maybe i am just reading it wrong?
#help-36
latent meteor
yes because
green ibex
you're subtracting 2 two times on one side, and one time on the other
theres an imbalance
latent meteor
So I am not supposed to -2 to both +4 and 3 i only choose one of them right?
latent meteor
how do i know which one to choose
latent meteor
okay cool how do close this
green ibex
.close
brazen stump
hello can someone help
final saddleBOT
hello can someone help
final saddleBOT
@brazen stump Has your question been resolved?
brazen stump
.close
rain sentinel
Hi, i hope you dont mind me asking this physics question here
final saddleBOT
Hi, i hope you dont mind me asking this physics question here
rain sentinel
but wouldnt the answer to question 1 be: The results would be less accurate because 100g is a light weight and the spring balance wouldn’t measure it accurately this is due to the fact a person must read the pointer’s position. This can be difficult if the pointer only moves a small distance. The measurement uncertainty becomes larger.
final saddleBOT
@rain sentinel Has your question been resolved?
urban spade
but wouldnt the answer to question 1 be: The results would be less accurate beca...
Your approach is on the right track but not reasoning
urban spade
Hi, i hope you dont mind me asking this physics question here
Definition of accuracy is what matters the most here
In the question,they have defined accuracy as the lowest percentage/numerical value of relative error,which decreases with decrease in absolute value
So when you actually lower the exact value,absolute error lowers down too,giving you more accuracy
final saddleBOT
@rain sentinel Has your question been resolved?
rain sentinel
So when you actually lower the exact value,absolute error lowers down too,giving...
so it should become more accurate with 500g
urban spade
so it should become more accurate with 500g
If we were to define a relation between mass of the sample and accuracy in this case,then it is inversely proportional
rain sentinel
If we were to define a relation between mass of the sample and accuracy in this ...
so im incorrect?
deft terrace
Hi, i hope you dont mind me asking this physics question here
what is experiment 1 here
like what are you trying to measure
If it's the mass then maybe ts could help
here i have taken the error for the weight to be 0 since we have been given precise weights
so the error in m is directly proportional to the m assuming the error in g is constant for both cases
so as we increase the weight the error increases as the mass increases
deft terrace
If you are calculating weight instead of the masses then ts statement would be valid as well
final saddleBOT
@rain sentinel Has your question been resolved?
rain sentinel
what is experiment 1 here
The aim of this lab is to calculate and predict the loading and reaction forces required to keep a real-life system in static equilibrium
im measuring the force not mass
deft terrace
deft terrace
If you are calculating weight instead of the masses then ts statement would be v...
ts is what would be applicable here then
rain sentinel
ts is what would be applicable here then
Experimental accuracy – if experiment 1 were repeated with 100g weights instead of 500g weights, would the results be less or more accurate? Why?
so this is the question
deft terrace
Experimental accuracy – if experiment 1 were repeated with 100g weights instead ...
ye so as you increase the. mass of the system you can see the error in weight or in turn force needed to be in equilibrium increases
which is not ideal so 100g would be more accurate
deft terrace
If you are calculating weight instead of the masses then ts statement would be v...
...
rain sentinel
so it would be more accirate with 100g
obviously i cant write what you said down
According to the results from experiment 1 and by comparing between thetheoretical and measured results for both 500g and 100g, it has been shownthat using the 100g weights gave out more accurate results. This clears out thatusing less weights produces more accurate results. This could be because theweights we used were just 10g, and using a large amount of them could causean increase in the errors or uncertainties giving out less accurate results.
so this is the valid answer
not this: Smaller weights produce smaller measurable effects, so the same measurement uncertainties have a greater impact on the final calculated results — making the experiment less accurate.
final saddleBOT
@rain sentinel Has your question been resolved?
final saddleBOT
@rain sentinel Has your question been resolved?
final saddleBOT
final saddleBOT
Available help channel!
jagged flare
5 people sat in a room with 5 chairs, and there are 2 of them, A and B. they were asked to come out of the room. how many ways could they sit upon their return, given A and B didnt sit in their own previous chair?
final saddleBOT
5 people sat in a room with 5 chairs, and there are 2 of them, A and B. they wer...
jagged flare
.close arghhh im stupid
final saddleBOT
jagged flare
why did i use 5!-7 when it shouldve been 5p2-7
final saddleBOT
Available help channel!
jagged flare
Let $p,q,r$ be roots to $x^3-7x^2+1=0$. Find
$$\sum_{\text{cyc}}\sqrt{\frac{(p^2+1)(q^2+1)}{r^2+1}}$$
final saddleBOT
Let $p,q,r$ be roots to $x^3-7x^2+1=0$. Find
$$\sum_{\text{cyc}}\sqrt{\frac{(p^2...
soft zealotBOT
ihave<skissue>
jagged flare
yeah idk status 1
final saddleBOT
@jagged flare Has your question been resolved?
fiery bluff
**ihave<skissue>**
hmm, maybe firstly you can consider making a polynomial where the roots are p^2+1, q^2+1 and r^2+1
@jagged flare
jagged flare
i thought of that, but wouldnt you get square roots
fiery bluff
well the easy way I would think is to produce the companion matrix $A$ for the polynomial, then computing $\det(xI-(A^2+I))$
soft zealotBOT
Element118
jagged flare
havent learnt anything that you said so this gonna take a while
fiery bluff
another way is probably whack vieta formulas
jagged flare
ew
fiery bluff
which I would think it is less easy
jagged flare
im open to new methods
fiery bluff
well ig companion matrix it is https://en.wikipedia.org/wiki/Companion_matrix
so basically the companion matrix is a matrix with eigenvalues equal to the roots, but using it, you can figure out the eigenvalues of A^2+I
jagged flare
whats an eigenvalue
mm ok so an eigenvalue is a scalar that multiplies the eigenvector such that it equals an n×n matrix multiplied by the eigenvector
tranquil pine
which I would think it is less easy
Wdym, if you find the polynomial in terms of x²+1 as you said then it's an immediate application of vieta
wait I guess I misunderstood
fiery bluff
Wdym, if you find the polynomial in terms of x²+1 as you said then it's an immed...
almost immediate
tranquil pine
yeah ok
Anyway you don't need those things to find that polynomial
I mean I know you know but here it's simple enough not to use them
deft terrace
i thought of that, but wouldnt you get square roots
you can just do it by finding the constants seperatly as well ,using the root relations
jagged flare
mm ok so an eigenvalue is a scalar that multiplies the eigenvector such that it ...
ok tldr, for a square matrix there exists an eigenvector and eigenvalue so that matrix×eigenvector=eigenvalue×eigenvector?
jagged flare
you can just do it by finding the constants seperatly as well ,using the root re...
wdym?
jagged flare
yeah how do you use that
deft terrace
so sum of roots of an equation with roots p^2+1 ..... would be p^2+q^2+r^2 + 3
which you can find out
fiery bluff
ok tldr, for a square matrix there exists an eigenvector and eigenvalue so that ...
not that there exists, but you need algebraically closed field for it to exist
jagged flare
whats a field and what does it mean for it to be algebraically closed 
fiery bluff
that's just how eigenvectors and eigenvalues are defined (though you need your eigenvector to be nonzero)
jagged flare
yeah ok im leaving linear algebra for college
deft terrace
fiery bluff
whats a field and what does it mean for it to be algebraically closed <:ded:1042...
field, like rationals, reals, complex, you can add subtract (commutative) multiply and divide. Basically complex numbers form an algebraically closed field, that is, every nonconstant polynomial has a root
jagged flare
so sum of roots of an equation with roots p^2+1 ..... would be p^2+q^2+r^2 + 3
ohhh like that
ok hold on thats interestinf
deft terrace
ok hold on thats interestinf
then do it
you can find p^4+q^4+r^4 using the newtons method
jagged flare
p^2+q^2+r^2+3=(p+q+r)^2-2(pq+pr+qr)+3=52
tranquil pine
it's a bit bashy but it works
jagged flare
you can find p^4+q^4+r^4 using the newtons method
err newton girard iirc?
deft terrace
err newton girard iirc?
i think ye
that one
basically x^4 = 7x^3 - x
better than learning matrices then the meaning of eigens on the fly tbh 
tranquil pine
hmm, maybe firstly you can consider making a polynomial where the roots are p^2+...
In case you wanted to see how to do this, notice that ||you expect it do be a 3rd degree polynomial in terms of y=x²+1 so you probably want to find it through squaring the equation you have and get an even polynomial. You can fix the odd degree terms using x³=7x²-1||
jagged flare
P_4=e_1P_3-e_2P_2+e_3p_1
tranquil pine
In case you wanted to see how to do this, notice that ||you expect it do be a 3r...
Actually ||if you square it in the right way you get an even polynomial right away||
jagged flare
e_1=7 e_2=0 e_3=-1
P_1=7 P_2=7×7-0×2=49 P_3=7×49-0×7+(-1)×3=340
p^4+q^4+r^4=P_4=7×340-0×49+(-1)×7=2373
err thats alot
jagged flare
you can find p^4+q^4+r^4 using the newtons method
actualy where does this even come into play
deft terrace
actualy where does this even come into play
to find p^2q^2+...... term
jagged flare
good point
(p^2+1)(q^2+1)+(p^2+1)(r^2+1)+(q^2+1)(r^2+1)=p^2q^2+p^2r^2+q^2r^2+2p^2+2q^2+2r^2+3=107+((p^2+q^2+r^2)^2-(p^4+q^4+r^4))/2=107+(331/2)
its not an integer??
fiery bluff
i'm thinking it has to be an integer
deft terrace
ye
fiery bluff
because my method would produce an integer
so that means the computation went wrong somewhere
jagged flare
hmm
deft terrace
**ihave<skissue>**
imo you should just simplify ts expression
jagged flare
p^4+q^4+r^4=P_4=7×340-0×49+(-1)×7=2373
prolly here somewhere then
deft terrace
you would get (p^2+1)(q^2+1) + ..... in the numerator
and the square root of the product of all three in the denominator right
fiery bluff
yeah that's one way to simplify it
deft terrace
(p^2+1)(q^2+1)+(p^2+1)(r^2+1)+(q^2+1)(r^2+1)=p^2q^2+p^2r^2+q^2r^2+2p^2+2q^2+2r^2...
your values are sorta wrong here
3 + 2(p^2+q^2+r^2) + (p^2q^2+....)
that should be what you get
jagged flare
whats what i got?
deft terrace
and all of these should be integers
deft terrace
shouldnt you do like (pq+pr+qr)^2-2pqr(p+q+r) and skip p^4+q^4+r^4
yup you can do that as well
deft terrace
3+2(49)+(331/2)
how did you get 52 shoudnt it be 49
jagged flare
(0)^2-2(-1)(7)=14=p^2q^2+p^2r^2+q^2r^2
jagged flare
how did you get 52 shoudnt it be 49
so the last bit would be ((49)^2-2373)/2=34
jagged flare
(0)^2-2(-1)(7)=14=p^2q^2+p^2r^2+q^2r^2
but why did i get 14 here
jagged flare
huhh 😭
jagged flare
3 + 2(p^2+q^2+r^2) + (p^2q^2+....)
so this would be 3+2(49)+(7)?
deft terrace
so this would be 3+2(49)+(7)?
yup
deft terrace
now for the bottom part take -1 common from each term
-(-1-p^2).... thats what you will get then right
it can just be seen as inserting -1 into a cubic polynomial with roots p^2,q^2 and r^2
then multiplying the result by -1
that should be it
jagged flare
huh sorry what
deft terrace
a cubic with roots a,b and c can be written as (x-a)(x-b)(x-c)
inserting -1 in it would give you (-1-a)(-1-b)(-1-c)
which is the same term as above
deft terrace
I'm talking about the denominator of the orignal problem yes
jagged flare
oh like after you combine them?
jagged flare
so (p^2+1)(q^2+1)+(p^2+1)(r^2+1)+(q^2+1)(r^2+1)=108
youll be left with this in the numerator
and sqrt((p^2+1)(q^2+1)(r^2+1)) in the denominator right?
deft terrace
ye
jagged flare
and how do you make the polynomial with roots p^2 q^2 r^2 again
oh wait
sum is 49
deft terrace
yup
jagged flare
the product sum thingy is 7 (?)
deft terrace
yup two at a time is 7
jagged flare
product is 1
deft terrace
yup
jagged flare
so t^3-49t^2+7t-1 has roots p^2, q^2, and r^2
deft terrace
yup
deft terrace
so t^3-49t^2+7t-1 has roots p^2, q^2, and r^2
t^3
jagged flare
subbing t=-1 we get -58
deft terrace
yup
jagged flare
since theres a - the denominator is sqrt(58)
deft terrace
now multiply that by the -1 we took out
jagged flare
so 108/sqrt(58)
jagged flare
-1-49-7-1=-58?
winter grail
This is very unordinary but I have a puzzle that works like in the video and I really need help.
I am looking for a password, 1 word or could be multiple
This is a tesseract, 5x5 Rubik's cube style. Where each letter moves according to the directions.
final saddleBOT
This is very unordinary but I have a puzzle that works like in the video and I r...
winter grail
Any help would be much appreciated
fiery bluff
This is very unordinary but I have a puzzle that works like in the video and I r...
it seems each face is an independent puzzle
winter grail
it seems each face is an independent puzzle
Yes and once the timer resets the letters get randomized
fiery bluff
idk what the solve state is
winter grail
This is more on the cryptography side
I was thinking a Polybius Square looks like it
That has the I/J letters
fiery bluff
idk, where is this on
fiery bluff
is this a webpage or something
fiery bluff
it's a webpage, you can view source or something
fiery bluff
the letters don't look important since they are random
so idk what's there to look at
winter grail
Well they should unlock the locked layer if you put them in the right order
And could give the English key for the password
final saddleBOT
@winter grail Has your question been resolved?
delicate sorrel
This is very unordinary but I have a puzzle that works like in the video and I r...
What program is that
final saddleBOT
@winter grail Has your question been resolved?
final saddleBOT
@winter grail Has your question been resolved?
winter grail
Any idea?
late rose
This is very unordinary but I have a puzzle that works like in the video and I r...
if this is a game you lowkey might have better luck decompiling it and inspecting its code
😭
late rose
if this is a game you lowkey might have better luck decompiling it and inspectin...
if this a web game you should try to use inspect element to analyse it more
late rose
Nope nothing in the source
can you send us the file?
winter grail
can you send us the file?
Its a webgame a puzzle hunt
winter grail
if this a web game you should try to use inspect element to analyse it more
Sadly not allowed to use the console
Considered "cheating"
Its an interactive js code cube
winter grail
if this is a game you lowkey might have better luck decompiling it and inspectin...
Looking to get a password out of this, if you sort the letters in the right way you can unlock the lock I believe
late rose
Sadly not allowed to use the console
well i was thinking about using a script to automate things
i guess that wouldnt be allowed?
nvm now that i think about it, it would just be easier to copy the faces into a program and then perform some scripting on it rather than interact with the webgame directly lol
can you show us a full set of 5 faces?
it's a little hard to tell from that video
it could be a playfair cipher https://en.wikipedia.org/wiki/Playfair_cipher
final saddleBOT
@winter grail Has your question been resolved?
winter grail
it could be a playfair cipher https://en.wikipedia.org/wiki/Playfair_cipher
Its polybius square confirmed
And the task is to sort the letters in abc format
But doing that for all 5 faces manually under 10 minutes is very hard
late rose
you could use a program/script to do it for you
winter grail
Got 10 minutes to solve all of this
winter grail
you could use a program/script to do it for you
Its got an anti cheat system
Unless you know a way to bypass it.. 👀
late rose
Its got an anti cheat system
input in each of the 25 characters in each face
and generate the optimal solving input
late rose
input in each of the 25 characters in each face
by hand
late rose
you could potentially optimise it further by reading each face with ocr
not sure how it would fare, given that there is a lot of 'noise' due to the cube's transparency
winter grail
Yea tried ocr tools but it couldnt read it
late rose
Even writing them down is slow
Idk i can type out one face in under a minute
winter grail
If you can make a script, how could you run that without the console or terminal
late rose
And the task is to sort the letters in abc format
oh yeah wdym "abc format" like alphabetical order?
winter grail
oh yeah wdym "abc format" like alphabetical order?
If I knew the exact steps for all faces I could do them under 10 minutes
late rose
what does that have to do with a polybius square btw
winter grail
Just the I/J I guess
late rose
bruh
winter grail
It's called the Polybius Cube in the source
winter grail
Yes
Once you solve one face it turns blue
5 faces under 10 minutes
You have about 2 minutes per face to solve
late rose
If anyone else wants to try this, here's the problem formalised a little:
You have a square face (say, the input square):
[ 0 1 2 3 4
0 ['D', 'U', 'Z', 'I', 'O'],
1 ['S', 'V', 'C', 'A', 'W'],
2 ['G', 'M', 'P', 'F', 'Y'],
3 ['N', 'R', 'T', 'E', 'Q'],
4 ['L', 'X', 'H', 'B', 'K'],
]
On each move, you can perform two things:
- Rotate a row i of the square by n places rightwards (positive direction) or leftwards (negative direction). Denote this as r(i, n) where i \in {0, 1, ... 4} and n \in {1, 2, 3, -2, -1}
- Rotate a column j of the square by n places rightwards (positive direction) or leftwards (negative direction). Denote this as _c(j, n) where j \in {0, 1, ..., 4} and n \in {1, 2, 3, -2, -1}.
A row, or column, will always wrap around itself.
Your goal is to get the square into the form
[
['A', 'B', 'C', 'D', 'E'],
['F', 'G', 'H', 'I', 'K'],
...
]
with as few moves as possible, while still maintaining a decent execution time.
Note that the letter 'J' is missing.
i know this is solvable by a greedy algorithm, however the move sequences will be pretty long
for a more optimal move sequence, i think maybe some sort of A* might help? but i'm not all too familiar with graph algortihms so...
winter grail
If anyone else wants to try this, here's the problem formalised a little:
You h...
Yes
Now do that for all 5
And it gets randomized each time
worthy radish
What is wrong?
final saddleBOT
What is wrong?
tired walrus
,rccw
soft zealotBOT
tired walrus
how did you simplify $\frac{a^{6/4}}{a^{-2/4}}$ to $a$?
soft zealotBOT
Ann
tired walrus
and for that matter how did $\frac{b^{5/4}}{b^{13/4}}$ become $b^{-8}$
worldly mesa
5/4 - 13/4 is also not 8
soft zealotBOT
Ann
worthy radish
a^4/4
worthy radish
It’s
2(a^2) now
It’s easier
To mark like that
5/4 - 13/4 = (-8)/4
Which is b^-2
@tired walrus
tired walrus
yes
worthy radish
Ok
I mistook it then and just pasted -8
I’m a little bit confused
Why isn’t it (2a^2)/b^2
@tired walrus
tired walrus
that two was in the denominator, and you weren't supposed to pull it into the numerator
worthy radish
tired walrus
also i admit i didn't notice what happened to the 2 because i was trying to help you get the exponents right
but the 2 was in the denominator and it stays in the denominator
worthy radish
How do you get a^2 from this equation?
I don’t seem to get a^2 but a @tired walrus
tired walrus
you're making the same sign error with the exponents on the a
a^(6/4 - (-2/4))
is a^(6/4 + 2/4)
worthy radish
So if it’s
1 -(-1)
It’s 1+1
Cause - is near the bracket
So you change the numbers
Just like
tired walrus
negative times negative gives positive, yes
tired walrus
no, 9(5+x) never happens in yours
5 - 9 - 5 - x
simplifies to -9 - x, and that's that
wheat zodiac
hii I need help on how to convert decimals as fractions
final saddleBOT
hii I need help on how to convert decimals as fractions
wheat zodiac
does that thumbs upmean youll help me? sorry im new
thin sequoia
yeah
wheat zodiac
ahh ok so like how do I do it. when im in class i get on the calclulator 0.75 but how am i supposed to know thats 3/4
thin sequoia
generally, post the question in mind here as a first msg to help with but yes
wheat zodiac
oh im sorry
tame vortex
I need help with this sum
thin sequoia
I need help with this sum
go read #❓how-to-get-help this channel is occupied
0.75 = 75/100, and can be simplified from there
tame vortex
Ok
thin sequoia
in general, it's good to know that certain numbers are divisible by other ones. like any number ending in a 5 or 10 is divisible by 5, like these oens
wheat zodiac
ohhhh
thin sequoia
so that you can divide bot 75 and 100 by 5
wheat zodiac
15 and 20
thin sequoia
yep
thin sequoia
wheat zodiac
.close
final saddleBOT
final saddleBOT
Available help channel!
final saddleBOT
Available help channel!
barren pebble
final saddleBOT
barren pebble
The given conditions are weird, ugh
barren pebble
Yep
bold igloo
what do you have so far
barren pebble
p(p-1/2)<0 so p>2p²
But that doesn't seem to help
As the equality occurs when variables equal 1/3 (i think)
barren pebble
I'm attempting a)
Haven't tried b) yet, because i thought maybe a) leads to b
rz=(1-x-y)(1-p-q), does that help?
bold igloo
Have you tried
- solve for Z in x+y+z=1
and solve for r in p+q+r=1
and plug that into px+qy+rz
bold igloo
Haven't tried b) yet, because i thought maybe a) leads to b
yea a leads to b
barren pebble
rz=(1-x-y)(1-p-q), does that help?
This?
barren pebble
I'm in grade 10
bold igloo
is this an AM GM question 😭
barren pebble
is this an AM GM question 😭
Probably
It's in the homework section of AM-GM, though some questions also use C-S and stuff
bold igloo
weighted am gm is roughh
bold igloo
Can you not just cheese this by moving the terms arround 😭
barren pebble
How so?
bold igloo
could re arrange it twice
like get
px+qy+rz-x/2-y/2 >= 0
then get by x and y terms
(p-1/2)x+(q-1/2)y+rz >= 0
then we know right p,q,r add up to 1, so we can write that as r=1-p-q
and sub that back into the above equation
(p-1/2)x+(q-1/2)y+(1-p-q)z>= 0 (edit : added the z)
barren pebble
It's rz, not just r btw
bold igloo
yea i am considering z
bold igloo
(p-1/2)x+(q-1/2)y+(1-p-q)**z**>= 0 (edit : added the z)
yea it works
then u would group by x and y
bold igloo
yea but u want p(x-z)+q(y-z)+ z -x/2 - y/2 >= 0
barren pebble
Are you trying to just factor it out?
bold igloo
yea
butt from the assumption that x<y<z
we can say that the coefficients of p and q must be 0 or lower
like (x-z) <= 0
same with (y-z) <= 0
barren pebble
But we're trying to get ≥0
bold igloo
yea but u want p(x-z)+q(y-z)+ z -x/2 - y/2 >= 0
no i mean thats the 2 terms in this equation
We are also given p > 0 and q > 0
p(x - z) <= 0
q(y - z) <= 0
and back to the question you know p,q <=1/2
and since p(x-z) and q(y-z) are both non positive
and we know p and q <= 1/2 we can have a lower bound
p(x - z) >= 1/2 (x-z)
same with q
barren pebble
Wait how?
bold igloo
q(y - z) >= 1/2 (y-z)
barren pebble
p(x - z) >= 1/2 (x-z)
This
bold igloo
so basically you know p and q are AT most 1/2
barren pebble
This just means p≥1/2 if x≠z, which is not true in general
barren pebble
so basically you know p and q are AT most 1/2
No?
bold igloo
multipying by negative switches sign
bold igloo
Yes at most
yeaaa
p(x-z) is least negative when p is smaller and most negative when p is 1/2
barren pebble
I meant it's q(y-z), not q(x-z)
bold igloo
yea but u want p(x-z)+q(y-z)+ z -x/2 - y/2 >= 0
into this equation
bold igloo
so it would be
p(x-z) + q(y-z) + z - x/2 - y/2 >= 1/2 (x-z) + 1/2 (y-z) + z - x/2 - y/2
barren pebble
And i'm assuming for b we try to prove (x+y)/2 ≥ 8xyz?
bold igloo
re arrange x and y and done
barren pebble
And i'm assuming for b we try to prove (x+y)/2 ≥ 8xyz?
Or is b) stronger than a)
barren pebble
LHS ≥x^p. y^q .z^r by W-AM-GM right?
bold igloo
yea this am gm
barren pebble
Wait...
barren pebble
Yeah
bold igloo
proving px+qy+rz>= (x+y)/2 >= 8xyz wont work
part a is under assumption that x<y<z
thats the core logic behind getting coefficients sub 0
barren pebble
Well b) is symmetric so we can assume the order of the variables
At least one set of them
Assuming both sets would lose generality
So we can still WLOG assume x≤y≤z if needed
bold igloo
So we can still WLOG assume x≤y≤z if needed
do idt we need to assume that
barren pebble
I don't even know, there were 30-ish questions, some in the IMO and IMO shortlist but i solved those easier
This question is weird
bold igloo
this is AM GM
what is this triangle man
oh
this is such a dumb way
but what if we say the region of pqr is
v1=<1/2, 1/2, 0>
v2 = <1/2,0,1/2>
v3 = <0,1/2,1/2>
barren pebble
8xyz≤(x+y)(y+z)(z+x)=(1-x)(1-y)(1-z)
barren pebble
but what if we say the region of pqr is
v1=<1/2, 1/2, 0>
v2 = <1/2,0,1/2>
v3 = ...
Oh hell nah
What in the linear algebra is this
bold igloo
if we prove the inequality works for those 3 corners
it should work for any point inside
so case1 case2 case 3 😭
barren pebble
Can we just use classical inequalities
bold igloo
im joking
there has to be a better way
but this is one way
waitttt
we need to prove the corner case
that (x+y)/2 > 8xyz
given that x+y+z=1
barren pebble
8xyz≤(x+y)(y+z)(z+x)=(1-x)(1-y)(1-z)
This is equivalent to 9xyz≤(xy+yz+zx) or 1/x + 1/y + 1/z ≥ 9
barren pebble
that (x+y)/2 > 8xyz
given that x+y+z=1
And x≤y≤z
bold igloo
why 7
barren pebble
If you don't assume x≤y≤z, we can't have LHS ≥ (x+y)/2
bold igloo
no its the other way (x+y)/2 >= 8xyz
barren pebble
I know, but where are you getting (x+y)/2 from
bold igloo
this is back to v1 btw
barren pebble
You need x≤y≤z to get LHS≥(x+y)/2
bold igloo
x/2 + y/2 + 0z >= 8xyz
testing the corner (1/2 , 1/2, 0)
could chose any corner tbf i just chose that
cus its v1
when i wrote it
bold igloo
u can AM GM that (x+y)/2 >= sqrt(xy)
so x+y >= 2 sqrt(xy)
and then u square both sides
assume idk T as x+y
so it would be T>=2 sqrt(xy)
barren pebble
Hold, i think i got it
bold igloo
T^2 >= 4xy meaning
xy <= T^2/4
bold igloo
Hold, i think i got it
yea prove for a corner and by symmetry it works for both the other
barren pebble
No i got it in a different way
bold igloo
how would u do without proving (x+y)^2 >= 4xy
barren pebble
Assume x≤y≤z. We need to prove 1-z≥16xyz, but xy≤(1-z)²/4, so we need 1≥4(1-z)z or (2z-1)²≥0
barren pebble
how would u do without proving (x+y)^2 >= 4xy
I did use that
bold igloo
Assume x≤y≤z. We need to prove 1-z≥16xyz, but xy≤(1-z)²/4, so we need 1≥4(1-z)z ...
how can you assume x<y<z
barren pebble
Because they have the same role
bold igloo
yea no mb for going into linear alg
idk why but i rlly like using linear alg to abuse symmetry
and just prove one case and be like cus symmetry its the same for other 2 so done
barren pebble
It's okay, i'm doing a research project in spectral graph theory, which involves linear alg so i get what you said
.close
final saddleBOT
final saddleBOT
Available help channel!
sudden pendant
final saddleBOT
tired walrus
!status
final saddleBOT
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
1
severe verge
what is the minimum value of the stuff in the parentheses
tired walrus
and at what x does it happen
sudden pendant
2+log2 5
tired walrus
not what was asked for
sudden pendant
i know,, but i need to find the x at which it happens right
tired walrus
you need to find the x at which the minimum happens, but this x isn't equal to 2+log_2(5)
sudden pendant
yeah i know that
sudden pendant
what is the minimum value of the stuff in the parentheses
i was answering this
tired walrus
if you know that then why give bs answers to mine or @severe verge's questions
final saddleBOT
final saddleBOT
Available help channel!
ember river
I'm using excel trying to do a loans question.
APR + interest rate
n = amt of times compounded; since this is monthly that would be 12
Y = years (I think)
PMT = the repeating payments (monthlies)
P is supposed to be down payment (Or anything invested initially, if any)
FV =FV(APR/n,n*y,-PMT,-P) That's basically the end result of all of the above put together in a loan plan (in this case I was trying to get to this number through x monthly interest within X years (which I was trying to find for this using goalseek) and with x monthly payments
What I'm not understanding is how I got such vastly different times, I really don't think a $40 monthly difference will amount to ~70yr loan length difference. It couldn't have accidentally translated it into months either because the monthly payment from a previous question for the left set ($50 less) I'm almost positive I got right pays off the loan in 3 years. 72.21 Months amounts to 6.02 years.
final saddleBOT
I'm using excel trying to do a loans question.
APR + interest rate
n = amt of ti...
soft zealotBOT
Piñata
ember river
**Piñata**
What is this
final saddleBOT
@ember river Has your question been resolved?
final saddleBOT
@ember river Has your question been resolved?
final saddleBOT
@ember river Has your question been resolved?
cedar obsidian
final saddleBOT
cedar obsidian
!status
final saddleBOT
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
From the usual graph of energy-time it shows for K(t) graph it starts from origin and then reaches max. According to this information option d suites
But (d) not the correct option choice as per answer key
fiery bluff
technically all 4 options are wrong since the KE graph should be sinusoidal [check this!]
cedar obsidian
Yes there isn't charcterstic of sine and consine graph
tiny kraken
But (d) not the correct option choice as per answer key
what's the answer key the book state then
cedar obsidian
Option c
It has explanation K is max at equilibrium position at t = T/2 and minimum at extreme t = T/4
fiery bluff
yes option c is "least wrong"
tiny kraken
yes option c is "least wrong"
yeah,but it's still not a sin/cosin function
cedar obsidian
yes option c is "least wrong"
Because it looks like a cosine?
tiny kraken
lmao
tiny kraken
Because it looks like a cosine?
not really
tiny kraken
It has explanation K is max at equilibrium position at t = T/2 and minimum at ex...
cuz this
and also at t=0 KE is at its max
autumn osprey
O)
autumn osprey
Why is C greater than A if they are both quaters of a circle
autumn osprey
Hi! Just a question but
Hello
main mirage
O)
Can you explain how the dog gets to these two points?
I.e explain (or draw) how the dog gets to the red point. (And then repeat for the yellow point).
autumn osprey
I.e explain (or draw) how the dog gets to the red point. (And then repeat for th...
It is because that is the area the dog is allowed to walk in
main mirage
Well yeah but like
I am interested to see physically how the dog could go to that point
main mirage
Can you explain how the dog gets to these two points?
Seeing as the dog is at the point (3, 2) and is attached to a 6m long rope, how does the dog go to the red point?
I know :)
autumn osprey
You're probably overcomplicating this
So...my original question
main mirage
You're probably overcomplicating this
To do this question, the most important bit is you understand how exactly this dog moves.
In particular,
The only restriction is that the dog starts at the point (3, 2) and can move anywhere that is 6m from that point, right?
autumn osprey
Bro wait
main mirage
(That is not phasing through the wall)
main mirage
My original question was why is c greater than a if they are both quarters
Yep, I know
In general, if two circles or sectors are different, that means the radius of each sector is different.
main mirage
Can you explain how the dog gets to these two points?
So the short answer is, the radius of circle C is larger than circle A. That means the area covered by circle C must be larger than the area covered by circle A.
Though, the most important thing to understand is why the radii are different, which was what I was trying to show with my question :)
autumn osprey
Though, the most important thing to understand is *why* the radii are different,...
Okay
main mirage
Okay
Indeed, so how does the dog move to the red point? I.e starting from (3, 2), how do you get to the point marked in red?
autumn osprey
Indeed, so how does the dog move to the red point? I.e starting from (3, 2), how...
Moving 4 units upwards
main mirage
Moving 4 units upwards
Uhuh, and?
autumn osprey
Opps
main mirage
Opps?
autumn osprey
Uhuh, and?
2 units to the left
autumn osprey
Yes
main mirage
Now, remember that we're trying to find the region that the dog can go to.
main mirage
The reason why A is then a sector in the first place is because we can imagine that the dog can rotate around the particular axis given by the point (3, 6).
So would it make sense then that, the region of shape A is a sector with radius 2?
autumn osprey
So would it make sense then that, the region of shape A is a sector with *radius...
What is the difference between it and c?
main mirage
What is the difference between it and c?
Good question!
Try and consider getting to the yellow point now :)
Using the exact same logic, how would the dog move to the yellow point?
autumn osprey
Using the exact same logic, how would the dog move to the yellow point?
2 units left and 2 units up
main mirage
2 units left and 2 units up
Mmm the thing is,
If you do this, you have only used 4m if the rope
That means the dog can actually go further!
autumn osprey
Ohhhh...
main mirage
My pleasure :)
autumn osprey
.close
uneven idol
Having trouble solving this integral. Supposed to only use the substitution method, this one here was my final attempt and i have no idea what went wrong
final saddleBOT
Having trouble solving this integral. Supposed to only use the substitution meth...
thorny canyon
Having trouble solving this integral. Supposed to only use the substitution meth...
oh this one's standard
1/sinx = cscx
multiply and divide by (cscx - cotx)
uneven idol
yeah im a bit of a beginner at this
thorny canyon
yeah im a bit of a beginner at this
ahh, bit tricky to think of the multiplication
uneven idol
I see that i tried this before and got stuck with two variables
oh i see what you meant
is there any way to solve it using only the basic trigonometric functions?
scarlet sequoia
I see that i tried this before and got stuck with two variables
this substitution will not help imo
wait
uneven idol
got any hint for which substitution to use please?
scarlet sequoia
you still have some cos(x) remaining
uneven idol
yeah thats why i got stuck on it
scarlet sequoia
you need to express it in terms of t
tiny kraken
maybe sinx/2
scarlet sequoia
tan(x/2) is better
tiny kraken
tan(x/2) is better
yeah that's also commonly used
scarlet sequoia
cos(x), sin(x) and tan(x) are all expressible in terms of tan(x/2) only
and so any rational function with trigs
can be changed into a rational function with regular "x" or "t"
depends on how you call your integration variable
scarlet sequoia
uh?
no
sin(x) = 2tan(x/2)/(1 + tan(x/2)^2) iirc
If you've never seen those formulas, better try to prove them
using the double angle formulas you already know
final saddleBOT
uneven idol
thanks for the help
final saddleBOT
Available help channel!
wide mason
the denominator isnt telescoping please help
final saddleBOT
the denominator isnt telescoping please help
true moth
maybe try to find a and b so that
1/(2n-2k+1)(2n-k+1) = a/(2n-2k+1) + b/(2n-k+1)
fiery bluff
the numerator also doesn't telescope
maybe it cancels out with the numerator somewhere
wide mason
maybe try to find a and b so that
1/(2n-2k+1)(2n-k+1) = a/(2n-2k+1) + b/(2n-k+1)
what will it do?
wide mason
the numerator also doesn't telescope
that's the problem
fiery bluff
what did you get for the denominator
wide mason
maybe it cancels out with the numerator somewhere
yeah but the thing is i have a thought about it like denominatior could be a bionomial or some standard expansion which cancels with numerator
true moth
what will it do?
do it and then you'll see
fiery bluff
yeah but the thing is i have a thought about it like denominatior could be a bio...
maybe you can show what you have tried so far
wide mason
what did you get for the denominator
no progress yet i just wrote it in telescoping form
fiery bluff
no progress yet i just wrote it in telescoping form
as in, you written the fraction $\frac{k}{(2n-2k+1)(2n-k+1)}$ as a difference of two fractions?
true moth
trust me it is telescopic
soft zealotBOT
Element118
wide mason
maybe you can show what you have tried so far
no i just gave a thought i didnt know hw to proceed bc it doesn't telescopes
wide mason
as in, you written the fraction $\frac{k}{(2n-2k+1)(2n-k+1)}$ as a difference of...
yes
$\frac{1}{(2n-2k+1)}-{1}(2n-k+1)}$
soft zealotBOT
DEXTER MORGAN
$\frac{1}{(2n-2k+1)}-{1}(2n-k+1)}$
```Compilation error:```! Extra }, or forgotten $.
l.49 $\frac{1}{(2n-2k+1)}-{1}(2n-k+1)}
$
I've deleted a group-closing symbol because it seems to be
spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.
Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2023/texmf-var/fonts/map/pdftex/updmap/pdftex.map}{/usr/l```
fiery bluff
try splitting into 2 summations and re-indexing the sum
wide mason
$\frac{1}{(2n-2k+1)}-\frac{1}{(2n-k+1)}$
wide mason
try splitting into 2 summations and re-indexing the sum
wdym by re-indexing
fiery bluff
"2n-2k+1" might be not very obvious to work with, but you can tell it's an odd number
wide mason
"2n-2k+1" might be not very obvious to work with, but you can tell it's an odd n...
yeah
fiery bluff
wdym by re-indexing
for instance say m=2n-k+1, and then write some of the terms as a summation over m
wide mason
oh i'll let yk if you get smt lmk too
soft zealotBOT
DEXTER MORGAN
fiery bluff
i think i have a good idea of the way forwards but i'm just trying to figure out how to explain
wide mason
trust me it is telescopic
how?
fiery bluff
don't look for it to telescope
wide mason
yeah it wont prolly
fiery bluff
but try to just write as summations of 1/m
fiery bluff
Try using the notation
$\sum_{m\in S}\frac{1}{m}$
where $S\subseteq\mathbb{N}$ is a subset of natural numbers
soft zealotBOT
Element118
true moth
how?
you see the two term you have ? if you have to sum them for k=1 to k=n, then part of the terms cancels, the - 1/(2n-k+1) half cancels with the 1/(2n-2k+1)
wide mason
you see the two term you have ? if you have to sum them for k=1 to k=n, then par...
yes but there is still some expansion which lives
fiery bluff
yes but there is still some expansion which lives
yeah there's a way to deal with that, I think the notation I suggested might be helpful in making it obvious what to do
true moth
yes but there is still some expansion which lives
yes but then you can try to write the expansion in another way for it to simplify with numeratir
wide mason
yes but then you can try to write the expansion in another way for it to simplif...
that's what I am stuck on
wide mason
yeah there's a way to deal with that, I think the notation I suggested might be ...
hmmm not sure either
fiery bluff
maybe I can step you through on how to use that notation
wide mason
yeah sure
fiery bluff
you have $\sum_{k\in{1,\dots,n}}\left(\frac{1}{2n-2k+1}-\frac{1}{2n-k+1}\right)$
wide mason
yea
soft zealotBOT
Element118
fiery bluff
first it probably helps if you split it up into two sums
wide mason
i did that
fiery bluff
$\sum_{k\in{1,\dots,n}}\frac{1}{2n-2k+1}-\sum_{k\in{1,\dots,n}}\frac{1}{2n-k+1}$
wide mason
did the same step
soft zealotBOT
Element118
fiery bluff
yeah so now we want to re-index - let m = 2n-2k+1 in the first sum, what set of values does m take
true moth
try to detail how to change the index on your sum so it's easier
for example if you want to change index of a sum indexed by k=1 to k=n with l = k+n
you know the new sum will start with l = n+1 and end with l = 2n
and then you can make the change in the sum expression
here you want something like l = 1 to l = n in the end
wide mason
yeah so now we want to *re-index* - let m = 2n-2k+1 in the first sum, what set o...
it takes weird fraction values if i express m in terms of n and k
fiery bluff
n is fixed in this sum, k varies
so m changes based on k
try plugging in k=1, k=2, and so on until k=n, do you see what values 2n-2k+1 would take?
wide mason
i'll let you guys know when I'll have some progress on this
steady wraith
what is the bay harbour butcher doing in a math discord server 😭
sorry thats the first thing that crossed into my mind when i saw the username of the person who needed help
final saddleBOT
@wide mason Has your question been resolved?
craggy plank
regal hamlet
final saddleBOT
fiery bluff
!status
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What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
fn is bounded pointwise, if we had it was defined on a numerable set, then we could get some sub-sequence at least
fiery bluff
2
what have you tried so far
let $q$ be a rational number. Can you find a subsequence such $\lim_{k\to\infty}f_{n_k}(q)$ exists?
soft zealotBOT
Element118
regal hamlet
fn is bounded pointwise, if we had it was defined on a numerable set, then we co...
this, I got this theorem that says we can get a converging pointwise sub sequence is the sequence of function is bounded and defined on a numerable set E
fiery bluff
iirc there's arzela-ascoli, I think the proof can be quite similar
regal hamlet
let $q$ be a rational number. Can you find a subsequence such $\lim_{k\to\infty}...
ah, the rationals is a numerable set right, so ${ f_{n_k} (q })$ converges pointwise for every $q \in \mathbb{Q}$
soft zealotBOT
Halex
regal hamlet
iirc there's arzela-ascoli, I think the proof can be quite similar
do you have its proof somewhere, I'd like to see it too but our text book do not include it afaik (Rudin)
fiery bluff
ah, the rationals is a numerable set right, so $\{ f_{n_k} (q \})$ converges poi...
yeah you can have that
though now you need to argue that it converges on the irrationals too
which actually... it's not so clear
regal hamlet
yes, how do i do 😭
fiery bluff
oh wait
it's increasing so there's only countably many jump discontinuities
think about how that can help you select a further subsequence so it converges on the "bad" irrationals
regal hamlet
it's increasing so there's only countably many jump discontinuities
I don't get why at all
fiery bluff
I don't get why at all
each jump discontinuity excludes at least one rational from its image
regal hamlet
each jump discontinuity excludes at least one rational from its image
Every monotone function on $\mathbb{R}$ has at most countably many discontinuities, and all of them are jump discontinuities?
soft zealotBOT
Halex
fiery bluff
Every monotone function on $\mathbb{R}$ has at most countably many discontinuiti...
yeah you should be able to prove it
Let $f$ be monotone increasing, and say $f$ is not continuous at $x$. Consider $\sup{f(t):t<x}$.
soft zealotBOT
Element118
fiery bluff
yeah the left limits and right limits exist for monotone functions
regal hamlet
Is it the limit to the Left? $$\text{sup} { f(t):t<x } = \lim_{t \to x^-} f(t);$$ if $f$ was continuous at $x$ then we would have $\text{sup} { f(t):t<x } = f(x)$
soft zealotBOT
Halex
regal hamlet
yeah the left limits and right limits exist for monotone functions
we would then have $$\text{inf} { f(t):t>x } \leq f(x) \leq \text{sup} { f(t):t<x }$$ as $f$ is increasing
soft zealotBOT
Halex
regal hamlet
But nout sure why its has countably many discontinuities
fiery bluff
But nout sure why its has countably many discontinuities
if the left and right sides are the same, then it cannot be discontinuous
the only way it can be discontinous at x is if the left and right limits are different
fiery bluff
but that means there has to be a jump at x
regal hamlet
yep, but why can we assume f discontinous at x at first?
fiery bluff
you can structure the argument
- say f and x, no assumption whether f is discontinuous
- talk about the left and right limits
- notice left and right limits equal iff f is continuous at x
regal hamlet
If they were equal then there would not be a discontinuity, hence it is still countable
fiery bluff
If they were equal then there would not be a discontinuity, hence it is still co...
from here, you can notice your countably many discontinuities and you can do something to target each of them
regal hamlet
from here, you can notice your countably many discontinuities and you can do som...
if f(a) < f(b), then the possible jump has lenght f(b)-f(a)>0. If there were infinite discontinuities, then the sum would be infinite, which would exceed f(b)-f(a)
did I get it right
fiery bluff
if f(a) < f(b), then the possible jump has lenght f(b)-f(a)>0. If there were inf...
consider $f(x)=\left\lfloor\frac{1}{x}\right\rfloor^{-1}$
soft zealotBOT
Element118
fiery bluff
of course it's only increasing on $[0, 1]$, but you can easily modify it to have infinitely many jumps
soft zealotBOT
Element118
regal hamlet
consider $f(x)=\left\lfloor\frac{1}{x}\right\rfloor^{-1}$
oh
there are infinite jumps and the sum if finite
fiery bluff
yeah which means your argument is wrong
what you need to show is that there are countably many jumps
which includes countably infinite
regal hamlet
If $f(a) < f(b)$, the total increase of the function on $[a,b]$ is $f(b)-f(a) > 0$. Each jump consumes a positive part of this total. Therefore, there can be only finitely many jumps larger than any fixed positive number
soft zealotBOT
Halex
fiery bluff
okay, then...
regal hamlet
Although there may be infinitely many smaller jumps, their total sum cannot exceed $f(b)-f(a)$
soft zealotBOT
Halex
regal hamlet
if there were infinite jumps bigger than 1/n, then the total sum of these jumps would exceed the total increase, f(b)-f(a)
so there can only be a finite amount of jumps bigger than 1/n
if we consider the points x such that their jump is bigger than 1/n, n=1,2,...; then the union of these points is an union of finite sets, which is at most countable
<@&268886789983436800> ^
regal hamlet
if we consider the points x such that their jump is bigger than 1/n, n=1,2,...; ...
and the union is the union of discontinuities
final saddleBOT
@regal hamlet Has your question been resolved?
fiery bluff
if we consider the points x such that their jump is bigger than 1/n, n=1,2,...; ...
yeah
so there are at most countably infinitely many jumps
now you need to deal with each of them
since that's where all the discontinuities are
fiery bluff
**Element118**
think about this
final saddleBOT
@regal hamlet Has your question been resolved?
regal hamlet
let $q$ be a rational number. Can you find a subsequence such $\lim_{k\to\infty}...
yes, we can, by the theorem I mentioned earlier
so can we define $f(x) = \text{sup} { f (q) : q \leq x }$
soft zealotBOT
Halex
regal hamlet
only where f is continuos at
final saddleBOT
@regal hamlet Has your question been resolved?
final saddleBOT
@regal hamlet Has your question been resolved?
final saddleBOT
@regal hamlet Has your question been resolved?
fiery bluff
so can we define $f(x) = \text{sup} \{ f (q) : q \leq x \}$
not really, you need to use a similar trick
without loss of generality you can choose a subsequence converging at every rational
that's the first step
but then you now have countably many jump discontinuities which you need to handle, and to resolve that, you need to use a similar trick
regal hamlet
🤔
regal hamlet
so can we define $f(x) = \text{sup} \{ f (q) : q \leq x \}$
Might I have forgotten to mention q is rational?
regal hamlet
but then you now have countably many jump discontinuities which you need to hand...
I'm not sure how to proceed then
fiery bluff
I'm not sure how to proceed then
look at what the trick on rationals got you - you can guarantee convergence on countably many points
which points are you left where you can't guarantee convergence, and how many are there?
regal hamlet
The Irrationals?
fiery bluff
The Irrationals?
[try proving that continuity of f(x) at x implies convergence at x]
final saddleBOT
@regal hamlet Has your question been resolved?
final saddleBOT
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vague anchor
,rccw 18th please
final saddleBOT
,rccw 18th please
soft zealotBOT
thorny canyon
,rccw 18th please
exactly in the interior as in we ignore the boundary
vague anchor
the half it
But
thorny canyon
But
we can find all points including boundary
and subtract boundary points
do you wanna try? or shd I give some hints?
thorny canyon
don't worry, you don't have to count
its simply sum of first 22 natural numbers
-boundary
ionic venture
There is a better and faster way.
vague anchor
Proof?
ionic venture
It is called Pick's theorem.
vague anchor
There is a better and faster way.
Expected
vague anchor
It is called Pick's theorem.
Well, I ll ask for written proof
ionic venture
Well, I ll ask for written proof
For the theorem?
thorny canyon
It is called Pick's theorem.
that still includes half the boundary points tho
doesn't really save calculation no?
vague anchor
For the theorem?
Yes
ionic venture
I mean it is the easiest way to solve this; you do need some adjustments.
ionic venture
But making a square seems more appropriate for such problem.
vague anchor
Yes
thorny canyon
each line has (22-y) integer points on it
vague anchor
each line has (22-y) integer points on it
Oh wow
thorny canyon
so summing up directly gives you the total points including boundary
vague anchor
so summing up directly gives you the total points including boundary
Actually man
vague anchor
x+y=42
thorny canyon
x=0 has 22 points, y = 0 has 22 points, x+y = 21 has 22 points
but the corners are repeated
so boundary = 22x3 - 3
vague anchor
x=0 has 22 points, y = 0 has 22 points, x+y = 21 has 22 points
Oh i see
thorny canyon
I think that shd work
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tranquil pine
solution for x in R
the final result i got was
cos(pi . e^x - pi/4) = root2/2
echo plover
Which one are we doing?
tranquil pine
well i worked on the cos + sin one
sorry i ended with
2^1/2 . cos(pi . e^x - pi/4) = 1
cos(pi . e^x - pi/4) = 2^1/2 / 2
ill try latex
$sqrt{2}cos(pie^x - frac{pi}{4}) = 1$
soft zealotBOT
Cat 404
atomic moon
Just \ before the sqrt, cos, pi and frac and you good
Also since its a product you can either do \cdot or write nothing
final saddleBOT
@tranquil pine Has your question been resolved?
tranquil pine
<@456226577798135808> Has your question been resolved?
i actually solved it thanks a lot
1st and 4th correct
2nd and 3rd are not
bcs when u try em it will give = -1 rather than the correct answer = 1
- k should be R*+
rain sentinel
is this correct?
final saddleBOT
is this correct?
graceful geyser
ye
versed crater
Would be helpful if you posted what the question was
versed crater
First derivative?
rain sentinel
yes
versed crater
Or second derivative of it
rain sentinel
First
versed crater
What is happening with the derivative of (x² + 1)²
rain sentinel
I found it using the chain rule
versed crater
(f/g)’ = (f’g - fg’)/g²
versed crater
What is happening with the derivative of (x² + 1)²
But chain rule for this just says (x²)’2(x² + 1)¹
Which is 4x(x² + 1)
versed crater
Your dv/dx is very wrong
versed crater
Isn’t that the same picture
rain sentinel
versed crater
Where is the x² + 1
rain sentinel
versed crater
You’ve added an extra x
Second term on the top
At this point I’d say the question is done
You can wiggle the form around but that won’t really change the answer
rain sentinel
ik but I've been trying to simplify
versed crater
There’s not really a point to doing so
rain sentinel
Becus I want to
versed crater
Have you learned about stationary points and how they relate to the derivative?
rain sentinel
yh
versed crater
You could turn this into something - something else
But if you wanted to solve for the critical points of 4x/(x² + 1)²
You want to find when its derivative is 0
Which means leaving the derivative as a quotient is more helpful
Since you can just say look at when the top is 0 (and the bottom not 0)
versed crater
You could probably factor and cancel one of the x² + 1 terms I suppose
rain sentinel
versed crater
But there’s not much to do beyond that
rain sentinel
im trying to figure out how to get it to this form
versed crater
Factor out a x² + 1