# X_(n+1)-aX_n+2=0 Which are the set values of "a" for the string "Xn" is descending?

## ${X}_{n + 1} - a {X}_{n} + 2 = 0$

Mar 18, 2017

See below.

#### Explanation:

${X}_{n + 1} - a {X}_{n} + 2 = 0$ Is a linear non homogeneous difference equation.

It's solution can be composed of a solution for the homogeneous

${X}_{n}^{h}$ such that ${X}_{n + 1}^{h} - a {X}_{n}^{h} = 0$

plus a particular solution ${X}_{n}^{p}$ for the non homogeneous equation such that

${X}_{n + 1}^{p} - a {X}_{n}^{p} + 2 = 0$

For the homogeneous solution the proposal is

${X}_{n}^{h} = C {a}^{n}$. Substituting

$C {a}^{n + 1} - a C {a}^{n} = C {a}^{n + 1} - C {a}^{n + 1} = 0$

Now for the particular we propose

${X}_{n}^{p} = {C}_{n} {a}^{n}$ then

${C}_{n + 1} {a}^{n + 1} - a {C}_{n} {a}^{n} = {a}^{n + 1} \left({C}_{n + 1} - {C}_{n}\right) = 2$

so

${C}_{n + 1} - {C}_{n} = \frac{2}{a} ^ \left(n + 1\right)$

then beginning with ${C}_{0}$ we have

${C}_{1} = {C}_{0} + \frac{2}{a}$
${C}_{2} = {C}_{1} + \frac{2}{a} ^ 2 = {C}_{0} + \frac{2}{a} + \frac{2}{a} ^ 2$
$\ldots$

and then

${C}_{n} = {C}_{0} + 2 {\sum}_{k = 1}^{n} {a}^{- k}$ and finally

${X}_{n} = {X}_{n}^{h} + {X}_{n}^{p} = \left({C}_{0} + 2 {\sum}_{k = 1}^{n} {a}^{- k}\right) {a}^{n} = {C}_{0} {a}^{n} + 2 {\sum}_{k = 1}^{n} {a}^{k}$

If we need that ${X}_{n} < {X}_{n - 1}$ then

${C}_{0} {a}^{n} + 2 {\sum}_{k = 1}^{n} {a}^{k} < {C}_{0} {a}^{n - 1} + 2 {\sum}_{k = 1}^{n - 1} {a}^{k}$

or

${C}_{0} {a}^{n} + 2 {a}^{n} < {C}_{0} {a}^{n - 1}$ or

$a < {C}_{0} / \left({C}_{0} + 2\right)$

Here ${C}_{0}$ is the initial condition.