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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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    What do you want to compute exactly? (i.e. what approximation are you asked to use?)2012-10-14
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    hahah sorry about that. Find f(2.5)2012-10-14
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    Oh. Then do you know any $R_{m,n}$ that involves $f(2.5)$? Perhaps that if you find that $R_{m_0,n_0}$ you can try using the recurrence relation involving the $R_{m,n}$'s and deduce the value of $f(2.5)$.2012-10-14
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    Nope I do not know.2012-10-14
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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...2012-10-14
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    Have not seen any of that style equations in my book. I am not exactly sure.2012-10-14
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    Doesn't look like anyone's going to go through the pain of trying. I can try.2012-10-14
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    Hey, you are awesome :) I just post my questions here and then research to see if I can come up with any clues. So far I haven't found anything that was helpful2012-10-14
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    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.2012-10-14
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    Is it possible you're lacking information? I did a quick sketch of how I would be supposed to find this thing and something shows me I might be lacking data, like I would need an extra $R_{m,n}$ or something.2012-10-14
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    No this is all the info I have been provided2012-10-14
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    hmmm maybe [this](http://www.google.ca/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&ved=0CDEQFjAC&url=http%3A%2F%2Ffacstaff.cbu.edu%2F~wschrein%2Fmedia%2FANA%2Fromberg.pdf&ei=4EZ6UPuZGOu00QH2y4DIAg&usg=AFQjCNFAxZc7T4GemwEr2jmeUBtM7FUBPA&sig2=uwQ0DZ_aFl21me86Hfy7HA) will be of help. You have to go to page #4 starting with pg 136 number 52012-10-14
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    The problem seems to be similar to the one, which is solved in this [exam sheet](http://www.lailafouad.name.eg/lailafouad/userdownloads/My%20Courses/Postgraduate/Exam_Answer_July2011.pdf).2012-10-14
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    YA IT DOES. Let me check it out :)2012-10-14
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    Oh this is my lucky day. This is the exact question. I really appreciate the time you took to research it for me. It seemed like I had reached a dead end. Thanks2012-10-14
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    @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.2012-10-14
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    @math101 In fact all I did was that I tried entering [romberg integral "51342"](http://www.google.com/search?q=romberg+integral+%2251342%22) into Google.2012-10-14
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    Thanks.I guess I gotta learn how to google :) @MartinSleziak2012-10-14
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    @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 answer2012-10-14

1 Answers 1

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$R_{1,1}$

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

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

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 $f(2.5)=.43457$