no your overcomplicating well tbh maybe i chose a poor diagram i could see why you would think its infinte but no assume its finite
#help-41
pseudo crescent
where does it end then?
nocturne swan
okay lets assume it only repeats 3 times so one loop including the voltage source and then two other loops containing the same pattern of resistors
oh wait nvm
i know how to do that but how would i do it if it was like this lemme draw it
one sec
just pretend everythings a resistor
how do i find the total resistance excluding the one on the very right?
.close
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hexed lintel
amber waspBOT
hexed lintel
i keep getting an x on its own when i integrate w respect to x
i cant do surface volumes unless i eliminate x and proceed to integrate the remaining w respect to y
honest dagger
i keep getting an x on its own when i integrate w respect to x
!show
amber waspBOT
i keep getting an x on its own when i integrate w respect to x
Show your work, and if possible, explain where you are stuck.
amber waspBOT
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.
hexed lintel
what are you trying to do btw? No question is being asked here
volume under the surface
hexed lintel
Please show the original problem, **exactly as it was stated to you**, with the ...
that is the question
honest dagger
!show
amber waspBOT
Show your work, and if possible, explain where you are stuck.
hexed lintel
Getting an x
Think I should be able to eliminate x before integrating w respect to y
honest dagger
Getting an x
Think hard about why you left that x in there. What are we integrate with respect to?
amber waspBOT
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hexed lintel
Think hard about why you left that x in there. What are we integrate with respec...
we integratign w respect to x
honest dagger
yes
and this is a definite integral with wrt x
So what do we do with definite integrals?
amber waspBOT
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acoustic fox
Is my proof correct?
amber waspBOT
Is my proof correct?
acoustic fox
For context I am proving a corollary of the archimedean property here, which I have already proven. It says that (\textbf{N}) is not bounded above in (\textbf{R}).
grizzled pagodaBOT
usopper.
acoustic fox
I don't think I am losing generality here either because every real number can be represented as a negative of another real number
indigo cloud
well properly you would choose s first, set r=-s and then do the argument
acoustic fox
Makes sense
Wait
For any real number (r), we know that there exists a natural number (n > r). Let (s = -r) so (n > -r). Taking the negative on both sides, (-n < r) where (-n \in \textbf{Z}).
grizzled pagodaBOT
usopper.
indigo cloud
let s in R be arbitrary. you want to find an integer m with m<s. you define r=-s. then you know there exists a natural number n with n>r. define m=-n. then m<s
acoustic fox
For any real number (r), let (s = -r). Since (s \in \textbf{R}), we know that there exists a natural number (n > s) because of the Archimedean property of real numbers. Taking the negative on both sides, (-n < -s) where (-n \in \textbf{Z}). Since (-s = r), we have (-n < r).
grizzled pagodaBOT
usopper.
acoustic fox
Good?
indigo cloud
yes
quiet drum
What is the integral of the function f(x) = 2x^2 + 3x - 1?
amber waspBOT
What is the integral of the function f(x) = 2x^2 + 3x - 1?
ancient raft
You could make the integral into multiple integrals, if that will help $$\int 2x^2+3x-1;dx$$
grizzled pagodaBOT
Good
patent raptor
What is the integral of the function f(x) = 2x^2 + 3x - 1?
Kennst du die Integrationsregeln?
quiet drum
nein, ich bin momentan in der 8 Klasse also weiß ich dies leider nicht
patent raptor
ok
also imprinzip ehöhst du die hochzahl um 1
und dividierst durch die neue hochzahl
also mal beispiel
grizzled pagodaBOT
bacc (unhelpful)
quiet drum
Ähhhh Gesundheit?
patent raptor
was
quiet drum
du ich bin 14 ich verstehe nicht mal was das F vorne bedeutet
lethal steppe
Wie kommt’s eigentlich dass du in der 8ten integrale machst
patent raptor
du ich bin 14 ich verstehe nicht mal was das F vorne bedeutet
ok lol
Ich ging davon aus du wüsstest wenigstens was du überhaupt fragst lol
quiet drum
extra kurs halt, ich habs halt nicht verstanden und hab auch keine Lust nach zufragen
quiet drum
also ich verstehe das ganze Thema nicht
lethal steppe
Ohne nachfragen wird schwierig langfristig 😅
Wir wissen halt nicht was genau du da machst im extra Kurs
quiet drum
ich frage nach den Ferien einfach nach
lethal steppe
Integrale kommen ja erst in der Oberstufe. Ich kann mir vorstellen dass das bei euch stark vereinfacht wird
patent raptor
Kannst ja in den Ferien dich trzdm damit befassen
wenns dich wirklich interessiert
Also du kennst ja Funktionen, zumindest sehr einfache wie Geraden, oder Parabeln... ein Integral ist schlicht und einfach ein Werkzeug um die Fläche unter einer Kurve auszurechnen
Die Grundidee besteht darin die Fläche zu schätzen in dem Rechtecke unter der Kurve zeichnet, die Fläche von denen ausrechnet und summiert
also so in etwa kann man sich das vorstellen
je mehr rechtecke man ausrechnet, die gleichzeitig dünner werden, desto besser kann man die fläche abschätzen
und am Präzisten gehts dann mit dem Integral
amber waspBOT
@quiet drum Has your question been resolved?
raven magnet
Hi. For anyone who speaks french I have no idea how to solve question 4-a. Can you help please
raven magnet
Tried recurrence tried using the function but nothing works
Or I made a mistake somewhere
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raven magnet
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gusty plaza
Can someone guide me on how to start this problem please?
amber waspBOT
Can someone guide me on how to start this problem please?
gusty plaza
I’m really not sure where to start can someone give hints
I know car dV/dt
And I’m given distance pl
Wait I got it
.close
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gusty plaza
Am I setting up my problem for number 4 here correct?
amber waspBOT
Am I setting up my problem for number 4 here correct?
@gusty plaza Has your question been resolved?
amber waspBOT
still pecan
.reopen
still pecan
Am I setting up my problem for number 4 here correct?
It might close the channel but not quite correct with the diagram
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fervent oasis
aren't all of them corect?
jaunty shale
no, matrices are not like real numbers
right multiplication and left multiplication do not yield same result
"matrices multiplication is not commutative, unlike with real numbers"
dull ivy
no, matrices are not like real numbers
You are undergraduate
amber waspBOT
You are undergraduate
amber waspBOT
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full elk
I think one big issue is that as you've defined it, Y can only be all 0's.
And in particular, I could reply and say that you divide it as x = "", y = 0^p and z = 1^p and then I can pump y and it will always be a multiple of p.
amber notch
I think one big issue is that as you've defined it, Y can **only** be all 0's.
is that not the case? i got the impression that was the only case possible (also hello again lol)
amber notch
And in particular, I could reply and say that you divide it as x = "", y = 0^p a...
huh
full elk
In particular, I think your first case breaks down if k = p
I would suggest the string 0^(4p)1^(2p) with pumping length p
amber notch
In particular, I think your first case breaks down if k = p
am i correct in thinking that the other 2 cases do not exist?
full elk
am i correct in thinking that the other 2 cases do not exist?
Yes. |xy| must be at most p, so if you have 0^k where k>p in the first character, you can't have any 1's in there
I think it's good, although you did write "n must be a multiple of n", which I suppose is a typo
If it's graded, maybe take the time to explain why 2p does not divide 4p+k
amber notch
Yes. |xy| must be at most p, so if you have 0^k where k>p in the first character...
okay nice, thats what i figured
amber notch
I think it's good, although you did write "n must be a multiple of n", which I s...
oops lol
full elk
Hmm, I wouldn't be convinced. k is any number between 1 and p, so it could be 2.
The reason why it can't be a multiple of 2p is that it can't be 0, and the next multiple of 2p is 2p, which is too large (k<= p)
amber notch
so i'd say "4p+k cannot be a multiple of 2p because it cannot be 0 and the next multiple of 2p is 2p, which is too large (k<= p)"? im a little confused on that tho, why is the next multiple of 2p 2p and how does that prove anything
full elk
We want 4p + k to be a multiple of 2p
4p is already a multiple of 2p
So if that is the case, k must be a multiple of 2p
So what are the multiples of 2p? We have 0, 2pi, 4pi, ...
But k >= 1, so it can't be 0
And k <= P, so it can't be 2p
amber notch
so i'd say "4p+k cannot be a multiple of 2p because it cannot be 0 and the next ...
would it just be like this? i dont fully get why my initial explanation doesnt work tho
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split sail
yo
split sail
need help with this last one
number d
We need a 5 digit number
That starts with either 2 or 6
That is odd and no number can be repeated
So we can write this as A B C D E
Yo uhvve 2 possible choices for A
And E must be an odd number
The middle letters can be any remaining letter apart from A and E
Then I think you know how to find the answer
amber waspBOT
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grim isle
amber waspBOT
shadow stump
the first bullet point
shadow stump
"reflected light ray is in the same plane as the incident ray and the normal to the mirror"
grim isle
how can i use this statement to prove question 3
shadow stump
question 3 asks you to prove that the statement above (the one about a and n) satisfies the two physical laws (bullet points)
grim isle
and what am i suppose to write
for angle i managed to do some calculations
but the first bullet point i have no idea
shadow stump
think about how you can show 3 vectors are coplanar
grim isle
their cross product is same?
vague thorn
or 0
grim isle
?
0 as in wut
dot produc of three points with normal vector is 0
ion got any points to prove lol
also what does arbitiary constant means
amber waspBOT
@grim isle Has your question been resolved?
humble onyx
a vector x is in the same plane as the plane formed by vectors y and z if a can be expressed as ay+bz where a, b are scalars
yo can do cross product to obtain normal to plane, then dot product that normal to plane with your vector x to check (you want the dot product to be 0)
molten stag
hi
molten stag
The grocery store earns 30% of the sale price for each box of biscuits sold. During the Thadingyut Festival, the store decides to take 20% off the sale price for one day. By what percentage must the number of boxes sold increase to earn the same amount of money as the previous day?
this is a very ez question but
smth is off
?
like the answers that i got it
is quite
different from the answer
so
like
can anyone help me with this
grim isle
a(b.c)
where abc are vectors
amber waspBOT
a(b.c)
where abc are vectors
grim isle
would it become abc or ab+ac
primal holly
Neither really
grim isle
. is dot product
primal holly
b.c is now a scalar, and you're scaling a by this new scalar
primal holly
so i can only simplify b.c
yep
grim isle
ya
primal holly
maybe you could dot product the whole thing with $\hat{\vb n}$
grizzled pagodaBOT
krypton
grizzled pagodaBOT
krypton
primal holly
this is what it becomes
hmm
still no
this is kind of weird hold on
n a_r and a_i need to be coplanar
grim isle
do u want the full question?
primal holly
yes
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still kelp
skibidi toilet
amber waspBOT
skibidi toilet
fluid wren
Don't troll
split sail
skibidi toilet
?
still kelp
No brainrots here
-1000 aura
still kelp
Don't troll
im demour
fluid wren
.close
amber waspBOT
split sail
gen alpha
split sail
There is no word called "rizzler"
still kelp
There is no word called "rizzler"
oh what a sigma
split sail
hi kaicenat follower
sigma is a mathematical expression
stop dying because of brainrot
split sail
What is a bussy
pseudo crescent
dont ask
latent schooner
bro u r brainrotted
pseudo crescent
this channel is gonna close anyway
still kelp
bro u r brainrotted
ur salty
split sail
Ok
pseudo crescent
just give it few mins
split sail
k
still kelp
no point in speaking with a non living thing
skill issue
split sail
brain issue for you
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Available help channel!
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woeful dagger
Has collatz conjecture been attempted to be disproven using quantum computers?
woeful dagger
And if not, why not?
cloud scarab
Quantum computers are better at solving some specific types of problems than classical computers - they are not just generally faster (and actually sometimes are slower).
I assume Collatz conjecture is not the type of problem that quantum computers solve efficiently (I mean bruteforcing the counterexample)
Although maybe there's a way
indigo cloud
the collatz conjecture (like most interesting mathematical problems) are not things you can just throw a computer at and wait long enough for them to solve it
quantum or not
lethal steppe
the collatz conjecture (like most interesting mathematical problems) are not thi...
Unless it’s false I assume?
indigo cloud
yes ok but how likely is that
woeful dagger
yes ok but how likely is that
A try cannot hurt?
indigo cloud
I mean people are probably still running shit on computers but I doubt anyone thinks they will actually find stuff
and like said before, quantum computers arent just straight up faster than classic computers, they are just better at certain tasks
heavy hollow
guys whats the command to get helpers to my room i forgot
indigo cloud
also its not exactly easy to have a quantum computer in the first place and then you dont necessarily want to waste computation time for this
sterile nymph
I strongly doubt it's false, mostly because if there is a counterexample it would be due to a very power of two being very close to a power of 3, which would say something very strong about rational approximations of ln(2)/ln(3), which seems to obey continued fraction structure that is common to a general real number. (and there's no reason to think it would be an exception.)
split sail
maybe when i will know enough maths i will prove it without needing a quantum computer
cloud scarab
I strongly doubt it's false, mostly because if there is a counterexample it woul...
Also Terence Tao published a paper titled "Almost all orbits of the Collatz map attain almost bounded values."
split sail
sterile nymph
and that's a great breakthrough, but it doesn't prove that there is no exceptional case.
cloud scarab
yes, but it kind of hints that there might be no counterexample
sterile nymph
but yeah, I doubt you'd get much quantum speedup, because the key to this computation is that it's iterative. And because it's an essentially iterative process, there isn't anything that a superposition of states can really do here.
sterile nymph
yes, but it kind of hints that there might be no counterexample
I would say that it just hints that there probably isn't a divergent orbit. A bounded orbit that isn't 1, 2, 4 is not at all ruled out by his paper, to my understanding.
Proving that there are measure 0 counterexamples to divergent orbits is important, but it's one of those things that everyone already expected due to heuristic arguments. Every time you multiply by 3 and add 1, there is on average 2 division by 2 steps, (if the last few bits of the result are taken to be bernoulli random variables with p=1/2), so the long term expected trend is (3/4)^i with i being the number of iterations.
of course, they aren't random variables.
cloud scarab
Yes, that's fair, but unfortunately there isn't much more progress than that
heavy hollow
can someone help me
amber waspBOT
@woeful dagger Has your question been resolved?
misty bobcat
amber waspBOT
misty bobcat
did i cook? I got r= 37.6 for part b and r=95.7 for partc
mint nacelle
how did you get that angle of 11.19 degrees?
oh wait using right angled trig, I can check that
,w (2pi - 5pi/4 - 2 * arctan(150/100))/2 * 180/pi
grizzled pagodaBOT
mint nacelle
ah perfect
then yeah your method for b is correct
your working for c however doesn't make any sense
you can't just assume 200pi * 225/360
first things first, 200r pi * 225/360 + 2pi r * 135/360 + 2y = 850
ah, but you have r/sin(11.19) = sqrt(100^2 + y^2)/sin(112.5), similar to what you did in part b
this must be a GDC paper or something cause you can just sub in r from the second equation straight into the 1st equation
weird
oh shit it's not 11.9 anymore
it's now (3pi/4 - arctan(y/100))/2
,w solve 200r pi * 225/360 + 2pi r * 135/360 + 2y = 850 and r/sin( (3pi/4 - 2 arctan(y/100))/2) = sqrt(100^2 + y^2)/sin(112.5 deg)
grizzled pagodaBOT
Requested by sooouth.
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midnight chasm
Not sure what to do with b+c here?
amber waspBOT
Not sure what to do with b+c here?
pallid canopy
if you know A and D you can plug them in and match coefficients after expanding
x^2 coefficient = 3, etc.
amber waspBOT
@midnight chasm Has your question been resolved?
amber waspBOT
@midnight chasm Has your question been resolved?
midnight chasm
if you know A and D you can plug them in and match coefficients after expanding
I thought I did do that?
midnight chasm
where do you solve for the x coefficient on the right side ?
I have to factor everything out first right?
amber waspBOT
@midnight chasm Has your question been resolved?
midnight chasm
where do you solve for the x coefficient on the right side ?
I'm not sure what you mean by this, I expanded it out but I'm not sure how to match the coefficients here
pallid canopy
there are no cubic terms on the left, but there's a Bx^3 term on the right, implying B = ?
amber waspBOT
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midnight chasm
.close
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charred onyx
I hate this type of questions, can somebody help me?
Fill in the 4 × 4 checkered board with the black top left box
with ones and zeros. In each 2×2 square, which has a black top left square, there is the same
number of zeros as the number of ones. How many different ways can the table be filled?
charred onyx
There can be only one number in each square, either a one or a zero
So this is a possible way to fill in the board
I thought it could be that a 2x2 square can be filled in by 6 different ways
Going like this, 1100,1010,1001,0011,0110,0101
Therefore a 4x2 table would be filled in 36 different ways?
And the 4x4 in 36 squared ways
But there is still the square in the middle thats ruining it all and thats my problem I need help with
<@&286206848099549185>
balmy sorrel
amber waspBOT
@charred onyx Has your question been resolved?
silk sonnet
so
what if you fill the middle 2x2 first
and then fill out the rest of the board
if one of the numbers in a 2x2 is fixed
then you can fill it 3 different ways
so
red thicket
I thought it could be that a 2x2 square can be filled in by 6 different ways
correct
silk sonnet
it would be 486
red thicket
then you can fill it 3 different ways
you sure?
red thicket
Thanks guys
alr
charred onyx
.close
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raw prawn
im doing completing the square right now. If the coefficient of x is higher than 1, what would I do?
formal yacht
Factor it out and continue as usual
formal yacht
im doing completing the square right now. If the coefficient of x is higher than...
Wait, x term or x^2 term?
raw prawn
so by factoring out the coefficient i would end up with lots of fractions?
formal yacht
Yeah probably, depends of course
raw prawn
thank you
amber waspBOT
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tough mica
amber waspBOT
tough mica
3.- Let $P(x) = 2x^5 - 5x^4 + 4x^3 - 37x^2 + ax + b \in \mathbb{R}[x]$. Find the values of $a$ and $b$ such that $3i$ is a root of $P$, and for the values found, determine all the roots of $P$.
grizzled pagodaBOT
938c2cc0dcc05f2b68c4287040cfcf71
amber waspBOT
@tough mica Has your question been resolved?
tough mica
,, \begin{cases} 2(-3i)^5 - 5(-3i)^4 + 4(-3i)^3 - 37(-3i)^2 + a(-3i) + b &= 0 \ 2(3i)^5 - 5(3i)^4 + 4(3i)^3 - 37(3i)^2 + a(3i) + b &= 0 \end{cases}
grizzled pagodaBOT
938c2cc0dcc05f2b68c4287040cfcf71
tough mica
,, \begin{cases} -72-378i -3ai + b = 0 \ -72 + 378i + 3ai + b = 0 \end{cases}
grizzled pagodaBOT
938c2cc0dcc05f2b68c4287040cfcf71
tough mica
-144 = -2b
b = 72
,, \begin{cases} -72-378i -3ai + 72 = 0 \ -72 + 378i + 3ai + 72 = 0 \end{cases}
grizzled pagodaBOT
938c2cc0dcc05f2b68c4287040cfcf71
tough mica
,, \begin{cases} -378i -3ai = 0 \ 378i + 3ai = 0 \end{cases}
grizzled pagodaBOT
938c2cc0dcc05f2b68c4287040cfcf71
grizzled pagodaBOT
938c2cc0dcc05f2b68c4287040cfcf71
tough mica
<@&286206848099549185>
simple quartz
amber waspBOT
@tough mica Has your question been resolved?
untold lake
patent raptor
rational root theorem
twin helm
do you know how to find em
see if -1 or 1 are factors of thi
and use synthetic divison if they are. if not try for -2 and +2
grizzled pagodaBOT
tough mica
$\polylongdiv{2x^5-5x^4+4x^3-37x^2-126x+72}{x^2+9}$
grizzled pagodaBOT
938c2cc0dcc05f2b68c4287040cfcf71
tough mica
@twin helm
tough mica
factor what
twin helm
the result
quotient
the question is find all the roots right?
so they are 3i, -3i and the roots of the quotient
twin helm
rational root theorem
strange spear
how do I factorize
test x=-2
twin helm
take the factors of the leading coefficent and divide it by the factors of the constant
strange spear
then factor out (x+2)
twin helm
I'd reccomend learning it if you don't know it alread
tough mica
,w divisors 8
grizzled pagodaBOT
tough mica
test x=-2
you are giving me the answer to one of the roots I think
because
otherwis I would need to brute force the roots until some number works
tough mica
like proper way of reducing guesses is rational root theorem but idk how it works
tough mica
I mean if you divide 2 that is a factor of 8 with 1 that is a factor of 2 and try 2/1 that is one root
$\polylongdiv{2x^3 - 5x^2-14x+8}{x+2}$
grizzled pagodaBOT
938c2cc0dcc05f2b68c4287040cfcf71
tough mica
I see
strange spear
yeah
tough mica
,w 2x^2-9x+4 =0
grizzled pagodaBOT
tough mica
,w 0 = 2x^5 -5x^4 + 4x^3 - 37x^2 -126x + 72
grizzled pagodaBOT
amber waspBOT
tough mica
bro too bored to write it down
only issue was with long division
because your intuition was on point but you need to justify that you found x=-2 using rational root theorem
strange spear
because your intuition was on point but you need to justify that you found x=-2 ...
yeah
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keen pawn
amber waspBOT
amber waspBOT
amber waspBOT
✅
earnest compass
take a basis
keen pawn
Rephrased into an implication this is : Suppose $V$ is a finite dimensional vector space, such that $U$ is a subspace of $V$, $V,U$ being distinct.If $n=dim(V) and m=dim(U)$ then there are (n-m) subspaces of $V$ , each of dimension $n-1$ and the interesection of these subspaces is $U$
grizzled pagodaBOT
A dense set
keen pawn
Let's start by constructing a subspace of $V$ of the desired dimension
grizzled pagodaBOT
A dense set
keen pawn
As the dimension of $U$ is $n-1$, let's assume a list of linearly independent vectors in $U$. Namely, $e_1,e_2,\dots e_{n-1}$
grizzled pagodaBOT
A dense set
earnest compass
i dont see how this route is helpful
id start by:
taking a basis b of U
taking a basis b' of V that contains b
keen pawn
Let's start by constructing a subspace of $V$ of the desired dimension
\
As the dimension of the subspaces of $V$ are $n-1$, let's assume a list of linearly independent vectors in $V$. Namely, $e1,e_2,\dots e_{n-1}$
\
Now, we wish to, perhaps , reduce this to a basis of $U$, we do this by deleting unnecessary vectors from the basis. To do this, we remove $n-1-m$ vectors from the basis.
\
It's evident there are n-m ways in which we can do this
grizzled pagodaBOT
A dense set
earnest compass
alr gl, my flights taking off so i wont be able to help
keen pawn
I think this shoudl work
keen pawn
alr gl, my flights taking off so i wont be able to help
Have a safe flight
amber waspBOT
@keen pawn Has your question been resolved?
keen pawn
Let's start by constructing a subspace of $V$ of the desired dimension
\
As the dimension of the subspaces of $V$ are $n-1$, let's assume a list of linearly independent vectors in $V$. Namely, $e1,e_2,\dots e_{n-1}$
\
Now, we wish to, perhaps , reduce this to a basis of $U$, we do this by deleting unnecessary vectors from the basis. To do this, we remove $n-1-(n-m)$ vectors from the basis.
grizzled pagodaBOT
A dense set
keen pawn
.close
amber waspBOT
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fathom notch
Hi guys I'm new
I want to be in my math Olympiad team And I'm 15 in 10th grade and in Morocco I just started grinding from today
I started seeing and uploading AOPS books that I needed to do so l can be better at problem solving ...
And started practicing a lot of maths alone
I don't know if we can call 3hourd per day a lot lmao
So anyone had an advice for me to be better and outperform everybody?
flat blaze
Hi guys I'm new
I want to be in my math Olympiad team And I'm 15 in 10th grade a...
It's awesome that you want to join your country's olympiad team and are putting in the effort to improve.
On a more personal note, while it’s natural to want to outperform others, relying on extrinsic motivation can lead to stress and burnout. Try to focus on internal motivation—finding joy in math itself, setting personal goals, and enjoying the challenge of solving problems. When you’re motivated by your own interest, you’ll find it easier to stay dedicated and improve over time.
being better than everybody is an unattainable goal, but being better than you were before is always possible. i recommend you focus on that, it'll be much better for you in the long run
inland pulsar
<@755365572388323378>
!noping
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fathom notch
Thanks guys
umbral osprey
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umbral osprey
im here after a bit of angle chasing
I think I have to use similar triangles
but idk which ones and I dont see enough information to prove similarity
<@&286206848099549185>
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tulip tapir
AHA get SNIPED anjali
(jk)
umbral osprey
bruh i put this question out for like 3 hours yesterday and noone helped
please help
tulip tapir
frankly, this question looks like hell
umbral osprey
frankly, this question looks like hell
please i beg
tulip tapir
i'm genuinely not sure what you're supposed to do lol
trying to figure out what's going on
umbral osprey
i'm genuinely not sure what you're supposed to do lol
it looks like it could be similarity
but we're given barely any angles
just the parallel lines
umbral osprey
<@&286206848099549185>
plz
<@&286206848099549185>
<@&286206848099549185>
<@&286206848099549185>
<@&286206848099549185>
<@&286206848099549185>
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tribal vessel
Let's say there exists smooth functions $\phi_x, \phi_y: \Omega \to \mathbb{C}$ such that $\frac{\partial \phi_x}{\partial y} = \frac{\partial \phi_y}{\partial x}$.
Is this enough to guarantee the existence of a smooth function $u: \Omega \to \mathbb{C}$ with $$\frac{\partial u}{\partial x} = \phi_x, \frac{\partial u}{\partial y} = \phi_y?$$
grizzled pagodaBOT
reking
tribal vessel
Oh, and Omega is a simply connected open set in C
the simply connected part should be important i suppose
pallid canopy
Try uhhhh #odes-and-pdes or #advanced-pdes
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@tribal vessel Has your question been resolved?
fervent oasis
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weak tangle
second one maybe
fervent oasis
why
amber waspBOT
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@fervent oasis Has your question been resolved?
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rancid parrot
Could someone please help me with this problem?
amber waspBOT
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loud field
How do you guys read something like :∠BAC
How can I read this and visualize this in my head?
I know A is the vertex
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hybrid sedge
I dont understand this approach?
amber waspBOT
I dont understand this approach?
shadow stump
,rcw
grizzled pagodaBOT
shadow stump
we have that ln(x) = 1*ln(x)
so we are choosing to use ln(x) and 1 as our two functions
shadow stump
if you have two functions $f(x)$ and $g'(x)$ then you have [ \int f(x) g'(x) \odif x = f(x) g(x) - \int f'(x) g(x) \odif x ] note that $g'(x)$, not $g(x)$, is the function appearing in our original integral
grizzled pagodaBOT
cloud
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lethal bridge
Why do we need to do pfd instead of u sub for this problem?
amber waspBOT
Why do we need to do pfd instead of u sub for this problem?
shadow stump
you had du = -2x dx, where did the x go?
lethal bridge
you had du = -2x dx, where did the x go?
Ooo good catch
I see now
Can’t use usub for it
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humble tusk
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neon token
is this correct?
amber waspBOT
is this correct?
pallid canopy
Why 0.95x instead of 0.05x
neon token
Why 0.95x instead of 0.05x
cuz its water
neon token
.close
long shadow
How do I get (5x-7)(5x+7)=0
amber waspBOT
How do I get (5x-7)(5x+7)=0
hard ermine
a^2-b^2 = (a-b)(a+b)
long shadow
a^2-b^2 = (a-b)(a+b)
Sorry I didn't understand
white lake
formula for the difference of square
latent schooner
there are some formulae for factorization
long shadow
What's was it
latent schooner
a^2-b^2 = (a-b)(a+b)
we can simply prove this by expanding the right hand side
white lake
Maybe you should recite it
long shadow
Im so stupid I couldn't understand
hard ermine
(a+b)(a-b) try expanding this
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split sail
amber waspBOT
split sail
is this step valid?
coral wigeon
yajat
split sail
no
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patent raptor
is this step valid?
looks weird, you just put a sum infront of the entries in the first column?
split sail
yea 💀
patent raptor
At least I am not aware of such rule
split sail
damn ok i was using the logic of the scalar multiplication logic
like you know when you multiply a scalar it gets mulitplied by all the elements of a row or a column
grizzled pagodaBOT
bacc (unhelpful)
patent raptor
Also "sum" is not a scalar
fierce edge
didn't i already 😭 resolve this
coral wigeon
<:kannawave:1261733746961289246>
split sail
.close
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patent raptor
n times d_k means we can use that logic right?
Yup, you could multiply the n into the first column, then your first row would be of n's to give a good start to simplify this
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rough turret
amber waspBOT
rough turret
i dont see how part a is relevant to part b
i did part a and im still stuck
(n+1)(4n²+ 14n + 9)
mint nacelle
(n+1)(4n²+ 14n + 9)
,w (4(n + 1)^2 + 6(n + 1) - 1) simplify
grizzled pagodaBOT
mint nacelle
well, yeah
for the inductive step you just need to show that $(2(n + 1) - 1)(2(n + 1) + 1) = \frac{1}{3} (n + 1) (4(n + 1)^2 + 6(n + 1) - 1) - \frac{1}{3} n(4n^2 + 6n - 1)$
grizzled pagodaBOT
south's secret twin brother
mint nacelle
from part a, (n + 1)(4(n + 1)^2 + 6(n + 1) - 1) has been expanded for you already
rough turret
😭
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sharp depot
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sharp depot
part iii pls
whats wrong with my way
8C3 to choose three spots out of the 8
then 3 * 2 *1 to represent ways u can arrange A B C within those 3 spots
so now u have a b c somehwere in the pass
and the rest of the 5 spots can be whatever
so its 3^5 cause u have 3 choices for everything else
so 8C3 * 3 * 2 * 1 * 3^5
the answer is meant to be 5796
mint nacelle
,w 3^5 * (8 choose 5) * 3!
grizzled pagodaBOT
mint nacelle
something about overcounting, cause that approach is logical
sharp depot
where
mint nacelle
I think the strategy here would be to subtract from the total, so only AB, only BC, only CA, only A, B, C
mint nacelle
where
you just can't guarantee that with your method, there is only 1 way to get to that sequence
sharp depot
i get the -3*256 cause thjere are 256 ways to have a pass with only B and C so timesing that by 3 is for AB BC CA and A B C
but wh ydid the add the 3
which cases did they add
mint nacelle
oh cause of inclusion-exclusion
sharp depot
cause teh 256 which is the answer of part ii
mint nacelle
the sequence with all As is subtracted twice, in AB and AC
sharp depot
ohhh
mint nacelle
so you need to add it once back, to only be subtracting once
yeah damn that's 2 marks, that's criminally little
mint nacelle
probably like 4 in A levels
mint nacelle
.close
if you were to add everything, you would need to find how many ways 5 objects can go into 3 bins using stars and bars, so that would be (5 + 3 - 1) choose (3 - 1) = 21 ways
but each of the 21 ways has a different number of repetitions, so you'd be dividing by different factorials every single time
you'd need to consider each of the 21 cases individually
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keen pawn
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keen pawn
We proceed by induction
We can trivially see that $dim(V_1) \leq dim(V_1)$
\
We now assume that $dim(V_1+V_2 + \dots + V_m) \leq dim(V_1)+ dim(V_2)+ \dots dim(V_m)$
\
We now prove. that from this it follows that $dim(V_1+V_2+ \dots V_{m+1} \leq dim(V_1)+ dim(V_2)+ \dots + dim(V_m)$
\
It's trivially true that $dim(V_1+V_2+ \dots + V_n+ V_{m+1}) \geq dim(V_1+V_2+ \dots + V_m)$
\
We however, know that $dim((V_1+V_2+ \dots +V_m)+V_{m+1})= dim(V_1+V_2+ \dots + V_m) + dim(V_{n+1}- dim[(V_1+V_2+ \dots + V_n) \cap V_{m+1}$.
\
We also know that $dim(V_1+V_2+\dots + V_m) \leq dim(V_1)+ dim(V_2)+ \dots dim(V_m)$.
This gives us that $dim(V_1+V_2+\dots V_{m+1}) \leq dim(V_1)+ dim(V_2)+ \dots dim(V_m)+ dim(V_{m+1}) -dim[(V_1+V_2+ \dots + V_n) \cap V)]_{m+1}$.
\
From this the desired result is evident
Hoes does this look?
indigo cloud
is it evident?
the logic is ok, still some bloat
not even mentioning the word induction is bad
dont write "attempt to prove", either you prove it or not
grizzled pagodaBOT
A dense set
keen pawn
not even mentioning the word induction is bad
How do mention induction in my proof
keen pawn
We proceed by induction
We can trivially see that $dim(V_1) \leq dim(V_1)$
\
We now assume that $dim(V_1+V_2 + \dots + V_m) \leq dim(V_1)+ dim(V_2)+ \dots dim(V_m)$
\
We now prove. that from this it follows that $dim(V_1+V_2+ \dots V_{m+1} \leq dim(V_1)+ dim(V_2)+ \dots + dim(V_m)$
\
It's trivially true that $dim(V_1+V_2+ \dots + V_n+ V_{m+1}) \geq dim(V_1+V_2+ \dots + V_m)$
\
We however, know that $dim((V_1+V_2+ \dots +V_m)+V_{m+1})= dim(V_1+V_2+ \dots + V_m) + dim(V_{n+1}- dim[(V_1+V_2+ \dots + V_n) \cap V_{m+1}$.
\
We also know that $dim(V_1+V_2+\dots + V_m) \leq dim(V_1)+ dim(V_2)+ \dots dim(V_m)$.
This gives us that $dim(V_1+V_2+\dots V_{m+1}) \leq dim(V_1)+ dim(V_2)+ \dots dim(V_m)+ dim(V_{m+1}) -dim[(V_1+V_2+ \dots + V_n) \cap V)]_{m+1}$.
\
From this the desired result is evident
grizzled pagodaBOT
A dense set
keen pawn
or literally any other phrase like that
Is this better?
keen pawn
We proceed by induction
We can trivially see that $dim(V_1) \leq dim(V_1)$
\
We now assume that $dim(V_1+V_2 + \dots + V_m) \leq dim(V_1)+ dim(V_2)+ \dots dim(V_m)$
\
We now prove. that from this it follows that $dim(V_1+V_2+ \dots V_{m+1} \leq dim(V_1)+ dim(V_2)+ \dots + dim(V_m)$
\
It's trivially true that $dim(V_1+V_2+ \dots + V_n+ V_{m+1}) \geq dim(V_1+V_2+ \dots + V_m)$
\
We however, know that $dim((V_1+V_2+ \dots +V_m)+V_{m+1})= dim(V_1+V_2+ \dots + V_m) + dim(V_{n+1}- dim[(V_1+V_2+ \dots + V_n) \cap V_{m+1}$.
\
We also know that $dim(V_1+V_2+\dots + V_m) \leq dim(V_1)+ dim(V_2)+ \dots dim(V_m)$.
This gives us that $dim(V_1+V_2+\dots V_{m+1}) \leq dim(V_1)+ dim(V_2)+ \dots dim(V_m)+ dim(V_{m+1}) -dim[(V_1+V_2+ \dots + V_n) \cap V_m+1})]$.
\
From this the desired result follows as $dim[(V_1+V_2+ \dots + V_m) \cap V_{m+1}] \geq 0$. This tells us $dim(V_1+V_2+ \dots + V_{m+1}) \leq dim(V_1)+dim(V_2)+ \dots + dim(V_n)$
grizzled pagodaBOT
A dense set
We proceed by induction
We can trivially see that $dim(V_1) \leq dim(V_1)$
\\
We now assume that $dim(V_1+V_2 + \dots + V_m) \leq dim(V_1)+ dim(V_2)+ \dots dim(V_m)$
\\
We now prove. that from this it follows that $dim(V_1+V_2+ \dots V_{m+1} \leq dim(V_1)+ dim(V_2)+ \dots + dim(V_m)$
\\
It's trivially true that $dim(V_1+V_2+ \dots + V_n+ V_{m+1}) \geq dim(V_1+V_2+ \dots + V_m)$
\\
We however, know that $dim((V_1+V_2+ \dots +V_m)+V_{m+1})= dim(V_1+V_2+ \dots + V_m) + dim(V_{n+1}- dim[(V_1+V_2+ \dots + V_n) \cap V_{m+1}$.
\\
We also know that $dim(V_1+V_2+\dots + V_m) \leq dim(V_1)+ dim(V_2)+ \dots dim(V_m)$.
This gives us that $dim(V_1+V_2+\dots V_{m+1}) \leq dim(V_1)+ dim(V_2)+ \dots dim(V_m)+ dim(V_{m+1}) -dim[(V_1+V_2+ \dots + V_n) \cap V_m+1})]$.
\\
From this the desired result follows as $dim[(V_1+V_2+ \dots + V_m) \cap V_{m+1}] \geq 0$. This tells us $dim(V_1+V_2+ \dots + V_{m+1}) \leq dim(V_1)+dim(V_2)+ \dots + dim(V_n)$
```Compilation error:```! Extra }, or forgotten $.
l.1434 ... -dim[(V_1+V_2+ \dots + V_n) \cap V_m+1}
)]$.
I've deleted a group-closing symbol because it seems to be
spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.
Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2023/texmf-var/fonts/map/pdftex/updmap/pdftex.map}]```
quick spoke
still bloat
i.e. what is the point of
dim(V_1 + V_2 + ... + V_(m+1)) >= dim(V_1 + V_2 + ... + V_m)
keen pawn
I thought it would be useful later
quick spoke
I thought it would be useful later
it's fine to do stuff in ur working that don't end up being useful, but in ur final write-up you should delete stuff that u don't use to avoid clutter
keen pawn
We proceed by induction
We can trivially see that $dim(V_1) \leq dim(V_1)$
\
We now assume that $dim(V_1+V_2 + \dots + V_m) \leq dim(V_1)+ dim(V_2)+ \dots dim(V_m)$
\
We now prove. that from this it follows that $dim(V_1+V_2+ \dots V_{m+1} \leq dim(V_1)+ dim(V_2)+ \dots + dim(V_m)$
\
We know that $dim((V_1+V_2+ \dots +V_m)+V_{m+1})= dim(V_1+V_2+ \dots + V_m) + dim(V_{n+1}- dim[(V_1+V_2+ \dots + V_n) \cap V_{m+1}$.
\
We also know that $dim(V_1+V_2+\dots + V_m) \leq dim(V_1)+ dim(V_2)+ \dots dim(V_m)$.
This gives us that $dim(V_1+V_2+\dots V_{m+1}) \leq dim(V_1)+ dim(V_2)+ \dots dim(V_m)+ dim(V_{m+1}) -dim[(V_1+V_2+ \dots + V_n) \cap V_m+1})]$.
\
From this the desired result follows as $dim[(V_1+V_2+ \dots + V_m) \cap V_{m+1}] \geq 0$. This tells us $dim(V_1+V_2+ \dots + V_{m+1}) \leq dim(V_1)+dim(V_2)+ \dots + dim(V_n)$
grizzled pagodaBOT
A dense set
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
quick spoke
How do mention induction in my proof
also yeah as was mentioned above, if ur doing a proof by induction, a very common thing to do is to begin with the words "we proceed via induction/we proceed by induction"
sometimes you want to mention explicitly what variable ur inducting on cus it's not always clear but here it's fine
indigo cloud
and it never hurts to also say "where we use our induction hypothesis" or something like that
quick spoke
i think personally, i would write it something like
We proceed via induction. The case $m=1$ is clear. Now, for $m \geq 2$,
\begin{align*}
dim(V_1 + V_2 + \cdots + V_{m+1}) &= dim(V_1 + V_2 + \cdots + V_m) + dim(V_{m+1}) - dim((V_1 + \cdots + V_m) \cap V_{m+1}) \
& \leq dim(V_1 + V_2 + \cdots + V_m) + dim(V_{m+1}) \
& \leq dim(V_1) + dim(V_2) + \cdots + dim(V_{m+1})
\end{align*}
by the induction hypothesis, and hence, we are done.
grizzled pagodaBOT
LY
indigo cloud
test $\dim$
grizzled pagodaBOT
Denascite
indigo cloud
oh this server even has it
indigo cloud
and now you can also try the hint given to you by bezier in your lin algebra thread
that proof is even easier
keen pawn
that proof is even easier
Do I attempt that proof here on in my thread
indigo cloud
up to you
keen pawn
keen pawn
We start off with a basis of $V$, say ${e_1,e_2,\dots, e_n}$.
\
We now reduce this to the basis of a $1$ dimensional vector space. To do so , we need to delete $n-1$ elemenents from the original basis . This can be done in $^{n}C_{n-1}=n$ ways. We thus have atleast $n$ one dimensional spaces of $V$.
\
As this is an existence proof, an example suffices. let $V_i = span{e_i}$. As all $e_is$ are LI, it follows that $V_i \cap V_j = {0}$ if $i \neq j$.
\
It's also evident by construction that $V =\sum_{i=1}^{n} V_i$.
\
As the only way to write $0$ is to set each $v_k$ to be zero, this sum is a direct sum, thus concluding our proof
grizzled pagodaBOT
A dense set
amber waspBOT
@keen pawn Has your question been resolved?
indigo cloud
wdym with v_k
keen pawn
wdym with v_k
Any arbitrary element belonging to a given vector space
keen pawn
$v_i \in V_i$
grizzled pagodaBOT
A dense set
amber waspBOT
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Available help channel!
unreal hollow
In an answering changing the subject questions would I loose marks for not including this sign in questions?
slender blaze
really depends on your teacher
but I guess most would
it’s kinda really important
unreal hollow
How can you tell where to put the sign because some awnsers have it and some don’t
slender blaze
ohh
that sign means there can be a positive and negative answer
so for square roots usually
x^2 = 16
the answer can be (+-)4
since both squared are 16
4, and -4
unreal hollow
So if the awnser is squat roots I should put that sign
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red pagoda
amber waspBOT
red pagoda
why does the 3rd row expand to 9(x+7)
quick ridge
$(3\sqrt{x+7})^2 = 9(x+7)$
grizzled pagodaBOT
knief
strange field
squared both sides
amber waspBOT
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sharp rapids
$\lim_{x \to c} \sqrt{x} = \sqrt{c}$
amber waspBOT
$\lim_{x \to c} \sqrt{x} = \sqrt{c}$
grizzled pagodaBOT
Halex
sharp rapids
Could someone teach me how to calculate these limits
$\forall \epsilon > 0, \exists \delta > 0 : |x - c| < \delta \implies |\sqrt{x} - \sqrt{c}| < \epsilon$
grizzled pagodaBOT
Halex
patent raptor
I think you could start multiplying sqrt(x)-sqrt(c) by the conjugate
sharp rapids
I know that $$|\sqrt{x} - \sqrt{c}| = \frac{|x-c|}{\sqrt{x} + \sqrt{c}|}$$
grizzled pagodaBOT
Halex
sharp rapids
my problem is setting the \delta in these kind of situations
because most of times we find that we have to choose a minimum and I don't get why
sharp rapids
I think you could start multiplying sqrt(x)-sqrt(c) by the conjugate
like this?
sharp rapids
you did that already lol
but as I said above I have problems finding delta
patent raptor
but as I said above I have problems finding delta
this may help
https://math.stackexchange.com/questions/1934477/prove-lim-x-to-a-sqrtx-sqrta-by-definition-stewart-solution-error
amber waspBOT
@sharp rapids Has your question been resolved?
full sinew
amber waspBOT
covert bear
This is
pallid canopy
can i just multiply limits with whatever i want
With legal algebra yes
full sinew
wdym with legal algebra
covert bear
Factorisation
lethal steppe
so generally i can multiply limits with whatever i want ?
Nothing was multiplied here
lethal steppe
They expanded the fraction by |x|
full sinew
is it like multiplying x/x
lethal steppe
Yeah
amber waspBOT
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brisk steeple
can someone help me figure out how to get the variable y? I got x which is 28, but idk how to get y
slender blaze
Isn’t 3y - 1 = 5x?
inner kernel
yes because they are External alternates
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rough totem
can you construct a regular nonagon?
amber waspBOT
can you construct a regular nonagon?
rough totem
(only using a straightedge and compass)
stoic ivy
maybe if you start with a triangle and somehow construct a parallelogram on each edge that might be a simpler problem
That would yield a nonagram I think
Well actually I guess you would need the parallelogram to have equal sides
So maybe it’s just as hard 😦
little widget
(only using a straightedge and compass)
a regular nonagon is not constructible with compass and straightedge (as 9 = 3², which is not a product of distinct Fermat primes)
see "Construction" section
little widget
so how do i remember doing it
it's not possible, so you didn't
little widget
ill try making one
bro is about to disprove the gauss wantzel theorem
little widget
is trisecting an angle impossible?
its impossible for most angles
little widget
alrigth, you have to tell me whats wrong
there is a part that takes 60° specifically as example
rough totem
are the red angles not actually equal?
little widget
idk
probabilly not, idk about this subject, i just know its impossible because of the theorem
rough totem
uhh prove that they re nt
pliant copper
the proof is in the wikipedia page
pliant copper
well the shape is symmetrical so the two outside are surely the same, so I suppose the middle is different. why do you think its the same?
pliant copper
are the red angles not actually equal?
okay I constructed it, the interior angle is larger than the outer two
its about 21.79 degrees
each outer angle is about 19.11
assuming you agree i drew your shape correctly
@rough totem does that answer your question
amber waspBOT
@rough totem Has your question been resolved?
stoic ivy
Wait if you construct an equilateral triangle and then on each side of the triangle you construct a hexagon, wouldn’t the envelope of those hexagons be a nonagon? What am I missing here
warm burrow
nice cicles
stoic ivy
Oh nvm I guess it wouldn’t have equal interior angles that’s the issue
pulsar coyote
what rule is this?
amber waspBOT
what rule is this?
warm burrow
hahaaa
shadow stump
basically we have [\lim_{x \to 0} 3x = 0] which means [ \lim_{x \to 0} = \lim_{3x \to 0} ]
grizzled pagodaBOT
cloud
amber waspBOT
amber waspBOT
Available help channel!
balmy flame
I need help with PART C of an AP Calculus BC FRQ question. It involves Macluarin Series and Intergals. I don’t think I’m on the right track with my work, so ignore what I put down so far.
amber waspBOT
I need help with PART C of an AP Calculus BC FRQ question. It involves Macluarin...
@balmy flame Has your question been resolved?
balmy flame
<@&286206848099549185>
balmy flame
<@&286206848099549185>
balmy flame
<@&286206848099549185> please 🙏
balmy flame
.close
strong brook
Hello