User:EGM6341.s11.TEAM1.Sujin.HW5

Problem 7: TheTheorem of Higher Order Error(HOTRE) for Trapezoidal rule
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Given
,where $$\begin{align} x_k=a+kh, \; \ h=(b-a)/n, \; \ x(t)= t*\frac{h}{2} + \frac{x_i+x_{i+1}}{2}, \; \ t \epsilon [-1,1] \end{align}$$ ,where $$\begin{align} g_k(t):=f(x(t)) \; \ x\in[x_k,x_{k+1}] \end{align}$$

Show eqn.7.1 is equal to eqn.7.2
from eqn.7.3 we have $$\begin{align} x_k=-1, \; x_{k+1}=1 \end{align}$$ $$\begin{align} \Rightarrow f(x_k)=g(-1)\end{align}$$ $$\begin{align}, \; f(x_{k+1})=g(1)\end{align}$$ $$\begin{align}, \; \frac{dx(t)}{dt}=\frac{h}{2} \end{align}$$

Plugging above terms and eqn7.3 into eqn.7.1

Show HOTRE for Trap. rule Step 1 is true
From lecture note [[media:Egm6341.s11.mtg30.djvu|30-3]], First step of HOTRE for Trap. rule is as following; Using Integration by parts, we can rewrite LHS of eqn.7.5; $$\displaystyle\begin{align} \int u {v}^{'}= uv - \int v {u}^{'} \end{align}$$ Let $$\displaystyle\begin{align} u=t \ \ {v}^{'}=g(t) \end{align}$$

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

So, HOTRE for Trap. rule Step 1(eqn.7.5) is true

Show HOTRE for Trap. rule Step 2a is true
From lecture note [[media:Egm6341.s11.mtg30.djvu|30-3]], Step 2a of HOTRE for Trap. rule is as following; where $$\displaystyle \begin{align} \; g_k^{(i)}(t)=\frac{d^i}{dt^i} g_k^{(t)}, \;x \epsilon[x_k, x_{k+1}] \end{align}$$ From eqn.7.1 and eqn.7.2,