Problem 33 - Calculus: Early Transcendentals

Question

Use the binomial series to expand the function as a power series. State the radius of convergence. \( \frac {1}{(2 + x)^3} \)

Step-by-Step Solution

Answer

The power series is \(\frac{1}{8} - \frac{3x}{16} + \frac{3x^2}{32} - \frac{5x^3}{64} + \cdots\) with radius of convergence \(2\).

Step 1: Identify the Form for Binomial Series

The Binomial series is used for functions of the form \((1 + x)^n\). To use the binomial series, we need to express \(\frac{1}{(2 + x)^3}\) in a similar form. Rewrite \(\frac{1}{(2+x)^3}\) as \(\left(\frac{1}{2+x}\right)^3\).

Step 2: Transform the Function

Re-express \(\frac{1}{2+x}\) as a series by writing it in the form \(\frac{1}{2}\cdot\frac{1}{1+\frac{x}{2}}\). This step is achieved by factoring \(2\) out of \((2+x)\). Look at the simple form \(\left(\frac{1}{2}\right)^3\cdot\left(1+\frac{x}{2}\right)^{-3}\), which is now suited for applying the binomial series.

Step 3: Use the Binomial Series Formula

The binomial series formula is \((1 + u)^n = \sum_{k=0}^{\infty} \binom{n}{k} u^k\). Here, let \(u= \frac{x}{2}\) and \(n = -3\) (since our power is negative).- The general term will be \(\binom{-3}{k} \left(\frac{x}{2}\right)^k\).

Step 4: Compute Binomial Series Terms

The term \(\binom{-3}{k}\) is given by \(\frac{(-3)(-4)(-5)...(-3-k+1)}{k!}\). Write out the first few terms:1. First term \(k=0\): \(1\)2. Second term \(k=1\): \(-3\cdot\frac{x}{2} = -\frac{3x}{2}\)3. Third term \(k=2\): \(\frac{-3 \cdot -4}{2!}\cdot\left(\frac{x}{2}\right)^2 = \frac{3 \cdot 4 \cdot x^2}{8} = \frac{3x^2}{4}\)Continue similarly for additional terms if needed.

Step 5: Determine Full Power Series

The power series for \((2+x)^{-3}\) becomes:\(1 - \frac{3x}{2} + \frac{3x^2}{4} - \frac{5x^3}{8} + \cdots\).Therefore, multiply each term by \((\frac{1}{2})^3 = \frac{1}{8}\) to go back to the original expression \(\frac{1}{(2+x)^3}\):\(\frac{1}{8}\left(1 - \frac{3x}{2} + \frac{3x^2}{4} - \frac{5x^3}{8} + \cdots\right)\) yields the series.

Step 6: Determine Radius of Convergence

To determine the radius of convergence, observe \(1 + \frac{x}{2} = 0\). Here, \(|\frac{x}{2}| < 1\) implies \(|x| < 2\). Thus, the radius of convergence, \(R\), is \(2\).

Key Concepts

Power Series

A power series is an infinite series in the form

• \( \sum_{k=0}^{\infty} a_k (x-c)^k \)

There, each \(a_k\) is a constant coefficient, \(x\) is the variable, and \(c\) is the center around which the series is expanded.
This series plays a crucial role in calculus, as it allows complex functions to be expressed as an infinite sum of simpler terms.
In our exercise, we transformed the function \(\frac{1}{(2+x)^3}\) into a power series using the Binomial Theorem, which provided a way to write each term of the series based on binomial coefficients.
The resulting series for \(\frac{1}{(2+x)^3}\) was:

• \( \frac{1}{8} \left(1 - \frac{3x}{2} + \frac{3x^2}{4} - \frac{5x^3}{8} + \cdots\right) \)

This representation is powerful as it allows you to approximate the function with as many terms of the series as needed to reach the desired accuracy.

Radius of Convergence

The radius of convergence is a key concept when dealing with power series.
It tells us the range of \(x\) values for which the series converges to the function.In simpler terms, it's the distance from the center \(c\) to the point where the series stops providing reliable approximations.
This is vital because outside this interval, the series may diverge, meaning it won't accurately represent the function.
For the transformed series \(\frac{1}{(2+x)^3}\), the term \(\left(1 + \frac{x}{2}\right)^{-3}\) determined the radius of convergence.

• The condition \(|\frac{x}{2}| < 1\) ensures convergence and leads to \(|x| < 2\).

Therefore, the radius of convergence for this series is \(R = 2\). This means when \(|x| < 2\), the series will converge to the original function \(\frac{1}{(2+x)^3}\).

Binomial Coefficients

Binomial coefficients are the numerical factors which appear in the binomial theorem.
They are expressed as \(\binom{n}{k}\) and are integral in expanding functions into power series.For any two numbers \((n, k)\), the binomial coefficient \(\binom{n}{k}\) is calculated using:

• \( \binom{n}{k} = \frac{n(n-1)(n-2)...(n-k+1)}{k!} \)

In our power series expansion of \(\left(1 + \frac{x}{2}\right)^{-3}\), these coefficients defined each term of the series.
When \(n=-3\), the binomial coefficients became crucial, influencing the signs and values of the resulting expressions.

• The first few binomial coefficients were:
• \(\binom{-3}{0} = 1 \)
• \(\binom{-3}{1} = -3 \)
• \(\binom{-3}{2} = \frac{-3(-4)}{2} = 6 \)

These coefficients smoothly guided the formation of the power series by modifying each subsequent term.