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 inho- mogenous 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 Sup´erieurein 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 alge- bra 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 arith- metic 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 opera- tions, 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 oper- ations 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 prin- ciple 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 equa- tion 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 homoge- nous 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.