The Maclaurin series expansion of f(x) = 1/(1-x)² is
1 + 2x + 6x² + 24x³ +......
The general expression for the Maclaurin series is given by the formula:
f(x) = f(0) + xf'(0) + x²f''(0) + ..... + xⁿf⁽ⁿ⁾(0)+...
= ∑ f⁽ⁿ⁾ (0)xⁿ/n!
n= 0 to infinity
Specifying the convergence region completes the Maclaurin series formulation. Detailed five steps for determining the Maclaurin series of
f(x) = 1/(1-x)²
Step 1 : Find Derivatives for f(x)
f(x) = 1/(1-x)²
f'(x) = -2(1-x)⁽⁻²⁻¹⁾ (-1) = 2/(1-x)³
The second derivative of f(x) is the derivative of f''(x) :
f"(x) = 2(-3)(1-x)⁽⁻³⁻¹⁾(-1) = 6/(1-x)⁴
Differentiating again, f"'(x) = 6(-4)(1-x)⁽⁻⁴⁻¹⁾(-1)
= 24/(1-x)⁵
Thus, the nᵗʰ derivative is ,
f⁽ⁿ⁾(x) = (n+1)!/(1-x)⁽ⁿ⁻¹⁾
Step 2: Evaluate These Derivatives and f(x) at x=0 let x = 0 , f(0) = 1
f'(0) = 2 , f"(0) = 6 , ....... , f⁽ⁿ⁾(0) = (n+1)!
Step 3 : Now, the Sum of Products :
f(x) = f(0) + xf'(0) + x²f''(0) + ..... + xⁿf⁽ⁿ⁾(0)+...
xf'(0) = 2x
x²f"(0) = 6x²
x³f"'(0) = 24 x³
----------------------------
xⁿf⁽ⁿ⁾(0)= (n+1)!
Thus, the Maclaurin series for ln(1 + x) is this:
f(x) = (1−x)⁻² = 1 + 2x + 6x² + 24x³ +......
To learn more about Maclaurin series , refer:
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