(that said, i don't really get the definition of 'concurrence' here. half the lines in the family pass through one point, half through the other.)

@TedShifrin -+ 2

Thank you @Semiclassical @TedShifrin

Yeah, the problem is wrong. You need a single point as the solution.

@TedShifrin K. So , do you mean to say the option are wrong

Also , the Q

the point is that the problem doesn't really make a lot of sense as written

K.

so much for simple algebra...this isn't pleasant at the moment

some examples may help. some triplets: (a,b,c)=(1,1,2), (1,1,-2), (1,-1,0)

so the lines x+y=-2, x-y=0, and x+y=2

there's no points which are concurrent between all three of them, and there's no concurrences between the first and last

but the first two do concur at (x,y) = (1,1), and the last two concur at (1,-1)

i guess the point is that, if you pick two lines, then if they intersect then they must do so at either (1,1) or (-1,-1). but they may not intersect at all

except that's not true either

yeah, i dunno wtf this problem meant

i get the algebra they intended you to do, but the problem itself is nonsense

@Semiclassical Thanks a lot again. I’ll verify this Q

Another Q : Differentiation of$ x^2 * e^x$ = 2x*e^x . Right ?

remember the product rule

Yes

so what does $f(x)g(x)$ differentiate to?

Product rule. K. I got it. Let me show sth.

This was my ans with product rule but my options still don’t match

thanks, covid.

again, what's the derivative of $f(x)g(x)$

your options should match.

oh

yeah

what options did they give you?

@Semiclassical f’(x) * g(x) + f(x)*g’(x)

oh boy

Am I missing sth

we're all missing something. the right answer isn't there.

Wow.....

This test bank continues to be completely garbage.

Over and over.

@leslietownes Seriously. No one in my class complained !!

I hope that's not a test bank and is just a teacher making their own questions and solutions....

they're learning another lesson, which is not to object to authority.

There are two collections of lines.One collection corresponds to $(a+1)x_0+(b+1)y_0=0$ and this is possible only if: a=-1 and b=-1 so we get the equation: $x+y+2=0$

I don’t think so, dc3rd.

@dc3rd Yes. It is a worksheet .

@S.M.T wtf is that second part two

Oh, then fire the teacher.

@Semiclassical No solutions.

jeeze

Hello, if $ a \equiv b \bmod {m}$ then $m$ divides $a-b$, but is it legal to write it $m$ divides $b-a$?

our condolences

which gives $x_0+y_0+2=0$, where (x_0,y_0) is the point at which lines in this collection meet. Like wise the other collection of lines (corresponding to $c=-a-b$) satisfies $x+y=2$ and you can that d) satisfies these conditions.

i had some fairly bad instructors but at least i was never at the mercy of a test bank.

@TedShifrin it is like a DPP I.e Daily practice problem worksheet.

What do you think, Avra?

it reminds me of the scene in robocop.

@Avra take the trivial case a=b=0

Huh?

oh

I think that second question is doing with probability and all those values should sum to $1$

misread

So it seems that the question is not complete. We may observe that line in one collection is parallel to collection line 2) @SMT

k.

then clearly just solve for $k$....

@Koro the problem is that you can find lines which satisfy the constraints but which don't intersect at (1,1) or (-1,-1)

@TedShifrin. I think if $m$ divides $a-b$ then it should also divide $b-a$ :/

regardless of positive or negative signs

(to be clear, ignore my silly statement about a=b=0, i misread what you said)

if m divides 9-3 = 7 then it should divide 3 - 9 = -7

I understand it like this but algebrically confusing me

@semi: Hi. a or b can’t be zero. Please note that under the given constraints either both are 1 or -1

I think I should coin it with and keep mod in mind while playing with them algebricallY @TedShifrin. What do you think?

I’ll buy a new test series.😅

Thanks all still

@S.M.T: do you see why (a,b) is either (1,1) or (-1,-1)?

I've been getting some wild algebra that I don't think you intended on us doing for the solution Ted. I may be wrong, but it is getting wild in my notepad...

@Koro what?

there's no requirement in the problem that a,b=1 or -1

i'll clarify for the culturally ill informed. a company is demonstrating an automated robot cop. a lackey at the company who hopes to sell it is chosen to demonstrate it at a board meeting by pulling a gun on it. the robot demands that he drop the gun. he does. the robot still demands that he drop the gun, and when he can't comply with the machine's nonsensical instructions, it shoots him to death.

that's us and the test bank

Algebraically confusing, really? If $z=cm$, then $-z=(-c)m$.

:/

found it

@TedShifrin. Also understanding it with respect to mod also helps I guess

Damn, fire alarm sounding over the building. Quite deafening. Poor kitty freaked and hiding but can’t escape.

as mod will be same regardless of sign

that movie is far smarter than it pretends to be. so much of the last 30 years is in that one scene.

similar to absolute value

@leslietownes. What is your favourite movie!!

warning: blooooood

What mods are the same?

probably carol reed's "the third man."

@Koro not exactly

@leslietownes. I was expecting a match, but I relaized you I watched newer movies :/

classic orson welles

@leslietownes. Can I ask if you have newer one?

I mean movie came later than that?

i really like "the shining," i've probably seen it 10 times.

newer than that is tough.

@TedShifrin. I mean Prof I will try to understand them in terms of what mod means as it helps me a lot

@leslietownes. :( This is the one who eats people alive?

@Koro Did you get it ?

@leslietownes. I like the hero but not the content though :(

it's hard to describe the shining. a guy who is an alcoholic, or former alcoholic, agrees to be a caretaker for a snowbound hotel in the middle of winter and ends up trying to kill his family.

i liked the new david copperfield movie. it's not in my all-time top 10 but it is good, and funny. with dev patel.

@leslietownes. I would agree with you on the hero, he is unrepeated

there's a system of 3 equations in 3 unknowns on the blackboard in the background of the david copperfield movie. and yes, i solved it instead of paying attention to the scene.

as an example of what can go wrong, take 3 x+5 y+8=0,7 x+3 y-10=0

Happy Halloween guys anyway

Last Q

both of those satisfy the condition on coefficients, and they do intersect, but not at (1,1) or (-1,-1)

for halloween, my favorite movie might be "alien." we watched that a few years ago.

I need help with this Q.

my wife had never seen it and screamed during that one scene.

@S.M.T what are the possible events here?

@Semiclassical I did not understand the Q. I’ll explain

@semi, @SMT: I made a mistake. I was wrong. Please ignore my comments on the question.

Just do it brute force, SMT.

i watch all kinds of horror genre crap because it's what my degenerate friends are into, but it might be the best genre.

@TedShifrin learned a new work. Thanks

i mean: you can roll either 1,2,3,4,5,6. how many factors do these each have

@leslie: try post-apocalyptic genre. I think that might be the best genre out there.

@Semiclassical Ok. So 1 has 1 factor . 2 has 2 factors = 1,2 like that …

3 has 1,3 factors = 2 factors , 4 has = 1,2 i.e 2 factors

um

how does 3 have three factors?

Check 4, too.

5 has 1 factor , 6 has 1,2,3 i.e 3 factors.

the definition of "number of factors" is a bit ambiguous to me tho

Why?

@TedShifrin K sir. Shouldn’t 4 have 1 , 2 as number of factors

if factors mean "positive divisors", then that's fine

No, you’re counting wrong. The factors of a prime are 1 and the prime.

koro i watch those too. my favorite is "the road" from the book by cormac mccarthy.

@leslietownes that seems like a brutal book to be a favorite

@TedShifrin 1,2,4 I.e 3 factors

semi i know. i even had my wife watch it, i think last halloween. it was too much.

Right.

So, the numbers are, in order?

K. So , total factors = 12 @TedShifrin @Semiclassical

total factors is irrelevant

give the counts for each roll

@Semiclassical K.

also, keep in mind that 6 is also a factor of 6

@Semiclassical yes

Not irrelevant. We want the probability distribution.

??

28 days later is another good entry in the genre. my wife thought that one was OK.

@Semiclassical Each roll I.e total 6 rolls possible ?

yes. how many factors for each possible roll?

this is just asking you to write them all out at once. you've already written some of them above

For 6 rolls , total 12 factors. So , do I say 2 factors as probability for each roll ?

But it still depends , 6 has 4 factors.

@Semiclassical Not following you exactly

i don't know what you're doing. all I'm asking is: for each roll of the dice, you get 1,2,3,4,5, or 6. how many factors do these give respectively?

Die

oops

@Semiclassical 1 = 1 (1) , 2 = 2 ( 1,2) , 3 = 2 (1,3) , 4 = 3 ( 1,2,4) , 5 = 2 ( 1,5), 6 = 4 . Total 13

yes. just one die

@leslietownes. Alien is my favourite too!

Still error.

yeah, you've got one wrong

@leslietownes. Every studied Group Lasso theorem (just asking not question)

alien is a perfect movie.

Hi all

Alien is my favourite

I like fantasy

hi balarka

Found it @Semiclassical @TedShifrin

hi Sen

The algebra is still being a pain Semi.....something isn't working out. I would've thought it would be a little easier than this...

never heard of it. maybe my wife, but she's asleep.

that looks right. so you either get one factor, two factors, three factors, or four factors

@Semiclassical Yes

how many ways do you have to get each of these?

Hi, a Balarka.

@leslietownes I thought she left earlier today?

@Semiclassical Then , what to do ?

what to do is answer the question i just asked

depends on how you define 'today.' she's napping and going to the airport soon.

how many of your rolls lead to one factor, how many to two factors, etc

Ah.

and then nothing but candy, r-rated movies, and takeout food.

Don’t forget leeks!

and leeks.

@Semiclassical K. Probability distribution of X is either get one factor, two factors, three factors, or four factors

Much better.

@TedShifrin. @leslietownes. And some mathematics books and complex algebra just for fun

What does probability distribution mean?

still haven't answered my question

how many of the rolls (1,2,3,4,5,6) lead to one factor, two factors, etc

probability distribution gives the probabilities of occurrence of different possible outcomes for an experiment.@TedShifrin

@Semiclassical Yes.

So do that. Semi keeps asking you!

@Semiclassical 4 factors = 1 roll , 3 factors = 1 roll , 2 factors = 3 , 1 factor = 1 roll

right. so 1/6 of the rolls give four factors, for instance.

therefore: what are the probabilities?

@Semiclassical K.

@TedShifrin Almost done with the first half of stratified Morse theory. Excited for the second half, where the applications are.

Nice!

@Semiclassical 1/6 of the rolls give 3 factors , 3/6 of the rolls give 2 factors & 1/6 of the rolls give 1 factor.

right.

Missing one

@TedShifrin 1/6 of the rolls give four factors

@leslietownes. How much candy your kid got!

maybe 10 items. we went to three houses, then my kid was too afraid. one house gave us a handful of stuff.

@Semiclassical So , I wrote the answer for each part right.

we still have most of one costco bag left. the little kids are excruciatingly cute.

right.

@leslietownes. Cool. I got one item today. I saw a bunch of kids having many, then I said to myself you will get nothing from me :( I felt bad afterwards

Could one of you @TedShifrin or @Semiclassical help me with getting over this hurdle?

Just ask; don't ask to ask

@Semiclassical @TedShifrin Thanks a lot

i give some to the parents of little kids. i hate when they call me sir, i know its a sign of respect, but...

@BalarkaSen we've been over this, so it's not a new question

ah

Don’t worry, copper. We call you worse.

I was already discussing the question with them earlier Balarka so they already know what I'm talking about

Give me the three partials, @dc3rd.

yeah, no "sir" from me.

Oh Halloween

it's in the chat transcript, let me find it

WHy chat is not scrolling down with new comments....

I'll get it since it is my question

a lot of people were doing socially distanced stuff in their driveways. i hadn't seen that before but appreciated it. although i don't see the difference with just answering the door. my daughter began yelling after someone had a dog in their lap. it was all over at that point.

other than (covid) masks, i didn't notice any difference...

i miss having kids around.

well.....how did you do that so quickly Semi?

magic

here I was scrolling back......manually....

if i substitute $z=3xy-2x^3$ into $Df$ (using mathematica) i get $Df=[-24 x^5 + 48 x^3 y - 24 x y^2, 12 x^4 - 24 x^2 y + 12 y^2, 0]$

So $z=x(3y-2x^2)$, and factoring out terms when we can, what do the other equations give?

hmmm...I have an extra term in my substitution of $z$ into the $x$ partial

i prefer doing the algebra intelligently by hand, but whatever. ….

I'm here doing it by hand too....

start by getting rid of the 6 in the y partial.

ok

Insert our (factored) expression for $z$.

@leslietownes. No social distancing here!

doing that as we speak

$3x^{2}y - 2x^{4}-2y^{2}+xy$

now if I had kept it factored: $x^{2}(3y - 2x^{2}) - 2y^{2} + xy$ which alludes to a quadratic

now I see why Leslie was complaining about {}, he doesn't use chatjax......

You don’t need braces for exponents unless they have more than one digit.

Are signs and exponents right there?

Sloppy errors will kill you.

yes they are...I factored out a $-6$

Look at the last term.

what's wrong with it?........bloody hell....it is a squared....

Now things will combine. Get rid of a factor of 2. What happens?

they didn't combine cleanly: working with $3x^2 y - 2x^4-2y^2+xy^2 = x^2(3y-2x^2)-y^2(2+x)$

It’s $x^2y$ in the last term?

At this point you have to be ULTRA meticulous.

😭😭😭.... I see what you mean I took that last term from rushing and looking at the $x$ partial instead of the $y$ one....

Now I get $-x^4 + 2x^2y-y^2 = 0$ that should factor nicely which i'm going to get once I type this

so doing the "sneaky" quadratic substitution I let $t = x^2$ then set up a quadratic to use the quadratic formula. after solving that I arrived at $t = y$ which translates to $y = x^2$. I suppose I could plug this into my $z$ derivative to find what will eventually be $z = x^3$...

well, you already have $z=3xy-2x^3$

Come on, man. Factor the trinomial.

yes...just used that and got what I expected from it

You get $(y-x^2)^2$.

The x partial factors similarly.

truthfully I "saw" that coming about, but I couldn't rememebr the exact form

Oy.

$(a+b)^2=…$

a^2 + 2ab + b^2........:(

Yup.

the exponents were messing with my head.....I'll do better next time...

no bonus points for being "sneaky" though?...... 😊

one also has $f(x,y,3xy-2x^3)=-4 x^6+12 x^4 y-12 x^2 y^2+4 y^3=-4(x^6-3x^2y+3x^2 y^2-y^3)$ if you do the algebra right

seems I didn't do that algebra right either the first go round.....

do you spot the factorization in that case?

I do actually...factoring out an $x^2$ from the first two terms and then $y^2$ from the second two terms

so here's a question then.....I probably know the answer...... but how do you know "when to stop" if you're solution is going in the wrong direction? and on the other side how do you know when you have a solution if it is something that has no prior solution available?

rough day in algebra for me to day though........damn meticulousness of it.....

more to the point, $f(x,y,3xy-2x^3)=-4(x^2-y)^3$

When it’s a textbook problem you should expect things to work out if done correctly. If it’s a research problem, all bets are off.

Fair enough. Figured as much....I guess I was asking in terms of research problems because you are venturing into the "unknown" so what to do?

because you don't have a nice defintion you are trying to satisfy to reach a conclusion

Experience has to be a guide, but who knows …

i don't understand how one would come up with this example, mind

I see what you did there Ted...I like it

well you have the examples creator right here to ask him how he did do it Semi

How did I, in fact? And the parametrization of the hyperboloid of one sheet where both parameter curves are lines.

how does one do a technical math presentation, in the spirit of 'good presentation practices', especially if one is constrained to say a one hour presentation. It seems to me like unless the audience is already familiar with so much jargon and subtle but important things that will be used in the presentation, this is pretty much impossible without grossly deviating from 'good presentation practices' (not too many words on a slide, 'big picture' rather than finnicky details)

i do see that the $(x,y,z)=(t+s,t^2+2ts,t^3+3t^2 s)$ works

i ask because i will be presenting something ive been immersed in for the last few months to my advisor who has likely been doing other unrelated things, and i dont want to end up doing a 'bad presentation'

but only in the sense that the algebra works out

and sure, maybe when i become an expert I can convey some subtle technical argument in a few words, but at this stage that doesn't seem possible for me

Just for your adviser or for general audience?

advisor plus some of his other students

So they know all the background and fundamentals, a lot of notation.

(this is in analysis btw)

some of the background yeah

ive already tried to omit as much stuff that they probably know as possible but my presentation is still bloated

i also have this 'big picture' version which takes a lot of liberties omitting technicalities, but reading it through, i cannot imagine anyone would actually derive any meaning out of it, lol

@TedShifrin so ruled surfaces, eh.

I suggest starting with 10-20 minutes of broad outline and then picking one or two things to do in detail.

hmm, thats fair enoguh

For the one we were just doing, I wrote the parametrization and eliminated the parameters.

hmm

Be prepared to elaborate if they ask questions, porridge.

i guess one can generalize this to $(x,y,z)=F(t)+F'(t)s$ for various choices of $F(t)$

Right.

with $F(t)=(t,t^2,t^3)$ simply being an obvious choice

It’s a universally important curve.

SHould I know this form that Semi just quoted? I see it being the "linear" approximation.....but maybe I'm off

$(x,y,z)=F(t)+F'(t)s$

LHS makes no sense

I just copy and pasted what he wrote....

Parametric equations back in chap 2

Yup, he wrote garbage too.

lol

I just didn’t read.

Yea I see what you're talking about. It's just a parameterized form of the surface....same tools of analysis still apply

when is the day going to come that I could use these budding mathematical powers for evil though?

Dunno.

better put when will I feel ready?....because every time i feel a step closer i discover more that is out there that is unknown to me and I have to learn...

@Semiclassic Hint for the other question — projective geometry. The hyperboloid is projectively equivalent to the saddle $z=xy$.

yup, im well prepared for questions

well, maybe ill get thrown a curve ball i hadn't considered

This should not be an inquisition — more fact-finding.

yeah of course haha

id like to think i am ready for an inquisition though

hence certainly for some fact finding

@TedShifrin Hello sir , there

Quite a surprise I got today.

i) is correct with all of its options. My teacher confirmed

Won’t tell home how to solve it but she said yes , it’s correct

for (i) apply the product rule

@shintuku I know but my answer doesn’t match options

hm, the four options don't include my answer either

is there any previous context?

No

seems like there is a mistake

@shintuku Nope, I just confirmed

you can use your symbolic math system of preference to confirm that the derivative is not in the four options

@shintuku I don’t know about it. Would you please teach me

wolframalpha.com/input/?i=%28x%5E2e%5Ex%29%27

@shintuku thanks , I’ll check it.

@S.M.T just b/c your teacher confirms something, doesn't mean they're right

the nice thing about math is that people can be wrong :)

the derivative of x^2 e^x is 2x e^x+x^2 e^x, full stop

(this doesn't mean you should tell them they're wrong---life is too short to get stuck on such things---but it is wrong)

@Semiclassical k.

This has answer d)

Hey guys! I am kinda curious in somethin. When looking at a hexagon (6-sided entity) and wanting to calculate the area, could you in theory treat the hexagon as 4 triangles and one square in the middle like this?

ibb.co/QPXXDm8

@S.M.T Given any a and b in R, there are two such families: 1) ax+by+(a+b)=0 and 2) ax+by-(a+b)=0. All straight lines in the family 1 pass through (-1,-1) as a(-1)+b(-1)+(a+b)=0 and all straight lines in family 2) pass through (1,1).

I think that’s what the question wanted.

Answer is d)

@Koro

@S.M.T That is not technically correct. What Koro said is correct. The entire family is not concurrent at any point, but there are two families; one concurrent at $(1,1)$ and the other concurrent at $(-1,-1)$.

K.

@NemanjaVuksanovic no, it's not

derivative of x^2 e^x is (x^2+2x)e^x

nice to know that that's still going.

@Koro i agree with this reading of it. It does lead to an interesting question: suppose you randomly pick one member of the first family and one member of the second. what's the distribution of intersections?

the sticking point here is what "random" means, mind

some of them give the right answer for specific values of x. it is always a waste of time to speculate but i wonder if a draft of the question involved "at x = k" and the draft answers did too and then someone decided to 'generalize' the answers without thinking about the product rule.

yeah

i did something like that writing a final exam once. it really frightened me because i only discovered it after i'd turned grades in, and it's a huge paper trail and hassle to change anything after that.

thankfully, 95% of the class 'read my mind' and knew what i was asking for and the 5% who didn't, it didn't make a difference in the ultimate grade.

also it wasn't multiple choice, so the grading process involved the input of a sane human and not just comparison of letters.