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 number of times and your resulting number is bigger than , so we may write it as with
Then the next number:
So if we start with a number greater than the next number is still greater than but by only as much so the sequence of results is decreasing towards
Now if you repeat this starting with a value less than you will find the sequence increases towards with ever decreasing differesnce from .
RonL
Hello, badi6!
Here's another approach . . . but it's much looonger!
We have a recurrence relation: . .[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: . .[2]
Subtract [2] - [1]: .
Multiply by 3: .
Let: . . . We have: .
Divide by
Factor: .
Hence, the desired function, , is of the form:
. . .[3]
Let the first term be
Then the second term is: .
From the first two terms of the sequence, we have:
. .
Solve the system: .
Substitute into [3]: .
Therefore: .
The best way to solve such relations is difference equations as Soroban shown.
Let be the starting value (the initial condition), then we obtain the following relation
,
which has the following unique solution
If you are asked what you will get by repeating this recustion infinitely many times, you have to check .
That is,
.........
.........
.........
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 and to obtain a second-order homogeneous equation, of which solutions are of the type .
You just substitute and solve and and write , where 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 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.