0
$\begingroup$

Given the recurrence relation $a_n = 2a_{n-1} - a_{n-2}$ how can I find the initial terms if $a_9 = 30$?

For which $x$ are there initial terms which make $a_9 = x$?

I know that there is no solution to the recurrence relation if $a_0 = 1$ and $a_1 = 2$ using the Characteristic Root Technique:

$ x^2 -2x +1 = 0$ which results in $r_1 = 1$ and $r_2 = 1$.

$a_n = ar_1^n + br_2^n$

$a_n = a(1)^n + b(1)^n$

$a_0 = a(1)^0 + b(1)^0$ which results in $1 = a + b$.

$a_1 = a(1)^1 + b(1)^1$ which results in $2 = a + b$.

Obviously $1 != 2$ resulting in no solution.

$\endgroup$
0
$\begingroup$

$$a_9=2a_8-a_7=2(2a_7-a_6)-a_7=3a_7-2a_6=3(2a_6-a_5)-2a_6=4a_6-3a_5=\cdots$$

and in the end

$$a_9=pa_1+qa_0.$$

$$a_9=9a_1-8a_0.$$

This linear equation in two unknowns has an infinity of solutions.

$\endgroup$
0
$\begingroup$

With repeated roots, you're supposed to use $a_n = ar^n+bnr^n$. So if $a_0=1$ and $a_1=2$, then you have

$a_0 = 1a +0b = 1$

$a_1 = 1a + 1b = 2$

So $a=1$, $b = 1$ and $a_n = 1+n$.

You can check this with

$1+n = a_n = 2a_{n-1}-a_{a-2} = 2(1+n-1) - (1+n-2) = 2n-n+1= 1+n$

If you're just given $a_9=30$, that's insufficient to find the initial terms, as you have two unknowns but only one equation.

Also, if you're using MathJax, typing != will result in a space between ! and =. Use \neq instead.

$\endgroup$
  • $\begingroup$ Ahh yes I did use the wrong formula. Thank you for your response! Very helpful as I am preparing for my final. $\endgroup$ – Josh Garza May 8 '18 at 18:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.