# How do you find the taylor series for y=(e^x)cos(x)?

Sep 13, 2015

e^x cos x = sum_(k=0)^oo (-4)^k(x^(4k)/((4k)!) + x^(4k+1)/((4k+1)!) -(2x^(4k+3))/((4k+3)!) )

#### Explanation:

Let $f \left(x\right) = {e}^{x} \cos \left(x\right)$

Successive derivatives look like this:

{:(f^((0))(x) = e^x cos x, f^((0))(0) = color(blue)(1)), (f^((1))(x) = e^x (cos x - sin x), f^((1))(0) = color(blue)(1)), (f^((2))(x) = -2e^x sin x, f^((2))(0) = color(blue)(0)), (f^((3))(x) = -2e^x(cos x + sin x), f^((3))(0) = color(blue)(-2)), (f^((4))(x) = -4e^x cos x, f^((4))(0) = -4), (f^((5))(x) = -4e^x(cos x - sin x), f^((5))(0) = -4), (f^((6))(x) = 8e^x sin x, f^((6))(0) = 0), (f^((7))(x) = 8e^x(cos x + sin x), f^((7))(0) = 8):}

So we have a pattern that repeats every $4$ derivatives, but multiplied by a factor of $- 4$ each time.

The general formula for a Taylor series at $0$ is:

f(x) = sum_(n=0)^oo f^((n))(0)/(n!) x^n

In our case, capture the recurring pattern as follows:

f(x) = sum_(k=0)^oo (-4)^k(x^(4k)/((4k)!) + x^(4k+1)/((4k+1)!) -(2x^(4k+3))/((4k+3)!) )