rm WeaveSupport

quarto/derivatives/taylor_series_polynomials.qmd

@@ -16,15 +16,7 @@ using Unitful
 ```{julia}
 #| echo: false
 #| results: "hidden"
-using CalculusWithJulia.WeaveSupport
 using Roots
-
-fig_size = (800, 600)
-const frontmatter = (
-        title = "Taylor Polynomials and other Approximating Polynomials",
-        description = "Calculus with Julia: Taylor Polynomials and other Approximating Polynomials",
-        tags = ["CalculusWithJulia", "derivatives", "taylor polynomials and other approximating polynomials"],
-);
 nothing
 ```
 
@@ -1259,5 +1251,3 @@ scatter!(cps, h1.(cps), markersize=5, marker=:square)
 ```
 
 Again by Rolle's theorem, between any pair of adjacent zeros $\xi^1_i, \xi^1_{i+1}$ there must be a zero $\xi^2_i$ of $h''(x)$. So there are $n-1$ zeros of $h''(x)$. Continuing, we see that there will be $n+1-3$ zeros of $h^{(3)}(x)$, $n+1-4$ zeros of $h^{4}(x)$, $\dots$, $n+1-(n-1)$ zeros of $h^{n-1}(x)$, and finally $n+1-n$ ($1$) zeros of $h^{(n)}(x)$. Call this last zero $\xi$. It satisfies $x_0 \leq \xi \leq x_n$. Further, $0 = h^{(n)}(\xi) = f^{(n)}(\xi)  - g^{(n)}(\xi)$. But $g$ is a degree $n$ polynomial, so the $n$th derivative is the coefficient of $x^n$ times $n!$. In this case we have $0 = f^{(n)}(\xi) - f[x_0, \dots, x_n] n!$. Rearranging yields the result.
-
-