Fun with Functions

Note that $a+b = f^{2013}(1)$

Since we know $f(1) = 2(1) - 1 = 1$ , $f^k(1) = 1$ for all integers $k$ . Hence, $a+b = 1$

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

From

$f(n) =1,3,5...=2n+(-1)$

$f^{2}(n)=1,9,25...=8n+(-7)$

$f^{3}(n)=1,27,125=26n+(-25)$

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Get

$a+b=\boxed{1}$

Notice that $f^{x}(n)$ is equivalent to $2^{x} \cdot n - (2^{x}-1)$ . If $2^{x}=y$ , then this equation is equal to $xy+(1-y)$ . The question asks for the value of $y+(1-y) \rightarrow \fbox{1}$ .

Well, obviously $f(1) = 1$ and $f^{2013}(1) = [f(1)]^{2013} = 1 = a + b$ .

$f^k(n)=2^kn-(1+2+\cdots+2^{n-1})$ .

Given that $f^{2013}(n)=an+b$ We can get $a+b$ by simply substituting $n=1$ in the function.

Thus we have $a+b=f(f(f(...f(1))$ But $f(1)=2*1-1=1$

From here on it can be simply observed that $a+b=1$

a+b=f^2013(1)

f^2013(1)=(f(1))^2013

f(1)=2-1=1

$f^{2013}$ is same as $(2n-1)^{2013}$ .

So the last constant term (-1) is multiplied 2013 times which gives -1. So b = $\boxed{-1}$

The second last term is obtained on muliplying (-1) 2012 times and then with 2n . So the coefficient is $1(\times)\ 2$ = 2; a= $\boxed{2}$

a+b = $\boxed{ 1}$