twilit bramble
and was hoping if then angle between would be 90
#help-42
primal plume
why does it say thre are 2 solutons
calm coralBOT
why does it say thre are 2 solutons
swift laurel
what was the equation?
subtle osprey
why does it say thre are 2 solutons
probably signifies a quadratic
sick current
that's a quadratic. (multiply by x(x-10) and it becomes clear).
Every quadratic equation has 2 solutions. In some cases, those solutions are double (both are the same number) or they might be complex numbers.
potent lotusBOT
primal plume
so 50 and - 50
primal plume
.close
calm coralBOT
swift laurel
basically just treat it as kinetic friction
it's still (coefficient)*(normal force)
calm coralBOT
@errant knot Has your question been resolved?
hoary topaz
calm coralBOT
calm coralBOT
<@&286206848099549185>
Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.
viscid zenith
Do you know what is meant by reflect?
remote mural
i need to use the inverse function ?
yes
lavish cove
first you need to find the negative reciprocal of the slope
remote mural
okay so the inverse is 1/8x-6 right
no
remote mural
first you need to find the negative reciprocal of the slope
no that's for a perpendicular line
remote mural
okay so the inverse is 1/8x-6 right
switch x and y in the original equation and solve for y
remote mural
wait waut wait
hoary topaz
okay
remote mural
- find perpendicular function
- use point slope form to find perpendicular function that also passes through the given point
- solve for the intersection of the perpendicular function and the original
- use the intersection as the midpoint in the midpoint formula to find the reflected point
im on phone rn, so annoying
hoary topaz
all good no worries but okay perpendiculr function is 1/8x-6? original one is -8x +6
lavish cove
no
the slope is right
you need to use point slope form and plug in the slope and the given point to find the perpindicular equation
lavish cove
it would be y-12=1/8(x-1)
lavish cove
put it into y=mx+b form
hoary topaz
oh my bad i multiplyed -12 by y and diveded
divided, i had to subtract it
okay so after that I take the intersection of both equations >
?
remote mural
1. find perpendicular function
2. use point slope form to find perpendicular fun...
just follow these steps my good sir
remote mural
it would be y=1/8x+95/8
yes
remote mural
okay so after that I take the intersection of both equations >
yes
the intersection is the midpoint between the given one and it's reflection
hoary topaz
yeah i set them equal to each other and solve i think
lavish cove
yes
plug in the orginal equation for y
and solve for x
alr i gtg but basically once you get x plug it into one of the equtions to get the y point of the midsection
then just use the midpoint fourmula
look up a yt video if u need help
hoary topaz
yeah im working on it thank you appricaite it
calm coralBOT
@hoary topaz Has your question been resolved?
languid bone
apparently my solution to this question is wrong, and just wanted to know why it might be
languid bone
the chance of arranging the letters of the word "WEIRD" to form that word is 1/5! = 1/120
so shouldn't the chance that the letters appear in that order if "DUMBWAITERS" is rearranged also 1/120?
pseudo moss
No
languid bone
yeah, but what particular case would go against that
pseudo moss
what do you mean
languid bone
idk how i can say what i mean, but if you could explain why that isn't true that would be great
graceful dust
what do you mean
pseudo moss
You are trying to find the probability of a particular class of words arising from DUMBWAITERS (that being the class of words that contains the substring WEIRD)
pseudo moss
<a:waves:880516410948853840>
helo layler
graceful dust
lily green arc when
pseudo moss
You are trying to find the probability of a particular class of words arising fr...
so, to this end, your first step should be counting the total number of strings you can get from DUMBWAITERS
languid bone
yep, 11!
graceful dust
hm…
pseudo moss
now count the number of arrangements of DUMBWAITERS that contains the substring WEIRD
sorry, arrangements of the letters
in DUMBWAITERS
languid bone
so would it be something like
graceful dust
i like 1/5! 
languid bone
the number of times the string WEIRD appears not broken up by letters + the number of times WEIRD appears broken up by letters
graceful dust
😭
graceful dust
you are correct that you can ignore the other letters
languid bone
yeah, that's what i was thinking
graceful dust
for 4
pseudo moss
layla to the rescue
languid bone
the answers say 13/1584, which is really close to 1/120, but its a bit more
pseudo moss
thanks for noticing my spread of misinformation ;-;
graceful dust
oh
graceful dust
“not all together”
languid bone
OH
ARE YOU KIDDING
why did i not notice that
i just assumed they wanted it in order
graceful dust
i just read that like “not necessarily all together”
graceful dust
only after seeing the answer would i read it as MUST be not all together
languid bone
so the chances that they would all appear together is 1/7920, from the first question
so 1/120 - 1/7920 = 13/1584
okay thank god
i was going insane
pseudo moss
hooray!!
languid bone
i dont even know what they're even going on about here but they were overcooking it
ok anyway
thanks guys
graceful dust
i dont even know what they're even going on about here but they were overcooking...
this is overly complicated
they didn’t realize you can ignore the other letters
languid bone
this is a hsc trial paper (australian y12 final exam) for like one of the top schools in the state
so i figured that their answer would have just been more correct or something
graceful dust
i dont even know what they're even going on about here but they were overcooking...
like this is an official solution?
languid bone
yeah, this is the solution the teachers came up with to mark students responses
graceful dust
haha they didn’t use the @exotic cosmos trick
languid bone
what's that
graceful dust
ignoring the other letters
graceful dust
would if i could
languid bone
the way that they structure these tests is they have a question 14 section at the end of the test
which is specifically designed to ensure that nobody gets 100% on the test
the questions get absolutely stupid
but they end up doing weird stuff so that if you get a 65% it actually scales up to a 90% or something because the test is so hard
its weird
graceful dust
loll
languid bone
i am just guessing they thought this question was harder than it was, so they just chucked it in there
ok anyway, thank you
cya
.close
calm coralBOT
graceful dust
only after seeing the answer would i read it as MUST be not all together
if i had read iii i probably would have noticed earlier 
calm coralBOT
Available help channel!
umbral terrace
Hi everyone, I took SAT a few days ago and got stuck on a question. I don't remember exactly how the question was but it was something like this, (x-5)(x+7)(x+5)(x+a)(x-b) the sum is 9 so what would be the number a-b if a and b are positive integers. I butchered the signs but it was something like this. I won't get my score for a few weeks but this question has been on my mind nonstop. Thanks
golden sable
Hi everyone, I took SAT a few days ago and got stuck on a question. I don't reme...
well first of all congrats on taking the SAT
umbral terrace
thank you
umbral terrace
would this be unsolvable without that?
umbral terrace
i see, but if 9 is the sum of the roots?
golden sable
idk
umbral terrace
but its not needed for my uni
golden sable
I took it 2 years ago
SAT went a lil up than PSAT
I got the lowest acceptable score
umbral terrace
yea i am not the best with multiple choice question
maybe if they had free wrties
i might have done better but yea
thanks, gonna be waiting for my score ig
.solved
calm coralBOT
calm coralBOT
Available help channel!
wheat peak
Help please
calm coralBOT
Help please
amber bay
for the first one
the avg rate of change of a function
b/w interval a and b
is defined as
f(b)-f(a)/b-a
so plug in x = -1 and x=h-1
and simplify
wheat peak
okki
calm coralBOT
@wheat peak Has your question been resolved?
wheat peak
and simplify
I believe imstill lost
when i plugged them in i got 16 and 16h...im not sure if i went right or wrong
calm coralBOT
@wheat peak Has your question been resolved?
calm coralBOT
@wheat peak Has your question been resolved?
zinc falcon
I am not really sure if the graph is correct
calm coralBOT
I am not really sure if the graph is correct
zinc falcon
it seems correct but at the same time it doesn't really seem correct
if x or y = 0 then no matter what the value for y or x is the graph would be 0
remote mural
I am not really sure if the graph is correct
It's a 2D graph for a function mapping from 2D to 1D
A 3D graph would be necessary
calm coralBOT
@zinc falcon Has your question been resolved?
exotic cosmos
haha they didn’t use the <@1266602198020919368> trick
omggg 😭 another application of it
graceful dust
loll yep
zinc falcon
I tried using desmos 3d to see and its a very complicated graph
calm coralBOT
@zinc falcon Has your question been resolved?
zinc falcon
.close
calm coralBOT
calm coralBOT
Available help channel!
eternal shard
Does the following system of linear congruences have an integer solution $(x,y)$?
\begin{align} 4x+6y &\equiv 1 \mod 7 \ 3x+8y &\equiv 1 \mod 7 \end{align}
potent lotusBOT
bacc (unhelpful)
eternal shard
will send my work in abit
So I managed to find out that it must have solutions because of the gcd(1,2) = 1 can be a multiple of 7
but I am stuck on finding the solutions, because applying Euclidean algorithm I cannot applly it backwards
glass heart
how would you solve 4x+6y=1, 3x+8y=1 over the real numbers
eternal shard
I would have x-2y = 0 so y = x/2 or better here x = 2y
Wait was my equation set up wrong
x-2y ≡ 0 mod 7 that means if I divide x-2y by 7 remainder is 0, so that means x-2y = 7k is a multiple of 7
hmm i thought x-2y = 7k or x-2y = 7 would work
can it be given what I got for the gcd that x-2y=0 is possible
so then my only solution would be (2,1)
glass heart
I mean, can you find integers with x=2y
eternal shard
and (-2,-1)
glass heart
are those the only ones?
glass heart
ok
with that aside for a moment
after you subtract the equations, you still have two equations
4x+6y=1 and x-2y=0
is that actually what you would do for real numbers to solve a linear system?
eternal shard
hmm in that case with two linear function, i wouldn't actually but I guess i must since i am doing modular arithmetic
glass heart
well you dont have to
but its not a bad idea to actually do the thing that allows you to solve linear systems in general
eternal shard
Yupok I will try
tranquil wasp
Euclid oiled up good luck
glass heart
but 14=0, therefore?
glass heart
14y = 7 * 2 * y
not sure what you want to say with that equation
glass heart
there no solutions
why
eternal shard
that no matter y
it's always a multiple of 7
meaning we cannot find integer y so that we get remainder 1
eternal shard
hmm ok
so then
the whole system cannot have solutions, because the 14y ≡ 1 mod 7 has none?
glass heart
hmm ok
this comment seems like you are not quite satisfied with my argumentation. am I misreading that? maybe I can rephrase it
calm coralBOT
eternal shard
this comment seems like you are not quite satisfied with my argumentation. am I ...
oh apologies lol, i understood it well
the argumentation
calm coralBOT
Available help channel!
eternal shard
Does the following system of linear congruences have an integer solution $(x,y)$?
\begin{align} 4x+6y &\equiv 1 \mod 11 \ 3x+8y &\equiv 1 \mod 11 \end{align}
potent lotusBOT
bacc (unhelpful)
eternal shard
oh nvm
potent lotusBOT
bacc (unhelpful)
eternal shard
I will continue from here
calm coralBOT
@eternal shard Has your question been resolved?
eternal shard
Ok I am actually having difficulties
eternal shard
**bacc (unhelpful)**
I converted this into
[ \begin{cases} 1-14y = 11n \ x-2y = 11m \end{cases} ]
where $n,m \in \mathbb{Z}$.
potent lotusBOT
bacc (unhelpful)
runic jewel
I think I got something
slate field
we could just solve this entirely mod 11 without pulling it back to the integers
if you pull it back now you’ve done something really unwieldy right
eternal shard
if you pull it back now you’ve done something really unwieldy right
I don't understand this
eternal shard
**bacc (unhelpful)**
Are you saying I should work with this?
slate field
**bacc (unhelpful)**
i mean that this is nicer to work with
eternal shard
ah ok
slate field
there are only 11 cases to check, one for each x.
eternal shard
**bacc (unhelpful)**
I figured here that the last equation is equivalent to x-2y = 0 because of ggt(1,2) = 1 so I would get solutions span{(1,2)}
which I then can try on 1-14y = 11n
brb
slate field
actually, there’s only one case to check
i’m not gonna verify the first equation that you got
but if you got one independent of x then actually you only have to check one equivalence class mod 11
1-14(4)= -55
and the other solutions for y will be 4 mod 11 as well
eternal shard
,, \bar 1 = { y \in \mathbb{Z} : | : 14y \equiv 1 \mod 11 }
potent lotusBOT
bacc (unhelpful)
slate field
yeah
eternal shard
3y = 1 mod 11
slate field
yes
eternal shard
ok 1 q
eternal shard
**bacc (unhelpful)**
what is wrong with this
slate field
**bacc (unhelpful)**
well you wrote that 1 bar was the solution to your equation
which it isn’t
it’s 4 bar
eternal shard
oh i wasnt trying
slate field
yeah i know you didn’t wanna say that
eternal shard
I tried to apply this [ \bar a = { b \in \mathbb{Z} : | : a \equiv b \mod m } ] because you mentioned equivalence class.
potent lotusBOT
bacc (unhelpful)
slate field
yes. and this will be true, but the point is that you’re trying to actually figure out a in this case
eternal shard
ohhh
eternal shard
yup
so
ok what does the equivalence class have to do with all of that
(x,y) = (8,4)
[ \bar 4 = { b \in \mathbb{Z} : | : 4 \equiv b \mod 11 } ]
potent lotusBOT
bacc (unhelpful)
calm coralBOT
@eternal shard Has your question been resolved?
eternal shard
because 3x4=1
Ok after some time, I think what it is, is that all solutions \begin{align*} x &= 8+11k \ y &= 4+11n \end{align*} can be expressed as [ (x,y) = (\bar 8, \bar 4) ] which now makes sense after considering the definition over and over again...
potent lotusBOT
bacc (unhelpful)
eternal shard
all integers x that have remainder 8 and all integers y that have remainder 4
mod 11
arghh
It should be actually $x \in \bar 8$ and $y \in \bar 4$
potent lotusBOT
bacc (unhelpful)
slate field
yep
potent lotusBOT
yøung matr!x
raven echo
@rare canyon
raven echo
I know a method
limpid pilot
this will close abruptly
rare canyon
OK
raven echo
If you have access to pascal triangle
rare canyon
If you have access to pascal triangle
binomial theory
bro that's smart af
but there is a problem
calm coralBOT
Available help channel!
raven echo
A2 + B2 = (A+B)2 -2AB
calm coralBOT
A2 + B2 = (A+B)2 -2AB
limpid pilot
A2 + B2 = (A+B)2 -2AB
what about it
shy nova
calm coralBOT
calm coralBOT
Available help channel!
inland wadi
Hi there, I have the idea as to why the set of all polynomials is not isomorphic to the the R^N, where N is the natural set and R is the real set. However, I cannot figure out how to prove it
inland wadi
My idea lies on the fact that for every polynomial, it contains a finite quantity of information, whereas there are elements in R^N that contain an infinite amount of information. So maybe the way would be to show that no map P --> R^N can be surjective?
calm coralBOT
@inland wadi Has your question been resolved?
calm coralBOT
@inland wadi Has your question been resolved?
calm coralBOT
@inland wadi Has your question been resolved?
vital plume
In special relativity, the famous Einstein equation $E = mc^2$ for the energy of an object at rest, in its full form, becomes $E^2 = (mc^2)^2 + (pc)^2$. Show that, when $p$ is fixed to the average momentum of an unladen swallow, this leads to
$$S(E) E = mc^2$$
where $S(E) = \sqrt{1 - \frac{1.3 \times 10^{16} \text{J}^2}{E^2}}$ is the swallow function.
potent lotusBOT
Ginger
vital plume
my question may be obvious
is it an african or european swallow?
it doesn't specify
nvm, i tried to clarify and now i can cross the bridge freely
i no longer need an answer
.close
calm coralBOT
tribal jetty
calm coralBOT
Available help channel!
dusky crescent
Need some assistance in finishing this off : )
calm coralBOT
Need some assistance in finishing this off : )
dusky crescent
I know I need to use the principle of inclusion-exclusion but can't figure out what I need to subtract from |A| to get my answer. Hopefully that makes sense
pseudo moss
why did you close your other help channel?
dusky crescent
just because I felt like I was over confusing people, so I just closed it to start fresh
dusky crescent
why did you close your other help channel?
Sorry about that!
calm coralBOT
@dusky crescent Has your question been resolved?
golden sable
first restriction is trivial
you can make a bijection between x_0 >= 5 ... rest of conditions
and x_1...+...+x_5 = 16
so then you apply I.E with the two sets
x_2 >= 9 and x_3 >= 8
dusky crescent
first restriction is trivial
you can make a bijection between x_0 >= 5 ... rest ...
How can I make a bijection?
potent lotusBOT
>> 20 & 0b111
golden sable
with $x_5 \ge 5$
potent lotusBOT
>> 20 & 0b111
golden sable
is bijective to $x_1 + x_2 + \ldots + x_5 = 16$
potent lotusBOT
>> 20 & 0b111
golden sable
with $x_5 \ge 0$
potent lotusBOT
>> 20 & 0b111
golden sable
trivially no? @dusky crescent
so we have decomp the problem into number of solutions $x_1 + x_2 + \ldots + x_5 = 16$ with $x_5 \ge 0$ and $x_2 \le 8$ and $x_3 \le 7$
potent lotusBOT
>> 20 & 0b111
golden sable
now to get rid of $x_2$ and $x_3$ conditions
potent lotusBOT
>> 20 & 0b111
golden sable
we can just simply find the amount of solutions with $x_2 \ge 9$ and $x_3 \ge 8$
potent lotusBOT
>> 20 & 0b111
golden sable
and use I.E to make sure we don't have duplicates
yea this problem is prettyy trivial
dusky crescent
we can just simply find the amount of solutions with $x_2 \ge 9$ and $x_3 \ge 8$
I did that here
So all my other cases that I did were not necessary?
golden sable
So all my other cases that I did were not necessary?
yeah
i dont know what u were doing there chef
😭
dusky crescent
i dont know what u were doing there chef
Okay so everything else I did can be erased?
Just keep this part?
golden sable
So can I just do the |A|, |B|, |C|? One for each restriction
I mean
just do restrictions on A and C
and subtract that result from B
like it would be like |B| - |A| - |C| + |A intersect C|
or something ike that
golden sable
So is this good to start?
I don't like the way it's structured
calm coralBOT
golden sable
oops
dusky crescent
.reopen
calm coralBOT
✅
dusky crescent
I don't like the way it's structured
Okay np I can definitely change that
How do you think I should change it?
golden sable
How do you think I should change it?
I would be like
the bijection thingy
to handle the $x_5$
potent lotusBOT
>> 20 & 0b111
golden sable
then the problem becomes
so we have decomp the problem into number of solutions $x_1 + x_2 + \ldots + x_5 = 16$ with $x_5 \ge 0$ and $x_2 \le 8$ and $x_3 \le 7$
potent lotusBOT
>> 20 & 0b111
potent lotusBOT
>> 20 & 0b111
golden sable
and the intersection
dusky crescent
Oh i think I know where I am confused. I don't get how it turn into 16
golden sable
Oh i think I know where I am confused. I don't get how it turn into 16
oh I just added 5
WAIT
why did you do that?
I said
potent lotusBOT
>> 20 & 0b111
$x_1 + x_2 + \ldots + (x_5 + 5) = 21$
>> 20 & 0b111
to ensure that the $x_5 + 5$ is greater than or equal to 5 when $x_5$ is greater than or equal to 0
dusky crescent
I had a sample question that was like this. Would it be better to structure this question in that same manner?
golden sable
I had a sample question that was like this. Would it be better to structure this...
YES
just for the $x_5 \ge 5$ part
potent lotusBOT
>> 20 & 0b111
dusky crescent
Ohh OKAY. I actually did that before I did all this, but it wasn't working out since it gave me -1. But now I realize my mistake. I included the other restrictions as well when it should've just been for the >= 5 part
that makes a lot more sense now
thank you
golden sable
Ohh OKAY. I actually did that before I did all this, but it wasn't working out s...
um
golden sable
lets gooo
now consider having 16 balls
and you want to seperate those balls into 5 groups
how many dividers would you need?
dusky crescent
how many dividers would you need?
^
golden sable
4 dividers
mhm'
so since we have (16 + 4)c(4)
you can see this as permuting a string of 16 balls and 4 dividers
and we divide by the duplicates (4)! and (16)!
so we get (20)!/(4!16!) = 20c4
cool cool
dusky crescent
Okay that makes sense
kind axle
Ohh OKAY. I actually did that before I did all this, but it wasn't working out s...
Great handwriting
dusky crescent
Great handwriting
Thank you!
kind axle
Are you done with your problem?
dusky crescent
Nope there is still more
golden sable
Are you done with your problem?
I don't think so
kind axle
Ah okay
dusky crescent
I don't think so
I know need to find the intersection of the other two restrictions? Why can't I just simply subtract them separately
golden sable
I know need to find the intersection of the other two restrictions? Why can't I ...
can I give you a scenario?
dusky crescent
Sure!
golden sable
$x_1 + x_2 + \ldots + x_5 = 16$
potent lotusBOT
>> 20 & 0b111
potent lotusBOT
>> 20 & 0b111
golden sable
@dusky crescent hmm you might be right for this specific case
that the intersection is zero
since 9 + 8 > 16
dusky crescent
Yup
dusky crescent
but in general this is not true
okay
dusky crescent
yeah should be good
So does that mean I just consider them separately without the intersection?
golden sable
So does that mean I just consider them separately without the intersection?
I mean the intersection would be zero
so it really doesn't matter
dusky crescent
yeah
dusky crescent
I mean the intersection would be zero
This is what I have now. What's next?
Or is there anything that needs to be fixed
dusky crescent
Yeahh
golden sable
Yeahh
I remember in high school
they taught me "stars and bars"
but when I went to uni they replaced it with
"weak and strong compositions"
dusky crescent
Ohh really? I never did this in high school. I am in 2nd year uni
So now I just subtract everything from the 20 choose 4?
golden sable
So now I just subtract everything from the 20 choose 4?
mhm
😭 yikes from freshman year
I remember taking this course and thinking
"oh no math is not fun anymore; where did my stars and bars go"
dusky crescent
Aww
dusky crescent
mhm
I just realized is it probably better to talk about the other two restrictions in terms of stars and bars?
golden sable
I just realized is it probably better to talk about the other two restrictions i...
you could
but you alr know how it works
dusky crescent
but you alr know how it works
truee
Don't think there is really a need to let things be |B| or |C| though
i think I can remove it
I can just subtract directly with the permutations
unless maybe it makes it more clear?
golden sable
unless maybe it makes it more clear?
prob does no?
and B intersect C
isn't always 0
most of the times it's not
dusky crescent
isn't always 0
yup I know : )
dusky crescent
prob does no?
Would you be able to explain what the subtraction does specifically? Does it just avoid duplicates
I understand the process but just want to make sure I understand the "why" of everything
golden sable
Would you be able to explain what the subtraction does specifically? Does it jus...
oh the subtraction basically removes the stuff
that violates the conditions
like violations the condition $x_3 \le 8$
potent lotusBOT
>> 20 & 0b111
dusky crescent
Ohh okay that makes sense!!
golden sable
etc etc
golden sable
Would you be able to explain what the subtraction does specifically? Does it jus...
mhm
but most of the time we could have violations happening simultaneously
and we don't wanna "double subtract"
that's why we consider the intersection
but the intersection is zero
in this case if that makes sense
you should give it a minute and lmk if you have further qs
dusky crescent
you should give it a minute and lmk if you have further qs
Okay thank you!
kind axle
"oh no math is not fun anymore; where did my stars and bars go"
combinatorics was never fun to begin with
golden sable
combinatorics was never fun to begin with
oh I loved combinatorics
esp in high school I did lots of comp math
kind axle
i hated it
golden sable
and then I took combinatorics and graph theory and linear algebra my first sem in uni
kind axle
still do
kind axle
good for you
kind axle
oh so its like that
kind axle
i dont like partying
golden sable
sucks for you man
dusky crescent
😭 I was devastated
With an A- ??
kind axle
it's not fun for me
golden sable
With an A- ??
yeah I mean I just never had that in hs
4.0 gpa in high school
now I'm a sophomore
dusky crescent
yeah I mean I just never had that in hs
Ohh true though me too. I had all above 95 in hs. Uni has humbled me
golden sable
(too old now)
golden sable
Ohh true though me too. I had all above 95 in hs. Uni has humbled me
my parents were extra mad
kind axle
Ohh true though me too. I had all above 95 in hs. Uni has humbled me
got 99% in grade 10
got 74% in grade 11
golden sable
and I guess fair because they do fund me
golden sable
engineering is chill
math is prob harder
but I do a "chiller" engineering tbf
I do computer engineering
kind axle
thats very chill
golden sable
and a bit of EE
kind axle
im studying fluids right now and this stuff is flowing over my head
golden sable
and I guess it also depends where you go to uni
like mit/stanford prob way harder
just because they can do that with their student demographic
kind axle
and I guess it also depends where you go to uni
im in hs rn
golden sable
im studying fluids right now and this stuff is flowing over my head
fluids is chill
btw if ur indian and in engineering
there is this one really good uni my friend told me about
IIT or sm
kind axle
btw if ur indian and in engineering
spot on, mate
golden sable
im preparing for the IITJEE
focus up ur life is on the line here
calm coralBOT
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zealous stream
hey this is the question and my (attempt of an) answer. it's not really about the content of the question but rather if what i'm saying is correct mathematically.
i'm starting with a minimization problem on x, where x is in the krylov subspace span{b,Ab,...,A^(r-1)b}, that is a subspace of dimension r in R^n
and i'm transforming it into a minimization problem on y, where y is in R^r, via a transformation x = Qy
the first highlighted part says that any x has a corresponding y, however i'm not sure about the second highlighted part when i'm going back the other way. given y* is a minimizer of the y-minimization problem, it should be the case that x* = Qy* is a minimizer of the original x-minimization problem... but i'm struggling to see exactly why.
my intuition is that x* = Qy* is not necessarily in the original krylov subspace span{b,Ab,...,A^(r-1)b} so how is it guaranteed to be the minimizer? i feel like i'm missing something and would appreciate any help or hints, tysm! 🫶
calm coralBOT
hey this is the question and my (attempt of an) answer. it's not really about th...
@zealous stream Has your question been resolved?
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@zealous stream Has your question been resolved?
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remote mural
$$2^x + x = 5$$
Anyway to do this algebraically?? Or can I just approximate
potent lotusBOT
Edmund Cloudsley
tall moon
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runic wave
how is this done
calm coralBOT
how is this done
idle marten
this is interesting
I think the red and green pairs are more easy to see
green uses the arithmetic series sum formula
could you try to manipulate this into the other part
idle marten
ye
idle marten
im lowk working with you on this one
1 is a constant here
I'm breaking this into multiple sums
do you have any other formulas that may help
runic wave
i do know this one but im fairly sure thats not how you do this question
its an ib question and its not in the syllabus
tacit moat
could you try to manipulate this into the other part
probably easier if you notice that that's a telescoping series
runic wave
probably easier if you notice that that's a telescoping series
what is that
idle marten
ok i got the answer, yea what is a telescoping series lol
idle marten
i do know this one but im fairly sure thats not how you do this question
ok, I did it with this formula tho
tacit moat
if you combine it into 1, then write it out, it'll become (2^3 - 1^3) + (3^3 - 2^3) + ...
notice that the 2^3 will cancel out
runic wave
ok, I did it with this formula tho
yeah still not in the syllabus
think theres another way
runic wave
if you combine it into 1, then write it out, it'll become (2^3 - 1^3) + (3^3 - 2...
ohh
idle marten
if you combine it into 1, then write it out, it'll become (2^3 - 1^3) + (3^3 - 2...
combine what into 1
runic wave
combine what into 1
the two sums
idle marten
could you try to manipulate this into the other part
these?
runic wave
if you combine it into 1, then write it out, it'll become (2^3 - 1^3) + (3^3 - 2...
so that ends up only remaining is -1 + (n+1)^3
tacit moat
yea, all the middle terms will cancel
calm coralBOT
tacit moat
that's why it's called a telescoping series
it's like a telescope that you can retract
tacit moat
sepdron can you explain in another channel
yea sure, open up a channel
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hidden ferry
Find all of prime numbers p,q,r such that: p | q^r + 1; q | r^p + 1; r | p^q + 1
I have analysed the case when one of the numbers is equal to 2 and the equality of any two numbers
=> p,q,r ≠ 2; p ≠ q; q ≠ r; r ≠ p.
q^r = -1 mod p
q^2r = 1 mod => ord_q(p) = 1, 2, r, 2r = d1
d1 ≠ 1, r (Otherwise d1 | r => q^r = 1 mod p)
=> d1 = 2, 2r
I don't know what to do next
hidden ferry
<@&286206848099549185>
autumn edge
Find all of prime numbers p,q,r such that: p | q^r + 1; q | r^p + 1; r | p^q + 1...
answer: all prime numbers except thoose wich arent the solution
hidden ferry
answer: all prime numbers except thoose wich arent the solution
🙂
calm coralBOT
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modern peak
answer: all prime numbers except thoose wich arent the solution
please stop trolling in the help channels
autumn edge
im not trolling
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ancient violet
Note that [pqr \mid (p^q + 1)(q^r + 1)(r^p + 1) = p^q q^r r^p + p^q q^r + q^r r^p + p^q r^p + p^q + q^r + r^p + 1.]
Because $pqr$ already divides $p^q q^r r^p$, it follows that [pqr \mid p^q q^r + q^r r^p + p^q r^p + p^q + q^r + r^p + 1]
potent lotusBOT
timuko
ancient violet
Without loss of generality, let $p \le q \le r$
potent lotusBOT
timuko
hidden ferry
**timuko**
p < q < r
ancient violet
Find all of prime numbers p,q,r such that: p | q^r + 1; q | r^p + 1; r | p^q + 1...
If d1=2, then either $r$ is odd and $p \mid q+1$ or $r$ is even and $p=2$
potent lotusBOT
timuko
ancient violet
If $r$ is even, then $r=p=2$ and $q \mid 5$. Thus, $q=5$.
potent lotusBOT
timuko
ancient violet
But p is prime
So $r$ cannot be even. Variables $p$, $q$, $r$ are symmetrical, so it follows that $p$, $q$, $r$ are all odd
potent lotusBOT
timuko
hidden ferry
So $r$ cannot be even. Variables $p$, $q$, $r$ are symmetrical, so it follows th...
(2,3,5) solution
ancient violet
(2,3,5) solution
$3 \mid 5^2 + 1$
potent lotusBOT
timuko
hidden ferry
5 | 3^2 + 1; 3 | 2^5 + 1; 2 | 5^3 + 1
5 | 10; 3 | 33; 2 | 126
p = 5; q = 3; r = 2
ancient violet
(3, 2, 5), (2, 5, 3), (5, 3, 2)
hidden ferry
Yep
calm coralBOT
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rotund granite
why is the exponent on n -1/5
calm coralBOT
why is the exponent on n -1/5
rotund granite
1/n ^ (6/5)
n^ (-1 + (6/5))
shouldn't it be positive 1/5
can someone please tell me
calm coralBOT
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marble pendant
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cunning reef
Let ABCD be a square with side lengths 48. circles with center P,Q,R,S are inscribed as shown. find the area of quadrilateral PQRS
cunning reef
oops forgot to write R
anyways
i know that the area is 1/2 x QS x PR
so id find each radius
but im stuck on that😓
oh
i got radius Q = radius S = 8
radius R = 18
radius P = 3
so the area is 432?
<@&286206848099549185>
am i correct
cunning reef
Can you please tell me how this works, I can't see it
its n/(n^6/5) = n ^(1 - 6/5)
calm coralBOT
patent parrot
Let ABCD be a square with side lengths 48. circles with center P,Q,R,S are inscr...
This diagram is inaccurate if you're trying to express the constraints
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glossy beacon
Im solving differential equations with series, what would be the derivative of this sum?
glossy beacon
$\sum^{\inf}_{n=0}C_nt^{n+1}$
potent lotusBOT
MapleMoose
potent lotusBOT
MapleMoose
potent lotusBOT
MapleMoose
eternal stirrup
When u plug n = 0 you dont have a constant
so dont write n=1 after
the only reason we shift the index like that is when it makes sense to do so
i.e when the first term (or thereafter who knows) is a constant
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knotty cypress
Hi!
knotty cypress
where γ is the combined curve x^2+y^2=4 from the positive direction (2,0) to (-2,0) and then followed with the line (-2,0) to (0,-2)
how do I use greens theorem here?
if we see P as 2x+y
and Q as x
Q'x - P'y would give us 1-1..
what does that mean and why does this happen and how do I fix it? xd
marsh valley
It means that the vector field (2x+y, x) has curl 0 everywhere, and as such the line integral is 0
knotty cypress
curl being?
marsh valley
$curl(F) (x,y) = Q_x(x,y) - P_y(x,y)$
knotty cypress
ah
potent lotusBOT
Azyrashacorki
knotty cypress
and then how would I calculate that question?
marsh valley
Green's Theorem is equivalent to $\int_C \vec{F}\cdot d\vec{r} = \int\int_G curl(F) dxdy$
potent lotusBOT
Azyrashacorki
marsh valley
You have calculated it, it's 0
knotty cypress
but the answer is -4 according to our professor
lol
like the answer section says that "the curve integral can be decided in multiple ways, for example by directly calculating it with an appropriate parameterization or by using greens formula (after an appropriate complementation), we'll here choose a third way
so this is why i'm very confused if I'm using greens theorem in the wrong way, or how to solve this question to begin with xd
marsh valley
Ohh
I thought the line at the end was (-2,0) to (2,0)
That would've closed the curve up and yielded 0
knotty cypress
oh wait did i misread
ffs
ok so that's why we can't use green's theorem then? since it's not a closed curve?
marsh valley
Yeah
knotty cypress
and to fix it we'd have to go from 0,-2 to 2,0?
could you please walk me through how we'd do that because i'm so lost atm 😭
marsh valley
Yeah of course
What they are doing in the solution is calculating the part that is on the circle.
They use the fundamental theorem of line integrals for that, which says that $\int_C \nabla f \cdot d\vec{r} = f(b) - f(a)$ where $a$ and $b$ are the endpoints of the curve $C$.
potent lotusBOT
Azyrashacorki
marsh valley
To do this, you need to find for which function $f$ is $\nabla f = (2x+y, x)$
potent lotusBOT
Azyrashacorki
knotty cypress
oh i see
marsh valley
That's what they get with their U
knotty cypress
so essentially if we were to end up somewhere where it's really hard to "come back" via parameterization this could be a good strategy
and then we take where we got to - our starting point
marsh valley
Yes
Here for instance, the hard part is that on the circle, you'd want to use polar coordinates, but your vector field doesn't really have any nice symmetry on the disk, so it's doable but a bit more work
knotty cypress
ok and if we want to "patch" it up manually, is it enough to say that we do γ+γ2 where γ2 is (0,-2)->(2,0)?
marsh valley
For the part of the line that goes from (-2,0) to (0,-2), you can just parametrize it as usual
knotty cypress
and if we want to "make that bit of more work", what would it look like?
marsh valley
Idk why they don't show that in the solution, maybe I'm missing something
knotty cypress
because from my understanding the part where you add the "path" in.. that doesn't come until later at which point i already have 1-1...
marsh valley
I think you'd have some integration by parts to work with, since x = r cos(theta) and y = r sin(theta). Then your path on the circle looks like r = 2 and theta = 0 to pi.
Might not actual;y be that bad
knotty cypress
so break it up into three sections?
semi circle, triangle downwards, triangle upwards?
but it doesn't really help with the main issue I have with this which is that
this is a youtube video from one of our professors
so he has this
marsh valley
The total curve is like this
knotty cypress
and he adds gamma 1 afterwards
and then he does this
but Q'x - P'y is always going to be 0?
in our case that is
like him adding gamma1 afterwards shouldn't change anything at all since it doesnb't change P or Q?
marsh valley
I think what they are doing is adding the line on the x axis, that yields 0, so the red part is just 0 - (path integral on line from the x axis)
then you can add the green line to that
knotty cypress
from your image you mean?
but then we wouldn't be using greens theorem.. or?
or oh
marsh valley
Like you'd use green's theorem to say that Red + Purple = 0, so then Red = - Purple.
So altogether Red + Green = -Purple + Green
So then the computation is reduced to computing purple and green
knotty cypress
hmm
marsh valley
Which play fairly easily with your vector field
marsh valley
Sorry
knotty cypress
no worries
marsh valley
The purple line can be parametrized as like $(-2,0)(1-t) + (2,0)t = (-2 + 4t, 0)$, $t$ from 0 to 1.
So then $\frac{dx}{dt} = 4$ and $\frac{dy}{dt} = 0$ and we have to compute $$\int_{purple} 2x + y dx + x dy = \int_0^1 2(-2+4t) * 4 dt = 0$$
potent lotusBOT
Azyrashacorki
knotty cypress
yes ;D
calm coralBOT
@knotty cypress Has your question been resolved?
marsh valley
Oh yeah ok
The half circle actually gives 0 on its path
So purple path also gives 0
Their solution uses the fundamental theorem of line integrals to compute the line integral for the whole path that starts at (2,0) and ends at (0,-2)
marsh valley
Like you'd use green's theorem to say that Red + Purple = 0, so then Red = - Pur...
But this still works, it’s just that the integrals along purple and red are both 0
In the solution they computed the potential function to be x^2 + xy, and then evaluating at the endpoints gives -4
But you could also just compute the integral along the green line
knotty cypress
but 0 is not a correct answer or?
marsh valley
No it’s -4
But the integral along the red semi circle is 0
So it turns out that only the green bit matters
I haven’t computed it but I suppose it gives -4
r(t)= (-2,0) (1-t) + (0,-2)t = (-2-2t, -2t)
dr/dt = (-2, -2)
Anyways
knotty cypress
i see
and if I want to use standard parameterization right away instead of greens?
which they said was an option? xd
marsh valley
Then you parametrize the semicircle first as (2cos(t), 2sin(t))
And compute the line integrals directly
t from 0 to pi
And then parametrize the green line
Add the results
knotty cypress
what would that look like if we calculate it? :(
marsh valley
Hum
well r’(t) gives (-2sin(t), 2cos(t))
I’m sorta walking rn so I can’t really write it all out
But you’d have to use IbP eventually as you’ll get integrals of sin(t)cos(t)
But they should cancel out since the integral is 0 !!
marsh valley
It comes from the green line
The whole line integral is red + green
And red gives 0
marsh valley
r(t)= (-2,0) (1-t) + (0,-2)t = (-2-2t, -2t)
So you parametrize the line segment like this
marsh valley
That’s a way of parametrizing a segment between two points
knotty cypress
i mean if we want to go between don't we want
marsh valley
If you have (a,b) and (c,d) and want to parametrize the line segment vetween them an easy one is
(a,b) * (1-t) + (c,d) * t
knotty cypress
which would be?
(a+ct, bt+dt)?
oh wait
it's not a , in the second parenthesis
knotty cypress
(a-at + ct, -bt + dt)?
marsh valley
Bottom also has a standalone b
knotty cypress
i thought you just did end point minus starting point to get the direction vector
marsh valley
But you want a path not a vector
knotty cypress
but what would the final thing be then?
marsh valley
It’s equivalent though
You could write the same expression as (a,b) + tv
Where v is the direction vector between the endpoints
marsh valley
but what would the final thing be then?
Well you have your parametrizing of the green segment
Then you apply the definition of a line integral
knotty cypress
(a(1-t) + ct , b(1-t) + dt)?
marsh valley
Yeah
knotty cypress
so for us
(-2,0) and (0,-2)
(-2+2t, -2t)
?
and then we'd just replace the x and y's with that in the integral
and we'd have 0<= t <= 1?
knotty cypress
and what would that look like then?
ugh
what is x even equals to with this
-2+2t?
x = -2+2t
y = -2t
dx/dt = 2
dx = 2dt
?
dy/dt = -2
dy = -2dt
dxdy = -4dt^2 xd?
nah that has to be smth wrong from me
knotty cypress
.close
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leaden bramble
Am I right here?
calm coralBOT
Am I right here?
hasty fiber
,rccw
potent lotusBOT
hasty fiber
how did you get that answer?
topaz gulch
timber abyss
Its E
With questions like these, start by considering whether it’s defined, then ask about continuity and differentiability
For what values of x is f(x) defined and why?
timber abyss
Am I right here?
What is f equal to at 0?
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