### 金融代写|金融数学代写Financial Mathematics代考|Newton’s Method

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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 金融代写|金融数学代写Financial Mathematics代考|Also Called the Newton-Raphson Method

Isaac Newton (1643-1727), an English philosopher and mathematician, did important work in both physics and calculus. His method for approximating roots to polynomials is a very nice application of the tangent line. Joseph Raphson (1648-1715), also English, was made a member of the Royal Society prior to his graduation from Cambridge. See more about these two at the MacTutor History of Mathematics site: http://www-history.mes.standrews.ac.uk/index.html

Newton’s Method solves the equation $f(x)=0$ using an iteration technique. An iteration technique involves three stages:
1) Determining an initial guess (or approximation) called $x_{0}$,
2) Constructing an algorithm to compute $x_{i+1}$ in terms of $x_{i}$,
3) A proof that the sequence $x_{n}$ converges to the required value, in our case a solution of the equation $f(x)=0$.

The process starts with the initial approximation $x_{0}$ and then computes $x_{1}$, $x_{2}$, etc., until a desired degree of accuracy is attained. We will discuss how to make an educated guess (the $x_{0}$ ) in the context of specific problems ${ }^{4}$. At this point, we are interested only in describing how Newton’s Method generates the iteration sequence in 2). A proof that the method works is beyond the scope of this text – consult an Advanced Calculus text, if you would like to see a proof.

To create the sequence of approximations using the Newton-Raphson Method, we start with a reasonable first approximation, $x_{0}$. Often this is done by using a graphing calculator to graph the function and then reading off an estimate from the graph. To find $x_{1}$, we first construct the tangent line to the graph of $f$ at the point $\left(x_{0}, f\left(x_{0}\right)\right)$. The second estimate, $x_{1}$, is the $x$-intercept of this tangent line.

## 金融代写|金融数学代写Financial Mathematics代考|Approximations Using Taylor Series

If $f(x)$ is a function which has derivatives of all orders $\left(f^{\prime}, f^{\prime \prime}, f^{\prime \mu^{\prime}}\right.$ etc., all exist) it can be shown that (under certain restrictions) $f(x)$ can be computed as an infinite sum of terms involving its derivatives.
$$f(x)=f(a)+f^{\prime}(a)(x-a)^{2}+\frac{f^{(2)}(a)}{2 !}(x-a)^{2}+\cdots+\frac{f^{(n)}(a)}{n !}(x-a)^{n}+\cdots$$
In the expression above $f^{(n)}(a)$ refers to the $n^{\text {th }}$ derivative of $f$ evaluated at a. We can compute approximate values of $f(x)$ near a known value $f(a)$ by using the first few terms in $1.12$.

Example 1.6: Use the first four terms of Equation $1.12$ and $a=0$ to approximate $\sin (x)$
Solution:
\begin{aligned} \sin (x) &=\sin (0)+\sin ^{\prime}(0)(x-0)+\frac{\sin ^{\prime \prime}(0)}{2 !}(x-0)^{2}+\cdots+\frac{\sin ^{\prime \prime \prime}(0)}{3 !}(x-0)^{3} \ &=0+\cos (0)(x-0)+\frac{-\sin (0)}{2 !}(x-0)^{2}+\frac{-\cos (0)}{3 !}(x-0)^{2} \ &=x-\frac{x^{3}}{6} \end{aligned}
As the sketch below illustrates this approximation is quite good for values of $x$ close to 0 (Figure $1.3$ ).

## 金融代写|金融数学代写Financial Mathematics代考|Exponents and Logarithms

For convenience we state some of the basic properties of the exponential and logarithmic functions. Unless otherwise stated, we will use the logarithm to base $e$, indicated as $\ln (x)$ and often referred to as the natural logarithm The TI BA II Plus and TI-30XS both have a button devoted to $\ln$ – the $2 \mathrm{ND}$ function for this button is $e^{x}$.
BASIC IDENTITIES
\begin{aligned} \ln (a b) &=\ln (a)+\ln (b) \ \ln \left(a^{r}\right) &=r \ln (a) \ \ln \left(\frac{a}{b}\right) &=\ln (a)-\ln (b) \ \frac{d \ln (x)}{d x} &=\frac{1}{x}, \quad \frac{d \ln (1+i)}{d i}=\frac{1}{1+i} \ \ln \left(e^{x}\right) &=e^{\ln (x)}=x \ \frac{d e^{x}}{d x} &=e^{x} \ \int e^{u} d u &=e^{u}+c \ \int \frac{1}{u} d u &=\ln (|u|)+c \end{aligned}
Example 1.9: Solve $(1.05)^{n}=2$.
Solution: We take $\ln$ of both sides to obtain $n \cdot \ln (1.05)=\ln (2)$. Thus, $n=\frac{\ln (2)}{\ln (1.05)} \approx 14.21$.
Example 1.10: Solve for $i:$
$$(1+i)^{3}=1+3 \cdot(.05)=1.15 .$$
Solution: We take $\ln$ of both sides to obtain $3 \ln (1+i)=\ln (1.15)$. This gives us $\ln (1+i)=\frac{\ln (1.15)}{3}=0.04658731412$. As a result, $1+i=e^{0.04658731412}=$ 1.047689553. Hence $i=.047686553$. We could also solve this problem by taking the cube root of both sides of the equation. $(1+i)=\sqrt[3]{1.15}=(1.15)^{\frac{1}{3}}=$ 1.047689553.

Note: Don’t round intermediate values in your calculations. It is appropriate to round the final answer. For example $\$ 145.8802$would be reported as$\$145.88$.
The TI BA II Plus does not have an $n^{\text {th }}$ root button, so you need to use the $y^{x}$ button with $x=\frac{1}{3}$. If you don’t know the decimal value of $\frac{1}{x}$ use the $\frac{1}{x}$ key. Here is how the calculation looks on the TI BA II Plus for $1.08^{\frac{1}{7}}, 1.08$, $y^{x}, 7, \frac{1}{x}=$ Result: $1.011055 .$

## 金融代写|金融数学代写Financial Mathematics代考|Also Called the Newton-Raphson Method

1) 确定初始猜测（或近似值），称为X0,
2) 构造一个算法来计算X一世+1按照X一世,
3) 证明该序列Xn收敛到所需的值，在我们的例子中是方程的解F(X)=0.

## 金融代写|金融数学代写Financial Mathematics代考|Approximations Using Taylor Series

F(X)=F(一个)+F′(一个)(X−一个)2+F(2)(一个)2!(X−一个)2+⋯+F(n)(一个)n!(X−一个)n+⋯

## 金融代写|金融数学代写Financial Mathematics代考|Exponents and Logarithms

ln⁡(一个b)=ln⁡(一个)+ln⁡(b) ln⁡(一个r)=rln⁡(一个) ln⁡(一个b)=ln⁡(一个)−ln⁡(b) dln⁡(X)dX=1X,dln⁡(1+一世)d一世=11+一世 ln⁡(和X)=和ln⁡(X)=X d和XdX=和X ∫和在d在=和在+C ∫1在d在=ln⁡(|在|)+C

(1+一世)3=1+3⋅(.05)=1.15.

TI BA II Plus 没有nth 根按钮，所以你需要使用是X按钮X=13. 如果你不知道十进制值1X使用1X钥匙。以下是 TI BA II Plus 上的计算结果1.0817,1.08, 是X,7,1X=结果：1.011055.

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## MATLAB代写

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