a limit of recurrence relation












1












$begingroup$


I have and the following recurrence relaation:



(x_{n})_{ngeq 0}, x_{n+1}=x_{n}+frac{1}{a}cdot x_{n}^{1-a},ageq 1,ngeq 0,x_{0}=1



I need to solve lim_{nrightarrow infty }frac{x_{n}^{a}}{n}



I found that this string is increasing and I tried to find z and w from general form x_{n+1}=zcdot x_{n}+w



I tried to factorize xn but I didn't get too far.






recurrence-relations







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I have and the following recurrence relaation:



    (x_{n})_{ngeq 0}, x_{n+1}=x_{n}+frac{1}{a}cdot x_{n}^{1-a},ageq 1,ngeq 0,x_{0}=1



    I need to solve lim_{nrightarrow infty }frac{x_{n}^{a}}{n}



    I found that this string is increasing and I tried to find z and w from general form x_{n+1}=zcdot x_{n}+w



    I tried to factorize xn but I didn't get too far.






    recurrence-relations







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      0



      $begingroup$


      I have and the following recurrence relaation:



      (x_{n})_{ngeq 0}, x_{n+1}=x_{n}+frac{1}{a}cdot x_{n}^{1-a},ageq 1,ngeq 0,x_{0}=1



      I need to solve lim_{nrightarrow infty }frac{x_{n}^{a}}{n}



      I found that this string is increasing and I tried to find z and w from general form x_{n+1}=zcdot x_{n}+w



      I tried to factorize xn but I didn't get too far.






      recurrence-relations







      share|cite|improve this question









      $endgroup$




      I have and the following recurrence relaation:



      (x_{n})_{ngeq 0}, x_{n+1}=x_{n}+frac{1}{a}cdot x_{n}^{1-a},ageq 1,ngeq 0,x_{0}=1



      I need to solve lim_{nrightarrow infty }frac{x_{n}^{a}}{n}



      I found that this string is increasing and I tried to find z and w from general form x_{n+1}=zcdot x_{n}+w



      I tried to factorize xn but I didn't get too far.






      recurrence-relations




      recurrence-relations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 5 at 12:43









      Vali ROVali RO

      686




      686






















          1 Answer
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          $begingroup$

          You can prove by induction that $x_ngeq 1$. This implies that for all $ngeq 0$, $$x_{n+1}-x_ngeq frac 1 a$$
          Summing this up leads to $$x_ngeq 1 + frac n a$$
          So, clearly, $x_nrightarrow +infty$ as $nrightarrow +infty$.



          The trick is to look for an estimate of $u_n = x_n^a$.
          Note that just like $(x_n)_{ngeq 0}$, $(u_n)_{n geq 0}$ tends to $+infty$, so we can use the appropriate Taylor expansion:
          $$begin{split}
          u_{n+1} &= left (x_n + frac 1 a x_n^{1-a}right)^a\
          &= u_n left (1 + frac 1 a frac 1 {u_n}right)^{a}\
          &= u_n left (1 + frac 1 {u_n} + oleft (frac 1 {u_n}right)right)\
          &= u_n +1 +o(1)
          end{split}$$
          So summing this up leads to $u_n=n+o(n)$, which implies that
          $$lim_{nrightarrow +infty} frac {x_n^a}n = 1$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks a lot! :)
            $endgroup$
            – Vali RO
            Jan 7 at 12:34










          • $begingroup$
            You're welcome!
            $endgroup$
            – Stefan Lafon
            Jan 7 at 16:57












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          1 Answer
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          1 Answer
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          active

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          1












          $begingroup$

          You can prove by induction that $x_ngeq 1$. This implies that for all $ngeq 0$, $$x_{n+1}-x_ngeq frac 1 a$$
          Summing this up leads to $$x_ngeq 1 + frac n a$$
          So, clearly, $x_nrightarrow +infty$ as $nrightarrow +infty$.



          The trick is to look for an estimate of $u_n = x_n^a$.
          Note that just like $(x_n)_{ngeq 0}$, $(u_n)_{n geq 0}$ tends to $+infty$, so we can use the appropriate Taylor expansion:
          $$begin{split}
          u_{n+1} &= left (x_n + frac 1 a x_n^{1-a}right)^a\
          &= u_n left (1 + frac 1 a frac 1 {u_n}right)^{a}\
          &= u_n left (1 + frac 1 {u_n} + oleft (frac 1 {u_n}right)right)\
          &= u_n +1 +o(1)
          end{split}$$
          So summing this up leads to $u_n=n+o(n)$, which implies that
          $$lim_{nrightarrow +infty} frac {x_n^a}n = 1$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks a lot! :)
            $endgroup$
            – Vali RO
            Jan 7 at 12:34










          • $begingroup$
            You're welcome!
            $endgroup$
            – Stefan Lafon
            Jan 7 at 16:57
















          1












          $begingroup$

          You can prove by induction that $x_ngeq 1$. This implies that for all $ngeq 0$, $$x_{n+1}-x_ngeq frac 1 a$$
          Summing this up leads to $$x_ngeq 1 + frac n a$$
          So, clearly, $x_nrightarrow +infty$ as $nrightarrow +infty$.



          The trick is to look for an estimate of $u_n = x_n^a$.
          Note that just like $(x_n)_{ngeq 0}$, $(u_n)_{n geq 0}$ tends to $+infty$, so we can use the appropriate Taylor expansion:
          $$begin{split}
          u_{n+1} &= left (x_n + frac 1 a x_n^{1-a}right)^a\
          &= u_n left (1 + frac 1 a frac 1 {u_n}right)^{a}\
          &= u_n left (1 + frac 1 {u_n} + oleft (frac 1 {u_n}right)right)\
          &= u_n +1 +o(1)
          end{split}$$
          So summing this up leads to $u_n=n+o(n)$, which implies that
          $$lim_{nrightarrow +infty} frac {x_n^a}n = 1$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks a lot! :)
            $endgroup$
            – Vali RO
            Jan 7 at 12:34










          • $begingroup$
            You're welcome!
            $endgroup$
            – Stefan Lafon
            Jan 7 at 16:57














          1












          1








          1





          $begingroup$

          You can prove by induction that $x_ngeq 1$. This implies that for all $ngeq 0$, $$x_{n+1}-x_ngeq frac 1 a$$
          Summing this up leads to $$x_ngeq 1 + frac n a$$
          So, clearly, $x_nrightarrow +infty$ as $nrightarrow +infty$.



          The trick is to look for an estimate of $u_n = x_n^a$.
          Note that just like $(x_n)_{ngeq 0}$, $(u_n)_{n geq 0}$ tends to $+infty$, so we can use the appropriate Taylor expansion:
          $$begin{split}
          u_{n+1} &= left (x_n + frac 1 a x_n^{1-a}right)^a\
          &= u_n left (1 + frac 1 a frac 1 {u_n}right)^{a}\
          &= u_n left (1 + frac 1 {u_n} + oleft (frac 1 {u_n}right)right)\
          &= u_n +1 +o(1)
          end{split}$$
          So summing this up leads to $u_n=n+o(n)$, which implies that
          $$lim_{nrightarrow +infty} frac {x_n^a}n = 1$$






          share|cite|improve this answer









          $endgroup$



          You can prove by induction that $x_ngeq 1$. This implies that for all $ngeq 0$, $$x_{n+1}-x_ngeq frac 1 a$$
          Summing this up leads to $$x_ngeq 1 + frac n a$$
          So, clearly, $x_nrightarrow +infty$ as $nrightarrow +infty$.



          The trick is to look for an estimate of $u_n = x_n^a$.
          Note that just like $(x_n)_{ngeq 0}$, $(u_n)_{n geq 0}$ tends to $+infty$, so we can use the appropriate Taylor expansion:
          $$begin{split}
          u_{n+1} &= left (x_n + frac 1 a x_n^{1-a}right)^a\
          &= u_n left (1 + frac 1 a frac 1 {u_n}right)^{a}\
          &= u_n left (1 + frac 1 {u_n} + oleft (frac 1 {u_n}right)right)\
          &= u_n +1 +o(1)
          end{split}$$
          So summing this up leads to $u_n=n+o(n)$, which implies that
          $$lim_{nrightarrow +infty} frac {x_n^a}n = 1$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 7 at 5:41









          Stefan LafonStefan Lafon

          3,005212




          3,005212












          • $begingroup$
            Thanks a lot! :)
            $endgroup$
            – Vali RO
            Jan 7 at 12:34










          • $begingroup$
            You're welcome!
            $endgroup$
            – Stefan Lafon
            Jan 7 at 16:57


















          • $begingroup$
            Thanks a lot! :)
            $endgroup$
            – Vali RO
            Jan 7 at 12:34










          • $begingroup$
            You're welcome!
            $endgroup$
            – Stefan Lafon
            Jan 7 at 16:57
















          $begingroup$
          Thanks a lot! :)
          $endgroup$
          – Vali RO
          Jan 7 at 12:34




          $begingroup$
          Thanks a lot! :)
          $endgroup$
          – Vali RO
          Jan 7 at 12:34












          $begingroup$
          You're welcome!
          $endgroup$
          – Stefan Lafon
          Jan 7 at 16:57




          $begingroup$
          You're welcome!
          $endgroup$
          – Stefan Lafon
          Jan 7 at 16:57


















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