Re: Suppose this, show that:

Hey Lambin.

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

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

Re: Suppose this, show that:

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

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Re: Suppose this, show that:

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

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Re: Suppose this, show that:

Quote:

Originally Posted by

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

Attachment 27725

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.