Tom edits

src/Advanced/Fast Inverse Root/parts/03 approx.typ

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+#import "@local/handout:0.1.0": *
+#import "@preview/cetz:0.3.1"
+#import "@preview/cetz-plot:0.1.0": plot, chart
+
+= Integers and Floats
+
+#generic("Observation:")
+For small values of $x$, $log_2(1 + x)$ is approximately equal to $x$. \
+Note that this equality is exact for $x = 0$ and $x = 1$, since $log_2(1) = 0$ and $log_2(2) = 1$.
+
+#v(5mm)
+
+We'll add the _correction term_ $epsilon$ to our approximation: $log_2(1 + a) approx a + epsilon$. \
+This allows us to improve the average error of our linear approximation:
+
+#table(
+  stroke: none,
+  align: center,
+  columns: (1fr, 1fr),
+  inset: 5mm,
+  [$log_2(1+x)$ and $x + 0$]
+    + cetz.canvas({
+      import cetz.draw: *
+
+      let f1(x) = calc.log(calc.abs(x + 1), base: 2)
+      let f2(x) = x
+
+      // Set-up a thin axis style
+      set-style(axes: (stroke: .5pt, tick: (stroke: .5pt)))
+
+
+      plot.plot(
+        size: (7, 7),
+        x-tick-step: 0.2,
+        y-tick-step: 0.2,
+        y-min: 0,
+        y-max: 1,
+        x-min: 0,
+        x-max: 1,
+        legend: none,
+        axis-style: "scientific-auto",
+        x-label: none,
+        y-label: none,
+        {
+          let domain = (0, 1)
+
+          plot.add(
+            f1,
+            domain: domain,
+            label: $log(1+x)$,
+            style: (stroke: ogrape),
+          )
+
+          plot.add(
+            f2,
+            domain: domain,
+            label: $x$,
+            style: (stroke: oblue),
+          )
+        },
+      )
+    })
+    + [
+      Max error: 0.086 \
+      Average error: 0.0573
+    ],
+  [$log_2(1+x)$ and $x + 0.045$]
+    + cetz.canvas({
+      import cetz.draw: *
+
+      let f1(x) = calc.log(calc.abs(x + 1), base: 2)
+      let f2(x) = x + 0.0450466
+
+      // Set-up a thin axis style
+      set-style(axes: (stroke: .5pt, tick: (stroke: .5pt)))
+
+
+      plot.plot(
+        size: (7, 7),
+        x-tick-step: 0.2,
+        y-tick-step: 0.2,
+        y-min: 0,
+        y-max: 1,
+        x-min: 0,
+        x-max: 1,
+        legend: none,
+        axis-style: "scientific-auto",
+        x-label: none,
+        y-label: none,
+        {
+          let domain = (0, 1)
+
+          plot.add(
+            f1,
+            domain: domain,
+            label: $log(1+x)$,
+            style: (stroke: ogrape),
+          )
+
+          plot.add(
+            f2,
+            domain: domain,
+            label: $x$,
+            style: (stroke: oblue),
+          )
+        },
+      )
+    })
+    + [
+      Max error: 0.041 \
+      Average error: 0.0254
+    ],
+)
+
+
+A suitiable value of $epsilon$ can be found using calculus or with computational trial-and-error. \
+We won't bother with this---we'll simply leave the correction term as an opaque constant $epsilon$.
+
+
+
+#v(1fr)
+
+#note(
+  type: "Note",
+  [
+    "Average error" above is simply the area of the region between the two graphs:
+    $
+      integral_0^1 abs( #v(1mm) log(1+x)_2 - (x+epsilon) #v(1mm))
+    $
+    Feel free to ignore this note, it isn't a critical part of this handout.
+  ],
+)
+
+
+#pagebreak()
+
+#problem(label: "convert")
+Use the fact that $log_2(1 + a) approx a + epsilon$ to approximate $log_2(x_f)$ in terms of $x_i$. \
+Namely, show that
+$
+  log_2(x_f) = (x_i) / (2^23) - 127 + epsilon
+$
+#note([
+  In other words, we're finding an expression for $x$ as a float
+  in terms of $x$ as an int.
+])
+
+#solution([
+  Let $E$ and $F$ be the exponent and float bits of $x_f$. \
+  We then have:
+  $
+    log_2(x_f)
+    &= log_2 ( 2^(E-127) times (1 + (F) / (2^23)) ) \
+    &= E - 127 + log_2(1 + F / (2^23)) \
+    & approx E-127 + F / (2^23) + epsilon \
+    &= 1 / (2^23)(2^23 E + F) - 127 + epsilon \
+    &= 1 / (2^23)(x_i) - 127 + epsilon
+  $
+])
+
+#v(1fr)
+
+
+#problem()
+Using basic log rules, rewrite $log_2(1 / sqrt(x))$ in terms of $log_2(x)$.
+
+#solution([
+  $
+    log_2(1 / sqrt(x)) = (-1) / (2)log_2(x)
+  $
+])
+
+#v(1fr)