It means you divide f(x) by g(x).

>>8174679
yeah, sure

>>8174678
"Fux the gux" obviously

It means you didn't pay attention in math

>>8174678
Excellent thread my friend

>>8174678
The outer parentheses are superfluous. Remove those and the meaning of the expression should become clear. If not, you should probably retake precalc.

Could it be rewritten as f/g(x)? Or would that make f seem to be a variable?

>>8174678
for every x, you divide the number you get out of f(x) by the number you get out of g(x). x can't be 0 or you fuck shit up senpai. This gives you a new function.

>>8174678
>>8174678
Let [math]R[/math] be a commutative ring and [math]S[/math] a multiplicative subset of [math]R[/math].

Define the localization [math]{S^{ - 1}}R \equiv \frac{{R \times S}}{\sim}[/math] .

Let [math]\left( {f,g} \right) \in R \times S[/math] .

Then [math]\frac{f}{g} \equiv {\left[ {\left( {f,g} \right)} \right]_\sim} = \left\{ {\left( {f',g'} \right) \in R \times S\;|\;fg' - f'g = 0} \right\}[/math] .

>>8177684
you mean there's a u in S such that u(fg' - f'g) = 0

>>8177684
Shut up u autistic tit

A function of x is divided by another function of x.

Its like (2x + 1)/(6x^2 + 11x + 4) = 1/(3x + 4)

>>8174684
That's what it means. When the functions are polynomials, here is the approach to simplify:

https://en.wikipedia.org/wiki/Polynomial_long_division

>>8175507
(f/g)(x) is acceptable in lower division math classes.

High division classes will expect more exact notation.

>>8177700
what are you talking about? f/g is the function defined by (f/g)(x) = f(x)/g(x)

>>8177713
(f/g)(x) = h(x)

Get it right dude, this is precal.

>>8177713
Oops f(g(x)) = h(x)

You are correct
I have been drinking cheap beer

>>8177684
But how can we be sure that the manifolds are cohomological?