IB Mathematics AA AHL Maclaurin series Study Notes - New Syllabus
IB Mathematics AA AHL Maclaurin series Study Notes
LEARNING OBJECTIVE
- Maclaurin series
Key Concepts:
- Maclaurin series
- Use of simple substitution
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Maclaurin Series
Maclaurin Series
The Maclaurin series is a special case of the Taylor series expanded about \( x = 0 \). It provides a way to represent functions as infinite sums of powers of \( x \).
The Maclaurin series of a function \( f(x) \) is given by:
\(\boxed{ f(x) = f(0) + f'(0) x + \frac{f”(0)}{2!} x^2 + \frac{f^{(3)}(0)}{3!} x^3 + \cdots = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n }\)
where \( f^{(n)}(0) \) is the \( n^{th} \) derivative of \( f(x) \) evaluated at \( x=0 \).
Below are some common Maclaurin series expansions:
1. Exponential function \( e^x \):
\( e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = \sum_{n=0}^\infty \frac{x^n}{n!} \)
2. Sine function \( \sin x \):
\( \sin x = x – \frac{x^3}{3!} + \frac{x^5}{5!} – \frac{x^7}{7!} + \cdots = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!} \)
3. Cosine function \( \cos x \):
\( \cos x = 1 – \frac{x^2}{2!} + \frac{x^4}{4!} – \frac{x^6}{6!} + \cdots = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!} \)
4. Natural logarithm \( \ln(1+x) \), valid for \( -1 < x \leq 1 \):
\( \ln(1+x) = x – \frac{x^2}{2} + \frac{x^3}{3} – \frac{x^4}{4} + \cdots = \sum_{n=1}^\infty (-1)^{n+1} \frac{x^n}{n} \)
5. General binomial expansion \( (1+x)^p \), for any rational \( p \in \mathbb{Q} \), valid for \( |x| < 1 \):
$(1+x)^p = 1 + p x + \frac{p(p-1)}{2!} x^2 + \frac{p(p-1)(p-2)}{3!} x^3 + \cdots = \sum_{n=0}^\infty \binom{p}{n} x^n $
where the generalized binomial coefficient is \( \displaystyle \binom{p}{n} = \frac{p(p-1)(p-2) \cdots (p-n+1)}{n!} \), with \( \binom{p}{0} = 1 \).
Example
Find the Maclaurin series expansion of \( f(x) = e^x \) up to the term in \( x^3 \).
▶️ Answer/Explanation
Solution:
Maclaurin series for \( e^x \):
\( e^x = \sum_{n=0}^\infty \frac{x^n}{n!} \)
terms up to \( n=3 \):
\( e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots \)
This is the required expansion up to \( x^3 \).
\( e^x \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6} \)
Using Manipulation to Obtain Other Maclaurin Series
Using Manipulation to Obtain Other Maclaurin Series
Once we know the Maclaurin series for some basic functions, we can find series for more complicated functions by applying:
- Substitution: Replace \( x \) by some function \( g(x) \) inside a known series.
- Products: Multiply two known series term-by-term.
- Differentiation: Differentiate a Maclaurin series term-by-term to find the series for the derivative of the function.
- Integration: Integrate a Maclaurin series term-by-term to find the series for the integral of the function.
Example :
Find the Maclaurin series for \( e^{x^2} \) up to and including the term in \( x^6 \).
▶️ Answer/Explanation
Solution:
Maclaurin series for \( e^x \):
\( e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \)
Substitute \( x^2 \) for \( x \):
\( e^{x^2} = 1 + x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + \cdots \)
terms up to \( x^6 \):
\( e^{x^2} \approx 1 + x^2 + \frac{x^4}{2} + \frac{x^6}{6} \)
Example :
Find the Maclaurin series for \( e^x \sin x \) up to and including the term in \( x^4 \).
▶️ Answer/Explanation
Solution:
Maclaurin series for \( e^x \):
\( e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots \)
Maclaurin series for \( \sin x \):
\( \sin x = x – \frac{x^3}{6} + \frac{x^5}{120} – \cdots \)
Multiply the two series term-by-term,
$ e^x \sin x \approx (1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24})(x – \frac{x^3}{6}) $
\( 1 \times x = x \)
\( 1 \times \left(-\frac{x^3}{6}\right) = -\frac{x^3}{6} \)
\( x \times x = x^2 \)
\( x \times \left(-\frac{x^3}{6}\right) = -\frac{x^4}{6} \)
\( \frac{x^2}{2} \times x = \frac{x^3}{2} \)
\( \frac{x^3}{6} \times x = \frac{x^4}{6} \)
Sum the terms up to \( x^4 \):
$ e^x \sin x \approx x + x^2 + \left(-\frac{1}{6} + \frac{1}{2}\right) x^3 + \left(-\frac{1}{6} + \frac{1}{6}\right) x^4 = x + x^2 + \frac{1}{3} x^3 + 0 \cdot x^4 $
So up to \( x^4 \), the series is:
\( e^x \sin x \approx x + x^2 + \frac{x^3}{3} \)
Maclaurin Series from Differential Equations
Maclaurin Series from Differential Equations
Sometimes, the Maclaurin series for a function can be found by solving a differential equation using power series methods.
- Assume the solution \( y = \sum_{n=0}^\infty a_n x^n \) (a power series centered at 0).
- Substitute \( y \), its derivatives \( y’, y” \), etc., into the differential equation.
- Match coefficients of powers of \( x \) to find a recurrence relation for \( a_n \).
- Use initial conditions (like \( y(0) \), \( y'(0) \)) to find specific coefficients.
This process allows you to derive the Maclaurin series expansion of solutions to differential equations.
Example:
Finding the Maclaurin Series for \( y \) satisfying \( y” + y = 0 \) with \( y(0) = 1 \), \( y'(0) = 0 \)
▶️ Answer/Explanation
Assume a power series solution
$ y = \sum_{n=0}^\infty a_n x^n = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + \cdots $
$ y’ = \sum_{n=1}^\infty n a_n x^{n-1} = a_1 + 2 a_2 x + 3 a_3 x^2 + 4 a_4 x^3 + \cdots $
$ y” = \sum_{n=2}^\infty n (n-1) a_n x^{n-2} = 2 a_2 + 6 a_3 x + 12 a_4 x^2 + 20 a_5 x^3 + \cdots $
Substitute into the differential equation \( y” + y = 0 \):
$ \sum_{n=2}^\infty n(n-1) a_n x^{n-2} + \sum_{n=0}^\infty a_n x^n = 0 $
Change index on first sum to match powers of \( x^n \) by letting \( m = n – 2 \):
$ \sum_{m=0}^\infty (m+2)(m+1) a_{m+2} x^m + \sum_{n=0}^\infty a_n x^n = 0 $ Rename \( m \to n \) for clarity:
$ \sum_{n=0}^\infty \left[ (n+2)(n+1) a_{n+2} + a_n \right] x^n = 0 $
For this to hold for all \( x \), each coefficient must be zero:
$ (n+2)(n+1) a_{n+2} + a_n = 0 \quad \Rightarrow \quad a_{n+2} = – \frac{a_n}{(n+2)(n+1)} $
initial conditions:
\( y(0) = a_0 = 1 \), and \( y'(0) = a_1 = 0 \).
Calculate coefficients:
$ a_2 = -\frac{a_0}{2 \times 1} = -\frac{1}{2} $ $ a_3 = -\frac{a_1}{3 \times 2} = 0 $
$ a_4 = -\frac{a_2}{4 \times 3} = -\frac{-1/2}{12} = \frac{1}{24} $ $ a_5 = -\frac{a_3}{5 \times 4} = 0 $
$ a_6 = -\frac{a_4}{6 \times 5} = -\frac{1/24}{30} = -\frac{1}{720} $ and so on.
$ y = 1 – \frac{x^2}{2!} + \frac{x^4}{4!} – \frac{x^6}{6!} + \cdots $ which is the Maclaurin series for \( \cos x \).