# Thread: Suppose this, show that:

1. ## Suppose this, show that:

PROBLEM:

Suppose that a function $\displaystyle f$ satisfies the following conditions for all real values $\displaystyle x$ and $\displaystyle y$:

i. $\displaystyle f(x+y)=f(x)*f(y)$
ii. $\displaystyle f(x)=1+xg(x)$, where $\displaystyle \lim_{x\to\0}g(x)=1$

Show that the derivative $\displaystyle f'(x)$ exists at every value of $\displaystyle x$ and that $\displaystyle f'(x)=f(x)$.

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 $\displaystyle x$ and $\displaystyle y$ belong in the same set of real numbers without restriction, then everything said about $\displaystyle x$ is also true about $\displaystyle y$.

Therefore, $\displaystyle f(y)=1+yg(y)$, where $\displaystyle \lim_{y\to\0}g(y)=1$

I thought, maybe, if I could rewrite condition (i) with a substitution in terms of (ii), I might get somewhere.

Any thoughts?

2. ## Re: Suppose this, show that:

Hey Lambin.

Hint: Consider the exponential function: Can you use it to show the first property (as a start)?

3. ## Re: Suppose this, show that:

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

4. ## Re: Suppose this, show that:

I think that Lambin was looking for a more direct proof.

If you look at the definition $\displaystyle f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ 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

5. ## Re: Suppose this, show that:

Hollywood posted his reply while I was typing, but here it is anyway:

6. ## Re: Suppose this, show that:

Originally Posted by johng
Hollywood posted his reply while I was typing, but here it is anyway:

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.