Given $y_0\in(0,b)$ and $y_{k+1}=\frac{1}{2}y_k(3-y_k^2)$ converges to $1$. Find $b$.
I know the sequence converges quadratically. But I have no idea how to find $b$.
Asked
Active
Viewed 81 times
1
-
Replace $y_k,y_{k+1}$ in the equation by by the limit, say, $L$. Solve for $L$. Examine then what happens when you start with various values of $y_0$. – uniquesolution Sep 22 '15 at 18:55
-
@uniquesolution $L$ you are referring are the fixed points. But I want the interval of starting values which converges to $1$. Of course you can do it numerically using mathematical software but it is not what I want. – DDaren Sep 22 '15 at 19:01
-
Show me what can you do numerically, please. – uniquesolution Sep 22 '15 at 19:05
-
try $y_0 = 10.$ What happens? – Will Jagy Sep 22 '15 at 19:05
-
and then try $y_0=1/10$. What happens? – uniquesolution Sep 22 '15 at 19:06
-
1@uniquesolution Believe it or not, $b=sqrt(3)$ – DDaren Sep 22 '15 at 19:08
-
So you are saying $y_0$ can be any number in $(0,\sqrt{3})$ and the limit will be $1$??? – uniquesolution Sep 22 '15 at 19:11
-
@uniquesolution yes – DDaren Sep 22 '15 at 19:15
-
What does the sequence converge to when $y_0=2$? It's not $1$. – Marconius Sep 22 '15 at 21:53
-
@Marconius Looks like it does not converge. My question is to find $b$ such that it converges. – DDaren Sep 23 '15 at 10:55
-
If $y_k=-1$ then $y_{k+1}=\frac{1}{2}(-1)(3-(-1)^2)=-1$, so $-1$ is a fixed point. With $g(x)=\frac{1}{2}x(3-x^2)=\frac{3}{2}x-\frac{1}{2}x^3$ then $g'(x)=\frac{3}{2}-\frac{3}{2}x^2 \implies g'(-1)=0 \implies |g'(-1)|<1$ so $x_0=-1$ is an attractive fixed point (just like $1$ is). – Marconius Sep 23 '15 at 11:27
1 Answers
2
Let $f(x) = \frac{1}{2}x(3-x^2)$.
Graphical analysis tells us that $f$ sends $(0,1]$ to $(0,1]$ and all points in $(0,1]$ converge to $1$ under iteration of $f$.
Graphical analysis also tells us that $f$ sends $[1,\sqrt3)$ to $(0,1]$.
Thus, the interval you seek is $(0,\sqrt3)$ because it is the largest interval containing the fixed point $1$ that is mapped into itself (or into $(0,1]$). This interval is called the immediate basin of attraction of the fixed point $1$.
$\qquad$
lhf
- 221,500