Explicity vs implicit first order euler to approximate $e^{alpha x}, alpha > 0$












0












$begingroup$


Hello for the function



$$
y(x) = e^{alpha x} Leftrightarrow y' = alpha y, y(0)=1
$$



I wanted to evaluate the error of the two following iterations



$$
left{
begin{array}{l}
y_k = y_{k-1} + alpha y_{k-1} delta t \
y_0 = 1
end{array}
right. Rightarrow y_k = (1+alpha delta t)^k
$$



and



$$
left{
begin{array}{l}
y_k = y_{k-1} + alpha y_{k}delta t \
y_0 = 1
end{array}
right. Rightarrow (1 - alpha delta t)y_k = y_{k-1} Rightarrow y_k = (1 - alpha delta t)^{-1} y_{k-1} Rightarrow y_k = (1-alpha delta t)^{-k}
$$



Now I'd like to evaluate the the errors using



$$
y(kdelta t) = e^{alpha k delta t} = 1 + alpha k delta t + alpha^2 frac{k^2}{2} delta t^2 + ...
$$



I can figure the taylor expansion of the first iteration (assuming $delta t rightarrow 0$), indeed I would get



$$
y_k = sum_{l=0}^{k} binom{k}{l} (alpha delta t)^l
$$



In such a case
$$
y(kdelta t) - y_k = O(alpha delta t)
$$



In the second iteration though I was expecting less error (asymptotically at least, i.e. something like $O(alpha^2 delta t^2)$), Taylor expansion I managed to get, not sure if this is correct



$$
y_k = sum_{l=0}^{+infty} frac{Gamma(1-k)}{Gamma(1-k-l)Gamma(l+1)} (-1)^l(alpha delta t)^l = 1 + alpha k delta t + frac{k (k+1)}{2}alpha^2delta t^2 + ...
$$



however the entailed error in this case is the same, is there something I'm missing? or is what I've done correct?






real-analysis ordinary-differential-equations numerical-methods taylor-expansion







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Hello for the function



    $$
    y(x) = e^{alpha x} Leftrightarrow y' = alpha y, y(0)=1
    $$



    I wanted to evaluate the error of the two following iterations



    $$
    left{
    begin{array}{l}
    y_k = y_{k-1} + alpha y_{k-1} delta t \
    y_0 = 1
    end{array}
    right. Rightarrow y_k = (1+alpha delta t)^k
    $$



    and



    $$
    left{
    begin{array}{l}
    y_k = y_{k-1} + alpha y_{k}delta t \
    y_0 = 1
    end{array}
    right. Rightarrow (1 - alpha delta t)y_k = y_{k-1} Rightarrow y_k = (1 - alpha delta t)^{-1} y_{k-1} Rightarrow y_k = (1-alpha delta t)^{-k}
    $$



    Now I'd like to evaluate the the errors using



    $$
    y(kdelta t) = e^{alpha k delta t} = 1 + alpha k delta t + alpha^2 frac{k^2}{2} delta t^2 + ...
    $$



    I can figure the taylor expansion of the first iteration (assuming $delta t rightarrow 0$), indeed I would get



    $$
    y_k = sum_{l=0}^{k} binom{k}{l} (alpha delta t)^l
    $$



    In such a case
    $$
    y(kdelta t) - y_k = O(alpha delta t)
    $$



    In the second iteration though I was expecting less error (asymptotically at least, i.e. something like $O(alpha^2 delta t^2)$), Taylor expansion I managed to get, not sure if this is correct



    $$
    y_k = sum_{l=0}^{+infty} frac{Gamma(1-k)}{Gamma(1-k-l)Gamma(l+1)} (-1)^l(alpha delta t)^l = 1 + alpha k delta t + frac{k (k+1)}{2}alpha^2delta t^2 + ...
    $$



    however the entailed error in this case is the same, is there something I'm missing? or is what I've done correct?






    real-analysis ordinary-differential-equations numerical-methods taylor-expansion







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Hello for the function



      $$
      y(x) = e^{alpha x} Leftrightarrow y' = alpha y, y(0)=1
      $$



      I wanted to evaluate the error of the two following iterations



      $$
      left{
      begin{array}{l}
      y_k = y_{k-1} + alpha y_{k-1} delta t \
      y_0 = 1
      end{array}
      right. Rightarrow y_k = (1+alpha delta t)^k
      $$



      and



      $$
      left{
      begin{array}{l}
      y_k = y_{k-1} + alpha y_{k}delta t \
      y_0 = 1
      end{array}
      right. Rightarrow (1 - alpha delta t)y_k = y_{k-1} Rightarrow y_k = (1 - alpha delta t)^{-1} y_{k-1} Rightarrow y_k = (1-alpha delta t)^{-k}
      $$



      Now I'd like to evaluate the the errors using



      $$
      y(kdelta t) = e^{alpha k delta t} = 1 + alpha k delta t + alpha^2 frac{k^2}{2} delta t^2 + ...
      $$



      I can figure the taylor expansion of the first iteration (assuming $delta t rightarrow 0$), indeed I would get



      $$
      y_k = sum_{l=0}^{k} binom{k}{l} (alpha delta t)^l
      $$



      In such a case
      $$
      y(kdelta t) - y_k = O(alpha delta t)
      $$



      In the second iteration though I was expecting less error (asymptotically at least, i.e. something like $O(alpha^2 delta t^2)$), Taylor expansion I managed to get, not sure if this is correct



      $$
      y_k = sum_{l=0}^{+infty} frac{Gamma(1-k)}{Gamma(1-k-l)Gamma(l+1)} (-1)^l(alpha delta t)^l = 1 + alpha k delta t + frac{k (k+1)}{2}alpha^2delta t^2 + ...
      $$



      however the entailed error in this case is the same, is there something I'm missing? or is what I've done correct?






      real-analysis ordinary-differential-equations numerical-methods taylor-expansion







      share|cite|improve this question









      $endgroup$




      Hello for the function



      $$
      y(x) = e^{alpha x} Leftrightarrow y' = alpha y, y(0)=1
      $$



      I wanted to evaluate the error of the two following iterations



      $$
      left{
      begin{array}{l}
      y_k = y_{k-1} + alpha y_{k-1} delta t \
      y_0 = 1
      end{array}
      right. Rightarrow y_k = (1+alpha delta t)^k
      $$



      and



      $$
      left{
      begin{array}{l}
      y_k = y_{k-1} + alpha y_{k}delta t \
      y_0 = 1
      end{array}
      right. Rightarrow (1 - alpha delta t)y_k = y_{k-1} Rightarrow y_k = (1 - alpha delta t)^{-1} y_{k-1} Rightarrow y_k = (1-alpha delta t)^{-k}
      $$



      Now I'd like to evaluate the the errors using



      $$
      y(kdelta t) = e^{alpha k delta t} = 1 + alpha k delta t + alpha^2 frac{k^2}{2} delta t^2 + ...
      $$



      I can figure the taylor expansion of the first iteration (assuming $delta t rightarrow 0$), indeed I would get



      $$
      y_k = sum_{l=0}^{k} binom{k}{l} (alpha delta t)^l
      $$



      In such a case
      $$
      y(kdelta t) - y_k = O(alpha delta t)
      $$



      In the second iteration though I was expecting less error (asymptotically at least, i.e. something like $O(alpha^2 delta t^2)$), Taylor expansion I managed to get, not sure if this is correct



      $$
      y_k = sum_{l=0}^{+infty} frac{Gamma(1-k)}{Gamma(1-k-l)Gamma(l+1)} (-1)^l(alpha delta t)^l = 1 + alpha k delta t + frac{k (k+1)}{2}alpha^2delta t^2 + ...
      $$



      however the entailed error in this case is the same, is there something I'm missing? or is what I've done correct?






      real-analysis ordinary-differential-equations numerical-methods taylor-expansion




      real-analysis ordinary-differential-equations numerical-methods taylor-expansion






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 18 at 14:04









      user8469759user8469759

      1,5731618




      1,5731618






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          You get easier results by comparing the powers in question in the exponents, that is as arguments of an exponential function after applying the appropriate logarithms.
          begin{align}
          (1+αδt)^k=expleft(kln(1+αδt)right)
          &=expleft(αkδtleft(1-frac{αδt}2+frac{(αδt)^2}3-frac{(αδt)^3}4+...right)right)
          \&=expleft(αt_k-frac{α^2}2t_kδt+...right),
          \
          (1-αδt)^{-k}=expleft(-kln(1-αδt)right)
          &=expleft(αkδtleft(1+frac{αδt}2+frac{(αδt)^2}3+frac{(αδt)^3}4+...right)right)
          \&=expleft(αt_k+frac{α^2}2t_kδt+...right).
          end{align}
          So you see, in both cases you get nearly the same first-order error, only with opposite signs.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Ah, I was expecting something better for the second, since it is an implicit method, I was wrong.
            $endgroup$
            – user8469759
            Jan 18 at 14:52












          • $begingroup$
            No, they are both first order and time-reverse to each other. Alternating explicit and implicit steps you get the midpoint method which is of second order.
            $endgroup$
            – LutzL
            Jan 18 at 14:54










          • $begingroup$
            So one method might be more convenient than the other only for stability reasons I suppose
            $endgroup$
            – user8469759
            Jan 18 at 14:55






          • 1




            $begingroup$
            Yes, but except for educational purposes you would almost always use higher order implicit methods, one- or multi-step.
            $endgroup$
            – LutzL
            Jan 18 at 15:18












          Your Answer








          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%2f3078291%2fexplicity-vs-implicit-first-order-euler-to-approximate-e-alpha-x-alpha-0%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









          0












          $begingroup$

          You get easier results by comparing the powers in question in the exponents, that is as arguments of an exponential function after applying the appropriate logarithms.
          begin{align}
          (1+αδt)^k=expleft(kln(1+αδt)right)
          &=expleft(αkδtleft(1-frac{αδt}2+frac{(αδt)^2}3-frac{(αδt)^3}4+...right)right)
          \&=expleft(αt_k-frac{α^2}2t_kδt+...right),
          \
          (1-αδt)^{-k}=expleft(-kln(1-αδt)right)
          &=expleft(αkδtleft(1+frac{αδt}2+frac{(αδt)^2}3+frac{(αδt)^3}4+...right)right)
          \&=expleft(αt_k+frac{α^2}2t_kδt+...right).
          end{align}
          So you see, in both cases you get nearly the same first-order error, only with opposite signs.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Ah, I was expecting something better for the second, since it is an implicit method, I was wrong.
            $endgroup$
            – user8469759
            Jan 18 at 14:52












          • $begingroup$
            No, they are both first order and time-reverse to each other. Alternating explicit and implicit steps you get the midpoint method which is of second order.
            $endgroup$
            – LutzL
            Jan 18 at 14:54










          • $begingroup$
            So one method might be more convenient than the other only for stability reasons I suppose
            $endgroup$
            – user8469759
            Jan 18 at 14:55






          • 1




            $begingroup$
            Yes, but except for educational purposes you would almost always use higher order implicit methods, one- or multi-step.
            $endgroup$
            – LutzL
            Jan 18 at 15:18
















          0












          $begingroup$

          You get easier results by comparing the powers in question in the exponents, that is as arguments of an exponential function after applying the appropriate logarithms.
          begin{align}
          (1+αδt)^k=expleft(kln(1+αδt)right)
          &=expleft(αkδtleft(1-frac{αδt}2+frac{(αδt)^2}3-frac{(αδt)^3}4+...right)right)
          \&=expleft(αt_k-frac{α^2}2t_kδt+...right),
          \
          (1-αδt)^{-k}=expleft(-kln(1-αδt)right)
          &=expleft(αkδtleft(1+frac{αδt}2+frac{(αδt)^2}3+frac{(αδt)^3}4+...right)right)
          \&=expleft(αt_k+frac{α^2}2t_kδt+...right).
          end{align}
          So you see, in both cases you get nearly the same first-order error, only with opposite signs.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Ah, I was expecting something better for the second, since it is an implicit method, I was wrong.
            $endgroup$
            – user8469759
            Jan 18 at 14:52












          • $begingroup$
            No, they are both first order and time-reverse to each other. Alternating explicit and implicit steps you get the midpoint method which is of second order.
            $endgroup$
            – LutzL
            Jan 18 at 14:54










          • $begingroup$
            So one method might be more convenient than the other only for stability reasons I suppose
            $endgroup$
            – user8469759
            Jan 18 at 14:55






          • 1




            $begingroup$
            Yes, but except for educational purposes you would almost always use higher order implicit methods, one- or multi-step.
            $endgroup$
            – LutzL
            Jan 18 at 15:18














          0












          0








          0





          $begingroup$

          You get easier results by comparing the powers in question in the exponents, that is as arguments of an exponential function after applying the appropriate logarithms.
          begin{align}
          (1+αδt)^k=expleft(kln(1+αδt)right)
          &=expleft(αkδtleft(1-frac{αδt}2+frac{(αδt)^2}3-frac{(αδt)^3}4+...right)right)
          \&=expleft(αt_k-frac{α^2}2t_kδt+...right),
          \
          (1-αδt)^{-k}=expleft(-kln(1-αδt)right)
          &=expleft(αkδtleft(1+frac{αδt}2+frac{(αδt)^2}3+frac{(αδt)^3}4+...right)right)
          \&=expleft(αt_k+frac{α^2}2t_kδt+...right).
          end{align}
          So you see, in both cases you get nearly the same first-order error, only with opposite signs.






          share|cite|improve this answer











          $endgroup$



          You get easier results by comparing the powers in question in the exponents, that is as arguments of an exponential function after applying the appropriate logarithms.
          begin{align}
          (1+αδt)^k=expleft(kln(1+αδt)right)
          &=expleft(αkδtleft(1-frac{αδt}2+frac{(αδt)^2}3-frac{(αδt)^3}4+...right)right)
          \&=expleft(αt_k-frac{α^2}2t_kδt+...right),
          \
          (1-αδt)^{-k}=expleft(-kln(1-αδt)right)
          &=expleft(αkδtleft(1+frac{αδt}2+frac{(αδt)^2}3+frac{(αδt)^3}4+...right)right)
          \&=expleft(αt_k+frac{α^2}2t_kδt+...right).
          end{align}
          So you see, in both cases you get nearly the same first-order error, only with opposite signs.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 18 at 14:53

























          answered Jan 18 at 14:51









          LutzLLutzL

          60.8k42057




          60.8k42057












          • $begingroup$
            Ah, I was expecting something better for the second, since it is an implicit method, I was wrong.
            $endgroup$
            – user8469759
            Jan 18 at 14:52












          • $begingroup$
            No, they are both first order and time-reverse to each other. Alternating explicit and implicit steps you get the midpoint method which is of second order.
            $endgroup$
            – LutzL
            Jan 18 at 14:54










          • $begingroup$
            So one method might be more convenient than the other only for stability reasons I suppose
            $endgroup$
            – user8469759
            Jan 18 at 14:55






          • 1




            $begingroup$
            Yes, but except for educational purposes you would almost always use higher order implicit methods, one- or multi-step.
            $endgroup$
            – LutzL
            Jan 18 at 15:18


















          • $begingroup$
            Ah, I was expecting something better for the second, since it is an implicit method, I was wrong.
            $endgroup$
            – user8469759
            Jan 18 at 14:52












          • $begingroup$
            No, they are both first order and time-reverse to each other. Alternating explicit and implicit steps you get the midpoint method which is of second order.
            $endgroup$
            – LutzL
            Jan 18 at 14:54










          • $begingroup$
            So one method might be more convenient than the other only for stability reasons I suppose
            $endgroup$
            – user8469759
            Jan 18 at 14:55






          • 1




            $begingroup$
            Yes, but except for educational purposes you would almost always use higher order implicit methods, one- or multi-step.
            $endgroup$
            – LutzL
            Jan 18 at 15:18
















          $begingroup$
          Ah, I was expecting something better for the second, since it is an implicit method, I was wrong.
          $endgroup$
          – user8469759
          Jan 18 at 14:52






          $begingroup$
          Ah, I was expecting something better for the second, since it is an implicit method, I was wrong.
          $endgroup$
          – user8469759
          Jan 18 at 14:52














          $begingroup$
          No, they are both first order and time-reverse to each other. Alternating explicit and implicit steps you get the midpoint method which is of second order.
          $endgroup$
          – LutzL
          Jan 18 at 14:54




          $begingroup$
          No, they are both first order and time-reverse to each other. Alternating explicit and implicit steps you get the midpoint method which is of second order.
          $endgroup$
          – LutzL
          Jan 18 at 14:54












          $begingroup$
          So one method might be more convenient than the other only for stability reasons I suppose
          $endgroup$
          – user8469759
          Jan 18 at 14:55




          $begingroup$
          So one method might be more convenient than the other only for stability reasons I suppose
          $endgroup$
          – user8469759
          Jan 18 at 14:55




          1




          1




          $begingroup$
          Yes, but except for educational purposes you would almost always use higher order implicit methods, one- or multi-step.
          $endgroup$
          – LutzL
          Jan 18 at 15:18




          $begingroup$
          Yes, but except for educational purposes you would almost always use higher order implicit methods, one- or multi-step.
          $endgroup$
          – LutzL
          Jan 18 at 15:18


















          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%2f3078291%2fexplicity-vs-implicit-first-order-euler-to-approximate-e-alpha-x-alpha-0%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

          張江高科駅

          Image
          張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2
          Read more