and we have a bijection between numbers with $n$ "digits" and $2^n$

the result follows from the definition of $2^\omega$ which is $\displaystyle \cup_{n \in \Bbb N} 2^n$

how to deal with 0.11111... is left as an exercise to the reader

ugh these problems are so hard

@DHMO what does it mean again? [0,1)

[0,1] means

the zero is in but the 1 not?

$\{0\} \cup (0,1)$

or, $\{x \in \mathbb{R} : 0 \leq x < 1\}$

ah ok

@saturatedexpo yes

can french, or maybe linguistic be applied mathematics? (on a university)

Any good topology puzzles

submitted a second paper and preparing a third!

[direct sum]en.wikipedia.org/wiki/…

guys would someone help me figure out the differnce between internal and external direct sum?

It says they're isomorphic, but are they the same set? It seems to me like they are. If I take the direct sum of $V_1$ and $V_2$, and they are isomorphic to some $V$, is it true that if $V_1$ and $V_2$ are both submodules/substructures of $V$, then it's an internal direct sum? Is every internal direct sum an external direct sum?

@Jake1234 math.stackexchange.com/questions/928524/… i hope some answer clears it up.

(i myself have no knowledge about this topic)

Thanks, yeah I looked at that already.. sadly not really.

internal tho sounds like, the sum lies in something.

just saying

Yeah I understand it like that, just not completely sure I get it precisely enough.

to me the second answer of the thread says it. but i can't imagine it "really".

"The biggest distinction I've seen is that if $A,B \subset G $, and $A\times B \cong G$, we say $G$ is the internal direct product of $A$ and $B$. However, if $A,B$ are not subgroups of $G$ (rather, they are isomorphic to direct factors of $G$), we would say $G \cong A \times B$ is an external direct product. "

seems like you study linear algebra?

This is mostly for a basic course in ring/module theory

i would ask my professor to give you a good example on that. professors are good in that ;)

from the answer above: suppose A and B are sets of numbers and G is a set of alphabetics (with an infinite alphabet), then we'd have an external product

(but only if G is isomorph to $A\times B$ )

[title]math.ncku.edu.tw/~fjmliou/advcal/sumvspace.pdf I think this explains it for me well enough.

Internal product is actually what it 'generates' in the given space.

well then feel free to try to explain it to me :)

(then you understand)

@ForeverMozart do you mind talking about what topic?

I think there might be some slight differences depending on the writer, but it looks to me like: Internal sum of two sub-structures is defined for two sub structures that are 'independent', and it's all their combinations. External sum are the actual tuples from the cartesian product of the sub-structures. If the internal sum of two sub-structures is defined, it's isomorphic to the external sum, hence naming it 'internal direct sum'.

@saturatedexpo in topology

@ForeverMozart That was the funniest moment in all of Family Guy to me for some reason, really caught me off guard the first time

@Jake1234 So could you make a small example? With specific sets and groups.

There's one in [direct sum]en.wikipedia.org/wiki/Direct_sum

for $\mathbb{Z}_6$, it's an internal direct sum of those two sets.

But for example you can take the subspace $V= \{ (0,0), (0,1) \}$ of $\mathbb{Z}_2 \times \mathbb{Z}_2$, take the external direct sum of $V$ and $V$, and you get something isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$, but there's not an internal direct sum defined for $V$ and $V$. (I think... might be nice if someone who understands this stuff checked it out xD.. I think I'll just ask someone in school to make sure).

Or if it is defined, it's still just $V$, not isomorphic to the external direct sum of $V$ and $V$.

@Jake1234 i think you don't fully understand (me neither), so yes go ahead and simply ask someone who does.

reciting definitions doesnt count :P

@Jake1234 youtube.com/watch?v=M67ajwjNsBE and youtube.com/watch?v=NSO5OM6EskQ (especially the theorem) made it for me clearer, as a total newbee to the topic

@Kaumudi dont ask to ask just ask :)

hey guys

If a system of equations is dependent when an equation is linear combination of two other equations, why is the system of equations ac=6, ad=4, bc=9, bd=6 still dependent?

Is nobody here??

can someone help me with my q

@Kaumudi please simply say your function

it's just system of equations....

yes, but its super hard to give you advice

and which level are you? in university or highschool?

because in highschool you can get away with lots of things xd

does monotony say something to you?

my claim is, that the deriviative is monotonly decreasing.

if you show that, you are basicly done.

well the normal function goes first up, then down

so the derivative has to go always down

depends really on how rigorous you like it

well, the derivative takes only at one point 0, the maximum of the function

that means

before -0.5 the function rises and is therefore not at a maximum

y

well

no LOCAL minimum

but the global mimimum would be -$\infty$

@WidowMaven The actual definition of an independent equation is more general.

what value do you get for f(1000)?

what value for f(1 gazillion)?

so the what is the limit of f(x) as $x\to\infty$?

yes

since it can't really get smaller then $-\infty$ you have your global minimum.

you don't even need to check the other side^^

what is the answer?

Oh wait @saturatedexpo You still here? D:

@Kaumudi then the answer is wrong

google.de/…

@ShinKim yyyy :D

@Kaumudi ah...

look at the graph of google

it explains a lot if you zoom out^^

not really possible if you don't know about poles.

@Kaumudi I think you should first find the singular points

Since the function is piece-wisely defined

Then work out the derivatives in each intervals then find max and min values

@Kaumudi simply think of, where does the denominator get 0

That'll be a tidy way to handle such problems

@Kaumudi What is the function again? $$f(x)=\dfrac{x^2+x+1}{x^2+x-1}?$$

yes @Brody

@saturatedexpo @Kaumudi Depending on level/context, knowing about vertical and horizontal asymptotes should be sufficient for now

@Kaumudi What's your task... Graphing, finding minima/maxima, range, etc.?

He wants to find $\mathrm{ran}(f)$.

My bad

im no expert, but isnt the range then sinply everything?

(except 1 and - something)

That's co-domain @saturatedexpo

@Kaumudi Range cannot be determined analytically, for now you need the graph of the function, which will tell all

And the asymptotes along with minima/maxima will get you there

well give me x, where f(x)=0 @Kaumudi

(hint: you can't)

Which also means the range of a function is the image of its domain.

@Kaumudi i mean, you can't give me x, such that f(x)=0. Thefore all of $\mathbb{R}$ CAN'T be the range.

@Kaumudi actually, you might be able to determine range w/o graphing, but this also requires finding asymptotes and extrema (minima/maxima)

@Brody how to you call "vertical" asymtotes?

@Kaumudi Wherever the bottom polynomial is zero, the function will have a "hole" or a vertical asymptote. This is where $f(x)$ is undefined

@Kaumudi plugin a big number in f. that will be close enough to the asymtote. then plugin a really small number (-1000)

@Kaumudi no, the function will have either a hole or a vertical asymptote

i always wondered about highschool math. it doesn't get one anywhere :/

@Kaumudi holes appear when the function evaluates (zero)/(zero), and vertical asymptotes where we have (non-zero number)/(zero)

I think that argument only works in a high school math class

@Kaumudi Are you a high school student

@Kaumudi yep, they'll be needed. just worry about vertical asymptotes for now

@ShinKim a slightly better question might be: do you want to just survive highschool^^

@Kaumudi congratulations :)

@Kaumudi yes, those as well, along with maxima/minima. this will divide your function into "sections" where you can check how your function behaves there

@Kaumudi perhaps, I don't remember too well

hmm, I think @saturatedexpo resolved this question already :p

@Kaumudi you only had to mind other points IF the domain would be restricted, like the interval between -4 and 4

then -4 and 4 would be interesting too

well, think of $f(x)=x^2$, and we are only allowed to plug in values of $|x|\leq 2$

@Kaumudi then obviuosly the maximum would be different than if all numbers where allowed.

@Kaumudi look at the leading coefficients of the numerator and denominator. if they equal, you'll have a hor. asym. at y = (top leading coeff)/(bottom leading coeff)

@Kaumudi but in your excercise that's not the case. Thus all holes/poles/asymtotes are interesting, but apart from that not much.

@Kaumudi in this case, the leading coefficient up top is 1, and so is the bottom. so we have a horizontal asym. at $y=\frac{1}{1}=1$

@Kaumudi what is $(x^2+x)/(x^2-x)$ for really big numbers?

nope

just use your calculator ;)

yes it is

@Kaumudi When you plug in $x$ of really really large magnitude, the leading terms in the top and bottom polynomials dominate, so the function behaves like (leading term)/(leading term) for these big $x$

@Kaumudi In other words, $(x^2+x+1)/(x^2+x-1)$ basically acts like $(x^2)/(x^2)$ for $x$ of large magnitude, because the effect from squaring a big number is much much greater than adding that number and adding/subtracting $x$, we can ignore those non-leading terms

(1000+10)/(1000-10)=1010/990~1

oh

well that was for x^3

well for x^2: 110/90=11/9~1

@Kaumudi that even for relative small values, the leading coefficient is the most important

@Brody what does $a\sim b$ mean in terms of relations i guess? Does that mean that: there exists some function that maps a to b?

yes, it's impossible

@Kaumudi I don't know of any general methods that work for any function. There is a definition for range, but I don't think it's can be used to compute actual results

@Kaumudi purplemath.com/modules/fcns2.htm

you can take LIMITs tho

the limit of f(x) as x APPROACHES the hole-value

well, you can say where the hole is located at, which involves a $y$-value. so in some sense, yes you can. the function is just not defined there

a quick way would be: where is the denominator zero (0)

or other way: at what values of x is f(x) not defined.

but that would be the domain

simply skip such an assignment in examen, its tedious.

especially without graph/calc, no thanks haha

@Kaumudi you got to ask yourselves, are you a human or a calculator.

and if an exam asks me to be a calculator, then no thanks

@Kaumudi You're right. Consider $y=\dfrac{x^2}{x}$. It looks just like $y=x$, but there is a "hole" at $(0,0)$. The range is $\mathbb{R}\setminus\{0\}$

@Kaumudi sorry, I might have said something wrong earlier. if the top and bottom polynomials have the same degree then there is a horizontal asymptote at (leading coefficient)/(leading coefficient)

how do you call that?

@Kaumudi Is it alright if we use this to talk from now on? $$f(x)=\dfrac{P(x)}{Q(x)}$$ where $P(x),Q(x)$ are polynomials in $x$

@Kaumudi where $P(x)$ and $Q(x)$ share a common zero, we have a "hole". where $P(x)$ is non-zero but $Q(x)$ is zero, we have a vertical asymptote

For example, look at $\dfrac{x^2-1}{x^2-2x+1}$

factoring, we get $\dfrac{(x+1)(x-1)}{(x-1)^2}$

y

@Kaumudi we're just talking about vertical asymptotes vs. holes right now (don't worry about horizontals or other stuff)

yes

@Kaumudi other stuff comes later. we find potential vertical asymptotes and holes by inspecting the zeroes of $P(x)$ and $Q(x)$

because horzontal asymtotes are when P(x)/Q(x) has a limit

Hello, please can i prove that an even and convex function is increasing ?

it has not much to do with holes/poles

@Vrouvrou yes you can prove it, where is your problem?

@Kaumudi other way around. the vertical asymptote "overwrites" the hole

i don't know how to start @saturatedexpo

if you plug in big numbers and P/Q has a limit thats not infinity, we talk about horizontal asmtote

@Kaumudi sorry I misleaded you with an earlier comment. From $$f(x)=\dfrac{x^2-1}{x^2-2x+1}=\dfrac{(x+1)(x-1)}{(x-1)^2}$$ so the $0/0$ at $x=1$ would indicate a "hole"

@Kaumudi but then we simplify to check for vertical asymptotes, it becomes $$\dfrac{x+1}{x-1}$$ giving $2/0$ (undefined) so now we see it's a vertical asymptote at $x=1$, not a hole

@Vrouvrou please state the definition of convex, that YOU know about.

@Kaumudi Yes, that's the general process. Check for common factors in $P(x)$ and $Q(x)$, then simplify. Vertical asymptotes override/overwrite holes

@Kaumudi the material is not too bad, I'm just not presenting this very clearly

@saturatedexpo f is convex then $f(t x+(1-t)y)\leq t f(x)+(1-t) f(y) , \forall x,y\in \mthbb{R},\forall t\in [0,1]$

@Kaumudi what is the horizontal asymptote is for the above rational function?

You're right. Finding those is much easier

@Vrouvrou and do you know how a convex function looks like?

leading coefficients

@Kaumudi Your answer is correct. The horizontal asymptote is at $y=1$. Determining horizontal asymptotes is faster/easier than vertical asymptotes and holes

$\frac{5x^3+...}{4x^3+...}´=\frac{5}{4}$

@saturatedexpo yes like here en.wikipedia.org/wiki/Convex_function

@Kaumudi Don't confuse them. The vertical asymptote is at $x=1$ (it's vertical), and the horizontal asymptote is at $y=1$ (flat)

@Vrouvrou proofwiki.org/wiki/…

@Kaumudi Another quick example for reference, $y=\frac{1}{x}$ has $x=0$ and $y=0$ as asymptotes, the former vertical and the latter horizontal

@saturatedexpo but i need if f is convex and even then f is increasing , not it's derivative

@Vrouvrou IFF is much much stronger

@Kaumudi Yep. Sorry for confusing things at moments, but I hope the net result was positive :p

@Vrouvrou and if you know its deriviate is increasing you only need a small step

@Kaumudi Perhaps later on we'll talk about the third and final class of asymptotes, oblique asymptotes (just so you know there's a little more, but don't worry about them for now)

@Brody can you think of a function that has a growing slope but stagnates?

@saturatedexpo stagnates?

@Brody the derivative grows, but the function has a limit it approaches

@saturatedexpo i don't know this step

@saturatedexpo what sort of limit? a limit at infinity or a point?

Vrouvrou does your nickname mean anything? I remember some french player of a game I used to play using it.

@Brody at infinity it has a finite value as a limit. but that would contradict an everygrowing slope or?