functional equation...

[man111]

If [tex]\bf f:R\rightarrow R[/tex] and function [tex]\bf f(x)[/tex] satisfy the equation [tex]\bf f(x)=f(\frac{x}{2})+\frac{x}{2}.f^{'}(x)[/tex]. Then find [tex]\bf f(x)[/tex]

[portokal]

[tex]f(x)=ax[/tex] , where [tex]a\in R[/tex]

[man111]

can anyone explain it how can i solve it.......

Thanks.

[drago]

Very interesting! It seems to me, I've seen it somewhere...
Obviously [tex]f(x)=ax[/tex] satisfies conditions, but why not other possibilities?!

[exp]

Apply the mean value theorem for the function f in the interval [x/2,x]. Then you'll get f'(x)=const and so f(x)=ax+b.

[L.e.o]

f(x) = ax + b is working too.
[tex]\bf f(x) - f(\frac{x}{2})= \frac{x}{2}.f^{'}(x) = A(x)f^{'}(x)[/tex]
[tex]A(x) = A(x^1)[/tex] which means that f(x) looks like (there is only one parameter):
[tex]f(x) = ax^n + c[/tex]
Now needs to find n.
[tex]ax^n - \frac{ax^n}{2n } =\frac{x}{2 }.\frac{x^{n-1}}{n }[/tex]
[tex]ax^n(1-\frac{1}{2n}) = \frac{ax^n}{2n }[/tex]
2n-1 = 1 => n=1

Example:
[tex]\bf f(x)=f(\frac{x}{2})+\frac{3x}{2}.f^{'}(x)[/tex]
and the solution would be with n=2
[tex]f(x) = ax^2 + c[/tex]

[drago]

Very good problem!

Apply the mean value theorem for the function f in the interval [x/2,x]. Then you'll get f'(x)=const and so f(x)=ax+b.

You'll get that [tex]\exists \xi \in (\frac{x}{2}, x), f'(\xi)= \frac{f(x)-f(\frac{x}{2})} {\frac{x}{2}}=f'(x)[/tex], and then what ?!
Of course this idea could be appllied, but You must get through some technical obstacles to prove that [tex]f'(x)=const[/tex] . And it is the most interesting part of the issue !
And I don't understand what L.e.o wrote!!?

[mkmarinov]

[tex]\bf f(x) - f(\frac{x}{2})= \frac{x}{2}.f^{'}(x) = A(x)f^{'}(x)[/tex]
[tex]A(x) = A(x^1)[/tex] which means that f(x) looks like (there is only one parameter):
[tex]f(x)=ax^n+c[/tex]

I feel that this needs some more explaining.

[exp]

What's so interesting, nothing special. With the assumption that f satisfies the given equation, you can easily prove that f is convex and at the same time concave, i.e. prove that f(tx/2+(1-t)x)=tf(x/2)+(1-t)f(x) for all t E [0, 1]. It's obvious that f is then linear.

[man111]

Yes I also want some explanation....

How can I got [tex]f^{'}(x)=C[/tex] Where [tex]C=[/tex] Constant from Mean value Theorem.

[drago]

What's so interesting, nothing special. With the assumption that f satisfies the given equation, you can easily prove that f is convex and at the same time concave, i.e. prove that f(tx/2+(1-t)x)=tf(x/2)+(1-t)f(x) for all t E [0, 1]. It's obvious that f is then linear.

Write it down pls!

[L.e.o]

Possible simalar tasks can be:
[tex]eq1: \bf f(x) - f(\frac{x}{2}) = A(x)f^{'}(x) ,where A(x) = ax+b[/tex]
[tex]eq2: \bf f(x) - f(\frac{x}{2}) = B(x)f^{''}(x) ,where B(x) = ax^2+bx+c[/tex]
etc...
The degree of f(x) and f(x/2) is the same and the degree of f '(x) is lower with 1. Thus, so the eq1. to have a solution the degree of A(x) should be 1. Analog for eq.2 and B(x).

Let's observe the 1st task: where A(x) = x/2 and
[tex]y(x) = a_n.x^n + a_{n-1}x^{n-1} +....+a_1.x + a_0[/tex]
Then
[tex]y(\frac{x}{2 }) = \frac{a_n.x^n}{2^n} + \frac{a_{n-1}x^{n-1}}{ 2^{n-1}} +....+\frac{a_1.x}{ 2} + a_0[/tex] (I have made a mistake here in the above solution typing "2n" instead of "2^n")

[tex]y'(x) = a_n.nx^{n-1} + a_{n-1}(n-1)x^{n-2} +....+a_1[/tex]

eq3: [tex]y(x)-y(\frac{x}{2}) = a_n.x^n\frac{2^n-1}{2^n } + a_{n-1}x^{n-1}\frac{2^{n-1}-1}{2^{n-1} } +....+a_1.\frac{x}{ 2}[/tex]
eq4: [tex]\frac{x}{2 } . f'(x) = \frac{x}{2 }(a_n.nx^{n-1} + a_{n-1}(n-1)x^{n-2} +....+a_1) = \frac{a_n.nx^n}{2} + \frac{a_{n-1}(n-1)x^{n-1}}{2 } +....+\frac{a_1 x}{ 2}[/tex]
Comparing eq3 and eq4: to be equal, the constants in front every degree of x should be identical.
But it is identical only in front x (equal to a1/2). The solutions then for y(x) is: y(x) = ax+c

Example: If A(x) = 3x/4, then the identical constants would be in front of x^2 and the solution would: y(x) = ax^2+c
No matter what linear A(x) (degree of 1) , the possible solution for y(x) will look like y(x) =ax^n + c (1 part + const)
For eq2: the solution for y(x) , where B(x) is function with degree of 2, y(x) = ax^n + b.x^m + c (2 parts + const)
That's what I meant in the first post: higher degree for A(x) = more parts for y(x)

[mkmarinov]

The degree of f(x) and f(x/2) is the same and the degree of f '(x) is lower with 1. Thus, so the eq1. to have a solution the degree of A(x) should be 1. Analog for eq.2 and B(x).

Is it necessary that the function be polynomial?

[L.e.o]

The degree of f(x) and f(x/2) is the same and the degree of f '(x) is lower with 1. Thus, so the eq1. to have a solution the degree of A(x) should be 1. Analog for eq.2 and B(x).

Is it necessary that the function be polynomial?

A function y(x) for which:
f(x) ~ x.f'(x) .
I couldn't come up of any other type than polynomial. If you can, then share it with us.
By the way, why is this topic in english when the forum is in bulgarian?

[mkmarinov]

Well, [tex]f(x)=\sqrt{x}=2x \frac{1}{2sqrt{x}}=2xf'(x)[/tex]
And the topic is in english because the OP doesn't seem to speak Bulgarian.

[L.e.o]

Let's have the very same task, but for which A(x) = x(2-√2)
Generating eq3 and eq4 and comparing the constants in front the x^n will give:
[tex]\frac{2^n-1}{2^n} = n(2-\sqrt{2})[/tex] which solution is n=0.5. Thus the answer to that task is again y(x) = a.x^n+c = a√x + c

[drago]

Just a comment.
for now being there are no assumtions how good is f around 0.
Of course f(x) is differentiable for [tex]x \ne 0[/tex] (and so continuous) but it is not necessary to exist [tex]f'(0)[/tex] or even f(x) to be continious at 0. If at least the latter is true, than the only possibility is f(x)=ax+b.
If not, there are other options...
So, if You prove f(x)=ax+b, it is mandatory to use at least that f(x) is continious at 0.
Hmm, ..., Sorry, I forgot that the initial condition was :
[tex]f'(x) \frac{x}{2} = f(x)-f(\frac{x}{2})[/tex]
not
[tex]f'(x) = \frac{f(x)-f(\frac{x}{2})} {\frac{x}{2}}[/tex]

[drago]

Obviously f(x) is smooth function on [tex]\mathbb{R}\setminus 0[/tex]
We know that f is differentiable at 0 , but let also assume that [tex]f'(x)[/tex]is continuous at 0.
In that case the only option is f(x)=ax+b.
To prove it is sufficient to show [tex]f'(x)=const,[/tex] when [tex]x>0[/tex]. Let [tex]\exists x_1, x_2, a=f'(x_1) \ne b=f'(x_2).[/tex]
Let [tex]m_a= inf\{x| 0 \lt x, f'(x)=a\}, m_b= inf\{x| 0 \lt x, f'(x)=b\} .[/tex]
Then [tex]m_a=m_b=0.[/tex] For example let [tex]0\lt m_a.[/tex] Because [tex]f'(x)[/tex] is continious when [tex]x>0 => f'(m_a)=a[/tex], then applying mean v.theorem,
[tex]\exists \xi \lt m_a, f'(\xi)=f'(m_a) = a[/tex] , which contradicts that [tex]m_a[/tex] is inf of such points.
then [tex]m_a=m_b=0[/tex] which contradics continuity of [tex]f'(x)[/tex] at 0.

[Makelov]

Well, we do not actually need the additional assumption that f'(x) is continuous. First of all, for all [tex]x\neq0[/tex], [tex]f'(x)=\frac{f(x)-f(x/2)}{\frac{x}{2}}[/tex] is the quotient of two continuous functions, and is therefore continuous. Next, by definition, [tex]f'(0)=\displaystyle\lim_{x\to 0} \frac{f(x)-f(0)}{x}=\lim_{x\to 0} \frac{f(x/2)-f(0)}{x/2}[/tex], which after a straightforward manipulation yields
[tex]f'(0)=\displaystyle\lim_{x\to 0} \frac{f(x)-f(x/2)}{\frac{x}{2}}=\displaystyle\lim_{x\to 0} f'(x)[/tex] which means that f'(x) is continuous at 0, too.

[drago]

Thanks, Makelov,
Yes, the question was: is [tex]f'(x)[/tex] continuous at 0 ?
[tex]\frac{f(x)-f(0)}{x} = \frac{f(x)- f(x/2) + f(x/2) - f(0)}{x} = \frac{1}{2} \frac{f(x)-f(x/2)}{\frac{x}{2}} + \frac{1}{2} \frac{f(x/2)-f(0)}{\frac{x}{2}}=\frac{1}{2}f'(x) + \frac{1}{2} \frac{f(x/2)-f(0)}{\frac{x}{2}}[/tex]
which of course yields:
[tex]f'(0)=\displaystyle\lim_{x\to 0} f'(x)[/tex]