Nikita Sharma 2019W2 MATH 101 209
Assignment Homework 12 due 9/12/14 at 9pm
1. (1 point) Find the sum of the following series. If it is 3. (1 point) Find the first five non-zero terms of Taylor series
divergent, type ”Diverges” or ”D”. centered at x = −4 for the function below.
∞ 2n+1
n 1 1 1 1 1 1 f (x) = ex
∑ (−1) = − + − +···
n=0 2n + 1 2 2 24 160 896
Answer:
Answer: f (x) = + + + +
+···
Note: You cannot write a decimal number for the answer.
Answer(s) submitted:
Hint: What power series is equal to this sum? •
•
•
Answer(s) submitted:
•
• •
(incorrect) (incorrect)
Correct Answers: Correct Answers:
• atan(1/2)
• eˆ(-4)
• eˆ(-4)*(x--4)
2. (1 point) • eˆ(-4)/2*(x--4)ˆ2
Suppose g is a function which has continuous derivatives, • eˆ(-4)/6*(x--4)ˆ3
and that g(5) = 3, g0 (5) = −5, g00 (5) = 2, g000 (5) = 5. • eˆ(-4)/24*(x--4)ˆ4
What is the Taylor series for g near 5, up to and including the
term containing x3 ?
4. (1 point) Write the Maclaurin series for f (x) = 4 cos 7x2
P3 (x) =
∞
Use this part of the Taylor series to approximate g(5.1).
g(5.1) ≈
as ∑ cn xn .
n=0
Solution: Find the following coefficients.
SOLUTION c0 =
We have c2 =
g00 (5) g000 (5) c4 =
g(x) = g(5)+g0 (5)(x−5)+ (x−5)2 + (x−5)3 +· · ·
2! 3! c6 =
Substituting gives c8 =
2 5 Answer(s) submitted:
g(x) = 3 − 5(x − 5) + (x − 5)2 + (x − 5)3 + · · ·
2! 3! •
From the first four terms of the Taylor series, we obtain •
•
1 1
P3 (4.9) = 3+(−5)(5.1 − 5)+ ·2(5.1 − 5) + ·5(5.1 − 5) = 2.51083.•
2 3
2! 3! •
Answer(s) submitted: (incorrect)
• Correct Answers:
• • 4
(incorrect) • 0
Correct Answers: • -98
• 0
• 3+-5*(x-5)+1/2!*2*(x-5)ˆ2+1/3!*5*(x-5)ˆ3
• 400.167
• 2.51083
1
, • 4*x
cos 6x2 − 1
• -(4ˆ3/6)*xˆ3
5. (1 point) Let f (x) = . Evaluate the 6th deriv- • 4ˆ5/5!*xˆ5
x2
ative of f at x = 0. • -(4ˆ7/7!)*xˆ7
f (6) (0) = • 4ˆ9/9!*xˆ9
Hint: Build a Maclaurin series for f (x) from the series for
cos(x). 8. (1 point)
Answer(s) submitted: Find the Maclaurin series of the function f (x) =
• (2x) arctan(7x2 ).
(incorrect) ∞
Correct Answers: f (x) = ∑ cn xn
• 38880 n=0
6. (1 point) Write the Maclaurin series for f (x) = 10x2 e−9x Determine the following coefficients:
∞ c3 =
as ∑ cn xn .
n=0
Find the first six coefficients. c5 =
c0 =
c7 =
c1 =
c2 = c9 =
c3 =
c4 = c11 =
c5 =
Answer(s) submitted:
Answer(s) submitted:
•
• •
• •
• •
• •
•
• (incorrect)
Correct Answers:
(incorrect)
Correct Answers: • 7 * 2
• 0 • 0
• 0 • -(7ˆ3) *
• 10 • 0
• -90 • (7ˆ5) *
• 405
• -1215 9. (1 point)
The Taylor series of function f (x) = ln(x) at a = 4 is given
7. (1 point) Find the first five non-zero terms of Maclaurin
by:
series (Taylor series centered at x = 0) for the function below.
f (x) = sin 4x ∞
f (x) = ∑ cn (x − 4)n
n=0
Answer: f (x) = + + + +
+···
Find the following coefficients:
Answer(s) submitted: c0 =
•
• c1 =
•
• c2 =
•
(incorrect) c3 =
Correct Answers:
2
Assignment Homework 12 due 9/12/14 at 9pm
1. (1 point) Find the sum of the following series. If it is 3. (1 point) Find the first five non-zero terms of Taylor series
divergent, type ”Diverges” or ”D”. centered at x = −4 for the function below.
∞ 2n+1
n 1 1 1 1 1 1 f (x) = ex
∑ (−1) = − + − +···
n=0 2n + 1 2 2 24 160 896
Answer:
Answer: f (x) = + + + +
+···
Note: You cannot write a decimal number for the answer.
Answer(s) submitted:
Hint: What power series is equal to this sum? •
•
•
Answer(s) submitted:
•
• •
(incorrect) (incorrect)
Correct Answers: Correct Answers:
• atan(1/2)
• eˆ(-4)
• eˆ(-4)*(x--4)
2. (1 point) • eˆ(-4)/2*(x--4)ˆ2
Suppose g is a function which has continuous derivatives, • eˆ(-4)/6*(x--4)ˆ3
and that g(5) = 3, g0 (5) = −5, g00 (5) = 2, g000 (5) = 5. • eˆ(-4)/24*(x--4)ˆ4
What is the Taylor series for g near 5, up to and including the
term containing x3 ?
4. (1 point) Write the Maclaurin series for f (x) = 4 cos 7x2
P3 (x) =
∞
Use this part of the Taylor series to approximate g(5.1).
g(5.1) ≈
as ∑ cn xn .
n=0
Solution: Find the following coefficients.
SOLUTION c0 =
We have c2 =
g00 (5) g000 (5) c4 =
g(x) = g(5)+g0 (5)(x−5)+ (x−5)2 + (x−5)3 +· · ·
2! 3! c6 =
Substituting gives c8 =
2 5 Answer(s) submitted:
g(x) = 3 − 5(x − 5) + (x − 5)2 + (x − 5)3 + · · ·
2! 3! •
From the first four terms of the Taylor series, we obtain •
•
1 1
P3 (4.9) = 3+(−5)(5.1 − 5)+ ·2(5.1 − 5) + ·5(5.1 − 5) = 2.51083.•
2 3
2! 3! •
Answer(s) submitted: (incorrect)
• Correct Answers:
• • 4
(incorrect) • 0
Correct Answers: • -98
• 0
• 3+-5*(x-5)+1/2!*2*(x-5)ˆ2+1/3!*5*(x-5)ˆ3
• 400.167
• 2.51083
1
, • 4*x
cos 6x2 − 1
• -(4ˆ3/6)*xˆ3
5. (1 point) Let f (x) = . Evaluate the 6th deriv- • 4ˆ5/5!*xˆ5
x2
ative of f at x = 0. • -(4ˆ7/7!)*xˆ7
f (6) (0) = • 4ˆ9/9!*xˆ9
Hint: Build a Maclaurin series for f (x) from the series for
cos(x). 8. (1 point)
Answer(s) submitted: Find the Maclaurin series of the function f (x) =
• (2x) arctan(7x2 ).
(incorrect) ∞
Correct Answers: f (x) = ∑ cn xn
• 38880 n=0
6. (1 point) Write the Maclaurin series for f (x) = 10x2 e−9x Determine the following coefficients:
∞ c3 =
as ∑ cn xn .
n=0
Find the first six coefficients. c5 =
c0 =
c7 =
c1 =
c2 = c9 =
c3 =
c4 = c11 =
c5 =
Answer(s) submitted:
Answer(s) submitted:
•
• •
• •
• •
• •
•
• (incorrect)
Correct Answers:
(incorrect)
Correct Answers: • 7 * 2
• 0 • 0
• 0 • -(7ˆ3) *
• 10 • 0
• -90 • (7ˆ5) *
• 405
• -1215 9. (1 point)
The Taylor series of function f (x) = ln(x) at a = 4 is given
7. (1 point) Find the first five non-zero terms of Maclaurin
by:
series (Taylor series centered at x = 0) for the function below.
f (x) = sin 4x ∞
f (x) = ∑ cn (x − 4)n
n=0
Answer: f (x) = + + + +
+···
Find the following coefficients:
Answer(s) submitted: c0 =
•
• c1 =
•
• c2 =
•
(incorrect) c3 =
Correct Answers:
2
