The old round off problem?

sam samschul at pacbell.net
Sat Mar 4 14:00:26 EST 2006


Hello all, I am taking a class in scientific programming at the local
college. My problem is that the following difference produces round off
errors as the value of x increases. For x >= 19 the diference goes to
zero.I understand the problem, but am curious as to whether their
exists a solution. I have tried various graphing programs,and they all
exihibit this problem.

thanks in advance Sam Schulenburg
f(x)  = cosh^2(x) - sinh^2(x)  = 1

>>> from math import *
>>> for x in range(20):
	print "x= %2d  Sinh^2(x) = %20.3f f(x) = %2.10f
"%(x,pow(cosh(x),2),pow(cosh(x),2)- pow(sinh(x),2))


x=  0  Sinh^2(x) =                1.000 f(x) = 1.0000000000
x=  1  Sinh^2(x) =                2.381 f(x) = 1.0000000000
x=  2  Sinh^2(x) =               14.154 f(x) = 1.0000000000
x=  3  Sinh^2(x) =              101.358 f(x) = 1.0000000000
x=  4  Sinh^2(x) =              745.740 f(x) = 1.0000000000
x=  5  Sinh^2(x) =             5507.116 f(x) = 1.0000000000
x=  6  Sinh^2(x) =            40689.198 f(x) = 1.0000000000
x=  7  Sinh^2(x) =           300651.571 f(x) = 0.9999999999
x=  8  Sinh^2(x) =          2221528.130 f(x) = 1.0000000000
x=  9  Sinh^2(x) =         16414992.784 f(x) = 1.0000000037
x= 10  Sinh^2(x) =        121291299.352 f(x) = 1.0000000298
x= 11  Sinh^2(x) =        896228212.033 f(x) = 0.9999998808
x= 12  Sinh^2(x) =       6622280532.961 f(x) = 1.0000019073
x= 13  Sinh^2(x) =      48932402357.710 f(x) = 0.9999923706
x= 14  Sinh^2(x) =     361564266073.369 f(x) = 0.9999389648
x= 15  Sinh^2(x) =    2671618645381.616 f(x) = 1.0000000000
x= 16  Sinh^2(x) =   19740740045670.668 f(x) = 0.9921875000
x= 17  Sinh^2(x) =  145865435631864.219 f(x) = 1.0000000000
x= 18  Sinh^2(x) = 1077807886778799.250 f(x) = 1.0000000000
x= 19  Sinh^2(x) = 7963982939278438.000 f(x) = 0.0000000000 
>>>




More information about the Python-list mailing list