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












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:



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










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    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:



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










    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:



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










      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:



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







      sequences-and-series






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      share|cite|improve this question













      share|cite|improve this question




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      edited 5 secs ago









      TheSimpliFire

      13.2k62464




      13.2k62464










      asked 1 hour ago









      Jwan622Jwan622

      2,41111632




      2,41111632






















          2 Answers
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          2












          $begingroup$

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



          Use directly




          • classical Taylor expansion :


          $$ln(1+x)=x-tfrac{x^2}{2}+tfrac{x^3}{3}-tfrac{x^4}{4}-...tag{1}$$



          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



          $$tfrac{0.4^n}{n}leq 0.001$$



          Due to the fact that $tfrac{0.4^6}{6} < 0.001 < tfrac{0.4^5}{5}$, we have $n=6$.




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




          Verification : $ln(1.4)= 0.3364722366...$ whereas
          $0.4-tfrac{0.4^2}{2}+tfrac{0.4^3}{3}-tfrac{0.4^4}{4}+tfrac{0.4^5}{5}=0.3369813333...$. 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 $frac1{1+x}$, and from the geometric series
            $$ frac1{1+x}=sum_{k=0}^infty (-1)^kx^k,$$
            you ought to see that $a_k=frac{(-1)^{k-1}}k$ 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 $frac{0.4^n}{n}<0.001$ and then take $sum_{k=0}^{n-1}frac{-(-0.4)^k}k$ 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









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              2 Answers
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              active

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              2 Answers
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              2












              $begingroup$

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



              Use directly




              • classical Taylor expansion :


              $$ln(1+x)=x-tfrac{x^2}{2}+tfrac{x^3}{3}-tfrac{x^4}{4}-...tag{1}$$



              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



              $$tfrac{0.4^n}{n}leq 0.001$$



              Due to the fact that $tfrac{0.4^6}{6} < 0.001 < tfrac{0.4^5}{5}$, we have $n=6$.




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




              Verification : $ln(1.4)= 0.3364722366...$ whereas
              $0.4-tfrac{0.4^2}{2}+tfrac{0.4^3}{3}-tfrac{0.4^4}{4}+tfrac{0.4^5}{5}=0.3369813333...$. 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 :


                $$ln(1+x)=x-tfrac{x^2}{2}+tfrac{x^3}{3}-tfrac{x^4}{4}-...tag{1}$$



                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



                $$tfrac{0.4^n}{n}leq 0.001$$



                Due to the fact that $tfrac{0.4^6}{6} < 0.001 < tfrac{0.4^5}{5}$, we have $n=6$.




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




                Verification : $ln(1.4)= 0.3364722366...$ whereas
                $0.4-tfrac{0.4^2}{2}+tfrac{0.4^3}{3}-tfrac{0.4^4}{4}+tfrac{0.4^5}{5}=0.3369813333...$. 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 :


                  $$ln(1+x)=x-tfrac{x^2}{2}+tfrac{x^3}{3}-tfrac{x^4}{4}-...tag{1}$$



                  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



                  $$tfrac{0.4^n}{n}leq 0.001$$



                  Due to the fact that $tfrac{0.4^6}{6} < 0.001 < tfrac{0.4^5}{5}$, we have $n=6$.




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




                  Verification : $ln(1.4)= 0.3364722366...$ whereas
                  $0.4-tfrac{0.4^2}{2}+tfrac{0.4^3}{3}-tfrac{0.4^4}{4}+tfrac{0.4^5}{5}=0.3369813333...$. 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 :


                  $$ln(1+x)=x-tfrac{x^2}{2}+tfrac{x^3}{3}-tfrac{x^4}{4}-...tag{1}$$



                  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



                  $$tfrac{0.4^n}{n}leq 0.001$$



                  Due to the fact that $tfrac{0.4^6}{6} < 0.001 < tfrac{0.4^5}{5}$, we have $n=6$.




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




                  Verification : $ln(1.4)= 0.3364722366...$ whereas
                  $0.4-tfrac{0.4^2}{2}+tfrac{0.4^3}{3}-tfrac{0.4^4}{4}+tfrac{0.4^5}{5}=0.3369813333...$. 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 9 mins ago

























                  answered 59 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 $frac1{1+x}$, and from the geometric series
                      $$ frac1{1+x}=sum_{k=0}^infty (-1)^kx^k,$$
                      you ought to see that $a_k=frac{(-1)^{k-1}}k$ 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 $frac{0.4^n}{n}<0.001$ and then take $sum_{k=0}^{n-1}frac{-(-0.4)^k}k$ 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 $frac1{1+x}$, and from the geometric series
                        $$ frac1{1+x}=sum_{k=0}^infty (-1)^kx^k,$$
                        you ought to see that $a_k=frac{(-1)^{k-1}}k$ 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 $frac{0.4^n}{n}<0.001$ and then take $sum_{k=0}^{n-1}frac{-(-0.4)^k}k$ 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 $frac1{1+x}$, and from the geometric series
                          $$ frac1{1+x}=sum_{k=0}^infty (-1)^kx^k,$$
                          you ought to see that $a_k=frac{(-1)^{k-1}}k$ 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 $frac{0.4^n}{n}<0.001$ and then take $sum_{k=0}^{n-1}frac{-(-0.4)^k}k$ 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 $frac1{1+x}$, and from the geometric series
                          $$ frac1{1+x}=sum_{k=0}^infty (-1)^kx^k,$$
                          you ought to see that $a_k=frac{(-1)^{k-1}}k$ 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 $frac{0.4^n}{n}<0.001$ and then take $sum_{k=0}^{n-1}frac{-(-0.4)^k}k$ 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 53 mins ago









                          Hagen von EitzenHagen von Eitzen

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