update condition

quarto/derivatives/condition.qmd

@@ -11,25 +11,25 @@ This sense of "small" is quantified through a *condition number*.
 
 If we let $\delta_x$ be a small perturbation of $x$ then $\delta_f = f(x + \delta_x) - f(x)$.
 
-The *forward error* is $||f(x+\delta_x) - f(x)||$, the *relative forward error* is  $||f(x+\delta_x) - f(x)||/ ||f(x)||$.
+The *forward error* is $\lVert\delta_f\rVert = \lVert f(x+\delta_x) - f(x)\rVert$, the *relative forward error* is  $\lVert\delta_f\rVert/\lVert f\rVert = \lVert f(x+\delta_x) - f(x)\rVert/ \lVert f(x)\rVert$.
 
-The *backward error* is $||\delta_x||$, the *relative backward error* is $||\delta_x|| / ||x||$.
+The *backward error* is $\lVert\delta_x\rVert$, the *relative backward error* is $\lVert\delta_x\rVert / \lVert x\rVert$.
 
- The *absolute condition number* $\hat{\kappa}$ is worst case of this ratio $||\delta_f||/ ||\delta_x||$ as the perturbation size shrinks to $0$.
-The relative condition number $\kappa$ divides $||\delta_f||$ by $||f(x)||$ and $||\delta_x||$ by $||x||$ before taking the ratio.
+ The *absolute condition number* $\hat{\kappa}$ is worst case of this ratio $\lVert\delta_f\rVert/ \lVert\delta_x\rVert$ as the perturbation size shrinks to $0$.
+The relative condition number $\kappa$ divides $\lVert\delta_f\rVert$ by $\lVert f(x)\rVert$ and $\lVert\delta_x\rVert$ by $\lVert x\rVert$ before taking the ratio.
 
 
 A *problem* is a mathematical concept, an *algorithm* the computational version. Algorithms may differ for many reasons, such as floating point errors, tolerances, etc. We use notation $\tilde{f}$ to indicate the algorithm.
 
-The absolute error in the algorithm is $||\tilde{f}(x) - f(x)||$, the relative error divides by $||f(x)||$. A good algorithm would have smaller relative errors.
+The absolute error in the algorithm is $\lVert\tilde{f}(x) - f(x)\rVert$, the relative error divides by $\lVert f(x)\rVert$. A good algorithm would have smaller relative errors.
 
 An algorithm is called *stable* if
 
 $$
-\frac{||\tilde{f}(x) - f(\tilde{x})||}{||f(\tilde{x})||}
+\frac{\lVert\tilde{f}(x) - f(\tilde{x})\rVert}{\lVert f(\tilde{x})\rVert}
 $$
 
-is *small* for *some* $\tilde{x}$ relatively near $x$, $||\tilde{x}-x||/||x||$.
+is *small* for *some* $\tilde{x}$ relatively near $x$, $\lVert\tilde{x}-x\rVert/\lVert x\rVert$.
 
 > "A *stable* algorithm gives nearly the right answer to nearly the right question."
 
@@ -41,14 +41,14 @@ $$
 \tilde{f}(x) = f(\tilde{x})
 $$
 
-for some $\tilde{x}$ with $||\tilde{x} - x||/||x||$ is small.
+	for some $\tilde{x}$ with $\lVert\tilde{x} - x\rVert/\lVert x\rVert$ is small.
 
 > "A backward stable algorithm gives exactly the right answer to nearly the right question."
 
 
-The concepts are related by Trefethen and Bao's Theorem 15.1 which says  for a backward stable algorithm the relative error $||\tilde{f}(x) - f(x)||/||f(x)||$ is small in a manner proportional to the relative condition number.
+The concepts are related by Trefethen and Bao's Theorem 15.1 which says  for a backward stable algorithm the relative error $\lVert\tilde{f}(x) - f(x)\rVert/\lVert f(x)\rVert$ is small in a manner proportional to the relative condition number.
 
-Applying this to the zero-finding we follow doi:10.1137/1.9781611975086.
+Applying this to the zero-finding we follow @doi:10.1137/1.9781611975086.
 
 To be specific, the problem is finding a zero of $f$ starting at an initial point $x_0$. The data is $(f, x_0)$, the solution is $r$ a zero of $f$.
 
@@ -63,7 +63,7 @@ Suppose for sake of argument that $\tilde{f}(x) = f(x) + \epsilon$ and $r$ is a
   &= 0 + f'(r)\delta + \epsilon
 \end{align*}
 
-Rearranging gives $||\delta/\epsilon|| \approx 1/||f'(r)||$ leading to:
+Rearranging gives $\lVert\delta/\epsilon\rVert \approx 1/\lVert f'(r)\rVert$ leading to:
 
 > The absolute condition number is $\hat{\kappa}_r = |f'(r)|^{-1}$.
 
@@ -83,4 +83,4 @@ Practically these two observations lead to
 
 For the first observation, the example of Wilkinson's polynomial is often used where $f(x) = (x-1)\cdot(x-2)\cdot \cdots\cdot(x-20)$. When expanded this function has exactness issues of typical floating point values, the condition number is large and some of the roots found are quite different from the mathematical values.
 
-The second observation helps explain why a problem like finding the zero of $f(x) = x * exp(x)$ using Newton's method starting at $2$ might return a value like $5.89\dots$. The residual is checked to be zero in a *relative* manner which would basically use a tolerance of `atol + abs(xn)*rtol`. Functions with asymptotes of $0$ will eventually be smaller than this value.
+The second observation helps explain why a problem like finding the zero of $f(x) = x \cdot \exp(x)$ using Newton's method starting at $2$ might return a value like $5.89\dots$. The residual is checked to be zero in a *relative* manner which would basically use a tolerance of `atol + abs(xn)*rtol`. Functions with asymptotes of $0$ will eventually be smaller than this value.