Mathematics - 2018-05-11 (page 1 of 5)

skull

12:12 AM

$Mathematics$ JEE Maths Zone

Related Rooms: [Physics](chat.stackexchange.com/rooms/54160/p...

All are welcome :-)

Mathematics is part 3

Daminark

1:57 AM

@Mathein Did you also do the business about it being a transitive non-abelian subgroup of $S_4$, leaving only those options? And then the quadratic stuff was because $A_4$ has no Index 2 subgroups, $S_4$ only has $A_4$, etc?

MatheinBoulomenos

yeah

Daminark

Nice

MatheinBoulomenos

knowing your subgroups of $S_4$ really helps with some Galois theory exercises

I tried to classify the subgroups of $S_5$ myself at some point, but after 6 pages of calculations with permutations, I gave it up

maybe it's easier if you have more theory

Daminark

Yeah I find 4 in particular to be common. 3 is often too easy, 5 loses a lot of what you can say

Oof yikes

MatheinBoulomenos

I tried an elementary approach, maybe I just didn't have the right ideas

I don't know why, but I actually like those finite group exercises even when you need some brute force. Like classify all groups of a certain order etc.

Fargle

2:06 AM

classify all groups

good luck

MatheinBoulomenos

lol

Daminark

Perhaps. I never actually classified the subgroups of S_4, so far I've been riding on knowing that the transitive ones are S_4/A_4/D_4/Z_4/Z_2^2, and that S_3 doesn't live in an index 2 subgroup of S_4

Fargle

not even up to isomorphism, just, straight up, all groups

Daminark

@Fargle every group is a subgroup of (S_n)! Surely you can just induct!

(avoided unexpected factorial but also sniped... :/)

Fargle

gottem

also good luck when doing $S_{\Bbb R^{\Bbb R}}$

MatheinBoulomenos

2:08 AM

@Daminark knowing actually all of them can be helpful if you need to compute all intermediate fields of say a Galois extension with $D_4$, but you could also work that out by hand

B. Mehta

from my Galois theory lecture notes

I don't think a single one of us did the exercise

MatheinBoulomenos

okay, transitive ones are probably not that bad

for $n \leq 4$ it's not too hard

I just tried to classify all subgroups

Fargle

What does it mean for a subgroup to be transitive again?

I'm betraying how little I know about algebra, and yet how much I care, in that one question.

MatheinBoulomenos

@Fargle it only makes sense for a subgroup of $S_n$, it means that the action on the set of $n$ elements is transitive, i.e. has only one orbit

Fargle

gotcha

B. Mehta

2:10 AM

which is important in Galois theory because irreducible polynomials have transitive galois groups

Fargle

Ah, nice. I need to actually go through field/Galois stuff again because that never stuck the first time around. Group and ring stuff I did just fine with, but I was burned out by the time we got to field extensions.

MatheinBoulomenos

the fact that $S_3$ and $S_5$ are transitive subgroups of $S_6$ still feels weird

I mean for $S_3$ it's kinda obvious

B. Mehta

wait what!

MatheinBoulomenos

of course it will depend on the embedding

B. Mehta

okay good

Fargle

2:12 AM

Wait which embedding of $S_3$ is transitive?

MatheinBoulomenos

you get one from Cayley's theorem

B. Mehta

$\langle (12)(34)(56),(135)(246) \rangle$ works i think?

Fargle

oh dang it you're right that's AAAAAA

Daminark

Ah clever

Fargle

that's just friggin bonkers

algebra is officially for mad lads (or just mad people, really, no need to be exclusionary 'bout it)

Daminark

2:14 AM

And yeah $S_5$ was some weird stuff, I think that had to do with the outer automorphism group of S_6 being non-trivial?

MatheinBoulomenos

yeah, it's related

Fargle

yeah I remember seeing something about that

MatheinBoulomenos

you can construct the embedding from Sylow theory: $S_5$ has $6$ 5-Sylow subgroups and acts transitively and faithfully on them by conjugation

Fargle

Is it a solved problem to determine which $S_k$'s embed transitively in $S_n$?

B. Mehta

@Fargle Yes, all of them do for $k \leq n$

unless you mean embed transitively :P

Fargle

2:16 AM

:P

I'm too tired to type cogent questions

Daminark

Get dunked on

MatheinBoulomenos

the Cayley trick always gives you a transtive embedding $S_n \to S_{n!}$

Daminark

Actually hmm, I think $S_4$ surjects onto $S_3$ by conjugating double transpositions, right? Does something go wrong if we compose that with the transitive embedding into $S_6$?

MatheinBoulomenos

that works

Daminark

Lol it seems like literally everyone and their kitten act transitively on 6 elements

MatheinBoulomenos

2:22 AM

$S_4 \to S_3$ and $S_n \to S_2$ via the sign map are the only surjections $S_n \to S_k$ for $n>k$ though

Fargle

Is there a classification theorem for this question in the literature somewhere, or is it not fully known but pretty well-understood for "small" $n$?

Daminark

Hmm, that's because the only normal subgroup of S_n from 5 up is A_n, right?

MatheinBoulomenos

yeah

you can also get a transitive action of $S_4$ on a $6$ element set by letting it act by conjugation on $2$-cycles

Daminark

Group theory is finally coming back to me. When rings came along groups just kinda disappeared into the void.

Actually are there contexts where (finite) group theory comes up in ring theory?

MatheinBoulomenos

if you do finite rings, then sure

all the proofs of Wedderburn's little theorem (a finite division ring is a field) that I know use some nontrivial group theory

Fargle

2:27 AM

Proof (Wedderburn's little theorem). Let $R$ be a finite division ring. By convention, $R$ is commutative. $\blacksquare$

B. Mehta

not sure if this counts but classifying finite abelian groups comes up in modules

Daminark

Get dunked on Weddenburn

Fargle

makes sense

MatheinBoulomenos

oh and the proof that all finite subgroups of the unit group of an integral domain are cyclic (so as a special case, the multiplicative group of a finite field) is a lot easier if you know that finite abelian groups are producrs of cyclic groups

Daminark

I actually intend to try some non-commutative stuff. My combo prof told me about Artin-Weddenburn once and it seemed real slick, though I do get the vibe that mostly thinking about commutative rings is gonna make transitioning cell tricky

MatheinBoulomenos

2:29 AM

Artin-Wedderburn is one of the theorems in ring theory that came up for me in so much different stuff

Daminark

Hmm, yeah I remember at the end of last quarter we had a problem about showing that finite abelian groups where there are at most n elements of order dividing n are cyclic, using the structure theorem

Fargle

the truly deep connection: every ring (with id) is just a subring of the ring of endomorphisms of its underlying abelian group

Daminark

This quarter I was talking to my prof once and he asked me if I knew that, I was like yeah we did structure theorem, at which point he was like 0_0

MatheinBoulomenos

@Daminark you can actually prove that without the structure theorem, but it's more work

you do some computations with the $\varphi$ function

Daminark

Told me to try to prove the fact that finite subgroups of F^{\times} are cyclic in an elementary way, and our TA had proven it soon after

He did the Euler totient business you speak of, so that was nifty

Fargle

2:33 AM

the totient function proves that Euler was a reality hacker

it's literally too good

MatheinBoulomenos

@Daminark I think it's better to start earlier with noncommutative stuff if you want to do it eventually

Daminark

Yeah the guy just saw the source code of existence for a second there

Also true

MatheinBoulomenos

a lot of results on noncommutative are still interesting if you specialize them for commutative rings, though the proofs may be easier

sometimes it's not even easier for commutative rings

a lot of basic results on modules are like that

you rarely use commtutativity of the base ring if you prove stuff of the form "modules with X and Y are Z"

GettingNifty

have you gusys ever had voices

like real voices

MatheinBoulomenos

I still have a voice right now

it's useful for communication

GettingNifty

2:37 AM

no i mean like god jesus and satan

Daminark

Sadly my voice box has been replaced with one made by Walmart

GettingNifty

they think it's funny i don't get it

i really don't get it

Daminark

Not physical voices, no

GettingNifty

no i hear seriously vibrant voices

there's a chance i can move on their commands too

they're really smart and really funny

MatheinBoulomenos

if you're serious, you should talk to a therapist

GettingNifty

2:39 AM

no I mean like I talk to God

MatheinBoulomenos

do you often talk with god about math?

GettingNifty

Im serious

LIke he'll tell me when a red car is going to come at me

i'm not joking i speak to God

MatheinBoulomenos

good for you

GettingNifty

Jesus too sometimes but more of a joke

Daminark

Yeah, I don't know of too many people that God communicates with directly while still alive since Moses (depending on who you ask, others will say differently, but ultimately it's not a daily occurrence) so maybe you should be 100% sure, and if not... Follow Mathein's advice

GettingNifty

2:41 AM

yeah i think only four people ever communicated with God directly in the bible

would you be terrified

i mean it's not like it stops

eh

MatheinBoulomenos

I would question my sanity (which seems like a healthy reaction). I probably wouldn't talk about it in a math chatroom

I acn prove it.

in ZFC?

what's that bro

it's like KFC but with more choice

Daminark

2:45 AM

Anyway we should probably get this back to math. I doubt there's much we can do about this.

MatheinBoulomenos

yeah

Daminark

Okay good thing that saga is over

Thomas Ward

→ 14 messages moved to Trash

quartata

:44540136 (to be fair it was counterflagged the first time)

Thomas Ward

well it's handled now, let's all move on.

Fargle

2:49 AM

So, uh, yeah, finite group theory as realized in ring theory.

MatheinBoulomenos

@ThomasWard thanks for the intervention

Thomas Ward

yep.

MatheinBoulomenos

yeah, apart from the result on finite subgroups of multiplicative groups and Wedderburn's little theorem, I don't know much applications in that direction

Daminark

@MatheinBoulomenos do you know some common "rookie mistakes" people make when they first play with non-commutative rings?

I dunno why I replied to that message

Fargle

Has the debt of ideas ever been paid in the other direction? That is, have ring-theoretic results ever been used to prove group-theoretic ones?

MatheinBoulomenos

2:53 AM

@Fargle yes, a lot actually

Mike Miller

One thing to keep in mind is that while you might be used to it, ab is NOT necessarily ba

MatheinBoulomenos

a lot of stuff in representation theory becomes a lot simpler if you think about it as modules over the group algebra

and Artin-Wedderburn etc. comes up

Daminark

:O

Re ab and ba

Mike Miller

Ty I thought it was a good bad joke

MatheinBoulomenos

really a lot of representation theory is about understanding the ring-theoretic properties of the group algebra really well

@MikeMiller big if true

Fargle

2:55 AM

@MikeMiller In turn, ba might not equal ab, either.

Mike Miller

Allow me to play devil's advocate and ask what you can actually do that sheds new light? To me it sounds like a dictionary rather than a magnifying glass

Daminark

@Fargle but if how do you commute the $\ne$ with the terms?

I can feel the grammar in the depths of my soul

Fargle

@Daminark $\neq$ is in the center.

MatheinBoulomenos

@MikeMiller you can probably do most things without it, but I don't find for example the written-out definition of an induced representation very natural

Daminark

Okay that was good

MatheinBoulomenos

2:58 AM

doing a base change along the inclusion $k[H] \hookrightarrow k[G]$ seems obvoius to do

and Frobenius reciprocity is just Hom-Tensor adjunction

Mike Miller

The last bit is a somewhat satisfying response, even though I don't know if it adds much for me personally

I can see why this feels like the right place for representation theory to you

MatheinBoulomenos

I agree that you can probably do the basic theory over $\Bbb C$ without mentioning noncommutative rings

but why prove the same theorems multiple times

I'd guess it's more important if you don't have a semisimple group algbera anymore

I don't know what the description of tensoring with $k[G]/J(k[G])$ with $J(k[G])$ being the Jacobson radical is without ring theory

but ring theory actually comes up in other ways: I remember one proof in Serrre where you show something is an integer by showing that it is rational and integral over $\Bbb Z$

Fargle

Forget the fly and the bazooka, that's killing a paramecium with the Big Crunch

MatheinBoulomenos

There's a proof of Artin's theorem on virtual characters that's very elegant where you want to show that the cokernel of a map between $\Bbb Z$ modules is finite and you just tensor with $\Bbb C$ and use Frobenius reciprocity. So the right-exactness of tensor products is essential to that proof

Mike Miller

I don't know why I want to mod out by the Jacobson radical

That might be sufficient reason to care if you convince me of that

MatheinBoulomenos

3:11 AM

you get something semisimple after that

Artinian (e.g. finite-dimensional over a field) + Jacobson radical is 0 = semisimple

Fargle

Jacobson is intersection of maximal ideals?

MatheinBoulomenos

yes

of all maximal left sided ideals or all maximal right sided ideals

so after you mod out the radical, you have unique decomposition into simple modules and the ring is a product of matrix algebras etc.

Mike Miller

Which tells you that modules are completely reducible?

MatheinBoulomenos

yes

Mike Miller

So this is something you might do in modular representation theory?

MatheinBoulomenos

3:14 AM

yeah

you also want to understand the jacobson radical itself though

Mike Miller

Can you give me a description of the Jacobson radical for like Fp-bar [G] for some simple cases?

Is this the universal quotient with that property or something?

MatheinBoulomenos

if you're interested in representations, a neat description of the Jacobson radical might be that that the elements of the Jacobson radical are those that act trivially on all simple modules

so you can define it as the intersection of all annulators of simple modules

if $k$ has characteristic $p$, then $g \in G$ is contained in the interection of all $p$-Sylow subgroups (so the largest normal $p$-subgroup) iff $g-1$ is contained in the radical of $k[G]$

Mike Miller

I'm not sure I care about annihalators of simple modules - part of the interest is that you can't describe all things from the simple things, right?

Or rather I should say I'm not sure I understand the logic behind the restriction to simple modules

MatheinBoulomenos

yeah, but this makes precise that the Jacobson radical is exactly that stuff which you can't describe from simple things

Mike Miller

Hmm, I see

Because it won't act trivially on like the regular rep

Let's say the modular representation ring is defined by the following construction. First take the Grothendieck group on Fp-bar representations; then quotient by the ideal generated by [V] - [W] + [W/V]. Can you recover this from the Jacobson radical stuff?

(The reason I'm asking so many questions is this seems p interesting!)

MatheinBoulomenos

3:27 AM

hmm not sure

one thing you can get from this Jacobson radical stuff is that over an algebraically closed field of characteristic $p$ the number of irreducible represenations is equal to the number of $p$-regular conjugacy classes, where a conjugacy class is called $p$-regular if the elements don't have order divisible by $p$

Prototank

Any manifold topologists awake?

Mike Miller

Sure

@Mathei Cool!

MatheinBoulomenos

if you don't asume algebraically closed, you still get an upper bound

Daminark

What with all the time zones, surely someone doing manifold topology is awake!

Prototank

the frustrating thing is that I understand everything in this book so far

except for the well-definedness of $df_x$

what is frustrating, also, is that defining the derivative of $f:X\to Y$, a smooth map of smooth manifolds, is done SEVERAL ways

But none, as far as I can tell, have decided to do it the way G&P did it

MatheinBoulomenos

3:39 AM

you can also ask if you have a mod $p$ representation if you can lift it to a characteristic $0$ representation, say over $\Bbb Z_p$. That's usually done by lifting certain idempotents, which is a very common problem/technique in noncommutative algebra

I'm taking a course on deformations of Galois representations right now, that's basically all about starting with a representation over some finite field and then asking how you can lift it to certain local rings of characteristic 0 which have that field as a residue field. The way this works is really hard to imagine without ring theory (though the rings we deal with are the (commutative) coefficient rings)

Mike Miller

@Prototank Can you say self contained what your question is here? Too lazy to engage with main

Prototank

yes, I can

The derivative of a map $f$ defined on open euclidean space at a point $x$ is the jacobian matrix of $f$ evaluated at $x$.

It is written $df_x$

G&P want to extend this definition to arbitrary smooth functions on manifolds. So if $f:X\to Y$ is one such map, then for local parametrizations $\phi$ and $\psi$ which fit into this diagram $$\begin{array}{ll}
X & \overset{f}{\to} &Y\\
\uparrow\phi & & \uparrow\psi\\
U & \overset{h}{\to}& V
\end{array}$$

we simply define $df_x=d\psi_0\circ dh_0\circ (d\phi_0)^{-1}$

Mike Miller

3:57 AM

Sorry I forgot about you

Prototank

the open question is this: what if we chose $\eta$ instead of $\phi$, and what if we chose $\rho$ instead of $\psi$? In other words, how do we know that $df_x$ is well defined?

Mike Miller

Any two parameterization that send 0 to a fixed point p, differ by a diffeomorphism of the domain

At least if you restrict enough

This is more or less the inverse function theorem

Once you do that, equating the two notions of df_x should be the chain rule for multivariable calculus on domains of Euclidean space

Prototank

also, just so the cart isn't being thrown before the horse, the terms on the right hand side of $df_x$ are all well defined since the functions are defined on open subsets of euclidean space

@MikeMiller, so yeah if we had, say, $\eta$ instead of $\phi$, they differ by $\phi\eta^{-1}$, right?

fyi, I'd be greatly pleased if you answered my question on the general :D

notice the date on it btw

Daminark

4:30 AM

@Mathein so the other algebra section didn't do too hot on their midterm and the professor gave them the pset of death

it's real nasty: math.uchicago.edu/~fcale/Algebra/HW7.pdf

MatheinBoulomenos

there's a lot to do, but you don't really need any clever ideas

Daminark

Oh I know, it's mostly just because he felt they needed a lot of practice after that last test. Which is fair, but also wow, that just seems really monotonous

MatheinBoulomenos

but computing 30 primitive elements and their minimal polynomials is a bit ridiculous

we had an exercise like that with a degree 5 polynomial where the Galois group had order 20 and a lot of subgroups

Daminark

Yeah I remember a few psets ago my prof gave us one problem outta Dummit and Foote which was just listing all the intermediate field extensions of something whose Galois group was $\mathbb{Z}/2\mathbb{Z}^3$ and I was just like, Emerton plz

Now I honestly don't feel like I have the right to complain about tedious homework anymore

MatheinBoulomenos

but that's still reasonable

you can classify all subgroups by linear algebra

Daminark

4:37 AM

Tru

MatheinBoulomenos

and every subfield is probably generated by some square roots with a Galois group like that

Daminark

Oh yeah it was something real easy, $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$

MatheinBoulomenos

who did this exercise? they had to work out a minimal polynomial polynomial of degree 24 for a primitive element for the whole thing

good luck

Daminark

I have no idea, their pset is due on Monday so I'll ask my friends taking that section how they fared

MatheinBoulomenos

I think the highest degree polynomials I worked with explicitly by hand was like degree 9 or something like that

and that was already really annoying

Daminark

4:44 AM

I imagine

MatheinBoulomenos

5:02 AM

@Daminark good lord

your professor must be trolling

the minimal polynomial of a possible choice for a primitive element is $x^{24} - 90*x^{21} - 70*x^{20} + 5695*x^{18} + 18690*x^{17} + 34895*x^{16} - 225900*x^{15} -
1544060*x^{14} - 3867780*x^{13} + 18840027*x^{12} + 62876100*x^{11} + 228621050*x^{10}
+ 222888810*x^9 + 999415025*x^8 - 9907474500*x^7 - 24575577355*x^6 -
34467394920*x^5 + 232838692457*x^4 + 705674357100*x^3 + 2030693398335*x^2 +
2155371295770*x + 1779496656001$

the magma algorithms usually try to keep these coefficients small ...

Anonymous

7:02 AM

Daminark

@Mathein jesus

Daminark

7:20 AM

Now that I think about I could totally be down for an April fool's pset where I do something like that

MatheinBoulomenos

7:39 AM

@Daminark here's a fun exercise: classify all groups of order 1024

spoiler: there are 49487365422 up to isomorphism

Alessandro Codenotti

8:01 AM

Can I classify the abelian ones instead?

Silent

@MatheinBoulomenos, Let $G=\left\{\begin{pmatrix}a&b\\0&a^{-1}\\ \end{pmatrix}:a,b\in \Bbb R, a>0\right\}$ and $N=\left\{\begin{pmatrix}1&b\\0&1\\ \end{pmatrix}:b\in \Bbb R\right\}$, how do we know that $G/N$ is isomorphic to $\Bbb R$ under addition?

mercio

8:14 AM

@Daminark that's some insane homework

imagine grading it by hand

Daminark

Yeah that sounds painful

Balarka Sen

Lel, this Olympiad problem wanted me to prove that $\lfloor n/1 \rfloor + \lfloor n/2 \rfloor + \cdots + \lfloor n/n \rfloor + \lfloor \sqrt{n} \rfloor$ is even

Downright trivial if you interpret it differently

mercio

8:31 AM

lol

that's a surprisingly cute thing

Balarka Sen

yeah

more so because you can actually tell what the number counts

mercio

oh I did it the hard way

I see the super easy way now lol

Balarka Sen

ah.

mercio

but I have to change the last $+$ into a $-$ for it to be pretty

Balarka Sen

well i take the last term of the sum to be a source of 1's i can plug in somewhere else

mercio

8:40 AM

with a $-$ it counts the size of $\{(a,b) \mid 1 \le a,b ; ab \le n ; a \neq b \}$ and it's even because switching $a$ and $b$ is an involution with no fixed points.

Balarka Sen

that works. I just noted that the first bit is $\sum_{k = 1}^n \tau(k)$, and $\tau(k)$ is even unless $k$ is a perfect square in which case take a 1 from the last term and make it even

same argument

yours is more geometric

mercio

Induction is what I thoguht of first

Balarka Sen

mhm.

mercio

with the number of divisors of numbers yeah

Nehal Samee

Hello ... I have a query ... How can I get the period of $|sin x|+|cos x |$ and $sin 4x . cos3x$ ? .... I wanted a mathematical approach ... I have seen the graphical considerations ...

If I convert the second function into first , then I think providing help for the first can facilitate solving of the second ...

Silent

8:49 AM

@LeakyNun,

46 mins ago, by Silent

@MatheinBoulomenos, Let $G=\left\{\begin{pmatrix}a&b\\0&a^{-1}\\ \end{pmatrix}:a,b\in \Bbb R, a>0\right\}$ and $N=\left\{\begin{pmatrix}1&b\\0&1\\ \end{pmatrix}:b\in \Bbb R\right\}$, how do we know that $G/N$ is isomorphic to $\Bbb R$ under addition?

Balarka Sen

Nehal Samee

9:31 AM

@BalarkaSen now , how do I find it for the second function ... ?

Btw , @BalarkaSen ... Can you provide support for the period of $ sin x + tan x$ ... From graph , I get $\pi$ ... But how do I prove it ?

konoa

Hi chat

Alessandro Codenotti

10:00 AM

Hi @konoa @Dami

mercio

o..o

Akiva Weinberger

11:31 AM

@IceInkberry What about $xe^x$, which gets bigger with every differentiation! :P ($(xe^x)'=xe^x+e^x$; $(xe^x)''=xe^x+2e^x$; $(xe^x)'''=xe^x+3e^x$; etc.)

Szabolcs

11:56 AM

Since my question was thoroughly ignored, I'm just going to drop it here ...

0

Q: Upper bound on size of largest hole in a graph

Are there any known easy-to-compute upper bounds on the size of the largest hole (induced cycle) in a graph? I am trying to check if a graph is perfect using the strong perfect graph theorem and a bit of brute forcing: I search for successively larger induced cycles of odd length in both the gra...

graph-theory

Semiclassical

12:12 PM

@AkivaWeinberger yeah, though the Taylor series in that case is pretty easy

The case of repeatedly acting by x(d/dx) is even more dramatic in that regard

B. Mehta

@Silent: Define $\varphi$ a homomorphism from $G$ to the reals under addition by $\begin{pmatrix} a & b \\ 0 & a^{-1}\end{pmatrix} \mapsto \ln a$. The kernel is $N$, and it's clearly surjective so we have an isomorphism $G/N \cong \mathbb{R}$.

Semiclassical

Also, in the case you suggest you’ve got $e^{-x}D^n(xe^x)$

Which is nice since you can write down the power series for both e^-x and the nth derivative of xe^x, and thereby express the resulting polynomial in x by a Cauchy product

Fun stuff

Silent

@B.Mehta thank you!

skull

93

Q: Student caught cheating when leaving class after handing me the exam

Usually I am quite clear that cheating and plagiarism is unacceptable. Although every semester I have to deal with several cases of plagiarism, I had not expected students to brazenly cheat in exams. A few seconds after one student handed me the exam, as she was leaving, I noticed that she had wr...

cheating

Akiva Weinberger

12:32 PM

> I remember exams where people were allowed to bring one hand-written sheet of A4 paper (in the U.S., you could declare one sheet of "legal" legal)

eye roll

Semiclassical

I have a worse example

skull

please share

Semiclassical

One prof I TAd for had that and his problems were very similar to previous discussion/HW problems

So ofc people included solutions of such on their crib sheet

(And it was occasionally very obvious that they had done so)

skull

hmmm

Semiclassical

IMO you can do one of the two: have problems similar to prior practice, or allow them to bring a crib sheet

Should not do both

skull

12:39 PM

you don't advocate "open book" exams?

Semiclassical

This was in an intro physics context to be precise

Akiva Weinberger

The eye roll was for the pun, to be clear

Secret

I prefer exams that focus on understanding, not recognising problem patterns

Semiclassical

@skull not really, no

Secret

Because as we all knew, Nature is the strictest examiner in existence

Semiclassical

12:40 PM

@Secret depends what you mean by patterns

Secret

have problems similar to prior practice to the point of basically minor differences

Akiva Weinberger

The best test for a thing-building class is, "Can you build the thing"

but you can't do that it an exam room

Semiclassical

The trouble is that the notion of ‘pattern’ that an intro physics student has can be pretty shallow

Akiva Weinberger

and also I guess this only applies to engineering/design subjects

Secret

I once have undergrad exams where they literally reuse last year's questions

Semiclassical

12:42 PM

Especially if it’s the version of intro physics intended for pre-Med and bio majors

Akiva Weinberger

Fun story: My high school's history department usually releases a practice exam so students have what to study off of and know what to expect

and one year they accidentally released the actual exam

Semiclassical

The notion of pattern matching for them is to memorize formulas and problem descriptions

Akiva Weinberger

This is a final, I mean

skull

lol

Akiva Weinberger

so they had to suddenly create a new final exam

Secret

12:44 PM

Ignore that, my internet hiccup sniped

Akiva Weinberger

The teacher who released the test accidentally was not the same as the teacher who wrote the test (and was quite proud of the test questions, in fact)

Semiclassical

Which is fine in the context of a bio course, where you’re asked to absorb a huge amount of info and recall it on command

skull

bio majors have incredible memories

Akiva Weinberger

History and language courses are also info dumps

Semiclassical

Less so in the context of intro physics, where you can change the wording of written problems without chabging the principles you’d apply

There is a notion of pattern there, but it’s at the level of problem-solving technique rather than the problems themselves

My own preference for such exams, at any rate, is to provide an equation sheet for the class

skull

12:47 PM

not make your own?

Semiclassical

That alleviates the need to memorize so much stuff, while still requiring that they understand said formulas enough to apply them correctly

Right

Secret

I like that better, exams back in hong kong and china basically are testing student's memories on the problems themselves

Exams in aust, they can be formulae sheet, no formulae sheet or A4 single sided paper of notes, and they don't make the same mistakes as the chinese ones (except they sometimes like to reuse last year's problems)

Semiclassical

Give everyone the same toolset, and test them on their ability to use those tools

Secret

What do you think of an exam having a formula sheet, but some problems are reused from last year's?

Semiclassical

Depends on the level of the course.

For an intro course, the problems are necessarily simple enough that using the same problems verbatim is problematic

For upper division courses it makes more sense to me, since the solutions are more involved and therefore just seeing the problem ahead of time doesn’t mean you’ll automatically be able to solve it

For upper division courses in general I think the calculus is different.

Secret

12:52 PM

I see

Akiva Weinberger

$\frac{\rm d}{{\rm d}x}e^{2x}=2e^{2x}$ thisisgettingoutofhand.jpg

philmcole

Hi chat

skull

hi

$Mathematics$ JEE Maths Zone

Transcript for

$Mathematics$ Mathematics