Using Newton's Method to Find f(x)=x^5+x-1's Zeros to 5 Decimal Places

[Telemachus]

Homework Statement

Hi there. I'm having some trouble with this exercise, which says: Use the Newton's method for determining the zeros of f(x) to 5 decimal places for f(x)=x^5+x-1.

So f'(x)=5x^4+1

I thought of iterating till I get x_n-x_{n+1}<0.000001. The thing is that I get to a point where the calculus become too difficult because of the numbers dimensions.

I've started with x=1
x_1=1\Rightarrow{x_2=1-\displaystyle\frac{1}{6}}=\displaystyle\frac{5}6{}

x_3=\displaystyle\frac{5}{6}-\displaystyle\frac{1829}{11526}=\displaystyle\frac{1296}{1921}

From here it becomes to difficult to continue.
f(x_3)=\displaystyle\frac{1296^5}{1921^5}+\displaystyle\frac{1296}{1921}-1
f'(x_3)=5\displaystyle\frac{1296^4}{1921^4}

And I still too far from the five decimal places that it asks me for.

I have this formula too, which I don't know how to use

It says:

Being f:[a,b]\longrightarrow{\mathbb{R}} a function two times derivable on the compact interval [a,b] such that 1º) f(a) and f(b) have different sign; 2º) exists k_1<0 such that |f'(x)|\geq{k_1} for all x\in{I}; and 3º) exists k_2\in{\mathbb{R}} such that |f''(x)|\leq{k_2} for all x\in{I}. Then it verifies:

1.º In ]a,b[ there is only one root r of the equation f(x)=0
2.º If r is enclosed on an interval [r-\delta,r+\delta]\subset{[a,b]} with \delta<2(k_1/k_2) and if we take x_1 on the interval, the succession x_n defined by recurrence by

x_{n+1}=x_n-\displaystyle\frac{f(x_n)}{f'(x_n)}, (n\in{\mathbb{N}}) (Newton iteration)

converges to the root r, and it verifies

|x_{n+1}-r|<\displaystyle\frac{k_2}{2k_1}|x_n-r|^2 y |x_{n+1}-r|<\displaystyle\frac{2k_1}{k_2}(\delta/\displaystyle\frac{2k_1}{k_2})^{2n}Can I use this last to know how many iterations I'll have to use to get the error that it asks me for?

Bye there, and thanks.

 

[vela]

Why don't you just use a calculator to evaluate each iteration? Newton's method is, after all, a way to estimate roots numerically. You'll find it converges quickly to the desired accuracy. (By the way, you messed up somewhere in calculating x₃.)

 

[Telemachus]

My calculator can't do much more than helping me with the divisions and multiplications :P

Nice observation by the way, I've made one mistake with x_3

 

[vela]

That's all you need it to do.