From: chenyh@memstvx1.memst.edu
Newsgroups: comp.compression
Subject: Re: Math Question
Date: 16 Nov 92 06:32:41 GMT
Organization: Memphis State University

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.