles triangle ABC, we’ve created two right angles… This is because the line L is perpendicular to the line BC. Remember that the line L cuts the isosceles triangle down its centre. Let’s name these right angles big R…

If you look at the diagram above carefully, what you will notice is that the radius of the circle is equal in length to the line AB and also the line AC. Let’s name the lines AB and AC… We’ll call them r.

Let’s also name the line BC… We’ll call it x. This means that the line BE is equal to half of x, and because of this, the line CE must also be equal to half of x…

Finally (I know you must be tired of drawing), let’s call the angle ABC alpha and the angle ACB beta…

With our diagram complete, we can now prove that alpha and beta are equivalent to each other.

*You will need to know a bit of trigonometry to pass this point. SOH CAH TOA rules to be precise.

It turns out that:

$\cos { \left( \alpha \right) } =\frac { A }{ H } =\frac { \frac { x }{ 2 } }{ r } =\frac { x }{ 2r }$

And also:

$\cos { \left( \beta \right) } =\frac { A }{ H } =\frac { \frac { x }{ 2 } }{ r } =\frac { x }{ 2r }$

This means that:

$\cos { \left( \alpha \right) } =\cos { \left( \beta \right) } \\ \\ \therefore \quad \alpha =\beta$

Hence, we’ve proven that an isosceles triangle has two angles (below its apex) equal to one another.

Proof: Opposite angles formed when two lines intersect, are equal to one another

How can we prove that opposite angles (when two lines intersect) are in fact equal to one another?

Well, first of all – let’s draw a circle…

We know that in a full circle, there are 360 degrees. This is an indisputable fact. Now, what happens if we split this circle in two with a straight line (going through its centre)?

Well, each half of the circle (top and bottom) – will now contain 180 degrees. We know this because:

$\frac { 360 }{ 2 } =180$

Ok, so far so good… Now, let’s draw another line through the circle (going through its centre) which intersects the first line we have drawn…

As we can see, because we have done this, we now have 4 different angles. Let’s name the two angles which are situated in the top half of the circle α and β

Earlier in this demonstration, we remarked that the top half of the circle (when it was split in two) contained 180 degrees. Mathematically and logically speaking, as this is the case, we must say that:

$\alpha +\beta =180$

Great, now let’s name the angles in the bottom half of the circle x and y

It follows, because the angles in the top half of the circle add up to 180 degrees, we must deduce that:

$x+y=180$

So, it turns out we now have two useful equations:

$\alpha +\beta =180$

$x+y=180$

Do we have enough to form our proof though? Unfortunately, not quite… We have to look at our most recent figure again, but this time from a different perspective…

You see, there are different top halves and bottom halves…

• There exists top halves α+β and also… β+y
• There exists bottom halves x+y but also α+x

You may ask, why is this important? Well, here’s what’s crucial:

$\beta +y=180$

$\alpha +x=180$

And now we have 4 different equations, 3 of which – will help us finally complete our proof.

$\alpha +\beta =180$

$x+y=180$

$\beta +y=180$

$\alpha +x=180$

Here’s why we need these equations…

α+β and α+x are equivalent (180 degrees), so we can deduce that:

$\alpha +\beta =\alpha +x\\ \\ \therefore \quad \beta =x$

*Subtract α from both sides of the equation.

α+β and β+y are equivalent (180 degrees), so we can deduce that:

$\alpha +\beta =\beta +y\\ \\ \therefore \quad \alpha =y$

*Subtract β from both sides of the equation.

Hence, we’ve proven that: Opposite angles (when two lines intersect) are equal to one another.

β=x and α=y:

How to quickly double the area of a square (simple geometry lesson)

In this post, I’ll be demonstrating how you can quickly double the area of a square using a simple geometrical trick.

Let’s say you have an ordinary square, like the one below…

Firstly, what you have to do is name the area of this square “A”…

Then, what you do next is divide this square (diagonally) into 4 equal parts…

After you have done this, you then name each part of this square “1/4 x A”…

Notice now, that to double the area of this square, all you have to do, is double the number of the 1/4 x A right angled triangles which currently exist – then configure them – like this…

As you can see, you’ve now got eight of these 1/4 x A right angled triangles neatly configured…

Not only are you left with a new square, double the size of your original square (follow the lines on the outside of the shape), but a handy equation, which proves that you doubled the area of the square you started off with…

$8\times \frac { 1 }{ 4 } A=2A$

How about that? 🙂

How to derive the formula for a circle from scratch

If you’d like to derive the formula for a circle from absolute scratch, then your best option would be to draw a diagram such as the one below:

If you look at this diagram carefully, what you will notice is:

• A circle exists and each point on this circle has the coordinate (x, y).
• The centre of the circle can be found at (a, b).
• The circle has a radius ‘r’.
• The right angled triangles in the diagram each have an adjacent length, opposite length and hypotenuse (r).

Once you’ve prepared a similar diagram, your next aim should be to turn your attention towards the right angled triangles which exist within the circle. You should also think about the many different right angled triangles which could fit within the circle provided they emanate from the centre point (a, b).

The reason I’ve mentioned these right angled triangles is because according to Pythagoras’ theorem, when you have a right angled triangle – its adjacent length squared plus its opposite length squared is equal to the length of its hypotenuse squared:

Adjacent²+Opposite²=Hypotenuse²

Now, in this case – the adjacent lengths of the right angled triangles which can fit within the circle on the diagram can be described using the expression:

$\left( x-a \right)$ or $\left| x-a \right|$

The opposite lengths can be described using the expression:

$\left( y-b \right)$ or $\left| y-b \right|$

Also, very interestingly:

• Each of the right angled triangles you can think of has a hypotenuse ‘r’.
• ${ \left( x-a \right) }^{ 2 }={ \left| x-a \right| }^{ 2 }$
• ${ \left( y-b \right) }^{ 2 }={ \left| y-b \right| }^{ 2 }$

When you combine all the information above, what you get is a neat formula which looks like this:

${ \left( x-a \right) }^{ 2 }+{ \left( y-b \right) }^{ 2 }={ r }^{ 2 }$

And it turns out… This is the formula for a circle on the x, y plane, whereby, (a, b) is the centre of the circle and ‘r’ is the length of its radius. How spectacular is that? 🙂

Derive the formula to find areas underneath curves

In this post I’ll be revealing how you can derive the formula which can be used to find areas underneath curves, from absolute scratch. Now, just below, what you will find is the diagram that will help us produce this formula…

In this diagram what you will discover is that:

*Please read the following contents carefully

• A length a exists, which starts at the origin O and ends at a;
• A length x exists, which starts at the origin O and ends at x;
• A length x+?x exists, which starts at the origin O and ends at x+?x;
• A length ?x exists, which starts at x and ends at x+?x;
• A height y exists, which starts at the origin O and ends at y;
• A height y+?y exists, which starts at the origin O and ends at y+?y;
• A height ?y exists, which starts at y and ends at y+?y;
• There is a curve called y=f(x);
• There is an area underneath the curve called A which commences at a and ends at x;
• There is an area underneath the curve called ?A which commences at x and ends at x+?x (Note: If you extend the distance from a to x what you get is a larger area, and the change in area can be measured. This change or difference is called ?A);
• There is a rectangle that exists called QRUT. It has an area which is y?x;
• There is a rectangle that exists called PRUS. It has an area which is (y+?y)?x;
• ?A has an area larger than that of the rectangle QRUT, but smaller than that of the rectangle PRUS.

Producing the formula with the information we’ve discovered…

Ok, so we want to produce the formula which will help us find areas underneath curves from absolute scratch. At our disposal we have a helpful diagram (which we’ve looked at and analysed carefully) and we’ve been able to discover a few facts about it. I think we can now get to work…

Let’s start off by saying that:

Area QRUT < ?A < Area PRUS

Which is something we already discovered.

If this is the case, we can say that:

y?x < ?A < (y+?y)?x

Now, check out what happens when we divide all the elements of this expression by ?x:

What we end up with is…

Alright, now you may be saying to yourself, why do I need to know this? Well, it turns out that:

This is because as ?x approaches 0,  ?y approaches 0 leaving (?A)/(?x) sandwiched between y and y+0.000000000000000001 which is virtually y.

And, also…

As a consequence, this ultimately means that:

$y=\frac { dA }{ dx }$

This is incredibly significant, because if we then integrate both sides of this equation, we get:

$\int { ydx=\int { \frac { dA }{ dx } } } dx\quad \Rightarrow \quad A=\int { ydx }$

And…

$A=\int { ydx } =F\left( x \right) +C$

Now, this equation can actually be used to find the area A underneath the curve from a to x. What we’re basically saying is that this area is equal to some function of x plus a constant. This ‘some function of x’ occurs when we integrate y which is a function of x.

Finalising the formula…

Alright so we’ve managed to latch on to something incredibly significant… We’ve got an important equation:

$A=\int { ydx } =F\left( x \right) +C$

However, it is not complete. We need to know what the constant C is. So…

If we say that at x=a the area A underneath the curve is 0, watch what happens… Look at what we get…

$O=F\left( a \right) +C$

Which means that:

$C=-F\left( a \right)$

Hence, we can conclude that:

$A=\int { ydx=F\left( x \right) } -F\left( a \right)$

And this formula can be transformed into something more fancy if we are measuring an area underneath a curve from x=a to x=b

This is probably the formula you’re most familiar with…

$A=\int _{ a }^{ b }{ ydx=F\left( b \right) } -F\left( a \right) ={ \left[ F\left( x \right) \right] }_{ a }^{ b }$

Which is the formula which can be used to find areas underneath curves.

If you are still confused and would like to go through this proof once again, please watch my video below…

You can also leave your comments below.

Related:

Trapezium Rule Formula – Derivation