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Re: How to show 1+2+3+ ... = -1/12 using Mathematica's symbols?
*To*: mathgroup at smc.vnet.net
*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
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