Question 373003
x= 4, f(4) = 0.0625,
x =3, f(3) = 0.125,
x = 2, f(2) = 0.25,
x = 1, f(1) = 0.5,
x = 0, f(0) = 1.
Notice that {{{(f(x))/(f(x-1)) = 1/2}}}.  Hence
{{{(f(x))/(f(x-1))*(f(x-1))/(f(x-2))*(f(x-2))/(f(x-3))}}}*...*{{{(f(2))/(f(1))*(f(1))/(f(0)) = (1/2)*(1/2)}}}*...*{{{(1/2)*(1/2)}}}, the product having x factors.  The left most expression gives way to a chain-reaction of cancellations, giving {{{(f(x))/(f(0)) = (1/2)^x = 1/2^x}}}.  But f(0) = 1, thus
{{{f(x) = 1/2^x}}}.

And that my STUDENT, is how it's done....