How do I generate a polynomial from roots that are repeated?

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Guillem Campins 2020-11-28

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评论： Star Strider 2020-11-28

采纳的回答： Star Strider

So I've got the roots of a polynomial from which I want to design the polynomial.

The thing is that I've got 8 roots, 4 of which are repeated. These are the roots I have:

-0.7437 - 1.4178i

-1.1031 + 0.1267i

-0.4571 + 0.9526i

-0.0977 + 1.0976i

When I do the polynomial by using the function poly, I get the polynomial that corresponds to the 8 roots but I'd like the polynomial for only the 4, non-repeated roots.

How could I do this? (I know I could probably eliminate the repeated, but I'd like a function that considers repeated roots or something).

Thanks in advance

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Answer 1

Star Strider 2020-11-28

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编辑：Star Strider 2020-11-28

Iinteresting. I would expect no repeats, and complex conjugate values instead.

Try this:

rts = [ -0.7437 - 1.4178i
        -0.7437 - 1.4178i
        -1.1031 + 0.1267i
        -1.1031 + 0.1267i
        -0.4571 + 0.9526i
        -0.4571 + 0.9526i
        -0.0977 + 1.0976i
        -0.0977 + 1.0976i];
Urts = unique(rts);
p1 = poly(Urts);
Ucrts = [Urts; conj(Urts)];
p2 = poly(Ucrts);

.

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Guillem Campins 2020-11-28

Alright, that works! Thanks very much :)

Star Strider 2020-11-28

As always, my pleasure!

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Answer 2

Image Analyst 2020-11-28

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Did you try just multiplying them all together?

x = linspace(-8, 5, 1000);
r1 =  -0.7437 - 1.4178i
r2 =  -0.7437 - 1.4178i
r3 =  -1.1031 + 0.1267i
r4 =  -1.1031 + 0.1267i
r5 =  -0.4571 + 0.9526i
r6 =  -0.4571 + 0.9526i
r7 =  -0.0977 + 1.0976i
r8 =  -0.0977 + 1.0976i
y = (x-r1).*(x-r2).*(x-r3).*(x-r4).*(x-r5).*(x-r6).*(x-r7).*(x-r8)
plot(x, real(y), 'r-', 'LineWidth', 2);
hold on;
plot(x, imag(y), 'b-', 'LineWidth', 2);
grid on;
legend('Real', 'Imaginary');

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Guillem Campins 2020-11-28

Hmm the problem with that is that I need the value of the polynomial in a straight form (as in x^3+2x^2+1) more than I need the plot, and the polynomial is expressed in a very non-straight form by doing that.

Thanks for the effort, though :)

Image Analyst 2020-11-28

Maybe might have to use the symbolic toolbox, which I don't have, to do the multiplication.

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