align fix; theorem style; condition number

quarto/integrals/area.qmd

@@ -230,7 +230,7 @@ To successfully compute a good approximation for the area, we would need to choo
 For Archimedes' problem - finding the area under $f(x)=x^2$ between $0$ and $1$ -  if we take as a partition $x_i = i/n$ and $c_i = x_i$, then the above sum becomes:
 
 
-
+$$
 \begin{align*}
 S_n &= f(c_1) \cdot (x_1 - x_0) + f(c_2) \cdot (x_2 - x_1) + \cdots + f(c_n) \cdot (x_n - x_{n-1})\\
 &= (x_1)^2 \cdot \frac{1}{n} + (x_2)^2 \cdot \frac{1}{n} + \cdots + (x_n)^2 \cdot \frac{1}{n}\\
@@ -238,6 +238,7 @@ S_n &= f(c_1) \cdot (x_1 - x_0) + f(c_2) \cdot (x_2 - x_1) + \cdots + f(c_n) \cd
 &= \frac{1}{n^3} \cdot (1^2 + 2^2 + \cdots + n^2) \\
 &= \frac{1}{n^3} \cdot \frac{n\cdot(n-1)\cdot(2n+1)}{6}.
 \end{align*}
+$$
 
 
 The latter uses a well-known formula for the sum of squares of the first $n$ natural numbers.
@@ -301,13 +302,18 @@ The general statement allows for any partition such that the largest gap goes to
 Riemann sums weren't named after Riemann because he was the first to approximate areas using rectangles. Indeed, others had been using even more efficient ways to compute areas for  centuries prior to Riemann's work. Rather, Riemann put the definition of the area under the curve on a firm theoretical footing with the following theorem which gives a concrete notion of what functions are integrable:
 
 
-> **Riemann Integral**: A function $f$ is Riemann integrable over the interval $[a,b]$ and its integral will have value $V$ provided for every $\epsilon > 0$ there exists a $\delta > 0$ such that for any partition $a =x_0 < x_1 < \cdots < x_n=b$ with $\lvert x_i - x_{i-1} \rvert < \delta$ and for any choice of points $x_{i-1} \leq c_i \leq x_{i}$ this is satisfied:
->
-> $$
-> \lvert \sum_{i=1}^n f(c_i)(x_{i} - x_{i-1}) - V \rvert < \epsilon.
-> $$
->
-> When the integral exists, it is written $V = \int_a^b f(x) dx$.
+::: {.callout-note icon=false}
+## Riemann Integral
+
+A function $f$ is Riemann integrable over the interval $[a,b]$ and its integral will have value $V$ provided for every $\epsilon > 0$ there exists a $\delta > 0$ such that for any partition $a =x_0 < x_1 < \cdots < x_n=b$ with $\lvert x_i - x_{i-1} \rvert < \delta$ and for any choice of points $x_{i-1} \leq c_i \leq x_{i}$ this is satisfied:
+
+$$
+\lvert \sum_{i=1}^n f(c_i)(x_{i} - x_{i-1}) - V \rvert < \epsilon.
+$$
+
+When the integral exists, it is written $V = \int_a^b f(x) dx$.
+
+:::
 
 
 
@@ -370,13 +376,14 @@ The area is invariant under shifts left or right.
 
 Any partition $a =x_0 < x_1 < \cdots < x_n=b$ is related to a partition of $[a-c, b-c]$ through $a-c < x_0-c < x_1-c < \cdots < x_n - c = b-c$. Let $d_i=c_i-c$ denote this partition, then we have:
 
-
+$$
 \begin{align*}
 f(c_1 -c) \cdot (x_1 - x_0) &+ f(c_2 -c) \cdot (x_2 - x_1)  + \cdots\\
  &\quad + f(c_n -c) \cdot (x_n - x_{n-1})\\
  &= f(d_1) \cdot(x_1-c - (x_0-c)) + f(d_2) \cdot(x_2-c - (x_1-c)) + \cdots\\
  &\quad + f(d_n) \cdot(x_n-c - (x_{n-1}-c)).
 \end{align*}
+$$
 
 
 The left side will have a limit of $\int_a^b f(x-c) dx$ the right would have a "limit" of $\int_{a-c}^{b-c}f(x)dx$.
@@ -471,7 +478,7 @@ Using the definition, we can compute a few definite integrals:
 This is just the area of a trapezoid with heights $a$ and $b$ and side   length $b-a$, or $1/2 \cdot (b + a) \cdot (b - a)$. The right sum   would be:
 
 
-
+$$
 \begin{align*}
 S &= x_1 \cdot (x_1 - x_0) + x_2 \cdot (x_2 - x_1) + \cdots x_n \cdot (x_n - x_{n-1}) \\
 &= (a + 1\frac{b-a}{n}) \cdot \frac{b-a}{n} + (a + 2\frac{b-a}{n}) \cdot \frac{b-a}{n} + \cdots  (a + n\frac{b-a}{n}) \cdot \frac{b-a}{n}\\
@@ -480,6 +487,7 @@ S &= x_1 \cdot (x_1 - x_0) + x_2 \cdot (x_2 - x_1) + \cdots x_n \cdot (x_n - x_{
 & \rightarrow a \cdot(b-a) + \frac{(b-a)^2}{2} \\
 &= \frac{b^2}{2} - \frac{a^2}{2}.
 \end{align*}
+$$
 
 
 > $$
@@ -502,7 +510,7 @@ This is similar to the Archimedes case with $a=0$ and $b=1$ shown above.
 Cauchy showed this using a *geometric series* for the partition, not the arithmetic series $x_i = a + i (b-a)/n$. The series defined by $1 + \alpha = (b/a)^{1/n}$, then $x_i = a \cdot (1 + \alpha)^i$. Here the bases $x_{i+1} - x_i$ simplify to $x_i \cdot \alpha$ and $f(x_i) = (a\cdot(1+\alpha)^i)^k = a^k (1+\alpha)^{ik}$, or $f(x_i)(x_{i+1}-x_i) = a^{k+1}\alpha[(1+\alpha)^{k+1}]^i$, so, using $u=(1+\alpha)^{k+1}=(b/a)^{(k+1)/n}$, $f(x_i) \cdot(x_{i+1} - x_i) = a^{k+1}\alpha u^i$. This gives
 
 
-
+$$
 \begin{align*}
 S &= a^{k+1}\alpha u^0 + a^{k+1}\alpha u^1 + \cdots + a^{k+1}\alpha u^{n-1}\\
 &= a^{k+1} \cdot \alpha \cdot (u^0 + u^1 + \cdot u^{n-1}) \\
@@ -510,6 +518,7 @@ S &= a^{k+1}\alpha u^0 + a^{k+1}\alpha u^1 + \cdots + a^{k+1}\alpha u^{n-1}\\
 &= (b^{k+1} - a^{k+1}) \cdot \frac{\alpha}{(1+\alpha)^{k+1} - 1} \\
 &\rightarrow \frac{b^{k+1} - a^{k+1}}{k+1}.
 \end{align*}
+$$
 
 
 > $$
@@ -541,9 +550,12 @@ Certainly other integrals could be computed with various tricks, but we won't pu
 ### Some other consequences
 
 
-  * The definition is defined in terms of any partition with its norm bounded by $\delta$. If you know a function $f$ is Riemann integrable, then it is enough to consider just a regular partition $x_i = a + i \cdot (b-a)/n$ when forming the sums, as was done above. It is just that showing a limit for just this particular type of partition would not be sufficient to prove Riemann integrability.
-  * The choice of $c_i$ is arbitrary to allow for maximum flexibility. The Darboux integrals use the maximum and minimum over the subinterval. It is sufficient to prove integrability to show that the limit exists with just these choices.
-  * Most importantly,
+* The definition is defined in terms of any partition with its norm bounded by $\delta$. If you know a function $f$ is Riemann integrable, then it is enough to consider just a regular partition $x_i = a + i \cdot (b-a)/n$ when forming the sums, as was done above. It is just that showing a limit for just this particular type of partition would not be sufficient to prove Riemann integrability.
+
+* The choice of $c_i$ is arbitrary to allow for maximum flexibility. The Darboux integrals use the maximum and minimum over the subinterval. It is sufficient to prove integrability to show that the limit exists with just these choices.
+
+
+* Most importantly,
 
 
 > A continuous function on $[a,b]$ is Riemann integrable on $[a,b]$.
@@ -553,13 +565,13 @@ Certainly other integrals could be computed with various tricks, but we won't pu
 The main idea behind this is that the difference between the maximum and minimum values over a partition gets small. That is if $[x_{i-1}, x_i]$ is like $1/n$ is length, then the difference between the maximum of $f$ over this interval, $M$, and the minimum, $m$ over this interval will go to zero as $n$ gets big. That $m$ and $M$ exists is due to the extreme value theorem, that this difference goes to $0$ is a consequence of continuity. What is needed is that this value goes to $0$ at the same rate – no matter what interval is being discussed – is a consequence of a notion of uniform continuity, a concept discussed in advanced calculus, but which holds for continuous functions on closed intervals. Armed with this, the Riemann sum for a general partition can be bounded by this difference times $b-a$, which will go to zero. So the upper and lower Riemann sums will converge to the same value.
 
 
-  * A "jump", or discontinuity of the first kind, is a value $c$ in $[a,b]$ where $\lim_{x \rightarrow c+} f(x)$ and $\lim_{x \rightarrow c-}f(x)$ both exist, but are not equal. It is true that a function that is not continuous on $I=[a,b]$, but only has discontinuities of the first kind on $I$ will be Riemann integrable on $I$.
+* A "jump", or discontinuity of the first kind, is a value $c$ in $[a,b]$ where $\lim_{x \rightarrow c+} f(x)$ and $\lim_{x \rightarrow c-}f(x)$ both exist, but are not equal. It is true that a function that is not continuous on $I=[a,b]$, but only has discontinuities of the first kind on $I$ will be Riemann integrable on $I$.
 
 
 For example, the function $f(x) = 1$ for $x$ in $[0,1]$ and $0$ otherwise will be integrable, as it is continuous at all but two points, $0$ and $1$, where it jumps.
 
 
-  * Some functions can have infinitely many points of discontinuity and still be integrable. The example of $f(x) = 1/q$ when $x=p/q$ is rational, and $0$ otherwise is often used as an example.
+* Some functions can have infinitely many points of discontinuity and still be integrable. The example of $f(x) = 1/q$ when $x=p/q$ is rational, and $0$ otherwise is often used as an example.
 
 
 ## Numeric integration
@@ -799,7 +811,7 @@ While such bounds are disappointing, often, when looking for specific values, th
 The Riemann sum above is actually extremely inefficient. To see how much, we can derive an estimate for the error in approximating the value using an arithmetic progression as the partition. Let's assume that our function $f(x)$ is increasing, so that the right sum gives an upper estimate and the left sum a lower estimate, so the error in the estimate will be between these two values:
 
 
-
+$$
 \begin{align*}
 \text{error} &\leq
 \left[
@@ -809,6 +821,7 @@ f(x_1) \cdot (x_{1} - x_0)  + f(x_2) \cdot  (x_{2} - x_1) + \cdots + f(x_{n-1})(
 &= \frac{b-a}{n} \cdot (\left[f(x_1) + f(x_2) + \cdots + f(x_n)\right] - \left[f(x_0) + \cdots + f(x_{n-1})\right]) \\
 &= \frac{b-a}{n} \cdot (f(b) - f(a)).
 \end{align*}
+$$
 
 
 We see the error goes to $0$ at a rate of $1/n$ with the constant depending on $b-a$ and the function $f$. In general, a similar bound holds when $f$ is not monotonic.
@@ -853,11 +866,12 @@ This formula will actually be exact for any 3rd degree polynomial. In fact an en
 The formulas for an approximation to the integral $\int_{-1}^1 f(x) dx$ discussed so far can be written as:
 
 
-
+$$
 \begin{align*}
 S &= f(x_1) \Delta_1 + f(x_2) \Delta_2 + \cdots + f(x_n) \Delta_n\\
   &= w_1 f(x_1) + w_2 f(x_2) + \cdots + w_n f(x_n).
 \end{align*}
+$$
 
 
 The $w$s are "weights" and the $x$s are nodes. A [Gaussian](http://en.wikipedia.org/wiki/Gaussian_quadrature) *quadrature rule* is a set of weights and nodes for $i=1, \dots n$ for which the sum is *exact* for any $f$ which is a polynomial of degree $2n-1$ or less. Such choices then also approximate well the integrals of functions which are not polynomials of degree $2n-1$, provided $f$ can be well approximated by a polynomial over $[-1,1]$. (Which is the case for the "nice" functions we encounter.) Some examples are given in the questions.