×

Inspired by Abhishek Sharma

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.

Inspiration

Note by Calvin Lin
1 year, 9 months ago

Sort by:

For 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. · 1 year, 9 months ago

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. Staff · 1 year, 9 months ago

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? · 1 year, 9 months ago

What is the value of $$f ( \pi )$$, if $$f(1) = 4$$? Staff · 1 year, 9 months ago

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

$\boxed{f(x)={ x }^{ 2 }+\left( \frac { f(r) }{ r } -r \right) x}$ · 1 year, 9 months ago

It is true that "If the function is continuous, then $$f(x) = x^2 + 3x$$.

However, 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? Staff · 1 year, 9 months ago

I don't know... Any clues? :D · 1 year, 9 months ago

Hint: What is $$f(3), f(3.1), f(3.14), f(3.141), f(3.1415), f(3.14159), ...$$? Staff · 1 year, 9 months ago

So we just have to approximate as we get closer and closer to $$\pi$$? Probably finding the upper and lower boundaries? · 1 year, 9 months ago

Yes, and no.

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". Staff · 1 year, 9 months ago