6.5. A Summation Method Asume that:

Slides:



Advertisements
Similar presentations
Solving Linear Programming Problems Shubhagata Roy.
Advertisements

The factor theorem The Factor Theorem states that if f(a) = 0 for a polynomial then (x- a) is a factor of the polynomial f(x). Example f(x) = x 2 + x -
MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 19, Friday, October 17.
MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 26, Monday, November 3.
MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 39, Monday, December 8.
MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 38, Friday, December 5.
Splash Screen. Lesson Menu Five-Minute Check (over Lesson 1–3) Then/Now New Vocabulary Key Concept: Distributive Property Example 1: Real-World: Distribute.
Equivalent Fractions Equivalent Fractions Using Cross-Multiplication.
Splash Screen. Lesson Menu Five-Minute Check CCSS Then/Now New Vocabulary Key Concept: Order of Operations Example 1:Evaluate Algebraic Expressions Example.
Over Lesson 1–3. Then/Now Understand how to use the Distributive Property to evaluate and simplify expressions.
Quadratics Solving equations Using “Completing the Square”
Splash Screen. Lesson Menu Five-Minute Check CCSS Then/Now New Vocabulary Key Concept: Order of Operations Example 1:Evaluate Algebraic Expressions Example.
Splash Screen. Example 1 Solve a Logarithmic Equation Answer: x = 16 Original equation Definition of logarithm 8 = 2 3 Power of a Power Solve.
Splash Screen. Lesson Menu Five-Minute Check (over Lesson 8–3) Then/Now New Vocabulary Example 1: Solve a Logarithmic Equation Key Concept: Property of.
MA.912.A.4.2: Add, subtract, and multiply polynomials. Which of the following expressions is equivalent to (5x − 3) 2 ? A. 25x 2 − 30x + 9 B. 25x 2 −
Splash Screen. Lesson Menu Five-Minute Check (over Lesson 1–3) CCSS Then/Now New Vocabulary Key Concept: Absolute Value Example 1:Evaluate an Expression.
Lesson 5 Menu Five-Minute Check (over Lesson 1-4) Main Ideas and Vocabulary Targeted TEKS Key Concept: Distributive Property Example 1: Distribute Over.
WARM UP EVALUATING EXPRESSIONS Evaluate the expression for the given value of the variable (Lesson 2.5). 1.-3(x) when x = 9 2.4(-6)(m) when m = (-n)(-n)
Splash Screen. Lesson Menu Objectives Vocabulary Example 1 Example 2 Quick Quiz.
REVIEW FOR QUIZ 3 ALGEBRA II. QUESTION 1 FACTOR THE FOLLOWING QUADRATIC 3N 2 + 7N + 4 Answer: (3n + 4)(n + 1)
POLYNOMIALS 9/20/2015. STANDARD FORM OF A POLYNOMIAL ax + bx + cx … mx + k nn-1n-2 The degree is ‘n’
Over Lesson 1–3 5-Minute Check 1 A.Multiplicative Property of Zero B.Multiplicative Inverse C.Commutative Property D.Identity Which property is demonstrated.
Splash Screen. Over Lesson 5–4 5-Minute Check 2 Solve the inequality. Show your work and 2 checks. 3t – 5 ≤ 13.
Bell Ringer: Simplify each expression
PreCalculus Section 2.6 Solve polynomial equations by factoring and the Rational Roots Theorem. Solve by factoring: x 3 + 5x 2 – 4x – 20 = 0 x 6 – x 3.
Factoring Factoring in the format ax2+bx+c Problem 1 X 2 + 7x + 10.
1.4 Identity and Equality Properties The sum of any number and 0 is equal to the number. So, 0 is called the additive identity. a + 0 = 0 + a = a
Splash Screen. Lesson Menu Five-Minute Check (over Lesson 1–3) Then/Now New Vocabulary Key Concept: Absolute Value Example 1:Evaluate an Expression with.
Splash Screen. Lesson Menu Five-Minute Check CCSS Then/Now New Vocabulary Key Concept: Order of Operations Example 1:Evaluate Algebraic Expressions Example.
PreCalculus Section 2.7 Apply general theorems about polynomials Fundamental Theorem of Algebra In the complex number system, a polynomial of degree n.
Distributive Property, Combining Like Terms (1-4) Objective: Use the Distributive Property to evaluate and simplify expressions. Combine like terms.
Splash Screen. Lesson Menu Five-Minute Check (over Lesson 1–3) CCSS Then/Now New Vocabulary Key Concept: Absolute Value Example 1:Evaluate an Expression.
Last Answer LETTER I h(x) = 3x 4 – 8x Last Answer LETTER R Without graphing, solve this polynomial: y = x 3 – 12x x.
SAT/ACT MATH UNIT 10 Equation with More Than One Variable.
Systems of Linear Differential Equations
Binomial Coefficients: Selected Exercises
Aim: What are the properties of a quadratic equation?
7.3 Solving Equations Using Quadratic Techniques
Name:__________ warm-up 5-6
Remainder Theorem What’s left over?.
Do Now – Take out your homework. Take You have 5 minutes.
Counting III: Pascal’s Triangle, Polynomials, and Vector Programs
Sum and Product of Roots
Lesson 7.6 EQ: How do you factor a polynomial when leading coefficient is not 1? Topic/Objective: To factor trinomials in the form ax2 +bx + c   Factor.
Linear Combination of Two Random Variables
Another Example -#x(Ax&Bx) 1) -#x(Ax&Bx) A $x(Ax>-Bx) The 2 ways to
Solving Quadratic Equations by Factoring
Notes 5.6 (Day 1) Find Rational Zeros.
Polynomials and the FFT(UNIT-3)
Factoring Quadratics using X-Box method
The Rational Zero Theorem
Cubic functions As we have seen a function in which the highest power of x is 3 is called a cubic function. The general form of a cubic function is: y.
Advanced Algorithms Analysis and Design
Polynomial Functions Equations and Graphs Characteristics
Great Theoretical Ideas in Computer Science
Evaluate the expression.
Linear Algebra Lecture 3.
Non-Homogeneous Systems
Factoring Quadratics using X-Box method
Counting III: Pascal’s Triangle, Polynomials, and Vector Programs
Practice Quiz Quadratics and Rationals
5.5 – Completing the Square
X-box Factoring.
5.4 Completing the Square.
Practice Problems ax2 + bx + c
A, b and c can be any numbers
BASIC FEASIBLE SOLUTIONS
A, b and c can be any numbers
A, b and c can be any numbers
Presentation transcript:

MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 30, Wednesday, November 12

6.5. A Summation Method Asume that: A(x) = S anxn, B(x) = S bnxn, and C(x) = S cnxn. If bn = dan, then B(x) = dA(x). If cn = an + bn, then C(x) = A(x) + B(x). If cn = a0bn + a1bn-1 + ... + an-1b1 + anb0, then C(x) = A(x)B(x). If bn = an-k, exept bi = 0 for i < k, then B(x) = xk A(x).

Example 1 Build a generating function h(x) for ar = 2r2. Answer: h(x) = 2x(1 + x)/(1 – x)3.

Example 2 Build a generating function h(x) with ar = (r + 1)/(r – 1) Answer: h(x) = 6x2(1 – x)-4.

Theorem 1 If h(x) is a generating function for ar, then h*(x) = h(x)/(1 – x) is the generating function for the sums of the ars. That is, h*(x) = a0 + (a0 + a1)x + (a0 + a1 + a2)x2 + ...

Example 1 (continued) Evaluate the sum: h*(x) = h(x)/(1 – x) = 2x/(1 - x)4 + 2x2/(1 – x)4. Answer: 2C(n+2,3) + 2C(n + 1, 3).

Example 2 (continued) Evaluate the sum 3 £ 2 £ 1 + 4 £ 3 £ 2 + ... + (n+1)n(n-1). Answer:h*(x) = 6x2/(1 – x)5. The coefficient at xn-2 is 6C(n + 2, 4).