what grade are you in tho
#help-23
hard sparrow
mossy lotus
someone else might help you in that case
mossy lotus
what grade are you in tho
im beyond school
normal moss
Amazing, used like 30% of the otherwise required letters
mossy lotus
!done btw
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If you are done with this channel, please mark your problem as solved by typing .close
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thin lion
Hi for this
safe radishBOT
Hi for this
solar hazel
maybe
there is not much written beyond the problem statement and a definition though
so, hard to say
thin lion
Iād assume that would be enough since, if yn is greater than L then for some n large enough, yn will also be greater then L?
Hm
solar hazel
as a hint you might want to take a max of Nās somewhere
thin lion
Wait do I consider like N = max(N1,N2) here?
Alright š was on the right lines š„
Sorry Iāve occupied this channel
!occupied
lean otter
Sorry Iāve occupied this channel
Oh that's okay š
Then should I ask this on a different channel??
thin lion
š
thin lion
Oh that's okay š
Then should I ask this on a different channel??
Yea on the ones which arenāt used by others
lean otter
Yea on the ones which arenāt used by others
Okiii
thin lion
Gl ā¤ļø
lean otter
š
solar hazel
yea that looks pretty good
thin lion
yea that looks pretty good
š tysm
Low-key I shouldāve got that cause I realised max N after a bit
.close
sterile pilot
safe radishBOT
keen tulip
What have you tried?
lone tangle
keen tulip
What
keen tulip
numerically
Oh
Itās so silly
Find the left side lower bound and right side upper bound
digital sparrow
that is still not easy to do
keen tulip
Not too hard
digital sparrow
without calc
keen tulip
Thereās a trick
digital sparrow
oh
formal pollen
I think for finding k it may be useful to consider ranges with inputs in [0, 2pi], then multiple the amount of solutions by 5? i.e. that of sin, then sin(sin), etc. and the same for cos. Letting u = sin(x) and then considering sin(u), and iterating that. Maybe a graphical approach for finding k but that seems like a pain.
For a and b, maybe use that sec^2(x) = tan^2(x) + 1? Alternatively, use some substitution which allows you to manipulate sqrt(a) - sqrt(b) where a = sec^2(x) + tan(x) and b = sec^2(x) - tan(x). They definitely want you to cancel something out since the tan's have opposite signs within the roots
safe radishBOT
@sterile pilot Has your question been resolved?
formal pollen
Yeah, setting $s = \sqrt{\sec^2{x} + \tan{x}} - \sqrt{\sec^2{x} - \tan{x}}$ and then squaring might lead somewhere
flat frigateBOT
ericbernefan
digital sparrow
Yeah, setting $s = \sqrt{\sec^2{x} + \tan{x}} - \sqrt{\sec^2{x} - \tan{x}}$ and ...
||you just need to find the range of root (sec^2 + tan^2) bro||
formal pollen
I'm pretty sure what I suggested isn't mutually exclusive with that
regardless I think the sign difference likely means that they want you to use some kind of difference / sub / algebra
safe radishBOT
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quasi panther
Hi! Does someone knows about Hill cipher?
safe radishBOT
Hi! Does someone knows about Hill cipher?
safe radishBOT
No need to ask āCan I askā¦?ā or āDoes anyone know aboutā¦?āāitās faster for everyone if you just ask your question!
quasi panther
But I need to do a task for Monday and gemini/ChatGPT can not help me, and I don not find the information I need for my task anywhere
quasi panther
my teacher gave me an encrypted text and the original text, and I have to find the key matrix which is necessary to encrypt the text
quasi bison
do you know what the size of the key matrix is supposed to be
like is it 3, 4, bigger, unknown?
also do you know linalg
maybe post the full task in here @quasi panther
quasi panther
like is it 3, 4, bigger, unknown?
the size of the matrix n x n is 4<n<10
quasi bison
best as an image
so that we don't have to pull every little bit of info out of you like rotten teeth
quasi bison
Im spanish so I think you wont understand my task
post it anyway
if i need a translation i will ask
quasi panther
Thats the original text and the explanation of the task
quasi panther
And thats the encrypted text
quasi panther
And thats the encrypted text
Maybe its better to send it by message instead of photo
quasi bison
why is the ciphertext a whole bunch of numbers that aren't even in [1, 27]
quasi bison
yes, this makes perfect sense to me
quasi panther
quasi bison
quasi panther
and this is the encrypted text
quasi bison
hm ok
so i guess we're not even taking any modulo stuff but rather just working with integers raw
this actually kinda makes things easier, i think...
so it says your encryption matrix is n Ć n with 5 ā¤ n ā¤ 9, yes?
i wonder why they said 4 < n < 10 and not 5 ā¤ n ā¤ 9... is it to make things more difficult to read or what
quasi panther
quasi panther
this is the unencrypted text
quasi panther
Im studying informatics engineering
quasi bison
that doesn't answer my question either way
quasi panther
Yes, Im studying linear algebra
quasi panther
but thats a big and difficult task and I dont know how to solve it
quasi bison
give me a second. i am about to show you how to reduce it to a linalg problem
quasi panther
ok
quasi bison
for example, for n=3 (i know it goes against the problem statement but just for example's sake) you would start with
[ 12 16 4 | 127 -60 -212 ]
[ 1 12 1 | -178 -37 -169 ]
[ 20 5 4 | 49 -115 75 ]
for this i pulled the first 9 numbers from message text
and the first 9 from ciphertext
and packaged tem as i showed
@quasi panther does this make sense to you
quasi panther
okay, and whit that matrix what would I have to do?
quasi bison
what matrix. the one i wrote?
quasi panther
for example, for n=3 (i know it goes against the problem statement but just for ...
this one
quasi panther
so I transform the left matrix into the identity and what would I have to do with the right one?
quasi bison
you apply row operations to the entire "double" matrix
wait
ah shit i fucked this up
you need to write the text in ROW-first, not column-first. my bad.
and then at the end, you need to transpose whatever you end up with in the right half
and THAT will be your K
quasi bison
you apply row operations to the _entire_ "double" matrix
but this still stands, so they both get transformed in the same way
[ 12 1 20 | 127 -178 49 ]
[ 16 12 5 | -60 -37 -115 ]
[ 4 1 4 | -212 -169 75 ]
it would be like this
quasi bison
yes, but you perform your row operations on the entire double matrix
so the right matrix transforms in sync with it
quasi panther
yes, i understand that
quasi bison
then, when you get to the identity, what ends up on the right will be K^T
quasi panther
and I transpose K and that would be the key matrix?
quasi panther
okay
I will try it, thank you very much
Ive been trying all day to do this task and I havent found any help
quasi bison
Ive been trying all day to do this task and I havent found any help
i wonder what your attempts were tbh
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graceful lichen
$$
\text{The vertices of a triangle are }
A(x_1,; x_1\tan\alpha),;
B(x_2,; x_2\tan\beta),;
C(x_3,; x_3\tan\gamma).
$$
$$
\text{If the circumcentre of } \triangle ABC \text{ is the origin and }
H(a,b) \text{ is the orthocentre, then } \frac{a}{b} \text{ is equal to:}
$$
$$
\text{(A)}\quad
\frac{\cos\alpha+\cos\beta+\cos\gamma}
{\cos\alpha\cos\beta\cos\gamma}
$$
$$
\text{(B)}\quad
\frac{\sin\alpha+\sin\beta+\sin\gamma}
{\sin\alpha\sin\beta\sin\gamma}
$$
$$
\text{(C)}\quad
\frac{\tan\alpha+\tan\beta+\tan\gamma}
{\tan\alpha\tan\beta\tan\gamma}
$$
$$
\text{(D)}\quad
\frac{\cos\alpha+\cos\beta+\cos\gamma}
{\sin\alpha+\sin\beta+\sin\gamma}
$$
safe radishBOT
$$
\text{The vertices of a triangle are }
A(x_1,\; x_1\tan\alpha),\;
B(x_2,\; x_...
flat frigateBOT
BlackidoZĪ£
graceful lichen
as H(ortho) G(centroid) and O' (circumcenter) are collinear and the ratio 2:1
i found G(x,y) = a/3 , b/3
quasi bison
jsyk you dont need to put a million \text blocks inside dollar signs
you make your own LaTeX life difficult this way
quasi bison
then write on paper and send picture lol
graceful lichen
already i am short on paper
the rough copy is about to end
further i found x = sigma(x1)/3 and y = sigma(x1 tan(alpha)) / 3 using formula of centriod coordinates
and dividing it i get a/b = sigma(x1)/sigma(x1 tan(alpha))
which doesn't match to any of the option
quasi bison
The vertices of a triangle are $A(x_1, x_1\tan(\alpha)$, $B(x_2, x_2\tan(\beta)$, $C(x_3, x_3\tan(\gamma))$.
If the circumcenter of $\triangle ABC$ is the origin and $H(a,b)$ is the orthocenter, then $\frac{a}{b}$ is equal to:
\begin{enumerate}[(A)]
\item $\frac{\cos\alpha+\cos\beta+\cos\gamma}{\cos\alpha\cos\beta\cos\gamma}$
\item $\frac{\sin\alpha+\sin\beta+\sin\gamma}{\sin\alpha\sin\beta\sin\gamma}$
\item $\frac{\tan\alpha+\tan\beta+\tan\gamma}{\tan\alpha\tan\beta\tan\gamma}$
\item $\frac{\cos\alpha+\cos\beta+\cos\gamma}{\sin\alpha+\sin\beta+\sin\gamma}$
\end{enumerate}
flat frigateBOT
Ann
graceful lichen
quasi bison
at least now the problem is visible
quasi bison
fucked up bracketing on the point specification but i have run out of shits to give
flat frigateBOT
quasi bison
mine is set to light
graceful lichen
,textconfig color
flat frigateBOT
BlackidoZĪ£
graceful lichen
bruh
graceful lichen
,texconfig color
flat frigateBOT
graceful lichen
how i can change the config to white background and black text
quasi bison
,texconfig color white
or if you want it to look like mine,
then ,texconfig color light
graceful lichen
,texconfig color light
flat frigateBOT
You have switched to the light colourscheme.
graceful lichen
$$
\text{The vertices of a triangle are }
A(x_1,; x_1\tan\alpha),;
B(x_2,; x_2\tan\beta),;
C(x_3,; x_3\tan\gamma).
$$
$$
\text{If the circumcentre of } \triangle ABC \text{ is the origin and }
H(a,b) \text{ is the orthocentre, then } \frac{a}{b} \text{ is equal to:}
$$
$$
\text{(A)}\quad
\frac{\cos\alpha+\cos\beta+\cos\gamma}
{\cos\alpha\cos\beta\cos\gamma}
$$
$$
\text{(B)}\quad
\frac{\sin\alpha+\sin\beta+\sin\gamma}
{\sin\alpha\sin\beta\sin\gamma}
$$
$$
\text{(C)}\quad
\frac{\tan\alpha+\tan\beta+\tan\gamma}
{\tan\alpha\tan\beta\tan\gamma}
$$
$$
\text{(D)}\quad
\frac{\cos\alpha+\cos\beta+\cos\gamma}
{\sin\alpha+\sin\beta+\sin\gamma}
$$
flat frigateBOT
BlackidoZĪ£
graceful lichen
okay its understood
quasi bison
but i guess we give no shits about solving except to like bashbashbash
who needs to think
who needs to know what is asked
nobody right??
graceful lichen
yes
quasi bison
bashbashbash
graceful lichen
lmao
graceful lichen
hehe
quasi bison
im not solving anything together with you if you demonstrate such disrespect towards questions
YOUR OWN questions
graceful lichen
GOD please solve this question i am already irritated by it
lol
there is even no solution available for it on the iternet
can i get help pelase
$$
\text{The vertices of a triangle are }
A(x_1,; x_1\tan\alpha),;
B(x_2,; x_2\tan\beta),;
C(x_3,; x_3\tan\gamma).
$$
$$
\text{If the circumcentre of } \triangle ABC \text{ is the origin and }
H(a,b) \text{ is the orthocentre, then } \frac{a}{b} \text{ is equal to:}
$$
$$
\text{(A)}\quad
\frac{\cos\alpha+\cos\beta+\cos\gamma}
{\cos\alpha\cos\beta\cos\gamma}
$$
$$
\text{(B)}\quad
\frac{\sin\alpha+\sin\beta+\sin\gamma}
{\sin\alpha\sin\beta\sin\gamma}
$$
$$
\text{(C)}\quad
\frac{\tan\alpha+\tan\beta+\tan\gamma}
{\tan\alpha\tan\beta\tan\gamma}
$$
$$
\text{(D)}\quad
\frac{\cos\alpha+\cos\beta+\cos\gamma}
{\sin\alpha+\sin\beta+\sin\gamma}
$$
flat frigateBOT
BlackidoZĪ£
safe radishBOT
graceful lichen
Show your work
i typed it
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nimble vine
Where?
graceful lichen
as H(ortho) G(centroid) and O' (circumcenter) are collinear and the ratio 2:1
Starting from here !
graceful lichen
which doesn't match to any of the option
ending here
nimble vine
Ic
mossy lotus
you speak like any random triangle with coordinates like that can have their circumcenter at origin
you should get another condition where you match the coordinates of the circumcenter and the origin
(or atleast you can get a set of equations considering the fact that OA, OB, OC are all equal)
graceful lichen
what to do now!
mossy lotus
and then simplify to eliminate the x1 x2 x3
graceful lichen
i didn't get what you text
graceful lichen
yes
mossy lotus
good
then can you tell me whatis the significance of that point?
what special characteristics it exhibits?
graceful lichen
by the definition it is intersection point of the perpendicular bisector
graceful lichen
yes that also
mossy lotus
coz they are all circumradius distance away from the circumcenter
feral linden
If you assume x_i/cos(a_i) (a_1, a_2, a_3 are Ī±,Ī²,Ī³) are all positive then your problem is solvable, it becomes one of the options
Two facts you can use
One: |OA|=|OB|=|OC|
Two: OH=OA+OB+OC
nimble vine
If you assume x_i/cos(a_i) (a_1, a_2, a_3 are Ī±,Ī²,Ī³) are all positive then your ...
in vector form
feral linden
Yes
(And (x, xcos(a))=(x/cos(a)) (cos(a), sin(a)), so it is reasonable to assume this perimeter x/cos(a) represents length, that is positive)
(I am not saying itās true. I am saying itās not unreasonable to assume that and it let your problem be solvable)
nimble vine
(I am not saying itās true. I am saying itās not unreasonable to assume that and...
Given that there are no x terms in options
feral linden
Yeah x we just choose it to align with the symbol of cos(a), his question wasnāt given in full anyway
x_i :=radius of the outer circle times cos(a_i)
Instead of the - of that
iron estuary
yeah the idea works the only thing is the sign assumption isnāt essential the vector identity already forces the answer
feral linden
Not sure. I used x_i =R cos(a_i) for all i. If some of them are x_j=-Rcos(a_j) it will be really messy
Instead of +++/+++ any combination like +-+/+-+ appears. No option is of this form thatās why I think we need to assume that
Anyway, after we assume this the answer is clear
iron estuary
the sign choice just fixes a parametrization it doesnāt affect a/b since the common factor cancels
safe radishBOT
@nimble vine Has your question been resolved?
buoyant bear
ABOC is a square, and is equal to the area of ADC It's asking for A's coordinates
safe radishBOT
ABOC is a square, and is equal to the area of ADC It's asking for A's coordinate...
zealous ingot
So CO and OD are equal
right
buoyant bear
I did some math idk if its correct
kind seal
So CO and OD are equal
If the areas are given equal and two angles is equal, the corresponding sides are equal. So you're correct
zealous ingot
what does the question want you to find?
buoyant bear
The coordinates of A
So y and 2x are both 5 since they are equal
Wait thats K's coordinates mb
hardy panther
i think the ans is c
zealous ingot
i think the ans is c
!noans
safe radishBOT
i think the ans is c
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.
safe radishBOT
i think the ans is c
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.
zealous ingot
nosols rather shit
kind seal
ooh lol. mb
buoyant bear
I dont really care about the answer tbh I'm studying for my exam
hardy panther
oops srry
zealous ingot
I dont really care about the answer tbh I'm studying for my exam
ok, can you tell me the coordinates of D and A, in terms of x and y
zealous ingot
I use the butterfly ratio thing?
thats for stuff in circles
I was thinking more along the lines of equation of line
zealous ingot
Wait I thought x was 2,5 and y is 5
you can't say that, we don't have information to conclude that
kind seal
ABCD is a square right? Just confirming
zealous ingot
bingo
buoyant bear
m = 1/2?
buoyant bear
This is all I could do
kind seal
Try putting these numbers in the line equation. The slope precisely
buoyant bear
So x is 1,5?
kind seal
y=mx+c, m is known and you have a point (x,y)
zealous ingot
This is all I could do
whats the y coordinate of D?
zealous ingot
W
kind seal
Can I ask another one?
Yes but try it one more time. Now that you see the techniques you used, you might be able to get it
zealous ingot
sure
buoyant bear
The only thing I could find is that this angle is 90 degrees
buoyant bear
Yes but try it one more time. Now that you see the techniques you used, you migh...
I already tried these thats why im sending them
kind seal
It is 90 degrees but can you reason that. You're right though
buoyant bear
yeah because -1 Ć 1 is -1
kind seal
You mean the product of slopes?
buoyant bear
Yeah when you multiply them
kind seal
$-\frac1{k} \times k = -1$, if you mean this by 1 x -1, then youre right
flat frigateBOT
Adhi
buoyant bear
Yeah
buoyant bear
The slope?
zealous ingot
The slope?
don't bother with that
you got intersection as being 90 deg right? that's sufficient for the slope part
can you think of some relation between perpedicular lines ||hint: semicircle related||
buoyant bear
Yeah I have no idea
zealous ingot
angles subtended on circumference by diameter of a semicircle/circle is always 90 degree
buoyant bear
Yeah but how's that gonna help
zealous ingot
you can translate the problem onto that right?
buoyant bear
I try forming a triangle?
zealous ingot
(a,b) is the point on the circumference
buoyant bear
It's just the point where they intercept
zealous ingot
the points on the x-axis where the lines pass through are endpoints of the diameter
buoyant bear
I don't know how I can use that we havent learned the formulae for circles
kind seal
You don't need formulae exactly
You should just know that the diameter is twice the radius
buoyant bear
Oh
So they are equal
I still dont get how you could form a semi-circle right there
Do we know that those sides are equal
nimble vine
I still dont get how you could form a semi-circle right there
if you check the x intercepts they are (-4,0) and (4,0)
yes they are equal
kind seal
Even if they weren't though you could find the diameter and half it. A definite number is enough
buoyant bear
You inserted 0 for y to find where they intercept?
kind seal
Yes that's what he did
nimble vine
and when you take the angle to be 90 you know that the itercepts form diameter
buoyant bear
Mb
kind seal
b does not lie on the y axis
b would be if it were on the y axis
Do you know the Pythagorean theorem? @buoyant bear
buoyant bear
=1?
nimble vine
Now a^2+b^2 is the pythagorean distance
you need this
nimble vine
how come 36?
buoyant bear
Mb 64
kind seal
?
nimble vine
OA is a radius
buoyant bear
?
nimble vine
yes semicircle should pass through A but ok
buoyant bear
Oh yeah mb
nimble vine
np
buoyant bear
.close
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Available help channel!
fathom plume
pls 2a help š
safe radishBOT
pls 2a help š
fathom plume
so $x\cdot x^{-1}=e$ and $x\cdot x^{n-1}=e$ hence ik that $x^{n-1}=x^{-1}$
flat frigateBOT
Cera
fathom plume
but idk what that implies abt the order of $x^{n-1}$ if im thinking in the right direction at all
flat frigateBOT
Cera
cedar widget
You can take the case of Z/nZ to see what the answer should be
fathom plume
so i come up w any group and try it out?
cedar widget
Yes, and a cyclic group is the simplest
fathom plume
i was unsure abt what you meant by z/nz so i used a generator of $e^{\frac{i\pi}{2}}$ , $\abs{e^{\frac{i\pi}{2}}}$ being 4 and inverse being 4, thus suggesting n?
flat frigateBOT
Cera
fathom plume
solution seems a bit diff tho
that first step looks like an abuse of notation idk how they put the n and -1 together
since -1 is just notation for inverse
cedar widget
i was unsure abt what you meant by z/nz so i used a generator of $e^{\frac{i\pi}...
Z/nZ as a group is cyclic of order n
fathom plume
Z/nZ as a group is cyclic of order n
ooh ok
cedar widget
Its integers modulo n
fathom plume
Its integers modulo n
so smt like $(G,+_{3})$?
flat frigateBOT
Cera
fathom plume
mb if its rlly obv im not g w notation rn
cedar widget
I don't know what is this notation
fathom plume
I don't know what is this notation
group of integers under addition modulo 3
cedar widget
(x^-1)^n=e so x^-1 has order at most n
fathom plume
I don't know what is this notation
my textbook seems to rlly like it
so id imagine thats how im expected to write it
fathom plume
(x^-1)^n=e so x^-1 has order at most n
thats what i dont get tho why does x^-1 have at most order n?
cedar widget
More common notation is Z/nZ, Zn and Cn
cedar widget
thats what i dont get tho why does x^-1 have at most order n?
By definition of order
Its the minimal integer 0<m such that (x^-1)^m=e
fathom plume
but order of x is also n?
so we saying that order of inveres of x has to be smaller than order of x?
or eq
cedar widget
Its not an assumption
fathom plume
theres def some math to prove it but i rlly cba and prob cant understand it anyway š
since rn it looks like they pulled the 'at most n' out of nowhere
cedar widget
Its really just the definition, as the order is just the minimal natural number satisfying some condition. So if you know n satisfies that condition, then the minimum must be smaller
fathom plume
i get where thats coming from, but we are imposing the coniditon on a unique element
so idk why it holds for x^-1 given x
also isnt 2b just based off associativity?
cedar widget
Its pretty straightforward from the axioms of a group
mortal sandal
You could say that yeah
fathom plume
Its pretty straightforward from the axioms of a group
is that referencing 2b?
because textbook wants me to do induction for that
š
aight tysm yall
.close
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carmine tulip
How to draw a line with the length of $\sqrt[3]{2}$ units on a paper?
flat frigateBOT
fort craft šµšø
brave wolf
How to draw a line with the length of $\sqrt[3]{2}$ units on a paper?
using what?
Straightedge and compass?
carmine tulip
Straightedge and compass?
Sorry English isn't my first language. What is a straightedge?
brave wolf
Sorry English isn't my first language. What is a straightedge?
ruler
carmine tulip
sure with a ruler and that thing used for drawing circles.
quasi bison
that one is called a compass
carmine tulip
I remember that my teacher taught us how to draw square root of 2 on a paper but not the cube root of 2.
carmine tulip
that one is called a compass
I thought compass was the thing which shows directions like north, south etc.
quasi bison
it's kinda both
sometimes the circle-making tool is called a "pair of compasses"
but yes compass is also the word for the tool that shows you where north is
quasi bison
I remember that my teacher taught us how to draw square root of 2 on a paper but...
cbrt(2) is impossible with compass & straightedge
brave wolf
I remember that my teacher taught us how to draw square root of 2 on a paper but...
it's impossible to do it with just straightedge and compass, using just standard "rules"
the wikipedia contains this construction, which however involves moving around the ruler until it fits (which isnt really allowed in standard geometry constructions, you can only connect 2 points with a line)
I'd suggest u to read the wikipedia page, it probably has everything u need to know
and it might be available in your language
carmine tulip
it's impossible to do it with just straightedge and compass, using just standard...
For the square root we drew a triangle with the sides having a length of 1 and the angles being 90 and two 45s and the big side was square root of 2. we did this on the number line and it worked.
brave wolf
For the square root we drew a triangle with the sides having a length of 1 and t...
Yes, but there isnt any construction like this for the cube root
in fact, the only lengths you can construct are those, which can be made only using +, -, *, / and sqrt
carmine tulip
and it might be available in your language
My english ain't that bad but I am just not familiar with math terms. Especially with those that are about geometry.
brave wolf
so for example you can make sth with length
sqrt(3^2 / 5 + 6 * sqrt(4 / 5))
but not
4*pi or cbrt(2)
carmine tulip
so for example you can make sth with length
sqrt(3^2 / 5 + 6 * sqrt(4 / 5))
but...
I mean if you take pi as 3 you should be able to do it.
brave wolf
Yes, but pi isnt 3
quasi bison
pi is pi, it is not 3, it is not 22/7, it is not 314/100, it is pi
carmine tulip
pi is pi, it is not 3, it is not 22/7, it is not 314/100, it is pi
ik. PI is irrational.
brave wolf
you are right that you can draw a pretty good approximation
quasi bison
ik. PI is irrational.
irrational doesn't mean not constructible
quasi bison
Pi has no explicit formula unlike e
hmmmmm what do you mean by explicit formula
if you allow infinite series then both pi and e can be expressed
brave wolf
Pi has no explicit formula unlike e
and how does e have one
carmine tulip
irrational doesn't mean not constructible
Can you elaborate?
quasi bison
if you don't then neither can
regal badge
hmmmmm what do you mean by explicit formula
e can be written as an infinite fraction, as an infinite sum, as a limit
Pi can't
quasi bison
Can you elaborate?
sqrt(2) is also irrational but very easy to construct
carmine tulip
and how does e have one
With limits I guess.
quasi bison
4 - 4/3 + 4/5 - 4/7 + ...
brave wolf
look up "pi series", youll find a bunch of them
carmine tulip
sqrt(2) is also irrational but very easy to construct
Yeah that is a good example.
quasi bison
that indeed you were
brave wolf
and e cant be constructed either btw, because again, it cant be written only using rationals, +, -, *, / and sqrt (as a finite combination, so no infinite sums)
carmine tulip
So are you guys pulling a read the manual card to me? That is, by not telling me how to draw it and just sending a Wikipedia article.
brave wolf
If you are familiar with the circle equation x^2 + y^2 = r^2, i could give you a rough explanation of why it's not possible
carmine tulip
the wikipedia contains this construction, which however involves moving around t...
@quasi bison doesn't this photo contradict what you are saying?
quasi bison
kinda bc im omitting the words 'compass and straightedge'
you need SOME kind of more advanced tool for it
carmine tulip
If you are familiar with the circle equation x^2 + y^2 = r^2, i could give you a...
I am not that familiar. I have seen it before tho.
carmine tulip
you need SOME kind of more advanced tool for it
Such as?
quasi bison
the wikipedia contains this construction, which however involves moving around t...
marked ruler, as per this image
brave wolf
<@183668144404037632> doesn't this photo contradict what you are saying?
the 6th step requires moving the ruler around, until the marked length falls on something, which is not allowed in classical geometry.
In classical geometry, straightedge can be used only to connect 2 points with a line
carmine tulip
marked ruler, as per this image
Isn't that just a ruler?
carmine tulip
classical constructions allow an unmarked one only
Wait what?! How can you draw a specific length using an unmarked ruler?
quasi bison
Wait what?! How can you draw a specific length using an unmarked ruler?
you don't
you start with one segment that you treat as having length 1 by declaration
faint seal
you can copy lengths using the compass
quasi bison
everything else is reckoned in relation to that
carmine tulip
We always used marked ones. (To be more accurate we didn't use any rulers at all but we were told that the proper way to do it, was by using a marked ruler)
brave wolf
Wait what?! How can you draw a specific length using an unmarked ruler?
U can start with a line segment of unit length. That segment is said to have length "1". Then you can for example double it to get 2, triple it to get 3, then you can use similar triangles to achieve length 1/3... and this way you can make any rational length. And by adding some clever constructions, you can make square root lengths as well
carmine tulip
you start with one segment that you treat as having length 1 by declaration
that is obvious. that is what we mean by unit or units
carmine tulip
everything else is reckoned in relation to that
But how can you say this is twice that without having a marked ruler? By eye?
quasi bison
But how can you say this is twice that without having a marked ruler? By eye?
no lmao
you can construct a segment that's 2 times your unit easily
then it will be provably so
carmine tulip
you can construct a segment that's 2 times your unit easily
How so?
maybe you use the compass. idk
quasi bison
start with point A and line l.
draw a circle of radius 1 unit centered at A.
see it intersect l at two points. call one of them B.
then draw a circle of radius 1 unit centered at B. see it intersect l at A and another point C.
then AC=2
carmine tulip
start with point A and line l.
draw a circle of radius 1 unit centered at A.
see...
that makes sense. But why there is an arbitrary restriction of using an unmarked ruler in classical constructions?
faint seal
because it's the simplest possible model for constructing geometric diagrams
carmine tulip
because it's the simplest possible model for constructing geometric diagrams
ok. I can kinda see why that would be something desirable.
quasi bison
blame euclid
brave wolf
It might be that the point of classical geometry was to construct theoretically perfect constructions and straightedge and compass were probably the best tools for that. E.g. the cbrt construction involves placing the marked ruler on the piece of paper, and then moving it around until the marks align nicely, this step can create a pretty large error, and so they probably rejected this
Classical geometry started in ancient greece, so you can probalby imagine that their tools werent perfect either, so it was prolly necessary to make the tools as simple as possible
carmine tulip
Classical geometry started in ancient greece, so you can probalby imagine that t...
If I am not mistaken, they even lacked a proper way to represent numbers unlike us who have the decimal system.
brave wolf
Well, they had a way to represent numbers, but it was certainly not as nice as our decimal system
and iirc, they had basically no algebra. So geometry was pretty much the only thing they did
carmine tulip
and iirc, they had basically no algebra. So geometry was pretty much the only th...
I mean geometry is cool. You can see stuff visually.
oblique thistle
How can I find the angle of AED? I need it to find the length of AD
quasi bison
you mean arc AD?
oblique thistle
Yes
quasi bison
consider that triangle ADE is isosceles
and the other angles in it are findable by basic trig
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cedar widget
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ā Original question: #help-23 message
frigid spruce
you can't reopen a channel that wasn't yours unless you have helpful role bibbly blud
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oblique thistle
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ā Original question: #help-23 message
cedar widget
xD
oblique thistle
Can someone explain to me why the answer is wrong when I use the chord formula to find the angle of AEB?
cedar widget
What formula are you using
finite igloo
${2 r \sin \left( \frac{\theta}{2}\right)}$
flat frigateBOT
k
finite igloo
ts?
oblique thistle
Yeah
finite igloo
how did u find the angle
oblique thistle
Which angle?
oblique thistle
I used that formula to find the angle AEB but the answer is wrong
No since I already know the chords length
I'm using the formula to find the angle
oblique thistle
Wait why I got 97.17
oblique thistle
This gives 112Ā°
That's the answer
cedar widget
12 = 2 * 7.21 sin(a/2)
oblique thistle
Oh wait nvm I made a common mistake
finite igloo
oh
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analog coral
hello
keen tulip
Hello
magic junco
but i dont know how to say it so ill just write it
Do you have a pic of it?
safe radishBOT
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.
thin bridge
what are those things in the denominator?
both 4?
analog coral
what are those things in the denominator?
both 4?
yes
faint hornet
how did you get -6?
thin bridge
no, the answer isn't -6
analog coral
no, the answer isn't -6
i knew chatgpt was wrong
analog coral
how did you get -6?
5-11
keen tulip
I think ChatGPT would have gotten this one right
thin bridge
5 - 11 is indeed -6
but the rest of the expression doesn't just vanish into thin air
faint hornet
okay, but the question is not 5-11 exactly
analog coral
I think ChatGPT would have gotten this one right
it did the same thing on the other question i asked
analog coral
okay, but the question is not 5-11 exactly
then what do i have to do?
thin bridge
$\frac ac - \frac bc = \frac{a-b}{c}$
flat frigateBOT
ĻĪ±Ī¼Omeganato5
analog coral
thin bridge
yes
quasi bison
!noai and all but i do really wonder how you could get it to fuck sth like this up
safe radishBOT
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).
thin bridge
nowhere did that say what you wrote was -6
quasi bison
Š° Š“ŃŠ¾Š±Š½ŃŃ ŃŠµŃŃŃ ŠŗŃŠ¾ ŃŃŠ°Š²ŠøŃŃ Š±ŃŠ“ŠµŃ?
Š“Š° Šø ŃŠøŃŠ»Š° ŃŃŃ Š½Šµ ŃŠµ
analog coral
Š° Š“ŃŠ¾Š±Š½ŃŃ ŃŠµŃŃŃ ŠŗŃŠ¾ ŃŃŠ°Š²ŠøŃŃ Š±ŃŠ“ŠµŃ?
idk why it said it in bulgarian im macedonian
quasi bison
oh wait
nimble vine
lmao
nimble vine
lmao
analog coral
oh shit
quasi bison
i forgor you were macedonian
analog coral
i didnt notice
primal bone
idk why it said it in bulgarian im macedonian
Because it's not sufficient information to determine the language; this would be the same in Russian
primal bone
Yh, it'll default to what it suspects the language would be
nimble vine
but gpt's answer is in Russian
quasi bison
"and who's gonna write the slash for the fractions?"
"and the numbers aren't the same ones here either"
primal bone
yh that's the other thing tbh
quasi bison
this is what i said
quasi bison
https://tenor.com/view/russia-communism-communist-cat-cat-meme-gif-1077514437313...
i am literally from Russia
nimble vine
ic
primal bone
In any case, this isn't the same as the question you'd asked here
analog coral
In any case, this isn't the same as the question you'd asked here
yea ik
i just followed those steps
primal bone
"followed"
quasi bison
anyway
fierce kernel
!noai and all but i do really wonder how you could get it to fuck sth like this ...
AI is really, really bad with fractions. chatgpt especially
primal bone
yes
quasi bison
sure
thin bridge
and 6/4 can be simplified further
analog coral
no I mean you edited the misspelling, and still misspelt it
IK
how would this be?
austere goblet
if in doubt convert everything into improper fractions
analog coral
if in doubt convert everything into improper fractions
quasi bison
convert all mixed numbers like $2 , \frac{3}{4}$ into improper fractions, then do the operations one at a time
flat frigateBOT
Ann
austere goblet
LCD is correct
but 10/6 is a yikes for the first fraction
the latest pic is correct though
analog coral
this one is right?
nimble vine
this one is right?
yes
quasi bison
analog coral
https://youtu.be/8ZbIrASBj54 btw
bro i asked my teacher abt this and she said it was wrong
quasi bison
don't call me "bro"
analog coral
don't call me "bro"
mb im used to saying that im sorry
analog coral
did she say anything SPECIFIC was wrong?
no she said its wrong and showed me the right way
quasi bison
ok so she just said the entire video is wrong.
that's strange and not very informative from her but ok.
analog coral
but i think the fractions need to specific to be right
analog coral
ok so she just said the entire video is wrong.
i just showed her for + not the other ones
she told me to use smth we call ŠŠŠ”
its basicly this
primal bone
You're naming how to add shit, not a governmental body
analog coral
analog coral
_You're naming how to add shit, not a governmental body_
tf does that even mean
primal bone
tf does that even mean
Calling this something like H3C feels like overkill for what is a fairly simple thing, is what I was getting at
quasi bison
Calling this something like H3C feels like overkill for what is a fairly simple ...
nah it's Cyrillic
NZS
which stands for something in Macedonian
primal bone
https://youtu.be/8ZbIrASBj54 btw
In any case, neither nor is wrong
Both involve putting the fractions under SOME common denominator
Whether it's the lowest or not is not that important as you think, while you're in the middle of computation
safe radishBOT
@analog coral Has your question been resolved?
nimble vine
Whether it's the lowest or not is not that important as you think, _while you're...
it matters when you are being graded
cerulean socket
no she said its wrong and showed me the right way
The thing is, let's say you want to add a/b+c/d.
That means you must multiply the numerator and denominator by something by something to make the denominators common, so you can factor it out and add the numerators.
So, that means you can either:
Make both denominators bd, so it becomes:
ad/bd + bc/bd = 1/bd(ad+bc)=(ad+bc)/bd
Or, you can multiply by LCM(b,d). The LCM is the smallest number such that b|LCM(b,d) and d|LCM(b,d), so then let, for some integers k1,k2,
bk1=LCM(b,d) -> equation 1
dk2=LCM(b,d)[this must be possible because LCM is a multiple of both b and d by definition. That is why it is called least common multiple]. -> equation 2
So, now, our fractions becomes:
ak1/LCM(b,d)+ck2/LCM(b,d)
[note that by equation 1 and 2, k1/LCM(a,b)=1/b, so ak1/LCM(b,d)=a/b, which gives us back our fraction, so doing this doesn't change our equation].
=(ak1+ck2)/(LCM(b,d)
This is why both approaches work.
nimble vine
The thing is, let's say you want to add a/b+c/d.
That means you must multiply t...
pov tldr, it says both right
primal bone
POV this is not what POV means
cold ridge
yeah, habit worth breaking
it ain't that deep honestly
unless you're a she
then yea
analog coral
Well I don't mind getting called princess or or sister honestly
primal bone
unless you're a she
I mean, the ā¦ļø next to her name?
nimble vine
unless you're a she
nah this gonna spark a debate
analog coral
.Close
cold ridge
I mean, _the ā¦ļø next to her name_?
oh true lol
analog coral
.close
quasi bison
unless you're a she
. i am, in fact, a she.
. that said, the simple golden rule of "if someone tells you not to call them [word], you RESPECT THAT and don't call them [word]" applies to anybody of any gender.
versed wave
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faint hornet
it was just an easy value to use to find C
versed wave
so any n works?
faint hornet
yes
mossy lotus
where did n = 2 come from?
its not that they used n=2, its that they divided both sides by 4^(n-2) and the next step is the simplified terms
but yea, in the end you can treat it as using some specific value
versed wave
ah
mossy lotus
its jus tthat stuff cancels out even then
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heady moss
Question 65
safe radishBOT
Question 65
heady moss
I need help with question 65 please. Iāve simplified the original function and did a u substitution tan x = t and tried solving the integral I1. I changed the intervals to 0-1 after taking said substitution and it simplified to 0. Tried the same thing with I2 and that also simplified to 0. Wondering what to do here.š
open wedge
Question 65
,rccw
flat frigateBOT
feral linden
Substitute u=tan(x), du=dx/cos^2(x)=(dx)(1+u^2)
This can let you calculate I1, though I2 I guess you have to use recursion on integral of arctan(u) u^n du
heady moss
uhhh yeah as I said it let me solve I1
I2 aswell but it was tedious
and it still simplified to 0
do u wanna see the work?
feral linden
Sure. Maybe I can spot where the error occurs
heady moss
okay! hold on
dusky harbor
just poking by, have you remembered adapting the bounds?
open wedge
,rccw
flat frigateBOT
heady moss
just poking by, have you remembered adapting the bounds?
yes! although please do check for any errors š
heady moss
okay let me redo I2
safe radishBOT
@heady moss Has your question been resolved?
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iron wolf
Can someone please explain to me why they didnt just put these equations from the question into a simplex table right away? instead of doing all that substituting at the start of the solution
safe radishBOT
Can someone please explain to me why they didnt just put these equations from th...
@iron wolf Has your question been resolved?
fathom jewel
Can someone please explain to me why they didnt just put these equations from th...
Because when you assume that the basic variables of the optimal basic feasible solution are x2 and x3 then the non-basic variables are automatically 0
iron wolf
Because when you assume that the basic variables of the optimal basic feasible s...
That makes sense, but the solution starts off with x1 and x2 as 0? Not x1 and x4
safe radishBOT
fathom jewel
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ā Original question: #help-23 message
fathom jewel
That makes sense, but the solution starts off with x1 and x2 as 0? Not x1 and x4
You could have put it right a way into a simplex, but it seems to me they chose some feasible basis by choosing x1=x2=0 and did the a "simplex iteration" algebraically
iron wolf
You could have put it right a way into a simplex, but it seems to me they chose ...
I guess that makes sense. Thank you
.close
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eternal breach
can the exponential integral function be calculated ?
safe radishBOT
can the exponential integral function be calculated ?
nimble vine
can the exponential integral function be calculated ?
Yes
eternal breach
do you mean like e^x?
e^x/x
faint hornet
no
faint hornet
or at least, not in the way you are thinking at least
there is no "nice way" to write it
you can approximate it but that's it
eternal breach
you can approximate it but that's it
yeah that's the thing i couldn't find any exact solution
faint hornet
yeah you will not find one
eternal breach
well thanks tho
eternal breach
.close
opal cypress
I'm having some trouble understanding eigenvectors and eigenvalues, consider a matrix with columns [3 0] and [1 2], using the "circle to ellipse" analogy, the axes of the ellipse should correspond to the eigenvectors of the linear transformation however it's not the case for this particular matrix and I don't quite understand why
safe radishBOT
I'm having some trouble understanding eigenvectors and eigenvalues, consider a m...
opal cypress
The dashed lines are the ellipse axes and the continuous lines are the eigenvectors
peak estuary
your matrix should be symmetric for this analogy
opal cypress
oh so it only works for symmetric matrices I see
then I need to revise my understanding of SVD, thanks!
.close
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frosty musk
Just advice on making a game to have transitional stuff in 4 dimensional+ graphics, I know I can easily do locked to dimension axes but just curious what anyone would like to add like how to translate that into more views? Matrix multiplication? Looking for transitions for when a player changes directions without going loopy, suggestions? If I have a translation matrix, is it possible to have the translated coordinates relate exactly as the dimension axes lock would give? Also, not needed but any suggestions on how to do like outside 3d of view plane (screen) stuff? Just ideas or educated opinions. X E.
safe radishBOT
Just advice on making a game to have transitional stuff in 4 dimensional+ graphi...
frosty musk
We are not warping space, so that comes with advantages. X E.
If I multiply out a matrix for all points, does the result behave basically like locked to dimension axes is the main question. X E.
Should I just test it myself?
safe radishBOT
@frosty musk Has your question been resolved?
frosty musk
I tested and the matrix works. X E.
fathom jewel
Great. X E.
frosty musk
Thanks. X E.
So yeah, this is easy now. X E.
Transitions do take Nā¢N but worth it, and I guess I can do z based still. X E.
Better than Nā¢N for almost nothing. X E.
And I can scale space with matrices. X E.
I tried it. X E.
So next question, how to do like xyz but w is off?
Broad question I know. X E.
I can use the matrix and z based in view finder and extra coordinate finder. X E.
This just got a lot easier. X E.
Any advice on how to find when an equation is true? Simplex method?
I eliminated x, y, and z now. X E.
fathom jewel
Any advice on how to find when an equation is true? Simplex method?
Can you elaborate what you mean?
Are you trying to solve a linear problem?
frosty musk
Are you trying to solve a linear problem?
Yes. X E.
fathom jewel
Can you send a picture of it? Generally speaking, you can apply Simplex.
frosty musk
Okay, so far I have found that in any dimensional, the z I speak of is in any moving dimension, ((end point)-(view point))/((view plane center)-(view point)) in any dimensional space iff the view plane is axis aligned. X E.
What I want is like a rotated P+(O-P)z+(X-O)x+(Y-O)y=Q+sum((V_i)s_i). X E.
Do I need an inverse matrix?
And x and y be like as long as view at point is zero, (projection-(view center))/z. X E.
In that dimension. X E.
Literally, this works. X E.
I think sticking to boxes. X E.
frosty musk
What I want is like a rotated P+(O-P)z+(X-O)x+(Y-O)y=Q+sum((V_i)s_i). X E.
When this is true with a matrix basically with x, y, and z set like I was doing before. X E.
Basically like z>=1, 0<=x<=constantā¢z, same as x for y, and constant bounds with no z in other parts. X E.
Should I just simplex method solve or what? This is a feasibility thing. X E.
Linear feasibility set. X E.
Actually, ask other people willing to implement, but this is awesome. X E.
.close X E.
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cunning kindle
how to do this?
safe radishBOT
how to do this?
fiery merlin
It looks like an integral problem.
quasi bison
integrate 2pi r N(r) dr from 0 to 20, by the looks of it?
cunning kindle
integrate 2pi r N(r) dr from 0 to 20, by the looks of it?
Why 2pi?
cunning kindle
It's like length times N(r).
So N(r) is basically the radius?
quasi bison
So N(r) is basically the radius?
no the radius is r
fiery merlin
No, N(r) is the density at that radius.
cunning kindle
Oh
quasi bison
2pi r dr is the area of a thin "shell" of radius r and thickness dr
cunning kindle
2pi r dr is the area of a thin "shell" of radius r and thickness dr
Ohhhh
Thanks @quasi bison and @fiery merlin
And @nimble vine
.close
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nova parcel
In this proof , at last it is written
" Since Q contains N it must be a denumerable set ."
Well if Q contains N . Then
N is a subset of Q.
Then how can the bijection exist
f: Q->N
opaque fern
An infinite set can have a bijection with a proper subset of itself
quasi bison
that's the thing with infinite sets
opaque fern
Finite set logic doesnt apply here
As an example, you can create a bijection between the set of naturals and the set of even numbers f: N -> E with f(n) = 2n
nova parcel
An infinite set can have a bijection with a proper subset of itself
Recently I proved a theroem :
There exists no injection from N to Nn.
where Nn={1,2,3.....n}
Doesn't this contradicts
feral linden
Itās simply that (countable) = (denumerable or finite). Since it contains N so it canāt be finite thus denumerable
Thatās what that sentence means
nova parcel
As an example, you can create a bijection between the set of naturals and the se...
Does it apply to only bijection from one infinite set to another infinite set...
opaque fern
Yes
opaque fern
Recently I proved a theroem :
There exists no injection from N to Nn.
where Nn...
This is about finite ones still
nova parcel
Okay then, ...I haven't encountered that theroem yet
opaque fern
Finite stuff you sre working with pigeonhole
feral linden
p<-> q or r
Then (p and not q)->r
p: Q is accountable
q: Q is denumerable
r: Q is finite
First paragraph gave you p true
The sentence you are asking about is saying (not q) is true
opaque fern
Infinite stuff is dedekind
austere forge
In this proof , at last it is written
" Since Q contains N it must be a denume...
for bijection, see https://en.wikipedia.org/wiki/Pairing_function
nova parcel
Itās simply that (countable) = (denumerable or finite). Since it contains N so i...
Denumerable itself means ..... bijection onto N
feral linden
Yeah
nova parcel
Infinite stuff is dedekind
Dedekind ?
nova parcel
Yeah
Then if Q is larger set than N . How can the bijection exist
Anyways .lex told a new theroem
opaque fern
Dedekind ?
https://en.wikipedia.org/wiki/Dedekind-infinite_set was talking about this
feral linden
Your first paragraph says Q is countable
being countable =being denumerable or being finite
Since it contains N, thus itās not finite, thus itās denumerable
I am explaining your picture. Your original question. How I would prove it comes later if you ask me
nova parcel
Sir my argument is ....if Q is denumerable.
Then by definition there exist a bijection from Q onto N
But since Q is larger to N.
How can the bijection exist
feral linden
Q is not larger than N
nova parcel
"Q contains N"
feral linden
{1,2,3,4,ā¦} is ālargerā than {2,4,6,8,ā¦}
Still there exists a bijection mapping k to 2k
A bijection mapping a set to its proper subset
Dedekind-infinite by definition shown in Lexās link
nova parcel
That example helpedn
Thanks ...
Are u alsk talking about that dedekind set
.close
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hybrid heath
does anyone know how to find elimination matrix, please teach me, i don't understand this
hybrid heath
topic - matrices
silent kayak
does anyone know how to find elimination matrix, please teach me, i don't under...
By elimination matrix do you mean a matrix that is used to transform another matrix into an upper triangular matrix?
hybrid heath
no
i mean , finding an identity matrix or reduced echelon form by multiplying with a matrix(an elimination matrix)
silent kayak
So you want to find a matrix E for matrix A so that EA=U where U is in REF?
(Also a reduced echelon form is a special type of upper triangular matrix!)
safe radishBOT
safe radishBOT
Available help channel!
warm fractal
Exercise and answer attached
safe radishBOT
Exercise and answer attached
warm fractal
I don't understand how the author constructed the rectangle with the dimensions based on the dot products
Maybe a hint would be nice
silent kayak
I don't understand how the author constructed the rectangle with the dimensions ...
Do you mean why can we think about it only in two dimensions?
warm fractal
No I understand that they all lie on a plane
But why does the rectangle have to be 2x3 based on their dot products
silent kayak
The dot product is a projection of a vector onto another
warm fractal
Oh
silent kayak
This basically means that here you can draw a line downwards from v in the diagram
warm fractal
Right
silent kayak
And you would get to 2a
warm fractal
Yeah so it is basically finding scalar component
silent kayak
and you could draw a line across and left and get 3 times b
silent kayak
no problem
warm fractal
.close
nova parcel
Kindly provide a proof of this theorem
safe radishBOT
Kindly provide a proof of this theorem
feral linden
Union of these sets is a subset of disjoint union of these sets
And you can prove N Cartesian product N has a bijection to N right?
nova parcel
Yes
feral linden
Then solved
nova parcel
Second part I understand
feral linden
Disjoint union ?
Let Aā={(a, A): a in A}
Disjoint union of these A_m I meant union of Amā
nova parcel
Ok
feral linden
Union of Am -> Union of Amā
mapping a in Am to (a, Am) in Amā is injective, and notice that those Amā, any two of them have empty intersection
(Itās the same as saying: you can assume that those Am are disjoint)
nova parcel
Okay
feral linden
Each Am is a subset or is bijective to N, thus this union is a subset of disjoint union of N many N, which is N Cartesian product N
nova parcel
U mean , I just create a set having elements of type (a,b) such that it contains information about all elements of all sets Am
feral linden
Yeah, which is a subset of N times N
nova parcel
Actually....that method is discussed is book
I was searching for some other method ...if any
feral linden
I see
nova parcel
Is there any other way ...
feral linden
Idk
short dawn
need help
quasi bison
need help
do you know how to find the sum of ALL coeffs in an expansion
without finding the expansion itself
short dawn
u mean sum of only binomial coefficient?
neat burrow
hi
short dawn
hey
mighty mango
need help
expand it then add! 
surely wont take you a longer amount of years than option C.
feral linden
For a polynomial f, consider f(1) and f(-1)
somber oriole
need help
The short answer is evaluate the expression at x =1 and x = -1 and then take the average of the two values
If you want an explanation of why this works I'm willing to elaborate
feral linden
You asked a similar question in channel 28
Why you ask an almost identical question
somber oriole
To be fair you never explained it to him