Solve this functional equation

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In summary, the conversation is about the function f(x) and its properties, specifically in relation to the equation $$f(xf(y) + f(x)) + f(y^2) = f(x) + yf(x + y)$$. Several possible solutions are discussed, including f(x) = x, f(x) = 0, and f(x) = cx. It is suggested to use Taylor expansion and the properties of f(0) to find the only possible solution. However, it is noted that this approach is not valid as f may not be continuous. The conversation ends with the suggestion to explore the function f(x) further.
  • #1
LCSphysicist
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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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  • #2
I see an obvious example of solution
[tex]f(x)=x[/tex].
 
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  • #3
anuttarasammyak said:
I see an obvious example of solution
[tex]f(x)=x[/tex].
And how do you prove this is the only function possible?
 
  • #4
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
anuttarasammyak said:
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
anuttarasammyak said:
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.
 
  • #7
LCSphysicist said:
Maybe @PeroK can think in something to help us.
Not immediately!
 
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  • #8
LCSphysicist said:
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
haruspex said:
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
LCSphysicist said:
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
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
anuttarasammyak said:
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
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
anuttarasammyak said:
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
haruspex said:
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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  • #16
anuttarasammyak said:
f(1)=1, f(-1)=-1, f(x)=0 otherwise
is a solution.
Are you sure? What about x=1, y=-2?
 
  • #17
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
anuttarasammyak said:
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
haruspex said:
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
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
William Crawford said:
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
anuttarasammyak said:
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
Excellent! but
[tex]f(f(x))=f(x)[/tex]
puts us
[tex]a^2=a[/tex]?
 
  • #25
anuttarasammyak said:
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##.
 
  • #26
Thanks. That meet with my #17. Can you tell me if other non analytic solutions exist?
 
  • #27
I believe that ##f(x) = ax## for ##a\in\{0,1\}## are the entire family of continuous solutions (consisting of elementary functions) to the original functional equation. As any continuous solution also has to be a solution to the Cauchy's functional equation. However there exist non-continuous solutions to the Cauchy's functional equation and there might therefore also exist non-continuous solutions to the original functional equation.
 
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  • #28
haruspex said:
If f(u) = u then f(nu) = nu for all integers n.
f(x)+f(y)=f(x+y) tells us f(nx)=f(x)+f((n-1)x)=... = n f(x)

Is "If f(u) = u" is necessary to say f(nu) = nf(u) and why n is limited to integer not any real number ?
 
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  • #29
anuttarasammyak said:
Thanks. That meet with my #17. Can you tell me if other non analytic solutions exist?
The function ##f(x) = \mathrm{max}(x,0)## (i.e. the positive part) is also a solution to Cauchy as well and though it is continuous it isn't differentiable at ##x=0## and thus not analytic. Check for yourself if also satisfy the original functional equation.
 
  • #30
Hi. Say f(x)=max(x,0)
[tex]f(x)+f(-x)=|x|[/tex]and

[tex]f(x)+f(-x)=0[/tex]
seem incompatible.
 
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  • #31
Okay, so we almost have come to the conclusion that ##f(x) = x## for all ##x##, but not quite. Instead, we have the facts that:
  1. ##f(p) = p## for all rational numbers ##p##
  2. ##f(ax + by) = af(x) + bf(y)## for all rational numbers ##a## and ##b##
  3. ##f(f(x)) = f(x)##
These are all consistent with ##f(x) = x##, but I don't see that they imply it, either. Is it possible, for example, to have 1-3 hold and to also have ##f(\pi) = 1##?
 
  • #32
Hi.
stevendaryl said:
Okay, so we almost have come to the conclusion that f(x)=x for all x, but not quite.
Another conclusion is f(x)=0 for all x, Anyway I will follow you.
stevendaryl said:
Instead, we have the facts that:
  1. ##f(p) = p## for all rational numbers ##p##
  2. ##f(ax + by) = af(x) + bf(y)## for all rational numbers ##a## and ##b##
  3. ##f(f(x)) = f(x)##
I have questions on 1. and 2.
As for 2. first, ##f(x)+f(y)=f(x+y)## that William Crawford identified tells ##f(nx+my)=nf(x)+mf(y)## for integer n,m. How would I develop it so that they are rational numbers ?
2. and ##f(1)=1## gives 1. How can I get ##f(1)=1##?
 
  • #33
anuttarasammyak said:
Another conclusion is f(x)=0 for all x, Anyway I will follow you.

If you look at the original functional equation, ##f(x) = x## is a solution.

I have questions on 1. and 2.
As for 2. first, ##f(x)+f(y)=f(x+y)## that William Crawford identified tells ##f(nx+my)=nf(x)+mf(y)## for integer n,m. How would I develop it so that they are rational numbers ?

Well, ##f(m * x/m) = m * f(x/m)##. So ##f(x/m) = 1/m f(x)##.

2. and ##f(1)=1## gives 1. How can I get ##f(1)=1##?

I'm saying that that is one solution, ##f(x) = x##. I'm asking if there are other solutions besides that one (and the trivial one, ##f(x) = 0##).
 
  • #34
Thanks. Now I understand :

The relation ##f(x+y)=f(x)+f(y)## gives ##f(ax+by)=af(x)+bf(y)## for rational numbers a,b, not for irrational numbers. We add ##f(1)=1## to the conditions of the solution. These two give ##f(x)=x## for all rational numbers x.

I assume @stevendaryl suggests this f(x) for rational number x together with a different definition of f(x) for irrational number x would satisfy the original equation in post #1.

From the relation ##f(x^2)=xf(x)##
[tex]f(2)=\sqrt{2} f(\sqrt{2})=2[/tex],
[tex]f(\sqrt{2})=\sqrt{2} [/tex]
Similarly
[tex]f(a\sqrt{2}+b\sqrt{3})=a\sqrt{2}+b\sqrt{3}[/tex] for rational numbers a,b. We observe f(x)=x for some irrational numbers.

Inputting x=1 into the equation in post #1, as f(1)=1, we get f(y)=y for any real number y including irrational numbers in spite of the above expectation.
 
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  • #35
anuttarasammyak said:
The relation f(x+y)=f(x)+f(y) gives f(ax+by)=af(x)+bf(y) for rational numbers a,b, not for irrational numbers.
Let ##c## be a rational number (possibly irrational), then there exist a sequence ##\{q_n\}_{n\in\mathbb{N}}## of purely rational numbers that converges to ##c##. Therefore (assuming ##f## is continuous)
$$
\begin{align*}
f(cx) &= \lim_{n\rightarrow\infty}f(q_nx) \\
&= \lim_{n\rightarrow\infty}q_nf(x) \\
&= cf(x)
\end{align*}
$$
and thus ##f(ax+by) = af(x) + bf(y)## is true even when ##a## and ##b## are irrational.
 
<h2>1. How do I approach solving a functional equation?</h2><p>Solving a functional equation involves finding a function that satisfies the given equation. The first step is to identify the type of functional equation and then use appropriate techniques such as substitution, elimination, or iteration to find the solution.</p><h2>2. What are some common types of functional equations?</h2><p>Some common types of functional equations include linear, quadratic, exponential, logarithmic, and trigonometric equations. Each type requires a different approach for solving.</p><h2>3. Can I use algebraic manipulation to solve a functional equation?</h2><p>Yes, algebraic manipulation is a common technique used to solve functional equations. This involves simplifying the equation by rearranging terms and applying algebraic rules to isolate the variable and find the solution.</p><h2>4. Are there any special cases or restrictions when solving functional equations?</h2><p>Yes, some functional equations may have restrictions on the domain or range of the function. It is important to identify these restrictions and consider them while solving the equation to ensure the solution is valid.</p><h2>5. Is there a specific method or formula for solving all functional equations?</h2><p>No, there is no one-size-fits-all method or formula for solving functional equations. Each equation may require a different approach and it is important to understand the properties and characteristics of each type of functional equation to find the appropriate solution.</p>

1. How do I approach solving a functional equation?

Solving a functional equation involves finding a function that satisfies the given equation. The first step is to identify the type of functional equation and then use appropriate techniques such as substitution, elimination, or iteration to find the solution.

2. What are some common types of functional equations?

Some common types of functional equations include linear, quadratic, exponential, logarithmic, and trigonometric equations. Each type requires a different approach for solving.

3. Can I use algebraic manipulation to solve a functional equation?

Yes, algebraic manipulation is a common technique used to solve functional equations. This involves simplifying the equation by rearranging terms and applying algebraic rules to isolate the variable and find the solution.

4. Are there any special cases or restrictions when solving functional equations?

Yes, some functional equations may have restrictions on the domain or range of the function. It is important to identify these restrictions and consider them while solving the equation to ensure the solution is valid.

5. Is there a specific method or formula for solving all functional equations?

No, there is no one-size-fits-all method or formula for solving functional equations. Each equation may require a different approach and it is important to understand the properties and characteristics of each type of functional equation to find the appropriate solution.

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