How did feynman compute $e^x$ with the accuracy he wanted?

Question

Excerpt from the book Surely You're Joking, Mister. Feynman!

Here Feynman calculates $e$ to a couple of powers. I understand that he luckily knew a couple of logs by heart. What I don't understand is the part where he adjusts the numbers to the accuracy he wanted to have. The accuracy he got in the first calculation e^3.3 was 27.1126. In the first example it could be : $$(e^{2.3026} \times e) / e ^ {0.0026} = e^{3.3}$$ But then he still had to use the power series for the $e^{0.3026}$ AND divide by it afterwards and I can't see how even Feynman would be able to do that. Maybe by a linear approximation? Thanks in advance.

score 2 · Answer 1 · answered Jul 27 '17 at 22:53

2

Regarding the adjustment - don't forget that the derivative of the exponential function is itself. So you have your first approximate answer, and that's good enough to use as the derivative to step to your refined answer.

So $e^{3.3}$ is about $27.1828$, but that's actually $e^{3.3026}$, so we want to adjust down by $0.0026\times 27.1828 \approx 0.07$ so call it $27.11$.

answered Jul 27 '17 at 22:53

Joffan

39,627

Thanks a lot for the answer! But i still don't get two things:1) Here we got an accuracy of 27.11 but he got the accuracy of 27.1126 . 2) How would you multiply 27.1828 * 0.0026 in your head? – delivosa Jul 28 '17 at 12:45
Feynmann was cleverer than me 2) I multiplied 28 by 1/4 by 0.01

Joffan

Jul 28 '17 at 13:52

Weird i've tried everything and can't get his accuracy but thanks for the answer i understand it much better now. – delivosa Jul 29 '17 at 16:37

Well, the accuracy is not really justified by the digits available. $e^{3.3026}\approx 27.18322$ so adjusting as if it were $27.1828$ should leave you out in the 4th dp. Maybe the story gained a little in the telling. – Joffan Jul 29 '17 at 17:37

score 2 · Answer 2 · answered Jul 27 '17 at 22:57

2

$e^{3.3} = e^{2.3026\dots} \times e \times e^{-0.0026\dots} = 10 \times e \times e^{-0.0026\dots}$. At this point you could use the linear approximation I guess, $$ e^{-0.0026\dots} =1 - 0.0026{\dotsc} + \frac{0.0026{\dotsc}^2}{2} - \dotsb \approx 1-0.0026{\dotsc} \approx 0.9974 . $$ I don't know if this is what Feynman actually did.

answered Jul 27 '17 at 22:57

Zach Teitler

3,020

That's likely what he did, and what he meant by adjusting. – DanielWainfleet Jul 28 '17 at 05:08
Thanks a lot for the answer! But i still don't get two things:1) When you multiply 0.9974 with 27.18 you get 27.109332 ≈ 27.11 but he got an accuracy of 27.1126. 2) How would you multiply 27.18 * 0.9974 in your head anyway? – delivosa Jul 28 '17 at 12:45

How did feynman compute $e^x$ with the accuracy he wanted?

2 Answers2