User:LudicrousTripe/sandbox

It is straightforward to show the continuity of the polynomials, and so we hate that silly [citation needed] on 🇦🇹.

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

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



is continuous, and then use the standard result that the sum of two continuous functions on some domain, in this case $ax^k : { a, x \in \mathbb{R} }, \ {n \in \mathbb{N} }$, 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 $(δ = |&#8239;ε&#8239;|^{1/n})$ it's not worth worrying about, so he says to get your shit in gear and forget about $x_{0} = 0$ already:



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



First we note that,
 * $$ \begin{align}


 * ax^n - ax_0^n | &= | ((ax - ax_0) + ax_0)^n - ax_0^n |. \end{align} $$

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{align}
 * 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| \\

&\le \sum_{k=1}^n {n \choose k} \left | (ax - ax_0)^k \ ax_0^{n-k} \right |. \end{align} $$

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 |a x_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)