The Principle of Linearity – Applications in the Areas of Algebra and Analysis

Proseminar Mathematisches Probleml¨osen University of Karlsruhe SS 2006

The Principle of Linearity – applications in the areas of algebra and analysis –

Franziska H¨afner

Contents

1 Converting complex problems into linear structures 2 1.1 General structure of the appendant vector space using arithmetic operations with a hexagon ...... 2 1.2 Speciﬁc example for a calculation using the linear structure ...... 4

2 Theory of linear recursion 5 2.1 Deﬁnition of linear recursion and diﬀerence between homogenous and inhomogenous linear recurrent functions ...... 5 2.2 Homogenous linear recursions ...... 5 2.2.1 Linear Shiftregister ...... 6 2.2.2 Derivation of the function R(a), which depicts the linear recursion, with the aid of eigenvalues and eigenvectors ...... 7 2.2.3 Application of the theory: Fibonacci Numbers ...... 8 2.3 Inhomogenous linear recursion ...... 12 2.3.1 Theory ...... 12 2.3.2 The Towers of Hanoi ...... 13

Sources 14 1 Converting complex problems into linear structures

“Structures are the weapons of the mathematicians” This is a quote by Bourbaki, a group of French mathematicians who came together in order to rearrange and structure mathematical cognitions. Nicolas Bourbaki is the pseudonym under which a group of mainly French 20th-century mathematicians wrote a series of books presenting an exposition of modern advanced mathematics, beginning in 1935. With the goal of founding all of mathematics on set theory, the group strove for utmost rigour and generality, creating some new terminology and concepts along the way. While Nicolas Bour- baki is an invented personage, the Bourbaki group is officially known as the Association des collaborateurs de Nicolas Bourbaki (“association of collaborators of Nicolas Bourbaki”), which has an office at the Ecole´ Normale Supérieurein Paris. According to their doing, the principle of linearity shall demonstrate how structures can help to solve tasks that appear to be quite complex.

The following paper is dedicated only to the use of the principle in the areas of algebra and analysis. The application in the area of geometry can be found on the handout of Jessica M¨uller.

1.1 General structure of the appendant vector space using arithmetic operations with a hexagon The ﬁrst example is to be an introduction to the kind of thinking needed to solve tasks with the help of the principle of algebra and analysis.

Task 1: Some number is positioned at each corner of a regular hexagon, a1 . . . an. One is allowed to use the following three operations:

1. Adding the same numer to two opposed numbers, e.g.

A1(1, 5, b):( a1, a2, a3, a4, a5, a6 ) → (a1 + b, a2, a3, a4 + b, a5, a6)

2. Adding the same number to all numbers which form an equilateral triangle, e.g.

A2(1, 3, 5, b):( a1, a2, a3, a4, a5, a6 ) → (a1 + b, a2, a3 + b, a4 + b, a5, a6)

3. Adding a speciﬁc number to one of the numbers and substracting this number from the adjacent numbers, e.g.

A2(1, 2, 3, b):( a1, a2, a3, a4, a5, a6 ) → (a1 − b, a2 + b, a3 − b, a4, a5, a6) Question: Which ﬁnal constellations can be achieved by applying only those three operations, given an initial constellation?

Solution: Application of the principle of linearity: 1. Using the model of a 6-dimensional vector space 2. Deﬁning the vectors of this vector space in the following way:   a1  a2     a3  ~a =    a4     a5  a6

3. The final combination can be found as a combination of (a) the initial constellation (b) a linear combination of the vectors representing the three operations i. The first operation is represented by the vectors:  1   0   0   0   1   0         0   0   1  ~x =   , ~y =   , ~z =    1   0   0         0   1   0  0 0 1 ii. The second operation is represented by the vectors:  1   0   0   1       1   0  ~a =   ,~b =    0   1       1   0  0 1 iii. The third operation can be constructed using a vector of the first operation and substraction one of the second operation. Example:  1   1   0   0   0   0         0   1   −1  ~x − ~a =   −   =    1   0   1         0   1   −1  0 0 0 All in all we now get the simple equation:

 1   0   0   1   0   0   1   0   0   1             0   0   1   1   0  afinal = ainit + b1   + b2   + b3   + b4   + b5    1   0   0   0   1             0   1   0   1   0  0 0 1 0 1 The solution is now to be seen in the representation of the task with the help of vectors. It has to be taken into consideration, however, that the vectors concerning the three operations are not linearly independent as it can easily be deduced looking at 3.iii . The sum of the ﬁrst three vectors hence equals the sum of the last two vectors. As there is no other combination that adds up to zero it can be deduced that the vector space is 4-dimensional. P6 ~ As a 2-dimensional invariant is existent (f(a) = j=1 ajOAj), it is obvious that there are no further restrictions and that any possible ﬁnal constellation can be achieved.

Summarizing the result it can be said that the principle of linearity is, in this example, used to structure the mass of information based on the innumerable possibilities to apply the three allowed operations. Hence it can explicitly be seen that complex problems that include many possible operations with and variations of the initial constellation can be solved in an easier way when transforming the operations into vectors – applying the principle of linearity.

1.2 Speciﬁc example for a calculation using the linear structure Fibonacci Numbers

Have you ever wondered where we got our decimal numbering system from? The Roman Empire left Europe with the Roman numeral system which we still see, amongst other places, in the copyright notices after TV programmes (1997 is MCMXCVII). The Roman numerals were not displaced until the 13th Century AD when Fibonacci pub- lished his Liber abaci which means “The Book of Calculations”. Fibonacci, or more correctly Leonardo da Pisa, was born in Pisa in 1175 AD. He found it quite useful to calculate using the fingers in order to have a record of the intermediate result. Did you know that you might have Fibonacci fingers? In 1200 he returned to Pisa and used the knowledge he had gained on his travels to write Liber abaci in which he introduced the Latin- speaking world to the decimal number system. Fibonacci is perhaps 1 hand best known for a simple series of numbers, introduced in Liber abaci 1 thenar 2 knuckles separating and later named the Fibonacci numbers in his honour. 3 parts of the 5 fingers The series starts with the terms F0=0, F1=1. After that the simple rule is used: Add the last two numbers to get the next. Thats for the easy part of the Fibonacci Numbers. Considering the linear recurrent equation per which the Fibonacchi Numbers are defined: Fn = Fn−1 + Fn−2 for all n ∈ N it seems obvious that it could be a hardship to compute F246224566 without using a robotized system. Therefore it is aimed to develop a formula which enables people to calculate any Fibonacci Number using only one calculation step.

2 Theory of linear recursion

As theory is the basis for any undertaking, the next paragraph shall elucidate the theory of linear recursion with reference to the principle of linearity.

2.1 Definition of linear recursion and difference between homogenous and inhomogenous linear recurrent functions Definition A linear recursion (over the field of K) of kth order is defined by a linear recurrent equation. Linear recurrent equations have the characteristic property that subsequent members are related to the preceding members by linear equations. Example: (R) xn = ck xn−1 + ... + c1 xn−k + g(n) in which c1, . . . , ck ∈ K and g : N → K is a function. A linear recursion is named homogenous when g(n) = 0, otherwise inhomogenous.

2.2 Homogenous linear recursions The set of solutions of the linear recursion may as well be depicted as the set of solutions of a linear system of equations. The reason for this is that you can imagine that for T some given vector a with the components (a1, . . . , an) , a solution in the form of x = T T ( x1, . . . , xn ) is existent. Now one is able to deﬁne a function R, which maps (a1, . . . , an) , T onto (x1, . . . , xn) . Theorem (homogenous) Given the homogenous linear recursion of kth order

(Rh) xn = ck xn−1 + ... + c1 xn−k for all n ≤ k + 1 N N The solutions of (Rh) form a subspace U of K , dim(K ) = k and the function R : KK → KN , a → R(a) an isomorphism in L.

Proof: This equation can be seen as a system of linear equations. Due to that a matrix can be thought of that depicts the linear function R. The kernel of this matrix gathers all the solutions of the system of linear equations. Meaning we just have to show that the kernel is a subspace. This is easily shown, as the kernel is not empty due to the fact that the zerovector is in the core. Furthermore it can be said that assuming that a and b were part of the core, due to the linearity of the linear function, we can see that α a + β b are also part of the kernel (proof page 85. Scriptum Aumann) As to the fact that K and L are ﬁnite-dimensional it can be deduced that an inverse function is existent. It has to be proven that this inverse function is injective and surjective. The inverse function is to be deﬁned as follows:

Φ: LN → Kk

(xnn>k ) → (x1, . . . , xk)

Injectivity can be proven as follows. It is assumed that ck 6= 0. Hence if xk+1 = 0 all the other xk are bound to be zero, which means that the kernel of the matrix does only contain the zerovector and the function is therefore injective. Surjectivity follows easily due to the fact that a linear recursion is looked at, as this means that all the precedent elements are used, to depict the ﬁnal one, meaning that all the elements of the value set do also lie within the target area. As the inverse function does have those qualities. It can be deduced that the original function does also have those qualities. Thus it can be deduced that the dimension of the subspace U equals k and therefore, as the dimension of the original vector space equals the dimension of the ﬁnal vector space the function can be considered to be an isomorphism.

In order to describe the system of a linear recursion more plastic the concept of a

2.2.1 Linear Shiftregister is looked at. According to a formal deﬁnition, a shiftregeister is a logical control unit and with every working cycle, the stored information is transferred to the next block. It applies the method of ﬁfo. ++++

· ck · c2 · c1

xn−1 xn−(k−1) xn−k

The problem of a linear recursion is easily transferred to the system of a shiftregister. For example, calculating x2:

xn−k(n=2,k=1) = x1 =⇒ x2 = x1 c1.

The next step would then be calculating x3:

First working cycle (k=2): xn−k(n=3,k=2) = x1 +

Second working cycle (k=1): xn−k(n=3,k=1) = x2 =⇒ x3 = x2 c2. As it can easily be seen, you need to shift n-times until you get the desired linear recurrent equation xn+1 = ...

2.2.2 Derivation of the function R(a), which depicts the linear recursion, with the aid of eigenvalues and eigenvectors

The function R is, hence, intended to depict the an onto the xn by shifting. Applying the function to the vector ( x1, . . . , xn ) has two eﬀects: Firstly, x1 vanishes and secondly, xk+1 is introduced, which is a linear combination of the preceding values x1, . . . , xk. 2.2.3 Application of the theory: Fibonacci Numbers For further explanation and evaluation the Fibonacci Numbers are to be used as a repre- sentative example instead generally proving and calculating the following.

The following matrix maps the initial values ~a of the Fibonacchi sequence onto the subsequent element ~b = R ~a 0 1 R = 1 1

An example would be: Let a1 be 5 and a2 be 8:

0 1 5 8 ~b = = 1 1 8 13

=⇒ b2 = a3 = 13.

Explanation of how the matrix was generated:

• b1 = a2

• b2 = a1 + a2 =⇒ Applying the matrix n times creates the nth Fibonacci Number.

Principal axis transformation

A diagonalization of the matrix is necessary with the objective of ﬁnding an explicit formula to calculate the nth Fibonacci Number. The function R maps vectors of the R2 onto vectors of the R2. This means that the subsequent elements are still dependent on the values of the two preceding elements. This is not adjuvant as the subsequent elements should not depend on a sequence of other elements. Hence, a principal axis transformation is essential:

1. Calculation of the eigenvalues:

−λ 1 0 = det(A − λ E) = 1 1 − λ

= λ2 − λ − 1 √ √ 1+ 5 1− 5 =⇒ λ1 = 2 , λ2 = 2 . 2. Calculation of the eigenvectors: First eigenvector: =⇒ (vx) = 1 (optional) √ ! − 1+ 5 1 v 0 2 √ x 1+ 5 = vy 0 1 1 − 2 √ 1+ 5 =⇒ (vy): − 2 + vy = 0 1 √ =⇒ ~v(λ1) = 1+ 5 2 Second eigenvector:

=⇒ (v ) = −1 (optional) y √ 1+ 5 =⇒ (vx): 2 √ 1+ 5 =⇒ ~v(λ ) = 2 2 −1 3. Creation of the transition matrix T . √ ! 1 1+ 5 √ 2 T = (~v(λ1), ~v(λ2)) = 1+ 5 2 −1

Why do we need this matrix? Lets just go back to the initial idea. We want to ﬁnd a matrix that maps the vector ~a onto the vector ~b, meaning we want to ﬁnd a matrix R with the qualities that

~b = R~a

Multiplication with T −1 (from the left)

T −1 ~b = T −1 R~a E z }| { T −1 ~b = T −1 R TT −1 ~a T −1 ~b = T −1 RT T −1 ~a | {z } (diagonalized) matrix Q

=⇒ Q is a representation of the mapping matrix R with respect to another basis.

(R − λ E)~v = 0 =⇒ R~v = λ~v In this case (calculated for the ﬁrst eigenvalue, the second is calculated analogical): √ ! 0 1 1 1+ 5 R~v(λ ) = √ = 2 √ 1 1 1 1+ 5 1+ 5 2 1 + 2 √ ! √ 1+ 5 1 + 5 1 = 2√ = √ 6+2 5 2 1+ 5 4 2

= λ1 ~v(λ1)

R (R~v(λ)) equals λ2 ~v(λ). 2 =⇒ ∃ new basis of sequences w(λi) = (1, λi, λi ,...), (i = 1, 2). These sequences can be gained by applying R 0-times, once, twice, . . . , n-fold.

T −1 ~a = ~s,(~s ∈ R2) is depiction of the vector ~a in the new basis. =⇒ ~a = T ~s.

n X R(a) = R( si vi) i=1 n X R(a) = si R(vi) (due to linearity of R) i=1

R(vi) is the depiction of the eigenvector in the new basis

=⇒ R(vi) = wi

n X R(a) = si wi. i=1

Theorem (Solution to linear recursions)

Given the homogenous linear recursion

(R): xn = ck xn−1 + ... + c1 xn−k for all n ≤ k + 1 over K

k k−1 1. Be λ ∈ K a zero of the polynomial f(t) = t − ck t − ... − c2 t − c1. This implies that w(λ) = (1, λ, λ2,...) is a solution to (R).

2. In the case of differing zeros, hence different eigenvalues, the sequences w(λj) = 2 (1, λj, λj ,...), j = 1, . . . , k form a basis to the solution space of (R). T n T 3. Be ( a1, . . . , an ) ∈ R a variable vector of initial values and ~s = ( s1, . . . , sn ) the Pn definit solution to the system of equations of i=1 si vi = a. The solution R(a) to R for the initial values a1, . . . , an is then given by Pn R(a) = i=1 si wi.

Determination of the si

Condition: F0 = 0 and F1 = 1 √ √ 2 √ √ 2 1+ 5 1+ 5 1+ 5 1− 5 First premiss: w1 = (1, 2 , 2 ,...) and w2 = (1, 2 , 2 ,...) Second premiss: Only s1 and s2 are to be considered in this case.

2 2 =⇒ (F0,F1,F2,...) = s1 (1, λ1, λ1 ,...) + s2 (1, λ2, λ2 ,...)

⇓

1. F0 = 0 = s1 + s2

√ √ 1+ 5 1− 5 2. F1 = 1 = s1 2 + s2 2

Solving this system of equations, we get s1 = −s2

1 1 s 0 √ √ 1 = 1+ 5 1− 5 s 1 2 2 2

0 1 √ 1 1− 5 1 s = 2 = √ 1 1 1 √ √ 5 1+ 5 1− 5 2 2

1 0 √ 1+ 5 1 1 s = 2 = −√ 2 1 1 √ √ 5 1+ 5 1− 5 2 2

We need the apply the function R n times, so we have to use the nth component of our new basis vectors ~w. Thus it can be deduced that

√ !n √ !n 1 1 + 5 1 1 − 5 F (n) = √ − √ 5 2 5 2 2.3 Inhomogenous linear recursion

2.3.1 Theory

Extract from the deﬁnition of linear recursion

(R): xn = ck xn−1 + ... + c1 xn−k + g(n)

in which c1, . . . , ck ∈ K and g : N → K is a function.

Assumed type of function g(n):

g(n) = a rn a, r, 6= 0

“Guessing” a solution to the inhomogenous system (mr, mr2, mr3,...) for some m ∈ K. Substitution of the solution into the lienar recurrent equation:

n n−1 n−k n mr = ck m r + ... + c1 m r + a r for all n ≥ k + 1. =⇒ Solution vector = (m r, . . . , m rn)

Division by r(n−k), extraction of m

k k−1 k =⇒ m (r − ck r − ... − c1) = a r | {zc } c being the characteristic polynomial of the apt homogenous system’s matrix (c 6= 0).

a =⇒ m = rk. c

Theorem (Structure of the solution space)

Given the linear recursion n (R): xn = ck xn−1 + ... + c1 xn−k + a r for all n ≤ k + 1 k k−1 in which c1, . . . , ck, a, r ∈ K and a, r 6= 0. Be c = r − ck r − ... − c1 6= 0, then 2 3 a k w = (mr, mr , mr ,...) is a solution to (R) with m = c r . All the solutions to (R) are thus appointed by the set of w + L, whereas L is the solution space to the adjacent homogenous recursion. 2.3.2 The Towers of Hanoi

The Tower of Hanoi puzzle was invented by the French mathematician Edouard Lucas in 1883. We are given a tower of eight disks (initially four in the applet below), initially stacked in increasing size on one of three pegs. The objective is to transfer the entire tower to one of the other pegs (the rightmost one in the applet below), moving only one disk at a time and never a larger one onto a smaller. Recurrent solution to the problem of determining the minimum number of transfers:

an = 2 an−1 + 1

Determining the solution to the homogenous system

Be a1 = 1. an = 2 an−1 (homogenous) =⇒ L = (s (1, 2, 22, 23,...): s ∈ K). (cmp. theorems)

“Guessing” the solution to the inhomogenous system Considering a rn. This expression equals 1 in this case =⇒ chose a = 1 and r = 1 k k−1 Considering the polynomial c : c = r − ck r − ... − c1 =⇒ c = r − c1 (Only the last two terms of the polynomial are to be taken into consideration.) c(1) 6= 0 =⇒ r = 1 has been a suitable choice.

a k 1 k =⇒ c = 1 − 2 and from m = c r =⇒ m = −1 1 . = −1 =⇒ (mr, mr2,...) becomes (−1, −1, −1,...) as m = −1 and r = 1 (speciﬁc solution). As we got the hint that every solution of R must be structured like speciﬁc solution + general solutuion, we can hence deduce that the solution space must look like this:

Extract of the theorem of the structure of the solution space: All the solutions to (R) are thus appointed by the set of w + L, whereas L is the solution space to the adjacent homogenous recursion.

According to this theorem, the solution can be written as:

(s − 1, 2 s − 1, 22 s − 1,...)

a1 being 1 ⇒ s − 1 = 1 ⇒ s = 2 ⇒ Composition of the solution vector: (2 − 1, 22 − 1,...) The solution to the problem of the Towers of Hanoi can hence be summarized in this equation: n−1 an = 2 for all integers n. This has been a short introduction into the application of the priciple of linearity in the ares of algebra and analysis. For more information, please refer to my sources. Sources books

Skriptum zur Vorlesung Lineare Algenbra und Analytische Geometrie I, Prof. Dr. H. Kuhnle, Prof. Dr. G. Aumann, Dr. U. Schober, WS 96/97 Repetitorium der Linearen Algebra, Dr. D. Wille Differentialgeometrie, Kühnel Analysis I, Forster Repetitorium der gewöhnlichen Differentialgleichungen, S. Timmann Das gelbe Rechenbuch 3, gewöhnliche Differentialgleichungen webpages

http://www.cut-the-knot.org/recurrence/hanoi.shtml use of the source of wikipedia http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/ﬁbnat.html http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/phi.html http://www.dgps.de/fachgruppen/methoden/mpr-online/issue4/art5/node2.html