Q16E

Question

In Problems 16, find a Taylor series for f(t) about t=0. Assuming the Laplace transform of f(t) can be computed term by term, find an expansion for \(\mathcal{L}\{ f\} (s)\)in powers of 1 / s. If possible, sum the series.

\({\rm{\;16}}{\rm{.\;}}f(t) = \sin t\)

Step-by-Step Solution

Verified

Answer

1Step 1: Define Laplace transform

A transformation of a function f(x) into the function \(g(t) = \smallint \infty 0e - xtf(x)dx\) that is useful especially in reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation.

2Step 2: Find expansion for \(\mathcal{L}\{ f\} (s)\)

Given

\(\begin{array}{c}f(t) = \sin t\\f(0) = \sin 0\\ = 0\end{array}\)

Also,

\(\begin{array}{c}f'(t) = \cos tf'(0)\\ = \cos 0\\ = 1\end{array}\)

\(\begin{array}{c}f''(t) = - \sin tf''(0)\\ = - \sin 0\\ = 0\end{array}\)

\(\begin{array}{c}f'''(t) = - \cos tf'''(0)\\ = - \cos 0\\ = - 1\end{array}\)

Now Taylor series for f(t) is given by

\(\begin{array}{c}f(t) = f(0) + \frac{{f'(0)t}}{{1!}} + \frac{{f''(0){t^2}}}{{2!}} + \ldots \\ = 0 + t - \frac{{{t^3}}}{{3!}} + \ldots \end{array}\)

Taking laplace transform we get.

\(\begin{array}{c}f(s) = \frac{1}{{{s^2}}} - \frac{{3!}}{{3!{s^4}}} + \ldots \\ = \frac{1}{{{s^2}}} - \frac{1}{{{s^4}}} + \ldots \end{array}\)

'This is Geometric Progression (GP) with