help-33

[quaint elm]

yes... he subtracted x from both sides

[gaunt lynx]

omgggg

im so stupid

[quaint elm]

so 1/2 x + 1/6 y = x
became
= 1/2 x + 1/6 y = 0

[gaunt lynx]

yeah shit

im dumb

[uneven prawn]

U german?

[marsh citrus]

@gaunt lynx Has your question been resolved?

[marsh citrus]

@gaunt lynx Has your question been resolved?

[marsh citrus]

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[steady fjord]

Hello i am trying to prove this iff statement

[steady fjord]

For all sets $a, b [ a \subseteq \bigcup b \iff (\forall z \in a)(\exists w \in b)(z \in w)]$

[elfin berry]

toast

[steady fjord]

$a, b [ a \subseteq \bigcup b \iff (\forall z \in a)(\exists w \in b)(z \in w)]$ \
Assume $\forall z(z \in a \implies z \in \bigcup b)$. \
This is equivalent to $\forall z (z \in a \implies \exists w \in b(z \in w))$

[elfin berry]

toast

[steady fjord]

but i am not sure how to proceed from here really

do i expand the implication to not p or q?

or am i done

[marsh citrus]

@steady fjord Has your question been resolved?

[brave marsh]

Well from there you can conclude that $(\forall z\in a)( \exists w \in b) z \in w$

[elfin berry]

Azyrashacorki

[marsh citrus]

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[marsh citrus]

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[gaunt lynx]

yo did i do itproperly

the deriviate

[brave marsh]

Yes

[gaunt lynx]

thank god

[waxen dust]

I will

[late geode]

2nd last line is dodgy

[night relic]

i think you wanted to write $0 = 4(e^x)^2 - 2b\cdot e^x$ here, but you have the next line corret

[elfin berry]

حسیب ♥

[night relic]

aw man ram beat me to it

[gaunt lynx]

oh

what did i do wrong here

[late geode]

here's everything in the last to lines coming from

[gaunt lynx]

i put in ln(1/2b) in gb(x)

[late geode]

where's the stuff on the right coming from

[gaunt lynx]

-2b timess e^ln(..)

[late geode]

can you show full work.

it seems like you're skipping a ton of steps

and something bad happened along the way

[gaunt lynx]

okay my bad

[late geode]

switching from ln(1/2 b) to (ln 1/2 b) is also bad
ideally you'd want to write something like ln(b/2)

[gaunt lynx]

this

ou

yeah i forgot the bracket thing

so many rules wtfff

basically its ln(1/2b)^2 right and not 2*ln(1/2b)

[late geode]

you want more () to make it less ambiguous and again b/2 is miles better than 1/2b

ln( (b/2)^2)

[gaunt lynx]

wait

[gaunt lynx]

so this is correct?

[late geode]

missing () aside,
both are ok

its just another way of express it

what you did seemed to have ignored that 2 completely

[gaunt lynx]

i mean

2 * 1/2b

isnt that b just b

[late geode]

2 * 1/2 b is indeed b

[gaunt lynx]

so i was correct?

[late geode]

however how are you turning $e^{2\ln(\frac b2)}$ or $e^{\ln((\frac b2)^2)}$ into $\frac b2$

[gaunt lynx]

i mean e^ ln

it cancels it both out

so we are left with just 2 * b/2

[late geode]

you don't explicitly have e^ln here

[gaunt lynx]

yes but e^x

where x is ln(b/2)

[elfin berry]

ραμOmeganato5

[late geode]

you don't have e^(ln(b/2)) here

[gaunt lynx]

look

this is what i understand

[late geode]

as mentioned

you can't just ignore that 2

[gaunt lynx]

oh

is thjat like a rule

[late geode]

yes, you can't just arbitrarily ignore components

you're essentially trying to say
$$a^{bc} = ba^c$$

[elfin berry]

ραμOmeganato5

[late geode]

which is NOT a thing

[gaunt lynx]

yeah thats not true

so how else do we do it

[late geode]

power law for logs
or exponent laws

power law for logs actually gives you something of the form e^(ln(stuff))

[gaunt lynx]

so

[late geode]

or you could apply exponent laws a^(bc) = (a^b)^c

[gaunt lynx]

ln(1/2b^2)

damn okay

[late geode]

missing ()

() are mega important, especially here

[gaunt lynx]

okay so ln ((1/2b)^2)

[late geode]

yes

[gaunt lynx]

ohhh

okay damnn

[late geode]

and write b/2 instead of 1/2 b to make it less ambiguous/ugly

[gaunt lynx]

alright

[turbid mica]

2^-1 b

[marsh citrus]

@gaunt lynx Has your question been resolved?

[gaunt lynx]

il keep this channel open cuz im practicing nr

[glossy flint]

But after some amount of time it will close up automatically

Unless you click the usual red cross

[prime ivy]

yea if you want you can close it and reopen another one later if you need to

[gaunt lynx]

oh my god

how do i derive this

[prime ivy]

pretend a is a constant

[gaunt lynx]

like

yes but

is it product rule here

or no

because we have e^x -ax * e^x

[prime ivy]

split up the derivative $\frac{d}{dx}e^x - a\frac{d}{dx}xe^x$

[elfin berry]

MarcoMa210

[prime ivy]

wait do you use d/dx notation or nah

[gaunt lynx]

no

confused me

😭

[prime ivy]

ight just derive e^x and axe^x

[gaunt lynx]

no product rule here?

[prime ivy]

there is

between x and e^x

[gaunt lynx]

shit yeah

okay

[prime ivy]

you can ignore a since it's a constant

[gaunt lynx]

here

?

[prime ivy]

correct

[gaunt lynx]

yes but

[prime ivy]

what's this symbol though

[gaunt lynx]

now factorize

[gaunt lynx]

its a 1

[prime ivy]

ah

[prime ivy]

just pull e^x out

since it's a common factor

[gaunt lynx]

but we have 3

e's

so

maybe

(e^x)^2?

and only leave one e^x in the bracket

why

[prime ivy]

we have 3 e^x's here

just pull that out

[gaunt lynx]

wait why did you remove the 1

we still have to add that

[prime ivy]

it's * 1

[gaunt lynx]

oh

yes

no but

[prime ivy]

you wrote * 1 which we can omit

[gaunt lynx]

it turns into a plus

[late geode]

what turns into a plus

[gaunt lynx]

its so confusing man sorry

sometimes you do +1

sometimes its * 1

[late geode]

+1 appears when you differentiate a lone x

[gaunt lynx]

okay ill show an example after this

[late geode]

this isn't that

[gaunt lynx]

okay so here

can we just pull all of them out

then do e^x^3

[late geode]

no

[gaunt lynx]

shit

[late geode]

this again falls under poor algebra foundations

how would you factorise
ab + ac

[gaunt lynx]

a (b+c)

right?

but here we have 3 e's that kinda confuses me

[late geode]

same principle

you didn't think of doing stuff like a^2 right?
even though you initially saw 2 as there

[gaunt lynx]

wait yes it makes sense

i confused it with factorising x's

so

e^x * (-ax -a)

[glossy flint]

Good

But there's also another e^x

[gaunt lynx]

thats what i meant

shit

so

[glossy flint]

So you can factor out e^x from all the three terms

[gaunt lynx]

yeah i did right?

[glossy flint]

With this you mean?

[gaunt lynx]

yes

[glossy flint]

Well, this is only for the last two terms 😅

[gaunt lynx]

so then e^x(e^x-ax-a)

no other possibility

[glossy flint]

No...

You're definitely overthinking

How would you factor ab + ac + ad?

[gaunt lynx]

a(b+c+d)

[glossy flint]

There you go

[gaunt lynx]

so

[glossy flint]

So if a = e^x what do you have?

[gaunt lynx]

yes but there is no d

in our case

or do you mean the 1

[glossy flint]

Huh? You have 3 terms and all of them have e^x in common
So THERE IS d

[gaunt lynx]

wait

oh my god

its 1?

[glossy flint]

Nope

[gaunt lynx]

can you tell me

[glossy flint]

Ok let me write this

How would you factor AB + AC + AD?

[gaunt lynx]

A(B+C+D)

[glossy flint]

Good, now can you write how this expression becomes with these values?

A = e^x
B = 1
C = -ax
D = -a

[gaunt lynx]

BRO

e^x(-ax-a+1)

this is theonly logical thing

[glossy flint]

There you go!

[glossy flint]

Indeed

[gaunt lynx]

yes but where did you pull the 1 from

it makes sense but

where

[glossy flint]

$$e^x = e^x \cdot 1$$

[elfin berry]

Alberto Z.

[glossy flint]

I hope you know this

[gaunt lynx]

oh

[glossy flint]

The same as this: A + AB = A(1 + B)

[gaunt lynx]

this guy just multiplied it

[glossy flint]

And?

[gaunt lynx]

ohh

[gaunt lynx]

yes so the 1 is gone

but

i get what u mean

it doesnt make sense otherwise

[glossy flint]

No?🤔

[gaunt lynx]

yes omg

no it isnt

ur right

its still 1 * e^x

yes

[glossy flint]

Indeed 👍

[gaunt lynx]

also is this correct for the 2nd derivative

[glossy flint]

Yup awesome 💪

[gaunt lynx]

wait but yk what i dont understand

the -a gets added into the bracket

[glossy flint]

Sure

[gaunt lynx]

because theres a +?

[glossy flint]

Yes

[gaunt lynx]

oh shit

[glossy flint]

Again, this is something you should really remember/know way before doing things with exponentials and derivatives

[gaunt lynx]

yes

i just got confused

do

so here

does the -infinity influence the infinity of the e‘s

[glossy flint]

You have to use the factored form

Otherwise you get the indeterminate form ∞ - ∞

[gaunt lynx]

oh so do e^x(-ax+1)?

[glossy flint]

Exactly

[gaunt lynx]

okay yeah so then its infinity

also i dont need a case differentiation here right

[glossy flint]

If x → +∞, then you have -∞

[glossy flint]

Of course you need. When you have an exponential, in particular

[gaunt lynx]

but e is stronger than x

[glossy flint]

Correct

[gaunt lynx]

so its infinity

but that also suggests kinda that

[glossy flint]

But you have +∞ • (-∞) which yields -∞

[gaunt lynx]

ohhh

[glossy flint]

It is infinity, but negative

[gaunt lynx]

okay thats what i needed

yes okay

[glossy flint]

• times - = -

[cyan flame]

does anybody here have comfortability with probability and statistics? I'm having trouble understanding this problem:

[glossy flint]

!occupied

[marsh citrus]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[gaunt lynx]

okay so then if a is negative

it makes it be just infinity

stuff like that

[glossy flint]

Aren't you given it is positive?

[gaunt lynx]

yes i sm 😭 😭

thanks

so we dont need cases then

[glossy flint]

But yes, if you were given a is < 0, then you would indeed have +∞

[glossy flint]

Yeah luckily 👍

[gaunt lynx]

nice okay

[gaunt lynx]

also for lim x->-infinity it would be 0+ right

[marsh citrus]

@gaunt lynx Has your question been resolved?

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[glossy flint]

.close

[marsh citrus]

Channel closed

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[glossy flint]

,Usually 0+ and 0- are used only for the argument (like lim as x → 0+) and not the output of the limit.
But yes, it would be a 0+ if you'd like (but just 0 is correct and enough)

[marsh citrus]

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[gaunt lynx]

can someone explain me how this results in that

[gaunt lynx]

,rotate

[gaunt lynx]

like idk what he did here

[kind solar]

Two matrices are equal if and only if their corresponding entries are equal

[gaunt lynx]

so just compare line by line?

[kind solar]

yeah

[gaunt lynx]

but how did he get ab=1

[kind solar]

they assume that a is nonzero

[gaunt lynx]

yes its fo

[kind solar]

Then a^2 b = a implies ab = 1

[gaunt lynx]

cyclic process

of 2

[gaunt lynx]

wait and that comes out from every equation?

[kind solar]

Well yes all the other equations can also be used to deduce that ab = 1

[gaunt lynx]

ohh

but one equation is enough

okay nice

[kind solar]

and going the other way as well from ab = 1 you can deduce any one of the three equations

so the last two statements are equivalent

[kind solar]

but one would've been enough

[gaunt lynx]

also if i had to prove that a matrix got a cycle of 3

abc = 1

right

[kind solar]

I'm not familiar with graph theory sorry

[gaunt lynx]

i got an exam in like an hour

im so scared lowkey

[kind solar]

aww

good luck

you got this

[marsh citrus]

@gaunt lynx Has your question been resolved?

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[marsh citrus]

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[main swan]

Super simple question I am just brainfarting. I want to specify that if an interval goes from one range of values to another range of values, from -infinity to 0[ to ]0 to infinity, how would I write that?

[main swan]

Cause it's an interval within an interval, would it be ]]-inf;0[;]0;inf[[?

That looks crazy to me :P

[lusty pond]

(-inf,0) union (0,inf)

[brave marsh]

If you specifically want interval notation, then (-infinity, 0)u(0, infinity) is ok

But really it's R - {0}

[hexed creek]

R \ {0}

If you want to combine two disjoint intervals then you have to write them as a union

[whole sleet]

R ; {0}

[night relic]

i.e. ]-inf;0[ U ]0;inf[

[main swan]

I want to specify that it is from a range in the negative to a range in the positives, relating it to an interval of a;b, so would it be nicer to say a<0 and b>0, or is there a fancier way?

[main swan]

Oh so this looks apt

[hexed creek]

you can't have square brackets on an infinite sided interval

[night relic]

]a;b[ = (a,b), it's a french/european notation

[main swan]

^ this

[hexed creek]

Never seen that 🙁 British maths education

[main swan]

It is so very nice when the standards are different, isn't it :P

[hexed creek]

I would prefer the notation (-inf,0) U (0,inf)

[main swan]

But yeah, so, in an interval a;b, where a is negative infinity to 0 excluding 0, and b is ]0 to infinity, this is how it would be written?

[hexed creek]

If you want that specific interval then R \ {0} (or whatever the euro equiv is)

but you cannot simplify beyond a union

[night relic]

as long as a < b, then ]a;b[ is fine regardless of positive or negative values

if that's what you're asking there?

[whole sleet]

It's not written. That's two intervals

[hexed creek]

By definition there can't be a hole in an interval

[main swan]

I want to use this as a visual example of why the mean value theorem wouldn't work on a discontinuous function, I wanted to write and specify that the interval [a;b] is specifically on these lines excluding 0

[hexed creek]

Use R ; {0}

[main swan]

I am not trying to define the functions values, I am saying if we specifically put a point anywhere on the left and a point anywhere on the right, the secant does not have a tangent in the interval with the same value

[hexed creek]

You say $f(x) = -1 for x \in (-inf,0) , 1 for x \in (0,inf)$

[elfin berry]

WeAreIngram

[hexed creek]

I can't remember how to do it properly

undefined otherwise

[main swan]

You misunderstand I think, I know the values of the function, I am using the function for something else, or am I confused here?

[hexed creek]

What do you want out of us

[main swan]

To specify, any value on the function with x < 0 to any value on the function with x > 0 would have a secant, I want to say that

But I want to exclude x=0

I think I've come to the conclusion I just separately specify that a > 0 and b < 0, seems so simple now

I have not been sleeping much ngl, sorry about the confusion y'all

[night relic]

how about: for any $x \in A = ]-\infty;0[$, and for $y \in B = ]0;\infty[$, then there is a secant line between $(x,f(x))$ and $(y,f(y))$

[elfin berry]

حسیب ♥

[hexed creek]

You don't need to set the set to the interval you can just use the interval

[main swan]

But the interval needs to be specified for my example, so I just say a < 0, b > 0?

[night relic]

i was going to say: you really only need A and B if you plan on referring to these intervals later, but in that case you can write out the intervals explicitly

[night relic]

you can either do this, or you can just pick a specific interval

[main swan]

I do like making it more generalised, so I will be doing it like that I think

Thank you guys for your time! And sorry for the confusion

[night relic]

np / pdp

[hexed creek]

Use .close if you're done

[main swan]

.close

[marsh citrus]

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[marsh citrus]

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[pine swan]

could someone explain to me this process? i get how to test whether it converges with the p-test, but im not sure why this process is being used to find where the integral converges?

[fervent rampart]

they used the limit comparison test to compare it with the integral [ \int_1^\infty \frac{dx}{x^{3/2}} ]

[elfin berry]

cloud ☁

[pine swan]

sorry what's the limit comparison test 😭

[vocal hinge]

When a polynomial or so approaches infinity, the numerator/denominator behave like their highest power term, i.e √x and x²

[fervent rampart]

https://websites.umich.edu/~mconger/dhsp/lct.pdf

the idea is, find an integral that you know the behavior of and where you expect the limit to exist and be finite, then since they both have the same behavior that tells you about your original integral

usually this looks like comparing it to the ratio of the highest power terms (in this case, sqrt(x)/x^2) because the convergence of the comparison function is easily found by p-test

[pine swan]

thanks! sorry, im still a bit lost on why we multiply that with the by the original integral?

[fervent rampart]

the idea is, we want to find out whether f(x) converges, so we make up a g(x) where we expect the limit of f(x)/g(x) to be finite and where the convergence of the integral of g(x) is easy to determine

because if the limit of f(x)/g(x) exists and is finite. we know either both integrals converge or both diverges

[pine swan]

OH

wait ok i got it tysm

[vocal hinge]

Isn't that g(x) technically 1/f(x)

[pine swan]

.close

[marsh citrus]

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[marsh citrus]

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[jagged orchid]

yooo i just finished acceleration and am skipping to geometry next year, is there anything important that is good to know prior to taking the course?

[vagrant gull]

which course?

geometry?

you mainly need to understand the basics of algebra before you start applying geometry to it. like solving equations, working with variables, and rearranging formulas that stuff helps a lot when the geometry gets harder

[marsh citrus]

@jagged orchid Has your question been resolved?

[jagged orchid]

yeah, i mean i just passed the algebra 1 final to get to geometry, so i know all of that stuff

[marsh citrus]

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[spark nest]

$x_1=3, x_2=7, x_{n+2}=x_{n+1}^2-x_n^2+x_n$. Find $lim_{n \rightarrow \infty} \sum_{k=1}^n 1/x_n$

[elfin berry]

Nerdyasianguy

[cunning fiber]

<@&268886789983436800>

[marsh citrus]

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

[spark nest]

I'm trying to split it into some telescoping sum

So that the adjacent numbers cancel out

But haven't got that

The limit should be 1/2 if i'm not mistaken

[brittle shuttle]

Correct

Can’t you prove this by strong induction on the sum?

[spark nest]

What is the induction statement?

[brittle shuttle]

Also I think you made a small mistake. Shouldn’t the denominator be x_k not x_n? Also I’m working on it but a strong induction proof should work

[marsh citrus]

@spark nest Has your question been resolved?

[spark nest]

I solved it

[spark nest]

Oh yeah

[marsh citrus]

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[spark nest]

Using telescoping sum

[marsh citrus]

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[sinful ocean]

Woah whay math is this. I am currently doing Algebra 1 in 7th.

[high raven]

This is a channel for helping with math and not discussion. Please type .close to close the channel.

[sinful ocean]

sorry. but one question, can I ask any type of math question like how to solve sin 350?

[main idol]

Yes you can ask here.

[marsh citrus]

@sinful ocean Has your question been resolved?

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[marsh citrus]

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[cobalt sedge]

do you have a question? if not, please close the channel so other people can use this

!done

[marsh citrus]

If you are done with this channel, please mark your problem as solved by typing .close

[lavish pelican]

😈

[cobalt sedge]

am I crazy or did OP leave the server? I can see that he has roles, but I dont share any servers with him???

ok, thats cached stuff

[lavish pelican]

ye bro has dipped

[cobalt sedge]

.close coz OP left

[marsh citrus]

Channel closed

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[reef fable]

maybe he got muted

[lavish pelican]

what does op stand for

overpowered

[cobalt sedge]

original poster

[lavish pelican]

o ok

[reef fable]

i thought it was Observation Post

[marsh citrus]

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[elfin berry]

Screaming Iced Tea

[blissful moat]

what have you tried so far

[elfin berry]

Screaming Iced Tea

Screaming Iced Tea

[blissful moat]

ok

why

[glossy sapphire]

3^x = 1

1 * 13 = 13

or you could just divide both LHS and RHS

[blissful moat]

write $3^x\cdot 13 = \frac {13}{3}$

[elfin berry]

ghost

[glossy sapphire]

oh shoot its 13/3 XD

[blissful moat]

non

[glossy sapphire]

so its not 0 haha

fraction

negative

[blissful moat]

as said you can cancel 13

right

also use \times for multiplication

[elfin berry]

Screaming Iced Tea

[blissful moat]

and \frac{}{} for fractio

yes

[elfin berry]

Screaming Iced Tea

[blissful moat]

if you are done with the thread and do not have any other query, you can close it by typing .close or .solved

[marsh citrus]

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[marsh citrus]

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[open meadow]

need help again i made this problem

[marsh citrus]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[quaint yoke]

what problem

[open meadow]

Players A and B each begin with one fair coin the start of every round both players simultaneously flip their coins The rules are If both coins show the same face the players exchange their coins if the coins show different faces each player keeps their own coin after the exchange keep step each player secretly chooses whether to Flip their coin again next round or replace their coin with a fresh fair coin from an infinite supply but they may only do this replacement if the other players coin showed heads this round. as the game goes on forever wht is the probability tht player a will hold their original coin at infinitely many different rounds

can u tell me answer

[drowsy plover]

A will hold his coin infinitely many round, with a probability of 1

[open meadow]

but the rule states tht the first coin he started with disappears from his hand

[drowsy plover]

per the second borel-cantelli lemma

if the events are independent and the sum of their probabilities is infinite, then the probability that infinitely many of them occur is 1.

[open meadow]

but i didnt assume it always disappear at the first chance

[open meadow]

but borel-cantelli lemma cant be used here bc it depends on previous rounds and the original coin disappears

well ur right i did not specify how players choose to replace their coin or not or if if they replace always or never

.close

[marsh citrus]

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[open meadow]

ty

[marsh citrus]

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[red nimbus]

Let $p$ be a prime such that $x^2 + 67x + 1$ is irreducible mod $p$. Show that the order of $x$ divides $p + 1$ in $\Z[x]/(p, x^2 + 67x + 1)$. \

any hints would be cool. People say it's trival hmm I tried using $(a+b)^p=a^p+b^p$ tho

[turbid mica]

six seven

[devout mauve]

whats the order of the field you are constructing

[red nimbus]

ok i we know that x²+67x+1=0 which means x²=-1-67x mod R where R is the ring in the task

so i can construct every monomial and by that polynomial with a linear function

so the order is 2?

[devout mauve]

R is the ring which is a field. how many elements does it have

[red nimbus]

ok i only remember it is some prime power

[devout mauve]

how many linear polynomials are there

[red nimbus]

1

like ax+b

[devout mauve]

how many choices for a and b

[red nimbus]

oh

so p * p = p^2

Z but with mod p

|R| = p²

[devout mauve]

so how many elements does the multiplicative group have

[red nimbus]

p²-1?

[devout mauve]

and what else can you tell me about the multiplicative group?

[red nimbus]

it should be periodic?

[devout mauve]

what do you mean by that

[red nimbus]

wasnt it for finite fields that the elements repeat themselves?

[main idol]

if you mean g^k = e, there's a special word for that in group theory

[red nimbus]

cyclic

so we can use lagrange?

x^(p²-1)=1

x^{(p-1)(p+1)}=1 => ord(x) | (p-1)(p+1)

[marsh citrus]

@red nimbus Has your question been resolved?

[marsh citrus]

@red nimbus Has your question been resolved?

[marsh citrus]

@red nimbus Has your question been resolved?

[marsh citrus]

@red nimbus Has your question been resolved?

[red nimbus]

my pity progres

[reef fable]

why no one helps bro ☠️

[red nimbus]

because i manifested it

[drowsy plover]

ya know you can ping helpers right?

[red nimbus]

i aint gon ping pre unis

[warm lake]

real helpers took off the helpers role years ago

[red nimbus]

frobenius it is

.solved

[marsh citrus]

Channel closed

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[hushed egret]

💀

blud asked my question in the help channels

[crystal lintel]

[crystal lintel]

it’s like that time you gave chris a question in mg then he asked it in a help channel and copy pasted the answer someone posted

[devout mauve]

I may have fallen asleep yesterday. anyway, the problem is that in general if you do this process modulo some quadratic polynomials x^2+ax+b then it can easily happen that the order of x is p^2-1. hence you will have to use some specific property of x^2+67x+1. but I dont know which. for starters, you know that b^2-4ac cannot be a quadratic residue mod p (otherwise the poly would not be irreducible), so maybe by quadratic reciprocity you can find out something about p. not sure

@red nimbus

[hushed egret]

you use the fact that x has norm 1 with respect to F_p

so that powers of x are actually in the subgroup of norm 1 elements, which is smaller

[devout mauve]

ah yes I thought about the 1 but didnt spot this. damn I'm really out of practice

[marsh citrus]

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[novel juniper]

I know this is simple but I have the feeling I messed up

$L\left(\theta \mid x_1 \dots x_n \right) = \theta^{\sum_{i_1}^{n} x_i} \left(1 - \theta \right)^{n- \sum_{i=1}^{n} x_i}$
\
Then the log-likelihood function is $\ln \left[L\left(\theta \mid x_1 \dots x_n \right) \right] = \sum x_i \ln(\theta) + (n- \sum x_i) \ln( 1- \theta)$

[elfin berry]

wai

[novel juniper]

oky

.close

[marsh citrus]

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[marsh citrus]

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[kind mulch]

what is this question trying to say? isnt y = f(x) already defined so why is it asking for me to find it

[hollow blaze]

it should be x = f(y) and then when you invert it , it becomes y = finverse (x)

[sleek lake]

you need to swap, so it becomes x = y³

and then you need to find y

[kind mulch]

oh thats weird

[dense sun]

It’s the result after the change.

[kind mulch]

so, y= x^3 ==> x = y^3 ==> y = x^1/3 ?

[sleek lake]

or find x and then swap, i guess

yes

[dense sun]

Exactly.

[kind mulch]

hmm ok, that makes sense although it seems useless

[dense sun]

Do you have more questions?

🙂

[kind mulch]

this question makes sense now ty, not for the time being

[dense sun]

You are welcome.

Have a nice day! You can close typing .close.

[kind mulch]

.close

[marsh citrus]

Channel closed

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[kind mulch]

.reopen

[marsh citrus]

✅ Original question: #help-33 message

[kind mulch]

actually, what do we do with y=2/x? because swapping y and x then solving for y just gives back y=2/x right?

so does it not have an inverse thing

.close

[marsh citrus]

Channel closed

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[sleek lake]

it has the thing, it just looks the same

inverse is reflecting the graph diagonally

2/x is inverse of 2/x

[marsh citrus]

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[low scaffold]

how do we solve this

[graceful mountain]

Do you remember the theorem?

[low scaffold]

What was it?

[glossy flint]

You don't have it in your notes/textbook?

[red nimbus]

,w max min theorem

[tame arch]

Extreme value thrm

a function is guaranteed to have both an absolute maximum and an absolute minimum if two conditions are met: 1. f(x) is continuous 2. there is a closed interval @low scaffold

if this is not in your notes, you are gonna want to put it in there asap

[low scaffold]

ahh i remember

also my other question is

this question

I did the workking ot and got

$x\geq11$

[elfin berry]

ø

[low scaffold]

i know its

,11]

I dont know what to put in the blank

[indigo nest]

I'm doing to assume that your working was correct

[late geode]

your result is incorrect,
can you show your work

[indigo nest]

Oh oops thanks ramonov

[indigo nest]

Also it's not that, x is greater than or equal to 11

Try drawing the region on a number line

[tame arch]

you should be getting an interval no?

[late geode]

x>=11 itself was already wrong

[exotic valley]

feels like you're gonna need to do some casework

[indigo nest]

Showing what you've done would help

Also too many cooks, I'm stepping away

[exotic valley]

same, sorry for interrupting

[tame arch]

it cant be x>=11 at all, for a fraction to be positive, the numerator and denominator must have the same sign. if you go larger than 11 you get a negative numerator and pos denom

oops my bad

[odd agate]

,rcw

,rccw

[indigo nest]

Why did the sign reverse on the first line

[tame arch]

i made this mistake so many times when i was younger lol

[low scaffold]

It didn’t

It’s the same as the sign in the question

[indigo nest]

Well I see a ≥ and then a ≤ in the next line

[tame arch]

it didnt flip

the mistake was that it didnt flip

[indigo nest]

Also no, not how inequalities work

You can't just multiply stuff to inequalities without being careful

[tame arch]

the best way to solve these is to never multiply by the variable. Instead, move everything to one side so the other side is zero

thats how i learned

[indigo nest]

I'm gonna let you take over, we're gonna keep talking over each other

[tame arch]

sorry didnt mean to

[low scaffold]

Because I heard someone say if u multiply by x, it will change the sign

[indigo nest]

That's not how it works at all

Unlike equalities where you can do the same thing to both sides without worrying about it

Inequalities are a bit more involved

Addition and subtraction work the same

But for multiplication, what happens depends on the sign of what you multiply by

Yes Celine I am coming to cases

[tame arch]

[indigo nest]

So when you're multiplying by (x-6), you get two outcomes depending on the value of x

For what region is x-6 positive

And for what region is it negative

[tame arch]

Great explanation actually

[low scaffold]

So we don’t change the signs

When multiplying by variables

[tame arch]

I would avoid it entirely, the entire issue can be solved by just moving everything to one side

[indigo nest]

Not true, you'd need to get rid of the denominator eventually

[turbid sinew]

Hope I'm not bothering... I was gonna say flip both the fractions (you can do that right?)

[indigo nest]

And you will run into the same issue

[glossy flint]

Stick to this, it's the most common way @low scaffold

[indigo nest]

Also it's a very bad idea to simply avoid stuff if you can't understand it

[low scaffold]

So how’s it meant to be done

[indigo nest]

That would also change the signs

[indigo nest]

Well I asked you a question

[glossy flint]

#help-33 message

[tame arch]

Alright im gonna leave this to Xavier

[low scaffold]

X-6 is negative in for values lesss than -6

[tame arch]

I think all of us trying to help is confusing him more

[indigo nest]

...no

With all due respect man, you can't be doing calculus and struggle at stuff like this

[low scaffold]

I’m a bit rusty tn

Rn

[indigo nest]

when is x-6 positive

What does it mean to be positive

[glossy flint]

Rusty on FIRST degree inequalities? That's a VERY SERIOUS problem!

[indigo nest]

Also genuinely asking both Alberto and Noctorin

[indigo nest]

Wouldn't this give you an x-6 in the denominator

That you would then need to get rid of?

Thus running into the exact same problem in the end?

Please correct me if I am misunderstanding what you're suggesting

[tame arch]

i think you are misunderstanding

[low scaffold]

Man am I cooked

[tame arch]

The way i learnt it, you dont need to get rid of it

[indigo nest]

Interesting

[glossy flint]

I usually do:
Num > 0 (or ≥)
Denom > 0

And then a sign scheme

[tame arch]

I think they called it boundary point rule

[glossy flint]

Probably it's a different method than some others that are taught

[indigo nest]

Ah right sign schemes

Forgot about those

Fair, but then we'd have to explain sign schemes down the line

[exotic valley]

either way sign analysis is inevitable

[glossy flint]

I mean, I hope he has been taught them 😅

[tame arch]

atp im down to stay up as late as i have to in order to help

[low scaffold]

Alr alr so

How do I do this again

[tame arch]

Alright lets start from the beginning

My first step is always to set the inequality = to 0

[low scaffold]

Ok

And then

[tame arch]

combine it into a single fraction

[exotic valley]

there's us as well

you don't have to if you are not available. just a gentle, friendly reminder

[tame arch]

of course, i have a passion for helping though lol

[reef fable]

good guy

[subtle shadow]

||good guys finish last, kidding||

[low scaffold]

And then

[tame arch]

find critical points

[low scaffold]

Set top and bottom

= 0

And solve the variables?

[tame arch]

yes

i was scared you didnt know what a critical point was lol

[low scaffold]

lol

[tame arch]

when you are done solving, let me know what you get

[red nimbus]

that would be critical

[tame arch]

LOL

[low scaffold]

Yeah and then

[low scaffold]

x=-1 and x=6

[tame arch]

um

hmm

show me ur work

i agree with 6

not -1

[tame arch]

you should have 11-x in numerator

let me see where you went wrong

its not 5-x-6

its 5-(x-6)

distribute the negative

5-x+6

11-x

if i had a dollar for every time i made this mistake i would probably own a porsche lol

[low scaffold]

X=11

[next ivy]

steel yourself for more.

[tame arch]

okay good

so we have (6, 11)

picture a number line

this gives us 3 intervals

what would those be?

[low scaffold]

Wdym

[tame arch]

Ok let me try to word this better

(6, 11) divides the number line into three intervals.

what would those three be

btw quick side note that you HAVE TO REMEMBER

6 in the denominator makes it 0

dividing by 0 is a HUGE NO in math

so you MUST remember to not include it in the interval by using a parenthesis instead of a bracket

[low scaffold]

Hm

Ik

Yeah

But what does the numerator

Tell us

[tame arch]

it gives us one of the critical points

alright so

what are the three intervals

if you need me to tell you that is okay

this might help you understand better, what is the domain?

[low scaffold]

All accepted values by the function

[tame arch]

yuh but like

what is it lol

like numerically

[low scaffold]

Wait

Can it be less than 6?

Because it’s not equal to 6

[tame arch]

yeah of course

[low scaffold]

But if we do

(6, 11]

That’s saying start from x >6

[tame arch]

nope

x=6 is a discontinuity thats all

[low scaffold]

Oh

So 5,4 works

As an example

[tame arch]

well no

you dont go high to low

in interval notation

[low scaffold]

Why

Wdym

[turbid sinew]

Could also mean 5.4?

[low scaffold]

I mean

As an example

The function can accept 5

Since it’s not 6

[tame arch]

yeah

alright im going to tell you the intervals

(-inf, 6) ∪ (6, 11) ∪ (11, inf)

[low scaffold]

But look

How is this accepted

[tame arch]

yeah we hadn't gotten to the answer lol

[low scaffold]

Ab

ah

[turbid sinew]

Cuz that's >= not just strictly >

[tame arch]

alright

so

we know

(-inf, 6) ∪ (6, 11) ∪ (11, inf)

are three intervals

so we test a value in each interval

if its true

then yuh

if not

then nuh

[low scaffold]

So we can do 11

That doesn’t matter tho because it’s in the numerator

It won’t be undefined

[tame arch]

well, no

you wouldnt want to

you would want a number between the interval

not on the endpoints

anyways, now we pick a number in the interval, plug it into the equation (11-x)/(5(x-6))

the important thing is if its positive or negative

because we are looking at >=, we want a positive value

because anything >= 0 is obviously positive

therefore true

[low scaffold]

how many steps does this require 😭

I thought it would be simple

[turbid sinew]

Same as how many intervals there are I guess?

[tame arch]

i promise its really easy

the more you practice, the faster you get at it

[low scaffold]

ohhhh

I think ik how to do this

It didnt click in my mind

[tame arch]

alright im glad you figured it out

[low scaffold]

i just lack practice

[tame arch]

practice is important

anyways im snoozing, goodnight fam

[low scaffold]

thanks fam

good night

[tame arch]

np g

gn

[marsh citrus]

@low scaffold Has your question been resolved?

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[marsh citrus]

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[alpine hull]

how do i find the area?

[marsh citrus]

@alpine hull Has your question been resolved?

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[marsh citrus]

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[misty hinge]

How do I check convergence of number 3

[misty hinge]

I tried D’alambert on that one as well as a major thingy

1/(k+2)

But I keep getting 1

[restive hedge]

@misty hinge umm convergence means ?? Ig u wanna solve questions right ?

My english weak

[dire basalt]

partial fractions

[misty hinge]

It meeans the sum exists

[sharp grove]

you can do a limit test or idk what it's called in your program

[dire basalt]

$$\dfrac{1}{k(k+2)} = \dfrac{1}{2} \left( \dfrac{1}{k} - \dfrac{1}{k+2} \right)$$

[misty hinge]

Oh that’s smart but is there another way because I think we haven’t seen that yet for numerical analysis

[elfin berry]

robin.dabanc_

[sharp grove]

he wants to prove convergence

[dire basalt]

exactly, the thing i wrote above is a telescoping so the infinite sum is finite

[restive hedge]

That what i tho convergence means solving questions?

[misty hinge]

Also I don’t think you can rarange order because you haven’t proven yet that is concergest absolutely

[sharp grove]

well yeah ig

[dire basalt]

idk what youre saying bruv ;

[restive hedge]

Nvm 😭 you handle this my english weak

[dire basalt]

convergence means showing that the sum is finite

[dire basalt]

ah i see

[misty hinge]

Yup

[tired ore]

this is 1/(k^2+2k) which is < 1/k^2

for large enough k

[misty hinge]

Also I don’t think you can do partial fractions because it isn’t asociative unless you prove absolute convergenxe

[dire basalt]

wdym? i can sum it up to n terms which gives some function of n, and then limit n to fininiity to get some finite number

[misty hinge]

Oh damn true hahaha how did I look over that

[dire basalt]

but then youll have to prove 1/k^2 is convergent

[sharp grove]

you can solve suppose $V_k = \frac{1}{k\sqrt{k}} and then take $$\lim_{k\to \infty} \frac{\frac{1}{k(k+2)}}{\frac{1}{k\sqrt{k}}=0$$

and $\sum V_k$ converges thus the original sum will converge

[elfin berry]

darky
Compile Error! Click the reaction for more information.
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[tired ore]

thats a basic fact

[misty hinge]

I mean you can’t rearange an infinite amount of terms in a sum because it’s not asociative unless you prove absolute convergence

[dire basalt]

its actually true for every positive k, so youve just lower bounded the sum

[dire basalt]

ik but idk if a "proof" question would accept that

[misty hinge]

It’s 0

[dire basalt]

no its pi^2/6

[misty hinge]

How?

[dire basalt]

we're talking about the sum not each term

1+1/4+1/9+1/16+...

[misty hinge]

No I meant that limit

So that tells me the sum is convergent

[dire basalt]

thtas not true

[restive hedge]

Can't u use this
First limit tend to 1 give 1/3 and last limit tend to infinity that will give zero so sum exist hence proved ?

[dire basalt]

take $\sum 1/k$

[elfin berry]

robin.dabanc_

[dire basalt]

1/k tends to 0, but the sum is divergent

[misty hinge]

Oh wait you didn’t prove the sum is convergebt you just proved the series is

[elfin berry]

darky

[sharp grove]

you can do it like this with the comparison test

[misty hinge]

Oh and how do you know that’s larger than the original

[dire basalt]

no we did

[elfin berry]

robin.dabanc_

[misty hinge]

I thought hyperharmonic converged to e 😭

[elfin berry]

robin.dabanc_

[sharp grove]

$$k^2 < k^2 + 2k \quad (\forall k \in \mathbb{N}^*)$$

$$\frac{1}{k^2+2k} < \frac{1}{k^2}$$

(summing up the inequality gives you)

$$\sum^{\infty}{k=1} \frac{1}{k^2+2k} < \sum^{\infty}{k=1} \frac{1}{k^2}$$

$$\sum^{\infty}_{k=1} \frac{1}{k^2+2k} < \frac{\pi^2}{6}$$

Thus $\sum^{\infty}_{k=1} \frac{1}{k^2+2k}$ converges.

[elfin berry]

darky

[misty hinge]

How do you prove that de sum for 1/(k•k^(1/2)) is convergent

[sharp grove]

it's a reiman series

or a P series

i think that's what you call it

[misty hinge]

Don’t know those

[sharp grove]

p = 3/2

[misty hinge]

Numarical analysis sis so hard 😭

[sharp grove]

$\sum \frac{1}{n^p}$

[elfin berry]

darky

[sharp grove]

do you recall seeing this

?

[misty hinge]

Nope

[restive hedge]

How u proved other questions?

[misty hinge]

Just for context I am in my last year of highschool

[dire basalt]

a person who doesnt know what convergence is should really not be talking here

[tired ore]

do partial fractions then

[dire basalt]

i suggested that

he said its not allowed

[sharp grove]

partial fractions it is

[restive hedge]

Mb man why u so mean 😭

[sharp grove]

huh

how do you usually solve these questions

in class?

[misty hinge]

Yeah I think so but I’m not sure

[sharp grove]

what tools have you studied

[dire basalt]

just show partial fractions, your teacher will be impressed either ways

[misty hinge]

D’alambert cauchy liebnitz integral and comparison test

[restive hedge]

I have right to learn man

[tired ore]

you know these?

[dire basalt]

yeah, not to poke into other's learning

[misty hinge]

I really think they won’t it’s supposed to be an easy excercise

[misty hinge]

Yes

[restive hedge]

I know how to solve but don't know how to prove i just asking how he proved other Question that will help

[restive hedge]

@dire basalt

[tired ore]

they are useless if you dont know that 1/n^p converges for p > 1

[dire basalt]

alright man, you confused me cuz you asked what convergence was in the beginning

[misty hinge]

That’s not true

And I do know that harmonic series is divergent and the hyperharmonic is convergent

And the alternating harmonic is also convergent

[tired ore]

just do partial fractions

[dire basalt]

its the simplest tbh

no bs

[sharp grove]

yeah

any other method is prob out of reach

since you didn't study reiman series

[dire basalt]

bounding is still valid

[misty hinge]

Alr thanks

[dire basalt]

squeezing

[misty hinge]

.close

[marsh citrus]

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[marsh citrus]

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[shell galleon]

h_j is positive semidefinite and h_j = e_jX(X'X)^-1X'e_j and by definition of pos semidefinite this is >= 0
however i dont know how to proof it is smaller equal to 1

[spark otter]

maybe "h_j that is positive semidefinite" is not what you meant

maybe you were talking about (X'X)^-1

[devout mauve]

without more context this is certainly not solvable

[spark otter]

I infer that x_j = X'e_j
What more do we know about e_j and X?

[shell galleon]

H is positive semidefinite

[spark otter]

what's H

[shell galleon]

X(X'X)^-1X'

[spark otter]

what are the dimensions of X

[shell galleon]

n * k

[spark otter]

and what are the e_j

[shell galleon]

e_j is a zero vector with the j'th element being a 1

[spark otter]

isn't this matrix H an orthogonal projection then

[shell galleon]

Yes

[spark otter]

so

what do we know about the norm of Hx compared to the norm of x

[shell galleon]

it is equal to each other

Hmm wait that is if H is orthogonal matrix

I am not sure tbh

[spark otter]

no

An orthogonal projection will not preserve norms in general

but it will affect it

[shell galleon]

maybe only if x is in X?

[spark otter]

Idk what "in X" would mean, but we're not interested in that

H is an orthogonal projection

so it's a projection onto some space S, parallel to S^orthogonal

then $x = x_S + x_{S^\perp}$

[elfin berry]

Rafilouyear2026

[spark otter]

(x_S being Hx)

x_s and x_S^orthogonal are... orthogonal

pythagoras

$|x|^2 = |x_S|^2 + |x_{S^\perp}|^2$

[elfin berry]

Rafilouyear2026

[spark otter]

thus... $|Hx|\leq |x|$

[elfin berry]

Rafilouyear2026

[shell galleon]

How can we determine |x| ?

or |Hx|

[spark otter]

well... it depends on the x

||Hx|| will appear on its own

[spark otter]

As long as you keep this in mind

Rewrite $h_j = e_j'He_j$ into something with norms...

[elfin berry]

Rafilouyear2026

[spark otter]

(final hint: ||you need a specific property of projections...|| ||H^2 = H||)

[shell galleon]

Aaaah okay

Ngl matrix algebra has been a long time ago so I needed this

thank you very much :)

.close

[marsh citrus]

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[marsh citrus]

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