02 - Polynomials and Conjugate Roots.pdf
Polynomials and Conjugate Roots. Name___________________________________. Date________________ Period____. A polynomial function with rational coefficients has
unit 3 - writing a polynomial equation given the roots
POLYNOMIAL FUNCTIONS. WRITING A POLYNOMIAL EQUATION GIVEN THE ROOTS. COMPLEX CONJUGATE ROOTS THEOREM. Review of solving a simple quadratic equation like y = x²+
Polynomials Complex Conjugate Root Theorem Worksheet 2
Polynomials Complex Conjugate. Root Theorem. Worksheet 2. Answer each of the following without using a calculator and using the boxes provided for your answers
13.5 Notes - Conjugate roots and descartes rule of signs
The conjugate roots theorem: Let f(x) be a polynomial all of whose coefficients are real numbers. Suppose that is a root
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3.6 Roots of Polynomials. Rational Root Theorem: to find the possible rational Conjugate Root Theorem: irrational roots occur in conjugate pairs. For ...
Conjugate Reciprocal Polynomials with all Roots on the Unit Circle
Number Theory and Polynomials. Conjugate Reciprocal Polynomials with all. Roots on the Unit Circle. Christopher D. Sinclair sinclair@math.ubc.ca. PIMS SFU
WARM-UP 5-5 THEOREMS ABOUT ROOTS OF POLYNOMIAL
What is a quartic polynomial equation that has roots 2-3i 8
Irreducible polynomials with many roots of equal modulus
hold between conjugates where the ni's are integers but no quotient of two roots is a root of unity. In Lemma 1 of [2] Smyth gives a different proof of the
Mathematical Focus 1
Jun 30 2013 A student then asks
Roots of Polynomials
1 positive real root. I negative real rout. 2 complex roots as a conjugate pair. Bounds (generalization of root bracketing from the real line to the complex
02 - Polynomials and Conjugate Roots.pdf
Polynomials and Conjugate Roots A polynomial function with rational coefficients has the follow zeros. Find all additional zeros. 1) -1 1 + 3i.
Irreducible polynomials with many roots of maximal modulus
complex conjugate roots. A Pisot polynomial is a monic polynomial with integer coefficients with a single positive root outside the unit circle and.
Polynomials Complex Conjugate Root Theorem Worksheet 2
Polynomials Complex Conjugate. Root Theorem. Worksheet 2. Answer each of the following without using a calculator and using the boxes provided for your.
Littlewood polynomials
a plot of the zeros of Littlewood polynomials with degree up to 26. This plot Let ? be its complex conjugate. l(?)=0 ?.
Understanding Poles and Zeros 1 System Poles and Zeros
It is often convenient to factor the polynomials in the numerator and denominator A system has a pair of complex conjugate poles p1
4.4.2 - The Conjugate Root Theorem
If z = a + bi is a root of the polynomial f (z) with real coefficients then. ¯z = a - bi is also a root
Mathematical Focus 1
polynomials may produce complex solutions. solving polynomial equations. ... The Complex Conjugate Root Theorem states that complex roots always.
Zeros of a Polynomial Function
Conjugate Zeros Theorem: If the polynomial P has z is also a zero of P. olynomial with integer coefficients that tisfies the given conditions
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A polynomial f ? C[x] is conjugate reciprocal (CR) if correspond to degree N CR polynomials with all roots on the unit circle. PIMS SFU
Lecture 5: Algebra 3: Irreducible Primitive and Minimal Polynomials
f(X) = X+X2 has 0 as a root therefore f(X) = X(1+X). (as Thus ? and ?2 are roots of 1+X+X2 in GF(4). ... Minimal Polynomials and Conjugate Elements.
Polynomials and Conjugate Roots Date Period - Kuta Software
that has two imaginary roots ©f e2X0_1n6i cKFuWtzad GS]o]fZtmwSavrke_ fLuLACT M f pAGlslz trSiBglhItvsM hrteesJelrKvBe[dC K E nMFaIdUeW BweiitJht oIJnTfIiEn`iPtPe KPorceCcwa[lVcHu^lKuBsJ Worksheet by Kuta Software LLC
Complex Conjugate Roots - Mechamath
Roots of Polynomials Ch 7 Roots of Polynomials General form: n = order of the polynomial ai = constant coefficients Roots – Real or Complex 1 For an nth order polynomial – n real or complex roots 2 If n is odd ÆAt least 1 real root 3 If complex roots exist they are in complex conjugate pairs ( ) 2 0 = 0 + 1 + 2 +???+ = n f x a a
Lecture 1: Real Rooted Polynomials - University of Washington
properties of real rooted polynomials and we use them to study properties of the above polynomials 1 2 Real-rooted Polynomials We start by recalling some properties of real-rooted polynomials In the following simple lemma we show that imaginary roots of univariate polynomials come in conjugate pairs Lemma 1 2
Lecture 2: Real Stable Polynomials - University of Washington
is real-rooted The roots of the above polynomial are the eigenvalues of the matrix M0= M 1=2(B+b 1A 1 + +b nA n)M 1=2 Since B;A 1;:::;A nare symmetric M0is symmetric So its eigenvalues are real and the above polynomial is real-rooted If A 1;:::;A n 0 i e if the matrices have zero eigenvalues then we appeal to the following theorem
Symmetries and Polynomials - Harvard University
distinct real roots and D(P) < 0 if and only if P has two complex conjugate roots and one real root Compare your answer to Exercise 1 3 Your homework is to complete up through Exercise 1 10 If you ?nish that and still have time try the following questions 2
Searches related to polynomials and conjugate roots filetype:pdf
The classical formulas for the roots of low degree polynomials give some clues The Quadratic Formula: The roots of ax2 +bx+c? Q[x] are: x= ?b± ? b2 ?4ac 2a where ± ? b2 ?4acare the square roots of b2 ?4ac Proof: Divide through by aand complete the square: x2 + b a x+ c a = (x+ b 2a)2 + c a ? b2 4a2 = 0 The solutions are then
What is conjugate roots theorem?
- Conjugate roots theorem If the complex number is a root of the polynomial in a variable with real coefficients, then the complex conjugate is also a root of that polynomial. This theorem is very useful for finding roots of polynomials.
What is the root of a polynomial?
- Roots of Polynomials General form: n= order of the polynomial ai= constant coefficients Roots – Real or Complex 1. For an nthorder polynomial –nreal or complex roots 2. If n is odd ÆAt least 1 real root 3.
How do you find complex roots in an nthorder polynomial?
- For an nthorder polynomial –nreal or complex roots 2. If n is odd ÆAt least 1 real root 3. If complex roots exist, they are in complex conjugate pairs ( )20 = 0 + 1 + 2 +???+ = n f x a a x a x anx
What is the conjugate of a polynomial?
- As has been pointed out in the comments, the conjugate of a polynomial has different meanings, depending on the context. In the expression the quantity q ( x) ¯ is the complex conjugate of q ( x), i.e., the complex conjugate of the number that you obtain by evaluating p at x. Note that x ? R, which we shall assume in the following.
Lecture 2: Real Stable Polynomials
Lecturer: Shayan Oveis Gharan Jan 10thDisclaimer:These notes have not been subjected to the usual scrutiny reserved for formal publications.
A multivariate polynomialp2C[z1;:::;zn] isH-stable(or stable for short) ifp(z1;:::;zn)6= 0 whenever (z1;:::;zn)2 HnwhereH=fc2C:=(c)>0g
is the upper-half of then-dimensional complex plane. We saypisrealstable if all coecients ofpare real.
Unless otherwise specied, all polynomials that we work with in this course have real coecients. Fact 2.1.A univariate polynomialp2R[t]is real rooted i it is real stable. This simply follows from the fact that the roots ofpcome in conjugate pairs. So, ifphas a roottwith =(t)<0, we havetis also a root with=(t)>0. The above denition can be hard to understand; so, instead we discuss an equivalent denition. Lemma 2.2.A multivariate polynomialp2R[z1;:::;zn]is real stable i for every pointa2Rn>0and b2Rn, the univariate polynomialp(at+b)is not identically zero and is real rooted. For example,z1z2is not real stable because fora= (1;1) andb= (0;0) Proof.): Fixa2Rn>0andb2Rn. Ifp(at+b) is identically zero, then forzj=aji+bj,p(z1;:::;zn) = 0 sopis not real stable. Otherwise, sayp(at+b) has a roottwith=(t)6= 0. Then, sincep(at+b) has real coecients by Lemma 1.2 (see rst lecture), we can assume=(t)>0. Writet=ci+d; then for z j=ajt+bj=ajci+bj+daj p(z1;:::;zn) = 0 sopis not real stable. (: Supposepis not real stable; then there exists (z1;:::;zn)2 Hnthat is a root ofp. Setaj==(zj) and b j=<(zj) thenaj>0 for alljsop(at+b) is not identically zero and it must be real rooted. Butt=iis a root ofp(at+b) which is a contradiction.SeeFigure 2.1 for applications of the ab ovelemma. Let us discuss several examples of real stable polynomials Linear Functions:A linear polynomialp=a1z1++anznis real stable ia1;:::;an0. To see thisnote that if allzihave positive imaginary value then any positive combination also has a positive imaginary
value and thus is non-zero. Elementary Symmetric Polynomial:For anynand anykthe elementary symmetric polynomial e k(z1;:::;zn) =P S2(n k)Q i2Sziis real stable. I leave this as an exercise. 2-1Lecture 2: Real Stable Polynomials2-2Figure 2.1: Left diagram shows zeros of the polynomial 1xyand the right diagram shows zeros of 1 +xy
in the planeR2. Note that in the left gure any line pointing to the positive orthant crosses the zeros at two
points so 1xyis real stable but this does not hold in the right gure so 1 +xyis no real stable. Non-exampleThe polynomialz21+z22is not real stable; for example letz1=ei=4andz2=e3i=4. One of the most important family of real-stable polynomials is the determinant polynomial. Lemma 2.3.Given PSD matricesA1;:::;An2Rddand a symmetric matrixB2Rdd, the polynomial p(z) = det B+nX i=1z iAi! is real stable. Proof.ByLemma 2.2 , it is enough to show that for anya2Rn>0andb2Rn p(b+ta) = det B+nX i=1b iAi+tnX i=1a iAi! is real-rooted. First, assume thatA1;:::;Anare positive denite. Then,M=Pn i=1aiAiis also positive denite. So, the above polynomial is real-rooted if and only if det(M)det M 1=2 B+nX i=1b iAi! M1=2+tI!
is real-rooted. The roots of the above polynomial are the eigenvalues of the matrixM0=M1=2(B+b1A1+ +bnAn)M1=2. SinceB;A1;:::;Anare symmetric,M0is symmetric. So, its eigenvalues are real and the above polynomial is real-rooted. IfA1;:::;An0, i.e., if the matrices have zero eigenvalues, then we appeal to the following theorem.This completes the proof of the lemma. In particular, we construct a sequence of polynomial with matrices
Ai+I=2k. These polynomials uniformly converge topand each of them is real-stable by the above argument;
sopis real-stable.Lecture 2: Real Stable Polynomials2-3
Lemma 2.4(Hurwitz [Hur95]).Letfpkgk0be a sequence of -stable polynomials overz1;:::;znfor a connected and open set Cnthat uniformly converge topover compact subsets of . Then,pis -stable. Denition 2.5(d-homogeneous).A polynomialp2R[z1;:::;zn]isd-homogeneous ifp(z1;:::;zn) = dp(z1;:::;zn)for any2R. How general are these real stable polynomials and where should we look for them? Theorem 2.6(Choe, Oxley, Sokal, Wagner [COSW02]).The support of any multi-ane homogeneous realstable polynomial corresponds to the set of bases of a matroid (more generally, the support corresponds to a
jump system). For example, the support of a an elementary symmetric polynomial correspond to the set of bases of a uniform matroid whereas the non-stable polynomialz1z2+z3z4does not correspond to bases of a matroid.2.1 Closure Properties
In general, it is not easy to directly prove that a given polynomial is real stable or a given univariate
polynomial is real rooted. Instead, one may use an indirect proof: To show thatq(z) is (real) stable we can
start from a polynomialp(z) where we can prove stability usingLemma 2.3 , then we apply a sequence of
operators that preserve stability top(z) and we obtainq(z) as the result.In a brilliant sequence of papers Borcea and Branden characterized the set of linear operators that preserve
real stability [BB09a;BB09b;BB10]. We explain two instantiation of their general theorem and weuse them to show that many operators that preserve real-rootedness for univariate polynomials preserve
real-stability for of multivariate polynomials.We start by showing that some natural operations preserve stability and then we highlight two theorems of
Borcea and Branden.
The following operations preserve stability.
ProductIfp;qare real stable so ispq.
SymmetrizationIfp(z1;z2;:::;zn) is real stable then so isp(z1;z1;z3;:::;zn). SpecializationIfp(z1;:::;zn) is real stable then so isp(a;z2;:::;zn) for anya2R. First, note that for any integerk,pk=p(a+{2k;z2;:::;zn) is a stable polynomial (note thatpkmay have com- plex coecients). Therefore by Hurwitz theorem 2.4 , the limit offpkgk0is a stable polynomial, so p(a;z2;:::;zn) is stable. External FieldIfp(z1;:::;zn) is real stable then so isq(z1;:::;zn) =p(1z1;:::;nzn) for any positive vectorw2Rn0. Ifq(z1;:::;zn) has a root (z1;:::;zn)2 Hnthen (1z1;:::;nzn)2 Hnis a root of psopis not real stable. InversionIfp(z1;:::;zn) is real stable and degree ofz1isd1thenp(1=z1;z2;:::;zn)zd11is real stable. This is because the mapz17! 1=z1is a bijection betweenHand itself.DierentiationIfp(z1;:::;zn) is real stable then so isq=@p=@z1. This follows from Gauss-Lucas theorem.
Ifq(z1;:::;zn) is not real stable it has a root (z1;:::;zn). Denef(z1) =p(z1;z2;:::;zn). Then,f0(z1)has a root inH. But the roots off0(z1) are in the convex hull of the roots off(z1) we get a contradiction
because the complement ofHis convex.Lecture 2: Real Stable Polynomials2-4
In the rest of this course we write@z1as a short hand for@p=@z1.We can continue this list and try to discover more and more closure properties. Borcea and Branden proved
a remarkable result characterizing all linear operators that are stability preserving. Here, we don't discuss
their theorem in full generality but we discuss one of the main applications of their theorem which is being
used in almost all applications. LetT:R[z1;:::;zn]!R[z1;:::;zn] be an (dierential) operator on polynomials with real coecients dened as follows:X ;2N0c ;z@For example, 1z1@z2. In the abovez=Qn
i=1ziiand@=Qn i=1@izi.DeneFT2R[z1;:::;zn;w1;:::;wn]
FT(z1;:::;zn;w1;:::;wn) =X
;2N0c ;z(w):Note thatFTis a polynomial with 2nvariables.
Theorem 2.7.Tis an stability preserver operator, i.e., it maps any real stable polynomial to another real
stable polynomial, iFTis real stable. For example, the operator 1@z1is stability preserver, because 1 +w1is real stable. Also, 1 +z2@z1and1@z1@z2are stability preserver.
For a non-example, 1+@z1@z2and 1@z1@z2@z3are not stability preserver. This is because (1+@z1@z2)(z1z2) =
z1z2+ 1 is not real stable.
As a direct consequence we show that the Multi-Ane-Part (MAP) operator is stability preserving. Given
a polynomialp, MAP(p) zeros out every monomial ofpwith a square and keeps all multilinear monomials.For example,
MAP(1 + 2x+xy+x2y+z3) = 1 + 2x+xy:
Lemma 2.8.MAP is a stability preserving operator.
Proof.Given a polynomialp2R[z1;:::;zn]. We can write MAP as the following dierential operatorMAP(p) =nY
i=1(1z0i@zi)jz1==zn=0:Now, observe that (1z0i@zi) is stability preserving. Furthermore, setting allzi= 0 is also stability preserving.2.2 Applications
Given a graphG= (V;E), our rst application is to show that the matching polynomialG(z1;:::;zn) =X
M(1)jMjY
isat inMz i(2.1)Lecture 2: Real Stable Polynomials2-5
is real stable. For observe that p(z1;:::;zn) =Y fi;jg2E(1zizj) is real stable. Now observe thatG= MAP(p). So,Gis real stable. As a consequence the univariate matching polynomial XM(1)jMjtjMj
is real rooted. This follows by symmetrizing the polynomial in ( 2.1 ), i.e., setzi=tfor alliand noting that any univariate real stable polynomial is real rooted.quotesdbs_dbs19.pdfusesText_25[PDF] polyphemus pronunciation
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