[Tutor] Re: Tutor digest, Vol 1 #2302 - 11 msgs

[Magnus Lycka]

Mic Forster wrote:
>TypeError: fun() takes exactly 2 arguments (1 given)
> >>>

I think the error message is very clear. :)

The bisect function provides one variable to the function
call. What I did in my original example, if I remember
coerrectly, was to do like this (although the values were
different then):

s = 33
j = 1. / 7861
def fun(k):
     return f(k, s) - j
delta = 1e-9
bisect(delta, 1-delta, delta, fun)

You could design the bisect function so that it takes more
parameters, but it gets more complicated.

If you don't like these global variables, you could use
lambda I guess... I'll give you two versions, without and
with lambda...

def bisect(min, max, delta, function):
     fMax = function(max)
     fMin = function(min)
     assert fMax * fMin < 0, locals()
     while max - min > delta:
         newX = 0.5 * (min + max)
         fNew = function(newX)
         if fNew * fMax > 0:
             max, fMax = newX, fNew
         else:
             min, fMin = newX, fNew
     return newX, fNew

def f(k, s, j):
     # I'll stick to the old equation
     return k / (1-k) * (1-k)**s / (1-(1-k)**s) - j

delta = 1e-9

# No lambda
for s, jInv in [(11, 862), (33, 7861)]:
     j = 1./jInv
     print "j = %f s = %i" % (j,s)
     def fun(k):
         return f(k, s, j)
     print "k = %.17f +/- %e" % bisect(delta, 1-delta, delta, fun)

print

# With lambda
for s, jInv in [(11, 862), (33, 7861)]:
     j = 1./jInv
     print "j = %f s = %i" % (j,s)
     print "k = %.17f +/- %e" % bisect(delta, 1-delta, delta,
                                       lambda k: f(k, s, j))


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Magnus Lycka, Thinkware AB
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