Light Math Challenge Thread #1

>>8740449
Shameless bump

[math]f(x)lim_{k\rightarrow \infty}\sum_{i=1}^{k-1}\frac{x-i/k}{x-i/k}[/math]

You mean a function that is discontinuous at rationals right?

>>8740500
No.

The question is find a function that is discontinuous everywhere in (0,1). Rationals AND Irrationals.

But continuous everywhere else.

>>8740494
http://www.wolframalpha.com/input/?i=lim+as+k+approaches+infinity+of+the+sum+from+i%3D1+to+k+of+(1+-+i%2Fk)%2F(1+-+i%2Fk)

Try harder my man. Your function is undefined (infinite) for x=1.

It has to be defined everywhere.

>>8740500
>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.

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

>>8740494
>>8740508
[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

> 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

>>8740449
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

>>8740531
Dirichlet's function is not continuous at 0 or at 1.

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.

>>8740524
http://www.wolframalpha.com/input/?i=Graph+f(x)+%3D+lim+as+k+approaches+infinity+of+the+product+from+i%3D1+to+(k-1)+of+(x+-+i%2Fk)%2F(x+-+i%2Fk)

Check a bit deeper.

>>8740531
Your function is then not continuous on 1 or 0.

For example:

If you approach one for the right then you are good to go.

But what happens if you approach it from the left?

>>8740538
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.

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.

>>8740538
not continuous at 0 and 1 tho

[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<x<1 \\
1
\end{cases}
[/math]

>>8740449
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.

>>8740564
I meant to put an else after that last part, but you get the idea. I got carried away with TeXing that thing.

>>8740549
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.

>>8740564
>>8740567

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.

>>8740580
Actually, such functions do exist.

>>8740566 is one of them.

And before making this thread I constructed a solution too, very similar to that poster but slightly different.

But there are many more solutions.

>>8740589
>Could you explain why it works? Specially, show why it is continuous at 0 and 1.
https://en.wikipedia.org/wiki/Nowhere_continuous_function#Dirichlet_function

>>8740449
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.

Now that the first correct solution is on, I will share my original solution:

[math]
f(x) =
\left\{
\begin{array}{lll}
0 & \mbox{if } x \in (0,1)^c \\
|x-1| & \mbox{if } x \in [ \frac{1}{2} , 1) \cap \mathbb{Q} \\
-|x-1| & \mbox{if } x \in [ \frac{1}{2} , 1) \cap \mathbb{I} \\
|x| & \mbox{if } x \in (0, \frac{1}{2}) \cap \mathbb{Q} \\
-|x| & \mbox{if } x \in (0, \frac{1}{2}) \cap \mathbb{i} \\

\end{array}
\right.
[/math]

I am sure there are smarter ones out there.

function that is the dirichlet function on (0,1) and 2 at x=0 and x=1

>>8740599
You are either bad at memes or bad at math.

>>8740596
>https://en.wikipedia.org/wiki/Nowhere_continuous_function#Dirichlet_function

If that is just the dirichlet function then wrong.

Discontinuous at 0 and 1, and the constraints of the problem say that it has to be continuous at those points.

[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(1-x)}\right \rfloor & : x \in \left[\frac{1}{2},1\right) \\
0 & : x \in \left(0,1\right)^c
\end{array}
\right.

[/math]

Graph [math] {\left \lfloor \frac{x}{y} \right \rfloor} = {0} [/math]

>>8740716
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]

>>8740737
What? Isn't that limit just unbounded?

>>8740836
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?

>>8740449
in 0<x<1:
{
1 for rationals
0 for irrationals
}
in any other area on the real line:
y=x

File: hm.png (2KB, 246x27px) Image search: [Google]

2KB, 246x27px

Bernoulli inequality: prove it for alpha > -1

>>8740449
f(x) on x=(0,1) = the total count of all odd perfect numbers in N
f(x) everywhere else = 12

>>8741233
[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].

[math](1 + \alpha)^{k + 1} = (1 + \alpha)(1 + \alpha)^k \geq (1 + \alpha)(1 + k\alpha) = 1 + k\alpha + \alpha + k\alpha^2 = 1 + k\alpha^2 + (k + 1)\alpha \geq 1 + (k + 1)\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

File: 1489093663073.jpg (66KB, 1170x653px) Image search: [Google]

66KB, 1170x653px

>>8740599
really makes you think

>>8740449
Retard here who never did math
Hows about
0 if x is irrational on the open interval (0,1)
1 everywhere else

>>8742053
That function isn't continuous at 0 or 1, which was a condition specified in the OP.

File: spivak lmct.png (10KB, 464x37px) Image search: [Google]

10KB, 464x37px

A nice question from Spivak's calc, chapter 3 that I think would be suitable for this thread.

>>8740449
Can't we just use the indicator function for rationals on [0,1]?

>>8742212
does it have to do with the fact that g(x) can be whatever for x<0?

>>8742212
Nice try, troll...

>>8742242
Why?

>>8742233
by that I mean that let g(x) = f(x) for x>=0, and for x<0 let it be whatever.

File: IMG_7082.jpg (2MB, 4032x3024px) Image search: [Google]

2MB, 4032x3024px

here is an interesting one

>>8742247
Because that's clearly wrong. Any injective f won't be representable as g(|x|) for ANY g...

>>8742273
>injective f
wat

>>8742273
Ah nevermind. Didn't read the part that required f to be even. In that case it's trivially true...

>>8742265
What the fuck?

Can you Tex that?

>>8742301

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]

Show that measure of E = 0