Math Is Fun Forum

{7/3}

Member

Registered: 2013-02-11

Posts: 210

Function

What is the function f(x) if this equation holds for each real x,y:  f(x+y)=f(x)+f(y)+2xy  [i know f(x) is x^2,but how do i formally obtain it?]

---

There are 10 kinds of people in the world,people who understand binary and people who don't.

SteveB

Member

Registered: 2013-03-07

Posts: 595

Re: Function

A few hints that I gave were:

Let y = 0, substitute this into your equation.

Use the fact that (x+0) = x
Simplify where possible. (You can fill in the intermediate steps.)
(.......)
Deduce that f(0) = 0

Start back with the original equation.
Let y = -x
Use the fact that (x-x) = 0
Then substitute the fact that f(0) = 0
Simplify where possible.
[As I have now discovered having checked my work and read Avon's post this does not lead to a solution.]

Try to get a formula for f(x) in terms of x and f(-x)
Then use this result to find what f(-x) in terms of x and f(x).
Make use of things like the fact that (-x)(-x) = x^2
Simplify your answer. Rearrange the equation. Try to make f(x) the subject.
[Actually we cannot get f(x) in terms of x this way. The result is true, but does not solve the problem.
The term of f(-x) cannot be eliminated because it does not always equal f(x).
However if we were allowed to assume that f(x) = f(-x) we could use this result to solve the problem.]

[I discovered later that my proof had a flaw caused by an accidental sign change in my notes.]

Last edited by SteveB (2013-05-18 22:26:37)

{7/3}

Member

Registered: 2013-02-11

Posts: 210

Re: Function

I have found these
f(0)=0
2x^2=f(x)+f(-x)
f(x-y)=f(x)+f(-y)-2xy
f(x+1)=f(x)+f(1)+2x
f(2x)=2( f(x)+x^2 )
f(-2x)=2( f(-x)+x^2 )
what do i do now?

---

There are 10 kinds of people in the world,people who understand binary and people who don't.

SteveB

Member

Registered: 2013-03-07

Posts: 595

Re: Function

It appears that you have done the bit to get f(0)=0 which is good, but the second bit where
you need to do a substitution like y = -x you appear not to have done correctly.

I will give you a few more lines of it to give you the idea of what I meant:

Let us assume that the result f(0)=0 is proven.
Now let y = -x

Using the original equation  f(x+y) = f(x) + f(y) + 2xy

We get:  f(x-x) = f(x) + f(-x) + 2(x)(-x)

Since x-x = 0 the bit in the left hand side can be simplified to f(0)
The bit where we have 2(x)(-x) can be written as -2x^2
So f(0) = f(x) + f(-x) - 2x^2
Using the result from earlier that f(0) = 0 we get:
0 = f(x) + f(-x) - 2x^2
This can be rearranged to give us:
f(x) = 2x^2 - f(-x)
[This was an interesting result, but we cannot eliminate f(-x).]

I am not sure at this point whether it is good style to change the letter that
is used like u = -x (as a substitution) or -u = x
[I now realise that this does not work. Sorry about that bit.]

Sorry about this. I have spotted a flaw in the final part of my proof.
This substitution does not appear to work, although it is confusing and
it is easy to accidentally flip the sign when writing it and think that it
proves that f(x) = f(-x) which would solve the problem.

Edit: Thanks Avon, also thanks to {7/3} for an interesting problem.
I have now read Avon's post, checked it to see that I agree, and yes it does work.
My result f(x) = 2x^2 - f(-x) is not needed so that was a blind alley.
However it is needed to do my y=-x substitution with the function g(x + y) to make the method that follows work.

Last edited by SteveB (2013-05-18 22:51:51)

Avon

Member

Registered: 2007-06-28

Posts: 80

Re: Function

I'm going to assume that f is a real valued function (not that it really matters).

It's easy to check that for any real number a, the function f(x) = x^2 + ax is a solution. I will show that these are the only continuous solutions.

Let g(x) = f(x) - x^2. Then g(x+y) = g(x) + g(y) and g(0) = 0.
Also 0 = g(0) = g(x) + g(-x), so g(-x) = -g(x) for all real x.

Let a = g(1).
We can prove by induction that g(nx) = ng(x) for all positive integers n and real x, and then we can deduce that g(x/n) = g(x)/n for all positive integers n and real x.
It follows that g(rx) = rg(x) for all rational r and real x, and so in particular g(r) = rg(1) = ar for all rational r.

If f is continuous then g is also continuous and so g(x) = ax for all real x.

{7/3}

Member

Registered: 2013-02-11

Posts: 210

Re: Function

Thanks to both of you.

---

There are 10 kinds of people in the world,people who understand binary and people who don't.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Function

hi {7/3}

Firstly, well doe to SteveB and Avon for two excellent contributions.

It would seem there are a whole 'family' of solutions of the form

Proof:

{7/3}: Were you expecting x^2 to be the only solution?  Or did you just spot this one and think it was the only one?

Maybe, there's more information?

eg If you were also told that f(x) is an even function then

is the only solution using SteveB's partial proof above.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

{7/3}

Member

Registered: 2013-02-11

Posts: 210

Re: Function

Oh,thanks,i thaught x^2 was the only one.

---

There are 10 kinds of people in the world,people who understand binary and people who don't.