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<h1>All about e</h1>
I recently read Richard Feynman’s <i>Lectures on Physics, vol. I</i>, and he explained something that none of my professors ever did— an equation he calls an "amazing jewel".
<p>So I thought I’d tell you about it. You can read Feynman yourself, of course— it’s chapter 22— but he was talking to college physics students, and I’m not going to assume you know anything more than high school math. Plus it’s no longer 1963, so I’ll assume you have a calculator, or access to <a href="http://www.wolframalpha.com">Wolfram Alpha</a>.
<h2>The derivative of e<sup>x</sup></h2>
Let’s start with the number e, which starts
<blockquote>
2.718281828459045235360287471352662497757247093699959574966967...
</blockquote>
You’ll probably remember e as the base of natural logarithms. But before we get to logarithms, e has some remarkable characteristics.
<p>To see one of them, let’s look at the function y = e<sup>x</sup>.
<center><img src="trig-yex.png"></center>
What’s its derivative? We can guess by figuring out the slope at a few points, for instance at x = 1. The slope for a straight line between points x<sub>1</sub>, y<sub>1</sub> and x<sub>2</sub>, y<sub>2</sub> is (y<sub>2</sub>-y<sub>1</sub>)/(x<sub>2</sub>-x<sub>1</sub>).
<center><img src="trig-slope.png"></center>
We can approximate the slope for the graph by picking two points very close together— say 1.0 and 1.0001. e<sup>1</sup> is of course e ≅ 2.7182818, while e<sup>1.0001</sup> is ≅ 2.7185537.
<p><i>Note: We’re going to be seeing a lot of irrational numbers here. If you see ≅ (approximately equal), that’s a reminder that I’ve truncated the number, which really goes on forever.</i>
<p>So the slope is
<blockquote>
(2.7185537 - 2.7182818) / (1.0001 - 1) ≅ 2.718417747
</blockquote>
Which is close to e. Let’s try a better approximation of the slope, by choosing x<sub>2</sub>, y<sub>2</sub> closer to 1:
<blockquote>
(e<sup>1.000001</sup> - e<sup>1</sup>)/ .000001 ≅ 2.718283188
</blockquote>
Which is even closer to e. It’s beginning to look like the slope at e<sup>1</sup> is e<sup>1</sup>, and in fact it is. You can verify all of these at Wolfram Alpha; e.g. copy and paste <tt>((e^1.000001) - (e^1))/.000001</tt>. See what happens if you use closer and closer points.
<p>Let’s try e<sup>0</sup>, whose value is 1.
<blockquote>
(e<sup>0.000001</sup> - e<sup>0</sup>)/.000001 ≅ 1.000000500
</blockquote>
How about e<sup>2</sup> ≅ 7.3890560989?
<blockquote>
(e<sup>2.000001</sup> - e<sup>2</sup>)/.000001 ≅ 7.389059793
</blockquote>
How about e<sup>π</sup> ≅ 23.14069263?
<blockquote>
(e<sup>π+.000001</sup>-e<sup>π</sup>)/.000001 ≅ 23.140704203
</blockquote>
As these examples suggest, the slope of e<sup>x</sup> at any point is e<sup>x</sup>. That is, <b>the derivative of e<sup>x</sup> is itself</b>.
<blockquote>
d/dx e<sup>x</sup> = e<sup>x</sup>
</blockquote>
This is the only function of x whose derivative is itself. That makes e<sup>x</sup> very nice to work with in calculus— we’ll get back to that later.
<h2>Interested in defining e</h2>
e was originally calculated by Jacob Bernoulli by evaluating the limit
<blockquote><img src="trig-bernoulli.png"></blockquote>
You have a dollar in an account that pays 100% interest in one year; after the year you’ll have $2.
<p>Suppose it’s 50% every 6 months, instead. Now you get $1 × 1.5 × 1.5 = $2.25.
<p>If there are n intervals, the interest is 100%/n... that is, 1/n. Plus you keep the principal, so you have 1 + 1/n. E.g. for one month it’s 1 + 1/12 = 108.33% and after the year you have
<blockquote>
(1 + 1/12)<sup>12</sup> ≅ 2.61303529
</blockquote>
<p>Compounding daily, at 0.2739726%, gives us
<blockquote>
(1+(1/365))<sup>365</sup> ≅ 2.714567482
</blockquote>
That’s starting to look like e, and if we keep making smaller and smaller divisions, that’s exactly where we end up.
<h2>Logarithms</h2>
To most people logarithms probably mean "mumble math something", so think of them as <b>exponents</b>, or <b>order of magnitude</b>. If you ask what order of magnitude a million is, you are in fact asking for log<sub>10</sub> 1,000,000 = 6.
<p>We use <b>logarithmic scales</b> all the time; they’re great for reducing enormous ranges of huge numbers to easy-to-grasp small numbers. Some common examples:
<ul>
<li>decibels: if a sound is ten times louder than another one, it’s 10 decibels higher
<li>the Richter scale: an earthquake 1.0 higher on this scale has 10 times the amplitude
<li>stellar magnitude: if one star is 100 times brighter than another, that’s 5 magnitudes
<li>the computing abbrevations kilo, mega, giga, tera, peta are a logarithmic scale— each step is 1024 times the last one
</ul>
As a refresher, if a<sup>b</sup> = c, then a = <sup>b</sup>√c, and b = log<sub>a</sub> c. For instance, 10<sup>3</sup> = 1000, so 10 = <sup>3</sup>√1000, and 3 = log<sub>10</sub> 1000.
<p>Since we normally use base 10, base 10 logarithms are particularly easy.
<ul>
<li>Log<sub>10</sub> 1000 = 3— just count the zeroes.
<li>Log<sub>10</sub> 10<sup>45</sup> = 45.
<li>Log<sub>10</sub> .001 = -3; count the places after the decimal point.
</ul>
For centuries the most useful thing about logs was that they made for easy multiplication of large numbers. E.g. you want to multiply 3.14159265 by 2.7182818. That works out to
<blockquote><tt>
3.14159265
<br> ×2.7182818
<br>-----------------
<br> 2513274120
<br> 3141592650
<br> 251327412000
<br> 629318530000
<br> 25132741200000
<br> 31415926500000
<br> 2199114855000000
<br> 6293185300000000
<br>-----------------
<br>8.539734123508770
</tt></blockquote>
Not difficult, but tedious and error-prone. But we can take advantage of the fact that
<blockquote>
log ac = log a + log c
</blockquote>
We can look up the logs in a table... tables to 14 decimal places have been available since 1620.
<blockquote>log<sub>10</sub> 3.14159265 ≅ .497149872
<br>log<sub>10</sub> 2.7182818 ≅ .434294477
<br>sum ≅ .93144434955
</blockquote>
Now we can use the same table to find out what number this is the log of, which turns out to be ≅ 8.53973412.
<p>Of course, these days we do nothing of the sort— we just use a calculator. But it was a huge time saver until the computer was invented.
<p>(Some of you retro-geeks may say "No, no, people used slide rules." What do you think a slide rule is? Among other things, it’s a couple of logarithmic scales.)
<h2>One weird trick</h2>
There’s an interesting trick for taking powers of 1/n with higher and higher n. Let’s look what happens with e<sup>1/n</sup>, doubling n each time. (Or to put it another way, take e, then √e, then the square root of that, and so on.)
<blockquote>
<table>
<tr>
<td>1/n</td> <td>e<sup>1/n</sup></td> <td>1 + 1/n</td>
</tr>
<tr><td>1 </td><td>2.7182818 </td><td>2 </td></tr>
<tr><td>1/2 </td><td>1.648721271 </td><td>1.5 </td></tr>
<tr><td>1/4 </td><td>1.2840254 </td><td>1.25 </td></tr>
<tr><td>1/8 </td><td>1.133148453 </td><td>1.125 </td></tr>
<tr><td>1/16 </td><td>1.06449446 </td><td>1.0625 </td></tr>
<tr><td>1/32 </td><td>1.0317434075 </td><td>1.03125 </td></tr>
<tr><td>1/64 </td><td>1.015747709 </td><td>1.015625 </td></tr>
<tr><td>1/128 </td><td>1.007843097 </td><td>1.0078125 </td></tr>
<tr><td>1/256 </td><td>1.003913889 </td><td>1.00390625 </td></tr>
<tr><td>1/512 </td><td>1.00195503 </td><td>1.001953125 </td></tr>
<tr><td>1/1024 </td><td>1.000977039 </td><td>1.0009765625 </td></tr>
<tr><td>1/2048 </td><td>1.000488400 </td><td>1.00048828125 </td></tr>
</table>
</blockquote>
If you’re thinking "Oooh, the limit of e<sup>1/n</sup> is 1", that’s correct, but that’s not the weird trick. The trick is that a good guess for e<sup>1/n</sup> is 1 + 1/n, and it’s an increasingly good guess as n gets huge.
<p>We’ll use this trick in a clever way in a moment, but I’ll also point out two things.
<ol>
<li>This particular series is important if you want to calculate logarithms by hand. See Feynman’s lecture if you want to know how.
<li>You may hear that e is a "convenient" base for logarithms. One thing that means is that this particular neat trick is made easier if your base is e. If it’s 10 instead, the trick involves another step: a good guess for 10<sup>1/n</sup> is 1 + 2.3026 * (1/n). We get rid of the constant by using base-e logarithms.
</ol>
<h2>What about e<sup>ix</sup>?</h2>
Now, all of the algebraic operations— addition, multiplication, exponentiation, and their inverses subtraction, division, taking a root, finding a logarithm— should work on any number, so long as we’ve defined ‘number’ broadly enough. It turns out that complex numbers are as far as we need to go.
<p>All of which means that a function like y = e<sup>ix</sup> should be fine. Now, we can understand raising a number to a power n as multiplying it by itself n times. This is stretched by such ideas as raising it to a fraction or a negative number, and seems almost to break if the number is imaginary. But the algebra all works, so we’ll let the meaning take care of itself.
<p>Still, how <b>do</b> we evaluate an expression like e<sup>ix</sup>? Where do we even start?
<p>We actually <i>have</i> a place to start— the observation from the last section that a good guess at e<sup>i/n</sup> is 1 + 1/n. Let’s assume that it’s true for complex numbers as well.
<p>Let’s set n to 1024. That is, a good guess for e<sup>i/1024</sup> is 1 + i/n = 1 + i/1024. That of course is the complex number 1 + .0009765625i.
<p>That gives us <i>one</i> exponential of an imaginary number. (Or rather, an approximation; the actual value ≅ .999999523 + 0.00097656234i.
<p>Given one value of e<sup>ix</sup>, here are ways to get more; here’s one. Let’s try for i/512. We could use the 1 + 1/n trick, but it gets less accurate as 1/n gets bigger. Instead, let’s take
<blockquote>
e<sup>i/1024</sup> = 1 + .0009765625i
</blockquote>
and square both sides. Now, (a<sup>b</sup>)<sup>c</sup> is always a<sup>bc</sup>, so
<blockquote>
(e<sup>i/1024</sup>)<sup>2</sup> = e<sup>(i/1024)×2</sup> = e<sup>i/512</sup>
</blockquote>
For the other side, the general rule for complex numbers is
<blockquote>
(a × bi)<sup>2</sup> = (a × bi)(a × bi)
<br> = a<sup>2</sup> + 2abi + b<sup>2</sup>i<sup>2</sup>
<br> = (a<sup>2</sup> - b<sup>2</sup>) + 2abi
</blockquote>
Here a = 1 and b = .0009765625, so we get
<blockquote>
(1×1 - .0009765625×.0009765625) + 2×1×.0009765625×i
<br> = 0.9999990463 + 0.001953125 i
</blockquote>
Let’s see what e<sup>in</sup> looks like, as a function. (I’m just going to let Wolfram Alpha calculate it.) Here’s how it goes for several values of n.
<blockquote>
<table>
<tr><td width="100px">0 </td><td>1</td><tr>
<tr><td>.2 </td><td>0.9800666 + 0.1986693 i</td><tr>
<tr><td>.4 </td><td>0.921061 + 0.389418 i</td><tr>
<tr><td>.6 </td><td>0.825336 + 0.564642 i</td><tr>
<tr><td>.8 </td><td>0.696707 + 0.717356 i</td><tr>
<tr><td>1 </td><td>0.540302 + 0.841471i</td><tr>
<tr><td>1.2 </td><td>0.362358 + 0.932039 i</td><tr>
<tr><td>1.4 </td><td>0.169967 + 0.985450 i</td><tr>
<tr><td>1.6 </td><td>-0.0291995 + 0.999574 i</td><tr>
<tr><td>1.8 </td><td>-0.227202 + 0.973848 i</td><tr>
<tr><td>2 </td><td>-0.416147 + 0.909297 i</td><tr>
<tr><td>2.2 </td><td>-0.588501 + 0.808496 i</td><tr>
<tr><td>2.4 </td><td>-0.737394 + 0.675463 i</td><tr>
<tr><td>2.6 </td><td>-0.856889 + 0.515501 i</td><tr>
<tr><td>2.8 </td><td>-0.942222 + 0.334988 i</td><tr>
<tr><td>3 </td><td>-0.989992 + 0.141120 i</td><tr>
<tr><td>3.2 </td><td>-0.998295 - 0.0583741 i</td><tr>
<tr><td>3.4 </td><td>-0.966798 - 0.255541 i</td><tr>
</table>
</blockquote>
It looks like there’s some patterns here. All the values we’ve seen so far, both for the real part and the imaginary part, live between -1 and 1. Moreover, looking down the real column, we see the value smoothly going down; when it reaches -1, toward the end of our list, it inches back up again. The imaginary column goes up, reaches 1, and then declines.
<h2>Circling round the subject</h2>
Feynman points out something else of note. Simply from the definition of a complex number, we know that
<blockquote>
e<sup>in</sup> = x + iy
</blockquote>
Now, a square root always has two solutions, positive and negative. We usually say that √2 is 1.41421356, but we could equally well use -1.41421356. Likewise, if i is √(-1), so is -i. Replacing i with -i in an equation is called taking the <i>complex conjugate</i>, and it gives you another true equation:
<blockquote>
e<sup>-in</sup> = x - iy
</blockquote>
Now let’s multiply these equations together:
<blockquote>
e<sup>in</sup> e<sup>-in</sup> = (x + iy)(x - iy)
</blockquote>
Simplifying the left side:
<blockquote>
e<sup>in</sup> e<sup>-in</sup> = e<sup>in - in</sup> = e<sup>0</sup> = 1
</blockquote>
And the right side:
<blockquote>
(x + iy)(x - iy) = x<sup>2</sup> +xyi -xyi - i<sup>2</sup>y<sup>2</sup> = x<sup>2</sup> + y<sup>2</sup>
</blockquote>
So for any complex number x + iy that we found as a value of e<sup>in</sup>, we know that
<blockquote>
x<sup>2</sup> + y<sup>2</sup> = 1
</blockquote>
Does this ring any bells yet? That’s the formula for a circle, so circles are involved in this somehow.
<h2>The big reveal</h2>
<p>Let’s just get a plot of e<sup>in</sup>. The red line is the real part, the blue line is the imaginary part.
<center><img src="trig-waves.png"/></center>
Those look like sine waves. And they <i>are</i> sine waves! In fact we can write
<blockquote>
e<sup>in</sup> = cos n + i sin n
</blockquote>
This is the equation Feynman calls the "amazing jewel" of mathematics. We’re just messing around with exponential functions, and we suddenly arrive at the trigonometric functions.
<p>The equation was first published by Leonhard Euler in 1748.
<p>Rather than plotting e<sup>in</sup>, we can also plot x and y as n varies. We get this graph:
<center><img src="trig-circle.png"/></center>
That shouldn’t be surprising, since x<sup>2</sup> + y<sup>2</sup> = 1 is how you plot a circle. However, it gives us an alternate way of thinking about e<sup>in</sup>— we can think of n as an angle, as in the picture. The radius r is also shown, and of course it’s 1.
<p>We can’t represent <i>all</i> complex numbers with e<sup>in</sup>... but we <i>can</i> represent them as re<sup>in</sup>. That is, an alternative representation of a complex number x + iy is re<sup>in</sup>.
<h2>Proofs</h2>
Maybe it looks like a sine wave but it really isn’t? Well, naturally, there are proofs of Euler’s formula. You can look <a href="http://en.wikipedia.org/wiki/Euler’s_formula#Proofs">here</a>, or <a href="https://proofwiki.org/wiki/Euler’s_Formula">here</a>, or <a href="http://math.stackexchange.com/questions/3510/how-to-prove-eulers-formula-eit-cos-t-i-sin-t">here</a>.
<p>The problem is that to understand the proofs, you need far more math than is needed just to understand e and sin/cos! Euler himself used the limit expansions of e<sup>n</sup>, of sine, and of cosine... but even though it’s true, it’s not intuitively obvious that cos x is the sum of -n/(2n)! × x<sup>2n</sup>.
<p>Let’s take another approach, along the lines of "use Wolfram Alpha to evaluate limits for us until we’re satisfied we know where this is going". Above I mentioned that Bernoulli’s definition of e was
<blockquote><img src="trig-bernoulli.png"></blockquote>
Now e = e<sup>1</sup>, so the general formula is
<blockquote><img src="trig-gen-bernoulli.png"></blockquote>
x can be a complex number, so we can also write
<blockquote><img src="trig-ix-bernoulli.png"></blockquote>
Let’s try a particular value, namely x = π.
<p>For n = 1, this is (1 + iπ/1)<sup>1</sup> = 1 + i π.
<p>You can enter <tt>(1 + (i*pi)/2)^2</tt> into Wolfram Alpha, and keep going with higher and higher n’s. Wolfram Alpha will helpfully give you a plot of the results. Here’s how it goes:
<blockquote><img src="trig-ipi.png"></blockquote>
It looks like the limit is simply -1, and so it is. Wolfram Alpha will even do limits for you— cut and paste <tt>lim((1+(i*pi)/n)^n)</tt>.
<p>But does Euler’s formula
<blockquote>
e<sup>ix</sup> = cos x + i sin x
</blockquote>
work for x = π? Cos π = -1 and sin π = 0, so yes. π is of course halfway round the unit circle.
<p>Let’s try another number, say π/6, or 30 degrees. Give Wolfram Alpha <tt> lim((1+(i*pi/6)/n)^n)</tt> and you get ½√3 + ½ i. And this checks out: if you take sin(π/6) you get ½, and cos(π/6) = ½√3.
<p>Now all you have to do is try out every other possible value of x from 0 to 2π. Any number you try, it’ll work.
<h2>An example</h2>
The good thing about learning your math from Feynman’s lectures is that you almost immediately get a practical application. The bad thing is that to understand the application requires not just following the physical analysis, but having a firm grounding in differential equations.
<p>So to demonstrate the use of complex exponents of e, I’m just going to show <i>one step</i> in a longer derivation. Feynman doesn’t even bother to show the step— he says it’s evident "by inspection"— but I’ll take it a bit slower.
<p>We have a harmonic oscillator (such as a weight held by a spring), driven by a force F, and resisted by friction. The equation is
<blockquote>
<font color="red">d<sup>2</sup>x/dt<sup>2</sup></font>
+ <font color="blue">γ(dx/dt)</font> +
<font color="green">ω<sub>0</sub><sup>2</sup>x</font> = F/m
</blockquote>
F is itself defined as F<sub>0</sub> cos(ωt + Δ). This makes for a pretty rough differential equation, but here’s where our trick comes in. We pretend that F isn’t a cosine but an exponential, and just ignore the imaginary part. So we use <u>F</u>e<sup>iωt</sup> for the force and <u>x</u>e<sup>iωt</sup> for the displacement. (The underlines are reminders that these are complex numbers.) Now the differential equation is
<blockquote>
<font color="red">d<sup>2</sup><u>x</u>e<sup>iωt</sup> /dt<sup>2</sup></font> +
<font color="blue">γ(d<u>x</u>e<sup>iωt</sup> /dt)</font> +
<font color="green">ω<sub>0</sub><sup>2</sup><u>x</u>e<sup>iωt</sup></font>
= <u>F</u>e<sup>iωt</sup>/m
</blockquote>
Recall that the derivative of e<sup>t</sup> is itself. The derivative of e<sup>iωt</sup> is almost as easy: bring down the constants, giving iωe<sup>iωt</sup>. So the second term
<blockquote>
<font color="blue">γ(d<u>x</u>e<sup>iωt</sup> /dt)</font>
</blockquote>
is immediately converted to
<blockquote>
<font color="blue">γiω<u>x</u>e<sup>iωt</sup> </font>
</blockquote>
The first term is a double differentiation, which becomes a double multiplication by iω. That is,
<blockquote>
<font color="red">d<sup>2</sup><u>x</u>e<sup>iωt</sup> /dt<sup>2</sup></font>
</blockquote>
becomes
<blockquote>
<font color="red">(iω)<sup>2</sup> <u>x</u>e<sup>iωt</sup></font>
</blockquote>
which reduces to
<blockquote>
<font color="red">-ω<sup>2</sup> <u>x</u>e<sup>iωt</sup></font>
</blockquote>
So the differential equation vanishes, replaced by
<blockquote>
<font color="red">-ω<sup>2</sup> <u>x</u>e<sup>iωt</sup></font> +
<font color="blue">γiω<u>x</u>e<sup>iωt</sup> </font>+
<font color="green">ω<sub>0</sub><sup>2</sup><u>x</u>e<sup>iωt</sup></font>
= <u>F</u>e<sup>iωt</sup>/m
</blockquote>
If your math is rusty this may not have seemed like an improvement, but again, the clever bit is that we’ve replaced a differentiation with a multiplication, which is far less hairy.
<p>Might as well take the next step: we can divide out e<sup>iωt</sup> from both sides:
<blockquote>
<font color="red">-ω<sup>2</sup> <u>x</u> </font> +
<font color="blue">γiω<u>x</u></font> +
<font color="green">ω<sub>0</sub><sup>2</sup><u>x</u></font> = <u>F</u>/m
</blockquote>
Solving for <u>x</u>:
<blockquote>
<u>x</u> = <u>F</u>/m(<font color="green">ω<sub>0</sub><sup>2</sup></font> <font color="red">- ω<sup>2</sup></font> + <font color="blue">γiω</font>)
</blockquote>
<p>You can see <a href="http://farside.ph.utexas.edu/teaching/315/Waves/node12.html">an example of doing all the equations with sine and cosine here</a>, while <a href="http://isites.harvard.edu/fs/docs/icb.topic251677.files/notes23.pdf">this page does the same thing with exponentials</a>.
<p>This e<sup>iωt</sup> business isn’t just a quirk of weights on springs; it turns up all over. It’s a basic part of analyzing electric circuits; it comes up in tidal effects on the atmosphere; it pops up in nuclear physics. Nature is lazy, and uses the same damn equations for everything.
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