help-10

[wicked crown]

you cant have more edges than a complete graph

[timid silo]

I think the correct way to express that would be O(n^2/m) even if n=2m worst case

if you know it's 2 you can substitute and get O(m)

wait

and every other case is better than that

so yeah it is O(n)

or m rather

[wicked crown]

I am kind of unsure rn whether my bound is even correct

oh no it should be fine

I know it is for sure greater than when n=2m

anyways

thank you very much for your help !

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[limber portal]

Why is there a uniqe function U(t) s.t. for all t, U(t) is invertible and solves this linear differential equation (U, A are nxn Matrices for all t)

[obtuse pebble]

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[obtuse pebble]

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[pine temple]

Question

[limber portal]

.

[obtuse pebble]

@limber portal Has your question been resolved?

[slim cove]

I'm guessing you're missing some conditions here. Probably that U(0) is fixed, and there's some sort of continuity of A with respect to time. The existence and uniqueness of solutions to differential equations is a deep theorem that's difficult to prove, but most arguments use something called the Banach contraction principle.

Here's some more information if you're curious. https://www2.math.upenn.edu/~kazdan/202F13/notes/ODE1.pdf

The proof is not for the faint of heart though.

[obtuse pebble]

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[jagged arrow]

could someone explain this integration to me, how did they get -ln(1+(1-s)k)

[slim cove]

it's 1 / (some linear funciton involving s), so the natural thing to guess is ln(that function involving s)

If you take the derivative of ln(1 + (1-s)k), you get -k/(1 + (1-s)k) by the chain rule

That's almost exactly what you want except it has an extra negative sign, so you get as an antiderivative -ln(1 + (1-s)k)

That's one way to do it

Another way you could do it is to substitute in u = 1 + (1-s)k

Then the integral will turn into -du/u

So after you integrate that, you get -ln(u), or -ln(1 + (1-s)k)

[teal turret]

YO i just did that problem yesterday!

are u taking FM ?

[jagged arrow]

yep

[teal turret]

broo look at what im doing rn lol:

[jagged arrow]

cool

what website is that?

[teal turret]

Coaching actuaries

highly recommend

[jagged arrow]

k thx

[teal turret]

did u understand eric's explanation

[jagged arrow]

kind of

[jagged arrow]

how do you chain rule this exactly? Im on summer break and brain isnt working well

[slim cove]

So, the outside function here is ln

And the inside function is 1 + (1-s)k

When you take the derivative of ln(something), you get 1/(something)

So that gives you 1/(1 + (1-s)k)

But then you gotta mutiply by the derivative of the inside function

1 + (1-s)k is the same as 1 + k - ks, and 1 and k are constants, so the derivative of that is just -k

[jagged arrow]

oh i get it now

[slim cove]

Therefore the chain rule tells us the derivative is 1/(1 + (1-s)k) * (-k)

[jagged arrow]

i didnt expand and that was confusing me

thanks

[slim cove]

no problem!

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[cerulean tulip]

hi guys! I'm having a little trouble understanding what a negative lagrange multiplier in constrained optimization means intuitively

[cerulean tulip]

like ik the lagrange multiplier is how much the output changes wrt a change in the constraint but im still having trouble visualizing what positive, negative, and zero lagrange multiplier might mean

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[native portal]

the probability of choosing 4th best person among 100 people, in the secretary problem, in general probability of choosing kth best person.

I derived this formula, but there is something wrong with it, the probabilities do not add up to 1

[native portal]

the additional part says the if kth best is to be selected, then k-1th, k-2th best should all be after i, and sorry there is a typo, the summation starts from i+1 to k, not n

[obtuse pebble]

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<@&286206848099549185>

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[rain oriole]

what have i done? is that correct? is that how that works?

[plain vapor]

looks like it is

[rain oriole]

i dont rly get the (-1)^n

[plain vapor]

wdym?

you want to understand why cosine can be written by that summation?

and yes i just checked it is correct

[rain oriole]

the cos() thing has the (-1) on top

[plain vapor]

oh oh ok

it doesnt change

[rain oriole]

but my sum has it on the bottom

[plain vapor]

if you multiply by -1 or divide by -1 you get the same result

[rain oriole]

ah I c

[plain vapor]

doesnt change dont worry about it

you could see it as (1/-1)^n, which can be rewritten into (-1/1)^n, which is just (-1)^n

[rain oriole]

nice

[obtuse pebble]

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[dawn vale]

i am stuck with

the

e^(x+10)

[queen jungle]

u sub

let u equal x+10

multiply by du for chain rule

[dawn vale]

not u's 😭

my teacher didnt teach us the u method

[queen jungle]

its fine

its the same thing as doing it w out

its just chain rule

[dawn vale]

so its (x+10)e^(x+10)?

[queen jungle]

no

whats the derivative of x+10

with respect to x

[dawn vale]

1

[queen jungle]

so then it would be 9e^(x+10)

[dawn vale]

so its just 9e^(x+10)?????

[queen jungle]

yup

try this one

[dawn vale]

i hate this question

[dawn vale]

thank you

[queen jungle]

9e^(2x+10)

try doing that problem

[dawn vale]

18e^(2x+10)

[queen jungle]

with the u sub method i showed u

gooodd job

great now you know how to do derivatives :3

[dawn vale]

😃

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[sick crystal]

Hey guys i got this probability problem from a friend:
There are giving raises to employees in a company and there are 18 employees that could get the raise. Only 5 of them will be promoted randomly. Between those employees there are two friends, Ana and María.
A is Maria gets a raise
B is Ana gets a raise

Describe AuB and calculate the probability
Describe An-B and calculate the prob

[sick crystal]

So, I am not sure on how to do this, I was about to use tree diagram but i dont see how to do it? Are those independent things?

[teal turret]

5c2 / 18c5 ?

IDK BRO

I QUIT

[obtuse pebble]

@sick crystal Has your question been resolved?

[sick crystal]

xD

<@&286206848099549185>

[keen quarry]

instead of trying to calculate A & B, try calculating not A and not B instead

[sick crystal]

No idea either, I guess 16/18?

[keen quarry]

what's the probably Maria is not the first person selected?

[sick crystal]

17/18

[keen quarry]

what's the probability she's not the 2nd person selected

[sick crystal]

16/17

[keen quarry]

not the 3rd?

[sick crystal]

15/16

and so on

[keen quarry]

so, the probability that maria is not selected for a promotion is equal to the probability maria is not the 1st, 2nd, 3rd, 4th, or 5th person selected

[sick crystal]

oh okay

makes sense

so it would be 17/18*16/17*15/16*14/15*13/14 = 13/18?

[keen quarry]

yep

[sick crystal]

and the same for ana

right?

[keen quarry]

exactly

[sick crystal]

okay yeah makes total sense thanks

I think i can work from here

[keen quarry]

and remember to subtract them from 1

[sick crystal]

so P(A) = 5/18 = P(B)

P(AnB) would be then 25/324 right?

so P(AUB) = 1 - P(AnB)

[keen quarry]

oh no

i forgot they're not independent

oh that makes this hard

cuz then you'll have to count every case individually

[sick crystal]

so?

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[mild sandal]

Question 12

I need some guidance

And if possible 13 as well

[nova bane]

three cubes of equal spheres?

[mild sandal]

i think the question is quite blurry

[nova bane]

Ok ill just wing it

Do you know how to find the volume of the gold bar

[mild sandal]

i would assume its lengthxwidthxheight

[nova bane]

yes

[mild sandal]

ayy

[nova bane]

so go get it

[mild sandal]

volume is 24m3?

[nova bane]

ok

[mild sandal]

o

k

[nova bane]

its going to be melted into 3 cubes

[mild sandal]

yes

[nova bane]

do you know the volume of a cube

[mild sandal]

side cubed?

[nova bane]

good

[mild sandal]

ty

[nova bane]

so you know the 3 cubes will be equal because "spheres are equal" or whatever

so it will be 3s^3

[mild sandal]

ye ye

[nova bane]

3s^3 = 24

[mild sandal]

okay

so s is gonna be 2?

[nova bane]

no

nvm yea

[mild sandal]

ok

[nova bane]

wait

then the answer is not in the choices

[mild sandal]

it asked for radius

[nova bane]

give me a moment to think about it

[mild sandal]

oh okay take ur time

[nova bane]

nvm i dont get the question

sorry

[mild sandal]

oh okay

can i ping for any helpers

or do i need to wait 15 mins from start?

[nova bane]

wait 15 min

[mild sandal]

<@&286206848099549185>

[mild sandal]

.

[nocturne bronze]

!15min

[obtuse pebble]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[mild sandal]

Damnnnn

[obtuse pebble]

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[obtuse pebble]

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[brave hatch]

Claim

[brave hatch]

In this question there are 2 normal distributions

Group 1: average of 7.4. Standarddeviation of 2

Group 2: average of 8 standarddeviation of 2

Question: the chance of someone from group 1 to score higher than someone from group 2

What I did:

Subtract average of group 2 from group 1: 7.4 - 8 = -0.6

Add up the squares of the standarddeviations: sqrt(2² + 2² ) = sqrt(8)

And then use normalcdf to get the chance of it being above 0

I would expect this to be right

But the actual answer they give is the following

[smoky vigil]

it would be below zero in this case

since you want the chance that group 1 scores higher than group 2

[smoky vigil]

and you computed the distribution of scores of group 2-group 1

[brave hatch]

No group 1-2

[brave hatch]

.

[smoky vigil]

ah

[brave hatch]

This is what they did tho. And I checked it all. I did everything the same except they use 4 instead of 2 for calculating the standarddeviation

And I have no idea where they get rhe 4 from

[smoky vigil]

ah

[brave hatch]

Are they just wrong or am I missing something

[smoky vigil]

you add square of variances and take square root, not standard deviations.

wai

t

[brave hatch]

Whaaaaat

Since when

I want to get a new standarddeviation

For the diff between the two distributions

I thought that should be the root of the squares of their deviations

Thats def what it says in my book

[smoky vigil]

did the question give 4 as standard deviation or 2?

[brave hatch]

2

[smoky vigil]

okay. Then its probably an error in the books answer.

[brave hatch]

They both have 2

It’s an exam 😭

It was a pilot exam

But still

[smoky vigil]

I would ask about the exams answer then.

[brave hatch]

I can’t

This was basically the exam for a previous year

And I have my actual exam tomorrow

This was practice

But my teacher didn’t make this it’s like for the whole country the same exam

[smoky vigil]

well, you could send an email to the one who made the answers, or the organization, surely they have some method of reporting wrong answers

[brave hatch]

So I am right then?

I mean I’m not gonna stress about it being the wrong answer in therr I just want to know if I did it the right way for my actual exam

[smoky vigil]

the way you did it is good, yes.

[brave hatch]

Ok nice

.close

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[timid silo]

is there a method that uses quadratic functions

the only methid that i can think of is since asinx + bcosx can be expressed as ksin(x+p), at the midpoint of alpha and beta, the derivative of sqrt3 acosx + 2bsinx would be 0

@nova vale original question by this dude btw

[obtuse pebble]

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[nova vale]

@timid silo

<@&286206848099549185>

[tiny cipher]

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[vague ibex]

for this problem do I just solve the matrix equations and then use row reduction to see if its gives me a solution ?

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[onyx crescent]

wich one is correct?

[onyx crescent]

i think the second one is correct

[nova bane]

whats the question.?

[onyx crescent]

wich of the affirmations is correct by using the graphic

.close

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[edgy needle]

um

[edgy needle]

how do i do question 1…

[sage geode]

Recall that the column vectors of a matrix are the images of the basis vectors

E.g., to determine the matrix A, start by applying the rotation to the vectors (1, 0) and (0, 1)

[edgy needle]

1,0 0,-1

is that how it works

wait now

No

0,-1 1,0

idk

cuz if a point was on 0,1 a transformation of 90 degrees would put it on 1,0

and 1,0 would go to 0,-1

[sage geode]

Right, so the column vectors of A will be (0, -1) and (1, 0)

[edgy needle]

what?!?!?!

is that sctually how it works

[sage geode]

Yes

[edgy needle]

oh

well that was easy

why is it the points

1,0 and 0,1

can it be any other point

[sage geode]

They are the standard basis vectors, it's safe to assume that those are picked as the basis

[edgy needle]

ohh ok

thx

[edgy needle]

Do u know the purpose of them trying to make me square the numbers and stuff’

[sage geode]

To 1) practice matrix multiplication and 2) see that the matrices are indeed related the way they are geometrically defined (e.g., rotating by 90 degrees clockwise twice is equivalent to rotating by 180 degrees clockwise, hence why A^2 = B)

[edgy needle]

ohhhh

OHHHHHHH

ok so

i can literally

multiply that 90 degree matrice by any matrice

to get a 90 degree rotation

[sage geode]

Multiplying any vector by A will give you the result of rotating that vector by 90 degrees clockwise, yes

[edgy needle]

i dont understand the order of the vectors

when theres 2 2x2 vectors

which goes first

?

[sage geode]

In a product of matrices, the one on the right should be applied first

[edgy needle]

wat?

so say i have matrice D and i want to rotate it 90 degrees clockwise

is it matrice D x matrice A

or matrice A x matrice d

[sage geode]

Ah

Well, depends on what you mean by rotating a matrix; If you mean rotating the space after D has been applied, then you should do AD

[edgy needle]

omg

this is confuse

[obtuse pebble]

@edgy needle Has your question been resolved?

[obtuse pebble]

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[long jasper]

HEL PLS

[ruby path]

factorize it

[long jasper]

so u^2 +sqrt3u

[stoic yacht]

that's a substitution, sure let u = tan(theta), then you want to solve u^2 + sqrt3 * u = 0

[long jasper]

so

u (u+sqrt3)

[stoic yacht]

yes

[long jasper]

ohh

then u = -sqrt3, 0

then sub it back into tantheta?

[ruby path]

yes

[long jasper]

so thjen it'kkl be

just 1/3pi

[ruby path]

uhh

that would give you sqrt(3) not -sqrt(3)

[long jasper]

wait im kinda confused when u sub it in

oh

oops

[ruby path]

all the way at the end

[long jasper]

so then what

[ruby path]

you did the sub right

[long jasper]

i sub that value into tha theta

uyeahj

but uis that all since its within 0 and 2pi

[ruby path]

you just need to find theta now

[ruby path]

you will have 4 solutions in all

2 for each

[long jasper]

wtff

im so ass at trig

im kinda confused tho

[ruby path]

hmm

do you know when tan is negative

[long jasper]

nope

idk how the tan graphs work

liek when u

havbe it between 0 to 2pui

its not liek cos, sin

[ruby path]

quadrants?

[long jasper]

oh

ik those yeah

[ruby path]

have you heard about them

yeah

[long jasper]

so when tan is negative

[ruby path]

so in what quadrants is tan negative

[long jasper]

its 2nd and 3th

4th*

[ruby path]

yes good

so you know your value for theta will lie in those two quadrants

[long jasper]

yeah

[ruby path]

I assume you can do tan x = 0 btw

[long jasper]

yeah

but then tan 1/'3pi?

idk abt that one

[ruby path]

there are 3 values for it

[ruby path]

right

well

[long jasper]

oh i kinda ssumed it was js 0

[ruby path]

tan (pi - x) = -tan x

do you know about this

[ruby path]

nope

[long jasper]

oh heck guys

[long jasper]

ah i remember it

ah dw i'll js figuire it out another time

[ruby path]

yeah

so if you sub pi/3 into it

you get your first answer

which is 2pi/3

[long jasper]

oh

and then 2pi - pi/3?

[ruby path]

yes

[long jasper]

ahh okok

i get it

[ruby path]

so what are the 5 solutions?

[long jasper]

wait how is there 5 tho

i thought it wss bounded by 2pi

and 0

[ruby path]

inclusive

when is tan 0

[long jasper]

pi?

[ruby path]

and 0 and 2pi in your case

remember they're included

[long jasper]

ohghh

okok thanks

[obtuse pebble]

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[obtuse pebble]

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[gloomy vector]

use SSS im p sure

[rugged cape]

they want me to use ASA/AAS :/

[stoic yacht]

are you allowed to use the fact that the opposite angles are congruent?

[high lily]

probably not

[rugged cape]

I'm not sure

[high lily]

should only be allowed to apply basic parallel line theorems

[stoic yacht]

can you use the properties regarding angles for transversals between parallel lines? opposite adjacent angles and stuff?

[rugged cape]

I don't quite understand you sorry 😭

i guess I'll give an example on how I solve my questions

BAC = DAC
angle B = angle D = 90
AC is common side
conclusion: ABD ~= ABC (ASA)

[stuck bough]

thats not asa, thats aas criterion

[rugged cape]

thanks for correcting me

would you mind helping me with my question too

[stuck bough]

which question

[rugged cape]

this

[stuck bough]

thus ASA criterion

[rugged cape]

thank you for the help man

have a good day

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[frozen kindle]

not sure how to approach question

[plush ginkgo]

Do you know how to find the distance between two 3-dimensional points?

[frozen kindle]

yea

but what does it mean a point closest to the origin?

[plush ginkgo]

It's the point on the line whose distance to the origin is the smallest

[frozen kindle]

i get that but how do i solve it bc i cant make it equal to 0 and get a same value for lander

[frozen kindle]

um

[plush ginkgo]

do you understand the question but are unsure how to find it?

[frozen kindle]

yeah

i just dont know how to solve the point closest to the origin

[plush ginkgo]

For example, you could express the distance as a function of λ. Then you can minimize that function

[trim portal]

idt thats the intended solution honestly

unless luvlin knows calculus

[frozen kindle]

i do but i doubt ur using calculus

its a vectors only question

[trim portal]

yeah, thats what i thought

[plush ginkgo]

yeah, in the 2d example above you can just draw the perpendicular from the origin to the line

[trim portal]

Note that this is perpendicular

[frozen kindle]

sooooo i-??

bc the answer just says 7 root 2 / 2

[trim portal]

Okay so basically you need to find a point on line M, whose position vector is perpendicular to the direction vector of line M

[trim portal]

sorry for late response

Here is a 2D illustration

the green vector is direction vector of the purple line

the red vector iis position vector of some point on the line

And the point is closest when red and green are perpendicular

So you'll basically need to find the position vector in terms of lambda, then the direction vector

and then set their dot product equal to 0

and solve for lambda

plugging it in the position vector should give you coordinates of P

and then just find its length

[obtuse pebble]

@frozen kindle Has your question been resolved?

[plush ginkgo]

Do you know how to find a perpendicular from a point to a line (in 3d)?

or in 2d for that matter

[obtuse pebble]

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[white python]

Can someone show me the steps of solving this

[hidden compass]

First, use the definition of abs value, so that you can get a 2nd degree polynomial

(remember that you want the derivative at x = -4)

[white python]

Could you show me how to solve the absolute value part of the equation please?

@hidden compass

[high lily]

do you know the (piecewise) definition of the abs val

[white python]

You mean where we'll do one side the positive abs and the negative abs?

[high lily]

$|\text{this}| = \begin{cases} \text{this}\ &\text{if } \text{this} > 0 \
-\text{this} \ &\text{if\ } \text{this}\leq 0 \end{cases}$

[warm shale]

ℝαμΩℕωⅤ

[white python]

Yea

I don't know how to solve it with this particular equation tho

[high lily]

for what you're considering
will the inside of |3x+4|
i.e. will 3x + 4 be positive or negative when x=-4

[white python]

It'll be negative

[high lily]

yes

and from the definition how would you express
|3x+4|
without absolute value bars when 3x+4 is negative

[white python]

-(3x+4) no?

[ruby path]

monster

[high lily]

yes

the problem then converts to finding $f'(-4)$ for $f(x) = -(3x+4) + 2x^2$

[white python]

And then just plug in the value of x and solve right?

[warm shale]

ℝαμΩℕωⅤ

[high lily]

differentiate first

[white python]

I saw this guy solve it in a weird way which confused me I can send u how he solved it @high lily

[high lily]

sure,

[high lily]

they're applying the derivative of the abs value directly (+chain rule)
their methods is more ideal if you're being asked to evaluate multiple things

$\dv{x} |x| = \frac{x}{|x|}$

[warm shale]

ℝαμΩℕωⅤ

[high lily]

by initially considering the sign around what we're interested in, that can be avoided

its up to personal preference which approach to take

[white python]

But where did he 3 outside the bracket come from @high lily

Oh nevermind

[high lily]

chain rule

[white python]

If we're using chain rule shouldn't we put a power of -1 on the bracket?

[high lily]

wdym

[white python]

Isn't this chain rule?

[high lily]

that's chain rule being applied in the specific case of a power function composition

more generally its
[f(g(x))]' = f'(g(x)) * g'(x)

[obtuse pebble]

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[wraith barn]

its a simple question i just forgot what to do when you state smth is an inf pt

[fossil pendant]

Ill give an example first and then say why im confused.

[wraith barn]

you draw the table and then what do you say?

[wraith barn]

sorry occupied

[fossil pendant]

We sent at the same time 🙂

Good luck

[wraith barn]

i write it like this but apparently this isnt enough so im wondering what more i should put

[obtuse pebble]

@wraith barn Has your question been resolved?

[fossil pendant]

I didnt learn how to graph yet wait for a few months or smth

Anyways

<@&286206848099549185>

[wraith barn]

:,)

[fossil pendant]

.

(Also idk what youre talking abt)

[wraith barn]

inflexion points

2nd derivative

[fossil pendant]

Ooooh wait im actually going to study that real soon 😭 sorry for not being helpful

[wraith barn]

allg dw

[lilac bloom]

so what's the problem asking? to list the inflection points?

@wraith barn

[wraith barn]

its just asking to find inflexion pts

i came do that

but the way i write it out is apprently wrong

[lilac bloom]

I see

[wraith barn]

like i have to SHOW they are inflexion pts

[lilac bloom]

so I can obviously see the y'' but what's u?

[wraith barn]

thats an x sorry

[lilac bloom]

ahhh

I see

Write out the second derivative of the function and show your work proving where y''(x)=0; then list like y''(0)=0, y''(2)=0

While a table is effective, it doesn't really say much as to what you actually did

[wraith barn]

yea

i have that down as well

sorry ill send it all

[lilac bloom]

If you have your work proving the inflection points then i'm confused why a math prof wouldn't accept that

[wraith barn]

idk man

my country’s syllabus is also that you have to properly sub in your points into a table to show if its an inflexion point

and so i did

but they didnt accept?

[lilac bloom]

that's so weird

I'm sorry but I'm not sure how to help you

[wraith barn]

:,) my country has issues

[lilac bloom]

To me, you've clearly done all that's asked in an elegant and concise manner

proving the inflection points and providing a table to demonstrate

[wraith barn]

yolo ig

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[jolly raptor]

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[silk bear]

4 3 5
Q2) Simplify √81 − 8 √216 + 15 √32 + √225.

[silk bear]

the 4 is power of √81, 3 is of √216 and 5 is of √32

that is complete question

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[karmic hedge]

Is this a mistake from the program I use?

[karmic hedge]

I tried a < 0 and it said it was wrong

I tried a <= 0 (based on the hint) and it said it was correct

But I dont see why my first answer was wrong

[trim portal]

it wants all the values of a probably

[karmic hedge]

But a = 0 gives 10/9 which doesnt have a HA, no?

[trim portal]

you sure

i think it gives -5/9

[karmic hedge]

Wait sry I mean it gives -5/9

yeye

Which doesnt have a HA

right?

[trim portal]

How precisely do you define HAs?

[karmic hedge]

Where a function approaches a horizontal line forever but never end up touching it

[rigid plaza]

I think this is a definition thing

[trim portal]

definitely

[trim portal]

with this definition, your answer would be right

but i think their definition was closer to this one ^

[karmic hedge]

Sry could you translate this a little bit I dont really understand 😅

Or show an example or something sry

[trim portal]

it basically means that the function approaches a finite constant as x -> infinity / -infinity

and -5/9 approaches -5/9 as x -> infinity

[rigid plaza]

This is a bad definition

Since it excludes oscillating asymptotes too

Which are less debatably asymptotes than constant functions

[karmic hedge]

but it gets to -5/9 in the end right? I thought thats the whole point of an asymptote is that they never intersect

[karmic hedge]

Idk what that is sry

I dont think ive covered that yet

[rigid plaza]

Picture level is that it keeps crossing the line it is asymptotic to

[trim portal]

something like this e.g.

[rigid plaza]

But with smaller and smaller magnitude

[trim portal]

well it mustn't be

it depends on the definition

you need to use the definition they give you

[karmic hedge]

Hmm ok I think I get it

But I need a break from maths now cause this is a mindfuck lol

Thank you guys for your help!

❤️

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[oblique pebble]

How do you solve this?

8. Assume that ( A ) is an invertible ( n \times n ) matrix and that ( B ) is an invertible ( m \times m ) matrix. Create an ( (n+m) \times (n+m) ) matrix ( C ) by placing ( A ) in the upper left corner, ( B ) in the lower right corner, and filling the rest with zeros. Symbolically, we write:

[
C = \left(\begin{array}{cc}
A & \mathbf{0} \
\mathbf{0} & B
\end{array}\right)
]

Show that ( C ) is invertible and that

[
C^{-1} = \left(\begin{array}{cc}
A^{-1} & \mathbf{0} \
\mathbf{0} & B^{-1}
\end{array}\right)
]

[warm shale]

Michael

[kind hawk]

what does it mean for a matrix to be invertible. what does it mean for a matrix to be the inverse of another matrix

[obtuse pebble]

@oblique pebble Has your question been resolved?

[oblique pebble]

That $C^{-1} C = I_{n+m}$

Oh, is that what I have to show

[warm shale]

Michael

[kind hawk]

yes

[oblique pebble]

Makes sense, how do I show that it's invertible in the first place? Or is that kind of proved if I calculate that both $C^{-1}C$ and $CC^{-1}$ equals the identity matrix?

[warm shale]

Michael

[fossil crag]

so you don't know yet that C^-1 is indeed C^-1

so you have to call it something else, like D

and show that CD = identity

or DC = identity

(left or right invertible is enough)

[oblique pebble]

yeah right

Well thanks

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[celest musk]

i think i dont understand compactness

[celest musk]

why can't every set be compact?

just pick the space the set resides in

that space covers the set, hence the set is compact

for example in R^2 any set can be covered by R^2 itself, which is an open set hence an open cover too

[polar fossil]

what's the definition of compactness?

[celest musk]

a set C is compact if every open cover of C has a finite subcover

[polar fossil]

yep good

so let's look at [0, 1)

this set is not compact, as you've probably been told

this is because there is an open cover that does not have a finite subcover

[celest musk]

it must be closed and bounded to be compact, and [0, 1) is neither closed nor bounded

[polar fossil]

ok yes you're right, but that's not the definition of compactness you're working with

[celest musk]

why's that?

[polar fossil]

(-1, 1/2)

(0, 1/2)
(0, 3/4)
(0, 7/8)
(0, 15/16)
...

is an open cover of [0, 1)

does it have a finite subcover?

[celest musk]

no

okay i get it

[celest musk]

anyway here, [0,1) is not closed but it is bounded

am i right

?

[polar fossil]

that's correct

[celest musk]

okay thanks

so going back to my original question

if we just pick a singleton like R^2

wouldn't R^2 be a finite subcover of C for every C in R^2 ?

[polar fossil]

pay attention to what it's saying

and let's forget about R^2 here, R works just as well

given an open cover of a set C (which is a set of sets like i listed earlier), you need to come up with a finite subset of those sets that still includes all the points in C

maybe an example of one that does work is useful

looking at [0, 1] -- an example open cover might be
{
(-1, 2)
(-1/2, 1)
(-1/4, 1/2)
(-1/8, 1/4)
...
}

that infinite set

[celest musk]

oh ok so FOR EVERY open cover of C there exists a finite subcover, which means that if i pick C and you give me any open cover, i can always find a finite subcover

[polar fossil]

yes

[celest musk]

i did it the opposite way

[celest musk]

just (-1, 2) is enough

hence it is compact

[polar fossil]

yep

[celest musk]

makes me think that every closed interval is compact

[polar fossil]

well

yea

remember what you said about closed and bounded sets

[celest musk]

uhm yea makes sense now

thanks very much 🙂

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[misty bear]

It's not about Math but more about Boolean Logic of Venn Diagrams. In the book "A concise introduction to a logic by J. Hurley 14th edition" I learning chapter about Categorical Proposition. There is section called "Conversion, Obversion, Contrapositive" and I learning this section now
I found that Contrapositive of statement "No A are B" looks like in a first screenshot bellow.

But here are the things:
Both transformations are correct, right?

No A are B
No non-B are non-A - Contrapositive
No non-A are non-B - Conversed
All non-A are B - Obversed

No A are B
No B are A - Conversed
No non-A are non-B - Contrapositive
No non-B are non-A - Conversed
All non-A are B - Obversed

But here is another transformation

No A are B
All A are non-B - Obversed
All B are non-A - Contrapositive
No B are A - Obversed
No A are B - Conversed

Have I done this transformation correctly and if Yes. How is that possible?

[misty bear]

<@&286206848099549185>

[timid silo]

yup

[misty bear]

Could you help me with this, pls?

[timid silo]

What Is The Question

I Think It Is Based On Sets

[misty bear]

I think I should put this question in another channel. Opened this question in #1256631400128446566

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[dapper robin]

how to show

[dapper robin]

this simplifies to

this using inductive hypothesis

<@&286206848099549185>

[obtuse pebble]

@dapper robin Has your question been resolved?

[dapper robin]

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[surreal jewel]

what is 5 + 54????

[median dome]

,calc 5+54

[warm shale]

Result:

59

[surreal jewel]

oh ok

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[dreamy cove]

#❓how-to-get-help

!da2a

[obtuse pebble]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

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[timid silo]

.reopen

The design of a house describes for its structure a quadrangular prism with a front length of 9 m, a depth of 12 m and a height of 2.50 m. This house will have:

1 front door, measuring 1.6 m x 2.20 m;
1 rear door, measuring 0.90 m x 2.20 m;
2 windows on each of the four sides of the house, measuring 1.20 x 1.10 m;

To build the house it is necessary to know the square footage and, therefore, buy the necessary quantities of materials, without wasting it. Calculate how many square meters of walls will be built.

[dapper robin]

do you know the answer?

[obtuse pebble]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

@timid silo Has your question been resolved?

[timid silo]

I've watched a 3 blue1brown video where he said that the first thing you could do to attack any problem of math that you're not getting the answer is try to simplify the problem

So, if there were no doors and no windows, what would the total area be?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

Thanks for your help, but I think I can solve it

Answer: 88.94 m².

Step 1: calculate the total area.

The area in front and back of the house are two rectangles measuring 9 m in front and 2.5 m in height.

2 space. space 9 space. space 2 comma 5 space equals space 45 straight space m squared

The side area is two rectangles 12 m deep by 2.5 m high.

2 space. space 12 space. space 2 comma 5 space equals space 60 straight space m squared

The total area is:

60 space plus space 45 space equals space 105 straight space m²

Step 2: calculate the area of ​​the windows and doors.

Door area.
1 front door, 1.6 m x 2.20 m = 3.52 m²
1 rear door, measuring 0.90 m x 2.20 m = 1.98 m²

Total door area: 3.52 + 1.98 = 5.5 m²

Window area.
There are 2 windows per side, with four sides, there are 8 windows. Each window is: 1.20 m x 1.10 m = 1.32 m².

Total window areas: 1.32 x 8 = 10.56 m².

Adding the areas of the doors and windows: 5.50 + 10.56 = 16.06 m².

Step 3: subtract the area of ​​the windows and doors from the total area.

105 space minus space 16 comma 06 space equals space 88 comma 94 space m²

Conclusion
88.94 m² of walls will be built.

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[timid silo]

Little tired so missing something obvious but where did the /2 go?

[median thunder]

10/2 = 5

its the same as sayingh

[timid silo]

I get 10/5 = 5 but then does the /2 not apply to the sqrt(2) as well?

[median thunder]

(10/2)*sqrt(2)

[meager glade]

$\frac{10\sqrt2}2=\frac{10}2\cdot\frac{\sqrt2}1=5\cdot\sqrt2$

[warm shale]

Flappie

[median thunder]

think of it as just a product

it wouldnt apply to both

thats like saying 3x(2x2) = 6x6

[timid silo]

@meager glade This makes sense, I was thinking square roots where it applies to everything the same.

@median thunder thanks 🙏

.close

[obtuse pebble]

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[obtuse pebble]

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[true urchin]

Please someone help me in solving it using Harmonic functions method.

[stoic yacht]

why not just use cauchy-riemann equations so you don't have to take as many derivatives?

[true urchin]

Naah
It's actually the demand
That's why

I am sharing my solved as well
Check if there is any misatake

@stoic yacht pls

@stoic yacht
are you reviewin it?

[obtuse pebble]

@true urchin Has your question been resolved?

[obtuse pebble]

@true urchin Has your question been resolved?

[obtuse pebble]

@true urchin Has your question been resolved?

[obtuse pebble]

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[sterile basin]

Idk how I got that u sub wrong

[half silo]

i think its coreect

[high lily]

your bounds aren't consistent

[sterile basin]

?

[high lily]

sqrt(15) turned into 15

[half silo]

nah at the end he corrected it

[high lily]

where?

[half silo]

since he put u back

[sterile basin]

I tried it twice

Yeah I corrected it

[half silo]

here

[sterile basin]

It's meant to be root 15

Online says the answer is 21 and I got way off

/ marking scheme

[sterile basin]

Updated

[high lily]

show your end result

[sterile basin]

I did it in my calculator and didn't get 21

[high lily]

show your calculations

what do you get when evaluating at x=sqrt(15)

[sterile basin]

8/3

[high lily]

also if you want to keep those bounds as sqrt(15) and 0,
you'd need to explicitly write x=0 and x=sqrt(15) while the variable of integration is u

that seems a bit low

can you show what you're typing into the calculator

[sterile basin]

Yeah

[high lily]

why sqrt(1)

this is for when x=0, you'll still need this value, but this isn't what I asked for

also this doesn't give 8/3

[obtuse pebble]

@sterile basin Has your question been resolved?

[obtuse pebble]

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[grizzled flume]

I was reviewing ODE's with the class: https://www.youtube.com/watch?v=VA6polQJToA&list=PLO1y6V1SXjjO-wHEYaM-2yyNU28RqEyLX&index=13
At 37:03 the result of the integral was ln|v-1| - 1/(v-1). I used integration by parts and got ln|v-1| - v/(v-1). When you derive those functions, both of them give you v/(v-1)^2. I have no idea what this means or how to interpret it.

[meager glade]

its a constant

$\frac{v}{v-1}=\frac{1+v-1}{v-1}=\frac{1}{v-1}+\frac{v-1}{v-1}=\frac{1}{v-1}+1$

[warm shale]

Flappie

[meager glade]

so when you take the derivative, itll disappear

[trail musk]

just a quick comment: 6/5 = 1 + 1/5

that's basically what you're doing here

[meager glade]

yeah, youre using $\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$

[warm shale]

Flappie

[grizzled flume]

Wow, I didn't saw it like that, really interesting, thanks. Maybe the only problem with it is that when I solved the DE, if it was on a test I would get and slightly different result, there'd be always that extra x when you undo the substitution, and maybe got invalidated because of that. And I had no way of knowing I was required to use a different integration method

[obtuse pebble]

@grizzled flume Has your question been resolved?

[meager glade]

if the reviewers are any bit competent, they would see that your solution is right within a constant

[obtuse pebble]

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[gilded falcon]

Hi, this is a problem from the book The Probabilistic Method by Alon and Spencer. How is the red statement true? I got that the expected value should actually be

[gilded falcon]

$$\mathbb{E}(X) = \sum_{k=1}^{n} \left((k)\binom{n}{k}p^n\right)$$

[warm shale]

sets

[gilded falcon]

and this is not equal to np
as stated in the textbook
pls help lol

@jaunty oasis maybe u can take a look? i know u said that u read some of this book 🙂

[bronze mica]

The probability that $|X|=k$ is not $\binom nkp^k$ (I assume the $p^n$ in your sum is a typo), but rather $\binom nkp^k(1-p)^{n-k}$: in order for $|X|=k$ you need to not only have picked $k$ vertices with probability $p$, but also have not picked each of the remaining $n-k$ vertices which happens with probability $1-p$ per vertex

[warm shale]

Edward II

[gilded falcon]

oh yeah

im dum

lol

[bronze mica]

The 'clearly' the book means is not, however, to actually use the definition of expectation but rather the intuitive "if I pick with probability p, I expect to pick a proportion p, i.e. np vertices"

[gilded falcon]

OH

I see

[random canyon]

hi

[bronze mica]

I mean you can do the rigorous treatment if you want, but really it's just reproving the expectation of a binomial distribution

[gilded falcon]

oh wait

could you please explain this argument in a little bit more detail

i don't get how its "obviously" np

[bronze mica]

If I were to roll a dice 600 times, how many times would you expect there to be a 1

[gilded falcon]

100

[bronze mica]

aka 600 * 1/6

[gilded falcon]

yea

[bronze mica]

it's essentially the same situation

[gilded falcon]

o yea

i get it

i think with stuff like this i should take examples to make it easier to understand

[bronze mica]

eh

[gilded falcon]

bc as soon as i thought of like n = 100 and p = 1/4 it made sense

[bronze mica]

examples can easily lead you astray because probability is often unituitive

[gilded falcon]

but it was confusing with only variables

o

anyway

thanks

.close

[obtuse pebble]

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[round oak]

Hi this is a problem from the book mathematical proof a transition to advanced mathematics. Here is what I have done so far. I don’t know how do I precede further and I wonder if there is a smarter way to do it.

[meager glade]

this channel is closed

please goto an open channel

[round oak]

Oh sry

[gilded falcon]

oh idk what to do

my question has been answered

how do i make the channel available to others now

[meager glade]

nothing, youve just got to wait for it to reset

[gilded falcon]

oki

[obtuse pebble]

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[heavy steppe]

For this question does there being a g and g' shown mean there are two functions at work here?

[meager glade]

you are given the slope of the tangent line, and a point on the tangent line

thenhow would oyu construct the tangent line?

[heavy steppe]

run the above info through point slope form? The main thing confusing me is g prime. I think I'm over emphasizing the g' part specifically.

[meager glade]

How would you normally get a line in point slope form?

[obtuse pebble]

@heavy steppe Has your question been resolved?

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[cerulean swallow]

Can you solve this

[dark vector]

which question do you want help with

[obtuse pebble]

@cerulean swallow Has your question been resolved?

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[valid oxide]

need help finding if this converges or not

it's a sequence, so i'm kinda lost on what to do

[gilded needle]

try rewriting (n+2)/(n+5) to get it into the form of 1 + something

[valid oxide]

how though?

yo

[gilded needle]

hint: n+2 = (n+5) - 3

[valid oxide]

what the

where did the -3 come from???

[gilded needle]

2 = 5 - 3 right?

[slim cove]

Hey, please don't post in help channels if you're not helping the person

[timid silo]

Oh

Sorry

[valid oxide]

bro

i'm so lost

are you trying to get things in terms of n to plug it in?

[gilded needle]

i'm saying to rewrite (n+2)/(n+5) to get it closer to the form displayed in the "recall that..." hint in your screenshot

[valid oxide]

I understand that

but idk how to

[gilded needle]

i gave a hint, did you try it?

[valid oxide]

cause my first instinct was to just take the limit of the original sequence

[gilded needle]

use n+2 = (n+5) - 3

[gilded needle]

ok, and how will you do that?

what is the limit?

[valid oxide]

ohhhh

[valid oxide]

it doesn't work

cause of the nth power

[gilded needle]

right

that's why the hint they gave is useful

[valid oxide]

right

are you able to call or are you strictly only allowed to help here?

[gilded needle]

we don't have voice channels here

[valid oxide]

can I add you if you're cool with it and call?

[gilded needle]

no, i prefer to stick to the help channels, is there something you're not sure how to typeset?

[valid oxide]

all good, it's just easier to explain myself over call

back to it

so I replace the n+2 = (n+5)-3

and then cancel out the n+5?

[tame python]

yea: $\frac{n+2}{n+5}=\frac{n+5-3}{n+5}=1+\frac{-3}{n-5}$

[warm shale]

mtt

[tame python]

so your limit is $\lim_{n\to\infty}\qty(1+\frac{-3}{n-5})^n$

[warm shale]

mtt

[valid oxide]

I really apprecaite this

can you explain to me how you did this?

I wanna make sure I understand

[tame python]

1. I followed your instructions
2. I typed what I did

[tame python]

nvm theres a typo here

let me redo 2.

you said to replace the n+2 with (n+5)-3

so $\frac{n+2}{n+5}=\frac{(n+5)-3}{n+5}$

[warm shale]

mtt

[tame python]

then to cancel the n+5s out

which is $\frac{(n+5)-3}{n+5}=1+\frac{-3}{n+5}$

[warm shale]

mtt

[tame python]

the original limit was $\lim_{n\to\infty}\qty(\frac{n+2}{n+5})^n$

[warm shale]

mtt

[tame python]

with what you were doing above, the inside of the parentheses has been changed to look different

you changed it to be $\lim_{n\to\infty}\qty(1+\frac{-3}{n+5})^n$

[warm shale]

mtt

[tame python]

that is what happened

[valid oxide]

I see

and so now, with the hint given, I can change it to e^-3

[valid oxide]

with r being -3

[tame python]

one issue

its not n in the denominator, its n + 5

so you cant directly use the limit

youll need a workaround

do you know how to do a limit substitution?

[valid oxide]

nope

[tame python]

youll notice that as n -> inf, n + 5 also -> inf

[valid oxide]

ohhhh

[gloomy vector]

could you set s=(n+5)/n? doe that work or no

[valid oxide]

nvm

yeah

[tame python]

no you cant

[gloomy vector]

oh

[tame python]

s is assumed to be a constant

if you see the hint, its used that way

[valid oxide]

oh s is a constant?