Let E be a function satisfying E(0)=E'(0)=1.?

Question

Prove that if E(a+b)=E(a)E(b) for all a and b, then E is differentiable and E'(x)=E(x) for all x. Find an example of a function satisfying E(a+b)=E(a)E(b).

Answer 1

The problem statement tells us that E(x) is differentiable at x = 0 ... We wish to show E is differentiable everywhere else and on top of that E'(x) = E(x). The proof is as follows:



Let a be fixed and arbitrary, then via definition E'(a) = lim (E(a + h) - E(a)) / h , as h --> 0



But E(a + h) = E(a) E(h) according to the given property for E(x), hence the above equals



lim [E(a) E(h) - E(a)] / h =factoring out the constant E(a) in the numerator= E(a) lim [(E(h) - 1)] / h , as h --> 0 (*)



Now since E(0) = 1, the limit of (E(h) - 1) / h as h--> 0 is of the ambiguous form 0/0. So the L'Hopitale Rule applies and taking derivatives from top and buttom it tells us the aforementioned limit equals lim (E'(h) - 0) / 1 as h --> 0 which equals E'(0) / 1 = 1/1 or simply 1.



Hence the limit exists, which proves E is differentiable and besides that, from (*) we conclude that E'(a) = E(a) times 1 = E(a). Since the latter is true for every a in the domain, it follows that E'(x) = E(x) for all x.



The example of course is Exp(x) or any exponential function with base b > 1 or 0 < b < 1. I believe there is a rather deep theorem stating that any continuous function f(x) satisfying the given functional equation with f(0) = 1 must be an exponential function. The result is due to Euler if I recall correctly.



****

I just saw the comment below by the other contributor. We do know the limit of E(h) as h --> 0: E'(0) exists so E is differentiable at x = 0 which means E is continuous at zero. Hence lim E(h) = E(0) = 1 as h --> 0, according to the definition of continuity; the two-sided limit of the function equals the value of the function at that point.

Answer 2

One nitpick about Mr. Rebound's answer. E(0) = a million tells you no longer something about what E(h) is as h approaches 0. you have not any thanks to understand even if that's an indeterminate style. You do recognize that E is differentiable at 0 and that that's a similar as a million notwithstanding. Use this alongside with the reduce definition of a spinoff to get: E'(0) = a million lim [E(0+h) - E(0)] / h = a million h?0 lim [E(h) - a million] / h = a million h?0 that's strictly the reduce you had before utilising l'Hopital's rule. ***** ok. i comprehend how the lim E(h) approaches a million as h ?0. that promises the indeterminate style 0/0. I initially became going to write down that you could not use l'Hopital's rule because you don't recognize what the by-product is close to 0 or perhaps even if that's differentiable everywhere else except 0. that's what you try to tutor after all. I left this out because i theory that the different 'blunders', which got here first, made the later one unnecessary to tutor. yet now that you've cleared up the single blunders is there a fashion to sparkling up the different? you could not use l'Hopital's rule without already understanding E is differentiable.

Answer 3

One nitpick about Mr. Rebound's answer.



E(0) = 1 tells you nothing about what E(h) is as h approaches zero. You have no way to know if it is an indeterminate form.



You do know that E is differentiable at zero and that it is equal to 1 though. Use this along with the limit definition of a derivative to get:



E'(0) = 1



lim [E(0+h) - E(0)] / h = 1

h→0



lim [E(h) - 1] / h = 1

h→0



Which is precisely the limit you had before using l'Hopital's rule.



*****

Ok. I understand how the lim E(h) approaches 1 as h →0. That gives you the indeterminate form 0/0. I originally was going to write that you can't use l'Hopital's rule because you don't know what the derivative is near 0 or even if it is differentiable anywhere else other than zero. That's what you're trying to prove after all. I left this out because I thought that the other 'error', which came first, made the later one unnecessary to mention.



But now that you've cleared up the one error is there a way to clear up the other? You can't use l'Hopital's rule without already knowing E is differentiable.

Answer 4

E(x) = ℯˣ

____________