![Page 1: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/1.jpg)
Linear Homogeneous Recurrence Relations
![Page 2: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/2.jpg)
The Fabled Beginnings
King Shirham of India
Grand Vizier Sissa Ben Dahir
The invention of chess
The prize: grains of wheat?
DefinitionA recurrence relation is an equation that recursively defines asequence, once one or more initial terms are given: each further termof the sequence is defined as a function of the preceding terms.
![Page 3: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/3.jpg)
The Fabled Beginnings
King Shirham of India
Grand Vizier Sissa Ben Dahir
The invention of chess
The prize: grains of wheat?
DefinitionA recurrence relation is an equation that recursively defines asequence, once one or more initial terms are given: each further termof the sequence is defined as a function of the preceding terms.
![Page 4: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/4.jpg)
Solving the Problem
let sk be the sum of the grains on the first k squares
let tk be the number of grains on the kth square
So, we know that tk+1 = 2tk. What we want is a general expression.
![Page 5: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/5.jpg)
Solving the Problem
let sk be the sum of the grains on the first k squares
let tk be the number of grains on the kth square
So, we know that tk+1 = 2tk. What we want is a general expression.
![Page 6: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/6.jpg)
Let’s Take a Look At What’s Going On
t1 = 1
t2 = 2t1t3 = 2t2 = 22t1
t4 = 2t3 = 22t2 = 23t1
and in general,tk = 2tk−1 = . . . = 2k−1t1
Using the initial condition, we have
tk = 2k−1
![Page 7: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/7.jpg)
Let’s Take a Look At What’s Going On
t1 = 1
t2 = 2t1
t3 = 2t2 = 22t1t4 = 2t3 = 22t2 = 23t1
and in general,tk = 2tk−1 = . . . = 2k−1t1
Using the initial condition, we have
tk = 2k−1
![Page 8: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/8.jpg)
Let’s Take a Look At What’s Going On
t1 = 1
t2 = 2t1t3 = 2t2 = 22t1
t4 = 2t3 = 22t2 = 23t1
and in general,tk = 2tk−1 = . . . = 2k−1t1
Using the initial condition, we have
tk = 2k−1
![Page 9: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/9.jpg)
Let’s Take a Look At What’s Going On
t1 = 1
t2 = 2t1t3 = 2t2 = 22t1
t4 = 2t3 = 22t2 = 23t1
and in general,tk = 2tk−1 = . . . = 2k−1t1
Using the initial condition, we have
tk = 2k−1
![Page 10: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/10.jpg)
Let’s Take a Look At What’s Going On
t1 = 1
t2 = 2t1t3 = 2t2 = 22t1
t4 = 2t3 = 22t2 = 23t1
and in general,tk = 2tk−1 = . . . = 2k−1t1
Using the initial condition, we have
tk = 2k−1
![Page 11: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/11.jpg)
Let’s Take a Look At What’s Going On
t1 = 1
t2 = 2t1t3 = 2t2 = 22t1
t4 = 2t3 = 22t2 = 23t1
and in general,tk = 2tk−1 = . . . = 2k−1t1
Using the initial condition, we have
tk = 2k−1
![Page 12: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/12.jpg)
So What About the Sum?
What we really want is sk, the sum of the number of grains on ksquares.
sk+1 = sk + tk+1
which is another form of a recurrence relation.
sk+1 = sk + 2k
![Page 13: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/13.jpg)
So What About the Sum?
What we really want is sk, the sum of the number of grains on ksquares.
sk+1 = sk + tk+1
which is another form of a recurrence relation.
sk+1 = sk + 2k
![Page 14: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/14.jpg)
So What About the Sum?
What we really want is sk, the sum of the number of grains on ksquares.
sk+1 = sk + tk+1
which is another form of a recurrence relation.
sk+1 = sk + 2k
![Page 15: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/15.jpg)
Again With Iteration
s2 = s1 + 2
s3 = s2 + 22 = s1 + 2 + 22
...
sk = sk−1 + 2k−1 = . . . = s1 + 2 + 22 + . . .+ 2k−1
Using initial condition that s1 = 1, we have
sk = 1 + 2 + 22 + . . .+ 2k−1
![Page 16: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/16.jpg)
Again With Iteration
s2 = s1 + 2
s3 = s2 + 22 = s1 + 2 + 22
...
sk = sk−1 + 2k−1 = . . . = s1 + 2 + 22 + . . .+ 2k−1
Using initial condition that s1 = 1, we have
sk = 1 + 2 + 22 + . . .+ 2k−1
![Page 17: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/17.jpg)
Again With Iteration
s2 = s1 + 2
s3 = s2 + 22 = s1 + 2 + 22
...
sk = sk−1 + 2k−1 = . . . = s1 + 2 + 22 + . . .+ 2k−1
Using initial condition that s1 = 1, we have
sk = 1 + 2 + 22 + . . .+ 2k−1
![Page 18: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/18.jpg)
Recall Series ...
What type of series is this?
It is a finite geometric series.
Formula for the partial sum of a geometric series?
n∑k=0
xk =1− xn+1
1− x
So, we have
sk =1− 2k
1− 2= 2k − 1
![Page 19: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/19.jpg)
Recall Series ...
What type of series is this?
It is a finite geometric series.
Formula for the partial sum of a geometric series?
n∑k=0
xk =1− xn+1
1− x
So, we have
sk =1− 2k
1− 2= 2k − 1
![Page 20: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/20.jpg)
Recall Series ...
What type of series is this?
It is a finite geometric series.
Formula for the partial sum of a geometric series?
n∑k=0
xk =1− xn+1
1− x
So, we have
sk =1− 2k
1− 2= 2k − 1
![Page 21: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/21.jpg)
Bad Decision King ...
For a chess board, this would give
264 − 1 = 18, 446, 744, 073, 709, 551, 615
grains. So, the King made a bad mistake ...
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Observations
We want an, the minimum number of moves
To move n rings, we have to play game with n− 1 rings
To move n− 1 rings, we have to play game with n− 2 rings
So this is a recursive relation.
an = an−1 + 1 + an−1 = 2an−1 + 1
with an initial condition of a1 = 1.
Closed form: an = 2n − 1
![Page 24: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/24.jpg)
Observations
We want an, the minimum number of moves
To move n rings, we have to play game with n− 1 rings
To move n− 1 rings, we have to play game with n− 2 rings
So this is a recursive relation.
an = an−1 + 1 + an−1 = 2an−1 + 1
with an initial condition of a1 = 1.
Closed form: an = 2n − 1
![Page 25: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/25.jpg)
Observations
We want an, the minimum number of moves
To move n rings, we have to play game with n− 1 rings
To move n− 1 rings, we have to play game with n− 2 rings
So this is a recursive relation.
an = an−1 + 1 + an−1 = 2an−1 + 1
with an initial condition of a1 = 1.
Closed form: an = 2n − 1
![Page 26: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/26.jpg)
Observations
We want an, the minimum number of moves
To move n rings, we have to play game with n− 1 rings
To move n− 1 rings, we have to play game with n− 2 rings
So this is a recursive relation.
an = an−1 + 1 + an−1 = 2an−1 + 1
with an initial condition of a1 = 1.
Closed form: an = 2n − 1
![Page 27: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/27.jpg)
Observations
We want an, the minimum number of moves
To move n rings, we have to play game with n− 1 rings
To move n− 1 rings, we have to play game with n− 2 rings
So this is a recursive relation.
an = an−1 + 1 + an−1 = 2an−1 + 1
with an initial condition of a1 = 1.
Closed form: an = 2n − 1
![Page 28: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/28.jpg)
Linear Homogeneous Recurrence Relations
Another method for solving these relations: using characteristic roots
an = c1an−1 + c2an−2 + . . .+ cpan−p
n ≥ p
all the ci’s are constants with cp 6= 0
called linear because all of the ap terms are to the first power
called homogeneous because all terms on the right hand sideinvolve some ap
![Page 29: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/29.jpg)
Linear Homogeneous Recurrence Relations
Another method for solving these relations: using characteristic roots
an = c1an−1 + c2an−2 + . . .+ cpan−p
n ≥ p
all the ci’s are constants with cp 6= 0
called linear because all of the ap terms are to the first power
called homogeneous because all terms on the right hand sideinvolve some ap
![Page 30: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/30.jpg)
Linear Homogeneous Recurrence Relations
Another method for solving these relations: using characteristic roots
an = c1an−1 + c2an−2 + . . .+ cpan−p
n ≥ p
all the ci’s are constants with cp 6= 0
called linear because all of the ap terms are to the first power
called homogeneous because all terms on the right hand sideinvolve some ap
![Page 31: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/31.jpg)
Linear Homogeneous Recurrence Relations
Another method for solving these relations: using characteristic roots
an = c1an−1 + c2an−2 + . . .+ cpan−p
n ≥ p
all the ci’s are constants with cp 6= 0
called linear because all of the ap terms are to the first power
called homogeneous because all terms on the right hand sideinvolve some ap
![Page 32: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/32.jpg)
Characteristic Equations and Characteristic Roots
TheoremSuppose that a linear homogeneous recurrence relation with constantcoefficients has characteristic roots α1, α2, . . . , αp. Then ifλ1, λ2, . . . , λp are constants then every expression of the form
an = λ1αn1 + λ2α
n2 + . . .+ λpα
np
is a solution to the recurrence relation. Moreover, if the roots aredistinct, every solution has this form for some constantsλ1, λ2, . . . , λp.
![Page 33: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/33.jpg)
Distinct Roots
ExampleSolve the linear homogeneous recurrence relation
an = 5an−1 − 6an−2
with a0 = a1 = 1.
First we rewrite as a polynomial
an = 5an−1 − 6an−2 ⇒ xn = 5xn−1 − 6xn−2
Then, divide by xk where k is smallest present power
x2 = 5x− 6⇒ x2 − 5x + 6 = 0
![Page 34: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/34.jpg)
Distinct Roots
ExampleSolve the linear homogeneous recurrence relation
an = 5an−1 − 6an−2
with a0 = a1 = 1.
First we rewrite as a polynomial
an = 5an−1 − 6an−2 ⇒ xn = 5xn−1 − 6xn−2
Then, divide by xk where k is smallest present power
x2 = 5x− 6⇒ x2 − 5x + 6 = 0
![Page 35: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/35.jpg)
Distinct Roots
ExampleSolve the linear homogeneous recurrence relation
an = 5an−1 − 6an−2
with a0 = a1 = 1.
First we rewrite as a polynomial
an = 5an−1 − 6an−2 ⇒ xn = 5xn−1 − 6xn−2
Then, divide by xk where k is smallest present power
x2 = 5x− 6⇒ x2 − 5x + 6 = 0
![Page 36: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/36.jpg)
Distinct Roots
Now, factor
x2 − 5x + 6 = 0⇒ (x− 2)(x− 3) = 0
characteristic roots: α1 = 2 and α2 = 3
So, two solutions are an = 2n and an = 3n - but not simultaneously.To find when that would happen, we need to solve
an = λ12n + λ23n
using the initial conditions.
![Page 37: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/37.jpg)
Distinct Roots
Now, factor
x2 − 5x + 6 = 0⇒ (x− 2)(x− 3) = 0
characteristic roots: α1 = 2 and α2 = 3
So, two solutions are an = 2n and an = 3n - but not simultaneously.To find when that would happen, we need to solve
an = λ12n + λ23n
using the initial conditions.
![Page 38: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/38.jpg)
Distinct Roots
Now, factor
x2 − 5x + 6 = 0⇒ (x− 2)(x− 3) = 0
characteristic roots: α1 = 2 and α2 = 3
So, two solutions are an = 2n and an = 3n - but not simultaneously.To find when that would happen, we need to solve
an = λ12n + λ23n
using the initial conditions.
![Page 39: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/39.jpg)
Distinct Roots
Using the initial conditions, we see
a0 = 1 : λ120 + λ230 = 1⇒ λ1 + λ2 = 1
a1 = 1 : λ121 + λ231 = 1⇒ 2λ1 + 3λ2 = 1
This becomes {λ1+λ2=1
2λ1+3λ2=1
When we solve, we get λ1 = 2 and λ2 = −1.This gives
an = 2 · 2n − 3n = 2n+1 − 3n
So, every solution has this form.
![Page 40: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/40.jpg)
Distinct Roots
Using the initial conditions, we see
a0 = 1 : λ120 + λ230 = 1⇒ λ1 + λ2 = 1
a1 = 1 : λ121 + λ231 = 1⇒ 2λ1 + 3λ2 = 1
This becomes {λ1+λ2=1
2λ1+3λ2=1
When we solve, we get λ1 = 2 and λ2 = −1.This gives
an = 2 · 2n − 3n = 2n+1 − 3n
So, every solution has this form.
![Page 41: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/41.jpg)
Distinct Roots
Using the initial conditions, we see
a0 = 1 : λ120 + λ230 = 1⇒ λ1 + λ2 = 1
a1 = 1 : λ121 + λ231 = 1⇒ 2λ1 + 3λ2 = 1
This becomes {λ1+λ2=1
2λ1+3λ2=1
When we solve, we get λ1 = 2 and λ2 = −1.This gives
an = 2 · 2n − 3n = 2n+1 − 3n
So, every solution has this form.
![Page 42: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/42.jpg)
Distinct Roots
Using the initial conditions, we see
a0 = 1 : λ120 + λ230 = 1⇒ λ1 + λ2 = 1
a1 = 1 : λ121 + λ231 = 1⇒ 2λ1 + 3λ2 = 1
This becomes {λ1+λ2=1
2λ1+3λ2=1
When we solve, we get λ1 = 2 and λ2 = −1.
This gives
an = 2 · 2n − 3n = 2n+1 − 3n
So, every solution has this form.
![Page 43: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/43.jpg)
Distinct Roots
Using the initial conditions, we see
a0 = 1 : λ120 + λ230 = 1⇒ λ1 + λ2 = 1
a1 = 1 : λ121 + λ231 = 1⇒ 2λ1 + 3λ2 = 1
This becomes {λ1+λ2=1
2λ1+3λ2=1
When we solve, we get λ1 = 2 and λ2 = −1.This gives
an = 2 · 2n − 3n = 2n+1 − 3n
So, every solution has this form.
![Page 44: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/44.jpg)
Distinct Roots
ExampleSolve the linear homogeneous recurrence relation
bk = −7bk−1 + 18bk−2
if b0 = 0 and b1 = 8.
![Page 45: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/45.jpg)
Solution
characteristic equation:
xk + 7xk−1 − 18xk−2 = 0
which becomesx2 + 7x− 18 = 0
that has roots
(x + 9)(x− 2) = 0⇒ x = −9, 2
so our solutions are all of the form
bk = λ1(−9)k + λ2(2)k
![Page 46: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/46.jpg)
Solution
characteristic equation:
xk + 7xk−1 − 18xk−2 = 0
which becomesx2 + 7x− 18 = 0
that has roots
(x + 9)(x− 2) = 0⇒ x = −9, 2
so our solutions are all of the form
bk = λ1(−9)k + λ2(2)k
![Page 47: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/47.jpg)
Solution
characteristic equation:
xk + 7xk−1 − 18xk−2 = 0
which becomesx2 + 7x− 18 = 0
that has roots
(x + 9)(x− 2) = 0⇒ x = −9, 2
so our solutions are all of the form
bk = λ1(−9)k + λ2(2)k
![Page 48: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/48.jpg)
Solution
characteristic equation:
xk + 7xk−1 − 18xk−2 = 0
which becomesx2 + 7x− 18 = 0
that has roots
(x + 9)(x− 2) = 0⇒ x = −9, 2
so our solutions are all of the form
bk = λ1(−9)k + λ2(2)k
![Page 49: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/49.jpg)
Solution
and when we apply the initial conditions we get
0 = λ1(−9)0 + λ2(2)0
8 = λ1(−9)1 + λ2(2)1
which leads to the system {λ1+λ2=0
-9λ1+2λ2=8
which has solutions λ1 = − 811 and λ2 = 8
11so all solutions of our recurrence relation are of the form
bk = −811
(−9)k +8
11(2)k
![Page 50: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/50.jpg)
Solution
and when we apply the initial conditions we get
0 = λ1(−9)0 + λ2(2)0
8 = λ1(−9)1 + λ2(2)1
which leads to the system {λ1+λ2=0
-9λ1+2λ2=8
which has solutions λ1 = − 811 and λ2 = 8
11so all solutions of our recurrence relation are of the form
bk = −811
(−9)k +8
11(2)k
![Page 51: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/51.jpg)
Solution
and when we apply the initial conditions we get
0 = λ1(−9)0 + λ2(2)0
8 = λ1(−9)1 + λ2(2)1
which leads to the system {λ1+λ2=0
-9λ1+2λ2=8
which has solutions λ1 = − 811 and λ2 = 8
11
so all solutions of our recurrence relation are of the form
bk = −811
(−9)k +8
11(2)k
![Page 52: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/52.jpg)
Solution
and when we apply the initial conditions we get
0 = λ1(−9)0 + λ2(2)0
8 = λ1(−9)1 + λ2(2)1
which leads to the system {λ1+λ2=0
-9λ1+2λ2=8
which has solutions λ1 = − 811 and λ2 = 8
11so all solutions of our recurrence relation are of the form
bk = −811
(−9)k +8
11(2)k
![Page 53: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/53.jpg)
Distinct Roots
ExampleSolve the linear homogeneous recurrence relation
fn+1 = 2fn + 3fn−1
if f0 = f1 = 2.
Our characteristic equation here is
xn − 2xn−1 − 3xn−2 = 0
and dividing by xn−2 gives
x2 − 2x− 3 = 0.
with characteristic roots are α1 = 3 and α2 = −1.Since the roots aredistinct we know all solutions of fn are of the form
fn = λ1(3)n + λ2(−1)n
![Page 54: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/54.jpg)
Distinct Roots
ExampleSolve the linear homogeneous recurrence relation
fn+1 = 2fn + 3fn−1
if f0 = f1 = 2.
Our characteristic equation here is
xn − 2xn−1 − 3xn−2 = 0
and dividing by xn−2 gives
x2 − 2x− 3 = 0.
with characteristic roots are α1 = 3 and α2 = −1.Since the roots aredistinct we know all solutions of fn are of the form
fn = λ1(3)n + λ2(−1)n
![Page 55: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/55.jpg)
Distinct Roots
ExampleSolve the linear homogeneous recurrence relation
fn+1 = 2fn + 3fn−1
if f0 = f1 = 2.
Our characteristic equation here is
xn − 2xn−1 − 3xn−2 = 0
and dividing by xn−2 gives
x2 − 2x− 3 = 0.
with characteristic roots are α1 = 3 and α2 = −1.Since the roots aredistinct we know all solutions of fn are of the form
fn = λ1(3)n + λ2(−1)n
![Page 56: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/56.jpg)
Distinct Roots
ExampleSolve the linear homogeneous recurrence relation
fn+1 = 2fn + 3fn−1
if f0 = f1 = 2.
Our characteristic equation here is
xn − 2xn−1 − 3xn−2 = 0
and dividing by xn−2 gives
x2 − 2x− 3 = 0.
with characteristic roots are α1 = 3 and α2 = −1.Since the roots aredistinct we know all solutions of fn are of the form
fn = λ1(3)n + λ2(−1)n
![Page 57: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/57.jpg)
The Rest of the Solution
using our initial conditions, we have
2 = λ1(3)0 + λ2(−1)0
2 = λ1(3)1 + λ2(−1)1
which gives the system of equations{λ1+λ2=23λ1-λ2=2
Solving gives λ1 = 1 and λ2 = 1, and so all solutions are of the form
fn = 3n + (−1)n
![Page 58: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/58.jpg)
The Rest of the Solution
using our initial conditions, we have
2 = λ1(3)0 + λ2(−1)0
2 = λ1(3)1 + λ2(−1)1
which gives the system of equations{λ1+λ2=23λ1-λ2=2
Solving gives λ1 = 1 and λ2 = 1, and so all solutions are of the form
fn = 3n + (−1)n
![Page 59: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/59.jpg)
The Rest of the Solution
using our initial conditions, we have
2 = λ1(3)0 + λ2(−1)0
2 = λ1(3)1 + λ2(−1)1
which gives the system of equations{λ1+λ2=23λ1-λ2=2
Solving gives λ1 = 1 and λ2 = 1, and so all solutions are of the form
fn = 3n + (−1)n
![Page 60: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/60.jpg)
Multiple Roots
ExampleSolve the linear homogeneous recurrence relation
an = 7an−1 − 16an−2 + 12an−3
with a0 = 1, a1 = 2 and a2 = 0.
We proceed as before and write the characteristic equation.
xn − 7xn−1 + 16xn−2 − 12xn−3
and then dividing by xn−3 gives
x3 − 7x2 + 16x− 12 = 0
![Page 61: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/61.jpg)
Multiple Roots
ExampleSolve the linear homogeneous recurrence relation
an = 7an−1 − 16an−2 + 12an−3
with a0 = 1, a1 = 2 and a2 = 0.
We proceed as before and write the characteristic equation.
xn − 7xn−1 + 16xn−2 − 12xn−3
and then dividing by xn−3 gives
x3 − 7x2 + 16x− 12 = 0
![Page 62: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/62.jpg)
Solution
This factors to (x− 2)2(x− 3) = 0. Let α1 = α2 = 2 and α3 = 3.
So, the solutions to this relation are of the form
an = λ1αn1 + λ2nαn
1 + λ3αn3
When we apply the initial conditions, we see
a0 = λ1 + λ3 = 1
a1 = 2λ1 + 2λ2 + 3λ3 = 2
a2 = 4λ1 + 8λ2 + 9λ3 = 0
Solving gives λ1 = 5, λ2 = 2 and λ3 = −4. So, our solution is
an = 5 · 2n + 2 · n2n − 4 · 3n
oran = (5 + 2n)2n − 4 · 3n
![Page 63: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/63.jpg)
Solution
This factors to (x− 2)2(x− 3) = 0. Let α1 = α2 = 2 and α3 = 3.So, the solutions to this relation are of the form
an = λ1αn1 + λ2nαn
1 + λ3αn3
When we apply the initial conditions, we see
a0 = λ1 + λ3 = 1
a1 = 2λ1 + 2λ2 + 3λ3 = 2
a2 = 4λ1 + 8λ2 + 9λ3 = 0
Solving gives λ1 = 5, λ2 = 2 and λ3 = −4. So, our solution is
an = 5 · 2n + 2 · n2n − 4 · 3n
oran = (5 + 2n)2n − 4 · 3n
![Page 64: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/64.jpg)
Solution
This factors to (x− 2)2(x− 3) = 0. Let α1 = α2 = 2 and α3 = 3.So, the solutions to this relation are of the form
an = λ1αn1 + λ2nαn
1 + λ3αn3
When we apply the initial conditions, we see
a0 = λ1 + λ3 = 1
a1 = 2λ1 + 2λ2 + 3λ3 = 2
a2 = 4λ1 + 8λ2 + 9λ3 = 0
Solving gives λ1 = 5, λ2 = 2 and λ3 = −4. So, our solution is
an = 5 · 2n + 2 · n2n − 4 · 3n
oran = (5 + 2n)2n − 4 · 3n
![Page 65: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/65.jpg)
Solution
This factors to (x− 2)2(x− 3) = 0. Let α1 = α2 = 2 and α3 = 3.So, the solutions to this relation are of the form
an = λ1αn1 + λ2nαn
1 + λ3αn3
When we apply the initial conditions, we see
a0 = λ1 + λ3 = 1
a1 = 2λ1 + 2λ2 + 3λ3 = 2
a2 = 4λ1 + 8λ2 + 9λ3 = 0
Solving gives λ1 = 5, λ2 = 2 and λ3 = −4. So, our solution is
an = 5 · 2n + 2 · n2n − 4 · 3n
oran = (5 + 2n)2n − 4 · 3n
![Page 66: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/66.jpg)
One More with Multiple Roots
ExampleSolve the linear recurrence relation
dn = 4dn−1 − 4dn−2
where d0 = 1 and d1 = 2.
Our characteristic equation is
xn − 4xn−1 + 4xn−2 = 0
and dividing by xn−2 gives
x2 − 4x + 4 = 0
which factors to(x− 2)2 = 0
which has roots x = 2 with multiplicity 2
![Page 67: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/67.jpg)
One More with Multiple Roots
ExampleSolve the linear recurrence relation
dn = 4dn−1 − 4dn−2
where d0 = 1 and d1 = 2.
Our characteristic equation is
xn − 4xn−1 + 4xn−2 = 0
and dividing by xn−2 gives
x2 − 4x + 4 = 0
which factors to(x− 2)2 = 0
which has roots x = 2 with multiplicity 2
![Page 68: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/68.jpg)
One More with Multiple Roots
ExampleSolve the linear recurrence relation
dn = 4dn−1 − 4dn−2
where d0 = 1 and d1 = 2.
Our characteristic equation is
xn − 4xn−1 + 4xn−2 = 0
and dividing by xn−2 gives
x2 − 4x + 4 = 0
which factors to(x− 2)2 = 0
which has roots x = 2 with multiplicity 2
![Page 69: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/69.jpg)
One More with Multiple Roots
ExampleSolve the linear recurrence relation
dn = 4dn−1 − 4dn−2
where d0 = 1 and d1 = 2.
Our characteristic equation is
xn − 4xn−1 + 4xn−2 = 0
and dividing by xn−2 gives
x2 − 4x + 4 = 0
which factors to(x− 2)2 = 0
which has roots x = 2 with multiplicity 2
![Page 70: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/70.jpg)
One More with Multiple Roots
which means that our solutions are of the form
dn = λ12n + λ2n2n
and using the initial conditions we see
d0 = λ120 + λ2(0)20 = 1
d1 = λ121 + λ2(1)21 = 2
which gives the system of equations{λ1=1
2λ1+2λ2=2
with solution λ1 = 1 and λ2 = 0and finally, this gives that all solutions are of the form
dn = 2n
![Page 71: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/71.jpg)
One More with Multiple Roots
which means that our solutions are of the form
dn = λ12n + λ2n2n
and using the initial conditions we see
d0 = λ120 + λ2(0)20 = 1
d1 = λ121 + λ2(1)21 = 2
which gives the system of equations{λ1=1
2λ1+2λ2=2
with solution λ1 = 1 and λ2 = 0and finally, this gives that all solutions are of the form
dn = 2n
![Page 72: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/72.jpg)
One More with Multiple Roots
which means that our solutions are of the form
dn = λ12n + λ2n2n
and using the initial conditions we see
d0 = λ120 + λ2(0)20 = 1
d1 = λ121 + λ2(1)21 = 2
which gives the system of equations{λ1=1
2λ1+2λ2=2
with solution λ1 = 1 and λ2 = 0and finally, this gives that all solutions are of the form
dn = 2n
![Page 73: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/73.jpg)
One More with Multiple Roots
which means that our solutions are of the form
dn = λ12n + λ2n2n
and using the initial conditions we see
d0 = λ120 + λ2(0)20 = 1
d1 = λ121 + λ2(1)21 = 2
which gives the system of equations{λ1=1
2λ1+2λ2=2
with solution λ1 = 1 and λ2 = 0
and finally, this gives that all solutions are of the form
dn = 2n
![Page 74: Linear Homogeneous Recurrence Relations - Dr. …btravers.weebly.com/.../recurrence_relations_slides.pdfSolving the Problem let s k be the sum of the grains on the first k squares](https://reader031.vdocument.in/reader031/viewer/2022020412/5af9b64b7f8b9a44658e16aa/html5/thumbnails/74.jpg)
One More with Multiple Roots
which means that our solutions are of the form
dn = λ12n + λ2n2n
and using the initial conditions we see
d0 = λ120 + λ2(0)20 = 1
d1 = λ121 + λ2(1)21 = 2
which gives the system of equations{λ1=1
2λ1+2λ2=2
with solution λ1 = 1 and λ2 = 0and finally, this gives that all solutions are of the form
dn = 2n