How do you find the Maclaurin Series for #e^x * sinx#?

1 Answer
Oct 29, 2016

# e^xsinx = x + x^2 + 1/3x^3 + ... #

Explanation:

Let # f(x) = e^xsinx #

The Maclaurin series is given by
# f(x) = f(0) + (f'(0))/(1!)x + (f''(0))/(2!)x^2 + (f'''(0))/(3!)x^3 + ... (f^((n))(0))/(n!)x^n + ...#

# f(0) = e^0sin0=0 #

Differentiating wrt #x# using the product rule;
# d/dx(uv)=u(dv)/dx+v(du)/dx #

# :. f'(x) =(e^x)(d/dxsinx) + (d/dxe^x)(sinx) #
# :. f'(x) =e^xcosx + e^xsinx # ............................... [1]
# :. f'(0) =e^0cos0 + e^0sin0 = 1 #

Differentiating [1] again wrt #x#:
# f''(x) = d/dx(e^xcosx) + d/dx(e^xsinx) #
# :. f''(x) = -e^xsinx + e^xcosx + e^xcosx + e^xsinx #
# :. f''(x) = 2e^xcosx # ............................... [2]
# :. f''(0) = 2e^0cos0 = 2 #

Differentiating [2] again wrt #x#:

# f'''(x) = 2d/dx(e^xcosx) #
# :. f'''(x) = 2{(e^x)(d/dxcosx) + (d/dxe^x)(cosx)} #
# :. f'''(x) = 2{-e^xsinx + e^xcosx} #
# :. f'''(x) = 2e^xcosx -2e^xsinx #
# :. f'''(0) = 2e^0cos0 -2e^0sin0 = 2 #

.... etc for higher derivatives

So, the power series is given by
# f(x) = f(0) + (f'(0))/(1!) + (f''(0))/(2!) + (f'''(0))/(3!) + ... #
# :. f(x) = 0 + (1)/(1)x + (2)/(2)x^2 + (2)/(6)x^3 + ... #
# :. f(x) = x + x^2 + 1/3x^3 + ... #