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# 数学代写|数值分析代写Numerical analysis代考|MATH/CS514 The bisection method

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## 数学代写数值分析代写Numerical analysis代考|The bisection method

Suppose that $f$ is a real-valued function defined and continuous on a bounded closed interval $[a, b]$ of the real line and such that $f(\xi)=0$ for some $\xi \in[a, b]$. A very simple iterative method for the solution of the nonlinear equation $f(x)=0$ can be constructed by beginning with an interval $\left[a_0, b_0\right]$ which is known to contain the required solution $\xi$ (e.g., one may choose $\left[a_0, b_0\right]$ as the interval $[a, b]$ itself, with $a_0=a$ and $b_0=b$ ), and successively halving its size.

More precisely, we proceed as follows. Let $k \geq 0$, and suppose that it is known that $f\left(a_k\right)$ and $f\left(b_k\right)$ have opposite signs; we then conclude from Theorem $1.1$ that the interval $\left(a_k, b_k\right)$ contains a solution of $f(x)=0$. Consider the midpoint $c_k$ of the interval $\left(a_k, b_k\right)$ defined by
$$c_k=\frac{1}{2}\left(a_k+b_k\right),$$
and evaluate $f\left(c_k\right)$. If $f\left(c_k\right)$ is zero, then we have located a solution $\xi$ of $f(x)=0$, and the iteration stops. Else, we define the new interval $\left(a_{k+1}, b_{k+1}\right)$ by
$$\left(a_{k+1}, b_{k+1}\right)= \begin{cases}\left(a_k, c_k\right) & \text { if } f\left(c_k\right) f\left(b_k\right)>0, \ \left(c_k, b_k\right) & \text { if } f\left(c_k\right) f\left(b_k\right)<0,\end{cases}$$
and repeat this procedure.

## 数学代写|数值分析代写Numerical analysis代考|Existence, Uniqueness, and Continuous Dependence

We have already seen how an iteration will often converge to a limit if the starting value is sufficiently close to that limit. The behaviour of the iteration, when started from an arbitrary starting value, can be very complicated. In this section we shall consider two examples. No theorems will be stated: our aim is simply to illustrate various kinds of behaviour.

First consider the simple iteration defined by
$$x_{k+1}=g\left(x_k\right), \quad k=0,1,2, \ldots, \quad \text { where } g(x)=a x(1-x), \quad(1.33)$$
which is often known as the logistic equation. We require the constant $a$ to lie in the range $0<a \leq 4$, for then if the starting value $x_0$ is in the interval $[0,1]$, then all members of the sequence $\left(x_k\right)$ also lie in $[0,1]$. The function $g$ has two fixed points: $x=0$ and $x=1-1 / a$. The fixed point at 0 is stable if $0<a<1$, and the fixed point at $1-1 / a$ is stable if $1<a<3$. The behaviour of the iteration for these values of $a$ is what might be expected from this information, but for larger values of the parameter $a$ the behaviour of the sequence $\left(x_k\right)$ becomes increasingly complicated.

For example, when $a=3.4$ there is no stable fixed point, and from any starting point the sequence eventually oscillates between two values, which are $0.45$ and $0.84$ to two decimal digits. These are the two stable fixed points of the double iteration
$$x_{k+1}=g^\left(x_k\right), \quad g^(x)=g(g(x))=a^2 x(1-x)[1-a x(1-x)] .(1.34)$$
When $3<a<1+\sqrt{ } 6$, the fixed points of $g^*$ are the two fixed points of $g$, that is 0 and $1-1 / a$, and also
$$\frac{1}{2}\left(1+\frac{1}{a} \pm \frac{1}{a}\left[a^2-2 a-3\right]^{1 / 2}\right) .$$

## 数学代写数值分析代写Numerical analysis代考|The bisection method

$$c_k=\frac{1}{2}\left(a_k+b_k\right),$$

$$\left(a_{k+1}, b_{k+1}\right)=\left{\left(a_k, c_k\right) \text { if } f\left(c_k\right) f\left(b_k\right)>0,\left(c_k, b_k\right) \text { if } f\left(c_k\right) f\left(b_k\right)<0,\right.$$

## 数学代写|数值分析代写Numerical analysis代考|Existence, Uniqueness, and Continuous Dependence

$$x_{k+1}=g\left(x_k\right), \quad k=0,1,2, \ldots, \quad \text { where } g(x)=a x(1-x),$$

$$\left.x_{k+1}=g^{\left(x_k\right)}, \quad g^{\prime} x\right)=g(g(x))=a^2 x(1-x)[1-a x(1-x)] .(1.34)$$

$$\frac{1}{2}\left(1+\frac{1}{a} \pm \frac{1}{a}\left[a^2-2 a-3\right]^{1 / 2}\right) .$$

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