Welcome to Maclaurin and Taylor Series!
Hello future mathematician! This chapter, often found in Further Pure Mathematics 2 (FP2), introduces one of the most powerful tools in calculus: Series Expansions.
What we are learning is how to transform complicated functions—like \(e^x\), \(\sin x\), or \(\ln(1+x)\)—into simple, infinite polynomials. Why is this important? Because polynomials are incredibly easy to differentiate, integrate, and calculate! This technique allows us to approximate function values with amazing accuracy.
Don't worry if this looks tricky at first; we will break down the formulas step-by-step, using clear analogies to make sure you master this essential topic!
Quick Review: Tools You Will Need
Before diving into the series, make sure you are comfortable with two key concepts:
1. Repeated Differentiation: You must be able to find the first, second, third, and subsequent derivatives of common functions quickly and accurately.
2. Factorials: The notation \(n!\) (read as "n factorial") means multiplying all positive integers less than or equal to \(n\).
Example: \(4! = 4 \times 3 \times 2 \times 1 = 24\). Remember that \(0! = 1\).
The Maclaurin Series: Centred at Zero
The Maclaurin series is a special case of the Taylor series. It approximates a function \(f(x)\) using a polynomial whose derivatives match the function's derivatives exactly at the point \(x=0\).
The Maclaurin Formula
If a function \(f(x)\) has derivatives of all orders at \(x=0\), its Maclaurin series is given by:
$$ f(x) = f(0) + x f'(0) + \frac{x^2}{2!} f''(0) + \frac{x^3}{3!} f'''(0) + \dots + \frac{x^n}{n!} f^{(n)}(0) + \dots $$In summation notation, this is written as:
$$ f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!} f^{(n)}(0) $$Breaking Down the Formula (The DNA Analogy)
Think of the function \(f(x)\) as a complex organism. The Maclaurin series is like scanning its 'DNA' (its derivatives) at \(x=0\) to create an identical copy in polynomial form.
- The Constant Term \(f(0)\): This ensures the polynomial approximation starts at the correct height when \(x=0\).
- The Coefficient \(\frac{f^{(n)}(0)}{n!}\): This is the crucial part. We are scaling the \(n\)-th derivative evaluated at zero by the factorial of \(n\). Why the factorial? The factorial cancels out the factors that naturally appear when you differentiate the polynomial terms repeatedly, ensuring the derivatives match perfectly at \(x=0\).
Step-by-Step Guide to Finding a Maclaurin Series
Let's find the first four non-zero terms for \(f(x) = e^x\).
Step 1: Write down the general formula (or the skeleton terms).
\(f(x) \approx f(0) + x f'(0) + \frac{x^2}{2!} f''(0) + \frac{x^3}{3!} f'''(0) + \dots\)
Step 2: Find the derivatives of the function.
\(f(x) = e^x\)
\(f'(x) = e^x\)
\(f''(x) = e^x\)
\(f'''(x) = e^x\)
Step 3: Evaluate the function and its derivatives at \(x=0\).
\(f(0) = e^0 = 1\)
\(f'(0) = e^0 = 1\)
\(f''(0) = e^0 = 1\)
\(f'''(0) = e^0 = 1\)
Step 4: Substitute these values back into the Maclaurin formula.
$$ e^x = 1 + x(1) + \frac{x^2}{2!}(1) + \frac{x^3}{3!}(1) + \dots $$ $$ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots $$Common Mistake Alert: Always remember to divide by the factorial \(n!\). It is the most common omission error!
Key Takeaway for Maclaurin Series
The Maclaurin series is a polynomial centered at \(x=0\). It requires repeated differentiation and careful evaluation at zero, followed by dividing by the appropriate factorial.
Standard Maclaurin Series Expansions
In exams, you are expected to be familiar with (and often required to derive) the Maclaurin series for several standard functions. It is highly recommended to memorize these patterns!
1. Exponential Function
$$
e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots = \sum_{n=0}^{\infty} \frac{x^n}{n!}
$$
2. Trigonometric Functions
Did You Know? Sine is an odd function (\(\sin(-x) = -\sin x\)), so its expansion only contains odd powers of \(x\). Cosine is an even function, so its expansion only contains even powers of \(x\). This is a great memory check!
$$ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} $$ $$ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!} $$3. Logarithmic Function
When finding the Maclaurin series for \(\ln x\), we run into a problem: \(\ln(0)\) is undefined! Therefore, we use the function \(\ln(1+x)\), which is defined at \(x=0\).
$$ \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^{n}}{n} $$Note: There are no factorials in the denominator here!
4. Binomial Series (A Quick Recap)
While the Binomial expansion is often covered earlier, it is included here as its structure is similar. It is valid for \(\lvert x \rvert < 1\).
$$ (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots $$Note: If \(n\) is a positive integer, the series terminates. If \(n\) is fractional or negative, it is an infinite series.
The Taylor Series: Centred at \(x = a\)
The Maclaurin series is incredibly useful, but it only gives a good approximation near \(x=0\). What if we need a highly accurate approximation near \(x=3\) or \(x= -1\)?
This is where the Taylor series comes in. The Taylor series is the generalized form of the Maclaurin series, allowing us to center the polynomial approximation around any point \(x=a\).
The Taylor Formula
If a function \(f(x)\) has derivatives of all orders at \(x=a\), its Taylor series is given by:
$$ f(x) = f(a) + (x-a) f'(a) + \frac{(x-a)^2}{2!} f''(a) + \frac{(x-a)^3}{3!} f'''(a) + \dots $$In summation notation:
$$ f(x) = \sum_{n=0}^{\infty} \frac{(x-a)^n}{n!} f^{(n)}(a) $$Taylor's Shift: Notice the shift from \(x^n\) to \((x-a)^n\), and the evaluation point shifting from \(x=0\) to \(x=a\).
Example: Finding the Taylor Series for \(\ln x\) about \(a=1\)
Since \(\ln 0\) is undefined, we must use a Taylor series centered around a point where the function is defined, for example, \(a=1\). This means our series will be in powers of \((x-1)\).
Step 1: Set \(a=1\) and find derivatives.
\(f(x) = \ln x\)
\(f'(x) = 1/x\)
\(f''(x) = -1/x^2\)
\(f'''(x) = 2/x^3\)
Step 2: Evaluate at \(x=a=1\).
\(f(1) = \ln 1 = 0\)
\(f'(1) = 1/1 = 1\)
\(f''(1) = -1/1 = -1\)
\(f'''(1) = 2/1 = 2\)
Step 3: Substitute into the Taylor formula.
$$ \ln x = f(1) + (x-1) f'(1) + \frac{(x-1)^2}{2!} f''(1) + \frac{(x-1)^3}{3!} f'''(1) + \dots $$ $$ \ln x = 0 + (x-1)(1) + \frac{(x-1)^2}{2!}(-1) + \frac{(x-1)^3}{3!}(2) + \dots $$ $$ \ln x = (x-1) - \frac{(x-1)^2}{2} + \frac{2(x-1)^3}{6} + \dots $$ $$ \ln x = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \dots $$Notice how similar this is to the \(\ln(1+x)\) Maclaurin series, but with \(x\) replaced by \((x-1)\)!
Key Takeaway for Taylor Series
The Taylor series generalizes Maclaurin by centering the approximation at \(x=a\). The process is identical to Maclaurin, but we evaluate the derivatives at \(x=a\) and use powers of \((x-a)\).
Advanced Techniques: Using Standard Expansions
Calculating high-order derivatives can be time-consuming and error-prone. Often, the quickest and safest way to find a series expansion is by using substitution, multiplication, or combination of the standard series you have memorized.
Technique 1: Substitution (The Power of Replacement)
If you know the series for \(f(u)\), you can find the series for \(f(g(x))\) by replacing \(u\) with \(g(x)\).
Example: Find the Maclaurin series for \(e^{2x}\) up to the term in \(x^3\).
We know: \(e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \dots\)
Let \(u = 2x\). Substitute \(2x\) everywhere you see \(u\):
$$ e^{2x} = 1 + (2x) + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \dots $$ $$ e^{2x} = 1 + 2x + \frac{4x^2}{2} + \frac{8x^3}{6} + \dots $$ $$ e^{2x} = 1 + 2x + 2x^2 + \frac{4}{3}x^3 + \dots $$Technique 2: Combining Series (Multiplication and Addition)
To find the expansion of a product of functions, say \(f(x)g(x)\), you multiply their individual series expansions, treating them as polynomials.
Example: Find the Maclaurin series for \(e^x \cos x\) up to the term in \(x^3\).
\(e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots\)
\(\cos x = 1 - \frac{x^2}{2} + \dots\)
We multiply the two polynomials, ignoring any resulting terms higher than \(x^3\):
$$ e^x \cos x = \left( 1 + x + \frac{x^2}{2} + \frac{x^3}{6} \right) \left( 1 - \frac{x^2}{2} \right) $$Term-by-term multiplication (only keeping power up to \(x^3\)):
- \(1 \times 1 = 1\)
- \(1 \times (-\frac{x^2}{2}) = -\frac{x^2}{2}\)
- \(x \times 1 = x\)
- \(x \times (-\frac{x^2}{2}) = -\frac{x^3}{2}\)
- \(\frac{x^2}{2} \times 1 = \frac{x^2}{2}\)
- \(\frac{x^3}{6} \times 1 = \frac{x^3}{6}\)
Now, combine the like terms:
$$ e^x \cos x = 1 + x + \left(-\frac{x^2}{2} + \frac{x^2}{2}\right) + \left(\frac{x^3}{6} - \frac{x^3}{2}\right) + \dots $$ $$ e^x \cos x = 1 + x + (0)x^2 + \left(\frac{x^3 - 3x^3}{6}\right) + \dots $$ $$ e^x \cos x = 1 + x - \frac{2x^3}{6} + \dots $$ $$ e^x \cos x = 1 + x - \frac{x^3}{3} + \dots $$Technique 3: Integration and Differentiation
If you need the series for a function \(f(x)\) but it's hard to differentiate, see if its derivative or integral is easier to expand.
Example: To find the series for \(\arctan x\), it's easier to find the series for its derivative:
\(\frac{d}{dx} (\arctan x) = \frac{1}{1+x^2}\).
You can expand \(\frac{1}{1+x^2}\) using the binomial series \((1+u)^{-1}\) where \(u=x^2\). Once expanded, integrate the resulting polynomial term-by-term to get the series for \(\arctan x\). (Remember the constant of integration, found by evaluating at \(x=0\)).
Quick Review Box: Tips for Success
- Maclaurin = Evaluate at \(x=0\).
- Taylor = Evaluate at \(x=a\).
- Always check for the Factorial \(n!\) in the denominator unless it's \(\ln(1+x)\).
- Use Substitution whenever possible to save time on differentiation.
You've covered the core concepts of series expansions! This is a foundational topic for much of higher mathematics. Keep practicing those standard derivations, and you will find these questions straightforward!