16.2 Completing the square


Table 16.1: Properties of the Laplace Transform. ($k,\tau,\omega>0$, $n=1,2,\ldots$)

\begin{picture}(402,593)(1,0)
% put(1,0)\{ framebox(402,593)\{\}\}
\put(0,295)...
... &
$\displaystyle e^{-\tau s}$ \\
\hline
\end{tabular*}}}
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Completing the square simply means that you write a quadratic

\begin{displaymath}
a x^2 + b x + c
\end{displaymath}

as

\begin{displaymath}
a \bigg[
\bigg(
\underbrace{x+\frac{b}{2a}}_{\textstyl...
...rac{b^2}{4a^2}}_{\textstyle \rm {new\atop constant}}
\bigg]
\end{displaymath}

If you multiply out, you see that it is the same.

One place where you often need this is in Laplace transforms. Laplace transforms are often given in terms of a quadratic $s^2+K$ where $K$ is an arbitrary constant. But you might have $as^2+bs+c$ instead of something of the form $s^2+K$. However, you can write the part inside the square brackets above as $s^2+K$ if you use the shift theorem to account for the $b/2a$ inside the parentheses. And the additional factor $a$ is trivial to account for.