Find all functions $f(x) : \mathbb{R} \rightarrow \mathbb{R}$ which satisfy

$f(x+y) = f(x) + f(y)+2xy.$

Hint: If $r$ is a rational number, what can we say about $f(rk)$ for any $k$?

**Prove that these are the only possible ones.**

Note:

1. It is not sufficient to just find a family of solutions.

2. You may not assume that $f(x)$ is continuous or differentiable.

3. There is more than 1 function that satisfies those conditions.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestFor any additive function $h(x)$ the function $f(x)=h(x)+x^2$ satisfies the equation. So there can be infinitely many wild solutions without additional constraints.

Log in to reply

Right, in particular, let $\{ 1, \pi, v_3, v_4, \ldots \}$ be a rational basis for the reals, then for $x = r_1 + r_2 \pi + \sum r_i v_i$, we could define

$f(x)= r_1 + 2r_2 + x ^2$

Such a function is neither differentiable, nor continuous.

Log in to reply

By definition, $f(r(k+1))=f(kr)+f(r)+2r^{2}k\quad---(1)$

Through pattern recognition of ${f(kr)}_{k=2}^{k=5}$ in terms of $f(r)$, it seems to follow the relation:

$f(kr)=kf(r)+k(k-1)r^{2} \quad --- (2)$ of which is directly derived from $(1)$

If this is the only solution, $f(n)$ has to have only $1$ value, where $n$ is any real number, which is dependent on the definition of the question. For instance, in the inspiration question, $f(1)$ is defined to be $4$ and only $4$.

Is this complete?

Log in to reply

What is the value of $f ( \pi )$, if $f(1) = 4$?

Log in to reply

For that, Chew-Seong Cheong has already given the answer, which is $\pi^{2} + 3\pi$, assuming the function is continuous.

For a general case, if given $f(r)$, $f(x)$ can be found if it is continuous.

Using Chew-Seong Cheong's method,

$f(x+r)=f(x)+f(r)+2rx\\ f(x+r)-f(x)=f(r)+2rx$

So, $f(x+r)=\sum _{ k=1 }^{ \frac { x }{ r } }{ (f(r)+2k{ r }^{ 2 }) } +f(r)$

Therefore, $f(x+r)=\frac { xf(r) }{ r } +x\left( x+r \right) +f(r)\\ ={ x }^{ 2 }+\left( r+\frac { f(r) }{ r } \right) x+f(r)\\ ={ (x+r) }^{ 2 }+\left( \frac { f(r) }{ r } -r \right) \left( x+r \right)$

$\boxed{f(x)={ x }^{ 2 }+\left( \frac { f(r) }{ r } -r \right) x}$

Log in to reply

$f(x) = x^2 + 3x$.

It is true that "If the function is continuous, thenHowever, since your argument never uses the condition that the function is continuous, hence it is flawed. The error is that you made the assumption that $\frac{x}{r}$ is an integer, otherwise your summation is meaningless. It could be adjusted to the case where $\frac{x}{r}$ is a rational number, but cannot apply to the case of irrational numbers. In particular, we don't know what $f(\pi )$ is.

So, how do we use the condition of continuity (but not differentiability) to prove it?

Log in to reply

Log in to reply

Hint:What is $f(3), f(3.1), f(3.14), f(3.141), f(3.1415), f(3.14159), ...$?Log in to reply

$\pi$? Probably finding the upper and lower boundaries?

So we just have to approximate as we get closer and closer toLog in to reply

If a function is continuous, then $f( \pi ) = \lim f(x_i)$ for any series of points that converge to $\pi$. We can pick \( x_i = 10^{-i} \lfloor 10^i \pi \rfloor as I did above.

The idea of upper and lower boundaries would apply for "increasing functions", which doesn't require the assumption of continuity. In this case, we have $f(3) \leq f(3.1) \leq f(3.14) \ldots \leq f( \pi ) \leq \ldots f(3.15) \leq f(3.2) \leq f(4) .$

Because the inner inequalities converge to each other, we get the result (without assuming continuity).

The take home is that for such functional equations, you have to be careful to work with exactly what you are given, instead of adding additional assumptions because it makes your working simpler.

@Abhishek Sharma See the above and it's relevance to "assume function extends to real numbers and is differentiable".

Log in to reply