User:LudicrousTripe/sandbox

This is the user sandbox of LudicrousTripe. A user sandbox is a subpage of the user's user page. It serves as a testing spot and page development space for the user and is not an encyclopedia article. Create or edit your own sandbox here.

Other sandboxes: Main sandbox | Template sandbox

Finished writing a draft article? Are you ready to request review of it by an experienced editor for possible inclusion in Wikipedia? Submit your draft for review!

It is straightforward to show the continuity of the polynomials, and so we hate that silly ^{[citation needed]} on Polynomial » Polynomial functions.

Weierstrass? Great choice! Great choice! Let's go!

Unfolding Weierstrass's continuity, one observes that any monomial,

⁠

ax^{k}:{a,x\in \mathbb {R} },\ {n\in \mathbb {N} }

⁠

is continuous, and then use the standard result that the sum of two continuous functions on some domain, in this case ⁠ $\mathbb {R}$ ⁠, is also continuous on that domain. Iteratively applying the result permits the conclusion that the sum of a finite number of monomials,

\sum _{i=0}^{n}a_{i}x^{i}

is continuous, i.e. that any polynomial is continuous.

Doc's not fucking having it. He says the fucking result for the case $x 0 = 0$ is so obvious $(δ = | ε | 1/ n)$ it's not worth worrying about, so he says to get your shit in gear and forget about $x 0 = 0$ already:

⁠

\mathrm {Let} \ x_{0}\in \mathbb {R} \backslash \{0\},\ \mathrm {and\ let} \ \varepsilon >0\ \mathrm {be\ given.} \ \mathrm {We\ need} \ \delta >0:

⁠

Well, we're talking monomials, so we have:

f(x)=ax^{n}

So let $ε > 0$ be given. We need to show $\exists δ > 0$ such that

⁠

First we note that,

{\begin{aligned}|ax^{n}-ax_{0}^{n}|&=|((ax-ax_{0})+ax_{0})^{n}-ax_{0}^{n}|.\end{aligned}}

Next we use the binomial expansion:

((ax-ax_{0})+ax_{0})^{n}=\sum _{k=0}^{n}{n \choose k}\ (ax-ax_{0})^{k}\ ax_{0}^{n-k}.

So we get,

{\begin{aligned}|ax^{n}-ax_{0}^{n}|&={\Big |}\sum _{k=0}^{n}{n \choose k}\ (ax-ax_{0})^{k}\ ax_{0}^{n-k}-ax_{0}^{n}{\Big |}\\&\leq \sum _{k=1}^{n}{n \choose k}\left|(ax-ax_{0})^{k}\ ax_{0}^{n-k}\right|.\end{aligned}}

There are $n$ terms in this sum, so we can say that

If,\ \forall k\in \{1,\dots ,n\},\ {\binom {n}{k}}|(ax-ax_{0})^{k}\ ax_{0}^{n-k}|<\varepsilon /n\ ,\ then\ |ax^{n}-ax_{0}^{n}|<\varepsilon .

To guarantee the above, we just pick

\delta =\min \left\{\ \left({\frac {\varepsilon }{{n \choose k}n|ax_{0}|}}\right)^{\frac {1}{k}},\ \ k\in \{1,\dots ,n\}\ \right\}.

OK, so we've just shown any monomial is continuous. Now, as we said at the outset, we just use the fact that, loosely speaking, $f + g$ is continuous if $f$ and $g$ are cts. And we're done. LudicrousTripe (talk) 01:30, 12 November 2013 (UTC)