International Baccalaureate Mathematics

Syllabus Content

Maclaurin series to obtain expansions for e^x, sinx, cosx, ln(1+x), (1+x)^p, p ∈ Q
Use of simple substitution, products, integration and differentiation to obtain other series.
Maclaurin series developed from differential equations.

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"(a) Write down the first three terms of the binomial expansion of $ (1 - a)^{-1} $ in ascending powers of $ a $." ... more

"(a) Find the first two non-zero terms in the Maclaurin series of" ... more

"The function $f$ is defined by $f(x) = e^x \sin{x} $, where $x \in \mathbb{R}$." ... more

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Calculus It is said that the word calculus comes from the Latin word for the small pebble used for counting and calculations. The two major branches, differentiation and integration, are studied by pupils only towards the end of their school days but does then form a major part of their studies. A course in calculus is a prerequisite for other, more advanced courses in mathematical analysis. Isaac Newton is best known for developing calculus. He was born in England in 1642 and is considered one of the most influential scientists and mathematicians in history.

Furthermore

Official Guidance, clarification and syllabus links:

Example: for substitution: replace $x$ with $x^2$ to define the Maclaurin series for $e^{x^2}$.

Example: the expansion of $e^x \sin{x}$.

Formula Booklet:

Maclaurin series

Maclaurin series for special functions

$$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots$$ $$e^x \approx 1 + x + \frac{x^2}{2!} + \cdots$$ $$\ln(1+x) \approx x - \frac{x^2}{2}+\frac{x^3}{3}-\cdots$$ $$\sin(x) \approx x - \frac{x^3}{3!} + \frac{x^5}{5!}-\cdots$$ $$\cos(x) \approx 1 - \frac{x^2}{2!} + \frac{x^4}{4!}-\cdots$$ $$\arctan(x) \approx x - \frac{x^3}{3} + \frac{x^5}{5}-\cdots$$

The Maclaurin series is a special case of the Taylor series, which is a way of representing a function as an infinite sum of terms involving its derivatives. The Maclaurin series specifically represents a function as a power series centered at x = 0, and is useful for approximating the value of a function near x = 0.

The formula for the Maclaurin series is:

$$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n$$

where $f^{(n)}(0)$denotes the n-th derivative of $f(x)$ evaluated at 0, and $n!$ denotes the factorial of $n$. In other words, the Maclaurin series represents the function $f(x)$ as an infinite sum of terms involving its derivatives evaluated at $x = 0$.

For example, let us find the Maclaurin series for the function $e^x$:

We start by computing the derivatives of $e^x$:

$$f(x) = e^x$$ $$f'(x) = e^x$$ $$f''(x) = e^x$$ $$f'''(x) = e^x$$

and so on. Evaluating each of these derivatives at x = 0, we get:

$$f(0) = e^0 = 1$$ $$f'(0) = e^0 = 1$$ $$f''(0) = e^0 = 1$$ $$f'''(0) = e^0 = 1$$

Substituting these values into the formula for the Maclaurin series, we get:

$$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$

which is the Maclaurin series for $e^x$ and can also be written as:

$$e^x \approx \sum_{n=0}^5 \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} \dotsb$$

The Maclaurin series can be used to approximate the value of a function near x = 0 by truncating the infinite sum at a certain point. The accuracy of the approximation depends on the number of terms included in the sum.

The following series are also provided in the formula booklet.

$$\ln(1+x) \approx \sum_{n=1}^6 (-1)^{n+1} \frac{x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6}$$ $$\sin(x) \approx \sum_{n=0}^5 (-1)^n \frac{x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \frac{x^{11}}{11!}$$ $$\cos(x) \approx \sum_{n=0}^5 (-1)^n \frac{x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \frac{x^{10}}{10!}$$ $$\arctan(x) \approx \sum_{n=0}^5 (-1)^n \frac{x^{2n+1}}{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \frac{x^{11}}{11}$$

The nth degree MacLaurin polynomial approximation to $f$ is:

$$M_n(x) = \sum_{k=0}^n \frac{f^{(k)}(0)}{k!} x^k$$

Finding the values of $x$ for which a Maclaurin Series converges involves determining the interval of convergence. The radius of convergence can be found using the ratio test. Just like with a geometric sequence, the series converges if the common ratio is between plus one and minus one. As there isn't always a 'common' ratio we can look at the ratio of a pair of consecutive terms as $n$ tends to infinity.

Example 1: Find the interval of convergence for the Maclaurin series of $ f(x) = \frac{1}{1 - x} $.
The Maclaurin series for this function is the geometric series: $$ f(x) = 1 + x + x^2 + x^3 + \cdots $$ To find the interval of convergence, use the ratio test: $$ \lim_{n \to \infty} \left| \frac{x^{n+1}}{x^n} \right| = |x| $$ The series converges when $ |x| < 1 $, so the interval of convergence is $ -1 < x < 1 $.

Example 2: Determine the values for which the Maclaurin series for $ e^{-x^2} $ converges.
The Maclaurin series for $ e^{-x^2} $ is: $$ e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots $$ Using the ratio test: $$ \lim_{n \to \infty} \left| \frac{(-1)^{n+1} \frac{x^{2n+2}}{(n+1)!}}{(-1)^n \frac{x^{2n}}{n!}} \right| = \lim_{n \to \infty} \frac{x^2}{n+1} = 0 $$ Since this limit is zero for all $ x $, the series converges for all real numbers.

Example 3: Find the convergence of the Maclaurin series for $ \ln(1 + x) $.
The Maclaurin series for $ \ln(1 + x) $ is: $$ \ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots $$ To find its interval of convergence, we examine the function's domain. The function $ \ln(1 + x) $ is defined for $ x > -1 $. Also, using the ratio test or analyzing the series as an alternating series shows that it converges when $ -1 < x \leq 1 $. Therefore, the interval of convergence for the Maclaurin series of $ \ln(1 + x) $ is $ -1 < x \leq 1 $.

Composite Maclaurin Series involve forming a new series by substituting another Maclaurin series into a given one. For instance, substituting the Maclaurin series of $ \sin x $ into that of $ e^x $ yields: $$ e^{\sin x} = 1 + \sin x + \frac{(\sin x)^2}{2!} + \frac{(\sin x)^3}{3!} + \cdots $$ This is a composite series where the inner series $ \sin x $ is expanded as $ x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots $.

Adding and subtracting Maclaurin Series can be done term-by-term. Consider the Maclaurin series for $ e^x $ and $ \cos x $: $$ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots $$ $$ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots $$ Adding these series, we get: $$ e^x + \cos x = 2 + x - \frac{x^4}{4!} + \frac{x^5}{5!} - \frac{x^6}{3!} + \cdots $$ Note that terms with the same power of $ x $ are combined.

Maclaurin Series Introduction. A good video from Mr Flynn showing how the Maclaurin Series approximates some functions very accurately.

Desmos demo of the Maclaurin series for sin(x)

Jim Simons, Twitter

Tuesday, May 21, 2024

Everyone knows e^(-1/x²) as an example of when the Maclaurin series doesn't work (all it's derivatives are 0 at the origin, but it isn't the 0 function). But have you ever looked at its graph to see just how extraordinarily flat it is at the origin? pic.twitter.com/MBrfn0RGrn
— Jim Simons (@pippinsboss) May 19, 2024

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