Problem 1 - The Calculus with Analytic Geometry

Question

Use a binomial series to find the Maclaurin series for the given function. Determine the radius of convergence of the resulting series.\(f(x)=\sqrt{1+x}\)

Step-by-Step Solution

Answer

The Maclaurin series for \( \sqrt{1+x} \) is \[ 1 + \frac{1}{2} x - \frac{1}{8} x^2 + \frac{1}{16} x^3 - ... \]. The radius of convergence is \ |x| < 1 \.

Step 1: Recall the Binomial Series Expansion

Recall that the binomial series expansion for \( (1+x)^n \) is given by: \[ (1+x)^n = \sum_{k=0}^{\infty} \binom{n}{k} x^k \], where \ \binom{n}{k} = \frac{n!}{k!(n-k)!} \.

Step 2: Set Up the Maclaurin Series for \( \sqrt{1+x} \)

In the given function \( f(x)=\sqrt{1+x} \), set \ n = \frac{1}{2} \ to apply the binomial series expansion: \[ \sqrt{1+x} = (1+x)^{1/2} = \sum_{k=0}^{\infty} \binom{1/2}{k} x^k \].

Step 3: Calculate the Binomial Coefficients

The binomial coefficient for \( \binom{1/2}{k} \) is calculated as: \[ \binom{1/2}{k} = \frac{(1/2)(1/2-1)(1/2-2)...(1/2-k+1)}{k!} \]. For example, the first few coefficients are: \binom{1/2}{0} = 1, \binom{1/2}{1} = \frac{1}{2}, \binom{1/2}{2} = \frac{1/2 (-1/2)}{2} = -\frac{1}{8}, \dots \.

Step 4: Write the Series

Using the calculated binomial coefficients, the Maclaurin series for \( \sqrt{1+x} \) becomes: \[ \sqrt{1+x} = 1 + \frac{1}{2} x - \frac{1}{8} x^2 + \frac{1}{16} x^3 - ... \].

Step 5: Determine the Radius of Convergence

The radius of convergence for a binomial series \( (1+x)^n \) with \ |n| < 1 \ is given by \( |x| < 1 \). Therefore, for \ f(x) = \sqrt{1+x} \ which uses \ n = \frac{1}{2} \, the radius of convergence is \( |x| < 1 \).

Key Concepts

Binomial Series Expansion

The binomial series expansion helps us to express functions like \( (1+x)^n \) in a series form. This is very useful in calculus and other areas of mathematics. The general form of the binomial series is: \[ (1+x)^n = \sum_{k=0}^{\infty} \binom{n}{k} x^k \].
Here, \( \binom{n}{k} \) represents the binomial coefficients which are computed as \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \).
The binomial coefficients tell us the weights of the terms in the series. By substituting values of \( k \) from 0 to \( \infty \), we can expand the function into an infinite series that approximates the function. This is particularly powerful because it allows us to simplify and work with complex functions in polynomial form.

Radius of Convergence

The radius of convergence determines how far from \( x = 0 \) our series will converge to the function. For a binomial series expansion, \( (1+x)^n \), the radius of convergence \( R \) depends on the exponent \( n \).
For example, if \( |n| < 1 \), the series will converge for \( |x| < 1 \).
So, if we have \( f(x) = \sqrt{1+x} \) which is the same as \( (1+x)^{1/2} \), the radius of convergence would be \( |x| < 1 \).
This means the series expansion is valid for \( x \) values between -1 and 1.

Binomial Coefficient Calculation

In our Maclaurin series, we need to calculate the binomial coefficients \( \binom{1/2}{k} \). These coefficients can be calculated using the formula: \[ \binom{1/2}{k} = \frac{(1/2)(1/2-1)(1/2-2)...(1/2-k+1)}{k!} \].
For example:

• \( \binom{1/2}{0} = 1 \)
• \( \binom{1/2}{1} = \frac{1}{2} \)
• \( \binom{1/2}{2} = \frac{1/2 (-1/2)}{2} = -\frac{1}{8} \)

Through this process, we generate the coefficients needed to construct the series. These coefficients can be a bit tricky to compute manually, so double-checking your calculations can be helpful.

Maclaurin Series

The Maclaurin series is a special case of the Taylor series, centered at \( x=0 \). It allows us to represent functions as an infinite sum of terms calculated from the derivatives at a single point (zero).
To find a Maclaurin series for \( \sqrt{1+x} \), we use the binomial series expansion with \( n = \frac{1}{2} \). This results in: \[ \sqrt{1+x} = 1 + \frac{1}{2} x - \frac{1}{8} x^2 + \frac{1}{16} x^3 - ... \].
Each term can be understood as a combination of the function's values and its derivatives at \( x=0 \). Using Maclaurin series, we can approximate the function \( \sqrt{1+x} \) for values within the radius of convergence.