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PolynomialTools

  

Hurwitz

  

decide whether a polynomial has all its zeros strictly in the left half plane

 

Calling Sequence

Parameters

Description

Examples

References

Calling Sequence

Hurwitz(p, z,'s','g')

Parameters

p

-

polynomial with complex coefficients

z

-

variable of the polynomial p

's'

-

(optional) name

'g'

-

(optional) name

Description

• 

The Hurwitz(p, z) function determines whether the the polynomial pz has all its zeros strictly in the left half plane.

• 

A polynomial is a Hurwitz polynomial if all its roots are in the left half plane.

• 

The parameter p is a polynomial with complex coefficients. The polynomial may have symbolic parameters, which evalc and Hurwitz assume to be real.  The paraconjugate  p* of p is defined as the polynomial whose roots are the roots of p reflected across the imaginary axis.

• 

The parameter 's', if specified, is a name to which the sequence of partial fractions of the Stieltjes continued fraction of pp&ast;p&plus;p&ast; will be assigned. The first element of the sequence returned in 's' is special. If it is of higher degree than 1 in z, p is not Hurwitz. If it is of the form bz+a, where a0orb<0, p is not Hurwitz, either. If each subsequent polynomial in the sequence returned is of the form bz+a, where a=0and0<b, then p is a Hurwitz polynomial.

  

This is useful if p has symbolic coefficients. You can decide the ranges of the coefficients that make p Hurwitz.

• 

If the Hurwitz function can use the previous rules to determine that p is Hurwitz, it returns true. If it can decide that p is not Hurwitz, it returns false. Otherwise, it returns FAIL.

• 

The parameter 'g', if specified, is a name to which the gcd of p and its paraconjugate  p&ast; will be assigned. The zeros of this gcd are precisely the zeros of p which are symmetrical under reflection across the imaginary axis.

• 

If the gcd is 1 while the sequence of partial fractions is empty, the conditions for being a Hurwitz polynomial are trivially satisfied. A manual check is recommended, though a warning is returned only if infolevel[Hurwitz] >= 1.

Examples

withPolynomialTools&colon;

p1z2+z+1

p1z2+z+1

(1)

Hurwitzp1&comma;z

true

(2)

p23z3+2z2+z+c

p23z3+2z2+c+z

(3)

Hurwitzp2&comma;z&comma;s2&comma;g2

FAIL

(4)

s2

3z2&comma;4z3c2&comma;3z2+zc

(5)

g2

1

(6)

The elements of s2 are all positive if and only if 0<c<23, by inspection. Thus, you can use the information returned even when the direct call to Hurwitz fails.

Separate calls to Hurwitz in the cases c=0 and c=23 give nontrivial gcds between p2 and its paraconjugate. Thus, the stability criteria are satisfied only as above.

p34z4+z3+z2+c

p34z4+z3+z2+c

(7)

Hurwitzp3&comma;z&comma;s3&comma;g3

FAIL

(8)

s3

0&comma;4z&comma;z&comma;zc&comma;z

(9)

Notice that the last term has coefficient −1. Thus, you can say unequivocally that p3 is not Hurwitz, for any value of c.

p4z5+5z4+4z3+3z2+2z+c

p4z5+5z4+4z3+3z2+c+2z

(10)

Hurwitzp4&comma;z&comma;s4&comma;g4

FAIL

(11)

s4

z5&comma;25z17&comma;289z55c+1&comma;5c+12z17c2+48c2&comma;c248c+2z5c+1c

(12)

By inspecting s4, notice that p4 is Hurwitz only if 15<c, and c2+48c<2, and 0<c. This can be simplified to the conditions 0<c<24&plus;172&equals;0.04... 

p5p2+Id

p53z3+2z2+c+z+Id

(13)

evalc and the Hurwitz function assume that symbolic parameters have real values.

Hurwitzp5&comma;z&comma;s5&comma;g5

FAIL

(14)

s5

3z2&comma;4z3c28Id3c22&comma;3c23z29c312c28d2+4c+Id3c229c312c28d2+4c

(15)

The coefficients of s5 can be inspected according to rules, but it is a tedious process.

p6expandx1x2+2xc

p6cx3+x4+cx2x32cx+2x2+2c2x

(16)

Hurwitzp6&comma;x&comma;s6&comma;g6

false

(17)

g6

x2+2

(18)

p7x+sqrt2

p7x+2

(19)

Hurwitzp7&comma;x

true

(20)

p8x3+cx2+c21x+1

p8x3+cx2+c21x+1

(21)

Hurwitzp8&comma;x&comma;s8&comma;g8

FAIL

(22)

s8

xc&comma;c2xc3c1&comma;c2xxxc

(23)

Examination of the above for real values of c is a way to determine whether the polynomial is Hurwitz.

p9expandcz2+1z+1z2+2z+2

p9cz5+3cz4+4cz3+2cz2+z3+3z2+4z+2

(24)

Hurwitzp9&comma;z&comma;s9&comma;g9

FAIL

(25)

s9

(26)

g9

cz2+1

(27)

In the previous example, c might be zero. Thus, Hurwitz cannot determine whether all the zeros are in the left half plane.

References

  

Levinson, Norman, and Redheffer, Raymond M. Complex Variables. Holden-Day, 1970.

See Also

evalc

expand

fsolve

Hurwitz Zeta Function

PolynomialTools

sqrt

subs