@KajHansen is there at all any rigorously studied concept of finding the piecewise constant function that when subtracted from a piecewise continuous function results in a continuous function?

I don't know @TheGreatDuck

@TheGreatDuck I think it might not always be possible

how so?

I can generate digits of $\pi$ with nothing more than coin flips and a huge amount of spare time

if, for example, it's discontinuous at $-1/n$ for all $n$ as well as at $0$, and the "jumps" are the same as the $x$-coordinate

hey guys I need help, this notation always confuses me

The leftmost and rightmost pieces would have to differ from each other by the harmonic series

ok so?

So there would be no piecewise constant function to add to it to make it continuous

you have yet to show why that would be the case...

Because after making it continuous the leftmost and rightmost pieces would have to differ from each other by the harmonic series, i.e., infinity!

@CausingUnderflowsEverywhere You know how to enable LaTeX, right?

a^3*b^3 \obolus (a^2*b^2)

but then the function you claim is an example is unbounded

therefore it is not piecewise continuous

the bracket eh.

Try $$\frac{a^3 \cdot b^3}{a^2 \cdot b^2}$$

an interesting though, sure.

@TheGreatDuck Not the original function

uh yeah

it is

just what you get after making it continuous

The pieces aren't flat

They're sloped

We just need piecewise continuity here

are you sure the limit would exist for such a function?

My professor simplifies the top using the bottom though, as if the brackets were not there. is \obolus (a^2*b^2) the same as \obolus a^2 \obolus b^2 ?

that sounds like an issue of heavy oscillation.

cause I see what you mean now

x*floor(1/x) is a decent example.

I was just about to say, x*floor(1/x)

Yeah

Thanks Kaj by the way! :D

Yes @CausingUnderflows

wolframalpha.com/input/?i=x*floor(1%2Fx) here's the graph

interesting.

(1/6) = (1/2)/3

and give me one moment

@CausingUnderflowsEverywhere 1/(a^2 b^2) = 1/a^2 * 1/b^2, yes

I got it @TheGreatDuck

I have a graphing app

@CausingUnderflowsEverywhere Example: 100÷(2*5)=(100÷2)÷5, as you can check

Instead of dividing by 10 you can divide by 2 and then divide by 5

I thought that was the example one gave in an answer

regardless

I am pretty sure the limit anywhere around 0 for that function is undefined

just like how sin(1/x) has undefined limits

wait whats the answer

= 10

100 * 0.4 = 40 or something

uuuh

yeah

thanks I gotta head in now, Ill jump to my last message tomorrow to review this

@TheGreatDuck $\lim_{x\to0}x\lfloor1/x\rfloor=1$ I think

.....

no

its undefined

best we can assume is 1

but even if not at 0

I don't think so, look at the graph

anything around that region is undefined limits

No it doesn't

Squeeze theorem

for instance 0.1

what's the limit at 0.1?

Sure, at 0.1 there's no limit, but that's irrelevant. @TheGreatDuck, it's possible for a function to be discontinuous everywhere except for at one point, even!

Continuous at only one point.

ummm

(That's not this)

dude

I'm surprise you haven't heard of it

I said Piecewise continuous

Right. That's not this.

But I brought it up because it shows why discontinuity at 0.1 has nothing to do with discontinuity at 0

then why are you using it as an example?

but dude

Look:

I said

"is there at all any rigorously studied concept of finding the piecewise constant function that when subtracted from a piecewise continuous function results in a continuous function?"

and as I kept saying, you're example is meaningless in proving that cannot be found

So? Is this function not piecewise continuous?

how can you say it is?

$$\begin{cases}x\lfloor\frac1x\rfloor,&x\ne0\\1,&x=0\end{cases}$$

to deal with the 0 problem

Now are you happy?

It's just a bunch of lines

That should count as piecewise continuous

do you even know what piecewise continuous means?

no it doesnt

Do you require finitely many pieces??

you're not even making sense

piecewise continuity means the right and left hand limits exist everywhere

They do!

you just said the limits at 0.1 dont exist

make up your mind

Oh, limits

I thought you just meant the limit

Yes, the right- and left-hand limits exist there

but not the limit

umm.... how so?

there's oscillation...

(because the right- and left-hand limits don't equal each other)

yeah, obviously

but how do either of them exist.

@TheGreatDuck The left-hand limit is $0$. The right-hand limit is $0.9$.

Look at the graph.

sorry

0.01

stupid scale

XD

At $0.01$, the left-hand limit is $0$ and the right-hand limit is $0.99$

that doesn't even make sense at all

which function are you graphing again?

cause there isn't even anything near 0 at 0.01

Oh! Sorry!

what function is that?

The LH limit is 1!

$x\lfloor\frac1x\rfloor$

I meant 1, not 0

you can zoom in close enough to make it out?

Yeah. Also I changed the vertical scale

oh

well on the wolram alpha there is no zoom

Hm

so it looks about as dense as sin(1/x)

In any case it's just a lot of lines

that get steeper and steeper

yeah, but that doesn't always mean a limit exists

:p

and also why exactly does that not work with a piecewise constant function anyway?

ah i see

@TheGreatDuck Because, if $g(x)$ was the "corrected" continuous version, what would the difference between $g(2)$ and $g(-2)$ be

depends on how you connect them

should be finite

since the graph descreases

and then increases

The size of each jump is $1/x$, right?

ok?

(Or the difference between $g(-2)$ and $g(0)$, doesn't matter)

it would be infinity

So the total sum of all of the jumps is $\infty$

alright fine fair enough

you cannot make it continuous

but isn't the asymptotic point in a way a point of continuity? Both sides are approaching the same thing?

I guess, if you make the codomain $\Bbb R\cup\{-\infty\}$ it would be continuous

yeah Idk really

but assuming it's left as $\Bbb R$ we can't get it defined everywhere

i assumed a function that messy were continuous

how about I say this then

What?

getting there...

instead of piecewise continuous functions all multivariate continuous compositions of continuous functions with functions of the form f(floor(g(x))) such that f and g are continuous.

and boy was that a mouthful.

:p

That would be piecewise constant, I think

(so subtract it from itself to get a continuous function)

umm

x*floor(x) is not piecewise constanct

That's not of the form f(floor(g(x))), though

I think you want f(x)*floor(g(x))

no

you didnt read my post

I think I misinterpreted something, yeah

I said all compositions of continuous functions with that form

h(j(x),f(floor(g(x))))

Ohh

although with any number of f terms...

Got it.

So what were you trying to do with this new class of functions

same thing...

i was just restricting it

But if xfloor(1/x) counts

then doesn't that mean that this doesn't work in general either

is 1/x continuous...?

Ohh, oh. True.

got you in a bind now, didn't i?

wow, it's way past DogAteMy's bedtime!

if you come up with a counter-example would you let me know?

'Tis. I need to go to bed.

@TedShifrin We were discussing how there is no piecewise-constant function that can be subtracted from $x\lfloor1/x\rfloor$ to make it continuous

and I think, possibly because of my tiredness, it went on for a lot longer than it needed to

you can't make a function with jump discontinuities continuous by adding a piecewise continuous function

(Corrected)

oh, piecewise-constant

i.e., step function

yeah

now I need to think slightly more

His definition of piecewise continuous was left and right limits exist everywhere

Night, DogAteMy :P

G'night.

Hasta mañana.

@AkivaWeinberger that's the only definition. They're all equivalent.

Still can't work, @TheGreatDuck. You can't fix both endpoints' jumps at once. And if you try to patch in the middle, you create another discontinuity.

huh? You're not making sense.

I always make sense.

i highly doubt that. nobody is perfect.

Well, fine, then.

so you're not going to explain yourself?

hello?

are you ok?

I had to tend to baking in the kitchen.

ah

I bet there's some yummy stuff in that oven

^^^

I thought maybe you got upset because I said nobody is perfect. XD

Well, I've always been perfect, but that's a different issue.

So how do you propose to add a function to make this continuous?

are you referring to x*floor(1/x) specifically still?

Yes.

cause that we did agree wouldn't work

ah ok

we had already switched subjects

That's what DogAteMy mentioned.

So what are you talking about now that I know nothing about?

instead of piecewise continuous functions all multivariate continuous compositions of continuous functions with functions of the form f(floor(g(x))) such that f and g are continuous.

i.e. like this h(j(x),f(floor(g(x))))

But a special case of that is the one we've already agreed can't work.

really?

Hi @Kaj

h(a,b) = a*b; j(x) = x; f(floor(g(x))) = floor(1/x)

Sure, @TheGreatDuck: Huh?

$j(x)=x$, $f(x)=x$, $g(x)=1/x$, $h(x,y)=xy$.

1/x isnt continuous

It sure is away from $0$. I don't care about $0$ at the moment.

Hey hey

The problems with your original question happen at every $1/n$. So I don't care about $0$.

no...

@Kaj: almond-orange cake :P

it's because the entire thing is the harmonic series

Huh?

You're the one who makes no sense.

the other guy said it wouldnt work as the "piecewise constant function" would have leaps totaling the harmonic series and would diverge.

I don't care about what the leaps total to.

then why exactly are you saying it wouldn't work?

I can't even make it continuous on $[1/3,1]$. Can you?

why wouldn't there be a piecewise constant function that when added to the function results in a continuous function for that period?

Well, you tell me one that works.

You have to define it on two intervals ... 1/3 to 1/2, 1/2 to 1 ... you do it.

-1.5 from [1/3 1/2)

and -1 from [1/2 to 1)

if I looked at it right, that should be continuous on 1/3 to 1

Not continuous from the right at 1/2 ... I haven't checked more.

well the left hand limit at 1/2 is 1

and the right hand limit is 0.5

if I lower both sides by 1 that has no effect

and then lowering the left side by 0.5 should make them match up

The original function is 1 at 1/2, but the original right hand limit was 1/2. So your function approaches -.5 from the left and has value 1-1=0 at 1/2 ...

ok so it's -1.5 from (1/3 to 1/2]

and -1 from (1/2 to 1]

i just did the limits wrong

OK, so I agree you can fix this. Now what happens when we go from 1/4 to 1/3? How're you going to make it work at 1/3?

just add the period -11/6 from (1/4 to 1/3]

and obviously this cant go on forever

cause then the resulting function diverges at 0

No, you have already defined it at 1/3, so you can't change it.

yeah, I fixed that

it's going to be right hand limits only being equal

Is there actually something you're getting at with this cause it feels like this will just keep repeating back...

Oh, I finally see what you guys are doing.

you misunderstood, didn't you?

:p

TheGreatDuck doesn't always make sense

only reason floor(1/x) doesn't work is cause the "continuous" version is asymptotic at 0

(and hence cannot be continuous)

It still is fine away from 0, but can't extend to 0, I see.

exactly

hence the restriction g(x) must be continuous in what I gave above

"h(j(x),f(floor(g(x))))"

j is continuous

I doubt you need continuity, though. If we were doing $\sqrt x[1/\sqrt x]$, your objection wouldn't hold any more. Interesting.

f is continuous

hello

well to be fair that would just generally not exist for negative values

Put in absolute values for ****'s sake.

or we would we would be dealing with the complex plane

I don't see any absolute values

hi @TedShifrin

hi @Ali ... you're either up all night or very early.

why didn't u say hi to me @Ali

@TedShifrin You might actually be interested in why I'm looking at this. It's a bit weird mind you and probably not useful. Mostly just something to mess around with.

@Mike: MikeMiller asked you to add another letter to your name.

@Mike_ because I don't think I've seen you before

Hi anyway

Is that better?

Did it not add the letter?

It won't change for a bit

ok I changed it to Mike El Jackson @TedShifrin

@TedShifrin I have been doing both things

LOL, great, @Ali :P

I had to finish some coding for uni

but then I found langs algebra in the library

what language u code in?

the language specified by the task

which is unfortunately matlab

I like matlab but not many people are using it in industry because of its cost :/ students get it free which is great

check out the juila language

you might like that (if u don't like maltab)

I wrote the whole thing in C but my instructor told me no

So here I am

@TedShifrin I'm just trying to think how to explain it clearly without confusing you. You know how in tables of integrals and things there's all those arbitrary constant parameters? Basically, I'm considering an operator where it is the same as the integral as has those identities but where those no longer need to be constants but can instead be any piecewise constant. No way for me to know for sure, but I have a pretty string feeling that continuous solutions to such an operator...

...are the same as solutions to the indefinite integral

Oh, you were talking about this weeks ago ... The only time that'll give you something reasonable is when the original function has discontinuities.

what makes you say that?

Because a continuous function can't have a discontinuous antiderivative

"I'm considering an operator where it is the same as the integral and has those identities but where those no longer need to be constants but can instead be any piecewise constant."

You could consider a family of operators for a given constant

and of course a function cannot have a discontinuous antiderivative.

@AliCaglayan not quite sure what you mean by that. Could you please elaborate a bit?

I think I give up for now.

fair enough

thought it might interest you

@TheGreatDuck It doesn't really make sense to talk about an operator and then be vague about a constant

I've heard that you tend to be the guy everyone always asks stuff to.

@AliCaglayan not quite sure how I was vague there...

Well if I have two functions A and B

and my magical operator O

OA doesn't always equal OB which doesn't really make it an operator

if A = B

well the integral varies by a constant does it not?

Yes

so why wouldn't mine vary by a piecewise constant

because then you don't have equality?

I probably didn't make that explicit

I am saying you have to specify which constant

@AliCaglayan that doesn't make any sense. By your reasoning the antiderivative is not an operator.

Well operators preserve equality

Your 'operator' does not

if D is the antiderivative

then DA =/= DB

exactly

cause DA = DA + 5

so D is not an operator

well then what is the integral classified as?

What I am saying is consider $D_s$ for a $s\in\Bbb R$

now you have an operator for each $s$

but im not choosing to consider a particular one...

where $s$ is the constant

The antiderivative is not an operator

then what is the antiderivative classified as?

anyway

I have to go

ok

still though. That's odd that the antiderivative sits in a bubble by itself

does it have no classification?

Well depends on what you mean by an antiderivative

there are many objects that differentiate to give a function

im taking about the general antiderivative

like how there's a general solution to a differential equation

Thinking of an antiderivative operator is the wrong way

yeah I see that now

Integration is more subtle than that

Believe me because I have done it before

alright fine

is indefinite integration/anti differentiation it's own class of thingy?

Its not really considered

ah fair enough

i guess one could think of it as a set in a weird way

if you want to learn why

take up a multivariable calculus course

you can get a sense for why integration does what it does

The fact that the antiderivative looks like an operator is kindof a special case of some more general theorem

(Stokes theorem) but do the multivariable calc first

oper8r

@KajHansen you are too modern

haha

the tangents in my brain have now scrambled that into oprah hater

:D