Factoring x + y

Question

I'm trying to find functions $f(x)$ and $g(y)$ such that $$f(x)\cdot g(y) = x + y$$

I can't seem to find a single solution to this problem. Anything I try becomes of the form $f(x,y) \cdot g(y) or f(x) \cdot g(x,y)$

Here is my work so far:

$$f(x)g(y) = x + y$$

$$f'(x)g(y) + f(x)g'(y)\frac{dy}{dx} = 1 +\frac{dy}{dx} $$

$$f'(x)g(y) - 1 = \frac{dy}{dx} - f(x)g'(y)\frac{dy}{dx} $$

$$f'(x)g(y) - 1 = \frac{dy}{dx} (1 - f(x)g'(y))$$

$$\frac{f'(x)g(y) - 1}{1 - f(x)g'(y)} = \frac{dy}{dx} $$

But since $g(y)$ doesn't have a known or restricted order, I have no idea what $f(x)$ would be:

I am considering taking higher derivatives and then using substitutions (ex: $g(y) = (x+y)/f(x))$ but I'm not sure if that will work.

Pedro · Accepted Answer · 2013-05-02T19:55:29.357

25

You have that $$\tag 1 g(0)f(0)=0$$ while $$f(1)g(0)=1$$ $$f(0)g(1)=1$$ The first equations says that either $g(0)=0$ or $f(0)=0$, or both. But this contradicts the last two equations.

ADD Note that what I wrote holds also when $x=-y$, and when $x=0$, $y=\alpha$ a constant, and vice-versa. In fact, the last observation means your function cannot be defined for any value of $x$ or $y$. Indeed, we have that for any $\alpha,\beta\in\Bbb R$

$$f(0)g(\alpha)=\alpha$$ $$f(\beta)g(0)=\beta$$

But beacuse of $(1)$, the last relations are impossible, so we cannot define either $f$ or $g$ for any $x\in\Bbb R$.

edited May 02 '13 at 19:55

answered May 02 '13 at 19:34

Pedro

122,002

The last line can be simplified to "multiplying the last two relations you get $0=1$." :) – N. S. May 02 '13 at 19:40
But can the following possible? – Sidharth Ghoshal May 02 '13 at 19:41
@N.S. "Simplified"? Is it really that complicated? =) – Pedro May 02 '13 at 19:41
1

@frogeyedpeas The point of my answer is that no such decomposition can exist. – Pedro May 02 '13 at 19:41
@sorry Peter, give me 15 seconds – Sidharth Ghoshal May 02 '13 at 19:42
lim x -> 0 f(x)g(x) = 0, lim x -> 0, y -> 1 f(x)g(y) = 1, lim x-> 1 y -> 0 f(x)g(y) = 1, does the contradiction still hold? I'm guessing it does but I don't see it – Sidharth Ghoshal May 02 '13 at 19:44
@frogeyedpeas You don't need to use limits. Just plug in numbers. In fact, I don't really see what you can deduce from your limits, really. You don't need derivatives either. – Pedro May 02 '13 at 19:45
I see your argument, I'm just curious if utilizin limits creates a potential solution, where f(x) and g(x) themselves are not well defined at 0 and 1 but when put together become defined – Sidharth Ghoshal May 02 '13 at 19:45
@frogeyedpeas I will add something – Pedro May 02 '13 at 19:50
1

15 seconds of posting delay is worth commenting about, which is why I'm doing it too! And it's taking me 30+ seconds to do so, too. :) – Kaz May 02 '13 at 23:20

score 2 · Answer 2 · answered May 02 '13 at 19:40

2

This would mean that $\frac{a+y}{b+y}$ would have to be a constant function of $y$ for all $a,b$. But $\frac{a+y}{b+y}=1-\frac{b-a}{b+y}$ is only constant when $a=b$.

answered May 02 '13 at 19:40

Thomas Andrews

177,126

Factoring x + y

2 Answers2