IB Maths AA HL 5.19 Maclaurin series

[Maximum number: 11]

The following question uses Maclaurin series to investigate approximations of mathematical constants and the accuracies of such approximations.

(a)

Using $x=\frac{1}{\sqrt{3}}$ and the first three (non-zero) terms of the Maclaurin series of $\arctan x$ , find an approximation for $\pi$ to three decimal places.

The Maclaurin series of $\arctan x$ is an example of an alternating series, ie a series where consecutive terms are positive and negative. Consider the following theorem.

Theorem: For alternating series with terms of decreasing magnitude, the error obtained in using a finite number of terms is less than or equal to the absolute value of the next term in the sequence.

Using the theorem, the maximum error in using the first three (non-zero) terms as an approximation to $\arctan x$ is given by $\left|-\frac{x^{7}}{7}\right|$ . In other words, $\left|\arctan x-\left(x-\frac{x^{3}}{3}+\frac{x^{5}}{5}\right)\right| \leq\left|-\frac{x^{7}}{7}\right|$ .

[ 3 ]

$\quad \arctan \frac{1}{\sqrt{3}}=\frac{\pi}{6}\left(\approx \frac{1}{\sqrt{3}}-\frac{1}{3}\left(\frac{1}{\sqrt{3}}\right)^{3}+\frac{1}{5}\left(\frac{1}{\sqrt{3}}\right)^{5}\right)$
attempt to evaluate $\frac{1}{\sqrt{3}}-\frac{1}{3}\left(\frac{1}{\sqrt{3}}\right)^{3}+\frac{1}{5}\left(\frac{1}{\sqrt{3}}\right)^{5}$ $\pi \approx 3.156$

(b)

Determine how many (non-zero) terms of the series would need to be used, such that the error in approximating $\arctan \left(\frac{1}{\sqrt{3}}\right)$ is less than 0.0001 .

[ 3 ]

METHOD 1
recognition that in the Maclaurin expansion of $\arctan x$ , the error term is $\frac{x^{2 m-1}}{2 m-1}$
attempt to solve $\frac{1}{2 m-1} \times\left(\frac{1}{\sqrt{3}}\right)^{2 m-1}<0.0001$ OR $m=6.60583 \ldots(=7)$

6 (non-zero) terms are needed

Note: Accept the error term in an absolute value.
Give marks as above for candidates who use 2 m+1 instead of 2 m-1.
This gives $m=5.60583 \ldots(=6)$ and leads to needing 6 terms.
Give marks as above for candidates who use n as the exponent and divisor. This gives $n=12.211675 \ldots$ and leads to needing 6 terms.

METHOD 2
recognition that in the Maclaurin expansion of $\arctan x$ , the error term is $\frac{x^{2 m-1}}{2 m-1}$
attempt to find the value of the error term for at least two consecutive values of m
error terms are
$\frac{\left(\frac{1}{\sqrt{3}}\right)^{7}}{7}=0.003054 \ldots, \frac{\left(\frac{1}{\sqrt{3}}\right)^{9}}{9}=0.0007919 \ldots, \frac{\left(\frac{1}{\sqrt{3}}\right)^{11}}{11}=0.0002159 \ldots, \frac{\left(\frac{1}{\sqrt{3}}\right)^{13}}{13}=0.00006092 \ldots$

6 (non-zero) terms are needed

Note: Accept the error term in an absolute value.

Accept 2 m+1 or n in place of 2 m-1.

METHOD 3
attempt to find the actual error using $\arctan x-$ (first m non-zero terms)
attempt to find the error for at least two consecutive values of m
$(m=4 \Rightarrow 0.000623, m=5 \Rightarrow 0.000169, m=6 \Rightarrow 0.0000473)$

6 (non-zero) terms are needed

(c)

Determine the smallest number of (non-zero) terms of the Maclaurin series for $\arctan x$ that should be used.

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METHOD 1
recognition error is of the form $\int_{0}^{\frac{1}{\sqrt{3}}}\left(\frac{x^{n}}{n}\right) \mathrm{d} x$

Note: Condone absence of limits of integration for (M1)

\frac{\left(\frac{1}{\sqrt{3}}\right)^{n+1}}{n(n+1)}<1 \times 10^{-6}

attempt to solve using GDC
$n=14.3336 \ldots$ $\Rightarrow n=15$
(hence $8^{\text {th }}$ term is the error term so) we require 7 (non-zero) terms

METHOD 2
recognition error is of the form $\int_{0}^{\frac{1}{\sqrt{3}}} \frac{x^{2 m-1}}{2 m-1} d x \operatorname{OR} \int_{0}^{\frac{1}{\sqrt{3}}} \frac{x^{2 m+1}}{2 m+1} d x$

Note: Condone absence of limits of integration for (M1)

\frac{\left(\frac{1}{\sqrt{3}}\right)^{2 m}}{2 m(2 m-1)}<1 \times 10^{-6} \text { OR } \frac{\left(\frac{1}{\sqrt{3}}\right)^{2 m+2}}{(2 m+1)(2 m+2)}<1 \times 10^{-6}

attempt to solve using GDC

\Rightarrow m=7.66681 \ldots \text { OR } \Rightarrow m=6.66681 \ldots

(hence $8^{\text {th }}$ term is the error term so) we require 7 (non-zero) terms

METHOD 3
recognition error is of the form $\int_{0}^{\frac{1}{\sqrt{3}}}\left(\frac{x^{n}}{n}\right) \mathrm{d} x$

Note: Condone absence of limits of integration for (M1)

\int_{0}^{\frac{1}{\sqrt{3}}}\left(\frac{x^{13}}{13}\right) \mathrm{d} x=2.51234 \ldots \times 10^{-6} \text { OR } \int_{0}^{\frac{1}{\sqrt{3}}}\left(\frac{x^{15}}{15}\right) \mathrm{d} x=6.35065 \ldots \times 10^{-7}

attempt to find the value of the integral for at least two consecutive odd values of n, n>9
$\int_{0}^{\frac{1}{\sqrt{3}}}\left(\frac{x^{13}}{13}\right) \mathrm{d} x=2.51234 \ldots \times 10^{-6}, \int_{0}^{\frac{1}{\sqrt{3}}}\left(\frac{x^{15}}{15}\right) \mathrm{d} x=6.35065 \ldots \times 10^{-7}$

Note: Award this A1 for the second correct value.
(hence $8^{\text {th }}$ term is the error term so) we require 7 (non-zero) terms

METHOD 4
attempt to find the actual error using $\int_{0}^{\frac{1}{\sqrt{3}}} \arctan x d x-$ (first m non-zero terms)
error with 6 terms is $2.00830 \ldots \times 10^{-6} \mathbf{OR}$ error with 7 terms is $5.04041 \ldots \times 10^{-7}$
attempt to find the error for at least two consecutive values of m
error with 6 terms is $2.00830 \ldots \times 10^{-6}$ and error with 7 terms is $5.04041 \ldots \times 10^{-7}$

Note: Award this A1 for the second correct value.
we require 7 (non-zero) terms

IB Maths AA HL Ahl 5 19 Hl Maclaurin Series Questions

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IB Maths AA HL Ahl 5 19 Hl Maclaurin Series Questions question 1

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Question (b)

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