I need help in quadratic equations
#help-41
amber waspBOT
@opal aurora Has your question been resolved?
opal aurora
Dawg what is my luck
amber waspBOT
@opal aurora Has your question been resolved?
inner zenith
@opal aurora what do u need help with?
amber waspBOT
@opal aurora Has your question been resolved?
loud lava
fierce hearth
Tried to find some info on this but without success so figured i'd ask here
fierce hearth
So eulers formula
what would it look like if we had cos^2 (x) ?
is that a thing?
is it just to square the expression in the RHS?
and wouldn't that turn into a real mess?
shadow stump
not hugely, it can be simplified a lot using exponent rules
fierce hearth
i mean wouldn't it be (e^ix+e^-ix)^2 / 4?
and then you get like (e^2ix + 2e^ix*e^-ix \ * e^-2ix) / 4
shadow stump
and what is e^(ix) e^(-ix)?
fierce hearth
yeah i guess 1
e^0
so then it's only really (e^2ix + 2 + e^-2ix) / 4
and what if I want to go back at a later point after doing various calculations (to cos / sin expressions) ? do I have to see that trend or is there an easier way?
shadow stump
you can get that back in terms of cos
fierce hearth
but I would have to see the pattern?
hmmm ok think I'm with you then, thanks!
have a nice day/evening :D
.close
burnt pumice
prove det(A) = 0 ⇐⇒ ∄A−1 with only the fact that det(A) = A1,1∀A ∈ R1×1 and that [det(EA) = det(E) det(A), det(AE) = det(A) det(E), and det(E) = det(E^T)] as well as simple matrix rules
burnt pumice
as in this:
and we also have typical matrix rules (not determinant rules) at our disposal
we can not use the fact that det(AB) = det(A)det(B) or that det(A)det(B) = det(AB)
for matrices A,B ∈ R
amber waspBOT
@burnt pumice Has your question been resolved?
amber waspBOT
@burnt pumice Has your question been resolved?
amber waspBOT
@burnt pumice Has your question been resolved?
amber waspBOT
@burnt pumice Has your question been resolved?
amber waspBOT
@burnt pumice Has your question been resolved?
amber waspBOT
amber waspBOT
Available help channel!
fierce hearth
amber waspBOT
fierce hearth
If I want to integrate this, I get that there's multiple ways to do it, but let's assume that I'm going for the direction of
sin(2u) = 2cos(u)sin(u)
right?
ancient jolt
yes
ancient jolt
yes
fierce hearth
okay so with that logic we have (1/2) * sin(2*theta)
and then we can take out the constant 1/2?'
so we end up with "just" integrating sin(2*theta)?
ancient jolt
(-cos 2theta)/2
fierce hearth
wouldn't that be -cos(2*theta)/2
ancient jolt
yuup
fierce hearth
and then /4
ancient jolt
yes
fierce hearth
which is the same as -(cos^2(theta)-sin^2(theta)/4?
ancient jolt
yes
ancient jolt
keep typing
fierce hearth
like in the video the professor says that we can think of it like sin^1(theta) and that cos is the inner derivative and then you get 1/2 sin^2(theta)
but another way to think of it is that cos(theta)sin(theta) is half of sin(2theta)
but i don't really end up with the same expression
like yes plotting it it probably will be the same, but I'm more trying to understand how I'd algebraically could get there
ancient jolt
(-1+2sin^2(theta))/4 = -1/4 + 1/2(sin^2(theta))
fierce hearth
yes but then aren't we -1/4 off?
fierce hearth
ah
yeah nevermind
seems like it doesn't matter
since we get - (-) when we put in the boundaries
aight, thank you!
have a nice day :D
.close
amber waspBOT
ancient jolt
i didn't help you btw
fierce hearth
i didn't help you btw
having someone reaffirming me that i'm doing the right thing does help XD
pretty much sums up 90% of my threads here ^^
dull pike
Plot twist, he lied 
ancient jolt
okay
amber waspBOT
Available help channel!
gritty holly
amber waspBOT
gritty holly
ive got this sin graph right
when i sketched i was gonna label the min and maximums right
so i tried doing pi/4 + 1/4 cycle (pi/4)
but thats wrong
i search it on desmos and the next 1/4 cycle is 3pi/4
why cant i start at the first intercept and plus a 1/4 cycle to get to the maximum
omg im silly
1/4 x period is half cycle
.close
amber waspBOT
amber waspBOT
Available help channel!
fallow kernel
Hello everyone. Can someone explain to me how the first term (1 + 1/x-1/2)^((x-1)/2) evaluates to e^2 here? I know why the 2nd term evaluates to 1, but this first term is really confusing for me to understand. Any help would be appreciated, thank you!
craggy sundial
Hello everyone. Can someone explain to me how the first term `(1 + 1/x-1/2)^((x-...
simplify the fraction, then do a substitution so that the inside looks like 1+1/t
fallow kernel
But why does that then evaluate to e^2? I haven't done much math involving Euler's number
craggy sundial
But why does that then evaluate to `e^2`? I haven't done much math involving Eul...
the definition of e is $\lim_{x\to\infty} (1 + \frac1x)^x$
grizzled pagodaBOT
Sepdron
fallow kernel
Okay, that makes more sense, but why is it e^2 and not just e?
craggy sundial
Hello everyone. Can someone explain to me how the first term `(1 + 1/x-1/2)^((x-...
is that (1/x-1)/2 or 1/((x-1)/2)?
fallow kernel
The latter
craggy sundial
Okay, that makes more sense, but why is it `e^2` and not just `e`?
yea I don't understand either, I got e
fallow kernel
What's the difference between the former and the latter?
Don't both evaluate to 1/(x-1) * 1/2?
craggy sundial
the first one is 1/(2x-2) and the second is 2/(x-1)
olive venture
yes
fallow kernel
the first one is 1/(2x-2) and the second is 2/(x-1)
Then it's 2/(x-1) definitely
craggy sundial
,w lim as x -> infinity of (1+2/(x-1))^((x-1)/2)
grizzled pagodaBOT
grizzled pagodaBOT
craggy sundial
nope, still e
fallow kernel
Hm, maybe my math professor just made a mistake then
proper nymph
is someone available=
amber waspBOT
is someone available=
proper nymph
pls
vast spade
?what is the question?
proper nymph
pliant parcel
Do you need to calculate y?
ancient jolt
uh
pliant parcel
(4*5)/2
split sail
20/2 = 10
split sail
10
proper nymph
how
split sail
there is no variable
proper nymph
whats the steps
split sail
4 and 2 have common factors therefore try to simplify them so 4 x 5/2 = (4/2) * 5 = 2 * 5
proper nymph
do you know the systems guys?
proper nymph
so if there's a fraction you multiply the number by the upper number then divide it to the denominator
yes?
vast spade
so if there's a fraction you multiply the number by the upper number then divide...
wdym
proper nymph
this
split sail
yes
proper nymph
you multiply the upper number and divide it by the down one
split sail
so if there's a fraction you multiply the number by the upper number then divide...
(a/b) * (c/d) = (ac)/(bd)
proper nymph
(a/b) * (c/d) = (ac)/(bd)
idk whats supposed to be a and b
proper nymph
you multiply the upper number and divide it by the down one
this is right?
vast spade
(a/b) * (c/d) = (ac)/(bd)
assuming that b and d are not zero
split sail
assuming that b and d are not zero
yesah
proper nymph
wait can i go to eat ?
vast spade
wait can i go to eat ?
close the channel then go and have your food
proper nymph
then i come back?
vast spade
yes
tulip tapir
.close
amber waspBOT
amber waspBOT
Available help channel!
proper nymph
guys
amber waspBOT
true jackal
What doubt do you have?
amber waspBOT
Available help channel!
proper nymph
calm flame
add and sub both the equations
add one will be like 3/2x + 3/2y = 0 so x+y=0 then x=-y
substract one will be like y/2-x/2=2 so y-x=4
proper nymph
can these both work?
calm flame
solve them and u get something like x=y=-2
calm flame
can these both work?
hmm should work since u just shifted some terms
amber waspBOT
@proper nymph Has your question been resolved?
keen pawn
amber waspBOT
keen pawn
$0_v \in U+U$ as $0_v+0_v = 0_v$.
\
let $u,v \in U \implies u+v \in U$ as $U$ is a vector space. Thus $U$ is closed under addition.
\
let $t \in U+U$. But $U+U=U$ so $U$ is a subspace too
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
indigo cloud
what are you even trying to show
your second line uses that U is a vector space to show that it is closed under addition
keen pawn
I'm trying to show $U+U=U$ and thus $U$ ia subspace of $V$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
$U+U = { v+w \mid v,w \in U}$
\
but $v+w \in U$ as $U$ is a vector space , so $U+U = U$
keen pawn
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
indigo cloud
ok good now with the different letters
keen pawn
$U+U = { v+w \mid v,w \in U}$
\
but $v+w \in U$ as $U$ is a vector space , so $U+U = U$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
ok good now with the different letters
Is this fine?
indigo cloud
good enough
amber waspBOT
amber waspBOT
Available help channel!
keen pawn
amber waspBOT
keen pawn
Now the vector space containing $0_v$ is an example
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
We now employ the fact that if a vector space has the field $\R$ it either has 1 element $0_v$ or a non-finite amount
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
fluid wren
**Veni, vidi, perii is $\\R \- \\Q$**
Yeah, would be better to call it the zero subspace since the problem is about an operation on subspaces
fluid wren
**Veni, vidi, perii is $\\R \- \\Q$**
Bad wording, but yeah
Why is that relevant though?
keen pawn
My bad, thought that the question asked about additive inverse in both parts
Which subspaces have additive inverses
hmm
Let me try something
fluid wren
Hint: When adding subspaces, their dimension (or cardinality if you haven't covered dimensions yet) shouldn't go down
keen pawn
If such an additive inverse exists $U+W= {0}$ By definition $U+V = {x+y \mid x+y =0 , x \in U , y \in W}$. From this we can conclude that $x=-y$.When both subspaces are the zero subspace, this is trivially satisfied.
\
\
We now use the following lemma :- If a subspace has an element other that $0$, it has infinitely any elements, including 0, which follows trivially from the vector space axioms.
\
From the definition of vector space addition, every possible element of $U$ is added to every possible element of $W$ piecewise. Thus $2v \in U+W$, where $v \neq 0$ and neither $U$ nor $W$ are the zero subspaces.
\
\
Thus the only subspace that has an additive inverse when added to another subspace. Is the zero subspace
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
fluid wren
Looks good
Or now that I think about it you could just use the fact that V is a always subspace of V+W
amber waspBOT
amber waspBOT
Available help channel!
keen pawn
amber waspBOT
keen pawn
I have some ideas
limpid sun
it is not true ... use U=V the whole vector space, and V_1 and V_2 whatever spaces
fluid wren
!nosols
amber waspBOT
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.
keen pawn
so I've just proven that additive inverses only exist under certain conditions
pseudo crescent
so I've just proven that additive inverses only exist under certain conditions
what?
keen pawn
so unless $U = {0} ; \not\exists U st V_1=V_2$
pseudo crescent
um, U, V1 and V2 arent vectors
keen pawn
I know
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
pseudo crescent
i dont even get how additive inverses are related
pseudo crescent
**Veni, vidi, perii is $\\R \- \\Q$**
and where did you get this from
keen pawn
fluid wren
i dont even get how additive inverses are related
They are talking about the addition of subspaces
pseudo crescent
okay, i see
keen pawn
Is that enough?
pseudo crescent
you should ideally provide a counterexample if you're disproving it
keen pawn
Why ?
fluid wren
If you've shown that cancellation (in a monoid) implies existence of inverses, yeah, this is sufficient
keen pawn
if $V_1+U = V_2+U$ \implies $V_1=V_2$. Then $\exists W \mid W+U= {0}$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
pseudo crescent
**Veni, vidi, perii is $\\R \- \\Q$**
Compile Error! Click the <:errors:50426655...
this doesnt convince me tbh
i get your idea, but I had to add my own thoughts into it to truly see why it works. I wouldn't be satisfied just by reading your proof.
counterexample is the simplest way to disprove it
- the question even asks you for one...
keen pawn
The thing is though, a counter examples just says this isn't always true. I've specified when exactly it's true, and when it's not
pseudo crescent
That's true but you have to actually write a proof for that
and it's unnecessary complication
pseudo crescent
**Veni, vidi, perii is $\\R \- \\Q$**
Compile Error! Click the <:errors:50426655...
- this isnt the best way to phrase what you've found
keen pawn
What would that be?
pseudo crescent
(forall V1, V2 (V1 + U = V2 + U -> V1 = V2)) iff (U = {0})
pseudo crescent
oh wait, it actually seems to be false
keen pawn
huh?
pseudo crescent
(forall V1, V2 (V1 + U = V2 + U -> V1 = V2)) iff (U = {0})
now it should be fine
the forall is important
pseudo crescent
huh?
if you had V1 = V2, the LHS would be true vacuously
while you could make the RHS false
keen pawn
if you had V1 = V2, the LHS would be true vacuously
So counter example is the way to go
uh
can't think of any counter example
pseudo crescent
So counter example is the way to go
use what you found to construct any counterexample
it can be almost anything
keen pawn
Hmm
amber waspBOT
@keen pawn Has your question been resolved?
keen pawn
use what you found to construct any counterexample
Let $U=\R^4, V_1 = {(x,0,0,0)|{; V_2 = {(0,x,0,0}$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
pseudo crescent
e.g. 1
amber waspBOT
amber waspBOT
Available help channel!
elder jay
Am I allowed to ask for help regarding uncertaintys (such as ±0.01) and stuff
elder jay
Ping me if responding
split kraken
Am I allowed to ask for help regarding uncertaintys (such as ±0.01) and stuff
sure, but you should just ask instead of asking to ask 
mint nacelle
sure, but you should just ask instead of asking to ask <:Sobbingcrying:118493825...
!da2a
amber waspBOT
sure, but you should just ask instead of asking to ask <:Sobbingcrying:118493825...
No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/
mint nacelle
useful command
split kraken
useful command
have you seen the discussion in #『meta-discussion』 and #1207832908438568990 lately?
I'm not a fan of the factoids
elder jay
Yea so I'm currently doing a physics lab ok where I vary the radius of a stopper which is moving in horizontal circular motion. I had to calculate the time it takes for one to occur. The way I did this is I had a video and a stopwatch inside the frame of the video and I would have the initial time of a period and then a final time of the period And subtract those. My question is how can I calculate the uncertainty for the frames per second or the video. It's because the video won't catch every single frame and they may miss when the period starts or ends so I'm a little confizzled
mint nacelle
Yea so I'm currently doing a physics lab ok where I vary the radius of a stopper...
great question, so how did you record the video?
elder jay
Use my phone, shot at 60fps
mint nacelle
Use my phone, shot at 60fps
ah great, so 60 frames per second
so you should assume that the uncertainty is 1 frame for both the start and the end
that's 2 frames in total, so 2/60 = 1/30 seconds
mint nacelle
So would I write 60 ± 1/30 fps?
no, the unit is seconds
$\frac{\text{2 frames}}{\text{60 frames/sec}} = \frac{1}{30} \text{sec}$
grizzled pagodaBOT
south's secret twin brother
mint nacelle
so I did 1/(1/sec)) = sec
elder jay
So it would be 60s ± 1/30s
mint nacelle
So it would be 60s ± 1/30s
yep!
if 60s is what you got ofc
wait that seems awfully high for the period
elder jay
No no don't worry
mint nacelle
oh wait it's like 10 periods or something right
elder jay
I got like 0.19s to about 0.34 seconds
mint nacelle
that makes sense
elder jay
I got like 0.19s to about 0.34 seconds
So it's just this ± 1/30s
mint nacelle
So it's just this ± 1/30s
yep
so another point, cause your values seem to be to 2 decimal places always
then you would round 1/30 to 2 dp as well
,calc 1/30
grizzled pagodaBOT
Result:
0.033333333333333
so 0.03
elder jay
Ah
mint nacelle
yeah and your uncertainty is also to 1 sf now so that's great
amber waspBOT
Available help channel!
tribal vapor
This is my problem:
When I solve it I get this as my answer, the issue I have is when I cross verify with desmos it seems incorrect.
what did i do wrong?
tribal vapor
Also, my work for the problem:
https://media.discordapp.net/attachments/978738131308261477/1295503558501863484/IMG_8914.jpg?ex=670ee335&is=670d91b5&hm=fc83bc240148ec48a93a369555a45df65520c1c6e53909a0e6f72bd9a45e0849&=&format=webp&width=868&height=1158
nevermind i inputted the wrong question in desmos
i put 3x^2 when i should of put 3x^3
waxen verge
If Im trying to find the number of unique combinations with limits, how do i got abt finding it?
how many different and unqiue combinations could I make by pulling out 10 balls from a bag containing
10 red balls
4 blue balls
10 green balls
10 yellow balls
2 pink balls
2 turquise balls
1 sapphire ball
6 viollet balls
as theres different number of balls, i realised i couldnt use plain nCr. I do have the number which i got from coding but I want to mathmatically check to make sure im correct
waxen verge
did your answer end in 15?
yes
yes
i misread as ur answer was 15
I apologise
it was ||7315||
coral wigeon
yep that's what i got
waxen verge
how did you go about it?
waxen verge
ah
i tried doing it by hand
so I did this
and then found coeffeicent of x^10
which was 2x^10/(1-x)^7
coral wigeon
ic. yea that's basically what's going on here
waxen verge
how do you go from 2x^10/(1-x)^7 to 7315?
and so what exactly is going on?
based on my research, im assuming it has to do with generating functions? for context im in final year of HS
coral wigeon
yes this is generating functions
coral wigeon
which was 2x^10/(1-x)^7
something about this is sus to me
coral wigeon
which was 2x^10/(1-x)^7
well wait wdym by this. what is 2x^10/(1-x)^7?
waxen verge
so I did this
i apolgoise, so I used chatgpt to try teach me how to solve it and it told me to this. so i then proceeded to expand it
coral wigeon
waxen verge
😭 no one else to ask
waxen verge
i apolgoise, so I used chatgpt to try teach me how to solve it and it told me to...
and so when i expanded it, i got 2 as the coeffeincet of x^10
however, the entire thing would be divided by (1-x)^7
waxen verge
ohh
coral wigeon
what is this problem from btw
waxen verge
its something i came up with myself
trying to calculate possibilitie of a game i happen to play
coral wigeon
oh cool
waxen verge
i just used balls as its easier for others to understand
coral wigeon
let's look at a simpler version: pulling 7 balls total from 4 red balls and 6 green balls
for the red balls our options are 0, 1, 2, 3, 4 balls
and similarly for green balls we can take 0, 1, 2, 3, 4, 5, or 6
hollow condor
trying to calculate possibilitie of a game i happen to play
What game
coral wigeon
now kind of ignore the fact that we need to pull exactly 7 for now
if we choose r red balls and g green balls, then we pull r + g balls in total
waxen verge
yes, i am following
coral wigeon
now consider the expression (1 + x + x^2 + x^3 + x^4)(1 + x + x^2 + x^3 + x^4 + x^5 + x^6)
coral wigeon
let's look at a simpler version: pulling 7 balls total from 4 red balls and 6 gr...
i just pulled it out of my hat but we'll connect it back to this
waxen verge
now consider the expression (1 + x + x^2 + x^3 + x^4)(1 + x + x^2 + x^3 + x^4 + ...
im assuming the first one is from red and the seond is green
coral wigeon
yep
now what if you expanded it? we're not actually gonna do it but just imagine first distributing the 1 from the first polynomial into (1 + x + x^2 + x^3 + x^4 + x^5 + x^6)
we would have 1*1 = x^0 which corresponds to the case where we take 0 red balls and 0 green balls
then 1*x = x which corresponds to the case where we take 0 red balls and 1 green ball
and 1*x^2 = x^2 which corresponds to the case where we take 0 red balls and 2 green balls
and so on
waxen verge
yes, i am still following
coral wigeon
when we're done with that we'll have 1 + x + x^2 + x^3 + x^4 + x^5 + x^6. which means like... when we choose 0 red balls, we have 1 way to select 0 balls total, 1 way to select 2 balls total, and so on
waxen verge
yes, make sense
coral wigeon
now when we distribute x to (1 + x + x^2 + x^3 + x^4 + x^5 + x^6) that will correspond to the cases where we choose 1 red ball
similar for x^2, x^3, and x^4
waxen verge
just to clarify, when you say distribute, do you mean multiply?
coral wigeon
yea
waxen verge
okay, I understand
coral wigeon
so then to find the ways to choose exactly 7, we just need to look at the coefficient of x^7
waxen verge
got this aswell
waxen verge
so then to find the ways to choose exactly 7, we just need to look at the coeffi...
okay that makes alot of sense
coral wigeon
like if you imagine what happens when we multiply this out, we get an x^7 term when we choose a summand from the first polynomial and a summand from the second polynomial whose exponents add to 7
waxen verge
what is a summand?
waxen verge
mhmm
coral wigeon
in the process of multiplying them together, we go through each way of choosing one summand from each polynomial and multiplying them together
waxen verge
ah, so like you go thru each step by step to multiply
coral wigeon
we multiply 1 and 1, then 1 and x, then 1 and x^2 and so on
then x and 1, then x and x, then x and x^2
and so on
coral wigeon
so in that case, 5*7 multiplications
coral wigeon
and similarly for green balls we can take 0, 1, 2, 3, 4, 5, or 6
just like 5*7 ways here to choose an amount of red balls and an amount of green balls
multiplying x^r and x^g is like choosing r red balls and g green balls
because x^r * x^g = x^(r+g)
waxen verge
ohhhh
coral wigeon
and choosing r red balls and g green balls makes r+g balls
waxen verge
yes, that is logical and understandable
coral wigeon
gr8
coral wigeon
so here we have one generating function for each color
just like the 2 ball example we just went through
coral wigeon
so I did this
you can rewrite them like this but i can't really recommend doing this by hand
waxen verge
could I quickly skip ahead and ask something before i forget?
waxen verge
If Im trying to find the number of unique combinations with limits, how do i got...
then for this would I do
(1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^10)^3 (1+x+x^2+x^3+x^4+x^5+x^6) (1+x+x^2+x^3+x^4)(1+x+x^2)^2 (1+x)
?
waxen verge
also i will have to leave soon
yup, thank you for the explaination, I believe I understand now
coral wigeon
then for this would I do
(1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^10)^3 (1+x+x^2+...
yes exactly
,w expand (1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^10)^3 (1+x+x^2+x^3+x^4+x^5+x^6) (1+x+x^2+x^3+x^4)(1+x+x^2)^2 (1+x)
grizzled pagodaBOT
waxen verge
YAYYY
waxen verge
i gtg but i can answer more later if you have questions
im done for now, thank you so much layla
this is stupendos and spectaculor
someone can close this now, Id say
.close
amber waspBOT
coral wigeon
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stone bloom
How do I get rid of X in this equation?
cos(Φ) = 1 - cos(X-α) - cos(X-β) + cos(α-β)
pallid canopy
Did you try difference identity for cosine on all of the terms with X
stone bloom
I am trying to show that Φ = θ/2
stone bloom
Did you try difference identity for cosine on all of the terms with X
that's how I simplified it lol
was I not supposed to simplify it?
amber waspBOT
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signal furnace
spam sum to product
stone bloom
~~spam sum to product~~
ok I tried that and got this
amber waspBOT
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stone bloom
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stone bloom
How do I show that angle α and β add up to 90 degrees in this image?
weak zinc
Maybe try and "fill angles in" in the picture you have: if I call that yellow unlabelled point on the circumference of the circle C, then maybe give some label to CBO, and then from there, fill in other angles in terms of those variables only?
stone bloom
Maybe try and "fill angles in" in the picture you have: if I call that yellow un...
I tried but I don't think it got me anywhere
my main goal was actually trying to show that Φ = θ/2
and it turns out that this is true if α + β = π/2
so that's what I am trying to find
weak zinc
Seems like you have a few more angles than you really need
For example, if you know angle AOB, you should know what angle ACB (C as per how I mentioned it before) is immediately?
stone bloom
For example, if you know angle AOB, you should know what angle ACB (C as per how...
how so?
weak zinc
There's a theorem that tells you about that 
stone bloom
There's a theorem that tells you about that <:catGiggle:668948717772079128>
inscribed angle is half of the central angle, right?
that's actually the theorem I am trying to derive, and it turns out I can derive it if I can show α + β = π/2
so I need to show this somehow without using the theorem I guess, lol
weak zinc
Oh 
amber waspBOT
@stone bloom Has your question been resolved?
stone bloom
Oh <:kek:1152776097847574578>
turns out it is pretty easy to solve if you assume that C is directly opposite A and B
but I have no idea how to do the general case
(also it should be AOBC ig)
amber waspBOT
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amber waspBOT
@stone bloom Has your question been resolved?
amber waspBOT
@stone bloom Has your question been resolved?
keen pawn
I'm trying to prove this definition is equivalent to saying that the sum of $n$ vector spaces, is said to be their direct sum if $ \bigcap V_i = {0}$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
I mean the fact that $\bigcap V_i = {0}$ follows trivially from the fact that $0_v \in V_i \forall 1 \leq i \leq n$. If each element can be written in only one way, it follows that ${0_v} \subseteq \bigcap V_i$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
robust isle
I'm trying to prove this definition is equivalent to saying that the sum of $n$...
you won't be able to prove this
because it is false
keen pawn
Huh?
robust isle
consider the x- and y-axes in R^2, and also the line passing through the origin and (1, 1)
these subspaces do not form a direct sum
but their intersection is 0
keen pawn
ah, so if a given set of subspaces, form a direct sum, their intersection is ${0}$
robust isle
that's true
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
In that case I'd like to prove the following lemma
(i) If the sum of $n$ vector spaces, forms a direct sum , then $V_1 \cap V_2 \cap \dots \cap V_n = {0}$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
I feel like the contrapositive may be easier to prove here
We wish to prove the following lemma :- (i) If the sum of $n$ vector spaces, forms a direct sum , then $V_1 \cap V_2 \cap \dots \cap V_n = {0}$ \
\
This is logically equivalent to proving if $V_1 \cap V_2 \cap \dots \cap V_n \neq {0}$, then the sum of those $n$ vector spaces isn't a direct sum
\
Let's start off with the assumption that the intersection is , say ${\alpha v \forall \alpha \in \R}$, which follows trivially from the vector space axioms. Now from the definition, we can conclude that say a vector $3v$, can be obtained in more than one way( how would I rigorously justify this), which means it isn't a direct sum
\
Which ends our proof
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
How's this?
robust isle
We wish to prove the following lemma :- (i) If the sum of $n$ vector spaces, for...
this doesn't seem to be a complete proof
why is the assumption that intersection is the set of all alpha v sufficient for this proof?
keen pawn
why is the assumption that intersection is the set of all alpha v sufficient for...
Say it's just this for now
robust isle
okay
keen pawn
The intersection will either be singletons or infinite
robust isle
well you said it's not 0
keen pawn
Oh, I meant not just 0
robust isle
anyway, fine just go with that assumption that you've got a bunch of alpha v
why do you fail to have a direct sum
keen pawn
I can express say 3v in more than 1 way
robust isle
which ones
keen pawn
All 0s except for one which is 3v
robust isle
right
keen pawn
And one 1v one 2v , the rest being 0
keen pawn
Yeah
robust isle
then 2u = u + u = 2u + 0
but U is in a direct sum
it's the direct sum consisting of just one subspace
robust isle
you can apply the definition of direct sum to only one subspace
every element of U can be uniquely written as an element of U
keen pawn
I don't follow
robust isle
okay if you don't like the fact that there's only one subspace
then do U direct sum with {0}
in U + {0} i have things like
2u = u + u = 2u + 0
you need to resolve why U + {0} is still a direct sum
keen pawn
Im saying I have n distinct vector spaces with this being the intersection, then a vector say 2v can be written in more than one way
robust isle
can you tell me what they are, while also keeping the example i gave you in mind
keen pawn
you can apply the definition of direct sum to only one subspace
I don't follow
robust isle
so in the case where m = 1
[0.7\textwidth]we have a single subspace $V_1 \subsete V$
\begin{itemize}
\ii the sum $V_1$ is a direct sum because each element $v \in V_1$ can be written in only one way $v = v_1$, where $v_1 \in V_1$
\end{itemize}
grizzled pagodaBOT
keen pawn
Yes
robust isle
therefore, V1 is a direct sum consisting of only one subspace
keen pawn
sure
robust isle
okay, nevertheless, if i have $2v \in V_1$, i can write it as
[ 2v = v + v = 2v + 0 ]
grizzled pagodaBOT
robust isle
why is this not a contradiction?
keen pawn
You can only take one element from each $V_i$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
robust isle
okay
so now in your case
argue why 2v or 3v or whatever can be written 2 different ways
keen pawn
say I have $2$ vector spaces for the sake of argument
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
I could have $v$ from $V_1$, $2v$ from $V_2$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
or vice versa
robust isle
and v is in V1 because? 2v is in V2 because?
keen pawn
$v \in V_1$ as $\alpha v\in V_1 \cap V_2$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
robust isle
okay and alpha v is in V_1 intersect V_2 because?
keen pawn
okay and alpha v is in V_1 intersect V_2 because?
I first assumed that ${0,v}$ is the intersection, so then as $V_1, V_2$ are vector spaces, $\alpha v \in V_1$ and $V_2$
robust isle
0,v?
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
robust isle
you should probably write v \ne 0
you can use the fact that V1 intersect V2 is a subspace
if v \ne 0 is in V1 intersect V2, then so is alpha v for any alpha
keen pawn
you can use the fact that V1 intersect V2 is a subspace
But this means $V_1 \subseteq V_2$ or vice versa
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
robust isle
wdym
robust isle
**Veni, vidi, perii is $\\R \- \\Q$**
this one?
keen pawn
yes
robust isle
we just went through and filled in all the missing details
theres nothing wrong with the proof per se
it just wasnt fully fleshed out
keen pawn
I mean I could consider arbitrary many vectors and their scalars to be in the direct sum
robust isle
but we only needed one nonzero vector to be in the intersection
and that gives you all alpha v
keen pawn
yeah
robust isle
amber waspBOT
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Available help channel!
jaunty orbit
I know I am wrong and stuck, but I tried.. any help?
amber waspBOT
I know I am wrong and stuck, but I tried.. any help?
@jaunty orbit Has your question been resolved?
amber waspBOT
@jaunty orbit Has your question been resolved?
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@jaunty orbit Has your question been resolved?
grand rivet
just wanted to check my answers is this correct?
(a) The mean age of all 24 cars is 4.02 years.
(b) The standard deviation of the ages of all 24 cars is 2.19 years.
grand rivet
<@&286206848099549185> someone please lol i resubmit this twice no one helping
i been waiting like an hr for one question someone help ples
pliant copper
problem kinda sucks to solve and it doesnt seem trivial to check without solving myself, if you show work its more likely to be checked
grand rivet
ok ill show u more than gimme a sec
grand rivet
problem kinda sucks to solve and it doesnt seem trivial to check without solving...
(3.6 * 9 + 64)/24 for a and for b ex^2/9 -3.6^2 = 1.925^2 so ex
twilit cradle
correct for b
pliant copper
yeah looks good
grand rivet
aight thx
pliant copper
the "ex" messed me up
pliant copper
$\sum$
grizzled pagodaBOT
grand rivet
yea i meant that symbol
it looks like an e
kinda ig
anyways thx for the help both of u much appreciated
.close
amber waspBOT
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Available help channel!
split aspen
im stuck
spare belfry
Show what you have tried so far
spare belfry
Move 3 to the right and divide both sides by 3
spare belfry
3k + 3 = log(base18) 53/7
spare belfry
Isn’t this your first attempt?
spare belfry
You’re pretty close to the answer
split aspen
chatgpt just converted everything to decimals so i couldnt understand it anymore
spare belfry
Oh, so was that generated by gpt?
split aspen
yeah the first attempt
spare belfry
3k + 3 = log(base18) 53/7
I’m referring to this
split aspen
yes
spare belfry
I see
split aspen
this was
split aspen
3k+3log18 = log(53/7)
im only here
spare belfry
What have you tried on your own?
split aspen
yes
split aspen
yes
spare belfry
3k+3log18 = log(53/7)
What did you do from my step to this step?
split aspen
ok so
spare belfry
18^(3k+3) = 53/7
do you agree with this?
My step
spare belfry
Yes, what’s next?
split aspen
18^(3k+3) = 53/7
do you agree with this?
then rhat
and then
i just put it in log form
because i was taught that way
and plus its smart
because through log form
you can get 3k+3 infornt
so
you can get the value of k
spare belfry
i just put it in log form
Cease here, do you know the log you applied was based on 10 log?
spare belfry
Now, you can apply either base 10 or base 18
split aspen
wait i can just
spare belfry
As long as the base on the both side are the same
spare belfry
Why?
split aspen
how does it go from 10 to 18
spare belfry
10 = 10
log 10 = log 10
log base18 10 = log base18 10
spare belfry
Yes
spare belfry
Np
spare belfry
No problem. Have a good one
split aspen
so now it makes sense
because log 18_18 = 1
and thats why its 3k+3 = log_18(53/7)
.close
amber waspBOT
split aspen
.reopen
amber waspBOT
✅
tropic wadi
what's the question btw?
tropic wadi
and thats why its 3k+3 = log_18(53/7)
everything was fine until this
split aspen
wait um
tropic wadi
but this is incorrect
split aspen
why
tropic wadi
what are the like terms?
split aspen
53/7 and -3?
tropic wadi
53/7 is in logarithm
tropic wadi
yeah
split aspen
like this?
tropic wadi
seems alright
split aspen
tysmm
tropic wadi
however there is also a process in which you can combine the whole thing in a single logarithm
split aspen
oh like
empty frost
so do i just leave it
either you leave it or $\log_kx-t = \log_kx-\log_k k^t = \log\frac{x}{k^t}$
grizzled pagodaBOT
xd_senBugha
split aspen
where did t come from..
empty frost
its just an example constant
empty frost
so in your case itll be $\log_{18}\frac{53}{7}-3 = \log_{18}\frac{53}{7}-\log_{18} 18^3 = \log_{18}\frac{\frac{53}{7}}{18^3}$
grizzled pagodaBOT
xd_senBugha
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split sail
going over review problems right now, the set of {2} is incorrect right, an error?
amber waspBOT
going over review problems right now, the set of {2} is incorrect right, an erro...
@split sail Has your question been resolved?
Available help channel!
proven vapor
Let $S={1,2,3,4,5,6}$. Then find the number of functions\ $f: S\mapsto P(S)$ which satisfy $$f(m)\subset f(n)\text{ whenever }m<n$$
grizzled pagodaBOT
kheerii
patent raptor
!occupied
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violet kraken
Sorry!
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grave dawn
amber waspBOT
grave dawn
bruh
im so stuck
i have no idea im going in circles
i know to to do dy/dx = dy/dt * dt /dx
so i get dy/dx = dy/dt(-2x^-3)
but then how would i compute the seconderitive toget d^2y/dx^2
$\frac{d^2y}{dx^2} = \frac{d}{dx}(-2x^{-3} \frac{dy}{dt})$
grizzled pagodaBOT
Mortta
grave dawn
would that be it?
junior jewel
amber waspBOT
quick ridge
what’s the issue
quick ridge
look at the hint
quick ridge
notice the right denominator is t(t+1)
so you only need to multiply the left fraction by (t+1)/(t+1)
then you can combine
amber waspBOT
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Available help channel!
frank basalt
amber waspBOT
frank basalt
Hi I am very lost when it comes to this problem
I don't really know where to start
I have to find a natural domain for this function
keen topaz
I think you should look at the arcsine.
Then you can look at the other one too.
To be clear you want to find the domain?
frank basalt
y
keen topaz
Is that yes
frank basalt
yes
keen topaz
Okay so what is the domain of arcsine u
frank basalt
Well, <-1, 1> is the domain of arcsine
keen topaz
So only natural numbers allowed?
No I mean for the question
frank basalt
I think by natural domain they just mean domain
So any real numbers without complex ones
keen topaz
Because ⁴√-n
frank basalt
x>=0
keen topaz
But |x+1|≤1
frank basalt
The only reason I am asking is becuase I recall that it had to do something with the roots
Depending if the base is even or odd something was done differently
I am extremely rusty when it comes to math and I am trying my best to get back in form
keen topaz
If it was for eample `2/3` instead of `3/4`
Okay so if it were ⅔ negative values are allowed
frank basalt
Knowing that I could do:
- `-1 <= |x+1| <= 1`
Let me expand this
keen topaz
So that means
Compare the two domains
keen topaz
For the second x≥½
frank basalt
ye ye
i was using a thought shortcut
i lsot my pen 💀
so the arcsine thing
its 2 in one
so I gotta split it up
keen topaz
but for the arcsine
0≤x≤-1
frank basalt
Can you just take off the modulus?
its
-2 <= x <= 0
or wait
no
0 on the right
this way
So -2 is not in domain of arcsine
keen topaz
0 on the right
No
Wait
arcsine|u| domain ranges from -1 to 1
keen topaz
`-1 <= |x+1| <= 1`
Yes
frank basalt
-1 <= |x+1| and |x+1| <=1
keen topaz
Okay.
frank basalt
And now, what the heck happens with the modulus
keen topaz
`-1 <= |x+1| <= 1`
remove the absolute value
keen topaz
without the modulus
Yes
keen topaz
Okay
frank basalt
And then the square root thing was x >= 1/2
keen topaz
Yeah
keen topaz
Do I plot this on a x-axis?
Huh, just find the intersection of the first and second one.
Since x≥½ and -2≤x≤0
Then...
keen topaz
Yeah
frank basalt
Thanks
keen topaz
If imaginary is not allowed
keen topaz
,w plot y=arcsin|x+1|-(2x-1)^¾
grizzled pagodaBOT
keen topaz
Huh
frank basalt
crazy innit
keen topaz
Not even sure how they got this
keen topaz
,w (-1.6)^¾
grizzled pagodaBOT
frank basalt
I have more of stuff like this to do 💀
keen topaz
frank basalt
Thanks my dude
I lack a lot from high school
I went to a trade school and our math teacher was abusing us instead of teaching
I barely passed my finals
keen topaz
Are you in uni
frank basalt
First year yes
3rd week in
We are going through elementary math + cyclometric functions
Tomorrow I have my first test and I am lowkey doomed
keen topaz
Good luck
frank basalt
No need
keen topaz
!done
amber waspBOT
If you are done with this channel, please mark your problem as solved by typing .close
amber waspBOT
keen topaz
I need time