User:Egm6341.s10.Team4.roni/HW4/4.5

= Problem 5: Theorem of Higher Order Error for Trap. rule Step1 =

Given
Higher Order Error for Trap. rule equation (5) on slide [[media:Egm6341.s10.mtg21.djvu|21-1]], which is:


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle Eq. (5)\ 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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Find
That using eq. (1) on slide [[media:Egm6341.s10.mtg21.djvu|21-2]],


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle g_k(t):=f(x(t)) \  such \  that \  x \   \epsilon  \left[ x_k,x_{k+1}\right]

For one panel Step 1 :



Eq.(2) \int_{-1}^{1}(-t) {g}^{(1)}(t)dt = \int_{-1}^{1}g_k(t)dt-[(g_k(-1)+g_k(+1))] $$


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Solution
Using integration by Parts :



\int u {v}^{'}= uv - \int v {u}^{'}\ \ where \ \ u=t \ \ {v}^{'}=g(t) $$

Writing the left side of Eq. (2):



\int_{-1}^{1}(-t) {g}^{(1)}(t)dt = -\int_{-1}^{1}(t) {g}^{(1)}(t)dt $$



= -\Bigg[(t \, g(t)) \Bigg|^{1}_{-1} -\int_{-1}^{1}g(t)dt \Bigg] $$



= -(1*g(1)-(-1)g(-1)-\int_{-1}^{1}g(t)dt) $$



= \int_{-1}^{1}g(t)dt)-(g(-1)+g(1)) $$


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$$\displaystyle \int_{-1}^{1}(-t) {g}^{(1)}(t)dt = \int_{-1}^{1}g(t)dt)-(g(-1)+g(1)) $$
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