Thoughts of a grumpy old man

You CAN divide by zero

Rate this Entry
10 Comments
by
Chimpy
, 3rd April 2012 at 01:14 AM (260 Views)
For starters, let's take the following function:
f(x)=ln(x)
Differentiating this gives:
f'(x)=(1*x')/(x*ln(e)=1/x
Let's integrate AND revert the function at the same time. This is possible because a differential is the opposite of an integral
F'(x)=ln(x)+c
However, integrating 1/x the normal way gives:
1/x=x(-1)
F'(x)=1/(-1+1)*x(-1+1)=1/0 * x0 = 1/0 * 1 = 1/0
Now that you have two values of the integral, let's make an equation:
1/0=ln(x)+c
Therefore you can divide by zero
Q.E.D
Categories
Uncategorized

Comments

  1. Mars's Avatar
    Error: Blog not found.
  2. aaaaaa123456789's Avatar
    You know that that's the whole reason why you can't integrate 1/x the "normal way", right?
    The "normal way" is proven by derivating, and it's known the proof works when you assume n =/= -1.
    So, using it for n = -1 is absurd, since it was assumed that n wasn't -1.
  3. Wolframium's Avatar
    Quote Originally Posted by aaaaaa123456789
    You know that that's the whole reason why you can't integrate 1/x the "normal way", right?
    The "normal way" is proven by derivating, and it's known the proof works when you assume n =/= -1.
    So, using it for n = -1 is absurd, since it was assumed that n wasn't -1.
    That makes sense to me. I was pretty sure there was something wrong with this, I just couldn't put my finger on it.
  4. PrinceWhite's Avatar
    What the...
  5. ~Twisted's Avatar
    This kind of makes sense to me.
  6. Chimpy's Avatar
    Quote Originally Posted by aaaaaa123456789
    You know that that's the whole reason why you can't integrate 1/x the "normal way", right?
    The "normal way" is proven by derivating, and it's known the proof works when you assume n =/= -1.
    So, using it for n = -1 is absurd, since it was assumed that n wasn't -1.
    It was intentionally wrong, but the intentional part was not the part you pointed out.

    The only thing I see wrong with it is that ln(x)+c is variable, and that you can create any number with it that way, but 1/0 does not have a single value that satifies it.

    Humor me, what exactly IS the problem with taking n = -1 so long as you don't actually perform the division by zero?
  7. Wolframium's Avatar
    Quote Originally Posted by Chimpy
    It was intentionally wrong, but the intentional part was not the part you pointed out.

    The only thing I see wrong with it is that ln(x)+c is variable, and that you can create any number with it that way, but 1/0 does not have a single value that satifies it.

    Humor me, what exactly IS the problem with taking n = -1 so long as you don't actually perform the division by zero?
    Simply because you end up with an impossible result for that one case. There are other, more general proofs of the theorem that are defined for n=-1.
  8. aaaaaa123456789's Avatar
    Quote Originally Posted by Chimpy
    It was intentionally wrong, but the intentional part was not the part you pointed out.

    The only thing I see wrong with it is that ln(x)+c is variable, and that you can create any number with it that way, but 1/0 does not have a single value that satifies it.

    Humor me, what exactly IS the problem with taking n = -1 so long as you don't actually perform the division by zero?
    The theorem that allows you to integrate as you want to clearly states that
    integral(xn * dx) = xn+1/(n+1) if and only if n isn't -1.
  9. skibå's Avatar
    Amen.
  10. Εclipse Midir's Avatar