User:LudicrousTripe/sandbox

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 x₀ = 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 x₀ = 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 ∃δ > 0 such that

⁠${\displaystyle }$⁠

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)