Expand log (1+x) as a Maclaurin's series upto 4 non-zero terms for x ≤ 1.

To expand the function \( \log(1+x) \) as a Maclaurin series, we start by recalling that the Maclaurin series is a Taylor series centered at \( x = 0 \). The general formula for the Maclaurin series is:

Maclaurin Series Formula

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 + \ldots

Finding Derivatives

We need to calculate the derivatives of \( \log(1+x) \) at \( x = 0 \):

• f(x) = log(1+x)
• f(0) = log(1+0) = log(1) = 0
• f'(x) = \frac{1}{1+x} → f'(0) = 1
• f''(x) = -\frac{1}{(1+x)^2} → f''(0) = -1
• f'''(x) = \frac{2}{(1+x)^3} → f'''(0) = 2

Constructing the Series

Now, we can substitute these values into the Maclaurin series formula:

• First term: \( 0 \)
• Second term: \( 1 \cdot x = x \)
• Third term: \( -\frac{1}{2}x^2 \)
• Fourth term: \( \frac{2}{6}x^3 = \frac{1}{3}x^3 \)

Final Series Expansion

Combining these terms, the Maclaurin series expansion of \( \log(1+x) \) up to four non-zero terms is:

log(1+x) ≈ x - \frac{1}{2}x^2 + \frac{1}{3}x^3

This series is valid for \( |x| \leq 1 \). Thus, you can use this approximation for values of \( x \) within that range.