help-13

[craggy haven]

but it’s wrong

somewhere…

[silk gust]

ok can we start over bc my brain refuses to comprehend anything rn

alr lemme ask you this

how far is kali from the school when matt leaves?

[craggy haven]

erm

idk

[silk gust]

ok so kali is traveling 40 km/h right

so how far would she travel in an hour

[craggy haven]

40

[silk gust]

so the answer to this would be...?

[craggy haven]

40

Ohhhh

[silk gust]

mhm!

[craggy haven]

w400-40?

is the next step right

[silk gust]

yep

alr now answer this: once matt leaves, how much distance is gained between him and kali each hour?

[craggy haven]

what

waittt

360

for kali

bout liek what do i do after

[silk gust]

wait so

lets say matt and kali both start from the school at the same exact time

ok?

but they ride in opposite directions like in the problem

[craggy haven]

okok

uh huh

[silk gust]

lets say matt travels west and kali travels east

so matt travels 50 km west in an hour, and kali travels 40 km east

you follow?

[craggy haven]

ya

time is distance /speed right

[silk gust]

yes

so how far apart would they be after an hour?

[craggy haven]

130

[silk gust]

how did you get that?

[craggy haven]

cuz matt left an hour ltr

so like

50+40+40

[silk gust]

oh wait mb you went back to the original problem

yes thats riht

right

so basically you keep adding 90 to 40 right? until you get 400

[craggy haven]

okok thx

u got it

[silk gust]

u see the answer?

[cedar kiln]

@craggy haven Has your question been resolved?

[cedar kiln]

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[charred oriole]

Not sure if anyone here understands dimensional analysis, but I'm having a hard time figuring this particular problem out

[charred oriole]

It's not hard math, it's the units that are tripping me up

[oblique flare]

Just find the good conversion
1 lb= 453.59g
1=453.59g/1lb or 1= 1lb/453.59g one of these is the useful one

[charred oriole]

they want the setups though

They don't want the answer

They want the full equation

I already answered this question but here are the answer choices anyway

This one was the right answer

I just am having a hard time trying to understand it

[oblique flare]

It’s really just multiplying this conversion out
but the numbers seem off

[charred oriole]

I work with whatever they tell me to

If I question anything else, I get the question wrong and I fail the test

[oblique flare]

$1.5g/ml \cdot \frac{1lb}{453.59g} \cdot \frac{3.785.41mL}{1gallon}$

[charred oriole]

I should explain why I am stuck, sorry.

I am stuck because they're asking me to convert g/mL and lbs/g. It's for that reason that I am stuck and don't know where to even start.

[wraith dagger]

qimmah

[charred oriole]

So why the hell is the 453.59 off to the side like that?

Wtf?

[split ice]

[charred oriole]

I KNEW it was weird

Fucking ATI and their buggy as hell app holy shit

This is hell

[charred oriole]

Thank you for the help! I appreciate it a lot <3

.close

[cedar kiln]

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[oblique flare]

Hehe told ya

:3

[charred oriole]

You did, it was a skill issue on my part 😅

ATI just loves formatting errors, their practice app is full of it

[oblique flare]

Hope it doesn’t happen again

[charred oriole]

It will. It's ATI.

[oblique flare]

😭😭😭😭 hope it isn’t that bad then

[charred oriole]

No, it's pretty awful xd. my only consolation is that my test is tomorrow and i'll be freed from ATI's bullfuckery

[oblique flare]

You gonna demolish that test king 🔥🔥

[cedar kiln]

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[torn marsh]

im kind of stuck on how to solve this question

[bitter gazelle]

You can use trigonometry

[honest field]

!nosols

[cedar kiln]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[honest field]

also it's wrong

[livid hound]

especially wrong solutions

[spring forum]

don't edit tho

nosols

just del it

[spring forum]

ty

[wicked mantle]

yeah, try to provide assistance in a “guidance”

[spring forum]

!status

[cedar kiln]

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

[spring forum]

then try this

[torn marsh]

its asking for

exact value

[spring forum]

yes

[torn marsh]

not in decimal approximation

[spring forum]

you do know sin 60º tho, right?

[livid hound]

that doesn't conflict with using trig

[spring forum]

its exact value

[bitter gazelle]

srry

[spring forum]

you're good! it happens :3

[torn marsh]

yea i know the 30 - 60 triangle

[spring forum]

right, so what is sin 60º?

[torn marsh]

sqrt3/2

[spring forum]

you also know that sin is opposite/hypotenuse

what is opposite/hypotenuse in this case?

[torn marsh]

5/x

[spring forum]

mhm

so you can set an equation

because 5/x is also supposed to be sin 60º

[torn marsh]

so sqrt3/2=5/x

[spring forum]

can you take it from here?

[torn marsh]

I should be

x=10/sqrt3?

[spring forum]

yes, but can you simplify that?

(rationalize the denominator)

[torn marsh]

10root3/3

[spring forum]

perfect

do you have any more questions?

[torn marsh]

il let u know

[spring forum]

!done

[cedar kiln]

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[torn marsh]

.close

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[pine laurel]

can anyone help me some maths

[pine laurel]

i could use some help in leslie matrix

[wicked mantle]

Hiiii, can you give us your question?

[pine laurel]

ok sure

It just asks to find the missing fertility rate for 20 - 24 and 30 - 34

i am not really sure how it works

[cedar kiln]

@pine laurel Has your question been resolved?

[cedar kiln]

@pine laurel Has your question been resolved?

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[sonic fossil]

find the 5 digit number formed by 1,2,3,4,5 without repetition, I would first find how many nos. can be formed

_ _ _ _ _ = 5! = 120 numbers

But if its asked to find 5 digit number obtained by permuting 0,1,2,3,4 then it comes there are zero number 4! x 0

[wicked mantle]

I don't quite get it

[sonic fossil]

maybe I have to tread it is 0!

treat*

[wicked mantle]

Are you looking for how many 5 digits number can be formed with 0~4?

[sonic fossil]

The actual question is I have to find the sum of 5 digit number obtained by permiting 0 1 2 3 4

If I know the 5 digit number I would find the sum

nvm I got it

[wicked mantle]

Mind sharing your method? I can verify it 4 you

[sonic fossil]

I'll share later on dm

.close

[cedar kiln]

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[wicked mantle]

Why don't just keep everything in this channel?

[sonic fossil]

I'll share in the channel afterwards

[wicked mantle]

well, don't close it in this case

[tropic oxide]

.reopen

[cedar kiln]

✅

[tropic oxide]

it might happen that the channel times out though

[sonic fossil]

I am giving the reasoning

[blazing zephyr]

shouldnt you exclide like 01234 cause its not a 5 digit

or do you want to count it

[sonic fossil]

Question says sum of 5 digit number obtained by permiting 01234

Permuting

[blazing zephyr]

fair

[sonic fossil]

I should subtract number of numbers starting from 0 otherwise it would not be 5 digit right?

[blazing zephyr]

how did you get 266660?

[sonic fossil]

I think my audition is wrong

[blazing zephyr]

yeah probably

you have the right idea but the answer is a bit off

[sonic fossil]

I did this suppose
716
234
123
+-------
(10 )(5)(13)

I tried addition like this

If you understand good otherwise it's omay

[blazing zephyr]

what did you add exactly?

[tropic oxide]

there are not 24 numbers that end with a 1 sadly

there are also not 24 numbers ending with 2, and the same for 3 and 4

[sonic fossil]

bye bye then

Lol

[tropic oxide]

though your idea of adding the numbers vertically without carrying is going to cook

[sonic fossil]

This is homework question, teacher will explain it today

I will explain after class how he did because he told do like this

.close

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[sacred tusk]

the sum of all permutations - the sum of all which starts with 0

[wicked mantle]

It's up to you, but here's a postgraduate to help you figure it out prior

[sonic fossil]

i am less on time today

It's like i have to go in 1 hour

[wicked mantle]

it's alr, have a good one 🧡

[sacred tusk]

my method is like there are a total of 5! = 120 permutations and you can pair each permutation in the way that each corresponding digit sums up to 4 like for example 23104+21340=44444 like for the first digit is 2 and 4-2=2 and the second digit is 3 and 4-3=1 and so on

so the sum is 44444 × (120/2)

and do the same for the sum of all which starts with 0

1342+4213=5555 etc

total permutation is 4! = 24 (cuz the first position is already placed by 0)

[wicked mantle]

Just drop it, he's not coming back

the channl will timeout soon

[sacred tusk]

oh okay

[cedar kiln]

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[coral jackal]

does anyone want to be my math tutor? for info, im in grade 10 going into grade 11 and i really need help with my math, im not bad but im also not good, i would really appreciate it

[wicked mantle]

We do not encourage paid service here.

But it’s up to you. If you insists, it would be better to demonstrate your current level of math such as showing your syllabus, demo questions etc.

[coral jackal]

how?

[wicked mantle]

Just post a few questions that you are dealing with recently.

[coral jackal]

.close

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[honest field]

if you guessed the scoreline for Manchester United vs West Ham correctly we will tutor you for free

[twilit escarp]

I suppose you are supporting West Ham

[honest field]

true...

mu will prob win anyway

[hallow pelican]

are you carbonite

[honest field]

no

[hallow pelican]

hmm

[honest field]

?

[cedar kiln]

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[verbal quest]

This is my question i was able to calculate the value of the integral but i am unable to convert it into the desired form

My integral value came out to be: -root2(ln (cosx + (root (cos2x/2))))

[cedar kiln]

@verbal quest Has your question been resolved?

[verbal quest]

<@&286206848099549185>

[sleek stump]

Hi Leo

Can you provide your question please

[verbal quest]

..

This is my question i was able to calculate the value of the integral but i am unable to convert it into the desired form

My integral value came out to be: -(ln (cosx + (root (cos2x/2))))

[sleek stump]

And what's the desired form?

[edgy spire]

The RHS

[verbal quest]

uhh its given in the Question

[sleek stump]

Oh sorry

Can you quickly go through what you've tried so far?

[verbal quest]

converted , sec2x - 1 into , 1-cos2x/cos2x

then converted it into, root2 sinx/ root(2(cosx)^2- 1)

substituted root2cosx = t

applied 1/root (t^2 - 1 ) form

[edgy spire]

I mean, one quick way to do it would be to differentiate their function and then find the values of alpha and beta by comparing the result with the integrand

[verbal quest]

go ahead

it filled my page to just differentiate right side

[sleek stump]

Yeah, that's what I suggest

[cedar kiln]

@verbal quest Has your question been resolved?

[verbal quest]

.close

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[cedar kiln]

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[deft brook]

Where does the 3/2 come from? And how does the (1+cos24X)/(2) become (1/48) (sin24)?

[neon kraken]

What strategy should I follow?

[deft brook]

https://tenor.com/view/moral-orel-meme-get-real-orel-puppington-clay-puppington-gif-22211944

[tawny sage]

They're combing the constants. So there was already a 1 at the start of the integral, and the fraction part of the integral was split up so the 1/2 part got separated out. Combine these you get 3/2, and then integrate them with respect to theta

[slender ginkgo]

what the hell is going on in this chat

[deft brook]

forgive me but I dont understand where that fraction part of the integral was split up from.

I see integral 1 and I think it just become theta.

[tawny sage]

Later in the integral there's a fraction like (1+cos(...))/2

They split that fraction into 1/2 + cos(...)/2

and added that 1/2 to the 1 at the start

[deft brook]

i seeeeeee nowwwwww

[deft brook]

Ok so how does that then allow 1/48 sin24x?

[tawny sage]

Do you know how to integrate cos(ax) for some constant "a"?

[deft brook]

sin(ax)/a right?

wait

am idiot

not realising it splits to become 1/2 + cos24x/2

[deft brook]

https://tenor.com/view/dr-disrespect-disrespect-unfaithful-transparent-omegalul-gif-13464158

[tawny sage]

yeah. That's where the 48 comes from. Dividing by 24 and 2

[deft brook]

I literally gotta break down the problem like this or am lost as shit.

[slender ginkgo]

oh u did it like that

[deft brook]

yee

https://tenor.com/view/moral-orel-clay-puppington-goodbye-slam-door-door-shut-gif-23642600

off to the next problem

.close

[cedar kiln]

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[cedar kiln]

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[royal zinc]

@slender ginkgo

[nova snow]

u got this k

[slender ginkgo]

fucking hell

is this some math club type shi

[royal zinc]

No buddy matter of fact these sinners threw this shi in my test😭

[slate lintel]

pre university this would be criminal

i'm wondering if it's some contour magic

[royal zinc]

2x in the denominator from 0 to 1

[slender ginkgo]

-1/2x is trivial

but the fucking square root tho

[royal zinc]

I guess good way to start would to combine first and last term. Ignore middle term

Then rationalisation

Doing that gives you

[slate lintel]

(not really; if you do that on its own over those bounds you get a divergent integral)

[nova snow]

true you cant split

[slender ginkgo]

i see

[royal zinc]

No but it is zero by zero form

The 2x in denominator cuts off

After rationalisation

[royal zinc]

You get this

But I'm pretty sure this is the point where you get stuck

Now what gang

Halp me

Where have the math graduates gone. Please lead the way 🙏🏿

[fast osprey]

Im glad im still in highschool

[royal zinc]

Don't worry. The damnation will come soon enough for you to

[fast osprey]

😭

[royal zinc]

<@&286206848099549185>

[lament badger]

hi

[royal zinc]

Not hi dawg solve this shi😭🙏

<@&286206848099549185>

@slender ginkgo is you there bruh?

[lament badger]

it's kind of hard to understand 🧐

[royal zinc]

😞

[lament badger]

You can maybe type it out here and I can see if I can help you

Trying to figure out

[royal zinc]

Type of what tho

[maiden narwhal]

Sqaure root gonna be the main problem

[viscid dust]

I feel some symetry here firstly we can try to calculate primitives without worrying about existence by calculating the integral between epilon >0 and 1

[maiden narwhal]

Simplify the middle root term

The last and first are done

[royal zinc]

Nothing works

Ever😞

[maiden narwhal]

Now what

[slender ginkgo]

wait

its from mit integration bee

[royal zinc]

Lmao what

[slender ginkgo]

there must be some trickery

i finna cheat

imma be fr i cant solve it

[royal zinc]

Noooo

[royal zinc]

No buddy

1/x can't be integrated

[maiden narwhal]

.

[royal zinc]

In those limits

[maiden narwhal]

Blud it can

Yea

I get u

[royal zinc]

It just gives infinity

[maiden narwhal]

Theres a discontonuity

We can bream the integral

[slender ginkgo]

its mit integration bee last final problem 💀

[royal zinc]

No fucking way they decided to give me that question

[slender ginkgo]

https://www.youtube.com/watch?v=uy5NirhE3jw

[maiden narwhal]

.

[royal zinc]

Yoooooooo

Nice

.close

[cedar kiln]

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[slender ginkgo]

@royal zinc if they actually gave u this problem, u can prolly ask them to solve it themselves and see if they can solve it :D

10-minute solution involving inverse

[royal zinc]

Lol

[cedar kiln]

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[white rover]

can someone help me understand what the sum is representing. Because what I don't understand is if the partition length is more than 1. Then there is more than there are multiple values of x that will give 1 for the characterstici function. say our partition length is 4. And the value is 5. Then we have 4 x's that will give 1 for the characterstic function. When taking the sum we would be adding 5 four times which is not equal to f(x).

[idle tusk]

can you write out your example a bit more explicitly?

[white rover]

in what sense?

[idle tusk]

i am not sure what exactly is confusing you

[sinful bay]

Here's what the graph of the characteristic function of the interval (0,1) looks like.

[white rover]

ok so I don't think I am understanding the sum correctly

[sinful bay]

Here's the graph of 2 chi_(0,1)

[white rover]

these are of partiton length of 1 tho

[sinful bay]

I'm getting to that but my desmosing is slow

Sorry

[idle tusk]

let's say the interval is [0,1] and the partition is (0, 1/5, 2/5, 3/5, 4/5, 1)

[white rover]

ok

[idle tusk]

explain where you think the contradiction arises

in your example

[white rover]

so one partition would be from 0 to 3/5 with a value of 5 and the other from 3/5 to 1 with a value of 7

[sinful bay]

Here's chi_(1,2) and (1/2)chi_(1,2).

[idle tusk]

just to clarify, when you say "value of 5", you mean f(x)=5 there?

[white rover]

so the sum is saying that I plug in the values from 0 to 3/5 and multiply each of them by one since each of them are on the line. that means I will ad 5 four times

[white rover]

yes

[sinful bay]

Finally, here's 2*chi_(0,1)+(1/2)*chi_(1,2)

[idle tusk]

c_{k} are arbitrary

so you are not adding 5 four times in a row

[white rover]

what is this?

[idle tusk]

sorry i mean c_k

[sinful bay]

Importantly, for any $x$, the sum $\sum_{n\leq k} c_k \chi_{I_k}(x)$ will have at most one of the summands be nonzero

[wraith dagger]

Buzzing Hornet

[idle tusk]

it's the coefficients in front of chis

[white rover]

I was thinking they would only be non zero on the partition

[white rover]

but can't it be the same ?

[sinful bay]

Yes, the sum will be nonzero only on the partition

But the things that are being summed will be nonzero only on the corresponding interval

[sinful bay]

So in this case, the 2*chi_(0,1) part of the sum is nonzero only on (0,1)

[white rover]

but then it isn't possible for that sum to equal a c_k corresponding to f(x)

[sinful bay]

And its value is 2 on that interval

[sinful bay]

I'm not sure what you mean by this

Like, ok, take any x in the partition

There is exactly one interval I_n in the partition which x is contained in

[white rover]

by this you mean that x is in a sub interval?

[sinful bay]

Thus, the sum $\sum_{n\leq k} ck \chi{I_k}(x)$ is in fact the sum $c_10+c_20+\dots c_n \chi_{I_n}(x)+\dots c_k*0$

[wraith dagger]

Buzzing Hornet

[sinful bay]

Because x is in I_n, the c_n*chi_{I_n} term is just c_n times 1

And so this evaluates to c_n

[white rover]

I want to make sure that I am following. We have n partitions. Each partition can only have 1 x in it? If so then for each partition you have c_k * 1. Then you add those up

[sinful bay]

We have one partition into n intervals

[white rover]

sorry

ok but on each interval there is only 1 x? If so how?

[sinful bay]

No, for any x there is one interval

Because the intervals are disjoint

[sinful bay]

Like let's do this visually really quick

[white rover]

so we can have an interval (0,4) right?

[sinful bay]

The partition includes the points in (0,1) and the points in (1,2)

[white rover]

ok

[sinful bay]

This is our c_1 chi_{I_1}

If we add this function to another function, it will only contribute to that sum when x is in (0,1)

[white rover]

ok so your saying that for our first interval (0,1) the c_k is 2 and then we multiply by 1

[sinful bay]

We multiply by 1 only if x is in (0,1)

[sinful bay]

Here, we have c_2 = 1/2 and I_2 = (1,2)

[sinful bay]

And when we take the sum of those two functions, they don't interfere with each other, so to speak

Because when one of them is nonzero, the other one must be 0

And vice versa

[white rover]

isn't our step function only defined on these open sub intervals. Thus the characterstis function is defined on each sub interval for x

oh wait

[sinful bay]

The characteristic function is defined on all of R

But it's 1 when we're on (0,1), and it's 0 everywhere else

[white rover]

I think I am starting to see it. so your saying that if we plug in any x then the characterstic function will only be 1 on one of the intervals in the partition?

[sinful bay]

So yes, it's defined on our other interval (1,2)

[sinful bay]

Yes!

Well, ok, kind of

There are n different characteristic functions at play here

Each individual interval gets its own characteristic function

[white rover]

ok so your are adding a bunch of c_k * 0 then one c_k * 1

correct?

[sinful bay]

Yes exactly

[white rover]

ok but here is where my issue comes up. Say in our example we know how one interval that can take multiple x. By this I mean we have our partion from [0,6] then from (2,6) c_k = 3.

wait

[sinful bay]

I'm not sure what "can take multiple x" means

[white rover]

by that I mean that f(3) = 3 f(3) =3 etc. but is the sum saying that we we are just taking the c_k value from that interval. We don't care for how many x, x = c_k

[sinful bay]

x = c_k doesn't matter at all

The only thing that decides the value of f(x) is which interval x is in

[white rover]

ok

[sinful bay]

Ok perhaps making this more explicit might help

Take the function $g: (I_1\cup \dots I_n)\to {1,\dots n}$ where $x\mapsto k$ if $x\in I_k$

[wraith dagger]

Buzzing Hornet

[sinful bay]

There's nothing deep going on here, g just takes an element in the partition and tells us which part of the partition it's in

Now, we know that this is well-defined, right?

Because if I take $x\in I_1\cup \dots I_n$, the $I$'s are disjoint, and so there's exactly one $I_k$ such that $x\in I_k$

[wraith dagger]

Buzzing Hornet

[sinful bay]

Are you with me up to there?

[white rover]

just making sure. so we have the function g which is the unions of all the subintervals in the partition. Where each k is mapped to a certian value k.

[sinful bay]

No

[white rover]

this is saying that for any k you plug in it can only be in one sub interval

[sinful bay]

The domain of g is the union of the subintervals

[sinful bay]

You aren't plugging in k's, you're plugging in real numbers

Specifically real numbers that are in at least one of the intervals in the partition

[sinful bay]

Let's use our toy example up here

I_1 = (0,1), I_2 = (1,2)

For any x in I_1, g(x) is 1

So g(1/2) is 1, for example

For any x in I_2, g(x) is 2

So g(3/2) is 2, for example

[white rover]

sorry I meant x

[sinful bay]

It might help to convince yourself of the things I said after this

Ok cool thumbs up

Great

So by definition, if $x$ is in $I_1\cup \dots I_n$, then for every $k\leq n$, $\chi_{I_k}(x)=1$ if and only if $g(x)=k$

[wraith dagger]

Buzzing Hornet

[sinful bay]

And $\chi_{I_k}(x)=0$ if and only if $g(x)\neq k$

[wraith dagger]

Buzzing Hornet

[sinful bay]

Does that make some sense?

[white rover]

I have one question about k. So k is the about for g(x). but then we also use k as an index for the sub intervals?

[sinful bay]

Yes

[white rover]

but how do we know that the output will be the same as the sub interval index?

[sinful bay]

Remember, up here, g(x) is the index of the interval that x is included in

If $x\in I_1$, $g(x)=1$

[wraith dagger]

Buzzing Hornet

[sinful bay]

If $x\in I_4$, $g(x)=4$

[wraith dagger]

Buzzing Hornet

[white rover]

ok ye so you defined our function as be the function that takes in x for sub interval k and outputs k

[sinful bay]

If $x\in I_k$, g(x)=k$

[wraith dagger]

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

[sinful bay]

Yeah basically

[sinful bay]

So all that this is saying

[white rover]

does that make our function to specfic though and not general enough ? because coulnd't there be a function say x is in the I_1 but g(x) = 6

[sinful bay]

Ok I'm not sure you understand how we defined g

I'm saying that we're literally defining the function so that if $x$ is in the interval $I_1$, then $g(x)=1$

[wraith dagger]

Buzzing Hornet

[sinful bay]

Like, that is (part of) the definition of g

[white rover]

but doesn't that then limit what we can apply this theorem to? Doesn't that limit us to only step functions of this defintion. ins't the oringal question says that it takes a value c_k on the kth interval it doesn't say that c_k msut equal k

[sinful bay]

Yes g is not f

I'm using g to build up to how f works

[white rover]

ok

[sinful bay]

Like, what I want to build up to is that if $x$ is in the partition, then $f(x)=c_{g(x)}$

[wraith dagger]

Buzzing Hornet

[white rover]

ok then I think I understand how g works. G is saying that if we have g(x) = k. then x is in the k sub interval and then the characterstic function is only 1 on that interval

[sinful bay]

Yes!

If $x$ is in the partition, then $\chi_{I_k}(x)=1$ is exactly the same statement as $g(x)=k$

[wraith dagger]

Buzzing Hornet

[sinful bay]

Because chi_{I_k}(x)=1 says that x is in I_k

And g(x)=k says that x is in I_k

So $f$ is defined as the sum $c_1\chi_{I_1}(x)+c_2\chi_{I_2}(x)+\dots c_n\chi_{I_n}(x)$, right?

[wraith dagger]

Buzzing Hornet

[sinful bay]

Well, the $g(x)$th term in that sum is $c_{g(x)}\chi_{I_{g(x)}}(x)=c_{g(x)}\cdot 1$

[wraith dagger]

Buzzing Hornet

[white rover]

this statement I am a bit confused about because we don't know if k= 1. woudln't we have to say k * X_ik(x) = f(x) iff X_ik(x) =1

[sinful bay]

It doesn't matter whether or not k=1

[white rover]

oh wait nvm I misread

ye g(x) = k only if it is in the interval so the charactersifc function is 1

[sinful bay]

Mhm

[sinful bay]

So this you probably get

[sinful bay]

But tell me if you need some more explanation of this

[white rover]

isn't this the opposite of what we just said. isn't the charactersteif function 1 if it is in the sub interval I_k and 0 if it is not

[white rover]

havent got there yet going down

[sinful bay]

Should have said 1, sorry

[white rover]

ok then let me continue reading down

[white rover]

yes since f(x) = k and this will equal k plus a bunch of 0's

[sinful bay]

What

No, this is just the definition of f

[white rover]

ok so we know that x can ony be in one of the intervals k

so then we have k*1

[sinful bay]

That's just me expanding out $\sum_{k\leq n} c_k \chi_{I_k}(x)$

[wraith dagger]

Buzzing Hornet

[white rover]

since x is not in any of the other intervals the characterstic funciton for those intervals is 0

[sinful bay]

Not k

c_k

[white rover]

yes sorry

so my only question now is that for this we started off with our function g. And then used it to build up to our function f. However I am confused on how we can act upon the prinicple of going from something specifc to something more general. Isn't it the otherway around?

[sinful bay]

There's no generality or specificity here

[sinful bay]

g was a tool so that I could say this

[white rover]

but we define g as the step function where g(x) = k for each k sub interval.

[sinful bay]

Basically, I wanted to be able to specify in symbols exactly which term in that sum was nonzero

[sinful bay]

g shouldn't really be thought of as the same kind of object as f

We should think of it as returning indices

[white rover]

what do you mean by returning indicies?

[sinful bay]

(The index of the interval which contains the input)

[sinful bay]

To restate the definition, it returns the index k of the interval I_k that contains the input x

[white rover]

yes

[sinful bay]

And that means that g(x) tells us exactly which term in the sum we care about

[white rover]

which is the k interval

[sinful bay]

It's the g(x) term

Which if x is in I_k is the k term, yeah

[white rover]

but can't their be step functions that are not defined this way?

[sinful bay]

I don't understand the question

[white rover]

say we have a step function where g(x) = 5 on the first sub interval

that means it is not a step funciton like g

[sinful bay]

Stop thinking about g as a step function

[sinful bay]

See here

[sinful bay]

And here

[white rover]

ok

[sinful bay]

Literally all g does is it tells us which term in this sum we care about

[white rover]

ok say g is not a step function. shouldn't what we are trying to prove also apply to a step function f where like i said earlier f(x) = 5 on the first sub interval. how could we use the sum of g for that?

[sinful bay]

There aren't any sums involving g

[sinful bay]

This expression is literally all that g is for

Notice that g only appears in the indices

Because what g does is it returns indices

[white rover]

so we don't need g at all then for defining f wiht the sum

[sinful bay]

Yes, f is defined independent of g

But $f(x)=c_{g(x)}$

[wraith dagger]

Buzzing Hornet

[sinful bay]

Which is the whole point

[white rover]

ins't it just c_k

[sinful bay]

It's c_k if x is in I_k

But you were having a lot of trouble with that idea earlier

Because of category errors and quantifier order

[white rover]

but c_k value does not at all correspond to the k sub interval

[sinful bay]

It depends on what you mean by "correspond to the k sub-interval"

[white rover]

by that i mean c_k can be 5 and it can be on the first sub interval

[sinful bay]

Yes that's true

But that's fine

That means that c_1=5

And then if x is in I_1, then f(x)=c_1=5

[white rover]

yes

[sinful bay]

Yes

That's all f does

c_k doesn't have to be related to k at all

Think of f sort of like a computer program

First, f determines the unique number k such that x is in I_k

And then f returns c_k

[white rover]

ok

[sinful bay]

That's it

That's really all that f does

[white rover]

ok makes sense

thx

.close

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[torn marsh]

confused in the second line

[stiff brook]

divide both sides by sinA

[torn marsh]

how did it transition into the second line

[twilit escarp]

and then by sinC

[torn marsh]

so thats gonna be

fraction over a fraction at one point

[idle tusk]

maybe $\frac{c}{\sin C} = \frac{a}{\sin A}$ would look more familiar to you?

[wraith dagger]

artemetra

[twilit escarp]

not neccessarly

[idle tusk]

if you imagine it's that instead of the second line

[slender ginkgo]

\begin{align*}
c\sin A &= a\sin C\
\frac{c\cancel{\sin A}}{\cancel{\sin A}} &= \frac{a\sin C}{\sin A}\
c &= \frac{a\sin C}{ \sin A}\
\frac{c}{\sin C} &= \frac{a\cancel{\sin C}}{\cancel{\sin C}\sin A}\
\frac{c}{\sin C} &= \frac{a}{\sin A}\
\frac{a}{\sin A}&= \frac{c}{\sin C}
\end{align*}

[torn marsh]

im confused in the highlighted part

[wraith dagger]

k

[torn marsh]

cuz im just wondering how that is possible

[dusky panther]

devide both sides by sin(c)

assuming it's not 0

[slender ginkgo]

[ \frac{\frac{a\sin C}{\sin A}}{\sin C} = \frac{a\sin C}{\sin A} \cdot \frac{1}{\sin C} = \frac{a\cancel{\sin C}}{\sin A} \cdot \frac{1}{\cancel{\sin C}} = \boxed{\frac{a}{\sin A}}]

[wraith dagger]

k

[slender ginkgo]

@torn marsh

does this make sense to u

[torn marsh]

Same process with sin C right

YES

caps

[slender ginkgo]

yup

[slender ginkgo]

ur working is similar to mine

[torn marsh]

but is this owrking still corect?

working out*

[slender ginkgo]

u skipped a lot of steps

[rapid mauve]

my brother in whichever god you pray to, i think you might wanna practice rearranging formulas

[wicked mantle]

@torn marsh

[rapid mauve]

i think he intended to divide both sides by sin C

[slender ginkgo]

ye

[slender ginkgo]

from this

but its uhh

ye..

[wicked mantle]

unclear, better him explain himself.

[rapid mauve]

fair

[slender ginkgo]

jump lots of steps

[cedar kiln]

@torn marsh Has your question been resolved?

[cedar kiln]

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[vital lodge]

Let N denote the set of positive integers. A function f : N → N is said to be bonza if
f (a) divides ba
− f (b)f (a)
for all positive integers a and b.
Determine the smallest real constant c such that f (n) ⩽ cn for all bonza functions f and all positive
integers n.

[vital lodge]

i mean , not to take this one very seriously though

[cedar kiln]

@vital lodge Has your question been resolved?

[vital lodge]

<@&286206848099549185>

[astral barn]

let me take a look

[vital lodge]

please do

[astral barn]

so i would first want to try the constant function where f(n)=1, and obviously here f(n) is bonza since (b-f(b))/1 works.
next i would want to try f(n)=n. here we have (b^a-b^a)/a=0 and this works too.
now lets suppose f(n)>cn and look for contradictions. lets say f(n)=n+1. then we have (b^a-(b+1)^(a+1)/(a+1). trying a=1 we get (-b^2-b-1)/2 which is not bonza.
this leads us to believe no bonza function grows faster than f(n)=n
now we just need to prove that f(n)>n leads to contradiction.
lets take a=n, f(n)=k, and k>n. we then have (b^n-f(b)^k)/k for all b. letting b=1, we get (1-f(1)^k)/k
suppose f(1)=1. then we have (1-1)/k which works
suppose f(1)=2. then we have (1-2^k)/2 which doesnt work since if we try trying k=4 leads to failure
therefore, growth beyond f(n)=n violates the divisibility condition for some values. hence, f(n)<=n must hold. so, c=1

[vital lodge]

i mean u are on very hard road

[astral barn]

?

[vital lodge]

i mean do u know about international mathematics olympiad

[astral barn]

ive heard of it

[vital lodge]

this is one of the problems asked in that

which happened this year

[astral barn]

oh cool

[vital lodge]

no need to worry about this problem

[astral barn]

is my solution wrong?

[vital lodge]

it just needs some fermat lil theorm

[vital lodge]

no, its just you were trying very hard and thats kinda very good of you

[astral barn]

👍

[vital lodge]

thanks for trying.

[astral barn]

ofc

[vital lodge]

.close

[cedar kiln]

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[tawdry sonnet]

hi can someone explain to me why it’s negative? i did the math but i got a positive answer so i’m really confused

[cerulean sail]

Do you happen to have what you did, and are happy to share it?

[tawdry sonnet]

yes!

it’s a little messy

[crimson delta]

you subtracted them in the wrong order

[tawdry sonnet]

ohhh

okay that makes sense

thank you

[cedar kiln]

@tawdry sonnet Has your question been resolved?

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[normal creek]

i need itsy bitsy help
the question says without using the calculator work out 6/7 ➗ 5/3 and show all the working blah blah and give your ans in lowest form what is meant by lowest form = i got ans 18/35 is that all or should i do something else

[cedar kiln]

Please don't occupy multiple help channels.

[wicked mantle]

Stick to your original channel

[dull oxide]

#help-46

[wicked mantle]

.close

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[sinful bay]

Which question are you asking for help on?

[alpine siren]

Question 19

[oblique flare]

log rules esp. on the left side

[oblique flare]

just a question though log here is base 10 or e ?

[alpine siren]

10

[alpine siren]

<@&286206848099549185>

[dire geode]

Use log and exponent rules

,tex .log rules

[wraith dagger]

riemann

[dire geode]

,tex .exp rules

[wraith dagger]

riemann

[alpine siren]

But there’s a + in between the value of log

I was not able to solve it

[topaz crow]

try using one of the log rules to the rhs so that it becomes the form of
log(something)=log(something)
from which you can remove the log from both sides

[alpine siren]

Okay I’ll try again

What to do after this ?

[cedar kiln]

@alpine siren Has your question been resolved?

[pastel vault]

somehow $x$ must satisfy $x^2 - x/2 - 7 = 0$

[wraith dagger]

south

[pastel vault]

I'm trying to reverse-engineer the answer

if $x^2 = ax + b$ then $4^{(ax + b)/2} + 4^{x + 15} = 4^{(a + 1/2) x + (b + 8)}$

but then this isn't a quadratic in $4^{x/2}$ cause $4^{x + 15}$ is not a constant

[wraith dagger]

south

[topaz crow]

divide the equation by 4^(x+15)

[pastel vault]

and then?

[topaz crow]

let u= (quadratic in terms of x)
then solve the quadratic in terms of 4^u

i think this works

[pastel vault]

yeah okay this actually isn't true: the roots are close but not quite

you get 4^(x/4 + 7/2) + 4^(x + 15) = 4^(x + 15)

[alpine siren]

Not able to solve

What to do with this one?

[pastel vault]

oh wait, so this is a quadratic in 4^(x^2 - x/2)

[pastel vault]

let u = 4^(x^2 / 2 - x/2)

[alpine siren]

That’s not equal to equation or am I missing ? Can you write on paper and show next steps

[pastel vault]

wait...

okay so you have x^2 / 2 - x and x^2 - x/2 in the exponents

yeah that doesn't match up

I think the question is faulty then

like if you let u = 4^(x^2 / 2 - x), you have u^2 = 4^(x^2 - 2x) which is different

unsolvable

[topaz crow]

yeah i also thought the exponents would match up but i got careless sorry
then i think it's a typo of the question (should be times instead of + in the lhs)

[pastel vault]

yeah

of course the intended solution is like this

when I checked on Desmos though it didn't match up so I was questioning everything

[alpine siren]

So do this even have solution?

Answer is -30 but idk how

[topaz crow]

there is an answer but is probably impossible to find without computer aid

also even if we assumed lhs is times instead of + it wouldn't be -30 (it would be -14)
so i'm not sure what the question was intended to be like initially

[alpine siren]

I don’t know.. but people solved it somehow

without computer and in 2 min

Can someone explain ?

[topaz crow]

i think that's wrong (not sure what he/she meant by possibility=1,2,4,8...)

wolfram alpha suggests that the product for the original question is very close to -7 (but not exactly -7 i believe)

[alpine siren]

Ahh

So there’s no solution

[alpine siren]

This is as per book. Arun Sharma - CAT

[topaz crow]

it was probably some kind of typo for the original question
not sure what the solution above means though

[alpine siren]

Maybe

[alpine siren]

I’ll close it then

.close

[cedar kiln]

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[true sapphire]

hello

[true sapphire]

x intercepts were 0 as discriminant was -4.

to fing turning point, i did completing square

method

-(x-2)^2 +1

[tropic oxide]

mmm think you screwed up the signs somewhere?

[true sapphire]

so x is +2 and y is +1

[tropic oxide]

-(x-2)^2 + 1 expands out to -x^2 + 4x - 3

you should have had y = -(x-2)^2 - 1

[true sapphire]

yess

-1

[tropic oxide]

so the vertex is at (2, -1) not (2, 1) as you just wrote.

[true sapphire]

but im confused

[tropic oxide]

what's confusing you

[true sapphire]

this is what they gave

[tropic oxide]

yeah?

the vertex is at (2, -1). that's where it's gonna land.

[true sapphire]

why is it saying (0, -5)

[tropic oxide]

0 not o.

(0, -5) is the y-intercept.

not the vertex.

the mark scheme checks for correct placement of the vertex AND correct marking of the y-intercept

giving a separate mark for each

[true sapphire]

thank you

[tropic oxide]

also you should always always always put parentheses when writing down the coordinates of a point

[true sapphire]

i did it

I will close this now

.close

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[edgy spade]

Define a set A on Rational numbers. If there exists an upper bound C belonging to Q, does that guarantee an existence of Supremum belonging to Q as well?

[edgy spade]

Also, if i were do change the definition of set A on real numbers instead of rational numbers and everything else with real numbers as well, does that make the statement true?

[raven shard]

yes: https://en.wikipedia.org/wiki/Least-upper-bound_property

[civic pumice]

you can try with some examples

[edgy spade]

Yes to what

[raven shard]

your second question

the first is fairly simple; think on it for a few minutes

[edgy spade]

Ohh

Ahh

Okay

[civic pumice]

try your first question with the set {x< sqrt(2)} for example

[edgy spade]

A = {x: x^4 < 7}

[edgy spade]

Yeah

Okay so its extreme bounds are irrationals

So lub property isnt satisfied here

[civic pumice]

its supremum is irrational

[edgy spade]

So what criteria is required for a set to satisfy the supremum property

[civic pumice]

but you can find a rational upper bound in Q like 2 for example

[edgy spade]

Yes

Wait is the supremum property only defined for sets on real numbers?

For any other set, we cannot say for sure?

[civic pumice]

its defined for ordered sets

if you can find an order, you can switch out < for your order and the definition holds

[edgy spade]

What are orders

Isn't Q an ordered set as well then?

[civic pumice]

you can order Q with <

[civic pumice]

I was answering this

you can define a supremum if you have an order

[jovial anchor]

yeah there's a difference between defining supremums and asking whether or not a set has a supremum, i suppose

[civic pumice]

yeah

[jovial anchor]

there are subsets of Q that do have a supremum in Q

[civic pumice]

but to answer your question, if you want your original question to apply for a certain set, the set must be closed under supremums

as in every subset of that set must have its supremum in the original set

[edgy spade]

Well then shouldn't the answer be yes to my first question

[civic pumice]

no because Q isnt closed under supremums

[edgy spade]

?

[jovial anchor]

as in, "supremum" is not generally a function that takes in a set and spits out a number

[civic pumice]

for example {x<sqrt(2)} where you take all x in Q is a subset of Q but its supremum isnt in Q

[jovial anchor]

it's more like a property that a number can have wrt a subset

so you can ask the question "given a subset A, is x a supremum of it?"

[edgy spade]

Yes we're already past this point

[jovial anchor]

but generally until you show that there is a supremum, "sup A" doesn't necessarily make sense

[civic pumice]

I was answering your question

I know we already found an answer but you said "shouldnt the answer be yes", and I said no because...

oh wait I misread what you answered to

my bad my bad

[edgy spade]

I based my statement on you saying that supremum property applies pn ordered sets and that Q is ordered set

[edgy spade]

Np

[jovial anchor]

i think the idea is that you can define the notion of "supremum" once you have an ordered set

[civic pumice]

I wasnt saying if you can define a supremum then it verifies your original question with two different sets

[jovial anchor]

but this doesn't guarantee that an ordered set will have all suprema

[civic pumice]

yes that is what I was trying to say

[jovial anchor]

if you just have an arbitrary set $X$ and a subset $A \subseteq X$, then it doesn't really make sense to ask whether $A$ has a supremum

[wraith dagger]

Pseudo (Cat theory #1 Fan)

[edgy spade]

So theres no specific criteria for a set to satisfy the supremum property?

[jovial anchor]

to illustrate the point, "the colour of an apple is white" is false, because the actual colour is red or green typically

but "the colour of 3 is white" is also false

but they are false for different reasons

[edgy spade]

Like a set can be ordered( like Q) but still not satisfythe property
But a set can be unordered but still satisfy the property?

[jovial anchor]

this is false not because the actual colour of 3 is something else, but rather that 3 doesn't have a colour

[jovial anchor]

in this case, asking about the supremum property for an unordered set is like asking whether the colour of 7 is green

it's not the kind of thing the question applies to

whereas asking whether Q has the supremum property is like asking whether a fruit is green

[edgy spade]

But im asking it because I want to know, not because im trying to prove its a non sensical thing to ask

[jovial anchor]

it may be true or false, but it definitely is the kind of thing the question applies to

[jovial anchor]

only ordered sets can satisfy the supremum property

[edgy spade]

Hhmm okay

[jovial anchor]

unordered sets can't in the same way that numbers can't be green

[edgy spade]

Oh

I see

I understand

[jovial anchor]

it's not that their actual colour is red, it's that they don't have a colour

[edgy spade]

Unordered sets dont have a supremum?

[jovial anchor]

they don't have a notion of supremum

in the same way that numbers don't have a notion of colour

[edgy spade]

Isn't it the same wrt unordered sets

If they dont have a notion of supremum, why define it

Why would it exist

[jovial anchor]

i don't understand your question

the notion of supremum is defined for ordered sets only

[edgy spade]

Having a notion of supremum vs having a supremum, what is the difference between these wrt unordered sets

[jovial anchor]

unordered sets don't have a notion of supremum

[jovial anchor]

.

[edgy spade]

I kind of get it

[jovial anchor]

as i said before

[edgy spade]

I think we're heading more into a linguistic depth than a mathematical one

[jovial anchor]

sure but this is an important mathematical point

sometimes the answer to a question is "true" or "false"

but sometimes, it doesn't make sense to ask the question

and that is different to the answer being "false"

[edgy spade]

Ok ok

Thanks a lot

[jovial anchor]

asking "is this fruit green?" is something that could be true or false

asking "is this number green?" is one where it doesn't make sense to ask the question

[edgy spade]

[jovial anchor]

💖

[edgy spade]

Thanks a lot both of you

.close

[cedar kiln]

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Use .reopen if this was a mistake.

[cedar kiln]

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[spark mist]

So what I have done is: first, i assume there is another set of integers (x', y', z') that satisfy the equation, and let x'+y'+z' be minimal we are going to find that there is another set of interger that also satisfy the equation but being smaller hence we find the contradiction

[spark mist]

wlog x and y and z are all positive

since if only 1 are negative it cant work

then assume that x > y > z

construct the quadratic in term of x

to find the other root

[spark mist]

f(t) = t^2 -2yz(t) +y^2 +z^2

the first root is x let the second root be a

we get 2yz - x = a

and (y^2+z^2 )/x = a

then i need to prove that a<x

[spark mist]

which i couldnt do it

[feral sapphire]

fair but you need to justify the wlog

[spark mist]

is this part not sufficient ?

i mean im practicing to write proof so i could still be wrong

[feral sapphire]

what if 2 are negative, what if three, and what if some are 0?

[spark mist]

if it is zero then all of three will be zero

hence satisfying the equation

i mean if suppose x and y are negative

then it will be same as if x and y are positive

if 3 are negative then it still cant work since the lhs would be positive while the rhs will be negative

[feral sapphire]

perfect, its good to write down these details sometimes for practicing

especially if u r practicing proof

[spark mist]

thanks

for the tips

[feral sapphire]

you mean a<x?

[spark mist]

yeah im sorry

[feral sapphire]

you extracted the conditions 2yz-x=a and (y^2+z^2)/x=a. Are these conditions sufficient to deduce a<x? One way to see is to plug in some values, simple values like x=1, y=1,z=1.

it might be helpful to think about the discriminant, which is 4y^2z^2-4y^2-4z^2

[cedar kiln]

@spark mist Has your question been resolved?

[spark mist]

so 1 proof i try is that

first

2yz - x < x we get 2yz<2x -> yz < x

then (z^2+y^2)/x < x

z^2+y^2 < x^2

and also (yz)^2 < x^2

z^2 + y^2 < (yz)^2 <x^2

z^2+y^2 < (yz)^2 would be false if

z or y = 0, 1

but since they cant be 0 we can ignore the case

so lets tst y or z = 1

pluggin it in

2y-x < x

2y < 2x

y < x

since we wlog z < y < x

man im confused

[feral sapphire]

are you saying "for the sake of contradiction assume x<=a"?

in which case it should be 2yz-x<=x

[spark mist]

uhh 2yz-x = a right ??

[feral sapphire]

oh wait yeah

[feral sapphire]

why are you using a<x when you haven't proved it though

[spark mist]

huff

i forgot

😭

[feral sapphire]

this should be x>=y>=z

it is possible they are equal

mb i haven't pointed that out

[spark mist]

wait i forgot

this question is like an extenstion

for another question

the proof is that let y=z

then we have x^2+2y^2 = kxy^2

[feral sapphire]

and what I am trying to say here, the three conditions "2yz-x=a, (y^2+z^2)/x=a, and x>=y>=z" for positive integers x,y,z,a is insufficient to deduce that a<x

[spark mist]

oh..

[feral sapphire]

because you can take x=y=z=a=1

all condition holds but yet a is not less than x

[spark mist]

i see

[feral sapphire]

This simply means that you have not extracted enought information from the original condition that x^2+y^2+z^2=2xyz

[spark mist]

oj

oh..

i mean here

[spark mist]

repeat all the step that we have done

then we need to deduce that a<x

what are going to do is

we are going to prove that y is between x and a

and since y < x

we are going to prove that a < y hence a < x

f(y) = y^2-ky^2z+y^2+z^2 < 3y^2-ky^2z = y^2(3-kz)

here i assume that k > 3

then the equation will hold since

f(y) <=0

then y will be between x and a

hence proving it

but here i havent prove for case k =2

which is the same as the first question

[feral sapphire]

y<=x

[feral sapphire]

it is possible x=a=y

[spark mist]

no it isnt

so the proof is

suppose y = z

then the equation become x^2 + 2y^2 = kxy^2

[feral sapphire]

aha ok

[spark mist]

hence y | x

then x = yL

plugging back and rewriting we will get

L^2 -kyL +2 = 0

the discriminant would be some square number to make the equation have interger root

so we get k^2y^2 - 8 = u^2

checking 1 by 1 we found that the only posible solution would be k = +- 3 or +-1

since we cant have negative

we have k = 3,1

which we also cant have

hence the three number must be all distinct

so thats why we can wlog x > y > z

[feral sapphire]

this step is unclear

you need more detail

[spark mist]

uh

since

x^2 + 2y^2 = kxy^2

then

it would have the same modulo residue

x^2 + 2y^2 = x^2 mod y

[feral sapphire]

mod what? y?

ok

[spark mist]

kxy^2 = 0 mod y

[blazing zephyr]

mod y isnt enough

[spark mist]

mod y^2 ??

wait

[feral sapphire]

u stopped strong inducting?

[blazing zephyr]

dunno how to continue

[spark mist]

yeah we still proved that y | x

so how should i prove that k = 2 work 😭

[feral sapphire]

so you get x^2=0 mod y^2 and y^2|x^2. can you prove then y|x?

[spark mist]

x^2 = ky^2