Calculate $sin(0.2)$ to 4 decimal places with its Taylor series and the Lagrange error bound












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$begingroup$


So, I have to estimate the value of sin(0,2) with at least 4 correct decimal places by using its Taylor series and the Lagrange error bound, but I am stuck in what I believe is a trivial step.



I understand the following:



$$sin(x)=sum_{n=0}^∞(-1)^nfrac{x^{2n+1}}{(2n+1)!}$$



$$f(x)=P_{n}(x) + R_{n}(x)$$ so, $$sin(0,2)=P_{n}(0,2)+R_{n}(0,2)$$



By having that, I assume that I should find the value for n so that $$f^{(n+1}(x)leq M$$ where M (I guess?) is 1, the maximum value for $sin(x)$



Knowing that, My Lagrange error should be:



$$R_n(x)leqfrac{M(x)^{n+1}}{(n+1)!}$$
so:
$$R_n(0,2)leqfrac{(0,2)^{n+1}}{(n+1)!}leq 0,0001$$
By checking values for n, I found that for n=3, its value is 0,00006..., and so I chose 3 as the degree of the Taylor polinomial that should give me an error smaller than 0,0001



Now I am stuck at calculating sin(0,2), which I assume it should be like this, where I do not know how to express Rn(x), nor if the n I chose is correct, since:
$$sin(0,2)=frac{(0,2)^{2n+1}}{(2n+1)!}=frac{(0,2)^{(2*3)+1}}{((2*3)+1)!}=2.539682539682542*10^{-9}$$
which is nowhere near its "correct" value, which I found that it's best approximated by making n=0:
$$sin(0,2)=frac{(0,2)^{2n+1}}{(2n+1)!}=frac{(0,2)}{1!}=0,2$$



If I take n=0, then I do not know how to even start writing Rn(x):
$$sin(0,2)=frac{(0,2)^{2n+1}}{(2n+1)!}=frac{(0,2)}{1!}=>0,2-R_n(x)=0,2-what?$$



Any help and corrections are greatly appreciated, I am also unsure if I chose the value for n correctly.






calculus taylor-expansion







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    1












    $begingroup$


    So, I have to estimate the value of sin(0,2) with at least 4 correct decimal places by using its Taylor series and the Lagrange error bound, but I am stuck in what I believe is a trivial step.



    I understand the following:



    $$sin(x)=sum_{n=0}^∞(-1)^nfrac{x^{2n+1}}{(2n+1)!}$$



    $$f(x)=P_{n}(x) + R_{n}(x)$$ so, $$sin(0,2)=P_{n}(0,2)+R_{n}(0,2)$$



    By having that, I assume that I should find the value for n so that $$f^{(n+1}(x)leq M$$ where M (I guess?) is 1, the maximum value for $sin(x)$



    Knowing that, My Lagrange error should be:



    $$R_n(x)leqfrac{M(x)^{n+1}}{(n+1)!}$$
    so:
    $$R_n(0,2)leqfrac{(0,2)^{n+1}}{(n+1)!}leq 0,0001$$
    By checking values for n, I found that for n=3, its value is 0,00006..., and so I chose 3 as the degree of the Taylor polinomial that should give me an error smaller than 0,0001



    Now I am stuck at calculating sin(0,2), which I assume it should be like this, where I do not know how to express Rn(x), nor if the n I chose is correct, since:
    $$sin(0,2)=frac{(0,2)^{2n+1}}{(2n+1)!}=frac{(0,2)^{(2*3)+1}}{((2*3)+1)!}=2.539682539682542*10^{-9}$$
    which is nowhere near its "correct" value, which I found that it's best approximated by making n=0:
    $$sin(0,2)=frac{(0,2)^{2n+1}}{(2n+1)!}=frac{(0,2)}{1!}=0,2$$



    If I take n=0, then I do not know how to even start writing Rn(x):
    $$sin(0,2)=frac{(0,2)^{2n+1}}{(2n+1)!}=frac{(0,2)}{1!}=>0,2-R_n(x)=0,2-what?$$



    Any help and corrections are greatly appreciated, I am also unsure if I chose the value for n correctly.






    calculus taylor-expansion







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      2



      $begingroup$


      So, I have to estimate the value of sin(0,2) with at least 4 correct decimal places by using its Taylor series and the Lagrange error bound, but I am stuck in what I believe is a trivial step.



      I understand the following:



      $$sin(x)=sum_{n=0}^∞(-1)^nfrac{x^{2n+1}}{(2n+1)!}$$



      $$f(x)=P_{n}(x) + R_{n}(x)$$ so, $$sin(0,2)=P_{n}(0,2)+R_{n}(0,2)$$



      By having that, I assume that I should find the value for n so that $$f^{(n+1}(x)leq M$$ where M (I guess?) is 1, the maximum value for $sin(x)$



      Knowing that, My Lagrange error should be:



      $$R_n(x)leqfrac{M(x)^{n+1}}{(n+1)!}$$
      so:
      $$R_n(0,2)leqfrac{(0,2)^{n+1}}{(n+1)!}leq 0,0001$$
      By checking values for n, I found that for n=3, its value is 0,00006..., and so I chose 3 as the degree of the Taylor polinomial that should give me an error smaller than 0,0001



      Now I am stuck at calculating sin(0,2), which I assume it should be like this, where I do not know how to express Rn(x), nor if the n I chose is correct, since:
      $$sin(0,2)=frac{(0,2)^{2n+1}}{(2n+1)!}=frac{(0,2)^{(2*3)+1}}{((2*3)+1)!}=2.539682539682542*10^{-9}$$
      which is nowhere near its "correct" value, which I found that it's best approximated by making n=0:
      $$sin(0,2)=frac{(0,2)^{2n+1}}{(2n+1)!}=frac{(0,2)}{1!}=0,2$$



      If I take n=0, then I do not know how to even start writing Rn(x):
      $$sin(0,2)=frac{(0,2)^{2n+1}}{(2n+1)!}=frac{(0,2)}{1!}=>0,2-R_n(x)=0,2-what?$$



      Any help and corrections are greatly appreciated, I am also unsure if I chose the value for n correctly.






      calculus taylor-expansion







      share|cite|improve this question











      $endgroup$




      So, I have to estimate the value of sin(0,2) with at least 4 correct decimal places by using its Taylor series and the Lagrange error bound, but I am stuck in what I believe is a trivial step.



      I understand the following:



      $$sin(x)=sum_{n=0}^∞(-1)^nfrac{x^{2n+1}}{(2n+1)!}$$



      $$f(x)=P_{n}(x) + R_{n}(x)$$ so, $$sin(0,2)=P_{n}(0,2)+R_{n}(0,2)$$



      By having that, I assume that I should find the value for n so that $$f^{(n+1}(x)leq M$$ where M (I guess?) is 1, the maximum value for $sin(x)$



      Knowing that, My Lagrange error should be:



      $$R_n(x)leqfrac{M(x)^{n+1}}{(n+1)!}$$
      so:
      $$R_n(0,2)leqfrac{(0,2)^{n+1}}{(n+1)!}leq 0,0001$$
      By checking values for n, I found that for n=3, its value is 0,00006..., and so I chose 3 as the degree of the Taylor polinomial that should give me an error smaller than 0,0001



      Now I am stuck at calculating sin(0,2), which I assume it should be like this, where I do not know how to express Rn(x), nor if the n I chose is correct, since:
      $$sin(0,2)=frac{(0,2)^{2n+1}}{(2n+1)!}=frac{(0,2)^{(2*3)+1}}{((2*3)+1)!}=2.539682539682542*10^{-9}$$
      which is nowhere near its "correct" value, which I found that it's best approximated by making n=0:
      $$sin(0,2)=frac{(0,2)^{2n+1}}{(2n+1)!}=frac{(0,2)}{1!}=0,2$$



      If I take n=0, then I do not know how to even start writing Rn(x):
      $$sin(0,2)=frac{(0,2)^{2n+1}}{(2n+1)!}=frac{(0,2)}{1!}=>0,2-R_n(x)=0,2-what?$$



      Any help and corrections are greatly appreciated, I am also unsure if I chose the value for n correctly.






      calculus taylor-expansion




      calculus taylor-expansion






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      edited Dec 30 '18 at 13:35









      AmateurMathPirate

      1,336721




      1,336721










      asked Dec 30 '18 at 12:49









      LightsongLightsong

      246




      246






















          3 Answers
          3






          active

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          0












          $begingroup$

          It looks like you're using the 3rd-order term, but you need to use the whole series up to the 3rd-order term. Since the 0th- and 2nd- order terms are $0$, you just need to add the 1st-order term, which is $x$.



          Second, to get 4 correct decimal places, the error really needs to be less than $0.00005.$



          Third, if you're using $2n+1$ as the index in your series, be sure to use $2n+1$ in the remainder term, rather than just $n$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you! I had not realized that I had to use all the terms up to the nth one, hence the confusion when I was checking other examples on how to do it.
            $endgroup$
            – Lightsong
            Dec 30 '18 at 16:20



















          1












          $begingroup$

          Since you found that $n=3$ is the degree of the Taylor polynomial that satisfies the error bound, your approximation should be
          $$sin(0.2)=0.2-frac{0.2^3}{3!},$$
          in which you use $P_3(x)$ to approximate $sin(x)$, based on your notation.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            $$sin x=x-frac{x^3}{3!}+R_n$$
            You calculated that for $n=3$ $R_n<varepsilon$ where $varepsilon$ is the error bound you have. Then $n=3$ is the degree of the Taylor polynomial that satisfies that error bound, so put $x=0.2$ into $P_n$ since that is what approximates the function, while $R_n$ tells you how good is the approximation of $P_n$.



            You'll get
            $$sin(0.2)=0.198669ldotssimeq P_3=0.1986666ldots$$






            share|cite|improve this answer









            $endgroup$













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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              0












              $begingroup$

              It looks like you're using the 3rd-order term, but you need to use the whole series up to the 3rd-order term. Since the 0th- and 2nd- order terms are $0$, you just need to add the 1st-order term, which is $x$.



              Second, to get 4 correct decimal places, the error really needs to be less than $0.00005.$



              Third, if you're using $2n+1$ as the index in your series, be sure to use $2n+1$ in the remainder term, rather than just $n$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Thank you! I had not realized that I had to use all the terms up to the nth one, hence the confusion when I was checking other examples on how to do it.
                $endgroup$
                – Lightsong
                Dec 30 '18 at 16:20
















              0












              $begingroup$

              It looks like you're using the 3rd-order term, but you need to use the whole series up to the 3rd-order term. Since the 0th- and 2nd- order terms are $0$, you just need to add the 1st-order term, which is $x$.



              Second, to get 4 correct decimal places, the error really needs to be less than $0.00005.$



              Third, if you're using $2n+1$ as the index in your series, be sure to use $2n+1$ in the remainder term, rather than just $n$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Thank you! I had not realized that I had to use all the terms up to the nth one, hence the confusion when I was checking other examples on how to do it.
                $endgroup$
                – Lightsong
                Dec 30 '18 at 16:20














              0












              0








              0





              $begingroup$

              It looks like you're using the 3rd-order term, but you need to use the whole series up to the 3rd-order term. Since the 0th- and 2nd- order terms are $0$, you just need to add the 1st-order term, which is $x$.



              Second, to get 4 correct decimal places, the error really needs to be less than $0.00005.$



              Third, if you're using $2n+1$ as the index in your series, be sure to use $2n+1$ in the remainder term, rather than just $n$.






              share|cite|improve this answer









              $endgroup$



              It looks like you're using the 3rd-order term, but you need to use the whole series up to the 3rd-order term. Since the 0th- and 2nd- order terms are $0$, you just need to add the 1st-order term, which is $x$.



              Second, to get 4 correct decimal places, the error really needs to be less than $0.00005.$



              Third, if you're using $2n+1$ as the index in your series, be sure to use $2n+1$ in the remainder term, rather than just $n$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 30 '18 at 14:12









              B. GoddardB. Goddard

              19.8k21442




              19.8k21442












              • $begingroup$
                Thank you! I had not realized that I had to use all the terms up to the nth one, hence the confusion when I was checking other examples on how to do it.
                $endgroup$
                – Lightsong
                Dec 30 '18 at 16:20


















              • $begingroup$
                Thank you! I had not realized that I had to use all the terms up to the nth one, hence the confusion when I was checking other examples on how to do it.
                $endgroup$
                – Lightsong
                Dec 30 '18 at 16:20
















              $begingroup$
              Thank you! I had not realized that I had to use all the terms up to the nth one, hence the confusion when I was checking other examples on how to do it.
              $endgroup$
              – Lightsong
              Dec 30 '18 at 16:20




              $begingroup$
              Thank you! I had not realized that I had to use all the terms up to the nth one, hence the confusion when I was checking other examples on how to do it.
              $endgroup$
              – Lightsong
              Dec 30 '18 at 16:20











              1












              $begingroup$

              Since you found that $n=3$ is the degree of the Taylor polynomial that satisfies the error bound, your approximation should be
              $$sin(0.2)=0.2-frac{0.2^3}{3!},$$
              in which you use $P_3(x)$ to approximate $sin(x)$, based on your notation.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Since you found that $n=3$ is the degree of the Taylor polynomial that satisfies the error bound, your approximation should be
                $$sin(0.2)=0.2-frac{0.2^3}{3!},$$
                in which you use $P_3(x)$ to approximate $sin(x)$, based on your notation.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Since you found that $n=3$ is the degree of the Taylor polynomial that satisfies the error bound, your approximation should be
                  $$sin(0.2)=0.2-frac{0.2^3}{3!},$$
                  in which you use $P_3(x)$ to approximate $sin(x)$, based on your notation.






                  share|cite|improve this answer









                  $endgroup$



                  Since you found that $n=3$ is the degree of the Taylor polynomial that satisfies the error bound, your approximation should be
                  $$sin(0.2)=0.2-frac{0.2^3}{3!},$$
                  in which you use $P_3(x)$ to approximate $sin(x)$, based on your notation.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 30 '18 at 13:18









                  KittyLKittyL

                  13.8k31534




                  13.8k31534























                      1












                      $begingroup$

                      $$sin x=x-frac{x^3}{3!}+R_n$$
                      You calculated that for $n=3$ $R_n<varepsilon$ where $varepsilon$ is the error bound you have. Then $n=3$ is the degree of the Taylor polynomial that satisfies that error bound, so put $x=0.2$ into $P_n$ since that is what approximates the function, while $R_n$ tells you how good is the approximation of $P_n$.



                      You'll get
                      $$sin(0.2)=0.198669ldotssimeq P_3=0.1986666ldots$$






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        $$sin x=x-frac{x^3}{3!}+R_n$$
                        You calculated that for $n=3$ $R_n<varepsilon$ where $varepsilon$ is the error bound you have. Then $n=3$ is the degree of the Taylor polynomial that satisfies that error bound, so put $x=0.2$ into $P_n$ since that is what approximates the function, while $R_n$ tells you how good is the approximation of $P_n$.



                        You'll get
                        $$sin(0.2)=0.198669ldotssimeq P_3=0.1986666ldots$$






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          $$sin x=x-frac{x^3}{3!}+R_n$$
                          You calculated that for $n=3$ $R_n<varepsilon$ where $varepsilon$ is the error bound you have. Then $n=3$ is the degree of the Taylor polynomial that satisfies that error bound, so put $x=0.2$ into $P_n$ since that is what approximates the function, while $R_n$ tells you how good is the approximation of $P_n$.



                          You'll get
                          $$sin(0.2)=0.198669ldotssimeq P_3=0.1986666ldots$$






                          share|cite|improve this answer









                          $endgroup$



                          $$sin x=x-frac{x^3}{3!}+R_n$$
                          You calculated that for $n=3$ $R_n<varepsilon$ where $varepsilon$ is the error bound you have. Then $n=3$ is the degree of the Taylor polynomial that satisfies that error bound, so put $x=0.2$ into $P_n$ since that is what approximates the function, while $R_n$ tells you how good is the approximation of $P_n$.



                          You'll get
                          $$sin(0.2)=0.198669ldotssimeq P_3=0.1986666ldots$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 30 '18 at 13:32









                          Lorenzo B.Lorenzo B.

                          1,8602520




                          1,8602520






























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