help-23

[jaunty spire]

just noticed that this is slighty wrong

[steep magnet]

Isnt that the same for x^2?

[jaunty spire]

no

x^2 is always positive

because a negative number times a negative is positive

i see now what you did wrong here

there is a minus infron of x^2 in the denominator

[steep magnet]

What was it? im still confused sry

[jaunty spire]

that means that if you multiply with x^(-2), the denominator will be negative in the limit

[steep magnet]

Ahhh I think I get it now

Ok I will try some more examples

[jaunty spire]

so after multiplying both num and den with x^(-k) you will be left with some x^(r-k) in the numerator, where r is the degree of the numerator

if x^(r-k) is odd and r > k, for instance if it's x^3, then you have to be careful when x goes to negative infinity, because in that case x^(r-k) will also go to negative infinity,

[steep magnet]

I got it down now. Just did like 20 examples and got them all right 😄

[jaunty spire]

great!

[steep magnet]

Thank you soooooooo much you are an amazing teacher

And thank you for being patient with me hahaha

[jaunty spire]

you're welcome ❤️

[steep magnet]

❤️

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[delicate trellis]

doing PLU decomp
i get the correct pivot L and U matrix
but i dont get how when i multiply all of the matrices
i do not get A

[robust drift]

i did not really understend what u r asking for but to multipliy matrice u have to multiply l1 by c1 and so on

just look at the color and u wil get the logic behind the multpication between matrix @delicate trellis

[delicate trellis]

PLU decomp

im using a online calc to multiply i do not get the same result

P*A

L*U

[robust drift]

sorry i can't help u with that i don't really get it <@&286206848099549185>

[lean otter]

<@&268886789983436800> they spamming across every channel

[polar lynx]

Don't spam this.

Muting you for the day for this. In the future please dm the modmail account to ask for permission before doing this kind of thing.

[safe radish]

@delicate trellis Has your question been resolved?

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[blazing peak]

i'm confused the way they solved this (the answer is a) but im getting different answers

[obsidian oracle]

what is the answer you got and how did you get it?

[flat frigate]

kebesque

[vapid pebble]

No she's here

@blazing peak

PR = X

NP = 2

2 = 2x - 4

X = 3

3+2 = 5

Still shows roles for me

[safe radish]

@blazing peak Has your question been resolved?

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[amber jay]

I’m confused on how to solve this problem; Solve the triangle (find all side lengths and angle measures)

[rough storm]

law of sines :3

ah wait

[vapid pebble]

Either of sines or cosines

[rough storm]

law of *cosines

[amber jay]

I’ll try it

[rough storm]

definitely law of cosines

[amber jay]

also my TI-30 cacultor sometimes bugs out on me

I’ve never done law of cosines with SSA I’ve only done it with SSS

[vapid pebble]

It was always one or the other

[rough storm]

think about formula of law of cosines

[amber jay]

hmm i think i got it thanks for the help :)

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[olive vault]

how do i prove the ranks of 2 matrices are equal?

[olive vault]

generally

[long oracle]

Without knowing their actual rank?

[olive vault]

yeah

well i know that the homogeneous systems represented by them have the same set of solutions

[long oracle]

Hm then u might know that they have the same nullity or whatever it’s called

[olive vault]

whats that

[long oracle]

And then maybe use rank nullity theorem to conclude about rank

[long oracle]

dim of null space

Maybe I’m using the wrong term here

[olive vault]

we havent learned that

so i dont think im supposed to use it

[long oracle]

Huh but rank u have?

[olive vault]

ok the full question is

[long oracle]

Hm too trivial in such case

[olive vault]

the only solution to Ax = 0 is the trivial one

and

i need to prove:

[olive vault]

wdym

[long oracle]

Like I meant maybe that’s why u haven’t learnt it yet for this exercise; otherwise it might be like almost directly true form the above I said

[olive vault]

ah

[long oracle]

And B has this too?

[olive vault]

B is what?

[long oracle]

Oops I meant to reply to this

[olive vault]

no

B can have any set of solutions, its just that its the same set of solutions to AB

[olive vault]

proving this was the first part of the question actually

[olive vault]

now i need to prove this

[long oracle]

What definition of rank do you have?

[olive vault]

the number of rows the arent 0s in the row reduced version of the matrix

[long oracle]

I see

Was there a term such as pivot row?

[olive vault]

well i also dont know all the terminology in english

so i might know what that is but not its name

[long oracle]

I don’t even know if it’s a thing, but trying to think of ways to translate this to something more tangible

Have you done proofs before using this definition?

[olive vault]

not proving that 2 ranks are equal no

only using the rank to prove something else

or the other way around

[long oracle]

I see, would you directly use this definition

[olive vault]

i mean intuitively it makes sense

[long oracle]

Or maybe something else?

[olive vault]

i could use something else ig?

[long oracle]

Well idk depends if you did so, I guess we can stick with it

[long oracle]

Ok so this is given?

[olive vault]

i proved it yeah

[long oracle]

And we also know this?

[olive vault]

yes

[long oracle]

Ah

Hm what’s V

Lol

[olive vault]

V(A) is the set of solution for the homogeneous system represented by A

ie the set of solutions to Ax = 0

[long oracle]

Oh that’s the null space!

[olive vault]

oh

[long oracle]

Cool cool

Or differently the kernel

Anyways okay let me think

How far did you come?

[olive vault]

well i didnt know how to start to prove a rank

so i didnt come anywhere lol

[long oracle]

I see, damnit if I just remembered a short proof of rank nullity theorem I might of been able to apply it here lol

[olive vault]

welp

[long oracle]

I can at least start by describing what the end goal is, we want by your definition to show that the number of rows that aren’t 0’s in row reduced AB is equal to the ones in B

Or even alternatively, that the number of rows that are 0’s in row reduced AB is equal to the ones in B

If that helps

[olive vault]

right

but we dont know the matrix to know how the reduced version would look like

lol can i just say that the dimention of the set of solutions is equal to the rank?

i dont see why not but thats obviously wrong

[long oracle]

Yes right, so the information given and the fact you proved in this case will hopefully illuminate us

[olive vault]

because that would prove it in one line

[long oracle]

Hm what exactly does this mean?

[olive vault]

lets say the rank of B is n

V(B) belongs to F^n

so every element of V(B) is a vector with n elements

and V(B) = V(AB)

so every solution to AB is a vector of n elements

and if all the solutions to AB are vectors of n elements then the number of non 0 rows in the row reduced version of AB must also be n

so the rank of AB is n

actually that might work

right?

[long oracle]

How does this follow?

[gusty trench]

are you trying to show that rank(AB) = rank(B)?

[olive vault]

well that means that there are n linearly independent rows

so if a vector of n elements is a solution

then

well yeah i dont know how to prove that part lol

[olive vault]

yeah

[gusty trench]

what information do you have about A and B?

[olive vault]

didnt you help me with a combinatorics question a couple of weeks ago?

[long oracle]

There’s a bit of information if you scroll up

[olive vault]

.

[gusty trench]

i might have

[long oracle]

Hm I’m not sure

[olive vault]

different ways to groups kids into groups of 3 or something like that

anyway

thanks for that if that was you

[gusty trench]

iirc rank(AB) = rank(B) isn't always true

[olive vault]

well theyre asking to prove it so it must be

[gusty trench]

what is V?

[olive vault]

.

[gusty trench]

oh the null space?

[olive vault]

apparantly except we dont call it that

(yet?)

[gusty trench]

wait why is this not trivially easy by rank nullity?

[long oracle]

They’re not allowed

[olive vault]

because i dont know what that is

[gusty trench]

bru

[long oracle]

Read the discussion lmao

[olive vault]

out of curiosity how would you prove this with rank nullity?

[gusty trench]

rank nullity says that rank(A) + nullity(A) = # cols of A

[olive vault]

oh i cant do that

[gusty trench]

you can technically prove this using rank-nullity without officially using rank nullity or calling it that

rank nullity is kind of one of those "no shit" results

[olive vault]

that works then

[long oracle]

Yeah

[gusty trench]

a matrix that has n columns has n variables

clearly a variable is either pivoted or free

the number of pivoted variables is the rank

[olive vault]

what does pivoted mean?

ah

[gusty trench]

it is a pivot when you take the rref

[olive vault]

whats rref?

[gusty trench]

row reduced echelon form

like gaussian reduction?

[olive vault]

oh

[gusty trench]

you've done that, right?

like solving Ax = b

[olive vault]

yeah

[gusty trench]

by row reducng

oaky

[olive vault]

yes

[gusty trench]

so you know how you parametize solutions

by parametrizing the free variables

[olive vault]

yes

[gusty trench]

and this reasoning makes sense?

[olive vault]

yes

[gusty trench]

either a variable is free or pivoted

great

so consider B which is nxn

then B has n variables

we know that V(AB) = V(B) so let's call this k

some k between 0 and n

sure?

[olive vault]

whats k?

[gusty trench]

just some number

the number of free variables

[olive vault]

ah

[gusty trench]

the dimension of V(AB)

for example, if B was a 3x3 matrix

and it was rank 2, then 2 variables are pivoted and 1 is free

[olive vault]

wouldnt that be the number of pivoted variables?

[gusty trench]

wouldn't what be?

[olive vault]

k

[olive vault]

yeah

so k is also the rank

[gusty trench]

i'm saying to call the rank (n-k)

k is the number of free variables

just work with me here

[olive vault]

how is it the dimention of V(B) then?

[gusty trench]

B has n variables, i'm saying that k of them are free and the remaining n-k are pivoted

[olive vault]

right

[long oracle]

I think you need to motivate the connection between free variables and nullity

[olive vault]

but isnt the dimention of V(B) the number of pivoted variables?

[long oracle]

Maybe that’s what’s confusing atm

[gusty trench]

i haven't seen this notation but i thought you said it was the set of homogenous solutions

so V(B) is the set of solutions to Bx = 0

right?

[olive vault]

yes

[gusty trench]

great now think of V(B) as a space that has a particular dimension

the dimension of that space corresponds to the number of free variables

[olive vault]

we havent learned vector spaces yet

but what im thinking is

[gusty trench]

i'm solving Ax = 0, A is a 3x3 matrix with 2 pivoted variables and 1 free variable. then the dimension of V(A) is 1

[long oracle]

I think there might be a more intuitive connection, in respect to his terminology

[olive vault]

maybe i misunderstood what pivoted and free means

[long oracle]

Like have you gone over dimension of vector spaces? I probably guess no if you haven’t talked about vector spaces

[gusty trench]

pivoted variables are the ones that are nonzero in the row reduced form, these are the variables that contribute to the rank

[olive vault]

right

so a solution so a matrix like this

[gusty trench]

oh here's a good example

consider this matrix

1 0 0
0 1 0
0 0 0

[olive vault]

would be a vector with the-number-of-pivoted-variables elements

[olive vault]

right

[gusty trench]

call this A

then A has 2 pivoted variables and one free varibale

now consider Ax = 0

think about what this means as a system of equations

call the first variable x, the second y, and the third z

then the solution set (x,y,z) to the system Ax = 0 satisfies 1x = 0, 1y = 0, 0 = 0

[olive vault]

ohhhhhhh

righttt

got it

yeah

[gusty trench]

so since x and y are pivoted

they have fixed values here

x = 0, y = 0

z can be anything

[olive vault]

so the number of solutions would be equal to the number of possible values for the free variable

[gusty trench]

whcih means the space of solutions to Ax = 0 is given by all vectors of the form (0,0,c) where c is any real number

[olive vault]

yeah

[gusty trench]

well technically in this case there are infinitely many solutions

[olive vault]

i got confused earlier i shouldve known this

my bad

[olive vault]

yeah

[gusty trench]

you can think of it as the number of degrees of freedom to describe solution vectors

in this case A has one free variable

and the solution set to Ax = 0 is vectors of the form (0, 0, c)

c is free to be anything

[olive vault]

right

[gusty trench]

great so we say that the dimension of the null space

the null space is V(A) btw

according to your notation

[olive vault]

but wait

[gusty trench]

the dimension of that is 1

[olive vault]

(0, 0, c) still has 3 elements

[gusty trench]

do you know what a basis is?

[olive vault]

no

[gusty trench]

do you know what a span is?

[olive vault]

no

well kind of but i cant use that

[long oracle]

Neil about to teach atleast 3 weeks of material lmao

[olive vault]

i know its all the possible vectors represented by something

[gusty trench]

the reason that the space of solutions (0,0,c) is of dimension 1 is because you can describe every vector in this space as a linear combination of only one vector, like (0,0,1)

[olive vault]

and i guess that something is the basis

or something like that

[gusty trench]

the dimesnion of the solution space has nothing to do with the fact that (0,0,c) has three elements

[olive vault]

ah

[gusty trench]

it has to do with the fact that every vector there can be described as a linear cobmination of only one vector

[olive vault]

ok

got it

[gusty trench]

so every vector there could be thought of as a scalar multiple of the vector (0,0,1)

[olive vault]

yeah

[gusty trench]

we say that (0,0,1) would be a basis vector for that space

since you only need one basis vector

[olive vault]

right

[gusty trench]

we say the space has dimension 1

you could think of it as degrees of freedom

if instead i gave you A =

1 0 0
0 0 0
0 0 0

and you consider Ax = 0

[olive vault]

dim would be 2

[gusty trench]

you have x = 0, 0 = 0, 0 = 0, meaning that your solution space is (0, c, d)

two degrees of freedom

two free variables in the matrix

[olive vault]

yeah

right

[gusty trench]

and the basis for that space would be (0,1,0) and (0,0,1)

since you can describe each vector of (0, c, d) as a linear combination of those two vectors

so diemnsion 2

[olive vault]

right

[long oracle]

Btw if you’d like to gain more intuition about this there is s great video by 3bue1brown

[gusty trench]

really recommend his LA playlist

great visualization, especially regarding rank and nullity in R^n

[olive vault]

noted :)

[gusty trench]

so do you understand why the dimension of the null space is the number of free variables?

[olive vault]

yeah

[gusty trench]

great then let's try again

B is an nxn matrix

that means n variables

[olive vault]

yeah

[gusty trench]

A is an mxn matrix

so AB also has dimension mxn

right

[olive vault]

yes

[gusty trench]

so AB also has n variables

[olive vault]

yes

[gusty trench]

now you know that V(AB) = V(B)

their null spaces are the same

this means the dimensions of those null spaces are the same

let's call that dimension k

[olive vault]

but i cant really use the term dimention

[gusty trench]

in other words, if you consider the matrix B, k out of those n variables will be free

[olive vault]

yeah

[gusty trench]

you can say that since V(AB) and V(B) are the same, AB and B have the same number of free variables

[olive vault]

ah

right

[gusty trench]

that might be a good enough explanation

so

let's call that number of free variables k

[olive vault]

right

[gusty trench]

that means that the remaining n-k variables are pivoted?

[olive vault]

yeah

[gusty trench]

so AB and B both have n variables, and they both have k free variables, so they both have n-k pivoted variables

so they both have the same number of pivoted variables, and therefore the same rank

this idea gets more formalized with the rank-nullity theorem of course but all it really is is a statement about how the number of free variables plus the number of pivoted variables equals the total number of variables

does that all make sense?

and is this enough of a proof without using something you haven't learned?

[olive vault]

i think thats enough yeah

[long oracle]

Perfect

[olive vault]

thanks a lot

[gusty trench]

happy to help lol

[long oracle]

I learned I’m rusty as hell so this was useful for me too

[olive vault]

while youre here

another question that i didnt know

given

does there eixst a

[gusty trench]

tf is a Z_7

[olive vault]

such that rk((A|c)) = n

[olive vault]

Z mod 7

F_7

[gusty trench]

matrices over fields other than R and C are a bit out of my comfort zone lol

good luck though

[olive vault]

no worries

thanks

[long oracle]

Hm I wonder if this just becomes a tedious exercise of checking

But also no nvm

[olive vault]

i dont think so

alright its 4:30 am i should probably sleep lmao

thanks for the help both of you

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[devout shale]

I don't know how to interpret the limit in question 4, because while the probability of single outcome with X=n with large n will be small, there will be more outcomes with that large n

[solar hazel]

well what do you think the probability is that game lasts 1000000000000000 rounds

approximately

[devout shale]

small

0

[solar hazel]

great, problem 4 is done

[devout shale]

But what if that is the wrong interpretatino

[solar hazel]

[devout shale]

How are you sure

[gusty trench]

a nice hint is also that it says "without calculating the probability"

kinda reinforces your intuition here

[solar hazel]

it says without calculating it, what do you think it is
so i don't think you need to be sure for 0 to be an acceptable answer

[devout shale]

I guess

[obtuse plover]

Recurrent MC moment 🙈🙉

[devout shale]

[gusty trench]

also since you obviously expect to lose money in the long run, the odds that you stay in for more and more rounds with this biased coin have to vanish

[solar hazel]

yea

you only expect the game to last so many rounds

so you can expect longer games to be unlikely

that's not 'rigorous' but it doesn't have to be lol

[devout shale]

Alrighty, thank you sirs and maam

.close

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[hazy tundra]

this is the question

i should show that thi slimit is 1/4

but i am getting 3/4

can i know where i made the mistake?

[jaunty spire]

do you use an "x" for multiplication?

[hazy tundra]

yes

[jaunty spire]

don't do that lol

it makes your writing kinda confusing

[hazy tundra]

srry

anyone?

[lime dust]

I can’t read well what did you do in your first step

[safe radish]

@hazy tundra Has your question been resolved?

[hazy tundra]

ignore this

[hazy tundra]

what i have done is i have taken cosec x out

[safe radish]

@hazy tundra Has your question been resolved?

[hazy tundra]

??

[hazy tundra]

<@&286206848099549185>

[heady lava]

hospital?

[hazy tundra]

bruh

u guys are useless

[safe radish]

@hazy tundra Has your question been resolved?

[safe radish]

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[broken nimbus]

How do i use WolfAlpha for row reduction of 2 matrices? like [A|B]

[broken nimbus]

For instance, how would I input this into WolfAlpha

[kind forge]

try image input and see if tht helps

[broken nimbus]

Ok i got it now, thanks

.close

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[barren bronze]

13d

[barren bronze]

how do i solve 3 + ln x> e^x. should i exponentiate everything or???

[faint seal]

Consider how the functions on both sides grow as x goes towards infinity

Particularly I want you to notice that the graph for 3+logx would start to grow much, much slower than the graph for e^x

[barren bronze]

ok how do i go from there?

how do i get theese numbers

[faint seal]

Hmm, the question seems to be much out of place of the others

The others are simple algebraic manipulations to solve the inequalities, while for this you will most likely need some sort of functional analysis using complex numbers

[barren bronze]

Ya agree

[faint seal]

First try to draw rough graphs for each of the curves $y=3+\ln(x)$ and $y=e^x$

[flat frigate]

kheerii

[faint seal]

Try to input some values to see where these two graphs intersect (roughly)

For example, at x=1, 3+logx > e^x but at x=0.5, 3+logx<e^x

[barren bronze]

but why is the lowest value 0.161. the intersect point seems alot more bigger than that

[faint seal]

0.161 sounds good to me

Why do you think it should be larger?

[barren bronze]

but why is the interval 0.1 - 1.1 the right? isnt e^x bigger every time when x is lower than their intersect compared to the natural log?

[faint seal]

Yeah, e^x is bigger when x is less than the first intersection point or when x is greater than the second intersection point

Which means the region between the two intersection points is the region you require

If you make a rough graph you’ll see what I mean

[barren bronze]

This is how i though it

[faint seal]

Hmm, not exactly like this

There should be a definite region between the two intersection points

Something of this sort

So the region between the two graphs is what you require

[barren bronze]

now i understand what the questions want. yeah the region between the blue and the green is where e^x is smlaler

[faint seal]

Indeed

[barren bronze]

but it just seems odd how this question cant be solved by akgebray

[faint seal]

Kinda hard to visualise this though but the best way to realise it is to try out random x values

[barren bronze]

because the other ones i used logs and stuff to simplify but this one was weird

[faint seal]

Yeah, there isn’t a simple algebraic way to solve for the intersection points unfortunately

[faint seal]

I agree the last part seems very out of place

The answer given is correct though, so there isn’t a mistake in the question

[barren bronze]

yeah true but the first question 13a had wrong awnser

so maybe the director were a bit trippy when they wrote this chapter lol

[faint seal]

Yeah that could be a possibility

[barren bronze]

but thanks for the help man, i appreciate it

🫡

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[eternal cloud]

my only question is for the sentence write an inequality for how far from the left you can shoot to get least 6 points

do I write my answer in {} or []

[marsh walrus]

what do you interpret those symbols to mean

[eternal cloud]

their for different functions, {} this i think would be for set notation

but its alright now

i found 3<=x<=7

[marsh walrus]

that sounds reasonable

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[primal plaza]

okay I’ve been stuck on this for 30 minutes if anyone could assist me I would be very grateful

[primal plaza]

No this entire problem is just hurting my brain

[short jackal]

the definition is given in the screenshot

[primal plaza]

How do I use it

[short jackal]

you look at the graph and see for which values of x the left limit does not agree with the right limit or the value of f(x)

that is the point of discontinuity

the function is also piecewise linear

so if you look at it it's definition which is given

there is only one possible candidate x value for a discontinuity

which you need to check to see if it satisfies or does not satisfy the given definition

[primal plaza]

Is it 3.5 or is that wrong

[short jackal]

how did you get 3.5?

[primal plaza]

Idk I set them equal to each other 😭

[short jackal]

that actually answers another interesting problem

[primal plaza]

But not this one ?

[short jackal]

which would be where should we switch between the two functions when defining a piecewise function to remain continuous

so all linear functions are continuous

then the only point where a discontinuity can be located is where we switch from one linear function to another

what point is that?

[primal plaza]

2?

[short jackal]

yes

now check the definition there

[primal plaza]

How do I do that plug it in ?

[short jackal]

a=2

[primal plaza]

So is that the answer cause a = x

[primal plaza]

So is that the answer ? 2

[short jackal]

I'm not giving the answer, that is the correct candidate to check if the definition is satisfied or not

If you think 2 is the answer, what part of the definition does it fail

That's what the question is asking

[primal plaza]

Oh it needs to fail ?

[short jackal]

you are giving the definition for continuity at point a

you are looking for points of discontinuity

which means NOT continuous

[primal plaza]

And is there work to do this

Or you just look at the graph

[short jackal]

either works

looking at the graph is easiest when given

but you also given the function explicitly and should know how to be able to use it to answer the problem

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[ocean tapir]

Can someone tell me of this is right

[long oracle]

what defintion of the inverses are you using

[dreamy perch]

give the question

[long oracle]

they have specific output ranges

its not any angle

usally arccos([-1,1])=[0,pi]

while arcsin([-1,1])=[-pi/2,pi/2]

by convention

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[frosty grail]

im confused on how you go from -cotx(-cscxcotx) to cscx(csc^2x -1)

[still charm]

sin^2 + cos^2 = 1
divide through by sin^2:
1 + cot^2 = csc^2, so cot^2 = 1 - csc^2

[frosty grail]

ok so would you multiply -cot x by (-cscxcotx) to get cotxcscxcot^2x. then you substitute cot^2x for 1-csc^2x. what do you do with the cotxcscx after that?

[still charm]

well first you want to learn how to multiply

multiply -cotx and -cscx cotx together to get csc x cot^2 x

then replace cot^2 with 1-csc^2

[frosty grail]

oh ok thank you.

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[marble tartan]

what does this look like in LaTeX ?

[thin bridge]

$\int_0^{10} \pi (3x-1)^2 \dd{x}$

[flat frigate]

ℝαμΩℕωⅤ

[nocturne elm]

I think the integral becomes {\tiny smol} when in "inline mode" $\int_0^{10} \pi (3x-1)^2 \dd{x}$

[flat frigate]

fix error

[nocturne elm]

Hmm, it doesn't

$\smallint_0^{10} \pi (3x-1)^2 \dd{x}$ is a bit too small

[flat frigate]

fix error

[marble tartan]

huh cool, thanks yall

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[lean otter]

if x^2+10x-3 is reducible over Q, then
3(2k+1)^2-(2p+1)^2 must be a multiple of 10 and [3(2k+1)^2]/[2p+1] must be an integer. (2k+1) and (2p+1) must be co prime

Prove there is no k,p∈Z that satisfies all conditions

[spice wing]

2k+1 and 2p +1 are coprime

Again 3*(2k+1)²/(2p+1) is integer

So 2p +1 = 3

Or p = 1

Also , 3*(2k+1)² - 3² = 10t where t is an integer

Or (2k+1)² = (10t+9)/3

Or (2k+1)² = 10t/3 + 3

1² = 1
2² = 4
3² = 9
4² = 6
5² = 5
6² = 6
7² = 9
8² = 4
9² = 1
You see no square number ends in 3

[spice wing]

which one ?

[true flame]

Helppp

[spice wing]

The formula for diagonals of a polygon given by n(n-3)/2

[true flame]

12

In the 13

[spice wing]

So n(n-3)/4 = n

This is 12

[spice wing]

Solve this

[true flame]

Okii

Resolutiom

[spice wing]

?

[spice wing]

^

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[upbeat ginkgo]

don't know how to start this

is it a power series?

[gusty trench]

a power series has a variable in it

like x

[upbeat ginkgo]

so it is?

[gusty trench]

it is not

[upbeat ginkgo]

it has x though

[gusty trench]

pretty sure that's a multiplication symbol

this question makes no sense if that's just an x

[upbeat ginkgo]

nah

if it was, it'd be the dot

haven't seen a multiplication symbol like that since middle school

i'd even say farther back

but nah, it's an x

[terse lichen]

nah the font for the x doesnt match the variable form of it

it definitely is a multiplication symbol

[gusty trench]

i'm like 99% sure it's just bad notation for multiplication

[upbeat ginkgo]

breh

lol

alrighty then!

[gusty trench]

you can make this into a geometric series

[upbeat ginkgo]

so 1/1-r?

[gusty trench]

more like ar^n

[gusty trench]

which you'll eventually simplify using this

but first you want it in the right form

so firstly pull out everything that doesn't depend on n

you'll get 7 * sum ((-1)^(n-1) * 3^n / 2^(2n))

can you write 2^(2n) as just a single number being raised to n?

[upbeat ginkgo]

uh

I don't think so

cause whatever value is n, you'd multiply 2 to it

which would drastically affect the value of the function when raised to that power

compared to simplifying it

unless i'm mistaken

but now how am I gonna get it into 1/1-r form for the geometric series?

[gusty trench]

x^(ab) = (x^a)^b

so 2^(2n) = (2^2)^n

is there a way you can write (-1)^(n-1) as something being raised to just n?

[upbeat ginkgo]

hmm

lemme think about that

one moment

I have no clue

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[pseudo scroll]

Does the series look familiar

[lean otter]

Summation 1/n! ?

[pseudo scroll]

[ f(x) = \sum_{n = 0}^{\infty} \f{x^n}{n!} ]

[flat frigate]

neon

[lean otter]

Was it the power series ?

[pseudo scroll]

yess

of what

[lean otter]

e^x ?

Left hand side = e^(loglog2)

Log2 > 3/5 [ True/False]

You need a calculator to check this

(to find value of log2)

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[hollow elm]

how does sin (kpi+x+pi) = -sin(kpi+x)

and how does that equal (-1)^k sinx

[burnt notch]

You can verify it using the formula for sine of a sum

sin(a+b) = sin(a)cos(b) + sin(b)cos(a)

Use a = kπ+x and b = π

Otherwise you can use a graphical way, this one: https://youtu.be/x36tRrVGtX0?feature=shared

[lime dust]

Also this is false

There is a - in front

[hollow elm]

i did it didn’t work mb i did something wrong

no it’s right

[lime dust]

-(-1)^ksinx

This is right

[hollow elm]

yea i did that?

[lime dust]

Not here

I read your writing

[hollow elm]

oh i meant that

[lime dust]

Did you do what alberto suggested?

[hollow elm]

how does -sin(kpi+.
x) = -(-1)^ksinx

how did he get rid of the x

k*

wait no go get it

.thanks

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[half canyon]

Find the equations of the tangent and the normal to 𝑦=9−𝑥2
at the point 𝐾(1,8)
im stuck with this question im not sure how to get the correct answer
i originally got mtan = -2x-4 and mn = 1/2x + 15/2

[burnt notch]

Is the function y = 9 - x² ?

[quick crater]

you mean $y=9-x^2$?

[flat frigate]

Skill_Issue

[half canyon]

yea

[quick crater]

do you know derivatives?

[half canyon]

yes

i gt -2x

[burnt notch]

Yep

[quick crater]

ok

use the formula for finding the ewuation of a line where you know a point and a slope

$y-y_1=m(x-x_1)$

[flat frigate]

Skill_Issue

[quick crater]

x1=1, y1=8, m you get as -2x

wait a minute

[half canyon]

isnt m just -2

since -2x = -2(1)

[burnt notch]

Yes exactly

[quick crater]

oh ye oop

[half canyon]

ok i understand where i went wrong

but i used a different formula that normaly works

thats just y = mx + b

[burnt notch]

You can use that too, no problem

[half canyon]

it just didnt work when i tried it idk why

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[lone arch]

My question is at #linear-algebra message

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[lone arch]

.close

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[worthy drift]

In Euler's formula,
When I take e^(2κπi),
when I take e^i(2κπ+π),
when I take e^πi

[low vector]

!original

[worthy drift]

In Euler e^i(x) when
x = 2kπ
or
x = 2kπ+π
or
x = π
??

[safe radish]

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.

[peak estuary]

you havent asked a question

when you do (thing), then (...) ?

[brave wolf]

Are you asking about the values of e^(2κπi), e^i(2κπ+π) and e^πi?

[worthy drift]

Yes

How I know what from the 3 should I take

[brave wolf]

e^(2κπi) = cos(2κπ) + i*sin(2κπ)

right?

[worthy drift]

No, when should I take each from the 3

[brave wolf]

?

[worthy drift]

The starting problem is 32x^6 + 2x^2 = 0

[brave wolf]

okay and you are solving for x?

[worthy drift]

Yes

[brave wolf]

first of all, factor out x^2

what will you get?

(Or 2x^2, that works too)

[worthy drift]

I found x=0, x⁴=-1/16

[brave wolf]

right

[worthy drift]

Then for the second x I need Euler's formula

[brave wolf]

indeed

-1/16 = (-1) * 1/16

and -1 = e^(πi)

[worthy drift]

Why πi and not i(2kπ+π)

[brave wolf]

You could use both. But πi is simpler

[worthy drift]

So it's the same?

[brave wolf]

or you can just use e^(πi + 2kπi)

which will probably be the best option if you arent entirely familiar with roots of unity

[brave wolf]

both are equal to -1, so they are essentially the same

[worthy drift]

Oh, now I get it, it is the same

Thank you

[brave wolf]

To get all the solutions, use this one though

it's the most general one

[worthy drift]

👍

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[trim vine]

can someone explain this step

[trim vine]

(the integral at the top is the question)

[vital tangle]

they've used the fact that 8 is 2^3 and 4 is 2^2 meaning you can then divide by 2^x

[trim vine]

yeah I got that

wait but

okay hold on

[trim vine]

okay so i'm getting [{2^(3+3x) + 2^(2+2x)}/2^x]

[vital tangle]

yea

[trim vine]

which gives

[vital tangle]

is that from integral calculator cause imo the way they've simplified it is kind of an unnatural way of doing it

[vital tangle]

you could instead just subtract one from each of the x terms and then use substitution

[trim vine]

yeah that's why i was really confused

[trim vine]

ah shite it's not clear is it

[vital tangle]

that looks like awht integral calculator gets

[trim vine]

this is what i got on me onw

answer isn't it the options tho

the closest I'm getting is (A)

[vital tangle]

It's B I think

[vital tangle]

using this step, doing the division and then use substitution u = 3+2x and u = 2+x

[trim vine]

oh yeah

[vital tangle]

which leads to (2x+2)/ln(2) + 2^x+2/ln(2) which simplifies to B

[trim vine]

how could i miss substitution

dagnabbit

[trim vine]

i see

[vital tangle]

well i'm pretty sure it simplifies to b i would verify that last step

multiple choice maths is stupid half the difficulty sometimes is putting it in some arbitrary form

[trim vine]

how do I apply the first u to the first term tho

sorry i'm really new to substitution lmao most of this just flies over my head

i'm getting du = 2dx yea

and du = dx in the second u

[vital tangle]

yes

[trim vine]

how do i apply it to the integration doe

that's where i always get stuck

[vital tangle]

so you end up with 2^u in the integral which is 2^u / ln(2)

[trim vine]

ah right

[vital tangle]

if you don't know the rule

[trim vine]

yeah i know of this one it's just that using it in questions is really confusing

like i don't even know where to put in the du i get half the time lmao

yeah okay so

[trim vine]

does 2^(3 + 2x)dx become 2^(u)du?

cus du is 2dx

OH WAIT
dx = 1/2 du

yeah i got it now

thanks a bunch

.close

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[fathom jewel]

$\textbf{Context.}\$
I am given the field $$H : 0 \leq z \leq \sqrt{9-x^2-y^2}$$ and the vector field $$\vec{v}(x,y,z) = \begin{pmatrix} \ln(e^x+e^y)+\ln(e^z+e^1) \ \ln(e^x+e^y)+z+1 \ x+y+z+1\end{pmatrix}$$

I had to calculate the flux $\Phi_B$ of $\vec{v}$ of the boundary of $H$ that lies in the $xy$-plane (ground area) from bottom to top. $$\Phi_B = 9\pi$$

Then I had to calculate with Gauss the flux $\Phi_R$ of $\vec{v}$ through the whole boundary are $R$ of $H$ from the inside to the outside. $$\Phi_R = 36\pi$$

$\textbf{Problem.}\$
The last task and my issue was to deduce the flux $\Phi_D$ of $\vec{v}$ through the ceiling area $D$ of the field $H$ from bottom to top and the solution was $$\Phi_D = -45\pi$$ and I didn't understand really why, I'll post the solution of my professor too.

[flat frigate]

𝔸dωn𝓲²s

[long oracle]

Are you sure you’re using the words “through”, “from bottom to top”, etc carefully here?

As that is 90% of the time the reason for the different signs

Also it could be the case that H induced some orientation

[fathom jewel]

I just tried to translate the context at best

[long oracle]

Hm okay, so geometrically you only have the fluxes for the boundary of the disk and then the open disk?

And want to find the flux for the dome, put differently Patrick’s ceiling from SpongeBob?

Well for shorthand let me just use R,B and D to go correspond to the fluxes

Is my interpretation correct so far?

[safe radish]

@fathom jewel Has your question been resolved?

[fathom jewel]

yes

[fathom jewel]

yes

[fathom jewel]

I mean here "boundary area" not "boundary are"

[long oracle]

Oh I see!

So we essentially have the flux boundary and you want the flux of the volume?

[fathom jewel]

I don't if it's equivalent to volume

but basically the flux of patricks ceiling + the circle gound

[long oracle]

Well I’m now having a hard time parsing the meaning of your use of boundary

[fathom jewel]

Rand

Let me ask chatgpt

[long oracle]

Uh no don’t that

[fathom jewel]

boundary surface

[long oracle]

Okay, what’s interesting is that that is already Patrick’s ceiling

[fathom jewel]

and what I meant by only "boundary"

basically

the half sphere has an intersection

with the xy-plane

this circle intersection

I call boundary

like a border or edge

[long oracle]

Oh okay, so it my was former interpretation

[fathom jewel]

yea

[long oracle]

I always use the topological boundary for some reason

When it’s the manifold one here

Okay

Let’s maybe write what the answer is saying, so using shorthand notation i mentored earlier

[fathom jewel]

i dont know what a manifold or topology is

[long oracle]

Doesn’t matter don’t worry!

[long oracle]

And if the orientation here is infact not the problem

Then we have that R+B=-D

Let’s try to interpret this

Does it even make sense

If we add the circle and the open disk flux then we get the negative flux of the ceiling

[fathom jewel]

yea basically

[long oracle]

And this is assuming all the orientations are the same

Which it seems it was

[long oracle]

So my question then is, do you think this makes sense?

[fathom jewel]

do you mean by orientation where the normal vector points?

[long oracle]

Yeah

Or rather also in which direction the flux is calculated

Which is indirectly determined by that too

[fathom jewel]

Yea basically it's always outwards

from my understanding, this is what was meant by "bottom to top" or "inwards to outwards"

[long oracle]

Yeah I would too

And so if our fluxes are positive it means the field was flowing that direction too as you may already know

[fathom jewel]

yea

[long oracle]

I guess one weird thing here is that, if you say that R was a circle, i.e a curve

How did you use Gauss?

[fathom jewel]

R was not the circle

it's the boundary surface

[long oracle]

Hm I thought that was the very thing we concluded

Okay, so let’s stop

[fathom jewel]

it was B what you mean

[long oracle]

Boundary surface of what?

[fathom jewel]

of the field H

I think I actually just got it

[long oracle]

H is already a surface too

There’s nothing about H that says it’s a volume

It could be interpreted as a shell

[fathom jewel]

well it says it's a field or literally translatede a "body"

unless I used the wrong word and field is something else

[long oracle]

Oh that’s the translation confusion lmao

Yeah body was the key here

We use that too in Swedish

Kropp

[fathom jewel]

jag vet

ok

[long oracle]

0_0

[long oracle]

Oh nice!

[fathom jewel]

well I got it like 50/50 I think

So the flux of the ground is -9pi geometrically speaking

[long oracle]

How?

[fathom jewel]

Probably cause the normal vector points down?

that's my assumption

[long oracle]

Oh it’s calculation from bottom to top?

[fathom jewel]

yea

[long oracle]

Then I think that should be reflected by the value used in Phi_B

[fathom jewel]

yeah idk why it isnt

[long oracle]

This might be an error in such case

So could you explain again what D was geometrically?

Just in case

I think I get it now too if my interpretation of D is the same

Might be an error of them to not write minus

[fathom jewel]

ok

hold on

[long oracle]

Since I may be underestimating the use of ceiling here for D

As otherwise D and R are the same

[fathom jewel]

It says Deckfläche D of the body K

Deckfläche is such area that you can see when you look at object from up

[long oracle]

Hm doesn’t sound like any nearby Swedish word

Oh’

Yeah

Maybe it’s that then

The normal is pointing opposite

And since R contains the floor

We have to remove it

Okay so I think that’s that

[fathom jewel]

,w -45 = -9 + x

[flat frigate]

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[long oracle]

Sign error for Phi_B

(Or so it seems)

D is just the same without the floor and opposite normal to R

Which explains the sign change

Oh or maybe I’m being stupid here

There is no sign error for Phi_B maybe?

[fathom jewel]

I don't think so

The funny thing is I asked my professor this already but I didn't understand anything 😭

[long oracle]

The B thing will have a normal point upwards right?

[fathom jewel]

yea

it literally says in the solution pointing upwars

[long oracle]

But for R, they are pointing outwards, so the floor for R will be opposite to B

[fathom jewel]

yes

so it's negative then actually

Actually shouldn't it be ground flux + ceiling flux = body flux

[long oracle]

Yeah but since D is like looking downwards we have that R is opposite to D without the floor, then just think carefully here what that means for B

[fathom jewel]

ground flux + ceiling flux = body flux
-9 + x = 36

,w -9 + x = 36

[flat frigate]

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Requested by adoniszgks

[fathom jewel]

I think it was like 45 going in and 9 going out

idk

😭

[long oracle]

Well think about it, if the normal of D is pointing downwards and we know that the vector field is flowing outwards

It has to be negative

Since we are calculating the flux for D from bottom to top

[fathom jewel]

hmm ok

[long oracle]

But the ceiling is pointing against that

According to how you described D atleast

Draw some pictures

It really helps

[fathom jewel]

I did

[long oracle]

Could you share?

[fathom jewel]

ye

,,\iint_R \vec{v} = \iint_B \vec{v} + \iint_D \vec{v} = \iiint_K \text{div}(\vec{v})

[long oracle]

This is how im interpreting it right now

[flat frigate]

𝔸dωn𝓲²s

[long oracle]

We know the vector field is flowing outwards becuase of Phi_R being positive

[fathom jewel]

ok

[long oracle]

and here R is closed, and D has no floor

[fathom jewel]

YEA

[fathom jewel]

SEE

[long oracle]

But Phi_D has to be negative

no?

[fathom jewel]

yea

it'S -45pi

[long oracle]

Oh so this confirms it for you?

[fathom jewel]

no

[long oracle]

Okay, look at this, we add R and B

[fathom jewel]

Oh why is D going inwards

[long oracle]

then we get D but pointing outwards

[long oracle]

because of your description from before?

[fathom jewel]

I am so dumb

[long oracle]

maybe its me being dumb here lmao