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#