User:Egm6341.s10.team4.anandankala/HW6-9

Prove That

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 * $$z_{i+1}=z_i+ \dfrac {h/2}{3}[f_i+ 4f_{i+1/2}+ f_{i+1}]$$


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Solution

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 * $$z^'_{i+1/2}= z^'(s=1/2)= \dfrac {-3}{2}(z_i-z_{i+1})- \dfrac{1}{4}(z^'_i +z^'_{i+1})$$


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 * We have $$\displaystyle z^'= hz^.$$


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 * Applying collocation at $$\displaystyle t_i$$ and $$\displaystyle t_{i+1}$$, we have $$z^._i= f_i$$ and $$z^._{i+1}=f_{i+1}$$


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 * $$\displaystyle \Rightarrow z^._{i+1/2}= -\dfrac {3}{2h}(z_i-z_{i+1})- \dfrac {1}{4}(f_i+ f_{i+1}) $$


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 * $$\displaystyle \bigtriangleup := z^._{i+1/2}- f_{i+1/2} $$


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 * Applying collocation at $$\displaystyle t_{i+1/2}$$, we have $$\displaystyle \bigtriangleup= 0$$ and $$\displaystyle z^._{i+1/2}= f_{i+1/2}$$


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 * $$\displaystyle \Rightarrow f_{i+1/2}= -\dfrac {3}{2h}(z_i-z_{i+1})- \dfrac {1}{4}(f_i+ f_{i+1}) $$


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 * $$\displaystyle \Rightarrow \dfrac {1}{4}(f_i+ 4f_{i+1/2}+ f_{i+1})= -\dfrac {3}{2h}(z_i-z_{i+1})= \dfrac {3}{2h}(z_{i+1}-z_i) $$


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 * $$\displaystyle \Rightarrow z_{i+1}-z_i= \dfrac {h}{6} (f_i+ 4f_{i+1/2}+ f_{i+1}) $$


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$$
 * $$\displaystyle \Rightarrow z_{i+1}= z_i+ \underbrace{\dfrac {(h/2)}{3} (f_i+ 4f_{i+1/2}+ f_{i+1})}_{Simpson's Rule}
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