# 数学代写|数值方法作业代写numerical methods代考|APM5333

## 数学代写|数值方法作业代写numerical methods代考|The root finding method

Solve $f(x)=0$ for $x$, when an explicit analytical solution is impossible. Sometimes, the solution of the problems sometime is “exact” in a fixed amount of time. The solution of other problems may have error tolerances, and our algorithms may have to iterate to compute them. Generally, the shortfall for the root finding system is its illogical termination of problems despite the sophistication of the algorithms and the duration at which convergence would be achieved at lower error tolerances. However, the advantages far outweigh the aforementioned disadvantages. In this chapter, three main root finding method will be discussed, i.e., Bisection, Secant, and Newton’s method. In practice, the duration of converge of these methods is in descending order, i.e., Newton’s method, Secant, and Bisection. By principle, the three methods, i.e., Bisection, Secant, and Newton’s method are defined in Eqs. (5.1) $-(5.3)$, respectively. Generally, a sequence $x_0, x_1, x_2, \ldots$ are constructed such that it converges to the root $x=r$.
$$\begin{gathered} x_{n+1}=x_{n-1}+\frac{x_n-x_{n-1}}{2} \ x_{n+1}=x_n-\frac{\left(x_n-x_{n-1}\right) f\left(x_n\right)}{f\left(x_n\right)-f\left(x_{n-1}\right)} \ x_{n+1}=x_n-\frac{f\left(x_n\right)}{f_{\left(x_n\right)}^{\prime}} \end{gathered}$$

## 数学代写|数值方法作业代写numerical methods代考|Euler method

Euler method is a common method for solving first-order numerical ordinary differential equations (ODEs) with a given initial value. It is also referred to as forward Euler method or simplest Runge-Kutta method (i.e., First-Order Runge-Kutta) and the simplest method for numerical integration of ordinary differential equations. The procedural steps to solving ODE via Euler method is prone to local error (error per step) which is proportional to the square of the step size.

The theory of the Euler’s method starts with the assumption that $y\left(x_1\right)$ denotes exact solution and $y_1$ the computed solution at $x_i=x_0+h$ such that
$$y\left(x_1\right)=y\left(x_0+h\right)$$
The solution can be derived from the Taylor’s series given as:
$$y\left(x_0+h\right)=y\left(x_0\right)+h y^{\prime}\left(x_0\right)+\frac{h^2}{2} y^{\prime \prime}\left(x_0\right)+\frac{h^3}{3} y^{\prime \prime \prime}\left(x_0\right)+\ldots$$
To apply the Taylor’s series here means that Eq. (6.4) will be truncated after second term
$$y\left(x_1\right)=y\left(x_0\right)+h y^{\prime}\left(x_0\right)$$
to give the Euler’s formula so that,
$$y_1=y_0+h f\left(x_0, y_0\right)$$
$f\left(x_0, y_0\right)$ is a known function and the values in the initial condition are also known numbers. Function is assumed to be continuous so that a unique solution to the initial value problem (IVP) will be obtained. The error in the formula is the third term of Eq. (6.4) i.e., $\frac{h^2}{2} y^{\prime \prime}(\xi)$ where $x_0 \leq \xi \leq x_1$.
When $x=x_{n+1}$, then the Euler’s formula becomes,
$$y_{n+1}=y_n+h f\left(x_n, y_n\right)$$
Hence, the error in the formula would be $\frac{h^2}{2} y^{\prime \prime}(\xi), x_n \leq \xi \leq x_{n+1}$. The error in the Euler’s method depends on the magnitude of the step size. Euler’s method uses the idea of local linearity or linear approximation; hence, the disadvantage of the Euler’s method is the tendency to sometimes fail or cumbersome to solve manually.
A typical application of Euler’s is presented below to solve the equation:
$$\frac{d y}{d x}=\frac{y \ln y}{2 x}$$
where the initial condition is $y(2)=e$.

# 数值方法代考

## 数学代写|数值方法作业代写numerical methods代考|The root finding method

$$x_{n+1}=x_{n-1}+\frac{x_n-x_{n-1}}{2} x_{n+1}=x_n-\frac{\left(x_n-x_{n-1}\right) f\left(x_n\right)}{f\left(x_n\right)-f\left(x_{n-1}\right)} x_{n+1}=x_n-\frac{f\left(x_n\right)}{f_{\left(x_n\right)}^{\prime}}$$

## 数学代写|数值方法作业代写numerical methods代考|Euler method

$$y\left(x_1\right)=y\left(x_0+h\right)$$

$$y\left(x_0+h\right)=y\left(x_0\right)+h y^{\prime}\left(x_0\right)+\frac{h^2}{2} y^{\prime \prime}\left(x_0\right)+\frac{h^3}{3} y^{\prime \prime \prime}\left(x_0\right)+\ldots$$

$$y\left(x_1\right)=y\left(x_0\right)+h y^{\prime}\left(x_0\right)$$

$$y_1=y_0+h f\left(x_0, y_0\right)$$
$f\left(x_0, y_0\right)$ 是一个已知函数，初始条件下的值也是已知数。假设函数是连续的，因此将获得初始值问题 (IVP) 的唯一解。公式中的误差是等式的第三项。(6.4) 即， $\frac{h^2}{2} y^{\prime \prime}(\xi)$ 在哪里 $x_0 \leq \xi \leq x_1$. 什么时候 $x=x_{n+1}$ ，则欧拉公式变为，
$$y_{n+1}=y_n+h f\left(x_n, y_n\right)$$

$$\frac{d y}{d x}=\frac{y \ln y}{2 x}$$

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