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infinitely many negative and infinitely many positive numbers


Is $k^2+k+1$ prime for infinitely many values of $k$?Diverging to Positive and Negative InfinityHow prove a sequence has infinitely many square numbers.infinitely descending natural numbersUse this sequence to prove that there are infinitely many prime numbers.Can decreasing sequence of sets with $A_i$ containing infinitely less elements than $A_i-1$ have finite limit?Show that this sequence contains infinitely many composite numbers.How do I rigorously show a sequence of positive real numbers converges to a non-negative real number?Series and positive sequenceDoes some Lucas sequence contain infinitely many primes?













2












$begingroup$


Suppose that
$$x_1=frac14, x_n+1=x_n^3-3x_n.$$



Show that the sequence has infinitely many negative and infinitely many positive numbers.



My idea: Suppose that it has finitely many negative numbers. then all the numbers after some index, must be larger than $sqrt3$. I want to show that the sequence cannot escape some interval.










share|cite|improve this question









$endgroup$
















    2












    $begingroup$


    Suppose that
    $$x_1=frac14, x_n+1=x_n^3-3x_n.$$



    Show that the sequence has infinitely many negative and infinitely many positive numbers.



    My idea: Suppose that it has finitely many negative numbers. then all the numbers after some index, must be larger than $sqrt3$. I want to show that the sequence cannot escape some interval.










    share|cite|improve this question









    $endgroup$














      2












      2








      2


      2



      $begingroup$


      Suppose that
      $$x_1=frac14, x_n+1=x_n^3-3x_n.$$



      Show that the sequence has infinitely many negative and infinitely many positive numbers.



      My idea: Suppose that it has finitely many negative numbers. then all the numbers after some index, must be larger than $sqrt3$. I want to show that the sequence cannot escape some interval.










      share|cite|improve this question









      $endgroup$




      Suppose that
      $$x_1=frac14, x_n+1=x_n^3-3x_n.$$



      Show that the sequence has infinitely many negative and infinitely many positive numbers.



      My idea: Suppose that it has finitely many negative numbers. then all the numbers after some index, must be larger than $sqrt3$. I want to show that the sequence cannot escape some interval.







      sequences-and-series polynomials






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 7 hours ago









      S_AlexS_Alex

      21219




      21219




















          4 Answers
          4






          active

          oldest

          votes


















          3












          $begingroup$

          You have essentially the right idea. Here are some hints to help you complete your proof.



          Let $f(x) = x^3 - 3x $.



          1. Show that $x_n in ( -2, 2 ). $


          2. Show that if $ x in (0, sqrt3)$, then $f(x) < 0 $.


          3. Show that if $x in ( sqrt3 , 2 )$, then $ 0 < f(x) < x$.

            This tells us that the values will decrease. However, do they decrease enough?


          4. Show that if $ x in ( sqrt3 ,2 )$, then there eixsts an $n$ such that $ f^n(x) < sqrt3$.

            This tells us that the values decrease enough to force a negative value, $f^n+1 (x)$.



          Note: There are multiple ways of doing 4. If you are stuck, consider $ frac 2 - f(x) 2 - x $. This tells you how quickly you're moving away from 2 (and hence will be less than $sqrt3$ eventually.)






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            For part (1), if $x_n=-1$ then $x_n+1=2$ and $x_i=2$ for all $i>n+1$. We need also to show $x_n$ does not get the value $-1$.
            $endgroup$
            – S_Alex
            5 hours ago


















          2












          $begingroup$

          Let $f(x) = x^3 - 3x$ and explore, for which $x$ one has
          $$
          |f'(x)| < 1.
          $$






          share|cite|improve this answer









          $endgroup$




















            0












            $begingroup$

            Notice that for $alpha>0$:



            $$(2+alpha)^3-3(2+alpha)=2+9alpha+6alpha^2+alpha^3>2+alpha$$
            while for $betain[0,2]$:
            $$(2-beta)^3-3(2-beta)=2-9beta+6beta^2-beta^3<2-beta$$



            The opposite arguments can be made around $-2$. Since $x_1in[-2,-2]$, your iteration is restricted to this range






            share|cite|improve this answer









            $endgroup$




















              0












              $begingroup$

              The desired claim follows from the following two observations:




              Claim. If $x_n in -1/4, 1/4$, then $x_n neq 0$ for all $n geq 1$.




              Proof. Let $p_1 = operatornamesign(x_1)$ and $p_n+1 = p_n^3 - 3p_n 4^2 cdot 3^n-1$. Then we inductively check that $p_n$ is always odd and $ x_n = p_n / 4^3^n-1$. Since the numerator is always odd integer, it cannot vanish, and the claim follows.




              Claim 2. If either $x_n geq 0$ for any sufficiently large $n$ or $x_n leq 0$ for any sufficiently $n$, then we actually have $x_n = 0$ for any sufficiently large $n$.




              Proof. Write $x_n = 2cos(2 pi f_n)$. Then



              $$ cos(2 pi f_n+1) = fracx_n+12 = fracx_n^3 - 3x_n2 = 4cos^3(2 pi f_n) - 3cos(2 pi f_n) = cos(2 pi cdot 3f_n). $$



              So it follows that $cos(2pi f_N+n) = cos(2pi cdot 3^n f_N)$. Now, replacing $(x_n)$ by $(-x_n)$ if necessary, we may assume that $x_n geq 0$ for all sufficiently large $n$. In other words, there exists $N$ so that $x_N+n geq 0$ for all $n geq 0$. Then each $3^n f_N$ must avoid the sets $(frac14, frac34) + mathbbZ$. So it follows that



              $$ f_N in mathbbR setminus bigcup_n=0^infty bigcup_kinmathbbZ left( frac4k+14 cdot 3^n, frac4k+34 cdot 3^n right) = left k pm frac14cdot 3^n : n geq 0 right. $$



              This implies that $3^n_0 f_N = pm 1$ for some $n_0 geq 0$, and hence $x_N+n = 0$ for all $n geq n_0$. This proves the desired claim.






              share|cite|improve this answer









              $endgroup$













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                4 Answers
                4






                active

                oldest

                votes








                4 Answers
                4






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                3












                $begingroup$

                You have essentially the right idea. Here are some hints to help you complete your proof.



                Let $f(x) = x^3 - 3x $.



                1. Show that $x_n in ( -2, 2 ). $


                2. Show that if $ x in (0, sqrt3)$, then $f(x) < 0 $.


                3. Show that if $x in ( sqrt3 , 2 )$, then $ 0 < f(x) < x$.

                  This tells us that the values will decrease. However, do they decrease enough?


                4. Show that if $ x in ( sqrt3 ,2 )$, then there eixsts an $n$ such that $ f^n(x) < sqrt3$.

                  This tells us that the values decrease enough to force a negative value, $f^n+1 (x)$.



                Note: There are multiple ways of doing 4. If you are stuck, consider $ frac 2 - f(x) 2 - x $. This tells you how quickly you're moving away from 2 (and hence will be less than $sqrt3$ eventually.)






                share|cite|improve this answer











                $endgroup$












                • $begingroup$
                  For part (1), if $x_n=-1$ then $x_n+1=2$ and $x_i=2$ for all $i>n+1$. We need also to show $x_n$ does not get the value $-1$.
                  $endgroup$
                  – S_Alex
                  5 hours ago















                3












                $begingroup$

                You have essentially the right idea. Here are some hints to help you complete your proof.



                Let $f(x) = x^3 - 3x $.



                1. Show that $x_n in ( -2, 2 ). $


                2. Show that if $ x in (0, sqrt3)$, then $f(x) < 0 $.


                3. Show that if $x in ( sqrt3 , 2 )$, then $ 0 < f(x) < x$.

                  This tells us that the values will decrease. However, do they decrease enough?


                4. Show that if $ x in ( sqrt3 ,2 )$, then there eixsts an $n$ such that $ f^n(x) < sqrt3$.

                  This tells us that the values decrease enough to force a negative value, $f^n+1 (x)$.



                Note: There are multiple ways of doing 4. If you are stuck, consider $ frac 2 - f(x) 2 - x $. This tells you how quickly you're moving away from 2 (and hence will be less than $sqrt3$ eventually.)






                share|cite|improve this answer











                $endgroup$












                • $begingroup$
                  For part (1), if $x_n=-1$ then $x_n+1=2$ and $x_i=2$ for all $i>n+1$. We need also to show $x_n$ does not get the value $-1$.
                  $endgroup$
                  – S_Alex
                  5 hours ago













                3












                3








                3





                $begingroup$

                You have essentially the right idea. Here are some hints to help you complete your proof.



                Let $f(x) = x^3 - 3x $.



                1. Show that $x_n in ( -2, 2 ). $


                2. Show that if $ x in (0, sqrt3)$, then $f(x) < 0 $.


                3. Show that if $x in ( sqrt3 , 2 )$, then $ 0 < f(x) < x$.

                  This tells us that the values will decrease. However, do they decrease enough?


                4. Show that if $ x in ( sqrt3 ,2 )$, then there eixsts an $n$ such that $ f^n(x) < sqrt3$.

                  This tells us that the values decrease enough to force a negative value, $f^n+1 (x)$.



                Note: There are multiple ways of doing 4. If you are stuck, consider $ frac 2 - f(x) 2 - x $. This tells you how quickly you're moving away from 2 (and hence will be less than $sqrt3$ eventually.)






                share|cite|improve this answer











                $endgroup$



                You have essentially the right idea. Here are some hints to help you complete your proof.



                Let $f(x) = x^3 - 3x $.



                1. Show that $x_n in ( -2, 2 ). $


                2. Show that if $ x in (0, sqrt3)$, then $f(x) < 0 $.


                3. Show that if $x in ( sqrt3 , 2 )$, then $ 0 < f(x) < x$.

                  This tells us that the values will decrease. However, do they decrease enough?


                4. Show that if $ x in ( sqrt3 ,2 )$, then there eixsts an $n$ such that $ f^n(x) < sqrt3$.

                  This tells us that the values decrease enough to force a negative value, $f^n+1 (x)$.



                Note: There are multiple ways of doing 4. If you are stuck, consider $ frac 2 - f(x) 2 - x $. This tells you how quickly you're moving away from 2 (and hence will be less than $sqrt3$ eventually.)







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 6 hours ago

























                answered 6 hours ago









                Calvin LinCalvin Lin

                36.6k349116




                36.6k349116











                • $begingroup$
                  For part (1), if $x_n=-1$ then $x_n+1=2$ and $x_i=2$ for all $i>n+1$. We need also to show $x_n$ does not get the value $-1$.
                  $endgroup$
                  – S_Alex
                  5 hours ago
















                • $begingroup$
                  For part (1), if $x_n=-1$ then $x_n+1=2$ and $x_i=2$ for all $i>n+1$. We need also to show $x_n$ does not get the value $-1$.
                  $endgroup$
                  – S_Alex
                  5 hours ago















                $begingroup$
                For part (1), if $x_n=-1$ then $x_n+1=2$ and $x_i=2$ for all $i>n+1$. We need also to show $x_n$ does not get the value $-1$.
                $endgroup$
                – S_Alex
                5 hours ago




                $begingroup$
                For part (1), if $x_n=-1$ then $x_n+1=2$ and $x_i=2$ for all $i>n+1$. We need also to show $x_n$ does not get the value $-1$.
                $endgroup$
                – S_Alex
                5 hours ago











                2












                $begingroup$

                Let $f(x) = x^3 - 3x$ and explore, for which $x$ one has
                $$
                |f'(x)| < 1.
                $$






                share|cite|improve this answer









                $endgroup$

















                  2












                  $begingroup$

                  Let $f(x) = x^3 - 3x$ and explore, for which $x$ one has
                  $$
                  |f'(x)| < 1.
                  $$






                  share|cite|improve this answer









                  $endgroup$















                    2












                    2








                    2





                    $begingroup$

                    Let $f(x) = x^3 - 3x$ and explore, for which $x$ one has
                    $$
                    |f'(x)| < 1.
                    $$






                    share|cite|improve this answer









                    $endgroup$



                    Let $f(x) = x^3 - 3x$ and explore, for which $x$ one has
                    $$
                    |f'(x)| < 1.
                    $$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 7 hours ago









                    avsavs

                    4,4751515




                    4,4751515





















                        0












                        $begingroup$

                        Notice that for $alpha>0$:



                        $$(2+alpha)^3-3(2+alpha)=2+9alpha+6alpha^2+alpha^3>2+alpha$$
                        while for $betain[0,2]$:
                        $$(2-beta)^3-3(2-beta)=2-9beta+6beta^2-beta^3<2-beta$$



                        The opposite arguments can be made around $-2$. Since $x_1in[-2,-2]$, your iteration is restricted to this range






                        share|cite|improve this answer









                        $endgroup$

















                          0












                          $begingroup$

                          Notice that for $alpha>0$:



                          $$(2+alpha)^3-3(2+alpha)=2+9alpha+6alpha^2+alpha^3>2+alpha$$
                          while for $betain[0,2]$:
                          $$(2-beta)^3-3(2-beta)=2-9beta+6beta^2-beta^3<2-beta$$



                          The opposite arguments can be made around $-2$. Since $x_1in[-2,-2]$, your iteration is restricted to this range






                          share|cite|improve this answer









                          $endgroup$















                            0












                            0








                            0





                            $begingroup$

                            Notice that for $alpha>0$:



                            $$(2+alpha)^3-3(2+alpha)=2+9alpha+6alpha^2+alpha^3>2+alpha$$
                            while for $betain[0,2]$:
                            $$(2-beta)^3-3(2-beta)=2-9beta+6beta^2-beta^3<2-beta$$



                            The opposite arguments can be made around $-2$. Since $x_1in[-2,-2]$, your iteration is restricted to this range






                            share|cite|improve this answer









                            $endgroup$



                            Notice that for $alpha>0$:



                            $$(2+alpha)^3-3(2+alpha)=2+9alpha+6alpha^2+alpha^3>2+alpha$$
                            while for $betain[0,2]$:
                            $$(2-beta)^3-3(2-beta)=2-9beta+6beta^2-beta^3<2-beta$$



                            The opposite arguments can be made around $-2$. Since $x_1in[-2,-2]$, your iteration is restricted to this range







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 6 hours ago









                            Rhys HughesRhys Hughes

                            7,1381630




                            7,1381630





















                                0












                                $begingroup$

                                The desired claim follows from the following two observations:




                                Claim. If $x_n in -1/4, 1/4$, then $x_n neq 0$ for all $n geq 1$.




                                Proof. Let $p_1 = operatornamesign(x_1)$ and $p_n+1 = p_n^3 - 3p_n 4^2 cdot 3^n-1$. Then we inductively check that $p_n$ is always odd and $ x_n = p_n / 4^3^n-1$. Since the numerator is always odd integer, it cannot vanish, and the claim follows.




                                Claim 2. If either $x_n geq 0$ for any sufficiently large $n$ or $x_n leq 0$ for any sufficiently $n$, then we actually have $x_n = 0$ for any sufficiently large $n$.




                                Proof. Write $x_n = 2cos(2 pi f_n)$. Then



                                $$ cos(2 pi f_n+1) = fracx_n+12 = fracx_n^3 - 3x_n2 = 4cos^3(2 pi f_n) - 3cos(2 pi f_n) = cos(2 pi cdot 3f_n). $$



                                So it follows that $cos(2pi f_N+n) = cos(2pi cdot 3^n f_N)$. Now, replacing $(x_n)$ by $(-x_n)$ if necessary, we may assume that $x_n geq 0$ for all sufficiently large $n$. In other words, there exists $N$ so that $x_N+n geq 0$ for all $n geq 0$. Then each $3^n f_N$ must avoid the sets $(frac14, frac34) + mathbbZ$. So it follows that



                                $$ f_N in mathbbR setminus bigcup_n=0^infty bigcup_kinmathbbZ left( frac4k+14 cdot 3^n, frac4k+34 cdot 3^n right) = left k pm frac14cdot 3^n : n geq 0 right. $$



                                This implies that $3^n_0 f_N = pm 1$ for some $n_0 geq 0$, and hence $x_N+n = 0$ for all $n geq n_0$. This proves the desired claim.






                                share|cite|improve this answer









                                $endgroup$

















                                  0












                                  $begingroup$

                                  The desired claim follows from the following two observations:




                                  Claim. If $x_n in -1/4, 1/4$, then $x_n neq 0$ for all $n geq 1$.




                                  Proof. Let $p_1 = operatornamesign(x_1)$ and $p_n+1 = p_n^3 - 3p_n 4^2 cdot 3^n-1$. Then we inductively check that $p_n$ is always odd and $ x_n = p_n / 4^3^n-1$. Since the numerator is always odd integer, it cannot vanish, and the claim follows.




                                  Claim 2. If either $x_n geq 0$ for any sufficiently large $n$ or $x_n leq 0$ for any sufficiently $n$, then we actually have $x_n = 0$ for any sufficiently large $n$.




                                  Proof. Write $x_n = 2cos(2 pi f_n)$. Then



                                  $$ cos(2 pi f_n+1) = fracx_n+12 = fracx_n^3 - 3x_n2 = 4cos^3(2 pi f_n) - 3cos(2 pi f_n) = cos(2 pi cdot 3f_n). $$



                                  So it follows that $cos(2pi f_N+n) = cos(2pi cdot 3^n f_N)$. Now, replacing $(x_n)$ by $(-x_n)$ if necessary, we may assume that $x_n geq 0$ for all sufficiently large $n$. In other words, there exists $N$ so that $x_N+n geq 0$ for all $n geq 0$. Then each $3^n f_N$ must avoid the sets $(frac14, frac34) + mathbbZ$. So it follows that



                                  $$ f_N in mathbbR setminus bigcup_n=0^infty bigcup_kinmathbbZ left( frac4k+14 cdot 3^n, frac4k+34 cdot 3^n right) = left k pm frac14cdot 3^n : n geq 0 right. $$



                                  This implies that $3^n_0 f_N = pm 1$ for some $n_0 geq 0$, and hence $x_N+n = 0$ for all $n geq n_0$. This proves the desired claim.






                                  share|cite|improve this answer









                                  $endgroup$















                                    0












                                    0








                                    0





                                    $begingroup$

                                    The desired claim follows from the following two observations:




                                    Claim. If $x_n in -1/4, 1/4$, then $x_n neq 0$ for all $n geq 1$.




                                    Proof. Let $p_1 = operatornamesign(x_1)$ and $p_n+1 = p_n^3 - 3p_n 4^2 cdot 3^n-1$. Then we inductively check that $p_n$ is always odd and $ x_n = p_n / 4^3^n-1$. Since the numerator is always odd integer, it cannot vanish, and the claim follows.




                                    Claim 2. If either $x_n geq 0$ for any sufficiently large $n$ or $x_n leq 0$ for any sufficiently $n$, then we actually have $x_n = 0$ for any sufficiently large $n$.




                                    Proof. Write $x_n = 2cos(2 pi f_n)$. Then



                                    $$ cos(2 pi f_n+1) = fracx_n+12 = fracx_n^3 - 3x_n2 = 4cos^3(2 pi f_n) - 3cos(2 pi f_n) = cos(2 pi cdot 3f_n). $$



                                    So it follows that $cos(2pi f_N+n) = cos(2pi cdot 3^n f_N)$. Now, replacing $(x_n)$ by $(-x_n)$ if necessary, we may assume that $x_n geq 0$ for all sufficiently large $n$. In other words, there exists $N$ so that $x_N+n geq 0$ for all $n geq 0$. Then each $3^n f_N$ must avoid the sets $(frac14, frac34) + mathbbZ$. So it follows that



                                    $$ f_N in mathbbR setminus bigcup_n=0^infty bigcup_kinmathbbZ left( frac4k+14 cdot 3^n, frac4k+34 cdot 3^n right) = left k pm frac14cdot 3^n : n geq 0 right. $$



                                    This implies that $3^n_0 f_N = pm 1$ for some $n_0 geq 0$, and hence $x_N+n = 0$ for all $n geq n_0$. This proves the desired claim.






                                    share|cite|improve this answer









                                    $endgroup$



                                    The desired claim follows from the following two observations:




                                    Claim. If $x_n in -1/4, 1/4$, then $x_n neq 0$ for all $n geq 1$.




                                    Proof. Let $p_1 = operatornamesign(x_1)$ and $p_n+1 = p_n^3 - 3p_n 4^2 cdot 3^n-1$. Then we inductively check that $p_n$ is always odd and $ x_n = p_n / 4^3^n-1$. Since the numerator is always odd integer, it cannot vanish, and the claim follows.




                                    Claim 2. If either $x_n geq 0$ for any sufficiently large $n$ or $x_n leq 0$ for any sufficiently $n$, then we actually have $x_n = 0$ for any sufficiently large $n$.




                                    Proof. Write $x_n = 2cos(2 pi f_n)$. Then



                                    $$ cos(2 pi f_n+1) = fracx_n+12 = fracx_n^3 - 3x_n2 = 4cos^3(2 pi f_n) - 3cos(2 pi f_n) = cos(2 pi cdot 3f_n). $$



                                    So it follows that $cos(2pi f_N+n) = cos(2pi cdot 3^n f_N)$. Now, replacing $(x_n)$ by $(-x_n)$ if necessary, we may assume that $x_n geq 0$ for all sufficiently large $n$. In other words, there exists $N$ so that $x_N+n geq 0$ for all $n geq 0$. Then each $3^n f_N$ must avoid the sets $(frac14, frac34) + mathbbZ$. So it follows that



                                    $$ f_N in mathbbR setminus bigcup_n=0^infty bigcup_kinmathbbZ left( frac4k+14 cdot 3^n, frac4k+34 cdot 3^n right) = left k pm frac14cdot 3^n : n geq 0 right. $$



                                    This implies that $3^n_0 f_N = pm 1$ for some $n_0 geq 0$, and hence $x_N+n = 0$ for all $n geq n_0$. This proves the desired claim.







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                                    answered 3 hours ago









                                    Sangchul LeeSangchul Lee

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