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Code:from scipy import integrate def myfunc(x, a, b): return (x**b) + a args = (1.0, 2.0) results = integrate.quad(myfunc, 0.5, 1.5, args) print 'Integral = ', results[0], ' with error = ', results[1]
Hi carol33;
Hi calccrypto;
i try to teach myself here and there and i have learned quite a bit since 2 years ago, but ive had limited experience with polynomials other than using polynomial division to write the AES algorithm in python (a big jump from having a lookup table). i still dont understand so much about polynomials that i would love to know, but are too advanced to learn by skipping around to something interesting that i see
Hi calccrypto;
oh deary me... ive gotten myself into more complicated stuff than ive attempted to learn. oops. i think i'll hold off asking more about this for a few more months, when i finally get into college and have a better background in calc
Hi calccrypto;
im just trying to get a more accurate answer, to at least the thousandths place since for p values, they are usually between .01 and .1 ...
Hi calccrypto;
thanks bobbym!
Hi calccrypto; As you can see from above that your cutoff of 100 for  infinty is introducing a small but noticeable error in the CDF Below is a closed form for the CDF, If you put 1.6 in there you will get the value produced by the second integral 0.945200708300442. But for the life of me, I cannot determine how he is getting those 2 numbers.
from 100 to the z scores, like normalcdf(100,1.6) (using 100 because infinity is a little too much), except the formula is a sum of normalcdfs
Hi calccrypto;
no. 1.6/1.9 are the 2 values to use for z in that big mess i have in the code block
Hi cal; Is this what you want? 