Mathematics - 2020-02-19

Leaky Nun

12:55 AM

@Astyx bon-minuit

Astyx

1:12 AM

Plus deux heures du matin en ce moment

Leaky Nun

@Astyx alors on dit quoi?

bon-presque-matin?

mdr

Astyx

On dit pas vraiment

geocalc33

bonjour

Jourbon

Sorbonne

bonnesoir

Lukas Heger

1:31 AM

hi @Leaky

Leaky Nun

@LukasHeger hoj

Lukas Heger

How are you doing?

Akiva Weinberger

5:35 AM

Pyramid schemes are recursive scams

Balarka Sen

6:08 AM

@AkivaWeinberger The Hopf fibration is unit circle bundle of a complex line bundle over $S^2$, so in particular fiberwise orientable.

adesh mishra

6:31 AM

@TedShifrin I don’t know why but I’m getting some strange doubts in Vector Calculus, it may be because I learned Vector Calculus not from a Mathematics book but from a Physics Book. I came across an argument in my book which says $$\oint dU =0$$ for any scalar $U$. I asked someone about how was it true and he told me that $$\oint dU = U \bigg |_x ^ x = U(x) - U(x) = 0 $$

Because of this I got a doubt of why for any vector field $$\oint \mathbf A \cdot d\mathbf l =0$$ is not true. If we find the anti-derivative of $\mathbf A$ and then apply Fundamental Theorem of Calculus Part 2 wouldn’t we gonna get zero?

Ted Shifrin

6:52 AM

@adesh: The statement $\oint dU=0$ is your latter statement only for $\mathbf A = \nabla U$. Nonconservative force fields do not have "antiderivatives." You should know Green's Theorem. But even if not, try $\mathbf A = (-y,x)$ in the plane.

Leaky Nun

@AkivaWeinberger Que dijo el numero 1 al numero 10?

Akiva Weinberger

Qué

Leaky Nun

@AkivaWeinberger "para ser como yo, tienes que ser sincero"

Akiva Weinberger

rolls eyes

Leaky Nun

:P

Akiva Weinberger

6:57 AM

Night

Leaky Nun

noches

@TedShifrin is there a bound to the number of singularities of a plane curve of degree d?

adesh mishra

7:38 AM

@TedShifrin Can you please explain how $$\oint dU =0$$ ? By the way how do you know that $\oint dU =0$ is the latter statement of $\mathbf A = grad~ U$ and then doing $$ \oint \mathbf A \cdot d\mathbf l = \oint \A_x dx + A_y dy + A_z dz = \oint dU + dU +dU = \oint dU$$

@TedShifrin Can you please explain how $$\oint dU =0$$ ? By the way how do you know that $\oint dU =0$ is the latter statement of $\mathbf A = grad~ U$ and then doing $$ \oint \mathbf A \cdot d\mathbf l = \oint A_x dx + A_y dy + A_z dz = \oint dU + dU +dU = \oint dU$$

Do you got some magic tricks :) ?

dsm

10:33 AM

@AkivaWeinberger I recall talking with you about some representation theory. I'm trying to show that if two representations $(\Pi_1,V_1)$ and $(\Pi_2,V_2)$ are equivalent then there exists appropriate bases $\mathcal{B}_1$ and $\mathcal{B}_2$ such that $[\Pi_1(g)]_{\mathcal{B}_1}=[\Pi_2(g)]_{\mathcal{B}_2}$. I'm confusing myself. If they are equivalent, there exists an intertwiner $\phi:V_1\rightarrow V_2$ such that

$$\phi\hspace{1mm}\circ\hspace{1mm}\Pi_1(g)=\Pi_2(g)\hspace{1mm}\circ\hspace{1mm}\phi\hspace{3mm}\Rightarrow\hspace{3mm}[\Pi_2(g)]=[\phi][\Pi_1(g)][\phi]^{-1}.$$

To have $[\Pi_1(g)]_{\mathcal{B}_1}=[\Pi_2(g)]_{\mathcal{B}_2}$ doesn't this just require $[\phi]=1$, in which case the basis $\mathcal{B}_2$ should be the same basis used for $\mathcal{B}_1$ represented in the standard orthogonal basis of $V_2$? Is that correct?

Archer

If I an given rank(A^2) what can I say about rank(A)?

Specifically, if rank(A^2) for 5*5 A is 2, then what will be the rank(A) ?

Howard

I have a doubt regarding bounties

Suppose I started a bounty today

Got the required answer

Now, can I award my bounty today to that answer

Archer

10:52 AM

Hi @LeakyNun !

geocalc33

11:11 AM

0

Q: curves on a plastic container using sunlight and blinds

What is going on in this image mathematically, and are there some neat underlying mathematical themes relating to this image? I did a little experiment by holding a translucent plastic container to some closed blinds on a bright sunny day. I think the even discretization of the light ...

inc.com/geoffrey-james/…

meditation can reverse-age a brain? cool.

MadSpaceMemer

11:24 AM

Can someone help me with the following problem. I wrote it down and defined what is what so no misunderstandings arise

This is in my script . However i do not see how is this true? Since a vector for an Eigenwert is defined to be inequal to the null vector, However the null vector is truly in the other set. How could the sets equal eachother since one of the sets contain a vector that is not in the other set?

I should probably mention that $f$ is a linear function from $V \to V$

So should it not be left side $\subset$ right side but right side $\not\subset$ left side

Leaky Nun

just add zero

MadSpaceMemer

huh?

Leaky Nun

define Eig(f) to be { v | f(v) = λv } U {0}

MadSpaceMemer

But that would break my professors Definition.....

Leaky Nun

then just claim that Eig(f) U {0} = ker(λid-f)

the point is

0 is the only difference

other than that the two sets are exactly the same

MadSpaceMemer

11:39 AM

Yes i understand I agree

I added now Zero and its correct

I knew that zero is only difference but it is wrong to write equality of sets withtout adding Zero I though I am missing something out but maybe he just did not notice

Leaky Nun

oh right

Eig(f) = { v | f(v) = λv } is alright

because 0 is in that set also

the problem is that Eig(f) is not the set of Eigenwerts of f lol

it's Zero mit die Menge Eigenwerts von f

MadSpaceMemer

I did not write full definition, mz apologies

V should not be the null vector

Leaky Nun

lol

MadSpaceMemer

Well i copied it, but it is defined the previous page to be strictly none null vector

Leaky Nun

anyway

it doesn't matter

the nullvector is not the point

NoName

12:00 PM

Does anyone have access to Mathematica/Maple?

I'm interested in the value of this integral:

$\displaystyle \int_0^{\pi/2}\frac{e^{-x}\sqrt{\cos{x}}}{\sqrt{\cos{x}}+\sqrt{\sin{x}}}\,dx$

Leaky Nun

0.452394, says wolframalpha

NoName

I was gonna say I'm not sure if Wolfram is getting it wrong after 2/3 decimal places.

Would the Mathematica do better do you think @LeakyNun? I don't know a lot about these softwares.

Leaky Nun

no idea

northerner

In follow up to a chat the other day regarding PEDMAS and order of operation (youtube.com/watch?v=y9h1oqv21Vs)... if it's arbitrary why not just always evaluate left to right? Wouldn't that be easier? So then $a+bc \neq bc + a$

@LeakyNun

Leaky Nun

no clue

northerner

12:07 PM

In other words why do we have order of operation?

Leaky Nun

because sums of products are useful i guess

and we wouldn't want to write (ab)+(cd)+(ef)+(gh)+(ij)

if it occurs so often

polynomials for examples

northerner

hmm

Thorgott

12:22 PM

turns out that taking an exam involving ODEs without knowing how to solve a linear system of differential equations was not the best idea I've ever had

skullpatrol

12:42 PM

@northerner there are more than enough details here :P

northerner

+1 for "The internet has been so bored lately"

skullpatrol

:D

user

Are fields closed under arbitrary products?

mick

1

Q: $ \int_R | \frac{ \cos(x) + \cos( \sqrt 2 x) + \cos( \sqrt 3 x)}{1 + x^2} | dx = ? $

Integrals where the integrand contains an absolute value can be very hard or impossible to express in closed form. Computing the zero’s of the integrand might help. But what if the zero’s have no closed form or there are infinitely many ? We can express any single zero by a contour integral but...

real-analysis calculus integration closed-form absolute-value

Im still puzzled

Also this one is still a mystery

2

Q: $ \sum_{n = 1}^{\infty} ( \frac{p_{2n-1}}{p_{2n}} - \frac{p_{2n}}{p_{2n +1}} ) = ?? $

Let $ p_n $ be the $n$ th prime. I was confused about the following idea. $$A = \sum_{n = 1}^{\infty} ( \frac{p_{2n-1}}{p_{2n}} - \frac{p_{2n}}{p_{2n +1}} ) $$ Very confused actually. Does this even converge ? Do we need a summability method ? Does its converge depend strongly on conjectures...

sequences-and-series limits prime-numbers summation-method

And no answer to this yet

1

Q: New commutative hyperoperator?

After reading about Ackermann functions , tetration and similar, I considered the commutative following hyperoperator ? $$ F(0,a,b) = a + b $$ $$ F(n,c,0) = F(n,0,c) = c $$ $$ F(n,a,b) = F(n-1,F(n,a-1,b),F(n,a,b-1)) $$ I have not seen this one before in any official papers. Why is this not con...

reference-request commutative-algebra tetration hyperoperation

Thorgott

1:08 PM

I'm not sure what "arbitrary" means, but for any sensible definition of arbitrary, the answer it yes

user

@Thorgott For me countable is enough.

so for countable it works, yes?

But: Q is a field and not closed under countable products?

Since every irrational number is an infinite product of rationals.

Thorgott

1:23 PM

I don't think there's a sensible definition of countable products on arbitrary fields

and indeed, it isn't true for Q either, despite the definition at least potentially making sense

user

Thanks alot anyway

user193319

2:34 PM

Let $x$ be an element in a $C*$-algebra. Is it true that $x$ is invertible if and only if $x^*x$ is invertible?

The forward direction is trivial.

courge9

Consider the square root of a function f on the complex plane. I understand that the branch points for sqrt(f) are given by the zeros of f. What now if f is holomorphic and does not vanish anywhere, like f(z)=exp(z)? It seems that then sqrt(exp(z)) has no branch point in 0, so it should be well defined... but I'm aware that there are two square roots, namely plus or minus exp(z/2). Where is the mistake in my line of thought?

Leaky Nun

2:50 PM

@courge9 each branch is well-defined

courge9

@LeakyNun Yes, I see that.

But zeros of f are branch points of sqrt(f), and f=exp has no zeros.

Which would mean that sqrt(exp(z)) has no branch points

but I think the mistake here is that while every zero of f is a branch point of sqrt(f), the converse doesn't hold

So sqrt(exp(z)) still has a branch point in 0... but why?

Semiclassical

An idea: there are two branches, but there’s no way to analytically continue from one to the other

As in, if you analytically continue sqrt(exp(z)) along any loop in z, then you’ll just end up with the same value you started with

Rather than possibly picking up a minus as you would for sqrt(z)

courge9

ok, so sqrt(exp(z)) is entire?

adesh mishra

@Semiclassical Can you tell me why $$\oint dU =0$$ for a scalar field $U$ ?

Semiclassical

Right.

courge9

3:04 PM

But then what's the matter with -exp(z/2) as square root of exp(z)? This looks like sqrt(exp(z)) still has some ambiguity which is only resolved after choosing a branch

How does this need for choosing a branch come into play, when there is no branch point at zero?

Thorgott

This doesn't look more ambiguous to me than sqrt(x) itself is ambiguous (in the sense that there are two square roots of any complex number)

Semiclassical

@courge9 sure. But that’s not the issue. The issue is that, if you analytically continue f(z)=sqrt(z) around zero, then you’ll end up with f(z)=-sqrt(z)

With sqrt(exp(z)), you never have the opportunity to reach that other definition

You can start with that other definition instead , but you aren’t forced to consider both simuy

courge9

So if I understand you correctly, once one chooses a branch cut for sqrt (e.g. the negative real axis), thus defining sqrt as well-defined function, sqrt(exp(z)) is an entire function (without any branch cuts)?

(which equals exp(z/2) upon this choice of branch cut for sqrt)

Semiclassical

@adeshmishra suppose you parametrize your path as $\gamma(t)$ for $t\in[0,1]$ Then, upon pulling back to integration over $t$, you just have a 1D integral with antidetrivative $U(\gamma(t))$. But $\gamma(0)=\gamma(1)$ (it’s a closed path) so the fundamental theorem of calculus yields $U(\gamma(1))-U(\gamma(0))=0$.

(In other words, you do substitutions to turn the line integral into a standard 1D integral, and argue that said integral vanishes)

The other workhorse in this computation is the multivariable chain rule, though

adesh mishra

@Semiclassical Is there any other way? Are you saying $$\oint dU = \int_{x}^{x} dU = 0$$ ?

Semiclassical

3:18 PM

No, though there’s a kernel of truth in that presentation as well

adesh mishra

Can you explain me in a way so that I wouldn’t get the doubt that for any vector field $\oint \mathbf A \cdot d\mathbf l $ is not necessarily zero?

Semiclassical

I’m saying that, once you use the parametrization, you end up with a scenario like $\int_0^1 f’(t)\, dt $ with $f0)=f(1)$.

@adeshmishra because $dU=(\nabla U)\cdot d\mathbf{l}$

adesh mishra

@Semiclassical Yea I agree with that, then?

Semiclassical

So it’s the special case $\mathbf{A}=\nabla U$

That is, your line integral does vanish if the vector field A is the gradient of a scalar field U

But that’s not always true

adesh mishra

Please help me in seeing that how $$\oint \nabla U \cdot d\mathbf l =0$$.

Leaky Nun

3:22 PM

Stokes theorem

no I mean what's wrong with what he said?

10 mins ago, by Semiclassical

@adeshmishra suppose you parametrize your path as $\gamma(t)$ for $t\in[0,1]$ Then, upon pulling back to integration over $t$, you just have a 1D integral with antidetrivative $U(\gamma(t))$. But $\gamma(0)=\gamma(1)$ (it’s a closed path) so the fundamental theorem of calculus yields $U(\gamma(1))-U(\gamma(0))=0$.

expanding the definition is the way to go

adesh mishra

I know Stokes’s Theorem in this way $$ \int \mathbf A \cdot d\boldsymbol{\sigma} = \oint \mathbf A \cdot d\mathbf l$$

@LeakyNun I couldn’t understand what he said.

Leaky Nun

how do you define $\oint$?

or $\oint \ \mathrm dU$ for that matter

courge9

@Semiclassical Have I understood this correctly (see my last message)? (Sorry for interrupting the current discussion, and thanks for your answers)

anakhro

Hi all.

adesh mishra

@LeakyNun I think we have to integrate from some $(x,y,z)$ and through some path we have to come back again at $(x,y,z)$

Leaky Nun

3:26 PM

and how is that defined?

let's say the path is $\gamma(t)$ for $t \in [0,1]$

so $\gamma(0) = \gamma(1) = (x,y,z)$

now how is $\oint \mathrm dU$ defined?

adesh mishra

@LeakyNun I’m not getting what is $\gamma (t)$ and why it belongs to an interval of zero and one.

Leaky Nun

give the path a name

adesh mishra

okay let’s call it gamma

Leaky Nun

what now

adesh mishra

What is t?

Leaky Nun

3:29 PM

ok let's say $U = x^2+yz$, and the path is a unit circle centred at the origin

can you calculate $\oint \mathrm dU$?

adesh mishra

Yeah

Leaky Nun

do it

calculate $\oint \mathrm dU$

adesh mishra

$$U = x^2 + yz \\
dU= 2x~dx + z~ dy+ y~dz$$
Am I right this far?

Leaky Nun

yeah

adesh mishra

$$\oint dU = \int_{0}^{0} 2x dx + \int_{0}^{0} z dy + \int _{0}^{0} y dz = 0$$

Leaky Nun

3:35 PM

why?

what is 0?

adesh mishra

You told me to integrate along the unit circle

Leaky Nun

and?

how does that give you $\oint dU = \int_{0}^{0} 2x dx + \int_{0}^{0} z dy + \int _{0}^{0} y dz$?

adesh mishra

for unit circle x goes from 1 to 0 to -1 to 0 to 1

y also goes from 0 to 1 to 0 to -1 to 0

and z doesn't go anywhere

Leaky Nun

now I can see why you would be confused

adesh mishra

:)

That's nice

Leaky Nun

3:38 PM

@Semiclassical I would appreciate if you would join in

courge9

me too, but regarding my question, lol ;)

adesh mishra

Leaky are you not going to help me ?

Leaky Nun

eh...

that isn't how path integral is defined lol

you can't just separate into the x y z components

the third integral doesn't even make sense

what's $y$ in $\int_0^0 y \ \mathrm dz$

so go back to the definition of path integral

from your textbook or whatever source you're learning from

adesh mishra

Okay I will wait for @TedShifrin to come for rescue :)

Leaky Nun

you don't have a textbook?

adesh mishra

3:43 PM

I have

hahahhah

Leaky Nun

now find the definition of path integral and tell me

adesh mishra

just a moment

Leaky Nun

in the meantime integrate dU along the straight line path from (0,0,0) to (1,1,1)

adesh mishra

It's straightforward, I just have to plug lower and upper limits

Leaky Nun

show me

Semiclassical

3:46 PM

@LeakyNun had to deal with something else and now I’m busy

adesh mishra

$$ \int_{0}^{1} 2x dx = x^2 \bigg|_{0}^{1} = 1 \\
\int_{0}^{1} z dy = z y\bigg|_{0}^{1} = z \\
\int_{0}^{1} y dz = y z\bigg|_{0}^{1} = y $$

Leaky Nun

yeah... you won't get a number that way

adesh mishra

yeah

Leaky Nun

the path integral should give you a number

adesh mishra

Yeah

Leaky Nun

3:47 PM

not some expression in terms of z and y

adesh mishra

why?

Leaky Nun

because it's integrating a known function along a known path

naturally the answer is a known number

adesh mishra

Should I quote the definition of line integral?

Leaky Nun

sure

adesh mishra

>If $f$ is defined on a smooth curve $C$, then the line integral of $f$ along $C$ is $$ \int_C f(x,y) ds = \lim_{n \to \infty} \sum_{I=1}^{n} f(x_i, y_i) \Delta s_i$$ if this limit exists.

Leaky Nun

3:50 PM

ok that isn't very helpful (well that's a Riemann sum)

are there any formulas that help you compute a line integral?

adesh mishra

I don't think so. I learned line integrals from Physics Textbook Feynman Lectures on Physics Vol 2 and the definiton that I have given you is from Stewart's Calculus

Leaky Nun

give me the exact title

adesh mishra

Of what?

Leaky Nun

Stewart's Calculus

adesh mishra

That was the exact text that I quoted above

Leaky Nun

3:53 PM

the title of the book

which edition?

adesh mishra

Calculus Early Transcendentals Sixth edition

Should I send images of section Line Integrals ?

Leaky Nun

no need

Alessandro Codenotti

Do the singular (co)homology functors $H_n\colon\mathrm{Top}\to\mathrm{Ab}$ have adjoints?

adesh mishra

Leaky are you thinking about a way to explain me or do I need to re-study and then come back to you (if you don't mind terribly)

?

Leaky Nun

@adeshmishra you should read through and understand this example on P.1038

@AlessandroCodenotti so you want a functor $\mathcal F: \mathbf{Ab} \to \mathbf{Top}$ such that $\operatorname{Hom}(H_n(X), G) = \operatorname{Map}(X, \mathcal F(G))$ (or reverse)

adesh mishra

4:03 PM

@LeakyNun Okay, I will come back later

Leaky Nun

@AlessandroCodenotti that reminds me of Eilenberg--MacLane spaces

$[X, K(G,n)] \cong H^n(X;G)$ @AlessandroCodenotti

$H_n$ cannot be a right adjoint because it doesn't preserve limits (Künneth formula)

now it does preserve coproducts

but it doesn't preserve quotients

$S^n = D^n/S^{n-1}$

@AlessandroCodenotti so I don't think it has adjoints on either side

Alessandro Codenotti

Oh, right, homology preserves very little

Makes sense

Leaky Nun

@AlessandroCodenotti but Eilenberg--MacLane spaces might be what you want

user10478

5:11 PM

I'm having trouble figure out what type of variable $z$ is here: math24.net/reduction-order-page-2/#example6 Their $y'$ doesn't seems to work out whether $z$ is an independent variable or a function of $x$, and if it's a function of $y$, the integral is trivial and $y = e^{xz}$.

Ted Shifrin

6:45 PM

@user10478 Since $'$ denotes differentiation with respect to $x$, everything is a function of $x$ here. To be more precise, they are setting $$y(x) = e^{\int_a^x z(t)\,dt}.$$ Then the fundamental theorem of calculus gives $y'(x) = z(x)y(x)$ and $y''(x)=z'(x)y(x)+z(x)^2y(x)$.

You could have defined $z(x)$ directly by taking the derivative of $\log(y(x))$, but that's not any easier to work with than this implicit construction.

@adeshmishra: You need to relearn the basics. Line integrals (path integrals) are defined by parametrizing the curve (path). Only when you know the integral is path-independent can you find a potential function and use the fundamental theorem of calculus.

user10478

7:29 PM

@TedShifrin If we use the substitution $Z(x) = \int z(x)\ dx$ (for simplicity, taking the constant of integration to be $0$), then by the chain rule $(e^Z)' = Ze^ZZ' =\int z\ dx\ e^{\int z\ dx}z$, right?

Ted Shifrin

No. There's no $Z$ in the front there.

$\int z(x)\,dx$ is just a symbol for a function whose derivative is $z(x)$.

Unless you put limits on the integral, it is not a well-defined function.

Anyhow, your chain rule is wrong.

user10478

How so?

Ted Shifrin

The derivative of $e^Z$ is $e^Z\,Z'$.

user10478

Ahh, right, I guess I was doing $(e^{Zx})'$

Ted Shifrin

Derivative with respect to what?

I don't think you were doing anything that makes sense.

user10478

7:44 PM

Right

Ted Shifrin

You certainly didn't have the product rule.

Anyhow.

user10478

So that makes their derivative work out then.

Ted Shifrin

Not surprisingly.

@Leaky: With regard to your question from a while ago, yes, the degree gives some bounds on singularities. For example, a conic can have a double point (two lines) but nothing more or worse. A cubic can have a double point or a cusp, but nothing more or worse. There are pretty Plücker formulas relating singularities and singularities of the dual curve.

Leaky Nun

8:03 PM

@TedShifrin nice

Jack Ohara

9:34 PM

How many numbers satisfy this relation

x^y == x mod(N)

for fixed y and N

for the simple case we might assume N is a product of two primes and that y is invertible modolo phi (N) eulers function

anakhro

11:15 PM

Given that the torus is a 2-covering space of the Klein bottle, can we determine the fundamental group of the Klein bottle using this?

Ted Shifrin

The torus is also a 2-sheeted covering of itself.

So you need to use (a lot) more information.

anakhro

In the same way, indeed.

This question just went "show that the torus is a 2-sheeted covering of the klein bottle. compute the fundamental group of the klein bottle.".

So I was very unsure how it wanted me to compute it.

topologicalmagician

Hi guys

does the Hausdorff dimension always exist?

Ted Shifrin

Presumably by considering the group of covering (deck) transformations, @anakhro.

anakhro

@topologicalmagician I think so.

Ted Shifrin

11:21 PM

@topologicalmagician: No. What if your space is the union of a line and a plane?

anakhro

With what metric, Ted?

Disjoint union?

Ted Shifrin

Sitting in $\Bbb R^3$ ... yeah.

I guess it doesn't have to be disjoint. Just make sure the line isn't contained in the plane.

I guess I could take unions of Cantor-type sets of different Hausdorff dimensions, too.

But isn't Hausdorff dimension defined as the inf of the $r$ so that $\mathscr H^r(X)=0$?

So I guess this would pick up the maximum of my dimensions in the cases I've described.

anakhro

Yes, isn't that always defined?

Ted Shifrin

So as long as $X$ isn't infinite-dimensional (so a countable union of things of increasing dimensions?) ... I guess ...

anakhro

Are you interpreting his question as "is it always finite"?

Maybe that's what he meant, but I consider infinite Hausdorff dimension to still "exist".

Ted Shifrin

11:30 PM

We don't know what his course is taking as definitions.

Now that you're teaching calculus, what do you think of the conversation I just had here?

anakhro

A classic example of a student who doesn't yet understand that a "proof" is only something which is sufficiently evident for the reader, not the writer.

Ted Shifrin

Well, the student doesn't want to be responsible for convincing the teacher he actually "knows" anything :P

anakhro

Heh, then I guess no one should be surprised that the teacher doesn't want to be responsible for giving him full marks.

Ted Shifrin

But part of the point, I guess, is that students have to learn that some precision in language is important. Saying $f(x)=0$ is a function isn't a good thing.

But since they've been sloppy for 15 years already, why have to get less sloppy now? :D

anakhro

Indeed, there is little incentive.

Ted Shifrin

11:36 PM

Why learn to write sentences and paragraphs correctly, too?

LOL, look at the student's final retort. But he thanked me :D

anakhro

With that response, I am not sure what he actually got out of the exchange.

Jack Ohara

@TedShifrin Hi

Ted Shifrin

Indeed, @anakhro. I wasn't an a***ole in the discussion, either :P

hi @JackOhara

anakhro

So in that situation, how do you think the teacher should have given more incentive to write properly, @TedShifrin?

Ted Shifrin

How do we know what the teacher said? I suspect the teacher was clear about what was expected in a solution.

anakhro

11:45 PM

Presumably, and the marks (not) given by the teacher provide a basic incentive to not be so sloppy "next time". However, the student did log on to stackexchange, and did post a question about it, and did continue to be sloppy, and did insist he was not that sloppy.

Ted Shifrin

Well, next time he posts, you can be more blunt than I was.

Most of the time in the 8 years I was associate department head, when a student brought me an instructor's grading to complain about, I went through it carefully and showed the student question by question what marks I would have assigned. Most of the time, my grading (which the student agreed was fair as I explained it) came out lower than the instructor's.

Heh. Golden.

Hi

Hi, @Astyx.

Who are you talking about ?

Ted Shifrin

11:50 PM

I linked it up 20 lines or so.

anakhro

Like I mean to say, it might be completely out of the hands of the teacher at that time. That is, the teacher did the best thing, and it is a problem of the history of the student's learning.

Indeed, I'd lean that way myself.

But I wonder if you have anything that you think that a teacher can do on top of that?

One of my favourite professors would write "come see me" instead of an explanation of why the answer was wrong.

Of course this leads to a lot of students not coming to see him. :P

But for those who did, I think it made a big diffeence.

Ted Shifrin

When I taught courses like this, when I returned the exams, I was careful to make explicit comments to the class about issues I had seen. And I often posted solutions to the exams and told them they should understand those carefully.

My big gripe over my 35+ years of teaching is that I spent countless hours grading homeworks (in upper-level courses and graduate courses) and writing lots of comments, and very few students even read (let alone digested) my comments. But I continued to put in the effort for all the years. Some students in each class definitely improved and thanked me.

anakhro

So that's your gripe, but would you do it differently if you got to do it again?

Ted Shifrin

LOL, you mean for another 35 years?

I was a rapid and fastidious grader up until the last homework and the last exam.

anakhro

Heh. More like if you got to go back in time and change your marking strategy. :P

(I absolutely forbid you to answer the question with "hire a grad student")

Ted Shifrin

11:54 PM

No, I did this where most other people did use grad students.

And in my honors calculus (and multivariable math) classes, I did hire a star undergrad who had taken the course from me and knew it well.

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