help-39

[turbid minnow]

How do you prove this?

[devout wagon]

hello

[red hedge]

Oh boy

[devout wagon]

Holy jesus

My brain is gonna do a explode

is that l or I

[turbid minnow]

hello

anyone done this before by any chance?

[red hedge]

Just wondering what level of maths is this?

[devout wagon]

I've tried.

[turbid minnow]

Year 11 3U NSW curriculum

[devout wagon]

I'm gonna use Michealsoft paint

[sharp smelt]

try re-writing everything in terms of sin and cos

[turbid minnow]

I was just sent the answers but I'm struggling to make sense of them

[turbid minnow]

It doesn't work

[devout wagon]

Did I write this correctly?

[turbid minnow]

yes

[devout wagon]

oki

[abstract vine]

$\frac{1+\frac{1}{sin^2(A)}\frac{sin^2(C)}{cos^2(C)}}{1+\frac{1}{sin^2(B)}\frac{sin^2(C)}{cos^2(C)}} = \frac{1+\frac{cos^2(A)}{sin^2(A)}sin^2(C)}{1+\frac{cos^2(B)}{sin^2(B)}sin^2(C)}$

[jolly parrot]

clonesolopros

[sharp smelt]

,w is (1+cosec^2(x)tan^2(y))/(1+cosec^2(z)tan^2(y))= (1+tan^2(x)sin^2(z))/(1+cot^2(y)sin^2(z))$

[sharp smelt]

ooh

yeah

this works

[devout wagon]

This is the writen one no?

[turbid minnow]

whats next?

[devout wagon]

As it's shown

[turbid minnow]

yep

[devout wagon]

nice

[sharp smelt]

use this

[abstract vine]

maybe now you do this:

[devout wagon]

I'm slightly brain dead so I'm still trying

[cunning laurel]

Looks like it's grade ten in india

[devout wagon]

I'm cooked I don't know how

Sorry

[abstract vine]

$\frac{1+\frac{1}{sin^2(A)}\frac{1+cos^2(C)}{cos^2(C)}}{1+\frac{1}{sin^2(B)}\frac{1+cos^2(C)}{cos^2(C)}} = \frac{1+{\frac{1+sin^2(A)}{sin^2(A)} sin^2(C)}}{1+\frac{1+sin^2(B)}{sin^2(B)}sin^2(C)}$

let me cook

[jolly parrot]

clonesolopros

[pearl pond]

@turbid minnow Has your question been resolved?

[pearl pond]

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[pearl pond]

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

[silk solar]

1

[brave schooner]

ok, so you want $\frac{x}{y}$ in your equation, do you see any way to get that?

[jolly parrot]

caspar

[silk solar]

no

[brave schooner]

what can you do to a fraction but still have it be the same?

[silk solar]

im unable to think it

[brave schooner]

you can divide and multiply by tha same value in the numerator and the denominator

[silk solar]

we can divide with y

[brave schooner]

yes!

[silk solar]

ok got it

thanks

.close

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[pearl pond]

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[river aurora]

A cannon ball is fired from x=0, y=0 with an angle theta and a speed v0. Find two angles that can be used to hit a target at ground level that is a distance of 50% the maximum range of the cannon ball

I am really stuck on this question

[river aurora]

I tried solving for the max range, then dividing by 2, but I don't know how to use thay

I get v0/2g=cos(theta)t

[elfin walrus]

So to find max range you have v0 vf and a right

So how would you solve for d

Remember at the apex the ball has a vertical velocity of 0

[river aurora]

How do I get a formula for range without t in it?

[elfin walrus]

Do you remember the big 5 formulas

There’s one without t

[river aurora]

Oh from physics?

Right

I'm in a math mindset

Do you think I can just use those

Or do I need to somehow derive it

[elfin walrus]

You can just use it

Here its

Vf^2=Vo^2+2ad

[river aurora]

Ok thank you

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[mystic fog]

I want to show that a sequence of real numbers of the form: $x_n = \frac{1}{n} \log(a_n)$ converges, for some other sequence $a_n$. What properties would be enough to prove about $a_n$ that would imply the convergence of $x_n$

[pearl pond]

Please don't occupy multiple help channels.

[jolly parrot]

Eduude

[mystic fog]

I already know that for all $n$, $a_n \leq e^n$ and $_a{n + m} \leq a_n a_m$

[jolly parrot]

Eduude

[pearl pond]

@mystic fog Has your question been resolved?

[cinder flower]

if there exists a polynomial P with positive leading coefficient such that a_n < P(n) eventually, that’s good enough i think

also assuming a_n > 0 for all n but that’s necessary for this question to make sense in the first place

[cinder flower]

i’m a little confused what you’re claiming here

[pearl pond]

@mystic fog Has your question been resolved?

[mystic fog]

This comes from a problem I'm working on, and on the setting of the problem, the sequence $a_n$ is always positive and bounded by $e^n$ and $a_{n+m} \leq a_n a_m$

[jolly parrot]

Eduude

[cinder flower]

oh

[dusky glen]

I mean, there's clearly the trivial ones (eventually strictly increasing or eventually strictly decreasing, since we're bounded above and below)

You could also show that a_n converges or that it's upper bounded

Or if you find a strictly decreasing subsequence of a_n I think that'll do the trick

[pearl pond]

@mystic fog Has your question been resolved?

[pearl pond]

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[simple dirge]

i need calculus books

[simple dirge]

can anyone send me a pdf

basic calculus limits, differentiqtion and stuff

[summer sundial]

Ever tried cengage?

[simple dirge]

nvr heard of it

[summer sundial]

its a good book

used mostly for jee prep

[simple dirge]

ok

ill try

.cloae

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[simple dirge]

.reopen

[pearl pond]

✅

[simple dirge]

@summer sundial its PAID

i was looking for free pdfs

[summer sundial]

yeah its a paid book but u usually get a pirated copy online

its very common nowadays

you could even try telegram

[sharp smelt]

I can suggest a cheap book

[quiet tendon]

https://openstax.org/details/books/calculus-volume-1/

openstax usually has a good series of books

[sharp smelt]

around 80 inr

[summer sundial]

u dont need to pay

[quiet tendon]

and mit has an opencourseware book from strang too
https://ocw.mit.edu/courses/res-18-001-calculus-fall-2023/pages/textbook/

[summer sundial]

wait imma share its pdf idk if its allowed tho

https://drive.google.com/drive/folders/1fRBgNouKce_KHerAiPk3yappaYW24M4k?usp=drive_link

[quiet tendon]

if you haven't used libgen it's also good

[summer sundial]

here are my math learning resources

it contains a copy of cengage also

[sharp smelt]

try a problem book in mathematical analysis

only around 1usd

[summer sundial]

@simple dirge

[simple dirge]

thanks people

🫰

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[harsh fern]

!help

[pearl pond]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[harsh fern]

hi

i dont understand

shouldnt it shift 3 units to the right

so the x coordinate should be at 3 not at 4

[light helm]

you're ignoring the other transformations

[harsh fern]

well

how do the other transformations

affect the x coordinate

the + 2 means it goes up vertically 2 units

[light helm]

yes, which will change the location of intercepts

[pearl pond]

@harsh fern Has your question been resolved?

[pearl pond]

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[compact summit]

would this be convergent or divergent?

[eternal tulip]

use alternating series test

i.e. check if (2n+1/(3n-2))^2n goes to 0 as n goes to infinity, if it does, then the series will converge

[compact summit]

I see, thank you

I believe it goes to 0

[eternal tulip]

so the series is convergent!

[compact summit]

many thanks

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[drifting root]

Ignore the names of the sets if they confuse you.

My question is as simple as: Whats the probable time to complete any of the starred sets considering that i can get 1 random piece from any of the remaining 13 sets once per week.
(One full set has 4 pieces)

🌟Bannaret | Legs

🌟Serpent | Gloves

🌟Doom | Gloves

🌟Relish | Chest Gloves

🌟Cobalt |

Arugant | Boots Legs

Luna | Boots

Blood rite | Boots Legs Chest Gloves ✅

Regal | Gloves

Mercenary | Chest

Marauder | Legs

Buccaneer | Legs

Outlander | Legs

Alchemist |

17/56 Pieces/Weeks
(1 random piece per week)

24 irrelevant Pieces(AKA "Without emojis")
15 Relevant Pieces (AKA "🌟" )
39 Total Remaining Pieces
14 Possible sets (1 Completed)
Duplicates Is not possible*

Date of completion is going to be "14 Mar 2025" But i only need one of the starred sets to be satisfied.

Thank you in advance

[drifting root]

<@&286206848099549185>

:( i gtg to bed <@&286206848099549185> close this when ive received an answer

[simple jasper]

This will close by itself due to inactivity. But it would depend on drop rates, I'm assuming this is a game?

[drifting root]

Yes its a game, and we're talking 1 piece of a random set every week

All i am wondering is how long it would take for me to get one of theese full sets based on the amount of items that i do not want, of course this is RNG, so it may vary a lot

You could also provide me a way to calculate this myself if that is easier for you.

I can get any of 56 items every week, never duplicate items, there are sets of four. I want a complete set from any of the 5/14 chosen ones, in due time it will happen, odds wise, when?

[simple jasper]

(# of items needed)/(# total items). That gives the probability per week. But to be honest there isn't really a good way to calculate how long it will take since it's random chance. You could finish in the least amount of weeks (# of items needed) or the most amount of weeks possible (# of items not needed first than # of items needed.

[simple jasper]

Can you choose the items you get?

[drifting root]

I obviously cannot

Completely random, only 1 per week

[drifting root]

If i wondered how long it would take for me to get 4 specific items i could probably calculate it myself, but i am satisfied with getting any 4 items of 5 different sets, as long as the 4 items is from the same one, the odds should massively increase when counting more sets as valid?

[simple jasper]

I mean, you're increasing the number of items wanted, but once it gets down to like 1 item from 1 set and are like 2 items from multiple than so on increasing. You would probably start wanting that 1 item and that 1 items only. But yes it would 'increase' your odds

[drifting root]

That's the thing, i still need a full set, just instead of 1/14 sets being the right one, i would be happy with any of 5/14 sets

[simple jasper]

How many items per sets?

[drifting root]

To get 1 out of theese 5 sets when there is 9 wrong sets, how long would it take on average, There is 4 items in one set

15 correct pieces
24 incorrect pieces

Considering that ive found 17 already

[drifting root]

I thank you for the help, i will just have to hope for the best, maybe this was a stupid question.

.close

[pearl pond]

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[simple jasper]

It's not a stupid question, it's just hard to quantify a plausible answer

[pearl pond]

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[trail bough]

Can someone please check my work

[midnight haven]

Looks perfect to me

[trail bough]

thank you

.close

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[chrome flame]

here i differentiate the function

and got D

e^sinx
e^sinx.cosx at x=0 it will be 1

and e^(1-cosx)sinx at x=0 it will be 0

so 0=1

[midnight haven]

f(x) is continous but not differentiable at x = 0

[chrome flame]

yes how did you check

like i did?

[midnight haven]

no

instead of differentiating g(f(x)) , i directly checked it for f(x)

[chrome flame]

how?

sinx=1-cosx?

x=pi/4

[midnight haven]

differentiate sinx and 1-cosx

thats how you do it for piece wise functions

[chrome flame]

they will equal only at pi/4

not at 0

[midnight haven]

this is wrong

[chrome flame]

i got it

what if the fnction is this

[midnight haven]

there is no need to equate both of them

[midnight haven]

since there is no differentation involved,you just have to check if f(x) is continous at 0

[chrome flame]

yes it is

1-cosx give 0 and sinx also

[midnight haven]

yes

you can either take e^sinx or e^(1-cosx)

and substitute 0

[midnight haven]

actually there is no need to check for continuity,i thought that was limit

[chrome flame]

no it is derivative

e function always continuous

[midnight haven]

are you talking about this ?

[chrome flame]

.

[midnight haven]

since x is exactly equal to 0 , g(f(0)) would be e^(1-cos0)

I didn't get you,what was your question ?

[chrome flame]

function is continuous at x=0

how to check for derivative

[cunning comet]

draw it and you will see it.

[chrome flame]

should i differentiate directly like i did above

or i have to use definition

[cunning comet]

you do not need to differentiate. draw the function and you will see its not differentiable at x = 0

[chrome flame]

.

i am asking about this

e^sinx

e^(1-cosx)

opps there is derivative also in the question

[cunning comet]

thats a different question as the the one you posted at the beginning. what is asked? g(f(0)) or the derivative of g(f(0))?

[chrome flame]

derivative

now i am asking about the second one derivative

which i posted

where equal sign is different

[cunning comet]

well again. my answer is: draw g(f(x)) and you will see its not differentiable.

[chrome flame]

.

[cunning comet]

is the change from x< 0 to x > 0 a smooth one or is there an abrupt change in direction?

[opal breach]

if it helps

[pearl pond]

@chrome flame Has your question been resolved?

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[edgy otter]

Can someone help me figure out where to start with this please? 😭

[sharp smelt]

is this a test or just practice ?

also, !show

!show

[pearl pond]

Show your work, and if possible, explain where you are stuck.

[edgy otter]

just cengage homework

[sharp smelt]

well, what do you think the answer is?

[edgy otter]

I don't even know where to start 😭

[sharp smelt]

what's sec(pi/2) as you approach it from the right

[edgy otter]

how do i figure that out?

[sharp smelt]

what's the defn of sec(x)

in terms of cos(x)

[edgy otter]

opposite over hypotenuse?

oh

wait

[waxen condor]

+∞ , can use L Hopitals first

[edgy otter]

1/cos(x)

[sharp smelt]

what, no

[edgy otter]

What's L Hopitals?

[sharp smelt]

yes

[waxen condor]

umm why bro ?

[sharp smelt]

what's (cos(\pi/2)) as we approach it from the right

[sharp smelt]

it's in the infty/ a constant form

[edgy otter]

How do I find that?

[sharp smelt]

what's cos(pi/2)

ok, I've got to go now

sorry

[edgy otter]

it's alright!

thank you for your help

.close

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[pearl pond]

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[midnight haven]

Is it correct?

[midnight haven]

simplying this gave me the above result which is useless

[midnight haven]

z represents a complex number

so how do i solve it ?

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

.close

[pearl pond]

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[pearl pond]

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[steep patrol]

so for this problem I have obtained my A and C but getting my B is hard

like this is what I have

I dont know what to set B(x+2)(x-1) equal to, in order to get the answer for B

because what I am doing seems right but I just cant further go about getting B

[runic quail]

why don't you expand everything out to get a system of equations

[steep patrol]

expand like how

oh wait systems of equation?

[runic quail]

yeah

if you expand and distirbute everything

[steep patrol]

Lemme try that then

[runic quail]

ok once you do that ping me

[steep patrol]

k

[runic quail]

now group the x^2 terms the x terms and the constant terms

youll get something like (but with different numbers) $3x-9=(2A-B)x^2+(A+4B-C)x+3A-B+C+5$

[jolly parrot]

The د

[runic quail]

since in the LHS there is no x^2 term

you know that 2A-B must equal 0

similarly yk that A+4B-C=3

and finally 3A-B+C+5=-9

now you have a system of equations

you can solve by hand or let the calculatutr solve it for you

(of course these numbers are just arbitary and youll have smthn different)

but do you get the idea? @steep patrol

[steep patrol]

I get the idea but I always distrubite it wrong

[runic quail]

ah ok lets go through it

whats the expansion of $A(x+2)^2$

[jolly parrot]

The د

[steep patrol]

Its just A(x^2+4x+4) right?

[runic quail]

yup

now just distribute the A to each term

[steep patrol]

and then it'll be Ax^2+4Ax+4A

[runic quail]

perfect

[steep patrol]

Nice

so Im assuming the expansion of B(x+2)(x-1) would look something like B(x^2-x-2) and then this would be Bx^2-Bx-2B correct?

[runic quail]

not exactly

look at (x+2)(x-1) mor closely

[steep patrol]

yeae

[runic quail]

foil it

wait mb

[steep patrol]

yea so if I foil it wouldn't it be x^2-x+2x-2?

[runic quail]

you have it correct

[steep patrol]

oh ok

[runic quail]

yeah yeah mb

ok for C you've made a slight mistake

when you have a 2nd degree denominator

you must account for the numerator bieng a linear term

(numerator always only 1 degree less than denominator)

so instead of C, use Cx

[steep patrol]

Ok so it would Cx(x-1) right?

then it would Cx^2-Cx right?

[runic quail]

wait one sec

why do you have (x+2)^2 in the denominator?

since you already have a B/(x+2), you only need a C/(x+2)

alternatively, you caould have had $\frac{Bx+C}{(x+2)^2}$

[jolly parrot]

The د

[runic quail]

but $\frac{B}{x+2}+\frac{C}{x+2}$

[jolly parrot]

The د

[runic quail]

works fine

[steep patrol]

I dont think thats right becuz the book teaches that if a function to the power of 2 ur supposed to split it but still add the power sign to it

here's what Prof Leonard also teaches

[runic quail]

ah i see i see yeah

mb mb

continue with what you were doing from the beginning

[runic quail]

from here, this is correct yeah

[steep patrol]

ok so when combining all formula's together u get x^2(A+B+C)+x(4A-B-C)+(4A-2B)

correct?

@runic quail

<@&286206848099549185>

[runic quail]

yes thats correct

now set the coefficients on the left equal to those on the right and solve the system

[steep patrol]

Im setting it up but not getting a right answer

[runic quail]

how did you set it up?

[steep patrol]

4A-B-C=0

[runic quail]

why 0

it should be 3

since you have 3x on the left hand side

[steep patrol]

omd

[runic quail]

A+B+C should be 0 since you have no x^2 term on the left

[steep patrol]

this prblm got me fucked

[steep patrol]

ok

that makes sense bcuz since we have a x term 4A-B-C has to equal=3x

right?

[runic quail]

well, yeah you need (4A-B-C)x = 3x

so you can just divide both by x

[steep patrol]

divide both sides by x?

wouldn't that equal (4A-B-C)= 3?

[runic quail]

yes

that's one of your three equations

you'll have a system with A,B,C to solve

[runic quail]

this is another equation

[steep patrol]

so 4(-2/3)-B-5=3

-8/3-B=8

[runic quail]

where did you get A and C from?

[steep patrol]

from here

man Im so fucked rn I just wanna understand why b=-2/3 so I can sleep it off

[runic quail]

bro I'm so sorry I messed you up

[runic quail]

it's supposed to be Cx-C here

and the final system should look smthn like

[steep patrol]

Man I fucking hate math

[steep patrol]

Is that wolf?

[runic quail]

yeah

[pearl pond]

@steep patrol Has your question been resolved?

[pearl pond]

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[ember spear]

,calc 9^(9^(9^(9^9)))

[jolly parrot]

Result:

Infinity

[ember spear]

shift + 6 on windows

[jolly parrot]

The following error occured while calculating:
Error: Value expected (char 3)

[ember spear]

parallel in geometry

[ember spear]

I don't think any calculator wants to calculate that monstrosity

∪

yes

yes

[pearl pond]

@midnight haven Has your question been resolved?

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[jolly parrot]

l'agit

l'agit

[pearl pond]

@cold gate Has your question been resolved?

[pearl pond]

@cold gate Has your question been resolved?

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[long bough]

<@&286206848099549185> If the height of a prism is doubled, the volume will increase by a factor of what

[ember spear]

what is it's formula, do you know it?

@long bough

Area of a prism = Base Area * Height

[long bough]

yes

so its either 2 or 4 then

except both work

when I plug in the question numbers

[ember spear]

umm they shouldn't

2 is the right answer

[long bough]

that is strange

[ember spear]

what type of prism is this?

[long bough]

the numbers given were 2,3, and 4.

It looks to be to a triangular??

its weird

its not a triangle cause its 2d, but this is 3d

[ember spear]

wdym

[light helm]

do you have a pic of what's coming up

[ember spear]

can you share a screenshot

[long bough]

yes ofc

this is geometry btw so its a bit challenging for me (im in 9th grade)

[light helm]

is the ss on its way?

[warm current]

!15m

[pearl pond]

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.

@long bough Has your question been resolved?

[pearl pond]

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[waxen sequoia]

I dont understand how to get B for ii)

[waxen sequoia]

<@&286206848099549185>

[turbid kestrel]

!15min

[pearl pond]

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.

[quiet tendon]

the eigenspace associated with an eigenvalue lambda of a matrix A is Null(A - lambda*I) where I is the identity matrix of appropriate size

what part is confusing?

you subtract your eigenvalue from the diagonal and then find a basis for the null space of that new matrix

[waxen sequoia]

like are eigenspace and eigenvector the same thing?

[waxen sequoia]

I did that and im not getting the same answer

even with an online calculator

[quiet tendon]

an eigenvector of a matrix A is a nonzero vector v such that Av = lambda * v for some lambda, which we call the associated eigenvalue. the eigenspace associated with an eigenvalue is the space spanned by all eigenvectors associated with that eigenvalue

for example, let's say i had a matrix where one of the eigenvalues was 3. then the space spanned by all eigenvectors of the matrix with eigenvalue 3 is called the eigenspace associated with the eigenvalue 3

[waxen sequoia]

this doesnt match with ii)

[waxen sequoia]

I did exactly that and got the answer above but that doesnt match with the posted solutions. Is it possible the solution is incorrect?

[quiet tendon]

i suppose it's always possible that the solutions are incorrect

,w rref {{-1 - 3i, 5},{-2, 1 - 3i}}

[jolly parrot]

Results provided by WolframAlpha

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Requested by nightshade5107

[waxen sequoia]

Then iii) is also incorrect. Only i) is the correct solution

[quiet tendon]

this tells me that any vector (x,y) in the eigenspace associated with 2 + 3i for ii satisfies x = (1/2 - 3i/2)y, so (1/2 - 3i/2, 1) is a basis vector for the eigenspace

which is equivalently that (1 - 3i, 2) is a basis for the eigenspace

so far ii is correct

for the other eigenspace associated with 2 - 3i

you have

,w rref {{-1 + 3i, 5},{-2, 1 + 3i}}

[jolly parrot]

Results provided by WolframAlpha

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

Requested by nightshade5107

[waxen sequoia]

how is this equivalent?

[quiet tendon]

which tells me that (1/2 + 3i/2, 1) is a basis vector for the other eigenspace

[quiet tendon]

we know that since (1/2 - 3i/2, 1) is a basis for the eigenspace, every vector in that space is some scalar c multiplied by this vector

so every eigenvector in the space is given by c(1/2 - 3i/2, 1)

every scalar c is exactly twice some other scalar d

so let c = 2d

then every eigenvector in the space is given by 2d(1/2 - 3i/2, 1) = d(1 - 3i, 2)

which means that (1 - 3i, 2) is also a basis for the space

if you have a basis for a vector space, you can always scale any vector in that basis and the result is also a basis for that space

[waxen sequoia]

ahh i see

[quiet tendon]

but yeah based on the calculations i did, their solution for ii is correct

[quiet tendon]

what you're getting here is exactly what i got before i scaled the vector by 2

that online calculator gave you an equivalent and correct answer

just not as "simplified" as the answer key you have

[waxen sequoia]

although both solutions should be acceptable

[quiet tendon]

yes

it's still a right answer

[waxen sequoia]

Alright, thank you.

.close

[pearl pond]

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[pearl pond]

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[deft kayak]

i dead

[deft kayak]

so i've got that E(R) = 5.5 and Var(R) = 3.85

pretty simple

i'm stuck on the laws of total expectation and variance, partly because the notation is kinda confusing

this is what we were told

\begin{gather*}
\sum E(N|R=y_2)f_2(y_2)\\
\frac{1}{10} \sum_{i= 1}^{10} n \cdot \frac{i}{10}\\
\frac{55}{100} n
\end{gather*}

[jolly parrot]

omgatriple

[deft kayak]

i'm pretty sure this is right, but i have no idea how to approach 3d?

[spiral pivot]

just use the definition of variance

Var(X) = E((E(X) - X)^2)

ugh, parenthesis

oh sorry

it specifies to use the law of total variance

which is

[deft kayak]

its this right

[spiral pivot]

Var(Y) - E(Var(Y | X)) + Var(E(Y | X))

Yes

[deft kayak]

i think i kinda got it, you find the inner E(Y|X) and Var(Y|X) using the binomial formulas

and then u have a function of r, then you just use definition of expecation and varaince using sums from r = 1 to 10?

[spiral pivot]

It's just a matter of applying the definition of E and Var. You don't need to do anything fancy

yup

[deft kayak]

For the second term, it would be $E(E(Y|X)^2) - E(E(Y|X))^2$ right

[jolly parrot]

omgatriple

[spiral pivot]

That is one way you can calculate it, yes

[deft kayak]

is there an easier way?

[spiral pivot]

It just depends on what is easiest.

for the situation

Var(X) = E((E(X) - X)^2) is equivalent to Var(X) = E(X^2) - E(X)^2

and depending on how your data is organized the second or the first might be easier to calculate.

[deft kayak]

i see, i think i'm getting the way now. i was a bit confused with the composition of the functions but it's making more sense now

thanks

[spiral pivot]

yw

[deft kayak]

.close

[pearl pond]

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[pearl pond]

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[final briar]

What did I do wrong

[glass salmon]

your diagonal length is incorrect

[heavy snow]

you said r = sqrt (42) but then drew the diagonal as sqrt(42)

[final briar]

Oh 💀

It should be 2 sqrt(42)

Thanks

.close

[pearl pond]

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[pearl pond]

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[stoic imp]

let $f(x)$ be the solution of the problem with initial values $\\y' + \frac{x}{1+x^2}y = \frac{2x}{1+x^2}, y(0) = -1\\$ find the positivity set of f

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[stoic imp]

I tried translating it but idk if its correct translation, this is original exercise in spanish

[snow sail]

what is a positivity set

[zealous shore]

I'm not sure if positivity set is a term I haven't heard of or if it's a bad translation. Is it where f(x)>0? If so, this can be solved with an integrating factor and then we can find what values of x give the needed condition

[snow sail]

yea, ODE is just IF

idk what a positivity set is

[snow sail]

so, yea

[stoic imp]

is the subset of the domain with positive images

[snow sail]

can you solve for y?

in the ODE

thats part 1

$y' + \frac{x}{1+x^2}y = \frac{2x}{1+x^2}$

[jolly parrot]

jan Niku

[snow sail]

https://mathworld.wolfram.com/IntegratingFactor.html

[stoic imp]

integrating factor?

the thing with the e

omg

can you help me with this ode

[stoic imp]

I cant I need more handholding

this is not even separable

🤯

[snow sail]

no, its not

that page kind of spells it out

we ultimately want to rewrite the LHS as a product rule

$\dv{y}{x} + p(x) y(x) = q(x)$

[jolly parrot]

jan Niku

[snow sail]

can you identify p and q

[stoic imp]

Im trying

[snow sail]

sure, yea

so $y' + py = 2p$

[jolly parrot]

jan Niku

jan Niku

[stoic imp]

Mmm

[snow sail]

no, you already got it screwy lol

[jolly parrot]

jan Niku

[stoic imp]

Shit mb

[jolly parrot]

jan Niku

[snow sail]

yea, so, at the moment, you need that integral

[stoic imp]

Thiss

[snow sail]

the exponent

dont worry about the e just yet

[stoic imp]

How to solve integral

Does it involve tangent im already screwed

[snow sail]

youre doing differential equations, i think you can handle a u-sub integral

write it this way

$v = \frac 12 \int \frac{2x}{1+x^2} \dd x$

[jolly parrot]

jan Niku

[stoic imp]

Ok

[snow sail]

does that help make it more clear?

theres no reason to waste time struggling with the integral really

not that you need to know how to do it, i dont think its useful to distract yourself with a small subcomponent of the larger problem

well, you do need to know, but maybe you know what i mean, lets bust out this integral 💪 so we can keep trucking

[snow sail]

look here

$u = x^2 + 1$ and $\dd u = 2x \dd x$

[jolly parrot]

jan Niku

[snow sail]

so yea

#1 i want you to do a quick mental verification if the absolute value bars are necessary

[stoic imp]

It ismt

[snow sail]

#2 you can either trust me that the +C isnt necessary, or we can track it through the entire problem annoyingly and find out later it doesnt matter

[stoic imp]

Ok

[snow sail]

maybe its easier to come back later when you understand the flow more, and see what happens if you decide to maintain it

[stoic imp]

Now wat

[snow sail]

okay so, we started here

$y' +py = 2p$

[jolly parrot]

jan Niku

[snow sail]

we found $v = \int p$ and we know v

[jolly parrot]

jan Niku

[snow sail]

now, you can multiply through by e^v

first, do you believe me that $v' = p$?

[jolly parrot]

jan Niku

[stoic imp]

Im so confused

[snow sail]

which part?

[stoic imp]

You are using curly v and regular v to refer to the same thing

[snow sail]

im trying to just use letters so we dont get too mixed up on the exact crazy looking functions

[snow sail]

curly v?

[stoic imp]

Is tripping me up

[snow sail]

do you mean y?

[stoic imp]

.

[snow sail]

oh, sorry, i forgot to latex

theyre the same, i'll be more careful

lets start here

$y' + py = 2p$

[jolly parrot]

jan Niku

[snow sail]

and we know that v' = p, you say is okay?

[stoic imp]

Ye

[snow sail]

so lets rewrite it slightly

$y' + v' y = 2p$

[jolly parrot]

jan Niku

[snow sail]

this will work for now

[stoic imp]

Rhs aswell

[snow sail]

well, maybe

lets see what happens

its not necessary just yet

now we multiply through by the integrating factor, thats e^v

$e^v y' + v' e^v y = 2p e^v$

[jolly parrot]

jan Niku

[snow sail]

can you recognize the application of the product rule on the LHS?

[stoic imp]

I camt

Recognize

Product rule

[snow sail]

how about this.... can you solve for $\dv x \qty( e^{v(x)})$

[jolly parrot]

jan Niku

[snow sail]

how do we take the derivative here?

[stoic imp]

Function composition rule

[snow sail]

chain rule, yea

well, the name isnt important

[stoic imp]

v’ x e^v

[snow sail]

$v'(x) e^v$?

[jolly parrot]

jan Niku

[snow sail]

that x in your answer has me worried

[stoic imp]

X is multiplication mb

[snow sail]

oh

so yea

okay so lets stare at

$e^v y' + v' e^v y = 2p e^v$

[jolly parrot]

jan Niku

[snow sail]

you actually just computed the derivative, ill go ahead and use it to rewrite slightly

$e^v y' + \qty(e^v)' y = 2p e^v$

[jolly parrot]

jan Niku

[snow sail]

do you see the product rule now?

[stoic imp]

Ye

[snow sail]

the derivative hits y in the first term, but leaves e^v alone

and in the second, it hits e^v, but leaves y alone

so, we can rewrite the LHS

as a derivative

of what?

how can we 'undo' the product rule. what product is the LHS a derivative of?

[stoic imp]

Unsure

[snow sail]

no

its a product rule right

[jolly parrot]

jan Niku

jan Niku

[snow sail]

seem reasonable?

[stoic imp]

Rhs PROD8CT RULE/?

[snow sail]

LHS product rule

[stoic imp]

Dont u mean lhs

[snow sail]

where did i say rhs

[stoic imp]

Lhs i see thr product rule yes

[snow sail]

okay

rewrite the lhs as the derivative of a product

[stoic imp]

e^v y

[snow sail]

okay

$\dv x \qty( e^v y) = 2pe^v$

[jolly parrot]

jan Niku

[stoic imp]

Yes

[snow sail]

lets go ahead and take your suggest now, about p = v'

on the RHS

$\dv x \qty( e^v y) = 2v'e^v$

[jolly parrot]

jan Niku

[stoic imp]

Sure

[snow sail]

can you solve for y from here?

its much more regular looking now

[stoic imp]

I cant

[snow sail]

how can we get rid of the derivative

[stoic imp]

Integration

[snow sail]

yup

so, lets integrate both sides

the LHS is easy

what is $\int 2 v' e^v \dd x$?

[jolly parrot]

jan Niku

[stoic imp]

What is v

Btw

[snow sail]

we defined it a long time ago

its not really necessary to know exactly what it is to do this integral

but, its $v(x) = \frac 12 \ln \qty( x^2 +1 )$

[jolly parrot]

jan Niku

[stoic imp]

Omg this looks horrible

[snow sail]

yea, this is why you shouldnt sub it in lol

its not important what v is

we can solve $\int 2 v'e^v \dd x$ by u sub

[jolly parrot]

jan Niku

[snow sail]

its not dv, its dx

remember that we had $\dv x \qty(ye^v)$ on the rhs

so we want to integrate wrt x to get rid of that derivative

[jolly parrot]

jan Niku

[snow sail]

what?

[stoic imp]

Idk how to dolve tbh

[snow sail]

think about it like this

you said $\dv x e^v = v' e^v$, right?

[jolly parrot]

jan Niku

[stoic imp]

2e^v is integral then

[snow sail]

i mean, just rewrite it

$\int 2 v' e^v \dd x = 2 \int \dv x (e^v) \dd x$

[jolly parrot]

jan Niku

[snow sail]

what happens?

(im just writing usub in a different way)

[stoic imp]

Im already lost but we know this integral is 2e^v

[snow sail]

well, what makes you feel lost?

[stoic imp]

We are jumping all over the place with differentiation u sub and integrations adn I still dunno how are we going to use Q(x)

[snow sail]

the idea is that we rewrote the differential equation substantially

we had before $y' + py = 2p$ which you said initially you didnt know how to solve

[jolly parrot]

jan Niku

[snow sail]

we got all the way to $\dv x \qty( ye^v ) = 2 \dv x \qty( e^v )$

[jolly parrot]

jan Niku

[stoic imp]

We multiplied by the integrating factor both sides of the equation aswell

[snow sail]

yea, its accounted for here

e^v is the integrating factor

[stoic imp]

Ohh

[snow sail]

we are super close here

the whole trick is like

integrating factor allows to usually rewrite a lot of stuff as derivatives

then, we can just integrate to eliminate the derivatives

sometimes this leaves you an easy integral sometimes it doesnt

here, BOTH sides are derivatives

so our life is pretty easy

[stoic imp]

Okay

[snow sail]

i dont want to throw too many symbols at you

but you could think of how wed solve something like....

[jolly parrot]

jan Niku

[snow sail]

its pretty straightforward

wed integrate both sides, and pick up a constant on both

lets group it on the right

[jolly parrot]

jan Niku

[snow sail]

we can confirm, does differentiating both sides give us the original differential equation?

[jolly parrot]

jan Niku

[snow sail]

what do you think? does that help?

[stoic imp]

Still stuck here

What black magic did you do to get x^2?

[snow sail]

this is just some differential equation

its not related to your current problem

[stoic imp]

Okay

[snow sail]

except that its kind of an example of how close we are

$\dv x \qty( ye^v ) = 2 \dv x \qty( e^v )$

[jolly parrot]

jan Niku

[snow sail]

were actually here

what happens when we integrate both sides?

[snow sail]

.

[snow sail]

yea

lets redefine quickly C = C2 - C1

and rewrite $ye^v = 2e^v + C$

[jolly parrot]

jan Niku

[snow sail]

can you solve for y

[stoic imp]

Mmm

[snow sail]

i like ... $y = 2 + Ce^{-v}$

[jolly parrot]

jan Niku

[stoic imp]

Okay

Can we say y = 2+ c

[snow sail]

no, how would you get that?

[stoic imp]

Yeah sorry

[snow sail]

no worries

[stoic imp]

I just wanted to ask

Now what?

We find c

[snow sail]

yea

which is fine

you have an initial condition

its so long ago i forgot it

was it y(0) = -1?

yea

do you remember $v(x) = \frac 12 \ln \qty( x^2 + 1)$?

[jolly parrot]

jan Niku

[snow sail]

so whats $y(0) = 2 + Ce^{-v(0)} = -1$

[jolly parrot]

jan Niku

[stoic imp]

f(x) = 2 and C = 0?

Fuck i messed up

[snow sail]

yea, try going it steps

[stoic imp]

Y = -1

[snow sail]

find v(0)

[snow sail]

then plug it in here.

[stoic imp]

Lowkey got c = -3

[snow sail]

yea, this seems fine

[stoic imp]

Cant believe we solved this crap

[snow sail]

this is wrong

you got C correct

but thats not y

[stoic imp]

But e^ln

[snow sail]

its not e^ln

its $\exp \qty(\frac 12 \ln \qty( x^2 + 1 ))$

[stoic imp]

Dammit

[snow sail]

this is a lil easier to read

[jolly parrot]

jan Niku

[snow sail]

you can simplify this, if you want

[snow sail]

yea

[stoic imp]

What about the positive set

[snow sail]

so heres the thought

well, this is just an algebra problem now

my thought is: for some x, we can make sqrt(x^2+1) pretty big

that will shrink the negative term

and we'll suddenly become positive

if you cant solve it, try looking at a graph

do you need other help with this part of the problem?

[stoic imp]

The image needs to be positive

[snow sail]

right

so solve $0 < 2 - \frac{3}{\sqrt{x^2+1}}$

[jolly parrot]

jan Niku

[stoic imp]

Mmm

[snow sail]

i think you made a mistake here

can you spot it?

idk if you followed my babbling earlier but we figured that the equality should be true for large enough x

[snow sail]

your inequality suggests small x as a solution though

did the direction get incorrect flipped or not flipped somewhere?

when does an inequality need to be flipped?

[stoic imp]

Im stupid

A moment

[snow sail]

no stress

[snow sail]

wait what

did we screw up somewhere

nah

[stoic imp]

Wdym

Where and when

[snow sail]

\begin{align*}
0 < 2 - \frac{3}{\sqrt{x^2+1}}
&\to \frac 32 < \sqrt{x^2+1} \
&\to \frac 94 - 1 < x^2
\end{align*}

[jolly parrot]

jan Niku

[stoic imp]

Waaat

[snow sail]

okay

youre halfways there

[snow sail]

sorry meant to reply to the image

[snow sail]

okay

lets simplify

[stoic imp]

X^2 + 1 is positive no matter what

[snow sail]

$x > \frac{\sqrt 5 }{ 2 }$

[jolly parrot]

jan Niku

[stoic imp]

True

[snow sail]

were missing solutions

but we found some

[stoic imp]

Wdym missing solutions

Wtff

[snow sail]

were missing some

[snow sail]

the last step here

you have a small problem

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

[jolly parrot]

jan Niku

[stoic imp]

,, \pm

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[stoic imp]

Mmm

[snow sail]

so actually $|x| > \frac{\sqrt 5}{ 2}$

[jolly parrot]

jan Niku

[snow sail]

,w plot 2 - 3 / ( sqrt(x^2+1 ) ) from x=-3 to x=3

[jolly parrot]

Results provided by WolframAlpha

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

Requested by janniku

[snow sail]

seems about right, this function looks like itd be symmetric about 0

[stoic imp]

So confusing

[snow sail]

so if we find some solutions off to the right, we should probably pick some up on the left, too

what a fun problem

the only thing its missing is partial fractions

well, its good for review

it hit a lot of things that people tend to struggle with

[stoic imp]

I.F?

[snow sail]

yea

and just general like

you know integration and differentiation

especially when we just have letters and not nice looking functions

using FTC a bit

solving inequalities without losing solutions

or making mistakes

[stoic imp]

Okay

[snow sail]

any remaining questions?

[stoic imp]

No and thank you for thr help, gonna go relax right now, thanks

[snow sail]

we didnt even do this

so you could do some review of the whole process since it was kind of long

and if you feel up to it, maybe see what happens

thanks for sticking with me

[stoic imp]

Review of the topics that were discussed

Like a summary or list

[snow sail]

i mean, there were a lot of details

so maybe you lost the big-picture somewhere in there

reading over it a few times now that you know a lot of the answers might help

or, you know, you do you

maybe you just move on

[stoic imp]

What would be the positive set in english

Wtf

[snow sail]

probably just 'find where y > 0'

idk it has a word or im too dumb to know it

[stoic imp]

Im gonna review the integrating factor method tbh it caught me off guard

[snow sail]

the wolfram site i linked is pretty concise

[stoic imp]

I was feeling naked when tackling the problem

[snow sail]

https://www.ncl.ac.uk/webtemplate/ask-assets/external/maths-resources/core-mathematics/calculus/integrating-factor.html

this ones good too

[snow sail]

integrating factor is like ... a trick

either you know it or, you mess around a lot, and maybe if you are lucky you figure out the trick for yourself

so its reasonable to not know it or forget it

[stoic imp]

Okay thank u ima scroll tiktok rn

Sorry

[snow sail]

[stoic imp]

My head hurts

.close

[pearl pond]

Channel closed

Closed by @stoic imp

Use .reopen if this was a mistake.

✅

[stoic imp]

find the monic equation that has the same roots as $\\f(x) = x^5 -13x^4 + 68x^3 -176x^2 +220x -100\\$ but each of them as simple roots

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[pearl pond]

@stoic imp Has your question been resolved?

[pearl pond]

@stoic imp Has your question been resolved?

[pearl pond]

@stoic imp Has your question been resolved?

[cunning comet]

determine the roots.

[pearl pond]

@stoic imp Has your question been resolved?

[stoic imp]

how

also please ping me

[cunning comet]

if you have no idea then start with drawing the graph.

[stoic imp]

but without using geogebra

is it possible to solve this exercise?

how do I determine the roots of a quintic

[cunning comet]

yes of course. draw the graph.

[stoic imp]

mmm

nah thats messed up let me open ggb real quick

[cunning comet]

come on. take x = -1, x = 0, x = 1, x = 2, ... calc y.

[stoic imp]

table method?

mmm

okay a moment

,calc -1-13-68-176-220-100

[jolly parrot]

Result:

-578

[stoic imp]

,calc 1-13+68-176+220-100

[jolly parrot]

Result:

0

[stoic imp]

,calc 2^5

[jolly parrot]

Result:

32

[stoic imp]

,calc 2^4

[jolly parrot]

Result:

16

[stoic imp]

,calc 32-1316+688-176*4+440-100

[jolly parrot]

Result:

4

[stoic imp]

no way im doing more than 4 values

x = -1 y = -578