Maximize $x_1^3+x_2^3+cdots + x_n^3$












5














This is from a Brazilian math contest for college students (OBMU):



Given a positive integer $n$, find the maximum value of



$$x_1^3+x_2^3+ cdots + x_n^3$$



where $x_j$ is a real number for all $j in {1,2,cdots, n}$ such that $x_1 + x_2 + cdots + x_n = 0$ and $x_1^2 + x_2^2 + cdots + x_n^2 = 1$ .










share|cite|improve this question


















  • 2




    What have you tried?
    – Frpzzd
    Nov 19 '18 at 23:49










  • What have you tried?
    – stuart stevenson
    Nov 19 '18 at 23:49






  • 1




    @stuartstevenson Jinx. XD $space$
    – Frpzzd
    Nov 19 '18 at 23:49










  • Nothing much. I just realized that the maximum value is less than 1, which is obvious.
    – Rafael Deiga
    Nov 19 '18 at 23:51






  • 1




    Are you familiar with en.wikipedia.org/wiki/Lagrange_multiplier?
    – Federico
    Nov 20 '18 at 0:37
















5














This is from a Brazilian math contest for college students (OBMU):



Given a positive integer $n$, find the maximum value of



$$x_1^3+x_2^3+ cdots + x_n^3$$



where $x_j$ is a real number for all $j in {1,2,cdots, n}$ such that $x_1 + x_2 + cdots + x_n = 0$ and $x_1^2 + x_2^2 + cdots + x_n^2 = 1$ .










share|cite|improve this question


















  • 2




    What have you tried?
    – Frpzzd
    Nov 19 '18 at 23:49










  • What have you tried?
    – stuart stevenson
    Nov 19 '18 at 23:49






  • 1




    @stuartstevenson Jinx. XD $space$
    – Frpzzd
    Nov 19 '18 at 23:49










  • Nothing much. I just realized that the maximum value is less than 1, which is obvious.
    – Rafael Deiga
    Nov 19 '18 at 23:51






  • 1




    Are you familiar with en.wikipedia.org/wiki/Lagrange_multiplier?
    – Federico
    Nov 20 '18 at 0:37














5












5








5


4





This is from a Brazilian math contest for college students (OBMU):



Given a positive integer $n$, find the maximum value of



$$x_1^3+x_2^3+ cdots + x_n^3$$



where $x_j$ is a real number for all $j in {1,2,cdots, n}$ such that $x_1 + x_2 + cdots + x_n = 0$ and $x_1^2 + x_2^2 + cdots + x_n^2 = 1$ .










share|cite|improve this question













This is from a Brazilian math contest for college students (OBMU):



Given a positive integer $n$, find the maximum value of



$$x_1^3+x_2^3+ cdots + x_n^3$$



where $x_j$ is a real number for all $j in {1,2,cdots, n}$ such that $x_1 + x_2 + cdots + x_n = 0$ and $x_1^2 + x_2^2 + cdots + x_n^2 = 1$ .







inequality optimization






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 19 '18 at 23:47









Rafael Deiga

662311




662311








  • 2




    What have you tried?
    – Frpzzd
    Nov 19 '18 at 23:49










  • What have you tried?
    – stuart stevenson
    Nov 19 '18 at 23:49






  • 1




    @stuartstevenson Jinx. XD $space$
    – Frpzzd
    Nov 19 '18 at 23:49










  • Nothing much. I just realized that the maximum value is less than 1, which is obvious.
    – Rafael Deiga
    Nov 19 '18 at 23:51






  • 1




    Are you familiar with en.wikipedia.org/wiki/Lagrange_multiplier?
    – Federico
    Nov 20 '18 at 0:37














  • 2




    What have you tried?
    – Frpzzd
    Nov 19 '18 at 23:49










  • What have you tried?
    – stuart stevenson
    Nov 19 '18 at 23:49






  • 1




    @stuartstevenson Jinx. XD $space$
    – Frpzzd
    Nov 19 '18 at 23:49










  • Nothing much. I just realized that the maximum value is less than 1, which is obvious.
    – Rafael Deiga
    Nov 19 '18 at 23:51






  • 1




    Are you familiar with en.wikipedia.org/wiki/Lagrange_multiplier?
    – Federico
    Nov 20 '18 at 0:37








2




2




What have you tried?
– Frpzzd
Nov 19 '18 at 23:49




What have you tried?
– Frpzzd
Nov 19 '18 at 23:49












What have you tried?
– stuart stevenson
Nov 19 '18 at 23:49




What have you tried?
– stuart stevenson
Nov 19 '18 at 23:49




1




1




@stuartstevenson Jinx. XD $space$
– Frpzzd
Nov 19 '18 at 23:49




@stuartstevenson Jinx. XD $space$
– Frpzzd
Nov 19 '18 at 23:49












Nothing much. I just realized that the maximum value is less than 1, which is obvious.
– Rafael Deiga
Nov 19 '18 at 23:51




Nothing much. I just realized that the maximum value is less than 1, which is obvious.
– Rafael Deiga
Nov 19 '18 at 23:51




1




1




Are you familiar with en.wikipedia.org/wiki/Lagrange_multiplier?
– Federico
Nov 20 '18 at 0:37




Are you familiar with en.wikipedia.org/wiki/Lagrange_multiplier?
– Federico
Nov 20 '18 at 0:37










3 Answers
3






active

oldest

votes


















4














One could indeed use Lagrange multipliers, but I thought I'd try to get geometric intuition in three dimensions.



Constraint sphere



This shows the constraint sphere $x_1^2 + x_2^2 + x_3^2=1$ and the constraint plane $x_1 + x_2 + x_3 = 0$, and the locus of their intersection, a (black) ring. The plane is perpendicular to the normalized vector ${bf a} = (1/sqrt{3}, 1/sqrt{3}, 1/sqrt{3})$. One (normalized) vector perpendicular to ${bf a}$ is ${bf b} = left( frac{1}{sqrt{6}},frac{1}{sqrt{6}},-frac{2}{sqrt{6}} right)$. Another (normalized) vector perpendicular to both is ${bf c} = left( -frac{1}{sqrt{2}},frac{1}{sqrt{2}},0right)$, determined by ${bf c} = {bf a} times {bf b}$.



Thus any potential solution point can be described as $(x_1, x_2, x_3) = cos (theta) {bf b} + sin (theta) {bf c}$.



Direct substitution, cubing the components and simplification yields $x_1^3 + x_2^3 + x_3^3 = -frac{cos (3 theta )}{sqrt{6}}$.



It is a simple matter to maximize this function of a single variable $theta$ and find that the solution is $1/sqrt{6}$.



enter image description here



Not surprisingly there are three equivalent solutions, corresponding to the permutation of the three variables.



As a check, I find the solution vector with $theta = 1$ (from the graph) to be ${bf s} = (-0.374432, 0.815587, -0.441155)$ indeed obeys the constraints and leads to the criterion sum-of-cubes to be $0.404163$, as visible on the graph.






share|cite|improve this answer























  • For $n=4$ taking $x_1=x_2=x_3=-frac1{sqrt{12}}$ and $x_4=frac3{sqrt{12}}$ gives $x_1^3+x_2^3+x_3^3+x_4^3=1/sqrt3$ perhaps this is helpful.
    – kingW3
    Nov 20 '18 at 2:03










  • The maximum should be an increasing sequence with $n$: let $f_k(vec{x})=sum x_i^k$. If for $vec{a}=(a_1,a_2,ldots ,a_n)$, $f_1(vec{a})=0$ and $f_2(vec{a})=1$, then for $vec{b}=(a_1,a_2,ldots , a_n, 0)$, we also have $f_1(vec{b})=0$, $f_2(vec{b})=1$, and $f_3(vec{a})=f_3(vec{b})$.
    – Jacob
    Nov 20 '18 at 2:42





















2














Hint: Try to use the technique of Fourier Analysis. We can reasonably guess that the maximum occurs when $x_i = -frac{1}{sqrt{n(n-1)}}, forall i<n$ and $x_n = sqrt{1-frac{1}{n}}.$



View $x_{cdot}$ as a function defined on the group $mathbf{Z}/nmathbf{Z}$. Let $zeta = exp(frac{2pi i}{n})$ be $n$-th root of unity and define its $r$-th Fourier coefficient as $s^{}(r) = frac{1}{n}sum_{k=1}^{n}x_kzeta^{-rk}.$ We note that
$$x_j = sum_{r=0}^{n-1}s(r)zeta^{rj}, quad forall j=1,ldots,n,$$ and
$$ sum_{1leq jleq n} |x_j|^2 = nsum_{1leq rleq n}|s(r)|^2,
$$
since $frac{1}{n}sum_{1leq rleq n} zeta^{rm} =1_{{m=0}}$. Express the constraints in the language of $s(r)$ (including $x$ is real-valued.) Then we get
$$s(r) =overline{s(n-r)},; s(0) = 0,; sum_{1leq rleq n}|s(r)|^2 = frac{1}{n}. $$
What is left is to express $Q = x_1^3 + cdots + x_n^3$ as a function involving $s(r)$'s. From the first identity above, we have
$$Q = nsum_{1leq r,r'leq n}s(r)overline{s(r')}s(r-r'). $$ Note that the support is
$$R = {(r,r');| ;1leq r,r'leq n-1, rneq r'}.$$ By applying Cauchy-Schwarz, we have
$$begin{eqnarray}|sum_{1leq r,r'leq n}s(r)overline{s(r')}s(r-r')|^2&leq& sum_{(r,r')in R}|s(r)|^2|s(r-r')|^2sum_{(r,r')in R}|s(r')|^2 \
&=&sum_{1 leq rleq n-1}|s(r)|^2left(frac{1}{n} - |s(r)|^2right)cdotsum_{1 leq rleq n-1}frac{1}{n} - |s(r)|^2 \
&=&left(frac{1}{n^2} - sum_{1 leq rleq n-1} |s(r)|^4right)frac{n-2}{n}\
&leq& frac{(n-2)^2}{n^3(n-1)},
end{eqnarray}$$
since $frac{1}{n^2} leq (n-1)sum_{1 leq rleq n-1} |s(r)|^4$.
Thus, we have
$$Q leq frac{n-2}{sqrt{n}sqrt{n-1}},$$ as desired.






share|cite|improve this answer































    1














    I come up with a solution using Lagrange Multipliers (one of the comments and one of answers suggests to use this). First, we need to realize that the domain



    $$ S ={ (x_1,x_2,cdots,x_n) in {mathbb{R}}^n | x_1 + x_2 + cdots + x_n = 0 text{ and } x_1^2 + x_2^2 + cdots + x_n^2 = 1} $$



    is closed (since the functions $(x_1,x_2,cdots,x_n ) mapsto x_1 + x_2 + cdots + x_n$ and $(x_1,x_2,cdots,x_n ) mapsto x_1^2 + x_2^2 + cdots + x_n^2$ are continuous) and bounded (since $x_1^2 + x_2^2 + cdots + x_n^2 = 1$). Then S is compact.



    Define
    $$f(x_1,x_2,cdots,x_n ) = x_1^3 + x_2^3 + cdots+ x_n^3 $$
    $$g(x_1,x_2,cdots,x_n ) = x_1^2 + x_2^2 + cdots+ x_n^2 -1 $$
    $$h(x_1,x_2,cdots,x_n ) = x_1 + x_2 + cdots+ x_n $$



    Since $S$ is compact and $f$ is continuous, then $f$ reaches a maximum value in
    $S$. Using Lagrange Multiplier, we should have



    $$nabla f = lambda_1 nabla g + lambda_2 nabla h$$



    Thus,



    $$ 3(x_1^2,x_2^2,cdots,x_n^2) = 2lambda_1(x_1,x_2,cdots,x_n ) + lambda_2(1,1,cdots,1)$$



    Then,



    $$3x_i^2 = 2lambda_1 x_i+ lambda_2 text{ } forall i in {1,2,cdots,n} label{1}tag{1}$$



    If we add all the equations and use the constraints equations, we obtain $lambda_2= frac{3}{n}$. Then,



    $$3x_i^2 = 2lambda_1 x_i+ frac{3}{n} $$



    Solving for $x_i$, we obtain



    $$ x_i = frac{lambda_1 pm sqrt{lambda_1^2 + frac{9}{n}}}{3} $$



    Let $k$ be a natural number less or equal to $n$. Therefore we can suppose



    $$ x_i = frac{lambda_1 + sqrt{lambda_1^2 + frac{9}{n}}}{3} forall i in {1,2,cdots,k}$$



    $$ x_i = frac{lambda_1 - sqrt{lambda_1^2 + frac{9}{n}}}{3} forall i in {k+1,k+2,cdots,n}$$



    Firtly, notice that $k$ is different of $0$ and $n$, because of the constraint $x_1+ x_2 + cdots + x_n =0$. Using the constraint $x_1^2 + x_2^2 + cdots + x_n^2 = 1$ and after some simplifications, we obtain



    $$lambda_1 left(nlambda_1 + (2k-n) sqrt{lambda_1^2 + frac{9}{n}} right) = 0label{2}tag{2}$$



    And using $x_1+ x_2 + cdots + x_n =0$, we get



    $$ nlambda_1 + (2k-n) sqrt{lambda_1^2 + frac{9}{n}} = 0 label{3}tag{3}$$



    So, we just need to satisfy ref{3}, because with that ref{2} is automatically satisfied. Then, we get



    $$lambda_1 = frac{3(n-2k)}{2sqrt{(n-k)nk}} $$



    Notice that we should consider the other solution with the other sign, but in the end we obtain the same maximum value for both cases. Multiplying ref{1} by $x_i$:



    $$ 3x_i^3 = 2lambda_1x_i^2 + frac{3x_i}{n} $$



    Adding in all i's:



    $$3f = 2lambda_1 $$



    Thus,



    $$ f(k) = frac{n-2k}{sqrt{(n-k)nk}} = frac{1}{sqrt{n}}left(sqrt{frac{n-k}{k}}-sqrt{frac{k}{n-k}}right)$$



    Remember that $k in {2,3,cdots, n-1}$. Making $x = sqrt{frac{k}{n-k}}$ and analyzing the derivative of



    $$y: (0,1) rightarrow mathbb{R}$$
    $$ x mapstofrac{1}{x}-x $$



    We see that the maximum value of $f$ is when $k=1$, that is,



    $$ frac{n-2}{sqrt{(n-1)n}} $$



    However, the Lagrange Multipliers just find the local extrema. How can I guarantee (if it's possible) that the above value is indeed the maximum value of $f$?






    share|cite|improve this answer























      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      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%2f3005707%2fmaximize-x-13x-23-cdots-x-n3%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4














      One could indeed use Lagrange multipliers, but I thought I'd try to get geometric intuition in three dimensions.



      Constraint sphere



      This shows the constraint sphere $x_1^2 + x_2^2 + x_3^2=1$ and the constraint plane $x_1 + x_2 + x_3 = 0$, and the locus of their intersection, a (black) ring. The plane is perpendicular to the normalized vector ${bf a} = (1/sqrt{3}, 1/sqrt{3}, 1/sqrt{3})$. One (normalized) vector perpendicular to ${bf a}$ is ${bf b} = left( frac{1}{sqrt{6}},frac{1}{sqrt{6}},-frac{2}{sqrt{6}} right)$. Another (normalized) vector perpendicular to both is ${bf c} = left( -frac{1}{sqrt{2}},frac{1}{sqrt{2}},0right)$, determined by ${bf c} = {bf a} times {bf b}$.



      Thus any potential solution point can be described as $(x_1, x_2, x_3) = cos (theta) {bf b} + sin (theta) {bf c}$.



      Direct substitution, cubing the components and simplification yields $x_1^3 + x_2^3 + x_3^3 = -frac{cos (3 theta )}{sqrt{6}}$.



      It is a simple matter to maximize this function of a single variable $theta$ and find that the solution is $1/sqrt{6}$.



      enter image description here



      Not surprisingly there are three equivalent solutions, corresponding to the permutation of the three variables.



      As a check, I find the solution vector with $theta = 1$ (from the graph) to be ${bf s} = (-0.374432, 0.815587, -0.441155)$ indeed obeys the constraints and leads to the criterion sum-of-cubes to be $0.404163$, as visible on the graph.






      share|cite|improve this answer























      • For $n=4$ taking $x_1=x_2=x_3=-frac1{sqrt{12}}$ and $x_4=frac3{sqrt{12}}$ gives $x_1^3+x_2^3+x_3^3+x_4^3=1/sqrt3$ perhaps this is helpful.
        – kingW3
        Nov 20 '18 at 2:03










      • The maximum should be an increasing sequence with $n$: let $f_k(vec{x})=sum x_i^k$. If for $vec{a}=(a_1,a_2,ldots ,a_n)$, $f_1(vec{a})=0$ and $f_2(vec{a})=1$, then for $vec{b}=(a_1,a_2,ldots , a_n, 0)$, we also have $f_1(vec{b})=0$, $f_2(vec{b})=1$, and $f_3(vec{a})=f_3(vec{b})$.
        – Jacob
        Nov 20 '18 at 2:42


















      4














      One could indeed use Lagrange multipliers, but I thought I'd try to get geometric intuition in three dimensions.



      Constraint sphere



      This shows the constraint sphere $x_1^2 + x_2^2 + x_3^2=1$ and the constraint plane $x_1 + x_2 + x_3 = 0$, and the locus of their intersection, a (black) ring. The plane is perpendicular to the normalized vector ${bf a} = (1/sqrt{3}, 1/sqrt{3}, 1/sqrt{3})$. One (normalized) vector perpendicular to ${bf a}$ is ${bf b} = left( frac{1}{sqrt{6}},frac{1}{sqrt{6}},-frac{2}{sqrt{6}} right)$. Another (normalized) vector perpendicular to both is ${bf c} = left( -frac{1}{sqrt{2}},frac{1}{sqrt{2}},0right)$, determined by ${bf c} = {bf a} times {bf b}$.



      Thus any potential solution point can be described as $(x_1, x_2, x_3) = cos (theta) {bf b} + sin (theta) {bf c}$.



      Direct substitution, cubing the components and simplification yields $x_1^3 + x_2^3 + x_3^3 = -frac{cos (3 theta )}{sqrt{6}}$.



      It is a simple matter to maximize this function of a single variable $theta$ and find that the solution is $1/sqrt{6}$.



      enter image description here



      Not surprisingly there are three equivalent solutions, corresponding to the permutation of the three variables.



      As a check, I find the solution vector with $theta = 1$ (from the graph) to be ${bf s} = (-0.374432, 0.815587, -0.441155)$ indeed obeys the constraints and leads to the criterion sum-of-cubes to be $0.404163$, as visible on the graph.






      share|cite|improve this answer























      • For $n=4$ taking $x_1=x_2=x_3=-frac1{sqrt{12}}$ and $x_4=frac3{sqrt{12}}$ gives $x_1^3+x_2^3+x_3^3+x_4^3=1/sqrt3$ perhaps this is helpful.
        – kingW3
        Nov 20 '18 at 2:03










      • The maximum should be an increasing sequence with $n$: let $f_k(vec{x})=sum x_i^k$. If for $vec{a}=(a_1,a_2,ldots ,a_n)$, $f_1(vec{a})=0$ and $f_2(vec{a})=1$, then for $vec{b}=(a_1,a_2,ldots , a_n, 0)$, we also have $f_1(vec{b})=0$, $f_2(vec{b})=1$, and $f_3(vec{a})=f_3(vec{b})$.
        – Jacob
        Nov 20 '18 at 2:42
















      4












      4








      4






      One could indeed use Lagrange multipliers, but I thought I'd try to get geometric intuition in three dimensions.



      Constraint sphere



      This shows the constraint sphere $x_1^2 + x_2^2 + x_3^2=1$ and the constraint plane $x_1 + x_2 + x_3 = 0$, and the locus of their intersection, a (black) ring. The plane is perpendicular to the normalized vector ${bf a} = (1/sqrt{3}, 1/sqrt{3}, 1/sqrt{3})$. One (normalized) vector perpendicular to ${bf a}$ is ${bf b} = left( frac{1}{sqrt{6}},frac{1}{sqrt{6}},-frac{2}{sqrt{6}} right)$. Another (normalized) vector perpendicular to both is ${bf c} = left( -frac{1}{sqrt{2}},frac{1}{sqrt{2}},0right)$, determined by ${bf c} = {bf a} times {bf b}$.



      Thus any potential solution point can be described as $(x_1, x_2, x_3) = cos (theta) {bf b} + sin (theta) {bf c}$.



      Direct substitution, cubing the components and simplification yields $x_1^3 + x_2^3 + x_3^3 = -frac{cos (3 theta )}{sqrt{6}}$.



      It is a simple matter to maximize this function of a single variable $theta$ and find that the solution is $1/sqrt{6}$.



      enter image description here



      Not surprisingly there are three equivalent solutions, corresponding to the permutation of the three variables.



      As a check, I find the solution vector with $theta = 1$ (from the graph) to be ${bf s} = (-0.374432, 0.815587, -0.441155)$ indeed obeys the constraints and leads to the criterion sum-of-cubes to be $0.404163$, as visible on the graph.






      share|cite|improve this answer














      One could indeed use Lagrange multipliers, but I thought I'd try to get geometric intuition in three dimensions.



      Constraint sphere



      This shows the constraint sphere $x_1^2 + x_2^2 + x_3^2=1$ and the constraint plane $x_1 + x_2 + x_3 = 0$, and the locus of their intersection, a (black) ring. The plane is perpendicular to the normalized vector ${bf a} = (1/sqrt{3}, 1/sqrt{3}, 1/sqrt{3})$. One (normalized) vector perpendicular to ${bf a}$ is ${bf b} = left( frac{1}{sqrt{6}},frac{1}{sqrt{6}},-frac{2}{sqrt{6}} right)$. Another (normalized) vector perpendicular to both is ${bf c} = left( -frac{1}{sqrt{2}},frac{1}{sqrt{2}},0right)$, determined by ${bf c} = {bf a} times {bf b}$.



      Thus any potential solution point can be described as $(x_1, x_2, x_3) = cos (theta) {bf b} + sin (theta) {bf c}$.



      Direct substitution, cubing the components and simplification yields $x_1^3 + x_2^3 + x_3^3 = -frac{cos (3 theta )}{sqrt{6}}$.



      It is a simple matter to maximize this function of a single variable $theta$ and find that the solution is $1/sqrt{6}$.



      enter image description here



      Not surprisingly there are three equivalent solutions, corresponding to the permutation of the three variables.



      As a check, I find the solution vector with $theta = 1$ (from the graph) to be ${bf s} = (-0.374432, 0.815587, -0.441155)$ indeed obeys the constraints and leads to the criterion sum-of-cubes to be $0.404163$, as visible on the graph.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Nov 20 '18 at 2:45

























      answered Nov 20 '18 at 0:53









      David G. Stork

      9,82021232




      9,82021232












      • For $n=4$ taking $x_1=x_2=x_3=-frac1{sqrt{12}}$ and $x_4=frac3{sqrt{12}}$ gives $x_1^3+x_2^3+x_3^3+x_4^3=1/sqrt3$ perhaps this is helpful.
        – kingW3
        Nov 20 '18 at 2:03










      • The maximum should be an increasing sequence with $n$: let $f_k(vec{x})=sum x_i^k$. If for $vec{a}=(a_1,a_2,ldots ,a_n)$, $f_1(vec{a})=0$ and $f_2(vec{a})=1$, then for $vec{b}=(a_1,a_2,ldots , a_n, 0)$, we also have $f_1(vec{b})=0$, $f_2(vec{b})=1$, and $f_3(vec{a})=f_3(vec{b})$.
        – Jacob
        Nov 20 '18 at 2:42




















      • For $n=4$ taking $x_1=x_2=x_3=-frac1{sqrt{12}}$ and $x_4=frac3{sqrt{12}}$ gives $x_1^3+x_2^3+x_3^3+x_4^3=1/sqrt3$ perhaps this is helpful.
        – kingW3
        Nov 20 '18 at 2:03










      • The maximum should be an increasing sequence with $n$: let $f_k(vec{x})=sum x_i^k$. If for $vec{a}=(a_1,a_2,ldots ,a_n)$, $f_1(vec{a})=0$ and $f_2(vec{a})=1$, then for $vec{b}=(a_1,a_2,ldots , a_n, 0)$, we also have $f_1(vec{b})=0$, $f_2(vec{b})=1$, and $f_3(vec{a})=f_3(vec{b})$.
        – Jacob
        Nov 20 '18 at 2:42


















      For $n=4$ taking $x_1=x_2=x_3=-frac1{sqrt{12}}$ and $x_4=frac3{sqrt{12}}$ gives $x_1^3+x_2^3+x_3^3+x_4^3=1/sqrt3$ perhaps this is helpful.
      – kingW3
      Nov 20 '18 at 2:03




      For $n=4$ taking $x_1=x_2=x_3=-frac1{sqrt{12}}$ and $x_4=frac3{sqrt{12}}$ gives $x_1^3+x_2^3+x_3^3+x_4^3=1/sqrt3$ perhaps this is helpful.
      – kingW3
      Nov 20 '18 at 2:03












      The maximum should be an increasing sequence with $n$: let $f_k(vec{x})=sum x_i^k$. If for $vec{a}=(a_1,a_2,ldots ,a_n)$, $f_1(vec{a})=0$ and $f_2(vec{a})=1$, then for $vec{b}=(a_1,a_2,ldots , a_n, 0)$, we also have $f_1(vec{b})=0$, $f_2(vec{b})=1$, and $f_3(vec{a})=f_3(vec{b})$.
      – Jacob
      Nov 20 '18 at 2:42






      The maximum should be an increasing sequence with $n$: let $f_k(vec{x})=sum x_i^k$. If for $vec{a}=(a_1,a_2,ldots ,a_n)$, $f_1(vec{a})=0$ and $f_2(vec{a})=1$, then for $vec{b}=(a_1,a_2,ldots , a_n, 0)$, we also have $f_1(vec{b})=0$, $f_2(vec{b})=1$, and $f_3(vec{a})=f_3(vec{b})$.
      – Jacob
      Nov 20 '18 at 2:42













      2














      Hint: Try to use the technique of Fourier Analysis. We can reasonably guess that the maximum occurs when $x_i = -frac{1}{sqrt{n(n-1)}}, forall i<n$ and $x_n = sqrt{1-frac{1}{n}}.$



      View $x_{cdot}$ as a function defined on the group $mathbf{Z}/nmathbf{Z}$. Let $zeta = exp(frac{2pi i}{n})$ be $n$-th root of unity and define its $r$-th Fourier coefficient as $s^{}(r) = frac{1}{n}sum_{k=1}^{n}x_kzeta^{-rk}.$ We note that
      $$x_j = sum_{r=0}^{n-1}s(r)zeta^{rj}, quad forall j=1,ldots,n,$$ and
      $$ sum_{1leq jleq n} |x_j|^2 = nsum_{1leq rleq n}|s(r)|^2,
      $$
      since $frac{1}{n}sum_{1leq rleq n} zeta^{rm} =1_{{m=0}}$. Express the constraints in the language of $s(r)$ (including $x$ is real-valued.) Then we get
      $$s(r) =overline{s(n-r)},; s(0) = 0,; sum_{1leq rleq n}|s(r)|^2 = frac{1}{n}. $$
      What is left is to express $Q = x_1^3 + cdots + x_n^3$ as a function involving $s(r)$'s. From the first identity above, we have
      $$Q = nsum_{1leq r,r'leq n}s(r)overline{s(r')}s(r-r'). $$ Note that the support is
      $$R = {(r,r');| ;1leq r,r'leq n-1, rneq r'}.$$ By applying Cauchy-Schwarz, we have
      $$begin{eqnarray}|sum_{1leq r,r'leq n}s(r)overline{s(r')}s(r-r')|^2&leq& sum_{(r,r')in R}|s(r)|^2|s(r-r')|^2sum_{(r,r')in R}|s(r')|^2 \
      &=&sum_{1 leq rleq n-1}|s(r)|^2left(frac{1}{n} - |s(r)|^2right)cdotsum_{1 leq rleq n-1}frac{1}{n} - |s(r)|^2 \
      &=&left(frac{1}{n^2} - sum_{1 leq rleq n-1} |s(r)|^4right)frac{n-2}{n}\
      &leq& frac{(n-2)^2}{n^3(n-1)},
      end{eqnarray}$$
      since $frac{1}{n^2} leq (n-1)sum_{1 leq rleq n-1} |s(r)|^4$.
      Thus, we have
      $$Q leq frac{n-2}{sqrt{n}sqrt{n-1}},$$ as desired.






      share|cite|improve this answer




























        2














        Hint: Try to use the technique of Fourier Analysis. We can reasonably guess that the maximum occurs when $x_i = -frac{1}{sqrt{n(n-1)}}, forall i<n$ and $x_n = sqrt{1-frac{1}{n}}.$



        View $x_{cdot}$ as a function defined on the group $mathbf{Z}/nmathbf{Z}$. Let $zeta = exp(frac{2pi i}{n})$ be $n$-th root of unity and define its $r$-th Fourier coefficient as $s^{}(r) = frac{1}{n}sum_{k=1}^{n}x_kzeta^{-rk}.$ We note that
        $$x_j = sum_{r=0}^{n-1}s(r)zeta^{rj}, quad forall j=1,ldots,n,$$ and
        $$ sum_{1leq jleq n} |x_j|^2 = nsum_{1leq rleq n}|s(r)|^2,
        $$
        since $frac{1}{n}sum_{1leq rleq n} zeta^{rm} =1_{{m=0}}$. Express the constraints in the language of $s(r)$ (including $x$ is real-valued.) Then we get
        $$s(r) =overline{s(n-r)},; s(0) = 0,; sum_{1leq rleq n}|s(r)|^2 = frac{1}{n}. $$
        What is left is to express $Q = x_1^3 + cdots + x_n^3$ as a function involving $s(r)$'s. From the first identity above, we have
        $$Q = nsum_{1leq r,r'leq n}s(r)overline{s(r')}s(r-r'). $$ Note that the support is
        $$R = {(r,r');| ;1leq r,r'leq n-1, rneq r'}.$$ By applying Cauchy-Schwarz, we have
        $$begin{eqnarray}|sum_{1leq r,r'leq n}s(r)overline{s(r')}s(r-r')|^2&leq& sum_{(r,r')in R}|s(r)|^2|s(r-r')|^2sum_{(r,r')in R}|s(r')|^2 \
        &=&sum_{1 leq rleq n-1}|s(r)|^2left(frac{1}{n} - |s(r)|^2right)cdotsum_{1 leq rleq n-1}frac{1}{n} - |s(r)|^2 \
        &=&left(frac{1}{n^2} - sum_{1 leq rleq n-1} |s(r)|^4right)frac{n-2}{n}\
        &leq& frac{(n-2)^2}{n^3(n-1)},
        end{eqnarray}$$
        since $frac{1}{n^2} leq (n-1)sum_{1 leq rleq n-1} |s(r)|^4$.
        Thus, we have
        $$Q leq frac{n-2}{sqrt{n}sqrt{n-1}},$$ as desired.






        share|cite|improve this answer


























          2












          2








          2






          Hint: Try to use the technique of Fourier Analysis. We can reasonably guess that the maximum occurs when $x_i = -frac{1}{sqrt{n(n-1)}}, forall i<n$ and $x_n = sqrt{1-frac{1}{n}}.$



          View $x_{cdot}$ as a function defined on the group $mathbf{Z}/nmathbf{Z}$. Let $zeta = exp(frac{2pi i}{n})$ be $n$-th root of unity and define its $r$-th Fourier coefficient as $s^{}(r) = frac{1}{n}sum_{k=1}^{n}x_kzeta^{-rk}.$ We note that
          $$x_j = sum_{r=0}^{n-1}s(r)zeta^{rj}, quad forall j=1,ldots,n,$$ and
          $$ sum_{1leq jleq n} |x_j|^2 = nsum_{1leq rleq n}|s(r)|^2,
          $$
          since $frac{1}{n}sum_{1leq rleq n} zeta^{rm} =1_{{m=0}}$. Express the constraints in the language of $s(r)$ (including $x$ is real-valued.) Then we get
          $$s(r) =overline{s(n-r)},; s(0) = 0,; sum_{1leq rleq n}|s(r)|^2 = frac{1}{n}. $$
          What is left is to express $Q = x_1^3 + cdots + x_n^3$ as a function involving $s(r)$'s. From the first identity above, we have
          $$Q = nsum_{1leq r,r'leq n}s(r)overline{s(r')}s(r-r'). $$ Note that the support is
          $$R = {(r,r');| ;1leq r,r'leq n-1, rneq r'}.$$ By applying Cauchy-Schwarz, we have
          $$begin{eqnarray}|sum_{1leq r,r'leq n}s(r)overline{s(r')}s(r-r')|^2&leq& sum_{(r,r')in R}|s(r)|^2|s(r-r')|^2sum_{(r,r')in R}|s(r')|^2 \
          &=&sum_{1 leq rleq n-1}|s(r)|^2left(frac{1}{n} - |s(r)|^2right)cdotsum_{1 leq rleq n-1}frac{1}{n} - |s(r)|^2 \
          &=&left(frac{1}{n^2} - sum_{1 leq rleq n-1} |s(r)|^4right)frac{n-2}{n}\
          &leq& frac{(n-2)^2}{n^3(n-1)},
          end{eqnarray}$$
          since $frac{1}{n^2} leq (n-1)sum_{1 leq rleq n-1} |s(r)|^4$.
          Thus, we have
          $$Q leq frac{n-2}{sqrt{n}sqrt{n-1}},$$ as desired.






          share|cite|improve this answer














          Hint: Try to use the technique of Fourier Analysis. We can reasonably guess that the maximum occurs when $x_i = -frac{1}{sqrt{n(n-1)}}, forall i<n$ and $x_n = sqrt{1-frac{1}{n}}.$



          View $x_{cdot}$ as a function defined on the group $mathbf{Z}/nmathbf{Z}$. Let $zeta = exp(frac{2pi i}{n})$ be $n$-th root of unity and define its $r$-th Fourier coefficient as $s^{}(r) = frac{1}{n}sum_{k=1}^{n}x_kzeta^{-rk}.$ We note that
          $$x_j = sum_{r=0}^{n-1}s(r)zeta^{rj}, quad forall j=1,ldots,n,$$ and
          $$ sum_{1leq jleq n} |x_j|^2 = nsum_{1leq rleq n}|s(r)|^2,
          $$
          since $frac{1}{n}sum_{1leq rleq n} zeta^{rm} =1_{{m=0}}$. Express the constraints in the language of $s(r)$ (including $x$ is real-valued.) Then we get
          $$s(r) =overline{s(n-r)},; s(0) = 0,; sum_{1leq rleq n}|s(r)|^2 = frac{1}{n}. $$
          What is left is to express $Q = x_1^3 + cdots + x_n^3$ as a function involving $s(r)$'s. From the first identity above, we have
          $$Q = nsum_{1leq r,r'leq n}s(r)overline{s(r')}s(r-r'). $$ Note that the support is
          $$R = {(r,r');| ;1leq r,r'leq n-1, rneq r'}.$$ By applying Cauchy-Schwarz, we have
          $$begin{eqnarray}|sum_{1leq r,r'leq n}s(r)overline{s(r')}s(r-r')|^2&leq& sum_{(r,r')in R}|s(r)|^2|s(r-r')|^2sum_{(r,r')in R}|s(r')|^2 \
          &=&sum_{1 leq rleq n-1}|s(r)|^2left(frac{1}{n} - |s(r)|^2right)cdotsum_{1 leq rleq n-1}frac{1}{n} - |s(r)|^2 \
          &=&left(frac{1}{n^2} - sum_{1 leq rleq n-1} |s(r)|^4right)frac{n-2}{n}\
          &leq& frac{(n-2)^2}{n^3(n-1)},
          end{eqnarray}$$
          since $frac{1}{n^2} leq (n-1)sum_{1 leq rleq n-1} |s(r)|^4$.
          Thus, we have
          $$Q leq frac{n-2}{sqrt{n}sqrt{n-1}},$$ as desired.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 20 '18 at 4:11

























          answered Nov 20 '18 at 4:05









          Song

          4,630317




          4,630317























              1














              I come up with a solution using Lagrange Multipliers (one of the comments and one of answers suggests to use this). First, we need to realize that the domain



              $$ S ={ (x_1,x_2,cdots,x_n) in {mathbb{R}}^n | x_1 + x_2 + cdots + x_n = 0 text{ and } x_1^2 + x_2^2 + cdots + x_n^2 = 1} $$



              is closed (since the functions $(x_1,x_2,cdots,x_n ) mapsto x_1 + x_2 + cdots + x_n$ and $(x_1,x_2,cdots,x_n ) mapsto x_1^2 + x_2^2 + cdots + x_n^2$ are continuous) and bounded (since $x_1^2 + x_2^2 + cdots + x_n^2 = 1$). Then S is compact.



              Define
              $$f(x_1,x_2,cdots,x_n ) = x_1^3 + x_2^3 + cdots+ x_n^3 $$
              $$g(x_1,x_2,cdots,x_n ) = x_1^2 + x_2^2 + cdots+ x_n^2 -1 $$
              $$h(x_1,x_2,cdots,x_n ) = x_1 + x_2 + cdots+ x_n $$



              Since $S$ is compact and $f$ is continuous, then $f$ reaches a maximum value in
              $S$. Using Lagrange Multiplier, we should have



              $$nabla f = lambda_1 nabla g + lambda_2 nabla h$$



              Thus,



              $$ 3(x_1^2,x_2^2,cdots,x_n^2) = 2lambda_1(x_1,x_2,cdots,x_n ) + lambda_2(1,1,cdots,1)$$



              Then,



              $$3x_i^2 = 2lambda_1 x_i+ lambda_2 text{ } forall i in {1,2,cdots,n} label{1}tag{1}$$



              If we add all the equations and use the constraints equations, we obtain $lambda_2= frac{3}{n}$. Then,



              $$3x_i^2 = 2lambda_1 x_i+ frac{3}{n} $$



              Solving for $x_i$, we obtain



              $$ x_i = frac{lambda_1 pm sqrt{lambda_1^2 + frac{9}{n}}}{3} $$



              Let $k$ be a natural number less or equal to $n$. Therefore we can suppose



              $$ x_i = frac{lambda_1 + sqrt{lambda_1^2 + frac{9}{n}}}{3} forall i in {1,2,cdots,k}$$



              $$ x_i = frac{lambda_1 - sqrt{lambda_1^2 + frac{9}{n}}}{3} forall i in {k+1,k+2,cdots,n}$$



              Firtly, notice that $k$ is different of $0$ and $n$, because of the constraint $x_1+ x_2 + cdots + x_n =0$. Using the constraint $x_1^2 + x_2^2 + cdots + x_n^2 = 1$ and after some simplifications, we obtain



              $$lambda_1 left(nlambda_1 + (2k-n) sqrt{lambda_1^2 + frac{9}{n}} right) = 0label{2}tag{2}$$



              And using $x_1+ x_2 + cdots + x_n =0$, we get



              $$ nlambda_1 + (2k-n) sqrt{lambda_1^2 + frac{9}{n}} = 0 label{3}tag{3}$$



              So, we just need to satisfy ref{3}, because with that ref{2} is automatically satisfied. Then, we get



              $$lambda_1 = frac{3(n-2k)}{2sqrt{(n-k)nk}} $$



              Notice that we should consider the other solution with the other sign, but in the end we obtain the same maximum value for both cases. Multiplying ref{1} by $x_i$:



              $$ 3x_i^3 = 2lambda_1x_i^2 + frac{3x_i}{n} $$



              Adding in all i's:



              $$3f = 2lambda_1 $$



              Thus,



              $$ f(k) = frac{n-2k}{sqrt{(n-k)nk}} = frac{1}{sqrt{n}}left(sqrt{frac{n-k}{k}}-sqrt{frac{k}{n-k}}right)$$



              Remember that $k in {2,3,cdots, n-1}$. Making $x = sqrt{frac{k}{n-k}}$ and analyzing the derivative of



              $$y: (0,1) rightarrow mathbb{R}$$
              $$ x mapstofrac{1}{x}-x $$



              We see that the maximum value of $f$ is when $k=1$, that is,



              $$ frac{n-2}{sqrt{(n-1)n}} $$



              However, the Lagrange Multipliers just find the local extrema. How can I guarantee (if it's possible) that the above value is indeed the maximum value of $f$?






              share|cite|improve this answer




























                1














                I come up with a solution using Lagrange Multipliers (one of the comments and one of answers suggests to use this). First, we need to realize that the domain



                $$ S ={ (x_1,x_2,cdots,x_n) in {mathbb{R}}^n | x_1 + x_2 + cdots + x_n = 0 text{ and } x_1^2 + x_2^2 + cdots + x_n^2 = 1} $$



                is closed (since the functions $(x_1,x_2,cdots,x_n ) mapsto x_1 + x_2 + cdots + x_n$ and $(x_1,x_2,cdots,x_n ) mapsto x_1^2 + x_2^2 + cdots + x_n^2$ are continuous) and bounded (since $x_1^2 + x_2^2 + cdots + x_n^2 = 1$). Then S is compact.



                Define
                $$f(x_1,x_2,cdots,x_n ) = x_1^3 + x_2^3 + cdots+ x_n^3 $$
                $$g(x_1,x_2,cdots,x_n ) = x_1^2 + x_2^2 + cdots+ x_n^2 -1 $$
                $$h(x_1,x_2,cdots,x_n ) = x_1 + x_2 + cdots+ x_n $$



                Since $S$ is compact and $f$ is continuous, then $f$ reaches a maximum value in
                $S$. Using Lagrange Multiplier, we should have



                $$nabla f = lambda_1 nabla g + lambda_2 nabla h$$



                Thus,



                $$ 3(x_1^2,x_2^2,cdots,x_n^2) = 2lambda_1(x_1,x_2,cdots,x_n ) + lambda_2(1,1,cdots,1)$$



                Then,



                $$3x_i^2 = 2lambda_1 x_i+ lambda_2 text{ } forall i in {1,2,cdots,n} label{1}tag{1}$$



                If we add all the equations and use the constraints equations, we obtain $lambda_2= frac{3}{n}$. Then,



                $$3x_i^2 = 2lambda_1 x_i+ frac{3}{n} $$



                Solving for $x_i$, we obtain



                $$ x_i = frac{lambda_1 pm sqrt{lambda_1^2 + frac{9}{n}}}{3} $$



                Let $k$ be a natural number less or equal to $n$. Therefore we can suppose



                $$ x_i = frac{lambda_1 + sqrt{lambda_1^2 + frac{9}{n}}}{3} forall i in {1,2,cdots,k}$$



                $$ x_i = frac{lambda_1 - sqrt{lambda_1^2 + frac{9}{n}}}{3} forall i in {k+1,k+2,cdots,n}$$



                Firtly, notice that $k$ is different of $0$ and $n$, because of the constraint $x_1+ x_2 + cdots + x_n =0$. Using the constraint $x_1^2 + x_2^2 + cdots + x_n^2 = 1$ and after some simplifications, we obtain



                $$lambda_1 left(nlambda_1 + (2k-n) sqrt{lambda_1^2 + frac{9}{n}} right) = 0label{2}tag{2}$$



                And using $x_1+ x_2 + cdots + x_n =0$, we get



                $$ nlambda_1 + (2k-n) sqrt{lambda_1^2 + frac{9}{n}} = 0 label{3}tag{3}$$



                So, we just need to satisfy ref{3}, because with that ref{2} is automatically satisfied. Then, we get



                $$lambda_1 = frac{3(n-2k)}{2sqrt{(n-k)nk}} $$



                Notice that we should consider the other solution with the other sign, but in the end we obtain the same maximum value for both cases. Multiplying ref{1} by $x_i$:



                $$ 3x_i^3 = 2lambda_1x_i^2 + frac{3x_i}{n} $$



                Adding in all i's:



                $$3f = 2lambda_1 $$



                Thus,



                $$ f(k) = frac{n-2k}{sqrt{(n-k)nk}} = frac{1}{sqrt{n}}left(sqrt{frac{n-k}{k}}-sqrt{frac{k}{n-k}}right)$$



                Remember that $k in {2,3,cdots, n-1}$. Making $x = sqrt{frac{k}{n-k}}$ and analyzing the derivative of



                $$y: (0,1) rightarrow mathbb{R}$$
                $$ x mapstofrac{1}{x}-x $$



                We see that the maximum value of $f$ is when $k=1$, that is,



                $$ frac{n-2}{sqrt{(n-1)n}} $$



                However, the Lagrange Multipliers just find the local extrema. How can I guarantee (if it's possible) that the above value is indeed the maximum value of $f$?






                share|cite|improve this answer


























                  1












                  1








                  1






                  I come up with a solution using Lagrange Multipliers (one of the comments and one of answers suggests to use this). First, we need to realize that the domain



                  $$ S ={ (x_1,x_2,cdots,x_n) in {mathbb{R}}^n | x_1 + x_2 + cdots + x_n = 0 text{ and } x_1^2 + x_2^2 + cdots + x_n^2 = 1} $$



                  is closed (since the functions $(x_1,x_2,cdots,x_n ) mapsto x_1 + x_2 + cdots + x_n$ and $(x_1,x_2,cdots,x_n ) mapsto x_1^2 + x_2^2 + cdots + x_n^2$ are continuous) and bounded (since $x_1^2 + x_2^2 + cdots + x_n^2 = 1$). Then S is compact.



                  Define
                  $$f(x_1,x_2,cdots,x_n ) = x_1^3 + x_2^3 + cdots+ x_n^3 $$
                  $$g(x_1,x_2,cdots,x_n ) = x_1^2 + x_2^2 + cdots+ x_n^2 -1 $$
                  $$h(x_1,x_2,cdots,x_n ) = x_1 + x_2 + cdots+ x_n $$



                  Since $S$ is compact and $f$ is continuous, then $f$ reaches a maximum value in
                  $S$. Using Lagrange Multiplier, we should have



                  $$nabla f = lambda_1 nabla g + lambda_2 nabla h$$



                  Thus,



                  $$ 3(x_1^2,x_2^2,cdots,x_n^2) = 2lambda_1(x_1,x_2,cdots,x_n ) + lambda_2(1,1,cdots,1)$$



                  Then,



                  $$3x_i^2 = 2lambda_1 x_i+ lambda_2 text{ } forall i in {1,2,cdots,n} label{1}tag{1}$$



                  If we add all the equations and use the constraints equations, we obtain $lambda_2= frac{3}{n}$. Then,



                  $$3x_i^2 = 2lambda_1 x_i+ frac{3}{n} $$



                  Solving for $x_i$, we obtain



                  $$ x_i = frac{lambda_1 pm sqrt{lambda_1^2 + frac{9}{n}}}{3} $$



                  Let $k$ be a natural number less or equal to $n$. Therefore we can suppose



                  $$ x_i = frac{lambda_1 + sqrt{lambda_1^2 + frac{9}{n}}}{3} forall i in {1,2,cdots,k}$$



                  $$ x_i = frac{lambda_1 - sqrt{lambda_1^2 + frac{9}{n}}}{3} forall i in {k+1,k+2,cdots,n}$$



                  Firtly, notice that $k$ is different of $0$ and $n$, because of the constraint $x_1+ x_2 + cdots + x_n =0$. Using the constraint $x_1^2 + x_2^2 + cdots + x_n^2 = 1$ and after some simplifications, we obtain



                  $$lambda_1 left(nlambda_1 + (2k-n) sqrt{lambda_1^2 + frac{9}{n}} right) = 0label{2}tag{2}$$



                  And using $x_1+ x_2 + cdots + x_n =0$, we get



                  $$ nlambda_1 + (2k-n) sqrt{lambda_1^2 + frac{9}{n}} = 0 label{3}tag{3}$$



                  So, we just need to satisfy ref{3}, because with that ref{2} is automatically satisfied. Then, we get



                  $$lambda_1 = frac{3(n-2k)}{2sqrt{(n-k)nk}} $$



                  Notice that we should consider the other solution with the other sign, but in the end we obtain the same maximum value for both cases. Multiplying ref{1} by $x_i$:



                  $$ 3x_i^3 = 2lambda_1x_i^2 + frac{3x_i}{n} $$



                  Adding in all i's:



                  $$3f = 2lambda_1 $$



                  Thus,



                  $$ f(k) = frac{n-2k}{sqrt{(n-k)nk}} = frac{1}{sqrt{n}}left(sqrt{frac{n-k}{k}}-sqrt{frac{k}{n-k}}right)$$



                  Remember that $k in {2,3,cdots, n-1}$. Making $x = sqrt{frac{k}{n-k}}$ and analyzing the derivative of



                  $$y: (0,1) rightarrow mathbb{R}$$
                  $$ x mapstofrac{1}{x}-x $$



                  We see that the maximum value of $f$ is when $k=1$, that is,



                  $$ frac{n-2}{sqrt{(n-1)n}} $$



                  However, the Lagrange Multipliers just find the local extrema. How can I guarantee (if it's possible) that the above value is indeed the maximum value of $f$?






                  share|cite|improve this answer














                  I come up with a solution using Lagrange Multipliers (one of the comments and one of answers suggests to use this). First, we need to realize that the domain



                  $$ S ={ (x_1,x_2,cdots,x_n) in {mathbb{R}}^n | x_1 + x_2 + cdots + x_n = 0 text{ and } x_1^2 + x_2^2 + cdots + x_n^2 = 1} $$



                  is closed (since the functions $(x_1,x_2,cdots,x_n ) mapsto x_1 + x_2 + cdots + x_n$ and $(x_1,x_2,cdots,x_n ) mapsto x_1^2 + x_2^2 + cdots + x_n^2$ are continuous) and bounded (since $x_1^2 + x_2^2 + cdots + x_n^2 = 1$). Then S is compact.



                  Define
                  $$f(x_1,x_2,cdots,x_n ) = x_1^3 + x_2^3 + cdots+ x_n^3 $$
                  $$g(x_1,x_2,cdots,x_n ) = x_1^2 + x_2^2 + cdots+ x_n^2 -1 $$
                  $$h(x_1,x_2,cdots,x_n ) = x_1 + x_2 + cdots+ x_n $$



                  Since $S$ is compact and $f$ is continuous, then $f$ reaches a maximum value in
                  $S$. Using Lagrange Multiplier, we should have



                  $$nabla f = lambda_1 nabla g + lambda_2 nabla h$$



                  Thus,



                  $$ 3(x_1^2,x_2^2,cdots,x_n^2) = 2lambda_1(x_1,x_2,cdots,x_n ) + lambda_2(1,1,cdots,1)$$



                  Then,



                  $$3x_i^2 = 2lambda_1 x_i+ lambda_2 text{ } forall i in {1,2,cdots,n} label{1}tag{1}$$



                  If we add all the equations and use the constraints equations, we obtain $lambda_2= frac{3}{n}$. Then,



                  $$3x_i^2 = 2lambda_1 x_i+ frac{3}{n} $$



                  Solving for $x_i$, we obtain



                  $$ x_i = frac{lambda_1 pm sqrt{lambda_1^2 + frac{9}{n}}}{3} $$



                  Let $k$ be a natural number less or equal to $n$. Therefore we can suppose



                  $$ x_i = frac{lambda_1 + sqrt{lambda_1^2 + frac{9}{n}}}{3} forall i in {1,2,cdots,k}$$



                  $$ x_i = frac{lambda_1 - sqrt{lambda_1^2 + frac{9}{n}}}{3} forall i in {k+1,k+2,cdots,n}$$



                  Firtly, notice that $k$ is different of $0$ and $n$, because of the constraint $x_1+ x_2 + cdots + x_n =0$. Using the constraint $x_1^2 + x_2^2 + cdots + x_n^2 = 1$ and after some simplifications, we obtain



                  $$lambda_1 left(nlambda_1 + (2k-n) sqrt{lambda_1^2 + frac{9}{n}} right) = 0label{2}tag{2}$$



                  And using $x_1+ x_2 + cdots + x_n =0$, we get



                  $$ nlambda_1 + (2k-n) sqrt{lambda_1^2 + frac{9}{n}} = 0 label{3}tag{3}$$



                  So, we just need to satisfy ref{3}, because with that ref{2} is automatically satisfied. Then, we get



                  $$lambda_1 = frac{3(n-2k)}{2sqrt{(n-k)nk}} $$



                  Notice that we should consider the other solution with the other sign, but in the end we obtain the same maximum value for both cases. Multiplying ref{1} by $x_i$:



                  $$ 3x_i^3 = 2lambda_1x_i^2 + frac{3x_i}{n} $$



                  Adding in all i's:



                  $$3f = 2lambda_1 $$



                  Thus,



                  $$ f(k) = frac{n-2k}{sqrt{(n-k)nk}} = frac{1}{sqrt{n}}left(sqrt{frac{n-k}{k}}-sqrt{frac{k}{n-k}}right)$$



                  Remember that $k in {2,3,cdots, n-1}$. Making $x = sqrt{frac{k}{n-k}}$ and analyzing the derivative of



                  $$y: (0,1) rightarrow mathbb{R}$$
                  $$ x mapstofrac{1}{x}-x $$



                  We see that the maximum value of $f$ is when $k=1$, that is,



                  $$ frac{n-2}{sqrt{(n-1)n}} $$



                  However, the Lagrange Multipliers just find the local extrema. How can I guarantee (if it's possible) that the above value is indeed the maximum value of $f$?







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 23 '18 at 13:58

























                  answered Nov 21 '18 at 2:00









                  Rafael Deiga

                  662311




                  662311






























                      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.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • 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.


                      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%2f3005707%2fmaximize-x-13x-23-cdots-x-n3%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

                      android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

                      SQL update select statement

                      WPF add header to Image with URL pettitions [duplicate]