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# 数学代写|数值分析代写Numerical analysis代考|MATH345 CONVERGENCE ANALYSIS OF EULER’S METHOD

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## 数学代写|数值分析代写Numerical analysis代考|CONVERGENCE ANALYSIS OF EULER’S METHOD

The purpose of analyzing Euler’s method is to understand how it works, to be able to predict the error when using it, and to perhaps accelerate its convergence. Being able to do this for Euler’s method will also make it easier to answer similar questions for other, more efficient numerical methods.

We begin by considering the error in approximation (9.26) that led directly to Euler’s method. Using Taylor’s theorem, write
$$Y\left(x_{n+1}\right)=Y\left(x_n\right)+h Y^{\prime}\left(x_n\right)+\frac{h^2}{2} Y^{\prime \prime}\left(\xi_n\right)$$
for some $x_n \leq \xi_n \leq x_{n+1}$. Using the fact that $Y(x)$ satisfies the differential equation, this becomes
$$Y\left(x_{n+1}\right)=Y\left(x_n\right)+h f\left(x_n, Y\left(x_n\right)\right)+\frac{h^2}{2} Y^{\prime \prime}\left(\xi_n\right)$$
The term
$$T_{n+1}=\frac{h^2}{2} Y^{\prime \prime}(\xi)$$
is called the truncation error for Euler’s method, and it is the error in the earlier approximation (9.26),
$$Y\left(x_{n+1}\right) \doteq Y\left(x_n\right)+h f\left(x_n, Y\left(x_n\right)\right)$$
To analyze the error in Euler’s method, subtract
$$y_{n+1}=y_n+h f\left(x_n, y_n\right)$$

from (9.31), to obtain
$Y\left(x_{n+1}\right)-y_{n+1}=Y\left(x_n\right)-y_n+h\left[f\left(x_n, Y\left(x_n\right)\right)-f\left(x_n, y_n\right)\right]$
$+\frac{h^2}{2} Y^{\prime \prime}\left(\xi_n\right)$

## 数学代写|数值分析代写Numerical analysis代考|Asymptotic Error Analysis

To obtain more accurate predictions of the error, we consider asymptotic error estimates, similar to what was done earlier in Chapter 7 with numerical integration. An asymptotic error formula is available for Euler’s method, provided that certain conditions are satisfied. If the two functions
$$\frac{\partial f(x, z)}{\partial z}, \quad \frac{\partial^2 f(x, z)}{\partial z^2}$$

are continuous for all values of $(x, z)$ near to $(x, Y(x)), x_0 \leq x \leq b$, then one can prove that the error in Euler’s method satisfies
$$Y\left(x_n\right)-y_h\left(x_n\right)=h D\left(x_n\right)+O\left(h^2\right), \quad x_0 \leq x_n \leq b$$
Assuming $y_0=Y_0$, the usual case, the function $D(x)$ satisfies an initial value problem for a linear differential equation,
$$D^{\prime}(x)=g(x) D(x)+\frac{1}{2} Y^{\prime \prime}(x), \quad D\left(x_0\right)=0$$
where
$$g(x)=\left.\frac{\partial f(x, z)}{\partial z}\right|_{z=Y(x)}$$
The conditions on (9.42) are almost always satisfied in practice, leading to (9.43), for all sufficiently small values of $h$. The notation $O\left(h^2\right)$ means a quantity proportional to $h^2$ or something smaller; it goes to zero more rapidly than the term $h D\left(x_n\right)$, which involves $h$ to the smaller power of 1 .

Obtaining $D(x)$ involves knowing the solution $Y(x)$, which we would not know in a practical situation. But the formula $(9.43)$ is still useful in obtaining practical methods to predict the error, as will be done following an example.

## 数学代写|数值分析代写Numerical analysis代考|CONVERGENCE ANALYSIS OF EULER’ S METHOD

$$Y\left(x_{n+1}\right)=Y\left(x_n\right)+h Y^{\prime}\left(x_n\right)+\frac{h^2}{2} Y^{\prime \prime}\left(\xi_n\right)$$

$$Y\left(x_{n+1}\right)=Y\left(x_n\right)+h f\left(x_n, Y\left(x_n\right)\right)+\frac{h^2}{2} Y^{\prime \prime}\left(\xi_n\right)$$

$$T_{n+1}=\frac{h^2}{2} Y^{\prime \prime}(\xi)$$

$$Y\left(x_{n+1}\right) \doteq Y\left(x_n\right)+h f\left(x_n, Y\left(x_n\right)\right)$$

$$y_{n+1}=y_n+h f\left(x_n, y_n\right)$$

\begin{aligned} &Y\left(x_{n+1}\right)-y_{n+1}=Y\left(x_n\right)-y_n+h\left[f\left(x_n, Y\left(x_n\right)\right)-f\left(x_n, y_n\right)\right] \ &+\frac{h^2}{2} Y^{\prime \prime}\left(\xi_n\right) \end{aligned}

## 数学代写|数值分析代写Numerical analysis代考|Asymptotic Error Analysis

$$\frac{\partial f(x, z)}{\partial z}, \quad \frac{\partial^2 f(x, z)}{\partial z^2}$$

$$Y\left(x_n\right)-y_h\left(x_n\right)=h D\left(x_n\right)+O\left(h^2\right), \quad x_0 \leq x_n \leq b$$

$$D^{\prime}(x)=g(x) D(x)+\frac{1}{2} Y^{\prime \prime}(x), \quad D\left(x_0\right)=0$$

$$g(x)=\left.\frac{\partial f(x, z)}{\partial z}\right|_{z=Y(x)}$$
(9.42) 的条件在实跕伐中几乎总是得到满足，导致 (9.43)，对于所有足够小的值 $h$. 符号 $O\left(h^2\right)$ 表示与 $h^2$ 或者更小的东西; 它比术语 更快地变为零 $h D\left(x_n\right)$, 这涉及 $h 1$ 的较小募。

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