User:GFauxPas/Sandbox

Welcome to my sandbox, you are free to play here as long as you don't track sand onto the main wiki. --GFauxPas 09:28, 7 November 2011 (CST)

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/existence

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/integrand

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/evaluation

User:GFauxPas/Sandbox/Zeta2/FourierSeries/

User:GFauxPas/Sandbox/Zeta2/Informal Proof

$\mathcal L \left\{{}\right\}$


 * $\mathcal L \left\{{e^{at}f\left({t}\right)}\right\} = F\left({s-a}\right)$

Theorem
Let $\mathcal Lf\left({x}\right) = F\left({x}\right)$ be the Laplace Transform of $f$.

Let $e^x$ be the exponential.

Then:


 * $\displaystyle \mathcal L \left\{{e^{ax} f\left({x}\right)}\right\} = F\left({s-a}\right)$

everywhere that $\mathcal Lf$ exists.

Theorem
Let $P_n$ be a real polynomial, of degree $n$.

Let $e^z$ be the complex exponential, where $z = x + iy$.

Then for every $n \in \N_{\ge 0}$:


 * $\displaystyle \lim_{x \mathop \to +\infty} \frac {P_n}{e^z} = 0$

Proof
The proof proceeds by induction on $n$, the degree of $P_n$.

Basis for the Induction
The case $n = 0$ is verified as follows:

This is the basis for the induction.

Induction Hypothesis
Fix $n \in \N$ with $n \ge 0$.

Assume:

holds for $n$.

This is our induction hypothesis.

Induction Step
This is our induction step:

The result follows by the Principle of Mathematical Induction.

Am I breaking any rules here? I feel uncomfortable about this for some reason. --GFauxPas (talk) 10:17, 12 May 2014 (UTC)


 * Why not just invoke Exponential Dominates Polynomial? Then there's no need to dig the heavy machinery of induction out of the garage. --prime mover (talk) 11:37, 14 May 2014 (UTC)