Finding the first non-zero terms of a power series












1












$begingroup$


I have the function:
$f(x) = frac{30}{(x^2 + 1)(x^2-9)}$



I need to find the first four non-zero terms of the power series centered at zero. I have not had much experience with power series so I am not sure how to start/complete this problem.






power-series







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$endgroup$

















    1












    $begingroup$


    I have the function:
    $f(x) = frac{30}{(x^2 + 1)(x^2-9)}$



    I need to find the first four non-zero terms of the power series centered at zero. I have not had much experience with power series so I am not sure how to start/complete this problem.






    power-series







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I have the function:
      $f(x) = frac{30}{(x^2 + 1)(x^2-9)}$



      I need to find the first four non-zero terms of the power series centered at zero. I have not had much experience with power series so I am not sure how to start/complete this problem.






      power-series







      share|cite|improve this question









      $endgroup$




      I have the function:
      $f(x) = frac{30}{(x^2 + 1)(x^2-9)}$



      I need to find the first four non-zero terms of the power series centered at zero. I have not had much experience with power series so I am not sure how to start/complete this problem.






      power-series




      power-series






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 7 '18 at 22:17









      ElijahElijah

      626




      626






















          2 Answers
          2






          active

          oldest

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          1












          $begingroup$

          $f(x)=frac{-3}{x^2+1}+frac{3}{x^2-9}$



          $frac{-3}{x^2+1}=-3(1-x^2+x^4-x^6+....)$



          $frac{3}{x^2-9}=frac{-1/3}{1-x^2/9}=(-1/3)(1+x^2/9+x^4/81+x^6/729+.....)$



          Combine to get $f(x)=-4/3+(80/27)x^2-........$



          I'll let you finish.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I am very sorry, this is a completely new topic to me. How would I go about solving this then? would the first term be the -4/3+(80/27)(x)^2 and substituting something in for x? Again this is a new topic for me so I don't know the proper techniques.
            $endgroup$
            – Elijah
            Dec 7 '18 at 22:46










          • $begingroup$
            @Elijah What I gave you are the first two terms of what you are looking for. To get the next two just add the coefficients of the $x^4$ and $x^6$ terms from the two given expressions. $-3-1/241$ and $3-1/2187$.
            $endgroup$
            – herb steinberg
            Dec 7 '18 at 23:10










          • $begingroup$
            I like this so much that I discarded my purely formal approach.
            $endgroup$
            – Lubin
            Dec 7 '18 at 23:17



















          0












          $begingroup$

          You can start by converting to rational parts.



          $displaystyle f(x)=frac a{x-3}+frac b{x+3}+frac c{x^2+1}$ in order to have a sum of usual power series :



          http://hardycalculus.com/calcindex/IE_powerseriestables.htm



          Then substitute in $displaystyle sum (pm 1)^nu^n=frac 1{1mp u}$ by factoring $3$ and get $u=frac x3$.



          For the other series substitute $u=x^2$ instead.



          Gather all the coefficients with the same power of $x$ to get the final power series $sum a_nx^n$ and determine its radius of convergence.



          Edit: I just saw your problem is just determining first coefficients, not all of them. But even though the methodology presented here allow to find the first coefficients quite quickly, just limit yourself to small powers of $x$.



          An alternative method is to calculate $f(0),f'(0),f''(0), ...$ and so on.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Separating $1/(x^2-9)$ adds unnecessary steps, as it's already in the form $1/(u+b)$.
            $endgroup$
            – eyeballfrog
            Dec 7 '18 at 22:40













          Your Answer





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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          $f(x)=frac{-3}{x^2+1}+frac{3}{x^2-9}$



          $frac{-3}{x^2+1}=-3(1-x^2+x^4-x^6+....)$



          $frac{3}{x^2-9}=frac{-1/3}{1-x^2/9}=(-1/3)(1+x^2/9+x^4/81+x^6/729+.....)$



          Combine to get $f(x)=-4/3+(80/27)x^2-........$



          I'll let you finish.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I am very sorry, this is a completely new topic to me. How would I go about solving this then? would the first term be the -4/3+(80/27)(x)^2 and substituting something in for x? Again this is a new topic for me so I don't know the proper techniques.
            $endgroup$
            – Elijah
            Dec 7 '18 at 22:46










          • $begingroup$
            @Elijah What I gave you are the first two terms of what you are looking for. To get the next two just add the coefficients of the $x^4$ and $x^6$ terms from the two given expressions. $-3-1/241$ and $3-1/2187$.
            $endgroup$
            – herb steinberg
            Dec 7 '18 at 23:10










          • $begingroup$
            I like this so much that I discarded my purely formal approach.
            $endgroup$
            – Lubin
            Dec 7 '18 at 23:17
















          1












          $begingroup$

          $f(x)=frac{-3}{x^2+1}+frac{3}{x^2-9}$



          $frac{-3}{x^2+1}=-3(1-x^2+x^4-x^6+....)$



          $frac{3}{x^2-9}=frac{-1/3}{1-x^2/9}=(-1/3)(1+x^2/9+x^4/81+x^6/729+.....)$



          Combine to get $f(x)=-4/3+(80/27)x^2-........$



          I'll let you finish.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I am very sorry, this is a completely new topic to me. How would I go about solving this then? would the first term be the -4/3+(80/27)(x)^2 and substituting something in for x? Again this is a new topic for me so I don't know the proper techniques.
            $endgroup$
            – Elijah
            Dec 7 '18 at 22:46










          • $begingroup$
            @Elijah What I gave you are the first two terms of what you are looking for. To get the next two just add the coefficients of the $x^4$ and $x^6$ terms from the two given expressions. $-3-1/241$ and $3-1/2187$.
            $endgroup$
            – herb steinberg
            Dec 7 '18 at 23:10










          • $begingroup$
            I like this so much that I discarded my purely formal approach.
            $endgroup$
            – Lubin
            Dec 7 '18 at 23:17














          1












          1








          1





          $begingroup$

          $f(x)=frac{-3}{x^2+1}+frac{3}{x^2-9}$



          $frac{-3}{x^2+1}=-3(1-x^2+x^4-x^6+....)$



          $frac{3}{x^2-9}=frac{-1/3}{1-x^2/9}=(-1/3)(1+x^2/9+x^4/81+x^6/729+.....)$



          Combine to get $f(x)=-4/3+(80/27)x^2-........$



          I'll let you finish.






          share|cite|improve this answer









          $endgroup$



          $f(x)=frac{-3}{x^2+1}+frac{3}{x^2-9}$



          $frac{-3}{x^2+1}=-3(1-x^2+x^4-x^6+....)$



          $frac{3}{x^2-9}=frac{-1/3}{1-x^2/9}=(-1/3)(1+x^2/9+x^4/81+x^6/729+.....)$



          Combine to get $f(x)=-4/3+(80/27)x^2-........$



          I'll let you finish.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 7 '18 at 22:42









          herb steinbergherb steinberg

          2,7182310




          2,7182310












          • $begingroup$
            I am very sorry, this is a completely new topic to me. How would I go about solving this then? would the first term be the -4/3+(80/27)(x)^2 and substituting something in for x? Again this is a new topic for me so I don't know the proper techniques.
            $endgroup$
            – Elijah
            Dec 7 '18 at 22:46










          • $begingroup$
            @Elijah What I gave you are the first two terms of what you are looking for. To get the next two just add the coefficients of the $x^4$ and $x^6$ terms from the two given expressions. $-3-1/241$ and $3-1/2187$.
            $endgroup$
            – herb steinberg
            Dec 7 '18 at 23:10










          • $begingroup$
            I like this so much that I discarded my purely formal approach.
            $endgroup$
            – Lubin
            Dec 7 '18 at 23:17


















          • $begingroup$
            I am very sorry, this is a completely new topic to me. How would I go about solving this then? would the first term be the -4/3+(80/27)(x)^2 and substituting something in for x? Again this is a new topic for me so I don't know the proper techniques.
            $endgroup$
            – Elijah
            Dec 7 '18 at 22:46










          • $begingroup$
            @Elijah What I gave you are the first two terms of what you are looking for. To get the next two just add the coefficients of the $x^4$ and $x^6$ terms from the two given expressions. $-3-1/241$ and $3-1/2187$.
            $endgroup$
            – herb steinberg
            Dec 7 '18 at 23:10










          • $begingroup$
            I like this so much that I discarded my purely formal approach.
            $endgroup$
            – Lubin
            Dec 7 '18 at 23:17
















          $begingroup$
          I am very sorry, this is a completely new topic to me. How would I go about solving this then? would the first term be the -4/3+(80/27)(x)^2 and substituting something in for x? Again this is a new topic for me so I don't know the proper techniques.
          $endgroup$
          – Elijah
          Dec 7 '18 at 22:46




          $begingroup$
          I am very sorry, this is a completely new topic to me. How would I go about solving this then? would the first term be the -4/3+(80/27)(x)^2 and substituting something in for x? Again this is a new topic for me so I don't know the proper techniques.
          $endgroup$
          – Elijah
          Dec 7 '18 at 22:46












          $begingroup$
          @Elijah What I gave you are the first two terms of what you are looking for. To get the next two just add the coefficients of the $x^4$ and $x^6$ terms from the two given expressions. $-3-1/241$ and $3-1/2187$.
          $endgroup$
          – herb steinberg
          Dec 7 '18 at 23:10




          $begingroup$
          @Elijah What I gave you are the first two terms of what you are looking for. To get the next two just add the coefficients of the $x^4$ and $x^6$ terms from the two given expressions. $-3-1/241$ and $3-1/2187$.
          $endgroup$
          – herb steinberg
          Dec 7 '18 at 23:10












          $begingroup$
          I like this so much that I discarded my purely formal approach.
          $endgroup$
          – Lubin
          Dec 7 '18 at 23:17




          $begingroup$
          I like this so much that I discarded my purely formal approach.
          $endgroup$
          – Lubin
          Dec 7 '18 at 23:17











          0












          $begingroup$

          You can start by converting to rational parts.



          $displaystyle f(x)=frac a{x-3}+frac b{x+3}+frac c{x^2+1}$ in order to have a sum of usual power series :



          http://hardycalculus.com/calcindex/IE_powerseriestables.htm



          Then substitute in $displaystyle sum (pm 1)^nu^n=frac 1{1mp u}$ by factoring $3$ and get $u=frac x3$.



          For the other series substitute $u=x^2$ instead.



          Gather all the coefficients with the same power of $x$ to get the final power series $sum a_nx^n$ and determine its radius of convergence.



          Edit: I just saw your problem is just determining first coefficients, not all of them. But even though the methodology presented here allow to find the first coefficients quite quickly, just limit yourself to small powers of $x$.



          An alternative method is to calculate $f(0),f'(0),f''(0), ...$ and so on.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Separating $1/(x^2-9)$ adds unnecessary steps, as it's already in the form $1/(u+b)$.
            $endgroup$
            – eyeballfrog
            Dec 7 '18 at 22:40


















          0












          $begingroup$

          You can start by converting to rational parts.



          $displaystyle f(x)=frac a{x-3}+frac b{x+3}+frac c{x^2+1}$ in order to have a sum of usual power series :



          http://hardycalculus.com/calcindex/IE_powerseriestables.htm



          Then substitute in $displaystyle sum (pm 1)^nu^n=frac 1{1mp u}$ by factoring $3$ and get $u=frac x3$.



          For the other series substitute $u=x^2$ instead.



          Gather all the coefficients with the same power of $x$ to get the final power series $sum a_nx^n$ and determine its radius of convergence.



          Edit: I just saw your problem is just determining first coefficients, not all of them. But even though the methodology presented here allow to find the first coefficients quite quickly, just limit yourself to small powers of $x$.



          An alternative method is to calculate $f(0),f'(0),f''(0), ...$ and so on.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Separating $1/(x^2-9)$ adds unnecessary steps, as it's already in the form $1/(u+b)$.
            $endgroup$
            – eyeballfrog
            Dec 7 '18 at 22:40
















          0












          0








          0





          $begingroup$

          You can start by converting to rational parts.



          $displaystyle f(x)=frac a{x-3}+frac b{x+3}+frac c{x^2+1}$ in order to have a sum of usual power series :



          http://hardycalculus.com/calcindex/IE_powerseriestables.htm



          Then substitute in $displaystyle sum (pm 1)^nu^n=frac 1{1mp u}$ by factoring $3$ and get $u=frac x3$.



          For the other series substitute $u=x^2$ instead.



          Gather all the coefficients with the same power of $x$ to get the final power series $sum a_nx^n$ and determine its radius of convergence.



          Edit: I just saw your problem is just determining first coefficients, not all of them. But even though the methodology presented here allow to find the first coefficients quite quickly, just limit yourself to small powers of $x$.



          An alternative method is to calculate $f(0),f'(0),f''(0), ...$ and so on.






          share|cite|improve this answer











          $endgroup$



          You can start by converting to rational parts.



          $displaystyle f(x)=frac a{x-3}+frac b{x+3}+frac c{x^2+1}$ in order to have a sum of usual power series :



          http://hardycalculus.com/calcindex/IE_powerseriestables.htm



          Then substitute in $displaystyle sum (pm 1)^nu^n=frac 1{1mp u}$ by factoring $3$ and get $u=frac x3$.



          For the other series substitute $u=x^2$ instead.



          Gather all the coefficients with the same power of $x$ to get the final power series $sum a_nx^n$ and determine its radius of convergence.



          Edit: I just saw your problem is just determining first coefficients, not all of them. But even though the methodology presented here allow to find the first coefficients quite quickly, just limit yourself to small powers of $x$.



          An alternative method is to calculate $f(0),f'(0),f''(0), ...$ and so on.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 7 '18 at 22:41

























          answered Dec 7 '18 at 22:33









          zwimzwim

          12k730




          12k730








          • 1




            $begingroup$
            Separating $1/(x^2-9)$ adds unnecessary steps, as it's already in the form $1/(u+b)$.
            $endgroup$
            – eyeballfrog
            Dec 7 '18 at 22:40
















          • 1




            $begingroup$
            Separating $1/(x^2-9)$ adds unnecessary steps, as it's already in the form $1/(u+b)$.
            $endgroup$
            – eyeballfrog
            Dec 7 '18 at 22:40










          1




          1




          $begingroup$
          Separating $1/(x^2-9)$ adds unnecessary steps, as it's already in the form $1/(u+b)$.
          $endgroup$
          – eyeballfrog
          Dec 7 '18 at 22:40






          $begingroup$
          Separating $1/(x^2-9)$ adds unnecessary steps, as it's already in the form $1/(u+b)$.
          $endgroup$
          – eyeballfrog
          Dec 7 '18 at 22:40




















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