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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
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