help-43

[carmine garden]

yea, it's free on his websiet

I love it because it avoids matrices and dets as much as possible

[upper carbon]

this is why i hate the book

[carmine garden]

Linear maps are so much nicer to think about

I mean at uni ( If I get this class), we'll probably use FIS

[upper carbon]

i think it is worth going back when you’re done with axler to some book and do a ton of determinant problems tbqh

[carmine garden]

yea, FIS probably

[upper carbon]

in any case, its good that you’re learning it and i don’t care what book you use as long as it gets you to do math

because that is one of the best things that you can do for your development as a mathematician

is to like it and do it often

[carmine garden]

yea, thanks

I'll probably be back soon, i have done isomorphisms and invertibility in LA 1, will just see if the theory is anything new, if nto i'll go straight to problems!

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[carmine garden]

I don't really get this proof

[carmine garden]

3.81 , and 3.82

I mean intutively it make sense

but the proof is confusing the hell out of me

Like my proof would be something like CBC^{-1} takes {v_1,v_2, \dot,v_n} \to {u_1,u_2,\dots ,u_n } \to {v_1,v_2,\dots, v_n}$ again, which is what A does

but that's just the intution

<@&286206848099549185>

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[deft tangle]

Three parallel and perpendicular forces act at the vertices of a triangle ABC. The resultant of these forces passes through the circumcentre of the triangle, if

[compact pewter]

Please don't occupy multiple help channels.

[plain smelt]

what is the entire question?

[kind viper]

"three parallel and perpendicular forces"...

[compact pewter]

@deft tangle Has your question been resolved?

[deft tangle]

@plain smelt

[unkempt sail]

Parallel and perpendicular?????

[plain smelt]

Does the question provide a diagram?

[deft tangle]

No

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[sick shard]

Does a sequence need to be infinite to call it's sum a series?

[sick shard]

I thought finite series exist

[mortal bay]

no series is just

sum of a sequence

[kind elm]

Generally we consider infinite terms yes

There is no point to a finite series, you can just call it a sum

[sick shard]

But I have worked with finite series in class 10th so a series can be finite or infinite but generally we work with infinite series in higher classes? Right?

[kind elm]

Uh, sure

Generally there is nothing interesting about "finite" series so it doesn't warrant a lot of study

[sick shard]

Ok

[kind elm]

In general though I will say, when you talk about a sequence, it has infinite terms, and when you talk about a series, it also has infinite terms

[stuck wigeon]

Hmm idk I remember using it for something in my real analysis class this year

Don't remember why

[stuck wigeon]

In all classes I've taken yes

[stuck wigeon]

Yes exactly

[silver canopy]

A series is simply the sum of the terms of a sequence. If the sequence is finite, the sum is called a finite series. If the sequence is infinite, the sum is called an infinite series

But anyways, there isn't really much point to worrying this much about semantics

[kind elm]

That's interesting

Unless you mean partial sums

[patent yoke]

I would never consider the phrase series to mean anything but an infinite sum. There is a good reason why the terms "partial sums" and "finite sum" exist.

[kind elm]

Which is cheating

[stuck wigeon]

Yes

I think it was to bound a limit or something irdk

[kind elm]

Yeah the limit of a series is defined as a limit of its partial sums

So you can peg it back to sequences and use the same machinery

[compact pewter]

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[carmine garden]

Chat could I have a hint

rank nullity comes to mind

I first want to show if blah blah blah , then dim(null(s)))= sum(Null(T))

[pine osprey]

If they are invertible there ker is 0

[carmine garden]

This is the right idea I suppose?

yea

but T's kernel may be non -zero

[pine osprey]

I can't a translation of this theorem

[pine osprey]

Yes

[carmine garden]

I suppose I could start by pre-multiplying by E_2^{-1}?

so $E_2^{-1}S=TE_1$

[boreal girder]

wai

[hushed magnet]

look at S =E2 T E1. that means that both sides are the same linear map

in particular they want to kill the same vectors and they want to reach the same vectors

[carmine garden]

Let $Q=e_1,e_2,\dots, e_m$ form a basis of a kernel of $S$. Thus $dim(Ker(S))=m$. We then have for some $v \in Span(Q)$, $E_2TE_1(v)=0$, as $E_2$ is one -one to be zero, $TE_1(v)$ must be zero, $E_1$ carries over a linearly indepndent set, to a linearly independent set, forming $E_1(e_1),\dots, E_M(e_m)$ as the basis of its kernel

[boreal girder]

wai

[carmine garden]

If a vector not in the span of this were to lie in the kernel of one, it would HAVE to lie in the kernel of the other

let's suppose $ v\ in Ker(T)$ isn't in the span of T(Q), then , it's input image would have to be E_1^{-1}(v), which would the lie in ker(S), which would make it lie in Q

and we're done

cooll

[carmine garden]

thanks for this push, was all I needed

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[carmine garden]

Give a combinatorial argument to establish $\binom{n}{k} = \sum_{i=k}^{n} \binom{i-1}{k-1}; n≥k$

[boreal girder]

wai

[carmine garden]

The left hand counts the number of ways to select k objects from n things , not too sure of what the right hand does

also the right is just zero forall i<k, is it not

oh right

but what does it count

$\binom{i-1}{k-1}$ counts the numebr of ways to select k-1 objects , from i-1 objects, done seperately each time, repeated $n-k$ times

[boreal girder]

wai

[carmine garden]

ncvm

nvm

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[steady grail]

Hi Need help on this chapter I am completely clueless on how to start This chapter is about integration

[subtle helm]

ok

so the density relates volume to mass (and vice versa)

so

from the problem, are u able to find the volume of the paraboloid?

hint: think about rotation

[steady grail]

I cant find the thing i need to derive

i mean

integrate

[subtle helm]

.

[steady grail]

volume of the paraboloid is er

30 x 60 - the semicircle

so 60x30 - ( pi x 30^2)

[subtle helm]

its not a semicircle

but a parabola, no?

[steady grail]

ye

idk what is a parabola

[subtle helm]

so u cant find area using a circle's formula?

[subtle helm]

have u taken precalc, before?

heres a more accurate picture

of a cross section

so the main idea of this problem

is that u would need to find the volume generated by a revolved parabola

for reference

this is what a paraboloid (aka the 'parabolic' surface in the quesiton) looks like

yes mb

to do this, we first need to derive the equation of the parabola

[subtle helm]

for simplicity, we select the following coordinate

[subtle helm]

to do this

we note that one
the equation is in the form of ${y = ax^2}$ and two ${30 = a(30)^2}$.

[boreal girder]

k

[subtle helm]

@steady grail from this, are u able to find the equation for the parabola?

[steady grail]

wait

let me process this

for a minute

[steady grail]

i know this is the like revolve around the y axis or x axis like turn 360 degrees form the volume
pi then integral

[steady grail]

the inner surface means?

[subtle helm]

?

[subtle helm]

the area that will form the surface for the rice in the container, i.e., the parabola?

[steady grail]

so the inside of the parabola

[subtle helm]

yes

have u found the equation for the parabola?

[steady grail]

sI stil dont get

i know it give y = ax^2

why x = 30

[subtle helm]

.

u see that the origin is at the vertex of the parabola, right?

[subtle helm]

so, from this the top right point should be the point (30,30), ya?

[steady grail]

yesss

origin

yes

wait

ya ya

Cus top left is -30,30

[subtle helm]

yes

so

u will find that the equation of that parabola is actualyl

${y = \frac{1}{30}x^2}$

[boreal girder]

k

[subtle helm]

from this we shall go on to find the volume

do u prefer disk or shell method?

[steady grail]

i use this

[subtle helm]

Yes

Ok

[subtle helm]

For our volume, which one are we using?

[steady grail]

we using the y-axis one

generated volume at the y-axis

[steady grail]

because the parabola is turn around at y axis

[subtle helm]

Yes

[subtle helm]

From this x^2 = 30y

And we want rotation from y=0 to y=30

What would be the formula?

[steady grail]

so we use the generated volume at the y -a xis so we use x so we need move x^2 = 30 y

so we just pi integrate 30 ~ 0 then the x^2 is 30 y dx

[subtle helm]

Yes

[steady grail]

so after i integrate that is the volume?

then i just x the density?

[subtle helm]

,w pi integral from 0 to 30 (30y) dy

[boreal girder]

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[steady grail]

42412 need divide density or times

[subtle helm]

.

[steady grail]

d = vxm?

[subtle helm]

d = m/v

dv = m

[steady grail]

ohh then it change to kg

[subtle helm]

,w 1.182(pi integral from 0 to 30 (30y) dy)

[boreal girder]

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[steady grail]

Okay i understand already thanksss

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[sweet aurora]

hi

[sweet aurora]

someone teach me recrusion relations

using a cas calculator for this

w,

pls

[latent ridge]

just calculate the values one by one

and in this case notice that the sequence formed is actually a geometric sequence

so you can also use that fact as a shortcut way

[compact pewter]

@sweet aurora Has your question been resolved?

[sweet aurora]

mind telling me more

[sweet aurora]

ye it is a gemotirc sequence

i just dont kow the formula for geomtric

[latent ridge]

for example for number 3 you are given the recurrence relation V(n+1) = 1.075V(n)

meaning that the current value is calculated by multiplying the previous value by 1.075

[sweet aurora]

hold on

[sweet aurora]

sorry im kinda confused

[latent ridge]

so V(n+1) is the value after V(n) right?

[sweet aurora]

yep

[latent ridge]

then what are you confused with?

[sweet aurora]

i know its the value but what do i input exactly into my CAS

[latent ridge]

you are given the initial value V(0) = 45000

from there you can calculate V(1)

and so on till V(12)

[sweet aurora]

so 1.075?

[latent ridge]

no

V(n+1) = 1.075V(n)

[sweet aurora]

wait

so

is that what i input

[latent ridge]

you are given V(0)

so n=0 right?

you can set n=0

[sweet aurora]

so n=0 is what i input?

[latent ridge]

V(0+1) = 1.075V(0)
V(1) = 1.075V(0)

right?

and V(0) is given

[sweet aurora]

yeah

[compact pewter]

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[midnight lark]

Test the convergence of the series 1+ (1.3/3!) +(1.3.5/5!) + (1.3.5.7/7!) + ... (7 Marks)
can u guys pls help me answer it

[midnight lark]

we can use ratio test here

[azure vault]

have you tried it?

[midnight lark]

i tried i got (2k-1)/(2k-1)! is it correct

[azure vault]

what is this supposed to represent?

[midnight lark]

as it's infinite number

[kind elm]

Is that the ratio?

[midnight lark]

ratio test = (Uk+1 )/(Uk) is the format

[kind elm]

Take the limit as k --> Inf

[azure vault]

btw that is not the correct ratio

[kind elm]

oof

[golden wyvern]

(1)(3)...(2n+1)=(2n+1)!/(2)(4)...(2n)=(2n+1)!/2^n n!

what is that reaction sir

[azure vault]

while that's cool to rewrite it that way, it's not completely required to find that formula

as in we can find the ratio without it

[midnight lark]

i need to find is whether it converge or diverge

[golden wyvern]

he is just asked to prove convergence of sum 1/2^n n!

[azure vault]

yes, so do the ratio test

[golden wyvern]

true

[azure vault]

and compute the ratio

(correctly)

[golden wyvern]

"correctly" lmao

[azure vault]

try at least to find a rough expression of the k-th term of your sequence

and find what do you multiply it with

to get the next term, the (k+1)-th term

[golden wyvern]

sqrt e is the limit i think if i did my headsolve correctly

[midnight lark]

what i did with the previous question was 1/k! then i got the answer Uk=1/k then Uk+1 =1/k+1 i did Uk+1/Uk= (1/(k+1)!)/(1/k!) ... at last i got 1/(infinity) = 0

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[honest merlin]

what's the difference between avg velocity and instantaneous velocity?

[white igloo]

the delta t

[honest merlin]

I don't follow

[unkempt sail]

Inst means for a very small time (infinitely small)

Average means over an interval of time

[honest merlin]

hmm

yeah makes sense

[nimble basalt]

is that all

[unkempt sail]

Average velocity is basically area of vt graph over interval of time

Meanwhile instantaneous velocity is the slope at any point of a - t graph

[honest merlin]

okay bet, I think I get it

thanks sm

[nimble basalt]

isnt that distance

[unkempt sail]

Over interval of time

Divided by delta t

[nimble basalt]

ah

[honest merlin]

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[carmine garden]

For each even $E$ in the sample space we define $n(E)$ to be the number of times in the first $n$ repetitions that $E$ occurs. The $P(E)$, the probability of the event $E$ is defined as $P(E)= \lim_{n \to. \infty} \frac{ n(E)}{n}$

[boreal girder]

wai

[carmine garden]

I don't get how we get, say the probability of a heads on flipping a coin once is 1/2 by this defn

[kind elm]

Well n(E) would be n/2 for your event

So (n/2)/n = 1/2

[carmine garden]

ah, okay

but the event is only done once

[kind elm]

Huh?

[carmine garden]

like the coin is only flipped once

[kind elm]

first n repetitions that E occurs.

[kind elm]

Well this corresponds to n being 1

[carmine garden]

yes

got it

thanks

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[young tinsel]

What symbol is this?

[solemn drum]

...you didn't post one?

[young tinsel]

please wait

[solemn drum]

oh mb

[young tinsel]

it's the weird "e" like thing

idk what that is

[solemn drum]

epsilon

[young tinsel]

what does it do

[solemn drum]

it denotes whether an element belongs to a set

so c ∈ {6,12,18,24...} means c is in the set {6,12,18,24...}

it's the equivalent of saying "belongs to"

[young tinsel]

oh okay

thanks for clarifying

[kind elm]

btw it's not written as a regular epsilon in handwriting

you stylize it just like that

[solemn drum]

ε here is the regular epsilon

[young tinsel]

how do you close the channel again

[uneven isle]

.close

[young tinsel]

.close

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[worn wyvern]

,rccw

[worn wyvern]

p is the set of points M of the plan with coordinates

Prove that p représente a part of conic

As far as I know I should write a relation between x and y

<@&286206848099549185>

[fleet frost]

Do you know compound angle formulas?

[muted elbow]

y = cos(t) = 2cos^2(t/2)-1
(y+1)/2 = cos^2(t/2)
x^2 = sin^2(t/2)
x^2 + (y+1)/2 = 1 which is the equation of a parabola

[compact pewter]

@worn wyvern Has your question been resolved?

[compact pewter]

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[next epoch]

is it a prerequisite that all vector spaces contain the zero vector?

[tranquil dawn]

yea

[next epoch]

does that also mean that all sub spaces cannot be considered sub spaces unless they contain the zero vector as well right?

subspace*

[tranquil dawn]

yea

[next epoch]

is this why lines and planes in R^3 are not considered subspaces of R^3 unless they intersect the origin?

[tranquil dawn]

yea

[next epoch]

thank you so much for your eloquent responses.

😃

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[teal badge]

is the answer to a) 1.5

[teal badge]

and for part b), do i just add up all the x multiplied by P(X=x). its just that right

and for c) is it just E(X) multiplied by 200, and substract that amount (570) from 3.50 x 200 (=700), hence answer being 130?

[compact pewter]

@teal badge Has your question been resolved?

[pine osprey]

Yes

Not sure about the c) so im not answering

[devout latch]

for c), i think this is valid because expected value is linear, however a probability bigger than 1 seems not right for a)

[compact pewter]

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[undone ether]

let the probability of me attempting a mcq in exam be x then probability that attempted question is correct is y .
in 100 mcq test , what is the probability that i got z questions correct .

sorry if i used so many variable , i wanna find the general case

[molten coral]

can you classify the kind of rv here

[sturdy forge]

wait my notation is whack

[molten coral]

you shouldnt give out the solution

[sturdy forge]

OH TRUE

[molten coral]

but its no biggie

[sturdy forge]

wait im a fool ive helped so many people here, and it's still my stupid bad habit

[molten coral]

they might even want that

[sturdy forge]

sorry about that

[molten coral]

or they have ghosted and we argue about nothing

youre all good no worries

[jagged blade]

I'm not sure this is a problem he's actually working on, rather than a curiosity

[sturdy forge]

true. I think it's also much more enlightening to see "why" a sum of iid bernoulli r.v. has the distribution of a binomial r.v. rather than me just saying "well, it's this"

[molten coral]

can you work out the probability of getting a single question right?

[jagged blade]

At least from the phrasing, I'm guessing he's guesstimating his performance on a recent exam

[sturdy forge]

nah idts

no way the exam has 100 mcqs

[molten coral]

idk thats not crazy is it

if theyre all mcq

we start with the probability that an exam is all mcq

then work with the distribution of # of questions

[jagged blade]

That would convert it to a percentage

[undone ether]

hey iam here

[sturdy forge]

well, I don't know. If he's in grade school, that's a lot of questions. If he's in university, he'd prob be able to figure it out

[winged lion]

wassup here

[sturdy forge]

a good point for sure

but that's an exhausting amount of questions to give

[jagged blade]

The point isn't that there are 100 questions, but that imagining we have 100 questions is equivalent to finding the percentage of questions that are correct

[sturdy forge]

yeah but I'm saying a real exam that he's taking probably won't actually have that many questions

cuz it's a lot of questions

y'know

[jagged blade]

But it doesn't need to

[molten coral]

@undone ether any thoughts on this? or how do you feel about starting here?

[jagged blade]

Sometimes I feel like these help channels are actually "suggest a topic for people to talk about" channels lol

[undone ether]

may be first lets find the probabilty of attemting some questions greater than z

[molten coral]

sure you can start here

If I remember right, we first have to agree that each question is independent

like, if you try or dont try a previous question, that doesn't change how likely you are to attempt the next one

or any other question

your probability y applies to all question

[jagged blade]

I think that's a given in this scenario

[molten coral]

hopefully

i think this is the only assumption we really need to nail down. Maybe its silly to focus on but yea

[undone ether]

guys but the probabilty of attempting a question correct is conditional probabilty

[molten coral]

Conditional? why is that

[molten coral]

I thought you wanted to ignore this for now?

[undone ether]

bro my discord broke not sending any messages

[molten coral]

I don't think this is true

[sturdy forge]

well then this suddenly makes the problem much much more difficult, if you're saying each question's attempt and answering is not independent to those of other questions

[molten coral]

unless you mean, x is defined to be the probability that you get a question right, given you attempt it

is that what you mean?

[undone ether]

yes
p(attempting correct / the question is attempted) = x
now this prob make sense

[molten coral]

yea this does not make your life hard

we could ignore it for now or incorporate it if you want

[undone ether]

like how will we approach it if we ignore this

[molten coral]

what we need to work out the z part is just to agree on

1. that there is some probability p of getting some question right
2. that p doesn't change question-to-question

whether p depends on x, or x and y, isn't important to get an answer for the z part in terms of p

[undone ether]

so basically this does'nt change even if the question is attempted or not

[molten coral]

if the previous question, is what I mean.

like, if you attempt 5 questions in a row, you're not less likely to attempt the 6th

Maybe it's not important to focus on this.

[molten coral]

it's a simplifying assumption, that doesn't try to model more complicated behavior, so it's easier to get an answer

[undone ether]

can you explain me why this ?

[sturdy forge]

we're assuming that it's true

because if you don't

it gets really ugly

[molten coral]

i kinda regret bringing this up haha I maybe gave you the idea it's more important than it is

[sturdy forge]

no I think it was important to bring up

there's just a disconnect here from the "realm of math" we're working in, and the "realm of reality" I think

[undone ether]

yup lets take this assumption and i think it is important

[molten coral]

okay

oops reply ping

[sturdy forge]

in reality, you could have this be true or not

but in how we're doing the math, assuming it to be true makes it much easier to do the math

because

[molten coral]

okay, the next step is to work out the answer for some z

[sturdy forge]

why did I change colors?!?!

[molten coral]

woah! ive never seen it happen in real time

[sturdy forge]

ok anyways, there's a very nice thing that happens

when you assume that the things are independent

[undone ether]

i may attempt z question and get z right , attempt z+1 and get z right , attempt z+k and get z right . so any idea to tackle this

[sturdy forge]

because then you can "add" the random variables together that represent trying to attempt and guessing one question

[molten coral]

yea

[sturdy forge]

and if they're independent (and identically distributed, which just means they're basically the same for all intents and purposes you're concerned about), they turn into something that behaves very nicely and that we understand very well

i.e. a binomial random variable

the big step here that really really is interesting is, why does this happen?

and that's something that I can probably find you some good reading on

[molten coral]

we include probability to attempt in probability to get it right

you cant get a question right without attempting it

we're just going to say for some question, you have p chance to get it right

that includes whether or not you attempt it

[undone ether]

so we can find binomial random variable for getting z correct of z+k attempted

[molten coral]

we don't care about # attempted

[undone ether]

how no of question attempted does'nt matter

[molten coral]

the idea is that we boil down your model into a single probability, p, that says the probability that you get a question right or not

this could be a function of a lot of things

[sturdy forge]

here's the thing

you wanted us to find the number ANSWERED CORRECTLY

[molten coral]

you want to have it be a function of both the chance you attempt a problem, and the prob of getting it right

[sturdy forge]

so to answer it correctly, you need to attempt it AND answer it correctly

[molten coral]

if the probability that you attempt a problem needs to be able to decrease p, that's fine

[sturdy forge]

the probability of such a thing happening is $P(attempt \cap correct)=P(attempt)P(correct)$ since we're assuming whether or not you attempt it is not going to affect the probability of getting it correct once you actually answer the question

[boreal girder]

00100000

[molten coral]

we can work out the p now, if you are doubtful

maybe it'd be helpful to show what were shooting for

Binomial Random Variable requirements:

[undone ether]

so we will boil down our model to single probabilty p , whether getting right or wrong . but in that case we should assume that all questions are attempted , lets solve in this way itself

[molten coral]

no we don't have to assume that

like say we want some probability that a person gets sick

we might include the probability that they are immune in that

which would be the probability that they can even get sick in the first place

[molten coral]

maybe we should just say p

it's ||xy||

[undone ether]

so this say that p = xy meaning we attempt the question and get it right

[molten coral]

it doesn't assume we attempt the question

it actually includes that theres only y probability to attempt the problem

we could write it out fully

[sturdy forge]

it's a probability, not an assumption

it means the probability that we attempt it and then get it right (assuming those are independent)

[jagged blade]

Well they are

[molten coral]

its a function of y, which is the prob that you attempt it

[sturdy forge]

they don't have to be

¯_(ツ)_/¯

[molten coral]

if you say that y = 0, there is no chance that yu attempt a problem, then p will just be 0

regardless of how likely it is you are right, given you do attempt it

[jagged blade]

Explain how to get a question right without attempting it

[sturdy forge]

I never said that that's a possibility

[jagged blade]

Then they're independent, by definition

[molten coral]

||it could include a memory effect, and be a function of the outcome of previous questions, so that if he has tried the previous two questions, hes more likely to skip the question he's on, for example||

[sturdy forge]

^

[jagged blade]

We've already established that isn't the case

[molten coral]

then ||the binomial coefficient doesn't work like you want in the distribution||

[jagged blade]

We already know that x and y apply to all questions

So we already know they're independent at this point

[molten coral]

okay

[sturdy forge]

well, at the very least it's not "by definition" that they're independent then

it just means they're identically distributed

[jagged blade]

It is at this point

[molten coral]

i think limbo agreed with this so it's not important to continue to discuss here

[sturdy forge]

sounds good

[undone ether]

yeh

[molten coral]

how do you feel about the xy thing

[undone ether]

i think it make more sense since , we cant get a question right without attempting it . so xy is correct , if we got a question wrong we may write like (1-x)*y

[molten coral]

yea, actually, thats close

i mean for p i think wed say like, your expected score is yx + (1-y)*0

(1-y) chance you dont attempt it, you get 0

y chance you do, and x chance to get it right

[undone ether]

what is expected scrore here

[molten coral]

if you did a lot of problems, what your average score would be

what you "expect" to happen, given you've done enough problems for the variation to even out

anyways, we have p

[undone ether]

so can we use binomial distribution to find the prob for getting z correct

[molten coral]

yea, thats one exit path now

if you dont want to rederive everything

we have created a binomial random variable

and the distribution is just known, we can look it up

the formula looks a little intense, but the idea is straightforward

we could track through it, or you might find a video to be faster, idk what you want to do from here

or if you had other questions

[undone ether]

so the final answer is \left( 100 CZ \right)(p^{z})(1-p)^{100-z}

[molten coral]

the final answer is $\binom{100}{z} (xy)^z (1-xy)^{(100-z)}$

[boreal girder]

jan Niku

[molten coral]

wow that was fast

sharpshooter mods

[undone ether]

super then we got the answer

[molten coral]

yea

any other questions?

@viral trout mods did you see that

[undone ether]

thanks man , for your time

[molten coral]

[undone ether]

thank you everyone for your time

[molten coral]

<@&268886789983436800>

[undone ether]

.close

[compact pewter]

Channel closed

Closed by @undone ether

Use .reopen if this was a mistake.

[glacial slate]

Was it something nsfw?

[molten coral]

it was a penis

[glacial slate]

Ah okay I'll ban

[compact pewter]

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• Show your work, and if possible, explain where you are stuck.
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[teal badge]

I posted this q before, but wasnt able to respond at the time how would I do question a). I got 1.5, but apparently its wrong. Is the answer just 0.3. Also, is the answer for c) 130

[teal badge]

For C btw, I just did (3.5 x 200) - (2.85 x 200)

[molten coral]

200?

oh, c

well okay, what did you do for a?

[teal badge]

i got 130 for c

[teal badge]

i did (4 x 0.1) + (5x0.1) + (6x0.1)

and added

should it just be 0.1 + 0.1 + 0.1

[molten coral]

you dont need to multiply by the value of x

[teal badge]

yeh thats what i think now

[molten coral]

when you do a question like this, P(x=4,5,6)

as long as theres no overlap between the outcomes, you can just add

since its not possible for the dice to come out as 4 and 5 at the same time, we just add

[molten coral]

another important thing is that its a probability, so the fact that you got an answer larger than 1 should have set alarm bells off

for what its worth

[teal badge]

yeh

true

so is the answer to a just 0.3

[molten coral]

yea

[teal badge]

is the answer to c) 130

[molten coral]

lemme work it out

[teal badge]

ight thx

[molten coral]

,calc 3.5 - (10.25 + 20.25 + 30.2 + 40.1 + 50.1 + 60.1)

[boreal girder]

Result:

0.65

[molten coral]

,calc 200 * 0.65

[boreal girder]

Result:

130

[molten coral]

yea

[teal badge]

whats this working out

oh nvm

[molten coral]

the expected value of a single roll

[teal badge]

i thoought it was 10.25

[molten coral]

discord eats some symbols

yea

[teal badge]

yeh ight thanks bro

.clsoe

[molten coral]

[teal badge]

.close

[compact pewter]

Channel closed

Closed by @teal badge

Use .reopen if this was a mistake.

[compact pewter]

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Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[vale perch]

I can't see my mistake, help to find where is the mistake

[kind viper]

what is the first term in y?

[vale perch]

lnx²

[kind viper]

is it ln(x^2) or is it (ln(x))^2?

[idle mountain]

your differenciation for ln(x^2) is wrong

[kind viper]

is the x squared, or is the logarithm squared?

@vale perch

[vale perch]

x squared

ln(x²)

[kind viper]

so it is ln(x^2)

or just 2 ln(x)

[vale perch]

Exactly

[kind viper]

in that case your chain rule is completely wrong

and the first factor should be d ln(x^2)/d(x^2)

and the second, d(x^2)/dx

bc x^2 is your inner function

(well that or you could just do it all without chain rule)

[vale perch]

And how

[idle mountain]

this is better actually

[vale perch]

do i just do 2ln(xl) then 2/x

[idle mountain]

$\log (a^b) = b \log (a)$

[boreal girder]

Bettim

[vale perch]

Oh

[idle mountain]

yea

[vale perch]

Got it

Thank you both

.close

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[compact pewter]

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Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[carmine garden]

Here, does this argument work

[carmine garden]

Let's assume S is not injective, then the rank of its null space is non-zero, which leads to a contradiction

[dusky nymph]

can you say what the contradiction is?

[old mica]

it is not obvious to me what the contradiction is

[carmine garden]

The contradiction is the range has dimension equal to that of V

[dusky nymph]

range of?

[carmine garden]

I mean the rank

[silver canopy]

You need to be more clear/rigorous in your proofs/explanations

[old mica]

it's kinda murky to me why that's the case?

it could be that I'm just not seeing something though

also, why is that a contradiction?

[carmine garden]

yea, I think RNT won't work here

[dusky nymph]

well it's part of the argument but not the whole thing

suppose in general that A,B are linear maps from V to V, can you say anything about how rank(AB) compares with rank(B)?

[carmine garden]

I think I'd first have to discuss about T

[carmine garden]

rank(AB)= rank(B) iff B is surjective to the domain of A

[dusky nymph]

is rank(AB) > rank(B) possible?

[carmine garden]

yes

[dusky nymph]

can you give an example?

[carmine garden]

uh, not off the top of my heaad

[dusky nymph]

maybe reconsider whether it's possible

[carmine garden]

hmm

[dusky nymph]

consider the null spaces of B and AB

how are they related?

[carmine garden]

The null space of B is that of AB too + maybe a bit more

[dusky nymph]

i think you have it reversed

null(B) is a subspace of null(AB) (because if Bv = 0 then ABv = 0)

now you can use rank nullity to relate the ranks

[carmine garden]

hmm

We have to start off with T

[dusky nymph]

well first let's focus on the simpler situation with just two maps A and B

then we can relate that to the RST situation

[carmine garden]

if I had two maps , A, B .Then AB is surjective iff B is injective as injectivity \implies surjectivity in a transformation from a VS ot itself

[dusky nymph]

why is AB automatically surjective if B is injective?

what if A is the zero map, for example

it is true that injectivity of B implies surjectivity of B since V is finite dimensional

[carmine garden]

Do you want me to prove it?

[dusky nymph]

prove what?

[carmine garden]

this

[dusky nymph]

that B injective ==> AB surjective?

sure, it should be a short proof

[carmine garden]

here A by itself has to be surjective though, does it not?

by rank nullity as B is injective , it rank is equal to the dim of the pre-image vector space

[dusky nymph]

yea you would have to assume something about A in order for this to hold

that's why i questioned it

[dusky nymph]

yea

ok now how about rank(A) and rank(AB)?

can you show that rank(AB) <= rank(B)?

[carmine garden]

Lemme try

[dusky nymph]

oops, meant rank(B), corrected

(it's also true with rank(A) on the right hand side, but i don't think we need that here)

[carmine garden]

hmm

[carmine garden]

what assumptions on B here

[dusky nymph]

A and B can be any linear maps from V to V

where V is a finite dimensional vector space

use the fact from above, null(B) is a subspace of null(AB)

[carmine garden]

got it

as $null(B)≤null(AB); dim(V)-rank(B)≤ dim(V)-rank(AB)$

[boreal girder]

wai

[dusky nymph]

yea and therefore..

[carmine garden]

dim(AB)≤ dim(B)

[dusky nymph]

ok good, now let's look at RST

[carmine garden]

we similarly have dim (RST)≤sim(ST)

[dusky nymph]

yes

so in particular, if dim(ST) < n, then RST can't be surjective

[carmine garden]

yes

so $dims(RST)≤dim(ST)≤dim(T)≤n$

[dusky nymph]

yea

well dim(T) at the end (but dim(S) is also true)

now the question is asking if S has to be injective

[carmine garden]

oops, yea,, lemme write dimT

[dusky nymph]

so a strategy might be to suppose S is not injective and then show that dim(RST) can't be n if that's the case

[boreal girder]

wai

[carmine garden]

well, we can start by removing the inequalities as dim(RST)=n

so $dims(RST)=dim(ST)=dim(T)=n$

[dusky nymph]

ok good

[boreal girder]

wai

[dusky nymph]

so focus on the middle one as it's the simplest one that involves S

if S is not injective, can dim(ST) be n?

[carmine garden]

no

no

[dusky nymph]

that is correct, but it will take a bit of arguing to explain why

any thoughts on how to start?

[carmine garden]

hmm

one minute

[carmine garden]

Let's suppose S were not injective, then the dimension of the null space won't be zero anymore, and thus , the dimension of the null space of RST won't be either

[dusky nymph]

it is true, but can you explain why?

(this part: "and thus , the dimension of the null space of RST won't be either")

[carmine garden]

As ST(v_1)=ST(v_2) for two vector v_1≠v_2, RST(v_1)= RST(v_2), for two vectors v_1≠v_2

[dusky nymph]

you seem to be using facts here which have not been proven:

[carmine garden]

yea, which are?

[dusky nymph]

oops let me back up

ok i guess just one thing

you're implicitly saying that ST is not injective

so you should explain why

[carmine garden]

Supposing S were not injective, then ST wouldn't be injective either and so on

[dusky nymph]

but this is the key fact, why is it true?

why does S not injective ==> ST not injective

somehow you have to use the fact that V is finite dimensional because that statement is false for infinite dimensions

[carmine garden]

yea, I get that

hmm

[carmine garden]

I'm not sure of how to rigoously justify this tbh

[dusky nymph]

here's one way you can use the finite dimensionality:

S not injective
==> S not surjective
==>(?) ST not surjective
==> ST not injective

where you would have to prove the (?) statement

(which is true even in infinite dimensions)

[carmine garden]

is ? a LT

[dusky nymph]

no ? is just intended to say "why is this implication true?"

the others being hopefully obvious

[carmine garden]

Can I just say the compsoition of a non surjective function with any other function is non-surjective

[dusky nymph]

well you should probably explain more carefully

if S is not surjective, why is ST not surjective?

[carmine garden]

Because S doesn't map to the entirety of V?

[dusky nymph]

ok and so?

[carmine garden]

so S composed with any function won't either

[dusky nymph]

true, but don't just state it as a fact

show it persuasively

if S doesn't map to the entirety of V, then there's some v in V such that Sw = v is always false, for all vectors w

[carmine garden]

I see

[dusky nymph]

if ST were surjective, then we would have STx = v for some vector x

thus Sw = v where w = Tx

contradicting the above

that's the kind of level of explanation i would expect if i were grading this

[carmine garden]

Hmm, I see. Okay , I'll work on building that level of rigour

[dusky nymph]

ok cool so are the rest of the implications clear?

S not injective
==> S not surjective
==> ST not surjective
==> ST not injective

(and do you see where we're using finite dimensionality of V?)

[carmine garden]

We're using that here?

[dusky nymph]

nope actually that's the only implication where we're not using it!

S not surjective ==> ST not surjective is true even in infinite dimensions

but the other ==> are not

injective ==> surjective
and surjective ==> injective
are both false in infinite dimensions

[carmine garden]

I see

[dusky nymph]

ok now the rest of your argument works

S not injective ==> ST not injective as we just showed
and therefore RST not injective, as you showed

so just one more thing to say after that and you're done!

[carmine garden]

Therefore S must be injective for RST to be Surjective

[dusky nymph]

right, because...

RST not injective ==> RST not surjective

so connecting all the arrows we get
S not injective ==> RST not surjective

[carmine garden]

Got it

[dusky nymph]

or taking the contrapositive,

RST surjective ==> S injective

voila!

this question was a bit trickier than it looked, eh?

[carmine garden]

Yea

Quite, to make it worse, I did't have any proof based math courses in sem 2, so I'm a bit out of touch with proof writing

[dusky nymph]

np, linear algebra is a good place to practice proof writing, probably better than real analysis or smth

[carmine garden]

Yea, I'm taking LA2 early(next sem!), hope that works out well

thanks agian!

*again

[dusky nymph]

yw, good luck!

[carmine garden]

.close

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[sour sun]

im working on a folio task on surge and logistic functions, and i need to use both these functions to relate to acceleration, ive found something i can do with the logistic function but cant figure out how to actually get it to work

im using the Mercedes-AMG F1 W11 EQ Performance because i thought it would be fun, ive got statistics of its 0-200km an hour and max speed, just dont really know how to start

[sour sun]

i also need to find a way to incooperate surge functions and i have no clue how imma do that

0-200km in 4.3~ seconds, and it has a max speed of 350km/phr. i was going to do the function based off of its 0-200 because i havent been able to find an accurate way to get its 350km/phr

would i be able to make L = 350 and fit A / b so that it passes y=200 at x = 4.3 to get an estimated time? would that work?

this is the logistic function im using (includes 1st and 2nd derivative incase they are useful, assuming i havent messed them up lol 😭)

[compact pewter]

@sour sun Has your question been resolved?

[sour sun]

i give up

.close

[compact pewter]

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[short ferry]

is this JEE or sth?

[short ferry]

firstly, you can safely remove the powers

[jagged lagoon]

where did u get stuck

[short ferry]

since they dont influence the sign (the ^5 and ^1/3)

[jagged lagoon]

or do u not know how to starrt

[warped oar]

No. I made this question. My mathematics teacher taught the wavy curve method today.

[short ferry]

this likely doesnt even have nice sols

I'm changing likely to surely

[short ferry]

isnt that for fractions of polynomials?

the denominator is always positive though

since (x^2-2x+2)^4 = ((x-1)^2+1)^4 >= 1^4, with equality only if x = 1

[sullen canopy]

wavy curve method lmao

[short ferry]

and 1^4 - cos^2(x) > 0 for all x other than multiples of pi

1 isnt a multiple of pi

so (x^2-2x+2)^4 - cos^2(x) > 0

and ((x^2-2x+2)^4 - cos^2(x))^1/3 > 0 as well

that means we can safely multiply both sides by this and the inequality reduces to

$\sin\left(x^{3}-x+7\right)e^{x^{2}}+\ln\left(x^{4}+1\right)<0$

and at this point, it's the right time to give up

[warped oar]

e^x^2 is always positive and so is ln(x^4 + 1)

[short ferry]

yeah, but sin isnt

[boreal girder]

MathIsAlwaysRight

[short ferry]

we should be able to prove that the equality case has infinitely many sols

but thats pretty much it

the solution is gonna be union of infinitely many ugly intervals without closed form

[kind viper]

can we get a source on that

[short ferry]

this is the source

[kind viper]

oh so bro made this question up

[warped oar]

Thanks MaethIsAlwaysRight for your help.

I may try myself more.

Thanks.

.close

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[carmine garden]

Stupid question. How does a quotient space differ from the sum of two vector spaces

[hushed magnet]

how are they similar in your opinion

except both being vector spaces

[carmine garden]

By definition they seem to be the same we add all vectors of one vector space to all vector spaces of the other, unless this means for some fixed v in V

[hushed magnet]

ok I can see where you are coming from

for starters, the elements of V/U are sets

instead of vectors

what this construction achieves is computing "modulo U"

[carmine garden]

I see

okay

[kind viper]

have you seen the quotient construction for rings...? you might not have

[carmine garden]

No, I've barely done group theory, ring theory is a while away

[hushed magnet]

or as a specific example, the quotient way to define modulo n

[carmine garden]

I suppose I've seen that, though I don't recall details

[hushed magnet]

look at it again

imo its the easiest quotient construction to understand

[carmine garden]

okay, will look at it

[carmine garden]

this defn?

[hushed magnet]

consider 5Z the set of all multiples of 5

define a relation on Z by a~b iff a-b in 5Z

show its an equivalence relation

write out all equivalence classes

[carmine garden]

Okay

a~a in 5Z, so it's reflexive. Let a~ b, then b~a as a-b=-(b-a) . Lastly if a~b, b~ c, then a~c

[carmine garden]

(....,-5,0,5); (....,-4,1,6..);(-3,2,7....),(-2,3,8..),(-1,4,9....).

[hushed magnet]

well it should be sets

but ok

and thats a very incomplete proof

anyway

we notice the first of those sets is 5Z or 0+5Z

the next is 1+5Z

then 2+5Z, 3+5Z, 4+5Z

what do you get if you add the second and third set together

(using set addition in the form A+B = {a+b: a in A, b in B})

[carmine garden]

we get the class 5 mod 3?

[hushed magnet]

yes

so (1+5Z) + (2+5Z) = 3+5Z

in general, (a+5Z) + (b+5Z) = (a+b) + 5Z

[carmine garden]

ah, which is simialr to quotient spaces

got it

Thanks!

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[coral iris]

how to do this

[acoustic yarrow]

well itd be nice if u gave some context

!xy

[compact pewter]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[coral iris]

that is the exact problem

[coral iris]

thats the whole problem

!xy

[compact pewter]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[acoustic yarrow]

it says "Now, we want to..."

what came before this?

[kind viper]

it is not part (b) or (c) of any bigger problem?

[kind viper]

ok so this is the context.

your earlier picture was cropped.

ok

now do you see the graph of f(x) in black?

[coral iris]

yes

[kind viper]

btw the domain and range should be given with [square brackets] for closed intervals not {curly brackets} for sets with only explicitly listed elements

anyway it looks like you were able to read f(-1), f(0) and f(1) from the graph

just continue like that

[coral iris]

so is it just based on the table

and if we are transalting x+2 shouldnt each value of x go 2 units to the left since it is horizontal

@kind viper

[kind viper]

yes

[kind viper]

mm kinda overthinking it but idk of an elegant way to word things at the minute

[coral iris]

i am still confused ebcuase we should be moving to the left when (x+2)

[tired bear]

yes

each point shows the value 2 units to the right of it

so the entire graph shifts left by 2 units

[compact pewter]

@coral iris Has your question been resolved?

[coral iris]

no but it shows it goes right accoring to the grpah but it should go left

[compact pewter]

@coral iris Has your question been resolved?

[sick shard]

Where does it show that?

I mean if it's in book or somewhere you can send pic

[coral iris]

it should be moving to the left but it moves to the right

[coral iris]

thats the whole worksheet for that problem

[sick shard]

Given: g(-4)=6

Look at (-2,6) in graph of f(x). That exact point must be placed at (-4,6) for g(-4)=6, so that point must be shifted 2 units left in graph of g(x).

Idk how do you see that as right shift

[coral iris]

becuase it is moving to the left to get to y=-6

you need to move form -4 thorugh -2 which is a +2 shift ot the right

[sick shard]

Oh man, ok leave it.
Just draw graph of g(x) by your own on a rough page

And see where it shifts

Ez

I see what you are trying to say but you are mistaking right shift of coordinate axes for right shift of graph

[sick shard]

This is best method for you

[coral iris]

i can do that and thats wroks to grraph but not onn here

@sick shard

[sick shard]

Idk nobody helping you cause you can't explain your problem better

[compact pewter]

@coral iris Has your question been resolved?

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[tropic agate]

I'm reading about permutations. Consider the symmetric group $S_n$. Why is the following identity true? $$(c\ c+1)(a\ c)(c\ c+1)=(a\ c+1).$$ I know the last two $2$-cycles on the left combine to $(a \ c\ c+1)$, and why does then $(c\ c+1)(a\ c\ c+1)=(a\ c+1)$?

[boreal girder]

psie

[hushed magnet]

what is (c c+1)(a c)(c c+1) (a)

wher the last () mean function argument

[tropic agate]

ah ok, right, (c c+1)(a c c+1) fixes c, and sends a to c to c+1 and c+1 to a to fixed. Ok, that helped, thanks! 😅

.close

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[strange pendant]

how do you

[strange pendant]

figure that the volume of a parelelepipid is the determinant of a 3x3 matrix

can someone explain step by step, I dont get the geometric idea

[muted elbow]

volume of parallelepiped = (area of a face)x(height between two parallel faces)

(axb) is the area of our base and .c is the perpendicular distance between the two faces

draw a diagram

the scalar triple product (axb).c is the determinant
[a]
[b]
[c]

[compact pewter]

@strange pendant Has your question been resolved?

[strange pendant]

can you do a drawing?

[muted elbow]

ok wait

[strange pendant]

yeah I know that, but thats exactly where my doubt arises, trying to understand the scalar triple product intuitively, also trying to understand the determinant of a 2x2 matrix is the area of a paralelogram

[strange pendant]

ping me if you manage to drew it please

[sullen canopy]

i think he might be drawing it

[strange pendant]

bro took ages to draw it

[strange pendant]

the fuck is this

[muted elbow]

ye my writing is shit tbf

[strange pendant]

where did the sin cos came from

[muted elbow]

trigonometry

right angled triangle

[strange pendant]

you skipped some steps

i dont remember soh cah toa

[muted elbow]

sin𝜃 = perpendicular/hypotenuse

|b| = length of the hypotenuse

so |b|sin𝜃 = perpendicular distance between two parallel edges

cos𝜃 = base/hypotenuse

[strange pendant]

perpendicular or opposite

the fuck you mean

[muted elbow]

opposite

same thing

ig its taught differently in different countries

[strange pendant]

sin=op/hyp

sin=op/|b|

|b|sin(theta) = op

[muted elbow]

yes

[strange pendant]

but where did cos came from

the letters are very tiny wtf?

[muted elbow]

cos𝜃 = base/hyp
sin(𝜋/2-𝜃) = cos𝜃

[strange pendant]

cos = adjacent/hypothenuse

[muted elbow]

sin(𝜋/2-𝜃) = opp/hyp
sin(𝜋/2-𝜃) = opp/|c|
|c|sin(𝜋/2-𝜃) = opp
|c|cos(𝜃) = opp

[strange pendant]

what is C here

|c|

[muted elbow]

third vector

[strange pendant]

upward is a

[muted elbow]

no

right is a

[strange pendant]

-> is A

[muted elbow]

forward is b

[strange pendant]

Why (pi/2-theta)

you are using trig identity i dont know

cos(x) = sin(pi/3-x)?

you skipped steps

[muted elbow]

pi/2

[strange pendant]

you skipped shit

[muted elbow]

wait how much trigono you know

[strange pendant]

only the basics

[strange pendant]

how am I supposed to know and why does it matter if you end up using cos anyways

[muted elbow]

ye idk how imma explain that

well dot product gives the projection of a vector along another vector

[strange pendant]

?

[muted elbow]

so geometrically you can see how that would give us the perpendicular