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Re: fractional derivative (order t) of (Log[x])^n and Log[Log[x]] etc.?

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  • Subject: [mg91284] Re: [mg91265] fractional derivative (order t) of (Log[x])^n and Log[Log[x]] etc.?
  • From: Curtis Osterhoudt <cfo at lanl.gov>
  • Date: Wed, 13 Aug 2008 04:41:01 -0400 (EDT)
  • Organization: LANL
  • References: <200808120847.EAA18835@smc.vnet.net>
  • Reply-to: cfo at lanl.gov

Hi, Michael, 

    If you go to the link you mention, and then click on the formula you're 
interested in, you should be taken to a page which contains the InputForm of 
that formula. For example, I clicked on the 6th one down, which takes me to 
http://functions.wolfram.com/ElementaryFunctions/Log/20/03/0006/. This page 
contains the InputForm of that formula, which I can then paste into 
Mathematica: D[Log[z]^n, {z, \[Alpha]}] == FDLogConstant[z, 0, n, 
\[Alpha]]/z^\[Alpha] /; Element[n, Integers] && n > 0 . 

   Hope that helps!

                 C.O.

On Tuesday 12 August 2008 02:47:10 hanrahan398 at yahoo.co.uk wrote:
> I'd be grateful if someone could tell me a nicely computable formula
> for the fractional derivative w.r.t. x (order t) of (Log[x])^n, where
> n is a positive integer.
>
> (Ideally I would like a formula where t can be any real number, but
> one for t>=0 would be most helpful!)
>
> The second thing I am seeking is a formula for the fractional
> derivative w.r.t. x (order t) of Log[Log[x]], Log[Log[Log[x]]], etc.,
> and more generally, of Log[...[Log[Log[x]]...], where there are n
> nested log functions, where n is of course a positive integer.
>
> (I have visited:
> <http://functions.wolfram.com/ElementaryFunctions/Log/20/03/>, and
> there is a formula there for the t-th (fractional) derivative of
> Log[x]^n, but I do not understand how to input it!!
>
> Basically I need formulae for the order-t fractional derivatives of
> (Log[x])^n and of Log[Log[x]], Log[Log[Log[x]]] (and generally with n
> nested logs), which I can use for variable x and given values of n and
> t, and can also evaluate at given values of x.
>
> Many thanks in advance.
>
> Michael



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