Fubini's theorem for series with dependent indices












2














As a part of a proposition about power series that I am trying to prove I have to show that
$$ sum_{m=0}^{infty} left(sum_{n=m}^{infty} frac{n!}{m!(n-m)!}(b-a)^{n-m}c_nright)(x-b)^m = sum_{n=0}^{infty}c_n(x-a)^n$$



Proof: What I have done so far is to show that
$$sum_{m=0}^{n} frac{n!}{m!(n-m)!}(b-a)^{n-m}(x-b)^m = (x-a)^n $$



So the next idea is to use Fubini's rearrangement theorem for infinite series (I have also shown that the series are absolutely convergent). However, the theorem does not accomodate for dependent indices. Intuitively (using triangular summation), I think that the proper way to interchange series would be
begin{equation*}
%begin{array}{ll}
sum_{m=0}^{infty} sum_{n=m}^{infty} frac{n!}{m!(n-m)!}(b-a)^{n-m}c_n(x-b)^m = sum_{n=0}^{infty} sum_{m=0}^{n}frac{n!}{m!(n-m)!}(b-a)^{n-m}c_n(x-b)^m
%end{array}
end{equation*}
and thus
$$=sum_{n=0}^{infty} c_n(x-a)^n$$
However, I do not know how to perform this step rigorously in case of infinite summation.






real-analysis sequences-and-series power-series







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    2














    As a part of a proposition about power series that I am trying to prove I have to show that
    $$ sum_{m=0}^{infty} left(sum_{n=m}^{infty} frac{n!}{m!(n-m)!}(b-a)^{n-m}c_nright)(x-b)^m = sum_{n=0}^{infty}c_n(x-a)^n$$



    Proof: What I have done so far is to show that
    $$sum_{m=0}^{n} frac{n!}{m!(n-m)!}(b-a)^{n-m}(x-b)^m = (x-a)^n $$



    So the next idea is to use Fubini's rearrangement theorem for infinite series (I have also shown that the series are absolutely convergent). However, the theorem does not accomodate for dependent indices. Intuitively (using triangular summation), I think that the proper way to interchange series would be
    begin{equation*}
    %begin{array}{ll}
    sum_{m=0}^{infty} sum_{n=m}^{infty} frac{n!}{m!(n-m)!}(b-a)^{n-m}c_n(x-b)^m = sum_{n=0}^{infty} sum_{m=0}^{n}frac{n!}{m!(n-m)!}(b-a)^{n-m}c_n(x-b)^m
    %end{array}
    end{equation*}
    and thus
    $$=sum_{n=0}^{infty} c_n(x-a)^n$$
    However, I do not know how to perform this step rigorously in case of infinite summation.






    real-analysis sequences-and-series power-series







    share|cite|improve this question



























      2












      2








      2


      1





      As a part of a proposition about power series that I am trying to prove I have to show that
      $$ sum_{m=0}^{infty} left(sum_{n=m}^{infty} frac{n!}{m!(n-m)!}(b-a)^{n-m}c_nright)(x-b)^m = sum_{n=0}^{infty}c_n(x-a)^n$$



      Proof: What I have done so far is to show that
      $$sum_{m=0}^{n} frac{n!}{m!(n-m)!}(b-a)^{n-m}(x-b)^m = (x-a)^n $$



      So the next idea is to use Fubini's rearrangement theorem for infinite series (I have also shown that the series are absolutely convergent). However, the theorem does not accomodate for dependent indices. Intuitively (using triangular summation), I think that the proper way to interchange series would be
      begin{equation*}
      %begin{array}{ll}
      sum_{m=0}^{infty} sum_{n=m}^{infty} frac{n!}{m!(n-m)!}(b-a)^{n-m}c_n(x-b)^m = sum_{n=0}^{infty} sum_{m=0}^{n}frac{n!}{m!(n-m)!}(b-a)^{n-m}c_n(x-b)^m
      %end{array}
      end{equation*}
      and thus
      $$=sum_{n=0}^{infty} c_n(x-a)^n$$
      However, I do not know how to perform this step rigorously in case of infinite summation.






      real-analysis sequences-and-series power-series







      share|cite|improve this question















      As a part of a proposition about power series that I am trying to prove I have to show that
      $$ sum_{m=0}^{infty} left(sum_{n=m}^{infty} frac{n!}{m!(n-m)!}(b-a)^{n-m}c_nright)(x-b)^m = sum_{n=0}^{infty}c_n(x-a)^n$$



      Proof: What I have done so far is to show that
      $$sum_{m=0}^{n} frac{n!}{m!(n-m)!}(b-a)^{n-m}(x-b)^m = (x-a)^n $$



      So the next idea is to use Fubini's rearrangement theorem for infinite series (I have also shown that the series are absolutely convergent). However, the theorem does not accomodate for dependent indices. Intuitively (using triangular summation), I think that the proper way to interchange series would be
      begin{equation*}
      %begin{array}{ll}
      sum_{m=0}^{infty} sum_{n=m}^{infty} frac{n!}{m!(n-m)!}(b-a)^{n-m}c_n(x-b)^m = sum_{n=0}^{infty} sum_{m=0}^{n}frac{n!}{m!(n-m)!}(b-a)^{n-m}c_n(x-b)^m
      %end{array}
      end{equation*}
      and thus
      $$=sum_{n=0}^{infty} c_n(x-a)^n$$
      However, I do not know how to perform this step rigorously in case of infinite summation.






      real-analysis sequences-and-series power-series




      real-analysis sequences-and-series power-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 6 at 11:38







      Cebiş Mellim

















      asked Jan 6 at 10:41









      Cebiş MellimCebiş Mellim

      12210




      12210






















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          Using the indicator function



          $$mathbf{1}_{{m leqslant n}}=mathbf{1}_{{n geqslant m}} = begin{cases}1 , & ngeqslant m \ 0, & n < mend{cases}$$



          we have



          $$sum_{m=0}^inftysum_{n=m}^infty f(m,n) = sum_{m=0}^inftysum_{n=0}^infty f(m,n)mathbf{1}_{{n geqslant m}}= sum_{m=0}^inftysum_{n=0}^infty f(m,n)mathbf{1}_{{m leqslant n}} $$



          With absolute convergence we can apply Fubini's theorem to interchange summations and obtain



          $$sum_{m=0}^inftysum_{n=m}^infty f(m,n) =underbrace{sum_{n=0}^inftysum_{m=0}^infty f(m,n)mathbf{1}_{{m leqslant n}}}_{text{after switching summation order}} = sum_{n=0}^inftysum_{m=0}^n f(m,n)$$






          share|cite|improve this answer





















          • When will I finally start to do the magic like this on my own ...
            – Cebiş Mellim
            2 days ago






          • 1




            @CebişMellim: Well, your intuition and observation about triangular summation was a good start and it allowed you to reach the correct conclusion.
            – RRL
            2 days ago











          Your Answer





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          active

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          1














          Using the indicator function



          $$mathbf{1}_{{m leqslant n}}=mathbf{1}_{{n geqslant m}} = begin{cases}1 , & ngeqslant m \ 0, & n < mend{cases}$$



          we have



          $$sum_{m=0}^inftysum_{n=m}^infty f(m,n) = sum_{m=0}^inftysum_{n=0}^infty f(m,n)mathbf{1}_{{n geqslant m}}= sum_{m=0}^inftysum_{n=0}^infty f(m,n)mathbf{1}_{{m leqslant n}} $$



          With absolute convergence we can apply Fubini's theorem to interchange summations and obtain



          $$sum_{m=0}^inftysum_{n=m}^infty f(m,n) =underbrace{sum_{n=0}^inftysum_{m=0}^infty f(m,n)mathbf{1}_{{m leqslant n}}}_{text{after switching summation order}} = sum_{n=0}^inftysum_{m=0}^n f(m,n)$$






          share|cite|improve this answer





















          • When will I finally start to do the magic like this on my own ...
            – Cebiş Mellim
            2 days ago






          • 1




            @CebişMellim: Well, your intuition and observation about triangular summation was a good start and it allowed you to reach the correct conclusion.
            – RRL
            2 days ago
















          1














          Using the indicator function



          $$mathbf{1}_{{m leqslant n}}=mathbf{1}_{{n geqslant m}} = begin{cases}1 , & ngeqslant m \ 0, & n < mend{cases}$$



          we have



          $$sum_{m=0}^inftysum_{n=m}^infty f(m,n) = sum_{m=0}^inftysum_{n=0}^infty f(m,n)mathbf{1}_{{n geqslant m}}= sum_{m=0}^inftysum_{n=0}^infty f(m,n)mathbf{1}_{{m leqslant n}} $$



          With absolute convergence we can apply Fubini's theorem to interchange summations and obtain



          $$sum_{m=0}^inftysum_{n=m}^infty f(m,n) =underbrace{sum_{n=0}^inftysum_{m=0}^infty f(m,n)mathbf{1}_{{m leqslant n}}}_{text{after switching summation order}} = sum_{n=0}^inftysum_{m=0}^n f(m,n)$$






          share|cite|improve this answer





















          • When will I finally start to do the magic like this on my own ...
            – Cebiş Mellim
            2 days ago






          • 1




            @CebişMellim: Well, your intuition and observation about triangular summation was a good start and it allowed you to reach the correct conclusion.
            – RRL
            2 days ago














          1












          1








          1






          Using the indicator function



          $$mathbf{1}_{{m leqslant n}}=mathbf{1}_{{n geqslant m}} = begin{cases}1 , & ngeqslant m \ 0, & n < mend{cases}$$



          we have



          $$sum_{m=0}^inftysum_{n=m}^infty f(m,n) = sum_{m=0}^inftysum_{n=0}^infty f(m,n)mathbf{1}_{{n geqslant m}}= sum_{m=0}^inftysum_{n=0}^infty f(m,n)mathbf{1}_{{m leqslant n}} $$



          With absolute convergence we can apply Fubini's theorem to interchange summations and obtain



          $$sum_{m=0}^inftysum_{n=m}^infty f(m,n) =underbrace{sum_{n=0}^inftysum_{m=0}^infty f(m,n)mathbf{1}_{{m leqslant n}}}_{text{after switching summation order}} = sum_{n=0}^inftysum_{m=0}^n f(m,n)$$






          share|cite|improve this answer












          Using the indicator function



          $$mathbf{1}_{{m leqslant n}}=mathbf{1}_{{n geqslant m}} = begin{cases}1 , & ngeqslant m \ 0, & n < mend{cases}$$



          we have



          $$sum_{m=0}^inftysum_{n=m}^infty f(m,n) = sum_{m=0}^inftysum_{n=0}^infty f(m,n)mathbf{1}_{{n geqslant m}}= sum_{m=0}^inftysum_{n=0}^infty f(m,n)mathbf{1}_{{m leqslant n}} $$



          With absolute convergence we can apply Fubini's theorem to interchange summations and obtain



          $$sum_{m=0}^inftysum_{n=m}^infty f(m,n) =underbrace{sum_{n=0}^inftysum_{m=0}^infty f(m,n)mathbf{1}_{{m leqslant n}}}_{text{after switching summation order}} = sum_{n=0}^inftysum_{m=0}^n f(m,n)$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 6 at 19:06









          RRLRRL

          49.4k42573




          49.4k42573












          • When will I finally start to do the magic like this on my own ...
            – Cebiş Mellim
            2 days ago






          • 1




            @CebişMellim: Well, your intuition and observation about triangular summation was a good start and it allowed you to reach the correct conclusion.
            – RRL
            2 days ago


















          • When will I finally start to do the magic like this on my own ...
            – Cebiş Mellim
            2 days ago






          • 1




            @CebişMellim: Well, your intuition and observation about triangular summation was a good start and it allowed you to reach the correct conclusion.
            – RRL
            2 days ago
















          When will I finally start to do the magic like this on my own ...
          – Cebiş Mellim
          2 days ago




          When will I finally start to do the magic like this on my own ...
          – Cebiş Mellim
          2 days ago




          1




          1




          @CebişMellim: Well, your intuition and observation about triangular summation was a good start and it allowed you to reach the correct conclusion.
          – RRL
          2 days ago




          @CebişMellim: Well, your intuition and observation about triangular summation was a good start and it allowed you to reach the correct conclusion.
          – RRL
          2 days ago


















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