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This question already has an answer here:

I don't understand the formula at all:

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

I've tried reading all sorts of webpages and answers on the subject but it's just not clicking with me. I don't understand how we can define things when these are already known quantities. We've already got an exact definition for $e$ as $\lim_{n \to 0} (1+\frac{1}{n})^n$, and we've got $i = \sqrt{-1}$, and we've got $\cos$ and $\sin$ as the $x$ and $y$ coordinates of where right triangles meet the unit circle.

So I don't understand how we get from those figures to Euler's formula. Then again I also don't understand why complex numbers are of the form $a + bi$ so that might have something to do with it. I have a hard time separating what identities are "emergent" from the previous mathematics and which pieces are "by definition" and why they are defined that way. I don't understand why we start talking about "rotations" as if it's obvious that's what's happening. I don't even know why $e$ is involved in any of this to begin with.

Is $e^{ix}$ simply being written in the slightly rearranged complex form $a + ib$ where $a = \cos(x)$ and $b = \sin(x)$? Is this to be interpreted as $\cos(x)$ being the $x$-coordinate and somehow $i \sin(x)$ is the $y$ coordinate?

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marked as duplicate by Jyrki Lahtonen Feb 23 '18 at 7:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @Arthur I've used them before but without understanding how or why they work. To my memory it's sort of like taking better and better estimates to infinity (which ultimately equals the function), representing it in the form of a summation. But I don't know how it works or how it applies here. $\endgroup$ – user525966 Feb 21 '18 at 15:14
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    $\begingroup$ I don't think it's something you understand as much as it's a thing you accept and learn to use. If you actually want a reason, the best reasoning I know lies in algebra and calculus and uses Taylor series. Inserting the standard series expansions of $e^{ix}$, $\cos(x)$ and $\sin(x)$, it is quite obvious that the two sides are the same. $\endgroup$ – Arthur Feb 21 '18 at 15:14
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    $\begingroup$ That limit is not the only (I think it's not even the most common) definition of $e$. The definition of $\cos$ and $\sin$ using right triangles are not particularly useful when you want to allow complex arguments. I would suggest you try to understand complex numbers before trying to understand this formula. $\endgroup$ – Henrik Feb 21 '18 at 15:16
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    $\begingroup$ @user525966 The coefficients of the Taylor series for $f$ at $x = 0$ are $\frac{1}{k!} f^{(k)}(0)$. But we know that the derivatives of $\sin$ and $\cos$ follow a simple repeating pattern. For example, the derivatives of $\sin$ at $x = 0$ starting with the zeroth are $0, 1, 0, -1, 0, 1, 0, -1, \ldots$. So, the Taylor series is $x - \frac{1}{3!} x^3 + \frac{1}{5!} x^5 - \frac{1}{7!} + \cdots$. Now, substitute $y = i x$ in the Maclaurin series for $e^y$ and compare. $\endgroup$ – Travis Willse Feb 21 '18 at 15:19
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    $\begingroup$ You seem to be asking at least 3 different good questions, each of which is probably answered to your satisfaction elsewhere on the site: 1. Why can complex numbers all be written in the form $a+bi$? 2. Given that understanding of complex numbers, how can we connect this to a plane in a way that rotations some up naturally? 3. Given a good understanding of both of those things, and the limit definition of $e$, where does Euler's formula come from? For #3, I gave one answer at math.stackexchange.com/a/1839484/26369 $\endgroup$ – Mark S. Feb 21 '18 at 17:37

11 Answers 11

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Euler’s identity

$$e^{ix}=\cos x+i\sin x$$

is related to the following geometric interpretation

enter image description here

and can be proved by series

$$\begin{align} e^{i\theta} &{}= 1 + i\theta + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \frac{(i\theta)^4}{4!} + \frac{(i\theta)^5}{5!} + \frac{(i\theta)^6}{6!} + \frac{(i\theta)^7}{7!} + \frac{(i\theta)^8}{8!} + \cdots \\[8pt] &{}= 1 + i\theta - \frac{\theta^2}{2!} - \frac{i\theta^3}{3!} + \frac{\theta^4}{4!} + \frac{i\theta^5}{5!} - \frac{\theta^6}{6!} - \frac{i\theta^7}{7!} + \frac{\theta^8}{8!} + \cdots \\[8pt] &{}= \left( 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \frac{\theta^8}{8!} - \cdots \right) + i\left( \theta- \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + \cdots \right) \\[8pt] &{}= \cos \theta + i\sin \theta . \end{align} $$

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    $\begingroup$ (To the OP): Euler discovered it by considering infinite series definitions for $e^x$, $\sin x$ and $\cos x$ and by using $i = \sqrt{-1}$, so this identity is indeed emergent $\endgroup$ – Yuriy S Feb 21 '18 at 15:20
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    $\begingroup$ Why does it have that geometric interpretation? Why is it true? It almost seems exactly like the unit circle where $\cos$ is the $x$-coordinate and $\sin$ is the $y$ coordinate again except the $y$ axis represents imaginary numbers here. $\endgroup$ – user525966 Feb 21 '18 at 15:21
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    $\begingroup$ @YuriyS How did Euler know the infinite series expansions for those quantities? $\endgroup$ – user525966 Feb 21 '18 at 15:22
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    $\begingroup$ @user525966 The correspondence between z complex on the unitary circle and $\cos \theta + i\sin \theta $ is only matter of trigonometry, the equivalence with $e^x$ emerges out by series. $\endgroup$ – gimusi Feb 21 '18 at 15:24
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    $\begingroup$ @user525966 complex numbers have different representation we can write them in cartesian form $z=x+iy$ or in trigonometric form $z=|z|(\cos \theta + i \sin \theta)$ and this is solely a trigonometry fact since there is a correspondence between $z$ and the point $(x,y)$. Is this concept clear to you? From here we can move forward and by series show that $(\cos \theta + i \sin \theta)=e^{i\theta}$ and thus we can also express z in polar form $z=|z|e^{i\theta}$. $\endgroup$ – gimusi Feb 21 '18 at 15:39
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I like to think of Euler's formula as an algebraic result which can then be interpreted geometrically. The fact that the power series for $e^{ix}$, $\cos{x}$ and $\sin{x}$ match up has already been mentioned. Here is (I hope) some more convincing evidence:

  1. if $f(x)=e^{ix}$ and $g(x)=\cos{x}+i\sin{x}$, $f(0)=1$ and $g(0)=1$.
  2. $f'(x)=if(x)$ and $g'(x)=ig(x)$. (This is pretty easy to verify). Since $f$ and $g$ are both differentiable everywhere, they have the same value at $x=0$ and they are solutions to the same first order linear differential equation, they must be the same function.

So that's one way of looking at it, but it suffers from having to know about the existence and uniqueness of solutions to differential equations.


Another way to look at it algebraically is using trig identities:

  • $e^{i(a+b)}=e^{ia}e^{ib}$, or equivalently $f(a+b)=f(a)f(b)$

We can show that $g(x)$ has the same property by using the identities $\cos{(a+b)}=\cos{a}\cos{b}-\sin{a}\sin{b}$, and
$\sin{(a+b)}=\sin{a}\cos{b}+\cos{a}\sin{b}$.

  • $g(a+b)=\cos{(a+b)}+i\sin{(a+b)}=\cos{a}\cos{b}-\sin{a}\sin{b}+i\sin{a}\cos{b}+i\cos{a}\sin{b}$
  • $g(a)g(b)=\left(\cos{a}+i\sin{a}\right)\left(\cos{b}+i\sin{b}\right)$,

If we multiply this out it becomes the same thing as $g(a+b)$.

You can try this with any set of trig identities you like, and it will always work out that $f$ and $g$ have the same properties.


If you buy all the algebraic evidence that $e^{ix}$ and $\cos{x}+i\sin{x}$ are really the same thing, then the geometric interpretation falls out for free.

  • $\cos{\theta}+i\sin{\theta}$ is a point on the unit circle in the complex plane, at an $\theta$ counterclockwise from the positive real axis.
  • By Euler's identity, $e^{i\theta}$ is the exact same thing, and can be interpreted in the exact same way.
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    $\begingroup$ I think this should be the accepted answer. $\endgroup$ – 3.14159 Feb 22 '18 at 6:16
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Maybe this is not answering your question, but I remember reading that Euler discovered some of the connection between $e$ and $i$ through this integral:

$$\int_{0}^{\infty}\frac{dx}{x^2+1}=\frac{\pi}{2}$$

We can also write $\frac{1}{x^2+1}=\frac{1}{2i}\left (\frac{1}{x-i}-\frac{1}{x+i}\right )$. This yields: $$\frac{1}{2i}\int_{0}^{\infty}\frac{1}{x-i}-\frac{1}{x+i}dx=\left [\frac{1}{2i}\ln\left (\frac{x-i}{x+i}\right)\right ]^{\infty}_{0}=\frac{1}{2i}\ln(-1)=\frac{\pi}{2}$$ So $\ln(-1)=\pi i$, or $$e^{\pi i}=-1$$

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    $\begingroup$ I don't understand how you get from evaluating it at $\infty$ and $0$ to getting $-1$, nor do I understand how you go from $\frac{1}{2i}\ln(-1)$ to $\pi/2$. How do you take the log of a negative number? I thought it was only defined for positive numbers. $\endgroup$ – user525966 Feb 21 '18 at 15:34
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    $\begingroup$ Also, math back in Euler's day was not as rigourous as it is now. Basically, Euler had enough intuition and talent that he was able to do calculations like the one above and be convinced that they were right. $\endgroup$ – Nick Guerrero Feb 21 '18 at 16:49
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    $\begingroup$ @user525966 "How do you take the log of a negative number? I thought it was only defined for positive numbers." It's only defined for positive numbers if you restrict yourself to the real numbers. The same is true of square-roots. But once you allow for complex numbers, it's easy to define an analytic continuation of ln whose domain includes the entire real line and whose range includes the complex plane. You might then ask "why allow for complex numbers?", the entire story of complex numbers is one of allowing them because it lets you get the right answer much more easily. $\endgroup$ – Shufflepants Feb 22 '18 at 16:15
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$e^{i\theta}$ is approximately $\left(1+\frac{i\theta}{n}\right)^n$ for large values of n.

The complex number $(1+\frac{i\theta}{n})$ has a radius close to $1$ and an angle close to $\frac{\theta}{n}$. Since multiplying complex numbers multiplies the radii and adds the angles, then $\left(1+\frac{i\theta}{n}\right)^n$ should have a radius of about $1$ and an angle of about $i\theta$. In other words, it should be very close to $\cos(\theta)+i\sin(\theta)$.

All of this can be made rigorous, but this is the way I think about it morally.

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    $\begingroup$ +1. It's important here that the modulus of $1+i\theta/n$ is $1+O(1/n^2)$, so that $|(1+i\theta/n)|^n \to 1$. If it were, say, $1+O(1/n)$ then this would not happen. That is, just having a radius close to $1$ is not good enough for the radius part to work out. $\endgroup$ – Ian Feb 22 '18 at 5:17
  • $\begingroup$ @Ian Exactly. Robjohn's excellent answer below (maybe soon above?) shows exactly why. This is the rigorous answer I alluded to. $\endgroup$ – Steven Gubkin Feb 22 '18 at 11:44
  • $\begingroup$ @Ian It is also worth noting that $(1+\frac{i\theta}{n})^n$ is exactly the expression that you obtain by applying Euler's method to the differential equation $y'=y$ along the line segment from $z=0$ to $z=i\theta$. $\endgroup$ – Steven Gubkin Feb 22 '18 at 11:46
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I'll try to answer this:

I don't see why anything involving imaginary numbers is a matter of trigonometry.

What is $\sin\theta$ and $\cos\theta$? You may have learned that, for a right triangle, one must have:

$$\sin \theta = \frac{opposite}{hypothenuse} \quad \quad \cos \theta=\frac{adjacent}{hypothenuse}$$

Here comes an interesting question: Do $\sin\theta, \cos\theta$ depend only on the angle of the triangle or do they depend on the lengths of the sides of the triangle? The answer is that they depend only on the angle $\theta$ (why?). This have the following meaning: I can simplify things a little bit just taking all triangles with $hypothenuse=1$ which are similar to the the triangle I am measuring $\sin\theta, \cos\theta$ originally, and then I'll get:

$$\sin \theta = \frac{opposite'}{1}= opposite' \quad \quad \cos \theta=\frac{adjacent'}{1}=adjacent'$$

Now, if we translate this triangle to the origin, we'll have:

enter image description here

Now we know what is $\sin \theta$ and $\cos \theta$. The coordinates of the point $X$ must be $(\cos \theta, \sin \theta)$. Now, we are using an idea: There is a "basis" and we are expressing points as multiples of the elements of this basis. This means that:

$$X=(\cos \theta, \sin \theta)= \cos \theta (1,0) + \sin \theta (0,1)$$

What is this "basis"? It is a set with elements which can not be constructed with any combination of the other elements. Look at the figure, intuitively, you can't reach that point only walking upwards, you need to be able to walk to the right too. It means that you can't go to the right walking to the top and vice-versa. If I want to reach that point, I need to walk $\cos \theta$ upwards and then $\sin \theta$ to the right. Each element of the basis can't be constructed with the other, this is called "linear independence". Isn't this similar to complex numbers? A complex number is a number of the form:

$$z=a+bi$$

In which you can't get $a$ by multiplying $i$ by some real number and vice-versa. This means that we also have a basis here, we just need to switch $(1,0)$ by $1$ and $(0,1)$ by $i$:

$$X=\cos \theta (1,0) + \sin \theta (0,1)=\cos \theta (1) + \sin \theta (i)=\cos \theta + i \sin\theta $$

Obviously, it's not as simple as that: We must also show that geometry done with multiples of $\{(1,0),(0,1)\}$ is the same of geometry done with multiples of $\{1,i\}$, as it turns out, they are usually defined to be very closely related. We could have done the same with polynomials of degree $\leq1$ too, for example:

$$X=\cos \theta (1) + \sin \theta (x)=\cos \theta + i \sin\theta x$$

But the most interesting case would be the case in which these polynomials can give you a geometry fairly similar to the geometries of the other cases I mentioned. Why did they went to a geometry of multiples of $\{(1,0),(0,1)\}$ to a geometry of multiples of $\{1,i\}$? Because the complex numbers have a lot of pleasing properties which - perhaps - you couldn't have derived the other way around. If we "translate" the geometry from one kind of basis to another, you gain every interesting property they have and also, you can find a translation from this pleasing property to the previous system you were working with! Now ask yourself: What properties were gained from this "translation"?

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My first response would have been along the lines given by gimusi above. Since that's already been well covered, I'll offer a non-rigorous perspective that might help make the connection between $exp$ and $sin/cos$ easier to accept.

Forget about triangles and circles for a moment, and forget about imaginary numbers as well. Let's start out with just the reals.

Suppose that we have some quantity that changes over time, whose second derivative is proportional to its value but in the opposite direction. For instance, consider a marble rolling back and forth in a round-ish bowl: the further it moves away from the centre, the steeper the slope, and the stronger the resultant force pushing it back towards the centre. If we shape the bowl just right, the acceleration will look something like:

$d^2x/dx^2 = ax$, with $x$ being the displacement and $a$ some negative constant.

We know that this results in simple harmonic motion, and a little bit of calculus shows that this motion can be described in terms of $sin$ and $cos$ functions.

Now flip that bowl upside-down, and place the marble on top of it. Instead of experiencing a restoring force (proportional and opposite to displacement), it now experiences a force that acts to magnify its displacement, proportional to displacement and in the same direction.

In mathematical terms: $d^2x/dx^2 = ax$, with $a$ now being some positive constant.

This is an exponential growth relationship, and a little bit of calculus shows that it can be described in terms of $exp$ functions.

So the same basic rule, $d^2x/dx^2 = ax$, gives us either an exponential function or a sine/cosine function, depending on the sign of $a$. Even with no knowledge of complex numbers, this suggests that $exp$ and $sin/cos$ are intimately related.

Now let's write that slightly differently. Instead of having to stipulate the sign of $a$, we can take advantage of the fact that the square of a real number is always non-negative. For real $k$:

$d^2x/dx^2 = -k^2x$ -> simple harmonic motion, described by $sin$ and $cos$

$d^2x/dx^2 = k^2x$ -> exponential growth, described by $exp$

At this point, we can reintroduce complex numbers. Note that $-k^2$ = $(ki)^2$ and $k^2$ = $-(ki)^2$. This suggests that an exponential function based on a real-valued $k$ should be the same thing as a sine/cosine function based on imaginary $k$, and an exponential function with imaginary $k$ should look like a sine/cosine function with real $k$.

It's a bit like the parable of the elephant: we tend to introduce $sin$ and $cos$ functions in a geometric context, because it's easy to visualise and doesn't require a lot of prior knowledge, and also because the geometric applications have a lot of practical use. But they're actually much more general, useful for analysing a huge variety of non-triangle-related concepts: heat flow, sound waves, etc. etc.

Meanwhile, we introduce exponential functions in terms of bank balances and compound interest, maybe phenomena like radio-decay. We don't tell students at the start that these are all really just different aspects of the same deeper concept, because it requires a lot more knowledge to understand them as a whole. By the time people do have that requisite knowledge, it's hard to rethink one's understanding of these functions.

Another analogy that can be used in place of the marble-and-bowl version is population dynamics: in a simple growth model with just one species and no capacity limits you get exponential growth, but in a predator-prey model you tend to get sine/cosine behaviours.

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  • $\begingroup$ +1, Euler's formula is straightforward when you realize that every solution of $f''=-f$ can be written as a linear combination of the sine and cosine functions, and that $e^{ix}$ is solution of this differential equation. $\endgroup$ – TonioElGringo Feb 22 '18 at 10:49
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Copied from this answer:

Starting with this formulation of $e^x$ $$ e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n\tag{1} $$ and extending this definition to $e^{ix}$: $$ e^{ix}=\lim_{n\to\infty}\left(1+\frac{ix}{n}\right)^n\tag{2} $$ For a complex number $z$, let $|z|$ be its magnitude and $\arg(z)$ be its angle. If it is not already known, only a small amount of algebra and trigonometry is needed to show that $$ \begin{align} |wz|&=|w|\cdot|z|\tag{3a}\\ \arg(wz)&=\arg(w)+\arg(z)\tag{3b} \end{align} $$ Induction then shows that \begin{align} |z^n|&=|z|^n\tag{4a}\\ \arg(z^n)&=n\arg(z)\tag{4b} \end{align} Let us take a closer look at $1+\dfrac{ix}{n}$. $$ \begin{align} \left|\,1+\frac{ix}{n}\,\right|&=\sqrt{1+\frac{x^2}{n^2}}\tag{5a}\\ \tan\left(\arg\left(1+\frac{ix}{n}\right)\right)&=\frac xn\tag{5b} \end{align} $$ Using $(4a)$, $(5a)$, and $(2)$, we get $$ \begin{align} |e^{ix}| &=\lim_{n\to\infty}\left|\,1+\frac{ix}{n}\,\right|^n\\ &=\lim_{n\to\infty}\left(1+\frac{x^2}{n^2}\right)^{n/2}\\ &=\lim_{n\to\infty}\left(1+\frac{x^2}{n^2}\right)^{\frac{n^2}{2n}}\\ &=\lim_{n\to\infty}e^{\frac{x^2}{2n}}\\[12pt] &=1\tag{6} \end{align} $$ It can be shown that when $x$ is measured in radians $$ \lim_{x\to0}\frac{\tan(x)}{x}=1\tag{7} $$ Using $(4b)$, $(5b)$, and $(7)$, we get $$ \begin{align} \arg(e^{ix}) &=\lim_{n\to\infty}n\arg\left(1+\frac{ix}{n}\right)\\ &=\lim_{n\to\infty}n\arg\left(1+\frac{ix}{n}\right) \frac{\tan\left(\arg\left(1+\frac{ix}{n}\right)\right)}{\arg\left(1+\frac{ix} {n}\right)}\\ &=\lim_{n\to\infty}n\frac xn\\ &=x\tag{8} \end{align} $$ Using $(6)$ and $(8)$, we get that $e^{ix}$ has magnitude $1$ and angle $x$. Thus, converting from polar coordinates: $$ e^{ix} = \cos(x) + i\sin(x)\tag{9} $$ We get the rotational action from $(9)$ and $(3)$.

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I'll try to address these pieces

I don't understand how we can define things when these are already known quantities. We've already got an exact definition for $e$ as $\lim_{n \to 0} (1+\frac{1}{n})^n$, and we've got $i = \sqrt{-1}$, and we've got $\cos$ and $\sin$ as the $x$ and $y$ coordinates of where right triangles meet the unit circle.


I have a hard time separating what identities are "emergent" from the previous mathematics and which pieces are "by definition" and why they are defined that way.

You're right, we do have those definitions. And yes, someone can't come up to you and define $e$ as $3.7$, because $e$ already has a definition.

We've had a definition of what $e^x$ is when $x$ is a real number. What we didn't have was a definition of $e^x$ when $x$ is an imaginary number. Before we give the definition, $e^{bi}$ is some meaningless writing. You could define it to to be whatever you like, ex. $b$, or $5$, or $\log 10b$, etc. But people chose to define it as $\cos(x) + i \sin(x)$ because it would have some really cool propreties (ex. defining it this way preserves some of the propreties of the real $e^x$ we are used to).

Another common example is infinite sums. We have defined $\sum_{i=1}^n a_i$ as $a_1+a_2+ \cdots + a_n$. But again this does not tell us anything about the meaning of the symbol $\sum_{i=1}^{\infty} a_i$. We could again assign anything we want to it, such as $7$, or $a_4$, or $\pi$. But we have decided to assign it $\lim_{n \to \infty} \sum_{i=1}^n a_i$

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    $\begingroup$ Is it really meaningless though? From the Taylor expansions it doesn't seem like we could have defined it otherwise. $\endgroup$ – user525966 Feb 23 '18 at 5:37
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    $\begingroup$ @user525966 Let $f(x)$ be the Taylor expansion of $e^x$. Since $e^i$ is a new symbol, previously undefined, there is no requirement for $e^i$ to be defined as $f(i)$. It would be nice if that turns out to be true (and this is one of the things that is preserved by defining $e^{ix} = \cos(x) + i \sin(x)$) but nothing requires it. We could just as simply defined another power series $g(x)$ and decreed that $e^x = f(x)$ when $x$ is real (this part we are forced to because we already have a definition for $e$ and real exponents) and $e^x = g(x)$ when $x$ is imaginary (This part we are free) $\endgroup$ – Ovi Feb 23 '18 at 5:50
  • $\begingroup$ Ahhh that makes a lot of sense when you put it that way, thanks $\endgroup$ – user525966 Feb 23 '18 at 6:49
  • $\begingroup$ @user525966 Glad I could help! $\endgroup$ – Ovi Feb 23 '18 at 7:13
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Proving the formula with Taylor Series

The Power series and the Taylor Series:

First, let's see the definition of a Power series at 0:

$$f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + ... + a_\infty x^\infty$$

Which is:

$$f(x) = \sum_{n=0}^\infty a_n x^n$$

How to find a Taylor series?

In functions whose derivative eventually end up resulting in the function itself (maybe changed by constants), we can create lots of equations of different orders, by derivating our known function and the sum.

By taking advantage of the fact that making x=0 in those equations will always result in a single constant (all other constants will vanish or be multiplied by 0), we will then find the values for the constants and indentify a pattern that will lead us to know all the values of the a constants up to infinity.

See:

$$\ f(x) = a_0 + a_1 x +\ a_2 x^2 + \ a_3 x^3 \ + ... + a_n x^n \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$ $$f'(x) = 0 + \ \ a_1 \ + 2 a_2 x + 3a_3 x^2 + ... + \ \ \ na_n x^{n-1}\ \ \ \ \ \ \ \ \ $$ $$f''(x) = 0 + \ \ 0\ \ + \ \ 2 a_2\ + 6a_3 x \ + ... + \ n(n-1)a_n x^{n-2}$$

And if we make x=0:

$$f(0) = a_0$$ $$f'(0) = a_1$$ $$f''(0) = 2a_2$$ $$f'''(0) = 6a_3$$ $$\frac{d^nf(0)}{dx^n} = (n!)a_n$$

Finding the series of the exponential

The easiest one, where its derivative is itself:

$$f(x) = f'(x) = f''(x) = ... = e^x$$

From where you get for x=0:

$$a_0 = a_1 = 2a_2 = 6a_3 = ... = 1$$

Then:

$$a_n = 1/n! \ \ \ \Longrightarrow \ \ \ e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$

Exponential with an imaginary expoent:

$$f(0) = e^{0i} = +1 = a_0$$ $$f'(0) = ie^{0i} = +i = a_1$$ $$f''(0) = i^2e^{0i} = -1 = 2!a_2$$ $$f'''(0) = i^3e^{0i} = -i = 3!a_3$$ $$f''''(0) = i^4e^{0i} = +1 = 4!a_4$$

Notice the pattern: +1; +i; -1; -i; +1; +i; -1; -i; ...

We see we've got shifting signs and a real and an imaginary parts: the real part taking even values of n and the imaginary part taking odd values of n.

$$e^{ix} = \sum_{n=0}^\infty \frac{(-1)^n x^{2n} }{(2n)!} + i \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}$$

Sinus and cosinus:

Now, here is the magic!

Follow the same process to get the Taylor series for these two.

For the sinus:

$$f(0) = +sin(0) = 0 = a_0$$ $$f'(0) = +cos(0) = +1 = a_1$$ $$f''(0) = -sin(0) = 0 = 2!a_2$$ $$f'''(0) = -cos(0) = -1 = 3!a_3$$ $$f''''(0) = +sin(0) = 0 = 4!a_4$$

Look at the shifting pattern! And look at the skipped constants (the zeros).

$$sin(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}$$

And the cosinus:

$$f(0) = +cos(0) = +1 = a_0$$ $$f'(0) = -sin(0) = 0 = a_1$$ $$f''(0) = -cos(0) = -1 = 2!a_2$$ $$f'''(0) = +sin(0) = 0 = 3!a_3$$ $$f''''(0) = +cos(0) = +1 = 4!a_4$$

Exactly the complement of the sinus:

$$cos(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}$$

Conclusion

And voilà!!!! Like a charm:

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

This conclusion does make me wonder and wonder. And I really can't understand what it means, but it's certainly one of the most magical and misterious things about Math I've seen....

I mean... I don't even know what an imaginary number is supposed to be, and then you make it exponential....

But the scret lies in the fact that, derivating all the functions, a cycling pattern appears.

The exponential generates increasing powers of i, which will create the +1, +i, -1, -i cycle.

The sinus and the cosinus also cycle when derivated: +sin, +cos, -sin, -cos.

It happens that both cycles match.

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  • $\begingroup$ Why do you say that none of the answers demonstrates the formula? More than one of them does. $\endgroup$ – Gareth McCaughan Mar 1 '18 at 20:36
  • $\begingroup$ My bad.... updated my answer. $\endgroup$ – Daniel Möller Mar 1 '18 at 22:42
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Let's start with $e=\lim_{n\rightarrow\infty}\left(1+\frac1n\right)^n$ and get to Euler's formula, hopefully with not too much handwaving. Each step here is meant to be "obvious" in some sense, and once you can fit them all into your brain at the same time Euler's formula should also be "obvious" :-).

Step 1: from $e$ to $e^x$

We have $e^x=\lim_{n\rightarrow\infty}\left(1+\frac1n\right)^{nx}=\lim_{m\rightarrow\infty}\left(1+\frac xm\right)^m$. (Write $m=nx$ or, if you're worried about keeping it integral, the nearest integer.)

Step 2: small exponents

Suppose $x$ is very small. The binomial theorem says that $\left(1+\frac xm\right)^m=1+m\frac xm+\cdots$ where the remaining terms all have $x^2$ or higher as a factor. Then for small $x$ those terms are small; and so $e^x\simeq1+x$ for small $x$.

Step 3: the derivative

What happens if we increase $x$ a little? $e^{x+h}=e^xe^h\simeq e^x(1+h)$ when $h$ is small, from which it follows that $\frac d{dx}e^x=e^x$.

Step 4: imaginary arguments, differential equations

So now let's look at $f(t)=e^{it}$. We have $f'(t)=ie^{it}=if(t)$.

Think of $t$ as time, so that this differential equation describes how $f(t)$ is moving around the complex plane.

Step 5: geometry

In the complex plane, $f'$ is perpendicular to $f$ (treating both as vectors from the origin) because $f'$ is just $f$ rotated by a right angle.

This means that $|f|$ is constant, because $\frac d{dt}|f|^2=\frac d{dt}\left(f\cdot f\right)=2f'\cdot f=0$ (here the $\cdot$ is the scalar product, thinking of $f(t)$ as a vector in ${\Bbb R}^2$). Since obviously $f(0)=1$, this means that $|f|=1$ always. So $f(t)$ is always on the unit circle.

This in turn means that $|f'|=1$ because $f'=if$. So $f(t)$ is moving with constant speed around the unit circle. The constant speed is 1 (i.e., 1 unit of distance moved when $t$ changes by 1) and it turns out that the motion is always anticlockwise.

Step 6: we're done

Well, what does moving with constant unit speed around the unit circle mean? Angles (in radians) are just distances around the unit circle, so at time $t$ this means that $f(t)$ is at angle $t$ on that circle. (Measured anticlockwise from the positive $x$-axis, which is exactly what the argument in step 5 actually gives us.)

And this is exactly what we need, because the correct definition of $\cos$ and $\sin$ is that the point at angle $t$ on the unit circle is $(\cos t,\sin t)$.

Remarks on handwaving

Since the idea is to make the path to Euler's formula intuitive, I've omitted details in a few places. Here, for the sake of honesty, are a few words about those places. 1 In step 1, the first equality assumes that we know what "raising to the power $x$" means and that it's a continuous function. Of course it is ... but, depending on how you define these things, that may not be a triviality. 2 In step 2, when I say "for small $x$ those terms are small" I'm implicitly interchanging two limiting operations (one where $n\rightarrow\infty$ and one where $x\rightarrow0$). That isn't always safe. I don't think making it rigorous is hard. 3 In step 3, I should really be keeping track of epsilons and deltas and being more precise about what that $\simeq$ means. This is just a matter of bookkeeping. 4 In step 4, I'm now raising things to imaginary powers. The cleanest way to define what that means actually uses the exponential function. So to make all this rigorous, what we should really do is to write $\exp x$ instead of $e^x$, prove that $\exp(x+h)=\exp x\exp h$, etc. None of that's difficult. 5 In step 5, I should really say "constant velocity" rather than "constant speed" and take care about which sort of right-angle rotation multiplication by $i$ does, so as to make it clear that $f(t)$ is always moving anticlockwise around the unit circle.

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Ultimately, what we want to do is extend real functions to the complex domain. In principle, this can be done in any arbitrary way, but there is at most only one unique analytic continuation of real functions to the complex plane, see here. As pointed out here, complex differentiable functions are analytic and vice versa, so all we need to do is check if Euler's formula yields a complex differentiable definition of the exponential function.

Any function of the complex variable $z = x + i y$, can be spit up in its imaginary and real parts:

$$f(x + i y) = u(x,y) + i v(x,y)$$

where $u$ and $v$ are real functions of the real variables $x$ and $y$. As explained here, complex differentiability is equivalent to $u$ and $v$ satisfying the Cauchy-Riemann equations. In case of the exponential function, Euler's formula implies:

$\exp(z) = \exp(x)\left[\cos(y) + i \sin(y)\right]\tag{1}$

This obviously reduces to the real exponential function on the real line. It's also easy to check that the pair $u(x,y) = \exp(x)\cos(y)$ and $v(x,y) = \exp(x)\sin(y)$ satisfy the Cauchy-Riemann equations, therefore Eq. (1) is the unique analytic continuation of the real exponential function.

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  • $\begingroup$ +1 for C-R equations. That $\cos x +i\sin x$ is the only way to describe $e^x$ on the complex plane while maintaining analyticity is, in my opinion, the best way to view the formula. Indeed, the many of other techniques to explain the formula---like the series method---assume analyticity to begin with! $\endgroup$ – Argon Feb 23 '18 at 5:32

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