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February 4th, 2008, 02:13 AM   #1
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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
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February 4th, 2008, 07:07 AM   #2
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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
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February 4th, 2008, 07:09 AM   #3
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Quote:
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
thanks for the answer,
but we haven't learned ln yet....
then it should be something else
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February 4th, 2008, 07:27 AM   #4
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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
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February 4th, 2008, 07:35 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.
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February 4th, 2008, 10:38 AM   #6
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Using Google calculator, 100! = 9.3... × 10^157, whereas 50.5^100 = 2.1... × 10^170.
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February 6th, 2008, 08:36 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
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February 6th, 2008, 11:49 AM   #8
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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
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February 7th, 2008, 05:36 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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