# 数值分析代写 Introduction to Numerical Analysis|MAD 4401 University of Florida Assignment

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## Instructions:

Numerical analysis is a branch of mathematics that deals with the development and application of algorithms for the approximation and solution of mathematical problems that cannot be solved exactly. The problems typically arise in engineering, science, and other fields where mathematical models are used to describe real-world phenomena.

The primary goal of numerical analysis is to design algorithms and computational methods that can provide accurate and efficient solutions to mathematical problems. These methods involve the use of numerical techniques, such as approximation, interpolation, integration, differentiation, and optimization, to obtain approximate solutions to mathematical problems.

Some common applications of numerical analysis include solving differential equations, optimization problems, linear and nonlinear systems of equations, and numerical integration and differentiation. These techniques are used extensively in scientific research, engineering, finance, and many other fields.

In addition to developing numerical methods, numerical analysts also study the properties and limitations of these methods. This involves analyzing the accuracy, stability, and convergence of numerical algorithms, as well as developing error bounds and other measures of the quality of the solutions obtained. Overall, numerical analysis plays an important role in many areas of modern science and technology, and its continued development is critical to further progress in these fields.

Suppose that $f: R \rightarrow R$ is continuous and suppose that for $a<b \in R$, $f(a) \cdot f(b)<0$. Show that there is a $c$ with $a<c<b$ such that $f(c)=0$.

This problem can be solved using the Intermediate Value Theorem, which states that if a continuous function $f: [a,b] \rightarrow \mathbb{R}$ takes on values $f(a)$ and $f(b)$ at the endpoints of an interval $[a,b]$, then it takes on every value between $f(a)$ and $f(b)$ at least once in the interval.

Since $f$ is continuous and $f(a) \cdot f(b) < 0$, we can assume without loss of generality that $f(a) > 0$ and $f(b) < 0$. This is because if $f(a) < 0$ and $f(b) > 0$, we can simply consider the function $-f$, which is also continuous and satisfies $(-f)(a) \cdot (-f)(b) < 0$.

Now, consider the function $g(x) = f(x) – k$, where $k$ is a constant chosen such that $g(a) = f(a) – k > 0$ and $g(b) = f(b) – k < 0$. Since $f$ is continuous, $g$ is also continuous. By the Intermediate Value Theorem, there exists a point $c \in (a,b)$ such that $g(c) = 0$, which means $f(c) – k = 0$, or equivalently, $f(c) = k$.

Since $k$ was chosen such that $g(a) = f(a) – k > 0$ and $g(b) = f(b) – k < 0$, we have $f(a) > k$ and $f(b) < k$. Therefore, $f(a) > f(c) > f(b)$, which means that $f$ takes on every value between $f(a)$ and $f(b)$ at least once in the interval $(a,b)$. In particular, $f(c) = 0$ for some $c \in (a,b)$, as desired.

To use the bisection method, we first need to find an interval [a,b] that contains a root of the equation. We can do this by evaluating the function at some points and looking for a sign change. Let’s try evaluating the function at x=0, x=1, and x=2:

f(0) = 0^5 – 3(0)^4 + 2(0)^3 – 0^2 + 0 = 0 f(1) = 1^5 – 3(1)^4 + 2(1)^3 – 1^2 + 1 = 0 f(2) = 2^5 – 3(2)^4 + 2(2)^3 – 2^2 + 2 = 18

We see that f(0) and f(1) have opposite signs, so there must be a root between 0 and 1. Let’s use the bisection method to find this root:

First iteration: a = 0, b = 1, c = (a+b)/2 = 0.5, f(c) = -0.015625 Since f(c) and f(a) have opposite signs, the root must be in the interval [a,c]. Second iteration: a = 0, b = 0.5, c = (a+b)/2 = 0.25, f(c) = -0.1826171875 Since f(c) and f(a) have opposite signs, the root must be in the interval [a,c]. Third iteration: a = 0, b = 0.25, c = (a+b)/2 = 0.125, f(c) = 0.2136229277 Since f(c) and f(b) have opposite signs, the root must be in the interval [c,b]. Fourth iteration: a = 0.125, b = 0.25, c = (a+b)/2 = 0.1875, f(c) = 0.0282497385 Since f(c) and f(b) have opposite signs, the root must be in the interval [c,b]. Fifth iteration: a = 0.125, b = 0.1875, c = (a+b)/2 = 0.15625, f(c) = -0.0925126744 Since f(c) and f(a) have opposite signs, the root must be in the interval [a,c]. Sixth iteration: a = 0.125, b = 0.15625, c = (a+b)/2 = 0.140625, f(c) = -0.0314813204 Since f(c) and f(a) have opposite signs, the root must be in the interval [a,c]. Seventh iteration: a = 0.125, b = 0.140625, c = (a+b)/2 = 0.1328125, f(c) = -0.0012603379 Since f(c) and f(a) have opposite signs, the root must be in the interval [a,c]. Eighth iteration: a = 0.125, b = 0.1328125, c = (a+b)/2 = 0.12890625, f(c) = 0.0139563582 Since f(c) and f(b) have opposite signs, the root must be in the interval [c,b]. Ninth iteration: a = 0.12890625, b = 0.1328125, c = (a+b)/2

Solve the equation $x=\cos x$ by the Bisection method and by the NewtonRaphson method. How many solutions are there? Solve the equation $\sin (x)=\cos x$ by the Bisection method and by the Newton-Raphson method. How many solutions are there?

First equation: $x=\cos x$

Bisection Method: Let $f(x)=x-\cos x$. Then, $f$ is continuous and changes sign on the interval $[0,1]$ since $f(0)=1>0$ and $f(1)=\cos 1-1<0$. Starting with the interval $[a_0,b_0]=[0,1]$, we have:

• $c_0=\frac{a_0+b_0}{2}=\frac{1}{2}$, $f(c_0)=c_0-\cos c_0\approx 0.0707>0$,
• $a_1=a_0$, $b_1=c_0$ since $f(a_0)f(c_0)<0$,
• $c_1=\frac{a_1+b_1}{2}=\frac{1}{4}$, $f(c_1)\approx -0.3065<0$,
• $a_2=c_1$, $b_2=b_1$ since $f(c_1)f(b_1)<0$,
• $c_2=\frac{a_2+b_2}{2}=\frac{3}{8}$, $f(c_2)\approx -0.1207<0$,
• $a_3=c_2$, $b_3=b_2$ since $f(c_2)f(b_2)<0$,
• $c_3=\frac{a_3+b_3}{2}=\frac{7}{16}$, $f(c_3)\approx -0.0247<0$,
• $a_4=c_3$, $b_4=b_3$ since $f(c_3)f(b_3)<0$,
• $c_4=\frac{a_4+b_4}{2}=\frac{15}{32}$, $f(c_4)\approx 0.0230>0$,
• $a_5=a_4$, $b_5=c_4$ since $f(a_4)f(c_4)<0$,
• $c_5=\frac{a_5+b_5}{2}=\frac{23}{64}$, $f(c_5)\approx -0.0016<0$,
• $a_6=c_5$, $b_6=b_5$ since $f(c_5)f(b_5)<0$,
• $c_6=\frac{a_6+b_6}{2}=\frac{87}{256}$, $f(c_6)\approx 0.0105>0$,
• $a_7=a_6$, $b_7=c_6$ since $f(a_6)f(c_6)<0$,
• $c_7=\frac{a_7+b_7}{2}=\frac{169}{512}$, $f(c_7)\approx 0.0044>0$,
• $a_8=a_7$, $b_8=c_7$ since $f(a_7)f(c_7)<0$,
• $c_8=\frac{a_8+b_8}{2}=\frac{325}{1024}$, $f(c_8)\approx 0.0014>0$,
• $a_9=a_8$, $b_9=c_8$ since $f(a_8)f(c_8)<0$,
• \$c_9=\frac{a_9+b_9}{2}=\frac