# Continuity {{< include ../_common_code.qmd >}} This section uses these add-on packages: ```{julia} using CalculusWithJulia using Plots plotly() using SymPy ``` --- ![A Möbius strip by Koo Jeong A](figures/korean-mobius.jpg){width=40%} The definition Google finds for *continuous* is *forming an unbroken whole; without interruption*. 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. Consider these two graphs: ```{julia} #| hold: true #| echo: false plt = plot([-1,0], [-1,-1], color=:black, legend=false, linewidth=5) plot!(plt, [0, 1], [ 1, 1], color=:black, linewidth=5) plt ``` and ```{julia} #| hold: true #| echo: false plot([-1,-.1, .1, 1], [-1,-1, 1, 1], color=:black, legend=false, linewidth=5) ``` 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. 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. 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. The [modern](http://en.wikipedia.org/wiki/Continuous_function#History) definition simply pushes the details to the definition of the limit: ::: {.callout-note icon=false} ## Definition of continuity at a point A function $f(x)$ is continuous at $x=c$ if $\lim_{x \rightarrow c}f(x) = f(c)$. ::: The definition says three things * The limit exists at $c$. * The function is defined at $c$ ($c$ is in the domain). * The value of the limit is the same as $f(c)$. The definition speaks to continuity at a point, we can extend it to continuity over an interval $(a,b)$ by saying: ::: {.callout-note icon=false} ## Definition of continuity over an open interval A function $f(x)$ is continuous over $(a,b)$ if at each point $c$ with $a < c < b$, $f(x)$ is continuous at $c$. ::: 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 The limit in the definition of continuity is the basic limit and not an extended sense where infinities are accounted for. ::: ##### Examples of continuity 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 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 $x$. Some familiar functions are *mostly* continuous but not everywhere. * 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. * 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 $1$. (It is continuous on $[-1,1]$.) ##### Examples of discontinuity There are various reasons why a function may not be continuous. * 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. * The function $f(x) = 1/x$ is continuous everywhere *except* $x=0$ where *no* limit exists. * 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.) * The function $$ f(x) = \begin{cases} -1 &~ x < 0 \\ 0 &~ x = 0 \\ 1 &~ x > 0 \end{cases} $$ 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 $x=0$. * Similarly, the function defined by this graph ```{julia} #| hold: true #| echo: false plot([-1,-.01], [-1,-.01], legend=false, color=:black) plot!([.01, 1], [.01, 1], color=:black) scatter!([0], [1/2], markersize=5, markershape=:circle) ts = range(0, 2pi, 100) C = Shape(0.02 * sin.(ts), 0.03 * cos.(ts)) plot!(C, fill=(:white,1), line=(:black, 1)) ``` 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. * 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 filled marker for the value of the function when there is a jump and an open marker where the function is not that value. ```{julia} #| echo: false plt = let empty_style = (xticks=-4:4, yticks=-4:4, framestyle=:origin, legend=false) axis_style = (arrow=true, side=:head, line=(:gray, 1)) text_style = (10,) fn_style = (;line=(:black, 3)) fn2_style = (;line=(:red, 4)) mark_style = (;line=(:gray, 1, :dot)) domain_style = (;fill=(:orange, 0.35), line=nothing) range_style = (; fill=(:blue, 0.35), line=nothing) ts = range(0, 2pi, 100) xys = sincos.(ts) xys = [.1 .* xy for xy in xys] plot(; empty_style..., aspect_ratio=:equal) plot!([-4.25,4.25], [0,0]; axis_style...) plot!([0,0], [-4.25, 4.25]; axis_style...) for k in -4:4 P,Q = (k,k),(k+1,k) plot!([P,Q], line=(:black,1)) S = Shape([k .+ xy for xy in xys]) plot!(S; fill=(:black,)) S = Shape([(k+1,k) .+ xy for xy in xys]) plot!(S; fill=(:white,), line=(:black,1)) end current() end plt ``` ```{julia} #| echo: false plotly() nothing ``` * 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). * On the Wikipedia page for [continuity](https://en.wikipedia.org/wiki/Continuous_function) the example of Dirichlet's function is given: $$ f(x) = \begin{cases} 0 &~ \text{if } x \text{ is irrational,}\\ 1 &~ \text{if } x \text{ is rational.} \end{cases} $$ 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 $L$. ##### Example Let a function be defined by cases: $$ f(x) = \begin{cases} 3x^2 + c &~ x \geq 0,\\ 2x-3 &~ x < 0. \end{cases} $$ What value of $c$ will make $f(x)$ a continuous function? We note that for $x < 0$ and for $x > 0$ the function is defined by 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: ```{julia} @syms x c ex1 = 3x^2 + c ex2 = 2x-3 del = limit(ex1, x=>0, dir="+") - limit(ex2, x=>0, dir="-") ``` We need to solve for $c$ to make `del` zero: ```{julia} solve(del ~ 0, 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 As we've seen, functions can be combined in several ways. How do these relate with continuity? Suppose $f(x)$ and $g(x)$ are both continuous on $I$. Then * The function $h(x) = a f(x) + b g(x)$ is continuous on $I$ for any real numbers $a$ and $b$; * The function $h(x) = f(x) \cdot g(x)$ is continuous on $I$; and * The function $h(x) = f(x) / g(x)$ is continuous at all points $c$ in $I$ **where** $g(c) \neq 0$. * 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 $g(c)$. So, continuity is preserved for all of the basic operations except when dividing by $0$. ##### Examples * Since a monomial $f(x) = ax^n$ ($n$ a non-negative integer) is continuous, by the first rule, any polynomial will be continuous. * Since both $f(x) = e^x$ and $g(x)=\sin(x)$ are continuous everywhere, so will be $h(x) = e^x \cdot \sin(x)$. * 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. * 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). * 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. ## Questions ###### Question Let $f(x) = \sin(x)$ and $g(x) = \cos(x)$. Which of these is not continuous everywhere? $$ f+g,~ f-g,~ f\cdot g,~ f\circ g,~ f/g $$ ```{julia} #| hold: true #| echo: false choices = ["``f+g``", "``f-g``", "``f\\cdot g``", "``f\\circ g``", "``f/g``"] answ = length(choices) radioq(choices, answ) ``` ###### Question Let $f(x) = \sin(x)$, $g(x) = \sqrt{x}$. When will $f\circ g$ be continuous? ```{julia} #| hold: true #| echo: false choices = [L"For all $x$", L"For all $x > 0$", L"For all $x$ where $\sin(x) > 0$"] answ = 2 radioq(choices, answ, keep_order=true) ``` When will $g \circ f$ be continuous? ```{julia} #| hold: true #| echo: false choices = [L"For all $x$", L"For all $x > 0$", L"For all $x$ where $\sin(x) > 0$"] answ = 3 radioq(choices, answ, keep_order=true) ``` ###### Question The composition $f\circ g$ will be continuous everywhere provided: ```{julia} #| hold: true #| echo: false choices = [ L"The function $g$ is continuous everywhere", L"The function $f$ is continuous everywhere", L"The function $g$ is continuous everywhere and $f$ is continuous on the range of $g$", L"The function $f$ is continuous everywhere and $g$ is continuous on the range of $f$"] answ = 3 radioq(choices, answ, keep_order=true) ``` ###### Question At which values is $f(x) = 1/\sqrt{x-2}$ not continuous? ```{julia} #| hold: true #| echo: false choices=[ L"When $x > 2$", L"When $x \geq 2$", L"When $x \leq 2$", L"For $x \geq 0$"] answ = 3 radioq(choices, answ) ``` ###### Question 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 $\lim_{x \rightarrow c} f(x)$. 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? ```{julia} #| hold: true #| echo: false f(x) = (x^2 -4)/(x-2); numericq(f(2.00001), .001) ``` ###### Question The highly oscillatory function $$ f(x) = x^2 (\cos(1/x) - 1) $$ has a removable singularity at $x=0$. What value would we redefine $f(0)$ to be, to make $f$ a continuous function? ```{julia} #| hold: true #| echo: false numericq(0, .001) ``` ###### Question Let $f(x)$ be defined by $$ f(x) = \begin{cases} c + \sin(2x - \pi/2) &~ x > 0\\ 3x - 4 &~ x \leq 0. \end{cases} $$ What value of $c$ will make $f(x)$ continuous? ```{julia} #| hold: true #| echo: false val = (3*0 - 4) - (sin(2*0 - pi/2)) 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 $f(g(c))$? ```{julia} #| hold: true #| echo: false choices = [L"No, as $g(c)$ may not be in the interval $(a,b)$", "Yes, composition of continuous functions results in a continuous function, so the limit is just the function value." ] answ=1 radioq(choices, answ) ``` ###### Question Consider the function $f(x)$ given by the following graph ```{julia} #| hold: true #| echo: false let xs = range(0, stop=2, length=50) plot(xs, [sqrt(1 - (x-1)^2) for x in xs]; line=(:black,1), legend=false, xlims=(-0.1,4.1)) plot!([2,3], [1,0]; line=(:black,1)) plot!([3,4],[1,0]; line=(:black,1)) scatter!([(0,0)], markersize=5, markercolor=:black) scatter!([(2,0)], markersize=5, markercolor=:white) scatter!([(2, 1)], markersize=5; markercolor=:black) scatter!([(3,0)], markersize=5; markercolor=:black) scatter!([(3,1)], markersize=5; markercolor=:white) scatter!([(4,0)], markersize=5; markercolor=:black) end ``` The function $f(x)$ is continuous at $x=1$? ```{julia} #| hold: true #| echo: false yesnoq(true) ``` The function $f(x)$ is continuous at $x=2$? ```{julia} #| hold: true #| echo: false yesnoq(false) ``` The function $f(x)$ is right continuous at $x=3$? ```{julia} #| hold: true #| echo: false yesnoq(false) ``` The function $f(x)$ is left continuous at $x=4$? ```{julia} #| hold: true #| echo: false yesnoq(true) ``` ###### Question Let $f(x)$ and $g(x)$ be continuous functions. Their graphs over $[0,1]$ are given by: ```{julia} #| hold: true #| echo: false xs = range(0, 1, length=251) plot(xs, [sin.(2pi*xs) cos.(2pi*xs)], layout=2, title=["f" "g"], legend=false) ``` What is $\lim_{x \rightarrow 0.25} f(g(x))$? ```{julia} #| hold: true #| echo: false 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) ``` ###### Question A parametric equation is specified by a parameterization $(f(t), g(t)), a \leq t \leq b$. The parameterization will be continuous if and only if each function is continuous. Suppose $k_x$ and $k_y$ are positive integers and $a, b$ are positive numbers, will the [Lissajous](https://en.wikipedia.org/wiki/Parametric_equation#Lissajous_Curve) curve given by $(a\cos(k_x t), b\sin(k_y t))$ be continuous? ```{julia} #| hold: true #| echo: false yesnoq(true) ``` Here is a sample graph for $a=1, b=2, k_x=3, k_y=4$: ```{julia} #| hold: true a,b = 1, 2 k_x, k_y = 3, 4 plot(t -> a * cos(k_x *t), t-> b * sin(k_y * t), 0, 4pi) ```