# How to derive Euler’s Identity using the Maclaurin Series

In this post I’ll be showing you how to derive Euler’s identity using the Maclaurin Series. It turns out that the Maclaurin series looks like this:

And expanded, it looks like this:

[*A larger version of this image can be found here.]

Now, since we want to derive Euler’s identity, we first have to find out what the formula for e^x looks like. In order to get this formula we must use the table below:

Derivatives of e^x When x=0
${ f }^{ \left( 0 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 0 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 1 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 1 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 2 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 2 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 3 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 3 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 4 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 4 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 5 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 5 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 6 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 6 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 7 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 7 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 8 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 8 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 9 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 9 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 10 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 10 \right) }\left( 0 \right) ={ e }^{ 0 }=1$

Ok. So we’ve got a useful table just above. Let’s write out the function of e^x in its Maclaurin Series form:

$f\left( x \right) ={ e }^{ x }\\ \\ =\frac { 1 }{ 0! } { f }^{ \left( 0 \right) }\left( 0 \right) { x }^{ 0 }+\frac { 1 }{ 1! } { f }^{ \left( 1 \right) }\left( 0 \right) { x }^{ 1 }+\frac { 1 }{ 2! } { f }^{ \left( 2 \right) }\left( 0 \right) { x }^{ 2 }\\ \\ +\frac { 1 }{ 3! } { f }^{ \left( 3 \right) }\left( 0 \right) { x }^{ 3 }+\frac { 1 }{ 4! } { f }^{ \left( 4 \right) }\left( 0 \right) { x }^{ 4 }+\frac { 1 }{ 5! } { f }^{ \left( 5 \right) }\left( 0 \right) { x }^{ 5 }\\ \\ +\frac { 1 }{ 6! } { f }^{ \left( 6 \right) }\left( 0 \right) { x }^{ 6 }+\frac { 1 }{ 7! } f^{ \left( 7 \right) }\left( 0 \right) { x }^{ 7 }+\frac { 1 }{ 8! } { f }^{ \left( 8 \right) }\left( 0 \right) { x }^{ 8 }\\ \\ +\frac { 1 }{ 9! } { f }^{ \left( 9 \right) }\left( 0 \right) { x }^{ 9 }+\frac { 1 }{ 10! } { f }^{ \left( 10 \right) }\left( 0 \right) { x }^{ 10 }+...$

Now, let’s replace

with the values from the table. If we do this, the formula for e^x will become:

Alright, so far, so good. We are certainly on the right track. Our next goal will be to discover what e^(i*x) is. This is because to produce Euler’s identity, we need to come up with:

${ e }^{ ix }=\cos { \left( x \right) } +i\sin { \left( x \right) }$

To come up with the formula above, we will need the table below, because our latest e^(x) formula will have to be transformed. x will be turned into i*x.

Imaginary Numbers Exponentiated
$\sqrt { -1 } =i$
$i\cdot i={ i }^{ 2 }=-1$
${ i }^{ 3 }={ i }^{ 2 }\cdot i=-i$
${ i }^{ 4 }={ i }^{ 3 }\cdot i=-i\cdot i=-{ i }^{ 2 }=1$
${ i }^{ 5 }={ i }^{ 4 }\cdot i=i$
${ i }^{ 6 }={ i }^{ 5 }\cdot i=i\cdot i={ i }^{ 2 }=-1$
${ i }^{ 7 }={ i }^{ 6 }\cdot i=-i$
${ i }^{ 8 }={ i }^{ 7 }\cdot i=-i\cdot i=-{ i }^{ 2 }=1$
${ i }^{ 9 }={ i }^{ 8 }\cdot i=i$
${ i }^{ 10 }={ i }^{ 9 }\cdot i={ i }^{ 2 }=-1$
${ i }^{ 11 }={ i }^{ 10 }\cdot i=-i$

As we’ve got the table above, we can figure out what the formula e^(i*x) would look like:

Since:

This means that:

${ e }^{ ix }=\cos { \left( x \right) } +i\sin { \left( x \right) }$

And finally, when x=π:

${ e }^{ i\pi }=\cos { \left( \pi \right) } +i\sin { \left( \pi \right) } \\ \\ \therefore \quad { e }^{ i\pi }=-1\\ \\ \therefore \quad { e }^{ i\pi }+1=0$

This is because:

$\cos { \left( \pi \right) } =-1\\ \\ \sin { \left( \pi \right) } =0$

You have produced Euler’s identity from almost absolute scratch. Give yourself a pat on the back! 🙂

# Deriving the Taylor Series from scratch

[Please note: In order to derive the Taylor Series, you will need to understand how to differentiate. If you know how to differentiate, finding the Taylor Series won’t be much of a problem. You also need to know that 0!=1, 1!=1, 2!=2, 3!=6, x^0=1, x^1=x.]

In this post I will be demonstrating how one can produce the Taylor Series from absolute scratch.

First of all, let’s look at the diagram above. Now, let’s suppose that the equation of the function above is:

$f\left( x+a \right) ={ C }_{ 0 }+{ C }_{ 1 }x+{ C }_{ 2 }{ x }^{ 2 }+{ C }_{ 3 }{ x }^{ 3 }+...$

Ok, so we have the equation for the function, however, it isn’t complete. C_0, C_1, C_2, C_3 etc are hidden constants. This means that our second task will be to discover these constants. We need to discover these constants to find the complete equation of the function so that we can arrive at the Taylor Series. Fortunately, this task won’t be too difficult. Let me show you how C_0, C_1, C_2, C_3 etc can be found fairly easily…

When x=0:

${ f }^{ \left( 0 \right) }\left( a \right) ={ C }_{ 0 }\quad \therefore \quad \frac { 1 }{ 0! } { f }^{ \left( 0 \right) }\left( a \right) ={ C }_{ 0 }$

Now:

${ f }^{ \left( 1 \right) }\left( x+a \right) ={ C }_{ 1 }+2{ C }_{ 2 }x+3{ C }_{ 3 }{ x }^{ 2 }+...$

When x=0:

${ f }^{ \left( 1 \right) }\left( a \right) ={ C }_{ 1 }\quad \therefore \quad \frac { 1 }{ 1! } { f }^{ \left( 1 \right) }\left( a \right) ={ C }_{ 1 }$

Also:

${ f }^{ \left( 2 \right) }\left( x+a \right) =2{ C }_{ 2 }+6{ C }_{ 3 }x+...$

When x=0:

${ f }^{ \left( 2 \right) }\left( a \right) =2{ C }_{ 2 }\quad \therefore \quad \frac { 1 }{ 2! } { f }^{ \left( 2 \right) }\left( a \right) ={ C }_{ 2 }$

And, finally:

${ f }^{ \left( 3 \right) }\left( x+a \right) =6{ C }_{ 3 }+...$

When x=0:

${ f }^{ \left( 3 \right) }\left( a \right) =6{ C }_{ 3 }\quad \therefore \quad \frac { 1 }{ 3! } { f }^{ \left( 3 \right) }\left( a \right) ={ C }_{ 3 }$

Alright, so now that we have discovered the hidden constants C_0, C_1, C_2 and C_3, our third task is to write down the complete equation of the function f(x+a). Thanks to the information we have above, the fact that x^0=1 and x^1=x, plus our ability to spot patterns, we will be able to do this quite quickly…

[*Image can be seen here if it appears to be too small on this page.]

And it turns out that the equation we have just above is the Taylor Series function. It can be simplified to look like this…

What is also interesting is that if we transform a=0, we get the Maclaurin Series function which can be used to discover formulas for things such as e^x.

If you have any questions regarding this post, please leave your comments below. Once again, thanks for stopping by! 🙂

# Latest Mathematics Proofs: August 2016

Recently I discovered a few more proofs, some related to A Level mathematics.  You can access these proofs by clicking on the links below.

Derive the formula of an ellipsehttps://plus.google.com/+mathsvideosforweb/posts/ZF8D3ghSNRD

Discover the distance between point A and B on the edge of a circle: https://plus.google.com/+mathsvideosforweb/posts/9fmuEJ4qeXY

Find the area of a sector within a circle: https://plus.google.com/+mathsvideosforweb/posts/cHt9t9PeSfg

Formula for a torus, derived from scratch: https://plus.google.com/+mathsvideosforweb/posts/TS84TL44BYd

Create the mathematical singularity shown in movie documentaries: https://plus.google.com/+mathsvideosforweb/posts/7p2onSzNYte

Parameterised formula for a torus derived from scratch: https://plus.google.com/+mathsvideosforweb/posts/AZrVMiPVdbv

Derive the formula for an ellipsoid (normal and parameterised) from scratch: https://plus.google.com/+mathsvideosforweb/posts/Wijf94ztn3v

Other news:

If you love mathematical proofs, please feel free to join my ‘mathematics proofs’ Google community at: https://plus.google.com/communities/106007058741903558109

I’ve also got a new Facebook page related to stuff about the universe and also mathematics. You can add me as a friend by accessing this link: http://www.facebook.com/tiago.hands

# Hannah Sweets Maths Problem – Edexcel (June 2015)

In this post I’ll be demonstrating how to solve the “Hannah Sweets” maths problem which was posed in an Edexcel exam paper around June 2015. This problem is in fact one which is related to probabilities and goes like this:

There are (n) sweets in a bag;

6 of the sweets are orange;

The rest of the sweets are yellow;

Hannah takes at random a sweet from the bag… She eats the sweet;

Hannah takes at random another sweet from the bag… She eats the sweet;

The probability that Hannah eats two orange sweets is 1/3;

Show that: n^2 – n – 90 = 0.

[Source: Edexcel]

At first sight, this question could look a bit tedious, but upon further inspection – we notice that it can be broken into parts and solved fairly easily. Let me explain how…

Firstly, because of the information provided above, we can say that the probability of finding an orange sweet in the bag (under initial conditions) is 6/n. Note that we were told that there are 6 orange sweets in the bag (to begin with) and (n) sweets in the bag in total.

Now, as the rest of the sweets in the bag are yellow, what we have under initial conditions is an (n-6)/n chance of finding a yellow sweet in the bag. This expression happens to be irrelevant though, because we are told: 1. The probability that Hannah eats two orange sweets is 1/3. 2. Show that n^2 – n – 90 = 0.

What we really need is to find the chances of finding a second orange sweet after taking one orange sweet out of the bag… The chances of this happening would be 5/(n-1) because if you take one orange sweet out of the bag, 5 will be left and (n-1) sweets will be left.

Using our most basic probabilistic theorems we can deduce that the probability of selecting two orange sweets out of the bag in a row would be:

(6/n) * (5/(n-1))

Since we’ve been told that the probability that Hannah eats two orange sweets in a row is 1/3, we can say that:

(6/n) * (5/(n-1)) = 1/3

This would give us the equation:

30/(n(n-1)) = 1/3

Which would in turn give us:

90/(n(n-1)) = 1

Then:

n(n-1) = 90

Therefore we’d get:

n^2 – n = 90

And finally:

n^2 – n – 90 = 0.

From here on we’d only have to solve this quadratic equation to figure out how many sweets were in the bag to begin with. It turns out that n=10.

# Multiplying Even and Odd Numbers

Today I’m going to be showing you what would happen if you were to multiply:

a) An even number by an even number;

b) An odd number by an odd number;

c) An even number by an odd number.

Firstly, let us define what an even number is:

An even number can be described using the expression $2n$, whereby (n) would be a whole number ranging from 0 upwards.

Next, let us define what an odd number is:

An odd number can be described using the expression $2n+1$, and similarly (as is the case with even numbers), (n) would be a whole number ranging from 0 upwards.

Now, since we’ve defined how both even numbers and odd numbers can be described in terms of mathematical expressions, let’s focus our attention on multiplying even numbers by even numbers, odd numbers by odd numbers and even numbers by odd numbers…

Multiplying even numbers by even numbers:

Let’s produce two whole numbers which could be equal to one another or not equal to one another… Let’s call these numbers ${ n }_{ 1 }$ and ${ n }_{ 2 }$.

Using these two whole numbers we can multiply two unknown even numbers by each other in such a manner:

$2{ n }_{ 1 }\cdot 2{ n }_{ 2 }$

This would invariably give us the result $4{ n }_{ 1 }{ n }_{ 2 }=2\cdot 2{ n }_{ 1 }{ n }_{ 2 }$. Now, the product of ${ n }_{ 1 }{ \cdot n }_{ 2 }$ would be a whole number and since this is the case, you would have to say that an even number multiplied by an even number would produce an even number. Let’s not forget that even numbers are multiples of 2.

Multiplying odd numbers by odd numbers:

Once again, let’s come up with two whole numbers which could be equal to one another or not equal to one another… These whole numbers will be ${ n }_{ 3 }$ and ${ n }_{ 4 }$.

This would mean that two odd numbers being multiplied by one another would produce an expression as such:

$\left( 2{ n }_{ 3 }+1 \right) \left( 2{ n }_{ 4 }+1 \right)$

And if we expand the expression above, we’ll get:

$4{ n }_{ 3 }{ n }_{ 4 }+2{ n }_{ 3 }+2{ n }_{ 4 }+1$

Now if we re-arrange the expression above, we can get:

$2\left( 2{ n }_{ 3 }{ n }_{ 4 }+{ n }_{ 3 }+{ n }_{ 4 } \right) +1$

Since the expression $2{ n }_{ 3 }{ n }_{ 4 }+{ n }_{ 3 }+{ n }_{ 4 }$ must be a whole number, you would be forced to conclude that an odd number multiplied by an odd number would produce an odd number.

Multiplying even numbers by odd numbers:

We will for the last time come up with two whole numbers ${ n }_{ 5 }$ and ${ n }_{ 6 }$.

An even number and an odd number being multiplied by one another could be shown using the mathematical expression below:

$2{ n }_{ 5 }\cdot \left( 2{ n }_{ 6 }+1 \right)$

Lazily, we could conclude that an even number multiplied by an odd number would produce an even number. This is because even numbers are all multiples of 2.

Ok… So, let’s summarise what we’ve discovered:

i) An even number multiplied by an even number would produce an even number;

ii) An odd number multiplied by an odd number would produce an odd number;

iii) An even number multiplied by an odd number would produce an even number.

Knowing this we can further strengthen our mathematical reasoning. 🙂

# Binomial Expansions on Python

The other day I was asked to solve a complex probability problem. This probability problem was related to a vocabulary quiz which was being circulated around the web. It is a vocabulary quiz which contains 15 questions with 2 two possible answers to each question. My goal was to simply state the chances of someone (with a terrible vocabulary) answering at least 7/15 questions in this quiz correctly without referring to Google or any other search engine… Their answers to these questions were going to be completely random.

Now, in order solve this probability problem I first had to acknowledge seven essential facts:

1) The person taking this quiz has a terrible vocabulary and just wouldn’t be able to answer any of the questions in this quiz with confidence;

2) The person answering these questions would be providing random answers to them;

3) The person taking this quiz isn’t allowed to use any tools/resources which would help him/her answer these questions correctly;

4) There were only two possible answers to each question (right (R) or wrong (W) answers);

5) Each answer to these questions were mutually exclusive, in other words, selecting the right answer to one question wouldn’t increase your chances of answering any other question correctly ;

6) There were 15 questions in total;

7) A binomial distribution  would be required to solve the problem as there were two mutually exclusive outcomes to each question being answered.

With these facts in mind I knew exactly what I had to do to solve this problem, however, expanding the expression (R+W)^15 was going to be a very tedious task indeed and would require plenty of manual labour.

Was there another way to solve this problem?? Could a computer program help me solve the problem more easily??

Well it turned out that I would be able to solve the problem more easily, but that I’d have to use Python (a high level general purpose programming language) including an “add on” called Sympy to achieve such a feat. So what I essentially did was install both Python(x, y) and Sympy on my Windows laptop.

*You can find out how to download both Python and Sympy by watching the video below:

I then searched “Python IDLE” on my laptop and opened up Python’s shell application. This is what I typed into the application:

If you look at the image above carefully you will see that this Python shell application (thanks to the help of the “add on” Sympy, including some handy code) was able to expand (R+W)^15 for me… In fact, it did it in a split second.

From the moment (R+W)^15 had been expanded, to solve the problem I had been asked to solve, all I had to do was:

i) Sum up the coefficients of the variables (R, W) included in the expansion;

ii) Place this sum (a) under the value of the sum of coefficients sitting beside R variables with exponentials greater or equal to 7 (b) – in a fraction b/a.

This fraction (b/a) would give me the probability / chances of someone randomly being able to answer at least 7/15 questions correctly in the vocabulary test under the conditions which had been set.

a = 2(1 + 15 + 105 + 455 + 1365 + 3003 + 5005 + 6435) = 32,768

b = 1 + 15 + 105 + 455 + 1365 + 3003 + 5005 + 2(6435) = 22,819

b/a = 22,819/32,768 = 0.70 (to 2 decimal places)

Solution to problem: 70% chance

Now the irony of the story is this… I gave myself this problem to solve because my “posh” vocabulary is horrendous. It turned out that I managed to get 9/15 questions in the vocabulary test correct and all my answers were guesses. 🙂

# Mathematical Art Work, including Isometric Drawings (Visualising Maths)

Over the past couple of months I’ve been drawing mathematical figures and logos. Quite recently I discovered that art can help me understand complex mathematics a little better, especially three dimensional problems. It turns out that infinitely many complex sketches can emerge out of isometric fields which can be created using pencils, rulers and compasses. Isometric fields are used by engineers and architects who model three dimensional structures. Isometric fields allow you to represent cubes and various different 3 dimensional shapes on 2 dimensional surfaces in quite spectacular fashion. Here are some pieces of work I produced using such fields…

Follow Mathematics Videos & Proofs’s board Mathematical Art & Beauty on Pinterest.

If you’d like to see more work that I’ve been able to produce, please visit my mathematics Pinterest page. I’ve thoroughly enjoyed drawing optical illusions and logos using isometric paper, and I could indeed start designing isometric logos and pieces of text to make some extra money whilst studying mathematics. It’s not that often you bump into sound business ideas, especially business ideas related to both mathematics and art.

If you are a mathematics student but have never used isometric paper before, I’d recommend downloading isometric paper via the links below and producing mathematical sketches. Isometric paper could potentially be very useful to those interested in learning more about vector geometry and extra dimensions.

And by the way, thanks for stopping by. Enjoy your new year celebrations! 🙂

# THE IRRATIONAL NUMBER VALUE INDEX

Want to derive the value of specific irrational numbers from scratch?

Just click on the surd you’d like to discover more about, and you will be directed to a page which will give you instructions on how to derive its value.

What we haven’t included on this list are the square roots of numbers which can easily be found. We’ve only found the value of surds up to the square root of 11.

 √2 √3 √5 √7 √11

# 1.618 as a continued fraction

The value 1.618 is an approximation to the golden ratio, a number which is found extensively in nature. As it’s a very interesting number, let’s find out what it would look like as a continued fraction…

$1.618\\ \\ =1+\frac { 618 }{ 1000 } \\ \\ =1+\frac { 1 }{ \frac { 1000 }{ 618 } } \\ \\ =1+\frac { 1 }{ \frac { 618 }{ 618 } +\frac { 382 }{ 618 } } \\ \\ =1+\frac { 1 }{ 1+\frac { 382 }{ 618 } } \\ \\ =1+\frac { 1 }{ 1+\frac { 1 }{ \frac { 618 }{ 382 } } } \\ \\ =1+\frac { 1 }{ 1+\frac { 1 }{ \frac { 382 }{ 382 } +\frac { 236 }{ 382 } } }$

$\\ \\ =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 236 }{ 382 } } } \\ \\ =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ \frac { 382 }{ 236 } } } } \\ \\ =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ \frac { 236 }{ 236 } +\frac { 146 }{ 236 } } } }$

$\\ \\ =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 146 }{ 236 } } } } \\ \\ =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ \frac { 236 }{ 146 } } } } } \\ \\ =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ \frac { 146 }{ 146 } +\frac { 90 }{ 146 } } } } } \\ \\ =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 90 }{ 146 } } } } } \\ \\ =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ \frac { 146 }{ 90 } } } } } }$

$\\ \\ =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ \frac { 90 }{ 90 } +\frac { 56 }{ 90 } } } } } } \\ \\ =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+... } } } } }$