# Math Help - Newton method

1. ## Newton method

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.
$f(x_3)=\displaystyle\frac{1296^5}{1921^5}+\display style\frac{1296}{1921}-1$
$f'(x_3)=5\displaystyle\frac{1296^4}{1921^4}$

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.

2. Since you want to find x to 5 decimal places, why not write the numbers as decimals?
You have $x_2= \frac{5}{6}= 0.833333$ (to 6 decimal places), $x_3= \frac{1296}{1921}= 0.674649$, $x_4= 1.091162$.

Do you see what is happening? The "oscillations" are getting larger and larger, not smaller! That is a warning that this sequence is NOT going to converge. I recommend choosing a new starting value. And use decimals to more than 5 decimal places. Also, I would not worry about a formula for the error. What should happen is that eventually, you will find that the first 5 decimal places of your iterations do not change. Once that happens you have the solution to 5 decimal places.

3. Ok, the formula only says when it converges, right? so I thought of using it to get the conditions to guarantee the convergence, but I don't know how to do that. I've choose to start with one, because f(0)=-1, and f(1)=1, then the root is between 0 and 1, and as f''(1)>0 and f(1)>0 I thought that it should work.

I've made a mistake with x3 :P

4. Originally Posted by Ulysses
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.
$f(x_3)=\displaystyle\frac{1296^5}{1921^5}+\display style\frac{1296}{1921}-1$
$f'(x_3)=5\displaystyle\frac{1296^4}{1921^4}$
You first iterate is OK but then you go wrong. I have:

Code:
                      1
0.8333333333333
0.7643821156601
0.7550248672318
0.7548777017701
0.7548776662467
0.7548776662467
0.7548776662467
0.7548776662467
You need to have a systematic method of doing the calculation, you should lay out a table with columns for x_n, f(x_n), f'(x_n) and x_{n+1} and fill these in as you go, and you should do this with decimals, this is an approximation method and not the place for exact arithmetic.

You may find it worth while using a spreadsheet to do this.

CB