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19 August, 19:59

Consider the function f (x) = xcos (x) sin (x) where x is in radians. Derive the taylor series

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  1. 19 August, 21:35
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    f (x) = x cos (x) sin (x) + (sin (2x) / 2) + xcos (2x) (x-x) + ((2 cos (2x) - 2xsin (2x)) / 2) * (x-x) ^2 + ((-6sin (2x) - 4xcos (2x)) / 6) * (x-x) ^3 + ...

    Step-by-step explanation:

    Taylor series formula

    f (x) = f (a) + (f' (a) / 1!) * (x-a) + (f'' (a) / 2!) * (x-a) ^2 + (f''' (a) / 3!) * (x-a) ^{3} + ...

    f (x) = x cos (x) sin (x)

    put 'a' instead of x

    f (a) = a cos (a) sin (a)

    take first, second, third derivative respectively, it comes

    f' (a) = (sin (2a) / 2) + acos (2a)

    f'' (a) = 2 cos (2a) - 2xsin (2a)

    f''' (a) = - 6sin (2a) - 4xcos (2a)

    put values in Taylor formula

    f (x) = a cos (a) sin (a) + (sin (2a) / 2) + acos (2a) (x-a) + ((2 cos (2a) - 2asin (2a)) / 2) * (x-a) ^2 + ((-6sin (2a) - 4acos (2a)) / 6) * (x-a) ^3 + ...

    put x instead of a

    f (x) = x cos (x) sin (x) + (sin (2x) / 2) + xcos (2x) (x-x) + ((2 cos (2x) - 2xsin (2x)) / 2) * (x-x) ^2 + ((-6sin (2x) - 4xcos (2x)) / 6) * (x-x) ^3 + ...
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