Strong induction - prove that $n \le 3^\frac{n}3$ for every integer $n \ge 0$.

Question

This question was asked somewhere else, but I am having trouble with the algebra in the inductive step. And if you don't mind, let me know if anything else seems blatantly wrong as well.

So far I have :

Assume the predicate $P(n)$, where $n \le 3^\frac{n}3$. We will prove this is true for every $n\ge 0$ via strong induction. Basis: $P(0)= 0 \le 1$, $P(1)= 1\le 3^\frac13$, $P(2)= 2\le3^\frac23$, $P(3)= 3=3$ holds for first four numbers

Inductive Step: Let $n=k$. Assume that $P(k)$ is true where $k$ is $3 \le i \le k$ and $i$ being some integer less than $k$. We will prove this also holds for the $k+1$ case: $4 + 5 + ... + k + (k+1) \le 3^\frac{k+1}3$.

Using algebra: $-(k+1) \le 3^\frac13 * 3^\frac{k}3$
$-3^{-\frac13}*(k+1) \le 3^\frac{k}3$
$-3^{-\frac13}*(k+1) \le k$ (substituting $P(k)$ back in)
stuck here...

"This question was asked somewhere else" could you maybe provide a link to the original? — Surb, Sep 25 '18 at 19:28
Sorry, it was ask here, but i was still a little confused. https://math.stackexchange.com/questions/1025734/use-strong-induction-to-prove-that-n-leq3n-3-for-every-integer-n-geq0 — aBitwise, Sep 25 '18 at 19:33

score 2 · Accepted Answer · answered Sep 25 '18 at 19:23

2

Assume you know it is true up to $k$. For $k \ge 4$ we have $$3^{\frac {k+1}3}=3\cdot 3^{\frac {k-2}3}\ge 3\cdot (k-2)\gt k+1$$

answered Sep 25 '18 at 19:23

Ross Millikan

383,099

I still a little unsure of your reasoning for this. So, you split 3^((k+1)/3) up into 3 * 3^((k-2)/3) and set it greater than equal to 3 * (k-2). Why? – aBitwise Sep 25 '18 at 19:41
In an induction proof you want to relate the new case to an old one that has been proven. The first step brings in the same expression for $k-2$ which we assume has already been proven. The second step makes use of what has been proven. I think this answer is a great discussion – Ross Millikan Sep 25 '18 at 19:50
Ahh, I see. In strong induction, we assume two things. For this example k, but also all its predecessors. I think I understand. Appreciate the link as well, definitely will give it a read. – aBitwise Sep 25 '18 at 20:13
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Yes, that is the difference between strong induction and normal induction. In this case we really have three chains going upward, one for numbers $0 \bmod 3$, one for $1 \bmod 3$ and one for $2 \bmod 3$. We could formulate it as normal induction for each of those. We could also do normal induction by splitting off a factor $\sqrt[3]3$ instead of a factor $3$. There are many routes to the solution here. – Ross Millikan Sep 25 '18 at 20:17

score 1 · Answer 2 · answered Sep 25 '18 at 19:19

1

Hint: I would prove that $$n^3\le 3^n$$ then we have to Show that $$(n+1)^3\le 3^{n+1}$$ Multiplying the first inequality by $3$ we get $$3^{n+1}\geq 3n^3\geq (n+1)^3$$

answered Sep 25 '18 at 19:19

Dr. Sonnhard Graubner

97,058

So you think that $3 = 3\times1^3 \ge (1+1)^3 = 8 $? – gammatester Sep 25 '18 at 19:31
All examples for which this inequality not hold can you check, for all other numbers this inequality is true. – Dr. Sonnhard Graubner Sep 25 '18 at 19:34
This inequalitiy is true for all $$n\geq n_0$$ – Dr. Sonnhard Graubner Sep 25 '18 at 19:34
Again: Why do you think that $3n^3 \ge (n+1)^3?$ This is complete non-sense. – gammatester Sep 25 '18 at 19:36
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So we have $$3n^3\geq (n+1)^3$$ if $$n(\sqrt[3]{3}-1)\geq 1$$ this is $$n\geq \frac{1}{\sqrt[3]{3}-1}$$ – Dr. Sonnhard Graubner Sep 25 '18 at 19:36
What you says is nonsense, read about the induction principle – Dr. Sonnhard Graubner Sep 25 '18 at 19:37
$$??? $$ Substitute $n=1$ in your inequality $3^{n+1}\geq 3n^3\geq (n+1)^3$ and get $9 \ge 3 \ge 8$. Do you agree that this is complete non-sense? – gammatester Sep 25 '18 at 19:41
@gammatester $3n^3- (n+1)^3= 2 n^3 - 3 n^2 - 3 n - 1\geq 0$ for $n>2$ isn't it? – Surb Sep 25 '18 at 19:41
I will say it again, your statement is nonsense at all. – Dr. Sonnhard Graubner Sep 25 '18 at 19:42
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Every induction prove must NOT start for $n=1$!!!! – Dr. Sonnhard Graubner Sep 25 '18 at 19:43
It is your statement. Again: substitue $n=1$ in your inequality. – gammatester Sep 25 '18 at 19:43
Again. read about the induction principle! – Dr. Sonnhard Graubner Sep 25 '18 at 19:44
@gammatester $\mathbf{n > 2}$ – Surb Sep 25 '18 at 19:45
Why don't you try the substitution? – gammatester Sep 25 '18 at 19:45
@Surb: No $n\ge 0$ in the original question. I do not read that $n>2$ in the Doctor's answer. – gammatester Sep 25 '18 at 19:47
@gammatester Well, OP has checked the claim for $n=0,1,2,3$. – Surb Sep 25 '18 at 19:49
@Surb: Yes, and I do not say the statement is wrong. But the answer, as it stands, is missing this detail. In the comments the Dr. admitted that the inequality is valid for $n\ge n_0.$ This should be added to the answer. – gammatester Sep 25 '18 at 19:56
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The answer of Dr. SG is a Hint. It is clearly written there. A hint has not to contain all details. – Viera Čerňanová Sep 25 '18 at 21:21

score 0 · Answer 3 · answered Sep 25 '18 at 19:23

0

It is clear for $n = 1,2,3,4$. (Base cases). Assume it is true for $n$, now notice that \begin{equation} n+1\leq 3^{\frac{n}{3}} + 1 = 3^{-\frac{1}{3}}3^{\frac{n+1}{3}} + 1 \end{equation} But the function $f(x) = 3^{-\frac{1}{3}}x + 1 \leq x$ for $x \in [x_0,\infty]$ where $x_0 = \frac{1}{1-3^{-\frac{1}{3}}} \simeq 3.2612$. So we have proved it for $n \geq 4$.

answered Sep 25 '18 at 19:23

Ahmad Bazzi

12,238

how do you find $x_0$? – Surb Sep 25 '18 at 19:30
1

Find the intersection $3^{-\frac{1}{3}}x + 1 = x$ and solve for $x$. @Surb – Ahmad Bazzi Sep 25 '18 at 19:31
How did you get 3^(n/3) + 1 in the first part? Shouldn't it just be 3^((n+1)/3)? – aBitwise Sep 25 '18 at 19:43
By induction hypothesis, you have that $n \leq 3^{n/3}$. So just add $1$ on both sides. – Ahmad Bazzi Sep 25 '18 at 19:52

Strong induction - prove that $n \le 3^\frac{n}3$ for every integer $n \ge 0$.

3 Answers3