[SYRFox]

Take the function:

f(x) = (1/x)(x^3 - x^2)

So: f(x) = (x^3/x) - (x^2/x)
= x² - x.

We proved that: (1/x)(x^3 - x^2) = x² -x.

But if x= 0:

-> x² - x = 0² - 0 = 0
-> (1/x)(x^3 - x^2) = (1/0)(0^3 - 0^2) : this is impossible (division per zero).

So these two functions are the same, but they're not the same (O_O)

OMG OMG WTF?!?

[Tokolosh]

Originally Posted by SYRFox

Take the function:

f(x) = (1/x)(x^3 - x^2)

So: f(x) = (x^3/x) - (x^2/x)
= x² - x.

We proved that: (1/x)(x^3 - x^2) = x² -x.

But if x= 0:

-> x² - x = 0² - 0 = 0
-> (1/x)(x^3 - x^2) = (1/0)(0^3 - 0^2) : this is impossible (division per zero).

So these two functions are the same, but they're not the same (O_O)

OMG OMG WTF?!?

If f(x) = x² - x
= f(x) = 0² - 0
then the division is the same. But 0² = 0.
You're dividing 0 by 0. Not possible unless 0 = x

OMG OMG WTF?!?

[A Thousand Threads]

Originally Posted by SYRFox

Take the function:

f(x) = (1/x)(x^3 - x^2)

So: f(x) = (x^3/x) - (x^2/x)
= x² - x.

We proved that: (1/x)(x^3 - x^2) = x² -x.

But if x= 0:

-> x² - x = 0² - 0 = 0
-> (1/x)(x^3 - x^2) = (1/0)(0^3 - 0^2) : this is impossible (division per zero).

So these two functions are the same, but they're not the same (O_O)

OMG OMG WTF?!?

a pole at x=0
what's so strange about that?

edit: Just realized that it isn't a pole.
Anyways, it's late

[!@#$%!]

tit, you dividing by 0, what's new?

--
the moral: do not solve math equations while stoned.

wow, man, did you see that? it's like... like... wow...

[!@#$%!]

ps-

youre doing [x^3 - x^2]/x, if x= 0 then it's 0/0

The expression 0/0 is an "indeterminate form". That does not simply mean that it is undefined; rather, it means that the limit of f(x)/g(x) is determined by the particular functions f and g as they both approach 0. As x approaches some number, the limit may approach any finite number, 0, ∞, or −∞, depending on the specific behavior of the functions. See l'Hôpital's rule.

you stated it was IMPOSSIBLE, which was wrong, hence the logical loophole.

[HECKLER SPRAY]

Please, can you stop speaking chinese, I don't understand...
Logical ? What does it mean ?

[golden child]

what he is struggling with is 1/0, not 0/0. meaning that they are not equal because y is undefined when x=0. can the first one even be considered a parabola since it has the break in it?

[!@#$%!]

Originally Posted by golden child

what he is struggling with is 1/0,

yeah but in the case of x=0 that formula amounts to 0/0. he's ignoring the value of the numerator, sez eye.

[HECKLER SPRAY]

??????????????????????????????????????????:confuse d:

[alyasa]

What !@#$%! said.

[finding nobody]

Should be able to make sense out of that? Cause I can't

[!@#$%!]

Originally Posted by HECKLER SPRAY

Please, can you stop speaking chinese, I don't understand...
Logical ? What does it mean ?

sorry for my bad english, what i meant was that his logic was short-circuiting. yes? he was reaching a self-contradictory conclusion. but it is because his premises were wrong.

[alyasa]

Basically, he created a self-contradictory loop, which feeds back onto itself, thereby creating an infinite paradox, which if questioned begins to melt down the logic circuits of the brain...

[!@#$%!]

Originally Posted by alyasa

Basically, he created a self-contradictory loop, which feeds back onto itself, thereby creating an infinite paradox, which if questioned begins to melt down the logic circuits of the brain...

^^ what he said. 2 thumbs up.

[StevOK]

The function is undefined at x=0. I forget what those are called, but they always have to be looked out for when you do this kind of thing.

[alyasa]

I think you might have to factor in limits in this equation. I think it's not so much undefined, but the proper term is indeterminate, I believe. If you factor in limits in the above equation and let the limits of x approach zero, then you effectively illustrate that the equation is unsolvable algebraically, thereby removing the conundrum that has been put in place by attempting to solve the equation further.

"...To avoid this little inconsistency the epsilon-delta limit device is used to establish that the LIMIT of the ratio dy/dx is 2ax + b as dx APPROACHES zero."

[golden child]

this function never crosses the y axis because when x=0 y is undefined, the reduced version is a full parabola so to speak that actually crosses the y axis at (0,0)

yeah but in the case of x=0 that formula amounts to 0/0. he's ignoring the value of the numerator, sez eye.

(1/0)(0^3 - 0^2)

[5Against1]

I think you forgot to carry the 1.

[_slavo_]