Answer to Problem 7E The McLaurin series is T = (−1) (2)²(2n+1) 2n+1 and series always converges for [-1, 1]. Explanation of Solution Given: The given series f(x) = tan¯¹x². Formula used: tan ¹x=x- Calculation: Given the given function is f(x) = tan¹². By replacing x = x², tan¯¹² = x² - +1/+ +(-1) (2)²+1 2n+1 +(-1) (2)+2 2n+1 The first three non-zero terms of the series are T₁ = x²,T₂ = −3²º‚T3 = x²º. The general term of the series is given by π = (-1) (2) (2n+1) The interval of convergence uses the ratio test. lim 004-u lim 004-u Tn Tn-1 T T-1 = = lim (-1) (2)2(2+1) 004-2 lim 004-2 (2)4(2n-1) 2n-1 2n+1 (-1)-(z)2(2+1) 2n+1 lim 004-2 2n+1 Tn =x x4 lim T 2n-1 2n+1 004-2 = x².1 lim 004-2 Tn T T lim n+oo T So the ratio tells us that if <1 the series will converge and if > 1 the series will diverge.We don't know that what will happen so we have |x| < 1, −1 < x < 1. The way to determine the convergence at end points is to simply plug them into the power series and see if the convergence or divergence using any test necessary. The series converges between -1 ≤ x ≤1.
Answer to Problem 7E The McLaurin series is T = (−1) (2)²(2n+1) 2n+1 and series always converges for [-1, 1]. Explanation of Solution Given: The given series f(x) = tan¯¹x². Formula used: tan ¹x=x- Calculation: Given the given function is f(x) = tan¹². By replacing x = x², tan¯¹² = x² - +1/+ +(-1) (2)²+1 2n+1 +(-1) (2)+2 2n+1 The first three non-zero terms of the series are T₁ = x²,T₂ = −3²º‚T3 = x²º. The general term of the series is given by π = (-1) (2) (2n+1) The interval of convergence uses the ratio test. lim 004-u lim 004-u Tn Tn-1 T T-1 = = lim (-1) (2)2(2+1) 004-2 lim 004-2 (2)4(2n-1) 2n-1 2n+1 (-1)-(z)2(2+1) 2n+1 lim 004-2 2n+1 Tn =x x4 lim T 2n-1 2n+1 004-2 = x².1 lim 004-2 Tn T T lim n+oo T So the ratio tells us that if <1 the series will converge and if > 1 the series will diverge.We don't know that what will happen so we have |x| < 1, −1 < x < 1. The way to determine the convergence at end points is to simply plug them into the power series and see if the convergence or divergence using any test necessary. The series converges between -1 ≤ x ≤1.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
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Question
This way the ratio test was done in this conflicts what I learned which makes it difficult for me to follow.
I was taught with the limit as n approaches infinity for (an+1)/(an) = L
I need to find the interval of convergence for the series tan-1(x2). (The question has a table of Maclaurin series which I followed as well)
https://www.bartleby.com/solution-answer/chapter-92-problem-7e-advanced-placement-calculus-graphical-numerical-algebraic-sixth-edition-high-school-binding-copyright-2020-6th-edition/9781418300203/2c1feea0-c562-4cd3-82af-bef147eadaf9
![Answer to Problem 7E
The McLaurin series is T = (−1) (2)²(2n+1)
2n+1
and series always converges for [-1, 1].
Explanation of Solution
Given:
The given series f(x) = tan¯¹x².
Formula used:
tan ¹x=x-
Calculation:
Given the given function is f(x) = tan¹².
By replacing x = x²,
tan¯¹² = x² - +1/+
+(-1) (2)²+1
2n+1
+(-1) (2)+2
2n+1
The first three non-zero terms of the series are T₁ = x²,T₂ = −3²º‚T3 = x²º.
The general term of the series is given by π = (-1) (2) (2n+1)
The interval of convergence uses the ratio test.
lim
004-u
lim
004-u
Tn
Tn-1
T
T-1
=
=
lim (-1) (2)2(2+1)
004-2
lim
004-2
(2)4(2n-1)
2n-1
2n+1 (-1)-(z)2(2+1)
2n+1
lim
004-2
2n+1
Tn
=x x4 lim
T
2n-1
2n+1
004-2
= x².1
lim
004-2
Tn
T
T
lim
n+oo T
So the ratio tells us that if <1 the series will converge and if > 1 the series will diverge.We don't know that what
will happen so we have |x| < 1, −1 < x < 1. The way to determine the convergence at end points is to simply plug
them into the power series and see if the convergence or divergence using any test necessary.
The series converges between -1 ≤ x ≤1.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3632f992-2a17-4bac-8a23-761c010bf45a%2F912841ee-bc1a-4f7d-952b-b8a16793cff6%2Fpbx96nl_processed.png&w=3840&q=75)
Transcribed Image Text:Answer to Problem 7E
The McLaurin series is T = (−1) (2)²(2n+1)
2n+1
and series always converges for [-1, 1].
Explanation of Solution
Given:
The given series f(x) = tan¯¹x².
Formula used:
tan ¹x=x-
Calculation:
Given the given function is f(x) = tan¹².
By replacing x = x²,
tan¯¹² = x² - +1/+
+(-1) (2)²+1
2n+1
+(-1) (2)+2
2n+1
The first three non-zero terms of the series are T₁ = x²,T₂ = −3²º‚T3 = x²º.
The general term of the series is given by π = (-1) (2) (2n+1)
The interval of convergence uses the ratio test.
lim
004-u
lim
004-u
Tn
Tn-1
T
T-1
=
=
lim (-1) (2)2(2+1)
004-2
lim
004-2
(2)4(2n-1)
2n-1
2n+1 (-1)-(z)2(2+1)
2n+1
lim
004-2
2n+1
Tn
=x x4 lim
T
2n-1
2n+1
004-2
= x².1
lim
004-2
Tn
T
T
lim
n+oo T
So the ratio tells us that if <1 the series will converge and if > 1 the series will diverge.We don't know that what
will happen so we have |x| < 1, −1 < x < 1. The way to determine the convergence at end points is to simply plug
them into the power series and see if the convergence or divergence using any test necessary.
The series converges between -1 ≤ x ≤1.
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