Solved

Proof by induction -- Inequalities

Posted on 2009-04-03
9
807 Views
Last Modified: 2012-05-06
Ok I can't figure out how to inductively prove the following inequalities, I have the base case and inductive hypothesis, I just don't see how to prove this stuff. I've already turned in the assignment, I'm just curious cause they never tell us. At least there's points in it for you guys.

1.) Prove that n! > n^3 when n is large enough. (n has to be greater than or equal to 6)

2.) Prove that n! < n^n for all positive integers by induction.

I just can't seem to prove these inequalities for P(K+1). Any help would be appreciated.

-Jeff   P.S. (If you solve one I'll give you half the points, unless only one is ever solved than I'll give you them all! HAHAHA)
0
Comment
Question by:jeffiepoo
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
9 Comments
 
LVL 84

Expert Comment

by:ozo
ID: 24058772

Assuming that k! < k^k, what can you then say about (k+1)! and (k+1)^(k+1) ?

what do you know about 1! and 1^1
0
 
LVL 84

Expert Comment

by:ozo
ID: 24058841
would it be any easier to prove that
(n-1)! > n^2 ?
0
 
LVL 6

Author Comment

by:jeffiepoo
ID: 24059363
I don't get what you're getting at
0
The Ultimate Checklist to Optimize Your Website

Websites are getting bigger and complicated by the day. Video, images, custom fonts are all great for showcasing your product/service. But the price to pay in terms of reduced page load times and ultimately, decreased sales, can lead to some difficult decisions about what to cut.

 
LVL 25

Expert Comment

by:InteractiveMind
ID: 24059442
You first assume that k!<k^k.
Now consider (k+1)! = (k+1)k! < (k+1)k^k,
it's easy to show that (k+1)^(k+1) > (k+1)k^k, in which case you're done.

As for (n-1)! > n^2, note that this is equivalent to n!>n^3. However, you may find it easier to prove that instead..
0
 
LVL 22

Accepted Solution

by:
NovaDenizen earned 250 total points
ID: 24061768
1.  n! > n^3 for all n >= 6.

First prove P(6), which is 6! > 6^3.  This is just easy arithmetic.

Then you have to prove P(k+1) given P(k) and k >= 6.
Given k! > k^3, and k >= 6, prove that (k+1)! > (k+1)^3.

Have you gotten this far?  The first thing you should do is expand the factorial and the polynomial.

2.  n! < n^n for all n >= 1.
This is actually not true.  1! < 1^1 reduces to 1 < 1, which is false.
Maybe you meant n! <= n^n for all n >= 0, or maybe n! < n^n for all n >= 2.
0
 
LVL 31

Expert Comment

by:GwynforWeb
ID: 24062376
n^n > n!

Assume true for n

(n+1)^(n+1) =  (n+1)(n+1)^n > (n+1)n^n

                 >  (n+1)n!   ( by inductive assumption)

                 =  (n+1)!

Hence if true for n true for n+1

Clearly true for n=2.  Hence by the principle of induction true for all n.
0
 
LVL 31

Expert Comment

by:GwynforWeb
ID: 24065015
 n^3  <  n!     (for n > 5)

Assume true for n

(n+1)^3 = n^3 + 3n^2 + 3n + 1
 
             <  n!  + 3n^2 + 3n + 1     (by inductive assumption)

             <  n!  + n^3 + n^3 +n^3    (if n > 3)

             <   n!  + n! + n! + n!        (by inductive assumption)

             <  4n!

             <  (n+1)n!  =  (n+1)!        (if n >4)
               
Hence if true for n true for n+1.

Since true for n=6 by the principle of induction true for all n.
0
 
LVL 31

Assisted Solution

by:GwynforWeb
GwynforWeb earned 250 total points
ID: 24065816
...a bit tidier is

(n+1)^3 = n^3 + 3n^2 + 3n + 1
 
             <  n^3 + n^3 + n^3 + n^3 = 4n^3     (if n > 3)

             <  4n!    (by inductive assumption)

             <  (n+1)n!  =  (n+1)!        (if n >4)  

 Hence if true for n true for n+1.
0

Featured Post

Independent Software Vendors: We Want Your Opinion

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Displaying an arrayList in a listView using the default adapter is rarely the best solution. To get full control of your display data, and to be able to refresh it after editing, requires the use of a custom adapter.
This article provides a brief introduction to tissue engineering, the process by which organs can be grown artificially. It covers the problems with organ transplants, the tissue engineering process, and the current successes and problems of the tec…
Viewers will learn how to properly install Eclipse with the necessary JDK, and will take a look at an introductory Java program. Download Eclipse installation zip file: Extract files from zip file: Download and install JDK 8: Open Eclipse and …
With the power of JIRA, there's an unlimited number of ways you can customize it, use it and benefit from it. With that in mind, there's bound to be things that I wasn't able to cover in this course. With this summary we'll look at some places to go…

696 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question