help-39

[burnt ibex]

right

https://tenor.com/view/yhwach-bleach-gif-13682097564047371169

[tropic saddle]

!occupied

[pearl pond]

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).

[burnt ibex]

what

[chrome river]

"An associative R-algebra, A, is a vector space over R with a prodct s.y. for all u,v,w in A a,b in R we have: associativity, distributivity, etc"

@burnt ibex if you need help, ask in another channel, #❓how-to-get-help

[chrome river]

so i guess the ring is with scalar multiplication and vector addition

so yea componentwise multiplication

what would be the mapping?

from R(beta) to R^2

but them being isomorphic still doesnt mean they are the same thing right?

here i have another doubt aswell.
Why do we say A = C for beta^2 < 0 but A is only isomorphic to R^2 when beta^2 > 0. From my understanding isomorphic means preserving structure. But if they preserve structure isnt it pretty much the same thing

atleast in abstract nonsense when we say two categories are isomorphic we can treat one as the other with no issue

[tropic saddle]

well when I have two groups like Z_4 and Z_5*, they are isomorphic but still different

different operations and so on

if beta^2 > 0, then A is R[beta]/(beta^2-x) iso R[beta]/(beta+sqrt(x)) x R[beta]/(beta-sqrt(x)) iso R x R

[chrome river]

right i get it

[tropic saddle]

I should have texed that

[chrome river]

ohhhh yea makes sense

okay yea thanks!

i think i get it now

:)

.close

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[pearl pond]

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[odd mason]

i got a logical question

[pearl pond]

Please don't occupy multiple help channels.

[odd mason]

(why doesnt integral of sin(x), upper limit pi , lower=0 )= (pi*1)/2

[prime bramble]

I'm not sure how to interpret this question; why would you expect the integral to evaluate to pi/2?

[odd mason]

because the integral means the area under a curve and sin(pi) is a semi circle

[prime bramble]

it isn't

[odd mason]

i meant sin(0) to sin(pi)

[prime bramble]

this isn't a semicircle

if it were, then we'd see a circle here with the other "hemisphere" added

but we don't

[odd mason]

wait i saw a question like this i am not sure if it was only sin(pi)

but thanks anyways

[prime bramble]

no problem

you can close the channel if you're done here

or... I can

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[brittle onyx]

If a numebr to the power of 2 is divisible by a number, then is the original number also divisible by that number?

[plush bramble]

Wut

Can you be more precise

"a number" refers to two numbers?

[sharp quest]

that's not true, a^2 divisible by b does not imply a is divisible by b

[regal herald]

2 isnt divisible by 4 as a simple counter

[brittle onyx]

Would by working for 2 be right then?

,rccw

[dapper kraken]

i dont think so for the second part

[brittle onyx]

Damn alright what do u think I should do for it then?

[dapper kraken]

personally id do modulo but i have no idea if thats applicable here (my learning road is fucked)

[brittle onyx]

Ok

[pearl pond]

@brittle onyx Has your question been resolved?

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[upper drum]

I am getting 15/36 which 5/12 as ans

condition would be alpha < beta

so
1/6 x 5/6 + 1/6 x 4/6 ................ 1/6 x 1/6

but no correct option

[tardy reef]

This condition is wrong. If you put $\alpha = 3, \beta = 4$, then your equation becomes,
$$x^2 + 6x + 4$$ which has value of -4 for x = -2

[jolly parrot]

Facter10Br4g

[tardy reef]

|| think of completing thesquare ||

[upper drum]

oh

[upper drum]

cannot we just do D < 0

[tardy reef]

that tells you about the roots

not the polynomial itself

[upper drum]

nop

when c > 0 and D < 0

it gives positive value

[tardy reef]

Yea. You need c>0 as well was my point

Also, you put the condition wrong

D<0 gives $\alpha^2 < \beta$

[jolly parrot]

Facter10Br4g

[upper drum]

aha yeah gotcha

.close

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[upper drum]

thanks

[upper drum]

it would be always positive

as dice cannot give negative

i just kept condition wrong

[tardy reef]

yea, brainfart moment

mb

[upper drum]

yeah np

thanks tho

[pearl pond]

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[exotic gale]

Can someone please help me on this question:

[leaden wadi]

Th total probability of picking a blue, red, or green ball has to equal 1.

[pearl pond]

@exotic gale Has your question been resolved?

[pearl pond]

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[dusk kelp]

how do you turn a quadratic expression into vertex form from 3x^2+1

[pearl pond]

Please don't occupy multiple help channels.

[compact ridge]

.close

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[sand robin]

why these two expressions are not the same

[west sapphire]

because log(a+b) does not equal log(a)log(b)

log(ab) equals log(a) + log(b)

[sand robin]

ahaa yes im stupid

thanks

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[west sapphire]

yw

[pearl pond]

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[west sapphire]

duplicate of #help-25

.close

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[carmine forum]

I can reach this point, I found put of that 1/siny is equal to ln|tan(y/2)| but I cant find how to reach that conclusion. I see my professor is doing like in the second photo but cant understand how he thinks about that substitution in first place

[drowsy adder]

tanx=t is quite a useful substitution as anywhere there is sec^2x you can write it as 1+t^2 another common thing is to multiply and divide by sec^2x or sec^2(x/2) like the later works in this case

you can also proceed with integrals of 1/(asinx+bcosx) in a similar way, putting tanx = t

[carmine forum]

.close

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[lime haven]

https://tenor.com/ssrfKELVEsE.gif

https://tenor.com/heK6nGTyEVU.gif

/imgine

[pearl pond]

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[lime haven]

gif

[feral sedge]

.close

[pearl pond]

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[lime haven]

https://tenor.com/cWvQcFMBK6A.gif

midjourney

[solar compass]

HI

I AM NEW IN THIS SERVER

[feral sedge]

#❓how-to-get-help

or #discussion

[solar compass]

2+2 = ?

[reef marlin]

2 + 2 = 4.

[pearl pond]

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[pulsar stump]

anyone has a correct definition for a k-ramsey graph? couldnt find one online

[pearl pond]

@pulsar stump Has your question been resolved?

[pearl pond]

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[pearl pond]

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[left dock]

why is this considered inversely proportional ?? (y=k/x)

when yx doesnt add up to 12 (5*2)

[mellow idol]

it's either an error in the table or the answer

i think they meant to write 2.4 not 2

[left dock]

if k isnt a constant

its NOTHING proportional to right???????/

[mellow idol]

what

[mellow idol]

if u mean to say y isn't inversely proprtional to x here ur correct

but i think theyu just forgot to write the .4 after 2

[left dock]

they marked me wrong

[mellow idol]

did u write 2 or 2.4

[left dock]

i did a wild guess cause the choices were just direct/inverse/joint/combined

[left dock]

the teachers were the one who wrote that

it was a quiz

[left dock]

they showed the table and these were the choices

[mellow idol]

ur not wrong cause the last values aren't consistent with inverse proprtion and if they wrote the wrong values thats on whoever wrote it not u

[left dock]

so im guessing MY TEACHER IS WRONG

[leaden marsh]

It has to be a writing error

[left dock]

thank you bro

i knew i was right

they were the ones who were wrong

/close

closed

close

[leaden marsh]

close after a dot

[mellow idol]

.close

like this

[pearl pond]

@left dock Has your question been resolved?

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[midnight haven]

write this without the root if x<-2

[mellow idol]

sqrt(x^2)=|x| for all real x

[mellow idol]

can you tell me how the x>-2 would affect the answer?

[midnight haven]

well

[spare lark]

I think he supposed to not have abs either

[midnight haven]

x has to be less than -2 so

[mellow idol]

so?

[spare lark]

In that case its just abs(2+x) but if you must have no abs also, it will be -(2+x) since 2+x < 0 for x < -2

[midnight haven]

when we get to |2+x|= 2+x
we need to use a - in front of the right side to make it true right?

|2+x|= -(2+x )

no idea

because if (2+x)<0

[mellow idol]

yup!

[midnight haven]

yeah so if I write this:
|2+x|= -(2+x )

am I done with the task? I eliminated the root

and that was the task

am I overcomplicating it

[spare lark]

Seems done

[midnight haven]

is this the correct final answer? @spare lark

[spare lark]

Yeah its good

[midnight haven]

thanks

its done then

[spare lark]

I let u close it

[midnight haven]

how do i

.close

thats how

.close

[pearl pond]

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[pearl pond]

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[alpine glacier]

Hello, please help with this question, i find it extremely difficult and the method i am trying isnt ont he mark scheme :((

[alpine glacier]

Here is the mark scheme

[neon void]

Hey guys can somebody help me, im not in school or anything this is just beyond my knowledge

[alpine glacier]

<@&286206848099549185> please help 🙏

[inland gate]

What help u need ??

[alpine glacier]

its not him, its the question above, that user is just confused

[still bane]

i would use casework tbh

1P2C3B4 where 1,2,3,4 are slots where other papers can go or can be empty

multiply final value by 3! because pcb can be swapped around

[meager grotto]

Yo

[still bane]

the rest is trivial i think

[meager grotto]

You need help ?

[alpine glacier]

wait but how are you dealing with the 18 possible slots?

[still bane]

you place blocks

blocks of subjects

[alpine glacier]

but P could also be in the first slot, no?

[still bane]

then slot 1 is empty

the slots can be empty

[alpine glacier]

ohh

[still bane]

def not the most efficient way to do it here

better with groups of indistinguishable objects

[alpine glacier]

honestly I thought I was cooking up a storm in the moment but I feel like I forogt to substract something that I

*that i overcounted

[still bane]

prolly

no actually your answer is smaller

4.39 * 10^15

you got 12.1 * 10^14

which is 1.21 * 10^15

your answer is 4x smaller

[alpine glacier]

well that answer is actually not the answer

[still bane]

oh

[alpine glacier]

i would have to substract that from 18!

[still bane]

ah

[alpine glacier]

although I also think I miscounted the zeros and that its supposed to be 12.1 * 10^15

[still bane]

okay

[alpine glacier]

oh wait nvm i didnt miscount

hot hugely far off, but also not right lmao

[still bane]

true

method 2 is similar to the method i wouldve used

but they did like the contrapositive

they placed the 15 non-scinece papers first

like 1A2A3A...

and there are 15! orders of those papers

and then there are 16 slots in between to put the science papers in

16*15*14*15!=4.39*10^15

[alpine glacier]

oh yeah

honestly I find that hard to wrap my head around...

i think the closest to my attempt was method 3

[still bane]

yeah

[pearl pond]

@alpine glacier Has your question been resolved?

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[cobalt hinge]

ill post my working in a few mins

[cobalt hinge]

gg

[jaunty jacinth]

??

[cobalt hinge]

actually you can close this one open a new one and i can help u there 😄

[jaunty jacinth]

oops

[open rivet]

unlucky guys

[jaunty jacinth]

my bad

i didn't see it 😭

[cobalt hinge]

alg

[jaunty jacinth]

.close

[cobalt hinge]

i have the post

just delet the msg

[cobalt hinge]

so i realized that $a_1+a_2+a_3...a_{20}=6(a_1+a_2...a_{10})$

[jolly parrot]

;(

[cobalt hinge]

then i denoted d to be the common difference so i got $20a_1+210d=6(10a_1+45d)$

[jolly parrot]

;(

[cobalt hinge]

this implies that $40a_1+60d=0\implies 2a_1+3d=0\implies a_1=-\frac{3d}{2}$

[jolly parrot]

;(

[open rivet]

wait how did you get 20a1?

nvm

the 210d is wrong

[cobalt hinge]

oh ok

[open rivet]

a20=a1+19*d

[cobalt hinge]

o hshit

lol

[open rivet]

added one term too many

[cobalt hinge]

im trying to work without arithmetic formula

[open rivet]

😄

[cobalt hinge]

i forgot a_1 doesnt have a d

🤨

anyways

[cobalt hinge]

$20a_1+190d=6(10a_1+45d)$

[jolly parrot]

;(

[cobalt hinge]

thats so odd cause i did it correct on the RHS

[open rivet]

yup

[hot canyon]

20(2a_1 + 19d) != 20a_1 + 210d

[cobalt hinge]

breh my mind is fucked i think

i stayed up till 230 yesterday playing games and doing this shit

[hot canyon]

You should be getting: (2a + 19d) = 3(2a + 9d) => 4a + 8d = 0 => a = -2d

[cobalt hinge]

yeah

so a_1/a_2=2 😄

thx

.close

[pearl pond]

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[cobalt hinge]

wait wait wait

.reopen

[pearl pond]

✅

[cobalt hinge]

this is the julia fractal right?

or is it mandelbrot

[open rivet]

mandelbrot is z_n+1=z_n^2+c i think

[cobalt hinge]

yeah

interesting

[rancid depot]

julia sets are of this kind of form

[cobalt hinge]

hmm

yeah ill figure this later

.close

[pearl pond]

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[pearl pond]

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[hot star]

I have been looking for a while but I can't find anything good on volumes of revolution about a line

[hot star]

if you can share a resource or tell me how to do it I would be very pleased

[cobalt hinge]

khan academy

[hot star]

anything specifically?

[cobalt hinge]

https://www.khanacademy.org/math/ap-calculus-bc/bc-applications-of-integration-new/bc-8-7/e/volumes-of-solids-of-known-cross-section

start from here

and work your way up till the end of unit 8

[boreal raptor]

:0

sal khan

my absolute goat

[hot star]

thanks 🙂

[cobalt hinge]

actually wait use lesson 9 to end of unit

[hot star]

coolio, will do!

[cobalt hinge]

you can also ask in what ever channel they share resources in

i cant link it for some reason

[hot star]

they?

[hot star]

XD

[cobalt hinge]

the server

[hot star]

ahhhh

gotcha!

coolio

just checking this is what I want before closing the channel

[jagged cape]

I Don,t understand this and tommorrow is my exam

[pearl pond]

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).

[hot star]

man, chuck this in a free channel XD

yup that was exactly what I was after! thanks man

.close

[pearl pond]

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[pearl pond]

Please don't occupy multiple help channels.

[iron stream]

20(x+1) + 130

@sharp sonnet

[cobalt hinge]

did u mistype?

[iron stream]

.close

[pearl pond]

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[midnight haven]

Can somebody explain how we get the highlighted inequality?

[open rivet]

they are using induction

so assuming 2<=2z_k<2_k+1<4

and then they show that z_k+1<z_k+2

[midnight haven]

But how does the root come in the inequality?

[pearl pond]

@midnight haven Has your question been resolved?

[plush bramble]

from the first sentence

[midnight haven]

Okay I got it

.close

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[pearl pond]

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[west bluff]

So, this problem is from a topology book, and while I've somewhat gotten to a solution, I'm not sure it's right, so I'd like to get it checked:
Show that there exists no f:R->Q such that f is continuous and non-constant.

[west bluff]

Essentially, my idea was to show that if this were true, there would be some irrational value contained in the image of f(which is Q, hence it cannot happen)

And to do that:
I assumed the existence of such an f and considered a generic open set A in Q, from which I considered a point x

Since A is open and f is continuous, f-(A) is open in R and therefore exists some open ball of radius δ>0 for f(x)

However, the image of all those points is also in A⊆Q, so I returned to the classical definition of continuity in R:
x∈(a-δ,a+δ)->f(x)∈(f(a)-ε,f(a)+ε) for some a, and some ε,δ>0, hence, I considered a succession of points {xn} contained in the open ball within the preimage I considered

All points in that succession must also be in A, and since f is continuous, they must satisfy that condition

Considering points from {xn} I then said that they form some interval (a-δ,a+δ) for some δ>0(which can't happen if f is constant)

From there, since a-δ and a+δ are reals, there exists some irrational value between them for any delta chosen, which would also be in A, which was chosen to be contained in Q

Hence, absurd and f does not exist

(does it look good?)

[autumn fossil]

are you sure about this?

all points of {xn} could be contained in that interval, but they could be all equal to a

[west bluff]

That's the thing I'm doubting the most

[autumn fossil]

i actually didnt fully grasp your argument

have you proved the intermediate value theorem already btw?

[west bluff]

Wouldn't it be constant then?

[autumn fossil]

not necessarily

[west bluff]

Someone mentioned that before, if you mean the intermediate value theorem for R, then yes

[autumn fossil]

it could have the same value on all points of xn, but a different value somewhere else

[west bluff]

If you mean some generalization to connected spaces they mentioned, then no

[autumn fossil]

not mean value

mean value is for derivatives

[west bluff]

The one that states that if there exist values greater and lower than 0 for a continuous function over R there exists a root in the interval?

[autumn fossil]

yeah, this one is good

see if you can apply it

[west bluff]

I cannot assure the existence of sone value greater and lower than 0

[autumn fossil]

you can slightly modify your function though

[west bluff]

Fairly sure, and even then, don't see how it'd help me, 0 is still rational

[autumn fossil]

like shift it up and down

[west bluff]

Which was the composition that shifted functions upwards? I'm thinking (x-a) but fairly sure it's a translation

[west bluff]

And still don't really see how it'd work out if I'm being honest

[autumn fossil]

Let's start from scratch

[west bluff]

The only thing I've got to work with is that it's continuous and goes from R to Q

[autumn fossil]

so you somehow figured out the proof would be by contradiction

[west bluff]

Most of these kind of statements are proved that way, really

[autumn fossil]

So assume there is a function f:R->Q such that f is continuous and non-constant.

what does it mean to be non-constant?

Can you give me a statement that defines it?

[west bluff]

That there exists some a,b such that f(a)≠f(b)

[autumn fossil]

okay, perfect

and this will be our assumption

notice that in your proof you actually never used this, you just took a random open set in Q

that's probably the main reason why it didnt lead anywhere

[autumn fossil]

anyway, we now have a continous function and 2 points on its graph which are unequal (in y-coords)

and we need to find some point with irrational y-value

where on the gaph do you think it would be reasonable to search for it?

think in terms of a and b

it can be general

[west bluff]

What do you mean by “where in the graph“?

[autumn fossil]

say that these are our points

and we have some continous function passing through them

what would be the irrational value in range of the function?

[west bluff]

Oh, well, within [f(a),f(b)]

[autumn fossil]

yeah

and this already hella looks like the theorem above

we are finding certain value between 2 bounds

so we have some irrational value there, say sqrt(2) for now

how do we guarantee in this specific example that sqrt(2) is indeed in the range

[west bluff]

We still cannot assure anything if the function crosses 0

[autumn fossil]

Yeah, but we can modify the function

[west bluff]

if f(b)>✓2 and f(a) is lower?

[autumn fossil]

in our specific example, say f(b) = 2 and f(a) = 1

we can generalize later

now I want you to use the theorem above to prove that sqrt(2) is somewhere in the image

[west bluff]

Oh, so we assume some irrational value c , compose the function with a traslation of the graph c units downwards

[autumn fossil]

Yeah, that's the idea in the right direction!

[west bluff]

And show that since f(a) is either higher or lower than c and so is f(b), f reaches c?

[autumn fossil]

yep, that's the idea

[west bluff]

But they can both be above or both be below, right?

Actually, nvm, we can choose c so that doesn't happen

[autumn fossil]

yep

[autumn fossil]

as you said here

[west bluff]

Yeah, alright, guess that makes sense

Only thing now

Is that I don't remember which composition mapped the plane upwards 💀

[autumn fossil]

Let's assume (without loss of generality) that f(a) < c < f(b)

how do you guarantee that g(a) < 0 < g(b)

where g is some newly defined function

[west bluff]

Subtracting c?

[autumn fossil]

there is only one step to do here

perfect

you'll be applying the theorem on f(a) - c and f(b) - c basically

[west bluff]

Can't this also be used to prove that if a real function is continuous on an interval [a,b] it reaches all the intermediate values between its maximum and minimum?

(which is quite the useful proposition, really)

[autumn fossil]

intermediate values... hmm.... that could be the "Intermediate value theorem" lol

but yeah, its a pretty useful proposition

[west bluff]

Actually, guess that makes sense 💀

[autumn fossil]

i was actually quite surprised when you came up with that root thing

intermediate value theorem is a simple corollary of that

[west bluff]

Which root thing?

[autumn fossil]

that theorem

it's often only used to prove the IVT and after that IVT is the main tool

proving that theorem and stopping there is like stopping 10 meters before the finish line

[west bluff]

Well, that tends to happen to me

I sometimes come up with some really strange ideas I later don't know how I thought of

And sometimes can't see the most obvious implications of things

[autumn fossil]

That sometimes happens to all of us ig

[west bluff]

Well, regardless, thanks for your help, it ended up being an easier problem than I was making it out to be

Didn't even need to use topology, which is somewhat weird, but oh well

.close

[pearl pond]

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[autumn fossil]

topology and analysis are strongly related

[pearl pond]

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[midnight haven]

My brain hurts just by looking at it can someone help solving it😭

[rough forge]

do you know FOIL?

[midnight haven]

yes

[rough forge]

do you know power rules?

[midnight haven]

not for roots

or is it the same

[rough forge]

roots are basically also powers

[midnight haven]

ohh yea so do I rewrite all the roots into powers after i did the foil?

[rough forge]

yea

[midnight haven]

makes sense give me some minutes i’ll say if i did it

@rough forge ok i got:

(a^3)^1/2 * (a^2) ^1/2 +(a^3)^1/2 * -(a^5)^1/4 +a^1/6 * (a^2)^1/2+ a^1/6 * (a^5) ^1/4

can I do something with those that have the same power?

[rough forge]

I am a bit confused

[midnight haven]

is it wrong?

[rough forge]

wait no

[jolly parrot]

anti-algebraist 𝔸dωn𝓲²s

[midnight haven]

where did I mess up

[rough forge]

How did you get (a^2) ^1/2 and -(a^5)^1/4

[midnight haven]

When doing the foil i took the right side as
-(4root)a5

cuz

[jolly parrot]

anti-algebraist 𝔸dωn𝓲²s

[plush sundial]

it comes in as a^4

[midnight haven]

im confused

[rough forge]

you have the 4-th root not square root

[jolly parrot]

anti-algebraist 𝔸dωn𝓲²s

[midnight haven]

ohhh

I forgot about that one

yea thats where I messed up

ok so after we did the foil coorectly and wrote it to all powers what’s next

[rough forge]

You'd simply the powers

[jolly parrot]

anti-algebraist 𝔸dωn𝓲²s

[midnight haven]

so at the left* starting from (a^3)^1/2= a^3*1/2 ?

[rough forge]

yea

and then since the terms are multiplied with the same base you can add the powers

[jolly parrot]

anti-algebraist 𝔸dωn𝓲²s

anti-algebraist 𝔸dωn𝓲²s

[rough forge]

yea so now just add the powers where you multiply

[midnight haven]

oh yea

long task

I hope it wont be in the test

it probably will be

anyways thanks for the help!

.close

[pearl pond]

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[pearl pond]

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[brave sluice]

this exercise is killing me

i'm working on number 16

[pearl pond]

@brave sluice Has your question been resolved?

[brave sluice]

okay we know that everything is G-sets already, so i think this question is only about defining an isomorphism

that's what we needed 14(a) for; not for an isomorphism, but to know that the disjoint union of left cosets is a G-set

still not sure what 14(b) is for

[pearl pond]

@brave sluice Has your question been resolved?

[pearl pond]

@brave sluice Has your question been resolved?

[pearl pond]

@brave sluice Has your question been resolved?

[pearl pond]

@brave sluice Has your question been resolved?

[pearl pond]

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[bright kindle]

how do i integrate this

[neon ermine]

Is that 2+sint

[unborn abyss]

multiply by conjugate

[iron stream]

Well first apply kings rule

1/(2+cost)
2+cost = 2cos²t/2 -1 + 2 = 3cos²t/2 +sin²t/2

[bright kindle]

okay let me try thanks

[iron stream]

Now just consider tant/2 = z

[bright kindle]

whats that

[bright kindle]

is my handwriting that bad

[unborn abyss]

yes

you're also missing dt

[rough forge]

If you have a+b then the conjugate is a-b

[bright kindle]

it was a long ques I don't really write it when i practice

[unborn abyss]

what is it actually

[bright kindle]

ohh ok

[iron stream]

What is the original question?

[iron stream]

1/1+sinxcosx can be integrated
By writing 1 = sin²x+cos²x and considering the sub tanx = z

[bright kindle]

thankyouu

.close

[pearl pond]

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[craggy crypt]

Is this correct

[rough forge]

what did you do in the second step?

[craggy crypt]

Made the bottom positive?

Oh I realised

.close

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[bright kindle]

uh

[bright kindle]

but by doing this i am getting tan²x in numerator

it will be harder to integrate ths

[sand robin]

https://www.integral-calculator.com

try this its pretty good

[pearl pond]

@bright kindle Has your question been resolved?

[pearl pond]

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[spark rock]

Hi I was here yesterday and a user showed me that I can reindex the sequence $(1 - \frac{1}{n+1})^n$ (which converges to 1/e) like this $(1 - \frac{1}{n+1})^{n+1-1}$ now let m = n+1: $(1 - \frac{1}{m})^{m-1} = \frac{(1-\frac{1}{m})^m}{1 - \frac{1}{m}}$ to show that $(1 - \frac{1}{m})^m$ converges to 1/e (since the denominator converges to 1)

[jolly parrot]

p1za

[spark rock]

now I was trying to do the same to prove that $(1 - \frac{1}{n+2})^{n}$ converges to 1/e aswell but Im not sure how i can apply the index shift here

[jolly parrot]

p1za

[gleaming moss]

m = n+1 --> (1 - 1 / (m+1))^(m-1)

[spark rock]

can I say $(1 - \frac{1}{n+2})^n = (1 - \frac{1}{n+2})^{(n+2) \cdot \frac{n}{n+2}}$ and since $\lim\limits_{n \to \infty} \frac{n}{n+2} = 1$ we have a sequence that is shifted by 2 so it also converges to 1/e?

[jolly parrot]

p1za

[spark rock]

ohhhhh

thats smart

[gleaming moss]

it's the same thing as before

[spark rock]

u think the way i described is also ok?=

[gleaming moss]

I think so, to rewrite it as the other example you showed it would look like:
Let ( m = n+2 )
[ \implies \left( 1 - \frac 1m \right)^{m-2} ]
[ \implies \frac{ \left( 1 - \frac 1m \right)^m }{ \left( 1 - \frac 1m \right)^2 } ]

[jolly parrot]

shsgd

[gleaming moss]

hope that helps

[spark rock]

yep definitely does thanks

[gleaming moss]

!done

[pearl pond]

If you are done with this channel, please mark your problem as solved by typing .close

[spark rock]

.close

[pearl pond]

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[cosmic reef]

hi

someone explain me 1+1=2

[pearl pond]

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[odd mason]

i bet u can't solve this

[pearl pond]

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/

[drowsy adder]

Just ask if you have a question

[odd mason]

dis

its been 6 hours

[hot canyon]

Find the first and second differences.

[odd mason]

so i am kinda sure that i am gonna burn this world down

for all this time i thought it was geometric

but turns out its gay

it could be anything

quadratic, cubic, quartic

[hot canyon]

Yeah it's not geometric but you'll realize the second differences have a pattern

[autumn trellis]

it could be anything, which is why these questions are not good questions. constant n-th differences mean an nth degree polynomial

[odd mason]

bruh we can't , we need atleats 3 2nd diffrences

[vestal tapir]

it's interleaved

3x and 2x

hm no

[autumn trellis]

infact there are a bunch of polynomials that will interpolate this. there is no real answer for this stuff

[odd mason]

i hate this shi

so when people give us 3,6,9,12 how do we know that it won't change after it and how do we know that the formula is 3n

[odd mason]

my life hurts by feeling my mind explode

[odd mason]

i though since there are bunch of polynomials that will interpolate this thus its an exponential

but turns out it still remains gay

do my questions have answer or maths just hates me

[vestal tapir]

(-1/24)x³ + (1/8)x² + (17/3)x − 1

[odd mason]

bro i dont think u are understanding the problem

[vestal tapir]

what why

[odd mason]

for cubic there was only 1 difference

[odd mason]

thus u can can't confirm that the sequence goes like this

[vestal tapir]

yes, just like we can't confirm 3,6,9,12

it's not about confirming

[odd mason]

thats the issue , how do we know the answer to a sequence

[vestal tapir]

like, hoping

you use your luck

[odd mason]

if its not about that , that should mean its indefenite meaning infinte answers

[vestal tapir]

it's a valid guess, maybe you're right and that's what it says in the answer
my guess is different

[odd mason]

luck in maths is a bit weird

[odd mason]

the cubic is natural to u , quadratic might be natural to me so does that mean answer of it is different relative to the person solving?

[vestal tapir]

it's different relative to the person who writes what counts as right

"the asker"

[odd mason]

everything is getting complex

that should also mean that an infinite series should have 1 answer?

i don't wanna be that guy

but maths is rigged

[odd mason]

convergent and divergent shouldn't exist because of this

[vestal tapir]

@hot canyondid you find something else?

or just that?

[hot canyon]

I dunno. I got 25 T-T

[vestal tapir]

mine is 25.25

[odd mason]

5th term?

mine is 25

[hot canyon]

:c if you're considering -0.5 = 2 x (-0.25) so the next should be -1

[odd mason]

but by geometric/exponential sequence

[vestal tapir]

ok i don't get it

[hot canyon]

but ow what froggie is doing is -0.5 = -0.25 - 0.25 so next is -0.75? :p

[vestal tapir]

yeah i see

[odd mason]

but i am thinking 6-0.25x2^(n-1)

this is a geometric for first difference

[vestal tapir]

it could be 0.25 0.5 1/sqrt2 btw

anyway i think it's a cubic

just i like it, like you said

[odd mason]

yeah so me and my friend got alot of answers for it, cubic,quadratic,quartic,geometric

[vestal tapir]

but can you write them as one expression

as they become more complicated they become less likely to be what was meant

and cubic is short

[hot canyon]

25 should work ✅

[vestal tapir]

they aren't asking for 25

read the problem

[odd mason]

in part b they asked

[hot canyon]

Yes, I'm saying we'd be safer to assume whichever general "a_n" gives 25 for the next place

[odd mason]

so thats why i wanted to detroy earth

[hot canyon]

Can you send a screen shot of the question? Or a pic?

[odd mason]

in the end it was the friends we made along the way

[odd mason]

can't , it came in a exam today they still have'nt given the exam paper

[vestal tapir]

maybe it's not enormous, and you;re right

i can;t write it

[hot canyon]

Did you write $a_n = 6n - 2^{n-3} - 1$ ?

[jolly parrot]

Arya

[odd mason]

no

[hot canyon]

;-;

[odd mason]

i am pretty sure in exam i wrote the worng answer for part a

but i think my part b was correct = 25

[pearl pond]

@odd mason Has your question been resolved?

[pearl pond]

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[stoic imp]

Let $\mathbb{L}{1}: X = \lambda(0,1,1) + (1,-3,0)$ and
$\mathbb{L}{2}: X = \mu(-1,-2,0) + (4,4,1)$.
Let $\Pi$ be the plane containing $\mathbb{L}{1}$ and $\mathbb{L}{2}$.
Find a line $\mathbb{L}{3}$ such that
[
\mathbb{L}{3} \perp \Pi,
\quad \mathbb{L}{3} \cap \mathbb{L}{1} \neq \varnothing,
\quad \text{and} \quad
\mathbb{L}{3} \cap \mathbb{L}{2} \neq \varnothing.
]

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[rancid depot]

theres a special vector product that could really help here

[stoic imp]

,w (0,1,1)x(-1,-2,0)

[jolly parrot]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[stoic imp]

this is the direction of L3

L3 : X = λ(2,-1,1) + (a,b,c)

[stoic imp]

hopefully L1 ∩ L2 is non empty

if we find a point of intersection we use it for L3 and we are done

[rancid depot]

it is expected given that the problem sttates theres a plane made by L1 and L2

[stoic imp]

wdym

[rancid depot]

if the two lines dont intersect (and we know they arent parallel) they will not have a plane passing through them

they will be skew

[stoic imp]

,w a(0,1,1) + (1,-3,0) = b(-1,-2,0) + (4,4,1)

[rancid depot]

you can also set up a system like a(0)+1=b(-1)+4, a(1)+(-3)=b(-2)+4

[stoic imp]

if they are parallel then what happens with the direction of L1 and L2? they are orthogonal? or they have same direction

[rancid depot]

the third equation should follow

[rancid depot]

there will be a plane that passes through the lines, but their intersection will be empty

[stoic imp]

true

yeah but direction of L1 is (0,1,1) is not parallel to (-1,-2,0) forrunately

otherwise this exercise will be hars

[rancid depot]

so, by the statement of the problem, we should expect the two lines to intersect

and they do

which can be shown by solving the system of equations

[stoic imp]

ye

1. 1 = -b + 4
2. a -3 = -2b + 4
3. a = 1

[stoic imp]

very nice observation sir

[gleaming moss]

it would be impossible; you cannot intersect both parallel lines and be orthogonal to them unless l1 = l2

[pearl pond]

@stoic imp Has your question been resolved?

[pearl pond]

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[craggy torrent]

ive been trying to study for a uni exam and i just dont understand what functions are and how to do them. the source they have provided makes no sense and i still dont understand some practise questions when put through chat gpt.

[open rivet]

!nogpt

[pearl pond]

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).

[open rivet]

please never use chatgpt for maths questions

[craggy torrent]

i understand but i just need help with whats happening

i apologise

[open rivet]

!original

[pearl pond]

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.

[craggy torrent]

the way my lecturer explains it doesnt make sense

[open rivet]

so you need to find $\dv{y}{x}$?

[jolly parrot]

Bonk

[craggy torrent]

yes

i think so

[open rivet]

do you know the product rule?

,tex .diff rules

[jolly parrot]

Bonk

[craggy torrent]

i dont understand :(

[open rivet]

which part?

[craggy torrent]

what youre asking me

[open rivet]

have you done derivatives?

[craggy torrent]

i am not very good with termanology

[craggy torrent]

this is all new to me

i havent done math since secondary and this is my first semerster at uni

[open rivet]

ok...

thats a big jump

have oyu ever seen $f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$?

[jolly parrot]

Bonk

[craggy torrent]

i dont think so

so far ive done graph theory, calculating deltas, finding gradients of straight lines and boolean expressions

which i understnad

[open rivet]

i suggest you study up on derivatives then

have you had limits?

i could explain them to you but it might take some time

[craggy torrent]

sorry im just stressed rn and i feel bad having to come here and ask for help

if you want to it would help i think

[open rivet]

okay so, this is calculus

what do you know about calculus so far

calculus includes, limits, derivaties, integrals

[craggy torrent]

can you give examples

uhm

i dont thinkk ive learnt any of that

but rn im only doing derivaties

it seems

[open rivet]

okay

and you havent had limits?

[craggy torrent]

i havent seen anything like that in my resource

[open rivet]

consider this function

this is a discontinuous function

as you can see by the "jump" at x=2

yes?

@craggy torrent

can i ping you btw?

[craggy torrent]

yes thats fine

[craggy torrent]

that makes sense yes

[open rivet]

a limit is approaching it slowly from one side

you have a left limit

and a right limit

which approach the point from the left and right respectively

$\lim_{x\to 2^-}f(x)$ is the left limit and $\lim_{x\to 2^+}f(x)$ is the right limit

[craggy torrent]

are those the black dots

[jolly parrot]

Bonk

[open rivet]

i will tell you about the dots later

[craggy torrent]

okay

[open rivet]

do you see what is happening here?

one has a - and one has a +

[craggy torrent]

ye i see that

[open rivet]

the - is from the left, the + is from the right

now, we define $\lim_{x\to x_0}f(x)$ to be the limit at $x_0$

[jolly parrot]

Bonk

[open rivet]

and this exists only if the left and right limit exist and are equal

so at this, lets look at $\lim_{x\to 0}f(x)$

[jolly parrot]

Bonk

[craggy torrent]

is that

5?

[open rivet]

indeed it is

[craggy torrent]

oh

yay

[open rivet]

because approaching it from the left is 5

and approaching it from the right is 5

now, those dots are where the function is or isnt

if you look at point a, you see a white dot

[craggy torrent]

yeas

[open rivet]

and its at x=-2

[craggy torrent]

okay yea

[open rivet]

that means that f(-2) does not exist

because there is a "hole"

but, lets take the limit

what is the left-sided limit?

remember, we approach it, but never actually reach it

[craggy torrent]

uh

3

idk

[open rivet]

i think its closer to 3.5

but sure

lets assume its 3

[craggy torrent]

oh okay

so we can use decimals

[open rivet]

and now what about the right-hand side limit?

[open rivet]

yeah it can be any number

[craggy torrent]

is ther one for right since its disconnected

[open rivet]

we are looking at point a

point a has a right-sided limit

(yes, its as easy as you think just like the left-sided limit)

so, whaddya think

[craggy torrent]

whats the difference between right and left

are they always the same number?

i know the - and +

[open rivet]

red is left sided, blue is right sided

[craggy torrent]

oh i see

that helps

ok

[open rivet]

nice

so, what is the right sided limit?

[craggy torrent]

3.5 again?

[open rivet]

indeed

and now, the left and right sided limits are equal

and are 3.5

thus

$\lim_{x\to a}f(x)=\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=3.5$

[jolly parrot]

Bonk

[open rivet]

this seems a bit strange at first right? because the value at a doesnt actually exist

beacuse of the hole

yes?

[craggy torrent]

sorry its just i dont think im doing these

theres nothing about limits

that is a whole new thing to me

[open rivet]

thats okay

but do you see whats happening here?

[craggy torrent]

yea

[open rivet]

so the limit exists

because if we approach it slowly from either side, they agree

but if we actually look the function at that point

it doesnt exist

[craggy torrent]

yes that makes sense

[open rivet]

now lets look at point b

[craggy torrent]

theres no right

it stops on left at ~6.5

[open rivet]

and the right?

[craggy torrent]

cos its a jump

is the right

9?

[open rivet]

yup

[craggy torrent]

are they meant to be

met

[open rivet]

the left limit is 6.5

the right limit is 9

[craggy torrent]

ye

[open rivet]

so what can we say about the limit as x approaches b?

[craggy torrent]

x stays teh same

isnt it y that changes?

[open rivet]

im not sure what you mean

[craggy torrent]

wait are you talking abt the

formula thingy

not the axis

[open rivet]

$\lim_{x\to b}f(x)$

im talking about this

[jolly parrot]

Bonk

[craggy torrent]

x-6.5 x+9?

[open rivet]

what?

[craggy torrent]

idk 😭

like

yea

i see what youre talking about

uhm

the - side is 6.5

igjht

then the + side is 9

?

wait

i

sorr

y

xd

[open rivet]

remember that the limit exists only if the left and right limit exist and are equal

[craggy torrent]

ye

it jumps up though

[open rivet]

are they equal?

[craggy torrent]

no

[open rivet]

thus the limit does not exist

$\lim_{x\to b}f(x)=$DNE

[jolly parrot]

Bonk

[craggy torrent]

yea tha makes sense

[open rivet]

so, now what is the value at f(b)?

[craggy torrent]

it wont have one?

[open rivet]

it does

look at the dots

white dot is not there

black dot is there

[craggy torrent]

so 9?

[open rivet]

indeed

[craggy torrent]

how come

the white means theres nothing there

but the black means an end or soething>?

alsos orry for my typign

my keyboard is displaced

[open rivet]

its to tell apart which values it does and does not attain

so at b it means f(b)=9

[craggy torrent]

and its only like that because its not equal?

[open rivet]

the f(b)=9 part?

its like that because there is a discontinuation

[craggy torrent]

yes

[open rivet]

it couldve also been that the lower one was black

then f(b)=6.5

but the important part here is that the limit does not exist

now, lets look at c

[craggy torrent]

how come the black dot takes priority

is it just because it does

[open rivet]

wdym priority?

its just how we define it

[craggy torrent]

okay

is the white dot there just to give it identity?

[open rivet]

it to show that its not there

like if you look at a

[craggy torrent]

but there is nothing there

yea

[open rivet]

its a gap there

[craggy torrent]

and the black dot means the opposite?

[open rivet]

black dot means it is there

white dot means it is not there

thats just how we define it

now, lets look at c

whats $\lim_{x\to c}f(x)$ and $f(c)$?

[jolly parrot]

Bonk

[craggy torrent]

on the line its ~5 for boht

but the black dot gives me a feeling otherwise

[open rivet]

thats correct

[craggy torrent]

does the black dot below c come into play here at all

[open rivet]

the limit at a point does not necessarily have to equal the function value at that point

[open rivet]

thats f(c)

[craggy torrent]

f(x) = 5 f(c) = 2

[open rivet]

\begin{align*}\lim_{x\to c}f(x)&=5\f(c)=2\end{align*}

[jolly parrot]

Bonk

[open rivet]

dont forget to write the limit

[craggy torrent]

lim

okay

[open rivet]

this is how you would write it down during a test or smth

[craggy torrent]

i see

[open rivet]

do you kinda understand what a limit means now?

[craggy torrent]

yea

that makes sense

[open rivet]

i know its a bit small but bear with me

[craggy torrent]

its fine

[open rivet]

what is $\lim_{x\to\infty} f(x)$?

[craggy torrent]

idk

[jolly parrot]

Bonk

[open rivet]

what do you think it is

this graph is 1/x btw

as x gets bigger and bigger

what does f(x) approach?

[craggy torrent]

infinity

?

idk

[open rivet]

not quite

x approaches infinity

but what does f(x) approach

like, fill in 1/x with increasingly bigger numbers x

1/10, 1/100, 1/1000, 1/10000, etc

,calc 1/10

[jolly parrot]

Result:

0.1

[open rivet]

,calc 1/100

[jolly parrot]

Result:

0.01

[open rivet]

,calc 1/100000

[jolly parrot]

Result:

1e-5

[open rivet]

what does this sequence approach?

[craggy torrent]

a repeated number of zeros with a 1 at the end

[open rivet]

it approaches 0

[craggy torrent]

ye