Questions regarding differentiation

[MathematicalPhysicist]

1)the function f(x) satisfies: f(x+y)=f(x)f(y)
for every x,y>0.
prove that if f(x) is differentiable then f(x)=0 or f(x)=e^(ax).

2)prove that if f(x) is differnetiable and satisfies f(xy)=f(x)+f(y)
for every x,y>0 then: f(x)=alnx.

in the first one i tried to put x=y, and then getting:
f(2x)=f^2(x), f'(2x)*2=2f(x)*x'
which means f'(2x)=f(x)
which happens when f'(2x)=(e^(2x)'/2
therefore f(x)=e^(2x) or if we want to generalise:
y=a
f(ax)=f(x)f(a)
a*f'(ax)=f'(x)f(a)+f(x)f'a)
f(x)=(af'(ax)-f'(x)f(a))/f'(a)
but how do i go out from here?

thanks in advance.

 

[dextercioby]

You may want to check your differentiations...

Daniel.

 

[VietDao29]

1)the function f(x) satisfies: f(x+y)=f(x)f(y)
for every x,y>0.
prove that if f(x) is differentiable then f(x)=0 or f(x)=e^(ax).

2)prove that if f(x) is differnetiable and satisfies f(xy)=f(x)+f(y)
for every x,y>0 then: f(x)=alnx.

in the first one i tried to put x=y, and then getting:
f(2x)=f^2(x), f'(2x)*2=2f(x)*x'
which means f'(2x)=f(x)
which happens when f'(2x)=(e^(2x)'/2
therefore f(x)=e^(2x) or if we want to generalise:
y=a
f(ax)=f(x)f(a)
a*f'(ax)=f'(x)f(a)+f(x)f'a)
f(x)=(af'(ax)-f'(x)f(a))/f'(a)
but how do i go out from here?

thanks in advance.

Uhmm, I haven't done this kind of problem before, but here's how I am going to tackle it:
f(x + y) = f(x) f(y)
$$f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \rightarrow 0} \frac{f(x) f(h) - f(x) f(0)}{h} = f(x) \lim_{h \rightarrow 0} \frac{f(h) - f(0)}{h} = f(x) \times f'(0)$$.
Now, I think you can go from here, right? :)
The bolded part in my quote is some errors that you made.
Someone may come up with a better apprach, though. :)

 

[MathematicalPhysicist]

ok, from your approach i get that f'(0)=0 cause f(0)=1, but then f'(x)=0
which means either f(x) equals some constant or zero, but i don't see how can i infer that f(x)=e^(ax)?

 

[MathematicalPhysicist]

i coule ofcourse claim that when f(0)=e^(a*0), but i could do it also for e^x or a^x, nothing so special about e here, is it?

 

[dextercioby]

$$f(2x)=f^2 (x)$$

By logarithmic differentiation, it follows

$$\frac{f'(2x)}{f(2x)}=\frac{f'(x)}{f(x)}$$

Denote 2x=u and you'll have

$$\frac{f'(u)}{f(u)}=\frac{f'(x)}{f(x)}$$

The thing above can't be true unless x=u from which it follows x=0 which is not allowed (initially we have x,y>0) or that the ratio is equal to a constant, let that be "a"

By integration

$$f(x) = C e^{ax}$$ with $$C^2 = C$$

therefore one gets the solutions

$$f(x) =0$$

or

$$f(x)=e^{ax}$$


Daniel.

 

[MathematicalPhysicist]

but it's not given to you that the function is a logarithm function, so how can you use the logarithmic differentiation if you don't know the idntity of the function?

 

[dextercioby]

Well, for part a) you'd have to prove the solution is the exponential. For doing that, my solution assumes the existence of the "ln" function along with all its known properties. This assumption does not enter conflict with the problem's data...

For the second part, i won't post a full solution for your problem, but instead i'll state and solve a problem which could prove (or not, it depends on you) itself useful for your solution to point b).

Solve the following functional equation

$$f(xy)=f(x)+f(y)$$ (1)

for $x,y >0$ and knowing the function f is differentiable at each point of its domain, the real positive semiaxis. Prove the solution you found is unique...

MY SOLUTION

*Let's pick $y=1$ in equation #1. It follows simply that

$$f(1) =0$$ (2).

*Let's pick $y=\frac{1}{x}$ in equation #1. Using #2 it follows that

$$f(x)=-f\left(\frac{1}{x}\right)$$ (3).

*Let's pick $y\rightarrow \frac{1}{y}$ in equation #1. Using equation #3, it follows simply that

$$f\left(\frac{x}{y}\right) =f(x)-f(y)$$ (4).

*Let's pick $x=y+\Delta x$ in equation #4. One gets simply

$$f\left(1+\frac{\Delta x}{y}\right) =f(y+\Delta x) -f(y)$$ (5).

*Let's divide both sides of the equation #5 by $\Delta x$ and take the limit $\Delta x \rightarrow 0$.

The RHS of #5 will become the LHS of #6 and the RHS of #5 will become the LHS of #6.

$$\left(\frac{df}{dy}=\right) f'(y)=\frac{1}{y}\lim_{\Delta x \rightarrow 0}\frac{f\left(1+\frac{\Delta x}{y}\right)}{\frac{\Delta x}{y}}$$ (6).

For the RHS of (6), since the limit is typically [itex]\frac{0}{0} [/tex], one can use l'Hôpital's rule (the function f is differentiable at each point of its domain and thus the number which results -$f'(1)$- exists and is finite) and thus, by the definition <br /> <br /> $$C=: f'(1)$$ (7),<br /> <br /> we find the following linear I-st order & separable ODE<br /> <br /> $$\frac{df}{dy} =C \frac{1}{y}$$(8)<br /> <br /> subject to the initial condition #2. From the theorem of the existence and unicity of the solution(s) of well-posed problem for an ODE, it follows that the obvious solution to the equation #8 subject to the condition #2, i.e.<br /> <br /> $$f(y)=C\ln y$$ <br /> <br /> , exists and is unique...<br /> <br /> Daniel.[/itex]

 

[dextercioby]

Well, for part a) you'd have to prove the solution is the exponential. For doing that, my solution assumes the existence of the "ln" function along with all its known properties. This assumption does not enter conflict with the problem's data...

For the second part, i won't post a full solution for your problem, but instead i'll state and solve a problem which could prove (or not, it depends on you) itself useful for your solution to point b).

Solve the following functional equation

$$f(xy)=f(x)+f(y)$$ (1)

for $x,y >0$ and knowing the function f is differentiable at each point of its domain, the real positive semiaxis. Prove the solution you found is unique...

MY SOLUTION

*Let's pick $y=1$ in equation #1. It follows simply that

$$f(1) =0$$ (2).

*Let's pick $y=\frac{1}{x}$ in equation #1. Using #2 it follows that

$$f(x)=-f\left(\frac{1}{x}\right)$$ (3).

*Let's pick $y\rightarrow \frac{1}{y}$ in equation #1. Using equation #3, it follows simply that

$$f\left(\frac{x}{y}\right) =f(x)-f(y)$$ (4).

*Let's pick $x=y+\Delta x$ in equation #4. One gets simply

$$f\left(1+\frac{\Delta x}{y}\right) =f(y+\Delta x) -f(y)$$ (5).

*Let's divide both sides of the equation #5 by $\Delta x$ and take the limit $\Delta x \rightarrow 0$.

The RHS of #5 will become the LHS of #6 and the RHS of #5 will become the LHS of #6.

$$\left(\frac{df}{dy}=\right) f'(y)=\frac{1}{y}\lim_{\Delta x \rightarrow 0}\frac{f\left(1+\frac{\Delta x}{y}\right)}{\frac{\Delta x}{y}}$$ (6).

For the RHS of (6), since the limit is typically [itex]\frac{0}{0} [/tex], one can use l'Hôpital's rule (the function f is differentiable at each point of its domain and thus the number which results -$f'(1)$- exists and is finite) and thus, by the definition <br /> <br /> $$C=: f'(1)$$ (7),<br /> <br /> we find the following linear I-st order & separable ODE<br /> <br /> $$\frac{df}{dy} =C \frac{1}{y}$$(8)<br /> <br /> subject to the initial condition #2. From the theorem of the existence and unicity of the solution(s) of well-posed problem for an ODE, it follows that the obvious solution to the equation #8 subject to the condition #2, i.e.<br /> <br /> $$f(y)=C\ln y$$ <br /> <br /> , exists and is unique...<br /> <br /> Daniel.[/itex]

$<br /> I hope one can see the Tex displays now.<br /> <br /> Daniel.$

 

[dextercioby]

Hooray, sorry about the multiple posting.

Maybe you didn't like the high school-type solutions that I've proposed.

Here are shorter ones, which use partial derivatives, which are not taght in HS.

For the first problem

$$f\left(x+y\right)=f(x)f(y)$$ (1)

Partial differentiate #1 wrt "x" and get

$$\frac{df(x+y)}{d(x+y)}\frac{\partial (x+y)}{\partial x}= f(y)\frac{df(x)}{dx}$$ (2)

Partial differentiate #2 wrt "y" and get

$$\frac{df(x+y)}{d(x+y)}\frac{\partial (x+y)}{\partial y}= f(x)\frac{df(y)}{dy}$$ (3)

Equate the RHS's from #2 and #3 and get

$$f(x)\frac{df(y)}{dy} =f(y)\frac{df(x)}{dx}$$ (4)

From which

$$\frac{1}{f(y)}\frac{df(y)}{dy}=\frac{1}{f(x)}\frac{df(x)}{dx} =C$$

,where C is a (real) constant. The reason i put that constant is that the "x" and "y" are assumed independent variables.

By simple integration the $f(x)=Be^{Cx}$ is found. Requiring for it to verify the initial equation #1, we find either B=0, or B=1.

Daniel.

 

[dextercioby]

For the second problem, first prove f(1)=0 (which is done above), then again partial differentiate wrt to "x" first and "y" then (or viceversa), but this time divide the 2 equations you get. Separate the variables, like I've done above, equate to a constant and then you'll get a separable I-st order linear ODE to which you can separate the variables and use the f(1)=0 condition to get the needed result.

Daniel.