Mathematics - 2018-02-04 (page 2 of 5)

jublikon

11:05

Never mind. I got it now :)

Secret

11:47

[Random]

Update notation to :O:

$\underbrace{[...]}_{n} := [...]_n$

Therefore:

Let _ be the empty list

The sequence goes as follows:

_

$[0]_0$

$[0,1,2,...,n]_n$

$[0,1,2,..._{\text{Huge}}$

$[0,1,2,...]_{\omega}$

For a foundation where ordinals are defined:

$[0,1,2,...,\alpha]_{\alpha+1}$

$[0,1,2,...]_{\lambda}$

where $\lambda$ is a limit ordinal and $\alpha$ is any ordinal

Continue:

$[0,1,2,...]_m$

In general $[0,1,2,...]$ is a countable unbounded list. The subscript gives a notion on its "size" in terms of how many steps is needed before an entry arbitrary large is reached. In other words, the subscript gives a measure on how many steps before it effectively becomes something similar to a proper class in set theory

Alex K Chen

12:25

Of course not intended for anyone, but some guy needs to stop posting crackpot theories related to set theory in Mathematics chatroom.

2

Secret

I never said what I said is true. The above is more like a brainstorming process than an actual theory, thus it is not even wrong

Everybody in this chatroom knew what I wrote is legitimate and what is nonsensical rambles

otherwise why would I use a [random]

Alex K Chen

I never said my comment was intended to anyone. Also, I don't think anybody reads what someone writes. (Again not indended towards anyone)

Secret

Of course, whether all of the above "response" is visible to any of you and whether any of you care is not my concern. As long The Plan remains unhindered, those who choose does not have not have a choice

(To the whole chatroom) I "respond" because I know what I wrote is nonsense and you guys knew it, as long you know when I am sane and the flooding controls is intact, then it will be ok

someone

Hello anyone know how can 1.052 x 12.504 x 0.53 = 6.7280 ? The result should be 6.9717 which is way higher. I've seen this same calculation in a lot of lectures and in the chemistry book but what the heck?

Secret

Anyway, as a cautionary measure, better to once again reiterate that anything that follows after [random] does not necessary have to make sense. This is because they only make sense after an interaction with the community

Alex K Chen

12:35

Anyway, again not indended to anyone, this site is for generally discussing math, not crackpot theories and random thoughts which occurs to brain. Someone may open a new room for that. Of course, if getting people's attention is the actual motto, then opening a room and talking there would work ...

Secret

Traffic of a very narrow topic in a new room is too low to be of significance

but perhaps:

4 hours ago, by The Great Duck

stop wasting them away in chat threads nobody is gonna read

someone

Can someone tell me please why it is 6.7280 is it a mistake? It's driving me insane :)

Secret

I should figure out how to make those stream of consciousness more coherent

@someone We need to know where the 3 numbers came from first. What quantities are they representing

Alex K Chen

@Secret So why don't try thought dumping in The Nineteenth byte ? A lot of people frequent that room, so I think it's a fantastic idea !

Secret

uh, do you think the stuff I am talking about is even remotely related to programming and coding puzzles?

someone

12:39

@Secret It is just an example given to show how to calculate the number of significant figures in multiplication e.g 6.7208 = 6.7 which takes the least number of significant figures (0.53)

Alex K Chen

not related to math too. That's the point

someone

@Secret that's the exact snippet from the book i.imgur.com/HNrOre5.png

Secret

Yeah I think I agree with you: 1.052*12.504*0.53=6.97173024 which to 2 sig fig should be 7.0

someone

It is probably a mistake but the exact numbers are mentioned way often even in practice questions and power point slides :) so weird

Secret

you might want to check with your instructor. I cannot think of any round up nor round down that can produce 6.7208

someone

12:55

The book is insanely inconsistent in the second example of multiplication they did not round but only count the significant figures, but in subtraction they did in one example and not in the other smh.

Tuki

hmm $$ \int \int_D 1-6x^2y dA, \quad D=[0,2]\times [-1,1] $$

Secret

you want to convert that to a double integral?

Tuki

yes thats right

i don't understand the notation

$$\int \int_{-1}^{1} \int_{0}^{2}1-6x^2ydxdy$$ ??

also $[0,2]$ denotes closed interval ?

is it $[x]\times[y]$ ?

$0\le x \le 2$ and $-1 \le y \le 1$ is this same ?

Secret

Yup, square commonly denote that edge of the interval is closed

So basically you are integrating this function over a rectangular region

someone

@Tuki this explains it pretty well tutorial.math.lamar.edu/Classes/CalcIII/DoubleIntegrals.aspx see R=[a,b] x [c,d]

Tuki

13:04

yes this is exactly what i wanted thanks @someone

Secret

13:17

Just like how reality cannot handle infinite entities, the point is the chat will be plunged into unusability, and this is more than enough for the adversories to achieve their goal.

3

Alex K Chen: Yeah fine, I will deal with the flooding problem more seriously, so as not to accidentally flood someone else's screen

(uh, accidentally caught balarka and co in the middle of my rambling. I need to time more carefully since the chemistry event means he and his group are to be avoided from getting involved in these rambles which sounds like anti-normie stuff)

point is, I ramble a lot less when the topology people are here now, thanks to them helping me to better grasps the subject so I can actually said something more sensible

I will soon figure out manifolds, and after that, GR will no longer be so unintuitive to me

IP Address

hi

I have question

How would I prove the sequence i,-1,-i,1 where the nth term is given by i^n

Would I use proof by mathematical induction?

Anderson Felipe Viveiros

13:35

@IPAddress Just use that $i^{4n+k}=i^{4n}i^k=(i^4)^ni^k=1^ni^k=i^k$

How do I solve this homework-like question?

Let $V$ be a vector space with inner product $\langle \, ,\,\rangle$ over the field $\Bbb K$ ($\Bbb R$ or $\Bbb C$). Suppose that the linear operator $T:V\to V$ is NORMAL, (i.e. there exists $T^\ast$ adjoint operator of $T$ and $T\circ T^\ast=T^\ast\circ T$).

If $T(v)=\lambda v\neq 0$, $T(u)=\mu u\neq 0$ and $\lambda\neq \mu\in \Bbb K$, then prove that $\langle u,v\rangle =0$.

IP Address

@AndersonFelipeViveiros thanks

Anderson Felipe Viveiros

@ip

* @IPAddress You're welcome

Tuki

14:19

so this notation $$ \int \int_D f(x,y) dA \quad D=[a,b]\times[c,d]=\int_{-1}^{1}(\int_{0}^{2}f(x,y)dy)dx $$ ??

hmm is false and i cannot remove this

@Shobhit you there ?

Shobhit

yes, but i dont have much time, exam tommorow @Tuki

Tuki

ok

Dattier

15:27

Hi,

$a$ is an integer, such write in basis $b$ with only $\{0,1\}$ and the size is $n<b$.
Is it true that $a^2$ is a palindrom in basis $b$ ?

sorry a mistake

$a$ is an integer, such write in basis $b$ (is palindrom) with only $\{0,1\}$ and the size is $n<b$.
Is it true that $a^2$ is a palindrom in basis $b$ ?

William Oliver

why are bilinear maps represented as the tensor product of the dual space of vector spaces as opposed to just the tensor product of non dual vector spaces?

Isn't there an isomorphism between both? Is it just convention?

Silent

What guarantees strict inequality here: $$\frac1{(n+1)!}+\frac1{(n+2)!}+\frac1{(n+3)!}+\ldots < \frac 1{(n+1)!} \left(1+\frac 1{n+1}+\frac1{(n+1)^2}+\ldots\right)$$? I see why weak inequality should hold.

Akiva Weinberger

Put a space somewhere in there

Silent

Well, it looks good in main site!

Akiva Weinberger

LaTeX gets annoyed if it's too long without spaces, it puts a line break somewhere arbitrarily

0celo7

15:42

@AkivaWeinberger you mean chatjax

Abcd

@Secret What's "The Plan"?

Tuki

what latex editor you use ?

Akiva Weinberger

and in this case it seems to have put the line break between "\fra" and "c"

Silent

@AkivaWeinberger Thank you!

Akiva Weinberger

@Silent $\frac1{(n+2)!}= \frac1{(n+1)!}\frac1{n+2}<\frac1{(n+1)!}\frac1{n+1}$

and similarly for all later terms

Silent

15:45

@AkivaWeinberger Yes, but that guarantees that $a_n<b_n$, but all we know from that is $\lim_\limits {n\to \infty} a_n\le \lim_\limits {n\to \infty} b_n$, right?

Abcd

$\ {\left(\dfrac {1+i}{1-i}\right)}^m= 1$, find the least positive integral value of $m$.
Attempt:

Akiva Weinberger

@Silent Is $a_n$ the partial sum, or the term?

Silent

Partial sum. @AkivaWeinberger

Abcd

$\implies e^{mi\dfrac{\pi}{2}}=1 \implies \sin (\dfrac{m\pi}{2})= -i $

Akiva Weinberger

What we know is that $a_{n+1}-a_n>b_{n+1}-b_n$, in that case

which is stronger

Abcd

15:52

Which is not possible, hence there's no solution.

Akiva Weinberger

In fact, it's equivalent to $a_{n+1}-b_{n+1}>a_n-b_n$, if we want to go down this route

Abcd

But another method (rationalisation and checking) yields he right answer i.e. m =4.

Akiva Weinberger

and thus by induction $a_n-b_n>a_1-b_1$ by induction, and we know $a_1-b_1>0$

Abcd

Why is my 1st method incorect?

Akiva Weinberger

And so $\lim a_n-\lim b_n\ge a_1-b_1>0$ and so $\lim a_n-\lim b_n>0$

but that's all ugly

so let's do that again but with the terms

If $A_n>B_n$, then $\sum A_n>\sum B_n$

Silent

15:55

I think i got that!

You will have to use too much effort!

@AkivaWeinberger

Akiva Weinberger

and that's just 'cause $A_n-B_n>0$, and so $\sum(A_n-B_n)>0$ (since adding positive stuff can only get you positive stuff)

and so $\sum A_n>\sum B_n$.

But yeah, from the first perspective, the gap between the partial sums increases, so it can never go down to zero

Silent

you are persistent!

Akiva Weinberger

@Abcd Don't you want $\sin(m\pi/2)=0$?

And $\cos(m\pi/2)=1$

Abcd

@AkivaWeinberger well, no.

its 0.

Akiva Weinberger

$\cos(4\pi/2)=1$

You said $e^{mi\frac\pi2}=1$, right?

Abcd

15:59

But $\cos3\pi/2 = 0$

Akiva Weinberger

$e^{mi\frac\pi2}=\cos(m\frac\pi2)+i\sin(m\frac\pi2)$

and you want that to equal $1$

Abcd

@AkivaWeinberger Yeah.

And I took cos mpi/2 = 0

Akiva Weinberger

Why?

If $\cos(m\frac\pi2)+i\sin(m\frac\pi2)=1$, then

Abcd

Okay, so we have to consider two cases.

m is odd and m is even.

Akiva Weinberger

looking at the real parts, we want $\cos(m\frac\pi2)=1$

and looking at the imaginary parts, we want $\sin(m\frac\pi2)=0$

Abcd

16:02

Ah! Simple!

Thanks.

Akiva Weinberger

That's not the way I'd do that, anyway

Abcd

how'd you do it?

Akiva Weinberger

From $e^{mi\frac\pi2}=1$

We know that $e^{ix}=1$ when $x$ is a multiple of $2\pi$.

So $m\frac\pi2$ is a multiple of $2\pi$.

And that first happens when $m=4$.

Alternatively, of course, you can write $\frac{1+i}{1-i}=i$ through rationalization (or through the geometry of the complex plane)

(double-checking by seeing that $(1-i)i=1+i$)

Abcd

@AkivaWeinberger hmm, i mentioned that above.

0celo7

@Semiclassical Did you have a way to access Soviet Math Dokl translations? I can't find it in the transcript or my downloads anywhere but I need to access that paper we once talked about.

Akiva Weinberger

16:06

Sure

Tuki

if i have $\int \int_{D} xy^2dA$ and is bounded with $x=y^2,y=x^2$

Akiva Weinberger

If you do it through $e^{mi\frac\pi2}$, you can see from that that $m\frac\pi2$ needs to be a multiple of $2\pi$, which is how I'd do it.

Tuki

How do i define more definite bounds ?

Akiva Weinberger

Draw it

Tuki

yes i'm on it

Akiva Weinberger

16:09

$y=x^2$ and $x=y^2$ both contain the points $(0,0)$ and $(1,1)$

Tuki

now these two defines area

which bounds the $xy^2$

Akiva Weinberger

@Tuki Something like that

0celo7

@AkivaWeinberger that looks suspicious

Akiva Weinberger

So $x$ goes from what value from what value? (Not involving $y$)

Tuki

so i want to find out $\begin{cases} y^2-x=0 \\ y-x^2=0 \end{cases}$ ?

Akiva Weinberger

16:13

We already know that that's solve at $(0,0)$ and $(1,1)$

Tuki

true

Akiva Weinberger

From the picture, what range does $x$ go over?

Not involving $y$

Tuki

goes from 0 to 1 ?

and y goes from 0 to 1

if you mean these to dots that are points where these lines intersect ?

Akiva Weinberger

Yeah, $x$ goes from $0$ to $1$

I'm thinking of having $x$ as the outer limit and $y$ as the inner limit

so the range for $y$ can involve $x$

$\int_0^1\int_?^?(xy^2)dy~dx$

The inner limits can involve $x$

(If I had the limits the other way around, then it would be the reverse- $y$'s range wouldn't involve $x$ and $x$'s range would involve $y$)

Tuki

y goes from 0 to 1 also ?

but then this would be bounded by rectangle ?

Akiva Weinberger

16:20

@Tuki Here's a picture of what we're doing.

The inner integral finds the area of one of those tiny rectangles (something times $dx$).

The outer integral adds up the areas of all those tiny rectangles, integrating from $x=0$ to $x=1$.

Note that the inner integral only finds the area of one of those tiny rectangles, for any particular $x$.

Let's say, as an example, that $x=\frac14$. That corresponds to one of the tiny rectangles towards the left.

What would be the height of that tiny rectangle?

(The picture I drew loaded, right?)

Tuki

$y=x^2$

so $(\frac{1}{4})^2$

$=1/16$ ?

Akiva Weinberger

This picture is getting complicated

1/16 is the distance from the bottom of the rectangle to the x-axis

Tuki

i think i see where this is going

Akiva Weinberger

The top of the rectangle touches the curve $x=y^2$

Tuki

$$ \int_{0}^{1}(\int_{0}^{\sqrt{x}}xy^2dy)dx $$ ??

Akiva Weinberger

16:31

Not quite. You didn't use the $x=y^2$ curve

Tuki

but now it leaves variable

Akiva Weinberger

Hint: $x=y^2$ is the same curve as $y=\sqrt x$

Tuki

yes makes sense

but integrals are basically Riemann sums when $dx\rightarrow 0$ ?

Akiva Weinberger

Yeah

Tuki

the rectangle x values approaches zero

Akiva Weinberger

16:34

The bottom of the red rectangle I drew is $x^2$ from the $x$-axis. The top of the rectangle is how far from the $x$-axis?

Remember that it touches the $x=y^2$ curve, aka the $y=\sqrt x$ curve

Kasmir Khaan

@MatheinBoulomenos How did it go with your exam ?

Tuki

16:51

$\int_{0}^{1}\int_{y}^{1}e^{-2x^2}dxdy$

like in normal distribution

Akiva Weinberger

$$\int_0^1\int_{x^2}^{\sqrt x}xy^2~\mathrm dy~\mathrm dx$$

@Tuki

is the answer

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$Mathematics$ Mathematics