Bro
#help-23
deep matrix
stiff aurora
hmm
deep matrix
I also got same
stiff aurora
my ans or book?
stiff aurora
I also got same
chalo koi na
stiff aurora
I also got same
btw kya tum ye kisi aur se puch k bta skte ho plz
deep matrix
Bro
stiff aurora
tum to In ho tum smj hi sakoge ki yr real life tutions vale kitne bekar honge mere and scholl to rehne do
deep matrix
How you got cosinverse?
stiff aurora
where?
stiff aurora
pic bhejna
stiff aurora
achaa
ty
1 thing more wont ake much time
bhai yr domain range k waqt aap kya approach rakhte ho
mujhe boht difficult lgta hai
any tips plz
stiff aurora
question to ma particularly ni keh rha but mujhe overall sochne ma difficult lagte hai
ye mtlb hai mera
stiff aurora
For 180±theta use kar
half cycle jise kehte hai?
stiff aurora
90 karega to function badal Jaye ga na
ok
deep matrix
Half cycle waicle nhi pata bro
deep matrix
Kaha?
stiff aurora
vaise yr aapko koi self study book ya online free course pta math ka
deep matrix
vaise yr aapko koi self study book ya online free course pta math ka
Khan academy
Best
It's very basic
But you should have patience
stiff aurora
.close
red sandal
hey, im really confused on this question:
safe radishBOT
hey, im really confused on this question:
red sandal
Ive done this so far:
sin^-1(42/78)
then
90 - ANS but it seems to be wrong?
Can anyone help?
heres my work, sry if its messy, i was bored:
The smiley face is the owl, the sad face is the mouse
misty bay
safe radishBOT
green flare
Please help
I keep getting D as the answer
The way I solved it was picking numbers
spiral moat
Were you able to find combinations of numbers that disprove a, b and c?
green flare
So for the 5 numbers I picked
They gave 1 and 10
So I picked 2, 3, 4
So the mean would be 4
m=4
And then I just plugged in each x value
And every value when plugged into x was less than 5 except 10
10-4=6
Not sure if this is wrong
spiral moat
Well I'm not entirely sure what do they mean by range
spiral moat
So for your numbers the range would be 6-0= 6
As the greatest deviation is 6 and smallest 0
That doesn't disprove a)
As 1 is the smallest ans 10 the greatest value
1<m<10
green flare
Hmmm why are you subtracting by 0
spiral moat
The smallest deviation is 4-4=0
green flare
Confused
spiral moat
As 4 is the mean and 4 is one of the values
spiral moat
6 is the greatest deviation from the mean for given values
10-4
And 0 is the smallest deviation for given values 4-4
So the range is 6
Which is greater than 5
So it doesn't disprove a)
Lets take these numbers: 1 2 4 9 10
green flare
Ok
spiral moat
Now m=5.2
green flare
How about 1
spiral moat
5.2 - 1 = 4.2
green flare
Ok
spiral moat
So deviation for x = 10 is still greater
green flare
Got it
spiral moat
And the smallest deviation is for x=4
green flare
Yes
spiral moat
b) is disproven with your numbers where the range was 6 which is greater than 5
Both examples disprove c)
So d) is correct
spiral moat
Maybe by range they mean something else
green flare
Yeah maybe
spiral moat
For our interpretation of range we found an example where range can be less than 5
green flare
Can you help me with that second question
I'll figure out the first one and ask about what they mean about range
spiral moat
Have you tried to do this one?
spiral moat
I would do it by rewriting a50 in terms of a49, then a49 in terms of a48 and so on until you get to a representation of a50 in terms of a46
Then put those those values for quant. A and quant. B into an inequasion
green flare
Would A be: a49+49
spiral moat
Yes
spiral moat
Would A be: a49+49
Try to rewrite this a49 in terms of a48
spiral moat
You're going the wrong way
spiral moat
a50 = a49 + 49 =
a48 + 48 + 49 =
a47 + 47 + 48 + 49 =
a46 + 46 + 47 + 48 + 49
So finally a50 = a46 + 190
Which means the two quantities are equal
lean otter
The circle contains point M(2,1) and it touches both coordinate axis.
Find the equation of the circle.
I need help, starting point, ideas how to solve. A tip.
safe radishBOT
The circle contains point M(2,1) and it touches both coordinate axis.
Find the e...
lean otter
I have a point M, point A(0,y) and point B(x, 0)
distance MC = r
distance CA = r
distance CB = r
3 equations, but 5 unknowns
fair hound
Think about the center of the circle
Finding the equation is knowing where the center is
lean otter
Finding the equation is knowing where the center is
Yeah, no problem, I need to find center and radius, putting equation is easy when I find these two
fair hound
Let's call the points A(0, a) and B(b, 0) to avoid any possible confusion
Your equation of the circle is (x-u)²+(y-v)² = r²
It's true for A:
(-u)² + (a-v)² = r²
But also for B and M
Also, the radius becomes more obvious once you plot the circle in your head: the distance between M and y-axis is very limited
(Because you know it touches x-axis too, so it can't be too far up)
You're necessarily in a case of symmetry
lean otter
radius is basically b
fair hound
Yeah, and in fact, a too
lean otter
then a and b are equal
lean otter
Also, the radius becomes more obvious once you plot the circle in your head: the...
yeah... visually some things do become more obvious, you are right
fair hound
Do you see how C(1, 1) is the only point at distance 1 of A, B and M ?
If you try to change that, because A and B are on the axis, you don't touch M
When the circle has to touch the two axis it makes sense that A and M have same y coordinate
Because of how M is positionned
lean otter
got it... but how do I tackle this problem algebraically?
I think I have 6 equations, and 5 unknowns now.. huh.. one more than enough 👀
fair hound
You can also check the equation for M
(2-u)²+(1-v)² = r²
4-2u+u² + 1-2v+v² = r²
If you sub that for A and B you should get that a = 1
Another way to see it's obvious is geometrically, because it's really supposed to be obvious on a drawing than a = b
And it helps a lot
lean otter
If you sub that for A and B you should get that a = 1
i am not sure what you mean? sub A and B where👀
fair hound
Sorry, I mean since r² has then 3 expressions
You can equate them
But since a = b obviously, doesn't leave too much choice for r and a = b
You can also use equations of lines if you're willing to
C is on the perpendicular bisector of every pair of {A, B, M}
There really is a lot of ways to do it as the conditions are very restrictive
lean otter
You can equate them
how do I equate 3 equations?
I know how to do that with 2.
Should I equate 1st and 2nd, and 2nd and 3rd?
(1st and 3rd would be another option), right..
fair hound
(2-u)²+(1-v)² = r²
(b-u)²+(0-v)² = r²
(0-u)²+(a-v)² = r²
so (2-u)²+(1-v)² = (b-u)²+(0-v)² = (0-u)²+(a-v)²
lean otter
I think that's what bothers me...
We wrote 3 equations, and we have 5 unknowns (a, b, u, v, r)
how is this solvable?
fair hound
One of the reasons is geometric, a = b
lean otter
One of the reasons is geometric, a = b
hmm.. that leaves us with 4 unknowns, still unsolvable
fair hound
so b² - 2bu + u² + v² = u² + a² - 2av + v²
so b² - 2bu = a² - 2av
but since a = b, u = v
and in fact, it was geometry again, we could deduce geometrically that since axis are orthogonal, C has necessarily the same x and y coordinates
the other keypoint is that a = b = u = v is very obvious too, u isn't equal to v for nothing
lean otter
what if we have 3 points M(x1, x2), A(x3, a), B(b, x4), that belong to a circle and where a, b are unknowns.
A more general example
lean otter
(2-u)²+(1-v)² = r²
(b-u)²+(0-v)² = r²
(0-u)²+(a-v)² = r²
so (2-u)²+(1-v)² = (b-...
I assume we would start with these equations
fair hound
yeah, and that would be a pain, a long system
in this case it doesn't necessarily touch the axis once and only once
so it's clearly not as restrictive and the equation will be x1, 2... dependant
lean otter
yeah, and that would be a pain, a long system
would it be solvable? Since we have more unknowns than equations
fair hound
3 non-aligned points are always concyclic
So there exists a circle that is a solution
And in such a case it's the only one because you can draw the perpendicular bisector of each segment
So yeah it would be solvable, there would be only one possibility for a and b
fair hound
3 non-aligned points are always concyclic
(it means you can always draw a circle going through 3 points that are not on a same straight line)
The problem is when you have a number of points different than 3
In case this number is 2, there isn't a unique solution
And in case this number is > 3, a solution doesn't always exist
(Still depends on x1, x2, x3, x4, note that if A, B, M are aligned, it's a problem)
lean otter
I actually can imagine that visually, but putting it all together algebraically is a bit difficult
fair hound
The easiest algebra for this is to be ok with the use of some analytical geometry.
You have 3 points A, B, M. If there are on a same straight line: no circle is going through all of them.
If they aren't, there is one (and only one):
Let's call the center C. Since C is at same distance of A, B, M, C belongs to the perpendicular bisectors of [AM] and [BM]. These perpendicular bisectors have equations you can write, so you can find the point where they cross which is necessarily C.
Once you know C, you know the equation of the circle.
Note that it doesn't determine the location of A and B, as they could be on either side of the circle and it wouldn't be relevant.
Your equation would in fact depends on a and b for obvious reasons.
(Since contrary to what happens in your exercise, they wouldn't necessarily be the same)
In your exercise, it's obvious C has positive and equal coordinates for "perpendicular reasons" since the circle touches each axis only once, axis are tangents
And now, the information about M fixes this unique coordinate to know
Which is obv the radius too bc same perpendicularity reason
lean otter
@fair hound
Appreciate your help!
I guess I'll need more exercise for this topic.
Thanks again!
safe radishBOT
@lean otter Has your question been resolved?
fair hound
@lean otter I'm sorry, after reading again, I think the exercise itself is wrong...
C(1, 1) works as I said, but C(5, 5) seems to work too
lean otter
there are two solutions
fair hound
yeah: here is why
Once you get that a = b, it follows that a = b = r
So C(r, r)
But then, r is not necessarily 1 as
(2-r)²+(1-r)² = r²
r²-4r+4 + r² -2r + 1 = r²
r² - 6r + 5 = 0
(r-5)(r-1) = 0
The trick is that as I said above, A and B are not fixed and the equation is a and b dependant
And in your exercice, this dependancy doesn't give a unique solution but two, as there is a symmetry about where M is on the circle
So even in the case of your exercise, we still doesn't have enough restrictions on a and b: the equation depends on if they are 1 or 5
lean otter
could we say that the solution is both, two circles
they both satisfy given conditions
fair hound
it gives two equations that work
(x-1)²+(y-1)² = 1
and (x-5)²+(y-5)² = 25 are both valid
Your exercise has two solutions
unreal kindle
Hello, how to show that $2^{n} - 1$ is a prime number only when n is prime
flat frigateBOT
lilisworld
unreal kindle
n>=2
untold cobalt
What have you tried?
stoic dune
Try factoring it
clever stag
Hello, how to show that $2^{n} - 1$ is a prime number only when n is prime
reasoning by absurd suppose $2^{n} - 1$ is a prime number and n not prime will have easily a contradiction
untold cobalt
it is not generally helpful to just tell people what to do in mathematics
stoic dune
Can be helpful sometimes
flat frigateBOT
clint00
untold cobalt
Can be helpful sometimes
i would argue it is never helpful
stoic dune
But yeah, that is just the problem statement
unreal kindle
reasoning by absurd suppose $2^{n} - 1$ is a prime number and n not prime will h...
easily?
clever stag
if i factor it what should i get?
a contradiction if u reasoning by absurd or if u have some arithmétic result
unreal kindle
i tried by absurd but it just didnt lead me anywhere
because i think they're something missing in my reasoning
untold cobalt
what was your reasoning?
unreal kindle
what was your reasoning?
the problem is that i don't know how to translate: "2^n - 1" is prime into arithmetics
clever stag
because i think they're something missing in my reasoning
if u suppose n not a prime number then u can write n=ab with pcgd(a,b) not 1
unreal kindle
so you're asking me to use gcd properties?
stoic dune
pcgd?
plucky elk
i would argue it is never helpful
definitely wrong
untold cobalt
definitely wrong
example?
clever stag
xd sorry im french we write pcgd
plucky elk
you claim it's never helpful, so the onus is on you to prove so
unreal kindle
if i assume 2^(ab) - 1 is prime
then where is the contraiction
is that that obvious?
clever stag
is that that obvious?
y
untold cobalt
you claim it's never helpful, so the onus is on you to prove so
In my experience, mathematics problems are usually designed to push people to understand material. The way that they achieve this is by discomfort, which forces one to put the pieces into place over time, from there they don't move too easily. Ideas have a tendency not to fix in the brain without this experience. I would be interested to hear in which scenario you think this isn't true.
plucky elk
Ah yes proof by anecdote. Very rigorous
plucky elk
it is not generally helpful to just tell people what to do in mathematics
If you have a better method of explaining, just do it rather than put kaynex down.
untold cobalt
Ah yes proof by anecdote. Very rigorous
I wasnt even referring to kaynex in the first place. This is a subject of genuine debate, I'm not sure why you feel the need to respond like this. There is a reason that no textbooks, for example, provide solutions and authors explicitly request that such manuals not be produced.
unreal kindle
where is the obvious?
plucky elk
I wasnt even referring to kaynex in the first place. This is a subject of genuin...
And yet you claim it's never helpful. Direct contradiction
unreal kindle
anyway it didnt work for me sooo
untold cobalt
And yet you claim it's never helpful. Direct contradiction
I said I would argue that it is never helpful. That is my opinion on it.
plucky elk
Your arguments are now opinions?
untold cobalt
I don't think I want to keep up the childishness. If you disagree with me, thats fine.
plucky elk
Anyway, keep them to yourself unless you're actively helping the helpee. All you did was interrupt Kaynex unnecessarily and accomplished nothing
unreal kindle
Anyway, keep them to yourself unless you're actively helping the helpee. All you...
ok sometimes it work but here it doesnt work so i still need help
plucky elk
ok sometimes it work but here it doesnt work so i still need help
Sorry I overtook your channel for the discussion.
clever stag
enlighten me
u have to proof "$2^{n} - 1$ prime then n prime" btw so let's put n=ab then $2^{n}-1$ divisible by $2^{b}-1$ and $2^{a}-1$ so u have 2 choices : $2^{a}-1=2^{n}-1$ and $2^{b}-1=1$ or $2^{b}-1=2^{n}-1$ and $2^{a}-1=1$ so what u can conclude ?
flat frigateBOT
clint00
unreal kindle
u have to proof "$2^{n} - 1$ prime then n prime" btw so let's put n=ab then $2^{...
if n=ab, then 2^b - 1 and 2^a - 1 divide 2^n - 1?
clever stag
if n=ab, then 2^b - 1 and 2^a - 1 divide 2^n - 1?
yep u can proof it it's just a Euclidean division
unreal kindle
u have to proof "$2^{n} - 1$ prime then n prime" btw so let's put n=ab then $2^{...
a=n or n=b?
stoic dune
if n=ab, then 2^b - 1 and 2^a - 1 divide 2^n - 1?
That's actually really cool, I didn't know this
clever stag
another proof by contrapositive if you want I hope more clear ^^ : u can put n=ab and write $2^{ab} – 1 = (2^{a} – 1) (2^{a (b-1)} + … + 1)$.Thus, we expressed $2^{n} - 1$ as the product of two integers greater than 1, which means it is not prime
flat frigateBOT
clint00
unreal kindle
and ?
it means that 2^(ab) - 1 can't be prime because 2^a - 1 and the other one devide it anyway and a and b is supposed to be different from n
clever stag
u have a=n and b=1 or b=n and a=1
unreal kindle
another proof by contrapositive if you want I hope more clear ^^ : u can put n=...
yeah but how did you write this
clever stag
$2^{ab}=2^{(a)b}$
flat frigateBOT
clint00
unreal kindle
u have a=n and b=1 or b=n and a=1
yes but wasnt the fact that 2^a - 1 and b divide it enough?
hard crest
yeah that's enough. but here's a question, why doesn't that argument work for prime n?
clever stag
yes but wasnt the fact that 2^a - 1 and b divide it enough?
y and ?
unreal kindle
ok so we proved that it doesnt work for n non prime
clever stag
if $2^{n}-1$ divisible by $2^{b}-1$ and $2^{a}-1$ u don't have another choice $2^{a}-1=2^{n}-1$ and $2^{b}-1=1$ or $2^{b}-1=2^{n}-1$ and $2^{a}-1=1$
flat frigateBOT
clint00
clever stag
ok so we proved that it doesnt work for n non prime
lol no it's not a contradiction a=n and b=1 or b=n and a=1 mean directly that n is prime ^^
unreal kindle
lol no it's not a contradiction a=n and b=1 or b=n and a=1 mean directly that n ...
ok now it's clearer
clever stag
ok now it's clearer
cool than
clever stag
**clint00**
this proof is better honestly by contrapositive but if u don't have the factorisation the other works
unreal kindle
but how do you even come up with that factorisation
clever stag
$a^{n} - b^{n}$
flat frigateBOT
clint00
clever stag
bc u can write $1=1^{n}$
flat frigateBOT
clint00
clever stag
Xd
safe radishBOT
clever stag
$a^n - b^n = (a - b) \sum_{k=1}^{n} a^{n-k} b^{k-1}$
flat frigateBOT
clint00
safe radishBOT
Available help channel!
mortal herald
safe radishBOT
mortal herald
Anyone knows how I could get rid of the -4/5 without a calculator?
So I could fuse the first fraction with the second
lean otter
So I could fuse the first fraction with the second
where did it come from
pure agate
Invert the expression to cancel out the negative.
lean otter
Invert the expression to cancel out the negative.
do what he siad
mortal herald
Invert the expression to cancel out the negative.
That will add more fractions, ill try though
lean otter
honestly i think you just leave the fractional exponent
i dont think you can simplify that 5th root
mortal herald
I didnt do what he asked to do
lean otter
(a/b)^-1 = b/a
lean otter
u mean -6/5?
mortal herald
Wdym
lean otter
there are 2 terms there
one with 4/5 exponent and one with 6/5 exponent
you can join those
lean otter
then alright sure
full fog
for part c I got the answer (7+√21)/2 which the mark scheme says is right but I rlly can't even understand the part d. which says a = (7+✓21)/2 is the answer. can someone explain a little pls
safe radishBOT
for part c I got the answer (7+√21)/2 which the mark scheme says is right but I ...
low canyon
If $g(x) = x$, then $g^{-1}(g(x)) = g^{-1}(x)$, and therefore $x = g^{-1}(x)$, so you have $x = g(x) = g^{-1}(x)$
flat frigateBOT
all matrices are invertible
tidal atlas
Really confused on this problem
worthy hemlock
safe radishBOT
safe radishBOT
Available help channel!
short rover
I'm working with projections in a Hilbert space $\frak{H}$. We define $P_Mf$ to be the projection of $f \in \frak{H}$ onto the subspace $M \subset \frak{H}$. \
\
They say "$M$ must be closed because $M = {f \in \frak{H} \vert f = P_Mf}$."\
\
I understand this definition of $M$, but I don't get how that implies $M$ is closed. Is it because it's defined by an equality ?
safe radishBOT
I'm working with projections in a Hilbert space $\frak{H}$. We define $P_Mf$ to ...
flat frigateBOT
burnpink
short rover
Intuitively strict inequalities define open sets, whereas equalities and nonstrict inequalities define closed sets. Can this be made more rigorous?
toxic stratus
you can rewrite that as
[
M = \set {f \in \mathfrak H \where (1 - P_M) f = 0} = \ker (1 - P_M) = (1 - P_M)^{-1}(\set 0)
]
flat frigateBOT
toxic stratus
and then $P_M : \mathfrak H \to \mathfrak H$ is continuous so that $1 - P_M : \mathfrak H \to \mathfrak H$ is continuous
flat frigateBOT
toxic stratus
so the inverse image of 0 is closed
toxic stratus
this assumes that you actually get a continuous projection ig
which might not be the case if you try to project onto a subspace that isnt closed
short rover
yeah in my notes they assumed P_M to be continuous
toxic stratus
safe radishBOT
toxic stratus
you too 
safe radishBOT
Available help channel!
pale hamlet
hey, can anyone help me understand how to find the first two x values that work for this equation?
safe radishBOT
hey, can anyone help me understand how to find the first two x values that work ...
pale hamlet
i know i can get 74/-197 = cos(pix/15)
oops i inputted the first question wrong
and first 2 as in first two values that work greater than 0
<@&286206848099549185>
safe radishBOT
@pale hamlet Has your question been resolved?
minor spruce
i have a problem... i don't think i know subtracting
3306
-1938
6 - 8, you need to borrow. you can't take from 0 so you take from 3 and 3 becomes 2. 0 - 3 you need to borrow again. where did i go wrong here
steep pewter
this is how i learnt it:
since i know even the first 3 isnt safe
cut it and put a 2 above it
minor spruce
yep i did that
steep pewter
then cut the next 3 and put 12 over it
now cut the 0 and put 9
and 16 instead of the 6
im skiiping a step everytime but i thinkk u understand where it comes from right?
minor spruce
ohhh i think i know what you're talking about
let me try it
okay nvm i don't lol. why does 0 become 9
and not 10
hard crest
it becomes 10 first when you borrow from the 3
then it becomes 9 when you borrow from it
minor spruce
Oh
so what i did was i did 10 -> 0
that's where i messed up
and not 10 - 9
ty
.close
sick quail
Hello!
I am trying to solve this system of linear congruences
I have done so via row reduction by turning it into the matrix
4 1 5
1 3 4 (mod 11)
But I end up with
1 3 4
0 0 0
I'm not sure what this means, y is then a free variable? Is this possible within systems of congruences?
Also, what does this mean for x?
hard crest
I'd recommend doing it with the x and y symbols for this one to see what that means
sick quail
I've done so like this, but then both variables are cancelled out, so I don't really have anything to work with
Though the equation is still valid
hard crest
yep the equations are equivalent! so yeah it's underconstrained
sick quail
Cool! So just to confirm, it's similar to having a free variable in linear algebra, so y can take on any value and the system will still hold? Regardless of it being modulo 11?
hard crest
well... it's one dimension of freedom
x and y are still related
but there are 11 solutions
in linear there would be infinite but we only have 11 elements in our field
sick quail
Oh I see, then I must plug values into x to solve for the 11 solutions, right?
flat frigateBOT
Hayley
sick quail
Got it!! Thank you so much 🙂
prisma glacier
safe radishBOT
prisma glacier
Im not sure how to do this question
devout shale
google maximum area of rectangle inscribed in a circle
prisma glacier
okay thanks
hard crest
google maximum area of rectangle inscribed in a circle
no don't
prisma glacier
.close
safe radishBOT
safe radishBOT
Available help channel!
lime nexus
how do u find q16a the surface area bit? i somehow managed to do all of them besides getting that bit wrong
safe radishBOT
how do u find q16a the surface area bit? i somehow managed to do all of them bes...
low canyon
What did you get?
lime nexus
uhh sorry it looks rlly wrong but the closest i gotten was SA= 2h^2+ 2000h/root2 +2000h 😅
the front triangle with h, i got the diagonal side to equal h/root2, and the base length (is that what u call it im not too sure) as 2h
low canyon
The diagonal side should be sqrt(2)*h, not h/sqrt(2)
lime nexus
oh-
tulip stag
hey
lime nexus
oh rlly?
tulip stag
can you help me
lime nexus
not the entire surface area of the triangle prism?
tulip stag
<#488120190538743810> <@770170323034177587>
ok thanks
low canyon
not the entire surface area of the triangle prism?
Exactly, not the entire surface, just the top
lime nexus
how do would yk that
low canyon
Because the ''surface area'' of the water can be interpreted as the area for the top, but I get what you wanted to do
The question should have been more clear about it
magic cove
I just can't find a sequence for this
safe radishBOT
I just can't find a sequence for this
deep matrix
Nvm
viscid crow
1-(1/2+1/4+1/6+1/8...
magic cove
the denominator is easy but I can't think of something that doesn't effect the numerator and effects the 2 to become a 3
Sorry forgot to add that it requested n=1 at the beginning
viscid crow
what is n
magic cove
like this
viscid crow
i think it is asking for the value at infinity
magic cove
I've got that the denominator is 2^n but didn't get what the numerator could be
viscid crow
do you know what the basel problem is?
viscid crow
I've got that the denominator is 2^n but didn't get what the numerator could be
its not a power of 2
magic cove
i think it is asking for the value at infinity
no it's asking for the sequence term
viscid crow
no it's asking for the sequence term
sorry im not very good at english can you explain what its asking, also this sum is solved by the basel problem, do you know what that it?
magic cove
the bottom isnt 2^n
my bad I read it wrong it's 2n
I've wrote the n weirdly in my notebook
jade gust
(2n-1)/2n?
magic cove
oh yeah that's right
jade gust
(a_n)^pos infty_n=1, a_n=that
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jade gust
\(\left(a_n\right)_{n\in\mathbb N_1},\quad a_n=\frac{2n-1}{2n}\)
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wraith prism
I didn't understand this question
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I didn't understand this question
wraith prism
viscid crow
p is a prime. Lets say p = a^2+b^2, what are the possible remainders when p is divided by 4?
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quasi bison
you've either misunderstood what riptechno asked you or confused yourself.
@wraith prism forget about the "prime" thing for a moment.
what remainders can a square give when divided by 4?
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wraith prism
<@788270795532533791> forget about the "prime" thing for a moment.
what remainde...
Square of any number?
quasi bison
square of an integer.
quasi bison
2?
give me an integer n such that n^2 ≡ 2 (mod 4).
@wraith prism
please do not wait until the bot pings you again
it's a bit annoying to have to do that
wraith prism
Opps. I rechecked
Accidentally divided 4/2
So I found only 0,1 as remainder
@quasi bison
quasi bison
5 was extraneous
but yes, ok
so a square can only ever give 0 or 1 as remainder
what remainders are possible for the sum of two squares?
wraith prism
Let me check again
Can I add the remainder of it?
It gives only 1,2
As remainder
Ohh so option 4 correct
Thanks ann
quasi bison
It gives only 1,2
0, 1 and 2.
but 0 is impossible for any prime to have, and 2 is only possible for the number 2 itself.
wraith prism
0, 1 and 2.
0 how?
quasi bison
0 + 0 = 0
wraith prism
but 0 is impossible for any prime to have, and 2 is only possible for the number...
I took 25+9 =34
Which gives 2 remainder
quasi bison
0 is impossible for any prime to have
let me repeat myself more explicitly
the possible remainders mod 4 of the sum of two squares (without regard to it being prime or not) are 0, 1 and 2.
when you also require said sum to be prime, 0 stops being an option,
and 2 is an option only as the number 2 itself.
wraith prism
and 2 is an option _only_ as the number 2 itself.
What does it mean?
quasi bison
the only prime number that is congruent to 2 mod 4 is the number 2.
there are no others.
wraith prism
the only prime number that is congruent to 2 mod 4 is the number 2.
For prime yes
quasi bison
i think there is some miscommunication here
or you are reading things into my words that i am NOT saying
wraith prism
I didn't understand 4th option
quasi bison
you don't understand what option 4 says?
it says "p = 2, or p ≡ 1 (mod 4)"
i.e.
p is congruent to 1 mod 4, or p just is 2.
wraith prism
Sum of prime number squares/4 will gives 1 or 2 only
Right?
I understand now in terms of mod
quasi bison
Sum of prime number squares/4 will gives 1 or 2 only
incorrect.
the sum of two squares can give remainders 0, 1 or 2 mod 4.
quasi bison
the sum of two squares can give remainders 0, 1 or 2 mod 4.
if this sum is also prime, it can only give remainders 1 and 2.
wraith prism
Sum of two prime numbers square(exclude 0)
Correct?
quasi bison
no
wraith prism
What is wrong with it😪?
wraith prism
the sum of two squares can give remainders 0, 1 or 2 mod 4.
if this sum is also ...
If the sum is not prime which also gives remainder 2
25+9 =34÷4 =2
quasi bison
now your notation is bad too
you cannot use / to refer to the remainder of a division
quasi bison
What is wrong with it😪?
it's a string of words with no coherence to it.
i cannot correct it except by offering my own wording
which i did twice
wraith prism
I don't know why my statement is wrong
It's okay we can move next question
I read the question again yes it says the sum of two squares and the sum is prime
wraith prism
X=3mod(9), x=7mod13 how to solve it?@quasi bison i have solved it but I would like to know once again from you sorry for tagging you
quasi bison
would have been much better if you closed this channel and opened a new one.
anyway, chinese remainder theorem
languid fable
Would like to ask if these are correct this is just free practice btw
languid fable
Hello
Are these right
This was done yesterday
Just sent it in
Today
Anyone out there
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twilit torrent
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pulsar pecan
can you send the entire question
twilit torrent
can you send the entire question
this is entire question
"from example above"
but example above is a video
i have taken the picture from the video
wait
twilit torrent
can you send the entire question
based on what i have shown, am i wrong?
could you please check first 1-2 minutes if the picture i have sent is not clear
lean otter
the product of the slopes should be -1
twilit torrent
the product of the slopes should be -1
yes
slope is y/x, is it not?
so negative reciprocal would be -x/y
so 2* -x/y
lean otter
okay then 2y/x -> -x/2y
yes, correct
twilit torrent
dumb question but how?
here
lean otter
its the same
lean otter
$-\frac{x}{2y}=-\frac12\cdot\frac xy$
flat frigateBOT
Toby
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proper maple
Hi
proper maple
I'm having a slight issue figuring out what should the variance and deviation be for this problem
`Simulate each game a million times and print out the wins and losses, assuming each bet was for $1.
- Simulate 1000000 plays of the first game: You win if you get one six in four rolls of one dice, and the round stops.
Given a payout of $1 when a win is hit calculate the:
● Mean
● Variance
● Standard deviation
For each game's payouts.`
I know mean would be the number of payouts (i.e. wins) divided by 1_000_000
But then I'm not too sure what the variance formula should look like.
From what I've googled, I need take a value, and subtract it from the mean.
But does this mean I need to record each number of the dice thrown in one round that's not a 6?
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@proper maple Has your question been resolved?
hard crest
you can also use the other formula for variance E[X^2] - E[X]^2
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kinglacto
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solemn vault
Hello again
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Hello again
solemn vault
I'm stuck on 'b'
So I basically have to prove this function is bijection
Well the first case is clearly bijective
In second case i proved that it is injective but not surjective...
It clearly means that second case is not bijective, doesn't that make the whole function non bijective....
Like
When I was proving cardinality of integers and natural numbers are equal... I proved for both case that the function are both surjective and injective ... that's how the whole function became bijective
olive tide
Isn’t the codomain just naturals tho
solemn vault
solemn vault
Isn’t the codomain just naturals tho
Of the main question, yes
olive tide
What was your approach to finding a q that works for 108
olive tide
Maybe you could prove surjective by constructing values for a genetic x in naturals
solemn vault
For the second case??
olive tide
For the whole thing?
solemn vault
Alr
olive tide
If there exists a set q for all natural numbers x then the function is surjective, doesn’t matter where q lies in the cases tbh
And since you proved injective for both cases
You’re done
olive tide
Soo we know that f is injective
solemn vault
Yes
olive tide
For both cases
solemn vault
Yes
olive tide
So you could think about it as the range generated by f is a subset of Naturals
Which means all we need to prove is that the range generated by f is also a superset (I.e surjective)
So if we can find a correct q for any x in naturals, then the range generated by f is a superset of naturals (I.e covers all natural numbers)
Does that help?
solemn vault
So you could think about it as the range generated by f is a subset of Naturals
So, basically if I've a positive 'q' that means I can get a natural number from the last two cases depending upon what my 'q' is
olive tide
Well that’s to be proven by you
solemn vault
If that's it then, the range of the function will be equal to co domain...that means it's surjective
Alr then
olive tide
Yeah
solemn vault
I get a unique natural number for all q > 0, q € Q
Any hints on how can i do so??
olive tide
Proof by contradiction
Assume that 2 different qs can get you to the same value
And then show that those 2 qs must be the same
I.e only exist one unique prime factorisation
solemn vault
What's qs??
solemn vault
So basically, f(q1)=f(q2) the I've to prove q1=q2
olive tide
Yeh
olive tide
The surjective stuff comes trivially
olive tide
Yeah
But you’ve shown that all q in natural don’t mix and all q not in natural don’t mix
But show that no q in first case hits any values that are also hit by a q in the second case
solemn vault
You mean, I've to show that I don't get 1 as an output in the second case??
olive tide
Oh thats fine then
olive tide
Yup same strategy
olive tide
Use the fact that prime factorisation is unique
flat frigateBOT
BasuDev
olive tide
kinda
we're not sure that the order is the same or anything
but left side and right side must include the same values
so x_1 must equal x_2
quasi bison
So I basically have to prove this function is bijection
dunno if it was mentioned here but you've got the wrong slashes here
solemn vault
quasi bison
you want \setminus for a backslash, not forward slash
solemn vault
Ohk
quasi bison
also wait you're like
required to show f is bijective
without a specific requirement of separately showing inj and surj
right
because i think it may be easier to simply construct the inverse...?
olive tide
that wouldnt automatically prove bijective, still need to show injective anyways
olive tide
e.g saying this inverse must be unique for reasons x,y,z
quasi bison
??
no?
constructing an inverse is enough, which of course involves also verifying that it is an inverse
you know that if an inverse exists then it is unique
unless you want to reprove that every time too
solemn vault
Alr alr we've that the function is injective and we've to show its bijective
*surjective
quasi bison
let n ∈ N \ {1}, decompose n into prime factors, see which ones have even exponents and which have odd exponents...
olive tide
normally just show that for all natural numbers there exist a q that returns that natural number
but there is a trick here
the domain is rational numbers which is the same size as natural numbers, and you know that f is injective so it should cover natural numbers
quasi bison
no, just because a function is injective and its domain and codomain are both countable does not by itself mean f is surjective automatically
olive tide
sorry, i thought it would be (is it true if domain is uncountable and codomain is countable)
solemn vault
let n ∈ N \ {1}, decompose n into prime factors, see which ones have even expone...
2=2
3=3
4=2²
5=5
6=3×2
7=7
8=2³
.....
quasi bison
that... is not really what i had in mind there
quasi bison
sorry, i thought it would be (is it true if domain is uncountable and codomain i...
do you want the smartass answer or the helpful answer?
solemn vault
.
olive tide
2=2
3=3
4=2²
5=5
6=3×2
7=7
8=2³
.....
you want a general inverse, dont technically prove anything if you cant cover all cases
olive tide
so think about how to go from some natural number to a q in domain, the step you used for f(q) = 108 should give you a hint
solemn vault
we proved that for every q>0, q€Q there's a unique natural number....
and our last two case gives us a natural number.. our Co domain is natural number and our range is also natural number doesn't that basically proves our function is surjective
thats what i said here, but its not super rigorous
infinity gets weird
solemn vault
Alr alr
solemn vault
so think about how to go from some natural number to a q in domain, the step you...
I can do prime factorisation
olive tide
yeah!!
flat frigateBOT
BasuDev
olive tide
well theres probably a few ways but imma give the one i found
if we have a natural number thats also a squared number
solemn vault
Yes
solemn vault
Alr we can
olive tide
otherwise we can use case 3
solemn vault
Ye
olive tide
if our natural number x is prime, then we let q = x/1
solemn vault
OK...
olive tide
if our natural number x isnt prime, then we need to find the p_1^2{r_1} * ... * p_n^2{r_n} * q_1 ^2{s_1-1} ... q_n^2{s_n-1}
such that x = p_1^2{r_1} * ... * p_n^2{r_n} * q_1 ^2{s_1-1} ... q_n^2{s_n-1}
which is just an argument saying that x isnt prime so there exist these numbers kind of
solemn vault
Your saying that for evry x>0 there's a number
Alr alr I gotta go now, I'll do the surjective part at night
@olive tide
Thanks for the help though I appreciate the time you have given
ThNks
solemn vault
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quasi bison
feels like a printing thing
it may have blended in with a grid line
you don't "calculate" it at all
all you can say is it must lie between Q1 and Q3 but beyond that you cannot say anything
quasi bison
dunno what you mean by "exactly between"
it lies between them, NOT necessarily halfway between them
solemn vault
Ye
quasi bison
@lean otter
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quasi bison
it was referring to basudev
anyway
why not simply multiply both sides by (33+m)?
well i mean like
it becomes "the same as" crossmultiplying if you insist on writing 3 as 3/1 and then writing the multiplication by 1 on the right explicitly
which i find a bit silly
it's just an ordinary multiplication of both sides by the same thing
also if you want to improve your factorization skills then that ought to include knowing when an expression is not factorable
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green flare
I keep getting quantity A is greater
For a put 6 for a and 3 for b and then 3/9 gives a remainder of 3 which is greater than 1
And when I plug in another number I get A as the greater value
Am I correct?
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crude bear
Quantity A is a?
green flare
Yes
crude bear
So you can have a = 18, b = 1 and b =9*0+ 1 => r = 1. We have a/r = 18 and b/1 = 1
Hmmm
We can have a = 9 and b = 1000.
a/r will be less than b/1
crude bear
1000 = 999+1
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nimble hinge
I should calculate the real fourrier series for f(x) = cos^nx. And based of the formular I came here. But i dont know how to really solve this. Can someone help me of starting solving these?
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I should calculate the real fourrier series for f(x) = cos^nx. And based of the ...
nimble hinge
I struggle with the cos^nx function
bold aurora
Not if n is even
quasi bison
are you not allowed to refer to complex stuff at all
bc if you could do that it would greatly simplify your life actually
bold aurora
Of note is that cos^n is always even, so you can get away with integrating from 0 to pi instead. Doing this for a generic n without complex numbers is a little beyond me 😛
nimble hinge
bc if you could do that it would greatly simplify your life actually
I am allowed to. I was thinking about it, but never done it before, we've been just introduced to it
quasi bison
let z = e^ix then cos(x) = (z + z^-1)/2
expand that with binomial
you basically get the entire fourier series for free
nimble hinge
this sounds hard and easy at the same time. I'll try, thanks
Am i plugging this in my integral then?
@quasi bison
nimble hinge
okay well, i dont know how to actually solve this with complex numbers
quasi bison
works out to
1/16 (z^4 + z^-4 + 4z^2 + 4z^-2 + 6)
= 1/8 cos(4x) + 1/2 cos(2x) + 3/8
some tinkering skipped over
but that's the general idea
you just get it directly as a linear combination of cosines
bold aurora
(You can also shove it in your integral and solve, it'll just be a little tedious, but the antiderivative of any power of z is trivial (and what you get is a linear combination of powers of z, that compliments the linearity of the integral well))
nimble hinge
Okay well, I think I get the idea and the handiness of using the complex numbers here. But is that already my solution? Like just forming an series of that and I am done?
quasi bison
yes
nimble hinge
This feels too easy ngl
bold aurora
((What Ann did is to notice that you already have a linear combination of cosines when you expand cos^4(x), so no need to go through the integrals to do the projections))
(((Afterall the idea of a fourier series is to deconstruct into a linear combination of sines and cosines))) 😄
nimble hinge
Ye idk, i never did this before really, so I had no clue "when im done" with the series
bold aurora
If you want the a_0 a_k and b_k for any n I can imagine the skipped over some tinkering might get a bit hairy? idk I didn't try it on paper
nimble hinge
hmm idk but quite happy to have an quite smart solution to this
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nimble hinge
Okay, sorry for asking again, but this is what i got this now. But I want to go back to the cosine form, right?
Because I feel like I am doing something wrong here, or not coming to a clean solution.
@quasi bison
bold aurora
$$z = e^{ix}$$ $$cos(x) = \frac{z + z^{-1}}{2}$$ $$cos(nx) = \frac{z^n + z^{-n}}{2}$$
hard crest
z^{-n}
flat frigateBOT
learath2
bold aurora
One day I'll learn how to use latex properly 
nimble hinge
Ah okay, thanks a lot
bold aurora
It is a little dicey to see that you can always pair a z^n with a z^-n when expanding
nimble hinge
It is a little dicey to see that you can always pair a z^n with a z^-n when expa...
Ah, I did not remember the formular below. It's not that hard as soon as you have it. I turned the last part to
cos((n-k)x) now. Is that right?
so something like this
bold aurora
I still don't have pen and paper, but it kinda looks wrong, I'd expect (n-2k), the range of the sum being halved since we are pairing the terms, and an issue about even and odd n causing one extra term
nimble hinge
Ye it is indeed wrong, I made a mistake
nimble hinge
I still don't have pen and paper, but it kinda looks wrong, I'd expect (n-2k), t...
but i dont know how to solve this. this -k is annoying
bold aurora
Let me give it a go
nimble hinge
Please note that the solution needs to be real
bold aurora
I got something like $$\frac{1}{2^n} \sum_{k = 0}^{\frac{n - 1}{2}} \binom{n}{k} cos(\binom{n}{k}x)$$
flat frigateBOT
learath2
bold aurora
I'm not completely sure about it, kinda rushed through it and then suddenly don't feel like looking more into it anymore
bold aurora
**learath2**
For odd n btw
I only did n=3 so I can visualize it. There should be another simplification there but I'm not in the vibe to math anymore
nimble hinge
Oh oof. Lemme see
bold aurora
For even ones you'll end up with a lone term in the middle, I don't remember how to get rid of it, you'll have to come up with something
Oh those (nCk) above the zs are wrong
They should be 2k - n and -2k + n
and you get a 2k+n in the cos instead
Ah, the lone term ends up being a z^0 by symmetry, so it's just constant
No more looking at this, I'll go sleep, gn
nimble hinge
Thank you a lot, I will also need some time to look at it now, haha
nimble hinge
Honestly, I feel like all of this is not the correct way. When I plug this in wolfram alpha it does not fit.
bright carbon
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bright carbon
how do you know which x term to make the subject to get a converging sequence?
is it trial and error or is there a method to know which x term to make the subject to achieve a converging iterative sequence
ember sphinx
wouldnt you just go with newton raphson?
lean otter
Hi
bright carbon
hi