Azzera filtri
Azzera filtri

express erf() function as qfunc()

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stephen williams
stephen williams il 18 Nov 2020
I have this script that is looking at intersections between probability density functions, and it gives the answer in term of erf(), which I believe is the correct answer, but I want to have it return the answer in terms of qfunc(). Any ideal how to do that?
Here is an example of an output, including the pretty() version.
ans =
-(1125899906842624*2^(1/2)*pi^(1/2)*(erf(1/(2*n^(1/2))) - 1))/5644425081792261
/ / 1 \ \
sqrt(2) sqrt(pi) | erf| --------- | - 1 | 1125899906842624
\ \ 2 sqrt(n) / /
- ----------------------------------------------------------
The erf() gets into the stream of results as the result of an integration of an exponential upstream. So in theory I could convert this way
erf(x)= 2*(1-qfunc(x*sqrt(2)))-1
but the erf() is built in, and the integration does nto recognize it being redefined ahead of time.
Any ideas?

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