help-4

[ruby sleet]

how can you simplify by "a" in your last line

[cloud coral]

are you sure you aren't missing any extra conditions?

[ruby sleet]

that requires at least a and n being coprime

otherwise your proof is not worth anything

[cloud coral]

well strictly, it's worth a lesson

[ruby sleet]

well I was saying the proof is not valid

[cloud coral]

fair

[ruby sleet]

here's an even tamer result

taking n = 10 and a = 4

say b = 0 just for the sake of it

then x = 1 and x = 6 result in the same value of ax+b mod n

as soon as a and n are not coprime

the statement is null

now the reason why your proof doesn't work: you try to simplify by "a"

in modular arithmetic, that makes no sense unless a has a modular inverse mod n

meaning some u such that ua = 1 mod n

that only happens when a and n are coprime

because when they are, then bezout theorem =>

$au + nv = 1$ for some $u,v$ integers

[rocky lotus]

rafilou is not not born in 2003

[ruby sleet]

then $au \equiv 1 \mod n$

[rocky lotus]

rafilou is not not born in 2003

[ruby sleet]

(this is even an equivalence)

[gritty carbon]

yeah i probably misunderstood a properties of modular arithmetic

[ruby sleet]

yeah... as you can see

$a \equiv b \mod n$ IMPLIES $ka \equiv kb \mod n$

[rocky lotus]

rafilou is not not born in 2003

[gritty carbon]

but the converse is not true

[ruby sleet]

the converse is indeed not always true

which is why we need the existence of $u$ such that $uk \equiv 1 \mod n$

if we do, then having $ka \equiv kb \mod n$

[rocky lotus]

rafilou is not not born in 2003

[ruby sleet]

then $uka \equiv ukb \mod n$

[rocky lotus]

rafilou is not not born in 2003

[ruby sleet]

so $a \equiv b \mod n$

[rocky lotus]

rafilou is not not born in 2003

rafilou is not not born in 2003

[gritty carbon]

Ok, understood, sorry for the stupid question

.close

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[ruby sleet]

there are no stupid questions

and no worries

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[small crystal]

do I use arc length here or smth? I don't know how sum is gon give the length

[sharp knot]

Do you remember the formula for arc lenght?

[small crystal]

yea

wait im given f'

dawg 😭

im blind

[sharp knot]

Do you see it now?

[small crystal]

so i do sum to find f'(x) over the interval?

[sharp knot]

The arc length formula is in terms of an integral. You estimate the integral with a reimann sum.

[small crystal]

i know the int a to b sqrt (1 + (dy/dx)^2 ) dx

but i dont get how im solving this problem

[sharp knot]

You're given the dy/dx at the points you need. It is the same as other reimann sum problems, you just use sqrt (1 + (dy/dx)^2 ) as the function you're integrating.

[small crystal]

do i have to do sum for all three f' and add that together?

[sharp knot]

Can you try setting it up? I'll let you know if I agree with what you have.

[small crystal]

would it be like this? I used the interval instead of integral for sum

[sharp knot]

Yes, exactly

wait

why do you have f'(7) for the last one? It should be f'(5).

[small crystal]

I wrote it wrong 😭 but I plugged correct value

[sharp knot]

Ah, I think you see the idea, though.

[small crystal]

yea

thanks for helping 🙂

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[sharp knot]

np

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[balmy iris]

i need some help with how to start this

[daring basalt]

current student in quantum physics coop placemnet needing imediate help on fixing this quantum wave equation:

[stark wedge]

!occupied

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[daring basalt]

but mine needs attension pls

[stark wedge]

this is someone else's channel. get your own.

[dapper osprey]

cold

[balmy iris]

thanks ann

[daring basalt]

I am sorry about the difficult situatino however these tought times call for important leadership decisions one of qhich should be adressing myproblem imediatly.

[stark wedge]

<@&268886789983436800> this is a troll.

[balmy iris]

its okay you guys can help him

ill open a new ticket

[stark wedge]

@balmy iris am i off base or like... doesn't the fourier sine transform simply involve taking the integral of f(x) sin(xi*x) dx

[stark wedge]

no

this channel has your name on it

[balmy iris]

okay

[daring basalt]

anna is right

[stark wedge]

@daring basalt can and should open his own channel

[daring basalt]

let me find a different one

[balmy iris]

oh wait yeah ur right

thank you ann idk why im slow

[dapper osprey]

some people are just born that way

☹️

[stark wedge]

....

what was the point of that comment again

[balmy iris]

yeah im born slow

!close

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[sturdy niche]

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[small crystal]

how do I solve this? I'm not given int f(x)

[safe fulcrum]

Integration by parts

[small crystal]

oh

lemme try

im getting f(x)g'(x) - int f'(x)g'(x)

is this correct?

do i plug in the bounds

[safe fulcrum]

Yes

[leaden cobalt]

you also could find f(x) if you wanted lol

[small crystal]

huh?

wdym

[leaden cobalt]

f'(x) = 3 for all x, so f is linear
and you have a point (x,f(x)) = (2,7)

ri?

[small crystal]

hmmm yea for IBP do i plug in bound for both parts?

[safe fulcrum]

Yep

[small crystal]

for f'(x) inside the integral i js treat it as constant and pull out right?

[leaden cobalt]

yeah, and if you want you can just replace it with "3"

[small crystal]

yea i did that

i got the right answer

thanks yall 🙂

[leaden cobalt]

ri short for right 💀

[small crystal]

huh?

wdym

oh yea i never seen someone use that but

ok

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[stone aspen]

!help

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[blazing rover]

Can f(x) ^ g(x) ever diverge as x goes to infinity if the limit as x goes to infinity of g(x) is 0?

[blazing rover]

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[mild parcel]

I'm having an extremely tough time figuring out this problem

[mild parcel]

I only got to the poiint where I found the intersection points

but I don't understand which method i should use

[safe fulcrum]

the outer volume would be given by x + y = 2 or y = 2 - x

and the inner volume by x = y^2 or y = sqrt(x)

one sec

ah that's nasty: okay so you need one integral for 0 < x < 1

and another integral for 1 < x < 4

[mild parcel]

wutdahel

i think i found the lower and upper bounds

[safe fulcrum]

but you need to split it into two integrals

and actually yeah disk/washer method works better

[mild parcel]

that means theres an inner and outer radius

[safe fulcrum]

yes

so what you need to do is to find the outer distance, so the distance between the bottom part of the red curve and y = 1

and then the inner distance

for example, the inner distance between the blue line and y = 1 is (2 - x) - 1 = 1 - x

(or x - 1, it doesn't matter if you square it)

(technically it's |x - 1| = |1 - x|)

https://www.desmos.com/calculator/bxecezh2wf

okay I'll just post this

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[spare pawn]

can anyone help me with this question

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[heady pawn]

For this I always remember AB, so we have to do b - a

like switch em around

[slow stone]

me and my friend are having a conflict wether we subtract them or not since they arent points but vectors

[slow stone]

Isnt that in the case if theyre points

[crimson yoke]

Points r vectors and vice versa

Points r nothing but 2d vectors

[slow stone]

So do I just solve it as if theyre 2 points?

[crimson yoke]

Yep

[slow stone]

100% sure?

[crimson yoke]

Yes

[heady pawn]

not entirely

[slow stone]

alr thank youu

[slow stone]

what

[heady pawn]

a - b is not the same as b - a

[slow stone]

i know that part i meant subtraction or addition

[crimson yoke]

That's not what she meant

[heady pawn]

just be carefull XD

[slow stone]

Haha alright dont worry, thank you!

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[pine spire]

Can someone explain example 2? How did they evaluate it?

[crimson yoke]

Try applying this there too

[pine spire]

Got that but how did they evaluate (x+1) and (x-1)

[crimson yoke]

x² - 1 = (x+1)(x-1)

[dusky pulsar]

(a+b)(a-b) = a^2 -b^2

this?

[crimson yoke]

There's an identity

[pine spire]

Okai got it thanks guys

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[pine spire]

.reopen

[vale dock]

✅

[pine spire]

pls explain

[crimson yoke]

Which step?

[pine spire]

the 3rd step

[crimson yoke]

$(x^2-x+1)(x+1)$

[rocky lotus]

Suika

[crimson yoke]

Expand it

$(x^2-x+1)x+(x^2-x+1)1$

[rocky lotus]

Suika

[pine spire]

alr okai

wont -x^2 + x^2 get canceled?

[crimson yoke]

Yes

They will

[pine spire]

Cant I cancel x^2 over here?

[crimson yoke]

Yes u can

[pine spire]

alr thanks

[crimson yoke]

But there is another x² here, it remains

[pine spire]

yep got the answer thanks

[crimson yoke]

Okay

[pine spire]

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[atomic pewter]

can someone explain how to do this question please

idk what to do

mb wait

wrong one

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[atomic pewter]

can someone explai how to do this question please

[mild kernel]

where'd you get stuck

[atomic pewter]

I tried to do an iscoles trianglew

but i dk what to do

[mild kernel]

Can you get the area of the pentagon?

Let's say the inner grey circle wasn't there

are you allowed to look up the formula?

[atomic pewter]

well we could just devide it into 3 triangles find the area of on and times it by 3

[mild kernel]

three?

[mild kernel]

why 3?

how about 5

alternatively, we could just use the formula

if you're allowed

[atomic pewter]

cuz there are 3 triangles in the pentagon

[mild kernel]

can you draw what you mean

[atomic pewter]

idk if am allowed

[atomic pewter]

alright

thats what I mean

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[mild kernel]

oh, geez

[mild kernel]

I didn't see that

[mild kernel]

have you found these areas?

this seems suboptimal to me

i think youre going to need to further split things here to make progress

[atomic pewter]

yea

I got 30.433.......

so just 30

for the area of one triangle

so the area of the pentagon is 90

[mild kernel]

I can tell you this is wrong

[atomic pewter]

how

[mild kernel]

because I know the answer

[atomic pewter]

how do you do it then

[mild kernel]

split it into 5 triangles

that are all identical

[atomic pewter]

how

[nova thicket]

hey! you are on the right track 😄 the area of the two of the 108° triangles combined is 8x8xsin(108)
but the middle triangle is not the same area as the other two!

[nova thicket]

draw lines from the center, to each vertex 😄

[mild kernel]

yea

then draw a line from the center to the center of each side

you get 10 identical triangles

[atomic pewter]

ohhh I see

[mild kernel]

they're all right

[atomic pewter]

that makes more sense

[nova thicket]

you can correct your approach, by finding the length of the unknown sides of the middle triangle by using the cosine rule 😄

the cosine rule says that cos(108) = (8^2 + 8^2 - EC^2)/(2x8x8). So now you can find EC, and similarly EB. then you can calculate the area of the middle triangle, since you know all three of its sides!!

[atomic pewter]

but dont you need an angle for that

[nova thicket]

what do you mean,?

[atomic pewter]

to find the area you need 2 sides and one angle

[nova thicket]

if you know all three sides, you dont need any angle 😄

[atomic pewter]

how

[nova thicket]

are you aware of herons formula?

[atomic pewter]

no I havnt heard of it before

the one highlited in red?

[nova thicket]

yes

first find EC and EB 😄 (using cosine rule)

[atomic pewter]

alright

so each side is 12.9

so B and C are 12.9 and A is 8

the angle is 36

[nova thicket]

yes 😄

[atomic pewter]

I meant like the letters in the formula

of the cosine rule

so now ill find out the area

48.9

so 108.9 for the area of the pentagon

[nova thicket]

its supposed to be around 110, but i think its because you rounded the two triangle areas to 30

[atomic pewter]

yea I did

[nova thicket]

dont do that, first add up the areas and then round it

[atomic pewter]

alright

so is it

109.8

[nova thicket]

🤔 still not sure how you got that, because its approximately 110.11

[atomic pewter]

you can round up to 110 for mine and round down to 110 for yours

I think I rounded the triangle in the middle

[nova thicket]

dont do any rounding after calculating the areas of the triangles!! add them up without rounding

[atomic pewter]

alright

so is it overall 110

[nova thicket]

the answer is supposed to be upto 3 sigfigs, why are you rounding at all in the first place?

lets keep everything unrounded. lets move onto finding the radius of the circle so we can calculate its area

[atomic pewter]

alright

so now we can use area of secter to calculate one of this

[nova thicket]

but then you should have followed jan-nikus suggestion from the beginning

[atomic pewter]

then times it by 5

[nova thicket]

all this area calculation was useless

[atomic pewter]

oh

what do we do then

[nova thicket]

we can simply find the radius of the big circle, and subtract the area of the pentagon from the big circle

[atomic pewter]

how do we find the radius of the big circle

[nova thicket]

cosine rule, again

this time draw lines from the center, connecting to each vertex of the pentagon

[atomic pewter]

ohhh

[nova thicket]

so the more efficient approach to this problem would have been:
calculate radius of the big circle
find sector area
multiply by 5

[atomic pewter]

oh we can use the circle therom to work out the angle in the middle

andgle in the middle twice angle at circumference

so 36 x 2

so the angle in the middle is 72

[nova thicket]

well, more simple would be that 360/5 = 72

but this is also fine

now find the radius using cosine rule

find the area of the sector

multiply by 5

[atomic pewter]

I used the sine rule

to work out the radius

which is 6.6 ........

[nova thicket]

why? cos(72) = 2r^2 - 8^2/(2r^2)

[atomic pewter]

it seemed easier

[nova thicket]

okay, sure 😄

now find the area of the sector and multiply by 5

or, find the area of the circle and subtract the pentagon area

[atomic pewter]

yea ill do the area of the circle

is quicker

cuz am doing this to practice for my exams

cuz I got paper 2 and 3 gcses in 2 weeks

paper 1 was abit hard so am practicing more

[nova thicket]

then what we just did was not the best method! i continued with it because you had already done some work in it, but let us keep in mind the fastest method:
find radius
find area of sector
multiply by 5

[atomic pewter]

ohhh

I need to do more of these questions to get to think more of finding the fastest methods cuz I dont have much time in the exam

[nova thicket]

anyways, let us get back to the problem

[atomic pewter]

so the area of the circle is

136.847776

so 136.85

136.85-110.11

26.74

what do I do now

[nova thicket]

so you need to find the radius of the small circle

dont forget what the problem asked for

[nova thicket]

by the way, this is supposed to be around 6.805

anyways, you know the area of the smaller circle, its 26.74

so you can find the radius

by making pi * r^2 = 26.74

then once you find r, you are finished 😄

[vale dock]

@mild kernel Has your question been resolved?

[atomic pewter]

alright

is it 2.92

@nova thicket

[spark sluice]

The correct answer is closer to 3.34, but this error seems to be because of the small errors which have been pointed out in your estimates for the areas. The method is fine, though

[atomic pewter]

alright thank you very much

[spark sluice]

For example you found the pentagon area = 109.8, but in reality its closer to 110.11

The big circle radius is closer to 6.80, but you got 6.6

[atomic pewter]

yea

[spark sluice]

If you find the radius of the small circle using these correct values, you get around 3.34

,calc sqrt((pi6.86.8-110.11)/pi)

[rocky lotus]

Result:

3.345280023003

[spark sluice]

As you can see

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[knotty musk]

i have done a really similar question where the degree of P(x) was specified and p(x) was x/x+1

so i was able to define a g(x) = (x+1)P(x) - x and then figure out P(-1)

but in this case i dont know how to determine the degree

[knotty musk]

i tried this same approach with g(x) = (x+1)(x+2)P(x) - x

and i assumed g(x) is a 11 degree polynomial

but then i realised that it doesnt satisfy g(-1) = 1 and g(-2) = 2

so thats where i got stuck

<@&286206848099549185> help pls

[vale dock]

@knotty musk Has your question been resolved?

[tame wolf]

Are you allowed to use a calculator or some software?

This is extremely tedious otherwise

[knotty musk]

no but i can leave the answer as a sum of fractions

thats what is shown in the answer key

[tame wolf]

Oh alright

You can use Lagrange interpolation for this

https://en.wikipedia.org/wiki/Lagrange_polynomial#Definition

So at each point (0, 1, 2, ...) there is a polynomial which evaluates to 1 at that point and 0 at the others

[knotty musk]

is there an easier way to do it? because this is meant to be solved with high school math

[tame wolf]

For example for 2 it is the following polynomial:
$$\frac{x-0}{2-0}\times\frac{x-1}{2-1}\times\frac{x-3}{2-3}\times\frac{x-4}{2-4}\times\dots\times\frac{x-10}{2-10}$$
Notice how it is a polynomial (with degree 10), such that when $x=2$, the product is exactly 1 (because each fraction becomes 1), but when it is any other number, one of the numerators is 0 and so the whole product becomes 0

[rocky lotus]

d

[tame wolf]

I'm not aware of an easier way

[knotty musk]

yeah this makes sense

so we write those polynomials for 0, 1, .. 10 and then add them?

[tame wolf]

Actually the answer is a fraction of five-digits numbers so I wouldn't expect an easier solution to exist

[tame wolf]

Well, there's one more detail

[knotty musk]

this is the answer key

[tame wolf]

That would evaluate to 1 at each point

[tame wolf]

That's interesting

[tame wolf]

But is it a high school math question, or a math olympiad question which only requires high school math?

[knotty musk]

high school math question

as a matter of fact, this is from our first math chapter which is called fundamentals of math

[tame wolf]

Maybe you have a book which contains some relevant piece of theory

A theorem or something

[knotty musk]

there was a question before this which was similar, perhaps that could help in solving this

[knotty musk]

nope

[knotty musk]

can i send it here?

[tame wolf]

Yes

[knotty musk]

this was solved by taking g(x) = (x+1)f(x) - x

[knotty musk]

heres the solution for this (sorry for the bad handwriting)

[tame wolf]

So that g is a 51 degree polynomial with 51 zeroes, hence g = 0 (?)

[tame wolf]

Wait this makes no sense, you need 52 zeroes

[knotty musk]

why? isnt g 51 degree

and 0,1,2, .... , 50 are zeroes

so u can write g as Kx(x-1)(x-2)..(x-50)

[tame wolf]

Yes, I mean if g had 52 zeroes then g would be 0

But I tried to apply that property with 51 instead of 52

[knotty musk]

ah ok

[knotty musk]

i meant g(u-1) wherever g(u) is written

[tame wolf]

In this case g(x) would have degree 12

So g(x) = kx(x-1)...(x-10)(x-a)

For some k,a

[knotty musk]

how did you determine that?

[tame wolf]

Well maybe we have to prove it

But what is clear is that it is at most 12

[knotty musk]

can you tell me your reasoning though? this was the part where i was stuck so i dont get it

[tame wolf]

There are 11 points so you can find a polynomial P with degree 10 or less

If you choose n numbers (here n=11) and want a polynomial which evaluates to some specific values at those numbers (in this case the values were 0/(0^2 + 3 0 + 2), 1/(1^2 + 3 1 + 2),...), then there is a polynomial with degree n-1 or less

[tame wolf]

This is a well known theorem and I don't think you aren't allowed to use it

[tame wolf]

To see why you can actually use these things

[tame wolf]

But just assume it

So in this case deg(P) <= 10 and g(x) = (x^2+3x+2)P(x)+x so deg(g) = 2 + deg(P) <= 12

[knotty musk]

hmm ok

[tame wolf]

Actually g(0), g(1), g(2),... g(10) must all be zero

So we can do something

If deg(g) < 11 then g = 0

So that means (x^2 + 3x + 2) P(x) = x, this is impossible

Therefore deg(g) is 11 or 12

[knotty musk]

ok that makes sense

[tame wolf]

Oh wait

If we define g like that, when we choose x = -1 the equation becomes g(-1) = 0 P(-1) + (-1), g(-1) = -1 and we lose the P(-1)!

So we want another g

[knotty musk]

g(-1) becomes 1 but yes we lose the P

but i think we can work around that using the same logic as that other question

the main problem was determining the degree

[tame wolf]

I think there is something more, because now the denominator vanishes at the points where you want to find P

[tame wolf]

That's why this happens

Maybe choosing g(x) = (x+2) P(x) + some fraction helps

[knotty musk]

the sub u = x + 1 helped with that last time

because the constant terms would cancel out and u can collect the coefficient of u

and the rest would become zero

but i guess itll be harder to apply that here since we're dividing by u(u+1)

[tame wolf]

I see

[tame wolf]

And there's also the (x-a)

[tame wolf]

(This one)

[knotty musk]

i found a to be -11/5

[tame wolf]

How?

[knotty musk]

im working it out right now

[knotty musk]

we know g(-1) is 1 and g(-2) is 2

and we have 2 unknowns, k and a

so we can solve for both

[tame wolf]

Right

Oh we also have to discard the case deg(g)=11

[knotty musk]

yeah

[tame wolf]

Just do this, but now there are 2 equations and 1 unknown (k)

Hopefully there is no solution

[knotty musk]

k is -10/12!

i think from here it is just tedious calculation

[tame wolf]

You mean in the case deg(g)=11 or deg(g)=12?

[knotty musk]

deg(g) = 12

for 11 there's no solutions

[tame wolf]

But why is g(-1) = 1?

[knotty musk]

g(x) = (x+1)(x+2)P(x) - x

[tame wolf]

Shouldn't it be -1?

Oh -

Yes

[tame wolf]

It's correct, I get the same

[tame wolf]

Yes it seems so

[knotty musk]

alright i still cant work it out

i guess ima give up on this one for now

thanks a lot for your help though @tame wolf

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[small crystal]

for part (a) i dont get why my setup are not giving the right answer

[vale dock]

@small crystal Has your question been resolved?

[wraith heart]

your limits are incorrect

[small crystal]

wait why? idk where i got it wromg

[wraith heart]

in general for areas of polar curves, your angle should never be outside of [0, 2pi]

[small crystal]

but this is also wrong

[wraith heart]

your limits are still incorrect

$\int_a^b = - \int_b^a$

[rocky lotus]

riemann

[wraith heart]

have you learned that yet?

your limits on the first integral is correct. it should be the same for the second integral

[small crystal]

yea i know this

then what would be the bounds for r=3

[wraith heart]

the bounds on the second integral should be the same as the first integral's bounds

[small crystal]

i did that but the area is wrong

[wraith heart]

you typed cosine instead of sine for some reason

[small crystal]

got the right answer

thanks 🙂

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[quick tartan]

Can someone help me with this

I used chebyshev's but then realized it was the lower bound

P(40<X<60) >= .75

Is it just 1 then?

[viscid spade]

May we see your work

[quick tartan]

P(40<X<60) = 1 - P(|X -50| >= 10) >= 1 - (25/100)

-> P(40 <X <60) >= 0.75

[viscid spade]

I see nothing wrong

[quick tartan]

It wants the upper

[viscid spade]

If yes the question is really useless

Thats not chebychev at start

[jaunty briar]

oh wait nvm

[viscid spade]

Chenychev is <= V(x) / eps²

[jaunty briar]

my bad

[quick tartan]

Okay i think it might just be 1 then

lol

idk what else it migh tbe

Cuz this hw unit is literally markov and cheby

[viscid spade]

Maybe they ask for lower bound and they typo also

[quick tartan]

yehh maybe

[viscid spade]

Ask them next time

[quick tartan]

aight

thx

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[real summit]

So

[real summit]

About limits and shiz

[plush crater]

yuh what ab em

[real summit]

I read about the epsilon delta definition

And it say something like, it a limit if you can go far enough out into the sequence so that the distance between ANY (?) number you find (after some particular N) and the proposed limit, is smaller than that epsilon

Wait

Oh wait

[pine prairie]

that's pretty close

the definition is talking about the existence of such an N

but you start with the epsilon. Given any positive epsilon, there exists an N such that bla bla bla

we need to actually be able to find this N

we need to show that for any epsilon > 0, we can find a suitable N such that so and so holds

the so and so is the ANY (?) number you find (after some particular N) and the proposed limit, is smaller than that epsilon

also ANY (?) number you find should be replaced with any term after the Nth one

[real summit]

So if i keep coming across a number in the sequence whose distance to the proposed limit is NOT less than epsilon, call em s_big, and i keep on finding such numbers, EVEN THOUGH most numbers that surround those s_big ARE less than epsilon, and in fact they get smaller and smaller-- because i cant find an appropriate N such that ANY term with n>N yields an appropriate distance, i can't call the sequence converging and it diverges

[pine prairie]

yes

so for example, if i have a sequence 1/n

clearly we know this converges to 0 right?

[real summit]

Yes

[pine prairie]

okay if i tell you

when n is even, the sequence is 1/n

when n is odd, the sequence is 2

this will not converge

because no matter how big you look for this N

all the odd terms after will not be less than 2 away from 0

so if you picked any epislon to be less than 2

you will fail to find convergence to 0

which means there is an epsilon you can pick that breaks it, meaning you don't have "for all epsilon > 0"

it could still possibly converge to something else, not 0, but we know that isn't the case

[real summit]

Is that kind of the same as this?

:

What i drew is like

So you do keep finding numbers that do get closer and closer to the limit, but ALSO sporadically there are numbers far away

So then you cant say it converges

[pine prairie]

yep

from the perspective of a drawing, convergence means that if you label those points, any (open) shape you draw around that limit point will contain the entire sequence, save a finite amount at the beginning

[real summit]

Oh yeah

[pine prairie]

so maybe you need to throw away the first 5 points or first 100 points

but everything afterwards will be inside the shape

for any shape you draw

well, any open shape containing the limit point

open as in dotted lines for the boundary (an open set)

[real summit]

So if the points that are far away is infinite, then that means the entire sequence isnt contained within like say a circle around the limit point

[pine prairie]

that too

[real summit]

But then what does it even mean to converge because an infinite part of the sequence does get arbitrarily close right or no

[pine prairie]

you need so that any circle you draw around the limit point, however small, will contain all but a finite number of points in the sequence

[real summit]

Oh yeah

[pine prairie]

you can say that a subsequence converges to that point

[real summit]

Oh

Interesting

[pine prairie]

but the sequence itself will not because there are circles that exclude infinitely many points

[real summit]

Rigt

[real summit]

I'm not at that paragraph yet

[pine prairie]

there is a cool theorem about bounded sequences always having convergent subsequences but that's further down the line

[real summit]

But a subsequence converging does have the same 'weight' as a sequence converging?

[pine prairie]

it's a completely different thing

[real summit]

Okok

Thankyou thankyou

Merci beaucoup

[pine prairie]

👍

[real summit]

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[small crystal]

why C? idk how to deal with limit with composite function

[storm perch]

Check left and right hand limit

Notice what happens

[small crystal]

i tried -0.11111 and +0.11111 and it does give same value but the graph is different ?

[frozen ledge]

don’t do that

[small crystal]

wot 😭

[storm perch]

Wdym

[frozen ledge]

just notice that whether you approach from the right or left that 1 - x^2 approaches 1 from below

[small crystal]

isnt that how you check it 😭

[frozen ledge]

so you’re really looking for lim x -> 1^- of f(x)

[storm perch]

That was what I was trying to point at

[frozen ledge]

don’t use numbers

it’s a waste of time

and leads to mistakes honestly

[storm perch]

Not really, you can "plug in" 0^- and 0^+ or just check for evenness of the function

[small crystal]

ive been taught to use acutal numbers when its conusing 😭

[glass kelp]

1-x^2 < 1 for x ≠ 0

[frozen ledge]

note that x^2 is always positive so if x -> 0^+ or 0^-, -x^2 will always approach 0^-

[frozen ledge]

this is a bad sign

[small crystal]

so we ignoring x and we js looking at f(1)?

[frozen ledge]

no

why ignore x?

wdym ignore x

[small crystal]

cuz its gonna be same doenst matter we going from right or left

[frozen ledge]

be careful

[frozen ledge]

^

[small crystal]

test in is thurday and im hella cooked

[frozen ledge]

this implies 1 - x^2 approaches 1^-

[small crystal]

so it approach 0 in same sides?

[frozen ledge]

what

[small crystal]

ummm

[frozen ledge]

maybe it’s easier to do what k said

1 - x^2 < 1 for x ≠ 0

so when you approach 0, 1 - x^2 approaches 1

but from below

[small crystal]

oh so its like its not actually 0 and somewhat zero so if we do when its not zero its gonna approach from less then that to 0

[glass kelp]

That’s the point of a limit 😭

I approach something close to that thing

But not actually be there

[small crystal]

but i didnt thought of its gonna approach from one side

[frozen ledge]

it only does that here because of how x^2 behaves

[small crystal]

im failing this exam 🔥

[frozen ledge]

there’s like one limit question tops

[small crystal]

yea cuz it cancels the sign

[frozen ledge]

you need like a 65% for a 5

[small crystal]

i suck at mcq

im really slow at solving problem

anyways

thank for helping!

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[hoary needle]

Is a marginal distribution of a bivariate distribution = to the actual distribution of one of the variables

[hoary needle]

let me phrase this better, if I have p(x,y) and p(y) is some function defined over 0 to 5 and p(x) is somefunction defined over 0-10 now if we compute the marginal which is often written as
p(x)=∫p(x,y)dy
p(x)=∫p(x∣y)p(y)dy
right so even if i integrate over x's domain (0-10) the values (5-10) would just go to 0 bcz p(y) is not defined there
so we sort of "lose" p(x) from 5-10 right so the marginal p(x) is NOT equal to the actual p(x) yes or no?

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@hoary needle Has your question been resolved?

[wraith heart]

the first equation is the right definition.

https://www.probabilitycourse.com/chapter5/5_2_1_joint_pdf.php

where

the slides here cover when to handle the p(y) = 0 cases
https://math.mit.edu/~sheffield/440/Lecture24.pdf

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[hybrid torrent]

im familiar with the inclusion-exclusion principle but not in this form, I would love some clarity

[vocal tusk]

N_n = {1, ..., n}

I is a subset of this

@hybrid torrent

[hybrid torrent]

what is i_1, i_2 then

?

are they just elements of N_n?

[vocal tusk]

numbers from 1 to n

[hybrid torrent]

what is r then

[vocal tusk]

the I is basically to pick sets from A_1, ..., A_n

I formalizes which of these elements to pick

[vocal tusk]

r is just any number 1 to n

which will be the size of I

like u can't write I = {1, ..., r}

cuz the numbers could be different

example:
N_n = {1, ..., n}
I = {1, 2, n}
i1 = 1, i2 = 2, i3 = n
r = 3

[hybrid torrent]

ok so I is an random subset of N_n, and we use the elements of I to pick sets to find an intersection with?

[vocal tusk]

yes

[hybrid torrent]

ok that part makes sense now

thank u

as for the second line

oh ok

it makes sense now that i fully understand that

thank you so much

[vocal tusk]

np

[hybrid torrent]

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[hybrid torrent]

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✅

[hybrid torrent]

@vocal tusk sorry to ask smth again but

for the base case for the proof

n=1

only subset of N_n, I, would be just a single element

so its one set A_1, for example, that makes up A_I

nvm typing it out

made me figure it out

lol

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[crystal wing]

ok, im trying to figure out a critical region for a test X~B(50,0.25).
null hypothesis is p=0.25, and alt hypothesis is p is not 0.25.
now here's where i'm confused because im supposed to get P(X<=a)<=0.025 (since it's a two tailed test), and i got that area under the curve from looking at a cumulative binomial distribution table and finding n=50, and p=0.25, which is 0.237
but if i try to get P(X<=b-1)>=0.975 (a value that's barely larger than 0.975, i don't know where to look because the table i was looking at only has one value for n=50 and p=0.25, which is the critical value for the lower tail.

[crystal wing]

here's the table i was using

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@crystal wing Has your question been resolved?

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@crystal wing Has your question been resolved?

[uneven oracle]

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[jade verge]

my current thouht process was the sum of the two small triangles have a "base" of 10 and so i could merge them to form a triangle with base 10 height 10 giving me an area of 50. Idk if thats right or not but anyway i cannot figure out how to get the big triangle, i didnt know whether or not to try a huge triangle and just minus the area of the square or split it up

[drifting hornet]

the big triangle has base 20, can you figure out the height?

[jade verge]

yeah thats what i was thinking first

geometry doesnt come very intuivtively to me though so i couldnt see anything instantly on how to find the height

[drifting hornet]

it takes 10 units to get from "height" 20 to "height" 10

how many units do you think it will take to get from "height" 10 to the vertex ("height" 0)?

you can also use similar triangles if you want to

the big triangle is similar to the small triangle (which is 2x smaller)

these 2 triangles are similar to be exact

@jade verge you here?

[jade verge]

oh damm sorry, that makes sense ill give it a go thanks

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[normal swan]

Hello, id like help understanding this part of a proof, its a part of a proof that uniform convergence "carries over" riemann integrability, i know how to continue after this point, its just this one part
Maybe theres some clever trick with suprema you can do that i dont quite get

[normal swan]

the closest i got was that the abs of the difference of their suprema would be smaller than the suprema of the abs of all their possible differences on that interval, which isnt enough, im assuming it will use the supremum of the abs of their differences at each point as thats what we know will be less than epsilon

[wraith heart]

can you just show the whole proof and explain which line you don't understand first

[normal swan]

well, its this one, how do we show that the abs of the differences of their suprema is less than or equal to epsilon if, at each point, the abs of their difference is less than epsilon
once i have that, i can show that for any partition the difference of upper integral sums of f(x) and fn(x) will be smaller than epsilon times the lenght of the interval, and a similar inequality will have to be shown for the difference of their infima if i want to compare the lower integral sums, but i might be able to do that one on my own once i understand this one, for the suprema

[wraith heart]

can you just show the whole proof

[stark wedge]

well he ain't doing it so

[wraith heart]

[karmic bloom]

you have fn-f < epsilon
first, using that just try proving sup(fn) - sup(f) <= epsilon

no modulus

[hard panther]

Just show the entire proof, people here can better understand with the full context. Its rarely possible to only have parts of a proof and understand the logic from that.

[normal swan]

This is it, i know how to do the rest

[hard panther]

even if its a sketch

[midnight pier]

Hint:
|f_n(x) - f(x)|<e <=> -e < f_n(x) - f(x) < e <=> f(x) - e < f_n(x) < f(x) + e

[normal swan]

i see
so if i were to apply the supremum to this inequality, id get that sup(f(x) - e) <= sup(f_n(x)) <= sup(f(x) + e) right?

and then i should be able to break apart the outer suprema

and then just subtract sup of f(x)

i think i see

alright that makes sense, thank you
and i think if i were to apply the infimum to that inequality i should be able to get the same thing but for the infima

[midnight pier]

Yes

[normal swan]

yep yep, got it

thank you once again

[midnight pier]

Welcome. Have a great day

[normal swan]

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[jolly stream]

need help with series

n cannot be a incomplete number so how am i wrong?

[crisp flare]

yo guys pppllllllssssssss dm me help me to understand powers must talk frensh but its ok if its english

[wraith heart]

!original

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Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[jolly stream]

if i solve the equation this way then i get the correct answer but how do the two different ways make a difference? they are both correct ways to solve equations😑

[jolly stream]

these are my earlier calculations if it helps

[wraith heart]

,calc (201 - 13)/4

[rocky lotus]

Result:

47

[wraith heart]

you did 4(n-1) = 4n - 1

[jolly stream]

ooh it was supposed to be -4

but still, in that case i get n=48

but the answer is 47

[rancid oxide]

maybe the answer is 47 because you need to add 47 steps to d

[wraith heart]

yea if the first term is indexed by n=0 and last term is n=47, then there are n+1 = 48 terms.

13, 13 + 4, 13 + 4 * 2, ..., 13 + 4 * (48 - 1)

,calc 13 + 4 * (48 - 1)

[rocky lotus]

Result:

201

[rancid oxide]

i would answer the same, i think the provided answer is wrong tbh

the instruction was "Find how many terms in the series if known"

[jolly stream]

interesting
its only a video who gave me the problem and answer

also chatgpt calculated n=48

[wraith heart]

show the video

[jolly stream]

oh wait no i'm an idiot,
the video at the end shows n=48
i only focused on the 47 before the final answer

so sorry guys but thanks for helping

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[wet tundra]

what does b mean by yz plane

[peak edge]

the y-z plane is the set of all points where the x coordinate is 0

[wet tundra]

oh ok

[peak edge]

so if you evaluate |x(t)|

[wet tundra]

do u have anhy idea holw to do dis question

[wet tundra]

wat

[peak edge]

so the x component of $s(t)$ should be $t^2-8t+18$ right

[rocky lotus]

nilo

[peak edge]

if you find the minimum of x(t) you should get the minimum distance to the y-z plane

[wet tundra]

wait how u know dis

[peak edge]

you wrote it in ur answer?

where you wrote out s(t) and the constants

[wet tundra]

ph

oh

wait minimum distance from y-z plane

i would thinkk we have to use the yz plane

not the x

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[safe fulcrum]

@wet tundra can't wait for you to trash our server with uni maths

next year

[sleek nebula]

fr

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[zenith pier]

based on what i know, you do the opposite of what is being done to x. so why is it 6x-6x, and done to the opposite side too? yes you have to make that disappear but i dont get it, why its 6x-6x

[lyric sundial]

Because you need to bring the things containing x on one side (of the equation) only, in this case the LHS

In other words, you should rearrange your first degree equation in this form:

something with x = something without x

[cold salmon]

in order to solve this question, just solve it

[lyric sundial]

Formally ax = b

[lyric sundial]

How else would you have solved it?

Did you want to bring the -11x to the RHS? That could have been a perfectly valid option, if you were wondering about that

[cyan jacinth]

This happened because in the explanation 6x was subtracted from both the sides to maintain equality

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[wanton sinew]

hi, im trying to solve a 2nd order differential equation, i found the roots from the characteristic equation and they are both imaginary, i and -i. so my general homogenous equation would be Ce^(ix) + De^(-ix)

[wanton sinew]

but how would I go about finding the particular solution?

this is the first time i get imaginary roots

the problem looks like this:

y'' + 100y = e^x

should i just guess that the particular solution is Ae^x?

[hot tide]

yes

maybe Ae^(Bx) but i'm not sure

my ODEs are rusty

[wanton sinew]

ok if i try Ae^x first I would get Ae^x(101) = e^x

weird way to write it

101Ae^x = e^x

so A would have to be 1/101

[torn jungle]

hallooooo i am new here

[wanton sinew]

hi new here

this channel is occupied but there are other help channels if you look above, it will say Math Help (Available)

you can ask your questions there

also welcome

[torn jungle]

well i am here to improve my math cuz maths in my country is extra dumb and i love maths for that o am here

[wanton sinew]

nice this is definitely the right place to improve your maths i would say!

[torn jungle]

thanks 🌟

[wanton sinew]

i see what went wrong

in my problem now

i didnt multiply by 100 XD

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[karmic wave]

hi, im trying to get a remainder for something, but the number is too big, i can solve it with exponentiation with powers of 2 but the format of the number isnt right and i have no clue what to do, for context i was solving a RSA decryption example, which was

(25760^26633) mod 41567

i used exponentiation for this and at the end got

(27539 × 24167 × 565 × 35935 × 25760) mod 41567

but as u can see this number is also too big, im stuck here. what im thinking of is shrinking this number to a power and redo exponentiation on it, but idk how and if that would even work

the answer to this should be 28.

[karmic wave]

,w 348083040473900732000 mod 41567

[rocky lotus]

Results provided by WolframAlpha

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

Requested by g9v

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[lapis meteor]

Numerical Integration (my clg taught this in the class when i was outside of the country and when i returned back they were on a new chapter😭) need help learning it.

[maiden isle]

Whats the question

[polar oxide]

if you have a question , go ahead and ask it , i dont think channels are meant for teaching someone an entire chapter

[maiden isle]

Send here a question we will help

[lapis meteor]

alr

Ahhh

No.2

[odd jackal]

,rccw

[lapis meteor]

Is this correct?

[maiden isle]

Yes

[lapis meteor]

OKKK

( I am solving by composite trapezodial rule )Alr so i wrote vakues of y0 y1 and y3 and i am thinking it is because it is divied into 3 equal parts and the difference (h) is 1?

is there a formula or u can just assume

Its also saying estimate the error how do i do that?

i got 4

its saying 9 😭

Wait

The equal value difference is supposed to be for x

mb

i got it

still havent found error yet

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@lapis meteor Has your question been resolved?

[lapis meteor]

.closed

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[noble saffron]

Hello all, I'm new in the server!

I just posted this on Computational Science Stack Exchange, but decided to try my luck here as well.

I am solving the heat-equation $\frac{\partial u}{\partial t} = \alpha\nabla^2 u$ on the domain $\Omega = [0,1]^2$ and interval $t\in[0,1\times10^{-2}]$ with homogeneous Dirichlet boundary conditions in 2D. I am using a central finite-difference method to discretize in space and the forward Euler method to discretize in time, and end up with a scheme like:

\begin{equation}
u^{n+1} = u^n + \delta t\alpha L u^n,
\end{equation}
where $u^n$ are the nodal values of the numerical solution and $L$ is the discrete analog of the Laplacian operator in 2D.

If I compute the relative global error as $e \colon = \frac{||u^T-u^\text{exact}(t_f)||_2}{||u^\text{exact}(t_f)||_2}$ and plot it against the mesh spacing $h$, I get the following plots for $\delta t = 5\times 10^{-4}$ and $\delta t = 1\times 10^{-4}$ (plots in the attached figures)

The red dots correspond to simulations done outside the CFL stability condition $dt \leq \frac{1}{4\alpha h^2}$.

I have two questions about these graphs.

Question 1
Why isn't the error blowing up sooner (for larger $h$) in the case of $\delta t = 5\times 10^{-4}$, since the points in red all violate the CFL condition? Might it be a sign of a problem in my implementation?

Question 2
I find this to be weird behaviour since I would expect the error to be a monotonic increasing function of $h$, instead it seems that there is an optimal $h$ for which the error is minimal.

An explanation I found for this behaviour is that, as $h$ decreases, the condition number of $L$ increases, namely $\frac{\lambda_\text{max}}{\lambda_\text{min}} \approx h^{-2}$. This causes the ODE system to become increasingly stiffer, and thus the local truncation error for an explicit method, such as the Euler method, becomes increasingly large. Is this the correct explanation for this behaviour? If so, can someone give me a more quantitative explanation? In specific, can I predict where the "inverted peak" will happen?

[rocky lotus]

Rui Martins

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[formal storm]

Hi! I was wondering if someone could help me solve for x?

[vestal gyro]

We will guide you.

[formal storm]

I'm pretty sure it has to do with pythagorean theorem but don't know exactly how to solve

[vestal gyro]

Do thou know similarity?

[formal storm]

pardon?

[vestal gyro]

Similarity.

[formal storm]

Ah yes

[vestal gyro]

Triangle ADE is similar to triangle ABC.

[formal storm]

Oh okay !

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[polar oxide]

trying to solve :
$n\geq 2 , (E)_n = (z-1)^n-e^{2i\theta}=0$

[rocky lotus]

<rajel />

[polar oxide]

$z = e^{\left(\frac{2\theta}{n} + \frac{2k\pi}{n}\right)} + 1$

thats what i did , and im not sure if its correct

[rocky lotus]

<rajel />

[ruby sleet]

you're missing some i in the exponent

but looks ok

[polar oxide]

alr , the qustion is to find the solutions to the equation

so i should take some values for n ?

[ruby sleet]

no

n is fixed

it's the k that you introduced

that can take multiple values

[polar oxide]

alr , so i keep taking values for until it get periodic

like the values start to repeat

[ruby sleet]

sure, but you should already know that without "trying values until it repeats"

[polar oxide]

0<k<n-1

so for k=n-1 it would repeat ?

im just guessing , how would to know it without plugin values

<@&286206848099549185>

@ruby sleet sorry to disturb

is what i said correct

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[summer heath]

is there any property or trick to solve this or i have to multiply them and subtract only?

[worn hemlock]

Do you know about row transformations

[next vortex]

hm, my first instinct might be to use row operations

[dusky pulsar]

You could use row manipulation

[viscid spade]

Apply the formula 🗿

[summer heath]

yeah but i don't think that should be used as it will take lot of time

[summer heath]

no

i don't know i mean

[viscid spade]

There is probably manipulation such as factorisation

Ima try

[autumn lantern]

after using the det formula yes

[dusky pulsar]

C1 to c1-c2

?

[ebon glade]

I would have written 7579 as x and then everything else as x+(something small), then multiply out and it probably simplifies nicely

[dusky pulsar]

Same thing tho yeah works

[summer heath]

Oh Thanks

[viscid spade]

Yeah factorisation is working after det formula

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[errant parrot]

hi, i dont understand why there are i and j values for the normal force and friction?

oh this channel is bugged

[errant parrot]

and why the normal force is cos60?

[cinder creek]

i and j represents the x- and y-axis (horisontal and vertical). Since normal force and friction is not vertical to the ground (if so, there are no horisontal component, only vertical), you need to decompose the forces into components.

[errant parrot]

how come normal force is not vertical to the ground?

[lyric sundial]

Normal means perpendicular to the plane of contact

In this case the inclined plane surface

[cinder creek]

If you see the gravity, it's vertical to the ground. If the normal force is vertical to the ground, it need to point opposite to gravity.

[clever sentinel]

Normal force is vertical to the base at which the object is placed.
Here, the base is elevated 30° above ground so the normal force's angle with ground is also increased by 30°.
So angle between normal force and ground = 90°+30°=120°

[twilit elk]

.