Taylor's coefficients keep being zero, but the function is not zero, why?

[Golabi Doon]

Hello everyone,

Consider f(x)=exp(-1/x^2)/(erf(1/x)*x) from R+ to R, where R+ is
defined as R+={x|x in R and x>0} (note that x>0 is strict). I need to
approximate f by a polynomial such that the approximation is most
accurate and tight around extremely small x's.

I tried to obtain Taylor's expansion coefficients, around the point
x=0, by computing the derivatives of f(x) up to 10th order, and then
finding the limit of the derivatives as x goes to 0. So far, all the
derivatives turned out to be zero so I stopped doing this for higher
order derivatives.

Is any thing is going to change if I continue this in higher orders?
If not, then how should I approximate f(x)?

Regards

Golabi

[Ray Vickson]

f is not analytic at x = 0. A plot of f(x) for small, positive x
reveals the problem you will encounter: f is very, very small in a
sizeable interval from zero; then it starts to grow quickly. There is
no way you can mimic such behaviour using a *polynomial*. You might be
better off setting f = 0 in some interval [0,a], then applying the
Taylor expansion to f(x) at x = a. This would make the approximation a
bit discontinuous at x = a. You could fix this by setting f(x) =~=
x*f(a)/a for 0 <= x <= a, then letting f(x) =~= Taylor expansion of f
for x > a. That would make the approximation continuous at x = a, but
now with a discontinuous first derivative at x = a. You could do even
better, making the approximations to f and f' continuous at x = a,
etc., by using a quadratic for 0 <= x <= a.

Using Maple 11 set to 50-digit precision, here are some of the values
of f(x) for small x > 0 (rounded to 10 digits in printout):
x f(x)
0.0020 1.19810310e-108571
0.0040 9.83605742e-27142
0.0100 1.13548386e-4341
0.0200 9.17836334e-1085
0.0300 9.40707175e-482
0.0500 3.83033919e-173
0.1000 3.72007598e-43
0.1500 3.32607294e-19

R.G. Vickson

[Robert Israel]

Golabi Doon <golab...@gmail.com> writes:

No, all derivatives are 0 at 0.

> If not, then how should I approximate f(x)?

Not with a polynomial, unless you want the polynomial to be just 0.

You might try

f(x) = exp(-1/x^2)/x + exp(-2/x^2)/sqrt(Pi)*(1-1/2*x^2+3/4*x^4-15/8*x^6+...)
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

[Axel Vogt]

Ray Vickson wrote:
> On Jan 6, 12:05 pm, Golabi Doon <golabid...@gmail.com> wrote:
>> Hello everyone,
>>

f(x)=exp(-1/x^2)/(erf(1/x)*x) from R+ to R, where R+ is
>> defined as R+={x|x in R and x>0} (note that x>0 is strict).

I think up to x=1/t and the multiplication by x it is s.th.
like the inverse of Mills ratio or a survival probability,
so x ~ 0 is asking for the latter with t ~ infinity.

May be the OP wants to say, why he needs it and what he is
going to do with it (being not aware, that rational*exp may
be good enough for his purpose as well - just guessing).

[Golabi Doon]

Thank you for all the responses. So now I am sure polynomial is not an
option for the approximation. I probably just use what Robert
suggested, although it still has exp, at least it gets rid of the less
elementary function erf !

Thanks

Golabi