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 February 4th, 2008, 01:13 AM #1 Newbie   Joined: Feb 2008 Posts: 8 Thanks: 0 prove question hey , it's first year calculus i need to prove next : 100!<=(50.5)^100 i thought about Bernoulli's inequality but i dont know how to continue (actually how to start ) 10X
 February 4th, 2008, 06:07 AM #2 Senior Member   Joined: Oct 2007 From: France Posts: 121 Thanks: 1 By the concavity of the function ln, we can write: ln [(x1+x2+..+xn)/n]>=(1/n)*[ln(x1)+..+ln(xn)] Put xk=k (1<=k<=n) and recall that 1+2+..+n=n(n+1)/2 to get: ln[(n+1)/2]>=1/n*[ln(1)+..+ln(n)]=1/n*ln(n!) This implies that ln[ [(n+1)/2]^n]>=ln(n!) and that: n!<=[(n+1)/2]^n If n=100, we obtain 100!<=50,5^100
February 4th, 2008, 06:09 AM   #3
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 Originally Posted by Richard André-Jeannin By the concavity of the function ln, we can write: ln [(x1+x2+..+xn)/n]>=(1/n)*[ln(x1)+..+ln(xn)] Put xk=k (1<=k<=n) and recall that 1+2+..+n=n(n+1)/2 to get: ln[(n+1)/2]>=1/n*[ln(1)+..+ln(n)]=1/n*ln(n!) This implies that ln[ [(n+1)/2]^n]>=ln(n!) and that: n!<=[(n+1)/2]^n If n=100, we obtain 100!<=50,5^100
but we haven't learned ln yet....
then it should be something else

 February 4th, 2008, 06:27 AM #4 Senior Member   Joined: Oct 2007 From: France Posts: 121 Thanks: 1 An elementary proof: 100!=(product for k=0...49)(50-k)(50+k+1) (50-k)(50+k+1)=[50,5-(k+0.5)]*[50.5+(k+0.5)] =50.5²-(k+0.5)²<=50.5² Thus: 100!<=(product for k=0...49)50.5²=50.5^100
February 4th, 2008, 06:35 AM   #5
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Re: prove question

Quote:
 Originally Posted by igalep132 100!<=(50.5)^100
100! = (100 * 1) * (99 * 2) * ... * (51 * 49) * 50 = sqrt(100 * 1) * sqrt(100 * 1) * sqrt(99 * 2) * sqrt(99 * 2) * ... * sqrt(51 * 49) * sqrt(51 * 49) * 50

Now you have 100 factors and you can apply whatever inequalities you can find to them.

 February 4th, 2008, 09:38 AM #6 Global Moderator   Joined: Dec 2006 Posts: 19,702 Thanks: 1804 Using Google calculator, 100! = 9.3... × 10^157, whereas 50.5^100 = 2.1... × 10^170.
February 6th, 2008, 07:36 AM   #7
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Re: prove question

Quote:
Originally Posted by CRGreathouse
Quote:
 Originally Posted by igalep132 100!<=(50.5)^100
100! = (100 * 1) * (99 * 2) * ... * (51 * 49) * 50 = sqrt(100 * 1) * sqrt(100 * 1) * sqrt(99 * 2) * sqrt(99 * 2) * ... * sqrt(51 * 49) * sqrt(51 * 49) * 50

Now you have 100 factors and you can apply whatever inequalities you can find to them.

10X

 February 6th, 2008, 10:49 AM #8 Newbie   Joined: Feb 2008 Posts: 8 Thanks: 0 i did it different way using average inequality. in the right side there is a need for Arithmetic progression formula part b to prove 100!<=(3408.5)^50
February 7th, 2008, 04:36 AM   #9
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Quote:
 Originally Posted by igalep132 part b to prove 100!<=(3408.5)^50
How is that any harder? sqrt(3408.5) > 58.38 > 50.5.

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