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work on limits section

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jverzani
2024-07-02 16:16:39 -04:00
parent 3f206753ec
commit 9c7c1d6d4c
8 changed files with 137 additions and 43 deletions
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52
quarto/limits/limits.qmd
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@@ -238,13 +238,13 @@ annotate!([(0,0+Δ,"A"), (x-Δ,y+Δ/4, "B"), (1+Δ/2,y/x, "C"),
annotate!([(.2*cos(θ/2), 0.2*sin(θ/2), "θ")])
imgfile = tempname() * ".png"
savefig(p, imgfile)
caption = "Triangle ``ABD` has less area than the shaded wedge, which has less area than triangle ``ACD``. Their respective areas are ``(1/2)\\sin(\\theta)``, ``(1/2)\\theta``, and ``(1/2)\\tan(\\theta)``. The inequality used to show ``\\sin(x)/x`` is bounded below by ``\\cos(x)`` and above by ``1`` comes from a division by ``(1/2) \\sin(x)`` and taking reciprocals.
caption = "Triangle ``ABD`` has less area than the shaded wedge, which has less area than triangle ``ACD``. Their respective areas are ``(1/2)\\sin(\\theta)``, ``(1/2)\\theta``, and ``(1/2)\\tan(\\theta)``. The inequality used to show ``\\sin(x)/x`` is bounded below by ``\\cos(x)`` and above by ``1`` comes from a division by ``(1/2) \\sin(x)`` and taking reciprocals.
"
plotly()
ImageFile(imgfile, caption)
```
To discuss the case of $(1+x)^{1/x}$ it proved convenient to assume $x = 1/m$ for integer values of $m$. At the time of Cauchy, log tables were available to identify the approximate value of the limit. Cauchy computed the following value from logarithm tables.
To discuss the case of $(1+x)^{1/x}$ it proved convenient to assume $x = 1/m$ for integer values of $m$. At the time of Cauchy, log tables were available to identify the approximate value of the limit. Cauchy computed the following value from logarithm tables:
```{julia}
@@ -337,13 +337,14 @@ Let's return to the function $f(x) = \sin(x)/x$. This function was studied by Eu
```{julia}
#| hold: true
f(x) = sin(x)/x
xs, ys = unzip(f, -pi/2, pi/2) # get points used to plot `f`
plot(xs, ys)
scatter!(xs, ys)
plot(f, -pi/2, pi/2;
seriestype=[:scatter, :line], # show points and line segments
legend=false)
```
The $y$ values of the graph seem to go to $1$ as the $x$ values get close to $0$. (That the graph looks defined at $0$ is due to the fact that the points sampled to graph do not include $0$, as shown through the `scatter!` command – which can be checked via `minimum(abs, xs)`.)
The $y$ values of the graph seem to go to $1$ as the $x$ values get close to $0$. (That the graph looks defined at $0$ is due to the fact that the points sampled to graph do not include $0$.)
We can also verify Euler's intuition through this graph:
@@ -432,7 +433,7 @@ The `NaN` indicates that this function is indeterminate at $c=2$. A quick plot g
```{julia}
#| hold: true
c, delta = 2, 1
plot(x -> (x^2 - 5x + 6) / (x^2 + x - 6), c - delta, c + delta)
plot(f, c - delta, c + delta)
```
The graph looks "continuous." In fact, the value $c=2$ is termed a *removable singularity* as redefining $f(x)$ to be $-0.2$ when $x=2$ results in a "continuous" function.
@@ -585,13 +586,11 @@ g(x) = (1 - cos(x)) / x^2
g(0)
```
What is the value of $L$, if it exists? A quick attempt numerically yields:
What is the value of $L$, if it exists? Getting closer to $0$ numerically than the default yields:
```{julia}
xs = 0 .+ hs
ys = [g(x) for x in xs]
[xs ys]
lim(g, 0; n=9)
```
Hmm, the values in `ys` appear to be going to $0.5$, but then end up at $0$. Is the limit $0$ or $1/2$? The answer is $1/2$. The last $0$ is an artifact of floating point arithmetic and the last few deviations from `0.5` due to loss of precision in subtraction. To investigate, we look more carefully at the two ratios:
@@ -606,7 +605,7 @@ y2s = [x^2 for x in xs]
Looking at the bottom of the second column reveals the error. The value of `1 - cos(1.0e-8)` is `0` and not a value around `5e-17`, as would be expected from the pattern above it. This is because the smallest floating point value less than `1.0` is more than `5e-17` units away, so `cos(1e-8)` is evaluated to be `1.0`. There just isn't enough granularity to get this close to $0$.
Not that we needed to. The answer would have been clear if we had stopped with `x=1e-6`, say.
Not that we needed to. The answer would have been clear if we had stopped with `x=1e-6` (with `n=6`) say.
In general, some functions will frustrate the numeric approach. It is best to be wary of results. At a minimum they should confirm what a quick graph shows, though even that isn't enough, as this next example shows.
@@ -633,23 +632,20 @@ A plot shows the answer appears to be straightforward:
```{julia}
#| echo: false
h(x) = x^2 + 1 + log(abs(11*x - 15))/99
plot(h, 15/11 - 1, 15/11 + 1)
```
Taking values near $15/11$ shows nothing too unusual:
Taking values near $15/11$ shows nothing perhaps too unusual:
```{julia}
#| hold: true
c = 15/11
hs = [1/10^i for i in 4:3:16]
xs = c .+ hs
[xs h.(xs)]
lim(h, c; n = 16)
```
(Though both the graph and the table hint at something a bit odd.)
(Though the graph and table do hint at something a bit odd -- the graph shows a blip, the table doesn't show values in the second column going towards a specific value.)
However the limit in this case is $-\infty$ (or DNE), as there is an aysmptote at $c=15/11$. The problem is the asymptote due to the logarithm is extremely narrow and happens between floating point values to the left and right of $15/11$.
@@ -726,7 +722,7 @@ For example, the limit at $0$ of $(1-\cos(x))/x^2$ is easily handled:
limit((1 - cos(x)) / x^2, x => 0)
```
The pair notation (`x => 0`) is used to indicate the variable and the value it is going to.
The pair notation (`x => 0`) is used to indicate the variable and the value it is going to. A `dir` argument is used to indicate ``x \rightarrow c+`` (the default), ``x \rightarrow c-``, and ``x \rightarrow c``.
##### Example
@@ -736,7 +732,7 @@ We look again at this function which despite having a vertical asymptote at $x=1
$$
f(x) = x^2 + 1 + \log(| 11 \cdot x - 15 |)/99.
h(x) = x^2 + 1 + \log(| 11 \cdot x - 15 |)/99.
$$
We find the limit symbolically at $c=15/11$ as follows, taking care to use the exact value `15//11` and not the *floating point* approximation returned by `15/11`:
@@ -744,8 +740,8 @@ We find the limit symbolically at $c=15/11$ as follows, taking care to use the e
```{julia}
#| hold: true
f(x) = x^2 + 1 + log(abs(11x - 15))/99
limit(f(x), x => 15 // 11)
h(x) = x^2 + 1 + log(abs(11x - 15))/99
limit(h(x), x => 15 // 11)
```
##### Example
@@ -764,16 +760,18 @@ We have for the first:
```{julia}
limit( (2sin(x) - sin(2x)) / (x - sin(x)), x => 0)
limit( (2sin(x) - sin(2x)) / (x - sin(x)), x => 0; dir="+-")
```
(The `dir = "+-"` indicates take both a right and left limit and ensure both exist and are equal.)
The second is similarly done, though here we define a function for variety:
```{julia}
#| hold: true
f(x) = (exp(x) - 1 - x) / x^2
limit(f(x), x => 0)
limit(f(x), x => 0; dir="+-")
```
Finally, for the third we define a new variable and proceed:
@@ -781,7 +779,7 @@ Finally, for the third we define a new variable and proceed:
```{julia}
@syms rho::real
limit( (x^(1-rho) - 1) / (1 - rho), rho => 1)
limit( (x^(1-rho) - 1) / (1 - rho), rho => 1; dir="+-")
```
This last limit demonstrates that the `limit` function of `SymPy` can readily evaluate limits that involve parameters, though at times some assumptions on the parameters may be needed, as was done through `rho::real`.
@@ -842,7 +840,7 @@ limit(f(x), x => PI/2)
Right and left limits will be discussed in the next section; here we give an example of the idea. The mathematical convention is to say a limit exists if both the left *and* right limits exist and are equal. Informally a right (left) limit at $c$ only considers values of $x$ more (less) than $c$. The `limit` function of `SymPy` finds directional limits by default, a right limit, where $x > c$.
The left limit can be found by passing the argument `dir="-"`. Passing `dir="+-"` (and not `"-+"`) will compute the mathematical limit, throwing an error in `Python` if no limit exists.
The left limit can be found by passing the argument `dir="-"`. Passing `dir="+-"` (and not `"-+"`), as done in a few examples above, will compute the mathematical limit, throwing an error in `Python` if no limit exists.
```{julia}
@@ -938,7 +936,7 @@ as this is the limit of $f(g(x))$ with $f$ as above and $g(x) = x^2$. We need $
##### Example: products
Consider this complicated limit found on this [Wikipedia](http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule) page.
Consider this more complicated limit found on this [Wikipedia](http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule) page.
$$