154 lines
3.5 KiB
Typst
154 lines
3.5 KiB
Typst
#import "@local/handout:0.1.0": *
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#import "@preview/cetz:0.4.2"
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#import "@preview/cetz-plot:0.1.2": chart, plot
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= Integers and Floats
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#generic("Observation:")
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If $x$ is smaller than 1, $log_2(1 + x)$ 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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[$log_2(1+x)$ 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(f1, domain: domain, label: $log(1+x)$, style: (
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stroke: ogrape,
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))
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plot.add(f2, domain: domain, label: $x$, style: (stroke: oblue))
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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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[$log_2(1+x)$ 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(f1, domain: domain, label: $log(1+x)$, style: (
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stroke: ogrape,
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))
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plot.add(f2, domain: domain, label: $x$, style: (stroke: oblue))
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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(type: "Note", [
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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)_2 - (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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#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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#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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])
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#solution([
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Let $E$ and $F$ be the exponent and float bits of $x_f$. \
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We then have:
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$
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log_2(x_f) & = log_2 ( 2^(E-127) times (1 + (F) / (2^23)) ) \
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& = E - 127 + log_2(1 + F / (2^23)) \
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& approx E-127 + F / (2^23) + epsilon \
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& = 1 / (2^23)(2^23 E + F) - 127 + epsilon \
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& = 1 / (2^23)(x_i) - 127 + epsilon
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$
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])
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#v(1fr)
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#problem()
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Using basic log rules, rewrite $log_2(1 / sqrt(x))$ in terms of $log_2(x)$.
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#solution([
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$
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log_2(1 / sqrt(x)) = (-1) / (2)log_2(x)
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$
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])
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#v(1fr)
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