“Surely You’re Joking, Mr. Feynman”: Adventures of a Curious Character

One day at Princeton I was sitting in the lounge and overheard some mathematicians talking about the series for e^x, which is 1 + x + x^2/2! + x^3/3! Each term you get by multiplying the preceding term by x and dividing by the next number. For example, to get the next term after x^4/4! you multiply that term by x and divide by 5. It’s very simple.

When I was a kid I was excited by series, and had played with this thing. I had computed e using that series, and had seen how quickly the new terms became very small.

I mumbled something about how it was easy to calculate e to any power using that series (you just substitute the power for x).

“Oh yeah?” they said. “Well, then what’s e to the 3.3?” said some joker—I think it was Tukey.

I say, “That’s easy. It’s 27.11.”

Tukey knows it isn’t so easy to compute all that in your head. “Hey! How’d you do that?”

Another guy says, “You know Feynman, he’s just faking it. It’s not really right.”

They go to get a table, and while they’re doing that, I put on a few more figures.: “27.1126,” I say.

They find it in the table. “It’s right! But how’d you do it!”

“I just summed the series.”

“Nobody can sum the series that fast. You must just happen to know that one. How about e to the 3?”

“Look,” I say. “It’s hard work! Only one a day!”

“Hah! It’s a fake!” they say, happily.

“All right,” I say, “It’s 20.085.”

They look in the book as I put a few more figures on. They’re all excited now, because I got another one right.

Here are these great mathematicians of the day, puzzled at how I can compute e to any power! One of them says, “He just can’t be substituting and summing—it’s too hard. There’s some trick. You couldn’t do just any old number like e to the 1.4.”

I say, “It’s hard work, but for you, OK. It’s 4.05.”

As they’re looking it up, I put on a few more digits and say, “And that’s the last one for the day!” and walk out.

What happened was this: I happened to know three numbers—the logarithm of 10 to the base e (needed to convert numbers from base 10 to base e), which is 2.3026 (so I knew that e to the 2.3 is very close to 10), and because of radioactivity (mean-life and half-life), I knew the log of 2 to the base e, which is.69315 (so I also knew that e to the.7 is nearly equal to 2). I also knew e (to the 1), which is 2. 71828.

The first number they gave me was e to the 3.3, which is e to the 2.3—ten—times e, or 27.18. While they were sweating about how I was doing it, I was correcting for the extra.0026—2.3026 is a little high.

I knew I couldn’t do another one; that was sheer luck. But then the guy said e to the 3: that’s e to the 2.3 times e to the.7, or ten times two. So I knew it was 20. something, and while they were worrying how I did it, I adjusted for the.693.

Now I was sure I couldn’t do another one, because the last one was again by sheer luck. But the guy said e to the 1.4, which is e to the.7 times itself. So all I had to do is fix up 4 a little bit!

They never did figure out how I did it.

When I was at Los Alamos I found out that Hans Bethe was absolutely topnotch at calculating. For example, one time we were putting some numbers into a formula, and got to 48 squared. I reach for the Marchant calculator, and he says, “That’s 2300.” I begin to push the buttons, and he says, “If you want it exactly, it’s 2304.”

The machine says 2304. “Gee! That’s pretty remarkable!” I say.

“Don’t you know how to square numbers near 50?” he says. “You square 50—that’s 2500—and subtract 100 times the difference of your number from 50 (in this case it’s 2), so you have 2300. If you want the correction, square the difference and add it on. That makes 2304.”

A few minutes later we need to take the cube root of 2Ѕ. Now to take cube roots on the Marchant you had to use a table for the first approximation. I open the drawer to get the table—it takes a little longer this time—and he says, “It’s about 1.35.”

I try it out on the Marchant and it’s right. “How did you do that one?” I ask. “Do you have a secret for taking cube roots of numbers?”

“Oh,” he says, “the log of 2Ѕ is so-and-so. Now one third of that log is between the logs of 1.3, which is this, and 1.4, which is that, so I interpolated.”

So I found out something: first, he knows the log tables; second, the amount of arithmetic he did to make the interpolation alone would have taken me longer to do than reach for the table and punch the buttons on the calculator. I was very impressed.

After that, I tried to do those things. I memorized a few logs, and began to notice things. For instance, if somebody says, “What is 28 squared?” you notice that the square root of 2 is 1.4, and 28 is 20 times 1.4, so the square of 28 must be around 400 times 2, or 800.