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  
Newbie Joined: Feb 2008 Posts: 8 Thanks: 0  Quote:
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)(50k)(50+k+1) (50k)(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  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: prove question Quote:
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: 18,845 Thanks: 1567 
Using Google calculator, 100! = 9.3... × 10^157, whereas 50.5^100 = 2.1... × 10^170.

February 6th, 2008, 07:36 AM  #7  
Newbie Joined: Feb 2008 Posts: 8 Thanks: 0  Re: prove question Quote:
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  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
 

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