help-38

[bright quarry]

think of like "largest" accumulation value vs "smallest" accumulation value

[trim lichen]

hold on tho

i dont think you got that right for the exponent then

[bright quarry]

🤔

[trim lichen]

wait

oh yeah ok

[bright quarry]

the root test says that if the limsup |a_n|^(1/n) < 1 then the series converges absolutely

diverges if it’s > 1

[dusky thunder]

yes

[bright quarry]

no information if it equals 1

ratio test would be $\limsup \biggr|\frac{a_{n+1}}{a_n}\biggr| < 1$

[solid kiln]

knief

[bright quarry]

for absolute convergence

and liminf > 1

for divergence

so in this example you can just break this up into even/odd n and use the root test to show that the limsup is 1/2 so it’s absolutely convergent but using the ratio test after splitting even and odd n youd get 1/8 for liminf and 2 for limsup

so limsup is not less than 1 and liminf is not greater than 1

no information

but in most normal looking series you come across it would probably just come down to what’s more convenient to use

and it wouldn’t make a difference

[dusky thunder]

that is so interesting. thank you for showing me this example!

[bright quarry]

like when the limits both exist for the ratio and root i don’t think it makes a difference

[bright quarry]

you’re welcome

[dusky thunder]

yeah i think in my case thats gonna be the series we tackle

.close

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[grand wave]

I want to ask about scalar projection. Im not sure i understand how it works. I understand that scalar projection is to show how much object is in front of you or behind you, i know it returns a single number and if number is positive its in front of you and if negative is behind you. Im just curious cant this be done by getting magnitude/distance between 2 vectors to game same thing?

[ionic pendant]

well a projection is a way to get the component of one vector in the direction of another vector

[grand wave]

what does that mean, can it be simplified a bit

[ionic pendant]

so in this example we are interested in the projection of y onto x

the vector projection of y onto x is the purple vector

the scalar projection of y onto x is the magnitude of that projection with a sign, being positive if it points in the direction of x and negative if it points in the opposite direction

so if x is "forward", then it is positive if y points "ahead" and negative if y points "behind"

[grand wave]

With my understanding that scalar projection returns a single number, which means its not a vector. Its just confusing for me.

It sounds simple, its magnitude between 2 vectors, so the key difference is that we know if its forward or behind?
But not sure how the distance is correct if the x line is straight and its not poining straight to y? Its like never reaches the y

Is it just showing how far ahead or behind it is rather showing the distance between 2 objects?

because for me it just seems that these both shows magnitude but in a sense its different and magnitude for me what is the distance directly from one object to another

[heady furnace]

magnitude can't be negative, in this case it just shows the distance

(this is because magnitude is calculated via pythagoras' theorem, which always gives a positive number)

while a scalar projection uses a frame of reference that can have negative values

[grand wave]

ok but does scalar projection shows the magnitude directly between 2 object or just how far ahead or behind its is in the directions they facing?

[heady furnace]

scalar doesn't show the magnitude of the distance vector, no

[grand wave]

Ok but what if happens when the scalar projections gets used on this case?

the distance clearly is not 0 but would the product be 0?

[heady furnace]

the scalar projection would be 0

it would not be either ahead or behind it

the magnitude of the distance vector would not be 0

[grand wave]

but the guides i read states that scalar projections shows the distance betwen 2 object, so in this case it only shows how far ahead or behind it is and not direct distance between 2 of them

[heady furnace]

it is not totally correct in telling the distance between 2 objects

with 2 scalar projections (like x and y axis) you can calculate the distance between 2 objects

but a scalar projection alone without any other information does not really show the distance between two objects unless they are aligned along the projection axis

[grand wave]

Ok so i understand there is 2 types of magnitudes?

[heady furnace]

not quite

the magnitude of the distance vector is just "how long the vector is"

but you can also express vectors if you have a frame of reference

like a x and y axis

[grand wave]

wait isint magnitue is a length between 2 vectors and its the same? magnitue == distance

[heady furnace]

in this particular case yes

it's the distance between the tail and the head of a vector if you want to see it that way

but vectors aren't always used for distances

[grand wave]

i understand that, im just a bit confused.
So scalar projection shows what?

[heady furnace]

have you worked with points in x and y axis before?

[grand wave]

I understand the x and y axis i just dont understand the concept of the scalar projection.

[heady furnace]

it's like the x and y coordinates of the object

you project the object along one axis

[grand wave]

So it show magnitude only perpendicular to the 2 objects?

[heady furnace]

kind of, but it could be negative

[grand wave]

so it cant be parallel?

[heady furnace]

you always have a way to find a perpendicular from a point to a line

[grand wave]

I know that there is math to found all solutions to the problems. Im just confused on the rules that follow scalar projection

[heady furnace]

this is basically scalar projection
P_y is projection of P along the y axis, and P_x is the projection along the x axis

it can be confusing because you could rotate the x and y axis but scalar projection requires a frame of reference

[grand wave]

what, i never seen 2 projections

the whole definition of scalar projection just doesnt make sense anymore

That looks more like Vector Projection

[heady furnace]

scalar projection is a projection onto another vector

i forgot to put the arrows on x and y

my bad

[grand wave]

Simply saying scalar projection is like 90 degree shadow and result is the distance between the one vector and where the shadow projected stops

[heady furnace]

if the shadow is perpendicularly to where you are projecting, yes

[grand wave]

isint it always perpendicular?

[heady furnace]

it usually depends on where you shine the light from but logically speaking yes it is perpendicular

[grand wave]

i just thought of that , so it can come into play later when i learn the degree of which shadow is casted

[heady furnace]

it does

[grand wave]

Thanks a lot for struggling with me

[heady furnace]

no prob

[grand wave]

!close

[heady furnace]

i think it is .close

.close

[grand wave]

.close

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[valid hamlet]

This week i learned propagation of errors, absolute uncertainty and etc. I would like to ask if you guys could check if i did this question right

[valid hamlet]

So f= 530^2.

So delta_f = 3.8% x 530^2

I cleaned it up a bit

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[formal jasper]

hello! can anyone help me spot my mistake

[formal jasper]

im stuck :(

oh, the E corner is the bottom right corner of the yellow square

the translator messed it up

[drifting blaze]

Wait a minute

The 8/7 information is given to you?

[formal jasper]

yeah

the translator messed it up so i had to write it

[drifting blaze]

Oh ok then ignore what I said

[formal jasper]

ohh okayy

[drifting blaze]

Someone else would help you.

[formal jasper]

thanks ^^

<@&286206848099549185>

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[dreamy pewter]

Hello, could someone please help me with this exercise?

[trim jolt]

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

[sturdy fiber]

yeah what step are you on man?

[dreamy pewter]

4

I got an answer and would like my work checked.

[sturdy fiber]

ok

[dreamy pewter]

i solved the inequality by the method <=-3 and => 3, and then i solved it as a rational inequality in both cases, and i got [-6/5,-1/2)U(-1/2,0]

but i dont know if it is ok

and the answer doesnt match

[sturdy fiber]

what do you mean?

which answer did you get?

[dreamy pewter]

in the first case ( => 3) i got [-6/5,-1/2) and in the second one (<=-3) i got (-1/2,0]

i thought it was option 1 or 3 but it doesnt match

idk if i did something wrong

[sturdy fiber]

are you willing to share your notes so that we can see your attempt to solve the exercise?

[dreamy pewter]

okay, give me a sec

[sturdy fiber]

-1/2 is excluded

yeah it's option 3

[dreamy pewter]

and where is the zero in -1/2

isn't it supposed to be -1/2,0?

Look my notes btw

lol i sent it in 144p

im sorry, its my phone

[sturdy fiber]

it's correct

[-6/5,-1/2)U(-1/2,0] and [-6/5,0]-{-1/2} are the same cause you subtract -1/2 from the range, since you cannot include it

[sturdy fiber]

yes that is correct

[dreamy pewter]

ohh i got it

[dreamy pewter]

yes yes

thank you man

so its option 3

perfect

[sturdy fiber]

yes

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[marsh forum]

Trying to determine if this converges or not

[marsh forum]

to start, we check for absolute convergence

so we check $\sum \frac{ \sqrt{n}}{n+1}$ for convergence

[solid kiln]

What a wonderful world !

[simple haven]

You don't need to check for absolute convergence

[marsh forum]

Yeah, just realised

divergence test works, I think

[simple haven]

An alternating series is convergent if it's decreasing (in abs, and the sequence converges to 0)

[marsh forum]

so this works too, I suppose

[simple haven]

This series is not absolutely convergent btw

[trim lichen]

we still might want to distinguish absolute vs conditional though

[marsh forum]

Which is why I mentioned divergence test

[simple haven]

Yup

[marsh forum]

the limit of a_n diverges to infty

[ionic pendant]

does it?

[marsh forum]

no

[trim lichen]

a_n itself does not approach infty

[marsh forum]

or a comparision would work then

[trim lichen]

sqrt(n)/(n+1) behaves roughly as 1/sqrt(n)

[marsh forum]

$\frac{1}{n+1}< \frac{ \sqrt{n}}{n+1}$

[solid kiln]

What a wonderful world !

[marsh forum]

As the harmonic series diverges , so does this

[ionic pendant]

for pretty much all algebraic functions the limit comparison test is the most straightforward method

[marsh forum]

This next

[simple haven]

Which wins, exponential growth or cubic growth?

Or I guess really the question should be exponential growth or quartic growth (because the denominator needs to be more than one exponent bigger to converge)

[marsh forum]

My first thought was to write 5^n as a power series , but we haven't done power series in class yet

[trim lichen]

no

that is way overcomplicated

[marsh forum]

root test should work, no?

that or the ratio test

ratio may be easier here

$\lim_{n \to \infty} 5 \frac{n^3}{(n+1)^3}$

[solid kiln]

What a wonderful world !

[marsh forum]

the limit is 5, which is greater than 1, thus it diverges

[trim lichen]

mmm

can you show your work and the exact wording of the test you applied

cause i think you may have gotten some things backwards.

[marsh forum]

$\frac{\frac{(n+1)^3}}{5^{n+1}}{\frac{n^3}{5^n}}$

[trim lichen]

$\frac{ (n+1)^3/5^{n+1} }{ n^3/5^n}$

[solid kiln]

What a wonderful world !
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

Ann

[trim lichen]

this is what you should've had

[marsh forum]

yea

[trim lichen]

ok

no backwardness so far

but your simplification of it was wrong

[marsh forum]

yea, the limit is 1/5

not 5

[trim lichen]

yes

do you know the spirit of the ratio test?

imo it's as important (if not more so) as knowing its literal statement

[marsh forum]

Yea, it's a pseudo-GP test basically

We are approximating it as a GP

[trim lichen]

yes

yours is like a GP with common ratio 1/5

you should expect it to converge

[marsh forum]

I'd like to solve a couple more problems, I'll post 'em here?

[trim lichen]

more series shit?

yeah whatever

[marsh forum]

Root test

so $\lim_{n \to \infty} \sqrt[n] {a_n} = \lim_{n \to \infty} \frac{n^2+1}{2n^2+1} =\frac{1}{2}<1 \implies \text{ convergence }$

[solid kiln]

What a wonderful world !

[marsh forum]

This works?

one more question

This series can be re-wriiten as $\sum a_n + \frac{a_n}{n}$

[solid kiln]

What a wonderful world !

[marsh forum]

$a_n$ converges, so we just have to show $\sum \frac{a_n}{n}$ converges

[solid kiln]

What a wonderful world !

[marsh forum]

Can I have a hint

A comparison test, will likely not work here

or it may actually

$a_n< \frac{1}{n^p} \text{ for some p ≥2 , thus } a_n/n< 1/n^{p+1}$, which converges, and we're done

[solid kiln]

What a wonderful world !

[marsh forum]

yea, probably wrong

for sufficently large n

okay, I'm gong insame

am I not

$\abs{a_n/n}< {a_n}$, $\abs{a_n}$ converges, so $\abs{a_n/n}$ converges too, thus $a_n/n$ converges

[solid kiln]

What a wonderful world !

[marsh forum]

.close

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[lean kraken]

Unit digit

[lean kraken]

How can I apply mod 10 here

I tried to do it separately but i got -7?

2^23^24 mod 10 is 2

9^24^23 mod 10 is 1

[half vigil]

2-1 = 1

[trim lichen]

well none of these answers could POSSIBLY be correct because you are subtracting even minus odd

this much you can see even without any calculations

the number will be odd and so end in an odd digit, so there is no correct option

[trim lichen]

how did 2 and 1 result in -7 though

[lean kraken]

My bad

I only got 2 and 1

[trim lichen]

assuming you got that correct (which it looks like you did), the answer would be 1.

[lean kraken]

thanks ann

Power of any even number would be even? And power of odd number will be odd always?

Yes it is

+close

[half vigil]

it's .

[half vigil]

.close

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[wraith hinge]

How do i get e ?

[sacred sapphire]

triangle ACE

you know what d is

[wraith hinge]

Yep its 60

[wraith hinge]

Isosceles truanglw?

[sacred sapphire]

no

[wraith hinge]

Oki

[sacred sapphire]

look closer

[wraith hinge]

Mm not sure

[wraith hinge]

Ohh angles on a straight line?

[sacred sapphire]

yes

[wraith hinge]

Ok that would be 120

So 120 +40 + e =180?

[sacred sapphire]

yes

[wraith hinge]

Oo ok so its 20

How do we know triangle ACE is not a isoscelec triangle

Incase i accidentally take it as a isosceles triangle?

[sacred sapphire]

because its not stated anywhere?

[wraith hinge]

Ohh oki

[sacred sapphire]

and you shouldnt randomly assume stuff

[wraith hinge]

Oo oki

Thank u very much

Have a nice day

.close

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[frosty nebula]

What's the formula for vertical asymptote?

[woven python]

depends but think of it as a abscissa or ordinate where the function or curve diverges to infinity

[frosty nebula]

chatgpt didn't help me

[woven python]

for example, $y=\frac{1}{x-a}$ has vertical asymptote at $x=a$ because the functiob diverges to $\pm \infty$ there

[frosty nebula]

I mean like I have the equation of x=1 is vertical asymptote of the graphic f:(1, Infinite)->R, f(x)=2x/(x-1)

[solid kiln]

parabolicinsanity

[woven python]

yeah, so?

[frosty nebula]

Shouldn't ne a limit? Like />?

[woven python]

i mean, yeah i guess but what i meant to say was to be taken more as intuition

[frosty nebula]

Like yhis

[woven python]

what

i mean sure

if $\abs{\lim_{x\to a^{\pm}}f(x)}=\infty$ then there's a vertical asymptote

[solid kiln]

parabolicinsanity

[woven python]

i guess

at x=a

[woven python]

yeah so there's an asymptote there

[frosty nebula]

And.... How do I find it?

[woven python]

literally find the point where there is divergence to infinity

thats all

thats what an asymptote is

so in that problem the asymptote is x=1

[frosty nebula]

Oh

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[wraith hinge]

Cn someone help me with this

[wraith hinge]

Can someone explain it to me cus i cant understand it

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[lusty token]

https://youtu.be/75dMcyCUo2g?si=1BWMLi7T-6KdNB1g

https://youtu.be/V5ArB_GFGYQ?si=IsGr4WuPjjZLWRZR

[lusty token]

This one is better

No need to watch the whole video

@wraith hinge

[shrewd hollow]

Is there something you specifically you don't understand about it? otherwise your probably better off just watching the videos.

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[livid bolt]

I'm going over

Ellipses

In class

And honestly I don't know what the fuck the teacher's talking about

I'm super confused

Does someone have any reccomendations for this

I am on calculus if that helps?

Like any videos or pages to check info about this specifically... maybe? I dunno I am lost

Or Hell just an explanation of what they did would be thankful

[hidden dew]

seems like deriving the standard form where the center is (0,0)

[livid bolt]

Which one is the standard form here? I think I passed this previously but I'm kinda overwhelmed and stuff is not coming up

[hidden dew]

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

[solid kiln]

yoboiqimmah

[hidden dew]

does this look familiar ? @livid bolt

[livid bolt]

Yes

This is the standard form

Okay

[hidden dew]

yep they just deriving this form where a>b

[livid bolt]

What do you mean with this?

[hidden dew]

https://youtu.be/BHtIBLsGdNU?si=NsTA-WLqjkxhmu52

[livid bolt]

Neat

Gonna close for now since I have more classes and need to watch and understand this anyways

Gonna pop up later again

.close

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[marsh forum]

I'd like to test this for convergence

[marsh forum]

we can use the divergence test

nvm

bad idea

root test

$\frac{ \sqrt[n]{n!}}{e^n}< \frac{n!}{e^n}$

[solid kiln]

What a wonderful world !

[marsh forum]

This doesn't work either

[nova spire]

you could do that

[marsh forum]

hmm

[nova spire]

what about ratio test

[marsh forum]

I did consider that, lemme try it

$\frac{(n+1} e^{{(n+1)^2}}{e^{n^2}}$

[solid kiln]

What a wonderful world !

[marsh forum]

oops

[nova spire]

$\frac{(n+1) e^{{(n+1)^2}}}{e^{n^2}}$

[solid kiln]

rafilou is not not born in 2003

[marsh forum]

yea, thanks

[nova spire]

but that seems wrong

what are you doing

a_(n+1)/a_n?

[marsh forum]

yea

[nova spire]

check your exps then

[marsh forum]

okie

got it

$\frac{(n+1) e^{n^2}}{e^{(n+1)^2}}$

[solid kiln]

What a wonderful world !

[marsh forum]

I then divide the num and denom by $e^{(n+1)^2}$

[solid kiln]

What a wonderful world !

[marsh forum]

ncm

*nvm

$lim_{n \to \infty} (n+1)e^{n^2- (n+1)^2}$

[solid kiln]

What a wonderful world !

[marsh forum]

which is $\lim_{n \to \infty} (n+1) e^{-(2n+1)}$

[solid kiln]

What a wonderful world !

[marsh forum]

which goes to 0

So it converges?

[nova spire]

yeah

[marsh forum]

Thanks

This next

This is calling for a comparison test imo

or nvm, that won't work

I was thinking of comparing it with 1/√n

[nova spire]

there is comparison at play

but 1/sqrt(n) is too bad

[marsh forum]

yeah, I realise, I'm thinking

[nova spire]

the sqrt(n) we won't touch

think about what you can do with the sin thingy

[trim lichen]

when comparing against some series you always have to understand the convergence status of the reference series

and thus the direction of comparison that produces a conclusive result

[marsh forum]

I can probably sandwich it

[trim lichen]

if the reference series is convergent, then the only fruitful direction is yours <= ref

if the reference series is divergent, then the only fruitful direction is yours >= ref

[marsh forum]

I have a feeling this diverges

nvm, it converges

sin(1/n) goes to 0

[trim lichen]

proof?

[marble wharf]

and?

1/sqrt(n) goes to 0 aswell

[trim lichen]

the general term of your series does approach 0, but that does not a convergent series make

@worldly cargo hi, do you have any more insightful and valuable commentary to add?

[marsh forum]

1/n< 1/√n

<@&268886789983436800>

[trim lichen]

any more? this is all so extremely relevant to this series question after all...

[marsh forum]

also, congrats rafilou

[chilly bobcat]

the entire expression becomes zero at infinity

[nova spire]

welp wasn't fast enough to be my first, thanks eric

[marsh forum]

I'm wondering if I can use this somehow

1/n≤sin(1/n)/√n<1/√n

[chilly bobcat]

nsin(1/n) approaches 1?

[marble wharf]

comparing it to 1/n would still not be enough

[marsh forum]

If true, this would mean divergence

[trim lichen]

you can use it if you are OK with using falsehoods

[marsh forum]

okay, so no

[marsh forum]

I can multiply and divide by √n

?

[nova spire]

I don't see much the relevance

[marble wharf]

lets pretend we are physicists for a moment. what would we do?

[chilly bobcat]

sin x / x limit..

[marsh forum]

sin(1/n)~1/n

[nova spire]

there you go

[marble wharf]

and what would we get then?

[marsh forum]

so I compare it to 1/n^{3/2}

[marble wharf]

1/

[marsh forum]

which converges

so this converges too?

[marble wharf]

good so now we know what we want to show

and clearly the key will be to rigorously compare sin(1/n) and 1/n

[marsh forum]

Even for a calc 2 course?

[nova spire]

yeah

every step needs to be rigorous

[marsh forum]

okay

[nova spire]

but there's a simple reason

why sin(1/n) is equivalent to 1/n

when n-> infinity

[marsh forum]

1/n \to 0

so we can replace 1/n with x as x \to 0

[nova spire]

it's the other way around but yes

sin(x)/x -> 1 as x-> 0

so "x = 1/n"

[marsh forum]

okie

thanks

[marble wharf]

a much easier solution is to say that sin(1/n) < 2/n

[marsh forum]

oh, fair

yes

[nova spire]

uh

just sin(1/n) <= 1/n

why the 2

[marble wharf]

I never wanna bother thinking about whether they intersect

[nova spire]

?

they don't

sin(x) < x when x > 0

[marble wharf]

I'm not saying that the 2/n is optimized

but depending on the level its much easier to argue

although I guess at calc 2 level it doesnt matter anymore

we had the same problem some months ago and werent able to show sin(x) < x yet iirc

[nova spire]

how about just

sin(x) <= x

[trim lichen]

yeah we do not need a strict comparison here

[marsh forum]

I would like to solve a few more problems

This is pretty easy $\frac{k \ln(k)}{(k+1^3)}< \frac{k^2}{(k+1)^3}$\
To show the bigger series converges, we use the ratio test \
$\lim_{k \to \infty} \frac{ (k+1)^2 (k+2)^3}{ k^2 \cdot (k+1)^3}$

[solid kiln]

What a wonderful world !

[trim lichen]

yeah?

and what's that give you then huh

[marsh forum]

the limit is 1, so inconclusive

[trim lichen]

easy
inconclusive

i believe this is what is known in the industry as a flop face first

maybe you should not have bounded the logarithm so crudely

also once again

tell me: what kind of familiar behavior does the ratio test capture

[marsh forum]

it treats the series as a GP

[trim lichen]

ok

does your series even remotely smell like a GP

[marsh forum]

no

[trim lichen]

ok then maybe trying to shoehorn ratio test is gonna be a bad idea

[marsh forum]

Well, then I want a comparison test, because the root test isn't going to be pleasent here

neither is the integral test

[trim lichen]

root test also captures GP-like behavior

anyway you're right that you want some kind of comparison

note that log(x) is dominated by any power function x^p as long as p > 0

[marsh forum]

yea

[nova spire]

so

you don't have to go as big as p = 1

[trim lichen]

$\log(k) < k^{0.42069}$

[solid kiln]

Ann

[trim lichen]

[marsh forum]

so I can do somethiing like

[nova spire]

p = 0.9999999999 is good

[marsh forum]

$k^{1.5}/(k+1)^3}$

[solid kiln]

What a wonderful world !
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[nova spire]

yeah sure

[trim lichen]

now you have you justify why that one converges.

[marsh forum]

in the long run teh denom behaves as k^3

1/k^{1.5} converges

[trim lichen]

can you turn this into a rigorous comparison

or is this 'long run behavior' thing already acceptable in your class

[marsh forum]

Well, our tutor used it

more rigourlsy though, I'd probably do a ratio or root test

[trim lichen]

no you wouldn't

and you should be able to tell me why you wouldn't

not least because we discussed it a mere fifteen minutes ago

[marsh forum]

Because it doesn't look much like a GP

k^{1.5}/(k+1)^3< k^{1.5}/k^3

that's what I'd do

[trim lichen]

is there a "quod licet Jovi non licet bovi" thing at your uni or is it agreed upon that what's good enough for your tutor is also good enough for you

[trim lichen]

yes, that's good.

now you have an honest to goodness p-series

[marsh forum]

Can I do a couple more

[marble wharf]

no, we will arrest you

what do you think

[marsh forum]

no, like can I get help with a few more

[marsh forum]

You're right, stupid question

I need something smaller than ln(n) for an effective comparison

like i want to comapre it with something bigger first to see if it converges

[trim lichen]

$(\log(n))^{\log(n)} = n^{\log\log(n)}$

[solid kiln]

Ann

[marsh forum]

yea

[trim lichen]

loglog(n) is quite the sluggish function but it does outgrow any constant, e.g. 2

[marsh forum]

hmm

[marsh forum]

I don't follow

[marble wharf]

when n is big enough then loglogn is bigger than 2

[marsh forum]

so it does converge

[marble wharf]

why?

[marsh forum]

as when n is big enough, it decreases faster than any n^p series

[marsh forum]

hmm, if this is wrong, I'm not too sure

[trim lichen]

make it rigorous

do a proper comparison test thing again

[marsh forum]

ln(ln(n))>n

Honestly, is it fine if I do this later, just realised I have to read a few ethnographies for my exam this friday

I forgot about that exam

[marble wharf]

you may want to let n_0 be the number for which loglogn>2 and split the sum

[trim jolt]

@marsh forum Has your question been resolved?

[trim jolt]

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[wraith hinge]

how is my answer wrong

[split chasm]

how are you getting the 36

and the -0 s

try to look more carefully at the image

[wraith hinge]

umm is it 49

are the 0s not correct

(x-0)^2 + (y+1)^2 = 49?

[ionic finch]

yeah

do you actually have 9 attempts

for one question

[wraith hinge]

lol yea

[ionic finch]

lmfao

insanity

[wraith hinge]

ik im making simple mistakes

tyy

.close

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[modern spear]

where should i start with this question

[modern spear]

the second part of the question is confusing

[red loom]

note that a geometric series is always of the form (a+ar+ar^2+ar^3+\ldots) this question is asking us to find an (x) such that (a+arx+ar^2x^2+ar^3x^3+\ldots=T_3(x)=-\frac54)

[solid kiln]

PajamaMamaLlama

[modern spear]

ah

so would r be (x/5) ?

.solved

[trim jolt]

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[random tapir]

help

.stop

.solved

[trim jolt]

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[daring bough]

Exercice 2 n3.a

J'ai trouvé x= 0,47 y=0,46 et z= 0,07 jsp si c'est correct

[trim jolt]

@daring bough Has your question been resolved?

[trim jolt]

@daring bough Has your question been resolved?

[daring bough]

<@&286206848099549185>

[nova spire]

C'est quoi la matrice T que t'as trouvé ?

[daring bough]

Celle la

Mais enft j'ai trouvé un résultat a cette question mais on me dit c'est pas bon

Surtout qu'il y a marqué l'ensemble des solutions

[clear cloud]

C'est pas pour autant qu'il y en a plusieurs si ?

Quoique

[daring bough]

Bah jsp justement

Comment je fais pour savoir si il y en a plusieurs

[clear cloud]

Jpeux voir?

[daring bough]

????

De

[clear cloud]

Comment tu as eu ça

Ah tu as juste donné à la machine

[daring bough]

Non

J'ai fais

Jsp si t'arrivera a me relire 🥲

[clear cloud]

Ça me paraît correct oui

[daring bough]

Mais deepseek me sors un autre résultat

C'est bizarre

[clear cloud]

Qu'est ce qu'il dit

[daring bough]

Il me sors ča

[clear cloud]

Bah il manque une ligne non?

[daring bough]

Mais il a dis on peut l enlevé elle sers a rien

En gros

Mais juste le soucis c'est on me dit déterminer l'ensemble des équations

[clear cloud]

Mouais, j'ai x = 1763/3745

y comme toi

z = 54/749

Donc ouais

!nogpt

[trim jolt]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[daring bough]

Donc ma réponse est bonne?

[clear cloud]

Avec cette matrice oui

[daring bough]

La matrice t oui

Mais dcp il y a qu'une solution ?

[clear cloud]

Ce qui paraît assez normale

[daring bough]

J'avoue ça serait bizarre d avoir plusieur x

Fin ça voudrai dire qu on a pas la même probabilité

Avec la même matrice de transition

J'écris en fraction ou en arrondi ?

[clear cloud]

Ça peut se justifier avec de la géométrie dans l'espace qu'il y a qu'une solution dailleurs

[clear cloud]

Les deux

[daring bough]

Daccc

[daring bough]

A bon?

Genre vecteur

[clear cloud]

Genre intersections de trois plans c'est un point ou vide

[daring bough]

Je crois qu'on l'a pas encore vue ça

Mais la toute dernière question j'ai pas compris ce qu'on nous demandait

Normalement c'est bon

Manque que la dernière question

[clear cloud]

C'est une interprétation d'une solution en particulier

alors x, y ou z ?

[daring bough]

X c'est les abeilles essaiment

Y les abeilles font du miel

Et z elles contractent une maladie

[clear cloud]

Et il demande quoi eux ?

[clear cloud]

Au moins l'intersection de deux plans non ?

[daring bough]

Oui??

Mais les matrices et les plans c'est pas pareil

[clear cloud]

Alors non, parcontre le système que tu résoud si

Tu as quatre eqn de plans la

[daring bough]

Si l'on observe l essaim sur une période quelle proportion représente les années où cet essaim produit du miel

[clear cloud]

Donc faut regarder lequel?

[daring bough]

J'avoue

Mais on se complique la vie

Y

[clear cloud]

Pas tant que ça, c'est deux phrases

Mais il y a pas besoin

[daring bough]

Oui t'as raison

[clear cloud]

Ok

[daring bough]

Mais enft jsp si c'est vraiment ça

Fin je crois

Normalement

[clear cloud]

Je viens à peine de vérifier P je suis une fraude mais normalement elle est correcte

[daring bough]

😭 tkttt

Normalement

Bah en plus la calculatrice donne ça

Donc c'est bon

[clear cloud]

Enfaite, ce que tu as trouvé dans ton système c'est ta distribution invariante et c'est dans le nom, ça bouge pas. Donc la proportion sur le long terme c'est la proba que tu as trouvé pour y associé, donc la val de y

[daring bough]

La proportion c'est 45,714%

[clear cloud]

Ça me paraît bon

[daring bough]

Mais j'ai pas compris pk deepseek trouve autre chose

[clear cloud]

Pour l'instant, considère les comme des élèves qui passent à l'oral sans connaître le sujet et qui brodent en regardant le diapo

Ça paraît correct mais dans les faits ils disent des bêtises

Et en plus comme c'est boîte noir (tu sais pas vraiment comment il fait) bah c'est compliqué d'avoir un vrai regard

[daring bough]

Bonne comparaison 🥲

Bon bah je pense avoir terminé sauf si j'ai fais une faute

[clear cloud]

D'ailleurs

[clear cloud]

Tu as la propriété qui dit pourquoi ça marche ?

[daring bough]

??

Comment ça

[clear cloud]

Enfaite

Si ta matrice de transition ne contient pas de 0

[daring bough]

A oui

Ça oui

Mais c'était la question 2

[clear cloud]

Alors la suite de matrices lignes convergent vers la solution de ton système qui est unique

[daring bough]

C'était la question 2

[clear cloud]

Ah ok

Mais ducoup cette propriété assure l'unicité

Comme tu te demandais

[daring bough]

Stv je peut te montrer ce que j'avais marqué

Dacc

[clear cloud]

Non j'pense que c'est ok ça

[daring bough]

Dacc

Sinon une autre question

[clear cloud]

Oui

[daring bough]

Mais enft c'est sur le grand oral

[clear cloud]

Dis moi

[daring bough]

En maths dcp

C'est compliqué ou pas a expliqué car il y aura quelqu'un qui connait pas grands chose en maths

Comment faire pour qu'elle comprenne sans qu'on puisse écrire

[clear cloud]

C'est quoi ton sujet ?

Ça depends beaucoup du sujet

[daring bough]

Fin moi en théorie j'ai comment peut t'on modélisé l evolution de population/ maledie etc grace au maths

Et dcp j'avais pris les équations différentiel

[clear cloud]

Pcq dans tout les cas tu as le droit et c'est fortement conseillé à un support ou tu peux leur marquer des trucs et leur montrer

[clear cloud]

Ah verlhust

[daring bough]

Oui mais faut choisir et jsp qu'est qui faut vrmt mettre

[clear cloud]

Choisir ?

[daring bough]

Bah on peut pas tout mettre

[clear cloud]

Bah c'est des graphique majoritairement

Pour le coup l'oral tu le fais rigoureusement mais sans trop vraiment rentrer dans les détails super précis et l'idée c'est que tu "donnes" des questions à te poser

[daring bough]

Mais juste j'ai peur un peu d'être trop vague ou trop precis

[clear cloud]

Bah ils te poseront des questions dessus

Et si tu réponds correctement c'est tant mieux

[nova spire]

Théorème de Perron Frobenius: 1 est valeur propre simple

mais bon on l'a pas

[clear cloud]

Si seulement tu l'avais

[daring bough]

Mais ça c'est le problème

Faut anticiper les question

[nova spire]

Enfin c'est facile à voir avec la similarité aussi

[daring bough]

Pour?

[nova spire]

on a une matrice de transition stochastique

donc somme des lignes = 1

donc 1 valeur propre

[daring bough]

Oui je suis d'accord

[clear cloud]

¯_(ツ)_/¯

[nova spire]

(vecteur propre colonne = (1 1 1))

[daring bough]

Je vois pas?

[nova spire]

$T \begin{pmatrix}1\1\1\end{pmatrix} =$ ?

[solid kiln]

rafilou is not not born in 2003

[nova spire]

Avec le fait que la somme des lignes de T ça fait 1...

c'est rapide

[daring bough]

Oui?

A ok

Mais tkt pas j'ai fini

[nova spire]

PT = P c'est une affaire légèrement différente

P ce sera pas un multiple de (1 1 1)

parce que là c'est plutôt les vecteurs lignes qu'on regarde

et pas colonnes

[daring bough]

Oui mais c'est une matrice ligne

[nova spire]

oui

Mais T est semblable à sa transposée

encore un résultat que je sens que tu connais pas

bref

PT = P a des solutions non nulles

[clear cloud]

Ils sont pas forcément expert non plus, et reviennent sur des points que tu as abordé, saches que si il te parle d'un truc un peu plus loin, c'est que tout ce que tu as dit était très bien

Et oui le "jury innocent" aura forcément des questions moins avancé que le professeur dont c'est la matière

[daring bough]

Dacc mercu

Merci

J'espère que j'y arriverai

[clear cloud]

Parcontre verlhust est extrêmement pris comme sujet

Afaik

[daring bough]

Afaik??

[clear cloud]

As far as i know

[daring bough]

Ah je croyais que c'était quelqu'un...

[clear cloud]

Ducoup les profs de maths sont habitués à le voir et ça peut être plus compliqué pour toi

Enfin c'est comme ça que je le vois

[daring bough]

Mais après tout les mathématicien sont connus

[clear cloud]

Ouais mais au niveau lycée il y en a des plus connus que d'autre

Et pour le GO ça se bouscule pas

[daring bough]

C'est grave ???

[clear cloud]

Non

[daring bough]

Après sinon la prof pourra me dire ce qu'elle en pense

Je pense

[clear cloud]

Si tu as l'occasion oui c'est le mieux

[daring bough]

D'acc

[clear cloud]

Après c'est beaucoup pour voir si tu es à l'aise à l'oral

[daring bough]

Oui cv je me débrouille

[clear cloud]

Connaît bien ton sujet et c'est tranquille

Puis en plus c'est une chance sur deux

Soit l'un soit l'autre

Sinon le programme de spe normal c'est ok ?

[daring bough]

Normalement oui

Juste les équations différentiel il y a des trucs un peu bizarre faut que je m entraine dessus

Primitiver aussi

[clear cloud]

Ok, en vrai spam les anciens sujets de bac

[daring bough]

Mais il y a pas les équations différentiel

[clear cloud]

Du style ?

[daring bough]

Apres il y a les intégral on a pas vue

[daring bough]

Fin il y a des choses que j'ai pas trop bien compris en cours att

[clear cloud]

Ah je me souviens d'un vilain sujet Asie avec des equa diff

[daring bough]

A bon?

[clear cloud]

Ouais sur l'apmep

2023 peut être ou 2022

[daring bough]

Dacc je vais voir

La faut que je fasse ma géographie

Mais c'est vrai que les intégrales c'est la partie la plus dure?

[clear cloud]

Ça reste toujours le même type d'exercice, si tu connais les propriétés et que tu t'entraîne assez c'est tranquille

[daring bough]

J'avoue c'est ce que j'ai fais pour le bac blanc en maths et physique

Des exo en boucle pour comprendre

Ča a marché

[clear cloud]

Le bac c'est pareil, il change pas fondamentalement des exos que tu fais en cours

[daring bough]

Je trouve c'est plus dure

[clear cloud]

Le bac ?

[daring bough]

Oio certain truc

Oui

Fin sauf les probabilités c'est super simple

[clear cloud]

Il y a encore tchebychev ?

[daring bough]

Non

Je crois pas

[clear cloud]

Pas plus mal

[daring bough]

Sûrement

Bref je vais pas t'embêter plus que ça 😅

[clear cloud]

Ouah tkt hein, tu fermes le truc quand tu veux

[clear cloud]

C'est vrai que faire markov avec si peu c'est dommage

[daring bough]

On vien juste de commencer qussi

Aussi

Peut être on verra plus de chose

[clear cloud]

Ouais mais ça c'est des trucs de matrices que tu auras pas en terminale je pense

[daring bough]

Jsp

Il reste que 1 mois après

[clear cloud]

Tant que ça ?

[daring bough]

Bah on va bientot être en juin

Avec le bac

[clear cloud]

C'est vrai ça

Tu fais quoi en géographie?

[daring bough]

L Union Européenne 😅

Pk ça supprime?

????

Ça veut pas

[clear cloud]

De quoi ?

Ahhh

Mdr

[daring bough]

Sinon je vais etre pas a jour

[clear cloud]

Je sais

[daring bough]

Pk le mots supprime?

Retar

[clear cloud]

Le mot que tu veut marquer est une insulte en anglais

[daring bough]

Retardé oui

A ok je vois

Et toi t'es en vacances ?

[clear cloud]

Les vacances dans le superieur kekw

[daring bough]

Ah 🥲

J'ai l'impression que dans le supérieur on a plus de vie

[clear cloud]

Juste moins de vacances

En université au moins

[daring bough]

J'avais prévu d'aller en prepa on dirait que j'aurai plus de vie

[daring bough]

.close

[trim jolt]

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[topaz musk]

how come for part A we dont add 20 ??

wait i got it

.close

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[urban copper]

(-3,1,3) - (0,-1,1) = (-3,2,2)

,, \mathbb{L}_1 : (x,y,z) = \lambda(-3,2,2) + (-3,1,3) \ \mathbb{L}_2 : (x,y,z) = \lambda(-3,2,2) + (2,-3,-1) \ \mathbb{L}_3 : (x,y,z) = \lambda(1,0,-1) + (0,1,-1)

[solid kiln]

938c2cc0dcc05f2b68c4287040cfcf71

[urban copper]

,, \mathbb{L}_1 \cap \mathbb{L}_2 = \emptyset

[solid kiln]

938c2cc0dcc05f2b68c4287040cfcf71

[urban copper]

because they are parallel and not the same line

for L1 n L3 we have

i) -3x - 3 = y
ii) 2x + 1 = 1
iii) 2x + 3 = -y - 1

from ii) we have 2x = 0 ==> x = 0

there is no intersection between L1 and L3 either because we get an absurd

for L2 n L3 we have

i) -3x + 2 = y
ii) 2x - 3 = 1
iii) 2x -1 = -y - 1

from 2 we get 2x = 4 ==> x = 2

from 1 we get -6 + 2 = y ==> -4 = y

so there is an intersection

and it is a point I think

L2 n L3 = 2(-3,2,2) + (2,-3,-1) = (-6,4,4) + (2,-3,-1) = (-4,1,3)

-4(1,0,-1) + (0,1,-1) = (-4,0,4) + (0,1,-1) = (-4,1,3)

[trim jolt]

@urban copper Has your question been resolved?

[urban copper]

someone help for fucks sake

[trim jolt]

@urban copper Has your question been resolved?

[cyan frost]

could you translate this

[solid kiln]

Waes (Wires)

[cyan frost]

thanks

@pseudo goblet

what do you need help on

[urban copper]

@pseudo goblet can I get some help

[urban copper]

2

[leaden nacelle]

you want to find the line perpendicular to L that goes through P

that way all conditions are met

[urban copper]

can you elaborate

[leaden nacelle]

you are given a line L

and a point P

and conditions for L'

the third condition is that L' is perpendicular to L

[leaden nacelle]

the second condition is that P is part of L'

and the first condition is that L and L' intersect

[urban copper]

yeah L' needs to pass through P

(1,-1,1) - (-5,7,4) = (6,-8,-3)

,w (6,-8,-3)x(-1,1,3)

[solid kiln]

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Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[leaden nacelle]

$\mathbb{L}' : \lambda (a, b, c) + (d, e, f)$ where for some $\lambda_1$, $\mathbb{L}'(\lambda_1) = P$

[solid kiln]

Katharine

[urban copper]

do you follow

[leaden nacelle]

i'm not sure i follow your reasoning

[urban copper]

we know L' passes through P

and we know L' intersects L

one point that is in L is (0,-1,1)

if we subtract them we get a vector that is in L

maybe I am mistaken doe

for get it, is stupid

[leaden nacelle]

how would you go about finding a vector that is perpendicular to a line

[urban copper]

I am mistaken, forget it

[urban copper]

I did a mistake

[leaden nacelle]

Seperate from what you did before

we need to find the plane that is perpendicular to L because L' lies in that plane

[urban copper]

(x,y,z).(-1,1,3) = 0
use dot product we get a plane of vectors orthogonal to the direction vector of L

-x + y + 3z = 0

now find the intersection between L and this plane I guess, idk

[urban copper]

yes, but I am stuck

[leaden nacelle]

We can constrain this plane further

as P needs to lie in it

[urban copper]

?

[urban copper]

?

WTF are you talking about

[leaden nacelle]

You've given the plane perpendicular to L going through the origin

but is that the correct one?

[urban copper]

wdym?

[leaden nacelle]

There are technically infinite planes perpendicular to the line L

[urban copper]

ok

[leaden nacelle]

the one you've given is the one that goes through the origin

(0, 0, 0)

[urban copper]

Ax + Bx + Cz + D = 0
Ax + Bx + Cz = -D

[leaden nacelle]

you've put D to 0

[urban copper]

ok so I think I get what you mean

-x + y + 3z + D = 0
P = (-5,7,4)
5 + 7 + 12 = -D
12 + 12 = -D
24 = -D
-24 = D

[leaden nacelle]

yes

so now we know the exact plane in which L' lies

[urban copper]

-x + y + 3z - 24= 0
-x + y + 3z = 24

[urban copper]

very interesting problem tho

now find the intersection of L with this plane, no?

I havent checked with GPT what does AI say

[leaden nacelle]

thas nice

[urban copper]

what is M

midpoint

[lime sphinx]

nope but we want to get M to specify L’

[urban copper]

M is (0,-1,1)

[lime sphinx]

M is the intersection between the plane you just found and L

[urban copper]

ohh

right

(x,y,z) = (-k, k-1, 3k + 1)
-x + y + 3z = 24
k + k -1 + 3(3k+1) = 24

2k-1+9k+3 = 24
11k+2 = 24
11k = 22
k = 2

(x,y,z) = (-2, 2-1, 6 + 1) = (-2,1,7)

so M = (-2,1,7)

now you want me to find PM x the normal of the plane

?

or we are done?

[lime sphinx]

yeah PM is a vector in the plane normal to L but containing P and it will specify L’

[lime sphinx]

modulo some verifications if you wish this should construct L’ with the desired properties

[urban copper]

lets verify after we find L'

[urban copper]

I got lost again

wtf you mean?

[lime sphinx]

i haven’t verified your computations, but that should be the right idea to construct L’

[urban copper]

what idea?

we did some stuf with kat aswell

we did a BUNCH of stuff lmfao

[lime sphinx]

get the normal plane contaning P then find M then construct L’

that’s the chain of ideas to construct L’

[urban copper]

what M did you got?

doesnt matter, we will verify later lmfao

my M = (-2,1,7)

[urban copper]

now what?

I got the plane containing P

-x + y + 3z = 24

P = (-5,7,4)

just find PM x N

N being the normal of the plane

no PM is already the direction of L'

we dont need cross product

so we can use the vector PM as the direction vector of L' and use either P or M as one of the points for the parametric equation of the line L'

[lime sphinx]

bingo

[urban copper]

OK

this exercises are always like this crap

[lime sphinx]

at least this one is in R^3

[urban copper]

https://tenor.com/view/i-have-played-these-games-before-seong-gi-hun-squid-game-i've-played-these-games-already-netflix-gif-16480153790313233198

[urban copper]

?

is it possible in R4 or Rn to have an exercise like this or wdym