A somewhat different Newton solver (HP35s)

[Thomas Klemm]

(08-07-2016 08:47 PM)Dieter Wrote:  What about limitations of this method? As far as I get it both f(x) as well as f'(x) are only approximate. I wonder how accurate especially f(x) is evaluated using this method. Maybe the math experts here can say more about this. ;-)

Using the Taylor series we can separate the real and imaginary part of \(f(x+ih)\):

\(\begin{align*}
\Re[f(x+ih)]&=f(x)-\frac{f''(x)}{2!}h^2+O(h^4)\\
\Im[f(x+ih)]&=h(f'(x)-\frac{f'''(x)}{3!}h^2+O(h^4))
\end{align*}\)

Thus by choosing a small \(h\) we can make the difference as small as we want.
You can verify this by using \(h=10^{-50}\) and the \(\sin\) function:

sin(2 + 1e-50 i) = 9.09297426826e-1 - 4.16146836547e-51 i

sin(2) = 9.09297426826e-1
cos(2) = -4.16146836547e-1

Both values are exact. But for a calculator with 12 significant digits something smaller like \(10^{-10}\) might be enough. It really depends on the value of the 2nd and 3rd derivative at the root.

We could go further and try to use \(h=10^{-99}\). But then we may end up with 0.
Thus we have to choose \(h\) small enough to avoid errors but not too small.

Kind regards
Thomas