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For given an expression like
$-\frac{e^{-\frac{\text{d1} t}{\text{c1}}} \left(-\text{a1} \text{a2} \text{d1}^2+\text{a1} \text{b2} \text{c1} \text{d1}+\text{a2} \text{b1} \text{c1} \text{d1}-\text{b1} \text{b2} \text{c1}^2\right)}{\text{c1} \text{d1} (\text{c2} \text{d1}-\text{c1} \text{d2})}-\frac{e^{-\frac{\text{d2} t}{\text{c2}}} \left(-\text{a1} \text{a2} \text{d2}^2+\text{a1} \text{b2} \text{c2} \text{d2}+\text{a2} \text{b1} \text{c2} \text{d2}-\text{b1} \text{b2} \text{c2}^2\right)}{\text{c2} \text{d2} (\text{c1} \text{d2}-\text{c2} \text{d1})}+\frac{\text{b1} \text{b2}}{\text{d1} \text{d2}}$ The Mathematica code of this expression is

(b1 b2)/(d1 d2) - ((-b1 b2 c1^2 + a2 b1 c1 d1 + a1 b2 c1 d1 - 
a1 a2 d1^2) E^(-((d1 t)/c1)))/(
c1 d1 (c2 d1 - c1 d2)) - ((-b1 b2 c2^2 + a2 b1 c2 d2 + a1 b2 c2 d2 - 
a1 a2 d2^2) E^(-((d2 t)/c2)))/(c2 d2 (-c2 d1 + c1 d2))

How to expand it into $\sum _{i=0}^{\infty } a_i e^{b_i}$ within Mathematica?

I know $a_i$ and $b_i$ for the given expression without writing any code, but how to write a personal general function to do this work?

Thank you very much!

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    $\begingroup$ I suggest you that you post something in Mathematica format. Otherwise it is rather impossible that you will get a reply. $\endgroup$
    – Dimitris
    Feb 24, 2017 at 13:05
  • $\begingroup$ @dimitris Thanks for your advice. I have modified my question. $\endgroup$
    – tanghe2014
    Feb 24, 2017 at 14:22
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    $\begingroup$ Surely you need to put some constraints on a_i and b_i. Otherwise you could trivially write every function f[t]=f[t]*E^0. Are you expecting this expansion to be unique? $\endgroup$
    – jjc385
    Feb 24, 2017 at 15:02
  • $\begingroup$ perhaps something along the lines of $b_i\to\log\beta_i$, and Taylor expand in $\beta_i$? $\endgroup$ Feb 25, 2017 at 13:43

2 Answers 2

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I would try something like this:

f = a E^x +b + c E^(7y) + d;
Replace[f, b_ Exp[a_] -> g[b, a], 1];
Replace[%, a_Plus :> List @@ a];
% /. g -> List;
Replace[%, b_ /; ! ListQ[b] -> {b, 0}, 1];
% //. {a___, {x_, e_}, b___, {y_, e_}, c___} -> {a, {x + y, e}, b, c};
{as, bs} = Transpose[%];
as
bs
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I find some code (What is the correct way to use specific rule for specific pattern match when parsing an expression?) to do some similar work.
This code is

Clear[coeff, z]

(*Applies to plus expressions*)
coeff[expr_Plus] := 
 Cases[expr,HoldPattern[(c : _?(FreeQ[#, z] &) : 1) (zVar : z^e_. : z^0)] :> {c,e}]

(*Applies to standalone expressions:nest them in a list first before using Cases*)
coeff[expr_] := 
 Cases[{expr},HoldPattern[(c : _?(FreeQ[#, z] &) : 1) (zVar : z^e_. : z^0)] :> {c,e}]

{#, coeff@#} & /@ {1, z, -z, 1/z, 1 + z^-1, 1/z^3 + 3/z + 3 z + z^3, 
 6 + 1/z^4 + 4/z^2 + 4 z^2 + z^4} // 
 Grid[Prepend[#, {"expr", "coeff[expr]"}], Frame -> All] &

The return of it is Out of this code. To get the coefficient and exponent I think one can modify this code to get the correct answer of my question, but I don't know how to finish it.

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