Euler’s Formula Presentation

Slide -1 (can use arrow keys to navigate; press "t" for timer or "h" to show/hide this UI and additional notes)

Introduction

Hello! This is a presentation about Euler’s formula that I presented at my high school’s math club on February 6, 2024.

Instead of making a boring slideshow, I decided to make an interactive website for my presentation!

These slides won’t make too much sense if you view them by themselves, so I’ve added some additional notes at the bottom of some slides (such as this one!) These additional notes are similar to the things I actually said while I was presenting.

Notes:

This presentation is designed for viewing on a computer rather than a mobile device.
Yes, this slide is slide -1. That’s because I added two slides [this one and the title slide that’s next] and I was too lazy to renumber all the slides.

Euler’s Formula Explained Simply

(and a preview of calculus)

By Eldrick Chen, calculus enjoyer and creator of unnamed calc website

Notes:

That is a picture of Leonhard Euler, who this formula is named after.

Exponentiation: It’s repeated multiplication!

\[ \large{2^3 = 2 \times 2 \times 2 = 8} \] \[ \large{3^4 = 3 \times 3 \times 3 \times 3 = 81} \]

Sounds simple, right?

Right.....?

Notes:

This basic definition of exponentiation only works if the exponent is a positive integer. But as we all know, there are numbers that aren’t positive integers.

Except it’s not so simple... we can expand the definition of exponentiation:

Zero exponent:

\[ \large{2^0 = 1} \]

Negative exponent:

\[ \large{2^{-1} = \frac{1}{2}} \]

Fractional exponent:

\[ \large{2^{1/2} = \sqrt{2}} \]

Irrational exponent:

\[ \large{2^\pi = 8.824977...} \]

Notes:

The general idea behind calculating \(2^\pi\) is that the sequence of numbers \(2^{3.14}\), \(2^{3.141}\), \(2^{3.1415}\), etc. approaches a specific value, and that value (8.824977...) is defined as \(2^\pi\).

But just how far can we push the definition of exponentiation?

\[ \large{i = \sqrt{-1}} \] \[ \large{i^2 = -1} \] \[ \large{2^i = \text{???}} \]

Can we make any sense of imaginary exponents?

Exponential Functions

The graph of \(a^x\) and the line tangent to \((0, 1)\)

\(a = \)

Slope at \(x = 0\) (Derivative):

Notes:

The tangent line to a point on a curve is the line that barely touches the curve at that point.
The slope of the tangent line is important because it tells us the rate of change of the function at that point (i.e. how fast it’s increasing).
The slope of the tangent line to a point on a function is known as the derivative of the function at that point.
In order to get the derivative of \(a^x\) at \(x = 0\) to equal 1, we must set \(a\) to a number that’s about 2.72. This number is called \(e\) and is irrational.

The Beauty of \(e^x\), the Natural Exponential Function

The graph of \(e^x\) and the line tangent to \((x, e^x)\)

\(x = \)

\(e^x \approx \)

Slope at \(x\) (first derivative):

Notes:

The derivative of \(e^x\) at any point is equal to the value of \(e^x\) at that point (i.e. the \(y\)-coordinate of that point). That’s what makes this function so important!
Higher-order derivatives:
- A first derivative describes how a function changes. A second derivative describes how the first derivative changes, a third derivative describes how the second derivative changes, and so on.
- The more derivatives you take, the better idea you have of how the function behaves.

How can we find an alternate representation of \(\sin(x)\)?

The graph of \(f(x) = \sin(x)\) and its Taylor polynomial \(p(x)\)

Degree:

Notes:

This is related to imaginary exponents, I promise.
We are trying to find Taylor polynomials for \(\sin(x)\). The idea is to approximate the sine function as well as we can with a polynomial around \(x = 0\) (the green point in the graph).

Review of polynomial degrees:
- 0th degree: Constant function (horizontal line)
- 1st degree: Any linear function
- 2nd degree: Quadratic (parabolic) function
The higher degree the Taylor polynomial is, the more accurately it approximates \(\sin(x)\).
The \(n\)th degree Taylor polynomial is designed in a way such that the first \(n\) derivatives of the polynomial evaluated at \(x = 0\) are the same as the first \(n\) derivatives of \(\sin(x)\) evaluated at \(x = 0\).
\(f'(x)\) denotes the derivative of \(f(x)\). \(f''(x)\) denotes the 2nd derivative, \(f'''(x)\) denotes the 3rd derivative, and so on. For higher derivatives, \(f^{(n)}(x)\) is often used to denote the \(n\)th derivative.
The Taylor polynomials for \(\sin(x)\) include odd powers and the factorials of odd integers. In addition, the signs between each term alternate between addition and subtraction. This will be important, so keep this in mind!
To summarize, Taylor polynomials are trying to imitate the behavior of the sine wave as best as they can. They do this by using the derivatives of \(\sin(x)\) evaluated at \(x = 0\), since derivatives tell you how a function behaves.

What if we extended our polynomial to infinity?

We get the Taylor series for \(\sin(x)\)!

\[ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \]

\(\sin(1)\) calculated with the Taylor series:

Degree:

True value of \(\sin(1)\)
Polynomial value at \(x = 1\)
Approximation error

Notes:

This Taylor series is exactly equal to \(\sin(x)\)!
The approximation error is the absolute value of the difference between the polynomial value and the true value of \(\sin(1)\).
A Taylor series is a type of infinite series.
- What this means is that as we add up more terms of the series, we get closer and closer to the true value of \(\sin(x)\) (i.e. the approximation error approaches 0).

How can we find an alternate representation of \(\cos(x)\)?

The graph of \(f(x) = \cos(x)\) and its Taylor polynomial \(p(x)\)

Degree:

Notes:

The Taylor polynomials for \(\cos(x)\) are similar to those of \(\sin(x)\), except the exponents are even and the denominators are the factorials of even integers.

Extending that last polynomial to infinity gives us the Taylor series for \(\cos(x)\):

\[ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \]

\(\cos(1)\) calculated with the Taylor series:

Degree:

True value of \(\cos(1)\)
Polynomial value at \(x = 1\)
Approximation error

The Taylor Polynomials for \(e^x\)

The graph of \(f(x) = e^x\) and its Taylor polynomial \(p(x)\)

Degree:

Notes:

We’re finally talking about exponentiation again!
The Taylor polynomials for \(e^x\) involve all of the positive integers (instead of just even or odd integers)! In addition, there are no minus signs anymore.

A new definition of exponentiation?

If we take our Taylor polynomial for \(e^x\) and add infinitely many terms, we can get the Taylor series for \(e^x\):

\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots \]

\(e\) calculated with the Taylor series:

Degree:

True value of \(e\)
Polynomial value at \(x = 1\)
Approximation error

Notes:

The Taylor series for \(e^x\) gives us a new definition of exponentiation. Instead of saying that \(e^x\) is \(e\) multiplied by itself \(x\) times, we can use this Taylor series to evaluate \(e^x\) instead.

The Major Breakthrough

What if we put an imaginary number into the Taylor series for \(e^x\)?

\[ e^i = 1 + i + \frac{i^2}{2!} + \frac{i^3}{3!} + \frac{i^4}{4!} + \frac{i^5}{5!} + \cdots \]

\(e^i\) calculated with the Taylor series:

Degree:

Polynomial value at \(x = i\):

What exactly is \(e^i\)?

\[ \begin{flalign} e^i &= 1 + i + \frac{i^2}{2!} + \frac{i^3}{3!} + \frac{i^4}{4!} + \frac{i^5}{5!} + \frac{i^6}{6!} + \frac{i^7}{7!} + \cdots \\ &= \class{red}{1} \class{blue}{+ i} \class{red}{- \frac{1}{2!}} \class{blue}{- \frac{i}{3!}} \class{red}{+ \frac{1}{4!}} \class{blue}{+ \frac{i}{5!}} \class{red}{- \frac{1}{6!}} \class{blue}{- \frac{i}{7!}}\cdots\\ &= \left(\class{red}{1 - \frac{1}{2!} + \frac{1}{4!} - \frac{1}{6!}} + \cdots\right) + \left(\class{blue}{i - \frac{i}{3!} + \frac{i}{5!} - \frac{i}{7!}} + \cdots\right)\\ &= \left(1 - \frac{1}{2!} + \frac{1}{4!} - \frac{1}{6!} +\cdots\right) + i\left(1 - \frac{1}{3!} + \frac{1}{5!} - \frac{1}{7!} + \cdots\right) \end{flalign} \]

Notes:

The exact reason why we are allowed to rearrange the terms like this is because the series \(1 - \frac{1}{2!} + \frac{1}{4!} - \frac{1}{6!} +\cdots\) and \(1 - \frac{1}{3!} + \frac{1}{5!} - \frac{1}{7!} + \cdots\) are absolutely convergent: that is, they converge (have a finite sum) even if you take the absolute value of every term. One property of absolutely convergent series is that you can rearrange their terms however you want without changing their sum.

Do those series seem familiar?

\[ e^i = \left(\class{red}{1 - \frac{1}{2!} + \frac{1}{4!} - \frac{1}{6!} +\cdots}\right) + i\left(\class{blue}{1 - \frac{1}{3!} + \frac{1}{5!} - \frac{1}{7!} + \cdots}\right) \] \[ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \] \[ \class{red}{\cos(1) = 1 - \frac{1}{2!} + \frac{1}{4!} - \frac{1}{6!} + \cdots} \] \[ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \] \[ \class{blue}{\sin(1) = 1 - \frac{1}{3!} + \frac{1}{5!} - \frac{1}{7!} + \cdots} \] \[ e^i = \class{red}{\cos(1)} + i \class{blue}{\sin(1)} \]

Generalizing The Major Breakthrough

\[ \begin{flalign} e^{ix} &= 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \cdots\\ &= 1 + ix + \frac{i^2x^2}{2!} + \frac{i^3x^3}{3!} + \frac{i^4x^4}{4!} + \frac{i^5x^5}{5!} + \cdots\\ &= \class{red}{1} \class{blue}{+ ix} \class{red}{- \frac{x^2}{2!}} \class{blue}{- \frac{ix^3}{3!}} \class{red}{+ \frac{x^4}{4!}} \class{blue}{+ \frac{ix^5}{5!}} \class{red}{- \frac{x^6}{6!}} \class{blue}{- \frac{ix^7}{7!}} + \cdots\\ &= \left(\class{red}{1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!}} + \cdots\right) + \left(\class{blue}{ix - \frac{ix^3}{3!} + \frac{ix^5}{5!} - \frac{ix^7}{7!}} + \cdots\right)\\ &= \left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots\right) + i\left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots\right) \end{flalign} \]

Finally, we arrive at Euler’s Formula:

\[ \begin{flalign} e^{ix} &= \left(\class{red}{1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots}\right) + i\left(\class{blue}{x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots}\right)\\ &= \class{red}{\cos(x)} + i\class{blue}{\sin(x)} \end{flalign} \]

Notes:

This is the formula that this entire presentation is about!

Geometric meaning

\(e^{ix}\) plotted on the complex plane. The red circle is a unit circle.

\(x =\)

\(e^{ix} \approx\)

Notes:

The complex plane is a way of visualizing a complex number by plotting its real part on the \(x\)-axis and its imaginary part on the \(y\)-axis.

Euler’s Formula: An Example

\[ \begin{flalign} e^{2i} &= \cos(2) + i\sin(2)\\ &\approx -0.416 + 0.909i \end{flalign}\]

Notes:

The angle that this point makes with the positive \(x\)-axis is 2 radians. The angle is 2 radians because we are evaluating \(e^{\class{red}{2}i}\), which Euler’s formula breaks down into a complex number involving \(\cos(2)\) and \(\sin(2)\).

Our original question: what is \(2^i\)?

\[ \begin{flalign} 2^i &= (e^{\ln(2)})^i\\ &= e^{i\ln(2)} \quad \text{(by definition)}\\ &= \cos(\ln(2)) + i\sin(\ln(2))\\ &\approx \cos(0.693) + i\sin(0.693)\\ &\approx 0.769 + 0.639i \end{flalign}\]

Notes:

The property \((a^b)^c = a^{bc}\) doesn’t always work when complex numbers are involved. However, we can do this here because \(2^i\) is defined to be equal to \(e^{i\ln(2)}\). More generally, if \(z\) is complex, \(a^z\) is defined as \(e^{z\ln(a)}\).

Euler’s identity

\[ \begin{flalign} e^{i\pi} &= \cos(\pi) + i\sin(\pi)\\ e^{i\pi} &= -1 + 0i\\ e^{i\pi} &= -1\\ e^{i\pi} + 1 &= 0 \end{flalign}\]

If we let \(\tau = 2\pi\)...

\[ \begin{flalign} e^{i\tau} &= \cos(\tau) + i\sin(\tau)\\ e^{i\tau} &= 1 + 0i\\ e^{i\tau} &= 1\\ e^{i\tau} - 1 &= 0 \end{flalign}\]

Notes:

Euler’s identity is often called a beautiful equation because it relates five important mathematical constants: 0, 1, \(e\), \(i\), and \(\pi\).
The symbol \(\tau\) (tau) is sometimes used to represent \(2\pi\) in a single constant, and some people prefer to use \(\tau\) over \(\pi\). Luckily, Euler’s identity works perfectly fine with \(\tau\)!

My Calculus Website (if you want to learn/review calculus!)

https://chaddypratt.org/calculus

Notes:

Yes, I shamelessly promoted my website after this presentation.
I have more details about the calculus behind Euler’s formula on this website! In fact, a good amount of this presentation was adopted from its section on Euler’s formula.