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align fix; theorem style; condition number

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jverzani
2024-10-31 14:22:21 -04:00
parent 3e7e3a9727
commit 18aae2aa93
61 changed files with 1705 additions and 819 deletions
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12
quarto/derivatives/linearization.qmd
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@@ -510,21 +510,23 @@ $$
Suppose $f(x)$ and $g(x)$ are represented by their tangent lines about $c$, respectively:
$$
\begin{align*}
f(x) &= f(c) + f'(c)(x-c) + \mathcal{O}((x-c)^2), \\
g(x) &= g(c) + g'(c)(x-c) + \mathcal{O}((x-c)^2).
\end{align*}
$$
Consider the sum, after rearranging we have:
$$
\begin{align*}
f(x) + g(x) &= \left(f(c) + f'(c)(x-c) + \mathcal{O}((x-c)^2)\right) + \left(g(c) + g'(c)(x-c) + \mathcal{O}((x-c)^2)\right)\\
&= \left(f(c) + g(c)\right) + \left(f'(c)+g'(c)\right)(x-c) + \mathcal{O}((x-c)^2).
\end{align*}
$$
The two big "Oh" terms become just one as the sum of a constant times $(x-c)^2$ plus a constant time $(x-c)^2$ is just some other constant times $(x-c)^2$. What we can read off from this is the term multiplying $(x-c)$ is just the derivative of $f(x) + g(x)$ (from the sum rule), so this too is a tangent line approximation.
@@ -533,7 +535,7 @@ The two big "Oh" terms become just one as the sum of a constant times $(x-c)^2$
Is it a coincidence that a basic algebraic operation with tangent lines approximations produces a tangent line approximation? Let's try multiplication:
$$
\begin{align*}
f(x) \cdot g(x) &= [f(c) + f'(c)(x-c) + \mathcal{O}((x-c)^2)] \cdot [g(c) + g'(c)(x-c) + \mathcal{O}((x-c)^2)]\\
&=[f(c) + f'(c)(x-c)] \cdot [g(c) + g'(c)(x-c)] + (f(c) + f'(c)(x-c)) \cdot \mathcal{O}((x-c)^2) + (g(c) + g'(c)(x-c)) \cdot \mathcal{O}((x-c)^2) + [\mathcal{O}((x-c)^2)]^2\\
@@ -541,6 +543,7 @@ f(x) \cdot g(x) &= [f(c) + f'(c)(x-c) + \mathcal{O}((x-c)^2)] \cdot [g(c) + g'(
&= f(c) \cdot g(c) + [f'(c)\cdot g(c) + f(c)\cdot g'(c)] \cdot (x-c) + [f'(c)\cdot g'(c) \cdot (x-c)^2 + \mathcal{O}((x-c)^2)] \\
&= f(c) \cdot g(c) + [f'(c)\cdot g(c) + f(c)\cdot g'(c)] \cdot (x-c) + \mathcal{O}((x-c)^2)
\end{align*}
$$
The big "oh" notation just sweeps up many things including any products of it *and* the term $f'(c)\cdot g'(c) \cdot (x-c)^2$. Again, we see from the product rule that this is just a tangent line approximation for $f(x) \cdot g(x)$.
@@ -803,13 +806,14 @@ numericq(abs(answ))
The [Birthday problem](https://en.wikipedia.org/wiki/Birthday_problem) computes the probability that in a group of $n$ people, under some assumptions, that no two share a birthday. Without trying to spoil the problem, we focus on the calculus specific part of the problem below:
$$
\begin{align*}
p
&= \frac{365 \cdot 364 \cdot \cdots (365-n+1)}{365^n} \\
&= \frac{365(1 - 0/365) \cdot 365(1 - 1/365) \cdot 365(1-2/365) \cdot \cdots \cdot 365(1-(n-1)/365)}{365^n}\\
&= (1 - \frac{0}{365})\cdot(1 -\frac{1}{365})\cdot \cdots \cdot (1-\frac{n-1}{365}).
\end{align*}
$$
Taking logarithms, we have $\log(p)$ is