Solution verification: finding a Maclaurin series for $f$, interval of convergence, and $f^(10)(0)$Question about Maclaurin Series for $cos x$Taylor and Maclaurin Series for $f(x)=e^x$Maclaurin series for $e^x +2e^-x$Maclaurin series of $sin(2pi x)$Finding interval of convergence for seriesMaclaurin Series expansion intervalMaclaurin Series for a natural logarithmMaclaurin series - Approximation and interval of convergenceMaclaurin series for lnMaclaurin Series from sin(x) to cos(x) using derivative

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Solution verification: finding a Maclaurin series for $f$, interval of convergence, and $f^(10)(0)$


Question about Maclaurin Series for $cos x$Taylor and Maclaurin Series for $f(x)=e^x$Maclaurin series for $e^x +2e^-x$Maclaurin series of $sin(2pi x)$Finding interval of convergence for seriesMaclaurin Series expansion intervalMaclaurin Series for a natural logarithmMaclaurin series - Approximation and interval of convergenceMaclaurin series for lnMaclaurin Series from sin(x) to cos(x) using derivative













2












$begingroup$


I have to find Maclaurin series for function $f(x)$ = $2x^2over16+x^4$, it's interval of convergence and $f^(10)(0)$. I managed to calculate Maclaurin series and $10^th$ derivative, but I'm not sure if it's done in proper way and if solution is correct.



On determining the series,



$$beginalign
f(x) &= frac2x^216+x^4 \
&= frac2x^216 cdot frac11-frac-x^416 \
&= frac2x^216 cdot sum_i=0^infty left(-fracx^416 right)^n \
&= frac2x^216 cdot sum_i=0^infty frac(-1)^n cdot x^4n16^n \
&= sum_i=0^infty frac(-1)^n cdot x^4n+22^4n+3
endalign$$



for $vert-x^4over16vert<1 implies xin(-2;2)$



On determining $f^(10)(0)$,



$$f^(10)(0)cdot frac x^10 10! = fracx^422^43 implies f^(10)(0) = x^32cdot frac10!2^43$$



Looking for feedback and opinion if the way I solved it is correct.






calculus proof-verification taylor-expansion







share|cite|improve this question











$endgroup$
















    2












    $begingroup$


    I have to find Maclaurin series for function $f(x)$ = $2x^2over16+x^4$, it's interval of convergence and $f^(10)(0)$. I managed to calculate Maclaurin series and $10^th$ derivative, but I'm not sure if it's done in proper way and if solution is correct.



    On determining the series,



    $$beginalign
    f(x) &= frac2x^216+x^4 \
    &= frac2x^216 cdot frac11-frac-x^416 \
    &= frac2x^216 cdot sum_i=0^infty left(-fracx^416 right)^n \
    &= frac2x^216 cdot sum_i=0^infty frac(-1)^n cdot x^4n16^n \
    &= sum_i=0^infty frac(-1)^n cdot x^4n+22^4n+3
    endalign$$



    for $vert-x^4over16vert<1 implies xin(-2;2)$



    On determining $f^(10)(0)$,



    $$f^(10)(0)cdot frac x^10 10! = fracx^422^43 implies f^(10)(0) = x^32cdot frac10!2^43$$



    Looking for feedback and opinion if the way I solved it is correct.






    calculus proof-verification taylor-expansion







    share|cite|improve this question











    $endgroup$














      2












      2








      2





      $begingroup$


      I have to find Maclaurin series for function $f(x)$ = $2x^2over16+x^4$, it's interval of convergence and $f^(10)(0)$. I managed to calculate Maclaurin series and $10^th$ derivative, but I'm not sure if it's done in proper way and if solution is correct.



      On determining the series,



      $$beginalign
      f(x) &= frac2x^216+x^4 \
      &= frac2x^216 cdot frac11-frac-x^416 \
      &= frac2x^216 cdot sum_i=0^infty left(-fracx^416 right)^n \
      &= frac2x^216 cdot sum_i=0^infty frac(-1)^n cdot x^4n16^n \
      &= sum_i=0^infty frac(-1)^n cdot x^4n+22^4n+3
      endalign$$



      for $vert-x^4over16vert<1 implies xin(-2;2)$



      On determining $f^(10)(0)$,



      $$f^(10)(0)cdot frac x^10 10! = fracx^422^43 implies f^(10)(0) = x^32cdot frac10!2^43$$



      Looking for feedback and opinion if the way I solved it is correct.






      calculus proof-verification taylor-expansion







      share|cite|improve this question











      $endgroup$




      I have to find Maclaurin series for function $f(x)$ = $2x^2over16+x^4$, it's interval of convergence and $f^(10)(0)$. I managed to calculate Maclaurin series and $10^th$ derivative, but I'm not sure if it's done in proper way and if solution is correct.



      On determining the series,



      $$beginalign
      f(x) &= frac2x^216+x^4 \
      &= frac2x^216 cdot frac11-frac-x^416 \
      &= frac2x^216 cdot sum_i=0^infty left(-fracx^416 right)^n \
      &= frac2x^216 cdot sum_i=0^infty frac(-1)^n cdot x^4n16^n \
      &= sum_i=0^infty frac(-1)^n cdot x^4n+22^4n+3
      endalign$$



      for $vert-x^4over16vert<1 implies xin(-2;2)$



      On determining $f^(10)(0)$,



      $$f^(10)(0)cdot frac x^10 10! = fracx^422^43 implies f^(10)(0) = x^32cdot frac10!2^43$$



      Looking for feedback and opinion if the way I solved it is correct.






      calculus proof-verification taylor-expansion




      calculus proof-verification taylor-expansion






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago









      Eevee Trainer

      7,80721339




      7,80721339










      asked 2 days ago









      MichaelMichael

      246




      246




















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Your computation of the MacLaurin series of $f$ looks right, but remember that $f^(10)(0)$ is a number. You know that$$fracf^(10)(0)10!=frac(-1)^22^11.$$Therefore,$$f^(10)(0)=frac10!2^11.$$



          Also, $leftlvertfrac-x^416rightrvert<1$ doesn't just imply that $xin(-2,2)$; it is actually equivalent to it.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            What happened to x is substituting it with 2 and -2 (what I got from interval of convergence)?
            $endgroup$
            – Michael
            2 days ago






          • 1




            $begingroup$
            The series diverges at those two points.
            $endgroup$
            – José Carlos Santos
            2 days ago










          • $begingroup$
            I’m ahead with the program and doing it Maclaurin on my own, so sorry for tedious questions. To get 10th derivative you input -2 or 2? Is that the way to calculate it?
            $endgroup$
            – Michael
            2 days ago










          • $begingroup$
            To get the $10^textth$ derivative, I put $n=2$, so that $x^4n+2$ becomes $x^10$.
            $endgroup$
            – José Carlos Santos
            2 days ago







          • 1




            $begingroup$
            I'm glad I could help.
            $endgroup$
            – José Carlos Santos
            2 days ago










          Your Answer





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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Your computation of the MacLaurin series of $f$ looks right, but remember that $f^(10)(0)$ is a number. You know that$$fracf^(10)(0)10!=frac(-1)^22^11.$$Therefore,$$f^(10)(0)=frac10!2^11.$$



          Also, $leftlvertfrac-x^416rightrvert<1$ doesn't just imply that $xin(-2,2)$; it is actually equivalent to it.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            What happened to x is substituting it with 2 and -2 (what I got from interval of convergence)?
            $endgroup$
            – Michael
            2 days ago






          • 1




            $begingroup$
            The series diverges at those two points.
            $endgroup$
            – José Carlos Santos
            2 days ago










          • $begingroup$
            I’m ahead with the program and doing it Maclaurin on my own, so sorry for tedious questions. To get 10th derivative you input -2 or 2? Is that the way to calculate it?
            $endgroup$
            – Michael
            2 days ago










          • $begingroup$
            To get the $10^textth$ derivative, I put $n=2$, so that $x^4n+2$ becomes $x^10$.
            $endgroup$
            – José Carlos Santos
            2 days ago







          • 1




            $begingroup$
            I'm glad I could help.
            $endgroup$
            – José Carlos Santos
            2 days ago















          2












          $begingroup$

          Your computation of the MacLaurin series of $f$ looks right, but remember that $f^(10)(0)$ is a number. You know that$$fracf^(10)(0)10!=frac(-1)^22^11.$$Therefore,$$f^(10)(0)=frac10!2^11.$$



          Also, $leftlvertfrac-x^416rightrvert<1$ doesn't just imply that $xin(-2,2)$; it is actually equivalent to it.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            What happened to x is substituting it with 2 and -2 (what I got from interval of convergence)?
            $endgroup$
            – Michael
            2 days ago






          • 1




            $begingroup$
            The series diverges at those two points.
            $endgroup$
            – José Carlos Santos
            2 days ago










          • $begingroup$
            I’m ahead with the program and doing it Maclaurin on my own, so sorry for tedious questions. To get 10th derivative you input -2 or 2? Is that the way to calculate it?
            $endgroup$
            – Michael
            2 days ago










          • $begingroup$
            To get the $10^textth$ derivative, I put $n=2$, so that $x^4n+2$ becomes $x^10$.
            $endgroup$
            – José Carlos Santos
            2 days ago







          • 1




            $begingroup$
            I'm glad I could help.
            $endgroup$
            – José Carlos Santos
            2 days ago













          2












          2








          2





          $begingroup$

          Your computation of the MacLaurin series of $f$ looks right, but remember that $f^(10)(0)$ is a number. You know that$$fracf^(10)(0)10!=frac(-1)^22^11.$$Therefore,$$f^(10)(0)=frac10!2^11.$$



          Also, $leftlvertfrac-x^416rightrvert<1$ doesn't just imply that $xin(-2,2)$; it is actually equivalent to it.






          share|cite|improve this answer









          $endgroup$



          Your computation of the MacLaurin series of $f$ looks right, but remember that $f^(10)(0)$ is a number. You know that$$fracf^(10)(0)10!=frac(-1)^22^11.$$Therefore,$$f^(10)(0)=frac10!2^11.$$



          Also, $leftlvertfrac-x^416rightrvert<1$ doesn't just imply that $xin(-2,2)$; it is actually equivalent to it.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          José Carlos SantosJosé Carlos Santos

          167k22132235




          167k22132235











          • $begingroup$
            What happened to x is substituting it with 2 and -2 (what I got from interval of convergence)?
            $endgroup$
            – Michael
            2 days ago






          • 1




            $begingroup$
            The series diverges at those two points.
            $endgroup$
            – José Carlos Santos
            2 days ago










          • $begingroup$
            I’m ahead with the program and doing it Maclaurin on my own, so sorry for tedious questions. To get 10th derivative you input -2 or 2? Is that the way to calculate it?
            $endgroup$
            – Michael
            2 days ago










          • $begingroup$
            To get the $10^textth$ derivative, I put $n=2$, so that $x^4n+2$ becomes $x^10$.
            $endgroup$
            – José Carlos Santos
            2 days ago







          • 1




            $begingroup$
            I'm glad I could help.
            $endgroup$
            – José Carlos Santos
            2 days ago
















          • $begingroup$
            What happened to x is substituting it with 2 and -2 (what I got from interval of convergence)?
            $endgroup$
            – Michael
            2 days ago






          • 1




            $begingroup$
            The series diverges at those two points.
            $endgroup$
            – José Carlos Santos
            2 days ago










          • $begingroup$
            I’m ahead with the program and doing it Maclaurin on my own, so sorry for tedious questions. To get 10th derivative you input -2 or 2? Is that the way to calculate it?
            $endgroup$
            – Michael
            2 days ago










          • $begingroup$
            To get the $10^textth$ derivative, I put $n=2$, so that $x^4n+2$ becomes $x^10$.
            $endgroup$
            – José Carlos Santos
            2 days ago







          • 1




            $begingroup$
            I'm glad I could help.
            $endgroup$
            – José Carlos Santos
            2 days ago















          $begingroup$
          What happened to x is substituting it with 2 and -2 (what I got from interval of convergence)?
          $endgroup$
          – Michael
          2 days ago




          $begingroup$
          What happened to x is substituting it with 2 and -2 (what I got from interval of convergence)?
          $endgroup$
          – Michael
          2 days ago




          1




          1




          $begingroup$
          The series diverges at those two points.
          $endgroup$
          – José Carlos Santos
          2 days ago




          $begingroup$
          The series diverges at those two points.
          $endgroup$
          – José Carlos Santos
          2 days ago












          $begingroup$
          I’m ahead with the program and doing it Maclaurin on my own, so sorry for tedious questions. To get 10th derivative you input -2 or 2? Is that the way to calculate it?
          $endgroup$
          – Michael
          2 days ago




          $begingroup$
          I’m ahead with the program and doing it Maclaurin on my own, so sorry for tedious questions. To get 10th derivative you input -2 or 2? Is that the way to calculate it?
          $endgroup$
          – Michael
          2 days ago












          $begingroup$
          To get the $10^textth$ derivative, I put $n=2$, so that $x^4n+2$ becomes $x^10$.
          $endgroup$
          – José Carlos Santos
          2 days ago





          $begingroup$
          To get the $10^textth$ derivative, I put $n=2$, so that $x^4n+2$ becomes $x^10$.
          $endgroup$
          – José Carlos Santos
          2 days ago





          1




          1




          $begingroup$
          I'm glad I could help.
          $endgroup$
          – José Carlos Santos
          2 days ago




          $begingroup$
          I'm glad I could help.
          $endgroup$
          – José Carlos Santos
          2 days ago

















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