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How many terms of the Maclaurin series for ln(1+x) do you need to use to estimate ln(1.4) to within 0.001?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Finding the first 5 terms of a Maclaurin Series using division?Estimate error using Taylor SeriesStrategy to recognize and solve sequence and series problems?Use the Maclaurin series for $cos(x)$to compute the value of $cos(5^circ)$ correct to five decimal places.Number of terms to estimate the integral within the indicated accuracy using series!User remainder theorem to estimate $ sum^infty_k=1 frac12(-1)^k+1k^2 $ within $frac25$How do you find nth maclaurin/taylor series or polynomial?Solving an integral using first two terms of the Mclaurin series for $f(x)$Show $log(1-x)=-sum_n=1^inftyfracx^nn,forall xin(-1,1)$.Calculating the approximation of an integral using the first four terms of a Maclaurin series.










1












$begingroup$


My question:




How many terms of the Maclaurin series for ln(1+x) do you need to use to estimate ln(1.4) to within 0.001?




I feel like I'm lost and my textbook that I'm using doesn't really go into how to solve problems like these in the section Taylor series and Maclaurin. Even worse, I'm a bit confused about all of this so I'm not sure where to even proceed.



So my process is this:



  • find the general formula for this series

  • If it's alternating, find the term so that the error is less than 0.001

  • I can also use the Taylor Inequality to estimate the error.

Where in the Stewart book does it even go into problems like these?



I have the derivatives of each:



$$f'(x) = frac11+x$$
$$f''(x) = frac-1(1+x)^2$$
$$f'''(x) = frac2(1+x)(1+x)^4 = frac2(1+x)^3$$
$$f^iv(x) = frac-6(1+x)^4$$










share|cite|improve this question









$endgroup$
















    1












    $begingroup$


    My question:




    How many terms of the Maclaurin series for ln(1+x) do you need to use to estimate ln(1.4) to within 0.001?




    I feel like I'm lost and my textbook that I'm using doesn't really go into how to solve problems like these in the section Taylor series and Maclaurin. Even worse, I'm a bit confused about all of this so I'm not sure where to even proceed.



    So my process is this:



    • find the general formula for this series

    • If it's alternating, find the term so that the error is less than 0.001

    • I can also use the Taylor Inequality to estimate the error.

    Where in the Stewart book does it even go into problems like these?



    I have the derivatives of each:



    $$f'(x) = frac11+x$$
    $$f''(x) = frac-1(1+x)^2$$
    $$f'''(x) = frac2(1+x)(1+x)^4 = frac2(1+x)^3$$
    $$f^iv(x) = frac-6(1+x)^4$$










    share|cite|improve this question









    $endgroup$














      1












      1








      1





      $begingroup$


      My question:




      How many terms of the Maclaurin series for ln(1+x) do you need to use to estimate ln(1.4) to within 0.001?




      I feel like I'm lost and my textbook that I'm using doesn't really go into how to solve problems like these in the section Taylor series and Maclaurin. Even worse, I'm a bit confused about all of this so I'm not sure where to even proceed.



      So my process is this:



      • find the general formula for this series

      • If it's alternating, find the term so that the error is less than 0.001

      • I can also use the Taylor Inequality to estimate the error.

      Where in the Stewart book does it even go into problems like these?



      I have the derivatives of each:



      $$f'(x) = frac11+x$$
      $$f''(x) = frac-1(1+x)^2$$
      $$f'''(x) = frac2(1+x)(1+x)^4 = frac2(1+x)^3$$
      $$f^iv(x) = frac-6(1+x)^4$$










      share|cite|improve this question









      $endgroup$




      My question:




      How many terms of the Maclaurin series for ln(1+x) do you need to use to estimate ln(1.4) to within 0.001?




      I feel like I'm lost and my textbook that I'm using doesn't really go into how to solve problems like these in the section Taylor series and Maclaurin. Even worse, I'm a bit confused about all of this so I'm not sure where to even proceed.



      So my process is this:



      • find the general formula for this series

      • If it's alternating, find the term so that the error is less than 0.001

      • I can also use the Taylor Inequality to estimate the error.

      Where in the Stewart book does it even go into problems like these?



      I have the derivatives of each:



      $$f'(x) = frac11+x$$
      $$f''(x) = frac-1(1+x)^2$$
      $$f'''(x) = frac2(1+x)(1+x)^4 = frac2(1+x)^3$$
      $$f^iv(x) = frac-6(1+x)^4$$







      sequences-and-series






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 1 hour ago









      Jwan622Jwan622

      2,41111632




      2,41111632




















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          You don't have to compute the derivatives as you do.



          Use directly



          • classical Taylor expansion :

          $$ln(1+x)=x-tfracx^22+tfracx^33-tfracx^44-...tag1$$



          with $x=0.4$ and



          • the theorem about alternating series saying that the error done by truncating this series is less that the module of the first discarded term (i.e., the first one not present in the truncated series).

          The condition is that this first discarded $n$th term is such



          $$tfrac0.4^nnleq 0.001$$



          Due to the fact that $tfrac0.4^66 < 0.001 < tfrac0.4^55$, we have $n=6$.




          Thus we need $n-1=5$ terms in expansion (1).




          Verification : $ln(1.4)= 0.3364722366...$ whereas
          $0.4-tfrac0.4^22+tfrac0.4^33-tfrac0.4^44+tfrac0.4^55=0.3369813333...$. The error $approx 0.0003$ is indeed less than $0.001$.






          share|cite|improve this answer











          $endgroup$




















            1












            $begingroup$

            First of all, you do need the MacLaurin series $$ tag1ln(1+x)=sum_k=0^infty a_kx^k.$$
            Given that the derivative of $ln(1+x)$ is $frac11+x$, and from the geometric series
            $$ frac11+x=sum_k=0^infty (-1)^kx^k,$$
            you ought to see that $a_k=frac(-1)^k-1k$ for $k>0$ (and $a_0=0$).
            In particular, for $0<x<1$, $(1)$ is an alternating series with secreasong (in absolute value) terms. Hence we can just stop when the further summands would be smaller than the desired error bound, without checking any specific error formula.
            In other words, we find $n$ with $frac0.4^nn<0.001$ and then take $sum_k=0^n-1frac-(-0.4)^kk$ as our approximation of $ln(1.4)$.



            Explicitly,
            $$ ln(1.4)approx 0.4-0.08+0.021overline3-0.0064+0.002048$$
            with the next summand $-0.000682overline6$ already small enough.






            share|cite|improve this answer









            $endgroup$













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              2 Answers
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              2 Answers
              2






              active

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              active

              oldest

              votes






              active

              oldest

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              2












              $begingroup$

              You don't have to compute the derivatives as you do.



              Use directly



              • classical Taylor expansion :

              $$ln(1+x)=x-tfracx^22+tfracx^33-tfracx^44-...tag1$$



              with $x=0.4$ and



              • the theorem about alternating series saying that the error done by truncating this series is less that the module of the first discarded term (i.e., the first one not present in the truncated series).

              The condition is that this first discarded $n$th term is such



              $$tfrac0.4^nnleq 0.001$$



              Due to the fact that $tfrac0.4^66 < 0.001 < tfrac0.4^55$, we have $n=6$.




              Thus we need $n-1=5$ terms in expansion (1).




              Verification : $ln(1.4)= 0.3364722366...$ whereas
              $0.4-tfrac0.4^22+tfrac0.4^33-tfrac0.4^44+tfrac0.4^55=0.3369813333...$. The error $approx 0.0003$ is indeed less than $0.001$.






              share|cite|improve this answer











              $endgroup$

















                2












                $begingroup$

                You don't have to compute the derivatives as you do.



                Use directly



                • classical Taylor expansion :

                $$ln(1+x)=x-tfracx^22+tfracx^33-tfracx^44-...tag1$$



                with $x=0.4$ and



                • the theorem about alternating series saying that the error done by truncating this series is less that the module of the first discarded term (i.e., the first one not present in the truncated series).

                The condition is that this first discarded $n$th term is such



                $$tfrac0.4^nnleq 0.001$$



                Due to the fact that $tfrac0.4^66 < 0.001 < tfrac0.4^55$, we have $n=6$.




                Thus we need $n-1=5$ terms in expansion (1).




                Verification : $ln(1.4)= 0.3364722366...$ whereas
                $0.4-tfrac0.4^22+tfrac0.4^33-tfrac0.4^44+tfrac0.4^55=0.3369813333...$. The error $approx 0.0003$ is indeed less than $0.001$.






                share|cite|improve this answer











                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  You don't have to compute the derivatives as you do.



                  Use directly



                  • classical Taylor expansion :

                  $$ln(1+x)=x-tfracx^22+tfracx^33-tfracx^44-...tag1$$



                  with $x=0.4$ and



                  • the theorem about alternating series saying that the error done by truncating this series is less that the module of the first discarded term (i.e., the first one not present in the truncated series).

                  The condition is that this first discarded $n$th term is such



                  $$tfrac0.4^nnleq 0.001$$



                  Due to the fact that $tfrac0.4^66 < 0.001 < tfrac0.4^55$, we have $n=6$.




                  Thus we need $n-1=5$ terms in expansion (1).




                  Verification : $ln(1.4)= 0.3364722366...$ whereas
                  $0.4-tfrac0.4^22+tfrac0.4^33-tfrac0.4^44+tfrac0.4^55=0.3369813333...$. The error $approx 0.0003$ is indeed less than $0.001$.






                  share|cite|improve this answer











                  $endgroup$



                  You don't have to compute the derivatives as you do.



                  Use directly



                  • classical Taylor expansion :

                  $$ln(1+x)=x-tfracx^22+tfracx^33-tfracx^44-...tag1$$



                  with $x=0.4$ and



                  • the theorem about alternating series saying that the error done by truncating this series is less that the module of the first discarded term (i.e., the first one not present in the truncated series).

                  The condition is that this first discarded $n$th term is such



                  $$tfrac0.4^nnleq 0.001$$



                  Due to the fact that $tfrac0.4^66 < 0.001 < tfrac0.4^55$, we have $n=6$.




                  Thus we need $n-1=5$ terms in expansion (1).




                  Verification : $ln(1.4)= 0.3364722366...$ whereas
                  $0.4-tfrac0.4^22+tfrac0.4^33-tfrac0.4^44+tfrac0.4^55=0.3369813333...$. The error $approx 0.0003$ is indeed less than $0.001$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 5 mins ago

























                  answered 55 mins ago









                  Jean MarieJean Marie

                  31.6k42355




                  31.6k42355





















                      1












                      $begingroup$

                      First of all, you do need the MacLaurin series $$ tag1ln(1+x)=sum_k=0^infty a_kx^k.$$
                      Given that the derivative of $ln(1+x)$ is $frac11+x$, and from the geometric series
                      $$ frac11+x=sum_k=0^infty (-1)^kx^k,$$
                      you ought to see that $a_k=frac(-1)^k-1k$ for $k>0$ (and $a_0=0$).
                      In particular, for $0<x<1$, $(1)$ is an alternating series with secreasong (in absolute value) terms. Hence we can just stop when the further summands would be smaller than the desired error bound, without checking any specific error formula.
                      In other words, we find $n$ with $frac0.4^nn<0.001$ and then take $sum_k=0^n-1frac-(-0.4)^kk$ as our approximation of $ln(1.4)$.



                      Explicitly,
                      $$ ln(1.4)approx 0.4-0.08+0.021overline3-0.0064+0.002048$$
                      with the next summand $-0.000682overline6$ already small enough.






                      share|cite|improve this answer









                      $endgroup$

















                        1












                        $begingroup$

                        First of all, you do need the MacLaurin series $$ tag1ln(1+x)=sum_k=0^infty a_kx^k.$$
                        Given that the derivative of $ln(1+x)$ is $frac11+x$, and from the geometric series
                        $$ frac11+x=sum_k=0^infty (-1)^kx^k,$$
                        you ought to see that $a_k=frac(-1)^k-1k$ for $k>0$ (and $a_0=0$).
                        In particular, for $0<x<1$, $(1)$ is an alternating series with secreasong (in absolute value) terms. Hence we can just stop when the further summands would be smaller than the desired error bound, without checking any specific error formula.
                        In other words, we find $n$ with $frac0.4^nn<0.001$ and then take $sum_k=0^n-1frac-(-0.4)^kk$ as our approximation of $ln(1.4)$.



                        Explicitly,
                        $$ ln(1.4)approx 0.4-0.08+0.021overline3-0.0064+0.002048$$
                        with the next summand $-0.000682overline6$ already small enough.






                        share|cite|improve this answer









                        $endgroup$















                          1












                          1








                          1





                          $begingroup$

                          First of all, you do need the MacLaurin series $$ tag1ln(1+x)=sum_k=0^infty a_kx^k.$$
                          Given that the derivative of $ln(1+x)$ is $frac11+x$, and from the geometric series
                          $$ frac11+x=sum_k=0^infty (-1)^kx^k,$$
                          you ought to see that $a_k=frac(-1)^k-1k$ for $k>0$ (and $a_0=0$).
                          In particular, for $0<x<1$, $(1)$ is an alternating series with secreasong (in absolute value) terms. Hence we can just stop when the further summands would be smaller than the desired error bound, without checking any specific error formula.
                          In other words, we find $n$ with $frac0.4^nn<0.001$ and then take $sum_k=0^n-1frac-(-0.4)^kk$ as our approximation of $ln(1.4)$.



                          Explicitly,
                          $$ ln(1.4)approx 0.4-0.08+0.021overline3-0.0064+0.002048$$
                          with the next summand $-0.000682overline6$ already small enough.






                          share|cite|improve this answer









                          $endgroup$



                          First of all, you do need the MacLaurin series $$ tag1ln(1+x)=sum_k=0^infty a_kx^k.$$
                          Given that the derivative of $ln(1+x)$ is $frac11+x$, and from the geometric series
                          $$ frac11+x=sum_k=0^infty (-1)^kx^k,$$
                          you ought to see that $a_k=frac(-1)^k-1k$ for $k>0$ (and $a_0=0$).
                          In particular, for $0<x<1$, $(1)$ is an alternating series with secreasong (in absolute value) terms. Hence we can just stop when the further summands would be smaller than the desired error bound, without checking any specific error formula.
                          In other words, we find $n$ with $frac0.4^nn<0.001$ and then take $sum_k=0^n-1frac-(-0.4)^kk$ as our approximation of $ln(1.4)$.



                          Explicitly,
                          $$ ln(1.4)approx 0.4-0.08+0.021overline3-0.0064+0.002048$$
                          with the next summand $-0.000682overline6$ already small enough.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 49 mins ago









                          Hagen von EitzenHagen von Eitzen

                          283k23273508




                          283k23273508



























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