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I am trying to solve this example:

The following recursive function defines a linear affine difference equation $$x(n+1) = 1.4*x(n) + 0.2$$ $$x(0) = -1$$

  1. Find the first three values of the iteration?
  2. Which initial value y(0) would cause the iteration to remain constant?

Can you give me a hint how to calculate this quickly?(For example by using wolfram alpha)

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up vote 1 down vote accepted

Suppose that $1.4^ny(n)=x(n)$, then the equation becomes $$ 1.4^{n+1}y(n+1)=1.4^{n+1}y(n)+0.2\tag{1} $$ Then you get $$ y(n+1)=y(n)+0.2(1.4)^{-n-1}\tag{2} $$ which becomes $$ y(n)=0.2\sum_{k=1}^n1.4^{-k}+y(0)\tag{3} $$ We can use the standard sum of a geometric series to get $$ \begin{align} y(n) &=0.2\frac{1.4^{-1}-1.4^{-n-1}}{1-1.4^{-1}}+y(0)\\ &=0.2\frac{1-1.4^{-n}}{0.4}+y(0)\\ &=\frac12(1-1.4^{-n})+y(0)\tag{4} \end{align} $$ Therefore, we get $$ \begin{align} x(n) &=\frac12(1.4^n-1)+1.4^nx(0)\\ &=1.4^n\left(\frac12+x(0)\right)-\frac12\tag{5} \end{align} $$

  1. We can easily compute the first few terms using $(5)$ and $x(0)=-1$.

  2. If $x(0)=-\frac12$, then we get from $(5)$ that $x(n)=-\frac12$ for all $n\ge1$.

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For part 2):

If $x(0)=x(1)$ then $x(0)=x(1)=1.4x(0)+0.2$, thus $x(0)=\frac{-1}{2}$

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it is correct now. – ashley Dec 15 '12 at 11:25

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