Mathematics - 2016-06-25 (page 1 of 4)

in principle, yeah. but i suspect you can truncate it eventually since 1) as a convergent series, its terms have to get smaller and smaller, and 2) the result has to be an integer

David Wheeler

Hello, all-I have a question..?

Semiclassical

so if you do enough terms you can presumably deduce what integer it actually is

Leaky Nun

@DavidWheeler tell us

David Wheeler

It has to do with this question on the site: math.stackexchange.com/questions/1836699/…

Leaky Nun

@Semiclassical What if it is $4+0.9+0.09+...$?

Semiclassical

12:07 AM

then you conclude that it's converging to 5?

"enough terms" is presumably not just the first one.

David Wheeler

One answer answers part c) by saying $K = \Bbb Q[X]/(X^4 - 2)$ is a real field, but I am not convinced of this.

Leaky Nun

@Semiclassical alright

What does $\mathbb Q[X]/(X^4-2)$ actually mean?

David Wheeler

The quotient of rational polynomials by the ideal generated by $X^4 - 2$

Leaky Nun

never mind

David Wheeler

I understand the coset of $X$ is "a" fourth root of 2, but I think to blithely assume it is the real one is somewhat facile.

Semiclassical

12:11 AM

by comparison, $\mathbb{R}[X]/(X^2+1)$ is isomorphic to $\mathbb{C}$.

David Wheeler

I understand that it is isomorphic to $\Bbb Q(\sqrt[4]{2})$, I just think that there's, I don't know exactly how to say it...a "couple steps missing".

Semiclassical

the only field theory is know in detail is physics, not abstract algebra, so i can't help

but, what's the definition of a 'real field'?

David Wheeler

The answerer goes on to say that $i \not\in \Bbb R$, but I think one has to show that in any field isomorphic to a subfield of $\Bbb R$ no square root of $-1$ exists.

@Semiclassical I mean a subfield of $\Bbb R$

Semiclassical

ah

can't say i know enough to prove that

David Wheeler

It's not a major issue, another answer shows we have an injection $K \to \Bbb R$, which is almost the same thing.

Semiclassical

12:19 AM

(judging from Wikipedia's page on formally real fields, it's not quite enough to show that no square root of -1 exists. one has to also rule out that any sum of squares would give -1.)

Leaky Nun

@Semiclassical may I ask a question?

David Wheeler

It's just that I think we could identify $\Bbb Q[X]/(X^4 - 2)$ with $\Bbb Q(i\sqrt[4]{2})$ (adjoining a different fourth root of 2), and then you have a non-real field.

anon

take for instance Q(sqrt(-2))

Semiclassical

there's probably a brute force way to proceed, e.g. take a generic element of $\mathbb{Q}[X]/(X^4-2)$ and square it

@leaky as per the chat description: just ask, don't ask to ask

Leaky Nun

@Semiclassical What's an ideal?

David Wheeler

12:21 AM

Oh, I'm sure there is-I just don't think its quite kosher to insist $\Bbb Q[X]/(X^4 - 2)$ is the real field.

Semiclassical

@david or at least not without some other legwork

David Wheeler

An ideal is the pre-image of zero under a ring-homomorphism.

Leaky Nun

@DavidWheeler and in English?

Semiclassical

now that's a helpful definition :p

Ideal (ring theory)

In ring theory, a branch of abstract algebra, an ideal is a special subset of a ring. Ideals generalize certain subsets of the integers, such as the even numbers or the multiples of 3. Addition and subtraction of even numbers preserves evenness, and multiplying an even number by any other integer results in another even number; these closure and absorption properties are the defining properties of an ideal. An ideal can be used to construct a quotient ring similarly to the way that, in group theory, a normal subgroup can be used to construct a quotient group. Among the integers, the ideals correspond...

David Wheeler

A ring is an algebraic structure that has addition and multiplication

So its an abelian group with a "compatible" semi-group operation

This compatability is usually stated as a distributive law

A ring-homomorphism is a map between rings that preserves sums AND products

Leaky Nun

12:25 AM

and what does pre-image of zero mean?

David Wheeler

So, in practical terms, an ideal is a subset of a ring that "acts sort of like zero"

Semiclassical

it means that if you act with the homomorphism, the entire ideal gets mapped to zero

David Wheeler

the pre-image of a set $Y \subseteq B$ for a function $f: A \to B$ is the set $X = \{x \in A: f(x) \in Y\}$

Semiclassical

you can also find that definition on said Wikipedia page, albeit phrased as 'kernel of a ring homomorphism' rather than pre-image

Leaky Nun

how does the even numbers get mapped to zero?

Semiclassical

12:28 AM

take everything modulo 2

Leaky Nun

is "modulo" defined in a ring?

Semiclassical

sure. that's the whole notion of a quotient ring.

David Wheeler

Suppose one takes the integers-even numbers are of the form $2k$, odd numbers are of the form $2k+1$. It turns out that "taking the remainder after dividing by 2" respects addition and multiplication.

Semiclassical

formally, the details are more involved. but the principle is the same.

Leaky Nun

alright

David Wheeler

12:29 AM

So the operation integer$\to$remainder is a ring-homomorphism.

Semiclassical

there's a bit in that Wikipedia article that talks about 'motivation' for the definition of an ideal, btw

Leaky Nun

then what is the ideal generated by $X^4-2$?

David Wheeler

The pre-image of the "zero remainder" is the even numbers.

Formally, it's all polynomials with rational coefficients that have $X^4 - 2$ as a factor.

You can find "polynomial remainders" in much the same way as you do with integers, by long division.

Semiclassical

and then $\mathbb{Q}[X]/(X^4-2)$ is what you get by taking the rational polynomials and acting as though $X^4$ can be replaced by $2$. (2 is the remainder of $X^4$ when divided by $X^4-2$)

Leaky Nun

And how can a set of polynomials contain $i$?

Semiclassical

12:32 AM

that's a handwavey explanation, mind

take $\mathbb{R}[X]/(X^2+1)$.

David Wheeler

Well, if you take remainders "modulo" $X^2 + 1$, then that is what $X + (X^2 + 1)$ becomes.

Semiclassical

$X^2$ has a remainder of -1 when divided by $X^2+1$, so within $\mathbb{R}[X]/(X^2+1)$ you could say that $X$ itself acts like $i$

David Wheeler

Because $X$ when squared, and added to $1$, then gives you a multiple of $X^2 + 1$, which under the suitable ring-homomorphism gets "modded (zeroed) out"

Semiclassical

more generically: if you've got a linear term $a+b X$, then squaring it gives $(a+bX)^2=a^2+2ab X+b^2 X^2$ which has remainder $(a^2-b^2)+2ab X$ when divided by $X^2+1$

David Wheeler

In other words, if you "tweak" the polynomial algebra so that $X^2 + 1 = 0$ as a rule

Semiclassical

12:36 AM

by comparison, $(a+i b)^2=(a^2-b^2)+(2ab)i$

Leaky Nun

@DavidWheeler becomes $X$?

David Wheeler

then $X^2 = -1$.

Semiclassical

that's not the same as saying $X=i$, though. that's a bridge too far within this context

it's better to say that there's an isomorphism between $\mathbb{R}[X]/(X^2+1)$ and $\mathbb{C}$

so that you can replace $X$ with $i$ everywhere in the above equations and everything still makes sense.

more formally, one should probably not say $X^2=-1$ but rather $[X^2]=[-1]$

Leaky Nun

So the question would be does there exist an isomorphism between $\mathbb Q[X]/(X^4-2)$ and $\mathbb C$

David Wheeler

Right, you have an extension of an isomorph of $\Bbb R$ (namely, the cosets $a + (X^2 + 1)$, cosets of constant polynomials), in which "a" square root of $-1$ exists.

Semiclassical

12:39 AM

which signals that one is working with elements of the quotient ring, not elements of $\mathbb{R}[X]$ itself

but dropping those square brackets is a pretty common thing

David Wheeler

But it is typical (although a bit lazy) to consider field isomorphisms as identifications

Semiclassical

right.

in much the same way as one typically would write the elements of $\mathbb{Z}/2\mathbb{Z}$ as 0 and 1 rather than [0], [1]

David Wheeler

And, no, there isn't a surjective ring-homomorphism of $\Bbb Q[X]/(X^4 - 2)$ onto $\Bbb C$, although there are at least 4 into it.

Leaky Nun

Is there an isomorphism between Q[X]/(X^4-2) and Q[X]/(X^4-1)?

Semiclassical

it won't be all complex numbers, just some of them

on that, i have no idea. i'd guess not, but...???

Leaky Nun

12:43 AM

I really don't get it

David Wheeler

Well, to get a field the polynomial you form the ideal with has to be "irreducible" (non-factorable into non-units)

Leaky Nun

how are they different?

Semiclassical

i last took abstract algebra back in...sheesh, 2008 or 2009

David Wheeler

$X^4 - 1$ has the factorization $(X - 1)(X + 1)(X^2 + 1)$.

Semiclassical

right. so one is a field and the other isn't, hence they're not the same

but if you've got something like X^4-3 then i'm back to ignorance

David Wheeler

12:44 AM

Proving a polynomial is in fact irreducible is not always a trivial task-factoring is known to be a "hard problem"

Leaky Nun

0

Q: $1+1=2$...but Why?

A study that was carried on recently showed that even babies at the age of few months know that $1+1=2$. My question is : is this a fact that can be proved, or is it a just a postulate as those in geometry ? Also what about the saying "A given statement can be either true or false." ? Why are t...

Oh, come on...

Semiclassical

which probably means that establishing whether two quotients of the same ring are isomorphic is even harder

@LeakyNun I'm not even going to look

David Wheeler

$X^4 - 3$ is indeed irreducible, but it yields a non-isomorphic field, which you can show by taking elements to the fourth power, and doing some messy algebra

Leaky Nun

I do not understand

Semiclassical

gotcha. i'm guessing that said messy algebra gets worse and worse the higher of degrees you consider

Leaky Nun

12:47 AM

How is X^4-2 and X^4-3 different?

David Wheeler

One gives a fourth root of 2, and one gives a fourth root of 3, these are not algebraically equivalent.

Semiclassical

I think this is one of those results you're best off accepting that 'it can be shown that.'

It does suggest the following harder question. Suppose someone gives you a ring $R$ with two distinct ideals $I_1,I_2$

anon

@LeakyNun they're obviously different polynomials. what you mean to ask is why they don't give isomorphic fields

Semiclassical

When is $R/I_1\cong R/I_2$?

David Wheeler

Hai anon!

Leaky Nun

12:50 AM

@anon yes

@Semiclassical when?

David Wheeler

Can you answer my question?

Semiclassical

I didn't say that I knew the answer to it :p

anon

@Semiclassical when they are. too broad of a question in that generality

Semiclassical

Fair enough.

David Wheeler

I know that $I_1 \cong I_2$ is not sufficient.

Semiclassical

12:51 AM

not sufficient, or not necessary?

wait, derp. no, not sufficient.

David Wheeler

1

Q: Does $K = \mathbb Q/(X^4 - 2)$ contain the imaginary unit $i$?

Let $P(X) = X^4 - 2 \in \mathbb Q[X]$. a) Prove that $P(X)$ is irreducible. b) Prove that the field $K = \mathbb Q/(P(X))$ is an algebraic extension of $\mathbb Q$ and find a generator of it. c) Considering $K$ as a subfield of $\mathbb C$, determine if it contains the complex num...

abstract-algebra polynomials vector-spaces

anon

@DavidWheeler you understand why Q[x]/(x^4-1) can't have a sqrt of -1 right?

Semiclassical

a narrower version would be, say, $R=\mathbb{Q}[X]$ and $I_1,I_2$ two different polynomial ideals in that setting

but even that's probably too hard

David Wheeler

@anon Uh...that's not even a field...

arctic tern

okay switched devices cant type on kindle

meant x^4-2 of course

David Wheeler

12:55 AM

I just want to know if I have this wrong-when you adjoin a root of a polynomial to a field, I don't think you get to say "which" root.

arctic tern

so what's your question?

David Wheeler

Well, I don't think it's enough to say "$K = \Bbb Q[x]/(X^4 - 2) \cong \Bbb Q(\sqrt[4]{2})$ and this field is real, and doesn't contain $i$".

arctic tern

Suppose $K=\mathbb{Q}[x]/(x^4-2)$ has an element $f$ which is a square root of $-1$. Let $\phi:K\to\Bbb R$ be the obvious field embedding taking $x$ to $\sqrt[4]{2}$. Then $\phi(f)$ would be a real number which squares to $-1$, a contradiction. Therefore $K$ has no element which squares to $-1$.

David Wheeler

It's my contention you have to show (or have previously shown) any field isomorphic to a subfield of $\Bbb R$ has no square root of $-1$.

arctic tern

different people have different thresholds for what they consider too trivial to mention

David Wheeler

1:00 AM

I agree, but "identification" is not quite the same thing as "equality".

arctic tern

then object to the equality

(which I do too)

Ted Shifrin

hi tern :) you less wise, as well? :P

heya @David

David Wheeler

I find your response perfectly satisfactory, I think even though it's a small step, it matters

arctic tern

@TedShifrin if ignorance is bliss, my wisdom seems to be increasing

Ted Shifrin

:P

I hope you're feeling no pain.

arctic tern

1:02 AM

was what I was referring to

David Wheeler

I tried to articulate what I felt was disturbing in a comment, but my longer one got deleted. I'll live.

Ted Shifrin

tern: I was concerned that you told me that $\Bbb Q[x]/(x^4-1)$ could contain no $\sqrt{-1}$.

Then I read you can't type on a kindle. I hate chatting on iPad. So I get it.

David Wheeler

Anon blames non-kindle.

Ted Shifrin

non- ?

David Wheeler

Non-non-kindle.

arctic tern

1:04 AM

just mention to Deidrich that for a noobie like OP, one shouldn't write K=Q(sqrt(2)) but rather \cong, and that an argument should be present as to why exactly a square roof -1 would lead to a contradiction

Ted Shifrin

Algebra, despite being "easier" (fewer quantifiers) than analysis, still has plenty of subtleties that confound.

David Wheeler

His reply was off-putting, as if I was stupid. I started to doubt what I knew.

Ted Shifrin

well, @David, these days I am stooopid.

David Wheeler

I should get my revenge by upvoting another answer. But, that's petty.

Ted Shifrin

I am ok with being downvoted so long as the downvoter gives me some rationale. I'd rather fix the thing.

David Wheeler

1:07 AM

Meh, the whole "rep" thing is a game for younger men than me.

Ted Shifrin

I'm older than you :D

David Wheeler

Then I shall give you copious rep-gimme a sec

Ted Shifrin

ha ha

I use up- and downvotes as means of signaling approbation or that there's something really wrong.

@David: Stop that. The computer's going to reverse all that.

You're in violation of MSE ethics.

David Wheeler

I'm too tired to continue. Old age.

Ted Shifrin

Should I buy you some Geritol?

David Wheeler

1:10 AM

Deal!

Ted Shifrin

How's your new life in the new place?

David Wheeler

I received an IRS scam call today.

Ted Shifrin

Lovely.

Perhaps Trump is behind it.

David Wheeler

I was suppose to send some Indian guy named "Steve Anderson" $6800 immediately or the sheriff would be dispatched to arrest me.

arctic tern

the best outcome is if trump wins, formally announces "psyche" and then we all scramble for a re-election

David Wheeler

1:12 AM

The donuts I made the sheriff are now cold.

Ted Shifrin

ROFL

arctic tern

was it a real person calling?

Ted Shifrin

@arctictern You need more drugs :)

David Wheeler

It wasn't a recording. They were even ready with a badge ID and everything. I filed a complaint with the attorney general's office, they can sort it out.

Ted Shifrin

Good for you, @David.

I would have just hung up.

I almost never answer my phone at this point.

David Wheeler

1:15 AM

Remind me not to call you.

arctic tern

it doesn't make sense to invest personal energy in obvious scams. usually they put out deliberately obvious scams en masse so it weeds out the ones that would waste any further time they invest in the scam after following up.

Ted Shifrin

OK.

Well, see you guys later. I hope you feel better soon, anon, tern, et al.

hi

bye hi

bye @MikeM

1:16 AM

Fargo is bland. The big issue here is: Lutheran, or Catholic?

Ted Shifrin

Or whose body ends up in the dumpster?

David Wheeler

Oh no-bodies go in the river, doncha know.

I actually had a bit of fun, week before last-went to a college reunion.

I've started reading books again. It's...weird. No hyperlinks.

and....I killed chat again.

user97303

1:36 AM

Hey, I have a question. If I have an uncountable set of irrational numbers, can I always construct a sequence from them that converge in the rationals?

Leaky Nun

@chell pi, pi/2, pi/4, pi/8, ... (it converges to 0)

additionally: pi+a/b, pi/2+a/b, pi/4+a/b, ... (it converges to a/b)

if a/b is rational, then all of the terms there are irrational.

user97303

I'm asking whether this always happens with any uncountable set of irrationals

Leaky Nun

if you have an uncountable set of irrational numbers, can you construct a sequence that converge in anything at all?

user97303

For instance, what I know is since there are uncountably many, the distance between terms in this set will be arbitrarily close (otherwise if d>0 is a lower bound on the distance between numbers in my set then I can just inject my set into the integers by dividing the real line into intervals of length d and it would be countable, contradiction)

Leaky Nun

@chell nice.

user97303

1:41 AM

@LeakyNun That converges in the rationals. I want to know whether it converges in the rationals.

Leaky Nun

in any rational?

user97303

@LeakyNun Yes, any rational

Leaky Nun

Does the fact that there is always a rational between two irrationals help?

user97303

You can assume that my set it bounded: ie, there are uncountably many numbers in my set $S \cup [n, n+1]$

user97303

@LeakyNun Well I know the rationals are dense in R and R\Q is also dense in R

Leaky Nun

1:45 AM

Alright, you should stop wasting your time with me, because I don't have an answer.

user97303

:( I don't think I'm wasting my time

user97303

I mean, unless you do

user97303

also maybe someone else will show up. but if not maybe I'll post a question

Leaky Nun

@MikeMiller Do you have anything to say about this question?

@chell But the point is, must there exist a region which is dense in the set?

user97303

I'm not sure

Leaky Nun

1:52 AM

If the answer is yes, then we're done

user97303

Yeah, that's a good way to look at it

user97303

Wait, I think I have a counterproof

Leaky Nun

@chell the Cantor set is nowhere dense and has a cardinality of uncountable

user97303

Yeah

user97303

I was going to construct a sequence (1/(n+1), 1/n]

user97303

2:03 AM

and then subdivide each of these intervals into 2^n intervals

user97303

and then pick out something in the irrationals in each of those subintervals to add to my set

user97303

then my set would be uncountable

user97303

but yours actually proves it, so nevermind

user97303

damn and I thought I was so close to having a nice proof

Leaky Nun

But can you still construct a sequence from it that converges to any rational?

arctic tern

2:34 AM

@chell Let $l_1,l_2,\cdots,$ be any sequence of positive lengths whose sum converges, and let $x_1,x_2,\cdots$ be an enumeration of the rational numbers. Let $X$ be the complement of $U=\bigcup_{n=1}^\infty (x_n-l_n,x+l_n)$. Since $U$ has finite measure, $X$ has infinite measure, in particular it's uncountable. Plus it has no subsequence converging to a rational since it's disjoint from a nbhd of every rational.

(it's a pretty paradoxical set)

user97303

I don't actually know what measure means but this seems to be the proof that someone else gave on my question just now

user97303

is there a way to change the notification sound on these chat rooms? I like being notified but this sound sounds like a dart thrown at me, scares me half to death sometimes

anon

2:53 AM

Click the volume button up top

You never said in chat that you posted the question on the main site

user97303

3:45 AM

2 hours ago, by chell

also maybe someone else will show up. but if not maybe I'll post a question

Prince M

4:36 AM

Can anyone tell me what the "swooping down function arrow" means?

well y naught

4:59 AM

huh? Pic?

Prince M

I just posted a question with the picture on main page

titled "inclusion mapping?"

Mike Miller

How is it not an inclusion?

anon

5:15 AM

The hook arrow indicates embedding, which doesn't even require one to be a subset of the other a priori (which from the sound of it is how you think of inclusion)

Although you do say «an» inclusion, so maybe I'm misreading your misunderstanding

Ted Shifrin

@DavidW: As predicted, the computer took away all your votes.

Wow, both anon and tern! Duelling IP addresses :)

anon

MAC addresses maybe

Ted Shifrin

:)

Axoren

5:35 AM

Do @anon and @arctictern share an office?

arctic tern

savor the mystery

2

Leaky Nun

7:43 AM

How is it that 20 rep is needed to post in chat?

Mike Miller

prevents spammers

Leaky Nun

I see

Andrew Thompson

@MikeMiller Good morning. Go to sleep.

Mike Miller

no

Andrew Thompson

>: (

Henry

7:57 AM

Does anyone know about the user Cleo and why don they not post anymore answers on MSE?

Mithlesh Upadhyay

flag as spam " math.stackexchange.com/questions/873565/… "

Henry

I'm talking about this user math.stackexchange.com/users/97378/cleo

Martin Sleziak

@MithleshUpadhyay In this case the spam flag is IMHO incorrect. The word spam has a special meaning in the SE network clearly defined in FAQ.

I'd say that "not an answer" of "offensive" could be correct flags in this case.

Leaky Nun

@MartinSleziak whether it's "offensive" is really subjective.

Martin Sleziak

The FAQ explicitly says that: "It should NOT be marked as spam when: A) The answer contains no useful information, ... B) It contains only gibberish, ..."

Leaky Nun

8:07 AM

It depends on whether you link it with terrorism, because "allahu akbar" is a regular Muslim saying.

Martin Sleziak

@LeakyNun Again, I would refer you to the FAQ - the same link as above. "If it contains only gibberish, such as "fsdguejgkfdlk". Use the 'offensive' flag for these cases, or flag 'for moderator attention' with a custom explanation if it requires more detail.

Leaky Nun

I see.

Martin Sleziak

Being a Muslim or a nun or whatever religion is not something to be offended by.

But even if the faq says to use offensive, I somehow felt more comfortable with not an answer. Which is what I chose to do.

Mithlesh Upadhyay

@MartinSleziak, yes you are correct. Thanks.

Henry

Does anyone know about the user Cleo and why don they not post anymore answers on MSE?

I'm talking about this user math.stackexchange.com/users/97378/cleo

Apparently they post marvelous results without proof. Does anyone know who are they?

They post mostly for integrals questions.

Leaky Nun

8:15 AM

@Henry geniuses.

Martin Sleziak

If you believe the information in the profile, you can use she instead of they. And the information in profile could also give a partial explanation for omitting proofs.

Leaky Nun

Now I am both amazed and intrigued by all of her answers.

Leaky Nun

8:28 AM

38

A: Calculating $\int_{\pi/2}^{\pi}\frac{x\sin{x}}{5-4\cos{x}}\,\mathrm dx$

$$\begin{align}\int_{\pi/2}^\pi\frac{x\sin x}{5-4\cos x}dx&=\pi\left(\frac{\ln3}2-\frac{\ln2}4-\frac{\ln5}8\right)-\frac12\operatorname{Ti}_2\left(\frac12\right)\\&=\pi\left(\frac{\ln3}2-\frac{\ln2}4-\frac{\ln5}8\right)-\frac12\Im\,\chi_2\left(\frac{\sqrt{-1}}2\right),\end{align}$$ where $\opera...

A step!!!!!!!!!!!!!!!

Leaky Nun

8:43 AM

Integral of the day:

$$\int_0^\infty\operatorname{arccot}(ax)\,\operatorname{arccot}(bx)\,\operatorname{arccot}(cx)\,dx=\\
\frac1{24 a b c}\left(2 a (b+2 c) \ln ^3(a)-3 \left(2 b (a+c) \ln (b)+(4 a c-2 b c) \ln (c)+\\
b \left(-2 c \operatorname{artanh}\left(\frac{b}{c}\right)+(c-a) \ln
(c-a)-(a+c) \ln (a+c)+a \ln \left(c^2-b^2\right)\right)\right) \ln ^2(a)\\
+3 \left(b c \left(\ln ^2(b)+2 \left(2 \ln (a+c)+\ln
\left(-\frac{c (b+c)}{(a-b) (a-c) (b-c)}\right)\right) \ln (b)+\ln ^2(b-a)\\
-3 \ln ^2(c)+\ln ^2(c-a)-\ln ^2(c-b)+\ln ^2(b+c)-4 \operatorname{artanh}\left(\frac{a}{c}\right) \ln (a+b)+\\

From here

19

A: Integrals of the form ${\large\int}_0^\infty\operatorname{arccot}(x)\cdot\operatorname{arccot}(a\,x)\cdot\operatorname{arccot}(b\,x)\ dx$

I finally managed to find a general solution to this problem. Although I put a lot of effort into simplification of the result, it is still not as pretty and symmetric as I would like it to be. I hope to improve it later. Sorry for poor typesetting. Assuming $0<a<b<c,$ $$\int_0^\infty\operatorna...

Ilan Aizelman WS

10:31 AM

Help is appreciated: math.stackexchange.com/questions/1839085/…

Leaky Nun

10:46 AM

@IlanAizelmanWS In general, if you can express a function in terms of $g(X)$ where $X=ax+by$, then you would have straight contour lines.

Ilan Aizelman WS

10:58 AM

@LeakyNun I see.

Thank you.

Another question: math.stackexchange.com/questions/1839111/…

Leaky Nun

@IlanAizelmanWS but D2 is wrong....

it should be similar to D1 with 0<=y<=x instead of 0<=x<=y

Mithlesh Upadhyay

11:14 AM

@robjohn , chat.stackexchange.com/rooms/11540/charcoal-hq

robjohn

11:32 AM

@MithleshUpadhyay Why are you posting in that room about a problem on SO? Either flag the actual question or ask someone on SO about it.

Ilan Aizelman WS

@LeakyNun Hi. Well, I'm trying to paint it on my papar, those two rectangles. and I was successful doing only the RHS one. how to do it with LHS?

@LeakyNun I mean, when I paint them, they look the same. like this: |_\

Mithlesh Upadhyay

@robjohn, I was unable to join chat on SO, since I've deleted my SO account then back on.

Ilan Aizelman WS

@LeakyNun I've solved it now. they're of curse triangulars. my English sux. I thought triangular is called rectanlge T_T

Leaky Nun

@ilan the first should be |/, while the second should be /_|

robjohn

12:05 PM

@MithleshUpadhyay you or someone else can flag the question asking that it be migrated.

Ilan Aizelman WS

12:26 PM

@LeakyNun Yup. Exactly.

Balarka Sen

1:00 PM

Morning @Semiclassical

Semiclassical

hey @balarka. any new physics problems? :)

Balarka Sen

Nothing pressing, but the pulley problems from Newtonian mechanics are a bit tiresome.

Saladin

0

Q: How to keep result of calculation to be in particular range?

I have data as :- Group A Element A score(3) Element B score(1) score 4/2 Group B Element A score(2) Element B score(3) Element C score(1) score 6/3 What I want in end, for Group A RATING (1-5) Group B RATING (1-5) The rating would be result of score, what the ...

modular-arithmetic

Semiclassical

ah

multiple uses of $F_{net}=ma$ etc?

Balarka Sen

Right

Semiclassical

1:03 PM

one thing that's useful to note in those are the internal forces in the problem

e.g. the force the pulley exerts on the object it's holding and vice versa

because a lot of the time, they'll cancel out and you can treat the pulley and object as one system.

(of course, if you want those internal forces, you don't have much option)

Balarka Sen

right

Semiclassical

have you done anything with oscillations yet?

Balarka Sen

not sure what you mean by oscillations.

Semiclassical

say, an object attached to a hanging spring. pull it down, see it move up and down

Balarka Sen

ohh. nope, not yet.

Semiclassical

1:07 PM

mmkay

Balarka Sen

i have been told about the hooke's law tho

Semiclassical

gotcha.

my pet peeve with hooke's law is the way it's usually presented

namely, F=-kx

Balarka Sen

The version I know is strain is proportional to stress.

Semiclassical

That works.

For force problems you typically convert strain to displacement and stress to force.

Balarka Sen

hmm, yeah

Semiclassical

1:09 PM

The way I prefer to write Hooke's law, anyways, is $\Delta F = - k \Delta x$

to emphasize that there need not be anything special about $x=0$. you can certainly arrange things so that $F=0$ when $x=0$, but that needn't be the case

Balarka Sen

That makes sense

Semiclassical

and to emphasize that Hooke's law isn't really a statement about the total force exerted by a spring, but rather a statement about how the force required to stretch it changes when the stretch is changed

Balarka Sen

Right.

Semiclassical

Which has the following practical implication. Suppose someone were to hook a spring to a ceiling and attach an object to it, then use a cardboard tube to disguise where the spring attaches to the ceiling (so you can't actually directly measure the total stretch)

You can nevertheless measure the spring constant, since you can still measure how much the stretch changes when you add another small weight

which is convenient because it means that if you're stretching a spring, there's really no need to fuss about where you take $x=0$ to be. put it somewhere, measure everything according to that coordinate system, and then fit the results with a line.

blah blah blah straight-line analysis blah blah blah

Balarka Sen

Yep.

Semiclassical

1:16 PM

Anyways, that's a mindset that comes in very handy if, say, you've got an object with springs attached to both sides

(which i've had students do in lab. not a bad problem at all measurement-wise.)

makes the derivation a heck of a lot easier.

but i'm rambling now, woops

Balarka Sen

heh, no, it's ok.

Semiclassical

mmkay. hmm, does your physics class actually have labs or just the text?

Balarka Sen

Yes, we have a lab. The practicals started today.

Semiclassical

ah, good

The thing which my students had to do for intro physics lab, which was always really really tedious

Balarka Sen

We'll probably do some measurements of the Young's constant at some point.

Or so we were told.

Semiclassical

1:20 PM

yeah, and that's exactly what i was yammering on about in the context of springs

because strain = dx/x0 and stress = dF/A, so k = dF/dx = stress.A/strain.x0 =Y.A/x0

Balarka Sen

Uh-huh.

Semiclassical

yadda-yadda-yadda

Anyways, the tedious thing my students had to do

Suppose you want to record the motion of some object. There's a few ways to do that. One way is using a stopwatch, but that doesn't really let you track the object.

There are some instruments that let you use sound to track the position of an object, and those are really nice. But my students didn't have that.

What we instead had were small video cameras. So the idea was to record a video of the motion, then analyze the video to get the motion from that

Balarka Sen

that doesn't sound so bad

Semiclassical

If the computer could do the analysis itself, it wouldn't be.

Balarka Sen

actually, what do you precisely mean by recording the motion?

Semiclassical

1:26 PM

Heh. Therein lies the problem.

What the video records is simply a black-and-white movie.

What the user has to do is walk through the video, going a few frames at a time, and mark the location of the object each time.

so that the computer gets a set of xy-coordinates in the video as a function of time.

Balarka Sen

yikes, that's going to be harder

Semiclassical

yeah. really quite tedious.

especially since you have to click and drag a cursor to the location each time. (you'd think they could just use the mouse to click the new spot or something)

The other thing they have to do ahead of time is include, in the actual video, some reference object whose length is known

so that they computer will know, say, that a distance of 20 pixels in the image is 1 cm of actual displacement

Balarka Sen

right, ack

Semiclassical

Combine that with an interface that's a real arse, and the fact that this is what they spend the first few weeks of first semester intro physics using, and...

yeah, kind've shitty for them

Balarka Sen

:S

Semiclassical

1:31 PM

intro physics lab, to be sure. the lecture doesn't use that

Balarka Sen

admittedly our physics lab is considerably less ~~pain in the~~ ah, how shall i say it, fascinating.

Semiclassical

snerk

For the actual description in the lab manual, see page 221-229 of this pdf

that's part of why TAing for first semester is so much more of a pain than second semester physics

no more MotionLab, woo

Balarka Sen

hehehe

Semiclassical

you use similar things in second semester, but none of them are nearly so onerous

plus, in second semester physics (the one geared towards pre-med/biology students, anyways) they get to use lasers for one or two labs

oh, and magnets. gotta love that

Balarka Sen

@Semiclassical lasers? like?

Semiclassical

1:44 PM

well, same idea as a laser pointer, but probably a bit more focused, and mounted in such a way as to be adjustable in direction

Balarka Sen

not space-lasers, surely? :P

@Semiclassical what kind of things do you do with them?

Semiclassical

the idea is to use the laser light to illuminate a thin slit (or a pair of such)

if it's smaller enough, then that'll produce a diffraction pattern owing to the finite wavelength of light

Balarka Sen

oh. double slit experiment?

i have done that with a pen-laser :P

Semiclassical

double or single

same ideas in either case, of course.

Balarka Sen

Uh-huh.

Semiclassical

1:47 PM

though my favorite bit of that is pointing out that you can use the same principle to estimate the size of a human hair

Balarka Sen

hmm?

Semiclassical

sure. if you illuminate a human hair with laser light, then you also get a diffraction pattern

and from that, and some other measurments (say, how far the hair being illuminated is from the surface you project the pattern on to) you can use the wavelength of the laser light to get a decent estimate of the width of the strand.

Balarka Sen

i see

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