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Re: Taylor series of the zeta function

• To: mathgroup at smc.vnet.net
• Subject: [mg103344] Re: [mg103317] Taylor series of the zeta function
• From: DrMajorBob <btreat1 at austin.rr.com>
• Date: Wed, 16 Sep 2009 05:48:00 -0400 (EDT)
• References: <200909150824.EAA12723@smc.vnet.net>
• Reply-to: drmajorbob at yahoo.com

Series used the same algorithm as always:

Series[f[x], {x, 85, 30}]

SeriesData[x, 85, {
f[85],
Derivative[1][f][85], Rational[1, 2] Derivative[2][f][85],
   Rational[1, 6] Derivative[3][f][85],
   Rational[1, 24] Derivative[4][f][85],
   Rational[1, 120] Derivative[5][f][85],
   Rational[1, 720] Derivative[6][f][85],
   Rational[1, 5040] Derivative[7][f][85],
   Rational[1, 40320] Derivative[8][f][85],
   Rational[1, 362880] Derivative[9][f][85],
   Rational[1, 3628800] Derivative[10][f][85],
   Rational[1, 39916800] Derivative[11][f][85],
   Rational[1, 479001600] Derivative[12][f][85],
   Rational[1, 6227020800] Derivative[13][f][85],
   Rational[1, 87178291200] Derivative[14][f][85],
   Rational[1, 1307674368000] Derivative[15][f][85],
   Rational[1, 20922789888000] Derivative[16][f][85],
   Rational[1, 355687428096000] Derivative[17][f][85],
   Rational[1, 6402373705728000] Derivative[18][f][85],
   Rational[1, 121645100408832000] Derivative[19][f][85],
   Rational[1, 2432902008176640000] Derivative[20][f][85],
   Rational[1, 51090942171709440000] Derivative[21][f][85],
   Rational[1, 1124000727777607680000] Derivative[22][f][85],
   Rational[1, 25852016738884976640000] Derivative[23][f][85],
   Rational[1, 620448401733239439360000] Derivative[24][f][85],
   Rational[1, 15511210043330985984000000] Derivative[25][f][85],
   Rational[1, 403291461126605635584000000] Derivative[26][f][85],
   Rational[1, 10888869450418352160768000000] Derivative[27][f][85],
   Rational[1, 304888344611713860501504000000] Derivative[28][f][85],
   Rational[1, 8841761993739701954543616000000] Derivative[29][f][85],
   Rational[1, 265252859812191058636308480000000] Derivative[30][f][
    85]}, 0, 31, 1]

The Rationals above are just 1/n! terms. The (x-5)^n terms don't appear  
explicitly in SeriesData, but Normal brings them out.

For Log[Zeta[x]], the chain rule would be used for the Derivatives.

Normal@Series[f[g[x]],{x,85,3}]
f[g[85]]+(-85+x) (f^\[Prime])[g[85]] (g^\[Prime])[85]+1/2 (-85+x)^2  
((g^\[Prime])[85]^2 (f^\[Prime]\[Prime])[g[85]]+(f^\[Prime])[g[85]]  
(g^\[Prime]\[Prime])[85])+1/6 (-85+x)^3 (3 (g^\[Prime])[85]  
(f^\[Prime]\[Prime])[g[85]] (g^\[Prime]\[Prime])[85]+(g^\[Prime])[85]^3  
(f^(3))[g[85]]+(f^\[Prime])[g[85]] (g^(3))[85])

Bobby

On Tue, 15 Sep 2009 03:24:44 -0500, Guy Verhofstadt  
<guy.verhofstadt at gmail.com> wrote:

> Hi
> I would like to compute the Taylor series of the (logarithm) of the
> Riemann zeta function at various integral points, up to high order and
> with high precision.
> Mathematica does quite well at this:
> N[Series[Log[Zeta[x]], {x, 85, 30}], 100]
> gives a result, for instance.
> I would like to know which algorithm is used to compute this, or how I
> could find out.
> Thank you
>


-- 
DrMajorBob at yahoo.com

• References:
  • Taylor series of the zeta function
    • From: Guy Verhofstadt <guy.verhofstadt@gmail.com>