Re: How to show 1+2+3+ ... = -1/12 using Mathematica's symbols?
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- Subject: [mg132278] Re: How to show 1+2+3+ ... = -1/12 using Mathematica's symbols?
- From: "Dr. Wolfgang Hintze" <weh at snafu.de>
- Date: Thu, 30 Jan 2014 23:03:39 -0500 (EST)
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Am Sonntag, 19. Januar 2014 08:52:36 UTC+1 schrieb Matthias Bode: > Hola, > > > > I came across this video (supported by the Mathematical Sciences Research Institute* in Berkeley, California): > > > > http://www.numberphile.com/videos/analytical_continuation1.html > > > > Could the method shown in this video be replicated using Mathematica symbols such as Sum[] &c.? > > > > Best regards, > > > > MATTHIAS BODES 17.36398=B0, W 66.21816=B0,2'590 m. AMSL. > > > > *) http://www.msri.org/web/msri [Hello sysop! I sent this message on 19th of January 23:34 to the group, and I don't know why it was not published. Probably my news-Server makes the problem. Therefore I send my response again] "Dr. Wolfgang Hintze" schrieb im Newsbeitrag news:... No problem: In[65]:= S = Limit[Sum[n^x, {n, 1, Infinity}], x -> 1] Out[65]= -(1/12) But I suggest reading my following remarks: The video is of course very funny, but as this physicist says himself his "derivation" is mathematical hocuspocus. Come on, let's instead "prove" that 1=0. Writing S = A - B with A = (1+3) + (2+4) + (3+5) + (4+6) + (5+7) + ... B = 3+4+5+... Consider A first: we leave the first two brackets unchanged. In the third bracket the 5 is okay but we have one 3 too much. It must be subtracted, and this will bei done in B. In the fourth bracket the 6 is okay but we have one 4 too much which will be subtracted in B. And so on. Now we have A = 4+6+8+10+... = 2(2+3+4+...) = 2 (S-1) B = S-1-2 = S-3 Hence S = 2(S-1) - (S-3) = S -2+3 = S+1 or, subtracting S on both sides, leaves us with 1=0 QED: This is true a fortiori since we alredy "know" that S is finite (= - 1/12). End of hocuspocus. Seriously, the result -1/12 must of course be obtained by analytic continuation of the appropriate function. In Mathematica the desired proof is immediately done by infact exchanging limits: via Zeta[] In[65]:= S = Limit[Sum[n^x, {n, 1, Infinity}], x -> 1] Out[65]= -(1/12) instead of In[96]:= Sum[Limit[n^x, x -> 1], {n, 1, Infinity}] During evaluation of In[96]:= Sum::div:Sum does not converge. >> Out[96]= Sum[n, {n, 1, Infinity}] S1 and S2 are easily calculated using the geometric series In[64]:= S1 = Sum[q^n, {n, 0, Infinity}] /. q -> -1 Out[64]= 1/2 In[69]:= Limit[Sum[n*q^(n - 1), {n, 1, Infinity}], q -> -1] Out[69]= 1/4 It's fun to play around a bit. For example, look at this In[94]:= T3g = Limit[Table[n^2*q^(n - 1), {n, 1, 5}], q -> -1] S3g = Limit[Sum[n^2*q^(n - 1), {n, 1, Infinity}], q -> -1] Out[94]= {1, -4, 9, -16, 25} Out[95]= 0 In[90]:= T3z = Limit[Table[n^(2*x), {n, 1, 5}], x -> 1] S3z = Limit[Sum[n^(2*x), {n, 1, Infinity}], x -> 1] Out[90]= {1, 4, 9, 16, 25} Out[91]= 0 That is both sums, 1-4+9-16+-... and 1+4+9+16+... give 0. The interesting question of the whole topic here is: in what sense are the results of divergent infinite sums well defined? Let me put it this way: is there anybody who finds another reasonable value for S = 1+2+3+4+ ...? Best regards, Wolfgang