perplexus.info :: Algorithms : Eleven Square Roots in a Logarithm

Eleven Square Roots in a Logarithm (Posted on 2006-09-05)

Presto the Mathematical Magician says, quite correctly, that ln(x), the natural logarithm (to the base e=2.718...) of x, is magically well-approximated by 2047(x^1/2048 - 1). Hence logarithms can be calculated with fair accuracy using a primitive calculator that only does square roots along with basic arithmetic.

What is behind Presto's magic?

By the same token, log(x), the common (base 10) logarithm of x, may be approximated by the similar formula K(x^1/2048 - 1) for a suitable value of K. For values of x between 1 and 10, explore the accuracy of this approximation, and that of similar formulas of the type K(x^1/N-1) where N=2ⁿ, under the assumption that a 10-digit calculator is being used to compute the repeated square roots. What values for K and n would you recommend when a 10-digit calculator is being used?

Submitted by Richard

Rating: 3.3333 (3 votes)

Solution: (Hide)

Both ln(x) and its approximation are zero when x=1. The derivative of ln(x) is 1/x and that of the approximation is 1/x times the factor (2047/2048)x^1/2048 which is nearly unity. So ln(x) and the approximation are the integral from 1 to x of nearly equal integrands. Another way to look at this is to consider y=N(x^1/N-1) and solve for x, which gives x=(1+y/N)^N which for large N is close to exp(y).
K(x^1/2048 - 1) approximates the common log well when K=889. Using the appropriate K and a higher power of 2 instead of 2¹¹=2048 would give a more accurate result theoretically, but with a limited-precision calculator, the accuracy of the computed value of the quantity x^1/N - 1 would eventually suffer when N=2ⁿ increases too far, because the difference of values that are becoming nearly equal is being taken. On a 10-digit calculator, 10^1/N computes as exactly 1 when N=2³³, so we can't let n get anywhere near as large as 33. Based on numerical experiments, K=889 and n=11 are good choices when a 10-digit calculator is being used -- for x between 1 and 10, the maximum error is about .0001 in absolute value. Charlie showed in his post that the best results are obtained (five accurate digits) by using 17 square roots and K a little less than 2¹⁷/ln(10).

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
Puzzle Answer	K Sengupta	2024-01-21 00:06:15
re(2): Not quite a spoiler	JLo	2006-09-24 15:25:53
re: Not quite a spoiler	vswitchs	2006-09-24 04:49:53
Few other thoughts	JLo	2006-09-10 13:51:06
some thoughts	Art M	2006-09-06 17:30:49
For the second part.	Charlie	2006-09-05 15:55:38
Not quite a spoiler	vswitchs	2006-09-05 14:52:28

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