Re: Highest precision of finite difference, given machine's epsilon

hmobahi@xxxxxxxxx wrote:

Computing derivative using finite difference is
df(x)/dx=(f(x+h)-f(x))/h . Of course, the smaller the h is the higher
precision is achieved, but if h gets too small, there is a risk of
losing precision due to machine's limitation in representing very small
numbers.

You're to be commended for the astute observation that even n e t l i b
ode codes apparently *missed*. e.g. look at their clumsy attempts to
estimate initial steps involving finite difference eqns.

Somewhere I read a rule of thumb for choosing h is as
h=|x|*sqrt(epsilon) , where epsilon is the smallest number that the
machine can represent.

Correct, although you have to be mindful of the variations applied
to higher order difference schemes. The formula is derived from a
general error model first postulated by Henrici in his classic textbook.

1. The first question: is this rule of thumb correct? I just read it in
a web page and I cannot trust it so easily.

You can trust, but you must verify, as you've done. btw, it's not a rule
of thumb, it's numerical analysis equivalent of "mc^2".

2. What if I am interested in the second derivative, i.e. (
f(x-h)-2*f(x)+f(x+h) ) / (h^2) ... now how the rule works? Should it be
h=|x|*sqrt(epsilon) or (h^2)=|x|*sqrt(epsilon) ?

Actually, it's neither. If it was, then the former squared would be
correct, that is, one formula, one h. For actual h in this case take a
look at, http://www.uc.edu/sashtml/ormp/chap5/sect28.htm

--
sdx
http://www.sdynamix.com

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