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











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      asked Jan 18 at 14:04









      user8469759user8469759

      1,5731618




      1,5731618






















          1 Answer
          1






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          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








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          1 Answer
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          active

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          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


















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          張江高科駅

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          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...
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