The Maclaurin series for sin−1(x) is given by sin−1(x) = x + [infinity] n = 1 1 · 3 · 5 (2n − 1) 2 · 4 · 6 (2n) x2n+1 2n + 1 . Use the first

Question

The Maclaurin series for sin−1(x) is given by sin−1(x) = x + [infinity] n = 1 1 · 3 · 5 (2n − 1) 2 · 4 · 6 (2n) x2n+1 2n + 1 . Use the first five terms of the Maclaurin series above to approximate sin−1 3 7 . (Round your answer to eight decimal places.)

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Maya 3 months 2021-10-18T07:03:43+00:00 1 Answer 0 views 0

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    2021-10-18T07:04:57+00:00

    Answer:

    0.44290869

    Step-by-step explanation:

    The Maclaurin series for sin⁻¹(x) is given by

    sin⁻¹(x) = x + x^{\alpha } _{n=1} \frac{1.3.5...(2n - 1)}{2.4.6...(2n)} * \frac{x^{2n + 1} }{2n + 1}

    Use the first five terms of the Maclaurin series above to approximate sin⁻¹ \frac{3}{7}. (Round your answer to eight decimal places.)

    Answer

    sin⁻¹(x) = x + x^{\alpha } _{n=1} \frac{1.3.5...(2n - 1)}{2.4.6...(2n)} * \frac{x^{2n + 1} }{2n + 1}

    in the above equation x^{\alpha } _{n=1}  summation from n=1 to ∞

    we are estimating this for the first 5 terms as follows

    sin⁻¹(x) = x +   \frac{1}{2} * \frac{x^{3} }{3}  +  \frac{1*3}{2*4} * \frac{x^{5} }{5}  +  \frac{1*3*5}{2*4*6} * \frac{x^{7} }{7}  +  \frac{1*3*5*7}{2*4*6*8} * \frac{x^{9} }{9}

    sin⁻¹(x) = x +  \frac{x^{3} }{6}  +  \frac{3x^{5} }{40}  +\frac{15x^{7} }{336}  +  \frac{105x^{9} }{3456}  

    now to get

    sin⁻¹(\frac{3}{7}) substitute

    hence,

    sin⁻¹(\frac{3}{7}) = \frac{3}{7}  + \frac{\frac{3}{7} ^{3} }{6}  + \frac{3 * \frac{3}{7} ^{5} }{40} + \frac{15* \frac{3}{7} ^{7} }{336} + \frac{105* \frac{3}{7} ^{9} }{3456}  

    sin⁻¹(\frac{3}{7}) = 0.42857142 + 0.01311953 + 0.00108437 + 0.00011855 + 0.00001482

               =  0.44290869

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