1. where is power series ∑_{k = 1}^∞ c_k(x - 3)^k centered at? could its interval of convergence be (-2,8)?
1. where is power series ∑_{k = 1}^∞ c_k(x - 3)^k centered at? could its interval of convergence be (-2,8)?
Answer
Explanation:
Step1: Determine the center of the power - series
For a power series of the form $\sum_{k = 1}^{\infty}c_{k}(x - a)^{k}$, the center is $a$. In the given power series $\sum_{k = 1}^{\infty}c_{k}(x - 3)^{k}$, the center is $a = 3$.
Step2: Analyze the interval of convergence
The radius of convergence $R$ for a power series centered at $a$ is related to the interval of convergence $(a - R,a+R)$. If the center $a = 3$ and the proposed interval of convergence is $(-2,8)$, then the radius of convergence $R$ is calculated as follows. The distance from the center $a = 3$ to either endpoint of the interval can be used. Using the left - hand endpoint: $R=|3-(-2)|=5$, and using the right - hand endpoint: $R = |8 - 3|=5$. So, it is possible for the interval of convergence to be $(-2,8)$.
Answer:
The power series $\sum_{k = 1}^{\infty}c_{k}(x - 3)^{k}$ is centered at $3$, and it is possible for its interval of convergence to be $(-2,8)$.