# The Equation

Part of the peterjamesthomas.com Maths and Science archive

 This article is essentially a coda to my previous piece, Euler’s Number. In the earlier text, I introduced the constant $e \approx 2.71828$ via its relation to The Exponential Function $\exp(x)$, or equivalently, $e^x$, this being the only function [1] whose derivative is the function itself, i.e.: $\dfrac{d}{dx}e^x=e^x$

In the short Sherlock Holmes story, A Scandal in Bohemia [2], Holmes meets, and is bested by, a lady of the name Irene Adler. While Holmes’s general perspective on the opposite sex is both ambiguous and ambivalent, he goes to far as to grant Ms Adler the honorific sobriquet “The Woman”, keeping a photo of her as a reminder of how she outwitted him.

The equation which appears at the top of this article is rightly extremely famous. Like Ms Adler, it has a claim to the definite article and thus the title of, The Equation.

The Equation also appears as a omnipresent candidate in “what is the most elegant/fundamental/important equation in Mathematics?” beauty parades. Indeed it has acquired a – wholly inappropriate – mystical quality. People feel it is imbued with the meaning of the Universe [3]. The Equation is of course important, but not for mystical reasons. Indeed when its meaning is properly understood, the equality becomes almost self-evident (at least to 21st Century eyes).

As a follow-on to our earlier exploration of Euler’s Number, here we will derive his famous Identity and probe its meaning.

Triangle Inequality

We will be looking at the definition of $\exp(x)$ that we established in the first article and comparing this to some other infinite series, ones relating to two trigonometric functions.

To start, let’s have a brief refresher on Trigonometry:

 Aside: Consider a generic right-angled triangle as in the figure below: Here the bottom left-hand angle has a value of $\theta$, the hypotenuse has length $c$, the adjacent side has length $a$ and the opposite side has a length of $b$. We then have the following definitions: \begin{aligned}\sin\theta &= \dfrac{b}{c}\hspace{5mm}\Rightarrow\hspace{5mm} b = c\sin\theta\hspace{5mm}(1) \\ \\\cos\theta &= \dfrac{a}{c}\hspace{5mm}\Rightarrow\hspace{5mm} a = c\cos\theta\hspace{5mm}(2) \\ \\\tan\theta &= \dfrac{b}{a}\hspace{5mm}\Rightarrow\hspace{5mm} b = a\tan\theta\end{aligned} Last time we focussed on $\tan\theta$, here $\sin\theta$ and $\cos\theta$ will be to the fore. We will also pause to note that – by Pythagoras’s Theorem – we have: $a^2+b^2=c^2$ Substituting for $a$ and $b$ from equations $(1)$ and $(2)$ above, we get: $(c\cos\theta)^2+(c\sin\theta)^2=c^2$ or: $c^2\cos^2\theta+c^2\sin^2\theta=c^2$ As the hypotenuse is of non-zero length, then we can divide both sides by $c^2$ to get: $\cos^2\theta+\sin^2\theta=1$ We will use this result later.

With these definitions fresh in our minds, we are now going to go back to our expression for $\exp(x)$. We have:

$\exp(x)=\displaystyle\sum_{n=0}^{\infty}\dfrac{x^n}{n!}$

We going to create infinite series expressions for both $\cos(x)$ and $\sin(x)$. To do this, we employ a technique called Taylor Series. The details of how to generate the Taylor Series of these two functions are covered as an aside, which can be skipped if desired.

 Aside: Taylor Series (often called a Maclaurin Series in the context I will use them) [4] employ another tool we introduced in the last article, derivatives. Taylor showed [5] that any function can be split into an infinite sum using its derivatives of different order at a specific point. For our purposes, this means that we can describe a function, $f(x)$ as: $f(x)=\dfrac{f(0)}{0!}+\dfrac{f^\prime(0)}{1!}x+\dfrac{f^{\prime\prime}(0)}{2!}x^2+\dfrac{f^{\prime\prime\prime}(0)}{3!}x^3+\ldots$ To be clear, $f(0)$ is the value of the function at the point $0$, i.e. a number. $f^\prime(0)$ is the value of the function’s first derivative at the point $0$, i.e. a second number. Using more economic notation, we can write: $\displaystyle f(x)=\sum_{n=0}^{\infty}\dfrac{f^{(n)}(0)}{n!}x^n$ Of course the general form of these equations probably looks rather familiar, more on this later. If we want to apply this approach to $\cos(x)$ and $\sin(x)$, we need to understand the derivatives of these trigonometric functions. I’ll state without proof that: \begin{aligned}\dfrac{d}{dx}\cos(x)&=-\sin(x) \\ \\\dfrac{d}{dx}\sin(x)&=\cos(x)\end{aligned} It also helps to know that $\sin(0)=0$ and $\cos(0)=1$, things that can be easily derived from the definitions of these functions provided above. If we consider the Taylor Series for $\cos(x)$, we have [6]: $\cos(x)=\dfrac{\cos(0)}{0!}+\dfrac{\cos^\prime(0)}{1!}x+ \dfrac{\cos^{\prime\prime}(0)}{2!}x^2+\dfrac{\cos^{\prime\prime\prime}(0)}{3!}x^3+\ldots$ which gives us: $\cos(x)=\dfrac{\cos(0)}{0!}+\dfrac{-\sin(0)}{1!}x+\dfrac{-\cos(0)}{2!}x^2+\dfrac{\sin(0)}{3!}x^3+\ldots$ Substituting our values for $\sin(0)$ and $\cos(0)$, we get: $\cos(x)=\dfrac{1}{0!}+\dfrac{0}{1!}x+\dfrac{-1}{2!}x^2+\dfrac{0}{3!}x^3+\ldots$ Dropping the zero terms (and adding in a couple more higher order ones), we have: $\cos(x)=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+\ldots$ Using the same logic, it is evident that we also have: $\sin(x)=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+\ldots$ Perhaps already some sense of how the Euler Identity emerges is beginning to become apparent from considering just these two results.

In the box above, we used Taylor Series to determine that:

$\cos(x)=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+\ldots$

and:

$\sin(x)=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+\ldots$

This means that:

$\cos(x)+\sin(x)=\dfrac{1}{0!}+\dfrac{x}{1!}-\dfrac{x^2}{2!}-\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}-\dfrac{x^6}{6!}-\dfrac{x^7}{7!}\ldots$

Which begins to look tantalisingly close to:

$\exp(x)=\dfrac{1}{0!}+\dfrac{x}{1!}+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}+\dfrac{x^6}{6!}+\dfrac{x^7}{7!}+\ldots$

How can we get the two expressions to match up? This is the problem to which we will now direct our energies.

The ‘i’s have it

So the issue is with pairs of negative terms appearing periodically. We need some factor that will make terms like $x^2$ and $x^3$ negative, but terms like $x^4$ and $x^5$ positive. That is we need a modifier that exhibits the pattern:

$+,+,-,-,+,+,\hspace{2mm}\ldots$

which seems to have a period of four. We also need this modifier not to change the absolute value of any terms, so we are looking for an $x$ such that $|x|=1$. This is the problem that we tackle next.

When considering how we can modify our series for $\cos(x)+\sin(x)$, an obvious first idea is to consider incorporating $-1$. However, this approach falls short, we get:

\begin{aligned}(-1)^0&=1 \\(-1)^1&=-1 \\(-1)^2&=1 \\(-1)^3&=-1 \\&\vdots\end{aligned}

The pattern is:

$+,-,+,-,\hspace{2mm}\ldots$

which has a period of two.

How to resolve this issue with a number whose modulus is one?

Here we introduce another of Mathematics’ fundamental numbers, $i$, which is defined as being the square root of $-1$ (so $i^2=-1$). The number $i$ is at the heart of another fundamental Mathematical concept, that of Complex Numbers. I only propose to skim this subject here, for readers who are new to the area, references to my writings about Complex Numbers appear in the footnotes [7].

Succinctly, Complex Numbers are ones of the form $a+bi$ [8], where $a$ and $b$ are Real Numbers and $i^2=-1$. Complex Numbers can be added, subtracted, multiplied and divided just the same as more familiar numbers. We can also – pertinent to the subject of this article, raise numbers to the power of Complex Numbers (see box).

 Aside: Let’s pick a number at random, maybe Euler’s Number will suffice [9]. Then what are we to make of an expression like: $e^{a+ib}$ Here the series formulation of $e^x$ comes to the rescue, we simply define: $e^{a+ib}=\exp(a+ib)=\displaystyle\sum_{n=0}^{\infty}\dfrac{(a+ib)^n}{n!}$ This expression involves only multiplying and adding Complex Numbers, something that – as mentioned above – is easy to carry out.

We will come back to Complex Numbers again a little later, introducing the idea of the Complex Plane. For now, let’s consider $i$ in isolation.

If we raise $i$ to successive powers, we have:

\begin{aligned}i^0&=1 \\i^1&=i \\i^2&=-1 \\i^3&=-i \\i^4&=1 \\i^5&=i \\i^6&=-1 \\&\vdots\end{aligned}

The pattern is:

$+,+,-,-,+,+,-\hspace{2mm}\ldots$

which has the required period of four.

If we plug this into our expression for $\exp(x)$ we get:

$\exp(xi)=\dfrac{1}{0!}+\dfrac{xi}{1!}+\dfrac{x^2i^2}{2!}+\dfrac{x^3i^3}{3!}+\dfrac{x^4i^4}{4!}+\dfrac{x^5i^5}{5!}+\ldots$

Which gives us:

$\exp(xi)=\dfrac{1}{0!}+\dfrac{xi}{1!}-\dfrac{x^2}{2!}-\dfrac{x^3i}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5i}{5!}+\ldots$

This matches our pattern and is very close to our expression for $\cos(x)+\sin(x)$. However, we note that every odd term is multiplied by $i$. These are precisely the contributions from $\sin(x)$, so if we replace $\sin(x)$ with $i\sin(x)$, we get an exact match and can say that:

$\exp(xi)=\cos(x)+i\sin(x)$

This result, which is more general than Euler’s Identity, is called Euler’s Formula. It is very useful as it allows us to express any Complex Number in exponential / polar form [10].

From Formula to Identity

We spent some time deriving Euler’s Formula:

$\exp(xi)=\cos(x)+i\sin(x)$

If we now set $x=\pi\hspace{2mm}\text{radians}=180^o$, we get:

$e^{\pi i}=\exp(\pi i)=\cos(\pi)+i\sin(\pi)=-1+0$

Which we rearrange to give:

$e^{\pi i}+1=0$

and we have derived Euler’s Identity. It’s as simple as that.

A Geometric Interpretation

When briefly introducing Complex Numbers above, I mentioned something called the Complex Plane. We will now spend a little while considering this and how it relates to Euler’s Formula.

 Aside: We can plot a Complex Number $z=a+bi$ on what is known as the Complex Plane. This has the Real part of $z$ (written $\mathfrak{Re}(z)=a$) plotted on the horizontal (or x-) axis and the Imaginary part of $z$ (written $\mathfrak{Im}(z)=b$) plotted on the vertical (or y-) axis. Part of the Complex Plane is shown below, together with the position of the Complex Number $\sqrt{2}+3.75i$: Plotting $\exp(xi)=\cos(x)+i\sin(x)$ on the Complex Plane we get the following: The circle is of radius $1$; because of both Pythagoras and the trigonometric identity $\cos^2\theta+\sin^2\theta=1$ that we established earlier in the article. Again, let’s recall that we are working with radians and $360^o=2\pi \hspace{2mm}$ radians so $90^o=\dfrac{\pi}{2} \hspace{2mm}$ radians. Let’s also recall that while we are talking about an angle $x$ in the Complex Plane, we are – as per Euler’s Formula – also working with a parameter of $xi$ for the Exponential Function. We can see that as the value of the angle $x$ increases, it moves up and round the circle. When $x$ exceeds $\dfrac{\pi}{2}$ (i.e. $90^o$) the point moves from the top right quadrant to the top left one. When $x$ exceeds $\pi$ (i.e. $180^o$) the point moves from the top left quadrant to the bottom left one. When $x$ exceeds $\dfrac{3\pi}{2}$ (i.e. $270^o$) the point moves from the bottom left quadrant to the bottom right one. Finally, when $x$ exceeds $2\pi$ (i.e. $360^o$) the point goes “round the clock” and moves from the bottom right quadrant back to to the top right one. Again where we talk about an angle $\dfrac{\pi}{2}$ here, the related parameter of the Exponential Function would be $\dfrac{\pi i}{2}$. The above goes to show that the Exponential Function with a Complex Number argument is periodic [i.e. $\exp(x + A)=\exp(x)$ for some constant $A$, called the period] and further that its period is $2\pi i$, another fundamental observation.

In the box covering plotting Complex Numbers on the Complex Plane, we saw that the expression $\exp(xi)=\cos(x)+i\sin(x)$ describes a circle of unit radius, centered at the origin. We also saw how increasing the value of the angle in the Complex Plane took us “round the clock” every $2\pi$ radians. This observation helps us understand Euler’s identity geometrically.

$\pi\hspace{2mm}$ radians is simply $180^o$ and represents the point on our unit circle diametrically opposed to $x=1$ ($0^o$). Of course this is simply $x=-1$. So if we start at $x=1$ and travel half way round our unit circle (sweeping out $\pi\hspace{2mm}$ radians) we end up at $x=-1$.

If we recall (for the third time) that we need to multiply the angle in the Complex Plane by $i$ to get the parameter of the Exponential Function, then going “half-way round the clock” in the Complex Plane is equivalent of a parameter of $\pi i$ for the Exponential Function. It is this observation which forms the geometric basis of Euler’s Identity:

$\exp(\pi i)=-1$

Euler’s Identity may be beautiful. It is certainly important. However, I hope that in this brief artilce, I have shown that – viewed through the lens of the Complex Plane – The Equation is not mysterious at all.

 Consider Supporting Us   Like all of the content on peterjamesthomas.com, this article is free. However, if you enjoyed reading it, you might consider helping to support the creation of new content by making a small contribution to defray our costs. Pay as much or as little as you want. Of course this is entirely optional. Peter James Thomas

 << Euler’s Number Finches, Feathers and Apples >>> Part of the peterjamesthomas.com Maths and Science archive.

Notes

 [1] If you don’t count multiples of it, like $2e^x$ that is. [2] Originally published in The Strand Magazine and then in the collection The Adventures of Sherlock Holmes. [3] To be fair often the same people who are obsessed by finding the words of Shakespeare in the decimal representation of $\pi$. [4] Maclaurin Series are a specific type of Taylor Series, it just happens that these are the type we will be using. [5] It is somewhat unclear who actually came up with the idea of Taylor Series and who definitively proved that they work. Taylor offered the first at least semi-rigourous proof, which is why his name attaches to them. A wholly rigourous proof appeared several years later. [6] It is important to note here that when we consider the functions $\cos(x)$ and $\cos(x)$, the $x$ is not in degrees, but instead in radians, where $\displaystyle A^o = \dfrac{2\pi A^o}{360^o}\hspace{2mm}\text{radians}$ So we have: \begin{aligned}\displaystyle 360^o &= 2\pi\hspace{2mm}\text{radians} \\\displaystyle 180^o &= \pi\hspace{2mm}\text{radians} \\\displaystyle 90^o &= \dfrac{\pi}{2}\hspace{2mm}\text{radians}\end{aligned} For a fuller explanation of radians, see a section of Glimpses of Symmetry, Chapter 11 – Root of the Problem. [7] For a brief introduction to Complex Numbers, see the relevant section of my article A Brief Taxonomy of Numbers. For a fuller review, see Glimpses of Symmetry, Chapter 7 – Imaginary Battleships. [8] Or indeed $a+ib$, the order of multiplication is immaterial. [9] Of course this choice might not be entirely random. [10] Consider again a generic right-angled triangle representing a Complex Number, $z$: We can define the size of $z$, written $|z|$ as being the length of the line we have drawn. Using Pythagoras, we can see that $|z|=\sqrt{a^2+b^2}$. Then Trigonometry gives us $a=|z|\cos\theta$ and $b=|z|\sin\theta$. So we can write $z=|z|\cos\theta + i|z|\sin\theta$. But now we can now Euler’s Formula states that $e^{ix}=\cos x+i\sin x$. If we use this in the above, we can see that $z=|z|e^{i\theta}$, so we have a way of expressing any Complex Number using the Exponential Function Also, this manner of expressing $z$ relies on what angle ($\theta$) it makes with the x-axis and how far away from the origin the point is ($|z|$); this is the essence of a polar representation of a number.
 Text & Images: © Peter James Thomas 2018. Published under a Creative Commons Attribution 4.0 International License.