Spathi Posted July 9, 2023 Report Share Posted July 9, 2023 In mathematics, there is the Ramanujan summation: 1+2+3+4…=-1/12 https://en.wikipedia.org/wiki/1_+_2_+_3_+_4_+_⋯ This sum is used in physics for predicting the Casimir effect: https://en.wikipedia.org/wiki/Casimir_effect I have also heard that this sum was used in the string theory (more precisely, in the original bosonic string theory). Then, in mathematics the p-adic numbers are used: https://en.wikipedia.org/wiki/P-adic_number https://youtu.be/tRaq4aYPzCc My question is: can the Ramanujan summation be relatively easily obtained using the p-acid numbers? Quote Link to comment Share on other sites More sharing options...

OceanBreeze Posted July 15, 2023 Report Share Posted July 15, 2023 (edited) On 7/9/2023 at 10:54 PM, Spathi said: My question is: can the Ramanujan summation be relatively easily obtained using the p-adic numbers? That depends on what you mean by "relatively easily"🤣 Edited July 21, 2023 by OceanBreeze Quote Link to comment Share on other sites More sharing options...

OceanBreeze Posted July 21, 2023 Report Share Posted July 21, 2023 P-adic functions exist in an exoteric area of mathematics seldom breached by engineers. So, you may take what I, a marine engineer, have to say about this with an atom of salt. P-adic numbers once were the domain of pure numbers-theory but have found their way into more and more Physics, thanks to some brilliant mathematical physicists. Here is an interesting paper on their many applications in Physics. Reverting back to the question asked in the OP, I can only find one equation which ties together p-adic functions ( actually only closely related L-adic functions ) and the Ramanujan function . I believe by solving this sigma summation in terms of an L-function of the Deligne motive: L(s) = ∑_{n}τ (n)n^{−s}, will yield the Ramanujan function τ (n). Based on that, I have to say that obtaining the Ramanujan function by means of p-adic numbers cannot be said to be “relatively easy”. But again I stress that the answer depends on your definition of “relatively easily” and that depends greatly on who is doing the calculating! Quote Link to comment Share on other sites More sharing options...

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