Plotting a Maclaurin series












1












$begingroup$


Can anyone plot the degree-8 Maclaurin polynomial for the function $(x+1)/((x^2)-9)$ and function $(x+1)/((x^2)-9)$ on the same plot.



Please help, I've tried everything I could think of.






plotting series-expansion







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  • $begingroup$
    Possible duplicates: (7559), (130464), (158080), (160684)
    $endgroup$
    – Michael E2
    21 mins ago
















1












$begingroup$


Can anyone plot the degree-8 Maclaurin polynomial for the function $(x+1)/((x^2)-9)$ and function $(x+1)/((x^2)-9)$ on the same plot.



Please help, I've tried everything I could think of.






plotting series-expansion







share|improve this question









New contributor




nik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Possible duplicates: (7559), (130464), (158080), (160684)
    $endgroup$
    – Michael E2
    21 mins ago














1












1








1





$begingroup$


Can anyone plot the degree-8 Maclaurin polynomial for the function $(x+1)/((x^2)-9)$ and function $(x+1)/((x^2)-9)$ on the same plot.



Please help, I've tried everything I could think of.






plotting series-expansion







share|improve this question









New contributor




nik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Can anyone plot the degree-8 Maclaurin polynomial for the function $(x+1)/((x^2)-9)$ and function $(x+1)/((x^2)-9)$ on the same plot.



Please help, I've tried everything I could think of.






plotting series-expansion




plotting series-expansion






share|improve this question









New contributor




nik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









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nik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|improve this question




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edited 1 hour ago









m_goldberg

89k873200




89k873200






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asked 2 hours ago









niknik

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nik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





nik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






nik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    Possible duplicates: (7559), (130464), (158080), (160684)
    $endgroup$
    – Michael E2
    21 mins ago


















  • $begingroup$
    Possible duplicates: (7559), (130464), (158080), (160684)
    $endgroup$
    – Michael E2
    21 mins ago
















$begingroup$
Possible duplicates: (7559), (130464), (158080), (160684)
$endgroup$
– Michael E2
21 mins ago




$begingroup$
Possible duplicates: (7559), (130464), (158080), (160684)
$endgroup$
– Michael E2
21 mins ago










1 Answer
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$begingroup$

The problem is probably that Series returns a SeriesData object. It must be converted to a polynomial with Normal before it can be plotted.



f = (x + 1)/((x^2) - 9);

p = Normal[Series[f, {x, 0, 8}]]

Plot[{f, p}, {x, -2, 2}]


enter image description here






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    $begingroup$

    The problem is probably that Series returns a SeriesData object. It must be converted to a polynomial with Normal before it can be plotted.



    f = (x + 1)/((x^2) - 9);
    
p = Normal[Series[f, {x, 0, 8}]]
    
Plot[{f, p}, {x, -2, 2}]


    enter image description here






    share|improve this answer









    $endgroup$


















      2












      $begingroup$

      The problem is probably that Series returns a SeriesData object. It must be converted to a polynomial with Normal before it can be plotted.



      f = (x + 1)/((x^2) - 9);
      
p = Normal[Series[f, {x, 0, 8}]]
      
Plot[{f, p}, {x, -2, 2}]


      enter image description here






      share|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        The problem is probably that Series returns a SeriesData object. It must be converted to a polynomial with Normal before it can be plotted.



        f = (x + 1)/((x^2) - 9);
        
p = Normal[Series[f, {x, 0, 8}]]
        
Plot[{f, p}, {x, -2, 2}]


        enter image description here






        share|improve this answer









        $endgroup$



        The problem is probably that Series returns a SeriesData object. It must be converted to a polynomial with Normal before it can be plotted.



        f = (x + 1)/((x^2) - 9);
        
p = Normal[Series[f, {x, 0, 8}]]
        
Plot[{f, p}, {x, -2, 2}]


        enter image description here







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 1 hour ago









        Henrik SchumacherHenrik Schumacher

        60.7k585171




        60.7k585171






















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