help-4

[hollow mica]

though that's where Im coming from

I was trying to prove catalan numbers (so (2n!)/(n+1)!n! are.. numbers

and this is the last step of my reasoning

[light saddle]

btw type .reopen or this will close suddenly

[hollow mica]

.reopen

[vale dock]

✅ Original question: #help-4 message

[hollow mica]

thanks!

[light saddle]

this is probably not what you want but you can show that using the induction formula

[hollow mica]

ah yeah (strong?) induction probably does something! I wanted to take a look at that too

but yeah I wanted to finish this reasoning before

[light saddle]

okay I think I got something: what can you say about (2n)!/(n! * n!) and (2n)!/((n-1)! * (n+1)!)

[vapid folio]

Haven't worked out the complete proof, but if you assume n is even, then (n+1)! = 2^k (sum of first n integers)P = 2^kNP where P is a product of odd numbers; you could probably work in cases show if d divided N, then it accounts for certain factors of the RHS and if d divides P, then it accounts for the remaining factors (I do not believe each factor needs to be distinct but this could be wrong)

[hollow mica]

I'll write that down to keep it in mind for later, thanks!

[hollow mica]

not sure I understand your rewriting of n+1! ?

[vapid folio]

you can factor out a 2 from the even integers which will leave an n/2 * (n+1) at the end of the product which turns into sum of first n integers (which is odd if n is even); you then can continue factoring 2 until you are only left with odd terms in the product

[hollow mica]

this is what I was using

[vapid folio]

Yeah, I haven't worked out the proof myself so I can't be certain but that may/may not be useful

Though I agree it's a cool problem 👍

[hollow mica]

god I'm slow with latex

sorry

with this

[hollow mica]

ah yeah! hadnt seen it that way of factoring

thanks for the answers! Guess I'll try to think about it a bit more 😅

.close

[vale dock]

Channel closed

Closed by @hollow mica

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[clear kelp]

can someone explain to me direct sum generally

[clear kelp]

and direct product

[spark sluice]

Both direct sums and direct products of two objects X,Y usually have as underlying sets X×Y (Cartesian product) such that X×Y is another object of the same kind. The word "sum" is used when your object is abelian , the word product is used when your object is generally not or when you are taking an infinite Cartesian product where direct sum and direct product may not agree
Cf. Direct sum of vector spaces, direct product of groups

[clear kelp]

my current understanding is that u can write it as an order pair or u can write it as a sum, with elements of both sets over some.. field? or something but im not sure how to say it well or if im understanding it well. or which is which. also they are analagous in finite dimension but not in infinite dimensions? im not sure

so just say with the set of arbitrary objects with no structure, we would use direct product? say the elements of the set were x1, x2, ..., xn with no clearly defined structure

and how would the direct product of two such sets look like, say X = {x1, x2, ..., xn} and Y = {y1, .., yn}.

if we did direct sum instead what happens or is it nonsense

what if we replaced it with an infinite set, X = {x1, x2, ..} . or if both were infinite sets

and thank u for helping <3

[spark sluice]

What I'm saying is, direct sum and direct product are somewhat the same

If you have vector spaces X,Y, you take X × Y and turn it into a vector space

And you say directsum to refer to this construction

If you have groups X,Y, you take X×Y and turn it into a group

And you say direct product to refer to this construction

[clear kelp]

yea.. what is this construction 😭

i tried going to wikipedia but its hard to understand what they are saying

so um... im trying to ask about like, precisely the definitions and so on

[spark sluice]

So your question of "what if I take direct product instead of direct sum" doesn't make much sense

[clear kelp]

words? in the category?

[spark sluice]

For vector spaces X,Y, their direct sum is the vector space X×Y where you define addition of (x,y) and (a,b) as (x+a,y+b) and scalar multiplication by c(x,y)=(cx,cy)

[clear kelp]

sorry, im sure ive seen all these before but my foundations are shaky so i would rather, use simpler terms or be told all terms and their definitions

and according to wikipedia, direct sum is different for different structures, so its hard to write a general definition for direct sum?

[spark sluice]

For abelian groups X, Y, their direct sum is the group X×Y where you define the group operation as (g,h)*(a,b)=(ga,hb)

As you can see the definition for direct sum is different between groups and vector spaces

[clear kelp]

is it not the same, only that the abelian operation changed from addition to multiplication and that vector spaces have a scalar multiplication which still works

[spark sluice]

Well so its not the same

There are similarities like you are pointing out

[clear kelp]

i guess

so

is there a general way to describe or understand the two, direct product and sum

[spark sluice]

The very general idea is that direct sums and direct products of some family of structures {Xs}_{s \in S} have the aim to put the same type of structure on the cartesian product of the family Xs. The wiki article goes into detail

[clear kelp]

family is similar to set? and different structures, can we use direct sums or products on them?

[spark sluice]

Yeah like family just means a collection of those structures. And no we don't usually take direct sums or products of stuff like groups with vector spaces

[clear kelp]

i see thanks

ill go read the wikipedia article a few times

[spark sluice]

If you want to look at this from a category theoretic viewpoint, the notion of products in a category can be taken as a generalization of direct sums and products

[clear kelp]

i never understood category theory and all the category stuff :(

[spark sluice]

There it is defined via a universal property, and in a lot of cases some cartesian product construction happens to satisfy the properties

[clear kelp]

yea, ill try to understand it sometime, and yes i do know the definitions of universal properties and so on. i just dont think i have a good understanding of them yet.

thank you aster

.close

[vale dock]

Channel closed

Closed by @clear kelp

Use .reopen if this was a mistake.

[vale dock]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[granite trellis]

Hi! Could someone help me verify this proof? I couldnt think of a function for A and B since the question doesnt mention whether it’s a number or an alphabet or any other element.

[spice topaz]

Do you know that NxN is countably infinite?

[granite trellis]

N x N has a bijective function with Q and Q is countably infinite, but i havent shown that over here

[spice topaz]

So if there is a bijection between NxN and AxB ，then AxB is countably infinite.

[granite trellis]

should i go about it that way, take N^2 and show that its countably infinite and the same thing would apply to A x B

[spice topaz]

In you proof, is |A| x |B| defined ?

[granite trellis]

no, im just given A and B

and i just assumed A x B would have same cardinality as N x N

[spice topaz]

Or is |N|x|N| defined? I mean, |N| is not a number, is multiplication defined on it ?

[spice topaz]

Then you should find a bijection between them.

[granite trellis]

oh wait i didnt think of that, then can i say |A| = |N| and |B| = |N| so |A x B| = |N x N|

[spice topaz]

If this is a theorem, you can use it directly.

If multiplication is not defined on |N| (which is not a number), the you proof is not correct, as it uses |N| x |N| .

[granite trellis]

ill look to check if thats a theorem, if not then if i prove that thing i can finish my proof just by using that right?

[spice topaz]

Yes

provided that you have theorem saying NxN is countably infinite

[granite trellis]

also i wont be forming a bijection, rather i would do this by double injection, but the problem is i dont know what kind of set A and B are so that i could create a function between it and Natural numbers

[spice topaz]

Because A is countably infinite, there must exist a bijection between A and N. This should be theorem. You don't have know the exact form.

Or maybe it is the definition of countably infinite.

[granite trellis]

that means even in A x B and N x N i dont really need an actual proof then, since i know A maps to N and B also maps to N so there is already a bijection by definition

the only proof would be to show N x N maps to N

[spice topaz]

You can list all elements in NxN, that makes it countable infinite.

like
(0,0)
(1,0) (0,1)
(2,0) (1,1) (0,2)
....
(n,0) (n-1,1) (n-1,2) ... (0,n)
...

Maybe there is already a theorem saying NxN is countable infinite.

[granite trellis]

wait so is this another approach to this proof? by showing we can list the elements?

[granite trellis]

yes there was an exercise in one of my readings

[spice topaz]

listing elements is an approach to prove countably infinity

You must show that every element appear at a finite position.

[granite trellis]

like at a number that could be defined by n?

[spice topaz]

what do you mean?

If you list NxN as (0,0), (0,1),(0,2),(0,3)....(0,n)....(1,0) (1,1), (1,2) .... , then (1,0) is not at a finite position, there infinite elements before it.

Sorry, have to leave now.

[granite trellis]

Thank you for your help! Really appreciate it

[vale dock]

@granite trellis Has your question been resolved?

[granite trellis]

.close

[vale dock]

Channel closed

Closed by @granite trellis

Use .reopen if this was a mistake.

[vale dock]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[fleet stirrup]

How do I prove the blue statement by using the bijection between the two green items?

[fleet stirrup]

The bijection is clear
Every tuple that can be formed with {0,1}^|C| can be associated with one subset of C

How does this lead to the statement in the blue box?

[pearl nacelle]

Are you asking why the cardinality of the set is 2^|C|?

[fleet stirrup]

I’m asking how to prove the blue statement

[pearl nacelle]

What us the blue statement I cant see it

[fleet stirrup]

Sorry the colors are bad

[pearl nacelle]

Right

So that’s the size of the set

It’s called the power set of C I’ll denote it by P(C) okay

Do you know that it’s every possible subset of C?

[fleet stirrup]

Yes

[fleet stirrup]

Yes that too

I know what it is

[pearl nacelle]

Alright good

[fleet stirrup]

But I don’t know how to determine its size

Because the only way to calculate that we got was 2^|C|

[pearl nacelle]

So imagine the elements of the set C are c1, c2, c3….,cn

So any subset either contains c1 or doesn’t contain c1, contains c2 or doesn’t contain c2 etc all the way up to cn

So we have 2 options for each element

2x2x2x….x2 n times

Where n is |C|

[fleet stirrup]

Yes

So we can portray this by using the tuples from {0,1} instead

Oh wait nvm

You explained the other thing

[pearl nacelle]

Yes but you’re right

[fleet stirrup]

The task explicitly asks us to use the results from a and c tho

Task a states that bijection implies |A|=|B|

[pearl nacelle]

Mhm

[fleet stirrup]

Task c was a proof for |A^n| = |A|^n

[pearl nacelle]

Oh so it wants to use a and c to prove that |P(C)|=2^|C|?

[fleet stirrup]

Yes

That was quick

[pearl nacelle]

Yeah lol

Well, we can do that

First let’s use part c

We can say that |{0,1}^n|=|{0,1}|^n=2^n right?

[fleet stirrup]

That one is taking me a moment

I think so?

[pearl nacelle]

Well you can see that |{0,1}|=2 yeah?

Two elements

[fleet stirrup]

Yes

Sorry I’m very slow

This homework has been a lot
I need to write it on paper right now and my concentration isn’t the best

[pearl nacelle]

That’s okay

If you read part c, it says that |A^n|=|A|^n

So using this, we can say |{0,1}^n|=2^n

[fleet stirrup]

Correct

[pearl nacelle]

Cool

So if |C|=n and we find a injection between P(C) and {0,1}^n, we’re done

[fleet stirrup]

How did you come up with the injection?

[pearl nacelle]

Well there’s a few ways of doing it

There’s one clever way which goes something like this

All of the subsets of {0,1}^n look something like (0,1,1,0,0…) right?

Some combination of 0’s and 1’s

[fleet stirrup]

Yes

They do

[pearl nacelle]

Ill write this out because it’s easier

[fleet stirrup]

Alright thank you

[pearl nacelle]

Hopefully this makes some sense

[fleet stirrup]

Yes this is the bijection I’ve mentioned earlier right

[pearl nacelle]

Yup

So we said that we’re happy that |{0,1}^n| is 2^n, that’s part c

We’ve found a bijection between that and our set P(C)

So using part (a) we can conclude what was inside of the blue box

[fleet stirrup]

Im not sure what part A‘s job is here

[pearl nacelle]

We know that P(C) and {0,1}^n are in bijection

We know that |{0,1}^n|=2^n

Part a tells us that if two sets are in bijection, they have the same size

So P(C) and {0,1}^n are the same size

2^n

[fleet stirrup]

Oh

Yeah that makes sense actually

[pearl nacelle]

Cool

[vale dock]

@fleet stirrup Has your question been resolved?

[fleet stirrup]

Sorry for disappearing I just wrote 4 pages

Thank you for your help it was clear in the end

May I ask for help with one more task?

[pearl nacelle]

Sure

[fleet stirrup]

M N are both sets
Not empty
A and B are subsets of M
While C and D are subsets of N
There’s also the function M->N

The first task wants me to figure out the mistake (there is a wrong equivalence)

This is a wrong proof for a wrong statement but once you correct the mistake the proof itself will become right and the statement of the proof will change

[fleet stirrup]

Im not sure if I was supposed to respond to a message to ping you or something but I’ll do it just in case

[pearl nacelle]

Do you hve any suggestions as to what it might be?

[fleet stirrup]

I don’t even understand the statement that they’re trying to prove

Im not talking about it being wrong
I know that much but I don’t understand what it’s saying

[pearl nacelle]

You don’t understand what the statement means?

[fleet stirrup]

The second part is confusing me

Sorry it’s 2 am and i haven’t slept enough in days this is killing me

[pearl nacelle]

Part b you mean right

[fleet stirrup]

No

The second part of the statement

f(A) and f(B)

I don’t know how to picture it

[pearl nacelle]

It’s saying that there is at least 1 x that is in A and B s.t f(x)=y

[fleet stirrup]

I guess it’s saying that there are values of the function that both A and B can imply?
That’d mean not having a unique pair
So it can’t be bijective

[fleet stirrup]

s.t.?

[pearl nacelle]

Such that

[fleet stirrup]

It can’t be injective either
Is it surjective?

[pearl nacelle]

I don’t think surjectivity/injectivity are really relevant here

[fleet stirrup]

Alright

Im currently interpreting the statement as
There’s a value y from a value x that both A and B share
But i don’t know how the right side is any different

[pearl nacelle]

Right side?

[pearl nacelle]

Oh

Think of it in terms of images of sets

[fleet stirrup]

I’m trying

[pearl nacelle]

Dont get too caught up on visualisation

Sometimes you just have to get comfortable in the abstract

[fleet stirrup]

So the image of a value that A and B share is supposed to be the same as… an image they both share?

I just can’t even tell how to phrase it

They seem like the same thing

Oh wait

Does it mean they share an x is supposed to be the same as they share a y

[fleet stirrup]

I thought about it more
Is it them supposedly sharing a value x for every y they have in common?

In that case this statement would be proven wrong by two different values of x that they don’t share that result in a common y

Sorry if you were going to come back but it’s almost 3am and I need to hurry up
I’ll open a new channel so more people can respond
@pearl nacelle
I’d appreciate seeing you there tho

.close

[vale dock]

Channel closed

Closed by @fleet stirrup

Use .reopen if this was a mistake.

[vale dock]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[fiery pelican]

Given a cube ABCD.A'B'C'D'. M and N are two point on AC and DC' which satisfy MN//BD'. Choose True or False
a) Set AM/AC = x, C'N/C'D = y [x, y ∈ (0,1)] -> Vct MN = (1 - x -y)vct BA + (1-y)BB' + (1-x)vct BC
b) Set MN/BD' = m/n (m,n ∈ N*) -> 2m + 3n = 13

Can someone prove those two statements false? (The answer sheet only show both of them are false, but i don't know why)

[fiery pelican]

I appreciate it :D

[vale dock]

@fiery pelican Has your question been resolved?

[fiery pelican]

<@&286206848099549185> sorry for disrupting you all but eh... can yall help me for a bit?

[pearl chasm]

Ques?

[fiery pelican]

here

its a true or false question
those two are false but they dont explain it clearly

[chilly violet]

Have you drawn a picture?

[fiery pelican]

i did

but after i drew it i just have no idea how to prove it false

[chilly violet]

Can you send the picture?

[fiery pelican]

okay

[fiery pelican]

sorry i drew without a ruler

[chilly violet]

Perf. I think it’s pretty obvious that (b) is false given the number of points M or N could be

[fiery pelican]

huh..?

[chilly violet]

you can draw MN in lots of different ways

[fiery pelican]

yeah but.. how is that related to B?

[chilly violet]

B is making a claim about the ratio of MN to the diagonal

ooh wait nvm i can’t read

[fiery pelican]

aw its okay its okay

[chilly violet]

my gut is saying false but i can’t prove it. it’s near midnight where i live and my brain is not currently equipped for geometry

so i may have to bail

[fiery pelican]

oh if that so then uhm, get some rest

thank you for helping me

[chilly violet]

i’d try drawing MN in more than one way

and see if you can make a similar triangles-esque argument with DCC’ or something

[fiery pelican]

okay

[chilly violet]

again sorry for not being able to do much

[fiery pelican]

no its okay

i really appreciatei t

https://tenor.com/view/spongebob-gif-5787005613059506418

[vale dock]

@fiery pelican Has your question been resolved?

[vale dock]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[vale dock]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[ivory stag]

Need help on circled b

[ivory stag]

And c, inplicit differentiation

Zero clue what im doing

The steps dont make any sense to me

[lyric jewel]

Hi! The entire principle behind implicit differentiation is that you just want to "take the derivative of both sides".

For example,
x^2 - y^2 = 1
What happens if you take the derivative of both sides (with respect to x)?

[ivory stag]

I learned a bit in class

Let me show u where i get stuck 1 sec

Wait

Deritive of y is always y’ right

[lyric jewel]

Yep.

[ivory stag]

I feel good i think until substituting y’ back in

Common denom of y

Oh wait

If x^2 - y^2 = 1, the original form

Can u make it -1/y^2

[lyric jewel]

Woah woah okay, hold on

Actually

It seems like your working out is right in a way:
x^2 - y^2 = 1
Taking the derivative of both sides
2x - 2y * y' = 0
y' = x/y, this is good.

Then you take the derivative of both sides, which is good too (using quotient rule).

[lyric jewel]

Yeah actually it seems like the only problem here is making common denominators.

How do you simplify y - x^2/y?

[ivory stag]

Multiply y by y/y

To get y^2/y

Then combone into one fraction

[lyric jewel]

Good; so: (y^2 - x^2)/y right?

[lyric jewel]

Now divide this by y^2, as per your expression.

[ivory stag]

Yea

Like

1 - x^2?

Is this equation of important

Oh

I got it

Multiply by the reciprocal to cancel out double fraction

[lyric jewel]

Yeah but there should not be a y' in front

Like it should just be 2x - 2y * y' = 0

[vale dock]

@ivory stag Has your question been resolved?

Channel closed

Closed by @ivory stag

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[heavy ridge]

Let K be the set of all real values of x where the function \ $f(x) =\sin \abs{x} -\abs{x}+2(x-\pi)\cos \abs{x}$ is not differentiable. Then the set K is: \ (A) $\phi$\ (B) $\left{ \pi \right}$ \ (C) $\left{ 0 \right}$ \ (D) $\left{ 0,\pi \right}$ \
I checked differentiability at x=0 because of the presence of the absolute value function. with that I'm getting different LHD and RHD but the solution manual says the answer is (A) null set

[rocky lotus]

T&C

[heavy ridge]

oops spotted my mistake. sorry

.close

[vale dock]

Channel closed

Closed by @heavy ridge

Use .reopen if this was a mistake.

[vale dock]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[modern estuary]

need help w dis

[modern estuary]

need to factorise by taking out common factors

[neat brook]

!status

[vale dock]

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

[modern estuary]

i dont know where to begin

[neat brook]

do you know what common factors are?

[modern estuary]

yeah

[neat brook]

okay, so - do you think we could take a common factor of a?

[modern estuary]

yeah

it'd be a square right

[neat brook]

and if we did that, we'd go from $a^2bc^3 + c^3ba^2 - a^3b^2c^2$ to $a( ??? )$

[rocky lotus]

ꙮ PPK ꙮ

[modern estuary]

a square right

[neat brook]

a square of what

[modern estuary]

A

[neat brook]

oh, you mean we could take a common factor of a^2

yeah

[modern estuary]

yeah yeah

[neat brook]

so we'd have $a^2( ???)$

[rocky lotus]

ꙮ PPK ꙮ

[modern estuary]

b?

[neat brook]

I was going one step at a time - so if we took a common factor of a^2 out, what would we have left

and then from what we have left, we can work out the common factor of b to take out

[modern estuary]

i didnt understand, sorry

[neat brook]

okay - so we've taken out the common factor of $a^2$. so what would we fit in the missing space such that $a^2 * (????) = a^2bc^3 + c^3ba^2 - a^3b^2c^2$

[rocky lotus]

ꙮ PPK ꙮ

[modern estuary]

it'd a square b and then c square right

[neat brook]

those would be some common factors, but you'd then need to work out what you're left with

if you want to do it all at once then that's fine

[modern estuary]

ok lets do it then

all at onnce ig

once

[neat brook]

$a^2bc^2 * (????) = a^2bc^3 + c^3ba^2 - a^3b^2c^2$

[rocky lotus]

ꙮ PPK ꙮ

[pearl nacelle]

The intersection of their images is the inage of their intersection

[modern estuary]

@neat brook its okay i got it

[neat brook]

you're 9 hours late lol

[modern estuary]

thanks for the help

you can close this

[neat brook]

.close

[vale dock]

Channel closed

Closed by @neat brook

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[pearl nacelle]

Ah well

[keen tundra]

.close

[vale dock]

Channel closed

Closed by @keen tundra

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[chrome karma]

Hey so i have this question here

[unborn sable]

whats up

[clear kelp]

whatve u tried

[chrome karma]

rnight so I got this

bUt the answer is wrong

[clear kelp]

what is the answer?

[safe fulcrum]

it's correct but they want a different form

they want 4 * e^(j * something)

[chrome karma]

the answer is this

[safe fulcrum]

oh that's even more interesting

[chrome karma]

isnt it

[safe fulcrum]

that's also correct, yes

[chrome karma]

so were both right?

[safe fulcrum]

you can just do 17/12 pi - 2pi

[lyric sundial]

Yep

It depends on the convention of the argument range, whether it's from -π to π or from 0 to 2π

[clear kelp]

its the same but 24pi/12 away, which is the same as e^(2pi i) = 1 = e^(24pi/12 i)

[chrome karma]

or right but the answer does show the range

[safe fulcrum]

the convention I've usually seen is that r > 0 so that surprised me

[chrome karma]

i mean the question

[lyric sundial]

Indeed, I also thought they wanted a positive mod

[clear kelp]

it doesnt matter too much unless its specified tho

[safe fulcrum]

yeah that's just saying that 'one possible answer is....'

of course you should pay attention in class to figure out which forms of the answer are wanted

[chrome karma]

alright thanks

.close

[vale dock]

Channel closed

Closed by @chrome karma

Use .reopen if this was a mistake.

[chrome karma]

.reopen

[vale dock]

✅ Original question: #help-4 message

[chrome karma]

wait

so then what aobut this here

how do u add two imaginary exponents since weve been only toguth multiply and divison now

[safe fulcrum]

oh, compute z1 z2 and z1 / z2 in polar form first

then convert both of those to Cartesian form

now you can add

[safe fulcrum]

$(x, y) = (r \cos \theta, r \sin \theta)$

[rocky lotus]

south

[vale dock]

@chrome karma Has your question been resolved?

[chrome karma]

so like this?

[cedar gazelle]

How do I use Pi in real life activities?

[chrome karma]

this is teh answer

either we both right or im wrong

[vale dock]

@chrome karma Has your question been resolved?

[left ruin]

is this the working for the second part of c

[chrome karma]

Yea

[vale dock]

@chrome karma Has your question been resolved?

[vale dock]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[vale dock]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[cobalt ledge]

hi, i was wondering if my proof would be valid, or if there are better ways or doing this. i feel like my ways feels a little dubious. ty😀

[cobalt ledge]

sorry, the image looks weird, im trying to prove that there are infinitely many non trivial triples, i.e. not a multiple of a different pythagoren triple

[vale dock]

@cobalt ledge Has your question been resolved?

[vale dock]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[regal stag]

Hi, I'm currently working through this question on a representation theory example sheet. I feel like I'm not understanding the concepts that have been taught in lectures at a foundational level - case in point I'm struggling to prove the first part of this question which feels like it should be not hard.

I'm pretty sure I know what I want to be using - the fact that the sum of the squares of the dimensions of the representations should add to |Q_8| = 8. To that end I'm pretty sure there should be exactly 4 1-D representations and 1 2-D representation.

But what is stopping it from having 2 2-D representations and no 1-D representations? or 8 1-D representations? I feel like there should be something nice to be said about how to find the exact number of 1-D representations (since they're so nice and come from abelian groups and are always irreducible for obvious reasons) but afaia we haven't covered anything on that in lectures (or I am conveniently blanking on that part of the course.)

[vale dock]

@regal stag Has your question been resolved?

[regal stag]

First time posting a question so I'm not 100% on how this works but I think it's been 15 minutes so im ok to ping <@&286206848099549185> ?

[wraith heart]

you can also ask in #advanced-algebra which is where more people who are familiar with rep theory hang out

[spice topaz]

I'm not very good at this, but I remember there is a theorem that says the number of irreducible representations equals the number of conjugate classes.

[regal stag]

i think that's not useful since again it gives a total and not dimensions pecific numbers

[spice topaz]

You three options listed all have different total numbers of representation.

There should be only one of them equals the number of conjugate classes.

[regal stag]

you're so right omg

wasnt actually thinkiung

ok sorry yeah back

[swift magnet]

so a few things to keep in mind

a group always has the trivial representation

[regal stag]

and yeah Q8 has 5 conjugacy classes of like 1, -1, +-i, +-j and +-k

[swift magnet]

so there's always at least one 1D rep

also yeah, number of irreps = number of conjugacy classes

in fact, a group is abelian iff all its irreps are 1D

[regal stag]

abelian group conjugacy classes are all singletons

that makes sense

[swift magnet]

mhm

[regal stag]

ok cool, i think im happy with the answer to this first part now. i'll close this and give the rest of the question a good bash. thanks guys o7

.close

[vale dock]

Channel closed

Closed by @regal stag

Use .reopen if this was a mistake.

[swift magnet]

though

oh, ok

[regal stag]

oop

[swift magnet]

i was going to ask whether they expect you to use character theory here

[regal stag]

no def not

wouldnt be covered on sheet 1, that's later in the course

[swift magnet]

wait then how do you know the sum of squares thing

[regal stag]

i think we proved it some other way

[swift magnet]

interesting

[vale dock]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[rigid whale]

I've been trying to help my gf to do this one problem for so long now but I can't seem to find any other way to do it other than through logging both sides, but she has an exam where it's JUST exponential equations and no logarithm whatsoever, so I'm wondering if my mind is going blank or is there just no way to solve this without the help of a logarithm??

The question goes like so:

10^(x-2) + 10^(x+2) = 99

[rigid whale]

I've tried so many things, only way I've gotten to a possible solution is by approximating the answer but given how simple her school is (and it's not a mathematical school) where exponential equations are learnt on the basics, I don't see why they would do so.

[solid tree]

let u=10^x

[rigid whale]

I don't think that does anything

10^x/10^2 + 10^x • 10^2 = 99

u = 10^x

u/100 + 100u = 99

It still requires to use logarithm.

[solid tree]

did she not learn about the logarithm?

[rigid whale]

They did, but this exam is just exponential equations without any logarithms.

[solid tree]

you use logarithms to solve certain exponential equations

its as you said, the only way in this circumstance

[sage gust]

Um no logs at all?

or can you stack onto exps

[rigid whale]

yes, i know and hence why i said i only see logarithm being the solution, but every other example they've done is without them, and this is just one that seems to be off, idk if it's a mistake or if they really want them to use logarithms on a test where they specifically said just pure exponential equations without any use of inverses (logarithms)

[sage gust]

oh..

[solid tree]

maybe its one of the 'harder' ones

idk

[rigid whale]

i'll just tell her the logarithm is the only way i can think of

ty regardless

[solid tree]

the answer must be in terms of a log anyways so there isn't really any going around it

[rigid whale]

.close

[vale dock]

Channel closed

Closed by @rigid whale

Use .reopen if this was a mistake.

[sage gust]

I can only think of logs but here’s sln

[vale dock]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[misty helm]

hey can someone help me I am learning about linear algebra

lets say I have bases b1 and b2, now lets take two scenarios

1. we have a vector v1 lets say 2i + 3j now I want to transform into the bases b1 and b2 subspace I would do that using [b1, b2]*v right? if not what is this video talking about https://www.youtube.com/watch?v=kYB8IZa5AuE
2. Now lets say I have a vector v1 which is already in the subspace of bases b1 and b2, now to convert v1 back into my i cap and j cap i would do the same multiply [b1, b2]*v right?

[vale dock]

@misty helm Has your question been resolved?

[vale dock]

@misty helm Has your question been resolved?

Channel closed

Closed by @misty helm

Use .reopen if this was a mistake.

[vale dock]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[merry hearth]

Whats the first step 😭

[merry hearth]

.closed

.close

[vale dock]

Channel closed

Closed by @merry hearth

Use .reopen if this was a mistake.

[stark wedge]

tf is "number of integers"

[ruby sleet]

My bet is on number of digits

[stark wedge]

but why not say digit if digit is meant

[ruby sleet]

Bc bad

[dawn vale]

hmm

we can find log 10 (5^200) to find the number of int in 5^200

bc is log 10

[dawn vale]

[vale dock]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[tired aspen]

Can you help me how to solve this problem?

[stark wedge]

do you know vieta

[tired aspen]

?

[dawn vale]

what is X

[tired aspen]

I dunno

[dawn vale]

what is root =)))

[ebon burrow]

this stuff?

[dawn vale]

i only know this is a part of tree

[pliant nova]

basically x intercepts of a polynomial function

[fleet burrow]

yes but for cubics

[shy geode]

is the answer 6?

[ebon burrow]

yeah, I thought he might know it without the name vieta

[shy geode]

sum of roots is -b/a which is zero in this case, therefore using the identity

[dawn vale]

i don't understand that. That is ( if ax+b=0 and x=-b/a => x is root)??

[pliant nova]

you could say that

[shy geode]

if b+c+d=0 then b^3+c^3+D^3=3dbc

then just use product of roots

-(-2)/1

[fleet burrow]

this is exactly it

[shy geode]

dbc=2 3dbc=6

[dawn vale]

tysjm

[pliant nova]

ur supposed to let OP solve it not lay it out in front of them

[shy geode]

wdym?

[dawn vale]

i don;t understand what "root" mean =))

[ebon burrow]

values of x for which f(x)=0

[shy geode]

values of x for which the plot y=f(x) intersects the x axis

[dawn vale]

okey tysm im only know a litte in english

[stark wedge]

this isn't your channel though. if you don't understand OP's question then what are you doing here

[dawn vale]

i can solve some problem =))

[dawn vale]

😭

[stark wedge]

you say you don't understand this specific problem. so in this channel specifically you cannot do anything

[dawn vale]

and i love math

[stark wedge]

not talking about the whole server

only this one help channel

[stark wedge]

do you know sum and product of roots

maybe that'll be more familiar

[dawn vale]

sorry im stupid

but now i can understand that problem require

[tired aspen]

Yea i know

[stark wedge]

ok so do you know the version of that for cubics

[tired aspen]

Yeap

I found out that

b+c+d equals to 0

and bc+cd+db equals to A

[stark wedge]

good

[tired aspen]

and bcd equals to 2

[stark wedge]

yeah now try to find a way to express b^3+c^3+d^3 in terms of these

[tired aspen]

Hmm

Just (b+c+d)(b^2+c^2+d^2) - something

Is it right?

[junior ore]

whats that?

what r u gonna equate it to?

[tired aspen]

To make

b^3+c^3+d^3

[junior ore]

then what about the other terms?

how r u gonna find the other terms

[tired aspen]

Oh

Other terms are..

[junior ore]

think of another way

[tired aspen]

Okay

Hmm

[junior ore]

do u know (a+b+c)^3 formula?

@tired aspen

[tired aspen]

?

Isn’t it

Oh no

Teach me bruh

[junior ore]

just replace a b c with b c d as given in the question

also b+c+d = 0 so its whole cube will also be 0

now try to solve this

[tired aspen]

Ohhhh

Yea that’s

(a+b+c)(a^2+b^2+c^2) is )t it?

[junior ore]

no

its not that

wait

no

its not equal

[dawn vale]

is a^3 .... =)))

and a^2....

[tired aspen]

Aha

Wait so

it’s just 6

6 is the answer?

[junior ore]

6?

[tired aspen]

Wow

Yeap

[junior ore]

yeah

[dawn vale]

nt

[tired aspen]

No

Thx

.close

[vale dock]

Channel closed

Closed by @tired aspen

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[dawn vale]

I have used it quite a lot =))

[viral idol]

!original

[vale dock]

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.

[junior ore]

there is no question

[viral idol]

ik

if h can

[junior ore]

he was just saying something related to previous question

[viral idol]

if he can't show the question

ohhhhh

[vale dock]

@dawn vale Has your question been resolved?

[dawn vale]

=)) i don't have question

[slate folio]

then close the channel

[river shale]

.solved

[vale dock]

Channel closed

Closed by @river shale

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[obtuse bluff]

this question has been really confusing me, i'm unsure of how to set up what i need to get the answer

[obtuse bluff]

my current point is having some sort of integrals along these lines, $\int^45_15 \int^{y+5}{y} f(x, y)dxdy + \int^60_0 \int^{x+5}{x} f(x, y)dydx$

[rocky lotus]

diego

[obtuse bluff]

augh

[wraith heart]

{}

[safe fulcrum]

search up 'geometric probability': the idea is that you plot person 1's arrival time on the x-axis, and plot person 2's arrival time on the y-axis

you can do it with integrals, but the shapes you get are triangles and trapezoids

[obtuse bluff]

$\int^{45}{15} \int^{y+5}{y} f(x, y)dxdy + \int^{60}{0} \int^{x+5}{x} f(x, y)dydx$

[rocky lotus]

diego

[obtuse bluff]

where $f(x, y) = f_X(x)f_Y(y) = \frac{1}{60}\frac{1}{45}$

[rocky lotus]

diego

[obtuse bluff]

since the probabilities are independent

[toxic ridge]

This info will help immensely

[safe fulcrum]

diagram spoiler

[obtuse bluff]

hmm i'll look into it but my prob course hasn't really gone over anything of the sort sooooooooooo idk if that's where i'm supposed to go with it

[bright wind]

-# Tbh I want to see what have he cooked

[obtuse bluff]

is the idea that i'm supposed to like.. find the probability by figuring out the relevant section that corresponds to the event i'm looking for and then like

figuring out the area of said section?

[safe fulcrum]

yep!

and then you divide by the total area of the sample space, so that would be the rectangle here

[obtuse bluff]

okay that makes sense

[safe fulcrum]

https://brilliant.org/wiki/1-dimensional-geometric-probability/
also there's this article if you want to see a few more examples

[obtuse bluff]

is using a joint density function possible though?? the geometric solution seems a lot simpler but we haven't gone over anything like that in class so idk if my professor will love it

[safe fulcrum]

I know this is probably not how your class wants you to do it, but it's honestly the simplest

[obtuse bluff]

ig my bounds are probably weird but i'll look into it

thank you for the help!

.close

[vale dock]

Channel closed

Closed by @obtuse bluff

Use .reopen if this was a mistake.

[safe fulcrum]

if you understand that the condition is |x - y| < 5

then you could do $\int_{15}^{45} \int_{x - 5}^{x + 5} f(x, y) \ dy \ dx$

[rocky lotus]

south

[safe fulcrum]

something like that