Like

Report

Show that $ \frac{1}{3} T_n + \frac{2}{3} M_n = S_{2n} $.

Integration Techniques

You must be signed in to discuss.

Campbell University

Baylor University

University of Nottingham

Boston College

Okay, So this is another question that wants us to show the relation between a couple of different approximations. So this one's to show that a Simpson approximation can be formed from the weighted average of a trap is oId in the mid point. So remember what each of these are so a trap. Inside approximation is Delta X over too times half of a plus, two times F A plus Delta X plus two times all the other mid interior points times F A B and then a midpoint. Some is just Delta X times F of A plus Delta X over too plus f of a plus three. Delta acts over too, plus all the way out. So you have an album. And so let's just compute this left hand side they gave us. So this is Delta X, over six times half of a plus two times F of April still toe acts, plus two times off of a plus two Delta X plus that all the way up to F. O. B. So that's a trap is laid part plus 2/3 Delta X Times. The midpoint part, which is half of a plus, Delta X over too, plus f of a plus three Delta X over too. So now you can see that we can factor out Delta X over three from both of these. And now we get half of a over too, plus f off a plus Delta X all the way up to F B over too, plus our midpoint terms, which get a factor of two. Okay, so where do we go from here? Well, to turn this into a form wheel, recognize we can actually pull out another factor of 1/2. So we write. This is Delta X, over six times half of a plus two f of A plus Delta X plus 2/2 of a plus $2 tax, plus all the way up to effort be plus Delta X over six times. Well, if we pulled out a factor of two 1/2 that is, if the multiply this by 24 f of A plus still tax over too, plus four times f of A plus three Delta X over too. And that continues on. So now let's group thes based on the order of what's in the parentheses. So we get Delta X over six times half of a plus four a plus Delta X plus two terms F of A plus Delta X plus four times F of A plus three Delta X all the way up to effort be, huh? Now look at this. After a moment, my factors to in every place. So we have this factor off 14 to 4, two for two except for the end points. So this is just Simpsons roll with twice as many some intervals. Because remember, Delta X of the two end is equal to Delta X of the end, divided by two. So the Delta X over three in the normal Simpsons rule becomes Delta X over six if we double the amount of Southern rebels. So again, looking at this final sum, we'll just rewrite it one more time. We get Delta X over six times F of a plus four F of A plus Delta X over too, plus two terms F of A plus Delta X plus four times F of A plus three Delta X, all the way up to F B. And again we have a factor of six instead of three in front because Delta X is now Delta X over to as seen from the parentheses, so have twice as many sub intervals, so we can conclude that the sum it is just equal to Simpsons rule with two n sub intervals, which is exactly what we wanted to prove.

University of Michigan - Ann Arbor

Integration Techniques