help-49

[stone sapphire]

ok thansk

[next rover]

yeah the other options are way wrong, there's no real choice

[cyan pendant]

definitely correct

[midnight plank]

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[last slate]

why do we need to take a determinant for evaluating eigen values?

[rose trout]

Eigenvalues/eigenvectors are such that $A\vec{v} = \lambda \vec{v}$. Rearranging, this means that $$A\vec{v} - \lambda I \vec{v} = \vec{0}.$$
And factoring we get $$(A-\lambda I)\vec{v} = \vec{0}.$$

[grand pond]

Azyrashacorki

[last slate]

yes

[rose trout]

Notice that since we usually want to find nonzero eigenvectors, this means that the matrix given by $A - \lambda I$ is not injective, so if there is such a pair of e-value/vector, the determinant of this matrix has to be 0.

[grand pond]

Azyrashacorki

[last slate]

why

[rose trout]

What part is not clear?

[last slate]

why determinant

[rose trout]

A - lambda I is not injective, meaning its kernel is not just {0}.

When the determinant of a matrix is 0, it means specifically that the kernel it not just {0}.

[last slate]

kernel?

[rose trout]

The set of vectors that get sent to 0 is called the kernel

i.e. if Ax = 0, then x is in the kernel. If the matrix is injective, the only such vector is 0

[last slate]

hm

okay

so why determinant

[midnight plank]

@last slate Has your question been resolved?

[rose trout]

Well you might recall that a matrix is invertible iff its determinant is nonzero.

And it's invertible iff it's bijective, and thus injective, which means its kernel must be just {0}.

So if in our equation we know the matrix can't be invertible, its determinant has to be 0.

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[real pivot]

$\int_{-1}^{1} \int_{x^2}^{2-x^2} f(x, y) , dy , dx$

Should the reversed order for this integral be:

$\int_{0}^{1} \int_{\sqrt{y}}^{\sqrt{2 - y}} f(x, y) , dx , dy$

[grand pond]

Calc III Victim

[real pivot]

Thing is I tried comparing it on desmos and it doesnt seem correct

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[real pivot]

<@&286206848099549185>

[real pivot]

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[viral dagger]

ABCD is a rectangle, E is an extension of CD and F is the point where BC and AE intersect, if AB=18, BC=24, area of CEF is 144 greater than ABF, find BF

[viral dagger]

let BF=x

[viral dagger]

we can see ABF≈CEF, let BF/FC=y

BF×18/2=BF×y×18×y/2-144
9x=9xy^2-144
x=xy^2-16

BF+CF=24
x+xy=24

idk how to solve this system of eqyatuon

[midnight plank]

@viral dagger Has your question been resolved?

[midnight plank]

@viral dagger Has your question been resolved?

[midnight plank]

@viral dagger Has your question been resolved?

[summer terrace]

i solved it without resulting in a system of eqs

using the intercept theorem, i found a relation between BF and CE

and then using the provided info that A(CEF)=145 A(ABF), i found another different relation between BF and CE

i got CE expressed in terms of BF from one of the two relations, and i substituted it in the other relation

i ended up with a quadratic equation in BF etc...

[viral dagger]

hey, how did you get A(CEF)=145 A(ABF)?

[summer terrace]

Area of CEF is 144 times greater than the area of ABF

[viral dagger]

its not 144 times greater

[summer terrace]

CEF = ABF + 144 ABF

[viral dagger]

its 144 greater

CEF=144+ABF

[summer terrace]

ummm idk, language here 😛

im not native english speaker

[viral dagger]

fair enough, probably couldve worded that better

[summer terrace]

I interpreted it as 144 times..

but but but, im pretty sure it's like I said

The quadratic I got was like

beautiful, you know what i mean

3x²+x-12=0

if I didn't treat it like that, idk what woulda happened

[viral dagger]

whats the relation between bf and ce btw?

not seeing it

[summer terrace]

there's 2, one from the intercept theorem and one from the areas relation

which one do you not see!

[viral dagger]

both :3

[summer terrace]

okay, from the intercept theorem, you get that

$\frac{BF}{FC}=\frac{AB}{CE}$

[grand pond]

Emily

[summer terrace]

and FC = 24 - BF

therefore

[viral dagger]

smort

[summer terrace]

$\frac{BF}{24-BF}=\frac{AB}{CE}$

$\frac{BF}{FC}=\frac{18}{CE}$

[viral dagger]

use double $$

[summer terrace]

$\frac{BF}{24-BF}=\frac{AB}{CE}$

hahaha

$\frac{BF}{24-BF}=\frac{18}{CE}$

[grand pond]

Emily

[summer terrace]

thats what i wanted to say :p

[viral dagger]

:p

[summer terrace]

and the other..

its simple, you can figure it out

find area of ABF, find area of CEF

set CEF = 145 ABF

you get in the end, after expressing CE in terms of BF:

$CE = \frac{2610 , BF}{24-BF}$

[grand pond]

Emily

[viral dagger]

$\frac{CE\cdot (24-BF)}{2}=9BF+144$

[grand pond]

Skill_Issue

[summer terrace]

and when you sub, it works out nicely

[viral dagger]

but thats wrong :<

[summer terrace]

idk, language/wording got me

anyways, i think i gave you the way though

[viral dagger]

$CE=\frac{18BF+288}{24-BF}$

[grand pond]

Skill_Issue

[summer terrace]

you can do it!

yeah yeah, i guess it works

whether you interpret it as i did, or as you said 144+ABF

nice!

[viral dagger]

$$BF=324BF+5128$$

uh wtf

[grand pond]

Skill_Issue

[summer terrace]

hahaha

[viral dagger]

:(

is tjis actually coreect

no way

dont believe

[summer terrace]

okay wait ill try it myself

[viral dagger]

ans chouces are 9 10 12 16 none of the above iirc

[summer terrace]

its 9

its correct, you surely made a little mistake, thats all i think 😛

do it again slowly

one of the steps: 18 BF² + 288 BF = 18 ( BF² -48 BF + 576)

[viral dagger]

alr thanks

.close

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[last slate]

answer is a and d

idk how to prove a

My working

Nvm

.close

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[cobalt dragon]

i read a point about skew symmetric matrix, that all diagonal elements must be zero. whys that

[nova yoke]

what's the definition of a skew symmetric matrix? start with that

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[paper whale]

can anyone tell me what's wrong here?
( i got 0/7 because it should be a disproof)

[runic hamlet]

well I mean even if its not unitary it could still be < at that point

which would make the whole thing <=

[paper whale]

so ur saying it's only the first direction?

[runic hamlet]

yes

[paper whale]

ok ty

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[glad merlin]

hello

[glad merlin]

im really confused on how we can interchange taylor series at a point with mclaurin series of shifted functions

is there more justification than this

[glad merlin]

was there a message here?

i swear i just saw something

[flint rain]

it was me but it wasn't really relevant

[humble torrent]

g(x) = f(x+a) is not enough?

since g(0) = f(a)

[flint rain]

It's just odd that you were introduced McLaurin series before Taylor series

[humble torrent]

not that odd, was for me too

makes more sense tbh

[flint rain]

A horizontal translation just corresponds to another function, if it still converges centered at 0 you can use the McLaurin expansion

[glad merlin]

i don't see how this relates to f(x). Like the main issue is that with the above statement is that i can't see (or maybe understand) the algebraic statement that links f(x) centred at a to f(x+a) centred at 0

[humble torrent]

you can retrieve f(x), by the fact that g(x-a) = f(x)

[glad merlin]

okay so let me just get the main stuff clear

we have f(x) centred at a point a

[humble torrent]

what do you mean by that?

[glad merlin]

sorry i was talking about finding the taylor expansion of f(x) centred at point a

okay so the above statement is saying that we can instead find the mclaurin series of f(x+a)

[humble torrent]

indeed

[glad merlin]

would i be able to prove using the expansion (i.e. f^(k)/factorial(k) * (x-a)^k) that f(x) at point a= f(x+a) at point 0

i know my notation is not very good, but hopefully the essence of my query was conveyed

[midnight plank]

@glad merlin Has your question been resolved?

[humble torrent]

im not sure i understand you, but yeah

just set g(x) = f(x+a)

apply maclaurin expansion

then you can get f(x) back by doing g(x-a)

[midnight plank]

@glad merlin Has your question been resolved?

[humble torrent]

@glad merlin ?

[midnight plank]

@glad merlin Has your question been resolved?

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[blissful drum]

im very confused as to how to graphically represent the additions/subtractions of vectors

[noble mural]

suppose that i want to go from A to C

but if i write a + b it is wrong as the 2 vectors are in opposite direction

when im asked to write vector AC

it means that how would i go if i want to travel from A to C

go throuhg B right?

[blissful drum]

are you saying distance vs displacement

[noble mural]

the direction

To go from A to C: i d go from A to B and then from B to C

[blissful drum]

sorry i forgot to phrase my concern. my concern is the tip to tail method

like you apparently need to

rearrange the vectors?

[noble mural]

ye thats what im demonstrating

[blissful drum]

and draw a new vector from the tail of the first vector to the tip of the second vector

[noble mural]

hmm sounds pretty strange

maybe i cant help on that :(

[blissful drum]

like this

the green is vector A + vector B

[light beacon]

yup

i always visualize it by flipping the second vector

and adding

[light beacon]

nope

[blissful drum]

ahhh

[light beacon]

$\vec{AB}=\vec{B}-\vec{A}$

[grand pond]

Hamdy Hisham

[blissful drum]

?

[blissful drum]

how do you prove that (ive heard of it before tho)

thats only if A and B are position vectors right

meaning they face different directions

like they spread apart

like this

[light beacon]

well right from your drawing $\vec{AB}+\vec{A}=\vec{B}$

[grand pond]

Hamdy Hisham

[light beacon]

that isn't necessary it can work at any vector

[blissful drum]

why is it ab + b

[light beacon]

look at your drawing and you will see

[blissful drum]

i dont follow

[blissful drum]

how is that possible

[light beacon]

.

[blissful drum]

lso for this

cant it be A + AB = -B

[light beacon]

nope

if that were true

that means that AB=-A-B

[blissful drum]

yeah but like

based on the direction of the vector

to represent vector B in terms of vector A and vector AB

[blissful drum]

wait whats the '-' and '+' signs based on

[light beacon]

note that $\vec{AB}$ is defined as the vector from the tip of $\vec{A}$ to the tip of $\vec{B}$

[grand pond]

Hamdy Hisham

[blissful drum]

yes

[light beacon]

the minus sign changes direction but with the same magnitude

[blissful drum]

ye so like for this one to express vector B, following the blue arrows, vector A + vector AB (a to b) = -vector B because the blue arrow is the opposite direction to the actua vector?

[light beacon]

nope

when adding vectors

the resulting vector is from the tail of the first to the tip of the last

so it will be +B

[blissful drum]

so the tail of the first meaning

the tail of A?

am i overcomplicating

is just remembering what vector AB enough

[light beacon]

yup

[light beacon]

yup

[blissful drum]

alr

thanks man

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[last slate]

How do i do the 1st question

[grim vector]

Isole y

[twilit field]

This ?

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[glad merlin]

hello

[glad merlin]

does anyone know what i have done wrong for the first part

as in the matrix representation of A2

i thought it would just be the matrix that was given to us

but apparently not

okay figured it out

basis means it must be linearly independent

not all of the vectors in the matrix are linearly independent

so ig i just take the ones that are

thanks

actually i'll keep this channel open just for a little bit more time

i may have another question real soon

[midnight plank]

@glad merlin Has your question been resolved?

[glad merlin]

how would you go about determining the radius of convergence here

i know its by the ratio test, but im getting a weird limit

which is approaching inf

that doesn't seem correct

[humble torrent]

Are you posting multiple questions?

Oh i see

[humble torrent]

What did you try?

[glad merlin]

the (1+1/n)^n is the e identity

but the (1+n) part makes it diverge to inf

i thought perhaps the radius was inf

but its no bueno

[humble torrent]

Do you recall the theorems related to power series and radius of convergence?

In any case, you used the ratio test and (correctly as far as i can see) the limit is greater than 1 (for most x)

what can you conclude in such case?

[glad merlin]

oh it diverges

[humble torrent]

Yup!

But what happens if x = -2 ?

[glad merlin]

converges

[humble torrent]

In any case the radius of convergence should be 0 if i havent missed anything

[glad merlin]

meaning that R = 0

[humble torrent]

yup!

[glad merlin]

cause its just a single point

that is so smart

[humble torrent]

well, it's just the ratio test

you did it all correctly aswell

it happens!

[glad merlin]

one more question

[humble torrent]

What was your work here?

[glad merlin]

i got this weird abs((x-4)/(n+1)^2)

and 1/(n+1)^2 also diverges to Inf

except when x = 4

so i thought that R will also be 0 in this case

but once again no bueno

have i perhaps done my factorials incorrectly

[humble torrent]

So the reason i mentioned theorems related to power series is because you can mostly ignore the (x-4)^n bit, as the coefficients here is what matters

i.e try the ratio test soley on the coffiencts c_n = 1/(n!)^2

if that is easier

[glad merlin]

ok i will try that again

just want to confirm one thing tho, ((n+1)!)^2 can be expressed as (n+1)! * (n+1)! right

[humble torrent]

sure, just depends on what youre doing i suppose, in our case we just "factor" the power 2 out of the quotient

if that is easier

i.e it feels reasonable to rather do a^2/b^2 = (a/b)^2 in this case, no?

[glad merlin]

ah yeah

im still getting 1/(n+1)^2

[humble torrent]

so as n approaches infinity?

[glad merlin]

yeah

[humble torrent]

oh i didnt catch that

[glad merlin]

it goes to Inf

[humble torrent]

Are you sure?

[glad merlin]

bro

[humble torrent]

just plug some really big number

[glad merlin]

im so dumb

its 0

[humble torrent]

no youre not dumb, again; all the right calcualtions but just a little mistake id do aswell

[glad merlin]

wait so how does that work then

you have 0

[humble torrent]

So this is the ratio test remember?

and in this case it seems this happens no matter what x we have

[glad merlin]

ah

Inf

[humble torrent]

What is inf?

[glad merlin]

as in like plus or minus infinity

or like the radius of convergence is the entire number line

[humble torrent]

Yea

Also, im not sure if youve seen it, but one theorem says that this limit you find, in this case 0—well the radius of convergence is just the reciprocal of that, and we're being a bit lazy here and letting 1/0 = inf (but this is not what happens in the proof of course)

[glad merlin]

ah very interesting

i always found 1/0 to be very ambiguous

[humble torrent]

yup and it is, and so its undefined

but in this rare instance its just a standard convention and theres theory to support the shortcut here, its just a way of symbolically expressing it; but its not the same as litrelaly doing 1/0

[glad merlin]

makes sense

also in regards to that previous question, sorry for just randomly ditching it

should have told you before

[humble torrent]

well usally its preferable if you use a channel per question

[glad merlin]

i was talking the mclaurin one

[humble torrent]

it was just confusing is all

Oh i see

Well i dont mind, i only help because i like to

[glad merlin]

thanks for your help aslan

really appreciate it

[humble torrent]

np!

[glad merlin]

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[humble torrent]

[midnight plank]

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[languid dagger]

I did the first part pretty easily, but I can't seem to get anywhere with the second part.

[languid dagger]

we can write expression for f(1) f(0) and f(-1)

but Im not sure how that really gets anywhere

f(-1) |a-b+c| <=1
f(0) |c| <= 1
f(1) |a+b+c| <=1

[runic hamlet]

use the expression for f'(x) you just obtained

[languid dagger]

yeah I was thinking I could try to find the maximum of it

[runic hamlet]

overthinking

[languid dagger]

but the negative at the end makes it a little trickier doesnt it?

[runic hamlet]

how big can the first term be at most

[languid dagger]

3/2

[runic hamlet]

what about the next one

[languid dagger]

surely 3/2 as well

[runic hamlet]

well I should be slightly more precise. how big in absolute value

uh wait made a mistake in my head. not as nice as I thought

ok, so obviously if the first one is 3/2, then the second one couldnt be 3/2 aswell

so we have to make that more precise

[languid dagger]

yea

[runic hamlet]

thankfully we can get rid of the f(1) and f(-1) by bounding them from above

so we are just left with |x+1/2| + |x-1/2|

[languid dagger]

oh as we make them 1 and -1 or smth?

f(1) and f(-1)^

[runic hamlet]

we want to bound |f(1)(x+1/2)+f(-1)(x-1/2)-2f(0)x|, yes?

[languid dagger]

yes

[runic hamlet]

by triangle inequality, thats <= |f(1)| |x+1/2| + |f(-1)| |x-1/2| + 2 |f(0)| |x|

[languid dagger]

sorry whats triangle inequality?

[runic hamlet]

|x+y| <= |x| + |y|

most useful inequality ever

abs of sum is annoying. sum of abs is better

[languid dagger]

alr

well if x = 1 then the max value of
f(1)(x+1/2) is 3/2
f(-1)(x-1/2) is 1/2
-2f(0)x is 2
and 3/2 + 1/2 + 2 = 4 which works

idk if it holds for -1

[runic hamlet]

well but what about all the other values of x

[runic hamlet]

keep working from here

[runic hamlet]

and then, like you wanted to at the beginning, find the max of this

[languid dagger]

um, the max of that would be 3/2 + 1/2 or |-1/2 - 3/2| so 2 in both cases

[runic hamlet]

ok why is it the max

but yes

and then finish the problem

[languid dagger]

well cause x is either -1 or 1

[runic hamlet]

why cant x be 0.71396

[languid dagger]

hmm if a were positive, the max would have to be at either -1 or 1

but im not too sure tbh

[runic hamlet]

make cases depending on whether x+-1/2 is positive or negative

[languid dagger]

so if x+1/2 is positive then the max is obviously x = 1
and if x + 1/2 is negative then the absolute max at x = -1
if x - 1/2 is positive then x = 1 is max
if x - 1/2 is negative then x = -1 is max

[runic hamlet]

good enough

so now, |x+1/2| + |x-1/2| <= 2

[languid dagger]

oh yes so if we take the max value lets say 2

then max of 2 - 2f(0)x, when x is 1

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[grim vector]

Area of a hemisphere ?

Is it a triangle or a conic under ?

[uncut topaz]

total surface area

i think its a hemisphere with a cone

[grim vector]

Ok so what is the area of a hémisphère?

[uncut topaz]

around 461?

[grim vector]

I meant the formula

[grim vector]

But indeed

3pi*49

[uncut topaz]

yea

[grim vector]

And what about the cône ?

[uncut topaz]

what would the radius of th cone be?

[grim vector]

The same as hemisphere

[uncut topaz]

then

it should be 517?

[grim vector]

How ?

[uncut topaz]

idk

[grim vector]

Wait we ll have a problem

[uncut topaz]

uhoh

what happened

nvm i got it

[grim vector]

Nah it will be 1/2* total surface area of a sphere + pi*r times sides length

Should be 637 or 638

[grim vector]

Counting the base, which is not part of area surface

[uncut topaz]

ok

is this right

@grim vector

[grim vector]

All good

[uncut topaz]

oh bet

thanks

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[grave locust]

can someone explain how to do this?

im used to making a matrix and solving

but i cant do it here

all i have right now is $0 = asin^2(x)+bcos^2(x)$

[grand pond]

marco.py

[grave locust]

not sure how to make a matrix out of that

[tired vapor]

for a function to be in $\text{span}{f(x),g(x)}$, they must be a linear combination, so $c_1f(x)+c_2g(x)$

[grave locust]

yes

[grand pond]

fish

[tired vapor]

wait you already had that step

[grave locust]

yes

[tired vapor]

since $c_1, c_2 \in \mathbb R$, i assume you can set both to 0 to confirm the first one

[grand pond]

fish

[grave locust]

so u just kind of guess

because 0 * anything = 0

[tired vapor]

i think so, not really sure how to set up a matrix for it

[grave locust]

ok that makes sense

[tired vapor]

otherwise, it would just be the product of 2 vectors, which gives about the same information as writing it as an equation

[grave locust]

what about for the sin(x) + cos(x)

i wrote

$sin(x) + cos(x) = asin^2(x) + bcos^2(x)$

[grand pond]

marco.py

[grave locust]

then i factored out

so

$0 = sin(x)(asin(x)-1) + cos(x)(bcos(x)-1)$

[grand pond]

marco.py

[grave locust]

how did u find that value?

[tired vapor]

wait nevermind

i guess i could say $a=1/sin(x)$ and $b=1/cos(x)$, but idk if that works

[grand pond]

fish

[robust isle]

no

it has to be an actual constant

no dependency on x

[tired vapor]

yeah, assumed so

[robust isle]

this is essentially an infinite system of equations, you plug in a value of x, you get a linear equation that a and b must satisfy

if you can extract a set of inconsistent equations for a and b (by plugging x's that simplify nicely with the sin's and cos's for example), it shows that sin(x) + cos(x) can't be in the span

[grave locust]

ok i think i get it

thank you

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[viral dagger]

hey, im just curious, how many times does 2^n and 3^m get 1 away from eachother?

[viral dagger]

i can count for small numbers, (2,3) and (8,9), but is there any more?

[merry pewter]

those are the only instances

in fact, a^b - c^d = 1 for a,b,c,d>1 has at most 1 solution for given a,b (unless a,b=2,3)

[viral dagger]

wow ok

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[viral dagger]

is there a proof for that?

actually it looks complicated nvm

[merry pewter]

ye should be a paper due to leveque

[merry pewter]

https://sci-hub.ru/10.2307/2371997

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[granite smelt]

my progress right now:
n=2 left, all other cases solved

anyone offer me some help to finish it off?

[midnight plank]

@granite smelt Has your question been resolved?

[merry pewter]

howd you get the n>2 case?

flt seems like the best tool here

[granite smelt]

No

N>2 was taken care of by LTE

[merry pewter]

i see

[granite smelt]

Firstly odd numbers are trivial, then just n>2 even numbers can be taken care ofby LTE

The reason why LTE doesnt fucking work for n=2 is cuz it’s p|n-1

No prime factors

And now i’m malding qwq

[granite smelt]

Some more elaboration:

Basically i just assumed what’s left was 1

Like

n^k = 1 (mod k)

In other words k|n^k-1

Then now if p|n-1, then n-1|n^k-1, so if i pet k=n-1, it’ll work:
n-1|n^(n-1)-1

To show that there are infinitely many cases, take k=(n-1)^m

[merry pewter]

yeah i get you

[granite smelt]

Yeah

And now i’m malidng over n=2

😭

[merry pewter]

wait wait im thinking

my thing ran out of battery

[granite smelt]

Dont directly spoil if you do get it tho

[merry pewter]

ye

[midnight plank]

@granite smelt Has your question been resolved?

[merry pewter]

ok bruh i think i got it

somewhat convoluted though

take k=2^m * p

been too long, im out of shape

[granite smelt]

uh

i read the solution already

SORRY LOL

but

[merry pewter]

rip

[granite smelt]

3*4^i was enough

i wouldve never found that though

[merry pewter]

yikes

[granite smelt]

what motivated this

[merry pewter]

stupid problem

[granite smelt]

(its 2014 N4

[merry pewter]

flt thing

[granite smelt]

yours align with sol 1 i suppose

[merry pewter]

yeah basically my solution

[granite smelt]

i woudlve mever found sol 1

sol 2 looks more approachable

dude what is this

💀💀💀💀

[merry pewter]

LOL

a_i\to infty

[granite smelt]

not in a million years is that going on my mind

i wonder what the motivation behind 3*4^i is

[merry pewter]

yeah that seems pretty silly

[granite smelt]

3 probably comes from the fact that the remainder of powers of 2 mod 3 form a seq with period 3

well that goes for pretty much all primes

[merry pewter]

maybe, i think it's just that 2^n*p is a solution frequently

for n large enough

the flt thing is kind of cool, so im glad i found that

you just go 2^k equiv 2^(2^m)

so you want to find p so that 2^(2^m-m)/p is odd

[granite smelt]

that's kind of insane

[merry pewter]

for many of these kinds of problems its safe to assume that flt is hiding somewhere

[granite smelt]

Usually

if there is an exponent

and there is a mos

mod

😱😱😱😱😱fermat: ever heard of prnis theorem

i mean little theorem

anyway cool

@merry pewter have you been to the imo?

not a lot of people can take on N4s easily

[merry pewter]

no

[merry pewter]

not easy

lucky because i am best at this kind of problem

[granite smelt]

wait

where are you from

usa?

[merry pewter]

ye

[granite smelt]

no wonder

have you been to mop

[merry pewter]

no

only got to aime twice 💀

not even to amo

are you preparing for imo

[midnight plank]

@granite smelt Has your question been resolved?

[granite smelt]

well

it's a far goal

i'd like to if i could make it onto team

[granite smelt]

dude but

jesus cheist

bro solves n4s under an hour

damn

are you still under 18

[merry pewter]

well i am in college, im 19

[granite smelt]

like are you elligble for team

oh

whoopsies

putnam?

[merry pewter]

yeah

[granite smelt]

ooooooo

youre preparing for putnam?

damnnnn

orz

anyway thanks bye

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[last slate]

problem 7

idk what to do

[fallen aurora]

I think using Vieta's formula is an approach to this problem

[last slate]

what

what's that

[fallen aurora]

uh like sum of roots = -b/a
and product of roots = c/a

[last slate]

so i take write both and simultaneous equation?

[fallen aurora]

Just apply Vieta's formula separately in those and try to eliminate the roots and m

[last slate]

ok ty i try

[last slate]

Would this work

[fallen aurora]

yeah I see no error

[last slate]

okay thank you very much

this was very helpful ty

[fallen aurora]

np

[last slate]

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[clever karma]

Why is a basis for A also a basis for B = (A)(A^t) ?

[clever karma]

the original problem was: construct a matrix with (1,0,1) and (1,2,0) as a basis for its row space and its column space.

they answered it by multiplying $\begin{bmatrix}
1 & 1\
0 & 2\
1 & 0
\end{bmatrix}

\begin{bmatrix}
1 & 0 & 1\
1 & 2 & 0
\end{bmatrix}
$

to get $\begin{bmatrix}
2 & 2 & 1\
2 & 4 & 0\
1& 0 & 1
\end{bmatrix}
$

[grand pond]

johnseymour20
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[clever karma]

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[icy cypress]

I am just asking why the continuity correction is addition rather then subtraction here

for p(z <= 2.2) I understand it would be addition because it has to "wrap around" the 2.2 value inclusive

but for what is p(z < 1.5) shouldnt it be that after the correction it is p(z < 1.5 - (0.5/50))

where 50 is the sample size?

Please ping me if you know the answer, thanks.

[midnight plank]

@icy cypress Has your question been resolved?

[icy cypress]

<@&286206848099549185>

[summer terrace]

but P(1.5 < X <= 2.2) is not equal to P(X<=2.2) - P(X**<**1.5)

it's actually: P(1.5 < X <= 2.2) = P(X<=2.2) - P(X <= 1.5)

[icy cypress]

wait wait

like after correction or before?

[summer terrace]

before

Notice the X range, 1.5 < X <= 2.2

using your thinking, which is

P(X<=2.2) - P(X<1.5)

X can be 1.5 right?

😛

[icy cypress]

WOAH

WAIT UP

WHAT THE

[summer terrace]

But it mustn't

[icy cypress]

wait doesnt this mean its just like any value to neg infinite

im confuzzled

[summer terrace]

it's an interval thing, that's all

[icy cypress]

so you just have to look at this questions and like think big brain style

while doing it

[summer terrace]

exactly
X must be strictly greater than 1.5

can't be exact 1.5

if you use < 1.5

[icy cypress]

because when p(x<=2.2) it has to be every value below

[summer terrace]

X can be 1.5

yes

[icy cypress]

okok

and without the

2.2

its like the same?

[summer terrace]

i didnt understaaand

[icy cypress]

or is it like the weird thing where when u sign flip with the normal distribution coz the alligator (<>) is the wrong way for your z table?

does it turn from > to <= because of the x <= 2.2 or because of the ^^

[summer terrace]

i dont know how you are thinking about it 😛

[icy cypress]

hm

[summer terrace]

the way im thinking about it is...

[icy cypress]

neither

[summer terrace]

in order to have the required interval ]1.5 ; 2.2]

you grab ]-oo ; 2.2]

and you remove from it

]-oo ; 1.5]

1.5] not 1.5[

[icy cypress]

sorry, I am not familiar with this notation

[summer terrace]

it's okay, it's okayy

[icy cypress]

mbbb

i feel bad now

[summer terrace]

no, dont be

its kinda hard to explain

[icy cypress]

so its nothing to do with the normal distribution

[summer terrace]

yesss, just the interval

[icy cypress]

you could just consider the equality 15 < x <= 22?

im not really familiar with like messin round with equalities like that

maybe I should be

[summer terrace]

are we on the same page, or not yet 😛

[icy cypress]

HMMMM WAIT ITS LIKE SAYING THE X>15 AND THEN THE PLUS X<=22

THEN IT FLIPS VIA NORMAL STUFF TO X<= 15

[summer terrace]

about why it is this: P(1.5 < X <= 2.2) = P(X<=2.2) - P(X <= 1.5)
and not this: P(1.5 < X <= 2.2) = P(X<=2.2) - P(X < 1.5)

[icy cypress]

?????

not question marking ur statement btw just my questiony thing

[icy cypress]

uhm yeah coz the 2.2 is like above it

and for x<=2.2 x must also be <= 1.5

[summer terrace]

yes, 2.2 and less than 2.2 are all included

[icy cypress]

not just skip the 1.5 value

[summer terrace]

now just remove 1.5 and what's less than it

you get the required interval!

[summer terrace]

yessss

you gotta erase 1.5 as well

[icy cypress]

wait so it just becomes x<=22??

this erasing and removin got me trippin ngl

[summer terrace]

okay, wait

i'll try to draw a quick thing

[icy cypress]

kk thx

[summer terrace]

good old paint app

[icy cypress]

no worries if u dont wanna'

[summer terrace]

its okayy

[icy cypress]

👍

[summer terrace]

okay, so here's the "X axis"

Look at the given interval

15 < X <= 22

[icy cypress]

yep

so blue means inclusive

and red means

not inclusive?

[summer terrace]

i dont know what that means

just follow with me

[icy cypress]

nvm

okok

[summer terrace]

we need this 15 < X <= 22

its in green

that's the needed interval

[icy cypress]

yes

[summer terrace]

notice it includes 22

[icy cypress]

mhm

[summer terrace]

but doesnt include 15

stops close to it

cuz its < 15

not <= 15

[icy cypress]

yeah

[summer terrace]

okay, so far so good

[icy cypress]

yes

[summer terrace]

to have this interval, we start with this interval, i will draw it in purple, wait

and we remove from it this one

in orange

[icy cypress]

my marker bout to have the full rainbow

[icy cypress]

O WAIT

O WTH

WTF

THATS ENLIGHTENING

WH

[summer terrace]

thats how you get the needed green interval

[icy cypress]

WTH

[summer terrace]

you remove 15 too

!!!!

[icy cypress]

WOT

HOW

blud needs to teach

[summer terrace]

😂

[icy cypress]

tysm

fr

[summer terrace]

you got it now?

!!

[icy cypress]

yeah !!

😜

[summer terrace]

what you were doing is... you were grabbing the purple interval

[icy cypress]

uhuh

[summer terrace]

and you were removing the orange interval, yes

[icy cypress]

take big daddy purple and remove small orange

[summer terrace]

but without removing 15 too

[icy cypress]

to get green

[summer terrace]

which is the required

[icy cypress]

yas

https://tenor.com/view/monday-blessings-gif-5947373921499176036

what is this 💀

[summer terrace]

and as you see, orange is P(X<=15) and not P(X<15)

[icy cypress]

yeah ]

the graph

[summer terrace]

i know you understood, just wanted to make it clear

[icy cypress]

like

awoke something in e

imma mathematics beast now

so tysm

[summer terrace]

Hahahhahaha

[icy cypress]

can i like rep u

[summer terrace]

you are welcome

[icy cypress]

smwhere

to be a helper or summin

[summer terrace]

noo, i dont think its possible

even if it was, no need

im just a random

[icy cypress]

<@&286206848099549185> make dis person helper

[summer terrace]

hovering around and helping when possible

😂

[icy cypress]

more helpfuil then yall 👀

aight laters

tysm

again

/close

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[last slate]

how what why

[noble mural]

have y tried to simplify it

[last slate]

i moved -1/tx-4 to the right

now i have 36x^2+16x-20/tx-4

should i multiply tx-4 by -9x+5

?

[worthy wing]

Factor numerator 36x^2+16x-20

[last slate]

4(9x^2+4x-5)

oh makes sense

[worthy wing]

You can factor better

[last slate]

how

[worthy wing]

Factor inside the parenthesis

[last slate]

u mean solve the quadratic equation?

[worthy wing]

Factor the polynomial 9x^2+4x-5

[last slate]

how tho

there is nothing left

[worthy wing]

Factor like (ax-c)(bx+d)

[last slate]

(9x-5)(1x+1)

[worthy wing]

So u have 4(9x-5)(x+1)

[last slate]

right

[worthy wing]

Now the exercise is easy

Check right side

U will see something there

[last slate]

-9x^2t+36

[worthy wing]

?

[last slate]

-9(x^2-4)

wait wai t

[worthy wing]

Where did you get this

[last slate]

-9x+5

[worthy wing]

Good

If you multiply both sides by -1 u will get 9x-5

And in the left side

[last slate]

right

then divide it

[worthy wing]

-4(9x-5)(x+1)

[last slate]

with the other 9x-5

[worthy wing]

Yes

U get 1 in the right side

[last slate]

-4(x+1)/tx-4 = 1

-4(x+1) = tx-4

-4x-4=tx-4

-4x=tx

t=-4?

[worthy wing]

Looks good

[last slate]

ayy thanks

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[strong pebble]

if ϕ : V -> R^n is a linear isomorphism, then does V=R^n?

[runic hamlet]

no

V is isomorphic to R^n

but they are not the same thing

[strong pebble]

oh ok ty

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[drifting root]

i have no idea what to do

ahhh wait

is it that cause its a to the power of 3, technically the most amount of intersections is 3?

so d jusy equals k

but then its a bit misleading to put 4 times if its infinite intersections

[fresh sparrow]

yeah the wording is weird

[fresh sparrow]

but it doesn't say exactly 4 so i guess it's fine (?)

[drifting root]

fair fair

thanks for clarification

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[inner timber]

A prism whose base is a 9 -sided polygon is intersected by a plane. In which case could the cross section be a 10 -sided polygon?

A. when the plane is parallel to the base of the prism

B. when the plane is perpendicular to the base and contains two edges of the prism

C. when the plane is perpendicular to the base and contains one edge of the prism

D. when the plane makes a narrow angle with the base

[inner timber]

how do you answer this question

I can't visualize it

[midnight plank]

@inner timber Has your question been resolved?

[inner timber]

I don't think it is A because then it would be a 9 sided polygon since it's parallel to the base

I have no idea bout the other 3 however

[inner timber]

@stiff heart

[midnight plank]

@inner timber Has your question been resolved?

[midnight plank]

@inner timber Has your question been resolved?

[delicate helm]

B and C would make a rectangle

D can make a bunch of stuff, but the idea is that if you just clip one angle, then you will add an aditional side there

should i model it?

[main cloud]

yes pls

[midnight plank]

@inner timber Has your question been resolved?

[inner timber]

Yes

Please

[inner timber]

Woah those are so good models

Thanks I can see it

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[midnight plank]

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[midnight plank]

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[wild cloak]

I don't understand what this question wants from me. I am in Calculus II and this question is part of a Taylor Series assignment. I am sure there is supposed to be a way to use my knowledge of Taylor Series to solve this question, but I am not seeing it.

[fossil knot]

Do you know binomial theorem

[wild cloak]

i'll look into it one sec

oh yeah ive seen that

alright thanks @fossil knot

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[fossil knot]

It's not super relevant here

o

[wild cloak]

Oh

.repoen

.reopen

[midnight plank]

✅

[fossil knot]

But you can see that when you expand the expression, its degree is 144 with leading term x^144

Binomial theorem tells you there's nothing with x^143

[wild cloak]

Why is that

Wouldn't there be one

Nevermind i see

Wait I think I see now

But once I expand it, won't there the leading term be (x^8)^18

cause its (x^8) * (x^8) * (x^8) * (x^8) * (x^8) * (x^8) * (x^8) * (x^8) * (x^8) * (x^8) * (x^8) * (x^8) * (x^8) * (x^8) * (x^8) * (x^8) * (x^8) etc. til its 18 times

[fossil knot]

yes

which is x^144

[wild cloak]

woops

my brain is fried

sorry

[fossil knot]

np

[frail carbon]

(so close, you copied it 17 times)

[wild cloak]

Oh i didnt think i was even close to 18 times

I just said etc

so basically

the factorial of the leading thing's exponent

so 144!

[fossil knot]

what

[wild cloak]

cause every other term below 144 would go to 0

by the 143rd derivative

and it would be 144 * 143 * 142 etc... all the way down to 1?

Or am i wrong

i just dont get why x going to 0 would be there either

maybe im lost

[fossil knot]

Each time you take the derivative of $ax^n$, the exponent decreases by 1

[grand pond]

Ari

[fossil knot]

oh i see what you're saying

yeah the coefficient is 144!

[wild cloak]

So

Would it go all the way down to

(144)!x

but then its x=0 so isnt it 0

i have 2 tries left so idk if i should check

[fossil knot]

exactly

[wild cloak]

omg

THank you

But no way it was just a GIMMICK QUESTION

thank u for ur help tho

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[midnight plank]

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[fossil knot]

lol

[midnight plank]

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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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• 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.

[drifting root]

I have no idea where to start

[fossil knot]

Try adding them

or that's my initial idea but idk if it works

[drifting root]

if i add them why does a common factor appear?

[fossil knot]

If something's a factor of A and B, it must also be a factor of A+B

So if you find factors of A+B, you can test and see if they are also factors of A and B

[drifting root]

ahhhhh

fair fair

makes sense