2
$\begingroup$

Use Newton’s method to find the solution of $$2x^3+5x-6=0$$ by using the following steps

a) Find, $f(0)$ where $$f(x)=2x^3+5x-6$$

b) Find, $f(1)$ where $$f(x)=2x^3+5x-6$$

c) Choose a guess point $$x_0$$

d) Perform the iterations, until the values stabilize to the 8-digits after the decimal sign.

my work

for a)

newton formula

$$x_{n+1} = x_n -\frac{f(x_n )}{f'(x_n )} $$ , for $n = 0,1,2,3,...$

$$f' (x)=6x^2+5$$

$$x_1=x_0-\frac{-6}{5} $$

$$ F(0)= x_0- \frac{-6}{5} $$

for b)

$$x_2=x_1-\frac{-6}{5} $$

for c)

Let $$x_0=1$$

$$x_1=1-\frac{-6}{5}=\frac{11}{5}$$

for d)

$$x_1=x_0-\frac{-6}{5}$$

$$x_2=x_1-\frac{-6}{5}$$

$$x_3=x_2-\frac{-6}{5}$$

$$x_4=x_3-\frac{-6}{5}$$

$$x_5=x_4-\frac{-6}{5}$$

is that correct ???

$\endgroup$
  • $\begingroup$ I would like to add that Newton's method does not, in general, give the roots of a polynomial. It approximates it to any desired degree of accuracy (doubling in decimal accuracy roughly each iteration) but it won't find the EXACT solution. $\endgroup$ – Trogdor Jun 15 '15 at 7:03
  • $\begingroup$ The goal of (a) and (b) is to see that $f(0)<0$ and $f(1)>0$, so there is a root of the polynomial in the interval $[0,1]$. Once you know this, you can choose a point $x_0$ accordingly, (c). $\endgroup$ – hjhjhj57 Jun 15 '15 at 7:07
1
$\begingroup$

As you wrote, Newton method updates the initial guess $x_0$ according to $$x_{n+1} = x_n -\frac{f(x_n )}{f'(x_n )}$$ So, if $f(x)=2 x^3+5 x-6$, $f'(x)=6 x^2+5$, this gives, after simplification $$x_{n+1} =\frac{4 x_n^3+6}{6 x_n^2+5}$$ As AjmalW answered, choose $x_0=1$ and start repeating the iterations. You will find $x_1=\frac{10}{11}$, $x_2=\frac{11986}{13255}$ and so on until the convergence criteria is met.

$\endgroup$
0
$\begingroup$

a) You have to determine $f(0)$, that is $f(0) = 2(0)^3+5(0)-6=\cdots$.

b) You have to determine $f(1)$. See above.

c) You have to guess a number, say $x_{0}=1$, since it's close to the zero point.

d) You have to solve $f(x)=0$ using the Newton's method. Since $f'(x_{0})=6x^{2}+5$, you have $$x_{1} = x_0 -\frac{f(x_0 )}{f'(x_0 )} = 10/11$$ where $x_{0}=1$. Find also $x_{2}$, $x_{3}$ etc.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.