2

So if I want a function $f:\mathbb R \times \mathbb R\to \mathbb R$ for which

$$f(a,b)+f(b,a)=0$$

holds, then in a most generic way I can take almost any $g(a,b)$ function, and construct $f$ as

$$f:= g(a,b) - g(b,a)$$

So now I want an $f$ that can do

$$f(a,b)+f(b,c)+f(c,a) = 0$$

or even

$$f(x_1,x_2) + f(x_2,x_3) +...+ f(x_i,x_j) +...+ f(x_n,x_1) = 0$$

This reminds of an integral over a closed path that must be zero. Such functions can be constructed as a potential difference. So by taking an arbitrary $g:\mathbb R\to \mathbb R$ "potential" function f can be constructed as

$$f:= g(a)-g(b)$$

But is this the most generic form for $f$? Why?

Shaun
  • 44,997
JohnDeeDoe
  • 21
  • 4
    Hey @JohnDeeDoe, your questions would receive better attention if you upgrade your typesettings. Take a look at https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Melquíades Ochoa Jan 27 '20 at 23:28
  • ah perfect, thank you for the edit! – JohnDeeDoe Jan 28 '20 at 00:08
  • Hint: $f(0, 0) + f(0, 0) + f(0, 0) = 0$; $f(0, a) + f(a, 0) + f(0, 0) = 0$; $f(0, a) + f(a, b) + f(b, 0) = 0$. I think if you define $g(x) := f(x, 0)$ and combine those things above, you'll necessarily get that $f(a, b) = g(a) - g(b)$. – Daniel Schepler Jan 28 '20 at 00:09
  • $f(x,y) = x-y$, then $f(a,b)+f(b,c)+f(c,d) = 0$ – fGDu94 Jan 28 '20 at 00:11

0 Answers0