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- Newsgroups: sci.math
- Path: sparky!uunet!snorkelwacker.mit.edu!bloom-picayune.mit.edu!athena.mit.edu!dmsilev
- From: dmsilev@athena.mit.edu (Daniel M Silevitch)
- Subject: Re: Help wanted in integration.
- Message-ID: <1992Nov9.140618.5960@athena.mit.edu>
- Sender: news@athena.mit.edu (News system)
- Nntp-Posting-Host: w20-575-79.mit.edu
- Organization: Massachusetts Institute of Technology
- References: <1992Nov6.144234.16137@rz.uni-karlsruhe.de> <1992Nov9.104113.17807@sun0.urz.uni-heidelberg.de>
- Date: Mon, 9 Nov 1992 14:06:18 GMT
- Lines: 38
-
- In article <1992Nov9.104113.17807@sun0.urz.uni-heidelberg.de>, afm@trillian (Andreas Mueller) writes:
- |> UKJW@DKAUNI2.BITNET ("Christoph Schlenker") writes:
- |> : Hello,
- |> :
- |> : I have just typed your integral into a math-prg and the solution
- |> : of it is just 0.
- |> <stuff deleted>
- |> : I hope thats right... Christoph
- |> It cannot possibly be right: the integrand is a function >=0,
- |> so the integral cannot vanish unless the integrand is identically
- |> zero, which it is obviously not.
- |> Andreas Mueller
- |>
- |> ---------------------------------------------------------
- |> Andreas Mueller afm@vogon.mathi.uni-heidelberg.de
- |> Mathematisches Institut
- |> Im Neuenheimer Feld 288
- |> W - 6900 Heidelberg 1
- |> ---------------------------------------------------------
-
- I missed the initial post, but I can offer some insight on why the math
- program gave an incorrect result. If the function inside the integral is
- strongly periodic (ex. sin^2(32*X) ), and the two limits contain an
- integer number of cycles, it is very likely that incorrect results will
- be returned, due to the algorithm used by most programs to evaluate
- integrals. The most common numeric algorithms are variants of Simpson's
- rule. They usually start with 2 or 4 trapezoids, and doube the number of
- subdivisions until the answer converges. With the function given above,
- integrating from 0 to 2*pi, the first 6 subdivisions will all have the
- integral evaluated at roots of the function. The system will see that 6
- consecutive estimates are 0, and therefore assume that the integral has
- value 0. To get around this, split the integral into two integrals of
- differing limits (1/3, 2/3 of the range works well) and sum the results.
- Note that this is a consequence of numerical integration algorithms, and
- is not present in symbolic math programs.
-
- Daniel Silevitch dmsilev@athena.mit.edu
- Massachusetts Institute of Technology
-