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1. Functional equations in one variable

1. Functional equations in one variable

Solution. x + 1 Let t =

Solution. x + 1 Let t = , then x 1 x = t − 1 . Directly substitution yields, () f t 2 ⎛ 1 ⎞ ⎜ ⎟ + 1 ⎝t −1⎠ 1 ⎛ 1 ⎞ 1 ⎜ ⎟ ⎝t −1⎠ t −1 2 = + = − 2 t t+ 1. Mathematical Database Thus 2 f ( x) = x − x+ 1. Example 1.3. 2 If f (ln x) = x + x+ 1, where x > 0 , find f ( x ). Solution. Let t = ln x, then x t = e . Directly substitution yields, t ( ) 2 t 2x x f() t = e + e + 1. Thus f( x) = e + e + 1. In general, if we have f ( g( x)) = h( x) and g( x) has an inverse function, then we may replace g 1 ( x) and get f ( x) = h( g ( x)) . − −1 x by Solving equation Sometimes after performing transformation of variables, we can arrive at simultaneous equations. We can find the unknown function after solving the simultaneous equations. We may also treat the unknown function as a variable in ordinary equations and solve it. Example 1.4. If f ( x) 4+ x = 2 3 + f ( x) x 2 , find f ( x ). Solution. It is equivalent to 2 2 x f x x f x ( ) = (4 + )(3 + ( )) . Simplifying it, we have Page 2 of 10

2 2 2 x f x = + x + + x f x ( ) 3(4 ) (4 ) ( ) − = + 2 4 f( x) 3(4 x ) f( x) =− 2 3(4 + x ) 4 Mathematical Database Example 1.5. If ⎛ x −1⎞ f ( x) + f ⎜ ⎟= 1+ x, find f ( x ). ⎝ x ⎠ Solution. Let x −1 t = , then x 1 x = 1 − t . Direct substitution yields f ⎛ 1 ⎞ 1 ⎜ ⎟ + f() t = 1 + ⎝1− t⎠ 1− t (1.1) ⎛ 1 ⎞ 1 f ⎜ ⎟ + f( x) = 1 + ⎝1− x ⎠ 1− x Let 1 t = 1 − x , then t −1 x = . Direct substitution yields t (1.2) ⎛t−1⎞ ⎛ 1 ⎞ t−1 f ⎜ ⎟+ f ⎜ ⎟= 1+ ⎝ t ⎠ ⎝1− t⎠ t ⎛ x −1⎞ ⎛ 1 ⎞ x −1 f ⎜ ⎟+ f ⎜ ⎟= 1+ ⎝ x ⎠ ⎝1− x⎠ x On the other hand, by substituting it into (1), we have ⎛ x −1⎞ (1.3) f ( x) + f ⎜ ⎟= 1+x ⎝ x ⎠ ((1.1) + (1.2) + (1.3))/2 : ⎛ x−1⎞ ⎛ 1 ⎞ 1⎛ 1 x−1 ⎞ (1.4) f ( x) + f ⎜ ⎟+ f ⎜ ⎟= ⎜3+ + + x ⎟⎠ ⎝ x ⎠ ⎝1−x⎠ 2⎝ 1−x x By subtracting (1.2) from (1.4), we have 3 2 1⎛ 1 x−1 ⎞ ⎛ x−1⎞ 1⎛ 1 x−1⎞ − x + x + 1 f( x) = ⎜3+ + + x⎟− ⎜1+ ⎟= ⎜1+ + x− ⎟= 2⎝ 1−x x ⎠ ⎝ x ⎠ 2⎝ 1−x x ⎠ 2 x(1 −x) Page 3 of 10

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