数值方法 [英] Numerical methods

问题描述

你好我只是新来的。我只是想问你是否可以用newton-rhapson'的方法来解决这个问题，虽然规定一个是割线方法。

问题是，f（x）= sinx + cos（1 + x ^ 2）-1

其中x是弧度并且进行四次迭代，初始值为

a）Xi-1 = 1且Xi = 3


并绘制结果图。

解决方案

你好我只是新来的。我只是想问你是否可以用newton-rhapson'的方法来解决这个问题，虽然规定一个是割线方法。

问题是，f（x）= sinx + cos（1 + x ^ 2）-1

其中x是弧度并且进行四次迭代，初始值为

a）Xi-1 = 1且Xi = 3


并绘制结果图表。

如果我没记错的话，割线方法是找根; Newton / Raphson

方法近似为积分。牛顿根发现方法很简单：


首先找到f w.r.t的第一个导数f''（x）。 x并重复应用以下内容：


x（n + 1）= x（n）-delta

delta = f（x（n））/ f ''（x（n））


直到| delta | < epsilon，其中epsilon是一个小的posititve值，而x（0）是你的第一个估计的
。


亲切的问候，


Jos

如果我没记错的话，割线方法是找根; Newton / Raphson

方法近似为积分。牛顿根发现方法很简单：


首先找到f w.r.t的第一个导数f''（x）。 x并重复应用以下内容：


x（n + 1）= x（n）-delta

delta = f（x（n））/ f ''（x（n））


直到| delta | < epsilon，其中epsilon是一个小的posititve值，而x（0）是你的第一个估计的
。


亲切的问候，


Jos

i忘了我们要求找到第一个正根。什么是三角洲？你能不能告诉我如何找到衍生物？

我忘了我们要求找到第一个正根。什么是三角洲？你能不知怎的告诉我如何找到衍生物？

要找到正数，请确保输入正数作为根的近似值。


函数的导数是通过基本微积分找到的，不确定它是否属于脚本域。你必须知道区别才能使用Newton-Rhapson方法。


delta定义为f（x）与其导数f''（x）的比值，在第i个近似值处计算得出。


如果你不能区分一个函数我建议你使用割线方法或二分法。


这些算法的代码可以在 www.nr.找到。 COM

Hello im only new here. i just wanna ask if you could solve this problem in newton-rhapson''s method although the prescribe one is secant method.
the problem is, f(x)= sinx + cos(1+x^2)-1
where x is in radians and make four iterations w/ initial values of
a)Xi-1 = 1 and Xi=3

and graph the results.

解决方案

Hello im only new here. i just wanna ask if you could solve this problem in newton-rhapson''s method although the prescribe one is secant method.
the problem is, f(x)= sinx + cos(1+x^2)-1
where x is in radians and make four iterations w/ initial values of
a)Xi-1 = 1 and Xi=3

and graph the results.

If I recall correctly, the secant method is for root finding; the Newton/Raphson
method approximates an integral. The Newton root finding method is easy:

first find the first derivative f''(x) of f w.r.t. x and apply the following repeatedly:

x(n+1) = x(n)-delta
delta= f(x(n))/f''(x(n))

until |delta| < epsilon, where epsilon is a small posititve value and x(0) is your
first estimation.

kind regards,

Jos

If I recall correctly, the secant method is for root finding; the Newton/Raphson
method approximates an integral. The Newton root finding method is easy:

first find the first derivative f''(x) of f w.r.t. x and apply the following repeatedly:

x(n+1) = x(n)-delta
delta= f(x(n))/f''(x(n))

until |delta| < epsilon, where epsilon is a small posititve value and x(0) is your
first estimation.

kind regards,

Jos

i forgot we are ask to find the 1st positive root. what is delta? and can you pls somehow tell me how to find the derivative?

i forgot we are ask to find the 1st positive root. what is delta? and can you pls somehow tell me how to find the derivative?

To find a positive make sure that you input a positive number as an approximation to the root.

The derivative of the function is found by basic calculus, not sure if this falls under the domain of the scripts. You must know differentiation to use the Newton-Rhapson method.

delta as defined is the ratio of f(x) and its derivative, f '' (x), evaluated at the ith approximation to the root.

If you can''t differentiate a function I suggest you use the secant method or the bisection method.

Code for these algorithms can be found at www.nr.com

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