Mixing
Limit $ a_{n+1} = frac{1}{3}(2a_n + frac{5}{a_n^{2}}), a_1 > 0 $
Let $ a_{n+1} = frac{1}{3}(2a_n + frac{5}{a_n^{2}}), a_1 > 0 $. Show the sequence converge and find it's limit.
I did the following:
We see that $a_n$ is positive for every n obviously (considering $a_1 > 0$). Then:
$$ a_{n+1} - a_n = frac{2a_n^{3} + 5}{3a_n^{2}} - a_n = frac{5 - a_n^{3}}{3a_n^{2}}$$
We see that for each $a_n > ^3sqrt{5}$ the sequence is decreasing.
That means the proof collapses to prove $a_n > ^3sqrt{5}$ in order for $a_n$ to be decreasing. Suppose $ a_n > ^3sqrt{5}$ then obviously $a_{n+1} = frac{2a_n^{3} + 5}{3a_n^{2}} > frac{2(sqrt[3]{5})^{3} + 5}{3(sqrt[3]{5})^{2}} = sqrt[3]{5}$ so we deduced $a_{n+1} > ^sqrt[3]{5} $.
(It's important to note $a_1$ might be less then $sqrt[3]{5}$, hence the sequence will decrease from the second element.)
So we observed $a_n$ is always positive and decreasig hence it's bounded below so it's bounded. As for it's limit:
$$begin{align} L = lim_{n to infty} a_{n+1} = lim_{n to infty} frac{1}{3}(2a_n + frac{5}{a_n^{2}}) = frac{1}{3}(2L + frac{5}{L^{2}}) = sqrt[3]{5}end{align}$$
I'm not sure as of the correctness of the solution. I'm aware that it's not a contradiction because it only implies it's in the epsilon-environment of $sqrt[3]{5}$, or rather in this case $(sqrt[3]{5}, sqrt[3]{5}+epsilon)$.
Is it correct?
EDIT: If anyone encounter a question in the likes, to sum it up you obviously should show the sequence is montonic and bounded. In this case for example we can show it using AM-GM inequality to see the sequence is bounded by $sqrt[3]{5}$ and then in the difference of two consequective elements we see the difference is negative due to every element being greater then the third root of 5. From here we know the sequence is bounded below and decreasing hence it's bounded. If we plug in the limit of the sequence to itself we will see it's the third root of five. And that's it.
calculus sequences-and-series analysis recursion
edited Nov 22 '18 at 10:18
Christian Blatter
172k7112326
asked Nov 21 '18 at 16:28
Moshe
376
add a comment |
Let $ a_{n+1} = frac{1}{3}(2a_n + frac{5}{a_n^{2}}), a_1 > 0 $. Show the sequence converge and find it's limit.
I did the following:
We see that $a_n$ is positive for every n obviously (considering $a_1 > 0$). Then:
$$ a_{n+1} - a_n = frac{2a_n^{3} + 5}{3a_n^{2}} - a_n = frac{5 - a_n^{3}}{3a_n^{2}}$$
We see that for each $a_n > ^3sqrt{5}$ the sequence is decreasing.
That means the proof collapses to prove $a_n > ^3sqrt{5}$ in order for $a_n$ to be decreasing. Suppose $ a_n > ^3sqrt{5}$ then obviously $a_{n+1} = frac{2a_n^{3} + 5}{3a_n^{2}} > frac{2(sqrt[3]{5})^{3} + 5}{3(sqrt[3]{5})^{2}} = sqrt[3]{5}$ so we deduced $a_{n+1} > ^sqrt[3]{5} $.
(It's important to note $a_1$ might be less then $sqrt[3]{5}$, hence the sequence will decrease from the second element.)
So we observed $a_n$ is always positive and decreasig hence it's bounded below so it's bounded. As for it's limit:
$$begin{align} L = lim_{n to infty} a_{n+1} = lim_{n to infty} frac{1}{3}(2a_n + frac{5}{a_n^{2}}) = frac{1}{3}(2L + frac{5}{L^{2}}) = sqrt[3]{5}end{align}$$
I'm not sure as of the correctness of the solution. I'm aware that it's not a contradiction because it only implies it's in the epsilon-environment of $sqrt[3]{5}$, or rather in this case $(sqrt[3]{5}, sqrt[3]{5}+epsilon)$.
Is it correct?
EDIT: If anyone encounter a question in the likes, to sum it up you obviously should show the sequence is montonic and bounded. In this case for example we can show it using AM-GM inequality to see the sequence is bounded by $sqrt[3]{5}$ and then in the difference of two consequective elements we see the difference is negative due to every element being greater then the third root of 5. From here we know the sequence is bounded below and decreasing hence it's bounded. If we plug in the limit of the sequence to itself we will see it's the third root of five. And that's it.
calculus sequences-and-series analysis recursion
edited Nov 22 '18 at 10:18
Christian Blatter
172k7112326
asked Nov 21 '18 at 16:28
Moshe
376
You can typeset $sqrt[3]{5}$ assqrt[3]{5}
– Martin R
Nov 22 '18 at 7:59
add a comment |
Let $ a_{n+1} = frac{1}{3}(2a_n + frac{5}{a_n^{2}}), a_1 > 0 $. Show the sequence converge and find it's limit.
I did the following:
We see that $a_n$ is positive for every n obviously (considering $a_1 > 0$). Then:
$$ a_{n+1} - a_n = frac{2a_n^{3} + 5}{3a_n^{2}} - a_n = frac{5 - a_n^{3}}{3a_n^{2}}$$
We see that for each $a_n > ^3sqrt{5}$ the sequence is decreasing.
That means the proof collapses to prove $a_n > ^3sqrt{5}$ in order for $a_n$ to be decreasing. Suppose $ a_n > ^3sqrt{5}$ then obviously $a_{n+1} = frac{2a_n^{3} + 5}{3a_n^{2}} > frac{2(sqrt[3]{5})^{3} + 5}{3(sqrt[3]{5})^{2}} = sqrt[3]{5}$ so we deduced $a_{n+1} > ^sqrt[3]{5} $.
(It's important to note $a_1$ might be less then $sqrt[3]{5}$, hence the sequence will decrease from the second element.)
So we observed $a_n$ is always positive and decreasig hence it's bounded below so it's bounded. As for it's limit:
$$begin{align} L = lim_{n to infty} a_{n+1} = lim_{n to infty} frac{1}{3}(2a_n + frac{5}{a_n^{2}}) = frac{1}{3}(2L + frac{5}{L^{2}}) = sqrt[3]{5}end{align}$$
I'm not sure as of the correctness of the solution. I'm aware that it's not a contradiction because it only implies it's in the epsilon-environment of $sqrt[3]{5}$, or rather in this case $(sqrt[3]{5}, sqrt[3]{5}+epsilon)$.
Is it correct?
EDIT: If anyone encounter a question in the likes, to sum it up you obviously should show the sequence is montonic and bounded. In this case for example we can show it using AM-GM inequality to see the sequence is bounded by $sqrt[3]{5}$ and then in the difference of two consequective elements we see the difference is negative due to every element being greater then the third root of 5. From here we know the sequence is bounded below and decreasing hence it's bounded. If we plug in the limit of the sequence to itself we will see it's the third root of five. And that's it.
calculus sequences-and-series analysis recursion
edited Nov 22 '18 at 10:18
Christian Blatter
172k7112326
asked Nov 21 '18 at 16:28
Moshe
376
Let $ a_{n+1} = frac{1}{3}(2a_n + frac{5}{a_n^{2}}), a_1 > 0 $. Show the sequence converge and find it's limit.
I did the following:
We see that $a_n$ is positive for every n obviously (considering $a_1 > 0$). Then:
$$ a_{n+1} - a_n = frac{2a_n^{3} + 5}{3a_n^{2}} - a_n = frac{5 - a_n^{3}}{3a_n^{2}}$$
We see that for each $a_n > ^3sqrt{5}$ the sequence is decreasing.
That means the proof collapses to prove $a_n > ^3sqrt{5}$ in order for $a_n$ to be decreasing. Suppose $ a_n > ^3sqrt{5}$ then obviously $a_{n+1} = frac{2a_n^{3} + 5}{3a_n^{2}} > frac{2(sqrt[3]{5})^{3} + 5}{3(sqrt[3]{5})^{2}} = sqrt[3]{5}$ so we deduced $a_{n+1} > ^sqrt[3]{5} $.
(It's important to note $a_1$ might be less then $sqrt[3]{5}$, hence the sequence will decrease from the second element.)
So we observed $a_n$ is always positive and decreasig hence it's bounded below so it's bounded. As for it's limit:
$$begin{align} L = lim_{n to infty} a_{n+1} = lim_{n to infty} frac{1}{3}(2a_n + frac{5}{a_n^{2}}) = frac{1}{3}(2L + frac{5}{L^{2}}) = sqrt[3]{5}end{align}$$
I'm not sure as of the correctness of the solution. I'm aware that it's not a contradiction because it only implies it's in the epsilon-environment of $sqrt[3]{5}$, or rather in this case $(sqrt[3]{5}, sqrt[3]{5}+epsilon)$.
Is it correct?
EDIT: If anyone encounter a question in the likes, to sum it up you obviously should show the sequence is montonic and bounded. In this case for example we can show it using AM-GM inequality to see the sequence is bounded by $sqrt[3]{5}$ and then in the difference of two consequective elements we see the difference is negative due to every element being greater then the third root of 5. From here we know the sequence is bounded below and decreasing hence it's bounded. If we plug in the limit of the sequence to itself we will see it's the third root of five. And that's it.
calculus sequences-and-series analysis recursion
calculus sequences-and-series analysis recursion
edited Nov 22 '18 at 10:18
Christian Blatter
172k7112326
asked Nov 21 '18 at 16:28
Moshe
376
edited Nov 22 '18 at 10:18
Christian Blatter
172k7112326
asked Nov 21 '18 at 16:28
Moshe
376
edited Nov 22 '18 at 10:18
Christian Blatter
172k7112326
edited Nov 22 '18 at 10:18
Christian Blatter
172k7112326
edited Nov 22 '18 at 10:18
Christian Blatter
172k7112326
172k7112326
asked Nov 21 '18 at 16:28
Moshe
376
asked Nov 21 '18 at 16:28
Moshe
376
asked Nov 21 '18 at 16:28
Moshe
376
376
You can typeset $sqrt[3]{5}$ assqrt[3]{5}
– Martin R
Nov 22 '18 at 7:59
add a comment |
You can typeset $sqrt[3]{5}$ assqrt[3]{5}
– Martin R
Nov 22 '18 at 7:59
You can typeset $sqrt[3]{5}$ as
sqrt[3]{5}– Martin R
Nov 22 '18 at 7:59
You can typeset $sqrt[3]{5}$ as
sqrt[3]{5}– Martin R
Nov 22 '18 at 7:59
add a comment |
2 Answers
2
active
oldest
votes
First observe that
$$
a_{n+1}=frac{1}{3}left(2a_n+frac{5}{a_n^2}right)=frac{1}{3}left(a_n+a_n+frac{5}{a_n^2}right)ge left(a_ncdot a_ncdot frac{5}{a_n^2}right)^{1/3}=5^{1/3}.
$$
Hence, $a_nge 5^{1/3}$, for all $n>1$.
Next observe that
$$
f(x)=frac{1}{3}left(2x+frac{5}{x^2}right)<x,
$$
for all $x$ in $[5^{1/3},infty)$, since
$$
x-f(x)=frac{1}{3}left(x-frac{5}{x^2}right)=frac{x^3-5}{3x^2}ge 0.
$$
Hence
$$
5^{1/3}le a_{n+1}=f(a_n)le a_n, quad text{for all $n>1$.}
$$
Thus ${a_n}$ is bounded and decreasing, and therefore it is convergent.
Also, if $a_nto a$, then, as $f$ is continuous in $(0,infty)$, then
$$
a=lim a_n=lim a_{n+1}=lim f(a_n)=lim f(a)
$$
and hence the limit $a$ satisfies
$$
a=f(a)=frac{1}{3}left(2a+frac{5}{a^2}right)
$$
and thus $a^3=5$, which implies that $a_nto 5^{1/3}$.
answered Nov 22 '18 at 10:44
Yiorgos S. Smyrlis
62.8k1383163
Thank you, that's that exact solution I intended.
– Moshe
Nov 22 '18 at 11:11
add a comment |
In order to for the sequence to converge:
$$lim_{ntoinfty}(a_{n+1}-a_{n})=0$$
Then, we can plug in the recursion formula for $a_{n+1}$ and we get
$$lim_{ntoinfty}(a_{n+1}-a_n)=lim_{ntoinfty}(frac{2}{3}a_n+frac{5}{3a_n^2}-a_n)=lim_{ntoinfty}(frac{1}{3}+frac{5}{3a_n^2})=3 lim_{ntoinfty}(-a_n+frac{5}{a_n^2})=3 lim_{ntoinfty}frac{5-a_n^3}{a_n^3}=0$$
As such, $lim_{ntoinfty} a_n=sqrt[3]5$
answered Nov 21 '18 at 17:20
MoKo19
1914
The first statement is false.
– lhf
Nov 22 '18 at 10:13
It is true that $a_2geqroot3of 5$ in any case, but you have not shown this.
– Christian Blatter
Nov 22 '18 at 10:27
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
First observe that
$$
a_{n+1}=frac{1}{3}left(2a_n+frac{5}{a_n^2}right)=frac{1}{3}left(a_n+a_n+frac{5}{a_n^2}right)ge left(a_ncdot a_ncdot frac{5}{a_n^2}right)^{1/3}=5^{1/3}.
$$
Hence, $a_nge 5^{1/3}$, for all $n>1$.
Next observe that
$$
f(x)=frac{1}{3}left(2x+frac{5}{x^2}right)<x,
$$
for all $x$ in $[5^{1/3},infty)$, since
$$
x-f(x)=frac{1}{3}left(x-frac{5}{x^2}right)=frac{x^3-5}{3x^2}ge 0.
$$
Hence
$$
5^{1/3}le a_{n+1}=f(a_n)le a_n, quad text{for all $n>1$.}
$$
Thus ${a_n}$ is bounded and decreasing, and therefore it is convergent.
Also, if $a_nto a$, then, as $f$ is continuous in $(0,infty)$, then
$$
a=lim a_n=lim a_{n+1}=lim f(a_n)=lim f(a)
$$
and hence the limit $a$ satisfies
$$
a=f(a)=frac{1}{3}left(2a+frac{5}{a^2}right)
$$
and thus $a^3=5$, which implies that $a_nto 5^{1/3}$.
answered Nov 22 '18 at 10:44
Yiorgos S. Smyrlis
62.8k1383163
Thank you, that's that exact solution I intended.
– Moshe
Nov 22 '18 at 11:11
add a comment |
First observe that
$$
a_{n+1}=frac{1}{3}left(2a_n+frac{5}{a_n^2}right)=frac{1}{3}left(a_n+a_n+frac{5}{a_n^2}right)ge left(a_ncdot a_ncdot frac{5}{a_n^2}right)^{1/3}=5^{1/3}.
$$
Hence, $a_nge 5^{1/3}$, for all $n>1$.
Next observe that
$$
f(x)=frac{1}{3}left(2x+frac{5}{x^2}right)<x,
$$
for all $x$ in $[5^{1/3},infty)$, since
$$
x-f(x)=frac{1}{3}left(x-frac{5}{x^2}right)=frac{x^3-5}{3x^2}ge 0.
$$
Hence
$$
5^{1/3}le a_{n+1}=f(a_n)le a_n, quad text{for all $n>1$.}
$$
Thus ${a_n}$ is bounded and decreasing, and therefore it is convergent.
Also, if $a_nto a$, then, as $f$ is continuous in $(0,infty)$, then
$$
a=lim a_n=lim a_{n+1}=lim f(a_n)=lim f(a)
$$
and hence the limit $a$ satisfies
$$
a=f(a)=frac{1}{3}left(2a+frac{5}{a^2}right)
$$
and thus $a^3=5$, which implies that $a_nto 5^{1/3}$.
answered Nov 22 '18 at 10:44
Yiorgos S. Smyrlis
62.8k1383163
Thank you, that's that exact solution I intended.
– Moshe
Nov 22 '18 at 11:11
add a comment |
First observe that
$$
a_{n+1}=frac{1}{3}left(2a_n+frac{5}{a_n^2}right)=frac{1}{3}left(a_n+a_n+frac{5}{a_n^2}right)ge left(a_ncdot a_ncdot frac{5}{a_n^2}right)^{1/3}=5^{1/3}.
$$
Hence, $a_nge 5^{1/3}$, for all $n>1$.
Next observe that
$$
f(x)=frac{1}{3}left(2x+frac{5}{x^2}right)<x,
$$
for all $x$ in $[5^{1/3},infty)$, since
$$
x-f(x)=frac{1}{3}left(x-frac{5}{x^2}right)=frac{x^3-5}{3x^2}ge 0.
$$
Hence
$$
5^{1/3}le a_{n+1}=f(a_n)le a_n, quad text{for all $n>1$.}
$$
Thus ${a_n}$ is bounded and decreasing, and therefore it is convergent.
Also, if $a_nto a$, then, as $f$ is continuous in $(0,infty)$, then
$$
a=lim a_n=lim a_{n+1}=lim f(a_n)=lim f(a)
$$
and hence the limit $a$ satisfies
$$
a=f(a)=frac{1}{3}left(2a+frac{5}{a^2}right)
$$
and thus $a^3=5$, which implies that $a_nto 5^{1/3}$.
answered Nov 22 '18 at 10:44
Yiorgos S. Smyrlis
62.8k1383163
First observe that
$$
a_{n+1}=frac{1}{3}left(2a_n+frac{5}{a_n^2}right)=frac{1}{3}left(a_n+a_n+frac{5}{a_n^2}right)ge left(a_ncdot a_ncdot frac{5}{a_n^2}right)^{1/3}=5^{1/3}.
$$
Hence, $a_nge 5^{1/3}$, for all $n>1$.
Next observe that
$$
f(x)=frac{1}{3}left(2x+frac{5}{x^2}right)<x,
$$
for all $x$ in $[5^{1/3},infty)$, since
$$
x-f(x)=frac{1}{3}left(x-frac{5}{x^2}right)=frac{x^3-5}{3x^2}ge 0.
$$
Hence
$$
5^{1/3}le a_{n+1}=f(a_n)le a_n, quad text{for all $n>1$.}
$$
Thus ${a_n}$ is bounded and decreasing, and therefore it is convergent.
Also, if $a_nto a$, then, as $f$ is continuous in $(0,infty)$, then
$$
a=lim a_n=lim a_{n+1}=lim f(a_n)=lim f(a)
$$
and hence the limit $a$ satisfies
$$
a=f(a)=frac{1}{3}left(2a+frac{5}{a^2}right)
$$
and thus $a^3=5$, which implies that $a_nto 5^{1/3}$.
answered Nov 22 '18 at 10:44
Yiorgos S. Smyrlis
62.8k1383163
edited Nov 22 '18 at 11:45
answered Nov 22 '18 at 10:44
Yiorgos S. Smyrlis
62.8k1383163
answered Nov 22 '18 at 10:44
Yiorgos S. Smyrlis
62.8k1383163
answered Nov 22 '18 at 10:44
Yiorgos S. Smyrlis
62.8k1383163
62.8k1383163
Thank you, that's that exact solution I intended.
– Moshe
Nov 22 '18 at 11:11
add a comment |
Thank you, that's that exact solution I intended.
– Moshe
Nov 22 '18 at 11:11
Thank you, that's that exact solution I intended.
– Moshe
Nov 22 '18 at 11:11
Thank you, that's that exact solution I intended.
– Moshe
Nov 22 '18 at 11:11
add a comment |
In order to for the sequence to converge:
$$lim_{ntoinfty}(a_{n+1}-a_{n})=0$$
Then, we can plug in the recursion formula for $a_{n+1}$ and we get
$$lim_{ntoinfty}(a_{n+1}-a_n)=lim_{ntoinfty}(frac{2}{3}a_n+frac{5}{3a_n^2}-a_n)=lim_{ntoinfty}(frac{1}{3}+frac{5}{3a_n^2})=3 lim_{ntoinfty}(-a_n+frac{5}{a_n^2})=3 lim_{ntoinfty}frac{5-a_n^3}{a_n^3}=0$$
As such, $lim_{ntoinfty} a_n=sqrt[3]5$
answered Nov 21 '18 at 17:20
MoKo19
1914
The first statement is false.
– lhf
Nov 22 '18 at 10:13
It is true that $a_2geqroot3of 5$ in any case, but you have not shown this.
– Christian Blatter
Nov 22 '18 at 10:27
add a comment |
In order to for the sequence to converge:
$$lim_{ntoinfty}(a_{n+1}-a_{n})=0$$
Then, we can plug in the recursion formula for $a_{n+1}$ and we get
$$lim_{ntoinfty}(a_{n+1}-a_n)=lim_{ntoinfty}(frac{2}{3}a_n+frac{5}{3a_n^2}-a_n)=lim_{ntoinfty}(frac{1}{3}+frac{5}{3a_n^2})=3 lim_{ntoinfty}(-a_n+frac{5}{a_n^2})=3 lim_{ntoinfty}frac{5-a_n^3}{a_n^3}=0$$
As such, $lim_{ntoinfty} a_n=sqrt[3]5$
answered Nov 21 '18 at 17:20
MoKo19
1914
The first statement is false.
– lhf
Nov 22 '18 at 10:13
It is true that $a_2geqroot3of 5$ in any case, but you have not shown this.
– Christian Blatter
Nov 22 '18 at 10:27
add a comment |
In order to for the sequence to converge:
$$lim_{ntoinfty}(a_{n+1}-a_{n})=0$$
Then, we can plug in the recursion formula for $a_{n+1}$ and we get
$$lim_{ntoinfty}(a_{n+1}-a_n)=lim_{ntoinfty}(frac{2}{3}a_n+frac{5}{3a_n^2}-a_n)=lim_{ntoinfty}(frac{1}{3}+frac{5}{3a_n^2})=3 lim_{ntoinfty}(-a_n+frac{5}{a_n^2})=3 lim_{ntoinfty}frac{5-a_n^3}{a_n^3}=0$$
As such, $lim_{ntoinfty} a_n=sqrt[3]5$
answered Nov 21 '18 at 17:20
MoKo19
1914
In order to for the sequence to converge:
$$lim_{ntoinfty}(a_{n+1}-a_{n})=0$$
Then, we can plug in the recursion formula for $a_{n+1}$ and we get
$$lim_{ntoinfty}(a_{n+1}-a_n)=lim_{ntoinfty}(frac{2}{3}a_n+frac{5}{3a_n^2}-a_n)=lim_{ntoinfty}(frac{1}{3}+frac{5}{3a_n^2})=3 lim_{ntoinfty}(-a_n+frac{5}{a_n^2})=3 lim_{ntoinfty}frac{5-a_n^3}{a_n^3}=0$$
As such, $lim_{ntoinfty} a_n=sqrt[3]5$
answered Nov 21 '18 at 17:20
MoKo19
1914
answered Nov 21 '18 at 17:20
MoKo19
1914
answered Nov 21 '18 at 17:20
MoKo19
1914
answered Nov 21 '18 at 17:20
MoKo19
1914
1914
The first statement is false.
– lhf
Nov 22 '18 at 10:13
It is true that $a_2geqroot3of 5$ in any case, but you have not shown this.
– Christian Blatter
Nov 22 '18 at 10:27
add a comment |
The first statement is false.
– lhf
Nov 22 '18 at 10:13
It is true that $a_2geqroot3of 5$ in any case, but you have not shown this.
– Christian Blatter
Nov 22 '18 at 10:27
The first statement is false.
– lhf
Nov 22 '18 at 10:13
The first statement is false.
– lhf
Nov 22 '18 at 10:13
It is true that $a_2geqroot3of 5$ in any case, but you have not shown this.
– Christian Blatter
Nov 22 '18 at 10:27
It is true that $a_2geqroot3of 5$ in any case, but you have not shown this.
– Christian Blatter
Nov 22 '18 at 10:27
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You can typeset $sqrt[3]{5}$ as
sqrt[3]{5}– Martin R
Nov 22 '18 at 7:59