How can I expand polynomials with matlab?

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Sachi
Sachi 2015 年 1 月 14 日
回答済み: Star Strider 2015 年 1 月 14 日
Can matlab expand something like the following and represent it in terms of powers of 'x'?
x(x-7)(x-6) + (x+4)(x-9)(x-2) + x(x-8)(x+1) ----> k_1 x^3 + k_2 x^2 + k_3 x + k_4
  1 件のコメント
Star Strider
Star Strider 2015 年 1 月 14 日
It can if you have the Symbolic Math Toolbox.

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採用された回答

John D'Errico
John D'Errico 2015 年 1 月 14 日
Trivial. With or without the symbolic toolbox. First, with...
syms x
P = expand(x*(x - 7)*(x - 6) + (x + 4)*(x - 9)*(x - 2) + x*(x - 8)*(x + 1))
P =
3*x^3 - 27*x^2 + 8*x + 72
Without the symbolic toobox, but with my sympoly toolbox , as found on the file exchange. Its free, but less capable than the symbolic one.
sympoly x
P = x*(x - 7)*(x - 6) + (x + 4)*(x - 9)*(x - 2) + x*(x - 8)*(x + 1)
P =
72 + 8*x - 27*x^2 + 3*x^3
In either case, it is easy enough to extract the numeric coefficients alone, if that is what you wanted.
And if you have neither toolbox and you like to do your own work (hey the sympoly toolbox is FREE after all)
coef = conv(conv([1 0],[1 -7]),[1 -6]) + ...
conv(conv([1 4],[1 -9]),[1 -2]) + ...
conv(conv([1 0],[1 -8]),[1 1])
coef =
3 -27 8 72

その他の回答 (1 件)

Star Strider
Star Strider 2015 年 1 月 14 日
An alternative using poly (assuming I got the maths correct) is:
% f(x) = x(x-7)(x-6) + (x+4)(x-9)(x-2) + x(x-8)(x+1)
rts1 = [ 0 7 6];
rts2 = [-4 9 2];
rts3 = [ 0 8 -1];
trm1 = poly(rts1);
trm2 = poly(rts2);
trm3 = poly(rts3);
ply = trm1 + trm2 + trm3;
produces:
ply =
3 -27 8 72

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