help-13

[cedar kiln]

@cinder jay Has your question been resolved?

[cedar kiln]

@cinder jay Has your question been resolved?

[cedar kiln]

@cinder jay Has your question been resolved?

[cedar kiln]

@cinder jay Has your question been resolved?

[cinder jay]

does no one know how to help me

[queen stirrup]

wait why do they talk about y

we are only told about k,v,a

also

when they say solutions

[cinder jay]

sin(k*v) = a and using the image to deterine the constants within the interval 0 to 360

[queen stirrup]

how do the coordinates correspond to the solution

cus if a is constant

the coordinates shoudnt change

im kinda confused sorry

[cinder jay]

Well this is the answer the book provided so im clueless too

[queen stirrup]

hmm

the solutions are probably in the form (sin(kv_i),cos(kv_i))

so

since a is constant

there must be two values kv1,kv2

whose sine is equal

rather four

cus there are four solutions

in the range 0 to 360

like sin(x) = sin(pi-x)

and sin(-x) = sin(x)

[cinder jay]

could be cause. the previous problems were like so then right under it there was the other one

[queen stirrup]

hm

but if u look at the diagram

the heights are different

so the sine values are different

have they told you how the diagram really represents the solution?

@cinder jay

*solutions

cus its not making any sense to me

by solution one would think the values of v

like if u take k=1,a=1/4; then sin(k.30 degrees) = 1/4 and sin(k.50 degrees)= 1/4

[cinder jay]

I mean it could be a misprint maybe? because its the only figure on the page

but the question above is the only one asking for k and a

[queen stirrup]

i did some calculation

looks like the solutions of v are actually the first number in each coordinate

so we get

sin(k*0.24) = a

sin(k*-0.24) = a

sin(k*0.97) = a

sin(k*-0.97=a)

but idk

at least the firsst equation

works with the solution u gave

[cinder jay]

thanks

[queen stirrup]

no no im not sure

cus even if u grant these are the equations

we have sin(-x) = -sin(x) => a=-a => a=0

this is messing with my head lol

[cinder jay]

yeah, I think this might been a misprint or something considering I had to ask how to even start doing when it was near the end of the problems on the page

anyway thanks

.close

[cedar kiln]

Channel closed

Closed by @cinder jay

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[queen stirrup]

sorry that i couldnt help

[cinder jay]

all good

[cedar kiln]

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[eager hemlock]

A and B are Orthogonal matrices, and det A + det B=0 . Then is A+B invertible?

[azure swift]

Want hints or?

[eager hemlock]

Sure

Sry my internet delayed so much

Damn

[azure swift]

Well what is ur gut feeling telling u?

i.e what do you think?

[eager hemlock]

Ok I feel like that one is 1 and other is -1, i mean the det

[azure swift]

As in what?

Oh, right, indeed

if det A = 1, then det B has to be -1

and vice versa

thats all fine, what else though, do you think the statment is true or?

[eager hemlock]

I'm not sure

Dunno anything about A+B

Thus I asked

[cedar kiln]

@eager hemlock Has your question been resolved?

[eager hemlock]

Why😭 What's wrong with my question

[azure swift]

Consider a few examples

try and play around with it

to get a better feel for it

before you decide to prove or not

actually, as a hint

start with the lowest possible dimension

1x1

what happens there?

[cedar kiln]

@eager hemlock Has your question been resolved?

[cedar kiln]

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[round frost]

heyyy

[round frost]

how do I find the equation of a expoential

from a graph

[oblique flare]

check at x=-1 ,x=0 , x=1
if the function is not shifted

[round frost]

heres the graph

[round frost]

its shifted

I just know that itll be y= b(x) + 4 rn

[oblique flare]

Check where the point (0,1) got shifted to then

[round frost]

to 0,8?

[oblique flare]

By how many units ?

[round frost]

7 units

[oblique flare]

Yep

So it’ll be something like
y=b(x)+7

[round frost]

but I thought it would be + 4, isnt C the asymptote?

[oblique flare]

Looks like a shifting to the left as well damn

[round frost]

yeah

[oblique flare]

Then yea you did
b(x) +4 right

[round frost]

yes

[oblique flare]

Is parent function given ?

[round frost]

idk what that is

[oblique flare]

Like the function without its transformations

[round frost]

uh no, I just know its an expoential

[oblique flare]

then wait I had an idea
-4 (so we get the exponential only)
we get a^(x+b)=4

[round frost]

well its y= b^(x-h) + 4

[cedar kiln]

@round frost Has your question been resolved?

[cedar kiln]

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[young flume]

I have a question, why isnt this -315? It does land on the same exact line at 45 and 405 but a "terminal" is the other side thats free to move around, but -315 cant move backwards?

[young flume]

or can it

[oblique prawn]

well if u do -315+360=45

so 45 and -315 are coterminal

and 45+360=405 so 45 and 405 are coterminal

[young flume]

o ok

.close

[cedar kiln]

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[cedar kiln]

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[cobalt scarab]

It won’t take anything I try

[cobalt scarab]

How do I write the domain

I tried x < 27

(0,27

)

[0,27]

(0,27]

Nothing works

The length can’t be 0

So y=-4x+108 can’t be 0

So -4x+108 > 0

Ohh wait

I wrote smaller than💀

Oh wait no I’m correct

Cuz yea if -4 is multiplied by anything bigger than or equal to 27 it’ll be 108 which will make y 0 or less

[woeful isle]

So do you have the answer or....

[cobalt scarab]

So x has to be smaller than 27

I do but it won’t take it

Domain should be x < 27 right?

[woeful isle]

I don't think so

[cobalt scarab]

How so?

Where was I wrong?

[woeful isle]

x = 27 - y/4

27 -y/4 > 0

y> 108

108 > y > x

[cobalt scarab]

Wait what’s happening

Why are you solving for y

[woeful isle]

Because y> x is given

If I know y

I know x

[cobalt scarab]

I don’t understand what’s that

Isn’t x the length

And the domain is the possible length

[woeful isle]

Yes

And....

[cobalt scarab]

So we just need to know what x can be

[woeful isle]

Exactly

[cobalt scarab]

So x can be anything below 27

[woeful isle]

If we knew what y could be

[cobalt scarab]

Why is y involved

[woeful isle]

We could also understand what x could be

[cobalt scarab]

But x can’t be 0 right

Cuz length can’t be 0

[cobalt scarab]

So you just make this inequality

[woeful isle]

Brb

[cobalt scarab]

At least that’s how we learned it

[woeful isle]

Back

Anyways where were we

Yes domain

[cobalt scarab]

Hahaha yup

[woeful isle]

So for volume

-4x^3 + 108x^2>0

x<27

[cobalt scarab]

That’s what I found

[woeful isle]

Hmm

[cobalt scarab]

That’s what I’m saying lol

[crimson sedge]

I didn't read the question but I'm assuming it doesn't have to do with the answer

[woeful isle]

Lemme read the question again

[crimson sedge]

It says unexpected character <

In the message

[cobalt scarab]

Yea idk why

[crimson sedge]

Could be the issue

[woeful isle]

(0,27)

This could work?

[crimson sedge]

It just doesn't get the notation for sow reason

Some *

[cobalt scarab]

I tried that as well

[woeful isle]

You have how many attempts

[cobalt scarab]

Unlimited

[woeful isle]

The upper bound isn't included

Open bracket at 27

[cobalt scarab]

For this that’s what it says

I know it was the message after

[cobalt scarab]

.

[cobalt scarab]

Tried this

[cobalt scarab]

And this

[cobalt scarab]

.

[woeful isle]

[0,27)

Only this left

Lol

[cobalt scarab]

Still says right endpoint incorrect

🥲

[woeful isle]

This is sad

[cobalt scarab]

So 27 is incorrect? That makes no sense

Stupid website

[woeful isle]

Wait

I am an idiot

.-.

[cobalt scarab]

Wut why xd

[woeful isle]

Remember 4x+y=108

Put x as 27

What do you get for y

-_-

0

Wasn't y > x

Omg

[cobalt scarab]

But then ) should work no?

[woeful isle]

No

[cobalt scarab]

Cuz then 27 is not included

[woeful isle]

Put it as 26 then

[cobalt scarab]

Only when it’s ]

[woeful isle]

You get y =4

Stil y<x

This violates our condition

[cobalt scarab]

(0,26) still doesn’t work💀

[woeful isle]

It doesn't ik

That was example

[cobalt scarab]

Ohh

[crimson sedge]

(0,21.6)

Can you try that for me

[cobalt scarab]

Wth

That worked how

[woeful isle]

This will work

[cobalt scarab]

Wtf

[woeful isle]

Put x = y

5x = 108

x =21.6

[crimson sedge]

^:)

[woeful isle]

So if x is anything lower than this

y>x

[cobalt scarab]

Wait so 4(27)+y=108?

[woeful isle]

And volume will be positive

Where did 27 come from

[cobalt scarab]

.

[woeful isle]

That was example

[cobalt scarab]

So how do you solve it I’m confused

[woeful isle]

To show how it violated the y>x condition

We calculate the limiting

Where x and y become equal

And then we say if x is anything lower

y will always be greater

[cobalt scarab]

But what is the equation to solve it

[woeful isle]

4x+y=108

[cobalt scarab]

Yea but what do you do

[woeful isle]

Put x=y

[cobalt scarab]

Or do you just plug infinite x values until it’s correct

[woeful isle]

Put x=y

[crimson sedge]

Just ignore all that you need a positive value for both x and y where y will still be greater than x. So let's assume y=x and do 4y+y =108 you'll find that y is 21.6 but x isn't equal to y it's less than it so x<21.6

[cobalt scarab]

So 4x=-y+108?

[woeful isle]

.-.

4x + x =108

Make sense

????

[cobalt scarab]

I’m so confused

[woeful isle]

Read this

[cobalt scarab]

Where did 4y+y come from

Or 4x+x

[woeful isle]

You know what substitution is

[crimson sedge]

It's just theoretical so we can find the value that satisfies y>x

[cobalt scarab]

Ohh holy shit that’s so weird

[woeful isle]

If you are too confused

We can start again

[cobalt scarab]

So you forget that you have two values basically and you make them of the same value

Same variable

[woeful isle]

Yes

[crimson sedge]

Pretty much

[cobalt scarab]

It doesn’t make sense in my head tho cuz won’t both be the same number

Cuz if y=8 and both y will have to be 8 for example

How does that work

[woeful isle]

They will be

That's the point

[crimson sedge]

We only make them equal so we can find the interval then they're no longer equal and we substitute accordingly

If that makes sense

[woeful isle]

Look at this

The black region is where x>y

[cobalt scarab]

That’s mind breaking that it works but pretty cool

[woeful isle]

The red region is where y>x

What's the limiting case

Where y and x become equal

[cobalt scarab]

The linear function?

[woeful isle]

Represented by dashes

[cobalt scarab]

Damn that’s cool

[woeful isle]

It's the y=x line

It's the boundary of the 2 regions

So what we tried to do is

[cobalt scarab]

And then you put 21.6 back in the equation,

[crimson sedge]

You can just trial and error if you want but takes more time lol

[woeful isle]

If know what the boundary is

Anything below it is x>y

[cobalt scarab]

4x+21.6=108

[woeful isle]

And anything above is y>x

This make sense now?

That why we looked for boundary

[cobalt scarab]

Yes it actually makes sense haha

[woeful isle]

And put x=y

[woeful isle]

Cool

So we found out

[cobalt scarab]

It works🥲

X=21.6

[woeful isle]

That at x = y

We get x= 21.6

[cobalt scarab]

If that didn’t work my brain would’ve malfunctioned rn lol so thank god

[woeful isle]

So if x is anything lower

[cobalt scarab]

Thank you guys😄

[woeful isle]

y>x

That's it

[cobalt scarab]

Then it wouldn’t work

[woeful isle]

What we wanted

[cobalt scarab]

Cuz x has to be smaller

[woeful isle]

Yup

[cobalt scarab]

So 21.6 is the limit

[woeful isle]

Yes

X has to be smaller than this

[cobalt scarab]

Thank you🥲

[woeful isle]

Y has to be greater then this

[crimson sedge]

Yep you exceed or reach that limit then you violate y>x

[woeful isle]

It's alright

This is what not carefully reading the question does. I literally got so confused

Lol

[cobalt scarab]

Hahaha

I mean I didn’t know this method at all🤣

I just thought it’s 27

[woeful isle]

It's alright

Remember the graph

[cobalt scarab]

I’ll try, visual math is my weak side

[crimson sedge]

You can think of a simpler version if you can't visualise it x+y = 10 for example where y>x

[cobalt scarab]

So 2x=10

X=5

[crimson sedge]

It's clear x would have to be less than 5 otherwise it'd be greater than y in this scenario

[cobalt scarab]

Y=5?

[crimson sedge]

The boundary is 5

[cobalt scarab]

Oh if it has to be less than ye

Y<5

[crimson sedge]

So x<5<y

[crimson sedge]

Other way around

[cobalt scarab]

Ohh I see

[crimson sedge]

If y<5 it'd be less than x

[cobalt scarab]

So they both can’t be 5

[crimson sedge]

Yep

[cobalt scarab]

One possibility could be x=4.9 and y=5.1

[crimson sedge]

Because it's not y<=x

[crimson sedge]

That's right

[woeful isle]

Yes

[cobalt scarab]

I see that makes sense

[woeful isle]

This what we wanted

Cool to look at lol

Anyways

@cobalt scarab anything else ??

Or is that all for today

[cobalt scarab]

Oh boy I have like 21 more math homework questions so I might open another channel soon🤣

But that’s it for that question

[woeful isle]

Well it's time to say good bye for now @cobalt scarab

[cobalt scarab]

Hahaha

[woeful isle]

Have a Great day

[cobalt scarab]

Thank you once again

[woeful isle]

Ur welcome

.close

[cedar kiln]

Channel closed

Closed by @woeful isle

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[void sand]

I have so many questions already

why can we make this identification?
why is the subspace k-dim?

[void sand]

here's the context

we're trying to put a smooth structure on G_k(V), but that isn't even my concern atm

at the moment, all I need are the answers to the two questions above

[cedar kiln]

@void sand Has your question been resolved?

[void sand]

<@&286206848099549185>

[cedar kiln]

@void sand Has your question been resolved?

[hidden hemlock]

V decomposes as $P \oplus Q$. Maps into a direct sum $A \oplus B$ correspond to a pair of a map into $A$ and a map into $B$.\
So in this case, given $X : P \to Q$, the pair $(id_P : P \to P, X: P \to Q)$ determined a map
$$(id,X):P \to P \oplus Q = V \qquad v \mapsto v + Xv$$
(Or if it helps you to think of $P \oplus Q$ as the product, $v \mapsto (v, Xv)$)\
Because on the first component $id_P$ it's injective, $(id,X)$ is injective hence $P$ can be identified with its image, the graph $\Gamma(X)$. You can also recover $X$ from the second component of $(id,X)$.\
Really the only essential fact we are using through all of this is $V = P \oplus Q$.

[wraith dagger]

Ryxie

[void sand]

why does id inj imply (id, X) inj?

[solid juniper]

[void sand]

oh hi Layla

[hidden hemlock]

if (v,Xv) = (v', Xv') then in particular v=v'

[void sand]

welcome back

[solid juniper]

hi

[void sand]

what does this mean

[solid juniper]

can't be explained in words

you just gotta look at the emote and if you get it you get it

[void sand]

no, I know what it means but like, why did you use it

[hidden hemlock]

This is the one and only time I enter help channels.

[solid juniper]

sorry ><

[void sand]

hold on.

okay, I see it

I'm a moron

that took too long

@hidden hemlock sorry for the ping... why is gamma(X) k dim?

it's the sum of a vector in P (dimP = k) and a vector in Q (dimQ = n - k), right? why is this gonna be k dim'l

hi Fungus

[hidden hemlock]

this is uhh rank nullity

[void sand]

fml

[hidden hemlock]

on the injection (id,X) : P -- > V

[solid juniper]

oh i thought the username typing was some random girl

[void sand]

let me see

[solid juniper]

but it is just fungus with a girly name

[hidden hemlock]

(Alternatively if you want: this is the first isomorphism theorem P \iso \Gamma(X) since ker(id,X) = 0)

[void sand]

ah okay hold on

nullity of this map is 0, right?

[hidden hemlock]

yeah

[void sand]

so the rank is k dim'l

[hidden hemlock]

yeah

[void sand]

wait, but the image is gamma((id, X))

do we just identify this with gamma(X)?

wait no

I'm being stupid

hold on.

the image is (P, X(P))

which looks like a graph

[hidden hemlock]

The image is what they call \Gamma(X), which looks like the graph yeah

[void sand]

okay I just spent the last 5 minutes checking, double checking, and triple checking the def

I get it now

thank you so much ryx @hidden hemlock

[hidden hemlock]

[void sand]

and thank you @solid juniper for the emotional support

[hidden hemlock]

@solid juniper

[void sand]

even if I don't know what your emote is supposed to be referencing

[solid juniper]

[void sand]

like to me, this seems like you're questioning my question

like it's a stupid question, which tbh, it could totally be

[solid juniper]

don't worry i don't know any of this stuff

[void sand]

I'm getting destroyed by linear algebra

this is actually exposing me as a fraud

[hidden hemlock]

you're not a fraud :3

[void sand]

maybe I'm just unfamiliar with using LA in other contexts

[solid juniper]

:3

[void sand]

but this is kinda scary for me

anyways, thank you ryx and Layla

welcome back to mathcord Layla!

[solid juniper]

#1152723433508057161 message you didn't do it

[hidden hemlock]

Ok.

[void sand]

new ryx banner

aaaaand the very next sentence is more linear algebra that I can't understand at first glance

time to pull out the paper

this one

okay, let's see

if $v \in Q \cap \Gamma(X)$, then $v \in Q$ and $v = w + Xu \in \Gamma(X)$, so $w + Xu \in Q$...

[wraith dagger]

higher!

[void sand]

then...

w is in P, Xu is in Q

w + Xu is in Q

ack, how am I gonna how that w + Xu = 0

smth smth complement, obviously

ah, hm

if w is nonzero, then w + Xu is not necessarily in Q, since w is outside of Q... I think?

lemme give this some thought

[hidden hemlock]

roughly, yes
I'd try the contrapositive: try to show that w = 0
(also note: v = w + Xw, i.e. your u and w are the same)

[void sand]

ah.

okay

I can do this >.<

[void sand]

ah. does this work?

if u + Xu is in Q, then u + Xu - Xu = u is in Q. but u is also in P, and Q \cap P = 0

Xu is in Q, so -Xu is too

[void sand]

so u = 0

and thus Xu = 0

and so u + Xu = v = 0?

am I yapping

[hidden hemlock]

Yes that's what I had in mind

[void sand]

that's... one paragraph of this construction understood

it's so over for me

thank you, again, ryx

I'll close this for now

sorry for all the questions

you don't go into this section of mathcord often

.solved

[cedar kiln]

Channel closed

Closed by @void sand

Use .reopen if this was a mistake.

[vague birch]

@everyone

[cedar kiln]

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

How do you compute the limit of (x+1)/(x²-1) as x approaches 1?

[maiden fractal]

If i simplify to 1/x-1, its still 1/0

[dire halo]

Can you write it on paper for me

[weak gull]

So that means it approaches infinity

Or negative infinity if approaching from the left

[dire halo]

Camr approach 1

Well

It approaches 1 since there is an asymtope at 1 and -1

[cedar kiln]

@maiden fractal Has your question been resolved?

[maiden fractal]

Oh so the limit is DNE?

[dire halo]

Graph it and see

[dire halo]

It’s -1 > x > 1

[cedar kiln]

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[cedar kiln]

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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[wanton jetty]

Someone pls help im genuinelly tweaking is my rounding wrong or something

[clear ember]

,w integral of k/sqrt(4 - 3x^2)) from 0 to 1

[wraith dagger]

Results provided by WolframAlpha

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

Requested by mmmm70177

[clear ember]

how are u supposed to be rounding these?

[wanton jetty]

To 3 sig figs

Did i round wrong 😭

[clear ember]

0.52 is 3 sig figs?

[clear ember]

i guess...

[wanton jetty]

Yeah

[clear ember]

the first one is right

[wanton jetty]

sike nah

Damn

[clear ember]

the last one surely can't be true

the median is where the area is 0.5 from the starting point

integral from 0 to 0 of f(x) is definitely not 0 as u can tell

[wanton jetty]

Im just wondering about the 3rd one because it should be 0.55 but it wasnt

[proper junco]

thats 2 sig figs

[clear ember]

i know

i'm asking them

[clear ember]

0.55 is not 3 sig figs

,calc sqrt(3)/pi

[wraith dagger]

Result:

0.55132889542179

[proper junco]

oh mb

[wanton jetty]

ohh

[clear ember]

the rest of your answers look right

just fix the sig figs

and the last question

[wanton jetty]

Ok

[wraith dagger]

mmmm7

[clear ember]

that's the objective of the last question

to find m

[wanton jetty]

Ok sounds good to me

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[blazing carbon]

I need to find this limit

[blazing carbon]

I think I can use squeeze theorem, from below its 0 and from above it should be 0

idk how to bound from above

it seems that the last term will be largest for sufficiently large n

so ideally I would bound the first terms by something times the last term

ok I can bound the middle term by 1/2 (0.999)^n (for sufficiently large n)

the first one is problematic

okay I got it its simple

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[lofty bluff]

Could someone help me with Q5

[lofty bluff]

This is all i could do for part a

😭

[azure herald]

well i forgot these actually..

lol turns out im not very smart

[lofty bluff]

Bruh

😭

[azure herald]

lol

[lofty bluff]

How long do i have to wait until im allowed to ping helpers

[azure herald]

like 15 minutes or so

[lofty bluff]

Okok

Ty

<@&286206848099549185>

<@&286206848099549185> pls

[cedar kiln]

@lofty bluff Has your question been resolved?

[lofty bluff]

<@&286206848099549185>

[hollow valve]

heya, here to help

you know how to represent a complex number in its exponential form, right?

[lofty bluff]

Yeh

Can i just say that PM=i cot(alpha/2) x AM

@hollow valve

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[glacial obsidian]

Hey, I'm stuck on this exercise on sequences.
In particular, I don't know how to do the induction (b). I have the base case but not the induction step itself.

[glacial obsidian]

Here's pretty much the entirety of my work so far

[cedar kiln]

@glacial obsidian Has your question been resolved?

[glacial obsidian]

<@&286206848099549185>

Update: I have been able to do the c) just fine with the result that I couldn't prove from b)

[cedar kiln]

@glacial obsidian Has your question been resolved?

[crimson sedge]

yo

uh you finished

okiiii

[glacial obsidian]

No I didn't

I still don't know how to do b)

@crimson sedge

[crimson sedge]

wha

oh oki

yeah nah im lost mb g

im js a beggining a level student 😭

<@&286206848099549185>

[glacial obsidian]

It's fine lmao, also you don't have to ping helpers I already did once it should be enough

[crimson sedge]

k

[steel crest]

let us call the RHS of b) as g_n show that v_n+1/v_n < g_n+1/g_n

u can do this by algebraic manipulation

[glacial obsidian]

I have tried that but I have found the contrary

[glacial obsidian]

That V_n+1/V_n > W_n+1/W_n

[steel crest]

hm

maybe you flipped an inequality sign?

v_n+1/v_n = root(n+1/n)*(2n+2)(2n+1)/4(n+1)^2

=root(n+1/n)*(2n+1)/(2n+2)

g_n+1/g_n = root(n+1/n)*root(2n+1/2n+3)

since root is positive

we have to show (2n+1)/(2n+2) < root(2n+1/2n+3)

squaring both sides

we have to show

(2n+1)^2/(2n+2)^2 < (2n+1)/(2n+3)

[glacial obsidian]

I'm trying to decrypt what you are writing lol

[steel crest]

is it understood yet?

[steel crest]

this is the same as showing
(2n+1)(2n+3) < (2n+2)^2

and this is true

as (2n+1)(2n+3) = (2n+2-1)(2n+2+1) =(2n+2)^2-(1)^2 < (2n+2)^2

[glacial obsidian]

Wait ok

I can't read this but you just made me realise that 4(n+1)^2 = (2n+2)^2 and I didn't realize that sooner so all of my computations have crazy exponents. I'll try to redo them with this simplification in mind

[steel crest]

would bracketing help or unhelp?

[glacial obsidian]

Hmmm I think I'll try to redo the steps for now so I think it's okay, if I can't find the correct result I'll tell you then

Ahhh I got it

I should have thought something was up when I originally had:

$\frac{32n^7+178n^6+392n^5+452n^4+284n^3+92n^2+12n}{32n^7+178n^6+352n^5+312n^4+122n^3+19n^2+n}$

[wraith dagger]

Dalilou

[glacial obsidian]

Thanks @steel crest

[steel crest]

welcome

[glacial obsidian]

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[thorny mantle]

so basically for the given system i need to find the value of a for which the system has only one answer

the answers are eather

1. a is negative

2. a is 0

3. a is positive

4 ) a is eather positive or 0

[random swift]

You want to find an a such that there exists only one x satisfying your system?

[thorny mantle]

yes

[random swift]

Look at the first equation - for what values of x satisfies this equation?

[fading garden]

wait hold on, the first equation has to equal a number less than 0?

[thorny mantle]

from -1 to 0

i guess

[random swift]

yep

Now, in terms of a, what values of x satisfy the second equation?

[thorny mantle]

yes

from 0 to a?

[random swift]

yep

So, what values of a do these two intervals intersect at exactly one point?

hint: or a to 0 if a < 0

[thorny mantle]

0?

idk😭

[random swift]

eqn 1. -1 <= x <= 0
eqn 2. a <= x <= 0 or 0 <= x <= a

look first at a <= x <= 0

[thorny mantle]

dont we need to use the determinant

?

[random swift]

nope

-1 <= x <= 0 and a <= x <= 0 means a <= x <= 0, right?

because they overlap there

[thorny mantle]

yes if a is greater than -1 right?

oh ye i understand now

[thorny mantle]

yup

and than what do we do

[random swift]

-1      -a       0
|----------------|  -1 < x < 0
         |-------|  -a < x < 0
=
         |-------|  -a < x < 0

-a      -1       0
|----------------|  -a < x < 0
         |-------|  -1 < x < 0
=
         |-------|  -1 < x < 0

in either case, we have multiple values for x, and we only want one!

So, clearly a cannot be less than 0.

[thorny mantle]

so its eather 0 or positive right

[random swift]

could be.
how many solutions are there if a = 0?

how many solutions are there if a > 0?

[thorny mantle]

if a is 0 than x is 0 too right?

and if a is positive than x is also 0 since the intersect at 0 right?

[random swift]

yep 🙂

[thorny mantle]

sooo

a is eather positive or 0 right

and thats the answer?

[random swift]

that's the answer

[thorny mantle]

thaaaankksssss

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[undone epoch]

Find if the integral converges or diverges $$\int_0^1 \frac{1}{\sqrt{x+x^3}};dx$$

[wraith dagger]

Good

[undone epoch]

Since it's not an improper integral, comparing and limit-comparing aren't for use. How do I solve it?

[cedar kiln]

@undone epoch Has your question been resolved?

[undone epoch]

<@&286206848099549185>

[summer otter]

are you sure about that it is not an improper integral?

[undone epoch]

Isn't the definition of an improper integral that it contains a infinite end?

[summer otter]

no that is one type of improper integral there is another type of improper integral that involves a finite interval

[undone epoch]

Right, but how do I solve it?

[summer otter]

$\int_{0}^{1}\frac{1}{\sqrt{x+x^3}} \dd x=\lim_{b\to 0^{+}}\int_{b}^{1}\frac{1}{\sqrt{x+x^3}} \dd x$

[wraith dagger]

convergence

[undone epoch]

Do I need to solve the integral, or could I know the answer by something?

[summer otter]

solving the integral is too tedious

[undone epoch]

I think I'll just keep on going.

Thanks for assistant.

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[summer otter]

you want to slove the integral?

[undone epoch]

No, I'll just solve other questions.

[summer otter]

ah you can use comparison test here

[undone epoch]

Then you need type 1.

[summer otter]

$\int \frac{1}{\sqrt{x}} \geq \int \frac{1}{\sqrt{x+x^3}}$

[wraith dagger]

convergence

[undone epoch]

I think I just do this question another day.

[summer otter]

alright

[cedar kiln]

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[small elbow]

I'm confused why this is fx(a - y)fy(y) instead of Fx(a-y)

Because I thought the integral of the pdf is the CDF

[dire geode]

yea that's a little inconsistent

these two are not equal as you've pointed out

the bottom one is the derivative wrt to a of the top one

$\frac{d}{da} F_{X+Y}(a) = f_{X+Y}(a)$

[wraith dagger]

riemann

[dire geode]

so the bottom one is equal to the right side

[small elbow]

thats what i thoguht

so probably a typo in this book

bc later on they do provide the other version

[dire geode]

yea they do exactly what you said here

[small elbow]

ah okay thx

that makes more sense now

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[drowsy rock]

Hey guys, the question for this problem was to find out which statement is true and which one is false. But here, it seems that both are true, I'm confused

[dusk goblet]

||

[drowsy rock]

OH

I see now

Tq

[dusk goblet]

you’re welcome

[drowsy rock]

There is no absolute value in the ln

Thanks

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[obsidian cosmos]

Need help with my logic course specifically with a book called logic in action if anyone is familiar. I’m trying to do the tableau method with predicate logic and I can’t seem to get it in the way the book teaches

[humble karma]

Do you have an example?

[obsidian cosmos]

Yes

Here’s the rules of the logic textbook

I’m working with quantifiers universal and existential

Here’s two I attempted

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[unkempt hearth]

Hey guys, anyone can help me? The exercise says to use the fundamental theorem of integral calculus

[dull oxide]

Note that, as x approaches zero, the numerator and denominator will both approach zero.

So you can use lhopital to find the limit by differentiating the numerator and differentiating the denominator.

The denominator should be simple for you (if not then you're going to need a lot of review).

For the numerator, you need to calculate the derivative of an integral. This is where the fundamental theorem of calculus (FTC) will be useful

[cedar kiln]

@unkempt hearth Has your question been resolved?

[static whale]

<@&268886789983436800>

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[crimson sedge]

Hi, can someone help me?

[dire geode]

what's the question asking you to show

[crimson sedge]

basically prove that inequality holds for this values

[dire geode]

do you know induction

[crimson sedge]

Yes, i tried to solve it through induction, but i think i got it wrong

in case u can show me how to solve it through induction - would be great

[humble karma]

If you show how you did your induction, then we can point out where you messed up, if you did.

[crimson sedge]

Well , dont think I have it🫤 . Did it - saw it is wrong and and erased it

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[mystic wren]

idk 😭

[wheat jewel]

We're trying to find the expectation of Y right?

so we can use this formula:

$E[Y] = E[E[Y \mid \lambda]]$

[wraith dagger]

DW1

[wheat jewel]

and we're given that Y is a poiisson distribution with parameter lambda, so the conditional expectation would be"

$E[Y \mid \lambda] = \lambda$

[wraith dagger]

DW1

[wheat jewel]

and the expected value of lambda, given that its a gamma distribution is ...

[cedar kiln]

@mystic wren Has your question been resolved?

[cedar kiln]

@mystic wren Has your question been resolved?

[mystic wren]

im sorry, i dont understand what this means

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[violet owl]

please help T-T

[violet owl]

i tryed 0.0833 and it didnt work?

also 0.0417

[warped coyote]

That's (1/6) * (1/2) and (1/6)*(1/2)^2 yeah? How does the n affect your result?

[solid juniper]

T-T

[violet owl]

no idea tbh english isnt my first langauge and im struggling with the wording of the question

i thought it would just be 1/12?

i know it cant be that simple tho ://

[solid juniper]

i think i have a way with probability generating functions

[violet owl]

ohh do tell

[solid juniper]

needing k dice rolls to get a six means we roll not a six k-1 times, then a six. so (5/6)^(k-1) * 1/6 probability
needing m coing flips to a get a heads means m-1 tails then a heads. so 1/2^m probability

then i wanna just convolute those 2 distributions' generating functions but lemme see if this works

ok i think that works??

[violet owl]

ohhh do you have an answer?

[solid juniper]

yea

[violet owl]

could i see it?

[violet owl]

very helpful x

[solid juniper]

[violet owl]

thats discusting

lets focus on the question please

english is my 5th language x

[solid juniper]

[violet owl]

@frozen rivet can you remove them i just want help on the question

oh wooooowwwww

[solid juniper]

[violet owl]

??????

thats not the answer???

[slate lintel]

ugh. @inland widget go away. if you type in here again i'm muting you

[violet owl]

thank you

obv not ????

your not your being horid

no it isnt

fuck off

@slate lintel can you mute him?

[slate lintel]

already did

[violet owl]

thank you <33

[violet owl]

i still need help on this question :((

[dull oxide]

https://tenor.com/view/chalmers-simpsons-may-i-see-it-talking-gif-11870196

[solid juniper]

i'm dying

[violet owl]

ewwwwww

[violet owl]

im desperate T-T

[solid juniper]

let's repost this for shame purposes

[slate lintel]

put it in the other server, not in here

[violet owl]

then he deleted his messages 💀

[dull oxide]

Not as desperate as vc05 it seems

[violet owl]

honestly you know youve fucked up when you try to hide the evidence

[dull oxide]

But ok I'll be srs now

[slate lintel]

i think the maximum value of pn would be rolling a six on the first try and getting heads on the first try so n=2

[solid juniper]

WTF HE DMED ME BEFORE 2 YEARS AGO

[violet owl]

trust i am

[violet owl]

WOAH

[violet owl]

yes please

[solid juniper]

hm the probabilities for each n look like this i think

[raven shard]

f.coefficients()

[solid juniper]

starting with n = 2

[dull oxide]

Oh dang this problem hard. I think slayla will be better help but I'll give it some thought and come back if i find anything and y'all haven't gotten a conclusion yet

[solid juniper]

my specialty

[raven shard]

indeed

[dull oxide]

God my best friend sends me that same dog every time i say something profoundly stupid to him. It's giving me flashbacks

[solid juniper]

taking 2 total moves being 1/12 probability makes sense so i trust this somewhat

[violet owl]

i put 0.0833 and it doesnt work :((

am i being dumb?

[solid juniper]

yea that's not the answer

well not yea to you're being dumb

but that's not the answer

[violet owl]

oh ok

[raven shard]

I haven’t looked any further but indeed according to layla’s image at the very least 1/9 is greater than 1/12

[sacred grail]

now decide whether the other coefficients are larger

[solid juniper]

easy for me

we want the largest coefficient of $$\left(\sum_{k=1}^{\infty} \left(\frac{5}{6}\right)^{k-1}\cdot \frac{1}{6}\cdot x^k\right)\left(\sum_{m=1}^{\infty} \frac{1}{2^m}\cdot x^m\right)$$

well

[wraith dagger]

slayla

[solid juniper]

i'm not sure how to do this by hand

[slate lintel]

what i would do is calculate the values out to like 10 or so dice rolls and like 5 flips

and just figure out which combination makes the most sense

actually probably 6 dice rolls and 2 flips is sufficient

[raven shard]

🫵

[solid juniper]

hm the (or at least my, which is hopefully the) answer might surprise you then

,calc 49/432

[wraith dagger]

Result:

0.11342592592593

[solid juniper]

@violet owl can you try that

[raven shard]

that does indeed appear to be the highest coefficient

[violet owl]

OH MY GOD IT WORKED

thank you T-T

[solid juniper]

so the right n was 4

[raven shard]

you’re probably just supposed to manually compute the cases for up to sum = 5 or something and deduce it’ll only grow smaller

[solid juniper]

let's all take a moment to appreciate the beauty of probability generating functions

[solid juniper]

yea the answer just happens to be small enough that you could do it by hand naively

[sacred grail]

maybe you can

[violet owl]

probably

[raven shard]

maybe you can

[sacred grail]

i on the other hand

[violet owl]

i think i did it and miswrote it as 0.013

[raven shard]

you on the other hand

[solid juniper]

ez

[sacred grail]

gf

[violet owl]

thank you guys so much

[sacred grail]

this is almost valid python you could've fooled me

[solid juniper]

R.<x> = QQ[]
snow: this is almost valid python you could've fooled me

[sacred grail]

that's why it's not

and the (1/2)^

[solid juniper]

actually it was 1/2^

[sacred grail]

i typed it from memory

[raven shard]

now prove it can’t be higher than 100 🫵

[sacred grail]

unfortunately my memory wasn't very good

[solid juniper]

i did all the work so far. someone else can do that

i could never use python because sage just makes working with generating functions so easy

treating lists as polynomials is for losers

[sacred grail]

,w generating function of geometric distribution

[raven shard]

nice moment generating function

L

[sacred grail]

L

[solid juniper]

moment generating function moment

[wraith dagger]

Results provided by WolframAlpha

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Requested by tearfox

[solid juniper]

another moment generating function moment

[sacred grail]

now does this start at 0 or 1 is the question

[raven shard]

compromise. start it at 1/2

[sacred grail]

brilliant

it appears to start at 0

,w apart px/(1-(1-p)x) * qx/(1-(1-q)x)

wow

[wraith dagger]

Results provided by WolframAlpha

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Requested by tearfox

[solid juniper]

[raven shard]

seems promising

[sacred grail]

okay this is getting too much to do on my phone

[violet owl]

.close

[cedar kiln]

Channel closed

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[solid juniper]

rip snow

[sacred grail]

dw i wasn't doing anything

[cedar kiln]

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[strange furnace]

substitution rule says its applicable if the integrand is f(g(x))*g'(x), but what if the integrand isnt so convienient?

[humble karma]

In general it might not be the case that you get exactly f(g(x)) * g'(x), but you can sometimes rearrange what is left in terms of g.
As you get more familiar with it, you can anticipate what results you'll have under certain substitutions, and learn how to pick the one that helps. Substitution isn't always the way to go either...

[strange furnace]

how else would i solve something that u substitution usually is needed for

specifically composite integrands

[humble karma]

I'm not saying it's never useful, I'm saying it sometimes isn't.
Regardless, you don't need the integrand to be exactly of the form f(g(x)) * g'(x) in order to use substitution.
If you use u = g(x), then du = g'(x) dx and you can always use that in your integrand. It'll generally help if you can rearrange the integrand in terms of u.

It's just that $\int f'(g(x)) g'(x) dx = f(g(x)) + C$ is a case in which it works well.

[wraith dagger]

Azyrashacorki

[cedar kiln]

@strange furnace Has your question been resolved?

[cedar kiln]

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[crimson sedge]

Not rlly a math question more of a idea collective of what do you think is the best way to grasp and understand a math concept if you feel like your struggling or lagging behind somewhat?

[astral leaf]

1. revise
2. practise
3. teach

first study your textbook to get a grasp of the concept then do some practice to passively remember it

if possible, try to teach it to people who are weaker than you so you actually retain a lot more information rather than just memorising from the textbook

[crimson sedge]

Thank you for the feedback

.close

[cedar kiln]

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[cedar kiln]

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[tawny flare]

I have a question in numerical analysis. We have a project to compare between integration methods and talk about their convergence, but one of them is Monte carlo. Is it fair to make Monte carlo run multiple trials and take the approximation with the highest probability density?

[tawny flare]

Because it might get bad points and if I just compared that would be worst case vs good cases

[grand stone]

wdym highest probability density

you mean the best monte carlo result?

[tawny flare]

I ran multiple trials for Monte carlo here

[grand stone]

okay

you don't really say "highest probability density"

you're referring to the mean

[tawny flare]

Ah, sorry.

[grand stone]

and that is the most natural thing to do here

🙂

[grand stone]

its ok

looks like the mean is very close to your true answer

[tawny flare]

yeah, but this is for N=100000. Which isn't ideal I will take smaller values for the comparasion.

it took one minute to run.

[grand stone]

im sure itll converge for smaller values too

the distribution looks normal 🙂

[tawny flare]

for smaller value it won't be a normal distribution. It will get close to the answer but with a higher error than other methods.

[grand stone]

it's never going to be normal

just approximately

[tawny flare]

This is for 1000 approximations. It's a really good approximation, but compared to other method the error is a bit high.

Anyway, thank you so much for the help. I really appreciate it.

.close

[cedar kiln]

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[vapid rivet]

am i trippin? or is my answer correct?

[modern dragon]

idk bud

[vapid rivet]

thanks

[grand stone]

LMAO

is that the solution at the bottom

[vapid rivet]

yes

[grand stone]

if so, this site is absolute trash

that value is incredibly close to zero

notice that?

[vapid rivet]

ah

tru lol

[grand stone]

ur right

the site is suffering from floating point error

hope ur teacher doesnt take off points for that

[vapid rivet]

happened on a quiz too

making me go crazy

thanks for verifying im not lol

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[cedar kiln]

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[cedar kiln]

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[zealous patio]

If I'm looking for the residues of z^n sin(1/z) where n is an integer, then when n is odd since -1 power term in the Laurent series will vanish, that means a residue of zero, not that there is no residue right

[zealous patio]

or i guess not just at the odd n's, also at all the negative n's