Given that f (x) satisfies f (x + y ^ 2) = f (x) + 2 (f (y)) ^ 2 for any natural number x and y, and f (1) is not equal to 0, then f (2012) =?

Given that f (x) satisfies f (x + y ^ 2) = f (x) + 2 (f (y)) ^ 2 for any natural number x and y, and f (1) is not equal to 0, then f (2012) =?

When y = 0, f (x) = f (x) + 2F & # 178; (0)
The solution is f (0) = 0
When x = 0, y = 1
f(1)=f(0)+2f²(1)=2f²(1)
Because f (1) is not equal to 0
So f (1) = 1 / 2
When y = 1, f (x + 1) = f (x) + 2 * F & # 178; (1) = f (x) + 2 * (1 / 2) &# 178;
So f (x + 1) = f (x) + 1 / 2
So f (2012) = f (2011 + 1) = f (2011) + 1 / 2
=f(2010)+1/2+1/2=f(2010)+(1/2)*2
=.=f(1)+(1/2)*2011
=1/2+(1/2)*2011
=(1/2)*2012
=1006

In this paper, we define a function y = f (x) on the set of natural numbers n. We know that f (1) + F (2) = 5. When x is odd, f (x + 1) - f (x) = 1. When x is even, f (x + 1) - f (x) = 3 (2) find the analytic expression of F (x)

(1) From F (1) + F (2) = 5F (2) − f (1) = 1, the solution is f (1) = 2, f (2) = 3. So f (2n + 1) - f (2n-1) = [f (2n + 1) - f (2n)] + [f (2n) - f (2n-1)] = 3 + 1 = 4, so f (1), f (3), f (5),..., f (1), f (3) and (2) when x is odd, f (x) = [f (x) - f (x-1)] + [f (x-1) - f (X-2)] + +When x is even, f (x) = [f (x) - f (x-1)] + [f (x-1) - f (X-2)] + +[f (2) - f (1)] + F (1) = 12.1 + X − 22.3 + 2 = 2x − 1, so f (x) = 2x, X is odd, 2x − 1, X is even

Given the function f (x) = {when x is greater than or equal to 10, it is x-3; when x is less than 10, it is f [f (x + 5)], where x is a natural number, find f (8) =?
I just learned to be confused~

Answer: 7
It comes step by step from the inside to the outside
According to the meaning of the title, it can be concluded that:
f(x)=x-3 x>=10
f(x)=＝f(f(x+5)) x