Logarithm's and study of 'e'

[Savage]

So, I have done extensive research on logarithms and to be honest I have not been able to find a formal definition of a logarithm.

I am not looking for 'well, log base a of b means a^(some power) = b" because this explanation completely disregards what I am looking for, let alone where logarithm's originated from.

I took a loot at  John Napier's 1614 book titled Mirifici Logarithmorum Canonis Descriptio and that did not help much.

Also, most google results resort to this book when talking about the discovery of 'e' but did not find it in there. Can someone please explain where the heck 'e' was first started to be noticed (like did someone notice the constant 2.7... from research or something?)

Lastly, can someone please explain:

1) (1+1/n)^n = e as n goes to infinity using elementary calculus or algebra?

2) How does one calculate the log of something if you know the base? I have heard the base creates an infinite series of polynomial expressions that one could plus the argument into to get a somewhat accurate result.

[Aves]

If the literal definition of logarithm isn't what you're looking for, then what ARE you looking for? That is the exact simple definition of what a logarithm is. It is like saying for addition, I am not looking for "well, a+b means the total quantitative amount of those two numbers a and b combined." because this explanation completely disregards what I am looking for, let alone where addition originated from. That equation is the exact relationship that a logarithm describes. Are you looking for the etymology of the word or something?

With regards to your question for e, the number is, again, literally DEFINED to be the summation of that series; any specific decimal is only an approximation of what e actually is. The concept of e originates from a search, possibly from that book you mentioned, looking for certain number with such and such properties (the summation of that series, for one), which, oh, we will call e from now on. No, the actual modern calculation of '2.718...' etc is not going to be in there, but the approximation is not significant to the definition of e. Similar to the definition of pi and how it is related to the radius and circumference of a circle, not to 3.1415926. The concept of 'e' came before the concept of '2.718...'

If you want a basic teaching aid, here's one:

[dragonsdemesne]

I think I understand what you're asking for in terms of a formal definition of a logarithm, other than the simple mathematical equation you mention, but I'm not sure exactly what that would be.  I'd have to think on that and see if I come up with anything.  Wikipedia says it is (if that helps any)

"Definition     The logarithm of a positive real number x with respect to base b, a positive real number not equal to 1[nb 1], is the exponent by which b must be raised to yield x. In other words, the logarithm of x to base b is the solution y to the equation b^y = x"

If memory serves, 'e' was known at least back in Newton/Leibnitz' day.  I'm not sure whether some concept of it was known in ancient times the way 'pi' was.  Ancient peoples didn't formally know what pi was, but they did know that there was a number that let them figure out circle areas and perimeters, and not having any concept of irrational numbers, they tried to express it with fractions that were close, like 22/7; some cultures had better approximations than others.

I think that in the early days of calculus, 'e' was first found when people were looking at derivatives, and discovered that the exponential function e^x was its own derivative.  (it is true that it is its own derivative, but I am not sure whether that's where they first found it)  If you read about the natural logarithm, which is the logarithm with base e, that might help explain it better than I could.

1) I have seen a definition of this, and I know it works, but I can't do it off the top of my head.  I'm pretty sure it was in my university calculus textbook.  I'm pretty sure you need calculus to do it; I don't think algebra is enough.  I think I recall the squeeze theorem being used to prove it, but that's about as far as my memory goes.  I hate to say it, but I think it was almost 15 years ago that I saw this...

2) For the first part of this, I'm not entirely sure what you're asking.  You can use a formula for 'change of base' that will convert the logarithm from something weird (like base 7.43 log or pi/666 base log or whatever it is) into something that you can deal with on a calculator easily, like base 10 or base e, but I'm not sure how to explain it better than that, or if I'm even understanding the question.  You still need logarithms to solve it, and if you mean 'calculate' to refer to doing it by hand as opposed to a log function on a calculator/computer, then I don't know the answer.  I'm not sure how logarithms were calculated historically.  I do know it was a giant pain in the ass, which is why they literally had tables and books full of logarithms to many decimal places back in the pre-computer days.  (they had such things for trig functions and stuff like that as well)  The book would have lookup tables and you'd go down lists of tables and find the one you wanted.  They also used slide rules for doing it mechanically in an approximate manner if they didn't want to do it that way.

The series you are referring to is called the Taylor series.  (you might also see the term Maclaurin series, which is one particular type of Taylor series)  Being an infinite series, the more terms you add up, the closer you get to the actual result, but being infinite, it'll also literally take forever to get to the actual result (without calculus, anyway) 

[andretimpa]

iirc the original motivation for logarithms was to find a way to make long multiplications using addition and the aid of comprehensive logarithm tables (so they were originally a computational aid in a world without calculators). from this point of view the original definition of what a logarithm is would be a function, log, obeying the property:

log(a.b) = log(a) + log(b)

This property parallels the property that exponentials have:

(a^b).(a^c) = a^(b+c)

so any function that for a given a, takes a^b and returns b for every real value of b obeys that definition of logarithm. Once you give a proper definition of what it means to calculate a^b when b is not an integer (you can use roots, that is, inverse exponentials, to define it for rational values of b and extend the definition for irrational values by imposing continuity) it follows that any function such that

log(a^b) = b

is a bona fide logarithm, which is what motivates the modern definition and the concept of a base for a logarithm.

Back in those days, the most important thing to make good use of logarithms was to have acurate logarithm tables, one thing that people started to notice was that if x ~ 0, then a^x ~ 1 + kx, with k depending only on a and not x. Not only that, but k behaved like a logarithm. These misterious logarithms are what Napier called the natural logarithms and he called the base that they'd use (since every logarithm must have a base) the number e (note that he didn't calculate e, but gave a definition for it). More than 100 years later, Euler studied the number e and the relationships we nowadays learn in calculus. The main point is that

a^x ~ 1 + k(a).x => k(a) ~ (a^x - a^0)/x, so that k is the derivate of a^x at x=0. If we want to calculate the derivate of a^x at x=z, that would give

lim h->0 (a^(z+h) - a^z)/h = a^z lim h->0 (a^h - 1)/h = k(a).a^z

To see that k(a) must be a logarithm, take b = a^c, then the derivate of b^x is k(b).b^x by the definition, but it is also the derivate of a^(c.x) so by the chain rule it equals c.k(a).b^x. As c = log_a (b), that means

k(b)/k(a) = log_a (b), so k obeys the logarithm base change formula and must be some logarithm in some base (e using Napier's definition) and we denote k(a) = ln(a). It follows that e must have the special property that the derivate of e^x is e^x itself (since ln(e) = 1 by definition). This leads to what was likely the first way of calculating e. Using (e^x)' = e^x and e^0 = 1, we can build a Taylor series for e^x. The series converges in all of R so substituting x = 1 leads to the following formula for e (I am skipping a lot here):

e = 1/0! + 1/1! + 1/2! + 1/3! + ...

To calculate lim n -> inf (1 + 1/n)^n (you need calculus for this one as limits are already outside the boundaries of what algebra can do) you do the following

lim n-> inf [ln ((1 + 1/n)^n)] = lim n -> inf [n .ln(1 + 1/n)] = lim h -> 0 [ln(1+h)/h] = ln '(1)

that is, the derivate of ln(x) at x = 1. Using the chain rule you can get that ln'(x) must be equal to 1/exp'(ln(x)) = 1/x  (exp'(x) is the derivate of e^x). Substituting,

lim n-> inf [ln ((1 + 1/n)^n)] = 1. Since e^x is continuous, we take the exponential in both sides and get    lim n -> inf (1 + 1/n)^n = e.

[Arum]

I haven't scanned through the rest of the thread, but there is a small table that helps me with inverting fractions or otherwise converting.

a^b=c Find the answer.

c^(1/b) or brt(c) both = a Find the base.

log_a c = b Find the exponent.

Not sure the use it'll be, but perhaps you could prove some theorems by converting the equation.

[MyNameIsJoey]

logarithms were invented to turn multiplications into additions, to simplify problems of high numbers espescially in the domain of astronomy.

What I've been unable to find whne i was searching on it is : how did people find number e? Find a number that when you derive e^x or when you integrate it it gives itself, thats pretty awesome.

[Savage]

one thing that people started to notice was that if x ~ 0, then a^x ~ 1 + kx, with k depending only on a and not x. Not only that, but k behaved like a logarithm. These misterious logarithms are what Napier called the natural logarithms and he called the base that they'd use (since every logarithm must have a base) the number e (note that he didn't calculate e, but gave a definition for it).

a^x ~ 1 + k(a).x => k(a) ~ (a^x - a^0)/x, so that k is the derivate of a^x at x=0. If we want to calculate the derivate of a^x at x=z, that would give

lim h->0 (a^(z+h) - a^z)/h = a^z lim h->0 (a^h - 1)/h = k(a).a^z

To see that k(a) must be a logarithm, take b = a^c, then the derivate of b^x is k(b).b^x by the definition, but it is also the derivate of a^(c.x) so by the chain rule it equals c.k(a).b^x. As c = log_a (b), that means

k(b)/k(a) = log_a (b), so k obeys the logarithm base change formula and must be some logarithm in some base (e using Napier's definition) and we denote k(a) = ln(a). It follows that e must have the special property that the derivate of e^x is e^x itself (since ln(e) = 1 by definition). This leads to what was likely the first way of calculating e. Using (e^x)' = e^x and e^0 = 1, we can build a Taylor series for e^x. The series converges in all of R so substituting x = 1 leads to the following formula for e (I am skipping a lot here):

Can you elaborate on how (1) a^x = 1 + k*x as x goes to 0? (2) How does k(b)/k(a) = log_a (b) imply k must be a logarithm?

To calculate lim n -> inf (1 + 1/n)^n (you need calculus for this one as limits are already outside the boundaries of what algebra can do) you do the following

lim n-> inf [ln ((1 + 1/n)^n)] = lim n -> inf [n .ln(1 + 1/n)] = lim h -> 0 [ln(1+h)/h] = ln '(1)

When you make the change from n to h, what happened to the 'n' on the outside of the ln?

that is, the derivate of ln(x) at x = 1. Using the chain rule you can get that ln'(x) must be equal to 1/exp'(ln(x)) = 1/x  (exp'(x) is the derivate of e^x). Substituting,

lim n-> inf [ln ((1 + 1/n)^n)] = 1. Since e^x is continuous, we take the exponential in both sides and get    lim n -> inf (1 + 1/n)^n = e.

If the literal definition of logarithm isn't what you're looking for, then what ARE you looking for?

Example: we define slope as (y2-y1)/(x2-x1) but even yet we can see that the slope is how steep something is. I meant by getting a proper definition, something more intuitive.

[Aves]

Hmm, so not a definition, but a real world analogy, you mean. A common real world example of logarithms in action is the Richter scale, used to measure the magnitude of earthquakes. A magnitude 2 earthquake is ten times stronger than a magnitude 1 earthquake, a magnitude 3 earthquake is 10 times stronger than a magnitude 2, and so forth. Thus a magnitude 3 earthquake is also 100 times stronger than a magnitude 1 earthquake, a magnitude 4 is 1000 times stronger than magnitude 1, etc.

A simple way of explaining the equation would be to say that all a logarithm does is counting the number of times the base would have to multiply by itself to reach a given number.

Also, have you looked over the add-multiply property of exponents and the multiply-add property of logarithms?

[andretimpa]

When I said people noticed that a^x ~ 1 + k(a)*x it was originally an empirical finding. A pattern that was noticed during the building of logarithm tables. Nowadays we understand it as a consequence of a^x ~ a^0 + k(a)*x with k(a) the derivate of a^x at x=0, which is just the first order Taylor approximation for a^x, but this was originally noticed before the development of calculus which is crucial to understand why the approximation is true.

For k(a) being a log, suppose we have

k(b)/k(a) = log_a(b) for all a and b. If we take c=b^(1/k(b)), then c^k(b) = b and k(b) = log_c(b). Substituting we get

k(a) = log_c(b)/log_a(b)

Now we use the base change formula log_x(y) = log_z(y)/log_z(x) and the special case when y=z: log_x(y) = 1/log_y(x):

k(a)=log_b(a)/log_b(c) = log_c(a).

Since this holds for all a and b then k is a logarithm in a base c that can be calculated as a^(1/k(a)) for any given a.


In the limit, the n that was multiplying the ln became an h dividing the ln. Essentially I did the variable change h =1/n.

[ProBacon]