User:Egm6341.s10.team3.heejun/Meeting 8 transcript

 '''MTG 8: Thu. 14 Jan 10'''

[[media: Egm6341.s10.mtg8.djvu | Page 8-1]] 

'''Trap. rule (simple)'''    fig.   [[media:Egm6341.s10.mtg7.djvu|p.7-1]] * as a particular case of Newton Cotes * approximated by function of straight line $$ \displaystyle \Rightarrow $$ Two points and One interval


 * {| style="width:100%" border="0" align="left"


 * $$ \displaystyle [a,b],\,\,\,\, x_{0}=a\,\, and\,\, x_{1}=b\,\,\,\, when\,\, n=1 $$
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"

&= \sum_{i=0}^{1}l_{i}(x)f(x_{i}) \\ &= l_{0}(x)\underbrace{f(x_{0})}_{\color{blue}{x_{0}=a}} + l_{1}(x)\underbrace{f(x_{1})}_{\color{blue}{x_{1}=b}} \\\end{align}\ $$
 * $$ \begin{align} \displaystyle f_{1}(x) &= p_{1}(x) \\
 * $$ \begin{align} \displaystyle f_{1}(x) &= p_{1}(x) \\
 * }
 * }

If recall the formula of Lagrange Interpolation, then $$ l_{0}\,\, and\,\, l_{1}$$ will be
 * {| style="width:100%" border="0" align="left"

&= 1\,\,\,\, for\,\, \color{blue}{x=x_{0}} \\ & \color{blue}{or}\\ &= 0\,\,\,\, for\,\, \color{blue}{x=x_{1}} \\\end{align}\ $$
 * $$ \begin{align} \displaystyle l_{0} &= \frac{x-x_{1}}{x_{0}-x_{1}} \\
 * $$ \begin{align} \displaystyle l_{0} &= \frac{x-x_{1}}{x_{0}-x_{1}} \\
 * }
 * }


 * {| style="width:100%" border="0" align="left"

&= 0\,\,\,\, for\,\, \color{blue}{x=x_{0}} \\ & \color{blue}{or}\\ &= 1\,\,\,\, for\,\, \color{blue}{x=x_{1}} \\\end{align}\ $$
 * $$ \begin{align} \displaystyle l_{1} &= \frac{x-x_{0}}{x_{1}-x_{0}} \\
 * $$ \begin{align} \displaystyle l_{1} &= \frac{x-x_{0}}{x_{1}-x_{0}} \\
 * }
 * }

* Ref. for Lagrange Interpolation:    (2)   [[media:Egm6341.s10.mtg7.djvu|p.7-3]]



[[media: Egm6341.s10.mtg8.djvu | Page 8-2]] 


 * {| style="width:100%" border="0" align="left"

&= 1\,\,\,\, for\,\, \color{blue}{i=j} \\ & \color{blue}{or}\\ &= 0\,\,\,\, for\,\, \color{blue}{i \neq j} \\\end{align}\ $$
 * $$ \begin{align} \displaystyle l_{i}(x_{j}) &= \underbrace{\delta_{ij}}_{\color{blue}{\delta = Kronecker\,\, delta}} \\
 * $$ \begin{align} \displaystyle l_{i}(x_{j}) &= \underbrace{\delta_{ij}}_{\color{blue}{\delta = Kronecker\,\, delta}} \\
 *   (1)
 * }.
 * }.

In this case, $$ \displaystyle l_{0}\,\, and\,\, l_{1}\,\, $$ are linear functions.

Therefore, $$ \displaystyle p_{1}(x) $$ is a linear combination of $$ \displaystyle l_{0}\,\, and\,\, l_{1}\,\, $$, and must be linear.

Integration of $$ \displaystyle f(x): $$


 * {| style="width:100%" border="0" align="left"


 * $$ \displaystyle I= \int_{a}^{b}f(x)dx \cong I_{1} = \int_{a}^{b}\color{blue}{p_{1}} \color{black}{(x)dx} = (\int_{a}^{b}l_{0}(x)dx)f(x_{0}) + (\int_{a}^{b}l_{1}(x)dx)f(x_{1}) $$
 *   (2)
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"


 * * p = polynomial and 1 = 1st order (linear)
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"

HW:
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style="width:100%; padding:10px; border:2px solid #8888aa" |

Use   (2)  $$\displaystyle \int\limits_{a}^{b} \sum_{i=0}^{n=1} l_i(x)f(x_i)\, dx $$ to obtain    (1)  $$\displaystyle I=\dfrac{b-a}{2}(f(a)+f(b)) $$     [[media:Egm6341.s10.mtg7.djvu|p.7-1]]
 * style = |
 * }.
 * }.

Simpson's rule (simple) * approximated by function of parabola (2nd order polynomial)

$$ \displaystyle \Rightarrow $$ Three points and Two intervals


 * {| style="width:100%" border="0" align="left"


 * $$ \displaystyle [a,b],\,\,\,\, x_{0}=a,\,\, x_{1}= \frac{a+b}{2}\,\, and\,\, x_{2}=b $$
 * }
 * }
 * }

[[media: Egm6341.s10.mtg8.djvu | Page 8-3]] 

How many methods do we have to get $$ \displaystyle f_{2}(x) $$ ?

Basically, 2 methods.

Method 1:


 * {| style="width:100%" border="0" align="left"


 * $$ \displaystyle f_{2}(x) = p_{2} (x) = c_{2}x^{2} + c_{1}x^{1} + c_{0} $$
 *   (1)
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"


 * $$ \displaystyle c_{0},\,\, c_{1}\,\, and\,\, c_{2} = 3\,\, unknowns $$
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"


 * $$ \displaystyle p_{2}(x_{i}) = f_{x_{i}}\,\,\,\, where\,\, i=0,1\,\, and\,\, 2 $$
 *   (2)
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"


 * $$ \displaystyle 3\,\, equations\,\, to\,\, slove\,\, for\,\, 3\,\, unknowns,\,\, c_{0},\,\, c_{1}\,\, and\,\, c_{2} $$
 * }.
 * }.
 * }.

Method 2:

Use the Lagrange polynomial,   (1)  and    (2)     [[media:Egm6341.s10.mtg7.djvu|p.7-3]], to get $$ \displaystyle p_{2}(x) $$


 * {| style="width:100%" border="0" align="left"


 * $$ \displaystyle p_{2}(x) = \sum_{i=0}^{n=2} l_{i}(x)f(x_{i}) $$
 *   (3)
 * }
 * }
 * }

Equavalent of Method 1 and 2:


 * {| style="width:100%" border="0" align="left"


 * $$ \displaystyle p_{2}(x_{j}) = \sum_{i=0}^{n=2} \underbrace{l_{i}(x_{j})}_{\color{blue}{\delta _{ij}}} f(x_{i})=f(x_{j})\,\,\,\, where\,\, j=0,1 \,\, and\,\, 2 $$
 *   (4)
 * }.
 * }.
 * }.

*    (4)  $$ \displaystyle \equiv $$    (3)


 * {| style="width:100%" border="0" align="left"

HW: Expand   (4)      p.8-3  to obtain $$ P_2 (x_j)= \sum_{i=0}^2  l_i(x_j) f(x_i) = f(x_j)$$
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }.
 * }.


 * {| style="width:100%" border="0" align="left"


 * $$ \displaystyle \underbrace{l_{0}(x)}_{\color{blue}{i=0}} = \prod_{j=0, j \neq i=0}^{n=2} \frac{x-x_{j}}{x_{0}-x_{j}} = \frac{(x-x_{1})(x-x_{2})}{(x_{0}-x_{1})(x_{0}-x_{2})} $$
 *   (4)
 * }.
 * }.
 * }.

[[media: Egm6341.s10.mtg8.djvu | Page 8-4]] 

It can be verified that:

$$ \displaystyle l_{0}(x_{0})=1, \,\, l_{0}(x_{1})=l_{0}(x_{2})=0, \,\, l_{i}(x_{j})= \,\, \delta_{ij} \,\, \,\, where\,\, i,j = 0,1 \,\, and \,\, 2 $$