Section 2.6 Zeros of polynomials and Horner`s method

2.6 Zeros of Polynomials and
Horner’s Method
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Zeros of Polynomials
β€’ Definition: Degree of a Polynomial
A polynomial of degree 𝒏 has the form
𝑃 π‘₯ = π‘Žπ‘› π‘₯ 𝑛 + π‘Žπ‘›βˆ’1 π‘₯ π‘›βˆ’1 + β‹― + π‘Ž1 π‘₯ + π‘Ž0 , π‘Žπ‘› β‰  0
Where the π‘Žπ‘–β€² 𝑠 are constants (either real or complex)
called the coefficients of 𝑃 π‘₯
β€’ Fundamental Theorem of Algebra
If 𝑃 π‘₯ is a polynomial of degree 𝑛 β‰₯ 1, with real or
complex coefficients, 𝑃 π‘₯ = 0 has at least one root.
β€’ Corollary
There exists unique constants π‘₯1 , π‘₯2 , … , π‘₯π‘˜ and unique
positive integers π‘š1 , π‘š2 , … , π‘šπ‘˜ such that π‘˜π‘–=1 π‘šπ‘– = 𝑛
and
𝑃 π‘₯ = π‘Žπ‘› (π‘₯ βˆ’ π‘₯1 )π‘š1 (π‘₯ βˆ’ π‘₯2 )π‘š2 … (π‘₯ βˆ’ π‘₯π‘˜ )π‘šπ‘˜
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Remark:
1. Collection of zeros is unique
2. π‘šπ‘– are multiplicities of the individual zeros
3. A polynomial of degree 𝑛 has exactly 𝑛 zeros, counting
multiplicity.
β€’ Corollary
Let 𝑃(π‘₯) and 𝑄(π‘₯) be polynomials of degree at most 𝑛.
If π‘₯1 , π‘₯2 , … , π‘₯π‘˜ with π‘˜ > 𝑛 are distinct numbers with
𝑃 π‘₯𝑖 = 𝑄(π‘₯𝑖 ) for all 𝑖 = 1,2, … , π‘˜, then 𝑃 π‘₯ = 𝑄(π‘₯)
for all values of π‘₯.
Remark:
If two polynomials of degree 𝑛 agree at at least (n+1) points,
then they must be the same.
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Horner’s Method
β€’ Horner’s method is a technique to evaluate polynomials
quickly. Need 𝑛 multiplications and 𝑛 additions to evaluate
𝑃 π‘₯0 .
β€’ Assume 𝑃 π‘₯ = π‘Žπ‘› π‘₯ 𝑛 + π‘Žπ‘›βˆ’1 π‘₯ π‘›βˆ’1 + β‹― + π‘Ž1 π‘₯ + π‘Ž0 .
Evaluate 𝑃 π‘₯0 .
Let 𝑏𝑛 = π‘Žπ‘› , π‘π‘˜ = π‘Žπ‘˜ + π‘π‘˜+1 π‘₯0 ,
for π‘˜ = 𝑛 βˆ’ 1 , 𝑛 βˆ’ 2 , … , 1,0
Then 𝑏0 = 𝑃 π‘₯0 .
β€’ At the same time, we also computed the
decomposition:
𝑃 π‘₯ = π‘₯ βˆ’ π‘₯0 𝑄 π‘₯ + 𝑏0 ,
Where 𝑄 π‘₯ = 𝑏𝑛 π‘₯ π‘›βˆ’1 + π‘π‘›βˆ’1 π‘₯ π‘›βˆ’2 + β‹― + 𝑏2 π‘₯ + 𝑏1
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Remark:
1. 𝑃′ π‘₯0 = 𝑄(π‘₯0 ), which can be computed by
using Horner’s method in (n-1) multiplications
and (n-1) additions.
2. Horner’s method is nested arithmetic
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β€’ Example. Use Horner’s method to evaluate
𝑃 π‘₯ = 2π‘₯ 4 βˆ’ 3π‘₯ 2 + 3π‘₯ βˆ’ 4 at π‘₯0 = βˆ’2.
Solution:
Step 1: π‘Ž4 = 2, 𝑏4 = 2
Step 2: π‘Ž3 = 0, 𝑏3 = π‘Ž3 + 𝑏4 π‘₯0 = 0 + 2 βˆ’2 = βˆ’4
Step 3: π‘Ž2 = βˆ’3, 𝑏2 = π‘Ž2 + 𝑏3 π‘₯0 = βˆ’3 + βˆ’4 βˆ’2 = 5
Step 4: π‘Ž1 = 3, 𝑏1 = π‘Ž1 + 𝑏2 π‘₯0 = 3 + 5 βˆ’2 = βˆ’7
Step 5: π‘Ž0 = βˆ’4, 𝑏0 = π‘Ž0 + 𝑏1 π‘₯0 = βˆ’4 + βˆ’7 βˆ’2 = 10
Thus 𝑃 βˆ’2 = 𝑏0 = 10
Remark:
π‘Ž4 π‘₯ 4 + π‘Ž3 π‘₯ 3 + π‘Ž2 π‘₯ 2 + π‘Ž1 π‘₯ + π‘Ž0 =
π‘Ž4 π‘₯ + π‘Ž3 π‘₯ + π‘Ž2 π‘₯ + π‘Ž1 π‘₯ + π‘Ž0
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