Naughty Functions
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Naughty Functions
I'm thinking about writing something about the role of functions in mathematics. One part of it will be dedicated to "weird" functions. I'm asking xkcd to help me come up with some good examples. That is, I'm looking for any function that is misbehaved or atypical or just plain cool that most people don't think of as a function. Nonelementary functions. Functions which are continuous but nowhere smooth. Functions that have any counterintuitive properties. Real functions are best, but easytoexplain functions on other domains work too.
Re: Naughty Functions
The Weierstrass function is your standard example.
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Re: Naughty Functions
The indicator function function of the rationals. Nowhere continuous.
The popcorn function, continuous only on the irrationals.
The Busy Beaver function.
Bijections from N to N^2. One that is given by a formula is f(n,m)=(n+m)*(n+m+1)/2+n (it's enumerates the elements on the diagonals). I haven't seen any neat bijections from R to R^2.
Spacefilling curves.
The devil's staircase: a continuous, monotonous function with zero derivative almost everywhere, but it increases from 0 to 1 on [0,1].
A wellordering on the reals.
[imath]f(x)=\begin{cases}\exp(1/x^2), & x\geq 0\\0,&x\leq 0\end{cases}[/imath] is infinitely differentiable but not analytic.
Additive (i.e. f(x+y)=f(x)+f(y)) functions that aren't linear.
The popcorn function, continuous only on the irrationals.
The Busy Beaver function.
Bijections from N to N^2. One that is given by a formula is f(n,m)=(n+m)*(n+m+1)/2+n (it's enumerates the elements on the diagonals). I haven't seen any neat bijections from R to R^2.
Spacefilling curves.
The devil's staircase: a continuous, monotonous function with zero derivative almost everywhere, but it increases from 0 to 1 on [0,1].
A wellordering on the reals.
[imath]f(x)=\begin{cases}\exp(1/x^2), & x\geq 0\\0,&x\leq 0\end{cases}[/imath] is infinitely differentiable but not analytic.
Additive (i.e. f(x+y)=f(x)+f(y)) functions that aren't linear.
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Re: Naughty Functions
I recommend, when you talk about "weird functions," that you point out the notquiteintuitive fact that, in the sense of cardinality, most functions are poorly behaved.
Here are some obvious starters, all with domain and codomain [imath]\mathbb{R}[/imath]:
Here are some obvious starters, all with domain and codomain [imath]\mathbb{R}[/imath]:
 The indicator function [imath]1_\mathbb{Q}[/imath] for [imath]\mathbb{Q}[/imath] in [imath]\mathbb{R}[/imath] (1 if rational, 0 if irrational) is defined everywhere but continuous nowhere.
 The Weierstrass function (http://en.wikipedia.org/wiki/Weierstrass_function) is continuous everywhere but differentiable nowhere.
 The function [math]f(x) = \begin{cases}\ \ x^2 & x \in \mathbb{Q} \\ \ \ 0 & x \not\in \mathbb{Q}\end{cases}[/math] is continuous only at 0and it is differentiable at 0.
 Thomae's function (http://en.wikipedia.org/wiki/Popcorn_function) is continuous at all irrational numbers and discontinuous at all rational numbers.
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Re: Naughty Functions
There's a book called "Counterexamples in Analysis" that's just full of this kind of stuff; you may find it useful if you can find a copy.
Re: Naughty Functions
Hix wrote:There's a book called "Counterexamples in Analysis" that's just full of this kind of stuff; you may find it useful if you can find a copy.
I thought of that book when I was writing this. However, why go out of my way to browse through it in the bookstore if I can bum the more familiar examples off of bored xkcd forumgoers
everyone else wrote:(stuff about continuity)
Thank you for your help. It's good to have the names of those functions so I can look into them. Can anyone think of any more weird functions. Ones that don't necessarily have to do with continuity. I'm looking for variety =)

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Re: Naughty Functions
SunAvatar wrote:The indicator function 1Q for Q in R (1 if rational, 0 if irrational)
[math]1Q(x) = lim_{n \to \infty} lim_{m \to \infty} cos^{m}(2n \pi x)[/math]
Do your functions have to be constructive? I remember an exercise which proved nonconstructively that there exists a smooth function f: [0,1] > R whose Taylor series around any point t diverges in any open interval containing t. Crazy stuff.
Generally I try to make myself do things I instinctively avoid, in case they are awesome.
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Re: Naughty Functions
The Banach–Tarski paradox has a few functions that are interesting, and require choice to exist.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision  BR
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Re: Naughty Functions
Most of the ways that functions in general can be 'nice' are rather more esoteric than continuity or differentiabilitythat is, even though they're nice things to have, a naïve observer wouldn't necessarily expect them by default. No one really expects all functions to be oneone or onto.
You could point out that the derivative is itself a function, from the set of differentiable realvalued functions to the set of realvalued functions.
You could consider two ways of looking at a binary operation like addition: as a function from [imath]\mathbb{R} \times \mathbb{R}[/imath] to [imath]\mathbb{R}[/imath], as is usually done, or in its curried form as a function from [imath]\mathbb{R}[/imath] to the set of functions from [imath]\mathbb{R}[/imath] to [imath]\mathbb{R}[/imath]. (That is, '+' is a function that, applied to 3, gives '3+', a function that, applied to 4, gives 7.) In either case, a binary operation is a function.
You could give a cardinality argument that, within any given fixed language, 'most' realvalued (or indeed even integervalued) functions have infinite Komolgorov complexity, and so can never actually be written down specifically. (This helps clarify what is imprecise about the informal "A function is a rule...." statement.)
You could point out that, for any set [imath]S[/imath], there is exactly one function from the empty set to [imath]S[/imath]. (Depending on how one formally defines a function, this function may itself be the empty set.)
You could point out that the derivative is itself a function, from the set of differentiable realvalued functions to the set of realvalued functions.
You could consider two ways of looking at a binary operation like addition: as a function from [imath]\mathbb{R} \times \mathbb{R}[/imath] to [imath]\mathbb{R}[/imath], as is usually done, or in its curried form as a function from [imath]\mathbb{R}[/imath] to the set of functions from [imath]\mathbb{R}[/imath] to [imath]\mathbb{R}[/imath]. (That is, '+' is a function that, applied to 3, gives '3+', a function that, applied to 4, gives 7.) In either case, a binary operation is a function.
You could give a cardinality argument that, within any given fixed language, 'most' realvalued (or indeed even integervalued) functions have infinite Komolgorov complexity, and so can never actually be written down specifically. (This helps clarify what is imprecise about the informal "A function is a rule...." statement.)
You could point out that, for any set [imath]S[/imath], there is exactly one function from the empty set to [imath]S[/imath]. (Depending on how one formally defines a function, this function may itself be the empty set.)
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Re: Naughty Functions
Some easy examples from calculus:
sin(1/x)  has an essential discontinuity at 0.
xsin(1/x)  is continuous (if you remove the removable discontinuity at 0), but not differentiable at 0.
x^{2}sin(1/x)  is everywhere differentiable, but has a noncontinuous derivative.
One example that hasn't yet been mentioned is a function which is differentiable but whose derivative is not Riemann integrable. A construction can be found in, e.g., Abbott, Understanding Analysis, section 7.6. The idea behind his construction is this:
Let C be a fat Cantor set. The complement of C is a union of disjoint open intervals. We define f by setting f=0 on C, and for each of the open intervals that makes up the complement of C, paste in something that is differentiable but has a scaled copy of x^{2}sin(1/x) at each end. The resulting function is everywhere differentiable, but the derivative has discontinuities at every point of C. Since the derivative is discontinuous on a set of positive measure, it is not Riemann integrable (by Lebesgue's criterion for Riemann integrability).
sin(1/x)  has an essential discontinuity at 0.
xsin(1/x)  is continuous (if you remove the removable discontinuity at 0), but not differentiable at 0.
x^{2}sin(1/x)  is everywhere differentiable, but has a noncontinuous derivative.
One example that hasn't yet been mentioned is a function which is differentiable but whose derivative is not Riemann integrable. A construction can be found in, e.g., Abbott, Understanding Analysis, section 7.6. The idea behind his construction is this:
Let C be a fat Cantor set. The complement of C is a union of disjoint open intervals. We define f by setting f=0 on C, and for each of the open intervals that makes up the complement of C, paste in something that is differentiable but has a scaled copy of x^{2}sin(1/x) at each end. The resulting function is everywhere differentiable, but the derivative has discontinuities at every point of C. Since the derivative is discontinuous on a set of positive measure, it is not Riemann integrable (by Lebesgue's criterion for Riemann integrability).
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Re: Naughty Functions
skeptical scientist wrote:sin(1/x)  has an essential discontinuity at 0.
That reminds me  Picard's great theorem is seriously weird and cool. It says that if an analytic function f(z) has an essential singularity at z = w, then f takes all possible complex values (with possibly one exception) infinitely often, in any open set containing w.
You can apply it to something simple like sin(1/z) or e^{1/z}, with w = 0.
Generally I try to make myself do things I instinctively avoid, in case they are awesome.
dubsola
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Re: Naughty Functions
Ended wrote:skeptical scientist wrote:sin(1/x)  has an essential discontinuity at 0.
That reminds me  Picard's great theorem is seriously weird and cool. It says that ifan analytica meromorphic function f(z) has an essential singularity at z = w, then f takes all possible complex values (with possibly one exception) infinitely often, in any open set containing w.
You can apply it to something simple like sin(1/z) or e^{1/z}, with w = 0.
Fix'd. Analytic functions can't technically have singularities.
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Re: Naughty Functions
These are interesting functions: I've never really looked into "weird" functions, although I've wondered about them.
I have a question about the Weierstrauss function: does its graph have infinite length between any 2 xordinates? If so, is this true for any nowheredifferentiable functions?
(I'm not sure how you'd even deal with the length of a function like that  I'm guessing you could take a line of any length and bend it so it's a smoothed version of the graph, and that the graph therefore has to have infinite length)
I have a question about the Weierstrauss function: does its graph have infinite length between any 2 xordinates? If so, is this true for any nowheredifferentiable functions?
(I'm not sure how you'd even deal with the length of a function like that  I'm guessing you could take a line of any length and bend it so it's a smoothed version of the graph, and that the graph therefore has to have infinite length)
Re: Naughty Functions
Huh? Meromorphic functions aren't allowed to have essential singularities. Sure, "analytic" is sometimes used to mean "analytic on C" (the proper term for this is "entire", I think), but here it's clear from context that that's not what skeptical means.Token wrote:Ended wrote:skeptical scientist wrote:sin(1/x)  has an essential discontinuity at 0.
That reminds me  Picard's great theorem is seriously weird and cool. It says that ifan analytica meromorphic function f(z) has an essential singularity at z = w, then f takes all possible complex values (with possibly one exception) infinitely often, in any open set containing w.
You can apply it to something simple like sin(1/z) or e^{1/z}, with w = 0.
Fix'd. Analytic functions can't technically have singularities.
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Re: Naughty Functions
Bah. I forgot the condition that meromorphic functions can have at worst poles. I shall stop being so pedantic about areas of mathematics I tended to sleep through. Not my fault if Complex Analysis is a hugely boring subject. "Analytic on D(w,r) \ {w} for some r > 0" then, if you insist. And it wasn't skeptical.
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Re: Naughty Functions
troyp wrote:I have a question about the Weierstrauss function: does its graph have infinite length between any 2 xordinates? If so, is this true for any nowheredifferentiable functions?
The length of a curve is defined in terms of the derivative, so I would say that the lenght of the weierstrass function is undefined, but you could probably make an argument for it being infinite.
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Re: Naughty Functions
How about a function that is unbounded in any open interval, but has a finite integral over all of R.
Let [imath]\{q_n\}[/imath] be an enumeration of the rational numbers. Let [math]g(x) = \left\{ \begin{array}{cc} \sqrt x 1 & \;\;\; 0<x \leq 1 \\ 0 & \textrm{else}\end{array}\right. .[/math] Finally, let [imath]f(x) = \sum_n 2^{n}g(xq_n)[/imath]. The function [imath]f[/imath] has a pole (from the right) at each rational number, yet [math]\int_{\mathbb R} f = \sum_n 2^{n} \int_0^1 g = \sum_n 2^{n} 1 =1.[/math]
Let [imath]\{q_n\}[/imath] be an enumeration of the rational numbers. Let [math]g(x) = \left\{ \begin{array}{cc} \sqrt x 1 & \;\;\; 0<x \leq 1 \\ 0 & \textrm{else}\end{array}\right. .[/math] Finally, let [imath]f(x) = \sum_n 2^{n}g(xq_n)[/imath]. The function [imath]f[/imath] has a pole (from the right) at each rational number, yet [math]\int_{\mathbb R} f = \sum_n 2^{n} \int_0^1 g = \sum_n 2^{n} 1 =1.[/math]
Re: Naughty Functions
Ackermann's Function
Not really fun unless you try to evaluate it using Graham's Number
Infinity doesn't stand a chance...
Not really fun unless you try to evaluate it using Graham's Number
Infinity doesn't stand a chance...

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Re: Naughty Functions
Counterexamples in Analysis is on that google book thing (google the title and it'll be at the top)
One cool example if a function that's finite everywhere but everywhere unbounded
f(x) = 0 if x is irrational, m if x is rational and of the form m/n when reduced
One cool example if a function that's finite everywhere but everywhere unbounded
f(x) = 0 if x is irrational, m if x is rational and of the form m/n when reduced
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Re: Naughty Functions
Most awesome function ever:
[math]f(x) =
\begin{cases}
0, & \mbox{if }x \notin \mathbb{Q}\\
1/q & \mbox{ if }x= \frac{p}{q} \in \mathbb{Q}
\end{cases}[/math]
It's awesome because it's continuous at every irrational in (0,1), but discontinuous at every rational in that set.
Crazy
[math]f(x) =
\begin{cases}
0, & \mbox{if }x \notin \mathbb{Q}\\
1/q & \mbox{ if }x= \frac{p}{q} \in \mathbb{Q}
\end{cases}[/math]
It's awesome because it's continuous at every irrational in (0,1), but discontinuous at every rational in that set.
Crazy
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Re: Naughty Functions
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

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Re: Naughty Functions
jestingrabbit wrote:http://en.wikipedia.org/wiki/Minkowski's_question_mark_function
Hehe, all the question marks in the formulae make it look like they're just guessing things.
Generally I try to make myself do things I instinctively avoid, in case they are awesome.
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Re: Naughty Functions
How about a continuous function f with finite integral as x > infinity but f(x) doesn't > 0 as x > infinity? An example would be f(x) = 0 everywhere except for intervals of length (1/2)^n centered on n. On those intervals, f rises to 1 and returns to 0 in series of evernarrower triangular spikes. I'm too lazy to write down the formula.
Re: Naughty Functions
Based on my previous post, how about this (easy?) exercise:
I described a function identically = 0 on an interval that then rises continuously at the interval's end. Now let's try smoothing it out at the corners. So... imagine any old function f and a real number a such that
f(x) = 0 for x ≤ a
f(x) > 0 for x > a
Show that you can construct such a function with an arbitrary number of derivates at a. That is, given n, I can make an ntimes differentiable function f with those properties.
But no such function can be infinitely differentiable.
I described a function identically = 0 on an interval that then rises continuously at the interval's end. Now let's try smoothing it out at the corners. So... imagine any old function f and a real number a such that
f(x) = 0 for x ≤ a
f(x) > 0 for x > a
Show that you can construct such a function with an arbitrary number of derivates at a. That is, given n, I can make an ntimes differentiable function f with those properties.
But no such function can be infinitely differentiable.
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Re: Naughty Functions
lgonick wrote:Based on my previous post, how about this (easy?) exercise:
I described a function identically = 0 on an interval that then rises continuously at the interval's end. Now let's try smoothing it out at the corners. So... imagine any old function f and a real number a such that
f(x) = 0 for x ≤ a
f(x) > 0 for x > a
Show that you can construct such a function with an arbitrary number of derivates at a. That is, given n, I can make an ntimes differentiable function f with those properties.
But no such function can be infinitely differentiable.
Bull.
f(x) = 0 for x<= a
= exp(1/(xa)^{2})
The guts of f were described earlier on this very page.
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Re: Naughty Functions
Personal favorite:
[math]f(x,y) = \left\{\begin{array}{ll}\frac{x^2y}{x^4 + y^2}&(x,y)\neq(0,0)\\0&(x,y)=(0,0)\end{array}\right.[/math]
Continuous along every straight line path, but discontinuous at (0,0); the limit as [imath](x,y)\rightarrow(0,0)[/imath] along the path [imath]y = \pm x^2[/imath] is [imath]\pm\frac12[/imath].
[math]f(x,y) = \left\{\begin{array}{ll}\frac{x^2y}{x^4 + y^2}&(x,y)\neq(0,0)\\0&(x,y)=(0,0)\end{array}\right.[/math]
Continuous along every straight line path, but discontinuous at (0,0); the limit as [imath](x,y)\rightarrow(0,0)[/imath] along the path [imath]y = \pm x^2[/imath] is [imath]\pm\frac12[/imath].
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Re: Naughty Functions
Not really naughty but more anecdotal, but not something I see every day.
[math]y= x^\frac {\left x \right }{x} + \frac {1}{x}[/math]
Function with all three asymptotes.
[math]y= x^\frac {\left x \right }{x} + \frac {1}{x}[/math]
Function with all three asymptotes.
Re: Naughty Functions
Right you are, and wrong I was!
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Re: Naughty Functions
lgonick wrote:How about a continuous function f with finite integral as x > infinity but f(x) doesn't > 0 as x > infinity? An example would be f(x) = 0 everywhere except for intervals of length (1/2)^n centered on n. On those intervals, f rises to 1 and returns to 0 in series of evernarrower triangular spikes. I'm too lazy to write down the formula.
The function sin(x^{2}) works. Use the alternating series test.
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Re: Naughty Functions
How about generating functions? Which, I suppose, are not functions in the usual sense. Or perhaps it's better to say is that the functions themselves may not be weird, it's how they are used that seems weird when you first encounter them. The best reference on these, for beginners anyway, is Herb Wilf's "generatingfunctionology" (yes, it's all one word) which is freely available as a PDF file on his website.
Re: Naughty Functions
skeptical scientist wrote:lgonick wrote:How about a continuous function f with finite integral as x > infinity but f(x) doesn't > 0 as x > infinity? An example would be f(x) = 0 everywhere except for intervals of length (1/2)^n centered on n. On those intervals, f rises to 1 and returns to 0 in series of evernarrower triangular spikes. I'm too lazy to write down the formula.
The function sin(x^{2}) works. Use the alternating series test.
I thought the absolute value of alternating series had to be constantly decreasing for that to work?
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Re: Naughty Functions
Charlie! wrote:skeptical scientist wrote:lgonick wrote:How about a continuous function f with finite integral as x > infinity but f(x) doesn't > 0 as x > infinity? An example would be f(x) = 0 everywhere except for intervals of length (1/2)^n centered on n. On those intervals, f rises to 1 and returns to 0 in series of evernarrower triangular spikes. I'm too lazy to write down the formula.
The function sin(x^{2}) works. Use the alternating series test.
I thought the absolute value of alternating series had to be constantly decreasing for that to work?
Well, he didn't spell it out, but the "alternating series" that skeptical is talking about is probably:
sin(x^{2}) is positive from 0 to sqrt(pi), and encloses a certain area;
sin(x^{2}) is negative from sqrt(pi) to sqrt(2pi), and clearly encloses a smaller area;
sin(x^{2}) is positive from sqrt(2pi) to sqrt(3pi), and encloses an area that's even smaller than that;
...
so the sum of all the (signed) areas from 0 to infinity will converge.
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Re: Naughty Functions
Hix wrote:Charlie! wrote:skeptical scientist wrote:lgonick wrote:How about a continuous function f with finite integral as x > infinity but f(x) doesn't > 0 as x > infinity? An example would be f(x) = 0 everywhere except for intervals of length (1/2)^n centered on n. On those intervals, f rises to 1 and returns to 0 in series of evernarrower triangular spikes. I'm too lazy to write down the formula.
The function sin(x^{2}) works. Use the alternating series test.
I thought the absolute value of alternating series had to be constantly decreasing for that to work?
Well, he didn't spell it out, but the "alternating series" that skeptical is talking about is probably:
sin(x^{2}) is positive from 0 to sqrt(pi), and encloses a certain area;
sin(x^{2}) is negative from sqrt(pi) to sqrt(2pi), and clearly encloses a smaller area;
sin(x^{2}) is positive from sqrt(2pi) to sqrt(3pi), and encloses an area that's even smaller than that;
...
so the sum of all the (signed) areas from 0 to infinity will converge.
Right. That tells you that if you consider the sequence [imath]\int_0^\sqrt{n\pi} \sin{x^2} \, \text{d}x[/imath], it converges to some limit L. Furthermore, since the integral can't change very much between [imath]\sqrt{n\pi}[/imath] and [imath]\sqrt{(n+1)\pi}[/imath] (since the function is bounded by 1, and [imath]\lim_n (\sqrt{(n+1)\pi}\sqrt{n\pi})=0[/imath]), this tells you that the indefinite integral converges.
Also, you can use methods of complex analysis to prove the limit is [imath]\sqrt{\frac{\pi}{8}}[/imath]  see Wikipedia.
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Re: Naughty Functions
OK, then, a nonnegative continuous function that doesn't > 0 but has finite integral as x > infinity.
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Re: Naughty Functions
I can't think of a way to define it nicely in one equation, but I'll describe such a function for you:
Let f(x) be zero for 0 <= x <= 1. Between x = 1 and x = 2, have it jump up and then down (in two straight lines if you don't mind it being nondifferentiable, or in a bump if you want it smooth), so that f(1.5) = 1. Now, have it do the exact same thing between x = 2 and x = 2.5, except squished by half in the x direction so that it fits in an interval half as big. (Don't squish it in the y direction, so f(2.25) = 1.) Have f(x) = 0 for 2.5 <= x <= 3. Have it jump up to 1 and back again between 3 and 3.25, and then be zero until x = 4.
Continue in this way, with a jump up to f(x) = 1 starting at each integer, but each such bump being half as wide as the bump before, and then f(x) is zero until the next integer. Each bump has an integral equal to half the previous bump, so the integral from zero to infinity of this function will be twice that of your original bump. And since the function keeps jumping up to the value of 1, it does not approach zero as x > infinity.
Let f(x) be zero for 0 <= x <= 1. Between x = 1 and x = 2, have it jump up and then down (in two straight lines if you don't mind it being nondifferentiable, or in a bump if you want it smooth), so that f(1.5) = 1. Now, have it do the exact same thing between x = 2 and x = 2.5, except squished by half in the x direction so that it fits in an interval half as big. (Don't squish it in the y direction, so f(2.25) = 1.) Have f(x) = 0 for 2.5 <= x <= 3. Have it jump up to 1 and back again between 3 and 3.25, and then be zero until x = 4.
Continue in this way, with a jump up to f(x) = 1 starting at each integer, but each such bump being half as wide as the bump before, and then f(x) is zero until the next integer. Each bump has an integral equal to half the previous bump, so the integral from zero to infinity of this function will be twice that of your original bump. And since the function keeps jumping up to the value of 1, it does not approach zero as x > infinity.
Let's have a fervent argument, mostly over semantics, where we all claim the burden of proof is on the other side!
 Cosmologicon
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Re: Naughty Functions
Here's an explicit description using the smooth bump idea:
[math]\sum_{n=1}^{\infty}\exp{\left({n^4(xn)^2}\right)}[/math]
which is greater than 1 on all positive integers. If I did this right, the integral goes like the sum of 1/n^2.
[math]\sum_{n=1}^{\infty}\exp{\left({n^4(xn)^2}\right)}[/math]
which is greater than 1 on all positive integers. If I did this right, the integral goes like the sum of 1/n^2.
 skeptical scientist
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Re: Naughty Functions
I thought sin x^{2} was a nice example because the formula is so simple. What's so great about positive functions anyways?
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Re: Naughty Functions
Cosmologicon wrote:Here's an explicit description using the smooth bump idea:
[math]\sum_{n=1}^{\infty}\exp{\left({n^4(xn)^2}\right)}[/math]
which is greater than 1 on all positive integers. If I did this right, the integral goes like the sum of 1/n^2.
Awesome! Not only is that function positive and does it have a a finite integral without going to zero, it's even strictly positive and differentiable.
I was actually wondering if such a function was possible. I guess you answered my question!
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Re: Naughty Functions
There was a function that my prof showed that was not well behaved anywhere and it was a summation. does anyone know an example?
 Yakk
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Re: Naughty Functions
samspotting wrote:There was a function that my prof showed that was not well behaved anywhere and it was a summation. does anyone know an example?
In what way was it not well behaved?
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Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
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