Hard to believe isn't it? Here is how some physicists proved it: http://www.nytimes.com/2014/02/04/science/in-the-end-it-all-adds-up-to.html?ref=science&_r=0
- from the link : - the ^^above quoted^^ from : http://www.nytimes.com/2014/02/04/science/in-the-end-it-all-adds-up-to.html?ref=science&_r=1 It does not seem "hard to believe", when taken in the context in which it was intended. Not to me, at least.
Can you please explain this in layman's terms? I got bored 15 seconds into the video and the article makes little sense to me. Does all this mean I'll be making more or less money?
Analytic continuation. http://en.wikipedia.org/wiki/Analytic_continuation The general idea: You have an infinite series which is convergent in some domain. In this domain there is a closed form equal to the series. The closed form is valid over a larger domain. The gimmick is calling the sum of the series (which is infinite) equal to the closed form. Example: \( \frac{1}{1-x} = 1 + x + x^2 + x^3 ...\) Let x = 2 and you have: -1 = 1 + 2 + 4 + 8 ...
http://en.wikipedia.org/wiki/1_+_2_+_3_+_4_+_·_·_· http://en.wikipedia.org/wiki/Riemann_zeta_function Assume \(S = 1 - 2 + 3 - 4 + 5 - \dots \; \textrm{and} \; Z = 1 + 2 + 3 + 4 + 5 + \dots\). Then \(Z - 4 Z = 1 + 2 - 4\times 1 + 3 + 4 - 4\times 2 + 5 + 6 - 4\times 3 + \dots = 1 - 2 + 3 - 4 + 5 - 6 + \dots = S\) Assume \(\frac{1}{(1 + x)^2} = 1 - 2 x + 3 x^2 - 4 x^3 + 5 x^4 - 6 x^5 + \dots\) converges even when x = 1, then \(\frac{1}{4} = \frac{1}{(1 + 1)^2} = 1 - 2 + 3 - 4 + 5 - 6 + \dots = S\) So if \(Z - 4 Z = S = \frac{1}{4} \) then it follows that \(Z = \, - \frac{1}{12}\) This is breaking a lot of the rules, but it makes sense in a more-or-less rigorous manner. ---- If instead we replace \(Z = 1 + 2 + 3 + 4 + 5 + \dots\) with \(\zeta(s) = 1^{-s} + 2^{-s} + 3^{-s} + 4^{-s} + 5^{-s} + \dots = \sum_{k=1}^{\infty} n^{-s} \) this happens to be what is analytically continued to be the Riemann zeta function. We can write \(\zeta(s) = \frac{2^{s-1}}{s-1} - 2^s \int_{0}^{\infty} \frac{ \sin \left( s \, \tan^{\tiny -1} \, t \right)}{\left( 1 + e^{\pi t}\right) \, \left( 1 + t^2 \right)^{\tiny \frac{s}{2}} } \, dt\) which converges for all s except s = 1, at s = -1 we have \(Z = \zeta(-1) = \frac{2^{-2}}{-2} - 2^{-1} \int_{0}^{\infty} \frac{ \sin \left( - \tan^{\tiny -1} \, t \right)}{\left( 1 + e^{\pi t}\right) \, \left( 1 + t^2 \right)^{\tiny - \frac{1}{2}} } \, dt = - \frac{1}{4} \times \frac{1}{2} - \frac{1}{2} \, \int_{0}^{\infty} \frac{ - t}{1 + e^{\pi t}} \, dt = - \frac{1}{4} \times \frac{1}{2} - \frac{1}{2} \, \left( \frac{1}{\pi^2} \, \sum_{k=1}^{\infty} \frac{ \left( -1 \right)^k}{k^2} \right) = - \frac{1}{8} - \frac{1}{2} \left( - \frac{1}{12} \right) = - \frac{1}{8} + \frac{1}{24} = - \frac{1}{12}\) So this method breaks less rules, and the only "sin" is equating a diverging series with the analytic continuation of a family of converging series. --- http://www.youtube.com/watch?v=w-I6XTVZXww http://www.youtube.com/watch?v=E-d9mgo8FGk Also \(\left( 1 - 2 \cdot 2^{\tiny -s} \right) \zeta(s) = 1^{-s} - 2^{-s} + 3^{-s} - 4^{-s} + 5^{-s} - 6^{-s} + \dots\) so for \(s = -1\), \(\frac{1}{4} = (-3) \, \left(- \frac{1}{12} \right) = \left( 1 - 2 \cdot 2^{\tiny 1} \right) \zeta(-1) = 1 - 2 + 3 - 4 + 5 - \dots = S\)
There are perhaps two problems with this so called proof: First: The sum of 1 -1 + 1 -1 . . . . . is either one or zero: Declaring it to be 1/2 is nonsense. Second: If a series does not converge, it is considered verbotten to attempt arithmetic manipulation of it.
u=1-1+1-... u-1=-u 2u=1 u=1/2 It oscillates at the mean value, like a tossed coin in the air is H and T.
[video=youtube_share;PCu_BNNI5x4]http://youtu.be/PCu_BNNI5x4[/video] http://en.wikipedia.org/wiki/1_+_2_+_3_+_4_+_⋯ http://terrytao.wordpress.com/2010/...tion-and-real-variable-analytic-continuation/ http://books.google.com/books?id=0pZJeQ2lEmkC&pg=PA249#v=onepage&q&f=false
Phyti: Your proof from Post #13 is an example of the reason that arithmetic manipulation of non-convergent series is verboten, as mentioned in my Post #11. The series is: 1 -1 +1 -1 +1 . . . . . . Depending on where you truncate it, the sum is either zero or one, never 1 / 2 Your proof that the sum is 1 /2 is an example of erroneous results due to manipulation of series which do not converge.
He's using this topic to promote his book, but what he says is interesting, and he's a pretty good professor. http://math.berkeley.edu/~frenkel/ [video=youtube_share;Vsvy8Ko2-YM]http://youtu.be/Vsvy8Ko2-YM[/video]
If you truncate it then it's not the same series any more. All kinds of weird stuff happens when you introduce infinities.
James R: From my Post #11: From your Post #17: Weird perhaps, but not magical. If you truncate after an odd number of terms, the sum is one. After terms: 1, 3, 5, 7,9 . . . . Truncation results in a sum of one. If you truncate after an even number of terms, the sum is zero. After terms: 2, 4, 6, 8 . . . . . truncation results in a sum of zero. The above is valid for any number of terms. It would require belief in magic, not logic, to claim otherwise. While introducing infinities often causes in counterintuitive results, it does not cause magical results. BTW: Your Post surprises me. From various others at this forum, it would not be surprising.
You're assuming that what holds for a finite number of terms also holds for an infinite number of terms. But that assumption requires justification. The basic problem with the given infinite series, of course, is that its sum diverges, which means that with an appropriate rearrangement of the order of terms you can make it sum to anything you like.
I remember in undergrad quantum physics, we had to solve a semi-realistic problem involving a calculation such as \(\int^\infty_{-\infty}sin^2x\ \rm{d}x\). So our professor says: "What do you do when you have a quantity that's not defined? Well, you define it." And the trick here was to define the integral as the limit of a series of physics situations approaching the one we were dealing with, in which the integral is unambiguously calculable. Specifically, we defined the integral as \(\lim_{k\to 0^+}\int^\infty_{-\infty}e^{-kx}sin^2x\ \rm{d}x\). If I remember correctly, they claimed that it can be mathematically shown that any similar scheme for defining the integral will yield the same result. In graduate quantum field theory, I remember the series sum \(\sum_{n=1}^\infty n\) came up in calculating the Casimir zero point energy force between two conducting plates, and for that problem we did something like define it as \(\lim_{k\to 0^+}\sum_{n=1}^\infty ne^{-kn}\). Specifically though, to calculate the measurable force, it didn't involve performing any infinite series sums, but rather subtracting the difference between two sums, with the result looking something like \(\infty - (\infty - \frac{1}{12})\). Anyhow, it's perfectly fine to define an otherwise undefined quantity in physics, as long as you remain conscious of the rule by which the quantity is defined and then calculated, the circumstances under which similar definitions will give the same result, and it matches with the results of experiment.