- #1
pat devine
- 9
- 0
The following is a repost from 2008 from someone else as there was no solution offered or provided I thought id post one here
Homework Statement neither my professor nor my TA could figure this out. so they are offering fat extra credit for the following problem
Let n be a positive integer greater than 1 and let p1,p2,...,pt be the primes not exceeding n.
show that p1p2...pt<4^n- An mathematically rigourous proof of this would take a long time but I would suggest the following outline which is logically correct I believe:
First we need to consider a lowest bound value of n and for that purpose is can be stated as pt above
so we want to show that p1*p2*p3...*pt < 4^pt
consider the natural log of both sides
log (p1*p2...) = log(4^pt)
consider the left hand side
log(p1*p2*p3...) = log p1+ logp2 + log p3 ...
Now consider the prime number theorem which can be restated such that:
logp1+logp2+logp3...+logpt ~ pt
another way to think about using a well known restatement of the prime number theorem is that the density of primes at any point n (ie the approximate prime gap) is approximately equal to log n.
eg logp1 is ~ the gap between p1 and p2
log22 is ~ the gap between p2 and p3 and so on
therefore is you sum up all the gaps between the first n primes the sum will be (approx.) the value of the nth prime pn.
Therefore the Left hand side = pt (approx within the order +/- 1% for large p)
Now the right hand side log(4^pt) = pt* log 4 = 1.386 * pt
Therefore this proves the original statement since pt < 1.386 * pt
Any comments would be welcome
Pat
Homework Statement neither my professor nor my TA could figure this out. so they are offering fat extra credit for the following problem
Let n be a positive integer greater than 1 and let p1,p2,...,pt be the primes not exceeding n.
show that p1p2...pt<4^n- An mathematically rigourous proof of this would take a long time but I would suggest the following outline which is logically correct I believe:
First we need to consider a lowest bound value of n and for that purpose is can be stated as pt above
so we want to show that p1*p2*p3...*pt < 4^pt
consider the natural log of both sides
log (p1*p2...) = log(4^pt)
consider the left hand side
log(p1*p2*p3...) = log p1+ logp2 + log p3 ...
Now consider the prime number theorem which can be restated such that:
logp1+logp2+logp3...+logpt ~ pt
another way to think about using a well known restatement of the prime number theorem is that the density of primes at any point n (ie the approximate prime gap) is approximately equal to log n.
eg logp1 is ~ the gap between p1 and p2
log22 is ~ the gap between p2 and p3 and so on
therefore is you sum up all the gaps between the first n primes the sum will be (approx.) the value of the nth prime pn.
Therefore the Left hand side = pt (approx within the order +/- 1% for large p)
Now the right hand side log(4^pt) = pt* log 4 = 1.386 * pt
Therefore this proves the original statement since pt < 1.386 * pt
Any comments would be welcome
Pat
Last edited by a moderator: