1. ## Exonential function

How to show that if f is continuous ,f(0)=1,f(1)=e,f(x+y)=f(x)f(y) then f(x)=e^x

2. Originally Posted by drpawel
How to show that if f is continuous ,f(0)=1,f(1)=e,f(x+y)=f(x)f(y) then f(x)=e^x
Try using Power series

3. Originally Posted by drpawel
How to show that if f is continuous ,f(0)=1,f(1)=e,f(x+y)=f(x)f(y) then f(x)=e^x
Do it in stages. First, $\displaystyle f(n) = e^n$ for a positive integer (by induction starting with n=1). Then you'll have to think about why it follows for a negative integer, and then for any rational number. Finally, use continuity to get it for any real number.

4. I was trying to prove it but it seems that I just cannot handle this problem.Can you help me more.

5. Originally Posted by drpawel
I was trying to prove it but it seems that I just cannot handle this problem.Can you help me more.
First step was to use induction to show that $\displaystyle f(n)=e^n$ if n is a positive integer. I hope you can manage that (you're certainly not getting any more hints on that part).

Now think about f(-n). The defining relation says that f(-n)f(n) = f(-n+n) = f(0) = 1, so $\displaystyle f(-n) = 1/f(n) = 1/e^n = e^{-n}$.

Next, use induction on n to prove that $\displaystyle f(nx) = \bigl(f(x)\bigr)^n$, for any x and any positive integer n. It follows that if m and n are integers (with n positive) then $\displaystyle e^m = f(m) = f(n\tfrac mn) = \bigl(f(\tfrac mn)\bigr)^n$. Take n'th roots to see that $\displaystyle e^{m/n} = f(\tfrac mn)$. In other words, $\displaystyle f(r) = e^r$ for any rational number r.

Finally, use continuity to deduce that the result is true for all real numbers.