Prove: f(x) = alnxProving Derivatives and ln Homework Statement

[JG89]

Homework Statement

If a differentiable function f(x) satisfies the equation f(xy) = f(x) + f(y), prove then that f(x) = alnx.

Homework Equations

The Attempt at a Solution

I have proved that if f satisfies f(x + y) = f(x)f(y) then f(x) = 0 or f(x) = e^(ax)

and I also know that if a function f satisfies f'(x) = af(x) for some constant a, then f(x) = ce^(ax) for some constant c and a.

 

[Dick]

Define f'(1)=a. Now write down f'(x) as a difference quotient and mess around with it. Maybe make a change of variables on the h. Can you show f'(x)=a/x?

 

[JG89]

[tex] f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \rightarrow 0} \frac{f(x+h) + (-f(x))}{h}[/tex].

I think the negative sign in front of f(x) messes things up because I don't think I can write f(x+h) + (-f(x)) as f(x(x+h)).

EDIT: I just saw your edit, I will continue to work on it.

 

[Dick]

You should be able to show f(1/x)=-f(x) pretty easily.

 

[JG89]

0 = f(1) = f(x(1/x)) = f(x) + f(1/x) and so f(1/x) = -f(x).
So then I have [tex]f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) + f(1/x)}{h} = \lim_{h \rightarrow 0} \frac{f(1 + h/x)}{h}[/tex]

But I can't see where to go from here to show that f'(x) = a/x

 

[NoMoreExams]

Isn't this a candidate for L'Hopital's now or is that a step in the wrong direction?

 

[JG89]

I was thinking that, so using l'hospital I have f'(x) = f'(1+x/h), no? But to me this doesn't make sense because if I want to show that the derivative is a/x then this cannot be true.

Plus in my textbook we haven;t covered l'hospital yet so I assume I'm expected to solve the question without using it.

 

[NoMoreExams]

No using L'Hopital's you have

[tex] \lim_{h\rightarrow 0} \frac{1}{x} f'\left(1 + \frac{h}{x} \right) = \frac{1}{x} [/tex]

 

[JG89]

Isn't the derivative of f(1 + x/h) = f'(1 + x/h)? Plus where did the 1/x come from before the f'(1+ h/x)?

 

[NoMoreExams]

Chain rule? And you keep writing x/h when you mean h/x, be careful.

 

[JG89]

Sorry if I'm being difficult, but how did you use the chain rule there?..

 

[JG89]

Okay, I think I have it.

Letting [tex]\phi(h) = h/x [/tex], we have [tex]d\dx (f(1+\phi)) = f'(1 + \phi)*\phi'(h) = f'(1 + \phi) * (1/x)[/tex].

So we have [tex] f'(x) = \lim_{h \rightarrow 0} f'(1 + h\x)*(1/x) = a/x [/tex]

 

[JG89]

Ugh, I messed up some of the latex in the previous post and whenever I try to edit it, it won't save the changes, so I will just retype what I wanted to type..

letting g(h) = h/x, we have d/dx(f(1+g)) = f'(1+g)*g' = f'(1 + h/x)*1/x.

Now, since f'(x) = lim as h approaches 0 of f'(1 + h/x)*1/x, we have f'(x) = f'(1)/x = a/x.

Dick, how did you see right away to define f'(1) = a?

 

[NoMoreExams]

If anything I would let [tex] \phi(h) = 1 + \frac{h}{x} [/tex] but you have the right idea there.

Next, you wrote [tex] f'(1+h) [/tex], you probably meant [tex] f'\left(1 + \frac{h}{x} \right) [/tex]?

 

[Dick]

0 = f(1) = f(x(1/x)) = f(x) + f(1/x) and so f(1/x) = -f(x).
So then I have [tex]f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) + f(1/x)}{h} = \lim_{h \rightarrow 0} \frac{f(1 + h/x)}{h}[/tex]

But I can't see where to go from here to show that f'(x) = a/x

You probably also showed that f(1)=0. So your difference quotient is
(1/x)*((f(1+h/x)-f(1))/(h/x))). Since if h->0, so does h/x->0, the second factor is a difference quotient for f'(1). So it's (1/x)*f'(1)=a/x.

 

[JG89]

If anything I would let [tex] \phi(h) = 1 + \frac{h}{x} [/tex] but you have the right idea there.

Next, you wrote [tex] f'(1+h) [/tex], you probably meant [tex] f'\left(1 + \frac{h}{x} \right) [/tex]?

Yup, that's what I meant. And of course, since f'(x) = a/x, integrating gives us f(x) = alnx.

Thanks guys

 

[JG89]

You probably also showed that f(1)=0. So your difference quotient is
(1/x)*((f(1+h/x)-f(1))/(h/x))). Since if h->0, so does h/x->0, the second factor is a difference quotient for f'(1). So it's (1/x)*f'(1)=a/x.

How are you able to so easily spot these things?

 

[Dick]

Practice, I guess. But it wasn't all THAT easy. I did have to think about it. I knew the answer was f(x)=a*ln(x), so f'(1)=a. Now you just have to figure out how to turn the difference quotient at a general point x into a difference quotient at x=1 times something using that f(xy)=f(x)+f(y). Forget the solution, get a blank piece of paper and try it. It's not super hard either.

 

[JG89]

Do you know of any questions similar to the one that I posted? There aren't anymore in my textbook and I'd love to try another one.

 

[Dick]

Not exactly. There aren't a lot of functions that have such a simple relation as f(xy)=f(x)+f(y) or f(x+y)=f(x)f(y). Try redoing the proof that the latter is c*exp(ax) where a=f'(0) for the latter case, if you can forget how it was proved before. It's even easier.

 

[JG89]

I will try redoing it with another proof. Thanks :)