At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Join our platform to connect with experts ready to provide precise answers to your questions in various areas. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
The terms through degree four of the Maclaurin series is [tex]f(x)=x+x^{2} +\frac{5x^{3} }{6} +\frac{5x^{4} }{6} +.....[/tex].
In this question,
The function is f(x) = [tex]\frac{sin(x)}{1-x}[/tex]
The general form of Maclaurin series is
[tex]\sum \limits^\infty_{k:0} \frac{f^{k}(0) }{k!}(x-0)^{k} = f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^{2} +\frac{f'''(0)}{3!}x^{3}+......[/tex]
To find the Maclaurin series, let us split the terms as
[tex]f(x)=sin(x)(\frac{1}{1-x} )[/tex] ------- (1)
Now, consider f(x) = sin(x)
Then, the derivatives of f(x) with respect to x, we get
f'(x) = cos(x), f'(0) = 1
f''(x) = -sin(x), f'(0) = 0
f'''(x) = -cos(x), f'(0) = -1
[tex]f^{iv}(x)[/tex] = cos(x), f'(0) = 0
Maclaurin series for sin(x) becomes,
[tex]f(x) = 0 +\frac{1}{1!}x +0+(-\frac{1}{3!} )x^{3} +....[/tex]
⇒ [tex]f(x)=x-\frac{x^{3} }{3!} +\frac{x^{5} }{5!}+.....[/tex]
Now, consider [tex]f(x) = (1-x)^{-1}[/tex]
Then, the derivatives of f(x) with respect to x, we get
[tex]f'(x) = (1-x)^{-2}, f'(0) = 1[/tex]
[tex]f''(x) = 2(1-x)^{-3}, f''(0) = 2[/tex]
[tex]f'''(x) = 6(1-x)^{-4}, f'''(0) = 6[/tex]
[tex]f^{iv} (x) = 24(1-x)^{-5}, f^{iv}(0) = 24[/tex]
Maclaurin series for (1-x)^-1 becomes,
[tex]f(x) = 1 +\frac{1}{1!}x +\frac{2}{2!}x^{2} +(\frac{6}{3!} )x^{3} +....[/tex]
⇒ [tex]f(x)=1+x+x^{2} +x^{3} +......[/tex]
Thus the Maclaurin series for [tex]f(x)=sin(x)(\frac{1}{1-x} )[/tex] is
⇒ [tex]f(x)=(x-\frac{x^{3} }{3!} +\frac{x^{5} }{5!}+..... )(1+x+x^{2} +x^{3} +......)[/tex]
⇒ [tex]f(x)=x+x^{2} +x^{3} - \frac{x^{3} }{6} +x^{4}-\frac{x^{4} }{6} +.....[/tex]
⇒ [tex]f(x)=x+x^{2} +\frac{5x^{3} }{6} +\frac{5x^{4} }{6} +.....[/tex]
Hence we can conclude that the terms through degree four of the Maclaurin series is [tex]f(x)=x+x^{2} +\frac{5x^{3} }{6} +\frac{5x^{4} }{6} +.....[/tex].
Learn more about Maclaurin series here
https://brainly.com/question/10661179
#SPJ4
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
