Solve this functional equation

  • #1
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Homework Statement:

"Find all functions f: R in R such that x and y belong to R" (R is obviously real space) (continuation in the blank frame

Relevant Equations:

\n
$$f(xf(y) + f(x)) + f(y^2) = f(x) + yf(x + y)$$

A tricky question, i think.
First fact i found was:
f(f(0)) = 0
So i separate it in two types of functions
f(0) = 0 and f(0) = u.

I was trying to analyzing both cases, with the cases where x = y and x = -y but is is rather extended way, so i believe there is a better attempt to solve the question
Anyway i am not sure if we have a solution with explicit functions (f(x) = x²) or if the answer end to be something like (f(x) + f(y) + ...)

Any tips?

@fresh_42
 
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Answers and Replies

  • #2
anuttarasammyak
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I see an obvious example of solution
[tex]f(x)=x[/tex].
 
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  • #3
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I see an obvious example of solution
[tex]f(x)=x[/tex].
And how do you prove this is the only function possible?
 
  • #4
anuttarasammyak
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I see another solution [tex]f(x)=0[/tex].

There may be more but I am not sure. How about writing f(x) in Taylor expansion form and taking a look at the coefficients of the same ##x^n y^m## term ,i.e
[tex] f(x)=f(0)+xf'(0)+...[/tex]
Hence we know ## f(0)=0, f'(0)=0,1##, f"(0)=0,...
 
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  • #5
member 587159
I see another solution [tex]f(x)=0[/tex].

There may be more but I am not sure. How about writing f(x) in Taylor expansion form and taking a look at the coefficients of the same ##x^n y^m## term ,i.e
[tex] f(x)=f(0)+xf'(0)+...[/tex]
Hence we know ## f(0)=0, f'(0)=0,1##, f"(0)=0,...
Who says a solution must be differentiable?
 
  • #6
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I see another solution [tex]f(x)=0[/tex].

There may be more but I am not sure. How about writing f(x) in Taylor expansion form and taking a look at the coefficients of the same ##x^n y^m## term ,i.e
[tex] f(x)=f(0)+xf'(0)+...[/tex]
Hence we know ## f(0)=0, f'(0)=0,1##, f"(0)=0,...
I imagined in open in Taylor, but i believe it would be so troublesome, we have not just x, we have f as function of y and other variables too.

I was thinking if the only possible answer happens to be just a lot of guess, i would be disappointed

. Maybe @PeroK can think in something to help us.
 
  • #8
haruspex
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So i separate it in two types of functions
f(0) = 0 and f(0) = u.
Follow that path. What general equation does x=0 give you? What about y=0 instead?
 
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  • #9
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Follow that path. What general equation does x=0 give you? What about y=0 instead?
Yes, i tried a lot and i found that the only functions which satisfy all the conditions is ##f(\gamma) = \gamma##
I checked and confirmed that the angular coefficiente need to be 1, and can not be any linear coefficient.
@anuttarasammyak made a nice guess #2.
 
  • #10
haruspex
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i found that the only functions which satisfy all the conditions is ##f(\gamma) = \gamma##
Are you saying you have proved that, or just that it is the only solution you have found?
 
  • #11
anuttarasammyak
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Say x=0
[tex]f(f(0))+f(y^2)=f(0)+yf(y)[/tex].
Say y=0
[tex]f(xf(0)+f(x))=f(x)[/tex],further say x=0
[tex]f(f(0))=f(0)[/tex]
So
[tex]f(y^2)=yf(y)[/tex]
f(0)=0. For ##y \neq 0##
[tex]\frac{f(y^2)}{y^2}=\frac{f(y)}{y}=c[/tex]
We can easily know allowed values of c by inputting f(y)=cy, which satisfy f(0)=0, to the original equation or the above second equation.

EDIT
I add on the features
[tex]f(f(x))=f(x)[/tex]
which shows nature of projection.
 
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  • #12
stevendaryl
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Say x=0
[tex]f(f(0))+f(y^2)=f(0)+yf(y)[/tex].
Say y=0
[tex]f(xf(0)+f(x))=f(x)[/tex],further say x=0
[tex]f(f(0))=f(0)[/tex]
So
[tex]f(y^2)=yf(y)[/tex]
f(0)=0. For ##y \neq 0##
[tex]\frac{f(y^2)}{y^2}=\frac{f(y)}{y}=c[/tex]
We can easily know allowed values of c by inputting f(y)=cy, which satisfy f(0)=0, to the original equation or the above second equation.
From
[tex]\frac{f(y^2)}{y^2}=\frac{f(y)}{y}[/tex]

we can't conclude that both sides are constant, can we? We can define a new function
[tex]g(x) = f(x)/x[/tex]

Then we have:
[tex]g(x) = g(x^2)[/tex]

Does that imply that g is a constant function?
 
  • #13
anuttarasammyak
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Hi.
[tex]\frac{f(x^2)}{x^2}=\frac{f(x)}{x},\ f(0)=0[/tex]
So f(x) is odd function of x. Let us expand it in series
[tex]f(x)=\sum_{n=0}^{+\infty}a_{2n+1}\ x^{2n+1}[/tex]
The above relation shows
[tex]\sum_{n=0}^{+\infty}a_{2n+1}\ (x^{4n}-x^{2n})=0[/tex]
All the coefficients except ##a_1## are zero.
 
  • #14
haruspex
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Hi.
[tex]\frac{f(x^2)}{x^2}=\frac{f(x)}{x},\ f(0)=0[/tex]
So f(x) is odd function of x. Let us expand it in series
[tex]f(x)=\sum_{n=0}^{+\infty}a_{2n+1}\ x^{2n+1}[/tex]
The above relation shows
[tex]\sum_{n=0}^{+\infty}a_{2n+1}\ (x^{4n}-x^{2n})=0[/tex]
All the coefficients except ##a_1## are zero.
As has been pointed out, this approach is utterly invalid. We are not given that f is continuous, let alone differentiable.
 
  • #15
anuttarasammyak
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We are not given that f is continuous, let alone differentiable.
Thanks. I admit my proof #13 works only if f(x) can be expressed in series.

For example
f(1)=1, f(-1)=-1, f(x)=0 otherwise
is a solution.
 
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  • #17
anuttarasammyak
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Thanks. #15 is a solution of equation f(x)/x=f(x^2)/x^2, f(x)=0. I see it does not satisfy the equation of post #1 , as you pointed out. As I should find the allowed values of c in post #11, not all the solutions of the equation f(x)/x=f(x^2)/x^2, f(x)=0 are the solution of the equation of post #1.

Now I am much interested to know whether we have more solutions other than analytic f(x)=cx, c={0,1}.
 
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  • #18
haruspex
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Thanks. #15 is a solution of equation f(x)/x=f(x^2)/x^2, f(x)=0. I see it does not satisfy the equation of post #1 , as you pointed out. As I should find the allowed values of c in post #11, not all the solutions of the equation f(x)/x=f(x^2)/x^2, f(x)=0 are the solution of the equation of post #1.

Now I am much interested to know whether we have more solutions other than analytic f(x)=cx, c={0,1}.
So far I have only managed to show
f(0) = 0
f(f(x)) = f(x)
f(x2) = x f(x)
If f(u) = u then f(nu) = nu for all integers n.

Edit:
Oh, and I forgot
f(-x)=-f(x)
 
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  • #19
So far I have only managed to show
f(0) = 0
f(f(x)) = f(x)
f(x2) = x f(x)
If f(u) = u then f(nu) = nu for all integers n.
That is also tremendous progress. This allow you to simplify the functional equation to $$xf(y) + yf(y) = yf(x+y)$$
 
  • #20
anuttarasammyak
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Re:#19
Your result toghether with the equation suggest the relation
[tex]f(xf(y)+f(x))=xf(y)+f(x)[/tex]
I am not sure it stands.
 
  • #21
haruspex
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That is also tremendous progress. This allow you to simplify the functional equation to $$xf(y) + yf(y) = yf(x+y)$$
If so, doesn't it immediately follow that yf(x )=xf(y)?
 
  • #22
Re:#19
Your result toghether with the equation suggest the relation
[tex]f(xf(y)+f(x))=xf(y)+f(x)[/tex]
I am not sure it stands.
No, you are right. What I wrote in 19 wasn't correct.
 
  • #23
From ##f(x^2)=xf(x)## conclude that ##f## is an odd function i.e. ##f(-x)=-f(x)##. If we take ##(x,y) = (-x,y)## then we get that
$$ f(-xf(y)+f(-x)) + yf(y) = f(-x) + yf(y-x) $$
or, by using the odd property of ##f##, that
$$ -f(xf(y)+f(x)) + yf(y) = -f(x) + yf(y-x). $$
Now, add this equation with the original functional equation, to get that
$$ 2f(x) = f(x+y) + f(x-y).$$
Next, use that ##f(x+y)## is symmetric under ##(x,y)\mapsto(y,x)## to derive that
$$ f(x)-f(y) = f(x-y)$$
or equivalently that
$$ f(x)+f(y) = f(x+y).$$
This is nothing more than Cauchy's functional equation. Thus, if it is assumed that ##f:\mathbb{R}\rightarrow\mathbb{R}## is continuous, then ##f(x) = ax## for ##a\in\mathbb{R}## are all the solutions.
 
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  • #24
anuttarasammyak
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Excellent! but
[tex]f(f(x))=f(x)[/tex]
puts us
[tex]a^2=a[/tex]?
 
  • #25
Excellent! but
[tex]f(f(x))=f(x)[/tex]
puts us
[tex]a^2=a[/tex]?
True, however that just imply that either ##a=1## or ##a=0##.
 

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