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Consider

F=1/((x-a)(x-b)) ; 
G= 1/((x-α)*(x-β));  
H=1/((x-r[1])(x-r[2])) ;

Using Apart[], the partial fraction decomposition is done only for $F$.

I am using version 13.1.0

What is the difference ?

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1 Answer 1

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sometimes you need to help Mathematica and tell it which is the variable to apply Apart on

F = 1/((x - a) (x - b))
G = 1/((x - α)*(x - β))
H = 1/((x - r[1]) (x - r[2]))

Compare


Mathematica graphics


Mathematica graphics

Basically, change Apart[G] to Apart[G, x] to make it work.

I am not sure if this is a feature or a bug but I've seen this before so may be it is a feature. If in doubt, always use the form Apart[expr,var]

enter image description here

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  • $\begingroup$ Thank you very much ! Being blind, there are many things that I miss $\endgroup$ Jan 7 at 9:24
  • $\begingroup$ @Nasser With symbolic coefficients Mathematica has no way to decide which symbols are variables and which should be understood as coefficients. Variables ordering is essential in Apart and documentation mentions that (can' t find now). $\endgroup$
    – Acus
    Jan 7 at 15:45

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