How to extract powers of a symbolic polynomial?

45 vues (au cours des 30 derniers jours)
Alina Rossi-Conaway
Alina Rossi-Conaway le 18 Mar 2021
Commenté : Walter Roberson le 25 Sep 2023
I'm working with a symbolic polynomial
y = 0.96*z^3500 + 0.04*z^0
I can extract the coefficients easily with
coeffs(y)
but I cannot figure out a way to pull off the corresponding powers of z into a vector. I've tried doing some wonky stuff with logs, but nothing so far. Am I SOL?
Thank you!!

Réponses (3)

Steven Lord
Steven Lord le 22 Mar 2021
syms z
y = 0.96*z^3500 + 0.04*z^0
y = 
[coefficients, powers] = coeffs(y)
coefficients = 
powers = 
syms y positive
exponents = simplify(subs(log(powers)./log(z), z, y))
exponents = 
  1 commentaire
Walter Roberson
Walter Roberson le 19 Juil 2021
Different formulation for finding the exponents.
syms z
y = 0.96*z^randi(9999) + 0.04*z^randi(9999)
y = 
[coefficients, powers] = coeffs(y)
coefficients = 
powers = 
exponents = mapSymType(powers, 'power', @(Z) children(Z,2));
if powers(end) == 1; exponents(end) = 0; end
exponents
exponents = 
This particular code relies on an enhancement to children() that was made a small number of releases ago. A workaround is possible for older releases.

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Shubham Rawat
Shubham Rawat le 22 Mar 2021
Hi Alina,
You may first find coefficients of all variables like this:
coef = sym2poly(y);
Then you can find all the index of all the non zero elements and -1 as indexing start from 1 in MATLAB:
polyPowers = find(coef) - 1;
Hope this Helps!
  1 commentaire
Davy Figaro
Davy Figaro le 19 Juil 2021
This is good, but because symbolic polynomial powers are presented in descending order, you need to double flip to get the polyPowers to line up to the original polynomial:
polyPowers = flip(find(flip(coef))) - 1;

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sil das
sil das le 25 Sep 2023
(x+1)^2
  1 commentaire
Walter Roberson
Walter Roberson le 25 Sep 2023
Could you explain how that line of code solves the problem mentioend by @Alina Rossi-Conaway ?

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