Standard crank division by zero thread - Page 22

CRGreathouse · 2011-04-19, 14:17

Quote:

Originally Posted by Don Blazys

If z=1,

then (T/T)*c^z = T*(c/T)^((z*ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1))

becomes (T/T)*c^1 = T*(c/T)^1

You removed a singularity when you transformed the first equation to the second. Since when T = c we're at the singularity and so can't remove it. It's just like transforming

x^2 / x

into

x

when x = 0. The analytic continuation of the first is indeed the second, but the two are not the same since the second is defined at x = 0 and the first is not.

Condor · 2011-04-19, 20:12

Quote:

Originally Posted by CRGreathouse

You removed a singularity when you transformed the first equation to the second. Since when T = c we're at the singularity and so can't remove it. It's just like transforming

x^2 / x

into

x

when x = 0. The analytic continuation of the first is indeed the second, but the two are not the same since the second is defined at x = 0 and the first is not.

It's worse than that. If you let $X=\frac{ln(c)}{ln(T)}-1$ , then the exponent in Don's equation is $\frac{z*X+z-1}{X}$ . If z!=1, then as T->c, X will approach 0, and the denominator is unbounded. This is what makes the whole equation indeterminate, of the form 1^inf. We can do this with Don's equation even though real numbers are not in his domain, because "any true equation, whether it be 2 + 3 = 5 or [Don's], is simply an actuality." In more meaningful words, its valid interpretation is independent of any arbitrary restrictions Don wants to place on the values. Just like he insisted it has to be independent of any actual restrictions that the laws of math require.

But when z=1, the exponent is $\frac{X}{X}$ . Don is literally saying he can replace this $\frac{X}{X}$ with 1 regardless of what X is, just because he expanded the "0" into a form that masks the fact that it is 0. Now it is this subexpression that is indeterminate, not he whole term, and Don's substitution only works if he allows the "analytic continuation" to be used by taking the limit as T->c (or X->0).

CRGreathouse · 2011-04-20, 01:02

Right, I pointed out the indeterminate form above. (You explain it more thoroughly -- I didn't bother, since I knew Don doesn't understand.)

Yes, the analytic continuation does get rid of the issues. Of course the analytic continuation is just c^z...!

Don Blazys · 2011-04-20, 11:14

Quoting CRGreathouse:

Quote:

You removed a singularity when you transformed the first equation to the second.

I didn't "remove a singularity".

Nor did I "transform the first equation into the second".

I simply evaluated

(T/T)*c^z = T*(c/T)^((z*ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1))

at z = 1, and the result was

(T/T)*c^1 = T*(c/T)^1

where we can let T = c.

Don.

science_man_88 · 2011-04-20, 11:59

Quote:

Originally Posted by Don Blazys

Quoting CRGreathouse:

I didn't "remove a singularity".

Nor did I "transform the first equation into the second".

I simply evaluated

(T/T)*c^z = T*(c/T)^((z*ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1))

at z = 1, and the result was

(T/T)*c^1 = T*(c/T)^1

where we can let T = c.

Don.

I haven't seen a place yet that I know we can't.

Don Blazys · 2011-04-20, 13:04

Thanks science man 88 !

CRGreathouse · 2011-04-20, 13:11

Quote:

Originally Posted by Don Blazys

I didn't "remove a singularity".

Nor did I "transform the first equation into the second".

I simply evaluated

Then you evaluated incorrectly. In a high school class, you'd lose points for an answer like that. In a college class, you'd probably get no credit for the problem at all.

It wouldn't matter much, except that your whole flawed proof is based on this mistake.

science_man_88 · 2011-04-20, 14:08

Quote:

Originally Posted by Don Blazys

Thanks science man 88 !

the problem with that response is:

Quote:

Originally Posted by Don Blazys

where we can let T = c.

Don.
[/COLOR]

but early ( though I now can't find it) you said they can't be equal but but here you allow the unallowable ?

Don Blazys · 2011-04-20, 21:02

Quoting CRGreathouse:

Quote:

Then you evaluated incorrectly.

No, I evaluated correctly.

If z=1, then

(T/T)*c^z = T*(c/T)^((z*ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1))

results in

(T/T)*c^1 = T*(c/T)^1

where letting T = c results in

(c/c)*c^1 = c*(c/T)^1

Don.

Condor · 2011-04-20, 21:58

Quote:

Originally Posted by science_man_88

the problem with that response is:

but early ( though I now can't find it) you said they can't be equal but but here you allow the unallowable ?

Don contradicts himself frequently, but I don't recall him doing it here. CRGreathouse and I said it, because the laws of math require it. T can't equal c in any expression that is predicated on the derivation of Don's "identity" :

$\frac{T}{T}C^z = T $\frac{C}{T}$^{\frac {ln( \frac{C^z}{T})}{[ln(\frac{C}{T})]}}=T $\frac{C}{T}$^{\frac {\frac{ln( \frac{C^z}{T})}{ln(T)}}{\frac{ln(\frac{C}{T})}{ln(T)}}} =T $\frac{C}{T}$^{\frac {\frac{ln( \frac{C^z}{T})}{ln(T)}-\frac{ln(T)}{ln(T)}}{\frac{ln(\frac{C}{T})}{ln(T)}-\frac{ln(T)}{ln(T)}}}=T $\frac{C}{T}$^{\frac {\frac{{z}*{ln(C)}}{ln(T)}-1}{\frac{ln(C)}{ln(T)}-1}}$

When c=T, the term I put inside [] is zero, and we all know that you can't divide by zero. Anything that follows from that point in the derivation is also technically undefined at c=T. I said "technically" because there are ways around it, but they depend on the assumption that you can substitute $\lim_{c \to T}F(c,T)$ for $F(T,T)$ , where $F(c,T)$ is the undefined part of the expression.

So, Don can't set z=1 in $\frac{\frac{{z}{ln(C)}}{ln(T)}-1}{\frac{ln(C)}{ln(T)}-1}$ , "divide out" the now identical terms, and replace it with 1 as he wants to do. It's still undefined at c=T whether or not you can still see the division by zero. He can, however, make the assumption and replace it with $\lim_{c \to T}\frac{\frac{{ln(C)}}{ln(T)}-1}{\frac{ln(C)}{ln(T)}-1}$ , which is 1. (Also note that he could do the same thing when z <>1 as he does when z=1, and rearrange it so that there is no division by zero. Just reverse the derivation! That "results in" c^z.)

Once he intentionally puts the zero in his derivation, he has to make the assumption about limits to use c=T. And once he does that, the same assumption allows replacing $T (\frac{C}{T}\)^{\frac {\frac{{z}*{ln(C)}}{ln(T)}-1}{\frac{ln(C)}{ln(T)}-1}}$ with $\lim_{c \to T}T (\frac{C}{T}\)^{\frac {\frac{{z}*{ln(C)}}{ln(T)}-1}{\frac{ln(C)}{ln(T)}-1}}$ . Since that limit is T^z, the entire main premise of his flawed proof disappears in a puff of smoke.

akruppa · 2011-04-20, 23:05

For those who haven't tried, quote Don and marvel at his text coloring prowess.

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2011-04-20, 01:02	#234
CRGreathouse Aug 2006 3×1,993 Posts	Right, I pointed out the indeterminate form above. (You explain it more thoroughly -- I didn't bother, since I knew Don doesn't understand.) Yes, the analytic continuation does get rid of the issues. Of course the analytic continuation is just c^z...!

2011-04-20, 13:04	#237
Don Blazys Feb 2011 163 Posts	Thanks science man 88 !

2011-04-20, 23:05	#242
akruppa "Nancy" Aug 2002 Alexandria 2,467 Posts	For those who haven't tried, quote Don and marvel at his text coloring prowess.