∂/∂x q^x = q^x .... solve for q... a hypothetical question for math geniuses?

Question

I ask the question, what is the value q of the exponential function q^x such that its derivative with respect to x is equal to itself?

Okay... dont just spout out the mindless answer "oh, q equals Euler's constant, 'e'"... not good enough.

'e' was originally defined as the base of an exponential function ( e^x ) such that it had a slope of 1 at x=0. Thats is what e is based on... its only coincidence that its derivative equals itself throughout its domain. That was proven as a consequence of its definition.

My proposition is, what if ∂/∂x e^x did not equal e^x... regardless of the fact that it has a slope of 1 at x=0.

How would mathematicians have determined (or proven to exist) that one special number, the base to an exponential function, whose derivative was equal to itself?

gudspelling... not a correct answer. Your so called proof relies on the preexisting fact that e^x is its own derivative... you are not adhering the the criteria of the hypothetical.

Answer 1

I've found this to be rather interesting myself, the definition of e. Just recently I've been able to get a better handle on things. The first thing to understand, which I didn't for a long while, is that there are various different, equivalent definitions for the exponential function and the constant e. You've likely seen the definition of e as the limit as n->infinity of (1+1/n)^n, and then defining the function as simply powers of e (which is tough to do for irrational powers).

An alternative definition of exp(x) is the function whose derivative is itself. That makes the question easy to answer, doesn't it? :)
I've wondered though whether exp(x) is really unique in this regard...I suppose it is, but I haven't seen proof of it.

Yet another definition (from my current Real Analysis class) is to define e^x as the familiar Taylor series expansion. This one is nice because all the properties of the exponential fall out of it rather quickly.

My book also notes the "rigorous calculus" approach, defining the natural log as an integral, then defining the exponential as the inverse of this function.

Okay, now to actually give you an idea of how you might do this, assuming we've defined the exponentiation over reals (note that we get some properties of exponentiation out of this), and we want to find q such that d/dx(q^x) = q^x. Well, using the definition of derivative (limits have h->0),
d/dx(q^x)
= lim [q^(x+h) - q^x] / h
= lim [q^h*q^x - q^x] / h
= q^x * lim [q^h-1]/h.

Now if this is to equal q^x, we need the remaining limit to be 1. So we need to find the value of q for which the limit is 1. Since we know that e is irrational, we would expect not to be able to tell much about this limit, other than using numerical methods to find the value of q for which this works. I was trying to perhaps re-shape this limit equation into the familiar limit definition of e, but I haven't had any luck.

(I was trying to find additional things to give you, and apparently most of what I've said is on wikipedia...trounced again! But you may want to visit there to see a couple of additional definitions for e I didn't know about.)

An excellent question, by the way.

EDIT: Looking at the limit that we wanted to equal 1, I've just realized that that is equivalent to requiring that we have a slope of 1 at x=0. So I'm not sure I've answered any part of your question at all...I've still only really shown that the derivative property is equivalent to the slope restriction. Let me know if there's something I'm missing here...

EDIT#2: Wow. Pascal, you're my newest hero.
(However, I must refute that 0^0=1. It can be defined as such if you'd like, but I prefer to leave it as indeterminate since the limiting behavior is such. Anyway, it won't matter for your exponential proof, as you could easily define exp(x) to be 1 + the sum of remaining terms, since x^0/0!=1 for any [nonzero by me] x.)

Answer 2

There are two questions here -- the first is how mathematicians found e, and the second is how to prove the existence and uniqueness of e.

How mathematicians found e is quite simple. Attempting to differentiate a general exponential function yields:

d/dx (a^x)
= [h→0]lim (a^(x+h) - a^x)/h
= a^x [h→0]lim (a^h - 1)/h

So assuming that the limit of (a^h - 1)/h as h→0 exists, the derivative of any exponential function must be a constant times the value of the function itself. But what constant? After failing to resolve the problem by a direct approach, mathematicians decided to go at the problem backwards, namely to decide what they wanted that limit to be, and then try to figure out what base of the exponential function would have to be in order to get that limit. Now, since they got to decide what they wanted the limit to be, they decided to make it as simple as possible -- namely, to try for a limit of 1. So mathematicians simply assumed that somewhere in the world, there was a number they could find (let's call it e) such that:

[h→0]lim (e^x - 1)/h = 1

Now, from the considerations above, it's obvious that IF such a number exists, that d/dx (e^x) = 1. So IF such a number exists, applying Taylor's theorem to the function e^x yields that:

e^x = [k=0, ∞]∑x^k/k!

And then of course, IF e exists, one may find e by simply evaluating the above function at 1. But note that I keep saying IF e exists -- nowhere have we established that it does! We could be talking about a nonexistent function.

Or not. Because whether or not e exists, the function [k=0, ∞]∑x^k/k! certainly exists, and is easily seen to converge everywhere. So if we can prove independently of any assumptions that this IS an exponential function that is equal to its own derivative, it will follow that such a function exists, and therefore that the number e exists. I outline a proof based on this technique below. There is one technical theorem that I'll leave unproved, which is this one:

Theorem: On the interior of the radius of convergence, power series may be differentiated term by term. That is, d/dx [k=0, ∞]∑(c_k x^k) = [k=0, ∞]∑(d/dx (c_k x^k)).

The above theorem is completely obvious for finite series, but for infinite series requires additional proof. Basically, the issue is to show that where f_n(x) = [k=0, n]∑(c_k x^k), that [h→0]lim [n→∞]lim (f_n(x+h) - f_n(x))/h = [n→∞]lim [h→0]lim (f_n(x+h) - f_n(x))/h. I'm going to leave this theorem unproved, since it requires a bit of experience in real analysis to understand, and I'm not sure whether you have any. If you do , a proof sketch can be found at http://homepages.ius.edu/WCLANG/m414/spring2007/notes26.pdf . Another simpler proof (that may be more accessible to you) can be found at http://links.jstor.org/sici?sici=0002-9890%28195205%2959%3A5%3C323%3ATDOPS%3E2.0.CO%3B2-6 , but requires that you be at a university that subscribes to JSTOR.

Anyway, assuming that theorem, we may prove the existence of e as follows:

Lemma 1: There exists a function which is 1 when evaluated at 0 and is equal to its own derivative

Proof: We define the function exp(x) = [k=0, ∞]∑x^k/k!. Evaluating at 0 yields exp(0) = 0^0 + [k=1, ∞]∑0^k/k! = 1 + 0 = 1. (btw, I read your profile. You're simply wrong on that point -- 0^0 IS equal to 1. Yes, that does mean the exponential function is discontinuous at (0, 0)). To find the derivative of exp(x), we note that the series given converges everywhere, so every point is on the interior of the radius of convergence, so for every x, we may apply the above theorem, so:

d/dx (exp (x))
= [k=0, ∞]∑d/dx (x^k/n!)
= d/dx (1) + [k=1, ∞]∑d/dx (x^k/k!)
= [k=1, ∞]∑(kx^(k-1)/k!)
= [k=1, ∞]∑(x^(k-1)/(k-1)!)
= [k=0, ∞]∑x^k/k!
= exp (x)

Thus exp (0) = 1 and exp (x) is equal to its own derivative.

Lemma 2a: If f is a function defined on all of R such that ∀x, y, f(x+y) = f(x)f(y), then for any natural number n, f(nx) = f(x)^n.

Proof: This is trivial in the case n=1. Suppose this holds for some n. Then f((n+1)x) = f(nx + x) = f(nx)f(x) = f(x)^n f(x) = f(x)^(n+1), so it holds for n+1. Thus by induction, it holds for all n.

Lemma 2b: If f is a positive function defined on all of R such that ∀x, y, f(x+y) = f(x)f(y), then for all natural numbers p and q, f(p/q) = f(1)^(p/q)

Proof: Since f will also satisfy the hypotheses of lemma 2a, we have that f(1) = f(q/q) = f(1/q)^q, thus taking qth roots of both sides yields f(1)^(1/q) = (f(1/q)^q)^(1/q) = f(1/q) (we had to assume that f(1/q) is positive in order to ensure that (f(1/q)^q)^(1/q) = f(1/q) instead of -f(1/q)). Using this result and lemma 2a again, we have, f(p/q) = f(1/q)^p = (f(1)^(1/q))^p = f(1)^(p/q), as required.

Lemma 2c: If f is a positive function defined on all of R such that f(0) = 1 and ∀x, y, f(x+y) = f(x)f(y), then for any rational number r, f(r) = f(1)^r

Proof: If r is positive, then it can be represented as the quotient of two positive integers p and q. Since f will satisfy the hypotheses of lemma 2b, we have f(r) = f(p/q) = f(1)^(p/q) = f(1)^r. If r is 0, then we note that per hypothesis, f(0) = 1 = f(1)^0. So finally suppose that r is negative. We note that 1 = f(0) = f(-r + r) = f(-r)f(r) = f(1)^(-r)f(r) (the last equality holds because if r is negative, then -r is positive). Multiplying both sides by f(1)^r then yields that f(r) = f(1)^r, as required.

Lemma 2d: If f is a positive continuous function defined on all of R such that f(0) = 1 and ∀x, y, f(x+y) = f(x)f(y), then for any real number x, f(x) = f(1)^x

Proof: Here we need the definition of exponentiation to an irrational number. One common definition is to take limits of rational exponents -- to wit, where x is irrational, we define a^x = [r→x, r∈ℚ]lim a^r, provided that this limit exists. We'll assume this definition is being used (other definitions essentially involve defining a^x as exp (x ln a), which although perfectly sound, would probably strike you as somewhat circular). Now, since f satisfies the conditions of lemma 2c, if x is rational, then f(x) = f(1)^x. If x is irrational, then since f was stipulated to be continuous, f(x) = [r→x, r∈ℚ]lim f(r) = [r→x, r∈ℚ]lim f(1)^r = f(1)^x. So this holds for all x, and f is an exponential function.

Lemma 3: The function exp (x) defined in lemma 1 satisfies the hypotheses of lemma 2d

Proof: Since exp (x) is differentiable everywhere, it is of course continuous. We already know that exp (0) = 1. So it remains to be shown that exp (x+y) = exp (x) exp (y) and that exp (x) is everywhere positive.

Consider the function g(z) = exp (x+y - z) exp (z). Since d/dz (exp (z)) = exp (z), taking the derivative of g yields:

d/dz (g(z))
= d/dz (exp (x+y - z) exp (z))
= d/dz (exp (x+y - z)) exp (z) + exp (x+y - z) d/dz (exp (z))
= exp (x+y - z) d/dz (x+y - z) exp (z) + exp (x+y - z) exp (z)
= - exp (x+y - z) exp (z) + exp (x+y - z) exp (z)
0

So g is in fact a constant function. Thus g(0) = g(y), and g(0) = exp (x+y - 0) exp (0) = exp (x+y) and g(y) = exp (x+y - y) exp (y) = exp (x) exp (y). Therefore, exp (x+y) = exp (x) exp (y), and as x and y were arbitrary, this holds for all x, y.

Now, to show that exp (x) is always positive -- suppose the contrary, that there is some x such that exp(x) is nonpositive. Then either exp (x) = 0 or exp (x) < 0 -- in the latter case, then since exp (0) = 1 is positive, then per the intermediate value theorem there is a number c in either (x, 0) or (0, x) (depending on whether x is greater or less than 0) such that exp (c) = 0. But then that means that 0*exp (-c) = exp (c) exp (-c) = exp (c+(-c)) = exp (0) = 1, contradicting the fact that 0*anything is 0. Therefore, exp (x) is always positive.

Main theorem: there exists a unique positive real number e such that d/dx (e^x) = e^x.

Proof: Let e = exp (1), where the exp function is defined as in lemmas 1 and 3. Per lemma 2d, exp (x) = exp (1)^x = e^x. Therefore, d/dx (e^x) = d/dx (exp (x)) = exp (x) = e^x, so e satisfies the given requirements.

To show uniqueness, suppose a is some other positive real number such that d/dx (a^x) = a^x. Then since a is positive, ln a exists so d/dx (a^x) = d/dx (e^(x ln a)) = e^(x ln a) d/dx (x ln a) = a^x ln a. But d/dx (a^x) = a^x, thus by transitivity, a^x = a^x ln a. Since a^x≠0, dividing by it yields 1 = ln a, so a=e, contradicting the assumption that a is a different positive real number. Q.E.D.

By the way, I've noticed that several of your previous posts involve questions related to the foundations of analysis (e.g. your question on how to prove π exists), so that's why I'm wondering whether you have any analysis background. If not, I would be happy to try to teach you some of what I know via e-mail (although obviously I cannot substitute for the services of an actual professor at a university). If you're interested, feel free to e-mail me.

Edit: Thanks for the complement Ben. Okay, maybe it's a bit strong for me to say Cogito is simply wrong about 0^0, since many textbooks do leave it undefined, but I generally feel that defining 0^0=1 is much more elegant algebraically. Plus, it serves the useful function of reinforcing the difference between the value of a function at a point and the limiting behavior as the function approaches the point, which I don't think is sufficiently stressed in calculus classes.

Answer 3

Use a power series:

q^x = a + bx + cx^2 + dx^3 + ex^4 + .....
d(q^x)/dx = b + 2cx + 3dx^2 + 4dx^3 + ....

a = b
b = 2c
c = 3d
d = 4e ....

Rearranging the equations:
b = a
c = b/2 = a/2
d = c/3 = b/(2*3) = a/3!
e = d/4 = ...... a/4!
f = a/5!

q^x = a + bx + cx^2 + dx^3 + ex^4 + .....
q^x = a + ax + (a/2!)x^2 + (a/3!)x^3 + (a/4!)x^4 + ....
q^x = a(1 + x + (x^2)/2! + (x^3)/3! + (x^4)/4! + ....)

Note that this is equal to a*e^x

q = q^1 = a(1 + 1/1! + 1/2! + 1/3! + 1/4! + ......)


Note: This is effectively a solution of the differential equation y' = y by the power-series method.

Edit:
This is a solution of the differential equation: dy/dx = y,
which is effectively what you are asking for in your question.

Answer 4

As Ben said, use (q^x) ' = lim(h->0){ (q^(x+h)-q^x)/ h }

= q^x lim(h->0) (q^h-1)/h

for which we need lim(h->0) (q^h-1)/h = 1 and that is

q = lim(h->0) (1+h)^(1/h)

This uniquely determined number, which approximates
2.718 is then called e.