Taylor Polynomial e^x * cos(x)

**deezy** · August 24th 2012, 12:41 PM

Question: Find the Taylor Polynomial $P_2 (x)$ for $e^x cos(x)$ about $x_0 = 0$ .

Here's what I did:

$P_2 (x) = e^x cos(x) + (e^x cos(x) - e^x sin(x)) (x) - \frac{2e^x sin(x) (x)^2 }{2!}$
$P_2 (x) = e^x cos(x) + (e^x cos(x) - e^x sin(x)) (x) - e^x sin(x) (x)^2$
$P_2 (x) = e^x (cos(x) + x cos(x) - x sin(x) - x^2 sin(x))$

The answer the book has is:

$P_2 (x) = 1 + x$

I'm not seeing how it simplifies to this?

**HallsofIvy** · August 24th 2012, 12:51 PM

You are not distinguishing between $x_0= 0$ and the variable x. The Taylor's series about $x_0$ is $f(x_0)+ f'(x_0)(x- x_0)+ \frac{f''(x_0)}{2!} (x- x_0)^2$ or, with $x_0= 0$ ,
$f(0)+ f'(0)x+ \frac{f''(0)}{2}x^2$

**Prove It** · August 24th 2012, 05:18 PM

Originally Posted by deezy

Question: Find the Taylor Polynomial $P_2 (x)$ for $e^x cos(x)$ about $x_0 = 0$ .

Here's what I did:

$P_2 (x) = e^x cos(x) + (e^x cos(x) - e^x sin(x)) (x) - \frac{2e^x sin(x) (x)^2 }{2!}$
$P_2 (x) = e^x cos(x) + (e^x cos(x) - e^x sin(x)) (x) - e^x sin(x) (x)^2$
$P_2 (x) = e^x (cos(x) + x cos(x) - x sin(x) - x^2 sin(x))$

The answer the book has is:

$P_2 (x) = 1 + x$

I'm not seeing how it simplifies to this?

Since you only want a Taylor Polynomial up to order 2, the easiest thing to do is to write the first few terms of the MacLaurin series for $\displaystyle \begin{align*} e^x \end{align*}$ and $\displaystyle \begin{align*} \cos{x} \end{align*}$ and multiply them together.

$\displaystyle \begin{align*} e^x &= 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \dots + \frac{x^n}{n!} + \dots \\ \\ \cos{x} &= 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots + \frac{(-1)^n x^{2n}}{(2n)!} + \dots \\ \\ e^x\cos{x} &= \left( 1 + x + \dots \right) \left( 1 - \frac{x^2}{2} + \dots \right) \\ &= 1 + x - \frac{x^2}{2} + \dots \end{align*}$

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