polynomial in matlab roots

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mar vouz 2016-1-6

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评论： mar vouz 2016-1-7

Hey ,I have a matrix

X= 0 0,142857142857143 0,285714285714286 0,428571428571429 0,571428571428571 0,714285714285714 0,857142857142857 1

and a matrix

so I created the polynomial by p=polyfit(X,Y,7) and I was asked to find the real solutions of this so I used r=roots(p) and all i get are imaginary numbers. Why?

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Roger Stafford 2016-1-6

It is not clear what "real solutions" you are trying to find. Solutions to what? The expression "roots(p)" would give you the points where p has a zero value and presumably there are none that are real-valued.

Since you have used degree 7 with 8 points, your polynomial p should match the values of Y at the points X precisely except for tiny round-off errors. Is that the "solution" you are seeking? In that case you already have it.

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Answer 1

Walter Roberson 2016-1-6

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Polynomials with an odd number of roots always enter through one side of the Y axis and exit through the other side of the Y axis (because -infinity to an odd power is still negative infinity, opposite sign of +infinity to an odd power, whereas for even number of roots, -infinity to an even power is positive infinity same as +infinity to an even power so with an even number of roots the graph will start and end on the same side.)

The graph plot(X,Y) enters from the lower left so it has to exit through the upper right. It does not cross 0 anywhere in the range of values so there are no internal real roots. The first value is positive. Therefore there must be a single real root that is to the lower left of the graph at an X smaller than any given value.

There would have to be some visible zero crossings within the defined range for there to be multiple real roots.

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Answer 2

John D'Errico 2016-1-6

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编辑：John D'Errico 2016-1-6

在 MATLAB Online 中打开

A 7th degree polynomial (with real coefficients) has at least ONE real root.

p = polyfit(X,Y',7);
roots(p)
ans =
     1.0951 +    0.20415i
     1.0951 -    0.20415i
    0.63639 +    0.47627i
    0.63639 -    0.47627i
    0.10224 +    0.36777i
    0.10224 -    0.36777i
   -0.15631 +          0i

Note that the last root is NOT complex. It has a zero imaginary part. It is the only real root, at x=-0.15631...

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John D'Errico 2016-1-7

编辑：John D'Errico 2016-1-7

在 MATLAB Online 中打开

Polyfit will always return the coefficients of a 7th degree polynomial.

Will some of them be essentially zero? They will probably never be exactly zero, but MAY be floating point trash. And given this data, the probability is nil that they will ever be exactly zero.

Even for the extreme case that Roger poses,

p = polyfit(X,ones(1,8),7)
p =
 3.3186e-12  -1.1719e-11    1.633e-11  -1.1339e-11    4.049e-12  -6.7884e-13   3.8415e-14            1

we see that the polynomial lacks zero coefficients, although they are all floating point trash except for the constant term, which was and should be 1.

roots(p)
ans =
     39.822 +     18.933i
     39.822 -     18.933i
     10.215 +     42.543i
     10.215 -     42.543i
    -26.704 +     34.116i
    -26.704 -     34.116i
    -43.135 +          0i

Even there, roots does not recognize this as essentially a constant function, although arguably it should. Long ago some code I wrote to solve for the roots of a spline function catered to this case. There, we had to compute the roots of a cubic polynomial, that in some cases might degenerate into a quadratic, linear, or even constant function but with trash for one or more leading coefficient(s). So the equivalent of a direct roots computation could generate spurious roots.

Our solution was to reduce the order of the polynomial before the roots were computed, by looking at the leading coefficient(s). We even got tricky there as I recall, in how we threw away those high order terms, but that is a long story and would make this a quite long comment.