Chapter 2: Q7P (page 59)

: Find the disk of convergence for each of the followingcomplex power series.
$\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} z^{2 n}}{(2 n)!}$

Short Answer

Expert verified

The required disk of convergence is . $| z | < \infty$

Step by step solution

Disk of Convergence

For any power series $\sum a_{n} z^{n}$ where z is a complex numbers, then disk of convergence is given by: . $ρ = \underset{n \to \infty}{l i m} | \frac{z \times n}{n + 1} | = | z |$

Step 2:Find the disk of Convergence

The given power series is: $\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} z^{2 n}}{(2 n)!}$ , where $a_{n} = \frac{{(- 1)}^{n} z^{2 n}}{(2 n)!}$

Now, let us evaluate the ratio as:

$\begin{matrix} ρ = \lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | \\ = \lim_{n \to \infty} | \frac{\frac{{(- 1)}^{n + 1} z^{2 (n + 1)}}{(2 (n + 1))!}}{\frac{{(- 1)}^{n} z^{2 n}}{(2 n)!}} | \\ = \lim_{n \to \infty} | \frac{(- 1) z^{2}}{(2 n + 1) (2 n + 2)} | \end{matrix}$

Now, for the series to be convergent, we have $ρ < 1$ . So,

$\begin{matrix} ρ = \lim_{n \to \infty} | \frac{(- 1) z^{2}}{(2 n + 1) (2 n + 2)} | < 1 \\ {| z |}^{2} < \lim_{n \to \infty} | (2 n + 1) (2 n + 2) | \\ {| z |}^{2} < \infty \\ | z | < \infty \end{matrix}$

Hence, the required disk of convergence is . $| z | < \infty$

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: Find the disk of convergence for each of the followingcomplex power series.
$\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} z^{2 n}}{(2 n)!}$

Short Answer

Step by step solution

Disk of Convergence

Step 2:Find the disk of Convergence

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: Find the disk of convergence for each of the followingcomplex power series.∑n=0∞(−1)nz2n(2n)!

Short Answer

Step by step solution

Disk of Convergence

Step 2:Find the disk of Convergence

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Conservation of Energy and Momentum

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: Find the disk of convergence for each of the followingcomplex power series.
$\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} z^{2 n}}{(2 n)!}$