# Math Help - recursive sequence ratio

1. ## recursive sequence ratio

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?

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?
Well have you done what the question asks?

Did you notice anything about then squences of numbers you found??

RonL

3. Originally Posted by CaptainBlack
Well have you done what the question asks?

Did you notice anything about then squences of numbers you found??

RonL

yes i notice that what ever number you stare with you end up with 7.5, but i dont know why?

would you please tell me why?

Thank you very much

yes i notice that what ever number you stare with you end up with 7.5, but i dont know why?

would you please tell me why?

Thank you very much
Suppose you have done this $n$ number of times and your resulting number is bigger than $7.5$, so we may write it as $x_n=7.5+\varepsilon,$ with $\varepsilon>0$

Then the next number:

$x_{n+1}=\frac{7.5+\varepsilon}{3}+5=2.5 + \frac{\varepsilon}{3}+5=7.5 + \frac{\varepsilon}{3}$

So if we start with a number greater than $7.5$ the next number is still greater than $7.5$ but by only $1/3$ as much so the sequence of results is decreasing towards $7.5$

Now if you repeat this starting with a value less than $7.5$ you will find the sequence increases towards $7.5$ with ever decreasing differesnce from $7.5$.

RonL

Here's another approach . . . but it's much looonger!

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?
We have a recurrence relation: . $a_n \;=\;\frac{1}{3}a_{n-1} + 5$ .[1]

Write the "next" term in the sequence: . $a_{n+1} \;=\;\frac{1}{3}a_n + 5$ .[2]

Subtract [2] - [1]: . $a_{n+1} - a_n \;=\;\frac{1}{3}a_n - \frac{1}{3}a_{n-1} \quad\Rightarrow\quad a_{n+1} - \frac{4}{3}a_n + \frac{1}{3}a_{n-1} \;=\;0$

Multiply by 3: . $3a_{n+1} - 4a_n + a_{n-1} \;=\;0$

Let: $X^n \,=\,a_n$ . . . We have: . $3X^{n+1} - 4X^n + X^{n-1} \;=\;0$

Divide by $X^{n-1}\!:\quad 3X^2 - 4X + 1 \;=\;0$

Factor: . $(X - 1)(3X-1) \:=\:0 \quad\Rightarrow\quad X \;=\;1,\:\frac{1}{3}$

Hence, the desired function, $f(n)$, is of the form:
. . $f(n) \;=\;A(1^n) + B\left(\frac{1}{3}\right)^n \quad\Rightarrow\quad f(n) \;=\;A + \frac{B}{3^n}$ .[3]

Let the first term be $a$
Then the second term is: . $\frac{1}{3}a + 5$

From the first two terms of the sequence, we have:

. . $\begin{array}{ccccc}
n=1\!: & A + \frac{1}{3}B &=& a \\ \\[-4mm] n=2\!: & A + \frac{1}{9}B &=& \frac{1}{3}a + 5 \end{array}$

Solve the system: . $A \:=\:\frac{15}{2},\;\;B \:=\:3a-\frac{45}{2}$

Substitute into [3]: . $f(n) \;=\;\frac{15}{2} + \frac{3a-\frac{45}{2}}{3^n}$

Therefore: . $\lim_{n\to\infty} f(n) \;\;=\;\; \lim_{n\to\infty}\left[\frac{15}{2} + \frac{3a -\frac{45}{2}}{3^n}\right] \;\;=\;\;\frac{15}{2} + 0 \;\;=\;\;\boxed{\frac{15}{2}}$

6. The best way to solve such relations is difference equations as Soroban shown.

Let $x(0)$ be the starting value (the initial condition), then we obtain the following relation
$x(n)=\frac{1}{3}x(n-1)+5\text{ for }n\in\mathbb{N}$,
which has the following unique solution
$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 $x(\infty)$.
That is,
$x(\infty)=\bigg(\frac{1}{3}\bigg)^{\infty}x(0)+5\s um\limits_{j=0}^{\infty}\bigg(\frac{1}{3}\bigg)^{k }$
......... $=0+5\frac{1}{1+\frac{1}{3}}$
......... $=5*\frac{3}{2}$
......... $=\frac{15}{2}.$
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 $5$ and to obtain a second-order homogeneous equation, of which solutions are of the type $\lambda^{n}$.
You just substitute $x(n)=\lambda^{n}$ and solve $\lambda_{1}$ and $\lambda_{2}$ and write $x(n)=c_{1}\lambda_{1}^{n}+c_{2}\lambda_{2}^{n}$, where $c_{1},c_{2}$ 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 $5$ 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.