em dash; sentence case

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jverzani
2025-07-27 15:26:00 -04:00
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@@ -1,4 +1,4 @@
# Taylor Polynomials and other Approximating Polynomials
# Taylor polynomials and other approximating polynomials
{{< include ../_common_code.qmd >}}
@@ -104,7 +104,7 @@ $$
tl(x) = f(c) + f'(c) \cdot(x - c).
$$
The key is the term multiplying $(x-c)$ -- for the secant line this is an approximation to the related term for the tangent line. That is, the secant line approximates the tangent line, which is the linear function that best approximates the function at the point $(c, f(c))$.
The key is the term multiplying $(x-c)$---for the secant line this is an approximation to the related term for the tangent line. That is, the secant line approximates the tangent line, which is the linear function that best approximates the function at the point $(c, f(c))$.
This is quantified by the *mean value theorem* which states under our assumptions on $f(x)$ that there exists some $\xi$ between $x$ and $c$ for which:
@@ -194,7 +194,7 @@ function divided_differences(f, x, xs...)
end
```
In the following--even though it is *type piracy*--by adding a `getindex` method, we enable the `[]` notation of Newton to work with symbolic functions, like `u()` defined below, which is used in place of $f$:
In the following---even though it is *type piracy*---by adding a `getindex` method, we enable the `[]` notation of Newton to work with symbolic functions, like `u()` defined below, which is used in place of $f$:
```{julia}
@@ -215,7 +215,7 @@ Now, let's look at:
ex₂ = u[c, c+h, c+2h]
```
If multiply by $2$ and simplify, a discrete approximation for the second derivative--the second order forward [difference equation](http://tinyurl.com/n4235xy)--is seen:
If multiply by $2$ and simplify, a discrete approximation for the second derivative---the second order forward [difference equation](http://tinyurl.com/n4235xy)---is seen:
```{julia}
simplify(2ex₂)
@@ -794,7 +794,7 @@ This is re-expressed as $2s + s \cdot p$ with $p$ given by:
```{julia}
cancel((a_b - 2s)/s)
p = cancel((a_b - 2s)/s)
```
Now, $2s = m - s\cdot m$, so the above can be reworked to be $\log(1+m) = m - s\cdot(m-p)$.
@@ -807,7 +807,7 @@ How big can the error be between this *approximations* and $\log(1+m)$? The expr
```{julia}
Max = (v/(2+v))(v => sqrt(2) - 1)
Max = (x/(2+x))(x => sqrt(2) - 1)
```
The error term is like $2/19 \cdot \xi^{19}$ which is largest at this value of $M$. Large is relative - it is really small: