Limit of a sequence of integrals involving continued fractions












16














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?










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  • Can you prove it is valid to change the position of $lim$ and $int$?
    – Kemono Chen
    Nov 19 '18 at 5:59






  • 1




    @Szeto: Could you post your comment as an answer (with details)? ( ̄▽ ̄)
    – Tianlalu
    Nov 19 '18 at 6:17






  • 1




    I think this is meant to be solved without that dominated convergence theorem as it is meant for first-year students.
    – Jimmy Sabater
    Nov 19 '18 at 6:17






  • 2




    @JimmySabater Although first-year students have never heard of the dominated convergence theorem, they often act as though it were obvious.
    – Andreas Blass
    Nov 19 '18 at 16:35
















16














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?










share|cite|improve this question
























  • Can you prove it is valid to change the position of $lim$ and $int$?
    – Kemono Chen
    Nov 19 '18 at 5:59






  • 1




    @Szeto: Could you post your comment as an answer (with details)? ( ̄▽ ̄)
    – Tianlalu
    Nov 19 '18 at 6:17






  • 1




    I think this is meant to be solved without that dominated convergence theorem as it is meant for first-year students.
    – Jimmy Sabater
    Nov 19 '18 at 6:17






  • 2




    @JimmySabater Although first-year students have never heard of the dominated convergence theorem, they often act as though it were obvious.
    – Andreas Blass
    Nov 19 '18 at 16:35














16












16








16


10





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?










share|cite|improve this question















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






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edited Nov 19 '18 at 18:12









200_success

669515




669515










asked Nov 19 '18 at 5:52









Jimmy Sabater

1,924219




1,924219












  • Can you prove it is valid to change the position of $lim$ and $int$?
    – Kemono Chen
    Nov 19 '18 at 5:59






  • 1




    @Szeto: Could you post your comment as an answer (with details)? ( ̄▽ ̄)
    – Tianlalu
    Nov 19 '18 at 6:17






  • 1




    I think this is meant to be solved without that dominated convergence theorem as it is meant for first-year students.
    – Jimmy Sabater
    Nov 19 '18 at 6:17






  • 2




    @JimmySabater Although first-year students have never heard of the dominated convergence theorem, they often act as though it were obvious.
    – Andreas Blass
    Nov 19 '18 at 16:35


















  • Can you prove it is valid to change the position of $lim$ and $int$?
    – Kemono Chen
    Nov 19 '18 at 5:59






  • 1




    @Szeto: Could you post your comment as an answer (with details)? ( ̄▽ ̄)
    – Tianlalu
    Nov 19 '18 at 6:17






  • 1




    I think this is meant to be solved without that dominated convergence theorem as it is meant for first-year students.
    – Jimmy Sabater
    Nov 19 '18 at 6:17






  • 2




    @JimmySabater Although first-year students have never heard of the dominated convergence theorem, they often act as though it were obvious.
    – Andreas Blass
    Nov 19 '18 at 16:35
















Can you prove it is valid to change the position of $lim$ and $int$?
– Kemono Chen
Nov 19 '18 at 5:59




Can you prove it is valid to change the position of $lim$ and $int$?
– Kemono Chen
Nov 19 '18 at 5:59




1




1




@Szeto: Could you post your comment as an answer (with details)? ( ̄▽ ̄)
– Tianlalu
Nov 19 '18 at 6:17




@Szeto: Could you post your comment as an answer (with details)? ( ̄▽ ̄)
– Tianlalu
Nov 19 '18 at 6:17




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
Nov 19 '18 at 6:17




I think this is meant to be solved without that dominated convergence theorem as it is meant for first-year students.
– Jimmy Sabater
Nov 19 '18 at 6:17




2




2




@JimmySabater Although first-year students have never heard of the dominated convergence theorem, they often act as though it were obvious.
– Andreas Blass
Nov 19 '18 at 16:35




@JimmySabater Although first-year students have never heard of the dominated convergence theorem, they often act as though it were obvious.
– Andreas Blass
Nov 19 '18 at 16:35










4 Answers
4






active

oldest

votes


















18














As requested by the OP in a comment deleted by a moderator, switching of limit and integral sign is avoided as it requires a higher-than-expected level of knowledge for justification. Thus, a ‘simpler’ approach is presented.



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 $phi=frac{1+sqrt 5}2$.






share|cite|improve this answer



















  • 10




    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
    Nov 19 '18 at 6:44



















26














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.






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  • @zahbaz Is there a particular reason you choose the positive root to the quadratic equation?
    – Jonathan Chiang
    Nov 19 '18 at 21:12










  • @JonathanChiang Every term in the continued fraction $S_infty(x)$ is positive and all operations involve addition, so it must be positive.
    – zahbaz
    Nov 19 '18 at 21:27












  • @zahbaz tell that to $zeta(-1)$
    – Peter
    Nov 20 '18 at 12:57



















8














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 $lim_{n to infty} A_n = phi$.






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    2














    $sqrt{x}$
    is a McGuffin.



    More generally,
    let
    $f_1 = frac{1}{1 + g(x) }
    $

    where
    $g'(x) > 0,
    g(0) = 0
    $
    ,
    $f_n(x)
    =frac{1}{1+f_{n-1}(x)}
    $
    ,
    and
    $A_n = int_0^1 f_n(x) dx
    $
    .



    Then
    $f_n(x)
    to dfrac{sqrt{5}-1}{2}
    $
    .



    Note:
    I doubt that any of this
    is original,
    but this was all done
    just now by me.



    Proof.



    $begin{array}\
    f_n(x)
    &=frac{1}{1+frac{1}{1+f_{n-2}(x)}}\
    &=frac{1+f_{n-2}(x)}{1+f_{n-2}(x)+1}\
    &=frac{1+f_{n-2}(x)}{2+f_{n-2}(x)}\
    end{array}
    $



    Therefore,
    if $f_{n-2}(x) > 0$
    then
    $frac12 < f_n(x)
    lt 1$
    .



    Similarly,
    if $f_{n-1}(x) > 0$
    then
    $0 < f_n(x)
    lt 1$
    .



    $begin{array}\
    f_n(x)-f_{n-2}(x)
    &=dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}-f_{n-2}(x)\
    &=dfrac{1+f_{n-2}(x)-f_{n-2}(x)(2+f_{n-2}(x))}{2+f_{n-2}(x)}\
    &=dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{2+f_{n-2}(x)}\
    end{array}
    $



    $begin{array}\
    f_n(x)+f_n^2(x)
    &=dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}+(dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)})^2\
    &=dfrac{(1+f_{n-2}(x))(2+f_{n-2}(x))}{(2+f_{n-2}(x))^2}+dfrac{1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
    &=dfrac{2+3f_{n-2}(x)+f_{n-2}^2(x)+1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
    &=dfrac{3+5f_{n-2}(x)+2f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
    text{so}\
    1-f_n(x)-f_n^2(x)
    &=dfrac{4+4f_{n-2}(x)+f_{n-2}^2(x)-(3+5f_{n-2}(x)+2f_{n-2}^2(x))}{(2+f_{n-2}(x))^2}\
    &=dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
    end{array}
    $



    Therefore
    $1-f_n(x)-f_n^2(x)$
    has the same sign as
    $1-f_{n-2}(x)-f_{n-2}^2(x)$.
    Also,
    $|1-f_n(x)-f_n^2(x)|
    lt frac14|1-f_{n-2}(x)-f_{n-2}^2(x)|
    $

    so
    $|1-f_n(x)-f_n^2(x)|
    to 0$
    .



    Let
    $p(x) = 1-x-x^2$
    and
    $x_0 = frac{sqrt{5}-1}{2}
    $

    so
    $p(x_0) = 0$,
    $p'(x) < 0$ for $x ge 0$.



    Since
    $f_n(x) > 0$,
    $f_n(x)
    to x_0$
    .






    share|cite|improve this answer





















    • what do you mean by Mcguffin?
      – Jimmy Sabater
      Dec 28 '18 at 8:52










    • Any g(x) > 0 will do. MacGuffin g.co/kgs/wDd8KQ
      – marty cohen
      Dec 28 '18 at 14:26











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    4 Answers
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    4 Answers
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    18














    As requested by the OP in a comment deleted by a moderator, switching of limit and integral sign is avoided as it requires a higher-than-expected level of knowledge for justification. Thus, a ‘simpler’ approach is presented.



    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 $phi=frac{1+sqrt 5}2$.






    share|cite|improve this answer



















    • 10




      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
      Nov 19 '18 at 6:44
















    18














    As requested by the OP in a comment deleted by a moderator, switching of limit and integral sign is avoided as it requires a higher-than-expected level of knowledge for justification. Thus, a ‘simpler’ approach is presented.



    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 $phi=frac{1+sqrt 5}2$.






    share|cite|improve this answer



















    • 10




      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
      Nov 19 '18 at 6:44














    18












    18








    18






    As requested by the OP in a comment deleted by a moderator, switching of limit and integral sign is avoided as it requires a higher-than-expected level of knowledge for justification. Thus, a ‘simpler’ approach is presented.



    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 $phi=frac{1+sqrt 5}2$.






    share|cite|improve this answer














    As requested by the OP in a comment deleted by a moderator, switching of limit and integral sign is avoided as it requires a higher-than-expected level of knowledge for justification. Thus, a ‘simpler’ approach is presented.



    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 $phi=frac{1+sqrt 5}2$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 20 '18 at 9:20

























    answered Nov 19 '18 at 6:29









    Szeto

    6,4362926




    6,4362926








    • 10




      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
      Nov 19 '18 at 6:44














    • 10




      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
      Nov 19 '18 at 6:44








    10




    10




    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
    Nov 19 '18 at 6:44




    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
    Nov 19 '18 at 6:44











    26














    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.






    share|cite|improve this answer





















    • @zahbaz Is there a particular reason you choose the positive root to the quadratic equation?
      – Jonathan Chiang
      Nov 19 '18 at 21:12










    • @JonathanChiang Every term in the continued fraction $S_infty(x)$ is positive and all operations involve addition, so it must be positive.
      – zahbaz
      Nov 19 '18 at 21:27












    • @zahbaz tell that to $zeta(-1)$
      – Peter
      Nov 20 '18 at 12:57
















    26














    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.






    share|cite|improve this answer





















    • @zahbaz Is there a particular reason you choose the positive root to the quadratic equation?
      – Jonathan Chiang
      Nov 19 '18 at 21:12










    • @JonathanChiang Every term in the continued fraction $S_infty(x)$ is positive and all operations involve addition, so it must be positive.
      – zahbaz
      Nov 19 '18 at 21:27












    • @zahbaz tell that to $zeta(-1)$
      – Peter
      Nov 20 '18 at 12:57














    26












    26








    26






    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.






    share|cite|improve this answer












    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.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 19 '18 at 7:29









    zahbaz

    8,22421937




    8,22421937












    • @zahbaz Is there a particular reason you choose the positive root to the quadratic equation?
      – Jonathan Chiang
      Nov 19 '18 at 21:12










    • @JonathanChiang Every term in the continued fraction $S_infty(x)$ is positive and all operations involve addition, so it must be positive.
      – zahbaz
      Nov 19 '18 at 21:27












    • @zahbaz tell that to $zeta(-1)$
      – Peter
      Nov 20 '18 at 12:57


















    • @zahbaz Is there a particular reason you choose the positive root to the quadratic equation?
      – Jonathan Chiang
      Nov 19 '18 at 21:12










    • @JonathanChiang Every term in the continued fraction $S_infty(x)$ is positive and all operations involve addition, so it must be positive.
      – zahbaz
      Nov 19 '18 at 21:27












    • @zahbaz tell that to $zeta(-1)$
      – Peter
      Nov 20 '18 at 12:57
















    @zahbaz Is there a particular reason you choose the positive root to the quadratic equation?
    – Jonathan Chiang
    Nov 19 '18 at 21:12




    @zahbaz Is there a particular reason you choose the positive root to the quadratic equation?
    – Jonathan Chiang
    Nov 19 '18 at 21:12












    @JonathanChiang Every term in the continued fraction $S_infty(x)$ is positive and all operations involve addition, so it must be positive.
    – zahbaz
    Nov 19 '18 at 21:27






    @JonathanChiang Every term in the continued fraction $S_infty(x)$ is positive and all operations involve addition, so it must be positive.
    – zahbaz
    Nov 19 '18 at 21:27














    @zahbaz tell that to $zeta(-1)$
    – Peter
    Nov 20 '18 at 12:57




    @zahbaz tell that to $zeta(-1)$
    – Peter
    Nov 20 '18 at 12:57











    8














    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 $lim_{n to infty} A_n = phi$.






    share|cite|improve this answer


























      8














      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 $lim_{n to infty} A_n = phi$.






      share|cite|improve this answer
























        8












        8








        8






        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 $lim_{n to infty} A_n = phi$.






        share|cite|improve this answer












        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 $lim_{n to infty} A_n = phi$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 19 '18 at 10:51









        gandalf61

        7,693623




        7,693623























            2














            $sqrt{x}$
            is a McGuffin.



            More generally,
            let
            $f_1 = frac{1}{1 + g(x) }
            $

            where
            $g'(x) > 0,
            g(0) = 0
            $
            ,
            $f_n(x)
            =frac{1}{1+f_{n-1}(x)}
            $
            ,
            and
            $A_n = int_0^1 f_n(x) dx
            $
            .



            Then
            $f_n(x)
            to dfrac{sqrt{5}-1}{2}
            $
            .



            Note:
            I doubt that any of this
            is original,
            but this was all done
            just now by me.



            Proof.



            $begin{array}\
            f_n(x)
            &=frac{1}{1+frac{1}{1+f_{n-2}(x)}}\
            &=frac{1+f_{n-2}(x)}{1+f_{n-2}(x)+1}\
            &=frac{1+f_{n-2}(x)}{2+f_{n-2}(x)}\
            end{array}
            $



            Therefore,
            if $f_{n-2}(x) > 0$
            then
            $frac12 < f_n(x)
            lt 1$
            .



            Similarly,
            if $f_{n-1}(x) > 0$
            then
            $0 < f_n(x)
            lt 1$
            .



            $begin{array}\
            f_n(x)-f_{n-2}(x)
            &=dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}-f_{n-2}(x)\
            &=dfrac{1+f_{n-2}(x)-f_{n-2}(x)(2+f_{n-2}(x))}{2+f_{n-2}(x)}\
            &=dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{2+f_{n-2}(x)}\
            end{array}
            $



            $begin{array}\
            f_n(x)+f_n^2(x)
            &=dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}+(dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)})^2\
            &=dfrac{(1+f_{n-2}(x))(2+f_{n-2}(x))}{(2+f_{n-2}(x))^2}+dfrac{1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
            &=dfrac{2+3f_{n-2}(x)+f_{n-2}^2(x)+1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
            &=dfrac{3+5f_{n-2}(x)+2f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
            text{so}\
            1-f_n(x)-f_n^2(x)
            &=dfrac{4+4f_{n-2}(x)+f_{n-2}^2(x)-(3+5f_{n-2}(x)+2f_{n-2}^2(x))}{(2+f_{n-2}(x))^2}\
            &=dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
            end{array}
            $



            Therefore
            $1-f_n(x)-f_n^2(x)$
            has the same sign as
            $1-f_{n-2}(x)-f_{n-2}^2(x)$.
            Also,
            $|1-f_n(x)-f_n^2(x)|
            lt frac14|1-f_{n-2}(x)-f_{n-2}^2(x)|
            $

            so
            $|1-f_n(x)-f_n^2(x)|
            to 0$
            .



            Let
            $p(x) = 1-x-x^2$
            and
            $x_0 = frac{sqrt{5}-1}{2}
            $

            so
            $p(x_0) = 0$,
            $p'(x) < 0$ for $x ge 0$.



            Since
            $f_n(x) > 0$,
            $f_n(x)
            to x_0$
            .






            share|cite|improve this answer





















            • what do you mean by Mcguffin?
              – Jimmy Sabater
              Dec 28 '18 at 8:52










            • Any g(x) > 0 will do. MacGuffin g.co/kgs/wDd8KQ
              – marty cohen
              Dec 28 '18 at 14:26
















            2














            $sqrt{x}$
            is a McGuffin.



            More generally,
            let
            $f_1 = frac{1}{1 + g(x) }
            $

            where
            $g'(x) > 0,
            g(0) = 0
            $
            ,
            $f_n(x)
            =frac{1}{1+f_{n-1}(x)}
            $
            ,
            and
            $A_n = int_0^1 f_n(x) dx
            $
            .



            Then
            $f_n(x)
            to dfrac{sqrt{5}-1}{2}
            $
            .



            Note:
            I doubt that any of this
            is original,
            but this was all done
            just now by me.



            Proof.



            $begin{array}\
            f_n(x)
            &=frac{1}{1+frac{1}{1+f_{n-2}(x)}}\
            &=frac{1+f_{n-2}(x)}{1+f_{n-2}(x)+1}\
            &=frac{1+f_{n-2}(x)}{2+f_{n-2}(x)}\
            end{array}
            $



            Therefore,
            if $f_{n-2}(x) > 0$
            then
            $frac12 < f_n(x)
            lt 1$
            .



            Similarly,
            if $f_{n-1}(x) > 0$
            then
            $0 < f_n(x)
            lt 1$
            .



            $begin{array}\
            f_n(x)-f_{n-2}(x)
            &=dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}-f_{n-2}(x)\
            &=dfrac{1+f_{n-2}(x)-f_{n-2}(x)(2+f_{n-2}(x))}{2+f_{n-2}(x)}\
            &=dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{2+f_{n-2}(x)}\
            end{array}
            $



            $begin{array}\
            f_n(x)+f_n^2(x)
            &=dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}+(dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)})^2\
            &=dfrac{(1+f_{n-2}(x))(2+f_{n-2}(x))}{(2+f_{n-2}(x))^2}+dfrac{1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
            &=dfrac{2+3f_{n-2}(x)+f_{n-2}^2(x)+1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
            &=dfrac{3+5f_{n-2}(x)+2f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
            text{so}\
            1-f_n(x)-f_n^2(x)
            &=dfrac{4+4f_{n-2}(x)+f_{n-2}^2(x)-(3+5f_{n-2}(x)+2f_{n-2}^2(x))}{(2+f_{n-2}(x))^2}\
            &=dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
            end{array}
            $



            Therefore
            $1-f_n(x)-f_n^2(x)$
            has the same sign as
            $1-f_{n-2}(x)-f_{n-2}^2(x)$.
            Also,
            $|1-f_n(x)-f_n^2(x)|
            lt frac14|1-f_{n-2}(x)-f_{n-2}^2(x)|
            $

            so
            $|1-f_n(x)-f_n^2(x)|
            to 0$
            .



            Let
            $p(x) = 1-x-x^2$
            and
            $x_0 = frac{sqrt{5}-1}{2}
            $

            so
            $p(x_0) = 0$,
            $p'(x) < 0$ for $x ge 0$.



            Since
            $f_n(x) > 0$,
            $f_n(x)
            to x_0$
            .






            share|cite|improve this answer





















            • what do you mean by Mcguffin?
              – Jimmy Sabater
              Dec 28 '18 at 8:52










            • Any g(x) > 0 will do. MacGuffin g.co/kgs/wDd8KQ
              – marty cohen
              Dec 28 '18 at 14:26














            2












            2








            2






            $sqrt{x}$
            is a McGuffin.



            More generally,
            let
            $f_1 = frac{1}{1 + g(x) }
            $

            where
            $g'(x) > 0,
            g(0) = 0
            $
            ,
            $f_n(x)
            =frac{1}{1+f_{n-1}(x)}
            $
            ,
            and
            $A_n = int_0^1 f_n(x) dx
            $
            .



            Then
            $f_n(x)
            to dfrac{sqrt{5}-1}{2}
            $
            .



            Note:
            I doubt that any of this
            is original,
            but this was all done
            just now by me.



            Proof.



            $begin{array}\
            f_n(x)
            &=frac{1}{1+frac{1}{1+f_{n-2}(x)}}\
            &=frac{1+f_{n-2}(x)}{1+f_{n-2}(x)+1}\
            &=frac{1+f_{n-2}(x)}{2+f_{n-2}(x)}\
            end{array}
            $



            Therefore,
            if $f_{n-2}(x) > 0$
            then
            $frac12 < f_n(x)
            lt 1$
            .



            Similarly,
            if $f_{n-1}(x) > 0$
            then
            $0 < f_n(x)
            lt 1$
            .



            $begin{array}\
            f_n(x)-f_{n-2}(x)
            &=dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}-f_{n-2}(x)\
            &=dfrac{1+f_{n-2}(x)-f_{n-2}(x)(2+f_{n-2}(x))}{2+f_{n-2}(x)}\
            &=dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{2+f_{n-2}(x)}\
            end{array}
            $



            $begin{array}\
            f_n(x)+f_n^2(x)
            &=dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}+(dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)})^2\
            &=dfrac{(1+f_{n-2}(x))(2+f_{n-2}(x))}{(2+f_{n-2}(x))^2}+dfrac{1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
            &=dfrac{2+3f_{n-2}(x)+f_{n-2}^2(x)+1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
            &=dfrac{3+5f_{n-2}(x)+2f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
            text{so}\
            1-f_n(x)-f_n^2(x)
            &=dfrac{4+4f_{n-2}(x)+f_{n-2}^2(x)-(3+5f_{n-2}(x)+2f_{n-2}^2(x))}{(2+f_{n-2}(x))^2}\
            &=dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
            end{array}
            $



            Therefore
            $1-f_n(x)-f_n^2(x)$
            has the same sign as
            $1-f_{n-2}(x)-f_{n-2}^2(x)$.
            Also,
            $|1-f_n(x)-f_n^2(x)|
            lt frac14|1-f_{n-2}(x)-f_{n-2}^2(x)|
            $

            so
            $|1-f_n(x)-f_n^2(x)|
            to 0$
            .



            Let
            $p(x) = 1-x-x^2$
            and
            $x_0 = frac{sqrt{5}-1}{2}
            $

            so
            $p(x_0) = 0$,
            $p'(x) < 0$ for $x ge 0$.



            Since
            $f_n(x) > 0$,
            $f_n(x)
            to x_0$
            .






            share|cite|improve this answer












            $sqrt{x}$
            is a McGuffin.



            More generally,
            let
            $f_1 = frac{1}{1 + g(x) }
            $

            where
            $g'(x) > 0,
            g(0) = 0
            $
            ,
            $f_n(x)
            =frac{1}{1+f_{n-1}(x)}
            $
            ,
            and
            $A_n = int_0^1 f_n(x) dx
            $
            .



            Then
            $f_n(x)
            to dfrac{sqrt{5}-1}{2}
            $
            .



            Note:
            I doubt that any of this
            is original,
            but this was all done
            just now by me.



            Proof.



            $begin{array}\
            f_n(x)
            &=frac{1}{1+frac{1}{1+f_{n-2}(x)}}\
            &=frac{1+f_{n-2}(x)}{1+f_{n-2}(x)+1}\
            &=frac{1+f_{n-2}(x)}{2+f_{n-2}(x)}\
            end{array}
            $



            Therefore,
            if $f_{n-2}(x) > 0$
            then
            $frac12 < f_n(x)
            lt 1$
            .



            Similarly,
            if $f_{n-1}(x) > 0$
            then
            $0 < f_n(x)
            lt 1$
            .



            $begin{array}\
            f_n(x)-f_{n-2}(x)
            &=dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}-f_{n-2}(x)\
            &=dfrac{1+f_{n-2}(x)-f_{n-2}(x)(2+f_{n-2}(x))}{2+f_{n-2}(x)}\
            &=dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{2+f_{n-2}(x)}\
            end{array}
            $



            $begin{array}\
            f_n(x)+f_n^2(x)
            &=dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}+(dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)})^2\
            &=dfrac{(1+f_{n-2}(x))(2+f_{n-2}(x))}{(2+f_{n-2}(x))^2}+dfrac{1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
            &=dfrac{2+3f_{n-2}(x)+f_{n-2}^2(x)+1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
            &=dfrac{3+5f_{n-2}(x)+2f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
            text{so}\
            1-f_n(x)-f_n^2(x)
            &=dfrac{4+4f_{n-2}(x)+f_{n-2}^2(x)-(3+5f_{n-2}(x)+2f_{n-2}^2(x))}{(2+f_{n-2}(x))^2}\
            &=dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
            end{array}
            $



            Therefore
            $1-f_n(x)-f_n^2(x)$
            has the same sign as
            $1-f_{n-2}(x)-f_{n-2}^2(x)$.
            Also,
            $|1-f_n(x)-f_n^2(x)|
            lt frac14|1-f_{n-2}(x)-f_{n-2}^2(x)|
            $

            so
            $|1-f_n(x)-f_n^2(x)|
            to 0$
            .



            Let
            $p(x) = 1-x-x^2$
            and
            $x_0 = frac{sqrt{5}-1}{2}
            $

            so
            $p(x_0) = 0$,
            $p'(x) < 0$ for $x ge 0$.



            Since
            $f_n(x) > 0$,
            $f_n(x)
            to x_0$
            .







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 27 '18 at 20:26









            marty cohen

            72.5k549127




            72.5k549127












            • what do you mean by Mcguffin?
              – Jimmy Sabater
              Dec 28 '18 at 8:52










            • Any g(x) > 0 will do. MacGuffin g.co/kgs/wDd8KQ
              – marty cohen
              Dec 28 '18 at 14:26


















            • what do you mean by Mcguffin?
              – Jimmy Sabater
              Dec 28 '18 at 8:52










            • Any g(x) > 0 will do. MacGuffin g.co/kgs/wDd8KQ
              – marty cohen
              Dec 28 '18 at 14:26
















            what do you mean by Mcguffin?
            – Jimmy Sabater
            Dec 28 '18 at 8:52




            what do you mean by Mcguffin?
            – Jimmy Sabater
            Dec 28 '18 at 8:52












            Any g(x) > 0 will do. MacGuffin g.co/kgs/wDd8KQ
            – marty cohen
            Dec 28 '18 at 14:26




            Any g(x) > 0 will do. MacGuffin g.co/kgs/wDd8KQ
            – marty cohen
            Dec 28 '18 at 14:26


















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