help-41

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[cerulean tapir]

can anyone help me?

<@&286206848099549185>

.closr

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[amber wasp]

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[severe basin]

Thank you

.close

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[amber wasp]

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Remember:
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[keen pawn]

could I have a hint

[keen pawn]

*hint

[rotund oak]

I believe you are trying to show that this matrix AB will not have full rank, and it’s giving you a hint about the kernel

I maybe could have been more clear in that

That may be too much of a hint or not enough of one, idrk

[weak zinc]

What is theorem 1.3.1, anyway?

[rotund oak]

I’m assuming that it’s something about if there’s more columns than rows then there exists some non trivial solution to Ax = 0

[amber wasp]

@keen pawn Has your question been resolved?

[keen pawn]

Shit

Forgot i had ths channel open

Sorry chat

[amber wasp]

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[warped hemlock]

Is this correct if I want the k to start from 1

[coral wigeon]

yes

[warped hemlock]

thanks

.close

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[amber wasp]

Available help channel!

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Remember:
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[vale vigil]

how do i multiply this polynomial (factor) this

[plain coral]

back

how much time do you have to complete this?

[vale vigil]

im studying for

factoring quiz tomorrow

but im not sure if I'll have that problem on the quiz

but i just want to make sure i know how to do it

[plain coral]

you dont know how to factor x^2 + 2x + 1 though

[vale vigil]

i got the answer for that problem but its super long work

wait

[plain coral]

the answer you got which was (x + 2)(x - 1) is different, thats x^2 + x - 2

[vale vigil]

is that wrong?

ih

oh

umm

[plain coral]

so can you factor x^2 + 2x + 1

[vale vigil]

waitimma try again

[plain coral]

yea

[vale vigil]

(x+1)(x+1)?

[plain coral]

yea thats correct

now are you capable of factoring 3x^2 + 4x + 1

[vale vigil]

i was thinking about something else

okay ill try

i think its wrong

i got (3x+3)(3x+1)

😭

[plain coral]

did you notice that 3 + 1 = 4

[vale vigil]

yaa

[plain coral]

so you can write it as 3x^2 + 3x + 1x + 1?

[vale vigil]

ya

[plain coral]

so what does 3x^2 + 3x factor to

[vale vigil]

this?

hii??

[tawdry lintel]

It's just 3x(x+1)

[vale vigil]

oh yaa

i need help multiplying this

61

[tawdry lintel]

Sure, what are you struggling with

About 61

[vale vigil]

i got my answer but its super long work

but i have a timed factoring quiz tmr

idk if there is a shorter way to do it

the super long at the bottom is the answer

[tawdry lintel]

That's just one of those things that you will have a long answer and will take a moment to do

[vale vigil]

but the quiz tomorrow i only get 3m30s for 5 questions

idk if that question will be on the quiz

[tawdry lintel]

That's a little evil

[vale vigil]

if youre saying so then.. its maybe not on the quiz

[tawdry lintel]

My only recommendation would be to skip the 2nd step and just write out every term from the original

[crude kite]

It's the way you're showing the work, it's redundant at times

[vale vigil]

ohh..

yaa so i was wondering if there is a shorter way

[tawdry lintel]

You would still be doing the same thing, just going straight to writing out every term instead of breaking it apart first

[crude kite]

Instead of writing that long line, just go straight to what I have

[tawdry lintel]

Step 2 of your work can be skipped entirely

[crude kite]

Step 3 too

[tawdry lintel]

Yes, it would just require doing the math in your head for each term

Considering you only have 3.5 minutes, I would try doing that

[vale vigil]

so what do i do first

[crude kite]

If you followed what you drew here, and just do the math mentally, it'll be faster

[vale vigil]

im bad at foiling , those long steps i got from chat gpt 😭

[crude kite]

You do -x^2 * 3x^2 mentally to get -3x^4

[vale vigil]

okay

[tawdry lintel]

First term multiplied by every term, then the second term multiplied by every term and so on

[crude kite]

Like I said, just follow this process of the lines

[tawdry lintel]

Going one term at a time, you can keep track by drawing those lines above like you did

[vale vigil]

oh

okayy

like that??

[crude kite]

Yes

[vale vigil]

okay okay

[tawdry lintel]

Go one line at a time, draw a line from term to term once you have already calculated the answer

[vale vigil]

okay thank you so much

.close

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[glass ridge]

@tawdry lintel hey

do you know how to do this

[pallid canopy]

#❓how-to-get-help

[amber wasp]

Please do not ping individual helpers unprompted.

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Remember:
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[noble kestrel]

Hello

[strange field]

hi

[strange field]

bye

[noble kestrel]

<@&268886789983436800>

[strange field]

wut

[silent bobcat]

?

[strange field]

why

[noble kestrel]

You spam useless things

[strange field]

who

[silent bobcat]

a little hasty but yes please avoid posting unhelpful stuff in the help channels

just because people are hoping that someone can come and help them, seeing hi and then bye is a little jarring

thanks

[strange field]

i came to help but as soon as i saw question (i cant do it) i said "bye"

was joking, sorry

[silent bobcat]

icic

[noble kestrel]

for question a)

I don't understand why {1}^c ∉ R

[night jacinth]

{1}^c is {2,3}

and {2,3} isn't in R

@noble kestrel

nothing fancy

[noble kestrel]

Ok so for it to be in R I need to have only one element in the singleton
{2}^c and {3}^c is not in R too

And the simple example speaking of the empty element ?

[night jacinth]

indeed

yeah

well the empty set has 0 elements inside of it

so it's not in R (as you said, all the sets in R have 1 element)

[noble kestrel]

but why say that the fact that it is not in R is sufficient to say that R is not a tribu ?

[night jacinth]

cause the definition of a sigma-algebra requires its presence

recheck your notes if you're unsure

[noble kestrel]

Ok I understand

Thank you for help !

[amber wasp]

@noble kestrel Has your question been resolved?

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[amber wasp]

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[raw oyster]

how is it impossible?

[raw oyster]

is it cause of the fact that we stated that a/b is an irreducable fraction?

[half rover]

yes

[raw oyster]

oh

ok

thx

.close

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[amber wasp]

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[wraith shale]

Is question 3 DNE?

[wraith shale]

Thank you

[south stag]

why would it be?

[tulip tapir]

DNE is when the left- and right-hand side limits converge to different values

but here they both go to 3

so no, not DNE, just 3

[wraith shale]

It is correct

thank you you two

havea good day

.close

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[amber wasp]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[winged spindle]

1 divided by root of x^5 * root of x = 8

[winged spindle]

iom conufsed

wouldnt the laws of expoinents make it x^-6 = 64 if u simplify

[crisp stratus]

$\frac{1}{x^5 \sqrt{x}} = 8$

this?

[grizzled pagoda]

artemetra

[winged spindle]

this

[crisp stratus]

oh

[crisp stratus]

yes

or just x^-3 = 8

[winged spindle]

and u do 64^1/6

well -.6

[crisp stratus]

yes

[winged spindle]

it gives the wrong answer

though.

the answer is 1/2

[crisp stratus]

i am not sure what happened here

[crisp stratus]

if you go with this

you get x^-1 = 2

so x=1/2

[winged spindle]

so i did ^2 to get rid of all the root

roots

would that not work

then use laws of exponents x^-6

[crisp stratus]

huh?

okay i mean

yeah

64^(-1/6) is still 1/2

[winged spindle]

oh yeah

i got it now

complicated it for myself

with the ^2 since its no calculator allowed

.close

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[amber wasp]

Available help channel!

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• Show your work, and if possible, explain where you are stuck.
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[keen pawn]

Not sure how to start (a)

The given any two points is confusing me

I was thinking $\forall x [ P(x) \land P(y) \implies \exists ! L(m) [ O(x,m) \land O(y,m)]]$

where x is an object on the caretesian plane

[grizzled pagoda]

Veni, vidi, perii

[amber wasp]

@keen pawn Has your question been resolved?

[keen pawn]

<@&286206848099549185>

[cunning birch]

ok so there's some idea there

but what you wrote is "given any x, ..." when the sentence is supposed to be "given any 2 points..."

you also didn't specify that x and y are disjoint points

and last thing

you wrote 'there exists a unique property applied to m (...)' when we don't know what m is

when you should say "there exists a unique object that is a line and..."

[keen pawn]

Ah

I see

Thanks

. close

[rotund oak]

.close

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[wraith shale]

I am appreicated for any response

[solemn escarp]

squeeze theorem maybe?

[spare belfry]

Squeeze

[wraith shale]

oh thx

[amber wasp]

@wraith shale Has your question been resolved?

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[craggy horizon]

can someone pls solve the H1 problem i have an issue with it

[craggy horizon]

<@&286206848099549185>

[amber wasp]

@craggy horizon Has your question been resolved?

[craggy horizon]

<@&286206848099549185>

[craggy horizon]

<@&286206848099549185>

[cyan agate]

<@&286206848099549185>

[amber wasp]

@craggy horizon Has your question been resolved?

[neat wind]

use the fact the total area of a parallelogram is base times height

[craggy horizon]

I believe I got it let me check once again

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@craggy horizon Has your question been resolved?

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[umbral prairie]

im trying to solve the equation 3^x = 2y² + 1 for (x, y) in N²

[umbral prairie]

i found 4 solutions: (0, 0), (1, 1), (2, 2) and (5, 11) up to x = 3000000 using a computer search

[crisp stratus]

those are probably it ahah

but i'll check

[umbral prairie]

also, if we let a and -a be the 3-adic roots of the polynomial 2X² + 1, and consider a solution (x, y), if we suppose wlog y = a [3], then by hensel's lemma y = a [3^x], but y < sqrt(3^x / 2) so that means a must have y preceded by a load of zeros (almost as many digits as y) in its 3-adic expansion

(note that i don't exclude there being a much simpler way to solve the equation)

also a theorem about repeated zeros in the digit expansions in algebraic p-adic numbers might not give an actual bound on the solutions

ik there is such a theorem for algebraic real numbers, but it doesn't give a bound, only proves that there is a finite number of them

also according to wikipedia its based on the borel cantelli lemma

[amber wasp]

@umbral prairie Has your question been resolved?

[umbral prairie]

<@&286206848099549185>

[neat wind]

there exists such solution to this problem here
(posting directly bcz this is not smth i would ever come up with on my own)

[umbral prairie]

oh okok

wow

thx

[neat wind]

yw

[umbral prairie]

also

is there always a finite number of solutions to equations of type p^k = P(n) for a given p > 1 and a polynomial P of degree at least 2?

[neat wind]

there may need to be a few additional restrictions as 2^x = y^2 has infinitely many for example

[umbral prairie]

oh right

i think if p is prime and P only has simple roots in Z/pZ it should work

mb weaker restrictions even

[amber wasp]

@umbral prairie Has your question been resolved?

[umbral prairie]

im kinda new to the server, so idk really what i should click

the original question has been resolved

but the generalization hasn't

[amber wasp]

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[normal dove]

was solving a trigonometric equation

got stuck at this

[normal dove]

what do i do now?

binomial?

theorem or smth?

any other way that is really tedious 😭

[warm glacier]

Yes

It should work

All the odd powers of x will get cancelled

[normal dove]

yeah but

any other way

?

[warm glacier]

Idk

Binomial is pretty good

[normal dove]

hmm

ping if anyone finds

[amber wasp]

@normal dove Has your question been resolved?

[amber wasp]

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[split sail]

can someone help me how i get from the first circled part to the second
finding the derivative using the d/dx

[patent raptor]

do you know power rule

[split sail]

yes

[patent raptor]

that's literally it

[split sail]

im having like a hard time like

i dont even knwoi how to explain it

applying the rules

i understand them

[patent raptor]

try colors

[grizzled pagoda]

bacc

[split sail]

I see

Oh i so totally see

that is literally it wow

[grizzled pagoda]

bacc

[split sail]

thank you

[patent raptor]

yea

[split sail]

i just have a hard time visualizing i appreciate you

[patent raptor]

color

it

[split sail]

.close

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[amber wasp]

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[ocean torrent]

need help with 8c please

[quick ridge]

do you know the formula for the sum of a geometric series

[ocean torrent]

yup

[mint wraith]

$u_1 + u_2 = 20 \to u_1 + u_1 \cdot r = 20 \ S_4-S_2 = u_3 + u_4 = 180 \to u_1 \cdot r ^2+ u_1 \cdot r^3 = 180$

[grizzled pagoda]

MetuMortis

[mint wraith]

does that look right to you @ocean torrent

[ocean torrent]

yep

it makes sense

[mint wraith]

do you think you can do something with following equations to find u_1 or r?:

$u_1 + u_1 \cdot r = 20 \ u_1 \cdot r ^2+ u_1 \cdot r^3 = 180$

[grizzled pagoda]

MetuMortis

[amber wasp]

@ocean torrent Has your question been resolved?

[ocean torrent]

im not sure

[mint wraith]

$u_1 + u_1 \cdot r = 20 \ r^2(u_1 + u_1 \cdot r) = 180$

[grizzled pagoda]

MetuMortis

[ocean torrent]

ahaa so r^2 (20) = 180

[mint wraith]

yes

[ocean torrent]

thank you so much!

[mint wraith]

you're welcome

[amber wasp]

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[strong granite]

i was hinted to flip the bits somehow, but i cant quite wrap my head around it

[amber wasp]

Please don't occupy multiple help channels.

[strong granite]

so the previous one was the one i did earlier, i think u had a look already

but 12 here is what im trying to wrap my head around

[pallid canopy]

can you close one of these two
https://discord.com/channels/268882317391429632/486844829209198613

[strong granite]

yes, sorry

done, its closed

[amber wasp]

@strong granite Has your question been resolved?

[amber wasp]

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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sharp dock]

yo

[amber wasp]

Please don't occupy multiple help channels.

[neat wind]

yo

[strong granite]

any thoughts on my work here #help-41 message

[amber wasp]

@sharp dock Has your question been resolved?

[sharp dock]

I need help myself 😭

With these 2 questions

[amber wasp]

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[left sage]

can someoen tell me how to solve this

[left sage]

im stumped..

[tepid minnow]

I won't tell you how to solve it but I can walk you through it

If I asked you to solve $x^2+4x+3$ would you know how to do that?

[grizzled pagoda]

NeptuneRises

[left sage]

I would factor

But the issue is I’m not sure how to factor it in this

With the x^4 and x^2

[tepid minnow]

Okay so when we factor $x^4+4x+3$ we turn it into $(x+3)(x+1)$ right\
What would happen if we instead had $x^4+bx^2+c$ What would that factor to

[grizzled pagoda]

NeptuneRises

[left sage]

That’s where I guess I get. Confused hmm could we just factor it like

Sorry let me do it in a moment I’m in a car

Sec

[tepid minnow]

All good

[left sage]

Okay back

Let me take a look

Okay I’m not sure

@tepid minnow

So like

I’m trying to do the strategy where we find a number that adds up to -117 and multiples to 2196

But I’m not sure if that’ll be applicable at all here

Or it might be but I’m not sure the best way of finding it

[tepid minnow]

That would be applicable but it would have to be used a little differently

[left sage]

How would I use it here

[tepid minnow]

We can say $x^4+bx^2+c = (x^2+x_1)(x^2+x_2)$

[grizzled pagoda]

NeptuneRises

[tepid minnow]

So instead of it factoring into $(x+a)(x+b)$ its $(x^2+a)(x^2+b)$

[grizzled pagoda]

NeptuneRises

[left sage]

Uhm

So would I root the numbers? I’m not sure

This is so strange my teacher did not teach this at all

That’s my punishment for taking an online class

[tepid minnow]

So you would have to root the numbers but in this case the roots should be easy to calculate

Looking at a different question may make it easier to understand

If I had $x^2-6x+5$ what would this factor into

[grizzled pagoda]

NeptuneRises

[left sage]

Wait how would that be possible I thought it had to add up to -6 and multiply to -5 right?

But how would that be possible here

-5x1 would be -4

[tepid minnow]

$(x-5)(x-1)$

[grizzled pagoda]

NeptuneRises

[tepid minnow]

(Also thats my fault I misstyped the +5 as -5)

[amber wasp]

@left sage Has your question been resolved?

[left sage]

Ok yes that makes sense

That’s what I’ve been doing so far in algebra

[amber wasp]

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[stone coral]

.help

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Commands:

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[stone coral]

.reopen

[hallow vault]

I don't understand why prof said that y can take values 1,4,9 etc?
that's just not possible given x is negative integer and y=x^4
also our pmf is coming out to be completely different,
their pmf has one more term in it where they consider negative of quadratic root of y

[hallow vault]

<@&286206848099549185> i've got a test in the next hour, please help early if u can

[hollow raven]

Hi I was needing some help simplifying an expression. 2 over (7/6x)+7

[hallow vault]

in another help channel please

.close

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[nocturne sonnet]

Hi! I just wanted to make sure my answers worked out. Here is the question and here is the work.

[nocturne sonnet]

I did
No solutions: never
Infinitely many solutions: when a = -2
Unique solution: never

[weak zinc]

second row, third entry

[nocturne sonnet]

Uh oh

Kk then i think it’ll be
No solutions: never
Unique solution: when a ≠-2
Infinitely many solutions: never

[weak zinc]

You're right for no solutions, as the equation you started off was homogenous, the solution (0, 0, 0) always works

Why do you think you'll never get infinitely many solutions?

[nocturne sonnet]

Infinitely many happens when there’s a row of all zeros right

I don’t think it’s possible to get that

the 2 values of a are different and the second row has 13 in it 😅

[weak zinc]

Welllllll...

[weak zinc]

...remember this here?

[nocturne sonnet]

I think Im slow sorry that should’ve been a 1

Not 13

[weak zinc]

Anyways, even if it was 13, you could've divided that row by 13 to make it a 1 anyway

[nocturne sonnet]

So that means we have a free variable

Which means infinitely many solutions 😅

Always

[weak zinc]

Not always, no-

[nocturne sonnet]

Won’t that row always be 0 0 1 0

[weak zinc]

It would be, but the thing is-

You should have then gotten to, as the final part on your paper here,
[
\pmqty{1 & 3 & 2 \ 0 & 0 & 1 \ 0 & -6 - 3a & 1 + 2a}
]
(ignoring the right column cause it's all zeros and nothing ever changes with it)

[grizzled pagoda]

@weak zinc

[nocturne sonnet]

Yes

I have that rn

[weak zinc]

It really doesn't matter what the value of a is, you can use that second row to always get rid of that 1 + 2a (and also the 2 above it too, for what it's worth)

So you can e.g. get yourself to, say
[
\pmqty{1 & 3 & 0 \ 0 & 0 & 1 \ 0 & -6 - 3a & 0}
]

[grizzled pagoda]

@weak zinc

[weak zinc]

(you don't need to clear that 2, but you may as well, cause it's easy to do )

[weak zinc]

Anyways, as you were saying

[nocturne sonnet]

wait sorry how did u use row 2 to get rid of 1+2-

1+2a**

[nocturne sonnet]

But yes I guess once that’s done, you have a row of zeros which mean you have infinitely many solutions as long as as a=-2

[weak zinc]

Subtract (1 + 2a) * row 2 from row 3

[nocturne sonnet]

As there would be a row of all 0s

So that means:
No solutions: never
Infinitely many solutions when a=-2
One unique solution: when a ≠ -2 !? Because if a doesn’t =-2 we have a free variable and we don’t have like the row of 0s

[weak zinc]

Yep

a being 2 means you can get a free variable, a not being 2 makes (0, 0, 0) the only solution

[nocturne sonnet]

Thank you sm !!

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[strong granite]

hi

[strong granite]

can i get some guidance here please

im not quite sure how to approach this

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@strong granite Has your question been resolved?

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@strong granite Has your question been resolved?

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[split sail]

the pythagorean identity say sinx+cosx = 1 or sinx +sin(90-x) = 1 whichch is sin(x+(90-x)) = sin 90 =1

[split sail]

are my deductions correct?

[shadow stump]

the pythagorean identity says that [ \sin^2(x) + \cos^2(x) = 1 ]

[grizzled pagoda]

cloud

[shadow stump]

the sine of the sum of two angles is also more complicated: [ \sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y) ]

[grizzled pagoda]

cloud

[split sail]

no

sinx+cosx not sin(x+y)

[lusty pine]

$\sin(x)+\sin(90-x)\neq \sin(x+(90-x))$

[split sail]

oh i forgot square

[grizzled pagoda]

Skill_Issue

[split sail]

oh

got my errotrs

.chthanks

.close

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[livid bolt]

Can somebody please help me understand this proof?

[grizzled pagoda]

Tommy

[lethal oasis]

What part dont you understand

[livid bolt]

wait a minute, i'll write it on paper

[livid bolt]

This first part is correct, right?

Then why does it say that the sequence {x_i^(k)} is Cauchy?

Wouldn't we need to have ( |x_i^{(k)} - x_i^{(m)}| < \epsilon )?

[grizzled pagoda]

Tommy

[lethal oasis]

Yeah but because x^k is cauchy, d(xk,xm)<e and then |x_i^k-x_i^m| <= d(xk,xm) < e

So x_i^k is also cauchy

[livid bolt]

oh ok, right

[livid bolt]

can you please explain the last part? where it says "and thus d(x^(k), x) tends to zero as k-->infinity, so the sequence {x^(k)} converges to x in R^n"

i understand the inequality but not what it implies

[livid bolt]

Sorry to bother you, is this correct?

[lethal oasis]

Just some small details if youd want to write it down correctly/rigorously but essentially yes

[livid bolt]

Thanks very much, and does it matter that we get d(x^(k), x) < n * epsilon instead of d(x^(k), k) < epsilon?

Epsilon is arbitrary so we can make it sufficiently small anyway

Right?

[lethal oasis]

Yeah but it has to be the same epsilon

[livid bolt]

What if we write abs(x_i^(k) - x_i) < epsilon/n for some epsilon > 0 and k sufficiently large?

Then we add all the terms and we get that d(x^(k), x) < epsilon

Is that doable?

[lethal oasis]

Yeah

But not that nice

Since you’d have to explain that e/n also can be any arbitrary number

[livid bolt]

But I've seen proofs where you have two terms which you know can be made arbitrarily small and by summing them you get that their sum is less than epsilon. For example if you know that a-->0 and b-->0, then abs(a+b) <= abs(a) + abs(b) = epsilon/2 + epsilon/2 = epsilon

[lethal oasis]

Yeah I mean you can do that

Ah forget what I said

Yeah you can do that

Actually the preferred way 😅

[livid bolt]

Okay

Thanks very much for the help 🙂

I appreciate it!!

.close

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[dim belfry]

hi

am dead

[amber wasp]

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[dim belfry]

WHAT

.close

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[tawny harbor]

real anal

[tawny harbor]

can someone please explain why "0 otherwise"

for applying Cauchy-Hadamard theorem

wait am i rlly rlly rlly dumb

it's cuz it doesn't appear in the series right?

[amber wasp]

@tawny harbor Has your question been resolved?

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[regal gust]

How do I integrate this?

[river falcon]

Take the constant out, complete the square in the denomimator, use the standard integrals and you should get your answer

Are you stuck at a particular point?

[regal gust]

I stuck when I want to integrate for the numerator and denominator

[river falcon]

Ok, what have you tried?

[regal gust]

Should the 2 be taken out too?

[regal gust]

I did this by flipping the equation

[river falcon]

And what are you trying to do with that? Keep going?

[regal gust]

I don’t know, I just tried

[river falcon]

Ok

So you can't invert like that

[regal gust]

Oh

My bad

[crisp trellis]

you can invert the polynomial by changing its exponent to the negative of its initial value, but you cannot invert constants

[regal gust]

Will this work?

[split sail]

Hi

[river falcon]

no because integral laws say that thats $$2 \int 2 + (1-3x)^{-2} dx$$ which isnt the original integral

[grizzled pagoda]

RadMeerkat62445

[regal gust]

Ooh, so I can’t simply add them just like that?

[river falcon]

No

Rather, try taking the 2 outside the integrand. You should get
$$ 4 \int (1-3x)^{-2} dx $$

[grizzled pagoda]

RadMeerkat62445

[river falcon]

Can you try using power rule on this? How would you do that?

[regal gust]

Sure let me try it

[river falcon]

From step no. 4 onwards remove the integral sign and dx because you're solving the integral directly

[regal gust]

Here is my final answer

[river falcon]

This looks correct

Except for the constant

when taking $$4 \int \frac{(1-3x)^{-1}}{3} dx$$, changing the sign of the (1-3x) term doesnt need you to invert the 3 as well

[grizzled pagoda]

RadMeerkat62445

[regal gust]

So 3 will stay at the denominator?

[amber wasp]

@regal gust Has your question been resolved?

[river falcon]

yes

[amber wasp]

@regal gust Has your question been resolved?

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[mental dome]

I wanted to try to understand how to do this using polar form

[quick ridge]

x = rcos(t), y = rsin(t) , x^2 + y^2 = r^2

[patent raptor]

Dootud said it basically

This looks odd

[split sail]

what is sen?

[mental dome]

sin

[quick ridge]

that’s a e bro

[mental dome]

im brazilian

[quick ridge]

ahh

[mental dome]

here we write sen

[patent raptor]

Did you do x = rsinθ and y = rcosθ instead

[quick ridge]

not everyone can be right i guess

[mental dome]

better?

[quick ridge]

yea this is just very wrong

[mental dome]

😦

what do i do now

[split sail]

hi my name is Madi

[mental dome]

hi madi 🫡

[split sail]

a little bit about me is I'm am in 8th grade and I am in accelerated math

[patent raptor]

can you calculate flux integrals

[mental dome]

????

[split sail]

what

[patent raptor]

[mental dome]

hey madi, this chat is for math questions

[split sail]

Ik this is why I joined

[mental dome]

#❓how-to-get-help read this

[split sail]

k

[mental dome]

<@&286206848099549185>

[split sail]

?

[mental dome]

what do i do now

[split sail]

oof sry

[split sail]

myguy my brain is not functioning

maybe tmrw

[patent raptor]

are u supposed to solve this system of equations?

[mental dome]

Yep

[grizzled pagoda]

bacc

[patent raptor]

\begin{align}
5r^2\cos\theta +5\cos\theta &= 12r \
5r^2\sin\theta -5\sin\theta &= 4r
\end{align}

[grizzled pagoda]

bacc

[patent raptor]

We could solve one equation for r and plug that into the other equation

Say we do that for (1)

[grizzled pagoda]

bacc

[patent raptor]

One may use the quadratic formula

[mental dome]

we can use complex?

cos + isin

[grizzled pagoda]

bacc

[patent raptor]

Damn doesn even simplify nicely

[grizzled pagoda]

bacc

[patent raptor]

,, 5\left [ \frac{12 \pm \sqrt{36-25\cos^2\theta}}{10\cos\theta} \right ]^2\sin\theta -5\sin\theta = 4\left [ \frac{12 \pm \sqrt{36-25\cos^2\theta}}{10\cos\theta} \right ]

[grizzled pagoda]

bacc

[patent raptor]

no way you solve this for theta

[mental dome]

i?

[patent raptor]

,w 5r^2cos(theta)+5cos(theta)=12r and 5r^2sin(theta)-5sin(theta)=4r

[grizzled pagoda]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by adoniszgks

[patent raptor]

interesting

Maybe there is a smarter way

They look very similar after all

[grizzled pagoda]

bacc

bacc

[patent raptor]

yea that should be way easier

We can actually make use of cos(theta) = sin(theta+pi/2)

[grizzled pagoda]

bacc

[patent raptor]

Now we should be able to solve for theta and equate them and solve for r

altho hmm

[grizzled pagoda]

bacc

bacc

[mental dome]

,w (cos(-a)+isin(-a))/(cosa+isina)

[grizzled pagoda]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by vinibied

[mental dome]

,w Conjugatez/(z(cos-isin))

[grizzled pagoda]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by vinibied

[mental dome]

,w [{5x(1+1/(x²+y²)) = 12, 5y(1-1/(x²+y²)) = 4}]

[grizzled pagoda]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by vinibied

[mental dome]

But now how do I get the result?

[grizzled pagoda]

bacc

bacc

[patent raptor]

Now apply this and use sine on both sides

[grizzled pagoda]

bacc

[patent raptor]

😭

solving this for r

I am probably missing out on something

a trick as always

idk why are you even being quiet

[mental dome]

I'm doing it here

take a look at my idea

[patent raptor]

oh i am dumb it's sine that i apply on both sides

not arcsin

[patent raptor]

whats your idea

[mental dome]

wait a sec

[patent raptor]

If we look at this is basically a cubic

[patent raptor]

So we get 2 real solutions and 1 complex

[mental dome]

sen is sin

[patent raptor]

yea

writing sen is a sin

[patent raptor]

but where did the i come from

[mental dome]

I multiplied eq 2

[patent raptor]

by i?

[mental dome]

yeah

[patent raptor]

is this allowed

[mental dome]

I multiplied the entire equation by i
so that I can analyze it by complexes

[mental dome]

yeah

[mental dome]

we just need the two real ones

[patent raptor]

\begin{align}
5r^2\cos\theta +5\cos\theta &= 12r \
5r^2i\sin\theta -5i\sin\theta &= 4ir
\end{align}

[grizzled pagoda]

bacc

[mental dome]

then I'm adding the two equations

[patent raptor]

ok

[mental dome]

ah

[patent raptor]

what

[mental dome]

I got confused, sorry

[patent raptor]

,, 5(r^2+1) \cdot \cos\theta + 5(r^2-1) \cdot i\sin\theta = 4r(3i+1)

[grizzled pagoda]

bacc

[patent raptor]

damn r^2-1

It's basically

[split sail]

What is even happening

[patent raptor]

,, a\cos\theta + bi\sin\theta = z

[grizzled pagoda]

bacc

[mental dome]

,w (a-bi)(cosa+isina)

[patent raptor]

no

[split sail]

Is this a question?

[patent raptor]

no

[mental dome]

I'm just testing something

[split sail]

Oh ok have fun

[grizzled pagoda]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by vinibied

[mental dome]

,w -(a+b)(cosa+isina)

[patent raptor]

,, 5(r^2+1) \cdot \cos\theta + 5(r^2+1) \cdot i\sin\theta -10(r^2-1)i \sin\theta = 4r(3i+1)

[grizzled pagoda]

bacc

[patent raptor]

bro

,, 5(r^2+1)e^{i\theta} -10(r^2-1)i \sin\theta = 4r(3i+1)

[grizzled pagoda]

bacc

[patent raptor]

,w isin(theta) in polar form

[patent raptor]

If we multiply (1) by sin(theta) and (2) by cos(theta)

,w sin(x)cos(x) = 0.5sin(x)

[grizzled pagoda]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by adoniszgks

[patent raptor]

forgot about the right side

,, 5(r^2+1) \cdot \cos\theta + 5(r^2-1) \cdot i\sin\theta = 4r(3i+1)

[grizzled pagoda]

bacc

[patent raptor]

this is bad

this is literally bad

i dont know what to do

you make me hate math now

[mental dome]

hahahaha

i have a new ideia

[patent raptor]

messi is better

[mental dome]

cr7 is better

[patent raptor]

hahahahaha

[mental dome]

sirrrrrrrrrrrr

okok

[patent raptor]

,w 5(r^2+1)e^(ix) -10(r^2-1)isin(x) = 4r(3i+1)

[patent raptor]

We need slang math

I think real math is just too weak

Assume $r^2-1 \approxeq r^2+1$

[grizzled pagoda]

bacc

[patent raptor]

😂

[mental dome]

bro

[patent raptor]

what

[mental dome]

I sent it to my teacher and he sent the bomb here

[patent raptor]

what bomb

what is you bombing about

[mental dome]

I think bomb is an expression used more here in Brazil, I don't know an equivalent expression

[patent raptor]

fuck

the conjugate

why did i not think of that

oh

[mental dome]

I'm trying to do it based on his resolution

[patent raptor]

resolution

[mental dome]

[patent raptor]

should have stayed in cartesian

[mental dome]

I'm not so good at English

[patent raptor]

you think i am?

i am the worst, most outrageous english speaker

[mental dome]

probably better than me

I don't even know how to speak

[patent raptor]

shit

[mental dome]

lol

[patent raptor]

if you told me it was a task regarding complex numbers

i would have thought of it differently

you and your i

confused me

[mental dome]

this was not

[patent raptor]

you can argue with your mirror

[mental dome]

[patent raptor]

https://tenor.com/view/ishowspeed-sewey-offside-goal-sui-sidemen-charity-match-gif-26793327

messi and ronaldo aside

this is the true goat

[mental dome]

just truths to be told

,close

.close

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Channel closed

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Use .reopen if this was a mistake.

[mental dome]

.reopen

[amber wasp]

✅

[patent raptor]

what is you cooking

[mental dome]

@patent raptor thanks for the help

.close

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Channel closed

Closed by @mental dome

Use .reopen if this was a mistake.

[patent raptor]

🥹

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[junior sundial]

I have to prove LHS = RHS

.close

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[frosty bone]

I have no idea how to approach this

[trail lintel]

me neither

[weak zinc]

Consider the factorisation (and order degree) of the denominator

[split sail]

just gimme answer

[weak zinc]

We don't do that, and you're in someone else's help channel - if you want genine help and not the answer, then read #❓how-to-get-help

[frosty bone]

wait wym by order?

[weak zinc]

The higest power you find in the denominator

Ah, degree! That was the word

It escaped me for a moment

[split sail]

ur no help

[weak zinc]

Go ask for help somewhere else then

[frosty bone]

is there like a general way to rewrite it

becaues just looking at it the decomp itself is sort of

uh

disgusting

💀

i can see that you would get 4 and 8 but nothing asides from that

[weak zinc]

Well, you can consider how the powers would work, you "basically" have (x^3)^3 in the first term, and (x^4)^2 in the second

(remember how you'd generally do a partial fraction decomposition: if the degree of the numerator is not less than that of the denominator, carry out polynomial division until it is, then split the denominator into its factors and create your fractions that way)

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[jovial field]

Need help with these questions
Hint and discussion would be appreciated

[patent raptor]

15 was it?

@jovial field

[jovial field]

Yes

[patent raptor]

So good news we can express these pseudo factorials

[jovial field]

Woohoo

Wait imma bring my rough sheets

Imma get the correct ideas for all questions then I fair them out

[patent raptor]

[ 1 \cdot 3 \cdot ... \cdot (2n-1) = \frac{1 \cdot 3 \cdot ... \cdot (2n-1) \textcolor{cyan}{\cdot 2 \cdot 4 \cdot ... \cdot 2n} }{\textcolor{cyan}{2^nn!}} = \frac{(2n)!}{2^nn!} ]

[jovial field]

Ok I have it

We can begin

[grizzled pagoda]

bacc

[jovial field]

Ok

[patent raptor]

This is what we need

[jovial field]

What we need is the nth term

(I am talking specific to my assignment)

[patent raptor]

Yes

[jovial field]

Yes

[patent raptor]

So using this knowledge

[jovial field]

Yes

[patent raptor]

We can get the nth term

[jovial field]

Ok

[patent raptor]

,, \sum_{n=1}^{\infty} \left ( \frac{\frac{(2n)!}{2^nn!}}{2^nn!} \right )^2

[grizzled pagoda]

bacc

[jovial field]

The morning term?

[patent raptor]

We can reduce this

[jovial field]

Yes

(2n)!/(2^2n (n!)^2)

[patent raptor]

,, \sum_{n=1}^{\infty} \frac{[(2n)!]^2}{[2^nn!]^4}

[grizzled pagoda]

bacc

[jovial field]

Absolutely

Now the problem is

We checked

[patent raptor]

yea

[jovial field]

In the morning

[patent raptor]

ratio test is inconclusive

[jovial field]

Yes it fails

Root test would not benefit

[patent raptor]

ye

[jovial field]

Let's apply Raabe's test

[patent raptor]

I tested root test is also inconclusive

[jovial field]

Yes

Should we try logarithmic test?

[patent raptor]

I think that won't work either

[jovial field]

Cz I feel having root and ratio test inconclusive I don't feel raabe would work

[jovial field]

Yes

Logarithmic test

?

[patent raptor]

Let me look that up

[jovial field]

Let me give it

[patent raptor]

Yea

[jovial field]

In a positive term series

[patent raptor]

that might work

log test

[jovial field]

Okok

If lim n tends to infinity

[jovial field]

Good

You got it

[patent raptor]

I tested it should be divergent

[jovial field]

Ok let's see

[patent raptor]

Oh does the base play a role?

[jovial field]

Okok I believe on you

I just need the idea

[patent raptor]

Like with natural log we get 1

[jovial field]

I would do myself after all the ideas

[jovial field]

Its ln

[patent raptor]

but we log base 10 we get something smaller 1

[jovial field]

Right?

[patent raptor]

then that doesn't work too

shit

[jovial field]

Log test considers ln right?

[jovial field]

Shit

What if we try integral test

[patent raptor]

We could try comparison

[patent raptor]

☠️

[jovial field]

Like a pinch of gamma

And beta

💀☠️

[patent raptor]

Let's make that our last option

[jovial field]

Na na no I can't use it

I can't use it in the assignment

Okok let's try comparison

Let's take Vn as (n!)^4

Fair enough

[patent raptor]

Ok this won't work

[jovial field]

Wait too fair I think so

No no wait

[patent raptor]

if we cancel the n! this should converge 100 %

[jovial field]

But see

If u consider

Un>Vn

The case should be of a diverging

Cz let's say Vn converges

And Un>Vn

What you think

Let's try log test

[patent raptor]

No won't work

[jovial field]

Oof

Do we have any option left

[patent raptor]

hmm

[jovial field]

Tell

[patent raptor]

I am thinkiing

[jovial field]

Okok

But bro

I told

Wait na its real

This cooks

By p series we know

[patent raptor]

oh shoot

no doesnt work shit

[jovial field]

What option we have

Now

[patent raptor]

Damn

I am thinking of limit comparison test

[jovial field]

Limit form

What bout (n!)^4

[patent raptor]

Wait hold on

[jovial field]

Ok

[patent raptor]

Stirlings approximation

,, n! \sim \sqrt{2\pi n} \left (\frac{n}{e} \right )^n

[grizzled pagoda]

bacc

[patent raptor]

Very very similar

[jovial field]

Yes ik this

[patent raptor]

Then we can probably apply root test

[jovial field]

But we apply this in limit form?

Wait

We changed the nth term

[patent raptor]

For very big n

they are very similar