# Newton's method for a given polynomial

Let $f(x)=\frac{1}{5}x^5+\frac{1}{3}x^3+x-1$

• Show that $f$ has only one zero $r$ in interval $(0,1)$
• To find approximation of $r$ we apply Newton's method $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$. Show that for every $x_0\in(0,1)$ this method converges.
• Show that order of convergence is exactly quadratic

First one is trivial, but I don't know how to solve second and third. For second I was trying to show that error $|x_{n+1}-r|=e_{n+1}=e_n-\frac{f(x_n)}{f'(x_n)}$ is approaching $0$ when $n$ approaches $+\infty$ but it didn't lead me to anywhere.

Also showing the third seems very complicated. Can anybody help?

• Hi, welcome to SciComp! This is a fine question, but in the future please mark homework questions as such with a tag. I think that the consensus is that homework questions are OK as long as the asker is upfront and explicit about it.
– tel
Nov 6 '13 at 23:55
• There's a really easy proof for parts 2 and 3 using contraction mapping / fixed point theorem. Are you familiar with these?
– Paul
Nov 7 '13 at 0:01
• @Paul, yes I am. I used it couple of times before, but don't see how to use it here.
– xan
Nov 7 '13 at 0:16

Newton's Method can be interpreted as a type of fixed point iteration of the form

$$x_{n+1}=F(x_n)$$

where

$$F(x)=x-\frac{f(x)}{f'(x)}$$

Assume we know the root $r$ such that $F(r)=r$ (i.e. $r$ the answer to part 1 of your question). According to fixed point theory, if $F$ is a contractive mapping on a closed set $[a,b]$ and $F'$ exists and is continuous, then the error $e_{n+1}$ can be expressed (with the help of mean value theorem) as

\begin{eqnarray} e_{n+1}&=& x_{n+1}-r\\ &=& F(x_n)-F(r) \\ &=& F'(\xi_n)(x_n-r) \end{eqnarray}

for some $\xi_n$ between $x_n$ and $r$. Thus, to ensure convergence by contractive mapping (your question 2), we require that $|F'(x)|<1$ for all $x$ in some $[a,b]$.

To prove 2nd order convergence, notice that the error term $e_{n+1}$ can also be expressed (with the help of Taylor Series) as

\begin{eqnarray} e_{n+1}&=&x_{n+1}-r\\ &=& F(r+e_n)-F(r)\\ &=&F(r) + e_nF'(r) + \frac{1}{2!} e_n^2F''(r) + \frac{1}{3!}F'''(r)+... - F(r) \\ &=&e_nF'(r) + \frac{1}{2!} e_n^2F''(r) + \frac{1}{3!}F'''(r)+... \end{eqnarray}

This last expression tells us that the order of convergence depends on the multiplicity of the root $r$ of $F$. If $r$ has multiplicity $q$, then all the derivatives of $F$ up to the $q-1$th order (at the root $r$) are all zero and the $q$th derivative $F^{(q)}(r)\neq0$. Hence, a root $r$ of multiplicity $q$ yields a convergence of order $q$.

In your particular case, you need to show that the root $r$ of $F$ (note the CAPITAL $F$, not lower case $f$) has multiplicity $2$. That is, you'll need to show that $F'(r)=0$ and $F''(0)\neq0$

Because $f$ is a polynomial and $f'(x)\ge 1$ for all $x$, it has a unique simple real root $r$. So it can be factored $f(x) = (x-r)g(x)$, where $g(x)$ is a polynomial satisfying $g(x)>0$ for all $x$. Then if $x_n = r + e_n$,

$$f(x_n) = e_ng(x_n)\\ f'(x_n)=g(x_n)+e_n g'(x_n)$$ Plugging this in to Newton's method: $$r+e_{n+1}=r+e_n-\frac{e_n g(x_n)}{g(x_n)+e_n g'(x_n)}$$

$$e_{n+1} = e_n^2 \frac{g'(x_n)}{f'(x_n)}$$

I haven't worked out the details, but you should be able to prove quadratic convergence from this by bounding $|g'(x_n)/f'(x_n)|$.

You're on the right track for the second and third parts. You might try looking at a Taylor series expansion of $f$ around the root in $(0,1)$, where the remainder term is second order. If you can state conditions required for quadratic convergence, that should also give you insight into some of the conditions required for convergence in general.