MY-ASSIGNMENTEXPERT™可以为您提供math.tamu.edu MATH609 Numerical analysis数值分析课程的代写代考和辅导服务！
MATH609课程简介
I am teaching two versions of 609 this semester. 609-700 is for students in the math department’s distance masters program while this course, 609-600 is for students in residence at TAMU, College Station. The class homepage for this course is located at:http://www.math.tamu.edu/~pasciak/classes/609-Local
Prerequisites
Next week (the week starting Oct. 5) is the middle of the semester and as promised in the syllabus, we shall have a midterm. This exam will be a take home but with a fairly restrictive timeline, namely, the exam will be sent to you on Wednesday Oct. 7 at 9:00 am and must be completed and EMAILED back to me ([email protected]) by 3:00 PM that day. The six hour window should suffice as I expect that most will be able to essentially finish in under two. In lieu of the this, there will be no new homework assigned the week before and only one lecture during the week of the exam.
MATH609 Numerical analysis HELP（EXAM HELP， ONLINE TUTOR）
For stopping the iterations use any of the conditions:
$$
\left|x^{(m+1)}-x^{(m)}\right|_2 /\left|x^{(0)}\right|_2<T O L, \quad\left|r^{(m)}\right|_2 /\left|r^{(0)}\right|_2 /<T O L,
$$
where $r^{(m)}=b-A x^{(m)}$ is the residual of the $m$-th iterate, or any other criterion you consider appropriate. Set $T O L=10^{-6}$ or $T O L=10^{-12}$ if using double precision.
The given conditions are used to stop the iterations of an iterative method for solving a linear system $Ax = b$.
The first condition measures the relative change in the solution vector between consecutive iterations, where $\left|x^{(m+1)}-x^{(m)}\right|_2$ is the Euclidean norm of the difference between the current and previous iterates of $x$, and $\left|x^{(0)}\right|_2$ is the Euclidean norm of the initial iterate. If this relative change is less than a certain tolerance level $TOL$, then the iteration process is stopped.
The second condition measures the relative size of the residual vector $r^{(m)}$ at the $m$-th iteration, where $r^{(m)}=b-Ax^{(m)}$ is the residual vector, and $\left|r^{(0)}\right|_2$ is the Euclidean norm of the initial residual vector. If the ratio of the Euclidean norm of the current residual vector to the Euclidean norm of the initial residual vector is less than a certain tolerance level $TOL$, then the iteration process is stopped.
The choice of the tolerance level $TOL$ depends on the desired accuracy of the solution and the precision of the floating-point arithmetic used in the implementation. A tolerance level of $10^{-6}$ or $10^{-12}$ is commonly used in double precision arithmetic.
In all problems take a r.h.s. $b=h^2(1,1, \ldots, 1)^t$ and $x^0=(0,0, \ldots, 0)^t$ or $x^0=b$.
It seems that you are referring to the initialization of the starting point for solving a system of linear equations using an iterative method.
In this case, the system of linear equations can be written in the form $Ax=b$, where $A$ is the coefficient matrix, $b$ is the right-hand side vector, and $x$ is the solution vector. The goal is to find the solution vector $x$ that satisfies the equation.
Iterative methods are algorithms that generate a sequence of approximate solutions $x^k$ that converge to the exact solution $x$. The starting point for these methods is usually an initial guess for the solution vector $x^0$.
In your statement, you suggest two possible choices for the initial guess:
- $x^0=(0,0,\ldots,0)^t$: This is a common choice for the initial guess, especially if the matrix $A$ is sparse (i.e., has many zeros). In this case, the initial guess assumes that the solution vector is entirely composed of zeros.
- $x^0=b$: This choice assumes that the solution vector is equal to the right-hand side vector $b$. This may be a good choice if the matrix $A$ has a dominant diagonal, meaning that the diagonal elements are much larger than the off-diagonal elements. In this case, the solution vector is approximately proportional to the right-hand side vector, and setting the initial guess to $x^0=b$ can accelerate the convergence of the iterative method.
In either case, the right-hand side vector $b$ is set to $b=h^2(1,1,\ldots,1)^t$, where $h$ is a scalar constant. This choice of $b$ is arbitrary and depends on the specific problem being solved.
MY-ASSIGNMENTEXPERT™可以为您提供UNIVERSITY OF ILLINOIS URBANA-CHAMPAIGN MATH2940 linear algebra线性代数课程的代写代考和辅导服务！ 请认准MY-ASSIGNMENTEXPERT™. MY-ASSIGNMENTEXPERT™为您的留学生涯保驾护航。
