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# 数学代写|数值分析代写Numerical analysis代考|MATH2722 Propagated Error in Function Evaluation

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## 数学代写|数值分析代写Numerical analysis代考|Propagated Error in Function Evaluation

Consider evaluating $f(x)$ at the approximate value $x_A$ rather than at $x_T$. Then consider how well does $f\left(x_A\right)$ approximate $f\left(x_T\right)$ ? There will also be an additional error introduced in the actual evaluation of $f\left(x_A\right)$, resulting in the noise phenomena described in Section 3.1. But we will be concerned here with just comparing the exact value of $f\left(x_A\right)$ with the exact value of $f\left(x_T\right)$.
Using the mean-value theorem (A.3) of Appendix A,
$$f\left(x_T\right)-f\left(x_A\right)=f^{\prime}(c)\left(x_T-x_A\right)$$
with $c$ an unknown point between $x_A$ and $x_T$. Since $x_A$ and $x_T$ are generally very close together, we have
$$f\left(x_T\right)-f\left(x_A\right) \doteq f^{\prime}\left(x_T\right)\left(x_T-x_A\right) \doteq f^{\prime}\left(x_A\right)\left(x_T-x_A\right)$$
In most instances, it is better to consider the relative error,
$$\operatorname{Rel}\left(f\left(x_A\right)\right) \doteq \frac{f^{\prime}\left(x_T\right)}{f\left(x_T\right)}\left(x_T-x_A\right)=\frac{f^{\prime}\left(x_T\right)}{f\left(x_T\right)} x_T \operatorname{Rel}\left(x_A\right)$$

## 数学代写|数值分析代写Numerical analysis代考|Rounding Versus Chopping

A more important difference in the errors shown in the tables is that which exists between rounding and chopping. Rounding results in a far smaller error in the calculated sum that does chopping. To understand why this happens, return to the formuia (3.33). As a typical case from it, consider the first term on the right side:
$$T \equiv-a_1\left(\epsilon_2+\cdots+\epsilon_n\right)$$
Assume that we are using rounding with the four-digit decimal machine of the above example. Using (2.22) with $\beta=10$ and $n=4$ for our decimal machine, we have that all $\epsilon$, satisfy
$$-0.0005 \leq \epsilon_j \leq 0.0005$$
Rounding errors can usually be treated as random in nature, subject to this bounding interval. Thus, the positive and negative values of the $\epsilon_j$ ‘s in (3.35) will tend to cancel, and the sum $T$ will be nearly zero. By using advançed methods from probability theory, it can be shown that (3.35) is very likely to satisfy
$$|T| \leq(1.49)(0.0005) \cdot \sqrt{n}\left|a_1\right|$$
Thus, (3.35) tends to be small until $n$ becomes quite large, and the same is true of the total error on the right of (3.33).
For our decimal machine with chopping, (2.22) becomes
$$-0.001 \leq \epsilon_j \leq 0$$

and the errors are all of one sign. Again, the chopping errors will vary randomly in this interval. But now the average value of the $\epsilon_j$ ‘s will be $-0.0005$, the middle of the interval, and the likely value of $(3.35)$ will be
$$-a_1(n-1)(-0.0005)$$
This grows with $n$, where the corresponding result for the case of rounding was zero. Thus the error (3.35) and (3.33) will grow much more rapidly when chopping is used rather than rounding.

## 数学代写|数值分析代写数值分析代考|函数评估中的传播误差

. 0$$f\left(x_T\right)-f\left(x_A\right)=f^{\prime}(c)\left(x_T-x_A\right)$$
with $c$ 之间的一个未知点 $x_A$ 和 $x_T$。自从 $x_A$ 和 $x_T$ 通常非常接近，我们有
$$f\left(x_T\right)-f\left(x_A\right) \doteq f^{\prime}\left(x_T\right)\left(x_T-x_A\right) \doteq f^{\prime}\left(x_A\right)\left(x_T-x_A\right)$$在大多数情况下，最好考虑相对误差
$$\operatorname{Rel}\left(f\left(x_A\right)\right) \doteq \frac{f^{\prime}\left(x_T\right)}{f\left(x_T\right)}\left(x_T-x_A\right)=\frac{f^{\prime}\left(x_T\right)}{f\left(x_T\right)} x_T \operatorname{Rel}\left(x_A\right)$$

## 数学代写|数值分析代写数值分析代考|舍入与切取

$$T \equiv-a_1\left(\epsilon_2+\cdots+\epsilon_n\right)$$

$$-0.0005 \leq \epsilon_j \leq 0.0005$$

$$|T| \leq(1.49)(0.0005) \cdot \sqrt{n}\left|a_1\right|$$
。因此，(3.35)趋于小，直到$n$变得相当大，(3.33)右侧的总误差也是如此。对于我们的斩式十进制机器，(2.22)变成
$$-0.001 \leq \epsilon_j \leq 0$$

，错误都是一个符号。同样，截取错误将在这个区间内随机变化。但是现在$\epsilon_j$的平均值将是$-0.0005$，即区间的中间值，而$(3.35)$的可能值将是
$$-a_1(n-1)(-0.0005)$$
。这随着$n$增长，其中舍入情况的相应结果为零。因此，当使用斩取而不是舍入时，错误(3.35)和(3.33)的增长要快得多

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