Understanding a part of the proof that sequence that converges to square root of two is decreasing.How to...

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Understanding a part of the proof that sequence that converges to square root of two is decreasing.


How to show that a sequence is positive, monotonically decreasing and converges to 0Showing the sequence converges to the square rootProve the following sequence is decreasingProve that the sequence ${{a_n}}$ converges.How do I finish my proof that the sequence converges?Prove that sequence is decreasing. Is my proof correct?Prove the sequence $x_{n+1}=frac{1}{4-x_n}, x_1=3$ converges.show that the recursive sequence converges to $sqrt r$Understanding a Cauchy Sequence ProofShow that $x_{n+1} = frac{2+x_n^2}{2x_n}$ is a decreasing sequence.






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$begingroup$


I have a very specific question about a part of the proof that the sequence that converges to $sqrt{2}$ given by: $x_1 = 1, x_{n+1} = frac{1}{2}(x_n + frac{2}{x_n})$ is monotonically decreasing.



While I do understand "how" it converges and why showing that $x_{n+1}-x_n leq 0$ proves that the sequence is monotonically decreasing, I don't understand how I get to $x_{n+1} - x_n = frac{1}{2}(frac{2}{x_n}-x_n)$ without knowing what $x_n$ looks like. Thanks for any quick hints and sorry if I'm missing the obvious.






calculus sequences-and-series algebra-precalculus







share|cite|improve this question









$endgroup$














  • $begingroup$
    The formula for $x_{n+1}-x_n$ just came from manipulating the definition in the first paragraph to pull over a single term of $x_n$. Was that what you were asking?
    $endgroup$
    – Matthew Daly
    Aug 16 at 12:04










  • $begingroup$
    Yes, it was. Wow, now that I see the solutions I feel stupid.
    $endgroup$
    – psyph
    Aug 16 at 12:06






  • 1




    $begingroup$
    Absolutely not. Stupid people don't ask questions when they are confused, which is why they stay stupid. ^_^
    $endgroup$
    – Matthew Daly
    Aug 16 at 12:07






  • 1




    $begingroup$
    Note that because $1 < sqrt2$ you have $x_1 < x_2$ so it is only monotonically decreasing for later steps
    $endgroup$
    – Henry
    Aug 16 at 12:12


















2












$begingroup$


I have a very specific question about a part of the proof that the sequence that converges to $sqrt{2}$ given by: $x_1 = 1, x_{n+1} = frac{1}{2}(x_n + frac{2}{x_n})$ is monotonically decreasing.



While I do understand "how" it converges and why showing that $x_{n+1}-x_n leq 0$ proves that the sequence is monotonically decreasing, I don't understand how I get to $x_{n+1} - x_n = frac{1}{2}(frac{2}{x_n}-x_n)$ without knowing what $x_n$ looks like. Thanks for any quick hints and sorry if I'm missing the obvious.






calculus sequences-and-series algebra-precalculus







share|cite|improve this question









$endgroup$














  • $begingroup$
    The formula for $x_{n+1}-x_n$ just came from manipulating the definition in the first paragraph to pull over a single term of $x_n$. Was that what you were asking?
    $endgroup$
    – Matthew Daly
    Aug 16 at 12:04










  • $begingroup$
    Yes, it was. Wow, now that I see the solutions I feel stupid.
    $endgroup$
    – psyph
    Aug 16 at 12:06






  • 1




    $begingroup$
    Absolutely not. Stupid people don't ask questions when they are confused, which is why they stay stupid. ^_^
    $endgroup$
    – Matthew Daly
    Aug 16 at 12:07






  • 1




    $begingroup$
    Note that because $1 < sqrt2$ you have $x_1 < x_2$ so it is only monotonically decreasing for later steps
    $endgroup$
    – Henry
    Aug 16 at 12:12














2












2








2





$begingroup$


I have a very specific question about a part of the proof that the sequence that converges to $sqrt{2}$ given by: $x_1 = 1, x_{n+1} = frac{1}{2}(x_n + frac{2}{x_n})$ is monotonically decreasing.



While I do understand "how" it converges and why showing that $x_{n+1}-x_n leq 0$ proves that the sequence is monotonically decreasing, I don't understand how I get to $x_{n+1} - x_n = frac{1}{2}(frac{2}{x_n}-x_n)$ without knowing what $x_n$ looks like. Thanks for any quick hints and sorry if I'm missing the obvious.






calculus sequences-and-series algebra-precalculus







share|cite|improve this question









$endgroup$




I have a very specific question about a part of the proof that the sequence that converges to $sqrt{2}$ given by: $x_1 = 1, x_{n+1} = frac{1}{2}(x_n + frac{2}{x_n})$ is monotonically decreasing.



While I do understand "how" it converges and why showing that $x_{n+1}-x_n leq 0$ proves that the sequence is monotonically decreasing, I don't understand how I get to $x_{n+1} - x_n = frac{1}{2}(frac{2}{x_n}-x_n)$ without knowing what $x_n$ looks like. Thanks for any quick hints and sorry if I'm missing the obvious.






calculus sequences-and-series algebra-precalculus




calculus sequences-and-series algebra-precalculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 16 at 11:59









psyphpsyph

898 bronze badges




898 bronze badges















  • $begingroup$
    The formula for $x_{n+1}-x_n$ just came from manipulating the definition in the first paragraph to pull over a single term of $x_n$. Was that what you were asking?
    $endgroup$
    – Matthew Daly
    Aug 16 at 12:04










  • $begingroup$
    Yes, it was. Wow, now that I see the solutions I feel stupid.
    $endgroup$
    – psyph
    Aug 16 at 12:06






  • 1




    $begingroup$
    Absolutely not. Stupid people don't ask questions when they are confused, which is why they stay stupid. ^_^
    $endgroup$
    – Matthew Daly
    Aug 16 at 12:07






  • 1




    $begingroup$
    Note that because $1 < sqrt2$ you have $x_1 < x_2$ so it is only monotonically decreasing for later steps
    $endgroup$
    – Henry
    Aug 16 at 12:12


















  • $begingroup$
    The formula for $x_{n+1}-x_n$ just came from manipulating the definition in the first paragraph to pull over a single term of $x_n$. Was that what you were asking?
    $endgroup$
    – Matthew Daly
    Aug 16 at 12:04










  • $begingroup$
    Yes, it was. Wow, now that I see the solutions I feel stupid.
    $endgroup$
    – psyph
    Aug 16 at 12:06






  • 1




    $begingroup$
    Absolutely not. Stupid people don't ask questions when they are confused, which is why they stay stupid. ^_^
    $endgroup$
    – Matthew Daly
    Aug 16 at 12:07






  • 1




    $begingroup$
    Note that because $1 < sqrt2$ you have $x_1 < x_2$ so it is only monotonically decreasing for later steps
    $endgroup$
    – Henry
    Aug 16 at 12:12
















$begingroup$
The formula for $x_{n+1}-x_n$ just came from manipulating the definition in the first paragraph to pull over a single term of $x_n$. Was that what you were asking?
$endgroup$
– Matthew Daly
Aug 16 at 12:04




$begingroup$
The formula for $x_{n+1}-x_n$ just came from manipulating the definition in the first paragraph to pull over a single term of $x_n$. Was that what you were asking?
$endgroup$
– Matthew Daly
Aug 16 at 12:04












$begingroup$
Yes, it was. Wow, now that I see the solutions I feel stupid.
$endgroup$
– psyph
Aug 16 at 12:06




$begingroup$
Yes, it was. Wow, now that I see the solutions I feel stupid.
$endgroup$
– psyph
Aug 16 at 12:06




1




1




$begingroup$
Absolutely not. Stupid people don't ask questions when they are confused, which is why they stay stupid. ^_^
$endgroup$
– Matthew Daly
Aug 16 at 12:07




$begingroup$
Absolutely not. Stupid people don't ask questions when they are confused, which is why they stay stupid. ^_^
$endgroup$
– Matthew Daly
Aug 16 at 12:07




1




1




$begingroup$
Note that because $1 < sqrt2$ you have $x_1 < x_2$ so it is only monotonically decreasing for later steps
$endgroup$
– Henry
Aug 16 at 12:12




$begingroup$
Note that because $1 < sqrt2$ you have $x_1 < x_2$ so it is only monotonically decreasing for later steps
$endgroup$
– Henry
Aug 16 at 12:12










3 Answers
3






active

oldest

votes


















6












$begingroup$

Just plug in $x_{n + 1}=frac12(x_n+frac{2}{x_n})$, i.e.,
$x_{n+1}-x_n = frac12(x_n+frac{2}{x_n})-x_n$. Now you can simplify things.






share|cite|improve this answer











$endgroup$























    5












    $begingroup$

    $$x_{n+1} - x_n = frac 12 left(x_n+frac{2}{x_n}right) - x_n = frac 12 left(x_n+frac{2}{x_n}right) - frac 12 cdot(2x_n) = frac 12 left(x_n+frac{2}{x_n} - 2 x_nright)=frac 12 left(frac{2}{x_n}-x_nright).$$






    share|cite|improve this answer









    $endgroup$















    • $begingroup$
      Thanks, I didn't know I could just do that.
      $endgroup$
      – psyph
      Aug 16 at 12:07



















    0












    $begingroup$

    An easy way is that$$x_{n+1}={1over 2}left(x_n+{2over x_n}right)={sqrt 2over 2}left({x_nover sqrt 2}+{sqrt 2over x_n}right)gesqrt 2$$therefore$$x_{n+1}={1over 2}left(x_n+{2over x_n}right)={1over 2}x_n+{1over x_n}le {1over 2}x_n+{1over 2}x_n=x_n$$






    share|cite|improve this answer









    $endgroup$


















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

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      active

      oldest

      votes






      active

      oldest

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      6












      $begingroup$

      Just plug in $x_{n + 1}=frac12(x_n+frac{2}{x_n})$, i.e.,
      $x_{n+1}-x_n = frac12(x_n+frac{2}{x_n})-x_n$. Now you can simplify things.






      share|cite|improve this answer











      $endgroup$




















        6












        $begingroup$

        Just plug in $x_{n + 1}=frac12(x_n+frac{2}{x_n})$, i.e.,
        $x_{n+1}-x_n = frac12(x_n+frac{2}{x_n})-x_n$. Now you can simplify things.






        share|cite|improve this answer











        $endgroup$


















          6












          6








          6





          $begingroup$

          Just plug in $x_{n + 1}=frac12(x_n+frac{2}{x_n})$, i.e.,
          $x_{n+1}-x_n = frac12(x_n+frac{2}{x_n})-x_n$. Now you can simplify things.






          share|cite|improve this answer











          $endgroup$



          Just plug in $x_{n + 1}=frac12(x_n+frac{2}{x_n})$, i.e.,
          $x_{n+1}-x_n = frac12(x_n+frac{2}{x_n})-x_n$. Now you can simplify things.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 16 at 12:08









          0XLR

          4,1301 gold badge12 silver badges31 bronze badges




          4,1301 gold badge12 silver badges31 bronze badges










          answered Aug 16 at 12:04









          user406143user406143

          835 bronze badges




          835 bronze badges




























              5












              $begingroup$

              $$x_{n+1} - x_n = frac 12 left(x_n+frac{2}{x_n}right) - x_n = frac 12 left(x_n+frac{2}{x_n}right) - frac 12 cdot(2x_n) = frac 12 left(x_n+frac{2}{x_n} - 2 x_nright)=frac 12 left(frac{2}{x_n}-x_nright).$$






              share|cite|improve this answer









              $endgroup$















              • $begingroup$
                Thanks, I didn't know I could just do that.
                $endgroup$
                – psyph
                Aug 16 at 12:07
















              5












              $begingroup$

              $$x_{n+1} - x_n = frac 12 left(x_n+frac{2}{x_n}right) - x_n = frac 12 left(x_n+frac{2}{x_n}right) - frac 12 cdot(2x_n) = frac 12 left(x_n+frac{2}{x_n} - 2 x_nright)=frac 12 left(frac{2}{x_n}-x_nright).$$






              share|cite|improve this answer









              $endgroup$















              • $begingroup$
                Thanks, I didn't know I could just do that.
                $endgroup$
                – psyph
                Aug 16 at 12:07














              5












              5








              5





              $begingroup$

              $$x_{n+1} - x_n = frac 12 left(x_n+frac{2}{x_n}right) - x_n = frac 12 left(x_n+frac{2}{x_n}right) - frac 12 cdot(2x_n) = frac 12 left(x_n+frac{2}{x_n} - 2 x_nright)=frac 12 left(frac{2}{x_n}-x_nright).$$






              share|cite|improve this answer









              $endgroup$



              $$x_{n+1} - x_n = frac 12 left(x_n+frac{2}{x_n}right) - x_n = frac 12 left(x_n+frac{2}{x_n}right) - frac 12 cdot(2x_n) = frac 12 left(x_n+frac{2}{x_n} - 2 x_nright)=frac 12 left(frac{2}{x_n}-x_nright).$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Aug 16 at 12:04









              TZakrevskiyTZakrevskiy

              20.5k1 gold badge24 silver badges56 bronze badges




              20.5k1 gold badge24 silver badges56 bronze badges















              • $begingroup$
                Thanks, I didn't know I could just do that.
                $endgroup$
                – psyph
                Aug 16 at 12:07


















              • $begingroup$
                Thanks, I didn't know I could just do that.
                $endgroup$
                – psyph
                Aug 16 at 12:07
















              $begingroup$
              Thanks, I didn't know I could just do that.
              $endgroup$
              – psyph
              Aug 16 at 12:07




              $begingroup$
              Thanks, I didn't know I could just do that.
              $endgroup$
              – psyph
              Aug 16 at 12:07











              0












              $begingroup$

              An easy way is that$$x_{n+1}={1over 2}left(x_n+{2over x_n}right)={sqrt 2over 2}left({x_nover sqrt 2}+{sqrt 2over x_n}right)gesqrt 2$$therefore$$x_{n+1}={1over 2}left(x_n+{2over x_n}right)={1over 2}x_n+{1over x_n}le {1over 2}x_n+{1over 2}x_n=x_n$$






              share|cite|improve this answer









              $endgroup$




















                0












                $begingroup$

                An easy way is that$$x_{n+1}={1over 2}left(x_n+{2over x_n}right)={sqrt 2over 2}left({x_nover sqrt 2}+{sqrt 2over x_n}right)gesqrt 2$$therefore$$x_{n+1}={1over 2}left(x_n+{2over x_n}right)={1over 2}x_n+{1over x_n}le {1over 2}x_n+{1over 2}x_n=x_n$$






                share|cite|improve this answer









                $endgroup$


















                  0












                  0








                  0





                  $begingroup$

                  An easy way is that$$x_{n+1}={1over 2}left(x_n+{2over x_n}right)={sqrt 2over 2}left({x_nover sqrt 2}+{sqrt 2over x_n}right)gesqrt 2$$therefore$$x_{n+1}={1over 2}left(x_n+{2over x_n}right)={1over 2}x_n+{1over x_n}le {1over 2}x_n+{1over 2}x_n=x_n$$






                  share|cite|improve this answer









                  $endgroup$



                  An easy way is that$$x_{n+1}={1over 2}left(x_n+{2over x_n}right)={sqrt 2over 2}left({x_nover sqrt 2}+{sqrt 2over x_n}right)gesqrt 2$$therefore$$x_{n+1}={1over 2}left(x_n+{2over x_n}right)={1over 2}x_n+{1over x_n}le {1over 2}x_n+{1over 2}x_n=x_n$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 16 at 12:47









                  Mostafa AyazMostafa Ayaz

                  19k3 gold badges10 silver badges43 bronze badges




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