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Quick question - what is the integral of e^-(x^2) dx from 0 to z
- Thread starter Howard
- Start date
dullard
Elite Member
Then what do you need to know? There is no polynomial or other answer we can give you (well we can give you a Taylor's series approximation which would be a polynomial but lets not get into approximations). That is just an integral that when multiplied by a constant was given a name - the error function.
Uh like simpson's rule? This function is nice & smooth, so any old integration rule (even trapezoid) will handle it ok...just pick a small step size. If you want to be fancy, use some form of extrapolation.
But yeah this integral equals a "set" value from 0 to infinity. Otherwise it's only approximate by numeric means. Beyond that, matlab, maple, excel, and mathematica all have erf(x) built in.
Well up to a point (0.5?), erf is roughly linear...how accurate do you have to be?
You can look up simpson's rule online...esp by hand, this will probably be your best bet. Other than that I can imagine erf values being tabulated in books like the CRC book of tables/formulas or whatever they call it.
chuckywang
Lifer
That integral you wrote cannot be integrated indefinitely.
Yeah. Simpson's rule essentially fits a parabola to the function at hand and integrates that (because well, parabolas are easy to integrate). So yes, you need 3 function evals of exp(-x^2).
Now the more tightly the 3 points are packed together, the better your result will be. For even better results, "string" several simpson approximations together--like do an approx on the range 0 to 0.5, then 0.5 to 1, 1-1.5, etc, and you end up with what's called the "composite simpson rule." (dividing up the interval)
I think the weights go something like 1/3, 2/3, 4/3, 2/3, 4/3, ... 4/3, 2/3, 1/3, I think.
-Eric
jamesbond007
Diamond Member
If you can't get help here, I suggest you try http://www.physicsforums.com I really like that place as there are a lot of interesting discussions that take place. It's filled with brains and I bet they can answer your question.
Not saying no one here on AT can, but you will probably get a faster answer there. 😉
Not saying no one here on AT can, but you will probably get a faster answer there. 😉
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