help-26

[neon iron]

how am i supposed to check if a basis can exist for this?

i understand the answer but im not sure how i could get the "no basis" answer

[molten vine]

do you know what vect is ?, here's an example

$\mathbb{R}^3=vect((1,0,0),(0,1,0),(0,0,1))$

[thorny flame]

Adam Chebil

[neon iron]

yes. but idk how this is linearly independent

i thought you needed 3 vectors for it to be independent

[molten vine]

it's a subspace of R^3

[neon iron]

ah wait yeah its less than so you check ci

[molten vine]

??

[molten vine]

it's an equation of a plane so 2 linearly independent vectors is enough

[neon iron]

oh

thats cheap bruh

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[sleek needle]

ive shown c is regular

dont rly know where to start when finding a reg rep

ik that the reg rep must be orientation preserving

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[gritty vigil]

what did I do wrong here?

[gritty vigil]

algebraically speaking

[sweet shard]

What's f(x)?

[gritty vigil]

2x^2-3

[mellow arrow]

Why did you convert x^2 to x^3?

[sweet shard]

You set up your difference quotient wrong

f(x+h) - f(x) = ?

[gritty vigil]

ah

yeah

thanks

[gritty vigil]

oh yeah that too lol

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[opal vault]

first of, a relation is a subset of E*E. However, I have trouble understanding what E is. Is it {1,...,9}? Is is {1,...,10}?

ok

Proof of antisymmetry : suppose nRm and mRn.
Since nRm, ...
Since mRn, ...
Thus ...

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[crimson kestrel]

ok i am stuck on this question

[crimson kestrel]

this is teh asnwer to the question and how would you go from first line to the second line of part b

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[crimson kestrel]

thanks @neon iron

okay...

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[humble sonnet]

limit when x approaches 1 from the right

[humble sonnet]

sadly L'hopital's not an accepted solving method here

[mild hearth]

probably taking a log to start is the way to go

[humble sonnet]

can u walk me through it? it's a question given whilst we haven't studied logarithms yet

[mild hearth]

basically $\log(ab) = \log(a) + \log(b)$ \

so $\log \left( (x-1) e^{{(x-1)}^{-1}} \right) = \log(x-1) + \log \left( e^{{(x-1)}^{-1}} \right)$ \

and then $\log(e^y) = y$ so $\log \left( e^{{(x-1)}^{-1}} \right) = {(x-1)}^{-1}$

[neon iron]

are you allowed to use Taylor series?

or the fact that exponentials dominate polynomials

[thorny flame]

Kaisheng21

[humble sonnet]

I can use that

[humble sonnet]

what's the limit of log^(x-1)

the only think Ik of Logs is that they oppose exponentials

[mild hearth]

because it's like, what e^y = 0?

log(x-1) is y st. e^y = x-1

so log(1 - 1) is y st. e^y = 0

so it's like, -infinity

[humble sonnet]

alright

thanks man

[mild hearth]

but yeah also note that this gives you the log of the limit

since we took the log in the first place

so to get the actual original one you might need to do exp of the answer

maybe this works idek

[humble sonnet]

oh, thanks for reminding me

amma go try it out

how do I tell it that my question has been answered?

.close

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[steady grotto]

Determine if the series is convergent or divergent by comparing against a relevant integral.

[topaz sinew]

@steady grotto Has your question been resolved?

[sweet shard]

usually you just replace the index variable with x and integrate the function wrt to x

scroll down to integral test
https://tutorial.math.lamar.edu/classes/calcii/IntegralTest.aspx

[steady grotto]

Yeah I know i did that, but the question asked specifically to compare to a relevant function, which I have trouble finding

[sweet shard]

show the question

[sweet shard]

if you did this then you're done

[steady grotto]

Question is in a foreign language that's why I have written it out, but it asks as i wrote for a comparison to another integral

the series is in the image above

So im specifically wondering how to do comparison with this series and not the integral test

[sweet shard]

what else does "relevant integral" mean then?

[steady grotto]

Using the comparison test

Cauchy's convergence test

(I believe)

[sweet shard]

sounds wrong

if you keep saying comparing to an integral

[sweet shard]

it's just this

[steady grotto]

Well Cauchy's test lets you rewrite the series as an integral you can use for comparison

Which I have succeded in doing several times, but with this question if having a hard time finding an integral to compare with

[sweet shard]

read this

[steady grotto]

PLEASE LISTEN! I know that is a relevant test an i have solved using it!

But this question requires another method

[sweet shard]

why

[sweet shard]

.

[steady grotto]

Here it's in swedish lol

I translated it in my original message

[sweet shard]

...

[steady grotto]

word for wrod

[sweet shard]

where does it say this

[sweet shard]

and not just this

[steady grotto]

"genom att jämföre med en lämplig integral" - by comparing with a relevant integral

So i cant just integrate it 😭

[sweet shard]

this tells you how to find the relevant integral

have you read that link yet?

look for "integral test"

[steady grotto]

aha I see where the comparison ocours now, thanks! good to know that i did it right from the start

if you could be bothered how does it work for this integral:

I have a hard time finding out using the integral test if it is convergent or divergent

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[spark vortex]

can someone explain what a co vector is

[spark vortex]

i cant seem to understand the purpose of it

ive read many def but none of them acc make sense

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[spark vortex]

.reopen

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[untold sable]

@spark vortex
Imagine you're walking on a straight road. Your position can be represented by a vector. Let's say the vector is ( \text{Position} = [3, 4] ), meaning you're 3 steps east and 4 steps north from your starting point.

Now, let's introduce a covector that represents the cost of walking. The covector is ( \text{Cost} = [2, 1] ), meaning it costs 2 coins to move one step east and 1 coin to move one step north.

To find out the total cost of your walk, you'd multiply each component of your position by the corresponding component of the cost, and then add them up:

[
\text{Total Cost} = (3 \times 2) + (4 \times 1) = 6 + 4 = 10 \text{ coins}
]

So, the covector ( \text{Cost} ) takes your ( \text{Position} ) vector and turns it into a simple number: 10 coins

[thorny flame]

Jane

[untold sable]

They’re used in most modern physics

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[neon iron]

A company will pack peaches in 0.5L cylindrical cans. Let x be the radius of the can in cm.
(a) Express the surface area of the can as a function of x.
(b)Find the radius and height of the can if the surface area is 900 cm^2. Solve using a graphing calculator. Label the dimensions.

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<@&286206848099549185>

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[quick vale]

How do I solve this problem?

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<@&286206848099549185>

[junior acorn]

angle between the lines is the angle between their direction vectors

so find the angle between (3,2,5) and (-2, -3, -4)

[quick vale]

gotcha

thanks!

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[wet sun]

!status

[topaz sinew]

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

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[jovial bay]

im confused on what the 4 cases are

[reef fjord]

all primes greater than 3 are odd right? what are the possible forms of odd numbers in mod 6?

[azure parrot]

(unless its looking for you to apply a certain theorem, i think this means 3 cases?)

[jovial bay]

1,3,5

[reef fjord]

Yeah i'm pretty sure there's only 3 cases, 1,3,5

it should be pretty easy to eliminate why integers of the form 3 mod 6 can't be prime

[jovial bay]

thats what i thought 😭 thats why i got confused and came here when i saw that it was 4 cases lmao

[azure parrot]

oh

there is one specific way you can make it 4 cases

hang on

its a little silly to do it over this method though, haha

[jovial bay]

also in response to this, i dont believe our instructor is looking for any specific theorem

[azure parrot]

in that case just analyse the different cases of odd numbers mod 6 and you should see a proof

[jovial bay]

ok thanks for the help

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[maiden needle]

how do you solve this using first and second differences?

[strange girder]

Quadratic it's x² no ?

[maiden needle]

textbook says linear

i want to know how to get there though

[strange girder]

Y it's linear ?

Linear it's under the form Ax+b

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[graceful tinsel]

Question f

[graceful tinsel]

I don’t even know where to start

[wet sun]

What is the question asking you to do

[graceful tinsel]

We are solving the equation. Verifying the solution. and the restriction being shown

[strange girder]

You can't solve it ?

[topaz sinew]

@graceful tinsel Has your question been resolved?

[graceful tinsel]

Nope

[strange girder]

You develop for arrived with equation 2nd degree

And you calculate discriminant with 2 solutions

∆= 745 if I know how to count

[graceful tinsel]

We have not done any of that

Im doing grade 11 math

[strange girder]

I don't know I'm french

How old re u

[graceful tinsel]

16 haha

[strange girder]

What s the question of yours exercise

[graceful tinsel]

2a

[strange girder]

Okay thx

[neon iron]

the answer is 4

[strange girder]

U don't know the discriminant with b²-4ac and X=(-b+√∆)/2a ?

[neon iron]

its 1

🧠 💡

[strange girder]

Of what?

[graceful tinsel]

Actually it’s

[neon iron]

the gradient vector identity in respect to the homemorphism across maps A and B along inverse bijections

🧠 🧠 💡

https://tenor.com/view/think-about-it-use-your-brain-use-the-brain-think-brain-gif-7914082

[strange girder]

Xd

[neon iron]

get your topology up not your funny up

[graceful tinsel]

What the hell are you talking about 😭

[graceful tinsel]

Bro this got me spinning heads

[strange girder]

Troll xd

[graceful tinsel]

I need to solve this

[neon iron]

the restrictions between the transvicity of the fields

[strange girder]

He want do big brain xd

[neon iron]

which question

[graceful tinsel]

Bro I got an exam tommorow 😭

[neon iron]

i will solve it for you

[neon iron]

i finished my test today

[graceful tinsel]

Like I need explaining

[neon iron]

50/51

[graceful tinsel]

2a

[neon iron]

i got 98%

[graceful tinsel]

See bro

Your chiling

[neon iron]

what

i will stop trolling

let me help u

[strange girder]

How old re u

[neon iron]

imma get a paper

[graceful tinsel]

Me? Hehe the exam is worth 24%

[neon iron]

im in grade 10 but finished up to manifold topology

[graceful tinsel]

Say your swear you in grade 10

[neon iron]

i am too

[strange girder]

I don't know xdd I'm french how old re u

[neon iron]

15

[graceful tinsel]

Im grade 11

[strange girder]

Xdd okay

[neon iron]

guys enough chatting

im trynna do 2a

[graceful tinsel]

Bet

[neon iron]

what is it asking

send the image here

so i dont have to scroll

wait

[neon iron]

which 2a

ok that one

[neon iron]

i see now

okay give me a second to work on it

if x =2/3 then

sqrt of 0

tf?

dawg it says x is greater than or equal to 2/3

if you substitute 2/3 into the values it will end up as 2/3=0

[graceful tinsel]

Wait you got a mic? Me and my other friend both tripping on this question lol

[neon iron]

yes

add my cord

like what

[graceful tinsel]

Could I add you to a chat and we talk?🙏

[neon iron]

yes

[strange girder]

@graceful tinsel a exemple

[topaz sinew]

@graceful tinsel Has your question been resolved?

[graceful tinsel]

Disconnected? Lol

<@&286206848099549185>

<@&286206848099549185>

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[graceful leaf]

uh

This one is weird for me

cuz there is no base

[vernal vale]

whats up ac

typically its implied that logs are base 10

[graceful leaf]

oh

So if there is no base shown for log is always 10

Ok

[vernal vale]

at least until you get past calculus

then it means natural

[graceful leaf]

o

[vernal vale]

here youd probably see ln

anyways

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[velvet vortex]

In this proof for Euler's totient function, it is claimed that $\text{The number of integers from 1 to n that are divisible by n is simply} \frac{n}{p_i}, \text{since} p_i \text{divides} n,$ but I have a difficulty understanding why. Can somebody explain this to me? Thanks in advance!

[thorny flame]

Joshua David

[velvet vortex]

(please ping me when you respond :D, thanks again!)

[sweet shard]

Your latex question got cut off

[alpine mist]

pretty sure that's a typo, and it's supposed to say the 'the number of integers from 1 to n that are divisible by p_i...'

[alpine mist]

it's just a counting argument.
since $p$ divides $n$, you have $n=kp$ and can say the integers less than $n$ that are divisible by $p$ are:\
$p,\ 2p,\ 3p,\ 4p,\ \ldots, kp$\
So there are $k$ integers that are divisible by $p$, and $k$ is exactly $n/p$

[thorny flame]

Zybikron

[velvet vortex]

I see it now, thank you both @alpine mist and @sweet shard :D!

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[boreal juniper]

How do you solve this question?

[drifting swift]

do you know what properties a function must satisfy to be a density function?

[topaz sinew]

@boreal juniper Has your question been resolved?

[boreal juniper]

no

should there be any specific properties?

[drifting swift]

yes, it needs to be >=0 everywhere and its integral over the entire real line must equal 1

[boreal juniper]

we were just told to consider the ideal conditions

[drifting swift]

the what now

[boreal juniper]

🤷‍♂️ even i'm confused.... i only know that X is a continuous random variable

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@boreal juniper Has your question been resolved?

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@boreal juniper Has your question been resolved?

[boreal juniper]

<@&286206848099549185> How do i solve the (e) part?

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[dark stream]

Specific to my written question on dy/dx

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[dark stream]

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[topaz sinew]

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[vagrant field]

Why does ln|t| = ln(t) if t>0?

[mellow venture]

ln not defined for negative numbers and zero

[long stirrup]

what do you mean

that's every function

arccos|a| = arccos(a) if a > 0
|n|! = n! if n > 0

[topaz sinew]

@vagrant field Has your question been resolved?

[sweet shard]

,tex .abs def

[thorny flame]

riemann

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[cinder owl]

can someone xplain to me

what the row 4 cofactor formula means i guess

like im trying to understand the hint

[odd pagoda]

do you know what it means to expand a determinant along a row or column

[cinder owl]

no, i think we will learn this osmetime this week

we definitely did not go over this in lecture

[odd pagoda]

well then ignore the hint for now

[topaz sinew]

@cinder owl Has your question been resolved?

[topaz sinew]

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[cinder owl]

.reopen

[topaz sinew]

✅

[cinder owl]

im sorta confused what it means to pick a non zero entry? sorry that might be really dumb

[odd pagoda]

do you know the permutation formula for the determinant?

[cinder owl]

I do not

I did computer P - lambda I tho

[odd pagoda]

well then also ignore that part of the hint

[topaz sinew]

@cinder owl Has your question been resolved?

[earnest timber]

@cinder owl this is exactly what you did in the last question when you solved for A - lambda I and then tried to find the eigenvalues from there, except that they just want you to find the expression

[earnest timber]

so the thing is that when you calculate for determinants, the calculation for normal matrices can be usually long and tedious, if you just take the baseline row or col and you see that there is a non-zero entry you can calculate the determinant easily

it's hard to put it in words

i think this is what they mean

[topaz sinew]

@cinder owl Has your question been resolved?

[cinder owl]

thank you this was helpful

.close

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[upper umbra]

could someone explain to me why this assumption was made?

[sweet shard]

what assumption

[upper umbra]

like is the even odd parts making up a function true?

is it always 50-50?

[sweet shard]

yes

[upper umbra]

ok

[sweet shard]

who is saying 50-50 ?

[upper umbra]

cus of the 1/2?

im trying to undersatnd why the above is formatted in the way that the fe and the fo are the halves of the sums of the even-odd concepts

[strange whale]

the only f_e and f_o that work are those yes

but they proved nothing

they just stated the result

[upper umbra]

by why is it the result?

is there a formal proof for it?

[strange whale]

there's a standard way of going about these kind of existence/uniqueness proofs

[upper umbra]

like why add the f(x) on top of the even-odd parts

[strange whale]

we'll arrive to it

[upper umbra]

where even is f(x) = f(-x)

got that

odd if f(x) = -f(-x)

ok got that as well

but why add?

[sweet shard]

because they're adding

if you want other operations, try proving they exist

[upper umbra]

i see...

[strange whale]

which is,

1. suppose such a decomposition f = f_e + f_o exists
2. try to restrict what f_e and f_o can be (in the best case there's only one possible choice), with the help of the decomposition
3. check that what you found in 2. actually satisfies the hypotheses of the decomp

[upper umbra]

"if such and such is so and such and such is so, then what"

ok i got it

[strange whale]

yeah so let's suppose we have f(x) = f_e(x) + f_o(x) with f_e even, and f_o odd

what does f(-x) look like ?

[upper umbra]

just plug in the -x to the even and odd functions and check their results isnt it?

[strange whale]

sure

write it out

@upper umbra you here still ?

[upper umbra]

oh yeah

i think i got it

given f(x) = fe(x) + fo(x)

recall for even -> f(x) = f(-x)

for odd -> f(x) = -f(-x)

checking what f(-x) looks like:
f(-x) = fe(-x) + fo(-x)

= fe(x) - fo(x)

[strange whale]

yeah

[upper umbra]

from solution add the f(x) + f(-x) = 2fe(x)

and im assuming from solution again that itd be f(x) - f(-x) = 2fo(x)

just divide by 2 and isolate for fe(x) and fo(x)

oh what

[strange whale]

gg

[upper umbra]

HAHAHA

[strange whale]

you can check that (f(x) + f(-x))/2 is even

and the other one is indeed odd

[upper umbra]

ye i think i got it

thats perfect

btw could you confirm my strategies as to how i could solve these next problems?

[strange whale]

yea

[upper umbra]

for this one can i seperate the unit steps

for each bound?

[strange whale]

it's the idea yes

[upper umbra]

got it

for this one

its a t shift?

or just dirac

[strange whale]

well they say it's a dirac below

the delta

[upper umbra]

ok il review both then

for this one

(last one)

how can i find the connection between the t domain and tau domain

i have an idea il have to consult the convolution theorem

with the t - tau

but for it to be an exponential i can only guess what other strategies to consult

i think i got it

thanks again 😄

.close

[topaz sinew]

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[hardy steppe]

im gonna admit

i dont understand how to solve b

i have the derivative

which is y = 4.9 * -sin( (pi/6 ) * t ) * (pi/6)

[neon iron]

you just need to find points where the derivative crosses x line

aka equal to 0

[hardy steppe]

okay, but. why does it say maximu,

and min

?

it seems like contradictive

its asking for when x is 0

but then at the same time its saying when y has a max

[hardy steppe]

i guess my real question is

should i do this algebraically or just plug it into my calculator

because i know that based on whats inside of -sin its going to be only multiples of 6

so like

0, 0 6, 0 12 ,0 18, 0 24, 0

[neon iron]

it says find both i presume

[hardy steppe]

im confused...

[neon iron]

its not asking when x is 0

its asking when y is 0

[hardy steppe]

okay right

[neon iron]

when y is 0 it crosses x

[hardy steppe]

ah so its asking me to take derived equation set to 0 and solve then using that find all the points where its = 0

[neon iron]

yeah

[hardy steppe]

divide out the multiplication factors and your left with the -sin(f(x)) = 0

so i just use that to plug values of x in until i find all the points where y = 0

between 0 < T < 24

.close

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[wicked tundra]

hello again, what formula or equation should I use in this question? i already know the box will fit in the elliptical hole, i just don't know how to prove it. (part 2)

[wicked tundra]

.close

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[brisk geyser]

can someone explain to me how these are the eigenvectors

[strange whale]

you just have to choose a non-zero vector which satisfies the equations

any one of them will do

@brisk geyser

[brisk geyser]

so like

1 = 1

1 = 1

[strange whale]

like if you choose (2, 2) on the left side it doesn't matter

[brisk geyser]

therefore its 1

and then

2 = 1/2y

therefore y is 4

im confused

so how does the second eigenvector get a 1 and a 2

[strange whale]

no you reversed x and y here

1 = 1/2 * 2

that's true

[brisk geyser]

kk

is the eigenvector like

(x,y) at its most factored form or something

[topaz sinew]

@brisk geyser Has your question been resolved?

[strange whale]

ppl usually take vectors with a 1, or only with integers in it yes

but that's just a common choice

there's no single eigenvector no matter what

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[lethal otter]

hey does anyone know why you can find the value of csc(cos^-1(-1/7) without a calculator, but not the value of cos^-1(-1/7)? i thought you could find both by making triangles and doing pythagorean theorem so im a little confused

[sweet shard]

yea you can. did you draw triangles?

oh you're asking why?

uhh, short answer is it's not a special triangle

[lethal otter]

wait wdym theyre both like asking cos^-1(-1/7) so wouldnt it be the same?

[sweet shard]

no

[topaz sinew]

@lethal otter Has your question been resolved?

[lethal otter]

oh ok thanks

[topaz sinew]

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[lethal otter]

.close

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[heavy jewel]

could someone explain how to use chebyshev inequality here

[topaz sinew]

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[severe lichen]

how do i find the zeros of x^6 -2x^5 -13x^4 +70x^3 -160x^2 +184x - 80

[severe lichen]

they gave me -5, 1, 2

[unique crypt]

use long division or synthetic division to remove those terms from the polynomial

then use rational root theorem on the cubic you get from that

[severe lichen]

is my divisor -80

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@severe lichen Has your question been resolved?

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[neon iron]

was shown this statement and I do not understand what it means

[neon iron]

O.A is the oblique asymptote

for reference, the oblique asymptote is y=3x+3

and the function is $\dfrac{3x^3-6x^2+x-2}{x^2-3x+4}$

[thorny flame]

RecRio

[bitter hemlock]

I guess it's saying, as x -> inf, the value of f(x) is below the OA, and as x -> -inf, the value of f(x) is above the OA
but it's just stating that, not really asking you to solve anything so I am not sure if this is supposed to be a problem or what, maybe you are supposed to prove those statements?

[neon iron]

its a part of the solution for this graph

I just wanted to know if this is backed by any proof or anything

or if it's just an inference made based on the graph, and not what the graph is dependant on

<@&286206848099549185>

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[neon iron]

Not sure how to do either

[neon iron]

<@&286206848099549185>

[topaz sinew]

@neon iron Has your question been resolved?

[upper crest]

For a geometric sequence, the formula is a*r^n

so a+ar+ar^2=14/3, and a*ar*ar^2=-8

[neon iron]

and then a isolate a from each

i get that

but then what

[upper crest]

lemme see, one minute

for the multiplicative formula, you get this:

and assuming a and r are integers, you know that the only possible values are -1, 1, -2, and 2

[neon iron]

the answer is

6 or

2/3

[upper crest]

you cant group the a's in the multiplying one like that, you can only cancel out one of them

wait i misread my bad

wait no i didnt, yeah you cant do that

[neon iron]

wdym

why not

[upper crest]

if you have a*ar, you cant do a(1*r)

[neon iron]

i did

a x ar x ar^2

and changed to

a(1 x r x r^2)

whats wrong with that

[upper crest]

that's only allowed when you're adding

[neon iron]

what

bro

im factoring out a

[upper crest]

a x ar x ar^2 is equal to a x a x a x r x r^2

= a^3 x r^3

[neon iron]

okay so

can u write out

the steps

im not following

and send a pic pls

[upper crest]

yeah one sec

[neon iron]

ty

the answer has to be

6 or 2/3

[topaz sinew]

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[neon iron]

.reopen

[topaz sinew]

✅

[neon iron]

so?

@upper crest

[upper crest]

yeah, sorry

[neon iron]

do u not know how to do it or

lol

[upper crest]

What do you normally do for these?

[neon iron]

idk

first time

[upper crest]

oh

[neon iron]

getting that question

[upper crest]

You could try what you did but factor out the a on the multiply one right

but im not sure how I'd do this yet

What math are you in currently?

[neon iron]

pre cal 12

[upper crest]

oh ok

11 im stuck on for the time being

sorry

[neon iron]

its ok

thank for tryin

[coral tiger]

@neon iron
Consider the terms to be a/r, a, ar

Then a³=-8

(according to the q)

[upper crest]

ohhh thats clever

[neon iron]

huh

i dont understand

[coral tiger]

One sec I'll show you my working

Oh NVM it says first 3 terms

[upper crest]

aww right, bummer

I can get a^3r^3=-8, which is ar=-2

and then you plug everything into the addition one, getting a-2+r^2=14/3

and substitute r for -2/a

[coral tiger]

Oh yeah and then use the next equation

[neon iron]

guys im kinda slow so

im gonna need to see it written

for me to understand

[coral tiger]

@neon iron you get this one right?

[neon iron]

no

[coral tiger]

What are your first 3 terms of the G.P?

[neon iron]

idk

thats what im trying to find

[coral tiger]

It's a, ar, ar²

[neon iron]

yeah

i get that

[upper crest]

this make sense?

[neon iron]

ok but

then whats a

whats r

it can be those 4 possibilities u mentioned earlier but

the answer is

6 or 2/3

[upper crest]

then you can use a+ar+ar^2, and plug -2 in for ar

[neon iron]

show me

pls

[upper crest]

ok one moment

Do you see what I used the ar=-2 for?

[neon iron]

💀

there's no way thats how its done

how do u even get 6 from that

[upper crest]

i mean multiply both sides by a

and then its a quadratic eq

[neon iron]

show

💀

lmfaoo

[upper crest]

ok lmao

[neon iron]

whered u get this from

isnt it

a+a(-2/a)+a(-2/a)^2

[upper crest]

yeees, and that simplifies to what i put

[neon iron]

bro whattttttt

my brain bro

sorry

wait

[upper crest]

lol u good

[neon iron]

oh ok

yeah

got it

whats the rest

@upper crest

[upper crest]

6 is suprisongly the right answer

is this a math competition problem or smth? its like really hard for a first geometric series problem lmao

[neon iron]

nah its not

its a warm-up

actually

😭

[upper crest]

i would actually hate my teacher LOL

[neon iron]

no way that shows up on a test

[upper crest]

no, theres no way

[neon iron]

why doesnt 2/3 work again

[upper crest]

it does actually, its just not a whole number

so its kinda gross to use

[neon iron]

oh ok

well

one more question

thanks btw

[upper crest]

yeah of course

[neon iron]

converting 9 to base 3

and then doing 3^2x-2

doiesnt work

or maybe it does

but

not that i can see

[upper crest]

hmm give me a minute

since the next term in the series is r * (the previous term), try that

3^x * r = 3^(3-2x)

youll have to use log base 3

at some point

[neon iron]

oh ok

thanks

[upper crest]

no prob!

use .close if ur done, also

[noble laurel]

@neon iron

[neon iron]

.close

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[half rampart]

hi

[half rampart]

this coz idk whats happening

[drifting swift]

do you know how to graph straight lines from their equations?

[upbeat crypt]

if u want to find where they cut then u should do:

[half rampart]

idk how to subsitute thats the problem

@upbeat crypt @drifting swift

no one ever taught me

[upbeat crypt]

then how you are solving this kind of question?

kinda weird lol

[half rampart]

it was like a class task

[upbeat crypt]

idk its not easy thing to explain to you from zero now

[half rampart]

we didnt have a teacher and i tried searching it up and stuff but they all have an = ___ and that equation doesnt

[upbeat crypt]

just do this:
2x + 2 = 4 -3x

idk if u understand

find x and whoo

[half rampart]

alrr lets try

when i did everything and put it in the equation i got 7

thats not an answer tho

[upbeat crypt]

7?

u got x = 7?

u sure ?

5x = 2? no? ..

so x = 7 ??try again

[half rampart]

no i got 2.5 for that

[upbeat crypt]

u sure 2.5?

try again

[half rampart]

5x=2 so we wld have to 5/2

do we not?

[upbeat crypt]

try again?

[half rampart]

what do i do then

[upbeat crypt]

idk dont guess tho

think

lmao

we will get 2 /5

not 5/2

[half rampart]

oh mb wrong way

that wld be 0.4

[upbeat crypt]

ye

its the same

2/5

2/5 = 4/10 = 0,4

[half rampart]

so x = 0.4

[upbeat crypt]

yh

anything else?

[half rampart]

hence it wld be C

[upbeat crypt]

ye?

[half rampart]

Alr thanks

appreciate it

[upbeat crypt]

try to do this :
y = 2x + 2
then replace x = 0.4

and find y

u will get y = 2.8

[half rampart]

so it wldnt be C off the fat

BAT*

[upbeat crypt]

so the answer is B

not C

[half rampart]

oh

[upbeat crypt]

yes sir? or no

[half rampart]

ill have to see

we dont find out if its right or not yet

but i do have another question

lmk if u can solve this or not

[upbeat crypt]

u should solve it

try

[half rampart]

i got 0.3

[upbeat crypt]

hmm

its like :
(13.32 / 4.44) * 1.15

ofc 0.3 is wrong

it must be bigger then 1.15

[half rampart]

3.45?

thats what i got now

[upbeat crypt]

ye

[topaz sinew]

@half rampart Has your question been resolved?

[half rampart]

is that the answer?

[upbeat crypt]

calc :
(13.32 / 4.44) * 1.15

and u will find the answer

[half rampart]

the answer is 3.45

[upbeat crypt]

okay well done

[half rampart]

thank you!!!

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[pseudo hamlet]

I've got an exercise where I'm supposed to use both L'Hopital and then the Taylor Series to decide the limit of the function in the image. I've solved the limit with L'Hopital and got the correct answer, but I'm stuck on trying to solve it with the Taylor Series. Any time I try to solve it by re-writing the function with the taylor series, and then solving it, I get 0 which isn't correct. Any help would be greatly appreciated.

[mellow cypress]

you don't need to use l'hopital

e^x expressed as its power series is the definition itself

[pseudo hamlet]

Well the exercise said it, so I just followed but that was easy.

[mellow cypress]

anyways, e^x dominates x^2, so it diverges to infty, so its not 0

[pseudo hamlet]

but evaluating the limit with the taylor series is what I'm having difficulties grasping

yeah that's what I got using L'hopital

[mellow cypress]

there must be some mistake

[pseudo hamlet]

what do you mean?

[mellow cypress]

lim e^x/x^2= lim e^x/2x= lim e^x/2=+infty

going via lopital

[pseudo hamlet]

correct

that's what I did to evaluate the limit with l'hopital

[mellow cypress]

oh nvm, I didn't read what you wanted

[pseudo hamlet]

and the next part of the exercise is to evaluate the limit again, but now with the taylor series

which is what I'm having difficulties with

probably poorly worded by me lol

[mellow cypress]

[\frac{e^x}{x^2}=\frac{1+x+\frac{x^2}{2}+..\frac{x^n}{n!}...}{x^2}=\frac 1{x^2}+\frac{1}{x}+\frac{1}{2}+\frac{x}{3!}....]

[pseudo hamlet]

that's what I did yeah

[mellow cypress]

taking lim x->infty the first 2 terms vanishes, and the rest diverges to +infty

what did you not get?

[pseudo hamlet]

I read that only functions that are derivable close to x0 can be evaluated with a taylor series, so I initially derived the function, plugged 0, and got 0/0. From there I drew the conclusion that I couldn't use the taylor series to evaluate the limit.

[mellow cypress]

the taylor series expansion of e^x is the definition of e^x

ik it sounds weird

also that power series is analytic on R, so it converges everywhere

you can try finding the radius of convergence

[pseudo hamlet]

So is what I read only applicable if the limit of the function approaches 0, or am I missing something?

[mellow cypress]

no what

[pseudo hamlet]

I'm refering to my previous message

[mellow cypress]

x-> +infinity

[pseudo hamlet]

English isn't my native language, so I'll try my best go explain what I mean

[mellow cypress]

yea but we care about taylor series of e^x, not e^x/x^2

the function is not defined at x=0

[pseudo hamlet]

I read that when you have a limit that you want to evaluate, which in my case is the limit I posted originally, you need to check if the function is derivable close to x0 before you can evaluate the limit with the taylor series.

But that isn't necessary in this case?

[mellow cypress]

whats x0 here?

[pseudo hamlet]

Yeah so that's what I mean with the previous question. Is that only applicable when you have a limit where x approaches 0 (Which isn't the case here)?

[topaz sinew]

@pseudo hamlet Has your question been resolved?

[pseudo hamlet]

So basically by evaluating 1/x^2 + 1/x + 1/2 + x/3! + ... where the limit goes to infinity you get +infinity, which is the same result as with L'hopital?

[mellow cypress]

yea

[mellow cypress]

lim x->0 e^x/x^2 does not converge, so yea the function isn't defined at that point so naturally it means nothing to find the taylor series around that point

[pseudo hamlet]

Gotchu. So with this this limit you'd ultimately get 1/x^2 - 1/2x^3 + 1/3x^4 - 1/4x^4, and then when you evaluate it where the limit goes to infinity you get 0 right?

[mellow cypress]

no

ln(x) taylor series at any point(say x=1) is not analytic everywhere

but since its the inverse of the exponential function,
we see that ln(x)<=x for all x

ln(x)=3ln(x^{1/3})<=3x^{1/3}

now try squeeze theorem

[pseudo hamlet]

The exercise just says evaluate the expression with (1) L'hoptial and (2) Taylor Series

L'hopital was no issue, but since ln(x) isn't analytic anywhere I guess it's not solvable with the taylor series?

[mellow cypress]

if you're allowed to use l'hopital sure, go ahead

I did it via elementary methods

[pseudo hamlet]

Yeah, the exercise just states to evaluate with both methods

so then you can't evaluate it with taylor series if I get what you mean?

[mellow cypress]

yes

[topaz sinew]

@pseudo hamlet Has your question been resolved?

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[lucid junco]

Why wouldn’t there be another solution at pi - 0?

[opal vault]

cos(5pi/4) = -1/sqrt(2)

[lucid junco]

where are those numbers comin from

[opal vault]

you're asking whether x = pi is a solution right?

cos(pi + pi/4) = -cos(pi/4) = -1/sqrt(2)

[lucid junco]

so whats wrong with this?

[opal vault]

the minus

cos(pi + pi/4) is not 1/sqrt(2)

[lucid junco]

oh

well tbh im not too sure if my thinking process was correct for -pi/2 solution

basically i got the 7pi/4 after setting that equal to the phase shift as normal

and i just subtracted a period from it

is that the right way?

but how do we know that gets us 1/sqrt(2)

[opal vault]

cos(2pi - x) = cos(-x) = cos(x)

so x = pi/4 makes this true as well

cos(7pi/4) = cos(pi/4) = 1/sqrt(2)

[lucid junco]

7pi/4 is the same thing as pi/4?

im not understanding

[opal vault]

I wrote this down

Use this with x = pi/4

7pi/4 = 2pi - pi/4

[lucid junco]

im just confused

cause

normally

when i solve trig equations if i wanna find more solutions in something like

0,2pi

i find the base angles and then set equal to phase shift

then add a period to it to find more

but that doesnt work here?

[opal vault]

if you want to solve $\cos(x) = \cos(a)$, then $x\equiv a$ mod $2\pi$ OR $x\equiv -a$ mod $2\pi$

[thorny flame]

rafilou2003

[opal vault]

because cos(-x) = cos(x)

[lucid junco]

o im unfamiliar with that notation

[opal vault]

$a \equiv b$ mod $2\pi \Longleftrightarrow \exists k\in \bZ, a = b+k\cdot 2\pi$

[thorny flame]

rafilou2003

[lucid junco]

i meant with the mod stuff

[opal vault]

that's what I wrote as a definition didn't I?

a is congruent to b mod 2pi if they're equal, up to a multiple of 2pi

[lucid junco]

this is even more notation im unfamiliar with haha

[opal vault]

$a \equiv b$ mod $2\pi \Longleftrightarrow a = b$ UP to a multiple of $2\pi$

[thorny flame]

rafilou2003

[opal vault]

better?

[lucid junco]

slightly i guess

i just dont know what mod is

like what does it mean

[opal vault]

i just explained what it means

a = b "MOD x" means that a = b UP TO A MULTIPLE OF x

[lucid junco]

oh

[topaz sinew]

@lucid junco Has your question been resolved?

[topaz sinew]

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[terse trellis]

For some reason i am missing the equation y=0, but i cant find my mistake

[terse trellis]

I should have this

[hollow hedge]

not only are you missing y=0, you are also missing x=0. The reason being you divided by xy in the first step.

For this reason, you assumed that x and y could not be 0, but it very well could be.

[terse trellis]

Hmm

Okay

So i shouldnt /xy

What -should- i do then

[ocean terrace]

you can, just test for x,y=0 afterwards

[hollow hedge]

you have a product of multiple terms being 0, what possible values can these terms take on for the whole product to be 0?

[terse trellis]

xy = 0

or

x^4-y^4

=0

[hollow hedge]

yes

[terse trellis]

Okay yes

[terse trellis]

And what about the method this person said

[hollow hedge]

sure, but I would actually prefer you do that before. Basically, dividing by xy here is fine, in the aspect that you know what happens if x or y is 0.

Whenever I try to divide by some terms that can be 0. I always try to check what would actually happen if they were 0 first to know if I'm missing some details.

Once again though, I highly reccommend avoiding doing so if possible. Just for the reason of convenience

[terse trellis]

Okay will do

Thanks!

[topaz sinew]

@terse trellis Has your question been resolved?

[topaz sinew]

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[cinder oxide]

k = -4, infinite solutions
k = 0, infinite solutions

or is it k = 0, unique solution?

this operation is allowed?

[ruby tree]

You're still on this

[ruby tree]

Yes

[ruby tree]

Yes if you set k =/= -4

[cinder oxide]

k represents "unknown" amount of x, right?

whenever you see a matrix with a variable, it actually represents infinitely many matrices, right?

that's why most matrices just include constants, as it's only one matrix (system of equations)

these variable matrixes really trip me up

[ruby tree]

Yes unless the variable cancels out or something

[cinder oxide]

how so?

like having two variables on the matrix?

k and -k underneath of it?

wait even if it was k and 1 underneath of it

[ruby tree]

Just with a term "k-k+2" for example

[cinder oxide]

i could use row operation -k for the row with 1 to make it -k

[cinder oxide]

hmmm

OK

[cinder oxide]

so this would not apply, for cancelling

you still end up with a k

no matter what

because row operation says so

but what if you multiply by 1/k

that would cancel k

[ruby tree]

Then you need to set k =/= 0

[cinder oxide]

so it wouldn't actually cancel k, is what you are saying?

completely

k will always be a part of the system of equations, in this domain restriction

[ruby tree]

If you have a row like [k k k]

It means x+y=1

It doesn't mean 0+0=0

[cinder oxide]

😮 it does?

i though it meant xk + yk = k

[ruby tree]

Well yeah, and you divide by k, assuming k =/= 0

[cinder oxide]

hmmm...

but that still gives a domain restriction, can't ignore that