The Laplace transform from scratch

Here's a natural way to derive the Laplace transform via taking the concept of a taylor series and making it continuous. We will, of course, start with a taylor series whose coefficients are given by $a(n)$

$\qquad \qquad \displaystyle\sum_0^{\infty} a(n) x^n \qquad x<1$

Then, make this continuous by replacing the sum by an integral

$\qquad \qquad \displaystyle\int_0^{\infty} a(n) x^n dn \qquad x<1$

We can change $x^n$ to $e^n \ln x$ and make the substitution $s =- \ln x$ to give us

$\qquad \qquad \displaystyle\int_0^{\infty} a(n) e^{-sn} dn \qquad s > 0$

Which is just the definition for the Laplace transform of $a(n)$ , $\mathcal{L}(a)(s)$ .

#Calculus

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Comments

Nice!

Krishna Karthik - 1 year, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$