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# 数学代写|数值分析代写Numerical analysis代考|MTH405 Operations Count

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## 数学代写|数值分析代写Numerical analysis代考|Operations Count

It is important to know the length of a computation and, for that reason, we count the number of arithmetic operations involved in Gaussian elimination. For reasons that will be apparent in later sections, the count will be divided into three parts. Also, as a notational convenience, let $U$ and $g$ denote the coefficient matrices in (8.26).

The elimination step. We count the additions/subtractions, multiplications, and divisions in going from the system (8.22) to the triangular system (8.26). We consider only the operations for the coefficients of $A$ and not for the right-hand side $b$ in (8.22). Table $8.1$ contains the number of these operations for each of the steps in the elimination. The totals in the last row of the table are obtained by using the formulas
$$\sum_{j=1}^p j=\frac{p(p+1)}{2}, \quad \sum_{j=1}^p j^2=\frac{p(p+1)(2 p+1)}{6}, \quad p \geq 1$$

Generally, the divisions and multiplications are counted together, since they are about the same in operation time. Doing this gives us
\begin{aligned} A S(A \rightarrow U) &=\frac{n(n-1)(2 n-1)}{6} \ M D(A \rightarrow U) &=\frac{n(n-1)(2 n-1)}{6}+\frac{n(n-1)}{2}=\frac{n\left(n^2-1\right)}{3} \end{aligned}
$A S(\cdot)$ denotes additions and subtractions for the enclosed operation, and $M D(\cdot)$ denotes multiplications and divisions. The notation $A \rightarrow U$ denotes the elimination process that converts $A$ in (8.22) into $U$ in (8.26).

Modification of the right side b to $g$. Proceeding as before, we get
$$\begin{gathered} A S(b \rightarrow g)=(n-1)+(n-2)+\cdots+1=\frac{n(n-1)}{2} \ M D(b \rightarrow g)=(n-1)+(n-2)+\cdots+1=\frac{n(n-1)}{2} \end{gathered}$$

The back substitution step, find $x$ from (8.26). As before,
$$\begin{gathered} A S(g \rightarrow x)=0+1+\cdots+(n-1)=\frac{n(n-1)}{2} \ M D(g \rightarrow x)=1+2+\cdots+n=\frac{n(n+1)}{2} \end{gathered}$$

## 数学代写|数值分析代写Numerical analysis代考|Computer Programs

Much mathematical research and program development has been carried out on producing efficient and accurate programs for Gaussian elimination, and your computer center should possess such a code. Some of the best of these codes are contained in the book Linpack: User’s Guide by Dongarra, Bunch, Moler, and Stewart (1979). Good programs will carry out the elimination in the most efficient manner and will arrange operations to minimize rounding errors. They will also offer error estimates, and this is needed because many linear systems are not as well behaved as our examples have been.

We give a subroutine LINSYS for implementing Gaussian elimination as described in this section. It is intended as a pedagogical tool; actual production computing should use a code of the type described in the preceding paragraph. The comment statements in LINSYS describe the values of the input and output variables. The use of matrix notation is based on ideas introduced in the next section. C TITLE: THIS IS A DEMO PROGRAM FOR SUBROUTINE LINSYS.
C IT WILL SOLVE A LINEAR SYSTEM $A * X=B$, GIVEN BY THE USER.
PARAMETER $($ MAXN=20）

## 数学代写|数值分析代写数值分析代考|操作计数

$$\sum_{j=1}^p j=\frac{p(p+1)}{2}, \quad \sum_{j=1}^p j^2=\frac{p(p+1)(2 p+1)}{6}, \quad p \geq 1$$

\begin{aligned} A S(A \rightarrow U) &=\frac{n(n-1)(2 n-1)}{6} \ M D(A \rightarrow U) &=\frac{n(n-1)(2 n-1)}{6}+\frac{n(n-1)}{2}=\frac{n\left(n^2-1\right)}{3} \end{aligned}
$A S(\cdot)$表示附加运算的加减运算，$M D(\cdot)$表示乘法和除法运算。符号$A \rightarrow U$表示将(8.22)中的$A$转换为(8.26)中的$U$的消元过程

$$\begin{gathered} A S(b \rightarrow g)=(n-1)+(n-2)+\cdots+1=\frac{n(n-1)}{2} \ M D(b \rightarrow g)=(n-1)+(n-2)+\cdots+1=\frac{n(n-1)}{2} \end{gathered}$$

$$\begin{gathered} A S(g \rightarrow x)=0+1+\cdots+(n-1)=\frac{n(n-1)}{2} \ M D(g \rightarrow x)=1+2+\cdots+n=\frac{n(n+1)}{2} \end{gathered}$$

## 数学代写|数值分析代写数值分析代考|计算机程序

PARAMETER $($ MAXN=20)

## MATLAB代写

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