help-26

[winter wave]

wow

im sorry for being so stupid

[opal vault]

It's alright

[winter wave]

I have a lot of other problems to do, but i wont bother you anymore XD

thank you for the help!

i am applying for a scholarship, the math is hard

.close

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[ionic oar]

Does this suffice?

[gleaming thunder]

yes

[ionic oar]

Really?

No problem whatsoever?

[gleaming thunder]

well depends on how much detail is expected for your class

[ionic oar]

no class

would any reasonable teacher accept this

[gleaming thunder]

I would

[ionic oar]

you're very reasonable

thanks

[gleaming thunder]

an extra line between the assumption coming from xn -> 2 and what you conclude wouldn't hurt as far as writing is concerned

[ionic oar]

ehhh

you're very reasonable

:))

Thanks, though

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[carmine parrot]

i need help with this

[burnt swift]

do you know definition of union of two sets

there is listing method and set builder notation

[carmine parrot]

Can you show me an example of a union

[burnt swift]

I just did

[carmine parrot]

Can you explain how it’s a union

Everything inside a and b?

[carmine parrot]

So everything in a and b

Is the answer 7?

[burnt swift]

impossible

[carmine parrot]

idk man

[carmine parrot]

.close

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[cursive patrol]

bro just spammed images

[carmine parrot]

I’m saying

Making my head hurt I had to close it bro

.reopen

[topaz sinew]

✅

[carmine parrot]

<@&286206848099549185>

[cursive patrol]

this last one is actually very slightly useful if you read the examples at the bottom

[carmine parrot]

I saw the first example matches

[rigid ivy]

Union of two sets is a set whose elements exist in either of those two sets

E.g. ${1}\cup{2}={1,2}$

[thorny flame]

SWR

[carmine parrot]

Wouldn’t it be B then? Empty set

[burnt swift]

maybe you are confusing union with intersection

[rigid ivy]

Please explain you're thinking on why you believe it would be the empty set

[carmine parrot]

I’m not sure

I’m confused

[rigid ivy]

Do you understand this example?

[burnt swift]

venn diagram maybe can make u understand

[carmine parrot]

Yeah

[rigid ivy]

So can you do union of {1,2,3} and {5}?

[carmine parrot]

No

[rigid ivy]

It's {1,2,3,5}

You just take the elements from both sets, and combine them into one set

[carmine parrot]

Oh yeah I thought you were referring to the question I sent

I can do that

[burnt swift]

do u still need help

[carmine parrot]

Yes I still can’t get the answer to the question I don’t know if I’m missing something

[burnt swift]

okay

[carmine parrot]

Is it {1,7,9,1,5,7}

[burnt swift]

yes but the duplicates

[carmine parrot]

So {1,7,9,5}

Wait

[burnt swift]

almost

[carmine parrot]

Fixed it

[burnt swift]

also, I think in listing method it should be sorted ascending

[carmine parrot]

How do I know when I need to answer in the listing method

[burnt swift]

write the resulting set using listing method

resulting set of taking the union of those two sets

[carmine parrot]

The answer would be {1,5,7,9} right?

[burnt swift]

,w Union[{1,7,9},{1,5,7}]

[thorny flame]

Results provided by WolframAlpha

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

Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[carmine parrot]

alright i understand now

[burnt swift]

great

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[clever citrus]

Question Let $f(x, y) = cos(y) +y^6 e^x sin(x^2)$ and find $f_{yyyyxyyyy}$. This will not take long if you use an appropriate theorem

[thorny flame]

Calc III Victim

[clever citrus]

now the only theorem I recall covering from the partial derivative was the clairaut theorem

but im not sure how that will help here

I honestly dont even understand clairaut theorem properly.

I solved this by finding the 4th derivative wrt to y and then wrt to x and then wrt to y 4 more times.

but how can this be simplified using the theorem

[junior fable]

honestly I just broke it up into parts, knew that differentiating cos(y) 4 times gives back cos(y), and differentiating more than 6 times completely kills the y^6

oh but of course the cos(y) part dies after differentiating wrt x

[clever citrus]

right so at the end you get 0 right

[junior fable]

and in that case you get (y^6)'''' * junk(x)

which dies after 3 more bouts of differentiating wrt y

yeah you get 0

[clever citrus]

but I just dont know how the theorem would have helped for this question.

[junior fable]

idk lol

[clever citrus]

lmao

[worthy flax]

You can instead start with differentiating wrt x

Cause of the theorem

[clever citrus]

wait so I could have done f_{xyyyyyyyy} ?

[worthy flax]

Ye

[clever citrus]

damnnnn

[worthy flax]

That’s what the theorem lets ya do

[clever citrus]

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[topaz sinew]

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[clever citrus]

just wanted to confirm one more thing so the order doesnt matter right.

f_yyyyxyyyy = f_xyyyyyyyy = f_yyyyyyyyx right

[frail reef]

if you dont want your help request to be locked, open a new channel

[worthy flax]

Yes (under the correct conditions)

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[cobalt helm]

Hi I need help especially with part 1

[shell shard]

This is not math

[cobalt helm]

Oh it’s only math

K

[shell shard]

https://discord.gg/eexdsFw

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[cobalt berry]

428 influencers were asked which platform they enjoy the most. 205 influencers said tik tok while 267 said instagram. if 120 said they enjoy tik tok but not instagram, what percent of influencers said the following? round to the nearest tenth of a percent. complete the corresponding venn diagram, label your sets, show work.

is this correct

[cobalt berry]

i just need a simple yes or no cause im running out of time

[cursive patrol]

basically everything is wrong

[cobalt berry]

damn

can you help me set it up properly then

[topaz sinew]

@cobalt berry Has your question been resolved?

[untold matrix]

U = 428
T = 205
I = 267
T - (T int I) = 120

a) Both means T int I
205 - (T int I) = 120
200 + 5 - 100 - 20 = T int I
T int I = 85

For %
85/428 = 0.1985
Nearest Tenth
0.2
Or 20 %

b) nor T nor I means
COMP(T uni I)
T uni I = T + I - (T int I)
T uni I = 205 + 267 - 85
T uni I = 387
Now,
COMP(T uni I) = U - (T uni I)
= 428 - 387 = 41

For %
41/428 = 0.0957
Nearest Tenth
0.1
Or 10 %

Note:
Int means intersection
Uni means union
COMP() means complement

@cobalt berry

[cobalt berry]

can you do it visually, im not very good text wise

[icy sky]

stop giving out answers

[velvet ether]

bro spawned in the whole fandom

[velvet ether]

um so basically the 205 applies to EVERYTHING in tiktok, not just the exclusive one

and then the 120 is exclusive to tiktok

[untold matrix]

Wat? I thought we should help ppl

[velvet ether]

dont just

tell them the answer

[untold matrix]

Oh

Ok then from now

I will tell formula alone

[icy sky]

!noans

[untold matrix]

Or concept

[topaz sinew]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[untold matrix]

Right

[icy sky]

ok lets talk about this problem now

[untold matrix]

Yeah sure U go ahead

[icy sky]

so when it says x influencers enjoy tiktok, thats the whole circle for tiktok

the phrasing for this problem is bad because it says "enjoy the most" but lets brush over that

[velvet ether]

i think he might be afk

[icy sky]

this is the part where i stop and wait

great

[cobalt berry]

no im here

[cobalt berry]

uh huh

[icy sky]

you good so far?

[cobalt berry]

yes

[icy sky]

cool

so in the part, "120 said they enjoy tiktok but not instagram", which area is that?

i.e. left, middle, right, or outside?

[cobalt berry]

left

[icy sky]

good

so we know the tiktok circle contains the sections left and middle

we know the value of left and left+middle

so we can find the value of middle

we are filling the venn diagram in section by section now

whats the value of middle?

[cobalt berry]

85?

[icy sky]

yup

do you see how to get right?

[untold matrix]

I'm gonna copy ur teacher style

Muahahahahaha

[icy sky]

good

you still there?

[cobalt berry]

yes

sorry im just on a big time crunch

[icy sky]

im making this as quick as i can while trying to make you understand the numbers

[untold matrix]

Same here eating doritos

[icy sky]

yes or no?

[cobalt berry]

yes

[icy sky]

what did you get

[cobalt berry]

uhhh

267

[icy sky]

remember (middle)+(right)=(people who enjoy instagram)

[cobalt berry]

yes

[icy sky]

we know (middle) and we know (people who enjoy instagram)

so we can find (right)

[cobalt berry]

ran out of time but i got this question finished

!close

i forgor

[icy sky]

do you still want to figure it out?

if not, use .close

[outer portal]

.close

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[deep crow]

How do i dothis

[burnt swift]

do comparison test with 1/e^x

[deep crow]

why doesnt 1/x work?

[burnt swift]

is this a pauls online note example?

[deep crow]

yes

i dont understand the explanation very much tho

[burnt swift]

you want to compare 1/(x+e^x) to 1/x instead of comparing it to 1/(e^x)?

[deep crow]

yes

[burnt swift]

but e^x grows exponentially faster

why choose 1/x

[deep crow]

oh so we usually choose the one with mire weight to compare?

[deep crow]

I thought choosing 1/x would be easier cause using the p-series would be easier to determine that that integral fiverges

[topaz sinew]

@deep crow Has your question been resolved?

[raven sparrow]

The integrand is smaller than 1/x for any x. Comparing it to 1/x won't help, because all you'll know is that the integral is smaller than the integral of 1/x, which even though it diverges could mean anything.

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[deep crow]

thanks

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[prime nova]

Hi, I'm stuck with a dumb determinant problem.

[prime nova]

This is the problem. Let me first tel what I did, because I just can believe how utterly confused I am by this thing.

[serene atlas]

Do you know how to multiply matrices

[prime nova]

Yes, that's not the part that's confusing me. Let me first explain. In the end it is surely a dumb mistake, but for now I don't get it.

[serene atlas]

Yea alright

I'm listening

[prime nova]

Not problem findind det(AB), it is equal to 2530. No problem either with det(A), it is 46. Therefore, det(B) should be 55. The thing is that I'm probably applying this one method wrong for finding the determinant. It relies on this theorem that is in the book the problem is from:

And this is what I did. The thing is that for some reason there seems to be two possible ways to proceed from this point on. (1) I can multiply row 2 by -16 and add row 3: that leads to det(B)=-55, that's wrong. (2) I can mutiply row 2 by 16 and subtract row 3: that lead to det(B)=55, that's right.

My problem is that I don't get why one option gives a right answer, and the other doesn't.

[prime nova]

That's basically it.

[serene atlas]

I see

Let me see

[serene atlas]

What are the matrices you get by doing these operations?

(1) (-16)R2+R3 => R3
(2) R3 - (16)R2 => R3

[serene atlas]

Is this what you mean?

(2) 16R2-R3 => R3
Or this is what you mean

[prime nova]

yeah

[serene atlas]

Right

[prime nova]

Yeah, I mean (1) (-16)R2+R3 => R3 and (2) 16R2-R3 => R3

[serene atlas]

I think subtracting rows is not allowed

[serene atlas]

In this case you have Multiplyed R3 with (-1)

So your det will be (-1)|A|

Are you following?

[prime nova]

Yeah, I follow you.

[serene atlas]

You basically you did 2 Row Operations

(-1)R3 => R3

And

16R2+R3 => R3

[prime nova]

Ohhh, I see. Thank you very much. I did not notice I was doing two row operations at once.

[serene atlas]

Yea yw

[prime nova]

Thank you very much! ❤️ A silly mistake, haha.

[serene atlas]

Happens

[prime nova]

.close

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[serene atlas]

Yw

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[plain mantle]

Can anyone help me with physics? I need to know to determine the reaction force of R2 so i can solve for R3 and R1.

[neon iron]

stationary?

take moments around R1

n u can find Fr2

[plain mantle]

How? can you teach me

[neon iron]

do u know what moments are?

[plain mantle]

distance and moment? like pounds?

ft and pounds?

M=F×d?

[neon iron]

Moments are a measure of rotation in a system

yes

Force * perpendicular distance to centre of rotation

[plain mantle]

Is this the formula? M=F x d?

[neon iron]

yes

in a stationary system, the clockwise and anti-clockwise moments should be equal

[plain mantle]

Okay i will try to answer, Tell me if it's correct.

[topaz sinew]

@plain mantle Has your question been resolved?

[plain mantle]

R1

Moment for 240lb

Distance from R1 = 8ft + 9ft = 17ft

Moment = 240lb x 17ft = 4080lb-ft > counterclockwise = positive

Moment for 960lb

Distance R1 = 3ft

Moment = 960 x 3ft = 2880lb-ft > counterclockwise = positive

How should i determine the reaction of R1

[topaz sinew]

@plain mantle Has your question been resolved?

[plain mantle]

Help me pls

[topaz sinew]

@plain mantle Has your question been resolved?

[topaz sinew]

@plain mantle Has your question been resolved?

[sly sleet]

@plain mantle

What is that?

Are there hinges?

[plain mantle]

Equillibrium Static

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[plain mantle]

.reopen

[topaz sinew]

✅

[plain mantle]

Formula like this can you help

Can you solve the problem? please

[fickle sorrel]

There is a physics Discord as well: https://discord.gg/BYDez64V

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[high fjord]

why does the ln2 not matter here?

[high fjord]

why can it be simplified into an expression without ln2 i mean

[icy sky]

it gets absorbed by the +C

[simple knot]

C=c+ln2

[high fjord]

Ah okay thanks

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[clever citrus]

Is this correct?

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[solid fog]

For the first one, whats confusing you

Do you not understand the translations visually? Whats up

[mild hedge]

idk why i am wrong, and the class never taught it

[mild hedge]

i did it in my head and it made sense/worked in my head

@solid fog is it option 2

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[hot flame]

i ended up with tan (173.3333..) so -0.117

this key has something way different

[topaz sinew]

@hot flame Has your question been resolved?

[outer portal]

tan(173 1/3) being about -0.117 is correct

think the key just made a mistake

[hot flame]

thanks\

i thought i was missing something

.close

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[lunar maple]

hello im doing calc one problem solving for integration there is this question that is confusing me im not gonna lie im stuck on it for like more than an hour so if some one can tell me what the hell is going on ill be gratefull im using solvely to help me but i think it confused me more than helped me

[sterile tangle]

which part are you stuck on?

[lunar maple]

Part a

[sterile tangle]

Do you know how to find the area of A

B should be simple since you can calculate it from the values on the graph

[lunar maple]

Ok so do we use integration to find c then subtract it from the whole thing and when you have Ab you will know what Aa is

[sterile tangle]

Uh

Theres a simpler way

So you know how we can use integration with respect to x (which I will abbreviate as wrt) in order to find the area under the curve right

We can also integrate wrt y in order to find the area between the curve and the y-axis

So rewrite y = arcsin(x) in terms of y

[lunar maple]

I’m not gonna lie I forgot how

Do we use the rule of the inverse

[sterile tangle]

take the inverse is what I would say

yep

sin(y) = x

then integrate that with your bounds of integration (Make sure to use y-values and not x)

[lunar maple]

So just one

[lunar maple]

Is this right I hope I didn’t waste your time lol sorry I’m just tired that’s why

[sterile tangle]

I think so

lemme check

yeah its right

[lunar maple]

Yk that just makes sense cause we just flipped it and finding the area under the curve from left like if y was the bottom

[sterile tangle]

mhm

[lunar maple]

@sterile tangle I saw another answer where they solve it by finding the difference between the upper portion of the region of function and the upper portion of region B

[sterile tangle]

yeah ig you could do that

but thats kind of what we did with the integration

you basically take the area of the entire thing above the grapth

then subtract the area of B above the graph to get A

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@lunar maple Has your question been resolved?

[lunar maple]

@sterile tangle would you be able to explain it thanks

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[uneven vault]

im doing some previous IOQM papers

pls tell me how you do this

<@&286206848099549185>

damn

this server

.close

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[icy sky]

it was two minutes but ok

[neon iron]

lmfao

bro cant wait

[icy sky]

i didnt even have time to think about the problem

[uneven vault]

.reopen

[topaz sinew]

✅

[uneven vault]

im impatient

[acoustic tangle]

Looks like a telescope

[uneven vault]

lo

lol

[coarse herald]

write it as k + k+1 in the numerator

should help

[uneven vault]

ok

but how

[icy sky]

oh wait

[uneven vault]

uhh

[icy sky]

you should write it as (k+1)^2-k^2

[uneven vault]

hold up

[coarse tusk]

Jeez

[uneven vault]

but

[coarse herald]

oh this is basically 1/k^2 - 1/(k+1)^2

[uneven vault]

how will it help tho

[uneven vault]

bro

[coarse herald]

yes im slow

[coarse herald]

telescopic series the terms just cancel out

[uneven vault]

im impatient i told

[uneven vault]

idk that

[icy sky]

you arent ioqm is a contest with difficult problems

[uneven vault]

??

you arent?

[coarse herald]

uhh

[icy sky]

it was in response to this

[coarse tusk]

Then you can’t solve the question 🤭

[uneven vault]

bruh

i prolly know it

but dont know the naame

[uneven vault]

also if u want to annoy kids then annoy someone else

[icy sky]

write out the first few terms in the form 1/k^2-1/(k+1)^2

[uneven vault]

ok one sec

[icy sky]

see what happens

[coarse tusk]

Annoy kids?

[coarse herald]

yeah youll see they cancel out and only the first and last term remain

[uneven vault]

not helping

[uneven vault]

lemme see

[coarse herald]

after which it should be easy

[coarse tusk]

I would’ve helped if you didn’t act so entitled

[uneven vault]

ye ill try

-_-

oh

i got it

[icy sky]

so whats N?

[uneven vault]

99

i think

[coarse herald]

great

[uneven vault]

i got

[icy sky]

good

[coarse herald]

yeah

[uneven vault]

(N+1)^2=10000

ye

ok

thx :))

[icy sky]

be patient next time please

[uneven vault]

yeah

sure

.close

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[sage sonnet]

Need help with the question

[sage sonnet]

I just can't solve 'to prove' questions in inequalities

[deft granite]

is this for reals or integers?

[cursive patrol]

it is true for all reals (edit: equality case matters oops)

[neon iron]

missing an equality as well?

[deft granite]

Yeah, if a = b then this is false

[deft granite]

Do you understand why it's true intuitively/informally? (assuming a and b are different)

[topaz sinew]

@sage sonnet Has your question been resolved?

[sage sonnet]

No

[sage sonnet]

Idk, this is the whole question

[deft granite]

Is there a note somewhere that a and b are not equal?

[sage sonnet]

No

[deft granite]

well, let's assume it's an oversight and it was implied

[sage sonnet]

I guess a and b are different

[deft granite]

Let's rearrange a and b so that a < b. Does that make our job easier?

[sage sonnet]

Ya

[sage sonnet]

But there can also be the possibility that a>b

[wicked bolt]

yea but in that case we can just swap a and b

the inequality is symmetric

so without loss of generality we can assume one is larger (or equal to) the other

[sage sonnet]

Ok

What's the next step

<@&286206848099549185>

[wicked bolt]

what type of inequalities have you learned?

[sage sonnet]

What type??

Do they have types?

🥲

[wicked bolt]

no they're just theorems

like have you learned the rearrangement inequality?

[sage sonnet]

Ya

[wicked bolt]

yea, here you can apply that

[deft granite]

I would start by moving everything to one side so that there's a 0 on the other side

[trim owl]

Factorize both side first

And simplify

Assume a>b wlog

I think it will work out

[sage sonnet]

Can we use binomial expansion in this?

[deft granite]

that is possible, but there's an easier way

(what me and kirito said)

[wicked bolt]

theres a way you could use completing the squares here but weve already got through this other solution

[sage sonnet]

Idk what to to do after getting a zero on one side

I'm dumbo

[deft granite]

So you got $a^4 + b^4 - a^3 b - ab^3 > 0$, right?

[thorny flame]

fix error

[sage sonnet]

Ya

[deft granite]

Are you familiar with solving an inequality where one side is a product and another one is zero? like $(x - 1)(x - 3) > 0$

[thorny flame]

fix error

[sage sonnet]

Yess

By wavy curve method

[deft granite]

Right

Is there a way to turn the left side into a product we can work with?

[molten musk]

bruh wavy curve method is for inferiors

[sage sonnet]

So what to do then??

Which method to use?

[molten musk]

it's not wrong to use any method

[sage sonnet]

here i got the solution but can't understand what they did

can someone explain

[sick escarp]

a^4+b^4 > a^3b+ab^3 is the same as (a^4+b^4)-(a^3b+ab^3)>0

after that they proceeded to factorize it

they factored out b^2 as it is a square, and then completed the square of what was left inside the last term

assuming that b is non-zero, this term will always be greater than 3/4

so the whole product will always be greater than 0

[trim owl]

Take in 1 side and factorize

[trim owl]

U cam also divide b4 on both side

Assume t as a/b

Then take 1 to left and t to right

[topaz sinew]

@sage sonnet Has your question been resolved?

[quaint belfry]

As soon as you get to $\left(a - b \right) \left(a^3 - b^3 \right)$, you are done. If $a > b$, then $a - b > 0$ and $a^3 - b^3 > 0$ so the product is positive. Similarly, if $a < b$, then $ a - b < 0$ and $ a^3 - b^3 < 0$, so multiplying two negatives will give you a positive.

[thorny flame]

PowerUp

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[visual rain]

is my formula correct?

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@visual rain Has your question been resolved?

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[timid mica]

Yo guys I need help on how they rearranged the formula to turn it into tan^2a - 4tana + 3 = 0. Except that for me the original formula im using is y=xtan(a) - gx/2u^2Cos^2(a)

Since that’s what we were taught but I can’t rearrange it💀🙏

[placid lance]

$\frac1{\cos^2\alpha}=\sec^2\alpha$

[thorny flame]

nameless individual

[timid mica]

Omfg

Im so retarted

Ive been stuck on that for so long

The identities😭🙏

Tysm dude🙏

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[topaz sinew]

@neon iron Has your question been resolved?

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@neon iron Has your question been resolved?

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[coarse tusk]

!status

[topaz sinew]

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

[neon iron]

What properties should I use ?

[neon iron]

1

[coarse tusk]

You’ll need fermat’s little theorem

And knowledge about how to use modular arithmetic

$a^{23}\equiv a$ (mod 23)

[neon iron]

Okay I will come back

[thorny flame]

kheerii

[coarse tusk]

This is what fermat’s little theorem says, utilising the fact that 23 is prime

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@neon iron Has your question been resolved?

[neon iron]

.close

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[noble olive]

Nikish is going to install five new game apps on her phone. She has shortlisted two word games, 5 quizzes and 16 saga games. Nikish wants to have at least one of each type of game. How many different selections of apps could Nimish possibly choose?

[long stirrup]

you gotta manually find partitions of 5 in to 3

3 1 1
2 2 1

that's probably it

[neon iron]

Is it compulsory that she have to install 5 games ? Or it can be less ?

[noble olive]

can you elaborate?

[long stirrup]

i mean there's 2 cases

[noble olive]

she has to install 5

[long stirrup]

you identify them

and then start adding

it's like a stupid problem, just a ton of calculations

[noble olive]

well i still don't get it but i'll see what i can do

[long stirrup]

3S 1Q 1W
3Q 1W 1S
2S 2Q 1W
2W 2Q 1S
2S 2W 1Q

[noble olive]

ah, i see what you mean

i'll try solving it now

[fair thorn]

i think wolframalpha can do that

nvm

[noble olive]

that comes out to be 9080

which is the same as the markscheme

ty

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[neon iron]

What's my error?

[neon iron]

Calculate m + n + p
given that the remainder is 3x + 5 and the sum of coefficients of the quotient is 7

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[cosmic siren]

how do I know if the distribution of X is normal or not?

[cosmic siren]

oop wait

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[clear wing]

(cos20+isin20) ^5 / (cos30+isin30)^3

[clear wing]

Can anyone tell me how to solve this problem by using only De Movier and Polar Form

[stable rivet]

Remember that (cos(x)+isin(x))^n = cos(xn)+isin(xn)

[clear wing]

[cos5(20) + isin5(20)] / [cos3(30) + isin3(30)]

cos100 + isin100 / cos 90 + isin90

right?

[brave coral]

yes

[stable rivet]

Yea

Do you remember how to divide polar coordinates

[coarse tusk]

the easiest way is just to write the terms as polar coordinates

[clear wing]

cos (90 + 10) + i sin (90 + 10 ) / 0 + i(1)

[clear wing]

no

[coarse tusk]

$e^{i\theta}:=\cos\theta+i\sin\theta$

[thorny flame]

kheerii

[coarse tusk]

normal exponent rules work if you write it like this

[clear wing]

nooo

do not use elure form

[stable rivet]

Could be precalc tho and they don’t know that form

[clear wing]

i have not read it yet

[coarse tusk]

oh, weird

[stable rivet]

Divide the magnitudes and subtract the angle of the denominator from the angle of the numerator

[clear wing]

hmmm

[coarse tusk]

$\frac{\cos 100^\circ+i\sin 100^\circ}{\cos 90^\circ+i\sin 90^\circ}$

[thorny flame]

kheerii

[clear wing]

ok

[coarse tusk]

you got this far yes?

[clear wing]

ya

[coarse tusk]

$\frac1{\cos\theta+i\sin\theta}=\frac{\cos\theta-i\sin\theta}{(\cos\theta+i\sin\theta)(\cos\theta-i\sin\theta)}=...$

[thorny flame]

kheerii

[clear wing]

?

[coarse tusk]

multiply the numerator and denominator of 1/(cosx+isinx) by its complex conjugate

[coarse tusk]

that's what I did here

[clear wing]

ok

[coarse tusk]

now simplify this, you'll see a nice identity

[clear wing]

ok

[coarse tusk]

and then you can use de moivre to solve the rest

[clear wing]

but we already use it

[coarse tusk]

sorry, I meant you can just add the arguments after

in the product

[clear wing]

ok

wait please

i am solving

what result you guys got

because my is not matching

got it thanks

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@slow bloom Has your question been resolved?

[slow bloom]

<@&286206848099549185>

[sterile finch]

What type of class is this for?

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[steep nebula]

I have some questions about the Sucessor function

[steep nebula]

$S(n) = n + 1$

[thorny flame]

Max-Cat

[steep nebula]

if so would $S( \aleph_0 ) = \omega$

[thorny flame]

Max-Cat

[steep nebula]

since omega is the number that comes after aleph null

I mean, $\aleph_0$ is the set cointaining all natural numbers

[thorny flame]

Max-Cat

[steep nebula]

woudn't it mean that $S(\aleph_0) \in \aleph_0$ ?

[thorny flame]

Max-Cat

[carmine marten]

There's a few things going in here

[carmine marten]

This definition only makes sense for natural numbers

[steep nebula]

(I'm sorry if its actually something basic)

[carmine marten]

You can't apply it to the set of natural numbers

Secondly both aleph0 and ω are, as sets, the same (I believe this is standard)

The distinction we make is in terms of understanding infinity

aleph0 is a cardinality, that is the size of a set, while ω is an ordinal which is a way to continue counting after infinity (someone else could probably explain this a little better)

So we count 1, 2, 3, 4, ... infinitely far, ω, ω+1, ω+2 etc.

[steep nebula]

so Aleph0 is the amount of numbers in N

[carmine marten]

yes

BUT ω+1, ω+2 have the same cardinality as ω

[steep nebula]

omega times omega; omega^omega and so on

[carmine marten]

yes

[steep nebula]

I find it hard because any cardinal number can be used as an ordinal, "there are 3 apples; this one is the third apple"

[carmine marten]

Unfortunately for intuition yeah, when finite they coincide

[steep nebula]

OH!

you can say "there are aleph0 apples" but you can never point to a final one

[carmine marten]

yes

[steep nebula]

and you can't mention "the last apple" since the meaning of "aleph_0 apples" is that for any apple you choose there are still infinitely more!

[carmine marten]

really ordinals introduce a notion of order (hence the name? maybe?) where you are allowed to go infinitely far and still point to a 'infinity'th element, but cardinality is just about sets without this

[carmine marten]

regardless trying to apply S as 'adding one' you need to keep in mind what addition is supposed to mean here

because the definitions are different for cardinals and ordinals

[steep nebula]

yeah I see

[carmine marten]

Also why did you think ω was an element of aleph0?

[carmine marten]

.

[steep nebula]

I thought about numbers as sets, 1 = { o } ; 2 = { o ; o } ...
and S(n) as adding one more element to the set containing n elements.
If aleph0 was a set of infinitely many elements, adding one more would still make it infinite.

again, sorry if this painfully obvious

[carmine marten]

What is your definition of aleph_0 actually

That's all true

It's just I realised there's at least two used at least to introduce these ideas

In one case using $\in$ would be correct (I was just thinking of the other)

[thorny flame]

Edward II

[steep nebula]

could you explain me both? As to better understand what you're thinking?

[carmine marten]

What I was thinking, and I think this is the more common definition in set theory because it actually works the way we do it nowadays, is that cardinals are certain ordinals, in which case $\aleph_0=\omega={0,1,2,3,\dots}$ (i.e. the set of natural numbers) In this case, $S(\aleph_0)\in\aleph_0$ would be incorrect regardless of whether you say $S(\aleph_0)=\aleph_0$ or $S(\aleph_0)=\omega$ (as you originally did)

[thorny flame]

Edward II

[carmine marten]

The other way, which is the nicer way, is to say that a cardinal is a collection of sets that all have the same size

In which case $\aleph_0=\Big{{0,1,2,3,\dots}, \omega+1, {2,4,6,\dots}, \dots\Big}$

[thorny flame]

Edward II

[carmine marten]

And I suppose saying $S(\aleph_0)\in\aleph_0$ would probably still be wrong, but more based on misunderstanding and consistency with notation rather than saying 'the number after something infinite (whatever that means) is finite', which is what it would mean in the first definition

[thorny flame]

Edward II

[steep nebula]

so let me see if I got it

a number n is defined as a set containing all possible sets with n elements

[carmine marten]

I abused notation, because it's not actually a set, but that's on me for not explaining

[steep nebula]

$3 = \Big{{A, B, C}, {3, 1, 4,} \dots \Big}$

[carmine marten]

you need to use \{ to actually get them to show

[steep nebula]

annyways, you get the idea

[carmine marten]

yes (it's not a set which is why the other definition is also used)

[steep nebula]

N is the set containing all natural numbers, all diferent numbers that have diferent amounts of elements

since we can always put another element there is no limit to how big a number can get, therefore N is a set with infinitely many elements

in another way, it is impossible to point to the last element, with last meaning "no thing after"

[carmine marten]

I'm going to cut in here because I didn't actually explain this

[steep nebula]

please do

[carmine marten]

ω also has no last element

The reason we treat it differently is that ω+1 does, and is the same cardinality as ω

[steep nebula]

wait, I'll get there

I'll try to

the amount of diferent numbers in N is Aleph0, it is not the set cointaining all numbers, as that is N, but is the amount of numbers

it's not which elements, but how many elements, it's not a set of numbers, but a quantity

as the cardinal that it is, it can be represented as a set with Aleph0 elements

one element for each natural number

N = { 1, 2, 3, 4 ... }
A0 = { {1}, {2}, {3} ... }

am I right until here?

[carmine marten]

Maybe? I'm not entirely sure people who actually study this would be happy with A0 = { {1}, {2}, {3} ... }? I personally don't see anything wrong with that as long as you remember A0 is really something else, and that this choice of using elements {1}, {2}, ... was arbitrary

[carmine marten]

also 'not the set containing all numbers' is using this more intuitive (at least to me) definition, but that's fine

[steep nebula]

Now the stupid question that started this all: "what happens if we add more?"

How can we add another element?

after all, Aleph0 as no ending element that we can follow up with a new one

we could put it in the beggining tho

but it would mean that the new would be the first element, the first would become the second element

like, a wave, objects next to eachother pushing each other

this "wave" would go on forever

the infinity'th element would be pushed aside

that new place cannot be any number, as it would be before the inf'th elementh

this new place the inf'th got pushed to is omega

right? does this analogy make sense?

do you want me to try explaining my thoughts in another way?

[carmine marten]

I don't really get it

So we have A0= {1,2,3,4, ...} (say)

Then we add another element which I will label as ω for reasons

and have {1,2,3,4, ... ω}

Υοu're saying we could have instead done {ω, 1, 2,3, ...}

so 1 is now second, etc?

[steep nebula]

yes, and "omega" is 1 and so on

[carmine marten]

but then what does this mean?

The infitieth element was ω, there's no longer an infinitieth element to move

[steep nebula]

ok, let me try to explain it better (prob can't)

imagine we live in a flat never ending, plane

and we have a line of boxes, one next to the other

there are infinitely many, the line never stops, it just gets smaller and smaller into the horizon

if we add a new box to the beggining of the line and push the whole line of boxes, so that the new one stays in the place of the first, the first stays in place of the second and so on

what would happen to the box in the horizon?

[carmine marten]

in this example it stays where it is

there's no box right before it that gets moved to the horizon

any box you pick before it, there would be another box after it that's not at the horizon

(infinity is weird and confusing)

[steep nebula]

i need to diner now

be right back

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[dawn shell]

Hi, I was solving a calc problem yesterday and I checked the solution today and am a bit confused, the question was to calculate the area above the cone and beneath the sphere for the first quadrant

[thorny flame]

Mephisto

Mephisto

[dawn shell]

The question was to solve for the volume V using a triple integral and spherical coordinats
so I thought:

$\theta \in[0,\pi/2]\
\phi \in [\pi/4 , \pi/2]\
r \in [0 , 2sin(\phi)]$

[thorny flame]

Mephisto

[dawn shell]

but in the solution phi goes from 0 to pi/4
and I don't understand why, considering we're looking for the volume above the cone, not beneath it

and r goes from 0 to 2cos(phi) instead of 2sin(phi)

ohhh is it becuase they defined phi differently?

<@&286206848099549185> can anyone confirm?

[topaz sinew]

@dawn shell Has your question been resolved?

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@dawn shell Has your question been resolved?

[old sparrow]

you mean the volume?

[dawn shell]

yeah sorry

[old sparrow]

you definitely don't want to use a triple integral unless there is something else you're not telling me like the fact that there is a scalar field

also does "beneath the sphere" mean inside the sphere?

[dawn shell]

yeah I should've been more clear

the volume enclosed by the sphere and the cone

above the cone

[old sparrow]

enclosed by the sphere and above the cone

[dawn shell]

yeah that's a better way to put it

[dawn shell]

the question explicitly stated to use a triple integral in spherical coordinats

I assume using flux is a better way of solving this?

[old sparrow]

why do we even need flux?

can you just send the original question please

[dawn shell]

it's in dutch lemme translate it really quickly

jezus

Let ( D \subset \mathbb{R}^3 ) be the region in the first octant ((x, y, z \geq 0)), lying above the cone ( z^2 = x^2 + y^2 ) and below the sphere ( x^2 + y^2 + z^2 = 2z ). Denote by ( V ) the volume of ( D ).

(a) Write ( V ) as a triple integral in cylindrical coordinates.

(b) Write ( V ) as a triple integral in spherical coordinates.

(c) Calculate ( V ) using one of the above integrals.

[thorny flame]

Mephisto

[old sparrow]

tbh i have no idea why one would use spherical coordinates

it makes it a complete mess

[dawn shell]

I mean

[old sparrow]

not to mention triple integral when you only have three dimensions

[dawn shell]

you could just use cartesian and rotate it around the z axis

but it's one of the exam questions from previous years so I'm just making sure I can do it

[old sparrow]

flux only works on R^3 to R^3

I'll let someone else have a look because I am clearly missing something

[dawn shell]

I guess it is an inefficient way to solve this question

I'm guessing they just want to know you know how to use different coordinate systems and triple integrals

[old sparrow]

Done it in cylindrical if you can call it cylindrical

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[dawn shell]

hold on

[dawn shell]

I had this, with [pi/4 , pi/2] as bounds for phi and [0,2sin(phi)] for r

[topaz sinew]

@dawn shell Has your question been resolved?

[raven sparrow]

Triple integrals are standard for computing volumes, so it should look somewhat like that, yes.

Is that the solution you're having doubts about?

[dawn shell]

yeah so I had 2sin(phi) in the first integral

I mean third

and pi/4 to pi/2 for the second one

and I wasn't sure why, my hipothesis was that they defined phi in a different way

[old sparrow]

oh i was looking at R^2 -> R

mb

[raven sparrow]

Phi is usually the angle from the z axis, not the angle made with the XY plane.

[dawn shell]

ohh alright, is this the norm?

[raven sparrow]

I think so , yes

[dawn shell]

I usually take the angle with the xy-plane so z = rsin(phi)

[raven sparrow]

Although TBf it will yield the same result provided the bounds are actually right

[dawn shell]

alright that's good to know

one more question I'd have

There is an extra question provided linked with further chapters

where you should be able to calculate this volume using flux

but I'm not sure what vector field I should use to start calculating that

[raven sparrow]

I suppose you could try some vector field whose divergence is 1 everywhere (so that the dV integral calculates the volume)

But it seems more work to compute the flux on the surface

[dawn shell]

oh because we'd have to calculate the surface area of this volume first?

[raven sparrow]

Yeah well it depends what it was asking exactly.

You could compute the flux using the volume, but you could also compute the volume using the flux. It depends which you're looking for

[dawn shell]

the second one

just another way to calculate this volume

[raven sparrow]

Yeah. But again you'd then have to compute the flux manually, which might turn out to be fine I haven't really looked into it

The vector field (x,y,z) might be simple enough

Divergence is 3, so you can divide the flux by 3 and get the desired volume

[dawn shell]

alright I'

I'll try that

hold on

if I'm working in spherical coordinates

do I just (r,theta,phi) ?

[raven sparrow]

It doesn't matter since you'll be computing a double integral for the flux, not a triple one

[dawn shell]

ah okay

ohhh yeah that's right

the whole stokes theorem

alright, thanks for now, I'll try to compute everything on my own

.close

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[topaz sinew]

@neon iron Has your question been resolved?

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@neon iron Has your question been resolved?

[verbal crater]

i mean does it have to be an integer or floating

[glacial adder]

you can take the binomial theorem formula and plug in 1/2 for n

computing the first few terms should be accurate enough

[verbal crater]

! is repeated multiplication

so like 3! = 3 * 2 * 1

thats true

you could just google square root algorithms

and see which one you like best

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[keen night]

Can I improve my skills here

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@keen night Has your question been resolved?

[cedar wagon]

It is a help server so ot depends what do you mean by improving

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[verbal ginkgo]

When the question says X>-1, do I sub in the positive or negative square root of √(X+5) as U in the final line?

[cedar wagon]

I dont understand what you mean by " positive or negztive sqrt"

[verbal ginkgo]

Cause I used the sub u^2 is X +5

If I solve for U, it's +-√(X+5)

[cedar wagon]

Oh ic

Since there is an absolute value i think it wont be a problem

[verbal ginkgo]

Oh alright thx

[coarse tusk]

No, that’s not the reason

[verbal ginkgo]

Oh

[coarse tusk]

The real substitution you’re doing here is $u=\sqrt{x+5}$

[thorny flame]

kheerii

[coarse tusk]

If you chose $u=-\sqrt{x+5}$ then the $\sqrt{x+5}$ in the denominator would be replaced by -u and not u

[thorny flame]

kheerii

[coarse tusk]

Which would make the whole thing work

[verbal ginkgo]

Ohhhhh

Alright that makes sense thx

Yeah gotcha thx

[coarse tusk]

You should write your substitutions explicitly like u=sqrt(x+5) so you don’t have confusions like this

[verbal ginkgo]

Okok

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[burnt swift]

For how many values of the digit $A$ is it true that $63$ is divisible by $A$ and $273{,}1A2$ is divisible by $4$?

[thorny flame]

938c2cc0dcc05f2b68c4287040cfcf71

[old cloak]

100 is divisible by 4 so we can discard the rest

factor 63 and check

[burnt swift]

factor 63 and check?

,w factors of 63

[thorny flame]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[old cloak]

don't forget 1

[burnt swift]

1, 3 , 7

is the prime facotirzation or

[old cloak]

prime factorization and 1 up to 9

because it is a digit

[burnt swift]

prime factorization and 1 up to 9? because its a digit?

[restive inlet]

wrong to use prime here

[burnt swift]

1 is not a prime

but

[old cloak]

combinations of primes too

[burnt swift]

this is all the factors of 63

[restive inlet]

you want the one digit factors of 63

[burnt swift]

why one digit

I dont get it

A is one digit?

[old cloak]

it states that

yes

the digit A

[burnt swift]

sure

so A is 1 to 9 and div by 63

we need divisibility rules for 1 3 and 7 to match or

[old cloak]

divides into

not div by

[coarse tusk]

First recall the divisibility rule for 4

[burnt swift]

well

can u define it

I dont remember

[old cloak]

also for 9 because 63 is div by 3^2

[coarse tusk]

The last two digits must be divisible by 4 for the whole number to be divisible by 4

[burnt swift]

fair

last two digits div 4 implies whole num is div 4

ok

[old cloak]

hence

[burnt swift]

2731A2

fair

[old cloak]

so check for A = 1, 3, 7, 9

[burnt swift]

can someone define div 3 and div 7 and div 9

[old cloak]

div 3 is sum of digits

recursively

[burnt swift]

can u define it

[old cloak]

but like it's not needed here

[burnt swift]

sum of all digits?

why

[old cloak]

if the sum of all digits of a number are divisible by 3 then the number itself is also divisible

[burnt swift]

by 3

well

[old cloak]

yes

[burnt swift]

3 | 2+7+3+1+A+2

[old cloak]

but you don't need it, we already have the prime factors from which you can construct all possible digits A that divide into 63

no

[old cloak]

.

[burnt swift]

how

[old cloak]

.

we are checking for divisibility by 4

[burnt swift]

4 | A2

[old cloak]

yes

[old cloak]

try them and done

[burnt swift]

how

[median temple]

273100 is divisible by 4
so all you need is the numbers for which A2 is divisible by 4 are (1, 3, 5, 7, 9)
and the numbers that satisfy 63 is divisible by A are (1, 3, 7, 9)
so the numbers that satisfy both would be 1, 3, 7, 9

[old cloak]

substitute

[burnt swift]

1,3,5,7,9?

why

why

[old cloak]

can you divide 2 digit numbers by 4

[burnt swift]

100 is divisible by 4 so we can discard the rest

[burnt swift]

unsure

[old cloak]

yes the rest of the number, it has been discarded and we are left with A2

divisible by 4 is twice divisible by 2

can you divide by 2

[burnt swift]

The divisibility rule for 2 states that any number with the last digit of 0, 2, 4, 6, or 8 will be divisible by 2. Simply put, any even number (numbers that end in 0, 2, 4, 6, or 8) is divisible by 2. If the number is not an even number, it is not divisible by two.

[old cloak]

are we trolling

[burnt swift]

A needs to be even

[burnt swift]

wdym

[median temple]

i think this is all you need to understand

ion think u need divisibility rules and whatnot

[burnt swift]

273100 is divisible by 4
so all you need is the numbers for which A2 is divisible by 4 are (1, 3, 5, 7, 9)
and the numbers that satisfy 63 is divisible by A are (1, 3, 7, 9)
so the numbers that satisfy both would be 1, 3, 7, 9

[old cloak]

you can't tell me that you don't know how to divide 2 digit numbers by 4 when you know limits

[burnt swift]

well, divisibility is not taught in calculus

[burnt swift]

so all you need is the numbers for which A2 is divisible by 4 are (1, 3, 5, 7, 9)
how did you find (1, 3, 5, 7, 9)
btw

[coarse tusk]

You don’t have to accuse OP of trolling just because they don’t understand something…

[old cloak]

division is 3rd grade I'm sorry

[median temple]

because you can substitute numbers in for A2 that would be divisible by 4, and the first value for A that this works for is 1, so 12, and the next is 3, so 32, and then 5, and so on and so forth so theres a pattern so (1, 3, 5, 7, 9)

[coarse tusk]

Maybe they aren’t familiar with divisibility rules, maybe the A is tripping them up