User:Egm5526.s11.team4.zou/NM HW6

=Problem 6.4 Redo proof of HOTRE=

From the lecture slide Mtg 31-4

''' This problem was solved without referring to S10 homework. '''

Problem Statement
Redo the proof of the HOTRE by another approach: cancel all the odd order derivative of $$\displaystyle g$$, and try to find where this approach collapse.

Solution
We have known form the class that the error at the panel $$\displaystyle k$$ can be expressed as:

If we keep on this integration by part and using $$\displaystyle p(t)$$ to cancel all the odd order derivative of $$\displaystyle {{g}_{k}}(t)$$, we have,

Similar to the process in step 2ab and step 3ab on the lecture, when we use $$\displaystyle p(t)$$ to cancel all the odd order derivative of $$\displaystyle {{g}_{k}}(t)$$, i.e. using $$\displaystyle {{p}_{2r}}$$ to cancel all $$\displaystyle {{g}_{2r-1}}(t)$$ by letting $$\displaystyle {{p}_{2r}}(1)={{p}_{2r}}(-1)=0$$, we are actually ensuring that all $$\displaystyle {{p}_{2r+1}}$$ are odd functions.

Hence,

Then Eq (4-2) becomes,

And the total error is a summation of $$\displaystyle {{E}_{k}}$$,

Reason why this approach fail
Then we can see the critical weakness of the approach of cancelling odd order derivative of $$\displaystyle {{g}_{k}}(t)$$ lies in the summation $$\displaystyle \sum\limits_{k=0}^{n-1}{\left[ g_{k}^{(2r)}(1){\color{red}+}g_{k}^{(2r)}(-1) \right]}$$.

Because the sign between $$\displaystyle g_{k}^{(2r)}(1)$$ and $$\displaystyle g_{k}^{(2r)}(-1)$$ is "plus" rather than "minus" then it can ‘ t cancel terms when sum according to $$\displaystyle k$$ making the final expression too complex compared to the approach stated in the lecture and also making the computation time enlongated.

Note that the corresponding summation in the preferred approach is with a "minus" sign reducing the summation symbol from 2 to 1.

=Problem 6.5 Calculation using Recurrence Formula=

''' This problem was solved by referencing Team 2, S10 homework 5, Problem8. '''

From the lecture slide Mtg 32-3

Problem Statement
Given the recurrence relation for $$\displaystyle {{p}_{2i}}$$ and $$\displaystyle {{p}_{2i+1}}$$:

And the relation for the coefficient $$\displaystyle {{c}_{2i+1}}$$:

Solution
From the S10 work we have known,

Then by Eq(5.3) we have,

Then by the recurrence formula Eq(5.1) and (5.2) we have:

and also,