Step 1: Understanding the Concept:
Direct computation of the Laplace transform for \( \cos\sqrt{t} \) using standard tables is not feasible. The recommended method involves employing the Taylor series expansion of the cosine function, followed by term-by-term application of the Laplace transform.
Step 2: Key Formula or Approach:
1. The Taylor series representation for \( \cos(x) \) is given by \( \cos(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} \).2. The Laplace transform of \( t^k \), where \( k \) is an integer, is defined as \( \mathcal{L}\{t^k\} = \frac{\Gamma(k+1)}{s^{k+1}} = \frac{k!}{s^{k+1}} \).
Step 3: Detailed Explanation:
1. Expand \( \cos\sqrt{t} \) using its Taylor series:
By substituting \( x = \sqrt{t} \) into the series for \(\cos(x)\), we obtain:\[ \cos\sqrt{t} = \sum_{n=0}^\infty \frac{(-1)^n (\sqrt{t})^{2n}}{(2n)!} = \sum_{n=0}^\infty \frac{(-1)^n t^n}{(2n)!} \]This expansion can be written as:\[ \cos\sqrt{t} = \frac{t^0}{0!} - \frac{t^1}{2!} + \frac{t^2}{4!} - \frac{t^3}{6!} + \ldots \]2. Apply the Laplace transform term by term:
Assuming the interchange of summation and integration is permissible:\[ \mathcal{L}\{\cos\sqrt{t}\} = \mathcal{L}\left\{ \sum_{n=0}^\infty \frac{(-1)^n t^n}{(2n)!} \right\} = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} \mathcal{L}\{t^n\} \]Utilizing the Laplace transform of \(t^n\), which is \( \mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}} \):\[ \mathcal{L}\{\cos\sqrt{t}\} = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} \frac{n!}{s^{n+1}} = \sum_{n=0}^\infty \frac{(-1)^n n!}{(2n)! s^{n+1}} \]Step 4: Final Answer:
The computed Laplace transform is represented by the series \( \sum_{n=0}^\infty \frac{(-1)^n n!}{(2n)! s^{n+1}} \), which corresponds to option (B).