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Weregoose
Authentic INTJ
Super Elite (Last Title)
Joined: 25 Nov 2004
Posts: 3976
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 Posted: 15 Feb 2006 02:07:46 pm Post subject: |
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How do we perform nth-order integrals without explicitly carrying out the integration processes? :?
(Are there any shortcuts or identities that I'm missing?) |
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Jeremiah Walgren
General Operations Director
Know-It-All
Joined: 24 May 2003
Posts: 1937
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| Posted: 15 Feb 2006 02:53:21 pm Post subject: |
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| It's been a while since I've done any calculus. What're the chances of you elaborating further? |
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Weregoose
Authentic INTJ
Super Elite (Last Title)
Joined: 25 Nov 2004
Posts: 3976
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| Posted: 15 Feb 2006 04:36:31 pm Post subject: |
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For example, let's say I want to compute the following on the TI-83+:
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Jeremiah Walgren
General Operations Director
Know-It-All
Joined: 24 May 2003
Posts: 1937
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| Posted: 15 Feb 2006 05:13:49 pm Post subject: |
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| Changing the coordinate system comes to mind, but I don't think that would help in this situation. I wasn't taught many ways to actually avoid doing the integral - my teacher didn't care if we slaved for several hours or not. |
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Fr0stbyte124
Advanced Newbie
Joined: 26 Jan 2006
Posts: 98
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| Posted: 15 Feb 2006 05:24:00 pm Post subject: |
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That is a crazy integral  . I would suggest doing whatever you can on a regular computer with good graphing software. Wolfram's online integrator http://integrals.wolfram.com/index.jsp couldn't solve, it either. It's a good tool for most integrals, just not this one. |
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DarkerLine
ceci n'est pas une |
Super Elite (Last Title)
Joined: 04 Nov 2003
Posts: 8328
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| Posted: 15 Feb 2006 05:38:17 pm Post subject: |
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The problem here is not that, it's that fnInt(fnInt(...)) results in an ERR:ILLEGAL NEST error on a TI-83+ series calculator.
The best way I can think of is to write your own routine for definite integrals and use it on the outer integral.
Ex:
Code:
"fnInt((X-1)/(1-AX)/log(AX),A,0,1->Y1
sum(seq(.01Y1(X),X,0,1,.01
You then get domain errors which you have to fix, but that's the essential idea. Plus, it takes a long time and isn't terribly accurate so you'd do well to use another way of calculating definite integrals. |
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thornahawk
μολών λαβέ
Active Member
Joined: 27 Mar 2005
Posts: 569
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| Posted: 16 Feb 2006 01:22:12 pm Post subject: |
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On the subject of multiple integrals... :)
Quote: Integrals of functions of several variables, over regions with dimension greater
than one, are not easy. There are two reasons for this. First, the number of function
evaluations needed to sample an N-dimensional space increases as the Nth power
of the number needed to do a one-dimensional integral. If you need 30 function
evaluations to do a one-dimensional integral crudely, then you will likely need on
the order of 30000 evaluations to reach the same crude level for a three-dimensional
integral. Second, the region of integration in N-dimensional space is defined by
an N − 1 dimensional boundary which can itself be terribly complicated: It need
not be convex or simply connected, for example. By contrast, the boundary of a
one-dimensional integral consists of two numbers, its upper and lower limits.
(from here.)
Substitutions to transform the double integral into a normal one are always helpful.
If the need to compute cannot be avoided, there are formulae available, depending on what region you're considering (e.g. a rectangle, triangle, etc.) Otherwise, Google either "multiple integral product rules" or "Monte Carlo integration".
thornahawk |
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Tiberious726
Advanced Member
Joined: 07 Oct 2005
Posts: 284
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| Posted: 21 Feb 2006 05:08:25 pm Post subject: |
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| don't you just integate the inside w/ respect to y and treat x as a constant then integrate the inside with repect to x? |
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thornahawk
μολών λαβέ
Active Member
Joined: 27 Mar 2005
Posts: 569
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| Posted: 28 Feb 2006 09:18:42 pm Post subject: |
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Analytically, that's what you do. Numerically, your proposal is a true PITA to implement.
Keep in mind that in computing a one dimensional integral, you're essentially sampling function such-and-such at come strategic points, and then adding up all of that after appropriate weighting. The equivalent of your proposal, numerically, is that you have this one dimensional function being integrated that is actually being computed as an integral. Hence the text I quoted: "First, the number of function
evaluations needed to sample an N-dimensional space increases as the Nth power
of the number needed to do a one-dimensional integral."
The sad fact is that it's not the most efficient way of doing so. Much research has been done into this; unfortunately, except for Monte Carlo, none of the efficient methods are simple enough to be programmed into a mere calculator. :(
thornahawk |
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