help-19

[cold swift]

yep exactly

so now we can get back to the limit comparison

[echo mango]

so since that series diverges then the other one does as well?

[cold swift]

yep since the limit approaches 1

[echo mango]

wait how do we know it approaches 1😭

[clever fjord]

y0shi

[cold swift]

i just simplified it a bit

to that

[echo mango]

oh we multiply it because we origionally divided by the "origional" equation?

[cold swift]

yeah essentially keep change flip

[echo mango]

okayyy

[cold swift]

so now we just need to evaluate this limit

which would be just 1

[echo mango]

so

when we cancel all the terms out like 3n^2 and 2n^3 we stil have 2 left over

where would that go

[clever fjord]

y0shi

[cold swift]

do you remember how to take the limit to infinity of a rational function?

you can totally lhopital this 3 times

but it is not necessary and requires too much time

[echo mango]

not rlly

[cold swift]

so if the degree of the numerator is greater than the degree of the denominator, the limit approaches infinity

if the degree of the denominator is greater than the degree of the numerator, it approaches 0

and if the degree of the numerator and the denominator are the same, the limit approaches the ratio of the coefficients of the highest degree term

[echo mango]

OHHHHHH thatttt

[cold swift]

this definitely something that you should try to remember since it helps a lot when computing these limits

[echo mango]

okay i get that

[echo mango]

so how did you get to that? by canceling out terms?

[cold swift]

yeah 3n^2 cancels out

[echo mango]

ohhh ur just writing that

ok nvm nvm

i was just multiplying it out first

so i had 6n^5/6n^5 +2

[cold swift]

ah yeah either is fine

that works too

but be careful

the denominator should be 6n^5+6n^2

[echo mango]

oh ya sorry

[cold swift]

not that it changes the value of the limit, but if your teacher is picky it could cost you a point

[cold swift]

all good

[echo mango]

so now we know that it

converges ?

since it is 1

[cold swift]

well the series we compared it to diverges right

that means the original series must diverge as well

[echo mango]

oh ya the harmonic

[cold swift]

as long as the limit evaluates to a finite positive number, both series will either diverge together or converge together

[cold swift]

yep

[echo mango]

ohh we determined if the origional is divergent first

then we used the limit comparison to let the series be divergent and compared to the origional?

and we did a_n /b_n

which the a_n was the one we were given and the b_n was the origional that we found?

[cold swift]

b_n is the series we compared a_n to

and we found that b_n is divergent

and since the limit evaluates to a finite positive number, the original must also be divergent

[echo mango]

okayyy i kinda get it now

and for the direct comparison test its kinda similar to that right

except that it has to be 0<=a_n<=b_n

[cold swift]

yep

also the limit comparison as a few extra conditions that basically resemble the direct comparison test

but you don’t need to know them since you can just use the direct comparison when you have those conditions

[echo mango]

okayyyy

and we do the same thing as the limit comparison and compare it to another series

and that series has to be greater than or equal to the one we have?

[cold swift]

the limit comparison test doesn’t really care whether one series is bigger than the other

which is why it helps when the direct comparison test conditions fails

[echo mango]

okayy i see

so do we usually use direct comparison

so the b_n in the direct comparison is what we are comparing to?

[cold swift]

we use direct comparison when the series we are comparing to our original to converges and is bigger than the original, or when the series we are comparing to diverges and it is smaller than the original

if the series we compared our original to converges let’s say it is smaller than the original

then the direct comparison test is inconclusive

and that’s when we turn to the limit comparison

basically we use limit comparison when direct comparison fails

[echo mango]

okay i see

could we try an example of that

[cold swift]

like when direct comparison fails?

[echo mango]

when it doesnt fail

[cold swift]

oh alright

[echo mango]

like idk which one of these works with direct comaprison

i wanna pick one to try the direct comparison test with

[cold swift]

number 3 works

so what would we compare that series to

[echo mango]

1/n+1?

or 1/n

1/n(n+1)

maybe 1/n is easier?

[cold swift]

well we usually just compare it to the highest degree term there

so $\sum \frac{1}{n^2}$

[clever fjord]

y0shi

[echo mango]

ohhhh

[cold swift]

so does this series converge or diverge

[echo mango]

so we compare it to the highest degree term in the denominator?

well that would be lim n->infinity 1/n^2

which is 1/infinity

so

[cold swift]

yeah and if there was something in the numerator

we would take the highest degree term as well

[echo mango]

oh so how would that go

[cold swift]

which means that the divergence test fails

[echo mango]

like if we had n^3/ n^2+1

[clever fjord]

y0shi

[echo mango]

but doesnt 1/infinity approach 0?

[cold swift]

yep

and if the limit equals 0

the divergence test is inconclusive

only when the limit doesn’t equal 0 does the divergence test say that the series is divergent

[echo mango]

oh right right

just the basic divergence test

[cold swift]

yep

[clever fjord]

y0shi

[cold swift]

doesnt that look like a p series?

[echo mango]

YEP

1/n^p

and its 2>1

[cold swift]

yep

so it converges

[echo mango]

yep

[cold swift]

and now for direct comparison test

we have to check whether or not this series is bigger or smaller than our original

and since this series we compared it with is convergent, for the direct comparison test to work, we need the original to be smaller

[echo mango]

okayyy

so we would do

1/n^2+n times n^2/1

which would be 1 bc of the limits being the same right

[cold swift]

well thats limit comparison test

and yes it would work

but direct comparison also works

[echo mango]

how would i do that

[clever fjord]

y0shi

[echo mango]

the series converges?

[cold swift]

the original converges yes

because it is less than 1/n^2

which converges

so by the direct comparison test, the original converges

[echo mango]

wait how do we know its less?

like if i plug in 1 it would be less

[cold swift]

so if we look at the denominators

the one on the left has a bigger denominator

if it has a bigger denominator, the overall value of the fraction would be smaller

since 1/3<1/2

[echo mango]

oh okiii

like 1^1 +1 =1/2 is smaller than the 1/1

[cold swift]

yep

[echo mango]

okayyyy

thank you so muchhhh

that helped a lot

[cold swift]

yw!

[echo mango]

im gonna take a break now

THANK YOU A LOT

[cold swift]

alright

[echo mango]

I learned so much

[cold swift]

ywyw!

just do more practice and youll get used to it

[echo mango]

🙏 better than the teacher

[cold swift]

haha

[odd edge]

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[odd edge]

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[wintry viper]

can someone help me with B please

[wintry viper]

i have no idea what its asking me

[open crown]

Draw it out

[wintry viper]

oh yea i think i get it now i just forgot about pe thats why

.close

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[magic fox]

y+1=-4/3 (x-1) can someone explain how to do this

[foggy ibex]

what do we need to do here?

[tacit mica]

what do u have to find in this?

or what do u have to do in this

[magic fox]

The orginal problem is Find the equation of the line that passes through (-2,3)(1,-1)

[foggy ibex]

then use the standard formula to find slope

[magic fox]

i found the slop

[tacit mica]

then what

[magic fox]

it is -4/3

i dont know how to put it in the equation form

[foggy ibex]

the standard eqn is y-y1=slope(x-x1)

and slope =y2-y1/x2-x1

[magic fox]

o

k

thank you

[foggy ibex]

you might have confused between y1 and y2

[magic fox]

ye

[foggy ibex]

ok have a nice day!

[odd edge]

@magic fox Has your question been resolved?

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[wispy birch]

Guys I know this is a chemistry question but I couldn't find any similar server like this. Please be kind enough to answer this T_T

[wispy birch]

Chapter: solid states

[frigid canopy]

https://discord.gg/RAAef3ye

[wispy birch]

I did no ones there it's dead

[fallow tapir]

What do you not understand about the question?

[wispy birch]

I got something but that's not the ans

[fallow tapir]

show your working

[wispy birch]

Wait

[wispy birch]

Ok this one thing I don't get... How do we calculate for Be??

I know how to for oxide ion and Al here but there's no info for be?

[fallow tapir]

i asked for your working 💀

[wispy birch]

Al is 1/4th of TV so--> 1/4(1) =1/4
O is in ccp ---> 1
Be---? If 1
Then be4AlO4

But ans is BeAl2O4

[fallow tapir]

why did you do 1/4(1)?

how many atoms are there in CCP?

[wispy birch]

Yeah mb 1/4 (2)

Mbbb

[fallow tapir]

again why

[wispy birch]

2 atoms ryt?

[fallow tapir]

no

[wispy birch]

I mean TVs are 2

Total atoms are 1?

[fallow tapir]

no

[wispy birch]

What

[fallow tapir]

lol

[wispy birch]

1/8*(8)

Is 1?

Is it not like this?lol

[fallow tapir]

there are some atoms on faces too

[wispy birch]

It's not FCC tho ryt?

[fallow tapir]

its CCP yeah

[wispy birch]

There are no atoms in face of CCP ryt?

[fallow tapir]

there are lol

[wispy birch]

This is called FCC ryt?

Wait are FCC and CCP same??

Oh yeah I think so

[fallow tapir]

both have the same structure yeah

[wispy birch]

Damn till now I was thinking CCP is same as SCC oh god

So let me try again

[wispy birch]

Ok one info was missing here Be is 1/4 of TV.

[fallow tapir]

i have some work, gtg

sorry

will brb

[wispy birch]

Alr thank you!

Alr got it finally thank you so much actually @fallow tapir

.close

[odd edge]

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[fallow tapir]

oh np

[odd edge]

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[chrome jungle]

helppp i need to find the first 7 terms

[tidal idol]

what's your thought?

[chrome jungle]

i have no idea how to start

[tidal idol]

try to write sqrt in terms of power indice

[chrome jungle]

so like,, to the power of 1/2?

[tidal idol]

yes

[chrome jungle]

okay buttt what then 😭 is there a formula or something

[tidal idol]

there is no formula, this is just multiplication with indices

[tidal idol]

show me your work and we will do this step by step

[chrome jungle]

idk what those are we haven't learned that yet

it says its recursive

[tidal idol]

ok that's ok then, we will just do it in terms of $3\sqrt{3}$

[clever fjord]

lgkoo

[tidal idol]

it has a recursive pattern yeah

so $a_1 = \sqrt{3}$

then $a_2 = \sqrt{3 \cdot \sqrt{3}}$.

do you understand this?

[clever fjord]

lgkoo

lgkoo

[chrome jungle]

yesss i think

where does the n-1 go

[tidal idol]

I substituted n = 2

they want the first 7 terms

so a_1, a_2, ..., a_7

[chrome jungle]

right

[tidal idol]

you're given a_1, and the recursive formula

where a_n is the new term, and a_n-1 is the last term

[tidal idol]

so the idea is to simply substitute the previous term into this

until you have a_7

[chrome jungle]

so a2 is 3sqrt3? and then we put that into the a3

and then keep going

[tidal idol]

a2 is not 3sqrt3, it is sqrt of 3sqrt3, so sqrt(3sqrt3)

[tidal idol]

but yh you get the idea

[chrome jungle]

ohh okay

andddd a3 would be that x3

and sqrt

[tidal idol]

yup

[chrome jungle]

okayy i need 2 more problems if you think you can help maybe 😭 sort of the same topic

just 28

[tidal idol]

right, so we have a sequence
$a_1, 10, a_3, a_4, \frac{1250}{27}, a_6, ..., a_n, ...$

[clever fjord]

lgkoo

[tidal idol]

do you know what is the property of a geometric seq?

[chrome jungle]

noooo

[tidal idol]

the property of a geometric sequence is that each consecutive terms have a common ratio r

for example

1, 2, 4, 8, 16, ...

you see how each term is two times the previous term?

That means this is a geometric seq with common ratio 2

[chrome jungle]

yesss right

[tidal idol]

so that mean, in general we have

$a_n = r \cdot a_{n-1}$

[clever fjord]

lgkoo

[chrome jungle]

yes

[tidal idol]

with this hint, can you work what the common ratio of your sequence?

[chrome jungle]

well how do i find the first term

[tidal idol]

you don't need the first term

you can start from the second term

it's a common ratio

so shouldn't matter where you start if it is geometric

i.e. start from a_2 = 10

[chrome jungle]

and we solve that to find r

[tidal idol]

yes, you have to come up with an equation in terms of r and solve for r

[chrome jungle]

so 10 = r • a_1 ??

[tidal idol]

yes, but we don't know about a_1,

think about this

a_2 = 10

a_5 = 1250/27

so how many r did we multiply 10 by to get to 1250/27?

[chrome jungle]

3?

[tidal idol]

yup

so what equation can we you deduce?

[chrome jungle]

i'm not sureeee 😭 is it 1250/27 = 10 • 3r ??

[tidal idol]

no

here's an example

b_1 = 3, b_4= 24

so to get from b_1 to b_4, we have to multiply 3 by r three times

that means

b_2 = 3*r

b_3 = 3 * r * r = 3r^2

b_3 = 3 * r * r * r = 3r^3

and b_4 = 3r^3 * 3 = 3r^4

which we know is 24

so 3r^4 = 24

and you can solve for r

does this make sense?

[chrome jungle]

yes yes

[tidal idol]

now apply the same logic to this

[chrome jungle]

so 10r³ ?

[tidal idol]

yup

now can you solve for r?

[chrome jungle]

is it 5/3?

[tidal idol]

yes, well done

[tidal idol]

now generally even though we have this recursive formula, we can write the nth term a_n in terms of r and the first term a_1

so we know a_2 = 10, and r = 5/3, can you find a_1?

[odd edge]

@chrome jungle Has your question been resolved?

[chrome jungle]

SORRY i had to do a test on this just now 😭 but was it 6 • 5/3^n-1 ?

[odd edge]

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[rapid kestrel]

For a first order DE with dy/dx = f(x,y), df(x,y)/dy must be continuous to gurantee unique solution. Why is that not present in this definition for uniqueness of higher order DEs?

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@rapid kestrel Has your question been resolved?

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[pallid rock]

can i get help how to solve this problem?

[tidal nest]

what are guy wires?

[pallid rock]

i think they are the two blue ones on the sides going diagonal

[tidal nest]

oh alr

is it including the blue line on the floor

or

[pallid rock]

i dont think so because it just says 2

[tidal nest]

ok i get it

very weird question ngl

[pallid rock]

i think the angle from the base on the left is 95 degrees and the opposite one on the right is 85 degrees because of the 5 degrees slant

ya its weird i have done similar ones before but never like this one i dont get it

[tidal nest]

and the 200ft is to the dotted line im guessin

[pallid rock]

that part im not sure because on the right it looks like its only going to the blue line (the base)

[tidal nest]

im pretty confused xd

[pallid rock]

thats the part thats weird to me too tho because if its what im thinking then you would just find the side using the law of cosines easy with a=200 and b=150 trying to find c2=a^2+b^2-2ab cos c

[tidal nest]

wait my brain doin smth

[pallid rock]

but like if the 200 goes past the base then idk

[tidal nest]

cosine rule could work

[pallid rock]

even if its a slant like that

i hate this question

can't even really tell whats what i know what u mean confusing its very badly drawn

[tidal nest]

yeah if it was like a better drawing it would make sense

look

i think i figured out the left wire

u can try to get the right one

[pallid rock]

did u get 260.25 for the left?

[tidal nest]

249.28 :/

[pallid rock]

hmm

[tidal nest]

ill send what i did

last step should be A not A^2

typo from me

based on what i did for the left wire, I literally cant do the same for the right one, so maybe i've over complicated it

[pallid rock]

can u explain to me i dont understand why its 200-150sin5

i didnt learn that in my class yet whatever that is

[tidal nest]

which part

the sin or.?

[pallid rock]

like how u got 150 and why its why is the 150sin5 being subtracted from 200

[tidal nest]

hm ok, so the total height is 200 feet

but its on a slope yeah?

[pallid rock]

ya

[tidal nest]

so a bit of it is underground

sin(x) = opposite side / hypotonus side

i def spelt that wrong but i hope u get it

[pallid rock]

thats how u found the measure of the little bit thats underground like can u explain how thats done

we never did that yet

[tidal nest]

maybe im wrong tho

but from the image im assuming i should do this

[pallid rock]

idk why i was given this problem tbh like we never learned how to do this yet only the basic law of cosines not with like a slant lmao

[tidal nest]

i mean if we assume the building is 0m wide, i could get the other wire but this seems like the question is just weird instead xd

i think i get what ur saying

only use the slope for the angle not the lengths of anything

[pallid rock]

its rigged not in my favor

[tidal nest]

that way u can get 260.25 for the length for the first wire xd

ill do it rq

[pallid rock]

ya cuz i never learned any other way to do it just if its regular

i might just put that and see if they accept it, cuz fuq

[tidal nest]

@pallid rock

ig for the other wire just use 85 degrees lol

[odd edge]

@pallid rock Has your question been resolved?

[pallid rock]

thanks

[odd edge]

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[outer wadi]

Can someone teach me how to do this or send me something explaining, my teacher didn’t go over this and i can’t find anything on youtube about it

[outer wadi]

I know how to find the general solution

i’m not sure what to do with the limit or what y(0) > 0 means

[mystic forge]

what is ur general solution

[mystic forge]

um

[outer wadi]

is that wrong 😭

[mystic forge]

u get this by separation

[brittle beacon]

1/(a + b) is not 1/a + 1/b

[mystic forge]

then simple integration leads to the solution

which u did

but

yeah

[brittle beacon]

(also note that they talk about y(0), and that dividing by zero, as you would in 2/t^2, is not good for your health...)

[outer wadi]

I’m kinda confused on where I went wrong, was it my second step? In my first step I took the integral of 1/y^2 and got -1/y = integral of tdt which is t^2/2 + C right

[mystic forge]

when u took the reciprocal

of -1/y

[outer wadi]

It’s possible to take the reciprocal of both sides right? I just did it wrong?

[mystic forge]

yeah

[outer wadi]

oh should I combine them and then take the reciprocal

[mystic forge]

yuh

[outer wadi]

also when solving for C does it matter that I keep track of wether c is 2c or c^2 because I watched a video that said we can just make 2c = c since it's still a constant but wouldn't that affect the value of C when we solve for it since we won't be diving by two?

[mystic forge]

it wont change the final expression

just simplify till the end

and whatever constant u have

make it A or something

[brittle beacon]

[though in this case, you want to have the constant in terms of y(0) if possible!]

[outer wadi]

hmm back to that, what is y(0) > 0 saying again? I dont't understand what that information is for, are they saying that y(0) can be negative or positive and we want the limit when it's positive?

[brittle beacon]

That the initial value y(0), you're assuming positive (which is just as well, seeing that it's a denominator and in a square root in that sqrt{2/y(0)} part)

[outer wadi]

oh ok

I was able to get the right answer, thank you both for the help!

.close

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[rugged lantern]

Can someone just read over this and tell me if I’m wrong

[rugged lantern]

Trying to just double check the original statement is

“Disprove the statement: there are two integers m>=2 and n>=1 such that m^2+1 = 2^n”

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@rugged lantern Has your question been resolved?

[rugged lantern]

.close

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[wanton prawn]

To keep it brief; I need to do a couple of interviews with people who have work (or have worked) as software developers or similar (Researchers (B.S-Ph.D), Web Dev, AI Devs, Data Scientists & Admins, Computer Vision Engineer, Cloud Engineers, etc....) for a sociology class I had to take to graduate. I have to ask ~20 brief questions like these:

18. What kinds of things have encouraged you to stay with the ORGANIZATION NAME?
19. Have you received any on-the-job training?
20. How would you counsel someone on preparing for your current position at ORGANIZATION NAME?

4. Are you aware of any pending laws related to labor/employment that might change your job? That is, change the relationship between you and the firm, for good or bad. If so, can you describe what those changes might be? 
5) Are there any rules or policies that are specific to (Organization Name) that makes this a better (or worse) place to work than other firms who hire people to do what you do here? 

   Etc...

I do have to record the audio for my professor (for legitimacy purposes), so the interviews will be as brief as possible because I can listen back to the audio instead of taking notes during.

If anyone is willing to help me out, I can literally do the interview with you any day, at any time. Between April 10th and April 20th.
It would help me out a ton!
Thanks in advance.

[odd edge]

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[shrewd sky]

is it 2n/n!

[cold sage]

no

[shrewd sky]

there a geometric series in a series

[cold sage]

where is the geometric part

[shrewd sky]

numerator

[cold sage]

thats not geometric 2 4 6 8 ...

factorise it

(what if you pulled out 2 from every component of that product?)

[shrewd sky]

you mean divide by 2

[cold sage]

factorise* but no, its not just a single 2

[shrewd sky]

i dont understand

first of all its 2 * 4 * 6 * 8 * 10 * 234234902394

and the denominator is a factorial

so the 2n isnt there

[cold sage]

?

[shrewd sky]

idk what the question is

the series

[cold sage]

i dont know what youre talking about/what you mean

[shrewd sky]

how would i factorise a series

[brittle beacon]

You're happy that for each of those, they're multiples of 2 you're dealing with? That 2 is a multiple of 2, 4 is a multiple of 2, 6 is, same up to 2n?

[shrewd sky]

yes

[cold sage]

so you can factor out a 2 from each number
(2*1)*(2*2)*(2*3)*...*(2*(n-1))*(2*n)

[shrewd sky]

so its 1 * 2 * 3 * ... n

thats what i meant when i said divide by 2

[cold sage]

but its not just a single 2 was my point

how many are there

[shrewd sky]

infinite

[cold sage]

how so?

i think youre slightly confusing the sum itself with just looking at the expression being summed

[shrewd sky]

i am confused on everything

i mean the numerator

its 2 being multiplied by 2 + 2 and then multiplied by 2 + 2 + 2

[cold sage]

if i start pulling 2's out of this
(2*1)*(2*2)*(2*3)*...*(2*(n-1))*(2*n)

itll start doing this
2 [1*(2*2)*(2*3)*...*(2*(n-1))*(2*n) ]
2^2 [1*2*(2*3)*...*(2*(n-1))*(2*n) ]
2^3 [ 1*2*3*...*(2*(n-1))*(2*n) ]

and keep going

what would be the exponent of 2 once ive taken them all out?

[shrewd sky]

2^n something my brain is still processing

[cold sage]

yup, and what would be in the square brackets?

[shrewd sky]

2?

a constant

[cold sage]

not quite
2^n [1*2*3*...*n]

[shrewd sky]

oh so youre not diving by 2

[cold sage]

just factoring out all the 2s

any idea what the stuff in the [] could be rewritten as

[shrewd sky]

so if its (2^n[1 * 2 * 3 * ... * n])/n! what do i do now

ratio test?

[cold sage]

not so fast buster

[cold sage]

gotta address this fellow

[shrewd sky]

well if n is going to infinity

wouldnt that be infinity

[cold sage]

n is finite in every term of that sum

just think of the expression

[shrewd sky]

:(

[cold sage]

n going to infinity just means there will be infinite terms thats all

but we're looking at simplifying the general form of a term

[shrewd sky]

ok i simplify whats inside the []

but isnt that just n being multiplied by itself

[cold sage]

its n multiplied by every natural number less than itself

[shrewd sky]

what

the only thing i can think of is facotiral

[cold sage]

correct

[shrewd sky]

WHAT

[cold sage]

the stuff in the [] is n!

[shrewd sky]

oh my god so its juist 2^n

[cold sage]

bingo

[shrewd sky]

bet ty ty

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[stiff reef]

if you have a set of vectors and convert them to a matrix if a row of the matrix is all zeros does that mean they are linearly dependant

[mystic forge]

yeah

[nocturne belfry]

i dont believe so, no

[mystic forge]

no???

[stiff reef]

ooh a battle

[nocturne belfry]

lol its not a battle

[mystic forge]

if u have 3 variables

(its been a while since i did LA)

[nocturne belfry]

consider the first 3 basis vectors for R4

using the standard basis

theyre LI

but youll have a row of all zeros

[mystic forge]

oh thats true

[nocturne belfry]

$\mqty[ \imat{3} \ 0 & 0 & 0 ]$

[clever fjord]

jan Niku

[stiff reef]

like if you have
[x x2 x3]
[x x2 X3]
[0 0 0]

[mystic forge]

yeah i was thinkin its like the one above^

[stiff reef]

if its in r3 and theirs 3

[mystic forge]

2 equations 3 variables

one free variable

[stiff reef]

or 2 in r2 kinda thing

okok thanbk you

[nocturne belfry]

idk youd have to describe a bit more exactly what is happenin

[stiff reef]

ok wait give me one sec

[mystic forge]

im guessing its a square matrix

[stiff reef]

ill draw it

yes

[nocturne belfry]

if its square then yea

i mean you have a row of all zeros

[stiff reef]

thats a way better explination then what i was babout to do

[nocturne belfry]

say its dimension n, the matrix

so the best you can hope for is the columns span R(n-1)

but, then you only need n-1 vectors to do that

so n many vectors (embedded) in R(n-1) cant be LI

maybe its easier to talk about subspaces

[stiff reef]

okok i see

wait

i have one more question about this

if thats ok

[nocturne belfry]

okay

[stiff reef]

would that mean if any of the rows are zero they are ld for a square matrix

like if the x compont was all zeors

or does it have to be the newst one

[nocturne belfry]

newst?

[stiff reef]

sorry like y in r2 or z in r3

wait im stupid

now that i said it im really dumb

[nocturne belfry]

i wouldnt worry about that

[stiff reef]

it doesnt matter what row is zeros

[nocturne belfry]

im just trying to understand what youre asking

[nocturne belfry]

yea, it doesnt matter

do you understand what i mean subspace?

like, you collect these vectors as columns in a matrix

and you zero out a row

or, maybe the row is all zeros all ready

[stiff reef]

i think so its like if you have a vector space like r3 r2 is a subspace or r3 right

oh wait

[nocturne belfry]

yea, its like ... 'inside' R3 yea?

[stiff reef]

yes

i just think of it like sets and subsets

idk if that will hurt me but it makes the most sence to me that way

[nocturne belfry]

like $\mqty[ 1 & 0 & 1 \ 0 & 1 & 1 \ 0 & 0 & 0 ]$

[clever fjord]

jan Niku

[nocturne belfry]

the 3 vectors, assembled as columns

these vectors span a 2-dimensional subspace of r3

theyre in R3 for sure, but the subspace is only 2 dimensional

because you have a row of 0s

[stiff reef]

okok i see it would just span a plane then right?

if its r2 in r3

[nocturne belfry]

yea plane

its just R2 but inside of R3

[stiff reef]

okok thank you

[nocturne belfry]

so to your question like

problem is the span is 2 dimensional right

[stiff reef]

yes

[nocturne belfry]

but, you only need 2 vectors to make a basis for a dimensional space

[stiff reef]

yes

[nocturne belfry]

and in fact, you cant have more and keep them LI

so if the subspace is 2d

but we have 3 vectors in it

they cant be LI

thats all i meant

[stiff reef]

ohh yeah ok i get it now that would mean if all 3 arnt linearly dependant wich is imposible they are all linear combinations of the other two?

i think

[nocturne belfry]

assume there isnt some other degenerate case

because like uhh

[stiff reef]

idk what that means

[nocturne belfry]

$\mqty[1 & 1 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0]$

[clever fjord]

jan Niku

[nocturne belfry]

this could also happen

[stiff reef]

ohh then its just a line in r3?

[nocturne belfry]

i dont wanna say in general what you are describing is true is all

[stiff reef]

so its subspace is r1

[nocturne belfry]

yea

or alternatively one of the vectors is the 0 vector

but assume this screwy stuff doesnt happen

[stiff reef]

then would it just be a dot

[nocturne belfry]

then yes, you can always pick one, and write it as a linear combination of the others

[stiff reef]

okok thank you its all kinda starting to click now

[nocturne belfry]

no, i mean like $\mqty[1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 0 ]$

[clever fjord]

jan Niku

[stiff reef]

oh ok thats just i hat and j hat right

[nocturne belfry]

we cant pick the second and third columns as a basis of the 2 dimensional subspace

[stiff reef]

it doesnt matter how many more zeros you add

[nocturne belfry]

do you see why?

answer: ||we cant write the first column as a linear combination of the second and third column||

[stiff reef]

yes because that would be r1?

[nocturne belfry]

well

[stiff reef]

like the basis vectors for r1

[nocturne belfry]

the first and second are LI and a basis for R2

[stiff reef]

i havent clicked on the answer yet

yes

[nocturne belfry]

there are cases is all haha

i think you understand it

[stiff reef]

hahaha yeah thats what i dont like about linear algebra

theirs always somthing

thank you though i understand now i think

[nocturne belfry]

if you can make a row of zeros you are in good shape

youre just looking at a smaller subspace

[stiff reef]

like with the row echalon form

[nocturne belfry]

then you need to do some investigating to figure out the size of the subspace

[nocturne belfry]

yea

[stiff reef]

okok so a row and a colomn of zeros are the same they can just get "thrown out" to find the sub space

[nocturne belfry]

sure but you dont actually need a column of zeros

[stiff reef]

and the only sub spaces are R0-RN

[nocturne belfry]

remember this guy

[stiff reef]

yeah ig you right

[stiff reef]

ohh okok thank you

[nocturne belfry]

the problem is that

if you add the first and second columns

you get the third

here we can take any two columns as a basis

but the third must be LD

[stiff reef]

yes but wouldnt that not be ref

[nocturne belfry]

hmm

[stiff reef]

im tripping

dont mind that

[nocturne belfry]

no youre good i dont remember the precise defns of these things

i have to google

i believe it is rref

[stiff reef]

its ok

[nocturne belfry]

you have 2 pivots

right

so 2 dimensional

[stiff reef]

oh okok thats like all zeros except on the diagonal right?

rref

[nocturne belfry]

its leading 1's

if i remember rightly

[stiff reef]

okok so just 1s and 0s

[nocturne belfry]

you may have like $\mqty[ 1 & 0 & 4 & 3 \ 0 & 0 & 1 & 2 ]$

[clever fjord]

jan Niku

you may have like $\mqty[ 1 & 0 & 4 & 3 \\ 0 & 0 & 1 & 2 ]$

[nocturne belfry]

but i could be wrong

2 pivots here

[stiff reef]

okok so that would mean its a vector in r2

[nocturne belfry]

no this one is not square

[stiff reef]

i spans r2

[nocturne belfry]

so its not so clear

[stiff reef]

can i just make is square by adding two rows of 0s?

[nocturne belfry]

i guess

what are we talking about now, the columns?

the columns can only span a 2 dimensional subspace

$\mqty[ 1 & 0 & 4 & 3 \ 0 & 0 & 1 & 2 \ \zmat{2}{4} ]$

[clever fjord]

jan Niku

[stiff reef]

ok so would i have to cheak if the other colums are scalar multiples of eatchother

[nocturne belfry]

thank you god for the physics package 🙏

[nocturne belfry]

you dont need to check anything here there are 2 pivots

[stiff reef]

lmao yeah it looks alot better then
[x x x]
[x x x]

oh wait you right sorry

so the amount of pivots in ref is the span

or dimention

like the r

[nocturne belfry]

say its a square matrix

the square matrix is n x n

and you have m pivots

[stiff reef]

ok

[nocturne belfry]

then the columns span a m-dimensional subspace of Rn

[stiff reef]

okk i get it thank you very much can i write that as like this set of vectors spans Rm

for the m pivots

[nocturne belfry]

sort of

its not really Rm right

this is why we have to say subspace

[stiff reef]

oh yeah ok i get that

[stiff reef]

this sentence helped alot

[nocturne belfry]

because the vectors will have n-many entries right

so, they cant be in Rm

(its implicit here m < n, so they have too many entries)

but you can think of the subspace as Rm inside of Rn

like you said, a plane maybe

or a line in 3d space

something like that

but the line is still in 3d space

[stiff reef]

yes ok i get it it would only be rN if its an nxn matrix where all vectors are linearly dependent

otherwise its a subspan

of a dimension

[nocturne belfry]

otherwise the vectors span a subspace

[stiff reef]

sorry subspace not subspan

[nocturne belfry]

if you have a square matrix with at least one row of all 0s

well, we can write out a nice clean defn if you want

i get the feeling youre just chasing the idea out though

[stiff reef]

i think i have it but i just dont know the termanology 100% ill give it like a very short paragraph of what i think it is

if you have an nxn matrix and all vectors are linearly independent without any zero rows these vectors will span Rn. if you take out all of the zero rows when in rref and are left with an MxN matrix the vectors span an m dimentional subspace

thats the best way i could word it

[nocturne belfry]

have you ever seen invertible matrix theorem?

[stiff reef]

i dont know i know about inverse matrixers and stiff

stuff

[nocturne belfry]

https://mathworld.wolfram.com/InvertibleMatrixTheorem.html

so look at # 7

any other statement there is equivalent to # 7

I would say .... maybe 1 or 2 is what you mean?

[stiff reef]

7. The columns of A span R^n.

this one

[nocturne belfry]

right

equivalently

1. A is row-equivalent to the nxn identity matrix

or 2. A as n pivot positions

every numbered statement here is logically equivalent

so if you have one, you have all of them

i just point you at this because this is the strongest form of the 'nice' case

where the columns span R^n

if youre not in the nice case, then your subspace has equal dimension to the number of pivots

[stiff reef]

okok thank you i think i understand it now

thank you this helped an insane amount

[nocturne belfry]

no problem

[stiff reef]

sorry for being a lil dumb

[nocturne belfry]

im glad you have already 90% understanding

you only need another 10% from me

[nocturne belfry]

haha dont say that

youre not dumb

you just have questions

[stiff reef]

you made it all click though wich is the important part

[nocturne belfry]

having questions is good

[stiff reef]

yeah i just feel bad cuz i dont wana wast peoples time i relly apreciate the help though i was watching youtube videos all day and nothing was making it stick

[nocturne belfry]

hmmmmm well if u must know

I am helping other ppl because i do not understand my own work

[stiff reef]

hahaha what are you working on

[nocturne belfry]

fractional calculus

[stiff reef]

damn that sounds a lil wacky

[nocturne belfry]

have you taken calculus?

[stiff reef]

i like calculus alot but i havent gotten very far

[nocturne belfry]

me either

[stiff reef]

just to integrals

[nocturne belfry]

its such a big subject

https://www.youtube.com/watch?v=2dwQUUDt5Is

if u are curious

[stiff reef]

like i know integration by parts and volume of solds of revilution stuff

but i like i like calculus alot more then linear algebra its just alot more intuitive

[nocturne belfry]

linear algebra is more important

[stiff reef]

damn this looks tuff

[nocturne belfry]

are you engineering?

[stiff reef]

nope im in com sci

[nocturne belfry]

ah you will definitely need it lol

get good at linear algebra

[stiff reef]

lmao yeah like half of com sci is vectors and matrixes

[nocturne belfry]

its like eating your vegetables

[stiff reef]

hahaha its just so much work

[nocturne belfry]

consider it an investment

you make the investment early

it pays off much more

[stiff reef]

yeah im trying my best but it takes a long time for me to understand the concepts but i feel like calculus just comes alot easier

wbu are you in engeneering

[nocturne belfry]

computational mathematics

[stiff reef]

zhamm thats pretty cool sounds hard

do you like it though

[nocturne belfry]

its just comp sci from another starting spot

its okay

[stiff reef]

what year are you in

[nocturne belfry]

im about to graduate

[stiff reef]

AYYY nice im in second year rn so i still got a lil ways to go );

[nocturne belfry]

and you already found the math server

you will do great

[stiff reef]

hahaha thank you its a huge help

[nocturne belfry]

no problem

good luck

[stiff reef]

thank you you too and thank you very much for the help

.close

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[spring iris]

Hi

[spring iris]

I need help on this

I have to prove that 3x^3 + 9x - 9x^3 is 3/2x^2(3-4x)

X is just alpha

[nocturne belfry]

were ookay to use x instead?

[spring iris]

Yea

[nocturne belfry]

Your problem is: I have to prove $3x^3 + 9x - 9x^3 = \frac 32 x^2 (3-4x)$

[clever fjord]

jan Niku

[spring iris]

Ye

[nocturne belfry]

what have you tried?

[spring iris]

Factoring by x^2 then by 3/2 which got me close i got the 4x term but i couldn't figure out how im gonna get the 3

[nocturne belfry]

you know im not sure this is true

is it possible youve made a mistake somewhere?

,w plot 3x^3+9x-9x^3 and 3/2 x^2*(3-4x)

[clever fjord]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by janniku

[nocturne belfry]

Can you tell me about where this problem came from?

[spring iris]

I proved with integration by parts this

[nocturne belfry]

Want to show: $\int _0 ^a \ln(1-x) \dd x = - a^3$

[clever fjord]

jan Niku

[spring iris]

Now im gonna use that proof to show that integral of this is this

That

3/2

Expression

[nocturne belfry]

Okay so if i understand

you want to otherwise assume that this is true: $$\int _0 ^a \ln (1-x) \dd x = -a^3$$

[clever fjord]

jan Niku

[spring iris]

Yea

[nocturne belfry]

and You want to apply that result in order to integrate this.

[spring iris]

Yep

[nocturne belfry]

so you want to calculate $$\int _0 ^a \qty( x^2 + 1 + \ln (1-x) ) \dd x$$

[clever fjord]

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

jan Niku

so you want to calculate $$\int _0 ^a \qty( x^2 + 1 + \ln (1-x) ) \dd x$$

[spring iris]

Yep

[nocturne belfry]

then this should be $\qty( \frac{x^3}{3} + x ) \bigg \vert _0 ^a - a^3$, right?

[clever fjord]

jan Niku

[spring iris]

Yep

Same result as me

[nocturne belfry]

which is ...

a^3/3 + a - a^3

[spring iris]

Yep

[nocturne belfry]

this is -2/3 a^3 + a

[spring iris]

Yep but the unit is 3cm so 3cm^2 is 9

[nocturne belfry]

$9 \int _0 ^a \qty( x^2 + 1 + \ln (1-x) ) \dd x = - 6 a^3 + 9a$

[spring iris]

Multiply by 9

[nocturne belfry]

unit? what?

[spring iris]

No remember to multiply by 9

Theres an air unit

Which is 3cm

[clever fjord]

jan Niku

[spring iris]

Yeah i mean i founs that result after

Correct

Its that but how can it be this

3/2 expression

[nocturne belfry]

I have to say that I don't have access to the entire problem

Maybe you are trying not overwhelm me with context?

Granted, I don't speak french!

[spring iris]

Its just a simple integral part of the problem it's sperate

[nocturne belfry]

but I can't say for sure what is happening I don't have enough context

[nocturne belfry]

this result checks out to me

beyond that I am not sure

[spring iris]

Yeah this is correct

[nocturne belfry]

If you can somehow share the entire problem

I'm lacking context to help you understand the final answer

[spring iris]

Its in french

[nocturne belfry]

maybe i can find a french helper

or translate it

chatgpt is pretty good at translation

sometimes

[spring iris]

Chatgpt sucked at this

Couldn't help

Spitting out nonsense

[nocturne belfry]

it sucks at math

but its okay to translate

lemme see

[pastel orbit]

I can kind of read it but not really lmao

[spring iris]

The final integral is just to find the blue part

[pastel orbit]

a) "Verify that for x in [0,1], 1/(1-x) - 1 = x/(1-x)"

[spring iris]

From 0 to alpha

[nocturne belfry]

were on the lower part

[pastel orbit]

which part

[nocturne belfry]

c, I believe

[spring iris]

7a and below

Thats where the integral problem start

[nocturne belfry]

ooh chatgpt is translating

if only i knew french

[spring iris]

Nah i wouldn't wish that on u

[spring iris]

This is the 3cm trick

Remember the result is in cm squared

So u have to multiply by 9

Its hidden

[nocturne belfry]

no, this is not true

well, just hold on

i still dont understand the problem

C_f and T?

[spring iris]

Oh you dont need to worry about that T is just a slope and cf is the graph

[spring iris]

It wants u to find the blue area part

I showed here

[nocturne belfry]

but this says the region bounded by x=0 and x=a which isnt a region

what does C_f and T mean

is C f like the figure of f?

so whats f

[spring iris]

It wants to find the area thats limited by cf which is the graph t which is the tangante and x=0 which is the origin and x= alpha which is just between 0.6 and 0.7

Its that blue part

This is f

Yes figure of f

It doesn't matter cuz the real problem is proving somehow that 3x^3 +9x -9x^3 is this

[nocturne belfry]

wouldnt it be the area under the curve

[spring iris]

It is

The area under it

[nocturne belfry]

from x=0 to x=a

[spring iris]

Yep

[nocturne belfry]

so we did that already

its this

[spring iris]

Yeah

[nocturne belfry]

oops

wait

[spring iris]

Yeah but that has to be this

Somehow

This is last years national exam

[nocturne belfry]

its not lol

[spring iris]

Bro

[nocturne belfry]

i mean its just not it doesnt even have the right order

[spring iris]

So ur telling me this is not possible and that the exam that the whole country did is wrong?

[nocturne belfry]

my guess would be im missing information

i mean

[spring iris]

Nope thats all u need to know we both came to the same expression

[nocturne belfry]

this function is only defined up to 1 right

[spring iris]

Yup

[nocturne belfry]

so what if a>1

[nocturne belfry]

or T is not just 'some tangent'

[spring iris]

A isnt > 1 because i proved this

[nocturne belfry]

its some function or something

[spring iris]

Nope it is

[nocturne belfry]

this is a root funding problem

[spring iris]

The tangent equation is y= -x + 1

If you need to know

[spring iris]

I also found this by the way

So were on the same page

[nocturne belfry]

this changes the answer

this is what i was asking way back when

[spring iris]

Big language barrier sorry

It still wont solve that 3/2 problem though

[nocturne belfry]

if you want to find the area between two functions

say here f and T

then you probably want to be sure sayy

f > T on [0,a]

[spring iris]

Oh my god

[nocturne belfry]

and then look at $\int _0 ^a (f-T) (x) \dd x$

[spring iris]

OH MY GOD

[clever fjord]

jan Niku

[spring iris]

YES

[nocturne belfry]

this is it

[spring iris]

Im stupid

Thanks a lot

[nocturne belfry]

youre good lol

[spring iris]

I completely misunderstood the question and went on integrating f

Instead f - t

[nocturne belfry]

well you want the area between the curves

so all the area up to f

but remove the area under T

(assuming f>= T)

[spring iris]

It is yup, youre the man bro

Thanks

[nocturne belfry]

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[maiden knoll]

how would you verify this not know r?

[maiden knoll]

i would do a cofactor expansion since i dont know how i would put this in my calculator

r_1 r_2 r_3 are distinct real roots, so can you just pick arbitrary r values?

[regal leaf]

just calculate the det

?

and use that $e^x \neq 0$

[clever fjord]

tobi

[regal leaf]

for all x in the Reals

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[solid niche]

Hello! I need help understanding what i am doing wrong in the next expression:
Let $ E(x) = \arcsin ( \sin (2x)) ; \forall x \in \R =>\newline
\ sin (E(x))= \sin (2x) ; \forall x \in \R <=>\newline

• \sin (E(x))= - \sin (2x) ; \forall x \in \R <=>\newline
  \sin ( E(x)- \pi )= \sin (2x- \pi ) ; \forall x \in \R <=>\newline
  E(x) - \pi = \arcsin(\sin(2x- \pi)) + 2k \pi ; \forall x \in \R ,; k \in \Z ; <=>\newline
  E(x) = \arcsin(\sin(2x- \pi)) + \pi (2k + 1) ; \forall x \in \R ,; k \in \Z
  ; =>\newline
  \arcsin(\sin(2x))=\arcsin(\sin(2x- \pi)) + \pi(2k+1) ; \forall x \in \R , ; k \in \Z \newline
  (FALSE) $

[cursive field]

...

[solid niche]

huh?

Why is it not turning into latex?

$ E(x) = \arcsin ( \sin (2x)) ; \forall x \in \R =>\newline
\ sin (E(x))= \sin (2x) ; \forall x \in \R <=>\newline
\sin (E(x))= - \sin (2x) ; \forall x \in \R <=>\newline
\sin ( E(x)- \pi )= \sin (2x- \pi ) ; \forall x \in \R <=>\newline
E(x) - \pi = \arcsin(\sin(2x- \pi)) + 2k \pi ; \forall x \in \R ,; k \in \Z ; <=>\newline
E(x) = \arcsin(\sin(2x- \pi)) + \pi (2k + 1) ; \forall x \in \R ,; k \in \Z
; =>\newline
\arcsin(\sin(2x))=\arcsin(\sin(2x- \pi)) + \pi(2k+1) ; \forall x \in \R , ; k \in \Z \newline
(FALSE) $

[cursive field]

$E(x) = \arcsin ( \sin (2x)) ; \forall x \in \R =>\newline \sin (E(x))= \sin (2x) ; \forall x \in \R <=>\newline\sin (E(x))= \sin (2x) ; \forall x \in \R <=>\newline\sin (E(x)- \pi)= \sin (2x- \pi ) ; \forall x \in \R <=>\newline E(x) - \pi = \arcsin(\sin(2x- \pi)) + 2k \pi ; \forall x \in \R ,; k \in \Z ; <=>\newline E(x) = \arcsin(\sin(2x- \pi)) + \pi (2k + 1) ; \forall x \in \R ,; k \in \Z; <=>\newline\arcsin(\sin(2x))=\arcsin(\sin(2x- \pi)) + \pi(2k+1) ; \forall x \in \R , ; k \in \Z \newline (FALSE)$