help-26

[ruby tree]

Ok so then for what values x and y does xy = 1? And then xy = -1?

[potent nova]

(1, 1)

For the first one

[ruby tree]

Is that the only solution?

Are you sure these are the only two solutions? Could you prove that?

[flat kindle]

why

you don't need all solutions of xy=1

[ruby tree]

It says for all x,y, surely you want all solutions

[flat kindle]

???

we are trying to prove f(1)=f(-1)=0

not prove f(xy)=f(x)+f(y)

[ruby tree]

Yeah and the only thing we have a definition of is f(xy)

[flat kindle]

substitute (x,y) =(1,1) and (x,y)=(-1,-1) to the equation

[ruby tree]

Ok never mind, you're right

[flat kindle]

close

it's f(xy) on the left

not f(x,y)

[thorny flame]

Shubh

[flat kindle]

yep

[potent nova]

Oh so as there are no x's, anything in the function will be true

So f(-1) is also 0

[flat kindle]

correct

[ruby tree]

It doesn't mean f(x) = 0

[potent nova]

Why not?

[ruby tree]

You only proved f(1) = f(1) + f(1), not f(x) = f(x) + f(x)

[potent nova]

Them how will we go on to confirm f(-1)=0?

[ruby tree]

Substitute (x,y) = (1,-1)

[flat kindle]

bruh

no one asked if f(x) = 0

[potent nova]

f(1*-1)=f(-1)+f(1)?

[flat kindle]

you just get f(-1)=f(-1)

try (x,y)=(-1,-1)

[potent nova]

so f(-1*-1) = f(-1)+f(-1)

f(1 ) = f(-1)+f(-1)?

[flat kindle]

and we know f(1)=0

[potent nova]

then -f(-1) = f(-1)

[potent nova]

What would we get next?

[flat kindle]

writing as 0=2f(-1) might be easier

[potent nova]

Oh ok

[flat kindle]

either way it's clear f(-1) = 0

[potent nova]

thx

[ruby tree]

Not sure what you mean. I made no such claim, Shubh was under the impression that because f(1) = f(1) + f(1) then f(x) = 0

[potent nova]

.close

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[ruby tree]

@potent nova here is an example of f if it makes it clearer: c*ln(abs(x))
https://www.desmos.com/calculator/i0wbcjjj1j

[ruby tree]

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[wintry oar]

Need help

[drifting swift]

,rccw

[topaz sinew]

@wintry oar Has your question been resolved?

[wintry oar]

<@&286206848099549185>

[topaz sinew]

@wintry oar Has your question been resolved?

[ruby tree]

I assume the question is "can P,Q,R, and S all lie on a circle"

Can you find a relation between the lengths PQ and SR?

[topaz sinew]

@wintry oar Has your question been resolved?

[wintry oar]

Idk how

[wintry oar]

Yes

[ruby tree]

What is PN equal to?

[wintry oar]

Side of hexagon and 12 side

Also half of the square

[ruby tree]

Ok, what is NQB?

[wintry oar]

Idk

I tried using angles to solve

Doesn’t work

[ruby tree]

What's the angle (BQN)

[wintry oar]

60 degrees

[ruby tree]

And what are the other angles of NQB

[wintry oar]

Triangle NQB has 60,60,60 degrees making it right triangle

[ruby tree]

Not right

[wintry oar]

How

[ruby tree]

A right triangle has an angle of 90, aka a right angle

[wintry oar]

My bad

[ruby tree]

So what does that mean for the length NQ?

[wintry oar]

It means NQ=NB=MB

[ruby tree]

Yes

Now if you take the midpoint between N and B, let's call it G, what's GQ?

[wintry oar]

GQ= NB*sqrt3/2

[ruby tree]

Is this sqrt(3)/2 or sqrt(3/2)?

[wintry oar]

Sqrt(3) / 2

[ruby tree]

Ok yes

One moment

Ok sorry there's a simpler way to do this

The goal is going to be to find the angle PQM

You already know the angle NQB

Can you think of something to link the two? @wintry oar

[wintry oar]

Yea sure i can find that easily

[ruby tree]

Well, show me how

[wintry oar]

One sec

30 degrees

Angle PQM

[ruby tree]

Yes but how

[wintry oar]

First construct line NQ

U know that angle N is 120 degrees right?

When connected to center of the hexagon (Q) it gets divided by 2 making angle BNQ 60 degrees

After than there is PN line that is perpendicular to CB making angle PNQ 90+60=150 degrees

[ruby tree]

We already know BNQ=60 from here

Ok PNQ=150

[wintry oar]

Now in triangle PNQ PN=NQ makes angle NPQ=(180-150)/2

PQ is equal to MQ

Making angle PQM 180-75*2=30 degrees

[ruby tree]

NPQ=15 ok, but I don't follow how you use PQ=MQ

You're correct but I'm just curious

[wintry oar]

Isosceles triangles have 2 same angles

[ruby tree]

Ok yes so MPQ = 90 - 15 = 75

Alright

So we have the angle PQM = 30º

What about the angle SRM?

[wintry oar]

Im unable to find that angle

But we can describe it using alpha and beta

[ruby tree]

SM is one side of a 12-side polygon

Just like NB is one side of an hexagon

[wintry oar]

360/12=24 degrees

[wintry oar]

U are right

[ruby tree]

No that's not 24

[wintry oar]

30

[ruby tree]

Yes

[wintry oar]

My bad

[ruby tree]

So what does it mean for the triangles PQM and SRM?

[wintry oar]

There are same triangles

[ruby tree]

Yes

Now can you find something about the lines PS and QR?

The goal is to find what kind of quadrilateral PQRS is

[wintry oar]

They are similiar

[ruby tree]

Lines aren't similar, do you mean parallel?

[wintry oar]

I cant see parallels

One sec

[ruby tree]

You can just use the angles of the triangles PMS and QMR

[wintry oar]

I did.

Pms has 120,30,30 Qmr has 75 and 52.5 52.5

Am i wrong

Sorry

I got it now

[ruby tree]

QMR is not 75

[wintry oar]

Its 90

[ruby tree]

Yes

[wintry oar]

How do we know it’s parallel tho

[ruby tree]

Well SPM = 30º, PMQ = 75º, MQR = 45º

PMQ = SPM + MQR

[wintry oar]

So if this is true it must be parallel?

[ruby tree]

Alternatively SPQ = 30º+75º = 105º and PQR = 30º+45º = 75º so SPQ + PQR = 180º

[wintry oar]

I get it now they are parallel

It makes them trapez

[ruby tree]

Yes but a special kind of trapezoid

[wintry oar]

Now what I believe only isosceles trapez fits in circle

[ruby tree]

Yes

Actually now that we're here I don't know if it's admitted that an isosceles trapezoid can be inscribed in a circle

We can go further if you want

[wintry oar]

We only have to check if its isosceles trapez or not

[ruby tree]

Well technically you need to know if there is a point O such that OS=OP=OQ=OR

But it's rather quick

Construct the bisector of QMR

All points on this new line are equidistant to P and S and also equidistant to Q and R

Now there must be a point on this line that is equidistant to P and Q because PQ is not perpendicular to it

Well if that point is O, then by construction OP=OQ and because of what I said just above, OP=OS and OQ=OR

I guess I forgot to state that the bisector of QMR is also the bisector of PMS but it's kind of obvious

[wintry oar]

How do we know OS=OP and others

[ruby tree]

O is on the bisector of PMS, that means OP=OS by definition

[wintry oar]

Never heard of that definition before

[ruby tree]

Oh sorry you also need MP=MS

But we got that already

[wintry oar]

Yes

[ruby tree]

Hm maybe I'm mistaken saying "by definition"

[wintry oar]

Is this stating that center of the circle is in meeting with bisectors?

[ruby tree]

It's more like "the bisector of the main angle of an isosceles triangle is also the perpendicular bisector of the opposite side"

And a perpendicular bisector of AB is by definition all the points equidistant to A and B

[wintry oar]

Is this theorem or something?

[ruby tree]

Not really but it's easy to prove

Don't mind the 1's everywhere

ABC is isosceles, AD is the angle bisector

Then because AB = AC and DAC = DAB, the triangles ACD and ABD are similar

[wintry oar]

In isosceles triangle the main angle is the center of the circle. Am i right?

I get it so

So it fits in circle

[ruby tree]

So CDA = BDA = 90º since D is on the line BC

Yeah I call the "main" angle the one that isn't equal to the other two

[wintry oar]

Thank you very much for ur time and energy

[ruby tree]

You're welcome

[wintry oar]

.close

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[jaunty meadow]

for th

[jaunty meadow]

for this one

can i divide the row 1 by three directly? (im doing gauss jordan elimination here)

[neon iron]

yes

[ruby tree]

Probably easier to multiply the other by 3 instead

[jaunty meadow]

which one

[ruby tree]

Third one? I assume you planned on subtracting one from the other

[jaunty meadow]

dividing first row by 3, not third row

i meant

[ruby tree]

You can multiply any row by any non-zero number, including 1/3

[jaunty meadow]

and i must go left to right, right?

[ruby tree]

That's usually the idea yes

[jaunty meadow]

and there can be more than one solution then, right?

[ruby tree]

Solution to what?

[jaunty meadow]

like, i said divide by 3, you asid multiply the other by 3 so anyone can find any different way and can find the correct answer right?

multiple trajectories

i mean

[ruby tree]

yes

The end result of Gaussian elimination is the unique reduced row echelon form

[jaunty meadow]

i found this:

[ruby tree]

So in a sense yes there is only one "solution"

[jaunty meadow]

1 0 0 | -2/3
0 1 0 | -1/3
0 0 1 | -11/9

do i need to find the answer in integer? (the right ones)

[ruby tree]

No

[jaunty meadow]

or they can stay like this

[ruby tree]

I think you made a mistake though, hold on

[jaunty meadow]

hm

R1 / 3 → R1
R2 - 6 R1 → R2
R2 / -9 → R2
R1 - 2 R2 → R1
R1 + 1 R3 → R1

that was my progress

[ruby tree]

You're definitely wrong, but I can't tell you where you went wrong with just that

[jaunty meadow]

okay let me write it to a paper

i was doing it in paint

[ruby tree]

That might be the reason

Maybe try another way, by first swapping the first and third rows

[jaunty meadow]

im sending the photo now

@ruby tree

welp

umm should i ping helpers

<@&286206848099549185>

omg

im suffering here

i think its correct tho

anyways

.close

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[ruby tree]

.reopen

[ruby tree]

Sorry, something came up

[ruby tree]

@jaunty meadow are you there?

[jaunty meadow]

hi

[ruby tree]

Ok hold on

The third row is 1 2 3 2

You wrote -3

[jaunty meadow]

oh wait

its actually -3

[ruby tree]

Hm ok hold on

Still wrong though

[jaunty meadow]

[ruby tree]

R3 -> R3 - R1

Why is there a 1 in the middle

I mean 0 1 0

It should be 0 0 0

[jaunty meadow]

OH

so its inconsistent

[ruby tree]

Funny because it makes the system inconsistent whereas with +3 it was consistent

[jaunty meadow]

wdym

[ruby tree]

This system is consistent

[jaunty meadow]

ooh

[ruby tree]

I mean I guess there are infinitely more consistent systems

I don't remember exactly

[jaunty meadow]

so i need to say the system is inconsistent after showing the work, right?

[ruby tree]

I guess so

[jaunty meadow]

oh and i want to ask you one more thing to you

[ruby tree]

As soon as you have a row of 0s, it's inconsistent

[jaunty meadow]

in second question it says gaussian elimination

is it like:
1 x x
0 1 x
0 0 1 ?

x can be any number, right?

in gaussian

[ruby tree]

To me it's the exact same thing

If the bottom-left triangle is all 0 and the diagonal is all 1 then you can eliminate those x's

[jaunty meadow]

true

true

right

thank you

<3

[ruby tree]

You're welcome

.close

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[tame void]

Help

[tame void]

I am reading Hatcher when I encountered this

"RP^n is the quotient space S^n/(v~-v), the sphere with antipodal points identified. This is equivalent to saying that RP^n is the quotient space of a hemisphere D^n with the antipodal points of boundary of D^n identified."

I fail to imagine that for RP^2 and so on. Can anyone help me with this?

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@tame void Has your question been resolved?

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@tame void Has your question been resolved?

[hexed tendon]

what do you need help with? Imagining RP^2? It cannot be embedded in R^3 so the only way of really imagining it is with a sphere with antipodal points being the 'same'.

[tame void]

That wasn't the issue. I wanted to know why can I claim RP^n both as the quotient of S^n and D^n under given identifications.

I got it though

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[thorny flame]

Bob Goldham

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[topaz sinew]

@neon iron Has your question been resolved?

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@neon iron Has your question been resolved?

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@neon iron Has your question been resolved?

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[coarse ether]

hi

[coarse ether]

so

i'm coding a fluid simulation

and i need to find the divergence of the vector field at each vector origin

but i don't have a function that describes the vectors

i just have a grid of vectors with a separation

how do i find the divergence at a point in the vector field?

[topaz sinew]

@coarse ether Has your question been resolved?

[strange whale]

but i don't have a function that describes the vectors
i just have a grid of vectors with a separation
where do you get those vectors from then ? just testing with random numbers ?
@coarse ether

[coarse ether]

they start at 0

and then i apply gravity to them

which will create divergence

and then the flow is created by correcting that divergence

by finding the divergence and then using that to compute the divergence free part of the field

[strange whale]

well the divergence is just a bunch of partial derivatives (dF_x/dx + dF_y/dy + ...), you can approximate each of these terms at each point in your space

[coarse ether]

ok that makes sense

but what if i'm trying to get the divergence at the origin of a vector

how do i compute the partial derivative

[strange whale]

$$\pdv{F_x}{x}(x, y) \approx \frac{F_x(x+h, y)-F_x(x-h, y)}{2h}$$
$$\pdv{F_y}{y}(x, y) \approx \frac{F_x(x, y+h)-F_x(x, y-h)}{2h}$$

[thorny flame]

aPlatypus

[strange whale]

you start at a point (x, y) where you want to evaluate the partials

then you just look around in the x and y directions respectively

and compute the slope between the 2 neighboring points

[strange whale]

wdym here ?

[coarse ether]

like

i have a grid of vectors

each with a x and a y component

[strange whale]

yeah

[coarse ether]

and they also each have a position

like in a grid

[strange whale]

yes

[coarse ether]

like how do i compute the partial derivative

at the exact location of one vector

[strange whale]

you don't have a closed form for your vector field

you only know your vector field on your grid

the best you can get is an approximation

[coarse ether]

i know

i'm asking how do i compute the approximation

[strange whale]

I've posted an approximation up above, where $F(x, y) = \begin{bmatrix} F_x(x, y) \ F_y(x, y) \end{bmatrix}$

F(x, y) is your 2d vector field

F_x(x, y) corresponds to extracting the x-coordinate of the vector at (x, y)

similarly for F_y

[thorny flame]

aPlatypus

[coarse ether]

ok i get it

[coarse ether]

but wait this doesn't take into account the vector at F(x, y)

only the vectors next to it

[strange whale]

looking at the two neighboring positions gives you a more precise approx in general

you could do it using F(x, y) if you want

and do the bog-standard calc 1 thing

$$\pdv{F_x}{x}(x, y) \approx \frac{F_x(x+h, y)-F_x(x, y)}{h}$$
$$\pdv{F_y}{y}(x, y) \approx \frac{F_y(x, y+h)-F_y(x, y)}{h}$$

wait also I typo'd

the dangers of copy pasting

[thorny flame]

aPlatypus

[coarse ether]

ok i understand now

[strange whale]

$$\pdv{F_x}{x}(x, y) \approx \frac{F_x(x+h, y)-F_x(x-h, y)}{2h}$$
$$\pdv{F_y}{y}(x, y) \approx \frac{F_y(x, y+h)-F_y(x, y-h)}{2h}$$

[coarse ether]

thank you so much :)

[thorny flame]

aPlatypus

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[viscid furnace]

"Let A and B be n x n diagonal matrices. Show that AB = BA." How can I prove this statement?

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@viscid furnace Has your question been resolved?

[torpid matrix]

its not true, so you cannot prove it

[viscid furnace]

what counter example could you give?

[torpid matrix]

try picking like basically any two matrices and test for yourself

diagonal matrices i missed that

[viscid furnace]

yes I was about to say, because they are both diagonal they are commutative to my knowledge

[torpid matrix]

my mistake, it is communtative with diagonal matrices

[viscid furnace]

but I'm just struggling to prove it for all n x n matrices

[torpid matrix]

can you prove that the product of two diagonal matrices is the diagonal matrix from multiplying the elements on the diagonal

[civic nymph]

out of curiousity

the statement does not say "for all A and B"

so even if it was just "Let A and B be n x n matrices, show AB=BA"

[torpid matrix]

its implied

[civic nymph]

all u need to do is find any nxn matrices that are communitive

sure, but

[torpid matrix]

the original problem probably includes it, and if it does not then the grader will still expect it anyways

you can be technically right and not get the pts

[viscid furnace]

so the determinant?

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[twilit robin]

So

For some reason I got that the derivative of ^3sqrt(x) was 1/3^3sqrtx

and it turns into 1/6 once I plug in 8

I feel like I'm missing something simple here

[jade thunder]

$\frac{d}{dx}\sqrt[3]{x} = \frac{d}{dx}x^{1/3} = (1/3)x^{-2/3}$

[drifting swift]

please don't write the cube root of x as ^3sqrt(x)

[thorny flame]

Frosst

[drifting swift]

write it as x^(1/3)

[twilit robin]

Okay

Ohh I see

I just made a mistake when thinking about x^(1/3-1)

[jade thunder]

yes

[twilit robin]

Thank you idk why that happens sometimes

.close

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[topaz sinew]

Please don't occupy multiple help channels.

[sturdy oracle]

You can do some kind of sign analysis

You know that the domain excludes x = 21

And we know that the denominator is always positive for any other number but 21

So you just gotta look at 3 - x

When is that positive and when is that negative

Oh but I guess you gotta find the derivative anyways lol

[proven narwhal]

ye I got (x+3) / (x-21)^5

so it's increasing from -inf to 3 and decreasing from 3 to inf?

[topaz sinew]

@proven narwhal Has your question been resolved?

[restive inlet]

how are you reaching that conclusion from the derivative?

[topaz sinew]

@proven narwhal Has your question been resolved?

[topaz sinew]

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[neon iron]

Hello chat

[neon iron]

Say I have an N x M grid.

example: 4x4 grid

If you can only move down and right

Count the amount of paths to go from S to e

This problem is trivial

You have to make 6 moves to go from S to E

You have to move down three times and right three times

so the amount of paths is $\binom{6}{3}$

[thorny flame]

Several people

[neon iron]

however!

What if we block a random square on the board, how many paths are there now?

also trivial

[worthy storm]

isn't that the same as just starting with the lower half

[neon iron]

yeah

[worthy storm]

with you in the upper left of the lower half

[neon iron]

[keen venture]

Well, you said this is also trivial. What's the difficult question?

[neon iron]

This would just be $2 * \binom{4}{1}$ becuase obviously

[thorny flame]

Several people

[neon iron]

The difficult problem is

[neon iron]

this on arbitrary N x M grid of any size with Xs at any spot

but atleast one path from S to E is guranteed to exist

how would i use combinatorics to count the number of path on any N x M grid like this

[keen venture]

In general it's:
Number of paths - Number of paths that go through an X

How many paths go through an X? That's the number of ways to go from S to X, times the number of ways to go from X to E

In your above example, there's
2C1*4C2 ways to go through the X

So the total number of paths that don't go through the X is
6C3 - 2C1*4C2 = 8

[neon iron]

I see

Inclusion Exclusion brilliant

SO for N x M grid

It would be something like

[keen venture]

Gets tricky with more than one X though

As you don't want to double count paths that go through both

Yeah might be inclusion-exclusion time in that case

[neon iron]

I see ty

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[serene imp]

Hi, I'm trying to get the answer given in blue. Can someone tell me what I did wrong here? I'm so lost 😭.. Isn't this just multiplying a matrix with its own identity?

[neon iron]

matrix multiplication is not commutative, so AB and BA are different things. when you multiplied V, you must multiply it to the left of DV-1

^

[topaz sinew]

@serene imp Has your question been resolved?

[serene imp]

How silly of me! I got it, thank you!

[neon iron]

no problem!

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[amber yacht]

∀𝑛∈ℕ, if 𝑛 is even then 3^n−2 is prime.

[amber yacht]

answers say false but i cant find a counterexample

you have to give a counterexample to back ur answer

[verbal crater]

n = 3

3^3 = 27

27 - 2 = 25

[amber yacht]

oh sorry

[verbal crater]

25 is a factor of 5

not prime

[amber yacht]

i forgot to include that n can only be an even number

[verbal crater]

oh ok

[amber yacht]

nvm i didnt forget it says it in question

[verbal crater]

sorry lol didnt see

[amber yacht]

lol

[odd pagoda]

well the first counterexample isnt that big

[verbal crater]

@amber yacht n = 8

[odd pagoda]

somewhat doable by hand

dont just give solutions

[amber yacht]

sorry im back

[amber yacht]

6559?

how did u know it was not prime

[amber yacht]

its hard because you have to go through even numbers 2,4,6,8 and also check if they are prime

[odd pagoda]

well 2 and 4 are easy

checking that 727 is a prime takes a while but is doable

finding out that 6559 is not a prime turns out to be very nice because it is divisible by 7 already

[amber yacht]

oh right

so u just put it into ur calculator?

but realistically its hard to believe that there even is a point of continuing after 4 or 6

you feel like its irrational to go that far to check

[odd pagoda]

"that far"

its three steps

[amber yacht]

thats quite a lot

[odd pagoda]

no

[amber yacht]

especially considering that u have to check its a prime

imagine if instead of being divisble by 7, it was divisible by 11

[odd pagoda]

from the start you dont expect the statement to be true

[amber yacht]

you never know how far you have to go

[odd pagoda]

any statement of the form "this formula always gives a prime" is very rarely true

[amber yacht]

ye i guess

[odd pagoda]

well then trust in the problem that you arent asked to do too much

[amber yacht]

but when u see ur first few attempts of n-0,2,4,6 all fail you lose hope and dont test 8

[odd pagoda]

no

0,2,4 are trivial

[amber yacht]

i just think its an irrational question to ask in a test

[odd pagoda]

the only actual thing you have tested is 727

[amber yacht]

BUT even then

if n=6 fials, will you really test n=8?

[odd pagoda]

yes

[amber yacht]

ok if n=8 doesnt work

how far will u rly go

[odd pagoda]

if I had to do it by hand I would stop somewherr during the n=8 step probably. switch to n=10 and try the first few primes for that. and if that also fails then I would think about doing something else

[amber yacht]

i see

so u try ur hardest to believe that statement like that would be false

[odd pagoda]

its basically guaranteed to be false

[amber yacht]

ye that makes sense

[odd pagoda]

we dont know any easy formula for primes

this one would be way too easy

[amber yacht]

idk its a bit tedious to think of that in a test

[amber yacht]

exactly thats the issue

anyways

thanks for helping everyone

.close

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[void pewter]

,rotate

[void pewter]

I don't understand

[fluid belfry]

thats baisc division

,, \frac{\frac{9N+9}{M}}{\frac{3N}{M}}= \frac{9N+9}{M}\times \frac{M}{3N}

[thorny flame]

!Yajat!

[void pewter]

so..

how to simplify it then?

[fluid belfry]

,, \frac{\cancel{3}(N+1)}{\cancel{M}}\times \frac{\cancel{M}}{\cancel{3}N}

[thorny flame]

!Yajat!

[fluid belfry]

just like you simplify numeric terms

[void pewter]

isnt it 3(n+3)?

nvm

[fluid belfry]

sorry i messed up

basically you factor out 9 from it

[void pewter]

o

9(n+1)

[fluid belfry]

and then it gets cancelled by 3 in the denom by 3 times

[void pewter]

9 / 3 = 3

[fluid belfry]

yes

[void pewter]

3(n+1)/N

[fluid belfry]

yes

[void pewter]

i was thinking

9 x n + 9

then divided by

3 x n

[fluid belfry]

no

thats

w

rong

[void pewter]

ik

[fluid belfry]

yea

[void pewter]

why cant i do 9n = 9 x n

?

ik its wrong

[fluid belfry]

9n can we written as 9 x n

what do you

mean

[void pewter]

one sec

[fluid belfry]

haha no

you cannot do this

because 3 times 3 times N is in addition with 3 times 3

you first need to simplify that

you can only divide like this in the numer and denom when on

both side every

term

is in product

form

[void pewter]

if question was written as the thing i did it would be right?

[fluid belfry]

this way?

[void pewter]

ye

this way

[fluid belfry]

yea then it would be wrong

[fluid belfry]

read this

[void pewter]

product form

?

[fluid belfry]

yea like

,, \frac{3 \times 3 \times 3}{3}

[thorny flame]

!Yajat!

[fluid belfry]

not $\frac{3+3+3}{3}$

[thorny flame]

!Yajat!

[void pewter]

so everything has to be multiplying?

[topaz sinew]

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[topaz sinew]

@void pewter Has your question been resolved?

[topaz sinew]

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[verbal crater]

sin(x) = cos (90 - x)
cos(x) = sin(90 - x)
tan(x) = cot(90 - x)
cot(x) = tan(90 - x)

[desert sluice]

CORRECT

[desert sluice]

?

[topaz sinew]

@desert sluice Has your question been resolved?

[desert sluice]

PLS TELL

[neon iron]

did AI write this question?

[neon iron]

you use those identities. for example, the first term 5 sin 28° sec 62° = 5sin 28°/cos 62° = 5 sin 28°/sin 28° = 5. Now try the other terms. I am also ignoring those random symbols around the tan/cot term.

[desert sluice]

ok

thanks

[neon iron]

no problem

[topaz sinew]

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[quaint nest]

Hello I just need help figuring out how to do this kind of problem, all I know is it has something to do graphing and finding whether out the expressions are equal, greater, or less than than 0

[quaint nest]

3606

[neon iron]

,rccw

[quaint nest]

2606

[quaint nest]

Just 2606

[neon iron]

oh that's interesting

what have you tried

[quaint nest]

Well so in a earlier problem it looked like this

One sec

So for 2600

I did stuff like this

But now the question does not specify to find without computation

I. 2606

So I’m just a little lost

[neon iron]

visualising cot and tan using the unit circle will be a bit tricky

do you instead know about the graph of the function $\m fx= \m \cot x$?

[quaint nest]

No

[quaint nest]

I think that’s why it allows computation but idk

[neon iron]

you can't without a calculator

it's irrational

[quaint nest]

Oh

[neon iron]

you can guess the signs of course there are various ways

but having a general idea on how the function is is probably the best

,w graph f(x) = cot(x) from -10 to 10

[thorny flame]

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[quaint nest]

Alright

[neon iron]

that's the graph of cot(x)

[quaint nest]

Mhm

[neon iron]

there are asymptotes for every x= 180k for an integer k

so, as you can notice

in a particular strip of the graph

as the x values increase, the function becomes smaller

can you intuit what cot(153)-cot(157) would be from that

[quaint nest]

Negative?

[neon iron]

well like I said, the larger the x, the smaller the y

so which one is bigger

cot(153) or cot(157)?

[quaint nest]

153

[neon iron]

right

and big - smaller = ?

[quaint nest]

Positive

[neon iron]

yes

[quaint nest]

Oh my god

Thank you

Do I just then do this for every problem now

[neon iron]

do you want to continue with the rest?

[quaint nest]

Yes please

[neon iron]

the MAIN takeaway from this question is to understand how those functions work

[neon iron]

you should understand why cot(x) behaves like this for example

and you will be able to ace the rest

but anyways yea let's continue

with b you are presented a similar situation but with the tan function

do you know the graph of tan(x)?

[quaint nest]

No

[neon iron]

okay

,w graph f(x) = tan(x) from -10 to 10

[thorny flame]

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[neon iron]

cot(x) is a mirrored version of what tan(x) looks like, basically

[quaint nest]

Right

[neon iron]

there are asymptotes for every x = 90k for an integer k

since sin(x)/cos(x), cos(x) = 0 for every 90k as well

okay but anyways

notice this time, for a particular strip, the larger the x, the larger the y

the opposite of the previous problem

[quaint nest]

Right

[neon iron]

can you intuit what tan(319) -tan(327) would be in this case

[quaint nest]

Negative

[neon iron]

yes

c) is exactly the same

you are simply working with radians now

I'll leave you to it to try an answe that one

[quaint nest]

Okay

Negative

No

Positive

It is positive

[neon iron]

yeah

okay great

[quaint nest]

Okay I think I get it

But I do have one more

Question

[neon iron]

okay

[quaint nest]

On part e

2606

How does one compute cot - tan

[neon iron]

yeah

okay so what would you say is the relation between cot and tan?

[quaint nest]

Hmm

Opposit

Reciprocal

[neon iron]

yea

so like

[quaint nest]

Oh wait

[neon iron]

tan(90) is undefined

byu tan(100) is only a bit after that

tan(100) is not considerably a big number

so 1/small - small = ?

[quaint nest]

0

[neon iron]

no

[quaint nest]

Oh

[neon iron]

you are saying 1/2 - 2 = 0

with that

[quaint nest]

Oh my bad

One sec

Negative

[neon iron]

well no but idk how to describe it better rn sorry

[quaint nest]

Your good

Would it be positive

[topaz sinew]

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[tight finch]

Trying to solve this IVP with method of undetermined coefficients

[tight finch]

But dunno how to tackle it from here

[strange whale]

well you end up w/ an impossible equation

2A = 1 and 2A = 0

so that means you should rework your guess for the particular sol

@tight finch

maybe a t^2e^t term could help

[topaz sinew]

@tight finch Has your question been resolved?

[tight finch]

I end up with this

Since my DE is y''+y

[vagrant wing]

hello

[tight finch]

Which won't cancel either

[vagrant wing]

can i ask something ?

[strange whale]

you can ask in a channel that's not occupied yet

see #❓how-to-get-help @vagrant wing

[tight finch]

.close

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[coral fog]

What am I supposed to do make the series start from n = 0

[faint fog]

if you want it to start it from n=0

then replace 2(1/e)^n with 2(1/e)^n+1

[coral fog]

I'm just worried between 2(1/e)^n+1 and 2(1/e)^n-1

[faint fog]

try putting zero

[drifting swift]

hold on hold on

@coral fog what are you more concerned about?
a) re-indexing the series to start at n=0
b) finding the sum of the series

[coral fog]

"a"

[drifting swift]

so you are unconcerned about b

yes?

[coral fog]

Yes

[drifting swift]

ok

this is a geometric series with ratio 1/e as should be clear.

is that clear?

[coral fog]

Exactly

[drifting swift]

ok

and its first term is 2 * (1/e)^1, or 2/e.

so it rewrites as $\sum_{n=0}^{\infty} \frac{2}{e} \cdot e^{-n}$.

[thorny flame]

AnnGhost

[drifting swift]

or $\frac{1}{e^n}$ or $\paren{\frac{1}{e}}^n$ instead of $e^{-n}$, whichever you like better.

[thorny flame]

AnnGhost

[coral fog]

So what my professor did is 2(1/e)^n+1

then

2(1/e)^n . (1/e)

it's the same yes?

[drifting swift]

yes sure

[coral fog]

Thank you ann savior

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[uncut token]

A

[acoustic pecan]

B

[uncut token]

How to get line 3 1/n....

@acoustic pecan

<@&286206848099549185>

[topaz sinew]

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

[uncut token]

<@&286206848099549185>

[soft phoenix]

what is the upper limit of sigma?

[uncut token]

Undifined @soft phoenix

[gray ridge]

to understand this , treat 1/n as a constant

e.g. $\sum_{i≥1}a\frac1{2^i}=a\sum_{i≥1}\frac1{2^i}$

[thorny flame]

Biscuity

[uncut token]

1/(1-(n-1)/n)^2
@gray ridge where is this come from?

<@&286206848099549185>

[topaz sinew]

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[gray ridge]

geometric series

[uncut token]

So from i(n-1/n)^i-1 become this?

@gray ridge

[gray ridge]

lemme check

oh, i missed that i

it's an AGS

[uncut token]

What is ags

[gray ridge]

https://en.m.wikipedia.org/wiki/Arithmetico-geometric_sequence

[uncut token]

But we have power ^-1

@gray ridge

[gray ridge]

so?

em

it's for ease of calculation

don't worry

maybe I'm too sleepy lol

the i-1

we will have a very neat
((n-1)/n)^0 for the first term

just shift the terms

[uncut token]

Oh ok" Thankyou for your help @gray ridge

.close

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[river plaza]

Can someone walk me through this basic complex numbers problem?

When we are asked for the solutions:

z^3=27

I don't understand this explanation:

"We know that the arguement of z^3, is 3*theta. since z^3 = 27, we also know that its argument is 0 degrees.

[topaz sinew]

@river plaza Has your question been resolved?

[river plaza]

<@&286206848099549185>

[modest galleon]

Were did you get z³=27 from?

[river plaza]

thats just the question.

z^3 = 27, we know one solution is 3, and it asks for the 2 distinct complex solutions

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[royal sky]

how do i find x here

[neon iron]

[\m\cos x = -\f12 \implies x = \m{\arccos}{-\f12}]

[neon iron]

so, what values cos(x) give you 1/2? as a start

[royal sky]

2pi/3 right

[royal sky]

and im not too sure what after the 2pi/3

[royal sky]

is arccos cos-1?

[topaz sinew]

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[vivid frost]

Hi, i just need to know how tall is a step (idk the english word) and when will the feature will hit the 73cm height

[topaz sinew]

@vivid frost Has your question been resolved?

[vivid frost]

hi, <@&286206848099549185>

.close

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[junior pumice]

Does this mean the solution of the homogeneous equation of this matrix is a span of one vector?

[strange whale]

yep

[junior pumice]

the second part im not quite sure on, is it asking will there be a solution for every b in Ax = b given Dim Nul A = 1?

[strange whale]

indeed

[junior pumice]

so there will be a "line" in R^5 (co domain) which there can't be a solution because its the null space?

[strange whale]

really sure about that ?

[junior pumice]

actually no because column space is in a different vector space

nul doesnt share the same vector space

[strange whale]

the matrix isn't square

you know rank-nullity theorem ?

[junior pumice]

i know rank theorem. that is that Rank A + Dim Nul A = n given an m x n matrix.

[strange whale]

yeah that

[junior pumice]

hmmm im not sure

[strange whale]

you have a 5x6 matrix

and you know its nullity already

[junior pumice]

at least how big it is

so the co domain is R^5

4 dimensions of that is the column/row space

1 is the nullity

is that right?

[strange whale]

you're screwing up the theorem again

[junior pumice]

R^6

5 are col/row 1 is null

[strange whale]

yeah

so the rank is 5

[junior pumice]

when rank < n that means there are b not within the range?

[strange whale]

no

the b live in R^5 mate

5x6 matrix (6 is the dim of the input space, 5 is the dim of the output space)

[junior pumice]

right, R^6 domain, R^5 co domain AH

so there is an x for every b then?

[strange whale]

yup

[junior pumice]

so as long as Col A has the same dimension as the co-domain, it generates every b

that is, Col A generates R^5?

[strange whale]

well colA is the whole space generated already

[junior pumice]

it spans R^5

[strange whale]

here you can say colA = R^5

colA is indeed made of 5-dim vectors

[junior pumice]

sorry theres like 80 different ways to say things i wanna be sure what im thinking is correct

is that enough?

[strange whale]

the essence of it is there yes

[junior pumice]

this isn't a rigorous class so hopefully he'll be fine with that.

thanks again. you are a linear algebra god

.close

[topaz sinew]

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[brave aspen]

stupid ass trig shit

[brave aspen]

(this is for the people in discussy) @uncut crow @violet fractal

[bright trail]

Look, if a is your first angle in 0 to 2pi

You have a second perfectly good angle at a+pi/2 for the other solution

[brave aspen]

yep

that cos is negative tho

[bright trail]

Your original -theta angle can be shifted up to match this a+pi/2 angle is all

[brave aspen]

what

im lost

theta has to be positive in this problem

[bright trail]

Yes, the other solution is the negative of the the original solution.

[brave aspen]

0 to 2pi strictly positive

right

[violet fractal]

oo wrinkle to iron

[bright trail]

a+pi/2 is the positive theta value.

[brave aspen]

yep

[bright trail]

If a is the original theta

[brave aspen]

are you saying the answer does not have to be positive tho?

[bright trail]

No

[brave aspen]

okay

well that's negative

theta is still positive cool but the answer is negative

[bright trail]

a+pi/2 is not

Yeah the answer is negative

[brave aspen]

the cos value is literally negative

cos is in the second quadrant

its negative

[uncut crow]

yes

[bright trail]

Look, given any cos(t)

[brave aspen]

cos (a+pi/2) is negative despite theta being positive still

[bright trail]

Yes yes

Mmm lemme think

See the pt here?

[brave aspen]

yeah

[bright trail]

The top arrow corresponds to a positive theta

[brave aspen]

yep

[bright trail]

The bottom to a negative theta

[brave aspen]

mhm

[bright trail]

So there are two different angles we can take to get to the same point.

[brave aspen]

yep

ones negative tho

cant use it

[bright trail]

With different signs

[brave aspen]

yep

cant use it doot

[bright trail]

Yeah, but we can just go in the opposite direction to find the positive one if we find a negative one that works

That's the point people are trying to make.

[brave aspen]

what

are you talking about the other top arrow?

the only "opposite direction" we can go is that top arrow

[bright trail]

You found the bottom arrow/angle yeah?

[brave aspen]

yep

the one we cant use for the problem

[bright trail]

All people are saying is you can get the same cosine and sine values by using the top angle/arrow instead.

[brave aspen]

okay

i literally know

but the cos value

is negative

[bright trail]

Yeah

[uncut crow]

That's ok

[brave aspen]

what the fuck

[bright trail]

?

[brave aspen]

are you fucking serious 😭

if i knew that i could have solved the problem in like a minute

[violet fractal]

didn't I tell you this

[bright trail]

Well now you know for the future

[brave aspen]

no

[brave aspen]

you threw me off with the triangles

[violet fractal]

I did. I said

once you see it you're gonna be like OH.

[bright trail]

See the two triangles here

They have the same sine but different cosines

One is positive the other negative

[brave aspen]

let me speak 😭

please

i fucking know that 😭

so i thought the problem wanted me to find a second value of cos that yielded the same positive answer

i didnt realize it could be negative

because when jay said draw triangles, i was imagining like pure triangles

[bright trail]

It's intentionally left kind of vague so people run into this issue imo

[brave aspen]

so i couldnt use negative values cuz we cant have negative side lengths

[bright trail]

People who don't understand will only see the one solution when there are two.

[violet fractal]

(even tho I do these in class...)

[brave aspen]

thats fucked

[brave aspen]

tee hee im just stupid ig idk

[bright trail]

Yeah sort of but it's important to hit this snag and remember this issue.