User:Egm6341.s10.Team4.roni/HW4/4.4

= Problem 4: Proof of Higher Order Error for Trap. rule Theorem =

Given
Higher Order Error for Trap. rule equation (1) on slide [[media:Egm6341.s10.mtg21.djvu|21-1]], which is:


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle Eq. (1) \ E_n^1= \sum_{k=0}^{n-1}[ \int_{x_k}^{x_{k+1}}f(x)dx- \frac{h}{2}(f_k(x_k)+f(x_{k+1}))]
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Show
That eq. (1) can be transformed to the following: equation (5) on slide [[media:Egm6341.s10.mtg21.djvu|21-2]]


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle E_n^1=\dfrac{h}{2} \sum_{k=0}^{n-1}[ \int_{-1}^{1}g_k(t)dt-(g_k(-1)+g_k(+1))]

where, equation (2) on slide [[media:Egm6341.s10.mtg21.djvu|21-1]]

h= \frac{(b-a)}{n} \ \ \ x_i=a+ih $$ and, equation (3) on slide [[media:Egm6341.s10.mtg21.djvu|21-1]]

x(t)= t*\frac{h}{2} + \frac{x_i+x_{i+1}}{2} \ \ \ t\in[-1,1] $$
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Solution
Defining: equation (1) on slide [[media:Egm6341.s10.mtg21.djvu|21-2]]

g_i(t):=f(x(t)) \ where \ x\in[x_i,x_{i+1}] $$

For one Panel, we have:



x_i=-1 \ \ and \ \ x_{i+1}=1 $$



f(x)_i=g(-1) \ \ \ f(x)_{i+1}=g(1) $$

Also,



\frac{dx(t)}{dt}=\frac{h}{2} $$



\Rightarrow E_n^1= \sum_{k=0}^{n-1} \int_{-1}^{1}g_k(t)\dfrac{h}{2}dt-\dfrac{h}{2}(g_k(-1)+g_k(+1)) $$

Giving,


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$$\displaystyle E_n^1=\dfrac{h}{2} \sum_{k=0}^{n-1}[ \int_{-1}^{1}g_k(t)dt-(g_k(-1)+g_k(+1))] $$
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