help-10

[obtuse pebble]

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[calm rover]

Hi, How can I get latex form for this formula?

[calm rover]

http://www.mathwords.com/b/bernoulli_trials.htm

I am trying to format it here but I can't seem to get the spacing right

[timid silo]

for latex help you can write on #latex-help

but the syntax is just $\binom{n}{k}p^k q^{n-k}$

[warm shale]

Al3dium

but the syntax is just $\binom{n}{k}p^k q^{n-k}$

[calm rover]

Ty!

[obtuse pebble]

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[obtuse pebble]

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[long sinew]

$\frac{\frac{1}{x+1} + \frac{1}{x-1}}{\frac{x}{x^{2} - 1}}$

[warm shale]

Breeziboi

[fierce lagoon]

Helper moment

[long sinew]

Lol i rarely ever need help

But this

Hurts my head

[thick oracle]

common denominator

[long sinew]

??

[thick oracle]

the top fractions

[fierce lagoon]

Yeah and then divide the fractions

[thick oracle]

bring them to it

[long sinew]

Ohg

Difference of squares

X^2 - 1

[thick oracle]

yes

[kind hawk]

multiply top and bottom by x^2-1

not necessary to combine the fractions in the numerator

[fierce lagoon]

^

Since x+1 and x-1 are factors of x^2 - 1 you can just multiply by (x^2-1)/(x^2-1)

[long sinew]

I see

It comes out to be equal to

...

2

Everything cancels out to 2x/x

I hate math sometimes

.close

[obtuse pebble]

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[timid crow]

Hi everyone

[timid crow]

I initially posed this question on the math StackExchange, but I got no answers

https://math.stackexchange.com/questions/4520261/proving-the-existence-of-a-certain-term-in-the-k-1th-derivative-to-find-riem

Does anyone know if its possible to prove the presence of a certain term in a [k]th derivative ?

Thanks in advance!

[obtuse pebble]

@timid crow Has your question been resolved?

[timid crow]

<@&286206848099549185>

[timid silo]

Yes?

I'm sorry, this is not my major.

Hope you find someone that could help a bit,

[timid crow]

Alright, no worries 🙂

[obtuse pebble]

@timid crow Has your question been resolved?

[obtuse pebble]

@timid crow Has your question been resolved?

[tardy epoch]

not sure about the answer, but these calculations look relevant
https://arxiv.org/abs/1202.1205

[obtuse pebble]

@timid crow Has your question been resolved?

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[timid crow]

@tardy epoch thank you, I will look through it

[obtuse pebble]

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[warm ether]

what is this called in english? in dutch it's called a "symbol scheme/table" , literally translated

[fierce lagoon]

Sign analysis

Sign schema

Relations table

Piece of shit

All the same

[obtuse pebble]

@warm ether Has your question been resolved?

[warm ether]

ty

[obtuse pebble]

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[red osprey]

I'm not sure which value out of the two intervals to put into the equation

I haven't done math in so long so i'm a little rusty

[timid silo]

u know what the average velocity is?

or what it means

[obtuse pebble]

@red osprey Has your question been resolved?

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[obtuse pebble]

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[drifting portal]

First question only

Actually i want both now i don’t get what he means in the second one

[obtuse pebble]

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[drifting portal]

.reopen

[obtuse pebble]

✅

[obtuse pebble]

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[subtle violet]

I am not sure if this fits here, sorry if it doesn't.
Question: Would normal subgroups be abelian? If I go by their definition, then they should exhibit abelian property to be true. But I found conflicting answers and not a proper answer. I would appreciate if anyone could help me understand this.

[zenith raft]

a normal subgroup of a group does not need to be abelian

[tardy stratus]

No, they are typically not abelian.

[zenith raft]

a normal subgroup of an abelian group is abelian though

[tardy stratus]

Why do you think they would be? Please explain your reasoning and we may be able to correct your misunderstanding

[zenith raft]

as is any other subgroup

[gilded needle]

any subgroup of an abelian group is abelian

any subgroup of an abelian group is also normal

[zenith raft]

yea actually that's what I was trying to get at

[subtle violet]

The definition I followed is that, for any element g belonging to G and h belonging to H, H is a normal subgroup of G if ghg^-1 belongs to H.

[tardy stratus]

Yes, this is indeed the definition

Are you perhaps assuming that ghg^-1 belonging to H implies that ghg^-1 = h?

[subtle violet]

yes

would that be incorrect?

[tardy stratus]

That would be incorrect; typically ghg^-1 will not be equal to h, but merely some other element of H

Are you familiar with the dihedral group?

[subtle violet]

yes

[tardy stratus]

Ok, great

Consider the group of rotations within the dihedral group with 6 elements

Actually, I'm sorry -- this is a bad example. Let me try again.

Are you familiar with the alternating group? Apologies that I'm pulling your chain around a bit

[subtle violet]

Apologies, I haven't studied about alternating groups yet.

[tardy stratus]

OK, that's fine.

I can use the example of the dihedral group still, although it isn't perfect

So yes, this normal subgroup of the rotations of the dihedral group is in fact Abelian, which is why it isn't a great example

However, even in Dih(6) we can see that typically ghg^-1 is not equal to h

if you take any non-trivial rotation, and conjugate by a particular reflection, you will get a different rotation

I'm not sure what notation you're familiar with, but typically I'd write something like d^-1ρd = ρ^-1

So although this is a bad example again because it is true that this normal subgroup is Abelian, it still demonstrates that you've misinterpreted the definition.

[subtle violet]

Ah right, it makes sense now that it would be any element of H

I think I misunderstood it so I was pretty confused for a while. Thank you very much for taking your time to explain this.

[tardy stratus]

No worries. I hope you have fun learning more group theory!

[gilded needle]

maybe for an easy counterxample, let G = H x K where H is any nonabelian group and K is any group

then H is normal in G and is nonabelian

[tardy stratus]

Yes, nice example

[gilded needle]

even easier example, let G be any nonabelian group, then G is normal in itself but is nonabelian

but that one's kind of unsatisfying somehow

[tardy stratus]

Yeah I agree

[subtle violet]

Every group would be normal to itself, correct?

[tardy stratus]

That's right

[gilded needle]

yep

[tardy stratus]

Every group except the trival group has two guaranteed normal subgroups: itself, and {1}.

[subtle violet]

Right, thanks, abstract concepts are hard to understand but still fun to learn

[tardy stratus]

It will get easier

[subtle violet]

By 1 here, I assume you mean identity element?

[tardy stratus]

Yeah, I like to use 1 where some people prefer e

[subtle violet]

ah right, makes sense

.close

[obtuse pebble]

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[white depot]

How do i work out kernel span

pathfinder this channel is occupied

please can you delete your message

[timid silo]

Whats your question?

[white depot]

how do I get the kernel span?

im confused on how to get it to row reduced echelon form

surely its just (010 000)

[timid silo]

(1,0)^t is kernel right?

[white depot]

uh

im not sure

[timid silo]

Can you post full question

I didn't get your question

[white depot]

how to work out the kernel span from the matrice

[gilded needle]

well you know the kernel of that matrix must have dimension 1 (because the matrix is 2x2 and has rank 1), so you know the kernel is spanned by a single vector

so you just need to find any nonzero vector in the kernel

[white depot]

thats the bit im stuck on

[gilded needle]

to do that, just solve $$\begin{pmatrix}0 & 2 \ 0 & 2\end{pmatrix}\begin{pmatrix}a \ b\end{pmatrix} = \begin{pmatrix}0 \ 0\end{pmatrix}$$

[warm shale]

Bungo

[white depot]

Okay I did that 2

I row reduce it

and end up with

(0 1 0) (0 0 0)

how would I go from there?

[gilded needle]

well that row reduced form means b=0

so any vector $\begin{pmatrix}a \ b\end{pmatrix}$ with b=0 is in the kernel

[white depot]

yea

[warm shale]

Bungo

[white depot]

Ohh

so (0 0 is right

(0 0)

[gilded needle]

well (0,0) is always in the kernel of any matrix, that's not super interesting

but notice also (1,0) is in the kernel

which is the one mentioned in your screenshot

[white depot]

Ahh I see

thanks so much

[gilded needle]

sure

[white depot]

.close

[obtuse pebble]

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[hallow phoenix]

Hello! I was wondering why and how did the calculator reach to this? It doesnt show any soultion.

[earnest orbit]

Well that's the equation for a circle

[hallow phoenix]

no, I meant how did it solve it?

there are the points, but im not so sure

Ive seen doing matrices but I dont understand it

some said geometry

[earnest orbit]

$(x - h)^2 + (y - k)^2 = (r)^2$

[warm shale]

Gustofwind

[earnest orbit]

Its just the application of the formula

[hallow phoenix]

hm do i just substitute

[earnest orbit]

yes

[hallow phoenix]

with all 3 or is it fine with 2

[earnest orbit]

h is x co-ordinate, k is is y co-ordinate, and r is radius

sub those three

[hallow phoenix]

Ahh

I see

The last time i did it was when i had a radius and it had 2 points

so when we are finding for all with points, we need 3 to solve for h,k and r

is that right?

[obtuse pebble]

@hallow phoenix Has your question been resolved?

[hallow phoenix]

I have found h, though for k im not so sure

if i sub h to one of the equations i still have r^2

which is unknown

[obtuse pebble]

@hallow phoenix Has your question been resolved?

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[haughty lance]

Hi, I have a screenshot of a proof I'm trying to work on for a Real Analysis class. I understand what I am supposed to do, but I feel like my execution at the end is lacking. Is there any way I could make this read better/more concise?

[obtuse pebble]

@haughty lance Has your question been resolved?

[tardy epoch]

This looks more like scratch work than a proof

Read through any of the examples here to try to see what constitutes a proof

https://www.milefoot.com/math/calculus/limits/DeltaEpsilonProofs03.htm

[haughty lance]

I think the issue that I have is generalizing the proofs in an abstract manner.

[obtuse pebble]

@haughty lance Has your question been resolved?

[tardy epoch]

Your proof should be line by line parallel to any of the examples in the link. Treat scratch work separately

[obtuse pebble]

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[buoyant ore]

Jane is driving to visit a friend who lives 480 miles away. She is driving 60 miles an hour and has already driven 100 miles when she starts to time herself. Write an equation for how many miles she has left on her trip.

[buoyant ore]

I am confused.

I think by equation he really means function because the next question is the graph the function you wrote for the question above.

So the function i got was

f(x)=60x+100

But that was wrong

Technically 100 is the starting point of time

so theres a t for time somewhere in this equation

60t

and then 100 miles have been driven

so 380 is the technical goal

would it be

380=60t?

I don't knwo and have spent too long on this

<@&286206848099549185> 15 minutes have passed

[kindred oasis]

It asks for an equation that gives you the miles she still has to travel. Think about how you would find that: you have the total miles she needs to travel, 380, as well as the miles she travels as a function of time, 60t

So the distance she still has to travel should start at 380, then go down to 0, right?

[buoyant ore]

Yes

so would my equation be

or function or something

be

f(x)=-60t+380

[kindred oasis]

Yes, exactly

You can rewrite it as 380-60t, which gets rid of an unnecessary sign. But that's it

[buoyant ore]

OML thank you so much 🫂 ive spent like 1 hour on this and I think its because i didnt understnad

[buoyant ore]

Ok thank you again!!

[kindred oasis]

yw

[buoyant ore]

.close

[obtuse pebble]

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[craggy ferry]

Real simple question

[craggy ferry]

To draw v-w, I would have an my initial point on (1,-1) and my terminal on (2,-3) and then from the origin (0,0) to (2,-3) correct?

.close

[obtuse pebble]

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[tawny tree]

Hello, i need help in solving this

[tawny tree]

i mean proving

and x1,x2,x3,...,xn are positive

and i should probably use the qagh inequality but idk how

[compact shadow]

The left hand side

Two sums

Which has m terms which has n terms?

[tawny tree]

its all n terms

[compact shadow]

I can’t distinguish the writing

[tawny tree]

sorry

[compact shadow]

Okay got it:

We want to prove that

[tardy epoch]

that was fast

[compact shadow]

Too tired to use latex, I will take photos

[tawny tree]

oki

[compact shadow]

Thanks

[tawny tree]

damn thats an interesting soultion thanks for help

.close

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

find the area of the region bounded by the curve x=at^2, y=2at between t=1 and t=2

[timid silo]

did anyone else get 28a^2/3?

oh wait nvm I see my mistake

.close

[obtuse pebble]

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[marble hamlet]

I've been trying different pythag identities to rework with integral before solving and am coming up with nothing.... can anyone help with this?

should I be doing a substitution?

[viscid delta]

Uhh, I think you might be overthinking this one.

You can simplify that expression to sec² theta + 1.

[marble hamlet]

i did that also on one try - and then just integrate?

[viscid delta]

Well yeah, integral of sec²x is just tan x after all.

[marble hamlet]

ok - then i get sin/cos right? and sub from there ?

[viscid delta]

Huh.

[marble hamlet]

tanx = sin(x)/ cos(x)

[viscid delta]

$\int_0^{\pi/4} \frac{1 + \cos^2 x}{\cos^2 x} dx = \int_0^{\pi/4} (\sec^2 x + 1) dx = [\tan x + x]_0^{\pi/4}$

Why do you need to sub anything?

[warm shale]

PhenomPlasma

[marble hamlet]

ahh since it's definite just evaluate when I get a simple trig function?

[viscid delta]

Yea.

[marble hamlet]

true that - thank you much

.close

[obtuse pebble]

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[warm ridge]

@fiery hull sir ive got it

[warm ridge]

but i dont get tgus

the formula k=rs

Qnd the latter parts

[thick oracle]

You don't get what?

[obtuse pebble]

@warm ridge Has your question been resolved?

[obtuse pebble]

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[warm ridge]

the formula

[warm ridge]

and how were the radiis found

[thick oracle]

K=....

That?

[warm ridge]

ya

im not familiar with thta

could u pls breif smth sbt jt

[thick oracle]

Heron's Formula

For the area

[warm ridge]

of inscribed cricle?

[thick oracle]

Of any triangle

[warm ridge]

sorry, with the inscribed cricle?

oh

okay, then how was the larger radius foumd

[thick oracle]

TE wasn't found

[obtuse pebble]

@warm ridge Has your question been resolved?

[obtuse pebble]

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[vast willow]

Hi

[vast willow]

Is this correct?

[stable rain]

reposting after no1 answered :c

but its too many numbers lol :c

[vast willow]

I wasn't answered for more than 2hrs or something like that 😭

[stable rain]

yea i mean

having 20s.f. is intimidating

i too dont wanna answer

lol

[vast willow]

What is 20s.f. btw

[stable rain]

gl

significant figures

aka 3.14 is 3 s.f.

[vast willow]

But this isn't complicated man

[stable rain]

19999.9999 is 9s.f.

yea ur numbers look horrible tho

its hard

ok bye gl

[vast willow]

No man I just copied the number after I got the answer from calculator XD

Ok tysm

[stable rain]

bru

[vast willow]

Yea and the number under it is just the whole part of the number copied from the calc +1

Bc it would be greater than that number

That's why I am saying that its not complicated

[obtuse pebble]

@vast willow Has your question been resolved?

[stable rain]

lol

[obtuse pebble]

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[stable rain]

oh i was

just about to help

well

nvm

lol

[obtuse pebble]

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[short zodiac]

This is more appropriate for a physics channel

[icy cape]

oh true

forgot abt that

[short zodiac]

Nevertheless I'll help you

[icy cape]

oh

thaks\

thanks*

[short zodiac]

Are you aware of conservation of momentum?

[icy cape]

y7es

yes*

[short zodiac]

That's what you need here

[icy cape]

Alright I am going to try and solve it but can you see if I am correct if not please locate and inform me of my mistake

[short zodiac]

M1v1 +M2v2=0

[icy cape]

(0.5kg x -48 m/s) + (4.0kg x V) = 0
-24kg m/s + 4.0kgV = 0
4kgV = 24kg m/s
V = 6 m/s

[icy cape]

^

[short zodiac]

Yep

[icy cape]

Thank you!

👍

[short zodiac]

.close

[obtuse pebble]

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[broken lichen]

Hi, can anyone help me with this? the options are injective but not surjective, surjective but not injective , Bjective and none of the above

[compact shadow]

ii is surjective , iii is injective, not sure why you chose none of the above both

[broken lichen]

umm i think iii is bijecive? cause i remember i put injective for 3 it is wrong

[compact shadow]

ii: any w there exists exactly two roots of 9z^2+5=w, so surjective not injective
iii is clearly injective no surjective, because n->9n^2+5 strictly monotonically increase

Oh wait

Z to Z

I thought N to N

Neither surjective nor injective

Image of 1 and -1 are equal so not injective

[broken lichen]

ahhhhhh

[compact shadow]

-2 isn’t in the image so not surjective

[broken lichen]

ahh it make sense now

thanks a lot !

[compact shadow]

Np

[broken lichen]

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[quiet crag]

Is there a different way to rewrite $-\pi Si(\frac{\pi}{x})$

[warm shale]

Wes

[quiet crag]

$\int \frac{sin x}{x}$ was the question but I don't think our professor wants the answer like above

[warm shale]

Wes

[quiet crag]

(times -pi oops)

[obtuse pebble]

@quiet crag Has your question been resolved?

[weary delta]

Given bounds you can numerically approximate it, or write it as a taylor series

[quiet crag]

the bounds, of course!! thank you

.close

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[obtuse pebble]

Channel closed

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[forest spruce]

All those tutorial vids keep saying that the factors of the last term are supposed to

add/subtract to the middle term

[small thicket]

wdym

[forest spruce]

They always say that to find factors that add/subtract to the middle term

[devout sable]

It’s not factors of last terms btw

[forest spruce]

Also dont tell me the asnwer, i just want to know what im missing

[small thicket]

yea find factors of -5 which add/subratct to middle term

[devout sable]

Its factors of (1st)*(last term)

[small thicket]

but remember u have a 6x

6 infront of the x

so u have to keep that in mind

[forest spruce]

5 - 1 = 4 ,, 5 + 1 = 6

They wont end up with (1)x

[small thicket]

boi

[devout sable]

Hm?

[small thicket]

what did i just say

[forest spruce]

WAIT SO im allowed to use first terms?

[small thicket]

u have a 6

[forest spruce]

And im not obligated to always use the last term?

You mean?

[devout sable]

You need to use both 1st and last term

[small thicket]

u dont need it

if u r confident u can skip that step

[devout sable]

Hm?

[small thicket]

wait wait let me show

[forest spruce]

About that

Wait nvm

You cant do anything with unlike terms

[devout sable]

Idk what video you saw but this video has clearly explained stuff

[forest spruce]

Since 6x² isnt compatible with plain 5

[devout sable]

https://youtu.be/9MUDAr1PnlE

[forest spruce]

Oh yeah datz the one

[devout sable]

Have you followed it completely?

[small thicket]

it would if u used -5 and 1

it is a factor of -5 right

and u have a 6 infront of the x^2

whats 6 times 1

6 right? 6-5 = 1

so what would u get sir.

[forest spruce]

Okay thanks! Ill try to follow the video

Too

Wait hold on

Negative 5 - 1 is -4

So youd end up with 6 and 4?

[obtuse pebble]

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[small thicket]

nonononn

[small thicket]

ehhh whatever

multiply right

bc thats the factor of -5

[forest spruce]

Wait multiply instead of add oh i get

Opps

[small thicket]

$-1\multiply5$

right

and we also know we have a 6.

we can either split it into 2 and 3 or 1 and 6

[forest spruce]

Yeah 1 and 6

Thanks

[small thicket]

i want 6 and 1 bc

yea

[forest spruce]

I got it now thankd

[small thicket]

its just less steps ig

.close

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[forest spruce]

Okay but this time its too latge

Large

12 + 5 is too weak its only 17

5 x 12 is too strong its 60

[proven zephyr]

ok so

how do i explain this

ima just set 2 variables, m and n

find 2 integers where... nvm

[forest spruce]

Idk

[proven zephyr]

wait..

hmm how do i ask this

[forest spruce]

Im exhausted today

Just finishing the last bits of my homework

[proven zephyr]

ok so

ima rephrase something

do you know how to factor the standard form of a quadratics: ax^2 + bx + c?

[vestal marsh]

So for this example I'd think it as 5x^2+30x+2x+12

[forest spruce]

Kind of

[proven zephyr]

you can do this if you want

factor by grouping if im not wrong

[forest spruce]

Idk grouping 😨💀

i think im too tired for today 😞

[vestal marsh]

Have you done factors of ac? Or just factor by groups

[proven zephyr]

i think he kinda mixed up how to factor ax^2 + bx + c with x^2 + bx + c

https://youtu.be/qeByhTF8WEw?t=295

this video good

similar to what amason did

[vestal marsh]

Whatever is the most familiar to you tbh

If you want an explanation to how i got 5x^2+2x+30x+12 I could do that.

[forest spruce]

Being the video skipper doesent help 😞

I may or may not ADHD 😞

[timid silo]

What are 2 numbers that add to 32 and multiply to 12*5

12*5=60 I think

so add to 32 and multiply to 60

that should be your goal first

[proven zephyr]

ima just write down the equation for you
m * n = 60
m + n = 32

[timid silo]

And if you are wondering yes there are numbers that work

[forest spruce]

How do you group 😞

[timid silo]

First off

Do you know how they got there to begin with?

[forest spruce]

They just split off the 32x into 30 and 2

[timid silo]

Because that's a very important step

[forest spruce]

Creating another x

[timid silo]

Yes but why those numbers specifically

[forest spruce]

Idk

[timid silo]

Why not 15 and 15? Or 31 and -1?

It's because 30+2=32 and 30*2=60

[forest spruce]

Like do you just divide numbers randomly and go

[timid silo]

No

Those are the special numbers we were looking for

[timid silo]

When I asked you this

[forest spruce]

Wait so we gotta trial and error 😨

[timid silo]

We'll, yes sort of

You should look for the numbers that make that possible

[forest spruce]

So now we have

[timid silo]

Then break the middle part

[forest spruce]

5x² + 2x + 30x + 12

What next

[timid silo]

Write it the other way around

[vestal marsh]

Factor

[forest spruce]

Do i start looking for factors now

[vestal marsh]

No

It's the correct way

[forest spruce]

Theres two middle terms

[timid silo]

$5x^2+30x+2x+12$

[warm shale]

jswatj

[forest spruce]

And 5 isnt divisible

[timid silo]

look at it this way

$(5x^2+30x)+(2x+12)$

[warm shale]

jswatj

[timid silo]

factor each bracket

On its own

[forest spruce]

How do we know which belongs to which

Why did 30 go to 5? Why 2 to 12?

[timid silo]

5,2 have no common factors

There's nothing we can pull out of them

it wouldn't help us

[forest spruce]

Okay then what next

[timid silo]

you can trial and error that too

pull out ? On the left and ? On the right

what should we factor on each side

[forest spruce]

PULL OUT

WHAT?

[vestal marsh]

Nah, we do 5x^2+2x and bring out the x to get x(5x+2)+6(5x+2)

[timid silo]

Sorry

[forest spruce]

2

[timid silo]

Distribute

oh actually I guess it doesn't matter

5x(x+6)+2(x+6)

[vestal marsh]

Yeah

Both work

[timid silo]

I thought it did

sometimes it doesn't work the other way however

[forest spruce]

If it WASN'T for the dang 5 this would have been solved now

Well then

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[obtuse pebble]

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[twilit sparrow]

hey

[twilit sparrow]

what's the root of "4.41 square miles"

[full cloud]

2.1 miles

[twilit sparrow]

If you have a square one side would be 2.1 miles?

exactly?

[full cloud]

If you have a square of area 4.41 square miles, then one side of the square is 2.1 miles

[twilit sparrow]

thanks g

[obtuse pebble]

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[quaint lily]

How do I find 16c?

[dusk portal]

@quaint lily if you have a function that gives you the displacement at every second, t, how do you sum up all these displacements? Remember, you're summing a continuous function

[full cloud]

Integrate it from 0 to the time it takes

[dusk portal]

or just be spoonfed that works too

[quaint lily]

how do I do this without integration?

[dusk portal]

well, you don't, really

you want to take a continuous sum over a range. first, figure out the range of t. i.e., from t=0 to t=?. this should have been done in part a

[quaint lily]

7 seconds

[dusk portal]

perfect

so you want to take a continuous sum from t=0 to t=7 of the displacement.

[quaint lily]

continuous sum?

[dusk portal]

nvm i misinterpreted the Q

[stable rain]

why wouldnt newton know this

lol

[next reef]

This can just be interpreted as a displacement-time graph

[stable rain]

weve an imposter

[next reef]

The slope gives you velocity

[stable rain]

among us

[full cloud]

I think you find x at t=7 and subtract it from x at t=0

[next reef]

And you already know the time interval

[quaint lily]

the answer is 49cm

[stable rain]

@quaint lily what formulas do u know

send em

bru what r u replying to

[ruby hound]

idk 💀

[stable rain]

lol

[quaint lily]

what formulas?

[stable rain]

keep it relavent

[quaint lily]

a lot

[stable rain]

mostly~

[quaint lily]

but not relevant to this

[stable rain]

send a list or pics

[quaint lily]

don't worry

i got it

.close

[obtuse pebble]

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[quaint lily]

pi = e

[stable rain]

wow

[obtuse pebble]

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[hot tiger]

Heyy

[hot tiger]

I wanna be a helper

[quaint lily]

prosciencedude

[versed cave]

Message modmail

But you don't need to be helper to help

[hot tiger]

oh

[versed cave]

Helper means that you get pinged

[hot tiger]

Can i still help others?

[versed cave]

Yes of course ahah

A lot of the usual and awesome helpers don't have the helper role

[hot tiger]

anyways,

@quaint lily can you pls explain your 2nd law

[quaint lily]

f=ma

[hot tiger]

f = /t

is that correct?

[quaint lily]

what is /t

[hot tiger]

F is idirectly porportional to time

[quaint lily]

?

[hot tiger]

Force is indirectly proportional to time

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[stable rain]

hey lovelies
cant we use cauchys theorem to deform c2 into c1
or viceversal

[stable rain]

last one last one

[willow ravine]

Either direction would both still contain the origin and would be divergent

[willow ravine]

Look at the first question, 1/z is not holomorphic and thus you can’t deform. You must have a holomorphic function.

[willow ravine]

The anti derivative is Log(z) which is valid in the complex plane minus the negative real axis. It’s domain is not the same as 1/z.

[obtuse pebble]

@stable rain Has your question been resolved?

[stable rain]

so applying cauchys theorem strictly means integrand=0? using it to deform curves doesnt count as using it?

[stable rain]

my question was more on the last option, why is there no holomorphic requirement on it

[stable rain]

oh wait yea, thanks

[willow ravine]

What do you mean by deforming it? If you were to take c2 and say it falls into c1 (contained) and then try to prove c2 = 0 by cauchy’s?

[stable rain]

i mean

integrating over c1 gives the same as when integrating over c2

[willow ravine]

It’s not a question about equality though, it’s a question on being able to apply the theorem.

[stable rain]

yea i guess just deforming it is not exactly applying the theorem and more of an indirect result of it

[stable rain]

how about this, why is this true when there is no specification on holomorphism

[willow ravine]

Which question are you referring to? The second or third one?

The anti derivative one or that you can’t use this on any function it must be holomorphic?

[obtuse pebble]

@stable rain Has your question been resolved?

[stable rain]

2.d

anti derivative

2nd*

[willow ravine]

The path independence?

Sorry, I was at the gym.

@stable rain

[stable rain]

yea

[willow ravine]

It’s just saying it won’t matter if you travel in a clockwise or counterclockwise direction.

On the path of C

Like if you flipped the arrow on your diagram in your first picture

[stable rain]

something like these would have the same value when we take the integral right

but it never mentioned holomorphicity so how can we apply cauchys theorem

or does existence of anti derivative imply something

[willow ravine]

Holomorphic means it is differentiable in the contour

The entire contour

On each point in the domain

[wild swallow]

if a function has a complex antiderivative

then it is holomorphic automatically

wherever that complex antiderivative exists

if a function is once differentiable and holomorphic then it is infinitely differentiable

but cauchy-goursat specifically uses the antiderivative to complete the proof

@stable rain

[stable rain]

omg

i forgot

yea cuz if antiderivative exists in a region then the first derivative exist in said region

and that implies holo n whatnot

but does it work the other way

does holomorphic imply infinitely integrable

imma go slp n check back tho itll probably be closed

thanks @wild swallow @willow ravine

[obtuse pebble]

@stable rain Has your question been resolved?

[obtuse pebble]

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[timid silo]

am i correct?

[dusk portal]

Not quite

[timid silo]

alrighty

thank you

[dusk portal]

@timid silo do you want help solving it?

[timid silo]

yes please 😭

[dusk portal]

ok, so where did u start?

[timid silo]

ive already rearranged it

im just unsure if im supposed to sutract 2 or 12

im thinking 12

[dusk portal]

Ah you've written it out wrong

[warm shale]

clr

[dusk portal]

My first thought would be that I need y on its own, right? So, let's get it on its own

$4y = 3x-6$

[warm shale]

clr

[timid silo]

ohh

[timid silo]

OH OKAY

[hexed agate]

Wron

[dusk portal]

how is it wrong?

[hexed agate]

4y = 3x + 6

[dusk portal]

nope?

[hexed agate]

Lmao I'm tripping

Sorry

Continue.

[dusk portal]

np 🙂
@timid silo ok, so now we have this, what would u do next?

[timid silo]

i divided 4/4 and 3/4
i got 3/4x - 6
then i got 2/3

tysm!

[dusk portal]

yea that's it 🙂

[timid silo]

thank you!!

[obtuse pebble]

@timid silo Has your question been resolved?

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[sharp flare]

hey everyone, i've been stuck on this newton law of cooling problem (using differential equations). I found y(t) and k, so i plugged in my k value and 50 (for 50 minutes) as t into y(t) but I am only getting 58 degrees which doesn't seem right

[obtuse pebble]

@sharp flare Has your question been resolved?

[obtuse pebble]

@sharp flare Has your question been resolved?

[obtuse pebble]

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[sharp flare]

oops i went afk

[sharp flare]

i still need help

[timid silo]

with what?

[tardy epoch]

your question got lost due to inactivity. copy and paste it here with your updated progress to get help

[sharp flare]

Sure one sec

hey everyone, i've been stuck on this newton law of cooling problem (using differential equations). I found y(t) and k, so i plugged in my k value and 50 (for 50 minutes) as t into y(t) but I am only getting 58 degrees which doesn't seem right

for clarification i just need help with part a

[obtuse pebble]

@sharp flare Has your question been resolved?

[wary vigil]

so youve found k

and now you fill in 50 instead of 30 right?

do you get 58 F?

is that the value you get from y(t) or from T(t)

@sharp flare

[sharp flare]

from y(t)

so i go back to my definition of y and plug in my y(t) result and T(t) is my answer?

[wary vigil]

yes

at least that seems reasonable to me

[sharp flare]

you might be right i got 133 which seems within reason

[wary vigil]

if the value for time you get from b

is between 50 minutes and infinity

and seems reasonably exponential

then i think you're correct in your calculation

[sharp flare]

yup you were totally right

thanks person

.close

[obtuse pebble]

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[obtuse pebble]

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[limber narwhal]

i solved for B idk the answer for A, I tried putting in "ln2.7" but that wasnt the right answer

[warm ridge]

whats that

[limber narwhal]

<@&286206848099549185>

.close

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

Im not sure how to start this problem

[timid silo]

I think Im supposed to do x^2-y^2 = 9 as the limit but not sure how to find the "Largest set"

what happens when the denominator is equal to zero

[timid silo]

Undefined?

Exactly so it won't be continuous when x^2+y^2=9

[timid silo]

So I just put that as the largest set?

And that equation is equation of circle with radius 3 so largest set would be R-range of that circle

[timid silo]

Im dont quite understand. R-range? I understand that the radius is 3.

[rigid pine]

f(x, y) is only undefined when x^2 + y^2 = 9 so either, in terms of largest sets, you need to decide between the set of points within x^2 + y^2 = 9 or the set of points outside of x^2 + y^2 = 9.

[timid silo]

R is the set of real number we will delete all points from R where denominator is equal to zero

[timid silo]

so R E x^2+y^2 = 9?

So denominator is equal to zero when (x,y) lie on circle of radius 3

[timid silo]

Whats E

the special e character

idk how to uhh

it looks like a opposite 3

Did you get that ?function will be discontinuous when denominator is equal to zero ?

[timid silo]

Yes

Yeah so first of all you will search (x,y) points where that denominator is equal to zero

3 and 0?

No . 9-x^2-y^2=0
9=x^2+y^2

Got it

And then?

Which is the equation of circle with radius 3

So denominator will be zero when (x,y) lie on that circle

Got it.

So you will delete all those points from set of real numbers which lies on circle

[timid silo]

So all real numbers excluding 9=x^2+y^2

R excluding 9=x^2+y^2?

Exactly

Yeah

How do i put the excluding as a equation?

Is that a negative sign after R?

Wait i just messed up 🤯

???

R-{(x,+-√(9-x^2))}
Or
R-{(x,-√(9-x^2)) union (x,+√(9-x^2))}

[timid silo]

You can say that but more precisely use \ this instead of -

got it. Thank you!

Have you understood

Yes.

Okay

.close

[obtuse pebble]

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[fallow tusk]

So I did y'*y^3 = 18x^2

[timid silo]

Yep

[fallow tusk]

Do I just integrate both sides to get 1/4y^4 = 18x^3?

[timid silo]

then dy y^3 = dx 18x^2

Then integrate both sides

And u are done sir just adda +C

[fallow tusk]

So I get 1/4y^4=18/3x^3

[timid silo]

Just add a +c on RHS SIDE

[fallow tusk]

But do I have to multiply the right by 4 and take the 4th root?

[timid silo]

Yep

[fallow tusk]

hmm lemme try

@timid silo did I screw something up?

[timid silo]

Nope

Wait it should be root (24xcube + c)

[fallow tusk]

Ah, there we go

thanks

[obtuse pebble]

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[stark pike]

in cases like this, how do i know when should i change the smaller than symbol to greater than and vise versa?

[teal turret]

they didnt change anything

they just moved the 20p to the other side

of the <

and then wrote it in front

[stark pike]

oh my god

thanks for pointing it out again

[nocturne minnow]

Technically they did

[stark pike]

im so blind

[nocturne minnow]

Whenever you divide/multiply by a negative

[stark pike]

i see

thanks alott

.close

[obtuse pebble]

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[smoky bear]

greetings, how does one prove these properties? for the additive inverse property of complex arithmetic, is it as easy as transposing $\alpha$ to the other side of the equation and stating that since $\alpha \neq - \alpha$ and $\alpha$ is a unique element of $\mathbb{C}$, there does exist a $\beta$ such that $\beta = -\alpha$ and $\alpha +\beta =0$? (and a similar argument for the multiplicative inverse wherein $\alpha$ is divided on both sides of the equation, $\beta = \frac{1}{\alpha}$, and, therefore, there exists a unique $\beta \in \mathbb{C}$ such that $\alpha \beta = 1$?)

[warm shale]

Plvzfq_rit

[obtuse pebble]

@smoky bear Has your question been resolved?

[smoky bear]

apologies for disturbing but um <@&286206848099549185> ?

a nvm

.close

[obtuse pebble]

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[obtuse pebble]

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[faint ore]

Hi

[faint ore]

Nvm

.close

[obtuse pebble]

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[light moon]

why can this

[light moon]

only simplify to this

if x is not -2

I get that when we start simplifying we get to here

[rigid pine]

The original expression's denominator still cannot be zero.

[light moon]

mhm I thought if you could simplify you could do whatever you wanted

but I guess that makes sense

otherwise what would be the point of having non simplified equations

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[obtuse pebble]

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[rigid pine]

There shouldn't a ^2 here.

[light moon]

shouldn't there?

[rigid pine]

No.

[rigid pine]

This simplifies to it without the ^2.

[light moon]

without the square root of 2 it just expands to x^2 + 2x

[obtuse pebble]

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[cosmic carbon]

can someone tell me why lim(√(x²-9) / x) as x approaches +∞ is 1?

[distant moth]

Take the highest powers of x common from the numerator and denominator, then try to simplify

[obtuse pebble]

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[fiery parrot]

heeeeeeeeeelp

i have tried calculus, but i is difficult

[timid silo]

have you tried implicit differentiation?

[obtuse pebble]

@fiery parrot Has your question been resolved?

[timid silo]

14th term of 120,130,140

Help

[timid silo]

How can we do this without that

[timid silo]

I was asking more for his working rather than whether he'd actually tried it

[timid silo]

Answer 10920

[obtuse pebble]

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[blazing imp]

I am trying to solve this number theory puzzle but I feel completely clueless...could anybody help to offer some hints? I think I need to try to factor the number into primes but the problem is I cannot tell who got the sum and who got the product...here is the question

Given are two integers both greater than one, for example, 2 and 3. There are two people, A, and B. One of them knows the sum of the integers (e.g., 5), and the other knows the product (e.g., 6). Both A and B` know whether they know the sum or the product, and the rules of the game.

A says to B: You cannot infer the value of the two numbers B to A: But, now I know the two numbers. A to B: But, now I also know the numbers!!!<@&286206848099549185>

[obtuse pebble]

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@blazing imp Has your question been resolved?

[obtuse pebble]

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[timid silo]

Hi, i got a question says"if z in C his imaginary part is 164 and K real positive number 4i = z/(z+k) find k."

[timid silo]

I tried to let Z be h-164i² but is this really correct ?

$\frac{h}{h-164i^{2}+k}+\frac{164}{h-164i^{2}+k} = 4i$

[warm shale]

Shihab

[timid silo]

h ∈ R; k ∈ R+

How they ended up with 4i ?

[timid silo]

if i worked with that i will end up with k = (1/4)i and that is not in R+, i think i made a lot of mistakes but i don't know where.

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

@scenic sand You have an idea ?

[scenic sand]

Heya

[scenic sand]

No you may tag helper

Cause it have been more than 15 minute

Tag helper if u need

[timid silo]

Yeah

i just saw you typing so i thought you may have an idea

<@&286206848099549185> Anyone have time to help me ?

[scenic sand]

The statment is false

[timid silo]

yes

so when i said z = h+164

it's wrong

don't have idea what should i do

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

.close

[obtuse pebble]

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[spare crypt]

How would u do part b?

[rain flicker]

If you already have |PQ|, find |OP| and |OQ|, then you can use the cosine theorem to find the angle

[spare crypt]

What’s that?

[rain flicker]

$c^{2}=a^{2}+b^{2}-2ab\cdot cos(\gamma)$

[warm shale]

draganb

[rain flicker]

Gamma is the angle you're looking for and c is |PQ|

two other sides are a and b

[spare crypt]

Yeah but the question is asking for the magnitude

How will finding the angle even help?

[rain flicker]

You mean magnitude like the answer being represented as some constanc c*pi ?

[spare crypt]

Yeah

[rain flicker]

Have you found |PQ|, |OP| and |OQ| ?

[spare crypt]

No i dont know how to

[rain flicker]

Do you know how to find the distance between two points on xy graph?

[spare crypt]

Yeah the distance formula right?

[rain flicker]

Yes, you can just take the Real part of P as x, its Imaginary part as y

You can do the same to all the points on a complex plane

And you can calculate the distance same as you do with points on a normal xy graph

In your case P would be P(2,2)

and Q(1, sqrt(3))

[obtuse pebble]

@spare crypt Has your question been resolved?

[obtuse pebble]

@spare crypt Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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