Proof that a finite series expansion of $f(X)$ at $alpha$ exists iff $Q(X)$ is a power of $(X-alpha)$, in...












0












$begingroup$


I'm working through Gouvea's P-adic numbers book, and early on they give the problem




Write $f(X)=frac{P(X)}{Q(X)}$ in lowest terms, so that $P(X)$ and $Q(X)$ have no common zeros. Show that the expansion of $f(X)$ in powers of $(X-alpha)$ is finite if and only if $Q(X)=(X-alpha)^m$ for some $mgeq 0$




The solution/hint given is




"if the expansion is finite, it will certainly become a polynomial after we multiply by $(X-alpha)^m$, where $m$ is the biggest exponnent appearing in a denominator.




From what I read so far, we should get a finite series in the event where we have a "pole", for example, the series expansion of $frac{X}{X-1}$ at the point $alpha=1$, gives us the series $(X-1)^{-1}+1$. But i'm not sure how to prove that we are assure a finite series expansion where the denomiator polynomial can be expressed as a power of $(X-alpha)$ and I'm also unsure of going in the other direction.



Thanks






complex-analysis elementary-number-theory proof-writing p-adic-number-theory







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    0












    $begingroup$


    I'm working through Gouvea's P-adic numbers book, and early on they give the problem




    Write $f(X)=frac{P(X)}{Q(X)}$ in lowest terms, so that $P(X)$ and $Q(X)$ have no common zeros. Show that the expansion of $f(X)$ in powers of $(X-alpha)$ is finite if and only if $Q(X)=(X-alpha)^m$ for some $mgeq 0$




    The solution/hint given is




    "if the expansion is finite, it will certainly become a polynomial after we multiply by $(X-alpha)^m$, where $m$ is the biggest exponnent appearing in a denominator.




    From what I read so far, we should get a finite series in the event where we have a "pole", for example, the series expansion of $frac{X}{X-1}$ at the point $alpha=1$, gives us the series $(X-1)^{-1}+1$. But i'm not sure how to prove that we are assure a finite series expansion where the denomiator polynomial can be expressed as a power of $(X-alpha)$ and I'm also unsure of going in the other direction.



    Thanks






    complex-analysis elementary-number-theory proof-writing p-adic-number-theory







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I'm working through Gouvea's P-adic numbers book, and early on they give the problem




      Write $f(X)=frac{P(X)}{Q(X)}$ in lowest terms, so that $P(X)$ and $Q(X)$ have no common zeros. Show that the expansion of $f(X)$ in powers of $(X-alpha)$ is finite if and only if $Q(X)=(X-alpha)^m$ for some $mgeq 0$




      The solution/hint given is




      "if the expansion is finite, it will certainly become a polynomial after we multiply by $(X-alpha)^m$, where $m$ is the biggest exponnent appearing in a denominator.




      From what I read so far, we should get a finite series in the event where we have a "pole", for example, the series expansion of $frac{X}{X-1}$ at the point $alpha=1$, gives us the series $(X-1)^{-1}+1$. But i'm not sure how to prove that we are assure a finite series expansion where the denomiator polynomial can be expressed as a power of $(X-alpha)$ and I'm also unsure of going in the other direction.



      Thanks






      complex-analysis elementary-number-theory proof-writing p-adic-number-theory







      share|cite|improve this question









      $endgroup$




      I'm working through Gouvea's P-adic numbers book, and early on they give the problem




      Write $f(X)=frac{P(X)}{Q(X)}$ in lowest terms, so that $P(X)$ and $Q(X)$ have no common zeros. Show that the expansion of $f(X)$ in powers of $(X-alpha)$ is finite if and only if $Q(X)=(X-alpha)^m$ for some $mgeq 0$




      The solution/hint given is




      "if the expansion is finite, it will certainly become a polynomial after we multiply by $(X-alpha)^m$, where $m$ is the biggest exponnent appearing in a denominator.




      From what I read so far, we should get a finite series in the event where we have a "pole", for example, the series expansion of $frac{X}{X-1}$ at the point $alpha=1$, gives us the series $(X-1)^{-1}+1$. But i'm not sure how to prove that we are assure a finite series expansion where the denomiator polynomial can be expressed as a power of $(X-alpha)$ and I'm also unsure of going in the other direction.



      Thanks






      complex-analysis elementary-number-theory proof-writing p-adic-number-theory




      complex-analysis elementary-number-theory proof-writing p-adic-number-theory






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 7 '18 at 2:10









      john fowlesjohn fowles

      1,198817




      1,198817






















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          $begingroup$

          Assume that the expansion is finite, i.e,



          $$f(x) = sum_{n=-k}^m a_i (x-alpha)^i$$



          where we assume $k>0$, and we let $a_{-k}=0$ if needed (I assume it only for convenience. Also note that if the sequence actually start with $ngeq 0$, $f$ is a polynomial and there's nothing to prove). Note that after multiplying by $(x-alpha)^k$ we get:



          $$(x-alpha)^kf(x)=sum_{n=-k}^m a_i (x-alpha)^{i+k}=sum_{n=0}^{m+k}a_i (x-alpha)^{i+k}$$



          RHS is a sum of polynomials, hence a polynomial $g(x)$, so, we may write $(x-alpha)^k f(x)=g(x)$ which gives $f(x)=g(x)/(x-alpha)^k$. After canceling factors, this is still the form of the rational function $f$ (maybe $k$ will be changed).



          The other way around, if $f(x)=g(x)/(x-alpha)^k$ for some $kgeq0$, where $g$ is a polynomial, note that $g$ can be written as $g(x)=g((x-alpha)+alpha)$. Now if we write $g$ explicitly as $g(x)=sum_{i=0}^m b_i x^i$, we get-



          $$g((x-alpha)+alpha)=g(x)=sum_{i=0}^m b_i ((x-alpha)+alpha)^i$$



          using the binomial theorem we get that $g(x)=sum_{i=0}^m b_i' (x-alpha)^i$ ($b_i'$ are the modified coefficients; we can calculate them explicitly, but this is not interesting). This gives
          $$f(x)=frac{g(x)}{(x-alpha)^k}=sum_{i=0}^m b_i' (x-alpha)^{i-k}$$



          And so the expansion is finite.






          share|cite|improve this answer









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            $begingroup$

            Assume that the expansion is finite, i.e,



            $$f(x) = sum_{n=-k}^m a_i (x-alpha)^i$$



            where we assume $k>0$, and we let $a_{-k}=0$ if needed (I assume it only for convenience. Also note that if the sequence actually start with $ngeq 0$, $f$ is a polynomial and there's nothing to prove). Note that after multiplying by $(x-alpha)^k$ we get:



            $$(x-alpha)^kf(x)=sum_{n=-k}^m a_i (x-alpha)^{i+k}=sum_{n=0}^{m+k}a_i (x-alpha)^{i+k}$$



            RHS is a sum of polynomials, hence a polynomial $g(x)$, so, we may write $(x-alpha)^k f(x)=g(x)$ which gives $f(x)=g(x)/(x-alpha)^k$. After canceling factors, this is still the form of the rational function $f$ (maybe $k$ will be changed).



            The other way around, if $f(x)=g(x)/(x-alpha)^k$ for some $kgeq0$, where $g$ is a polynomial, note that $g$ can be written as $g(x)=g((x-alpha)+alpha)$. Now if we write $g$ explicitly as $g(x)=sum_{i=0}^m b_i x^i$, we get-



            $$g((x-alpha)+alpha)=g(x)=sum_{i=0}^m b_i ((x-alpha)+alpha)^i$$



            using the binomial theorem we get that $g(x)=sum_{i=0}^m b_i' (x-alpha)^i$ ($b_i'$ are the modified coefficients; we can calculate them explicitly, but this is not interesting). This gives
            $$f(x)=frac{g(x)}{(x-alpha)^k}=sum_{i=0}^m b_i' (x-alpha)^{i-k}$$



            And so the expansion is finite.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Assume that the expansion is finite, i.e,



              $$f(x) = sum_{n=-k}^m a_i (x-alpha)^i$$



              where we assume $k>0$, and we let $a_{-k}=0$ if needed (I assume it only for convenience. Also note that if the sequence actually start with $ngeq 0$, $f$ is a polynomial and there's nothing to prove). Note that after multiplying by $(x-alpha)^k$ we get:



              $$(x-alpha)^kf(x)=sum_{n=-k}^m a_i (x-alpha)^{i+k}=sum_{n=0}^{m+k}a_i (x-alpha)^{i+k}$$



              RHS is a sum of polynomials, hence a polynomial $g(x)$, so, we may write $(x-alpha)^k f(x)=g(x)$ which gives $f(x)=g(x)/(x-alpha)^k$. After canceling factors, this is still the form of the rational function $f$ (maybe $k$ will be changed).



              The other way around, if $f(x)=g(x)/(x-alpha)^k$ for some $kgeq0$, where $g$ is a polynomial, note that $g$ can be written as $g(x)=g((x-alpha)+alpha)$. Now if we write $g$ explicitly as $g(x)=sum_{i=0}^m b_i x^i$, we get-



              $$g((x-alpha)+alpha)=g(x)=sum_{i=0}^m b_i ((x-alpha)+alpha)^i$$



              using the binomial theorem we get that $g(x)=sum_{i=0}^m b_i' (x-alpha)^i$ ($b_i'$ are the modified coefficients; we can calculate them explicitly, but this is not interesting). This gives
              $$f(x)=frac{g(x)}{(x-alpha)^k}=sum_{i=0}^m b_i' (x-alpha)^{i-k}$$



              And so the expansion is finite.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Assume that the expansion is finite, i.e,



                $$f(x) = sum_{n=-k}^m a_i (x-alpha)^i$$



                where we assume $k>0$, and we let $a_{-k}=0$ if needed (I assume it only for convenience. Also note that if the sequence actually start with $ngeq 0$, $f$ is a polynomial and there's nothing to prove). Note that after multiplying by $(x-alpha)^k$ we get:



                $$(x-alpha)^kf(x)=sum_{n=-k}^m a_i (x-alpha)^{i+k}=sum_{n=0}^{m+k}a_i (x-alpha)^{i+k}$$



                RHS is a sum of polynomials, hence a polynomial $g(x)$, so, we may write $(x-alpha)^k f(x)=g(x)$ which gives $f(x)=g(x)/(x-alpha)^k$. After canceling factors, this is still the form of the rational function $f$ (maybe $k$ will be changed).



                The other way around, if $f(x)=g(x)/(x-alpha)^k$ for some $kgeq0$, where $g$ is a polynomial, note that $g$ can be written as $g(x)=g((x-alpha)+alpha)$. Now if we write $g$ explicitly as $g(x)=sum_{i=0}^m b_i x^i$, we get-



                $$g((x-alpha)+alpha)=g(x)=sum_{i=0}^m b_i ((x-alpha)+alpha)^i$$



                using the binomial theorem we get that $g(x)=sum_{i=0}^m b_i' (x-alpha)^i$ ($b_i'$ are the modified coefficients; we can calculate them explicitly, but this is not interesting). This gives
                $$f(x)=frac{g(x)}{(x-alpha)^k}=sum_{i=0}^m b_i' (x-alpha)^{i-k}$$



                And so the expansion is finite.






                share|cite|improve this answer









                $endgroup$



                Assume that the expansion is finite, i.e,



                $$f(x) = sum_{n=-k}^m a_i (x-alpha)^i$$



                where we assume $k>0$, and we let $a_{-k}=0$ if needed (I assume it only for convenience. Also note that if the sequence actually start with $ngeq 0$, $f$ is a polynomial and there's nothing to prove). Note that after multiplying by $(x-alpha)^k$ we get:



                $$(x-alpha)^kf(x)=sum_{n=-k}^m a_i (x-alpha)^{i+k}=sum_{n=0}^{m+k}a_i (x-alpha)^{i+k}$$



                RHS is a sum of polynomials, hence a polynomial $g(x)$, so, we may write $(x-alpha)^k f(x)=g(x)$ which gives $f(x)=g(x)/(x-alpha)^k$. After canceling factors, this is still the form of the rational function $f$ (maybe $k$ will be changed).



                The other way around, if $f(x)=g(x)/(x-alpha)^k$ for some $kgeq0$, where $g$ is a polynomial, note that $g$ can be written as $g(x)=g((x-alpha)+alpha)$. Now if we write $g$ explicitly as $g(x)=sum_{i=0}^m b_i x^i$, we get-



                $$g((x-alpha)+alpha)=g(x)=sum_{i=0}^m b_i ((x-alpha)+alpha)^i$$



                using the binomial theorem we get that $g(x)=sum_{i=0}^m b_i' (x-alpha)^i$ ($b_i'$ are the modified coefficients; we can calculate them explicitly, but this is not interesting). This gives
                $$f(x)=frac{g(x)}{(x-alpha)^k}=sum_{i=0}^m b_i' (x-alpha)^{i-k}$$



                And so the expansion is finite.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 7 '18 at 11:55









                MadarbMadarb

                346111




                346111






























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