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- Path: sparky!uunet!dtix!darwin.sura.net!nntp.msstate.edu!memstvx1!chenyh
- From: chenyh@memstvx1.memst.edu
- Newsgroups: comp.compression
- Subject: Re: Math Question
- Message-ID: <1992Nov16.003241.4092@memstvx1.memst.edu>
- Date: 16 Nov 92 06:32:41 GMT
- References: <24005@hacgate.SCG.HAC.COM> <1e75usINN31v@agate.berkeley.edu>
- Organization: Memphis State University
- Lines: 91
-
- In article <1e75usINN31v@agate.berkeley.edu>, bsmith@mickey.NoSubdomain.NoDomain (Brian Smith) writes:
- > In article <24005@hacgate.SCG.HAC.COM>, abirenbo@rigel.cel.scg.hac.com (Aaron Birenboim) writes:
- > |>
- > |> While working on an optimization problem related to compression
- > |> I bumped into the following nasty little problem :
- > |>
- > |> evaluate:
- > |>
- > |> { [(i+1)q }
- > |> d { [ }
- > |> --- { Integral [ (x - (i + 0.5) q)^2 * exp(-c * x) dx }
- > |> dq { [ }
- > |> { [i*q }
- > |>
- > |> I have evaluated the integral..... and it is terribly nasty. Even if I
- > |> didn't goof on the evaluation, i would surely goof on the differentiation.
- > |>
- > |> c and i are constants. I'm trying to minimize error for q.
- > |>
- > |> I seem to remember some terribly nasty formulas for handling the
- > |> differentiation before evaluating any integral.
- > |>
- > |> I think it involved the bounds, derivitives of the bounds, the integrand
- > |> and the derivitave of the integrand.
- > |>
- > |> My memory comes from a dim recolection of some stuff covered
- > |> by Dr. Sawchuk at USC (southern cal) in a class on random variables
- > |> based on the book by papoulis. I cannot find it mentioned in papoulis...
- > |> i think is was ONLY in Sawchuk's notes (which i cannot find).
- >
- > I believe the following is correct (this is the result of about 2 minutes
- > of calculation, but it seems to work). I decided to post it so other people
- > could check it, since I'm not perfectly confident in the answer (my calculus
- > was a long time ago!).
- >
- > Theroem:
- >
- > If f(x) is a function that is analytically integrable, then
- >
- > { [y }
- > d { [ }
- > --- { Integral [ f(x) dx } = f(y)
- > dy { [ }
- > { [0 }
- >
- > Proof:
- >
- > Let g(x) be the indefinate integral of f(x). Then, by the fundamental theorem
- > of integral calculus, g'(x) = f(x). Now, the definate intergral from 0 to y
- > of f(x) is g(y) - g(0). The derivative of this expression wrt y is g'(y) which
- > is f(y). QED
- >
- > Given this, and a few reductions, the expression
- >
- > { [(i+1)q }
- > d { [ }
- > --- { Integral [ (x - (i + 0.5) q)^2 * exp(-c * x) dx }
- > dq { [ }
- > { [i*q }
- >
- > evaluates to
- >
- > q^2 * exp(-c * q) * [ ({(i+1) - (i+0.5)}^2) - ({i - (i+0.5)}^2) ]
- >
- > which, if I didn't make any algebra errors, is zero.
- >
- > -----
- > Brian C. Smith arpa: bsmith@cs.Berkeley.EDU
- > University of California, Berkeley uucp: uunet!ucbvax!postgres!bsmith
- > Computer Sciences Department phone: (510)642-9585
-
- ===>I run mathcad symbolic evaluate,comes out as follow,
-
- (.25)
- (---) *(i+1)*exp(-1*c*(i+1)*q)*(4*q*c+8+q*q*c*c)
- (c^2)
-
- -.25*exp(-1*c*(i+1)*q)*[4*c+2*q*c^2]/c^3
-
- -(.25)
- (---)*(i)*exp(-1*c*(i)*q)*(-4*q*c+q*q*c*c+8)
- (c^2)
-
- +.25*exp(-1*c*i*q) * [-4*c+2*q*c^2]/c^3
-
-
- looks nasty allright.
-
- YiHUNG CHEN
- Memphis State Univ.
- Math. Dept.
-