Welcome to Series and Limits (FP2.6)!
Hello! This chapter takes some of the concepts you might have touched on in A-Level Pure Maths—like sequences and integration—and pushes them to their theoretical limits. Literally!
Why is this important? Series expansions (like Maclaurin series) allow us to approximate complicated functions using simple polynomials. This is crucial for everything from calculator programming to advanced physics. Meanwhile, understanding limits lets us analyze the long-term behaviour of functions and determine whether certain sums and integrals actually have a finite value.
Don't worry if this seems abstract at first. We will break down Maclaurin series, look at some powerful limit rules, and apply these ideas to solve tricky integrals. Let's dive in!
1. Maclaurin Series: Turning Functions into Polynomials
What is a Maclaurin Series?
A Maclaurin series is a special type of power series that represents a function \(f(x)\) as an infinite polynomial, centred around \(x=0\).
Analogy: Imagine you are standing at \(x=0\) on the graph of \(f(x)\). The Maclaurin series is a perfect map of the function built solely from information (the function's value and all its derivatives) gathered right at that point.
The general formula for a Maclaurin series is:
\[ f(x) = f(0) + xf'(0) + \frac{x^2}{2!}f''(0) + \frac{x^3}{3!}f'''(0) + \dots + \frac{x^n}{n!}f^{(n)}(0) + \dots \]
Key Standard Expansions You Must Know (and use)
The syllabus requires you to know and use the standard Maclaurin series expansions for several common functions. These are usually provided in your formulae booklet, but knowing them helps immensely!
- The Exponential Function: \[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \]
- The Natural Logarithm: \[ \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots \]
- Sine Function: (Only odd powers!) \[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots \]
- Cosine Function: (Only even powers!) \[ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots \]
- Binomial Expansion: (For rational powers \(n\)) \[ (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots \]
Range of Validity (Radius of Convergence)
A series only equals the function it represents for specific values of \(x\). This is the range of validity.
For $e^x$, $\sin x$, and $\cos x$, the series converges for all real $x$ (i.e., \(\vert x \vert < \infty\)).
For $\ln(1+x)$ and $(1+x)^n$, the series usually only converges if \(\vert x \vert < 1\).
How to Find the Range for Related Functions
If you substitute something into a standard series, the condition for validity must still hold true for that substitution.
Example: Find the range of validity for the expansion of \(\ln(1-2x)\).
The standard series is \(\ln(1+u)\), valid for \(\vert u \vert < 1\).
Here, the substitution is \(u = -2x\).
Therefore, we need \(\vert -2x \vert < 1\), which simplifies to \(2\vert x \vert < 1\), or \(\vert x \vert < \frac{1}{2}\).
Quick Review: Maclaurin Series
- They are polynomials centered at \(x=0\).
- Use the standard expansions given in the booklet.
- Always check the range of validity, especially when substituting functions of \(x\).
2. Using Series Expansion to Find Limits
One of the most powerful applications of Maclaurin series is finding limits, especially those of the indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) as \(x \to 0\).
If you tried to calculate \(\lim_{x\to 0} \frac{\sin x}{x}\) by plugging in \(x=0\), you would get \(\frac{0}{0}\). We use series to simplify the expression before evaluating the limit.
Step-by-Step Method for Limits
- Replace the functions in the limit expression with their Maclaurin series expansions. Keep terms until the denominator is cancelled out, or until you reach a non-zero constant term in the numerator.
- Simplify the numerator and denominator by cancelling out common factors of \(x\).
- Substitute \(x=0\) into the simplified expression.
Example: Finding a tricky limit
Find the limit: \(\lim_{x\to 0} \frac{x^2 e^x}{\cos(2x) - 1}\)
Step 1: Substitute Maclaurin Series
Use standard expansions, substituting \(2x\) into the cosine series:
\[ e^x = 1 + x + \frac{x^2}{2} + \dots \]
\[ \cos(2x) = 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \dots = 1 - 2x^2 + \frac{2}{3}x^4 - \dots \]
The expression becomes:
\[ \lim_{x\to 0} \frac{x^2 \left( 1 + x + \frac{x^2}{2} + \dots \right)}{\left( 1 - 2x^2 + \frac{2}{3}x^4 - \dots \right) - 1} \]
Step 2: Simplify
Numerator: \(x^2 + x^3 + \frac{x^4}{2} + \dots\)
Denominator: \(-2x^2 + \frac{2}{3}x^4 - \dots\)
We can cancel \(x^2\) from every term:
\[ \lim_{x\to 0} \frac{1 + x + \frac{x^2}{2} + \dots}{-2 + \frac{2}{3}x^2 - \dots} \]
Step 3: Evaluate the Limit
Substitute \(x=0\):
\[ \frac{1 + 0 + 0 + \dots}{-2 + 0 - \dots} = -\frac{1}{2} \]
Key Takeaway: Series expansions provide a robust algebraic tool to handle indeterminate limits, often avoiding the need for L'Hôpital's rule. Make sure you expand to a high enough order to clearly separate the numerator and denominator after cancelling \(x\).
3. Important Limits at Infinity and Zero
The syllabus requires knowledge of two critical limits concerning the relative "strength" of different types of functions as \(x\) approaches infinity or zero.
The Race to Infinity: Exponential Dominance
When \(x \to \infty\), the exponential function \(e^x\) grows faster than any polynomial function, \(x^k\), where \(k > 0\).
The important result is: \[ \lim_{x \to \infty} x^k e^{-x} = 0 \quad \text{for any } k > 0 \]
What this means: As \(x\) gets huge, the denominator \(e^x\) completely overwhelms the numerator \(x^k\), driving the entire fraction to zero. The polynomial term \(x^k\) could be \(x^{100}\), but the exponential \(e^x\) will still win the race to infinity!
The Race to Zero: Logarithmic Weakness
When \(x \to 0\), the logarithmic function \(\ln x\) goes to \(-\infty\), but it does so very slowly compared to any power of \(x\).
The important result is: \[ \lim_{x \to 0} x^k \ln x = 0 \quad \text{for any } k > 0 \]
Why this works: As \(x\) approaches zero (from the positive side), the term \(x^k\) approaches zero very quickly, and this powerful polynomial term "drags" the slower-changing logarithm down to zero.
Memory Aid: Exponentials (E) dominate everything at infinity. Logs (L) are the weakest near zero. (E>P>L)
4. Improper Integrals
What makes an Integral "Improper"?
An integral is improper if either:
1. One or both of its limits of integration is infinite (\(\infty\) or \(-\infty\)). (Type I)
2. The function being integrated (the integrand) has an infinite discontinuity within the integration interval. (Type II)
The Procedure: Using Limits to Define the Integral
We cannot simply substitute \(\infty\) into a function. We must replace the improper part with a variable (often \(t\)) and then take the limit as that variable approaches the problematic value.
Step 1: Replace the troublesome limit with a temporary variable \(t\).
Example (Type I): For \(\int_{0}^{\infty} f(x) \, dx\), write \(\lim_{t \to \infty} \int_{0}^{t} f(x) \, dx\).
Example (Type II): For \(\int_{0}^{1} \frac{1}{\sqrt{x}} \, dx\) (discontinuous at \(x=0\)), write \(\lim_{t \to 0^+} \int_{t}^{1} \frac{1}{\sqrt{x}} \, dx\).
Step 2: Evaluate the definite integral in terms of \(t\).
Step 3: Evaluate the limit as \(t\) approaches the problematic value.
Convergence and Divergence
If the limit in Step 3 exists and is a finite number, the improper integral converges. If the limit is \(\infty\), \(-\infty\), or does not exist, the integral diverges.
Example: Type I Improper Integral (Exponential)
Evaluate \(\int_{1}^{\infty} x e^{-x} \, dx\)
Step 1 & 2: Set up the limit and integrate
We use integration by parts (not shown here, but necessary):
\[ \int x e^{-x} \, dx = -x e^{-x} - e^{-x} + C \]
Now apply the limits using \(t\):
\[ \lim_{t \to \infty} \int_{1}^{t} x e^{-x} \, dx = \lim_{t \to \infty} \left[ -x e^{-x} - e^{-x} \right]_1^t \]
\[ = \lim_{t \to \infty} \left( \left[ -t e^{-t} - e^{-t} \right] - \left[ -1 e^{-1} - e^{-1} \right] \right) \]
Step 3: Evaluate the limit
We use the dominance result from Section 3: \(\lim_{t \to \infty} t e^{-t} = 0\).
Also, \(\lim_{t \to \infty} e^{-t} = \lim_{t \to \infty} \frac{1}{e^t} = 0\).
The limit simplifies to:
\[ = (0 - 0) - \left( -\frac{1}{e} - \frac{1}{e} \right) \]
\[ = - \left( -\frac{2}{e} \right) = \frac{2}{e} \]
Since the result is a finite number, the integral converges to \(\frac{2}{e}\).
Example: Type II Improper Integral (Discontinuity)
Evaluate \(\int_{0}^{1} \frac{1}{\sqrt{x}} \, dx\)
The function is undefined at \(x=0\), so we use \(t\) approaching \(0\) from the positive side (\(0^+\)).
Step 1 & 2: Set up the limit and integrate
Since \(\frac{1}{\sqrt{x}} = x^{-1/2}\), the antiderivative is \(2x^{1/2}\) or \(2\sqrt{x}\).
\[ \lim_{t \to 0^+} \int_{t}^{1} x^{-1/2} \, dx = \lim_{t \to 0^+} \left[ 2\sqrt{x} \right]_t^1 \]
\[ = \lim_{t \to 0^+} \left( 2\sqrt{1} - 2\sqrt{t} \right) \]
Step 3: Evaluate the limit
As \(t \to 0^+\), \(2\sqrt{t} \to 0\).
\[ = 2 - 0 = 2 \]
The integral converges to 2.
Did you know? Even though the function \(\frac{1}{\sqrt{x}}\) shoots up to infinity at \(x=0\), the area under the curve is still finite! This is a fascinating result of mathematical convergence.
Key Takeaway: Improper Integrals
- Identify the problematic limit or discontinuity.
- Replace it with a variable \(t\).
- Integrate normally, then apply \(\lim_{t \to \dots}\).
- Use the dominance rules (Section 3) when evaluating limits as \(t \to \infty\).
You have now covered the essentials of Series and Limits (FP2.6). Remember, this chapter is highly procedural. Practice applying the Maclaurin expansions correctly and always show the limiting process clearly when dealing with improper integrals!