Question

Use the methods of this section to find the first few terms of the Maclaurin series for each of the following functions. $$ \ln \frac{1+x}{1-x} $$

Short answer

The first few terms of the Maclaurin series are 2x + x^2.

Step-by-step solution

Understanding the Maclaurin Series

The Maclaurin series for a function f(x) is given by: \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \]. We need to find the derivatives of the function at x = 0.

Define the Function

The function is given as: \[ f(x) = \, \ln \left( \frac{1+x}{1-x} \right) \]. Identify the first few derivatives of this function.

Compute the First Derivative

Find the first derivative of the function: \[ f'(x) = \frac{d}{dx} \left[ \ln \left( \frac{1+x}{1-x} \right) \right] = \frac{1}{1+x} \cdot \frac{d}{dx}(1+x) - \frac{1}{1-x} \cdot \frac{d}{dx}(1-x) = \frac{1}{1+x} + \frac{1}{1-x} = \frac{2}{1-x^2} \cdot x \].

Compute the Second Derivative

Find the second derivative of the function: \[ f''(x) = \frac{d}{dx} \left[ \frac{2x}{1-x^2} \right] = 2 \cdot \frac{(1-x^2)-x(-2x)}{(1-x^2)^2} = 2 \cdot \frac{1+x^2}{(1-x^2)^2}.\]

Evaluate the Derivatives at x=0

Evaluate the derivatives at x=0: \[ f(0) = \, \ln \left( \frac{1+0}{1-0} \right) = \, \ln(1) = 0, \]. \[ f'(0) = 2, \]. \[ f''(0) = 2. \].

Construct the Maclaurin Series

Substitute the derivatives into the Maclaurin series formula: \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots \]. This yields: \[ \ln \left( \frac{1+x}{1-x} \right) = 0 + 2x + \frac{2}{2}x^2 + \cdots \].

Key concepts

Derivative Calculation

When constructing a Maclaurin series, it's essential to calculate derivatives of the function at the point x=0. For our function \( f(x) = \ln \left( \frac{1+x}{1-x}\right) \), we start by finding the first few derivatives.

The first derivative, written as \ f'(x) \, helps us understand the function's slope. We apply the chain rule and get:\( f'(x) = \frac{d}{dx} \left[ \ln \left( \frac{1+x}{1-x} \right) \right] = \frac{2}{1-x^2} \).

Next, we find the second derivative by differentiating \ f'(x) \ again. This time, we need the quotient rule for differentiation:
\ f''(x) = \frac{d}{dx} \left[ \frac{2x}{1-x^2} \right] = 2 \cdot \frac{1+x^2}{(1-x^2)^2} \.

Remember, the derivatives must be computed accurately as they directly influence the terms in our Maclaurin series.

Evaluating Series at x=0

Evaluating the function and its derivatives at \( x=0 \) is a crucial step. Each term in the Maclaurin series is determined by these values.

Start with evaluating the function itself at zero:
\( f(0) = \ln \left( \frac{1+0}{1-0} \right) = \ln(1) = 0 \).
Then, we find the first derivative at x=0:
\( f'(0) = 2 \).
Lastly, evaluate the second derivative at x=0:
\( f''(0) = 2 \).

These evaluations are plugged into the Maclaurin series formula to get the corresponding coefficients. Higher-order derivatives follow the same evaluation process.

Series Expansion

Finally, build the Maclaurin series using the calculated values. Start with the general form of the series:

\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots \].

For our function, using the values we calculated:\(f(0) = 0, f'(0) = 2, f''(0) = 2\), we substitute these into the series:
\[\ ln \left( \frac{1+x}{1-x} \right) = 0 + 2x + \frac{2}{2}x^2 + \cdots \].

Thus, the first few terms of the Maclaurin series for \( \ln \left( \frac{1+x}{1-x} \right) \) are:

\[ 2x + x^2 + \cdots \].

Each additional term comes from higher-order derivatives, evaluated at \(x=0\), and contributes to the series expansion's accuracy. The more terms you include, the more accurately the series represents the function near \(x=0\).