# Does any one know how to write this e^sqrt(lognloglog(n)) ,where n is biginteger

• 03-07-2011, 09:05 AM
ajay.eeralla
Does any one know how to write this e^sqrt(lognloglog(n)) ,where n is biginteger
how to write e^sqrt(lognloglog(n)) ,where n is biginteger
i wrote Double a;
a=Math.pow(e,Math.sqrt(Math.log(num.longValue())*M ath.log(Math.log(num.longValue())))));
but iam getting error
• 03-07-2011, 12:37 PM
JosAH
Quote:

Originally Posted by ajay.eeralla
how to write e^sqrt(lognloglog(n)) ,where n is biginteger
i wrote Double a;
a=Math.pow(e,Math.sqrt(Math.log(num.longValue())*M ath.log(Math.log(num.longValue())))));
but iam getting error

Is n very big? i.e. n > Double.MAX_VALUE. If so you have to approximate those logs. The disadvantage with approximating a log with, say, a Taylor or McLaurin series is that they become quite inaccurate for large values. A way to reduce the inaccuracy is:

log(x+y) == log(y)+log(1+y/x))

for suitable values of x and y when x+y == n. If you can elaborate on this, feel free to reply here.

kind regards,

Jos
• 03-08-2011, 04:56 AM
ajay.eeralla
Thank you josAH for quick reply...
n is very big,i.e n>>Double.maxvalue,
actually iam implementing quadratic sieve algorithm for integer factorization problem..
in that i have to find lognloglogn value..
• 03-08-2011, 06:15 AM