help-28

[weak pelican]

i like subtracting lambda

[fast peak]

lambda I

[weak pelican]

subtracting lambda I

so if you want to know if a vector is an eigenvector in the matrix

ya

[fast peak]

if you want to know whether a vector is an eigenvector then you should just multiply it by A

and check whether you get the same vector but scaled back

[weak pelican]

so A(v) - lambda(v) = 0

[fast peak]

note that you dont even need to know what lambda is yet

just Av

with the example above, if you do A*(3,2,1) you get (9,6,3) and it is easy to notice that this is 3 times the original vector

so therefore (3,2,1) is an eigenvector with eigenvalue 3

[weak pelican]

yea

[fast peak]

you did not need to know lambda before

[weak pelican]

oh so A(v) - lambda(v) = 0 is to find out the lambda

[fast peak]

are you fucking listening to me?

just Av

no Av-lambdav

[weak pelican]

oh sorry

sorry im going to go ahead and close this now

.close

[full forum]

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[young heart]

Hi, I am looking for help with DH parameters. I think I am close but something is not quite right. I found all of these paramaters and then checked them but I am not getting the correct output for forward kinematics so i don't think my format is correct

[young heart]

My layout looks like this but that table is not completely up to date, and the screenshot is what the arm looks like so I feel like those should be the correct DH values

Here is a more updated photo of the math

[full forum]

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[tranquil belfry]

need help after finding the area of the shaded region underneath the x axis

i get that to be 1/4 units

since I solved for when y = 0 of x^3-x and got the x int to be 1

so integrated the function over that interval (0 to 1) and got 1/4

since the area there is supposed to be equaul to the integral of x^3-x from (1 to b)

i did x^3-x = 0

since we subtracting 1/4 when we do the 1 part since thats already 1/4 as we calculated earlier

so the -1/4 and + 1/4 cancel

and we are left with [x^3-x] = 0 where x is some b value

so i am now stuck

cuz this was the first thing i solved and the three roots are -1,0 and 1

neither of which make sense given the picture

[proven dirge]

so it should intercept with the axis at $x(x-1)(x+1)=0$ so $x=1$

[glossy valve]

Gizmic

[proven dirge]

so you found the area by taking the integral from 0 to 1 for $x^3-x$ and got $\frac{1}{4}$ ?

[tranquil belfry]

yes

[glossy valve]

Gizmic

[tranquil belfry]

well -

but yeah

[proven dirge]

did you get -1\4 instead?

[tranquil belfry]

yes

but area is a positive number i thought

cuz how can the area above the x axis be -?

[proven dirge]

it is yeah so you take the absolute value in this case

[tranquil belfry]

from 1 to b

yeah

ok

[proven dirge]

well what you should do is take the integral from 1 to B and set that equal to 1/4

[tranquil belfry]

why 4?

i set it to 1/4

[proven dirge]

$\int_1^B(x^3-x )dx=\frac14$

[glossy valve]

Gizmic

[tranquil belfry]

ye

[proven dirge]

you're right

[tranquil belfry]

taths what i wrote

[proven dirge]

so what did you get when you did that

[tranquil belfry]

so then wouldn't it be js x^3-x = 0

cuz when we do the part where x^3-x where x = 1 thats the same as 1/4 but since we are subtracting that would cancel out?

ohwait

no

i am stupid

its adding

so 1/2

[jolly horizon]

what do you get when you take that integral

[tranquil belfry]

so x^3-x = 1/2

where x is some value b

[proven dirge]

you should get $F(x)=[\frac{x^4}{4}-\frac{x^2}{2}]_1^B$ right?

[glossy valve]

Gizmic

[tranquil belfry]

yea

[proven dirge]

if you take the integral

[tranquil belfry]

[(b^4)/4] - [(b^2/2)] = 1/2

[proven dirge]

yeah looks right

then solve for B

[tranquil belfry]

ohhh

wait i forgot to yeah

ok

[proven dirge]

note that it should only be positive and greater than 1

if you get multiple sols

[tranquil belfry]

im getting x^4 -2x^2 -2 = 0

[proven dirge]

factorise

[tranquil belfry]

isn't the descrim <0?

no

nvm

ok

idk how to factorize this though

quadratic right

well no its not a quadratic eq

what i do

[proven dirge]

it is a quadratic

[tranquil belfry]

oh

[upper valley]

Idk what r u doing but solve delta

[tranquil belfry]

1+- root 3

[proven dirge]

if you want to make it clearer you can substitute $x_2=x^2$ and you'll get $x_2^2-2x_2-2=0$

[glossy valve]

Gizmic

[tranquil belfry]

obv plus tho

so 1 + root 3

[proven dirge]

that's correct but we should have also set $x^2 = 1 \pm \sqrt{3}$

[glossy valve]

Gizmic

[tranquil belfry]

no its root 2

oh

yeah

true

wait no

why x^2?

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[clear galleon]

Hello, does anybody know what will happen on a Trig Function Graph if the period is negative?

[tight mortar]

the period being “negative” will have the same graph as it being positive

[clear galleon]

I've just done this question right here and I'm confused on why the period isn't negitave

[clear galleon]

Can you explain that to me?

[tight mortar]

except flipped

like

basically imagine the period as being a speed control

if the period is 2x the graph “takes twice the time”

if the period is 0.1 the graph “takes a tenth of the time”

[clear galleon]

I don't think of it like that but I get what you mean continue

[tight mortar]

now what happens if the period is negative?

[tight mortar]

you’re not meant to normally lol

[clear galleon]

the period will go backwards????

[tight mortar]

it’s like rewinding

YES

[clear galleon]

the graph will go backwards???

[tight mortar]

so it’ll be like

flipped

do you notice how like

the graph of tan normally goes to the top-right

but there it’s going to the top-left

!!

because it’s like it’s going backwards

but it takes the same “amount of time”

[clear galleon]

Bro's onto something....

But wait I don't understand

if that's so

then why the Period he wrote is 2pi

is it mistake

it should be -2pi

[tight mortar]

well

technically the period is the absolute value of the coefficient

[clear galleon]

Is it really?

I never knew

[tight mortar]

but when the period is negative it’s like the graph is backwards which is what u asked lol

note none of this is rigorous or like

[clear galleon]

Right so now is the period basically going to make it kind of turn into Cotan?

[tight mortar]

formally correct at all

well no

[clear galleon]

sorry

I meant

Cotan

[tight mortar]

umm

yea I think

in fact tan(pi/2-x)=cot(x)

[clear galleon]

is this what you're trying to say?

[tight mortar]

uhh that’s a form of it

[tight mortar]

this version is more common lol

but that form might be slightly more relevant

[clear galleon]

wait brother so let me get this straight

if A, B, and the X are negative then the tan graph will have the cotan shape?

[tight mortar]

umm wait wdym by a and b

[clear galleon]

Y= Atan (B(x-h)+d

That's how my teacher teaches it at least

[tight mortar]

ok

if ab is negative yea it looks like cotan

because tan/cotan are odd

[clear galleon]

Are what?

[tight mortar]

odd

which means tan(-x)=-tan(x)

[clear galleon]

what does odd mean

[tight mortar]

even functions would be like f(-x)=f(x)

like odd- and even-degree x^n

that’s a story for another day tho

[clear galleon]

Okay brother thank you

Do you mind if I close this?

@tight mortar

[tight mortar]

ya?

lol

[clear galleon]

.close

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[tame ivy]

need help with a tiling with perfect squares problem, cant show the image but i need to dm about it

[tame ivy]

for reference,

[short goblet]

Wdym tiling with perfect squares?

[tame ivy]

this is what it looks like

there are square tiles that make another square, and the tiles dont repeat

[short goblet]

Not sure I understand the problem😅

[tight mortar]

all the tiles are squares

they fit perfectly into another square

[tame ivy]

also about the stuff i need to dm

its not you doing the work its more you check my work because

something about it is wrong(because these perfect square problems can be solved algebraically)

[tight mortar]

why do you need to dm?

[tame ivy]

im not allowed to publicly show my work because its for a

in basic terms its for a competition(unrelated to math btw, but it would ruin the integrity if someone saw my work)

[safe trench]

So you are basically cheating?

[tame ivy]

no

im allowed to ask others for help

[acoustic forum]

Wow what am emoni

Emoji

[tame ivy]

the difference is consent basically

[safe trench]

I don't even know where it came from. Surprised myself

[tight mortar]

google emoji kitchen

[safe trench]

Wait what? It was an accident. Now I don't know how to do it

[placid sapphire]

17

[full forum]

@tame ivy Has your question been resolved?

[full forum]

@tame ivy Has your question been resolved?

[full forum]

@tame ivy Has your question been resolved?

[full forum]

@tame ivy Has your question been resolved?

[torn jolt]

@full forum whats the scientific notation?

[sudden dock]

i wanna know if this is right

[full forum]

@tame ivy Has your question been resolved?

[frail zinc]

someone help if i dont find the awnser im going to die

[amber robin]

It is a right angled triangle

Use pythagoras theorem

[soft dew]

Is it really?

[amber robin]

Yes

Always, the diameter and any point on the circle make a right angled triangle

Yk how if that angle is theta, that one would be 2 theta

[amber robin]

Similarly, 2 theta is 180° for that case making theta 90°

[tame ivy]

this is still my channel

[amber robin]

😅 sorry

[ember viper]

I guess you could use heron's formula....

[ember viper]

oh oop

[full forum]

@tame ivy Has your question been resolved?

[tame ivy]

this isnt directly related

but if you can prove or disprove 8a+16x+5a+12x=14x+8y+5x+2y+3x+y+a+4x+3a+8x=14x+8y+5x+2y+2x+y+x+y+x+2y+10x+8y=5a+12x+3a+8x+10x+8y

that would tell me if im wasting my time or not

[austere cove]

@tame ivy https://www.wolframalpha.com/input?i=+8a%2B16x%2B5a%2B12x%3D14x%2B8y%2B5x%2B2y%2B3x%2By%2Ba%2B4x%2B3a%2B8x%3D14x%2B8y%2B5x%2B2y%2B2x%2By%2Bx%2By%2Bx%2B2y%2B10x%2B8y%3D5a%2B12x%2B3a%2B8x%2B10x%2B8y

WA thinks the only solution is a = x = y = 0

[tame ivy]

it cant be 0

or 1

sadly

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@tame ivy Has your question been resolved?

[eternal dawn]

x = 306526303 ( can be anything)

if i add to x a number only having 1s and is as long as the x number has digits
will the result always be a number without any zeros?

(x can be any number)

[raven imp]

Like x = 9999 ?

[raven imp]

If I understand correctly

[onyx glen]

!occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

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@tame ivy Has your question been resolved?

[tame ivy]

it got resolved

i just bruteforced it

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[kind wraith]

Hi

[quaint prawn]

hi sir

what is your doubt

[kind wraith]

I want some help

From you

[quaint prawn]

with what sir

[kind wraith]

I am not a teacher

[quaint prawn]

Oh well what is your doubt

[kind wraith]

I am just a student

What mens doubt

[silver pecan]

What is your question

Doubt meaning something you are unsure of

[quaint prawn]

yes exactly

[kind wraith]

See it's vary easy

2×5+5-658÷28695

Here my question

Any one Indian here

[silver pecan]

I'm not

[quaint prawn]

i am from madras

[quaint prawn]

Why can you not solve this or use a calculator?

,w 2*5+5-658/28695

[glossy valve]

Results provided by WolframAlpha

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

Requested by austinu

[silver pecan]

Or if you need to not use a calculator make use of the order of operations

And long arithmetic

[kind wraith]

Ok Austin can you speak Hindi

[kind wraith]

I am just checking is this server is really good

[silver pecan]

This is wrong, it used a fraction bar instead of a division symbol

[kind wraith]

You have studied the set chapter

[quaint prawn]

ys maderchod

[kind wraith]

Ye betichod kahe bokradi kar rha ha

Ham Bihar se ha

[keen vector]

bro

what's going on

[quaint prawn]

😂

[keen vector]

ru actually indian

[quaint prawn]

ys

[keen vector]

yea i believe you 😂

[kind wraith]

He fucked offf

[keen vector]

how old ru

[kind wraith]

14 years

[quaint prawn]

ofc he from noth india

😂

that why he cant do arithmetic

[kind wraith]

Yes I am from

I am from Bihar

[kind wraith]

Bhai dekh ham bas chek kar rhe the samjha

[ocean jackal]

Bro racist much?

💀

[kind wraith]

Bro seee be friendly okk

Don't be abuse anyone

[quaint prawn]

ok

i apologize

i wont abuse

[kind wraith]

That's the friendship

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@kind wraith Has your question been resolved?

[kind wraith]

Ofcourse

[full forum]

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[kind wraith]

What does it mean

[full forum]

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[kind wraith]

I didn't understand

[kind wraith]

118520718180

[full forum]

@kind wraith Has your question been resolved?

[shrewd falcon]

what is this? your mother's number?

[kind wraith]

Just byy mistake it's has typed

I didn't understand 6th question

[shrewd falcon]

,rccw

[shrewd falcon]

What is the set W?

And do you know a finite vs. infinite set?

[kind wraith]

Yes it's the basic

[shrewd falcon]

k

so which one do you think it is

[kind wraith]

Can you solve 6th num all questions

[shrewd falcon]

nah

[kind wraith]

Why

[shrewd falcon]

but it should be clear enough as to which are finite vs. infinite

[shrewd falcon]

!noans

[full forum]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[kind wraith]

No I am just asking the procees

[shrewd falcon]

!nosols

!nosols

[full forum]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[kind wraith]

What does it mean

[shrewd falcon]

read it

[kind wraith]

It's not what are you thinking

Just 13 number question

[shrewd falcon]

,rccw

[kind wraith]

Bro what does it means

I don't understand your language

[shrewd falcon]

it means im gonna explain it step by step rather than giving a solution

[shrewd falcon]

then lowkey just go find another server

[amber robin]

It's a command to rotate the image

[full forum]

@kind wraith Has your question been resolved?

[pure basin]

Where is 13 number question?

[amber robin]

,rotate

[amber robin]

@kind wraith what don't u understand in the 6th question?

[kind wraith]

I haven't read the sentence properly that's why

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[full forum]

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[torn jolt]

this equation has 0 solutions, is there a way for me to determine quickly that this equation isn't solveeable?

[void nova]

I don't think so

You could factorize the numerator, if you want

And see that you get a first degree equation which has no solutions

But I don't know if that's what you wanted as an answer

[torn jolt]

well the way I solved it is that I multiplied the denom on both sides then factorized on the left and subtracted the terms from each other which leads to something like -3 is 0

actually now that I notice

if I am able to subtract two quadratic terms from each other that means it's unsolveable generally no?

[void nova]

Wait, by doing this you should get a first degree equation that has x = -3 as solution

But then you notice that -3 makes the denominator 0, hence it's not acceptable

[void nova]

You only get a first degree equation, which might or not have solutions

[torn jolt]

ah so I check if the solutions don't divide by zero

[void nova]

Wdym?

[torn jolt]

so I solved it normally then see if the solution doesn't lead to division by zero

[void nova]

Well, this is only for this equation of course

Not generally true

[torn jolt]

this is pretty confusing

[void nova]

I don't know how to explain it better, sorry 😐

Let's wait for some other helpers, I guess

[torn jolt]

let me show you how I did it wait

@void nova

[void nova]

You lost a (x + 3)

Or,in other words,why did you cancel that (x + 3)?

[torn jolt]

because this term exists on both sides and I subtracted the full expression on the right from the left

[void nova]

I believe you did some steps in your head wrongly

Try to write it again, showing all steps

[torn jolt]

okay

@void nova

bruh I tried a different method with subtracting both sides

[gritty flax]

3 - (-6) is 9

[torn jolt]

oh right

then I get -3 as answer

yeah

so it's invalid because division with it leads to 0

but what did I do wrong when subtracting both terms when factorized?

[void nova]

Yep

That you didn't do the factorisation correctly

You canceled an extra factor for no reason

[torn jolt]

can you plesse show me how its done on paper

ig not

.close

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[glossy valve]

devthemasked

[onyx glen]

is modular arithmetic allowed

[pure basin]

Uh yeah

I added and subtracted n

[onyx glen]

ok then look at the values of this for n=0,1,2,3,4

and verify that mod 5 they all reduce to 0

[shadow surge]

this bot helps with math?

[onyx glen]

what bot?

[pure basin]

No

Texit I think

[shadow surge]

texit

[onyx glen]

no it doesn't

[pure basin]

It only puts your question

[onyx glen]

it can only WRITE math

[covert heath]

yeah kinda

[onyx glen]

and sometimes call wolfram alpha

[shadow surge]

aa okay thanks

[onyx glen]

or do basic calculations

[covert heath]

it has access to the wolffram api

[onyx glen]

off topic tho

[shadow surge]

oh okay

[pure basin]

Also can we do it how I did

[onyx glen]

can also go for n = -2,-1,0,1,2 to make arithmetic a bit less scary

[pure basin]

$n^5-n+5n$

[glossy valve]

devthemasked

[pure basin]

$n(n^4-1^4)$

[glossy valve]

devthemasked

[onyx glen]

sure, then you will just be evaluating n^5 - n instead

or you could do some factorization stuff sure

[covert heath]

valid so far sure

[pure basin]

Yeah so finally I get $(n-1)(n)(n+1)(n^2+1)+5n$

[glossy valve]

devthemasked

[onyx glen]

you can even do a trick

[covert heath]

sheesh i dont think thats helping

[onyx glen]

n^2 + 1 = n^2 - 4 + 5

[covert heath]

do you know FLT

[pure basin]

What is FLT

[covert heath]

i think itd be helpful

[onyx glen]

so your thing becomes

n(n+1)(n-1)(n^2-4) + 5n(n-1) + 5n

[covert heath]

not last lol

little

[onyx glen]

fermat's little theorem

[pure basin]

nope

[covert heath]

i dont wanna spoil the whole thing but FLT does the trick

[onyx glen]

and this in turn becomes

(n-2)(n-1)n(n+1)(n+2) + 5n(n-1) + 5n

which is also a viable route

[covert heath]

$a^p \equiv a \pmod p$, where $p$ is a prime.

[onyx glen]

\pmod{p}

[covert heath]

dammnit.

[pure basin]

We both are still learning latex

[glossy valve]

Percy

[covert heath]

anyway yeah that

[covert heath]

im kinda decent at it i just dont do modular arithmetic much

[covert heath]

you can try flt-ing from here if your instructor is chill with that.

[onyx glen]

...

[covert heath]

the proof is kinda wonky but you'll be able to get it ig

[pure basin]

How did you get

[covert heath]

spoonfeeding perhaps, i just figured i might as well just tell em if they dont know flt

[onyx glen]

n^2 + 1 = n^2 - 4 + 5

[pure basin]

Yea got that

[onyx glen]

oh sorry

5n(n-1)(n+1)

not just 5n(n-1)

does not even matter much anyway since it's a multiple of 5 either way

[covert heath]

just use flt please

arithmetic is what we resort to when we wish to sell our soul to the devil

[pure basin]

If we need to use a theorem we have to prove it

I don't wanna remember all proofs

[slender onyx]

if OP didnt learn flt then we need other strats

shame tho, flt is indeed very fast here

[pure basin]

omg just realised your a senior moderator

[covert heath]

never too late to learn it though

[slender onyx]

imagine a pinker being in a help channel

[onyx glen]

op said he knows modular arithmetic and then proceeded not to follow my suggestion

[onyx glen]

https://tenor.com/view/gojo-jjk-pink-role-gif-701518254404567742

[covert heath]

if you're memorising proofs that's kinda concerning

[pure basin]

How did you get 5n you took the 5 out? from the n^2-4+5?

[onyx glen]

yes i took the 5 out

5*(rest of the factors)

[pure basin]

so 5n(n-1)(n+1)(n^2-4)

[slender onyx]

looks like theyre doing ur 2nd suggestion so all good ig

[pure basin]

Then I just take 5 common right

[pure basin]

?

[onyx glen]

let me get back home and i will show what i did in full detail

what are your pronouns btw

[pure basin]

how does it matter?

[hushed briar]

😆

[pure basin]

What

[hushed briar]

Sorry, I couldn't help laughing

[pure basin]

Why?

Like what is funny?

[onyx glen]

i referred to you with he/him here but then realized that i don't know whether it's correct

and you didn't correct me

but it does not hurt to ask

if you don't give a shit you can just say you don't give a shit :P

anyway

[slender onyx]

‘they’ never hurts

[onyx glen]

one moment

[slender onyx]

so i never have to ask 🙂

[pure basin]

Ok so just tell this one step is correct or not

Then I need to leave

[pure basin]

.

[onyx glen]

that's not how i would have called it

[pure basin]

There is a +5n also

[onyx glen]

oh that's what it was

i hid it in the ... bc i didn't remember nor care what it was

cause i didn't touch it

[pure basin]

oh ok

[covert heath]

We have \begin{align*}
n^5 + 4n \pmod 5\
&\equiv n^5 - n +5n \pmod{5}\
&\equiv n^5 - n \pmod 5 + 5n \pmod{5}\
&\equiv 0 \text{ by flt} + 0 \text{ because 5n \pmod{5} \equiv 0}
\end{align*}

QED.

But uh okay don't use flt ig /lh

kill me

[pure basin]

yea so all of these terms

Just don't try latex

Are divisible by 5

Ok got it

Thanks

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[glossy valve]

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

[covert heath]

yeah

whats up with that

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[sly prairie]

Hey

I'm not sure how to prove it

I translate

if int(f) converges and if f is decreasing then lim t f(t) =0

f has to be continuous and positive

And I have to prove it if it's true

since I haven't found any counterexamples I assume it's true but I don't see how to start

[rustic frigate]

Are you deducing that or is that one of the assumptions?

[amber robin]

t tends to 0 and f(t) also tends to 0. So, t f(t) should also tend to 0

Is that not enough?

Oh, wait no

Shit

[sly prairie]

I'm deducing

[amber robin]

T tends to infinity

My bad

[sly prairie]

maybe it's just false

[rustic frigate]

Why should f have to be continuous for example, what’s your argument?

[sly prairie]

oh sorry

No, f has to be continous and positive

i'm not deducing it ^^

[rustic frigate]

Oh so it’s one of the assumption in your problem, right?

[sly prairie]

yes

[tired quarry]

if f wasnt continous, it wouldnt be true

not sure if positive is necessary

[sly prairie]

f is continous

I'm not sure I was clear ^^

I have f, wich is continous and positive

and I have this assumption

[slate violet]

I'm jumping in to say that $f(t)$ must behave like $\frac{1}{x^p}$ for $p > 1$

[glossy valve]

south

[covert heath]

does that say 'f is decreasing'?

okay you translated it

nice

fuck my french

[rustic frigate]

the integration bound is from 0

[slate violet]

from the p-test for convergence (the summation and the integral must either both converge or both diverge), well the lower limit of the integral is 0 not 1 but that shouldn't matter

[sly prairie]

here the exercice

[slate violet]

just take t large enough since the limit is $t \to + \infty$

[glossy valve]

south

[covert heath]

i see

thanks

[slate violet]

like that's the idea

I'd probably mess up if I tried to point you on how to prove this more formally

[rustic frigate]

Oh you’re saying it behaves like it convergence wise

[sly prairie]

yes

so for you guys it's true

[unique wagon]

i dont remember the exact proof but my prof did cover this question once

im guessing its a popular result in analysis, i could find u a source

[sly prairie]

it would be very cool

but I had an idea with series

[unique wagon]

https://math.stackexchange.com/questions/562987/for-monotonic-f-if-the-improper-integral-int-0-infty-fxdx-converges-th

lol second google search

[slate violet]

the idea should be to bound $f(t)$ above by $k \cdot \frac{1}{t}$, $\forall k \in \mathbb R^+ \forall t \in [1, \infty)$

[sly prairie]

is that completly out of context or good?

[glossy valve]

south

[sly prairie]

thank you!

[sly prairie]

okay I think I have an idea

thank you all!

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[frigid crag]

If I have a converging geometric series C, where Cn=An+Bn (A and B are geometric series), is it true that both A and B converge?

[onyx glen]

the sum of two geometric series is not itself a geometric series in general

unless A_n and B_n happened to have the same common ratio

[frigid crag]

I first assume that C is a geometric series so it means that has to be correct

How does it help that they have the same common ratio?

[onyx glen]

hold on

[frigid crag]

Oh I think I got it

[onyx glen]

ok this is sounding sus

!xy

[full forum]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[frigid crag]

Alright one moment

[tired quarry]

Cn=0
Bn=(-1)^n
An=-1*(-1)^n

[covert heath]

xy??

[tired quarry]

Counter example

[onyx glen]

http://xyproblem.info/

[onyx glen]

oh yeah i guess

[covert heath]

AH lmao

[frigid crag]

Can=0 isn't a geometric series

Cn

[onyx glen]

waiting for you to show the original problem tho

[frigid crag]

Yes my bad

[tired quarry]

it is, with c1=0 and ratio equal to, let's say, 1

[frigid crag]

"Let A, B be infinite geometric series. Prove or disprove:
If (a_n+b_n) converges, then both A and B converge."

Couldn't provide a screenshot unfortunately

[covert heath]

hmmm

[onyx glen]

ok right so

at no point does it say that the sequence a_n + b_n itself has to be geometric.

anyway, tomzizek's counterexample still holds

[frigid crag]

But only geometric sequences can converge

[onyx glen]

what

what are you talking about

do you deny the existence of convergent series which AREN'T geometric

like at all

[covert heath]

right so thats super wrong

[frigid crag]

We haven't been taught that at least

[onyx glen]

if you want something more drastic: $a_n = 2^n, b_n = -2^n$

[glossy valve]

Ann

[onyx glen]

$\sum a_n$ and $\sum b_n$ both diverge but $a_n + b_n = 0$

[glossy valve]

Ann

[frigid crag]

We've also been taught that the series 0,0,0,0... Isn't a geometric series

[tired quarry]

question, what does it mean that sequances converge?

[onyx glen]

ok but 0+0+0+... is a convergent series

[frigid crag]

When the sum approaches a finite number

[onyx glen]

geometric or otherwise

[covert heath]

i mean i think it is, but if it isnt then yeah counter example

[onyx glen]

i think this is a misunderstanding on your part

[frigid crag]

Here's what I'm thinking, If the sum converges, either C1 (C being the series of the sum) or it's ratio are smaller than 1, and that can be used to prove either A1, B1 or their ratios (which must be equal) are less than one, meaning they both converve

[onyx glen]

a huge one

[onyx glen]

you are overthinking it

[frigid crag]

There's a good chance that's true

[onyx glen]

you are massively overthinking it my dude

[frigid crag]

Okay could you explain it one more time

[onyx glen]

but first off you DO realize that there is more to the world of series than just geometric ones

right?

[frigid crag]

Yes there's geometric and the addition one which idk the name of

[onyx glen]

there's more than just geometric and arithmetic ones.

[frigid crag]

Those ones I haven't learned and thus dont need to use in this question

[onyx glen]

ok

let me give you my counterexample once again

$a_n = 2^n$ and $b_n = -1 \cdot 2^n$

[glossy valve]

Ann

[onyx glen]

do you agree or disagree that these are both geometric series

[frigid crag]

Agree

[onyx glen]

do you agree or disagree that these both diverge

[frigid crag]

Yes

Agree

[onyx glen]

do you agree or disagree that $a_n + b_n = 0$ for each and every $n$

[glossy valve]

Ann

[frigid crag]

2^2 + -1 isn't 0 though

[onyx glen]

2^2 + (-1) is not 0 but where would you see that exact sum?

[frigid crag]

n=2

[onyx glen]

$a_2 + b_2 = 4 + (-4) = 0$

[glossy valve]

Ann

[frigid crag]

Why is bn -4?

[onyx glen]

$b_2 = -1 \cdot 2^2 = -4$...

[glossy valve]

Ann

[frigid crag]

Ah the image cut off I see

I understand this example but we were explicitly told that 0,0,0,.. is not a geometric sequence

[covert heath]

wait until you hear about this bloke called fibonacci

[onyx glen]

ok but so what.

[onyx glen]

so what.

the question does not say that (a_n + b_n) must itself be a geometric series

only that it must converge.

[frigid crag]

But the only sequence we've been taught that converges is a geometric sequence

[onyx glen]

do you agree or disagree that the series 0 + 0 + 0 + 0 + ..., however you wish to classify it, is still convergent?

[frigid crag]

Wait do arithmetic series converge aswell?

Ohhhhhh

[onyx glen]

in general, no.

[frigid crag]

Yes I agree it's convergent

[covert heath]

This one does

[onyx glen]

ok well in that case

theres your counterexample

without a care in the world as for geometricity

[covert heath]

https://en.wikipedia.org/wiki/Convergent_series#Examples_of_convergent_and_divergent_series

one google search is all it takes

i mean hell we find e and pi using a convergent series

[onyx glen]

surely your professor didn't do something crazy like "YOU SHALL NOT EVER DECLARE A SERIES CONVERGENT IF IT IS NOT GEOMETRIC" or sth

[frigid crag]

No but I believe we defined a convergent series as a geometric series that (...)

[covert heath]

sad $e$ noises.

[glossy valve]

Percy

[frigid crag]

Either way I understand now, thank you

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[onyx glen]

you did??

[frigid crag]

Yep

[onyx glen]

.reopen

[full forum]

✅

[onyx glen]

wait can i see your notes tho

cause like

[frigid crag]

They're not in English

[onyx glen]

what language are they in

[frigid crag]

Hebrew

[covert heath]

F.

[frigid crag]

Yea

[onyx glen]

rip

[covert heath]

I was like 'I can kinda get german and french!'

Freaking Hebrew

[onyx glen]

but i dont think your prof would have dismissed all non-geometric series as automatically disqualified for convergence.

unless they are crazy.

[frigid crag]

That's not out of the picture

[covert heath]

or just really really oversimplifing

idk

what grade are you in

[frigid crag]

Ninth but I take these university courses

[covert heath]

Hm interesting

[frigid crag]

Confusing mostly

[covert heath]

Well just for future reference

e literally is the sum of a convergent series

as is pi

[frigid crag]

Right, isn't e the sum of the series (n-1)!/n² or something similar

[onyx glen]

1/n!

for n from 0 to infinity

[frigid crag]

Close enough

[covert heath]

not geometric

[frigid crag]

And it does converge

Oof

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[lapis kite]

Why do we differentiate x then multiple it with y THEN integrate

To find area under the curve you integrate Y to the respect of x

Now both y and x equations are given in terms of t

So for some reason we differentiate x
Then multiple the diffrenatation of x with y then integrate

So instead of ∫ y dx
We basically solve for ∫ y*f'(x) dt

[robust slate]

You can Google this

https://tutorial.math.lamar.edu/classes/calcii/paraarea.aspx

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[lapis kite]

Tysm
Someone also sent me this

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[static fiber]

2cos3x trying to get the value but it shows 0. But desmos shows that x value 0 starts with y=2

[robust slate]

You’re doing (2\cos 3) \times 0$ and not $2\cos(3 \times 0)$

[glossy valve]

Civil Service Pigeon
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[static fiber]

Ohhh

ohhhh

Thanks

Wait so do i do this all the time

I kinda don't get how to input it on calculator sometimes

Like for example the given gets longer

[robust slate]

You’re taking the cosine of 3 times 0 in its entirety

So put that in parentheses

[static fiber]

OHH

So i just substitute the x value to the x

I just realized LOL

I didn't get it on our first discussion it wasn't even explained

Thanks, I get it now

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[static fiber]

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[amber whale]

How can I solve the cubic 45x^3+585x^2+2475x+3375 without using cardanos method?

[frozen fiber]

,w hcf(45,585,2475,3375)

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[slate violet]

then for x^3 + 13x^2 + 55x + 75

rational root theorem tells you that the only zeroes are the factors of 75/1, including the negative ones

you should be able to just guess a root r and then divide by x - r using your preferred method

[onyx glen]

did you miss the =0

[onyx glen]

in fact the negative ones are the only ones you could hope for, as this cubic is clearly positive for x > 0

[slate violet]

nice observation

the generalisation of course would be Descartes' rule of signs

[amber whale]

Thx

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[stuck fiber]

This is the derivative of f(X)

[thick hedge]

Awesome

[stuck fiber]

So sub in x = 0 and we get -1

[onyx glen]

f'(0) = -1?

[stuck fiber]

Yeye

[thick hedge]

Yea? (0-0-1)e^0 =-1 *1=-1

[stuck fiber]

My question is, I know we need to get arctan of both slopes to get the answer in degrees

I did arctan(-1) = 45 degrees and arctan(1/7) = ~8 degrees

But I did 45 - 8 = 36

[onyx glen]

need

[stuck fiber]

They wanted 45 + 8 = 53

[onyx glen]

first off arctan(-1) = **-**45°.

[stuck fiber]

Ohhhhhhh

That was my mistake

[onyx glen]

second we dont rly need to introduce arctan bs at this juncture imo

[stuck fiber]

How would you do it?

[onyx glen]

find the angle between the vectors (7, 1) and (1, -1)

via dot product

[stuck fiber]

Also you said 'bs' do you think this method is unreliable or something?

[stuck fiber]

Ok 1 moment

[onyx glen]

a little bit unreliable yes

like you already introduce the potential for rounding errors

[stuck fiber]

sqrt(7^2 + 1^2) = sqrt(50)
sqrt(1^2 + (-1)^2) = sqrt(2)

cos(theta) = ab/|a|*|b|

[dry coral]

um

read that again

[stuck fiber]

Oh oops

[onyx glen]

7^2 + 1^2 is not 8

also you should not indicate dot product with juxtaposition

[stuck fiber]

Im not really sure what you mean by this sry

cos(theta) = 6/(sqrt(2))(sqrt(50))

cos(theta) = 6/(sqrt(100))

theta = arccos(3/5)

[stuck fiber]

Ah I see

Oh cool

Ok thanks!

❤️

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[fossil wedge]

Isn't this wrong? Take Z2 x Z2 for example. Then g1 = 1 g2 = 1 and gcd(o(g1), o(g2)) = gcd(2, 2) = 2. Yet all the subgroups of Z2 x Z2 are of the form M x N where M, N are subgroups of Z2

[rapid rain]

I'm not sure how $<(1,1)>$ can be written as $M\times N$

[glossy valve]

rafilou is not not born in 2003

[rapid rain]

I didn't see your problem was with the other way

but yeah if you find g1, g2 such that gcd(o(g1),o(g2)) is not 1

then <(g1,g2)> is a counterexample

[fossil wedge]

Subgroups of Z2 x Z2 are: Z2 x Z2, <(1,1)>, Z2 x {0} = <(1,0)>, {0} x Z2 = <(0,1)>, <(0,0)> = {0,0}. Correct?

[fossil wedge]

Now I see, thanks!

Also, if I'm asked to find the subgroups of Z9 x Z2 x Z2 for example, is there a quick way to do this? Z9 x Z2 x Z2 is isomorphic to Z18 x Z2 and if we take for example g1 = 3, g2 = 1, gcd(o(g1), o(g2)) = gcd(6, 2) = 2, so not all subgroups are of the form M x N

We know that for all M <= Z18, N <= Z2, M x N is a subgroup of Z18 x Z2 (right?) but by the theorem above there must be more subgroups of Z18 x Z2 that are not of the form M x N

So how do you solve an exercise like this?

Or the subgroups of S3 x Z6, how do you do it?

In general how do you find the subgroups of G x H if the theorem doesnt hold? @rapid rain thanks if you can answer 🙃

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@fossil wedge Has your question been resolved?

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@fossil wedge Has your question been resolved?

[full forum]

@fossil wedge Has your question been resolved?

[fossil wedge]

Is this correct?

I tried tò find the subgroups of S3 x Z6 too by saying that S3 x Z6 is isomorphic to S3 x Z3 x Z2, and gcd(card(S3 x Z3), card(Z2)) = 1 so by using the same theorem I used above, the subgroups of S3 x Z3 x Z2 are direct products of subgroups of S3 x Z3 and Z2 respectively. I found the subgroups of S3 x Z3 that can be written as a direct product of a subgroup of S3 and a subgroup of Z3 respectively, which are: for order 18, S3 x Z3, for order 9 <(123)> x Z3, for order 6 <(12)> x Z3, <(23)> x Z3, <(13)> x Z3, {id} x Z3, for order 3 <(123)> x {0}, {id} x Z3, for order 2 <(12)> x {0}, <(23)> x {0}, <(13)> x {0}, for order 1 {id, 0}

But then there can be hybrid subgroups that cannot be written as the product of a subgroup of S3 and a subgroup of Z3, right? Because S3 and Z3 don't have coprime orders. If so, how do I find these "hybrid" subgroups?

[full forum]

@fossil wedge Has your question been resolved?

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@fossil wedge Has your question been resolved?

[fossil wedge]

<@&286206848099549185> please

[opaque pier]

God damn. Sorry, can't help.

Good luck

[fossil wedge]

Ok no problem, thanks

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@fossil wedge Has your question been resolved?

[winter rune]

well, for example, if you have GCD(o(g1),o(g2)) =\= 1, the the subgroup generated by (g1,g2) is an example of such a hybrid subgroup

[fossil wedge]

I'm sorry, why is that?

[winter rune]

because the order of (g1,g2) is less then the product of orders of g1, and g2

so the |subgroup generated by (g1,g2)| < |subgroup generated by g1| * |Subgroup generate by g2| = |subgroup generated by g1 X subgroup generated by g2|

[fossil wedge]

So if <g1> * <g2> is a subgroup of G1 X G2, then <(g1,g2)> is a subgroup of <g1> * <g2>, so <(g1,g2)> is a subgroup of G1 x G2

[fossil wedge]

In the example here, if we take <(123)> x Z3 = <(123)> x <1> for example, then <((123), 1)> is a subgroup of S3 x Z3. Similarly, from <(12)> x <1> we get that <((12), 1)> is also a subgroup and so on...

I'm not sure why these always form subgroups though... |<(g1,g2)>| divides |<g1>| * |<g2>| by the definition of least common multiple: |<(g1,g2)>| = o(g1, g2) = lcm(o(g1), o(g2)) = (o(g1) * o(g2)) / gcd(o(g1), o(g2)), and |<g1>| = o(g1), |<g2>| = o(g2).

But how can we be sure they form subgroups? Do we have to check the properties of closure and inverses every time?

[winter rune]

Well the possible elements of <(g1,g2)> is <(g1^i,g2^i)> this is a subset of <(g1^a,g2^b)>
And since it is generated, it must be a group

So it's a subgroup

[fossil wedge]

Okay yeah I understand, and are there other "hybrid" subgroups or are these all?

What if we have subgroups that are not cyclic products? For example in S3 x Z3 by using the theorem we got <(123)> x <1>, <(12)> x <1>, etc... which are all cyclic products and this lets us find the subgroups <((123), 1)>, <((12), 1)>, etc... but if we had non-cyclic product subgroups this wouldnt be possible

I'd like to know if there's a general rule to find hybrid subgroups of direct products

[stiff summit]

The gcd of card(S3 x Z3) and card(Z2) is the gcd of 18 and 2, which isn’t 1 though?

[fossil wedge]

yeah my mistake...

[stiff summit]

By the way, this isn’t a rule or something you can apply by hand, but you could use the software GAP to find all subgroups of a small group, if you want to check your answers with it

[fossil wedge]

So if we have two subgroups G and H, and gcd(|G|, |H|) is not equal to 1, there is no rule on how to find the "hybrid" subgroups of G x H?

[stiff summit]

I didn’t say there was no rule

[fossil wedge]

I mean rules I can apply by hand

I can't use any software, I need to be able to solve these problems in the exam too .-.

[winter rune]

there are others, you can use subgroups generated by multiple elemnets

every subgroup is generated by some elements

so let us say you are finding subgroups of G1 X G2, you just need to find <(g1,g2) (g3,g4) (g5,g6) ... (g2n-1,g2n)> such that there exists some combination of g2,g4,...,g2n, k such that (1,k) is not in the group

(assuming abelian, because i dont have much intuition on non abelian groups, probly you can generalise)
if you choose a minimal set of generators, that means
g2^(a1 o(g1))g4^(a2 o(g3))...g2n^(an o(g2n-1)) is not all posibilities

hence
gcd(o(g1),o(g2)) or gcd(o(g3),o(g4)) or ... or gcd(o(g2n-1),o(g2n)) is not 1
(assuming the set of generators is minimal, and we are talking about abelian groups)

[graceful arch]

(a+b)3

@graceful arch

[full forum]

@fossil wedge Has your question been resolved?

[woven pumice]

hey can anyone explain me trignometry

[amber robin]

Ask in #geometry-and-trigonometry

Or make a new help channel

!occupied

[full forum]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[full forum]

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[formal summit]

If i have a set of the 12 elements {2,2,1,1,1,1,1,1,1,0,0,0} and i draw 5 elements, then without putting them back draw 5 elements again, how do I calculate the chance that on at least one of the draws the sum of the elements is at least 5 (example: 0+1+1+1+2)

[onyx glen]

first off that's a multiset rather than a set

but. hm.

!nosols

[full forum]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[onyx glen]

you posted an answer before you edited that

...

i saw what you posted dude don't try to make a fool of me

anyway

[onyx glen]

i would consider finding the probability of the complement instead

cause it doesn't look like there are a lot of ways to draw 2 sets of 5 numbers without replacement with totals at or below 4...

[formal summit]

I already know about complements, I just dont know how to calculate any of this.

i mean i dont know how to calculate P(both draws<5)

I mean, I want to know how to calculate it. I already wrote a simulation to get the solution

This is just using complements when already having partial solutions. Maybe I am not explaining right what I am looking for...

[onyx glen]

considering the complement, it looks like the lowest that the total across both draws can POSSIBLY be is 7 (taking the ten lowest numbers), while the highest is 8 (bc each draw must be at most 4)

so it's either 3+4, 4+3 or 4+4, and i am pretty sure all these cases can be enumerated by hand

[pure basin]

Just being in the server for a lot of time

But most commands are related

[full forum]

@formal summit Has your question been resolved?

[formal summit]

Ok this is useful but what I am wondering is if the problem were more complicated, such that I couldn't enumerate the number of cases by hand, is there a formula to still solve it?

[onyx glen]

probably not

best way becomes crunching it on a computer

[formal summit]

Ok, cause right now I write simulations that just do the thing 100,000 times and assume that I am very close to the mathematical result. I was wondering if a more elegant approach is available

Thanks for the response!

[full forum]

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[stone mortar]

I have a question.
a relation is R transitive
if (a,b), (b,c) and (a,c) belongs to R
is that right?

[stone mortar]

?????????

[onyx glen]

that's not the definition of a transitive relation, no.

[stone mortar]

are you sure about that

[onyx glen]

yes.

[twilit leaf]

oops im wrong

[stone mortar]

and you would not like to give me the correct defination?

[earnest dagger]

what are the quantifiers on a, b, c here

[stone mortar]

a,b,c belongs to R'

R' is set of real numbers

and relation R is defined from R' to R'

[onyx glen]

uh now you are overusing the letter R

[stone mortar]

call it R'

[onyx glen]

anyway

a relation $R$ on a set $X$ is called transitive if it satisfies the following property:

for all $a, b, c \in X$, \textbf{if} $(a,b) \in R$ and $(b,c) \in \bR$, \textbf{then} $(a,c)\in R$.

[glossy valve]

Ann

[stone mortar]

hmmm

interesting

[rapid rain]

basically, as long as there is a "path of relations" from a to b, then a is in relation with b

for transitive relations

[stone mortar]

hmmmm so phi(empty set) is transitive afterall

[earnest dagger]

what is phi

[stone mortar]

empty set

[earnest dagger]

yes the empty set is a transitive relation

vacuously

[rapid rain]

yes

[stone mortar]

interesting

and...

who is rafilou

[rapid rain]

because "for all a,b,c such that (a,b) in empty set and (b,c) in empty set" the rest is vacuous

[rapid rain]

me?

[stone mortar]

and how do you know he was not born in 2003

[rapid rain]

he is not not born in 2003

because it's me

anyways

[void nova]

Are you trolling?

[rapid rain]

!done

[full forum]

If you are done with this channel, please mark your problem as solved by typing .close

[stone mortar]

no

[onyx glen]

you should probably not call the empty set "phi" unless you like being misunderstood

[stone mortar]

was curious about the name

[stone mortar]

what?

its called phi here

[rapid rain]

although $\emptyset$ looks like $\phi$, don't call it phi

[glossy valve]

rafilou is not not born in 2003

[rapid rain]

call it emptyset to avoid confusion

[stone mortar]

descrete is so confusing

my teacher call it phi

and its written as $\phi$ in books

[glossy valve]

Info

[rapid rain]

then they will be burned on the stake when it's their time

(/j)

[stone mortar]

in my books*

[earnest dagger]

it's not written $\phi$ the probably use something more like ∅

[rapid rain]

.

[onyx glen]

in badly typeset books

[earnest dagger]

it's usually a zero or an o with a slash through it, but almost always typeset differently from phi

[onyx glen]

but then again i know a lot of books out there are made with quite poor typesetting quality

[stone mortar]

call it whatever... why does it matter

[rapid rain]

for others to understand you

[onyx glen]

just saying that if you say "phi" on here and it turns out to mean empty set

[stone mortar]

hmm

[onyx glen]

then people will get confused

like i can speak about this server at least

[rapid rain]

I could call breakfast "glorgle" for all I care, but when I ask for my glorgle in the morning I might be surprised

[stone mortar]

so what is the symbol for phi really called(so I can show off in class 😎 )

[gritty rose]

$\emptyset$ also exists

[glossy valve]

riemann

[onyx glen]

it is just called "empty set"

bc that is its one and only meaning

[rapid rain]

I was there

[gritty rose]

oh yes

[stone mortar]

no but it is "denoted by a 0 with a slash"

what is the "0 with a slash" called? a 0 with a slash?

[earnest dagger]

I would call probably it the "empty set symbol". wikipedia has an article for https://en.wikipedia.org/wiki/Null_sign

[gritty rose]

$\nullsign$

[glossy valve]

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

[gritty rose]

Ohno

[earnest dagger]

$\varnothing$ vs $\emptyset$

[glossy valve]

zkzach

[onyx glen]

the symbol $\varnothing$ denotes one thing and one thing only, and so it does not have a separate name from the thing it denotes.