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anonymous
 5 years ago
prove induction n^2 < n! , for n>=4 ,
if it is easier n! > n^2
i didnt get far, the basis case is true
and (k+1)!= (k+1)k! > ...
anonymous
 5 years ago
prove induction n^2 < n! , for n>=4 , if it is easier n! > n^2 i didnt get far, the basis case is true and (k+1)!= (k+1)k! > ...

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watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1I am sorry did you just delete your earlier question with my answer in it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this program is strange , i didnt see anything in it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i apologize earlier, anything you have to say is helpful :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it doesnt have to be induction

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but i have a proof of my own, the basis case is true 4! > 4^2 (k+1)! = k!(k+1) > k^2 ( 2 + 1) = k^2( 1 + 1 + 1) = k^2 + k^2 + k^2 > k^2 + 2k + 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh wait, that doesnt work

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not the most elegant proof, but it might work

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Ok, if you want we can use induction here we go. So we assume that k^2 < k! Notice that that k>1 is equivalent to 2k+1 < 3k It follows that k>= 4 we have 2k+1 < 3k < (k)! k Now using the assumption k^2 < k! we have k^2 +(2k+1) < k! +(2k+1) + k! +(k)! k=(k+1)! So (k+1)^2 < (k+1)!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that looks, good,. one sec, what was your other proof . now im curious

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1So you haven't read it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im reading it , one sec

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1The statement \(n^2<n!\) is equivalent to \(n<(n1)!\) Notice that for \(n\geq 4\) we have \((n1)!\geq (n2)(n1)\) So it is enough to prove \((n2)(n1)>n\) The inequality is equivalent to \(n^23n+2>n\) \(n^24n+2>0\) \(n(n4)+2>0\) which is clearly true for \(n\geq 4\).

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Are you trying to use this for some statement about the series? if it is the case maybe it will be easier to prove the statement for some bigger n (not 4) to make it more obvious.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats clever, can you look over my proof the basis case is true 4! > 4^2 (k+1)! = k!(k+1) > k^2 ( 2 + 1) = k^2( 1 + 1 + 1) = k^2 + k^2 + k^2 > k^2 + 2k + 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not that i know of about serious

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh but then i need to show k^2 > 2k ?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1correct, that is what I was going to say :)

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1well actually you need 2k^2 > 2k+1

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Notice that \((k1)^2 \geq 0\) So \(k^22k+1 \geq 0\) So \(k^2\geq 2k1\) For \(k\geq 4, k^2 > 2\) So \(2k^2> k^2+2\geq (2k1)+2 =2k +1\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can i ask a quick question. im helping a student who thinks that you can form a solid of revolution , bounded by y=x^2, y = x+2, and revolve it about the y axis

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Yes I think you can but there is some overlap between the solid form by the region on the right of the yaxis and the solid form by the region on the left side of the yaxis
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