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Simple recurrence and contextual meaning of IncludeSingularTerm = false ?

Posted 2 months ago
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Hello, while playing with this recurrence relation:

g(n+1)=10*log(n)+g(n)  

Wolfram Alpha produces this answer:

g(n) = c_1 + 10 log(Γ(n)) (c_1 is an arbitrary parameter)

(by the way, a Step-by-Step option would be great).
When I simply change the sign of the LHS term, the proposed solution is instead:
enter image description here

I have checked the generalized Riemann Zeta function documentation, and then searched the web as well, but could not get a clue about the meaning and use of:
IncludeSingularTerm = False
Any suggestion about how to proceed to gain a thorough understanding of its meaning in this context?
I was hoping to get some hint from an examination of the "Open Code" output, but its generation appears to hang forever (I have a Premium Pro Subscription).

Many Thanks !

3 Replies
Posted 2 months ago

Evaluating ??IncludeSingularTerm in Mathematica sheds some light:

??IncludeSingularTerm
   "IncludeSingularTerm is an option for LerchPhi and Zeta. With
    IncludeSingularTerm -> True, terms involving ((k + a)^2)^(-s/2) with
    k + a == 0 are included. With IncludeSingularTerm -> False, they are not."

This is also mentioned in the docs of the more general Lerch transcendent, LerchPhi[] (of which Hurwitz zeta is a special case); unfortunately the docs for Zeta[] neglect to mention this setting.

Evaluating

Options[Zeta]
   {IncludeSingularTerm -> False}

shows that the singular term is excluded by default:

Zeta[2, -1, IncludeSingularTerm -> True]
   ComplexInfinity

Zeta[2, -1, IncludeSingularTerm -> False]
   1 + π^2/6
POSTED BY: J. M.
Answer

Thanks J.M.
I then also wonder whether I am correctly interpreting the meaning of (for clarity, I have replace n/2 with n)

enter image description here

as the following derivative with respect to the first variable, s, in enter image description here

enter image description here

a result listed at http://mathworld.wolfram.com/HurwitzZetaFunction.html
or am I rather mistaking this particular Wolfram Alpha syntax for derivatives?

Posted 2 months ago

Yes, your interpretation is correct.

POSTED BY: J. M.
Answer
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