# Numerical iterative method, estimating error

Given iterative method: $x_{n+1}=0.7\sin x_n +5 = \phi(x_n)$ for finding solution for $x=0.7\sin x +5$, I want to estimate $|e_6|=|x_6-r|$ as good as possible, with $x_0=5$, where $r$ is exact solution. This method obviously converges, because $\phi$ is contraction mapping, so $r=\phi(r)$ is a fixed point. So, with mean value theorem:

$|e_{n+1}|=|x_{n+1}-r|=|\phi(x_n)-\phi(r)|\le \max_{c\in\mathbb{R}}|\phi'(c)|\cdot |x_n-r|$

and we have:

$|e_n|\le 0.7^n \cdot |e_0|$

But I don't know how can I estimate $|e_0|$ without a computer? I suppose there is some simple way to finish it and with clever observation $|e_0|\le 0.7$. Can anybody help?

• If you knew the error exactly, that implies you know the solution, which makes the whole exercise trivial. I think your observation of e_0 < 0.7 is a fairly good one. And cosidering the true error is about 0.66, you likely won't get much better. – Godric Seer Nov 12 '13 at 22:34
• Yes, but that $e_0\le 0.7$ was only my guess. I don't know that. I'm looking for some trick that will help me with derivation of some decent upper bound for $|e_0|$ with given $x_0$. Without use of a computer. – xan Nov 12 '13 at 22:40
• Only thing I have so far is that $|e_n|\le 0.7^n \cdot |e_0|$ but this is useless without any approximation of $|e_0|$ which can be very big for many $x_0$. – xan Nov 12 '13 at 22:41
• Ok, I think substituting $x=4.3$ into $f(x)=x-0.7\sin x -5$ will suffice. It's easy to see that $f(x)$ is negative and for $x=5$ positive, so $|e_0|\le 0.7$ in this case. – xan Nov 12 '13 at 22:47

Given some $x_0$, you are looking for a bound on the error of the root for the equation:
$0 = 0.7 \mathrm{sin}(x) - x + 5$
Note that the function is bounded above by $5.7 - x$ and below by $4.3 - x$
Each of these have a trivial root at $5.7$ and $4.3$ respectively. You know that any roots of your function then fall in $[ 4.3, 5.7]$. Therefore, for any $x_0$
$e_0 = max\{|x_0 - 5.7|,|x_0 - 4.3|\}$
• @xan : $\max\{|x_0-5.7|,|x_0-4.3|\} = 0.7 + |x_0-5|$. – Stefan Smith Nov 28 '13 at 1:01