A Standard Integral Equation












3












$begingroup$


Consider the integral equation



$$phi(x) = x + lambdaint_0^1 phi(s),ds$$



Integrating with respect to $x$ from $x=0$ to $x=1$:



$$int_0^1 phi(x),dx = int_0^1x,dx + lambda int_0^1Big[int_0^1phi(s),dsBig],dx$$



which is equivalent to



$$int_0^1 phi(x),dx = frac{1}{2} + lambda int_0^1phi(s),ds$$



How can I go from here in order to solve the problem for the homogeneous case and find the corresponding characteristic values and associated rank?






linear-algebra integration matrix-equations







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What is $lambda$? What do you mean by "solve the problem"? I don't see what "the problem" is supposed to mean. Of which object do you want to find the characteristic values and ranks? Have you checked your definition of $phi$?
    $endgroup$
    – James
    8 hours ago










  • $begingroup$
    My apologies, $lambda$ is an arbitrary constant. In essence I want to obtain an expression of $phi(x)$ which does not contain a function of $s$, which the initial integral equation has.
    $endgroup$
    – LightningStrike
    8 hours ago
















3












$begingroup$


Consider the integral equation



$$phi(x) = x + lambdaint_0^1 phi(s),ds$$



Integrating with respect to $x$ from $x=0$ to $x=1$:



$$int_0^1 phi(x),dx = int_0^1x,dx + lambda int_0^1Big[int_0^1phi(s),dsBig],dx$$



which is equivalent to



$$int_0^1 phi(x),dx = frac{1}{2} + lambda int_0^1phi(s),ds$$



How can I go from here in order to solve the problem for the homogeneous case and find the corresponding characteristic values and associated rank?






linear-algebra integration matrix-equations







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What is $lambda$? What do you mean by "solve the problem"? I don't see what "the problem" is supposed to mean. Of which object do you want to find the characteristic values and ranks? Have you checked your definition of $phi$?
    $endgroup$
    – James
    8 hours ago










  • $begingroup$
    My apologies, $lambda$ is an arbitrary constant. In essence I want to obtain an expression of $phi(x)$ which does not contain a function of $s$, which the initial integral equation has.
    $endgroup$
    – LightningStrike
    8 hours ago














3












3








3





$begingroup$


Consider the integral equation



$$phi(x) = x + lambdaint_0^1 phi(s),ds$$



Integrating with respect to $x$ from $x=0$ to $x=1$:



$$int_0^1 phi(x),dx = int_0^1x,dx + lambda int_0^1Big[int_0^1phi(s),dsBig],dx$$



which is equivalent to



$$int_0^1 phi(x),dx = frac{1}{2} + lambda int_0^1phi(s),ds$$



How can I go from here in order to solve the problem for the homogeneous case and find the corresponding characteristic values and associated rank?






linear-algebra integration matrix-equations







share|cite|improve this question











$endgroup$




Consider the integral equation



$$phi(x) = x + lambdaint_0^1 phi(s),ds$$



Integrating with respect to $x$ from $x=0$ to $x=1$:



$$int_0^1 phi(x),dx = int_0^1x,dx + lambda int_0^1Big[int_0^1phi(s),dsBig],dx$$



which is equivalent to



$$int_0^1 phi(x),dx = frac{1}{2} + lambda int_0^1phi(s),ds$$



How can I go from here in order to solve the problem for the homogeneous case and find the corresponding characteristic values and associated rank?






linear-algebra integration matrix-equations




linear-algebra integration matrix-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 8 hours ago







LightningStrike

















asked 8 hours ago









LightningStrikeLightningStrike

455




455








  • 1




    $begingroup$
    What is $lambda$? What do you mean by "solve the problem"? I don't see what "the problem" is supposed to mean. Of which object do you want to find the characteristic values and ranks? Have you checked your definition of $phi$?
    $endgroup$
    – James
    8 hours ago










  • $begingroup$
    My apologies, $lambda$ is an arbitrary constant. In essence I want to obtain an expression of $phi(x)$ which does not contain a function of $s$, which the initial integral equation has.
    $endgroup$
    – LightningStrike
    8 hours ago














  • 1




    $begingroup$
    What is $lambda$? What do you mean by "solve the problem"? I don't see what "the problem" is supposed to mean. Of which object do you want to find the characteristic values and ranks? Have you checked your definition of $phi$?
    $endgroup$
    – James
    8 hours ago










  • $begingroup$
    My apologies, $lambda$ is an arbitrary constant. In essence I want to obtain an expression of $phi(x)$ which does not contain a function of $s$, which the initial integral equation has.
    $endgroup$
    – LightningStrike
    8 hours ago








1




1




$begingroup$
What is $lambda$? What do you mean by "solve the problem"? I don't see what "the problem" is supposed to mean. Of which object do you want to find the characteristic values and ranks? Have you checked your definition of $phi$?
$endgroup$
– James
8 hours ago




$begingroup$
What is $lambda$? What do you mean by "solve the problem"? I don't see what "the problem" is supposed to mean. Of which object do you want to find the characteristic values and ranks? Have you checked your definition of $phi$?
$endgroup$
– James
8 hours ago












$begingroup$
My apologies, $lambda$ is an arbitrary constant. In essence I want to obtain an expression of $phi(x)$ which does not contain a function of $s$, which the initial integral equation has.
$endgroup$
– LightningStrike
8 hours ago




$begingroup$
My apologies, $lambda$ is an arbitrary constant. In essence I want to obtain an expression of $phi(x)$ which does not contain a function of $s$, which the initial integral equation has.
$endgroup$
– LightningStrike
8 hours ago










4 Answers
4






active

oldest

votes


















4












$begingroup$

Relabelling the dummy variable $xmapsto s$ on the LHS of your final equation, $$int_0^1phi(s),ds-lambdaint_0^1phi(s),ds=frac12\implies int_0^1phi(s),ds=frac1{2(1-lambda)}$$



Thus $$phi(x)=x+fraclambda{2(1-lambda)}$$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Note $int_{0}^{1}{phi(s)ds}$ is a constant say $a$. Your functional equation (FE) can be rewritten as: $$phi(x)=x+alambda$$
    Putting into FE yields:




    $$x+alambda=x+lambdaint_{0}^{1}{(s+alambda )ds }iff alambda=lambdabig(frac{1}{2}+lambda abig)$$
    If $lambda=0$ then $phi(x)=x$



    if $lambdane 1$ $a=frac{1}{2}+lambda aiff ( 1-lambda)a=frac{1}{2}iff a=frac{1}{2-2lambda}$ and then $phi(x)=x+frac{lambda}{2-2lambda}$



    If $lambda=1$ there won’t besuch $phi$.







    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      If you are after finding $phi(x)$, one approach that comes to mind is to assume it is smooth enough to have a normally convergent (so we can interchange series summation and integration) Taylor expansion on $[0, 1]$:
      $$
      phi(x) = sum_{n geq 0} a_{n} x^{n}.
      $$
      Substituting it into your equation, we get:
      $$
      sum_{n geq 0} a_{n} x^{n} = x + lambda sum_{n geq 0}{a_{n} over n+1}.
      $$
      Matching up the coefficients of the difference powers of $x$, we get:
      $$
      a_{n} = 0 quad mbox{ for } n geq 2,
      $$
      $$
      a_{1} = 1,
      $$
      and
      $$
      a_{0} = lambda left(a_{0} + {a_{1} over 2}right).
      $$
      This gives a relationship between $a_{0}$ and $lambda$.






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        Note that since $lambdaint_0^1 phi(s),ds$ is a constant (with respect to $x$), then we can write$$phi(x)=x+a$$and by substitution we conclude that $$x+a=x+lambdaint _{0}^{1}x+adximplies\a=lambda({1over 2}+a)implies\a={lambdaover 2-2lambda}$$ and we obtain$$phi(x)=x+{lambdaover 2-2lambda}quad,quad lambdane 1$$The case $lambda=1$ leads to no solution.






        share|cite|improve this answer









        $endgroup$













          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%2f3162106%2fa-standard-integral-equation%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          Relabelling the dummy variable $xmapsto s$ on the LHS of your final equation, $$int_0^1phi(s),ds-lambdaint_0^1phi(s),ds=frac12\implies int_0^1phi(s),ds=frac1{2(1-lambda)}$$



          Thus $$phi(x)=x+fraclambda{2(1-lambda)}$$






          share|cite|improve this answer









          $endgroup$


















            4












            $begingroup$

            Relabelling the dummy variable $xmapsto s$ on the LHS of your final equation, $$int_0^1phi(s),ds-lambdaint_0^1phi(s),ds=frac12\implies int_0^1phi(s),ds=frac1{2(1-lambda)}$$



            Thus $$phi(x)=x+fraclambda{2(1-lambda)}$$






            share|cite|improve this answer









            $endgroup$
















              4












              4








              4





              $begingroup$

              Relabelling the dummy variable $xmapsto s$ on the LHS of your final equation, $$int_0^1phi(s),ds-lambdaint_0^1phi(s),ds=frac12\implies int_0^1phi(s),ds=frac1{2(1-lambda)}$$



              Thus $$phi(x)=x+fraclambda{2(1-lambda)}$$






              share|cite|improve this answer









              $endgroup$



              Relabelling the dummy variable $xmapsto s$ on the LHS of your final equation, $$int_0^1phi(s),ds-lambdaint_0^1phi(s),ds=frac12\implies int_0^1phi(s),ds=frac1{2(1-lambda)}$$



              Thus $$phi(x)=x+fraclambda{2(1-lambda)}$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 8 hours ago









              John DoeJohn Doe

              11.3k11239




              11.3k11239























                  2












                  $begingroup$

                  Note $int_{0}^{1}{phi(s)ds}$ is a constant say $a$. Your functional equation (FE) can be rewritten as: $$phi(x)=x+alambda$$
                  Putting into FE yields:




                  $$x+alambda=x+lambdaint_{0}^{1}{(s+alambda )ds }iff alambda=lambdabig(frac{1}{2}+lambda abig)$$
                  If $lambda=0$ then $phi(x)=x$



                  if $lambdane 1$ $a=frac{1}{2}+lambda aiff ( 1-lambda)a=frac{1}{2}iff a=frac{1}{2-2lambda}$ and then $phi(x)=x+frac{lambda}{2-2lambda}$



                  If $lambda=1$ there won’t besuch $phi$.







                  share|cite|improve this answer









                  $endgroup$


















                    2












                    $begingroup$

                    Note $int_{0}^{1}{phi(s)ds}$ is a constant say $a$. Your functional equation (FE) can be rewritten as: $$phi(x)=x+alambda$$
                    Putting into FE yields:




                    $$x+alambda=x+lambdaint_{0}^{1}{(s+alambda )ds }iff alambda=lambdabig(frac{1}{2}+lambda abig)$$
                    If $lambda=0$ then $phi(x)=x$



                    if $lambdane 1$ $a=frac{1}{2}+lambda aiff ( 1-lambda)a=frac{1}{2}iff a=frac{1}{2-2lambda}$ and then $phi(x)=x+frac{lambda}{2-2lambda}$



                    If $lambda=1$ there won’t besuch $phi$.







                    share|cite|improve this answer









                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$

                      Note $int_{0}^{1}{phi(s)ds}$ is a constant say $a$. Your functional equation (FE) can be rewritten as: $$phi(x)=x+alambda$$
                      Putting into FE yields:




                      $$x+alambda=x+lambdaint_{0}^{1}{(s+alambda )ds }iff alambda=lambdabig(frac{1}{2}+lambda abig)$$
                      If $lambda=0$ then $phi(x)=x$



                      if $lambdane 1$ $a=frac{1}{2}+lambda aiff ( 1-lambda)a=frac{1}{2}iff a=frac{1}{2-2lambda}$ and then $phi(x)=x+frac{lambda}{2-2lambda}$



                      If $lambda=1$ there won’t besuch $phi$.







                      share|cite|improve this answer









                      $endgroup$



                      Note $int_{0}^{1}{phi(s)ds}$ is a constant say $a$. Your functional equation (FE) can be rewritten as: $$phi(x)=x+alambda$$
                      Putting into FE yields:




                      $$x+alambda=x+lambdaint_{0}^{1}{(s+alambda )ds }iff alambda=lambdabig(frac{1}{2}+lambda abig)$$
                      If $lambda=0$ then $phi(x)=x$



                      if $lambdane 1$ $a=frac{1}{2}+lambda aiff ( 1-lambda)a=frac{1}{2}iff a=frac{1}{2-2lambda}$ and then $phi(x)=x+frac{lambda}{2-2lambda}$



                      If $lambda=1$ there won’t besuch $phi$.








                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 8 hours ago









                      HAMIDINE SOUMAREHAMIDINE SOUMARE

                      1,478211




                      1,478211























                          0












                          $begingroup$

                          If you are after finding $phi(x)$, one approach that comes to mind is to assume it is smooth enough to have a normally convergent (so we can interchange series summation and integration) Taylor expansion on $[0, 1]$:
                          $$
                          phi(x) = sum_{n geq 0} a_{n} x^{n}.
                          $$
                          Substituting it into your equation, we get:
                          $$
                          sum_{n geq 0} a_{n} x^{n} = x + lambda sum_{n geq 0}{a_{n} over n+1}.
                          $$
                          Matching up the coefficients of the difference powers of $x$, we get:
                          $$
                          a_{n} = 0 quad mbox{ for } n geq 2,
                          $$
                          $$
                          a_{1} = 1,
                          $$
                          and
                          $$
                          a_{0} = lambda left(a_{0} + {a_{1} over 2}right).
                          $$
                          This gives a relationship between $a_{0}$ and $lambda$.






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            If you are after finding $phi(x)$, one approach that comes to mind is to assume it is smooth enough to have a normally convergent (so we can interchange series summation and integration) Taylor expansion on $[0, 1]$:
                            $$
                            phi(x) = sum_{n geq 0} a_{n} x^{n}.
                            $$
                            Substituting it into your equation, we get:
                            $$
                            sum_{n geq 0} a_{n} x^{n} = x + lambda sum_{n geq 0}{a_{n} over n+1}.
                            $$
                            Matching up the coefficients of the difference powers of $x$, we get:
                            $$
                            a_{n} = 0 quad mbox{ for } n geq 2,
                            $$
                            $$
                            a_{1} = 1,
                            $$
                            and
                            $$
                            a_{0} = lambda left(a_{0} + {a_{1} over 2}right).
                            $$
                            This gives a relationship between $a_{0}$ and $lambda$.






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              If you are after finding $phi(x)$, one approach that comes to mind is to assume it is smooth enough to have a normally convergent (so we can interchange series summation and integration) Taylor expansion on $[0, 1]$:
                              $$
                              phi(x) = sum_{n geq 0} a_{n} x^{n}.
                              $$
                              Substituting it into your equation, we get:
                              $$
                              sum_{n geq 0} a_{n} x^{n} = x + lambda sum_{n geq 0}{a_{n} over n+1}.
                              $$
                              Matching up the coefficients of the difference powers of $x$, we get:
                              $$
                              a_{n} = 0 quad mbox{ for } n geq 2,
                              $$
                              $$
                              a_{1} = 1,
                              $$
                              and
                              $$
                              a_{0} = lambda left(a_{0} + {a_{1} over 2}right).
                              $$
                              This gives a relationship between $a_{0}$ and $lambda$.






                              share|cite|improve this answer









                              $endgroup$



                              If you are after finding $phi(x)$, one approach that comes to mind is to assume it is smooth enough to have a normally convergent (so we can interchange series summation and integration) Taylor expansion on $[0, 1]$:
                              $$
                              phi(x) = sum_{n geq 0} a_{n} x^{n}.
                              $$
                              Substituting it into your equation, we get:
                              $$
                              sum_{n geq 0} a_{n} x^{n} = x + lambda sum_{n geq 0}{a_{n} over n+1}.
                              $$
                              Matching up the coefficients of the difference powers of $x$, we get:
                              $$
                              a_{n} = 0 quad mbox{ for } n geq 2,
                              $$
                              $$
                              a_{1} = 1,
                              $$
                              and
                              $$
                              a_{0} = lambda left(a_{0} + {a_{1} over 2}right).
                              $$
                              This gives a relationship between $a_{0}$ and $lambda$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 8 hours ago









                              avsavs

                              3,779514




                              3,779514























                                  0












                                  $begingroup$

                                  Note that since $lambdaint_0^1 phi(s),ds$ is a constant (with respect to $x$), then we can write$$phi(x)=x+a$$and by substitution we conclude that $$x+a=x+lambdaint _{0}^{1}x+adximplies\a=lambda({1over 2}+a)implies\a={lambdaover 2-2lambda}$$ and we obtain$$phi(x)=x+{lambdaover 2-2lambda}quad,quad lambdane 1$$The case $lambda=1$ leads to no solution.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    Note that since $lambdaint_0^1 phi(s),ds$ is a constant (with respect to $x$), then we can write$$phi(x)=x+a$$and by substitution we conclude that $$x+a=x+lambdaint _{0}^{1}x+adximplies\a=lambda({1over 2}+a)implies\a={lambdaover 2-2lambda}$$ and we obtain$$phi(x)=x+{lambdaover 2-2lambda}quad,quad lambdane 1$$The case $lambda=1$ leads to no solution.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      Note that since $lambdaint_0^1 phi(s),ds$ is a constant (with respect to $x$), then we can write$$phi(x)=x+a$$and by substitution we conclude that $$x+a=x+lambdaint _{0}^{1}x+adximplies\a=lambda({1over 2}+a)implies\a={lambdaover 2-2lambda}$$ and we obtain$$phi(x)=x+{lambdaover 2-2lambda}quad,quad lambdane 1$$The case $lambda=1$ leads to no solution.






                                      share|cite|improve this answer









                                      $endgroup$



                                      Note that since $lambdaint_0^1 phi(s),ds$ is a constant (with respect to $x$), then we can write$$phi(x)=x+a$$and by substitution we conclude that $$x+a=x+lambdaint _{0}^{1}x+adximplies\a=lambda({1over 2}+a)implies\a={lambdaover 2-2lambda}$$ and we obtain$$phi(x)=x+{lambdaover 2-2lambda}quad,quad lambdane 1$$The case $lambda=1$ leads to no solution.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 8 hours ago









                                      Mostafa AyazMostafa Ayaz

                                      17.6k31039




                                      17.6k31039






























                                          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%2f3162106%2fa-standard-integral-equation%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

                                          GameSpot

                                          Image
                                          body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} GameSpot 戰地風雲：惡名昭彰2在Gamespot的評論 网站类型 新聞 持有者 CBS 创始人 Pete Deemer Vince Broady Jon Epstein 网站 http://www.gamespot.com/ 注册 Optional (free and paid) 推出时间 1996年5月1日 [1] GameSpot （中国大陆：游戏基地），於1996年5月由Pete Deemer和Vince Broady創立，是一個提供新聞、評論、預告片、下載及其他的相關資訊的電子遊戲網站。GameSpot被一間後來被CNET收購的企業ZDNet所收購。根據Alexa，GameSpot.com是200個網路擁擠最嚴重的網站之一。 除了由GameSpot員工創作的內容，網站還允許用戶寫評論、網誌、之後在網路論壇分享。一些在CNET旗下的GameFAQs分享。 2004年， GameSpot被Spike TV的觀眾選上「電子遊戲賞節目」贏得「最傑出遊戲網站。 [2] 其他的遊戲網站還有IGN、1UP.com、GameSpy是它最大的競爭對手。2008年，根據Compete.com的統計，「gamespot.com」吸引了最少6000萬人的點擊率。 [3] GameSpot的主頁鏈結了到最近新聞、評論、預告、和一些有關遊戲機的入口：Wii、任天堂DS、電腦遊戲、Xbox 360、PSP、PlayStation 2、PlayStation 3。它還有一列「最受歡迎遊戲名單」，還有給用戶快速獲得遊戲資訊的搜尋器。GameSpot 還包括一些小範圍的遊戲機：任天堂64、GameCube、Game Boy Color、Game Boy Advance、Xbox、PlayStation、SEGA Saturn、Dreamcast、Neo Geo Pocket Color、N-Gage、手機遊戲。 目录 1 歷史 1.1 國際歷史 1.2 著名的員工 2 評論和分
                                          Read more

                                          connect to host localhost port 22: Connection refused

                                          Image
                                          .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0; } 1 1 I created a Ubuntu Vagrant box and ssh'd into it. Now, created a ssh key and trying to ssh to my localhost but I get the following error. vagrant@vagrant-ubuntu-trusty-64:~$ ssh -vvvv localhost OpenSSH_6.6.1, OpenSSL 1.0.1f 6 Jan 2014 debug1: Reading configuration data /etc/ssh/ssh_config debug1: /etc/ssh/ssh_config line 19: Applying options for * debug2: ssh_connect: needpriv 0 debug1: Connecting to localhost [127.0.0.1] port 22. debug1: connect to address 127.0.0.1 port 22: Connection refused ssh: connect to host localhost port 22: Connection refused When I check the list of ports opened as follows: vagrant@vagrant-ubuntu-trusty-64:~$ s
                                          Read more

                                          Getting a Wifi WPA2 wifi connection

                                          Image
                                          .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0; } 0 I've been trying to get a simple wifi connection ! bu I have to admit I just can't get it done ! here is what I've tried: root@MYONE:/etc# clear root@MYONE:/etc# vi wpa_supplicant.conf ctrl_interface=/var/run/wpa_supplicant ctrl_interface_group=0 update_config=1 network={ ssid="Mi Engine" psk="7407352688e4" } root@MYONE:/etc# iw wlan0 link Not connected. root@MYONE:/etc# ifconfig eth0 Link encap:Ethernet HWaddr 74:FE:48:3A:55:AB inet addr:192.168.178.88 Bcast:192.168.178.255 Mask:255.255.255.0 inet6 addr: fe80::76fe:48ff:fe3a:55ab%4804152/64 Scope:Link UP BRO
                                          Read more