Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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)

share|improve this question

2 Answers 2

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 $n\ge1$.

share|improve this answer

For part 2):

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

share|improve this answer
    
it is correct now. –  ashley Dec 15 '12 at 11:25

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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