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Romberg Integration is used to approximate $$\int_2^3 \! f(x) \, \mathrm{d} x$$ If $f(2)=.51342, f(3)=.36788, R_{3,1}=.43687 , R_{3,3}=.43662$ Find $f(2.5)$

I am quite confused how to proceed since I don't even know what the function is. The only peice of information that may be helpful is $$R_{1,1}= { h \over 2}(f(2)+f(3))={1 \over 2}(.51342+.36788)=.44065$$
Maybe putting this into a piece-wise function may be helpful

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There must probably be one of the $R(n,0)$ of the form $$ \frac h2 \left( f(2) + f(2.5) + f(3) \right) $$ To be honest I didn't like Romberg integration when I learned it and I don't feel like diving back into this ugly recurrence relation... – Patrick Da Silva Oct 14 '12 at 3:41
You really want to try to backtrack the recurrence relation to find a linear combination of $f(2)$, $f(2.5)$ and $f(3)$ using $R_{3,3}$ and $R_{3,1}$. I'm going to try that now. – Patrick Da Silva Oct 14 '12 at 4:26
The problem seems to be similar to the one, which is solved in this exam sheet. – Martin Sleziak Oct 14 '12 at 6:15
@math101: If you've found the solution, you could write it up as an answer for posterity and accept it so the question doesn't remain unanswered. – joriki Oct 14 '12 at 7:00
@joriki I have a few questions that have not been answered. When I have the time I will answer them so that others can have the answer – math101 Oct 14 '12 at 14:32
up vote 1 down vote accepted



$R_{3,1}, R_{3,2}, R_{3,3}$

We need to use the recursive formulas. The first formula is simple and easy: $$R_{1,1}={h \over 2}( f(a)+f(b))={1 \over 2}(.51342+.36788)=.44065$$ $$R_{3,3}={16R_{3,2}-R_{2,2} \over 15}=.43662$$ $$R_{3,2}={4R_{3,1}-R_{2,1} \over 3}={4(.43687)-R_{2,1} \over 3}=.58249-{ 1 \over 3}R_{2,1}$$ $$R_{2,2}={4R_{2,1}-R_{1,1} \over 3}={4r_{2,1}-.44065\over 3}={4 \over 3}R_{2,1}-.14688$$

Now we have to subsitute $R_{3,2}$ and $R_{2,2}$ into $R_{3,3}$ $$R_{3,3}={1 \over 15}(16(.58249-{1 \over 3}R_{2,1})-({4 \over 3}R_{2,1}-.14688)=.43662$$ Now that we only have one variable its simple to solve and $R_{2,1}=.43761$ Now we can solve $R_{2,1}$ $$R_{2,1}={ 1\over 2}R_{1,1}+h_1f(a+h_2)={ 1\over 2}R_{1,1}+h_1f(2.5)$$ $$f(2.5)={2R_{2,1}-R_{1,1} \over 2h}={2(.43761)-(.44065) \over 2(.5)}=.43457$$ So we can conclude that $f(2.5)=.43457$

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