174 lines
3.7 KiB
Typst
174 lines
3.7 KiB
Typst
#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,
|
|
[$log(1+x)$ 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
|
|
],
|
|
[$log(1+x)$ 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) - (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)
|