> [!theorem] > > Let $f(x) = (1 - t)^{1/2}$, then > $ > \frac{d^nf}{dt^n}(0)= \prod_{k = 0}^{n-1}\paren{\frac{1}{2} - k} > $ > and let > $ > c_n = \frac{1}{n!}\frac{d^nf}{dt^n}(0) \quad p(t) = \sum_{n = 0}^\infty c_nt^n > $ > be the corresponding Maclaurin [[Power Series|series]]. > 1. $p$ has a radius of convergence of $1$. > 2. $p(t) = (1 - t)^{1/2}$. > 3. $p$ converges on the endpoints. > > *Proof*. The coefficients comes from the power rule, where > $ > \frac{d^nf}{dt^n}(t) = (1 - t)^{1/2 - n}\prod_{k = 0}^{n-1}\paren{\frac{1}{2} - k} > $ > From here, we can distribute the factorial as > $ > c_n = \frac{1}{n!}\prod_{k = 0}^{n - 1}\paren{\frac{1}{2} - k} = \prod_{k = 0}^{n - 1}\frac{1}{k+1}\paren{\frac{1}{2} - k} > $ > since $\abs{(1/2 - k)/(k + 1)} \le 1$ for all $k \ge 0$, $\abs{c_n} \le 1$ for all $n \in \nat$. On the other hand, shifting the coefficients yields > $ > c_n = C \cdot \frac{1}{n(n-1)} \cdot \prod_{k = 2}^{n - 1}\frac{1}{k - 1}\paren{\frac{1}{2} - k} > $ > therefore > $ > 1 = \limsup_{n \to \infty} \braks{n(n-1)}^{-1/n} \le \limsup_{n \to \infty}\abs{c_n}^{1/n} \le 1 > $ > and the radius of convergence is $1$. For any $t \in (0, 1)$, the [[Taylor's Formula|Taylor]] remainder can be estimated as > $ > R_n(t) \le \int_0^1 \frac{(1 - s)^{n - 1}}{(n - 1)!}(1 - st)^{1/2 - n}t^{n}ds > $ > If $\abs{t} < r < 1$, then there exists $M_r \ge 0$ such that $\abs{1 - st}^{1/2 - n} \le M_r^n$ for all $s \in [0, 1]$. So > $ > R_n(t) \le \int_0^1 \frac{M_r^nC^n}{(n-1)!}ds = \frac{M_r^nC^n}{(n-1)!} \to 0 > $ > as $n \to \infty$. Lastly, by the [[Monotone Convergence Theorem for Sequences|monotone convergence theorem]], > $ > \sum_{n = 1}^\infty c_n = \lim_{t \to 1}\sum_{n = 1}^\infty c_n t^n = 1 - \lim_{t \to 1}(1 - t)^{1/2} = 1 > $ > so the series converges absolutely on the endpoints as well. > [!theorem] > > For any $\eps > 0$, there exists a polynomial $p \in \real[x]$ such that $p(0) = 0$ and $\norm{\abs{\cdot} - p}_{u, [-1, 1]} < \eps$. > > *Proof*. Let $q \in \real[x]$ such that $\norm{(1 - (\cdot))^{1/2} - q}_{u, [0, 1]} < \eps/2$. Let $t = 1 - x^2$ and $p = q(1 - x^2)$, then $\norm{\abs{\cdot} - p}_{u, [-1, 1]} < \eps/2$. From here, offsetting $p$ such that $p(0) = 0$ yields the desired result.