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?

(In reply to Not quite a spoiler by vswitchs)

Sorry for my obscure first post.  I did think people would catch on, and didn't want to spoil all the fun.  But apparently nobody had heard of the following way of defining the exponential function:

exp(x) = lim_n->infinity (1 + x/n)^n

So for a given large n (for instance 2048), this means:

exp(x) ~ (1 + x/n)^n

(where ~ is meant to mean "approximately equal".)  Plugging in x=ln(y), we get:

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

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

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

Posted by vswitchs on 2006-09-24 04:49:53