help-42

[opaque yoke]

ok thaks

ok thanks i got it

i appreciate the help

.close

[calm coral]

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[tacit moat]

np!

[calm coral]

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[gusty lance]

can someone explain to me how this is done

im not sure what the first step even is

[edgy leaf]

have you taken the natural log on both sides?

[gusty lance]

oh no

ok

one sec

ive got lny = 2ln(x2+1) + 3ln(x2+x+1)

[edgy leaf]

indeed

now implicitely differentiate

[gusty lance]

ok

this is so long

6 terms x 5 terms

holy

[edgy leaf]

It should not be that bad

[gusty lance]

oh

wait i subbed y in for what we got in the q

maybe thats why

[edgy leaf]

oh you mean the final answer?

[gusty lance]

no i mean

one sec

[edgy leaf]

your steps are correct

now here is a neat trick mathematicians use when finding derivatives

you dont need to simplify

fully

[gusty lance]

oh

[edgy leaf]

like multiplying out y with the whole equation

that will take a while

[gusty lance]

so i could js leave it at this?

[edgy leaf]

yes but do write what y is in terms of x

[gusty lance]

oh okk

not too bad then

thank you

.close

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[wild marten]

Would the non-alternative backwords direction still work as n is a whole number

[calm coral]

@wild marten Has your question been resolved?

[wild marten]

<@&286206848099549185>

[thick sinew]

wdym by work

regarding the n = 2p/n step?

this step should be more elaborate

it's not immediately clear why p/n is in Z

likewise you can't assume that n isn't 0, so divisibility by n isn't given

[wild marten]

oh right

its a whole number not natural

thats a pain, alr tysm

.close

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[potent lotus]

ω

ω

ω

ω

ω

[upbeat venture]

I would agree that that's a good counterexample

that span preservation property only works for the rows, hence the name

[remote mural]

Awesome, thank you!

[calm coral]

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[exotic ruin]

@upbeat venture are you online?

[calm coral]

Please don't occupy multiple help channels.

[agile pilot]

!original

[calm coral]

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

[agile pilot]

Wait

[exotic ruin]

Ok

[agile pilot]

Post your question

!da2a

[calm coral]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[exotic ruin]

I need help with like 260 digits of pi and I only know like 5..

[agile pilot]

Ann can prob help though

@velvet osprey you got this 😭

[exotic ruin]

😂

[velvet osprey]

uh

[eternal shard]

wth

[velvet osprey]

you want to do... what exactly?

memorize 260 digits of pi?

[agile pilot]

Yeah

She does

[exotic ruin]

I only know like 3.1415926535 8979323846 2643383279 5028841971 6939937510 5820974944 5923078164 0628620899 8628034825 3421170679 8214808651 3282306647 0938446095 5058223172 5359408128 4811174502 8410270193 8521105559 6446229489 5493038196 4428810975 6659334461 2847564823 3786783165 2712019091 4564856692 3460348610 4543266482 1339360726 0249141273 7245870066 0631558817 4881520920 9628292540 9171536436 7892590360 0113305305 4882046652 1384146951 9415116094 3305727036 5759591953 0921861173 8193261179 3105118548 0744623799 6274956735 1885752724 8912279381 8301194912 9833673362 4406566430 8602139494 6395224737 1907021798 6094370277 0539217176 2931767523 8467481846 7669405132 0005681271 4526356082 7785771342 7577896091 7363717872 1468440901 2249534301 4654958537 1050792279 6892589235 4201995611 2129021960 8640344181 5981362977 4771309960 5187072113 4999999837 2978049951 0597317328 1609631859 5024459455 3469083026 4252230825 3344685035 2619311881 7101000313 7838752886 5875332083 8142061717 7669147303 5982534904 2875546873 1159562863 8823537875 9375195778 1857780532 1712268066 1300192787 6611195909 2164201989

[velvet osprey]

is that what ze wants?

[exotic ruin]

It’s my math teacher

[agile pilot]

Probably

[velvet osprey]

are you like... shitting us rn

[exotic ruin]

I skipped class like

5-6 times

[velvet osprey]

this sounds like a very unserious inquiry

[exotic ruin]

It’s h-op

[fickle wigeon]

Bro are you ok

[exotic ruin]

no

[velvet osprey]

<@&268886789983436800> this sounds like a potential troll but im not sure

[leaden thunder]

Yea definition of troll

[agile pilot]

I would just close the channel lol 😭

[leaden thunder]

Block and move on everyone

[exotic ruin]

.?

I ain’t trolling

[agile pilot]

If it’s memorization it’s just memorize digits in different groups

[fickle wigeon]

why don’t you start with the first 1000, so that the 260 won’t be so difficult

[exotic ruin]

1000?!?

[velvet osprey]

your math teacher is making you engage in stupid and completely useless memorization?

[torpid brook]

this clearly isnt productive

[exotic ruin]

yes

[torpid brook]

.close

[calm coral]

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[velvet osprey]

it is indeed not

[exotic ruin]

He strict asf

[torpid brook]

please dont abuse help channels

[fickle wigeon]

W moderator

[agile pilot]

Math teachers are supposed to teach you math

This is just… memorization

I would watch a video on pi songs

And learn it through there

If you’re actually sseuous, but realize that this inquiry is quite time wasteful

[calm coral]

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[wary plover]

Small question
X={(x,y): x^2+y^2<=1, x>=0, y>=0}
Y=R^2

I said the set A={(x,y): x^2+y^2<=1, x=0, y>0} is closed in X because A^c is open in X. Is this true?

[wary plover]

And A is open in Y because (0,0) is not included

[velvet osprey]

that... sounds like a misapplied heuristic

[wary plover]

Oops

(0,0) is not in A so can it really be closed in X😭

[velvet osprey]

it is not closed in X nor in R^2, no.

but it isn't open either

[wary plover]

So not open doesn’t always mean closed and vice versa?

I thought there was sth like Let Y be a subset of a metric space X. Y is closed iff its complement is open.

Why doesn’t it work here😭

Wow I understand now

Thanks

Is there a set closed in X but open in Y?

[calm coral]

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[calm coral]

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[vague crest]

the answer is supposed to be e. can someone tell me what i did wrong?

[harsh siren]

where'd the other 2 go

[vague crest]

wdym

[vague crest]

oh silly me

[calm coral]

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[outer sedge]

how to solve this? do i set x = 3-z and parametrize? r(y, z) = <3-z, y, z> and then we have a circle on the yz plane with radius = 2?

[calm coral]

@outer sedge Has your question been resolved?

[outer sedge]

<@&286206848099549185>

[vague bison]

i suggest having an image of the region of integration

[outer sedge]

is my approach correct

but im confused cuz there's two values of x

[vague bison]

the components in your r(y, z) satisfy the equation x = 3 - z, but not other equations defining S

[outer sedge]

do i need to parametrize it like a cylinder?

where x = x, y = rcos(theta), z = rsin(theta)

r = 2

[vague bison]

you may
then you need to describe the region of integration with your new parameter
also, note that a surface is parametrized with two parameters

now i see θ, and one more parameter needed

[outer sedge]

but what if i wanna parametrize it using the other method (cartesian), how would i do it

do i need 2 integrals

cuz x = 0 and x = 3-z

[vague bison]

i suggest having a mental image of the integration region first

then use each define/bounding equation to write the region of integration in set notation

to ensure that you're a clear image of that
before proceeding to applying

∬ f dS

S

For rectangular coordinates, you can apply the relevant formula on Wiki
https://en.wikipedia.org/wiki/Surface_integral#Surface_integrals_of_scalar_fields

[outer sedge]

what do i put as x, 0 or 3-z, when i write r(y, z)

[vague bison]

i suggest having a mental image of the region of intgration first

[outer sedge]

the 2D region?

it's a circle

[vague bison]

||you want x as a function of y and z (i.e. x = x(y, z)), but from the (mental) image (of the region of integration), that's clearly impossible by the definition of "function", which have one and only one output.||

[outer sedge]

we wanna break it into two integrals?

[vague bison]

that would complicate matters

[outer sedge]

one where x = 0 and one where x = 3-z

[vague bison]

again, what do you have as a mental image of the image of integration

[outer sedge]

does that work?

in ur parametrization

[vague bison]

having a mental image of the problem is often the first key step to solve the problem

[outer sedge]

is there a way to do it using cartesian coordinates or no

[vague bison]

wiki has the formula for rectangular coordinates

but it involves the region of integration

that's why you want to understand clearly the region of integration before you proceed

[outer sedge]

it's a cylinder that is divided between the plane x = 0 and x = 3-z

[vague bison]

then try to write this in set(-building notation)

if things are to complicated

start with simpler things first

before adding more constraints

say, the set notation for a cylinder in the question is ...

[outer sedge]

0<x<3-z

[vague bison]

that's not a cylinder

[outer sedge]

x is between 0 and plane 3-z

cylinder y^2 + z^2 = 4

the projection is a circle on the yz plane

[vague bison]

write them out clearly in set-builder notation, like

• "x btwn ...." ⇒ {(x, y, z) ∈ ℝ³ | ???}
• "cylinder ..." ⇒ {(x, y, z) ∈ ℝ³ | ???}

[outer sedge]

(0, 3-z)

for x

[vague bison]

• first constraint: {(x, y, z) ∈ ℝ³ | x ∈ [0, 3 − z]}, or even better (easier to manipulate), {(x, y, z) ∈ ℝ³ | 0 ≤ x ≤ 3 − z}
• what about the 2nd constraint?

[outer sedge]

y^2+z^2=4, it's an equation, does not have two values between each other

[vague bison]

write it in set-builder notation to make things clear

[outer sedge]

{(x, y, z) ∈ ℝ³ | y^2 + z^2 = 4}

[vague bison]

to find out the a set of point (in ℝ³ ) satisfying these two constraints, what (set operation) can you do?

[outer sedge]

im not sure

you add the constraints together?

but how

[vague bison]

1. points satisfying the first condition in first set
2. points satisfying the second condition in second set
3. points satisfying both conditions in both sets

[outer sedge]

1. all x between 0 and 3-z, 2. -2<y<2, -2<z<2

[vague bison]

that's the only condition? also, regions of integration are usually defined on closed regions for some theoretical reasons. that's why you see things like "Riemann integration defined on closed and bounded interval [a, b]". you don't want to exclude the boundary points of the region of integration by writing strict inequalities above

[outer sedge]

1. all x between 0 and 3-z, any z, any y

2. y's between -2 and 2, z's between -2 and 2

[vague bison]

1. first constraint: {(x, y, z) ∈ ℝ³ | 0 ≤ x ≤ 3 − z}
2. seond constraint: {(x, y, z) ∈ ℝ³ | y² + z² = 4}
3. both constraints: intersection of these two sets = "{(x, y, z) ∈ ℝ³ | "first condition" AND "second condition"}]

[outer sedge]

x could be anything in the second set

my bad

[vague bison]

you haven't really used the 2nd condition
from these two inequalities, we don't see the the point has to be on the cylinder surface (i.e. satisfy y² + z² = 4)
That's the symbolic expression (for the regional of integration) that I'm looking for
||{(x, y, z) ∈ ℝ³ | 0 ≤ x ≤ 3 − z AND y² + z² = 4}||

It's good to be able to pass from

if possible

1. verbal
2. visual
3. symbolic
   understanding
   here's a mental image that I'm looking for

the red "ring" is the region of integration

that has to be parametrized by r(?, ?)

[outer sedge]

how do we parametrize this thing

[vague bison]

in 2D plane, we have the vertical line test to test whether a planar curve represents the graph of some function
now we can do the same (in our mind) using the yz-plane as the candidate domain

to see whether r(y, z) make sense

[outer sedge]

i don't think we can parametrize it in cartesian

cuz x is variying

it's not x = f(y, z)

[vague bison]

we can see that any line parallel to the x-axis passing through the circle y² + z² = 4 on the yz-plane intersects with the region of integration at more than one (in fact uncountably many) points

that's why we can't use y and z to parametrize any point r on S

[outer sedge]

can we split it into multiple integrals?

btw, what is the projection of the surface onto the yz plane? the disk y^2 + z^2 <= 4?

[vague bison]

so if you wanna parametrize r using Cartesian coordinates, x shld b one of the parameters

[vague bison]

the circle y² + z² = 4

[outer sedge]

so i have r= (x, y, z)

i need it to be in terms of two variables

i can't

[vague bison]

we already know that x should be one of them

if you feel blocked, that often means that you haven't utilized some given condition

[outer sedge]

z = sqrt(4-y^2)

y = y

something like that?

but why don't we take the -sqrt(4-y^2)

[vague bison]

i suggest writing in complete sentences to make things clear

[vague bison]

if you feel stuck, passing from one understanding

to another understanding can often unblock you

1. verbal
2. visual
3. symbolic

here you feel stuck at symbols

so look back into the graph representing the region of integration

[vague bison]

.

[outer sedge]

it's in the first octant

i see why

wait maybe no, let me graph it

[vague bison]

oh i see yeah if you use Cartesian coordinates you need to split it into two integrals

if you use cylindrical parametrization you don't

[outer sedge]

what are the parametrizations of the two integrals, one being z= -sqrt(4-y^2) and the other being z = sqrt(4-y^2)?

[vague bison]

now you have z = z(x, y) (without x), so you can write r(x, y) = (x, y, z) = (x, y, z(x, y))

so r is parametrized by two variables

then you can directly apply

,,{\displaystyle \iint _{S}f,\mathrm {d} S=\iint _{T}f(\mathbf {r} (s,t))\left|{\partial \mathbf {r} \over \partial s}\times {\partial \mathbf {r} \over \partial t}\right|\mathrm {d} s,\mathrm {d} t}

[potent lotus]

vin100

[vague bison]

i copied the formula from wiki

s and t represents coordinates in a plane

here you can use s = x and t = y

[outer sedge]

what would z = in both cases

sqrt(4-y^2)?

[vague bison]

yes
so the T can be taken to be S

in the formula

[outer sedge]

why not -sqrt(4-y^2)

[vague bison]

ofc if you choose to parametrize with x and y/z then you need a couple of surface integrals

[outer sedge]

i see ok

[vague bison]

that's why i suggest cylindrical parametrization

[outer sedge]

for this, in the parametrization, why do we set y = 3x^2 + 3z^2 instead of y = 6

[vague bison]

with x and θ representing the counterclockwise angle between the projection of r on the yz-plane and the positive y-axis

[vague bison]

same technique to the question that we've gone through

have a mental image in mind
symbolic description

[outer sedge]

i think whenever they say "S is the portion of [equation]", then we use that specific equation, in the other question, it said S is the portion of y^2 + z^2 = 4", so we use that specific equation in the parametrization

[vague bison]

the "y = 6" can be regarded as some sort of "boundary condition"

[outer sedge]

yes

ok i think i'm done

thanks for ur time

.close

[calm coral]

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[gusty cliff]

The mark scheme says the only answer is -1+sqrt2

How do I know that -1-sqrt2 is invalid?

.close

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[cloud walrus]

cylinder area is

[cloud walrus]

am i correct?

.close

[calm coral]

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[eternal shard]

not even 1 minute time to reply

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[blazing coyote]

Can I have a hint

[blazing coyote]

I'm trying to formulate a dynamical system

what I noticed was the difference vareries as follows

a_2-a_1=150

a_3-a_2=555

a_4-a_3=370

a_5-a_4=550

a_6-a_5=-100

,w 555/150

[potent lotus]

Results provided by WolframAlpha

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

Requested by math_rocks

[blazing coyote]

.close

[calm coral]

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[lament sorrel]

combinatorics

why do we take "-" at the end of b)?

[lament sorrel]

plugging x=0 => B(x)=0/0

[glass heart]

you have to consider it as a limit x->0

[lament sorrel]

i dont get why

[tribal osprey]

well

ig they don't teach u that in high scool

[leaden thunder]

plot it and see if it's just a hole discontinuity

[calm coral]

@lament sorrel Has your question been resolved?

[lament sorrel]

why does it matter, it's GFs

[leaden thunder]

because you're questioning the limit

[lament sorrel]

i dont understand why it matters

[leaden thunder]

why what matters

[lament sorrel]

i just want to know + or - here

all this calc stuff

[leaden thunder]

well find what the value for + is

calculus is usually a prerequisite for generating functions

[lament sorrel]

2/0

[lament sorrel]

we dont care for x values and all that

it's just a represention

[leaden thunder]

right and limit as x goes to 0 of 2/x isn't an integer

[calm coral]

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[lament sorrel]

0/0 is also not an integer

[leaden thunder]

you're back here

[leaden thunder]

and here

[calm coral]

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[kindred imp]

if I have a function thats just a straight line with the value (f(q))^2. Would the integration of (f(q))^2 be (f(q))^2x since (f(q))^2 is a constant and if so, would the sum of the integration between 0 and q be (f(q))^2q?

[random adder]

Yes

[kindred imp]

okay nice. just needed to be 100% sure

.close

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[calm coral]

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[umbral onyx]

Hello, how do I integrate this?

[calm coral]

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[umbral onyx]

.close

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[lapis sage]

can someone help with this

[lapis sage]

l'hopital is possible 3 times but it gets very ugly

does anyone see any shortcuts or factorizations for this to be more elegant

[random adder]

You can do some subs

[lapis sage]

like?

[random adder]

Like t = cos(x)

[ivory pilot]

Yeah

[lapis sage]

thnx

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[wild marten]

Is the statment: if H is a subgroup of cyclic group G, then H is also cyclic an iff statement? Like does it wrok the other way round?

[marsh valley]

Hmmm, from the statement the converse holds simply because a group is a subgroup of itself (misinterpreted)

I think?

[wild marten]

Idk i was thinking about it

I cant find a counter example

but also how would u go about proving it

[teal coral]

are you asking whether if there exists a cyclic subgroup then the whole group is cyclic

or are you asking whether if all subgroups are cyclic then the whole group is cyclic

[wild marten]

Yes, like if i a G such that H is a subset of G

and G turns out to be a group

can I assume its cyclic if H is cyclic?

[teal coral]

no, the trivial subgroup is cyclic...

[wild marten]

ur a genius

but yeah holy shit

how did that not hit me

Ok excluding the trivil group

whats an example of it going the other way but not working

[marsh valley]

I think there are groups for which every proper subgroup is cyclic but the whole thing isn’t

[wild marten]

With an element other then e also in the group

[random adder]

Wouldnt it just be an element such that x² = e and e

[wild marten]

ok i might just need to spend more time on this then

[wild marten]

that would be e

[marsh valley]

See this

[wild marten]

wait wdym

then x^-2 also has to exist

[random adder]

Like -1 , 1 in Q \ {0} under multiplication

[wild marten]

ooh hmm

but in that case

Q* is not cyclic

i think

[random adder]

Well isnt that the whole point

[wild marten]

yeah ur right

Ok so lets assume (for this case say < means subgroup) H<D<S<R<F<G

And R is cyclic

then S D H are also cyclic

but we dont know if F and G are

[random adder]

I think i could construct something

Like imagine multiplication mod n

[wild marten]

ok

im new to abstract algebra btw

[random adder]

Ok no thats just cyclic

[wild marten]

wait for all n

I know addition is with <1>

but like what about U(3)

[random adder]

Oh yeah its only cyclic for like p^n , 2p , 2 , 4 - odd p

Condition such that primitive root exists

[wild marten]

{1,2} is not cyclic right

under multiplication

wiat

[random adder]

Under what operation?

[wild marten]

nvm lemme think of a better xample

[random adder]

Im guessing like multiplication mod 12 would be a counterexample

[wild marten]

U(5)= {1,2,3,4}

this would be non-cyclic

You cant get 3 with any number in there

actually nvm its in mod

2, 4, 1,3 come from <2>

[wild marten]

ok continue this lmao

[random adder]

Like n = 35 makes sense ig

So there should be 24 numbers coprime to it.

But the maximum order is 12

[wild marten]

whats coprime

relitively prime? like gcd=1?

[random adder]

Yes

[wild marten]

ok just making sure

[random adder]

And ig you can find like set with 3 elements aka (an element with order 3) then a set with 6 elements. So that it contains those 3 and some other 3 and so on to 12

[wild marten]

hmm

im gonna study a bit more into it

I think i understand what you are saying tho

[random adder]

So like for example
{1 , 4 , 16 , 29 , 11 , 9}
{1 , 2 , 4 , 8 , 16 , 32 , 29 , 23 , 11 , 22 , 9 , 18}
And the original group

So original group is not cyclic

Because multiplication mod 35 has no primitive root

And then you have a cyclic subgroup of length 12. And then a subgroup of that subgroup of length 6

[wild marten]

Its not a subgroup tho

because they dont share multiplication

right

@random adder

[random adder]

They do. Its same operation

[wild marten]

wait what is the first one and second one

[random adder]

Its just $2^n \pmod {35}$ and $4^n \pmod {35}$

[potent lotus]

casework

[random adder]

And then you just consider the original group that isnt cyclic

[wild marten]

wait which is the top group

im a little lost

gimme a sec let me write this down

[random adder]

The top is $2^n \pmod {35}$

[potent lotus]

casework

[wild marten]

im reading more on cyclic groups

what does it mean by order of b?

like order of a is jus the lowest int st it = e eight

right

[keen flower]

the order of b is the order of the cyclic subgroup generated by b

[wild marten]

oh

ohhhh

[keen flower]

i think it's equivalent to what you said

the order of b is the smallest positive integer n such that b^n=e

[wild marten]

yeah it is i just didnt realize they were making b a genirator of another group

[random adder]

Yes thats basically the thing i did before

2^n has order 12

4^n then has order 6

Pardon my bad notation

[wild marten]

just to be clear, you can take any element of a cyclic group and use it to genirate a cyclic subgroup

[wild marten]

ur good dw

im also lacking a large ammount of understanding lmao

[keen flower]

yes, in fact you can take any element of a group, whether or not the group is cyclic, and use it to generate a cyclic subgroup

[random adder]

It just might happen to be the same group

[wild marten]

i think i just proved this too

Closure is met as the powers just add and they remain in the group

Asocitivity is met through the parent group

identity is mean as a^0 = e

and inverses are met as if a^n is in the group a^-n is in the group

sorry all these definitions are building up lmao

[keen flower]

that sounds right

[wild marten]

im only like 4 days into abstract algerbra and i have so many things scrambling around in my head lmao

[random adder]

I mean for infinite groups its very easy to give a constuction for what you said. Just consider Z over addition.

[wild marten]

its very intutitive for an upper dib math class tho

[wild marten]

yeah, also the fact that groups use exponent laws

makes it really easy

[random adder]

Or exponents use group laws

Jk. But i dont really know what came first. I would guess exponents

[wild marten]

lmfao either way its intutive

if G is order N does that mean

|a| = n

or is it saying there are n elements in G

[wild marten]

for this

[keen flower]

G is order n iff there are n elements in G

that's the definition of the order

[wild marten]

so its the same as set notation

[keen flower]

yeah

if a generates G then a has order n as well

[wild marten]

Is this the proper way of looking at it

im a little lost by this one lmao

[keen flower]

sure
"the order of b is n/d" is equivalent to "|<b>| = n/d"

[calm coral]

@wild marten Has your question been resolved?

[wild marten]

Ok i need some help understanding this proof:

from the book for better legibility

Why can we just change k to d?

Like every step makes sense, but k is a set number and so it n

[random adder]

Which part exactly dont you get

[wild marten]

Second to last to the last step in my work

I skipped a proposition by mistake tho so im going back

[random adder]

This?

Oh one before.

[wild marten]

the one u circled and the one right after

I dont get that jump

Because n and k are set values

not abstract

right

[random adder]

You dont have km = n but rather n | km

[wild marten]

wdym

[random adder]

Just because $a^n = a^{km}$ doesnt mean $n = km$

[potent lotus]

casework

[wild marten]

hmm

[random adder]

But $n | km$

[potent lotus]

casework

[wild marten]

ok im a little confused

[random adder]

Like take for example $n = 9$ , $k = 8$

[potent lotus]

casework

[wild marten]

a^(2*3) = a^(6) right

ok

[random adder]

You cant find $m$ such that $mk = n$

[potent lotus]

casework

[wild marten]

ok

[random adder]

At least not integer

But that doesnt mean you cant get a multiple of n

[wild marten]

oh all multiples of n will give u identity

not just n itsself

[random adder]

Yes

[wild marten]

oh

lemme redo

I kept treating it as exponenets

forgetting there are multiple ns

[random adder]

You can. But like exponents with complex numbers

$e^{i2\pi} = e^0$ doesnt imply $2\pi i = 0$

[potent lotus]

casework

[random adder]

As you are working with finite cyclic groups (they cycle)

[wild marten]

ohhh thats why its called cyclic

(idk if thats true )

@random adder can you help me understand this proof a bit

im getting somewhat lost

[calm coral]

@wild marten Has your question been resolved?

[keen flower]

what part confuses you?

[calm coral]

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[pulsar owl]

We are given two equilateral triangles: ABC and XY Z. It is known that X is the
midpoint of side BC. It is also known that A lies on side Y Z (but A is not the
midpoint of Y Z. What is the maximum possible value of angle ∠XZC?

[pulsar owl]

The above is the picture

<@&286206848099549185>

[calm coral]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

@pulsar owl Has your question been resolved?

[pulsar owl]

.helopers

<@&286206848099549185>

[unique saffron]

uh

[pulsar owl]

Can you help?

<@&286206848099549185>

.close

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[prisma sapphire]

I’d like help understanding multiplying functions

[prisma sapphire]

The answer is A but I don’t know why

[torpid canopy]

I'm assuming the dot here denotes multiplication and not composition?

[prisma sapphire]

Yeah

[torpid canopy]

okay

so to be in the domain of f(x) * g(x)

it has to be in the domain of both

otherwise you can't multiply it

[prisma sapphire]

Got it

Range? I also calculated the range of f(x) wrong, it’s [0,5] I think

[torpid canopy]

for the range

[torpid canopy]

correct

and the range for g is just {-2}

what possible values can you obtain by multiplying a number in [0,5] by -2

[prisma sapphire]

0,-10

[torpid canopy]

right

and inclusive

since 0 and 5 are in the interval

so [-10,0]

[prisma sapphire]

Alr

Thanks 👍

[torpid canopy]

[prisma sapphire]

.close

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[prisma sapphire]

Uh

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[lament prawn]

can someone help me ques 1.89 im complete clueless

[distant trout]

assuming the cube has side unit length, can you identify the coordinates of the vectors?

for a) the coord of the edge vector and the diagonal vector

[lament prawn]

im sorry i dont understand

[frosty spruce]

assume a cube has length = 1

find coord of the vectors

[lament prawn]

b giải thích đề bài đc k mình đọc không hiểu lắm

[frosty spruce]

Cho khối lập phương, 1 đỉnh trùng với O (Origin là O(0,0)), 3 cạnh trùng với Ox, Oy, Oz. Tìm:
a) Góc tạo bởi Oz và đường chéo của khối lập phương
b) Góc tạo bởi đường chéo của 1 mặt và đường chéo của khối lập phương

nôm na đơn giản là như thế này

cái này chắc là lớp 12 phải tiếp cận r

[lament prawn]

mình hiểu r

done

.close

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[weary oasis]

Question 1
2 / 8 points

How would you explain binomial expansion in your own words to your peers?

What kind of problems can we solve using the binomial formula?

My answer:
Describes how a polynomial with two terms raised to a power can be expanded.
It is used to find probability in mathematics.

What did I do wrong?

[weary oasis]

i just got 2 points from that...

[random adder]

Grading of this question seems very subjective.

[weary oasis]

what?

[random adder]

How would you that is very personal question. (It depends on who you ask , doesnt have only 1 solution)

[weary oasis]

dont find loopholes...

[random adder]

You probably couldve written an essay about this question. And you couldve written 2 sentences. Idk how its graded.

[weary oasis]

use substantive thinking

[random adder]

Everything you said was right. Maybe you just needed to write more.

[weary oasis]

Really?

was i right?

everyone else confirm this

[calm coral]

@weary oasis Has your question been resolved?

[calm coral]

@weary oasis Has your question been resolved?

[weary oasis]

<@&286206848099549185>

[calm coral]

@weary oasis Has your question been resolved?

[calm coral]

@weary oasis Has your question been resolved?

[ivory lion]

what casework said was right

nothing you said was wrong but the teacher was probably expecting more

you answered the first question pretty literally

the second question is barely correct, it can be used to solve some probability problems

but it really doesnt even scratch the surface as to what the binomial theorem is good for

the grading is clearly subjective and contextless but i dont think the grade you got was unreasonable

at least not without more context

you could elaborate on question 1 by saying that the binomial expansion is just noticing a pattern in distributing in the expansion of a binomial raised to a power, which allows you to group terms such that the coefficients are combination numbers n choose k

you could elaborate on question 2 by saying that it can be used to split a combinatorial problem into cases or vice versa, such as proving that the sum of a row of Pascal's triangle is 2^n

or that it helps us pick out particular terms in a binomial expansion

you really could write a lot about it

[calm coral]

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[astral rain]

What is wrong with this graph?

[velvet osprey]

it's stretched vertically for no good reason

[astral rain]

yeah i have no idea where the upper dot should be placed in.

but i think i kept it for around the same as the one in before

[ancient grotto]

What is g(4)?

[astral rain]

4

[ancient grotto]

No

[astral rain]

4-2^2

[ancient grotto]

So 4-4 which is?

[astral rain]

0

[ancient grotto]

Poop

I thought you meant other way

[ancient grotto]

This would be?

[astral rain]

g(x-0)

[ancient grotto]

g(4)=(4-2)^2-4=2^2-4

[astral rain]

so is that the x or y cordinate?

i just go ask chatgpt

.close

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[ancient grotto]

y-coordinate

.reopen

[calm coral]

✅

[ancient grotto]

!nogpt

[calm coral]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[teal drift]

Are you really asking chatGPT this basic question?? You have to learn properly with books or class notes, otherwise you won't learn anything

[leaden thunder]

they gone to the almighty gpt

.close

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[deft trellis]

I can't figure this question out

[olive yacht]

do you know what continuous means?

@deft trellis i can help u rn

[deft trellis]

I believe it is when

the function and the limit result in the same output

like lim x-3 equals f(3)

[olive yacht]

ok yes

more simply tho, because you have the graph

the function is continuous where there is no holes, jumps, or breaks

or an infinite discontinuity where there's an asymptote

[deft trellis]

So would something like [-4, -2) be continuous?

Because there are none of those, and it excludes -2

[olive yacht]

well it is continuous there because it's not including -2
the parentheses ) means the interval doesn't include -2

everywhere on that interval is continuous

[deft trellis]

Yes

@olive yacht Thank you I think I got it

[olive yacht]

ofc!

[deft trellis]

.close

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[tall moon]

let $f(n)$ be the sums of $\frac{1}{a!b!c!}$ such that $a,b,c\geq 0$ and $a+b+c=n$, find $p+q$ if $f(2023)=\frac{p}{q}$ (yes its simplified)

[tall moon]

honestly my only thought is to multiply it all by (n!)^3 butt idrk

[olive yacht]

have you learned..taylor series perchance

[tall moon]

no

[olive yacht]

oknvm

im sure there is a simpler way

i just cant find it

[potent lotus]

skissue.in.a.teacup

[tall moon]

i dont think the last part is needed but ill just add it to be sure

[leaden thunder]

did you try partial fractioning 1/(stuff) = A/a! + B/b! + C/c!

[tall moon]

err

[leaden thunder]

and sub c= n-a-b

[tall moon]

A(b!c!)+B(a!c!)+C(a!b!)=1

[tall moon]

is this for doing pfd or something else

[leaden thunder]

i don't know when the best time to make that sub is

maybe after pfd

[tall moon]

you want A,B,C to be independent from a,b,c right

[leaden thunder]

they should be ?

[tall moon]

honestly idk how to find A,V,C hrre

[rustic osprey]

Isn’t it easier to say that the sum is $\frac{1}{n!} \sum_{a+b+c=n} \binom{n}{a,b,c}$

[potent lotus]

Civil Service Pigeon

[leaden thunder]

yea that's a better idea

[rustic osprey]

Yeah so it should be ||(3^n)/n!||

@tall moon you alive?

[tall moon]

ye

how does (n,(a,b,c)) work

like (n,a)(n,b)(n,c)?

[rustic osprey]

Prob should’ve just sent the whole theorem ngl

https://en.m.wikipedia.org/wiki/Multinomial_theorem

[tall moon]

lemme givr it a read

hm

is it like the sum (n,(a,b,c))×1^a×1^b×1^c=(3)^n

[rustic osprey]

Yeah

[tall moon]

so the ans would be 3^2023/2023! ?

ok ty

.close

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[tall moon]

.reopen

[calm coral]

✅

[tall moon]

wait hold on how do you add the simplified 3^2023/2023!

2023! has 1006 powers of 3

do you write it like 3^1017+(2023!)/3^1006

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[opaque yoke]

how could i do this

i multiplied both side by 10^x

[unkempt drift]

yep so you can let 10^x = u

you get a quadratic in u and you use the quadratic formula

[opaque yoke]

oh ok

thanks

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[opaque yoke]

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[calm coral]

✅

[opaque yoke]

.cllse

.close

[calm coral]

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[calm coral]

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[opaque yoke]

confused on how this is wrong

Pe^rt

2=e^rt

ln2=rt

compounded monthly so ln2=12rt

r=0.08

ln2/0.08*12

=.7220283

so .72

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[ivory pilot]

Where did I go wrong?

[ivory pilot]

My problem is that while using the negative velocity formula, the output velocity don't get to be 0 at t=3.470239

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[narrow harness]

how would i go about answering this pls

[potent smelt]

via integration, no?

[vocal pagoda]

You need to solve the DE.

[narrow harness]

how did you find out its differential equations pls

[narrow harness]

probably

[zealous mulch]

Fairly sure you're lacking some info

[potent smelt]

It's a differential equation because it's an equation involving the derivative of a variable.

[narrow harness]

its the whole question

[potent smelt]

It's a very simple kind of differential equation though

[zealous mulch]

You need a point to unequivocally define the function f(x)

[potent smelt]

as it doesn't involve the variable itself, just the derivative

[zealous mulch]

Guess they want you to figure out a point by the graph then?

[potent smelt]

have you tried just integrating it?

[narrow harness]

surely xe^x -e^x integrated is the same?

like e^x - xe^x

[potent smelt]

it's slightly different

[zealous mulch]

You're integrating a product

[sacred oyster]

Helpppp

Help

[potent smelt]

!occupied

[sacred oyster]

Plss

[calm coral]

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

[sacred oyster]

?

[potent smelt]

he expanded the product

[zealous mulch]

Still will be left with xe^x

[potent smelt]

it doesn't integrate to itself.

ah, that's what you were talking about

Yes, he'll need to use Integration by Parts

[zealous mulch]

Yeah, that's what I was trying to tell him, though Ideally I wanted him to come up with it(assuming he's covered it)

[potent smelt]

ah fair enough, sorry for the spoiler then

[narrow harness]

ive done all pure alevel maths now

so ive done IBP

[zealous mulch]

Even After that though there's still the problem of actually determining which function it is

[calm coral]

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[blazing hinge]

ok so you are given two points and it's a linear equation

do you know the slope formula

[calm coral]

@restive hare Has your question been resolved?

[restive hare]

no

[blazing hinge]

a = (y2-y1)/(x2-x1)

and you're given points (x1, y1) and (x2, y2)

[restive hare]

yes

a=7-4/-1-1?

[blazing hinge]

not quite

that's (y2-x2)/...

[blazing hinge]

we want this instead

[restive hare]

ok

sorry

4-1/7--1

so that is a

[blazing hinge]

flip the fraction and that's your a

we wants the y's on top and x's at the bottom

[restive hare]

ok

7--1/4-1

now what

do I switch it around for b?

[blazing hinge]

ok so you have

a, y and x now

so yeah, you can find b

using y = ax + b

where (x, y) is any of the point given in the question

(1, -1) will be easier

[restive hare]

wait so a is the fraction

y is 7--1

and x is 4-1

[blazing hinge]

yup

[restive hare]

so to find b

[blazing hinge]

that gives 8/5 right

[restive hare]

ax-y

8/3?

4-1

[blazing hinge]

don't forget y is -1, not 1

[restive hare]

ohhhh

yep

[blazing hinge]

so your b would be -1 - (8/5)

[restive hare]

is it not 8/5x5-8?

[blazing hinge]

umm idk how u got there

let's try again

[restive hare]

ok

[blazing hinge]

a is 8/5 correct?

[restive hare]

yes

[blazing hinge]

you have an x, -1

and a y, 1

[restive hare]

how is the x only -1 and not both?

[blazing hinge]

bc coordinates are in the form (x, y)

[restive hare]

ok

[blazing hinge]

and the q gives you (-1, 1) as a point

[restive hare]

but also 4,7 is a cord?

[blazing hinge]

alright so

yes

you can use that one too but -1 and 1 are easier to work with

so

[restive hare]

ok

wait for 8/5

[blazing hinge]

the line equation is in the form of y = a*x + b

[restive hare]

why do we use -1 for both

should it be

4-1

which is 3

there not two substraction signs

[blazing hinge]

yoo y is -1 sorry

and x is 1

[restive hare]

so it it 8/3?

[blazing hinge]

i mean it's still 8/5

[restive hare]

you better than me its ok

[restive hare]

whst

7--1=8

4-1=3

?

[blazing hinge]

woops

got the coordinates wrong again lol

ok

8/3

phew

[restive hare]

8/3?

[blazing hinge]

8/3 is a

so then, y = ax + b => -1 = 8/3*1 + b => b = -1 -8/3

and once you find b from there, you can multiply the whole equation by a number which would turn the fractions into nice integers...

[restive hare]

yesx

ty

ok

so b is -11/3

[blazing hinge]

yup

should be right

[restive hare]

ok

so

-1=8/3+-11/3

[blazing hinge]

you can keep -1 as y

and don't forget that it's a*x

so y = (8/3)x - 11/3

that's your line equation

[restive hare]

ok

[blazing hinge]

now you need to turn the fractions into integer

and change the form so that it's ax + cy = b

[restive hare]

ok

uhh

they are not nice numberss

[blazing hinge]

how would you normally turn 8/3 into an integer

[restive hare]

x 10

nvm sry

[blazing hinge]

• so that it gives 8

you'd multiply 8/3 by 3 right

the 3's cancel out, now you're left with an integer

[restive hare]

yes

8

[blazing hinge]

so

now multiply both sides of your equation by 3

see what happens

[restive hare]

ok

[blazing hinge]

this one

[restive hare]

y=8-11

8+y=11

[blazing hinge]

if you're gonna multiply one side by 3, you gotta multiply the other side by 3 as well to keep things equal

sooo that would be 3y

[restive hare]

ok

8x-3y=11

[blazing hinge]

yessir

[restive hare]

finally

ty

[blazing hinge]

np!

[restive hare]

.close

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[restive hare]

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[restive hare]

wait

how is it 3y

if its -11/3 x3

its -11?

oh wait nvm im dumb

.close

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[fickle wigeon]

please help

[velvet osprey]

there is a nice trick here that works

[fickle wigeon]

if you are talking about putting x as -1 and 1 that wont work here

[velvet osprey]

i and -i actually

[fickle wigeon]

fuck

i thought about w and w^2

crap

but as far as i remember that gave a relation between coefficients spaced by 2

anyway thanks

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[violet jackal]

been trying to understand analytical applied maths,,,

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[opaque palm]

can anyone tell me why i get the exact same graph with the exact same values no matter if i have "k" or "2k" as an upperbound in the second sum

[jolly pilot]

Coz the expression with the inner sum reduces to 2x^2k in both cases. The terms from r = k+1 onwards are all 0s

Even if you increase the upper bound the 3k, 4k, ... the expression would evaluate to the same value

[opaque palm]

thanks. however, ive tried to simplify it with the binomial theorem to (2x)^2k before but i get a different graph

[jolly pilot]

Oh wait, its all 2r, so theres no odd terms. my bad

it wont be exactly 2x^2k

but nonetheless, the point remains that all the terms beyond r=k+1 are all 0s

[opaque palm]

thank you. but that means i cant use the binomial theorem right

[jolly pilot]

thats why both answers are the same

[jolly pilot]

yea

[opaque palm]

thats sad

it looks so nice

[jolly pilot]

pretty sure theres some identity tho

[opaque palm]

but i didnt know there can not only be even terms for this to work

[sonic fern]

you can use the binomial theorem

first take the x out of the first summation

[jolly pilot]

Like Im 99% sure the even binomial coefficients sum out to 2^(2k-1)

[sonic fern]

yes they do