Answer:
-x¹⁴ / 5040
-½ < x < ½
Step-by-step explanation:
f(x) = e^(-x²)
The Taylor series for eˣ centered at 0 is:
eˣ = ∑ (1/n!) xⁿ
Substitute -x²:
e^(-x²) = ∑ (1/n!) (-x²)ⁿ
e^(-x²) = ∑ (1/n!) (-1)ⁿ x²ⁿ
The 14th degree term occurs at n=7.
(1/7!) (-1)⁷ x¹⁴
-x¹⁴ / 5040
ln(1 + x) = ∑ₙ₌₁°° (-1)ⁿ⁺¹ xⁿ / n
If we substitute 4x²:
ln(1 + 4x²) = ∑ₙ₌₁°° (-1)ⁿ⁺¹ (4x²)ⁿ / n
Using ratio test:
lim(n→∞)│aₙ₊₁ / aₙ│< 1
lim(n→∞)│[(-1)ⁿ⁺² (4x²)ⁿ⁺¹ / (n+1)] / [(-1)ⁿ⁺¹ (4x²)ⁿ / n]│< 1
lim(n→∞)│-1 (4x²) n / (n+1)│< 1
4x² < 1
x² < ¼
-½ < x < ½
Answer:
0.5,1.2,23,-34,-73
Step-by-step explanation:
Answer:
i) 28 - 30i
ii) 36 + 28i
Step-by-step explanation:
i) x = 6 + i ⇒2x = 2(6 + i) = 12 + 2i
z = 4 - 8i ⇒ 4z = 4(4 - 8i) = 16 - 32i
2x + 4z = (12 + 2i) + (16 - 32i) = 28 - 30i
ii) w = -1 + 5i and z = 4 - 8i
w × z = (-1 + 5i)(4 - 8i) = -4 + 8i + 20i - 40
⇒collect like terms
w × z = -4 + 28i - 40
∵ 
∴w × z = -4 + 28i - 40(-1) = -4 + 28i + 40 = 36 + 28i
Answer:
-50
Step-by-step explanation:
