Re: solving the equation [ spline way ]

Nimo wrote:
On Jan 31, 10:26 pm, "Greg Neill" <gneil...@xxxxxxxxxxxxxxxx> wrote:
Nimo wrote:
On Jan 31, 9:03 pm, "Greg Neill" <gneil...@xxxxxxxxxxxxxxxx> wrote:
Yes. But each spline s_i incorporates:
1. Information about the location of its endpoints
2. Information about the slope of the fitted curve
at its endpoints (which means information about
the points beyond that lie beyond its end points).

That information is encoded in the cubic equation of
each individual spline; a*x^3 + b*x^2 +c*x + d.
Four parameters for each spline, multiplied by the
number of splines, gives 4*(n-1) individual parameters
to fit your n data points. How does that jibe with your
idea of "complexity" for a solution?

so, what exactly is your point here ?

I too saw about 'the degree of freedom of the curve' from wiki
article, but the article GOES on saying that, we can very
easily over come that 'difficulty' by taking some
continuous differentiation values and etc,etc.....?

well, I didn't understand it clearly.
so I'm asking your clarification for that
http://en.wikipedia.org/wiki/Spline_interpolation
see sub-category 'cubic-spline'
Waiting..?

http://mathworld.wolfram.com/CubicSpline.html

Work out an example by hand to see what it's all about.

At the end T(S) is not a single cubic curve,
but a set of cubic curves.

Am I right ? in that sense

That's what I've been trying to point out...


If that is the case, we only over come
the difficulty of higher polynomial,
at lease it's better than having nothing for me.

It seems to me that, since you are not worried about the function
values at intermediate points (like F(1.5)), then you don't need
the complications of cubic splines. Linear splines would suffice.

Define an elementary function D(x) as follows:

D(x) = (0 if x<0; 1 otherwise) (The unit step function)

If d_i is the difference between the point values:

d_i = y_i - y_i-1 ; where y_0 = 0 by definition

Then you can write your function as:

F(x) = d_1*D(x - x_1) + d_2*D(x - x_2) + d_3*D(x - x_3) + .... d_n*D(x -
x_n-1)

So for your example,

F(x) = 2*D(x - 0) + 4*D(x - 1) + 5*D(x - 2) + 6*D(x - 3) + 8*D(x - 4)

F(x) will return your point values for each value of x.

You can also construct the derivative function which, when integrated,
will yield your point values:

f(x) = 2*D(x - 0) + 2*D(x - 1) + D(x - 2) + D(x - 3) + 6*D(x - 4)

so that F(x) = integral[0,x; f(x)*dx]


.

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