Mathematics | Binomial expansion

Welcome to the World of Infinite Series: The Binomial Expansion (P4)

Hey there, future mathematician! Welcome to Unit P4. You might remember the Binomial Expansion from P2, where we dealt with nice, tidy powers like \((a+b)^3\) or \((x+y)^5\). Those expansions always stopped eventually.

In P4, we’re leveling up! We are going to explore the General Binomial Expansion, which allows us to expand expressions with difficult powers—like fractions (\(\frac{1}{2}\) for square roots) or negative numbers (\(-1\) for fractions like \(\frac{1}{1+x}\)).

This is crucial because these expansions often lead to infinite series, which are powerful tools used in engineering, physics, and advanced mathematical analysis to approximate difficult functions accurately.

Ready to turn complicated expressions into simple polynomials? Let’s dive in!

Section 1: The General Binomial Theorem (The Core P4 Skill)

1.1 Prerequisite Quick Review (P2 Recap)

In P2, if \(n\) was a positive whole number (like 3 or 5), the expansion was finite and we used the formula:

\[(a+x)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}x + \binom{n}{2}a^{n-2}x^2 + \dots + \binom{n}{n}x^n\]

We used Pascal’s triangle or the \(\binom{n}{r}\) button on our calculator. This method does not work when \(n\) is a negative integer or a fraction, because the series never terminates—it’s infinite!

1.2 Introducing the P4 General Binomial Formula

For the General Binomial Theorem, we must ensure the expression is in the specific form \((1+x)^n\). Here, \(n\) can be any rational number (\(n \in \mathbb{Q}\)), meaning \(n\) can be a fraction, a negative integer, or both.

The formula for the expansion of \((1+x)^n\) is:

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

Key Observations:

• The expansion starts with 1 (since \(1^n = 1\)).
• The second term is always simply \(nx\).
• The terms continue indefinitely (the dots \(\dots\) are important!).

Memory Aid (Remembering the Coefficients):
Notice the pattern in the numerators: Start with \(n\), then \(n\) multiplied by one less (\(n-1\)), then \(n\) multiplied by two less (\(n-2\)), and so on. The denominator is always the factorial of the power of \(x\): \(2!\), \(3!\), \(4!\), etc.

Example: Expanding \(\frac{1}{\sqrt{1+x}}\)

First, rewrite the expression using index notation: \((1+x)^{-\frac{1}{2}}\). Here, \(n = -\frac{1}{2}\).

Term 1: \(1\)
Term 2: \(nx = (-\frac{1}{2})x = -\frac{1}{2}x\)
Term 3: \(\frac{n(n-1)}{2}x^2 = \frac{(-\frac{1}{2})(-\frac{3}{2})}{2}x^2 = \frac{\frac{3}{4}}{2}x^2 = \frac{3}{8}x^2\)
And so on...

Section 2: The Condition for Validity (Range of Convergence)

2.1 Why Convergence Matters

Since the expansion is an infinite series, we need to know if adding more terms actually gives us a value closer to the real answer. If the terms keep getting bigger, the series is useless! This is called divergence.

For the series to be accurate (to converge), the terms \(x^r\) must get smaller and smaller as \(r\) increases.

2.2 Determining the Range of \(x\)

For the standard P4 expansion of \((1+x)^n\), the terms will only shrink if the absolute value of \(x\) is less than 1.

The Fundamental Condition for Validity:

The expansion of \((1+x)^n\) is only valid if:

\[|x| < 1 \quad \text{or} \quad -1 < x < 1\]

Analogy: The Telescope Effect
Imagine the expansion is like looking at an object through a telescope. The expansion only works if the item you are looking at (\(x\)) is small enough to fit within the lens (\(|x| < 1\)). If \(x\) is too large (e.g., \(x=10\)), \(x^2=100\), \(x^3=1000\). The terms are blowing up, and the expansion fails completely!

2.3 When the Expansion is \((1+bx)^n\)

If the expression is \((1+bx)^n\), the term being raised to the power is \(bx\). We treat \(bx\) as our 'X' and apply the condition to it:

\[|bx| < 1\]

To find the range of \(x\), you must isolate \(x\):

\[|x| < \frac{1}{|b|}\]

Step-by-Step Example (Finding Validity):

Find the range for the expansion of \((1+3x)^{-4}\).

1. Identify the term being powered (the 'x' part): \(3x\).
2. Set the absolute value less than 1: \(|3x| < 1\).
3. Divide by 3: \(|x| < \frac{1}{3}\).
4. Range: \(-\frac{1}{3} < x < \frac{1}{3}\).

Section 3: Handling Expressions Not Starting with 1

This is often the trickiest part of the Binomial Expansion for P4 students. Remember, the general formula only works for \((1+x)^n\).

3.1 The Crucial Step: Factorisation

If you have an expression in the form \((a+bx)^n\) where \(a \neq 1\), you must factor out \(a\) first to create the required 1 inside the bracket.

The Factorisation Rule:

\[(a+bx)^n = \left[a\left(1 + \frac{b}{a}x\right)\right]^n\]

Using the index law \((XY)^n = X^n Y^n\):

\[(a+bx)^n = a^n \left(1 + \frac{b}{a}x\right)^n\]

Now, you treat the term \(\frac{b}{a}x\) as your new single variable, \(X\), and expand \(\left(1 + X\right)^n\) using the standard formula. Finally, you multiply the entire expansion by the factor \(a^n\).

3.2 Step-by-Step Expansion of \((4-2x)^{\frac{1}{2}}\)

We want the first three terms and the range of validity.

Step 1: Rewrite into the \((1+X)^n\) form.
Factor out the 4 from the bracket and apply the power \(\frac{1}{2}\) to both factors:

\[(4-2x)^{\frac{1}{2}} = \left[4\left(1 - \frac{2x}{4}\right)\right]^{\frac{1}{2}}\]
\[= 4^{\frac{1}{2}} \left(1 - \frac{1}{2}x\right)^{\frac{1}{2}}\]
\[= 2 \left(1 - \frac{1}{2}x\right)^{\frac{1}{2}}\]

Here, \(n = \frac{1}{2}\) and our new variable is \(X = -\frac{1}{2}x\).

Step 2: Expand \((1+X)^n\).
Use the formula \(1 + nx + \frac{n(n-1)}{2!}x^2 + \dots\)

Term 1: \(1\)
Term 2: \(n(X) = \left(\frac{1}{2}\right)\left(-\frac{1}{2}x\right) = -\frac{1}{4}x\)
Term 3: \(\frac{n(n-1)}{2!}(X)^2 = \frac{(\frac{1}{2})(\frac{1}{2}-1)}{2} \left(-\frac{1}{2}x\right)^2\)
\[= \frac{(\frac{1}{2})(-\frac{1}{2})}{2} \left(\frac{1}{4}x^2\right) = \frac{-\frac{1}{4}}{2} \left(\frac{1}{4}x^2\right) = -\frac{1}{8} \cdot \frac{1}{4}x^2 = -\frac{1}{32}x^2\]

The expansion inside the bracket is: \(\left(1 - \frac{1}{4}x - \frac{1}{32}x^2 + \dots \right)\)

Step 3: Multiply by the factor.
Multiply the entire result by \(4^{\frac{1}{2}} = 2\):

\[2 \left(1 - \frac{1}{4}x - \frac{1}{32}x^2 + \dots \right) = 2 - \frac{2}{4}x - \frac{2}{32}x^2 + \dots\]
\[= 2 - \frac{1}{2}x - \frac{1}{16}x^2 + \dots\]

Step 4: Find the range of validity.
The expansion is valid when \(|X| < 1\). Our \(X\) is \(-\frac{1}{2}x\).

\[|-\frac{1}{2}x| < 1\]
\[\left|\frac{1}{2}x\right| < 1\]
\[|x| < 2\]

Section 4: Applications and Advanced Techniques

4.1 Approximations using the Binomial Series

One of the most powerful uses of the infinite series is to calculate approximations of complex values without relying on a calculator.

Step-by-Step Approximation Example:

Use the expansion of \((1+x)^{\frac{1}{3}}\) up to the \(x^2\) term to estimate \(\sqrt[3]{1.06}\).

Step A: Find \(x\).
We need \((1+x)^{\frac{1}{3}} = 1.06^{\frac{1}{3}}\).
This means \(1+x = 1.06\), so \(x = 0.06\).

Check validity: Since \(|x| = 0.06\), and \(0.06 < 1\), the approximation will be good.

Step B: Perform the Expansion.
For \(n = \frac{1}{3}\):
\[(1+x)^{\frac{1}{3}} \approx 1 + nx + \frac{n(n-1)}{2}x^2\]
\[\approx 1 + \frac{1}{3}x + \frac{(\frac{1}{3})(-\frac{2}{3})}{2}x^2\]
\[\approx 1 + \frac{1}{3}x - \frac{2/9}{2}x^2\]
\[\approx 1 + \frac{1}{3}x - \frac{1}{9}x^2\]

Step C: Substitute \(x = 0.06\).

\[\sqrt[3]{1.06} \approx 1 + \frac{1}{3}(0.06) - \frac{1}{9}(0.06)^2\]
\[\approx 1 + 0.02 - \frac{1}{9}(0.0036)\]
\[\approx 1 + 0.02 - 0.0004\]
\[\approx 1.0196\]

The more terms you use, the more accurate the approximation is!

4.2 Using Expansion for Rational Functions (Partial Fractions Connection)

Sometimes you need to expand a complex fraction, especially if the denominator is irreducible (cannot be factored).

Example: Find the expansion of \(\frac{3+x}{1-x}\)

Method: Separate the fraction and use negative indices.

\[\frac{3+x}{1-x} = (3+x)(1-x)^{-1}\]

First, expand \((1-x)^{-1}\) where \(n=-1\) and \(X=-x\):

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

This expansion is valid for \(|-x| < 1\), or \(|x| < 1\).

Second, multiply the resulting series by \((3+x)\):

\[(3+x)(1 + x + x^2 + \dots) = 3(1 + x + x^2 + \dots) + x(1 + x + x^2 + \dots)\]
\[= (3 + 3x + 3x^2 + \dots) + (x + x^2 + x^3 + \dots)\]

Finally, collect like terms (up to \(x^2\)):

\[= 3 + (3x+x) + (3x^2+x^2) + \dots\]
\[= 3 + 4x + 4x^2 + \dots\]

Did You Know?
The series \(1 + x + x^2 + x^3 + \dots\) is actually a famous geometric progression. Since the common ratio is \(x\), it only converges when \(|x| < 1\). This connection reinforces why the validity condition is so important!

Section 5: Summary and Final Checks

Key Takeaways for P4 Binomial Expansion

1. Index \(n\) is Rational: If \(n\) is negative or fractional, you must use the P4 General Formula, resulting in an infinite series.

2. The 1 is Mandatory: The formula \((1+X)^n\) is the only starting point. If you have \((a+bx)^n\), you must factorise \(a\) out: \(a^n(1+\frac{b}{a}x)^n\).

3. Validity Check: The series is only accurate within a specific range. If the expanded term is \(X\), the condition is \(|X| < 1\).

Don't worry if this topic feels dense initially! The mechanics are the same every time: factor, expand, multiply, check the range. Practice makes perfect!

You’ve mastered a fundamental tool in advanced mathematics. Keep going!