Always bringing me back

Solution 1: According to the definition, $X_1 = \mathcal{L}(X_0) = aX_0 + b$

$X_2 = \mathcal{L}(X_1) = aX_1 + b = a(aX_0 + b) + b = a^2X_0 + b(1 + a)$

$X_3 = \mathcal{L}(X_2) = aX_2 + b = a(aX_1 + b) + b = a^3X_0 + b(1 + a + a^2)$

Observation: In general, we can write $X_n$ as

$X_n = a^n X_0 + b \sum _{i = 0} ^{n - 1} a^{i}$

Now $X_{\infty} = \displaystyle \lim _{n \to \infty} X_n = \displaystyle \lim _{n \to \infty} \left(a^n X_0 + b \sum _{i = 0} ^{n - 1} a^{i}\right)$

Since $|a| < 1, \displaystyle \lim_{n \to \infty} a ^n = 0$ . Also, $\displaystyle \lim_{n \to \infty} b \sum _{i = 0} ^{n - 1} a^{i} = b \times \frac{1}{1 - a}$

Therefore, $X_{\infty} = \frac{b}{1- a} = \frac{5}{1 - \frac{4}{7} } = \boxed{\frac{35}{3}}$ $_\square$

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Solution 2: Draw graphs of $L_1 : y = \mathcal{L}(x)$ and $L_2 : y = \mathcal{L}^{-1} (x)$

Start with the point $(X_0, \mathcal{L}(X_0))$ , i.e. $(X_0, X_1)$ . If we take a reflection of this point off the $y = x$ line, we obtain the point $( X_1, X_0)$ . Then we move parallel to the $y$ axis until we reach the line $y = \mathcal{L}(x)$ . We will meet at the coordinates $(X_1, X_2)$ . We then move to $(X_2, X_1)$ and then $(X_2, X_3)$ . This is shown in the graph below.

graph

If we continue this process infinitely many times, we can show that we will be converging to a point. Because after every iteration, the area of the triangle formed by the lines $L_1, L_2$ and line joining points $(X_n, X_{n+1})$ and $(X_{n+1}, X_{n})$ decreases and the triangle converges to a point. We then reach the point $(X_{\infty}, X_{\infty})$ . At this point, the $x$ coordinate is equal to the $y$ coordinate is equal to $X_{\infty}$ , since for very large $n$ , $X_{n} = X_{n+1}$ .

We substitute $x = y = X_{\infty}$ in $y = ax + b$ .

$X_{\infty} = aX_{\infty} + b$ . Therefore $X_{\infty} = \frac{b}{1-a}$ . $_\square$ .