The sum of all the positive integers is not -1/12(blog.rongarret.info)
blog.rongarret.info
The sum of all the positive integers is not -1/12
http://blog.rongarret.info/2014/01/no-sum-of-all-positive-integers-is-not.html
10 comments
Two additional details:
* Terence Tao got the Fields Medal in 2006 http://en.wikipedia.org/wiki/Terence_Tao
* ColinWright's post is now undead.
* Terence Tao got the Fields Medal in 2006 http://en.wikipedia.org/wiki/Terence_Tao
* ColinWright's post is now undead.
The really interesting point, or at least what struck me when I first heard about this, is that there are alternative, useful, counterintuitive notions of ideas things that folks might think they have some intuitive grasp of. There's more to math (and not only the parts you _know_ you don't know about) than is dreamt of in one's philosophy.
Math is mostly about exploring the consequences of alternative definitions for notions we take for granted. In this case, it is the equality notion that is used in a different fashion
> Step 1: Let S1 = 1 - 1 + 1 - 1 ...
Because it's not so clearly stated in the article, step 1 is already flawed.
> the algebraic rules that apply to regular numbers do not apply to non-converging infinite sums
I'm sure the author knows this but they don't even apply to all converging infinite sums. (Only absolutely converging ones.) E.g. 1 - 1/2 + 1/3 - 1/4 ... converges to a value, but you can also re-arrange the terms to get any value you want.
Because it's not so clearly stated in the article, step 1 is already flawed.
> the algebraic rules that apply to regular numbers do not apply to non-converging infinite sums
I'm sure the author knows this but they don't even apply to all converging infinite sums. (Only absolutely converging ones.) E.g. 1 - 1/2 + 1/3 - 1/4 ... converges to a value, but you can also re-arrange the terms to get any value you want.
And, strangely enough, the value that 1 - 1/2 + 1/3 - 1/4 + 1/5... converges to if evaluated left to right is the natural log of 2. It doesn't converge very quickly, though. It takes a billion terms to reach just six decimal places of precision.
Actually, it only takes a million terms to reach six decimal places.
Why is it strange that this sum is ln 2?
Why is it strange that this sum is ln 2?
It's strange that it converges to a number we can describe succinctly, as opposed to a random number that has no relation to anything else.
Oh, this is actually surprisingly easy to see with elementary calculus knowledge.
High school algebra tells us that
1 - x + x^2 - ... + (-1)^n x^n = 1 + (-x) + (-x)^2 + ... + (-x)^n = (1 - (-x)^{n+1})/(1-(-x)).
Thus, for the infinite sum we have:
1 - x + x^2 - x^3 + ... = lim_{n -> inf} (1 - x + x^2 - ... + (-1)^n x^n) = lim_{n -> inf} (1 - (-x)^{n+1})/(1-(-x)) = 1/(1+x).
But then if you integrate the infinite sum term by term, you'll get:
\int (1 - x + x^2 - x^3 + ...) = x - x^2/2 + x^3/3 - ...
On the other hand, if you integrate 1/(1+x), you'll get precisely log(1+x). Now, one would want to argue that:
log(1+x) = x - x^2/2 + x^3/3 - ...
and here this is actually true, but in general this may not be true that the integral of the sum of infinite series is equal to sum of the integrals of each term -- what you need here is the notion of uniform convergence, but fortunately 1 - x + x^2 - ... converges uniformly.
But, since Leonard Euler wouldn't really be pedantic about stuff like this, as long as the result is correct, neither should the occasional math hacker. Thus, we put x = 1 in both sides of the equality and we get 1 - 1/2 + 1/3 - ... = log(2).
Apart from integrating term by term, there's one more problem with above reasoning, and there are bonus points for people who notice that (hint: uggc://ra.jvxvcrqvn.bet/jvxv/Nory'f_gurberz )
High school algebra tells us that
1 - x + x^2 - ... + (-1)^n x^n = 1 + (-x) + (-x)^2 + ... + (-x)^n = (1 - (-x)^{n+1})/(1-(-x)).
Thus, for the infinite sum we have:
1 - x + x^2 - x^3 + ... = lim_{n -> inf} (1 - x + x^2 - ... + (-1)^n x^n) = lim_{n -> inf} (1 - (-x)^{n+1})/(1-(-x)) = 1/(1+x).
But then if you integrate the infinite sum term by term, you'll get:
\int (1 - x + x^2 - x^3 + ...) = x - x^2/2 + x^3/3 - ...
On the other hand, if you integrate 1/(1+x), you'll get precisely log(1+x). Now, one would want to argue that:
log(1+x) = x - x^2/2 + x^3/3 - ...
and here this is actually true, but in general this may not be true that the integral of the sum of infinite series is equal to sum of the integrals of each term -- what you need here is the notion of uniform convergence, but fortunately 1 - x + x^2 - ... converges uniformly.
But, since Leonard Euler wouldn't really be pedantic about stuff like this, as long as the result is correct, neither should the occasional math hacker. Thus, we put x = 1 in both sides of the equality and we get 1 - 1/2 + 1/3 - ... = log(2).
Apart from integrating term by term, there's one more problem with above reasoning, and there are bonus points for people who notice that (hint: uggc://ra.jvxvcrqvn.bet/jvxv/Nory'f_gurberz )
The natural logarithm of 2 is only described so succinctly because that number (and related ones) were so common and useful that we introduced a shorthand to denote the limiting process which tends to it. I think you'd find that, were the definition expanded completely, that its description is not so succinct after all. :)
Now that you mention it, I seem to remember us working through examples of this in our elementary calculus class. But I had also forgotten that they needed to be absolutely converging.
Actually, I didn't know that. And it surprises me, actually. Intuitively I would have guessed that converging infinite sums would be well-behaved under normal algebraic manipulations. But you learn something new every day :-)
Can you point me to a reference?
Can you point me to a reference?
The basic idea is pretty simple. If Σa_i is conditionally convergent — that is, Σa_i converges but Σ|a_i| does not — then there are an infinite number of both positive and negative a_i. See [1] below for a hint on how to prove this; it's straightforward.
Given that, let's say we want to rearrange the a_i in Σa_i to converge to some arbitrary real number M. Take all the positive a_i in order and call that set A_+ and take all the negative a_i in order and call that set A_-. From [1] we know that both A_+ and A_- have an infinite number of elements.
You then construct a new series like so. Take the fewest elements from A_+ (in the order they appear in the original series) such that their sum is greater than M. Next, add to that sum the fewest number of elements from A_- (also in order) such that this new sum is smaller than M. Go back to taking elements from A_+ until you're just larger M and then back to taking elements from A_- until you're just under M. Repeat this ping-pong maneuver ad infinitum to construct a new series.
You can prove that this new series converges (conditionally) to M.
[1]: Here's a hint. What can we say about Σ|a_i| if we know that Σa_i converges but has only a finite number of negative terms?
Given that, let's say we want to rearrange the a_i in Σa_i to converge to some arbitrary real number M. Take all the positive a_i in order and call that set A_+ and take all the negative a_i in order and call that set A_-. From [1] we know that both A_+ and A_- have an infinite number of elements.
You then construct a new series like so. Take the fewest elements from A_+ (in the order they appear in the original series) such that their sum is greater than M. Next, add to that sum the fewest number of elements from A_- (also in order) such that this new sum is smaller than M. Go back to taking elements from A_+ until you're just larger M and then back to taking elements from A_- until you're just under M. Repeat this ping-pong maneuver ad infinitum to construct a new series.
You can prove that this new series converges (conditionally) to M.
[1]: Here's a hint. What can we say about Σ|a_i| if we know that Σa_i converges but has only a finite number of negative terms?
> If Σa_i is conditionally convergent then there are an infinite number of both positive and negative a_i.
That's true, but that's not enough for your proof -- e.g. \sum (-1)^n/n^2 also has infinitely many positive and negative terms, but converges absolutely.
What you need is that sum of all elements in at least one of A_+, A_- (thus also the sum of all elements in the second one) diverges (the order you take the sum in doesn't matter here, as you surely know).
That's true, but that's not enough for your proof -- e.g. \sum (-1)^n/n^2 also has infinitely many positive and negative terms, but converges absolutely.
What you need is that sum of all elements in at least one of A_+, A_- (thus also the sum of all elements in the second one) diverges (the order you take the sum in doesn't matter here, as you surely know).
I left out lots of other details, too, because it was meant to be an outline for the original commenter, not a proof. In any case,
What's more, if Σa_i is conditionally convergent then the sums of both A_+ and A_- diverge. You're right that one has to use this fact in the full proof.
\sum_{i=1}^{\infty} \frac{(-1)^i}{i^2}
What's more, if Σa_i is conditionally convergent then the sums of both A_+ and A_- diverge. You're right that one has to use this fact in the full proof.
http://mathworld.wolfram.com/ConditionalConvergence.html gives a nice intuitive explanation:
"A series is said to be conditionally convergent iff it is convergent, the series of its positive terms diverges to positive infinity, and the series of its negative terms diverges to negative infinity. […] The Riemann series theorem states that, by a suitable rearrangement of terms, a conditionally convergent series may be made to converge to any desired value, or to diverge. The Riemann series theorem can be proved by first taking just enough positive terms to exceed the desired limit, then taking just enough negative terms to go below the desired limit, and iterating this procedure. Since the terms of the original series tend to zero, the rearranged series converges to the desired limit. A slight variation works to make the new series diverge to positive infinity or to negative infinity."
"A series is said to be conditionally convergent iff it is convergent, the series of its positive terms diverges to positive infinity, and the series of its negative terms diverges to negative infinity. […] The Riemann series theorem states that, by a suitable rearrangement of terms, a conditionally convergent series may be made to converge to any desired value, or to diverge. The Riemann series theorem can be proved by first taking just enough positive terms to exceed the desired limit, then taking just enough negative terms to go below the desired limit, and iterating this procedure. Since the terms of the original series tend to zero, the rearranged series converges to the desired limit. A slight variation works to make the new series diverge to positive infinity or to negative infinity."
The problem is that 1 - 1 + 1 - 1 ... is not convergent. The sum is not infinite, but it’s neither a finite number, simply “not exist”.
The correct method is to calculate the partial sum of the first N term: S_N=sum_{n=1}_N (-1)^n. When N is even the amount of 1 and -1 are equal and the partial sum is 0. When N is odd then there is an additional 1 and the partial sum is 1. So the partial sum has no limit and then 1 - 1 + 1 - 1 ... is not convergent.
Conditionally convergent series like 1-1/2+1/3-1/4+1/5-1/6+ ... =ln(2) are much better. (The problem is that 1+1/2+1/3+1/4+1/5+1/6+ ... = infinity.) You can operate with them almost freely, but you can’t rearrange the order of an infinite number of term.
The unconditionally convergent series like 1+1/4+1/9+1/16+...=pi^2/6 are almost like finite summations. You can do whatever sensible operation you want and the result will be correct.
The correct method is to calculate the partial sum of the first N term: S_N=sum_{n=1}_N (-1)^n. When N is even the amount of 1 and -1 are equal and the partial sum is 0. When N is odd then there is an additional 1 and the partial sum is 1. So the partial sum has no limit and then 1 - 1 + 1 - 1 ... is not convergent.
Conditionally convergent series like 1-1/2+1/3-1/4+1/5-1/6+ ... =ln(2) are much better. (The problem is that 1+1/2+1/3+1/4+1/5+1/6+ ... = infinity.) You can operate with them almost freely, but you can’t rearrange the order of an infinite number of term.
The unconditionally convergent series like 1+1/4+1/9+1/16+...=pi^2/6 are almost like finite summations. You can do whatever sensible operation you want and the result will be correct.
OK, but what you said was:
"they [normal rules of algebra] don't even apply to all converging infinite sums"
I know that 1 - 1 + 1... is not convergent, but that's a non-sequitur relative to the claim that you made.
> The unconditionally convergent series like 1+1/4+1/9+1/16+...=pi^2/6 are almost like finite summations. You can do whatever sensible operation you want and the result will be correct.
Yes, that's what I thought.
"they [normal rules of algebra] don't even apply to all converging infinite sums"
I know that 1 - 1 + 1... is not convergent, but that's a non-sequitur relative to the claim that you made.
> The unconditionally convergent series like 1+1/4+1/9+1/16+...=pi^2/6 are almost like finite summations. You can do whatever sensible operation you want and the result will be correct.
Yes, that's what I thought.
I guess any introduction to real analysis should cover that.
http://www.amazon.com/Calculus-Vol-One-Variable-Introduction...
Unfortunately, I don't know any online references out of the top of my mind.
http://www.amazon.com/Calculus-Vol-One-Variable-Introduction...
Unfortunately, I don't know any online references out of the top of my mind.
Look under "examples" here: http://en.wikipedia.org/wiki/Riemann_series_theorem
Oh for the love of - does this really need to be said?
Really?
I mean, of course it's false! It's an instructive example of how apparently reasonable things go wrong, and why you sometimes really, really need to pay attention to the details.
Do we really need to be told that the sum of all the positive integers is a negative fraction? Of course we don't.
I despair sometimes, I really do. I need to go away and spend some time in my happy place.
Really?
I mean, of course it's false! It's an instructive example of how apparently reasonable things go wrong, and why you sometimes really, really need to pay attention to the details.
Do we really need to be told that the sum of all the positive integers is a negative fraction? Of course we don't.
I despair sometimes, I really do. I need to go away and spend some time in my happy place.
Well... I don't understand the mindset of "of course it's false!" The first question that comes to my mind is "can I use normal-seeming algebra to make 1 + 2 + 3 + ... sum to a number other than -1/12"? If it's trivial to make it come out to anything at all, we can chalk it up to apparently reasonable things going wrong; if it seems to be mysteriously easier to add it all up to -1/12, it's worth investigating why that might be.
I mean, we can apply the good old formula for the geometric series,
I mean, we can apply the good old formula for the geometric series,
1 + x + x^2 + x^3 + ... = 1/(1-x)
1 + 2 + 4 + 8 + 16 + ... = -1
Ha, what a coincidence.
It won't work for decimal 10s complement numbers:
For x=10 we get
It won't work for decimal 10s complement numbers:
For x=10 we get
1 + 10 + 100 + 1000 + 10000 + ... = -1/9
9 + 90 + 900 + 9000 + 90000 + …
9*(1 + 10 + 100 + 1000 + 10000 + ...) = 9*(-1/9) = -1> Ha, what a coincidence.
> It won't work for decimal 10s complement numbers:
I don't get why you say this -- you go on to demonstrate the opposite?
This is just the other-side-of-the-decimal-point inverse of the well known 0.9999999... = 1 equality.
One of my favorite proofs for 0.9999... being equal to 1 goes like so: imagine subtracting it from 1. You'll get a number which has a 0 at every decimal place: 0.0000000.... Obviously, the number with a 0 in every decimal place is 0 itself.
Similarly, if you have your integer represented decimally as a bunch of coefficients (0 <= c < 10) of powers of 10, and all of the coefficients are 9, it's fairly straightforward to see that adding 1 will get you a new number for which all of the coefficients are 0. Since adding 1 to the original number gave us 0, we can treat the original number as -1.
> It won't work for decimal 10s complement numbers:
I don't get why you say this -- you go on to demonstrate the opposite?
This is just the other-side-of-the-decimal-point inverse of the well known 0.9999999... = 1 equality.
One of my favorite proofs for 0.9999... being equal to 1 goes like so: imagine subtracting it from 1. You'll get a number which has a 0 at every decimal place: 0.0000000.... Obviously, the number with a 0 in every decimal place is 0 itself.
Similarly, if you have your integer represented decimally as a bunch of coefficients (0 <= c < 10) of powers of 10, and all of the coefficients are 9, it's fairly straightforward to see that adding 1 will get you a new number for which all of the coefficients are 0. Since adding 1 to the original number gave us 0, we can treat the original number as -1.
I don't get why you say this
Just a narrative of my own experience thinking this through :-)
I thought "OMG" would tell you that, sorry.
Just a narrative of my own experience thinking this through :-)
I thought "OMG" would tell you that, sorry.
I think it needs to be said because the people promulgating the falsehood are legitimate mathematicians. And most of the time their blog has quality stuff. But they dropped the ball on this one.
Forgive me, but I really don't understand your point. Do you think they believe it?
This is proper geek fun with serious points underneath and a good smattering of why it's actually valid to consider these things. This isn't established people with proper reputations setting out to con the unsuspecting public. There is solid math going on here. See Terry Tao's recent blog post[0] about why we can and should play with these things.
And now it's late here and I'm going off-line for a while. As I said elsewhere, I need to detach for a bit.
[0] https://news.ycombinator.com/item?id=7078744
Added in edit: Just checking before signing off for the night and I see that the submission of Terry Tao's blog entry on this topic has been killed. By mods, flags, whatever, I don't know. here's the link[1] if you're interested, and this is me, signing off.
[1] http://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin...
This is proper geek fun with serious points underneath and a good smattering of why it's actually valid to consider these things. This isn't established people with proper reputations setting out to con the unsuspecting public. There is solid math going on here. See Terry Tao's recent blog post[0] about why we can and should play with these things.
And now it's late here and I'm going off-line for a while. As I said elsewhere, I need to detach for a bit.
[0] https://news.ycombinator.com/item?id=7078744
Added in edit: Just checking before signing off for the night and I see that the submission of Terry Tao's blog entry on this topic has been killed. By mods, flags, whatever, I don't know. here's the link[1] if you're interested, and this is me, signing off.
[1] http://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin...
> Do you think they believe it?
Depends on what you mean by "they" and "it". The numberphile folks undoubtedly know that they are playing fast and loose with the rules. The readers of Slate almost certainly don't know it.
> There is solid math going on here.
No, there isn't. There is solid math going on in complex analysis where you have concepts like analytic extensions that produce unintuitive but useful and (more importantly) consistent results. But that is NOT what this video is about. This video is about using high school math to "prove" a result that is simply not true under the rules of high school math. The only thing separating this from outright crackpottery is that the result they derive happens to look like one that can be legitimately derived under the rules of complex analysis and analytic extensions. But that's a mighty thin reed. It doesn't change the fact that they present the result as if it were true under the rules of high school math, and under those rules it isn't true.
EDIT: Terry Tao's article is excellent, and I am appalled (but, sadly, not surprised) that your submission was killed.
Hm, my own submission seems to have fallen off the front page awfully quickly.
EDIT2: I have been corresponding with an HN mod who informed me that my submission triggered the voting ring detector. (It was a false positive, which ought to worry someone at YC.) Also, the original Terry Tao submission has now been unkilled. I encourage you to upvote it.
Depends on what you mean by "they" and "it". The numberphile folks undoubtedly know that they are playing fast and loose with the rules. The readers of Slate almost certainly don't know it.
> There is solid math going on here.
No, there isn't. There is solid math going on in complex analysis where you have concepts like analytic extensions that produce unintuitive but useful and (more importantly) consistent results. But that is NOT what this video is about. This video is about using high school math to "prove" a result that is simply not true under the rules of high school math. The only thing separating this from outright crackpottery is that the result they derive happens to look like one that can be legitimately derived under the rules of complex analysis and analytic extensions. But that's a mighty thin reed. It doesn't change the fact that they present the result as if it were true under the rules of high school math, and under those rules it isn't true.
EDIT: Terry Tao's article is excellent, and I am appalled (but, sadly, not surprised) that your submission was killed.
Hm, my own submission seems to have fallen off the front page awfully quickly.
EDIT2: I have been corresponding with an HN mod who informed me that my submission triggered the voting ring detector. (It was a false positive, which ought to worry someone at YC.) Also, the original Terry Tao submission has now been unkilled. I encourage you to upvote it.
I hope that a good question makes you happy. I don’t know the details of the Ramanujan summation and don’t want to dig too deep in the subject now.
What is the Ramanujan summation of S_3 = 1+1+1+ ... ? (With a liberal interpretation of the Wikipedia article, I think it’s 0.)
Where is the fault in 0 = S_3-S_3 = S_3-shift(S_3)=1+0+0+0+...=1 ?
Is the shift a legal operation? When?
What is the Ramanujan summation of S_3 = 1+1+1+ ... ? (With a liberal interpretation of the Wikipedia article, I think it’s 0.)
Where is the fault in 0 = S_3-S_3 = S_3-shift(S_3)=1+0+0+0+...=1 ?
Is the shift a legal operation? When?
X - X = 0 only if X is finite. If X is infinite, X - X can be anything.
That’s what I secretly wished. If it where infinite then everything is explained.
But the Wikipedia article “said” that it is 0 (because 1^0+2^0+3^0+...=1+1+1, and 0 is even). But the [dead?!] article submitted by ColinWright from Tao’s blog says it’s -1/2. (I prefer not to disagree with Tao, just fixed Wikipedia.)
The problem is more deep. I should read the complete version of Tao’s article.
But the Wikipedia article “said” that it is 0 (because 1^0+2^0+3^0+...=1+1+1, and 0 is even). But the [dead?!] article submitted by ColinWright from Tao’s blog says it’s -1/2. (I prefer not to disagree with Tao, just fixed Wikipedia.)
The problem is more deep. I should read the complete version of Tao’s article.
> The problem is more deep.
You want to focus on the idea of analytic continuation. That's where the magic happens.
You want to focus on the idea of analytic continuation. That's where the magic happens.
the claim in the operations is that these sums have finite values.
Yes, but you cannot assume that the values are finite in the "proof" that they are finite. You have to prove that they are finite before you are allowed to manipulate them as finite quantities.
to prove by contradiction, that's exactly what you do. You assume they are finite, use allowed operations, show a contradiction.
G. Hardy had a hard time understanding it, and Ramanujan apparently knew it[1]. I'm not a mathematician, but if Hardy had a hard time understanding this at first glance, I don't think anyone should be ashamed for not knowing that the same of all numbers is -(1/12)
[1] http://books.google.cz/books?id=Of5G0r6DQiEC&pg=PA53&dq=grat...
[1] http://books.google.cz/books?id=Of5G0r6DQiEC&pg=PA53&dq=grat...
People believe witchcraft math because they don't understand things like e^pi*i - 1 = 0, so they have no faith in their own reasoning.
I didn't watch the numberphile video and just read this.
Can someone explain to me why I am allowed to add a padding zero at the start without taking it out at the end?
By adding it, I am giving the sum an offset, and I can kind of understand why adding zero is not a big deal, but they reasoned:
Then S1 + S1 = (1 - 1 + 1 - 1 ...) + (1 - 1 + 1 - 1 ...)
= (1 - 1 + 1 - 1 ...) + (0 + 1 - 1 + 1 - 1 ...)
= (1 + 0) + (1 - 1) + (1 - 1) ....
= 1 + 0 + 0 ... = 1
But I might as well leave the leading Zero out and argue that:
Then S1 + S1 = (1 - 1 + 1 - 1 ...) + (1 - 1 + 1 - 1 ...)
= (1 - 1 + 1 - 1 ...) + (1 - 1 + 1 - 1 ...)
= (1 + 1) + ( -1 - 1) + (1 + 1) ....
= (2 - 2) + (2 - 2) + (2 - 2)... = 0
Can someone explain to me why I am allowed to add a padding zero at the start without taking it out at the end?
By adding it, I am giving the sum an offset, and I can kind of understand why adding zero is not a big deal, but they reasoned:
Then S1 + S1 = (1 - 1 + 1 - 1 ...) + (1 - 1 + 1 - 1 ...)
= (1 - 1 + 1 - 1 ...) + (0 + 1 - 1 + 1 - 1 ...)
= (1 + 0) + (1 - 1) + (1 - 1) ....
= 1 + 0 + 0 ... = 1
But I might as well leave the leading Zero out and argue that:
Then S1 + S1 = (1 - 1 + 1 - 1 ...) + (1 - 1 + 1 - 1 ...)
= (1 - 1 + 1 - 1 ...) + (1 - 1 + 1 - 1 ...)
= (1 + 1) + ( -1 - 1) + (1 + 1) ....
= (2 - 2) + (2 - 2) + (2 - 2)... = 0
That's the point! Using the algebra from the article:
(1 - 1 + 1 - 1 ...) = 0 + (1 - 1 + 1 - 1 ...)
and therefore:
(1 - 1 + 1 - 1 ...) = (0 + 1 - 1 + 1 - 1 ...)
And as they're mathematically the same (again, in the article's algebra), why not replace one with the other?
The problem is that when you do, and you sum the two infinite series using the "zipping" method (as in the numberphile video) the two equations equate to different results.
(1 - 1 + 1 - 1 ...) = 0 + (1 - 1 + 1 - 1 ...)
and therefore:
(1 - 1 + 1 - 1 ...) = (0 + 1 - 1 + 1 - 1 ...)
And as they're mathematically the same (again, in the article's algebra), why not replace one with the other?
The problem is that when you do, and you sum the two infinite series using the "zipping" method (as in the numberphile video) the two equations equate to different results.
Alright, got it. Thanks!
> Can someone explain to me why I am allowed to add a padding zero at the start without taking it out at the end?
Because there is no "end". These are infinite series.
> But I might as well leave the leading Zero out and argue that:
Yes, exactly, that's the whole point. You cannot manipulate non-convergent series in this way if you want consistent results.
Because there is no "end". These are infinite series.
> But I might as well leave the leading Zero out and argue that:
Yes, exactly, that's the whole point. You cannot manipulate non-convergent series in this way if you want consistent results.
Cheers, thanks!
The answer is clearly 42. Monkeying with infinity allows you to derive any answer you like and I like 42.
I didn't feel like signing up for a Blogger account, but I still wanted to drop this here:
> My friend, Ray Sidney, with a PhD in mathematics from MIT explained away this nonsense rather succinctly.
> infinity + 7 = infinity
> If you minus infinity from both sides, you can tell the world 7 = 0!
http://en.wikipedia.org/wiki/Riemann_sphere#Arithmetic_opera...
infinity - infinity (along with a few others) is very often left undefined for this very reason--not to mention it lets you divide by zero without breaking math.
> My friend, Ray Sidney, with a PhD in mathematics from MIT explained away this nonsense rather succinctly.
> infinity + 7 = infinity
> If you minus infinity from both sides, you can tell the world 7 = 0!
http://en.wikipedia.org/wiki/Riemann_sphere#Arithmetic_opera...
infinity - infinity (along with a few others) is very often left undefined for this very reason--not to mention it lets you divide by zero without breaking math.
For the love of math, this!
= (1 - 1 + 1 - 1 ...) +
(0 + 1 - 1 + 1 - 1 ...)
= (1 + 0) + (1 - 1) + (1 - 1) ....
You're just adding the elements of the two series pair-wise (which "works" because addition is commutative and associative).
Here are a few additional intermediate steps:
Here are a few additional intermediate steps:
= (1 + -1 + 1 + -1 ...)
+ (0 + 1 + -1 + 1 ...)
= (1 + 0) + (-1 + 1) + (1 + -1) + (-1 + 1) ...[deleted]
Just add 0 in front of the second sum. 0 + x = x, so it is the same.
This is a non-proof based only off of notational styles.
Let s be a series , with s_n = sum (1 to n, 1) = n if S3 exists (i.e. lim n->infinity s_n exists), then the series (s-s) (where (s-s)_n = s_n - s_n) converges , and it converges towards S3 - S3 (=0).
> = (1 + 1 + 1 + 1 ...) - (0 + 1 + 1 + 1 ...) the series whose sum he's describing here is not (s-s), but another one entirely : u where u_1 = s_1, u_n+1= s_n+1-s_n
These are different series, so it is entirely reasonable for them to converge at different values, 1 doesn't equal 0.
The "real" reason this result isn't what we think it is(apart from "infinite sums are different" argument, which is a non-answer):
>Step 1: Let S = 1 + 2 + 3 + 4 ... What you're doing here firstly, is saying that "I assume that this sum converges, let S be the value it converges to".
So you end up proving (if the other steps weren't also flawed) that S = -1/12, all you're saying is that if the sum exists, then it is -1/12.
The issue here depends on what sort of thing you're working on. The "classic" definition of a limit (convergence of a series) does not work here, because you can prove that for all n, the sum of n numbers will always be at least -1/12 away from -1/12 (on account of it being positive), so it can't converge to -1/12 (hence S not existing).
However, there are techniques for assigning limits to divergent sums. these summability methods will give the same result as the classic analysis for convergent series, but will also give values for some divergent series.
It is similar to analytic continuation( f(x) doesn't exist, but a limit exists in x both from the left and to the right (named y), so we sometimes say that f(x)=y ), in that it allows us to extend the resolution domain slightly. But the classic definition no longer works.
The one used here is zeta function regularization, which consists in the following:
you have a series a , and a function Za(s) = (1/a1^s)+(1/a2^s)+(1/a3^s)+(1/a4^s)+.....
Za is only defined for certain values of s depending on the series. But for a serie representing a convergent sum, we know that Za(-1)= a1+a2+a3+... exists. So this method will give the same limit for convergent sums as the classical method.
For a series with a diverging sum (a_n)=n, Za(-1) doesn't exist, but the limit in -1 exists from both ends (by analytic continuation), so we extend the domain of Za to -1 by the value of this "limit" : -1/12
The article mentions Ramanujan summation, but Zeta regularization is actually a much more useful tool. In
Let s be a series , with s_n = sum (1 to n, 1) = n if S3 exists (i.e. lim n->infinity s_n exists), then the series (s-s) (where (s-s)_n = s_n - s_n) converges , and it converges towards S3 - S3 (=0).
> = (1 + 1 + 1 + 1 ...) - (0 + 1 + 1 + 1 ...) the series whose sum he's describing here is not (s-s), but another one entirely : u where u_1 = s_1, u_n+1= s_n+1-s_n
These are different series, so it is entirely reasonable for them to converge at different values, 1 doesn't equal 0.
The "real" reason this result isn't what we think it is(apart from "infinite sums are different" argument, which is a non-answer):
>Step 1: Let S = 1 + 2 + 3 + 4 ... What you're doing here firstly, is saying that "I assume that this sum converges, let S be the value it converges to".
So you end up proving (if the other steps weren't also flawed) that S = -1/12, all you're saying is that if the sum exists, then it is -1/12.
The issue here depends on what sort of thing you're working on. The "classic" definition of a limit (convergence of a series) does not work here, because you can prove that for all n, the sum of n numbers will always be at least -1/12 away from -1/12 (on account of it being positive), so it can't converge to -1/12 (hence S not existing).
However, there are techniques for assigning limits to divergent sums. these summability methods will give the same result as the classic analysis for convergent series, but will also give values for some divergent series.
It is similar to analytic continuation( f(x) doesn't exist, but a limit exists in x both from the left and to the right (named y), so we sometimes say that f(x)=y ), in that it allows us to extend the resolution domain slightly. But the classic definition no longer works.
The one used here is zeta function regularization, which consists in the following:
you have a series a , and a function Za(s) = (1/a1^s)+(1/a2^s)+(1/a3^s)+(1/a4^s)+.....
Za is only defined for certain values of s depending on the series. But for a serie representing a convergent sum, we know that Za(-1)= a1+a2+a3+... exists. So this method will give the same limit for convergent sums as the classical method.
For a series with a diverging sum (a_n)=n, Za(-1) doesn't exist, but the limit in -1 exists from both ends (by analytic continuation), so we extend the domain of Za to -1 by the value of this "limit" : -1/12
The article mentions Ramanujan summation, but Zeta regularization is actually a much more useful tool. In
Ugh. First, the title of this article is wrong because in the article itself the author says it IS -1/12, just not the way the video proved it.
And, yes, I know in math that HOW you get the answer is somehow seen as being more important than the answer itself, but really? You consider this attention damaging? This is like the people who complain about Mythbusters because it's not "real science".
Look, anything that gets people more interested in math is great, especially when it's a video as harmless as this one (it's not as if this video could actually impact someone's life or well-being).
So drop this "I was the cool form of uncool before uncool became a thing" attitude and just be happy that people are interested in math for a change.
And, yes, I know in math that HOW you get the answer is somehow seen as being more important than the answer itself, but really? You consider this attention damaging? This is like the people who complain about Mythbusters because it's not "real science".
Look, anything that gets people more interested in math is great, especially when it's a video as harmless as this one (it's not as if this video could actually impact someone's life or well-being).
So drop this "I was the cool form of uncool before uncool became a thing" attitude and just be happy that people are interested in math for a change.
No, the title is right. The sum is not -1/12, there is no sum because it diverges. The Ramanujan summation is not the real summation, its a way to assign a sum to divergent series which can be useful in some cases.
> just be happy that people are interested in math for a change
Sorry, can't do it. Mythbusters actually is real science, and anyone who thinks it isn't doesn't understand science. But this video is not real math, notwithstanding that the result it derives looks the same as one that can be derived using real math.
Sorry, can't do it. Mythbusters actually is real science, and anyone who thinks it isn't doesn't understand science. But this video is not real math, notwithstanding that the result it derives looks the same as one that can be derived using real math.
You are wrong and the title is correct. As proof, I present you my hypothesis:
"You cannot obtain a negative number by adding positive numbers."
Try and disprove me, I dare you.
"You cannot obtain a negative number by adding positive numbers."
Try and disprove me, I dare you.
It's a shame that ColinWright's post, linked by him below, got nuked by the moderators because it contains the most relevant information on the matter directly from an accredited mathematics professor, Terry Tao [0]. This is what should gratifies one's intellectual curiosity, not a series of posts that argue about the boring parts.
[0] http://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin...