📜  门| GATE-CS-2007 |问题 28

📅  最后修改于: 2021-09-26 04:07:10             🧑  作者: Mango

考虑从 Newton-Raphson 方法获得的系列 X n+1 = X n /2 + 9/(8 X n ), X 0 = 0.5。级数收敛到
(一) 1.5
(B)方格(2)
(三) 1.6
(四) 1.4答案:(一)
解释:

As per Newton Rapson's Method, 

Xn+1  = Xn − f(Xn)/f′(Xn)

Here above equation is given in the below form

Xn+1 = Xn/2 + 9/(8 Xn)

Let us try to convert in Newton Rapson's form by putting Xn as
first part.
Xn+1  = Xn - Xn/2 + 9/(8 Xn)
                 = Xn - (4*Xn2 - 9)/(8*Xn) 

So    f(X)  =  (4*Xn2 - 9)
 and  f'(X) =  8*Xn 

很明显 f(X) = 4X 2 − 9。我们知道它的根是±3/2 = ±1.5,但是如果我们从 X 0 = 0.5 开始,根据方程,我们在任何时候都不能得到负值,所以回答是 1.5 即选项 (A) 是正确的。
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