Fun(ctional) equation
f ( x ) + f ( y ) + f ( x y ) = 2 + f ( x ) f ( y )
If the above equation is true for all x , y ∈ R and f ( x ) is a polynomial function with f ( 4 ) = 1 7 and f ( 1 ) = 1 , then find the value of f ( 5 ) .
The answer is 26.
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1 solution
The eq can be written as ( f ( x ) − 1 ) ( f ( y ) − 1 ) = ( f ( x y ) − 1 ) .Now we set g ( x ) = f ( x ) − 1 .This yields g ( x ) g ( y ) = g ( x y ) ,since f ( x ) is a polynomial in x so is g ( x ) .So g ( x ) = a ( n ) x n + a ( n − 1 ) x ( n − 1 ) . . . . . . . . . . . . . . a ( o ) .Where a ( j ) denotes the cefficient of x j Now comparing coefficients its easy to show that g ( x ) = x n .Hence f ( x ) = x n + 1 .Now set x = 4 yields n = 2 .Hence f ( x ) = x 2 + 1
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Great algebraic solution, Spandan!
I'm actually going to apply a calculus approach here. Since f(x) is a polynomial function, let us assume it is differentiable everywhere on its domain. Differentiating the above functional equation with respect to x and y respectively gives:
f ′ ( x ) + y ⋅ f ′ ( x y ) = f ′ ( x ) f ( y )
f ′ ( y ) + x ⋅ f ′ ( x y ) = f ( x ) f ′ ( y )
and some rearranging of variables, we now obtain:
f ( x ) − 1 x ⋅ f ′ ( x ) = f ( y ) − 1 y ⋅ f ′ ( y ) = A ⇒ f ( x ) − 1 f ′ ( x ) = x A
and integrating with respect to x gives:
l n ( f ( x ) − 1 ) = A ⋅ l n ( x ) + B ⇒ f ( x ) = e B x A + 1 , where A , B ∈ R . Substitution of the initial condition f ( 4 ) = 1 7 now produces:
f ( 4 ) = e B ⋅ 4 A + 1 = 1 7 ⇒ A = 2 , B = 0
or f ( x ) = x 2 + 1 is the required polynomial. Thus, f ( 5 ) = 2 6 .