Explanation:
1) I believe you are correct about the exponent. f(x) = e^(-x²).
A Maclaurin series is a Taylor series about x=0:
f(x) = f(0) + f'(0)/1! x + f''(0)/2! x² + f'''(0)/3! x³ + ...
First, we find f''(x).
f(x) = e^(-x²)
f'(x) = -2x e^(-x²)
f''(x) = 4x² e^(x²) − 2 e^(x²)
Evaluate at x = 0:
f''(0) = -2
Find the coefficient:
f''(0) / 2! = -2/2! = -1
2) ∑ (xᵏ / k!) = eˣ
∑ 1 / (4ᵏ / k!)
= ∑ (¼ᵏ / k!)
= e^¼
= ∜e
3) The x⁴ term in the Maclaurin series will have a coefficient of f⁽⁴⁾(0) / 4!. From the pattern, that coefficient is 36. Therefore:
36 = f⁽⁴⁾(0) / 4!
f⁽⁴⁾(0) = 864
4) f(x) = x cos(x²)
cos(t) = ∑ (-1)ⁿ t²ⁿ / (2n)!
cos(x²) = ∑ (-1)ⁿ x⁴ⁿ / (2n)!
x cos(x²) = ∑ (-1)ⁿ x⁴ⁿ⁺¹ / (2n)!
When n = 0, the first term of the Maclaurin series is x.
When n = 1, the next non-zero term of the Maclaurin series is -x⁵/2.
So the 5th order Maclaurin polynomial is f(x) = x − x⁵/2.
5) e^x = ∑ (xⁿ / n!)
e^(x²) = ∑ (x²ⁿ / n!)
x² e^(x²) = ∑ (x²ⁿ⁺² / n!)