I'm not sure if the title is descriptive enough; feel free to change it if you come up with something better.

I've been reading through Feynman's Lectures on Physics. In the first volume, he dedicates a chapter to just math. He starts with the natural numbers and addition by 1, and builds his way to the complex numbers, with the purpose of proving Euler's formula $e^{ix} = \cos x + i \sin x$. It's a very nice read, but there's a part in the middle I'm having trouble understanding.

After having introduced irrational numbers, he begins to explain how to calculate (or define, rather) irrational powers as succesive approximations of rational powers and how to calculate logarithms, which is a related problem. In particular, he gives as an example how to find solutions to the equations $x = 10^\sqrt{2}$ and $x = \log_{10} 2$. To do that, he makes a table of successive square roots of ten, by calculating $10^s$ for $s = 1, \frac1{2}, \frac1{4}, \frac1{8}, \cdots , \frac1{1024}$. He remarks that this is enough to calculate some logarithms, because if we already calculated $10^s$ and we want its logarithm, it is simply $s$.

He also notices that as we take square roots of number that get closer and closer to $1$, there is a pattern: $\sqrt{1+\delta}$ is approximately $1+\delta/2$, so, for numbers that are already close to $1$, (such as $10^{1/1024}$, which is the last square root he calculated) instead of keeping on doing square roots we can just "guess" at the result with a pretty good accuracy.

Now, here's the part I don't understand: Having calculated $10^{1/1024}$ to be approximately $1.0022511$, he says the following:

[...] it is clear that, to an excellent approximation, if we take another root, we shall get 1.00112 something, and rather than actually take all the square roots, we guess at the ultimate limit. When we take a small fraction $\Delta$ of $1024$ as $\Delta$ approaches zero, what will the answer be? Of course it will be some number close to $0.0022511 \Delta$. Not exactly $0.0022511 \Delta$, however -- we can get a better value by the following trick: we substract the $1$, and then divide by the power $s$. This ought to correct all the excesses to the same value.

He then adds another column: for each $s$, in addition to $10^s$ there's $\frac{10^s-1}{s}$, and it looks like this converges to something as $s$ gets smaller. I recognize this as one of the usual formulas for the logarithm, but I don't follow why he introduced it. Later he uses this to make an approximation formula: $10^\Delta \approx 1+\Delta \ln 10$. I understand this, but I don't get where he got that from. Could someone clarify this?

I wasn't sure about asking this question because I thought it might be hard to understand if you've never read this chapter. If this is the case, let me know and I'll try to edit it a bit.


2 Answers 2


Thanks for bringing this to my attention. I checked the lecture tape, and found that what Feynman originally said was, "When I took a small fraction of 1024, as the fraction went to zero, what would the answer be here?" This was edited in 1961 by Robert Leighton into the form it now takes in The Feynman Lectures on Physics. I agree with you that the current version is a bit confusing, and will recommend that it be changed from, "When we take a small fraction Δ of 1024 as Δ approaches zero..." to, "When we take a small fraction Δ/1024 as Δ approaches zero..."

Mike Gottlieb [Editor, The Feynman Lectures on Physics New Millennium Edition]

  • 2
    $\begingroup$ I just saw in the feynmanlectures.info site that this has been added to the errata. Awesome! $\endgroup$
    – Javier
    Jul 21, 2012 at 14:06
  • $\begingroup$ Actually are the lecture tapes in any way accessible to the public? $\endgroup$
    – lalala
    Jan 29, 2019 at 20:09

I'm not surprised you didn't understand this; it's rather uncharacteristically badly written. It should say "When we take a small fraction $\Delta$ of $1/1024$", not $1024$. The idea is that, since when we halve the exponent the distance from $1$ is roughly halved, the distance from $1$ (which he calls the "excess") is roughly proportional to the exponent, so $10^s\approx1+\alpha s$ for some $\alpha$. The "excess" is $\alpha s$, and by dividing through by $s$ he gets increasingly accurate approximations of $\alpha\approx(10^s-1)/s$.

The reason for $\alpha=\log10$ is that $10^s=\mathrm e^{s\log10}\approx1+s\log10$ for small $s$.

  • $\begingroup$ I get it now, thank you! This paragraph is really weird, considering that the rest of the lectures is very well written and clear. $\endgroup$
    – Javier
    Jul 15, 2012 at 21:52
  • $\begingroup$ I follow what you said, except, where does $e^{s\log{10}} \approx 1 + s\log{10}$ come from? $\endgroup$
    – Jac Frall
    Oct 8, 2020 at 23:37
  • 1
    $\begingroup$ @JacFrall: This is from the exponential series, $\mathrm e^x=\sum_{k=0}^\infty\frac{x^k}{k!}=1+x+\cdots$. $\endgroup$
    – joriki
    Oct 11, 2020 at 8:16

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