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DP IB Maths: AA HL

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5.11 MacLaurin Series

Question 1a

Marks: 4

Consider the general Maclaurin series formula

 space f open parentheses x close parentheses equals f open parentheses 0 close parentheses plus x f to the power of apostrophe open parentheses 0 close parentheses plus fraction numerator x squared over denominator 2 factorial end fraction f to the power of apostrophe apostrophe end exponent open parentheses 0 close parentheses plus   horizontal ellipsis   plus fraction numerator x to the power of n over denominator n factorial end fraction f to the power of open parentheses n close parentheses end exponent open parentheses 0 close parentheses plus   horizontal ellipsis

(wherespace f to the power of open parentheses n close parentheses end exponent indicates the n to the power of t h end exponentderivative ofspace f).

a)
Use the formula to find the first five terms of the Maclaurin series for e to the power of 2 x end exponent.

 

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    Question 1b

    Marks: 2
    b)
    Hence approximate the value of e to the power of 2 x end exponent when x equals 1.
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      Question 1c

      Marks: 3
      c)
      (i)
      Compare the approximation found in part (b) to the exact value of e to the power of 2 x end exponent when x equals 1.

      (ii)
      Explain how the accuracy of the Maclaurin series approximation could be improved.

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        Question 1d

        Marks: 2
        d)
        Use the general Maclaurin series formula to show that the general term of the Maclaurin series for e to the power of 2 x end exponent is 

        fraction numerator open parentheses 2 x close parentheses to the power of n over denominator n factorial end fraction

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          Question 2a

          Marks: 3
          a)
          Use substitution into the Maclaurin series for sin space x

          sin space x equals x minus fraction numerator x cubed over denominator 3 factorial end fraction plus fraction numerator x to the power of 5 over denominator 5 factorial end fraction minus   horizontal ellipsis 

          to find the first four terms of the Maclaurin series for sin space open parentheses x over 2 close parentheses.
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            Question 2b

            Marks: 3
            b)
            Hence approximate the value of sin space pi over 2 and compare this approximation to the exact value.
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              Question 2c

              Marks: 2
              c)
              Without performing any additional calculations, explain whether the answer to part (a) would be expected to give an approximation of sin space pi over 4 that is more accurate or less accurate than its approximation for sin pi over 2.
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                Question 3a

                Marks: 4

                The Maclaurin series for e to the power of xand sin space x are

                e to the power of x equals 1 plus x plus fraction numerator x squared over denominator 2 factorial end fraction plus   horizontal ellipsis blank and          sin space x equals x minus fraction numerator x cubed over denominator 3 factorial end fraction plus fraction numerator x to the power of 5 over denominator 5 factorial end fraction minus   horizontal ellipsis

                a)
                Find the Maclaurin series for e to the power of x s i n space x up to and including the term in x to the power of 4.
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                  Question 3b

                  Marks: 3
                  b)
                  Use the Maclaurin series for sin space x, along with the fact that begin mathsize 16px style   fraction numerator straight d over denominator straight d x end fraction open parentheses sin space x close parentheses equals cos space x end style,  to find the first four terms of the Maclaurin series for cos space x.
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                    Question 4a

                    Marks: 4
                    a)
                    Use the general Maclaurin series formula to find the first four terms of the Maclaurin series for  begin mathsize 16px style fraction numerator 1 over denominator 1 plus x end fraction end style.
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                      Question 4b

                      Marks: 3
                      b)
                      Confirm that the answer to part (a) matches the first four terms of the binomial theorem expansion of begin mathsize 16px style ​ fraction numerator 1 over denominator 1 plus x end fraction end style .
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                        Question 4c

                        Marks: 2

                        The Maclaurin series for ln space left parenthesis 1 plus x right parenthesis is 

                        ln open parentheses 1 plus x close parentheses equals x minus x squared over 2 plus x cubed over 3 minus   horizontal ellipsis

                        c)
                        Differentiate the Maclaurin series for ln space left parenthesis 1 plus x right parenthesis up to its fourth term and compare this to the answer from part (a). Give an explanation for any similarities that are found.
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                          Question 5a

                          Marks: 3
                          a)
                          Use the Maclaurin series for sin space x and cos space x to find a Maclaurin series approximation for 2 space sin space x space cos space x up until the term in x to the power of 4.
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                            Question 5b

                            Marks: 3

                            The double angle identity for sine tells us that 

                            sin space 2 x equals 2 space sin space x space cos space x

                            b)
                            Use substitution into the Maclaurin series for sin space x to find a Maclaurin series approximation for sin space 2 x up until the term in x to the power of 4, and confirm that this matches the answer to part (a).
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                              Question 6a

                              Marks: 4
                              a)
                              Use the Binomial theorem to find a Maclaurin series for the functionspace f defined by 

                              space f open parentheses x close parentheses equals square root of 1 minus 2 x squared end root

                              Give the series up to and including the term in x to the power of 6.
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                                Question 6b

                                Marks: 2
                                b)
                                State any limitations on the validity of the series expansion found in part (a).
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                                  Question 6c

                                  Marks: 4
                                  c)
                                  Use the answer to part (a) to estimate the value of square root of 0.5 end root, and compare the accuracy of that estimated value to the actual value of square root of 0.5 end root.
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                                    Question 7a

                                    Marks: 4

                                    Consider the differential equation 

                                    y to the power of apostrophe equals 2 y squared plus x

                                    together with the initial condition y left parenthesis 0 right parenthesis equals 1.

                                    a)
                                    (i)
                                    Show that y double apostrophe equals 4 y y to the power of apostrophe plus 1.

                                    (ii)
                                    Use an equivalent method to find expressions for y to the power of apostrophe apostrophe apostrophe end exponenty to the power of left parenthesis 4 right parenthesis end exponent and y to the power of open parentheses 5 close parentheses end exponent. Each should be given in terms of y and of lower-order derivatives of y.
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                                      Question 7b

                                      Marks: 3
                                      b)
                                      Using the boundary condition above, calculate the values of y to the power of apostrophe open parentheses 0 close parentheses, y to the power of apostrophe apostrophe end exponent open parentheses 0 close parentheses, y to the power of apostrophe apostrophe apostrophe end exponent open parentheses 0 close parenthesesy to the power of open parentheses 4 close parentheses end exponent open parentheses 0 close parentheses and y to the power of open parentheses 5 close parentheses end exponent open parentheses 0 close parentheses.
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                                        Question 7c

                                        Marks: 4

                                        Let space f left parenthesis x right parenthesis be the solution to the differential equation above with the given boundary condition, so that y equals f left parenthesis x right parenthesis.

                                        c)
                                        Using the answers to part (b), find the first six terms of the Maclaurin series for space f left parenthesis x right parenthesis.
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                                          Question 7d

                                          Marks: 2
                                          d)
                                          Hence approximate the value of to 4 d.p. when x equals 0.1.
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                                            Question 8a

                                            Marks: 5

                                            Consider the differential equation

                                            y to the power of apostrophe equals 2 x y squared

                                            with the initial condition y left parenthesis 0 right parenthesis equals 1.

                                            a)
                                            (i)
                                            Find y to the power of apostrophe apostrophe end exponent.

                                            (ii)
                                            Hence show that y to the power of apostrophe apostrophe apostrophe end exponent equals 8 y y to the power of apostrophe plus 4 x open parentheses y to the power of apostrophe close parentheses squared plus 4 x y y to the power of apostrophe apostrophe end exponent  and
                                             
                                                     y to the power of open parentheses 4 close parentheses end exponent equals 12 open parentheses y to the power of apostrophe close parentheses squared plus 12 y y to the power of apostrophe apostrophe end exponent plus 12 x y to the power of apostrophe y to the power of apostrophe apostrophe end exponent plus 4 x y y to the power of apostrophe apostrophe apostrophe end exponent

                                             

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                                              Question 8b

                                              Marks: 4
                                              b)
                                              Use the results from part (a) along with the given initial condition to find a Maclaurin series to approximate the solution of the differential equation, giving the approximation up to the term in x to the power of 4.
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                                                Question 8c

                                                Marks: 4
                                                c)
                                                Use separation of variables to show that the exact solution of the differential equation with the given initial condition is 

                                                y equals fraction numerator 1 over denominator 1 minus x squared end fraction

                                                 

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                                                  Question 8d

                                                  Marks: 3
                                                  d)
                                                  Use the binomial theorem to find an approximation for begin mathsize 16px style ​ fraction numerator 1 over denominator 1 minus x squared end fraction end style up to the term in x to the power of 4, and verify that it matches the answer to part (b).
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