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:

$$\sum _{k=0}^{\mathrm{\infty }}{\displaystyle \frac{f^k(a)}{k!}}(x-a{)}^k$$

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.

Expert Answer:

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

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

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:

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

Simplifying the equation:

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

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

Numerical Result:

Finally, we have our Taylor Series Expansion:

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

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:

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

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:

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

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

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

Finally, we have our Taylor Series Expansion:

$$T3(x)=x–{\displaystyle \frac{1}{2}}{x}^3$$

Figure 2