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Home/ Questions/Q 7038727
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Editorial Team
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Editorial Team
Asked: May 28, 20262026-05-28T01:42:14+00:00 2026-05-28T01:42:14+00:00

The expressions I am working with are much too complicated to fully enter here,

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The expressions I am working with are much too complicated to fully enter here, but I’ve included a simple examples that highlights the problem I am running into. I am hoping there is someone out there with enough programming fortitude to help me around this issue. Let me preface this by saying I have little to no background in programming in general, but I know the basics of Mathematica. Any and all help is greatly appreciated. Suppose I have set up the following functions:

X[x_] := x Log[x]
X[0] := 0
Y[y_] := y Log[y]
Y[0] := 0
Z[z_] := z Log[z]
A[x_, y_, z_] := X[x] + Y[y] + Z[z]

In[7]:= A[x, y, z]

Out[7]= x Log[x] + y Log[y] + z Log[z]

In[8]:= B[x_, y_, z_] := 
 Evaluate[A[x, y, z] - x*D[A[x, y, z], x] - y*D[A[x, y, z], y] - 
   z*D[A[x, y, z], z]]

In[9]:= B[x, y, z]

Out[9]= x Log[x] - x (1 + Log[x]) + y Log[y] - y (1 + Log[y]) + 
 z Log[z] - z (1 + Log[z])

I have set up A[x,y,z] with the rules for X[x], Y[y], and Z[z] so that it can handle the case where x,y,z == 0, i.e. when x == 0 I want all expressions in A[x,y,z] with x to go to zero or be neglected including Log[x]. I’ve defined a function B[x,y,z] that involves the partial derivatives of A[x,y,z]. Now, I want the result so that B[0,y,z] yields

yLog[y]-y(1+Log[y])+zLog[z]-z(1+Log[z])

that is to basically go back and make A[x,y,z]:= Y[y]+Z[z] but instead I am currently running into the following, understandable, error:

Infinity::indet: Indeterminate expression 0 (-[Infinity]) encountered. >>

There must be some way around this with Mathematica and I am wondering if it will involve the Hold function or something related. Thank you all for the help.

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  1. Editorial Team
    Editorial Team
    2026-05-28T01:42:14+00:00Added an answer on May 28, 2026 at 1:42 am

    One way is to use dummy variables in the derivatives, and replace afterward with the actual values.

    Clear[X, Y, Z];
    X[x_] := x Log[x]
    X[0]   = 0;
    Y[y_] := y Log[y]
    Y[0]   = 0;
    Z[z_] := z Log[z]
    Z[0]   = 0;
    
    A[x_, y_, z_] := X[x] + Y[y] + Z[z]    
    B[x_, y_, z_] := 
     Module[{xx, yy, zz}, 
      A[x, y, z] - x*D[A[xx, y, z], xx] - y*D[A[x, yy, z], yy] - 
        z*D[A[x, y, zz], zz] /. {xx -> x, yy -> y, zz -> z}]
    
    B[0, y, z]
    
    (* Output: y Log[y] - y (1 + Log[y]) + z Log[z] - z (1 + Log[z]) *)
    

    This works because each derivative has a factor of 0. A more general solution might involve defining UpValues on X, Y, Z for handling derivatives at the origin (I don’t have time to check this at the moment).

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