Problems proving $e^{n+1} gt (1+n)^2$ [closed]
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I'm trying to prove by induction, but I'm getting stuck in the middle.
I'd use some help.
calculus
asked Nov 22 at 12:29
Bartosz
315
closed as off-topic by amWhy, José Carlos Santos, Cesareo, Chinnapparaj R, user302797 Nov 23 at 3:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Cesareo, Chinnapparaj R, user302797
If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
0
down vote
favorite
I'm trying to prove by induction, but I'm getting stuck in the middle.
I'd use some help.
calculus
asked Nov 22 at 12:29
Bartosz
315
closed as off-topic by amWhy, José Carlos Santos, Cesareo, Chinnapparaj R, user302797 Nov 23 at 3:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Cesareo, Chinnapparaj R, user302797
If this question can be reworded to fit the rules in the help center, please edit the question.
1
Take $n=4$. Then $e^{n+1}approx148.4$, whereas $(1+n)^n=5^4=625$. So, you will never be able to prove that.
– José Carlos Santos
Nov 22 at 12:31
@JoséCarlosSantos My mistake, it should be $(1+n)^2$
– Bartosz
Nov 22 at 12:33
Where did you get stuck?
– José Carlos Santos
Nov 22 at 12:36
I end up with $e^{k+2} = e cdot e^{k+1} > e * (1+k)^2 > ... $
– Bartosz
Nov 22 at 12:42
It is enough to show that $f(x)=(x+1)^2 e^{-x}>0$ is bounded by $e$ on $mathbb{R}^+$. On the other hand $f'(x)$ only vanishes at $x=pm 1$, hence $f(x)leq f(1) = frac{4}{e} < e$.
– Jack D'Aurizio
Nov 22 at 21:02
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to prove by induction, but I'm getting stuck in the middle.
I'd use some help.
calculus
asked Nov 22 at 12:29
Bartosz
315
I'm trying to prove by induction, but I'm getting stuck in the middle.
I'd use some help.
calculus
calculus
asked Nov 22 at 12:29
Bartosz
315
asked Nov 22 at 12:29
Bartosz
315
asked Nov 22 at 12:29
Bartosz
315
asked Nov 22 at 12:29
Bartosz
315
asked Nov 22 at 12:29
Bartosz
315
315
closed as off-topic by amWhy, José Carlos Santos, Cesareo, Chinnapparaj R, user302797 Nov 23 at 3:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Cesareo, Chinnapparaj R, user302797
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, José Carlos Santos, Cesareo, Chinnapparaj R, user302797 Nov 23 at 3:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Cesareo, Chinnapparaj R, user302797
If this question can be reworded to fit the rules in the help center, please edit the question.
1
Take $n=4$. Then $e^{n+1}approx148.4$, whereas $(1+n)^n=5^4=625$. So, you will never be able to prove that.
– José Carlos Santos
Nov 22 at 12:31
@JoséCarlosSantos My mistake, it should be $(1+n)^2$
– Bartosz
Nov 22 at 12:33
Where did you get stuck?
– José Carlos Santos
Nov 22 at 12:36
I end up with $e^{k+2} = e cdot e^{k+1} > e * (1+k)^2 > ... $
– Bartosz
Nov 22 at 12:42
It is enough to show that $f(x)=(x+1)^2 e^{-x}>0$ is bounded by $e$ on $mathbb{R}^+$. On the other hand $f'(x)$ only vanishes at $x=pm 1$, hence $f(x)leq f(1) = frac{4}{e} < e$.
– Jack D'Aurizio
Nov 22 at 21:02
add a comment |
1
Take $n=4$. Then $e^{n+1}approx148.4$, whereas $(1+n)^n=5^4=625$. So, you will never be able to prove that.
– José Carlos Santos
Nov 22 at 12:31
@JoséCarlosSantos My mistake, it should be $(1+n)^2$
– Bartosz
Nov 22 at 12:33
Where did you get stuck?
– José Carlos Santos
Nov 22 at 12:36
I end up with $e^{k+2} = e cdot e^{k+1} > e * (1+k)^2 > ... $
– Bartosz
Nov 22 at 12:42
It is enough to show that $f(x)=(x+1)^2 e^{-x}>0$ is bounded by $e$ on $mathbb{R}^+$. On the other hand $f'(x)$ only vanishes at $x=pm 1$, hence $f(x)leq f(1) = frac{4}{e} < e$.
– Jack D'Aurizio
Nov 22 at 21:02
1
1
Take $n=4$. Then $e^{n+1}approx148.4$, whereas $(1+n)^n=5^4=625$. So, you will never be able to prove that.
– José Carlos Santos
Nov 22 at 12:31
Take $n=4$. Then $e^{n+1}approx148.4$, whereas $(1+n)^n=5^4=625$. So, you will never be able to prove that.
– José Carlos Santos
Nov 22 at 12:31
@JoséCarlosSantos My mistake, it should be $(1+n)^2$
– Bartosz
Nov 22 at 12:33
@JoséCarlosSantos My mistake, it should be $(1+n)^2$
– Bartosz
Nov 22 at 12:33
Where did you get stuck?
– José Carlos Santos
Nov 22 at 12:36
Where did you get stuck?
– José Carlos Santos
Nov 22 at 12:36
I end up with $e^{k+2} = e cdot e^{k+1} > e * (1+k)^2 > ... $
– Bartosz
Nov 22 at 12:42
I end up with $e^{k+2} = e cdot e^{k+1} > e * (1+k)^2 > ... $
– Bartosz
Nov 22 at 12:42
It is enough to show that $f(x)=(x+1)^2 e^{-x}>0$ is bounded by $e$ on $mathbb{R}^+$. On the other hand $f'(x)$ only vanishes at $x=pm 1$, hence $f(x)leq f(1) = frac{4}{e} < e$.
– Jack D'Aurizio
Nov 22 at 21:02
It is enough to show that $f(x)=(x+1)^2 e^{-x}>0$ is bounded by $e$ on $mathbb{R}^+$. On the other hand $f'(x)$ only vanishes at $x=pm 1$, hence $f(x)leq f(1) = frac{4}{e} < e$.
– Jack D'Aurizio
Nov 22 at 21:02
add a comment |
3 Answers
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up vote
3
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Let consider
$$e^n>n^2$$
and we have
base case: $n=1$
induction step
$$e^{n+1}=ecdot e^n>ecdot n^2stackrel{?}>(n+1)^2$$
answered Nov 22 at 12:37
gimusi
92.5k94495
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up vote
1
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You can use the Taylor expansion of $e^{n+1}$.
begin{align}
e^{n+1}&=sum_{k=0}^inftyfrac{(n+1)^k}{k!}=1+1+n+frac{(n+1)^2}{2!}
+frac{(n+1)^3}{3!}+O((n+1)^4)\
&=1+n+1+frac{(n+1)^2}{2}+n^2/2+...=2+n+n^2+n+1/2+...\
&=n^2+2n+1+3/2+...>(n+1)^2
end{align}
answered Nov 22 at 12:39
minmax
48518
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up vote
0
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$n=1$ is ok.
Our assumption: $e^n>n^2$.
For $n>2ln(n)$ we must show that $n+1>2ln(n+1)$. $(n>1)$
$$n+1>2ln(n)+1 stackrel{?}>2ln(n+1)$$
$$1stackrel{?}>2ln(1+frac{1}{n})$$
$$e>(1+frac{1}{n})^2 text{Because maximum of } (1+frac{1}{n})^2=2.25 text{Big idea is there} n>1 . $$
answered Nov 22 at 13:09
1ENİGMA1
960316
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Let consider
$$e^n>n^2$$
and we have
base case: $n=1$
induction step
$$e^{n+1}=ecdot e^n>ecdot n^2stackrel{?}>(n+1)^2$$
answered Nov 22 at 12:37
gimusi
92.5k94495
add a comment |
up vote
3
down vote
Let consider
$$e^n>n^2$$
and we have
base case: $n=1$
induction step
$$e^{n+1}=ecdot e^n>ecdot n^2stackrel{?}>(n+1)^2$$
answered Nov 22 at 12:37
gimusi
92.5k94495
add a comment |
up vote
3
down vote
up vote
3
down vote
Let consider
$$e^n>n^2$$
and we have
base case: $n=1$
induction step
$$e^{n+1}=ecdot e^n>ecdot n^2stackrel{?}>(n+1)^2$$
answered Nov 22 at 12:37
gimusi
92.5k94495
Let consider
$$e^n>n^2$$
and we have
base case: $n=1$
induction step
$$e^{n+1}=ecdot e^n>ecdot n^2stackrel{?}>(n+1)^2$$
answered Nov 22 at 12:37
gimusi
92.5k94495
answered Nov 22 at 12:37
gimusi
92.5k94495
answered Nov 22 at 12:37
gimusi
92.5k94495
answered Nov 22 at 12:37
gimusi
92.5k94495
92.5k94495
add a comment |
add a comment |
up vote
1
down vote
You can use the Taylor expansion of $e^{n+1}$.
begin{align}
e^{n+1}&=sum_{k=0}^inftyfrac{(n+1)^k}{k!}=1+1+n+frac{(n+1)^2}{2!}
+frac{(n+1)^3}{3!}+O((n+1)^4)\
&=1+n+1+frac{(n+1)^2}{2}+n^2/2+...=2+n+n^2+n+1/2+...\
&=n^2+2n+1+3/2+...>(n+1)^2
end{align}
answered Nov 22 at 12:39
minmax
48518
add a comment |
up vote
1
down vote
You can use the Taylor expansion of $e^{n+1}$.
begin{align}
e^{n+1}&=sum_{k=0}^inftyfrac{(n+1)^k}{k!}=1+1+n+frac{(n+1)^2}{2!}
+frac{(n+1)^3}{3!}+O((n+1)^4)\
&=1+n+1+frac{(n+1)^2}{2}+n^2/2+...=2+n+n^2+n+1/2+...\
&=n^2+2n+1+3/2+...>(n+1)^2
end{align}
answered Nov 22 at 12:39
minmax
48518
add a comment |
up vote
1
down vote
up vote
1
down vote
You can use the Taylor expansion of $e^{n+1}$.
begin{align}
e^{n+1}&=sum_{k=0}^inftyfrac{(n+1)^k}{k!}=1+1+n+frac{(n+1)^2}{2!}
+frac{(n+1)^3}{3!}+O((n+1)^4)\
&=1+n+1+frac{(n+1)^2}{2}+n^2/2+...=2+n+n^2+n+1/2+...\
&=n^2+2n+1+3/2+...>(n+1)^2
end{align}
answered Nov 22 at 12:39
minmax
48518
You can use the Taylor expansion of $e^{n+1}$.
begin{align}
e^{n+1}&=sum_{k=0}^inftyfrac{(n+1)^k}{k!}=1+1+n+frac{(n+1)^2}{2!}
+frac{(n+1)^3}{3!}+O((n+1)^4)\
&=1+n+1+frac{(n+1)^2}{2}+n^2/2+...=2+n+n^2+n+1/2+...\
&=n^2+2n+1+3/2+...>(n+1)^2
end{align}
answered Nov 22 at 12:39
minmax
48518
edited Nov 22 at 13:24
answered Nov 22 at 12:39
minmax
48518
answered Nov 22 at 12:39
minmax
48518
answered Nov 22 at 12:39
minmax
48518
48518
add a comment |
add a comment |
up vote
0
down vote
$n=1$ is ok.
Our assumption: $e^n>n^2$.
For $n>2ln(n)$ we must show that $n+1>2ln(n+1)$. $(n>1)$
$$n+1>2ln(n)+1 stackrel{?}>2ln(n+1)$$
$$1stackrel{?}>2ln(1+frac{1}{n})$$
$$e>(1+frac{1}{n})^2 text{Because maximum of } (1+frac{1}{n})^2=2.25 text{Big idea is there} n>1 . $$
answered Nov 22 at 13:09
1ENİGMA1
960316
add a comment |
up vote
0
down vote
$n=1$ is ok.
Our assumption: $e^n>n^2$.
For $n>2ln(n)$ we must show that $n+1>2ln(n+1)$. $(n>1)$
$$n+1>2ln(n)+1 stackrel{?}>2ln(n+1)$$
$$1stackrel{?}>2ln(1+frac{1}{n})$$
$$e>(1+frac{1}{n})^2 text{Because maximum of } (1+frac{1}{n})^2=2.25 text{Big idea is there} n>1 . $$
answered Nov 22 at 13:09
1ENİGMA1
960316
add a comment |
up vote
0
down vote
up vote
0
down vote
$n=1$ is ok.
Our assumption: $e^n>n^2$.
For $n>2ln(n)$ we must show that $n+1>2ln(n+1)$. $(n>1)$
$$n+1>2ln(n)+1 stackrel{?}>2ln(n+1)$$
$$1stackrel{?}>2ln(1+frac{1}{n})$$
$$e>(1+frac{1}{n})^2 text{Because maximum of } (1+frac{1}{n})^2=2.25 text{Big idea is there} n>1 . $$
answered Nov 22 at 13:09
1ENİGMA1
960316
$n=1$ is ok.
Our assumption: $e^n>n^2$.
For $n>2ln(n)$ we must show that $n+1>2ln(n+1)$. $(n>1)$
$$n+1>2ln(n)+1 stackrel{?}>2ln(n+1)$$
$$1stackrel{?}>2ln(1+frac{1}{n})$$
$$e>(1+frac{1}{n})^2 text{Because maximum of } (1+frac{1}{n})^2=2.25 text{Big idea is there} n>1 . $$
answered Nov 22 at 13:09
1ENİGMA1
960316
answered Nov 22 at 13:09
1ENİGMA1
960316
answered Nov 22 at 13:09
1ENİGMA1
960316
answered Nov 22 at 13:09
1ENİGMA1
960316
960316
add a comment |
add a comment |
1
Take $n=4$. Then $e^{n+1}approx148.4$, whereas $(1+n)^n=5^4=625$. So, you will never be able to prove that.
– José Carlos Santos
Nov 22 at 12:31
@JoséCarlosSantos My mistake, it should be $(1+n)^2$
– Bartosz
Nov 22 at 12:33
Where did you get stuck?
– José Carlos Santos
Nov 22 at 12:36
I end up with $e^{k+2} = e cdot e^{k+1} > e * (1+k)^2 > ... $
– Bartosz
Nov 22 at 12:42
It is enough to show that $f(x)=(x+1)^2 e^{-x}>0$ is bounded by $e$ on $mathbb{R}^+$. On the other hand $f'(x)$ only vanishes at $x=pm 1$, hence $f(x)leq f(1) = frac{4}{e} < e$.
– Jack D'Aurizio
Nov 22 at 21:02