The idea here is to ask math questions that really make you think, but can be understood and solved by someone with even just high school level mathematics, as long as he is able to think it through enough. This way we can engage the entire board instead of just the minority who know a lot of number theory and geometry, as happened in the math challenge threads of the old days.
Here is the first question:
Construct a function, through any means possible, that is defined for every real number but is discontinuous at every point in [math] (0,1) [/math], but continuous everywhere else.
If you know analysis then adding a proof for why your function is satisfactory is a nice bonus, but not needed.
If you don't know rigorous calculus then just keep in mind that continuous means that there are no sudden jumps between points of the function.
Hint: Everywhere else includes 1 and 0.
There are obviously many possible solutions so try it even if someone else already gave an answer.
And if anyone has other problems that fit into the spirit of this Light Math Challenge then post them too.
Pic related is visual aid. It is called Thomae's function and if you are not familiar with the notion of continuity then keep in mind that Thomae's function is continuous at every rational point, but discontinuous at every irrational point. Pretty good.
Try harder my man. Your function is undefined (infinite) for x=1.
It has to be defined everywhere.
Jason Jones
>You mean a function that is discontinuous at rationals right?
Oh, if you meant my description of Thomae's function then yeah, I got it reversed. But the point still holds. Just an example of how discontinuous functions look.
Kevin Baker
Just hopping in to say that I like this thread OP. I'm sadly busy with other work but i might give it a try when im done later today
Samuel Hill
[math]f(x)lim_{k\rightarrow \infty}\prod_{i=1}^{k-1}\frac{x-i/k}{x-i/k}[/math] that was meant to be a product
Logan Cook
> Construct a function, through any means possible, that is defined for every real number but is discontinuous at every point in (0,1), but continuous everywhere else.
let f(x) be defined as:
the Dirichlet function from x = (0, 1), x everywhere else
Camden Flores
What about using some chaotic function in (0,1) like [math]f^{k}(x)[/math] for some fixed integer [math]k[/math] where f is the tent map or something like that
Nathan Rivera
Dirichlet's function is not continuous at 0 or at 1.
Dominic Adams
Isn't this super easy? The function f(x) is 0 for x in (-infty,0] and x in [0,infty); otherwise, if x is rational, it is 1; otherwise, it is -1.
If you approach one for the right then you are good to go.
But what happens if you approach it from the left?
Your function has the same problem. It is discontinuous at 0 and 1.
Think about it for a sec.
There is a reason I asked this. If it was THIS trivial then I wouldn't have bothered.
Benjamin Sanchez
Using squares, and math it is easy to construct using simple algebraic methods, infinite functions continuing forever that when condensed are discontinuous. Perhaps at every point.
Angel Peterson
not continuous at 0 and 1 tho
Easton Kelly
[math]f(x)= \begin{cases} \lim_{k\rightarrow\infty}\left(\lim_{j\rightarrow\infty}\left( \cos{\left(k!\pi x \right)} \right)^{2j}\right),& \text{if } 0
Lincoln Parker
I think you could do something like
in (0,1/2] f(x) = x if x is rational, 0 if x is irrational
in (1/2,1) f(x) = x if x is rational, 1 if x is irrational
outside of (0,1) f(x) = x
There is probably a slick answer that doesn't involve just frankensteining together functions though.
Dominic Morgan
I meant to put an else after that last part, but you get the idea. I got carried away with TeXing that thing.
Thomas Brooks
Yes, you're right, it is discontinuous there. I change my answer: there is no function satisfying your constraints. Suppose there was, and you can find a contradiction by using the triangle inequality.
Brayden Thomas
That looks nice but to be quite honest I have no idea at all on how to visualize it, let alone prove its continuity.
Could you explain why it works? Specially, show why it is continuous at 0 and 1.
Actually, such functions do exist.
is one of them.
And before making this thread I constructed a solution too, very similar to that poster but slightly different.
0.99999999.... is in the interval, yet if your function has to be undefined in that number, it would also be undefined at 1 since 0.9999....=1 thus there is no solution.
Oliver Gutierrez
Now that the first correct solution is on, I will share my original solution:
It's really going to bother me that I didn't just write it like this: [math]f(x)=\lim_{k\to\infty}\left\{ \begin{array}{lr} \left \lfloor{kx}\right \rfloor & : x \in \left(0,\frac{1}{2}\right) \\ \left \lfloor{k(x-1)}\right \rfloor & : x \in \left[\frac{1}{2},1\right) \\ 0 & : x \in \left(0,1\right)^c \end{array} \right.[/math]
Robert Sanders
What? Isn't that limit just unbounded?
Carson Bailey
I was thinking about it as the limit of the sequence [math]f_k[/math], but you're right that it doesn't work. How about this one? [math]k \in \mathbb{N} [/math] [math]f_k(x)=\left\{ \begin{array}{111} \left \lfloor{2^kx}\right \rfloor \bmod 2 & \text{if } x \in \left(0,1\right) \\ 0 & \text{if } x \le 0 \\ 1 & \text{if } x \ge 1 \end{array} \right.[/math] [math]f_k \rightarrow f[/math] I'm not sure if this is actually well-defined. Anyone know?
Justin Mitchell
in 0
Ryan Myers
Bernoulli inequality: prove it for alpha > -1
Michael King
f(x) on x=(0,1) = the total count of all odd perfect numbers in N f(x) everywhere else = 12
Ayden Morris
[math](1 + \alpha)(1 + \alpha)^n \geq (1 + \alpha)(1 + n\alpha)[/math] Divide through by [math]1 + \alpha[/math], which is possible since [math]1 + \alpha > 0[/math]. [math](1 + \alpha)^n \geq 1 + n\alpha[/math] Testing the base case, [math]n = 0[/math], it is obviously true. Assume it is true for some [math]n = k, k \in \mathbb{N}[/math], or [math](1 + \alpha)^k \geq 1 + k\alpha[/math].
Thus, the original inequality is proven by the principle of mathematical induction.
please forgive my tex mess, and please point out any mistakes
Isaac Rogers
really makes you think
Caleb Reed
Retard here who never did math Hows about 0 if x is irrational on the open interval (0,1) 1 everywhere else
Nicholas Evans
That function isn't continuous at 0 or 1, which was a condition specified in the OP.
Charles Allen
A nice question from Spivak's calc, chapter 3 that I think would be suitable for this thread.
Jackson Bell
Can't we just use the indicator function for rationals on [0,1]?
Brandon Howard
does it have to do with the fact that g(x) can be whatever for x
Owen Campbell
Nice try, troll...
Levi Parker
Why?
Camden Perry
by that I mean that let g(x) = f(x) for x>=0, and for x
Easton Jenkins
here is an interesting one
Elijah Bailey
Because that's clearly wrong. Any injective f won't be representable as g(|x|) for ANY g...
Hunter Myers
>injective f wat
Adam Rodriguez
Ah nevermind. Didn't read the part that required f to be even. In that case it's trivially true...
Adam Howard
What the fuck?
Can you Tex that?
Levi Roberts
Suppose there is a Set E in [math]\mathbb{R^3}[/math] such that for any x in [math]\mathbb{R^3}[/math] and any r in [math] \mathbb{R^3}[/math] there exists a point z in a ball of radius r around x [math] B_r(z) [/math] such that [math]B_r(z) \cap B_2r(x) \cap E = \0[/math]
