help-49

[fleet sonnet]

isnt vector used in physics

[obsidian wedge]

It is

[fleet sonnet]

so if i didnt use cosine law i would use vector for this question?

[flat plinth]

Yeah

[fleet sonnet]

oh

also

for vectors

what if they give u dot

wait

[flat plinth]

There's many ways to peel an apple

[obsidian wedge]

We are only concerned with the form or expression

It doesn't matter since the question doesn't ask you to find the resultant

[fleet sonnet]

what do u do

[obsidian wedge]

But you could use it since it gives the same form

[fleet sonnet]

im new to vectors like

i just found out what it is

[obsidian wedge]

Find the resultant force acting on the object

Force is a vector

Since acceleration is a vector

[fleet sonnet]

yes

wait but

do u solve that question?

like what do i do with it

[obsidian wedge]

Use the resultant

Formula

[fleet sonnet]

a-b?

[flat plinth]

This

[obsidian wedge]

Understand it intuitively.. You are applying force upwards and sideways

[fleet sonnet]

thats for paralellogram

[slim eagle]

just ask your teacher to help

[fleet sonnet]

its like a class tutor

[slim eagle]

i dont think its possible to understand vector addition at first so easily

[fleet sonnet]

idk what their contact is

[slim eagle]

whenever u meet

[fleet sonnet]

oh alr

wait but can u just do that question

[obsidian wedge]

Fr watch videos online

[fleet sonnet]

like what would i write

is it just 2N

i draw a middle line between them and state 2N?

[obsidian wedge]

Do you know trig

[fleet sonnet]

ye

[obsidian wedge]

Well then you can solve it

[fleet sonnet]

so we make that into a triangle?

oh ok

[obsidian wedge]

Yup

[fleet sonnet]

okay ty

.close

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[fleet sonnet]

with vectors if u have 2 that are perpendicular to each other

[fleet sonnet]

wait ima draw

will the answer for these be the same

even though like one of the arrors is following the other

[white gate]

what answer

dot product?

[fleet sonnet]

question says find the resultant vector for the following

[fleet sonnet]

idk

i wana know if they have the same answer

cuz for the right u can use trig to solve it

[white gate]

oh, resultant vector

[fleet sonnet]

yes

[white gate]

the vectors arent the same, but the modulus are the same

[fleet sonnet]

what would u write for the answer

would the left one be 0 and the right be 14

[white gate]

it says to find the resultant vector so surely you have to know the direction and length

[fleet sonnet]

doesnt it give the direction and length

oh ye i forgot to add N to it

but thats just like outward

[white gate]

sigh

!original

[midnight plank]

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.

[fleet sonnet]

okay

leme ss

:/ my phone died

[white gate]

rip

[fleet sonnet]

ok but the original

is

thats what it gives

and the question is find the resultant vector for the following

[white gate]

oh

yeah, first you draw the resultant vector to show the direction

then you find its length

[fleet sonnet]

so the green

is the resultant vector right

and for length

uh

is it just 10-10

how do we do it

[white gate]

what

[fleet sonnet]

uh

ok so this sheet says

eg a-b= a + (-b)

[fleet sonnet]

wait how do u do it

[white gate]

think of it like a very normal geometry problem without the vector

[fleet sonnet]

oh ok so its just pyth

[white gate]

ye

[fleet sonnet]

i thought it would be hard bruh

what does this mean then

wait i cant take pic so lemme draw

eg a-b= a + (-b)

turns into

which then turns into this

i was tryna draw dotted lines

[white gate]

well, yeah, that's true, a+b = a+(-b)
in this formula, the length doesn't always satisfy that

[fleet sonnet]

oh so thats a seperate formula?

[white gate]

the length can be found by treating it like a normal geometry problem

[fleet sonnet]

yes

[white gate]

that formula is only to determine the direction of resultant vector

[fleet sonnet]

oh the direction

[fleet sonnet]

so my final answer would be just draw that middle line

with it labeled as 14.14

[white gate]

,w 10sqrt(2)

[grand pond]

Results provided by WolframAlpha

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Requested by return.h3x.is.targetvn

[white gate]

yep that's right

[fleet sonnet]

oh okay

ty

.close

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[orchid grove]

Relation between statndard deviation and mean deviation?

[orchid grove]

My book says

5× mean deviation = 4 standard deviation

[silent elbow]

Idk, did you try googling it?

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[verbal pumice]

I'm kinding of having a hard time working out what's gonna happen with the 7 A current source in between

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[verbal pumice]

<@&286206848099549185>

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[wind oxide]

H

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[spice scroll]

how do i prove T(n)=Theta(n)?

[spice scroll]

also how do i solve when 9=a,13=b,18=c , and i a,b,c are different

as in different numbers from above

[wind oxide]

Is t(n) a sequence??

Or arbitrary function

[spice scroll]

RR

[midnight plank]

@spice scroll Has your question been resolved?

[wind oxide]

Bruh wth-

That rr makes no sense for n=1

Should be $T(18n) = T(2n) + T(13n) + \Theta(18n)$

[grand pond]

rak³en

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[tight bison]

Is my math correct? I feel like I didn’t do anything wrong

[tight bison]

Perhaps it was my algebra when cancelling out the sec^2x

It was in that, it was in my algebra I messed up

I’ll just wait for somebody to confirm I fucked up my algebra lol

Yea I figured it out

Wrong identity

And I should’ve distributed

[drifting birch]

Yea

[tight bison]

Thank you😂

[drifting birch]

Also you went from x to theta

and forgot all your dx dtheta and du

[tight bison]

Yea, in this specific problem I don’t think I’ll need to convert back into x

[drifting birch]

😭

[tight bison]

Yea😭

Man I’m tired, calc 2 is kicking my ass

[drifting birch]

you do because it's an indefinite integral

[tight bison]

Oh ok thank you

[drifting birch]

if I integrate cosx and get sinu something is wrong

😭

[tight bison]

Yea I totally get it

I completely whiffed on this one

.close

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[tight bison]

You alr know ya boy is back

How do I convert from theta to x again

.reopen

[midnight plank]

✅

[tight bison]

The stuff above the bot is my question

Lemme use a better picture so you aren’t craning your head actually

[hybrid crane]

$$sec^2 (x) = tan^2 (x) + 1$$

[grand pond]

StrangeQuarkAL

[drifting birch]

your theta sub is wrong

Your root is √(576+576tan²) but substituting tan does make 576 magically appear

😢

[tight bison]

Well it turned into 24

Sqrt of 576

[drifting birch]

no like

You have your integral right

and you have x=tan(theta)

If you substitute you get √(576+tan²)

Not √(576+576tan²)

[tight bison]

I thought the substitution was x=atan(theta)

And a would be 576

OHHHHHH

[drifting birch]

it is

you forgot the a

[tight bison]

x=24tan(theta)

Oh that makes much more sense now

So dx would be 24sec^2(theta)

[drifting birch]

yes

[tight bison]

Thank you very much

Twice you have helped me🙏🏾

I am in your debt oh wise one

[drifting birch]

[tight bison]

.close

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[steady trail]

But is a set X said minimal if there is only 1 element in it that satisfies a certain property?

[midnight plank]

@steady trail Has your question been resolved?

[midnight plank]

@steady trail Has your question been resolved?

[runic hamlet]

generally a set is called minimal with respect to some property if no proper subset of it also satisfies that property

so if a set with one element satisfies some property then its likely that its minimal, cause the empty set probably doesnt satisfy the property

but it depends highly on the specific property

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[bronze mason]

Could somebody check my proof? I'm not sure I fully get this whole 'choose-an-epsilon' business, whether it's continuity, limits or even something like this.
(oops used wrong sign in the middle of the second case should be \geq not \leq )

[uneven sandal]

Why is $s^3 - 7 \eps > 2$ should it not be greater or equals?

[gilded turtle]

unrelated but how do you get the filled in box in latex

[grand pond]

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

[bronze mason]

\blacksquare, I think it's in amsmath or amssymb or something

[gilded turtle]

ah

ty

[uneven sandal]

but shouldnt matter I guess

[bronze mason]

Why so?

Yes, I am not sure that changes the accuracy of the proof.

[uneven sandal]

just choose another epsilon tilde defined as epsilon/2

if thats the problem

usally

[bronze mason]

Well, that is indeed what I've done (for safety's sake).

But I am not sure why it would be a non-strict inequality anyway

[uneven sandal]

also

it should be less than not greater than right

[bronze mason]

the middle one should be greater than or equal I believe

the s^3 - 7\eps > 2 is right

[uneven sandal]

but s = 3rd root of 2

and that minus a tiny positive number

raised to the third power os smaller than 2

[bronze mason]

Well, the whole point is to get a contradiction.

Umm... actually, wait.

No, s isn't the third root of 2. I'm saying suppose s is greater than the third root of 2.

[uneven sandal]

ah yeah my bad

[bronze mason]

Then if I set this inequality to be the case, I'm getting that s - epsilon will actually be an upper bound too.

I think this proof is right, but I only got it after borderline copying a different proof. 😛

[uneven sandal]

it looks good

[bronze mason]

Excellent, thanks for your assistance... I am never sure with these choose-an-epsilon mind games.

,close

Wrong symbol.

/close

Nope.

[uneven sandal]

use a dot

[bronze mason]

Ah

.close

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[slate sandal]

Hi

.reopen

Can anyone help me with this?

[bronze mason]

Technically this channel is still assigned to me I think for a while so please post in one of the unoccupied ones

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[fleet flume]

Given that W=240N, what is Q?

[fleet flume]

Lamy's theorem

how should I start?

[steady jacinth]

Hint: F/sin(theta) is constant for a given object where F is a force acting on the object and theta is the angle that it makes with some fixed line

[fleet flume]

sine rule?

[fleet flume]

yeah, but I need help to apply it

[stiff lion]

Doing the free body diagram of each block. Then, draw the tension forces of the strings.

[steady jacinth]

Like @stiff lion said, an easier and more intuitive method is to draw the free body diagrams and use the fact that the object is in equilibrium so the net force on the object in any direction is 0.

[fleet flume]

is it ok if B points in different directions in the free body diagrams?

if not, it would be impossible that the forces are in equilibrium

[stiff lion]

Yes

[fleet flume]

thx

got it

thanks

.close

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[last arch]

So I mean a is clearly false

The mid points are not the same

So let's look at b,c and d

Which of these properties are unique to paralleograms

[iron pagoda]

I think C

I was thinking C was the answer but idk

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@iron pagoda Has your question been resolved?

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[orchid grove]

A delegation of four students is to be selected from a total of 12 students. In how many
ways can the delegation be selected

[orchid grove]

two particular students refuse to be together and two other particular students wish
to be together only in the delegation ?

[unique juniper]

for this one it's just 12C4

for the second criteria

in one case, one of the two students is picked, then the other, then neither

[midnight plank]

@orchid grove Has your question been resolved?

[midnight plank]

@orchid grove Has your question been resolved?

[midnight plank]

@orchid grove Has your question been resolved?

[fallen gust]

Are all of these congruent bcs I feel like 1 and 9 are not due to the angles not matching the other angle

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[viral dagger]

err did i not just post an image here

[viral dagger]

:/

ok im doing part b

hints are
"you can explicitly find K"
"what has equal power to both circles"

playing around it looks like NK is the diameter of circumcircle ANB, but i have no idea how to prove it

[modern sapphire]

tangents at A and B have something special

[viral dagger]

yeah its at K but idk how to prove it :(

[modern sapphire]

what can you say about AKB triangle?

focus on angles at A and B

[viral dagger]

not sure

[modern sapphire]

what can you say about small arc AM?

use tangent secant thm there at A

[viral dagger]

its 120?

[modern sapphire]

yea

[viral dagger]

whar

[modern sapphire]

draw tangent AK at A, and find angle BAK

same with ABK

[viral dagger]

hm

60?

[modern sapphire]

I may have named the wrong theorem 💀

I just realized it oops

[modern sapphire]

but yea

[viral dagger]

call center E, MAE=MAC/2=30
KAB=90-30=60

[modern sapphire]

you uhh dont need to do that

you found angle BAK = 60, similarly ABK is also 60

that makes ABK an equilateral triangle, or AK = BK

and ABK is unique coz AB is fixed

that makes K unique, and it lies on radical axis, which is all lines MN

[viral dagger]

wait how do you know AK is tangent at A hold on

im retracing my steps and im doing a silly?

[modern sapphire]

MAK = 60 right?

[viral dagger]

yea but the proof needs AK tangent at A no?

[modern sapphire]

I explicitly draw a tangent at A

I dont construct AK

I draw a line at A, such that it is tangent to the circle

[viral dagger]

oh phantom point?

[modern sapphire]

At A

A is not phantom

it is on the segment

and on the circle

[viral dagger]

i mean like call point K' as a phantom point of tangents A and B, everything above got ABK' to be equilateral, and K' would be proven to lie on the radical axis proving K'=K?

[modern sapphire]

yes

[modern sapphire]

also, not a phantom point, just point of intersection

[viral dagger]

whats the diffrence?

[modern sapphire]

idk whats phantom point

I am just assuming its something imaginary

while point of intersections is a concrete concept

[viral dagger]

ok i see

[modern sapphire]

I mean, you just jumped ahead here

[viral dagger]

oh that was just from drawing

i was hoping for that to be like an insightcor smth

[modern sapphire]

I just wanted to show the point of intersection lies on the radical axis coz tangents at A and B are equal in length

[viral dagger]

cause it seemed useful and it looks true so its worth tryung to prove?

[modern sapphire]

yea, it is. But while solving problems, its not good to find insights based on results

[viral dagger]

ok

sorry uhh i gtg ill try continuing this

.close

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[tall rapids]

$Prove that any two consecutive odd numbers are relatively prime (coprime).
$

[somber canopy]

MY GOD

[tall rapids]

Hhhhhhhhhhh

[somber canopy]

YOU JUST HAD TO CLAIM THIS

U COULDN'T SEE ME TYPING FOR THE LAST 5 MINUTES?

[tall rapids]

Guess what

Ok I'll help you

[somber canopy]

ok

wait i'll get my own thread

[tall rapids]

Tyt

[keen saddle]

@somber canopy i recommend first claiming the channel
then type your question
if its long

[tepid lodge]

let n be a prime number

assume n and n+2 are not comprime

so gcd(n, n+2) > 1

since they're odd, it can't be 2

and since their difference is 2, they can't both be divisible by something > 2

[tall rapids]

Make scenes

.close

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[inland patio]

I'm studying a theorem that states a continuous, bijective function from a compact metric space onto another metric space has a continuous inverse. The author provides a counterexample when the domain is not compact, namely f(t)=(cos t,sin t) on [0, 2pi). I wonder, what is the inverse of this function?

[next bramble]

it has lots of inverses, because the codomain (x,y) contains points where there doesn't exist a t such that x = cos t and y = sin t.

[inland patio]

ok 👍

[next bramble]

but you can take a representative, i.e. the unit circle and find an inverse from that

[inland patio]

it seems counterintuitive to me that there are lots of inverses when Rudin claims it is 1-1 and onto. Shouldn't this give us a unique inverse?

Rudin is the author, by the way.

[next bramble]

It's not injective because f(t) = f(t + 2n pi) for all integer n.

maybe I'm misunderstanding the context

oh, oops. It is injective on [0,2pi)

my bad

I suppose you could define this

$$f^{-1}(x,y) = \begin{cases} \arccos{x} & \qq{if} y \ge 0 \ 2 \pi - \arccos{x} & \qq{if} y < 0 \end{cases}$$

[grand pond]

Shuba

[next bramble]

But you'd need to restrict the domain of this function to the image of f, which is where x^2 + y^2 = 1

you could equally define g as an inverse also

$$g(x,y) = \begin{cases} \arcsin{y} & \qq{if} x \ge 0 \ 2 \pi - \arcsin{y} & \qq{if} x < 0 \end{cases}$$

[grand pond]

Shuba

[inland patio]

ok 👍

.close

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[stone spear]

hi

[full leaf]

hi what is your question

[stone spear]

so i was doing

linear programming

so the max value i got was

752 dollars

by putting in 12,38

in the model 12x+16y

however the value is outside the feeasible region

btw i shaded out to see the values better

[midnight plank]

@stone spear Has your question been resolved?

[midnight plank]

@stone spear Has your question been resolved?

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[gusty falcon]

sin^2(x) + cos^2(x) = 1 is an identity because x in R

[gusty falcon]

but like tan^2(x) + 1 = sec^2(x) is an identity(?)

technically it isn't right

i mean that begs the question what the equal sign means

it's an identity for all values for which tan(x) and sec(x) is defined

which happens to be the same set

so like we care about the domain?

but i guess there's a problem with that as well

😭

[lyric crag]

It will work fine until the value u substituting is not undefined

[gusty falcon]

the identity is also true for x = pi/2 in some sense

so no that doesn't answer my question

[visual tiger]

both sides are \textbf{defined} and equal for all $x \not \equiv \frac{\pi}{2}\mod \pi$

[grand pond]

rafilou is not not born in 2003

[visual tiger]

so it is an identity on that set

[gusty falcon]

oh

do we not specify what set something is a set on?

is it just uh retrieved from context?

[misty gorge]

whenever you have the identity f(x)=g(x), then the set on which that equation is an identity is usually the intersection of the domains of f and g

[misty gorge]

so here i would say the fact that this is true means that tan^2 + 1= sec^2is an identity at the level of functions because whenever it makes sense to talk about both tan^2(x) and sec^2(x) then that identity holds.

[gusty falcon]

right i guess that is straightforward enough

i had the impression that the domain of the f has to match the domain of g

like uh if you rewrite the "identity" as: 1 + tan^2(x) = sec^2(x)

now both sides have the same domain

and then if we have x in R then rewriting the above identity as sec^2(x) - tan^2(x) = 1 😭 isn't an identity

but yeah i guess uh we need x for which both are defined

[misty gorge]

you can just restrict the domain of both functions if you want it to be a “true” identity, since g|_Dom(f) is another perfectly good function.

lol

so take g(x) to be 1 if x is not pi/2 and don’t define it otherwise

well i would say g is still 1

but like you might want to make the distinction that g is 1|Dom(tan)

[lyric crag]

How?

[gusty falcon]

oh you mean like make some kind of continuous extension?

[misty gorge]

i mean the other way. restrict the domain of 1

like f(x) =1

[gusty falcon]

oh i thought u meant extend the domain of sec^2 - tan^2 such that it matches the domain of 1

whoops

[gusty falcon]

yeah i guess that's in some sense

the same 😭 as restricting the domain

[gusty falcon]

it's just like x/x = 1 isn't an identity for x in R

[misty gorge]

yeah it’s a bit silly. but this actually does come up sometimes in geometry (like, the keyword here i’m thinking of is “change of chart maps”) where you want to talk about the composition of two functions but some of the image of the first function is not in the domain of the second, and that’s okay, just take the intersection where it makes sense.

[misty gorge]

same thing here but we’re talking about a relationship between functions, we have to restrict our domain a little bit.

[gusty falcon]

yeah i guess i just i'll encounter that way later lol

[gusty falcon]

for now i can live with this

cuz it just makes sense intuitively at least

[lyric crag]

So sin²x + cos²x =1 is also not an identity?

[gusty falcon]

so basically an identity is just when f(x) = g(x) for all x in the intersection of both the domains, right?

[gusty falcon]

it is

that's true for all x in R

[lyric crag]

Ohk

But writing tan²x - sec²x =-1 doesn't changes its domain?

[lyric crag]

.

[gusty falcon]

huh

my point was that sec^2(x) - tan^2(x) is not an identity if x in R

but the other version 1 + tan^2(x) = sec^2(x) is an identity when x in R

in some sense at least

https://en.wikipedia.org/wiki/Partial_function

[lyric crag]

I don't know if I am getting your point but I am fine with it

[gusty falcon]

what do you not get though?

[lyric crag]

The thing I get is that sec²x -tan²x won't hold and just answer 1

Tell me if I am tru

[gusty falcon]

when x = (2n + 1)pi/2

your equation clearly doesn’t hold

[lyric crag]

Oh yes I get now

I have just started functions before my grade

Just drawing graphs and stuff.

[gusty falcon]

okay then, glad that’s sorted

anyway thanks everyone that helped

.close

[midnight plank]

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[oak granite]

There is a theorem (idk the name) says:
in a the plane
if you have 4 points A, B, C and D
where A ≠ B
AB = CD (distance)
and AB ≠ CD (vector)
there exist a unique rotation that rotates A in C and B in D (or rotates [AC] on [BD]) which it's center is the intersection of the medians of AB and CD, and the angle of rotation is (AB, CD) [2pi]
Can anyone tell me its proof or send me a wiki link of the proof ?

[blissful trench]

probably euclidean plane isometry?

cool portfolio btw

[oak granite]

I'll search and come back

[oak granite]

thank you

I didn't find 😦

[blissful trench]

I just guessed the name based on your problem but im not sure if thats what are you looking

is this related to geometry or linear algebra?

[oak granite]

geometry

the definition I am working with is that: R is a rotation application with center I and angle theta
R(A) = B means IA = IB (distance) and (IA, IB) = theta [2pi]

[lyric crag]

Can you give full info or pic?

[oak granite]

it is in french

wanna picture?

[oak granite]

?

[lyric crag]

The thing i understand is that diagonal intersection is the rotation axis (R) and because ab = cd by symmetry the AR = DR so after rotating A and D will rotate on a circle so D will take A's place

I can't give written proof right now

Try to Google it by translating something should show up cuz it's a theorem

[oak granite]

I tried for a long time searching but I didn't find

[midnight plank]

@oak granite Has your question been resolved?

[lyric crag]

https://www.doubtnut.com/qna/642545521

I am getting these results check this out

[midnight plank]

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[wispy osprey]

first is the question, second is my answer, third is the solution

[wispy osprey]

isn't it supposed to be (1,3]? why is the answer (2,3]?

[rose trout]

It's (2,3] because for i = 1 you have X_1 = (2,5)

You can't just take the limit of both endpoints and call it a day

In this case, the upper limit gets smaller, but so does the lower limit.

[midnight plank]

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[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

,, A \oplus B = C \iff A \subset C \quad B \subset C

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

i) a + 2 - 1 = 0
ii) a + 1 + 1 = 0

[graceful ferry]

Damn bro u study so much give me some motivation how

[tidal turret]

the motivation is to finish my degree brother, dont you want to finish too?

[graceful ferry]

Yes I do but the discipline is not the best🫡

[tidal turret]

you got this man, I have seen your homework is tough

but you need to persevere

[graceful ferry]

I take some your questions to solve haha

You're right

You got it too let's go!

[tidal turret]

thanks bro, gl u2 with algebra

[midnight plank]

@tidal turret Has your question been resolved?

[brisk iris]

Hey @tidal turret

Did you follow the steps ?

[tidal turret]

what steps

[brisk iris]

That I told you for the previous exercice

[tidal turret]

oh yeah

for the linear transformation?

[brisk iris]

There is no linear transformation in this exercice

But the steps remain valid

So, Step 1:

--> What are the dimensions of each space ? Is there an intersection ?

[tidal turret]

SnT is only trivial

dim S = 2

dim T = 2

[brisk iris]

Trivial as in null ?

[tidal turret]

yeah SnT = {0}

[brisk iris]

Good

So S+T is basically a direct sum that creates R4

[tidal turret]

S (+) T = R4

[brisk iris]

Nice

So on the other side, you should also find a basis of R4

Can you do that ?

[tidal turret]

I found a basis for R4

[brisk iris]

Tell me

[tidal turret]

,w Nullspace {{1,0,2,-2},{1,1,0,1}}

[grand pond]

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Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[brisk iris]

???

[tidal turret]

,w rref {{-2,2,1,0},{2,-3,0,1},{1,3,-1,-3},{0,1,1,0}}^T

[grand pond]

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[tidal turret]

this is my basis of R4

S n T = {0}

[brisk iris]

Yeah but your basis should include (a,1,1,1)

[tidal turret]

S = <(-2,2,1,0),(2,-3,0,1)>

[tidal turret]

right

[brisk iris]

You're getting lost again.

[tidal turret]

but that is not in S

[brisk iris]

Yeah but forget S+T

We already established that's just R4

Now we're looking at the other side of the equations

This side

If you want this thing to be R4

You should create a basis for R4 that includes (a,1,1,1)

[tidal turret]

okay

[brisk iris]

that would look like this <(a,1,1,1),w1,w2,w3>

[tidal turret]

R4 = {(a,1,1,1),(0,1,0,0),(0,0,1,0),(0,0,0,1)}

[brisk iris]

Yeah but compute that determinant ...

of the 4 vectors

[tidal turret]

,w det {{a,1,1,1},{0,1,0,0},{0,0,1,0},{0,0,0,1}} = 0

[grand pond]

Results provided by WolframAlpha

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Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[buoyant yoke]

cmon you couldve done this yourself

[brisk iris]

Why did you equate it to 0 ??

[tidal turret]

oh, it should be nonzero

[brisk iris]

Anyway, you need that determinant to be different than 0 no matter the value of a

So your basis doesn't work

Try to have a determinant that is always positive for example

[tidal turret]

is a triangular matrix the determinant is the diagonal elements multiplied, det = a

[brisk iris]

Yes, but what we want is for the determinant to be strictly positive (or negative)

So the vectors you chose for the base don't give you that

Can you think of ways to change this matrix in order to obtain a strictly positive determinant (like a²+1)

If you find the matrix, you find the right basis

[tidal turret]

we know (a,1,1,1) ∈ S + T

[brisk iris]

The first vector should not be touched

forget S+T, we know it's just R4, so it doesn't add any information

I want you to change this coefficients (as simply as possible), to obtain a matrix for which the determinant is strictly positive

Try this

[tidal turret]

what?

[brisk iris]

Try this, see what the determinant is

[tidal turret]

(-1,a,0,0)?

[brisk iris]

The determinant of the matrix I gave you

[tidal turret]

but (-1,a,0,0) from where?

[brisk iris]

I wanted to make sure that the determinant was positive, like : a²+1

So I tweaked the second vector to make sure that's what I get

Here's exactly how I thought of it

I made sure det (A1) = a²+1 , because I knew det (A2) = 1

[tidal turret]

why do we want determinant positive

[brisk iris]

In order to build a basis, we need the 4 vectors to be free

no matter what the value of a is

So a way to build the 3 vectors is by looking at them like a matrix

So that means det of the 4 vectors can't be 0

[tidal turret]

yeah determinant of 4 vectors is NOT 0 because then we dont generate R4

[brisk iris]

yes

[tidal turret]

it can be positive or negative tho

[brisk iris]

you could also look for 3 vectors that are free, and orthogonal to (a,1,1,1)

[tidal turret]

but why positive?

[brisk iris]

We don't care about positive

[tidal turret]

the orthogonal complement of <(a,1,1,1)>

[brisk iris]

Strictly positive allows me to avoid 0

[brisk iris]

yeah that would work aswell

[brisk iris]

yes

W could be : ax1+x2+x3+x4 = 0

It would work as well

[tidal turret]

yes with the = 0

[brisk iris]

careful, don't divide by a, it could be 0

so it's solved

I always prefer to use the determinant cause it allows you to instantly check if vectors are dependant or not. No need to worry about any peculiarities

[tidal turret]

why is this soo hard

[midnight plank]

@tidal turret Has your question been resolved?

[brisk iris]

Is this not solved ?

[midnight plank]

@tidal turret Has your question been resolved?

[tidal turret]

how do I find a?

[brisk iris]

He’s not asking you to find a

He’s asking you to find a W that works no matter the value for a

[tidal turret]

it says find a in R

and W

[brisk iris]

Oh we did better than he asked

For the vectors I gave you, you can choose any a and it would still work

However

You can simplify by using a=1

Then the vectors you gave me at the beginning would work

[midnight plank]

@tidal turret Has your question been resolved?

[midnight plank]

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[orchid grove]

A or B?

I am comfused in domain

[vast ginkgo]

b basically represents all answers $n\pi + \frac{\pi}{4}$ where n is any integer

[grand pond]

BuilderDolphin

[vast ginkgo]

if that's what was confusing you

[orchid grove]

Ahh no

[vast ginkgo]

(not saying it's the answer just clarifying it)

[orchid grove]

I meant is it working?

[vast ginkgo]

i mean

[orchid grove]

I understood its meaning

[vast ginkgo]

oh

[orchid grove]

But can it be our answer?

If yes then why and if no then why?

😬

[vast ginkgo]

well can a function have multiple outputs for a single input

[orchid grove]

No

[vast ginkgo]

then is b even an option

[orchid grove]

It violates our function property

You meant inverse function f(1)

Should have one value

So A is correct

[vast ginkgo]

yep

[orchid grove]

Hmm makes sense

[vast ginkgo]

if you plug in 1 into arctan

the only thing you possibly get is pi/4

[orchid grove]

Can I not get other particular value?

@vast ginkgo

[vast ginkgo]

i mean again can you get two outputs from one input

[orchid grove]

Not two

[vast ginkgo]

you're finding the value of the inverse tangent with input 1

[orchid grove]

I meant if my function is from (π/2,3π/2)

[vast ginkgo]

uh nope

the question is literally just asking what the value of $f^{-1}(1)$ is

[grand pond]

BuilderDolphin

[vast ginkgo]

$= arctan(1) = \frac{\pi}{4}$

[grand pond]

BuilderDolphin

[orchid grove]

They didn't write the R-->R

Because tanx is not define

at π/2

And other

(2n+1)π/2

[vast ginkgo]

i guess i can see what you mean

no domain was given for tan so i just assumed (-pi/2, pi/2)

for arctan

but again that would violate the definition of a function

having multiple outputs

[vast ginkgo]

in that case yeah the inverse of $tan(x)$ in $(\frac{(2n-1)\pi}{2}, \frac{(2n+1)\pi}{2})$ at x = 1 would be $n\pi + \frac{\pi}{4}$ where $n \in Z$

[orchid grove]

What?

[grand pond]

BuilderDolphin

[orchid grove]

So what option work?

[vast ginkgo]

the domain of tan would also depend on n

[orchid grove]

I see

So which option is suitable

[vast ginkgo]

i would probably go with A logically since no domain is given but im no expert

depends on what ur learning too

B is correct if domain is specified in terms of n

but there isn't really anything related to that on the problem so i not sure

[orchid grove]

Can we wait for another helpers?

[midnight plank]

@orchid grove Has your question been resolved?

[midnight plank]

@orchid grove Has your question been resolved?

[orchid grove]

<@&286206848099549185>

[orchid grove]

Could you explain it?

[spiral blaze]

bro, obv answer is b

u have to consider eery value of x that gives 1

since its an inverse

[orchid grove]

Thanks

.close

[midnight plank]

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[dusty aspen]

What happens in the pink area?

[dusty aspen]

inf/inf=1 ???

[lyric charm]

would you like me to show you that more general thing that i didn't get to show last time

[dusty aspen]

sure but I have 20 minutes till exam so...

[lyric charm]

$\frac{x^2 - 4}{x^2 - 1} = \frac{(x^2 - 4)x^{-2}}{(x^2-1)x^{-2}} = \frac{1 - 4x^{-2}}{1 - x^{-2}}$

[grand pond]

ann.in.a.teacup

[lyric charm]

the negative powers of x go to 0 and you are left with 1/1

because x once again goes to infinity

[dusty aspen]

why did you add x^-2 in the middle part

[lyric charm]

i didn't add it

[dusty aspen]

you multiplied by x^-2

[lyric charm]

i multiplied by it

the numerator and denominator are both quadratic

so i divided them both by x^2 so that their leading terms would instead be constants

and so that we don't have an infty/infty situation anymore

[dusty aspen]

you can't divide inf/inf and get 1?

[lyric charm]

no

$\frac{\infty}{\infty}$ is undefined --- as shorthand for a limit (of the ratio of two functions, each of which goes to $\pm\infty$ by itself) it could be any number or even not exist at all

[grand pond]

ann.in.a.teacup

[lyric charm]

case in point: if i slapped a coefficient of 2 on that x^2 term on the top (you'd get a different function), you'd get that the limit is 2 not 1

[dusty aspen]

So how to continue and get 1 from here:

[lyric charm]

the negative powers of x go to 0

[dusty aspen]

negative powers of x? X^-2 is something really small, like 0.0000...?

[lyric charm]

$x^{-2}$ approaches 0 as $x \to \pm\infty$.

[grand pond]

ann.in.a.teacup

[lyric charm]

so does x^(any negative number)

that is what i meant with "negative power of x"

nothing more

[dusty aspen]

oh I see

Could we split this fraction? Original one into two fractions? And then erase something?

[dusty aspen]

this one

[lyric charm]

not worth the effort bc the denominator isn't a pure power of x.

what i showed just now is how you do it generally

[dusty aspen]

can you turn x^2-4 and turn it into x^2-1-3?

then you could split and something would cancel

[lyric charm]

again

not worth the effort

you do not want to spend 17 years doing that with like idk 6th degree polynomials on the top and bottom both

[dusty aspen]

okay

thanks again

.solved

[midnight plank]

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[twilit field]

So I have $\int_{a}^{b} \norm{ f'(t)}dt$

[lyric charm]

\int

[grand pond]

What a wonderful world it is !

[twilit field]

Which is simply $\int _{a}^{b} \sqrt{2}dt$

[lyric charm]

are you sure?

[twilit field]

oops

no

[lyric charm]

what did you get for f'(t) itself?

[grand pond]

What a wonderful world it is !

[twilit field]

so $\sqrt{2} (b-a)$

[grand pond]

What a wonderful world it is !

[twilit field]

I'm interested in everything but part(a)

First to prove it isn't simple

$f'(t) = ( -3\cos^2(t)\sin(t), 3\sin^2(t)\cos(t))$

[grand pond]

What a wonderful world it is !

[twilit field]

The derivative is $(0,0)$ at $t = \frac{\pi}{2}$. It is thus not simple

[grand pond]

What a wonderful world it is !

[twilit field]

Now to come up with a piecewise paramatization

Could I define the piecewise paramatization as $(cos^3(t), sin^3(t))$ on $[0,\pi] - \frac{\pi}{2}$ and $(0,1)$ at $t=\pi/2$

[grand pond]

What a wonderful world it is !

[twilit field]

so the curve length would be $\int_{0}^{\pi/2} \sqrt{ (-3\cos^2(t)\sin(t))^2+(3\sin^2(t)\cos(t))^2}+ \int_{\pi/2}^{\pi}\sqrt{ (-3\cos^2(t)\sin(t))^2+(3\sin^2(t)\cos(t))^2}$

[grand pond]

What a wonderful world it is !

[twilit field]

Is that right

[midnight plank]

@twilit field Has your question been resolved?

[twilit field]

<@&286206848099549185>

.cloe

.close

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[lunar pivot]

if $S = {u_1, \dots, u_n}$ is an orthogonal subset of an inner product space $(V, \mathbb{F})$ then how do i prove $S$ is linearly independent?

[grand pond]

carburetor

[lunar pivot]

since $S$ is an orthogonal set, $\langle u_i, u_j \rangle = 0$ for $1 \le i, j \le n$

[grand pond]

carburetor

[lunar pivot]

so i have $\sum_{1\le i, j\le n} c_i \bar{c_j}\langle u_i, u_j\rangle = 0$ or $\langle\sum_{i=1}^n c_iu_i, \sum_{j=1}^n c_ju_j\rangle = 0$

[grand pond]

carburetor

[runic hamlet]

0 = <ui, 0> = <ui, (...)> = ...

[lunar pivot]

but doesn't this mean $\sum_{i=1}^n c_iu_i = 0$

[grand pond]

carburetor

[runic hamlet]

what are the c_i supposed to be if you didnt choose them specifically that way

you want to check lin independence

so given sum c_i u_i =0 you have to conclude c_i=0

[lunar pivot]

c_i ∈ F i guess?

[runic hamlet]

well yes of course the c_i are scalars

[lunar pivot]

yes but that's not what i'm getting

i started with <u_i, u_j> = 0 for all 1 <= i, j <= n

[runic hamlet]

for all i < j

not for all i,j

you have to take a step back

what is it that you want to prove

what is the definition of that thing

[lunar pivot]

oh right

[runic hamlet]

so therefore, how do you have to start the proof

then you can think about using what you know

[midnight plank]

@lunar pivot Has your question been resolved?

[lunar pivot]

i think i got it

i can use the fact that $\langle u_i, u_i\rangle \ge 0$ to get something like $c_j || u_j ||^2 = \langle\sum_{i=1}^n c_i v_i, v_j\rangle$

[grand pond]

carburetor

[lunar pivot]

but $\sum_{i=1}^n c_iv_i = 0$ and $||u_j||^2 \ge 0$ which implies $c_j = 0$

[grand pond]

carburetor

[lunar pivot]

@runic hamlet this is fine right?

[midnight plank]

@lunar pivot Has your question been resolved?

[runic hamlet]

what are the v_i now

but this sounds vaguely correct, yes

[midnight plank]

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[somber canopy]

Wasn't there some way of writing the taylor expansion as

[somber canopy]

$f(x)$ = $e^{\epsilon\frac{d}{dx}}f(a)$

[grand pond]

kronium_

[somber canopy]

something like this

I dont exactly remember

Thats why im asking if someone knows

[midnight plank]

@somber canopy Has your question been resolved?

[robust isle]

https://math.stackexchange.com/questions/185354/exponential-of-the-differential-operator
yeah you could do something like that

@somber canopy

[somber canopy]

yes

[robust isle]

yes

[midnight plank]

@somber canopy Has your question been resolved?

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[midnight plank]

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[ancient dragon]

How would I prove that i is the derivative of d? Would saying they both have a vertical asymptote be enough, and if not, what else could I use to justify my answer?

[subtle zinc]

you cannot give a proof as you do not know the functions exactly

[sharp coral]

you can also note where the function is increasing and tie that to the sign of the derivative

[ancient dragon]

Thank you. Is there a proper way to word that saying that?

[dusty portal]

If I am reading it correctly I disagree with the proposition you make

[dusty portal]

“Intervals of increase/decrease”

[ancient dragon]

Thank you

[dusty portal]

Autocorrect

[ancient dragon]

.close

[midnight plank]

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[midnight plank]

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[remote owl]

I understand how to do it when its in standard form with 3 variables but i dont understand when its just 2 variables

[dusty portal]

It's the same process. Factor, make a sign chart, and test it.

x^2-2x=<0

[remote owl]

huh]

[dusty portal]

How did you learn it?

[remote owl]

what is a sign chart

[remote owl]

we just put all variables on one side then you find 2 numbers that add to something and multiply to something

but I cant do that when there is just 2

[dusty portal]

Well, you literally only have to factor the x out with 2 variables.

That's it.

[last arch]

Maybe I can use desmos to demo

No pun intended

Here is a graph of the quadratic x^2-2x

[remote owl]

ok

[last arch]

The shaded region is where the quadratic is less than 0

Where would you say this region lies between

On the x axis at least

[remote owl]

I just dont understand how to factor x^2-2x=<0

[last arch]

oh

Well then uh

[remote owl]

I have only factored like stuff in standard form

[last arch]

You might need to revise a tiny bit

[buoyant yoke]

$x\cdot x-2\cdot x$ do they have a common term?

[grand pond]

Bonk

[last arch]

What is a common factor in both x^2 and 2x

[remote owl]

x

[dusty portal]

^

[last arch]

Yea so

Just factor the x out

[remote owl]

ok

[last arch]

You good from here?

[remote owl]

yes thanks

.close

[midnight plank]

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[midnight plank]

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[tidal turret]

I need help with 1)

translation will be provided in a second

[buoyant yoke]

youre german?

[tidal turret]

is my second name lmao

Renato German

[buoyant yoke]

chavez

[tidal turret]

yee

second midterm is a little bit harder

[tidal turret]

I finished first midterm, still thinking about second exam, hopefully I am strong enough

[tidal turret]

Last time I got a 2/10 😅

[buoyant yoke]

when you do the questions here it looks like you do fine

do you get stressed during the exam?

[tidal turret]

I got adhd i think so 1h30 is usually not enough time to think through, when I do them here Usually is not under the clock

[buoyant yoke]

are you allowed extra time?

here youd get 15 minutes extra if you have some sort of condition

[tidal turret]

also in the exam the desks are so uncomfortable and the professors talk with each other and i can't concentrate bro

[tidal turret]

extra time students

[buoyant yoke]

yeah then ask if youre allowed some kind of extra time

look into your unis policy

[tidal turret]

I don't think they give a fuck

the professors talk with each other bro loudly like to make distraction

they want us to fail

[buoyant yoke]

then let them know to keep quieter

[tidal turret]

this is why I am studying with loud music so I simulate the exam room

[tidal turret]

they dont care bro, but i will consider it

[dawn dagger]

is that a thing in your uni?

[tidal turret]

I think so lmao

at least in the foundation year

[dawn dagger]

yea we humans will go extinct one day

[tidal turret]

oof this one is hard as hell

I don't even know how to find a

[midnight plank]

@tidal turret Has your question been resolved?

[shadow schooner]

dim(T) = 2 probably. Then dim(T) = dim(Im(f)) must mean the kernel of f, Ker(f) is of dimension 2, but the second condition says we have a direct sum between S and Ker(f) = to R^4, meaning dim(S) = 2 because you can sum dimension with direct sums. Then there should a value for a such that S is of dimension 2 not 3. For the last condition idk.

[shadow schooner]

That's what I can think of without computing the things, I have to say it might not work. Yeah this does not look easy

[tidal turret]

right, you are gorgeous, thank u

[tidal turret]

you gave me a huge insight though thanks !!

[midnight plank]

@tidal turret Has your question been resolved?

[midnight plank]

@tidal turret Has your question been resolved?

[tidal turret]

@brisk iris

[brisk iris]

Helloo

Where is the problem ?

[brisk iris]

Y still need help with this ?

[tidal turret]

is tricky

[brisk iris]

Can I try to write it and send it ?

[tidal turret]

sure

I found a

[brisk iris]

You did most of the work

Did you find v ?

[tidal turret]

yes

[tidal turret]

I think so yeah

[brisk iris]

SO I think you solved it

what's the issue ?

[tidal turret]

can you check

looks okay ish

[brisk iris]

The last vectors should be a vector of S

[tidal turret]

wdym?

[brisk iris]

I'm not sayuing it's wrog

But it's better if you use a vector of S here

[tidal turret]

ok

let me change it

[brisk iris]

Now to check, just make sure that all conditions are met and you're good

[tidal turret]

should be correct, thank you shaki

[tidal turret]

thanks

. solved

.solved

[midnight plank]

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[midnight plank]

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[noble elbow]

bit of a silly question.
If we use two inputs, one to define the distance between each factor and one to define the start value (for example, x, 2 would be x double factorial) For convenience im going to just notate it like $F_{6}(x)$
is there a reasonable way to get something like... $F _ {4}(5) = F_ {5} (5)$ that isn't as cheesy 😭