Induction proof for expression $4^n > n^3$

Question

I'm trying to proof that expression $(4^n>n^3)$ for $n\in \mathbb{N}$ using the induction.

1.There is $n0 = 0 $ for what $L=4^0=1$ and $P=n^0=0$

That is why $L>P$

2.Let's see what happen for $n+1$

Assumption: $4^n > n^3$ Thesis: $4*4^n > (n+1)^3$

I was always doing it like that $L = 4*4^n$ due to the assumption is greater than $4 * n^3$

And I do not know how can I solve it.

$(n+1)^3$ could be develop as a binomial serie. – user1050421 Mar 27 '16 at 22:29 — user1050421, Mar 27 '16 at 22:29

score 1 · Accepted Answer · answered Mar 27 '16 at 22:31

1

The bases case $n = 1, 2$ are clear since $4^1 = 4 > 1 = 1^3, 4^2 = 16 > 8 = 2^3$. Assume the statement is true when $n = k\geq 3$, you prove its true for $n = k+1$. This means to prove: $4^{k+1} > (k+1)^3$. We have: $4^{k+1} = 4\cdot 4^k > 4\cdot k^3 = k^3 + k^3+k^3+k^3 \geq k^3+3k^2 + 3k+ 1 = (k+1)^3$. Thus its true for all $n \geq 1$.

answered Mar 27 '16 at 22:31

DeepSea

77,651

You also need to say why k^3 +k^3 ... > 3k^2 + 3k + 1 though, but this is a better start – Inazuma Mar 27 '16 at 22:46
How can i prove that k^3+k^3.. > 3k^2 +3k +1? – FieryCod Mar 27 '16 at 22:54
1

@Charllie You want to see that $3k^3\ge 3k^2+3k+1$; it is equivalent to $3k(k^2-k-1)>1$. Note that, for $k>1$, $k^2-k-1>0$ and you're done. – egreg Mar 27 '16 at 23:00
You can look at this, and the 'duplicate' it also suggests – Inazuma Mar 27 '16 at 23:00
1

If your inductive hypothesis is that the statement is true when $n=k\ge3$, then you need to include $n=3$ as a base case. – Barry Cipra Mar 27 '16 at 23:26

score 1 · Answer 2 · answered Mar 27 '16 at 22:37

1

You want to show that if $4^n > n^3$ then $4 * 4^n > (n + 1)^3$.

If n ≥ 2 then $(1 + 1/n)^3 ≤ 1.5^3 = 3.375 < 4$. This shows that $(n + 1)^3 ≤ 3.375 n^3$ for n ≥ 2.

If n ≥ 2 then $4^{n+1} = 4 * 4^n > 4 * n^3 ≥ 3.375 n^3 ≥ (n+1)^3$.

Since $4^n > n^3$ is true for n = 1 and n = 2, it is true for all n.

answered Mar 27 '16 at 22:37

gnasher729

10,113

It is really a good proof? I thought that proof is when we operate on n not on real numbers. – FieryCod Mar 27 '16 at 22:40
This is not a valid proof. It has to be shown for the (n+1)th case, not just numbers 1 and 2 (as the pattern could terminate at 3, or 4 etc.) – Inazuma Mar 27 '16 at 22:43
This proof is valid. It uses $n \ge 2$ to show that $n^3$ grows by less than a factor of 4. – marty cohen Mar 28 '16 at 00:11

Induction proof for expression $4^n > n^3$

2 Answers2