Any identity (involving addition, multiplication, substitution) between real/complex power series is an...












-2












$begingroup$


I need help to prove the following principle-




Any identity between real or complex power series, involving addition,
multiplication (possibly infinite sums and products), and substitution, is an identity in the ring of formal power series.




Answer:



Suppose $f(x)=sum_{n geq 0} a_nx^n$ and $g(x)=sum_{n geq 0} b_nx^n$ be two power series in $ mathbb{R}$.



Also let the identity $ f(x)=g(x)$ holds in $mathbb{R}$.



Then we have to show that same identity $f(X)=g(X)$ i.e., $ sum_{n geq 0} a_nX^n=sum_{n geq 0} b_nX^n$ holds in the ring of formal power series $mathbb{R}[[X]]$.



Let $ h(x)=f(x)-g(x)=sum_{n geq 0} a_nx^n-sum_{n geq 0} b_nx^n=0$.



We know that Taylor series of an analytic function is unique.




Does this conclude the proof?




Help me to prove the principle with any further requirements.






power-series formal-power-series







share|cite|improve this question











$endgroup$

















    -2












    $begingroup$


    I need help to prove the following principle-




    Any identity between real or complex power series, involving addition,
    multiplication (possibly infinite sums and products), and substitution, is an identity in the ring of formal power series.




    Answer:



    Suppose $f(x)=sum_{n geq 0} a_nx^n$ and $g(x)=sum_{n geq 0} b_nx^n$ be two power series in $ mathbb{R}$.



    Also let the identity $ f(x)=g(x)$ holds in $mathbb{R}$.



    Then we have to show that same identity $f(X)=g(X)$ i.e., $ sum_{n geq 0} a_nX^n=sum_{n geq 0} b_nX^n$ holds in the ring of formal power series $mathbb{R}[[X]]$.



    Let $ h(x)=f(x)-g(x)=sum_{n geq 0} a_nx^n-sum_{n geq 0} b_nx^n=0$.



    We know that Taylor series of an analytic function is unique.




    Does this conclude the proof?




    Help me to prove the principle with any further requirements.






    power-series formal-power-series







    share|cite|improve this question











    $endgroup$















      -2












      -2








      -2


      1



      $begingroup$


      I need help to prove the following principle-




      Any identity between real or complex power series, involving addition,
      multiplication (possibly infinite sums and products), and substitution, is an identity in the ring of formal power series.




      Answer:



      Suppose $f(x)=sum_{n geq 0} a_nx^n$ and $g(x)=sum_{n geq 0} b_nx^n$ be two power series in $ mathbb{R}$.



      Also let the identity $ f(x)=g(x)$ holds in $mathbb{R}$.



      Then we have to show that same identity $f(X)=g(X)$ i.e., $ sum_{n geq 0} a_nX^n=sum_{n geq 0} b_nX^n$ holds in the ring of formal power series $mathbb{R}[[X]]$.



      Let $ h(x)=f(x)-g(x)=sum_{n geq 0} a_nx^n-sum_{n geq 0} b_nx^n=0$.



      We know that Taylor series of an analytic function is unique.




      Does this conclude the proof?




      Help me to prove the principle with any further requirements.






      power-series formal-power-series







      share|cite|improve this question











      $endgroup$




      I need help to prove the following principle-




      Any identity between real or complex power series, involving addition,
      multiplication (possibly infinite sums and products), and substitution, is an identity in the ring of formal power series.




      Answer:



      Suppose $f(x)=sum_{n geq 0} a_nx^n$ and $g(x)=sum_{n geq 0} b_nx^n$ be two power series in $ mathbb{R}$.



      Also let the identity $ f(x)=g(x)$ holds in $mathbb{R}$.



      Then we have to show that same identity $f(X)=g(X)$ i.e., $ sum_{n geq 0} a_nX^n=sum_{n geq 0} b_nX^n$ holds in the ring of formal power series $mathbb{R}[[X]]$.



      Let $ h(x)=f(x)-g(x)=sum_{n geq 0} a_nx^n-sum_{n geq 0} b_nx^n=0$.



      We know that Taylor series of an analytic function is unique.




      Does this conclude the proof?




      Help me to prove the principle with any further requirements.






      power-series formal-power-series




      power-series formal-power-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 1 at 17:22









      Blue

      48.6k870156




      48.6k870156










      asked Jan 1 at 14:39









      M. A. SARKARM. A. SARKAR

      2,2671619




      2,2671619






















          1 Answer
          1






          active

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          1












          $begingroup$

          Hint: $a_n = dfrac{f^{(n)}(0)}{n!}$ holds in the ring of analytic functions on $mathbb R$ and in the ring of formal power series $mathbb{R}[[X]]$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            @@lhf, How does it work? I think it should be $c_n=a_n-b_n=frac{h^{(n)}(0)}{n!}=0$. But since $h(x)=0$, we must have $c_n=0$ implying $a_n-b_n=0 Rightarrow a_n=b_n$. Is not it?
            $endgroup$
            – M. A. SARKAR
            Jan 1 at 14:50












          • $begingroup$
            @M.A.SARKAR, exactly
            $endgroup$
            – lhf
            Jan 1 at 15:01










          • $begingroup$
            ok now $a_n=b_n$ implies $ sum a_nX^n=sum b_nX^n$. Is it?
            $endgroup$
            – M. A. SARKAR
            Jan 1 at 15:03











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          1 Answer
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          1 Answer
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          active

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          1












          $begingroup$

          Hint: $a_n = dfrac{f^{(n)}(0)}{n!}$ holds in the ring of analytic functions on $mathbb R$ and in the ring of formal power series $mathbb{R}[[X]]$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            @@lhf, How does it work? I think it should be $c_n=a_n-b_n=frac{h^{(n)}(0)}{n!}=0$. But since $h(x)=0$, we must have $c_n=0$ implying $a_n-b_n=0 Rightarrow a_n=b_n$. Is not it?
            $endgroup$
            – M. A. SARKAR
            Jan 1 at 14:50












          • $begingroup$
            @M.A.SARKAR, exactly
            $endgroup$
            – lhf
            Jan 1 at 15:01










          • $begingroup$
            ok now $a_n=b_n$ implies $ sum a_nX^n=sum b_nX^n$. Is it?
            $endgroup$
            – M. A. SARKAR
            Jan 1 at 15:03
















          1












          $begingroup$

          Hint: $a_n = dfrac{f^{(n)}(0)}{n!}$ holds in the ring of analytic functions on $mathbb R$ and in the ring of formal power series $mathbb{R}[[X]]$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            @@lhf, How does it work? I think it should be $c_n=a_n-b_n=frac{h^{(n)}(0)}{n!}=0$. But since $h(x)=0$, we must have $c_n=0$ implying $a_n-b_n=0 Rightarrow a_n=b_n$. Is not it?
            $endgroup$
            – M. A. SARKAR
            Jan 1 at 14:50












          • $begingroup$
            @M.A.SARKAR, exactly
            $endgroup$
            – lhf
            Jan 1 at 15:01










          • $begingroup$
            ok now $a_n=b_n$ implies $ sum a_nX^n=sum b_nX^n$. Is it?
            $endgroup$
            – M. A. SARKAR
            Jan 1 at 15:03














          1












          1








          1





          $begingroup$

          Hint: $a_n = dfrac{f^{(n)}(0)}{n!}$ holds in the ring of analytic functions on $mathbb R$ and in the ring of formal power series $mathbb{R}[[X]]$.






          share|cite|improve this answer









          $endgroup$



          Hint: $a_n = dfrac{f^{(n)}(0)}{n!}$ holds in the ring of analytic functions on $mathbb R$ and in the ring of formal power series $mathbb{R}[[X]]$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 1 at 14:42









          lhflhf

          166k10171396




          166k10171396












          • $begingroup$
            @@lhf, How does it work? I think it should be $c_n=a_n-b_n=frac{h^{(n)}(0)}{n!}=0$. But since $h(x)=0$, we must have $c_n=0$ implying $a_n-b_n=0 Rightarrow a_n=b_n$. Is not it?
            $endgroup$
            – M. A. SARKAR
            Jan 1 at 14:50












          • $begingroup$
            @M.A.SARKAR, exactly
            $endgroup$
            – lhf
            Jan 1 at 15:01










          • $begingroup$
            ok now $a_n=b_n$ implies $ sum a_nX^n=sum b_nX^n$. Is it?
            $endgroup$
            – M. A. SARKAR
            Jan 1 at 15:03


















          • $begingroup$
            @@lhf, How does it work? I think it should be $c_n=a_n-b_n=frac{h^{(n)}(0)}{n!}=0$. But since $h(x)=0$, we must have $c_n=0$ implying $a_n-b_n=0 Rightarrow a_n=b_n$. Is not it?
            $endgroup$
            – M. A. SARKAR
            Jan 1 at 14:50












          • $begingroup$
            @M.A.SARKAR, exactly
            $endgroup$
            – lhf
            Jan 1 at 15:01










          • $begingroup$
            ok now $a_n=b_n$ implies $ sum a_nX^n=sum b_nX^n$. Is it?
            $endgroup$
            – M. A. SARKAR
            Jan 1 at 15:03
















          $begingroup$
          @@lhf, How does it work? I think it should be $c_n=a_n-b_n=frac{h^{(n)}(0)}{n!}=0$. But since $h(x)=0$, we must have $c_n=0$ implying $a_n-b_n=0 Rightarrow a_n=b_n$. Is not it?
          $endgroup$
          – M. A. SARKAR
          Jan 1 at 14:50






          $begingroup$
          @@lhf, How does it work? I think it should be $c_n=a_n-b_n=frac{h^{(n)}(0)}{n!}=0$. But since $h(x)=0$, we must have $c_n=0$ implying $a_n-b_n=0 Rightarrow a_n=b_n$. Is not it?
          $endgroup$
          – M. A. SARKAR
          Jan 1 at 14:50














          $begingroup$
          @M.A.SARKAR, exactly
          $endgroup$
          – lhf
          Jan 1 at 15:01




          $begingroup$
          @M.A.SARKAR, exactly
          $endgroup$
          – lhf
          Jan 1 at 15:01












          $begingroup$
          ok now $a_n=b_n$ implies $ sum a_nX^n=sum b_nX^n$. Is it?
          $endgroup$
          – M. A. SARKAR
          Jan 1 at 15:03




          $begingroup$
          ok now $a_n=b_n$ implies $ sum a_nX^n=sum b_nX^n$. Is it?
          $endgroup$
          – M. A. SARKAR
          Jan 1 at 15:03


















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