Timeline for How does one motivate the analytic continuation of the Riemann zeta function?
Current License: CC BY-SA 4.0
8 events
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| Sep 5 at 15:37 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
http -> https (the question was bumped anyway)
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| Jan 6, 2021 at 7:49 | comment | added | shane.orourke | I like your use of the words 'elementary' and 'rigorously' ;) | |
| Jan 8, 2016 at 11:21 | comment | added | Gottfried Helms | Hmm, the question as well this answer is pretty old. It's worth to be noted here anyway, that the question asks for the motivation for the consideration of the "functional equation" which concerns the relation of zeta at negative and positive arguments, not for the motivation to consider the relation between the alternating to the non-alternating series. | |
| Mar 10, 2011 at 21:20 | comment | added | Frank Thorne | I wonder if it's possible to prove that $1 + 2 + 3 + \cdots$ is also equal to $-1/14$, or $\pi$, or $e$, or zero, or $\sqrt{-163}$, or ... | |
| Mar 10, 2011 at 2:01 | comment | added | Theo Johnson-Freyd | Anyway, I bring this up to emphasize that one must be very carefuly with "elementary algebraic manipulations" like whichever argument you use to conclude that $\sum (-1)^n(n+1) = \frac14$. | |
| Mar 10, 2011 at 2:00 | comment | added | Theo Johnson-Freyd | (continuation) a deeper example is closely related to your $1-2+3-4+\dots=\frac14$. Namely, the same argument gives $s=1-1+1-1+\dots=\frac12$ --- $s+s(\text{shifted})=1$, so $s=\frac12$. But $t=1-1+0+1-1+0+1-1+0+\dots$ satisfies $t+t(\text{shifted})+t(\text{shifted twice}) = 1$, so $t=\frac13$. This also illustrates that "associativity" is actually a continuous property. $a+(b+c)$ is the addition where $b$ and $c$ are infinitely close together compared to their distance to $a$, but there are other additions like $a\quad+b\;+\;c$, or $a\;+\;b\quad+\quad c$. | |
| Mar 10, 2011 at 1:58 | comment | added | Theo Johnson-Freyd | I feel compelled to emphasize the following. Just like conditionally convergent series are not commutative (for any $x\in\mathbb R$, there exists a permutation of $\{1,-\frac12,\frac13,-\frac14,\dots\}$ with sum $x$), divergent series are not even associative. The most basic example is $0=0+0+\dots=(1-1)+(1-1)+\dots=1+(-1+1)+(-1+)\dots=1+0+0+\dots=1$. But (continued) | |
| Mar 10, 2011 at 0:36 | history | answered | Frank Thorne | CC BY-SA 2.5 |