• Mar 15th 2008, 09:29 PM
xomichelleybelly
Use math induction to show that
1) n!>= n^2 for n>=6

2) n!>=n^3 for n>=6
• Mar 16th 2008, 12:24 AM
CaptainBlack
Quote:

Originally Posted by xomichelleybelly
Use math induction to show that
1) n!>= n^2 for n>=6

I will assume that you have done the base case and that we assume this is true for some $k \ge 6$

So we need to show that:

$(k+1)! \ge (k+1)^2$,

well:

$(k+1)! = k!~ k$

and by supposition $k! \ge k^2$, so:

$(k+1)! = k! k \ge k^3$

So if we can prove that $k^3 \ge (k+1)^2$ we are done, but $k^3 \ge (k+1)^2$ is equvalent to $k^3-k^2-2k-1 \ge 0$. So to compete the proof of the induction step you need to show that:

$k^3-k^2-2k-1 = 0$

has no real root greater than or equal to $6$ (this is sufficient as for $k$ greater than its largest real root the cubic is positive).

This you do by making the substitution $x=k-6$, then using Descartes rule of signs to show that this has no positive roots.

(In fact the only real root of $k^3-k^2-2k-1 = 0$ is near $k=2.1$)

RonL
• Mar 16th 2008, 12:54 AM
CaptainBlack
Quote:

Originally Posted by xomichelleybelly
Use math induction to show that

2) n!>=n^3 for n>=6

Exactly the same method as for part 1, but the equation that you need to
show has no root greater than 6 is:

$k^4-k^2-2k-1=0$

and exactly the same method will work with this.

RonL
• Mar 16th 2008, 01:01 AM
mr fantastic
Quote:

Originally Posted by xomichelleybelly
Use math induction to show that
1) n!>= n^2 for n>=6

2) n!>=n^3 for n>=6

By proving 2) you've automatically proved 1).

(And I reckon a lawyer could argue that induction has been used to prove both (Rofl) )