help-0

[lament basalt]

Then turning it clockwise assume that it's negative right?

Iike from 0 degrees to - 30 degrees?

Ohkay

Wait um last thing

So after the sine difference identity it showed me how to prove cosine sum identity

I don't get why cos (90-theta) = sin theta, sin (90 - 0) = cos theta and the rest

[tardy stag]

if you think about a right triangle

,tex .sohcahtoa

[ocean seal]

Hayley

[tardy stag]

if we call the other angle phi

think of how sin theta relates to cos phi

[lament basalt]

Sry is it on the angle of opposite side?

Angle between the hypotenuse and opposite

[tardy stag]

yeah the one on the top

\tikz {
\draw (0,0) -- (0,1) node [pos=0.6, right]{$\phi$} --
(3, 0) -- cycle node [pos=0.3, above]{$\theta$};
}

[lament basalt]

Wait how is phi found out again?

[ocean seal]

Hayley

[lament basalt]

Yes the phi is the one on the top

[tardy stag]

well what's the sum of the angles of a triangle?

[marsh night]

180

duh

[lament basalt]

They all have to be 180

[tardy stag]

yeh

so what's the relationship between theta and phi

[marsh night]

90

sum

[tardy stag]

i'm sorry, are you the one who asked this question?

[marsh night]

ok mb

[lament basalt]

So like angle phi + angle theta + right angle =180

[tardy stag]

yeah

and the right angle is 90

[lament basalt]

Im left with angle + angle theta = 90 bcs I subtracted the right angle on both sides

Is that right?

[tardy stag]

yeah

so we know that $\phi = 90^\circ - \theta$ and vice versa

[ocean seal]

Hayley

[lament basalt]

Ohhh

So like may I ask how is it related to the cos (90- theta) = sin theta earlier?

[tardy stag]

well let's look at that triangle again

[lament basalt]

Sin (90-theta) = cos theta
& tan (90-theta) = cot theta

[tardy stag]

,tex .sohcahtoa

[ocean seal]

Hayley

[lament basalt]

Okay

[tardy stag]

what's sin(phi)? Use the labels on this triangle

[marsh night]

[lament basalt]

The sin (phi) is definitely opposite/hypotenuse right?

[tardy stag]

well...

it's opposite/hypotenuse from phi's perspective

but the opposite angle to phi

is the adjacent angle to theta

[lament basalt]

Wait so earlier

It says cos (90-theta) = sin theta right

[tardy stag]

yeah

[lament basalt]

Does that mean that the cosine = sine talking about
is that it's actually the cosine of theta is equal to the sin of phi?

[tardy stag]

yep! cos(theta) = sin(phi) in this case

and we established that phi = 90º - theta

[lament basalt]

Ohh dayummm I see

One more thing

Why is it a theta sign as well?

Instead of phi, lemme reply onto it

[lament basalt]

Below the "to derive the sum..."

-of the photo

[tardy stag]

well phi is just something we made up

and defined to be 90 - theta

[tardy stag]

so you said $\cos(\theta) = \sin(\phi)$ and from that we can get that $\cos(\theta) = \sin(90^\circ - \theta)$

[ocean seal]

Hayley

[lament basalt]

Ohhh OK okayyy

[lament basalt]

Alright definitely appreciate ur help, thanksss

[tardy stag]

yep! hve fun, i never actually proved these myself

just accepted them on faith

[lament basalt]

Yeah, it has become tricky honestly

.close

[lone heart]

Channel closed

Closed by @lament basalt

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[crystal lotus]

i have three questions relating to eliminating parameters. i finished an assignment on it but there are a few stubborn ones that I cant get right. This is the first one. Im pretty sure this is the right equation. But it says otherwise

[crystal lotus]

Wait. Nevermind. That one I just used a instead of an alpha sin

thats good

For this problem I had the correct graph. But my elimination of the parameter was considered incorrect.

[lone heart]

@crystal lotus Has your question been resolved?

[crystal lotus]

.close

[lone heart]

Channel closed

Closed by @crystal lotus

Use .reopen if this was a mistake.

[lone heart]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[vapid root]

I'm having a trouble trying to understand how Xn distribution will be affected by Xi. Do I just simply use the pdf default of gamma distribution, If that is the case, would I just 1/n times the distribution?

[lone heart]

@vapid root Has your question been resolved?

[vapid root]

<@&286206848099549185>

[lone heart]

@vapid root Has your question been resolved?

[naive valley]

in general, if you take the sum of n iid random variables, then the pdf of the sum equals the convolution of n copies of the pdf for one of them

or, in terms of characteristic functions, the characteristic function of the sum of n of them equals (characteristic function of one of them) to the n'th power

presumably the characteristic function approach would be simplest here, assuming you are allowed to use it

[lone heart]

@vapid root Has your question been resolved?

[vapid root]

so if I'm understanding your explanation correct. The distribution of Xn is equal to n times the Gamma Distribution of Xi?

But why would it be to the power of n?

[naive valley]

where did you get that from what I said?

first, be explicit what you mean about "distribution", do you mean pdf?

if so, the pdf of the sum of n of them equals the convolution of n copies of the pdf of one of them

convolution is not that same as "multiply by n"

[vapid root]

I assume this is what you mean "in terms of characteristic functions, the characteristic function of the sum of n of them equals (characteristic function of one of them) to the n'th power"

[naive valley]

characteristic function = fourier transform of pdf

[vapid root]

i c

yeah, I think I got it now

Ty

.closed

.close

[lone heart]

Channel closed

Closed by @vapid root

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[icy violet]

Can someone help me start question e

[last ether]

By turning point do they mean an extrema

Like the vertex?

In that case you would know from the statements that:

• f'(-1) = 0
• f(-2) = 5

[icy violet]

Ahh

[icy violet]

So do I differentiate and sun in points?

[last ether]

Yeah

I mean

Turning point refers to the extrema

[icy violet]

?

[last ether]

Well yeah f'(x) = 2ax + b

[icy violet]

Since there’s 2 points do I have to make 2 equations

And then simultaneous equation

[last ether]

Yes

[icy violet]

Are the 2 points (-1,0) and (-2,5)

[last ether]

Literally just make the two equations from the info I gave you

There's no point in trying to obtain a second point

[icy violet]

Aight

.close

[lone heart]

Channel closed

Closed by @icy violet

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[sonic wharf]

Given that f(x) = x + 3, gf(x) = 3x^2 + 18x + 20. Find g(x)

[sonic wharf]

how do i do this?

[gray isle]

let u = x + 3

express x in terms of u
and substitute

[sonic wharf]

got it

.close

[lone heart]

Channel closed

Closed by @sonic wharf

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[hollow thicket]

.close

[lone heart]

Channel closed

Closed by @hollow thicket

Use .reopen if this was a mistake.

[hollow thicket]

why is the inverse of f(x): pi - sin^-1(x) instead of pi + sin^-1(x)?

[lone heart]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[hollow thicket]

why is the inverse of f(x): pi - sin^-1(x) instead of pi + sin^-1(x)?

[lone heart]

@hollow thicket Has your question been resolved?

[lone heart]

@hollow thicket Has your question been resolved?

[marsh rapids]

Think of some specific values. f^-1(1) = pi/2 = pi - pi/2

Also sin(pi - sin^-1(x)) = -sin(-sin^-1(x)) = sin(sin^-1(x)) = x, so - works and + doesn't

[hollow thicket]

ah makes sense

thanks

.close

[lone heart]

Channel closed

Closed by @hollow thicket

Use .reopen if this was a mistake.

[lone heart]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[ember needle]

I'm having a lot of trouble with this question

[ember needle]

I would be glad if someone could walk me through a)

and I'll try the rest myself

[somber eagle]

Start by squaring both sides

[ember needle]

alright, hold on a moment please

x - 2 = 9 + 6sqrt(2) + 8?

wait

oh sorry

okay hold on again

x - 2 = 9 + 6sqrt(x) + 4x?

[somber eagle]

Wouldn’t be 6sqrt(x)

[ember needle]

oh right, 12 right?

[somber eagle]

yeah

[ember needle]

what then?

[somber eagle]

theres a little trick you can use to be able to solve this like a quadratic equation

[ember needle]

how do I do that?

I would be glad if you could show me the standard way of doing it, and then that way of doing it

so I can learn properly

[somber eagle]

this is the standard way of doing it

im not telling you some weird random method

[ember needle]

oh alright, then go on please

[somber eagle]

try rewriting x as (sqrtx)^2

[ember needle]

like what?

[ember needle]

this?

[somber eagle]

you now have a quadratic in terms of sqrtx

yeah

and the other x term

[ember needle]

hmm alright, what then?

which one is the other x term?

12sqrt(x)?

[somber eagle]

the x from x-2

[ember needle]

hmm

then what do we do?

[somber eagle]

you should have something looking like that

[ember needle]

yeah

I just apply the quadratic formula?

[somber eagle]

would you know how to solve that using the quadratic formula

yeah

[ember needle]

well in that case, wouldn't this quadratic have a solution for x?

[somber eagle]

you have to find sqrt(x) first to find that out

[ember needle]

the discriminant comes out as positive

what do you mean?

[somber eagle]

find sqrt(x)

[ember needle]

how do I do that?

[somber eagle]

use the quadratic formula

your equation is in terms of sqrt(x)

using the quadratic formula finds you sqrt(x), not x

[ember needle]

with the quadratic formula, this question has a solution

[somber eagle]

no

[ember needle]

according to the answers page, it shouldn't

how

what

[somber eagle]

sqrt(x) has a solution

just find sqrt(x) first

[ember needle]

I'm so confused right now

[somber eagle]

i will tell you after

[ember needle]

I can't

this question was asked before the author wen t over the quadratic formula

I think he wants another solution

[somber eagle]

wdym?

[ember needle]

I couldn't solve this question, so I postponed it

the author hadn't yet talked about the quadratic formula when he first asked this

[somber eagle]

all im asking you to find is sqrt(x)

[ember needle]

then it's

[somber eagle]

are the solutions positive or negative

[ember needle]

they're negative

oh that's why this has no solutions

[somber eagle]

yeah

the square root of a number cant be negative

[ember needle]

ooh alright

but there probably still is a solution without the quadratic formula?

[somber eagle]

if theres no solution with the quadratic formula, there wont be solutions without it

using a different method wont get you a different answer

[ember needle]

I know the answer will be the same

I just want to know the method the author intended

[somber eagle]

how are you so sure the author used a different method

[ember needle]

he asked this before he ever spoke of the quadratic formula in the book

it's a basic mathematics book

he wouldn't presume the reader already knows about the quadratic formula, the book starts with addition

[somber eagle]

there is more than one way to solve a quadratic equation

[ember needle]

yeah, but this is the topic before quadratic equations

he hadn't shown quadratic equations before this at all

[somber eagle]

i really dont know any other way that this could be solved, but maybe theres a neat trick to doing it without quadratics

[ember needle]

unfortunately, he only gave answers to this question in the solution part

no method

this is all he gave

oh and, why does b have no answers?

with this method, I've found this for b)

[opal stone]

I'm dogwater at math and helping my lil bro. To estimate the thickness of one sheet of A4 paper, You measure a ream of paper, which consists of 500 sheets. You determine that the pile of paper is 55mm thick. How thick is one sheet of paper? Express your answer as a decimal number and also a fraction

[ember needle]

you've got to ask in an open question channel @opal stone

like #help-14

[somber eagle]

idk what that answer to b is

since im pretty sure it does have solutions

[ember needle]

this is confusing

[somber eagle]

the answer is wrong

there is exactly 1 solution to that equation

[ember needle]

the book says otherwise

[somber eagle]

the book is wrong

that is what im telling you

[ember needle]

well I don't know

anyways, I must go now

.close

[lone heart]

Channel closed

Closed by @ember needle

Use .reopen if this was a mistake.

[somber eagle]

please do not trust the book though

[lone heart]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[white karma]

is it
I. A and B are independent given S: False
II. A and C are independent given S: false
A and C are independent given B and S:true
IV. B and C are independent given A and S false

[white karma]

<@&286206848099549185>

[upbeat gorge]

<@&268886789983436800> ping spam

[white karma]

Accident

[gray isle]

don't MASS 👻 ping

[white karma]

keyboard glitched

sorry

While you are here

[white karma]

you may aswell help

[gray isle]

no

[white karma]

Plese

please

[lone heart]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[white karma]

Im sorry for mass pinging it was accident

I asked it before

no one answered

after ping

after 1 hour

Dude

stop

[upbeat gorge]

Bruh

[white karma]

they ain't gonna help me

[vernal latch]

dude shut up

[white karma]

my message can't ven be seen

[vale crag]

<@&268886789983436800> mute them ffs

[alpine sable]

wth

[upbeat gorge]

<@&268886789983436800> intentional ping spam

[vernal latch]

hes just trolling at this point

[wintry patrol]

SHUT THE FUCK UP

[waxen glade]

SHUT UP

[sick cedar]

why are u doing this

[vivid rampart]

bro stop

[alpine sable]

Stop

[vernal latch]

ah great

[zealous lichen]

<@&268886789983436800> help plz

[vernal latch]

someone pls kick him

[foggy nimbus]

stop

[alpine sable]

Where's the moderators when you need them?

[sick cedar]

this guy is seriously

[white karma]

STOP

PLEASE

[waxen glade]

<@&268886789983436800> pls kick this guy

[alpine sable]

Answer john

[white karma]

Not sure

[keen pagoda]

Oh hell no

[vale crag]

ayy

[fading citrus]

got it

[keen pagoda]

Ayyyy

[zealous lichen]

nice

[vernal latch]

yay

[keen pagoda]

W

[upbeat gorge]

oh my god

[white karma]

W

[keen pagoda]

Finally

[waxen glade]

thank god

[alpine sable]

Help john smith

[vernal latch]

thank you god

[keen pagoda]

I kept getting pinged

[alpine sable]

(op was also ping spamming)

[white karma]

on accident

[keen pagoda]

Bruh on accident

[white karma]

I sent

[keen pagoda]

You ping spammed on accident

[white karma]

three

cause I tried before

[alpine sable]

Let my man go it was an accident

[somber eagle]

"spam", literally 1 ping more than he should have

[white karma]

on another channel

no one

responded

[alpine sable]

108 pings frim this channel

[alpine sable]

Just help john

[upbeat gorge]

He pinged 7 times at first lol

[white karma]

Im sorry about that

[somber eagle]

did he delete them

[upbeat gorge]

He did delete 6 of them

[alpine sable]

I dont even know the answer

🗣️

[somber eagle]

as long as no one here has access to the audits of the server its fine

[fading citrus]

If you ping spam I will mute you so don't do it again

[white karma]

ok

[fading citrus]

I can see deleted messages

[somber eagle]

welp john its been a good run

[white karma]

?

[somber eagle]

he sees everything

[white karma]

I mean no one helps me anyways

[somber eagle]

true

[white karma]

some dude said he would help me yesterday

and just gave up

today I waited 4 hours

no response

so 😭

[alpine sable]

🗣️

[somber eagle]

wait another 4 and ask again

[white karma]

IG

[white karma]

minutes?

[somber eagle]

hog all the channels and get banned

[somber eagle]

weeks

[white karma]

Genius

[white karma]

Someone has to help me

[somber eagle]

yep

[white karma]

can I ping helper in another 15 minutes

[somber eagle]

and if someone asks, i didnt tell you to do it

probably

[white karma]

can you help

?

[somber eagle]

i hate probability so idk

[white karma]

yea

I have more

I believe Im right

but id like someone to help me check

[somber eagle]

ok

send some

[white karma]

the first ones up there

[white karma]

^

[somber eagle]

are all your questions about probability?

[white karma]

Next one is active paths and if they are d-seperated

so yea

[somber eagle]

😭

i havent done stats in a long time, any advice i give will probably be wrong

[white karma]

😭

do you know anyone

[somber eagle]

yeah, their name is helpers, but make sure you put an @ before it

[thorn monolith]

Uhh can you repost the question? It takes a lot of scrolling to get back to it

[upbeat gorge]

It’s pinned

[white karma]

^

?

any updates

<@&286206848099549185>

...

<@&286206848099549185>

[vapid shuttle]

!15mins

[lone heart]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[white karma]

sorry it was 3 minutes early

[somber eagle]

😭

told off again

[white karma]

dude chill

[thorn monolith]

I dont know much probability but i'd try to use the fact that:

If A and B are independent then:
$$P(A \cap B) = P(A). P(B) $$

And if A and B are dependent then:
$$P(A \cap B) = P(B).P(A|B) = P(A).P(B|A)$$

[somber eagle]

me when i ping 14 seconds too early

[vapid shuttle]

It wasn’t ‘early’ you need to read the once part of the message in bold

[ocean seal]

Cyrenux

[vapid shuttle]

Considering this channel has tagged helpers many times, each additional tag counts as more than ‘once’

[white karma]

Dude

Chill

[somber eagle]

very strict server

[thorn monolith]

@white karma focus on this

[somber eagle]

tbh it is about maths so it figures

no one fun

[white karma]

yea I got that

I just wanted to know if I was right

trying to see if anyone would check

[somber eagle]

trust yourself

[white karma]

No

can't

[upbeat gorge]

Do you know what independent events mean

[somber eagle]

if he already answered the question and is confirming, my guess is yes

[upbeat gorge]

Because our answers are different

So @white karma, can you explain your reasoning for each statement

[white karma]

Ok sure

[somber eagle]

😊 john smith can finally get the help he needs

[white karma]

let me scroll up

Ive changed them slightly

[upbeat gorge]

I just don’t want helpers to get pinged a 500th time by the same asker

[white karma]

I. A and B are independent given S: true
II. A and C are independent given S: true
A and C are independent given B and S:false
IV. B and C are independent given A and S false

although im a little lost

[somber eagle]

so what im hearing is that the constant pinging led to help being given, it does work after all

[upbeat gorge]

Explain why you think each is true

[white karma]

ok well

given samsple space s

[upbeat gorge]

Or false

[white karma]

the value of A doesn't effect the value of be

same for II

right?

[upbeat gorge]

Well, notice that we have b1 and b2

Let’s make a, b, and c actual events so that they’re not abstract

And since we only have two possibilities for each, let’s make them “happened” and “didn’t happen”

Let’s say A is “today is Monday”, then a1 is “today is Monday” and a2 is “today is not Monday”

Then let’s let b1 be “I rolled a 4”, so b2 would be “I didn’t roll a 4”

[white karma]

ok

[upbeat gorge]

Given it is Monday, what’s the probability I rolled a 4?

[white karma]

0.4

and not monday

is 0.6

[upbeat gorge]

Yes

Are they the same?

[white karma]

oh so false

[upbeat gorge]

yeah so A and B are not independent

[white karma]

yea thats what I originally

thought

then I confused myself

[upbeat gorge]

Let’s say c1 is “I bought ice cream”, so c2 is “I didn’t buy ice cream”

[white karma]

ok

[upbeat gorge]

Let’s do II first

If today is Monday, what’s the probability that I bought ice cream

[white karma]

0.3375

[upbeat gorge]

Why

[white karma]

0.45*0.4*0.75+0.45*0.6*0.75

[upbeat gorge]

The “today is Monday” is a “given” thing

[white karma]

oh

mb

0.4*0.75+0.6*0.75

0.75

[upbeat gorge]

Ok

Given today is NOT Monday, what’s the probability I buy ice cream?

[white karma]

0.25

[upbeat gorge]

The same?

[white karma]

no

[upbeat gorge]

So is statement II true?

[white karma]

no

[upbeat gorge]

Ok

Now verify statements III and IV

[white karma]

ok

A and C are independent given B and S: false

0.135

and 0.1375\

[upbeat gorge]

ok this is correct

[white karma]

and IV is false

[upbeat gorge]

Though the reasoning is more identical to statement 2

[white karma]

oh ok

thats weird

[upbeat gorge]

But IV is actually true

[white karma]

really

?

[upbeat gorge]

The key is we are given S and A

What that means is that we don’t need to know which A it is

We just pick 1, and examine the Bs and Cs from that branch

Let’s say today is Monday

Given I rolled a 4, what’s the probability I bought ice cream?

[white karma]

0.45

thanks

!close

/close

[thorn monolith]

Dot

[white karma]

.close

[lone heart]

Channel closed

Closed by @white karma

Use .reopen if this was a mistake.

[lone heart]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[dusty gull]

Hello my question is this: An n × n matrix C is skew-symmetric if CT = −C. Prove that every square matrix A can be written uniquely as A = B + C where B is symmetric and C is skew-symmetric

[dusty gull]

I know that by the definition of uniqueness, we must have A = B + C and another equation A = D + E, where B and D are symmetric matrices, and C and E are skew-symmetric matrices

$$ B + C = D + E \implies B-D = E-C \therefore B = D \text{and} E = C$$

[ocean seal]

April | Koi of Mahjong

[dusty gull]

I'm not sure where to go from here though

[upbeat hornet]

The conclusion does not follow from the premises

3 + 5 = 2 + 6

[vale crag]

how do you justify your final deduction ? @dusty gull

[dusty gull]

This was the given first step in the textbook, well the justification wasn't given, but it was given that B = D and E = C

I tried to deduce this, and then work from there

[vale crag]

it comes from the fact that if a matrix is both symmetric and anti-symmetric, then it has to be the zero matrix

[dusty gull]

My thinking was that if B and D are symmetric matrices, and E and C are skew-symmetric matrices, then B - D = 0 and E - C = 0 , so B - D = E - C, and B + C = E + D

[vale crag]

we don't care about this implication

it's the reverse we want

[dusty gull]

the reverse?

[vale crag]

you want to show B+C = D+E implies B=D and C=E

showing B=D and C=E implies B+C = D+E doesn't accomplish that

[upbeat hornet]

didn’t they do that in the second message

[vale crag]

sure

it falls from nowhere the way they stated their proof though

[dusty gull]

So we get to this step $$B + C = D + E \implies B - D = E - C$$

[ocean seal]

April | Koi of Mahjong

[dusty gull]

Which states that a symmetric matrix is equal to a skew-symmetric matrix

[vale crag]

alright alright

[dusty gull]

Zero is the only matrix that is both skew-symmetric and symmetric, yes? (if so, is this the only one?)

[vale crag]

symmetric = skew-symmetric so both have to be zero fine

[dusty gull]

So if $B-D = 0 = E-C$, then we have $B- D = 0$ and $E-C = 0$. or $B = D$ and $E = C$

[ocean seal]

April | Koi of Mahjong

[vale crag]

right

[dusty gull]

okay and what does this tell us about our expression $A = B + C$?

[ocean seal]

April | Koi of Mahjong

[vale crag]

tells you that if such an expression of A exists, then it is unique

[dusty gull]

So, do we also need to show that a square matrix is the sum of a symmetric and skew-symmetric matrix? or do we just need to show that its unique? ( which we did)

[vale crag]

nah you need to show it exists also

"every square matrix A can be written uniquely as"

[dusty gull]

alright

[vale crag]

one typical way to do that is to work with the condition you're given

"A = B + C, with B symmetric, and C skew-symmetric"

and try to find conditions on B and C such that this decomposition works

and ultimately check that what you have in the end is indeed a valid decomposition

[vale crag]

here I suggest you look at A^T, works well with symmetric and skew-symmetric matrices

@dusty gull

[dusty gull]

okay so $A^{T} = B - C$

[ocean seal]

April | Koi of Mahjong

[vale crag]

right, so you have 2 equations

A = B + C

A^T = B - C

[dusty gull]

$A - A^{T} = 2C \implies C = \frac{1}{2} (A - A^{T})$

[ocean seal]

April | Koi of Mahjong

[dusty gull]

Therefore $A = B + \frac{1}{2} ( A - A^{T}) \implies \frac{1}{2} (A + A^{T} ) = B$

[vale crag]

indeed

[ocean seal]

April | Koi of Mahjong

[vale crag]

now to be sure, you can check that indeed B is symmetric, C is skew-symmetric, and A = B+C

[dusty gull]

By properties of symmetry, we W.T.S $B^{T} = B$, so $\frac{1}{2} ( A + A^T)^{T} = \frac{1}{2} A^{T} + \frac{1}{2} A^{T}^{T} = \frac{1}{2} A^{T} + \frac{1}{2} A = \frac{1}{2} \left( A + A^{T}\righgt)$ Therefore B is symmetric

[ocean seal]

April | Koi of Mahjong
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[dusty gull]

$\frac{1}{2} \left( A - A^{T} \right)^{T} = \frac{1}{2} A^{T} - \frac{1}{2} A^{T}^{T} = \frac{1}{2} \left(A^{T} - A \right) = -\frac{1}{2}(A-A^{T})$ Therefore C is skew-symmetric =

[ocean seal]

April | Koi of Mahjong
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[dusty gull]

$A = B + C \implies A = \frac{1}{2} \left( A + A^{T} \right) + \frac{1}{2} \left( A - A^{T} \right) = \frac{1}{2} A + \frac{1}{2} A + \frac{1}{2} A^{T} - \frac{1}{2} A^{T} \implies A = A$

[ocean seal]

April | Koi of Mahjong

[dusty gull]

Does this show what we're trying to prove?

[vale crag]

yeah gg

[dusty gull]

Okay! Thanks 🙂

[vale crag]

and technically the thing we just did also takes care of the uniqueness

"if A = B+C, with B symm and C skew-symm, then B = ... and C = ..." that's what we did

there can only be one possible solution for a given A

@dusty gull

[dusty gull]

okay

[lone heart]

@dusty gull Has your question been resolved?

Channel closed

Closed by @dusty gull

Use .reopen if this was a mistake.

[lone heart]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[reef pebble]

can someone tell me how to solve this? how

[vale wigeon]

this is not an equation

there's nothing to solve

did you mean simplify?

[reef pebble]

my bad lmao

yeah simplify

[winter light]

First of all make common denominator

[reef pebble]

oh shit i think i got it

thanks everyoen

[limpid spade]

youre welcome

[reef pebble]

.close

[lone heart]

Channel closed

Closed by @reef pebble

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[ember dagger]

I don’t get this at all, I’ve been trying for a good 15 minutes now and I’ve gotten all the problems wrong

[ember dagger]

!status

[lone heart]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[ember dagger]

1

[somber eagle]

try and put each inequality with y as the subject

ie y < ax + b

[wary stream]

That looks like a khan academy problem, did you watch the videos?

[ember dagger]

6x-2y>-11
-2y>6x-11
y>-6/2x-6

[ember dagger]

Yeah, I did

[wary stream]

First, you can simplify 6/2

Second, when you multiply or divide by a negative, you have to flip the inequality sign

[winter light]

And you should get 11/2 not -6

[ember dagger]

my apologies, I got it mixed up

[winter light]

Dw

[ember dagger]

y>-11/2x-6

So I can graph this already

It’s saying that it is greater than

So it should be above

[somber eagle]

11/2 should be the constant

not the x

[ember dagger]

So it’s A or D

Oh

[somber eagle]

and the 3 should be the x

instead of making it y >

try and put it as y <

[wary stream]

.

[somber eagle]

instead of dividing by negative numbers, try just moving the y to one side and the other terms to the other

[ember dagger]

Ohhh

[ember dagger]

Got it

So it’s

Y>3x+6

[wary stream]

Using this inequality 6x - 2y > -11 ?

[ember dagger]

Wait I don’t get it

I thought that was wrong

[tardy tapir]

why were there 116 pings just from this channel

[wary stream]

6x - 2y > -11
Original problem

Y>3x+6
Your answer

I was confirming if that was that same inequality or a new problem

Because I'm trying to see how you got +6

When the problem had -11 to begin with

[ember dagger]

Yea I put it into slope intercept form so it’s easier to graph

[wary stream]

Because I'm trying to see how you got +6
When the problem had -11 to begin with

How did you get +6?

[somber eagle]

also it should be less than

[ember dagger]

Whattttttt

Ok

[wary stream]

Show your full working from 6x - 2y > -11 to y > 3x + 6

[ember dagger]

6x-2y>-11
-2y>6x-11
y>-6/2+6

[somber eagle]

you dont keep the 6x positive

[ember dagger]

Oops typo

[somber eagle]

also -11/2 doesnt equal 6

[wary stream]

6x - 2y = -11
If you were given that, could you set that equation equal to y?

[ember dagger]

no

[wary stream]

You don't know how to solve for y?

Like making it y = ...?

Because if you don't, then I suggest reviewing equations before going to inequalities

[ember dagger]

I Can

[ember dagger]

I don’t get why we are setting it to Y

[wary stream]

To get it into the slope intercept form

You can graph it much more easily if it's in slope intercept form, right?

Like what's easier to graph, 6x - 2y = -11 or y = mx + b?

[ember dagger]

y=mx+b of course

I was following by this example

[wary stream]

Now as asked, given 6x - 2y = -11 can you set it equal to y?

[ember dagger]

Yeah

[wary stream]

So do it

[ember dagger]

y=6x-2y=-11?

[wary stream]

Setting it equal to y means solve for y

6x - 2y = -11
Solve for y

[ember dagger]

Ok

6x - 2y = -11
-6 -6
-2y=-17
—— —-
-2 -2
-17/2

[wary stream]

No

What happened to the x?

It's 6x

Why did that x magically disappear?

[alpine sable]

What the hell, 100+ pings from this channel??

[ember dagger]

Huh 6-6=0

[wary stream]

6x - 2y = -11
You have 6x

You subtracted 6

6x - 2y = -11
-6 -6
-2y=-17

What happened to the x?

[ember dagger]

Oh

Wait hold on

[ember dagger]

Wdym

6x-6x

= 0

[wary stream]

Look at your work

[ember dagger]

I forgot to type in the z

X

[somber eagle]

you are not subtracting 6x

[wary stream]

You made the x disappear

6x is not the same as 6

6x - 2y = -11
-6 -6
-2y=-17

[ember dagger]

You subtracted 6

6x - 2y = -11
-6x -6x
-2y=-17

[wary stream]

No

-11 - 6x is not -17

[wary stream]

I'm still standing by this statemet

[ember dagger]

HUH

[ember dagger]

?????????

[wary stream]

Because if you don't, then I suggest reviewing equations before going to inequalities

You are failing to understand how to set equations equal to a variable

[ember dagger]

I don’t Even understand what you are trying to say

[wary stream]

reviewing equations before going to inequalities
Do you understand this part?

[ember dagger]

Yeah

[ember dagger]

Solving for Y?

[wary stream]

Yes you're solving for y

[ember dagger]

y = -11/2 - 3x

????????

[wary stream]

Almost

[ember dagger]

Ohnyvod

[wary stream]

Can you show the full work on how you got that?

[ember dagger]

Nevermind i got lost in my own math I don’t even know how or why I did that

[wary stream]

The question still stands, can you solve for y, in this equation, 6x - 2y = -11

[tardy stag]

this is close, instead of just dividing by 2 you need to divide by -2 since you had -2y on the left

[lone heart]

@ember dagger Has your question been resolved?

[solid stirrup]

sin of what is equal to pi

[wary stream]

Not your channel to ask help in

[solid stirrup]

ok sorry

[lone heart]

Please read #❓how-to-get-help

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[lone heart]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[torn isle]

Hi, can anyone guide me how to solve this?

The answer given is this

May I know how to solve from (b)

Like how we see it?

I got this

Why k can be directly equal to n(n+1) like that shown in the general equation of the legendre’s equation?

[lone heart]

@torn isle Has your question been resolved?

[torn isle]

<@&286206848099549185> anyone can help me see how to get the legendre equation from this form by suitable change of variable?

[junior idol]

Wouldn't the default substitution $x=\cos(\theta)$ work?

[ocean seal]

DAILI

[torn isle]

So no need change to cot theta like what I did?

[junior idol]

Depends on which derivatives are easier to convert I guess

[torn isle]

May I know how to get legendre form?

All the d theta change to dx as well?

[ocean seal]

DAILI

DAILI

[junior idol]

Could be wrong with this one

Been a while since I did stuff like this

Also

Is it jsut me or is your Legendre's diff eq wrong

[torn isle]

I see this question just 4 marks should be pretty easy but I just don’t know why ever since part b I don’t know how to do 🥲

[torn isle]

Which one?

[junior idol]

This is different no?

Like $(1-x^2)$ instead of $(1-x)^2$

[ocean seal]

DAILI

[torn isle]

Oh, I think that YouTube put wrongly

In my formula sheet for legendre’s, yours is correct

[junior idol]

Yea thought so

[junior idol]

Do you understand this formula

This is just the chain rule

[torn isle]

You mean side the first two terms?

[junior idol]

Yes it's like differentiating f(g(x)) where f is F and g is x and x is theta

[torn isle]

Something like this right?

[junior idol]

you use product rule first for this stuff

[junior idol]

Start from this its easier

Then let $x=\cos(\theta)$

So that

[ocean seal]

DAILI

[junior idol]

$\frac{\mathrm{d}x}{\mathrm{d}\theta} = -\sin(\theta)$

[ocean seal]

DAILI

[junior idol]

And also $\sin(\theta) = \sqrt{1-\cos(\theta)^2} = \sqrt{1-x^2}$

[ocean seal]

DAILI

[junior idol]

And then you should get the desired result

[torn isle]

I still don’t see how to change 🫠

May I know Second derivative how to change?

[tacit arch]

well did you plug in your expression for dF/dx * dx/dtheta into the (...) ?

[torn isle]

I got something like this

Got -x and not -2x

Or is there any mistakes I made?

[torn isle]

Here

[tacit arch]

where's your dx/dtheta explicitly?

it's muddled in with your other calculations

[torn isle]

this onky -sin theta

[tacit arch]

how did you get from the top line to the bottom one here

[torn isle]

This actually I don’t really know how to solve

[tacit arch]

you should be following known formulas and not making up your own

googling "second derivative chain rule" gives
https://math.stackexchange.com/a/865491

try following that

[torn isle]

Okay, I’ll search for that

[torn isle]

Okay I got it

So after I got the form in legendre equation, for the k, I just compare with the standard legendre equation and that’s why I got n(n+1)?

May I know what to do with considering the boundary condition at the poles and what it meant by state the restriction on k?

[tacit arch]

Did you plug in cosine at those endpoints

And maybe sine?

[torn isle]

No I didn’t

I just solve and then compare with the standard equation and assume that k=n(n+1)

[torn isle]

Okay, nvm, I think I got it

When plug in we’ll get <infinity and it’s must be finite then only we come out the conclusion that k=n(n+1)

For d May I know how to solve?

Also starting from du/d theta?

Or I should directly split them into F(theta) and G(t)?

[tacit arch]

observe that u is a product of a function of theta and a function of time.

[torn isle]

I solved for G which is w

But not for F(theta)

Let me continue to think of that

[torn isle]

Yea, I thought of this and solve using this for G(t)

[tacit arch]

You just read off what G is

[tacit arch]

G is not equal to w

[torn isle]

Not something like this?

I started with this also

[tacit arch]

Yea that looks right

[torn isle]

You mean which pic?

The one I start with?

I don’t know how to show they satisfy both DEs

[tacit arch]

Well plug u in to both

[torn isle]

What do you mean?

This?

[tacit arch]

You plug u into the two equations in 4a)

[torn isle]

I don’t know what you mean plug u into two equations

[torn isle]

Is it like the latest I did

But it’s long

I don’t think 6 marks question will require long workings

[tacit arch]

Yes this is the right start, but you forgot chain rule on F

[tacit arch]

Then don't do it?

[torn isle]

Not sure, only this questions already spent me 6 hours plus 🙂

Maybe in real exam

I’ll just give my lecturer this marks

[tacit arch]

Doesn't seem that long if this is your first PDE problem

[torn isle]

Maybe because I don’t know

I already solving d for hours 🙂

[tacit arch]

Didn't you only just finish part b

[torn isle]

Yes

But it’s talking about show that one (a)

Nvm, it’s already 3am in my area 🙂

Tomorrow I’ll ask again in this channel

.close

[lone heart]

Channel closed

Closed by @torn isle

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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 moon]

Might be a blurry so I can resend

[lone heart]

Channel closed

Closed due to the original message being deleted

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[dense wind]

If a set is countable, there exists a bijection between Y and the set of Natural Numbers

But I can’t tell if there is or isn’t a bijection between the two

And also how is part 2 of the question different from part 3?

[mortal trellis]

countable could also mean finite. depending on definition

[dense wind]

I think this class accepts infinite sets as countable too

[mortal trellis]

thats not what I mean

countable should always include countably infinite sets

the question is whether it also includes finite sets

in which case the second question is slightly different from the third (the second asks whether it is finite or countably infinite, the third just asks whether it is countably infinite)

[merry depot]

what do you mean there isn't an element associated to the second natural number?

[dense wind]

Like we skip 4 numbers everytime we plug in an Natural number in the function

[merry depot]

which is fine.
plug in 1 you get 5
plug in 2 you get 9
so there's your association.

[dense wind]

Oh wait that’s all is needed to prove countable?

That for each n we can get a distinct f(n)?

[wispy ibex]

Yep this defines a bijective map between the set of natural numbers and Y

[mortal trellis]

well and that each element in Y is equal to f(n) for some n

[lone heart]

@dense wind Has your question been resolved?

[lone heart]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[alpine sable]

what kinda maniac writes inbetwhen the lines like that 😭

[lone heart]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[alpine sable]

Trying to figure out this problem but I'm really stumped https://i.imgur.com/uAuwaYt.png
I'm not exactly sure how to get dy/x from this. I know about the chain rule applying to y to make dy/dx, but unless if I'm mistaken it wouldn't apply here since its simply ln(y) which should just be 1/y. If I go from 1/y, I would end up having to derive it again to get dy/dx but I'd need the 2nd derivative of f and g which we havent learned how to do yet.
Even once I find dy/dx, I'm not exactly sure how the f'g/fg' would end up dividing? Their derivatives should add, how does one end up dividing the other?
Just learned about explicit/implicit differentiation btw, ty for all the help!

[alpine sable]

Also just had the thought of f' and g' being dy/dx but I don't know how that'd work out

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

<@&286206848099549185> im supposed to ping now i think :)

still been trying to work on it but Ive honestly no clue

[near apex]

Differentiate w.r.t. x and d(lny)/dx = 1/y * dy/dx.

[near apex]

Just start and show us where you get stuck.

[alpine sable]

Okay I get that part now after thinking about it lol, it's in terms of x not in terms of y (still iffy on derivative notation) so now I can attempt the problem but still a bit stuck

Driving everything gets (I'ma be lazy writing the functions)
f'g + fg' = 2 + 1/y(dy/dx)
Subtract 2 and multiplying y gets me
y(f'g + fg' - 2) = dy/dx, but that doesn't even resemble one of the options

PS: Writing this in the car w/ motion sickness so lmk if my brain turned off somewhere lol

[near apex]

Your implicit differentiation of LHS is wrong. try again.

Remember with respect to which variable are you differentiating.

[alpine sable]

Wdym LHS?

[near apex]

Left hand side.

of f(x)g(y)

Basically,

$$\frac {\dd {\left(f(x)(g(y)\right)}}{\dd {x}}$$
Using multiplication rule,
$$ = \frac {\dd {\left(f(x)\right)}}{\dd {x}} \cdot g(y) + \frac {\dd {\left(g(y)\right)}}{\dd {x}} \cdot f(x) $$

Here, notice that you are differentiating g(y) with respect to x in second term so you need to use chain rule.

[ocean seal]

Enemagneto

[near apex]

@alpine sable

[alpine sable]

Is f(x)/dx different from f'(x)?

[alpine sable]

@near apex oop

[near apex]

No, those are same.

[near apex]

If you meant d(f(x))/dx.

[sudden burrow]

$g'(y)$ is equal to $\frac{d(g(y))}{dy}$. We want the derivative with respect to $x$, not $y$, so we need to use the chain rule to get $\frac{d(g(y))}{dy} \cdot \frac{dy}{dx}$.

[ocean seal]

Calculustache

[alpine sable]

Aight I think I got it
Driving brings
f'g + g'f(dy/dx) = 2 + 1/y(dy/dx)
Swag around a bit
f'g - 2 = 1/y(dy/dx) - g'f(dy/dx) or dy/dx(1/y - g'f)
d/dx = (f'g - 2)/(1/y - fg')

This look good? TYSM for all your help btw :)

[sudden burrow]

That looks right.

[alpine sable]

Ok had to eat lol ty!

.close

[lone heart]

Channel closed

Closed by @left siren

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[unborn hatch]

g

[unborn hatch]

.close

[lone heart]

Channel closed

Closed by @unborn hatch

Use .reopen if this was a mistake.

[lone heart]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[proven leaf]

I have a recursive definition of a series and am looking to find an explicit formula, this is the defintion:

$$S_{p}=\binom{p}{1}S_{p-1}+\binom{p}{2}S_{p-2}+\binom{p}{3}S_{p-3}+\cdots$$

I was recently pointed to the Bell Numbers, however they have a different recursive definition than this series and I'm beginning to beg the question is it even feasible to try to find an explicit definition?

[ocean seal]

XxMrFancyu2xX

[royal socket]

How is this different to the bell numbers?

[near apex]

Is that P choose 1 ?

[proven leaf]

it is slightly different I can type them side by side if you want :)

[proven leaf]

yep!

[royal socket]

So by your definition

$$S_p = \sum_{k=1}^p \binom{p}{k} S_{p-k}$$

[ocean seal]

Im you

[royal socket]

or is it to p-1

Whats the upper bound?

[proven leaf]

to p :)

oh shoot I didn't include the end mbmb

[royal socket]

What's S0 defined as

[proven leaf]

$S_{0}=\sum_{n=0}^{\infty}\frac{1}{2^n}=2$

[ocean seal]

XxMrFancyu2xX

[proven leaf]

oh I should also include $S_{p}=\sum_{n=0}^{\infty}\frac{n^p}{2^n}$

[ocean seal]

XxMrFancyu2xX

[proven leaf]

idk if it helps

¯_(ツ)_/¯

[royal socket]

Hmm interesting... can we compare it to the bell numbers

[proven leaf]

,w Bell Numbers

[ocean seal]

Results provided by WolframAlpha

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

Requested by XxMrFancyu2xX#6050

[proven leaf]

so bell numbers are 1,1,2,5,15,52,203, etc.

mine start 2,2,6,150,1082 etc.

I don't have a proof nor know of one but it seems like my guy grows faster than the Bell Numbers

[royal socket]

I mean I recognise that its similar to the lerch transcendent?

[proven leaf]

it can be expressed in terms of that

[royal socket]

Actually i put this into wolfram and it did come up with that

[proven leaf]

but I would prefer elementary

[proven leaf]

ye it's like LerchPhi[1/2,-p,0] or smthn like that

[royal socket]

not sure if this will help but

compare to the polylogarithm

Its urs but

the summand is reciprocal

And i dont think there would be an elementary representation

[proven leaf]

yea that also came up when I tried manipulating some stuff

[royal socket]

usually there isnt for yucky stuff like this

we can keep trying though ill try some stuff

[worn fox]

what exactly is it you're looking for that would be better than this

[proven leaf]

something that's doesn't include the sum and uses elementary functions ig

[royal socket]

Have you seen the oeis entry for this?

The formula section doesnt show any "elementary" formula youre looking for im afriad

[proven leaf]

yea but it does show the recursive one I derived :(

[royal socket]

Ah

[proven leaf]

I guess really something with all those binomial coefficients, hoping there was some pattern by substituting in S_{p-1}

the same was one derives the explicit arithmetic formula

[worn fox]

its 2^A000670(n) and A000670 has a ton equivalent definitions

[royal socket]

Your sequence is exactly twice the "ordered bell numbers"

[royal socket]

2*A000760

[worn fox]

we're both wrong

2*A000670

[proven leaf]

because it's my exact definition but the initial condition is 1 instead of 2 :(

I already tried Fubini numbers

[royal socket]

funny.
$$S_p = 2 \left( x\frac{d}{dx} \right) ^p \left( \frac{1}{2-x} \right) \Big|_{x=1}$$

[ocean seal]

Im you

[proven leaf]

woah where did that come from?

haven't seen that

[royal socket]

for the fubini numbers: a(n) = ((x*d/dx)^n)(1/(2-x)) evaluated at x=1. - Karol A. Penson, Sep 24 2001

[proven leaf]

interesting...

so that means $S_{p}=2x^{p}\frac{d^p}{dx^p}\left(\frac{1}{2-x}\right)\Big|_{x=1}$?

[ocean seal]

XxMrFancyu2xX

[royal socket]

idk why does that not look very right

surely then x^p is just 1 lol

what

[proven leaf]

p factorial? but yea idk

[royal socket]

nah but the x^p is outside the operator?

[proven leaf]

then what does x mean

if you solve the derivative at x=1 then we just have an x^p hanging around, no?

[worn fox]

its meant to be like $x\frac{d}{dx}\left(x\frac{d}{dx}\left(\frac{1}{2-x}\right)\right)$ etc

[ocean seal]

ΣΑC

[royal socket]

AH right yeah

that is so ugly

[proven leaf]

yeaa that's pretty bad

[royal socket]

yeah i dont think theres gonna be like an elementary sol cuz sums and integrals they can just be ugly like that

[proven leaf]

tuff well I guess the recursive is as nice as nice is going to get

[worn fox]

oh it has an exponential generating function

[royal socket]

or the original sum tbh

[proven leaf]

that's for like a power series I think

[worn fox]

$\frac{1}{1-\frac{e^x}{2}}$

[proven leaf]

I remember seeing a video on it I believe

[ocean seal]

ΣΑC

[proven leaf]

,w Table[1/(1-e^n/2),{n,1,10}]

[ocean seal]

Results provided by WolframAlpha

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

Requested by XxMrFancyu2xX#6050

[proven leaf]

ah yes thanks wolfram I didn't know that

[worn fox]

S_p will be the pth derivative evaluated at 0 iirc

oh maybe not, exponentials gen functions different to usual

[proven leaf]

My original thought was:
$$S{p}=\binom{p}{1}S{p-1}+\binom{p}{2}S{p-2}+\binom{p}{3}S{p-3}+\cdots+\binom{p}{p-1}S_{1}+\binom{p}{p}S_{0}$$\
$$S{p-1}=\binom{p-1}{1}S{p-2}+\binom{p-1}{2}S{p-3}+\binom{p-1}{3}S{p-4}+\cdots+\binom{p-1}{p-1}S_{0}$$\

Hence,
$S{p}=\binom{p}{1}\left(\binom{p-1}{1}S{p-2}+\binom{p-1}{2}S{p-3}+\binom{p-1}{3}S{p-4}+\cdots+\binom{p-1}{p-1}S_{0}\right)+\binom{p}{2}S{p-2}+\binom{p}{3}S{p-3}+\cdots+\binom{p}{p-1}S_{1}+\binom{p}{p}S_{0}$

however, it can be proven (shown is better word) that this method requires O(2^n) time so it takes really long to compute as so I couldn't find a pattern after even substituting S_{p-4} so gave up on the method :(

[ocean seal]

XxMrFancyu2xX

[proven leaf]

.close

[lone heart]

Channel closed

Closed by @proven leaf

Use .reopen if this was a mistake.

[lone heart]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[ember isle]

So here is the true/false exercise:

[ember isle]

Does it suffice to just assume c1=c2=1, then T(v1 + v2) = T(v1) + T(v2)
then assume c2 = 0, so c1T(v1) = T(cv1)
and say that this satisfies the definition for a linear transformation and thus is true?

i'm mostly asking because i'm not sure if that's alright to claim

[merry depot]

that's fine

[ember isle]

alright, thanks!

.close

[lone heart]

Channel closed

Closed by @ember isle

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[jolly plover]

How should I solve for $a_n$ given that $a_1=1$ and $a_n=\frac{n+1}{n-1}a_{n-1}+(n+4)/3$