Discover the magic of Maclaurin and Taylor series!
Maclaurin series is a Taylor series expansion around x = 0.
f(x) = eˣ ⇒ f(0) = e⁰ = 1
f'(x) = eˣ ⇒ f'(0) = 1
f''(x) = eˣ ⇒ f''(0) = 1
All derivatives at x=0 are 1!
Maclaurin series: f(x) = f(0) + f'(0)x + f''(0)x²/2! + f'''(0)x³/3! + ...
eˣ = 1 + x + x²/2! + x³/3! + x⁴/4! + ...
f(x) = sin(x) ⇒ f(0) = 0
f'(x) = cos(x) ⇒ f'(0) = 1
f''(x) = -sin(x) ⇒ f''(0) = 0
f'''(x) = -cos(x) ⇒ f'''(0) = -1
The pattern repeats every 4 derivatives.
Only odd powers remain (since even derivatives are 0 at x=0)
sin(x) = x - x³/3! + x⁵/5! - x⁷/7! + ...
f(x) = cos(x) ⇒ f(0) = 1
f'(x) = -sin(x) ⇒ f'(0) = 0
f''(x) = -cos(x) ⇒ f''(0) = -1
f'''(x) = sin(x) ⇒ f'''(0) = 0
The pattern repeats every 4 derivatives.
Only even powers remain (since odd derivatives are 0 at x=0)
cos(x) = 1 - x²/2! + x⁴/4! - x⁶/6! + ...
f(x) = log(1-x) ⇒ f(0) = log(1) = 0
f'(x) = -1/(1-x) ⇒ f'(0) = -1
f''(x) = -1/(1-x)² ⇒ f''(0) = -1
f'''(x) = -2/(1-x)³ ⇒ f'''(0) = -2
Pattern: f⁽ⁿ⁾(0) = -(n-1)!
log(1-x) = -x - x²/2 - x³/3 - x⁴/4 - ...
Valid for -1 ≤ x < 1
f(x) = tan⁻¹(x) ⇒ f(0) = 0
f'(x) = 1/(1+x²) ⇒ f'(0) = 1
f''(x) = -2x/(1+x²)² ⇒ f''(0) = 0
f'''(x) = (6x²-2)/(1+x²)³ ⇒ f'''(0) = -2
Pattern alternates between 0 and non-zero values
Only odd powers remain, alternating signs
tan⁻¹(x) = x - x³/3 + x⁵/5 - x⁷/7 + ...
Valid for -1 ≤ x ≤ 1
We know that cos²(x) = (1 + cos(2x))/2
cos(x) = 1 - x²/2! + x⁴/4! - x⁶/6! + ...
So cos(2x) = 1 - (2x)²/2! + (2x)⁴/4! - (2x)⁶/6! + ...
cos²(x) = ½[1 + (1 - (2x)²/2! + (2x)⁴/4! - ...)]
cos²(x) = 1 - x² + x⁴/3 - 2x⁶/45 + ...
f(x) = log(x) ⇒ f(1) = 0
f'(x) = 1/x ⇒ f'(1) = 1
f''(x) = -1/x² ⇒ f''(1) = -1
f'''(x) = 2/x³ ⇒ f'''(1) = 2
Pattern: f⁽ⁿ⁾(1) = (-1)ⁿ⁺¹(n-1)!
Taylor series about x=1: f(x) = f(1) + f'(1)(x-1) + f''(1)(x-1)²/2! + ...
log(x) = (x-1) - (x-1)²/2 + (x-1)³/3 - ...
First three non-zero terms shown
f(x) = sin(x) ⇒ f(π/4) = √2/2 ≈ 0.7071
f'(x) = cos(x) ⇒ f'(π/4) = √2/2
f''(x) = -sin(x) ⇒ f''(π/4) = -√2/2
f'''(x) = -cos(x) ⇒ f'''(π/4) = -√2/2
Pattern repeats every 4 derivatives
Taylor series about x=π/4:
sin(x) = √2/2 [1 + (x-π/4) - (x-π/4)²/2! - (x-π/4)³/3! + ...]
First three non-zero terms shown
f(x) = x² - 3x + 2 ⇒ f(1) = 1 - 3 + 2 = 0
f'(x) = 2x - 3 ⇒ f'(1) = -1
f''(x) = 2 ⇒ f''(1) = 2
Higher derivatives are 0
f(x) = 0 + (-1)(x-1) + 2(x-1)²/2! = -(x-1) + (x-1)²
Let y = x-1 ⇒ x = y+1
f(x) = (y+1)² - 3(y+1) + 2 = y² + 2y + 1 - 3y - 3 + 2
f(x) = y² - y = (x-1)² - (x-1)
Same result as Method 1!