Composite functions of domain natural numbers dealing with perfect squares












0












$begingroup$


The domain of a function f is the set of natural numbers. The function is defined as follows:
$$f(n) =n+[sqrt{n}] $$



Where $[x] $=nearest integer smaller than or equal to x. For example, $[$pi$]$ =3.



Prove that for every natural number m the following sequence contains at
least one perfect square
$$m, f(m), f^2 (m), f^3 (m)........ $$



The notation $f^k$ denotes the function obtained by composing f with itself k times, e.g.,$f^2 = f ◦ f$.



My attempt:
I don't know what to say but I have just used numbers to put in the equation to verify. Other than this I am not able to think anything about it.






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    consider what happens when n = 101. Then consider what happens when n = 102, n=103, ..., n=110. Then formulate a hypothesis (i.e. preliminary lemma) and prove it. Then use the lemma to prove the initial assertion.
    $endgroup$
    – user2661923
    Sep 8 '18 at 22:38












  • $begingroup$
    Could you be more specific as to which numbers you used "to put in the equation to verify" and how you did that? Did you use a computer program? Which range of numbers did you test (a computer could test a very large range)?
    $endgroup$
    – Rory Daulton
    Sep 9 '18 at 0:11










  • $begingroup$
    No did not use computer program. Just randomly put 1,2,3 or 10..
    $endgroup$
    – jayant98
    Sep 9 '18 at 4:27










  • $begingroup$
    @user2661923 I don't know about lemma. Can you provide me any other information on this lemma?
    $endgroup$
    – jayant98
    Sep 9 '18 at 4:30










  • $begingroup$
    @user2661923 can you provide any information about lemmas?
    $endgroup$
    – jayant98
    Sep 9 '18 at 19:46
















0












$begingroup$


The domain of a function f is the set of natural numbers. The function is defined as follows:
$$f(n) =n+[sqrt{n}] $$



Where $[x] $=nearest integer smaller than or equal to x. For example, $[$pi$]$ =3.



Prove that for every natural number m the following sequence contains at
least one perfect square
$$m, f(m), f^2 (m), f^3 (m)........ $$



The notation $f^k$ denotes the function obtained by composing f with itself k times, e.g.,$f^2 = f ◦ f$.



My attempt:
I don't know what to say but I have just used numbers to put in the equation to verify. Other than this I am not able to think anything about it.






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    consider what happens when n = 101. Then consider what happens when n = 102, n=103, ..., n=110. Then formulate a hypothesis (i.e. preliminary lemma) and prove it. Then use the lemma to prove the initial assertion.
    $endgroup$
    – user2661923
    Sep 8 '18 at 22:38












  • $begingroup$
    Could you be more specific as to which numbers you used "to put in the equation to verify" and how you did that? Did you use a computer program? Which range of numbers did you test (a computer could test a very large range)?
    $endgroup$
    – Rory Daulton
    Sep 9 '18 at 0:11










  • $begingroup$
    No did not use computer program. Just randomly put 1,2,3 or 10..
    $endgroup$
    – jayant98
    Sep 9 '18 at 4:27










  • $begingroup$
    @user2661923 I don't know about lemma. Can you provide me any other information on this lemma?
    $endgroup$
    – jayant98
    Sep 9 '18 at 4:30










  • $begingroup$
    @user2661923 can you provide any information about lemmas?
    $endgroup$
    – jayant98
    Sep 9 '18 at 19:46














0












0








0


1



$begingroup$


The domain of a function f is the set of natural numbers. The function is defined as follows:
$$f(n) =n+[sqrt{n}] $$



Where $[x] $=nearest integer smaller than or equal to x. For example, $[$pi$]$ =3.



Prove that for every natural number m the following sequence contains at
least one perfect square
$$m, f(m), f^2 (m), f^3 (m)........ $$



The notation $f^k$ denotes the function obtained by composing f with itself k times, e.g.,$f^2 = f ◦ f$.



My attempt:
I don't know what to say but I have just used numbers to put in the equation to verify. Other than this I am not able to think anything about it.






algebra-precalculus







share|cite|improve this question











$endgroup$




The domain of a function f is the set of natural numbers. The function is defined as follows:
$$f(n) =n+[sqrt{n}] $$



Where $[x] $=nearest integer smaller than or equal to x. For example, $[$pi$]$ =3.



Prove that for every natural number m the following sequence contains at
least one perfect square
$$m, f(m), f^2 (m), f^3 (m)........ $$



The notation $f^k$ denotes the function obtained by composing f with itself k times, e.g.,$f^2 = f ◦ f$.



My attempt:
I don't know what to say but I have just used numbers to put in the equation to verify. Other than this I am not able to think anything about it.






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 17:41







jayant98

















asked Sep 8 '18 at 22:02









jayant98jayant98

513116




513116












  • $begingroup$
    consider what happens when n = 101. Then consider what happens when n = 102, n=103, ..., n=110. Then formulate a hypothesis (i.e. preliminary lemma) and prove it. Then use the lemma to prove the initial assertion.
    $endgroup$
    – user2661923
    Sep 8 '18 at 22:38












  • $begingroup$
    Could you be more specific as to which numbers you used "to put in the equation to verify" and how you did that? Did you use a computer program? Which range of numbers did you test (a computer could test a very large range)?
    $endgroup$
    – Rory Daulton
    Sep 9 '18 at 0:11










  • $begingroup$
    No did not use computer program. Just randomly put 1,2,3 or 10..
    $endgroup$
    – jayant98
    Sep 9 '18 at 4:27










  • $begingroup$
    @user2661923 I don't know about lemma. Can you provide me any other information on this lemma?
    $endgroup$
    – jayant98
    Sep 9 '18 at 4:30










  • $begingroup$
    @user2661923 can you provide any information about lemmas?
    $endgroup$
    – jayant98
    Sep 9 '18 at 19:46


















  • $begingroup$
    consider what happens when n = 101. Then consider what happens when n = 102, n=103, ..., n=110. Then formulate a hypothesis (i.e. preliminary lemma) and prove it. Then use the lemma to prove the initial assertion.
    $endgroup$
    – user2661923
    Sep 8 '18 at 22:38












  • $begingroup$
    Could you be more specific as to which numbers you used "to put in the equation to verify" and how you did that? Did you use a computer program? Which range of numbers did you test (a computer could test a very large range)?
    $endgroup$
    – Rory Daulton
    Sep 9 '18 at 0:11










  • $begingroup$
    No did not use computer program. Just randomly put 1,2,3 or 10..
    $endgroup$
    – jayant98
    Sep 9 '18 at 4:27










  • $begingroup$
    @user2661923 I don't know about lemma. Can you provide me any other information on this lemma?
    $endgroup$
    – jayant98
    Sep 9 '18 at 4:30










  • $begingroup$
    @user2661923 can you provide any information about lemmas?
    $endgroup$
    – jayant98
    Sep 9 '18 at 19:46
















$begingroup$
consider what happens when n = 101. Then consider what happens when n = 102, n=103, ..., n=110. Then formulate a hypothesis (i.e. preliminary lemma) and prove it. Then use the lemma to prove the initial assertion.
$endgroup$
– user2661923
Sep 8 '18 at 22:38






$begingroup$
consider what happens when n = 101. Then consider what happens when n = 102, n=103, ..., n=110. Then formulate a hypothesis (i.e. preliminary lemma) and prove it. Then use the lemma to prove the initial assertion.
$endgroup$
– user2661923
Sep 8 '18 at 22:38














$begingroup$
Could you be more specific as to which numbers you used "to put in the equation to verify" and how you did that? Did you use a computer program? Which range of numbers did you test (a computer could test a very large range)?
$endgroup$
– Rory Daulton
Sep 9 '18 at 0:11




$begingroup$
Could you be more specific as to which numbers you used "to put in the equation to verify" and how you did that? Did you use a computer program? Which range of numbers did you test (a computer could test a very large range)?
$endgroup$
– Rory Daulton
Sep 9 '18 at 0:11












$begingroup$
No did not use computer program. Just randomly put 1,2,3 or 10..
$endgroup$
– jayant98
Sep 9 '18 at 4:27




$begingroup$
No did not use computer program. Just randomly put 1,2,3 or 10..
$endgroup$
– jayant98
Sep 9 '18 at 4:27












$begingroup$
@user2661923 I don't know about lemma. Can you provide me any other information on this lemma?
$endgroup$
– jayant98
Sep 9 '18 at 4:30




$begingroup$
@user2661923 I don't know about lemma. Can you provide me any other information on this lemma?
$endgroup$
– jayant98
Sep 9 '18 at 4:30












$begingroup$
@user2661923 can you provide any information about lemmas?
$endgroup$
– jayant98
Sep 9 '18 at 19:46




$begingroup$
@user2661923 can you provide any information about lemmas?
$endgroup$
– jayant98
Sep 9 '18 at 19:46










1 Answer
1






active

oldest

votes


















1












$begingroup$

Let
$n = m^2+k$
where
$1 le k le 2m$.



Then



$begin{array}\
f(n)
&=n+[sqrt{n}]\
&=m^2+k+[sqrt{m^2+k}]\
&=m^2+k+m\
end{array}
$



If $1 le k le m$ then



$begin{array}\
f^{(2)}(n)=f(f(n))
&=m^2+k+m+[sqrt{m^2+k+m}]\
&=m^2+k+m+m\
&=m^2+2m+k\
&=(m+1)^2+k-1\
end{array}
$



Therefore
$f(f(n))$
is one closer to a square than
$f(n)$.
In particular,
if $k=1$ then
$f(f(n))$ is a square.



If $m+1 le k le 2m$ then



$begin{array}\
f^{(2)}(n)=f(f(n))
&=m^2+k+m+[sqrt{m^2+k+m}]\
&=m^2+2m+1+k-(m+1)+[sqrt{m^2+2m+1+k-(m+1)}]\
&=(m+1)^2+k-(m+1)+(m+1)\
&=(m+1)^2+k\
end{array}
$



and
$k-(m+1)
=(k-m)-1
lt k-m$
so the difference between $k$
and the square below it
decreases by $1$.
Repeating this $m$ times
brings it down where
$k le m$
(where "$m$" is the current
square, not the original one)
and the paragraph above applies.



In both cases,
eventually
$f^{(j)}(n)
=lfloor sqrt{f^{(j)}(n)} rfloor ^2
$.






share|cite|improve this answer









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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Let
    $n = m^2+k$
    where
    $1 le k le 2m$.



    Then



    $begin{array}\
    f(n)
    &=n+[sqrt{n}]\
    &=m^2+k+[sqrt{m^2+k}]\
    &=m^2+k+m\
    end{array}
    $



    If $1 le k le m$ then



    $begin{array}\
    f^{(2)}(n)=f(f(n))
    &=m^2+k+m+[sqrt{m^2+k+m}]\
    &=m^2+k+m+m\
    &=m^2+2m+k\
    &=(m+1)^2+k-1\
    end{array}
    $



    Therefore
    $f(f(n))$
    is one closer to a square than
    $f(n)$.
    In particular,
    if $k=1$ then
    $f(f(n))$ is a square.



    If $m+1 le k le 2m$ then



    $begin{array}\
    f^{(2)}(n)=f(f(n))
    &=m^2+k+m+[sqrt{m^2+k+m}]\
    &=m^2+2m+1+k-(m+1)+[sqrt{m^2+2m+1+k-(m+1)}]\
    &=(m+1)^2+k-(m+1)+(m+1)\
    &=(m+1)^2+k\
    end{array}
    $



    and
    $k-(m+1)
    =(k-m)-1
    lt k-m$
    so the difference between $k$
    and the square below it
    decreases by $1$.
    Repeating this $m$ times
    brings it down where
    $k le m$
    (where "$m$" is the current
    square, not the original one)
    and the paragraph above applies.



    In both cases,
    eventually
    $f^{(j)}(n)
    =lfloor sqrt{f^{(j)}(n)} rfloor ^2
    $    .






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Let
      $n = m^2+k$
      where
      $1 le k le 2m$.



      Then



      $begin{array}\
      f(n)
      &=n+[sqrt{n}]\
      &=m^2+k+[sqrt{m^2+k}]\
      &=m^2+k+m\
      end{array}
      $



      If $1 le k le m$ then



      $begin{array}\
      f^{(2)}(n)=f(f(n))
      &=m^2+k+m+[sqrt{m^2+k+m}]\
      &=m^2+k+m+m\
      &=m^2+2m+k\
      &=(m+1)^2+k-1\
      end{array}
      $



      Therefore
      $f(f(n))$
      is one closer to a square than
      $f(n)$.
      In particular,
      if $k=1$ then
      $f(f(n))$ is a square.



      If $m+1 le k le 2m$ then



      $begin{array}\
      f^{(2)}(n)=f(f(n))
      &=m^2+k+m+[sqrt{m^2+k+m}]\
      &=m^2+2m+1+k-(m+1)+[sqrt{m^2+2m+1+k-(m+1)}]\
      &=(m+1)^2+k-(m+1)+(m+1)\
      &=(m+1)^2+k\
      end{array}
      $



      and
      $k-(m+1)
      =(k-m)-1
      lt k-m$
      so the difference between $k$
      and the square below it
      decreases by $1$.
      Repeating this $m$ times
      brings it down where
      $k le m$
      (where "$m$" is the current
      square, not the original one)
      and the paragraph above applies.



      In both cases,
      eventually
      $f^{(j)}(n)
      =lfloor sqrt{f^{(j)}(n)} rfloor ^2
      $      .






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Let
        $n = m^2+k$
        where
        $1 le k le 2m$.



        Then



        $begin{array}\
        f(n)
        &=n+[sqrt{n}]\
        &=m^2+k+[sqrt{m^2+k}]\
        &=m^2+k+m\
        end{array}
        $



        If $1 le k le m$ then



        $begin{array}\
        f^{(2)}(n)=f(f(n))
        &=m^2+k+m+[sqrt{m^2+k+m}]\
        &=m^2+k+m+m\
        &=m^2+2m+k\
        &=(m+1)^2+k-1\
        end{array}
        $



        Therefore
        $f(f(n))$
        is one closer to a square than
        $f(n)$.
        In particular,
        if $k=1$ then
        $f(f(n))$ is a square.



        If $m+1 le k le 2m$ then



        $begin{array}\
        f^{(2)}(n)=f(f(n))
        &=m^2+k+m+[sqrt{m^2+k+m}]\
        &=m^2+2m+1+k-(m+1)+[sqrt{m^2+2m+1+k-(m+1)}]\
        &=(m+1)^2+k-(m+1)+(m+1)\
        &=(m+1)^2+k\
        end{array}
        $



        and
        $k-(m+1)
        =(k-m)-1
        lt k-m$
        so the difference between $k$
        and the square below it
        decreases by $1$.
        Repeating this $m$ times
        brings it down where
        $k le m$
        (where "$m$" is the current
        square, not the original one)
        and the paragraph above applies.



        In both cases,
        eventually
        $f^{(j)}(n)
        =lfloor sqrt{f^{(j)}(n)} rfloor ^2
        $        .






        share|cite|improve this answer









        $endgroup$



        Let
        $n = m^2+k$
        where
        $1 le k le 2m$.



        Then



        $begin{array}\
        f(n)
        &=n+[sqrt{n}]\
        &=m^2+k+[sqrt{m^2+k}]\
        &=m^2+k+m\
        end{array}
        $



        If $1 le k le m$ then



        $begin{array}\
        f^{(2)}(n)=f(f(n))
        &=m^2+k+m+[sqrt{m^2+k+m}]\
        &=m^2+k+m+m\
        &=m^2+2m+k\
        &=(m+1)^2+k-1\
        end{array}
        $



        Therefore
        $f(f(n))$
        is one closer to a square than
        $f(n)$.
        In particular,
        if $k=1$ then
        $f(f(n))$ is a square.



        If $m+1 le k le 2m$ then



        $begin{array}\
        f^{(2)}(n)=f(f(n))
        &=m^2+k+m+[sqrt{m^2+k+m}]\
        &=m^2+2m+1+k-(m+1)+[sqrt{m^2+2m+1+k-(m+1)}]\
        &=(m+1)^2+k-(m+1)+(m+1)\
        &=(m+1)^2+k\
        end{array}
        $



        and
        $k-(m+1)
        =(k-m)-1
        lt k-m$
        so the difference between $k$
        and the square below it
        decreases by $1$.
        Repeating this $m$ times
        brings it down where
        $k le m$
        (where "$m$" is the current
        square, not the original one)
        and the paragraph above applies.



        In both cases,
        eventually
        $f^{(j)}(n)
        =lfloor sqrt{f^{(j)}(n)} rfloor ^2
        $        .







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 7 '18 at 18:13









        marty cohenmarty cohen

        73.2k549128




        73.2k549128






























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