From the Power Series to the Laplace Transform

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The Laplace Transform can be derived from the Powerseries. Let's consider the Powerseries \begin{align} \sum_{n=0}^{\infty} a(n)x^n=A(x) \end{align} where we use the notation $a_n=a(n)$ for the coefficients of the series. The function $a(n)$ with $n$ as variable is transformed ($\leadsto$) into a function $A(x)$ with $x$ as variable. \begin{align} a(n) \leadsto A(x) \end{align} We illustrate this by a few examples. The simple function $a(n)=1$ leads to a geometric series, which converges for $x < 1$ \begin{align} 1 \leadsto A(x)\\ A(x)=\frac{1}{1-x} \forall \, 0\leq x<1 \end{align} Another example \begin{align} \frac{1}{n!} \leadsto A(x)\\ \end{align} has the solution $A(x)=e^x$, since $\sum_n \frac{1}{n!}x^n$ is exactly the Taylorexpansion of $e^x$. If we now change the descrete variable $n=1,2,3,\dots$ to the continuous variable $t$ and replace the sum by an integral we have \begin{align} \int_0^{\infty} a(t) x^t dt = A(x) \end{align} We are allowed to write \begin{align} x=e^{\ln x}\\ x^t= \left( e^{\ln x} \right)^t \end{align} and just rename $\ln x =-s$, so we get \begin{align} x^t= e^{-s \cdot t} \end{align}

This leads us to the definition of the Laplace transform \begin{align} F(s)=\int_0^{\infty} f(t) e^{-st} dt \end{align}