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1. Functional equations in one variable

Method of undeterm**in**ed coefficients Mathematical Database When we know that the unknown function satisfies certa**in** conditions, say it is a quadratic or 2 cubic function, we can immediately set up **variable**s (e.g. let f ( x) = ax + bx+ c if f ( x ) is a quadratic polynomial) and solve for them. Example **1.**6. If f ( x) is a quadratic function such that f ( x+ 1) − f( x) = 8x+ 3 and f (0) = 5 , f**in**d f ( x ). Solution. Let 2 f ( x) ax bx c Simplify**in**g gives = + + , then ax ( + 1) 2 + bx ( + 1) + c−ax 2 −bx− c= 8x+3. After solv**in**g, we have a = 4 and b =−**1.** Putt**in**g x = 0, we have c = 5 . 2ax + a + b = 8x + 3. Therefore 2 f ( x) = 4x − x+ 5. 2. **Functional** **equations** **in** more than **one** **variable** For functional **equations** with more than **one** **variable**, we can also apply the methods menti**one**d above. Besides we can also try to substitute some special values, say x = y = 0 **in**to the given condition given to obta**in** some results. As it is very difficult to describe these techniques **in** words, we will try to see their application **in** various examples below. Example 2.**1.** If f : → satisfies **1.** f (1) = 2 , 2. For all xy∈ , , f( xy) = f( x) f( y) − f( x+ y) + 1, f**in**d f ( x ). Page 4 of 10

Solution. Putt**in**g y = 1, then Mathematical Database f( x) = f( x) f(1) − f( x+ 1) + 1 = 2 f( x) − f( x+ 1) + 1 f( x+ 1) = f( x) + 1 Therefore apply**in**g condition 1 and by mathematical **in**duction, for all **in**teger x, we have f ( x) = x+ **1.** m m For any rational number, let x = where m, n are **in**tegers and n is not zero. Putt**in**g x = , y = n, n n ⎛ then ( ) m ⎞ ⎛m ⎞ f m = f ⎜ ⎟( n+ 1) − f ⎜ + n⎟ 1 ⎝ n ⎠ ⎝ n ⎠ + . S**in**ce f( x+ 1) = f( x) + 1 for ∀x ∈ , we have ⎛m ⎞ ⎛m⎞ f ⎜ + n⎟= f ⎜ ⎟+n. ⎝ n ⎠ ⎝ n ⎠ ⎛m⎞ ⎛m⎞ Substitut**in**g this **in**to the orig**in**al equation, we have m+ 1 = f ⎜ ⎟( n+ 1) − f ⎜ ⎟−n+ **1.** Thus ⎝ n ⎠ ⎝ n ⎠ ⎛m⎞ m f ⎜ ⎟ = + **1.** ⎝ n ⎠ n So we have f ( x) = x+ 1 ∀x ∈ . Remark. In this question we can see that we first solve the functional equation for a special case (we found f ( x ) when x ∈ ), then we solve for the more general case. (we found f ( x ) when x ∈ ). Indeed, this method is used **in** many occasions. For f**in**d when f ( x ) when x∈ about this later. x ∈ . Then, f**in**d f ( x ) when x ∈ by substitut**in**g f : → , we can apply this method similarly. First, m x = . F**in**ally, f**in**d f ( x ) n by the density of rational numbers (for cont**in**uous functions only). We shall see more Example 2.2. If 2 2 ( x y) f( x y) ( x y) f( x y) 4 xy( x y ) − + − + − = − for all x, y , f**in**d f ( x ). Solution. The given condition is equivalent to Page 5 of 10

- Page 1 and 2: Mathematical Database FUNCTIONAL EQ
- Page 3: 2 2 2 x f x = + x + + x f x ( ) 3(4
- Page 7 and 8: If f ( x ) =−x, from (2.4) we hav
- Page 9 and 10: Mathematical Database Now, as f is