0
$\begingroup$

I am looking for the exponential function that describes the following behavior:

Start at the value 1. The next value of the sequence is calculated by the following formula:

Value of Previous + Value of Previous * (1/3)

The idea is that this is experience in a video game, where each level is harder to obtain than the last. Meaning the sequence would be:

1, 4/3, 16/9, 64/27 ...

After starring at this for so long, I couldn't think the equation to do it. It obviously needs to be an exponential function but I thought for sure I'd need to write 1/3 somewhere in the correct form.

The actual answer to this is (3/4)^(1-n) but I do not understand how I could have came to that conclusion if I didn't use Wolfram Alpha's Pattern Finder.

How can this be solved? I have to approach these kind of patterns all the time in my programming works and I rely on the Wolfram tool far too often.

  • 0
    Where did you take the word "holonomic" from? What you want is "recursive" - "holonomic" means a totally unrelated thing. – 2017-01-31
  • 1
    @AlexM. when you use Wolfram Alpha's Pattern calculator, if it cannot find a function that matches your series then it says "No holonomic function exists that matches your data". I assumed this means that I am looking for a holonomic function. – 2017-02-01

2 Answers 2

1

Let $x_{n-1}$ be the previous number. Then,

$$x_n=x_{n-1}+\frac13x_{n-1}=\frac43x_{n-1}$$

We then know this is in geometric progression, so $x_n=x_0\times\left(\frac43\right)^n$.

If you did not see this, my first attempt would be to write out the first few terms to see if you can spot a pattern. Once you think you have the right formula, you can use induction to verify your results.

Or, since you knew the result should be an exponential function, you could try substitution:

$$x_n=b^n\implies b^n=\frac43b^{n-1}$$

Divide both sides by $b^{n-1}$, you will find that $b=\frac43$.

  • 0
    Also note that $$(3/4)^{1-n}=x_0(4/3)^n$$where $x_0=3/4$. – 2017-01-31
0

Denote the $n^{th}$ term of the sequence as $a_n$.

Notice $$a_n=a_{n-1}+\frac13 a_{n-1}=(\frac43)^1 a_{n-1}=(\frac43)^2a_{n-2}=\cdots=(\frac43)^{n-1}a_1$$

Given $a_1=1$ and recall that $\displaystyle a^{-b}=(\frac1a)^b$, we have $$a_n=(\frac43)^{n-1}(1)=(\frac34)^{1-n}$$