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.)

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

which yields the polynomial:

$P(x)={\begin{cases}2x^{3}-6x^{2}+11x-6,&if\ x\leq 2\\2x^{3}-6x^{2}+11x-6,&if\ 2<x\leq 3\\-4x^{3}+48x^{2}-151x+156,&if\ 3<x\end{cases}}$

c.)

5.) Runge Interpolation and 6.) Judicious Interpolation

Uploading the 40 images for these two questions would be a headache and a chore. Therefore I provide 4 images - the interpolations at n=10 and n=20. If you'd like to generate more, then you can use these Maple worksheets I created:

Runge Interpolation

Judicious Interpolation

10th degree Runge

File:Runge10.jpg

20th degree Runge

File:Runge20.jpg

10th degree Judicious

File:Judicious10.jpg

20th degree Judicious

File:Judicious20.jpg

9.) Uniqueness of Interpolating Polynomial

a.) Power Form

${\begin{bmatrix}1&1&1&1\\8&4&2&1\\64&16&4&1\\512&64&8&1\\\end{bmatrix}}{\begin{bmatrix}a\\b\\c\\d\\\end{bmatrix}}={\begin{bmatrix}1\\2\\3\\4\\\end{bmatrix}}$

This yield $a=1/56$, $b=-7/24$, $c=7/4$, and $d=-10/21$. So $P(x)=x^{3}/56-7x^{2}/24+7x/4-10/21$.

b.) Lagrange Form

$L_{1}={\frac {(x-2)(x-4)(x-8)}{(1-2)(1-4)(1-8)}}$

$L_{2}={\frac {(x-1)(x-4)(x-8)}{(2-1)(2-4)(2-8)}}$

$L_{3}={\frac {(x-1)(x-2)(x-8)}{(4-1)(4-2)(4-8)}}$

$L_{4}={\frac {(x-1)(x-2)(x-4)}{(8-1)(8-2)(8-4)}}$

So $P(x)=1*L_{1}+2*L_{2}+3*L_{3}+4*L_{4}\,\!$. Substituting and simplifying everything down leaves us with: $P(x)=7/4*x-7/24*x^{2}-10/21+1/56*x^{3}$

c.) Newton Form

$P(x)=b_{0}+b_{1}(x-x_{0})+b_{2}(x-x_{0})(x-x_{1})+b_{3}(x-x_{0})(x-x_{1})(x-x_{2})$

$P(x_{0})=b_{0}$

$P(x_{1})=b_{0}+b_{1}(x_{1}-x_{0})$

$P(x_{2})=b_{0}+b_{1}(x_{2}-x_{0})+b_{2}(x_{2}-x_{0})(x_{2}-x_{1})$

$P(x_{3})=b_{0}+b_{1}(x_{3}-x_{0})+b_{2}(x_{3}-x_{0})(x_{3}-x_{1})+b_{3}(x_{3}-x_{0})(x_{3}-x_{1})(x_{3}-x_{2})$

$1=b_{0}$

$2=b_{0}+b_{1}(2-1)=1+b_{1}$

$1=b_{1}$

$3=b_{0}+b_{1}(4-1)+b_{2}(4-1)(4-2)=1+1(4-1)+b_{2}(4-1)(4-2)=4+6b_{2}$

$-1/6=b_{2}$

$4=b_{0}+b_{1}(8-1)+b_{2}(8-1)(8-2)+b_{3}(8-1)(8-2)(8-4)=1+7-1+168b_{3}$

$1/56=b_{3}$

So, $P(x)=1+(x-1)-(x-1)(x-2)/6+(x-1)(x-2)(x-4)/56$. If we expand all the terms and then simplify, we arrive at: $P(x)=7/4*x-7/24*x^{2}-10/21+1/56*x^{3}$

10.) The Method of Undetermined Coefficients

We expect that since we have four points, we'll at least have accuracy up to cubic functions. We then want to check if we can find values for $a_{0},\ a_{1},\ a_{2},\ a_{3}$ such that the we have exactness for quartic polynomials.

I claim that there are no $a_{0},\ a_{1},\ a_{2},\ a_{3}$ that fulfill the conditions. To show this, consider this counter-example:

Let $a=0,\ h=1$. In order to find $a_{0},\ a_{1},\ a_{2},\ a_{3}$, let's throw four specific functions at the equality, and see what $a_{0},\ a_{1},\ a_{2},\ a_{3}$ turn up.

$f_{0}(x)=x^{4}\,\!$

$f_{0}(x)=x^{4}+1\,\!$

$f_{0}(x)=x^{4}+2\,\!$

$f_{0}(x)=x^{4}+3\,\!$

$\int _{0}^{1}f(x)\,dy=a_{0}f(0)+a_{1}f(1/3)+a_{2}f(1/2)+a_{3}f(1)$. Plugging in the four functions above yields:

$1/5=0a_{0}+a_{1}/81+a_{2}/16+1a_{3}\,\!$

$6/5=1a_{0}+82a_{1}/81+17a_{2}/16+2a_{3}\,\!$

$11/5=2a_{0}+163a_{1}/81+33a_{2}/16+3a_{3}\,\!$

$16/5=3a_{0}+244a_{1}/81+49a_{2}/16+4a_{3}\,\!$

Or, in matrix form:

${\begin{bmatrix}0&1/81&1/16&1\\1&82/81&17/16&2\\2&163/81&33/16&3\\3&244/81&49/16&4\\\end{bmatrix}}{\begin{bmatrix}a\\b\\c\\d\\\end{bmatrix}}={\begin{bmatrix}1/5\\6/5\\11/5\\16/5\\\end{bmatrix}}$

${\begin{bmatrix}a\\b\\c\\d\\\end{bmatrix}}={\begin{bmatrix}0.0750651\\-0.5958984\\1.4010417\\0.1197917\\\end{bmatrix}}$

However, if we consider a slightly different set of functions

$f_{0}(x)=x^{4}+1\,\!$

$f_{0}(x)=x^{4}+2\,\!$

$f_{0}(x)=x^{4}+3\,\!$

$f_{0}(x)=x^{4}+5\,\!$

And try to find $a_{0},\ a_{1},\ a_{2},\ a_{3}$, we get:

${\begin{bmatrix}a\\b\\c\\d\\\end{bmatrix}}={\begin{bmatrix}0.7980469\\1.1707031\\-1.2312500\\0.2625000\\\end{bmatrix}}$

Which is different! So we can't find a solution for $a_{0},\ a_{1},\ a_{2},\ a_{3}$ that works for all quartic functions.

${\displaystyle }$