When the value is taken in what range, the iterative method takes the value and the convergence orde

Here's Newton

Method is the method of finding the root of the equation f(x)=0.

Use an iterative approach. Through a certain iterative formula, x(k+1)=g(xk) is obtained, if ek |xk－x*|Thereinto.

x* is the root of f(x) 0. ek is a measure of the distance between the iterative sequence and the true solution, and ek 0 indicates that the true solution has been obtained.

It can be proved that f(x) satisfies certain conditions, then the quadratic convergence to x*, roughly speaking.

ek is about e(k 1) 2, which is a method that converges quickly.

Because you think, e.g. e1=, then e2 is about, e3 is about 10 (4), e4 is about 10 (8), e5 is about 10 (16), and it only takes a few iterations to get a significant number of bits of the solution is about.

16-bit approximate solution, convergence is very fast.

Of course, it is generally difficult to do this fast, but the Newton method is generally considered to be a very effective way to solve the roots of nonlinear equations.

When the main diagonal is strictly dominant (i.e., the absolute value of the main diagonal element is greater than the sum of the absolute values of the rest of the elements in the line), the jacobi iteratively converges, so when |a|When > 4, it must be convergent. However, it should be noted that this is a sufficient condition for convergence, not a necessary condition.

Local convergence has the following theorem.

Let it be known that f(x) = 0 has a root, and f(x) is sufficiently smooth (derivatives of each order exist and are continuous).

If f'(a) != 0 (single-weight zero), then the initial value is taken in a certain neighborhood of a, and the iterative method x[n+1] = x[n] -f(x[n]) f'(x[n]) always converges to a, and the convergence speed is at least second-order.

If f'(a) == 0 (multiple zeros), then the convergence velocity is first-order when the initial value is taken in a certain neighborhood of a.

Note g(x)=x-f(x) f'(x), where"A neighborhood"Available from |g'(x)|

k=2。Calculate that x(n+1)- 2 (xn- 2) 2xn), (x(n+1)- 2) (xn- 2) 1 (2xn), the limit is 1 (2 2).

The convergence speed is 2nd order.

Local convergence has the following theorem.

Let it be known. f(x)

There are roots. a,f(x)

Sufficiently smooth (derivatives of all orders are present and continuous).

If. f'(a)

0 (a single mu rushes high and heavy zero point), then the initial value of the brother is taken.

a.

Internal, iterative method.

x[n+1]

x[n]f(x[n])/f'(x[n])

The resulting sequence.

x[n] total convergence gauge to.

a, and the convergence speed is at least second-order.

If. f'(a)

0 (multiple zeros), then the initial value is taken in.

a, the convergence velocity is of the first order.

Remember. g(x)=x-f(x)/f'(x), where"A neighborhood"Available from |

g'(x)|

Wrong. There are two types of series convergence, conditional convergence and absolute convergence of aluminium hail.

A convergent series, if its absolute value.

If the series also converges, then we call it the series of absolute convergence, otherwise, we call it the series of conditional convergence and laughter.

So absolute convergence is just a subset of convergence.

Example: Consider the series (sigma) n from 1 to positive infinity.

1)^(n-1)]/n^a]

a is a constant. When a<2 is conditional convergence.

When a>2 is absolute convergence.

Convert cosx with sinx at 0 point Taylor.

Because just consider the infinitesimal amount of the same order as x 5.

According to the original formula, we only need to guess Dan to make the following approximation:

cosx=1-x 2 2+x 4 24+o(x 4)sinx=x-x 3 6+x 5 120+o(x 5) substitution into the original Kaizhaofu, we have.

f(x)=x-[a+b(1-x^2/2+x^4/24+o(x^4))]x-x^3/6+x^5/120+o(x^5)]

1-a-b)*x - a 6 + 2b squint 3)*x 3 + a 120 + 2b 15)* x 5 + o(x 5).

So, 1-a-b=0, a 6 + 2b 3=0, a 120 + 2b 15≠0

solution, a=4 3, b=-1 3

The landlord's question should not be the landlord's own question, it should be your teacher's statement, right?

A few days ago, I was puzzled to see exactly the same statement, why did such a question arise?

1. Constant series, literally translated, is a constant series.

A constant series with only two outcomes, either convergent or non-convergent.

nonconvergent，or inconvergent；Non-convergence does not necessarily diverge divergent.

2. To calculate the sum of the series of several terms, it is generally to construct a function and then find the sum function.

For the sum function, it corresponds to an infinite number of points, and the number of terms in question is only a series.

is one of them.

The sentence "to meet the conditions for each fix" in the landlord's question seems to be incomprehensible.

I probably want to ask if this number series belongs to a fixed one.

Dot? That is, one of the infinite points in the branches of the convergence domain?

If you want to ask this question, the answer is: yes, you're right !

If the landlord's question is not what it means, please follow up and answer all questions.

It's not a question of whether it's right or not, it's a question that can't be understood by a rough question.

1. What is "sufficient conditions for each fixed order"?

2. The series of several terms can be constructed to obtain its convergence domain, and then substituted.

special x to get the sum of the series of terms;

Whether to converge or not, generally use the comparison test = comparison method.

Can the landlord add to the whole question? Zhen Heng looked confused, and didn't understand the real meaning of the landlord.

Looking forward to the landlord's addition and questioning.