Find the radius of convergence:
[tex]
\begin{array}{l}
\text { 3y using cachy } \varepsilon_{n=2}^\alpha(n-1)^a(z-3+2 i)^n \
\sum_{n=0}^\alpha n^2\left(\frac{i}{4}\right)^n(z-2 i)^n \
\sum_{n=0}^\alpha \frac{2+i n}{2^n} z^n \
\sum_{n=0}^\alpha \frac{(-1)^n}{2^{2 n}(n d)^2} z^{2 n}
\end{array}
[/tex]
Answer (2)
Series 1 has a radius of convergence of R = 1 .
Series 2 has a radius of convergence of R = 4 .
Series 3 has a radius of convergence of R = 2 .
Series 4 has a radius of convergence of R = 2 .
The radii of convergence for the given series are 1 , 4 , 2 , 2 .
Explanation
Problem Setup We are asked to find the radius of convergence for each of the given series. We will use the ratio test to find the radius of convergence for each series.
Radius of Convergence for Series 1 For the first series, ∑ n = 2 ∞ ( n − 1 ) a ( z − 3 + 2 i ) n , let a n = ( n − 1 ) a ( z − 3 + 2 i ) n . Then, we have
R = n → ∞ lim a n + 1 a n = n → ∞ lim n a ( z − 3 + 2 i ) n + 1 ( n − 1 ) a ( z − 3 + 2 i ) n = n → ∞ lim n a ( n − 1 ) a ⋅ z − 3 + 2 i 1 = z − 3 + 2 i 1 .
For convergence, we need ∣ z − 3 + 2 i ∣ < 1 . Thus, the radius of convergence is R = 1 .
Radius of Convergence for Series 2 For the second series, ∑ n = 0 ∞ n 2 ( 4 i ) n ( z − 2 i ) n , let a n = n 2 ( 4 i ) n ( z − 2 i ) n . Then, we have
R = n → ∞ lim a n + 1 a n = n → ∞ lim ( n + 1 ) 2 ( 4 i ) n + 1 ( z − 2 i ) n + 1 n 2 ( 4 i ) n ( z − 2 i ) n = n → ∞ lim ( n + 1 ) 2 n 2 ⋅ i 4 ⋅ z − 2 i 1 = z − 2 i 4 .
For convergence, we need ∣ z − 2 i ∣ < 4 . Thus, the radius of convergence is R = 4 .
Radius of Convergence for Series 3 For the third series, ∑ n = 0 ∞ 2 n 2 + in z n , let a n = 2 n 2 + in z n . Then, we have
R = n → ∞ lim a n + 1 a n = n → ∞ lim 2 n + 1 2 + i ( n + 1 ) z n + 1 2 n 2 + in z n = n → ∞ lim 2 + i ( n + 1 ) 2 + in ⋅ z 2 = z 2 .
For convergence, we need ∣ z ∣ < 2 . Thus, the radius of convergence is R = 2 .
Radius of Convergence for Series 4 For the fourth series, ∑ n = 0 ∞ 2 2 n ( n d ) 2 ( − 1 ) n z 2 n , let a n = 2 2 n ( n d ) 2 ( − 1 ) n z 2 n . Then, we have
R = n → ∞ lim a n + 1 a n = n → ∞ lim 2 2 ( n + 1 ) (( n + 1 ) d ) 2 ( − 1 ) n + 1 z 2 ( n + 1 ) 2 2 n ( n d ) 2 ( − 1 ) n z 2 n = n → ∞ lim ( n d ) 2 (( n + 1 ) d ) 2 ⋅ 2 2 n 2 2 n + 2 ⋅ z 2 1 = z 2 4 .
For convergence, we need ∣ z 2 ∣ < 4 , which means ∣ z ∣ < 2 . Thus, the radius of convergence is R = 2 .
Final Answer The radii of convergence for the given series are 1, 4, 2, and 2, respectively.
Examples
Understanding the radius of convergence is crucial in many areas of engineering and physics. For example, when designing filters in signal processing, engineers use power series to represent the filter's transfer function. The radius of convergence determines the range of frequencies for which the filter is stable and performs as expected. Similarly, in quantum mechanics, perturbation theory involves expanding solutions as power series, and the radius of convergence dictates the range of parameters for which the approximation is valid.
The radii of convergence for the given series are: 1, 4, 2, and 2 respectively. This means that the first series converges within a distance of 1 from a specific point, the second within 4, and the last two within a distance of 2. The analysis used the Ratio Test for each series to determine the convergence intervals.
;
