still not, but there are only 2 options, or the equation is wrong , or you made a mistake while computing the answer, so you should check if all the computations are correct
#help-19
mystic saffron
I checked it 2 times
Maybe I made the equation wrong
try to write it in the most simply way
Wait
for example , i think the first line should be l = 2g
How?
it says for the ones to failed only english
and only science
All those who failed ONLY in English are {(M intersection S)-(E intersection M intersection S)}
and also the 3 math
like, l = 2g + 3f
mystic saffron
All those who failed ONLY in English are {(M intersection S)-(E intersection M i...
yes
Doesn't "l" include all those who passed in English
oh yeah
I missread
mystic saffron
I missread
If you get the answer ping me
mystic saffron
All those who failed ONLY in English are {(M intersection S)-(E intersection M i...
ok
@mystic saffron ok, the problem must be for sure , in making the first equation, since it is the harder part of the problem
odd edgeBOT
@mystic saffron Has your question been resolved?
mystic saffron
also , why your set U is not defined as E∪M∪S + x = U?, since x is what are you trying to find
@mystic saffron Essentially , I dont know why did you write -65 , since you just should use the equations, to solve every a,b,c,d,f,g, then that summ be substracted from 260
by the way , is not possible you get like b as a fraction, maybe you did a mistake, since that fraction should be in some proportion to other set, i mean , cant be 5/2 students who did a test, it should be (5/2)*k, where k is the cardinality of some set there
glacial pollen
Hm
lethal wharf
?
glacial pollen
So do you just keep trying combinations until you get an answer?
There isn't a method?
glacial pollen
And how does that last question still equate to the full polynomial?
How does the ^2 get restored
I did the math ^2 doesn't return
Does 2x multiplied by 3x create ^2 somehow?
Or is it something to do with removing the parameters?
glacial pollen
No
lethal wharf
$-13x = 2x-15x$
clever fjordBOT
jandro
spring smelt
So do you just keep trying combinations until you get an answer?
how did you do the first 3 problems?
glacial pollen
Is 2x multiplied by 3x 6x^2
spring smelt
yes
glacial pollen
I didn't that's the answer sheet.
glacial pollen
Is 2x multiplied by 3x 6x^2
How
spring smelt
there is a method for this
because 2*3 is 6
and x*x is x^2
so together its 6x^2
glacial pollen
Kk
spring smelt
all good?
spring smelt
if that was all, remember to close the chat!
spring smelt
yes
glacial pollen
Do you do inverse operations or give input values for X to attain the value were it is zero now?
And what was the purpose of factoring it
spring smelt
to find the values of x
usually the equation is equals to 0
then you will have 2x-5=0 or 3x+1=0
you solve those individually and you will get 2 values of x
glacial pollen
So for both would you just add or subtract the terms or would you start by trying to multiply by 2x?
spring smelt
why multiply 2x?
glacial pollen
Order of operations
But to be fair this is inverse so
And I should've remembered inverse doesn't appear to have any rules
So adding and subtracting the terms?
spring smelt
what do you mean by inverse
glacial pollen
Inverse operations solving for X
spring smelt
like 2x-5+5=+5?
glacial pollen
2x +5 = 0 +5 = +5
spring smelt
yea
glacial pollen
Is that what you're suppose to do?
spring smelt
yes
glacial pollen
Oh
spring smelt
but
glacial pollen
Kk
glacial pollen
?
That is what I did xd
I just didn't write -5 turning into +5
And put +5
(Real)
spring smelt
2x +5 = 0 +5 = +5
oh cause you wrote 2x+5 here
glacial pollen
Real
Is that the solution?
Or do I have to divide by 2 or something (though that probably wouldn't work bc uneven)
spring smelt
no
glacial pollen
Isn't x suppose to be zero
spring smelt
no
glacial pollen
O_O
glacial pollen
oH
Ok
Literally the only time I managed to at least somewhat manage to understand lol
spring smelt
do you know how to factorise?
glacial pollen
It's not factorization "/
spring smelt
but the video you were watching was on factorisation
glacial pollen
We are taught factorization is 10 5 x2. Or. 20 5x4 10x2
What they did was entirely new
They were like
glacial pollen
They could literally delete 10 and turn it into 1 if the other numbers around it would equate to it
Oh
That explains alot
spring smelt
do you need to know factorisation or was that all?
glacial pollen
So once you solved the other one
0+1 = 1 div 3 0.3333333
glacial pollen
but theres two solutions cause you still have 3x+1=0
Error
spring smelt
error?
spring smelt
yea its correct
glacial pollen
Do I put these into the normal equation now?
spring smelt
but i think its neater to do like this
glacial pollen
O?
spring smelt
3x+1=0
3x=-1
x=-1/3
spring smelt
Do I put these into the normal equation now?
your answer is just x=-1/3 or 2.5
glacial pollen
That's it?
spring smelt
to check if youre correct, you can sub those numbers into the quadratic equation and see if its equals to 0
spring smelt
That's it?
yea
glacial pollen
to check if youre correct, you can sub those numbers into the quadratic equation...
Oh yessss
That
glacial pollen
Ok but for the last question again
Did they just try random combos until they got that?
spring smelt
no
glacial pollen
Did they narrow it down based on what they new the likely values were?
Like how would one guess 13x turns into 1?
And 5 moves
spring smelt
thats why i asked if you know how to factorise
glacial pollen
Oh I dont
spring smelt
let me draw out something for you
spring smelt
we'll start simple with the first question
you know its in the form of ax^2+bx+c right
glacial pollen
Yes
spring smelt
you draw this thing out first then write the bottom with ax^2 c | bx
can you tell its in that form?
glacial pollen
A doesn't have the coefficient the second one doesn't either nor does it have x and c isn't a term or y
Unless it doesn't have as much rules as one would think and so long as it has ^2 it is considered in the form.
spring smelt
this is what i do next
lets start on the first line
basically what im doing is think of what multiplies what to give x^2
the only scenario is x times x right?
spring smelt
likewise for -20, think of a way to multiply 2 numbers such that it gives -20, but before you pick any 2 multiples
the multiples have to add up to be -1
we can do 20x1, 10x2, 5x4
but the first 2 doent give -1 when added up, so it can only be 5x4
-1 because theres a negative sign in -20
to check if youre correct, cross multiply like the arrows and write the answer on the rightmost column
when added together, it should be equals to -x
and youre left with 2 equations if you read it horizontally
x-5 and x+4
glacial pollen
Oh!
If I am guessing correctly erm.
This is basically a multiplication table for solving which values will work and add properly and the 20 and -1 were parts of the equation you chose? (I thought it was related to the original one so I was confused)
spring smelt
somewhat yes
glacial pollen
What did I miss?
spring smelt
the left side is finding the multiples that will give the bottom number
and the right side is for you to check by adding them up
you can try to draw a table similar to mine for the second question and let me see your attempt
for negative numbers it can be tricky because you might mix up 5x-4 and -5x4
but here we need -1 so it has to be -5x4 cause -5+4=-1 as compared to 5-4=1
glacial pollen
These gnats.... (I swear they have a damn npc spawner block or something)
ANGERY:
Ok one second let me think
Can I come back in 10 minutes multiple friends dmed me I got raided by 2million gnats and I was trying to eat at the same time ._.
Welp see you then
odd edgeBOT
@glacial pollen Has your question been resolved?
alpine abyss
quartz pier
Im watching one of my lectures (there is no audio) and the prof is proving this question
quartz pier
Im confused about how
this is true?
is it because
3 | 3 +(a-b)
so we assume 3 | (a-b) to be true aswell?
and we know (a+1)-(b+1) = a+1-b-1 = a-b
oops im an idiot I just read the IH part again
.close
odd edgeBOT
quartz pier
.reopen
odd edgeBOT
✅
quartz pier
Im confused by the inductive step for 2)
what exactly is going on here and how does this help us prove
so they divided it into 2 cases where y = 0 or y > 0
for y = 0, we know 3|x-y => 3|x => x>= 3 but why did they say apply IH to (x-3, y)
same for second why did they say apply IH to (x-1, y-1)
ik that (x-3, y) is true coz of IH but im confused about why the -3
same for second part why the -1
oops im an acc idiot
.close
odd edgeBOT
signal oar
So how do we do this?
wise badge
if you have a set of vectors, you can always get an orthonormal basis for the span with gram schmidt
sleek moth
From the equation you know that (1, -1, 1) is orthogonal to the plane
so the problem reduces to finding two orthogonal unit vectors that are orthogonal to (1, -1, 1)
wise badge
and you can also find a vector orthogonal to two others with the cross product
sleek moth
so the problem reduces to finding two orthogonal unit vectors that are orthogona...
The most elegant solution would probably be to combine this observation with the Gram Schmidt algorithm
wise badge
then find a vector obviously orthogonal to (1,-1,1) and then cross
you're thinking too hard about it
signal oar
sleek moth
The most elegant solution would probably be to combine this observation with the...
although you'd have to come up with two other vectors such that the three vectors together form a basis, which I suppose should not be taken for granted
wise badge
although you'd have to come up with two other vectors such that the three vector...
cross product gives orthogonal, and orthogonal (and nonzero) means independent
signal oar
then find a vector obviously orthogonal to (1,-1,1) and then cross
Then how do we do this quicker?
sleek moth
then find a vector obviously orthogonal to (1,-1,1) and then cross
I don't think finding a vector like that is super obvious
wise badge
choose one of the constants zero, one of the constants 1, and then find the other
signal oar
Oh
wise badge
I don't think finding a vector like that is super obvious
(1,1,0), (1,0,-1), (0,1,1) are all very easy choices
signal oar
$\mrm{0 \ 1 \ 1}$
wise badge
yeah
sleek moth
choose one of the constants zero, one of the constants 1, and then find the othe...
sure fair enough
wise badge
yeah
wise badge
choose one of the constants zero, one of the constants 1, and then find the othe...
even this is thinking too hard about it imo
wise badge
sure
wise badge
that works
signal oar
Except we need scale them to be unit vectors I guess
signal oar
Yeah
odd edgeBOT
odd edgeBOT
Available help channel!
jolly zephyr
odd edgeBOT
odd edgeBOT
odd edgeBOT
Available help channel!
weak bluff
===========/============
Johnny has 10 stones in his ownership. Two of them have the same color.
Johnny puts them in a line like this:
🔴🟠🟡🟤⚪️⚫️🟢🟢🔵🟣
But he doesn't like that the two green ones are next to one another.
a) How many different ways are there to arrange the stones without the two green ones touching?
b) what about if Johnny had 3 or more of the same color stones? What if he had n stones of the same color?
c) what if Johnny had more than 10 stones? What if he had m stones of which 2 have the same color?
d) Can you come up with a way to express the amount of ways to arrange m stones of which n have the same color?
What is the function N(n,m), where N is the number of possible arrangements of m stones of which n have the same color such that the stoness with the same color do not touch?
Hint: configurations like this
🟢🟡🟠🔵🟣🟤⚪️⚫️🔴🟢 with the same color at the start and end of the line must be included!
Hint: symmetric configurations are indistinguishable!
🟢1🔴🟢2 = 🟢2🔴🟢1
Hint 2:
N(n,m) = N_edge(n,m) + N_center(n,m)
🟢 | 🔴⚫️⚪️🟤🟣🔵🟠🟡 | 🟢
Also, dont forget that this:
🟢🟡🟠🔵🟣🟤⚪️🟢⚫️🟢 (n = 3)
and this:
🟢⚫️⚪️🟢🟤🟣🔵🟢🟡🟢 (n = 4)
are valid configurations as well
BONUS QUESTION 1 (thanks to @lordvoldemort_0099993300)
e) what if the line was looped so that the ends would touch?
weak bluff
BONUS QUESTION 2
f) now Johnny wants to put his stones in a square like this:
🟢🟡🟠
🔵🟣🟢
⚪️⚫️🔴
This means he needs a total of m² stones. Only configurations with directly adjacent stones of the same color are invalid. Example:
🟢🟢🟣
🔴⚫️⚪️
🟠🟡🟤 invalid
🟢🟡🟠
🔵🟢🟤
🔴⚪️⚫️ valid
How many combinations are there?
g) What is N(m,n) if configurations with diagonally adjacent stones are not allowed?
Ultra bonus question: until here we assumed that the stones are indistinguishable, that is:
🟢1🔴🟢2 = 🟢2🔴🟢1
what is the expression for N(m, n) if indistinguishability is violated, that is:
🟢1🔴🟢2 =/= 🟢2🔴🟢1?
Apply the condition of distinguishability to all previous exercises, and provide the answer for both the distinguishable case and indistinguishable case for following questions
Ultra mega bonus question:
Johnny now has two pairs of stones with the same color! Actually he has l pairs of the same color. Note that this configuration is also not allowed:
🟢🔴🟢🔵🔴🔵 because blue touches green.
What is N(m, n, l)?
What is N(m, n, l) if 🟢🔴🟢🔵🔴🔵 configurations are allowed?
Hint: make sure to exclude configurarions that do not allow for placement of l pairs of the same color (l = 1 implies n >= 3 and l = 2 implies n >= 7 ).
Ultra giga bonus question:
Now Johnny wants to experiment with stone configurations of the same color that are longer than two. The length of the forbidden stone configuration is k.
What is N(m, n, l, k) in this case?
Note: if k = 3, then configurations such as 🔵🔵🔴🟢🟠 are allowed.
Mega gigachad ultra bonus question:
What if none of the combinations with 2 or more stones adjacent are allowed, that is 🔵🔵🔴🟢🟠, 🔵🔵🔵🔴🟢, are also not allowed?
Ultra epic mega giga Erdős bonus question
Generalize all of the above exercises to:
-
n dimensional cubic arrangements of stones
-
distill a geometric factor from the equations that allows for variation of the lattice type and make it valid for all known space groups
Erdos question 2:
Go back to question f). Here we did the first and second coordination shell for a square lattice. What if we need to consider that a stone of the same color within the pth coordination shell of another stone of the same color is not allowed?
Can you generalize the coordination shell problem for
- n dimensional cubic space groups, and,
- generalize it for arbitrary space group?
<@&286206848099549185>
.open
.reopen
.stayopen
<@&286206848099549185> has it been 15 minutes?
bitter lodge
<@&286206848099549185> has it been 15 minutes?
!15m
odd edgeBOT
<@&286206848099549185> has it been 15 minutes?
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.
bitter lodge
only ping once
weak bluff
Ok sry
bitter lodge
also no, it has been 8
weak bluff
Sorry
bitter lodge
nws
weak bluff
Maainly the Erdős questions
Actually
I'm mainly stuck on having this exercise be complete
So that it covers all of the combinatorics problems you can imagine regarding the pairing of the stones
So far its decent, especially if I add Erdős question 3 which is to now consider Erdos question 1 and 2 results and make the geometric objects closed
As in question e), which was remarked by a very smart discorder, where we go from line to ring
But then for the higher dimensional cases
As well as the case which is generalized for all space groups
Basically I am looking for feedback on if the exercise is complete in the sense that it covers all cases of the pairing distinguishable and indistinguishable like objects, as well as that all the subquestions are well-posed combinatorics problems
Is there some all-encompassing combinatorics theorem that I could use to test my problems for completeness and well-posedness?
Also, sorry for being a bit, what can be perceived as, rude
There is haste in me for some reason
Johnny needs help
Also sorry if the help question is not really clear in terms of 'i need help solving this or that' 😔
bitter lodge
might i ask what your current level of mathematical background is?
bitter lodge
ok then i recommend you just go pick up a book on combinatorics
weak bluff
Any recommendations on combinatorics in context of defect diffusion
In bulk solids and surfaces
I need time-dependent combinatorics
With creation and annihilation of defects and such
bitter lodge
youre likely better off asking in #discrete-math
weak bluff
And I can't find a comprehensive overview of the role of combinatorial factors in the time dependent defect species concentration per unit area/volume
bitter lodge
or just googling, seeing if people have recommendations on stackexchange or etc
weak bluff
Anywhere in literature
bitter lodge
is that to say you expect there is no literature or that you cannot find it
weak bluff
I have searched over and under and I have not found a such a function family yet which describes the configurations of defects in a material that would lead to destruction
Of the defects
It also applies to diffusion of atoms and molecules on surfaces
If one could find the probability of being in such an annihilating configuration then one can use it to simplify Fick's equation
Or even Levy diffusion
bitter lodge
then all i can recommend is that you try and find some professors working on such problems / in related fields and emailing them
weak bluff
Wel it involves many factors
How can I model the chance a combinatorics or mathematical physics professor is going to join discord
So it was more a shot in the dark I guess
Still, no shooting at all would never hit any target so better off to ask
In Discords larget Math community
And I'm too lazy to make account for reddit, etc, so there's a factor of convenience involved as well
Anyways I guess I know what I am looking for now
Last q
bitter lodge
i doubt reddit is the place to go either
weak bluff
Yeah
So how would you descrbe the branch of combinatorics where some combinations disappear from the set
According to some function f(t)
Or can get modified
bitter lodge
ive no clue, not a combinatorist haha
perhaps it might be worthwhile building up your combinatorial foundations first
such that you can proceed on to these more complex questions
weak bluff
I dont really care
bitter lodge
not only will it make it easier, but youll also know better what youre actually looking for
weak bluff
I just need to see the equation
See, I mostly care about the physics
But yeah you are right
Dude
Well thanks for the discussion
I needed some reflection on the problem I want to solve and also needed new literature and talking about it with you Helpers gave just that
Thanks
.closed
.close
odd edgeBOT
bitter lodge
glad you found what you were after
weak bluff
😁😄
bitter lodge
all the best
weak bluff
Peace brotha
odd edgeBOT
Available help channel!
chilly pond
Could somebody help me to understand this?
odd edgeBOT
Could somebody help me to understand this?
vernal yacht
Could somebody help me to understand this?
Which part are you having problem with?
proven beacon
do you have the original problem
chilly pond
I thought to calculate the points of the vertex, and calculate the segments
here
I dont have any problem with write the code... but I need to know the geometric and/or analytics to do that
I think there is a general formulae winch I can apply to any cases
maybe Olog(n) or less
cause if I will try to get the segments as limits, and brush ever lines inside the limits the complexity will be impracticable for large rectangles sizes.
<@&286206848099549185>
Does somebody is here?
@proven beacon @vernal yacht
Task
A rectangle with sides equal to even integers a and b is drawn on the Cartesian plane. Its center (the intersection point of its diagonals) coincides with the point (0, 0), but the sides of the rectangle are not parallel to the axes; instead, they are forming 45 degree angles with the axes.
How many points with integer coordinates are located inside the given rectangle (including on its sides)?
Example
For a = 6 and b = 4, the output should be 23
The following picture illustrates the example, and the 23 points are marked green.
This is the problem
.close
odd edgeBOT
odd edgeBOT
Available help channel!
pine drift
Hey guys
If I want to list the perpendicular angle would I write segment G is perpendicular with GF or VG with GF ?
pine drift
So I guess I’m wondering if I should include the whole G line or the little part of it that connects with GF
mental ridge
Hey guys
If I want to list the perpendicular angle would I write segment G is p...
if g denotes a line then the former
torpid owl
looks like v is the line?
pine drift
Wait I can show the whole thing
It’s like this
The answer shows its GF with v
Whyyyy
Does it have to be specifically like that?
mental ridge
this is a very confusing notation. You seem to be using G as both a line and a point
torpid owl
v is the line. and u and t
pine drift
Ok so G is a point got it
And I can’t say vG because one is a line and one is a point right
And I can’t just say G because that would be a point not a line
So I can only say v?
But that means including the wholleeee line
Not just until the G point
Right
And that’s ok to do?
mental ridge
yes
pine drift
yes
I have another question
Can I say angle UST is congruent to UTS ? Instead of angle S is congruent to T
Or does it have to be only 1 letter
torpid owl
I've seen both being used
mental ridge
since s is a point, the former is more correct
odd edgeBOT
odd edgeBOT
Available help channel!
pliant iris
how do i do (v)? The answer is 144
odd edgeBOT
how do i do (v)? The answer is 144
pliant iris
im thinking like 5x4x3*2x1(2x2)2x1
devided by 2!2!
cuz in the middle theres 20's avaible and 2L's avaible
so its 2x2
and then theres just 5 left
so its (2x2)*5!
devided by 2!2! for the 2L's and 2O's
which gives u 120
not quite 144
odd edgeBOT
@pliant iris Has your question been resolved?
pliant iris
<@&286206848099549185>
odd edgeBOT
@pliant iris Has your question been resolved?
mystic saffron
<@&286206848099549185>
i'm thinking there would be some more complications because the L can be to the left of the O or to the right of the O
pliant iris
Yah
But
That would be 120 to one side and then x2 for the other side
Which is way above 140
mystic saffron
hmmm
oh
are we counting for both l's next to it
because it says exactly one
also it might not be 120 for each side
pliant iris
Hmm
Yah
welp
i dont know
i cant do this
bruh
why is this so hard
everyone ive asked dosent know
and my teacher has decided to stop existing for the next couple of days to prepare for some other stuff
is it time to give up
pliant iris
i giveup
odd edgeBOT
Available help channel!
acoustic grotto
odd edgeBOT
acoustic grotto
how do I solve this
this is what I've done so far
this is the original question
mystic saffron
odd edgeBOT
mystic saffron
Is there a khan academy or similar practice where I can repeat these problems to memorize them
long chasm
I am pretty sure, but you can look up a few worksheets
Lemme pull up the link
I can’t seem to find it
grave mason
mystic saffron
Ty
odd edgeBOT
odd edgeBOT
Available help channel!
undone merlin
I solved this equation algebraically but I don’t understand how I would solve it graphically. Would someone be able to help?
eager crater
You can solve it graphically by putting the equation in to a graphing program
undone merlin
Like this?
Apparently this is what it should look like, but I don’t really understand why the graph would look this way
undone merlin
I entered it as two equations like this, I believe this is the way our teacher wants us to do it
eager crater
no because your calculator uses y. but there is no y
undone merlin
Yes, that is what I am confused about
Because my teacher told me to do it like this
But I don’t get why
undone merlin
Two vertical lines?
eager crater
yes
peak maple
odd edgeBOT
peak maple
How do you do b
peak maple
ah yh i j didnt see P was thousands so i didnt get the mark scheme
thanks tho
.close
odd edgeBOT
odd edgeBOT
Available help channel!
queen moat
odd edgeBOT
queen moat
.close
odd edgeBOT
odd edgeBOT
Available help channel!
broken oyster
part b
brittle beacon
What did you get in part (a)?
broken oyster
q=x^2-3x+5
r=-13x+12
brittle beacon
You could, e.g., write $x^4 - 2x^3 + x^2 + ax + b = {\color{green} (x^4 - 2x^3 + x^2 - 5x + 7)} + (a + 5)x + (b - 7)$, from where you know the first term from your division
clever fjordBOT
@brittle beacon
brittle beacon
At that point, you want the remainder of that to be identically zero
broken oyster
You could, e.g., write $x^4 - 2x^3 + x^2 + ax + b = {\color{green} (x^4 - 2x^3 +...
why do u write (a+5)x+(b-7)?
brittle beacon
why do u write (a+5)x+(b-7)?
the +5 and the -7 cancel out so that it doesn't change the original expression, and then you can use what they've told you
broken oyster
the +5 and the -7 cancel out so that it doesn't change the original expression, ...
oh i see
broken oyster
the +5 and the -7 cancel out so that it doesn't change the original expression, ...
wait what have they told me
brittle beacon
wait what have they told me
Well, what they made you find rather, that
[
{\color{green} x^4 - 2x^3 + x^2 - 5x + 7 = (x^2 - 3x + 5)(x^2 + x - 1) - 13x + 12}
]
clever fjordBOT
@brittle beacon
broken oyster
Well, what they made you find rather, that
\[
{\color{green} x^4 - 2x^3 + x^2 -...
so do i sub that into
broken oyster
<@788085606483361802>
this?
broken oyster
alr lemme do that rq
broken oyster
You should <:SCgoodjob2:750436894915166321>
i'm lost what do i do after this?
brittle beacon
When you do, you basically have that $x^4 - 2x^3 + x^2 - 5x + 7 = (x^2 - 3x + 5) {\color{red} (x^2 + x - 1)} + \text{stuff}$, and you want the stuff to be zero
broken oyster
When you do, you basically have that $x^4 - 2x^3 + x^2 - 5x + 7 = (x^2 - 3x + 5)...
ohhhhhhhhh i get u
clever fjordBOT
@brittle beacon
broken oyster
When you do, you basically have that $x^4 - 2x^3 + x^2 - 5x + 7 = (x^2 - 3x + 5)...
Alright i got it now thanks
.close
odd edgeBOT
broken oyster
.reopen
odd edgeBOT
✅
odd edgeBOT
odd edgeBOT
Available help channel!
hollow jolt
which courses should I look over to be able to integrate these 2?
hollow jolt
i think they use 2 different methods but I'm not sure of their names so I'm pretty much just shooting in the dark
yes, I have an exam tomorrow and there's around 30 courses on my university page, with 20-30 pages each, so unfortunately I don't have the time to skim through them all
someone said that i need to use the Residue theorem for the first one
and yes those 2 exercises appear in a mock test that our professor posted for us
not yet, i'm still working on learning some other stuff now, but I thought I'd put together a learning plan for those 2 kind of exercises in the mean time
odd edgeBOT
@hollow jolt Has your question been resolved?
modern sundial
You can do a problem like that using standard techniques from a first semester in Calculus. Using the Residue theorem is probably beyond the capability of someone that is asking a question like this to begin with.
hollow jolt
yes
hollow jolt
You can do a problem like that using standard techniques from a first semester i...
eh i'll still try, thanks for the answer though
.close
odd edgeBOT
odd edgeBOT
✅
odd edgeBOT
odd edgeBOT
Available help channel!
pale wharf
how can i normalise this?
vernal yacht
how can i normalise this?
What does it represent?
stable axle
Who came up with 2n-1 is odd, it's pretty legit
odd edgeBOT
Who came up with 2n-1 is odd, it's pretty legit
@stable axle Has your question been resolved?
odd edgeBOT
@stable axle Has your question been resolved?
south plume
Who came up with 2n-1 is odd, it's pretty legit
what is your question, exactly?
keen canyon
odd edgeBOT
summer river
okay let's start this off by making some comments
like b in terms of h,k, an dj
we can do this by just expanding (hx+k)(x+j)
since it's simple enough
@keen canyon would you like to do this?
odd edgeBOT
@keen canyon Has your question been resolved?
tulip nest
So the question is use synthetic division to test possible rational roots, until I'm able to find the complete factorization. But I forgot how to factor lmao. Polynomial is 3x⁴-17x³+9x+41x+12
odd edgeBOT
So the question is use synthetic division to test possible rational roots, until...
@tulip nest Has your question been resolved?
tulip nest
<@&286206848099549185> 
mystic saffron
So the question is use synthetic division to test possible rational roots, until...
What's tripping you up? You might need to continue using synthetic till you have a quadratic. Can't really do anything with the 3rd degree polynomial you found.
tulip nest
What's tripping you up? You might need to continue using synthetic till you have...
Maybe thats it, I started with 3x⁴-17x³+9x+41x+12, maybe I did my division wrong?
I'm not really fluent with mathematics, I am trying to get better though
mystic saffron
Your division is correct. It looks like you just need to find another root, you can do that by continuing using synthetic division and the possible roots you found from the original polynomial. You just use the new polynomial you found as the numerator/dividend
tulip nest
So just pick out the factors of 3 and 12 for this new polynomial?
mystic saffron
Yeah I believe so.
tulip nest
Thanks bro
mystic saffron
No problem. Let me know if that works. You should have a quadratic after that which can be factored using standard factoring methods.
dry plank
Hey can someone please check my answers :P.
odd edgeBOT
Hey can someone please check my answers :P.
@dry plank Has your question been resolved?
odd edgeBOT
@dry plank Has your question been resolved?
glacial pollen
the multiples have to add up to be -1
?
-10 x 2 = -20
10 x -2 = -20
i thought about that for a long time and i still couldnt figure out why you would need that since it didnt appear to connect contextually or logically. Why would you even add in the first place? That doesn't appear to occure in the problem and even if it did how would it affect -20? I can't see how this relates to the problem, especially when it seems all you would really do is just look at the factor list and choose factors that work when you can do -10 x 2 10 x -2 -5x4 5x-4
odd edgeBOT
@dry plank Has your question been resolved?
finite glade
Can someone help me solve 12 and explain the process to me please and thanks
dawn tiger
one approach might be finding the length of the hypotenuse using distance formula
then finding the distance between A and C in terms of t, and the distance between B and C in terms of t
set up the pythagorean theorem and you should be able to solve an equation for t
odd edgeBOT
@finite glade Has your question been resolved?
finite glade
Awesome thanks
odd edgeBOT
odd edgeBOT
Available help channel!
hollow jolt
can someone please help me understand cauchy-riemann equations?
modern sundial
What specifically do you not understand
modern sundial
I'll be honest, I was willing to explain something conceptual in nature but I'm not going to read that.
hollow jolt
I'll be honest, I was willing to explain something conceptual in nature but I'm ...
you were being rude on a question i asked earlier, now on this one too, at this point just don't bother.. nobody forces you to answer each of them if you don't want to
modern sundial
I answered in the way I did to your other question because you were being given bad advice. For one, that integral was like, 3 or 4 lines of work using a first semester techniques. Someone asking something like that usually is not prepared to suddenly arm themselves with a high powered and relatively technical theorem. At no point did you explain why you felt compelled to use that, all you said was that you were going to try anyway. It is made even more confusing when some other user gave you a now deleted and completely ridiculous computer generated answer that was like 20 lines long and that was somehow acceptable to you.
hollow jolt
I answered in the way I did to your other question because you were being given ...
well, i'm trying my best. I couldn't attend university a lot this semester because of personal issues, and i'm trying to catch up on stuff for an exam that is tomorrow. My bad for not mentioning that.
the exam is in 6 hours and i'm functioning on pure caffeine, so sorry if my answer felt snappy.
modern sundial
I'm not going to delve into your personal issues, but you should remember that there are resources available at every university if you need help dealing with complicated situations, and there are always options to get an exam deferred so you can put your best foot forward instead of rushing through excessive material.
hollow jolt
.close
lavish laurel
how do I graph complex numbers here ?
odd edgeBOT
how do I graph complex numbers here ?
lavish laurel
−1±2i
so the real part , a = -1 and the imaginary part b = 2
but my notes say that phi, the angle is ALWAYS counter clockwise
raw trail
Yes
lavish laurel
so would it look like this?
like the anlge phi is beginning at zero and ends there
so the angle in my triangle is - phi ?
lavish laurel
But if Arctan(b/a) = phi
then we have a negative angle
and the angle drawn is positive
arctan of -2 is -1.107etc
does that make sense
odd edgeBOT
@lavish laurel Has your question been resolved?
odd edgeBOT
@lavish laurel Has your question been resolved?
lavish laurel
<@&286206848099549185>
sullen ferry
But if Arctan(b/a) = phi
arctan only outputs angles between -pi/2 and pi/2
sullen ferry
phi=pi-the angle from the terminal arm to the negative x-axis
so you'd need to add pi to it, like this
odd edgeBOT
@lavish laurel Has your question been resolved?
turbid sleet
Hi, I am currently learning about the discrete wavelet transform. One thing I dont understand is that when I go down a level (split the signal frequencies) and I apply the transform, keeping the lower frequencies, the lenght of the orginial signal is cut in half. I am using the python library pywt to do this, but form my research this is not specific to the python library, but the wavelet transform in general. I dont know why my signal is cut in half, I know it has something to do with the fact that I keep the lower half of frequencies but I cant connect this with the signal lenght. Any help would be greately appreciated.
odd edgeBOT
Hi, I am currently learning about the discrete wavelet transform. One thing I do...
@turbid sleet Has your question been resolved?
odd edgeBOT
@turbid sleet Has your question been resolved?
odd edgeBOT
@turbid sleet Has your question been resolved?
odd edgeBOT
@turbid sleet Has your question been resolved?
reef forum
how do I start
odd edgeBOT
how do I start
reef forum
I just need to know how to start
odd edgeBOT
@reef forum Has your question been resolved?
stable axle
what is your question, exactly?
i guess like who came up with that it is true how come it isnt like documented
odd edgeBOT
i guess like who came up with that it is true how come it isnt like documented
@stable axle Has your question been resolved?
odd edgeBOT
odd edgeBOT
Available help channel!
reef sierra
Hi, I need some help with linear diophantine equations
Consider $ax + by = c$ where $c = k * gcd(a, b)$, that is, c is a multiple of the gcd.
\vspace{5px}
The way to solve this is to first obtain a solution $(x_0, y_0)$ of $ax + by = gcd(a, b)$.
\vspace{5px}
Then you scale it by c/g so the scaled solution is $(c/g * x_0, c/g * y_0)$ where g is the gcd. Let this be denoted by $(x_1, y_1)$.
\vspace{5px}
So, now you have $ax_1 + by_1 = c$.
\vspace{5px}
Then, to find the general solution, you'll add and subtract $ p * ab / g $ to the LHS of the above equation. You'll get $ a * (x_1+ pb/g) + b * (y_1 - pa/g) = c $ where p is an integer parameter
\vspace{5px}
Thus, the general solution would be $ ( x_1 + pb/g, y_1 - pa/g ) $
\vspace{5px}
Now, let's say that you want to find the number of solutions between $ [x_{min}, x_{max} ] $ and $ [y_{min}, y_{max} ]$
\vspace{5px}
So, you'll put the above value of the parameterised solution into the inequations $x_{min} \leq x \leq x_{max}, y_{min} \leq y \leq y_{max} $ . You'll get intervals for p. You can then find the intersection of these inequations.
\vspace{5px}
My Question: This is fine for positive coefficients a, b but what if they're negative?
reef sierra
(Sorry for the multiple deletes)
(Latex timed out and the formatting changes I made weren't being registered)
clever fjordBOT
DarkCharlotte
odd edgeBOT
@reef sierra Has your question been resolved?
reef sierra
<@&286206848099549185>
royal flame
Hell no, I’m getting a seizure reading the first problem
reef sierra
sorry, is it badly formatted?
reef sierra
ah, okay
bitter lodge
you just need to do it in cases
so that you know the sign of a and b
such that when you deal with the inequalities you know when to flip them
reef sierra
why would itn ot work for negative coefficients
So the solution to the equation $ax + by = gcd(a, b)$ is obtained by the Extended Euclidean Algorithm (the same as Euclid's GCD algorithm but you back-substitute the remainders). But the gcd is only defined for positive numbers. That's why I thought there might be some trouble
clever fjordBOT
DarkCharlotte
bitter lodge
the gcd of a negative is just the same as that of its psotiive
reef sierra
such that when you deal with the inequalities you know when to flip them
So, I ignore the negative sign. Then, I flip the inequality?
bitter lodge
yeah so when you get to xmin - x1 <= pb/g <= xmax - x1
you need to divide by b and so you flip the inequalities g/b (xmin - x1) >= p >= ...
reef sierra
the gcd of a negative is just the same as that of its psotiive
okay, so i can get the solution to $ |a| x + |b| y = gcd( |a|, |b| ) $ by this
where |.| is the absolute value function
but is that the solution of ax + by = gcd(a, b)?
gcd(a, b) = gcd( |a|, |b| ) so the RHS is fine
but the LHS don't match
bitter lodge
say x0, y0 is a solution to |a| x + |b| y = g
then see that |a| = sgn(a) a so then a (sgn(a) x0) + b (sgn(b) y0) = g
reef sierra
Ah okay, that makes sense. The solutions to the signed equation are merely the solutions to the unsigned equations multiplied by the appropriate sign.
So now we obtained a solution for ax + by = gcd(a, b) where a, b are signed
Then we can scale the equation by c/gcd to get the solution for ax + by = c
And then we can add-subtract p * ab/g on the LHS to get a parameterised solution
And then we can put it in our inequations to get intervals for k
Ah, is this where we'll need to take care of flipping signs?
reef sierra
yeah so when you get to xmin - x1 <= pb/g <= xmax - x1
you need to divide by b a...
If b is positive, no need to flip the inequations but if its negative then you have to. and i guess that's all we need to do!
thanks for the help :D
reef sierra
Hell no, I’m getting a seizure reading the first problem
btw, @royal flame , you could see these resources if you want:
the equation i gave is a slight variation of Bezout's lemma - the RHS constant is different (https://en.wikipedia.org/wiki/Bézout's_identity)
and Euclid's Extended Algorithm (extended because it builds on top of the original euclidean algorithm to find the GCD of two numbers) can be used to find any one solution (it has infinite) to this equation (https://en.wikipedia.org/wiki/Extended_Euclidean_algorithm)
this link describe the way to find the general solution of this equation https://math.stackexchange.com/questions/20717/how-to-find-solutions-of-linear-diophantine-ax-by-c
zinc oriole
odd edgeBOT
sterile blaze
cant the right angle be opposite:(
zinc oriole
yes
i dont know how
how do i do that
i already solved the squares
but idk where to go next
yes
subtract on both sides?
324
um
divide?
square root
so sqrt of 576
24
0.8
24/30
4/5
thank you so much
odd edgeBOT
If you are done with this channel, please mark your problem as solved by typing .close
zinc oriole
.close
odd edgeBOT
gusty pulsar
@zinc oriole please don't open multiple help channels at once
odd edgeBOT
Available help channel!
lone elbow
find differentiable $f: \mathbb{R} \to \mathbb{R}$ such that $\2f(x)f'(x)\left(\frac{x^2}{2} + 2\right) = x \left(f^2(x) + 1\right)$ and $f(0) = 5$
clever fjordBOT
studying_calc_real_analysis
haughty scaffold
usually when you get this type of problem you want to make it so that all f(x)'s are on one side and all x's or constants are on the other
then look for antiderivatives
so just as deepfriedpack said, dividing by those two accomplishes that
lone elbow
,, \frac{2f(x)f'(x)\left(\frac{x^2}{2} + 2\right)}{\left(f^2(x) + 1\right)\left(\frac{x^2}{2} + 2\right)} = \frac{x \left(f^2(x) + 1\right)}{\left(f^2(x) + 1\right)\left(\frac{x^2}{2} + 2\right)} \ \implies \frac{2f(x)f'(x)}{\left(f^2(x) + 1\right)} = \frac{x}{\left(\frac{x^2}{2} + 2\right)}
haughty scaffold
yup, and theres also a 2 you forgot to add on the left hand side, its 2f(x)f'(x)....
but you notice that a few things simplify
clever fjordBOT
studying_calc_real_analysis
lone elbow
I dont get it why are we dividing by $\left(f^2(x) + 1\right)\left(\frac{x^2}{2} + 2\right)$
clever fjordBOT
studying_calc_real_analysis
mystic saffron
you see no f'(x)/f(x) form?
lone elbow
usually when you get this type of problem you want to make it so that all f(x)'s...
ok.
now what?
haughty scaffold
now notice the 2f(x)f'(x) on the left hand sides numerator...do you see what its antiderivative is>
its a tricky one if you havent seen it before
lone elbow
,, \frac{2f(x)f'(x)\left(\frac{x^2}{2} + 2\right)}{\left(f^2(x) + 1\right)\left(\frac{x^2}{2} + 2\right)} = \frac{x \left(f^2(x) + 1\right)}{\left(f^2(x) + 1\right)\left(\frac{x^2}{2} + 2\right)} \ \implies \frac{2f(x)f'(x)}{\left(f^2(x) + 1\right)} = \frac{x}{\left(\frac{x^2}{2} + 2\right)}
clever fjordBOT
studying_calc_real_analysis
haughty scaffold
at this point what we want to do is isolate f(x) to find the function. since we also have an f'(x) we know that we want to find an antiderivative
lone elbow
how do I do that
haughty scaffold
so, the antiderivative of 2f(x)f'(x) is actually f^2(x). do you see that? try deriving f^2(x) with the chain rule
lone elbow
,w differentiate f^2(x)
woeful lynx
,, \frac{2f(x)f'(x)\left(\frac{x^2}{2} + 2\right)}{\left(f^2(x) + 1\right)\left(...
I dont do much calc, but youre a real one for this LaTeX, that too on discord
Did you do it in overleaf and then copy paste it here, or did you just straight up write this and send
lone elbow
write this and send, im am cs student, so coding is not too hard for me
woeful lynx
write this and send, im am cs student, so coding is not too hard for me
wow. I do cs asw, but this is a different level compared to what im used to
though to be fair, I dont do much LaTeX stuff
lone elbow
just familiarity, nothing clever
lone elbow
so, the antiderivative of 2f(x)f'(x) is actually f^2(x). do you see that? try de...
how did you integrated that?
woeful lynx
just familiarity, nothing clever
fair
haughty scaffold
when you see the product of f(x)f'(x) you should know that to integrate that you need to have f^2(x) in mind always
woeful lynx
anyways, back to the math now
haughty scaffold
with the chain rule, f^2(x) turns into 2f(x) * f'(x)
haughty scaffold
now we are going to do a trick here, instead of f^2(x) we are going to use f^2(x)+1. the +1 is inconsequential because under the derivative its just 0, but it helps when u notice that on the denominator you have exactly f^2(x)+1
clever fjordBOT
studying_calc_real_analysis
lone elbow
okay
haughty scaffold
now, do you remember what the antiderivative of a function of the form of g'(x)/g(x) is?
haughty scaffold
this one is ln(g(x))
regal leaf
haughty scaffold
deriving ln(g(x)) gives you 1/g(x) * g'(x) by the chain rule again,therefore g'(x)/g(x)
and now we have this
which is exactly of the form we want
the antiderivative of this therefore will be ln(f^2(x)+1), yes?
haughty scaffold
because g(x) = f^2(x)+1 in our case
thats why we also added the +1 on the numerator
to make them identical
we couldnt avoid it
lone elbow
you said derivative of f^2(x) + 1 is 2f(x) * f'(x)
haughty scaffold
it is, and we used that to transform the numerator
but now we are taking the antiderivative of the whole fraction
lone elbow
okay
haughty scaffold
okay does this help?
lone elbow
now what?
haughty scaffold
do you understand the steps i took?
i started off with the equality you have me and made sure to take the antiderivative of both sides
basically integrate both sides
lone elbow
I need to understand how did you do top part
lone elbow
deriving ln(g(x)) gives you 1/g(x) * g'(x) by the chain rule again,therefore g'(...
mmm
haughty scaffold
ah id actually made a mistake too, so it turns out simpler
lone elbow
where?
ln implications part looks ok
but dunno how did you got $f^2(x) + 1 = \left(x^2 + 4\right)$
clever fjordBOT
studying_calc_real_analysis
haughty scaffold
look at this
youre right, i shouldnt have
at this point you need to use the value u are given
lone elbow
haughty scaffold
it tells you that f(0) = 5. so now, if we put in our last relation as x=0 we get ln(26) = ln(4) + c
so c = ln(26)-ln(4) or c=ln(26/4) which is an odd number... kinda worrying but oh well
lone elbow
wait a moment
haughty scaffold
does that make sense?
yeah?
did you see what i did? i just turned the sum into one fraction
lone elbow
deriving ln(g(x)) gives you 1/g(x) * g'(x) by the chain rule again,therefore g'(...
mmm
haughty scaffold
deriving ln(g(x)) gives you 1/g(x) * g'(x) by the chain rule again,therefore g'(...
remember this
yeah
lone elbow
but suddenly its $\frac{d}{dx} \left(\ln\left(f^2(x) + 1\right)\right)$ and I thought it was only $\ln\left(f^2(x) + 1\right)$
clever fjordBOT
studying_calc_real_analysis
haughty scaffold
do keep in mind though, the ln's have absolute values in them
we just remove them later since their contents are positive anyway
haughty scaffold
(0, +inf) but yeah
thats why when we integrate we need absolute values inside the ln
but the expressions in the logs are already positive
so we can discard the absolutes soon after
lone elbow
deriving ln yields g'/g
you integrated both sides for last implication
but
how is that $\int \frac{\frac{d}{dx}\left(x^2 + 4\right)}{x^2 + 4} dx = \ln| x^2 + 4| + C$?
clever fjordBOT
studying_calc_real_analysis
haughty scaffold
we did the same thing on the right side
we have a form of g'(x)/g(x), which is just ln(g(x))'
whats confusing you?
how can i help you understand
lone elbow
"deriving ln(g(x)) gives you 1/g(x) * g'(x) by the chain rule again,therefore g'(x)/g(x)"
haughty scaffold
**studying\_calc\_real\_analysis**
this would be more accurate
clever fjordBOT
studying_calc_real_analysis
haughty scaffold
well at this point, as we said, we have to find the C
and for that you use the point that is given to you, f(0)=5
substitute x=0 in our equation above
what do you get for C?
lone elbow
,, \ln(26) - \ln(4) = C \ \implies \ln\left(\frac{26}{4}\right) = c?
clever fjordBOT
studying_calc_real_analysis
haughty scaffold
yup, so C=ln(26/4)
therefore the right hand side turns into ln(x^2+4) + ln(26/4), or ln(26/4 x^2 + 26)
right?
clever fjordBOT
studying_calc_real_analysis
haughty scaffold
well look
haughty scaffold
lone elbow
okay so how do I check my answer?
haughty scaffold
that'll be an annoying thing to do but youre welcome to do so
find the derivative f'(x) and see if our function satisfies the equation given
the whole thing that was in the start
lone elbow
,w differentiate sqrt((26(x^2))/4 + 25)
clever fjordBOT
lone elbow
<@&286206848099549185>
haughty scaffold
its right, i checked
lone elbow
according to who
haughty scaffold
the two graphs are on top of each other
i took the initial problem, reordered it a little to the form 2f(x)f'(x)/f^2(x)+1 = x/(x^2/2 + 2)
and i see that the graph of each side is the same
so they describe the same function
haughty scaffold
mmm no? this thing also computed it
and it just turns out to be the same thing
thats the answer
haughty scaffold
you mean when we took the antiderivative of 2f(x)f'(x) into f^2(x)?
why we instead added a +1 too?
lone elbow
you said something so that is not the same in numerator denomi
haughty scaffold
it was because we noticed the denominator. we saw that the denominator was f^2(x)+1, and if the numerator was just f^2(x) then wed have a problem...how would we take the antiderivative with the ln?
we wouldnt have identical things in the numerator and the denominator
right?
so to fix that we added the +1, which doesnt change our expression because the +1 just goes under the derivative
but its significant to be able to continue with the problem
the antiderivative of this cannot be found with the ln tactic
because we dont have the form g'(x)/g(x)
do u get it?
haughty scaffold
great
lone elbow
thank you! i am closing
haughty scaffold
nw! have a good one
odd edgeBOT
odd edgeBOT
Available help channel!
analog void
odd edgeBOT
frigid canopy
Is ac parallel to pr
analog void
They share a common angle at the vertex
this angle?
analog void
ok
frigid canopy
OR isn't given ?
analog void
OR isn't given ?
it isnt
frigid canopy
Well if the lines are parallel do you see why they are similar?
analog void
Well if the lines are parallel do you see why they are similar?
thats my stumbling block
so
if theyre both parallel
why does it mean they have same angles
on that line
so they have the same angle at the vertex
frigid canopy
Corresponding angles
As for why they are parallel it's because PQR is an enlargement of ABC with a common centre
analog void
As for why they are parallel it's because PQR is an enlargement of ABC with a c...
cool
analog void
Corresponding angles
doesnt there need to be a third line intersecting?
for coressponding angles to work?
frigid canopy
that would be line PO
analog void
that would be line PO
could u say RO as well?
frigid canopy
yes
ok, it turns out that any shape and its enlargemt are similar
keep that in mind
analog void
ok, it turns out that any shape and its enlargemt are similar
ohhh
ok so
they both share one angle
and if the two smaller triangles r similar due to enlargement
then the base angles must be the same
hence they all have the same angles
frigid canopy
well that proves PQR~ABC
