# Thread: Another Induction

1. ## Another Induction

Got another problem on induction, this time I don't even know who to start and what to do.

What I want to do is to show that $\displaystyle 3^n \geq n^3$ is true for all
$\displaystyle n \in$ N.
Here we define the natural set of numbers decluding 0.

From here I don't know much about what to do but show it true for n=1, witch I of course have done.
Anyone got any ideas?

2. Originally Posted by a4swe
Got another problem on induction, this time I don't even know who to start and what to do.

What I want to do is to show that $\displaystyle 3^n \geq n^3$ is true for all
$\displaystyle n \in$ N.
Here we define the natural set of numbers decluding 0.

From here I don't know much about what to do but show it true for n=1, witch I of course have done.
Anyone got any ideas?
Next time, please post thy questions in a new thread.
---
We note that,
$\displaystyle 3^0\geq 0^3$
$\displaystyle 3^1\geq 1^3$
Let S represent all N such that it is true.
We know that $\displaystyle 0,1\in \mathbb{N}$.

Assume, $\displaystyle k\in S$, i.e. that it is true for k.
Then,
$\displaystyle 3^k\geq k^3$
Multiply by 3,
$\displaystyle 3^{k+1}\geq 3k^3$

Note that,
$\displaystyle 2k^3\geq 3k^2+3k+1$
For $\displaystyle k\geq 2$
Add $\displaystyle k^3$ to both sides,
$\displaystyle 3k^3\geq k^3+3k^2+3k+1$
Gaze upon ye perfect cubical,
$\displaystyle 3k^3\geq (k+1)^3$
Transitivity,
$\displaystyle 3^{k+1}\geq (k+1)^3$
Thus,
$\displaystyle (k+1)\in S$

Thus, S contain all N, i.e. true for all integers.

3. Next time, please post thy questions in a new thread.
Excuse me, the others forums I am active on keep a high "one topic only" policy, witch means that all subjects about let's say induction goes under one topic. That was not the case here I have to and will check the rules and posting guide lines.
Once again, I am sorry.

We know that $\displaystyle 0,1\in \mathbb{N}$.
Well, that wasn't true for the definition of N I used, but ok it doesn't interfere with the soulution in general, I just don't care for the 0.
Thanks for the soultion, I will now go back to my pencils and papers.
Good night to you!

4. Originally Posted by a4swe
Excuse me, the others forums I am active on keep a high "one topic only" policy
Maybe it was okay to do that, I am just a moderator I like to boss people around, it makes me feel as a superior being (which of course I am).

5. Originally Posted by a4swe
Excuse me, the others forums I am active on keep a high "one topic only" policy, witch means that all subjects about let's say induction goes under one topic. That was not the case here I have to and will check the rules and posting guide lines.
Once again, I am sorry.
Frankly I don't care what other site's policies are. Posting a new
question into an existing thread confuses the thread, so very
quickly nobody can tell which question any new replies apply to.

If I find it happening I will split the thread, and if people persist
after having this explained to them they will be given negetive
reputation feedback.

RonL

6. Frankly I don't care what other site's policies are.
Off course you don't, I just wanted to explain (not to make excuses) my wrongfull actions, it won't happen again.

7. More on topic then:

Note that,
$\displaystyle 2k^3\geq 3k^2+3k+1$
For
Yes, I have noted that and yes it's true.
But how do I show it?

8. Originally Posted by a4swe
More on topic then:

Yes, I have noted that and yes it's true.
But how do I show it?
You may find it easier to show
$\displaystyle 3k^3 \ge (k + 1)^3$

or equivalently

$\displaystyle 3 \ge \frac{(k+1)^3}{k^3}$

This is easily done by considering different values of k and noting that $\displaystyle \frac{(k+1)^3}{k^3}$ monotonically decreases to a value of 1 as k increases indefinitely.

-Dan

9. A lot easier, thanks.
Off to bed now.

10. Originally Posted by a4swe
Yes, I have noted that and yes it's true.
But how do I show it?
Use another induction argument

11. Use another induction argument
Was afraid so.
Ok, we try that.