Ryan's function composition

It is matter of simple algebraic manipulation to find that

$f(x) = \frac{x - 1}{x}, f^2(x) = \frac{1}{1 - x},f^3(x) = x$

$g(x) = \frac{3x + 1}{3 - 9x}, g^2(x) = -\frac{1}{9x}, g^3(x) = \frac{3x - 1}{9x + 3}, g^4(x) = x$

Now for any $k \in \mathbb{N}$ and $a \in \{0,1,2\}$ we get $f^{3k + a}(x) = f^a\left(f^{3k}(x)\right) = f^a(x)$ which follows by simple induction (or intuition). Similarly for any $k \in \mathbb{N}$ and $a \in \{0,1,2,3\}$ we get $f^{4k + a}(x) = f^a\left(f^{4k}(x)\right) = f^a(x)$ Now note that $\frac{2520}{3} = 840$ and $\frac{2520}{4} = 630$ . Hence $$\sum_{k = 1}^{2520} \Big(f^k(2) - g^k(2)\Big) = 840\left(f(2) + f^2(2) + 2\right) - 630\left(g(2) + g^2(2) + g^3(2) + 2\right)$$ Simply computing each of $f(2), f^2(2), g(2), g^2(2), g^3(2)$ we get $\begin{aligned} \sum_{k = 1}^{2520} \Big(f^k(2) - g^k(2)\Big) &= 840\left(\frac{1}{2} - 1 + 2\right) - 630\left(\frac{-7}{15} + \frac{-1}{18} + \frac{5}{21} + 2 \right) \\ &= \boxed{179} \end{aligned}$

$f(x) = \frac {x-1}{x} = 1 - \frac {1}{x}$

$f^2 (x) = 1 - \frac {1}{1 - \frac {1}{x}} = 1 - \frac {x}{x-1} = - \frac {1}{x-1}$

$f^3 (x) = 1 - \frac {1}{- \frac {1}{x-1} } = 1 + x - 1 = x$

$f^4 (x) = x$

Since $f^4 (x) = f(x)$ , $f^n (x)$ has a period of $3$ , that is $f^n (x) = f^{n \pmod 3 } (x)$

$f(2) = f^4 (x) = f^7 (2) = \ldots = f^{2518} (2) = \frac {1}{2}$ , total of 840 of them

$f^2 (2) = f^5 (2) = f^8 (2) = \ldots = f^{2519} (2) = -1$ , total of 840 of them

$f^3(2) = f^6 (2) = f^9 (2) = \ldots = f^{2520} (2) = 2$ , total of 840 of them

$\Rightarrow \displaystyle \sum_{k=1}^{2520} f^k (2) = 840(\frac {1}{2} - 1 + 2) = 1260$

Similarly, $g^2(x) = \frac {3 (\frac {3x+1}{3-9x}) + 1 }{3 - 9 (\frac {3x+1}{3-9x}) } = - \frac {1}{9x}$

$g^3(x) = \frac {3 ( - \frac {1}{9x} ) + 1 }{3 - 9 ( - \frac {1}{9x} ) } = \frac {-3+9x}{27x+9}$

$g^4 (x) = g^2 (g^2 (x)) = - \frac {1}{9 (- \frac {1}{9x} ) } = x$

$g^5 (x) = x = g(x)$

$g(2) = g^5 (2) = g^9 (2) = \ldots = g^{2517} (2) = - \frac {7}{15}$ , total of 630 of them

$g^2(2) = g^6 (2) = g^{10} (2) = \ldots = g^{2518} (2) = - \frac {1}{18}$ , total of 630 of them

$g^3(2) = g^7 (2) = g^{11} (2) = \ldots = g^{2519} (2) = \frac {5}{21}$ , total of 630 of them

$g^4(2) = g^8 (2) = g^{12} (2) = \ldots = g^{2520} (2) = 2$ , total of 630 of them

$\Rightarrow \displaystyle \sum_{k=1}^{2520} g^k (2) = 630(- \frac {7}{15} - \frac {1}{18} + \frac {5}{21} + 2) = 1081$

Answer is $1260 - 1081 = \boxed{179}$

It is easy to check these functions for periodicity when evaluated at $2$ , and we find that they are. Namely,

$f(2) = \frac {1}{2}, f^2(2) = -1, f^3(2) = 2, f^4(2) = \frac{1}{2} \cdots$

$g(2) = -\frac {7}{15}, g^2(2) = -\frac {1}{18}, g^3(2) = \frac {5}{21}, g^4(2) = 2, g^5(2) = -\frac{7}{15} \cdots$

Thus, $\sum_{k = 1}^{2520} f^k(2) = (\frac{2520}{3})(\frac{1}{2} - 1 + 2) = 1260$ and $\sum_{k = 1}^{2520} g^k(2) = (\frac{2520}{4})(-\frac{7}{15} - \frac{1}{18} + \frac {5}{21} + 2) = 1081$

So the desired answer is $1260 - 1081 \rightarrow 179$ .

If we calculate $f(2)$ , we get that:

$f(2) = \frac{2-1}2 = \frac12$ $f^2(2) = \frac{\frac12 - 1}{\frac12} = -1$ $f^3(2) = \frac{-1-1}{-1} = 2$

Therefore, there is a repeating pattern of $2, 0.5, -1$ . We can see that $\frac{2520}{3} = 840$ , so the value of:

$\displaystyle \sum_{k=1}^{2520}{f^k(2)} = 840 \cdot (2 + 0.5 -1) = 1260$

Following the same process with $g(2)$ , we can see another repeating pattern, this time with the numbers $2, -\frac7{15}, -\frac1{18}, \frac5{21}$ . We can see that $\frac{2520}4 = 630$ so the value of:

$\displaystyle \sum_{k=1}^{2520}{g^k(2)} = 630 \cdot (2-\frac7{15}-\frac1{18}+\frac5{21}) = 1081$ .

Therefore, we can see that the value of:

$\displaystyle \sum_{k=1}^{2520}{(f^k(2)-g^k(2))} = 1260 - 1081 = \fbox{179}$

By inspection, it eventually becomes clear that $f^n(x)=\begin{cases} \frac{x-1}{x}, & n \equiv 1 \pmod 3 \\ \frac{1}{1-x}, & n \equiv 2 \pmod 3 \\ x, & n \equiv 0 \pmod 3 \end{cases}$ and $g^n(x)=\begin{cases} \frac{3x+1}{3-9x}, & n \equiv 1 \pmod 4 \\ \frac{-1}{9x}, & n \equiv 2 \pmod 4 \\ \frac{3x-1}{9x+3}, & n \equiv 3 \pmod 4 \\ x, & n \equiv 0 \pmod 4 \\ \end{cases}$ Breaking up the given sum, we have $\left ( \sum_{k=1}^{2520} {f^k(2)} \right ) - \left ( \sum_{k=1}^{2520} {g^k(2)} \right )$ $(\frac{2520}{3})(\frac{1}{2}+\frac{-1}{2}+2)-(\frac{2520}{4})(\frac{-7}{15}+\frac{-1}{18}+\frac{5}{21}+2)$ The above simplifies to $\boxed{179}$ .

We can solve the problem by looking at the repeating patterns of the functions and its compositions. f(x)= (x-1)/x, f^2 (x) = -1/(x-1), f^3 (x) = x, f^4 (x) = f(x), f^5 (x) = f^2 (x),... etc.It is repeating after 3 steps.So f(2) = 1/2, f^2 (2) = -1, f^3(2) = 2.So the total sum will be 3/2 (2520/3) = 1260. For the function of g(x) and its compositions, it is repeating after 4 steps, namely, g(x) = (3x+1)/(3-9x), g^2 (x) = -1/(9x), g^3(x) = (3x-1)/(3+9x), and g^4(x) = x,...etc.So total sum of g(x) and its compositions evaluated at x=2, are (-7/15 + (-1/18) + (5/21) + 2) (2520/4) = 1081 So Their total difference is 1260 - 1081 = 179

$f(x)={x-1 \over x}, f^2(x) = {-1 \over x-1 }, f^3(x) = x$ .

We also have:

$g(x) = -{1 \over 3}{3x+1 \over 3x - 1}, g^2(x) = -{1 \over 3}{1 \over 3x}, g^3(x) = {1 \over 3}{3x-1 \over 3x + 1}, g^4(x) = x$

So, The required sum is:

${2520 \over 3}\left(f(x) + f^2(x) + f^3(x)\right) - {2520 \over 4}\left(g(x) + g^2(x) + g^3(x) + g^4(x)\right)$

Substituting $x=2$ in the above, it results in $179$ .

A calculation that is slightly too tedious to write down here tells us that $f^4(x)=\frac{1}{x}$ and that $g^4(x)=x$ . This tells us that $g$ repeats itself every four terms, i.e. $g^5=g, g^6=g^2$ , etc. We may also see that after the third iteration, $f$ becomes $3$ -repetitive and concerning the outcome when we fill in $2$ , it even is 3-repetitive from the start. As a result, we may split the sum to give $\frac{2520}{3}\cdot(\text{sum of each three succesive terms of }f)-$

$\frac{2520}{4}\cdot(\text{sum of each four succesive terms of }g) =$

$\frac{2520}{3}\cdot(-1+2+\frac{1}{2}) - \frac{2520}{4}\cdot(2+\frac{5}{21}-\frac{7}{15}-\frac{1}{18}) =$

$\frac{2520}{3}\cdot\frac{3}{2}-\frac{2520}{4}\cdot\frac{1260+150-294-35}{630} =$

$1260 - 1260 -150 + 294 + 35 = 179.$