Exponent Combination Laws

[gfauxpas]

Hello everyone.

I'm investigating the different definitions of the exponential function exp x.

I noticed that all the proofs I've seen for exponent combination laws or logarithm laws for irrational indices/bases assumes the definition:

\exp y = x \iff \ln x = y, \ln x = \int_1^x \frac {\mathrm {d}t} t

How does one derive the exponent combination laws or log laws for other definitions of exp x, particularly for exp x = e^x?

[korona]

You can define exp through its power series. Most identities can be easily proven from this definition.

[Yakk]

The limit definition of e can give you both a definition for e, and a direct definition of e^x.

[legend]

You usually just show that all your definitions are equivalent and then you can prove the properties you need using any definition you like.

[gfauxpas]

Korona: can you elaborate on that a bit more, please?

And Yakk, the only way I know how to use the limit definition to go to e^x is to say:

\exp x = \lim_{n \to \infty}(1 + x/n)^n = \lim_{(n/x) \to \infty}((1 + 1/(n/x))^{(n/x)})^x = e^x

but that assumes exponent combination laws already.

Legend, I'm working on showing all the definitions are equivalent, but e^x seems isolated from all others, and I can't get the limit definition to imply any others.

[legend]

You might be interested this:
http://en.wikipedia.org/wiki/Characterizations_of_the_exponential_function
Also sorry, but I don't know what you mean by e^x. Usually for real numbers you define a^b:=exp(ln(a)*b). So using this to define the exp function gives you a circular definition.

[OverBored]

Ok, so in my experience, we normally derive e, and the exponential function by considering power series. In particular we define them and then a radius of convergence, and then we note that this particular power series happens to be everywhere convergent. It is a standard theorem in analysis the power series can be differentiated termwise, which gives us the standard identity d(e^x)/dx = e^x
From this we get e = exp(1) (which happily we know converges). Getting our standard laws are a little work in this, in particular we want
\Sigma \frac{(x+y)^n}{n!} = \Sigma \frac{(x)^n}{n!} \Sigma \frac{(y)^n}{n!}
There are a few ways of proving this (including just writing it out) but in my opinion the most slick way (so long as you have a couple of differentiation rules) is as follows:

Let F(x) = exp(a + b - x) * exp(x) , then F'(x) = exp(a + b - x)*exp(x) - exp(a + b - x)exp(x) = 0,
and so F(x) is constant.
But F(0) = exp(a + b), and F(b) = exp(a)*exp(b)
Done

The line where we state that derivative is 0 implies constant is non-trivial, but doesn't require too much in the way of tools.

I especially like this characterisation, as it follows over into the complex plane quite simply.

[gfauxpas]

You might be interested this:
http://en.wikipedia.org/wiki/Characterizations_of_the_exponential_function
Also sorry, but I don't know what you mean by e^x. Usually for real numbers you define a^b:=exp(ln(a)*b). So using this to define the exp function gives you a circular definition.

Okay, let me try to explain.

e^x for natural number indices: e multiplied by itself x times. Product of no numbers is defined as 1.
Negative integers: e^x = 1/e^(-x)
rational numbers: e^(p/n) = (the nth root of e)^p = the nth root of (e^p)
irrational numbers: .... ??

Is there a way to continue this pattern using, I dunno, limits? sequences?

[Yakk]

Yes. Non-rational numbers can be represented as a cauchy sequence of rationals. To generate the cauchy sequence that represents f(x), where f is defined on the rationals, and where x={a_i} is a cauchy sequence, simply calculate f(a_i), possibly dropping some "head" of the cauchy sequence for which f() is not always defined.

This works for defining e^x for x non-rational reasonably well. (Note that it requires certain things of f that I didn't mention -- continuity will do, I think.)

[gfauxpas]

Thanks everyone for your answers, it's been helpful.

Yakk I don't know enough analysis to prove continuity like that, but thanks for that insight. I imagine it's straightforward to prove exponent combination laws for integer indices, but can you prove it for rational indices without recourse to the existence of the logarithm? Would then taking the limit of the sequences prove that the properties hold for all real indices?

[korona]

e^x is continuous so if x is a real number, (x_1, x_2, x_3, ...) is a rational sequence converging to x then (e^x_1, e^x_2, e^x_3, ...) converges to e^x.
But if you define e^x that way then you have to show that the sequence converges and does not depend on the choice of the x_i.
Defining exp as a power series gives you differentiability on the radius of convergence for free.
EDIT: @yakk You cannot say "that holds if f is continuous" as you cannot really define continuity for a function defined on the rationals. In general not every function from Q can be extended to R that way.

[gfauxpas]

e^x is continuous so if x is a real number, (x_1, x_2, x_3, ...) is a rational sequence converging to x then (e^x_1, e^x_2, e^x_3, ...) converges to e^x.
But if you define e^x that way then you have to show that the sequence converges and does not depend on the choice of the x_i.
Defining exp as a power series gives you differentiability on the radius of convergence for free.

I hear what you're saying, thanks.

I see how you can jump from most of the definitions to the other, using the FToC or theorems for the existence of power series or whatever. But the definition as the limit of a sequence (1 + x/n)^n seems to be in the middle of nowhere. How does that one imply any other definition?

[legend]

Okay, let me try to explain.

e^x for natural number indices: e multiplied by itself x times. Product of no numbers is defined as 1.
Negative integers: e^x = 1/e^(-x)
rational numbers: e^(p/n) = (the nth root of e)^p = the nth root of (e^p)
irrational numbers: .... ??

Is there a way to continue this pattern using, I dunno, limits? sequences?

Ok, I see where this is going. But you're defining e as a limes and your real number x as a limes of rationals. You're finally ending up with a really ugly definition and I really don't see what it would be good for.
The usual approach is to show that exp(ln(a)*b)=a^b holds for rational b and use this to define a^b for rationals.

[tomtom2357]

EDIT: @yakk You cannot say "that holds if f is continuous" as you cannot really define continuity for a function defined on the rationals. In general not every function from Q can be extended to R that way.

Actually, you can define continuity on the rational numbers, because the rationals are dense, from which it follows that between any two rational numbers there is another rational number. All you have to do is tweak the epsilon delta definition slightly so that the only x values allowed are rational numbers.

[PM 2Ring]

I see how you can jump from most of the definitions to the other, using the FToC or theorems for the existence of power series or whatever. But the definition as the limit of a sequence (1 + x/n)^n seems to be in the middle of nowhere. How does that one imply any other definition?

You can do what Newton himself did: use binomial expansion on (1 + x/n)^n, which will give you the power series. It's not exactly rigorous, but hey.

Actually, you can define continuity on the rational numbers, because the rationals are dense, from which it follows that between any two rational numbers there is another rational number. All you have to do is tweak the epsilon delta definition slightly so that the only x values allowed are rational numbers.

* kicks tomtom2357 *

[korona]

Let f: Q -> Q; f(x) = 0 if x^2 < 2 and f(x) = 1 otherwise. This function is continuous in Q as the pre-image of every open set is open:
The sets f^-1( (-epsilon, epsilon) ) = ( (-sqrt(2), sqrt(2)) intersected with Q)
and f^-1( (1 - epsilon, 1 + epsilon) ) = ( (-infinity, -sqrt(2)) union (sqrt(2), infinity) intersected with Q)
are open in Q and form a partition of Q. But clearly this function cannot be extended to R.
EDIT: Uniform continuity would be sufficient but e^x is not uniform continuous on Q.

[jestingrabbit]

I see how you can jump from most of the definitions to the other, using the FToC or theorems for the existence of power series or whatever. But the definition as the limit of a sequence (1 + x/n)^n seems to be in the middle of nowhere. How does that one imply any other definition?

You can do what Newton himself did: use binomial expansion on (1 + x/n)^n, which will give you the power series.

Another approach: let f(x)= lim (1 + x/n)^n. Prove convergence for all x (don't have a simple argument for that atm, sorry). Note that f(a+b)<=f(a)f(b)<=f(a+b+e) for all positive e, and conclude that f(a+b) = f(a)f(b). You also have monotonicity (perhaps prove this after convergence) and expanding the brackets will give you a value for f(1). That there is only one function that satisfies these properties is the final result you need.

[mfb]

I see how you can jump from most of the definitions to the other, using the FToC or theorems for the existence of power series or whatever. But the definition as the limit of a sequence (1 + x/n)^n seems to be in the middle of nowhere. How does that one imply any other definition?

If you already know some rules about exp, then exp(x+1/n)=c*exp(x) and therefore exp(0+n/n)=c^(n)*exp(0) => exp(1)=c^n
But the taylor expansion of the first equation gives you c=(1+1/n+O(n^2))... so after you found out that you can neglect the higher orders in the limit, you have exp(1)=lim((1+1/n)^n,n->inf). It is possible to do the same with an additional x, just substitute stuff.

[gfauxpas]

Sorry, why does \exp (x + \frac 1 n) = c \exp x?

You can do what Newton himself did: use binomial expansion on (1 + x/n)^n, which will give you the power series. It's not exactly rigorous, but hey.

Oh whoa the binomial theorem works for all real powers? Cool, I'll look into that

[mike-l]

You can do what Newton himself did: use binomial expansion on (1 + x/n)^n, which will give you the power series. It's not exactly rigorous, but hey.

Oh whoa the binomial theorem works for all real powers? Cool, I'll look into that

\
It does, but you can evaluate the limit using any sequence that goes to infinity, in particular with just integers. (This doesn't prove the limit exists, but proves that if it does, then it's equal to what you get)

[gfauxpas]

It does, but you can evaluate the limit using any sequence that goes to infinity, in particular with just integers. (This doesn't prove the limit exists, but proves that if it does, then it's equal to what you get)

Oh, right.

I can prove the limit exists if I'm allowed to use logarithm laws. But, um, yeah.

[jestingrabbit]

I can prove the limit exists if I'm allowed to use logarithm laws. But, um, yeah.

Well, I trust that you can show that the limit is monotonically increasing for positive x, so all you need is an upper bound, so I would suggest that that is where to focus your energy. In particular, you know that 3^x is an upper bound, or 4^x, or those after you round x up etc etc. I suspect that if you pick a large enough upper bound for the sequence the proof will be doable.

Note that you might need to assume the rule is true for integer powers of integers, but that's not a real problem imo. To some extent, this result should come out of that other one.

[gfauxpas]

I can prove the limit exists if I'm allowed to use logarithm laws. But, um, yeah.

Well, I trust that you can show that the limit is monotonically increasing for positive x, so all you need is an upper bound, so I would suggest that that is where to focus your energy. In particular, you know that 3^x is an upper bound, or 4^x, or those after you round x up etc etc. I suspect that if you pick a large enough upper bound for the sequence the proof will be doable.

Note that you might need to assume the rule is true for integer powers of integers, but that's not a real problem imo. To some extent, this result should come out of that other one.

I'm willing to assume the power rules for integer powers even not of integer bases. Though I don't know the proofs I'm quite sure they're not Calculus proofs.

I'd have to think about how to prove something is an upper bound, I've never done that before for a case similar to this one. Can I have a hint?

[mfb]

Sorry, why does \exp (x + \frac 1 n) = c \exp x?

Follows directly from exp(a+b)=exp(a)*exp(b) with a=x and b=1/n. Define c as exp(1/n) and look at its series expansion.

[gfauxpas]

Oh, okay. But I need to know exponent combination laws for that, which goes back to my original problem

[mfb]

Well, you just need a good order in your proofs.

IIRC, my prof introduced it like this:
ln(x) gets defined as an integral over 1/x (->derivative obvious, power series calculatable)
exp(x) is the inverse function of ln(x) (derivative can be calculated -> power series follows -> convergence radius -> it is equivalent to power series definition of exp(x))
e can be defined as exp(1) which is 1/0!+1/1!+1/2!+...
ln(a*b)=ln(a)+ln(b) can be derived from the integral definition of ln(x), this can be translated to exp(a+b)=exp(a)*exp(b)
Now there might be some magic involved, but I think you can extend stuff like exp(ln(a)+ln(a))=a*a=a^2 to identify exp(ln(a)*b) with a^b for rational b (which was defined earlier using multiplication and integer roots). Define a^c for irrational c as the limit of rational b->c, proof that it is well-defined. This also justifies the notation exp(x)=e^x.

[gfauxpas]

Ok, so in my experience, we normally derive e, and the exponential function by considering power series. In particular we define them and then a radius of convergence, and then we note that this particular power series happens to be everywhere convergent. It is a standard theorem in analysis the power series can be differentiated termwise, which gives us the standard identity d(e^x)/dx = e^x
From this we get e = exp(1) (which happily we know converges). Getting our standard laws are a little work in this, in particular we want
\Sigma \frac{(x+y)^n}{n!} = \Sigma \frac{(x)^n}{n!} \Sigma \frac{(y)^n}{n!}
There are a few ways of proving this (including just writing it out) but in my opinion the most slick way (so long as you have a couple of differentiation rules) is as follows:

Let F(x) = exp(a + b - x) * exp(x) , then F'(x) = exp(a + b - x)*exp(x) - exp(a + b - x)exp(x) = 0,
and so F(x) is constant.
But F(0) = exp(a + b), and F(b) = exp(a)*exp(b)
Done

The line where we state that derivative is 0 implies constant is non-trivial, but doesn't require too much in the way of tools.

I especially like this characterisation, as it follows over into the complex plane quite simply.

What do you mean by writing it out? I tried playing with various expansions but couldn't get any to work.

Your second suggestion assumes the knowledge that D_x exp x = exp x though, and I'm pretending not to know that yet.

[OverBored]

Okay, I will write it out explicitly, but I am not sure how useful it is really.

Spoiler:

e^x e^y =(\sum_0^\infty {\frac{x^n}{n!}})(\sum_0^\infty {\frac{y^n}{n!}}) = (1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots)(1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \cdots)

We find the constant term, then the term linear in x and y, quadratic in x and y, etc

= 1 + x + y + \frac{x^2}{2!} + xy + \frac{y^2}{2!} + \frac{x^3}{3!} + \frac{x^2 y}{2!} + \frac{x y^2}{2!} + \frac{y^3}{3!} \cdots = 1 + (x+y) + \frac{(x+y)^2}{2!} + \frac{(x+y)^3}{3!} \cdots

or a little more carefully

(\sum_0^\infty {\frac{x^n}{n!}})(\sum_0^\infty {\frac{y^n}{n!}}) = \sum_{n=0}^\infty{\sum_{k=0}^n{\frac{x^k y^{n-k}}{k! (n-k)!}}}= \sum_{n=0}^\infty{\frac{(x+y)^n}{n!}}

The inner sum is just the binomial coefficients, and we get them just from adding the terms.

The other method does assume the derivative property, yes. If you don't want to, that is fair enough, but proving it can be useful. The way I was shown it was to prove that power series can be differentiated termwise (which actually looking back was a fairly long proof) and then just applying this to our definition of exp(x). It isn't the most elegant way to go about defining exp and e, but once you have done the work it gives you a nice solid definition, and quite a few properties.

[gfauxpas]

Cool, thanks for showing me that.

I'm thinking that, using the limit definition of exp, if I can show:

\left({1 + \frac{x + y}{n}} + \frac {xy}{n^2}\right)^n \to \left({1 + \frac{x + y}{n}}\right)^n as n \to +\infty and similar things, I can prove the exponent combination laws that way. My friend is examining this with me to see if it's feasible, feel free to chime in.

I appreciate everyone's help btw.

edit: oh, the reason I'm trying to avoid that knowledge is because f = Df, f(0) = 1 is one of the definitions, I'm trying to keep the definitions seperate.

[mfb]

\left({1 + \frac{x + y}{n}} + \frac {xy}{n^2}\right)^n = 1 + (x+y) + \frac{n-1}{n}(x+y) + ... + \frac {xy}{n} + c \cdot \frac{(x+y)^2}{n} + O(1/n^2)
Well, write it down in a strict mathematical way, and you get the usual exponential power series with (x+y).

[gfauxpas]

I'm bull-headed so I insisted on proving one law directly from the limit of a series definition (with a lot of help from an online friend). It's here if anyone wants to see it, it's proof 2.

edit: on further investigation, that is the hardest one to prove, I think.