| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Hi,
A question. Can anybody help me to integrate the expression:
e to the minus half x squared.
and also tell me how to type it in on one of these keyboards!
This question comes from my brother who wants to be able to find the
area under a normal distribution curve between N and infinity.
He and I both did Maths degrees but it was so long ago we can't
remember anything about it!
Oh how the memory fades!
Graham
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1458.1 | re .0 | EAGLE1::BEST | R D Best, sys arch, I/O | Mon Jun 17 1991 10:34 | 14 |
This sounds a little bit like
erf(x) = (2/sqrt(pi)) * integral(0, x, exp(-t^2), t)
^ ^
(the arguments of operator integral are respectively:
lower limit, upper limit, integrand, and variable of integration)
I don't think there is a closed form algebraic formula for this. This
is a table lookup or numerical integration problem. There may be approximation
formulae for certain ranges of abscissa values.
For a numerical technique, see 'Numerical Recipes in C', by Press, Flannery,
Teukolsky, and Vetterling, Cambridge University Press, p. 174-176.
If I have time later today, I might try to enter some code.
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| 1458.2 | No "pure" closed form solutions. | CADSYS::COOPER | Topher Cooper | Mon Jun 17 1991 12:34 | 6 |
There is definitely no closed form solution to the cumlative normal
integral. It can be solved in terms of the error function (which
then must be approximated to be useful) or approximated directly.
In other words, it is no more or less "solvable" than sin(x).
Topher
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| 1458.3 | Analytic solution for special case only | VMSDEV::HALLYB | The Smart Money was on Goliath | Mon Jun 17 1991 16:53 | 14 |
While there is no closed form solution for the integral
/\ B
| -x�/2
| e dx
|
\/ A
There is a fairly easy solution to the definite integral obtained by
letting A = -oo and B = +oo (oo == infinity). As I recall it involved
computing the double integral exp(-x�/2) exp(-y�/2) dy dx and switching
to a polar system with the ultimate answer being 1/sqrt(2 pi)
John
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