if you choose a starting number and divide it by 3 and add 5, divide the result by 3 and add 5. keep repeating the process. what do you notice? and why?
Suppose you have done this $\displaystyle n$ number of times and your resulting number is bigger than $\displaystyle 7.5$, so we may write it as $\displaystyle x_n=7.5+\varepsilon,$ with $\displaystyle \varepsilon>0$
Then the next number:
$\displaystyle x_{n+1}=\frac{7.5+\varepsilon}{3}+5=2.5 + \frac{\varepsilon}{3}+5=7.5 + \frac{\varepsilon}{3}$
So if we start with a number greater than $\displaystyle 7.5$ the next number is still greater than $\displaystyle 7.5$ but by only $\displaystyle 1/3$ as much so the sequence of results is decreasing towards $\displaystyle 7.5$
Now if you repeat this starting with a value less than $\displaystyle 7.5$ you will find the sequence increases towards $\displaystyle 7.5$ with ever decreasing differesnce from $\displaystyle 7.5$.
RonL
Hello, badi6!
Here's another approach . . . but it's much looonger!
We have a recurrence relation: .$\displaystyle a_n \;=\;\frac{1}{3}a_{n-1} + 5$ .[1]If you choose a starting number and divide it by 3 and add 5,
divide the result by 3 and add 5. keep repeating the process,
what do you notice? and why?
Write the "next" term in the sequence: .$\displaystyle a_{n+1} \;=\;\frac{1}{3}a_n + 5$ .[2]
Subtract [2] - [1]: .$\displaystyle a_{n+1} - a_n \;=\;\frac{1}{3}a_n - \frac{1}{3}a_{n-1} \quad\Rightarrow\quad a_{n+1} - \frac{4}{3}a_n + \frac{1}{3}a_{n-1} \;=\;0 $
Multiply by 3: .$\displaystyle 3a_{n+1} - 4a_n + a_{n-1} \;=\;0$
Let: $\displaystyle X^n \,=\,a_n$ . . . We have: .$\displaystyle 3X^{n+1} - 4X^n + X^{n-1} \;=\;0$
Divide by $\displaystyle X^{n-1}\!:\quad 3X^2 - 4X + 1 \;=\;0$
Factor: .$\displaystyle (X - 1)(3X-1) \:=\:0 \quad\Rightarrow\quad X \;=\;1,\:\frac{1}{3}$
Hence, the desired function, $\displaystyle f(n)$, is of the form:
. . $\displaystyle f(n) \;=\;A(1^n) + B\left(\frac{1}{3}\right)^n \quad\Rightarrow\quad f(n) \;=\;A + \frac{B}{3^n} $ .[3]
Let the first term be $\displaystyle a$
Then the second term is: .$\displaystyle \frac{1}{3}a + 5$
From the first two terms of the sequence, we have:
. . $\displaystyle \begin{array}{ccccc}
n=1\!: & A + \frac{1}{3}B &=& a \\ \\[-4mm] n=2\!: & A + \frac{1}{9}B &=& \frac{1}{3}a + 5 \end{array}$
Solve the system: .$\displaystyle A \:=\:\frac{15}{2},\;\;B \:=\:3a-\frac{45}{2}$
Substitute into [3]: .$\displaystyle f(n) \;=\;\frac{15}{2} + \frac{3a-\frac{45}{2}}{3^n} $
Therefore: .$\displaystyle \lim_{n\to\infty} f(n) \;\;=\;\; \lim_{n\to\infty}\left[\frac{15}{2} + \frac{3a -\frac{45}{2}}{3^n}\right] \;\;=\;\;\frac{15}{2} + 0 \;\;=\;\;\boxed{\frac{15}{2}} $
The best way to solve such relations is difference equations as Soroban shown.
Let $\displaystyle x(0)$ be the starting value (the initial condition), then we obtain the following relation
$\displaystyle x(n)=\frac{1}{3}x(n-1)+5\text{ for }n\in\mathbb{N}$,
which has the following unique solution
$\displaystyle x(n)=\bigg(\frac{1}{3}\bigg)^{n}x(0)+5\sum\limits_ {j=0}^{n-1}\bigg(\frac{1}{3}\bigg)^{k}\text{ for all }n\in\mathbb{N}.
$
If you are asked what you will get by repeating this recustion infinitely many times, you have to check $\displaystyle x(\infty)$.
That is,
$\displaystyle x(\infty)=\bigg(\frac{1}{3}\bigg)^{\infty}x(0)+5\s um\limits_{j=0}^{\infty}\bigg(\frac{1}{3}\bigg)^{k }$
.........$\displaystyle =0+5\frac{1}{1+\frac{1}{3}}$
.........$\displaystyle =5*\frac{3}{2}$
.........$\displaystyle =\frac{15}{2}.$
Soroban just illustrated how it can be done if you know nothing about the solutions of difference equations.
Please see the following tutorial, which may be very useful
Math tutorial: first-order difference equations
Note. Soroban just iterated the recursion twice to get rid of the forcing term $\displaystyle 5$ and to obtain a second-order homogeneous equation, of which solutions are of the type $\displaystyle \lambda^{n}$.
You just substitute $\displaystyle x(n)=\lambda^{n}$ and solve $\displaystyle \lambda_{1}$ and $\displaystyle \lambda_{2}$ and write $\displaystyle x(n)=c_{1}\lambda_{1}^{n}+c_{2}\lambda_{2}^{n}$, where $\displaystyle c_{1},c_{2}$ are arbitrary constants, which can be determinated when the initial conditions are given.
However, I have to point out that this is not always possible; for instance, when the forcing term is not a constant (it is $\displaystyle 5$ for this equation).
Moreover, higher-order difference equations with variable coefficients are always solveble (even for second-order).
This is why studying the qualitative (oscillation and stability) behaviour of such equations is important since these equations appear in many fields of real life modelings.