help-0

[leaden bobcat]

with what values?

[opaque shard]

so like this? I dont have a -5 on my graph is the problem

[leaden bobcat]

yes, but with undefined values for x=2 and x=3 since the function is not differentiable there

usually drawn as empty circles

[opaque shard]

so now i just have to figure this proabla out

[leaden bobcat]

yes

now for the last part where x < 2

[opaque shard]

and since its a proabla it has to be a diagonal line

[leaden bobcat]

yes

[opaque shard]

does the line start where f(x) is 0? aka -1,4?

no wait

it starts at -1

[leaden bobcat]

wdym starts? Lines don't really start
What I think you're trying to say is, to draw a line we need 2 points, and the point where it touches 0 is easy to find and that is when x = -1

[leaden bobcat]

yes

[opaque shard]

do the ends need a close circle or an open?

[leaden bobcat]

again undefined for x=2

[leaden bobcat]

if it were closed, we would be saying that the derivative exists for x=2, but this is not the case

the function is not differentiable for x=2 since it is not continuous

[leaden bobcat]

great

[opaque shard]

damn i really made that harder than it was

[leaden bobcat]

The way you replied at the beginning I thought you had no clue about this type of exercises

then you showed me you doing it right for 3 exercises already

Was shocked ngl

[opaque shard]

haha

i was overthinking this problem so much thats insane

thank you sm for your help and taking time out of your day it means alot

[leaden bobcat]

Btw, I suppose you jumped straight to derivatives without knowing limits, correct?

[opaque shard]

i know how to solve limits just not really why or how they are used

[leaden bobcat]

You're learning the computational part of this topic without a solid understanding of what's going on, at least this is what it seems to me

The fact that you're able to do these exercises alone is pretty solid though

[opaque shard]

when doing the review ill go back and try to understand the definetions and everything

[leaden bobcat]

imo learning why these things are used/studied is a big part of the subject, so I would suggest taking a book and starting from there, but ofc you decide what you need this stuff for

[opaque shard]

will do

thank you once again

.close

[lone heart]

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[ivory ivy]

So far I have the set L

where L = {l : a|l and b|l}

and ab is an element of L so its non empty

and by well ordering principle we have a smallest element, say m = ab/ gcd(a,b)

i feel like this shows existence

now i need to do the second part

where I assume n is an element of L

and prove m must divide n

not exactly sure how to go about this

[modern sedge]

why "say m = ab / gcd(a,b)"

that part is unnecessary

[ivory ivy]

i mean ok

then just say m

[mortal trellis]

you dont know that its true

and also why should m=lcm(a,b)

[ivory ivy]

cuz the lcm is the least common multiple

and m is the least element

of the set of common multiples

[mortal trellis]

least common multiple is not defined by being the smallest multiple

[ivory ivy]

smallest common multiple no?

[mortal trellis]

no

[ivory ivy]

wow

[mortal trellis]

its a multiple and all other common multiples are multiples of it

from your own definition

the definition does not talk about the size of it

[ivory ivy]

i mean sure but those r def the same

[mortal trellis]

why are they

[ivory ivy]

because multiples of a positive integer

are larger

than the original

2a > a

3a > a

etc.

[mortal trellis]

yes but thats only part of it

you also need to show that m divides every element in the set L

[ivory ivy]

yes

[mortal trellis]

and that does not follow from it being the smallest

[ivory ivy]

thats what i wasn't

sure how to go about

[ivory ivy]

agreed

[mortal trellis]

so you cant claim that m is the lcm

yet

[ivory ivy]

well

ok

i didn't in my actual thing i just said that here

so yall can see where im going with this

[mortal trellis]

"yeah I dont actually tell you what I'm actually doing, I'm telling you wrong things on purpose"

helpful

[ivory ivy]

bruh

[mortal trellis]

anyway

consider division with remainder

of n and m

[ivory ivy]

?

like say

n = mq + r

and prove r = 0

?

[mortal trellis]

yes

[ivory ivy]

suree

lemme try something

assume r > 0

r = n - mq

wait

omg

and r would be

in the set

which is a contradiction

to m

being least

so therefore r = 0

ok bet

i got it

since a and b divide both n and m

they divide r

so r would be a common divisor in this case

and by division algorithm it would be less than m

which is a contradiction

to m's definition

thank you denascite

great help

[mortal trellis]

common multiple

but yes

so this shows that m=lcm(a,b)

and this shows existence

for uniqueness, let m1 and m2 both be lcms

what do you know about them dividing each other?

[ivory ivy]

ok

lemme think

m1 | m2

and m2 | m1

so m1 = m2

thats it right?

QED

[mortal trellis]

the word nonnegative should be mentioned somewhere

but yes

[ivory ivy]

ah yes

well

a and b r non negative

so ig just say my set L

is nonnegative common multiples

and then all elements r nonnegative

i think that cleans things up

[mortal trellis]

the lcm is defined as the nonnegative integer satisfying the definition

[ivory ivy]

yes

so if the set L is non negative

the lcm that is an element in it

would be non negative

[mortal trellis]

the set L was just for showing that some lcm exists

that was the last part

[ivory ivy]

well-ordering principle

[mortal trellis]

this is now completely separate

[ivory ivy]

won't even work

[mortal trellis]

m1 and m2 have nothing to do with L

[ivory ivy]

oh ok

so what

say m1 and m2 r nonneg

[mortal trellis]

but we know they are both nonnegative and divide each other

[ivory ivy]

yeah

so im done right?

[mortal trellis]

and they are nonnegative because of the def of lcm

yes you are. but its important that m1 and m2 are nonnegative

so you need to mention it

[ivory ivy]

ok

[mortal trellis]

otherwise lcms are not unique

[ivory ivy]

thanks

[lone heart]

@ivory ivy Has your question been resolved?

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[bleak portal]

7x7 to the power of 2 diveded by 3 and subtracted by 4. 😭

[sage rain]

I’ve been trying to solve this using chain rule but difficult

[ivory ivy]

sirez

is the bottom

-1/3

an exponent?

[sage rain]

Yes

[ivory ivy]

ok

so

simplify denominator to x^5/3

then make it 6x^(-5/3)

the rest is trivial

[placid zinc]

I can't see this function as a composition, so chain rule doesn't work

[sage rain]

i would make it a negative exponent

one sec

So now you can chain rule

The inside exponent is still negative

[plucky widget]

hello

@sage rain

[sage rain]

Hello

[alpine sable]

hello

wanna learn maths

algebra especially

[desert raptor]

#prealg-and-algebra
I assume you're pre-uni

[alpine sable]

yeah

you are good at assuming

[desert raptor]

This is an occupied help channel

Please create a new one

@alpine sable all of these are unoccupied

[alpine sable]

i couldnt understand sir

but alright

[desert raptor]

!occupied

[lone heart]

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).

[sage rain]

Ok. Back to my question

[desert raptor]

do you know

[desert raptor]

x^a * x^b = x^(a+b)

[desert raptor]

not quite

[sage rain]

What’s wrong in mine?

This is where I’m lost

[cobalt oriole]

I think the problem is that taking the ` like theres 2 different functions

My english is not good I'm sorry

Like,, the thing you did is I think when its like (fog)'(x)

But theres only 1 function here

Please correct me if im wrong

[desert raptor]

is this clear

[sage rain]

Wait, how you get 6?

[cobalt oriole]

It's probably a typo

[sage rain]

@desert raptor how did you get a 6 in the denominator?

Would 2=6/3?

[desert raptor]

oh

it's a

mistake

it's 5/3

@sage rain

now can you differentiate this

[sage rain]

Ok. Let me give this a shot

Sorry for the messy handwriting

So it ends up -10/x^8/3

Can you leave it that way though?

With the 8/3?

[desert raptor]

yes

it's correct

[sage rain]

Thank you @desert raptor

That was brutal

[ivory ivy]

bro

that was what i said

originally

[sage rain]

Sorry. Didn’t understand how you said it

.close

[lone heart]

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[lone heart]

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[alpine sable]

MA is
AB is 12
Big radius is 10
Find small one

[alpine sable]

<@&286206848099549185>

[toxic verge]

Bro wait 15 mins before pinging

[alpine sable]

Plz i meed dast

[iron crystal]

ok.

[alpine sable]

Fast

[toxic verge]

Why dont you read #rules

[compact surge]

💀

[alpine sable]

I am sos sorry

[iron crystal]

6^2+a^1=100

100-36

64

a^2=64

a=8

r MA=8

[toxic verge]

!nosols

[lone heart]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[iron crystal]

the solution is so poorly solved and explained, im 99% sure he will be accused for copying it.

[toxic verge]

What if its an mcq

[alpine sable]

I dont understand

[iron crystal]

i gave you the answer enjoy.

you said you wanted it fast

here it is.

[alpine sable]

Answer is 3

In solutions

[iron crystal]

6^2+a^2=10^2

[tacit arch]

Stop doing that

[iron crystal]

ok ok sorry

[lone heart]

@alpine sable Has your question been resolved?

[warped wave]

guys, help me in something, X(2X+9)+1=46.

[lone heart]

@alpine sable Has your question been resolved?

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[blissful summit]

Hi guys, I recalled the
lim f(g(x)) = f(lim g(x))
rule, so during a proof for sequences I came to wonder is the equation legal:

[buoyant saddle]

take f(x) = |x|

[naive valley]

it's valid if the limit inside the abs values on the right hand side exists

[blissful summit]

.close

[lone heart]

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[ember lantern]

Elp

[ember lantern]

Help

<@&286206848099549185>

[digital magnet]

Post your question

And wait 15 mins before pinging helpers

[ember lantern]

it is urgent

they are going to kill me

[digital magnet]

?

[ember lantern]

the key

My parents are going to kill me

[digital magnet]

This is a math server

[balmy bloom]

can you press it back down into place?

[echo grotto]

Just check the nature of tge function and try if lhl =rhl.

[ember lantern]

No

returns

Help

@lavish cave

[shadow coral]

maybe look up a vid that tells you how to reattach keys for ur specific model of laptop?

if one exists

[ember lantern]

I need help

[solid raptor]

what u need help withN

[lone heart]

@ember lantern Has your question been resolved?

[ember lantern]

No

[ember lantern]

A keynoard

Key board

[solid raptor]

can you be more specific please

[solid raptor]

is it the W key that is loose?

[ember lantern]

Yep

[solid raptor]

try press it in again

[ember lantern]

I fixed it but

[solid raptor]

do not take it out fully, it might break the underlying parts

[ember lantern]

is submerged

[solid raptor]

submerged?

[ember lantern]

Ye

Look

[solid raptor]

where?

the picture above does not look submerged, it looks loose?

[solid raptor]

So you did take it out fully?

[ember lantern]

Yep

But

Idk why its doesnt work how the other work

@solid raptor

[solid raptor]

Hmmm but if a small part broke off, you cannot fix it anymore easily

[lone heart]

@ember lantern Has your question been resolved?

[lean depot]

Hi, I saw there was a mod ping here. Was it deleted? Please let me know

[stone orchid]

yea i didnt it on accident

[lean depot]

Ah OK no problem!

[lone heart]

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[lone heart]

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[serene vault]

does this involve adding any vectors?

[twin nimbus]

Naturally. |0> is a vector and so is |1>

[serene vault]

what does that actually mean tho?

|0> as a vector is (00)?

[twin nimbus]

If you have a quantum system with 2 states then |0> is a length 2 column vector, most commonly [1;0], and |1> is a length 2 column vector, most commonly [0;1]

In general, assuming these states are orthogonal, then their vectors are as well

[serene vault]

those natural vectors you just mentioned [1;0] and [0;1] what do those numbers mean?

[twin nimbus]

and then for the bras they are row vectors

[twin nimbus]

These are the basis vectors of the space in question

[serene vault]

can there be numbers > 1 in those vectors?

[twin nimbus]

They are generally going to need to be normalized, so no

[serene vault]

i see

ok so going back to the question - the first step is to find the "magnitude" which is just squaring those fracitions?

[lone heart]

@serene vault Has your question been resolved?

[lone heart]

@serene vault Has your question been resolved?

[twin nimbus]

@serene vault you find α*α-bar

[serene vault]

thats - a times a minus bar?

[twin nimbus]

a times a conjugate

[serene vault]

so i take the fraction times itself

then do something called a conjugate on it?

[twin nimbus]

These are real, so the complex conjugation is a no-op

But in general they can be complex values

[serene vault]

shoot... most of the vocab youre using im clueless to

what is a no-op?

[twin nimbus]

An operation that does nothing

[serene vault]

OHH

what would an example of a "non real" number be?

[twin nimbus]

2+3i

[serene vault]

ahh so anything with that "i" in it?

[twin nimbus]

Not technically true, but morally true

[serene vault]

ok so i was somewhat right? that to find the probability of ket0 here we would square the fraction?

64/289?

[twin nimbus]

Technically, you would do <0|ψ>

[serene vault]

ive never seen that before

[twin nimbus]

And square that

[serene vault]

i know what the psi ~ is

but wrapping 0 and "|" in "<>" i have never seen

[twin nimbus]

<0| is a bra

[serene vault]

what different from this to just squaring the fraction?

[twin nimbus]

It's more general of a treatment

|ψ> means you have a system in state ψ

<0| represents an observation

So <0|ψ> represents the probability amplitude that you observe the state ψ in state 0.

[serene vault]

OHHHH

[lone heart]

@serene vault Has your question been resolved?

[lone heart]

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[wheat isle]

how do i start? do i set the parametric equations equal to eachother or somth

[lone heart]

@wheat isle Has your question been resolved?

[toxic verge]

Try to convert it into this format like
x-1/1= y-2/1=z-3/1=s

This must be familiar

[wheat isle]

where does x-1/1 and y-2/1 come from

[toxic verge]

From l1

s=x-1

and s=y-2

There's brackets i around x-1, y-2 and z- 3 btw

Forgot to put

Doesnt really matter

But yeah

[wheat isle]

so u solve for s?

[toxic verge]

You get the parallel vector for l1 from this

i+j+k

From making the same eqn for l2 youll get its parallel vector

Then take cross product to find normal vector of the plane

And for finding a point on the plane just equate the x y and z parts

Understand?

[wheat isle]

we didnt learn i j k method

[toxic verge]

O

Well then just put two different values of s to get two points on the line

And find parallel vector by subtracting

[toxic verge]

And find the point by this only

[wheat isle]

euhm

[toxic verge]

What part having difficulty with?

[wheat isle]

approaching

so shoud we write

l1 and l2 as

vector equation form

x = p + td where p is a point and d is a direction

[toxic verge]

No need to do that if udk that parallel vector can be found from that

[toxic verge]

Just do this

Eg

Put s=0

And s=1

And subtract the points to find parallel vector

Just do that first then you'll see where im going with this

[wheat isle]

i dont remember doing like subtraction stuff before

if we choose s=0 and s=1 do we choose t as well

[toxic verge]

Yeah take two random values of t

[toxic verge]

Here d is parallel vector btw if udk what that is

[toxic verge]

How do you usually solve two lines for a plane then

[wheat isle]

we called d direction vector

[toxic verge]

Same thing

[toxic verge]

Did this yet?

[wheat isle]

so if s=0 and t=0

p1 = (1,2,3)

p2 = (1,2,3)

[toxic verge]

No dont do t first just do s

Lets work on l1 first

Put s 0 ans s 1

Also my messages might be coming really late since im travelling in like tunnels and the net is dogshit

Sorry for that

Someone else can take over if they're too slow btw i dont wanna inconvenience you

[toxic verge]

Though you did find a pt on the plane with this

[wheat isle]

so if s=0 then <x,y,z> = (1,2,3) and if s=1 then <x,y,z> = (2,3,4)

[toxic verge]

Ok cool

Now convert that into vector form

Those points

[wheat isle]

x = (1,2,3) + t<1,1,1>
x = (2,3,4) + t<1,1,1>

[toxic verge]

How did your direction vector change when its the same line

[wheat isle]

oh

i was looking at l2 i think

[toxic verge]

Ok so anyway

You have one direction vector now

Also you just found out direction vector by intuition?

[wheat isle]

i just looked at coefficient of s

[toxic verge]

Yeah exactly what i was trynna do with this lol

So anyway

Find direction vector of l2 now

[wheat isle]

<1,0,2> for l2

[toxic verge]

Ok cool

So what do you think we should do now for finding the normal vector (you might call this perpendicular vector idk) for the plane?

[wheat isle]

cross product direction vector of l1 and l2

[toxic verge]

Correct

Do that

[wheat isle]

<2,-1,-1>

[toxic verge]

Ok now

You have a pt on the plane (multiple, in fact) and its normal vector

What should be the planes eqn

Which form do u use btw, (r-a)•n=0?

[wheat isle]

we use this one (x-p) dot n = 0

[toxic verge]

Ah same thing

So you have the point and the normal vector

Put them in

[wheat isle]

i have two points for s which one do i use?

[toxic verge]

Any

Doesnt matter

You can use t points as well

Since normal vector is a free vector

[wheat isle]

wait so if i could use any whats the point of finding s=1 for example if i only use s=0

[toxic verge]

I was trying to teach you how to find the direction vector by subtracting two points on the line but u already knew that u had to look at coeffs lol

[wheat isle]

oh

so did i only need to choose s=0 for example

[toxic verge]

Yes

[wheat isle]

I see

[toxic verge]

Didnt even need to use t

[wheat isle]

Yes

does this approach work for all questions that give me 2 lines and i have to find a plane containing them?

[toxic verge]

Yes

Not for two parallel lines tho

Since you cant find the normal vector then

You cant find the planes eqn given two parallel lines by any method

[wheat isle]

ohh

I see

[toxic verge]

Is it clearer now

[wheat isle]

ya i think so

[toxic verge]

Cool, you can ask me about the other two methods i was explaining as well if u want btw

[wheat isle]

oh yea wait

i have another question

if i have a plane ax + by + cz = d

[toxic verge]

Yes

[wheat isle]

how do i find vector parallel to that

[toxic verge]

A,b,c is the coeffs of vec parallel to it

[wheat isle]

so thats just the normal vector?

[toxic verge]

Wait no i mean perpendicular

You cant find parallel

[toxic verge]

Yes

[jaunty olive]

Check options

[wheat isle]

how do i check

[jaunty olive]

That's the only way

[toxic verge]

Just a,b,c

For perp

[wheat isle]

but its asking for parallel isnt it

[toxic verge]

For parallel yeah check options

[wheat isle]

how do i check it

[toxic verge]

Take the dot product

And see if its 0

[toxic verge]

With the perpendicular vector

[wheat isle]

so <2,-1,5> dot with the options

[toxic verge]

-1

Yes

[wheat isle]

and if its 0 its parallel?

[toxic verge]

Yes

[toxic verge]

Parallel with the plane

[wheat isle]

okay

[toxic verge]

And perpendicular to the normal vector

[wheat isle]

thanks

[toxic verge]

Yw

[wheat isle]

i have another question but this one is more confusing Lol

wait nvm i got it

ok thanks

.solved

[lone heart]

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[alpine sable]

i have this recursive definition for sum

[alpine sable]

i wanna prove this fact for n>=1 using induction:

base step: at n=1 LHS=RHS so it holds for basis step

ok so the problem is that im stuck proving p(k) -> p(k+1)

the reason is that the upper limit wont match

although what i think happens is that n is length of the sequence, let me show

(i dont know why that would work)

[lavish cave]

ah so you have (sum of a(i) from k to (n + 1))
= (sum of a(i) from k to n) + a(n + 1)

then using the induction hypothesis, you get:
= a(k) + sum of a(i) from (k + 1) to n + a(n + 1)

so the sum will range from (k + 1) to (n + 1) as desired

[alpine sable]

ok my bad i forgot to say that i have to prove that the formula is valid for all n>=1 but considering the case k=1

[lavish cave]

yeah I left out the case where k = n cause that's pretty obvious

just working with the definition for 1 <= k < n

[lavish cave]

that shouldn't matter for what I wrote though

[alpine sable]

alr

[alpine sable]

thats the part i dont get. why only the "range" of the sum matters? (not the upperbound)

the purple part

from what i understand, its basically shifting i so it "lands" on a range of "length" m

but wouldn't that be problematic since the original assumption only talks about a_i?

not a_i'

[lavish cave]

I don't understand why you're working with m + 2

you should be replacing n with m + 1 instead

so you have (sum of a(i) from k to (n + 1)), if you let me reuse n again

[alpine sable]

wouldnt it be m+2?

ok let me make this more readable

[lavish cave]

WAIT I MISSED THIS PART

sorry

[alpine sable]

np what i sent is a mess

let me organize it

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

okay i think my question can be reduced to why is the change of index legal

in the yellow part

[lone heart]

@alpine sable Has your question been resolved?

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[alpine sable]

have i gone wrong somewhere

[median oar]

did you think you did something wrong?

[alpine sable]

i mean im getting the wrong answers

soo

lmao

13 = lambda is a correct root

but there's a lambda = 3

[median oar]

oh god

[alpine sable]

sooo

[median oar]

can you actually do row reduction with the determinant

do you know if that's a real thing

[hushed scroll]

a33 should be -1-lembda

[median oar]

ah

clever guy

[alpine sable]

wait

🤦‍♂️

🤦‍♂️

🤦‍♂️

🤦‍♂️

🤦‍♂️

🤦‍♂️ 🤦‍♂️ 🤦‍♂️ 🤦‍♂️ 🤦‍♂️ 🤦‍♂️ 🤦‍♂️ 🤦‍♂️ 🤦‍♂️

[hushed scroll]

[median oar]

nah arithmetic is hard

i hate doing matrix algebra

[hushed scroll]

now do it again

[alpine sable]

crying bro

let me do it again

[alpine sable]

actually though

i get so lost doing arithmetic with matrices

🥴

WELL WELL WELL

GUESS WHO SUDDENLY GOT THE RIGHT ANSWE

R

ALL CAUSE OF A DAMN SIGN

@hushed scroll you saved me 3 mental breakdowns

🙏

[hushed scroll]

ok good 👍

[alpine sable]

the answer to this is

yes

you can swap rows

multiply by a scalar k

or add a multiple of one row to another row

[median oar]

interesting

[alpine sable]

indeed

.close

[lone heart]

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[lone heart]

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[oak temple]

hello, i was wondering on how id be able to get this into cot 1/2

[oak temple]

i know thats the double angle formula for tan

but i really cant see any correlation here

is there a different formula i should know?

[raw linden]

sin(2(theta/2))=?

[oak temple]

i know sin2 theta is 2sincos

but i dont think ive ever seen what you asked before 😭

[raw linden]

2sin(theta/2)cos(theta/2)

And now do cos(2(theta/2))

[tribal verge]

Check if these help

[oak temple]

ive never seen those before

they definitely help tho

[tribal verge]

These are derived from double angle identities

[oak temple]

i get given these in the exam

[oak temple]

that would be a choice of 3!

[tribal verge]

These identities were used in some of the problems in my book so

I think you’ll easily find their derivations online

[raw linden]

One is exactly what you want tho

[tribal verge]

Or you could try to do it yourself using sin 2x and cos 2x identities

[oak temple]

im confused though because i dont know where im going to subsitute these double angle formulas into

or will it become obvious

when i get to the stage im at, should i try and make the right side (cot 1/2 theta) into something which looks more similar to what i have?

if that makes sense

[raw linden]

Do you see a way to simplify 2sin(theta/2)cos(theta/2)/(1-(1-2cos^2(theta/2))

[oak temple]

this?

sorry its not very neat

[raw linden]

Yes

[oak temple]

wait, so sin theta is equal to the top fraction?

numerator

is that what we did earlier

[raw linden]

Sure

[oak temple]

i never realised that

thats already very useful thanks

and then that would be -tan theta?

over 2

[raw linden]

A small sign error

[oak temple]

ohh, 1--2cs

[raw linden]

Now what’s the def of tan? Or cot?

[oak temple]

sin/cos for tan

if thats what you mean

[raw linden]

With theta/2?

[oak temple]

oh, is it not just tan 1/2 x

[tribal verge]

There’s a mistake here

[raw linden]

Easily rectified

[tribal verge]

Cos x = 2cos^2( x/2 ) - 1

[tribal verge]

This is the correct formula

Using this you will directly get cot theta/2

[raw linden]

Go through from the beginning now you know the gist of it

[oak temple]

i shall

woud this be good?

and thank you both very much for the help

[tribal verge]

Exceptions?

[oak temple]

huhhhhh

ohh

the other part

anything that makes it 0

the denominator

so 2 pi

[tribal verge]

2 . n . pi

I think the question requires generalised answers

[oak temple]

ah, okay

thank you for the help 😄

.close

[lone heart]

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[lone heart]

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[worldly sleet]

Like nCr(2,2)=1
Well, can it show those possibilities like?
ab
ba

[ornate lintel]

i think u mean nPr?

[worldly sleet]

mo

nCr

[ornate lintel]

oh

if u wanna show combinations like a,b and b,a u need to use nPr instead

[worldly sleet]

OK, let me rephrase it. Let’s say.
nCr(3,2)

ab or ba

ac

bc

@ornate lintel

[ornate lintel]

so basically you are asking why nCr cant show combinations like a,b and b,a but only show one?

or am i mistaken

[worldly sleet]

No

I know why it doesn’t show up I’m asking how can it show combinations

[ornate lintel]

i dont know if this will help cuz idrk what point u dont know
but like cause if you do a nCr(3,2) you will get a,b b,c a,c three combinations
each of them can be changed into a,b => (a,b and b,a) etc...
so you'll need to run nCr(3,2)x2! which gets nPr(3,2)

cause order doesnt matter in nCr

[alpine sable]

guyss

i need help

[worldly sleet]

Let me rephrase

I want to show the combinations of nCr

[alpine sable]

there is a question

[worldly sleet]

I don’t want the same equation

[alpine sable]

Find five rational numbers between 3/5 and 4/5.

what will be the answer

[ornate lintel]

oh you mean like for nCr(2,1) you want to show a ,b instead of 2 or?

[worldly sleet]

Yes

[ornate lintel]

i think u need to have a hypothesis first

like object a,b,c

[worldly sleet]

Correct

Meaning that there’s three possibilities

ab

bc

ac

[ornate lintel]

yes

[worldly sleet]

OK, yes

I’m aware of that

But when I write it as
nCr(3,2)

It only tells me how many combinations there are

Not with the combinations are

How do I do that?

[ornate lintel]

i think mainly we write it manually

but you can also get a code to write that

[worldly sleet]

Yes, that that is what I want

[ornate lintel]

https://www.dcode.fr/combinations

this website should help

[worldly sleet]

Thank you

But there’s one problem

That post doesn’t have a choose condition

[ornate lintel]

do you mean this?

[worldly sleet]

No this

[ornate lintel]

you mean you want choices like 4 choose 3?

[worldly sleet]

What does choose mean?

[ornate lintel]

can you rephrase what you asked before

sorry

[worldly sleet]

Like how does the calculator work? How does it do that? Show me the possible calculations?

[compact surge]

https://planetcalc.com/3757/

Is that what you're looking for?

You could also imagine implementing this in something like python

https://docs.python.org/3/library/itertools.html#itertools.combinations

[ornate lintel]

yea exactly

[worldly sleet]

Thank you

[lone heart]

@worldly sleet Has your question been resolved?

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[strange fractal]

guys i dont get it

[strange fractal]

they say x cannot be 1 or -3

but then they make x = 1 and -3

[median oar]

where

[vale wigeon]

yeah bit of a sneaky

[median oar]

oh yucky

[compact surge]

this isn't an equation in a field of numbers because the entire thing happens in a field of fractions of a polynomial ring

[median oar]

ok that's surely understandable to them

[alpine sable]

hello

[compact surge]

😭

[vale wigeon]

@strange fractal the idea is that if a polynomial equality is known to hold for infinitely many values of x, then in fact it also holds for all x ∈ R

[compact surge]

if you're trying to solve for A(x + 3) + B(x - 1) = 3x + 5 over all x in R then the values of A and B must also be the solution of that system on R \ {1,-3}

[gritty bramble]

what even is this explanation

[lone heart]

@strange fractal Has your question been resolved?

[lone heart]

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[fervent kestrel]

Given $\overrightarrow{F} = \nabla (4x - 8y), find \int_C \overrightarrow{F} * d\overrightarrow{r}$ where C is the quarter of the circle$ x^2 + y^2 = 9$ in the fourth quadrant, oriented counterclocwise

[ocean seal]

dragonbreath

[fervent kestrel]

Using Green's theorem, I should just be able to use $\phi (x,y) = 4x - 8y$

[ocean seal]

dragonbreath

[fervent kestrel]

and plug in the the start and end points, which I would believe to be (3, 0) and (0, -3) which gives

Using phi(end x, end y) - Phi(start x, start y) I get
(4(0) - 8(-3)) - (4(3) - 8(0)) = 24 - 12 = 12

but it is saying the answer is -12. Why is it -12?

wait

nvm on the wait, I originally had the message saying start - end

[dawn zephyr]

4th quadrant,
quarter of a circle,

oriented counter clockwise,

draw and check, your start should be at -pi/2 and end should be at 0

(0,-3) start and (3, 0) end

rest is all correct

[fervent kestrel]

So oriented counter clockwise means I trace the circle in a counter clockwise fashion and that dictates my start and end points?

[dawn zephyr]

yea pretty much

[fervent kestrel]

Okay, that makes more sense

and this would end up flipping it to 12 - 24 = -12

Thank you, this makes more sense to me now!

.close

[lone heart]

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[glossy wigeon]

Stuck on this question. I’ve split the integral but I’m not sure that my bounds are correct since f(x,y) dips below the xy plane. I’ve also tried different simplifications for the natural log and keep getting it wrong.

[glossy wigeon]

a better screenshot

[charred summit]

The question does not specify that the solid is on the positive side of the z axis

So you will integrate the given function as it is
On the boundaries 0 ≤ x ≤ 7 and 0 ≤ y ≤ 5

[glossy wigeon]

so no splitting the integral?

i believe that is also incorrect

i was told specifically: "You have to take the unsigned area, not the signed area"

it says "which lies between the xy-plane" in the question

[charred summit]

You are right

It seems correct to me

[glossy wigeon]

okay thanks for your help

.close

[lone heart]

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[charred summit]

What?
You found the solution?

[glossy wigeon]

no

ive been trying it all day

im not sure what else i should try

[charred summit]

.reopen

[lone heart]

✅

[charred summit]

Did you calculate the volume in the negative of f(x,y) @glossy wigeon ? while making it positive

I think what you did there only accounted for the positive volume of f(x,y)

[glossy wigeon]

how do you mean?

i tried to restrict the domain further by accounting for only when the solid is above the xyplane or touching

[toxic verge]

How can you reopen others channels yourself btw? Is it the green role

[glossy wigeon]

its fine

for me at least ie i would have used it

[toxic verge]

No im just curious lol

[glossy wigeon]

this is what it looks like in desmos

so if you integrate for y = 5, x = 7, f(x,y) <= 0

[charred summit]

There is some part of the function in the negative

For x<7 and y<5

[glossy wigeon]

I agree

[charred summit]

Like for f(x,y) <= 0
12-xy <= 0
12 <= xy
12/x<=y<=5
So now for 0 <= x <= 12/5
y wouldn't have a value
So we start with
12/5 <= x <= 7
And 12/x <= y <= 5

But we here will integrate for -f(x,y) to get positive volume
So xy-12

[glossy wigeon]

I believe this is close to my second integral?

[charred summit]

$$\int_\frac{12}{5}^7 \int_\frac{12}{x}^5 (xy-12) dy dx$$

[ocean seal]

Sherif Player

[charred summit]

Yeah it is close to it

[glossy wigeon]

and should this be added to my first integral?

[charred summit]

It will be added to the final result you had above

[glossy wigeon]

Oh okay

That integral is equal to 72ln(35/12) - 23/4 or something

[charred summit]

Hmm it gave me 71.something

[glossy wigeon]

71.32?

.3217864

[charred summit]

Yeah

[glossy wigeon]

alright

[charred summit]

Add it to the previous final answer and check it

[glossy wigeon]

Then add that to 108 + 72ln(35/12)?

[charred summit]

Yeah

So it will be
108 + 144ln(35/12) - 23/4

[glossy wigeon]

omg correct!

@charred summit wow amazing

this is fantastic work

[charred summit]

Nice

You are welcome

[glossy wigeon]

thank youuu

.close

[lone heart]

Channel closed

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[charred summit]

You understood what I have done right?,

[glossy wigeon]

which part was i missing?

[charred summit]

The additional part and integral
Where we accounted for the volume under xy-plane (f(x,y) <= 0)

You understood how we did it ?

[glossy wigeon]

i see that you took -f(x,y) but im still trying to understand the intuition

[charred summit]

We set 12-xy <= 0
To get the boundary condition of y and x

Which happened to be
12/x <= y <= 5

[glossy wigeon]

ohhhh i think im beginning to understand

[charred summit]

Then for 0 <= x <= 12/5
y would be bigger than 5 so we wouldn't take it

[glossy wigeon]

there was volume underneath the xy plane i was missing of the solid so we needed to account for that by taking -f(x,y)?

[charred summit]

Yeah

[glossy wigeon]

crazy

wow u are really good

[charred summit]

I was just making sure that you understood that so you would be able to solve future problems by yourself

[glossy wigeon]

for sure

i really appreciate that

[charred summit]

Wanna anything more?

[glossy wigeon]

i dont think so. ill have to probably redo this and make sure i get it all but thanks for everything!

[charred summit]

You are welcome, bye

[lone heart]

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[unique dune]

Just for sure, as ${n \to \infty}$, is [ \left| \frac{2e^nx^n}{e^{2n} - 1} \right| = \left| \frac{2e^nx^n}{e^{2n}} \right|]
Acceptable?

[ocean seal]

k

[grizzled mauve]

well the notation is wrong

[unique dune]

why

similar / approx?

[grizzled mauve]

because theyre not equal. they have equal limits

so you have choices:

a) put lim in front of both

[raw herald]

It should be approx

[unique dune]

[ \left| \frac{2e^nx^n}{e^{2n} - 1} \right| \approx \left| \frac{2e^nx^n}{e^{2n}} \right|]?

[grizzled mauve]

b) use ~ , which means, asymptotically equal to

[ocean seal]

k

[unique dune]

[ \left| \frac{2e^nx^n}{e^{2n} - 1} \right| \sim \left| \frac{2e^nx^n}{e^{2n}} \right|]?

[ocean seal]

k

[unique dune]

this one, right?

[grizzled mauve]

yes

if you want to be SUPER rigorous

[unique dune]

ok

another question

[grizzled mauve]

you write "as n->infty" but its implied if the exercise is about limits at infinity

option c)

[grizzled mauve]

multiply by e^-n top and bottom then you have literal equality with =

[unique dune]

the original question was finding radius of convergence for (cschn)x^n

i was making sure that my reasoning and notation in the response make sense

[unique dune]

plz 🥹

[grizzled mauve]

so if you can write this using regular factorials

you can use Stirling

[unique dune]

it seems to be double factorials..

[grizzled mauve]

and the top is triple yeah

write out double factorial with 2n see if you can find a pattern

2!!=2

4!!=4×2

6!!=6×4×2

8!!=8×6×4×2

notice anything?

[unique dune]

no?

they are iterated??

[grizzled mauve]

true

but all the numbers have a common factor

[unique dune]

2(n!)

[grizzled mauve]

almost!

waitzis it?

no not quite

close

[winter light]

You can factor out a 2 from EACH term

[unique dune]

so

n!! = 2(n!) in this case, no?

[winter light]

Here 2 appears only once...

But in your function how many terms are there in the denominator?

[unique dune]

n terms?

[winter light]

So why 2 here and not 2^n ? 😅

[unique dune]

im not following...

[grizzled mauve]

consider 8×6×4×2

=2(4×6×4×2)=2²(4×3×4×2)=2³(4×3×2×2)=2⁴(4×3×2×1)

[unique dune]

oh

i got confused with addition😭

[grizzled mauve]

me too at first its okay

for the numerator

can you show that

(3n-1)!!!~(3n)!!! ?

as n-> infty?

or maybe your prof will let you state it as obvious

why don't you treat it as obvious and then at the end of your assignment see if you have time and patience to go back and justify it

once you say theyre asymptomatically equivalent, can you simplify?

[unique dune]

ok

(3/2)^2 x^n

|x| < 2/3

[grizzled mauve]

ding ding

[unique dune]

because 3n-1 ~ 3n, so they grow at similar rate?

[grizzled mauve]

this is how youd formally prove it

[unique dune]

ive only learned big o, small o🥀

[grizzled mauve]

look at last column

ez notation for ~

ratio goes to one thats it

try it actually

(3n-1)!!!/(3n)!!! as n->infty

good exercise

$$\dfrac{2 \cdot 5 \cdot 8 \cdot 11 \cdot 14 \cdot \cdots \cdot (3n-1)}{3 \cdot 6 \cdot 9 \cdot 12 \cdot 15 \cdot \cdots \cdot (3n)}$$

[ocean seal]

gfauxpas

[unique dune]

can i say stuff like

At very large n, ${3n - 1 \sim 3n}$. Therefore, the ratio is 1?

[ocean seal]

k

[unique dune]

lets do

At very large n, ${3n + k \sim 3n}$ for any constant ${k}$. Therefore, the ratio is 1?

[ocean seal]

k

[grizzled mauve]

hmmm, kinda handwavey, maybe

[unique dune]

handwavey arugment is an actual term?? 😭

ive seen my teacher used it, thought he made that up

[unique dune]

could u give me more hint 😭

[grizzled mauve]

sure

once you write it this way

you see you can combine them into one product symbol

$\prod$

[ocean seal]

gfauxpas

[unique dune]

ok

gimme a sec

[grizzled mauve]

you might be overthinking it

[unique dune]

i got it

[grizzled mauve]

$$\frac{\prod(3n-1)}{\prod(3n)}$ you can combine the numerator and the denominato the same way you can combone sums like

[ocean seal]

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

[grizzled mauve]

$\sum (3k) + \sum (3k-1)$

[unique dune]

nope

[ocean seal]

gfauxpas

[unique dune]

ye its 1 - 1/3k

im trying to find the limit by taking ln of both sides

[grizzled mauve]

I think that's good enough actually

$\prod \left({1-\frac{1}{3k}}\right)$

[ocean seal]

gfauxpas

[unique dune]

and then

i use squeeze theorem

[grizzled mauve]

there's mroe than one way, but if i were the teacher and saw this id give you full credit iuf you said this tends to 1