• Status: Solved
  • Priority: Medium
  • Security: Public
  • Views: 867
  • Last Modified:

Proof by induction -- Inequalities

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
jeffiepoo
Asked:
jeffiepoo
2 Solutions
 
ozoCommented:

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
 
ozoCommented:
would it be any easier to prove that
(n-1)! > n^2 ?
0
 
jeffiepooAuthor Commented:
I don't get what you're getting at
0
Protect Your Employees from Wi-Fi Threats

As Wi-Fi growth and popularity continues to climb, not everyone understands the risks that come with connecting to public Wi-Fi or even offering Wi-Fi to employees, visitors and guests. Download the resource kit to make sure your safe wherever business takes you!

 
InteractiveMindCommented:
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
 
NovaDenizenCommented:
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
 
GwynforWebCommented:
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
 
GwynforWebCommented:
 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
 
GwynforWebCommented:
...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
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

Join & Write a Comment

Featured Post

Cloud Class® Course: Ruby Fundamentals

This course will introduce you to Ruby, as well as teach you about classes, methods, variables, data structures, loops, enumerable methods, and finishing touches.

Tackle projects and never again get stuck behind a technical roadblock.
Join Now