• 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 \$\displaystyle k \ge 6\$

So we need to show that:

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

well:

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

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

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

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

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

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

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

(In fact the only real root of \$\displaystyle k^3-k^2-2k-1 = 0\$ is near \$\displaystyle 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:

\$\displaystyle 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