Functional Equation Fun

Algebra Level 4

{ f ( a + c , b + c ) = f ( a , b ) + c f ( a c , b c ) = c f ( a , b ) f ( a , b ) = f ( b , a ) \begin{cases} f\left(a+c,b+c\right) = f\left(a,b\right) + c \\ f\left(ac, bc\right) = cf\left(a,b\right) \\ f\left(a,b\right) = f\left(b,a\right) \end{cases}

How many functions f : R 2 → R f: \mathbb{R}^2 \to \mathbb{R} satisfy the above conditions?

0 1 2 4

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Josh Banister by Josh Banister , 24, United Kingdom

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1 solution

Josh Banister
Nov 27, 2016

First we consider a = x , b = − x a = x, b = -x . We have f ( x , − x ) = − f ( − x , x ) = − f ( x , − x ) f(x,-x) = -f(-x,x) = -f(x,-x) meaning f ( x , − x ) = 0 f(x,-x) = 0 . This implies f ( x + x , − x + x ) = x f ( 2 x , 0 ) = x f ( x , 0 ) = x 2 \begin{aligned} f(x+x, -x + x) &= x \\ f(2x, 0) &= x \\ f(x,0) &= \frac{x}{2} \end{aligned} Now instead of x x , we substitute x = a − b x = a-b . we now get f ( a − b , 0 ) = a − b 2 f ( a − b + b , 0 + b ) = a − b 2 + b f ( a , b ) = a + b 2 \begin{aligned} f(a-b, 0) &= \frac{a-b}{2} \\ f(a-b + b, 0+b) &= \frac{a-b}{2} + b \\ f(a,b) &= \frac{a+b}{2} \end{aligned}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice sequence of substitution steps.

Note: We have only shown that f ( a , b ) = a + b 2 f(a, b) = \frac{ a+b}{2} is a necessary condition. We still need to show that is sufficient, namely by substituting it back into the equations and verify that they still hold.

Calvin Lin Staff - 4 years, 6 months ago

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