Question

I'm struggling with the sum of the logs problem in Problem Set 1. My ratio seems to approach 0, and very slowly. I'm not getting the output I expected from reading the problem.

Answer 1

from math import *
n = 10
oddinteger = 3
checkdigit = 2
primecount = 1
primelogsum = log(2)
while n < 10000:
while primecount < n:
if oddinteger%checkdigit == 0:
oddinteger += 2
checkdigit = 2
elif oddinteger%checkdigit != 0 and checkdigit*2 < oddinteger:
checkdigit += 1
else:
primecount += 1
primelogsum += log(oddinteger)
oddinteger += 2
checkdigit = 2
print "Sum of the logs of the primes: " + str(primelogsum)
print "n: " + str(n)
print "Ratio of sum of the logs of the primes to the value n: " + str(float(primelogsum)/float(n))
print ""
primelogsum = log(2)
n += 10

Answer 2

what does n represent and what is it meant to be doing?

Answer 3

Well, I'm a moron.

I was operating under the assumption that:

as n increases - primelogsum approaches 1.

when it now seems that I need to keep taking the sum of the logs each time I get a new prime number, and that ratio will approach 1.

Answer 4

while leaving n as a constant

Answer 5

actually now that I read it again, I think I was right - n is supposed to increase with each iteration. I set it to some value, then add up all the logs of prime numbers up to n, divide that result by n, and then increase n and have another iteration.

Answer 6

Are you actually adding up ALL the logs? Why does primelogsum keep getting reset to log(2)?

Answer 7

I'm doing that each time the while primecount < n condition becomes false, which begs the question, why do I care whether my primecount is < n? I should be checking my oddinteger that is being tested for primeness against n.

Answer 8

... and we're approaching 1 on a squiggly line :-)

Answer 9

cool!

Answer 10

Thanks for the help!