538 lines
14 KiB
Plaintext
538 lines
14 KiB
Plaintext
# Continuity
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```{julia}
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#| echo: false
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import Logging
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Logging.disable_logging(Logging.Info) # or e.g. Logging.Info
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Logging.disable_logging(Logging.Warn)
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import SymPy
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function Base.show(io::IO, ::MIME"text/html", x::T) where {T <: SymPy.SymbolicObject}
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println(io, "<span class=\"math-left-align\" style=\"padding-left: 4px; width:0; float:left;\"> ")
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println(io, "\\[")
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println(io, sympy.latex(x))
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println(io, "\\]")
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println(io, "</span>")
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end
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# hack to work around issue
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import Markdown
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import CalculusWithJulia
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function CalculusWithJulia.WeaveSupport.ImageFile(d::Symbol, f::AbstractString, caption; kwargs...)
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nm = joinpath("..", string(d), f)
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u = ""
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Markdown.parse(u)
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end
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nothing
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```
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This section uses these add-on packages:
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```{julia}
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using CalculusWithJulia
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using Plots
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using SymPy
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```
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```{julia}
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#| echo: false
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#| results: "hidden"
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using CalculusWithJulia.WeaveSupport
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const frontmatter = (
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title = "Continuity",
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description = "Calculus with Julia: Continuity",
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tags = ["CalculusWithJulia", "limits", "continuity"],
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);
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nothing
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```
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---
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The definition Google finds for *continuous* is *forming an unbroken whole; without interruption*.
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The concept in calculus, as transferred to functions, is similar. Roughly speaking, a continuous function is one whose graph could be drawn without having to lift (or interrupt) the pencil drawing it.
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Consider these two graphs:
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```{julia}
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#| hold: true
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#| echo: false
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plt = plot([-1,0], [-1,-1], color=:black, legend=false, linewidth=5)
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plot!(plt, [0, 1], [ 1, 1], color=:black, linewidth=5)
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plt
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```
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and
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```{julia}
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#| hold: true
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#| echo: false
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plot([-1,-.1, .1, 1], [-1,-1, 1, 1], color=:black, legend=false, linewidth=5)
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```
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Though similar at some level - they agree at nearly every value of $x$ - the first has a "jump" from $-1$ to $1$ instead of the transition in the second one. The first is not continuous at $0$ - a break is needed to draw it - where as the second is continuous.
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A formal definition of continuity was a bit harder to come about. At [first](http://en.wikipedia.org/wiki/Intermediate_value_theorem) the concept was that for any $y$ between any two values in the range for $f(x)$, the function should take on the value $y$ for some $x$. Clearly this could distinguish the two graphs above, as one takes no values in $(-1,1)$, whereas the other - the continuous one - takes on all values in that range.
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However, [Cauchy](http://en.wikipedia.org/wiki/Cours_d%27Analyse) defined continuity by $f(x + \alpha) - f(x)$ being small whenever $\alpha$ was small. This basically rules out "jumps" and proves more useful as a tool to describe continuity.
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The [modern](http://en.wikipedia.org/wiki/Continuous_function#History) definition simply pushes the details to the definition of the limit:
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> A function $f(x)$ is continuous at $x=c$ if $\lim_{x \rightarrow c}f(x) = f(c)$.
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This says three things
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* The limit exists at $c$.
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* The function is defined at $c$ ($c$ is in the domain).
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* The value of the limit is the same as $f(c)$.
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This speaks to continuity at a point, we can extend this to continuity over an interval $(a,b)$ by saying:
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> A function $f(x)$ is continuous over $(a,b)$ if at each point $c$ with $a < c < b$, $f(x)$ is continuous at $c$.
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Finally, as with limits, it can be convenient to speak of *right* continuity and *left* continuity at a point, where the limit in the defintion is replaced by a right or left limit, as appropriate.
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:::{.callout-warning}
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## Warning
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The limit in the definition of continuity is the basic limit and not an extended sense where infinities are accounted for.
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:::
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##### Examples of continuity
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Most familiar functions are continuous everywhere.
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* 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.
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* Similarly, the basic trigonometric functions $\sin(x)$, $\cos(x)$ are continuous everywhere.
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* So are the exponential functions $f(x) = a^x, a > 0$.
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* The hyperbolic sine ($(e^x - e^{-x})/2$) and cosine ($(e^x + e^{-x})/2$) are, as $e^x$ is.
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* The hyperbolic tangent is, as $\cosh(x) > 0$ for all $x$.
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Some familiar functions are *mostly* continuous but not everywhere.
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* For example, $f(x) = \sqrt{x}$ is continuous on $(0,\infty)$ and right continuous at $0$, but it is not defined for negative $x$, so can't possibly be continuous there.
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* Similarly, $f(x) = \log(x)$ is continuous on $(0,\infty)$, but it is not defined at $x=0$, so is not right continuous at $0$.
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* 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).
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* The hyperbolic co-tangent is not continuous at $x=0$ – when $\sinh$ is $0$,
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* 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 $1$.
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##### Examples of discontinuity
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There are various reasons why a function may not be continuous.
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* The function $f(x) = \sin(x)/x$ has a limit at $0$ but is not defined at $0$, so is not continuous at $0$. The function can be redefined to make it continuous.
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* The function $f(x) = 1/x$ is continuous everywhere *except* $x=0$ where *no* limit exists.
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* A rational function $f(x) = p(x)/q(x)$ will be continuous everywhere except where $q(x)=0$. (The function $f$ may still have a limit where $q$ is $0$, should factors cancel, but $f$ won't be defined at such values.)
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* The function
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$$
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f(x) = \begin{cases}
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-1 & x < 0 \\
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0 & x = 0 \\
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1 & x > 0
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\end{cases}
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$$
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is implemented by `Julia`'s `sign` function. It has a value at $0$, but no limit at $0$, so is not continuous at $0$. Furthermore, the left and right limits exist at $0$ but are not equal to $f(0)$ so the function is not left or right continuous at $0$. It is continuous everywhere except at $x=0$.
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* Similarly, the function defined by this graph
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```{julia}
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#| hold: true
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#| echo: false
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plot([-1,-.01], [-1,-.01], legend=false, color=:black)
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plot!([.01, 1], [.01, 1], color=:black)
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scatter!([0], [1/2], markersize=5, markershape=:circle)
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```
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is not continuous at $x=0$. It has a limit of $0$ at $0$, a function value $f(0) =1/2$, but the limit and the function value are not equal.
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* The `floor` function, which rounds down to the nearest integer, is also not continuous at the integers, but is right continuous at the integers, as, for example, $\lim_{x \rightarrow 0+} f(x) = f(0)$. This graph emphasizes the right continuity by placing a point for the value of the function when there is a jump:
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```{julia}
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#| hold: true
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#| echo: false
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x = [0,1]; y=[0,0]
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plt = plot(x.-2, y.-2, color=:black, legend=false)
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plot!(plt, x.-1, y.-1, color=:black)
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plot!(plt, x.-0, y.-0, color=:black)
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plot!(plt, x.+1, y.+1, color=:black)
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plot!(plt, x.+2, y.+2, color=:black)
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scatter!(plt, [-2,-1,0,1,2], [-2,-1,0,1,2], markersize=5, markershape=:circle)
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plt
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```
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* The function $f(x) = 1/x^2$ is not continuous at $x=0$: $f(x)$ is not defined at $x=0$ and $f(x)$ has no limit at $x=0$ (in the usual sense).
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* On the Wikipedia page for [continuity](https://en.wikipedia.org/wiki/Continuous_function) the example of Dirichlet's function is given:
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$$
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f(x) =
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\begin{cases}
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0 & \text{if } x \text{ is irrational,}\\
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1 & \text{if } x \text{ is rational.}
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\end{cases}
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$$
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The limit for any $c$ is discontinuous, as any interval about $c$ will contain *both* rational and irrational numbers so the function will not take values in a small neighborhood around any potential $L$.
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##### Example
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Let a function be defined by cases:
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$$
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f(x) = \begin{cases}
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3x^2 + c & x \geq 0,\\
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2x-3 & x < 0.
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\end{cases}
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$$
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What value of $c$ will make $f(x)$ a continuous function?
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We note that for $x < 0$ and for $x > 0$ the function is a simple polynomial, so is continuous. At $x=0$ to be continuous we need a limit to exists and be equal to $f(0)$, which is $c$. A limit exists if the left and right limits are equal. This means we need to solve for $c$ to make the left and right limits equal. We do this next with a bit of overkill in this case:
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```{julia}
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@syms x c
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ex1 = 3x^2 + c
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ex2 = 2x-3
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del = limit(ex1, x=>0, dir="+") - limit(ex2, x=>0, dir="-")
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```
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We need to solve for $c$ to make `del` zero:
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```{julia}
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solve(del, c)
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```
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This gives the value of $c$.
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## Rules for continuity
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As we've seen, functions can be combined in several ways. How do these relate with continuity?
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Suppose $f(x)$ and $g(x)$ are both continuous on $I$. Then
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* The function $h(x) = a f(x) + b g(x)$ is continuous on $I$ for any real numbers $a$ and $b$;
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* The function $h(x) = f(x) \cdot g(x)$ is continuous on $I$; and
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* The function $h(x) = f(x) / g(x)$ is continuous at all points $c$ in $I$ **where** $g(c) \neq 0$.
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* The function $h(x) = f(g(x))$ is continuous at $x=c$ *if* $g(x)$ is continuous at $c$ *and* $f(x)$ is continuous at $g(c)$.
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So, continuity is preserved for all of the basic operations except when dividing by $0$.
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##### Examples
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* Since a monomial $f(x) = ax^n$ ($n$ a non-negative integer) is continuous, by the first rule, any polynomial will be continuous.
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* Since both $f(x) = e^x$ and $g(x)=\sin(x)$ are continuous everywhere, so will be $h(x) = e^x \cdot \sin(x)$.
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* Since $f(x) = e^x$ is continuous everywhere and $g(x) = -x$ is continuous everywhere, the composition $h(x) = e^{-x}$ will be continuous everywhere.
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* Since $f(x) = x$ is continuous everywhere, the function $h(x) = 1/x$ - a ratio of continuous functions - will be continuous everywhere *except* possibly at $x=0$ (where it is not continuous).
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* The function $h(x) = e^{x\log(x)}$ will be continuous on $(0,\infty)$, the same domain that $g(x) = x\log(x)$ is continuous. This function (also written as $x^x$) has a right limit at $0$ (of $1$), but is not right continuous, as $h(0)$ is not defined.
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## Questions
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###### Question
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Let $f(x) = \sin(x)$ and $g(x) = \cos(x)$. Which of these is not continuous everywhere?
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$$
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f+g,~ f-g,~ f\cdot g,~ f\circ g,~ f/g
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$$
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```{julia}
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#| hold: true
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#| echo: false
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choices = ["``f+g``", "``f-g``", "``f\\cdot g``", "``f\\circ g``", "``f/g``"]
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answ = length(choices)
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radioq(choices, answ)
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```
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###### Question
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Let $f(x) = \sin(x)$, $g(x) = \sqrt{x}$.
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When will $f\circ g$ be continuous?
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```{julia}
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#| hold: true
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#| echo: false
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choices = [L"For all $x$", L"For all $x > 0$", L"For all $x$ where $\sin(x) > 0$"]
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answ = 2
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radioq(choices, answ, keep_order=true)
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```
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When will $g \circ f$ be continuous?
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```{julia}
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#| hold: true
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#| echo: false
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choices = [L"For all $x$", L"For all $x > 0$", L"For all $x$ where $\sin(x) > 0$"]
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answ = 3
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radioq(choices, answ, keep_order=true)
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```
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###### Question
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The composition $f\circ g$ will be continuous everywhere provided:
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```{julia}
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#| hold: true
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#| echo: false
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choices = [
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L"The function $g$ is continuous everywhere",
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L"The function $f$ is continuous everywhere",
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L"The function $g$ is continuous everywhere and $f$ is continuous on the range of $g$",
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L"The function $f$ is continuous everywhere and $g$ is continuous on the range of $f$"]
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answ = 3
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radioq(choices, answ, keep_order=true)
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```
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###### Question
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At which values is $f(x) = 1/\sqrt{x-2}$ not continuous?
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```{julia}
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#| hold: true
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#| echo: false
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choices=[
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L"When $x > 2$",
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L"When $x \geq 2$",
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L"When $x \leq 2$",
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L"For $x \geq 0$"]
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answ = 3
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radioq(choices, answ)
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```
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###### Question
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A value $x=c$ is a *removable singularity* for $f(x)$ if $f(x)$ is not continuous at $c$ but will be if $f(c)$ is redefined to be $\lim_{x \rightarrow c} f(x)$.
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The function $f(x) = (x^2 - 4)/(x-2)$ has a removable singularity at $x=2$. What value would we redefine $f(2)$ to be, to make $f$ a continuous function?
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```{julia}
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#| hold: true
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#| echo: false
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f(x) = (x^2 -4)/(x-2);
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numericq(f(2.00001), .001)
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```
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###### Question
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The highly oscillatory function
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$$
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f(x) = x^2 (\cos(1/x) - 1)
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$$
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has a removable singularity at $x=0$. What value would we redefine $f(0)$ to be, to make $f$ a continuous function?
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```{julia}
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#| hold: true
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#| echo: false
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numericq(0, .001)
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```
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###### Question
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Let $f(x)$ be defined by
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$$
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f(x) = \begin{cases}
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c + \sin(2x - \pi/2) & x > 0\\
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3x - 4 & x \leq 0.
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\end{cases}
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$$
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What value of $c$ will make $f(x)$ continuous?
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```{julia}
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#| hold: true
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#| echo: false
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val = (3*0 - 4) - (sin(2*0 - pi/2))
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numericq(val)
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```
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###### Question
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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 $f(g(c))$?
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```{julia}
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#| hold: true
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#| echo: false
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choices = [L"No, as $g(c)$ may not be in the interval $(a,b)$",
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"Yes, composition of continuous functions results in a continuous function, so the limit is just the function value."
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]
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answ=1
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radioq(choices, answ)
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```
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###### Question
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Consider the function $f(x)$ given by the following graph
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```{julia}
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#| hold: true
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#| echo: false
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xs = range(0, stop=2, length=50)
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plot(xs, [sqrt(1 - (x-1)^2) for x in xs], legend=false, xlims=(0,4))
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plot!([2,3], [1,0])
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scatter!([3],[0], markersize=5)
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plot!([3,4],[1,0])
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scatter!([4],[0], markersize=5)
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```
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The function $f(x)$ is continuous at $x=1$?
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```{julia}
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#| hold: true
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#| echo: false
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yesnoq(true)
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```
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The function $f(x)$ is continuous at $x=2$?
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```{julia}
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#| hold: true
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#| echo: false
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yesnoq(false)
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```
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The function $f(x)$ is right continuous at $x=3$?
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```{julia}
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#| hold: true
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#| echo: false
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yesnoq(false)
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```
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The function $f(x)$ is left continuous at $x=4$?
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```{julia}
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#| hold: true
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#| echo: false
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yesnoq(true)
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```
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###### Question
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Let $f(x)$ and $g(x)$ be continuous functions whose graph of $[0,1]$ is given by:
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```{julia}
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#| hold: true
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#| echo: false
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xs = range(0, 1, length=251)
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plot(xs, [sin.(2pi*xs) cos.(2pi*xs)], layout=2, title=["f" "g"], legend=false)
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```
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What is $\lim_{x \rightarrow 0.25} f(g(x))$?
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```{julia}
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#| hold: true
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#| echo: false
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val = sin(2pi * cos(2pi * 1/4))
|
||
numericq(val)
|
||
```
|
||
|
||
What is $\lim{x \rightarrow 0.25} g(f(x))$?
|
||
|
||
|
||
```{julia}
|
||
#| hold: true
|
||
#| echo: false
|
||
val = cos(2pi * sin(2pi * 1/4))
|
||
numericq(val)
|
||
```
|
||
|
||
What is $\lim_{x \rightarrow 0.5} f(g(x))$?
|
||
|
||
|
||
```{julia}
|
||
#| hold: true
|
||
#| echo: false
|
||
choices = ["Can't tell",
|
||
"``-1.0``",
|
||
"``0.0``"
|
||
]
|
||
answ = 1
|
||
radioq(choices, answ)
|
||
```
|
||
|