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

User Matheno
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Answer:

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 +.........................

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 +.........................

User YakovL
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