What do you call a function based on (w*x) / (y*z)
where w and y remain the same, and x always increases linearly with z?

For example
at one point on the graph it might look like

(60*1) / (25*1)

and on the next it equals

(60*2) / (25*2)

Constant.

You crackhead.

Does z always equal x? Then it can be rewritten as (w*x)/(y*z) == (w*x)/(y*x) == w/y and if w and y are fixed then it’s just one constant

It's a linear function.

it's called rational function, and I understand your love for maths, but you need to learn some more comprehensible version of English.

Due to the fact that x and z increase by the same amount, lets call it n, and n goes to positive infinity, the value of x(n)/z(n) = [x+n]/[z+n] will be approaching 1, as for larger n, x and z become negligible. x(n)/z(n) is not linear because the relationship of x(n) and z(n) changes with n from x/z to almost n/n=1.

w/y*[x(n)/z(n)]
x(n) = x0 + n
z(n) = z0 + n

limit n=>inf of [x(n)/z(n)]

applying De l'Hôpital's rule:
d[x(n)]/dn = x'(n) = n
d[z(n)]/dn = z'(n) = n

x'/z' = n/n = 1

The linearity of the function f(n) = x(n)/z(n) is tested by it's first derivative f'(n) being a constant:
f'(n) = d[x(n)/z(n)]/dn
f'(n) = [x'*z - x*z']/z² (Quotient rule)
f'(n) = [n*(z+n) - (x+n)*n]/[z+n]²
f'(n) = [n*(z-x)]/[z+n]²
as we can see, f'(n) is not a constand thus f(n) is not linear.

@PonySlaystation I'll take your word for it.

@theabbie "but you need to learn some more comprehensible version of English."

Thats fair.

And also

How DARE you. lol.

@Root "You crackhead."

This is pretty much all my posts these days.

If x=kz+n, then the overall function is actually hyperbolic

w/y * (n/z + k)

@PonySlaystation he said (w*c) / (y*c) where c is a constant.
Times, not plus!

@Root
1) "w and y remain the same" thus w and y are constant which leads to w/y=c, constant.

2) "x increases linearly with z" thus x0 and z0 are constant, n is my increment:
x(n)=x0+n
z(n)=z0+n

(w*x)/(y*z) = (w/y)*(x/z) = c*x(n)/z(n) = c*(x0+n)/(z0+n)

limit of c*(x0+n)/(z0+n) for n approaching infinity is c.
I omitted c in my previous comment because it was irrelevant. 🙂

@PonySlaystation ...what in the goddamn...

Sorry, I don’t get why this isn’t a constant. Z and x are the same and should cancel out, leaving x and w which are fixed numbers. So you’re just left with one constant divided by another constant

Oh wait...I get it now. Lol

@PonySlaystation Aha! Yes, you are right. Apologies for doubting :)

@Root no worries m8 ❤️