Proof by Induction of a Recursively Defined SequenceProof by Induction - Sequence of integersDivisible by 19...

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Proof by Induction of a Recursively Defined Sequence


Proof by Induction - Sequence of integersDivisible by 19 Induction ProofAnother quick induction question for a recursively defined sequence (with closed form formula given)Showing a sequence defined recursively is convergentProve sequence defined by recurrence relation using inductionProof by induction, inequalityAm I properly using induction (specifically the induction hypothesis)?Inductive proof on a sequenceInductive proof of sequence defined by recurrence relationprove that ${a_n}_{n=1}^{infty}$ is bounded and monotone increasing.













2












$begingroup$


The question I'm currently struggling with is:




Consider the following recursively defined sequence:



$$ a_1 = 0, a_{n+1}= frac{a^2_n+4}{5} $$



a) Show by induction that $0le a_n le 1 $ for all n



b) Prove that if $a_n$ converges, so that $lim_{nto infty}a_n=A$ then $A^2−5A+4 =0$, and thus $A = 1$.



c) Prove by induction that $a_n$ is strictly increasing, $a_{n+1}>a_n$ for all n.



d) Conclude that $lim_{n toinfty}a_n$ exists, and thus that $lim_{n to infty}a_n=1$




I've managed to complete part a) where I did:



Basis - at $n=1$ $a_{1+1}=frac{0+4}{5}$



Inductive Hypothesis - for k as an element of the natural numbers, it is always true that $0 le a_{k+1} le 1$ where $a_{k+1}=frac{a_k+4}{5}$



Inductive Step - Here I tried to prove it by making $n=k+1$ which meant



$a_{k+2}= frac{a_k+1^2+4}{5}$



which I split in to $frac{a_k+1^2}{5}+frac{4}{5}$



and from the inductive hypothesis I know that $0 le a_{k+1} le 1$ and therefore it's evident that $frac{a_k+1^2}{5}$ can take a minimum value of $frac{0}{5}$ and a minimum value of $frac{1}{5}$



Part b) seems relatively simple as you can prove



$A^2−5A+4 =0$ quite easily but I'm unsure how to get the equation in the first place. I think it has something to do with rearranging:



$a_{k+1}= frac{a^2_k+4}{5} $



Part c) is the part I'm struggling with most and the answer in my notes doesn't really explain to me how it's done.



I did the basis and my inductive hypothesis and at the inductive step I got to:



$a_{k+1}^2-5a_{k+1}+4>0$ but I'm not sure what to do after this



Would anyone mind helping me with part c) and d)?






real-analysis sequences-and-series induction







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  • $begingroup$
    @saulspatz I already wrote that
    $endgroup$
    – greedoid
    4 hours ago
















2












$begingroup$


The question I'm currently struggling with is:




Consider the following recursively defined sequence:



$$ a_1 = 0, a_{n+1}= frac{a^2_n+4}{5} $$



a) Show by induction that $0le a_n le 1 $ for all n



b) Prove that if $a_n$ converges, so that $lim_{nto infty}a_n=A$ then $A^2−5A+4 =0$, and thus $A = 1$.



c) Prove by induction that $a_n$ is strictly increasing, $a_{n+1}>a_n$ for all n.



d) Conclude that $lim_{n toinfty}a_n$ exists, and thus that $lim_{n to infty}a_n=1$




I've managed to complete part a) where I did:



Basis - at $n=1$ $a_{1+1}=frac{0+4}{5}$



Inductive Hypothesis - for k as an element of the natural numbers, it is always true that $0 le a_{k+1} le 1$ where $a_{k+1}=frac{a_k+4}{5}$



Inductive Step - Here I tried to prove it by making $n=k+1$ which meant



$a_{k+2}= frac{a_k+1^2+4}{5}$



which I split in to $frac{a_k+1^2}{5}+frac{4}{5}$



and from the inductive hypothesis I know that $0 le a_{k+1} le 1$ and therefore it's evident that $frac{a_k+1^2}{5}$ can take a minimum value of $frac{0}{5}$ and a minimum value of $frac{1}{5}$



Part b) seems relatively simple as you can prove



$A^2−5A+4 =0$ quite easily but I'm unsure how to get the equation in the first place. I think it has something to do with rearranging:



$a_{k+1}= frac{a^2_k+4}{5} $



Part c) is the part I'm struggling with most and the answer in my notes doesn't really explain to me how it's done.



I did the basis and my inductive hypothesis and at the inductive step I got to:



$a_{k+1}^2-5a_{k+1}+4>0$ but I'm not sure what to do after this



Would anyone mind helping me with part c) and d)?






real-analysis sequences-and-series induction







share|cite|improve this question









New contributor




king is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$












  • $begingroup$
    @saulspatz I already wrote that
    $endgroup$
    – greedoid
    4 hours ago














2












2








2





$begingroup$


The question I'm currently struggling with is:




Consider the following recursively defined sequence:



$$ a_1 = 0, a_{n+1}= frac{a^2_n+4}{5} $$



a) Show by induction that $0le a_n le 1 $ for all n



b) Prove that if $a_n$ converges, so that $lim_{nto infty}a_n=A$ then $A^2−5A+4 =0$, and thus $A = 1$.



c) Prove by induction that $a_n$ is strictly increasing, $a_{n+1}>a_n$ for all n.



d) Conclude that $lim_{n toinfty}a_n$ exists, and thus that $lim_{n to infty}a_n=1$




I've managed to complete part a) where I did:



Basis - at $n=1$ $a_{1+1}=frac{0+4}{5}$



Inductive Hypothesis - for k as an element of the natural numbers, it is always true that $0 le a_{k+1} le 1$ where $a_{k+1}=frac{a_k+4}{5}$



Inductive Step - Here I tried to prove it by making $n=k+1$ which meant



$a_{k+2}= frac{a_k+1^2+4}{5}$



which I split in to $frac{a_k+1^2}{5}+frac{4}{5}$



and from the inductive hypothesis I know that $0 le a_{k+1} le 1$ and therefore it's evident that $frac{a_k+1^2}{5}$ can take a minimum value of $frac{0}{5}$ and a minimum value of $frac{1}{5}$



Part b) seems relatively simple as you can prove



$A^2−5A+4 =0$ quite easily but I'm unsure how to get the equation in the first place. I think it has something to do with rearranging:



$a_{k+1}= frac{a^2_k+4}{5} $



Part c) is the part I'm struggling with most and the answer in my notes doesn't really explain to me how it's done.



I did the basis and my inductive hypothesis and at the inductive step I got to:



$a_{k+1}^2-5a_{k+1}+4>0$ but I'm not sure what to do after this



Would anyone mind helping me with part c) and d)?






real-analysis sequences-and-series induction







share|cite|improve this question









New contributor




king is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




The question I'm currently struggling with is:




Consider the following recursively defined sequence:



$$ a_1 = 0, a_{n+1}= frac{a^2_n+4}{5} $$



a) Show by induction that $0le a_n le 1 $ for all n



b) Prove that if $a_n$ converges, so that $lim_{nto infty}a_n=A$ then $A^2−5A+4 =0$, and thus $A = 1$.



c) Prove by induction that $a_n$ is strictly increasing, $a_{n+1}>a_n$ for all n.



d) Conclude that $lim_{n toinfty}a_n$ exists, and thus that $lim_{n to infty}a_n=1$




I've managed to complete part a) where I did:



Basis - at $n=1$ $a_{1+1}=frac{0+4}{5}$



Inductive Hypothesis - for k as an element of the natural numbers, it is always true that $0 le a_{k+1} le 1$ where $a_{k+1}=frac{a_k+4}{5}$



Inductive Step - Here I tried to prove it by making $n=k+1$ which meant



$a_{k+2}= frac{a_k+1^2+4}{5}$



which I split in to $frac{a_k+1^2}{5}+frac{4}{5}$



and from the inductive hypothesis I know that $0 le a_{k+1} le 1$ and therefore it's evident that $frac{a_k+1^2}{5}$ can take a minimum value of $frac{0}{5}$ and a minimum value of $frac{1}{5}$



Part b) seems relatively simple as you can prove



$A^2−5A+4 =0$ quite easily but I'm unsure how to get the equation in the first place. I think it has something to do with rearranging:



$a_{k+1}= frac{a^2_k+4}{5} $



Part c) is the part I'm struggling with most and the answer in my notes doesn't really explain to me how it's done.



I did the basis and my inductive hypothesis and at the inductive step I got to:



$a_{k+1}^2-5a_{k+1}+4>0$ but I'm not sure what to do after this



Would anyone mind helping me with part c) and d)?






real-analysis sequences-and-series induction




real-analysis sequences-and-series induction






share|cite|improve this question









New contributor




king is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this question









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share|cite|improve this question




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edited 2 hours ago









greedoid

45k1157112




45k1157112






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asked 4 hours ago









kingking

235




235




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  • $begingroup$
    @saulspatz I already wrote that
    $endgroup$
    – greedoid
    4 hours ago


















  • $begingroup$
    @saulspatz I already wrote that
    $endgroup$
    – greedoid
    4 hours ago
















$begingroup$
@saulspatz I already wrote that
$endgroup$
– greedoid
4 hours ago




$begingroup$
@saulspatz I already wrote that
$endgroup$
– greedoid
4 hours ago










4 Answers
4






active

oldest

votes


















2












$begingroup$

It is $$a_{k+1}^2-5a_{k+1}+4>0$$ and not $a_{k+1}^2-5a_{k+1}-4>0$. So you have $$(a_{k+1}-1)(a_{k+1}-4)>0$$



which is true by a).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ahh yes thank you that was a typo in my post but I see what you do with it now
    $endgroup$
    – king
    3 hours ago



















2












$begingroup$

The inductive step for the monotonicity (c) is just
$$
a_{n+1} > a_n implies frac{a^2_{n+1}+4}{5} > frac{a^2_n+4}{5} implies a_{n+2} > a_{n+1}
$$
The first implication holds because $a_n ge 0$.



The convergence (d) then follows from the monotone convergence theorem.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    b) The argument is that at the fixed point, which has the value A,
    $$a_{k+1}=a_{k}=A$$

    So
    $$a_{k+1}=frac{a_k^2+4}{5}$$
    becomes
    $$A=frac{A^2+4}{5}$$
    $$5A=A^2+4$$
    $$A^2+5A+4=0$$






    share|cite|improve this answer








    New contributor




    Martin Hansen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$





















      1












      $begingroup$

      a) By the assumption of the induction we obtain:
      $$0leq a_{n+1}=frac{a_n^2+4}{5}<frac{1^2+4}{5}=1.$$
      b),c) $$a_{n+1}-a_n=frac{a_n^2+4}{5}-a_n=frac{(a_n-1)(a_n-4)}{5}>0,$$
      which says that $a$ increases and bounded.



      Thus, there is $limlimits_{nrightarrow+infty}a_n$ and let this limit be equal to $A$.



      Thus, by theorems about limits we obtain: $$A^2-5A+4=0,$$ which gives $A=1$.



      I like the following way:
      $$|a_{n+1}-1|=left|frac{a_n^2+4}{5}-1right|=frac{a_n+1}{5}|a_n-1|<frac{2}{5}|a_n-1|<...<left(frac{2}{5}right)^n|a_1-1|rightarrow0,$$
      Which says $$lim_{nrightarrow+infty}a_n=1.$$






      share|cite|improve this answer











      $endgroup$













        Your Answer





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        4 Answers
        4






        active

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        4 Answers
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        2












        $begingroup$

        It is $$a_{k+1}^2-5a_{k+1}+4>0$$ and not $a_{k+1}^2-5a_{k+1}-4>0$. So you have $$(a_{k+1}-1)(a_{k+1}-4)>0$$



        which is true by a).






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          Ahh yes thank you that was a typo in my post but I see what you do with it now
          $endgroup$
          – king
          3 hours ago
















        2












        $begingroup$

        It is $$a_{k+1}^2-5a_{k+1}+4>0$$ and not $a_{k+1}^2-5a_{k+1}-4>0$. So you have $$(a_{k+1}-1)(a_{k+1}-4)>0$$



        which is true by a).






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          Ahh yes thank you that was a typo in my post but I see what you do with it now
          $endgroup$
          – king
          3 hours ago














        2












        2








        2





        $begingroup$

        It is $$a_{k+1}^2-5a_{k+1}+4>0$$ and not $a_{k+1}^2-5a_{k+1}-4>0$. So you have $$(a_{k+1}-1)(a_{k+1}-4)>0$$



        which is true by a).






        share|cite|improve this answer









        $endgroup$



        It is $$a_{k+1}^2-5a_{k+1}+4>0$$ and not $a_{k+1}^2-5a_{k+1}-4>0$. So you have $$(a_{k+1}-1)(a_{k+1}-4)>0$$



        which is true by a).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 4 hours ago









        greedoidgreedoid

        45k1157112




        45k1157112












        • $begingroup$
          Ahh yes thank you that was a typo in my post but I see what you do with it now
          $endgroup$
          – king
          3 hours ago


















        • $begingroup$
          Ahh yes thank you that was a typo in my post but I see what you do with it now
          $endgroup$
          – king
          3 hours ago
















        $begingroup$
        Ahh yes thank you that was a typo in my post but I see what you do with it now
        $endgroup$
        – king
        3 hours ago




        $begingroup$
        Ahh yes thank you that was a typo in my post but I see what you do with it now
        $endgroup$
        – king
        3 hours ago











        2












        $begingroup$

        The inductive step for the monotonicity (c) is just
        $$
        a_{n+1} > a_n implies frac{a^2_{n+1}+4}{5} > frac{a^2_n+4}{5} implies a_{n+2} > a_{n+1}
        $$
        The first implication holds because $a_n ge 0$.



        The convergence (d) then follows from the monotone convergence theorem.






        share|cite|improve this answer











        $endgroup$


















          2












          $begingroup$

          The inductive step for the monotonicity (c) is just
          $$
          a_{n+1} > a_n implies frac{a^2_{n+1}+4}{5} > frac{a^2_n+4}{5} implies a_{n+2} > a_{n+1}
          $$
          The first implication holds because $a_n ge 0$.



          The convergence (d) then follows from the monotone convergence theorem.






          share|cite|improve this answer











          $endgroup$
















            2












            2








            2





            $begingroup$

            The inductive step for the monotonicity (c) is just
            $$
            a_{n+1} > a_n implies frac{a^2_{n+1}+4}{5} > frac{a^2_n+4}{5} implies a_{n+2} > a_{n+1}
            $$
            The first implication holds because $a_n ge 0$.



            The convergence (d) then follows from the monotone convergence theorem.






            share|cite|improve this answer











            $endgroup$



            The inductive step for the monotonicity (c) is just
            $$
            a_{n+1} > a_n implies frac{a^2_{n+1}+4}{5} > frac{a^2_n+4}{5} implies a_{n+2} > a_{n+1}
            $$
            The first implication holds because $a_n ge 0$.



            The convergence (d) then follows from the monotone convergence theorem.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 4 hours ago

























            answered 4 hours ago









            Martin RMartin R

            29.4k33558




            29.4k33558























                1












                $begingroup$

                b) The argument is that at the fixed point, which has the value A,
                $$a_{k+1}=a_{k}=A$$

                So
                $$a_{k+1}=frac{a_k^2+4}{5}$$
                becomes
                $$A=frac{A^2+4}{5}$$
                $$5A=A^2+4$$
                $$A^2+5A+4=0$$






                share|cite|improve this answer








                New contributor




                Martin Hansen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.






                $endgroup$


















                  1












                  $begingroup$

                  b) The argument is that at the fixed point, which has the value A,
                  $$a_{k+1}=a_{k}=A$$

                  So
                  $$a_{k+1}=frac{a_k^2+4}{5}$$
                  becomes
                  $$A=frac{A^2+4}{5}$$
                  $$5A=A^2+4$$
                  $$A^2+5A+4=0$$






                  share|cite|improve this answer








                  New contributor




                  Martin Hansen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    b) The argument is that at the fixed point, which has the value A,
                    $$a_{k+1}=a_{k}=A$$

                    So
                    $$a_{k+1}=frac{a_k^2+4}{5}$$
                    becomes
                    $$A=frac{A^2+4}{5}$$
                    $$5A=A^2+4$$
                    $$A^2+5A+4=0$$






                    share|cite|improve this answer








                    New contributor




                    Martin Hansen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    $endgroup$



                    b) The argument is that at the fixed point, which has the value A,
                    $$a_{k+1}=a_{k}=A$$

                    So
                    $$a_{k+1}=frac{a_k^2+4}{5}$$
                    becomes
                    $$A=frac{A^2+4}{5}$$
                    $$5A=A^2+4$$
                    $$A^2+5A+4=0$$







                    share|cite|improve this answer








                    New contributor




                    Martin Hansen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    share|cite|improve this answer



                    share|cite|improve this answer






                    New contributor




                    Martin Hansen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                    answered 4 hours ago









                    Martin HansenMartin Hansen

                    687




                    687




                    New contributor




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                    New contributor





                    Martin Hansen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    Martin Hansen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.























                        1












                        $begingroup$

                        a) By the assumption of the induction we obtain:
                        $$0leq a_{n+1}=frac{a_n^2+4}{5}<frac{1^2+4}{5}=1.$$
                        b),c) $$a_{n+1}-a_n=frac{a_n^2+4}{5}-a_n=frac{(a_n-1)(a_n-4)}{5}>0,$$
                        which says that $a$ increases and bounded.



                        Thus, there is $limlimits_{nrightarrow+infty}a_n$ and let this limit be equal to $A$.



                        Thus, by theorems about limits we obtain: $$A^2-5A+4=0,$$ which gives $A=1$.



                        I like the following way:
                        $$|a_{n+1}-1|=left|frac{a_n^2+4}{5}-1right|=frac{a_n+1}{5}|a_n-1|<frac{2}{5}|a_n-1|<...<left(frac{2}{5}right)^n|a_1-1|rightarrow0,$$
                        Which says $$lim_{nrightarrow+infty}a_n=1.$$






                        share|cite|improve this answer











                        $endgroup$


















                          1












                          $begingroup$

                          a) By the assumption of the induction we obtain:
                          $$0leq a_{n+1}=frac{a_n^2+4}{5}<frac{1^2+4}{5}=1.$$
                          b),c) $$a_{n+1}-a_n=frac{a_n^2+4}{5}-a_n=frac{(a_n-1)(a_n-4)}{5}>0,$$
                          which says that $a$ increases and bounded.



                          Thus, there is $limlimits_{nrightarrow+infty}a_n$ and let this limit be equal to $A$.



                          Thus, by theorems about limits we obtain: $$A^2-5A+4=0,$$ which gives $A=1$.



                          I like the following way:
                          $$|a_{n+1}-1|=left|frac{a_n^2+4}{5}-1right|=frac{a_n+1}{5}|a_n-1|<frac{2}{5}|a_n-1|<...<left(frac{2}{5}right)^n|a_1-1|rightarrow0,$$
                          Which says $$lim_{nrightarrow+infty}a_n=1.$$






                          share|cite|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            a) By the assumption of the induction we obtain:
                            $$0leq a_{n+1}=frac{a_n^2+4}{5}<frac{1^2+4}{5}=1.$$
                            b),c) $$a_{n+1}-a_n=frac{a_n^2+4}{5}-a_n=frac{(a_n-1)(a_n-4)}{5}>0,$$
                            which says that $a$ increases and bounded.



                            Thus, there is $limlimits_{nrightarrow+infty}a_n$ and let this limit be equal to $A$.



                            Thus, by theorems about limits we obtain: $$A^2-5A+4=0,$$ which gives $A=1$.



                            I like the following way:
                            $$|a_{n+1}-1|=left|frac{a_n^2+4}{5}-1right|=frac{a_n+1}{5}|a_n-1|<frac{2}{5}|a_n-1|<...<left(frac{2}{5}right)^n|a_1-1|rightarrow0,$$
                            Which says $$lim_{nrightarrow+infty}a_n=1.$$






                            share|cite|improve this answer











                            $endgroup$



                            a) By the assumption of the induction we obtain:
                            $$0leq a_{n+1}=frac{a_n^2+4}{5}<frac{1^2+4}{5}=1.$$
                            b),c) $$a_{n+1}-a_n=frac{a_n^2+4}{5}-a_n=frac{(a_n-1)(a_n-4)}{5}>0,$$
                            which says that $a$ increases and bounded.



                            Thus, there is $limlimits_{nrightarrow+infty}a_n$ and let this limit be equal to $A$.



                            Thus, by theorems about limits we obtain: $$A^2-5A+4=0,$$ which gives $A=1$.



                            I like the following way:
                            $$|a_{n+1}-1|=left|frac{a_n^2+4}{5}-1right|=frac{a_n+1}{5}|a_n-1|<frac{2}{5}|a_n-1|<...<left(frac{2}{5}right)^n|a_1-1|rightarrow0,$$
                            Which says $$lim_{nrightarrow+infty}a_n=1.$$







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 1 hour ago

























                            answered 3 hours ago









                            Michael RozenbergMichael Rozenberg

                            105k1893198




                            105k1893198






















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