Mixing
prove this sequence is increasing
Clash Royale CLAN TAG#URR8PPP
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2
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The sequence $A_n$ is described as-
beginalign*
A_1 & = sqrt2\
A_n+1 & =(2+A_n)^0.5
endalign*
where $A_n$ is the $n$th term in the sequence.
I can prove by induction that this sequence is bounded by above by $2$ and then find the limit to be $2$. But in order to assume that the limit exists, I need to show that it is also increasing so I can use the monotone convergence theorem. It is easy to see why it is increasing, but how do I prove it?
real-analysis sequences-and-series
edited 1 hour ago
N. F. Taussig
42.1k93254
asked 2 hours ago
childishsadbino
395
add a comment |Â
up vote
2
down vote
favorite
The sequence $A_n$ is described as-
beginalign*
A_1 & = sqrt2\
A_n+1 & =(2+A_n)^0.5
endalign*
where $A_n$ is the $n$th term in the sequence.
I can prove by induction that this sequence is bounded by above by $2$ and then find the limit to be $2$. But in order to assume that the limit exists, I need to show that it is also increasing so I can use the monotone convergence theorem. It is easy to see why it is increasing, but how do I prove it?
real-analysis sequences-and-series
edited 1 hour ago
N. F. Taussig
42.1k93254
asked 2 hours ago
childishsadbino
395
Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
1 hour ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The sequence $A_n$ is described as-
beginalign*
A_1 & = sqrt2\
A_n+1 & =(2+A_n)^0.5
endalign*
where $A_n$ is the $n$th term in the sequence.
I can prove by induction that this sequence is bounded by above by $2$ and then find the limit to be $2$. But in order to assume that the limit exists, I need to show that it is also increasing so I can use the monotone convergence theorem. It is easy to see why it is increasing, but how do I prove it?
real-analysis sequences-and-series
edited 1 hour ago
N. F. Taussig
42.1k93254
asked 2 hours ago
childishsadbino
395
The sequence $A_n$ is described as-
beginalign*
A_1 & = sqrt2\
A_n+1 & =(2+A_n)^0.5
endalign*
where $A_n$ is the $n$th term in the sequence.
I can prove by induction that this sequence is bounded by above by $2$ and then find the limit to be $2$. But in order to assume that the limit exists, I need to show that it is also increasing so I can use the monotone convergence theorem. It is easy to see why it is increasing, but how do I prove it?
real-analysis sequences-and-series
real-analysis sequences-and-series
edited 1 hour ago
N. F. Taussig
42.1k93254
asked 2 hours ago
childishsadbino
395
edited 1 hour ago
N. F. Taussig
42.1k93254
asked 2 hours ago
childishsadbino
395
edited 1 hour ago
N. F. Taussig
42.1k93254
edited 1 hour ago
N. F. Taussig
42.1k93254
edited 1 hour ago
N. F. Taussig
42.1k93254
42.1k93254
asked 2 hours ago
childishsadbino
395
asked 2 hours ago
childishsadbino
395
asked 2 hours ago
childishsadbino
395
395
Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
1 hour ago
add a comment |Â
Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
1 hour ago
Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
1 hour ago
Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
1 hour ago
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
2
down vote
Prove by induction that $A_n leq 2$. Then note that $(A_n-frac 1 2)^2 leq frac 9 4$. When expanded this reduces to $A_n+1 geq A_n$.
answered 2 hours ago
Kavi Rama Murthy
37k31746
add a comment |Â
up vote
2
down vote
We need to prove that
$$A_n+1=sqrt2+A_nge A_n iff A_n^2-A_n-2 le 0 iff -1le A_n le2$$
then we have $A_n ge 0$ and by inducion we can prove that $A_nle 2$ indeed
- base case: $A_1=sqrt 2le 2$
- induction step: assuming true $A_nle 2$ we need to prove that $A_n+1le 2$ then
$$A_n+1=sqrt2+A_n lesqrt2+2=2$$
answered 1 hour ago
gimusi
82.6k74091
add a comment |Â
up vote
1
down vote
If we can assume $A_n, A_n+1 > 0$ then
$A_n+1 > A_n iff A_n+1^2 > A_n^2$
And $A_n+1^2n = ((2 + A_n)^frac 12) = 2 + A_n$
If we can assume $A_n < 2$ then $A_n+1^2 = 2 + A_n > A_n + A_n > 2*A_n > A_n*A_n = A_n^2$.
And that's that.
.....
We do have to prove that $A_n > 0$ but that's trivial:
By induction: Base case: $sqrt 2 > 0$. Induction step. If $A_n >0$ then $A_n + 2 > 0$ so $A_n+1 > sqrtA_n + 2 > 0$.
And we have to prove $A_n < 2$ but that's easy:
By induction: Base case: $sqrt 2 <2$. Induction step. If $A_n < 2$ then $A_n + 2 < 4$ and $A_n+1 = sqrt2+A_n < sqrt 4 = 2$.
=====
If we start from the begining we can to it is one fell swoop:
Proposition: $0 < A_1 < A_2 < .... A_n < A_n+1 <.... < 2$.
Base Case: $0 < A_1 = sqrt 2 < 2$
Induction case: Suppose $0 < A_n < 2$ then
$0 < A_n^2 < 2A_n=A_n + A_n < A_n + 2 < 2+2=4$
so $0< sqrtA_n^2 < sqrtA_n + 2 < sqrt4$
and $0< A_n < A_n+1 < 2$.
answered 1 hour ago
fleablood
64.8k22680
add a comment |Â
up vote
1
down vote
A proof by contradiction.
Suppose to the contrary that for some term, we have $A_k+1 leq A_k$. Suppose that this is the first instance for which this occurs. Now, we have that
$$A_k+1 = sqrt2+A_k,$$ so that by using our assumption,
$$ 2 + A_k= A_k+1^2 leq A_k^2.$$
Now, since $$A_k = sqrt2+A_k-1,$$
then also $$A_k^2 = 2+A_k-1.$$
So, plugging in above, we have that
$$2+A_k leq A_k^2 = 2+A_k-1.$$
In other words, $$A_k leq A_k-1.$$
This is a contradiction to the assumption that $A_k+1$ was the first instance for which this occurs. Thus, the sequence must be increasing.
answered 1 hour ago
Euler....IS_ALIVE
2,59311336
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Prove by induction that $A_n leq 2$. Then note that $(A_n-frac 1 2)^2 leq frac 9 4$. When expanded this reduces to $A_n+1 geq A_n$.
answered 2 hours ago
Kavi Rama Murthy
37k31746
add a comment |Â
up vote
2
down vote
Prove by induction that $A_n leq 2$. Then note that $(A_n-frac 1 2)^2 leq frac 9 4$. When expanded this reduces to $A_n+1 geq A_n$.
answered 2 hours ago
Kavi Rama Murthy
37k31746
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Prove by induction that $A_n leq 2$. Then note that $(A_n-frac 1 2)^2 leq frac 9 4$. When expanded this reduces to $A_n+1 geq A_n$.
answered 2 hours ago
Kavi Rama Murthy
37k31746
Prove by induction that $A_n leq 2$. Then note that $(A_n-frac 1 2)^2 leq frac 9 4$. When expanded this reduces to $A_n+1 geq A_n$.
answered 2 hours ago
Kavi Rama Murthy
37k31746
answered 2 hours ago
Kavi Rama Murthy
37k31746
answered 2 hours ago
Kavi Rama Murthy
37k31746
answered 2 hours ago
Kavi Rama Murthy
37k31746
37k31746
add a comment |Â
add a comment |Â
up vote
2
down vote
We need to prove that
$$A_n+1=sqrt2+A_nge A_n iff A_n^2-A_n-2 le 0 iff -1le A_n le2$$
then we have $A_n ge 0$ and by inducion we can prove that $A_nle 2$ indeed
- base case: $A_1=sqrt 2le 2$
- induction step: assuming true $A_nle 2$ we need to prove that $A_n+1le 2$ then
$$A_n+1=sqrt2+A_n lesqrt2+2=2$$
answered 1 hour ago
gimusi
82.6k74091
add a comment |Â
up vote
2
down vote
We need to prove that
$$A_n+1=sqrt2+A_nge A_n iff A_n^2-A_n-2 le 0 iff -1le A_n le2$$
then we have $A_n ge 0$ and by inducion we can prove that $A_nle 2$ indeed
- base case: $A_1=sqrt 2le 2$
- induction step: assuming true $A_nle 2$ we need to prove that $A_n+1le 2$ then
$$A_n+1=sqrt2+A_n lesqrt2+2=2$$
answered 1 hour ago
gimusi
82.6k74091
add a comment |Â
up vote
2
down vote
up vote
2
down vote
We need to prove that
$$A_n+1=sqrt2+A_nge A_n iff A_n^2-A_n-2 le 0 iff -1le A_n le2$$
then we have $A_n ge 0$ and by inducion we can prove that $A_nle 2$ indeed
- base case: $A_1=sqrt 2le 2$
- induction step: assuming true $A_nle 2$ we need to prove that $A_n+1le 2$ then
$$A_n+1=sqrt2+A_n lesqrt2+2=2$$
answered 1 hour ago
gimusi
82.6k74091
We need to prove that
$$A_n+1=sqrt2+A_nge A_n iff A_n^2-A_n-2 le 0 iff -1le A_n le2$$
then we have $A_n ge 0$ and by inducion we can prove that $A_nle 2$ indeed
- base case: $A_1=sqrt 2le 2$
- induction step: assuming true $A_nle 2$ we need to prove that $A_n+1le 2$ then
$$A_n+1=sqrt2+A_n lesqrt2+2=2$$
answered 1 hour ago
gimusi
82.6k74091
answered 1 hour ago
gimusi
82.6k74091
answered 1 hour ago
gimusi
82.6k74091
answered 1 hour ago
gimusi
82.6k74091
82.6k74091
add a comment |Â
add a comment |Â
up vote
1
down vote
If we can assume $A_n, A_n+1 > 0$ then
$A_n+1 > A_n iff A_n+1^2 > A_n^2$
And $A_n+1^2n = ((2 + A_n)^frac 12) = 2 + A_n$
If we can assume $A_n < 2$ then $A_n+1^2 = 2 + A_n > A_n + A_n > 2*A_n > A_n*A_n = A_n^2$.
And that's that.
.....
We do have to prove that $A_n > 0$ but that's trivial:
By induction: Base case: $sqrt 2 > 0$. Induction step. If $A_n >0$ then $A_n + 2 > 0$ so $A_n+1 > sqrtA_n + 2 > 0$.
And we have to prove $A_n < 2$ but that's easy:
By induction: Base case: $sqrt 2 <2$. Induction step. If $A_n < 2$ then $A_n + 2 < 4$ and $A_n+1 = sqrt2+A_n < sqrt 4 = 2$.
=====
If we start from the begining we can to it is one fell swoop:
Proposition: $0 < A_1 < A_2 < .... A_n < A_n+1 <.... < 2$.
Base Case: $0 < A_1 = sqrt 2 < 2$
Induction case: Suppose $0 < A_n < 2$ then
$0 < A_n^2 < 2A_n=A_n + A_n < A_n + 2 < 2+2=4$
so $0< sqrtA_n^2 < sqrtA_n + 2 < sqrt4$
and $0< A_n < A_n+1 < 2$.
answered 1 hour ago
fleablood
64.8k22680
add a comment |Â
up vote
1
down vote
If we can assume $A_n, A_n+1 > 0$ then
$A_n+1 > A_n iff A_n+1^2 > A_n^2$
And $A_n+1^2n = ((2 + A_n)^frac 12) = 2 + A_n$
If we can assume $A_n < 2$ then $A_n+1^2 = 2 + A_n > A_n + A_n > 2*A_n > A_n*A_n = A_n^2$.
And that's that.
.....
We do have to prove that $A_n > 0$ but that's trivial:
By induction: Base case: $sqrt 2 > 0$. Induction step. If $A_n >0$ then $A_n + 2 > 0$ so $A_n+1 > sqrtA_n + 2 > 0$.
And we have to prove $A_n < 2$ but that's easy:
By induction: Base case: $sqrt 2 <2$. Induction step. If $A_n < 2$ then $A_n + 2 < 4$ and $A_n+1 = sqrt2+A_n < sqrt 4 = 2$.
=====
If we start from the begining we can to it is one fell swoop:
Proposition: $0 < A_1 < A_2 < .... A_n < A_n+1 <.... < 2$.
Base Case: $0 < A_1 = sqrt 2 < 2$
Induction case: Suppose $0 < A_n < 2$ then
$0 < A_n^2 < 2A_n=A_n + A_n < A_n + 2 < 2+2=4$
so $0< sqrtA_n^2 < sqrtA_n + 2 < sqrt4$
and $0< A_n < A_n+1 < 2$.
answered 1 hour ago
fleablood
64.8k22680
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If we can assume $A_n, A_n+1 > 0$ then
$A_n+1 > A_n iff A_n+1^2 > A_n^2$
And $A_n+1^2n = ((2 + A_n)^frac 12) = 2 + A_n$
If we can assume $A_n < 2$ then $A_n+1^2 = 2 + A_n > A_n + A_n > 2*A_n > A_n*A_n = A_n^2$.
And that's that.
.....
We do have to prove that $A_n > 0$ but that's trivial:
By induction: Base case: $sqrt 2 > 0$. Induction step. If $A_n >0$ then $A_n + 2 > 0$ so $A_n+1 > sqrtA_n + 2 > 0$.
And we have to prove $A_n < 2$ but that's easy:
By induction: Base case: $sqrt 2 <2$. Induction step. If $A_n < 2$ then $A_n + 2 < 4$ and $A_n+1 = sqrt2+A_n < sqrt 4 = 2$.
=====
If we start from the begining we can to it is one fell swoop:
Proposition: $0 < A_1 < A_2 < .... A_n < A_n+1 <.... < 2$.
Base Case: $0 < A_1 = sqrt 2 < 2$
Induction case: Suppose $0 < A_n < 2$ then
$0 < A_n^2 < 2A_n=A_n + A_n < A_n + 2 < 2+2=4$
so $0< sqrtA_n^2 < sqrtA_n + 2 < sqrt4$
and $0< A_n < A_n+1 < 2$.
answered 1 hour ago
fleablood
64.8k22680
If we can assume $A_n, A_n+1 > 0$ then
$A_n+1 > A_n iff A_n+1^2 > A_n^2$
And $A_n+1^2n = ((2 + A_n)^frac 12) = 2 + A_n$
If we can assume $A_n < 2$ then $A_n+1^2 = 2 + A_n > A_n + A_n > 2*A_n > A_n*A_n = A_n^2$.
And that's that.
.....
We do have to prove that $A_n > 0$ but that's trivial:
By induction: Base case: $sqrt 2 > 0$. Induction step. If $A_n >0$ then $A_n + 2 > 0$ so $A_n+1 > sqrtA_n + 2 > 0$.
And we have to prove $A_n < 2$ but that's easy:
By induction: Base case: $sqrt 2 <2$. Induction step. If $A_n < 2$ then $A_n + 2 < 4$ and $A_n+1 = sqrt2+A_n < sqrt 4 = 2$.
=====
If we start from the begining we can to it is one fell swoop:
Proposition: $0 < A_1 < A_2 < .... A_n < A_n+1 <.... < 2$.
Base Case: $0 < A_1 = sqrt 2 < 2$
Induction case: Suppose $0 < A_n < 2$ then
$0 < A_n^2 < 2A_n=A_n + A_n < A_n + 2 < 2+2=4$
so $0< sqrtA_n^2 < sqrtA_n + 2 < sqrt4$
and $0< A_n < A_n+1 < 2$.
answered 1 hour ago
fleablood
64.8k22680
answered 1 hour ago
fleablood
64.8k22680
answered 1 hour ago
fleablood
64.8k22680
answered 1 hour ago
fleablood
64.8k22680
64.8k22680
add a comment |Â
add a comment |Â
up vote
1
down vote
A proof by contradiction.
Suppose to the contrary that for some term, we have $A_k+1 leq A_k$. Suppose that this is the first instance for which this occurs. Now, we have that
$$A_k+1 = sqrt2+A_k,$$ so that by using our assumption,
$$ 2 + A_k= A_k+1^2 leq A_k^2.$$
Now, since $$A_k = sqrt2+A_k-1,$$
then also $$A_k^2 = 2+A_k-1.$$
So, plugging in above, we have that
$$2+A_k leq A_k^2 = 2+A_k-1.$$
In other words, $$A_k leq A_k-1.$$
This is a contradiction to the assumption that $A_k+1$ was the first instance for which this occurs. Thus, the sequence must be increasing.
answered 1 hour ago
Euler....IS_ALIVE
2,59311336
add a comment |Â
up vote
1
down vote
A proof by contradiction.
Suppose to the contrary that for some term, we have $A_k+1 leq A_k$. Suppose that this is the first instance for which this occurs. Now, we have that
$$A_k+1 = sqrt2+A_k,$$ so that by using our assumption,
$$ 2 + A_k= A_k+1^2 leq A_k^2.$$
Now, since $$A_k = sqrt2+A_k-1,$$
then also $$A_k^2 = 2+A_k-1.$$
So, plugging in above, we have that
$$2+A_k leq A_k^2 = 2+A_k-1.$$
In other words, $$A_k leq A_k-1.$$
This is a contradiction to the assumption that $A_k+1$ was the first instance for which this occurs. Thus, the sequence must be increasing.
answered 1 hour ago
Euler....IS_ALIVE
2,59311336
add a comment |Â
up vote
1
down vote
up vote
1
down vote
A proof by contradiction.
Suppose to the contrary that for some term, we have $A_k+1 leq A_k$. Suppose that this is the first instance for which this occurs. Now, we have that
$$A_k+1 = sqrt2+A_k,$$ so that by using our assumption,
$$ 2 + A_k= A_k+1^2 leq A_k^2.$$
Now, since $$A_k = sqrt2+A_k-1,$$
then also $$A_k^2 = 2+A_k-1.$$
So, plugging in above, we have that
$$2+A_k leq A_k^2 = 2+A_k-1.$$
In other words, $$A_k leq A_k-1.$$
This is a contradiction to the assumption that $A_k+1$ was the first instance for which this occurs. Thus, the sequence must be increasing.
answered 1 hour ago
Euler....IS_ALIVE
2,59311336
A proof by contradiction.
Suppose to the contrary that for some term, we have $A_k+1 leq A_k$. Suppose that this is the first instance for which this occurs. Now, we have that
$$A_k+1 = sqrt2+A_k,$$ so that by using our assumption,
$$ 2 + A_k= A_k+1^2 leq A_k^2.$$
Now, since $$A_k = sqrt2+A_k-1,$$
then also $$A_k^2 = 2+A_k-1.$$
So, plugging in above, we have that
$$2+A_k leq A_k^2 = 2+A_k-1.$$
In other words, $$A_k leq A_k-1.$$
This is a contradiction to the assumption that $A_k+1$ was the first instance for which this occurs. Thus, the sequence must be increasing.
answered 1 hour ago
Euler....IS_ALIVE
2,59311336
answered 1 hour ago
Euler....IS_ALIVE
2,59311336
answered 1 hour ago
Euler....IS_ALIVE
2,59311336
answered 1 hour ago
Euler....IS_ALIVE
2,59311336
2,59311336
add a comment |Â
add a comment |Â
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Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
1 hour ago