3n The Taylor series for a function f(x) centered at a = 2 is given by > o L r – 2)". What is f(32) (2), that is, n=0 the 32nd derivative of f(x) evaluated at a = 2. 96
3n The Taylor series for a function f(x) centered at a = 2 is given by > o L r – 2)". What is f(32) (2), that is, n=0 the 32nd derivative of f(x) evaluated at a = 2. 96
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Title: Calculating the 32nd Derivative of a Function Using a Taylor Series**
**Introduction:**
The Taylor series for a function \( f(x) \) centered at \( a = 2 \) is expressed as:
\[
\sum_{n=0}^{\infty} \frac{3n}{(n+1)!}(x-2)^n.
\]
Our goal is to determine \( f^{(32)}(2) \), which represents the 32nd derivative of \( f(x) \) evaluated at \( a = 2 \).
**Options:**
(a) \(\frac{96}{33}\)
(b) \(\frac{96}{33!}\)
(c) \(\frac{96}{32! \cdot 33!}\)
(d) \(96 \cdot 32!\)
(e) None of these
**Explanation:**
To solve this problem, we need to identify the coefficient of \((x-2)^{32}\) in the given Taylor series to find \( f^{(32)}(2) \). Typically, the coefficient of \((x-a)^n\) in a Taylor series expansion for the function at a point \( a \) is given by
\[
\frac{f^{(n)}(a)}{n!}.
\]
The provided series is \(\sum_{n=0}^{\infty} \frac{3n}{(n+1)!}(x-2)^n\), indicating the coefficient must be compared directly to this form to find \( f^{(32)}(2) \).
**Conclusion:**
Based on matching coefficients and simplification, the choice that correctly applies the formula to obtain the 32nd derivative is:
**Answer: (c) \(\frac{96}{32! \cdot 33!}\).**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fadc447ea-4c16-4666-abc4-68a3d9e32223%2F2b99f9a0-7527-41bc-953e-3d8d690b78be%2Fx8rxmrg_processed.png&w=3840&q=75)
Transcribed Image Text:**Title: Calculating the 32nd Derivative of a Function Using a Taylor Series**
**Introduction:**
The Taylor series for a function \( f(x) \) centered at \( a = 2 \) is expressed as:
\[
\sum_{n=0}^{\infty} \frac{3n}{(n+1)!}(x-2)^n.
\]
Our goal is to determine \( f^{(32)}(2) \), which represents the 32nd derivative of \( f(x) \) evaluated at \( a = 2 \).
**Options:**
(a) \(\frac{96}{33}\)
(b) \(\frac{96}{33!}\)
(c) \(\frac{96}{32! \cdot 33!}\)
(d) \(96 \cdot 32!\)
(e) None of these
**Explanation:**
To solve this problem, we need to identify the coefficient of \((x-2)^{32}\) in the given Taylor series to find \( f^{(32)}(2) \). Typically, the coefficient of \((x-a)^n\) in a Taylor series expansion for the function at a point \( a \) is given by
\[
\frac{f^{(n)}(a)}{n!}.
\]
The provided series is \(\sum_{n=0}^{\infty} \frac{3n}{(n+1)!}(x-2)^n\), indicating the coefficient must be compared directly to this form to find \( f^{(32)}(2) \).
**Conclusion:**
Based on matching coefficients and simplification, the choice that correctly applies the formula to obtain the 32nd derivative is:
**Answer: (c) \(\frac{96}{32! \cdot 33!}\).**
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