The Maclaurin series for a function is a special case of the Taylor series centered at zero. For the function ln(x), we need to find the Maclaurin series around x = 1, since ln(x) is not defined at x = 0.
To derive the Maclaurin series for ln(x), we can use the fact that it can be expressed as a Taylor series around a point. The general formula for the Taylor series of a function f(x) around a point a is given by:
f(x) = f(a) + f'(a)(x-a) + f”(a)(x-a)²/2! + f”'(a)(x-a)³/3! + …
In our case, we use x = 1 as the center:
First, we compute the derivatives of ln(x):
- f(x) = ln(x) ⇒ f(1) = ln(1) = 0
- f'(x) = 1/x ⇒ f'(1) = 1/1 = 1
- f”(x) = -1/x² ⇒ f”(1) = -1/1² = -1
- f”'(x) = 2/x³ ⇒ f”'(1) = 2/1³ = 2
- f⁽⁴⁾(x) = -6/x⁴ ⇒ f⁽⁴⁾(1) = -6/1⁴ = -6
Continuing this process, we can see a pattern that leads to:
ln(x) = (x-1) – (1/2)(x-1)² + (1/3)(x-1)³ – (1/4)(x-1)⁴ + …
More formally, the Maclaurin series (centered at x = 1) for ln(x) can be expressed as:
ln(x) = ∑ (-1)^(n-1)(x-1)ⁿ/n, for n = 1 to ∞
This series converges for 0 < x ≤ 2. This means that the Maclaurin series for ln(x) will help us estimate the function near x = 1 accurately.