Find the Maclaurin series for the function. arcsin(x) f(x) = f(x) = 1, (2n)! 2² (n!) ² 1 42n X #0 X=0 x X #0 X = 0
Find the Maclaurin series for the function. arcsin(x) f(x) = f(x) = 1, (2n)! 2² (n!) ² 1 42n X #0 X=0 x X #0 X = 0
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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![**Maclaurin Series for a Function**
To find the Maclaurin series for the function:
\[
f(x) =
\begin{cases}
\frac{\arcsin(x)}{x}, & x \neq 0 \\
1, & x = 0
\end{cases}
\]
The Maclaurin series is expressed as:
\[
f(x) =
\begin{cases}
\sum_{n=0}^{\infty} \frac{(2n)!}{2^{2n}(n!)^2} \cdot x^{2n}, & x \neq 0 \\
1, & x = 0
\end{cases}
\]
The expression within the series has been verified for \( x \neq 0 \) with a cross indicating it needs further checking, while the expression for \( x = 0 \) has been verified correctly with a checkmark.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8530ca5f-6e8e-4cce-aaa4-95b39b1a4e45%2Fd3a6bea8-2773-4ad0-b689-ae06e42f1768%2Fi8x6l3e_processed.png&w=3840&q=75)
Transcribed Image Text:**Maclaurin Series for a Function**
To find the Maclaurin series for the function:
\[
f(x) =
\begin{cases}
\frac{\arcsin(x)}{x}, & x \neq 0 \\
1, & x = 0
\end{cases}
\]
The Maclaurin series is expressed as:
\[
f(x) =
\begin{cases}
\sum_{n=0}^{\infty} \frac{(2n)!}{2^{2n}(n!)^2} \cdot x^{2n}, & x \neq 0 \\
1, & x = 0
\end{cases}
\]
The expression within the series has been verified for \( x \neq 0 \) with a cross indicating it needs further checking, while the expression for \( x = 0 \) has been verified correctly with a checkmark.
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