Question about using fzero to find all real roots of a polynomial
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I have the polynomial y = x^7-4.75*x^6+10.875*x^5-20.125*x^4+20*x^3+1.75*x^2-30*x+25 and I want to find not just one real root, but all three of them. I know for a fact that they are: -1.00, 1.25, and 2.50. I DO know how to find these values in many other ways, but I want to see if I can display all three of these real roots with this following program:
ff=[1 -4.75 10.875 -20.125 20 1.75 -30 25];
fun = @f;
x0 = 0;
B = fzero(fun,x0)
In particular, this yields B = -1 with x0 = 0 as my guess, but I want to find and display all three of them. Now the basic question: Is it possible to display all of the real roots of this polynomial using the fzero command?
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回答 (2 件)
Azzi Abdelmalek
2014 年 11 月 9 日
編集済み: Azzi Abdelmalek
2014 年 11 月 9 日
ff=[1 -4.75 10.875 -20.125 20 1.75 -30 25];
out=roots(ff)
out=out(imag(out)==0)
%or
out=out(imag(out)<1e-5)
Roger Stafford
2014 年 11 月 9 日
Is it possible to display all of the real roots of this polynomial using the fzero command?
The answer to this question is yes provided 1) that you furnish a sufficiently close initial estimate (x0) for each root, and 2) that none of the roots is a double root - that is, a point where the curve becomes only tangent to the x-axis instead of crossing it. To accomplish 1) it is helpful to make a graph of the polynomial.
Your 'fun' is not properly defined. I hope you used code other than what you show.
2 件のコメント
Roger Stafford
2014 年 11 月 9 日
No, 'fzero' does not function that way. For each initial estimate it will give only one answer, so to get three roots you would have to call on it with at least three different estimates. The 'roots' function is what you need to get all roots with one run.
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