Exploring the Properties of Differentiable Functions: E, ln, and Differentiation

[MathematicalPhysicist]

i have three questions which go like this:
the function f(x) satisfies the equation f(x+y)=f(x)f(y)
1)if f(x) is differentiable, either f(x)=0 or f(x)=e^ax
2)if f(x) is continuous, either f(x)=0 or f(x)=e^ax

3)if a differentiable function f(x) satisfies the equation f(xy)=f(x)+f(y)
then f(x)=a*log(x)

for the first question i followed the definition:
f'(x)=lim (f(x+h)-f(x))/h=lim (f(x)(f(h)-1)/h)=f(x)*f'(0)
there's a theorem which states that if y'=ay then the y=ce^ax
so i think because f'(0) is a constant, it follows from this that f(x)=e^ax
about the second question I am not sure, perhaps i should use the intermediate theorem here, but I am not given any interval, and i thought perhaps using the definition of conintuity (at least one of them) lim (f(x+h)-f(x))=0 as h appraoches zero, but it didnt get me anywhere.

about the third question, here what i did:
if we put g(x)=e^f(x) then we have this equality:
g(xy)=g(x)g(y) obviously we have here a function of the form x^a, but how do i prove it, i started using the defintion of derivative but this also haven't got me anywhere.

your help is appreciated.

 

[StatusX]

For the second, you can easily show that for integers y, f(y)=f(1)^y. You can then extend this to rational numbers, and finally to all real numbers by continuity.

For the third, try taking a derivative of the equation f(xy)=f(x)+f(y).

 

[MathematicalPhysicist]

let me see, if i understand:
f(x)=f(x)*f(0)
if f(x) doenst equal 0 then f(0)=1
f(1)=f(f(0))
f(2)=f(1+1)=f(1)f(1)=f(1)^2
f(3)=f(2)f(1)=f(1)^3
i.e by induction we have f(x)=f(1)^x
and then i extend it as you said to the reals, but shouldn't i be showing that f(1)=e^a to finalise the proof?

for the third if i take the derivative i get:
f'(xy)=f'(x)+f'(y)
after that i should use the fact that g(xy)=e^f(xy) that g'(xy)=f'(xy)e^f(xy)=(f'(x)+f'(y))e^(f(x)+f(y)) but is this enough to show that g(x)=x^a, i feel that something is missing?

 

[StatusX]

For the first, e^a can be made to be any number by the right choice of a (or if you want to restrict f(x) to be real, argue why f(1) needs to be positive). For the second, you need to differentiate with respect to either x or y.

 

[MathematicalPhysicist]

so i should take either x or y as a constant and the other as a variable:
so it should be yf'(xy)=f'(x)?
and yg'(xy)=(f'(x)/y)*e^f(xy), but how from here i get that g(x)=x^a?

p.s
im not sure my differentiation is correct, i differntiated wrt x.

 

[StatusX]

Forget g! Play around with that first equation in your last post. I can't give you any more hints without giving it away.

 

[MathematicalPhysicist]

just some queries, for the second question in order to prove that f(1) i have assumed that it's negative and got a contradiction to the fact that it's continuous, is it the right idea here?

for the third question i got this far:
af'(ax)=f'(x)
so if we divide by f(ax) when it's different than 0, then af'(ax)/f(ax)=f'(x)/f(x) the lhs is (ln(f(ax)))', i tried to integrate this equation, but i didnt succeded in arrived at the equation i need to derive.

 

[StatusX]

f(1) could be negative and f could be continuous if f is complex, so I don't know how this would work. You can show what you've done, but there's a simpler proof that doesn't use continuity.

For the third, remember you're trying to show f(x)=a*log(x). What should f'(x) look like?

 

[MathematicalPhysicist]

it should be a/x, so i should divide by x?
af'(ax)/x=f'(x)/x
(aln(x))'f'(ax)=f'(x)/x
(aln(x))'=f'(x)/f'(ax)x
now i need to show that f'(x)/xf'(ax)=f'(x), how do i do that?

btw, for the other question, what is the other approach?
im not sure if my approach is correct but i assumed that f(1)<0 and thus because f(2)=f(1)^2>0 then for a suitable d, we have |1-0|<d |f(1)-f(0)|<e and for d' |2-1|<d' |f(2)-f(1)|<e', so we have for s=min(d,d') that |2-0|<s but because s is the smaller between them then it could be less than 2.
but as i said i don't think it's correct.

 

[StatusX]

it should be a/x, so i should divide by x?
af'(ax)/x=f'(x)/x
(aln(x))'f'(ax)=f'(x)/x
(aln(x))'=f'(x)/f'(ax)x
now i need to show that f'(x)/xf'(ax)=f'(x), how do i do that?

Your problem is with f'(ax). Differential equations usually don't involve things like y'(ax), only functions of single variables, like y'(x) (or y'(a) (hint)).

f(2)=f(1)^2>0

?? How could you write this and then keep going for another paragraph? Just modify this slightly.

 

[MathematicalPhysicist]

about your last remark, obviously the function isn't monotone if f(1)<0 but does it mean that it's not continuous? i can't find in my book a theorem that states so.

anyway, for the other question, a is a constant so we are still dealing here with one variable,x.
for x=1 we have f'(a)=f'(1)/a but how do i use it on af'(ax)/x=f'(x)/x?

 

[StatusX]

about your last remark, obviously the function isn't monotone if f(1)<0 but does it mean that it's not continuous? i can't find in my book a theorem that states so.

You just showed f(2)>0 without using any assumptions of continuity, monotonicity, etc. Can you modify this to show f(1)>0?

anyway, for the other question, a is a constant so we are still dealing here with one variable,x.
for x=1 we have f'(a)=f'(1)/a but how do i use it on af'(ax)/x=f'(x)/x?

Who says a is a constant? That equation is true for all a, so just integrate it with respect to a to get f(a), and thus f(x). (note a here is not the same as a in a*log(x). I think you plugged in this a originally for a different reason, but go back and see that you can replace a with x, y, or whatever letter you want.)

 

[MathematicalPhysicist]

f(2)=f(1)^2
f(1)=f(2)/f(1)
obviously we have here a geometric sequence, but i don't see how can i conclude that f(1) must be positive.
if f(1)<0 then f(1)<f(0)=1
f(f(1))<f(1)<f(0)
f(f(1))<f(2)/f(1)<f(0)
f(1)<f(2)<f(1)f(f(1))=f(1+f(1))=f(1)^(1+f(1))
but i don't how to procceed from here.

 

[StatusX]

You're overthinking this. f(1)=f(1/2+1/2)=f(1/2)^2>0.

 

[MathematicalPhysicist]

foolish me, forgot about this.
thanks, status.