Functional equations 1.

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1 Functional equations. What is a function? Technically speaking, it would be rather impossible to give a proper notion of a function. We will need to use other words, such as 'relation', 'map' or other, which are just synonyms of a word function and need to be properly dened. Question: What we can do? We will get understanding of not what the object is, but how we can work with it. Assume, we have two sets, call them A and B, and the aim is to describe a function, which maps A on B. We will just say, that for eery from set A there is a choice of an element y from set B. This element y we will denote as f, it's called the image of. Notation: f : A B; y = f or f : y Composition of functions. Denition. Let us have two functions y = f, g : A B and z = gy, f : B C. Then a function z = F = gf, F : A C is called a composition of functions g and f. Notation: F = g f. Remark. There can be a composition of more than two functions, i. e. : z = F = f f... f n = f f f 3... f n.... Remark. Note, that g f and f g are dierent functions. Eample. Let g =, fy = cos y.then F = f g = cos. Eample. Let ϕ = 3. Then ϕ ϕ = 3 3 = 9. Eample 3. Let ψ = Then ψ ψ = = Let f = +, g = +. Find f g.. Let ϕ = , [ 5; ]. Find ϕ ϕ. 3. Let f = +. Find: a f f; b f f f; c f f f f; d f f f f f. Homework 4. Find ϕ ϕ, if a ϕ = ; b ϕ = ; c ϕ = ; d ϕ =. 5. Let ϕ =. Find ϕ ϕ ϕ. 6. Let ϕ =. Find } ϕ ϕ {{... ϕ. } n times

2 The simplest functional equations Functional equations are equations where unknown is a function. Generally, those functions are connected with well-known functions using composition. Eamples of functional equations: f + 5 f f3 = 0 or f y = f + fy, where f is unknown, and the aim is to nd f. Denition. Function f is called a soultion of a functional equation on set M, if it satises this equation for all values of variable in the set M. Problems. Prove that a function f = A + B 3, where A and B some constant numbers, is a solution of f + 5 f f = 0.. Proof, that a function f = C, where C is some constant, is a solution of f + y = f + fy. 3. Prove, that f = α for any α is a solution of f y = f fy. 4. Prove, that f = sgn is a solution of f y = f fy. Remark. Last two tasks show, that dierent functions may be solutions of the same equation. To solve a functional equation means to nd all of the solutions or prove that they do not eist Prove, that f = g, where g is any function, is a solution of 6. Find all functions f dened on the set of real numbers, such that 3 f f +5 = 0 for any real. 7. Find all functions f and g satisfying net condition for all 0 f + g = + +, f + g = Find all functions f dened on the set of all real numbers, such that for all real and y this function satises sin + cos y = f + fy + g gy. Homework 9. Prove, that a function f = a for any a > 0 is a solution of f + y = f fy. 0. Find all functions f dened on real numberssatisfying for any real π f = f0 sin f cos +.. Find all functions f and g dened on real numbers, such that for any real numbers and y the following holds f + fy + g gy = y.. A function fn from integers to integers satises the condition: { n 0, if n > 00, fn = f f n +, if n 00. Prove that fn = 9 for all n 00.

3 Functional equations-3. Substituion in equations without free variables. Find all functions f, such that for every ± it satises f =. +. Find all functions f, such that for every 0 it satises f f =. 3. Find all functions f, such that for every, 0 it satises f + f =. 4. Find all functions f, such that for every it satises f + f + = Let a, b, c be some constant numbers, a b. Find all functions f satisfying af + bf = c. 6. Find all functions f, such that for every 0 it satises 3f + f + f =. 7. Find all functions f, such that for every 0, it satises f + f + f = Homework 8. Find all functions f, satisfying f + =. 9. Let a, c, k be some constant numbers and a c. Find all functions f, such that for a + every c, it satises f = k. c + 0. Let a ±. Find all functions f, such that for every 0 it satises af + f = a.. Find all functions f, such that for every 0 it satises + f + f =.. Let a ± and let g be some function. Find all functions f, such that for every it satises f af = g. 3. Find all functions f, such that for every 5 it satises f f =. 4. Let a 0 be some constant number. Find all functions f, such that for every 0, a it satises a f + f =. a

4 5. Find all functions f, such that for every ± it satises 3 + f + f = Let a and b be some constant numbers, a 0, a b. Find all functions f, such that for every ± it satises + af + + bf = Let a and b be some constant real numbers, a ±, and let n be a natural number. Find all functions f, satisfying af n + f n = b n. 8. Let a be some constant number, a ±. Find all functions f, such that for every 0 è ± it satises + f + af = +. +

5 Rules Setup. The board consists of a grid of 8 8 squares.. Squares are adjacent horizontally, vertically or diagonally. 3. The game is played by two or four players. 4. Each of the player has his own camp: a 3 3 square adjoint to a corner of the whole board. 5. Each player has a set of pieces in a distinct color, of the same number as squares in each camp. 6. The board starts with all the squares of each player's camp occupied by a piece of that player's color. Objective The winning objective is to be the rst player to race all one's pieces into the opposing camp the camp diagonally opposite one's own. Play sequence. Players randomly determine who will move rst.. Pieces can move in eight possible directions up, down, left, right and diagonally. 3. Each player's turn consists of moving a single piece of one's own color in one of the following plays: 3a. One move to an empty square: Place the piece in an empty adjacent square. This move ends the play. 3b. One or more jumps over adjacent pieces: An adjacent piece of any color can be jumped if there is an adjacent empty square on the directly opposite side of that piece. Place the piece in the empty square on the opposite side of the jumped piece. The piece which was jumped over is unaected and remains on the board. After any jump, one may make further jumps using the same piece, or end the play. 4. Then the other player makes his step in the same way. First set of tasks 0. Play as many as possible games, to understand game mechanics and simplest rules of the game. Try to see logical connections, and try to distinguish somehow good moves and bad moves. The more you play, the deeper you understand how to play this game.. To nish the rules you need to write down when the game ends.. To make the game proper you need to add up etra rules, so the game will always end up somehow one of the players winning or draw. Also you need to epress properly the winning conditions. 3. Find ways to measure moves of each player to describe which move sequence of moves is better. Can you do it numerically?

6 Symmetric polynomials variables case Theorem Vieta. Let, be roots of polynomial + p + q. Then { + = p; = q. Both + and are polynomials over variables and and moreover those polynomials do not change, if one replace with and with. We will study such polynomials. Denition. Polynomial f, y, which does not change if one echange variables and y are called symmetric Eample. Polynomial f, y = y + y is symmetric, while f, y = 3 3y is not. Denition. Polynomials σ = + y and σ = y are called elementary symmetric polynomials over two variables. Aside of σ and σ we will be using polynomials + y, 3 + y 3,..., n + y n, which are called power sums and we will denote them as S n : S n = n + y n. We will try to epress rst several of S n using σ and σ : S = + y = + y y = σ σ ; S 3 = 3 + y 3 = + y y + y = σ 3 3σ σ ; S 4 = 4 + y 4 = + y y = σ 4 4σ σ + σ. From that you may assume, that it will hold for any n, which is your rst task: Lemma. Every power sum S n = n + y n can be represented as a polynomial over σ and σ. I hope your soultion for the rst task was constructive so it gives you way to construct that polynomial over sigma and sigma, so you will easily show us such representation for S 5, S 6 and S 7. Etra task: There also direct formulas for S n, which you may try to prove or at least to show that those formulas are equal to each other: S n = n + y n = σ + σ 4σ n + σ σ 4σ n S n = n S n = n [ n ] m=0 [ n ] m=0 n m σ n m σ 4σ m. m n m! σ n m σ m, 3 m! n m! where [ ] n means the biggest integer less than n. The third formula was constructed by Edward Waring in 779. Using the proven lemma, one can etend it to the proof of the main theorem about symmetric polynomials

7 Theorem Main. Any symmetric plynomial over two variables can be represented as a polynomial over σ, σ. This the main task to prove the main theorem. Aso one can show that such representaion is unique it does not depend on the way we obtain this representation, we will always end up with the same thing.