Hey Lambin.
Hint: Consider the exponential function: Can you use it to show the first property (as a start)?
PROBLEM:
Suppose that a function satisfies the following conditions for all real values and :
i.
ii. , where
Show that the derivative exists at every value of and that .
ATTEMPT:
I honestly don't have much of a clue here. Instead, I've tried to ask myself questions to get some idea going but to no avail. First, I tried to make sense of what the conditions imply and discovered that because and belong in the same set of real numbers without restriction, then everything said about is also true about .
Therefore, , where
I thought, maybe, if I could rewrite condition (i) with a substitution in terms of (ii), I might get somewhere.
Any thoughts?
Lambin
Did you realize the important properties of your function?
1. f(0) = 1 : Indeed if you put x = y =0 then you will get it.
2. f(x) is different than zero for all real numbers x… Indeed if for a real number y we have
f(y) =0 then f(x+y) = 0 for all values of x .i.e for all real numbers ..that is absurd…
3. f(x-y) = f(x) /f(y) indeed substitute x = x-y and you get f(x)=f(x-y)f(y) then…
4. f(x) >0 for every real number x . indeed substitute x = y and you will get it….
The only known function from functional Analysis to have all these properties is the function f(x) = e^x which is differentiable for all real numbers x and f’(x) = f(x).
MINOAS
I think that Lambin was looking for a more direct proof.
If you look at the definition and work with the given properties of f, I think you'll get to the result pretty quickly: start by substituting f(x)f(h) for f(x+h).
- Hollywood
Oh, this is fantastic! You know, I actually did peer into the definition of the derivative but must have overlooked its resemblance to condition (i), otherwise I would have been able to figure it out. Intuitively, I thought it might be possible that f'(x) = f(x) because the first condition was simulating an exponential function.
Thanks again for all these replies.