Question

Find the Maclaurin series for the following functions. $$\cos [\ln (1+x)]$$

Short answer

The Maclaurin series for \(\cos[\ln(1 + x)]\) is \(1 - \frac{x^2}{2} + \frac{x^3}{2} - \frac{x^4}{8} + \ldots \).

Step-by-step solution

Understand the Maclaurin Series

The Maclaurin series for a function is its Taylor series expansion at 0. It is given by the formula: \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \ldots \]

Express the Inner Function

Recognize that the inner function \(\ln (1 + x)\) has its own Taylor series expansion. The expansion for \(\ln(1+x)\) around 0 is: \[ \ln (1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]

Substitute the Inner Function into Cosine

Substitute \(\ln (1 + x)\) into the cosine function. So, we need the Maclaurin series for \(\cos [\ln (1 + x)]\). Start with the known Maclaurin series of \(\cos u\): \[ \cos u = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \ldots \]

Combine the Series

Replace \(u\) with \(\ln (1 + x)\) in the cosine series: \[ \cos [\ln (1 + x)] = 1 - \frac{(\ln(1 + x))^2}{2!} + \frac{(\ln(1 + x))^4}{4!} - \ldots \]

Expand and Simplify

Using the expansion of \(\ln(1 + x)\), substitute it into the equation from Step 4 and expand it to collect terms in powers of \(x\):For example, substituting the first few terms:\[ \cos [\ln (1 + x)] \approx 1 - \frac{(x - \frac{x^2}{2} + \ldots)^2}{2!} + \ldots \]

Write the First Few Terms

Expanding this up to the first few terms, we get:\[ 1 - \frac{(x - \frac{x^2}{2})^2}{2} \]Calculating the squared term and simplifying gives:\[ 1 - \frac{x^2 - x^3 + \frac{x^4}{4}}{2} + \ldots = 1 - \frac{x^2}{2} + \frac{x^3}{2} - \frac{x^4}{8} + \ldots \]

Key concepts

Taylor Series

To understand the Maclaurin series, it's essential to grasp the broader concept of the Taylor series. A Taylor series is a way to represent any function as an infinite sum of terms calculated from its derivatives at a single point. The formula for the Taylor series expansion around a point \(a\) is:
\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \]
For a Maclaurin series, which is a special case of the Taylor series around \(a=0\), the formula simplifies to:
\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \]
This series allows us to approximate functions with polynomial terms, making calculations simpler and analyses more manageable.

Expansion

Expansion in terms of series means expressing functions as sums of potentially infinite terms. In the case of power series, we express functions as sums of polynomials.
The Maclaurin series for \( \ln(1+x) \) provides a clear example of this concept:
\[ \ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \]
Another familiar series is that of the cosine function:
\[ \cos u = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \cdots \]
Using these expansions, we can substitute one series into another to understand the behavior of more complex functions, such as \( \cos [\ln (1 + x)] \).

Logarithmic Function

The logarithmic function \( \ln(1 + x) \) has a well-known Taylor series expansion around 0. This series is:
\[ \ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \]
The expansion helps us approximate the logarithmic function with polynomials. This representation is particularly useful for integrating logarithms into other functions, such as in the exercise involving \( \cos [\ln (1 + x)] \). By substituting the expansion of \( \ln(1 + x) \) into the cosine series, we can then re-expand and simplify the resulting expression.

Cosine Function

The cosine function, \( \cos(x) \), is also commonly represented by its Maclaurin series. This is expressed as:
\[ \cos u = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \cdots \]
In our exercise, we utilized this series to determine \( \cos [\ln (1 + x)] \). By substituting \( \ln(1 + x) \) into each term of the cosine series and expanding, we approximate \( \cos [\ln (1 + x)] \) in terms of powers of \( x \). The combination of these series enables us to simplify complex functions into more manageable, understandable forms.