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Regarding this question, I found no answer that combines mine, so I want to know if my proof is valid as well:

Problem: Prove that:$\ 3^n \gt n^4 \ \forall n \ge 8$

Base step (n=8): is true because$$P(8): 6561 \gt 4096 \\$$ Inductive step ($P(n) \implies P(n+1)$): $\ 3^{n+1} \gt \ (n+1)^4 =n^4 +4n^3+6n^2+4n+1$

$$3^{n+1} = 3^n + 3 = 3^n + 3^n + 3^n = 3^n + 3^n + \frac{(n+1)^4}{3} \gt (n+1)^4 \\ \text{I found } \frac{(n+1)^4}{3} \text{ by dividing 3 on both sides of } 3^{n+1} \gt (n+1)^4 \text{ as soon as I suppose this to be true.}$$

Loris Simonetti
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    Whoa... hold up. $3^n\cdot 3$ and you say that this is equal to $3^n\color{red}{\times}3^n\color{red}{\times}3^n$? No. $3^n\cdot 3 = 3^n\color{red}{+}3^n\color{red}{+}3^n$. There is a big difference between multiplication and addition. – JMoravitz Apr 01 '21 at 16:25
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    As for the structure of your proof in general, what you suppose is true would just be $P(n)$... that $3^n>n^4$ for some $k\geq 8$. You want to show that $P(n)\implies P(n+1)$, but you can not suppose that $P(n)\implies P(n+1)$... that would be circular reasoning. – JMoravitz Apr 01 '21 at 16:28
  • @JMoravitz I correct that, sorry it was a very Huge typo, thanks for the notice – Loris Simonetti Apr 01 '21 at 16:29
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    So... yes, we can have $3^{n+1} = 3^n\cdot 3 = 3^n+3^n+3^n$ and using the induction hypothesis say that this is greater than or equal to $n^4+n^4+n^4$... It is not clear in your attempt at a proof how you got to $\frac{(n+1)^4}{3}$ however, and this is where the real meat of the challenge lies... showing that $n^4+n^4+n^4\geq (n+1)^4$ – JMoravitz Apr 01 '21 at 16:31
  • @JMoravitz Ok, thanks for the answer, so basically what I did doesn't make any sense, right? – Loris Simonetti Apr 01 '21 at 16:32
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    Quite. There was a huge leap in logic that took you to $(n+1)^4$ that was unjustified, there were several typos confusing addition symbols for multiplication symbols or vice versa (even after the edit), and the grammar/structure of the proof was off. I encourage reading How to write a clear induction proof. – JMoravitz Apr 01 '21 at 16:34
  • Are you claiming $3^n = \frac {(n+1)^4}3$? And your your reason is because you divided $3$ on both sides of $3^{n+1}$? What are the two sides of $3^{n+1}$? So $\frac {3^{n+1}}3 = 3^n$ but how on earth doe $\frac{3^{n+1}}3 = \frac {(n+1)^4}3$? I don't understand what you are thinking at all. – fleablood Apr 01 '21 at 16:54
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    also even if $3^n = \frac {(n+1)^4}3$. Wouldn't that mean $3^n + 3^n + \frac{(n+1)^4}3 = \frac {(n+1)^4}3+ \frac {(n+1)^4}3+ \frac {(n+1)^4}3 = (n+1)^4$ and that would be equal, not greater than. What are you trying to do? – fleablood Apr 01 '21 at 16:57
  • @fleablood for the first comment the > is in the next line, however, you're absolutely right! – Loris Simonetti Apr 01 '21 at 17:05
  • @fleablood As I understand, the OP tried to use $3^{n+1}>(n+1)^4$ as a hypothesis to get that $3^n>\dfrac{(n+1)^4}{3}$ and tried to use this to prove the inductive step, but as alluded to previously this is circular reasoning. The whole point is that all we know is that $3^k>k^4$ for some particular value of $k\geq 8$ and we try to use this to prove that it follows that $3^{k+1}>(k+1)^4$. We can't assume the desired conclusion in order to prove itself, we must put in some effort to get to the conclusion using only the stated hypotheses. – JMoravitz Apr 01 '21 at 17:42
  • I'm a little surprised how many attempts at proofs by induction where you need to prove: $P(n)\implies P(n+1)$ will start with a simple statement of: $P(n+1)$. – fleablood Apr 01 '21 at 22:03

2 Answers2

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Rewrite the expression as $e^{n \log 3 - 4 \log n}$. This is your induction step. Now you need to show it for $n+1$. $$ e^{(n+1)\log 3 - 4 \log (n+1)} = e^{n \log 3 - 4 \log n} \cdot e^{\log 3 - \log (1+\frac{1}{n})} $$ The first term is greater that $1$ by the inductive argument. All is left to show that the second term is greater than $1$. This is true (again, using the definition of logarithm) for $$ \frac{3n}{n+1} >e $$ which is true for $n>\frac{e}{3-e} \approx 1.5$

Alex
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Inductive step ($P(n) \implies P(n+1):) \ 3^{n+1} \gt \ (n+1)^4 =n^4 +4n^3+6n^2+4n+1$

You need to PROVE $ 3^{n+1} \gt \ (n+1)^4 =n^4 +4n^3+6n^2+4n+1$. You can NOT start by assumenting that.

But you start by assuming $3^n > n^3$.

So $3^{n+1} = 3\cdot 3^n > 3n^4$

So how does $3n^4$ compare with $(n^4 + 4n^3 + 6n^2 + 4n + 1)$?

Well $3n^4 = n^4 + 2n^4$.

SO how does $2n^4$ compare with $4n^3 + 6n^2 + 4n + 1$?

Well $n \ge 8$ so $2n^4 = (2n)n^3 \ge 8n^3$.

So

$n^4 + 2n^4 \ge n^4 + 8n^3 = n^4 + 4n^3 + 4n^3$.

So how does $4n^3$ compare with $6n^2 + 4n + 1$.

Well, we just keep doing this $n \ge 8$ so $n^k \ge 8n^{k-1}$ to get:

$3^{n+1} = 3\cdot 3^n >$

$3n^4 = n^4 + 2n^4 \ge $

$n^4 + 8n^3 = n^4 + 4n^3 + 4n^3 \ge$

$n^4 + 4n^3 + 32n^2= n^4 + 4n^3 + 6n^2 + 26n^2\ge$

$n^4 + 4n^3 + 6n^2 + 26\cdot 8 n=$

$n^4 +4n^3 + 6n^2 + 4n + (26\cdot 8 - 4)n \ge$

$n^4 + 4n^3 + 6n + 4n + (26\cdot 8 -4)\cdot 8 >$

$n^4 + 4n^3 + 6n + 4n + 1=(n+1)^4$

fleablood
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  • why $n^4 + 4n^3 + n^3 + n^2 + n^3=n^4+7n^3$? – Loris Simonetti Apr 01 '21 at 17:40
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    @LorisSimonetti you should be able to see that for $n\geq 8$ you have that $n^4>n^3>n^2>n$, yes? And since $n\geq 8$ you have $n^2\geq 8n>7n>6n>5n>4n>\dots$ yes? Recall your basic properties of inequalities... that if $0<a<b$ and $0<c$ then $ac<bc$. – JMoravitz Apr 01 '21 at 17:45
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    @LorisSimonetti as for the specific line you refer to, I don't see it, but $n^4+4n^3+n^3+n^{\color{red}{3}}+n^3=n^4+7n^3$ as a simple matter of algebraic simplification... you have $4n^3+n^3+n^3+n^3 = (4+1+1+1)n^3$ by factoring a common factor of $n^3$. – JMoravitz Apr 01 '21 at 17:49
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    Twas a typo. It was meant to bey $n^3 $and $4n^3 + n^3 + n^3 + n^3 = (4 + 1+1+1)n^3 = 7n^3$. But I didn't really think they way was clear. And going one direction (backwards) and converting it do go forward, was weirdly not as straightforward as I thought it would be. – fleablood Apr 01 '21 at 21:59