How to extract powers of a symbolic polynomial?

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Alina Rossi-Conaway
Alina Rossi-Conaway am 18 Mär. 2021
Kommentiert: Walter Roberson am 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!!

Antworten (3)

Steven Lord
Steven Lord am 22 Mär. 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 Kommentar
Walter Roberson
Walter Roberson am 19 Jul. 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 am 22 Mär. 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 Kommentar
Davy Figaro
Davy Figaro am 19 Jul. 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 am 25 Sep. 2023
(x+1)^2
  1 Kommentar
Walter Roberson
Walter Roberson am 25 Sep. 2023
Could you explain how that line of code solves the problem mentioend by @Alina Rossi-Conaway ?

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