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# 数学代写|数值分析代写Numerical analysis代考|MATH345 The Influence of Rounding Errors in One-Step Methods

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## 数学代写数值分析代写Numerical analysis代考|The Influence of Rounding Errors in One-Step Methods

If a one-step method
\begin{aligned} &\eta_0:=y_0 ; \ &\text { for } i=0,1, \ldots: \ &\eta_{i+1}:=\eta_i+h \Phi\left(x_i, \eta_i ; h\right), \ &x_{i+1}:=x_i+h \end{aligned}
is executed in floating-point arithmetic ( $t$ decimal digits) with relative precision eps $=5 \times 10^{-t}$, then instead of the $\eta_i$ one obtains other numbers $\tilde{\eta}i$, which satisfy a recurrence formula of the form \begin{aligned} \tilde{\eta}_0 &:=y_0 ; \ \text { for } i=0,1, \ldots: \ c_i &:=\operatorname{fl}\left(\Phi\left(x_i, \tilde{\eta}_i ; h\right)\right), \ d_i &:=\operatorname{fl}\left(h c_i\right), \ \tilde{\eta}{i+1} &:=\mathrm{fl}\left(\tilde{\eta}i+d_i\right)=\tilde{\eta}_i+h \Phi\left(x_i, \tilde{\eta}_i ; h\right)+\varepsilon{i+1}, \end{aligned}
where the total absolute rounding error $\epsilon_{i+1}$, in first approximation, is made up of threee components:
$$\varepsilon_{i+1} \doteq h \Phi\left(x_i, \tilde{\eta}i ; h\right)\left(\alpha{i+1}+\mu_{i+1}\right)+\tilde{\eta}{i+1} \sigma{i+1}$$

## 数学代写|数值分析代写Numerical analysis代考|Practical Implementation of One-Step Methods

In practice, initial-value problems present themselves mostly in the following form: What is desired is the value which the exact solution $y(x)$ assumes for a certain $x \neq x_0$. It is tempting to compute this solution approximately by means of a one-step method in a single step, i.e., choosing stepsize $\bar{h}=x-x_0$. For large $x-x_0$ this of course leads to a large discretization error $e(x ; \bar{h})$; the choice made for $\bar{h}$ would be entirely inadequate. Normally, therefore, one will introduce suitable intermediate points $x_i, i=1$, $\ldots, k-1, x_0<x_1<\cdots<x_k=x$, and, beginning with $x_0, y_0=y\left(x_0\right)$, compute successive approximate values of $y\left(x_i\right)$ : Having determined an approximation $\bar{y}\left(x_i\right)$ of $y\left(x_i\right)$, one computes $\bar{y}\left(x_{i+1}\right)$ by applying a step of the method with stepsize $h_i:=x_{i+1}-x_i$,
$$\bar{y}\left(x_{i+1}\right)=\bar{y}\left(x_i\right)+h_i \Phi\left(x_i, \bar{y}\left(x_i\right) ; h_i\right), \quad x_{i+1}=x_i+h_i .$$

There again arises, however, the problem of how the stepsizes $h_i$ are to be chosen. Since the amount of work involved in the method is proportional to the number of individual steps, one will attempt to choose the stepsizes $h_i$ as large as possible. On the other hand, they must not be chosen too large if one wants to keep the discretization error small. In principle, one has the following problem: For given $x_0, y_0$, determine a stepsize $h$ as large as possible, but such that the discretization error $e\left(x_0+h ; h\right)$ after one step with this stepsize still remains below a certain tolerance $\varepsilon$. This tolerance $\varepsilon$ should not be selected smaller than $K$ eps, i.e., $\varepsilon \geq K$ eps, where eps is the relative machine precision and $K$ a bound for the solution $y(x)$ in the region under consideration,
$$K \approx \max \left{|y(x)| \mid x \in\left[x_0, x_0+h\right]\right} .$$

## 数学代写数值分析代写Numerical analysis代考|The Influence of Rounding Errors in One-Step Methods

$$\eta_0:=y_0 ; \quad \text { for } i=0,1, \ldots: \eta_{i+1}:=\eta_i+h \Phi\left(x_i, \eta_i ; h\right), \quad x_{i+1}:=x_i+h$$

$\tilde{\eta}0:=y_0 ;$ for $i=0,1, \ldots: c_i \quad:=\operatorname{fl}\left(\Phi\left(x_i, \tilde{\eta}_i ; h\right)\right), d_i:=\operatorname{fl}\left(h c_i\right), \tilde{\eta} i+1 \quad:=\operatorname{fl}\left(\tilde{\eta} i+d_i\right)=\tilde{\eta}_i+h \Phi\left(x_i, \tilde{\eta}_i ; h\right)+\varepsilon i+1$, 其中总绝对舍入误差 $\epsilon{i+1}$ ，在第一个近似值中，由三部分组成:
$$\varepsilon_{i+1} \doteq h \Phi\left(x_i, \tilde{\eta} ; h\right)\left(\alpha i+1+\mu_{i+1}\right)+\tilde{\eta} i+1 \sigma i+1$$

## 数学代写|数值分析代写Numerical analysis代考|Practical Implementation of One-Step Methods

$$\bar{y}\left(x_{i+1}\right)=\bar{y}\left(x_i\right)+h_i \Phi\left(x_i, \bar{y}\left(x_i\right) ; h_i\right), \quad x_{i+1}=x_i+h_i .$$

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