Timeline for Optimal knot placement for fitting piecewise-continuous linear functions to a nonlinear function
Current License: CC BY-SA 2.5
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 15, 2010 at 4:39 | vote | accept | Gilead | ||
| Nov 14, 2010 at 16:10 | comment | added | Tracy Hall | The value $a^2c^5/120$ comes from direct calculation: either observe the invariance and reduce to the special case $\int_0^c (ax^2/2-axc/2)^2\ dx$, or take the fully general $(a/2)x^2+a_1x+a_0$, find the line that crosses it at $x_0$ and $x_0+c$, and integrate the squared error, proving the invariance by the cancellation of all parameters other than $a$ and $c$. Since the original problem is invariant with respect to addition of linear terms, and any horizontal shift of a parabola is equivalent to adding a linear term, equispaced knots are indeed optimal for a quadratic polynomial. | |
| Nov 12, 2010 at 20:20 | comment | added | Gilead | Thanks for the very insightful answer. I have a few questions: 1) could you provide a reference for the $a^2 c^5 /120$ error, or a clue as to how it was derived? 2) For any quadratic polynomial $f(x)=a_2 x^2+a_1 x+a_0$, your integral function $I(x)=\int_L^x f^{\prime\prime}(s)^{2/5}\ ds=(2a_2)^{2/5}(x-L)$ is a linear function. If we take equally spaced horizontal strips in the range of $I(x)$, all the crossing points correspond to equally spaced knots; somehow I don't think equispaced knots are optimal for all quadratic polynomials. Have I missed something? Thanks. | |
| Nov 12, 2010 at 5:29 | history | answered | Tracy Hall | CC BY-SA 2.5 |