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November 2nd, 2015, 12:22 PM   #21
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Quote:
 Originally Posted by Karma Peny You completely ignored the 'paradoxes' I mentioned and went off to talk about sets.
I thought I was denying any paradox and giving other examples of infinite objects (having infinitely many elements) that we routinely use as complete objects.

Quote:
 Originally Posted by Karma Peny So, either it is one huge coincidence that these figures match or the Extreme Finitism approach has correctly identified where these ‘mysterious’ values are coming from (& it surely isn’t infinity!).
The problem with claiming that your approach identifies where this value is coming from, is that you have just decided to integrate a function generated from partial sums. It doesn't actually explain very much at all because these rational numbers only appear in the integrals. I would be astonished if they aren't related in some way though.

I also have been looking at those functions and integrals, but they aren't all of the answer because these results are the ones that come from the analytic continuation of the Riemann Zeta function, and not from techniques such as Generic Summation. Other results remain unexplained by the integrals:
$$1-1+1-1+\ldots = \tfrac12 \\ 1+0-1+1-0+1+\ldots = \tfrac23 \\ 1+2+4+8+16+\ldots = -1$$
Note the the latter can be obtained by analytic continuation, and the two results agree.

Last edited by v8archie; November 2nd, 2015 at 12:26 PM.

November 2nd, 2015, 12:34 PM   #22
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Quote:
 Originally Posted by Karma Peny You can keep your mysteries and your belief in 'infinity'
What does belief have to do with anything here?

November 2nd, 2015, 06:01 PM   #23
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Quote:
Originally Posted by Karma Peny
Quote:
 Originally Posted by v8archie Why does there need to be a meaning?
Because if you then proceeed to use it, you will not understand what you are doing. And if you don't proceed to use it, then it is of no value.
Perhaps we should leave John von Neumann to answer that one:
Quote:
 Young man, in mathematics you don't understand things. You just get used to them.

November 3rd, 2015, 06:40 PM   #24
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Quote:
 Originally Posted by v8archie Perhaps we should leave John von Neumann to answer that one:
So, mathematics relies on faith? Is math a religion now, along with pdf?

November 3rd, 2015, 06:47 PM   #25
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Quote:
 Originally Posted by Monox D. I-Fly So, mathematics relies on faith? Is math a religion now, along with pdf?
And the Flying Spaghetti Monster. Never forget him. In sauce may he reside.

-Dan

Last edited by skipjack; November 3rd, 2015 at 10:59 PM.

November 3rd, 2015, 07:09 PM   #26
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Quote:
 Originally Posted by Monox D. I-Fly So, mathematics relies on faith? Is math a religion now, along with pdf?
In most, if not all, of my work, I have faithfully relied on the "fact" that 1 + 1 = 2.

 November 13th, 2015, 06:37 PM #27 Math Team     Joined: May 2013 From: The Astral plane Posts: 2,272 Thanks: 942 Math Focus: Wibbly wobbly timey-wimey stuff. I realize this topic is basically done and buried but I spent a few idle moments of research and though I'd tell y'all my results. As the topic has come up frequently in the last few months I thought it might be helpful. First: I have no idea how to derive this formula. But it is the simplest I could find that analytically continues the (Riemann) zeta function to negative s. I am uncertain of the region of convergence but $\displaystyle \zeta$ is meromorphic and it converges everywhere in the complex s plain with the exception of s = 1. The formula below provides a form that converges for all but even negative integer s (and, of course, s = 1) so far as I can tell. I also can't find a way to use this definition to produce a value for s = 0. $\displaystyle \zeta (s) = 2^s \pi ^{s - 1} ~ sin \left ( \frac{ \pi s}{2} \right ) ~ \Gamma (1 - s) ~ \zeta (1 - s)$ (This is called the "Riemann function equation.") For those who don't know, the Gamma function ( $\displaystyle \Gamma(n)$) is closely related to the factorial function and for our purposes all we need to know is that $\displaystyle \Gamma (n) = (n - 1)!$ for integer $\displaystyle n \geq 1$. The idea here is that we know the values of $\displaystyle \zeta (s)$ for any positive integer greater than 1. So to get $\displaystyle \zeta (-1)$: $\displaystyle \zeta (-1) = 2^{-1} \pi ^{-1 - 1} ~ sin \left ( \frac{ \pi (-1)}{2} \right ) ~ \Gamma (1 - -1) ~ \zeta (1 - -1)$ $\displaystyle = \frac{1}{2 \pi ^2} (-1) \Gamma (2) ~ \zeta (2)$ and since $\displaystyle \Gamma (2) = (2 - 1)! = 1! = 1$ and $\displaystyle \zeta (2) = \sum_{n = 1}^{\infty} \frac{1}{n^2} = \frac{ \pi ^2}{6}$ we get $\displaystyle \zeta (-1) = \frac{1}{2 \pi ^2} (-1) \frac{\pi ^2}{6}$ $\displaystyle \zeta (-1) = - \frac{1}{12}$ This has nothing to do with the sum 1 + 2 + 3 + ... -Dan Last edited by topsquark; November 13th, 2015 at 07:05 PM.
November 13th, 2015, 07:14 PM   #28
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Quote:
 Originally Posted by topsquark The idea here is that we know the values of $\displaystyle \zeta (s)$ for any positive integer greater than 1.
I believe that we need to (and do) know $\zeta (s)$ for a continuous region to continue it analytically without ambiguity.

November 13th, 2015, 07:21 PM   #29
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Quote:
 Originally Posted by v8archie I believe that we need to (and do) know $\zeta (s)$ for a continuous region to continue it analytically without ambiguity.
True. I was making that assumption for the sake of doing the calculation of the function version at -1 specifically. Continuity has been proven for s > 1 using the "usual" sum formula so I took that as a given.

-Dan

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