# How did Feynman produce solutions in under a minute?

As detailed here and elsewhere, Feynman and others at Los Alamos could calculate many problems to 10% accuracy in minutes:

When I was at Los Alamos I found out that Hans Bethe was absolutely topnotch at calculating... A few minutes later we need to take the cube root of $$2 \frac 1 2$$... and he says, "It's about $$1.35$$."... I had a lot of fun trying to do arithmetic fast, by tricks, with Hans... I announced, "I can work out in sixty seconds the answer to any problem that anybody can state in ten seconds, to 10 percent!"... People started giving me problems they thought were difficult, such as integrating a function like $$\frac 1 {1 + x^4}$$, which hardly changed over the range they gave me. The hardest one somebody gave me was the binomial coefficient of $$x^{10}$$ in $$(1 + x)^{20}$$; I got that just in time.... [Paul Olum] says, $$\tan 10^{100}$$. I was sunk: you have to divide by $$\pi$$ to $$100$$ decimal places!

What methods do Feynman, Bethe, Olum, use to do these? Or, since we can't really know the answer to that: What methods can we use to easily approximate calculations within 10% error?

Now, one might simply respond: Wolfram Alpha. But we do this not for lack of a calculator! For example, Sanjay Mahajan requires his students to show "number sense" by

Without a calculator, estimate $$\sqrt{1.3}, \sqrt[3]{1.6}, \sin 7,$$ and $$1.01^{100}$$.

What methods can we use to do this? I'll post my collection as an answer below. Surprisingly, while I've found good methods for estimating logs, exponents, and trig, approximate long division (in less steps than the real thing) proves to be the hardest!

• The binomial coefficient is perhaps the easiest of all if you know the formula. $29 \choose 19$ Commented Jan 3 at 1:35
• Keep in mind he likely only reported the successes. Von Neumann and Ulam have similar folklore. Commented Jan 3 at 2:08
• @Aruralreader That's fine - I'm not looking for bragging rights. I'm looking for the insight gained. Going from "$1.01^{100}$, hmm, hand me a calculator" to "$e^{100 \log 1.01}, \log 1.01 \approx 0.01$, so it's approximately $e$"! is a quantum leap. Commented Jan 3 at 2:50
• @SRobertJames: that’s not the point :) One might guess, in the heat of the moment, that ‘$1.01^{100}$’ is slightly more than one-ish, but who knows, yet, thinking exponentiation, maybe $e$. And if wrong, never re-share the story. Commented Jan 3 at 3:24
• This is like the story with Ramanujan with the number $1729$ which was also only an anecdote. I see no reason why this story is not also just a myth. Commented Jan 3 at 5:48

### Log

$$\log (1 + x) = x - \frac {x^2}2 + \frac{x^3} 3 +... \\ \log 2 \approx 0.69 \ \log 3 \approx 1.1 \log 10 \approx 2.3$$ allows easily computing any log to within 1%.

### Exponential

$$e^x = 1 + x + \frac {x^2}{2!}+...$$ along with the constants above allows easily computing any exp to within 1%.

### Trig

$$\sin x = x - \frac {x^3}{3!} + ... \\ \cos x = 1 - \frac {x^2} {2!} + ...$$ provide sufficient estimates of all trig functions. The cubic term is only needed near $$\frac \pi4$$. Near $$\frac \pi 2$$, use cofunctions near $$0$$. For inverse functions, use the above and algebra.

### Factorial

Stirling's approximation $$n! \approx \sqrt{2 \pi n} \left[ \frac n e \right]^n$$ along with the exp estimators allows computing good estimates. It may be easier to take the log: $$\log (n!) \approx n \log n - n + \frac 1 2 \log n + \frac 1 2 \log 2 \pi.$$

### Integrals

Numeric estimates via sampling are surprisingly accurate. For example, for Feynman's example $$\int \frac {dx} {x^4 + 1}$$ (with bounds $$[0,1]$$) it suffices to take three easy points $$0 \leadsto 1, 1 \leadsto \frac 1 2, \frac 1 2 \leadsto \frac {16} {17}$$, which average to $$\frac {83} {102}$$, within range of the true value $$0.867...$$.

Sometimes thought is required where and how to sample. For example, to solve Arnold's problem of $$\int_0^{\frac \pi 2}\cos^{100}x$$, notice that $$\cos^{100}x$$ is nearly always either close to $$1$$ or very close to $$0$$. We can therefore treat it as a square pulse which begins at $$0$$ and ends when $$\cos^{100} x$$ crosses $$\frac 1 2$$.

To find this threshold $$x$$, first solve $$t^{100} = \frac 1 2$$, $$100 \log t \approx -0.69 \\ t \approx \exp (-0.0069) \approx 0.9931$$ and then $$\arccos x = t \approx 0.9931 \\ x^2/2 \approx 0.0069 \\ x \approx 0.12$$ which is within about 4% of the true value $$0.12502...$$.

### Polynomials and Rational Functions

$$(1+x)^n \approx 1 + nx$$ for small $$x$$

$$\frac 1 {1+x} \approx 1 - x + x^2$$ for small $$x$$

### Division

Division within 10% is easy by rounding (and perhaps correcting), but often division errors are magnified, and therefore require closer to 2% accuracy.

No good method found!

Is there a method that can e.g. determine $$\frac {562} {2.3}$$ to within 2% error with less steps than long division?!

• Well structured ! A little detail : "Exponents" should be "Exponential" of "Exponential function" Commented Jan 3 at 8:55
• For $(1+x)^n$ you might consider using the rule of 72 or something sililiar. For division $562 \over 2.3$ is basically integer division It's easy to calcualte $562/23$ with an error of $1$ or $2$ in seconds. What about dividng fractions or even polynomials? Commented Jan 3 at 18:30
• To illustrate $(1+.10)^{50}$ equals about 6 using your method. A quick calculation using rule of 72 shows that $(1+.10)^{50.4}$ equals about $128$. Per wolfram the correct answer is about $117.4$ Commented Jan 4 at 15:30
• The series for $\log(1+x)$ only converges for $|x|<1$, so to say you can "compute any log" seems misleading. Commented Jan 11 at 2:29
• @angryavian We can indeed use to estimate any log, by using log identities and the few constants. For example, to estimate $\log 2.5$, we write $\log 2.5 = \log 2 + \log 0.5$. Similarly e.g. to estimate $\log 2.4 \cdot 10^6$. Commented Jan 11 at 3:49

Feynman once got into a showdown with a man with an abacus. Feynman explained how he was able to mentally compute the cube root of 1729.03:

The number was 1729.03. I happened to know that a cubic foot contains 1728 cubic inches, so the answer is a tiny bit more than 12. The excess, 1.03 is only one part in nearly 2000, and I had learned in calculus that for small fractions, the cube root's excess is one-third of the number's excess. So all I had to do is find the fraction 1/1728, and multiply by 4 (divide by 3 and multiply by 12). So I was able to pull out a whole lot of digits that way.

The full story can be found here: https://www.ecb.torontomu.ca/~elf/abacus/feynman.html

• What does "for small fractions, the cube root's excess is one-third of the number's excess." mean? Commented Jan 3 at 2:56
• The source you linked also has a great explanation of the value of approximation methods: "I started to explain that it was an approximate method, and had to do with the percentage of error.... I realized something: he doesn't know numbers. With the abacus... all you have to do is to learn to push the little beads up and down.... So [using approximation methods is] slower at basic arithmetic, but we know numbers" (emphasis added). Commented Jan 3 at 2:58
• @SRobertJames Incidentally, this example is discussed in Did Feynman mentally compute $\sqrt[3]{1729.03}$ by linear approximation?. Commented Jan 3 at 4:01