Welcome to Further Calculus!
Welcome to one of the most powerful chapters in Further Pure Mathematics 1 (FP1). In your standard A Level studies, you learned how to differentiate and integrate basic functions. Here, we are going to "supercharge" those skills. We will learn how to approximate any function as a simple polynomial, find limits that seem impossible, and use a "magic" substitution to solve the toughest trig integrals.
Don't worry if some of these formulas look a bit intimidating at first. We’ll break them down step-by-step, and you'll soon see that they are just clever extensions of things you already know!
1. Taylor Series
Imagine you have a very curvy, complicated function like \( \sin(x) \) or \( e^x \), but you really wish it was just a simple quadratic or cubic equation because those are easier to work with. A Taylor Series allows you to do exactly that! It approximates a function near a specific point using a polynomial.
The Big Idea
If we know everything about a function at one single point (its value, its gradient, its curvature, etc.), we can predict what the function looks like nearby. We call the point we are "centering" our approximation around \( a \).
The Formula
\( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \dots \)
Or, in sigma notation: \( f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n \)
Step-by-Step: Expanding \( \sin(x) \) at \( x = \pi \)
1. Find the derivatives: \( f(x) = \sin x \), \( f'(x) = \cos x \), \( f''(x) = -\sin x \), \( f'''(x) = -\cos x \).
2. Evaluate at the center (\( a = \pi \)):
\( f(\pi) = \sin(\pi) = 0 \)
\( f'(\pi) = \cos(\pi) = -1 \)
\( f''(\pi) = -\sin(\pi) = 0 \)
\( f'''(\pi) = -\cos(\pi) = 1 \)
3. Plug into the formula:
\( f(x) \approx 0 - 1(x-\pi) + 0 + \frac{1}{6}(x-\pi)^3 \dots \)
\( f(x) \approx -(x-\pi) + \frac{1}{6}(x-\pi)^3 \)
Quick Review: Taylor series are centered at \( x = a \). If the center is \( x = 0 \), it’s specifically called a Maclaurin Series (which you saw in Core Pure).
Common Mistake: Forgetting the factorial (\( n! \)) in the denominator. Always remember: "The power matches the factorial!" (Power of 3 goes over 3!).
Key Takeaway: The Taylor series lets us turn complex functions into polynomials centered around any point \( a \).
2. Using Series to Find Limits
Sometimes, when you try to find a limit by plugging in a number, you get \( \frac{0}{0} \). While you can often use L'Hospital's Rule (see below), using series expansions is often much faster for "nested" fractions.
How it works
Replace the tricky functions in your limit with their series expansions. Usually, the first few terms are enough to see what happens as \( x \) gets very small.
Example: Find \( \lim_{x \to 0} \frac{x - \arctan x}{x^3} \)
1. We know the Maclaurin series for \( \arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \dots \)
2. Substitute this into the limit: \( \lim_{x \to 0} \frac{x - (x - \frac{x^3}{3} + \dots)}{x^3} \)
3. Simplify the top: \( \lim_{x \to 0} \frac{\frac{x^3}{3} - \frac{x^5}{5}}{x^3} \)
4. Divide by \( x^3 \): \( \lim_{x \to 0} (\frac{1}{3} - \frac{x^2}{5}) \)
5. As \( x \to 0 \), the term with \( x^2 \) vanishes, leaving us with \( \frac{1}{3} \).
Key Takeaway: If a limit looks like a mess of trig and powers, try replacing the trig with its series expansion!
3. Leibnitz's Theorem
You already know the Product Rule for finding the first derivative of two functions multiplied together (\( uv \)). But what if you need the 10th derivative? Or the \( n \)-th derivative?
The Analogy
Leibnitz's Theorem is the "Binomial Expansion" of calculus. It looks almost exactly like the expansion of \( (a+b)^n \), but instead of powers, we use derivatives!
The Formula
\( \frac{d^n}{dx^n}(uv) = \sum_{r=0}^{n} \binom{n}{r} u^{(n-r)} v^{(r)} \)
Where \( u^{(k)} \) means the \( k \)-th derivative of \( u \).
A Handy Trick
When picking which function is \( u \) and which is \( v \), always pick the one that disappears after a few differentiations (like \( x^2 \)) to be your \( v \). This makes most of the terms in the long formula become zero!
Example: To find the 4th derivative of \( x^2 \sin x \), let \( v = x^2 \).
\( v = x^2 \)
\( v' = 2x \)
\( v'' = 2 \)
\( v''' = 0 \) (All terms after this will be zero!)
Key Takeaway: Use Leibnitz’s Theorem for the \( n \)-th derivative of a product. It follows the same pattern as the Binomial Theorem.
4. L'Hospital's Rule
If you have a limit \( \lim \frac{f(x)}{g(x)} \) that results in an indeterminate form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), L'Hospital's Rule is your best friend.
The Rule
\( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \)
Note: You differentiate the top and bottom separately! This is NOT the quotient rule.
Repeated Application
Don't worry if you use the rule once and still get \( \frac{0}{0} \). You can just do it again! Keep differentiating until you get a real number or a clear infinity.
The "Power" Limit: \( \lim_{x \to \infty} (1 + \frac{a}{x})^x \)
This is a classic exam favorite. To solve it:
1. Set \( y = (1 + \frac{a}{x})^x \)
2. Take natural logs: \( \ln y = x \ln(1 + \frac{a}{x}) \)
3. Rewrite as a fraction: \( \frac{\ln(1 + a/x)}{1/x} \)
4. Apply L'Hospital's Rule to find the limit of \( \ln y \).
5. Did you know? The result of this specific limit is always \( e^a \). It’s a great way to check your work!
Key Takeaway: L'Hospital's Rule turns "0/0" problems into derivative problems. Just remember to differentiate the top and bottom separately.
5. The Weierstrass Substitution (The \( t \)-Substitution)
Sometimes you encounter an integral involving \( \sin x \) and \( \cos x \) that seems impossible to solve using standard methods. The Weierstrass Substitution is a "magic" tool that turns trig integrals into algebraic ones.
The Substitution
Let \( t = \tan(\frac{x}{2}) \).
When you use this, the trig functions transform as follows:
\( \sin x = \frac{2t}{1+t^2} \)
\( \cos x = \frac{1-t^2}{1+t^2} \)
\( dx = \frac{2}{1+t^2} dt \)
Why use it?
It converts a "wiggly" trig integral into a "rational" integral (fractions with powers of \( t \)). You can then usually solve these using partial fractions or standard arctan/ln integrals.
Example: Integrating \( \int \csc x \, dx \)
1. Remember \( \csc x = \frac{1}{\sin x} \).
2. Substitute \( \sin x = \frac{2t}{1+t^2} \) and \( dx = \frac{2}{1+t^2} dt \).
3. The integral becomes: \( \int \frac{1+t^2}{2t} \cdot \frac{2}{1+t^2} dt \).
4. Everything cancels out except: \( \int \frac{1}{t} dt \).
5. This equals \( \ln|t| + C \).
6. Replace \( t \): \( \ln|\tan(\frac{x}{2})| + C \).
Don't worry if this seems tricky! You don't need to re-derive the \( \sin x \) and \( \cos x \) formulas every time; just memorize them or know how to use the reference triangle where the opposite is \( 2t \), adjacent is \( 1-t^2 \), and hypotenuse is \( 1+t^2 \).
Key Takeaway: Use \( t = \tan(x/2) \) as a last resort for difficult trig integrals. It turns trig into algebra.