How to relate region of convergence between a function and its Laplace Transform.

21 Views Asked by At

G'day,

I have a query related to convergence of the Laplace Transform of a function:

Consider we have a function $f(x)$ defined by the infinite series: $$ f(x) = \sum_{n = 0}^\infty a_n x^n $$ Suppose this is convergent via the Ratio Test with : $\left|x\right| < \frac{1}{K}$ where: $$ K = \lim_{n \rightarrow \infty} \left|\frac{a_{n + 1}}{a_n}\right| $$ Noting that if $K = 0$, then $\left|x\right| < \infty$.

Now, if we take the Laplace Transform of $f(x)$, then

$$ F(s) = \mathscr{L}\left[f(x)\right] = \sum_{n = 0}^\infty a_n \mathscr{L}\left[x^n\right] = \sum_{n = 0}^\infty a_n \frac{n!}{s^{n + 1}} $$
If we then apply the Ratio Test for the transform: $$ R =\lim_{n \rightarrow \infty} \left|\frac{b_{n + 1}}{b_n}\right| = \lim_{n \rightarrow \infty} \left|\left(n + 1\right) \frac{a_{n + 1}}{a_n}\frac{1}{s}\right| = \frac{1}{\left|s\right|}\lim_{n \rightarrow \infty}\left|n\frac{a_{n + 1}}{a_n} + \frac{a_{n + 1}}{a_n}\right| $$ Via the triangle inequality, this becomes: $$ R =\frac{1}{\left|s\right|}\left(\lim_{n \rightarrow \infty}\left|n\frac{a_{n + 1}}{a_n}\right| +\lim_{n \rightarrow \infty} \left| \frac{a_{n + 1}}{a_n}\right|\right) \leq \frac{1}{\left|s\right|}\left(\lim_{n \rightarrow \infty}\left|n\frac{a_{n + 1}}{a_n}\right| +K\right) $$ For convergence, we require $R < 1$, and so we require: $$ \frac{1}{\left|s\right|}\left(\lim_{n \rightarrow \infty}\left|n\frac{a_{n + 1}}{a_n}\right| +K\right) < 1 \longrightarrow \left|s\right| > K + \lim_{n \rightarrow \infty}\left|n\frac{a_{n + 1}}{a_n}\right| $$ And so, if that $L = \lim_{n \rightarrow \infty}\left|n\frac{a_{n + 1}}{a_n}\right|$ is finite, then the series converges for $s$ as defined above.

Now, I respect there are more tests to define convergence and thus this is not the sole method, but does this provide a set of conditions to define regions of convergence of $F(s)$ and $f(x)$??