help-13

[wraith dagger]

Ann

[tropic oxide]

??

[cedar kiln]

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[rustic flax]

what happens from the second last to the last line? Is it just by definition of total probability that intersection with all the parts of the partition renders the intersection E n B ?

[cedar kiln]

@rustic flax Has your question been resolved?

[vale gyro]

You mean the third line the the last?

[rustic flax]

Yeah

[vale gyro]

The total probability is a theorem not a definition (only btw)

But yes it is there are using this theorem. But there must be given that F_i is a partition of course

Partition of omega

[rustic flax]

Ah ok but is it actually correct to call that the law of total probability what they do on that line?

[vale gyro]

Yes. But you have to differentiate between definition and law/ theorem/… there exist many equivalents to theorem

But it means the same: a statement that derives of other statements or definitions

[rustic flax]

alright dude thanks but my point is, law of total probability is like P(A)=sum_k P(A | B_k)P(B_k) correct? What they do on the last line is not that

[vale gyro]

Well just use the definition of conditional probability to substitute this: P(A | B_k)

Than you can see it

I think

[rustic flax]

alright thanks

.close

[cedar kiln]

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[crimson sedge]

In the survey, each participant is asked: “Do you normally use drugs at least once a week?”. To achieve randomized responses, the participant is then instructed to respond as follows:
1. Flip a fair coin (i.e., 50% chance of heads and 50% chance of tails).
2. If tails, then respond truthfully.
3. If heads, then flip the coin again. Respond “Yes” if heads, and “No” if tails.

suppose b is the value of the true answer to the question, where b = 1 is “Yes” and b = 0 is “No”.
where b' = 1 means the participant responded “Yes” and b' = 0 means the participant responded “No” 

[crimson sedge]

To formalize this idea, let q [0,1] denote the true fraction of the population that frequently uses drugs. Compute the probability P(b = 0 | b' = 0) that the true answer is “No”, given that the survey response is “No”. Your answer should be in terms of q and constants.

Compute the probability P(b = 0 | b' = 0) that the true answer is “No”, given that the survey response is “No”. our answer should be in terms of q and constants.

my work so far:

P( b= 0 | b’ = 0) =  P( b’=0 | b = 0) * P(b = 0) / P(b’ = 0)
( ,75 * .5 ) / (.5) = 0.75

can someone verify this? I'm stuck because I'm not sure where q comes into play. How can I write my answer in terms of q

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

<@&286206848099549185>

I think i have the right answer. I just don't have it written out properly. I found the probability using baye's rule (i'm sure I got that part right) but I'm stuck with how to write my answer in terms of q

Please help

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

No

[crimson sedge]

😭

its a super simple question

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

how long can i have this chat

[main needle]

P(b=0) is q instead of 0.5

since it's the real one and not a coinflip

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

wow you king

sorry

i didn't see your messagge

P( b= 0 | b’ = 0) =  P( b’=0 | b = 0) * P(b = 0) / P(b’ = 0)
P( q = 0 | b’ = 0) =  P( b’= 0 | q = 0) * P(q = 0 ) / P(b’ = 0)

[crimson sedge]

so like this ^

[main needle]

no, I mean P(b=0)=q

[crimson sedge]

oh

okay

P( b= 0 | b’ = 0) =  P( b’=0 | b = 0) * P(b = 0) / P(b’ = 0)
P( q| b’ = 0) =  P( b’= 0 | q) * P(q ) / P(b’ = 0) where q is P(b = 0)

[crimson sedge]

like this

sorry im an idiot

[main needle]

uh q shouldn't show up inside P(...), it's just a value

P( b= 0 | b’ = 0) = P( b’=0 | b = 0) * P(b = 0) / P(b’ = 0) is fine

[crimson sedge]

im confused then

how do i use q then

it wants final answer in terms of q

[main needle]

one of those terms on the right is q

[crimson sedge]

gotchuuu

let me do it again

P( b= 0 | b’ = 0) =  P( b’=0 | b = 0) * P(b = 0) / P(b’ = 0)
P( q| b’ = 0) =  P( b’= 0 | b = 0) * q / P(b’ = 0) where q is P(b = 0)

?

[main needle]

yea

[crimson sedge]

okay thank you

[crimson sedge]

did i do the bayes rule correctly

for a previous question these were the answers

[main needle]

sorry idk

[crimson sedge]

if i did bayes rule correctly?

no worries plurmorant

ill just wait

i dont want to confuse you with the picture i sent

its somewhat unrelated. I had to find the b'|b propabilities of where it was 0 and where it was 1 before. for that one i had to use bernouli distribution

for this problem the distribution was bayes

[sick bison]

Украинская

[crimson sedge]

Привіт

I’m ukrainian yes but ukrainian american

[sick bison]

👋

Ah nice

I am Russian-Polish-American

[crimson sedge]

Nice always great to meet eastern European Russians

[crimson sedge]

I am Antarctican

its pretty hard here to live with Polar bears and the low temperature

[crimson sedge]

Lol I just want help

[drifting marlin]

@crimson sedge still need help?

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

Yes please

Sorry it’s quite late for me

[cedar kiln]

@crimson sedge Has your question been resolved?

[drifting marlin]

@crimson sedge still need help? That was at 4am for me :p

[crimson sedge]

yes i do

it was 3AM for me

[crimson sedge]

this is the context

[crimson sedge]

this is the question

[drifting marlin]

So how far did you get again?

[crimson sedge]

this is my answer so far

I did Baye's rule

[crimson sedge]

this was my original answer. i just don't know how to write my answer in terms of q

P( b= 0 | b’ = 0) =  P( b’=0 | b = 0) * P(b = 0) / P(b’ = 0)
P( q = 0 | b’ = 0) =  P( b’= 0 | q = 0) * P(q = 0 ) / P(b’ = 0)

my original answer

To formalize this idea, let q [0,1] denote the true fraction of the population that frequently uses drugs. Compute the probability P(b = 0 | b' = 0) that the true answer is “No”, given that the survey response is “No”. Your answer should be in terms of q and constants.

Compute the probability P(b = 0 | b' = 0) that the true answer is “No”, given that the survey response is “No”. our answer should be in terms of q and constants.

the question

[drifting marlin]

So P(b=0 | b'=0) = P(b'=0 | b=0)*P(b=0)/P(b'=0) yes?

And you subbed in q=P(b=0)

[crimson sedge]

yes that is my bayes rule

yes

is that enough?

[drifting marlin]

Well they want the answer in terms of constants, so next is to remove the other two P() pieces

[crimson sedge]

so would P(b' = 0) = q'

[drifting marlin]

q' doesn't really mean anything

You can expand the denominator using the law of total probability

ie P(b'=0) = P(b'=0 | b=0)P(b=0) + P(b'=0 | b=1)P(b=1)

You know P(b=0) and P(b=1), which is nice

[crimson sedge]

would P( b = 1) also equal to q since q is constrained within 0 and 1

[drifting marlin]

No

[crimson sedge]

its just 0.5?

[drifting marlin]

P(b=0) + P(b=1) = 1 since there are only two values b can take on

[crimson sedge]

okay and in this case its 0.5

[drifting marlin]

No it isn't

P(b=0) = q

[crimson sedge]

ohhh

1-q

[drifting marlin]

Good

[crimson sedge]

loool

[drifting marlin]

Now, think about P(b'=0 | b=0)

If the true value of b is 1, what would have to happen for b' to be 0?

[crimson sedge]

they would have to have flipped a coin then gotten heads then have gotten tails

[drifting marlin]

And that happens with probability...

[crimson sedge]

I solved for P(b'0 | b = 0) it was 0.75

.5 + (.5 * .5)

[drifting marlin]

I made a typo

But yes that works

I meant P(b'=0 | b=1)

[crimson sedge]

ohhh

it would be 1 - 0.75

so 0.25

[drifting marlin]

So you end up with $$P(b=0 | b'=0) = \frac{0.75q}{0.75q + 0.25(1-q)}$$

[wraith dagger]

Steakanator

[drifting marlin]

Simplify the denominator a little and you're good to go

[crimson sedge]

blesss thank you steakanator

you explained it so nicely

i appreciate you

[drifting marlin]

All good

Best of luck

[crimson sedge]

.close

[cedar kiln]

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[dense wing]

nope

cause Q is closed

[cedar kiln]

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[cedar kiln]

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[sudden timber]

can I have a hint for this integral: Integral of (ln(x))^2

[main needle]

there's a way to do integration by parts that makes a simple integral

[cedar kiln]

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[wintry steeple]

So I'm finding a SVD of a matrix A which is a 2x3, has one singular value (A'A has one non zero eigenvalue), and we want to find P and Q such that P'AQ=D where D is the diagonal matrix with the singular values on the diagonal.
I'm at the last step when I need to find P but since I only have one singular value, I need to use gram schmidt to extend the one vector I have for P to an orthonormal basis.
I'm confused though because I'm not sure what vectors to use with gram schmidt here?

[cedar kiln]

@wintry steeple Has your question been resolved?

[wintry steeple]

<@&286206848099549185>

.close

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[magic fossil]

The question says "Consider three consecutive positive integers. If the third number is subtracted from the sum of first two numbers, the difference is 10. Find the numbers" The problem Is that I don't quite understand what consecutive numbers means even after I searched it up.

[cedar kiln]

@magic fossil Has your question been resolved?

[magic fossil]

<@&286206848099549185>

[regal oak]

mind if I help ?

[magic fossil]

yea

its easy anyway

@regal oak

I dont understand

but its easy

i mean

[regal oak]

consecutive numbers mean the numbers next to each other

like

2 3

3 4

[magic fossil]

what about 2,3 , 13

[regal oak]

they are neighbours

neighboring numbers are called consecutive
eg 2, 3
3,4
4,5
understood what consecutive is ?? @magic fossil

[magic fossil]

Then how do you solve this?

The diffrence can't be 10

[regal oak]

ok now I'll help you solve it

[regal oak]

it is

[magic fossil]

the sum of 1st and 2nd numbers are always going to be bigger then the third number

[regal oak]

ya

3rd number is subtracted from

for ease lets take the middle number ( 2nd one ) as x
ok ?

[magic fossil]

oh

Okay

I get it

Can i ask another questoin

question

[regal oak]

sure

[magic fossil]

I have to write "8 more than a number" as an algebraic experssion

I don't understand the "more than"

[regal oak]

x-1, x , x+1
take these 3 are the consecutive numbers u solve for x u get the 2nd number find the remaining 2

[regal oak]

x+8

[magic fossil]

oh

Okay

order matters?

[regal oak]

for the 1st q ?

[magic fossil]

q?

[regal oak]

question

[magic fossil]

No, This

It doesn't

Right?

[regal oak]

x+8 = 8+x

[magic fossil]

Yea...

Okay

One more question

last one

Can I ask?

[regal oak]

sure

[regal oak]

order does matter for this

[magic fossil]

I have to write, "Twenty more than three times a number is 20. " As an equation

And 20 + 3x = 20 right?

[regal oak]

20+3x=20

ya

[magic fossil]

Okay

Thanks

[regal oak]

np

u can dm me if u get further doubts

on this

[magic fossil]

Okay

.close

[cedar kiln]

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[floral thistle]

polynomial long division

[cedar kiln]

@formal storm Has your question been resolved?

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[amber kiln]

how do i get to the second step

why is it e^y-1 and not e^y-x

[main needle]

they factor out an x from both

xe^y goes to e^y, and x goes to 1

[amber kiln]

ah ok

this separation of variables btw

so can I not just divide the e^y out of the right side and bring it to the left

and then multiply the right side by dx to get dy = dx

[main needle]

divide (e^y-1) yeah

[amber kiln]

no like before the factoring out

can you not just divide the e^y by itself?

[cedar kiln]

@amber kiln Has your question been resolved?

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[static harness]

i just end up getting 0

[static harness]

after integrating

am i doing something wrong?

ohh wait its not 0

would it just end up being 3/4 - 7/8

it seems like a property

[cedar kiln]

@static harness Has your question been resolved?

[cedar kiln]

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[quartz sedge]

find the exact value if sin(u) = 5/13 and cos(v)=-3/5, find sin(u+v) (both u and v are in quadrant 2)

[quartz sedge]

I got to -3/13 + cos(u)sin(v)

idk how to go from there

[livid hound]

determine cos(u) and sin(v) using the given information

[quartz sedge]

would that be with inverse sin and cos?

[livid hound]

with the aid of drawing a triangle / pythagorean trig identities and/or properties of the unit circle

no

[quartz sedge]

oh ok

[livid hound]

do not attempt to explicitly determine the values of u and v

[quartz sedge]

oh i got it ty

.close

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[crimson sedge]

80% of x is 12!

[regal oak]

12 factorial ?

[crimson sedge]

=12

80% of x =12

Find x

[regal oak]

80/100 x = 12
solve x

x=1200/80

@crimson sedge understood ?

[crimson sedge]

Ooo

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

Yep all thanks to youuuu! Imma do physics now!

[cedar kiln]

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[regal oak]

u can dm if u get doubt on it

[cedar kiln]

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[celest seal]

.close

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[blazing gust]

Could someone help me find the particular solution to this differential equation?

[cedar kiln]

@blazing gust Has your question been resolved?

[flint granite]

what are the initial conditions?

like does it say anywhere “y(0) = 1” or something like that

[cedar kiln]

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[feral haven]

Hi, I could do with some help breaking down points 1 and 2. Mainly point 2. I'm fine with the definition of a Laplace transform just not sure how to interpret the conditions for it

[feral haven]

.close

[cedar kiln]

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[azure cosmos]

Hey guys, i dont get this question, and i dont how to do it and dont know the workings, thank youuu

[graceful karma]

Well what's the perimeter in terms of x for the square

[azure cosmos]

Is it 4?

[graceful karma]

4 then what?

[azure cosmos]

So for a

Is perimeter the area?

[graceful karma]

The length of the edges

Combined

[dense wing]

What is perimeter in the most general sense?

[azure cosmos]

I dont know tbh, sorry

[dense wing]

So check your notes, or Google it

[azure cosmos]

Like is it cirumference?

[azure cosmos]

Or maybe thats for circle..?

[dense wing]

Circumference is perimeter of a circle.

[azure cosmos]

Yesss

Okayokay give me a moment

[dense wing]

Your first thing should also be googling it tbh... not asking here

[azure cosmos]

Nono i dont know the workings

I tried googling it

I got no results

[dense wing]

Right... but if you Google "what is perimeter" I highly doubt nothing comes up

[azure cosmos]

I know how to do a) which is 4x4x4x4 right?

[dense wing]

No

[azure cosmos]

We use the formula?

Which is

Area=bxh?

[dense wing]

Why are you using the area formula (and failing at it)?

When you're not asked for area

[azure cosmos]

.close

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[balmy loom]

I am currently trying to show that

[balmy loom]

can be written as

this is what i have tried so far. But i am not able to read much out of that last equation

does someone have an idea?

About the parameters:
A_x and A_y are real amplitudes and delta a real phase.
a_1 and a_2 are real vectors that are orthogonal

<@&286206848099549185> Maybe one of you?^^

[cedar kiln]

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[tender finch]

Help please

.close

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[split jungle]

Could somebody help me with this problem?

I got this far Volume of the jar is r²pih

surface of the jar= 2pir²+h2rpi

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[split jungle]

Then i wrote that the volume of the jar was 1liter so i rewrote the height in functie of the radius and subsituted it in the surface of the jar formula

[split jungle]

then i got this function surface of the jar=2pir² + 2/r

But kinda stuck here

[stone charm]

you can find the derivative of this function and then equate it to 0

$4pi*r - 2/r^2 = 0$

[wraith dagger]

Dzinski

[split jungle]

Why are we taking the derivative ?

sorry if it's a stupid question

[stone charm]

to find the minimum of the function

[stone charm]

https://www.onlinemath4all.com/how-to-find-maximum-and-minimum-value-of-a-function.html

check this article

[split jungle]

Aight thanks i think i got it

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[zinc pagoda]

can someone help me with this question please?

[zinc pagoda]

<@&286206848099549185>

.close

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[sly girder]

how would i find these sums

just a rough idea of how to do it would be appreciated

[bold vine]

For 8a, I think writing out the first few terms and seeing what each of the terms have in common will help

[sly girder]

i thought you could only find the sum of geometric or telescopic series

hm

[bold vine]

Since the denominator will always be even

[sly girder]

but they're always positive

nothing ever cancels

[bold vine]

b. I think that's a power series representation for a function you already know

[sly girder]

o

[bold vine]

You can probably still find a closed form for it

Try writing the first few terms out:

1/4, 1/10, 1/18, 1/28

What if we factor out a 1/2?

1/2(1/2 + 1/5 + 1/9 + 1/14 ....)

Hmmm

Not sure - good point

[sly girder]

i dont ever remember doing sums of these

[bold vine]

I was hoping things would work out nicely

[sly girder]

professor has never done these before

except for geometric or telescoping series

[bold vine]

For b and c tho, those are representations of familiar series

The first one I can't think of anything

[sly girder]

ok ill try b and c

thx

[bold vine]

np

[sly girder]

@bold vine ive got these

does it matter where the series starts

8(c) starts at n = 2

but the one i have recorded in this picture starts at n = 0

[bold vine]

Can't read that super easily

[bold vine]

You can probably reindex the sum so it starts at the point you'd like

[sly girder]

might be better

would reindexing mess it up tho

feel like it wouldnt represent that power series anymore

[cedar kiln]

@sly girder Has your question been resolved?

[crimson sedge]

if you haven't figured out (a) yet, I'd bet it telescopes

[sly girder]

not sure how

[crimson sedge]

if that is the case, a partial fraction decomposition will do the trick

[sly girder]

hm

[A / n] + [B / (n + 3)]

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[compact junco]

ayo where did the extra g' come from

[compact junco]

shouldn't the derivative of g^-1 be -g^-2?

i don't understand

why did they just add an extra g'

[dull dome]

it's called the chain rule

[livid hound]

also the notation is questionable

[compact junco]

wat

[dull dome]

I think the sentence is assuming that g is some kind of function

[compact junco]

how does it apply here

OHHH

wait what

[livid hound]

$\dv{x} (g(x))^{-1}$

[wraith dagger]

ℝamonov

[livid hound]

= ?

[compact junco]

but if g(x) is a function cant u just raise it to a power of -1

[livid hound]

wdym?

[compact junco]

and then differentiate it

[livid hound]

wdym?

[compact junco]

uh

oh yeah i havent learned the chain rule yet

i should probably go do that

thanks for the help

[livid hound]

it should be one of the first few things you learn

[compact junco]

.cose

[livid hound]

also the notation is bad as

[compact junco]

lol i literally just started yesterday

because i cant learn calculus

[livid hound]

the use of ^-1 for the exponent in functions usually indicates the inverse function

[compact junco]

okay

thanks for the help

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[worn cosmos]

.open

.ask How do i do this

?????

[glad kestrel]

well you wrote p/q right there

do that

[worn cosmos]

okie

i did that

[glad kestrel]

then you've answered the question

[bright surge]

https://brilliant.org/wiki/cardano-method/
(general method you can use to solve problems like this)

(also, i dont think that you should remember the entire cubic formula)

[glad kestrel]

^definitely not needed

[worn cosmos]

THATS IT ?? @glad kestrel

[glad kestrel]

yes

[worn cosmos]

OWWWOWOWO!!

[glad kestrel]

it just says to list possible roots

you don't need to solve it

[worn cosmos]

what about the 3rd one

[glad kestrel]

do you know descartes' rule of signs?

[worn cosmos]

nope 😄

kindaish

[glad kestrel]

i'll send a good video on it, one sec

[worn cosmos]

thank u!

[glad kestrel]

https://www.youtube.com/watch?v=YaU5JTe3cPU

[worn cosmos]

thank youuuuu

how do i close this

.close

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[worn cosmos]

oh wait

@glad kestrel

does thst mean #2 has no possible roots?

[glad kestrel]

it does have possible roots

p/q

[worn cosmos]

yea

oh

1 and 5

lol

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[ruby charm]

not really sure where to start here

[main needle]

it's a little tricky, but would you agree the bigger triangle ABC is just a scaled up ADE? (similar)

[ruby charm]

Yeah

[main needle]

and when you scale, the sides increase evenly

so when it increases the height by 3 and the base by 4, that means the height is 3/4 the base

and with that you can just call the base x, height 3x/4, and use pythag on the small triangle

[ruby charm]

So x^2 + (3x/4)^2 = 6^2?

[main needle]

yea

then you can do big triangle - small triangle area

[ruby charm]

So if I plug in the value of x, would the the height of small triangle be 12.96, and the base 23.04?

[cedar kiln]

@ruby charm Has your question been resolved?

[ruby charm]

@main needle

[main needle]

,w x^2 + (3x/4)^2 = 6^2

[wraith dagger]

Results provided by WolframAlpha

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

Requested by Plurmorant#1811

[main needle]

the base should be 4.8

[ruby charm]

ah ok

So I would do 1/2(4.8)(3.6) to find small triangle

and then 1/2(8.8)(6.6) to find the big triangle

And then subtract?

[main needle]

yea

[ruby charm]

Alright I got it

Thanks

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[tardy quail]

Question: \ Use the product-to-sum identities to find the exact value.\\$\sin(\frac{3\pi}{8})\sin(\frac{5\pi}{12})$

[tardy quail]

So my problem with this is that I end up with $\frac{1}{2} (\cos(\frac{\pi}{24}) - \cos(\frac{19\pi}{24}))$, and those aren't angles we've memorized on the unit circle.

[wraith dagger]

@tardy quail

@tardy quail

[tardy quail]

Did I do something wrong, then? D: Or is there some other step I'm missing?

I feel like if we had to use the half angle identities, the instructions would've mentioned it or at least have been less specific.

Especially considering how it would be needed twice for each cosine since it's over 24.

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[radiant scaffold]

hellppp TwT

[radiant scaffold]

<@&286206848099549185>

[cedar kiln]

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[radiant scaffold]

;-;

<@&286206848099549185>

Well if someone does come pls ping me

<@&286206848099549185>

HELPERSSSSS

[radiant scaffold]

<@&286206848099549185> ;-;

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[static eagle]

i would like to ask why use cosine rule?

i drew a right angle triangle

with 6 as a height

[livid hound]

can you show your diagram

labelled and clearly showing the ground, radar station, path of the plane

[static eagle]

Oh sorry i didnt label

[livid hound]

also that does not represent the problem presented in the problem

[static eagle]

So its not a right angle?

[livid hound]

try draw the correct diagram

first by starting the ground

placing something to mark the location of the radar station

[static eagle]

Ok

[static eagle]

Not sure

But when the plane is flying constant

Is it flying at a straight line

[livid hound]

yes.

[static eagle]

With a height of 6

[livid hound]

at an angle of 25°

[static eagle]

When it says climb

Its raised ye?

[livid hound]

With a height of 6
it was 6km above the radar station and then starts climbing at an incline

[static eagle]

Wait lemme change something

6km from the initial point to radar

[static eagle]

So is it the rate of distance above 25° that i should look for?

[livid hound]

no

unclear wording

the slanted line represents the path of the plane

you're interested in the distance between the plane and the radar
and its rate of change

[static eagle]

Like this?

[livid hound]

yes

[static eagle]

The distance b im quite confused

Should i use 6 as opposite angle? With sin25

To get b

[livid hound]

no

consider the speed of the plane to get an expression for b

[static eagle]

Oh ok

So does that mean 4*4 is b?

[livid hound]

you'd might use that later that at some point

but you'd also want the distance between the plane and the radar in terms of t

so that you can differentiate wrt t

[static eagle]

So

B is 4t

[livid hound]

lowercase b but yeh

[static eagle]

Ok

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@static eagle Has your question been resolved?

[static eagle]

.close

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[crimson sedge]

What real value of $t$ produces the smallest value of the quadratic $t^2 -9t - 36$?

i got $\frac{9}{2}$. is that right

[wraith dagger]

d-static

[dense wing]

No

[dense wing]

oh wait, yes, thought it asked for minimum value

[crimson sedge]

so is it correct or not?

[dense wing]

yes....

hence why I corrected myself to say yes...

[crimson sedge]

ok

i didn't understand

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[mortal saddle]

.OPEN

.open

[low ocean]

lol

[low ocean]

u have to type something to claim the cahnnel

.close

[cedar kiln]

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[mortal saddle]

ty

^^

[low ocean]

instead of ".open"

.close

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[crimson sedge]

How can I set up my equation for this problem?

[dapper raven]

@crimson sedge Sketch it

if that doesn't help i can explain why it could've helped

[dapper raven]

you see what region you need?

[crimson sedge]

I think I need the small region between the blue and red lines

[crimson sedge]

the 0 to 1

[dapper raven]

that was my thought too

but because we need to rotate around x =2

do you know integrals?

[crimson sedge]

Im familiar but Im not good at them

[dapper raven]

ok

because we need to rotate around x =2

we can change the graphics so that the area in between them stays the same but they are much easier to work with in terms of finding the volume

[crimson sedge]

How do I do that?

[dapper raven]

do you know inverse functions?

[crimson sedge]

When the x and y changes and solve for y?

[dapper raven]

yeah

[crimson sedge]

Yes

[dapper raven]

we can do that to make them vertical instead of horizontal

to make the rotation around the horizontal axis instead of a vertical axis

which is easier

at least in my opinion

[crimson sedge]

Then do I inverse both functions?

[dapper raven]

yes

[crimson sedge]

y=x^2 and y=x

[dapper raven]

exactly

can you plot those too?

[dapper raven]

without the previous :d

otherwise it might be confusing

😄

[dapper raven]

and the axis around which you rotate transformes too

what does it become?

[crimson sedge]

2 on the y axis?

[dapper raven]

yeah

[dapper raven]

now to make the rotation around the x axis instead of y = 2

we translate it down 2

all of them

[dapper raven]

so now we can calculate the integrals

[crimson sedge]

Im unsure how

[dapper raven]

one sec i'm gonna do it myself to make sure i'm not talking nonsense

😄

[crimson sedge]

Thank you!

[dapper raven]

ok i got it

[crimson sedge]

Ok

[dapper raven]

one thing though

volume is always positive

but the integrals we'll get are negative

[crimson sedge]

Got it

[dapper raven]

cuz we have x^2-2

and x-2

so we gotta look out for that

to get the area between x^2-2 and the x axis

[crimson sedge]

Ok

[dapper raven]

we do the integral of 0 - (x^2-2) dx

$\int_{0}^{1} 0 - (x^{2} - 2)dx$

[wraith dagger]

Katharine

[dapper raven]

but we wanna find the volume

so we do $\pi r^2$

[wraith dagger]

Katharine

[dapper raven]

so instead it becomes

$\pi \int_{0}^{1} (0 - (x^{2} - 2))^2 dx$

[wraith dagger]

Katharine

[dapper raven]

does that make sense?

[crimson sedge]

Yes

[dapper raven]

so we have a volume now

but it's too much

because it's the whole function

not the little bit inbetween the two functions

what we can do though

is find the volume of the other function and the x axis

and subtract that from this one

can you try setting up the integral of the other function

it doesn't have to be in the fancy text

[crimson sedge]

pi integral (0-(x-2))^2 dx

[dapper raven]

perfect

do you know how to solve integrals?

[crimson sedge]

Yes

[dapper raven]

or at least these integrals?

[crimson sedge]

I know what to do next

Thank you!

[dapper raven]

ok 😄

[crimson sedge]

.close

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[gleaming tartan]

I am unsure how to approach Q1

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[sleek smelt]

If O is open in $X\subseteq R^n$ and U is open in $Y\subseteq R^m$ is OxU open in $X\times Y\subseteq R^{n+m}$?

[wraith dagger]

alef0

[cedar kiln]

@sleek smelt Has your question been resolved?

[sleek smelt]

<@&286206848099549185>

[crimson sedge]

In general topology that would be from definition of product topology, so it would be trivial
But if you're building it up in terms of open sets in metric spaces then I think you could prove it that way as well

So you want to show that a general point in OxU has an open ball in XxY fully contained in OxU right

Tbh I'm not sure where to go from there lol, have you tried expanding on this?

[cedar kiln]

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[crimson sedge]

Ahh yeah I figured out the crux of the argument I think you can too

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[steady breach]

I would like some intutition on question c

I got b, and it was a negative binaomical distribution with big R as the number of successes did it with a little bit of mgf manipulattion

but for this question Im not so sure what I can do

thank you

<@&286206848099549185>

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[gusty lotus]

Hi

[gusty lotus]

I get how they condensed the first and second equation into one and the third and fourth into another
but I don't get how they combined the results from those two into the last one

[cedar kiln]

@gusty lotus Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

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[gusty lotus]

.reopen

[cedar kiln]

✅

[crimson sedge]

I think they just added them and divided by 2 since on the left hand side you would have 2x/2 = x

@gusty lotus

[gusty lotus]

sorry im confused

so we got pi/4 +kpi

and 3pi/4 +kpi

how would we combine them

[crimson sedge]

wait what I said doesnt work lmao 😭

[gusty lotus]

oh

[crimson sedge]

ah wait I got it

x being allowed to be either of those two means, that since they are just pi/2 away from each other + a multiple of pi, we can add any multiple of pi/2 to them

multiples of pi are also a multiples pi/2, so the additional multiples being added are just multiples of pi/2

so it's the same logic as the other combining that happened earlier on

[gusty lotus]

oh I see

cause pi/4 has a pattern with 3pi/4

in that they are pi away from each other

so we can just do pi/4 plus pi

thanks a lot

@crimson sedge

.close

[cedar kiln]

Channel closed

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[crimson sedge]

pi/2 away from each other

[cedar kiln]

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[opal epoch]

?

[bright surge]

???

[cedar kiln]

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@opal epoch Has your question been resolved?

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[gray vine]

How do I find angle u?

[gray vine]

I know that triangles with 2 equal sides are isosceles triangles, which means 2 of the three angles are equal. Which ones that'd be is difficult to tell.

I assume e.g. angle_1 = 14°, angle_2 = 180-2(14)

[livid hound]

repeated application of angle sums and properties of isosceles triangles

[gray vine]

I assume e.g. angle_1 = 14°, angle_2 = 180-2(14)

Is this correct?

[livid hound]

why assume?

there is no need to assume anything

[gray vine]

I am unsure if I get it mixed up, i.e. angle_2 = 14° and angle_1 = (180-2(14))°

[livid hound]

angle_1 = 14°, angle_2 = 180-2(14)
is fine

just remember that the angles opposite congruent sides (of a triangle) are congruent

[gray vine]

So angle_2 = 152°, angle_3 = 28°, angle_4 = 28°, angle_5 = (180-56)°

[gray vine]

How do we know they're congruent

I found this example

[livid hound]

its a properties of an isosceles triangle
(provable by triangle congruency)

[gray vine]

Are you referring to this?

[livid hound]

yes

[gray vine]

Does that mean the sides are equal? Like 5 and 5

[livid hound]

identical hatch markings on the sides indicate congruency

[gray vine]

Call it side x, would the other be the same

English is not my first language so I am not sure what congruency translates to

Is it size? Shape?

[livid hound]

yeh pretty much

[gray vine]

Two questions, which one

[livid hound]

depends on how its used

in this case it indicates the sides have the same length

(i.e. the reason you knew you had isosceles triangles)

[gray vine]

Angle_6 = 42°
Angle_7 = 42°
Angle_8 = 96°
Angle_9 = 180-(96+28) = 56°
Angle_10 = 56°
Angle_11 = 180-112 = 68°
u = 180-(68+42) = 180-110 = 70°

Is there a simpler way to get the answer than doing it this way?

[livid hound]

not really

you could save a few steps with exterior angle theorem but the idea is still pretty much the same

[gray vine]

Could you demonstrate?

So the last steps?

u = angle_9 + angle_10?

[livid hound]

at the very start

< 3 = 2 * <1 = 28

you could also do stuff like <9 = 14° + <7

[gray vine]

So what's the easiest or fastest method to find angle 9?

We'd need <8, <4 first?

[livid hound]

well < 3 = < 4

[gray vine]

< 3 = 14*2
<9 = 2*28?

[livid hound]

a lot of the angles are trivially congruent so you can pretty much reduce the amount of numbers to represent your angles by half

[gray vine]

How can we do that

That's interesting

[livid hound]

exterior angle theorem

[gray vine]

Yeah but <7 is in another triangle

With different interior angles

[livid hound]

there are more triangles in the diagram than you might think

[gray vine]

So <9 would depend on different interior angles

If we apply exterior angle theorem

[gray vine]

<u = 2*<9?

[livid hound]

fk

[gray vine]

Or <u = <9 + <10?

[livid hound]

that was wrong, didn't spot that error

u = 14° + <9 = 14° + <10

[gray vine]

what error

[livid hound]

<u = <9 + <10?
when you typed that earlier

[gray vine]

I'm confused because the theorem uses the triangle next to it

[livid hound]

using just these 4 labels.
<1 = 2 * 14 = 28
<2 = 14 + 28 = 42
<3 = 14 + 42 = 56
<4 = 14 + 56 = 70

[gray vine]

We use the exterior angle theorem on <2? How is it that <2 = 14+28?

[livid hound]

14 + <1 = <2

[gray vine]

I don't see why 14 is included

[livid hound]

consider bigger triangles

[gray vine]

Oh ok

And < 3 = 14 + this angle?

.close

[cedar kiln]

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[granite gate]

How would you express the fact that x and y can be permuted in the following function and not change the output?:

[wraith dagger]

fuguehoppy

[leaden otter]

$f(y,x)=\frac{yx}{y^2+x^2}=\frac{xy}{x^2+y^2}=f(x,y)$

[wraith dagger]

Euclid31415

[granite gate]

What about something like this, where it becomes impractical to write out every possibility:

[wraith dagger]

fuguehoppy

[leaden otter]

Commutativity of multiplication and addition

[granite gate]

Okay

I just feel like I remember seeing some sort of notation for this but I don't remember what it was called

[leaden otter]

You can show that exchanging any two terms doesn’t change the value of expression. Then, every permutation can be formed by swapping terms

[granite gate]

Oh, my bad, I was recalling the notation $\sum\limits _ {\text{sym}} a^2 b = a^2 b + a^2 c + b^2 c + b^2 a + c^2 a + c^2 b$ or $\sum\limits_{\text{cyc}} a^2 b = a^2 b + b^2 c + c^2 a$ if you are working with three variables but that doesn't apply here, at least directly, from what I can tell

[wraith dagger]

fuguehoppy

[leaden otter]

Or, you can notice that both the numerator and denominator are symmetric polynomials.

[granite gate]

Oh yea, so it can be expressed as

$\frac{\prod\limits_{0 \leq i \leq n} a_i}{\sum\limits_{0 \leq i \leq n} a_i^2}$

[wraith dagger]

fuguehoppy

[granite gate]

Thanks!

.close

[cedar kiln]

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[winged acorn]

i need help in general, idk what to do ive tried doing the ln thing but i think im getting the wrong answer

[trail whale]

did you try expressing it as a power of the exponential function @winged acorn ?

[winged acorn]

can i send you my work?

cause i feel like i did it wrong

[trail whale]

yep

[winged acorn]

tried doing 2 different ways

give me a sec

[trail whale]

instead of using L'Hopital, try to use the limit of the exponential function

expand cos x using Taylor's series, if you know it

[winged acorn]

Not sure if I ever learned that

I was told to evaluate it using L'Hopital rule

[granite gate]

L'Hopital can be a successful approach here; try bringing out the exponent 4/x instead of just the 1/x

[winged acorn]

wdym?

if possible couild you walk me through the whole thing? i'm just super confused

[granite gate]

In your approach on the left, I think you had accidentally left the exponent of 4 in the logarithm on the fourth line

Well not accidentally, per say

[winged acorn]

i tried to combine 4/x with ln(xcos(x))

and that's what i got

ln((xcos(x))^4)

[livid hound]

where's xcos(x) coming from

[winged acorn]

from the original problem

[livid hound]

does not have xcos(x)

[winged acorn]

oh

[livid hound]

x+cos(x) isn't the same as xcos(x)
being inside a log doesn't change that

[winged acorn]

oh right yeah

i just forgot to write the + is that where everything went wrong?

or

[granite gate]

Oh oops now that I see that I apologize for imposing my approach on you even though both should be equally valid

[trail whale]

is it a must to use L'Hopital here?

[winged acorn]

Yeah

[trail whale]

it gets messy with that

[winged acorn]

wait so was that all i had wrong? just forgetting to add the +

or am i missing steps

[granite gate]

Yes, it should be good now

[winged acorn]

okay so after that i would get 0/0 and then take derivative of both top and bottom because of L hopital

[granite gate]

Yup

[winged acorn]

so it would be (((4(1-sin(x)))/(cos(0)+0)))/(1)?

[granite gate]

Sorry just give me one moment to parse that

Is what you wrote this?

$\frac{\frac{4(1-\sin (x))}{\cos (0) + 0}}{1}$

[wraith dagger]

fuguehoppy

[winged acorn]

yeah

i wrote that

well replacing x with 0

mb

I got 4, but im not sure if that's correct

[granite gate]

Okay yeah in this case the denominator of (cos(0) + 0) happens to be 1 so it's okay

But for the future

Since you're taking the derivative of ln((cos(x)+x)^4) the exponent of ^4 should also be on the denominator

[granite gate]

That is correct

[winged acorn]

Okay, thank you!

[granite gate]

And then

That means ln(y) = 4

So you just need to solve for y

[winged acorn]

and that would be my answer?

[granite gate]

Yes, since y represents (x + cos(x))^(4/x)

[winged acorn]

this makes so much more sense now

tysm

[granite gate]

No problem!

[limber stag]

I'm trying to draw the path of point B, with point A being the velocity.

https://www.desmos.com/calculator/lrkrvtbomd

[winged acorn]

im gonna close this so you can re-post your question

[limber stag]

uhh k

[winged acorn]

.close

[cedar kiln]

Channel closed

Closed by @winged acorn

Use .reopen if this was a mistake.

[winged acorn]

or you can just repost in the available section

[cedar kiln]

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[gleaming cave]

prove that there exists objects in zf that are not definable in zf [does that make sense?]? how do you speedrun this proof?

[bright surge]

zf?

what;s that?

[gleaming cave]

zermelo frankel

perhaps the question is better formulated as "prove that there exists objects in zf that are not definable in set theory"

[cedar kiln]

@gleaming cave Has your question been resolved?

[cedar kiln]

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[silver fable]

Can you explain this function domain notation?

[silver fable]

Let's say f(x) =1/x what would the notation be in this case

[elfin hemlock]

The natural domain of f would be R\{0}

But you can define the function to have whatever domain you want (hence domain is a part of the definition of a function)

[silver fable]

What about the notation?

A--->B

Is it R---->R*??

[silent pawn]

It can be anything.

[elfin hemlock]

Just told you what the domain is

[elfin hemlock]

So no not R

[silver fable]

Yeah it's R*

[elfin hemlock]

What is R* supposed to mean?

[silent pawn]

probably real extended numbers

[silver fable]

So it's f:R*--->R

[silent pawn]

but as I said it can be anything

[silver fable]

R* is real numbers except 0

[elfin hemlock]

Like I have said you can define the domain and codomain how you wish. They are a part of the definition of the function

[silent pawn]

A and B can be anything. It is convenient to let $A=\mathbb R\setminus{0}$ and $B=\mathbb R$. Done.

[wraith dagger]

daniele_danielo

[silver fable]

How i wish?? So i can write it like this f : R{0} - - - -> (1,3)

[elfin hemlock]

No

Not a function then (if we still working with f(x)=1/x)

[silver fable]

This is so confusing 😅

Yes I'm taking about f(x) =1/x

[elfin hemlock]

No then