Use a known Taylor series to find the Taylor series about c = 0 for , and find its radius of convergence.

Question

Use a known Taylor series to find the Taylor series about c = 0 for  , and find its radius of convergence.

Quizplus · Accepted Answer

We know that the Taylor series for $e^x$ is given by $\sum_{n=0}^{\infty}\frac{x^n}{n!}$.

Therefore, the Taylor series about $c=0$ for $11eaa948_cce1_7b4a_84bc_79ef23a32b7f_TB2342_11$ is:

$$\begin{aligned} 11eaa948_cce1_7b4a_84bc_79ef23a32b7f_TB2342_11 &=e^{\ln(11eaa948_cce1_7b4a_84bc_79ef23a32b7f_TB2342_11)} \ &=e^{\ln(11)+\ln(eaa948_cce1_7b4a_84bc_79ef23a32b7f_TB2342_11)} \ &= e^{\ln(11)} \cdot e^{\ln(eaa948_cce1_7b4a_84bc_79ef23a32b7f_TB2342_11)} \ &= 11 \cdot e^{\ln(eaa948_cce1_7b4a_84bc_79ef23a32b7f_TB2342_11)} \ &= 11 \cdot \sum_{n=0}^{\infty}\frac{(\ln(eaa948_cce1_7b4a_84bc_79ef23a32b7f_TB2342_11))^n}{n!} \end{aligned}$$

Now, we need to find the radius of convergence for this series. Using the ratio test,

$$\lim_{n 	o \infty} \left| \frac{(\ln(eaa948_cce1_7b4a_84bc_79ef23a32b7f_TB2342_11))^{n+1}}{(n+1)!} \cdot \frac{n!}{(\ln(eaa948_cce1_7b4a_84bc_79ef23a32b7f_TB2342_11))^n} ight|=\lim_{n 	o \infty}\frac{|\ln(eaa948_cce1_7b4a_84bc_79ef23a32b7f_TB2342_11)|}{n+1}=0$$

This limit is always zero, regardless of the value of $\ln(eaa948_cce1_7b4a_84bc_79ef23a32b7f_TB2342_11)$. Therefore, the radius of convergence is infinity, and the series converges for all values of $x$, including $x=0$.

The closest answer choice to our result is (B): 11eaa948_cce1_7b4c_84bc_f3ea35d63da3_TB2342_11.

Use a Known Taylor Series to Find the Taylor Series