Definition of the natural logarithm

I know that there is a year old thread that has just been resurrected that is slightly on this topic, but I had a question that I thought might merit a new thread.

I'm working with my Calculus book here, and I'm working on the chapter called, "Logarithmic Functions from the Integral Point of View." It works through various proofs of logarithmic functions in a very explicit mathematical manner. However, they are all based on a definition of the natural logarithm that does not seem to match the mathematical rigor that is apparant in almost every other proof in the book. I'm sure this is simply due to my failure to understand a certain aspect of this definition, so I thought I'd ask on here and see if you can clear it up.

Basically, it starts out talking about the history of the natural logarithm. Newton, and a few others were trying to solve the problem of finding the values of x1, x2, x3, etc. for which the the areas A1, A2, A3, under the curve of y = [tex]\frac{1}{x}[/tex] would all be equal.

They found that the x values that satisfy this problem were exp(1), exp(2), exp(3) etc. Due to the relationship between the area under the curve, and the definite integral, we get the following:
[tex]\int_1^{x}{1/t}\,dt = lnx[/tex].

So, basically, it seems that they experimented around with the graph, found that there was this one number that kept coming up as a solution to this problem. This then led to the natural logarithm. The issue I have with this is that it seems it's all built on basically guesswork. They guessed around to find the right x value that would satisfy their area, and they found e. It just doesn't seem as rigorous as everything else. Is there something I'm missing? Or is this just the only way to "Discovery," a transcendental number like e?
 

HallsofIvy

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I'm not sure what you mean by "rigor". It is perfectly reasonable to "experiment" to see how you would like to define something. It is the proofs after the definition that have to be rigorous. There is an important difference between "motivation" for a particular definition and rigorous proofs.

It would have been perfectly reasonable to reason the following way:
We can show, easily, that [itex]\int x^n dx= x^{n+1}/(n+1)[/itex] for all n except n= -1. That means we still don't have an anti-derivative for 1/x! Okay, let's just define
[tex]\int_1^x (1/t) dt[/tex]
to be "ln(x)".

That is, don't bother talking about "Exp(1)", etc. yet. Just define this new function ln(x) as being a specific anti-derivative of 1/x. 1/x is continuous for x> 0 so this function is defined for all positive x. By the fundamental theorem of calculus, it is also differentiable for x> 0 and its derivative is 1/x. That's positive for x> 0 so we know ln(x) is strictly increasing. From the definition, [itex]ln(1)= \int_1^1 dt/t= 0[/itex] so ln(x) is positive for x> 1 and negative for x< 1.

I'm sure your text shows that ln(xy)= ln(x)+ ln(y), ln(1/x)= -ln(x) and ln(xy)= yln(x) so I won't show those here.

What we can do then is this: Since ln(x) is differentiable for all positive x, it is also continuous for all postive x and we can apply the mean value theorem to the interval [1, 2].
[tex]\frac{ln(2)- ln(1)}{2- 1}= 1/c[/tex]
for some c between 1 and 2. But
[tex]\frac{ln(2)- ln(1)}{2-1}= \frac{ln(2)- 0}{2- 1}= ln(2)= 1/c[/itex]
Since [itex]0\le 1/c \le 2[/itex], [itex]1/2\le c\le 1[/itex] so [itex]ln(2)\ge 1/2[/itex].

From that, for any X> 0, taking Y= 22X, ln(Y)= ln(22X)= 2Xln(2)> 2X(1/2)= X. That is, given any X> 0, there is Y such that ln(Y)> X. ln(x) is unbounded and, since it is an increasing function, [itex]\lim_{x\rightarrow \infty} ln(x)= \infty[/itex]. And, since ln(1/x)= -ln(x), [itex]\lim_{x\rightarrow 0} ln(x)= -\infty[/itex]. While the domain of ln(x) is x> 0, its range is all real numbers.

But ln(x) is an increasing function from [itex][0, \infty)[/itex] to R so it is invertible and its inverse, which, for no particular reason, I shall label "exp(x)", is from R to [itex][0, \infty)[/itex].

Now, here is the crucial point. If y= exp(x) then, by definition, x= ln(y) and, since x> 0, 1= (1/x)ln(y)= ln(y1/x. Going back to the inverse function, exp(1)= y1/x and so y= (exp(1))x.

That is, y= exp(x), defined simply as the inverse function to ln(x), can be written y= (exp(1))x, a number to the x power. Defining "e" to be the number exp(1), y= exp(x)= ex.

That is, by defining ln(x) as an integral, we arrive at the fact that its inverse function can be written ex, we do not start from ex as a separate function.
 
Definitions need no proof. They just help simplify things. Why is x^0 = 1? Because defining it that way makes exponent arithmetic work out. For example, x^n / x^n = x^(n-n) = x^0, which better be 1 since x^n/x^n = 1. (Maybe that looks like a proof that x^0 = 1, but it's not; it's really just extending another definition -- exponents.)
 
HallsofIvy, thank you very much for your proof. I managed to follow and understand it, but I had a small, and somewhat trivial issue here:

Since [itex]0\le 1/c \le 2[/itex], [itex]1/2\le c\le 1[/itex] so [itex]ln(2)\ge 1/2[/itex].
Basically, how'd you get from [itex]1/2\le c\le 1[/itex] to [itex]ln(2)\ge 1/2[/itex]?

Definitions need no proof. They just help simplify things. Why is x^0 = 1? Because defining it that way makes exponent arithmetic work out. For example, x^n / x^n = x^(n-n) = x^0, which better be 1 since x^n/x^n = 1. (Maybe that looks like a proof that x^0 = 1, but it's not; it's really just extending another definition -- exponents.)
That is very helpful. Thank you for the explanation
--
Just to add to all of this, I had something related to what DoctorBinary was saying. Am I correct in saying that there are some constants that are simply existent in nature and are not arbitrarily defined, like many of our definitions? For example, pi, is found in the circle, e is found all over the place, etc. So, we might define x^0 = 1, but e, pi, etc are defined by nature, so to speak.

Is this a correct way of looking at these constants?
 
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disregardthat

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We can show, easily, that [itex]\int x^n dx= x^{n+1}/(n+1)[/itex] for all n except n= -1.
I know a way to show that without using the properties of [tex]y=e^x[/tex], but it's not "easily" shown. Do you have a suggestion to how we easily can show this for all real n!=-1?
 
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HallsofIvy

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HallsofIvy, thank you very much for your proof. I managed to follow and understand it, but I had a small, and somewhat trivial issue here:


Basically, how'd you get from [itex]1/2\le c\le 1[/itex] to [itex]ln(2)\ge 1/2[/itex]?
My fault. I miswrote that. I had already written, from the mean value theorem, that [itex]ln(2)\ge 1/c[/itex]. Since [itex]1/2\le c[/itex] it follows that [itex]c\ge 2[/itex] and so [itex]ln(2)\ge 2[/itex], not 1/2.
 

Office_Shredder

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My fault. I miswrote that. I had already written, from the mean value theorem, that [itex]ln(2)\ge 1/c[/itex]. Since [itex]1/2\le c[/itex] it follows that [itex]c\ge 2[/itex] and so [itex]ln(2)\ge 2[/itex], not 1/2.
I think you're supposed to conclude ln(2)<2
 

gel

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I know a way to show that without using the properties of [tex]y=e^x[/tex], but it's not "easily" shown. Do you have a suggestion to how we easily can show this for all real n!=-1?
It follows from [tex]\frac{d}{dx}x^{n+1}=(n+1)x^n[/tex]
 

disregardthat

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It follows from [tex]\frac{d}{dx}x^{n+1}=(n+1)x^n[/tex]
It is obviously this formula I am referring to in the first place.
 

Office_Shredder

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It is obviously this formula I am referring to in the first place.
You just use the definition of the derivative and the binomial theorem to expand (x+h)n

If you're looking for an in depth discussion on how that works, look up the power rule on wikipedia
 

DrGreg

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You just use the definition of the derivative and the binomial theorem to expand (x+h)n

If you're looking for an in depth discussion on how that works, look up the power rule on wikipedia
You missed Jarle's point. He said "all real n", i.e. including non-integer n.
 
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morrobay

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Hello The Anomaly
I have been obsessed with this exact question;
See my two posts at Calculus and Analysis:
https://www.physicsforums.com/showthread.php?t=215045
https://www.physicsforums.com/showthread.php?t=317165
Yesterday I found a page at http://www.physclips.unsw.edu.au/jw/calculus.htm [Broken]
see log function
There is a graph with two plots on it:
y=ln x and y= 1/x
The slope of y= ln x is given by y=1/x
The curve of y=ln x is rising at the same rate as the area under cuve of y= 1/x is increasing,
ln x = integral 1/t dt
from 1 to x
 
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To paraphrase earlier posts, ask yourself what function of t satisfies the relation
d/dt f(t) = 1/t. In physics, we know that the field of a charge in 3-D space falls off as 1/R2, and the potential as 1/R. For a long line charge (like on a wire), the field drops off as 1/R. How does the potential fall off?
 

Hurkyl

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I should point out that that's how research usually goes.

(1) You experiment, solve problems, and whatnot.

(2) You conceive of / find something that 'appears' to be true, or 'should' be true.

(3) You prove a theorem that says it really is true, or at least what conditions need to be assumed to make it true

(4) You incorporate it into a good exposition of the subject


In rigorous expositions of real analysis, the definition [itex]\ln x = \int_1^x 1/t \, dt \qquad (x > 0)[/itex] is quite frequently taken as a starting point for presenting some of the elementary functions, probably because it is completely explicit and depends on very little theory.
 
Thank you for the responses, I see your points about the role of experimentation and definitions in mathematics.

To paraphrase earlier posts, ask yourself what function of t satisfies the relation
d/dt f(t) = 1/t. In physics, we know that the field of a charge in 3-D space falls off as 1/R2, and the potential as 1/R. For a long line charge (like on a wire), the field drops off as 1/R. How does the potential fall off?
I'm afraid I don't fully understand how the potential fall off fits in. If you integrate 1/R, which is the field drop off, then clearly we have the natural logarithm we're talking about. However, you're asking about the potential fall off for the wire, and I'm not sure how that fits in. For a charge in a 3d space, you said the potential fall off (1/R) was the square root of the electric field fall off (1/R2). Does the same apply here?

I'm obviously missing the point. The f(t) in your initial question is clearly ln(x), but I do not know how your physics example fits in, unless we integrate the 1/R for some reason.
 

Office_Shredder

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You missed Jarle's point. He said "all real n", i.e. including non-integer n.
Whoops, silly me. The 'n' confused me.

Ok, I sent him a PM with the details for the generic real case
 
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