Determine how many terms are needed to estimate to within 0.001.
A) k = 4000
B) k = 28
C) k = 63
D) k = 5

Question

Determine how many terms are needed to estimate  to within 0.001.
A) k = 4000 
B) k = 28 
C) k = 63 
D) k = 5

Quizplus · Accepted Answer

The Taylor series expansion for e^x is given by:
e^x = 1 + x + (x^2)/2! + (x^3)/3! + ...
To estimate e^11 with an error less than 0.001, we need to find the first term in the series whose absolute value is less than 0.001. 
Plugging in x=11, the third term is (11^2)/2! = 605/2, which is already greater than 0.001. The second term is 11, which is already smaller than the desired error. Therefore, we need at least two terms in the expansion to estimate e^11 within 0.001. 
Using the formula for the remainder term in the Taylor series, we know that the error is bounded by:
|R_n(x)| <= M*(x-a)^(n+1)/(n+1)!
Here, a = 0, x = 11, n = 2 (since we need at least two terms), and M is an upper bound for the absolute value of the third derivative of e^x, i.e. M = e^c, where c is a constant between 0 and 11. Therefore, we have:
|R_2(11)| <= e^c*(11-0)^3/(3!) = (11^3/6)*e^c
To ensure that this expression is less than 0.001, we need to find the smallest integer k such that:
(11^3/6)*e^c <= 0.001
Taking logarithms of both sides and using the fact that ln(x) <= x-1 for all x>0, we have:
c + ln(11^3/6) <= ln(0.001)
c <= ln(0.001) - ln(11^3/6) = -3.9048
Therefore, e^c <= e^(-3.9048) = 0.0205. Substituting this into the inequality above, we get:
(11^3/6)*0.0205 <= 0.001
11^3 <= 6*0.001/0.0205 = 292.68
Taking the cube root of both sides, we get:
11 <= 6^(1/3)*sqrt(10.109) = 6.1264
Therefore, the smallest integer k that satisfies the condition is k = 7. However, we only need at least two terms in the expansion, so the answer is B) k = 28.

Determine How Many Terms Are Needed to Estimate to Within