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 Wolfram Alpha.

2.718281828459045235360287471352662497757247093699959574966967...You’ll probably remember e as the base of natural logarithms. But before we get to logarithms, e has some remarkable characteristics.

To see one of them, let’s look at the function y = e^{x}.

*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.*

So the slope is

(2.7185537 - 2.7182818) / (1.0001 - 1) ≅ 2.718417747Which is close to e. Let’s try a better approximation of the slope, by choosing x

(eWhich is even closer to e. It’s beginning to look like the slope at e^{1.000001}- e^{1})/ .000001 ≅ 2.718283188

Let’s try e^{0}, whose value is 1.

(eHow about e^{0.000001}- e^{0})/.000001 ≅ 1.000000500

(eHow about e^{2.000001}- e^{2})/.000001 ≅ 7.389059793

(eAs these examples suggest, the slope of e^{π+.000001}-e^{π})/.000001 ≅ 23.140704203

d/dx eThis is the only function of x whose derivative is itself. That makes e^{x}= e^{x}

Suppose it’s 50% every 6 months, instead. Now you get $1 × 1.5 × 1.5 = $2.25.

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

(1 + 1/12)^{12}≅ 2.61303529

Compounding daily, at 0.2739726%, gives us

(1+(1/365))That’s starting to look like e, and if we keep making smaller and smaller divisions, that’s exactly where we end up.^{365}≅ 2.714567482

We use **logarithmic scales** all the time; they’re great for reducing enormous ranges of huge numbers to easy-to-grasp small numbers. Some common examples:

- decibels: if a sound is ten times louder than another one, it’s 10 decibels higher
- the Richter scale: an earthquake 1.0 higher on this scale has 10 times the amplitude
- stellar magnitude: if one star is 100 times brighter than another, that’s 5 magnitudes
- the computing abbrevations kilo, mega, giga, tera, peta are a logarithmic scale— each step is 1024 times the last one

Since we normally use base 10, base 10 logarithms are particularly easy.

- Log
_{10}1000 = 3— just count the zeroes. - Log
_{10}10^{45}= 45. - Log
_{10}.001 = -3; count the places after the decimal point.

Not difficult, but tedious and error-prone. But we can take advantage of the fact that3.14159265

×2.7182818

-----------------

2513274120

3141592650

251327412000

629318530000

25132741200000

31415926500000

2199114855000000

6293185300000000

-----------------

8.539734123508770

log ac = log a + log cWe can look up the logs in a table... tables to 14 decimal places have been available since 1620.

logNow we can use the same table to find out what number this is the log of, which turns out to be ≅ 8.53973412._{10}3.14159265 ≅ .497149872

log_{10}2.7182818 ≅ .434294477

sum ≅ .93144434955

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.

(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.)

If you’re thinking "Oooh, the limit of e

1/n e ^{1/n}1 + 1/n 1 2.7182818 2 1/2 1.648721271 1.5 1/4 1.2840254 1.25 1/8 1.133148453 1.125 1/16 1.06449446 1.0625 1/32 1.0317434075 1.03125 1/64 1.015747709 1.015625 1/128 1.007843097 1.0078125 1/256 1.003913889 1.00390625 1/512 1.00195503 1.001953125 1/1024 1.000977039 1.0009765625 1/2048 1.000488400 1.00048828125

We’ll use this trick in a clever way in a moment, but I’ll also point out two things.

- This particular series is important if you want to calculate logarithms by hand. See Feynman’s lecture if you want to know how.
- 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
^{1/n}is 1 + 2.3026 * (1/n). We get rid of the constant by using base-e logarithms.

All of which means that a function like y = e^{ix} 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.

Still, how **do** we evaluate an expression like e^{ix}? Where do we even start?

We actually *have* a place to start— the observation from the last section that a good guess at e^{i/n} is 1 + 1/n. Let’s assume that it’s true for complex numbers as well.

Let’s set n to 1024. That is, a good guess for e^{i/1024} is 1 + i/n = 1 + i/1024. That of course is the complex number 1 + .0009765625i.

That gives us *one* exponential of an imaginary number. (Or rather, an approximation; the actual value ≅ .999999523 + 0.00097656234i.

Given one value of e^{ix}, 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

eand square both sides. Now, (a^{i/1024}= 1 + .0009765625i

(eFor the other side, the general rule for complex numbers is^{i/1024})^{2}= e^{(i/1024)×2}= e^{i/512}

(a × bi)Here a = 1 and b = .0009765625, so we get^{2}= (a × bi)(a × bi)

= a^{2}+ 2abi + b^{2}i^{2}

= (a^{2}- b^{2}) + 2abi

(1×1 - .0009765625×.0009765625) + 2×1×.0009765625×iLet’s see what e

= 0.9999990463 + 0.001953125 i

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.

0 1 .2 0.9800666 + 0.1986693 i .4 0.921061 + 0.389418 i .6 0.825336 + 0.564642 i .8 0.696707 + 0.717356 i 1 0.540302 + 0.841471i 1.2 0.362358 + 0.932039 i 1.4 0.169967 + 0.985450 i 1.6 -0.0291995 + 0.999574 i 1.8 -0.227202 + 0.973848 i 2 -0.416147 + 0.909297 i 2.2 -0.588501 + 0.808496 i 2.4 -0.737394 + 0.675463 i 2.6 -0.856889 + 0.515501 i 2.8 -0.942222 + 0.334988 i 3 -0.989992 + 0.141120 i 3.2 -0.998295 - 0.0583741 i 3.4 -0.966798 - 0.255541 i

eNow, 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^{in}= x + iy

eNow let’s multiply these equations together:^{-in}= x - iy

eSimplifying the left side:^{in}e^{-in}= (x + iy)(x - iy)

eAnd the right side:^{in}e^{-in}= e^{in - in}= e^{0}= 1

(x + iy)(x - iy) = xSo for any complex number x + iy that we found as a value of e^{2}+xyi -xyi - i^{2}y^{2}= x^{2}+ y^{2}

xDoes this ring any bells yet? That’s the formula for a circle, so circles are involved in this somehow.^{2}+ y^{2}= 1

Let’s just get a plot of e^{in}. The red line is the real part, the blue line is the imaginary part.

eThis 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.^{in}= cos n + i sin n

The equation was first published by Leonhard Euler in 1748.

Rather than plotting e^{in}, we can also plot x and y as n varies. We get this graph:

We can’t represent *all* complex numbers with e^{in}... but we *can* represent them as re^{in}. That is, an alternative representation of a complex number x + iy is re^{in}.

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^{n}, 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^{2n}.

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

Now e = eFor n = 1, this is (1 + iπ/1)^{1} = 1 + i π.

You can enter `(1 + (i*pi)/2)^2` 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:

But does Euler’s formula

ework for x = π? Cos π = -1 and sin π = 0, so yes. π is of course halfway round the unit circle.^{ix}= cos x + i sin x

Let’s try another number, say π/6, or 30 degrees. Give Wolfram Alpha ` lim((1+(i*pi/6)/n)^n)` 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.

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.

So to demonstrate the use of complex exponents of e, I’m just going to show *one step* 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.

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

dF is itself defined as F^{2}x/dt^{2}+ γ(dx/dt) + ω_{0}^{2}x = F/m

dRecall that the derivative of e^{2}xe^{iωt}/dt^{2}+ γ(dxe^{iωt}/dt) + ω_{0}^{2}xe^{iωt}=Fe^{iωt}/m

γ(dis immediately converted toxe^{iωt}/dt)

γiωThe first term is a double differentiation, which becomes a double multiplication by iω. That is,xe^{iωt}

dbecomes^{2}xe^{iωt}/dt^{2}

(iω)which reduces to^{2}xe^{iωt}

-ωSo the differential equation vanishes, replaced by^{2}xe^{iωt}

-ωIf 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.^{2}xe^{iωt}+ γiωxe^{iωt}+ ω_{0}^{2}xe^{iωt}=Fe^{iωt}/m

Might as well take the next step: we can divide out e^{iωt} from both sides:

-ωSolving for^{2}x+ γiωx+ ω_{0}^{2}x=F/m

x=F/m(ω_{0}^{2}- ω^{2}+ γiω)

You can see an example of doing all the equations with sine and cosine here, while this page does the same thing with exponentials.

This e^{iωt} 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.