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Testing latest pari + WASM + node.js... and it works?! Wow.

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License: GPL3
ubuntu2004
Function: diffop
Section: polynomials
C-Name: diffop0
Prototype: GGGD1,L,
Description:
 (gen,gen,gen,?1):gen    diffop($1, $2, $3)
 (gen,gen,gen,small):gen diffop0($1, $2, $3, $4)
Help: diffop(x,v,d,{n=1}): apply the differential operator D to x, where D is defined
 by D(v[i])=d[i], where v is a vector of variable names. D is 0 for variables
 outside of v unless they appear as modulus of a POLMOD. If the optional parameter n
 is given, return D^n(x) instead.
Doc:
 Let $v$ be a vector of variables, and $d$ a vector of the same length,
 return the image of $x$ by the $n$-power ($1$ if n is not given) of the
 differential operator $D$ that assumes the value \kbd{d[i]} on the variable
 \kbd{v[i]}. The value of $D$ on a scalar type is zero, and $D$ applies
 componentwise to a vector or matrix. When applied to a \typ{POLMOD}, if no
 value is provided for the variable of the modulus, such value is derived
 using the implicit function theorem.

 \misctitle{Examples}
 This function can be used to differentiate formal expressions:
 if $E=\exp(X^2)$ then we have $E'=2*X*E$. We derivate $X*exp(X^2)$
 as follows:
 \bprog
 ? diffop(E*X,[X,E],[1,2*X*E])
 %1 = (2*X^2 + 1)*E
 @eprog
 Let \kbd{Sin} and \kbd{Cos} be two function such that
 $\kbd{Sin}^2+\kbd{Cos}^2=1$ and $\kbd{Cos}'=-\kbd{Sin}$. We can differentiate
 $\kbd{Sin}/\kbd{Cos}$ as follows,
 PARI inferring the value of $\kbd{Sin}'$ from the equation:
 \bprog
 ? diffop(Mod('Sin/'Cos,'Sin^2+'Cos^2-1),['Cos],[-'Sin])
 %1 = Mod(1/Cos^2, Sin^2 + (Cos^2 - 1))
 @eprog
 Compute the Bell polynomials (both complete and partial) via the Faa di Bruno
 formula:
 \bprog
 Bell(k,n=-1)=
 { my(x, v, dv, var = i->eval(Str("X",i)));

   v = vector(k, i, if (i==1, 'E, var(i-1)));
   dv = vector(k, i, if (i==1, 'X*var(1)*'E, var(i)));
   x = diffop('E,v,dv,k) / 'E;
   if (n < 0, subst(x,'X,1), polcoef(x,n,'X));
 }
 @eprog
Variant:
 For $n=1$, the function \fun{GEN}{diffop}{GEN x, GEN v, GEN d} is also
 available.