Los Angeles City College Chemistry 60/68 Mock Exam _2 ANSWER KEY ...

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Chemistry 60/68 Mock Exam _2 ANSWERS pp. 1 Los Angeles City College Chemistry 60/68 Mock Exam _2 ANSWER KEY Professor Torres Multiple Choice 1. …C10H12O4S(s) + …O2(g) → …CO2(g) + …SO2(g) + …H2O(g) When the equation above is balanced and all coefficients are reduced to their lowest whole-number terms, the coefficient for O2(g) is A. 6 B. 7 C. 12 D. 14 2.

• molecular weight of 78.05 amu
• h.a.
• h. a.
• aqueous sodium iodide
• hc hc
• h2o a. balance
• precipitate
• reaction
• a.
• c.
• 3 c.
• 2b.
• 3 b.
• b.

Sujets

• Exposé
• NutraSweet
• Aqueous solution
• Réaction
• Molecule
• Precipitation (chemistry)
• Aspartame
• Chloride

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• Publié par: istyo
• Nombre de lectures: 19
• Langue: English

Extrait

Lecture 18 MATH1904

Homogeneous Linear Recurrence Relations

Ak-th order linear recurrence relationfor the sequencex ,x , ∙ ∙ ∙has the form 0 1

x=a x n 1n +a x+∙ ∙ ∙+a x+f n , −1 2n−2 k n−k forn≥k, wherea,a,. . .,aare 1 2 k constants andf,f,f,. . .is some k k+1k+2 given sequence.

We shall restrict our attention to the homogeneouslinear recurrence relations: those for which all thef n are 0. These are the recurrence relations of the form

x=a x+a x+∙ ∙ ∙+a x , n 1n−1 2n−2 k n−k forn≥k, or equivalently,

x−a x− ∙ −∙ ∙ x= 0 x n −a n−1 2n−2 a k n−k , 1 forn≥k.

1

Example:

Consider the recurrence relation:

x= 0, x n + 2x−15 n−2 n−1 forn≥2, wherex= 0 andxLet= 1. 0 1 2 3 G(z) =x+x z+x z+x z+∙ ∙ ∙ 0 1 2 3 be the generating function of the sequence x,x,x,. . .multiplying by. After zand 0 1 2 2 zwe find that 2 3 zG(z) =x z+x z+x z+∙ ∙ ∙ 0 1 2 2 2 3 z G(z) =x z+x z+∙ ∙ ∙ 0 1

We now form the expression 2 G(z) + 2zG(z)−15z G(z). Because of the recurrence relation most of the terms cancel and so 2 G(z)(1 + 2z−15z) =x+ (x+ 2x)z. 0 1 0 Using the initial valuesx= 0 andx= 1 0 1 2 and dividing by (1 + 2z−15z) leads to the closed form z G(z) =. 2 1 + 2z−15z

2

2 We have 1 + 2z−15z= (1−3z)(1 + 5z) and we want to find numbersAandBsuch that z A B = +. 2 1 + 2z−15z1−3z1 + 5z 1 Solving this forAandBwe find thatA= 8 1 andB=−. Thus 8 1 1 1 1 G(z) =−. 8 1−3z8 1 + 5z −1 Using the formula for (1−zreplacing) , z by 3zfor the first term, andzby−5zfor the second term, we obtain 1 2 2 G(z(1 + 3) = z+ 3z+∙ ∙ ∙) 8 1 2 2 3 3 −(1−5z+ 5z−5z+∙ ∙ ∙) 8 n The coefficient ofzinG(z) isx n and from the expression forG(z) just found we have the solution 1 1 n n x n =∙3−(−5) 8 8 to our recurrence relation.

3

This example suggests that the solution to a homogeneous recurrence relation is a combination ofn-th powers. If we know this to be true we could solve the previous problem as follows.

Step 1Find the values ofλsuch that n x n =λis a solution to the recurrence relation.

Step 2Find out what combinations of the solutions found in Step 1 satisfy the initial conditions.

Here is how this approach works for the previous example

2x−15x= 0, x n + n−1n−2 forn≥2 with initial conditionsxand= 0 0 x= 1. 1

n First substitutex n =λin to the recurrence. After cancellation we see thatλ satisfies 2 λ+ 2λ−15 = 0.

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That is, (λ−3)(λand so the only= 0 + 5) solutions forλare 3 and−5.

n n The next step is to putx n =A3 +B(−5) and find the values ofAandBwhich satisfy the initial conditions. That is, we must solve

A+B= 0 3A−5B= 1 1 1 ThereforeA= andB=−. 8 8

Thus

as before.

1 1 n n x n =∙3−(−5), 8 8

5

First order linear recurrence relations

The simplest example of a recurrence relation is

x=λx n n −1

If 2 3 G(z) =x+x z+x z+x z+∙ ∙ ∙ 0 1 2 3 Then 2 3 zG(z) =x z+x z+x z+∙ ∙ ∙ 0 1 2 and therefore (1−λz)G(z) =x. Thus the 0 −1 generating function forG(z) isx(1−λz) 0 n and sox=x λ. n 0

Indeed we can also prove this by induction. Forn= 1 we havex=λxby assumption. 1 0 n On the other hand, ifn≥1 andx n =λ x 0 , n n+1 thenx=λx n =λ λ x=λ x 0 , n +1 0 completing the induction step.

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Second order linear recurrence relations

ax+bx, x n = n−1n−2

0 0 0 Ifx,x,x,. . . 0 1 2 sequences which relation and ifA 0 00 x Bx x n =A n + n

forn≥2

00 00 00 andx,x,x,. . .are two 0 1 2 satisfies the recurrence andBare constants, then also satisfies it.

0 00 Thus thelinear combination(Ax+Bx) n n 0 00 of the two solutions (x) and (x) is also a n n solution. This is a very important property and holds in general for all homogeneous linear recurrence relations.

The general solution to second order recurrence relations

We introduce the generating functionG(z) ofx,x,x,. . .and observe that 0 1 2 2 3 G(z) =x+x z+x z+x z+∙ ∙ ∙ 0 1 2 3 2 3 zG(z) =x z+x z+x z+∙ ∙ ∙ 0 1 2 2 2 3 z G(z) =x z+x z+∙ ∙ ∙. 0 1

7

Therefore 2 G(z)−azG(z)−bz G(z) =x+ (x−ax)z. 0 1 0 From this we obtain the closed form x+ (x−ax)z 0 1 0 G(z) =. 2 1−az−bz 2 The next step is to factorize 1−az−bz; i.e., write 2 1−az−bz= (1−λ z)(1−λ z) 1 2 for some numbersλandλ. We can now 1 2 find a partial fraction expansion forG(z). If λ6=λ, this will be of the form 1 2 A B G(z+) = , 1−λ z1−λ z 1 2 for some constantsAandB; whereas if λ=λ, it will be of the form 1 2 A B G(z) = +. 2 1−λ z(1−λ z) 1 1

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n In the first case, the coefficient ofzin G(z) is n n x n =Aλ+Bλ 1 2 n and in the second case, the coefficient ofz is n n n n n+ 1)λ= )λ x n =Aλ+B( 1 (A+B 1 +Bnλ . 1 1 To find the values ofAandBwe need to know someinitial conditionssuch as the values ofxandx. 0 1

The Characteristic Equation

λandλare the roots of the 1 2 characteristic equation 2 λ−aλ−b= 0.

TheoremIf 2 1−az−bz= (1−λ z)(1−λ z), then the 1 2 general solution to the recurrence relation

x n =ax+bx −2 ,forn≥2. n−1n n n is eitherx n =Aλ+Bλor 1 2 n n x n =Cλ+Dnλaccording to whether 1 1 λ6=λorλ=λ, respectively. 1 2 1 2 9