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Is the solution correct? Question: 8, E(-1)" (n + 1) x" (converges for -1 < x < 1) to find a series that converges to (for -1 < x < 1). Use the fact that %3D (1+2)* (1+2) "Solution": Line 1: Notice that if f (x) = y then f' (x) (1+x)* (1+2) Line 2: Therefore, to find a series for -2 we could differentiate the series for (1+2)3 (1+a) Line 3: To find the derivative of a convergent series, we can differentiate term-by-term (i.e. differentiate one term at a time, as if we were differentiating a polynomial). Line 4: Let's expand the given series: = 1- 2x + 3x2 – 4x3 + 5x4 (1+z) 6x5 +... Line 5: Line 6: Now differentiate both sides: Line 7: (1+2) = -2 + 3. 2 – 4. 3x2 + 5 . 4x3 – 6· 5x4 +... Line 8: Write using sigma notation: Line 9: = E (-1)" (n + 1) · n. x" (1+z) n=0 Line 10: The radius of convergence does not change when differentiating, so E (-1)" (n + 1) · n. x" is a power series that converges to -2 for -1

Is the solution correct? Question: 8, E(-1)" (n + 1) x" (converges for -1 < x < 1) to find a series that converges to (for -1 < x < 1). Use the fact that %3D (1+2)* (1+2) "Solution": Line 1: Notice that if f (x) = y then f' (x) (1+x)* (1+2) Line 2: Therefore, to find a series for -2 we could differentiate the series for (1+2)3 (1+a) Line 3: To find the derivative of a convergent series, we can differentiate term-by-term (i.e. differentiate one term at a time, as if we were differentiating a polynomial). Line 4: Let's expand the given series: = 1- 2x + 3x2 – 4x3 + 5x4 (1+z) 6x5 +... Line 5: Line 6: Now differentiate both sides: Line 7: (1+2) = -2 + 3. 2 – 4. 3x2 + 5 . 4x3 – 6· 5x4 +... Line 8: Write using sigma notation: Line 9: = E (-1)" (n + 1) · n. x" (1+z) n=0 Line 10: The radius of convergence does not change when differentiating, so E (-1)" (n + 1) · n. x" is a power series that converges to -2 for -1

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.2: Arithmetic Sequences
Problem 67E
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q4 6.2

Is the solution correct? Question: Use the fact that (1+2)? = E(-1)" (n + 1) 2" (converges for -1 < x < 1) to find a series that converges to (1+2) (for -1 < z < 1). れ=0 "Solution": Line 1: Notice that if f (x) = (1+2) then f' (x) (1+2) Line 2: Therefore, to find a series for we could differentiate the series for (1+2)3 (1+2)? Line 3: To find the derivative of a convergent series, we can differentiate term-by-term (i.e. differentiate one term at a time, as if we were differentiating a polynomial). Line 4: Let's expand the given series: Line 5: 1+2) 6x5+... = 1- 2x + 3x2 – 4x3 + 5x4 Line 6: Now differentiate both sides: Line 7: = -2 + 3. 2x – 4. 3x2 + 5 . 4x3 – 6- 5x4+... (1+2)3 Line 8: Write using sigma notation: Line 9: -2 E(-1)" (n + 1) - n - 2" n=0 Line 10: The radius of convergence does not change when differentiating, so E (-1)" (n + 1) · n· x" is a power series that converges to (1+a)' for -1 < a < 1. n=0 Is the above "solution" correct? If not, in which line is the first error? [ Select ] What is the correct final answer to the question? (Choose from the options below) [ Select] DO Option I: E (-1)" (n + 1) · n · x" n=0 Option II: E (-1)" (n +1) - n · r"-1 n=1 Option III: E(-1)" (n - 1) · n xn+1 n=0 Option IV: E (-1)"+1 (n + 1) · n · x"-1 n=1
Transcribed Image Text:Is the solution correct? Question: Use the fact that (1+2)? = E(-1)" (n + 1) 2" (converges for -1 < x < 1) to find a series that converges to (1+2) (for -1 < z < 1). れ=0 "Solution": Line 1: Notice that if f (x) = (1+2) then f' (x) (1+2) Line 2: Therefore, to find a series for we could differentiate the series for (1+2)3 (1+2)? Line 3: To find the derivative of a convergent series, we can differentiate term-by-term (i.e. differentiate one term at a time, as if we were differentiating a polynomial). Line 4: Let's expand the given series: Line 5: 1+2) 6x5+... = 1- 2x + 3x2 – 4x3 + 5x4 Line 6: Now differentiate both sides: Line 7: = -2 + 3. 2x – 4. 3x2 + 5 . 4x3 – 6- 5x4+... (1+2)3 Line 8: Write using sigma notation: Line 9: -2 E(-1)" (n + 1) - n - 2" n=0 Line 10: The radius of convergence does not change when differentiating, so E (-1)" (n + 1) · n· x" is a power series that converges to (1+a)' for -1 < a < 1. n=0 Is the above "solution" correct? If not, in which line is the first error? [ Select ] What is the correct final answer to the question? (Choose from the options below) [ Select] DO Option I: E (-1)" (n + 1) · n · x" n=0 Option II: E (-1)" (n +1) - n · r"-1 n=1 Option III: E(-1)" (n - 1) · n xn+1 n=0 Option IV: E (-1)"+1 (n + 1) · n · x"-1 n=1
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