Question

The Laplace transform of \( \cos\sqrt{t} \) is:

• A. \( \sum_{n=0}^\infty \frac{(-1)^{n-1} n!}{(2n)! s^n} \), s is the parameter of Laplace transform.
• B. \( \sum_{n=0}^\infty \frac{(-1)^n n!}{(2n)! s^{n+1}} \), s is the parameter of Laplace transform.
• C. \( \sum_{n=0}^\infty \frac{(-1)^{n+1} n!}{(2n)! s^n} \), s is the parameter of Laplace transform.
• D. \( \sum_{n=0}^\infty \frac{(-1)^n n!}{(2n)! (s+1)^{n+1}} \), s is the parameter of Laplace transform.

Hint:

When faced with a Laplace transform of a function for which you don't have a standard formula (especially involving compositions like \( \cos\sqrt{t} \) or \( \frac{\sin t}{t} \)), the Taylor series expansion is a powerful technique. Expand the function, then transform term-by-term using the basic formula \( \mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}} \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We cannot directly compute the Laplace transform of \( \cos\sqrt{t} \) using standard table entries. The best approach is to use the Taylor series expansion of the cosine function and then apply the Laplace transform term by term.
Step 2: Key Formula or Approach:
1. The Taylor series for \( \cos(x) \) is \( \cos(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} \). 2. The Laplace transform of \( t^k \) is \( \mathcal{L}\{t^k\} = \frac{\Gamma(k+1)}{s^{k+1}} = \frac{k!}{s^{k+1}} \) for integer k.
Step 3: Detailed Explanation:
1. Expand \( \cos\sqrt{t} \) using its Taylor series:
Substitute \( x = \sqrt{t} \) into the series for \(\cos(x)\): \[ \cos\sqrt{t} = \sum_{n=0}^\infty \frac{(-1)^n (\sqrt{t})^{2n}}{(2n)!} = \sum_{n=0}^\infty \frac{(-1)^n t^n}{(2n)!} \] \[ \cos\sqrt{t} = \frac{t^0}{0!} - \frac{t^1}{2!} + \frac{t^2}{4!} - \frac{t^3}{6!} + \ldots \] 2. Apply the Laplace transform term by term:
Assuming we can interchange the sum and the integral: \[ \mathcal{L}\{\cos\sqrt{t}\} = \mathcal{L}\left\{ \sum_{n=0}^\infty \frac{(-1)^n t^n}{(2n)!} \right\} = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} \mathcal{L}\{t^n\} \] Now, use the transform for \(t^n\): \( \mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}} \). \[ \mathcal{L}\{\cos\sqrt{t}\} = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} \frac{n!}{s^{n+1}} = \sum_{n=0}^\infty \frac{(-1)^n n!}{(2n)! s^{n+1}} \] Step 4: Final Answer:
The Laplace transform is the series \( \sum_{n=0}^\infty \frac{(-1)^n n!}{(2n)! s^{n+1}} \), which matches option (B).