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Name: Alan
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We know about the calculator log function, but my interest is in how the calculator arrives a the value. I searched through several textbooks (algebra 1 and algebra 2), none of which described how the calculation is accomplished without factoring the numerical value by its base. The problem still intrigues me. I guess that I have grown too dependent on Texas Instruments. In college chemistry and physics, we were taught the calculator tricks; so, the application could have a solution. This is >for an Earth Science class in a unit on Richter Scale.

There are many (probably hundreds) of algorithms for calculating the log of a number "x", i.e. ln(x). One class of algorithms is the sum of an infinite series, which can be calculated to as many terms as you desire the accuracy of ln(x). The "trick" is to use an infinite series that converges rapidly to the "answer" and has a "remainder formula" that lets you know what error you made by computing only "n" terms rather that the full infinite series. For example, one can prove the formula: ln{(1+x)/(1-x)}= 2*[x +(x^3)/3 + (x^5)/5 + ...+ (x^p)/p+...] for |x| < 1. For x = 1/3 one gets immediately: ln{2}= 2*[1/3 + 1/(3*3^3) + 1/(5*3^5) + ...] The typical general term in the [bracket] is: An = 1/((2*n +1)*3^(2*n+1)) and one can show further that the remainder, Rn, is bracketed by: 0< Rn < An/8. Using only 7 terms one can calculate ln(2) to an accuracy of 7 decimal places and Rn < 0.000000001. Then by a substitution "trick" in the original formula let: x = 1/(2p + 1) the formula for p becomes: ln(p+1) = ln(p) +2*[ 1/(2p+1) + 1/(3(2p+1)^3) + 1/(5(2p+1)^5)) + ...] This gives ln(3) in terms of ln(2) + a small correction term in the square bracket. So ln(3) is also now known to 7 places with only seven terms. I would be plagiarizing if I left you with the impression that I derived the above. Chapter 8 Section 34 of "Theory and Application of Infinite Series" by Konrad Kopp is devoted to the problem of the numerical evaluation of infinite series of all sorts. By the way in the original formula 'x' need not be an integer. The only requirement for this particular formula is: |x| < 1

I do not know the particular formula TI uses in their calculators, but it is undoubtedly some sort of variation on the method above.

Vince Calder

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