# Solving x = a x + b

Problem: Solving $x = ax +b$ for $x$.

Solution:

Choose $x_0$ arbitrarily, we generate the sequence $\{x_i\}$ recursively from $x_i = a x_{i-1} + b, i=1, 2, ...$: $x_1= a x_0 + b,$ $x_2 = a x_1 +b =a(a x_0+b)+b =a^2 x_0 +ab + b,$ $x_3 = a x_2 + b = a (a^2 x_0 + ab +b) +b = a^3 x_0 +a^2b +ab +b,$ $\ddots$ $x_n = a x_{n-1} +b = a^n x_0 + b \sum\limits_{i=0}^{n-1} a^i=a^n x_0 + b\cdot\frac{1-a^{n}}{1-a}.$

It follows that for $|a| <1$, $x = \lim\limits_{n \rightarrow \infty} x_n = \lim\limits_{n\rightarrow \infty}a^n x_0+\frac{b(1-a^n)}{1-a}= \lim\limits_{n\rightarrow \infty}a^n x_0+ \lim\limits_{n \rightarrow \infty}\frac{b(1-a^n)}{1-a}\overset{(\star)}{=}0\cdot x_0 + \frac{b(1-0)}{1-a}.$

i.e., $x = \frac{b}{1-a}.$

Fig. 1 shows a CAS -aided solution using Omega CAS Explorer:

Fig. 1

For $|a|> 1$, we rewrite $x=ax +b$ as $x = \frac{1}{a}x - \frac{b}{a} \overset{A=\frac{1}{a}, B=-\frac{b}{a}}{=} Ax +B.$

Since $|a| >1 \implies \frac{1}{|a|} = |A| <1\implies \lim\limits_{n\rightarrow \infty}A^n = 0$, $x =\lim\limits_{n \rightarrow \infty}A^n x_0 + B\cdot \frac{1-A^n}{1-A}\frac{}{} = \frac{B}{1-A} = \frac{-\frac{b}{a}}{1-\frac{1}{a}} = \frac{-b}{a-1} = \frac{b}{1-a}.$

We have used fact that $|a| < 1 \implies \lim\limits_{n \rightarrow \infty} a^n = 0.\quad\quad\quad(\star)$

A proof is as follows:

Since $|a^n| = |\underbrace{a\cdot a\cdot a ... a}_{n\;a's}| =\overbrace{|a||a|...|a|}^{n\;|a|'s} = |a|^n,$

if $a=0$ then $a^n = 0\cdot 0 ... 0 = 0.$ It means $(\forall \epsilon >0, \forall n \in N^+, |a^n-0|<\epsilon)\implies \lim\limits_{n \rightarrow \infty} a^n = 0.$

Otherwise ( $a \ne 0$), $\forall \epsilon >0, |a^n-0| =|a|^n < \epsilon \implies n \log(|a|)<\log(\epsilon).$

For $|a|<1,$ $n\log(|a|)<\log(\epsilon) \overset{\log(|a|) < 0}{\implies} n > \frac{\log(\epsilon)}{\log(|a|)}.$

And so, $(\forall \epsilon > 0, \exists n^* = \lceil\frac{\log(\epsilon)}{\log(|a|)}\rceil \ni n > n^*, |a^n-0| < \epsilon) \implies \lim\limits_{n \rightarrow \infty}a^n = 0.$

Exercise-1 Prove by mathematical induction: $(x_{k} = a x_{k-1} + b, k=1,2,...,n ) \implies x_n = a^n x_0 + b\sum\limits_{i=0}^{n-1}a^i.$

# “Instrumental Flying”

Consider the following set $S_1 = \{(x,y) | x=\frac{e^y-e^{-y}}{2}\}.\quad\quad\quad(1-1)$

For all $(x, y_1), (x, y_2) \in S_1$, we have $\frac{e^{y_1}-e^{-y_1}}{2}=x, \frac{e^{y_2}-e^{-y_2}}{2}=x.$

That is, $(x, y_1), (x, y_2) \in S_1 \implies \frac{e^{y_1}-e^{-y_1}}{2}- \frac{e^{y_2}-e^{-y_2}}{2}= 0.\quad\quad\quad(1-2)$

Since $\frac{e^{y_1}-e^{-y_1}}{2}- \frac{e^{y_2}-e^{-y_2}}{2}= 0$ if and only if $y_1 = y_2$ (Exercise-1), $\frac{e^{y_1}-e^{-y_1}}{2}-\frac{e^{y_2}-e^{-y_2}}{2}=0 \implies y_1 = y_2.\quad\quad\quad(1-3)$

From (1-2) and (1-3), we conclude: $\forall (x, y_1), (x, y_2) \in S_1 \implies y_1 = y_2.$

i.e., $S_1$ defines a function.

Alternatively, (1-1) can be expressed as $S_1 = \{(x, y) | x = \sinh(y)\}.$

It shows that $S_1$ is the inverse of $\sinh(x)$. Therefore, we name the function defined by (1-1) $\sinh^{-1}$ and write it as $y = \sinh^{-1}(x).$

Let’s look at another set: $S_2 = \{(x,y)|x=\frac{e^y+e^{-y}}{2}\}.\quad\quad\quad(2-1)$

For a pair $(x, y_1>0) \in S_2$, we have $x=\frac{e^{y_1}+e^{-y_1}}{2}.\quad\quad\quad(2-2)$

For another pair $(x, y_2=-y_1)$, $\frac{e^{y_2} + e^{-y_2}}{2} = \frac{e^{-y_1}+e^{-(-y_1)}}{2} = \frac{e^{-y_1} + e^{y_1}}{2} \overset{(2-2)}{=} x\implies (x, y_2=-y_1) \in S_2.$

Since $y_1>0, y_2 = -y_1 \neq y_1, (x, y_1), (x, y_2) \in S_1$ does not implie $y_1 = y_2$. It means $S_2$ does not define a function.

However, modification of $S_2$ gives $S_3 = \{(x, y) | x=\frac{e^y+e^{-y}}{2}, y \ge 0\}.\quad\quad\quad(2-3)$

It defines a function.

Rewrite (2-3) as $S_3 = \{(x, y) | x = \cosh(y), y \ge 0\}.\quad\quad\quad(2-4)$

Then $\forall (x, y_1), (x, y_2) \in S_3$, we have $x=\cosh(y_1), x=\cosh(y_2)\implies \cosh(y_1) = \cosh(y_2).\quad\quad\quad(2-5)$

Notice that by (2-3), $y_1, y_2 \ge 0.$

Cleary, (2-5) is true if and only if $y_1 =y_2$. For if $y_1 \ne y_2$, by $(\star)$, $\cosh(y_1) \ne \cosh(y_2).$

Therefore, $(x, y_1), (x, y_2) \in S_3 \implies y_1 = y_2.$

We name the function defined by (2-3) $\cosh^{-1}$ as (2-4) shows that it is the inverse of $\cosh(x).$

See “Deriving Two Inverse Functions” for the explicit expressions of $\sinh^{-1}$ and $\cosh^{-1}$.

Prove $t_1 \ge 0, t_2 \ge 0, t_1 \ne t_2 \implies \cosh(t_1) \ne \cosh(t_2).\quad\quad\quad(\star)$

Without loss of generality, we assume that $t_2 > t_1.$ By Lagrange’s mean-value theorem (see “A Sprint to FTC“), $\cosh(t_2) -\cos(t_1) =\cosh'(\xi) (t_2-t_1)$

where $\xi \in (t_1, t_2).$

We have $t_2 > t_1 \implies t_2 - t_1 >0$

and $\forall t > 0, (\cosh(t))' = (\frac{e^t+e^{-t}}{2})' = \frac{e^t-e^{-t}}{2} \overset{t>0\implies t >-t \implies e^t >e^{-t}}{>} 0.$

Consequently, $\cosh(t_2) - \cosh(t_1) = \cosh'(\xi) (t_2-t_1) > 0.$

i.e., $\cosh(t_2) \ne \cosh(t_1).$

Exercise-1 Show that $\frac{e^{y_1}-e^{-y_1}}{2}- \frac{e^{y_2}-e^{-y_2}}{2}= 0$ if and only if $y_1 = y_2$.