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@ -5,67 +5,141 @@
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= Integers and Floats
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#generic("Observation:")
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For small values of $a$, $log_2(1 + a)$ is approximately equal to $a$. \
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Note that this equality is exact for $a = 0$ and $a = 1$, since $log_2(1) = 0$ and $log_2(2) = 1$.
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#v(2mm)
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We'll add a "correction term" $epsilon$ to this approximation, so that $log_2(1 + a) approx a + epsilon$.
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#cetz.canvas({
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import cetz.draw: *
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let f1(x) = calc.log(calc.abs(x + 1), base: 2)
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let f2(x) = x
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// Set-up a thin axis style
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set-style(axes: (stroke: .5pt, tick: (stroke: .5pt)))
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plot.plot(
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size: (8, 8),
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x-tick-step: 0.2,
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y-tick-step: 0.2,
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y-min: 0,
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y-max: 1,
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x-min: 0,
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x-max: 1,
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legend: none,
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axis-style: "scientific-auto",
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{
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let domain = (0, 10)
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plot.add-fill-between(
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f1,
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f2,
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domain: domain,
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style: (stroke: none, fill: luma(75%)),
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)
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plot.add(
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f1,
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domain: domain,
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label: $log(1+x)$,
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style: (stroke: black),
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)
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plot.add(f2, domain: domain, label: $x$, style: (stroke: black))
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},
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)
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})
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TODO: why? Graphs.
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#problem(label: "convert")
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Use the fact that $log_2(1 + a) approx a + epsilon$ to approximate $log_2(x_f)$ in terms of $x_i$. \
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For small values of $x$, $log_2(1 + x)$ is approximately equal to $x$. \
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Note that this equality is exact for $x = 0$ and $x = 1$, since $log_2(1) = 0$ and $log_2(2) = 1$.
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#v(5mm)
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We'll add the _correction term_ $epsilon$ to our approximation: $log_2(1 + a) approx a + epsilon$. \
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This allows us to improve the average error of our linear approximation:
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#table(
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stroke: none,
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align: center,
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columns: (1fr, 1fr),
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inset: 5mm,
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[$log(1+x)$ and $x + 0$]
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+ cetz.canvas({
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import cetz.draw: *
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let f1(x) = calc.log(calc.abs(x + 1), base: 2)
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let f2(x) = x
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// Set-up a thin axis style
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set-style(axes: (stroke: .5pt, tick: (stroke: .5pt)))
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plot.plot(
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size: (7, 7),
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x-tick-step: 0.2,
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y-tick-step: 0.2,
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y-min: 0,
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y-max: 1,
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x-min: 0,
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x-max: 1,
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legend: none,
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axis-style: "scientific-auto",
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x-label: none,
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y-label: none,
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{
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let domain = (0, 1)
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plot.add(
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f1,
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domain: domain,
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label: $log(1+x)$,
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style: (stroke: ogrape),
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)
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plot.add(
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f2,
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domain: domain,
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label: $x$,
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style: (stroke: oblue),
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)
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},
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)
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})
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+ [
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Max error: 0.086 \
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Average error: 0.0573
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],
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[$log(1+x)$ and $x + 0.045$]
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+ cetz.canvas({
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import cetz.draw: *
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let f1(x) = calc.log(calc.abs(x + 1), base: 2)
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let f2(x) = x + 0.0450466
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// Set-up a thin axis style
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set-style(axes: (stroke: .5pt, tick: (stroke: .5pt)))
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plot.plot(
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size: (7, 7),
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x-tick-step: 0.2,
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y-tick-step: 0.2,
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y-min: 0,
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y-max: 1,
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x-min: 0,
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x-max: 1,
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legend: none,
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axis-style: "scientific-auto",
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x-label: none,
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y-label: none,
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{
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let domain = (0, 1)
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plot.add(
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f1,
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domain: domain,
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label: $log(1+x)$,
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style: (stroke: ogrape),
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)
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plot.add(
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f2,
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domain: domain,
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label: $x$,
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style: (stroke: oblue),
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)
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},
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)
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})
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+ [
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Max error: 0.041 \
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Average error: 0.0254
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],
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)
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A suitiable value of $epsilon$ can be found using calculus or with computational trial-and-error. \
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We won't bother with this---we'll simply leave the correction term as an opaque constant $epsilon$.
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#v(1fr)
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#note(
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type: "Note",
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[
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"Average error" above is simply the area of the region between the two graphs:
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$
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integral_0^1 abs( #v(1mm) log(1+x) - (x+epsilon) #v(1mm))
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$
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Feel free to ignore this note, it isn't a critical part of this handout.
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],
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)
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#pagebreak()
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#problem(label: "convert")
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Use the fact that $log_2(1 + a) approx a + epsilon$ to approximate $log_2(x_f)$ in terms of $x_i$. \
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Namely, show that
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$
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log_2(x_f) = (x_i) / (2^23) - 127 + epsilon
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$
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for some correction term term $epsilon$ \
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#note([
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In other words, we're finding an expression for $x$ as a float
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in terms of $x$ as an int.
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@ -4,6 +4,9 @@
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The following code is present in _Quake III Arena_ (1999):
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#v(5mm)
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```c
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float Q_rsqrt( float number ) {
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long i = * ( long * ) &number;
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@ -12,6 +15,8 @@ float Q_rsqrt( float number ) {
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}
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```
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#v(5mm)
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This code defines a function `Q_rsqrt` that consumes a float `number` and approximates its inverse square root.
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If we rewrite this using notation we're familiar with, we get the following:
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$
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@ -112,7 +117,7 @@ What is the exact value of $kappa$ in terms of $epsilon$? \
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#solution[
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This problem makes sure our students see that
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$kappa = (2^24)/3(epsilon - 127)$. \
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$kappa = 2^23 3/2 (127 - epsilon)$. \
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See the solution to @finalb.
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]
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