help-39

[cobalt hinge]

👍

[shell tapir]

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[pearl pond]

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[lean lotus]

pls explain this

why did we assume x is 3sin(Thera)

[verbal whale]

That 3 was just to make that 9 factor out

And then you'd want to make use of the fundamental trig relationship i.e. sin² + cos² = 1
So you can choose x = 3sin(θ)

[lean lotus]

oh

wait

still why do we do sin

[verbal whale]

You can pick cos, but sin is better because its derivative is cos, so we don't have extra minuses annoying us

[lean lotus]

hmm okay

do u have a sheet of all like trig identity

needed for calc

[shell tapir]

there is one pinned in the precalculus channel

[lean lotus]

alr

also for my questions

why is sqrt(9-x^2) = the next step

for the first example

[verbal whale]

Well, if you've just picked x = 3sin(θ) you'll have to calculate how your integral becomes, of course 😅

[pearl pond]

@lean lotus Has your question been resolved?

[pearl pond]

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[midnight haven]

Hello

[midnight haven]

could someone guide me through this problem

[west bluff]

Well, a bunch of things

First, what is n?

And second, what do they mean by “find the chance“?

[midnight haven]

im guessing n is an arbitrary number

and this question is on probability

[west bluff]

Oh, okay

[tardy reef]

AM GM tells you the highest product is n^2

But the minimum is 2n-1

[midnight haven]

ohhh

[west bluff]

Well, gotta admit probability is not my forte so I'll let someone else help ya

[tardy reef]

so, you gotta find all the products such that a * (2n-a) >= 0.75n^2

[midnight haven]

i did that

a range would be

n/2 =< a =< 3n/ 2

[tardy reef]

Oh, they are not naturals, so this should be 0 mbmb

[tardy reef]

Now, focus on the smaller of the two possible values

[midnight haven]

n/2

wait

we have infinite dont we ?

why just 2

[tardy reef]

what are the possible ranges to choose the smallest value in general

[tardy reef]

I meant, the sum a+b = 2n has a<b wlog

So, by the two I meant focus on only a and not b

[midnight haven]

oh

okay

[midnight haven]

i didnt get you

[tardy reef]

Can a be n+1? If a<b

[midnight haven]

no

[tardy reef]

So its bounded right?

[midnight haven]

yeah a is bounded ?

[tardy reef]

The range is upper bound - lower bound

What is the size of sample space for a?

[midnight haven]

n

3n/2 - n/2

[tardy reef]

Uhh, sample space is all the admissible values of a

It should be from 0 to n

Of that we only want n/2 to n

since a is smaller than n

[midnight haven]

ohh

[tardy reef]

What we want is $P(a| a \times (2n-a) > 0.75n^2)$

[jolly parrot]

Facter10Br4g

[midnight haven]

I get it

[tardy reef]

So, uhhh all the possible values of a are from 0 (because they only want non-negative) to n (because of what I said before)

And you want the subset of that so you satisfy the product condition

[tardy reef]

[midnight haven]

Thanks

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[pearl pond]

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[drifting pawn]

Hi i need some urgent help cause i have an exam in 2 hours and i am not able to solve this can someone help me please
So it is a cauchy problem the picture is in french but the question is to solve the cauchy problem below

[lime river]

U3k = U3k-3 + 1 = U3k-6 + 2 = U3k-9 + 3 ..... = U0 + k
Similarly U3k + 1 = U1 + k
U3k + 2 = U2 + k

[lime river]

A way to think about this can be if you had the recurrence Un = Un-3 how would have solved it if U0 U 1 and U2 were given

[lime river]

Here terms will be repeating.

[lime river]

Now if we add 1 to the recurrence here terms of the form U3k+r where r belongs to remainder class of 3 will follow a ap instead of repeating

[jolly parrot]

Arya

[drifting pawn]

Bu why do the 1 2 .... k increase

[lime river]

U(n-3) = U(n-3-3) + 1
So U(n-3) = U(n-6) + 1
U(n) = U(n-6) + 1 + 1

U(n) = U(n-6) +2

[pearl pond]

@drifting pawn Has your question been resolved?

[drifting pawn]

Can you please tell me if we are gonna get Un= [n/3]+1

[lime river]

Yea it's correct

[drifting pawn]

Thank you so much

[lime river]

Wlc

[pearl pond]

@drifting pawn Has your question been resolved?

[pearl pond]

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[torn sky]

In my book we've been taught this rule.

Now a question asked us to derivate 10^-x. I'd say that would be 10^-x * ln(10), but the book says that it's 10^-x * ln(10) * -1

[midnight haven]

10^-x = (10^-1)^x

[rose robin]

or chain rule, inner function is -x

[open rivet]

you are forgetting the chain rule

[torn sky]

So can we say that instead, generally for such type functions we can do a^x * x` * ln(a)?

I can rephrase if wished for

So if the exponent can be seen as a function, also that in this case -x = u(x) the derivative becomes a^x * u'(x) * ln(a)

[open rivet]

yup

u-sub with chain rule

[torn sky]

So do these rules apply for all kinds of situations? Thing I find difficult is knowing when to use all those other rules when a new rule is introduced

[open rivet]

yeah

so with the chain rule: $g(f(x))'=g'(f(x))f'(x)$

[jolly parrot]

Bonk

[open rivet]

so if we have f(x)=10^(-x)

and use a u-sub u(x)=-x

f(u(x))=10^u

then f(u(x))'=f'(u(x))u'(x)=log(u)10^u *-1

kinda make sense?

[torn sky]

Ah alright, yeah makes sense, thanks!

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[wild fable]

How do we know whether to use the limit as pi/2 approaches from the left or right hand side

[spare lark]

0 to pi/2 + would be sus since its not defined in pi/2

So pi/2 -

[wild fable]

wdym?

[spare lark]

Pi/2+ is above pi/2

[wild fable]

o

[spare lark]

So integral from 0 to pi/2+👎

[versed mica]

because pi/2 is the upper limit

[wild fable]

ok gotcha

How can I start with this problem?

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[pearl pond]

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[midnight haven]

Part b

[plush bramble]

the range of g is the domain of the inverse function

[compact ridge]

all A level students should engrave this into their brains

[compact ridge]

but also you know that cx + d = 0 gives you the asymptote

if that x-value is not in the domain x > k

g(x) will be a 1-to-1 function in this domain

one of the endpoints will be the horizontal asymptote, which is a/c

the other endpoint will be g(k)

also the form will be ... < g(x) < ....

[midnight haven]

Am I meant to find the inverse function for a single mark or am I missing something

[midnight haven]

I don't get the a/c thing

[compact ridge]

actually you shouldn't use that fact for this q

[midnight haven]

I figured

[midnight haven]

Mind explaining this to me tho

[compact ridge]

https://www.youtube.com/watch?v=miPOf5lXfpw

[plush bramble]

oh yes i missed that g was a special kind of function

[compact ridge]

(1 minute 30 seconds)

[plush bramble]

follow the first example here: https://www.cuemath.com/calculus/rational-function/

[compact ridge]

riemann you'd be way better at proof-based maths lol

leave the pre-rigorous grunt work to helpers like me, lmao

[plush bramble]

[midnight haven]

Ahaa

[midnight haven]

Okay so g<1/3

What about the other boundary

[compact ridge]

then did you find what g(k) equals

[midnight haven]

What was k again

[compact ridge]

ah right I can figure it out for you then

cause 1/t + 2 > 2 basically for all real t > 0

so k = 2 and you just need to find g(2)

[midnight haven]

Oh so k is the least possible value of x for t>0

So -2

Got it ty

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[pearl pond]

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[compact ridge]

nwnw!

[pearl pond]

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[midnight haven]

If we have two events A and B

[midnight haven]

and they are mutually independent events

we cant represent that in a venn diagram can we ?

like

[wet swallow]

Independent or mutually exclusive?

[midnight haven]

if A and B are mutally exclusive then we just draw 2 seperate circles with no intersection between them

is there a way to do smth similiar like that for mutually independent ?

[midnight haven]

independent

[midnight haven]

we say mutually independent when we have more than 2 events right ?

okay

thanks

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[quick star]

wait for two events, is there even a distinction between independent and mutually independent?

[midnight haven]

i dont think so

[midnight haven]

🤷‍♂️

[quick star]

and i'm guessing plainly saying independent is just pairwise independence

and doesn't account for stuffs like P(A n B n C) = P(A) P(B) P(C)

okay then

[pearl pond]

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[quiet cargo]

please do the last one someone

[cobalt hinge]

!answer

[cobalt hinge]

so part c)?

[quiet cargo]

yeah c ii

the extremely last question

[cobalt hinge]

k

[quiet cargo]

thank you

[cobalt hinge]

oh boy

i gotta rack my memory for this one

casework

so we have 44 boys and the given information in the table

[quiet cargo]

$$ \text{is there LaTeX here?} $$

[jolly parrot]

hades

[cobalt hinge]

lets imagine one of the boys likes athletics

[quiet cargo]

oh ok there is

[cobalt hinge]

what is that probability?

[quiet cargo]

nothing just continue

its a programming language

u can do cool shit

wait

$$ \int^7_3 3x^2 - 4x +3 \text{ dx} $$

[jolly parrot]

hades

[quiet cargo]

@cobalt hinge

just makes stuff look better

[warm current]

$\dd x$

[jolly parrot]

SWR

[quiet cargo]

wait whats that command?

$$ \dd x $$

[warm current]

An easy dx

[quiet cargo]

i didnt know thanks

[jolly parrot]

hades

[cobalt hinge]

yeah i know

#latex-testing or #latex-help

is suitable for this

[quiet cargo]

lmfao

just continue plz

@cobalt hinge

[cobalt hinge]

well i was asking you a question

[warm current]

You don't always need double dólares btw single dollars for inline like $x\ne3$ and whatever

[jolly parrot]

SWR

[quiet cargo]

yo what

how did u do it without the $

[warm current]

My keyboard went Spanish for a second...

[warm current]

I was saying without $$

[quiet cargo]

how did u do it without that

[warm current]

I didn't

[quiet cargo]

can u solve my didnt

doubt

extreme last part

<@&286206848099549185>

[cobalt hinge]

@quiet cargo

[quiet cargo]

yeah then...

i did 1 - when all 3 like the same

but got the wrong answer

[cobalt hinge]

4C3=4 possibilties

[quiet cargo]

wdym?

i know PnC

[cobalt hinge]

4 possiblities for different combinations

[quiet cargo]

but what do u mean by that

why pnc here

[warm current]

This doesn't exclude if two boys like the same

[somber epoch]

How old is this guy

@quiet cargo

[quiet cargo]

oh yeah i forgot about that ty

[quiet cargo]

old enough...

[somber epoch]

Just wondering no hate

❤️

[quiet cargo]

come math battle if u want

@somber epoch

[somber epoch]

What grade are you in

[quiet cargo]

wdyt

[somber epoch]

Where are you from then 😭

Idk what your system is

Year or grade idk

[somber epoch]

Sure ig

[quiet cargo]

bro

[somber epoch]

Crazy

[quiet cargo]

wydt = what do you think

[somber epoch]

Oh

[quiet cargo]

smh

[somber epoch]

I thought you misspelled wdym

[quiet cargo]

no

[somber epoch]

Fr

[quiet cargo]

wdyt

[somber epoch]

Uh

Maybe you are grade 8

Idk

[quiet cargo]

no

10th

[somber epoch]

Cold

[quiet cargo]

i was j conf

come math battle

[cobalt hinge]

???

[quiet cargo]

which grade r u in?

[cobalt hinge]

can we stay on topic, please?

[somber epoch]

10th too

[somber epoch]

Let me answer them all rq

[quiet cargo]

i got the q

[cobalt hinge]

dont.

[quiet cargo]

board or smtg?

come dms

[cobalt hinge]

i swear, just take this to #math-discussion or ANYTHING except here

[quiet cargo]

.close

[pearl pond]

Channel closed

Closed by @quiet cargo

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[cobalt hinge]

you're kidding.

[quiet cargo]

BRO NOBODY CARES

[somber epoch]

Real

[quiet cargo]

continue it here

oesnt matter

[quiet cargo]

whats ur board

[cobalt hinge]

nice thanks for wasting my precious time

[quiet cargo]

do uk calculus?

@somber epoch

[somber epoch]

Idk let's see

Dms

[versed mica]

#help-20 message

[cobalt hinge]

you were the one who was asking for help.

[quiet cargo]

holy fuck 💀

[versed mica]

yep lol

[pearl pond]

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[brazen vector]

given the sequence ${u_n}$ defined as follows:\ $u_0=\frac 12$ and $u_{n+1}=1+\frac 1{u_n}$\ Study the convergence of ${u_n}$

[warm current]

!status

[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

[brazen vector]

1

[warm current]

Can you bound the sequence?

[rough forge]

do you know the monotony criterion?

[brazen vector]

monotone convergence theorem? yes

[brazen vector]

it is bounded below by u_0

u_n>0 for all n in N

and u_{n+1}=1+1/u_n>1>1/2=u_0

[brazen vector]

i am not sure how i can show this tbh

[wraith river]

induction

[brazen vector]

ah yes

[wraith river]

if u_n > 0 then obviously 1 + 1/u_n > 0

[brazen vector]

i am starting to forget about induction since i havent used this in a while

[wraith river]

that being said, you can just consider the limit case directly

[brazen vector]

ok so this much is pretty straighforward

[brazen vector]

wdym by that

[wraith river]

if there is a limit u, then u = 1 + 1/u

if there is a solution, you have the limit value

if there isn't it diverges

[brazen vector]

hmmm

but this doesnt serve as a proof does it

[warm current]

But you need to prove there is a limit first

[brazen vector]

so how do i proceed now

[warm current]

bound your sequence above

and consider monotonicity

[brazen vector]

the sequence isnt monotonic

[warm current]

oh hmm

okie gimme a minute

[brazen vector]

take your time

btw in class there was a case like this so we were given a recurrent sequence and then the professor considered the subsequences consisting of odd numbered terms and even numbered terms separately

[warm current]

yeah the subsequences are monotone

[brazen vector]

but whats the idea behind this

[wraith river]

I think a simple ratio test works here

[warm current]

but that doesn't help you (I don't think) for the entire sequence

[brazen vector]

ie the idea behind considering these 2 subsequences

[unborn abyss]

do they each converge to the same number?

[warm current]

yes

[unborn abyss]

seems good to me

[brazen vector]

yes i mean if you want to work with subsequences then you should prove that all subsequences converge to the same limit right

[warm current]

oh. yeah. That would still imply that every subsequence converges to the same number

[brazen vector]

oh wait

each subsequence consists of a number of odd and even numbered terms

[warm current]

If the even and odd subsequences converge to the same number, then all of their sub-subsequence will converge to that number too. But even/odd partitions your sequence

[unborn abyss]

the important part is that it covers the whole thing

[brazen vector]

is this the idea behind considering these 2 subsequences in such cases ?

[warm current]

yeah true. "covers" is a better word than "partitions"

[unborn abyss]

go back to the original defn of limit of sequence to prove it

[brazen vector]

because they cover the entire sequence ?

[unborn abyss]

and there's a finite number of them, that's important as well

[brazen vector]

you asked me for an upper bound before

before doing that

[warm current]

that was my bad

I thought the sequence was monotone

[brazen vector]

i prove that the subsequence of odd numbered terms is decreasing

and that of the even numbers is increasing

since that of the odd numbered terms decreasing then u_1 serves as an upper bound of {u_{2n+1}}

oh but that doesnt mean it bound {u_n} for any n in N

but thats not necessary

oh wait it is

[warm current]

You don't need this necessarily

[brazen vector]

why not

i need an upper bound for {u_{2n}} to converge

[warm current]

I mconfused you a bit when I asked you to bound your sequence earlier. I mistakenly believe your original sequence was monotone

[warm current]

Yes. This you will need. And you will need a lower bound for your odd subsequence

[brazen vector]

for the odd subsequence it is u_0 because u_0 is a lower bound for the whole sequence {u_n}

but what about the upper bound of {u_{2n}}

[warm current]

It is, but I do not see how you figured that

[warm current]

Maybe you could apply similar logic for u_1 that you did for u_0 just above

[brazen vector]

u_n>0 by induction so 1/u_n>0 and then 1+1/u_n>1>u_0

which means that u_{n+1}>u_0 for all n in N

and so u_n>u_0 for all n in N

[warm current]

looks good

technically, $u_n\ge u_0$

[jolly parrot]

SWR

[brazen vector]

yes since n=0 is possible

[warm current]

Okay, try the same idea with $u_1$ now

[jolly parrot]

SWR

[warm current]

That would be the first thing I do, anyway

[brazen vector]

but the problem of the upper bound is the only thing bothering me

[brazen vector]

sure let me try that

i was planning to do this

[brazen vector]

after you said this

[warm current]

it may not be as simple, but it is certainly true

[brazen vector]

oh wait . For all n in N, u_n>= u_0=1/2_ so u_{n+1}=1+1/u_n=<1+1/u_0=3

since u_{n+1}=< 3 for all n in N then u_n=< 3 for all n in N

this proves it

without using induction

it is bad tex , how do can i fix the problem of _ making the text italic

lol i tried my best to fix it

but it is still not totally fixed

[brazen vector]

so first of all is this proof correct and valid

[warm current]

$\forall(n\in\bN)\left(u_n\ge u_0=\frac12\to u_{n+1}=1+\frac1{u_n}\le1+\frac1{u_0}=3\right)$

[jolly parrot]

SWR

[warm current]

Yeah. That looks good

[brazen vector]

this proves that both of these subsequences converge

but doesnt prove that they have the same limit

how do i do that

[warm current]

what are your even and odd subsequences?

[brazen vector]

$u_{2n+2}=\frac {2u_{2n}+1}{u_{2n}+1}$ and $u_{2n+1}=\frac {2u_{2n-1}+1}{1+u_{2n-1}}$

[warm current]

wow really? Those don't seem right

[brazen vector]

ah wait

[brazen vector]

you can also rewrite them as $u_{2n+2}=2-\frac 1{u_{2n}+1}$ and $u_{2n+1}=2-\frac 1{u_{2n-1}+1}$

[warm current]

you can?

how did you change the denominator?

[brazen vector]

mb i meant to write it like this

[warm current]

ah

I see I see

The easy way is to assume $u_{2n+2}=u_{2n}$ for $n\to\infty$

[jolly parrot]

SWR

[warm current]

The way to do that with rigor is

For any $\varepsilon>0$, there is some $N\in\bN$ such that $\abs{u_{2n+2}-u_{2n}}<\varepsilon$ for all $n>N$. Then you have some inequality equations to play with

[jolly parrot]

SWR

[brazen vector]

i thought about using triangular inequality

by adding and subtracting u_{2n+1}

oh no wait

what i tried was something like this

[warm current]

Well for your even subsequence, it is monotonic increasing, so $u_{2n+2}>u_{2n}$, or $u_{2n+2}-u_{2n}>0$. This can help you with your absolute value

[jolly parrot]

SWR

[brazen vector]

assume that ${u_{2n}}$ converges to $l$ then $\forall\varepsilon >0\ \exists\delta >0\ \exists n_0\in\mathbb{N}\ \forall\in\mathbb{N}$ such that $2n\geq n_0\implies |u_{2n}-l|<\varepsilon$ and also $|u_{2n}-l|\leq |u_{2n}-u_{2n-1}|+|u_{2n-1}-l|$

[brazen vector]

but what can i do from this

[brazen vector]

i dont see what i can do this either of these

[warm current]

okay yeah lemme help you out

We are assuming $\abs{u_{2n+2}-u_{2n}}<\varepsilon$ for sufficiently large $n$, yeah?

[jolly parrot]

SWR

[brazen vector]

yes after all this subsequence is cauchy

[warm current]

no I think you are misunderstanding

[brazen vector]

why no

isnt this subsequence convergent ?

[warm current]

yes, but not because it is cauchy

[brazen vector]

no

[warm current]

Technically, it is cauchy, but that's not how we showed that the subsequence is convergent

[brazen vector]

what i meant is that this is true since the sequence is cauchy

i didnt say that it is convergent since it is cauchy

[warm current]

oh right

yeah

[brazen vector]

what i meant was that since it is convergent then it is cauchy so this is true

[warm current]

The subsequence is monotonic increasing, so it is convergent, so it is cauchy

that's the whole picture

anyway...

[warm current]

We have this, so now we move on

the even subsequence is monotonic increasing, so $\abs{u_{2n+2}-u_{2n}}=u_{2n+2}-u_{2n}$, yeah?

[jolly parrot]

SWR

[brazen vector]

sure

[warm current]

So, $u_{2n+2}-u_{2n}<\varepsilon$ right?

[jolly parrot]

SWR

[brazen vector]

yes

oh wait

are you planning to write u_{2n+2} in terms of u_{2n+1}?

[warm current]

nope

[brazen vector]

ohh

[warm current]

We're gonna use this

[warm current]

We modify this to get $u_{2n+2}<u_{2n}+\varepsilon$. So, using your first subsequence, we get
$$2-\frac{1}{u_{2n}+1}<u_{2n}+\varepsilon$$

[jolly parrot]

SWR

[warm current]

Now, we have an inequality where we can solve for $u_{2n}$

[jolly parrot]

SWR

[brazen vector]

$u_{2n}^2+(\varepsilon-1)u_{2n}+\varepsilon-1>0$

[warm current]

looks good

[brazen vector]

so now do i find the roots in terms of ε and check the sign of this ?

[warm current]

yep

[brazen vector]

$u_{2n}=\frac{1-\varepsilon\pm\sqrt{(\varepsilon-1)(\varepsilon-5)}}2$

[warm current]

Inequality, not equal

[brazen vector]

u_{2n}>0 between these 2 values, = 0 at these these values and < 0 otherwise

[warm current]

Now, rule out one of the cases

[brazen vector]

u_{2n}< 0 is not possible as proven previously so_ u_{2n} must be in the closed interval with the boundaries of the intevral being the 2 values here

ah wait a second

[brazen vector]

isnt this here false

[brazen vector]

instead of u_{2n} the lhs of this should be written

[warm current]

wdym?

[warm current]

Yeah I didn't know what you were trying to get at here

basically, you should have this:
$$u_{2n}\in\left(\frac{1-\varepsilon-\sqrt{(\varepsilon-1)(\varepsilon-5)}}{2}, \frac{1-\varepsilon+\sqrt{(\varepsilon-1)(\varepsilon-5)}}{2}\right)$$

[jolly parrot]

SWR

[brazen vector]

so $u_{2n}^2+(\varepsilon-1)u_{2n}+\varepsilon-1>0$ when $u_{2n}$ is between these values, $= 0$ if $u_{2n}=$ one of these values and $< 0$ otherwise

[brazen vector]

in other words this is the solution of the inequality

[warm current]

yes

And for any epsilon, there are sufficiently large n to satisfy it

[brazen vector]

how is that

[warm current]

Because we started with the fact the subsequence converged
#help-39 message

[warm current]

And that means this must be true now

[brazen vector]

yes but this doesnt assume u_{2n}>0

we should look for this too no ?

[brazen vector]

for sufficiently large ε, at least a part of these interval will contain negative values

if we cant ensure that there will always be positive values in this interval

then we cant proceed

[warm current]

$$u_{2n}\in\left(\max\left(0, \frac{1-\varepsilon-\sqrt{(\varepsilon-1)(\varepsilon-5)}}{2}\right), \max\left(0, \frac{1-\varepsilon+\sqrt{(\varepsilon-1)(\varepsilon-5)}}{2}\right)\right)$$

[jolly parrot]

SWR

[violet bone]

how do you learn to write in that text

[warm current]

#latex-help and #latex-testing

[violet bone]

thanks

[warm current]

for sufficiently small epsilon, the inequality will be all positive values

[warm current]

putting you back to this

[brazen vector]

sure but what about the sufficiently large epsilon

[warm current]

then this is still valid

but we dont care about large epsilon. We want to find the limit, so we want small epsilon

[brazen vector]

if both ends here are < 0

[warm current]

it's not possible

[brazen vector]

then u_{2n}∈(0,0) so there isnt u_{2n}

[warm current]

right side is always positive. I just did max because I copy-pasted

[brazen vector]

how are you sure about that

thats my question

[warm current]

okay, let's get pathological here and I'll show you

[brazen vector]

this is > 0 only when epsilon < 1

[warm current]

,w plot y=sqrt((x-1)(x-5)

[jolly parrot]

Results provided by WolframAlpha

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

Requested by spelwerdzrite

[warm current]

it looks like x=3 will be imaginary

So, you might think that choosing $\varepsilon=3$ would be impossible

[jolly parrot]

SWR

[warm current]

but, remember, we want to satisfy $\abs{u_{2n+2}-u_{2n}}<\varepsilon$. If we satisfy $\abs{u_{2n+2}-u_{2n}}<\frac12$, then our $\varepsilon$ inequality will still be satisfied. For large $\varepsilon$, just bound it to satisfy anything smaller.

[jolly parrot]

SWR

[brazen vector]

this doesnt hold for epsilon > 1

[warm current]

read what I wrote here

You can do this:

To satisfy $\abs{u_{2n+2}-u_{2n}}<\varepsilon$, satisfy $\abs{u_{2n+2}-u_{2n}}<\min(\varepsilon, 1)$ instead.

[jolly parrot]

SWR

[brazen vector]

i didnt get the last sentence about large epsilon

[warm current]

basically this 👆

[brazen vector]

sure

but isnt the first inequality already satisfied

[warm current]

I think we're going in circles right now

[brazen vector]

probably

[warm current]

walk me through what you know right now and where you confused

[brazen vector]

sure

so first of all, ${u_{2n}}$ is convergent so it is cauchy and thus $\forall\varepsilon >0, |u_{2n+2}-u_{2n}|<\varepsilon$ for sufficiently large $n$. from here you can rewrite the inequality as $u_{2n}^2+(\varepsilon-1)u_{2n}+\varepsilon-1>0$ after using $u_{2n+2}=2-\frac 1{u_{2n}+1}$

[brazen vector]

now solving the inequality gives us $u_{2n}\in\left(\frac{1-\varepsilon-\sqrt{(\varepsilon-1)(\varepsilon-5)}}{2}, \frac{1-\varepsilon+\sqrt{(\varepsilon-1)(\varepsilon-5)}}{2}\right)$ must be satisfied in order to satisfy the inequality

[brazen vector]

now here comes my problem

my problem is that for ε>1 , none of the boundaries here is > 0

so for large enough ε, there will be no u_{2n} that satisfies this

so how can we deal with this

[warm current]

this is correct

[brazen vector]

also if i am wrong somewhere then where did i go wrong

[warm current]

one moment

[brazen vector]

sure take your time

[warm current]

Sorry, gotta do a work thing. I'll be a few minutes

[brazen vector]

no problem at all , when you come back please ping me

[warm current]

@brazen vector this inequality is backwards

[brazen vector]

wdym by that

[warm current]

<, not >

[brazen vector]

how did you get that

[warm current]

How did you get this?

[brazen vector]

$2-\frac 1{u_{2n}+1}<u_{2n}+\varepsilon\implies\frac 1{u_{2n}+1}>2-u_{2n}-\varepsilon\implies (u_{2n}+1)(2-u_{2n}-\varepsilon)-1>0\implies u_{2n}^2+(\varepsilon-1)u_{2n}+\varepsilon-1<0$

so yes you are right

mb

[warm current]

nw

[brazen vector]

in other words we need $u_{2n}\in (\frac{1-\varepsilon+\sqrt{(\varepsilon-1)(\varepsilon-5)}}{2},\infty)$

[brazen vector]

we can have u_{2n} in the interval (-∞,r_1) where r_1 is the other root from above

but we need to be careful to choose only the positive values of u_{2n} from there

if they exist

so what we need is $u_{2n}\in(-\infty,\frac{1-\varepsilon-\sqrt{(\varepsilon-1)(\varepsilon-5)}}{2})\cup (\frac{1-\varepsilon+\sqrt{(\varepsilon-1)(\varepsilon-5)}}{2},\infty)$ with $u_{2n}>0$

[brazen vector]

now what

what does that say about {u_{2n+1}} converging to the same limit

as {u_{2n}}

[warm current]

Good question. My idea was to take limit $\varepsilon\to 0$, but the interval changed, which makes things awkward

[jolly parrot]

SWR

[warm current]

you can rule one of these cases out since $u_{2n}>\frac12$ for every $n\in\bN$

[jolly parrot]

SWR

[brazen vector]

ohh yes

[brazen vector]

this remains

maybe some elements of this can also be cut off depending on epsilon but in general it can remain like this

[brazen vector]

so your idea now doesnt work anymore ?

[warm current]

I'm trying to figure out

[brazen vector]

i will try to do something too

[warm current]

I gotta do some more work

I'm still looking at this, but may be absent for a bit

[brazen vector]

np you can try it whenever you want

[pearl pond]

@brazen vector Has your question been resolved?

[violet bone]

dang ur lucky ur getting help

[pearl pond]

@brazen vector Has your question been resolved?

[violet bone]

i’m curious what math class is this?

[brazen vector]

this is real analysis

[violet bone]

ight

[pearl pond]

@brazen vector Has your question been resolved?

[brazen vector]

how did you get this closed form

[hot canyon]

Wait t.t

[hot canyon]

Can solve for $u_n = \frac{a_n + 2a_{n-1}}{a_{n-1} + 2a_{n - 2}}$, and prove convergence there.

[jolly parrot]

Arya

[brazen vector]

what is a_n

[hot canyon]

Fibonacci

[brazen vector]

and what do you mean by solve for

[brazen vector]

why does this work

[hot canyon]

Basically, by observation and if rigor is required, you can prove this via induction

[brazen vector]

i am not allow to rely on something like the convergence of the fibonacci sequence but i would like to see this way too if it works

but before that SWR and I were near the end of a proof

we only had one thing left

which is to prove that the subsequences of the odd and even numbered terms converge to the same limit

[hot canyon]

Once you know $a_n$'s are Fibonacci, you can manipulate your way to $u_n = \frac{L_n+2}{2L_n - 1}$ where $L_n = \frac{a_n}{a_{n-1}}$

[jolly parrot]

Arya

[hot canyon]

And it's easy to show $\lim_{n \to \infty} \frac{a_n}{a_{n-1}} = \phi$

[jolly parrot]

Arya

[hot canyon]

If you want you can try this way too, it's a lil rigorous but doable

[brazen vector]

i want it to be rigorous xD

[brazen vector]

how to do that

i am not so sure about it tbh

on how to proceed here i mean

[hot canyon]

Assume $L = \lim_{n\to \infty} \frac{a_n}{a_{n-1}} = 1 + \frac{a_{n-1}}{a_{n-2}} = 1+\frac{1}{L} \$ n solve

[jolly parrot]

Arya

[brazen vector]

but i dont even know if a_n/a_{n-1} converges or no

so am i assuming that the ratio of the consecuive terms of the fibonacci sequence is convergent ?

[hot canyon]

To show that this is indeed true, you can utilise this:

[brazen vector]

hmmm

i tried to prove the first one using induction

but i wasnt reaching anywhere

i will try again in a moment

but you see this gave me an idea for the place i was stuck in before

nvm this doesnt work

[pearl pond]

@brazen vector Has your question been resolved?

[naive hemlock]

@brazen vector Do you know about tests for convergence or Divergence?

[pearl pond]

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[pearl pond]

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[prime bramble]

[cobalt hinge]

[brave sluice]

donnie was here

[cobalt hinge]

well they aren't here anymore

hence was

[pearl pond]

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[silent marsh]

how they deduced that its gonna be 1/3^2 and so on? shouldnt it be 1/3f(2^nx)

[cunning comet]

use f(x) = (1/3) f(2x) iteratively, this means
first step f(x) = (1/3) f(2x), then use this formula in the same way for 2x and plug this in.

[pearl pond]

@silent marsh Has your question been resolved?

[pearl pond]

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[sharp smelt]

Trying to find the partial derivatives here, would $\pdv{z}{x}$ just be f'(x)g(y)$?

[jolly parrot]

What a wonderful world!
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[spare lark]

Yes

[pearl pond]

@sharp smelt Has your question been resolved?

[sharp smelt]

thanks

[pearl pond]

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[sharp smelt]

Trying to find $\pdv{z}{s}$
\
$ e^{st} (t+ \frac{dt}{dx})cos(\sqrt{s^2+t^2})- e^{st} sin(\sqrt{s^2+t^2})\frac{2s+ 2t \frac{dt}{ds}}{2\sqrt{s^2+t^2}}$

[sharp smelt]

Is this fine

[jolly parrot]

What a wonderful world!

[lunar knot]

Why do you have dt/dx terms

[sharp smelt]

oops

yeah

keep getting confused

Trying to find $\pdv{z}{s}$
\
$ e^{st} (t)cos(\sqrt{s^2+t^2})- e^{st} sin(\sqrt{s^2+t^2})\frac{2s}{2\sqrt{s^2+t^2}}$

[jolly parrot]

What a wonderful world!

[lunar knot]

Yes I have the same

One would refactor it

Consider anything except "s" to be a constant

[sharp smelt]

hmm?

$e^{st} (cos(\sqrt{s^2+t^2}) - sin(\sqrt{s^2+t^2})\frac{s}{\sqrt{s^2+t^2}})$

[jolly parrot]

What a wonderful world!

[lunar knot]

yes, you can even write it with theta then

[sharp smelt]

Cool

got it

thanks

$e^r(t cos(\theta)-e^r sin(\theta) \frac{s}{\theta}$

[jolly parrot]

What a wonderful world!

[lunar knot]

Depending on the next step of the exercise, if you study it for theta ->0, you may use sin(x) ~ x in 0

[sharp smelt]

The question was just find the derivative

thanks

.close

[pearl pond]

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[pearl pond]

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[gleaming moss]

!show

[pearl pond]

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

[gusty moss]

i did

x^3 + 1 + mx + m >= 0

thats it

idk where to continue

[light helm]

try factor by grouping

[gusty moss]

okay

[gusty moss]

grouping which

the x or the m

[light helm]

first two and last two

[gusty moss]

so x(x^2+1) + (m+1) >= 0

[light helm]

no

[gusty moss]

?

oh

u mean the x-1

factor

[light helm]

no

there is no x-1 factor here

[gusty moss]

now its case work, right?

i. both are negative

ii. both are positive

iii. x^2 -x + 1 + m = 0

?

or is it the discriminant is smaller than 0

smaller than or equal to

[light helm]

go back

what did you get after just factorising

[gusty moss]

(x+1)(x^2-x+1+m) >= 0

for this to be true

either x>=-1

wait no

case 1 is x = -1 which gives 0

case 2 is when the quadratic = 0

case 3 is when both functions are negative

and case 4 is when both functoins are positive

[light helm]

now try focus on analysing the sign of

x^2 -x + 1 + m

[gusty moss]

the discriminant has the condition that m >= -3/4 for one solution or none

but since the problem gives that m > 0

it means that this quadratic is always positive (no solution)

therefore the only condition that matters is x >= -1

[pearl pond]

@gusty moss Has your question been resolved?

[gusty moss]

why the hell is my name michael

[iron basin]

wasn't it already that

anyways

you now got the factorization

which says that the product of x+1 and x^2-x+1+m is positive

what can you say in general about two numbers if they have a positive product?

[pearl pond]

@gusty moss Has your question been resolved?

[gusty moss]

Greater than 0

[gusty moss]

I can’t even change it back

[sacred cypress]

i agree

[gusty moss]

okay so this line of thinking is correct?

[sacred cypress]

idk but is still having the same answer as you

i*

[iron basin]

what is greater than zero

[gusty moss]

the product

[iron basin]

i know

im asking about the numbers

their signs

[gusty moss]

?

both of them are + is

if x >= -1

[iron basin]

and another case

both of them are negative

[gusty moss]

x+1 is negative if x < -1

[gusty moss]

no the quadratic cant be negative

because m>0

thats the given condition

[iron basin]

yeah but in general

[gusty moss]

so the only way for their product to be greater or

oh

[iron basin]

oh nvm

[gusty moss]

then both are negative and you get a positive product too

[iron basin]

you said that alredy

[gusty moss]

so

my logic is correct?

[iron basin]

i guess so

then x >= -1

[gusty moss]

okay

thanks

.close

[pearl pond]

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[next ore]

a letter is chosen form the word sleeplessness
a. find no. of ways it can be done
b. find no. of ways it can be a vowel
c. find probability of it being a vowel

[next ore]

am i correct if my answer is
a. 5
b. 1
c. 4/13

[vestal tapir]

that makes sense

[next ore]

okay cool

i have another one

[urban briar]

simple

use nCr

I believe

so...

[next ore]

3 identical dice are roled simultaneously, find the probability for all 3 to show the same number

will the answer be 1/36 ?

cuz identical doesnt really matter in probability

[urban briar]

wait mb

[next ore]

for a. and b. part its basic combinatorics and for c. we gotta treat the like alphabets as distinct

[vestal tapir]

yeah this strains the metaphor too much

[next ore]

??

wdym

[urban briar]

ok

[vestal tapir]

i mean you're right, by the same logic as before the probability is not equal to number of ways divided

[next ore]

yeah

[next ore]

im asking cuz someone told me the answer to this is 3/28 which i believe is wrong

[pulsar schooner]

You’re hot today.

[pulsar schooner]

It’s 1/36

[next ore]

cool i get it

my friend confused me between combinatorics and probability

[iron stream]

Which one ?

[jolly parrot]

Steinar

[pulsar schooner]

Dice one

[iron stream]

Probably wrong answer for c

[next ore]

** here they treated the dice alike as they would do in combinatorics

[next ore]

dawg 😭

[next ore]

^2 ?

[pulsar schooner]

6 vowels
26 letters in totals

So

6/26

3/13

[vestal tapir]

in the first question it could be like one of three things, they all make sense, but dice are physicl things you throw, it's much harder to ignore the right answer

[iron stream]

Your answers seem correct!

[pulsar schooner]

C is incorrect @next ore

[next ore]

okayy

[next ore]

theres another quesiton i asked after that

[next ore]

so 1/36 is wrong for the 2nd one?

[vestal tapir]

no 1/36 is the only right answer

[pulsar schooner]

^^^^

[vestal tapir]

it's less ambiguous

[pulsar schooner]

!donw

!done

[pearl pond]

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

[pulsar schooner]

@iron stream

[next ore]

okay cool thanks

.close

[pearl pond]

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[pearl pond]

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[radiant whale]

what can I do here with u-sub

[radiant whale]

the only hint I got was “this one will require properties of exponents”

[unborn abyss]

pick some part of it to be u

[radiant whale]

wow 😔

do u think this looks right

[unborn abyss]

no

[radiant whale]

oh uh

[unborn abyss]

$\frac{2x+1}{2x} \neq 1$

[jolly parrot]

hayley (immeasurable)

[radiant whale]

it equals 2 doesn’t it

apart from that

is it fine…

[unborn abyss]

no

[radiant whale]

…

what is it then

[brave sluice]

combine the exponentials

then you can do the u-sub

[radiant whale]

as in make it e^2x^3

[brave sluice]

huh?

[radiant whale]

oh wait

idk

im just confusing myself atp

[brave sluice]

$e^{x^2}e^x e^2=?$

[jolly parrot]

Axe

[brave sluice]

these all have the same base so you can add the exponents

[radiant whale]

e^x^2 + x + 2?

oh wait yes it is

omg

I see it now

I see the vision

[brave sluice]

yes but please use parentheses 🙏

oh good

👍

[radiant whale]

mb mb

[brave sluice]

the answer looks right

[radiant whale]

thank you 🙏🏻

.close

[pearl pond]

Channel closed

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[pearl pond]

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[craggy crypt]

Is this right or

Is it not mathematically correct

[cobalt hinge]

is $y=6^{\log_5(t)}$?

[jolly parrot]

;(

[craggy crypt]

Yea

[cobalt hinge]

so $\ln(y)=\ln(6^{\log_5(t)})=\overbrace{\log_5(t)\ln(6)}^{\ln(a^b)=b\ln(a)}$

[jolly parrot]

;(

[craggy crypt]

Yes

[cobalt hinge]

oh you just wrote it weirldy

i thought i was tripping

[craggy crypt]

Sorry 😭

[old marsh]

ur answers good ali

[cobalt hinge]

it looks like the log_5 is in the exponent

[craggy crypt]

Yea I converted to ln

[cobalt hinge]

i approve

👍

[craggy crypt]

Okay sweeet

.close

[pearl pond]

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[cosmic reef]

hi

[pearl pond]

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[open rivet]

!status

[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

[true basin]

2

[open rivet]

!show

[pearl pond]

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

[true basin]

The ‘z’ looks like 2 sorry about that

[open rivet]

,rccw

[open rivet]

which question is this?

[true basin]

9

[open rivet]

9?

ok

[open rivet]

thats why its wrong

you made a mistake seeing your own z as a 2

[true basin]

Where

[open rivet]

actually, i think im confusing myself xd

nvm

[true basin]

Oh lol

[gleaming moss]

I believe the signs are wrong here

-(12a^2 - 3z^2) then distribute the -1

And please put a line through your zs

[true basin]

Is this right now?

[gleaming moss]

Think so

[true basin]

Alright thanks

.close

[pearl pond]

Channel closed

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[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[hot canyon]

!status

[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

[hot canyon]

!show

[pearl pond]

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

[hot canyon]

Hint: b = ak => the a-by-b piece is basically a piece with length a, breadth ak

[light orbit]

I know how to solve the second question(by partitioning each piece into a b-minoes and then by applying theorem that m-by-n board has perfect cover with b-nominoes iff b is a factor of m or n) but I got stuck with the first one

[hot canyon]

Wow, first was the easier one actually

[light orbit]

Is this an assumption that b is equal to ak or what?

if so, then what if a is not equal to ak

[hot canyon]

It's not assumption. It's given.

The question has given you the fact that a | b, or you can write b as ak

[light orbit]

Oh no

I havent read the question properly

[hot canyon]

Can you do it now?

[light orbit]

I think that we can think of each piece as a b/k by 1 piece

[hot canyon]

Assume the first row of the partition is covered by a-by-b pieces in such a way that x pieces are stacked along the length and y along the breadth

Then, xa + yb = m

Implies a|m

[light orbit]

Of what partition?

[hot canyon]

Nevermind the partition, visualize that if such a cover were to exist, the above would hold true

[light orbit]

Sorry, but I can't get it

[hot canyon]

Hmm.. partition into rows n columns of size a

[light orbit]

Yeah, that what I meant by saying that we can consider a by b pieces as 1 by a/k pieces

[hot canyon]

✓

[light orbit]

Thank you very much

You've been a big help

.close

[pearl pond]

Channel closed

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[pearl pond]

Available help channel!

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[chrome river]

Hi! I have a question regarding associative $\mathbb{R}$-algebras.
Suppose $\mathscr{A}$ is a 2-dimensional unital $\mathbb{R}$-algebra with basis $\mathcal{B} = {1, \beta}$ i.e. $$z \in \mathscr{A} \iff z = x+y \beta, \quad x,y \in \mathbb{R}$$ and that $\beta^2 \in \mathbb{R}$. \
I dont understand why when $\beta^2 < 0$, $\mathscr{A}$ is equal to the complex numbers. Wouldnt $\mathscr{A} = \mathbb{C}$ specifically when $\beta^2 = -1$?

[jolly parrot]

woomy

[tropic saddle]

if say beta=2i, then A is C but the "coordinates" are different

[chrome river]

so its just a translation of C?

Like C but everything is moved up by 2 i guess

[tropic saddle]

scaled by 2

[chrome river]

oh yea

but only vertically

[tropic saddle]

yes

[chrome river]

im still having a hard time understanding as to why beta must be a translation of i

[tropic saddle]

beta^2 is real and negative

so beta^2 = x*(-1) for some positive real x

so beta = sqrt(x)i

[chrome river]

sorry im back

yea makes sense

thanks!

if beta is positive then

A = |R^2?

cause then beta would just be equal to sqrt(x)*1

[pearl pond]

@chrome river Has your question been resolved?

Channel closed

Closed by @chrome river

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[tropic saddle]

no

[chrome river]

.reopen

[pearl pond]

✅

[tropic saddle]

what ring structure are you trying to put on R^2

the problem is that beta is explicitly not supposed to be in R

so sqrt(x)*1 is not a unique thing

its both beta and the real number sqrt(x)

there is another new solution to z^2=x

sqrt(x), -sqrt(x) and now also beta

[chrome river]

oh yea

wait so it isnt necessarily R^2 since beta doesnt have to be in R

beta^2 has to be

[tropic saddle]

but what ring do you even mean by R^2

componentwise multiplication?

[chrome river]

R x R

wait sorry my brain is melting im quite tired

[tropic saddle]

there is an easier description in terms of a factor ring of R[x]

[burnt ibex]

yo

[tropic saddle]

hmm I suppose that is indeed isomorphic to R^2

[burnt ibex]

dividing polynomials is basicly divide multiply subtract