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In the biopic "infinity" about Richard Feynman. (12:54) He computes $\sqrt[3]{1729.03}$ by mental calculation. I guess that he uses linear approximation. That is, he observe that $1728=12^3$. Let $f(x)=\sqrt[3]{x}$. Then $f'(x)=\frac{1}{3\sqrt[3]{x^2}}$ and $f'(1728)=\frac{1}{3\sqrt[3]{1728^2}}=\frac{1}{3\cdot 12^2}$. Therefore, $$\sqrt[3]{1729.03}=f(1729.03)\approx f(1728)+f'(1728)(1729.03-1728)=12+\frac{1.03}{3\cdot 12^2}=12.002384\overline{259}.$$

Question 1. If he used the linear approximation, how did he compute $\frac{1.03}{3\cdot 12^2}=0.002384\overline{259}$ by a mental calculation?

Question 2. If he didn't use the linear approximation, what is another method he might have used?

bfhaha
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  • Well how good was his approximation? $12$ is already a decent approximation. – ClassicStyle May 01 '16 at 17:51
  • His approximation is $12.002384$. – bfhaha May 01 '16 at 17:54
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    You're asking about how a mental computation was accomplished in _a scene from a fictional movie_...? (Just clarifying; I'm not willing to click a youtube link to get the content of the question.) – Andrew D. Hwang May 01 '16 at 18:04
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    @AndrewD.Hwang Yes. Although it is a fictinal movie, I think that maybe the scenarist had a method to approximate $\sqrt[3]{1729.03}$. And I hope this scene can motivate my student to learn linear approximation. – bfhaha May 01 '16 at 18:12
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    $1.03/3\approx.34333\approx.34332$ $.34332/144=.05722/24=.02861/12\approx.002384$. – Mark S. May 01 '16 at 18:30
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    You should mention what value was computed so that the question is self-contained. – Bill Dubuque May 01 '16 at 18:41
  • @MarkS. Thanks! – bfhaha May 01 '16 at 18:48
  • @BillDubuque Did you mean that I should mention the Feynman's approximation? – bfhaha May 01 '16 at 18:49
  • @MarkS. I think that it is an approximation which can be computed by a mental calculation. Please post the comment as an answer. – bfhaha May 01 '16 at 18:57
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    @Andrew D. Hwang It is mentioned in his book "Surely You're joking, Mr. Feynman!". Give it a shot. –  Jul 01 '18 at 13:09

2 Answers2

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Feynman tells the story in one of his books of anecdotes.

http://www.ee.ryerson.ca/~elf/abacus/feynman.html

$12$ is a very good first approximation and the linear term of the series expansion suffices to get high precision.

$$ \sqrt[3]{1728 + d} = 12\sqrt[3]{1+x} = 12 + 4x + O(x^2)$$

where $d = 1.03$ and $x = \frac{d}{1728}$ is, in Feynman's words, about 1 part in 2000, so that the error term is of order $10^{-6}$.

Feynman says that he computed $12 + \frac{4d}{1728}$ as the approximate value.

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.

He describes that as though $d=1$ for this part of the calculation, so maybe $12 + \frac{1}{432}$ was what he actually computed. By "adding two more digits" (to 12.002) he seems to mean working out the division in the fraction. It could also mean adding (0.03)/432 as two more decimal digits of accuracy to $(12 + 432^{-1})$, which requires only a multiplication by 3 of an already computed quantity 1/432.

Feynman's method is the one that would have been immediate for anyone familiar with the binomial series and with $12^3 = 1728$. He said that knew the latter as ft^3/in^3 and other people might know it from the Ramanujan 1729 story. The other ingredient, as Feynman says in the story, was being good at integer division.

zyx
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I upvoted the other answer as it comes from the other book of anecdotes, but one way to calculate this without Feynman's experience might be the following.

You can start with the linear approximation, and then you have to calculate $(\dfrac{1.03}{3})/12^2$. $\dfrac{1.03}{3}\approx .34333$, but that doesn't lend itself to division by $12^2$, so you can change the last digit to get $\dfrac{1.03}{3}\approx .34332$

(You know this will be helpful by the divisibility test for $3$, and can note it's also divisible by $2$ and $4$ if you're thinking ahead.)

Then $\dfrac{.34332}{12^2}=\dfrac{.05722}{12*2}=\dfrac{.02861}{12}\approx.002384$.

Mark S.
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  • "linear approximation" here means taking the linear part of $(12 + x)^3$ when solving $(12 + x)^3 = 12^3 + 1.03$. Then $3x(12^2) = 1.03$ and the rest is as in the answer. Thinking of it as 1 or 1.03 divided by 432, how to divide by 432 with minimum calculation? The error in approximating that 432 by 500 is (1/432 - 1/500) = (500- 432)/500*432 or about 68/200000. This gives 0.002 + 0.00034 = 0.00234 using only division by 2, 4*5=20, and 100-32=68. – zyx May 02 '16 at 02:49
  • Thanks MarkS. You and zyx gave the best answer for me. But it can only has one answer. So please forgive me to use upvote instead of mark it as an answer. – bfhaha May 02 '16 at 05:12