Q. 17 - Calculus

Question

Let $f\left(x\right)=a{x}^3+b{x}^2+cx+d$, where $a,b,c,$ and $d$ are constants. Find the first- through fourth-order Taylor polynomials, $P_1\left(x\right), P_2\left(x\right), P_3\left(x\right),$ and $P_4\left(x\right)$, for $f$ at $x_0$. Explain why $f\left(x\right)=P_3\left(x\right)=P_4\left(x\right)$.

Step-by-Step Solution

Answer

The Taylor polynomials are,

$$\begin{gathered} P_1\left(x\right)=a{x}_0^3+b{x}_0^2+c{x}_0+d+(3a{x}_0^2+2b{x}_0+c)(x-x_0) \\ {P}_2\left(x\right)=a{x}_0^3+b{x}_0^2+c{x}_0+d+(3a{x}_0^2+2b{x}_0+c)(x-x_0)+(6a{x}_0+2b){(x-x_0)}^2 \\ {P}_3\left(x\right)=a{x}_0^3+b{x}_0^2+c{x}_0+d+(3a{x}_0^2+2b{x}_0+c)(x-x_0)+(6a{x}_0+2b){(x-x_0)}^2+6a{(x-x_0)}^3 \\ {P}_4\left(x\right)=a{x}_0^3+b{x}_0^2+c{x}_0+d+(3a{x}_0^2+2b{x}_0+c)(x-x_0)+(6a{x}_0+2b){(x-x_0)}^2+6a{(x-x_0)}^3 \end{gathered}$$

Step 1. Given Information.

The function is,

$f\left(x\right)=a{x}^3+b{x}^2+cx+d$.

Step 2. Describing the Taylor polynomial.

The first-, second, third-, and fourth-, order Taylor polynomials at $x=x_0$ that is, $P_1\left(x\right),P_2\left(x\right),P_3\left(x\right),P_4\left(x\right)$ are given by,

$$\begin{gathered} P_1\left(x\right)=f\left(x_0\right)+f\text{'}\left(x_0\right)(x-x_0) \\ {P}_2\left(x\right)=f\left(x_0\right)+f\text{'}\left(x_0\right)(x-x_0)+\frac{f\text{'}\text{'}\left(x_0\right)}{2!}{(x-x_0)}^2 \\ {P}_3\left(x\right)=f\left(x_0\right)+f\text{'}\left(x_0\right)(x-x_0)+\frac{f\text{'}\text{'}\left(x_0\right)}{2!}{(x-x_0)}^2+\frac{f\text{'}\text{'}\text{'}\left(x_0\right)}{3!}{(x-x_0)}^3 \\ {P}_4\left(x\right)=f\left(x_0\right)+f\text{'}\left(x_0\right)(x-x_0)+\frac{f\text{'}\text{'}\left(x_0\right)}{2!}{(x-x_0)}^2+\frac{f\text{'}\text{'}\text{'}\left(x_0\right)}{3!}{(x-x_0)}^3+\frac{f\text{'}\text{'}\text{'}\text{'}\left(x_0\right)}{4!}{(x-x_0)}^4 \end{gathered}$$

Step 3. Finding the Taylor polynomials.

Step 3. Finding the Taylor polynomials.

The value at $x=x_0$ is,

$$\begin{gathered} f\left(x_0\right)=a{\left(x_0\right)}^3+b{\left(x_0\right)}^2+c\left(x_0\right)+d \\ =a{x}_0^3+b{x}_0^2+c{x}_0+d \end{gathered}$$

Finding the derivatives of the function,

$$\begin{gathered} f\text{'}\left(x\right)=\frac{d(a{x}^3+b{x}^2+cx+d)}{dx} \\ =3a{x}^2+2bx+c \\ f\text{'}\left(x_0\right)=3a{\left(x_0\right)}^2+2b\left(x_0\right)+c \\ =3a{x}_0^2+2b{x}_0+c \end{gathered}$$

Also,

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=\frac{d(3a{x}^2+2bx+c)}{dx} \\ =6ax+2b \\ f\text{'}\text{'}\left(x_0\right)=6a\left(x_0\right)+2b \\ =6a{x}_0+2b \end{gathered}$$

Also,

$$\begin{gathered} f\text{'}\text{'}\text{'}\left(x\right)=\frac{d(6ax+2b)}{dx} \\ =6a \\ f\text{'}\text{'}\text{'}\left(x_0\right)=6a \end{gathered}$$

Also,

$$\begin{gathered} f\text{'}\text{'}\text{'}\text{'}\left(x\right)=\frac{d\left(6a\right)}{dx} \\ =0 \\ f\text{'}\text{'}\text{'}\text{'}\left(x_0\right)=0 \end{gathered}$$

Hence, the Taylor polynomials are,

$$\begin{gathered} P_1\left(x\right)=a{x}_0^3+b{x}_0^2+c{x}_0+d+(3a{x}_0^2+2b{x}_0+c)(x-x_0) \\ {P}_2\left(x\right)=a{x}_0^3+b{x}_0^2+c{x}_0+d+(3a{x}_0^2+2b{x}_0+c)(x-x_0)+(6a{x}_0+2b){(x-x_0)}^2 \\ {P}_3\left(x\right)=a{x}_0^3+b{x}_0^2+c{x}_0+d+(3a{x}_0^2+2b{x}_0+c)(x-x_0)+(6a{x}_0+2b){(x-x_0)}^2+6a{(x-x_0)}^3 \\ {P}_4\left(x\right)=a{x}_0^3+b{x}_0^2+c{x}_0+d+(3a{x}_0^2+2b{x}_0+c)(x-x_0)+(6a{x}_0+2b){(x-x_0)}^2+6a{(x-x_0)}^3 \end{gathered}$$

Step 4. Explanation.

Here, $P_3\left(x\right)=P_4\left(x\right)$. This is because for any polynomial function $f$ of degree $n$ , the $mth$Taylor polynomial for $f$, $P_m\left(x\right)=f\left(x\right) when m\ge n$.