infinitely many negative and infinitely many positive numbers












2












$begingroup$


Suppose that
$$x_1=frac{1}{4}, x_{n+1}=x_{n}^3-3x_n.$$



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 $sqrt{3}$. I want to show that the sequence cannot escape some interval.










share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Suppose that
    $$x_1=frac{1}{4}, x_{n+1}=x_{n}^3-3x_n.$$



    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 $sqrt{3}$. I want to show that the sequence cannot escape some interval.










    share|cite|improve this question









    $endgroup$















      2












      2








      2


      2



      $begingroup$


      Suppose that
      $$x_1=frac{1}{4}, x_{n+1}=x_{n}^3-3x_n.$$



      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 $sqrt{3}$. I want to show that the sequence cannot escape some interval.










      share|cite|improve this question









      $endgroup$




      Suppose that
      $$x_1=frac{1}{4}, x_{n+1}=x_{n}^3-3x_n.$$



      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 $sqrt{3}$. 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, sqrt{3})$, then $f(x) < 0 $.


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

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


          4. Show that if $ x in ( sqrt{3} ,2 )$, then there eixsts an $n$ such that $ f^n(x) < sqrt{3}$.

            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 $sqrt{3}$ 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
            6 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 $p_1 = operatorname{sign}(x_1)$ 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



              $$ cos(2 pi f_{n+1}) = frac{x_{n+1}}{2} = frac{x_n^3 - 3x_n}{2} = 4cos^3(2 pi f_n) - 3cos(2 pi f_n) = cos(2 pi cdot 3f_n). $$



              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 $(frac{1}{4}, frac{3}{4}) + mathbb{Z}$. So it follows that



              $$ f_N in mathbb{R} setminus bigcup_{n=0}^{infty} bigcup_{kinmathbb{Z}} left( frac{4k+1}{4 cdot 3^n}, frac{4k+3}{4 cdot 3^n} right) = left{ k pm frac{1}{4cdot 3^n} : n geq 0 right}. $$



              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









              $endgroup$














                Your Answer








                StackExchange.ready(function() {
                var channelOptions = {
                tags: "".split(" "),
                id: "69"
                };
                initTagRenderer("".split(" "), "".split(" "), channelOptions);

                StackExchange.using("externalEditor", function() {
                // Have to fire editor after snippets, if snippets enabled
                if (StackExchange.settings.snippets.snippetsEnabled) {
                StackExchange.using("snippets", function() {
                createEditor();
                });
                }
                else {
                createEditor();
                }
                });

                function createEditor() {
                StackExchange.prepareEditor({
                heartbeatType: 'answer',
                autoActivateHeartbeat: false,
                convertImagesToLinks: true,
                noModals: true,
                showLowRepImageUploadWarning: true,
                reputationToPostImages: 10,
                bindNavPrevention: true,
                postfix: "",
                imageUploader: {
                brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
                contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
                allowUrls: true
                },
                noCode: true, onDemand: true,
                discardSelector: ".discard-answer"
                ,immediatelyShowMarkdownHelp:true
                });


                }
                });














                draft saved

                draft discarded


















                StackExchange.ready(
                function () {
                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3203826%2finfinitely-many-negative-and-infinitely-many-positive-numbers%23new-answer', 'question_page');
                }
                );

                Post as a guest















                Required, but never shown

























                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, sqrt{3})$, then $f(x) < 0 $.


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

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


                4. Show that if $ x in ( sqrt{3} ,2 )$, then there eixsts an $n$ such that $ f^n(x) < sqrt{3}$.

                  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 $sqrt{3}$ 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
                  6 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, sqrt{3})$, then $f(x) < 0 $.


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

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


                4. Show that if $ x in ( sqrt{3} ,2 )$, then there eixsts an $n$ such that $ f^n(x) < sqrt{3}$.

                  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 $sqrt{3}$ 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
                  6 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, sqrt{3})$, then $f(x) < 0 $.


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

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


                4. Show that if $ x in ( sqrt{3} ,2 )$, then there eixsts an $n$ such that $ f^n(x) < sqrt{3}$.

                  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 $sqrt{3}$ 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, sqrt{3})$, then $f(x) < 0 $.


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

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


                4. Show that if $ x in ( sqrt{3} ,2 )$, then there eixsts an $n$ such that $ f^n(x) < sqrt{3}$.

                  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 $sqrt{3}$ 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
                  6 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
                  6 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
                6 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
                6 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 $p_1 = operatorname{sign}(x_1)$ 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



                                $$ cos(2 pi f_{n+1}) = frac{x_{n+1}}{2} = frac{x_n^3 - 3x_n}{2} = 4cos^3(2 pi f_n) - 3cos(2 pi f_n) = cos(2 pi cdot 3f_n). $$



                                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 $(frac{1}{4}, frac{3}{4}) + mathbb{Z}$. So it follows that



                                $$ f_N in mathbb{R} setminus bigcup_{n=0}^{infty} bigcup_{kinmathbb{Z}} left( frac{4k+1}{4 cdot 3^n}, frac{4k+3}{4 cdot 3^n} right) = left{ k pm frac{1}{4cdot 3^n} : n geq 0 right}. $$



                                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 $p_1 = operatorname{sign}(x_1)$ 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



                                  $$ cos(2 pi f_{n+1}) = frac{x_{n+1}}{2} = frac{x_n^3 - 3x_n}{2} = 4cos^3(2 pi f_n) - 3cos(2 pi f_n) = cos(2 pi cdot 3f_n). $$



                                  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 $(frac{1}{4}, frac{3}{4}) + mathbb{Z}$. So it follows that



                                  $$ f_N in mathbb{R} setminus bigcup_{n=0}^{infty} bigcup_{kinmathbb{Z}} left( frac{4k+1}{4 cdot 3^n}, frac{4k+3}{4 cdot 3^n} right) = left{ k pm frac{1}{4cdot 3^n} : n geq 0 right}. $$



                                  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 $p_1 = operatorname{sign}(x_1)$ 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



                                    $$ cos(2 pi f_{n+1}) = frac{x_{n+1}}{2} = frac{x_n^3 - 3x_n}{2} = 4cos^3(2 pi f_n) - 3cos(2 pi f_n) = cos(2 pi cdot 3f_n). $$



                                    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 $(frac{1}{4}, frac{3}{4}) + mathbb{Z}$. So it follows that



                                    $$ f_N in mathbb{R} setminus bigcup_{n=0}^{infty} bigcup_{kinmathbb{Z}} left( frac{4k+1}{4 cdot 3^n}, frac{4k+3}{4 cdot 3^n} right) = left{ k pm frac{1}{4cdot 3^n} : n geq 0 right}. $$



                                    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 $p_1 = operatorname{sign}(x_1)$ 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



                                    $$ cos(2 pi f_{n+1}) = frac{x_{n+1}}{2} = frac{x_n^3 - 3x_n}{2} = 4cos^3(2 pi f_n) - 3cos(2 pi f_n) = cos(2 pi cdot 3f_n). $$



                                    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 $(frac{1}{4}, frac{3}{4}) + mathbb{Z}$. So it follows that



                                    $$ f_N in mathbb{R} setminus bigcup_{n=0}^{infty} bigcup_{kinmathbb{Z}} left( frac{4k+1}{4 cdot 3^n}, frac{4k+3}{4 cdot 3^n} right) = left{ k pm frac{1}{4cdot 3^n} : n geq 0 right}. $$



                                    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












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 3 hours ago









                                    Sangchul LeeSangchul Lee

                                    97k12173283




                                    97k12173283






























                                        draft saved

                                        draft discarded




















































                                        Thanks for contributing an answer to Mathematics Stack Exchange!


                                        • Please be sure to answer the question. Provide details and share your research!

                                        But avoid



                                        • Asking for help, clarification, or responding to other answers.

                                        • Making statements based on opinion; back them up with references or personal experience.


                                        Use MathJax to format equations. MathJax reference.


                                        To learn more, see our tips on writing great answers.




                                        draft saved


                                        draft discarded














                                        StackExchange.ready(
                                        function () {
                                        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3203826%2finfinitely-many-negative-and-infinitely-many-positive-numbers%23new-answer', 'question_page');
                                        }
                                        );

                                        Post as a guest















                                        Required, but never shown





















































                                        Required, but never shown














                                        Required, but never shown












                                        Required, but never shown







                                        Required, but never shown

































                                        Required, but never shown














                                        Required, but never shown












                                        Required, but never shown







                                        Required, but never shown







                                        Popular posts from this blog

                                        日野市

                                        GameSpot

                                        Tu-95轟炸機