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 $\displaystyle f(x)=x^5+x-1$.

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

I thought of iterating till I get $\displaystyle x_n-x_{n+1}<0.000001$. 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

$\displaystyle x_1=1\Rightarrow{x_2=1-\displaystyle\frac{1}{6}}=\displaystyle\frac{5}6{}$

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

From here it becomes to difficult to continue.

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

$\displaystyle f'(x_3)=5\displaystyle\frac{1296^4}{1921^4}$

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