User:SigmaJargon/Math5610Assignment4

1.) Spline vs. Cubic Hermite

a.) Data:

$s(0)=1$

$s'(0)=1$

$s(1)=2$

$s'(1)=y$

$s(2)=2$

$s'(2)=0$

The coefficients of the piecewise cubic function on the first interval (0-1) are found by solving:

${\begin{bmatrix}0&0&0&1\\0&0&1&0\\1&1&1&1\\3&2&1&0\\\end{bmatrix}}{\begin{bmatrix}a\\b\\c\\d\\\end{bmatrix}}={\begin{bmatrix}1\\1\\2\\y\\\end{bmatrix}}$

And on the second interval (1-2):

${\begin{bmatrix}1&1&1&1\\3&2&1&0\\8&4&2&1\\12&4&1&0\\\end{bmatrix}}{\begin{bmatrix}a\\b\\c\\d\\\end{bmatrix}}={\begin{bmatrix}2\\y\\2\\0\\\end{bmatrix}}$

b.) This becomes a cubic spline if the second derivative is continuous. Since we only have two intervals, this will hold if the second derivative for both equations is equal at 1. That is, if:

$6(y-1)-2(y-1)=6y-10y$

$4y-4=-4y$

$8y=4$

$y=1/2$

2.) More on Cubic Splines

Data:

$(x_{1},y_{1})=(1,1)$

$(x_{2},y_{2})=(2,8)$

$(x_{3},y_{3})=(3,27)$

$(x_{4},y_{4})=(4,64)$

a.) I'm not sure I should even bother solving the system of equations to find the coefficients - it's pretty obvious that $x^{3}$ is what we're looking for.

b.)

${\begin{bmatrix}1&1&1&1&0&0&0&0&0&0&0&0\\8&4&2&1&0&0&0&0&0&0&0&0\\0&0&0&0&8&4&2&1&0&0&0&0\\0&0&0&0&27&9&3&1&0&0&0&0\\0&0&0&0&0&0&0&0&27&9&3&1\\0&0&0&0&0&0&0&0&64&16&4&1\\12&4&1&0&-12&-4&-1&0&0&0&0&0\\0&0&0&0&27&6&1&0&-27&-6&-1&0\\12&2&0&0&-12&-2&0&0&0&0&0&0\\0&0&0&0&12&2&0&0&-18&-2&0&0\\6&2&0&0&0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0&24&2&0&0\\\end{bmatrix}}$

${\begin{bmatrix}a_{1}\\b_{1}\\c_{1}\\d_{1}\\a_{2}\\b_{2}\\c_{2}\\d_{2}\\a_{3}\\b_{3}\\c_{3}\\d_{3}\\\end{bmatrix}}={\begin{bmatrix}1\\8\\8\\27\\27\\64\\0\\0\\0\\0\\0\\0\\\end{bmatrix}}$