Finding the iterates

Question

Let $0 < r < 1$ and $a > 0$ and consider the mapping $f : R → R$ given by $f(x) = rx + a.$

Find, in terms of $a$ and $r$, the ﬁrst four iterates of $x_0 = 0$ under $f$

How would I go about this? I know that to find the iterates, I need to apply the function again, but I am not too sure what this would look like in this equation.

So far, I have that:
$X_0 = 0$
$X_1 = f(0) = r(o) + a = a$
$X_2 = f(a) = r(a) + a = ra + a$
$X_3 = f(ra + a) = r(ra + a) = r^2a + ra + a$
$X_4 = f(r^2 +ra + a) = r(r^2a + ra + a) + a = r^3a +r^2a + ra + a$

But I don't think this looks correct...

I'm not too sure if this is what you are talking about. But 0 < r < 1 and a > 0 — Roy Sheehan, Feb 03 '14 at 09:53
yes but you have only one condition,if we are talking about system,then you need two equation and two variable — dato datuashvili, Feb 03 '14 at 09:57
Oh, I think I might understand now... The equations are y = f(x) and y = x — Roy Sheehan, Feb 03 '14 at 10:00
if you want that instead of $x_1$,$x_2$,...$x_n$,put $a$ and $r$,then no different,your approaches look correct — dato datuashvili, Feb 03 '14 at 10:03

gammatester · Answer 1 · 2014-02-03T11:36:15.020

2

$$x_0 = 0\\ x_1 = r x_0+a = a\\ x_2 = r x_1+a = r a + a\\ x_3 = r x_2+a = r (r a + a) + a = r^2 a + r a + a\\ x_4 = r x_3+a = r (r (r a + a) + a) + a = r^3 a + r^2 a + r a + a$$

Can you continue?

Show with induction: $x_n=a\sum_{k=0}^{n-1}r^k=a\frac{1-r^n}{1-r}$

edited Feb 03 '14 at 11:36

answered Feb 03 '14 at 10:07

gammatester

18,827

but there is one issues,he has not mention that $f(x_i)=x_{i-1}$,instead of $x_i$,use $f(x_i)$ – dato datuashvili Feb 03 '14 at 10:10
@dato datuashvili: That is the standard meaning of iterates – gammatester Feb 03 '14 at 10:11
ok but my name is $dato$ :) – dato datuashvili Feb 03 '14 at 10:11

dato datuashvili · Answer 2 · 2014-02-03T09:56:12.790

1

if we have

$x_0=0$

then $f(x_0)=r*x_0+a$

or $f(x_0)=a$

now what is problem?ok let us try some more iteration

$f(x_1)=r*x_1+a=r*x_1+f(x_0)$

$f(x_2)=r*x_2+f(x_0)$

and so on,but about $r$,how can we determine?if we want to compute $r$,then we have following equation

$f(x_1)=r*x_1+f(x_0)$

$f(x_2)=r*x_2+f(x_0)$

now subtract

$f(x_1)-f(x_2)=r(x_1-x_2)$

and

$r=(f(x_1)-f(x_2))/(x_1-x_2)$

edited Feb 03 '14 at 09:56

answered Feb 03 '14 at 09:46

dato datuashvili

9,194

Finding the iterates

2 Answers2