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I have expression: $$f(x)=\sum_{i=0}^{K-1}\sum_{j=0}^{N-1}a_{i,j}e^{ix}x^j.$$ I want to extract coefficient matrix $A$ from expression $f(x)$ where $$A=\left( \begin{array}{ccccc} a_{0,0}& \cdots & a_{0,j} & \cdots & a_{0,N-1} \\ \vdots & \cdots & \vdots & \cdots & \vdots\\ a_{i,0} & \cdots & a_{i,j} & \cdots & a_{i,N-1} \\ \vdots & \vdots & \cdots & \vdots\\ a_{K-1,0} & \cdots & a_{K-1,j} & \cdots & a_{K-1,N-1} \\ \end{array} \right).$$ For example: $$f(x)=-\frac{1}{2} e^{-x} x^3+\frac{1}{2} e^{-2 x} x^2+\frac{1}{2} e^{-x} x^2+2 e^{-2 x} x-2 e^{-x} x+e^{-2 x}-2 e^{-x}+1$$ and its $A$ is $$A=\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ -2 & -2 & \frac{1}{2} & -\frac{1}{2} \\ 1 & 2 & \frac{1}{2} & 0 \\ \end{array} \right).$$ Does Mathematica have specific function for this, or else can someone help me to develop a technique using existing functions in Mathematica?

Frey
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2 Answers2

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A simple solution would be using Coefficient 

p = Sum[c[i, j] x^i Exp[I j x], {i, 0, 3}, {j, 0, 3}];

Table[a[i, j] = Coefficient[Coefficient[p, x, i], Exp[I x], j]
 , {j, 0, 3}, {i, 0, 3}];

MatrixForm[%]

$ \left( \begin{array}{cccc} c(0,0) & c(0,1) & c(0,2) & c(0,3) \\ c(1,0) & c(1,1) & c(1,2) & c(1,3) \\ c(2,0) & c(2,1) & c(2,2) & c(2,3) \\ c(3,0) & c(3,1) & c(3,2) & c(3,3) \\ \end{array} \right) $

General structure

The syntax is applicable for any arbitrary polynomial $\sum_{i,j} x^i y^j$

p = Sum[c[i, j] x^i y^j, {i, 0, 3}, {j, 0, 3}];

var1 = x;
var2 = y;
Table[a[i, j] = Coefficient[Coefficient[p, var1, i], var2, j]
 , {j, 0, 3}, {i, 0, 3}];

For your given example

p = -1/2 Exp[-x] x^3 + 1/2 Exp[-2 x] x^2 + 1/2 Exp[-x] x^2
 + 2 Exp[-2 x] x - 2 Exp[-x] x + Exp[-2 x] - 2 Exp[-x] + 1

var1 = x ;
var2 = Exp[-x] ;
Table[a[i, j] = Coefficient[Coefficient[p, var1, i], var2, j]
 , {j, 0, 3}, {i, 0, 3}];

MatrixForm[%]

$ \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ -2 & -2 & \frac{1}{2} & -\frac{1}{2} \\ 1 & 2 & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) $

You can easily extend it for more variable. In that case you would not be able to see it as a matrix, but the information will be stored in a[i,j,...].

Sumit
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  • Thanks @Sumit, How can I change the second line if my p = 1 - E^(-3 x) + 3 E^(-2 x) - 3 E^-x + 3 E^(-2 x) x^2 - 3 E^-x x^2 + E^(-2 x) x^3 + E^-x x^3 + 1/4 E^(-2 x) x^4 - 1/4 E^-x x^4 – Frey May 15 '15 at 08:24
  • just replace Exp[I x] by Exp[-x] in Coefficient. The idea is if you have $\sum_{i,j} x^i y^j$ then a[i, j] = Coefficient[Coefficient[p, y, j], x, i]. In your case $y=e^{-x}$. – Sumit May 15 '15 at 16:45
  • Thanks @Sumit. This is what exactly I needed, and this saves my time :) – Frey May 15 '15 at 20:17
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You can just use CoefficientList:

p = -1/2 Exp[-x] x^3 + 1/2 Exp[-2 x] x^2 + 1/2 Exp[-x] x^2 + 2 Exp[-2 x] x -
    2 Exp[-x] x + Exp[-2 x] - 2 Exp[-x] + 1;
CoefficientList[p, {Exp[-x], x}] //TeXForm

$\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ -2 & -2 & \frac{1}{2} & -\frac{1}{2} \\ 1 & 2 & \frac{1}{2} & 0 \\ \end{array} \right)$

Carl Woll
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