help-36

[faint edge]

You've got everything other than the antipode part

I guess a way to think of it is that you can imagine I=[0,1] wrapping around the equator of a sphere (or the earth). You just proved you can find a point where g(c)=0. This is equivalent to saying you can find a point where f(c)=f(c+1/2)

If you are at a pt c on the equator of a sphere and you walk half way around the equator you end up at the antipode of a c.

If f is measuring temperature this shows the temperature at c and at the antipode of c are the same.

[ripe jewel]

you can also consider the interval [0,2pi] if that helps

[ionic gate]

What was the formal proof for the question?

[final saddle]

@ionic gate Has your question been resolved?

[strong hill]

hi is it okay if i’m in middle school i’m barely gonna start 8th grade and i’m really struggling in 7th grade math. i got masters on my texas starr test for reading but did not meets on my math. any suggestions?

[craggy plank]

Claim your own channel

#❓how-to-get-help

[fierce finch]

Let I=[0,1] and imagine I is a circle (the equator), 0 represents a starting point (e.g. Greenwich) so 1/2 represents the diametrically opposite point and 1 brings you back to point 0 so when u are on 0 or 1 it’s like being in Greenwich so if f is the temperature u have f(0)=f(1) and the temperature is also continuous .

Two points are antipodal if they are opposite each other on the circle, i.e. separated by half a turn, or an arc of 1/2 so if i take a point c between 0 and 1/2 the opposite is c+1/2

I'll let you draw your own conclusions based on what we've proven.

[final saddle]

@ionic gate Has your question been resolved?

[ionic gate]

I meant for the first part.

[fierce finch]

g is continious on [0,1/2] bcs f is continuous on [0,1] in addition to that g(0)=f(0)-f(1/2) and g(1/2)=f(1/2)-f(1)=f(1/2)-f(0)=-g(0)

let's not live too much, we have two things to do here

First case : if f(1/2)=f(0) so c exists and it is even c=0

[faint edge]

Writing out the proof for them seems like doing too much.

[fierce finch]

Ok ill delete

[faint edge]

Summarizing might be a good middle ground?

[fierce finch]

I'll let you think about the second case maybe

[faint edge]

That seems fair.

[final saddle]

@ionic gate Has your question been resolved?

[final saddle]

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[still drum]

honestly been thinking abt this question for a day-ish, don't really know where to start? could i get a hint as to what i can do to make progress?

[tired walrus]

hard to give a hint without spoiling it

look for hidden assumptions made in the argument

[still drum]

thats fair

ok

i guess an obvious question is

can $x ~{} y $ even be true if $x \not ~{} x$

[soft zealot]

lifelong dumbass
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[still drum]

ok screw latex

uhhh

[tired walrus]

\sim

that's how you get the tilde

[onyx peak]

Maybe a better question would start with "must"

[still drum]

thank you

[still drum]

ok

im going to try sth

[still drum]

ok i have a question

the above logic assumes that x, y are in the same equivalence class

but what if the equivalence class just contains one element (ie x)

in which case y doesnt exist

and thus the symmetric and transitive properties cant be applied

[tired walrus]

i mean nobody said y has to be distinct from x

but you're kinda on the right track

namely that questioning whether the things you talk about exist is a good habit in general and also for this problem

[still drum]

uhh ok

[still drum]

well if y and x were the same then you've just assumed reflexive property already no?

[onyx peak]

What even allows you to talk about equivalence classes if its not an equivalence relation

[still drum]

isnt the point to assume its an equivalence relation and then show a contradiction?

[onyx peak]

I'd say that the point is finding a relation which is symmetric, transitive but not reflexive (i.e. not an equivalence relation)

showing that reflexivity isnt redundant

[still drum]

oh so thats what they mean by counter example

[still drum]

i guess an obvious example is x < y

[onyx peak]

Not symmetric

[still drum]

uhhh

oh ok

[onyx peak]

1 < 2 but not 2 < 1

[still drum]

so this was a typo?

[onyx peak]

no, it wasnt

im saying that x < y isnt symmetric

[still drum]

sorry im confused i thought you were asking me to find a relation that was

• symmetric, transitive, not reflexive

[onyx peak]

so how does it help you in finding a relation which is symmetric, transitive but not reflexive

[still drum]

oh wait

yea right

yea my bad

i cant read

sorry

[onyx peak]

you wont always find a familiar relation which serves as a counterexample

perhaps take some simple set, such as {a, b, c} and try constructing the counterexample on that

[still drum]

ok thank u

[onyx peak]

to make sure it's not reflexive, say that you'll disallow e.g. a ~ a

now you just gotta make it transitive and symmetric

[onyx peak]

btw this idea was very close, just rephrase it without equivalence classes stuff

[still drum]

mmm ok

[still drum]

by this you mean assume $a \not \sim a$

[soft zealot]

lifelong dumbass

[still drum]

or do you mean that the operation just isnt allowed

[onyx peak]

ye

[still drum]

oh ok

[onyx peak]

you can use a table to construct the relation ~, or draw some kind of graph with vertices a, b, c (since it's symmetric, every ~ works in both direction)

[versed crater]

Can you at least see why it’s not true?

[still drum]

you mean if $a \not \sim a$ or if the axiom is redundant

[soft zealot]

lifelong dumbass

[versed crater]

Second one

[onyx peak]

why the axiom is *not redundant

[still drum]

right yes my bad

[still drum]

somewhat

but ultimately not really

[versed crater]

Do you think it’s true

(True that it is redundant, that we don’t need it)

[still drum]

i dont think its true

but i dont see why its not

[versed crater]

I’m trying to give you this perspective because for counterexamples you usually need to start with some intuition

Okay, what is the exact implication that they claim to be true?

(What do you what to disprove)

[still drum]

that you can obtain $a\sim a$ by using the other 2 properties

[soft zealot]

lifelong dumbass

[versed crater]

Write it formally

[still drum]

so

the logic goes

suppose $\exists x, y $ s.t. $x\sim y$

[soft zealot]

lifelong dumbass

[onyx peak]

I'd say that their x is arbitrary

[still drum]

thats fair ok

[onyx peak]

since they are trying to prove that x ~ x for all x

[versed crater]

Reflexive + transitive => symmetric

Yeah?

[still drum]

suppose for all $ x, \exists y $ s.t. $x \sim y$

[versed crater]

What do these words mean in math symbols?

[soft zealot]

lifelong dumbass

[versed crater]

No that’s a function

Well, if you put unique then it’s a function

[versed crater]

But that’s neither of these 3 words

[still drum]

no i was trying to follow this

[versed crater]

Aren’t we trying to write this in more detail?

[onyx peak]

I think that the OP wrote out the ||faulty|| assumption they made here in the proof

[versed crater]

Im kinda lost on what the current goal is

[still drum]

wait just to ask shouldnt this be symmetric + transitive => reflexive

[versed crater]

Did I write it wrong

I did

Sorry you’re right

[still drum]

ok

[versed crater]

Write out what each of those words mean in math

[still drum]

$x\sim y \implies y \sim x \implies x\sim x$

[soft zealot]

lifelong dumbass

[versed crater]

Nope

Why are there 2 implications

Each of the words

Reflexive, symmetric and transitive

[still drum]

ok

reflexive implies that $x \sim x$, symmetric implies $x \sim y \implies y \sim x$, transitive implies $x \sim y$ and $y \sim z$ $\implies $ x \sim z$

[soft zealot]

lifelong dumbass
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[versed crater]

Not implies, is

[still drum]

oh ok

[versed crater]

Also you’re missing quantifiers

[still drum]

reflexive is that $x \sim x$, symmetric is $x \sim y \implies y \sim x$, transitive is $x \sim y$ and $y \sim z$ $\implies $ x \sim z$

[soft zealot]

lifelong dumbass
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[still drum]

ok let me add them

for an equivalence relation $R \subset X \times X$

reflexive is that $ x \sim x \forall x \in X$, symmetric is $ x \sim y \implies y \sim x \forall x, y \in X$, transitive is $x \sim y$ and $y \sim z$ $\implies $ $x \sim z \forall x,y,z \in X$

[soft zealot]

lifelong dumbass

[still drum]

i think

[versed crater]

Usually we put the quantifiers before though

But that’s right

So we need to show the last 2 imply the first (or that it doesn’t imply)

[still drum]

ok

[versed crater]

Notice that the last 2 statements are implications, whereas the first isn’t

They tell you more about the behaviour of relations when you pick 2 or 3 related elements

What about reflexivity?

[still drum]

reflexivity describes just the relation of an element with respect to itself

so

the last two are logical statements (transitive and symmetric)

whereas the first is simply an assumption you take to be true?

and without an assumption you take to be true, logical statements are meaningless?

is that the point you're getting at

[versed crater]

Yes here

The first statement is just always true, for every x in X

We don’t need any assumptions

The other 2 tells us more if we first pick some related elements

Notably, if there’s nothing to pick, they will be vacuously true

[still drum]

but by picking related elements, you've already assumed reflexivity?

[versed crater]

No I haven’t?

[still drum]

oh ok

never mind

i was just wondering where you were going with this

[versed crater]

I can just say suppose (a, b) ∈ R

If R is symmetric then we know (b, a) ∈ R

But first I need that (a, b) ∈ R before I can use symmetric to say more

If R is reflexive then I immediately have (a, a) ∈ R

[versed crater]

I don’t need this step

[still drum]

yes but how does that solve the problem

of the reflexive property being redundant

[versed crater]

Well, if R is empty, is R symmetric?

[still drum]

i mean a, b don't exist

[versed crater]

It’s a yes or no question

[versed crater]

.

[still drum]

so yes

[versed crater]

Right

Is it transitive

[still drum]

well theres nothing to pick so yes

[versed crater]

What about reflexivity?

[still drum]

i think yes because theres also nothing?

like no element exists so

[versed crater]

Okay, what elements are we picking in here

Like, from where do the elements that we pick come from

[still drum]

for the symmetric property?

[versed crater]

Yeah

[still drum]

the empty set

[versed crater]

From R

[versed crater]

What about this? From where do we pick the elements from?

[still drum]

we cant pick the elements from anywhere because they don't exist because R is empty?

[versed crater]

We’re still picking from R

There’s just nothing in R but we attempt to pick stuff from R anyway

[versed crater]

Okay what about

Where do these elements come from

[still drum]

i guess we still try to pick them from R

[versed crater]

Well, no

[still drum]

but why not

[versed crater]

We don’t pick from R because we don’t need an assumption for reflexitivity

[onyx peak]

What do you even mean by elements? R is a set of ordered pairs, are your elements ordered pairs?

[versed crater]

Yeah

Symmetric is (a, b) ∈ R => (b, a) ∈ R

Transitivity is (a, b), (b, c) ∈ R => (a, c) ∈ R

[onyx peak]

You might want to re-read this btw @still drum

[still drum]

i mean i think i do get it

like i understand that reflexivity is just a fact that you assume

[still drum]

like i think i understand what the words i said me

[versed crater]

Write reflexivity in this way

[still drum]

$(a,a) \in R \implies (a,a) \in R$

[soft zealot]

lifelong dumbass

[still drum]

which is the same statement

[versed crater]

Woah

Nope

That’s a tautology

That’s not reflexivity

[still drum]

ok then reflexivity is just that $ (a,a) \in R $

[versed crater]

This is if a ~ a then a ~ a, it doesnt say that a ~ a for all a ∈ X

[still drum]

ok

so reflexivity is $ (a,a) \in R \forall a \in X$

[versed crater]

So where do we pick out elements from

For this

[still drum]

X

[versed crater]

Ah

[still drum]

we pick a from X

[versed crater]

But for the other ones we picked ordered pairs from R

So somehow this is different

[still drum]

ok

[versed crater]

So here if R is empty then symmetric and transitive is always true

What about reflexivity

[still drum]

R is defined as $R \subset X \times X$

[soft zealot]

lifelong dumbass

[still drum]

R being empty says nothing abt X being empty

[versed crater]

That’s true

So can we know for sure that R is reflexive

[still drum]

hmm

X could be the empty set

[versed crater]

It could

[still drum]

but then R still wouldnt be empty

because then you could have the ordered pair $(\phi, \phi)$

[versed crater]

No we’re asserting here that R is the empty set

[soft zealot]

lifelong dumbass

[versed crater]

No that doesn’t work

[still drum]

oh

[versed crater]

X doesn’t contain the empty set

[versed crater]

X is the empty set

[still drum]

oh thats fair

right

my bad

[versed crater]

So suppose X was empty

Then when we try to find everything in X to test of (a, a) ∈ R, for a ∈ X

What happens

[still drum]

theres nothing to test

[versed crater]

So it’s true

If R is empty and X is empty then it’s true that R is symmetric and transitive => R is reflexive

Well, if R and X are both empty then it’s always all 3 anyway

What if X was not empty?

[still drum]

if X was not empty there is actually stuff to test

[versed crater]

And then…?

Let’s suppose X = {a}

[still drum]

ok well

R is empty

so

the ordered pair (a,a) cannot be in R

[versed crater]

So R is not…?

[still drum]

well if (a,a) was in R then R is not empty

and so you cant have both

[versed crater]

So R is not what

[still drum]

reflexive

hence the counter-example is constructed

[versed crater]

Yeah

[versed crater]

So my mind goes to R is empty because of this

[still drum]

so

you saw that

and considered the case for when R is empty but X non-empty?

[versed crater]

I realised that symmetric and transitive didn’t necessarily mean (a, a) ∈ R

Because what if a isn’t related to anything

What if every element in R doesn’t have a as one of it’s ordered pairs

If I find just 1 thing (call it a) in X that never appears in either entry of R, we can’t say that (a, a) ∈ R

[still drum]

but you also needed the other two statements to be true

[versed crater]

I’m assuming it’s true

I’m trying to show that the implication doesn’t hold

Suppose X is any set, and R a relation on X, and R is symmetric and transitive. Does this imply R is reflexive?

So then I need to show that for any a ∈ X, I can show using the assumptions that (a, a) ∈ R

[versed crater]

Then I get to here

If a is related to something, anything, then (a, a) is automatically in R as well

[still drum]

ohh

so like

suppose i had R such that

for X = {a, b, c}

R = {(a,a), (b,c), (c,b)}

then the line of logic doesnt hold because

[versed crater]

Well R isn’t transitive here

So your example is irrelevant

[still drum]

oh

[versed crater]

If R is transitive then (b, b) and (c, c) are in R by picking the last 2 elements in both orders and using transitivity

[still drum]

right

ok

[versed crater]

Then because every element of X appears in at least 1 of the pairs in R, it is true that R is also reflexive (note that we didn’t assume thing)

[still drum]

right ok

so that example doesnt work

[versed crater]

Well, suppose X = {a, b, c, d}

And R = {(a, a), (b, c), (c, b), (b, b), (c, c)}

Now R is transitive and symmetric

[still drum]

but its not reflexive

[versed crater]

Yup

[still drum]

because (d,d) isnt in there

[versed crater]

Exactly

[still drum]

ok that made more sense to me

[versed crater]

.

[still drum]

than the example with the empty set of R

alright

im sorry

this took way longer than it should've

[versed crater]

No no don’t be sorry

No this stuff takes time

[versed crater]

Yes but the empty set was a much simpler example

Once I realise the difference was in picking stuff, it turns out the empty set will work

[still drum]

i guess thats also fair but requires you to accept that vacuously true statements just dont matter

or rather

they do matter

but you can arbitrarily choose whether they're true or false

[versed crater]

No they are always true

They can’t be false

[still drum]

oh really?

[versed crater]

This has to do with how the implication sign logic works

[still drum]

oh ok

huh

sure

[still drum]

sorry im just trying to look this up now and misread a line on wikipedia

[versed crater]

p => q is always true if p is always false (I think)

[still drum]

i mean ik it was really frustrating so

[versed crater]

Nah it wasn’t

I wouldn’t be here if it were frustrating

[still drum]

well i felt really frustrated throughout all of it (this stuff really does require patience huh)

oh i think

i realised where

i went wrong

[still drum]

.

[versed crater]

That frustration is the learning part!! Embrace it!

[still drum]

here

this is wrong

because

instead of it should be reflexive is that $ x \sim x \forall x \in X$, symmetric is $ x \sim y \implies y \sim x \forall x, y \in X$, transitive is $x \sim y$ and $y \sim z$ $\implies $ $x \sim z \forall x,y,z \in X$

[soft zealot]

lifelong dumbass

[still drum]

it should be

reflexive is that $ x \sim x \forall x \in X$, symmetric is $ x \sim y \implies y \sim x \forall (x,y) \in R$, transitive is $x \sim y$ and $y \sim z$ $\implies $ $x \sim z \forall (x,y), (y,z) \in R$

[soft zealot]

lifelong dumbass

[versed crater]

No thats the same thing

[still drum]

oh

[versed crater]

Reflexivity doesn’t have an implication

Rather, it says a ∈ X => (a, a) ∈ R

Go google implication sign truth table

It always returns true when p is false

[versed crater]

So in the case where X is empty, a ∈ X is always false

So the implication always returns true

So any relation R on an empty set X is always reflexive

That’s what I meant by the “vacuously true” message

[still drum]

ok i see

alright

excuse me while my brain falls apart into a million pieces

[versed crater]

No no put it back together and learn something new!

[still drum]

i mean i just need a break thats all

its tired

[versed crater]

It do be like that

[still drum]

ok

like i just finished doing a bunch of exercises before this

so im already smoked

thanks again for all the help

alright ill close the channel now

.close

[final saddle]

Channel closed

Closed by @still drum

Use .reopen if this was a mistake.

[atomic moon]

yes it is cuz the negation will be always false then

not(p => q) = p and not q

so whatever q since p is false the and is false

[final saddle]

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[tranquil pine]

Okay, I have a very stupid question i need to ask. Why can i not use the concept of maxima-minima here? I just know we cant, but why not?

[tranquil pine]

Please tag me

[still drum]

wdym by the concept of maxima-minima

@tranquil pine

[tired walrus]

how are you planning to use the "concept of maxima-minima" an wdym by that?

[tranquil pine]

I somehow had it all figured out in my head, but now that i tried to type it out, i realised that i was wrong

I was thinking of using the derivative

brain fart

[tired walrus]

yeah derivative wrt a discrete variable is sus

[tranquil pine]

.close

[final saddle]

Channel closed

Closed by @raven blade

Use .reopen if this was a mistake.

[solar crest]

what’s the intended sol?

general form of binomial expansion formula and then.. idk

[tired walrus]

find the ratio between two adjacent terms and note when it switches from >1 to <1

[solar crest]

idk wym by that

[tired walrus]

the terms are C(9,k) 2^(9-k) (9/2)^k

find the ratio between the (k+1)st and k'th term

[solar crest]

wait
it’s consistent?

k+1th term and kth term have a ratio?

[tired walrus]

wdym "consistent"

i never said the ratio was constant

it'll depend on k obviously

[solar crest]

oh it’s dependent on k

[solar crest]

i see

[tired walrus]

ratio>1 means the next term is bigger, ratio<1 means the next term is smaller

[solar crest]

oh i see

so ur solving for ratio>1

and then you’ll get k

sub back in

thats final sol

ty

[final saddle]

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• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
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[tranquil pine]

I've been able to conclude that sum of coefficients would be 4^n. how do i proceed? i dont even fully understand the demand of the question after that

[tranquil pine]

Please tag

[desert mantle]

@tranquil pine for which n is that between 4000 and 10000?

for that n, what is the largest coefficient?

[tranquil pine]

n = 6

[errant basin]

yeah, its singular, i wonder about the strangely specific range

[tranquil pine]

so (1 + 3x)^6

[errant basin]

when is (n,i) the largest?

[tranquil pine]

its binomal theorem, so only positive integral powers

[tired walrus]

yes now you need to find the largest coeff in that

[errant basin]

i hate helping other people and discovering im wrong

[tired walrus]

which you can even do by just finding all the coeffs outright

[tired walrus]

c'est la vie

[errant basin]

i dont think thats the best approach 😦

[tired walrus]

takes about as much effort as the ratio thing

[errant basin]

c'est un crossiant

[tired walrus]

n=6 is p low

[errant basin]

anyway, i=3 right?

[desert mantle]

this isnt only about the binomial coefficient

[tired walrus]

the binomial coefficients are 1, 6, 15, 20, 15, 6, 1

[tranquil pine]

what if i had like n=10, what can i do then?

[tired walrus]

then do the ratio thing

[errant basin]

i hate helping other people and discovering im wrong due to a index shift or something

[tired walrus]

ie find the ratio of the k+1st term to the kth

and find at which k this ratio crosses 1

[tranquil pine]

wait let me try that

I got 1 = (3x)(7-k)/k

how do i proceed?

[desert mantle]

well the x shouldnt appear

solve for k

[tranquil pine]

i dont understand

did i make a mistake?

[desert mantle]

the coefficients dont contain x

[tranquil pine]

oh

then the 6th term should be greatest

right?

so the coefficient is 1458?

[vital surge]

that is correct

[tranquil pine]

okay!

thank you so much everyone!!

.close

[final saddle]

Channel closed

Closed by @raven blade

Use .reopen if this was a mistake.

[final saddle]

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Send your question here to claim the channel.

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

[silk fjord]

is there a difference between ≨ and <?

[thin cloud]

No...I guess? Maybe that's why I have never seen it before

[silk fjord]

I was just looking at unicode characters for something unrelated to math, and found that

and was just curious as to why it even exists

[thin cloud]

I did some google and it say ≨ mean "less than but not equal to", kinda pointless while we have <

[silk fjord]

yeah

[final saddle]

@silk fjord Has your question been resolved?

[royal tangle]

I mean, some people use ⊆ to mean “subset” and ⊂ for “subset but not equal to”
while other people use ⊂ to mean “subset” and ⊊ for “subset but not equal to”

so I guess ≨ exists for “completion” of possible combinations, idk if it has any real usage

[final saddle]

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[summer lynx]

this question

[tired walrus]

Lt = limit?

anyway:

!status

[final saddle]

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 jackal]

6

[summer lynx]

1

[summer lynx]

yes

i did something like putting the value of x=5 but thats undefined

i know i have to do something to make it defined to be able to put the limits

[tired walrus]

question was not addressed to you.

[tired walrus]

indeed

try adding and subtracting 5f(5).

and grouping the terms into two pairs

[summer lynx]

jus to the numerator

[tired walrus]

yes on the numerator.

[summer lynx]

ok

[tired walrus]

i take it you understood?

[summer lynx]

yea so i can cancel out the first term

and the socond one rules out

the one after the + sign in the numerator

[summer lynx]

i think

nah i didnt get the answer

[summer lynx]

no

what did they do in the second steo

[ebon agate]

L'hopital

[summer lynx]

🤡

oh

[summer lynx]

but any i dea what i can do here

@ebon agate

[ebon agate]

so f(5) + 5*(Lt x->5 (f(5)-f(x))/x-5)

Do you know the definition of a derivative

[summer lynx]

yes

[ebon agate]

,rccw

[ebon agate]

@summer lynx

The equation in the black box is the definition of a derivative

[summer lynx]

that f'(5)

Oh

[ebon agate]

Yes

[summer lynx]

thanks

.close

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[vivid jungle]

$$\lim_{x\to2}{(3-x)}^{\tan{(\frac{\pi}{4}x)}}$$

[soft zealot]

Koren

[vivid jungle]

i don't know what to do at all,
i suppose it's related to e^lnx=x

[vital crag]

Yes use this to rewrite your expression

$e^{\log(a^b)} = e^{b\log(a)}$

[soft zealot]

riemann

[vivid jungle]

$$e^{\tan{(\frac{\pi}{4}x)} * ln(3-x)}$$

[soft zealot]

Koren

[vital crag]

lim (e^thing) = e^(lim thing)

[vivid jungle]

ok, that's still \pm\infty times 0

[vital crag]

Yea you're not done manipulating

[vivid jungle]

ok, what else can i do?

subtitute t = x-2 so ill have t\to0

[vital crag]

What else have you tried/learned

[vivid jungle]

?

[vivid jungle]

i don't think i learned how to do this exrecise, it's from a test from 2014 and the subjects might have changed

so i want to see which tools there are to solve it

to see if i know those

[vital crag]

There's usually more than one way to do limit problems

L'hopital is one option here

[vivid jungle]

L'hopital is something that was teached every year for like 20 years in my course but this year it wasn't

so it may be it

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@vivid jungle Has your question been resolved?

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[kindred sail]

The question is written as below:
A study of 55 patients with low-back pain reported that the mean duration of the pain was 17.6 months, with a standard deviation of 5.1 months. Assuming that the duration of this problem is normally distributed in the population, determine a 99% confidence interval of the mean duration of low-back pain in the population.

[final saddle]

@kindred sail Has your question been resolved?

[kindred sail]

<@&286206848099549185>

[cyan kayak]

!status

[final saddle]

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

[kindred sail]

1

im on step 1

[dusty quarry]

oh dear hypothesis testing

[cyan kayak]

Do you know the difference between the sample variance and population variance?

[kindred sail]

yeah

[cyan kayak]

Do you remember how to relate them in an unbiased way?

[kindred sail]

no

[cyan kayak]

So if you have a sample variance of s^2, then your unbiased estimate of the population variance is n/(n-1) s^2

then you can just use z-scores to find the rest.

[kindred sail]

i dont know xi tho

the question didnt give the values im supposed to subtract the mean from

[cyan kayak]

xi meaning $\xi$ or $x_i$?

[soft zealot]

OmnipotentEntity

[kindred sail]

the second one, not the greek letter

[cyan kayak]

where x_i means the start of the interval we're seeking?

[kindred sail]

wait
given that my sample size is 55, the mean duration is 17.6 months and the standard deviation is 5.1 months, i just need to find the Z-score of the 99% confidence, and then use it to find the margin of error

and then after that i dont know what else

[cyan kayak]

So the 99% confidence interval is the smallest interval that contains 99% of the probability mass of the distribution, right?

[kindred sail]

yes

[cyan kayak]

the cdf ranges from (0, 1)

so what range of values will contain 99% of this range?

[kindred sail]

0 to 0.99?

[cyan kayak]

well, 0 and 1 shoot off to infinity

we want the smallest interval

[kindred sail]

oh right

then its 0 to 0.01 i think

[cyan kayak]

???

[kindred sail]

bruh im confused

[cyan kayak]

it's ok

[kindred sail]

wait let me show my steps

[cyan kayak]

let's do an easier one

let's say we want a confidence interval of 50%

[kindred sail]

uh huh

[cyan kayak]

our confidence interval should always contain the mean, because that's where the most probability mass is concentrated

[kindred sail]

yeah

[cyan kayak]

So our range, from (0, 1), we need to cover 50% of this, so we can have (0, 0.5) or (0.4, 0.9) or (0.2, 0.7). But which interval is uniquely situated right in the middle of the range and so covers the most mass in as small an area as possible?

[kindred sail]

id say (0.4, 0.9), if not (0, 0.5)

[cyan kayak]

do you remember where your mean is?

in this range?

[kindred sail]

for the question im asking about?
if thats what youre asking its 17.6

oh wait

[cyan kayak]

I'm asking about inside of the cdf interval

[kindred sail]

no i dont remember
well currently im taking grade 12 data management, i dont think i learned about CDF yet

[cyan kayak]

so a normal distribution is symmetric. The mean is located at 0.5

[kindred sail]

yeah

[cyan kayak]

We want to surround this as evenly as possible.

which will give the interval (0.25, 0.75)

[kindred sail]

okay

[cyan kayak]

do you see why?

[kindred sail]

because of the 50% confidence

[cyan kayak]

well, any of these intervals would be a 50% confidence interval.

but the (0.25, 0.75) interval is the smallest one.

[kindred sail]

you can divide 0.5 by 2 and just make the interval (0.5 - 0.25, 0.5 + 0.25)

[cyan kayak]

yup

so what's the 99% confidence interval?

[kindred sail]

okay
going back to my question, with 99% confidence we could say that the confidence interval is (0.005, 0.995)

[cyan kayak]

so you take the edges of the range and convert it into a z-score

using the z-score table working backwards.

then using the z-score you find the interval.

[kindred sail]

hmm
when i worked on it, there was something called a critical value, and i used a t-distribution table

also by taking the edges of the range, you mean i take 0.005 and 0.995 and convert them to Z-scores, then calculate the margin of error?

and then use the margin of error to find the interval

[cyan kayak]

with 55 samples you don't need to use the t-distribution, you're well estimated by a normal distribution (central limit theorem)

The switchover point is about 30 samples.

[kindred sail]

about 30?

let me show you what ive got as of now

from the looks of it the switchover point is about 17?

[cyan kayak]

sorry

I went afk

30 samples

you have 55 samples in your problem which is > 30

therefore you use a regular normal distribution instead of student's-T

[kindred sail]

ok

[cyan kayak]

I mean, you can still use the T distribution

but you don't need to

[kindred sail]

does my interval look correct tho

[cyan kayak]

let me actually work this problem through and see.

oh

no actually I can tell immediately that it's not correct.

You should get an interval of about (4, 31). @kindred sail

Obviously, you should work the problem to get more significant figures.

but you can use my estimate to check your work

[kindred sail]

can you explain your steps to getting this estimate

wait with that interval that would make the margin of error too large

[final saddle]

@kindred sail Has your question been resolved?

[final saddle]

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[covert girder]

Hello. After a bunch of calc 2, I forgot how to manipulate algebraically to get a remarkable limit

[covert girder]

What's the catch? I know I have 1 + 1/(3x+1)

But I got x+1 up there

[steep nest]

use e^ln

[tired walrus]

i was about to reply but then noticed you greeted with "hello gentlemen"

[terse crypt]

or defn of e

[tired walrus]

i wonder if that means you only want men replying or if this was unintentional

@covert girder

[covert girder]

Unintentional.

[steep nest]

this is a way better method

lol

[tired walrus]

ok then you could simply erase the word "gentlemen"

[terse crypt]

a cooler method 😎

[covert girder]

You got it

[tired walrus]

and not leave a girl wondering if she could even engage

[covert girder]

Math has no gender

But I got it

Just my way of addressing, sexist or not, I apologise

[tired walrus]

the inside stuff rewrites as 1 + 1/(3x+1)

so you want this raised to the power of 3x+1 to set up your e

[tired walrus]

yeah, consider changing it. even "hello everyone" has a lot less risk of misread

[covert girder]

So as far as I see I got 3 ways of solving it

[tired walrus]

the pro gamer move is to write $x+1 = (3x+1)\cdot\frac{x+1}{3x+1}$

[soft zealot]

Ann

[covert girder]

For sure. Don't want to cause trouble in a community I've been lurking for years

[terse crypt]

the defn of e 🔥 🔥

[covert girder]

Y'all are the best

What's the defn?

[shrewd nexus]

x^n / n!

sum from 0 to inf

[terse crypt]

not wrong😭

[covert girder]

Is that Taylor or what, seems similar

[terse crypt]

but TheVibe is jking here

[tired walrus]

not that one

[tired walrus]

limit of (1+1/n)^n at infinity

[thin cloud]

Not that useful in this casexd

[terse crypt]

that was mclaurin (taylor) expansion ye

[tired walrus]

in your case, 3x+1 plays the role of n

[covert girder]

I think I got it, no?

[shrewd nexus]

Yup

[covert girder]

Uuuu, well this is a pre uni test

[terse crypt]

https://tenor.com/view/chiikawa-absolute-cinema-gif-1155133254255634218

[covert girder]

🔥 🥳

[tired walrus]

yup

[covert girder]

Thanks everyone! I'll make sure to follow a more diverse etiquette

[thin cloud]

What a nice solution

[shrewd nexus]

It's how it was defined for me, but ye we do a lil trollin

[terse crypt]

a lil tomfoolery

[covert girder]

Thanks for the laughs and the help, all!

Saving my calc as always

Have a nice one!

.close

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[final saddle]

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[heavy lance]

im about to stress TF out

[final saddle]

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[heavy lance]

whoops

cut off is just "nt" ive redone this type of problem like 30 times now and i really need some help

ive tried calculator and multiple AI but i just cannot do it im stressing out HARD

please....can anyone help

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[jade iris]

hey, my specialist gave me another problem, how do i approach it

[final saddle]

@jade iris Has your question been resolved?

[final saddle]

@jade iris Has your question been resolved?

[slim creek]

Not 100% certain this is the correct solution, but first I’d recommend playing around with small values of a and b, evaluating a#b as you go, to try and find a pattern in what a#b is

From this, you can deduce that a#0=a, so a#1=a+1 (for a>0) and a#2=a+2 (for a>1). From this, a logical conjecture would be that a#b=a+b (which would work nicely with the value you want to evaluate) as 1#(2#(…(13#14)…)) is just the sum of the first 14 positive integers

[teal swallow]

Prove that the operator # is commutative and associative

Then you can do the calculations from left to right

And ignore all the parentheses

[muted prairie]

3#1 = 2

[jade iris]

ok i proved its commutative and associative but how do i evaluate the expressions. It's still so much work

[teal swallow]

here's an additional hint: # = XOR

[jade iris]

oh wow

why is that?

xor like the logic gate?

[teal swallow]

let's see if i write it like this, would u understand

3 # 1 = (convert to binary) 11 # 01 = 10 = (convert back to decimal) 2

[jade iris]

but how does the operation # have anything to do with binary

[teal swallow]

this is so complicated to explain gicahpfcrvnjkra

[jade iris]

how did you know it was xor

you can explain it however you want

[muted prairie]

out of curiousity, did you chatgpt this?

[jade iris]

my specialist will not accept chatgpt answers 😠

[teal swallow]

no

i dont even have money to chatgpt

smh

[muted prairie]

hmm

free ai?

[pale niche]

wouldn't doubt it

it's gotten good enough to answer these types of questions but its explanations are quite awful

[jade iris]

can someone please just explain how you get to xor

that is so confusing

i can try and solve it from there then

and no ai please

[teal swallow]

https://youtu.be/GRlGknQEOW8?si=k7wJ-adXFZ068lFd
bingo

it's not a coincidence

[jade iris]

ty!! i put this into my ai and it also showed the same vid, must be right then

[teal swallow]

damn

honestly the # = XOR guess took me a solid 30 minutes

to figure out

also didnt know ai now can generate correct https address lol

[muted prairie]

the ones that have been specifically programmed to search for sources can

i dont think a genai with no help could

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@jade iris Has your question been resolved?

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[snow lily]

help

[snow lily]

the red part is shaded only right

or am i tripping

<@&286206848099549185>

havent done this in so long

[edgy mauve]

Can u show the original question without the markings

And !15

[thin cloud]

!15m

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

[snow lily]

(ignore the shade)

[edgy mauve]

Yea so that's the shaded part

[snow lily]

this red part only?

[edgy mauve]

Wdym red part is shaded

[snow lily]

wait what

how

[edgy mauve]

Where else can u see shaded marks or whtever u call them

Sorry

I misunderstood ur question

U r ryt

Red is shaded

Along with this part

[thin cloud]

Uhh, I need to learn something here, what's the relation between set A and A'?

[edgy mauve]

A' means U-A

Everything except A

[snow lily]

its that without the middle no?

[thin cloud]

Danm, that's new to me, I have never been taught that, thanks

[edgy mauve]

Lemme walk u through this

So here everything except The marked part is E'

Do we agree

[snow lily]

ye

[edgy mauve]

And this is DUE'

Ryt?

[snow lily]

u forgot the outside and C

cuz its union

[edgy mauve]

Oh ryt

So basically (DUE')=E'

[snow lily]

this is DUE'

[edgy mauve]

Ryt ,u r ryt

Okay soo what would be (DUE')'

[snow lily]

the ones not shaded

[edgy mauve]

Yea

[inner acorn]

what is happening?

[edgy mauve]

Now this sounfs correct

This is correct

[snow lily]

yup thx

[edgy mauve]

This is happening

[inner acorn]

C but not D and E

ah ok seems like yall done

[snow lily]

.close

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[undone tiger]

What is commutative binary operation?

[fiery bluff]

,w commutative binary operation

[undone tiger]

[fiery bluff]

worth a shot i guess

[undone tiger]

G×G-->G

[fiery bluff]

https://en.wikipedia.org/wiki/Commutative_property

[undone tiger]

So A×A will have 25 elements

To 5 elements

Now my doubt is how can I make map?

And the map should be symmetric?

[fiery bluff]

try filling in a table

for each box, determine how many choices you have

[undone tiger]

I'm confused because elements are not equal

I can't make squares matrix

[fiery bluff]

to make a binary operation, it suffices to specify what's the output for every possible pair of inputs

[undone tiger]

The output will be like (a,b) form

The diagonal elements will have choices

I found something best explanation

https://math.stackexchange.com/questions/2353141/what-is-the-possible-number-of-commutative-binary-operations-that-can-be-defined

Wow brilliant people!! Explained easily hurrayy!!! Thanks everyone

.close

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[covert flower]

how do we solve min max problems?

[covert flower]

What are its steps

[fallen vigil]

be more specific? what problem

[covert flower]

min max word problems where you need to find its maximum/ minimum extrema

[fallen vigil]

find the minimum or maximum given an equation?

[atomic moon]

do you have a specific example

[fallen vigil]

differentiate it. when the derivative is 0, check if its a min or max. for all points where derivative is 0

[thin cloud]

You mean application of derivatives problem?

[fallen vigil]

some points are neither, u can find out by checking second derivative or a sketch

[final saddle]

@covert flower Has your question been resolved?

[covert flower]

Yea one sec

[covert flower]

this one

[fallen vigil]

differentiate it. thats what they did when they found S'. they set that =0. and found that is when r=cbrt(250/pi). they put that into S'' to find out that its >0. which implies that its a maximum. in this case, maximum volume.

S' = 0 denotes the change is zero. so it can be a maximum point, minimum point of inflexion point

with the use of the second derivative we can find out which it is

S''=0, weird behavior
S'' >0 minimum
S'' < 0 maximum

[thin cloud]

S"=0 isn't always inflexion point tho

[atomic moon]

indeed, you have to study the sign of f'' in the neighborhood of the point, if it changes at the point then it is

[fallen vigil]

oh sorry. what else can it be

[final saddle]

@covert flower Has your question been resolved?

[atomic moon]

it can be something, you can take x^4/12 and see that the second derivative is x^2 and so indeed 0 at x=0 but its not changing convexity cuz it does not changes sign

f'' = 0 is a neccessary condition for an inflexion point but its not sufficient

[covert flower]

sorry I was outside 😭

let me read

[fallen vigil]

thanks

[atomic moon]

welcome on the server btw

[fallen vigil]

thanks

[covert flower]

I’m confused

what the question is asking

save me

In order to do this question what do I need to know

This is basically word problems

[atomic moon]

esthesia wrote the method here, we were just talking about the case where f'' is equal to zero where this is a case you can't say its a max or a min. it probably won't happen in your exercices since you're asked to find a min or a max

[sonic crystal]

yakuuu

[covert flower]

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[bold dust]

help please

[thin cloud]

This problem relates to euler "candy division problem"

Which is a really powerful tool when it comes to counting solutions for sums

You should look it up on Google, tho. It's quite a lot

[bold dust]

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[whole verge]

Hello, so the dunford decomposition of a endomorphisme u is the unique couple (d,n) such that nd=dn and u=d+n where d is diagonalizable and n is nilpotent , I have proven the existence of d and n that satisfy this property and I have proven that they are a polynomials of u .
in the proof of the uniqness we have assumed that if any other couple (d',n') exists they must also be a polynomial of u, without providing any proof of it !
So my question is:
is the uniqness true only if the couples are a polynomial of u or can one prove that (d',n') must be a polynomial of u ?

[shrewd nexus]

if you found that d and n are polynomials of u, when you take another couple that verifies the original properties, they too will be polynomials of u

[whole verge]

How so ?

[shrewd nexus]

Well by doing what you did for d and n

because the uniqness is always true that's for sure when it exists

[whole verge]

I constructed d and n from projections on the characteristic spaces

[shrewd nexus]

yes

[whole verge]

I can't do that for d' and n'!

[shrewd nexus]

well you have d' + n

oops

you have d' + n' = u because d' n' = n' d' you have d' u = u d', same for n', and since d and n are polynomials of u, d' and n' commute with d and n

and i think that's all you need

you dont need that they are polynomials of u actually, just need the fact that they commute

[shrewd nexus]

(and yeah im sorry i just didnt remember the construction of d and n)

dunford decomp was a while ago

Does that answer your question?

[whole verge]

yeah that's much better actually

[shrewd nexus]

ok great:)

[lofty seal]

The uniqueness of the Dunford decomposition does not require the assumption that
𝑑 and n are polynomials in 𝑢.
Even if another decomposition
(𝑑′,𝑛′) is given with the same properties (commuting, one diagonalizable, one nilpotent), it must be equal to the original one. So:
You don’t need to assume
𝑑′,𝑛′∈𝐾[𝑢]— the uniqueness proof works without that

[whole verge]

the whole point of the polynomial of u was to show the commutivity

[whole verge]

yeah i see that now

[shrewd nexus]

you can use it to show it yes

not required tho

[whole verge]

yeah thnx

[shrewd nexus]

:)

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[tranquil pine]

how can i make the construction of this triangle using geogebra?

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[normal shuttle]

How do i solve this

[normal shuttle]

AH is 9 btw, not 12

<@&286206848099549185>

[tranquil pine]

What is your grade ?

[normal shuttle]

Why does that matter

[tranquil pine]

No just asking

[normal shuttle]

Oh

I don't wanna say

[tranquil pine]

Because I'm in grade 10

Okay as u like i asked for the question

[normal shuttle]

This is the question

Are you going to help?

[final saddle]

@normal shuttle Has your question been resolved?

[scenic notch]

🗿

bro really anounced his grade and left

[normal shuttle]

Fr bro

Also

I think that might be unsolveable

Im not sure

[edgy mauve]

What if 12 is the measurement of something else

[normal shuttle]

Its not i asked

My teacher said it wasnt

He said its a typo

I also have the answer but that doesnt make sense to me

This might jusy be wrong 💀

I dont think thats how euclidian works

[scenic notch]

My intuition says that AHB and CHB are similar triangles, but I can't really prove it. Edit* I'm so totally incredibly wrong

[normal shuttle]

They have 1 similar side and angle

But nothing more

Thats not enough 😕

[edgy mauve]

Dang this makes sense

[normal shuttle]

Does it?

Not to me

[edgy mauve]

Okay u see triangle ACD and trainfle ABC r congruent ryt?

[normal shuttle]

Yeah

[edgy mauve]

So we can say
Angle <DAC=<BAC

Similarly
<ACD=<ACB

[normal shuttle]

Uhmm

[edgy mauve]

Ryt?

[normal shuttle]

DAC and BCA are equal

Not DAC and BAC

I think

[edgy mauve]

Ryt ryt sorry

But well that doesnt get us anywhere now

[normal shuttle]

Lol

[edgy mauve]

Taking different approach

[normal shuttle]

This is like the 4th person whos struggled with it

Therefore

[edgy mauve]

BHC and ABC

[normal shuttle]

I declare it to be flawed

[edgy mauve]

Nah it's prolly possible the 4 person u met rnt qualified enough

[normal shuttle]

One of then was me

Thank you 🥲

[edgy mauve]

And me but lets try either way

[normal shuttle]

Yes sir

[dense badge]

BHC and ABC share two angles no?

[edgy mauve]

U can prove that BHC is congruent to ABC

[normal shuttle]

Abc is not 90

Bhc is 90

They arent congruent

[edgy mauve]

It is 90

[dense badge]

rectangle

[normal shuttle]

Haha

Yes it is

[edgy mauve]

Rectangle has 4 90s bro

[normal shuttle]

I was looking at abh

Oops

Continue

[edgy mauve]

Oh okay so they r congruent

[normal shuttle]

Yes

SIMILAR TRIANGLES?

[edgy mauve]

So u could say that
<HBC=<BAH

[edgy mauve]

Yeah triangle BAH AND BHC

[crisp spade]

are we still on this question?

[normal shuttle]

Yes

[crisp spade]

ok

three equations

from three triangles

[edgy mauve]

U can find that those two r similar ryt?

[crisp spade]

one hypotenuse AC=13 one hypothenuse AB other hypotenuse BC

Pythagoras

write down three equations

two of them involves BH

cancel the BH out

[normal shuttle]

I had a stroke reading this

[crisp spade]

compare with the third

[normal shuttle]

Compare AB and BC with AC?