User:Egm6341.s10.team4.anandankala/HW5-3

Given

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$$
 * $$\displaystyle x(t_k)= \frac{1}{2} (x_k+x_{k+1})+ t_k \frac{h}{2}


 * }.
 * }.

Find

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 * Find tk in terms of x.
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 * }.

Solution

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$$
 * $$\displaystyle x(t_k)= \frac{1}{2} (x_k+x_{k+1})+ t_k \frac{h}{2}


 * }.
 * }.


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$$
 * $$\Rightarrow \displaystyle t_k \frac{h}{2}=x-\frac{1}{2} (x_k+x_{k+1})


 * }.
 * }.


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$$
 * $$\Rightarrow \displaystyle t_k =\frac{2}{h} (x-\frac{1}{2} (x_k+x_{k+1}) )


 * }.
 * }.

Substituting x=xk.


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$$
 * $$\Rightarrow \displaystyle t_k =\frac{2}{h} (x_k-\frac{1}{2} (x_k+x_{k+1}) )


 * }.
 * }.


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$$
 * $$\Rightarrow \displaystyle t_k =\frac{2}{h} (\frac{1}{2} x_k- \frac{1}{2}x_{k+1})


 * }.
 * }.


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$$
 * $$\Rightarrow \displaystyle t_k =\frac{-1}{h} (x_{k+1}-x_k)


 * }.
 * }.

We know that xk+1-xk=h.


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$$
 * $$\Rightarrow \displaystyle t_k =\frac{-1}{h} (h)


 * }.
 * }.


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$$
 * $$\Rightarrow \displaystyle t_k =-1


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 * }.

Substituting x=xk+1.


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$$
 * $$\Rightarrow \displaystyle t_k =\frac{2}{h} (x_{k+1}-\frac{1}{2} (x_k+x_{k+1}) )


 * }.
 * }.


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$$
 * $$\Rightarrow \displaystyle t_k =\frac{2}{h} (\frac{1}{2}x_{k+1}-\frac{1}{2} x_k)


 * }.
 * }.


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$$
 * $$\Rightarrow \displaystyle t_k =\frac{1}{h} (x_{k+1}-x_k)


 * }.
 * }.

We know that xk+1-xk=h.


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$$
 * $$\Rightarrow \displaystyle t_k =\frac{1}{h} (h)


 * }.
 * }.


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$$
 * $$\Rightarrow \displaystyle t_k =1


 * }.
 * }.