Mathematics - 2016-11-20 (page 4 of 9)

Ramanujan

15:05

OK,iam not getting

So for first digit 4 we get 0100

DHMO

yes

Ramanujan

Then for 2nd digit 3 which is 48

DHMO

no

Ramanujan

?

16×3=?

DHMO

the point is to treat each digit separately

4 is 0100
3 is 0011
2 is 0010
1 is 0001

join them together and you get 0001-0010-0011-0100 (hyphens just for readibility)

so you get 1001000110100

heather

15:09

okay, I think I understand integration by parts now

Ramanujan

16×3=48 so it is 32+16 so in binary …?

DHMO

@heather nice

@Ramanujan $48_{10} = 110000_{2}$

heather

and I woke up early this morning and taught myself about limits

so I think I can do this pretty well now.

DHMO

@heather self-learning!

what did you learn about limits?

Ramanujan

@DHMO OK,so I got that in base 10

That's the point?

heather

15:10

@DHMO, what they meant and how to calculate them

DHMO

@Ramanujan ??

heather

and about right and left limits

DHMO

@heather so you learnt the formal definition of limits?

heather

@DHMO, I don't know, I don't think so? when I said what they mean, I meant intuitively

DHMO

@heather I see. That's nice. Building a intuition is a good thing to do.

Would you evaluate $\lim\limits_{x\to1}x^3$?

heather

15:12

well, that'd be 1, right?

DHMO

yes

what about $\lim\limits_{x\to1}\dfrac{x^2-1}{x-1}$?

Ramanujan

@DHMO ?double limits?

DHMO

@Ramanujan ??

Ramanujan

1 min ago, by DHMO

Would you evaluate $\lim\limits_{x\to1}x^3$?

DHMO

@Ramanujan "double limits" means right limit and left limit?

heather

15:14

@DHMO 2, I think

DHMO

@heather why?

Ramanujan

Yes,2

Mahmoud

Hiiii

DHMO

@Mahmoud hi. how is your revision?

heather

@DHMO, well, you can rewrite the top as (x+1)(x-1) and then the x-1's cancel, so you are left with x+1, where you can just plug in 1, and then you have 2

DHMO

15:15

@heather brilliant.

Mahmoud

It seems like whatever I do isn't enough @DHMO

DHMO

@heather what about $\lim\limits_{x\to1}\dfrac{\sqrt x-1}{x-1}$?

@Mahmoud have you encountered difficult problems?

Ramanujan

@DHMO that's my favorite one!

Let's see if Heather will figure out with himself for first time

Mahmoud

@DHMO No, but I'm still horrible at proof-writing.

DHMO

@Mahmoud what is your major difficulty in proof-writing?

Ramanujan

15:19

@Mahmoud can I know in which class you are?

Mary Star

@DHMO @robjohn Could you take a look at the edit part of my question? math.stackexchange.com/questions/2022192/limit-of-sequence

heather

@DHMO I'm not sure if I did one step right, is it 1/2, or do...hmm, no I think I have to switch to the third type of limit there, so I have to evaluate the left and right integral, and then those both give 2, so I think the answer is 2.

Yeah, I think the answer is 2.

DHMO

@MaryStar $a_{n_{k_j}}$ lol

@heather integral?

Ramanujan

@heather nope

heather

@DHMO, lol, I meant limit

Ramanujan

15:21

@DHMO I doubt he can't do for very first

DHMO

@heather can you explain how you get 2?

@Ramanujan let her try again

Mary Star

@DHMO Is this wrong?

DHMO

@MaryStar no, just that the amount of subscripts is funny

heather

@DHMO, well, okay, I rationalized the numerator, so I got x-1 over (x-1)($\sqrt{x}$+1), and then the x-1's cancel, so I got 1 over $\sqrt{x}$+1 and then I wasn't sure what to do. Either evaluate it by just plugging in, in which case you get 1/2, or by using left and right limits, because it is a different type of limit now. If you do the second way, you evaluate the right limit of 1 over sqrt x +1 and get two, and then left limit of 1 over sqrt x +1 and get 2, so it must be defined giving 2.

Astyx

@MaryStar what is it you're asking ?

Mahmoud

15:25

@DHMO I don't even know how to start the proof.

DHMO

@heather you're almost there. How does the right limit of $\dfrac1{\sqrt{x}+1}$ get 2?

Mary Star

I want to show that a sequence $(a_n)_{n=1}^{\infty}$ converges to $a\in \mathbb{R}$ iff each subsequence $(a_{n_k})_{k=1}^{\infty}$ of $(a_n)_{n=1}^{\infty}$ has a subsequence $(a_{n_{k_j}})_{j=1}^{\infty}$ that converges to $a$. @Astyx

DHMO

@Mahmoud well, then proof-writing isn't the problem. the problem is you don't know how to prove a statement

Can you tell me the definition of subset?

heather

@DHMO, well, we are basically taking sqrts of numbers just above 1, like 1.0001, and then adding 1. and this approaches 2 as x gets closer and closer to 1; I used a calculator to double check.

oh, wait, no that's wrong

DHMO

@heather you verified that $\sqrt x+1$ approaches $2$, but not $\dfrac1{\sqrt x+1}$

Astyx

15:27

@MaryStar I suggest showing the opposite

heather

@DHMO, okay, well then the right limit approaches 1/2, right?

DHMO

@heather yes

Mahmoud

A subset B of a set A is another set who"s elements are all contained in the set A. @DHMO ?

heather

and then I did the same thing wrong on the left side

so it would also be 1/2

DHMO

@Mahmoud you are not taught symbols?

heather

15:27

so the right answer would be 1/2? @DHMO

DHMO

@heather yes, so you knew how to do it all along

Mahmoud

$A \subset B$

heather

stupid errors haunt me wherever I go =P

DHMO

@Mahmoud the definition of subset using symbols

Mahmoud

Oh

DHMO

15:28

@heather what about $\dfrac{\sqrt[3]x-1}{x-1}$?

heather

what does x approach, @DHMO? 1 again?

DHMO

@heather oops, stupid mistakes on my account; yes.

heather

okay

Ramanujan

@heather @DHMO why I loved that problem was my teacher wrote denominator as (√x + 1 ) (√x - 1) and cancelled √x - 1 so by plugging in x we got 1/2

Mahmoud

$A \subset B=\{\forall x : x \in A \implies x \in B\}$

DHMO

15:30

@Mahmoud your curriculum does have a special way to avoid using $\forall$

Mary Star

@Astyx What do you mean?

DHMO

@Mahmoud no, that isn't correct.

Mahmoud

Now ?

Astyx

@MaryStar showing that a sequence does not converge to a iff each of its subsequence has no subsequence converging to a

DHMO

@Mahmoud still not correct

you used set comprehension notation

$A \subset B := \forall x : x \in A \implies x \in B$

Mary Star

15:33

@Astyx Is this what I shown at the direction $\Rightarrow$ ?

heather

@DHMO 1/2 again I think

DHMO

@heather why?

heather

well, same sort of thing, rationalized, canceled, got 1 over cube root of x +1, then the left and right limit, and it approaches 1/2 in both cases so the original is 1/2

did I get it wrong?

DHMO

@heather can you show the step of rationalization?

heather

yeah, sure I did the original times the cube root of x +1 over the cube root of x +1 and got x-1 over (x-1)(cube root of x +1)

Astyx

15:35

@MaryStar what I mean is that I would do the opposite of what you did : if $(a_n)$ converges to a then each of its subsequence converges to a, and then each subsequence oviously has a subsequence that converges to a, which prove $\implies$

Then suppose that $(a_n)$ does not converge to a, use what this means to create a subsequence that does not have any subsequence that converges to a

DHMO

@heather so $(\sqrt[3]x-1)(\sqrt[3]x+1) = x-1$?

heather

well, the two middle terms cancel out but I'm guessing i did the first term wrong, cube root of x * cube root of x $\neq$ x

Mahmoud

Let $E$ be an non-empty set, why is $E \times \emptyset =\emptyset$ ? @DHMO

heather

just put it in a calculator and it gave (cube root of x)^2

darn, so I did the whole thing wrong

DHMO

@Mahmoud use the definition of $\times$ and the definition of $\varnothing$

SteamyRoot

15:38

@Mahmoud can you give me an element in $E \times \emptyset$?

Mahmoud

No because there is no elements in $\emptyset$ to pair with @SteamyRoot

SteamyRoot

There you go.

DHMO

@Mahmoud so you know how to prove it intuitively. now can you write a formal proof?

Mahmoud

That's my issue @DHMO, maybe use the proof by contradiction ?

Mary Star

suppose that $(a_n)$ does not converge to a then there are at east two limit points, ie., two different subsequences converge to a different point, say a and b.
This means that the subsequence that converge to b does not have any subsequence that converge to a. Is this correct?

DHMO

15:39

@Mahmoud how would you proceed?

Can you recall the definitions of $\times$ and $\varnothing$?

Lozansky

@DHMO $\int sin^4 t \: dt$, Euler??

Mahmoud

Yes, $\emptyset$ is the set with no elements.

DHMO

@Mahmoud formally?

heather

@DHMO, I'm not completely sure how to solve that one then...I can't think of an algebraic manipulation that would work there.

DHMO

@heather do you know the difference of cubes?

@Lozansky that would be troublesome

consider using the double angle formula

heather

15:41

@DHMO, no, I don't, let me google it real quick

Mahmoud

Okay, $\emptyset=\{x| x \not= x\}$

DHMO

@Mahmoud brilliant.

now what is the definition of $\times$?

Evinda

Hello!!! We have that $c(x) \leq -c_0<0$.

I have shown that $u(x) \leq \frac{f(x_0)}{c(x_0)}$.

I think that it does not hold that $u(x) \leq \frac{\max f(x)}{-c_0}$ since $\frac{1}{c(x_0)} \geq \frac{1}{-c_0}$. Do you agree?

heather

@DHMO, okay, I've got some idea now I think

DHMO

@heather nice

Mahmoud

15:42

The set of all pairs $(x,y)$ such that $x\in E$ and $y\in \emptyset$

SteamyRoot

If you don't like cubes, just substitute them away :P

heather

@DHMO But maybe I don't understand it well enough, because I'm not sure how it helps

DHMO

@Mahmoud and formally?

@heather read the comment above you

Astyx

@MaryStar yes it is

DHMO

@heather I guess you would like to learn the substitution rule in limits

heather

15:44

@DHMO, that might be good, yeah. Okay, I'll look that up.

Evinda

Hi @Astyx
Did you see my question?

Astyx

No

Hi

Oh now I did

Evinda

We have that $c(x) \leq -c_0<0$.

I have shown that $u(x) \leq \frac{f(x_0)}{c(x_0)}$.

I think that it does not hold that $u(x) \leq \frac{\max f(x)}{-c_0}$ since $\frac{1}{c(x_0)} \geq \frac{1}{-c_0}$. Do you agree?

Mahmoud

$E \times \emptyset=\{(x,y)|x\in E \land y\in \emptyset\}$

But $y\in \emptyset \implies y\not=y$

Contradiction ?

DHMO

@Mahmoud therefore?

Mahmoud

15:46

So is $E \times \emptyset$ undefined or equal to $\emptyset$ ?

DHMO

@Mahmoud it isn't undefined

Mahmoud

But the contradiction ...

SteamyRoot

@Lozansky $\sin^4(t) = \frac{3 - 4\cos(2t) + \cos(4t)}{8}$

DHMO

you could argue that $\{x|x\ne x\}$ contains a contradiction as well? @Mahmoud

Astyx

@Evinda It does not mean it does not hold, only that what you have done is not sufficient to prove it (nor sufficient to disprove it)

Lozansky

15:48

@SteamyRoot Who remembers that?

:D

Astyx

@Lozansky don't you ?

SteamyRoot

@Lozansky You remember the simple formulas and use them to derive it :P

Mary Star

Ah ok... Thank you very much!! :-) @Astyx

Mahmoud

Oh that's right @DHMO

Astyx

@MaryStar My pleasure !

Use De Moivre to get it almost instantly

Mahmoud

15:49

It is the definition of the $\emptyset$ itself, thanks :D

DHMO

@Mahmoud you are welcome

so you learnt how to prove results using their very definitions

Mahmoud

The prof just wrote the definition down.

Evinda

@Astyx What other condition would be needed?

DHMO

@Mahmoud the point is that the condition would hold for no object

so that the set would contain no object

Mahmoud

Exactly @DHMO :)

Astyx

15:51

@Astyx such innequalities can never lead you to this result. But you could study $u(x)$ to get more info

DHMO

@Mahmoud an equivalent definition is of course $\varnothing := \{x|\bot\}$

Mahmoud

What is that symbol ?

Evinda

We have that $Lu=f(x) \text{ in } \Omega \\ u|_{\partial{\Omega}}=0$.
$L$ is an elliptic operator @Astyx

DHMO

@Mahmoud it means "contradiction", but you can just view it as the symbol for the truth value "false"

Astyx

And what do you want to prove ?

Mahmoud

15:54

Ok, @DHMO

Ramanujan

@Mahmoud can I know in which class you are?

DHMO

@Mahmoud Prove that there is no bijection between $X$ and $\mathscr P(X)$ where $X \ne \varnothing$.

heather

@DHMO I'm finding no explanation for this that makes sense.

Mahmoud

@Ramanujan I don't really known because I'm probably in a different country than you.

heather

just as a side question, is it okay to, for that limit, substitute 1 in for some spots there is x but not all of them? Or no, is that not okay?

Ramanujan

15:56

@Mahmoud are you from India?

SteamyRoot

In general, that's very dangerous

heather

@SteamyRoot, so...no?

Mahmoud

@DHMO What exactly is $\mathscr P(X)$

DHMO

@Mahmoud the power set of $X$

Mahmoud

@Ramanujan No.

DHMO

15:56

@heather that is not okay

Astyx

@heather no

heather

okay

Mahmoud

I know but how do we define it explicitly ? @DHMO

DHMO

@heather well, then don't use it?

Evinda

@Astyx That $\min \{ 0, \frac{\min_{\Omega} f(x)}{-c_0} \leq u(x) \leq \max \{ 0, \frac{\max_{\Omega} f(x)}{-c_0} \}$.

I tried to show the second inequality first.

If the maximum of u is achieved at the boundary, it is equal to 0.

Suppose that the maximum is achieved at some $x_0 \in \Omega$.

Then we have $Lu(x_0)=\sum_{i,j=1}^n a_{ij}u_{x_ix_j}(x_0)+cu(x_0)=f(x_0)$ and thus $c(x_0) u(x_0) \geq f(x_0)$.

From this we get that $u(x) \leq u(x_0) \leq \frac{f(x_0)}{c(x_0)} \leq \max_{\Omega} \frac{f(x)}{c(x)}$.

DHMO

15:57

@Mahmoud does your test include power sets?

Null

$\mathcal{P}(X)$ i saw more often

Balarka Sen

@Mahmoud The set of all subsets of $X$.

heather

@DHMO but then how do I solve that limit?

DHMO

@Null I see

Mahmoud

@DHMO Yes.

Evinda

15:57

But can we use somehow the fact that $c(x) \leq -c_0<0$ in $\overline{\Omega}$ ? @Astyx

DHMO

@heather substitute $x=u^3$

Mahmoud

Like the problem you saw yesterday.

Astyx

@Evinda Yes

DHMO

@Mahmoud $\mathscr P(X) := \{Y|Y\subseteq X\}$

Astyx

Because you have that these number are negative

DHMO

15:58

@Mahmoud the proof is quite beautiful

heather

@DHMO, which would be basically the cube root of 1^3, you mean? you can do that?

Mahmoud

Show me please !

DHMO

@heather I don't understand.

Null

would you prefer in multilineequations the "=" at the end of the lines or at the beginning?

DHMO

@Mahmoud proofwiki.org/wiki/Cantor%27s_Theorem

@Mahmoud This proves a stronger result, i.e. no surjection from set to power set

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