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
No, 0=0 has an infinite amount of solutions.
Answer:
Step-by-step explanation:
If two angles are supplement of each other then one of the angles must be acute
Answer:
$40.52
Step-by-step explanation:
You add the amount of all the t-shirts that Mike bought and subtract it from 100.
