# Maclaurin series HL (TN)

#### Teacher Notes

HL syllabus content:  Maclaurin series; use of simple substitution, products, integration and differentiation to obtain other series; Maclaurin series developed from differential equations (AHL 5.19)

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In the Analysis & Approaches HL course, Maclaurin series can be taught informally without covering some of the formal elements of power series that were in the previous syllabus for the ‘old’ Maths HL Calculus option topic but are not in the current syllabus such as convergence tests, radius of convergence, interval of convergence and error terms. Because of the condensed and informal nature of Maclaurin series in the Analysis & Approaches HL syllabus, it should not require a great deal of instructional time to cover. It is necessary to have covered sequences & series and all the other content in the calculus topic including differential equations before teaching Maclaurin series.

Deriving polynomial approximations to certain functions was investigated by the English mathematician Brook Taylor (1685-1731) and by the Scottish mathematicians Colin Maclaurin (1698-1756) and James Gregory (1638-1675).

A Maclaurin series is a type of power series. It is important for students to have an informal understanding that a power series is essentially a polynomial with an infinite number of terms; however, we construct power series with a suitable number of finite terms in order to approximate certain functions. Compared to many other types of functions, polynomials are not only easily evaluated (only requiring addition and multiplication) but they are also easily differentiated and integrated. Thus, power series are useful for approximating other functions which may not be easily evaluated, differentiated or integrated.

Students are already familiar with infinite geometric series and that if the common ratio, r, has a value such that $$- 1 < r < 1$$ then the infinite geometric series converges to the sum $${S_\infty } = \frac{{{u_1}}}{{1 - r}}$$, where $${u_1}$$ is the first term in the series. I build on this previous knowledge and introduce power series by asking students to consider the following infinite polynomial.

$$1 + x + {x^2} + {x^3} + \; \cdots \; + {x^n} + \; \cdots$$

Clearly this is an infinite geometric series such that $${u_1} = 1$$ and $$r = x$$. Using the formula for $${S_\infty }$$, we get $$1 + x + {x^2} + {x^3} + \; \cdots \; + {x^n} + \; \cdots = \frac{1}{{1 - x}}$$. However, since an infinite geometric series converges to a finite sum only if $$- 1 < r < 1$$ then the power series $$1 + x + {x^2} + {x^3} + \; \cdots \; + {x^n} + \; \cdots$$ is a valid representation of $$\frac{1}{{1 - x}}$$ only when $$- 1 < x < 1$$; that is, $$\frac{1}{{1 - x}} = \sum\limits_{n = 0}^\infty {{x^n}}$$ for $$- 1 < x < 1$$. This power series can be used to generate approximations to $$\frac{1}{{1 - x}}$$ when x is close to zero. The graphs below illustrate how polynomial functions in the form $$y = 1 + x + {x^2} + {x^3} + \; \cdots \; + {x^n} + \; \cdots$$become better approximations of the function $$y = \frac{1}{{1 - x}}$$ in the region around $$x = 0$$ as more terms are added to the polynomial.

While some power series are valid representations of a function for any real number value of $$x$$, many power series – such as the example above – converge to the values of a function only within a specified interval of convergence. The syllabus content on Maclaurin series is relatively narrow compared to the 'old' syllabus content on power series in the HL Calculus option topic for the former Maths HL course. Even though interval of convergence is not in the current syllabus, I think it is appropriate to briefly discuss with students the fact that many power series are only reasonable approximations of a particular function for a limited range of values of $$x$$, and this can be effectively demonstrated with graphs such as the ones above. However, the issue of an interval of convergence will not appear on exam questions. The interval of convergence is not given for the power series listed in the formula booklet for the functions $${{\rm{e}}^x},\;\ln \left( {1 + x} \right),\;\sin x,\;\cos x$$ and $$\arctan x$$.

The two primary skills that students need to develop with respect to Maclaurin series are: (1) to use an existing Maclaurin series (such as one of the five listed in the formula booklet) to derive another Maclaurin series by means of a simple substitution, products, integration or differentiation; and (2) to develop a Maclaurin series from a given differential equation. Here are a couple of examples to illustrate these two skills.

skill:  use an existing Maclaurin series to derive another Maclaurin series by means of a simple substitution, products, integration or differentiation

1.   Find a Maclaurin series for $$\frac{1}{{{{\left( {1 - x} \right)}^2}}}$$ by differentiating the Maclaurin series for $$\frac{1}{{1 - x}}$$.

worked solution:

Although it is not in the formula booklet, students should recognize that $$\frac{1}{{1 - x}}$$ is the sum of the infinite geometric series (power series) $$1 + x + {x^2} + {x^3} + \; \cdots \; + {x^n} + \; \cdots$$.

$$\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {\frac{1}{{1 - x}}} \right) = \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {{{\left( {1 - x} \right)}^{ - 1}}} \right) = - {\left( {1 - x} \right)^{ - 2}}\left( { - 1} \right) = \frac{1}{{{{\left( {1 - x} \right)}^2}}}$$

Hence, the power series for $$\frac{1}{{{{\left( {1 - x} \right)}^2}}}$$ will be the derivative of $$1 + x + {x^2} + {x^3} + \; \cdots \; + {x^n} + \; \cdots$$

$$\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {1 + x + {x^2} + {x^3} + \; \cdots \; + {x^n} + \; \cdots } \right) = 0 + 1 + 2x + 3{x^2} + \; \cdots \; + \left( {n + 1} \right){x^n} + \; \cdots$$

Therefore, $$\frac{1}{{{{\left( {1 - x} \right)}^2}}} = 1 + 2x + 3{x^2} + \; \cdots \; + \left( {n + 1} \right){x^n} + \; \cdots = \sum\limits_{n = 0}^\infty {\left( {n + 1} \right){x^n}}$$

Note: The result does not have to be expressed in sigma notation, but it is worthwhile for students to practice using sigma notation.

skill:  develop a Maclaurin series from a given differential equation and initial value for $$y\left( 0 \right)$$

2.   Given the differential equation $$\frac{{{\rm{d}}y}}{{{\rm{d}}x}} - xy = x$$ with the initial condition $$y\left( 0 \right) = 1$$, find the first three non-zero terms of the Maclaurin series for $$y$$.

worked solution:

Apply the formula for Maclaurin series found in the formula booklet.

$$f\left( x \right) = f\left( 0 \right) + x \cdot f'\left( 0 \right) + \frac{{{x^2}}}{{2!}} \cdot f''\left( 0 \right) + \; \cdots$$

A Maclaurin series is being developed for the unknown function $$y$$ in the differential equation. The value of the function $$y$$ at $$x = 0$$ is given to be 1. It is necessary to find successive derivatives for $$y$$ and evaluate them at $$x = 0$$, $$y = 1$$. Then the required number of terms of the Maclaurin series will be found by substituting into the formula.

$$\frac{{{\rm{d}}y}}{{{\rm{d}}x}} - xy = x\;\;\; \Rightarrow \;\;\;\frac{{{\rm{d}}y}}{{{\rm{d}}x}} = xy + x$$

At $$\left( {0,\;1} \right)$$$$\frac{{{\rm{d}}y}}{{{\rm{d}}x}} = 0 + 0 = 0$$

$$\frac{{{{\rm{d}}^2}y}}{{{\rm{d}}{x^2}}} = y + x\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + 1\;\;\; \Rightarrow$$    At $$\left( {0,\;1} \right)$$$$\frac{{{{\rm{d}}^2}y}}{{{\rm{d}}{x^2}}} = 1 + 0 + 1 = 2$$

$$\frac{{{{\rm{d}}^3}y}}{{{\rm{d}}{x^3}}} = \frac{{{\rm{d}}y}}{{{\rm{d}}x}} + \frac{{{\rm{d}}y}}{{{\rm{d}}x}} + x\frac{{{{\rm{d}}^2}y}}{{{\rm{d}}{x^2}}}\;\;\; \Rightarrow$$    At $$\left( {0,\;1} \right)$$$$\frac{{{{\rm{d}}^3}y}}{{{\rm{d}}{x^3}}} = 0 + 0 + 0 = 0$$

$$\frac{{{{\rm{d}}^4}y}}{{{\rm{d}}{x^4}}} = \frac{{{{\rm{d}}^2}y}}{{{\rm{d}}{x^2}}} + \frac{{{{\rm{d}}^2}y}}{{{\rm{d}}{x^2}}} + \frac{{{{\rm{d}}^2}y}}{{{\rm{d}}{x^2}}} + x\frac{{{{\rm{d}}^3}y}}{{{\rm{d}}{x^3}}}\;\;\; \Rightarrow$$    At $$\left( {0,\;1} \right)$$$$\frac{{{{\rm{d}}^4}y}}{{{\rm{d}}{x^4}}} = 2 + 2 + 2 + 0 = 6$$

Substituting into the formula gives $$y = 1 + 0 + \frac{{{x^2}}}{{2!}} \cdot 2 + 0 + \frac{{{x^4}}}{{4!}} \cdot 6$$

Therefore the first three non-zero terms of the Maclaurin series for $$y$$ are:  $$y = 1 + {x^2} + \frac{1}{4}{x^4}$$

#### ♦  teaching materials  ♦

AA_HL_5.19_Maclaurin_1_v1
A set of three exercises on Maclaurin series. Worked solutions are included.

AA_HL_5.19_Maclaurin_2_v1
A set of four exercises on Maclaurin series. Answers are included.

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