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# 数学代写|数值分析代写Numerical analysis代考|STAT434 Bracketing a root

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## 数学代写数值分析代写Numerical analysis代考|Bracketing a root

The function $f(x)$ has a root at $x=r$ if $f(r)=0$.
The first step to solving an equation is to verify that a root exists. One way to ensure this is to bracket the root: to find an interval $[a, b]$ on the real line for which one of the pair ${f(a), f(b)}$ is positive and the other is negative. This can be expressed as $f(a) f(b)<0$. If $f$ is a continuous function, then there will be a root: an $r$ between $a$ and $b$ for which $f(r)=0$. This fact is summarized in the following corollary of the Intermediate Value Theorem 0.4 :

Let $f$ be a continuous function on $[a, b]$, satisfying $f(a) f(b)<0$. Then $f$ has a root between $a$ and $b$, that is, there exists a number $r$ satisfying $a<r<b$ and $f(r)=0$.

In Figure $1.1, f(0) f(1)=(-1)(1)<0$. There is a root just to the left of 0.7 . How can we refine our first guess of the root’s location to more decimal places?

We’ll take a cue from the way our eye finds a solution when given a plot of a function. It is unlikely that we start at the left end of the interval and move to the right, stopping at the root. Perhaps a better model of what happens is that the eye first decides the general location, such as whether the root is toward the left or the right of the interval. It then follows that up by deciding more precisely just how far right or left the root lies and gradually improves its accuracy, just like looking up a name in the phone book. This general approach is made quite specific in the Bisection Method, shown in Figure 1.2.

## 数学代写|数值分析代写Numerical analysis代考|How accurate and how fast?

If $[a, b]$ is the starting interval, then after $n$ bisection steps, the interval $\left[a_n, b_n\right]$ has length $(b-a) / 2^n$. Choosing the midpoint $x_c=\left(a_n+b_n\right) / 2$ gives a best estimate of the solution $r$, which is within half the interval length of the true solution. Summarizing, after $n$ steps of the Bisection Method, we find that
Solution error $=\left|x_c-r\right|<\frac{b-a}{2^{n+1}}$
and
Function evaluations $=n+2$.
A good way to assess the efficiency of the Bisection Method is to ask how much accuracy can be bought per function evaluation. Each step, or each function evaluation, cuts the uncertainty in the root by a factor of two.

A solution is correct within $p$ decimal places if the error is less than $0.5 \times 10^{-p}$.
Use the Bisection Method to find a root of $f(x)=\cos x-x$ in the interval $[0,1]$ to within six correct places.

First we decide how many steps of bisection are required. According to (1.1), the error after $n$ steps is $(b-a) / 2^{n+1}=1 / 2^{n+1}$. From the definition of $p$ decimal places, we require that
\begin{aligned} \frac{1}{2^{n+1}} & <0.5 \times 10^{-6} \ n & >\frac{6}{\log _{10} 2} \approx \frac{6}{0.301}=19.9 \end{aligned}

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