Question
Asked Nov 13, 2019
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By manipulating a geometric series, determine a closed form power series for each function below. Find the interval of convergence.

b. f(x)-3 tan (2x)
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b. f(x)-3 tan (2x)

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Expert Answer

Step 1
The given function is f(x)3 tan(2x)
Obtain the power series expansion for 3tan (2x
.7
tan(x)x
3
5
7
n+1
-Σ-1).
(2n 1)
n0
2n+1
)" (2x)
3 tan (2x)-3-1
(2n 1)
n 0
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The given function is f(x)3 tan(2x) Obtain the power series expansion for 3tan (2x .7 tan(x)x 3 5 7 n+1 -Σ-1). (2n 1) n0 2n+1 )" (2x) 3 tan (2x)-3-1 (2n 1) n 0

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Step 2
Use the Ratio test to find the interval of convergence
3(-1)* (2x} ?»*
(2n 1)
2n+1
+2x)2n+1+1
3(1)* (2x)4)4
(2(n+1) 1)
(2x) 2e3
(2n 3)
(2.r)
3(-1)
=3-1)
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Use the Ratio test to find the interval of convergence 3(-1)* (2x} ?»* (2n 1) 2n+1 +2x)2n+1+1 3(1)* (2x)4)4 (2(n+1) 1) (2x) 2e3 (2n 3) (2.r) 3(-1) =3-1)

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Step 3
Apply the ratio test as follows
L lim
а,
2n+3
2x)
3(-1)
n+1
(2n 3)
(2x
=lim
2n+1
3(-1)'
(2n 1)
n
2n+3
(2.x)
(2n 1)
-lim 3(-1)
n+1
2n+1
(2n+3) 3(-1) (2x)
(2x)(2n 1)
-lim 3(-1)
(2n 2)
1
no
2n 1
(2x) lim
n2n2
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Apply the ratio test as follows L lim а, 2n+3 2x) 3(-1) n+1 (2n 3) (2x =lim 2n+1 3(-1)' (2n 1) n 2n+3 (2.x) (2n 1) -lim 3(-1) n+1 2n+1 (2n+3) 3(-1) (2x) (2x)(2n 1) -lim 3(-1) (2n 2) 1 no 2n 1 (2x) lim n2n2

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Math

Calculus