6.2.2-Numerical Integration: Romberg Integration and Richardson's Extrapolation
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- Опубликовано: 4 окт 2024
- These videos were created to accompany a university course, Numerical Methods for Engineers, taught Spring 2013. The text used in the course was "Numerical Methods for Engineers, 6th ed." by Steven Chapra and Raymond Canale.
Oh my god you explained this soooo much better than my boring engineer teacher. He took 2 hours of taylor series crap and missed the whole point. Props, and thanks!
At about 4:20, it should be pointed out he DOUBLED the number of function evaluations. The way he said it in the video sounded like he only needed 2 more evaluations to go from O(h2) to O(h4). It's a true statement provided you've only taken 2 evaluations to start with, but it's better to be more general when teaching.
4:49 General Romberg formula should have I(j+1),(k-1) not I(j+1),(k+1) as written in the video
not clear w.r.t I_{j,k}. You did not explain what I_{j,k} is so it is difficult to understand the table.
Yes, as Jacob mentioned in a comment below, it should be k-1 instead of k+1 (around 4:30-5:30 time mark).
At 5:10, the first integration of the general equation is supposed to be I_(j+1, k-1), not I_(j+1, k+1)
thanks for the lessons, it's helping me a lot. What do j and k stand for in the general Romberg algorithm?
Thanks for the video. I have one comment on the audio quality. I think you should lower the treble of your recordings, it's a little bit disturbing if you watch the video with a headphone.
Thank you for the video. I think you should explain in depth the meaning of the subindexes. I figured it out after a few minutes and the comments confirmed my suspicions but I think that would improve dramatically the understanding of the video. Anyhow, thank you very much.
It was awesome!! Thanks a lot
Thank you so much you that was really helpful
i think your general formula is wrong...... At 4:29 you formula says something like this : I(1,2) = [ 4 I(2,1) - I(1,1) ] / 3 , there is no I(1,2) , maybe you mean I(2,2) , because we're moving forward in the algorithm :)
Thanks for the comment. I just went back over this, and there is an error, but it's not what you might think. The formula I(1,2) = [ 4 * I(2,1) - I(1,1) ] / 3 is actually correct. As I mentioned in my previous comment, it's important to keep in mind what we mean by j and k. I'm not sure why you think there is no I(1,2). I(1,2) refers to the integral approximation accurate to O(h^4) for one interval segment. This is equal to a weighted combination of two separate O(h^2) approximations (each using the trapezoid rule): one assuming a single trapezoid for the entire region [ I(1,1) ] and another assuming two trapezoids for the region [ I(2,1) ].
The error in the general formula is on the second index of the first I in the numerator. It should be k-1 not k+1. The general formula is thus:
I(j,k) =~ [ 4^(k-1)*I(j+1,k-1) - I(j,k-1) ] / [ 4^(k-1) - 1]. This is given correctly on p. 636 of Chapra & Canale.
thought the same as @trollLoki. Thought they were just indexes or sthing.. Thx for explanation
@jacob 1st question please: 1) Do you mean, I(1,1) is like using a single trapezoid rule on the whole region, I(2,1) is like using a "composite" trapezoid on the region, and I(1,2) is like using two (normal / single) trapezoid-rules on the whole region? 2nd question: 2) What is more accurate, 2 "normal" (single) trapezoid-rule on 1 region, or a "composite" trapezoid rule? Thanks in advance!
5:58 , sir what is the name of that book ?
"Numerical Methods for Engineers, 6th ed." by Steven Chapra and Raymond Canale.
gokil