Show that $(exp(x)cdot exp(x')=exp(x+x')$ and $ exp(x)>exp(0),x>0)Rightarrow exp$ is strictly...












1












$begingroup$


$exp(x)=sum_{n=0}^{infty}frac{1}{n!}x^n$



My calculations so far:



Assume



$exists _{x,x'inmathbb{R}}:$
$x<x'$ and $exp(x)>exp(x')$



$x<x'Rightarrow x'-x>0 Rightarrow exp(x-x')>exp(0)$



Can I somehow invoke a contradiction with this?






real-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Try setting $x= y-x'$ where $y>x'$.
    $endgroup$
    – irchans
    Jan 12 at 19:52










  • $begingroup$
    I read this first as "Show that $A$, and $(BRightarrow C)$, not as the (presumably intended) "Show that $(A$ and $B)Rightarrow C$."
    $endgroup$
    – TonyK
    Jan 12 at 20:12












  • $begingroup$
    Oh sorry I will eddit the question
    $endgroup$
    – RM777
    Jan 12 at 20:13
















1












$begingroup$


$exp(x)=sum_{n=0}^{infty}frac{1}{n!}x^n$



My calculations so far:



Assume



$exists _{x,x'inmathbb{R}}:$
$x<x'$ and $exp(x)>exp(x')$



$x<x'Rightarrow x'-x>0 Rightarrow exp(x-x')>exp(0)$



Can I somehow invoke a contradiction with this?






real-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Try setting $x= y-x'$ where $y>x'$.
    $endgroup$
    – irchans
    Jan 12 at 19:52










  • $begingroup$
    I read this first as "Show that $A$, and $(BRightarrow C)$, not as the (presumably intended) "Show that $(A$ and $B)Rightarrow C$."
    $endgroup$
    – TonyK
    Jan 12 at 20:12












  • $begingroup$
    Oh sorry I will eddit the question
    $endgroup$
    – RM777
    Jan 12 at 20:13














1












1








1





$begingroup$


$exp(x)=sum_{n=0}^{infty}frac{1}{n!}x^n$



My calculations so far:



Assume



$exists _{x,x'inmathbb{R}}:$
$x<x'$ and $exp(x)>exp(x')$



$x<x'Rightarrow x'-x>0 Rightarrow exp(x-x')>exp(0)$



Can I somehow invoke a contradiction with this?






real-analysis







share|cite|improve this question











$endgroup$




$exp(x)=sum_{n=0}^{infty}frac{1}{n!}x^n$



My calculations so far:



Assume



$exists _{x,x'inmathbb{R}}:$
$x<x'$ and $exp(x)>exp(x')$



$x<x'Rightarrow x'-x>0 Rightarrow exp(x-x')>exp(0)$



Can I somehow invoke a contradiction with this?






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 12 at 20:15







RM777

















asked Jan 12 at 19:41









RM777RM777

38312




38312








  • 1




    $begingroup$
    Try setting $x= y-x'$ where $y>x'$.
    $endgroup$
    – irchans
    Jan 12 at 19:52










  • $begingroup$
    I read this first as "Show that $A$, and $(BRightarrow C)$, not as the (presumably intended) "Show that $(A$ and $B)Rightarrow C$."
    $endgroup$
    – TonyK
    Jan 12 at 20:12












  • $begingroup$
    Oh sorry I will eddit the question
    $endgroup$
    – RM777
    Jan 12 at 20:13














  • 1




    $begingroup$
    Try setting $x= y-x'$ where $y>x'$.
    $endgroup$
    – irchans
    Jan 12 at 19:52










  • $begingroup$
    I read this first as "Show that $A$, and $(BRightarrow C)$, not as the (presumably intended) "Show that $(A$ and $B)Rightarrow C$."
    $endgroup$
    – TonyK
    Jan 12 at 20:12












  • $begingroup$
    Oh sorry I will eddit the question
    $endgroup$
    – RM777
    Jan 12 at 20:13








1




1




$begingroup$
Try setting $x= y-x'$ where $y>x'$.
$endgroup$
– irchans
Jan 12 at 19:52




$begingroup$
Try setting $x= y-x'$ where $y>x'$.
$endgroup$
– irchans
Jan 12 at 19:52












$begingroup$
I read this first as "Show that $A$, and $(BRightarrow C)$, not as the (presumably intended) "Show that $(A$ and $B)Rightarrow C$."
$endgroup$
– TonyK
Jan 12 at 20:12






$begingroup$
I read this first as "Show that $A$, and $(BRightarrow C)$, not as the (presumably intended) "Show that $(A$ and $B)Rightarrow C$."
$endgroup$
– TonyK
Jan 12 at 20:12














$begingroup$
Oh sorry I will eddit the question
$endgroup$
– RM777
Jan 12 at 20:13




$begingroup$
Oh sorry I will eddit the question
$endgroup$
– RM777
Jan 12 at 20:13










3 Answers
3






active

oldest

votes


















2












$begingroup$

Let $y>x leftrightarrow y-x>0$. Then



$exp(y)=exp(y-x+x)=exp(y-x) exp(x)> exp(0) exp(x) =exp(x)$.



In the second and third step you use the two assumptions.






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    First let $x=x'=0$ therefore $$exp(0)cdotexp(0)=exp(0)$$which yields to $$exp(0)=0text{ or }1$$ if $exp(0)=0$ then we have $$exp(x){=exp(x+0)\=exp(0)cdotexp(x)\=0quad,quad forall xin Bbb R}$$which is non-sense. Therefore $exp(0)=1$. Now let $x'>x$. Then there exists $epsilon>0$ for which $x'=x+epsilon$. This leads to$$exp(x'){=exp(x+epsilon)\=exp(x)exp(epsilon)\>exp(x)exp(0)\=exp(x)}$$therefore$$exp(x')>exp(x)$$which completes our proof.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      You haven't used the definition of $exp$ at all, so how can you prove anything about it?
      $endgroup$
      – TonyK
      Jan 12 at 20:03










    • $begingroup$
      It needs not. Any function with such properties is strictly increasing.
      $endgroup$
      – Mostafa Ayaz
      Jan 12 at 20:04










    • $begingroup$
      @TonyK Everything on the left side of the implication can be taken for granted. We have proved this property with the Cauchy-Product already in class
      $endgroup$
      – RM777
      Jan 12 at 20:11










    • $begingroup$
      Perhaps I misinterpreted the question $-$ see my comment there.
      $endgroup$
      – TonyK
      Jan 12 at 20:12



















    0












    $begingroup$

    In general, for any $f(cdot)$ with $f(x+y)=f(x)f(y)$ where $f(y)>f(0)=1$ for every $y>0$, you can immediately prove that $f(cdot)$ is strictly increasing.



    For this, I start by proving $f(x)>0$ for every $x$. To see this, use $f(x+y)=f(x)f(y)$, take $y=-x$ for an $x>0$ and get $f(x)f(-x)=f(0)>0$. Since $f(x)>0$, we deduce $f(-x)>0$.



    Now, fix an $xinmathbb{R}$, and let $z>x$. Then, $z=x+epsilon$ for an $epsilon>0$. This gives, $f(z)=f(x)f(epsilon)>f(x)$, as $f(x)>0$ and $f(epsilon)>1$. Done.






    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%2f3071300%2fshow-that-expx-cdot-expx-expxx-and-expx-exp0-x0-rightar%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









      2












      $begingroup$

      Let $y>x leftrightarrow y-x>0$. Then



      $exp(y)=exp(y-x+x)=exp(y-x) exp(x)> exp(0) exp(x) =exp(x)$.



      In the second and third step you use the two assumptions.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Let $y>x leftrightarrow y-x>0$. Then



        $exp(y)=exp(y-x+x)=exp(y-x) exp(x)> exp(0) exp(x) =exp(x)$.



        In the second and third step you use the two assumptions.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Let $y>x leftrightarrow y-x>0$. Then



          $exp(y)=exp(y-x+x)=exp(y-x) exp(x)> exp(0) exp(x) =exp(x)$.



          In the second and third step you use the two assumptions.






          share|cite|improve this answer









          $endgroup$



          Let $y>x leftrightarrow y-x>0$. Then



          $exp(y)=exp(y-x+x)=exp(y-x) exp(x)> exp(0) exp(x) =exp(x)$.



          In the second and third step you use the two assumptions.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 12 at 19:59









          MaxMax

          586




          586























              4












              $begingroup$

              First let $x=x'=0$ therefore $$exp(0)cdotexp(0)=exp(0)$$which yields to $$exp(0)=0text{ or }1$$ if $exp(0)=0$ then we have $$exp(x){=exp(x+0)\=exp(0)cdotexp(x)\=0quad,quad forall xin Bbb R}$$which is non-sense. Therefore $exp(0)=1$. Now let $x'>x$. Then there exists $epsilon>0$ for which $x'=x+epsilon$. This leads to$$exp(x'){=exp(x+epsilon)\=exp(x)exp(epsilon)\>exp(x)exp(0)\=exp(x)}$$therefore$$exp(x')>exp(x)$$which completes our proof.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                You haven't used the definition of $exp$ at all, so how can you prove anything about it?
                $endgroup$
                – TonyK
                Jan 12 at 20:03










              • $begingroup$
                It needs not. Any function with such properties is strictly increasing.
                $endgroup$
                – Mostafa Ayaz
                Jan 12 at 20:04










              • $begingroup$
                @TonyK Everything on the left side of the implication can be taken for granted. We have proved this property with the Cauchy-Product already in class
                $endgroup$
                – RM777
                Jan 12 at 20:11










              • $begingroup$
                Perhaps I misinterpreted the question $-$ see my comment there.
                $endgroup$
                – TonyK
                Jan 12 at 20:12
















              4












              $begingroup$

              First let $x=x'=0$ therefore $$exp(0)cdotexp(0)=exp(0)$$which yields to $$exp(0)=0text{ or }1$$ if $exp(0)=0$ then we have $$exp(x){=exp(x+0)\=exp(0)cdotexp(x)\=0quad,quad forall xin Bbb R}$$which is non-sense. Therefore $exp(0)=1$. Now let $x'>x$. Then there exists $epsilon>0$ for which $x'=x+epsilon$. This leads to$$exp(x'){=exp(x+epsilon)\=exp(x)exp(epsilon)\>exp(x)exp(0)\=exp(x)}$$therefore$$exp(x')>exp(x)$$which completes our proof.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                You haven't used the definition of $exp$ at all, so how can you prove anything about it?
                $endgroup$
                – TonyK
                Jan 12 at 20:03










              • $begingroup$
                It needs not. Any function with such properties is strictly increasing.
                $endgroup$
                – Mostafa Ayaz
                Jan 12 at 20:04










              • $begingroup$
                @TonyK Everything on the left side of the implication can be taken for granted. We have proved this property with the Cauchy-Product already in class
                $endgroup$
                – RM777
                Jan 12 at 20:11










              • $begingroup$
                Perhaps I misinterpreted the question $-$ see my comment there.
                $endgroup$
                – TonyK
                Jan 12 at 20:12














              4












              4








              4





              $begingroup$

              First let $x=x'=0$ therefore $$exp(0)cdotexp(0)=exp(0)$$which yields to $$exp(0)=0text{ or }1$$ if $exp(0)=0$ then we have $$exp(x){=exp(x+0)\=exp(0)cdotexp(x)\=0quad,quad forall xin Bbb R}$$which is non-sense. Therefore $exp(0)=1$. Now let $x'>x$. Then there exists $epsilon>0$ for which $x'=x+epsilon$. This leads to$$exp(x'){=exp(x+epsilon)\=exp(x)exp(epsilon)\>exp(x)exp(0)\=exp(x)}$$therefore$$exp(x')>exp(x)$$which completes our proof.






              share|cite|improve this answer









              $endgroup$



              First let $x=x'=0$ therefore $$exp(0)cdotexp(0)=exp(0)$$which yields to $$exp(0)=0text{ or }1$$ if $exp(0)=0$ then we have $$exp(x){=exp(x+0)\=exp(0)cdotexp(x)\=0quad,quad forall xin Bbb R}$$which is non-sense. Therefore $exp(0)=1$. Now let $x'>x$. Then there exists $epsilon>0$ for which $x'=x+epsilon$. This leads to$$exp(x'){=exp(x+epsilon)\=exp(x)exp(epsilon)\>exp(x)exp(0)\=exp(x)}$$therefore$$exp(x')>exp(x)$$which completes our proof.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 12 at 19:57









              Mostafa AyazMostafa Ayaz

              15.8k3939




              15.8k3939












              • $begingroup$
                You haven't used the definition of $exp$ at all, so how can you prove anything about it?
                $endgroup$
                – TonyK
                Jan 12 at 20:03










              • $begingroup$
                It needs not. Any function with such properties is strictly increasing.
                $endgroup$
                – Mostafa Ayaz
                Jan 12 at 20:04










              • $begingroup$
                @TonyK Everything on the left side of the implication can be taken for granted. We have proved this property with the Cauchy-Product already in class
                $endgroup$
                – RM777
                Jan 12 at 20:11










              • $begingroup$
                Perhaps I misinterpreted the question $-$ see my comment there.
                $endgroup$
                – TonyK
                Jan 12 at 20:12


















              • $begingroup$
                You haven't used the definition of $exp$ at all, so how can you prove anything about it?
                $endgroup$
                – TonyK
                Jan 12 at 20:03










              • $begingroup$
                It needs not. Any function with such properties is strictly increasing.
                $endgroup$
                – Mostafa Ayaz
                Jan 12 at 20:04










              • $begingroup$
                @TonyK Everything on the left side of the implication can be taken for granted. We have proved this property with the Cauchy-Product already in class
                $endgroup$
                – RM777
                Jan 12 at 20:11










              • $begingroup$
                Perhaps I misinterpreted the question $-$ see my comment there.
                $endgroup$
                – TonyK
                Jan 12 at 20:12
















              $begingroup$
              You haven't used the definition of $exp$ at all, so how can you prove anything about it?
              $endgroup$
              – TonyK
              Jan 12 at 20:03




              $begingroup$
              You haven't used the definition of $exp$ at all, so how can you prove anything about it?
              $endgroup$
              – TonyK
              Jan 12 at 20:03












              $begingroup$
              It needs not. Any function with such properties is strictly increasing.
              $endgroup$
              – Mostafa Ayaz
              Jan 12 at 20:04




              $begingroup$
              It needs not. Any function with such properties is strictly increasing.
              $endgroup$
              – Mostafa Ayaz
              Jan 12 at 20:04












              $begingroup$
              @TonyK Everything on the left side of the implication can be taken for granted. We have proved this property with the Cauchy-Product already in class
              $endgroup$
              – RM777
              Jan 12 at 20:11




              $begingroup$
              @TonyK Everything on the left side of the implication can be taken for granted. We have proved this property with the Cauchy-Product already in class
              $endgroup$
              – RM777
              Jan 12 at 20:11












              $begingroup$
              Perhaps I misinterpreted the question $-$ see my comment there.
              $endgroup$
              – TonyK
              Jan 12 at 20:12




              $begingroup$
              Perhaps I misinterpreted the question $-$ see my comment there.
              $endgroup$
              – TonyK
              Jan 12 at 20:12











              0












              $begingroup$

              In general, for any $f(cdot)$ with $f(x+y)=f(x)f(y)$ where $f(y)>f(0)=1$ for every $y>0$, you can immediately prove that $f(cdot)$ is strictly increasing.



              For this, I start by proving $f(x)>0$ for every $x$. To see this, use $f(x+y)=f(x)f(y)$, take $y=-x$ for an $x>0$ and get $f(x)f(-x)=f(0)>0$. Since $f(x)>0$, we deduce $f(-x)>0$.



              Now, fix an $xinmathbb{R}$, and let $z>x$. Then, $z=x+epsilon$ for an $epsilon>0$. This gives, $f(z)=f(x)f(epsilon)>f(x)$, as $f(x)>0$ and $f(epsilon)>1$. Done.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                In general, for any $f(cdot)$ with $f(x+y)=f(x)f(y)$ where $f(y)>f(0)=1$ for every $y>0$, you can immediately prove that $f(cdot)$ is strictly increasing.



                For this, I start by proving $f(x)>0$ for every $x$. To see this, use $f(x+y)=f(x)f(y)$, take $y=-x$ for an $x>0$ and get $f(x)f(-x)=f(0)>0$. Since $f(x)>0$, we deduce $f(-x)>0$.



                Now, fix an $xinmathbb{R}$, and let $z>x$. Then, $z=x+epsilon$ for an $epsilon>0$. This gives, $f(z)=f(x)f(epsilon)>f(x)$, as $f(x)>0$ and $f(epsilon)>1$. Done.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  In general, for any $f(cdot)$ with $f(x+y)=f(x)f(y)$ where $f(y)>f(0)=1$ for every $y>0$, you can immediately prove that $f(cdot)$ is strictly increasing.



                  For this, I start by proving $f(x)>0$ for every $x$. To see this, use $f(x+y)=f(x)f(y)$, take $y=-x$ for an $x>0$ and get $f(x)f(-x)=f(0)>0$. Since $f(x)>0$, we deduce $f(-x)>0$.



                  Now, fix an $xinmathbb{R}$, and let $z>x$. Then, $z=x+epsilon$ for an $epsilon>0$. This gives, $f(z)=f(x)f(epsilon)>f(x)$, as $f(x)>0$ and $f(epsilon)>1$. Done.






                  share|cite|improve this answer









                  $endgroup$



                  In general, for any $f(cdot)$ with $f(x+y)=f(x)f(y)$ where $f(y)>f(0)=1$ for every $y>0$, you can immediately prove that $f(cdot)$ is strictly increasing.



                  For this, I start by proving $f(x)>0$ for every $x$. To see this, use $f(x+y)=f(x)f(y)$, take $y=-x$ for an $x>0$ and get $f(x)f(-x)=f(0)>0$. Since $f(x)>0$, we deduce $f(-x)>0$.



                  Now, fix an $xinmathbb{R}$, and let $z>x$. Then, $z=x+epsilon$ for an $epsilon>0$. This gives, $f(z)=f(x)f(epsilon)>f(x)$, as $f(x)>0$ and $f(epsilon)>1$. Done.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 12 at 20:22









                  AaronAaron

                  1,912415




                  1,912415






























                      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%2f3071300%2fshow-that-expx-cdot-expx-expxx-and-expx-exp0-x0-rightar%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

                      Questions related to Moebius Transform of Characteristic Function of the Primes

                      Image
                      2 $begingroup$ Consider the function defined in (1) below related to the fundamental prime counting function $pi(x)$ . Note that A143519(n) is not multiplicative. (1) $quad f(x)=sumlimits_{n=1}^{x}A143519(n)$ https://oeis.org/A143519 The following plot illustrates $f(x)$ defined in formula (1) above. Figure (1) : Illustration of $f(x)$ defined in formula (1) The integer zeros of $f(x)$ for $xle 10,000$ are listed in (2) below. (2) $quad$ {1,6,9,12,19,30,79,80,81,116,193,201,287,288,291,668,673,679,680,685,686,1109} The zero crossings of $f(x)$ for $xle 10,000$ where $f(x)$ doesn't settle at zero are listed in (3) below. (3) $quad$ {14,21,33,114,115,118,195,286,290,295,442,445,665,667,670,671,678,682} Question (1) : Does $f(x)$ have a finite number of integer zeros, and if so ...
                      Read more

                      List of scandals in India

                      Image
                      List of scandals in India From Wikipedia, the free encyclopedia Jump to navigation Jump to search This is a dynamic list and may never be able to satisfy particular standards for completeness. You can help by expanding it with reliably sourced entries. This article has multiple issues. Please help improve it or discuss these issues on the talk page . (Learn how and when to remove these template messages) This article has an unclear citation style . The references used may be made clearer with a different or consistent style of citation and footnoting. ( January 2014 ) (Learn how and when to remove this template message) This list is in reverse order. As a consequence it appears to be slanted towards recent events. Please try to keep recent events in historical perspective and add more content related to non-recent events. ( May 2013 ) (Learn how and when to remove this template message) ...
                      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