Mathematics - 2021-02-23 (page 1 of 2)

12:18 AM

@LeakyNun looking at another source, one systematic approach seems to be to compute $\gcd(x^p-x,f)$ in $\mathbb{F}_p[x]$

Leaky Nun

aha

@Semiclassical do you know that $\Bbb F_p^\times$ is cyclic?

Semiclassical

didn't

i mean, my knowledge of finite fields is pretty minimal

the context of said source was trying to count roots of polynomial f in Z/(p^t) for t>1

Leaky Nun

@Semiclassical anyway what that means is that for every $p$ there is some $a$ such that the powers of $a$ modulo $p$ goes through every nonzero modulo

e.g. for $5$, the powers of $2$ go like 2, 4, 3, 1

so 2 is a generator mod 5

Semiclassical

ah. that sounds like fermat's little theorem

Leaky Nun

Fermat's little theorem says that a^4 = 1 mod 5 for every a coprime to 5

so the 4th term must always be 1

but it doesn't say that you must go through every possible residue before reaching 1

Semiclassical

12:22 AM

i guess that's not enough to figure out which a are generators, yeah

Leaky Nun

so e.g. if I started with 4, I would go 4, 1, 4, 1

Semiclassical

right

it constrains things but doesn't determine them

Leaky Nun

so, how this relates to your question is this trick

Semiclassical

i do vaguely remember this stuff, it's just been a while

Leaky Nun

$x^2+x+1 = 0$ iff ($x \ne 1$ and $x^3 = 1$)

because $x^3-1 = (x^2+x+1)(x-1)$

Semiclassical

12:23 AM

of course

Leaky Nun

so suppose such $x$ exists

let $a$ be a generator mod $p$

Semiclassical

i think i begin to see, yeah

Leaky Nun

then $x = a^k$ for some $k$ between $1$ and $p-2$

the upper bound is because of Fermat's last theorem

Semiclassical

right

so $a^{3k}=1$

Leaky Nun

then $x^3=1$ means $a^{3k}=1$, so after you went through 3k steps of the cycle you must reach 1

but you only reach 1 every p-1 steps

(by Fermat little theorem + counting)

Semiclassical

12:26 AM

so p-1 has to be a multiple of 3

Leaky Nun

exactly

and for the converse, if p-1 is a multiple of 3, then $a^{(p-1)/3}$ is a solution

Semiclassical

i'm guessing p=3 is a weird exception because p-1 is already smaller than 3

Leaky Nun

of course this depends on you knowing a generator

Semiclassical

oh. p=3 means x=a^k for some k between 1 and 1, so x=a

Leaky Nun

p=3 is really weird

4 mins ago, by Leaky Nun

$x^2+x+1 = 0$ iff ($x \ne 1$ and $x^3 = 1$)

this doesn't even hold if $p=3$

Semiclassical

12:27 AM

right

Leaky Nun

because $x^2+x+1 = (x-1)^2$

Semiclassical

x=1 is a solution regardless

Leaky Nun

so you get a repeated root $x=1$

Semiclassical

wacky

so posing this as x^3=1 is a good idea from the "multiplicative group" perspective

Leaky Nun

right

Semiclassical

12:29 AM

even though x^2+x+1 is what's irreducible

i guess that already points to why p=3 is weird: x^2+x+1 simply stops being irreducible

Leaky Nun

yeah

Semiclassical

i wonder. are there any other p for which x^2+x+1 is reducible? seems like it shouldn't be

Leaky Nun

exactly for those p for which x^2+x+1 has a solution

Semiclassical

oh

i guess p=3 isn't that weird then

well. still weird because it's a perfect square

Leaky Nun

over 7, we know 2 is a solution, so x^2+x+1 = (x-2)(x-4)

Semiclassical

12:31 AM

but still reducible

right

Leaky Nun

the weirdness of p=3 is that it does not factorize into distinct factors

Semiclassical

right. huzzah for repeated roots

Leaky Nun

what we call separable

Semiclassical

okay, dinnertime

BigSocks

c*ck*d by a physics b**mer that puts his wikipedia in his math se bio

-1

Q: Is this the right notion of "relatively arithmetical"?

Say I want to tease out an arithmetical property from the ether: I get it's $\Pi^0_n$ $(\Sigma^0_n)$, and I get some indices for programs that compute structures (their (atomic) diagrams) with that property. So lets say I want to check for a certain property among those guys, and I manage to show...

clearly does not know logic but is is unafraid of this small inconvenience

Ted Shifrin

12:42 AM

@LeakyNun Do we know where yet?

anakhro

1:00 AM

@BigSocks tbh I have no idea what you are asking in that question, either.

Ted Shifrin

Nor I.

BigSocks

@anakhro that is ok

Mike Miller

Eh, nevermind. Not here.

BigSocks

the difference is that you two don't go on my question and pretend to know better and flame me and downvote me

anakhro

1:17 AM

I don't think he was flaming you.

BigSocks

flaming is a strong word, but he definitely was talking down to me

anakhro

I do agree that stackexchange somewhat suffers from a "let's say it in a way that makes the asker feel like they are an idiot" vibe.

Not sure how to fix that. In a society where people enjoy not only being right, but thinking someone else has their pants on backwards, it's kind of a large undertaking.

BigSocks

yeah absolutely that is a big part of the problem here. The objective this guy had was never to clarify anything, and you can tell because he has no clue what I am talking about even when cleared up. really sad

hilarious how one of the biggest dudes in computability shows up and gives him the same link and suddenly it's standard and ok

anakhro

But you do have an issue on your end, too. Be the change you see in the world.

BigSocks

what? that I could phrase things better? yeah I guess. There is always things you could improve for sure

I would rather have that problem $\aleph_0$ times than have his though

anakhro

1:23 AM

There is also how you treat your interlocutor.

BigSocks

elaborate

anakhro

"The "Huh?" after "...this set of indices I was talking about" betrays your lack of knowledge" <--- basically you just said "you dun goofed" to him, which doesn't exactly further your friendship.

BigSocks

Sure- that is definitely the one moment that I was not 100% level in the interaction, I can agree with that. OTOH, I wasn't even rude- if you know a smidge of computability you know what I said in that sentence. To not know is not bad at all. He did in fact goofed and I tried to put it in the nicest way.

compared to how people treat eachother in this chat and on this website... I always put my best foot forward and never stooped to his level. I answered each of his attitude-laden doubts pretty much calmly. I then diagnosed his actual issue. I would love it if people spoke to me that way wh…

And if I am being frank, I would not like to be friends with a guy that shows up to questions he cannot contribute more than nitpicks to.

anakhro

If you think you put your best foot forward, then I think you should take a step back from the conversation. This wasn't the only time you said something that only widened the valley of misunderstanding and discord.

BigSocks

absolutely- he has continued to comment and I am just not gunna respond to him

anakhro

1:30 AM

I guess that's a little bit more mature than making derisive comments.

user2103480

1:44 AM

@BigSocks man the SE logicians are so good

BigSocks

they really are

Noah is amazing

user2103480

but tbh the question could be formulated in a clearer way

BigSocks

absolutely, I basically wrote it as fast as I could and hit enter, no doubt about that

but is it wrong even a little? no

user2103480

makes the speed at which noah deciphered and answered it even more impressive :D

BigSocks

hyea

Sarabsri mt

1:47 AM

If T is a vector here

Semiclassical

ow

Sarabsri mt

then , have I written its components down there correct.

Semiclassical

with respect to what coordinate system?

Can't specify components without saying what the x-direction and y-direction are

Sarabsri mt

Ok

So , there was a force mg acting downwards and T was perpendicular to it.

So , angle between two vectors is 90 degree

So , i took its components

Semiclassical

Sure. So if your axes are +x to the right and +y upwards

then the components would be Tx = T and Ty=0

Sarabsri mt

1:50 AM

T cos 90 (+ve x axis )and T sin 90(-ve y axis)

is this right

@Semiclassical ok

Semiclassical

that'd work. I prefer to take x,y to fit with the right-hand rule usually

Sarabsri mt

T cos 90 should be up or down.

Either up or down. Both are perfect I think.

As you say sir

Semiclassical

Either will work, but you really do need to specify the coordinates

i mean, "specify" is fancy

Sarabsri mt

@Semiclassical Right hand rule? I don’t know about it.

Semiclassical

what I really mean is "draw coordinate axes"

Sarabsri mt

1:51 AM

@Semiclassical ok

@Semiclassical k

Semiclassical

well, when you think of x and y, you usually think of +y as counter-clockwise from +x

rather than clockwise

Sarabsri mt

K

Got it

Thanks a lot.

Semiclassical

that's a right-handed coordinate system. it doesn't matter much in first-semester physics

though it does show up when talking about rotation. but it's a big deal in second-semester physics (electromagnetism)

Sarabsri mt

Ohk

Semiclassical

if cross products show up, this matters. if not, it really doesn't

Sarabsri mt

1:54 AM

Ok

yeah then it will

Right

@Semiclassical one more Q

1min,I’ll paste it

Semiclassical

sure

Sarabsri mt

Why is it

that T sin 90 is right and T cos 90 is down or up

Semiclassical

it's definition of sine and cosine, really

Sarabsri mt

If i take 90 as theta

then , perpendicular would be hypotenuse

I can’t find it then

Semiclassical

if you're having trouble visualizing, take an angle slightly smaller than 90 degrees

Sarabsri mt

2:00 AM

Ok

one more thing. Sorry .

Semiclassical

if the hypotenuse is T, then sin(theta) = opp/hyp = Ty/T -> opposite = T*sin(theta). as theta->90 degrees, then Ty->T

and thus sin(90) = 1

similarly Tx->0 as theta->90 deg, so cos(90)=0

Sarabsri mt

Ok ok

user2103480

@Thorgott did your prof cover smooth submanifolds of R^n in analysis 2

Sarabsri mt

@Semiclassical ok

Semiclassical

you can also think of the unit circle.

Sarabsri mt

2:02 AM

So , how did you know before starting the Q where angle 90 should be.S

since you had 3 choices right

Semiclassical

i had to make a choice about my coordinates

@user2103480 yes

harsh

@Semiclassical Ok. But why specifically choose that. The angle 90 in my diagram is what you chose right sir

user2103480

2:03 AM

general topology as well?

Thorgott

nah

Sarabsri mt

@Semiclassical but how from coordinates were you so sure ?

Thorgott

I mean, we defined the term topology, but that was all

user2103480

ah okay

Thorgott

and we only defined it for the sake of remarking the metric space have a natural topology

we just did general metric space theory

Semiclassical

2:04 AM

i mean, you're the one who said 90 degrees. if T is supposed to be at a 90 degree angle, then it has to be with respect to the downward direction

Thorgott

and basic normed spaces stuff

Clarinetist

Got a question: suppose I have three vectors $(a, b, c), (d, e, f), (g, h, i)$. I want to solve for such values such these three vectors have unit length, are (pairwise) orthogonal, and satisfy
$$\begin{align}
&(a+c)(d+f) + (a+b)(d+e) + (b+c)(e+f) = 0 \\
&(a+c)(g+i) + (a+b)(g+h) + (b+c)(h+i) = 0 \\
&(d+f)(g+i) + (d+e)(g+h) + (e+f)(h+i) = 0
\end{align}$$
Besides just plopping this in Mathematica, is there a methodical way of doing this?

Semiclassical

but you could just as well take the angle to be 0 with respect to +x (to the right)

Clarinetist

(in $\mathbb{R}^3$)

Sarabsri mt

@Semiclassical Ok. Sir

Semiclassical

2:05 AM

in which case it's Tx=T cos(0)=T and Ty=T*sin(0)=0

Sarabsri mt

Ohk.

user2103480

we did a good overview of general topology (compactness, hausdorffness, closure/interior, connectedness in arbitrary topological spaces) but ended at implicit function theorem, banach fixed point and lagrange multipliers I think

Semiclassical

@Clarinetist i feel like this demands a matrix formulation

Sarabsri mt

@Semiclassical got it.

@Semiclassical thanks again.

Ted Shifrin

That's a sea of letters, @Clarinet. Did it come from something about the actual vectors?

Sarabsri mt

2:06 AM

Have a great day @Semiclassical

Semiclassical

you too

Ted Shifrin

In other words, is it meaningful?

Semiclassical

the labeling here is also not especially helpful

Thorgott

ah, so you didn't do any measure theory?

user2103480

normed spaces too, and for some reason I think derivatives in such a general form that it might as well couldve been frechet derivatives

Clarinetist

2:08 AM

Well, $\{(a, b, c), (d, e, f), (g, h, i)\}$ should be orthonormal. Other than that, if we consider the matrix
$$\mathbf{A} = \begin{bmatrix}
a + c & a + b & b + c \\
d + f & d + e & e + f \\
g + i & g + h & h + i
\end{bmatrix}$$
the above conditions are what make the off-diagonal entries of $\mathbf{A}\mathbf{A}^{T}$ equal $0$.

Semiclassical

that's a lot nicer to look at :P

user2103480

except that we assumed finite dimensional normed vector space

@Thorgott nope, that was analysis 3

Semiclassical

So $A=U+U^\top$ where $U$ is orthogonal

user2103480

measure theory and submanifolds of R^n, with stokes at the end

Semiclassical

wait, no

I'm being silly

user2103480

2:09 AM

@Thorgott what the heck did you cover in analysis 3 then

functional analysis or some fourier/distribution stuff?

sounds excessive

Ted Shifrin

What are those conditions in terms of the actual vectors, not their damn coordinates?

If no answer, I'm totally not interested.

Semiclassical

$A=(I+S)U$ where $S=\begin{pmatrix} 0 & 1 & 0\\0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}$

Thorgott

nah, we also have a separate measure/integration theory lecture (together with a complex analysis lecture, both of them are small, they make up for analysis 3), but we already constructed Lebesgue measure at the end of ana2

Clarinetist

Oh, I see what you're saying...

I'll have to give that some thought. It would be *so* nice if I could find a way to do that...

Semiclassical

where $U$ is orthogonal

user2103480

2:11 AM

ah okay we went through the whole tour of lebesgue measure, fubini tonelli and the likes

Semiclassical

So you're looking for orthogonal matrices $U$ such that $A^\top A$ is diagonal when $A=(I+S)U$.

user2103480

but tbh to me this whole stuff was a bit quick. The exams were generous so grades didnt suffer at all

Semiclassical

hmm, maybe that should be $A=U(I+S)$

user2103480

@Thorgott I've progressed to 1.8 speed now

Thorgott

great

2x is endgame

user2103480

2:14 AM

2x is impossible I've tried that

Semiclassical

@Clarinetist to clarify what I'm on about: $$A=\begin{bmatrix} a + c & a + b & b + c \\ d + f & d + e & e + f \\ g + i & g + h & h + i \end{bmatrix}=\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} + \begin{bmatrix} c & a & b \\ f & d & e \\ i & g & h \end{bmatrix}$$

user2103480

the words become unintelligible at times

Semiclassical

The first matrix is some orthogonal matrix $U$.

Thorgott

depends on the prof, I guess

BigSocks

it is more for your subconscious. so you can have unknown knowns

hmm maybe I should do that while I sleep

user 85795

2:16 AM

Yeah, speed kills.

Semiclassical

the second is just $U$ with columns permuted. so it can be written as $$\begin{bmatrix} c & a & b \\ f & d & e \\ i & g & h \end{bmatrix} = \begin{bmatrix} a & b & c \\ d & d & f \\ g & h & i \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}$$

Clarinetist

I really wish people would teach undergraduate-level probability with vectors, rather just sticking with scalars for everything. That would probably make my life easier with this question.

Semiclassical

Hence, $A=U+US=U(I+S)$. Therefore $AA^\top =U(I+S)(I+S)^\top U^\top$

@clarinest rip

Clarinetist

Oh, how interesting

Semiclassical

nicely, $(I+S)(I+S)^\top) = (I+S)(I+S^{-1})=I+S+S^{-1}+I$

not sure how helpful that actually is :P

Ted Shifrin

2:18 AM

@Clarinetist generally, stat majors know no math.

Semiclassical

but, we ultimately want $AA^\top = U(I+S)(I+S)^\top U^\top = D$ for diagonal $D$

so $(I+S)(I+S)^\top = U^\top D U$

Clarinetist

I don't deny that, unfortunately. Didn't realize how watered-down undergrad-level stats was until my MS core classes, and the MS core was with linear algebra and calc assumed.

user 85795

Yet, stats is math.

Semiclassical

stats can be made mathematical. that doesn't mean it's always true :P

Ted Shifrin

They teach it with high school math, not university math.

Clarinetist

2:20 AM

Lol, if you're talking freshman-level stats, that stuff is extremely useless, yet taught so often

Semiclassical

$$(I+S)(I+S)^\top = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix}=\begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix}=I +uu^\top$$

Clarinetist

I've TAd and tutored for a few of those classes. The only people who have an inkling of what is going on are those who know a bit of calculus.

Semiclassical

where $u=(1,1,1)$

Ted Shifrin

A bit of calculus is still high school.

Semiclassical

ahah. and the eigenvalues of $I+uu^\top$ are $1,1,2$

Ted Shifrin

2:22 AM

Pay attention to Semiclassic.

Thorgott

r8 this diagram

Semiclassical

so I think that means $D$ has diagonal entries 1,1,2 in some order

i guess the point now is to find some orthogonal matrix U. any rotation or permutation of such will preserve this property

so...just find the change of basis which makes $I+uu^\top$ diagonal.

and there's a pretty obvious basis for that :P

Clarinetist

Thanks for your help! Don't give me any more hints. Believe it or not, I haven't seen change of basis stuff since... 2012

Semiclassical

lol, okay

hmm, i think i may have made an error somewhere anyways

user2103480

@Thorgott d h u d h

Clarinetist

2:27 AM

No problem, I'll figure it out :)

Semiclassical

kk

Thorgott

h u h

Clarinetist

Stats is just odd at the undergraduate level.

You have one class which a lot of non-math majors (or some non-stats math majors) have to take, requiring only some pre-calculus.

You then have another class often taught at the sophomore level, requiring Calc. II as a prerequisite (not requiring any knowledge from the lower-level class), often taken only by stats or related (e.g., actuarial science) majors.

Semiclassical

i did use a statistical argument earlier, albeit a pretty obvious one

suppose you're measuring values $X_1,X_2,\cdots,X_n$

and then compute $X_2-X_1,X_3-X_2,\cdots,X_{n}-X_{n-1}$

assuming that the X's are all iid, then the variance of the differences are twice the original variances

moreover, it means that successive X's are negatively correlated

which is simple but neat

Clarinetist

I'm going to have a hard time teaching undergraduates probability if the first thing that comes to mind when computing a variance for me now is is "$X \in L^2$"

Semiclassical

2:33 AM

so statistically you tend to see oscillations when you take first differences

hah

i had in mind $X$ as a Gaussian in my head, not that it matters for the moments

yeah, i sorted my error. basically, dont' believe me when I stated the eigenvalues of $I+uu^\top$.

BigSocks

@Thorgott huh pretty good

Semiclassical

check that yourself and then realize why it's easy

BigSocks

@Thorgott you already did the thing

Thorgott

comedy has been solved

user 85795

comedy of errors?

Thorgott

2:37 AM

I have a diagram in this write-up that's twice as wide as the page

think I need to fix that

user2103480

@Thorgott mood

Clarinetist

Well, this question has done me some good... I opened Mathematica up for the first time, and realized I need to brush up on a ton of linear algebra

(The department chair was a Maple user in my undergrad.)

Semiclassical

shouldn't need to do mathematica, tho

Clarinetist

Oh yeah, I get that. But at least it gave me an excuse to finally use it.

Semiclassical

the annoying bit here is that the three vectors are highly non-unique

Clarinetist

2:42 AM

Do people still use Maple these days? Lol.

Semiclassical

any orthogonal transformation (including non-proper) will give you another set

user 85795

3:59 AM

Fred Armisen Does Every North American Accent | Standup For Drummers | Netflix Is A Joke

►

Semiclassical

@TedShifrin there does end up being something nicely geometric about that ugly-looking linear algebra problem from earlier. it's equivalent to finding 3 vectors at 60 degree angles to each other

Ted Shifrin

You mean 120?

So this should be seen somehow in terms of the vectors without coordinates.

Although starting with an orthonormal basis I don’t see it yet.

user 85795

4:17 AM

have you adjusted to the accent shift from Georgia to southern cal dr?

Semiclassical

@TedShifrin no, 60

direction cosines are 1/2 not -1/2

Clarinetist

So I've got to ask, considering that I will likely not pursue a PhD in pure/applied math...

Let's say you're a professor applying for tenure and promotion. How in the world do you get enough papers published so as to help your portfolio? Stats people have it easy because you can basically work with anyone who needs stats help and get listed as an author, but it seems a LOT more difficult in pure and applied math, pure math especially.

Ted Shifrin

4:40 AM

@Semiclassical Ah, right, 3D. So an equilateral pyramid.

Semiclassical

right

@Clarinetist what was the initial context for your problem, anyways

ery cerious matherfunker

8:38 AM

I have become a noob after a car accident

I cannot understand why regular number is divisible by 2^n and 2^m where n and m is positive interger

can any expert number theorist explain this

what is this theorem called?

Leaky Nun

9:08 AM

what is a regular number?

ery cerious matherfunker

9:49 AM

finite decimal number

ok I think first I figure out the proof and if I can't then I will ask it here

maths student

Each week a shop sells 5, 56-inch televisions and 10, 25-inch televisions. The sale prices vary from time to time, but in any single week the prices remain fixed. The price of 56-inch televisions has a mean of 25000 Rs and a standard deviation of 2900 Rs. The mean and standard deviation for the 25-inch televisions are 11000 and 600 Rs respectively. Assume the the prices are normally distributed with means and standard deviations given above.

What is the probability that in a given week the total sales of the company will exceed 191050 Rs. Go up to 4 decimals

Leaky Nun

@eryceriousmatherfunker your question doesn't make sense

what are n and m

could you give an example?

ery cerious matherfunker

10:06 AM

@LeakyNun well a/b=finite decimal number when b=2^m×5^n where a and b=/=0 are interger and m and n are positive interger

it is number theory i guess

Leaky Nun

well you wrote 2^n and 2^m

ery cerious matherfunker

hehe may be due to my brain damage

Leaky Nun

@eryceriousmatherfunker so are you trying to understand why a regular number can be written as a/(2^m x 5^n), or why every number of the form a/(2^m x 5^n) is regular?

ery cerious matherfunker

@LeakyNun yes capt

Leaky Nun

I said "or"

I'm asking you

which one of the two

ery cerious matherfunker

10:18 AM

first one

Leaky Nun

ok

so let x be a regular number

do you know that multiplying by 10 shifts the number leftwards with respect to the decimal point? e.g. 3.456 x 10 = 34.56

ery cerious matherfunker

I should have said finite decimal number instead of regular number

@LeakyNun yes

Leaky Nun

so since there are finitely many digits after the decimal points, 10^k x is an integer for some k

ok?

ery cerious matherfunker

ok

Leaky Nun

let's call the integer c

so 10^k x = c

so x = c/10^k = c/(2^k x 5^k)

maths student

10:27 AM

Each week a shop sells 5, 56-inch televisions and 10, 25-inch televisions. The sale prices vary from time to time, but in any single week the prices remain fixed. The price of 56-inch televisions has a mean of 25000 Rs and a standard deviation of 2900 Rs. The mean and standard deviation for the 25-inch televisions are 11000 and 600 Rs respectively. Assume the the prices are normally distributed with means and standard deviations given above.

What is the probability that in a given week the total sales of the company will exceed 191050 Rs. Go up to 4 decimals

@LeakyNun I tried to define variable X and Y for 56 inch and 25 inch televisons.

So by given condition we need to find P(5X+10Y > 191050)

ery cerious matherfunker

$a_1×10^{-1}+a_2×10^{-2}+...+a_n10^{-n}$

a_n is int for n={1,2,...,n}

Leaky Nun

@mathsstudent you need to how scalar multiplication of normal distributions work, as well as addition of two independent normal distributions

@eryceriousmatherfunker which is the form you seek

maths student

I want to know whether approach is correct or not ?

@LeakyNun

Leaky Nun

yes

ery cerious matherfunker

@LeakyNun yes I get the idea

now I am doing it in another form

it is nearly complete

Leaky Nun

10:33 AM

great

ery cerious matherfunker

oh so it is that there are many possible common multiples and canceling them out you get integer

so out goal is to make numerator a int

denominator 2^n×5^m

get it thanks

also apologies for ambiguous statements , haven't sleep for 3 days

the m and n might not be same because of different exponent

user 85795

@eryceriousmatherfunker not sleeping for 3 days will not help you in math

Astyx

^

Leaky Nun

maybe maths isn't the best activity to undertake in the occasion of a deprivation of sleep for a triduum

user 85795

^

maths student

10:43 AM

@LeakyNun just check whether this is correct or not

Leaky Nun

in which country do you write 125000 as 1,25,000?

you're missing the normal symbobls

user 85795

india

Leaky Nun

it's N(25000,2900^2) not just (25000,2900^2)

unless that's what you've been taught

also the second line should read 5X not X

unless that's what they teach you

ery cerious matherfunker

that hindu number system

Leaky Nun

and for Y you didn't square the s.d.

ery cerious matherfunker

10:47 AM

God bless you guys. Keep doing kind work

maths student

Somewhere I directly converted it into standard deviation as variance is preety large number.

@LeakyNun Approach is correct right ?

user 85795

@eryceriousmatherfunker try to get some well deserved sleep

maths student

@user85795 you can also check it.

whether I am going in right direction or not ?

Leaky Nun

@mathsstudent yes

user 85795

@mathsstudent Leaky is doing a fine job :-)

maths student

10:50 AM

thanks @LeakyNun

user 85795

Indian numbering system

The Indian numbering system is used in the Indian subcontinent (Bangladesh, Bhutan, India, Maldives, Nepal, Pakistan and Sri Lanka) to express large numbers. The terms lakh (1,00,000) and crore (1,00,00,000) are the most commonly used terms (even in English, such as in a local variety called Indian English) to express large numbers in the system. == System == There are words for numbers larger than 1 crore as well, but these are not commonly used and are unfamiliar to most. These include 1 arab (equal to 100 crore or 1 billion), 1 kharab (equal to 100 arab or 100 billion), 1 nil (sometime...

Leaky Nun

ok thanks

user 85795

np

porridgemathematics

if $\mu(k) > 0$ and $\mu(k) \rightarrow 0$ decreasingly in $k = 1,2,3...$, is it possible for $\frac{1}{n^2} \sum_{k=1}^n k \mu(k) $ to not go to $0$?

Astyx

no

porridgemathematics

11:00 AM

oh wait. i am an idiot, is it simply because $\sum_{k=1}^n k\mu(k) \leq C * \sum_{k=1}^n k$ for some constant $C$

Astyx

take n_0 and separate the sum in $\sum_{k=1}^{n_0}$ and $\sum_{k=n_0}^{n}$

no, because that converges to C when divided by $n^2$

porridgemathematics

oh right

Astyx

The point is you can take C to be as close to 0 as you want

porridgemathematics

ah yes I see, and we can prove that by separating and setting $n_0$ arbitrarily large?

Astyx

yes

porridgemathematics

11:02 AM

thanks!

alexmesa

Posting my problem here in hopes it gets some attention: math.stackexchange.com/questions/4035718/…

User 873110

1:26 PM

Suppose that we are given a finite collection $\Sigma$ of compact curves on a surface
M such that all intersections are transverse. If
M has a Riemannian metric, clearly cut and paste leaves the total length of $\Sigma$
unchanged. But rounding corners, if done carefully, strictly reduces the total
length of $\Sigma$.

What is the meaning of these lines? Any reference will be helpful.

Mike Miller

2:22 PM

Shouldn't you have a faculty member to talk about this stuff with? You have a lot of questions.

Akiva Weinberger

Cut an equilateral triangle into three pieces such that all are the same shape (similar), but two are the same size (congruent) and the third is a different size

Leaky Nun

2:33 PM

@AkivaWeinberger now what

Akiva Weinberger

You mutilated a perfectly good equilateral triangle, congratulations

Leaky Nun

thanks

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$Mathematics$ Mathematics