Welcome to Westonci.ca, where your questions are met with accurate answers from a community of experts and enthusiasts. Get detailed and accurate answers to your questions from a community of experts on our comprehensive Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
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 choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.
