How to extract powers of a symbolic polynomial?

Question

Alina Rossi-Conaway 2021-3-18

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https://ww2.mathworks.cn/matlabcentral/answers/777072-how-to-extract-powers-of-a-symbolic-polynomial

评论： Walter Roberson 2023-9-25

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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!!

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Answer 1

Steven Lord 2021-3-22

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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 =

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Walter Roberson 2021-7-19

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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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Answer 2

Shubham Rawat 2021-3-22

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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!

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Davy Figaro 2021-7-19

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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