MIT Slides on Diagonalisatiojn

8/7/2019 MIT Slides on Diagonalisatiojn

1/124

Introduction

The Three Proofs

Unifying the Proofs

Conclusion

The Fibonacci numbersProofs of Binets formula and other mysteries

Qiaochu Yuan

Department of MathematicsMassachusetts Institute of Technology

February 21, 2009 / HMMT

Qiaochu Yuan The Fibonacci numbers
http://find/http://goback/


2/124

Introduction

The Three Proofs

Unifying the Proofs

Conclusion

Outline

1 Introduction

Whats Going On?

Background

2 The Three ProofsProof 1: Linear Independence

Proof 2: Partial Fraction Decomposition

Proof 3: Diagonalization

3 Unifying the ProofsThe shift operator

4 Conclusion



3/124

Introduction

The Three Proofs

Unifying the Proofs

Conclusion

Whats Going On?

Background

What Were Going To Do

Definition: The Fibonacci numbers satisfy

F0 = 0, F1 = 1, Fn+2 = Fn+1 + Fn

Theorem (Binets formula): Fn =1

5

1+5

2

n

152

n

Were going to run through three proofs,

Were going to learn why theyre really the same,

We might even do other cool stuff!



4/124

Introduction

The Three Proofs

Unifying the Proofs

Conclusion

Whats Going On?

Background



F0 = 0, F1 = 1, Fn+2 = Fn+1 + Fn


5

1+5

2

n

152

n






5/124

Introduction

The Three Proofs

Unifying the Proofs

Conclusion

Whats Going On?

Background



F0 = 0, F1 = 1, Fn+2 = Fn+1 + Fn


5

1+5

2

n

152

n





I d i


6/124

Introduction

The Three Proofs

Unifying the Proofs

Conclusion

Whats Going On?

Background



F0 = 0, F1 = 1, Fn+2 = Fn+1 + Fn


5

1+5

2

n

152

n





I t d ti


7/124

Introduction

The Three Proofs

Unifying the Proofs

Conclusion

Whats Going On?

Background

Linear Algebra Is Important!

Definition: A (complex) vector space V is a set with:

Vector addition v + u,Scalar multiplication c v, c C,

A zero vector v + 0 = v.

Examples: Cn,C[x], the solutions to y + y = 0


Introduction


8/124

Introduction

The Three Proofs

Unifying the Proofs

Conclusion

Whats Going On?

Background







Introduction


9/124

Introduction

The Three Proofs

Unifying the Proofs

Conclusion

Whats Going On?

Background







Introduction


10/124

Introduction

The Three Proofs

Unifying the Proofs

Conclusion

Whats Going On?

Background







Introduction


11/124

Introduction

The Three Proofs

Unifying the Proofs

Conclusion

Whats Going On?

Background







Introduction


12/124

Introduction

The Three Proofs

Unifying the Proofs

Conclusion

Whats Going On?

Background


Definition: The dimension d of V is the smallest number of

vectors v1,...vd such that:

Every v is of the form c1v1 + ... + cdvd,

This representation is unique,

c1v1 + ... + cdvd = 0 if and only if c1 = ... = cd = 0 (linearindependence).

We call (vi) a basisof V and say that it spans V.

Examples:

dimCn = n, dimC[x] = , dim{y|y + y = 0} = 2


Introduction


13/124

The Three Proofs

Unifying the Proofs

Conclusion

Whats Going On?

Background








Examples:



Introduction


14/124

The Three Proofs

Unifying the Proofs

Conclusion

Whats Going On?

Background








Examples:



Introduction


15/124

The Three Proofs

Unifying the Proofs

Conclusion

Whats Going On?

Background








Examples:



Introduction


16/124

The Three Proofs

Unifying the Proofs

Conclusion

Whats Going On?

Background








Examples:



Introduction

Th Th P f Wh G i O ?


17/124

The Three Proofs

Unifying the Proofs

Conclusion

Whats Going On?

Background








Examples:



Introduction

Th Th P f Wh t G i O ?


18/124

The Three Proofs

Unifying the Proofs

Conclusion

Whats Going On?

Background


Definition: A linear transformation T : V1 V2 is a function

such that:T(v1 + v2) = T(v1) + T(v2),

T(cv1) = cT(v1).

Examples: n n matrices, T(P(x)) = xP(x), ddx


Introduction

The Three Proofs Whats Going On?


19/124

The Three Proofs

Unifying the Proofs

Conclusion

What s Going On?

Background




T(cv1) = cT(v1).



Introduction

The Three Proofs Whats Going On?


20/124

The Three Proofs

Unifying the Proofs

Conclusion

What s Going On?

Background




T(cv1) = cT(v1).



IntroductionThe Three Proofs Whats Going On?


21/124

The Three Proofs

Unifying the Proofs

Conclusion

What s Going On?

Background




T(cv1) = cT(v1).





22/124

The Three Proofs

Unifying the Proofs

Conclusion

What s Going On?

Background


Definition: An eigenvectorof T : V V is a v such thatT(v) = v.

is called an eigenvalue. If V has dimension n, T has at

most n eigenvalues.

Most T are characterized by their eigenvectors and

eigenvalues.




23/124

The Three Proofs

Unifying the Proofs

Conclusion

What s Going On?

Background




most n eigenvalues.


eigenvalues.




24/124

Unifying the Proofs

Conclusion

g

Background




most n eigenvalues.


eigenvalues.


IntroductionThe Three Proofs

Proof 1: Linear Independence



25/124

Unifying the Proofs

Conclusion



Motivation: Geometric Series

What are the solutions to sn+1 = asn?

sn = ans0! (Induction)

"First-order" version of what we want to look at.

Identity to keep in mind: an+1s0 = a(an)s0






26/124

Unifying the Proofs

Conclusion













27/124

Unifying the Proofs

Conclusion










U if i h P f




28/124

Unifying the Proofs

Conclusion










U if i th P f




29/124

Unifying the Proofs

Conclusion

oo a t a act o eco pos t o


Fibonacci-type Sequences

Definition: A sequence sn is of Fibonacci typeif

sn+2 = sn+1 + sn.Proposition: The set S of Fibonacci-type sequences forms a

vector space.

Proof.

(sn+2 + tn+2) = (sn+1 + tn+1) + (sn + tn),

csn+2 = csn+1 + csn,

0 = 0 + 0 (zero vectors are important!)



Unifying the Proofs




30/124

Unifying the Proofs

Conclusion

p





vector space.

Proof.

(sn+2 + tn+2) = (sn+1 + tn+1) + (sn + tn),





Unifying the Proofs




31/124

Unifying the Proofs

ConclusionProof 3: Diagonalization




vector space.

Proof.

(sn+2 + tn+2) = (sn+1 + tn+1) + (sn + tn),





Unifying the Proofs




32/124

Unifying the Proofs





vector space.

Proof.

(sn+2 + tn+2) = (sn+1 + tn+1) + (sn + tn),





Unifying the Proofs




33/124

Unifying the Proofs


Finding a Basis

Proposition: S has dimension 2; it has basis Fn and Fn1.Proof.

Suppose s0 = a, s1 = b. Then,s2 = a+ b, s3 = a+ 2b, s4 = 2a+ 3b, s5 = 3a+ 5b, s6 =5a+ 8b...

In fact sn = aFn1 + bFn, n > 1. (Induction)

The first two values determine the whole sequence, sothere arent any others.



Unifying the Proofs



P f 3 Di li i


34/124

Unifying the Proofs


Finding a Basis



In fact sn = aFn1 + bFn, n > 1. (Induction)

The first two values determine the whole sequence, sothere arent any others.



Unifying the Proofs



P f 3 Di li ti


35/124

y g


Finding a Basis



In fact sn = aFn1 + bFn, n > 1. (Induction)The first two values determine the whole sequence, so

there arent any others.



Unifying the Proofs





36/124

y g


Finding a Basis







Unifying the Proofs





37/124


Finding a Basis







Unifying the Proofs





38/124


A Fancy Word For Guessing

Now we use the ansatz sn = an:

sn+2 = sn+1 + sn if and only if an+2 = an+1 + an,

If and only if a2 = a+ 1,

If and only if a= 1

52 .

= 1+

52 is the golden ratio, =

152 is the othergolden

ratio.



39/124


Unifying the Proofs

C l i





40/124

Conclusionoo 3 ago a at o





If and only if a= 1

52 .

= 1+



ratio.



41/124


Unifying the Proofs

Conclusion





42/124

Conclusion





If and only if a= 1

52 .

= 1+



ratio.



Unifying the Proofs

Conclusion





43/124

Conclusion

Guessing Works!

Proposition: n, n form a basis for S.

Proof.

Suppose s0 = a, s1 = b and sn = c1n + c2

n.

Then s0 = c1 + c2 and s1 = c1 + c2.

Can we solve this system? Yes!

c1 =ba , c2 =

ba

So every Fibonacci-type sequencehas this form.

Corollary: Fn =nn



Unifying the Proofs

Conclusion





44/124

Conclusion

Guessing Works!


Proof.


n.

Then s0 = c1 + c2 and s1 = c1 + c2.


c1 =ba , c2 =

ba


Corollary: Fn =nn



Unifying the Proofs

Conclusion

Proof 1: Linear IndependenceProof 2: Partial Fraction Decomposition



45/124

Conclusion

Guessing Works!


Proof.


n.

Then s0 = c1 + c2 and s1 = c1 + c2.


c1 =ba , c2 =

ba


Corollary: Fn =nn



Unifying the Proofs

Conclusion




46/124

Guessing Works!


Proof.


n.

Then s0 = c1 + c2 and s1 = c1 + c2.


c1 =ba , c2 =

ba


Corollary: Fn =nn



Unifying the Proofs

Conclusion




47/124

Guessing Works!


Proof.


n.

Then s0 = c1 + c2 and s1 = c1 + c2.


c1 =ba , c2 =

ba


Corollary: Fn =nn



Unifying the Proofs

Conclusion




48/124

Guessing Works!


Proof.


n.

Then s0 = c1 + c2 and s1 = c1 + c2.


c1 =ba , c2 =

ba


Corollary: Fn =nn



Unifying the Proofs

Conclusion




49/124

Guessing Works!


Proof.


n.

Then s0 = c1 + c2 and s1 = c1 + c2.


c1 =ba , c2 =

ba


Corollary: Fn =nn



Unifying the Proofs

Conclusion




50/124

Generating Functions

Definition: The (ordinary) generating functionof sn is

s0 + s1x + s2x2 + ....

Examples:

1 + x + x2 + ... = 11xs0 + as0x + a

2s0x2 + ... = a01sx

1 + 2x + 3x2 + ... = 1(1x)2

n0

+

n1

x +

n2

x2 + ... = (1 + x)n



Unifying the Proofs

Conclusion




51/124



s0 + s1x + s2x2 + ....

Examples:

1 + x + x2 + ... = 11xs0 + as0x + a

2s0x2 + ... = a01sx

1 + 2x + 3x2 + ... = 1(1x)2

n0

+

n1

x +

n2

x2 + ... = (1 + x)n



Unifying the Proofs

Conclusion




52/124



s0 + s1x + s2x2 + ....

Examples:

1 + x + x2 + ... = 11xs0 + as0x + a

2s0x2 + ... = a01sx

1 + 2x + 3x2 + ... = 1(1x)2

n0

+

n1

x +

n2

x2 + ... = (1 + x)n



53/124

Introduction

The Three Proofs

Unifying the Proofs

Conclusion




54/124



s0 + s1x + s2x2 + ....

Examples:

1 + x + x2 + ... = 11xs0 + as0x + a

2s0x2 + ... = a01sx

1 + 2x + 3x2 + ... = 1(1x)2

n0

+

n1

x +

n2

x2 + ... = (1 + x)n


Introduction

The Three Proofs

Unifying the Proofs

Conclusion




55/124

A Very Special Function

So let F(x) = F0 + F1x + F2x2 + F3x

3 + ....

Then xF(x) = F0x + F1x2 + F2x

3 + ....

Then x2

F(x) = F0x2

+ F1x3

+ ....Then xF(x) + x2F(x) = F0x + F2x

2 + F3x3 + ....

But this is also F(x) F0 F1x + F0x = F(x) x!

It follows that (1 x x2)F(x) = x, so F(x) = x1xx2 .

Corollary: F

110

= 1089 = 0.0112358...


Introduction

The Three Proofs

Unifying the Proofs

Conclusion




56/124



3 + ....


3 + ....

Then x2

F(x) = F0x2

+ F1x3

+ ....Then xF(x) + x2F(x) = F0x + F2x

2 + F3x3 + ....



Corollary: F

110

= 1089 = 0.0112358...


Introduction

The Three Proofs

Unifying the Proofs

Conclusion



A V S i l F i


57/124



3 + ....


3 + ....

Then x

2

F(x) = F0x2

+ F1x3

+ ....Then xF(x) + x2F(x) = F0x + F2x

2 + F3x3 + ....



Corollary: F

110

= 1089 = 0.0112358...


Introduction

The Three Proofs

Unifying the Proofs

Conclusion



A V S i l F ti


58/124



3 + ....


3 + ....

Then x

2

F(x) = F0x2

+ F1x3

+ ....Then xF(x) + x2F(x) = F0x + F2x

2 + F3x3 + ....



Corollary: F

110

= 1089 = 0.0112358...


Introduction

The Three Proofs

Unifying the Proofs

Conclusion



A V S i l F ti


59/124



3 + ....


3 + ....

Then x

2

F(x) = F0x2

+ F1x3

+ ....Then xF(x) + x2F(x) = F0x + F2x

2 + F3x3 + ....



Corollary: F

110

= 1089 = 0.0112358...


Introduction

The Three Proofs

Unifying the Proofs

Conclusion



A V S i l F ti


60/124



3 + ....


3 + ....

Then x

2

F(x) = F0x

2

+ F1x

3

+ ....Then xF(x) + x2F(x) = F0x + F2x

2 + F3x3 + ....



Corollary: F

110

= 1089 = 0.0112358...


Introduction

The Three Proofs

Unifying the Proofs

Conclusion





61/124



3 + ....


3 + ....

Then x

2

F(x) = F0x

2

+ F1x

3

+ ....Then xF(x) + x2F(x) = F0x + F2x

2 + F3x3 + ....



Corollary: F

110

= 1089 = 0.0112358...


Introduction

The Three Proofs

Unifying the Proofs

Conclusion



Remember Those Partial Fractions?


62/124


1 = x + x2 1x2

= 1x + 1, so:

1 x x2 = (1 x)(1 x). Moreover,x

1xx2 =

A

1x+ B

1xfor some A, B,

Hence x = A(1 x) + B(1 x).

This gives A = 1 , B =

1 , so...

F(x) = 1

11x

11x

!

F(x) = 1

(1 1) + ( )x + (2 2)x2 + ...


Introduction

The Three Proofs

Unifying the Proofs

Conclusion





63/124


1 = x + x2 1x2

= 1x + 1, so:


1xx2 =

A

1x+ B

1xfor some A, B,

Hence x = A(1 x) + B(1 x).


1 , so...

F(x) = 1

11x

11x

!

F(x) = 1

(1 1) + ( )x + (2 2)x2 + ...


Introduction

The Three Proofs

Unifying the Proofs

Conclusion





64/124


1 = x + x2 1x2

= 1x + 1, so:


1xx2 =

A

1x+ B

1xfor some A, B,

Hence x = A(1 x) + B(1 x).


1 , so...

F(x) = 1

11x

11x

!

F(x) = 1

(1 1) + ( )x + (2 2)x2 + ...


Introduction

The Three Proofs

Unifying the Proofs

Conclusion





65/124


1 = x + x2 1x2

= 1x + 1, so:


1xx2 =

A

1x+ B

1xfor some A, B,

Hence x = A(1 x) + B(1 x).


1 , so...

F(x) = 1

1

1x 1

1x

!

F(x) = 1

(1 1) + ( )x + (2 2)x2 + ...


Introduction

The Three Proofs

Unifying the Proofs

Conclusion





66/124


1 = x + x2 1x2

= 1x + 1, so:


1xx2 =

A

1x+ B

1xfor some A, B,

Hence x = A(1 x) + B(1 x).


1 , so...

F(x) = 1

1

1x 1

1x

!

F(x) = 1

(1 1) + ( )x + (2 2)x2 + ...


Introduction

The Three Proofs

Unifying the Proofs

Conclusion





67/124


1 = x + x2 1x2

= 1x + 1, so:


1xx2 =

A

1x+ B

1xfor some A, B,

Hence x = A(1 x) + B(1 x).


1 , so...

F(x) = 1

1

1x 1

1x

!

F(x) = 1

(1 1) + ( )x + (2 2)x2 + ...


Introduction

The Three ProofsUnifying the Proofs

Conclusion





68/124


1 = x + x2 1x2

= 1x + 1, so:


1xx2 =

A

1x+ B

1xfor some A, B,

Hence x = A(1 x) + B(1 x).


1 , so...

F(x) = 1

1

1x 1

1x

!

F(x) = 1

(1 1) + ( )x + (2 2)x2 + ...


Introduction


Conclusion



Recurrences as Linear Transformations


69/124


(a, b) (b, a+ b) is a linear transformation on C2

Matrix form:

1 1

1 0

b

a

=

b+ a

b

(a, b) (b, a+b) (a+b, a+2b) (a+2b, 2a+3b) ...

In fact Fn =

1 1

1 0

n=

Fn+1 Fn

Fn Fn1

. (Induction)

Corollary: Fn+1Fn

1 F2n = (1)

n. (Determinant)

How do we quickly compute the powers of a matrix?


Introduction


Conclusion





70/124



Matrix form:

1 1

1 0

b

a

=

b+ a

b

(a, b) (b, a+b) (a+b, a+2b) (a+2b, 2a+3b) ...

In fact Fn =

1 1

1 0

n=

Fn+1 Fn

Fn Fn1

. (Induction)

Corollary: Fn+1Fn

1 F2n = (1)

n. (Determinant)



Introduction


Conclusion





71/124



Matrix form:

1 1

1 0

b

a

=

b+ a

b

(a, b) (b, a+b) (a+b, a+2b) (a+2b, 2a+3b) ...

In fact Fn =

1 1

1 0

n=

Fn+1 Fn

Fn Fn1

. (Induction)

Corollary: Fn+1Fn

1 F2n = (1)

n. (Determinant)



Introduction


Conclusion





72/124

ecu e ces as ea a s o at o s


Matrix form:

1 1

1 0

b

a

=

b+ a

b

(a, b) (b, a+b) (a+b, a+2b) (a+2b, 2a+3b) ...

In fact Fn =

1 1

1 0

n=

Fn+1 Fn

Fn Fn1

. (Induction)

Corollary: Fn+1Fn

1 F2n = (1)

n. (Determinant)



Introduction


Conclusion





73/124


Matrix form:

1 1

1 0

b

a

=

b+ a

b

(a, b) (b, a+b) (a+b, a+2b) (a+2b, 2a+3b) ...

In fact Fn =

1 1

1 0

n=

Fn+1 Fn

Fn Fn1

. (Induction)

Corollary: Fn+1Fn

1 F2n = (1)

n. (Determinant)



Introduction


Conclusion





74/124


Matrix form:

1 1

1 0

b

a

=

b+ a

b

(a, b) (b, a+b) (a+b, a+2b) (a+2b, 2a+3b) ...

In fact Fn =

1 1

1 0

n=

Fn+1 Fn

Fn Fn1

. (Induction)

Corollary: Fn+1Fn

1 F2n = (1)

n. (Determinant)



Introduction


Conclusion



Changing Basis


75/124

g g

x, y = xe1 + ye2; ei is the standard basis

x, y = zv1 + wv2; how do we compute z, w?

yx

=

v2 v1 w

z

v2 v1

1 yx

=

w

z

We can also write linear transformations in a different

basis!


Introduction


Conclusion



Changing Basis


76/124

g g



yx

=

v2 v1 w

z

v2 v1

1 yx

=

w

z


basis!



77/124

Introduction


Conclusion




Changing Basis


78/124



yx

=

v2 v1 w

z

v2 v1

1 yx

=

w

z


basis!


Introduction


Conclusion




Changing Basis


79/124



yx

=

v2 v1 w

z

v2 v1

1 yx

=

w

z


basis!


Introduction


Conclusion




Linear Transformations in a Different Basis


80/124

Suppose Av = u but we want to change to a basis P. Then:

If new coordinates are v, u then v = Pv, u = Pu

So APv = Pu, or P1APv = u

New matrix in new basis is P1AP, a conjugate of AGoal: find basis such that new matrix is easier to work with


Introduction


Conclusion






81/124






Introduction


Conclusion






82/124





Qiaochu Yuan The Fibonacci numbers Introduction


Conclusion






83/124







Conclusion






84/124







Conclusion




The Power of Eigenvectors


85/124

Proposition: , are the eigenvalues of F. The eigenvectors

are

1

,

1

Proof.

1 1

1 0

1 =

+ 1

2 = + 1 by definition (and same for )

Dimension 2, so there are no other eigenvalues or

eigenvectors.

Computation can be done in general: we can find acharacteristic polynomial.

With P =

1 1

, P1FP =

0

0

, a diagonal matrix



Conclusion






86/124


are

1

,

1

Proof.

1 1

1 0

1 =

+ 1



eigenvectors.


With P =

1 1

, P1FP =

0

0

, a diagonal matrix



87/124

Introduction


Conclusion






88/124


are

1

,

1

Proof.

1 1

1 0

1 =

+ 1



eigenvectors.


With P =

1 1

, P1FP =

0

0

, a diagonal matrix



Conclusion






89/124


are

1

,

1

Proof.

1 1

1 0

1 =

+ 1



eigenvectors.


With P =

1 1

, P1FP =

0

0

, a diagonal matrix



Conclusion






90/124


are

1

,

1

Proof.

1 1

1 0

1 =

+ 1



eigenvectors.


With P =

1 1

, P1FP =

0

0

, a diagonal matrix



Conclusion






91/124

Conjugation respects multiplication: P1FnP =

n 0

0 n

Back to old basis: Fn

= P n 0

0 n

P1

Computations give

Fn = 1

n+1 n+1 n n

n n n1 n1

as desired



Conclusion


Proof 2: Partial Fraction DecompositionProof 3: Diagonalization



92/124


n 0

0 n


= P n 0

0 n

P1

Computations give

Fn = 1

n+1 n+1 n n

n n n1 n1

as desired



Conclusion


Proof 2: Partial Fraction DecompositionProof 3: Diagonalization



93/124


n 0

0 n


= P n 0

0 n

P1

Computations give

Fn = 1

n+1 n+1 n n

n n n1 n1

as desired



Conclusion

The shift operator

Eigendecomposition as a General Principle


94/124

What do the three proofs have in common?

Geometric series n, n appeared naturally

General idea of decomposing into a sum of simpler partsappeared naturally

"Simpler parts" = geometric series = eigenvectors?

Identify linear transformations acting in the first two proofs



Conclusion

The shift operator



95/124








Conclusion

The shift operator



96/124







Introduction


Conclusion

The shift operator



97/124







Introduction


Conclusion

The shift operator



98/124







Introduction


Conclusion

The shift operator

Interesting Linear Transformations


99/124

Vector spaces: spaces of sequences (proof 1), spaces of

functions (proof 2)

We want an be to associated with an eigenvalue of a

Sequences: take T(sn) = sn+1; then an is an eigenvector

with eigenvalue a

Functions: take T(f(x)) = f(x)f(0)x ; then1

1ax is aneigenvector with eigenvalue a

Shift operators:

T : a0 + a1x + a2x2 + ... a1 + a2x + a3x

2 + ...

Eigenvectors are precisely the geometric series,eigenvalues are their common ratios

Caution: spaces are infinite-dimensional


Introduction


Conclusion

The shift operator



100/124


functions (proof 2)



with eigenvalue a



Shift operators:

T : a0 + a1x + a2x2 + ... a1 + a2x + a3x

2 + ...




Introduction

The Three Proofs

Unifying the Proofs

Conclusion

The shift operator



101/124


functions (proof 2)



with eigenvalue a



Shift operators:

T : a0 + a1x + a2x2 + ... a1 + a2x + a3x

2 + ...




Introduction

The Three Proofs

Unifying the Proofs

Conclusion

The shift operator


V f ( f 1) f


102/124


functions (proof 2)



with eigenvalue a



Shift operators:

T : a0 + a1x + a2x2 + ... a1 + a2x + a3x

2 + ...





103/124

Introduction

The Three Proofs

Unifying the Proofs

Conclusion

The shift operator


V t f ( f 1) f


104/124


functions (proof 2)



with eigenvalue a



Shift operators:

T : a0 + a1x + a2x2 + ... a1 + a2x + a3x

2 + ...




Introduction

The Three Proofs

Unifying the Proofs

Conclusion

The shift operator


Vector spaces: spaces of sequences (proof 1) spaces of


105/124


functions (proof 2)



with eigenvalue a



Shift operators:

T : a0 + a1x + a2x2 + ... a1 + a2x + a3x

2 + ...




Introduction

The Three Proofs

Unifying the Proofs

Conclusion

The shift operator

A Generic Description


106/124

A sequence s is Fibonacci-type if

T2s = Ts+ s (T2 T 1)(s) = 0

When T is the shift on sequences, this is the usual

definitionWhen T is division by x, this is the definition of the

generating function

When T = F, this is the identity F2 F I = 0

(characteristic polynomial)


Introduction

The Three Proofs

Unifying the Proofs

Conclusion

The shift operator



107/124


T2s = Ts+ s (T2 T 1)(s) = 0



generating function

When T = F, this is the identity F2 F I = 0

(characteristic polynomial)


Introduction

The Three Proofs

Unifying the Proofs

Conclusion

The shift operator



108/124


T2s = Ts+ s (T2 T 1)(s) = 0



generating function

When T = F, this is the identity F2 F I = 0(characteristic polynomial)


Introduction

The Three Proofs

Unifying the Proofs

Conclusion

The shift operator



109/124


T2s = Ts+ s (T2 T 1)(s) = 0



generating function

When T = F, this is the identity F2 F I = 0(characteristic polynomial)


Introduction

The Three Proofs

Unifying the Proofs

Conclusion

The shift operator

A Generic Proof


110/124

Operators like T can be added and multiplied:

T2 T 1 = (T )(T )

T a : a0 + a1x + a2x2 + ...

(a1 a0a) + (a2 a1a)x + (a3 a2a)x2 + ...

(T a)s = 0 a is a geometric series with common ratioa

Operators commute, so we just take everything that is

annihilated by both operators

So we expect solutions to be span of{s|(T )s = 0}, {s|(T )s = 0}


Introduction

The Three Proofs

Unifying the Proofs

Conclusion

The shift operator

A Generic Proof


111/124


T2 T 1 = (T )(T )

T a : a0 + a1x + a2x2 + ...

(a1 a0a) + (a2 a1a)x + (a3 a2a)x2 + ...






Introduction

The Three Proofs

Unifying the Proofs

Conclusion

The shift operator

A Generic Proof


112/124


T2 T 1 = (T )(T )

T a : a0 + a1x + a2x2 + ...

(a1 a0a) + (a2 a1a)x + (a3 a2a)x2 + ...






Introduction

The Three Proofs

Unifying the Proofs

Conclusion

The shift operator

A Generic Proof


113/124


T2 T 1 = (T )(T )

T a : a0 + a1x + a2x2 + ...

(a1 a0a) + (a2 a1a)x + (a3 a2a)x2 + ...






Introduction

The Three Proofs

Unifying the Proofs

Conclusion

The shift operator

A Generic Proof


114/124


T2 T 1 = (T )(T )

T a : a0 + a1x + a2x2 + ...

(a1 a0a) + (a2 a1a)x + (a3 a2a)x2 + ...






Introduction

The Three Proofs

Unifying the Proofs

Conclusion

The shift operator

Generalization


115/124

Theorem: Let (sn) satisfy sn+k = ak1sn+k1 + ... + a0sn andlet P(x) = xk ak1xk1 ... a0 = (x 1)m1 ...(x r)mr.Then:

There exist polynomials p1,...pr with deg pi < mi suchthat...

sn = p1(n)n1 + ... + pr(n)

nr

Example: sn+2 = 4sn+1 4sn sn = (c1n+ c0)2n


Introduction

The Three Proofs

Unifying the ProofsConclusion

The shift operator

Generalization


116/124



sn = p1(n)n1 + ... + pr(n)

nr



Introduction

The Three Proofs


The shift operator

Generalization


117/124



sn = p1(n)n1 + ... + pr(n)

nr



Introduction

The Three Proofs


The shift operator

Generalization


118/124



sn = p1(n)n1 + ... + pr(n)

nr



Introduction

The Three Proofs


Conclusion


119/124

Binets formula is best understood as eigendecomposition.

Eigendecomposition can appear in many situations.

Further reading: artofproblemsolv-

ing.com/Forum/weblog_entry.php?t=175257, 212490,215833, 217361

Even further reading:

www.math.upenn.edu/ wilf/DownldGF.html

GOOD LUCK ON GUTS!



120/124

Introduction

The Three Proofs


Conclusion


121/124







GOOD LUCK ON GUTS!


Introduction

The Three Proofs


Conclusion


122/124







GOOD LUCK ON GUTS!


Introduction

The Three Proofs


Conclusion


123/124







GOOD LUCK ON GUTS!


Introduction

The Three Proofs


Conclusion


124/124







GOOD LUCK ON GUTS!


Date post:	08-Apr-2018
Category:	Documents
Upload:	diana-jenkins
View:	219 times
Download:	0 times

MIT Slides on Diagonalisatiojn

Documents