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I'm trying to find a simple way to define a bilinear and Leibniz functional/map "$B(\cdot,\cdot)$", that eats two functions $f(x),f(y)$ or products theirof and produces the following, $$ B(f(x),f(y))=f(x**y-y**x) \,, $$

where "$**$" is a non-commutative product that's un-important for my question.

The problem I have is making the map $B(\cdot,\cdot)$ obey the Leibniz rule $$ B(f(x_1)f(x_2)...f(x_n),f(y))=\sum_if(x_1)...f(x_{i-1})B(f(x_i),f(y))f(x_{i+1})..f(x_n) \,. $$

I have read this question and tried the following code to implement just for Leibniz rule

B[f[x_], f[y_]] := f[x ** y - y ** x]
B[f[x_]*f[y_], f[z_]] := B[f[x], f[z]]*f[y] + f[x]*B[f[y], f[z]]

But the above code doesn't seem to understand how to handle power imputs of $f(x)$ with the same argument $x$, e.g. $f(x)^2$, and higher order products like $f(x)f(y)f(z)$. For example if I evaluate

B[f[a]^2, f[e]]
B[f[a]*f[b]*f[c], f[e]]

Mathematica doesn't seem to know what to do.

I guess it's because I only told Mathematica how to handle two imputs in one slot of $B(\cdot,\cdot)$. But since I am going to be manipulating higher order products of $f(x)$, how should I define the map $B$ without making one definition for each order of product/power of the imput function?

PS: The problem also comes up when I try to implement Bilinearity for the map $B$.

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Try the following code:

B[f[x_],f[y_]]:=f[x**y-y**x]; B[x_Times,y_]:=Sum[ReplacePart[x,{i}:>B[x[[i]],y]],{i,Length[x]}]; B[Power[x_,y_],z_]:=y*Power[x,y-1]*B[x,z];

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  • $\begingroup$ Thanks! I tried a few examples it worked out pretty well. May I ask what "x_Times" means in the 2nd line? And if you have the 2nd line, why do you still need the 3rd? I am trying to understand this well enough to impletement the bilinearity of B. Thanks! $\endgroup$ Apr 8 '20 at 22:49
  • $\begingroup$ @JDING x_Times is a pattern which represents an expression with head Times. Similarly, Power[x_, y_] is pattern represents an expression with head Power. For more detail, you may check the document for Pattern and Blank. Mathematically, x^2 can be identified with x*x. However, these are two different expressions in Mathematica. The former is of head Power, while the latter is of head Times (Mathematica would automatically replace x*x by x^2, but not vice versa). That's why we need the third line. $\endgroup$
    – Wen Chern
    Apr 9 '20 at 7:57

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