Q18E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 18, find a Taylor series for \(f\left( t \right)\) about t=0. Assuming the Laplace transform of \(f\left( t \right)\) can be computed term by term, find an expansion for \(\mathcal{L}\{ f\} (s)\)in powers of \(1/s\). If possible, sum the series.

\({\rm{\;18}}{\rm{.\;}}f(t) = {e^{ - {r^2}}}\)

Step-by-Step Solution

Answer

\(\mathcal{L}\left\{ {{e^{ - {t^2}}}} \right\}(s) = \sum\limits_{n = 1}^\infty   {\frac{{{{( - 1)}^n}(2n)!}}{{n!}}{{\left( {\frac{1}{s}} \right)}^{2n + 1}}} \)

Step 1: Define Laplace transform

A transformation of a function f(x) into the function  that is useful especially in reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation.

Step 2:Find expansion for \(\mathcal{L}\left\{ {{e^{ - {t^2}}}} \right\}(s)\)

Since the Taylor's series of \({e^t}\) is

\({e^t} = 1 + t + \frac{{{t^2}}}{{2!}} + \frac{{{t^3}}}{{3!}} -   \cdots  + \frac{{{t^n}}}{{n!}} +  \cdots \)

we have that the Taylor series of f(t) is

\({e^{ - {t^2}}} = 1 - {t^2} + \frac{{{t^4}}}{{2!}} - \frac{{{t^6}}}{{3!}} +   \cdots  + {( - 1)^n}\frac{{{t^{2n}}}}{{n!}} +  \cdots \)

Using linearity of Laplace transform we get

\(\begin{array}{c}\mathcal{L}\left\{ {{e^{ - {t^2}}}} \right\}(s) = \mathcal{L}\left\{ {1 - {t^2} + \frac{{{t^4}}}{{2!}} - \frac{{{t^6}}}{{3!}} +   \cdots  + {{( - 1)}^n}\frac{{{t^{2n}}}}{{n!}} +  \cdots } \right\}\\ = \mathcal{L}\{ 1\} (s) - \mathcal{L}\left\{ {{t^2}} \right\}(s) + \frac{1}{{2!}}\mathcal{L}\left\{ {{t^4}} \right\}(s) +  \cdots   + \frac{{{{( - 1)}^n}}}{{n!}}\mathcal{L}\left\{ {{t^{2n}}} \right\} +  \cdots \\ = \frac{1}{s} - \frac{2}{{{s^3}}} + \frac{{12}}{{{s^5}}} +  \cdots  + \frac{{{{( - 1)}^n}(2n)!}}{{n!{s^{2n + 1}}}} +  \cdots \\ = \sum\limits_{n = 1}^\infty  {\frac{{{{( - 1)}^n}(2n)!}}{{n!{s^{2n + 1}}}}} \end{array}\).

From where it follows that

\(\sum\limits_{n = 1}^\infty  {\frac{{{{( - 1)}^n}(2n)!}}{{n!{s^{2n + 1}}}}}  = \sum\limits_{n = 1}^\infty  {\frac{{{{( - 1)}^n}(2n)!}}{{n!}}{{\left( {\frac{1}{s}} \right)}^{2n + 1}}} \)

Therefore,

\(\mathcal{L}\left\{ {{e^{ - {t^2}}}} \right\}(s) = \sum\limits_{n = 1}^\infty   {\frac{{{{( - 1)}^n}(2n)!}}{{n!}}{{\left( {\frac{1}{s}} \right)}^{2n + 1}}} \)