ah mb
#help-49
hard pewter
tidal turret
(1,0,0,0) works tho
hard pewter
yh
tidal turret
(0,1,-1,0) in H
hard pewter
yes or 0,0,-2,1
tidal turret
yeah XD
grand pondBOT
hard pewter
same here
hard pewter
sure we can
tidal turret
is more of the same I think, but with a direct sum
oh, last one also had direct sum
yeah, thats why we checked linear independence
hard pewter
yeah its the same
almost
we get a,b in the similar manner as you did in the prev one
tidal turret
hard pewter
but wait
tidal turret
lets figure dimension of T first
hard pewter
yea just like the previous problem,
dim H1 = 3
like it didnt depend on k this time it doesnt depend on a and b
hard pewter
yes so dim T is 1 making this easier ig
tidal turret
yeah dim T = 1
tidal turret
yes so dim T is 1 making this easier ig
wdym?
$T = S^\perp \cap H1 \cap H2\$ btw, but too overkill using complements
grand pondBOT
renato
hard pewter
yh lets just avoid complements
tidal turret
better if we avoid it yeah
is killing a fly with a bazooka
lets find a,b ∈ R
1 + b = 0
b = -1
4 + 2a = 0
2a = -4
a = -2
hard pewter
yeah coo
l
now in the last one we could have picked any extension to get W, this time we just need to ensure what we pick belongs to H2 as well
tidal turret
H1nH2
hard pewter
yh i mean H1 is obvious
tidal turret
nah nvm it is
yeah
hard pewter
yh works
tidal turret
,w rank {{1,0,1,1},{1,0,1,4},{4,2,0,3}}
grand pondBOT
tidal turret
XD
we didnt even computed anything else than to check they are linearly independent
hard pewter
yeah lmao
hard pewter
it was fun
sage helm
I would like (b) checked
midnight plankBOT
I would like (b) checked
midnight plankBOT
@sage helm Has your question been resolved?
midnight plankBOT
@sage helm Has your question been resolved?
twilit field
Prove that the intersection $K \cap H$ of subgroups of a group $G$ is a subgroup of H, and that if K is a normal subgroup of G, then $K \cap H$ is a normal subgroup of $H$.
I would like to prove the second part.
\
As $K$ is a normal subgroup of $G$, it follows for $x \in G, y \in K$, $x^{-1}yx \in K$.
\
Let $u \in H , t \in K \cap H$. As $t \in K \implies x^{-1}tx \in K$. This implies $K \cap H$ is a normal subgroup of $G$, what am I missing
midnight plankBOT
> Prove that the intersection $K \cap H$ of subgroups of a group $G$ is a subgro...
grand pondBOT
What a wonderful world !
flat spire
wai
twilit field
hi
runic hamlet
you need u^-1 t u in K cap H
twilit field
so I have to figure out how to get that
As all $u \in G$, I can replace $x$ by $u$?
grand pondBOT
What a wonderful world !
twilit field
That's it?
twilit field
and I already know $u \in H$, so $u \in G \cap K$
grand pondBOT
What a wonderful world !
twilit field
Then as the intersection of two groups is a group, by closure, $u^{-1}tu \in H \cap K$
runic hamlet
all these typos really dont help
twilit field
I don't really see any typos, sorry /srs
grand pondBOT
What a wonderful world !
runic hamlet
I really dont know if you mean the correct thing
twilit field
we've established $u \in H \cap K$. And $t \in H \cap K$. Thus, by closure, $u^{-1}tu \in H \cap K$
Is tht better?
grand pondBOT
What a wonderful world !
runic hamlet
no, u in H
twilit field
you have only shown its in K
and here we showed it's in K too
cobalt nest
you need to show that u^-1 t u in H @twilit field
grand pondBOT
What a wonderful world !
runic hamlet
why?
runic hamlet
good
twilit field
as t \in K \cap H, t \in H
We then by closure have $u^{-1}t \in H$, and the once again by closure, we have $u^{-1}tu \in H$
grand pondBOT
What a wonderful world !
runic hamlet
yes
twilit field
It this follows it is a normal subgroup of $H$
grand pondBOT
What a wonderful world !
cobalt nest
It this follows it is a normal subgroup of $H$
did u prove its a subgroup first?
twilit field
did u prove its a subgroup first?
Yea, I've proven the first part already
So I'm done then?
Thanks a lot , both of you!
.close
midnight plankBOT
midnight plankBOT
Available help channel!
inland patio
Let V be a complex inner product space. I need to show that if T is a normal operator, then there exists a normal operator U such that U^2 = T.
My attempt; if T is normal, its spectral decomposition is T = a1T1 + a2T2 + ... +akTk. Now, can I just define U = sqrt(a1) T1 + sqrt(b1)T2 +.... Is this well-defined?
midnight plankBOT
Let V be a complex inner product space. I need to show that if T is a normal ope...
inland patio
A useful fact is that TiTj=delta_ijTi, so if U is well-defined, then in fact U^2 = T.
cobalt nest
Let V be a complex inner product space. I need to show that if T is a normal ope...
U = sum sqrt(a_i) T_i
U² = sum_i,j sqrt(ai)sqrt(aj) T_i T_j
u can use that useful fact u have
u can prove that the operator is normal quite trivial
inland patio
u can prove that the operator is normal quite trivial
ok, thanks 👍
stone fractal
my sum to this taylor series is equal to -(x - 2)^-n
midnight plankBOT
my sum to this taylor series is equal to -(x - 2)^-n
stone fractal
which is apparently wrong.
i don't understand what x = 1 means it would be easier if it was something like centered at c = 1
lyric charm
i don't understand what x = 1 means it would be easier if it was something like ...
that is exactly what it means though
the center should be 1
and so your series should be written in powers of (x-1)
stone fractal
and so your series should be written in powers of (x-1)
.
lyric charm
OHHH -1 is the c in ( x-c )
c is 1 not -1
stone fractal
right mb
so the answer is supposedly - ( x-1 ) ^n but shouldn't n be negatif because theyre fractions here ill send photo
oh wait i just figured it out thx twin on god 🙏
!done
midnight plankBOT
If you are done with this channel, please mark your problem as solved by typing .close
stone fractal
.close
midnight plankBOT
midnight plankBOT
Available help channel!
twilit field
Let $ƒ: R^{+} \to C^{\cross}$ be the map $f(x)=e^{ix}$. Prove $f$ is a homomorphism and find its kernel and image.
\
We wish to prove $f(ab)=f(a)fb)$. $f(ab) = e^i(ab)$. $f(a)ƒ(b)= e^{i(a+b)}$ . What am I doing wrong
midnight plankBOT
Let $ƒ: R^{+} \to C^{\cross}$ be the map $f(x)=e^{ix}$. Prove $f$ is a homomorph...
grand pondBOT
What a wonderful world !
hard umbra
the group operation on the domain is addition of reals
hard umbra
you should probably use additive notation on the domain
twilit field
as in?
hard umbra
a + b rather than ab
twilit field
a + b rather than ab
This would map to the entire complex plane , right
twilit field
no
1 is
so a circle of radius 1
centred at origon
*origin
excluding the origin
hard umbra
the unit circle contains the origin?
twilit field
this is a punctured disc
hard umbra
can you get 0.99
lyric charm
can you get i
midnight plankBOT
@twilit field Has your question been resolved?
molten bay
limit ( x ) tends to infinity ( \left(\frac{\sin x}{x}\right)^{\frac{1}{x}} ).
midnight plankBOT
limit \( x \) tends to infinity \( \left(\frac{\sin x}{x}\right)^{\frac{1}{x}} \...
grand pondBOT
Andy
lavish venture
ln
fallen sparrow
limit \( x \) tends to infinity \( \left(\frac{\sin x}{x}\right)^{\frac{1}{x}} \...
Yeah i think l hospital is the way to go after taking ln
lavish venture
l hospital
molten bay
so here is my hint is that i will apply log both sides and get ( e^{\frac{1}{x} \log\left(\frac{\sin x}{x}\right)} ).
grand pondBOT
Andy
lavish venture
why
molten bay
what why?
fallen sparrow
so here is my hint is that i will apply log both sides and get \( e^{\frac{1}{x}...
And then l hospital yes
molten bay
because we are given 1/x expontial
lavish venture
$\ln L = \lim_{x \to \infty} \frac{\ln\left(\frac{\sin x}{x}\right)}{x}$
fallen sparrow
It's the same thing
fallen sparrow
use lhopitals here
... it's the same thing
grand pondBOT
knief
fallen sparrow
Take limit inside e
fallen sparrow
what’s the point of doing e^ first
Sigh
midnight plankBOT
@molten bay Has your question been resolved?
cursive swan
Find a condition/relation between a, b and c so that the quadratic ax^2 + bx + c has a root twice the other.
midnight plankBOT
Find a condition/relation between a, b and c so that the quadratic ax^2 + bx + c...
lyric charm
!status
midnight plankBOT
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
!nosols
midnight plankBOT
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
midnight plankBOT
Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).
cursive swan
lol
lyric charm
and also you managed to ping me and not OP. triple fail.
<@&268886789983436800> can we get the wall of text cleaned up
astral talon
and also you managed to ping me and not OP. triple fail.
||why do i feel like they did it on purpose to spite you||
lyric charm
dunno.
astral talon
honestly i dont get why people try to gpt to "help" others, if you dont know something you dont have to answer the question. even if gpt is correct, you're doing no one a favour, you're not promoting good learning
fallen sparrow
Find a condition/relation between a, b and c so that the quadratic ax^2 + bx + c...
Vieta's works just fine here
Also note that they've mentioned a sizable comparison between roots so they have to be real
exotic stratus
Vieta's works just fine here
Not the issue though. The issue is you're not supposed to provide answers and on top of use gpt to do that
lyric charm
Also note that they've mentioned a sizable comparison between roots so they have...
no, "twice" means "two times"
nobody says anything about them being real or not
cursive swan
Yes, 2 + 2i = 2(i + 1)
exotic stratus
Also note that they've mentioned a sizable comparison between roots so they have...
You can in fact multiply a complex number with 2
formal blade
You should not give people direct solutions or use llms to answer questions here.
carmine sigil
The post in question reads like a human wrote it
cursive swan
I let one root be x. The other is, therefore, 2x.
Now x + 2x = -b/a
x = -b/3a
Now, x(2x) = 2x^2 = c/a
2 (-b/3a)^2 = c/a
c/a = 2b^2/9a^2
2b^2 = 9ac
lyric charm
mmm
carmine sigil
And an online detector says human written
lyric charm
ok issue #1 is you really shouldn't be reusing the letter x.
carmine sigil
It's just a lot of text, but probably not gpt.
lyric charm
but otherwise your algebra seems fine @cursive swan
fallen sparrow
no, "twice" means "two times"
Ok true. I should've specified we can't compare two complex numbers
lyric charm
Ok true. I should've specified we can't compare two complex numbers
no comparison is being made.
fallen sparrow
no comparison is being made.
Yes i understand that now
lyric charm
nosols is a pretty hardline policy.
exotic stratus
Not preferable \neq you shouldn't do it.
And you shouldn't do it
@cursive swan you're already known to not go the known way, I recommend that this time since you've been handed the answer, you go the long route and don't use Vieta's relations and solve this problem without them.
Ofc I am in no place to tell you what you must do, but I think you should try since it'd prolly be a worthwhile investment this time around.
lyric charm
what like explicitly start with f(r) = f(2r) = 0 and do some other BS?
like simplifying f(2r) - 4f(r) or something
exotic stratus
what like explicitly start with f(r) = f(2r) = 0 and do some other BS?
Honestly I was thinking starting from ||(kx - r)(kx - 2r)|| but there are a lot of things which can be done
fallen sparrow
what like explicitly start with f(r) = f(2r) = 0 and do some other BS?
That's honestly not a bad idea if he doesn't wanna do vietas
lyric charm
OP has already shown the algebra for vieta's
fallen sparrow
Which is why I said if.
midnight plankBOT
midnight plankBOT
Available help channel!
midnight plankBOT
Bro how to occupy this??
fast meadow
Nvm I got it while raging
A number is divided by any number we get "the number", but the product of the number times any number is "the number", if the number is added with with any number, we get the "any number", if we delete this number from "any number" we get "any number".
What is " any number" and what is "the number"?
exotic stratus
!xy
midnight plankBOT
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.
exotic stratus
No I am not able to do that
Are you in middle of an exam?
fast meadow
No
exotic stratus
then?
fast meadow
Bro this is a question
I will just write it on a paper and send the image, but what is the difference??
exotic stratus
It is, but a severly malformed one, is there a concept that you don't understand that you'd like help with or is this you outlining you though process while solving a question?
fast meadow
I am sorry for giving my question in this channel cuz this channel is for help, but I am not asking help I want u to try to get the solution from my question.
woeful turret
I am sorry for giving my question in this channel cuz this channel is for help, ...
thats not how this works
exotic stratus
A number is divided by any number we get "the number", but the product of the nu...
This sounds a lot like someone who's learning algebra for the first time trying to describe it
exotic stratus
Well are you familiar with algebra?
fast meadow
Ye
exotic stratus
Ok great, can you write this algebraically?
exotic stratus
Yea... that's kinda the point...
fast meadow
Ok fine
x/y = x, x*y = x, x+y = y, y-x= y, what is x?
Bro u will get this in 10 seconds
That's why I told u to just read what I gave
exotic stratus
Ok, so if you know the answer why did you post it here?
Also that's not literally what you sent
fast meadow
I am sorry for giving my question in this channel cuz this channel is for help, ...
☝🏽
midnight plankBOT
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
exotic stratus
!noans
midnight plankBOT
The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.
midnight plankBOT
exotic stratus
Also for future reference that sentence doesn't translate to the equations you sent @fast meadow
fallen sparrow
It's already closed
midnight plankBOT
✅
fast meadow
A number is divided by any number we get "the number", but the product of the number times any number is "the number", if the number is added with with any number, we get the "any number", if we delete this number from "any number" we get "any number". Make this sentences into equation.
exotic stratus
Ok, give me a bit
exotic stratus
$$\forall v, w, x, y, z \exists k$$
$$\frac x y = k$$
$$k \cdot z = k$$
$$k + w = w$$
$$v - k = v$$
@exotic stratus
exotic stratus
Exactly
fast meadow
Bro how is this?
exotic stratus
We have algebra for a reason, common language is confusing and easy to get wrong
Even this maybe wrong
lol
fast meadow
Bro "the number" is the same number in all of my sentences
lyric charm
A number is divided by any number we get "the number", but the product of the nu...
ngl this wording feels like it was deliberately made unreadable
fast meadow
The first one is x/y = k, I thought it was k/y = k
lyric charm
with references to "the number", "any number", etc.
fast meadow
ngl this wording feels like it was deliberately made unreadable
That's the whole point!
lyric charm
are you translating this from another language
fast meadow
Nah
lyric charm
That's the whole point!
ok and the point is to torture you? or what
fast meadow
I am english
fast meadow
ok and the point is to torture you? or what
Exactly
blissful pier
do you have the original question verbatim?
fast meadow
Bro I made the thing up
lyric charm
ok masochist lol
fast meadow
I made the question myself
fast meadow
The first one is x/y = k, I thought it was k/y = k
@exotic stratus
exotic stratus
<@946379454056386560>
No it isn't, because the number wasn't stated before that
and A number and any number just mean any possible number
exotic stratus
which BTW maybe you'd surprised to know but x * y =/= y * x for some of them
exotic stratus
yes
fast meadow
Wait how???
exotic stratus
Since you like pain
blissful pier
Then perhaps a rewording would be nice, "Property 1: For any nonzero real number a, dividing the number x by a yields x itself." as an example?
exotic stratus
Enjoy my gift to you @fast meadow
blissful pier
If I am understanding ur original sentence right?
fast meadow
Enjoy my gift to you <@1125585255919788132>
It's like getting coal for Christmas
Anyways do u know what "the number " is?
exotic stratus
I know what the number would be if we were dealing with real numbers, x = 0, I don't remember at which point we get multiple zeros in the number systems but I'd say from there this becomes ambiguous
fast meadow
x/y = x, x*y = x, x+y = y, y-x= y, what is x?
This can also be the answer right?
.close
midnight plankBOT
midnight plankBOT
Available help channel!
inland patio
I struggle with this exercise. I need to specifically use the spectral decomposition of a normal operator T to prove that T is invertible if and only if no eigenvalue equals 0. Recall the spectral decomposition of a normal operator is a1T1+a2T2+...+akTk, where a1,a2,...,ak are the distinct eigenvalues and T1,T2, ..., Tk are the orthogonal projections on the eigenspaces. I'm trying to prove this in one line sort of, if possible. Since T is an operator, it suffices to show that the null space N(T) = {0} if and only if no eigenvalue equals 0. I don't see how to make use of the spectral decomposition. Any help is very appreciated. 😔
midnight plankBOT
I struggle with this exercise. I need to specifically use the spectral decomposi...
nova yoke
side note, this is almost trivial to prove directly, without the spectral decomposition
(assuming we're in a finite dimensional space)
T is not invertible
<==> T is not injective
<==> there is a nonzero v such that Tv = 0
<==> 0 is an eigenvalue and v is an eigenvector
lyric charm
yeah but op needs to use specifically the spectral decomp thing.
nova yoke
understood, that's why i said side note, just wanted to point it out
inland patio
T is not invertible
<==> T is not injective
<==> there is a nonzero v such that ...
yeah, thanks for the alternative proof Bungo. For a spectral decomposition proof, here's my attempt (I feel like this is a bit of a roundabout way of doing things, but meeh). If N(T)={0} then simply pick v to be an eigenvector in W_i, the eigenspace corresponding to \lambda_i, and then 0\neq T(v)=\lambda_i v, so \lambda_i\neq 0 (yes, I didn't use the spectral decomposition in this direction, but whatever). But now, conversely, if no eigenvalue equals 0 and we are considering T(v)=0 for v in V, then v can be decomposed as sum of vectors in the respective eigenspaces (since V is a direct sum of those), and we obtain from the spectral decomposition that T(v) is a linear combination of linearly independent vectors with coefficients being the nonzero eigenvalues. It has to be the case that v=0.
nova yoke
"we obtain from the spectral decomposition that T(v) is a linear combination of linearly independent vectors with coefficients being the nonzero eigenvalues" - can you show the details for this?
midnight plankBOT
@inland patio Has your question been resolved?
inland patio
"we obtain from the spectral decomposition that T(v) is a linear combination of ...
actually, I don't think I can, as I think I was speaking gibberish 😅
inland patio
"we obtain from the spectral decomposition that T(v) is a linear combination of ...
here's the full exercise by the way:
nova yoke
Tv = 0 is equivalent to <Tv, Tv> = 0, maybe considering this inner product would be helpful?
inland patio
Tv = 0 is equivalent to <Tv, Tv> = 0, maybe considering this inner product would...
I don't see though how T acts on v, if say v = w1 + w2 + ... + wk, where wi belongs to the eigenspace corresponding to lambda_i
nova yoke
maybe work with the decomposition of T instead of decomposing v, just leave v as a generic vector
i'm feeding my cats and trying to do this in my head at the same time, so quite possibly i'm leading you down a dead end 😆
inland patio
no worries 😉
inland patio
Tv = 0 is equivalent to <Tv, Tv> = 0, maybe considering this inner product would...
from what you wrote, I gather we want to conclude that v = 0, so somehow we want to obtain the chain of equalities 0=<Tv, Tv>= ... =||v||^2. I'm just guessing now.
we know that T is normal, so TT*=T*T
nova yoke
$$\langle \lambda_1 T_1 v + \cdots + \lambda_k T_k v, \lambda_1 T_1 v + \cdots + \lambda_k T_k v\rangle$$
grand pondBOT
Bungo
nova yoke
what if you expand this out, then you'll get $k$ terms of the form $\langle \lambda_i T_i v, \lambda_i T_i v\rangle$ as well as a bunch of cross terms
grand pondBOT
Bungo
runic hamlet
why didnt you continue from the old approach? Tv = sum lambda_i T_i v and T_i v are eigenvectors
inland patio
what if you expand this out, then you'll get $k$ terms of the form $\langle \la...
right, and the cross terms vanish since $T_iT_j=\delta_{ij}T_i$, and $\langle \lambda_i T_i v, \lambda_i T_i v\rangle=|\lambda_i|^2\langle v,T(v)\rangle$. I'm not a 100% sure what these terms add up to
grand pondBOT
psie
inland patio
why didnt you continue from the old approach? Tv = sum lambda_i T_i v and T_i v ...
I did some mistakes in my old approach and just thought it was no good 😅
inland patio
why didnt you continue from the old approach? Tv = sum lambda_i T_i v and T_i v ...
So 0=Tv = sum lambda_i T_i v and T_i v are (linearly independent) eigenvectors, so lambda_i=0 for all i?
runic hamlet
all T_iv have to be zero
well ok, depends on direction
god I hate problems like this
which overcomplicate shit
inland patio
yeah, pain in the ass
inland patio
all T_iv have to be zero
basic question, but why is it true that if sum a_iv_i=0 and a_i are all nonzero plus the vectors vi are linearly independent, then v_i=0 for all i? I'm only familiar with the statement if sum a_iv_i=0 and vi are linearly independent, then a_i=0 for all i.
runic hamlet
well its more like, the v_i are from V_i which are in direct sum
and a linear combination of vectors from a direct sum that equals zero has to be the trivial sum
so more accurately, lambda_i T_i v =0 for all i
so if lambda_i nonzero, then T_i v= 0
inland patio
ok 👍
inland patio
so if lambda_i nonzero, then T_i v= 0
dumb question again, and probably my final one, but how does one go from T_iv = 0 for all i <=> v = 0. The <= is obvious, but => I don't see.
runic hamlet
T_i v is is the eigenvector decomposition of v
woeful turret
midnight plankBOT
fallen sparrow
!status
midnight plankBOT
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
Yeah okay my previous statement was correct
Try to relate between f(n) and f(1-n)
fallen sparrow
shit bro why didnt i think of that
Pretty common in these types of questions
fallen sparrow
i had even done a similar question before
Yeah they're asked quite frequently
woeful turret
wait bro arent u giving advanced tomorrow
woeful turret
all the best bro papers gonna be ez
fallen sparrow
Idk I'm very stressed out rn im reading chemistry
woeful turret
which part of chem
woeful turret
oh we havent done P block yet..but d block less questions are asked learn K2Cr2O7 and KMnO4 properly
and environmental chem i havent seen a single question from advanced
fallen sparrow
Yes i know that thank you
fallen sparrow
and environmental chem i havent seen a single question from advanced
There's still a chance
And I don't wanna waste it
My sir also said to do so
fallen sparrow
Kk
fallen sparrow
What
woeful turret
yeah i think i went wrong somewhere
fallen sparrow
Pretty sure
You should be getting f(x) + f(1-x) = 1/root (2021)
Oh no
Why would you do that
Just take root 2021 common from the denominator in the second term
fallen sparrow
woeful turret
in the advanced question there was nothing to take common so i kind of used the same logic
fallen sparrow
Never do that
Never use logic from the question onto another questi9h
Even if it looks similar
woeful turret
yeah ur right 😔
fallen sparrow
Always do a check inthe back of your mind
Anyways
Gn i don't think I'll be on tomorrow
toxic prism
Gl w your exam 🙏
fallen sparrow
Thanks
woeful turret
Gn i don't think I'll be on tomorrow
goodnight bro, yeah dont come online, atb
got the answer as well
ty
.close
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woeful turret
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@woeful turret Has your question been resolved?
woeful turret
<@&286206848099549185>
sudden heart
<@&286206848099549185>
you are over thinking the sin(a) part, just think of sin(a) as an arbitrary number between 0 and 1 exclusive
alpha is independent of x, so there is no need to graph sin
woeful turret
you are over thinking the sin(a) part, just think of sin(a) as an arbitrary numb...
ohh ok yeah ur right but how do i get number of solutions?
sudden heart
ohh ok yeah ur right but how do i get number of solutions?
you have it correctly graphed at the bottom of that paper, how many different inputs (x values) create the same y value for any given y value between 0 and 1?
woeful turret
4
sudden heart
yes
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last slate
does lny= lnx + lnw equal y=x+w?
midnight plankBOT
does lny= lnx + lnw equal y=x+w?
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surreal moon
@last slate what?
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kind ledge
!help
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kind ledge
need help with f(0) = -5 [0.25 (0-1)] + 1
rn i did
f(0) = -5[0.25(-1)] + 1
f(0) = -5 [-0.25] + 1
idk what to do next
livid python
f(0) = -5 [-0.25] + 1
[x] denotes greatest integer function ?
kind ledge
[x] denotes greatest integer function ?
x was 0
oh
yea
the [] is greatest integer functio
livid python
from the way it's defined [-0.25] should go to -1
livid python
alr so that means f(0) = -5 [-1] + 1
just -5(-1)
livid python
alr so that means f(0) = -5 [-1] + 1
was correcting this
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blazing moat
how do I find the expression for the area? I tried different bounds on desmos for ages but I still didn't find it
midnight plankBOT
how do I find the expression for the area? I tried different bounds on desmos fo...
twilit field
I think you should try computing it by hand
$\int_{0}^{2 \pi }2- 2 \cos( 3 \theta) d \theta$
grand pondBOT
What a wonderful world !
blazing moat
im not getting the right answer
twilit field
one minute
dusty portal
how do I find the expression for the area? I tried different bounds on desmos fo...
Try to find the area of one of the petals and multiply that by 3
Then subtract the area of the circle from it
blazing moat
$\frac{1}{2}\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}\left(2\cos3x\right)^{2}dx$
this is like one petal
dusty portal
**What a wonderful world !**
Idk why but for some reason the integral from 0 to 2pi just never works out nicely for petals...
dusty portal
this is like one petal
Yeah
blazing moat
dawg no half a petal
dusty portal
?
dusty portal
Hmm you forgot the 1/2 coefficient
twilit field
Idk why but for some reason the integral from 0 to 2pi just never works out nice...
yeah, wrong formula for area
my bad
blazing moat
yea
dusty portal
Idk why but for some reason the integral from 0 to 2pi just never works out nice...
Nevermind it is just not squared
grand pondBOT
ruby
dusty portal
But when I did this problem that’s how I did it
blazing moat
do i do *3 for petal and subtract from whole circle
blazing moat
yea
dusty portal
NAHHHH there’s no way you took AB instead of BC
blazing moat
wdym?
dusty portal
The hell happened for you to get a late exam
dusty portal
The only other exam was AB
^
blazing moat
i didnt take AB
blazing moat
nah I js signed up for late exam
dusty portal
Why
dusty portal
Also the AP Psych exam shenanigans were crazy 💀
dusty portal
regular is 8 late is 12
Nah.
Bro did you really sign up for the late exam just for that 😭
blazing moat
Also the AP Psych exam shenanigans were crazy 💀
i got a 4 on ap psych last year it was so embarassing 😭
blazing moat
Bro did you really sign up for the late exam just for that 😭
yea my apush is also on 5/20
dusty portal
DAWG
blazing moat
what 😭
dusty portal
i got a 4 on ap psych last year it was so embarassing 😭
I know someone who got a 1 💀
blazing moat
dawg how do you even get a 1
blazing moat
I didn’t even know you could sign up for the late exam
you can
blazing moat
polar 😭
dusty portal
Yeah
blazing moat
i want parametrics
dusty portal
Unfortunate
blazing moat
😭
dusty portal
Parametric are ass
dusty portal
im crashing out if i get polar
It isn’t even that hard 💀
Anyways
This problem should be straightforward
I think when I mocked this I got like a 106/108
💀
blazing moat
$\frac{1}{2}\int_{0}^{2\pi}\left(2\right)^{2}dx-\frac{3}{2}\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}\left(2\cos3x\right)^{2}dx$
grand pondBOT
ruby
blazing moat
i solved it
dusty portal
Yay
blazing moat
I think when I mocked this I got like a 106/108
😭
blazing moat
yea
blazing moat
which grade are you in
dusty portal
No you aren’t
blazing moat
im graduating next year
dusty portal
I had seniors in my class lol
Bro this one freshman was gonna take the exam in my school
Bro forgot his fucking password 💀
blazing moat
is bro good at calc
blazing moat
cuz 9th is kinda
dusty portal
Nah
blazing moat
My school is sweaty
bay area?
blazing moat
ok
dusty portal
I live in Georgia
blazing moat
atlanta?
dusty portal
Near
blazing moat
i live in cali
dusty portal
💀
blazing moat
then you see a cooked cali kids 💀
blazing moat
😭
dusty portal
An 8th grader finished BC 💀
blazing moat
dawg what
dusty portal
The only reason why he isn’t taking university math is because the Dual Enrollment laws here are holding him back 💀
At minimum you have to be a 10th grader
dusty portal
It’s fine
As long as you find something interesting
And don’t screw up your college apps
blazing moat
tbh i dont know what to write for college apps
dusty portal
Lock in 😭
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molten bay
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slender walrus
where's that coming from
dusty portal
Sus
I would also just simplify the left hand side in my opinion
But there is a domain restriction here
molten bay
( \log_2(x^2-2x) \log 2 = \log(x-2) )
grand pondBOT
Andy
molten bay
Is this correct?
( \log_2(x^2 - 2x) \log 2 = \log(x - 2) )
( \log_2(2(x^2 - 2x) = \log_2(x - 2) )
grand pondBOT
Andy
molten bay
2x^2-4x=x-2
2x^2-5x+2=0
grand pondBOT
ℝαμOmeganato5
molten bay
Is this not how we take log?@slender walrus
slender walrus
no
molten bay
2^t
t log 2?
slender walrus
log(2^t) would be, t log (2)
molten bay
They are we not taking both side?
slender walrus
yes
molten bay
$$
2^{(x^2-2x)} = x - 2
$$
$$
\log(2^{(x^2-2x)}) = \log(x - 2)
$$
grand pondBOT
Andy
slender walrus
but you're taking log_2 on both sides,
you'll end up with log_2(2) where you had the red part (which is just 1)
grand pondBOT
Andy
grand pondBOT
Andy
slender walrus
just reread what you did more carefully,
the initial step was technically valid, just inefficient
missing log on the right side
molten bay
Hang
On
( 2^{\log_2(x^2-2x)} = x-2 )
taking log both sides
( \log_2(x^2-2x) \log 2 = \log_2(x-2) )
slender walrus
the issue that actually made the work invalid was what you did afterwards
grand pondBOT
Andy
slender walrus
ok, now its invalid again
slender walrus
the log(2) again
molten bay
I have already given you proper reason
slender walrus
you need to be REALLY clear that you're taking the same log base on both sides
slender walrus
because here it seems you're taking common log on the left
but log_2 on the right
slender walrus
1 sec
molten bay
Bro you were making issues without any point for 15 minutes
molten bay
**Andy**
We can remove log now
x^2-2x=x-2
And yes x^2-2x>0 should be
x^2-3x+2=0
we get 3+-1/2=2,1
slender walrus
this is what you get if you apply common or unspecified log to the left side
$$\log(2^{\log_2(x^2-2x)}) = \log_2(x^2-2x)\log(2)$$
and if you wrote
$$ \log_2(x^2-2x)\log(2) = \log(x-2)$$
that would be fine, but you wrote
$$ \log_2(x^2-2x)\log(2) = \log_2(x-2)$$
(where log was applied on the left and $\log_2$ on the right
\
applying $\log_2$ to both sides will get you
$$\log_2(x^2-2x)\underbrace{\log_2(2)}_{1} = \log_2(x-2)$$
grand pondBOT
ℝαμOmeganato5
molten bay
And we can remove log now
slender walrus
or more simply: $\log_a(a^k) = k$ \
and similarly $a^{\log_a(k)} = k$ \
(where defined) and you could avoid a lot of this
grand pondBOT
ℝαμOmeganato5
slender walrus
Bro you were making issues without any point for 15 minutes
wdym
molten bay
You wrote the same as me and you said my term was invalid why?
slender walrus
i didn't write the same as you
slender walrus
that's what i'm saying was wrong and had iissues
$\log 2$ isn't 1. $\log_2(2)$ is 1.
grand pondBOT
ℝαμOmeganato5
slender walrus
your issue is with the omission of your log bases and how they kept appearing and disappearing
molten bay
that's what i'm saying was wrong and had iissues
We are taking log_2() not log_e both sides
slender walrus
i don't know what you wrote on your actual sheet, but i'm taking whatever gets sent here at face value
molten bay
x^2-3x+2=0
...
slender walrus
We are taking log_2() not log_e both sides
yes, that's what you intended but you just writing log 2 instead of log_2(2) indicated otherwise
from the thing i made earlier, you wrote the top one where you should be writing the bottom one, (or equivalent)
my point is that if you intended to apply $\log_{\red{2}}$ on both sides and wanted to include that braced component
$$\log_2(x^2-2x)\underbrace{\log_{\red{2}}(2)}_{1} = \log_2(x-2)$$
the red $\red{2}$ in the base isn't optional
grand pondBOT
ℝαμOmeganato5
slender walrus
having fixed that, it will lead to x=1 or 2
and you'll need to check if those satisfy the condition x^2-2x>0
as for one of the earlier issues, you did something like
log(p) * log(q) = log(pq)
which is invalid
midnight plankBOT
@molten bay Has your question been resolved?
midnight plankBOT
@molten bay Has your question been resolved?
molten bay
( 2^{\log_2(x^2-2x)} = x-2 )
Taking log
( \log_2(x^2-2x) \log_2(2) = \log(x-2) ) ab dono side se antilog le lo lekin condition dyan rakhna h
( x^2-2x = x-2 )
( x^2-3x+2 = 0 )
( x = 1, 2 )
( x^2-2x > 0 ) and ( x-2 > 0 )
( x(x-2) > 0 ) it will gives ( (-\infty, 0) \cup (2, \infty) )
( x > 2 )
So we get ( (2, \infty) ) so we get 0 solution@slender walrus
grand pondBOT
Andy
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
slender walrus
overall there will be no real solutions
but you are again having the same notation issues as before
now you're omitting the 2 in the base of the log on the right side
molten bay
Right side log_2(x-2)
Yes
Yes i know the understanding of log_e amd log_2 as you can see
It is just typo and mostly we forget to mention properly
That is why it is making issue.
.close
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neat silo
Solve for x in the interval [0, 2pi)
sin x - cos x - tan x < 1
midnight plankBOT
Solve for x in the interval [0, 2pi)
sin x - cos x - tan x < 1
lyric charm
exotic stratus
Solve for x in the interval [0, 2pi)
sin x - cos x - tan x < 1
!status
midnight plankBOT
Solve for x in the interval [0, 2pi)
sin x - cos x - tan x < 1
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
so like i split the tan into sin and cos
sin x - cos x - sin x/cos x < 1
then i dont have an idea
lyric charm
i'd imagine subtracting 1 from both sides and making the entire thing into one fraction can't hurt
grim vector
And studying an associated function
grim vector
Thats how i would do it
neat silo
obviously cos x could not be 0
grim vector
?
neat silo
so we exclude the values that make cos x 0
grim vector
It just exclude -pi/2 and pi/2
grim vector
?
neat silo
cos 3pi/2 = 0
neat silo
so i take the derivative?
grim vector
On [0,pi/2) U (pi/2, 3pi/2) U (3pi/2, 2pi)
neat silo
f(x) = sin x - cos x - tan x - 1
then the derivative would be
f'(x) = cos x + sin x - sec^2 x
neat silo
So you know limits and continuity
yes
neat silo
Study the sign of f(x)
oo so do i have to check the regions?
grim vector
Put it under the same denominator
neat silo
Put it under the same denominator
alright so its
(sin x cos x - cos^2 x - sin x)/cos x - 1
exotic stratus
(sin x cos x - cos^2 x - sin x - cos x)/cos x < 0
Take the denominator to right, remember to split the inequality based on sign of cos(x)
neat silo
assuming cos x is not 0
so we can divide by cos x
sin x cos x - cos^2x - sin x - cos x < 0
sin x cos x - sin x < cos^2 x + cos x
sin x (cos x - 1) < cos x (cos x + 1)
?
exotic stratus
I mean yea but I don't think you split anything
neat silo
how can i do so
exotic stratus
Well see the intervals for which the sign of cos(x) is negative
and there the sign will reverse
neat silo
oh my god its so complicated
neat silo
Well see the intervals for which the sign of cos(x) is negative
so by splitting the quadrants
cos x is only negative in the 2nd and 3rd quadrant
so do i have to do casework?
exotic stratus
You do, yes
neat silo
so its like
Case 1 for [0, pi/2) and (3pi/2, 2pi)
Case 2 for (pi/2, pi) dan (pi, 3pi/2)
neat silo
Solve for x in the interval [0, 2pi)
sin x - cos x - tan x < 1
<@&286206848099549185> can someone help me
alright i got 2 of the solutions
[0, pi/2), (pi, 3pi/2)
i am having trouble getting last one
midnight plankBOT
@neat silo Has your question been resolved?
barren loom
it's expressible as 2 arctan(real root of 2u^3-u^2+1)
barren loom
You can also think of it as solving the zero set of s c - c^2 - s - c, s^2+c^2-1, two conics giving (c+1)(2c^3-2c^2+3c-1)=0 and take arccos of (the real root of 2c^3-2c^2+3c-1)
midnight plankBOT
@neat silo Has your question been resolved?
neat silo
it's expressible as 2 arctan(real root of 2u^3-u^2+1)
how did u obtain 2u^3 - u^2 + 1
barren loom
The other answer is easier to follow
neat silo
what do you mean by solving the zero set
i know its related to some sort of poly bash but where can i start
barren loom
Think of s c - c^2 - s - c=0 as a conic
It intersects the unit circle in two real points
(One for c=-1)
neat silo
Think of s c - c^2 - s - c=0 as a conic
s and c are random variables right?
barren loom
They are sin(x) and cos(x) also secretly
barren loom
Since we intersect with the unit circle
neat silo
sin x cos x - cos^2 x - sin x - cos x = 0 as a conic that intersects the unit circle
barren loom
That’s the solution step
Letting one equation into the other
sin(x)^2+cos(x)^2=1 is thought of as parametrization
neat silo
do i plot it like this using desmos?
barren loom
No
neat silo
oops
barren loom
x=cos(theta), y=sin(theta), x y-x^2-x-y=0 and x^2+y^2=1
neat silo
oops yes
barren loom
I prefer keeping the theta x and y just so because of this
Theta is something angle
x,y algebraic
neat silo
alright
barren loom
Does it make sense now?
barren loom
That’s just since it happens to be a quadratic in sin and cos, allows this trick
neat silo
ah
barren loom
neat silo
accidentaly deleted it oof
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neat silo
.close
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last slate
midnight plankBOT
last slate
how did we get 1/2(-3 pi/4)
fallen sparrow
Arg works a lot like logarithm
subtle blaze
Use another form of the point
fallen sparrow
Cuz of the e^itheta form
subtle blaze
Like re^{i(θ + 2pi)}
And then do the calculation again you should get 2 different solutions
Square root should give 2 solutions
dusty portal
how did we get 1/2(-3 pi/4)
$\arg(z^n)=n\arg(z)$
grand pondBOT
;(
subtle blaze
Unless you’re at the branch point which is 0
dusty portal
And also we have multiples of 2pik for all the angles
subtle blaze
What
last slate
oh go it
subtle blaze
0 is the only number that has only 1 square root
last slate
got it
last slate
but why does the arg of u squared equal arg of u
dusty portal
?
last slate
wait
subtle blaze
but why does the arg of u squared equal arg of u
Look at this picture
last slate
nvm got it
subtle blaze
dusty portal
cuz arg is tan
subtle blaze
Both the blue and green angles when you double will reach z²
So then there’s 2 solutions
subtle blaze
Also you’d be looking to use atan2 not arctan
arctan has the issue that it suggests pi-periodicity but it’s actually 2pi-periodicity
subtle blaze
Yeah
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keen harbor
In linear algebra, a coordinate vector $[\vec{v}]_B$ is a tuple of numbers, where those numbers are the coefficients of a linear combination with some basis B. The result of that linear combination is some vector $\vec{v}$, whose components may or maynot be the same as those in coordinate vector $[\vec{v}]_B$.
Regarding the resulting vector $\vec{v}$:
- Does that vector continue to have a relation to the coordinate system of basis
B? Or is that relationship severed, and now this vector is just part of the original vector space and nothing more? Which would mean that this vector has no basis?
midnight plankBOT
In linear algebra, a coordinate vector $[\vec{v}]_B$ is a tuple of numbers, wher...
grand pondBOT
Vulkanoid
runic hamlet
presumably you think that v itself lives in R^n ?
keen harbor
yeah
runic hamlet
there are lots of other vector spaces v could be from
keen harbor
I'm not sure, actually. I guess it's part of the vector space, and I would think it's also part of R^n.
so part of both.
runic hamlet
well for example v could be some polynomial, eg v=1+x-18x^2
and if your basis if B=(1,x,x^2) then [v]_B might be (1,1,-18)
of course given B and [v]_B you can compute v
and given B and v you can compute [v]_B
so in that sense v and [v]_B are of course connected
but I dont think thats what you mean?
keen harbor
I'm not following what you're saying 100%.
I'm trying to think of what is the mathematical object that is the vector v (the result of the linear combination). So, things like "what relations to other math structures does it have", etc. Or, I'm wondering if it's just now a tuple of numbers with no further structure.
For example, [v]B is a tuple of numbers which, further, has a relationship to a coordinate system with basis B.
runic hamlet
well v is an element of a vector space
how v looks like will depend on how that vector space looks like
but the structure is that vector space
again, v could be a polynomial
polynomial form a vector space
(you can add polynomials and scale them by a number)
keen harbor
Now, I'm not sure if I'm explaining myself correctly.
Maybe this:
Does the resulting vector v has some type of relationship to the coordinate system with basis B?
runic hamlet
and given B and v you can compute [v]_B
except for this relationship, no
keen harbor
So, basically, once [v]B is evaluated and results in vector v, then v loses the connection from the coordinate system from where it came. Is correct?
Like, you can't get back to the coordinate system without knowing the basis.
But, the vector v itself does not carry that type of info embedded within itself.
runic hamlet
a coordinate space is useless without the basis, yes
keen harbor
And thus, at that point, it's just a tuple of numbers.
runic hamlet
you have to know what your coordinates mean
the point I am still trying to bring across is that not all vectors are just tuples of numbers
v does not have to be like that
keen harbor
I understand, but I'm trying to focus one a type, so as to make things as simple as possible for myself. I'm already finding this confusing as it.
The main confusion, for me, is that vector definition is somewhat recursive.
You have a vector, that's the precursor to some "true" vector.
So, I'm trying to see if this recursiveness stops at some point.
runic hamlet
[v]_B is not some "true" vector