Trouble with little o notation.












0












$begingroup$


Problem:
Compute the differential of $phi(t) = e^{A+tB}$ at $0$ where $A$ and $B$ are commuting square matrices of size $ntimes n.$



We proceed in the usual manner:
$$phi(0+h) = e^{A+hB} = e^{A}e^{hB}$$ since $A$ and $hB$ commute. Thus
$$phi(h) = e^{A}left(I+hB+o(hB)right)$$
$$ = e^A+h e^AB+e^{A}o(hB) = phi(0)+he^{A}B+o(h).$$
I am not sure why $e^{A}o(hB)$ is $o(h).$ Perhaps someone can explain?










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    0












    $begingroup$


    Problem:
    Compute the differential of $phi(t) = e^{A+tB}$ at $0$ where $A$ and $B$ are commuting square matrices of size $ntimes n.$



    We proceed in the usual manner:
    $$phi(0+h) = e^{A+hB} = e^{A}e^{hB}$$ since $A$ and $hB$ commute. Thus
    $$phi(h) = e^{A}left(I+hB+o(hB)right)$$
    $$ = e^A+h e^AB+e^{A}o(hB) = phi(0)+he^{A}B+o(h).$$
    I am not sure why $e^{A}o(hB)$ is $o(h).$ Perhaps someone can explain?










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Problem:
      Compute the differential of $phi(t) = e^{A+tB}$ at $0$ where $A$ and $B$ are commuting square matrices of size $ntimes n.$



      We proceed in the usual manner:
      $$phi(0+h) = e^{A+hB} = e^{A}e^{hB}$$ since $A$ and $hB$ commute. Thus
      $$phi(h) = e^{A}left(I+hB+o(hB)right)$$
      $$ = e^A+h e^AB+e^{A}o(hB) = phi(0)+he^{A}B+o(h).$$
      I am not sure why $e^{A}o(hB)$ is $o(h).$ Perhaps someone can explain?










      share|cite|improve this question









      $endgroup$




      Problem:
      Compute the differential of $phi(t) = e^{A+tB}$ at $0$ where $A$ and $B$ are commuting square matrices of size $ntimes n.$



      We proceed in the usual manner:
      $$phi(0+h) = e^{A+hB} = e^{A}e^{hB}$$ since $A$ and $hB$ commute. Thus
      $$phi(h) = e^{A}left(I+hB+o(hB)right)$$
      $$ = e^A+h e^AB+e^{A}o(hB) = phi(0)+he^{A}B+o(h).$$
      I am not sure why $e^{A}o(hB)$ is $o(h).$ Perhaps someone can explain?







      real-analysis asymptotics






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      asked Jan 8 at 17:12









      model_checkermodel_checker

      4,44821931




      4,44821931






















          1 Answer
          1






          active

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          1












          $begingroup$

          One definition of $o(h)$ is as follows:




          A matrix valued function $f(h)$ is $o(h)$ if $lim_{h to 0} frac{|f(h)|}{h} = 0$ (for some matrix norm $|cdot|$).




          With this definition, I think you should find the result easy to prove.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            But the function $f(h)= e^{A}o(hB).$ How do I deal with the $o(hB)$ term?
            $endgroup$
            – model_checker
            Jan 8 at 17:19












          • $begingroup$
            The $o(hB)$ term is a stand-in for a function $g(h)$ satisfying $$ lim_{h to 0} frac{|g(h)|}{|hB|} = 0 $$ However, this is equivalent to saying that $g$ is $o(h)$. So, we can simply replace $o(hB)$ with $o(h)$.
            $endgroup$
            – Omnomnomnom
            Jan 8 at 17:22










          • $begingroup$
            Typically, however, I wouldn't use the notation $o(hB)$. Instead, I would say that $$ e^{hB} = I + hB + o(h) $$
            $endgroup$
            – Omnomnomnom
            Jan 8 at 17:24












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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

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          active

          oldest

          votes









          1












          $begingroup$

          One definition of $o(h)$ is as follows:




          A matrix valued function $f(h)$ is $o(h)$ if $lim_{h to 0} frac{|f(h)|}{h} = 0$ (for some matrix norm $|cdot|$).




          With this definition, I think you should find the result easy to prove.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            But the function $f(h)= e^{A}o(hB).$ How do I deal with the $o(hB)$ term?
            $endgroup$
            – model_checker
            Jan 8 at 17:19












          • $begingroup$
            The $o(hB)$ term is a stand-in for a function $g(h)$ satisfying $$ lim_{h to 0} frac{|g(h)|}{|hB|} = 0 $$ However, this is equivalent to saying that $g$ is $o(h)$. So, we can simply replace $o(hB)$ with $o(h)$.
            $endgroup$
            – Omnomnomnom
            Jan 8 at 17:22










          • $begingroup$
            Typically, however, I wouldn't use the notation $o(hB)$. Instead, I would say that $$ e^{hB} = I + hB + o(h) $$
            $endgroup$
            – Omnomnomnom
            Jan 8 at 17:24
















          1












          $begingroup$

          One definition of $o(h)$ is as follows:




          A matrix valued function $f(h)$ is $o(h)$ if $lim_{h to 0} frac{|f(h)|}{h} = 0$ (for some matrix norm $|cdot|$).




          With this definition, I think you should find the result easy to prove.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            But the function $f(h)= e^{A}o(hB).$ How do I deal with the $o(hB)$ term?
            $endgroup$
            – model_checker
            Jan 8 at 17:19












          • $begingroup$
            The $o(hB)$ term is a stand-in for a function $g(h)$ satisfying $$ lim_{h to 0} frac{|g(h)|}{|hB|} = 0 $$ However, this is equivalent to saying that $g$ is $o(h)$. So, we can simply replace $o(hB)$ with $o(h)$.
            $endgroup$
            – Omnomnomnom
            Jan 8 at 17:22










          • $begingroup$
            Typically, however, I wouldn't use the notation $o(hB)$. Instead, I would say that $$ e^{hB} = I + hB + o(h) $$
            $endgroup$
            – Omnomnomnom
            Jan 8 at 17:24














          1












          1








          1





          $begingroup$

          One definition of $o(h)$ is as follows:




          A matrix valued function $f(h)$ is $o(h)$ if $lim_{h to 0} frac{|f(h)|}{h} = 0$ (for some matrix norm $|cdot|$).




          With this definition, I think you should find the result easy to prove.






          share|cite|improve this answer









          $endgroup$



          One definition of $o(h)$ is as follows:




          A matrix valued function $f(h)$ is $o(h)$ if $lim_{h to 0} frac{|f(h)|}{h} = 0$ (for some matrix norm $|cdot|$).




          With this definition, I think you should find the result easy to prove.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 8 at 17:17









          OmnomnomnomOmnomnomnom

          129k793187




          129k793187












          • $begingroup$
            But the function $f(h)= e^{A}o(hB).$ How do I deal with the $o(hB)$ term?
            $endgroup$
            – model_checker
            Jan 8 at 17:19












          • $begingroup$
            The $o(hB)$ term is a stand-in for a function $g(h)$ satisfying $$ lim_{h to 0} frac{|g(h)|}{|hB|} = 0 $$ However, this is equivalent to saying that $g$ is $o(h)$. So, we can simply replace $o(hB)$ with $o(h)$.
            $endgroup$
            – Omnomnomnom
            Jan 8 at 17:22










          • $begingroup$
            Typically, however, I wouldn't use the notation $o(hB)$. Instead, I would say that $$ e^{hB} = I + hB + o(h) $$
            $endgroup$
            – Omnomnomnom
            Jan 8 at 17:24


















          • $begingroup$
            But the function $f(h)= e^{A}o(hB).$ How do I deal with the $o(hB)$ term?
            $endgroup$
            – model_checker
            Jan 8 at 17:19












          • $begingroup$
            The $o(hB)$ term is a stand-in for a function $g(h)$ satisfying $$ lim_{h to 0} frac{|g(h)|}{|hB|} = 0 $$ However, this is equivalent to saying that $g$ is $o(h)$. So, we can simply replace $o(hB)$ with $o(h)$.
            $endgroup$
            – Omnomnomnom
            Jan 8 at 17:22










          • $begingroup$
            Typically, however, I wouldn't use the notation $o(hB)$. Instead, I would say that $$ e^{hB} = I + hB + o(h) $$
            $endgroup$
            – Omnomnomnom
            Jan 8 at 17:24
















          $begingroup$
          But the function $f(h)= e^{A}o(hB).$ How do I deal with the $o(hB)$ term?
          $endgroup$
          – model_checker
          Jan 8 at 17:19






          $begingroup$
          But the function $f(h)= e^{A}o(hB).$ How do I deal with the $o(hB)$ term?
          $endgroup$
          – model_checker
          Jan 8 at 17:19














          $begingroup$
          The $o(hB)$ term is a stand-in for a function $g(h)$ satisfying $$ lim_{h to 0} frac{|g(h)|}{|hB|} = 0 $$ However, this is equivalent to saying that $g$ is $o(h)$. So, we can simply replace $o(hB)$ with $o(h)$.
          $endgroup$
          – Omnomnomnom
          Jan 8 at 17:22




          $begingroup$
          The $o(hB)$ term is a stand-in for a function $g(h)$ satisfying $$ lim_{h to 0} frac{|g(h)|}{|hB|} = 0 $$ However, this is equivalent to saying that $g$ is $o(h)$. So, we can simply replace $o(hB)$ with $o(h)$.
          $endgroup$
          – Omnomnomnom
          Jan 8 at 17:22












          $begingroup$
          Typically, however, I wouldn't use the notation $o(hB)$. Instead, I would say that $$ e^{hB} = I + hB + o(h) $$
          $endgroup$
          – Omnomnomnom
          Jan 8 at 17:24




          $begingroup$
          Typically, however, I wouldn't use the notation $o(hB)$. Instead, I would say that $$ e^{hB} = I + hB + o(h) $$
          $endgroup$
          – Omnomnomnom
          Jan 8 at 17:24


















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