help-17

[heady sparrow]

Just differentiate it

[twin meteor]

Devil Wears Prada

[sinful pollen]

this is it

when x = -1

[viral copper]

-2 right

[sinful pollen]

so then the left dy/dx would cancel out the right

so its nonexistent

wait

[steel snow]

why would they cancel out

[sinful pollen]

y = -2

when they cancel

that doesnt solve problem

nvm

[viral copper]

one side is dy/dx and the other side is -dy/dx

they don't cancel

[sinful pollen]

asks for dy/dx

why not

i add dy\dx

to cancel it out

and do the same on the other side

[viral copper]

so you have 2dy/dx on the other side

[sinful pollen]

ohhh

so then dy\dx = y\2

so the answer is A

[steel snow]

dy/dx = (y+2)/2

[sinful pollen]

ydes

yes

oh wait

y

dy/dx = y

[viral copper]

,w y = xy + x^2 + 1; dy/dx at x = -1

[twin meteor]

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Requested by neonperseus

[viral copper]

smh

[steel snow]

ok so what do you need to find dy/dx now

[sinful pollen]

find y

[steel snow]

good thing you have this

what is y when x = -1

[sinful pollen]

-y + 2

[steel snow]

what

[sinful pollen]

yes

[steel snow]

y = (-1)y + 1 +1

[sinful pollen]

exactly

[steel snow]

so 2y = 2

y = 1

[sinful pollen]

what

no y = -y + 2

[steel snow]

add y to both sides

[sinful pollen]

omg

2y = 2

y = 1

[steel snow]

yuhh

now substitute it back in hnere

[sinful pollen]

ok now

3.2

3/2

[steel snow]

wait I think you didn't differentiate correctly then 😭

[sinful pollen]

oh

how

[gentle sleet]

yea the derivative isn't right, it should be 2x

[sinful pollen]

oh yeah

i forgot to diff x^2

i just left it as x^2

gime a sec

[steel snow]

my differentiation would be dy/dx = dy/dx * x + dx/dx * y + 2x + 0

dy/dx = dy/dx*x + y +2x

[sinful pollen]

dy/dx = -1

is the asnwer

[steel snow]

dy/dx - x*dy/dx = y+2x

[sinful pollen]

so C

is the choice

[steel snow]

dy/dx ( 1-x) = y+2x

dy/dx = (y+2x)/1-x

I got B

[sinful pollen]

is it C

what

[steel snow]

did you use product rule when you differentiated

[sinful pollen]

yes

i got the same diff as you

[steel snow]

dy/dx = (y+2x)/1-x ?

[sinful pollen]

no

[steel snow]

check my work

[sinful pollen]

according to this it would then be dy/dx = -dy/dx + y - 2

which would then be 2dy/dx = y - 2

oh

so tis then

-1/2

u jsut arranged it differently

i made a mistake

its B

thanks

[steel snow]

hmm

yes

thank you

[sinful pollen]

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[sinful pollen]

im stuck on $\sqrt{y}$

[twin meteor]

Devil Wears Prada

[sinful pollen]

what would that be

[steel snow]

what are you talking about

like what would sqrt(y) be when

[sinful pollen]

dy/dx

the diff of it

[steel snow]

you can do u sub

[sinful pollen]

whats that

[viral copper]

[ \dv{\sqrt y}{x} = \dv{\sqrt y}{y} \cdot \dv{y}{x} ]

[twin meteor]

neon

[viral copper]

chain rule

[steel snow]

wait stop im on integrals unit and I keep dying

use chain rule

[viral copper]

beautiful formatting I know

[sinful pollen]

i dont understand that chain rule formatting

[steel snow]

yeah im not trying to learn this texit stuff

[sinful pollen]

how do i do chain rule here

[steel snow]

u is sqrt(y)

[sinful pollen]

what is u

you just said the whole thing

[steel snow]

wait did i die

[sinful pollen]

no

u is sqrt

or 1/2

v = y

i think

wait

im not sure

[sinful pollen]

is it

$$\frac12y^{-\frac12} \cdot \frac{dy}{dx}$$

[twin meteor]

Devil Wears Prada

[steel snow]

i think so yes

[sinful pollen]

ok

well i got dy/dx

$$\frac{dy}{dx}=-\frac12x^{-\frac12}-\frac12y^{-\frac12}$$

[twin meteor]

Devil Wears Prada

[steel snow]

👍

[sinful pollen]

so is the asnwer D

[steel snow]

what do you think 😭

i didn't compute it

[gentle sleet]

it's not, it's in the wrong direction

orange should be the tangent line

[sinful pollen]

what

so im right?

4/3

is the slope

of the tangent line

[gentle sleet]

no, it should look like this
this is the correct tangent

[sinful pollen]

4/3

r u saying my diff is the opposite

like -4/3

[gentle sleet]

that eq isn't right
how did you get to that?

[sinful pollen]

its not?

[gentle sleet]

no, just that the slope should be -, and yours is positive

[gentle sleet]

no

[sinful pollen]

what is the correct diff

[gentle sleet]

how did you go from $-\frac12x^{-\frac12} - \frac12y^{-\frac12}\frac{dy}{dx} = 0$ to that?

[twin meteor]

Sepdron

[sinful pollen]

cant i just move the dy/dx to the right

then divide by -1

[gentle sleet]

you can't move just the dy/dx, it's attached to the -1/2 y^(-1/2)

[sinful pollen]

oh yeah

so how would i find dy/dx here

i cant simplify can i

[gentle sleet]

move the whole y term, then divide

[sinful pollen]

what

how

[gentle sleet]

$-\frac12x^{-\frac12} = \frac12y^{-\frac12}\frac{dy}{dx} $

uh

[sinful pollen]

$-\frac12x^{-\frac12} = \frac12y^{-\frac12}\frac{dy}{dx}$

[twin meteor]

Devil Wears Prada

[gentle sleet]

ty

[sinful pollen]

ohhh

so it would be that x component divided by that y component

[gentle sleet]

yup

[sinful pollen]

and in order to find the slope of dy/dx at a certain point can i just input the given coordinate values into dy/dx

or do i have to use the original function

[gentle sleet]

here you already have the x and y
so you can directly plug it in

[sinful pollen]

ok

[gentle sleet]

but if you don't have the y, you'd need to use the original equation

[sinful pollen]

so the answer is B

[gentle sleet]

mhm

[gentle sleet]

that's what I used there

[sinful pollen]

ok thanks

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[next meteor]

im asking about b

what would the integral bounds be if the triple integral is dzdydx

i can tell you what i came up with but im not sure if it's correct

[vocal sleet]

@next meteor Has your question been resolved?

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@next meteor Has your question been resolved?

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[fresh compass]

hey

[fresh compass]

wouldnt the remaning pens will be 40 instead of 20

[arctic mantle]

he sells 20 out of the 40 at a profit of 20% and the pens left over at a loss of 60%

[fresh compass]

he make profit from 20 pens but loss from 40 pen . he sold only 60

then why here it say 20 from loss too ? because 20x12 is the profit

[sour wadi]

No way its "Devil Wears Prada"

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@fresh compass Has your question been resolved?

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[silent wren]

how would you solve this:

[somber portal]

wtf is a pronumeral?

[silent wren]

the x and h

[somber portal]

okay. can we assume the base of the pyramid is a square?

[silent wren]

yea

[somber portal]

so x is the diagonal of the square

[silent wren]

yes

[somber portal]

can you compute that?

[silent wren]

OH

yeahhh

that was easy

thanks

[somber portal]

you are missing h yet

[silent wren]

once i find x

wait im confused

nvm

help pls

[somber portal]

do you have x?

[silent wren]

2.83?

[somber portal]

2sqrt2, but okay

h is the altitude of the pyramid

[silent wren]

ye

[somber portal]

it goes from the middle of x, perpendicularly

[silent wren]

yep

[somber portal]

so now you got a right triangle

[silent wren]

yeah

[somber portal]

you know the hypothenuse, 3, and one leg, x/2

can you get the other leg?

[silent wren]

whats a leg??

i got it

x=2.83 and h=2.65

.close

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[gleaming kelp]

.close

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[storm cedar]

Solve without using L'Hôpital's rule. I'm assumming using just analysis and set theory. My attempts to multiply by conjugate always end in (5-x) either in nominator or denominator, and 0 in the opposite in each case, so (0/0) no matter what. I can't get rid of the root to analyze it easily either.

[cosmic cloud]

multiply by conjugate

both numerator and denominator contain square root so

multiply/divide by conjugate of numerator and also with denominator

[storm cedar]

that's right, thanks

.close

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[tranquil linden]

I solved this matrix to find "a" if Delta = 0, the calculations are correct. My only question is, what if I tried to do it with a column and not a row? Do I have to select a row?

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[cobalt dagger]

hi

[tropic geode]

hi

[vocal sleet]

@cobalt dagger Has your question been resolved?

[cobalt dagger]

i never had a problem ...

[flat whale]

.close

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[fallow topaz]

Does my answer look correct

[vocal sleet]

Please don't occupy multiple help channels.

[flat whale]

can you close one of your channels

[fallow topaz]

i did it now but it doesnt close i think its a glitch

[flat whale]

thx bot's just slow.

[fallow topaz]

ok👍

[vocal sleet]

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@fallow topaz Has your question been resolved?

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[simple mason]

For a non-negative intgere k evaluate $\sum_{j=0}^{k} 2^{k-j} {{k+j}\choose{j}}$

[twin meteor]

Sherif Player

[simple mason]

What I see here is that this looks like a lot to binomial theorem of (2+1)^k
Except instead of K choose j we got k+j choose j

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[midnight crag]

is this 1/2sqrt(7) like 1/2sqrt(7) x (solution to 1 + solution to cos 2theta) or is it like 1/2sqrt(7) x solution to 1 x solution to cos 2theta

[midnight crag]

with x being multiplying

[silk osprey]

the first

[midnight crag]

a oki

[silk osprey]

but please write legibly

[midnight crag]

always get them mixed up lol

[midnight crag]

LMAO

im doing it with a mouse

[silk osprey]

makes sense

[flat whale]

you could use this input thingy
https://www.wolframalpha.com/

but latex would be the best

[midnight crag]

too much trouble typing it all out ngl...

[silk osprey]

it’s so simple

$\frac{1}{2\sqrt{7}}\int (1+\cos(2\theta)) d\theta$

[twin meteor]

knief

[midnight crag]

THAT IS SIMPLE?

[silk osprey]

yes

[midnight crag]

all of those curly brackets just confuse me lol

[flat whale]

just type one character at a time

[silk osprey]

i mean you can create shortcuts to not have to use those i guess

[midnight crag]

but if i have some free time ill look into it

[silk osprey]

takes less than a day

look up overleaf

[midnight crag]

huuh okay

ty

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[woven path]

So what do I do here?

[woven path]

do I use this on the original equation?

[silk osprey]

do you need to use this

[silk osprey]

^

[woven path]

awesome

[silk osprey]

or can you just use differentiation rules

[flat whale]

,tex .diff rules

[woven path]

differentiation rules?

[twin meteor]

riemann

[flat whale]

have you seen any of those

[woven path]

yeah, some

[silk osprey]

power rule?

[woven path]

im just trynna refresh before the test

[woven path]

yeah

[silk osprey]

then just use that

no need to use first principles

[woven path]

alr

I just hope that I dont have to do it the long way during the test

thanks guys 🔥

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[woven path]

y'all

this aint it

[flat whale]

the derivative only gives you the slope

tangent lines have the form (slope) * x + (y-intercept)

you still need to find y-intercept. that's when x=-5 comes in

[woven path]

so

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[keen grail]

Help

[covert python]

find the angle between segment x and segment 34

use the fact that its a complementary angle

[keen grail]

Complementary angle?

[covert python]

adds to 90°

is x parallel to segment 16?

[keen grail]

I think so, its text question transferred to a drawing and x is a person standing and 16 is a wall so i'm assuming that it is parallel

[covert python]

that seems like it would be

[keen grail]

how would that help me find x?

[covert python]

you could find all the angles in the triangle

[keen grail]

and then use trigonometry to find distance x?

[covert python]

yes

[keen grail]

makes sense, thank you

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[autumn idol]

Hi guys I need help with question 1

F continuous in IR ; limx->+ ♾ of f(x) = l And Limx->- ♾ of f(x) = l’

Show that if ll’>0 then there exist a,b in IR^2 in which f(a)*f(b)> 0

[hybrid flicker]

tu as soit l > 0, l' < 0 ou l'inverse

[autumn idol]

Oui

[hybrid flicker]

si l > 0, tu peux trouver un a tel que f(a) > 0

[autumn idol]

Je l’avais resolu en utilisant la definiton du limite mais je veux savoir s’il y a une autre methode

[autumn idol]

Comment ?

[hybrid flicker]

def de la limite, trouve a tel que |f(a)-l| < l/2

(epsilon= l/2)

[autumn idol]

C’est correct comme ca ?

[hybrid flicker]

oui, dis juste que le cas non traité se fait de manière analogue

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[raw quail]

its a simple question but im somehow struggling lmao

[silk rampart]

60 degree in standard position

Could you please clarify what is meant by "standard position"

@raw quail

[raw quail]

an angle where the initial side lies on the x axis and the vertex is the origin point in the cordinate plane

[silk rampart]

Great, alright

[raw quail]

so it’s 60?

also can you explain why so?

[silk rampart]

do you know how many degrees are in one revolution

[raw quail]

yup 360

[silk rampart]

no, i havent solved it yet

[raw quail]

ohh okay

[silk rampart]

alright, well we're rotating it by two and a quarter revolutions

two and a quarter, that's 2.25

so we're rotating it by 2.25 * 360 degrees

are you getting this

[raw quail]

yup

[silk rampart]

,w 2.25 * 360

[twin meteor]

Results provided by WolframAlpha

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Requested by _nofluxgiven

[silk rampart]

and we're rotating it by 810 degrees anticlockwise

well, by rotating an angle by 360 degrees, nothing changes

[raw quail]

ohh so its the quarter that makes a difference?

[silk rampart]

so in essence we're rotating it by 810 - 360*2 degrees anticlockwise

yep

which is 90 degrees, if i've done my arith right

[raw quail]

yup quarter is supposed to be 90 i think

but none of the choices are 810

or is it an equivalent angle?

[silk rampart]

we're getting there

810 is not the answer

it was the angle we have to rotate it by

which is just 90 degree

rotating the angle 90 degree anticlokwise would yield this

an angle of 150 degrees

[raw quail]

ahh i see, so if i ever get a question that mentions a rotation and a half, rotation and a quarter etc do i just ignore the whole rotation and solve it with the half or the quarter?

[silk rampart]

depends on what is asked, though when the opportunity arises now you should be able to save some time

can you do the 22nd question on your own now?

[raw quail]

yeah i think its the 3rd quadrant?

[silk rampart]

indeed

3.5 revolutions

is nothing but 0.5 revolutions

which is a 180 degree turn

so your angle was initially in the first quadrant

flip it around, we get the third

done in seconds

[raw quail]

alright thank you so much

[silk rampart]

anytime

[raw quail]

.close

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[blissful peak]

how do i factor the original fuction

[blissful peak]

nah u good

.close

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[teal flame]

can someone help me solve this?

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[desert glen]

.close

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[tulip peak]

So, how are you supposed to do these derivatives? Don't know math terms in English, I've asked my group about them, but they're not responding, so I'd like to know how to solve them

If possible, I'd like someone that knows Spanish to help

[mild rapids]

ola

en la mayoria se usan solo reglas de derivacion

se usan mayormente la regla de que la derivada es aditiva o algo asi, y la regla de potencia

[hidden turret]

hawk tuah differentiate that thang

[tulip peak]

Entonces solo las haría normales, en base a esas reglas?

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[opal quarry]

can i use componentwise contiunity

[opal quarry]

to prove this?

i tried to

consdier
$( (g \circ f) (p) )_i = g_i (f(p))$

[twin meteor]

rhurhuzz ☆

[opal quarry]

and i thought that should prove it but that doesnt use the conitnuity of f at all so im not really sure

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[sacred jewel]

can somebody help me with

[sacred jewel]

Ive done 2πS=rh+r^2

to get rid of the bracket

what do i do next

move the r squared to the other side?

[heavy yoke]

that would be a good next step

but your first step is not quite correct

[sacred jewel]

i mean like

S/2π

kinda hard to type it up

[vocal sleet]

@sacred jewel Has your question been resolved?

[sacred jewel]

this was the answer

is there like

something wrong

[woeful igloo]

thats the same answer as stockadvisor gave you

[sacred jewel]

oh

[opaque shell]

mhm

[woeful igloo]

You just need to bring that solitary -r over the denominator 2pi*r, and then they would look the same

[sacred jewel]

alr

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[lime kraken]

how do i 15

[heavy yoke]

now that you have access to the determinant you have a simple test for invertibility

[lime kraken]

i know if (k-3) and (k-2) = 2 or -2

[oak magnet]

Well det(A) = 0, A not invertible

[lime kraken]

det(A) can only be 0 if i know if (k-3) and (k-2) = 2 or -2

but how do i find the k here

[oak magnet]

whats det of a 2x2 ?

[lime kraken]

1/ad-bc

[heavy yoke]

no, that is 1/det(A)

[oak magnet]

Thats for invert calcul we don't actually need it here

But you got det(A) right so what does it gives if u equal it to zero ?

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[mellow wigeon]

find all the real numbers α where ∀ n ∈ ℕ* we have ⌊α⌋ + ⌊2α⌋ + ... + ⌊nα⌋ is a multiple of n

[river minnow]

Start by considering cases on floor(2a) and plugging in n = 2

[mellow wigeon]

so we should have ⌊α⌋ + ⌊2α⌋ ≡ 0 (mod 2)

[uneven flicker]

What is N*?

[mellow wigeon]

positive integers different to 0

[uneven flicker]

Ohh

[mellow wigeon]

so ⌊α⌋ and ⌊2α⌋ must both be odd or both even

[uneven flicker]

You can't say

[mellow wigeon]

why

[uneven flicker]

floor(1.9) = 1

floor(3.8) = 3

[river minnow]

Does 1.9 satisfy the condition of a?

floor(1.9) + floor(3.8) = 4

3 is not a multiple of 2

[mellow wigeon]

3.8 not 2.8

[river minnow]

,w floor(1.9) + floor(3.8) + floor(5.7) + floor(7.6) + floor(9.5) + floor(11.4) + floor(13.3) + floor(15.2) + floor(17.1) + floor(19) + floor(20.1)

[uneven flicker]

No condition on a

[mellow wigeon]

for α positive

α - ⌊α⌋ should be ≥ 0.5

is that true?

(in case of n = 2)

[river minnow]

Not a multiple of 11

[mellow wigeon]

missing a +

you did ×

between 11.4 and 13.3

[twin meteor]

Results provided by WolframAlpha

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Requested by alonerbean

[river minnow]

No, why?

[mellow wigeon]

hmm try for n is an even number

try 10

remove the last one

[river minnow]

10 does work

[mellow wigeon]

12?

[river minnow]

12 too

You mean n = 12?

[mellow wigeon]

yes

[river minnow]

Why does it matter?

It fails for one n, so it's not a solution

[mellow wigeon]

I want to see if it is working with n where n is an even number

[river minnow]

,w floor(1.9) + floor(3.8) + floor(5.7) + floor(7.6) + floor(9.5) + floor(11.4) + floor(13.3) + floor(15.2) + floor(17.1) + floor(19) + floor(20.9) + floor(22.8)

[warm vine]

why does value of n matter

[twin meteor]

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Requested by alonerbean

[river minnow]

The parity of n doesn't matter much

[mellow wigeon]

hmm

[river minnow]

.

[warm vine]

hint: let $\alpha = k + \epsilon$ where $k$ is integer and $0\le \epsilon < 1$

[mellow wigeon]

we got ⌊α⌋ + ⌊2α⌋ = 2k

[twin meteor]

Astar777

[river minnow]

What can floor(2a) equal in what cases?

[mellow wigeon]

depends on this inequality:
α - ⌊α⌋ ... 0.5

[river minnow]

Right, floor(2a) = 2floor(a) if floor(a) <= a < floor(a) + 1/2 and floor(2a) = 2floor(a) + 1 if floor(a) + 1/2 <= a < floor(a) + 1

[mellow wigeon]

like if α = 1.4
⌊2α⌋ = 2 = ⌊α⌋ + 1
but if α = 1.6
⌊2α⌋ = 3 = ⌊α⌋ + 2

this is where α is positive

[mellow wigeon]

I didn't get it

[warm vine]

,, \floor{x} = x - {x}

[twin meteor]

Astar777

[warm vine]

right?

[mellow wigeon]

what does {x} represent?

[warm vine]

fractional part function

[mellow wigeon]

after . ?

like 2.3
it is 0.3

[civic otter]

For positive numbers yes

[mellow wigeon]

yes that what I was going to say

[civic otter]

For negatives it's 1 - decimal part

[uneven flicker]

For -3.4 it's 0.6 lmao

[river minnow]

Consider these cases in floor(a) + floor(2a) = 0 mod 2

[mellow wigeon]

works for positive and negative α ?

right?

[river minnow]

Yes

[mellow wigeon]

in the first case ⌊2α⌋ = 2⌊α⌋ or 2 + ⌊α⌋ ???

[river minnow]

In the first case floor(2a) = 2floor(a) and in the second case floor(2a) = 2floor(a) + 1

[mellow wigeon]

so in the first case
⌊α⌋ should be even

I got 3⌊α⌋ ≡ 0 (mod 2)

[river minnow]

Right

[mellow wigeon]

so in this case
α ∈ ... U [-4, -3.5[ U [-2, -1.5[ U [0, 0.5[ U [2, 2.5[ U ...

[river minnow]

What about the second case?

[mellow wigeon]

wait

⌊α⌋ + 1 is even

or

wait

[river minnow]

That's correct

[mellow wigeon]

oh yes

so ⌊α⌋ is odd

[river minnow]

Now, for each case separately, plug in n = 3, 4, ...; Each time try narrowing the inequality of a down and at some point you should notice a pattern

[mellow wigeon]

α ∈ ... U [-5, -4.5[ U ....
combining the first and the second one
α ∈ ... U [-5, -4.5[ U [-4, -3.5[ U ...

[mellow wigeon]

👌

⌊3α⌋ = 3⌊α⌋ if ⌊α⌋ ≤ α < ⌊α⌋ + ⅓ ?

[river minnow]

Yes, but what if floor(a) + 1/3 <= a < floor(a) + 1/2?

[mellow wigeon]

you mean in the end + ⅔ ?

[river minnow]

No

[mellow wigeon]

huh

[river minnow]

Let's stay in the case floor(a) <= a < floor(a) + 1/2

[mellow wigeon]

ok

for exemple 3.4
⌊3 × 3.4⌋ = ⌊10.2⌋ = 10
so it is 3⌊α⌋ + 1 ?

[river minnow]

Yes

Right so you have floor(a) + floor(2a) + floor(3a) = 0 mod 3

[mellow wigeon]

mod 3

[river minnow]

We know that floor(2a) = 2floor(a)

What can be said about floor(3a) then?

[mellow wigeon]

yes

⌊α⌋ is odd?

[river minnow]

You should get floor(3a) = 0 mod 3

So floor(3a) must be 3floor(a)

[mellow wigeon]

oh wait I forgot it is mod 3 not 2

[river minnow]

Can you express this using an inequality?

[mellow wigeon]

I'll try my best

⌊3α⌋ = 3⌊α⌋
if ⌊α⌋ ≤ α < ⌊α⌋ + ⅓ ?

[river minnow]

Yes

You see how the inequality has narrowed down?

We had floor(a) <= a < floor(a) + 1/2 now it became that

[mellow wigeon]

yes

+1/n

[river minnow]

You may check yourself that the same happens at every n after that

Now prove this pattern (by induction) and conclude what a should be equal to

[mellow wigeon]

⌊nα⌋ = n⌊α⌋
if ⌊α⌋ ≤ α < ⌊α⌋ + 1/n ?

[river minnow]

Yes

[mellow wigeon]

and if we have this? then
⌊(n-1)α⌋ = (n-1)⌊α⌋ ?
since 1/n is less than 1/(n-1)

[mellow wigeon]

idk if this is helpful

[river minnow]

Yes, the same will hold for all positive integers <= n consequently

[mellow wigeon]

yes

[river minnow]

Okay, anyway, now use induction to show that floor(na) = nfloor(a) for all positive integers n

[mellow wigeon]

what does induction mean?

pattern?

sorry my english is not very good

[river minnow]

Proof by induction of a statement for positive integers is a proof involving two steps. You show that the statement holds for the number 1 and then you show that the statement being true for some positive integer k implies it to also be true for k+1

[mellow wigeon]

means
for n = 2:
...
for n = 3:
...
in general for any positive integer n different to 0:
...

is that what you meant?

[river minnow]

That's what we want to show, yes

If in ... you meant floor(na) = nfloor(a)

[mellow wigeon]

⌊nα⌋ = n⌊α⌋
if ⌊α⌋ ≤ α < ⌊α⌋ + 1/n

⌊nα⌋ = n⌊α⌋ + 1
if ⌊α⌋ + 1/n ≤ α < ⌊α⌋ + (n-1)/n

wait I didn't understand too much

back to n = 3

we said if ⌊α⌋ ≤ α < ⌊α⌋ + ⅓
so:
⌊α⌋ + ⌊2α⌋ + ⌊3α⌋ ≡ 0 (mod 3)
⌊α⌋ + 2⌊α⌋ + 3⌊α⌋ ≡ 0 (mod 3)
6⌊α⌋ ≡ 0 (mod 3)
which is always true
if ⌊α⌋ ≤ α < ⌊α⌋ + ⅓

[river minnow]

You don't need to assume anything here

[mellow wigeon]

means in the first case
α ∈ ... U [-3, -2.6666...[ U [-2, -1.6666...[ U [0, 0.3333...[ U [1, 1.3333[ U ...

[mellow wigeon]

why

[river minnow]

We know that floor(a) + floor(2a) + floor(3a) = 0 mod 3

We know that floor(2a) = 2floor(a)

Plugging in, you get floor(3a) = 0 mod 3, right?

[mellow wigeon]

or 2⌊α⌋ + 1?

[river minnow]

Let's stay in the first case for now

Doing both simultaneously will make things difficult

[mellow wigeon]

ok

[river minnow]

floor(3a) can only ever be equal to 3floor(a), 3floor(a) + 1, 3floor(a) + 2

Which one of these possibilities satisfies floor(3a) = 0 mod 3?

[mellow wigeon]

yes

[mellow wigeon]

3⌊α⌋

[river minnow]

Right, so floor(3a) = 3floor(a)

[mellow wigeon]

yes

[river minnow]

For practice before writing the proof, try showing floor(4a) = 4floor(a) with floor(a) + floor(2a) + floor(3a) + floor(4a) = 0 mod 4

[mellow wigeon]

ok hold on

6⌊α⌋ + ⌊4α⌋ ≡ 0 (mod 4)
so ⌊4α⌋ ≡ 0 (mod 2) ?

[river minnow]

You are missing some steps

[mellow wigeon]

before?

[river minnow]

Can't just right from 6floor(a) + floor(4a) = 0 mod 4 to floor(4a) = 0 mod 4

[mellow wigeon]

6⌊α⌋ ≡ 0 (mod 2)

2×3⌊α⌋ ≡ 0 (mod 2)

[river minnow]

Right, but what about mod 4?

[mellow wigeon]

6⌊α⌋ + ⌊4α⌋ ≡ 0 (mod 4)
since 6⌊α⌋ ≡ 0 (mod 2)
means ⌊4α⌋ ≡ 0 (mod 2)

[river minnow]

How does 6floor(a) = 0 mod 2 imply floor(4a) = 0 mod 2?

[mellow wigeon]

logicly?

[river minnow]

Ah that yes

[mellow wigeon]

no?

[river minnow]

But why are you doing mod 2

[mellow wigeon]

I just tried I thought it maybe lead to a result

to see ⌊4α⌋ = ..

[river minnow]

6foor(a) = 2floor(a) mod 4
Since floor(a) is even, think about possible values of 2floor(a) mod 4

[mellow wigeon]

didn't get it

first line

6⌊α⌋ + ⌊4α⌋ ≡ 0 (mod 4)

[river minnow]

6 = 2 mod 4, right?

[mellow wigeon]

yes

multiply by

⌊α⌋

[river minnow]

So 6floor(a) = 2floor(a) mod 4

[mellow wigeon]

yes

6⌊α⌋ ≡ 2⌊α⌋ (mod 4)

[river minnow]

Now think about the possible values of 2floor(a) mod 4 given that floor(a) is even

[mellow wigeon]

multiple of 6?

I mean ⌊α⌋

[river minnow]

Sure

And take remainder when dividing by 4

[mellow wigeon]

⌊α⌋ multiple of 2?

[river minnow]

Yeah

[mellow wigeon]

so α ∈ ... U [-2, -1[ U [0, 1[ U [2, 3[ U ...

?

[river minnow]

That's true, yes

[mellow wigeon]

so

[vocal sleet]

@mellow wigeon Has your question been resolved?

[vocal sleet]

@mellow wigeon Has your question been resolved?

[mellow wigeon]

bro I didn't get it

for n = 2
α ∈ ... U [-4.5 , -3.5[ U [-2.5 , -1.5[ U [-0.5 , 0.5[ U [2.5 , 3.5[ U ...

for n = 3
α ∈ ... U [-5.3333, -4.6666[ U [-4.3333, -3.66666[ U [-3.3333, -2.6666[ U ...

for n = 4
α ∈ ... U [-4.25, -3.75[ U [-3.5, -3.3333[ U [-2.666, -2.5[ U [-2.25, -1.75[ U [-1.5, -1.3333[ U [-0,6666, -0,5[ U ...

I see no relation between the values of n
please help me

<@&286206848099549185>

find all the real numbers α ∀ n ∈ ℕ*
⌊α⌋ + ⌊2α⌋ + ... + ⌊nα⌋ is a multiple of n

[mellow wigeon]

???

help please

I am cooked

[mellow wigeon]

I got something
if n is odd
α ∈ [⌊α⌋, ⌊α⌋+1/n[ U [⌊α⌋+(n-1)/n , ⌊α⌋+1[

I have no proof
just I realised that for n = 3
I got
α ∈ [⌊α⌋, ⌊α⌋+⅓[ U [ ⌊α⌋+⅔, ⌊α⌋+1[

ans for n = 5 I got
α ∈ [⌊α⌋, ⌊α⌋+⅕[ U [⌊α⌋+⅘, ⌊α⌋+1[

I didn't try n = 7 tbh but I guess it is the same

[mellow wigeon]

pleaseeeeeee helppp

find all the real numbers α ∀ n ∈ ℕ*
⌊α⌋ + ⌊2α⌋ + ... + ⌊nα⌋ is a multiple of n

pleaseeeee helpppppp
<@&286206848099549185>

[lofty vapor]

That’s imo last year

[dusk zodiac]

imo p1

[mellow wigeon]

how did you knoww

ok what is the result?

[mellow wigeon]

is this correct first?

[mellow wigeon]

hope you remember me

[prisma rain]

it's a imo problem, there are many resolutions on youtube

[dusk zodiac]

yeah

[mellow wigeon]

can you send me link

[dusk zodiac]

https://artofproblemsolving.com/wiki/index.php/2024_IMO_Problems/Problem_1

[vocal sleet]

@mellow wigeon Has your question been resolved?

[vocal sleet]

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[fading cove]

do you want to solve x ?

[tropic dock]

i think its -5

[queen root]

u can try verifying that

by substituting it into x in the equation

[fading cove]

look put 1 to the left side

so you get an equation equal to 0

then you put 1 in the fraction

[oak magnet]

If a/b = 1 then a = b neccessarly

So you solve for x² + x -30 = x-5

Which is quite forward to sol, despite a little trap to avoid falling in

[peak matrix]

alternatively, just factor the numerator

[tropic dock]

alright

.close

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[plush summit]

I have the function given in the picture

now my question is how can I shorten the function

is e^-4x²-1 (15sin(-3x + 1/6π) - 40xcos(-3x+1/6π)) correct?

[hybrid flicker]

yes

and you can still factor by 5

[plush summit]

is this more correct?

[median crane]

Looks good

[vocal sleet]

@plush summit Has your question been resolved?

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[vital latch]

so the body K is limited by the curves at the top

and we're supposed to calculate the volume

the only thing I'm struggling with is how they figured out what the value of theta had to be?

Like I get that we can describe the dxdy as a circle with x = rcos(phi) and y = rsin(phi) if we let phi go between 0 and 2pi and r = 1

but how do we know what value theta has to be for z to not "go outside" the boundaries when we integrate?

and how do we know what to integrate when we actually integrate as well? very confused :C

never really understood why/how you put a 1 in the first one for example

like I do see how we can get theta

if we set sqrt(x^2+y^2) = sqrt(1-x^2-y^2)

then we can get x^2+y^2=1-x^2-y^2 = 2(x^2+y^2) = 1

x^2+y^2 = 1/2

so we should be intersecting at a circle with the radius 1/sqrt(2)

[gaunt sparrow]

Theta represents the angle from the z axis to a given point.
Here it's delimited by 0 and pi/4 because the cone opens at angle pi/4.
In general, you could find that value by solving for the intersection.

[gaunt sparrow]

x^2 + y^2 = 1/2 gives you z^2 (from the first equation)

Then you can find r since you have a value for x^2 + y^2 + z^2

So now you have values for z and r and you can solve for theta in z = r cos(theta)

[vital latch]

uhhh wait

so assuming i don't know that a cone opens at angle pi/4

i can set the restrictions to be equal to each other to know the intersection, yeah?

and then I know that x^2+y^2=1/2

[gaunt sparrow]

Yep

1/2

[vital latch]

ah wait

yes

[gaunt sparrow]

z will be the sqrt of that

[vital latch]

and how would I know that? a lot of numbers in my head atm

from the first equation?

but in either case, how do I get theta from this?

or are you saying the idea is to do

[gaunt sparrow]

Yes. We have $x^2 + y^2 = \frac{1}{2}$ and $z = \sqrt{x^2 + y^2}$.

[twin meteor]

Azyrashacorki

[vital latch]

this equation system?

[gaunt sparrow]

Then the point is that $r = \sqrt{x^2 + y^2 + z^2}$ at the intersection

[vital latch]

and then find the numbers that I can know, like r and phi

[twin meteor]

Azyrashacorki

[vital latch]

and I'd get z=cos(theta)

hmmm

[gaunt sparrow]

And you know what $x^2 + y^2$ is and you know what $z^2$ is.

Then in the last equation you know r and z, and you can solve for theta

[twin meteor]

Azyrashacorki

[vital latch]

i'm just a bit afraid of reusing the same values :x

but essentially then

set equations equals to each other to find out where they intersect

and that value would be what z is at the intersection?

[gaunt sparrow]

You just need the last equatino

[vital latch]

then go back to z = r cos theta, and assuming i set it up like i did where 0<=r<=1 and 0<=phi<=2pi

i get 1/sqrt(2) = cos(theta) ?

which is 45 degrees or pi/4 radians

[gaunt sparrow]

Yes that's the gist of it

[vital latch]

ok i think i'm with you on that part then, could you explain how they then proceed to set up the triple integral?

where do we get the 1 from in the leftmost one?

[gaunt sparrow]

The full working looks like $\frac{1}{\sqrt{2}}= \sqrt{x^2 + y^2}=z = r\cos(\theta) = \sqrt{x^2 + y^2 + z^2} \cos(\theta) = \sqrt{\frac{1}{2} + \frac{1}{2}}\cos(\theta) = \cos(\theta)$

[twin meteor]

Azyrashacorki

[gaunt sparrow]

Anyways

[vital latch]

how did you get the rad to be what you just put? o_O

unit sphere?

[gaunt sparrow]

r is defined as sqrt(x^2 + y^2 + z^2)

i.e. the distance from the origine to (x,y,z)

[vital latch]

and then you say x^2+y^2+z^2 = r^2

so r = sqrt(x^2+y^2+z^2)?

[gaunt sparrow]

Yes

[vital latch]

ah

ok then im with u

[gaunt sparrow]

Okok

You put 1 in the integral just do show that you're just integrating dxdydz.

Really you're integrating dV, which is small pieces of volumes

But that's because dxdydz = dV

[vital latch]

ok so whenever im integrating volumes i should do that?

and for areas?

[gaunt sparrow]

For areas you're integrating dxdy, which is units of area dA.

[vital latch]

yes

[gaunt sparrow]

You can think of dxdy as small pieces of area made up of rectangles with length dx and width dy

Then dxdydz is a small piece of volume made of a prism with length dx, width dy and height dz

[vital latch]

i more meant if I should have the 1 in those cases as well

[gaunt sparrow]

Yeah

You could just not write 1 at all

[vital latch]

ok then i think i understand what you're trying to say

it's just the conventional way of expressing that we intend to calculate the volume

ok but then for the next step, r^2sintheta dr dtheta dphi, is that from the "functional determinant"?

[gaunt sparrow]

Yes. Literally, you are integrating the constant function 1 over a certain domain. That gives you the volume contained in there

[vital latch]

not sure if you have that term in english

[gaunt sparrow]

Yes that comes from the Jacobian determinant that pops out of transforming into spherical coordinates

[vital latch]

and then for the next part we just seperate them into their own to make it easier to integrate?

is there any case when it's not ok to split them up into their own integrals?

ok managed to do it now

thank you so much!

hope you have an amazing day

.close

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[bright nacelle]

Hello I have a question
If I want to find E(X) using g(X) I need to have g(X) = a function in terms of x only ?
Then E(X)= integral of x g(x) dx?
From negative infinity of positive infinity

[bright nacelle]

this is about probability and statistics

where E is the expected value or "mean"

[hard atlas]

what else should g(x) be

if not a function in terms of x only

[bright nacelle]

a real number?

[hard atlas]

well then g(X)=that real number and E(some real number) = that real number

so you mean that you have a random variable Y=g(X) and you know g and then want to find E(X) ?

hopefully g is invertible

[bright nacelle]

I have a probability density function

f(x,y)

I found g(x) and h(y)

but I want to find E(x) using g(x)

[hard atlas]

you presumably mean something by specifically using the notation g(x) but I dont know what you mean

do you mean that its the density of x?

so then its derivative is the density of x?

[bright nacelle]

the marginal distribution of x my bad

[bright nacelle]

g(x)

which is the integral from negative infinity to positive infinity of f(x,y) dy

[hard atlas]

ok at least the question is clearer now. I think yes but its been some time since I've done that kinda stuff

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[fickle apex]

the center of mass should have x=y=0 right?

[hexed needle]

it would be if the density was the same throughout

oh nevermind

you're correct in that aspect

I didn't realize it was in 3d

[fickle apex]

i just need to find z then

[hexed needle]

yep

[fickle apex]

so for the total mass, I should have 2 triple integrals right? The first with bounds 0<r<R. 0<theta<2pi. -R<z<h.
the second with 0<r<R. 0<theta<2pi. h<z<R?

[hexed needle]

well that makes it seem like you're using cylindrical coordinates, but this is a sphere

so you need spherical coords

[fickle apex]

hm

i dont think ive been taught that yet

I suppose i'll come back to this after it gets taught in lecture then

[hexed needle]

sounds good lol

[fickle apex]

i dont think we'll learn about it until tomorrow, so im just gonna close this channel. dont need to hog it for no reason lol

.close

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[fallen sandal]

why isn't the derivative defined on sharp turns like the middle part in the absolute value function

[peak matrix]

how would you define it?

[tawny hollow]

One intuitive reason is what would the tangent look like?

[peak matrix]

if there even is the tangent

[fallen sandal]

Umm i think it would be a straight line

[tawny hollow]

I dont

[grizzled vault]

any line that intersects that corner works

so there isn't any unique "tangent line"

[peak matrix]

https://www.desmos.com/calculator/uzsrs029de

any of these lines would work

[dull bear]

(Indeed, if you remember that the definition of the derivative is supposed to be the limit of the difference quotient [where it exists], you’d notice on either side the limits aren’t the same)

[tawny hollow]

So in that sense, if derivatives were defined in terms of the slope of its tangents, then the derivative at that point is not well defined

So even our intuitive notion of a derivative does not apply here

[fallen sandal]

Umm so what

[dull bear]

Remember you have the derivative of f at a being the limit of, say, (f(x) - f(a))/(x - a) as x gets closer to a

But if that limit exists, it needs to be the same from either side

[fallen sandal]

Umm I think they approach 0 from both directions

[dull bear]

From the left, you get -1, from the right, you get +1

Think about what the slope of the pairs of points you get for some negative x and the origin, and also some positive x and the origin

[fallen sandal]

But isn't the slope like always different from both directions when u find the derivative

Unless it's like a straight line idk

[dull bear]

In many of the exercises you’ve done, you’ve shown that they become the same from either side as you get closer and closer to the point you’re considering

Remember the first few here #help-7｜zen1thxyz message we found the limits actually existing

[dull bear]

Whereas here it doesn’t

[fallen sandal]

Ohh

[fallen sandal]

Wat Abt like extrema

[dull bear]

As in derivatives at extrema? You’d find the derivative to be zero if it exists there

[fallen sandal]

umm how come it doesn't work here

It it bcuz it doesn't like curve

To like get so close to each other

[dull bear]

Because of the “if it exists there” basically nothing guarantees the derivative exists at extrema, but if it does, it must be zero

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[velvet cove]

how to do

[flat whale]

yea you're supposed to integrate del f(x,y) wrt to a variable

https://math.libretexts.org/Bookshelves/Analysis/Supplemental_Modules_(Analysis)/Ordinary_Differential_Equations/2%3A_First_Order_Differential_Equations/2.7%3A_Exact_Differential_Equations

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[vocal sleet]

Please don't occupy multiple help channels.

[fading cove]

send your question

[flat whale]

closed your other channel

are the numbers just numbers in the base of the subscript?

$10_8 = 8_{10}$ ?

[twin meteor]

riemann

[flat whale]

yes

just convert all the numbers to the same base and add

no idea what F and C in base 16 are

[cyan shadow]

C is 12, F 15

[dark rune]

F = 15, C = 12

[cyan shadow]

haha sniped

[dark rune]

damn it

[flat whale]

https://www.youtube.com/watch?v=8YUILmUCHSc

oh hex okay

if you can do addition, you can evaluate f(x, y)

$f(3_{10}, 10_8) = 3_{10} + 10_8 + 0101_2$

[twin meteor]

riemann

[flat whale]

is the converting hard or the addition?

https://math.libretexts.org/Bookshelves/Applied_Mathematics/Contemporary_Mathematics_(OpenStax)/04%3A__Number_Representation_and_Calculation/4.05%3A__Addition_and_Subtraction_in_Base_Systems

i believe in you

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[frigid mason]

HELP

[frigid mason]

oops

how do i find someone good at geometry i can't do this anymore

[brittle cipher]

we can't do the work for you but we can help, where are you stuck?

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[twin meteor]

Ousel

Ousel

[livid void]

Kinetic converts to potential:

[twin meteor]

Ousel

Ousel

[frigid mason]

no

[livid void]

I already started a new room, sorry xP

[livid void]

Post your question in one of the help channels that are available. (Channel group above this one)

[frigid mason]

okk tysm

[livid void]

<@&286206848099549185>

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[livid void]

<@&286206848099549185> xP

[vast shale]

😭

[livid void]

Can't help? @vast shale

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✅

[livid void]

So a friend of mine somehow brute forced the answer just typing in integers. Apparently its 0.78 but the only way I know to work it leads to 0.52

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@livid void Has your question been resolved?

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@livid void Has your question been resolved?

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@livid void Has your question been resolved?

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@livid void Has your question been resolved?

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[livid void]

<@&286206848099549185>

[livid void]

😮‍💨

[empty delta]

im not that at home in these kinds of questions but i think i know the correct way of solving it using these formulas.

i used
u_s*m*g=kd
w_tot=K_f-K_i=1/2*m*v_2^2-1/2m*v_1^2
w_tot=w_spring+w_fric=-1/2*k*d^2-u_k*m*g*d

(idk how to use the bot to make these show as actual formulas)

@livid void

[livid void]

idk what wtot is, but this is what I did

unless that means total weight