Re: D vs. Derivative
- To: mathgroup at smc.vnet.net
- Subject: [mg15673] Re: D vs. Derivative
- From: "Allan Hayes" <hay at haystack.demon.co.uk>
- Date: Sat, 30 Jan 1999 04:29:05 -0500 (EST)
- References: <78pcjv$d3r@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Gianluca Gorni wrote in message <78pcjv$d3r at smc.vnet.net>...
>
>Hello!
>
>It seems that D[f[x],x] and f'[x] are not equivalent, and the latter can
>give useless outputs.
>
>Consider the following power series, that converges in the unit disk of
>the complex field:
>
> f[x_] = Sum[ x^(n-1)/(n^3+n), {n, 1, Infinity} ]
>
>With immediate assignment, f[x] is evaluated to a special function.
>Suppose now that I need the derivative of f[x]. If I do it with
>
> D[ f[x], x ]
>
>there is no problem: I get a regular-looking special function
>combination. But if I try to get the derivative with
>
> f'[x]
>
>the output is a formula containing DirectedInfinity. Moreover
>
> f'[x] // Simplify
>
>gives Indeterminate.
>
>By the way, the integral
>
> Integrate[ (1-Cos[y])/(E^y-x), {y, 0, Infinity} ]
>
>is left as it is by Mathematica, although it is equal to the special
>function f[x] above, at least for many values of x.
>
>
> +---------------------------------+
> | Gianluca Gorni |
> | Universita` di Udine |
> | Dipartimento di Matematica |
> | e Informatica |
> | via delle Scienze 208 |
> | I-33100 Udine UD |
> | Italy |
> +---------------------------------+
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> +---------------------------------+
>
>
Gianluca,
Clear["`*"]
f' is df& where df = D[f[#],#]
The problem that you mention seems to involve
h[x_] =HypergeometricPFQ[{I,1},{1+I},x];
D[h[#],#]
DirectedInfinity[((1/2 + I/2)*Sqrt[2]*
Sign[Gamma[2 + I]])/Sign[Gamma[1 + I]]]
which is equal to
h'[#]
DirectedInfinity[((1/2 + I/2)*Sqrt[2]*Sign[Gamma[2 + I]])/
Sign[Gamma[1 + I]]]
while
D[h[x],x]
-(1/((-1 + x)*x)*I*(1 -
HypergeometricPFQ[{I, 1}, {1 + I}, x] +
x*HypergeometricPFQ[{I, 1}, {1 + I}, x]))
Interestingly, from
feh[x_] = FunctionExpand[h[x]]
(I*Beta[x, I, 0])/x^I
We get
D[feh[#],#]
I/((1 - #1)*#1) + Beta[#1, I, 0]*#1^(-1 - I)
which is the same as
feh'[#]
I/((1 - #1)*#1) + Beta[#1, I, 0]*#1^(-1 - I)
and
D[feh[x],x]
I/((1 - x)*x) + x^(-1 - I)*Beta[x, I, 0]
Also
Sum[1/n^2,{n,Infinity}]
Pi^2/6
but
Sum[1/#^2,{#,Infinity}]
Sum[1/#1^2, {#1, 1, Infinity}]
Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
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