🚀 Welcome to Maclaurin and Taylor Series!

Hey there, math explorers! This chapter is one of the most powerful tools in Further Pure Mathematics. It might look a little intense at first because of all the derivatives, but don't worry—we'll break it down into simple, manageable steps.

What are we doing here? We are learning how to take incredibly complicated functions (like \(f(x) = e^x\) or \(f(x) = \sin x\)) and turn them into simple, manageable polynomials (functions with just \(x\), \(x^2\), \(x^3\), etc.).

Why is this important? Calculators and computers can't calculate \(\sin(1)\) exactly! They use Maclaurin and Taylor series expansions to approximate the value extremely accurately. You are learning the math behind the technology!

Let's dive in!

1. Understanding Series Expansion: The Mathematical Look-Alike

Imagine you have a curvy function, \(f(x)\). We want to find a polynomial \(P(x)\) that looks exactly like \(f(x)\) at a specific point, and also has the same slope, the same curvature, the same rate of curvature change, and so on.

By making the function and its derivatives match up at that central point, the polynomial \(P(x)\) becomes an excellent approximation of the function \(f(x)\) near that center.

Maclaurin Series: Centered at \(x=0\)

The Maclaurin series is the simpler form, where we choose the center point of our approximation to be the origin, \(x=0\).

The Maclaurin Formula

The Maclaurin series for a function \(f(x)\) is given by:

\[ f(x) = f(0) + x f'(0) + \frac{x^2}{2!} f''(0) + \frac{x^3}{3!} f'''(0) + \frac{x^4}{4!} f^{(4)}(0) + \dots \]

Or, written using summation notation (which is crucial for FP2):

\[ f(x) = \sum_{r=0}^{\infty} \frac{x^r}{r!} f^{(r)}(0) \]

Key Terminology Check:

  • \(f(0)\): The value of the function at \(x=0\) (the constant term).
  • \(f'(0)\): The value of the 1st derivative (slope) at \(x=0\).
  • \(f^{(r)}(0)\): The value of the \(r\)th derivative at \(x=0\).
  • \(r!\): The factorial of \(r\). (Remember: \(3! = 3 \times 2 \times 1 = 6\), and \(0! = 1\)).
Quick Review: The Maclaurin series is a polynomial designed so that its value and the values of all its derivatives match the original function's value and derivatives at \(x=0\).

2. Step-by-Step: Finding a Maclaurin Series

Finding a Maclaurin series involves a repetitive process of differentiation and evaluation. This is where most students make small algebraic errors, so be careful and organized!

Process: Finding the Maclaurin Expansion for \(f(x) = \ln(1+3x)\) up to the \(x^3\) term.

Step 1: Write down the derivatives. Find the required derivatives of \(f(x)\).

  • \(f(x) = \ln(1+3x)\)
  • \(f'(x) = 3(1+3x)^{-1}\)
  • \(f''(x) = -3(3)(1+3x)^{-2} \times 3 = -27(1+3x)^{-2}\)
  • \(f'''(x) = -27(-2)(1+3x)^{-3} \times 3 = 162(1+3x)^{-3}\)

Step 2: Evaluate the derivatives at \(x=0\). Substitute \(x=0\) into each derivative.

  • \(f(0) = \ln(1+0) = \ln(1) = 0\)
  • \(f'(0) = 3(1)^{-1} = 3\)
  • \(f''(0) = -27(1)^{-2} = -27\)
  • \(f'''(0) = 162(1)^{-3} = 162\)

Step 3: Substitute the values into the Maclaurin formula.

The formula is: \( f(x) = f(0) + x f'(0) + \frac{x^2}{2!} f''(0) + \frac{x^3}{3!} f'''(0) + \dots \)

\[ f(x) = 0 + x(3) + \frac{x^2}{2}(-27) + \frac{x^3}{6}(162) + \dots \]

Step 4: Simplify the expression.

\[ f(x) = 3x - \frac{27}{2}x^2 + 27x^3 + \dots \]
⚠️ Common Mistake Alert!

A very common error is forgetting to divide by \(r!\) (the factorials). Forgetting the factorials means you are generating a Taylor Polynomial, not a series expansion. Always remember the denominator: \(2!, 3!, 4!\) etc.

3. Standard Maclaurin Expansions (The Essential Five)

You must know, and be able to reproduce quickly, the expansions for these five fundamental functions. They are the building blocks for almost every FP2 problem involving series.

Note: These expansions are only valid for certain ranges of \(x\) (called the radius of convergence), but for most basic FP2 manipulations, we assume \(x\) is small enough for the expansion to be valid.

A. Exponential Function

For all \(x\):

\[ \text{e}^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots \]

B. Sine Function (Odd Powers Only)

For all \(x\):

\[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots \]

Memory Aid: Sine is an odd function, so it only has odd powers of \(x\). The signs alternate (plus, minus, plus, minus).

C. Cosine Function (Even Powers Only)

For all \(x\):

\[ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots \]

Memory Aid: Cosine is an even function, so it only has even powers of \(x\). The signs alternate.

D. Natural Logarithm

For \(-1 < x \leq 1\):

\[ \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots \]

Did you know? Notice the factorials are missing here! The denominator is just \(n\), not \(n!\). This is a crucial difference to remember.

E. Generalised Binomial Expansion

For \(-1 < x < 1\):

\[ (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots \]

This series is identical to the Maclaurin expansion of \(f(x)=(1+x)^n\). It works even when \(n\) is fractional or negative.

Key Takeaway: Memorize the standard five series. They save immense amounts of time and are required for the manipulation questions later on.

4. The Taylor Series: Moving the Center

Maclaurin series is great, but what if we want a highly accurate polynomial approximation of a function near \(x=5\)? The Maclaurin series (centered at \(x=0\)) would be very inaccurate there.

The Taylor Series is the generalization of the Maclaurin series. Instead of centering the expansion at \(x=0\), we center it at any point \(x=a\).

The Taylor Formula (Centered at \(x=a\))

The Taylor series for a function \(f(x)\) expanded about \(x=a\) is:

\[ f(x) = f(a) + (x-a) f'(a) + \frac{(x-a)^2}{2!} f''(a) + \frac{(x-a)^3}{3!} f'''(a) + \dots \]

In summation notation:

\[ f(x) = \sum_{r=0}^{\infty} \frac{(x-a)^r}{r!} f^{(r)}(a) \]

Simple Trick: Look at the Maclaurin formula and replace every \(x\) with \((x-a)\) and every \(0\) with \(a\).

Example: Finding a Taylor Series

Find the Taylor expansion of \(f(x) = \sin x\) centered about \(a = \frac{\pi}{2}\) up to the \( (x-\frac{\pi}{2})^2 \) term.

Step 1: Derivatives.

  • \(f(x) = \sin x\)
  • \(f'(x) = \cos x\)
  • \(f''(x) = -\sin x\)

Step 2: Evaluate at \(a = \frac{\pi}{2}\).

  • \(f(\frac{\pi}{2}) = \sin(\frac{\pi}{2}) = 1\)
  • \(f'(\frac{\pi}{2}) = \cos(\frac{\pi}{2}) = 0\)
  • \(f''(\frac{\pi}{2}) = -\sin(\frac{\pi}{2}) = -1\)

Step 3: Substitute into the Taylor formula.

\[ f(x) = f(a) + (x-a) f'(a) + \frac{(x-a)^2}{2!} f''(a) + \dots \] \[ f(x) = 1 + (x-\frac{\pi}{2})(0) + \frac{(x-\frac{\pi}{2})^2}{2}(-1) + \dots \]

Result:

\[ \sin x \approx 1 - \frac{1}{2}\left(x-\frac{\pi}{2}\right)^2 \]
Key Takeaway: The Maclaurin series (centered at 0) is a specific type of Taylor series. The Taylor series allows us to center the approximation wherever we need the most accuracy.

5. Advanced Applications in FP2

In Further Pure Mathematics, you are rarely asked just to differentiate three times. The challenge comes from using the standard series to solve more complex problems efficiently.

A. Using Manipulation (Substitution)

Instead of finding 5 derivatives for \(f(x) = e^{-3x^2}\), we can use the known series for \(\text{e}^u\) and substitute \(u = -3x^2\).

We know: \( \text{e}^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \dots \)

Substituting \(u = -3x^2\):

\[ \text{e}^{-3x^2} = 1 + (-3x^2) + \frac{(-3x^2)^2}{2!} + \frac{(-3x^2)^3}{3!} + \dots \]

Simplifying:

\[ \text{e}^{-3x^2} = 1 - 3x^2 + \frac{9x^4}{2} - \frac{27x^6}{6} + \dots \] \[ \text{e}^{-3x^2} = 1 - 3x^2 + \frac{9}{2}x^4 - \frac{9}{2}x^6 + \dots \]

This method is far quicker and less prone to differentiation errors than the traditional step-by-step Maclaurin process.

Other Manipulation Techniques:

  1. Differentiation: If you need the series for \(\frac{1}{1+x}\), you can differentiate the series for \(\ln(1+x)\).
  2. Integration: If you need the series for \(\arctan x\), you can integrate the series for \(\frac{1}{1+x^2}\) (which comes from the binomial expansion).

B. Evaluating Limits

This is a major FP2 use case. When you encounter a limit of the form \(\lim_{x \to 0} \frac{f(x)}{g(x)}\) that evaluates to \(\frac{0}{0}\) (an indeterminate form), the series expansion can help you resolve the ambiguity.

The Goal: Replace the complicated functions \(f(x)\) and \(g(x)\) with their polynomial expansions up to a high enough power (usually the lowest power that doesn't cancel out). This allows for easy cancellation of the problem term (often \(x\)).

Step-by-Step Example Limit:

Evaluate \( L = \lim_{x \to 0} \frac{x - \sin x}{x^3} \)

Step 1: Write out the expansion for the function. We need the expansion of \(\sin x\) up to the \(x^3\) term, as the denominator is \(x^3\).

\[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots \]

Step 2: Substitute the expansion into the numerator.

Numerator: \( x - \sin x = x - \left( x - \frac{x^3}{6} + \frac{x^5}{120} - \dots \right) \)

Numerator: \( x - \sin x = \frac{x^3}{6} - \frac{x^5}{120} + \dots \)

Step 3: Substitute the numerator back into the limit expression.

\[ L = \lim_{x \to 0} \frac{\frac{x^3}{6} - \frac{x^5}{120} + \dots}{x^3} \]

Step 4: Divide every term in the numerator by \(x^3\).

\[ L = \lim_{x \to 0} \left( \frac{1}{6} - \frac{x^2}{120} + \dots \right) \]

Step 5: Evaluate the limit. As \(x \to 0\), all terms involving \(x\) disappear.

\[ L = \frac{1}{6} \]

This technique provides a fast, algebraic way to solve limits without using L'Hôpital's rule (which is typically not an FP2 method).

C. Approximations

We can use truncated series (polynomials that stop after a certain power) to approximate values. The more terms we include, the better the approximation, especially when \(x\) is close to the center of expansion.

For example, for small angles, the Maclaurin expansion of \(\cos x\) gives us:

\[ \cos x \approx 1 - \frac{x^2}{2} \]

This shows why \(\cos x \approx 1\) is often a good approximation when \(x\) is tiny (since the \(x^2\) term becomes negligible).

Final Key Takeaway: Series expansions are versatile tools. Use the direct differentiation method for general functions, and use substitution/manipulation for functions based on the standard five. Always simplify before taking a limit!