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wjn
Member since Mar-19-02
Mar-19-02, 02:15 PM (EST)

"napier logs"

 If someone knows where to find it !!Long time ago, when i was at college, an uncle showed me how Napier get to the number "e" and the natural logs. The demonstration was really beautiful. My uncle died and i could never find the demo again, and never got to reproduce it.Can anyone help me?Thanks,Wagner

alexb
Charter Member
787 posts
Mar-19-02, 02:24 PM (EST)

1. "RE: napier logs"
In response to message #0

 It's most certain that Napier himself never discovered the number e, nor what we today call the natural logarithm. A very readable source on the matter is e: The story of a number by Eli Maor (Princeton University Press, 1994)

wjn
guest
Mar-19-02, 03:06 PM (EST)

2. "RE: napier logs"
In response to message #1

 Thanks for the info. I may be wrong, i just assumed it was Napier. Actually it doesn't matter who did it. I'm only interested in how someone, beginning with two geometric series got to the natural logs and the "e". I'll look for the book, but i would appreciate if someone knows a website where i could look up.wjn

dysprosia
guest
May-09-02, 07:15 AM (EST)

3. "RE: napier logs"
In response to message #2

 lim_n->inf((1+1/n)^n)=e. You can visualize this numerically.

Michael Klipper
guest
Aug-03-02, 05:33 PM (EST)

4. "RE: napier logs"
In response to message #2

 I don't know exactly what motivated the definitions of e and the natural logarithm, but I can provide a possible development that my analysis class did. It defines the natural logarithm and uses it to define e. This might not be what you are looking for, but it always fascinated me to see a way to develop logarithms from only some small properties.Let's say we want to make a function L(x) with the property that L(xy) = L(x) + L(y) for all x and y in the domain. Let's investigate its properties and then form a definition to encompass those properties.First of all, if 0 is in the domain of L(x), then we'd have L(x*0) = L(0) = L(0) + L(x) and L(x) = 0 for all x. Thus, to make a more interesting function, let's forbid 0 from being in the domain.Now, we see that L(x*1) = L(x) = L(x) + L(1), so L(1) = 0.Furthermore, L(-1*-1) = L(1) = L(-1) + L(-1), so L(-1) = 0 as well.In fact, for any x in the domain of L,L(x*x) = L(x^2) = L(-x*-x) = 2L(x) = 2L(-x), so L(x) = L(-x) and the function L is an even function.Now, let's consider a L(x) which is differentiable and examine its properties.We know that for any fixed y, we can differentiate with respect to x to get ' = ' = L'(x) = yL'(xy), where the second part comes from the Chain Rule. By now substituting x = 1, we get L'(1) = yL'(y), or L'(y) = L'(1) / y. Since L(x) is an even function, it has an odd derivative, so L'(y) = L'(1) / y holds for all y which are not 0.Let K = L'(1). We know from the Fundamental Theora of Calculus that the integral from 1 to y of Kdx / x is L(y) - L(1) = L(y), if y > 0.So for positive y, we have L(y) = K dx / x.If y < 0, then L(y) = L(|y|) = K dx / x, since L is an even function. This is the most general rule which describes ALL logarithms. If we let K=1, and we restrict our domain to positive numbers, then we get the natural logarithm function. However, note that if K = 1 / ln b for some positive b, then we get logarithms to the base b. It's interesting to see that, as a consequence of this reason, the function L(x) = K dt / t is the MOST GENERAL FUNCTION that fulfills L(xy) = L(x) + L(y). Hence, the only functions which do this are logarithms!e is defined as the unique number > 0 such that L(e) = 1. This is unique, because L(x) is strictly increasing for x > 0. Once you develop Taylor Polynomial approximations, then you can prove that0> (1 + ax)^(1/ax) = e^a, and this justifies the formula that most people know for e.