Mixing
Limit of a sequence of integrals involving continued fractions
up vote
11
down vote
favorite
The following question was asked in a calculus exam in UNI, a Peruvian university. It is meant to be for freshman calculus students.
Find $lim_{n to infty} A_n $ if
$$ A_1 = intlimits_0^1 frac{dx}{1 + sqrt{x} }, ; ; ; A_2 =
intlimits_0^1 frac{dx}{1 + frac{1}{1+sqrt{x}} }, ; ; ; A_3 =
intlimits_0^1 frac{dx}{1 + frac{1}{1+frac{1}{1+sqrt{x}}} },
...$$
First of all, I think this is a hard question for a midterm exam, but anyway, notice that we can calculate $A_1$ by making $t=sqrt{x}$
$$ A_1 = intlimits_0^1 frac{2 t dt }{1+t} = 2 intlimits_0^1 dt - 2 intlimits_0^1 frac{dt}{1+t}=2-2(ln2)=2-ln2^2 $$
Now, as for $A_2$ I would do $t = frac{1}{1+sqrt{x}}$ which gives $d t = frac{ dx}{2 sqrt{x} (1+sqrt{x})^2} = frac{t^2 dx}{2 (t-1)}$ thus upon sustituticion we get
$$ A_2 = - intlimits_1^{1/2} frac{2 (t-1) }{t^2(1+t) } dt $$
which can easily solved by partial fractions or so. But, apparently this is not the way this problem is meant to be solved as this exam contained 4 questions to be solved in an hour. What is the trick, if any, that can be used to solve this problem without doing the unellegant partial fractions?
calculus limits definite-integrals continued-fractions
edited 4 hours ago
200_success
669515
asked 17 hours ago
Jimmy Sabater
1,799218
add a comment |
up vote
11
down vote
favorite
The following question was asked in a calculus exam in UNI, a Peruvian university. It is meant to be for freshman calculus students.
Find $lim_{n to infty} A_n $ if
$$ A_1 = intlimits_0^1 frac{dx}{1 + sqrt{x} }, ; ; ; A_2 =
intlimits_0^1 frac{dx}{1 + frac{1}{1+sqrt{x}} }, ; ; ; A_3 =
intlimits_0^1 frac{dx}{1 + frac{1}{1+frac{1}{1+sqrt{x}}} },
...$$
First of all, I think this is a hard question for a midterm exam, but anyway, notice that we can calculate $A_1$ by making $t=sqrt{x}$
$$ A_1 = intlimits_0^1 frac{2 t dt }{1+t} = 2 intlimits_0^1 dt - 2 intlimits_0^1 frac{dt}{1+t}=2-2(ln2)=2-ln2^2 $$
Now, as for $A_2$ I would do $t = frac{1}{1+sqrt{x}}$ which gives $d t = frac{ dx}{2 sqrt{x} (1+sqrt{x})^2} = frac{t^2 dx}{2 (t-1)}$ thus upon sustituticion we get
$$ A_2 = - intlimits_1^{1/2} frac{2 (t-1) }{t^2(1+t) } dt $$
which can easily solved by partial fractions or so. But, apparently this is not the way this problem is meant to be solved as this exam contained 4 questions to be solved in an hour. What is the trick, if any, that can be used to solve this problem without doing the unellegant partial fractions?
calculus limits definite-integrals continued-fractions
edited 4 hours ago
200_success
669515
asked 17 hours ago
Jimmy Sabater
1,799218
Can you prove it is valid to change the position of $lim$ and $int$?
– Kemono Chen
16 hours ago
1
If we let the integrand of $A_n$ be $f_n$, then $f_{n+1}=1/(1+f_n)$. If $f_infty$ exists, then necessarily $$f_infty=1/(1+f_infty)implies f_infty^2+f_infty-1=0$$. Indeed, $f_inftyequiv frac{-1+sqrt 5}2$ is a constant function. The main problem is justifying the switching of order between integral and limit. Dominated convergence theorem surely will do the job.
– Szeto
16 hours ago
1
@Szeto: Could you post your comment as an answer (with details)? (￣▽￣)
– Tianlalu
16 hours ago
1
I think this is meant to be solved without that dominated convergence theorem as it is meant for first-year students.
– Jimmy Sabater
16 hours ago
1
@JimmySabater Although first-year students have never heard of the dominated convergence theorem, they often act as though it were obvious.
– Andreas Blass
6 hours ago
add a comment |
up vote
11
down vote
favorite
up vote
11
down vote
favorite
The following question was asked in a calculus exam in UNI, a Peruvian university. It is meant to be for freshman calculus students.
Find $lim_{n to infty} A_n $ if
$$ A_1 = intlimits_0^1 frac{dx}{1 + sqrt{x} }, ; ; ; A_2 =
intlimits_0^1 frac{dx}{1 + frac{1}{1+sqrt{x}} }, ; ; ; A_3 =
intlimits_0^1 frac{dx}{1 + frac{1}{1+frac{1}{1+sqrt{x}}} },
...$$
First of all, I think this is a hard question for a midterm exam, but anyway, notice that we can calculate $A_1$ by making $t=sqrt{x}$
$$ A_1 = intlimits_0^1 frac{2 t dt }{1+t} = 2 intlimits_0^1 dt - 2 intlimits_0^1 frac{dt}{1+t}=2-2(ln2)=2-ln2^2 $$
Now, as for $A_2$ I would do $t = frac{1}{1+sqrt{x}}$ which gives $d t = frac{ dx}{2 sqrt{x} (1+sqrt{x})^2} = frac{t^2 dx}{2 (t-1)}$ thus upon sustituticion we get
$$ A_2 = - intlimits_1^{1/2} frac{2 (t-1) }{t^2(1+t) } dt $$
which can easily solved by partial fractions or so. But, apparently this is not the way this problem is meant to be solved as this exam contained 4 questions to be solved in an hour. What is the trick, if any, that can be used to solve this problem without doing the unellegant partial fractions?
calculus limits definite-integrals continued-fractions
edited 4 hours ago
200_success
669515
asked 17 hours ago
Jimmy Sabater
1,799218
The following question was asked in a calculus exam in UNI, a Peruvian university. It is meant to be for freshman calculus students.
Find $lim_{n to infty} A_n $ if
$$ A_1 = intlimits_0^1 frac{dx}{1 + sqrt{x} }, ; ; ; A_2 =
intlimits_0^1 frac{dx}{1 + frac{1}{1+sqrt{x}} }, ; ; ; A_3 =
intlimits_0^1 frac{dx}{1 + frac{1}{1+frac{1}{1+sqrt{x}}} },
...$$
First of all, I think this is a hard question for a midterm exam, but anyway, notice that we can calculate $A_1$ by making $t=sqrt{x}$
$$ A_1 = intlimits_0^1 frac{2 t dt }{1+t} = 2 intlimits_0^1 dt - 2 intlimits_0^1 frac{dt}{1+t}=2-2(ln2)=2-ln2^2 $$
Now, as for $A_2$ I would do $t = frac{1}{1+sqrt{x}}$ which gives $d t = frac{ dx}{2 sqrt{x} (1+sqrt{x})^2} = frac{t^2 dx}{2 (t-1)}$ thus upon sustituticion we get
$$ A_2 = - intlimits_1^{1/2} frac{2 (t-1) }{t^2(1+t) } dt $$
which can easily solved by partial fractions or so. But, apparently this is not the way this problem is meant to be solved as this exam contained 4 questions to be solved in an hour. What is the trick, if any, that can be used to solve this problem without doing the unellegant partial fractions?
calculus limits definite-integrals continued-fractions
calculus limits definite-integrals continued-fractions
edited 4 hours ago
200_success
669515
asked 17 hours ago
Jimmy Sabater
1,799218
edited 4 hours ago
200_success
669515
asked 17 hours ago
Jimmy Sabater
1,799218
edited 4 hours ago
200_success
669515
edited 4 hours ago
200_success
669515
edited 4 hours ago
200_success
669515
669515
asked 17 hours ago
Jimmy Sabater
1,799218
asked 17 hours ago
Jimmy Sabater
1,799218
asked 17 hours ago
Jimmy Sabater
1,799218
1,799218
Can you prove it is valid to change the position of $lim$ and $int$?
– Kemono Chen
16 hours ago
1
If we let the integrand of $A_n$ be $f_n$, then $f_{n+1}=1/(1+f_n)$. If $f_infty$ exists, then necessarily $$f_infty=1/(1+f_infty)implies f_infty^2+f_infty-1=0$$. Indeed, $f_inftyequiv frac{-1+sqrt 5}2$ is a constant function. The main problem is justifying the switching of order between integral and limit. Dominated convergence theorem surely will do the job.
– Szeto
16 hours ago
1
@Szeto: Could you post your comment as an answer (with details)? (￣▽￣)
– Tianlalu
16 hours ago
1
I think this is meant to be solved without that dominated convergence theorem as it is meant for first-year students.
– Jimmy Sabater
16 hours ago
1
@JimmySabater Although first-year students have never heard of the dominated convergence theorem, they often act as though it were obvious.
– Andreas Blass
6 hours ago
add a comment |
Can you prove it is valid to change the position of $lim$ and $int$?
– Kemono Chen
16 hours ago
1
If we let the integrand of $A_n$ be $f_n$, then $f_{n+1}=1/(1+f_n)$. If $f_infty$ exists, then necessarily $$f_infty=1/(1+f_infty)implies f_infty^2+f_infty-1=0$$. Indeed, $f_inftyequiv frac{-1+sqrt 5}2$ is a constant function. The main problem is justifying the switching of order between integral and limit. Dominated convergence theorem surely will do the job.
– Szeto
16 hours ago
1
@Szeto: Could you post your comment as an answer (with details)? (￣▽￣)
– Tianlalu
16 hours ago
1
I think this is meant to be solved without that dominated convergence theorem as it is meant for first-year students.
– Jimmy Sabater
16 hours ago
1
@JimmySabater Although first-year students have never heard of the dominated convergence theorem, they often act as though it were obvious.
– Andreas Blass
6 hours ago
Can you prove it is valid to change the position of $lim$ and $int$?
– Kemono Chen
16 hours ago
Can you prove it is valid to change the position of $lim$ and $int$?
– Kemono Chen
16 hours ago
1
1
If we let the integrand of $A_n$ be $f_n$, then $f_{n+1}=1/(1+f_n)$. If $f_infty$ exists, then necessarily $$f_infty=1/(1+f_infty)implies f_infty^2+f_infty-1=0$$. Indeed, $f_inftyequiv frac{-1+sqrt 5}2$ is a constant function. The main problem is justifying the switching of order between integral and limit. Dominated convergence theorem surely will do the job.
– Szeto
16 hours ago
If we let the integrand of $A_n$ be $f_n$, then $f_{n+1}=1/(1+f_n)$. If $f_infty$ exists, then necessarily $$f_infty=1/(1+f_infty)implies f_infty^2+f_infty-1=0$$. Indeed, $f_inftyequiv frac{-1+sqrt 5}2$ is a constant function. The main problem is justifying the switching of order between integral and limit. Dominated convergence theorem surely will do the job.
– Szeto
16 hours ago
1
1
@Szeto: Could you post your comment as an answer (with details)? (￣▽￣)
– Tianlalu
16 hours ago
@Szeto: Could you post your comment as an answer (with details)? (￣▽￣)
– Tianlalu
16 hours ago
1
1
I think this is meant to be solved without that dominated convergence theorem as it is meant for first-year students.
– Jimmy Sabater
16 hours ago
I think this is meant to be solved without that dominated convergence theorem as it is meant for first-year students.
– Jimmy Sabater
16 hours ago
1
1
@JimmySabater Although first-year students have never heard of the dominated convergence theorem, they often act as though it were obvious.
– Andreas Blass
6 hours ago
@JimmySabater Although first-year students have never heard of the dominated convergence theorem, they often act as though it were obvious.
– Andreas Blass
6 hours ago
add a comment |
3 Answers
3
active
oldest
votes
up vote
13
down vote
accepted
By the substitution $t=sqrt x$,
$$A_n=int^1_0f_n(t^2)(2tdt)$$
$f_n(t^2)$ is of the form
$$f_n(t^2)=frac{a_n+b_nt}{c_n+d_nt}$$
We have the recurrence relation
$$a_{n+1}=c_n$$
$$b_{n+1}=d_n$$
$$c_{n+1}=a_n+c_n$$
$$d_{n+1}=b_n+d_n$$
Or
$$c_{n+1}=c_n+c_{n-1}$$
$$d_{n+1}=d_n+d_{n-1}$$
which are the Fibonacci recurrence with initial conditions
$$c_0=1, c_1=1$$
$$d_0=0, d_1=1$$
I think you can now proceed.
Also, the general term of Fibonacci sequence $0,1,1,cdots$ is
$$frac{phi^n-overlinephi^n}{sqrt5}$$ where $phi=frac{1+sqrt 5}2$.
answered 16 hours ago
Szeto
5,9442726
8
It is very nice, but I believe this is a hard question to ask in a midterm for beginners. I dont think there is a more elementary solution, correct?
– Jimmy Sabater
16 hours ago
add a comment |
up vote
15
down vote
If I were taking that exam, I'd speculate covergence and write the integrand for $A_infty$ as
$$ S_infty(x) = frac{1}{ 1 + frac{1}{1+frac{1}{1+frac{1}{ddots} }}} = frac{1}{1+S_infty(x)}$$
Solve the resulting quadratic for $S_infty^2(x) + S_infty(x) -1 = 0$ for $S_infty(x)=frac{-1+sqrt{5}}{2}$. Then we immediately have $A_infty = S_infty$.
Then, I'd sit there and wonder what they intended for me to actually show on a freshman calculus exam.
answered 15 hours ago
zahbaz
8,02121837
2
Well, this is already mentioned in my comment on the OP. I don’t see the point to repost it as an answer. Moreover, this answer does not mention the switching of integral and limit, which is highly non-trivial. Poorly written.
– Szeto
14 hours ago
15
@Szeto Comments to the question should be left to improve the question, not to answer it. Comments may also be deleted if they do not serve this purpose. You should have incorporated your remarks into an/your answer.
– Dubu
14 hours ago
9
@Szeto Are you suggesting that I copied your comment? Because I didn't pay attention to it. This answer is simply what would first cross my mind in a test-taking scenario. Your comment has strong points in it, and I encourage you to turn it into an answer.
– zahbaz
13 hours ago
@zahbaz The may also have "intended" you to show that the limit actually exists, which is easy enough to prove.
– alephzero
10 hours ago
@zahbaz Is there a particular reason you choose the positive root to the quadratic equation?
– Jonathan Chiang
1 hour ago
|
show 1 more comment
up vote
3
down vote
Let $A_n = intlimits_0^1 f_n(x) dx$. First show that each $f_n$ is a monotonically increasing or decreasing function in the range $[0,1]$, and hence $A_n$ lies between $f_n(0)$ and $f_n(1)$. Then note that
$f_n(0) = 1, frac{1}{2}, frac{2}{3}, frac{3}{5}, dots$
are the convergents of the continued fraction expansion of $phi$, and $f_n(1) = f_{n+1}(0)$. So we have
$frac{1}{2} le A_1 le 1$
$frac{1}{2} le A_2 le frac{2}{3}$
$frac{3}{5} le A_3 le frac{2}{3}$
and so on.
So the sequence $A_n$ is squeezed between consecutive convergents of the continued fraction expansion of $phi$. And so $lim_{n to infty} A_n = phi$.
answered 12 hours ago
gandalf61
7,157523
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
13
down vote
accepted
By the substitution $t=sqrt x$,
$$A_n=int^1_0f_n(t^2)(2tdt)$$
$f_n(t^2)$ is of the form
$$f_n(t^2)=frac{a_n+b_nt}{c_n+d_nt}$$
We have the recurrence relation
$$a_{n+1}=c_n$$
$$b_{n+1}=d_n$$
$$c_{n+1}=a_n+c_n$$
$$d_{n+1}=b_n+d_n$$
Or
$$c_{n+1}=c_n+c_{n-1}$$
$$d_{n+1}=d_n+d_{n-1}$$
which are the Fibonacci recurrence with initial conditions
$$c_0=1, c_1=1$$
$$d_0=0, d_1=1$$
I think you can now proceed.
Also, the general term of Fibonacci sequence $0,1,1,cdots$ is
$$frac{phi^n-overlinephi^n}{sqrt5}$$ where $phi=frac{1+sqrt 5}2$.
answered 16 hours ago
Szeto
5,9442726
8
It is very nice, but I believe this is a hard question to ask in a midterm for beginners. I dont think there is a more elementary solution, correct?
– Jimmy Sabater
16 hours ago
add a comment |
up vote
13
down vote
accepted
By the substitution $t=sqrt x$,
$$A_n=int^1_0f_n(t^2)(2tdt)$$
$f_n(t^2)$ is of the form
$$f_n(t^2)=frac{a_n+b_nt}{c_n+d_nt}$$
We have the recurrence relation
$$a_{n+1}=c_n$$
$$b_{n+1}=d_n$$
$$c_{n+1}=a_n+c_n$$
$$d_{n+1}=b_n+d_n$$
Or
$$c_{n+1}=c_n+c_{n-1}$$
$$d_{n+1}=d_n+d_{n-1}$$
which are the Fibonacci recurrence with initial conditions
$$c_0=1, c_1=1$$
$$d_0=0, d_1=1$$
I think you can now proceed.
Also, the general term of Fibonacci sequence $0,1,1,cdots$ is
$$frac{phi^n-overlinephi^n}{sqrt5}$$ where $phi=frac{1+sqrt 5}2$.
answered 16 hours ago
Szeto
5,9442726
8
It is very nice, but I believe this is a hard question to ask in a midterm for beginners. I dont think there is a more elementary solution, correct?
– Jimmy Sabater
16 hours ago
add a comment |
up vote
13
down vote
accepted
up vote
13
down vote
accepted
By the substitution $t=sqrt x$,
$$A_n=int^1_0f_n(t^2)(2tdt)$$
$f_n(t^2)$ is of the form
$$f_n(t^2)=frac{a_n+b_nt}{c_n+d_nt}$$
We have the recurrence relation
$$a_{n+1}=c_n$$
$$b_{n+1}=d_n$$
$$c_{n+1}=a_n+c_n$$
$$d_{n+1}=b_n+d_n$$
Or
$$c_{n+1}=c_n+c_{n-1}$$
$$d_{n+1}=d_n+d_{n-1}$$
which are the Fibonacci recurrence with initial conditions
$$c_0=1, c_1=1$$
$$d_0=0, d_1=1$$
I think you can now proceed.
Also, the general term of Fibonacci sequence $0,1,1,cdots$ is
$$frac{phi^n-overlinephi^n}{sqrt5}$$ where $phi=frac{1+sqrt 5}2$.
answered 16 hours ago
Szeto
5,9442726
By the substitution $t=sqrt x$,
$$A_n=int^1_0f_n(t^2)(2tdt)$$
$f_n(t^2)$ is of the form
$$f_n(t^2)=frac{a_n+b_nt}{c_n+d_nt}$$
We have the recurrence relation
$$a_{n+1}=c_n$$
$$b_{n+1}=d_n$$
$$c_{n+1}=a_n+c_n$$
$$d_{n+1}=b_n+d_n$$
Or
$$c_{n+1}=c_n+c_{n-1}$$
$$d_{n+1}=d_n+d_{n-1}$$
which are the Fibonacci recurrence with initial conditions
$$c_0=1, c_1=1$$
$$d_0=0, d_1=1$$
I think you can now proceed.
Also, the general term of Fibonacci sequence $0,1,1,cdots$ is
$$frac{phi^n-overlinephi^n}{sqrt5}$$ where $phi=frac{1+sqrt 5}2$.
answered 16 hours ago
Szeto
5,9442726
edited 16 hours ago
answered 16 hours ago
Szeto
5,9442726
answered 16 hours ago
Szeto
5,9442726
answered 16 hours ago
Szeto
5,9442726
5,9442726
8
It is very nice, but I believe this is a hard question to ask in a midterm for beginners. I dont think there is a more elementary solution, correct?
– Jimmy Sabater
16 hours ago
add a comment |
8
It is very nice, but I believe this is a hard question to ask in a midterm for beginners. I dont think there is a more elementary solution, correct?
– Jimmy Sabater
16 hours ago
8
8
It is very nice, but I believe this is a hard question to ask in a midterm for beginners. I dont think there is a more elementary solution, correct?
– Jimmy Sabater
16 hours ago
It is very nice, but I believe this is a hard question to ask in a midterm for beginners. I dont think there is a more elementary solution, correct?
– Jimmy Sabater
16 hours ago
add a comment |
up vote
15
down vote
If I were taking that exam, I'd speculate covergence and write the integrand for $A_infty$ as
$$ S_infty(x) = frac{1}{ 1 + frac{1}{1+frac{1}{1+frac{1}{ddots} }}} = frac{1}{1+S_infty(x)}$$
Solve the resulting quadratic for $S_infty^2(x) + S_infty(x) -1 = 0$ for $S_infty(x)=frac{-1+sqrt{5}}{2}$. Then we immediately have $A_infty = S_infty$.
Then, I'd sit there and wonder what they intended for me to actually show on a freshman calculus exam.
answered 15 hours ago
zahbaz
8,02121837
2
Well, this is already mentioned in my comment on the OP. I don’t see the point to repost it as an answer. Moreover, this answer does not mention the switching of integral and limit, which is highly non-trivial. Poorly written.
– Szeto
14 hours ago
15
@Szeto Comments to the question should be left to improve the question, not to answer it. Comments may also be deleted if they do not serve this purpose. You should have incorporated your remarks into an/your answer.
– Dubu
14 hours ago
9
@Szeto Are you suggesting that I copied your comment? Because I didn't pay attention to it. This answer is simply what would first cross my mind in a test-taking scenario. Your comment has strong points in it, and I encourage you to turn it into an answer.
– zahbaz
13 hours ago
@zahbaz The may also have "intended" you to show that the limit actually exists, which is easy enough to prove.
– alephzero
10 hours ago
@zahbaz Is there a particular reason you choose the positive root to the quadratic equation?
– Jonathan Chiang
1 hour ago
|
show 1 more comment
up vote
15
down vote
If I were taking that exam, I'd speculate covergence and write the integrand for $A_infty$ as
$$ S_infty(x) = frac{1}{ 1 + frac{1}{1+frac{1}{1+frac{1}{ddots} }}} = frac{1}{1+S_infty(x)}$$
Solve the resulting quadratic for $S_infty^2(x) + S_infty(x) -1 = 0$ for $S_infty(x)=frac{-1+sqrt{5}}{2}$. Then we immediately have $A_infty = S_infty$.
Then, I'd sit there and wonder what they intended for me to actually show on a freshman calculus exam.
answered 15 hours ago
zahbaz
8,02121837
2
Well, this is already mentioned in my comment on the OP. I don’t see the point to repost it as an answer. Moreover, this answer does not mention the switching of integral and limit, which is highly non-trivial. Poorly written.
– Szeto
14 hours ago
15
@Szeto Comments to the question should be left to improve the question, not to answer it. Comments may also be deleted if they do not serve this purpose. You should have incorporated your remarks into an/your answer.
– Dubu
14 hours ago
9
@Szeto Are you suggesting that I copied your comment? Because I didn't pay attention to it. This answer is simply what would first cross my mind in a test-taking scenario. Your comment has strong points in it, and I encourage you to turn it into an answer.
– zahbaz
13 hours ago
@zahbaz The may also have "intended" you to show that the limit actually exists, which is easy enough to prove.
– alephzero
10 hours ago
@zahbaz Is there a particular reason you choose the positive root to the quadratic equation?
– Jonathan Chiang
1 hour ago
|
show 1 more comment
up vote
15
down vote
up vote
15
down vote
If I were taking that exam, I'd speculate covergence and write the integrand for $A_infty$ as
$$ S_infty(x) = frac{1}{ 1 + frac{1}{1+frac{1}{1+frac{1}{ddots} }}} = frac{1}{1+S_infty(x)}$$
Solve the resulting quadratic for $S_infty^2(x) + S_infty(x) -1 = 0$ for $S_infty(x)=frac{-1+sqrt{5}}{2}$. Then we immediately have $A_infty = S_infty$.
Then, I'd sit there and wonder what they intended for me to actually show on a freshman calculus exam.
answered 15 hours ago
zahbaz
8,02121837
If I were taking that exam, I'd speculate covergence and write the integrand for $A_infty$ as
$$ S_infty(x) = frac{1}{ 1 + frac{1}{1+frac{1}{1+frac{1}{ddots} }}} = frac{1}{1+S_infty(x)}$$
Solve the resulting quadratic for $S_infty^2(x) + S_infty(x) -1 = 0$ for $S_infty(x)=frac{-1+sqrt{5}}{2}$. Then we immediately have $A_infty = S_infty$.
Then, I'd sit there and wonder what they intended for me to actually show on a freshman calculus exam.
answered 15 hours ago
zahbaz
8,02121837
answered 15 hours ago
zahbaz
8,02121837
answered 15 hours ago
zahbaz
8,02121837
answered 15 hours ago
zahbaz
8,02121837
8,02121837
2
Well, this is already mentioned in my comment on the OP. I don’t see the point to repost it as an answer. Moreover, this answer does not mention the switching of integral and limit, which is highly non-trivial. Poorly written.
– Szeto
14 hours ago
15
@Szeto Comments to the question should be left to improve the question, not to answer it. Comments may also be deleted if they do not serve this purpose. You should have incorporated your remarks into an/your answer.
– Dubu
14 hours ago
9
@Szeto Are you suggesting that I copied your comment? Because I didn't pay attention to it. This answer is simply what would first cross my mind in a test-taking scenario. Your comment has strong points in it, and I encourage you to turn it into an answer.
– zahbaz
13 hours ago
@zahbaz The may also have "intended" you to show that the limit actually exists, which is easy enough to prove.
– alephzero
10 hours ago
@zahbaz Is there a particular reason you choose the positive root to the quadratic equation?
– Jonathan Chiang
1 hour ago
|
show 1 more comment
2
Well, this is already mentioned in my comment on the OP. I don’t see the point to repost it as an answer. Moreover, this answer does not mention the switching of integral and limit, which is highly non-trivial. Poorly written.
– Szeto
14 hours ago
15
@Szeto Comments to the question should be left to improve the question, not to answer it. Comments may also be deleted if they do not serve this purpose. You should have incorporated your remarks into an/your answer.
– Dubu
14 hours ago
9
@Szeto Are you suggesting that I copied your comment? Because I didn't pay attention to it. This answer is simply what would first cross my mind in a test-taking scenario. Your comment has strong points in it, and I encourage you to turn it into an answer.
– zahbaz
13 hours ago
@zahbaz The may also have "intended" you to show that the limit actually exists, which is easy enough to prove.
– alephzero
10 hours ago
@zahbaz Is there a particular reason you choose the positive root to the quadratic equation?
– Jonathan Chiang
1 hour ago
2
2
Well, this is already mentioned in my comment on the OP. I don’t see the point to repost it as an answer. Moreover, this answer does not mention the switching of integral and limit, which is highly non-trivial. Poorly written.
– Szeto
14 hours ago
Well, this is already mentioned in my comment on the OP. I don’t see the point to repost it as an answer. Moreover, this answer does not mention the switching of integral and limit, which is highly non-trivial. Poorly written.
– Szeto
14 hours ago
15
15
@Szeto Comments to the question should be left to improve the question, not to answer it. Comments may also be deleted if they do not serve this purpose. You should have incorporated your remarks into an/your answer.
– Dubu
14 hours ago
@Szeto Comments to the question should be left to improve the question, not to answer it. Comments may also be deleted if they do not serve this purpose. You should have incorporated your remarks into an/your answer.
– Dubu
14 hours ago
9
9
@Szeto Are you suggesting that I copied your comment? Because I didn't pay attention to it. This answer is simply what would first cross my mind in a test-taking scenario. Your comment has strong points in it, and I encourage you to turn it into an answer.
– zahbaz
13 hours ago
@Szeto Are you suggesting that I copied your comment? Because I didn't pay attention to it. This answer is simply what would first cross my mind in a test-taking scenario. Your comment has strong points in it, and I encourage you to turn it into an answer.
– zahbaz
13 hours ago
@zahbaz The may also have "intended" you to show that the limit actually exists, which is easy enough to prove.
– alephzero
10 hours ago
@zahbaz The may also have "intended" you to show that the limit actually exists, which is easy enough to prove.
– alephzero
10 hours ago
@zahbaz Is there a particular reason you choose the positive root to the quadratic equation?
– Jonathan Chiang
1 hour ago
@zahbaz Is there a particular reason you choose the positive root to the quadratic equation?
– Jonathan Chiang
1 hour ago
|
show 1 more comment
up vote
3
down vote
Let $A_n = intlimits_0^1 f_n(x) dx$. First show that each $f_n$ is a monotonically increasing or decreasing function in the range $[0,1]$, and hence $A_n$ lies between $f_n(0)$ and $f_n(1)$. Then note that
$f_n(0) = 1, frac{1}{2}, frac{2}{3}, frac{3}{5}, dots$
are the convergents of the continued fraction expansion of $phi$, and $f_n(1) = f_{n+1}(0)$. So we have
$frac{1}{2} le A_1 le 1$
$frac{1}{2} le A_2 le frac{2}{3}$
$frac{3}{5} le A_3 le frac{2}{3}$
and so on.
So the sequence $A_n$ is squeezed between consecutive convergents of the continued fraction expansion of $phi$. And so $lim_{n to infty} A_n = phi$.
answered 12 hours ago
gandalf61
7,157523
add a comment |
up vote
3
down vote
Let $A_n = intlimits_0^1 f_n(x) dx$. First show that each $f_n$ is a monotonically increasing or decreasing function in the range $[0,1]$, and hence $A_n$ lies between $f_n(0)$ and $f_n(1)$. Then note that
$f_n(0) = 1, frac{1}{2}, frac{2}{3}, frac{3}{5}, dots$
are the convergents of the continued fraction expansion of $phi$, and $f_n(1) = f_{n+1}(0)$. So we have
$frac{1}{2} le A_1 le 1$
$frac{1}{2} le A_2 le frac{2}{3}$
$frac{3}{5} le A_3 le frac{2}{3}$
and so on.
So the sequence $A_n$ is squeezed between consecutive convergents of the continued fraction expansion of $phi$. And so $lim_{n to infty} A_n = phi$.
answered 12 hours ago
gandalf61
7,157523
add a comment |
up vote
3
down vote
up vote
3
down vote
Let $A_n = intlimits_0^1 f_n(x) dx$. First show that each $f_n$ is a monotonically increasing or decreasing function in the range $[0,1]$, and hence $A_n$ lies between $f_n(0)$ and $f_n(1)$. Then note that
$f_n(0) = 1, frac{1}{2}, frac{2}{3}, frac{3}{5}, dots$
are the convergents of the continued fraction expansion of $phi$, and $f_n(1) = f_{n+1}(0)$. So we have
$frac{1}{2} le A_1 le 1$
$frac{1}{2} le A_2 le frac{2}{3}$
$frac{3}{5} le A_3 le frac{2}{3}$
and so on.
So the sequence $A_n$ is squeezed between consecutive convergents of the continued fraction expansion of $phi$. And so $lim_{n to infty} A_n = phi$.
answered 12 hours ago
gandalf61
7,157523
Let $A_n = intlimits_0^1 f_n(x) dx$. First show that each $f_n$ is a monotonically increasing or decreasing function in the range $[0,1]$, and hence $A_n$ lies between $f_n(0)$ and $f_n(1)$. Then note that
$f_n(0) = 1, frac{1}{2}, frac{2}{3}, frac{3}{5}, dots$
are the convergents of the continued fraction expansion of $phi$, and $f_n(1) = f_{n+1}(0)$. So we have
$frac{1}{2} le A_1 le 1$
$frac{1}{2} le A_2 le frac{2}{3}$
$frac{3}{5} le A_3 le frac{2}{3}$
and so on.
So the sequence $A_n$ is squeezed between consecutive convergents of the continued fraction expansion of $phi$. And so $lim_{n to infty} A_n = phi$.
answered 12 hours ago
gandalf61
7,157523
answered 12 hours ago
gandalf61
7,157523
answered 12 hours ago
gandalf61
7,157523
answered 12 hours ago
gandalf61
7,157523
7,157523
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004575%2flimit-of-a-sequence-of-integrals-involving-continued-fractions%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Can you prove it is valid to change the position of $lim$ and $int$?
– Kemono Chen
16 hours ago
1
If we let the integrand of $A_n$ be $f_n$, then $f_{n+1}=1/(1+f_n)$. If $f_infty$ exists, then necessarily $$f_infty=1/(1+f_infty)implies f_infty^2+f_infty-1=0$$. Indeed, $f_inftyequiv frac{-1+sqrt 5}2$ is a constant function. The main problem is justifying the switching of order between integral and limit. Dominated convergence theorem surely will do the job.
– Szeto
16 hours ago
1
@Szeto: Could you post your comment as an answer (with details)? (￣▽￣)
– Tianlalu
16 hours ago
1
I think this is meant to be solved without that dominated convergence theorem as it is meant for first-year students.
– Jimmy Sabater
16 hours ago
1
@JimmySabater Although first-year students have never heard of the dominated convergence theorem, they often act as though it were obvious.
– Andreas Blass
6 hours ago