ohh
#help-36
woven ledge
well we have p sin theta/ q cos theta = p^2/q^2
if we subtract one from both sides
what do we get?
thin cloud
no space at the start
soft zealotBOT
CherryMan
woven ledge
!nosol
dry ledge
**CherryMan**
ye
dry ledge
what
craggy atlas
I divided both numerator and denominator by cos theta
And you have tan theta given as p/q
craggy atlas
Yeah
woven ledge
well i guess that method might be easier
dry ledge
what was the cherry man saying then
dry ledge
well i guess that method might be easier
whats your method
dry ledge
teach me every method cause im 100% sure ill forget either of wm
woven ledge
whats your method
basically my method makes more sense if you knew componendo dividendo
well ill continue
dry ledge
go slow 😭
woven ledge
we have $\frac{p \sin \theta}{q \cos \theta} -1 = \frac{p^2}{q^2} - 1$
soft zealotBOT
CherryMan
woven ledge
so $\frac{p \sin \theta -q \cos \theta}{q \cos \theta} = \frac{p^2-q^2}{q^2}$
soft zealotBOT
CherryMan
dry ledge
why
woven ledge
i made a common denominator
woven ledge
$\frac{p \sin \theta +q \cos \theta}{q \cos \theta} = \frac{p^2+q^2}{q^2}$
soft zealotBOT
CherryMan
woven ledge
similarly if i added 1 instead
dry ledge
these are two diff equations right
woven ledge
now if we divide one equation by the other we get the required identity
woven ledge
these are two diff equations right
yeah but both from the same original one
woven ledge
then divide
this whole fiasco is actually one step of componendo dividendo
dry ledge
ohhh
woven ledge
so if you were allowed you could skip to the required identity directly from this
dry ledge
is this like a theorem
oh identity
im not sure if im allowed this identity cause we've not really been taught this
ill try to derive it
or use the other method
thanks both of yall
woven ledge
welcome!
craggy atlas
componendo dividendo works on any problem not only trig btw
dry ledge
componendo dividendo works on any problem not only trig btw
yeah its an identity thats why
thankx for the help.
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final saddleBOT
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valid carbon
bot test
valid carbon
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✅
valid carbon
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odd seal
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Please don't repeatedly close and claim a new channel with the exact same question. This erases all previous progress made towards your problem and is confusing for helpers, making it more difficult to help you. Please be patient, even if your channel has not received much attention.
tired walrus
i literally said its a test
@ Moderators test cheater
final saddleBOT
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sturdy cloak
final saddleBOT
sturdy cloak
could someone plz explain why we do the opposite for the example below??
late gazelle
when you say the example below, which one are you referring to? or just in general how do transformations of functions work.
sturdy cloak
when you say the example below, which one are you referring to? or just in gener...
translste then reflect
why do we do the opposite here
ie in the top example we do rflect then translate
why do we do the other way round in the bottom one
late gazelle
ah k
so if the coefficient of f(x) is negative its a reflection about the x-axis, where as if its f(-x), its a reflection about the y axis, there isnt an order to how you should do it but it all happens at once, an example would be: consider the function f(x) = x^2, if i want f(-x) i would sub in -x for x, which would give f(-x) = (-x)^2 = x^2. so In this case the function is the same since the power is even. In the other example of -f(x), we put a negative outside the entire function: -f(x) = -[x^2] = -x^2, and if we wanted to horizontally translate: f(x-a) = (x-a)^2 which is a shift a units to the right since the coefficient of a is -1.
Tldr: f(x) is a function which takes an input x and outputs an answer, in the case of f(-x), we sub each x in for -x, in the case of -f(x), we have a factor of -1 outside the function, and f(x-a) is a shift along the x axis a units to the right if a is negative, and left if a is positive.
hopefully that makes sense, if it doesnt lmk
sturdy cloak
so if the coefficient of f(x) is negative its a reflection about the x-axis, whe...
ahh yep perfect makes sense now
thanks!!
late gazelle
and if you want an interactive: https://www.desmos.com/calculator/m8vtiyrlmc
ivory bear
sturdy cloak
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violet junco
violet junco
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heavy briar
Find a closed form of this sum:
sin(1/e) + sin(1/e^2) + sin(1/e^3) + sin(1/e^4) + ..........
final saddleBOT
Find a closed form of this sum:
sin(1/e) + sin(1/e^2) + sin(1/e^3) + sin(1/e^4) ...
heavy briar
Where e is Euler's number.
tired walrus
$\sum_{n=1}^{\infty} \sin(e^n)$
soft zealotBOT
Ann
tired walrus
e^{-n}
soft zealotBOT
Normie
tired walrus
$\sum_{n=1}^{\infty} \sin(e^{-n})$
soft zealotBOT
Ann
tired walrus
like this?
heavy briar
yes
tired walrus
may i know where you got this problem from?
seems quite difficult to find a closed form for this, unless i am missing something that can somehow miraculously be done with taylor series shenanigans.
not even WA gives one
plucky rover
You shouldn't be on discord
dusty quarry
<@&268886789983436800>
lethal totem
!da2a
final saddleBOT
No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/
nimble fiber
oh shi
tired walrus
<@&268886789983436800> underage
nimble fiber
please
tired walrus
sorry, you need to be at least 13 to be on discord at all
not just on here but to use the app
on this server we have to take this seriously
dusty quarry
now I have to check the logs.
i can dm you
tired walrus
plucky rover
It was? Still here for me lol
tired walrus
i see it here
dusty quarry
oh nvm
tired walrus
we got receipts
cyan kayak
amazingly, it's not showing up in logs-deleted.
cyan kayak
oh, it's not there for me, one second
plucky rover
Oh okay nvm it is deleted
I can't reply to it
But it shows up in the client
Wtf discord
autumn sage
guys, need clue
proper dagger
guys, need clue
!occupied
final saddleBOT
guys, need clue
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
@heavy briar Has your question been resolved?
fallen drum
Hello, i am struggling to find r and U1 in this geometric sequence
final saddleBOT
Hello, i am struggling to find r and U1 in this geometric sequence
fallen drum
mainly stuck on after the dividing part
dusty quarry
uhhh
fallen drum
whats wrong
slate narwhal
isn't U_n = U_1 * r^(n-1)
fallen drum
isn't U_n = U_1 * r^(n-1)
yes
slate narwhal
this looks more like U_6 than U_3, no?
final saddleBOT
fallen spruce
not even WA gives one
what is WA?
slate narwhal
wolfram alpha (most likely)
fallen spruce
is it some kind of tool?
slate narwhal
yes
final saddleBOT
Available help channel!
pale violet
final saddleBOT
edgy mauve
What have u tried
pale violet
guys help pls
pale violet
What have u tried
basic trigonometry
craggy plank
basic trigonometry
plz demo
edgy mauve
basic trigonometry
Ur work?
snow bear
asking for help and then saying you don't have time to get help is quite something
pale violet
asking for help and then saying you don't have time to get help is quite somethi...
i have to get phone first for that
any ways help plsss bro
edgy mauve
any ways help plsss bro
Can u find the two numbers for which the product is equal to the product of constant and leading coeffecient and sum if the coeffecient of middle term
pale violet
Can u find the two numbers for which the product is equal to the product of cons...
ya i should use calculator
edgy mauve
ya i should use calculator
U can find it without calculator as well
4,-3
Use these 2
pale violet
ok then
pale violet
Do u k how to do middle term splitting
ya
pale violet
ok
pale violet
ok
true pelican
I'm working on a theoretical framework for coordination in multi-agent systems. The core mathematical formulation is:
C = k * η^α * (1-η)^γ * S^β
Where:
C: coordination effectiveness
η ∈ [0,1]: efficiency parameter
S: system size (# of agents)
k, α, γ, β > 0: constants
From this, I derive an optimal efficiency point:
η* = α / (α + γ)
My claim is that this produces an inverted-U relationship with efficiency, a scaling law with system size, and that η* is independent of S.
Could anyone check whether the derivation of η* and the general mathematical reasoning here are correct? Are there any hidden assumptions or edge cases I might be missing?
Thanks!
edgy mauve
I'm working on a theoretical framework for coordination in multi-agent systems. ...
!occupied
final saddleBOT
I'm working on a theoretical framework for coordination in multi-agent systems. ...
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
pale violet
Show ur work once u r done
bro the value of A is 60 degree is that correct?
edgy mauve
bro the value of A is 60 degree is that correct?
That's 1 value yes
The other one would be the value of A in 2nd quadrant
pale violet
The other one would be the value of A in 2nd quadrant
ya the minus one
am i supposed to find that too?
edgy mauve
ya the minus one
The minus one wont give u a value
The minus one will give u undefined value
Cuz it exceeds the limit of sinA
edgy mauve
am i supposed to find that too?
SinA is positive in both 1st and 2nd quadrant right?
pale violet
ehh i find these stuffs tough
edgy mauve
A=60 in 1st quadrant, so u can find the value of A in 2nd quadrant right?
edgy mauve
ehh i find these stuffs tough
Which part
edgy mauve
Sin(theta) is valid if -1<sin(theta)<1
edgy mauve
like which value is acceptable and which is not😢
But -2/sqrt3 is -1.1547....
Which is not within the range that's y it's invalid and cant be acceptable
pale violet
But -2/sqrt3 is -1.1547....
Which is not within the range that's y it's invalid ...
tnx bro
edgy mauve
Np
karmic torrent
How would we go about q5?
final saddleBOT
How would we go about q5?
heavy briar
1/(1+2) + 1/(1+2+3) + ... + 1/(1+2 + ... + 51)?
karmic torrent
Urn sorry the numbers got cut of q5 is evaluate the expression
plucky rover
The question numbers aren't visible
heavy briar
okay
karmic torrent
karmic torrent
What I’ve tried was that summation $n^2/((n^2) - 10n + 50)$ where the lim n=1 -> 9
soft zealotBOT
YaBoiVGC
karmic torrent
But 10 and 50 dont factorize well
fiery bluff
yeah it seems pretty sus, completing the square for the denominator doesn't look too good either
unless it wants you to use complex numbers
karmic torrent
It’s Olympiad, so integration CANNOT be used
plucky rover
Okay dumb question, are the denominators actually symmetric here
Cuz we could do smth with that maybe
fiery bluff
oh yeah they are, just complete the square
1^2/(5^2 + (1-5)^2) + 2^2/(5^2 + (2-5)^2) +...
plucky rover
Right yes
karmic torrent
Pls use $$ i can’t read the expression
soft zealotBOT
YaBoiVGC
karmic torrent
.close
dense coral
there's an implied summation symbol here, yes? because we're writing \omega as a linear combination of the \lambda^i's?
final saddleBOT
there's an implied summation symbol here, yes? because we're writing \omega as a...
dense coral
but I'm confused as to why the summation isn't written out explicitly
it's probably the Einstein summation convention if I had to guess, but I was under the impression that there's an implied sum only when we have an upper index followed by the same lower index
while we have a lower index followed by an upper one here 
final saddleBOT
@dense coral Has your question been resolved?
shadow marlin
apparently you only need the index to appear twice for it to be a sum, and top/bottom distinction is related to contravariance/covariance
I just read that on wiki though
modest fog
17
soft zealotBOT
dapper hull
,rccw
soft zealotBOT
dapper hull
I’d suppose you did it right now try to find a pattern when differentiating g(x)
@modest fog
modest fog
I’d suppose you did it right now try to find a pattern when differentiating g(x)...
Yea I thought abt it
I differentiated like 4 times
Can't see a definitive pattern tho
dapper hull
I don't want to differentiate more <:cat_happycry:636407813144576010>
Yk e^(x+lnx)=xe^x
dapper hull
I m so stupid
Happens when shit is so complicated you forget the small things
It’s ok
modest fog
Thank you for ur time
dapper hull
Now I think the pattern is easier
final saddleBOT
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robust horizon
The overall run time complexity should be O(log (m+n)).``` third time stating this problem here but i definitely got stuff mixed up, so would like to start from scratchafk for a few minutes; please ping if you are replying
final saddleBOT
@robust horizon Has your question been resolved?
robust horizon
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robust horizon
0 <= a, b, c, d < n
a, b, c, and d are distinct.
nums[a] + nums[b] + nums[c] + nums[d] == target
You may return the answer in any order.``` my work: ```cpp
#include <iostream>
int main()
{
std::vector<int> a{1, 0, -1, 0, -2, 2};
int target = 0;
int n = 6;
std::sort(a.begin(), a.end());
std::vector<std::vector<int>> result;
for (int i = 0; i < n - 3; i++)
{
if (i > 0 && a[i] == a[i - 1])
{
continue;
}
for (int j = i + 1; j < n - 2; j++)
{
if (j > i && a[j] == a[j - 1])
{
continue;
}
int left = j + 1, right = n - 1;
while (left < right)
{
int sum = a[i] + a[j] + a[left] + a[right];
if (sum == target)
{
result.push_back({a[i], a[j], a[left], a[right]});
left++;
right--;
}
else if (sum < target)
{
left++;
}
else
{
right--;
}
}
}
}
}
``` this does not work properly; need help on how to fix it@robust horizon Has your question been resolved?
final saddleBOT
@robust horizon Has your question been resolved?
tiny gorge
first problem is that you're not actually returning the result
if this is leetcode, don't they give you a function (not main) that you're supposed to write?
robust horizon
What exactly doesn't work?
it did not pass the test cases
robust horizon
if this is leetcode, don't they give you a function (not `main`) that you're sup...
i have that version, let me paste that real-quick
robust horizon
it did not pass the test cases
(also sorry for replying late, i fell asleep)
class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
vector<vector<int>> result;
int n = nums.size();
sort(nums.begin(), nums.end());
for (int i = 0; i < n - 3; i++) {
if (i > 0 && nums[i] == nums[i - 1])
continue;
for (int j = i + 1; j < n - 2; j++)
{
if (j > i + 1 && nums[j] == nums[j - 1])
continue;
int left = j + 1, right = n - 1;
while (left < right)
{
sum = nums[i] + nums[j] + nums[left] + nums[right];
if (sum == target)
{
result.push_back({nums[i], nums[j], nums[left], nums[right]});
left++;
right--;
}
else if (sum < target)
{
left++;
}
else
{
right--;
}
}
}
}
return result;
}
};
i think the issue is here: cpp { result.push_back({nums[i], nums[j], nums[left], nums[right]}); left++; right--; } ?
opal plinth
Yes it is
opal plinth
It keeps the leading spaces
robust horizon
right
opal plinth
If the input ends in ... a, a, b, b), then you get a duplicate
opal plinth
Instead of incrementing left and decrementing right, you could just increment left until you get a different number
robust horizon
an if condition?
(more like a loop)
robust horizon
Instead of incrementing left and decrementing right, you could just increment le...
makes sense
that would be easier
i will try and write the code and show you
well, does this work in my previous approach, or no? cpp while (left < right && a[left] == a[left + 1]) left++; while (left < right && a[right] == a[right - 1]) right--;
this is easy to visualize
but well, is probably wrong
opal plinth
Hm that should work
robust horizon
this is O(n^3), right?
opal plinth
Yes
opal plinth
well, does this work in my previous approach, or no? ```cpp
while (left < right ...
Oh this doesn't work if you can get to the target with the same last two numbers
robust horizon
how do i fix that?
opal plinth
Hold on
robust horizon
okay
opal plinth
Never mind it depends where you put it
opal plinth
Ok yes that works
opal plinth
Note that you don't actually need the right-- part
robust horizon
can we do it in less time complexity?
robust horizon
Note that you don't actually need the `right--` part
right
opal plinth
Hm I don't think you can do better than O(n^3)
robust horizon
i see, thank you!
is it okay if we discuss the other problem as well? or if that is not allowed, that's okay too
the median one
opal plinth
Sure
robust horizon
let me paste the problem real-quick, again
The overall run time complexity should be O(log (m+n)).
I tried to find other solutions but nothing is lower, so yeah I think this is optimal as far as we know
robust horizon
fair enough, thank you
opal plinth
Right so let's take examples
(A) [0, 1, 2, 3, 4], [5, 6, 7, 8, 9, 10]
(B) [1, 2, 5, 6, 7], [0, 3, 4, 8, 9, 10]
opal plinth
Uh so any idea how to begin?
robust horizon
well
i thought about the question you asked me earlier
i am not sure if this is correct but
if we pick the middle element of the first array, it is the median if and only if the biggest number on the left is less than equal to the smallest number on the right? from both arrays
or does that not make sense
well i probably did not word that correctly
opal plinth
Ye I don't know if I understand
robust horizon
one thing i am getting confused about a lot in this problem is keeping track of when we are dealing with values and when with indices
opal plinth
The arrays are sorted, so everything on the left is always less than (or equal to) everything on the right
The main issue is that there are two arrays
Ok, in (B), if you pick the 5, how can you figure out that it is the global median?
And in (A), if you pick the 2, how can you figure out that it is not?
robust horizon
Ok, in (B), if you pick the 5, how can you figure out that it is the global medi...
is the answer in terms of indices?
opal plinth
No I'm only asking for some process you can apply
Let me ask something different first
What's the defining property of the median in this problem?
robust horizon
we do a split of the two arrays such that left and right halves are balanced, and the largest element on the left is less than or equal to the smallest element on the right?
because the arrays are sorted
opal plinth
What exactly is a split here, and what does it mean for the halves to be balanced?
robust horizon
equal number of elements in the left part of our median and in the right part of it?
opal plinth
Right
robust horizon
does that not make sense?
opal plinth
Just to be extra sure, let me ask you what the "parts" are
robust horizon
can you please tell me if i am wrong here?
robust horizon
what would that be?
opal plinth
The arrays have m and n elements
Can you use that as part of your definition of the median?
opal plinth
Can you elaborate
robust horizon
our first array has five elements
and the second one has six elements
assuming our array is sorted,
the median would be the 6th element
using this
(m + n + 1)/2
now you are going to say that the arrays are not merged
opal plinth
Ok I see
No I understand that you're talking about a combined array but there's a small issue with the even case
robust horizon
we will need to take average of the two central elements
opal plinth
[0, 1], [2, 3]
What would be the median here
robust horizon
1.5
opal plinth
Right, and so what is (m + n)/2 exactly?
Like, here it equals 2, but that's evidently not the median
opal plinth
The index of the first element in the central pair?
opal plinth
Ok that's one way to put it
robust horizon
is it wrong?
opal plinth
No. What I was expecting is: there are as many elements to the left as to the right of the median
opal plinth
"to the left" meaning less than of course, and right, greater than
robust horizon
yeah that is way easier to understand
opal plinth
So if the arrays have m and n elements, there should be floor((m+n)/2) elements less than the median
robust horizon
makes sense
opal plinth
So back to the examples
(A) [0, 1, 2, 3, 4], [5, 6, 7, 8, 9, 10]
(B) [1, 2, 5, 6, 7], [0, 3, 4, 8, 9, 10]
In either example, if you pick the middle of the first array, how can you tell whether it's already the median?
robust horizon
No. What I was expecting is: there are as many elements to the left as to the ri...
by checking this
for instance
in (A)
the middle element of the first element is '2'
we check if it has 6 elements on its left side or not
we don't
well no
that won't work for the second example
opal plinth
That's the problem: you do not know how many elements total are less than the one you picked
If you did, this would be a simple binary search
robust horizon
so basically
opal plinth
You do know one piece of information though, and with one additional step, you can know a second one; combining these two pieces of info, you should be able to tell if your pick was the median
robust horizon
can you please tell me those?
opal plinth
Try picking two adequate elements in the second array as well, and make some comparisons
robust horizon
can we do something like count and compare
opal plinth
In (B), if you had to place 5 in the second array, where would that be?
robust horizon
after 4
opal plinth
Ok, is that piece of information enough to know whether 5 is the median?
robust horizon
not yet
because we must also check the counts on both sides across both arrays, not just the local placement?
opal plinth
True, however this is not exactly the process we are going to use
robust horizon
right, sorry
opal plinth
In (A), where could you place 2 in the second array to immediately know whether it's the median?
opal plinth
Hm that wasn't a good question... rather, once you've picked 2 already, where could you then look inside the second array to immediately know that 2 is not the median?
robust horizon
um, the first element of the second array?
opal plinth
Why is that?
robust horizon
is that wrong?
opal plinth
I need to understand your reasoning first, I can't tell you whether you're on the right track if I don't know what track it is
What would you do knowing the value of the first element of the second array?
robust horizon
it's a little scary when you ask questions like that without telling if i am in the right direction or not, because if i am not then the whole thing just sounds plain stupid
my reasoning:
opal plinth
Ok but you need to stop thinking like that; you learn by making mistakes, and the time you take is not wasted
robust horizon
if the first element of the second array is larger than 2, then the second array adds nothing to the left count, and we can instantly conclude that 2 cannot be the median
opal plinth
Yes! That's a good rationale
robust horizon
really?
opal plinth
Now all you need to do is improve upon it
For example if you look instead at the second element, the 6
It also is greater than 2 (obviously, but let's say you didn't know the first was 5), but what could this add to the left count?
Brb
robust horizon
here's what i am thinking of:
- instead of checking the very first element of the second array, we can check any element in it at or after where 2 would be inserted
- if that element is already greater than 2, we immediately know that the second array contributes nothing to the left
- but if that element were less than or equal to 2, then all elements before it also count toward the left?
sorry if it does not make sense
opal plinth
Ok but you don't know where 2 would be inserted
robust horizon
right
opal plinth
It also is greater than 2 (obviously, but let's say you didn't know the first wa...
Answer this first
robust horizon
one element?
robust horizon
um, yes?
opal plinth
I mean would that be enough to make 2 the median
robust horizon
no
opal plinth
Right, so how much would be enough?
robust horizon
we require 6, right?
opal plinth
So back to the examples
(A) `[0, 1, 2, 3, 4]`, `[5, 6, 7, 8, 9, 10]`
(B) `[1, 2,...
Here m=5, n=6
How many elements are less than the median?
robust horizon
5
opal plinth
Yes, and how many do we know are less than 2 in (A)?
opal plinth
Yes, so how many more do we need
robust horizon
3
opal plinth
So where to look in the second array?
robust horizon
till 7?
a minute
we need 3 elements from the second array that are less than or equal to 2?
opal plinth
Yeah
robust horizon
well
opal plinth
If the arrays can have duplicates, you also need 3 elements greater than or equal to 2
Here we're only considering simple examples but duplicates just add a little bit of complexity
robust horizon
fair enough
opal plinth
So again, just making sure, how exactly do you check whether the second array has 3 elements less than 2?
robust horizon
- we look at the first element of the second array
- if it’s already greater than 2, no elements in that array can be less than 2
- if we need 3 elements from that array to reach the median, but the maximum possible contribution less than 3 implies that the element cannot be median
opal plinth
[0, 1, 3, 4, 5], [2, 6, 7, 8, 9, 10]
First, we pick 3
We want to know whether it's the median, what do we do?
robust horizon
we first check how many elements are less than 3 in the first array
opal plinth
1. we look at the first element of the second array
2. if it’s already greater t...
Hm well that's already a different process than what you said here
opal plinth
Uh I'm a little surprised, are you sure you got it?
robust horizon
yeah
opal plinth
Ok then
final saddleBOT
plucky rover
i do not really want to code it
Yeah writing the code for this one is a nightmare
final saddleBOT
Available help channel!
brittle breach
So I am given to differentiate f(x) = sqroot (4 - x²), x € (-2,2).
What to do when x has two value?
final saddleBOT
So I am given to differentiate f(x) = sqroot (4 - x²), x € (-2,2).
What to do wh...
brittle breach
My status is that I can differentiate but not solve for x or whatever
lilac moat
So I am given to differentiate f(x) = sqroot (4 - x²), x € (-2,2).
What to do wh...
Use the chain rule
lucid nymph
youre not supposed to solve for x, its only saying that x can only be between -2 and 2 since if its outside that domain you would have a sqrt of a negative number
final saddleBOT
@brittle breach Has your question been resolved?
brittle breach
.close
final saddleBOT
final saddleBOT
Available help channel!
gentle zephyr
final saddleBOT
@gentle zephyr Has your question been resolved?
final saddleBOT
@gentle zephyr Has your question been resolved?
daring lion
translate
gentle zephyr
FIND THE SET OF POINTS FOR WHICH THE FOLLOWING FUNCTIONS ARE CONTINOUS
daring lion
i think f=g/h is continuous at all points in domain where g,h are continuous
gentle zephyr
wdym?
worldly mesa
Everywhere both the denominator and numerator are continous the the fraction is continous
gentle zephyr
IS THAT A THEOREM OR A COROLLARY?
whole halo
renato, is there a reason you never translate these?
less people will immediately help you out if its just in spanish
gentle zephyr
less people will immediately help you out if its just in spanish
FIND THE SET OF POINTS FOR WHICH THE FOLLOWING FUNCTIONS ARE CONTINOUS
whole halo
no, I mean you shouldnt need to be asked to translate
most people in the server dont speak spanish
gritty drift
that's not what he's asking. he's asking why do you wait till someone asks you to translate for you to translate?
whole halo
so you should always give a translation right below the original image
gritty drift
very few people speak Spanish in the server, you'll get faster responses if you provide a translation with your question, rather than wait for someone to ask for it
gentle zephyr
i translated now, can I receive help or not?
whole halo
renato, this is an intervention
whole halo
i translated now, can I receive help or not?
if you make a mistake once, its a typo
if you make a mistake twice, its concerning
if you make a mistake multiple times, its a problem
so if you keep doing the same wrong thing, wouldnt you want to fix it?
gritty drift
this isn't about this question specifically, you never translate your questions until prompted. you will surely ask another question in the future. in order to get faster help in the future, you should translate your question when you send it, rather than wait for someone to ask for it
whole halo
if you leave the question untranslated, only people who speak spanish are more willing to help you
I dont speak spanish, most of us dont
imagine if you cut the image off halfway, then we had to ask to get the other half
thats what stitches is saying here, make it easier for us to know what you need to do
@gentle zephyr you need to translate the question the next time you post one, ok?
gentle zephyr
ok, now what?
the problem is that my english is very bad, sometimes I don't know how to translate accurately
can you guys put yourself in my shoes for a second?
@whole halo @gritty drift
whole halo
in most cases, a machine translation is better than no translation at all
gritty drift
yes but my spanish is worse than your english, whatever attempt you make at a translation (machine or manual) will be better than me trying to understand a statement in a language I do not know
whole halo
can you guys put yourself in my shoes for a second?
you havent even put yourself in our shoes
why should we?
when we ask for the translation, you still just give one without hesitation
gentle zephyr
im wearing shoes rn, wdym
gentle zephyr
when we ask for the translation, you still just give one without hesitation
true
whole halo
just go for the translation you think is correct
and include that below any images in the future
gentle zephyr
sometimes a bad translation changes the route of the exercise completely has happened to me in the past
whole halo
that is why you always include the original image
but here youve already done that
so what is there to worry? we cant blame you for an accident
whole halo
sometimes a bad translation changes the route of the exercise completely has hap...
what you can do is machine translate it and see if it lines up
since you speak spanish, youll have the most experience to validate the machine translation as well
that way, you can be sure of the translation
gentle zephyr
its about the mathematics terms that the machine makes mistakes sometimes
either way, can we continue with the exercise?
gentle zephyr
?
whole halo
you should be telling us what your answers are
are you really saying you never start on these exercises until we tell you to begin?
gentle zephyr
you should be telling us what your answers are
dude Prof believes in no answers
is better for learning
whole halo
that you read it that way is concerning
gentle zephyr
according to him
gritty drift
he means what you've tried so far, not the actual answer
whole halo
I asked for your answers... that means answers that belong to you
gentle zephyr
I tried to start but dont know how to begin
whole halo
youll have to keep asking us for help if you dont know how to begin
here, lets practice that
gentle zephyr
have you guys ever been there? not knowing how to start an exercise?
whole halo
yes, theres a few things I do to try to begin
first, I get a good sense of the thing Im working with
do you have a good sense of what a continuous function is?
gentle zephyr
kind a
whole halo
alright, tell me what a continuous function should be
gentle zephyr
in single variable is when both lateral limits coincide
no?
then we can say the func is continuous
@whole halo
whole halo
that is correct
but think about it: this problem is instead asking for every point where the function is continuous
we cant check every point individually you know
so we'll have to solve this a different way
whole halo
alright, tell me what a continuous function should be
a continuous function is a function where every point on it is continuous
gentle zephyr
check for points of discontinuity
assume is continuous outside those discontinuity points
whole halo
yes
most functions that you are working with are almost continuous
theres a few patterns about these continuous functions that can justify this assumption of "mostly continuous"
adding them keeps continuity
subtracting them keeps continuity
multiplying them keeps continuity
dividing them only keeps continuity when the denominator isnt 0
square rooting them keeps continuity when the inside is not negative
etc.
heres the first example
tell me which parts of this function are continuous
gentle zephyr
gentle zephyr
tell me which parts of this function are continuous
idk
everything
because e is monotonically increasing
e^x > 0 AFAIK
XY I ASSUME ASWELL IS CONTINUOUS
DONT KNOW HOW TO EXPLAIN IT
whole halo
renato, you need to learn not to press the caps lock key or whatever youre doing
gentle zephyr
denomi is never zero
whole halo
thats correct
gentle zephyr
whats your point?
whole halo
DONT KNOW HOW TO EXPLAIN IT
I can explain this
gentle zephyr
what?
whole halo
as I was saying
all polynomials are continuous
x and y are polynomials, so they are continuous
xy is a polynomial also, so it is continuous
so is x - y, the exponent in the denominator
1 + e^x is continuous also
you can put one continuous function in another and still get a continuous function
1 + e^(x - y) is continuous too
since e^x > 0, that means 1 + e^x is never 0, so that is why xy/(1 + e^(x - y)) is continuous everywhere
whole halo
because e is monotonically increasing
theres a typo here
e^x is positive is all thats required
that e^x is monotonically increasing is true, but not necessary for what we need
other than that, you have the right proof to solve (a)
gentle zephyr
dont want to argue about inecessary stuff, but doesnt e^x being monotonically increasing func directly implies that e^x > 0
whole halo
unfortunately it does not
x + 2 is also monotonically increasing and it isnt always positive
tired walrus
the problem is that my english is very bad, sometimes I don't know how to transl...
you try your best anyway. you give us both the original AND your english translation, regardless of its quality.
gentle zephyr
yes but in this case the derivative is equal to the same function
tired walrus
im wearing shoes rn, wdym
"put yourself in someone's shoes" is a metaphor
whole halo
its strange that renato used the metaphor first
Im going to assume renato not getting it when I used it was a joke
whole halo
yes but in this case the derivative is equal to the same function
so that means you need e^x is monotonically increasing & e^x is its own derivative to know that e^x is positive
that would work, yes
thats good, moving on to (b)
gentle zephyr
we already passed the stage about arguing about the translation issues, can we continue
whole halo
gentle zephyr
this are sum of polynomials in numerator and denominator
whole halo
keep going
gentle zephyr
is still possible the denominator is zero when -x^2 - y^2 = -1
whole halo
correct
worldly mesa
These are all of the dicontinuities
gentle zephyr
we dont know
whole halo
you, in fact, know
whole halo
is still possible the denominator is zero when -x^2 - y^2 = -1
the discontinuities happen when x^2 + y^2 = 1 here
mainly because the function isnt even defined when that happens
gentle zephyr
yes exactly
whole halo
but everywhere else, the function is continuous, right?
whole halo
that would be your answer, everywhere except when x^2 + y^2 = 1
worldly mesa
Then we are done
gentle zephyr
of center 0 with radius 1
whole halo
youll need to write this answer down properly with math
do you know the way to do that?
gentle zephyr
no
whole halo
try writing it using "set builder notation"
for example, ${x\mid x>0}$ represents all positive numbers
soft zealotBOT
mtt
tired walrus
${(x,y) \in \bR^2 \mid .......... }$
soft zealotBOT
Ann
whole halo
this is half of it done for you, above
tired walrus
fill in the dots appropriately to capture "all points except those on the circle x^2+y^2=1"
gentle zephyr
this circle has many other points other than (0,1) and (1,0), correct?
whole halo
yes
you arent just focusing on integers right now
you just need a way to say "all points except those on the circle x^2 + y^2 = 1"
gentle zephyr
but above you guys said the only points where (0,1) and (1,0), so there are more points where the func is discontinuous or what
gentle zephyr
you just need a way to say "all points except those on the circle x^2 + y^2 = 1"
but...after we find out all those points, how to prove its discontious at those points? or what are we supposed to do
tired walrus
but above you guys said the only points where (0,1) and (1,0), so there are more...
above you guys said the only points where (0,1) and (1,0)
where did any of us say that
link to the msg
whole halo
you already proved it was discontinuous earlier on
you said that the denominator would be 0 when -x^2 - y^2 = -1
so naturally if x^2 + y^2 = 1, the denominator would be 0, and the function wouldnt be defined
cant be continuous if its like that
so that proof is done
whole halo
but...after we find out all those points, how to prove its discontious at those ...
you just now need to write down "those not on the circle x^2 + y^2 = 1" in math notation
I can give you a hint if youre stuck on not knowing the right way to write it
gentle zephyr
,, \mbb{R}^2 \setminus {(x,y) \in R \times R \mid x^2 + y^2 = 1}
soft zealotBOT
Renato
whole halo
interesting approach
now theres a simpler way to go about this
,,{(x,y)\in\mathbb{R}^2\mid x^2+y^2\ne1}
soft zealotBOT
mtt
gentle zephyr
hahah
whole halo
you can use the not equals sign
powerful little tool when youre just constructing a set
thats (b) done
now for (c)
this I can say isnt immediately straightforward
gentle zephyr
this I can say isnt immediately straightforward
its a piecewise
whole halo
yes
this kind of pattern is common sometimes
theyll give you most of the function, and then a single point
whole halo
take a look at the function when (x, y) isnt (0, 0)
is it continuous?
gentle zephyr
the limit of the function needs to be equal to 1 at all the points
whole halo
theres only one point, but sure
worldly mesa
theyll give you most of the function, and then a single point
Usually that point "fills" the discontinuity, but we don't know that until we check.
whole halo
I sort of wanted renato to realize that but sure
gentle zephyr
depends if denominator is equal to zero
x^2 + xy + y^2 = 0 then its discontinuous if xy not zero
gentle zephyr
we need to find the limit
gentle zephyr
we need to find the limit of the function for when (x,y) approaches (0,0) and show that the limit is equal to 1
otherwise is not continuous
or what?
whole halo
renato, did you see the question I asked?
whole halo
is this function continuous?
lets assume that (x, y) cannot be (0, 0)
gentle zephyr
is this function continuous?
its continuous when 2x^2 + y^2 is not zero
that only happens when (0,0) that the denominator is zero
whole halo
there you go, you have to mention that first
now whats the proof that its continuous for everywhere else around (0, 0)?
gentle zephyr
I was sort of confusing things at first, didnt took a whole lot of time to think things through, mb
gentle zephyr
now whats the proof that its continuous for everywhere else around (0, 0)?
idk?
whole halo
for (x, y) not being (0, 0), take a look at the fraction here
why is x^2 y^3 continuous?
severe canyon
I was sort of confusing things at first, didnt took a whole lot of time to think...
Let me say, you have to improve your reading ability. It's not about math
gentle zephyr
patience, reading comprehension, are in my bucket list
whole halo
this reading ability has nothing to do with language - sometimes you need to know when you dont have the full picture
doesnt matter what language youre reading
gentle zephyr
why is x^2 y^3 continuous?
multiplication of polynomials
gentle zephyr
sum of poly
whole halo
you gotta mention that polynomial / polynomial, so its continuous when 2x^2 + y^2 isnt 0
you leave that out and the proof isnt complete
alr now for the limit we need to be solving
,,\lim_{(x,y)\to(0,0)}\frac{x^2y^3}{2x^2+y^2}
soft zealotBOT
mtt
whole halo
what do you think could work here?
gentle zephyr
x = y
whole halo
and?
gentle zephyr
and y = x?
whole halo
x = y and y = x are identical
gentle zephyr
jajaj
whole halo
go on try something else other than just x = y
oh right theres an important thing I left out
whole halo
you havent done everything yet for (c)
for the function to be continuous,
theyre saying $\lim_{(x,y)\to(0,0)}\frac{x^2y^3}{2x^2+y^2}=1$
soft zealotBOT
mtt
whole halo
you mentioned this a few times and I forgot to include that
what does the limit simplify to when you try out x = y?
gentle zephyr
,, \frac{y^5}{2y^2 + y^2} = \frac{y^5}{3y^2} = \frac{y^3}{3}
soft zealotBOT
Renato
gentle zephyr
this function is not continuous
we literally found a line for which the limit is equal to zero
whole halo
yep
gentle zephyr
so its not continuous
whole halo
thats how youd show it
whole halo
**mtt**
this limit btw is always 0 regardless of the path you pick
so we now know the function is continuous everywhere except (0, 0)
how do you write that down in math notation?
gentle zephyr
idk
whole halo
we need to write down, in notation, "every point in R^2 except (0, 0)"
try writing part of it first
gentle zephyr
,, {(x,y) \in \mbb{R}^2 \mid (x,y) \neq (0,0)}
soft zealotBOT
Renato
whole halo
very good
(c) is done
youll be repeating yourself for (d) here
just say everything you think is required, I think youve got this
gentle zephyr
poly/poly
whole halo
more details, what is this for?
gentle zephyr
the function when (x,y) is not (0,0) has a continuous denominator and continuous numerator, but the function when (x,y) itself is not (0,0) is continuous when x^2 + xy + y^2 is not zero
gentle zephyr
more details, what is this for?
I mean, if the numerator of a rational function is continuous and the denominator of the rational function is continuous then it can only be discontinuous when we divide by 0, that is when the denominator x^2 + xy + y^2 = 0
whole halo
thats good, keep going
gentle zephyr
idk how to continue
whole halo
when does x^2 + xy + y^2 = 0 happen?
gentle zephyr
it's not simple
whole halo
well at least you know what step is next
gentle zephyr
idk
gentle zephyr
when does x^2 + xy + y^2 = 0 happen?
there are many possibilities, but mainly when (0,0)
pliant parcel
x^2 + xy + y^2 = 0
-> x^2 + y^2 = -xy
gentle zephyr
x^2 + xy + y^2 = 0
-> x^2 + y^2 = -xy
what's your point?
pliant parcel
i was just thinking out loud for a bit
daring lion
it's not simple
-# quadratic formula exists
pliant parcel
this could be rewritten to $\left(\frac xy \right)^2 + \frac xy + 1 = 0$ \
let $t = \frac xy \rightarrow t^2 + t + 1 = 0$
$$\therefore t = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm i\sqrt3}2$$
$$\therefore \frac xy = \frac{-1 \pm i\sqrt3}2 \rightarrow x = y \cdot \frac{-1 \pm i\sqrt3}2$$
soft zealotBOT
creeperdoesredstone
gentle zephyr
the problem is that (x,y) ∈ R^2
gentle zephyr
-# quadratic formula exists
what's your point?
daring lion
**creeperdoesredstone**
like he showed, from here you can say there aren't any real solutions for x^2 + xy + y^2 = 0 except (0,0)
gentle zephyr
,w solve x^2 + xy + y^2 = 0
soft zealotBOT
daring lion
I know 2 more ways to prove this
- By some manipulations
x²+xy+y²=0
x²+2xy+y²=xy
(x+y)²=xy
x²+xy+y²=0
x²+y²=-xy
Adding them,
(x+y)²+x²+y²=0
(x+y)²=-(x²+y²)
LHS is always +ve due to squaring, RHS is always negative due to being negative of squared terms, so both can only be equal at 0
- By using quadratic formula on
x²+(y)x+(y²)=0
but you should be able to do this
gentle zephyr
I know 2 more ways to prove this
1. By some manipulations
x²+xy+y²=0
x²+2xy+y²...
wdym?
gentle zephyr
I know 2 more ways to prove this
1. By some manipulations
x²+xy+y²=0
x²+2xy+y²...
what are u proving here?
@daring lion @daring lion @daring lion
you here?
pliant parcel
they're proving that there are no (x, y) pairs in R^2 that satisfy x^2 + xy + y^2 = 0 except for (0, 0)
gentle zephyr
how?
bold turtle
final saddleBOT
@gentle zephyr Has your question been resolved?
severe canyon
<@811854308634001419> <@811854308634001419> <@811854308634001419>
Please STOP PINGING (especially MORE THAN ONCE)
It's EXTREMELY IRRITATING for every helper
Also, pinging multiple times has exactly the same effect as pinging just ONCE
final saddleBOT
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velvet wind
so uhh I don't really know how to use maple and the last answer that says RootOf(z^5+Z^3-2root3, index=1)^2 is just supposed to be 2 root 3 
Is it something wrong with the syntax or am I using the wrong command?
final saddleBOT
so uhh I don't really know how to use maple and the last answer that says RootOf...
@velvet wind Has your question been resolved?
velvet wind
.close
final saddleBOT
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quasi plume
final saddleBOT
quasi plume
literally at the tip of my tongue
yet I can't figure out how to solve it
is this right?
iron dagger
it would depend on what they use as the coordinate system here but for a traditional one then yeah
quasi plume
okay so how would I solve this?
tan(30)=x/y I tried. I tried 40=sqrt(x^2 + y^2). and I tried using 30 60 90 triangle rule
iron dagger
it would depend on what they use as the coordinate system here but for a traditi...
actually no for a traditional system the x and y's would be swapped in your image
soft zealotBOT
Result:
0.57735026918963
bruh
iron dagger
tan(30)=x/y I tried. I tried 40=sqrt(x^2 + y^2). and I tried using 30 60 90 tria...
you already have the angle and magnitude, there's no need for all that
just think about what sin30 and cos30 equal
quasi plume
just think about what sin30 and cos30 equal
well cos30 is sqrt(3)/2 and sin30 is 1/2
iron dagger
ok, and what ratios do they represent?
quasi plume
wdym?
iron dagger
like sine and cosine represent fractions of which sides of a triangle?
quasi plume
sin is the y in the drawing and cos is the x. after the positions got swapped in this pic
iron dagger
not necessarily true
that would only be true if the magnitude of the vector was 1
recall sohcahtoa
quasi plume
that would only be true if the magnitude of the vector was 1
oh, right
so do I needa apply scalar to <sqrt(3)/2,1/2> ?
iron dagger
yes
quasi plume
40<sqrt(3)/2,1/2> ?
iron dagger
yep
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jade matrix
final saddleBOT
jade matrix
taking 2 balls out at the same time is the same as taking out one ball not replacing then taking out another
we know 1 ball is red
so the asnwer is 1 - (1/2 * 1/3)
which is wrontg
runic tulip
we know 1 ball is red
2 balls are red?
I'm still not sure that do you mean by 1/2 * 2/3
runic tulip
So basically you first pick one red ball out of 4
The another one out of 3?
This is incorrect because the other does not matter
Let's say the red balls are R1 and R2
Say you first pick R1 and then R2
And in another case you first pick R2 and then R1
Both these are the same case as the balls are identical but you count them seperately
jade matrix
hmm
you can either pick RB RR BB BR
we know there is a red ball
so the options are RB RR BR
we only want RB OR BR
is it just 2/3
as out of the outcomes we want 2 of the 3
final saddleBOT
@jade matrix Has your question been resolved?
final saddleBOT
@jade matrix Has your question been resolved?
brazen stump
shouldnt this be 22/62
final saddleBOT
shouldnt this be 22/62
final tangle
how are you getting 22
brazen stump
because
in the 62 it contains the 44+6+4
acc nvm i dont understand why is 12/62
i know is over 62 but i dont know the numerator
final saddleBOT
@brazen stump Has your question been resolved?
final tangle
out of the 62, you want the ones that are also in B
can you identify those sections on the venn diagram
brazen stump
can you identify those sections on the venn diagram
but b also coaintins the other 10 people
why are they not included
final tangle
because those aren't in A
final saddleBOT
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Available help channel!
fallen marsh
hello, i watched some video's provided by the teacher but i still dont understand how to plug this into the formula to figure this out
sweet summit
Do you know which part of the quadratic formula is the Discriminant?
fallen marsh
i do not
sweet summit
it's the "b^2 - 4ac" portion of the quadratic formula.
So can you tell me something about square roots? are there certain values that can't be possible?
just to be clear, the orange part is the discriminant
fallen marsh
its the opposite of ^2 so only numbers multiplied by themselves?
i have that written down
already
sweet summit
No, think of the domain of square roots. Can square roots be negative?
fallen marsh
no?
sweet summit
bingo
so what that tells us is that if the discriminant is < 0, then we have no solutions
what about if the discriminant is 0?
fallen marsh
then there are no solutions
fallen marsh
isnt +- 0 the same thing? i know from the video that if there are 2 solutions they come from that +-
mb i am new to this haha
sweet summit
No worries, this is all about thinking logically about this. 🙂
so if b^2 - 4ac = 0, then the square root of b^2 - 4ac is 0, right?
so what's the square root of 0?
fallen marsh
0?
sweet summit
bingo
so if D = 0
then we get the solutions are -b / 2a
so there's 1 solution
so if D < 0, we have no solutions, if D = 0, we have 1 solution
so waht can we conclude if D > 0?
fallen marsh
2 solutions!
sweet summit
bingo!
fallen marsh
and can there be more then 2 solutions?
sweet summit
nope 🙂
fallen marsh
nice ok
steep nest
the amount of solution a polynomial has is the degree
sweet summit
that's what the +/- basically is telling us
sweet summit
the amount of solution a polynomial has is the degree
Yeah this
steep nest
and the amount of real solutions is using discriminant
which is beautifully stated in the picture above
fallen marsh
ax2 + bx + c = 0 , so what is D here?, 0 1 and 2 are all wrong in the websight?
i thought it was 0, so 1 solution
sorry i have to go, thank you for the help i apricate it
!close
.close
final saddleBOT
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scenic junco
final saddleBOT
scenic junco
We were learning to verify inverse functions
The right image is my work for problem 27
In problem 27, one of the inverses has a domain restriction
How does it play into the verifying process?
trail crest
How does it play into the verifying process?
It just means that f(g(x)) = x for x <= 0
As that is where g(x) is defined
scenic junco
Your inequality is flipped
The textbook solution shows the first part as x<=0
trail crest
trail crest
The textbook solution shows the first part as x<=0
That isnt what you wrote