Prove that the polynomial is $g(x,y)(x^2 + y^2 -1)^2 + c$












1












$begingroup$


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



Let $f(x,y)$ be a polynomial in two real variables such that the polynomials



$$frac{partial f}{partial x}(x,y)$$



$$frac{partial f}{partial y}(x,y)$$



are divisible by $x^2+y^2-1$. Prove that there's a polynomial $g(x,y)$ and a constant $c$ such that



$$f(x,y) = g(x,y)(x^2+y^2 -1)^2 +c$$






real-analysis multivariable-calculus contest-math multivariate-polynomial







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


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



    Let $f(x,y)$ be a polynomial in two real variables such that the polynomials



    $$frac{partial f}{partial x}(x,y)$$



    $$frac{partial f}{partial y}(x,y)$$



    are divisible by $x^2+y^2-1$. Prove that there's a polynomial $g(x,y)$ and a constant $c$ such that



    $$f(x,y) = g(x,y)(x^2+y^2 -1)^2 +c$$






    real-analysis multivariable-calculus contest-math multivariate-polynomial







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


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



      Let $f(x,y)$ be a polynomial in two real variables such that the polynomials



      $$frac{partial f}{partial x}(x,y)$$



      $$frac{partial f}{partial y}(x,y)$$



      are divisible by $x^2+y^2-1$. Prove that there's a polynomial $g(x,y)$ and a constant $c$ such that



      $$f(x,y) = g(x,y)(x^2+y^2 -1)^2 +c$$






      real-analysis multivariable-calculus contest-math multivariate-polynomial







      share|cite|improve this question











      $endgroup$




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



      Let $f(x,y)$ be a polynomial in two real variables such that the polynomials



      $$frac{partial f}{partial x}(x,y)$$



      $$frac{partial f}{partial y}(x,y)$$



      are divisible by $x^2+y^2-1$. Prove that there's a polynomial $g(x,y)$ and a constant $c$ such that



      $$f(x,y) = g(x,y)(x^2+y^2 -1)^2 +c$$






      real-analysis multivariable-calculus contest-math multivariate-polynomial




      real-analysis multivariable-calculus contest-math multivariate-polynomial






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 20 '18 at 0:01









      Jean Marie

      28.9k41949




      28.9k41949










      asked Nov 19 '18 at 23:34









      Rafael DeigaRafael Deiga

      662311




      662311






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Treating $f$ as a polynomial in $(mathbb{R}[y])[x]$, there exist polynomials $p(x,y) in (mathbb{R}[y])[x], q(y), r(y) in mathbb{R}[y]$, such that $f(x,y)=(x^2+y^2-1)p(x,y)+xq(y)+r(y)$.



          We have that $frac{partial f}{partial x}(x,y)=(x^2+y^2-1)frac{partial p}{partial x}(x,y)+2xp(x,y)+q(y)$ and $frac{partial f}{partial y}(x,y)=(x^2+y^2-1)frac{partial p}{partial y}(x,y)+2yp(x,y)+xq’(y)+r’(y)$ are divisible by $(x^2+y^2-1)$. Therefore $2xp(x,y)+q(y)$ and $2yp(x,y)+xq’(y)+r’(y)$ are divisible by $(x^2+y^2-1)$



          As we did earlier, there exist polynomials $s(x,y) in (mathbb{R}[y])[x], t(y),u(y) in mathbb{R}[y]$ such that $p(x,y)=(x^2+y^2-1)s(x,y)+xt(y)+u(y)$.



          We have that
          begin{align}
          2xp(x,y)+q(y)&=2x(x^2+y^2-1)s(x,y)+2x^2t(y)+2xu(y)+q(y) \
          &=(x^2+y^2-1)(2xs(x,y)+2t(y))+x(2u(y))+(q(y)-2(y^2-1)t(y))\
          end{align}
          is divisible by $x^2+y^2-1$. Thus $2u(y)=0,q(y)-2(y^2-1)t(y)=0$. Have $q’(y)=4yt(y)+2(y^2-1)t’(y)$



          Next we have
          begin{align}
          2yp(x,y)+xq’(y)+r’(y)&=2y(x^2+y^2-1)s(x,y)+2xyt(y)+2yu(y)+xq’(y)+r’(y)\
          &=2y(x^2+y^2-1)s(x,y)+x(2yt(y)+4yt(y)+2(y^2-1)t’(y))+r’(y)\
          end{align}
          is divisible by $x^2+y^2-1$. Thus $6yt(y)+2(y^2-1)t’(y)=0, r’(y)=0$. Thus $r(y)=c$ for some constant $c$.



          We shall show $t(y)=0$. Assume on the contrary $t$ is not identically $0$. Let $t$ have degree $n$ with nonzero leading coefficient $a$. Comparing the leading coefficient in $6yt(y)+2(y^2-1)t’(y)=0$, we get $6a=2an$, so $n=3$. Note $t$ is divisible by $y^2-1$, so $t(y)=a(y^2-1)(y+b)$, some $b in mathbb{R}$. Substituting $y=0$ in $6yt(y)+2(y^2-1)t’(y)=0$ gives $t’(0)=0$. However $t’(0)=-a$ is nonzero, a contradiction. Thus $t(y)=0$, and so $q(y)=2(y^2-1)t(y)=0$.



          Thus
          begin{align}
          f(x,y)&=(x^2+y^2-1)p(x,y)+xq(y)+r(y)\
          &=(x^2+y^2-1)^2s(x,y)+x(x^2+y^2-1)t(x,y)+(x^2+y^2-1)u(y)+xq(y)+r(y)\
          &=(x^2+y^2-1)^2s(x,y)+c\
          end{align}
          and we are done.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            When you gonna prove that $t(y) =0$, actually we have $6a + 2an =0$, which implies much more easier that $a =0$ xD. The rest is correct i think. Could you fix that?
            $endgroup$
            – Rafael Deiga
            Dec 30 '18 at 11:16











          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%2f3005690%2fprove-that-the-polynomial-is-gx-yx2-y2-12-c%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Treating $f$ as a polynomial in $(mathbb{R}[y])[x]$, there exist polynomials $p(x,y) in (mathbb{R}[y])[x], q(y), r(y) in mathbb{R}[y]$, such that $f(x,y)=(x^2+y^2-1)p(x,y)+xq(y)+r(y)$.



          We have that $frac{partial f}{partial x}(x,y)=(x^2+y^2-1)frac{partial p}{partial x}(x,y)+2xp(x,y)+q(y)$ and $frac{partial f}{partial y}(x,y)=(x^2+y^2-1)frac{partial p}{partial y}(x,y)+2yp(x,y)+xq’(y)+r’(y)$ are divisible by $(x^2+y^2-1)$. Therefore $2xp(x,y)+q(y)$ and $2yp(x,y)+xq’(y)+r’(y)$ are divisible by $(x^2+y^2-1)$



          As we did earlier, there exist polynomials $s(x,y) in (mathbb{R}[y])[x], t(y),u(y) in mathbb{R}[y]$ such that $p(x,y)=(x^2+y^2-1)s(x,y)+xt(y)+u(y)$.



          We have that
          begin{align}
          2xp(x,y)+q(y)&=2x(x^2+y^2-1)s(x,y)+2x^2t(y)+2xu(y)+q(y) \
          &=(x^2+y^2-1)(2xs(x,y)+2t(y))+x(2u(y))+(q(y)-2(y^2-1)t(y))\
          end{align}
          is divisible by $x^2+y^2-1$. Thus $2u(y)=0,q(y)-2(y^2-1)t(y)=0$. Have $q’(y)=4yt(y)+2(y^2-1)t’(y)$



          Next we have
          begin{align}
          2yp(x,y)+xq’(y)+r’(y)&=2y(x^2+y^2-1)s(x,y)+2xyt(y)+2yu(y)+xq’(y)+r’(y)\
          &=2y(x^2+y^2-1)s(x,y)+x(2yt(y)+4yt(y)+2(y^2-1)t’(y))+r’(y)\
          end{align}
          is divisible by $x^2+y^2-1$. Thus $6yt(y)+2(y^2-1)t’(y)=0, r’(y)=0$. Thus $r(y)=c$ for some constant $c$.



          We shall show $t(y)=0$. Assume on the contrary $t$ is not identically $0$. Let $t$ have degree $n$ with nonzero leading coefficient $a$. Comparing the leading coefficient in $6yt(y)+2(y^2-1)t’(y)=0$, we get $6a=2an$, so $n=3$. Note $t$ is divisible by $y^2-1$, so $t(y)=a(y^2-1)(y+b)$, some $b in mathbb{R}$. Substituting $y=0$ in $6yt(y)+2(y^2-1)t’(y)=0$ gives $t’(0)=0$. However $t’(0)=-a$ is nonzero, a contradiction. Thus $t(y)=0$, and so $q(y)=2(y^2-1)t(y)=0$.



          Thus
          begin{align}
          f(x,y)&=(x^2+y^2-1)p(x,y)+xq(y)+r(y)\
          &=(x^2+y^2-1)^2s(x,y)+x(x^2+y^2-1)t(x,y)+(x^2+y^2-1)u(y)+xq(y)+r(y)\
          &=(x^2+y^2-1)^2s(x,y)+c\
          end{align}
          and we are done.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            When you gonna prove that $t(y) =0$, actually we have $6a + 2an =0$, which implies much more easier that $a =0$ xD. The rest is correct i think. Could you fix that?
            $endgroup$
            – Rafael Deiga
            Dec 30 '18 at 11:16
















          2












          $begingroup$

          Treating $f$ as a polynomial in $(mathbb{R}[y])[x]$, there exist polynomials $p(x,y) in (mathbb{R}[y])[x], q(y), r(y) in mathbb{R}[y]$, such that $f(x,y)=(x^2+y^2-1)p(x,y)+xq(y)+r(y)$.



          We have that $frac{partial f}{partial x}(x,y)=(x^2+y^2-1)frac{partial p}{partial x}(x,y)+2xp(x,y)+q(y)$ and $frac{partial f}{partial y}(x,y)=(x^2+y^2-1)frac{partial p}{partial y}(x,y)+2yp(x,y)+xq’(y)+r’(y)$ are divisible by $(x^2+y^2-1)$. Therefore $2xp(x,y)+q(y)$ and $2yp(x,y)+xq’(y)+r’(y)$ are divisible by $(x^2+y^2-1)$



          As we did earlier, there exist polynomials $s(x,y) in (mathbb{R}[y])[x], t(y),u(y) in mathbb{R}[y]$ such that $p(x,y)=(x^2+y^2-1)s(x,y)+xt(y)+u(y)$.



          We have that
          begin{align}
          2xp(x,y)+q(y)&=2x(x^2+y^2-1)s(x,y)+2x^2t(y)+2xu(y)+q(y) \
          &=(x^2+y^2-1)(2xs(x,y)+2t(y))+x(2u(y))+(q(y)-2(y^2-1)t(y))\
          end{align}
          is divisible by $x^2+y^2-1$. Thus $2u(y)=0,q(y)-2(y^2-1)t(y)=0$. Have $q’(y)=4yt(y)+2(y^2-1)t’(y)$



          Next we have
          begin{align}
          2yp(x,y)+xq’(y)+r’(y)&=2y(x^2+y^2-1)s(x,y)+2xyt(y)+2yu(y)+xq’(y)+r’(y)\
          &=2y(x^2+y^2-1)s(x,y)+x(2yt(y)+4yt(y)+2(y^2-1)t’(y))+r’(y)\
          end{align}
          is divisible by $x^2+y^2-1$. Thus $6yt(y)+2(y^2-1)t’(y)=0, r’(y)=0$. Thus $r(y)=c$ for some constant $c$.



          We shall show $t(y)=0$. Assume on the contrary $t$ is not identically $0$. Let $t$ have degree $n$ with nonzero leading coefficient $a$. Comparing the leading coefficient in $6yt(y)+2(y^2-1)t’(y)=0$, we get $6a=2an$, so $n=3$. Note $t$ is divisible by $y^2-1$, so $t(y)=a(y^2-1)(y+b)$, some $b in mathbb{R}$. Substituting $y=0$ in $6yt(y)+2(y^2-1)t’(y)=0$ gives $t’(0)=0$. However $t’(0)=-a$ is nonzero, a contradiction. Thus $t(y)=0$, and so $q(y)=2(y^2-1)t(y)=0$.



          Thus
          begin{align}
          f(x,y)&=(x^2+y^2-1)p(x,y)+xq(y)+r(y)\
          &=(x^2+y^2-1)^2s(x,y)+x(x^2+y^2-1)t(x,y)+(x^2+y^2-1)u(y)+xq(y)+r(y)\
          &=(x^2+y^2-1)^2s(x,y)+c\
          end{align}
          and we are done.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            When you gonna prove that $t(y) =0$, actually we have $6a + 2an =0$, which implies much more easier that $a =0$ xD. The rest is correct i think. Could you fix that?
            $endgroup$
            – Rafael Deiga
            Dec 30 '18 at 11:16














          2












          2








          2





          $begingroup$

          Treating $f$ as a polynomial in $(mathbb{R}[y])[x]$, there exist polynomials $p(x,y) in (mathbb{R}[y])[x], q(y), r(y) in mathbb{R}[y]$, such that $f(x,y)=(x^2+y^2-1)p(x,y)+xq(y)+r(y)$.



          We have that $frac{partial f}{partial x}(x,y)=(x^2+y^2-1)frac{partial p}{partial x}(x,y)+2xp(x,y)+q(y)$ and $frac{partial f}{partial y}(x,y)=(x^2+y^2-1)frac{partial p}{partial y}(x,y)+2yp(x,y)+xq’(y)+r’(y)$ are divisible by $(x^2+y^2-1)$. Therefore $2xp(x,y)+q(y)$ and $2yp(x,y)+xq’(y)+r’(y)$ are divisible by $(x^2+y^2-1)$



          As we did earlier, there exist polynomials $s(x,y) in (mathbb{R}[y])[x], t(y),u(y) in mathbb{R}[y]$ such that $p(x,y)=(x^2+y^2-1)s(x,y)+xt(y)+u(y)$.



          We have that
          begin{align}
          2xp(x,y)+q(y)&=2x(x^2+y^2-1)s(x,y)+2x^2t(y)+2xu(y)+q(y) \
          &=(x^2+y^2-1)(2xs(x,y)+2t(y))+x(2u(y))+(q(y)-2(y^2-1)t(y))\
          end{align}
          is divisible by $x^2+y^2-1$. Thus $2u(y)=0,q(y)-2(y^2-1)t(y)=0$. Have $q’(y)=4yt(y)+2(y^2-1)t’(y)$



          Next we have
          begin{align}
          2yp(x,y)+xq’(y)+r’(y)&=2y(x^2+y^2-1)s(x,y)+2xyt(y)+2yu(y)+xq’(y)+r’(y)\
          &=2y(x^2+y^2-1)s(x,y)+x(2yt(y)+4yt(y)+2(y^2-1)t’(y))+r’(y)\
          end{align}
          is divisible by $x^2+y^2-1$. Thus $6yt(y)+2(y^2-1)t’(y)=0, r’(y)=0$. Thus $r(y)=c$ for some constant $c$.



          We shall show $t(y)=0$. Assume on the contrary $t$ is not identically $0$. Let $t$ have degree $n$ with nonzero leading coefficient $a$. Comparing the leading coefficient in $6yt(y)+2(y^2-1)t’(y)=0$, we get $6a=2an$, so $n=3$. Note $t$ is divisible by $y^2-1$, so $t(y)=a(y^2-1)(y+b)$, some $b in mathbb{R}$. Substituting $y=0$ in $6yt(y)+2(y^2-1)t’(y)=0$ gives $t’(0)=0$. However $t’(0)=-a$ is nonzero, a contradiction. Thus $t(y)=0$, and so $q(y)=2(y^2-1)t(y)=0$.



          Thus
          begin{align}
          f(x,y)&=(x^2+y^2-1)p(x,y)+xq(y)+r(y)\
          &=(x^2+y^2-1)^2s(x,y)+x(x^2+y^2-1)t(x,y)+(x^2+y^2-1)u(y)+xq(y)+r(y)\
          &=(x^2+y^2-1)^2s(x,y)+c\
          end{align}
          and we are done.






          share|cite|improve this answer











          $endgroup$



          Treating $f$ as a polynomial in $(mathbb{R}[y])[x]$, there exist polynomials $p(x,y) in (mathbb{R}[y])[x], q(y), r(y) in mathbb{R}[y]$, such that $f(x,y)=(x^2+y^2-1)p(x,y)+xq(y)+r(y)$.



          We have that $frac{partial f}{partial x}(x,y)=(x^2+y^2-1)frac{partial p}{partial x}(x,y)+2xp(x,y)+q(y)$ and $frac{partial f}{partial y}(x,y)=(x^2+y^2-1)frac{partial p}{partial y}(x,y)+2yp(x,y)+xq’(y)+r’(y)$ are divisible by $(x^2+y^2-1)$. Therefore $2xp(x,y)+q(y)$ and $2yp(x,y)+xq’(y)+r’(y)$ are divisible by $(x^2+y^2-1)$



          As we did earlier, there exist polynomials $s(x,y) in (mathbb{R}[y])[x], t(y),u(y) in mathbb{R}[y]$ such that $p(x,y)=(x^2+y^2-1)s(x,y)+xt(y)+u(y)$.



          We have that
          begin{align}
          2xp(x,y)+q(y)&=2x(x^2+y^2-1)s(x,y)+2x^2t(y)+2xu(y)+q(y) \
          &=(x^2+y^2-1)(2xs(x,y)+2t(y))+x(2u(y))+(q(y)-2(y^2-1)t(y))\
          end{align}
          is divisible by $x^2+y^2-1$. Thus $2u(y)=0,q(y)-2(y^2-1)t(y)=0$. Have $q’(y)=4yt(y)+2(y^2-1)t’(y)$



          Next we have
          begin{align}
          2yp(x,y)+xq’(y)+r’(y)&=2y(x^2+y^2-1)s(x,y)+2xyt(y)+2yu(y)+xq’(y)+r’(y)\
          &=2y(x^2+y^2-1)s(x,y)+x(2yt(y)+4yt(y)+2(y^2-1)t’(y))+r’(y)\
          end{align}
          is divisible by $x^2+y^2-1$. Thus $6yt(y)+2(y^2-1)t’(y)=0, r’(y)=0$. Thus $r(y)=c$ for some constant $c$.



          We shall show $t(y)=0$. Assume on the contrary $t$ is not identically $0$. Let $t$ have degree $n$ with nonzero leading coefficient $a$. Comparing the leading coefficient in $6yt(y)+2(y^2-1)t’(y)=0$, we get $6a=2an$, so $n=3$. Note $t$ is divisible by $y^2-1$, so $t(y)=a(y^2-1)(y+b)$, some $b in mathbb{R}$. Substituting $y=0$ in $6yt(y)+2(y^2-1)t’(y)=0$ gives $t’(0)=0$. However $t’(0)=-a$ is nonzero, a contradiction. Thus $t(y)=0$, and so $q(y)=2(y^2-1)t(y)=0$.



          Thus
          begin{align}
          f(x,y)&=(x^2+y^2-1)p(x,y)+xq(y)+r(y)\
          &=(x^2+y^2-1)^2s(x,y)+x(x^2+y^2-1)t(x,y)+(x^2+y^2-1)u(y)+xq(y)+r(y)\
          &=(x^2+y^2-1)^2s(x,y)+c\
          end{align}
          and we are done.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 30 '18 at 2:58

























          answered Dec 30 '18 at 2:53









          user630376user630376

          813




          813












          • $begingroup$
            When you gonna prove that $t(y) =0$, actually we have $6a + 2an =0$, which implies much more easier that $a =0$ xD. The rest is correct i think. Could you fix that?
            $endgroup$
            – Rafael Deiga
            Dec 30 '18 at 11:16


















          • $begingroup$
            When you gonna prove that $t(y) =0$, actually we have $6a + 2an =0$, which implies much more easier that $a =0$ xD. The rest is correct i think. Could you fix that?
            $endgroup$
            – Rafael Deiga
            Dec 30 '18 at 11:16
















          $begingroup$
          When you gonna prove that $t(y) =0$, actually we have $6a + 2an =0$, which implies much more easier that $a =0$ xD. The rest is correct i think. Could you fix that?
          $endgroup$
          – Rafael Deiga
          Dec 30 '18 at 11:16




          $begingroup$
          When you gonna prove that $t(y) =0$, actually we have $6a + 2an =0$, which implies much more easier that $a =0$ xD. The rest is correct i think. Could you fix that?
          $endgroup$
          – Rafael Deiga
          Dec 30 '18 at 11:16


















          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%2f3005690%2fprove-that-the-polynomial-is-gx-yx2-y2-12-c%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

          Human spaceflight

          Image
          Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e
          Read more

          Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

          Image
          0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
          Read more

          File:DeusFollowingSea.jpg

          Image
          File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
          Read more