Use Fixed Point iteration to find the root.
f(x)=x^4-x+0.2 near 1 , x0=1

Question

Use Fixed Point iteration to find the root.
f(x)=x^4-x+0.2     near 1 , x0=1

anonymous · Answer

well we want to solve $$x^4-x+0.2=0$$we must isolate one the $x$'s in that expression

anonymous · Answer

$$x^4-x+0.2=0$$$$x=x^4+0.2$$or$$x=\sqrt[4]{x-0.2}$$which one will converge? what do u think?

anonymous · Answer

i should take derivative to check ??

anonymous · Answer

emm... just plug some values and see what happens u have 2 equations

1
$$x_{n+1}=x_n^4+0.2 \ \ \ \ \ x_0=1$$
2
$$x_{n+1}=\sqrt[4]{x_n-0.2} \ \ \ \ x_0=1$$

anonymous · Answer

second one will converge i think.

anonymous · Answer

Yup...i think second one is better. lets check it

anonymous · Answer

n=0$$x_{1}=\sqrt[4]{x_0-0.2}=\sqrt[4]{1-0.2}=0.9457 $$n=1$$x_{2}=\sqrt[4]{x_1-0.2}=\sqrt[4]{0.9457-0.2}=0.9292 $$n=2$$x_{3}=\sqrt[4]{x_2-0.2}=\sqrt[4]{0.9292-0.2}=0.924 $$ oh converged very soon

anonymous · Answer

thanks :)

anonymous · Answer

welcome :)

anonymous · Answer

lol...u playin football there?

anonymous · Answer

yup :P

anonymous · Answer

messi or ronaldo?

anonymous · Answer

...messi is the best ....haha...

anonymous · Answer

yeah he is :).
can you help me with another problem ?

anonymous · Answer

np :)