s2dbaker
s2dbaker
Joined: Jun 10, 2010
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February 28th, 2014 at 9:48:36 PM permalink
Quote: AxiomOfChoice

Consider the following exchange:

Me: You will get wet if you stand outside in the rain
3-year-old: But I'm outside now and I'm not getting wet
Me: But it's not raining now. I only said that you would get wet if you stood outside when it was raining.
3-year-old: I don't know what you are talking about. I'm outside now and I'm not wet. lalalalala I'm a ninja!

That is basically the equivalent of the conversation that we are having now. Guess which part you are playing?

It's not difficult to see that you are calling me a three year old and that's fine as long as you don't mind violating the new non-Statler and Waldorf rules but I was pointing out that there's a body of evidence that shows that 1 + 2 + 3 + 4 + 5 + 6 + ... = -1/12

Have I said anything incorrect? Did I get any analysis of the Zeta function wrong? Is WolframAlpha calculating the Zeta function at -1 = -1/12?
If y = x and z = x doesn't it follow that y = z?
Someday, joor goin' to see the name of Googie Gomez in lights and joor goin' to say to joorself, "Was that her?" and then joor goin' to answer to joorself, "That was her!" But you know somethin' mister? I was always her yuss nobody knows it! - Googie Gomez
AxiomOfChoice
AxiomOfChoice
Joined: Sep 12, 2012
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February 28th, 2014 at 9:57:16 PM permalink
Quote: s2dbaker

Have I said anything incorrect?

Yes

Quote:

Did I get any analysis of the Zeta function wrong?

Yes

Quote:

Is WolframAlpha calculating the Zeta function at -1 = -1/12?

No, that part is right.


Quote:

If y = x and z = x doesn't it follow that y = z?

That is also right.

The part what you said that was incorrect was that zeta(-1) = 1+2+3+4+5+...

You only get wet by standing outside if it's raining, and the zeta function is only equal to that infinite sum when the sum converges.

Why is this so hard to understand?
24Bingo
24Bingo
Joined: Jul 4, 2012
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February 28th, 2014 at 10:12:05 PM permalink
Remember folks, when you insisted 0^0 = 1 because the indeterminate form so often went to 1?

This is the madness that lay on that road.
The trick to poker is learning not to beat yourself up for your mistakes too much, and certainly not too little, but just the right amount.
AxiomOfChoice
AxiomOfChoice
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February 28th, 2014 at 10:17:48 PM permalink
Quote: s2dbaker

It's not difficult to see that you are calling me a three year old and that's fine as long as you don't mind violating the new non-Statler and Waldorf rules



And, for the record, I was clearly insulting your argument and not you.
s2dbaker
s2dbaker
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March 1st, 2014 at 6:29:19 AM permalink
Quote: AxiomOfChoice

The part what you said that was incorrect was that zeta(-1) = 1+2+3+4+5+...

Well then I'll try to understand. Let me work through it. We'll take it one step at a time. Is the Zeta function defined as:

ζ(n) = 1/1n + 1/2n + 1/3n + 1/4n + 1/5n + 1/6n + ...
Someday, joor goin' to see the name of Googie Gomez in lights and joor goin' to say to joorself, "Was that her?" and then joor goin' to answer to joorself, "That was her!" But you know somethin' mister? I was always her yuss nobody knows it! - Googie Gomez
MangoJ
MangoJ
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March 1st, 2014 at 6:29:48 AM permalink
So what lacks in this discussion is a proper generalization of infinite sums - those that cover converging sums as well as diverging sums in a unified, consistent way.
Right ? Wrong ?
AxiomOfChoice
AxiomOfChoice
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March 1st, 2014 at 12:05:15 PM permalink
Quote: MangoJ

So what lacks in this discussion is a proper generalization of infinite sums - those that cover converging sums as well as diverging sums in a unified, consistent way.
Right ? Wrong ?



There is not one unique way to do that. You can use different methods to assign different values to divergent sums, but that is not the same thing as saying that the series sums to that value.

In other words, you can redefine the term "sum" to mean something completely different, which may be useful for some purposes, but it's important to realize that you are using a different definition of the word than anyone else in this case.
24Bingo
24Bingo
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March 1st, 2014 at 12:07:53 PM permalink
Quote: s2dbaker

Well then I'll try to understand. Let me work through it. We'll take it one step at a time. Is the Zeta function defined as:

ζ(n) = 1/1n + 1/2n + 1/3n + 1/4n + 1/5n + 1/6n + ...



No.

It's defined as the analytic extension of the function defined by that sum, which is itself defined only for numbers with real parts strictly greater than 1. That defines the function over enough points that there's a unique analytic function equal to it at every point, but it's not the same function because this one is defined at points where the other isn't.

Let's go back to the definition of an infinite sum: a limit. Namely, the limit of the partials as the number of terms summed goes to infinity. This means that "for every epsilon, there is a delta" - or rather, in this case, for every epsilon, there is an N - such that for any radius epsilon from the sum, there is an N such that any number of terms greater than N produces a partial sum within that radius. When the very first partial is over a full unit from the "sum," and the distance shoots up quadratically without end, something's gone wrong.

In just the same way that it's trivial that .999... = 1 once the definition of that ellipsis is properly understood, it's trivial that this sum diverges once you have a coherent idea of what an infinite sum is; math is all about such coherent, rigorous ideas, not silly tricks.

The extension to that family of series provided by the zeta function is useful in both string theory and (rather infamously) the study of primes, but that doesn't mean they're one and the same, only that some properties of one carry over to the other.
The trick to poker is learning not to beat yourself up for your mistakes too much, and certainly not too little, but just the right amount.
AxiomOfChoice
AxiomOfChoice
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March 1st, 2014 at 12:26:20 PM permalink
Quote: s2dbaker

Well then I'll try to understand. Let me work through it. We'll take it one step at a time. Is the Zeta function defined as:

ζ(n) = 1/1n + 1/2n + 1/3n + 1/4n + 1/5n + 1/6n + ...



Oh, I missed this post of yours. Bingo answered it perfectly. But, in short, the answer to your question is "no".

Do you really not understand the difference between the statements:

1. "The zeta function is equal to that sum", and
2. "The zeta function is equal to that sum when it converges"?

You are saying the first statement, but the first statement is incorrect. The 2nd statement is correct.
s2dbaker
s2dbaker
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March 2nd, 2014 at 5:49:06 AM permalink
Quote: AxiomOfChoice

Oh, I missed this post of yours. Bingo answered it perfectly. But, in short, the answer to your question is "no".

Do you really not understand the difference between the statements:

1. "The zeta function is equal to that sum", and
2. "The zeta function is equal to that sum when it converges"?

You are saying the first statement, but the first statement is incorrect. The 2nd statement is correct.

Then I'm just missing the second definition of the Zeta function. It might help me to understand better if I knew the definition of the Zeta function that WolframAlpha uses when n is less than 1. Anyone know what that is?
Someday, joor goin' to see the name of Googie Gomez in lights and joor goin' to say to joorself, "Was that her?" and then joor goin' to answer to joorself, "That was her!" But you know somethin' mister? I was always her yuss nobody knows it! - Googie Gomez

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