1

I am trying to prove that

$$e^{k+1} ≥ 3 + 3k + k^2$$ with, $$k>2$$

WhatI have done so far:

What we are trying to prove is that $$e^n≥1+n+n^2$$ is a true statement. Since $n=3$ holds, this is our base case. Then, the inductive hypothesis is: $$e^k≥1+k+k^2$$ Hence, I need to show $$e^{k+1}≥1+(k+1)+(k+1)^2=3+3k+3k^2$$ is a true statement.

Lerbi
  • 303
  • 2
  • 9
  • $e^{k} ≥ 1 + k + k^2$ may not be true because $e^{1}=e< 3=1 + 1 + 1^2$ – mike Nov 21 '14 at 01:23
  • Sorry, fixed the question! – Lerbi Nov 21 '14 at 01:33
  • If you're trying to prove $e^{k+1}\geq 3+3k+k^2$, shouldn't your inductive hypothesis be $e^{k}\geq 3+3(k-1)+3(k-1)^2$? – Randy E Nov 21 '14 at 01:45
  • No sir. What I'm trying to prove is that $e^n≥1+n+n^2$ is a true statement. Since $n=3$ holds, this is our base case. Then, the inductive hypothesis is: $e^k≥1+k+k^2$. Hence, I need to show $e^{k+1}≥1+(k+1)+(k+1)^2=3+3k+3k^2$ is a true statement. – Lerbi Nov 21 '14 at 06:41

3 Answers3

1

It does not hold.

What about $k=1$? Then $e=e^1 > 1+1+1=3$. It is a contradiction with $e \approx 2.7.$

ADDED:

If $k\ge 3$, then it holds.

Note that $e^x=1+x+\frac {x^2}{2}+\frac {x^3}{3!}+ \cdots+\frac {x^n}{n!}+\cdots$.

So $e^k-(1+k+k^2)>\frac{k^3}{3!}-\frac{k^2}{2}=k^2(\frac k6-\frac12)\ge 0$

Paul
  • 20,553
1

$e^{x}\geq (x^2+x+1)\\ \impliedby e^{x}\geq (x^2+x+\dfrac{1}{4})+\dfrac{3}{4}\\\impliedby e^x\geq \left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4} \\\impliedby e^{x+1}-e^x>\left(x+\dfrac{3}{2}\right)^2-\left(x+\dfrac{1}{2}\right)^2\\ \impliedby e^{x+1}-e^x> 2(x+1)\\ \impliedby f'(x)> 0 \ \ \forall x> \ln 2\ \left(f(x)=e^x-2(x+1)\right)$

0

To prove:

$$e \cdot[1+k+k^2] ≥ 3 + 3k + k^2\tag{1}$$

We can rearrange the terms in (1) (moving $k^2$ to the LHS and moving $e(1+k)$ to the RHS and prove:

$$(e-1)k^2 ≥ (3-e)(1+k).\tag{2}$$

This is true because $(e-1)≥(3-e)$ and $k^2≥(1+k),k>2$.

mike
  • 5,604
  • I'm not really following your answer..why is it necessary to show that $(e - 1)k^2 >= (3-e)(1+k)$? – Lerbi Nov 21 '14 at 01:55
  • please see my update... – mike Nov 21 '14 at 02:00
  • Oh whoops. Yes, I can prove it fairly easily IF I'm able to use both sides. But for these questions, my prof prefers it when we work from the left, until we eventually get to the right. So here I would show $e⋅[1+k+k^2] >= e⋅[1+k]+k^2]$... Or something like that. Until I can eventually get to $3 + 3k +k^2$. Some of these get pretty tricky. – Lerbi Nov 21 '14 at 06:32