If f(a + b) = f(a) + f(b) - 2f(ab) for all nonnegative integers a and b, and f(1) = 1, compute f(1986). ???????

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Answer 1 · 2017-05-24T14:09:32

$f(1986)=0$

$f(1986)=f(1985+1)=f(1985)+f(1)-2f(1985*1)=1-f(1985)$

$f(1985)=f(1984+1)=f(1984)+f(1)-2f(1984*1)=1-f(1984)$

$f(1984)=f(1983+1)=f(1983)+f(1)-2f(1983*1)=1-f(1983)$

$ldots$

$f(n)=f((n-1)+1)=f(n-1)+f(1)-2f((n-1)*1)=1-f(n-1)$

Now, $f(1)=1,f(2)=f(1+1)=f(1)+f(1)-2f(1)=0$

Thus, $f(1)=1,f(2)=0,f(3)=1,f(4)=0,ldots$ The function is $1$ for odd numbers and $0$ for even numbers. Since $1986$ is even, $f(1986)=0$

Answer 2 · 2017-05-24T17:09:38

See below.

Considering

$f(n+1)=f(n)+f(1)-2f(n)=-f(n)+1$

the difference equation

$f_(n+1)+f_n=1$ has a solution with the structure

$f_n = a (1)^n+b(-1)^n$ so we have

${(f_(n+1)=a+b(-1)^(n+1)+a+b(-1)^n=1),(f_n=a+b(-1)^n = 1):}$

solving for $a,b$ we have

$a=1/2, b = -1/2(-1)^n$ so

$f_n = 1/2(1-(-1)^n)$

so if $n$ is odd

$f_(2k+1)=1$

and if $n$ is even

$f_(2k)=0$

so

$f(1986)=0$