help-23

[lime dust]

up to you

[pure yew]

I just wanted to solve this equation using subtitution method:
x^2-4x+y^2=0
x-y=1
so i did like this: x = y+1
(y+1)^2-4(y+1)+y^2=0
2y^2-2y-3=0 from there i dont know what to do

[lime dust]

We were almost done.
2y^2-2y-3=0

And this
(y-1/2)^2 = 7/4

are the same

we only needed to square root both sides

sqrt((y-1/2)^2) = sqrt(7/4)

|y-1/2| = sqrt(7)/2

y-1/2==Sqrt[7]/2 or y-1/2==-(Sqrt[7]/2)

[pure yew]

what do u mean by |y-1/2|

[lime dust]

So y = (1 + sqrt(7)) / 2
or y = (1 - sqrt(7)) / 2

[pure yew]

what the hell

[slow sequoia]

There is another way like using delta

[pure yew]

in answer it says When x = 0 y = -1; x=1 y=0

[lime dust]

You should read the whole process from here #help-23 message

I am talking about your initial equation

Which is what we were doing the whole time.

[slow sequoia]

Have you guys tried using delta

[pure yew]

what is delta?

[lime dust]

Muta, you have to check the basics again.

[slow sequoia]

Delta id calculated with
b^2-4ac
Then if delta<0 there is no solution
If delta=0 there is one solution
And if delta>0 then there is multiple solution

[lime dust]

You can't solve 2nd degree equations if you have not studied them

[slow sequoia]

Yes

[pure yew]

i tried i got D = 28

but its wrong because i cant do sqrt(28) :/

[lime dust]

sqrt(28) is 2sqrt(7)

because 28 is 4*7

and 4 is 2^2

[pure yew]

but its not whole number like answer is in whole number form

[lime dust]

but that's not because of the equation you gave me

[slow sequoia]

Then you follow another formula after that
y1=-b+sqrt(b)/2a
y2=-b-sqrt(b)/2a

[lime dust]

anyways, i'm checking it and that's an error

x = 0 y = -1; x=1 y=0 This is not the answer

Where did you get that from?

[pure yew]

exercise book

[lime dust]

the book is wrong, or you copied it bad

x^2-4x+y^2=0; x-y=1

For this system of equations you can clearly see that your solutions are wrong

[slow sequoia]

The book is right lmao

[lime dust]

by replacing them in the equations

[lime dust]

I said either the book is wrong or the exercise is not written correctly here

[slow sequoia]

Yes

Muta can you rewrite the exercise or take a picture

[lime dust]

x=3/2-Sqrt(7)/2, y=1/2-Sqrt(7)/2

x=3/2+Sqrt(7)/2, y=1/2+Sqrt(7)/2

These are the solutions for his system of equations

[slow sequoia]

I scrolled up a bit, there is a lot

[lime dust]

not related

we were solving 1 single equation

then he said it came from a system of equations

x^2-4x+y^2=0; x-y=1 this one

[pure yew]

x^2-4x+y^2=0
x-y=1

When x = 0 y = -1 ; x = 1 y = 0

[lime dust]

lol

[pure yew]

its written in a book :/

[lime dust]

totally out of context

[slow sequoia]

Yes

They just asked you to find solution where x and y give 0 with the equation

You just have to determine a value to one of these and resolve

[pure yew]

no they asked to solve

[slow sequoia]

????

[pure yew]

this is written in the back of book of answers

[lime dust]

What does solve means to you?

I understand now

so in yout book of answers

they put this system x^2-4x+y^2=0; x-y=1

and they say the answers are 0, -1, 1 and 0?

i don't think it says that but

if that's what they say

the book is wrong

[pure yew]

:/

[lime dust]

can u take a picture

or better

can you read the exercise without solution?

i mean, in the book of questions

not the solution

[slow sequoia]

Its very confuse

[lime dust]

you can check yourself

When x = 0, y = -1
x^2-4x+y^2=0
0^2-4(0)+(-1)^2=0
1=0

wrong

as you can see, either the book is wrong, or you are not reading it properly

[pure yew]

Ok here is word by word:```System of equations:

x^2-4x+y^2=0
x-y=1

The number of solutions is

A. 0
B. 1
C. 2
D. No soliution```

[lime dust]

lol

so, you solve it, you get this
x=3/2-Sqrt(7)/2, y=1/2-Sqrt(7)/2

x=3/2+Sqrt(7)/2, y=1/2+Sqrt(7)/2
and you answer

C) 2

if the book is saying other thing, it's wrong, probably a typo or talking baout other exercise

[pure yew]

but why it says in answers when x = 0 y = -1

[lime dust]

because books are made by humans, and they can get confused

propably talking about other exercise

[pure yew]

nah this book is revised in printing there are no mistakes

[lime dust]

As I told you before.

1. The book is wrong
   or
2. You're not reading correctly.

The system of equations you gave me have the solutions I provided

Period

[pure yew]

1. I still dont undearstand how to solve this

[lime dust]

Then, read the whole explanation I gave before again until you understand it.

[pure yew]

i dont undearstand pizzas

random numbers popping

i dont know what to do

nah f this i cant spend more than hour

Thank you for your patience and explanations you provided

.close

[safe radish]

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[pale crag]

In how many ways can we arrange the letters murmur, such that no two alike letters are together?

[pale crag]

Ive been trying to solve this for the past hour..

[pseudo scroll]

Draw a Venn diagram

[pale crag]

MY MIND IS LAGGING RN

[pseudo scroll]

I don't blame you

I've solved something similar before

It's pretty hard for the first time

[pure yew]

4 ways?

[pale crag]

Tried that it didn’t work

[pale crag]

Idk tbh

Show me how

[pseudo scroll]

Lol

Are you sure

[pale crag]

Yes.

[pseudo scroll]

Three bubbles

[pale crag]

Bro ive seen smthn similar on quora

[pseudo scroll]

that's objectively the best way to do this

[pale crag]

Tried applying it here

Idk ugh

[idle cave]

Rip

[pure yew]

1.rmumur
2.rmrumu
3.rumrmu
4.rumurm

[pale crag]

U can start with m or u

Its gonna give u more than that

I dont think it can be solved like that i got lost while doing it

[pure yew]

1.mumurr
2.mumrur
3.umrumr
4.umrmur

ok 2 ways

[lime dust]

There are 30 ways

[pale crag]

Uh idts

[pale crag]

Pleaseee show me how

[pure yew]

damn

[pale crag]

I appreciate the effort man

[lime dust]

First, tell me how many ways to rearrange them.

[pale crag]

90

[lime dust]

ok

u have laready something done?

can i see all your job?

or u're just too fasT?

xD

[pale crag]

I tried counting two letters as one

[lime dust]

ok

how much would be that?

[pale crag]

But then didn’t know how to know the repetition of the rest

[pale crag]

30

[lime dust]

usually, when someone ask you for something, u look the opposite

[pale crag]

Yea thats what ive tried ig

[lime dust]

but

what is 30 exactly?

[pale crag]

The arrangements Where RR exists

[pure yew]

why number 30 exactly?

[pale crag]

I tried multiplying that by 3

[lime dust]

so u have 30 for MM, 30 for RR, 30 for UU

[pale crag]

But then i should know when did the letters appear twice right?

[pale crag]

Yup

[lime dust]

how many for MU together?

[pale crag]

30?

No wait

There wont be the same reputations

[lime dust]

right, so how much?

[pale crag]

Lemme count that

[pure yew]

oh yeah hes right

[pale crag]

5!/2! ?

[lime dust]

why 5?

[pure yew]

urmurm
umrurm
umurmr
murumr
urmrum
rumurm
rmurmu
mrmuru
rmumru
umrumr
rumrmu
rmurum
urumrm
umrmru
mrumru
mrumur
mururm
rumrum
umrmur
urmumr
mrurmu
murmru
rumumr
mrurum
murmur
rmrumu
rurmum
urmrmu
rmumur```

[pale crag]

Considering MU one letter

[pale crag]

Damn

[lime dust]

lol

sry, i don't want to spit the answer, but the last thing u said is wrong

is not 5!/2!

[pale crag]

What is it

Its fine pls do

My mind is dying

Brain

Idk

[lime dust]

mine too lol

[pale crag]

NOOO

U CAN DO IT

I believe in u

[lime dust]

it's fine, i already know how to do it

[pure yew]

wait so its not 30???

[lime dust]

it is

[pure yew]

good

[lime dust]

sry i think i asked the wrong question

[pale crag]

Its fine go on

[lime dust]

give me 1 second, i'm trying to find the good question

it is 4!/2!, i'm just exhausted

and can't find the way to ask u

[pure yew]

i just did it by hand

[lime dust]

these ones

are the ways to arrange the letters so that both

for example RU are kept together

but remember that you have to substract this from the ways to rearrange 3 letters

like RUM

which is

3! right?

[pale crag]

Hold on lemme process that

Oh okay

So ways to keep two different letters-the ways to keep 3 diff letters right?

[lime dust]

yes, but i'm stupid

there is a much easier way to solve this

[pale crag]

Youre nottt go on

[lime dust]

ok we finish with this solution

so it would be

90-30-30-30

that's 0

12+12+12 36

-6 = 30

but i have a better solution for this problem

we are gonna make this

MU _ _ _ _

now the next letter is an R that can be here

[pale crag]

Okay

[lime dust]

MURxRx
MURxxR
MUxRxR

starting the word with MU 2+2+1 = 5

right?

2! for the first

[lime dust]

1 for the second

2! for the third

[pale crag]

Okay

[lime dust]

There are three possible ways to choose the first letter of the word, and two possible ways to choose the second letter.

So the number of ways would be 3 * 2 * 5

which would be 30

[pale crag]

Wait

I actually get that

[pale crag]

But why is the second one 1

[lime dust]

this is only possible because the letters repeated wice

[lime dust]

let me re-check

sry it's 2! for the second

and 1 for the third

i said wrong

[pale crag]

Oh yah it makes sense now

[pure yew]

So for each pair of alike letters, calculate the number of permutations where those two letters are adjacent, then subtract the sum of all these values from the total number of permutations to get the final result, in your case the word murmur, there are 90 total permutations without any rules set, so there are 3 pairs of alike letters: mm, rr, and uu, following that each pair has 30 permutations where they are adjacent. So, the final result is 90 - 3*30 = 0.

[pale crag]

Cuz we cant put U

Its always gnna be M

[lime dust]

yes

[pale crag]

Wait imma try to understand this as well

[pale crag]

Omg tysffmmmmmm

[lime dust]

I'll make a draw

amazing drawing skills

[pale crag]

Check this out

Im a good student 🌹

[pure yew]

damn

[lime dust]

correct

muta, go close your question, i told u your book is wrong xDD

[pale crag]

Thank u so much guys

Really i would be sobbing rn if it weren’t for u

[lime dust]

you're welcome

[pseudo scroll]

I respect your dedication

[pure yew]

somehow inclusion exclusion principle is more easily undearstandable than my systems of equation

xd

[safe radish]

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[unreal magnet]

Looks like trig substitution

I forgot which one

[plucky osprey]

it should be sec theta

but i couldnt continue

[surreal stump]

yeah post work homie

[unreal magnet]

^

[plucky osprey]

wait doing it again

its all messed up

[surreal stump]

Gotcha, take your time for it to be legible. Also, symbolab is super helpful for finding the specific substitution sometimes for a quick check.

[plucky osprey]

could i do it like : tan^2 (x) + 1 = sec^2 (x)

so that would make the integral of sec^2 (theta) - 1
which equals tan(theta) - theta ?

.close

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[vale hollow]

Linear Algebra help

[drowsy moss]

they did the row reduction on [A | 0]

that's the first column after the | in the matrix

[vale hollow]

The row reduction on [A|0] is just this:?

[drowsy moss]

yes

[vale hollow]

Ok so from this we know that x_1 = -2

How did they get x_2 to be 1?

[drowsy moss]

and the the [4 8] is the next column, and [-2 -4] is the third column

[vale hollow]

How did they get this?

[drowsy moss]

So converting it back to an equation gives you x + 2y = 0, and x = -2y

[vale hollow]

There aren't any free variables so how it there a T?

[drowsy moss]

so your vector [x y] is [-2y y] = [-2 1]y

or t, doesn't matter what you call the variable

[vale hollow]

ah ok

.close

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[mossy wadi]

why is there a flat line between 1/2 and 8?

[mossy wadi]

when i try to remove the root and the power it gives a differen't function and im not sure why

[lean otter]

$\sqrt{a^2}=|a|$

[flat frigate]

Toby

[safe radish]

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[jovial fjord]

what is wrong with my answer?

[jovial fjord]

The above is my solution, anything wrong?

btw, I also tried replacing them with u and v, which is also not right

[misty bay]

sorry I can't really follow your work

[jovial fjord]

so z_x basically means dz/dx

[misty bay]

did you get dz/dx = e^y, dz/dy = (y+1)e^y + xe^y?

[jovial fjord]

yeah I got it wrong with the dz/dx

but shouldn't dz/dy be (y+1)e^y+xe^y?

[misty bay]

yes it should

[jovial fjord]

now I got the second one right

[misty bay]

what did you get for dx/du and dy/du?

[jovial fjord]

got it all right now, thanks man!

just gotta replace all of them with u and v

so pain

.close

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[signal ocean]

.

How do i solve 1 and 2 in IX)

[safe radish]

@signal ocean Has your question been resolved?

[signal ocean]

Not but my genius big brain solved it on it self

.close

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[unique valley]

For part b, how is gH the set of left cosets

[lean otter]

gH is a left coset by definition

wdym by "set"?

[unique valley]

In the question it refers to create a bijection between the set of left cosets and Y

[lean otter]

the symbol $\mapsto$ means "maps to"

[flat frigate]

Toby

[lean otter]

which means gH gets mapped to g(a)

the set of all possible gH is the set of all left cosets

[unique valley]

Isn’t this a bijection from a singular left coset to Y though?

[lean otter]

no, in this context, it defines a function on all possible cosets gH for any g in G

[unique valley]

I see, for the final statement can I just say by lagrange’s theorem |G|/|H|=number of left cosets =|Y|

[lean otter]

yes

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@unique valley Has your question been resolved?

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[junior wagon]

How does this prove the existence of i such that p^i(x) =x?

[junior wagon]

Is the contradiction literally just the fact that they have found that i = k-j+1?

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[visual aurora]

Why can't I do this?

[finite yacht]

Because lim(cos(x) + x^3 sin(1/x)) does not exist

[visual aurora]

So when one of the two limits does not exist alone, I can't use the property of limit multiplication to separate the limit.

[finite yacht]

Exactly

[visual aurora]

Ok, thanks.

[finite yacht]

However, now that I remember it, in the case that one of them is 0 you don't need the other one to exist, it is enough if it is bounded

But in this case the second one goes to infinity

[safe radish]

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[crude phoenix]

How do I graph y=-2^(x-3) without an ordered pair?

[devout shale]

What do you mean without an ordered pair?

Could you send your original question

[crude phoenix]

I mean without making a list of when x=1, y=...

Js there a way to graph this by just looking at the numbers like you would on an graph of y=-(x+3)^3

[devout shale]

hmmmm well we can probably tell a few things without making a list

like for example

any number raised to the 0th power is equal to 1

so looking at -2^(x-3)

we can see this will be raised to the 0 at x=3

and that would give y=1

[crude phoenix]

Thanks, how would I find two other points?

[devout shale]

I actually need to correct myself first

this equation is actually (-1)*(2)^(x-3)

so at x=3, we get (-1)*(2)^(0)

which is actually -1

not 1

Does that make sense?

[crude phoenix]

So it's reflected

Oh yeah

[devout shale]

We can try to find the y-intercept too

that would be nice

that's when x=0

-(2)^(-3)

[flat frigate]

AustinU

[devout shale]

so that would be the y-intercept

[crude phoenix]

thanks

I think two points would be enough

[devout shale]

hmmm maybe

[flat frigate]

AustinU

[devout shale]

Does this function have an x-intercept?

i.e, can it equal 0?

[flat frigate]

AustinU

[crude phoenix]

I have another question
f(x) = -2f(-2x-6)-3. Is the correct equation [-1/2x-6, -2y-3]?

[devout shale]

I won't help with another question until you respond to my help with your last one :p.... but anyways I'd need more info for that one

[crude phoenix]

Yes

The base graph doesn't touch the x axis

[devout shale]

does it get close to it?

[crude phoenix]

Sry, I was typing it

[devout shale]

no worries!

[crude phoenix]

Yeah it does bc as the x value increases, the fraction is smaller

[devout shale]

yup, but you'd say, never an actual x-intercept

seems good to me

[crude phoenix]

Okay 👍

Thanks

[crude phoenix]

Could you help with this one?

[devout shale]

send the original question please

[devout shale]

I'd put in the chat here what your confusions are about this one, and someone else will come along and help out

[crude phoenix]

Wdym?

[devout shale]

describe what you need help with about this question

[crude phoenix]

I think i did it right but comparing it with someone else's answer it's not correct

Actually, I found my mistake

When factoring at the beginning, I forgot to factor the 2 out of the 6

[crude phoenix]

Thanks for the help!

.close

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[junior wagon]

I'm a little confused by the notation of Stirling numbers of first kind

[junior wagon]

In this example, what is n and k?

[safe radish]

@junior wagon Has your question been resolved?

[safe radish]

@junior wagon Has your question been resolved?

[junior wagon]

<@&286206848099549185>

[safe radish]

@junior wagon Has your question been resolved?

[safe radish]

@junior wagon Has your question been resolved?

[spiral bane]

I have no idea what the stirling numbers are but n and k are just numbers

i dont understand why you are confused

[safe radish]

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[full girder]

y'(x) = x, y(0)=y(1)= 0

[full girder]

I know this ODE can be solved using Green's Functions and I already did

But the question asks for the verification

[quasi bison]

... are you sure you wrote it correctly

[full girder]

So, how to do it normally?

[quasi bison]

as written, the equation has no solution

[full girder]

I get your concern, that is also what is bugging me

I got the Green's Functions and intergrating it gives me

[quasi bison]

integrating it twice?

did they maybe mean y''(x) = x?

[ionic solstice]

probably

[full girder]

Bruh 😭

I hate this course and the TA so much

This is the 5th time they made atypo

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[lean otter]

how do you know that cost = -cost is the same as pi/2, and 3pi/2

i have a unit circle and i see that this is when both cos are = 0

[devout shale]

So you are trying to find when cos(t) is equal to -cos(t)

[flat frigate]

AustinU

[devout shale]

whattt ahhhaahah noooo

that can't be!

so obviously something went wrong when you divided by cos(t)

what could it have been, to stop you from being able to divide by it?

well, if cos(t)=0 then you can't divide by it

that would be dividing by 0

which you can't do

So it is clear that this equation is only true when cos(t)=0

and that is your new problem to solve

using your unit circle you should see that the two angles that satisfy that are pi/2 and 3pi/2 (within 0 to 2pi)

Does that make sense?

[lean otter]

one sec just processing this lol

[lean otter]

so are you saying that you cannot mulitply by reciprocal then distrubute the -1?

[flat frigate]

AustinU

[devout shale]

So the error must have occured when you divide by cos(t) as that is literally the only step that we took to produce this false statement

so from that you can see that cos(t) must equal 0, and that is why the error arises

Does that soothe your mind?

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

i think i get it but thats mind twisting honestly

since you would not be able to divide by cos only if it were equal to 0

and cos is equal to 0 at those two points listed in the answer

[devout shale]

yes

[lean otter]

thats crazy no way im coming up with that on the test

[lean otter]

well thank you again

[devout shale]

Well, I was just trying to reason it out for you

the explanation can be much simpler

so don't pysch yourself out

when does negative (number) equal positive (number)

?

never

well

[lean otter]

yea

[devout shale]

when that number is 0

so that's the main idea

is - cos(t) = positive cos (t)

then cos(t) is 0

[lean otter]

oh i get it more now

since zero isnt postive or negative

[devout shale]

yeah, negative 0 equals positive 0

although that really doesn't mean much

just it isn't true for anything other than 0

[lean otter]

bro thatz crazy

[devout shale]

so that is what is useful

[lean otter]

ay good helping man

[devout shale]

https://tenor.com/view/spongebob-maths-spongebobmaths-gif-19232863

No problem!

Have a good one

[lean otter]

you too

.close

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[crude phoenix]

If f(x) = g(x-1) and g(x) = √(x+1), find f(4)

[quasi bison]

!status

[safe radish]

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

[crude phoenix]

Do I sub x in g(x-1) for √(x+1)

[quasi bison]

f(x) = g(x - 1)
what happens when you replace x with 4 in this equation?

[crude phoenix]

g(4-1)

[quasi bison]

f(4) = g(4-1).

or, f(4) = g(3).

are you now able to calculate g(3) and hence also f(4)?

[crude phoenix]

Yes, thank you

-2f(-1/2x) is the transformation applied to √x. Would I have √(-2x) or √(-1/2x) under the radical?

[quasi bison]

f(-x/2) = sqrt(-x/2)

and -2 f(-x/2) = -2 sqrt(-x/2)

[crude phoenix]

Can you explain why? I thought when I do 1/b the number would be flipped

[quasi bison]

??

what's "1/b"?

i am simply showing you what happens on a purely symbolic level when you evaluate -2 f(-x/2) for f(x) = sqrt(x).

your question was purely symbolic in nature and so is my answer to it.

[crude phoenix]

y=f(bx) and there is a horizontal compression by 1/b or stretch by 1/b if 0<b<1 on a graph

[safe radish]

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[crude phoenix]

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[west lily]

Injective, Surjective, Bijective, for functions in the complex plane

[west lily]

do the same tests that work for real functions work for complex ones as well?

[peak estuary]

yes

but well now you can take roots of negative numbers and stuff

so results may vary

a function that may have been injective over R might now not be anymore

same for surjective

[west lily]

so how would i work out this specific example?

[peak estuary]

well what would you do over R

[west lily]

ngl i've just been graphing and then seeing what they are

im struggling to graph this tho

[peak estuary]

oh

yeah graphing over C is not a good idea

work with equations instead

[west lily]

hmm

ok so:

for the function f(z) with complex numbers z_1 and z_2, equating f(z1) and f(z2) gives the same images

so therefore they can't be injective

how do i test for surjective

.close

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[coarse kelp]

how do I show convergence of the sequence:

$\Bigg\{\frac{\log_b(n)}{n}\Bigg\}, b>1$

[flat frigate]

steamhahasteamhaha

[coarse kelp]

I have no idea where to start

I was thinking maybe squeeze theorem

but I don't know how exactly

I don't know how to "bound"

$\log_b(n)$

[flat frigate]

steamhahasteamhaha

[quasi bison]

a variety of ways exist

do you want the from-scratch version or a more refined version

[coarse kelp]

i wanna do refined version first

[quasi bison]

$\ln(n) \leq \frac{1}{p} n^p$

[flat frigate]

Ann

[quasi bison]

for any positive constant p

for your purposes, any value of p less than 1 fits; if you're buridan's ass, take p = 1/2

and of course, log_b(n) = ln(n)/ln(b) so we might as well take the base as e

[coarse kelp]

hmmmm

[quasi bison]

do you need me to clarify anything else or are you going to continue

(don't answer "yes" to this one)

[coarse kelp]

what about the from scratch version

[quasi bison]

$\ln(n) < n$

[flat frigate]

Ann

[quasi bison]

whence the one for n^p is derived by some trickery

[safe radish]

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[sand ember]

Let m be an integer, let X_m = {p = n^2 + m^2 | n is a natural number} be the set of primes of the form n^2 + m^2. Prove that the Dirichlet density of X_m is 0 for each fixed m.

[trim dune]

ask this is #advanced-number-theory or #elementary-number-theory

in*

[sand ember]

No one ever posts in those I asked a question and it got ignored for 2 days

[safe radish]

@sand ember Has your question been resolved?

[sand ember]

I know that usually due to Fermat we have that p = x^2 + y^2 iff p = 1 (mod 4), but that seems different to here

[safe radish]

@sand ember Has your question been resolved?

[safe radish]

@sand ember Has your question been resolved?

[sand ember]

<@&286206848099549185>

[trim dune]

You're more likely to get an answer in those channels

I got an answer in them recently

Again I have not learnt anything about Dirichlet density but it seems like It would be worth noting that the sum of X_m is 2* all primes congruent to 1 mod 4

So maybe you could say something about the density of X_m vs X_n for m not equal n and deduce that they're 0 by considering the sum of densities or something

[sand ember]

I've asked it but I'm keeping this open for now in case someone can help me with it

[sand ember]

I know this, but this seems different

[trim dune]

I mean the union of X_m is the set of such primes

[sand ember]

If I'm understanding you right, you think I should do the union. We do have that if X_i intersection X_j is the empty set then dens(X_i u X_j) = dens(X_i) + dens(X_j), but we don't have that they're empty, e.g. 5 is in X_1 and X_2

So I don't think I can break it up like this

[trim dune]

yeah but it's like twice the density

or maybe not

there may be something to consider in the number of ways a prime can be written as a sum of two squares

[sand ember]

Idk if this is going to be the way, I think it'll have to do with fermat's theorem of the sum of two squares, but I don't why it wouldn't hold here

[trim dune]

yeah - this is building on that

idk

[sand ember]

I guess we could look at the specific formula

The dirichlet density is going to be $$lim_{s\rightarrow 1^+} \frac{(\frac{1}{m^2 + 1})^s + (\frac{1}{m^2 + 2})^s + ...}{log(\frac{1}{s-1})}$$

[flat frigate]

LeftySam

[sand ember]

Note here the top should be the primes of that form that is being summed over, so for instance if m = 1, then m^2 + 3 isn't a prime so that isn't summed over. In addition, m is fixed

[safe radish]

@sand ember Has your question been resolved?

[safe radish]

@sand ember Has your question been resolved?

[safe radish]

@sand ember Has your question been resolved?

[raw scroll]

.close

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[autumn sable]

Need help

After differentiating siny=tanx implicitly and getting cosy dy/dx = sec^2x I don’t know the directory to take

Or if it’s even the right one

[safe radish]

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[dusty talon]

how should i set this problem up?

[granite idol]

what would you guess?

[dusty talon]

i feel like something has to = 81

[granite idol]

yes..

[grizzled shoal]

What have you tried?

[dusty talon]

im not sure where to start, if i dont have momentum from the beginning ill just be confused

[grizzled shoal]

Momentum?
Momentum is mass times velocity

You mean velocity?

[dusty talon]

i mean momentum on doing the problem

not actual momentum in math

[quasi bison]

when the diver hits the water, how far has he fallen

[dusty talon]

81

[quasi bison]

81 feet, correct.

you have a function that gives the distance fallen by the diver in terms of time

now you need to put the two together

[dusty talon]

81/16 ?

[granite idol]

81/16 equals what

[dusty talon]

5.0625

[quasi bison]

ew, decimals.

also, is this number suspended in a vacuum?

[dusty talon]

then sqrt?

[granite idol]

no, I meant the equation you're working with

[dusty talon]

81/16 = t^2

[granite idol]

good

[quasi bison]

"I started with the equation 81 = 16t^2 and divided both sides by 16."

[dusty talon]

yes

[quasi bison]

you should make it a habit to put everything you do into words.

anyway, continue.

you now have t^2 = 81/16.

your goal is to solve for t.

keep going.

[dusty talon]

i sqrt both sides to get t= sqrt5.0625

which is t = 2.25

[granite idol]

don't turn it into a decimal number...

[dusty talon]

why not?

[quasi bison]

decimals, in general, are ugly.

and for fractions whose denominators aren't powers of ten or divisors thereof, also imprecise.

[granite idol]

and you can lose precision

[quasi bison]

that, or you'll have to always write 0.124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857124857... whenever you want to talk about the number one-seventh.

you might be under the impression that fractions are less "numbery" than decimals. let go of such bias.

[dusty talon]

ya thats it

[quasi bison]

let go of this misconception.

it will do you no good whatsoever.

[granite idol]

is that a thing?

[quasi bison]

sure is.

[dusty talon]

except for when they are clean numbers

[quasi bison]

"cleanness" is subjective.

i'd take 1/2 over 0.5 (and especially over .5) any day.

[dusty talon]

ok so now i know my seconds to impact

81 ft in 2.25 seconds

[quasi bison]

ok, cool, yes, you know this.

now read (and perhaps type out here) the second question.

[dusty talon]

With what velocity (in ft/s) does the diver hit the water?

[quasi bison]

ok

now think

how does the answer to this question relate to your function, d(t)?

[dusty talon]

velocity is related to d(t)

we go 81 ft in 2.25 seconds

and i think velocity is distance over time

but 81 / 2.25 = 36

and that is incorrect

[dusty talon]

.

[quasi bison]

you are wrong, but in fairness i did kind of not give many pointers in the right direction

you want the velocity of the diver at the instant he hits the water.

does this clarify what is being asked for?

[dusty talon]

yes

but i dont know why it isnt 81 / 2.25

[quasi bison]

81/2.25 would give you the average velocity of the diver over the course of his fall.

which is not the same thing as his instantaneous velocity at the moment of impact.

you understand the difference between these two, do you not?

[dusty talon]

yes

we need his speed at the 2.25'th second

how we get there i dont know

[dusty talon]

.

do we use the average velocity of 36ft/s ??

somehow

i think multiply something with 81 or 2.25

i have no idea what or why tho

<@&286206848099549185>

i do not know

velocity = distance / time

we cant get derivative of a normal number like 81 or 2.25

its 0

im not sure what ur getting at

ok

[quasi bison]

no

you want d'(2.25)

[dusty talon]

but thats 0

[quasi bison]

??

[dusty talon]

derivative of 2.25

0

[quasi bison]

i did not say the derivative of 2.25

i said the derivative of d, evaluated at t=2.25

[dusty talon]

d is distance? 81?

32t

[quasi bison]

try to be a little less oblivious to the problem statement.

i've seen that a lot. you go into a problem and not more than 4 steps in you forget any and all details from the outset.

[dusty talon]

this the momentum i was talking about

like jumping to rocks in the water, if i miss a beat , im lost

we're looking for the velocity at the 2.25'th second

[quasi bison]

"2.25'th second" is a bit cringe as far as wording goes.

[dusty talon]

ok

[quasi bison]

you're not interested in a one-second interval, you're interested in an instant. a point in time.

[dusty talon]

instantaneous velocity

ok...

The distance the diver falls in t seconds is given by the function
d(t) = 16t^2 ft.

d(2.25) = 16(2.25)^2

?

[quasi bison]

no,

again,

the derivative of d, evaluated at t=2.25

[dusty talon]

do you mean distance when you say d?

d'(2.25) = 81

81'(2.25) = 81

[quasi bison]

when i

ugh.

when i say d, i mean d.

the function named d.

[dusty talon]

oh the function

[quasi bison]

i am using notation consistent with the problem.

[dusty talon]

d'(2.25) = 16(2.25)^2

[quasi bison]

no

you're confusing d' with d

you've done things with derivatives before, haven't you?

[dusty talon]

yes

[quasi bison]

ok then

given that i've been talking about the derivative of d(t) incessantly all this time,

was this not cue enough for you to work out d'(t)?

[dusty talon]

idk why i thought you meant d as in distance

d'(t) = 81

no d(t) = 81

[quasi bison]

no, bad, bad

d(t) = 16t^2 is the definition of the function itself. d(t) = 81 is an equation we needed to solve for the previous part.

[dusty talon]

ok so i should not think about d(t) = 81 anymore

that was for previous

[quasi bison]

indeed

[dusty talon]

we differentiate 16(t)^2

to get 32t

then 32(2.25)

wow

it would have helped to know that instantaneous velocity = derivative of velocity function

[quasi bison]

yeah, you're supposed to know that.

er,

no

instantaneous velocity = derivative of POSITION function

the derivative of velocity is acceleration.

[dusty talon]

.close

[safe radish]

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[young mantle]

Hi, I’m looking for help with writing a mathematical formula to calculate probability involving dice rolls. It’s for a tabletop RPG I’m going to be playing.

What I need this formula to be able to do is:

• Calculate the chance of rolling a double at least on X amount of 6 sided dice (by “at least” I mean that triples, quadruples, etc count as rolling a double)
• Double 6 needs to be excluded (it counts as a failure)
• The face number (not the total of the double) is greater than or equal to a number between 1 and 5

Any explanation for how said formula works would be appreciated!

[safe radish]

@young mantle Has your question been resolved?

[empty gyro]

The chance of rolling a double a better is the inverse of the chance of rolling all singles

If X>6, then there is a 100% chance of rolling a double or higher

[young mantle]

Right, but double 6 is to be excluded as a success (this is to stop players from getting a 100% chance to succeed with 7 or more dice)

[empty gyro]

So success is double or better of 1-5, and everything else is failure

[young mantle]

Correct

[empty gyro]

So we can write failure as rolling zero or one of all 1-5

X=0 and X=1 are trivial at 100% failure (so 0% success)

[young mantle]

Right

[empty gyro]

X=2 failure, we consider two cases:
Case 1: first dice is 1-5, then second dice can be any value except matching the first, so there are 5 choices for the second dice.
Case 2: if first dice is 6, then second dice can be anything.

[young mantle]

Yes

[empty gyro]

Number of outcomes in these two cases is 5*5+1*6=31

So probability of failure of 31/36, and success is 5/36 (which matches the naive way of considering 5 possible doubles)

[young mantle]

Right, makes sense

[empty gyro]

X=3 is when things start to get a little more tricky

Probably where you are stuck too

[young mantle]

Most likely

[empty gyro]

Kinda do the same thing, just more tedious.
Case 1: First dice is 1-5 (5 outcomes)
Case 1.1: Second dice is any non match 1-5 (4 outcomes)
Case 1.1.1: Third dice is any of the remaining 4 values (4 outcomes)
Case 1.2: Second dice is 6 (1 outcome)
Case 1.2.1: Third dice is any non match of first dice (5 outcomes)
Case 2: Second dice is 6 (1 outcome)
Case 2.1: Second dice is 1-5 (5 outcomes)
Case 2.1.1: Third dice is any non match of the second dice (5 outcomes)
Case 2.2: Second dice is 6 (1 outcome)
Case 2.2.1: Third dice is anything (6 outcomes)

Then just combine all the outcomes

5*(4*4+1*5)+1*(5*5+1*6)

,w 5*(44+15)+1*(55+16)

[flat frigate]

Results provided by WolframAlpha

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Requested by SWR#8588

[empty gyro]

probability of failure is 136/6³

Tedium remains until X>6, then it's just probability of successive 6

There may be a better way to do this, but that's the best I got

[young mantle]

This has been very helpful, thanks.

So once I get to 7 dice, does the probability of failure become the same as (probability for 6 dice + the inverse of probability of successive 6s) / 6 ^ 7 ?

[empty gyro]

Hmm.. hard to say now. I never was that great with stats. Maybe other <@&286206848099549185> are

[young mantle]

That’s fine, it’s quite the complex problem and I’m not the best at probability as it is

[lean otter]

if this is for a practical thing, maybe it would be easier to do it numerically via monte carlo

[safe radish]

@young mantle Has your question been resolved?

[empty gyro]

Ran some tests for you

@young mantle

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[turbid path]

for this

on the left

can i move the 2 back up and make it log(9-x^4)

[green bane]

no

[turbid path]

so

what do i do

[green bane]

it would be log ((9-x^2)^2)

[turbid path]

the goal is to turn it into one logarithm

right side is okay

okay perfect

so its just log(((9-x^2)^2)/((3+x)(3-x)))

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@turbid path Has your question been resolved?

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[honest terrace]

Can a double or hald angle be in two quadrants?

.close

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[craggy garden]

.Ask

[craggy garden]

I'm in trig. And I don't understand what this is

What symbol is that

[limber smelt]

what are we seeing here

[green bane]

angle?

[limber smelt]

the symbol above 180?

[craggy garden]

Yes

[limber smelt]

erm the symbol for measured angle

[craggy garden]

Oh.

Okkk

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[dusty bramble]

so i need help i know i need to use that equation to just plug in the years but the correct answer isn’t popping up

[obtuse plover]

Uh

Show what u plugged in ur calculator

@dusty bramble

[dusty bramble]

well i just did p(1)=12,520e^0.058(1)

[obtuse plover]

Good, and what did u get

[dusty bramble]

12,551.05179776

[obtuse plover]

Hmm

[dusty bramble]

rounded

12,551.05

[obtuse plover]

I’m not getting that answer

Can u show ur calculator

[dusty bramble]

one sec

im usiing an online calculator since i dont have a proper graphing one

[obtuse plover]

Gotcha

[dusty bramble]

but even if i use my normal oone

its not working

idk what im doing wrong

[obtuse plover]

Yea just show me and I’ll see

[obtuse plover]

Try just typing in exactly this: 12520 • e^0.058

[obtuse plover]

Why’d u do ln?

ln is the opposite of e^

[dusty bramble]

oh shit

i forgot to click shift e

[obtuse plover]

[dusty bramble]

hmm

[obtuse plover]

That’s what I got as well

[dusty bramble]

bruh

i was just clicking the wrong thing this whole time

[obtuse plover]

[dusty bramble]

what would i plug in for c?

[obtuse plover]

Yea use ur scientific for these kinds of things, not mathway

c?

[dusty bramble]

for doulbe time

double time

[obtuse plover]

So it’s asking u for the amount after 2 years right?

[dusty bramble]

yes

[obtuse plover]

So what does ‘t’ represent

[dusty bramble]

time

[obtuse plover]

So what would t be for us in this scenario

[dusty bramble]

so the doubling time is just 2 years

wait nvm

[obtuse plover]

Oh that’s what u meant

Doubling time means the amount of time it takes for us to get to double our original balance

So how do u think we could do that

[dusty bramble]

hmmmm

not too sure

[obtuse plover]

Well what would double our original balance be

[dusty bramble]

would the og balance b e after 1 yr?

[obtuse plover]

it means when we start at t=0

[dusty bramble]

wait nvm

2 x 12,520?

[obtuse plover]

Which is?

[dusty bramble]

25040

[obtuse plover]

So set 25040 = P(t) and solve

[obtuse plover]

Nice job

[dusty bramble]

so rounded would be 12.0?

[obtuse plover]

Ye

[dusty bramble]

thanks for ur help!

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[obtuse plover]

Np

[safe radish]

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[native pagoda]

I am not sure what I messed up with part a

the answer is 4Pi/3

[patent blaze]

I'm pretty sure you are supposed to multiply your result by 2 because you shaded half of the region because there is two values of y for the same x

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@native pagoda Has your question been resolved?

[patent blaze]

so your bounds for theta should've been -pi/2 to pi/2

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[native pagoda]

.reopen

[safe radish]

✅

[native pagoda]

pog

how so

wait why is it -Pi/2 and Pi/2

https://www.youtube.com/watch?v=I2Z6K_g5kpc

I just follow this guy method

11:47

he only shaded the x boundary

0=<x=<2

[patent blaze]

look at the given y bounds it shows a positive semi circle in quadrant 1 and 2 and a negative semi circle in quadrants 3 and 4. but you are given x =0 to x =2 which is only the right side of those regions put together which would be -pi/2 to pi/2 in theta. The person in the video wasn't doing a problem with the same type of bounds as you. You can tell by the graph he made and the region he shaded.

in the video he was looking at the area between y=x^2 and y=x which starts from theta = 0 to theta = pi/4. He got pi/4 because the top is y=x which is a line that is 45 deg above x axis which would be theta = pi/4. He got 0 because the functions intersect at the origin which is on the x axis and that's theta = 0.

[native pagoda]

I see

oh yeah fact bro

I have to shade

the lower part too

https://tenor.com/view/pewdiepie-bright-gif-21056137

have to account for the y-boundary too

[safe radish]

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[rancid fjord]

.reopen

[safe radish]

✅

[rancid fjord]

Hi

How do i solve this?

[fickle talon]

!status

[safe radish]

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

[rancid fjord]

1

[fickle talon]

also make a new channel

this is gonna close

[rancid fjord]

Ok

[native pagoda]

make a new one brother

.close

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[rocky gate]

Does anyone know how to solve this?
(\tan (\pi )/(15)+\tan (4\pi )/(15))

(1-\tan (\pi )/(15)\tan (4\pi )/(15))

[rocky gate]

its a double angle i believe

but im not completely sure

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[rocky gate]

i got it lol

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[peak depot]

Can someone explain what the hell homogeneous means

[topaz tree]

just google it

[peak depot]

So you know what it means?

[topaz tree]

i know in chemistry not inmaths

[fluid token]

chatGPT
google Bard

both will give you a more thorough explanation

[zenith lodge]

I know in biology

[topaz tree]

what is homogeneous in bio

[zenith lodge]

Actually it homogeneous pair

Of chromosomes

[topaz tree]

In mathematics, the term "homogeneous" refers to a property of a mathematical object or equation. A mathematical object is said to be homogeneous if it exhibits symmetry under scaling, i.e., if multiplying its values by a constant results in an object that is identical to the original object.

In the context of equations, a homogeneous equation is an equation in which all the terms have the same degree or order. In other words, the equation is homogeneous if each term can be expressed as a constant times a function raised to a power. For example, the equation:

ax^2 + bxy + cy^2 = 0

is homogeneous of degree 2, since each term can be expressed as a constant times a function of degree 2 (x^2, xy, and y^2, respectively).

Homogeneous equations and functions have many important applications in mathematics, particularly in areas such as linear algebra, differential equations, and geometry. In linear algebra, for example, homogeneous linear equations can be used to study the properties of vector spaces and linear transformations, while homogeneous differential equations arise frequently in the study of physical systems that exhibit scale-invariance.

there you go

[zenith lodge]

GPT

[topaz tree]

ya

[safe radish]

@peak depot Has your question been resolved?

[peak depot]

This doesnt explain why a first order ode dy/dx=f(x,y) is homogeneous if you can express f(x,y) as f(y/x)

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[opaque sorrel]

hi, can someone help me with this

[obtuse jackal]

Do you know about derivatives?

[opaque sorrel]

yea

[obtuse jackal]

Did you find <= 1 ?

[opaque sorrel]

uh nop

idek how to start

[obtuse jackal]

Notice that x^r <= x for all x in [0, 1]

(btw, another proof would be using the binomial theorem, but that would be much longer)

[opaque sorrel]

wait is it possible to prove the edge cases then treat everything inside as if it works>

like case for x = 0 and x = 1

[obtuse jackal]

They're not special in any way

[opaque sorrel]

oh

hmm