help-49

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[twilit field]

Let G be a cyclic group of order $6$ , how many of its elements generate G

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

I'm kind of confused here

so the elements are of the form $e^{-1}.1,e,e^2,e^3,e^4, e^5$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

Intutively, I suspect e^{-1} and e

[radiant roost]

e inverse?

[twilit field]

yeah, all integral powers lie in the cyclic group, right

[radiant roost]

o

[twilit field]

of the generator

Let me check the defn just to be safe

[radiant roost]

it should be fine

[obtuse basin]

Yeah, can you prove that though?

[twilit field]

I'm not too sure. I could try

essentially my argument is the only numbers whose span is Z are -1 and 1, where the scaling factor is an integer

all even numbers would generate only even powers

[radiant roost]

e^-1 is e^5 btw

[twilit field]

hmm, yeah

[obtuse basin]

Right, so an element of a cyclic group is a generator iff its power is coprime to the size of the group

[twilit field]

I guess I should prove that

[radiant roost]

they only ask you to work with a group of order 6 so it would be doing more work than necessary

u can just check e^-1 generates the group and 1, e^2, e^3, e^4 don't

[twilit field]

The next question asks for a group of arbitrary order n

so might as well do it now

[radiant roost]

ah

[twilit field]

hmm

[twilit field]

so I prove this theorm for that I suppose

[carmine sigil]

The theorem should be pretty straightforward. If the power isn't coprime you'll arrive back at 1 prematurely and therefore not generate the entire group

[twilit field]

I'm still not used to thinking about this kind of stuff, I guess that's why I'm finding this confusing

[twilit field]

hmm, yeah

makes sense

[pine thicket]

would you arrive at 1?

oh yes

i was thinking of e^1

[twilit field]

Suppose the group is of order $n$, let $e^k$, k<n, be a generator, then $e^kl$ should be multiples of 1 though n, however, this won't happen unless k is co-prime to l

[grand pond]

ƒ(Why am. I here)=I don't Know

[carmine sigil]

That's a little sparse on justification.

Maybe consider the least common multiple perhaps

[twilit field]

if $k,l$ are co-prime their lcm is kl

[carmine sigil]

If they are coprime

[grand pond]

ƒ(Why am. I here)=I don't Know

[carmine sigil]

Right, which means if we consider k, 2k, etc the first value which is divided by l is k

Because otherwise kl would not be the least common multiple, definitionally.

So as a result the smallest value of x, such e^(kx) = 1 is l.

[twilit field]

yes

[carmine sigil]

Then we can prove that the values e^k, e^(2k), etc are distinct up to the size of the subgroup they generate.

The above is probably already proven by your text

[twilit field]

Ah, I see

Thanks

[carmine sigil]

Yw

[twilit field]

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[last slate]

Hello.Doing LA
I understand that with any 2 perpendicular vectors we can define any point in that plane.

[last slate]

But why do they have to be perpendicular?

[twilit field]

They don't have to be perpendicular

just have to be linearly independent

Like (1,0), (1,1) aren't perpendicualr, but using them, I can obtain any vector on the Xy plane

Which book are you using

[last slate]

$a\vec{v} + b\vec{u} \neq 0$

[last slate]

This relation?

[twilit field]

well, almost

except when a and b are both 0

[last slate]

I'm just watching a video rn.
But I will be using Anton resse elementary linear algebra book

[last slate]

Give me some time

[twilit field]

sure

[last slate]

I get it

I remember a statement saying that we can build any vector using just 2 arbitrary vectors which are perpendicular

Got confused

Thanks

.close

[midnight plank]

Channel closed

Closed by @gleaming geode

Use .reopen if this was a mistake.

[twilit field]

That;s only if the vector space is 2 dimensional

[last slate]

We are talking about a plane here aren't we ?

2d?

.reopen

[midnight plank]

✅

[twilit field]

I mean here I assume you're talking about R^2 so yes

But there is something called a vector space too, and basis vector and more

for instance if two matrices are linearly independent, their span forms something called a 2d vector space

but I don;t think you've covered all that yet

sorry

[last slate]

Ah

[last slate]

So I hat j hat

[twilit field]

That's the standard basis, yes

[last slate]

Vector space means any arbitrary vectors which replace I hat and j hat

And don't have to be perpendicular

[twilit field]

uh, no

[last slate]

😔

[twilit field]

a vector space is a set satisfying certain axioms

you'll come to it soon

don't worry

[last slate]

Dang thought I got it

[twilit field]

you did

for R^2

[last slate]

Alright

[twilit field]

Sorry

[last slate]

Lol why sorry

Thanks

.close

[midnight plank]

Channel closed

Closed by @gleaming geode

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[viral dagger]

find all positive integers $(a,b)$ such that $3^a+3^b+a!+b!+1$ is a perfect square

[grand pond]

Skissue ping4response

[viral dagger]

i tried taking mod 3 and mod 4 and couldnt really get anything out of it

[pine thicket]

try bigger mods

[viral dagger]

like?

[pine thicket]

8

[viral dagger]

n^2 mod 8=0 or 1 iirc

[pine thicket]

or 4

[viral dagger]

eh?

oh yeah

0 1 4 on rhs

lhs is (1 or 3)+(1 or 3)+1 (assuming a,b>4)

oh

so do we guess and check for a,b<=4

[pine thicket]

4! = 0 mod 8, so you dont need to worry about a=b=4

[viral dagger]

aa

(1,1) (1,2) (2,1)

hype thanks

.close

[midnight plank]

Channel closed

Closed by @viral dagger

Use .reopen if this was a mistake.

[pine thicket]

you have to be careful

you could still have a<4<b

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[pine thicket]

@viral dagger

[pine thicket]

you have to check to make sure a<4<b cases dont work

[viral dagger]

oh god uhh

well take a<b

them (1 or 3)+(1 or 3)+1+(1 or 2 or 6)=0 or 1 or 4

move the 1 to get
(1 or 3)+(1 or 3)+(1 or 2 or 6)=0 or 3 or 7

(1,1,6) (1,1,1) (3,3,1) (3,3,2) works

hm

if the thid is 6 then the first would be 3, which has no sol

[pine thicket]

wait im silly, i think mod 8 might still work for a<4<b cases

[viral dagger]

if the third is 2, then rhe first is 1 (no sol)

if the third is 1, then the first one is 1 (has sols)

so it changes to $5+3^b+b!$

[grand pond]

Skissue ping4response

[pine thicket]

break it into cases of a=1, a=2, and a=3

mod 8 appears to take care of most cases for the last two

[viral dagger]

take mod 3 you get 2+b! being either 1 or 0 which cant happen if b>=4

right?

[pine thicket]

yes

in fact it wont work for b=3 either

[viral dagger]

yeah but the presumption was b>=4 at the start

ok is there any more edge cases that i missed

[pine thicket]

i was working the cases my way, but it seems like you got something to work

yeah it appears your solution works

[viral dagger]

nice

thank you!

[pine thicket]

so it looks like your solution set is just (1,1),(1,2),and(2,1)

youre welcome

.close

[midnight plank]

Channel closed

Closed by @pine thicket

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[graceful ferry]

Guys can you give me an example of a function $f:\mathbb{R} \rightarrow \mathbb{R}$ that's continous but not uniformly continous?

[grand pond]

prograce

[graceful ferry]

I'm just trying to better understand the difference

If there even exists a function like this ?

Maybe not on R

[frigid mountain]

sin x^2

it varies very fast when |x| is large

so it is not uniformly continous

[graceful ferry]

I see

[surreal moon]

I think even just y=x² if I'm not crazy

[tribal temple]

I was about to say this (you can't "bound the distance" independent of the points you choose)

[frigid mountain]

oh, yes

sin x^2 is a continuous, bounded but not uniformly continuous function

[surreal moon]

This example i like most since it is also bounded

,w plot sin(x)*sin(1/x)

[grand pond]

Results provided by WolframAlpha

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

Requested by spelwerdzrite

[surreal moon]

That one may be uniform

[graceful ferry]

I just cannot understand it intuitevely💀

[surreal moon]

Uniform continuity or my silly example function?

[graceful ferry]

Uniform continuity, why all the examples work😃

[surreal moon]

@frigid mountain gave you the best example, ill defer to to him to explain if he wants to

[graceful ferry]

I would appreciate it

[frigid mountain]

Uniform continuity is "stronger" continuity. It means that the $\delta$ does not depend on x. That is, give a positive $\epsilon$ for every interval, no matter where it is, as long as it's shorter than $2\delta$, the range of y will not be more than $2\epsilon$

[grand pond]

CPearl

[frigid mountain]

So, in the case of x^2, when the interval is going away from 0, the range of y tends to be infinity

and it will exceed $2\epsilon$

[grand pond]

CPearl

[frigid mountain]

so it is not uniformly continuous

[graceful ferry]

Yeye

I'm asking so much but what about sinx^2 ? It's intervals of y don't go to infinity it's bounded [-1,1]

[frigid mountain]

as for sinx^2, we can consider the case that $\epsilon=\frac{1}{2}$

[grand pond]

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

[frigid mountain]

,w plot sin(x^2)

[grand pond]

Results provided by WolframAlpha

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

Requested by cpearl0_32852

[frigid mountain]

for any fixed $\delta$, when the interval is going away from 0, the range of y tends to be [-1, 1]

[grand pond]

CPearl

[frigid mountain]

so it will finally exceed $2\epsilon=1$

[grand pond]

CPearl

[graceful ferry]

Hmmm

Thx!

.close

[midnight plank]

Channel closed

Closed by @graceful ferry

Use .reopen if this was a mistake.

[frigid mountain]

cya

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[last slate]

<@&268886789983436800>

[lunar ocean]

<@&268886789983436800>

[tawdry laurel]

<@&268886789983436800>

[twilit field]

!staus

!status

[midnight plank]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[twilit field]

well, you know 0.5mv^2=10 and ma=10

what can you get from that

hint : ||u^2=2as|| here

[midnight plank]

@last slate Has your question been resolved?

Channel closed

Closed by @humble meteor

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[balmy flume]

Does comparing coefficient only work when the L.H.S and R.H.S are identical
Or even if they are not identical?
We cannot say like
2x^2+4 identical to x+2
So comparing coefficients should not work if the L.H.S and R.H.S are not identical

[balmy flume]

I am asking this question based on partial fraction decomposition

Which can be done through two methods which are substitution and comparing coefficients

In comparing coefficients we could also make the L.H.S the same degree as the R.H.S and then use comparing coefficients

EASY QUESTION

when anybody answers please reply or tag me

[midnight plank]

@balmy flume Has your question been resolved?

Channel closed

Closed by @balmy flume

Use .reopen if this was a mistake.

[balmy flume]

.reopen

[midnight plank]

✅

@balmy flume Has your question been resolved?

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[balmy flume]

.reopen

[midnight plank]

✅

@balmy flume Has your question been resolved?

[polar mortar]

Yes, if you’re comparing two polynomials in standard form you can look at coefficients of corresponding terms, by degree.
If all the coefficients are the same, on the same degree terms, then the polynomials are equal.

[balmy flume]

Should the two functions on both sides of the equation be identical

?

or even if they are just equal it works

[polar mortar]

If you’re saying the two polynomials are equal, for all x, then yes they should be identical

[balmy flume]

I am asking for general application of comparing coefficients in polynomial factorization

[balmy flume]

that's if they are identical

Should they be identical for applying comparing coefficients

[polar mortar]

What do you mean? Give me an example

[balmy flume]

like this example?

If the degree of the L.H.S and R.H.S is different

[polar mortar]

Then the polynomials are different and have different coefficients.

[balmy flume]

Is this a yes or no?

[polar mortar]

You can compare coefficients of different polynomials, they’ll just be different somewhere.

[balmy flume]

So can we compare coefficients for x^2+2x^3=2x^4+x^3?

[polar mortar]

You can compare them, yes. And see that those polynomials cannot be equal for all x.

[balmy flume]

So being identical is not needed?

[polar mortar]

Because you compare coefficients of corresponding degree terms.

[polar mortar]

No, they do not need to be identical. Setting them equal means different things though, and you just reach different conclusions.

Like, saying x+4 = 3 is setting two polynomials equal to each other.
They are not identical, and cannot be equal for all x. But we can see that they are equal for x=-1.
Vs
x^2 - 1 = (x-1)(x+1) are identical. And if we distribute out the RHS to standard form we can compare coefficients to see they are all the same.

So those polynomials are equal for all x

[balmy flume]

So its nuanced and based on the structure of the polynomials

[polar mortar]

Yes

[balmy flume]

And interplays with the domain and range of the function

[polar mortar]

We set identical polys equal in partial fraction decomposition to get a system of equations and solve for unknown coefficients all the time.

We set different polys equal all the time to find roots and intersections.

[balmy flume]

If identical then domain and range is all real numbers

[balmy flume]

Is my understanding ok?

[polar mortar]

That’s not true.

[polar mortar]

Generally seems fine.

[balmy flume]

What would be a better way to conceptualize it?

[polar mortar]

What exactly are you trying to conceptualize?

[balmy flume]

Solid understanding of where comparing coefficients can be applied and where it cannot be applied

And also the relationship between a function being identical and its domain and range

[polar mortar]

The domain of any polynomial is all reals.
The range depends on the polynomials.
If the degree is odd then the range is all reals.
If the degree is even then there is some maximum or minimum.

There’s nothing required about equaling another polynomial.

And take x^2 and 2x^2. Different polynomials with identical domain and range.
So I’m not sure what that has to do with setting them equal.

[balmy flume]

Ah yes quartic quadratic then all real and cubic quintet then maximum minimum because of the graph structure

[balmy flume]

So your saying comparing coefficients has no relationship or you are not sure with the domain and range

[polar mortar]

That’s backwards, but yes.

[polar mortar]

None that I know of

[balmy flume]

They are mostly separate topics but if they are combined it would be a tough question

[midnight plank]

@balmy flume Has your question been resolved?

[balmy flume]

Ty so much I think these sub topics are way more clearer now

[midnight plank]

Channel closed

Closed by @balmy flume

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[graceful ferry]

Let V be a vector space and U be a non trivial subspace of V
Prove that there doesn't exist $v_0 \in V$ such that every $v \notin U$ , $v \in V$ satisfies: $v \in sp{v_0}$

[grand pond]

prograce

[graceful ferry]

Assume that there exists a $v_0 \in V$ like this, divide into two cases:\
If $v_0 \in U$ and if $v_0 \notin U$

[grand pond]

prograce

[graceful ferry]

How do I continue from here ?

[runic hamlet]

take u in U. what can you say about u+v_0 ?

[vague seal]

sp = span?

[graceful ferry]

That it's in U?

[graceful ferry]

Yes

[vague seal]

am i dumb because take R^3

V = R^3

then take U = <(1, 0, 0), (0, 1, 0)>

and v_0 = (0, 0, 1)

V \ U = <(0, 0, 1)> \ {0}

maybe that's the case

that 0 is the problem

[graceful ferry]

Yes

[vague seal]

but then again there's no indication that v has to BE the span(v_0)

[nova yoke]

V \ U is not just <(0,0,1)> \ {0}, for example (1,1,1) is in V \ U

[vague seal]

You're completely right

and that's also exactly why it's impossible

youre right :D

i was being dumb

[graceful ferry]

I wish I could immediately understand intuitevely like this🥹

[vague seal]

and that's what denascite was talking about

think of u + v_0

is that in U?

is that in span(v_0)?

let me explain bungos example by also taking denascites point

assume that my faulty example was a proper example

that would mean that all vectors in V that are not in U are elements of a span of a v_0

that means that for all vectors v

v is not an element of <(1, 0, 0), (0, 1, 0)>

but v is an element of <(0, 0, 1)>

now bungo gives a counter example

namely (1, 1, 1)

which is a linear combination of v_0 with some other vector

which is (1, 1, 0)

but since it is such a combination it is not an element of the span(v_0)

[graceful ferry]

I understand

[vague seal]

so the point is take some v_0

and some u in U

take their linear combination

and look at the result

is it part of U

if so then it's not interesting

if it is not

...

:D

[graceful ferry]

Oki I tried doing it I think thanks

.close

[midnight plank]

Channel closed

Closed by @graceful ferry

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[grim vector]

,rotate

[potent star]

What’s the range

[hearty rune]

you tell us

[potent star]

Smaller than 1

[grim vector]

🫵

[hearty rune]

that it?

[potent star]

But like there’s another

Part idk

[hearty rune]

the domain is restricted, what is x at its lowest points?

[potent star]

-1

[hearty rune]

then what is y

[potent star]

0

Ohhhh

[grim vector]

Wunderbar

[midnight plank]

@potent star Has your question been resolved?

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[lyric scroll]

guys what did i do wrong

,rotate

[lyric scroll]

i forgot the e^3x

my bad

[midnight plank]

@lyric scroll Has your question been resolved?

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[wild timber]

i don't understand a single part of the answer

1. how is the lower limit and upper limit of integration what they are
2. how is the integral what it is

for this second question too i'm even more lost on the components of the answer

1. how to get upper and a lower limit of integration?
2. the stuff inside the sum
3. the 2/n what does that even mean

[midnight plank]

@wild timber Has your question been resolved?

[midnight plank]

@wild timber Has your question been resolved?

[midnight plank]

Channel closed

Closed by @wild timber

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[viral dagger]

find all positive integer $(a,b,c)$ such that there exists positive integer $k$ that satisfies
$$k^2=2^a+2^b+2^c+3$$

[grand pond]

Skissue ping4response

[viral dagger]

taking mod 4 then if a,b,c>1 then it becomes k^2=3 mod 4 which is impossible

if all of them are 1 then its k^2=9=1 mod 4
if 2 of them are 1 then its
k^2=3 mod 4 (imposs)
if 1 of them are 1 then its
k^2=1 mod 4 (poss)

for all 1 if we check its obvious it works

say its the c thats 2, you get
$$2^a+2^b=k^2-5$$

[grand pond]

Skissue ping4response

[viral dagger]

im not sure hiw to continue this

[jaunty canopy]

does mod 8 work?

[viral dagger]

emmm

well a,b>2 doesent work

if none of them are greater than 2 and greater than 1 thst doesent work

if one of them is grearer thsn 2, you get 2^a+5=(0 or 1 or 4) which works for a=2 only

so it simplifies further to $2^a=k^2-9=(k-3)(k+3)$

[grand pond]

Skissue ping4response

[jaunty canopy]

i'm gonna go to bed now, gl with the problem!

[viral dagger]

oh okai

gn!

the onlt pair that works is k=5

so (1,2,3)

huh.. but that doesent give anything uhh

oh

its 124

.close thanks ly

[midnight plank]

Channel closed

Closed by @viral dagger

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[sharp coral]

,rcw

[sharp coral]

try to write down as many angle measurements you can find from the given information

please specify angles with 3 points, it's difficult to tell which angles you are referring to

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[sharp coral]

,rccw

[sharp coral]

i agree with all of the angles marked in the figure but i can't figure out how you would get x = 82

[rugged otter]

what grade math is this?

[sharp coral]

well at this point we can use the fact that BCD is isosceles

[viral fable]

180-48 = 132

since it is isosceles divide 132 by 2

angle bcd = 66, angle dbc = 66

therefore x has to equal 34

[sharp coral]

we have that angle BCD = angle DBC because triangle BCD is isosceles

and angles BCD, CDB, and DBC must add to 180 since they form a triangle

no, you don't divide by 2 twice

you have that x + 32 = angle BCD

yes

what 2?

[sharp coral]

the 32 is already marked here

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[wanton shore]

x²+2xy-y²= 6 find (x²+y²)min

[wanton shore]

In this qstn can we put x=rsinθ and y = rcosθ

?

[fallow scarab]

You can add 2y^2 to both sides then the left side will become a perfect square

Oh nvm that's not helpful

[twilit field]

sounds like a good idea

[wanton shore]

But i can't seem to understand why it worked

Because the eqn is not a circle

[twilit field]

you're converting it to polar form

basically

[wanton shore]

Then why this substitution worked?

[knotty frost]

maybe a hyperbolic sub?

[wanton shore]

What is hyperbolic sub

[knotty frost]

hyperbolic substitution

[wanton shore]

I haven't studied that

[wanton shore]

Got it

[knotty frost]

oh

I say this because it could help to simplify

[wanton shore]

Ok thanks

.close

[midnight plank]

Channel closed

Closed by @wanton shore

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[twilit field]

Let $x,y$ be elements of a group $G$. Assume that each of the elements x,y and xy are of order 2. Prove that the set $H = {1,x,y,xy}$ is a subgroup of order 4.

[grand pond]

ƒ(Why am. I here)= I don't know

[twilit field]

so here the order in the first line and second line mean different things

right

So here all I really have to do is prove closure, identity and inverses exist

$1$ belongs in the group, thus so does the identity. Every element is of order $2$ and thus its own inverse.

[grand pond]

ƒ(Why am. I here)= I don't know

[twilit field]

I'm not sure of how to concisely prove closure

I feel like proving xy=yx is sufficent

but not sure

[frigid mountain]

also xxy, yxy, xyx, xyy, I guess

[twilit field]

yeah, but xxy=y, and xyy=x

so yeah

xy=yx, yxy and xyx

[frigid mountain]

yxy=x can be shown by xyxy=xx

[twilit field]

fair

yes

and xyx=y

[frigid mountain]

and then xy=yx because yxy=yyx

[twilit field]

hmm

ooh

cool

thanks

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[twilit field]

I'm trying to find The directional derivative here
\
$\del{f(x,y)} = 2\sqrt{y} + \frac{x}{\sqrt{y}}$
\
$\del{f(3,4)} = 4 + \frac{3}{2}$
\
Thus the directional deriavtive is $5.5(4,-3)$

[grand pond]

ƒ(Why am. I here)= I don't know

[rough birch]

um

okay so what are $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$?

[grand pond]

Arnavutköy

[sharp coral]

the gradient is a vector

[twilit field]

2\sqrt{y} and x/\sqrt{y}

[sharp coral]

so by definition [ \nabla f = \ab(\pdv fx, \pdv fy) ]

[grand pond]

cloud

[twilit field]

oh right

I got confused

oops

[rough birch]

wait isn't it $2\sqrt{y}$?

[grand pond]

Arnavutköy

[twilit field]

yeah, that's what I wrote, right

[rough birch]

oh yeah

so evaluate these derivatives at $(3,4)$

[grand pond]

Arnavutköy

[twilit field]

👍

$(4,3/2) \cdot (4,-3)$

[grand pond]

ƒ(Why am. I here)= I don't know

[rough birch]

yeah thats what you do

[twilit field]

16-9/2 = 23/10

[sharp coral]

the direction vector must be a unit vector

[twilit field]

so (4/5,-3/5)

thanks!

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[rough birch]

wait no

[sharp coral]

that's been the convention in all the textbooks i've used

[twilit field]

let me confirm what my book says

[rough birch]

the gradient is something else though

wait what are you claiming is normalized?

[sharp coral]

the vector representing the direction of the directional derivative must be normalized

[rough birch]

according to wikipedia there isn't any such normalization

[sharp coral]

it is in my textbook and paul's online notes and wolfram mathworld. you should double check in your own textbook, though

and wikipedia does have it in some places but not in others (since sources vary somewhat), but every source on intro multivar seems to adopt that convention

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[paper hemlock]

Need help with this integral. I think I’ve done the partial fraction thing right but I have no idea how to get the values of A and C (I can find B because it doesn’t multiplicate to 0)

[paper hemlock]

Feel free to correct my working if it’s wrong

<@&286206848099549185>

[modern shard]

it would be Cx + D

since it goes in quad.

everything except it is correct

[midnight plank]

@paper hemlock Has your question been resolved?

[paper hemlock]

Wait why

[modern shard]

since it's type of x^2 + b^2

[paper hemlock]

How do you mean?

[modern shard]

see whenever there is a non-repeating quadratic factor in denominator we write as of type (CX+D)/quad

[paper hemlock]

So the denominator still stays the same but I just have to change the C to Cx + D?

[modern shard]

yup

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[paper hemlock]

.reopen

[midnight plank]

✅

[paper hemlock]

So how do you keep going after that

[modern shard]

compare coefficients

after simplifying

[paper hemlock]

I have

But it came out wrong

[modern shard]

show me your work

[paper hemlock]

But when you put that into solver then it spits out gibberish

Even if the 256 is 1 (1 being the coefficient of 256) it turns wrong

[modern shard]

A would be 0 i guess

my bad

[paper hemlock]

What have I done wrong here

[modern shard]

yeah a sec

B = 4

is correct

[paper hemlock]

Yes that’s the only thing I know for sure

Oh maybe I have to set up a third equation

[modern shard]

A+ C = 0

[paper hemlock]

Fourth I mean

Since 4 unknowns

[modern shard]

[modern shard]

A, B, C, D

yk B

so 3 eqns

[paper hemlock]

How come

[modern shard]

compare coefficients of x^3

since there is no x^3 on LHS

we took it as 0

same goes with x^2

[paper hemlock]

I am still getting wrong answer with that

[modern shard]

what are you getting

[paper hemlock]

The integrating part is easy btw it’s just this is making me trouble

[modern shard]

yeah yeah dw

[paper hemlock]

Not the same graph aka not the same

[modern shard]

i am getting A= -1

C=1

D=0

[paper hemlock]

Still wrong

[modern shard]

aha

[paper hemlock]

Did you find the issue?

[modern shard]

,w simplify (2x+4)/(16+x^2) + (-2)/(x-4) + 4/(x-4)^2

[grand pond]

Results provided by WolframAlpha

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

Requested by cool_vansh

[modern shard]

sure

[modern shard]

we did it correctly, i messed up in handwriting XD

[paper hemlock]

Huh how

[modern shard]

it was messy

uh?

[paper hemlock]

I put it wrong Into the calculator

One hour of losing my mind because i forgot a minus

….

But thanks for the help I got it now

.close

[midnight plank]

Channel closed

Closed by @paper hemlock

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[twilit field]

Let $U_n$ be the group of units of $\Z_n$. Find the multiplication table for $U(12)$ So It's the multiplication table for those numbers that have an inverse in $Z_{12}$. This would be that set of numbers that's relatively prime to $12$, so $5,7,11$ would be the elements

[grand pond]

ƒ( wai ina teacup)= I don't know

[twilit field]

Is that right

[quiet hinge]

yes but whats the multiplication table for U(12)

[twilit field]

I'm figuring out how to LaTeX it

$\begin{tabular}{c|cccc}
\cdot & 5 & 7 & 11&1 \ \hline
5&1&9&7 &5\
7&11&1&5 &7\
11&7&5&1&11\
1&5&7&11&1
\end{tabular}$

[quiet hinge]

you are missing a unit

namely the unit 1

[twilit field]

huh

Is 1 taken to be co-prime to 12 by convention

[quiet hinge]

no its by definition

[strong lava]

not just a convention

[quiet hinge]

gcd(1,12)=1

[twilit field]

fair

got it

[grand pond]

ƒ( wai ina teacup)= I don't know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[twilit field]

oops

[jaunty canopy]

also we want it to be a group so we'd like an identity

[twilit field]

oh right

[
\begin{tabular}{c|cccc}
\cdot & 5 & 7 & 11 & 1 \ \hline
5 & 1 & 9 & 7 & 5 \
7 & 11 & 1 & 5 & 7 \
9 & 7 & 5 & 1 & 11 \
1 & 5 & 7 & 11 & 1
\end{tabular}
]

[quiet hinge]

it is also convention to put the 1 at the beginning of the rows and columns of a multiplication table

[twilit field]

also I've made a mistake

7*5=9

[
\begin{tabular}{c|cccc}
\cdot & 5 & 7 & 11 & 1 \ \hline
5 & 1 & 9 & 7 & 5 \
7 & 11 & 1 & 5 & 7 \
9 & 7 & 5 & 1 & 11 \
1 & 5 & 7 & 11 & 1
\end{tabular}
]

[grand pond]

ƒ( wai ina teacup)= I don't know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[quiet hinge]

9?

afaik 9 not in U(12)

[twilit field]

Hmm

,w 35 mod 12

oops

11

[grand pond]

Results provided by WolframAlpha

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

Requested by physicsrocks

[quiet hinge]

gcd 9 12 = 3

[twilit field]

got it

yeah

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[twilit field]

thanks!

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[twilit field]

Let $S = \R \setminus {-1}$ Define a binary operation $\ast$ on $S$ by$a+b+ab$. Prove $(S, \ast)$ is an abelian group.
\
We first prove that $a \ast b = b\ast a$. $a+b+ab = a \ast b$, and $b \ast a = b+a + ba$ . $a+b+ab = b+a+ba$ as multiplication and addition over the reals is commutative.
\
We now prove the existence of an identity in this group. An identity is such an element such that $a \ast b = a$. We let this identity be $x$ . We thus have $a \ast x = a= a+x+ax$. We thus have $x(1+a)=0$ . Thus the identity in this group is $0$.
\
The inverse is defined to be that element that when composed with any other element gives us the identity element. We thus have $a \ast a^{-1} =0$, or $a^{-1}+aa^{-1}=-a$, or $a^{-1} = \frac{-a}{1+a}$.
\
\
I'm not too sure of how to prove closure though

[grand pond]

ƒ( wai ina teacup)= I don't know

[jaunty canopy]

oh this is kinda funny

this was a problem on our groups problem sheet too lol

there's a brute force way to do it, and a nice way to do it

nice way: ||a+b+ab = (a+1)(b+1)-1||

[twilit field]

ah, and that proves closure how?

[modern sapphire]

• is closed under R
  Only way * is not closed under S is if a*b = -1.

[twilit field]

ah yes

[modern sapphire]

This factorization makes it easy to show that

[twilit field]

the span doesn't have to be $\R$

[grand pond]

ƒ( wai ina teacup)= I don't know

[twilit field]

I forgot that

Thanks

[jaunty canopy]

as in this was a nice way to prove

• abelian
• associative
• invertible
• closed
  etc.

[twilit field]

Ah

I see

Nice

Thanks

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[left rover]

can somebody explain why the ray joining two homogeneous 3D points A and B is the vector span of A and B? i.e. aA + bB?

[rough birch]

wait i think you need to add a condition

that $a+b=1$

[grand pond]

Arnavutköy

[rough birch]

then this fact is true

[left rover]

https://tenor.com/view/dis-belief-surprised-oh-gif-10085527

but can you explain why that's the case?

[rough birch]

we can represent this using vectors

[runic hamlet]

and a,b >= 0

[rough birch]

if it is a segment, yes. if it is a line, no.

if its a ray, then depending on which direction we start in, its either $a>=0$ or $b>=0$

[grand pond]

Arnavutköy

[rough birch]

okay, so take a point $M$ on the segment connecting $A$ and $B$ okay.

[grand pond]

Arnavutköy

[rough birch]

We will let $O$ denot the origin

[grand pond]

Arnavutköy

[rough birch]

okay, so do you agree that $\rightarrow{AM}=c\rightarrow{AB}$ for some constant $c$

[grand pond]

Arnavutköy

[rough birch]

wait how do i do the thing again

[left rover]

[rough birch]

oh wait, okay, so do you agree that $\overrightarrow{AM}=c\overrightarrow{AB}$ for some constant $c$

[grand pond]

Arnavutköy

[left rover]

this seems intuitive to me but I cannot reason why

[left rover]

I meant this

[rough birch]

its because $A$ $M$ and $B$ are collinear

[grand pond]

Arnavutköy

[rough birch]

does that make sense?

[left rover]

ahh yes

[rough birch]

okay, so moreover, do you agree that $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$, where $O$ is the origin

[grand pond]

Arnavutköy

[left rover]

wait A, B, and O are all homogeneous yes?

[rough birch]

yes

[left rover]

yes

i agreezies

[rough birch]

okay, so we have that $\overrightarrow{AM}=c\overline{OB}-c\overline{OA}$ for some constant $c$

[grand pond]

Arnavutköy

[rough birch]

(those should be vectors not segments)

[left rover]

yes

[rough birch]

and additionally, we now use that $\overrightarrow{OM}=\overrightarrow{OA}+\overrightarrow{AM}$

[grand pond]

Arnavutköy

[rough birch]

and since $|overrightarrow{AM}=c\overrightarrow{OB}-c\overrightarrow{OA}$, this means that $\overrightarrow{OM}=(1-c)\overrightarrow{OA}+c\overrightarrow{OB}$

[grand pond]

Arnavutköy

[rough birch]

whoops

and since $\overrightarrow{AM}=c\overrightarrow{OB}-c\overrightarrow{OA}$, this means that $\overrightarrow{OM}=(1-c)\overrightarrow{OA}+c\overrightarrow{OB}$

[grand pond]

Arnavutköy

[left rover]

okay, yes

but the coefficients are related somehow and not free

[rough birch]

yeah, specifically, they sum up to $1$

[grand pond]

Arnavutköy

[left rover]

ahhhhh

got it. so any coefficients as long as they sum to one

thanks a lot!!

.close

[midnight plank]

Channel closed

Closed by @left rover

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[last slate]

Is the integral of |sin(x)| from 0 to x = (2[x/pi]) + (int from 0 to pi{x/pi - [x/pi]} of sinx)?

[last slate]

Where [.] is the floor function.

[rough birch]

um close

i think you want to take the integral from 0 to $x-\pi\left\lfloor\frac{x}{\pi}\right\rfloor$ of $\sin(x)$ instead

[grand pond]

Arnavutköy

[last slate]

(2[x/pi]) + (int from 0 to pi{x/pi - [x/pi]} of sinx)?

x is anything

idk

'{}' does not represent the fractional part; it's just a bracket.

[midnight plank]

@last slate Has your question been resolved?

Channel closed

Closed by @humble meteor

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[twilit field]

prove that the inverse of $g_1g_2 \dots g_n$ is $g_n^{-1} g_{n-1]^{-1} \dots g_1^{-1}$
\
Would induction be the right idea here?
\
Let the inverse of $g_1g_2 \dots g_n$ be $g_n^{-1} g_{n-1]^{-1} \dots g_1^{-1}$
\
We thus have $(g_1g_2 \dots g_n )\cdot (g_n^{-1} g_{n-1]^{-1} \dots g_1^{-1})=1 $
\
We then multiply across by 1 by post multiplying by (g_n g_n^{-1}$ after the nth position
\
This gives us $$(g_1g_2 \dots g_n g_{n+1} )\cdot (g_{n+1}^{-1}g_n^{-1} g_{n-1]^{-1} \dots g_1^{-1})=1 $
thus giving us the desired result

[grand pond]

ƒ( wai ina teacup)= I don't know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[strong lava]

induction works yes

[rough birch]

well we don't even need to do induction to prove that step

[strong lava]

but arguably it's just "through direct computation"

[twilit field]

fair

yeah

[rough birch]

we simply need to show that $g_1\cdots g_ng_n^{-1}\cdots g_1^{-1}=e$

[grand pond]

Arnavutköy

[rough birch]

and we can just keep on replacing the innermost terms with the identity and taking them out

and then we are left with the identity

[twilit field]

[rough birch]

the other direction around however

is that you need to show

$g_1^{-1}\cdots g_n^{-1}g_1\cdots g_n=e$

[runic hamlet]

well technically you are doing induction when you say "keep on replacing like that"

[grand pond]

Arnavutköy

[twilit field]

ah yes, induction would work similarly here too

Thanks

[runic hamlet]

its just such an obvious induction that no one really bothers writing it out

[runic hamlet]

wrong order

[rough birch]

oh yeah whoops

[twilit field]

Thanks

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[dry acorn]

I'm plugging it into my calculator but I think I'm messing some part of the equation up. Help pls

[dry acorn]

I figured it out nvm

.close

[midnight plank]

Channel closed

Closed by @dry acorn

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[oak adder]

how

[oak adder]

😢

I’ve been stuck on this problem for a while could someone help

[main owl]

theres no way u can use the sinx/x = 1 rule here right

[oak adder]

i don’t think so

[digital crag]

Comparison test

Are we just trying to find if it converges or diverges?

If so then we can just use comparison test

[oak adder]

Mm

I was trying to evaluate it

[digital crag]

What does the question ask?

[oak adder]

But o haven’t considered that yet

Holdon

[digital crag]

I don’t think u can evulare that

Yeah it’s not an elementary Antiderivative

[sharp wave]

after using by parts you get -sin(x^2)/x how will you plug in the limits there?

[digital crag]

U don’t get -sinx^2/x after using by parts

It’s not possible to integrate this

[oak adder]

Ah

[feral sedge]

u had 2 people tell u the same thing a few days ago

can u at least show ur work or smth

instead of reposting this same question

[oak adder]

Yeah I know

I wasn’t available then

[sharp wave]

one of the terms, i didnt write the one under the integral

[oak adder]

Also

I tried IBP

And got nothing from it

Someone suggested something to do with the gamma function and didn’t elaborate further

[feral sedge]

so this would have been nice

if u tried ibp and "got nothing"

[oak adder]

Here

this is what I got

[feral sedge]

i c

[sage olive]

what tools do you know? this can likely be done by residues

[oak adder]

I tried residue

[sage olive]

didn't work?

[feral sedge]

switch your choice of u and dv

[oak adder]

not really

[sage olive]

feynman's?

[oak adder]

I’ll write it

[oak adder]

Not yet

[sage olive]

maybe the function e^(iz^2) / z^2 over an indented semicircular contour

[oak adder]

Hollldon a minute

Hollllldon

I think I see something here

Unless I did this wrong

[feral sedge]

derivative of sin(x^2) is not sin(2x)

[oak adder]

Oh wait

I was thinking sin^2(x)

my bad

[last slate]

idk how to evaluate the integral tho 😢

[sage olive]

my proposal

[oak adder]

I see

But wait

I think the other idea works aswell

Just need to integrate a bit more

Or

[sage olive]

the ibp?

[oak adder]

Ok I don’t know why I keep mixing up powers of trig and input powers

[sage olive]

ibp isn't gonna work

[oak adder]

I just ended up with a special integral

yeah

[sage olive]

if ibp would work, the integral wouldn't be nonelementary

you'd ultimately have to resort to some sort of feynman or residue or other trick anyway

[sage olive]

that technically is a known result at this point

the fresnel integral

[oak adder]

I haven’t looked at fresnel integral

Let me try your approach

[sage olive]

i get to here for now

[oak adder]

I just got 0

hm

The calculator also gave me 0

[sage olive]

the integral from 0 to inf of sin(x²)/x² is not 0

[oak adder]

Then I didn’t approach it right

Holdon

How would you continue after combining the two integrals?

[sage olive]

parametrize little gamma

i'm trying rn

[feral sedge]

evaluate the left expression at the limits 0 and infinity

fresnel type integrals are evaluated using a contour involving pi/4

gl doing it any other way

[oak adder]

I have a headache

Thank you guys for helping by the way

[feral sedge]

$\int_0^{\infty} \frac{\sin(x^2)}{x^2} , dx = \underbrace{- \frac{\sin(x^2)}{x^2} \Bigr|0^{\infty}}{0} + 2\underbrace{\int_0^{\infty} \cos(x^2) , dx}_{\text{Fresnel C}} = 2\cdot\frac{1}{2}\sqrt{\frac{\pi}{2}} = \sqrt{\frac{\pi}{2}}$

[grand pond]

Mqnic_

[feral sedge]

read proof of central result here https://en.wikipedia.org/wiki/Fresnel_integral

[oak adder]

So the approach is to solve Fresnal C using analysis

[feral sedge]

it's probably the most standard way

[oak adder]

That makes sense

Thank you

Oop I forgot to close

How do I close channel

[sage olive]

.close

[midnight plank]

Channel closed

Closed by @sage olive

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[short zealot]

help!

[austere flume]

Well do you have any options eliminated or are u completely lost

[short zealot]

i'm lost entirely on this because i haven't seen content displayed in this fashion

@austere flume

[austere flume]

Ok lets start off with whether or not its Geometric or Binomial

What are the characteristics of each one

@short zealot

[short zealot]

i believe

binomial represents a more normal distribution as the n increases

and geometric has a discernible right skew

is that correct

[austere flume]

Yea, basically as the number of trials in a geometric distribution increases, the probability of achieving the first success on that trial decreases, and this distribution here doesn't represent that

[short zealot]

awesome

so definitely not d or e

so now a b and c

so n has to be 8 since the scale goes up to 8 but that doesn't narrow it much down

but i get that conceptually

[austere flume]

Now, how do you find the expected value of a distribution

[short zealot]

np

OH

and since it's approximately normal

2.4 looks almost like the mean if it was a denomination on the x-axis

so it's b

awesome

thank you robert downey jr laying down silly

[austere flume]

Np

[short zealot]

expected value typa

anyways bless you bro

.close

[midnight plank]

Channel closed

Closed by @short zealot

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[radiant roost]

what symbol can i use for an arbitrary element of the dihedral group D_4?

[fallow scarab]

g

[radiant roost]

i want to write a function from D_4->S_3

that takes flips to a transposition (1,2)

and non-flips to the identity

do i just explain this in words?

or can i write it with symbols

it is meant to be a homomorphism

like this?

[surreal moon]

Either is fine

If you are concerned about clarity, use words to explain the idea or motivation

[radiant roost]

Okay, thanks 😊

. close

.close

[midnight plank]

Channel closed

Closed by @radiant roost

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[cedar jungle]

Hey! So, I am trying to solve the & graph the system of inequalties here and whenever I am doing the corner points I it is saying that it is incorrect, I even used Desmos to help me out a bit seeing if the graph I am making I am wrong but it still says incorrect. It is fine if you can answer just one of the questions if you want, I am not sure what I am doing wrong. If you want I can send the graph from desmos on here to show.

[midnight plank]

@cedar jungle Has your question been resolved?

[midnight plank]

@cedar jungle Has your question been resolved?

[midnight plank]

@cedar jungle Has your question been resolved?

[midnight plank]

@cedar jungle Has your question been resolved?

[cedar jungle]

rip

[last slate]

maybe you are missing the corner points, the ones you get from intersection of lines

[midnight plank]

@cedar jungle Has your question been resolved?

[lethal path]

not all of the intersections between any 2 lines will be the corner points

you need to mark out the region on Desmos

ideally, you can flip $\ge$ to $\le$ and vice versa

[grand pond]

south

[lethal path]

then the white region will be the one you want

[young scaffold]

Ibrahim bro

Your name sounds like it came from thr bible

Ibrahim and his genealogy/descendants

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[twilit field]

Trying to find the equation of the tangent plane and normal line at this point

[dull ferry]

wut now

z + 1 = xw^y

whats cos

[twilit field]

a. function

Hmm

I first re-write it as $z=xe^{y} \cos(z)$

[grand pond]

ƒ( wai ina teacup)= I don't know

[twilit field]

oops

NO

I'm confused

this is a function of two variables, is it not?

We first differentiate wrt $z$ to get

[grand pond]

ƒ( wai ina teacup)= I don't know

[twilit field]

I think I'm tripping

how do I differentaite this

can someone remind me

Hmm

[lusty steppe]

differentiate implicitly i guess

[twilit field]

how do you mean

I'm really confused

ooh

z is just the output innit

so here is it

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[ancient wyvern]

how do i get the derivative of this

[midnight plank]

@ancient wyvern Has your question been resolved?

Channel closed

Closed by @ancient wyvern

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[dense arch]

hello, can anyone tell me why the magnitude of the cross product of 2 vectors gives the area of the parallelogram?

[dense arch]

like maybe a visualization or something

[sharp wave]

you know the area is bh right?

[dense arch]

yeah

[sharp wave]

you can write h as asin theta

where theta is the angle between the sides

so you get ab sintheta

which is nothing but the magnitude of the cross product of the vectors

[dense arch]

ahh becasue of the formula IIa x b II =IIaII IIbII sin(theta)

[sharp wave]

yess

[dense arch]

ahh okok

thank you very much

.close

.close

[midnight plank]

Channel closed

Closed by @dense arch

Use .reopen if this was a mistake.

[sharp anchor]

Can someone help with this pls

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[lofty scarab]

In the triangular land plot 𝑇𝑈𝑉, two sides are 𝑇𝑈=300m, TU=300m and 𝑇𝑉=500m, with the angle between them being 60 degrees. A straight road passes through the plot, with one end located 120m away from the vertex 𝑈, dividing the land into two equal areas. Find the length of this road in whole meters.

Answer: The length of the road is _ m

[lofty scarab]

can not figure this out

[delicate sage]

You can find the area of KTL triangle

Because you're given the angle

1/2 TK * TL * sin 60

Then you can just consider side KL to be x, and proceed to find the area with the formula of area for scalene triangles

Okk

[lofty scarab]

hmm

[lofty scarab]

i dont have TL

TV is given

[delicate sage]

Find the whole area

And divide by 2

[stark shale]

you can find TL using the fact that Area of triangle KTL is half the area of triangle UTV

[lofty scarab]

ooh

tysm

i didnt even realize that was an option

@delicate sage@stark shale legends

[midnight plank]

@lofty scarab Has your question been resolved?

Channel closed

Closed by @lofty scarab

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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