(-1)"x2n+ Given the power series > n=0 (2n + 1)n! |an+1 (a) Simplify an and find the limit. State the radius of convergence and the end- points. If applicable, substitute to show whether each endpoint is in the interval of convergence or not.
(-1)"x2n+ Given the power series > n=0 (2n + 1)n! |an+1 (a) Simplify an and find the limit. State the radius of convergence and the end- points. If applicable, substitute to show whether each endpoint is in the interval of convergence or not.
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.3: Geometric Sequences
Problem 49E
Related questions
Question
![(-1)"x²n+1
(2n + 1)п!
2. Given the power series >
n=0
An+1
(a) Simplify
and find the limit. State the radius of convergence and the end-
An
points. If applicable, substitute to show whether each endpoint is in the interval
of convergence or not.
(b) It can be shown that the series converges to f(x) =
dt on its interval of
convergence. To illustrate this, find s10, 820, and s30. Plot these three polynomials
and f on the same set of axes in the window x E [-10, 10], y E [-2, 2].
-t2
(c) Notice that e-t dt cannot be integrated using standard techniques, but the
series can be used to approximate values of the definite integral in f for any value
of x. Use s100 to obtain a decimal approximation for f(5) (NOTE from the graph
that this is a pretty good approximation for e-ť dt). Compare your answer
with the decimal approximation for
What do you notice?
2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc6d45b7c-770b-4dbe-b3cf-74fe1dfded8d%2F398fc581-de21-4b95-8cf1-fb6dcb62e90f%2F20tuewa_processed.png&w=3840&q=75)
Transcribed Image Text:(-1)"x²n+1
(2n + 1)п!
2. Given the power series >
n=0
An+1
(a) Simplify
and find the limit. State the radius of convergence and the end-
An
points. If applicable, substitute to show whether each endpoint is in the interval
of convergence or not.
(b) It can be shown that the series converges to f(x) =
dt on its interval of
convergence. To illustrate this, find s10, 820, and s30. Plot these three polynomials
and f on the same set of axes in the window x E [-10, 10], y E [-2, 2].
-t2
(c) Notice that e-t dt cannot be integrated using standard techniques, but the
series can be used to approximate values of the definite integral in f for any value
of x. Use s100 to obtain a decimal approximation for f(5) (NOTE from the graph
that this is a pretty good approximation for e-ť dt). Compare your answer
with the decimal approximation for
What do you notice?
2
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