Welcome to FP2 Series: Unlocking the Secrets of Sums and Approximations!
Hello Future Mathematicians! This chapter, Series, might look intimidating at first, but it is one of the most powerful tools in Further Mathematics. You will learn how to find the sum of incredibly long sequences (sometimes even infinite ones!) and, perhaps most excitingly, how to approximate complex functions like \(\sin x\) or \(e^x\) using simple polynomials.
Why is this important? Series expansions are the backbone of numerical analysis, physics simulations, and even the calculators you use every day! Let's dive in and make these concepts crystal clear.
1. Summation of Finite Series: The Method of Differences
Before tackling advanced methods, let’s quickly recall the standard summation formulas—these are often prerequisites or starting points for more complex problems.
1.1 Review: Standard Summation Results (Prerequisites)
You should be fluent in these results. They are given in the formula booklet, but knowing them helps speed up calculations significantly.
- The sum of the first \(n\) integers: $$\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$$
- The sum of the first \(n\) squares: $$\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$$
- The sum of the first \(n\) cubes: $$\sum_{r=1}^{n} r^3 = \left(\frac{n(n+1)}{2}\right)^2$$
1.2 The Method of Differences (Telescoping Sums)
When you encounter a series that doesn't fit the standard formulas (especially involving fractions or trigonometric terms), the Method of Differences is often the key.
The core idea is to express the general term \(u_r\) as the difference between two terms, \(f(r) - f(r+1)\) or \(f(r) - f(r-1)\). When you sum these terms, everything in the middle cancels out!
Analogy: The Collapsing Telescope
Imagine an old telescope. When you collapse it, only the first piece and the last piece remain visible; all the middle sections tuck away neatly. This is exactly what happens with telescoping sums!
Step-by-Step Process for Rational Functions
This method is most commonly applied to rational functions (fractions) in the form \(\sum_{r=1}^{n} \frac{P(r)}{Q(r)}\).
Step 1: Express the general term \(u_r\) using Partial Fractions.
If \(u_r = \frac{1}{r(r+1)}\), you must rewrite it as: $$u_r = \frac{1}{r} - \frac{1}{r+1}$$
Step 2: Write out the first few terms of the Sum, \(S_n\).
Substitute \(r = 1, 2, 3, \dots, n\) into your difference form:
- \(r=1\): \(\left(1 - \frac{1}{2}\right)\)
- \(r=2\): \(\left(\frac{1}{2} - \frac{1}{3}\right)\)
- \(r=3\): \(\left(\frac{1}{3} - \frac{1}{4}\right)\)
- ...
- \(r=n\): \(\left(\frac{1}{n} - \frac{1}{n+1}\right)\)
Step 3: Observe the Cancellation and find \(S_n\).
When you add them up: $$S_n = \left(1 - \cancel{\frac{1}{2}}\right) + \left(\cancel{\frac{1}{2}} - \cancel{\frac{1}{3}}\right) + \left(\cancel{\frac{1}{3}} - \cancel{\frac{1}{4}}\right) + \dots + \left(\cancel{\frac{1}{n}} - \frac{1}{n+1}\right)$$
The only terms remaining are the first part of the first bracket and the last part of the last bracket: $$S_n = 1 - \frac{1}{n+1}$$
Quick Tip: Spotting the Difference Pattern
Sometimes the difference is \(f(r) - f(r+2)\) or \(f(r) - f(r+3)\). If the terms are separated by more than 1, you will need to list out the first two or three terms, and the last two or three terms, to find the leftovers.
Common Mistake to Avoid: Always ensure the difference you create in Step 1 is equal to the original term \(u_r\). If \(u_r = \frac{2}{r(r+2)}\), then \(\frac{1}{r} - \frac{1}{r+2}\) actually equals \(\frac{2}{r(r+2)}\). If your partial fraction difference results in a constant \(k \ne 1\), you must factor out \(k\) or multiply by \(1/k\).
Key Takeaway (Finite Series)
The Method of Differences allows us to find exact closed-form expressions for \(S_n\) by forcing internal cancellation. This relies heavily on accurate partial fraction decomposition.
2. Sums to Infinity (\(S_{\infty}\)): Convergence
What happens if the series never ends? In FP2, once you have found the sum to \(n\) terms, \(S_n\), it is usually straightforward to find the sum to infinity, provided the series converges.
2.1 Defining Convergence
A series converges if the sum approaches a finite, fixed value as the number of terms \(n\) tends towards infinity. If the sum grows indefinitely, it diverges.
To find the sum to infinity, \(S_{\infty}\), you simply take the limit of \(S_n\) as \(n \to \infty\): $$S_{\infty} = \lim_{n \to \infty} S_n$$
Example using the Method of Differences Result
Using our earlier example where \(S_n = 1 - \frac{1}{n+1}\):
$$S_{\infty} = \lim_{n \to \infty} \left(1 - \frac{1}{n+1}\right)$$
As \(n \to \infty\), the term \(\frac{1}{n+1} \to 0\).
Therefore, \(S_{\infty} = 1 - 0 = 1\).
Don't worry if limits feel fuzzy! For the FP2 Series chapter, typically you just need to know that terms like \(1/n\), \(1/n^2\), or \(e^{-n}\) all tend to zero as \(n\) gets very large.
Key Takeaway (Infinite Series)
If a series has a defined \(S_n\), finding \(S_{\infty}\) is achieved by evaluating \(\lim_{n \to \infty} S_n\). If the limit is a finite number, the series converges.
3. Power Series: Maclaurin and Taylor Expansions
This is the heart of the FP2 Series chapter. We are now moving away from discrete sums and looking at how to represent functions themselves using polynomials. These representations are called Power Series.
3.1 The Core Idea: Polynomial Approximation
Imagine you are trying to calculate \(\sin(0.1)\) without a calculator. A polynomial, like \(f(x) = x - \frac{x^3}{6}\), is much easier to compute than the original function.
The idea of a Maclaurin or Taylor Series is to construct an infinitely long polynomial \(P(x)\) that perfectly matches the function \(f(x)\) at a specific point (and whose derivatives also match!). The more terms we use, the better the approximation.
3.2 Maclaurin Series (Centred at \(x=0\))
A Maclaurin Series is a special case of a Taylor Series centred at \(a=0\). It expresses a function \(f(x)\) as an infinite polynomial using its derivatives evaluated at \(x=0\).
The Maclaurin Formula
If \(f(x)\) can be differentiated repeatedly, its Maclaurin expansion is: $$f(x) = f(0) + x f'(0) + \frac{x^2}{2!} f''(0) + \frac{x^3}{3!} f'''(0) + \dots + \frac{x^n}{n!} f^{(n)}(0) + \dots$$
Step-by-Step Procedure for Finding a Maclaurin Series
Example: Find the Maclaurin expansion for \(f(x) = e^x\).
We need to calculate the derivatives and evaluate them at \(x=0\).
- Find the Function Value: $$f(x) = e^x \implies f(0) = e^0 = 1$$
- Find the First Derivative: $$f'(x) = e^x \implies f'(0) = 1$$
- Find the Second Derivative: $$f''(x) = e^x \implies f''(0) = 1$$
- Substitute into the Formula: $$e^x = f(0) + x f'(0) + \frac{x^2}{2!} f''(0) + \frac{x^3}{3!} f'''(0) + \dots$$ $$e^x = 1 + x(1) + \frac{x^2}{2!}(1) + \frac{x^3}{3!}(1) + \dots$$ $$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots$$
Important Reminder: Do not forget the Factorials (\(n!\)) in the denominators! \(2! = 2\), \(3! = 6\), \(4! = 24\), etc.
3.3 Taylor Series (Centred at \(x=a\))
The Taylor Series is the general form. It allows us to approximate a function around any point \(x=a\), not just \(x=0\). This is particularly useful if we are working with values far away from the origin, or if the function is undefined at \(x=0\) (like \(\ln x\)).
The Taylor Formula
The Taylor expansion of \(f(x)\) about the point \(x=a\) is: $$f(x) = f(a) + (x-a) f'(a) + \frac{(x-a)^2}{2!} f''(a) + \frac{(x-a)^3}{3!} f'''(a) + \dots$$
How to use the Taylor Formula
The process is identical to Maclaurin, but instead of setting \(x=0\), you evaluate all derivatives at \(x=a\), and the powers involve \((x-a)\) instead of just \(x\).
Did you know? If you are approximating \(f(x)\) near \(x=4\), the Taylor series will give a much better approximation near \(x=4\) than the Maclaurin series would.
3.4 Standard Maclaurin Expansions
In exams, you may be asked to derive these, but you should recognize and be able to use the results quickly. The interval of validity (the range of \(x\) for which the series converges to \(f(x)\)) is crucial.
Standard Result Table
- \(e^x\): $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$$ Valid for: All real \(x\).
- \(\sin x\) (Odd powers only): $$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$$ Valid for: All real \(x\).
- \(\cos x\) (Even powers only): $$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots$$ Valid for: All real \(x\).
- \(\ln(1+x)\): $$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$$ Valid for: \(-1 < x \le 1\).
- \((1+x)^k\) (Binomial Expansion, typically covered in FP1 or C4/P4, but crucial here): $$(1+x)^k = 1 + kx + \frac{k(k-1)}{2!}x^2 + \frac{k(k-1)(k-2)}{3!}x^3 + \dots$$ Valid for: \(-1 < x < 1\).
3.5 Combining and Substituting Series
Often, you won't need to differentiate a function from scratch. You can use the standard known series via substitution or multiplication.
Example: Find the series for \(e^{2x}\).
Since we know the expansion for \(e^y = 1 + y + \frac{y^2}{2!} + \dots\), we simply substitute \(y = 2x\): $$e^{2x} = 1 + (2x) + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \dots$$ $$e^{2x} = 1 + 2x + \frac{4x^2}{2} + \frac{8x^3}{6} + \dots$$
Example: Find the series for \(e^x \sin x\) up to the \(x^3\) term.
Multiply the required parts of the two series expansions: $$e^x \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$$ $$\sin x \approx x - \frac{x^3}{6}$$ Multiply: $$e^x \sin x \approx \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6}\right) \left(x - \frac{x^3}{6}\right)$$ Focus only on terms that result in \(x\), \(x^2\), or \(x^3\): $$ = (1)(x) + (x)(x) + (\frac{x^2}{2})(x) + (1)(-\frac{x^3}{6}) + \dots$$ $$ = x + x^2 + \frac{x^3}{2} - \frac{x^3}{6}$$ $$ = x + x^2 + \frac{3x^3 - x^3}{6}$$ $$ e^x \sin x \approx x + x^2 + \frac{2x^3}{6} = x + x^2 + \frac{x^3}{3}$$
Accessibility Tip for Multiplication
When multiplying series, draw lines to connect the terms you need. Remember, if you are looking for the \(x^4\) term, ignore any term that is \(x^5\) or higher!
3.6 Approximating Integrals and Limits
Series expansions are powerful because they allow us to deal with functions that are otherwise hard to integrate or simplify in limits.
1. Evaluating Limits: Use the series expansion to replace the complex function near \(x=0\).
Example: Find \(\lim_{x \to 0} \frac{\sin x - x}{x^3}\).
Replace \(\sin x\) with its series: $$\frac{\left(x - \frac{x^3}{6} + \dots \right) - x}{x^3} = \frac{-\frac{x^3}{6} + \text{Higher Order Terms}}{x^3}$$ $$\lim_{x \to 0} \left(-\frac{1}{6} + \text{terms involving } x, x^2, \dots \right) = -\frac{1}{6}$$
2. Integrating Series: Once a function is in polynomial form, integration becomes simple: $$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$$
Key Takeaway (Power Series)
Maclaurin and Taylor series give polynomial approximations by matching the function and its derivatives at a specific point. Remember the formula, the derivatives, and the factorials! Using standard expansions through substitution saves significant time.
Quick Review: Essential Formulas and Checks
Before tackling any problem, ask yourself:
- Is it a Summation Problem (\(S_n\))?
- If it involves \(r, r^2, r^3\): Use standard formulas.
- If it involves fractions/trig: Use the Method of Differences (Partial Fractions required).
- Is it an Approximation Problem (\(f(x)\))?
- If centre is \(x=0\): Use Maclaurin. \(f(0), f'(0), \dots\)
- If centre is \(x=a\): Use Taylor. \(f(a), f'(a), \dots\) and powers of \((x-a)\).
- Check: Did I include all the necessary factorials in the denominator?
Keep practicing those derivatives and partial fractions—they are the essential building blocks for success in the Series chapter! You've got this!