Degenerate kernel method to solve Fredholm integral equation of the second kind












2












$begingroup$


$$ f(x) + int_0^1 (xy+x^2y^2) f(y) dy = x^3 +frac16x^2+ frac15x
$$



I have this fredholm integral equation of the second kind and am not sure how to answer this equation.
I know that is has to be written in the form
$$ sum a_jx^{j-1} $$



I do not know how to integrate with the function $f(y)$ in the integral.



Any help would be appreciated.



Thanks






calculus linear-algebra integration integral-equations







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    $$ f(x) + int_0^1 (xy+x^2y^2) f(y) dy = x^3 +frac16x^2+ frac15x
    $$



    I have this fredholm integral equation of the second kind and am not sure how to answer this equation.
    I know that is has to be written in the form
    $$ sum a_jx^{j-1} $$



    I do not know how to integrate with the function $f(y)$ in the integral.



    Any help would be appreciated.



    Thanks






    calculus linear-algebra integration integral-equations







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      0



      $begingroup$


      $$ f(x) + int_0^1 (xy+x^2y^2) f(y) dy = x^3 +frac16x^2+ frac15x
      $$



      I have this fredholm integral equation of the second kind and am not sure how to answer this equation.
      I know that is has to be written in the form
      $$ sum a_jx^{j-1} $$



      I do not know how to integrate with the function $f(y)$ in the integral.



      Any help would be appreciated.



      Thanks






      calculus linear-algebra integration integral-equations







      share|cite|improve this question











      $endgroup$




      $$ f(x) + int_0^1 (xy+x^2y^2) f(y) dy = x^3 +frac16x^2+ frac15x
      $$



      I have this fredholm integral equation of the second kind and am not sure how to answer this equation.
      I know that is has to be written in the form
      $$ sum a_jx^{j-1} $$



      I do not know how to integrate with the function $f(y)$ in the integral.



      Any help would be appreciated.



      Thanks






      calculus linear-algebra integration integral-equations




      calculus linear-algebra integration integral-equations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 9 '18 at 19:27







      p s

















      asked Dec 2 '18 at 16:02









      p sp s

      358




      358






















          2 Answers
          2






          active

          oldest

          votes


















          0












          $begingroup$

          can you start with:
          $$f(x)+int_0^1(xy+x^2y^2)f(y)dy=x^3+frac16x^2+frac15x$$
          $$f'(x)+(x+x^2)f(1)=3x^2+frac13x+frac15$$
          $$f'(x)=(3-f(1))x^2+left(frac13-f(1)right)+frac15$$
          so:
          $$f(x)=intleft[(3-f(1))x^2+left(frac13-f(1)right)+frac15right]dx+C$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Once integrating that final integral would it be able to write it as a sum of a function. Also, how would I integrate f(1)
            $endgroup$
            – p s
            Dec 2 '18 at 21:00










          • $begingroup$
            f(1) is just a constant, but we would need a way to work out its value, which is an issue as the function is defined with the function itself
            $endgroup$
            – Henry Lee
            Dec 2 '18 at 21:02



















          0












          $begingroup$

          Note that we can write the equation as
          $$ f(x) = x^3 + left[frac{1}{6} - int limits_0^1 y^2 f(y) , mathrm{d} yright] x^2 + left[frac{1}{5} - int limits_0^1 y f(y) , mathrm{d} yright] x , , , x in [0,1] , . $$
          Obviously, we do not know the values of the integrals yet, but it turns out that we do not need to! It is sufficient to conclude that there are constants $a, b in mathbb{R}$ such that
          $$ f(x) = x^3 + a x^2 + b x , , , x in [0,1] , . $$
          We can then plug this form of $f$ into the previous equation and compute the integrals to obtain
          $$ x^3 + a x^2 + b x = x^3 - left(frac{a}{5}+frac{b}{4}right) x^2 - left(frac{a}{4}+frac{b}{3}right)x , , , x in [0,1] , . $$
          This yields the linear system
          $$ begin{pmatrix} frac{6}{5} && frac{1}{4}\ frac{1}{4} && frac{4}{3} end{pmatrix} begin{pmatrix} a \b end{pmatrix} = begin{pmatrix} 0 \0 end{pmatrix} , ,$$
          which only has the trivial solution $a=b=0$ . Therefore
          $$ f(x) = x^3 , , , x in [0,1] , , $$
          is the unique solution to the integral equation.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you, this gives me a final answer but I have to write it as a sum of functions in the form $$ sum a_jx^{j-1} $$ how would I change what you have solved above into this format?
            $endgroup$
            – p s
            Dec 2 '18 at 23:22












          • $begingroup$
            @ps You can write $f(x) = x^3 = sum_{j=2}^4 a_j x^{j-1} $ with $a_4 = 1$ and $a_3 = a_2 = 0$ .
            $endgroup$
            – ComplexYetTrivial
            Dec 2 '18 at 23:32










          • $begingroup$
            how did you get to $$ x^3 - left(frac{a}{5}+frac{b}{4}right) x^2 - left(frac{a}{4}+frac{b}{3}right)x , , , x in [0,1] , . $$
            $endgroup$
            – p s
            Dec 9 '18 at 19:00













          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%2f3022806%2fdegenerate-kernel-method-to-solve-fredholm-integral-equation-of-the-second-kind%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          can you start with:
          $$f(x)+int_0^1(xy+x^2y^2)f(y)dy=x^3+frac16x^2+frac15x$$
          $$f'(x)+(x+x^2)f(1)=3x^2+frac13x+frac15$$
          $$f'(x)=(3-f(1))x^2+left(frac13-f(1)right)+frac15$$
          so:
          $$f(x)=intleft[(3-f(1))x^2+left(frac13-f(1)right)+frac15right]dx+C$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Once integrating that final integral would it be able to write it as a sum of a function. Also, how would I integrate f(1)
            $endgroup$
            – p s
            Dec 2 '18 at 21:00










          • $begingroup$
            f(1) is just a constant, but we would need a way to work out its value, which is an issue as the function is defined with the function itself
            $endgroup$
            – Henry Lee
            Dec 2 '18 at 21:02
















          0












          $begingroup$

          can you start with:
          $$f(x)+int_0^1(xy+x^2y^2)f(y)dy=x^3+frac16x^2+frac15x$$
          $$f'(x)+(x+x^2)f(1)=3x^2+frac13x+frac15$$
          $$f'(x)=(3-f(1))x^2+left(frac13-f(1)right)+frac15$$
          so:
          $$f(x)=intleft[(3-f(1))x^2+left(frac13-f(1)right)+frac15right]dx+C$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Once integrating that final integral would it be able to write it as a sum of a function. Also, how would I integrate f(1)
            $endgroup$
            – p s
            Dec 2 '18 at 21:00










          • $begingroup$
            f(1) is just a constant, but we would need a way to work out its value, which is an issue as the function is defined with the function itself
            $endgroup$
            – Henry Lee
            Dec 2 '18 at 21:02














          0












          0








          0





          $begingroup$

          can you start with:
          $$f(x)+int_0^1(xy+x^2y^2)f(y)dy=x^3+frac16x^2+frac15x$$
          $$f'(x)+(x+x^2)f(1)=3x^2+frac13x+frac15$$
          $$f'(x)=(3-f(1))x^2+left(frac13-f(1)right)+frac15$$
          so:
          $$f(x)=intleft[(3-f(1))x^2+left(frac13-f(1)right)+frac15right]dx+C$$






          share|cite|improve this answer









          $endgroup$



          can you start with:
          $$f(x)+int_0^1(xy+x^2y^2)f(y)dy=x^3+frac16x^2+frac15x$$
          $$f'(x)+(x+x^2)f(1)=3x^2+frac13x+frac15$$
          $$f'(x)=(3-f(1))x^2+left(frac13-f(1)right)+frac15$$
          so:
          $$f(x)=intleft[(3-f(1))x^2+left(frac13-f(1)right)+frac15right]dx+C$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 2 '18 at 18:22









          Henry LeeHenry Lee

          1,865219




          1,865219












          • $begingroup$
            Once integrating that final integral would it be able to write it as a sum of a function. Also, how would I integrate f(1)
            $endgroup$
            – p s
            Dec 2 '18 at 21:00










          • $begingroup$
            f(1) is just a constant, but we would need a way to work out its value, which is an issue as the function is defined with the function itself
            $endgroup$
            – Henry Lee
            Dec 2 '18 at 21:02


















          • $begingroup$
            Once integrating that final integral would it be able to write it as a sum of a function. Also, how would I integrate f(1)
            $endgroup$
            – p s
            Dec 2 '18 at 21:00










          • $begingroup$
            f(1) is just a constant, but we would need a way to work out its value, which is an issue as the function is defined with the function itself
            $endgroup$
            – Henry Lee
            Dec 2 '18 at 21:02
















          $begingroup$
          Once integrating that final integral would it be able to write it as a sum of a function. Also, how would I integrate f(1)
          $endgroup$
          – p s
          Dec 2 '18 at 21:00




          $begingroup$
          Once integrating that final integral would it be able to write it as a sum of a function. Also, how would I integrate f(1)
          $endgroup$
          – p s
          Dec 2 '18 at 21:00












          $begingroup$
          f(1) is just a constant, but we would need a way to work out its value, which is an issue as the function is defined with the function itself
          $endgroup$
          – Henry Lee
          Dec 2 '18 at 21:02




          $begingroup$
          f(1) is just a constant, but we would need a way to work out its value, which is an issue as the function is defined with the function itself
          $endgroup$
          – Henry Lee
          Dec 2 '18 at 21:02











          0












          $begingroup$

          Note that we can write the equation as
          $$ f(x) = x^3 + left[frac{1}{6} - int limits_0^1 y^2 f(y) , mathrm{d} yright] x^2 + left[frac{1}{5} - int limits_0^1 y f(y) , mathrm{d} yright] x , , , x in [0,1] , . $$
          Obviously, we do not know the values of the integrals yet, but it turns out that we do not need to! It is sufficient to conclude that there are constants $a, b in mathbb{R}$ such that
          $$ f(x) = x^3 + a x^2 + b x , , , x in [0,1] , . $$
          We can then plug this form of $f$ into the previous equation and compute the integrals to obtain
          $$ x^3 + a x^2 + b x = x^3 - left(frac{a}{5}+frac{b}{4}right) x^2 - left(frac{a}{4}+frac{b}{3}right)x , , , x in [0,1] , . $$
          This yields the linear system
          $$ begin{pmatrix} frac{6}{5} && frac{1}{4}\ frac{1}{4} && frac{4}{3} end{pmatrix} begin{pmatrix} a \b end{pmatrix} = begin{pmatrix} 0 \0 end{pmatrix} , ,$$
          which only has the trivial solution $a=b=0$ . Therefore
          $$ f(x) = x^3 , , , x in [0,1] , , $$
          is the unique solution to the integral equation.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you, this gives me a final answer but I have to write it as a sum of functions in the form $$ sum a_jx^{j-1} $$ how would I change what you have solved above into this format?
            $endgroup$
            – p s
            Dec 2 '18 at 23:22












          • $begingroup$
            @ps You can write $f(x) = x^3 = sum_{j=2}^4 a_j x^{j-1} $ with $a_4 = 1$ and $a_3 = a_2 = 0$ .
            $endgroup$
            – ComplexYetTrivial
            Dec 2 '18 at 23:32










          • $begingroup$
            how did you get to $$ x^3 - left(frac{a}{5}+frac{b}{4}right) x^2 - left(frac{a}{4}+frac{b}{3}right)x , , , x in [0,1] , . $$
            $endgroup$
            – p s
            Dec 9 '18 at 19:00


















          0












          $begingroup$

          Note that we can write the equation as
          $$ f(x) = x^3 + left[frac{1}{6} - int limits_0^1 y^2 f(y) , mathrm{d} yright] x^2 + left[frac{1}{5} - int limits_0^1 y f(y) , mathrm{d} yright] x , , , x in [0,1] , . $$
          Obviously, we do not know the values of the integrals yet, but it turns out that we do not need to! It is sufficient to conclude that there are constants $a, b in mathbb{R}$ such that
          $$ f(x) = x^3 + a x^2 + b x , , , x in [0,1] , . $$
          We can then plug this form of $f$ into the previous equation and compute the integrals to obtain
          $$ x^3 + a x^2 + b x = x^3 - left(frac{a}{5}+frac{b}{4}right) x^2 - left(frac{a}{4}+frac{b}{3}right)x , , , x in [0,1] , . $$
          This yields the linear system
          $$ begin{pmatrix} frac{6}{5} && frac{1}{4}\ frac{1}{4} && frac{4}{3} end{pmatrix} begin{pmatrix} a \b end{pmatrix} = begin{pmatrix} 0 \0 end{pmatrix} , ,$$
          which only has the trivial solution $a=b=0$ . Therefore
          $$ f(x) = x^3 , , , x in [0,1] , , $$
          is the unique solution to the integral equation.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you, this gives me a final answer but I have to write it as a sum of functions in the form $$ sum a_jx^{j-1} $$ how would I change what you have solved above into this format?
            $endgroup$
            – p s
            Dec 2 '18 at 23:22












          • $begingroup$
            @ps You can write $f(x) = x^3 = sum_{j=2}^4 a_j x^{j-1} $ with $a_4 = 1$ and $a_3 = a_2 = 0$ .
            $endgroup$
            – ComplexYetTrivial
            Dec 2 '18 at 23:32










          • $begingroup$
            how did you get to $$ x^3 - left(frac{a}{5}+frac{b}{4}right) x^2 - left(frac{a}{4}+frac{b}{3}right)x , , , x in [0,1] , . $$
            $endgroup$
            – p s
            Dec 9 '18 at 19:00
















          0












          0








          0





          $begingroup$

          Note that we can write the equation as
          $$ f(x) = x^3 + left[frac{1}{6} - int limits_0^1 y^2 f(y) , mathrm{d} yright] x^2 + left[frac{1}{5} - int limits_0^1 y f(y) , mathrm{d} yright] x , , , x in [0,1] , . $$
          Obviously, we do not know the values of the integrals yet, but it turns out that we do not need to! It is sufficient to conclude that there are constants $a, b in mathbb{R}$ such that
          $$ f(x) = x^3 + a x^2 + b x , , , x in [0,1] , . $$
          We can then plug this form of $f$ into the previous equation and compute the integrals to obtain
          $$ x^3 + a x^2 + b x = x^3 - left(frac{a}{5}+frac{b}{4}right) x^2 - left(frac{a}{4}+frac{b}{3}right)x , , , x in [0,1] , . $$
          This yields the linear system
          $$ begin{pmatrix} frac{6}{5} && frac{1}{4}\ frac{1}{4} && frac{4}{3} end{pmatrix} begin{pmatrix} a \b end{pmatrix} = begin{pmatrix} 0 \0 end{pmatrix} , ,$$
          which only has the trivial solution $a=b=0$ . Therefore
          $$ f(x) = x^3 , , , x in [0,1] , , $$
          is the unique solution to the integral equation.






          share|cite|improve this answer









          $endgroup$



          Note that we can write the equation as
          $$ f(x) = x^3 + left[frac{1}{6} - int limits_0^1 y^2 f(y) , mathrm{d} yright] x^2 + left[frac{1}{5} - int limits_0^1 y f(y) , mathrm{d} yright] x , , , x in [0,1] , . $$
          Obviously, we do not know the values of the integrals yet, but it turns out that we do not need to! It is sufficient to conclude that there are constants $a, b in mathbb{R}$ such that
          $$ f(x) = x^3 + a x^2 + b x , , , x in [0,1] , . $$
          We can then plug this form of $f$ into the previous equation and compute the integrals to obtain
          $$ x^3 + a x^2 + b x = x^3 - left(frac{a}{5}+frac{b}{4}right) x^2 - left(frac{a}{4}+frac{b}{3}right)x , , , x in [0,1] , . $$
          This yields the linear system
          $$ begin{pmatrix} frac{6}{5} && frac{1}{4}\ frac{1}{4} && frac{4}{3} end{pmatrix} begin{pmatrix} a \b end{pmatrix} = begin{pmatrix} 0 \0 end{pmatrix} , ,$$
          which only has the trivial solution $a=b=0$ . Therefore
          $$ f(x) = x^3 , , , x in [0,1] , , $$
          is the unique solution to the integral equation.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 2 '18 at 22:57









          ComplexYetTrivialComplexYetTrivial

          3,7302629




          3,7302629












          • $begingroup$
            Thank you, this gives me a final answer but I have to write it as a sum of functions in the form $$ sum a_jx^{j-1} $$ how would I change what you have solved above into this format?
            $endgroup$
            – p s
            Dec 2 '18 at 23:22












          • $begingroup$
            @ps You can write $f(x) = x^3 = sum_{j=2}^4 a_j x^{j-1} $ with $a_4 = 1$ and $a_3 = a_2 = 0$ .
            $endgroup$
            – ComplexYetTrivial
            Dec 2 '18 at 23:32










          • $begingroup$
            how did you get to $$ x^3 - left(frac{a}{5}+frac{b}{4}right) x^2 - left(frac{a}{4}+frac{b}{3}right)x , , , x in [0,1] , . $$
            $endgroup$
            – p s
            Dec 9 '18 at 19:00




















          • $begingroup$
            Thank you, this gives me a final answer but I have to write it as a sum of functions in the form $$ sum a_jx^{j-1} $$ how would I change what you have solved above into this format?
            $endgroup$
            – p s
            Dec 2 '18 at 23:22












          • $begingroup$
            @ps You can write $f(x) = x^3 = sum_{j=2}^4 a_j x^{j-1} $ with $a_4 = 1$ and $a_3 = a_2 = 0$ .
            $endgroup$
            – ComplexYetTrivial
            Dec 2 '18 at 23:32










          • $begingroup$
            how did you get to $$ x^3 - left(frac{a}{5}+frac{b}{4}right) x^2 - left(frac{a}{4}+frac{b}{3}right)x , , , x in [0,1] , . $$
            $endgroup$
            – p s
            Dec 9 '18 at 19:00


















          $begingroup$
          Thank you, this gives me a final answer but I have to write it as a sum of functions in the form $$ sum a_jx^{j-1} $$ how would I change what you have solved above into this format?
          $endgroup$
          – p s
          Dec 2 '18 at 23:22






          $begingroup$
          Thank you, this gives me a final answer but I have to write it as a sum of functions in the form $$ sum a_jx^{j-1} $$ how would I change what you have solved above into this format?
          $endgroup$
          – p s
          Dec 2 '18 at 23:22














          $begingroup$
          @ps You can write $f(x) = x^3 = sum_{j=2}^4 a_j x^{j-1} $ with $a_4 = 1$ and $a_3 = a_2 = 0$ .
          $endgroup$
          – ComplexYetTrivial
          Dec 2 '18 at 23:32




          $begingroup$
          @ps You can write $f(x) = x^3 = sum_{j=2}^4 a_j x^{j-1} $ with $a_4 = 1$ and $a_3 = a_2 = 0$ .
          $endgroup$
          – ComplexYetTrivial
          Dec 2 '18 at 23:32












          $begingroup$
          how did you get to $$ x^3 - left(frac{a}{5}+frac{b}{4}right) x^2 - left(frac{a}{4}+frac{b}{3}right)x , , , x in [0,1] , . $$
          $endgroup$
          – p s
          Dec 9 '18 at 19:00






          $begingroup$
          how did you get to $$ x^3 - left(frac{a}{5}+frac{b}{4}right) x^2 - left(frac{a}{4}+frac{b}{3}right)x , , , x in [0,1] , . $$
          $endgroup$
          – p s
          Dec 9 '18 at 19:00




















          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%2f3022806%2fdegenerate-kernel-method-to-solve-fredholm-integral-equation-of-the-second-kind%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

          Ellipse (mathématiques)

          Image
          Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
          Read more

          Quarter-circle Tiles

          Image
          up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
          Read more

          Mont Emei

          Image
          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
          Read more