Re: Smoothest function passing through n points

In article <d72eg5$rhc$1@xxxxxxxxxxxxxxxxxxx>,
klewis@xxxxxxxxxxxxxxx (Keith A. Lewis) wrote:

> "XGR131" <XGR131@xxxxxxxxx> writes in article
> <1117032203.893719.167040@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> dated 25 May 2005
> 07:43:23 -0700:
> >let points x_1,y_1 x_2,y_2 x_3,y_3 .. x_n,y_n be given
> >let a=x_1 and b=x_n
> >We are sseking the function f(x) , where f(x_1)=y1 , f(x_2)=y_2 ...
> >f(x_n)=y_n and derivative of f exists on the interval (a,b)
> >
> >now if there is any other function g that also pass through all the
> >given points then f'(x)=<g'(x) on the interval (a,b)
>
> As somebody else pointed out, the average g'(x) on (a,b) must be equal to
> the average f'(x) but I will assume you're looking for the one with the
> lowest maximum absolute value. Then what you are describing is the function
> you get by connecting the dots using straight line segments.
>
> >But first : Does f always exist? if not what are the conditions for it
> >to exist
> >and for illustraion let points 0,0 1,1 2,4 be given , what is f if f
> >passes through all the given points and also if g too passes through
> >all the points 0,0 1,1, 2,4 then f'=<g'
>
> The connect-the-dots function always exists and always has the least
> variance in the first derivative, but most people wouldn't consider it
> "smooth" because f''(x) has Dirac delta functions in it.

OK, so, what's a good definition of smooth? Under what circumstances
would you say function f is smoother on interval (a, b) than function
g is? Assuming, of course, that both functions have continuous
derivatives up to the same order, or assuming that both functions are
infinitely differentiable, so the usual definition of smooth doesn't
help.

--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)
.