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Mercuria
Merc's Husband's Wife!

 Posted: Thu Jan 27, 2005 7:36 am    Post subject: 1 i'm trying to take the integral of t^(2n+2)/(1+t^2) dt from 0 to x, and i just can't do it. well, i have no confidence i'll get the answer. help, please? feel free to move this thread wherever...
kevinatilusa
Daedalian Member

 Posted: Thu Jan 27, 2005 9:58 am    Post subject: 2 Let's try a few small values of n. If n=-1, we've got 1/(1+t^2)=arctan(t)+C. This is just one to have memorized. if n=0, we've got t^2/(1+t^2). Looks ugly, but we can can fix it up by a trick: t^2/(1+t^2)=(t^2+1-1)/(1+t^2)=1-1/(1+t^2), so integrating we get t-arctan(t)+C if n=1, we've got t^4/(1+t^2). The initial instinct is to use the same trick, but this time we're looking to get a t^2 (the quotient of the leading terms of our numerator/denominator) as part of our quotient, so we'll want the numerator to look like t^2(1+t^2)=t^4+t^2. This means we'll write it like (t^4+t^2-t^2)/(1+t^2)=(t^2)-1/(1+t^2). We know how to do t^2, and we just did 1/(1+t^2) as the n=0 case, so we get t^3/3-(t-arctan(t))+C=t^3/3-t+arctan(t)+C if n=2, we've got t^6/(1+t^2). The same argument (but now we want a (t^6+t^4 up top, gives that t^6/(1+t^2)=(t^6+t^4-t^4)/(1+t^2)=t^4-(t^4)/(1+t^2) is the way to go. We know how to do t^4, and throwing in the n=1 case gives t^5/5-t^3/3+t/1-arctan(t)+C The same pattern continues, and we can always reduce from the n case to the n-1 case using a bit of clever division. If you're looking for rigor, this can probably also be transformed into an induction proof.
kevinatilusa
Daedalian Member

 Posted: Thu Jan 27, 2005 10:09 am    Post subject: 3 If you've seen Taylor series before, you may recognize the Taylor series for arctan starting to form in your answer here. If you remember how that Series was derived, you can get to a (very slick) way of doing this integral: 1/(1+t^2)=1-t^2+t^4-t^6+t^8-t^10+... Instead of writing the whole thing as an infinite series, stop at t^(2n) and write the rest as a geometric series: 1/(1+t^2)=1-t^2+t^4-t^6+...+(t^(2n+2)/(1+t^2))(-1)^(n+1) Now just integrate everything in sight, and you've got an equation relating a bunch of known integrals to an unknown one, and you can just solve for the one you want. The catch here is...I don't know if I could ever have thought of this "nice" method without doing the "ugly" one in the previous post first and seeing what the answer was in advance.
CrystyB
Misunderstood Guy

Posted: Thu Jan 27, 2005 2:46 pm    Post subject: 4

minor correction:
 kevinatilusa wrote: (t^4+t^2-t^2)/(1+t^2)=(t^2)-1/(1+t^2)
=(t^2)-(t^2)/(1+t^2)

My initial thought was: I could promote http://mathforum.org/dr.math/ask/ , but there are already lots of people using that, so the chances of one particular question being skipped made me reconsider posting that reference. I kept this paragraph in just for the "fwiw" of it all.
Guest

 Posted: Thu Jan 27, 2005 3:19 pm    Post subject: 5 Well, the Integrator gives: -[x (1-2n) /(1+2n)]*Hypergeometric 2 F 1 [(n+1/2), 1; (n+3/2), -x 2 ] Does that help?
CrystyB
Misunderstood Guy

 Posted: Thu Jan 27, 2005 3:29 pm    Post subject: 6 well, i get a slightly different thing
ZutAlors!
Daedalian Member

 Posted: Thu Jan 27, 2005 3:49 pm    Post subject: 7 Ooops. That would be because my copying skills aren't up to par. And now I can't edit it -- shame on me for being too lazy to log in. Ah well. It's not like mathematics requires exactitude.
Mercuria
Merc's Husband's Wife!

 Posted: Thu Jan 27, 2005 9:41 pm    Post subject: 8 sorry guys, i should have put in more background information... this is the remainder (taylor/lagrange remainder, i think) of the partial taylor polynomial for arctan. the homework was to find the partial and the remainder, but i just can't get a concise term for the remainder. i think i'm just going to leave it as an integral, since i don't think i actually have the background to integrate it. i was just wondering if i missed any special cases, since it's been years since i took calculus. thanks everybody =]
kevinatilusa
Daedalian Member

 Posted: Fri Jan 28, 2005 2:16 am    Post subject: 9 Even if you can't do the integral, you can sometimes get bounds on it by bounding the function you're integrating. For example, here we have that the remainder is the integral from 0 to x of t^(2n+2)/(1+t^2). Since 1+t^2 is always bigger than one, we can replace it by 1 and just make things bigger in absolute value, meaning that your error is at most integral(0 to x) of |t^(2n+2)|=|x^(2n+3)/(2n+3)|. Another way to see this is that the Taylor series you're trying to sum is an alternating series with decreasing terms (assuming |x|<=1...but if it isn't you're series isn't converging anyway), and the error when you're adding up such a series is always at worst the absolute value of the next term of the series after you stop, which would again be |x^(2n+3)/(2n+3)|
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