Sorry in advance as English is not my primary language.

I randomly thought of the following simple problem, and I coudn't solve it after one one hour trying. Maybe you guys can help.

Let $a_1$ and $a_2$ be positive real numbers. Let $a_n$ be the arithmetic average of the previous 2 numbers, i.e.:

$$a_n = \frac{a_{n-1}+a_{n-2}}{2}$$

If I draw this as points on a paper, I can obviously see that the limit of $a_n$ as $n\rightarrow\infty$ is a function of $a_1$ and $a_2$, but I can't solve it.

How do I proceed?

  • $\begingroup$ Did you solve the recurrence equation ? If you did, did you apply the conditions ? By the way, welcome to the site ! $\endgroup$ – Claude Leibovici Feb 8 at 11:42
  • $\begingroup$ You can try with, for example $a_0 = 0$ and $a_1 =100$ and plot it! It gives an interesting graphical solution ... $\endgroup$ – Matti P. Feb 8 at 11:42
  • $\begingroup$ Take a look at Solving homogeneous linear recurrence relations with constant coefficients. $\endgroup$ – robjohn Feb 8 at 12:05
  • $\begingroup$ @ClaudeLeibovici, Thank you! I didn't even know this was a "recurrence equation", I'm not a mathematician, just an Engineer. $\endgroup$ – Anderson Linhares Feb 8 at 15:33
  • $\begingroup$ @MattiP., You are right, that was helpful. I fount out that the limit as $n\rightarrow\infty$ is $\frac{a_1+2a_2}{3}$. However, I still can't find the solution for a general $a_n$. $\endgroup$ – Anderson Linhares Feb 8 at 15:33

If you rewrite the equation as

  • $2a_n -a_{n-1} - a_{n-2} = 0$

you obtain a so called homogeneous linear difference equation.

This type of equation can be solved.

A possible method is to check what happens if you plug in a "guessed" solution of the form $a_n = c\cdot \lambda^n$ where $c$ is a real constant.

You will find that the solution can be written as $$a_n = c_1\cdot 1^n + c_2\cdot \left(-\frac{1}{2} \right)^n = c_1 + c_2\cdot \left(-\frac{1}{2} \right)^n$$

The $1$ and $-\frac{1}{2}$ come from solving the quadratic equation $2\lambda^2 - \lambda - 1 = 0$ you will come across when you carry out the suggested approach $a_n = c\cdot \lambda^n$.

  • $\begingroup$ Thank you for this insight. This is not the solution though, right? Experimenting with the values, I'm certain that the limit as $n\rightarrow\infty$ is $\frac{a_1+2a_2}{3}$. I don't know what solution to "guess" to get an expression for $a_n$ though. $\endgroup$ – Anderson Linhares Feb 8 at 15:25
  • $\begingroup$ it is. it iis. you may use your two starting values to determine the constants $c_1$ and $c_2$. i thought to leave this last step for you so you can play around with the general formula. $\endgroup$ – trancelocation Feb 8 at 15:29

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