Re: D vs. Derivative
- To: mathgroup at smc.vnet.net
- Subject: [mg15646] Re: [mg15601] D vs. Derivative
- From: BobHanlon at aol.com
- Date: Sat, 30 Jan 1999 04:28:42 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
In a message dated 1/28/99 6:17:12 AM, gorni at dimi.uniud.it writes:
>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.
>
>My version is 3.0.1 for PowerMacintosh.
>
Gianluca,
f1[x_] := Evaluate[Simplify[Sum[ x^(n-1)/(n^3+n), {n, 1, Infinity} ]]];
f2[x_] := Evaluate[FullSimplify[Sum[ x^(n-1)/(n^3+n), {n, 1, Infinity}
]]]
f1[x]
-(1/(4*x)*(2*I - 2*I*HypergeometricPFQ[{I, 1}, {1 + I}, x] +
(1 + I)*x*HypergeometricPFQ[{1, 1 - I}, {2 - I}, x] +
4*Log[1 - x]))
f2[x]
-(1/2)*x^(-1 - I)*(Beta[x, I, 0] +
x^I*(I + x^I*Beta[x, 1 - I, 0] + 2*Log[1 - x]))
f1[x] - f2[x] == 0 // FullSimplify
True
D[f1[x], x]//Simplify
1/(2*x^2)*(-1 + I + (1 - I)*
HypergeometricPFQ[{I, 1}, {1 + I}, x] +
x*HypergeometricPFQ[{1, 1 - I}, {2 - I}, x] +
2*Log[1 - x])
f2'[x]//Simplify
1/x^2*(1/2 + I/2)*(Beta[x, I, 0]/x^I -
I*(-1 + x^I*Beta[x, 1 - I, 0] + (1 + I)*Log[1 - x]))
% - %% == 0 //FullSimplify
True
Since the form of f2 avoids the problem, it should be used.
Bob Hanlon