Q. 42 - Calculus

Question

In Exercises 41–48 find the fourth Taylor polynomial $P_4\left(x\right)$ for the specified function and the given value of $x_0$.

$e^{x },1$

Step-by-Step Solution

Answer

Ans: The fourth Taylor polynomial for the specified function is $=e\left[x+\frac{(x-1{)}^2}{2}+\frac{(x-1{)}^3}{6}+\frac{(x-1{)}^4}{24}\right]$

Step 1. Given information:

$e^{x },1$

Step 2. The fourth Taylor polynomial:

Since for any function $f$ with a derivative of order 4 at $x=1$, the fourth Taylor polynomial for $x=1$ is given by

$P_4\left(x\right)=f\left(1\right)+f^{\text{'}}\left(1\right)(x-1)+\frac{f^{\text{'}\text{'}}\left(1\right)}{2!}(x-1{)}^2+\frac{f^{\text{'}\text{'}}\left(1\right)}{3!}(x-1{)}^3+\frac{f^{\text{'}\text{'}\text{'}\text{'}}\left(1\right)}{4!}(x-1{)}^4$

Therefore, first find the value of the function along with$f^{\text{'}}\left(x\right),f^{\text{'}\text{'}}\left(x\right),f^{\text{'}\text{'}}\left(x\right)$ and $f^{\text{'}\text{'}\text{'}\text{'}}\left(x\right)$ at $x=1$

Step 3. Finding the fourth Taylor polynomial through derivative of order 4:

Thus, the value of the function at $x=1$ is

$f\left(1\right)=e^1=e$

Therefore, all the derivatives of the function $f\left(x\right)=e^x$ is $e^x$, so $f^n\left(1\right)=e$

Step 4. Substituting the derivative of order 4 in the fourth Taylor polynomial :

Therefore, the fourth Taylor polynomial for the function $f\left(x\right)=e^x$ is

$P_4\left(x\right)=e+e(x-1)+\frac{e}{2!}(x-1{)}^2+\frac{e}{3!}(x-1{)}^3+\frac{e}{4!}(x-1{)}^4$$$\begin{gathered} =e\left[1+x-1+\frac{(x-1{)}^2}{2}+\frac{(x-1{)}^3}{6}+\frac{(x-1{)}^4}{24}\right] \\ =e\left[x+\frac{(x-1{)}^2}{2}+\frac{(x-1{)}^3}{6}+\frac{(x-1{)}^4}{24}\right] \end{gathered}$$