New member, induction problem

May 2010
6
0
Hobart
I have just joined this forum. I studied maths 45+ years ago, and I have not used it since - so I have forgotten nearly everything! I decided to learn calculus again, but do not have access to tutors etc, which makes it hard when I want to check answers. Anyway ...

Show that n! > n^3 if n is large enough by induction? I get stuck.
 
Apr 2008
748
159
I have just joined this forum. I studied maths 45+ years ago, and I have not used it since - so I have forgotten nearly everything! I decided to learn calculus again, but do not have access to tutors etc, which makes it hard when I want to check answers. Anyway ...

Show that n! > n^3 if n is large enough by induction? I get stuck.
Have you done any induction problems before, do you know/understand the basic ideas behind the proof by induction?

Or is it just setting up the problem initially that you're struggling with?
 
May 2010
6
0
Hobart
Have you done any induction problems before, do you know/understand the basic ideas behind the proof by induction?
Or is it just setting up the problem initially that you're struggling with?
I should have been more precise. Yes, I understand induction and have solved other problems OK.

First, the smallest n such that \(\displaystyle n! > n^3\) is 6.

Second, assuming it is true for n = k > 6 then show that

\(\displaystyle (k+1)! > (k+1)^3\)

We know that \(\displaystyle (k+1)! = k!(k+1) > k^3(k+1)\)

At which point I get stuck!

PS: My first attempt at using LaTex and I don't know why a formulae is moved down relative to the rest of the line that it is in.
 
Nov 2009
485
184
It looks like you're doing fine.

I'd like to note that it suffices to show that \(\displaystyle k^3(k+1)\geq (k+1)^3\) for \(\displaystyle k\geq 6\). Can you do it?
 
May 2010
6
0
Hobart
Show \(\displaystyle k^3(k+1) \geq (k+1)^3\) for \(\displaystyle k \geq 6\)

Step 1 rearrange

\(\displaystyle k^3(k+1) \geq (k+1)^3 \Rightarrow k^4 + k^3 \geq k^3 + 3k^2 + 3k + 1\)
\(\displaystyle \Rightarrow k^4 - (3k^2 + 3k + 1) \geq 0\)

Step 2 induction

First it is true for \(\displaystyle k = 6\) by substitution.

Second, assume it is true for some \(\displaystyle k > 6\). Then

\(\displaystyle (k+1)^4 = k^4 + 4k^3 + 6k^2 +4k + 1\)
\(\displaystyle 3(k+1)^2 + 3(k+1) + 1 = 3k^2 + 9k + 7\)

So show \(\displaystyle k^4 + 4k^3 + 3k^2 -5k - 6 \geq 0\) for \(\displaystyle k > 6\)

Now for \(\displaystyle k \geq 6 \ \ 3k^2 > 18k > 17k + 6\)

So

\(\displaystyle k^4 + 4k^3 + 3k^2 -5k - 6 > k^4 + 4k^3 + 17k +6 -5k - 6 = k^4 + 4k^3 + 12k > 0\) for all k > 6

Is that OK?