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 January 30th, 2019, 05:29 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 514 Thanks: 80 Doubt on math web-pages What am I missing ? In every pdf or page it says : $\displaystyle 1+2+3+4+... =-\frac{1}{12}$ . How can the sum of all natural numbers be -1/12 ?
January 30th, 2019, 05:57 AM   #2
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 Originally Posted by idontknow What am I missing ? In every pdf or page it says : $\displaystyle 1+2+3+4+... =-\frac{1}{12}$ . How can the sum of all natural numbers be -1/12 ?
Short answer: it's not. The sum of all natural numbers is infinity.

 January 30th, 2019, 08:05 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,660 Thanks: 2635 Math Focus: Mainly analysis and algebra A longer, but still simplified answer is that there are no infinite sums that we can evaluate. If you imagine doing it on an infinitely precise calculator, you will never get to press the "=" button. What we can do (in some cases) is to assign a value to the infinite sum. Some of these cases are more obvious than others. Convergent series, for example, have pretty obvious values. Others have less obvious values that still seem to contain some kernel of truth. Typically these values are consistent across different methods of determining the value in question. Those methods also producing values for simpler (perhaps convergent) series that agree with simpler methods. It is in this way that we assign the value $$1+2+3+4+ \ldots =-\frac{1}{12}$$ The value also appears to have meaning in the real world, producing verifiable results in physical theories.
 January 30th, 2019, 08:15 AM #4 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 The most frequently seen "proof" of this idiocy on youtube involves outright deception. The narrator says something is true if x < 1. But that is false. What he says is true if |x| < 1. Later he says let x = - 1. That of course would be valid if the limitation were merely x < 1. But it is not valid because the limitation is |x| < 1 and it is false that |-1| < 1.
 January 30th, 2019, 08:19 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,644 Thanks: 2084 This question has been discussed here before. I'll find a link.
 January 30th, 2019, 09:18 AM #6 Math Team     Joined: May 2013 From: The Astral plane Posts: 2,162 Thanks: 879 Math Focus: Wibbly wobbly timey-wimey stuff. I'll do a quick run-down on the basics until skipjack can find a good link. I don't know of one, myself. This deals with something very useful known as the Riemann Zeta Function $\displaystyle \zeta (s) = \sum_{n = 1}^{\infty} \dfrac{1}{n^s}$ Now, there are values of s that are problematic here as they give nasty answers. s = -1 is one of these places. We then get $\displaystyle \zeta (-1) = 1 + 2 + 3 + \text{...}$ But. There is a trick known as "analytic continuation." It deals with complex variables (in this case we'd make s a complex number) and is really really messy. The long and short of it is that we can use this to extend the zeta function into regions where it ought not to go. In this case we can analytically continue the zeta function to give us a value at s = -1. The whole point here is that the value of the zeta function at s = -1 is no longer given by 1 + 2 + 3 + 4... It is given as a different function. (I have no idea what it is.) Anyway, analytic continuation gives $\displaystyle \zeta (-1) = -\dfrac{1}{12}$. Relatively easy to conceptualize, rather difficult to calculate. -Dan Addendum: v8archie: I don't know if it's an artifact of my String Theory text but it really doesn't use $\displaystyle \zeta (-1)$. I haven't looked it up to refresh my memory but text starts doing a count of something or other and it goes 1 + 2 + 3 + 4 +... It then says this is -1/12. But it's an actual count and does not reference the zeta function at all. The only reason I go along with it (and many many others, I suspect) is that it works! Thanks from greg1313
 January 30th, 2019, 11:59 AM #7 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,660 Thanks: 2635 Math Focus: Mainly analysis and algebra That's presumably because $\zeta(-1)$ isn't the only way to get there. Some proofs do involve stretching (or breaking) traditional rules, others (I think) don't. The point is that, proofs that do break the rules somehow miraculously reach the same answer as we get by analytic continuation of the $\zeta$ function. The methods also confirm results of other series values. So there seems to be something inherently reasonable about assigning the value $-\frac1{12}$ to that particular series. Note that the rule $|x| < 1$ cited above is a rule required for convergence of a series, but here we aren't talking about a convergent series, so there's no obvious reason why that rule should still hold. Thanks from topsquark Last edited by v8archie; January 30th, 2019 at 12:02 PM.
 January 30th, 2019, 04:12 PM #8 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,660 Thanks: 2635 Math Focus: Mainly analysis and algebra It's also worth pointing out that, just because some values of the $\zeta$ function can be obtained by calculating the value of convergent series, it doesn't follow that all values of the analytic continuation of the $\zeta$ function can be interpreted as the values of series. The analytic continuation of a function is distinct from the original function. Besides, as I was careful to state above. $\zeta(2)=\frac{\pi^2}6$ does not mean that the result of the infinite sum $1 + \frac14 + \frac19 + \ldots$ is equal to $\frac{\pi^2}6$. It indicates that the series of finite partial sums converges to $\frac{\pi^2}6$. We have decided as mathematicians to define the value of the infinite sum to be $\frac{\pi^2}6$, that is to say we have assigned the value $\frac{\pi^2}6$ to that infinite sum. Moreover, we have assigned that value in a particular way - by looking at the value to which the infinite series of finite partial sums converge. Clearly, we cannot do the same for the sum $1 + 2 + 3 + \ldots$ because the finite partial sums do not converge. But the evidence that the infinite sum should be assigned the value $-\frac1{12}$ comes from analytic continuation, other methods of proof and, as you alluded to, the fact that such a value works in physical contexts. (I believe another is the effect that makes ships at harbour drift together although I don't recall it's name). Thanks from topsquark
January 30th, 2019, 05:24 PM   #9
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 Originally Posted by v8archie That's presumably because $\zeta(-1)$ isn't the only way to get there. Some proofs do involve stretching (or breaking) traditional rules, others (I think) don't. The point is that, proofs that do break the rules somehow miraculously reach the same answer as we get by analytic continuation of the $\zeta$ function. The methods also confirm results of other series values. So there seems to be something inherently reasonable about assigning the value $-\frac1{12}$ to that particular series. Note that the rule $|x| < 1$ cited above is a rule required for convergence of a series, but here we aren't talking about a convergent series, so there's no obvious reason why that rule should still hold.
i am quite aware that different results arise from different axioms. My point was that the proof given in that video was deliberately deceptive.

There is a proof that if |x| < 1 the series has a finite limit. That video says that there is a sum, not a limit, if x < 1. Where is the proof? Indeed, where is the definition of an infinite sum? Moreover what other changes result by choosing an axiom that says

$a > 0 < b \text { is consistent with } a + b < 0.$

Alternate axiom systems do not invalidate the logical deductions of an axiom system.

It is one thing to say "if we choose the following axioms, we get results that both differ from the results of a different sets of axioms and that are useful in some branches of physics." It is an entirely different thing to say "conclusion A of this set of axioms is wrong because, if we lie about conclusion B of those axioms, we can deduce not A."

My comment was not about the free choice of axioms in mathematics. My comment was about the difference between honesty and dishonesty.

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