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work on limits section

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17
quarto/limits/continuity.qmd
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@@ -76,6 +76,8 @@ This speaks to continuity at a point, we can extend this to continuity over an i
Finally, as with limits, it can be convenient to speak of *right* continuity and *left* continuity at a point, where the limit in the definition is replaced by a right or left limit, as appropriate.
In particular, a function is *continuous* over $[a,b]$ if it is continuous on $(a,b)$, left continuous at $b$ and right continuous at $a$.
:::{.callout-warning}
## Warning
@@ -90,7 +92,7 @@ Most familiar functions are continuous everywhere.
* For example, a monomial function $f(x) = ax^n$ for non-negative, integer $n$ will be continuous. This is because the limit exists everywhere, the domain of $f$ is all $x$ and there are no jumps.
* Similarly, the basic trigonometric functions $\sin(x)$, $\cos(x)$ are continuous everywhere.
* Similarly, the building-block trigonometric functions $\sin(x)$, $\cos(x)$ are continuous everywhere.
* So are the exponential functions $f(x) = a^x, a > 0$.
* The hyperbolic sine ($(e^x - e^{-x})/2$) and cosine ($(e^x + e^{-x})/2$) are, as $e^x$ is.
* The hyperbolic tangent is, as $\cosh(x) > 0$ for all $x$.
@@ -103,7 +105,7 @@ Some familiar functions are *mostly* continuous but not everywhere.
* Similarly, $f(x) = \log(x)$ is continuous on $(0,\infty)$, but it is not defined at $x=0$, so is not right continuous at $0$.
* The tangent function $\tan(x) = \sin(x)/\cos(x)$ is continuous everywhere *except* the points $x$ with $\cos(x) = 0$ ($\pi/2 + k\pi, k$ an integer).
* The hyperbolic co-tangent is not continuous at $x=0$ – when $\sinh$ is $0$,
* The semicircle $f(x) = \sqrt{1 - x^2}$ is *continuous* on $(-1, 1)$. It is not continuous at $-1$ and $1$, though it is right continuous at $-1$ and left continuous at $1$.
* The semicircle $f(x) = \sqrt{1 - x^2}$ is *continuous* on $(-1, 1)$. It is not continuous at $-1$ and $1$, though it is right continuous at $-1$ and left continuous at $1$. (It is continuous on $[-1,1]$.)
##### Examples of discontinuity
@@ -210,6 +212,13 @@ solve(del, c)
This gives the value of $c$.
This is a bit fussier than need be. As the left and right pieces (say, $f_l$ and $f_r$) as both are polynomials are continuous everywhere, so would have left and right limits given through evaluation. Solving for `c` as follows is enough:
```{julia}
solve(ex1(x=>0) ~ ex2(x=>0), c)
```
## Rules for continuity
@@ -384,7 +393,7 @@ numericq(val)
###### Question
Suppose $f(x)$, $g(x)$, and $h(x)$ are continuous functions on $(a,b)$. If $a < c < b$, are you sure that $lim_{x \rightarrow c} f(g(x))$ is $f(g(c))$?
Suppose $f(x)$, $g(x)$, and $h(x)$ are continuous functions on $(a,b)$. If $a < c < b$, are you sure that $\lim_{x \rightarrow c} f(g(x))$ is $f(g(c))$?
```{julia}
@@ -453,7 +462,7 @@ yesnoq(true)
###### Question
Let $f(x)$ and $g(x)$ be continuous functions whose graph of $[0,1]$ is given by:
Let $f(x)$ and $g(x)$ be continuous functions. Their graphs over $[0,1]$ are given by:
```{julia}

26
quarto/limits/intermediate_value_theorem.qmd
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@@ -76,7 +76,7 @@ ImageFile(imgfile, caption)
In the early years of calculus, the intermediate value theorem was intricately connected with the definition of continuity, now it is a consequence.
The basic proof starts with a set of points in $[a,b]$: $C = \{x \text{ in } [a,b] \text{ with } f(x) \leq y\}$. The set is not empty (as $a$ is in $C$) so it *must* have a largest value, call it $c$ (this requires the completeness property of the real numbers). By continuity of $f$, it can be shown that $\lim_{x \rightarrow c-} f(x) = f(c) \leq y$ and $\lim_{y \rightarrow c+}f(x) =f(c) \geq y$, which forces $f(c) = y$.
The basic proof starts with a set of points in $[a,b]$: $C = \{x \text{ in } [a,b] \text{ with } f(x) \leq y\}$. The set is not empty (as $a$ is in $C$) so it *must* have a largest value, call it $c$ (this might seem obvious, but it requires the completeness property of the real numbers). By continuity of $f$, it can be shown that $\lim_{x \rightarrow c-} f(x) = f(c) \leq y$ and $\lim_{y \rightarrow c+}f(x) =f(c) \geq y$, which forces $f(c) = y$.
### Bolzano and the bisection method
@@ -87,8 +87,19 @@ Suppose we have a continuous function $f(x)$ on $[a,b]$ with $f(a) < 0$ and $f(b
We use this fact when building a "sign chart" of a polynomial function. Between any two consecutive real zeros the polynomial can not change sign. (Why?) So a "test point" can be used to determine the sign of the function over an entire interval.
The `sign_chart` function from `CalculusWithJulia` uses this to indicate where an *assumed* continuous function changes sign:
Here, we use the Bolzano theorem to give an algorithm - the *bisection method* - to locate the value $c$ under the assumption $f$ is continuous on $[a,b]$ and changes sign between $a$ and $b$.
```{julia}
f(x) = sin(x + x^2) + x/2
sign_chart(f, -3, 3)
```
The intermediate value theorem can find the sign of the function *between* adjacent zeros, but how are the zeros identified?
Here, we use the Bolzano theorem to give an algorithm - the *bisection method* - to locate a value $c$ in $[a,b]$ with $f(c) = 0$ under the assumptions:
* $f$ is continuous on $[a,b]$
* $f$ changes sign between $a$ and $b$. (In particular, when $f(a)$ and $f(b)$ have different signs.)
::: {.callout-note}
#### Between
@@ -339,6 +350,13 @@ find_zero(h, (0, 2))
### Solving `f(x) = g(x)` and `f(x) = c`
The above shows a means to translate a given problem into one that can be solved with `find_zero`. Basically to solve either when a function is a non-zero constant or when a function is equal to some other function, the difference between the two sides is formed and turned into a function, called `h` above.
If using symbolic expressions, as below, then an equation (formed by `~`) can be passed to `find_zero`:
```{julia}
@syms x
solve(cos(x) ~ x, (0, 2))
```
:::
##### Example: Inverse functions
@@ -493,7 +511,7 @@ Note that the function is infinite at `b`:
d(b)
```
From the graph, we can see the zero is around `b`. As `d(b)` is `-Inf` we can use the bracket `(b/2,b)`
From the graph, we can see the zero is around `b`. As `d(b)` is `-Inf` we can use the bracket `(b/2, b)`
```{julia}
@@ -569,7 +587,7 @@ find_zero(f, I), find_zero(f, I, p=2)
The second number is the solution when `p=2`.
The above used a *keyword* argument, but a positional argument allows for broadcasting:
The above used a *keyword* argument to pass in the parameter, but using a positional argument (the last one) allows for broadcasting:
```{julia}
find_zero.(f, Ref(I), 1:5) # solutions for p=1,2,3,4,5

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.
$$

4
quarto/limits/limits_extensions.qmd
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@@ -192,7 +192,7 @@ Describe the limits at $-1$, $0$, and $1$.
## Limits at infinity
The loose definition of a horizontal asymptote is "a line such that the distance between the curve and the line approaches $0$ as they tend to infinity." This sounds like it should be defined by a limit. The issue is, that the limit would be at $\pm\infty$ and not some finite $c$. This requires the idea of a neighborhood of $c$, $0 < |x-c| < \delta$ to be reworked.
The loose definition of a horizontal asymptote is "a line such that the distance between the curve and the line approaches $0$ as they tend to infinity." This sounds like it should be defined by a limit. The issue is, that the limit would be at $\pm\infty$ and not some finite $c$. This requires the idea of a neighborhood of $c$, $0 < |x-c| < \delta$, to be reworked.
The basic idea for a limit at $+\infty$ is that for any $\epsilon$, there exists an $M$ such that when $x > M$ it must be that $|f(x) - L| < \epsilon$. For a horizontal asymptote, the line would be $y=L$. Similarly a limit at $-\infty$ can be defined with $x < M$ being the condition.
@@ -284,7 +284,7 @@ That is, right limits can be analyzed as limits at $\infty$ or right limits at $
## Limits of infinity
Vertical asymptotes are nicely defined with horizontal asymptotes by the graph getting close to some line. However, the formal definition of a limit won't be the same. For a vertical asymptote, the value of $f(x)$ heads towards positive or negative infinity, not some finite $L$. As such, a neighborhood like $(L-\epsilon, L+\epsilon)$ will no longer make sense, rather we replace it with an expression like $(M, \infty)$ or $(-\infty, M)$. As in: the limit of $f(x)$ as $x$ approaches $c$ is *infinity* if for every $M > 0$ there exists a $\delta>0$ such that if $0 < |x-c| < \delta$ then $f(x) > M$. Approaching $-\infty$ would conclude with $f(x) < -M$ for all $M>0$.
Vertical asymptotes are nicely defined with, as with horizontal asymptotes, by the graph getting close to some line. However, the formal definition of a limit won't be the same. For a vertical asymptote, the value of $f(x)$ heads towards positive or negative infinity, not some finite $L$. As such, a neighborhood like $(L-\epsilon, L+\epsilon)$ will no longer make sense, rather we replace it with an expression like $(M, \infty)$ or $(-\infty, M)$. As in: the limit of $f(x)$ as $x$ approaches $c$ is *infinity* if for every $M > 0$ there exists a $\delta>0$ such that if $0 < |x-c| < \delta$ then $f(x) > M$. Approaching $-\infty$ would conclude with $f(x) < -M$ for $M>0$.
##### Examples