help-13

[royal loom]

quick example

what is the column space of $\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$

[wraith dagger]

Austan

[noble perch]

so like

(x,y)

cuz you can add em to get like a two dimensional plane

[royal loom]

I wouldn't write it like that, but I see what you mean. Say R2 instead

[noble perch]

oh

[royal loom]

all 2D vectors

[noble perch]

right

[royal loom]

(real 2D vectors)

Okay but what if I added another vector

what if it was

$\begin{bmatrix} 1 & 0 & 1\ 0 & 1 & 1 \end{bmatrix}$

[wraith dagger]

Austan

[royal loom]

what's the column space of that?

can we get anywhere new? is there any more linear combinations we can reach that we couldn't before?

[noble perch]

well the vector in column 3 is linearly dependent on the first two columns

[royal loom]

yup perfect

so no

we can't

the column space is unchanged if we ignore dependent columns

so when we row reduce a matrix and find its pivot columns

we find out what columns are dependent

we ignore these

and make our column space from the pivot columns

[noble perch]

oh okay

[royal loom]

$$A=\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$$
$$B=\begin{bmatrix} 1 & 0 & 1 \ 0 & 1 & 1 \end{bmatrix}$$
$$C(A)=c_1 \cdot \begin{bmatrix} 1 \ 0 \end{bmatrix} + c_2 \cdot \begin{bmatrix} 0 \ 1 \end{bmatrix} = C(B)=\mathbb{R}^{2}$$
Because the pivot columns of $A$ and $B$ are the same.

[wraith dagger]

Austan

[royal loom]

C(A) meaning column space of A

C(B) meaning column space of B

[noble perch]

yeah this makes a lot of sense

[royal loom]

now if you understand column space, row space should come easy. It is the same idea but for the rows

if we want to know the row space of our matrix

all we have to do is make the rows columns

and find the column space of that

[noble perch]

transpose

[royal loom]

yes

exactly

[noble perch]

right

[royal loom]

So are you feeling good about all of this?

[noble perch]

so what would we do if the question asked us for the basis of the column space

[royal loom]

do you know what a 'basis' is?

[noble perch]

pretty good actually this has been an enlightening discussion

[noble perch]

well i know it has two properties: linearly independent and they span the space

[royal loom]

a basis of a space, is a set of linearly independent vectors, that span a space

yes

so for R2

a basis can be

<1,0>, <0,1>

right?

[noble perch]

yeah

[royal loom]

those are linearly independent

they span R2

so they form a basis for R2

how about

<5,1>, <0,8>

do those form a basis for R2?

[noble perch]

id say they do

[royal loom]

yes

actually there can be infinitely different choices of basis's for any given space. Usually we like to choose the simplest one possible

for R2 we would normally choose <1,0>, <0,1>

right

okay

so earlier we said that

the column space, is the space spanned by the linear combinations of your columns

if we want a basis for the column space

we want to find linear independent vectors, in our column space, whose linear combinations fill the space

you should have a good idea of where to find these

[noble perch]

oh so in a test, i would probably be asked to tell them a basis or a few bases since there are infinitely many

[royal loom]

yes they will 99% likely only ask you for one basis

[noble perch]

ah okay

let me do a problem on bases and i'll be back

[royal loom]

okay, I have to go for now, but my final hint to you is that there was a reason we found the pivot columns

you can DM me if you need more help but I wont get to it until later

[noble perch]

because they are the basis?

[royal loom]

not in REF no

but the columns in the original matrix, that are pivot columns in REF

yes

[noble perch]

ah okay

[noble perch]

alright thanks man i actually have another small question, do you want me to dm it to you?

.close

[cedar kiln]

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[umbral dew]

!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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[crimson sedge]

1

1

[dire geode]

Differentiate the right side and use FTC

[crimson sedge]

FTC

?

[dire geode]

,tex .FTC1

[wraith dagger]

riemann

[dire geode]

F' = f

[crimson sedge]

Oh

Ok

How do you know that but are you an iitian

Ok thanks @dire geode

.close

[cedar kiln]

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[grizzled scarab]

hey, im trying to prove that +mod3 is associative and having some trouble

[grizzled scarab]

binary operation on the group {0,1,2}

I end up with versions of this and I can't really get it to be the same on both sides but my book says otherwise, also checking it seems like it is associative
Any help?

[crimson delta]

you cant just divide by 3 like that

[nimble mulch]

what specifically do you want to prove

• or *?

[grizzled scarab]

Yea I was a bit confused because I technically need to take the remainder so I am not really sure how to approach this

[nimble mulch]

ok you'd rather do ([a]+[b])+[c] ≡ ([a+b])+[c] ≡ [a+b]+[c] ≡ [a+b+c] ≡ [a]+[b+c] ≡ [a]+([b+c]) ≡ [a]+([b]+[c])

you usually never specifically apply mod n or division by n in the modulo set for n

[grizzled scarab]

not sure I really understand this

[nimble mulch]

I presume you're familiar with the notation [x] in a Residue Class Ring

[grizzled scarab]

I am not 😅

[nimble mulch]

e.g. in the Ring Z/5Z: [0] ≡ [5] ≡ [10] ≡ [15] ≡ ...

let me adjust that for the above proof as well

@grizzled scarab if you just take this general notation the proofs of the properties of the Ring become simpler

[grizzled scarab]

Idk what you mean by ring, I'd try to read English is not my native as well

But I probably can't use that in my course since I didn't learn it

[nimble mulch]

what's your native, possible they match

[grizzled scarab]

I speak Hebrew

[nimble mulch]

kk nvm

[grizzled scarab]

Is there another way to prove this? Kinda in the way I did it that makes sense at all?

[nimble mulch]

it's objectively better as it also circumvents you having to define what a,b,c are in your proof

[nimble mulch]

are a,b,c elements of the set {0,1,2} for your proof?

[grizzled scarab]

Yes

I forgot to define

[nimble mulch]

I mean if you work with modulo operations you'd usually also at least have to be able to use equivalence classes

which are denoted by [x]

[nimble mulch]

therefore I'm not sure what you're allowed to use

you need to be able to somehow intertwine addition and the modulo operation, which you'd do via equivalence classes

@grizzled scarab like could you provide what operations and properties you've already learnt / area allowed to use

[grizzled scarab]

Pretty much nothing, it's like first course of uni and this is the beginning but this course assume no previous background

[nimble mulch]

the approach of rewriting a modulo addition as an addition and integer division seems rather sloppy

[grizzled scarab]

I'll have to look into the stuff you said and ask my teacher

I appreciate the info

[nimble mulch]

If you have been introduced to modulo, you also at least need to know equivalence, otherwise you can't calculate :D

since within Z/3Z, the number 0 suddenly is equivalent to 3

as well as to 6, 9, etc.

so you need to define this behavior as equivalence

and then you end up with equivalence classes @grizzled scarab

{0,3,6,9,12,15,...}

{1,4,7,10,13,...}

{2,5,8,11,14,...}

all numbers within these sets are equivalent to each other within Z/3Z

[grizzled scarab]

Yea they didn't really do that I guess, just said you divide and take remaineder = modulo

If I understand what you mean

[nimble mulch]

yeah but once you actually want to prove anything, you can't apply the numeric method to the proof

with your half addition symbol

[grizzled scarab]

I think I have to ask my teacher for clarification

[nimble mulch]

this part here is clear right?

[grizzled scarab]

But you gave me a direction now

Yes

[nimble mulch]

k, because then the only missing step is

that you want to summarize these equivalence classes with a notation

because 0 would have the same meaning as 3 etc.

which is where the notation [n] comes in

so [0] stands for the equivalence class of 0

[grizzled scarab]

Ah I think I understand you now

[nimble mulch]

when you write [0]+[1]

you don't "add" sets together, but you add equivalence classes

[0]+[1] = [1]

[2]+[2]=[4]=[1]

[grizzled scarab]

I'll read about it some more and look for more examples too regardless

[grizzled scarab]

Yea I think it's kinda what I meant here

[nimble mulch]

kk, I'll again refer to the initial proof I provided, because that's how you'd prove it almost always

most definitely not with translating division and taking the remainder using symbols

[grizzled scarab]

Thanks again for all the info!

[nimble mulch]

np, the main reason it's easier is that you can combine brackets:

[2]+[2] = [2+2] = [4] = [1] (within Z/3Z)

so yeah, I'd just write down this proof, you'll quickly grasp it

[cedar kiln]

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[cedar kiln]

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[royal loom]

@coarse shadow Stick to one channel

post your question here

[cedar kiln]

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[jaunty mural]

[royal loom]

[coarse shadow]

with this

[royal loom]

this channel is going to close soon because you deleted the original message

you need to open a new channel for your questio

!help

[cedar kiln]

Please read #❓how-to-get-help

[coarse shadow]

how do I add images because I need to upload the math problem

[dire geode]

https://support.discord.com/hc/en-us/articles/211866427-How-do-I-upload-images-and-GIFs-

[cedar kiln]

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[bold totem]

$\frac{dy}{dx}=\frac{sin^3x}{tan^2y}$

[cedar kiln]

Please don't occupy multiple help channels.

[wraith dagger]

Xiyyr

[bold totem]

(differential equation)

[stuck walrus]

It's separable

[magic solar]

Multiply by tan ^2y to the LHS and move the dx to the RHS and integrate

[cedar kiln]

@bold totem Has your question been resolved?

[cedar kiln]

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[teal kraken]

what does « determine the roots unity » mean?

[stuck walrus]

Determine values of x for which x^n = 1

[teal kraken]

ok

so just

roots of unity?

[stuck walrus]

isn't that the question lol

[teal kraken]

ik that one just not sure what my teacher means 😭😭

[stuck walrus]

might have missed out an "of" probably a typo

[teal kraken]

that makes sense

thank you!

.close

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[cedar kiln]

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[young dome]

With what

[cedar kiln]

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[cedar kiln]

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[chrome atlas]

Can I get help with Statcrunch statistics software?

[flint plinth]

fyi, there is a #computing-software channel in case you don't get any luck here

(i don't know the software myself)

[chrome atlas]

ok thank you

.close

[cedar kiln]

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[cedar kiln]

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[spare narwhal]

I need help with this question

[tropic oxide]

!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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[tropic oxide]

@spare narwhal

[spare narwhal]

2

[tropic oxide]

show current progress

[tropic oxide]

right ok

so far so good except for the way you write your fives and sevens.

[spare narwhal]

Wym

[tropic oxide]

your 7's should have a stroke through them, and your 5's should not be looking like S's

[spare narwhal]

How should I write a 5 then

[tropic oxide]

two strokes

one moment, let me show you

[spare narwhal]

Ohh

Lemme try it and you tell me if it looks better

[tropic oxide]

ok

stroke your sevens also

[spare narwhal]

Okay

[tropic oxide]

once you fix all this:

solve for x from the equation 3x+c = 5x - 13/2, and solve for y from the equation 3y + c = 5y - 7.
both x and y will be expressed in terms of c.

[tropic oxide]

the leftmost 5 and 7 look the best

[spare narwhal]

I've always written my 5's like the other way so this is a change

[spare narwhal]

Okay

[tropic oxide]

well it won't do for math because you don't want to ever confuse two different symbols for one another

and those fives are VERY misreadable as S. and vice versa.

mistakes like that are very hard to track down later lol

[spare narwhal]

Ah

[tropic oxide]

ok yeah good now do it for y

[spare narwhal]

@tropic oxide

[tropic oxide]

??

[spare narwhal]

Okay

Okay what next

I got y=1/2c+7/2

@tropic oxide

[tropic oxide]

ok yeah that's correct

now calculate x-y and choose the right answer option based on that

[spare narwhal]

I got a as my answer

@tropic oxide

.close

[cedar kiln]

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[tropic oxide]

yeah checks out

[cedar kiln]

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[jolly sable]

I want to understand this formula

I guess it is dot product of two vectors

a,b,c are normal vectors and x-x1,y-y1,z-z1 are plane vectors

[crimson sedge]

you guess?

[jolly sable]

yes

[crimson sedge]

$$x-x_1$$ looks like $$\Delta x$$ to me

[wraith dagger]

One person

[jolly sable]

yes

so

[cedar kiln]

@jolly sable Has your question been resolved?

[knotty salmon]

Hey

I am very poor in maths

Can anyone help me

[cedar kiln]

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[jolly sable]

.reopen

[cedar kiln]

✅

[jolly sable]

.close

[cedar kiln]

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[jolly sable]

How to find the equation of a plane which passes through middle point (3,-2,1)(1,4,-3)

[lunar lynx]

You mean middle point of those two points?

[cedar kiln]

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[cursive path]

Can someone explain integration of area under graph

[cursive path]

for negative area questions?

Do I just add a negative sign in front of the whole equation?

[crimson sedge]

Example question?

[cursive path]

Do I split this into 2 parts

the negative area and positive area?

[slate lintel]

yeah

[crimson sedge]

I don't think you have to

[cursive path]

But that part is negative area sir

I will have to put a negative sign in front of the equation

[crimson sedge]

mmmmm kinda confusing question

[cursive path]

Like if I integrate area within sin x

[crimson sedge]

it could be either one because the area under the x axis would be considered negative in most cases. It never asked you to consider it not negative either

[cursive path]

I cant set the lower and upper bound to be 0 and 2pi

wont that give me an area of 0

since the 2 symmetrical areas will cancel out

[crimson sedge]

the area under the curve is zero

[cursive path]

Shouldnt it be positive area

[crimson sedge]

not unless you were asked to consider the area under the theta axis to be non-negative

[cursive path]

Im confused now

So we always say that areas under the x-axis is negative areas

and minus them off the total area?

[crimson sedge]

idk about always

but I learnt that if the area is under the x-axis it's negative

it should be negative unless the question says otherwise

[cursive path]

Coz I had questions before

that showed a graph purely under the x-axis

and they asked to find the area bounded

the answer gave a positive number

[cedar kiln]

@cursive path Has your question been resolved?

[cedar kiln]

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[strange horizon]

if i have integral(1/(-e^(-y))dy, whats the solution?

[urban orchid]

3cosy+coty=0 (solve for x)

[stiff totem]

if you mean $\int\frac{1}{-e^{-y}},dy$, you can simplify the fraction using the negative exponent

[wraith dagger]

Desync

[stiff totem]

then it should be a recognisable integral

[cedar kiln]

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[urban orchid]

<@&286206848099549185>

[cedar kiln]

Please don't occupy multiple help channels.

[royal loom]

stick to #help-26

.close

[cedar kiln]

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[crimson sedge]

$$ v_{pa} = v_{pb} + v_{ba}$$

[wraith dagger]

One person

[crimson sedge]

$$ 50 = v_{pb} + 50$$

[wraith dagger]

One person

[crimson sedge]

$$v_{pb} = 0$$

[wraith dagger]

One person

[crimson sedge]

The particle p relatiive b has a velocity of zero

so why does

(b) say the distance between the two is increasing?

huh

idk?

its just numbers

i assume that

is b is staionary

if a is staionary

then b is moving 50 km away

and p is moving 50 km away

no

no picture

nope

oh wait

picture i never seen

[cedar kiln]

@crimson sedge Has your question been resolved?

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[thin tusk]

A variant of Rod cutting problem, I know it's a Computer Science problem but it is also a Math problem and I want the mathematical approach for this question.
I have provided an image of the original question from the past ICPC problem, I have tried several things to find its best solution but I couldn't get to a valid solution, so I need someone who can help me understand this problem and find a solution.
the next three lines shows the result for each row
you can get to a maximum of 4 portions/pieces for the loaf of bread if the constant factor is 1.500000 (Float value in C++ which is up-to 6-to-7 decimal places)

[cedar kiln]

@thin tusk Has your question been resolved?

[thin tusk]

I can try and find the largest possible cut by simple brute force and with a little trick

0.500000 / 1.5 = 1 / 3 = 0.333333 // upto 6-7 decimal places max
1.000000 - 0.333333 = 2 / 3 = 0.666667
0.6666667 > 0.4999995 // So this cut is not possible

// Now I aim for 0.666667
(2 / 3) / 1.5 = 4 / 9 = 0.444444
1 - 4 / 9 = 5 / 9 = 0.5555556
0.5555556 < 0.6666667 // Repeat this and you get to the largest possible cut.
// But it won't give me the optimal cut, but I can keep this approach going for a few more steps

// Now I aim for 0.5555556
(5 / 9) / 1.5 = 10 / 27 = 0.3703704
1 - (10 / 27) = 17 / 27 = 0.6296296
0.6296296 > 0.5555556 // This cut is not possible

// Now I aim for 0.6296296
(17 / 27) / 1.5 = 34 / 81 = 0.4197531
1 - 34 / 81 = 47 / 81 = 0.5802691
0.5802691 < 0.6296296 // Acceptable
// If I keep repeating it for n-steps I'll eventually get to the solution but it would take alot of operations which I don't want, it's not an optimal approach I want to find make sort of formula which would get me either the best first cut or the end state cut which in this case (k = 1.5) gives the result 0.214285.
// If a formula cannot be deviced then the next possible option is to find best cut with minimal operations.

[cedar kiln]

@thin tusk Has your question been resolved?

[thin tusk]

Can i get some help?

<@&286206848099549185>

[cedar kiln]

@thin tusk Has your question been resolved?

[thin tusk]

<@&286206848099549185>

[cedar kiln]

@thin tusk Has your question been resolved?

[thin tusk]

<@&268886789983436800> waiting for a long time, still haven't gotten a single response

i need to close this ticket

[worldly chasm]

Many people on this server who can solve this were asleep when you asked, please be patient and only ping once.

Would a simple recursive process, where you cut the largest loaf such that you create a new smallest piece that is the (new largest piece)/k in size work? It's a greedy algorithm, so it might be suboptimal.

[wanton sail]

you can close it by typing .close

as it says in the instructions

please don't ping moderators about stuff like waiting for a long time

!volunteers

[cedar kiln]

Helpers are just people volunteering their time to help you. Be polite.

[thin tusk]

IK how to close it but i havent gotten a single response

the problem is the cutting part

[worldly chasm]

For instance, for k = 1.5

Loaf: 1.0
Let the size of the new loaf be n, Now we solve n = (1.0 - n)/k which gives n = 1.0/(k+1) or 0.4

Loaf: 0.6 0.4
We take the largest of these 0.6 and do the same trick, except we need to look at the 0.4 loaf or the 0.6 loaf depending on the size of the newest loaf, largest is better.

Looking at 0.4 we have n = 0.4/k = 0.2667

Looking at 0.6 we have n = (0.6 - n)/k => n = 0.6/(1+k) = 0.24

Because 0.2667 > 0.24 we take the former

Loaf: 0.3333 0.2667 0.4

Next, consider the two options for 0.4, which are 0.3333/1.5 = 0.2222 or 0.4/2.5 = 0.16, we find that 0.2222 is larger so we want to create that, however, doing so would create a loaf smaller than 0.2222 from the 0.4, so the best we can do is 0.2, which ends the iteration.

Loaf: 0.3333 0.2667 0.4

Given that a solution was provided that is better than this, a simple greedy algorithm seems like it cannot work.

[thin tusk]

need more accurate results than 0.6 and 0.4

how to reach those precise numbers

[worldly chasm]

I'm sorry?

Did you read what I wrote?

[thin tusk]

the first cut is 0.571429 and 0.428571

[worldly chasm]

Yes, I know. Please read all of the way through, it is not a solution, it is describing my attempt at a simple greedy solution that does not work

My guess is that the first cut is not unique, however

There are a range of values that will work.

And they are intentionally randomizing within this range to obscure the solution

Another attempt, let's find the smallest k such that you can make 3 loaves with any cuts.

If we make the first cut greedy to maximize the size difference between the two loaves, then we have

Loaf = 1/(k+1) k/(k+1)

Then we need a cut such that the larger loaf cut into half precisely is exactly equal to 1/(k(k+1)) which is k/(k+1) = 2/(k^2 + k), which implies k = sqrt(2)

So for k < sqrt(2) the best you can get is a single cut, for k >= sqrt(2) you can get at least 2 cuts (3 loaves).

And in the case where k exactly equals sqrt(2) this cut is unique. But for k = sqrt(2) + epsilon this cutting isn't unique, there is some "wiggle room".

I'll leave it to you to exactly determine how much wiggle room, it's just straightforward algebra. I'm going to move on to a minimal k for 3 cuts.

[worldly chasm]

So for three cuts we don't want to leave as large a second loaf as possible, but as large of a third(!) loaf as possible, which means our first cut should be less greedy.

If we have loaves of sizes a, b, c after two cuts, we want the largest loaf (a) to get cut exactly in half and be exactly b/k in size (where b is the second largest loaf).

c will come from the first cut, and a and b will be determined from the second cut.

First cut:

Loaf: a+b c

Subject to the constraint that a+b > c, a+b < ck.

Second cut: a b c

Subject to the constraint that ak = c, equality required for optimality.

Finally third cut: a b c/2 c/2

Subject to the constraint that ck/2 = b <= ak, equality is required for optimality here as well.

Using a computer algebra system, it looks as if, ironically, you can get a solution for 3 cuts with k = sqrt(2) as well???

It feels like I've made a mistake somewhere but I cannot see it.

@thin tusk I don't know whether or not this helps you see an approach, but it looks like kind of a mean problem to reason about.

at least mathematically

There is probably some recursive solution that, while it does a lot of work, should be fast enough to run in a single second because computers are good at this sort of thing?

[cedar kiln]

@thin tusk Has your question been resolved?

[thin tusk]

we are not sure how many total cuts we can make, so this assumption might not work

[worldly chasm]

@thin tusk we're starting from the premise that we can only make 2 cuts, and trying to find the minimum k that will get us there.

[thin tusk]

I liked this approach

[worldly chasm]

I did as well, unfortunately it is sub-optimal 😦

[thin tusk]

it didnt said I have to get that accurate result which was bothering me alot

i was thinking how they got to that result

those specific values which perfectly overlap with one another in smaller cuts

Loaf = 1/(k+1) plus k/(k+1)
How did you make this formula

[worldly chasm]

that's not really a formula, but a description of the size of the pieces in the loaf at this step

the first piece is 1/(k+1), the second piece is k/(k+1)

[thin tusk]

should there be a + sign in b/w?

[worldly chasm]

No, because it's not a formula

think of it as an array:

loaf = [1/(k+1), k/(k+1)]

[thin tusk]

ahh

[thin tusk]

Loaf = 1.0
k = 1.5
1/(k+1) = 1/2.5 = 0.4
k/(k+1) = 1.5/2.5 = 0.6
Loaf = 1/(k+1) + k/(k+1)
Loaf = 0.4 + 0.6 = 1.0
or [1.0] -> [0.4, 0.6]

[worldly chasm]

that's what that part is saying

but unfortunately, 0.4, being the absolute smallest cut you can make, is actually too small to work with the 3rd cut

[thin tusk]

the problem itself is solved with this simple approach but there's still a big mystery

how they got to that precise numbers

[worldly chasm]

I don't think that the problem is solved.

Because it's not optimal, and the problem specifies that you must cut the loaf into as many pieces as possible.

[thin tusk]

Any proof

atleast it gives a rough estimate

it might fail on some corner cases

[worldly chasm]

"indicating the maximal achievable"

and I think it probably fails in most cases.

because 1.5 doesn't seem like a special value

[thin tusk]

for 1.999, how many cuts can there be

lets assume the inputs are lower and upper bound

[worldly chasm]

a whole hell of a lot

[thin tusk]

1.000 and 1.999

[worldly chasm]

it's infinite for 2

[thin tusk]

they fixed it for 3 decimal places

not 1.9999999

[worldly chasm]

and for 1 <= k < sqrt(2) it's only 1.

[thin tusk]

so there might be a finite answer

[worldly chasm]

there is certainly a finite answer.

but my guess is that the value approaches infinite quite rapidly

[thin tusk]

the output size is limited as well

only 6 decimal places after decimal

so ig below 1 million max cuts

[worldly chasm]

while your output is limited in precision, you could still in theory come up with a scenario where you output mostly 0.000000 and have more than 1 million entries, provided that you're keeping track of your numbers with enough internal precision

[thin tusk]

float will keep the result upto 6 decimal places for me so my answer wont search beyong 6-7 decimal places

[worldly chasm]

a single precision float only does 6-7 decimal digits of precision, yeah. This is probably why they're saying not to output more than that.

[thin tusk]

also I got close to the actual result with my first cut
0.5849625 and 0.4150375

[worldly chasm]

It looks, strangely enough, now that I've double checked, that sqrt(2) gets you both a second and a third cut.

k = sqrt(2) that is

[thin tusk]

i did log2(k)

[worldly chasm]

I do not immediately see why that would be a good idea.

Can you share your thoughts?

[thin tusk]

i was messing around with random inputs and i thought maybe log has something to do with this problem and i tried different bases of log

like log base k, log 10 and log 2

which gave a close enough result to actual

yea when I cant find a pattern, I do some trial and error

[worldly chasm]

so in other words, if you have a loaf = [a, b, c] with c the largest loaf, you try to make a cut such that [a, b, c (1 - log2(k)), c (log2(k))]? and if the smaller loaf is too small to satisfy the constraint you?

[thin tusk]

it has the same issue as your first approach

it might or most probably will fail at some point

since its not giving a precise cut as the test case

[worldly chasm]

3/7

[thin tusk]

O_O

[worldly chasm]

I do not know why 3/7

and the k=1.6 starts off with 25/61

[thin tusk]

how in the what

can you make a table for values from 1.0 to 1.9

we can see a pattern emerge from it

also how is the first cut 25/61 for 1.6

[worldly chasm]

k=1.5 cuts are 3/7, 2/7, 3/14
k=1.6 cuts are 25/61, 16/61, 25/122, 10/61, 8/61

I'm not solving this, I'm determining a simple rational fraction pattern given the decimal numbers.

these are all quite close, and all have related denominators, so I'm confident that there is some underlying pattern that they are exploiting.

[thin tusk]

why even 61, thats such an absurd number to pick

[worldly chasm]

I do not know

[thin tusk]

surely if we dig deeper we can find their hidden pattern

[worldly chasm]

If you have any other examples of valid output from the oracle function, it would be appreciated.

[thin tusk]

they only provided 2 test cases

[worldly chasm]

So:

k=1.5

Loaf: [1]
Loaf: [4/7, 3/7]
Loaf: [2/7, 2/7, 3/7]
Loaf: [2/7, 2/7, 3/14, 3/14]

was the strategy

which seems more reasonable.

you make the initial cut, then cut each other in half.

k=1.6

Loaf: [1]
Loaf: [36/61, 25/61] (interesting that they're using perfect squares here)
Loaf: [20/61, 16/61, 25/61]
Loaf: [20/61, 16/61, 25/122, 25/122]
Loaf: [10/61, 10/61, 16/61, 25/122, 25/122]
Loaf: [10/61, 10/61, 8/61, 8/61, 25/122, 25/122]

[thin tusk]

perfect squares and a random prime as denominator (random to us for now)

with this pattern the next number should be
k: 1.7
49/85, 36/85

[worldly chasm]

that doesn't seem to follow

[thin tusk]

adding consecutive perfect squares

[worldly chasm]

well, there's no reason to assume that it's going to be linear (and every reason to assume it's not, as k=2 is literally infinite), and moreover the k = 1.5 case didn't feature consecutive perfect squares

[thin tusk]

maybe the jump is a little bigger b/w squares

ik the ICPC test questions have a really well constructed pattern, they always do

[worldly chasm]

oh, this is a programming competition

not homework

[thin tusk]

an old paper

[thin tusk]

nah i do my own homework

i dont want to look at the solution online, I would rather discuss it with someone else and approach the solution from scratch

[thin tusk]

they didnt cut 36/61 in half
instead they cut it into 20 and 16

[worldly chasm]

I noticed.

it looks like historically ICPC problems at least from where I found, are numbered as problem A, not problem 1.

[thin tusk]

instead of writing 4/7 and 3/7
we can write it as
8/14 and 6/14
16/28 and 12/28

trying to make them a perfect square

[worldly chasm]

Yes, but you'll be unable to make a perfect square, because you'd have to find a common factor such that you have 2^2 * 2^x 3^y = 2^2n 3^2m, and also 3 * 2^x 3^y = 2^2j 3^2k, this implies that 2n = 1 + x, but 2j = x, but both 1+x and x cannot be even.

[thin tusk]

I'll probably think of new patterns while counting sheeps in my sleep

[worldly chasm]

@thin tusk before you go, a potentially useful observation

16, 20, 25 have common differences of 4 and then 5

36 and 25 the difference is 11

The algorithm might have the following structure:

[thin tusk]

why does it look like (a + b)^2

[worldly chasm]

4^2, 4*5, and 5^2 ?

Dunno

That might be a coincidence or it might not

[thin tusk]

does the same apply for k=1.5?

4/7, 3/7
2/7, 2/7, 3/7

[worldly chasm]

Well in that case the common difference in the numerator is just 1

I'm looking at the algorithm in two steps, first we try to divide the bins in a specific way, then cut them all in half

Then it terminates unable to do more

[thin tusk]

i have to take a deeper look into Strassen multiplication algo after this one is solved

F

@worldly chasm

1/(k+1) = 1/(1.6+1) = 1/2.6 = 5/13

take square of numerator and you get 25

[worldly chasm]

hmmmm....

[thin tusk]

1/(1.5+1) = 1/2.5 = 2/5
take square of 2 u get 4

for k=1.7
i got 10/27 one half must be 100

is there a Pythagoras Theorem hidden in this

[worldly chasm]

I don't think it's quite that simple.

So one thing you've missed is that in the 1.5 case, you're taking the square of 2 to get 4, but 4 is the larger of the two numbers.

[thin tusk]

in this case its 25 the lower bound

[worldly chasm]

whereas in the 1.6 case, you're taking the square of 5 to get 25, but 25 is the smaller of the two numbers.

[thin tusk]

yea i noticed

[worldly chasm]

so I think this is just a coincidence

law of small numbers thing

[thin tusk]

I have been a victim of this so many times i lost count

well gn for now, i'll continue this struggle later today

[worldly chasm]

good luck

[cedar kiln]

@thin tusk Has your question been resolved?

[thin tusk]

.close

[cedar kiln]

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[tough fulcrum]

my answer for (b) is wrong and I'mm not sure why

pls help

[cedar kiln]

@tough fulcrum Has your question been resolved?

[wanton sail]

Hey! lemme take a look

[wanton sail]

you found the correct value of lambda (20/33) but you just made a typo when calculating y

so your answer is basically right, just a small arithmetic mistake at the end :)

[tough fulcrum]

Oh i see 🤦‍♂️

thanks for your help

.close

[cedar kiln]

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[wanton sail]

no problem!

[cedar kiln]

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[supple mantle]

how am I not sophisticated enough to see why this x just vanishes bruh: I wanna set 2x ln(x) +x = 0 to find critical points | Actually just got a grasp on it while writing this question, is it because we can just find the 0 of the factor and that will do? Are there other ways to find the 0 too?

[supple mantle]

cuz then it's just $ln(x) = \frac{-1}{2}$ then $ e^{-0.5}$

[wraith dagger]

Maggi

[glass sky]

i mean x can be 0 and e^-0.5 if im not wrong

[supple mantle]

yea, it's just that 0 is not that useful when finding the absolute minimum

[idle cedar]

but x cant really be 0 can it ?

because ln(0) is not defined

[supple mantle]

and that yea, 0 is not in the domain

[idle cedar]

therefore , as a product equals 0 iff one of the factors is 0 , x=0 is not an option

meaning the first factor has to be 0 , therefore you can drop x

[supple mantle]

yup!

thanks

.close

[cedar kiln]

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[unique isle]

how do i solve this one?

[unique isle]

is there a formula or something?

[proud arrow]

allowed to use calculator?

[unique isle]

yeah

[proud arrow]

do you see the two R's

which make the arc

dont you see a triangle

that too isoceles

[unique isle]

yea

what are the r s though

[proud arrow]

could you apply some trigono metric properties over here?

[proud arrow]

they wont matter

if you see closely

[unique isle]

ok

[proud arrow]

try going over basic definitions of

them functions

oh yes how about we drop a line from ground to the plane

straight line

that'll make it easier to apply trig

[unique isle]

ook

[proud arrow]

could you see how we are gonna apply trig

[cedar kiln]

@unique isle Has your question been resolved?

[unique isle]

no

i dont understand

i lost the plot

[crimson sedge]

https://images.app.goo.gl/yZg82T9edYdFktQQ9

You get r from this

Then use trigonometry by dropping an altitude from centre of plane to vertex

sin(0.5) ≈ 0.5

No wait that's radian

You're given degrees

sin(0.5°) = 0.008

And sin(0.25°) = 0.004

Use these to get answer

@unique isle

[unique isle]

okok

let me try

[cedar kiln]

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[crimson sedge]

How would I go about solving this without doing any actual "calculations"

[crimson sedge]

the if part seems weird

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

<@&286206848099549185> you don't have to read the question at all

Just asking

how I would get rid of the if statement

[worldly chasm]

You cannot 😦

At least not without using functions that are essentially if statements

Like the indicator function or floor function or similar

[crimson sedge]

oh yeah

I could

hmm

use that

do like

$$sign(\Delta x - d_1)$$

[wraith dagger]

One person

[crimson sedge]

but still

[worldly chasm]

Let u(x) be the function that gives 0 if x is negative and 1 of u(x) is positive (undefined at 0), u(x - Δx) will return 1 if plateau and 0 if ramp

And undefined if it lands on the corner

But I'm not sure what you're planning on using this information for?

[crimson sedge]

just learning

how not to

do math when doing math

[worldly chasm]

Are you trying to come up with a general formula for b) and c) without using cases?

[crimson sedge]

yes

[worldly chasm]

Probably not worth it imo. Cases are cleaner and clearer.

The purpose of writing down math as math is to communicate the result with yourself and anyone else who reads it

The less that person needs to think to understand it, the better, imo

There's obviously a lower bound of thinking effort, but the expression itself shouldn't hinder understanding but facilitate it as much as possible.

[crimson sedge]

What about making a statement like

wait nvm

literally exact same situation

I was gonna ask

How to answer True/False statement with math

I did this for a

Got kinda lazy

and didn't make variables for some things

[crimson sedge]

.close

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[shadow kite]

Can someone help me with this, I’m not sure how to proceed

[celest ledge]

It’s sqrt(log(-x))=log (|x|)

Now x<0 so sqrt(log(-x))=log(-x)

log(-x)=0 or 1

sqrt(x^2) is |x| not x

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[primal lodge]

https://cdn.discordapp.com/attachments/808388205867040778/1142084244961775657/image.png

[primal lodge]

How am i supposed to do this?

I guess i could plot the rectangle

Do i just reflect every point?

[upbeat forge]

Yes, and then find overlap of original and reflected rectangle

[primal lodge]

i did something wrong

i guess it was just cuz i sketched wrong

looks like a straight forward question

should probs try again

okay yeah i got it

bad diagram lol

.close

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[crimson sedge]

How???

[glossy mango]

any chance the question is wrong?

it says log 2 = m

and than asks us to find log 25 in terms of x

does it mean in terms of m?

[tropic oxide]

yes most likely it's just a typo

[crimson sedge]

Yes it means in terms on m

But I still can’t do it

That’s why I asked this question, just to clarify

@glossy mango

[glossy mango]

when the base of the logarithm isnt specified, it's assumed to be log base 10 in most scenarios

log 10 is 1, obviously

smilarly, log 100 is 2

[crimson sedge]

yea...

[glossy mango]

log 100 - 2 log 2 is the same as log 100 - log 4 = log 25

and log 2 is m

[crimson sedge]

oooooooooo

[glossy mango]

so log 25 is 2(1-m)

[crimson sedge]

i get it now!

thank you!

[glossy mango]

np <3

[crimson sedge]

.clos

c.lose

.close

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[crimson estuary]

Just wondering if someone could see if its correct since i dont have the answers

[ocean mural]

if a = 2
and b = a - 14
b cannot be 4

[crimson estuary]

ah shit

can u help me find where i went wrong

[ocean mural]

well its probably right there

if a = 2

and b = a - 14

what is b. really

[cinder venture]

Is this an exam?

[crimson estuary]

its a past paper

[crimson estuary]

i first got b from the 2a+ b = 8 so thats how i got 4

[ocean mural]

oh ok

[crimson estuary]

so i definitely got something wrong somewhere

[ocean mural]

that says 2a + b = -8

[crimson estuary]

oh wait

so b does = -12

ok well thank you for helping

.close

[cedar kiln]

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[spring coyote]

Guys

is this very small number?

below 0.05?

or very big

[crimson sedge]

What exactly?

7 is bigger than 0.5

So

[spring coyote]

h mmmm

[ocean mural]

70/100 = 0.7
70/1000 = 0.07
70/10000 = 0.007

[spring coyote]

oho

[ocean mural]

so 70/9999999 must be very small

[spring coyote]

like basically

[cinder venture]

Your calculator is very peculiar

[spring coyote]

i learn statistaks, and it say fraction of choice if its below 0.05 then use certain formula

[ocean mural]

I think it's google formula

I mean calculatro lol

[spring coyote]

and fraction of choice is base sample divided by total possible sample size

[crimson sedge]

Anyways e-6 Is basically 10^(-6)

[spring coyote]

ye

so i dont get why total number of surveyees in a theatre is theoretically infinite

its task where i dont know sample size

so i first need to deterimne n/N = fraction of choice translated

and textbook say

we can determine without calculating that its below 0.05 (which means we use special formula for below 0.05

because it is possible to survey infinite amout of people about what they think about theatre price

but humans are 8 billion

shiiiiiiiiiiii

so i calculate sample size is 70

[cedar kiln]

@spring coyote Has your question been resolved?

[cedar kiln]

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[primal lodge]

how would i do this?

[primal lodge]

not really sure where to start

[stiff totem]

you can fill in some of the missing angles, as they are opposite angles

as in, the angle where l and m meet is x degrees on both sides

so you can fill in one of the blanks at the bottom there

then, once you have all of them, do the replacements as given in the question (y=2w, v=110, etc.)

then they should add up to 360, which should give you something to work with

[primal lodge]

oh yeah and v = y + z= 110 right?

not sure why i didn't even think of even doing anything lol

[stiff totem]

yeah

[primal lodge]

can i also do w+z+x+y = 180

[stiff totem]

yes

[primal lodge]

okay thanks

answer should be 3/4 ig

i have another question

I'm just reviewing questions from a practice test

but the way i approached this iirc was

y is either in (-15, 15)

and the other one involves the (-20,10)

i guess i could combine it, but i don't have to right?

The answer is D? I don't really have an answer key so just ocnfirming

[stiff totem]

x=8, y=-16 satisfies those inequalities, but so does x=8, y=-14

so yeah, can be either way, so d

[primal lodge]

okay yeah i think it's trivial either ways, just confirming my answer

thanks

.close

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[cedar kiln]

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[velvet lynx]

Find Min

[sly sandal]

Min?

[velvet lynx]

minium

[slate lintel]

are there constraints on x and y?

[velvet lynx]

no

[south tundra]

I thought there should be too, but realised it might just be about completing the squares

Try doing that

[slate lintel]

you've got an xy term so that will be fun for you

[south tundra]

Hint: ||Complete the square in x and y first and then in y||

Huh wait

That -2x

[vague rapids]

Does a minimum even exist?

[south tundra]

Alright I found it

Yes it is possible

[agile wyvern]

just take the partial right?

[south tundra]

Actually

[velvet lynx]

of course

[south tundra]

No nvm

[agile wyvern]

i got an answer

[south tundra]

Is dungx familiar with calc?

[vague rapids]

you sure you are not missing any part of the question?

[agile wyvern]

oh

if you can use calc its simple

[south tundra]

Yeah can you show us the context?

[velvet lynx]

no

[agile wyvern]

oh

[velvet lynx]

NOoooooooooooooooooo

[south tundra]

Or does it just say "Find the minimum" and that's it

[velvet lynx]

the test said

cant use

calc

[south tundra]

By calc we mean calculus just in case

[velvet lynx]

its not english

[vague rapids]

Still, it would be better if you send an image

[velvet lynx]

ok

Tìm giá trị nhỏ nhất của biểu thức:M=...

use translate

Giá trị nhỏ nhất đó đạt được khi x,y bằng bao nhiêu?

[vague rapids]

So there was a condition

x=y

[junior dome]

Find the minimum value of the expression: M=x(x + y) ) +y(y - 3) – 2x + 2010 What is the minimum value that is reached when x, y equals ?

[vague rapids]

wait did I misinterpret

[junior dome]

yeah i read it like that too

but there's ? sign

[boreal ravine]

math

[velvet lynx]

no

when M reach min what is x,y

[junior dome]

i think it is x = y tbh

? is for what is the min. value

and not for x and y

[vague rapids]

Well, if the condition x=y is not given then there would be no minimum

[velvet lynx]

wat

[vague rapids]

,w minimum of x(x+y)+y(y-3)-2x+2010

[junior dome]

ahh

[agile wyvern]

thats what i got when i used calc but how do you do it without calc

[wraith dagger]

Results provided by WolframAlpha

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

Requested by beard420

[velvet lynx]

idk

.close

[cedar kiln]

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[crimson sedge]

making a circular candy braclet, you have four letters to choose from: a,b,c, (with only three candies for each letter), if you can create braclets of length 1-4, how many possible bracelets can you make?

[crimson sedge]

I said:
4 length = 3 * 3 * 3 * 2
3 length = 3 * 3 * 3
2 length = 3 * 3
1 lenght = 3

answer = 3 + 9 + 27 + 54 = 93
is this correct?
oh shoot i think thats wrong
because abc = bca =cab since its ciruclar

my next attempt:

length 1: 3

length 2: aa,ab,ac,bb,bc,cc = 6

im not sure how to do this recursively tho

[fair geyser]

it's not just circular it also has 2 sides

bca = bac

[crimson sedge]

nono

bca != bac

I dont think

[fair geyser]

in real life it is

depends if the intent of rthe question follows rl

[crimson sedge]

yea it does

it says specifically the LOOP of the chain

[fair geyser]

idk what you mean

[crimson sedge]

basically yea it follows real life

[ocean mural]

lol if it's circular then yes bca = bac

[crimson sedge]

such that abc = cab = bca

but abc != acb (=bac)

[fair geyser]

a real life bca bracelet is indistiguishable from bac

[crimson sedge]

oh

yea

i agree

so how many bracelets can she make

using up to 4 beads

that are UNIQUE

[sharp mantle]

Are we assuming the bracelet can be flipped

Or no

[crimson sedge]

no

doesnt state it so id assume not

[sharp mantle]

Then abc ≠ acb

[ocean mural]

well in real life bracelets can be flipped

[crimson sedge]

the only consideration is that you can loop one bead around

and abc -> cab

[sharp mantle]

So then they can't be flipped

[crimson sedge]

no

how would i approach this problemm

[fair geyser]

i don't know

[crimson sedge]

just writing them down?

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

<@&286206848099549185>

[twin marsh]

Burnside's Lemma may be helpful for resolving issues like this.

(Namely issues like abc = bca = cab).

[cinder venture]

Dont use ChatGPT for Math

May not always be correct

[dire geode]

<@&268886789983436800>

[wanton sail]

Please don't use ChatGPT, especially to answer questions from helpees @crimson sedge

[crimson sedge]

@dire geode you know how to do it?

[crimson sedge]

I couldn't solve it. So I Asked AI(Chat GPT). Sorry If I violated any rules.

[wanton sail]

If you can't solve it, then don't answer; answering using GPT and similar AI is against the rules

[crimson sedge]

Ok

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[oblique prawn]

yo

[oblique prawn]

is $\lim_{x\to 0^+}{\frac{1}{x}}$ equal to infinity or DNE

[wraith dagger]

Jash

[royal loom]

depends

what is your definition for a right sided limit converging to infinity?

[compact junco]

technically it doesnt exist

but you can say its infiniy

[oblique prawn]

idk is there like a formal definition

[royal loom]

yes

there can be

[oblique prawn]

so what would it be

Find $a$ and $b$ such that $\lim_{x\to 0}{\frac{\sqrt{ax+b}-3}{x}}=\frac{1}{3}$

[wraith dagger]

Jash

[royal loom]

well

for a normal right handed limit

it is something like

we say $f(x)\to L$ as $x\to c^{+}$ if $\forall \varepsilon >0:\exists \delta >0$ s.t $\forall x \in S$ with $c<x<c+\delta \implies |f(x)-L|<\varepsilon$

[wraith dagger]

Austin

[royal loom]

and then combine this with a definition for a limit going to infinity

and ya

[oblique prawn]

oh ok

ty

.close

[cedar kiln]

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[crimson sedge]

Open the brackets

(x-5)-x = -5

In numerator

That's your last step

[stark quest]

x-5-x=-5

[crimson sedge]

Yeah

-5

Not just 5

No worries

[cedar kiln]

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[candid lotus]

determine the exact value of:

[candid lotus]

what should i do here?

i need to apply formulas?

[gritty viper]

all of these have values you should know

like, every individual sin() and cos()

so you can just find those all

[candid lotus]

like for example cos(pi/3) is 1/2 ?

[gritty viper]

yeah

[candid lotus]

thats bcs i saw taht

i was confused

what is that for

[gritty viper]

well you could try using that but there's no need

those formulas are so cursed

it should be an exercise for very strong students to derive them from the normal ones (sin(a+b), etc) and then completely forget about them after

[candid lotus]

oh ok i found out

In bold and in blue, the new features at the start of sup

thats what im going to see this year

it has nothing to do with the revision exercises of last years im doing

[gritty viper]

ah ok

[candid lotus]

what studies are you doing or did you do btw? @gritty viper

[gritty viper]

nothing super focused yet

studying some mathematical logic on my own

[candid lotus]

what are you planning to do?

[gritty viper]

idk

PhD

probably

[candid lotus]

alright i see, thats really nice

[cedar kiln]

@candid lotus Has your question been resolved?

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[flint plinth]

como

[dire geode]

[crimson sedge]

yo como gatos

i ate a cat

[sharp ginkgo]

yo comí los gatos

bc that’s the past tense for it