# Find the Taylor polynomial T3(x) for the function f centered at the number a. f(x) = x + e^{−x}, a = 0

This problem aims to find the Taylor polynomials up to $3$ places for a given function $f$, centered at a point $a$. To better understand the problem, you must know about Power Series, as it forms the basis of the Taylor Series.

Taylor series of a function is defined as an infinite sum of derivative terms of that function at a single point. The formula for this Series is derived from the Power series and can be written as:

$\sum_{k=0}^{\infty} \dfrac{f^{k}(a)}{k!} (x-a)^k$

where $f(k)(a)$ denotes the nth derivative of $f$ evaluated at point $a$ and $k$ is the degree of the polynomial. If $a$ is set to 0, it’s known as Maclaurin Series.

###### But not every function has a Taylor Series expansion.

Firstly, expanding the series for $k = 3$ as $T3$

$T3(x) = f(a) + \dfrac{f(a)}{1!}(x-a) + \dfrac{f“(a)}{2!}(x-a)^ 2 + \dfrac{f“(a)}{3!}(x-a)^ 3$

Next, we are going to find the derivatives of $f(x)$ which will get plugged into $T3(x)$ equation:

$f(x) =x + e^{-x}, f(0) = 1$

First Derivative:

$f(x) = 1 – e^{-x}, f(0) = 0$

Second Derivative:

$f“(x) = e^{-x}, f“(0) = 1$

Third Derivative:

$f“(x) = – e^{-x}, f“(0) = -1$

Substituting the above derivatives into $T3(x)$ becomes:

$T3(x) = f(a) +\dfrac{f(a)}{1!}(x-a) + \dfrac{f“(a)}{2!}(x-a)^2 + \dfrac{f“(a)}{3!}(x-a)^ 3$

Simplifying the equation:

$= 1 +\dfrac{0}{1!}(x-0) + \dfrac{1}{2!}(x-2)^ 2 + \dfrac{-1}{3!}(x-0)^ 3$

$T3(x) = 1 +\dfrac{x^ 2} {2} – \dfrac{x^ 3} {6}$

## Numerical Result:

Finally, we have our Taylor Series Expansion:

$T3(x) = 1 +\dfrac{x^ 2} {2} – \dfrac{x^ 3} {6}$

Figure 1

## Example:

Find the taylor polynomial $t3(x)$ for the function $f$ centered at the number a. $f(x) = xcos(x), a = 0$

Expanding the series for $k = 3$ as $T3$ gives us:

$T3(x) = f(a) + \dfrac{f(a)}{1!}(x-a) + \dfrac{f“(a)}{2!}(x-a)^ 2 + \dfrac{f“(a)}{3!}(x-a)^ 3$

Next, we are going to find the derivatives of $f(x)$ which will get plugged into $T3(x)$ equation:

$f(x) =xcos(x), f(0) = 0$

$f(x) = cos(x) – xsin(x), f(0) = 1$

$f“(x) = -xcos(x) -2sin(x), f“(0) = 0$

$f“(x) = xsin(x) -3cos(x), f“(0) = -1$

Substituting the above derivatives into $T3(x)$ becomes:

$T3(x) = f(a) +\dfrac{f(a)}{1!}(x-a) + \dfrac{f“(a)}{2!}(x-a)^ 2 + \dfrac{f“(a)}{3!}(x-a)^ 3$

Plugging in the values in $T3(x)$ equation.

$= \dfrac{1}{1!}x + 0 + \dfrac{-3}{3!}x^ 3$

Finally, we have our Taylor Series Expansion:

$T3(x) = x – \dfrac{1}{2}x^ 3$

Figure 2

Images/mathematical drawings are created with GeoGebra.