1+2+3+4+...(to infinity) = -1/12
Here's the thing.
You can look at it, and think it's just a bunch of trickery, and that it shows how math and science are disconnected from the world. Or, you can look at it and believe it, and doubt your perception of the world. "We're not making physical infinities in nature.... Intuitively, you want to stop the sequence, and the minute you stop the sequence... you've got a big number."
Me, I believe it.
Starting from false premises allows any silly conclusion.
The sum 1 - 1 + 1 - 1 + 1 ... is NOT equal to .5 . It is something that does not exist or, in math parlance, is undefined.
I'm not a mathematician, you probably know a lot more about it than I do. It appears that both statements can be proven. However, if the equation is useful in physics, then there must be real world justification for the result.
Quote: MoscaThe sum of 1 + 2 + 3 + 4 + 5 + ... until infinity is somehow -1/12
Here's the thing.
You can look at it, and think it's just a bunch of trickery, and that it shows how math and science are disconnected from the world. Or, you can look at it and believe it, and doubt your perception of the world. "We're not making physical infinities in nature.... Intuitively, you want to stop the sequence, and the minute you stop the sequence... you've got a big number."
Me, I believe it.
1 + 2 + 3 + 4 + .... is undefined. It doesn't equal anything. That's all that there is to it. Infinite sums are only defined when they converge.
X = 1-2+4-8...
2X = 2-4+8-16...
3X = (1-2+4-8...)+(2-4+8...) = 1
X = 1/3.
However how can that be given all the values are integers!!
From my own work the best parallel is the Bayesian and Frequentist viewpoints in statistics. Both have value, but each side will only admit that about the other behind closed doors.
Quote: charliepatrickIf I remember correctly you can't reorder sequences which don't converge, otherwise you can prove lots of weird things (and once 1=2 you're kind of stuck!)
X = 1-2+4-8...
2X = 2-4+8-16...
3X = (1-2+4-8...)+(2-4+8...) = 1
X = 1/3.
However how can that be given all the values are integers!!
X = 1 + 2 + 3 + ... (all positive numbers)
2*X = 2 + 4 + 6 + ... (double them to get all even numbers)
2*X - X = X = 1 + 3 + 5 + 7 + ... (remove all even numbers from X)
1 + 3 + 5 + 7 + ... = 1 + 2 + 3 + 4 + ... (so sum of all odd numbers = sum of all numbers)
0 = 2 + 4 + 6 + 8 + ... (so sum of all even numbers = 0)
0 = 2*X (since 2*X = sum of all even numbers)
X = 0
Once again, it's undefined. If you make the incorrect assumption that it's defined, you can prove whatever you want. This isn't particularly interesting. False ==> True follows in one step from an axiom of logic.
Quote: charliepatrickIf I remember correctly you can't reorder sequences which don't converge, otherwise you can prove lots of weird things (and once 1=2 you're kind of stuck!)
X = 1-2+4-8...
2X = 2-4+8-16...
3X = (1-2+4-8...)+(2-4+8...) = 1
X = 1/3.
However how can that be given all the values are integers!!
Its worse then that you can't reorder any sequence which is not absolutely convergent. For instance the alternating harmonic series 1-1/2+1/3-+... converges but if you allow rearrangement it can be shown to converge to any real number.
Quote: AxiomOfChoiceX = 1 + 2 + 3 + ... (all positive numbers)
2*X = 2 + 4 + 6 + ... (double them to get all even numbers)
2*X - X = X = 1 + 3 + 5 + 7 + ... (remove all even numbers from X)
1 + 3 + 5 + 7 + ... = 1 + 2 + 3 + 4 + ... (so sum of all odd numbers = sum of all numbers)
0 = 2 + 4 + 6 + 8 + ... (so sum of all even numbers = 0)
0 = 2*X (since 2*X = sum of all even numbers)
X = 0
Once again, it's undefined. If you make the incorrect assumption that it's defined, you can prove whatever you want. This isn't particularly interesting. False ==> True follows in one step from an axiom of logic.
Yeah called the principle of explosion in logic. Decided to look it up on wikipedia to see how they explained it and I love the example they give. Argument follows if you assume both the statements "All lemons are yellow" and "Not all lemons are yellow"
We know that "All lemons are yellow" as it is defined to be true.
Therefore the statement that (“All lemons are yellow" OR "Santa Claus exists”) must also be true, since the first part is true.
However, if "Not all lemons are yellow" (and this is also defined to be true), Santa Claus must exist - otherwise statement 2 would be false. It has thus been "proven" that Santa Claus exists. The same could be applied to any assertion, including the statement "Santa Claus does not exist".
Not all values are integers, there's an infinity mixed in and that screws up everything.Quote: charliepatrickIf I remember correctly you can't reorder sequences which don't converge, otherwise you can prove lots of weird things (and once 1=2 you're kind of stuck!)
X = 1-2+4-8...
2X = 2-4+8-16...
3X = (1-2+4-8...)+(2-4+8...) = 1
X = 1/3.
However how can that be given all the values are integers!!
It's the same as:
S = 1-1+1-1+1-1+1-1..
1-S = 1-(1-1+1-1+1-1+1-1..)
or
1-S = 1-1+1-1+1-1+1-1..
or
1-S = S
Add S to both sides:
1=2S
Solve for S by dividing by 2
S = .5
I can understand that.
Quote: s2dbakerNot all values are integers, there's an infinity mixed in and that screws up everything.
It's the same as:
S = 1-1+1-1+1-1+1-1..
1-S = 1-(1-1+1-1+1-1+1-1..)
or
1-S = 1-1+1-1+1-1+1-1..
or
1-S = S
Add S to both sides:
1=2S
Solve for S by dividing by 2
S = .5
I can understand that.
You may think you understand it, but it's incorrect.
Again, it's undefined. It doesn't converge.
Quote: AxiomOfChoiceYou may think you understand it, but it's incorrect.
Again, it's undefined. It doesn't converge.
Yep interestingly enough pretty sure that series was on my first final in analysis show that 1-1+1-1+-... did not converge.
By averaging the sums, the result converges and gets you the pseudo limit of .5Quote: AxiomOfChoiceYou may think you understand it, but it's incorrect.
Again, it's undefined. It doesn't converge.
(please don't say that .333...*3<>1)
Quote: s2dbakerBy averaging the sums, the result converges and gets you the pseudo limit of .5
(please don't say that .333...*3<>1)
No, I am saying that 1-1+1-1+1-1... doesn't exist. It just doesn't. You may as well argue about 1/0 is.
This is mathematics. You can't just make stuff up and pretend that it's well defined. There are no axioms which define arbitrary infinite sums. Either your infinite sum converges or it's undefined. There is no middle ground.
Quote: s2dbakerBy averaging the sums, the result converges and gets you the pseudo limit of .5
(please don't say that .333...*3<>1)
No it doesn't converge. You can show it doesn't converge since convergence implies for every epsilon there exist an N such that if m>N |X_m-X|<epsilon where X_m is the mth partial sum. I set epsilon as 1/4. There is no such N then since the distance from the partial sums and 1/2 is always a distance of 1/2 away. Hence the partial sums don't converge and hence the sum doesn't converge.
I see math as an analog of the world. Sometimes when a result doesn't appear to make sense, it should be set aside until something hooks up to it from another direction. String theory doesn't make sense in helping to understand the left wing lock, but it's still there.
Quote: MoscaThe point is that the answer of -1/12 has a useful, practical application in string theory; it answers questions about the world. Therefore under some circumstances, at some level, -1/12 works.
I see math as an analog of the world. Sometimes when a result doesn't appear to make sense, it should be set aside until something hooks up to it from another direction. String theory doesn't make sense in helping to understand the left wing lock, but it's still there.
None of this changes the fact that the sum is undefined, and that using the same techniques it's possible to get other answers.
Quote: AxiomOfChoiceNone of this changes the fact that the sum is undefined, and that using the same techniques it's possible to get other answers.
Sometimes it is undefined, and sometimes it isn't.
WolframAlpha disagrees, Zeta(-1)Quote: AxiomOfChoiceNone of this changes the fact that the sum is undefined.
Quote: s2dbakerWolframAlpha disagrees, Zeta(-1)
That is not the value of 1+2+3+... the zeta function is only defined that way when the series converges. There is an analytic continuation of the zeta function and with the analytic continuation zeta(-1)=-1/12 but that is something entirely different. Similarly 1+3+9+...=/=-1/2 even though as a geometric series it would be 1/(1-3). The problem is these things are not defined if the series doesn't converge.
Quote: AxelWolfWill any of this help me beat casinos? Or disprove someones lame system?
No. But it's in the math forum, so it doesn't need to be about gambling.
Seriously, math is an interesting field of study. Useful, but also so much more. Don't treat it like a hammer :)
I have no doubt this is the place to discuss it, but it sounds like something I can't use, as do many other things, but this is not something, I would or could have an opinion about, not even one I would like.Quote: AxiomOfChoiceNo. But it's in the math forum, so it doesn't need to be about gambling.
Seriously, math is an interesting field of study. Useful, but also so much more. Don't treat it like a hammer :)
So Zeta(-1) doesn't equal -1/12?Quote: TwirdmanQuote: s2dbakerWolframAlpha disagrees, Zeta(-1)
That is not the value of 1+2+3+... the zeta function is only defined that way when the series converges. There is an analytic continuation of the zeta function and with the analytic continuation zeta(-1)=-1/12 but that is something entirely different. Similarly 1+3+9+...=/=-1/2 even though as a geometric series it would be 1/(1-3). The problem is these things are not defined if the series doesn't converge.
String theory dosn't make sense, period. No experimental result EVER in 40 years? Man, any other field of study would have dumped it long ago.Quote: MoscaThe point is that the answer of -1/12 has a useful, practical application in string theory; it answers questions about the world. Therefore under some circumstances, at some level, -1/12 works.
I see math as an analog of the world. Sometimes when a result doesn't appear to make sense, it should be set aside until something hooks up to it from another direction. String theory doesn't make sense in helping to understand the left wing lock, but it's still there.
String theory is like neoclassical microeconomics. Seductive as a pure math model, but with no practical relevance. So, excuse me, saying that -1/12 has practical application in string theory is akin to saying that a white unicorn has practical application in Greek mythology.
Quote: s2dbakerSo Zeta(-1) doesn't equal -1/12?Quote: TwirdmanQuote: s2dbakerWolframAlpha disagrees, Zeta(-1)
That is not the value of 1+2+3+... the zeta function is only defined that way when the series converges. There is an analytic continuation of the zeta function and with the analytic continuation zeta(-1)=-1/12 but that is something entirely different. Similarly 1+3+9+...=/=-1/2 even though as a geometric series it would be 1/(1-3). The problem is these things are not defined if the series doesn't converge.
No I saidly that zeta(-1)=-1/12 as an analytic extension of the zeta function. Just like gamma(3/2)=1/2*sqrt(pi) that does not mean (1/2)! =1/2*sqrt(pi) he gamma function is an analytic extension of the function (x-1)!. But factorial is still only defined for integers. Similarly the sum form of the zeta function is only defined for when the real part of the argument is greater then 1. You can do an analytic extension but again that does not mean 1+2+3+...=-1/12 it just means the analytic extension of the zeta function taken at an argument of -1.
Quote: kubikulannString theory dosn't make sense, period. No experimental result EVER in 40 years? Man, any other field of study would have dumped it long ago.
String theory is like neoclassical microeconomics. Seductive as a pure math model, but with no practical relevance. So, excuse me, saying that -1/12 has practical application in string theory is akin to saying that a white unicorn has practical application in Greek mythology.
Neoclassical microeconomics has no practical relevance? Whaaaaaaa?
Your signature discredits you for discussion on economics, sorry.Quote: bigfoot66Neoclassical microeconomics has no practical relevance? Whaaaaaaa?
No, not really.Quote: BuzzardRand Paul thinks..
Speaking of ressurecting threads from the dead..Quote: TwirdmanQuote: s2dbakerSo Zeta(-1) doesn't equal -1/12?Quote: TwirdmanQuote: s2dbakerWolframAlpha disagrees, Zeta(-1)
That is not the value of 1+2+3+... the zeta function is only defined that way when the series converges. There is an analytic continuation of the zeta function and with the analytic continuation zeta(-1)=-1/12 but that is something entirely different. Similarly 1+3+9+...=/=-1/2 even though as a geometric series it would be 1/(1-3). The problem is these things are not defined if the series doesn't converge.
No I saidly that zeta(-1)=-1/12 as an analytic extension of the zeta function. Just like gamma(3/2)=1/2*sqrt(pi) that does not mean (1/2)! =1/2*sqrt(pi) he gamma function is an analytic extension of the function (x-1)!. But factorial is still only defined for integers. Similarly the sum form of the zeta function is only defined for when the real part of the argument is greater then 1. You can do an analytic extension but again that does not mean 1+2+3+...=-1/12 it just means the analytic extension of the zeta function taken at an argument of -1.
ζ(n) = 1/1n + 1/2n + 1/3n + 1/4n + 1/5n + 1/6n + ...
ζ(-1) = 1/1-1 + 1/2-1 + 1/3-1 + 1/4-1 + 1/5-1 + 1/6-1 + ...
Given that 1/x-1 = x Then
ζ(-1) = 1 + 2 + 3 + 4 + 5 + 6 + ...
Yet WolframAlpha says that ζ(-1) = -1/12
Discuss
What Twirdman says still applies. You cannot just ignore the parts of the definition that you don't like.
WolframAlpha can so why can't I?Quote: AxiomOfChoiceWe have had this discussion before.
What Twirdman says still applies. You cannot just ignore the parts of the definition that you don't like.
Quote: s2dbakerWolframAlpha can so why can't I?
The problem is that you think that, for all s, zeta(s) = the sum from n = 1 to infinity of 1/n^s. And that is wrong. zeta(s) only equals that sum when that sum is defined, and that sum is only defined when it converges, because infinite sums are only defined when they converge.
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?
Quote: aceofspadesAnd if the sky was wet it would be the ocean
The sky is wet.
Ruh-roh.
... OOOPs forget that. Just remembered my Aunt Sally UGH !
Quote: FaceThe sky is wet.
Ruh-roh.
In Cali, sounds like they think it IS the ocean. Take care, left-coasters.
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 + ... = -1/12Quote: AxiomOfChoiceConsider 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?
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?
YesQuote: s2dbakerHave I said anything incorrect?
YesQuote:Did I get any analysis of the Zeta function wrong?
No, that part is right.Quote:Is WolframAlpha calculating the Zeta function at -1 = -1/12?
That is also right.Quote:If y = x and z = x doesn't it follow that y = z?
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?
This is the madness that lay on that road.
Quote: s2dbakerIt'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.
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:Quote: AxiomOfChoiceThe part what you said that was incorrect was that zeta(-1) = 1+2+3+4+5+...
ζ(n) = 1/1n + 1/2n + 1/3n + 1/4n + 1/5n + 1/6n + ...
Right ? Wrong ?
Quote: MangoJSo 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.
Quote: s2dbakerWell 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.
Quote: s2dbakerWell 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.
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?Quote: AxiomOfChoiceOh, 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.
