Inequality proof with sequences - showing $x_n<x_n+1$ for $x_n+1=x_n^2 + 1/4$, $x<1/2$

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the question



So I was doing part 2 of this question and I wanted to know if my approach is correct.



$x_n+1 - x_n = x_n^2 + 1/4 - x_n$



Now since it is a sequence of positive terms
$x_n$ > 0



Therefore $x_n^2 + 1/4 > 1/4 > x_n$



Hence $x_n^2 + 1/4 - x_n > 0$



$x_n+1 - x_n > 0$



$x_n < x_n+1$




sequences-and-series proof-verification inequality induction recurrence-relations




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  • I think your reasoning is not so good. See my post.
    – Michael Rozenberg
    Aug 13 at 6:26










  • The same idea from my answer to your other question works here as well: $;x_n+1-x_n=x_n^2-x_n-1^2,$.
    – dxiv
    Aug 13 at 7:03











  • In case it helps, here is another question about the same recurrence: $a_n+1=frac14+a_n^2$ is converging and have a limit. Drawing cobweb plot might help to get some intuition in problems like this, you can see some examples here and here. In this case, the function is $f(x)=x^2+frac14$.
    – Martin Sleziak
    Aug 21 at 2:22














up vote
1
down vote

favorite












the question



So I was doing part 2 of this question and I wanted to know if my approach is correct.



$x_n+1 - x_n = x_n^2 + 1/4 - x_n$



Now since it is a sequence of positive terms
$x_n$ > 0



Therefore $x_n^2 + 1/4 > 1/4 > x_n$



Hence $x_n^2 + 1/4 - x_n > 0$



$x_n+1 - x_n > 0$



$x_n < x_n+1$




sequences-and-series proof-verification inequality induction recurrence-relations




share|cite|improve this question






















  • I think your reasoning is not so good. See my post.
    – Michael Rozenberg
    Aug 13 at 6:26










  • The same idea from my answer to your other question works here as well: $;x_n+1-x_n=x_n^2-x_n-1^2,$.
    – dxiv
    Aug 13 at 7:03











  • In case it helps, here is another question about the same recurrence: $a_n+1=frac14+a_n^2$ is converging and have a limit. Drawing cobweb plot might help to get some intuition in problems like this, you can see some examples here and here. In this case, the function is $f(x)=x^2+frac14$.
    – Martin Sleziak
    Aug 21 at 2:22












up vote
1
down vote

favorite









up vote
1
down vote

favorite











the question



So I was doing part 2 of this question and I wanted to know if my approach is correct.



$x_n+1 - x_n = x_n^2 + 1/4 - x_n$



Now since it is a sequence of positive terms
$x_n$ > 0



Therefore $x_n^2 + 1/4 > 1/4 > x_n$



Hence $x_n^2 + 1/4 - x_n > 0$



$x_n+1 - x_n > 0$



$x_n < x_n+1$




sequences-and-series proof-verification inequality induction recurrence-relations




share|cite|improve this question














the question



So I was doing part 2 of this question and I wanted to know if my approach is correct.



$x_n+1 - x_n = x_n^2 + 1/4 - x_n$



Now since it is a sequence of positive terms
$x_n$ > 0



Therefore $x_n^2 + 1/4 > 1/4 > x_n$



Hence $x_n^2 + 1/4 - x_n > 0$



$x_n+1 - x_n > 0$



$x_n < x_n+1$





sequences-and-series proof-verification inequality induction recurrence-relations





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 21 at 2:17









Martin Sleziak

43.6k6113260




43.6k6113260










asked Aug 13 at 6:16









user122343

695




695











  • I think your reasoning is not so good. See my post.
    – Michael Rozenberg
    Aug 13 at 6:26










  • The same idea from my answer to your other question works here as well: $;x_n+1-x_n=x_n^2-x_n-1^2,$.
    – dxiv
    Aug 13 at 7:03











  • In case it helps, here is another question about the same recurrence: $a_n+1=frac14+a_n^2$ is converging and have a limit. Drawing cobweb plot might help to get some intuition in problems like this, you can see some examples here and here. In this case, the function is $f(x)=x^2+frac14$.
    – Martin Sleziak
    Aug 21 at 2:22
















  • I think your reasoning is not so good. See my post.
    – Michael Rozenberg
    Aug 13 at 6:26










  • The same idea from my answer to your other question works here as well: $;x_n+1-x_n=x_n^2-x_n-1^2,$.
    – dxiv
    Aug 13 at 7:03











  • In case it helps, here is another question about the same recurrence: $a_n+1=frac14+a_n^2$ is converging and have a limit. Drawing cobweb plot might help to get some intuition in problems like this, you can see some examples here and here. In this case, the function is $f(x)=x^2+frac14$.
    – Martin Sleziak
    Aug 21 at 2:22















I think your reasoning is not so good. See my post.
– Michael Rozenberg
Aug 13 at 6:26




I think your reasoning is not so good. See my post.
– Michael Rozenberg
Aug 13 at 6:26












The same idea from my answer to your other question works here as well: $;x_n+1-x_n=x_n^2-x_n-1^2,$.
– dxiv
Aug 13 at 7:03





The same idea from my answer to your other question works here as well: $;x_n+1-x_n=x_n^2-x_n-1^2,$.
– dxiv
Aug 13 at 7:03













In case it helps, here is another question about the same recurrence: $a_n+1=frac14+a_n^2$ is converging and have a limit. Drawing cobweb plot might help to get some intuition in problems like this, you can see some examples here and here. In this case, the function is $f(x)=x^2+frac14$.
– Martin Sleziak
Aug 21 at 2:22




In case it helps, here is another question about the same recurrence: $a_n+1=frac14+a_n^2$ is converging and have a limit. Drawing cobweb plot might help to get some intuition in problems like this, you can see some examples here and here. In this case, the function is $f(x)=x^2+frac14$.
– Martin Sleziak
Aug 21 at 2:22










1 Answer
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(i).



Because by assumption of the induction $$x_n+1=x_n^2+frac14<frac14+frac14=frac12.$$



$(ii).$



$$x_n+1-x_n=x_n^2-x_n+frac14=left(x_n-frac12right)^2>0.$$






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    (i).



    Because by assumption of the induction $$x_n+1=x_n^2+frac14<frac14+frac14=frac12.$$



    $(ii).$



    $$x_n+1-x_n=x_n^2-x_n+frac14=left(x_n-frac12right)^2>0.$$






    share|cite|improve this answer
























      up vote
      0
      down vote













      (i).



      Because by assumption of the induction $$x_n+1=x_n^2+frac14<frac14+frac14=frac12.$$



      $(ii).$



      $$x_n+1-x_n=x_n^2-x_n+frac14=left(x_n-frac12right)^2>0.$$






      share|cite|improve this answer






















        up vote
        0
        down vote










        up vote
        0
        down vote









        (i).



        Because by assumption of the induction $$x_n+1=x_n^2+frac14<frac14+frac14=frac12.$$



        $(ii).$



        $$x_n+1-x_n=x_n^2-x_n+frac14=left(x_n-frac12right)^2>0.$$






        share|cite|improve this answer












        (i).



        Because by assumption of the induction $$x_n+1=x_n^2+frac14<frac14+frac14=frac12.$$



        $(ii).$



        $$x_n+1-x_n=x_n^2-x_n+frac14=left(x_n-frac12right)^2>0.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 13 at 6:20









        Michael Rozenberg

        88.3k1579180




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