Taylor expansion for the exponential series












0












$begingroup$


Inside this question, a key derivation was made in which



$$x(a)=expBig(afrac{partial}{partial{t}}Big)x(t)Big|_{t=0}$$



I know that we can write $x(a)$ as a series by



$$x(a)=sum_{n=0}^inftyfrac{x^{(n)}}{n!}a^n$$



But, I don't see how



$$x(a)=sum_{n=0}^inftyfrac{x^{(n)}}{n!}a^n=cdots=expBig(afrac{partial}{partial{t}}Big)x(t)Big|_{t=0}$$



How do you derive $x(a)=expBig(afrac{partial}{partial{t}}Big)x(t)Big|_{t=0}$?






derivatives taylor-expansion exponential-function







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Related
    $endgroup$
    – A.Γ.
    Jan 6 at 19:41










  • $begingroup$
    There is also a nice historical chapter on the topic.
    $endgroup$
    – A.Γ.
    Jan 6 at 20:15
















0












$begingroup$


Inside this question, a key derivation was made in which



$$x(a)=expBig(afrac{partial}{partial{t}}Big)x(t)Big|_{t=0}$$



I know that we can write $x(a)$ as a series by



$$x(a)=sum_{n=0}^inftyfrac{x^{(n)}}{n!}a^n$$



But, I don't see how



$$x(a)=sum_{n=0}^inftyfrac{x^{(n)}}{n!}a^n=cdots=expBig(afrac{partial}{partial{t}}Big)x(t)Big|_{t=0}$$



How do you derive $x(a)=expBig(afrac{partial}{partial{t}}Big)x(t)Big|_{t=0}$?






derivatives taylor-expansion exponential-function







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Related
    $endgroup$
    – A.Γ.
    Jan 6 at 19:41










  • $begingroup$
    There is also a nice historical chapter on the topic.
    $endgroup$
    – A.Γ.
    Jan 6 at 20:15














0












0








0





$begingroup$


Inside this question, a key derivation was made in which



$$x(a)=expBig(afrac{partial}{partial{t}}Big)x(t)Big|_{t=0}$$



I know that we can write $x(a)$ as a series by



$$x(a)=sum_{n=0}^inftyfrac{x^{(n)}}{n!}a^n$$



But, I don't see how



$$x(a)=sum_{n=0}^inftyfrac{x^{(n)}}{n!}a^n=cdots=expBig(afrac{partial}{partial{t}}Big)x(t)Big|_{t=0}$$



How do you derive $x(a)=expBig(afrac{partial}{partial{t}}Big)x(t)Big|_{t=0}$?






derivatives taylor-expansion exponential-function







share|cite|improve this question











$endgroup$




Inside this question, a key derivation was made in which



$$x(a)=expBig(afrac{partial}{partial{t}}Big)x(t)Big|_{t=0}$$



I know that we can write $x(a)$ as a series by



$$x(a)=sum_{n=0}^inftyfrac{x^{(n)}}{n!}a^n$$



But, I don't see how



$$x(a)=sum_{n=0}^inftyfrac{x^{(n)}}{n!}a^n=cdots=expBig(afrac{partial}{partial{t}}Big)x(t)Big|_{t=0}$$



How do you derive $x(a)=expBig(afrac{partial}{partial{t}}Big)x(t)Big|_{t=0}$?






derivatives taylor-expansion exponential-function




derivatives taylor-expansion exponential-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 19:32









A.Γ.

22.9k32656




22.9k32656










asked Jan 6 at 19:04









Axion004Axion004

398313




398313








  • 1




    $begingroup$
    Related
    $endgroup$
    – A.Γ.
    Jan 6 at 19:41










  • $begingroup$
    There is also a nice historical chapter on the topic.
    $endgroup$
    – A.Γ.
    Jan 6 at 20:15














  • 1




    $begingroup$
    Related
    $endgroup$
    – A.Γ.
    Jan 6 at 19:41










  • $begingroup$
    There is also a nice historical chapter on the topic.
    $endgroup$
    – A.Γ.
    Jan 6 at 20:15








1




1




$begingroup$
Related
$endgroup$
– A.Γ.
Jan 6 at 19:41




$begingroup$
Related
$endgroup$
– A.Γ.
Jan 6 at 19:41












$begingroup$
There is also a nice historical chapter on the topic.
$endgroup$
– A.Γ.
Jan 6 at 20:15




$begingroup$
There is also a nice historical chapter on the topic.
$endgroup$
– A.Γ.
Jan 6 at 20:15










1 Answer
1






active

oldest

votes


















1












$begingroup$

Through the related question, I think I understand the derivation.



We are given that



$$x(a)=sum_{n=0}^inftyfrac{x^{(n)}}{n!}a^n$$



We also know that



$$exp(a)=sum_{n=0}^inftyfrac{a^n}{n!}$$



Therefore,



$$expBig(afrac{partial}{partial{t}}Big)=sum_{n=0}^inftyfrac{a^n}{n!}Big(frac{partial}{partial{t}}Big)^n$$



Hence if we evaluate this at $x(t)$ we have



$$expBig(afrac{partial}{partial{t}}Big)x(t)=sum_{n=0}^inftyfrac{a^n}{n!}Big(frac{partial{x(t)}}{partial{t}}Big)^n =sum_{n=0}^inftyfrac{a^n}{n!}x^{n}(t)=sum_{n=0}^inftyfrac{x^{n}(t)}{n!}a^n$$



It is necessary for $t=0$ in order to compute the Maclaurin expansion. If $tneq0$, then we would have



$$x(a)=sum_{n=0}^inftyfrac{x^{n}(t)}{n!}(a-t)^n$$



Therefore,



$$expBig(afrac{partial}{partial{t}}Big)x(t)Big|_{t=0}=sum_{n=0}^inftyfrac{x^{n}(t)}{n!}a^n=x(a)$$






share|cite|improve this answer









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    1












    $begingroup$

    Through the related question, I think I understand the derivation.



    We are given that



    $$x(a)=sum_{n=0}^inftyfrac{x^{(n)}}{n!}a^n$$



    We also know that



    $$exp(a)=sum_{n=0}^inftyfrac{a^n}{n!}$$



    Therefore,



    $$expBig(afrac{partial}{partial{t}}Big)=sum_{n=0}^inftyfrac{a^n}{n!}Big(frac{partial}{partial{t}}Big)^n$$



    Hence if we evaluate this at $x(t)$ we have



    $$expBig(afrac{partial}{partial{t}}Big)x(t)=sum_{n=0}^inftyfrac{a^n}{n!}Big(frac{partial{x(t)}}{partial{t}}Big)^n =sum_{n=0}^inftyfrac{a^n}{n!}x^{n}(t)=sum_{n=0}^inftyfrac{x^{n}(t)}{n!}a^n$$



    It is necessary for $t=0$ in order to compute the Maclaurin expansion. If $tneq0$, then we would have



    $$x(a)=sum_{n=0}^inftyfrac{x^{n}(t)}{n!}(a-t)^n$$



    Therefore,



    $$expBig(afrac{partial}{partial{t}}Big)x(t)Big|_{t=0}=sum_{n=0}^inftyfrac{x^{n}(t)}{n!}a^n=x(a)$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Through the related question, I think I understand the derivation.



      We are given that



      $$x(a)=sum_{n=0}^inftyfrac{x^{(n)}}{n!}a^n$$



      We also know that



      $$exp(a)=sum_{n=0}^inftyfrac{a^n}{n!}$$



      Therefore,



      $$expBig(afrac{partial}{partial{t}}Big)=sum_{n=0}^inftyfrac{a^n}{n!}Big(frac{partial}{partial{t}}Big)^n$$



      Hence if we evaluate this at $x(t)$ we have



      $$expBig(afrac{partial}{partial{t}}Big)x(t)=sum_{n=0}^inftyfrac{a^n}{n!}Big(frac{partial{x(t)}}{partial{t}}Big)^n =sum_{n=0}^inftyfrac{a^n}{n!}x^{n}(t)=sum_{n=0}^inftyfrac{x^{n}(t)}{n!}a^n$$



      It is necessary for $t=0$ in order to compute the Maclaurin expansion. If $tneq0$, then we would have



      $$x(a)=sum_{n=0}^inftyfrac{x^{n}(t)}{n!}(a-t)^n$$



      Therefore,



      $$expBig(afrac{partial}{partial{t}}Big)x(t)Big|_{t=0}=sum_{n=0}^inftyfrac{x^{n}(t)}{n!}a^n=x(a)$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Through the related question, I think I understand the derivation.



        We are given that



        $$x(a)=sum_{n=0}^inftyfrac{x^{(n)}}{n!}a^n$$



        We also know that



        $$exp(a)=sum_{n=0}^inftyfrac{a^n}{n!}$$



        Therefore,



        $$expBig(afrac{partial}{partial{t}}Big)=sum_{n=0}^inftyfrac{a^n}{n!}Big(frac{partial}{partial{t}}Big)^n$$



        Hence if we evaluate this at $x(t)$ we have



        $$expBig(afrac{partial}{partial{t}}Big)x(t)=sum_{n=0}^inftyfrac{a^n}{n!}Big(frac{partial{x(t)}}{partial{t}}Big)^n =sum_{n=0}^inftyfrac{a^n}{n!}x^{n}(t)=sum_{n=0}^inftyfrac{x^{n}(t)}{n!}a^n$$



        It is necessary for $t=0$ in order to compute the Maclaurin expansion. If $tneq0$, then we would have



        $$x(a)=sum_{n=0}^inftyfrac{x^{n}(t)}{n!}(a-t)^n$$



        Therefore,



        $$expBig(afrac{partial}{partial{t}}Big)x(t)Big|_{t=0}=sum_{n=0}^inftyfrac{x^{n}(t)}{n!}a^n=x(a)$$






        share|cite|improve this answer









        $endgroup$



        Through the related question, I think I understand the derivation.



        We are given that



        $$x(a)=sum_{n=0}^inftyfrac{x^{(n)}}{n!}a^n$$



        We also know that



        $$exp(a)=sum_{n=0}^inftyfrac{a^n}{n!}$$



        Therefore,



        $$expBig(afrac{partial}{partial{t}}Big)=sum_{n=0}^inftyfrac{a^n}{n!}Big(frac{partial}{partial{t}}Big)^n$$



        Hence if we evaluate this at $x(t)$ we have



        $$expBig(afrac{partial}{partial{t}}Big)x(t)=sum_{n=0}^inftyfrac{a^n}{n!}Big(frac{partial{x(t)}}{partial{t}}Big)^n =sum_{n=0}^inftyfrac{a^n}{n!}x^{n}(t)=sum_{n=0}^inftyfrac{x^{n}(t)}{n!}a^n$$



        It is necessary for $t=0$ in order to compute the Maclaurin expansion. If $tneq0$, then we would have



        $$x(a)=sum_{n=0}^inftyfrac{x^{n}(t)}{n!}(a-t)^n$$



        Therefore,



        $$expBig(afrac{partial}{partial{t}}Big)x(t)Big|_{t=0}=sum_{n=0}^inftyfrac{x^{n}(t)}{n!}a^n=x(a)$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 6 at 20:19









        Axion004Axion004

        398313




        398313






























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