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Eleven Square Roots in a Logarithm (Posted on 2006-09-05) Difficulty: 2 of 5
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(x1/2048 - 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(x1/2048 - 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(x1/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?

See The Solution Submitted by Richard    
Rating: 3.3333 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): Not quite a spoiler | Comment 6 of 7 |
(In reply to re: Not quite a spoiler by vswitchs)

Ah yes, actually using this definition of the exp-function crossed my mind, but then I got to the same second-last statement as you did, namely

y^(1/n) ~ 1 + ln(y)/n

and I thought, well that doesn't really mean much more than

1 ~ 1

and you could in fact replace the term "y" on the left-hand side with "y^2" or even with "exp(y)+y^y" if you wish. The almost-equation would still hold. But of course when solving after ln(y) you'd get a wrong result. Your deduction of Richard's approximation can probably be repaired (replace "~" by "=...+o(n)" or similar). However Art M has already given another convincing deduction that doesn't fall into this trap.

Edited on September 24, 2006, 3:28 pm
  Posted by JLo on 2006-09-24 15:25:53

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