Proof of de Moivre's Theorem

 The following post is based of how de Moivre's theorem can be shown through a matrix representation of complex multiplication and linear transformations. The 'proof' is not by any means rigorous, but instead is used to show an interesting connection. For those unfamiliar with complex numbers (with concepts such as modulus-argument form) and matrices, I suggest familiarising yourself first. De Moivre's Theorem is stated below:

Where theta is the argument and r is the modulus of a complex number. The expression in brackets is a complex number raised to the n power. For simplicity we will assume n is real. Note how instead of expanding the bracket using the binomial expansion, the theorem states that the n can just be multiplied by theta. This is quite unique and offers a route for exploration.

Complex Multiplication:

For multiplication of two complex numbers of the form a+ib and c+id, it can be seen as:

However, this can be represented in the form of a linear transformation of a vector by a matrix:

(note the top part of the vector is the real component and the bottom part is imaginary, also what complex number is a vector and what is the transformation is not important)

This can now be extended for a general complex number a+ib to the power of n:

The first step is used to make the following step more obvious, we will now replace the concept of repeated multiplication by a+ib with repeated linear transformations:

And therefore:

(note how the matrix is in the form of a rotation transformation)

Modulus-Argand Form

For a given complex number a+ib, it can be represented in modulus argument form as such:

r is the modulus and theta is the argument. This can be seen as like the polar form of complex numbers. Below is the number 2+i, in modulus argument form it is:

Rotation Transformation

Now back to the original theorem, this can be rewritten as:

r to the n can be cancelled out to give:

Now let:

Therefore, the equation above can be written as:

We now need to prove this statement. To do so we will look at this transformation geometrically, consider the point:

Under the rotation transformation the point is rotated by theta degrees and therefore after a single rotation the point becomes:

This can be proved by using matrix multiplication of the point and using the double angle identities. More generally by n-1 rotations the point is transformed to:

We can now see that:

If you are struggling with the last step check out rotation transformations. I hope you have enjoyed this 'proof' with thanks to desmos again.-naBla5040

Projectile Motion 2

Projectile Motion 2
In this post, I will expand on my previous one with things relating to the range, time and distance. Expanding on from before:

We will now derive an equation for the time of flight, from leaving the ground to hitting it. Consider the case below:

At the peak height (very top) the vertical velocity is 0 so:

From this we can see the time to the peak height is given by the formula above, now we can find the total time for the ball to reach the ground. If your intuition is right, you should have realised you need to multiply the formula above by 2. This is because the trajectory is a parabola (we will go more into that in part 3) and so symmetric about the peak; also because of the conservation of energy the velocity going up and the velocity going down will be equal and opposite for a certain height (as illustrated by the example at height H in the above diagram). As an extension see if you can show that fact from the equation at the bottom of the post. So:

(where T is the time to hit the ground, note this is where the ground is level)

From this, we can now calculate the maximum horizontal distance (the range) by using our formula for the constant horizontal velocity, multiplying the horizontal velocity by the total time of the flight you get the range as:

(Here I simplified the formula for range by using trigonometric identities, don't worry too much)

Now we can see to achieve the maximum horizontal range we need an angle of 45 degrees, as we multiply this by 2 to get 90 degrees which is where sine is at its peak value. This makes sense if you think about it.

To find the peak height we can use the conservation of energy:

(In this example capital V is the initial velocity and the lowercase v is the velocity at a certain height )

(I got the second equation by cancelling the mass and multiplying each term by 2)

The statement above comes from the fact that energy cannot be created nor destroyed and says that at any given point along the trajectory the sum of the objects kinetic and gravitational energy stores will equal the total kinetic energy store of the ball to begin with (the kinetic energy it had at the very beginning). Now, at the peak height, all the velocity is in the horizontal direction which we know from before is VcosO, and so rearranging and substituting gives:

(here I used the Pythagorean identity that cosx squared plus sinx squared equals 1, this is trivial to prove)

Now we have the formula for the peak height, as you can see to achieve the maximum height possible we need an angle of 90 degrees with the ground which should also be obvious when you think about it.

Thank you for reading part 2, especially if you struggled with part 1, and I hope you understood some if not all parts of this, and look out for part 3-Joe

Projectile Motion 1

I understand some of you dislike maths, so well done if you are one of those who are reading this. Most people were reluctant to study science and maths even when they had to, therefore I don't expect many to be interested in this. This is more of an experiment for me to see how easy my content is to understand, so if it all goes over your head don't be discouraged and look out for other posts.
Projectile Motion 1
I hope for this to be a simple post, however, I am unsure as to how some people who struggle with maths may find it, so I intend to split it over 2 or 3 posts with varying difficulty.
This topic appears in A-Level physics and A-Level maths
Projectile motion describes how an object launched behaves in flight and describes its corresponding trajectory: for example throwing a stone or kicking a ball. In its simplest case, we can look at how a ball that is launched with some angle and speed moves over the ground.
To begin with:

The Black object is our ball and the blue vector is the velocity of the ball at the very start. As it is a vector quantity it will have a magnitude (the speed of the ball) and a direction (the angle with the surface). In this example, we will be ignoring drag, but before it is important to go over vectors:

Our blue velocity vector can be split apart (resolved) into an upward speed and a side speed, and adding these two together will result in the blue velocity vector, which we gave the ball. The upward speed is our vertical velocity at the very start and the side speed is our horizontal velocity at the very start. If we know the angle separating v from the ground, we can then calculate both of these velocities using trigonometry:

Now we need to take into account gravity: since gravity is only acting in the downwards vertical direction it will have no effect on the horizontal velocity, but over time it will reduce the vertical velocity. We know the acceleration due to gravity is:

(it's negative because it is downwards and the triangle means change in)

From this:

I understand if you may be a little confused but all I have done here is rearranged the formula; remember that the change in time is just the time since you kicked the ball (t) with time starting at 0, and the change in vertical velocity being the vertical velocity after a certain time minus the starting vertical velocity, which is VsinO. So now:

(where the vertical velocity is given as a function of time, the starting angle and the starting speed)

We can do the same for the horizontal velocity, however since gravity only acts vertically the horizontal velocity will be a constant and it won't change, so it will just be what I calculated to begin with:

Now we have these two equations that describe the horizontal and vertical velocity of our ball after a certain time t:

I used the Pythagorean Theorem to calculate the magnitude of the velocity tangent to the trajectory as the vertical and horizontal components form a right-angled triangle. Don't confuse this with the vector V.

As you can see from the two equations above describing the components of the velocity, the horizontal velocity is constant throughout, and the vertical velocity is reduced by 9.81m/s every second. These two equations can be used to find more information about the trajectory, such as maximum range, speed, peak height, and duration. 

If you understood this well and enjoyed it please read the next posts, else don't worry and try to re-read it in case you missed something. Often not understanding something in maths comes from having shaky foundations in the things before. I hope you enjoyed reading and 'thank you for wanting to know more today than you did yesterday'(Neil deGrasse Tyson).

Infinite Series of Geometric Progression

Please note this post is quite advanced, so don't worry if you find this confusing. Remember that … means to carry on the pattern forever.
If you are unfamiliar with infinite series, then put simply it is the summation of an infinite number of terms, for example:

These terms form a sequence with a particular pattern. The patterns above should be quite obvious.
Infinite series either converge or diverge, if a series converges it approaches a specific value, if it diverges it continually grows or declines (approaches plus or minus infinity). This may seem odd but think about this particular series:

This series converges at a particular value, if you add the terms repeatedly you should see it approaches 1 (gets closer to 1 as you add more and more terms). Which is unlike: 

as this series diverges, as it doesn't approach a specific value. The series gets larger each time by a greater and greater amount, so we say the series approaches infinity (the more terms you add on, you should find the series doesn't get closer and closer to a specific value but rather continually increases).

Sigma Notation

We can represent these infinite series more concisely with proper mathematical notation.

Below, the Z-shape is called a sigma and is the Greek capital for the letter s. It means to add up the terms of a sequence for all integers from the bottom limit to the top limit. You can hopefully see this from the 2 examples below:

note: I didn't put the answer to the first equation, but you should see it is just some number

For a series to be infinite the upper bound has to be infinity. We can either say:

We will apply this notation to show the above convergent series:

remember the sigma is just a different way of representing the expression 

Proof of Convergence for the above Equation

To show the above example equals one:

when I divided x by 2, it was the same as subtracting one half

Geometric Progression

A geometric sequence involves multiplying the last term by the common ratio to find the next term. Examples include:

We can see for the first sequence the common ratio is 2, and for the second it is 3. These can be represented by the formulae:

I have only subtracted one from n to shift the sequence to start at 1-don't worry about it

Deriving the Formula for Convergence for the example below

Now we know about geometric progression, we can formalise the proof above. Suppose we have an infinite series, taking the form below:

k is just some real number

We can expand this out, as we know what the sigma represents:

We can divide both sides by k to get the expression below; the expression is just the above X excluding 1/k. So:

We can now solve:

The formula for the value the infinite series converges to as a function of some real number k:

(when k>1, else when k=1 or is smaller than 1 the formula gives illogical results)

When k=2, X is one, which is what we saw above

When k=3, X is one half, which is one third plus one ninth plus one twenty-seventh...

When k=4, X is on third, which is one fourth plus one sixteenth...

I hope you have enjoyed reading this, I feel this may be quite difficult for some, but after all it is a hard topic.

Differentiation

This is an extension of my earlier post on differentiation from first principles. I intend to go over the three main rules for differentiating functions, and important derivatives to know.
When you take a derivative of a function, you are essentially finding a formula for the gradient of the function (the gradient function). For example:

The gradient is clearly changing at different points, unlike linear graphs.

We can find the formula for the gradient by differentiating the equation for the graph:

the gradient function is m=2x

For the notation we can use either:

to represent the derivative of a function.

The formula above is the gradient equals two times x. So when x=0, the gradient =0; this makes sense if you look at the graph; when x=2, the gradient =4; but when x=-2, the gradient =-4. So now you know what differentiating gives us, we need to know how to do it. To differentiate a function you only need to focus on the x parts, anything else is a constant. The basic rule for polynomials of any degree:

For example to differentiate:

the number 5 is a constant in the equation above, so we ignore it

Important derivatives to know:

Now, to differentiate more complex stuff, we can use the product and quotient rule:

The product rule:

The quotient rule:

For example to differentiate:

remember the derivative of sine x is cosine x, and the above is from the quotient rule

And:

from the product rule

Finally the chain rule:

Now to use it in practise, and to differentiate the function below:

 remember that y is equal to the square root of u, and u is 16 minus x squared

 remember the square root of u is u to the power of 0.5

the equation I have derived here is the gradient function for a circle with a radius of 4

The three rules I have stated here allow you to differentiate most functions, to get the derivatives of the functions. Now we'll apply the rules on one big problem:

from the quotient rule

from the product rule

from the chain rule

I hope you have enjoyed reading-naBla5040