Although to penetrate into the intimate mysteries of nature and thence to learn the true causes of phenomena is not allowed to us, nevertheless it can happen that a certain fictive hypothesis may suffice for explaining many phenomena. — Leonhard Euler

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I will try to be fairly objective with this final blog and let the mathematics speak for itself. Thanks so much for your continued interest and response to these blogs! I greatly appreciate it, and I hope you learned a thing or two about mathematics and its eloquence.

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Recall that a complex number takes the form ** a + bi = r(cos(x) + i*sin(x))**, where

**.**

*x =**θ*Let us focus our attention on * cos(x) + i*sin(x)*.

From last blog, we can rewrite *sin *and *cos* as infinite sums.

*sin(x) = x – x ^{3}/6 + x^{5}/120 – x^{7}/5040 + x^{9}/362880 + … + (-1)^{n}x^{2n + 1}/(2n+1)! + …*

*cos(x) = 1 – x ^{2}/2 + x^{4}/24 – x^{6}/720 + x^{8}/40320 + … + (-1)^{n}x^{2n}/(2n)! + …*

Therefore, our expression *cos(x) + i*sin(x)* becomes the following.

* cos(x) + i*sin(x) = 1 – x^{2}/2 + x^{4}/24 – x^{6}/720 + x^{8}/40320 + … + i*(x – x^{3}/6 + x^{5}/120 – x^{7}/5040 + x^{9}/362880 + …)*

Let us now distribute *i *through the parenthesis.

*cos(x) + i*sin(x) = 1 – x ^{2}/2 + x^{4}/24 – x^{6}/720 + x^{8}/40320 + … + (ix – ix^{3}/6 + ix^{5}/120 – ix^{7}/5040 + ix^{9}/362880 + …)*

We can rearrange the right side of the equation by exploiting the commutative property of addition, which states that the order in which numbers are summed does not matter. We will rearrange the equation so that the exponent of *x* increases as one reads the expression from left to right.

*cos(x) + i*sin(x) = 1 + ix – x ^{2}/2 – ix^{3}/6 + x^{4}/24 + ix^{5}/120 – x^{6}/720 – ix^{7}/5040 + x^{8}/40320 + ix^{9}/362880 + …*

As it is currently written, this infinite sum does not convey much information, but after exploiting a property of the principle imaginary number *i*, we will be able to see something quite striking.

Recall ** i = sqrt(-1)**. Therefore, if we square both sides of the equation,

*i*= -1. If we multiply this equation again by

^{2}*i*, we get

*i*

^{3}*=*

*-i*. If we multiply this equation by

*i*one more time, we get

*i*

^{4}*=*

*1*. Notice that, if we multiply the subsequent equation again by

*i*, we come full circle in our answers, and the pattern repeats itself.

To avoid confusion, the pattern is written explicitly below.

*i ^{0}*

*=*

*1*

**i^{1}** = i

*i ^{2}*

*=*-1

*i ^{3}*

*=*-i

*i ^{4}*

*=*

*1*

*i ^{5}*

*=*i

*i ^{6}*

*=*-1

*i ^{7}*

*=*-i

*i ^{8}*

*=*

*1 …*

Let us now apply this property to the infinite sum.

**cos(x) + i*sin(x) = i ^{0}1 + i^{1}x + i^{2}x^{2}/2 + i^{3}x^{3}/6 + i^{4}x^{4}/24 + i^{5}x^{5}/120 + i^{6}x^{6}/720 + i^{7}x^{7}/5040 + i^{8}x^{8}/40320 + i^{3}x^{9}/362880 + …**

Notice that, if we went through each term we could see that *i ^{0}* goes to one,

*i*goes to

^{1}*i*,

*i*

^{2 }^{ }goes to negative one,

*i*goes to negative

^{3 }*i*,

*i*goes to one, and so on. Thus, you can see that the expression above is compatible with our previous expression for

^{4}*cos(x) + i*sin(x)*. We can write this symbolically below.

*1 + ix – x ^{2}/2 – ix^{3}/6 + x^{4}/24 + ix^{5}/120 – x^{6}/720 – ix^{7}/5040 + x^{8}/40320 + ix^{9}/362880 + … = i^{0}1 + i^{1}x + i^{2}x^{2}/2 + i^{3}x^{3}/6 + i^{4}x^{4}/24 + i^{5}x^{5}/120 + i^{6}x^{6}/720 + i^{7}x^{7}/5040 + i^{8}x^{8}/40320 + i^{9}x^{9}/362880 + …*

Let us now exploit a property of exponents to group *i* and *x* together.

*i ^{0}1 + i^{1}x + i^{2}x^{2}/2 + i^{3}x^{3}/6 + i^{4}x^{4}/24 + i^{5}x^{5}/120 + i^{6}x^{6}/720 + i^{7}x^{7}/5040 + i^{8}x^{8}/40320 + i^{9}x^{9}/362880 + … = (ix)^{0}1 + (ix)^{1} + (ix)^{2}/2 + (ix)^{3}/6 + (ix)^{4}/24 + (ix)^{5}/120 + (ix)^{6}/720 + (ix)^{7}/5040 + (ix)^{8}/40320 + (ix)^{9}/362880 + …*

To finish our rearragment, let us call make a quick substitution: ** ix = u**.

** (ix) ^{0}1 + (ix)^{1} + (ix)^{2}/2 + (ix)^{3}/6 + (ix)^{4}/24 + (ix)^{5}/120 + (ix)^{6}/720 + (ix)^{7}/5040 + (ix)^{8}/40320 + (ix)^{9}/362880 + … = u^{0} + u^{1} + u^{2}/2 + u^{3}/6 + u^{4}/24 + u^{5}/120 + u^{6}/720 + u^{7}/5040 + u^{8}/40320 + u^{9}/362880 + …**

The right side of the equation should look very familiar to you now, as we derived it in last week’s blog. You may recall the following infortmation.

**e ^{u} = u^{0} + u^{1} + u^{2}/2 + u^{3}/6 + u^{4}/24 + u^{5}/120 + u^{6}/720 + u^{7}/5040 + u^{8}/40320 + u^{9}/362880 + …**

*We can now substitute **ix* back in for *u* to get the following.

** e^{ix} = (ix)^{0}1 + (ix)^{1} + (ix)^{2}/2 + (ix)^{3}/6 + (ix)^{4}/24 + (ix)^{5}/120 + (ix)^{6}/720 + (ix)^{7}/5040 + (ix)^{8}/40320 + (ix)^{9}/362880 + … **

**After performing this substitution, let us pause for a moment to reflect on what we have just proven.**

** e^{ix} = cos(x) + i*sin(x)**

This equality, known as Euler’s Great Formula, serves as a basis for complex analysis and has many important applications. Right now, we will discuss one of those important outcomes, i.e. the most amazing formula in all of mathematics!

Recall *x* is just an angle. Let us say, for the sake of discussion, we choose the angle *180 degrees*. As you may recall from the first grand finale blog, mathematicians prefer to measure their angles in radians, so *180 degrees *really becomes *π *radians. Let us plug this value into Euler’s Great Formula and see what we find.

**e ^{iπ} = cos(π) + i*sin(π)**

From basic trigonometry, we know * cos(π) = -1* and

*.*

**sin(π) = 0****e ^{iπ} = -1**

If we add one to both sides of the equation, the following relationship ensues.

**e ^{iπ} + 1 = 0**

This, my friends, is the most amazing formula in mathematics. Titled Euler’s Great Identity, it combines the five fundamental constants with the three fundamental operations and has applications far reaching in mathematics. For example, have you ever thought to take the logarithm of a negative number? Is it really as blasphemous as your math teachers once told you?

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