- #1

pat devine

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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

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