Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that R₂(x) → 0.] f(x) = 10x - 2x³, a = -1

Algebra and Trigonometry (6th Edition)
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ChapterP: Prerequisites: Fundamental Concepts Of Algebra
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Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Find the Taylor series for \( f(x) \) centered at the given value of \( a \). [Assume that \( f \) has a power series expansion. Do not show that \( R_n(x) \to 0 \).]

\[ f(x) = 10x - 2x^3, \quad a = -1 \]

Choose the correct series expansion:

1. \[ \sum_{n=0}^{\infty} \frac{f^{(n)}(-1)}{n!} (x+1)^n = -8 + 4(x+1) + 6(x+1)^2 - 2(x+1)^3 \]

2. \[ \sum_{n=0}^{\infty} \frac{f^{(n)}(-1)}{n!} (x+1)^n = -8 + 4(x+1) + 2(x+1)^2 - 6(x+1)^3 \]

3. \[ \sum_{n=0}^{\infty} \frac{f^{(n)}(-1)}{n!} (x+1)^n = -8 - 4(x+1) + 6(x+1)^2 + 2(x+1)^3 \]

4. \[ \sum_{n=0}^{\infty} \frac{f^{(n)}(-1)}{n!} (x+1)^n = -8 + 6(x+1) + 4(x+1)^2 - 2(x+1)^3 \]

5. \[ \sum_{n=0}^{\infty} \frac{f^{(n)}(-1)}{n!} (x+1)^n = -8 - 4(x+1) + 2(x+1)^2 + 6(x+1)^3 \]

Find the associated radius of convergence \( R \).

\[ R = \underline{\hspace{2cm}} \]
Transcribed Image Text:Find the Taylor series for \( f(x) \) centered at the given value of \( a \). [Assume that \( f \) has a power series expansion. Do not show that \( R_n(x) \to 0 \).] \[ f(x) = 10x - 2x^3, \quad a = -1 \] Choose the correct series expansion: 1. \[ \sum_{n=0}^{\infty} \frac{f^{(n)}(-1)}{n!} (x+1)^n = -8 + 4(x+1) + 6(x+1)^2 - 2(x+1)^3 \] 2. \[ \sum_{n=0}^{\infty} \frac{f^{(n)}(-1)}{n!} (x+1)^n = -8 + 4(x+1) + 2(x+1)^2 - 6(x+1)^3 \] 3. \[ \sum_{n=0}^{\infty} \frac{f^{(n)}(-1)}{n!} (x+1)^n = -8 - 4(x+1) + 6(x+1)^2 + 2(x+1)^3 \] 4. \[ \sum_{n=0}^{\infty} \frac{f^{(n)}(-1)}{n!} (x+1)^n = -8 + 6(x+1) + 4(x+1)^2 - 2(x+1)^3 \] 5. \[ \sum_{n=0}^{\infty} \frac{f^{(n)}(-1)}{n!} (x+1)^n = -8 - 4(x+1) + 2(x+1)^2 + 6(x+1)^3 \] Find the associated radius of convergence \( R \). \[ R = \underline{\hspace{2cm}} \]
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