Further Pure Mathematics | The binomial series

🚀 The Binomial Series: Expanding Your Mathematical Horizon

Welcome to the fascinating world of The Binomial Series! If you have previously studied the standard Binomial Theorem, you might remember how useful it is for expanding expressions like \((a+b)^5\). However, that method only works if the power is a positive whole number.

In Further Pure Mathematics, we learn a powerful extension called the Binomial Series. This allows us to expand expressions where the power (\(n\)) is negative, fractional, or any rational number—like \((1+x)^{\frac{1}{2}}\) or \(\frac{1}{(1-x)}\)!

This skill is crucial because it allows us to approximate complicated expressions, especially those involving roots, using simple polynomials.

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1. Laying the Foundation: From Finite Expansion to Infinite Series

When you expanded \((1+x)^4\), you knew exactly how many terms you would get (five terms). This is a finite series.

The Binomial Series is different. When the power \(n\) is not a positive integer, the expansion never stops! It becomes an infinite series.
(Don't worry, in exams, you usually only need to calculate the first three or four terms.)

Did you know? The concept of using infinite series to approximate functions is a fundamental idea in advanced calculus and physics!

Prerequisite Check: Factorials (!!!)

You must remember what a factorial is, as it appears in the denominator of the formula:

• \(2! = 2 \times 1 = 2\)
• \(3! = 3 \times 2 \times 1 = 6\)
• \(r! = r \times (r-1) \times \dots \times 1\)

The factorial tells you which term coefficient you are calculating.

2. The General Binomial Series Formula

The General Binomial Series is only guaranteed to work reliably for expressions in the form \((1+x)^n\).

The formula for expanding \((1+x)^n\), where \(n\) is any rational number, is:

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

Let's break down the pattern:

• Term 1: Always 1.
• Term 2: \(nx\) (Power times the second term).
• Term 3: The numerator introduces one extra factor: \(n(n-1)\). The denominator is \(2!\). The power of \(x\) is \(x^2\).
• Term 4: The numerator introduces one extra factor: \(n(n-1)(n-2)\). The denominator is \(3!\). The power of \(x\) is \(x^3\).

Memory Aid (Numerator Trick): The number of factors in the numerator (e.g., \(n, n-1, n-2\)) always matches the number in the factorial denominator (e.g., 3 factors for \(3!\)) and the power of \(x\).

Example Walkthrough: Finding the First Four Terms

Let's find the expansion of \((1-x)^{-2}\) up to the term in \(x^3\).

Here, we have:
\(n = -2\)
\(x\) is replaced by \((-x)\)

Step 1: The first term (1)
$$ 1 $$

Step 2: The second term (\(nx\))
$$ n x = (-2)(-x) = +2x $$

Step 3: The third term (\(\frac{n(n-1)}{2!}x^2\))
$$ \frac{(-2)(-2-1)}{2!} (-x)^2 = \frac{(-2)(-3)}{2} (x^2) = \frac{6}{2}x^2 = +3x^2 $$

Step 4: The fourth term (\(\frac{n(n-1)(n-2)}{3!}x^3\))
$$ \frac{(-2)(-2-1)(-2-2)}{3!} (-x)^3 = \frac{(-2)(-3)(-4)}{6} (-x^3) = \frac{-24}{6}(-x^3) = (-4)(-x^3) = +4x^3 $$

Result:
$$ (1-x)^{-2} \approx 1 + 2x + 3x^2 + 4x^3 + \dots $$

3. The Crucial Condition for Validity: When Does It Work?

Since the Binomial Series is an infinite expansion, we need to know under what conditions the terms actually get smaller and smaller, ensuring the series sums up to the correct, finite value. This is called convergence.

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

$$ |x| < 1 $$

This means that \(x\) must be between -1 and 1:

$$ -1 < x < 1 $$

Analogy: Imagine using a cheap map to navigate. The map is accurate close to your starting point (where \(x\) is small), but the further you travel (as \(x\) gets close to or larger than 1), the map becomes useless, and your estimate is wrong.

Applying the Condition

If you expanded \((1+2y)^{-1}\), the \(x\) term in the formula is \(2y\). Therefore, the condition for validity becomes:

$$ |2y| < 1 $$ $$ 2|y| < 1 $$ $$ |y| < \frac{1}{2} $$ $$ -\frac{1}{2} < y < \frac{1}{2} $$

Always state the condition for validity after finding the expansion!

4. Generalising the Base: Handling \((a+x)^n\)

The General Binomial Series formula only applies directly to the form \((1+x)^n\). What if you need to expand \((2+x)^{-3}\)?

We must algebraically manipulate the expression so that the first term in the bracket is 1.

Step 1: Factor out the first term (\(a\))
Factor \(a\) out of the bracket. Since the bracket is raised to the power \(n\), the factored \(a\) also receives the power \(n\):

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

Step 2: Apply the series
Now, the expression is in the correct form, \(a^n (1 + X)^n\), where \(X = \frac{x}{a}\). You apply the standard Binomial Series formula to the bracket \(\left( 1 + \frac{x}{a} \right)^n\).

Step 3: Multiply and State Validity
Multiply the entire expansion by \(a^n\). The validity condition is now based on the new 'x' term, \(\frac{x}{a}\):

$$ \left| \frac{x}{a} \right| < 1 $$ $$ |x| < |a| $$

Example: Expanding \((4+x)^{\frac{1}{2}}\)

We want to find the first three terms of \((4+x)^{\frac{1}{2}}\).

1. Rewrite the expression:

$$ (4+x)^{\frac{1}{2}} = 4^{\frac{1}{2}} \left( 1 + \frac{x}{4} \right)^{\frac{1}{2}} = 2 \left( 1 + \frac{x}{4} \right)^{\frac{1}{2}} $$

Here, \(n = \frac{1}{2}\) and the 'x' term is \(X = \frac{x}{4}\).

2. Expand the bracket \(\left( 1 + \frac{x}{4} \right)^{\frac{1}{2}}\):

Term 1: \(1\)

Term 2 (\(nX\)):

$$ nX = \left(\frac{1}{2}\right) \left(\frac{x}{4}\right) = \frac{x}{8} $$

Term 3 (\(\frac{n(n-1)}{2!}X^2\)):

$$ \frac{\frac{1}{2}(\frac{1}{2}-1)}{2!} \left(\frac{x}{4}\right)^2 = \frac{\frac{1}{2}(-\frac{1}{2})}{2} \left(\frac{x^2}{16}\right) = \frac{-\frac{1}{4}}{2} \left(\frac{x^2}{16}\right) = -\frac{1}{8} \left(\frac{x^2}{16}\right) = -\frac{x^2}{128} $$

3. Combine and State Validity:

$$ (4+x)^{\frac{1}{2}} \approx 2 \left( 1 + \frac{x}{8} - \frac{x^2}{128} + \dots \right) $$ $$ (4+x)^{\frac{1}{2}} \approx 2 + \frac{2x}{8} - \frac{2x^2}{128} + \dots $$ $$ (4+x)^{\frac{1}{2}} \approx 2 + \frac{x}{4} - \frac{x^2}{64} + \dots $$

Validity condition: \(|X| < 1\)

$$ \left| \frac{x}{4} \right| < 1 \implies |x| < 4 \text{ or } -4 < x < 4 $$

5. Using the Binomial Series for Approximation

One primary use of these expansions is to find approximate values for numbers.

If you have a convergent series expansion for, say, \((1+x)^{\frac{1}{2}}\), and you choose a value for \(x\) that is small and within the validity range, the expansion gives a good numerical approximation.

Example: Approximating \(\sqrt{1.02}\)

Let's use the expansion for \((1+x)^{\frac{1}{2}} \approx 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \dots\)

We know that \(\sqrt{1.02} = (1.02)^{\frac{1}{2}}\). If we set \(1+x = 1.02\), then \(x = 0.02\).

Since \(x = 0.02\) is very small and satisfies the validity condition (\(|0.02| < 1\)), the approximation will be very accurate.

Substitute \(x = 0.02\) into the series:

$$ \sqrt{1.02} \approx 1 + \frac{1}{2}(0.02) - \frac{1}{8}(0.02)^2 $$ $$ \sqrt{1.02} \approx 1 + 0.01 - \frac{1}{8}(0.0004) $$ $$ \sqrt{1.02} \approx 1 + 0.01 - 0.00005 $$ $$ \sqrt{1.02} \approx 1.00995 $$

(A calculator gives \(\sqrt{1.02} \approx 1.00995049\), showing how close our three-term approximation is!)

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Summary: Essential Binomial Series Concepts

The binomial series is a powerful tool, but mastering it relies on two key things: correct substitution and understanding convergence.

• Formula: \((1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots\)
• Convergence: The expansion is only valid if \(|x| < 1\).
• General Form: If you have \((a+x)^n\), you must factor out \(a\) first: \(a^n \left(1 + \frac{x}{a}\right)^n\).
• Infinite Nature: If \(n\) is negative or fractional, the series never ends.

Don't worry if this seems tricky at first—practice calculating those fractional coefficients carefully, and always double-check your signs, especially when \(n\) is negative!

Good luck!