Question

In Exercises 49–56 find the Taylor series for the specified function and the given value of $x_0$. Note: These are the same functions and values as in Exercises 41–48.

$50.e^x,1$

Short answer

The Taylor series for the function$f\left(x\right)=e^x$at$x=1$is$P_n\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{e}{k!}(x-1{)}^k$.

Step-by-step solution

Step 1. Given data

We have the function$f\left(x\right)=e^x$

Step 2. Table of the taylor series

Any function $f$with a derivative of order $n$, the taylor series at localid="1649409912679" $x=1$is given by,

$P_n\left(x\right)=f\left(1\right)+f^{\mathrm{\prime }}\left(1\right)(x-1)+\frac{f^{\prime \prime }\left(1\right)}{2!}(x-1{)}^2+\frac{f^{\prime \prime }\left(1\right)}{3!}(x-1{)}^3+\frac{f^{\prime \prime \prime \prime }\left(1\right)}{4!}(x-1{)}^4+\dots$

we can write the general of the Taylor series of the function $f$is,

$P_n\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{f^k\left(x_0\right)}{k!}{(x-x_0)}^n$

So, let us first construct the table of the Taylor series for the function $f\left(x\right)=e^x$at $x=1$.

n	$f^n\left(x\right)$	$f^n\left(1\right)$	$\frac{f^n\left(1\right)}{n!}$
0	$e^x$	$e^1$	$e$
1	$e^x$	$e^1$	$e$
2	$e^x$	$e^1$	$\frac{e}{2!}$
. . .	. . .	. . .	. . .
k	$e^x$	$e^1$	$\frac{e}{k!}$
. . .	. . .	. . .	. . .

Step 3. Taylor series for the f(x)=ex

The Taylor series for the function $f\left(x\right)=e^x$at $x=1$is $P_n\left(x\right)=e+e(x-1)+\frac{e}{2!}(x-1{)}^2+\frac{e}{3!}(x-1{)}^3+\frac{e}{4!}(x-1{)}^4+\dots$

Or we can write this as$P_n\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{e}{k!}(x-1{)}^k$