giống như tìm đạo hàm điểm cực trị
#help-36
tropic crest
còn derive AB_1 nghĩa là sao
tức là làm cách nào suy ra AB_1
à
...
AB_1 = AB - V_B * t
rồi anh giải thích kĩ hơn phần này được ko
leaden moon
tức là làm cách nào suy ra AB_1
Good
timber leaf
d^2 = (v^2_A +v^2_B)t^2 - 2Iv_Bt+I^2 kìa, nó chính là dạng parabol
với a= (v^2_A +v^2_B), b = 2Iv_B, t =I^2
công thức đỉnh của parabol là (-b/2a, -delta/4a)
vì cái này là parabol với chiều hướng lên trên nên giá trị nhỏ nhất tại đỉnh của nó
mà câu hỏi cũng hỏi rằng tìm giá trị d sao cho nó là min, nên mình tìm giá trị y
d^2 đây chính là f(t)
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stark mauve
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stark mauve
Just estimated y will not effect ot 😂😭
timber leaf
do you think sin would work?
timber leaf
now, you check that, when x = 0, what is the derivative?
stark mauve
1??
timber leaf
when x approach infinity, what do you see the slope of the lines?
they tend to lay down, right?
stark mauve
Slopes r getting horizontal
timber leaf
have you tried to draw these functions out?
timber leaf
you should try draw graphs out, extremely helpful
one of my buddy on this sv told me to draw graph too
stark mauve
Graph the slope and match with each option??
solar crest
spam plug in random points
stark mauve
How?
solar crest
by inspection, is it A?
stark mauve
Hnm
solar crest
yea you would need to know what the graph of some of those functions look like
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@stark mauve Has your question been resolved?
woeful haven
is the answer D OPTION ?
strange pelican
Match graph with y or dy/dx
derivative
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@stark mauve Has your question been resolved?
brazen breach
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brazen breach
ok so like this is just 2^3+2^2x3+2x3^2+3^3+...+8x9^2+9^3 which sums up to be about 5170 but somehow the answer is in the 8000s
from 1 to n^4 there's like n^3+n^2(n-1)+n(n-1)^2+(n-1)^3 oh
ohhhhhh
yeah this is probably an arithmetic error on my part sigh
nope it is still nowhere near close
6259 🥀
brazen breach
from 1 to n^4 there's like n^3+n^2(n-1)+n(n-1)^2+(n-1)^3 oh
is there anything wrong w the formula I'm making up
onyx peak
how did you make that formula?
brazen breach
how did you make that formula?
I'm imagining base 10 as a start
from 1 to 10000, 9000-9999 has 10^3, all the 900s beforehand have 9x10^2, 90s other than the 900 ones is 9x9x10, and lastly just 9s in singular r 9x9x9, so I'm assuming [formula I've said]
brazen breach
some numbers can have more than one 9
oh n-1 appearances
I only counted the no of 9- hey wait a minute that makes this question 10x easier
:(
onyx peak
is it 8096 btw?
brazen breach
is it 8096 btw?
yeah
yeah I'm gonna close
I would lowkey not survive timed competitions
😔
holy misread
.close
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lime crest
Find all pairs of primes $p,q$ such that [ p^3 -q^5 = (p+q)^2 ]
final saddleBOT
Find all pairs of primes $p,q$ such that \[ p^3 -q^5 = (p+q)^2 \]
soft zealotBOT
Copter
lime crest
i only have (3,7) as a solution but idk how to find the others
unborn forge
Reducing the equation mod p and mod q would probably help
rugged merlin
i only have (3,7) as a solution but idk how to find the others
I think you mean (7,3)?
Alao its probably the case that thst is the only solution
tribal stag
rugged merlin
,, p^3- q^5 \equiv p-q \q (\op{mod}3)
soft zealotBOT
rugged merlin
You could probably work with this
lime crest
oh im helpful
lime crest
woah
wait why si this true?
assume p,q > 2 then we have p-q = 1 mod 3
then p,q mod 3 = (0,2),(1,0),(2,1) mod 3
can i just case check this
rugged merlin
can i just case check this
yes
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old kettle
How do I do this?
final saddleBOT
How do I do this?
old kettle
Like can someone explain what is the logic behind solving such equations
scarlet sequoia
Like can someone explain what is the logic behind solving such equations
we're close to a simple quadratic equation
it's just about working around the absolute value
as luck would have it, |y| can be simplified into y or -y, depending on the sign of y
so break into two cases
second method (that applies here only), you can solve for t = |x-3| first
but then remember that t must be non-negative
tame gyro
i think t = |x-3| is way better here
i mean that would be my go to. but 1st method works always
tame gyro
Like can someone explain what is the logic behind solving such equations
whole logic is getting it to a quadratic equation
tame gyro
we're close to a simple quadratic equation
as he said
if you want i can help you on pv with other examples, ive got plenty of time now
halcyon ether
Let t=x-3, if t=a is a root of that quadratic then so is t=-a
By that you can argue thier sum has to be 0
scarlet sequoia
By that you can argue thier sum has to be 0
the sum of the t's, not the x's
but it can circumvent the problem a bit
halcyon ether
the sum of the t's, not the x's
Yes but then you can work out the sum of x's with that
scarlet sequoia
Yes but then you can work out the sum of x's with that
well if it weren't an MCQ we would be stuck
scarlet sequoia
yes, we can have 0 (no solutions), 6 (two) and 12 (four) as possible answers
and 3 or 9 if x = 3 itself were a solution
(which it isn't here)
halcyon ether
Let say x_1-3=a and x_2-3=-a then x_1+x_2-6=a-a=0
scarlet sequoia
It's nice to have tricks like this, but I think OP is more interested in a general way to solve those problems
tame gyro
are you guys just guessing values based on the posibilities rather than just solving the equation?
halcyon ether
It's nice to have tricks like this, but I think OP is more interested in a gener...
I mean it works w/o the choices
halcyon ether
are you guys just guessing values based on the posibilities rather than just sol...
No? I'm not guessing anything, but I'm definitely not solving quad either
tame gyro
No? I'm not guessing anything, but I'm definitely not solving quad either
ig bro, shit isnt really helpful either
halcyon ether
ig bro, shit isnt really helpful either
It does wdym
That's like the quick way, we skip the part we solve quadratic
tame gyro
explain it the easitest way you can
halcyon ether
explain it the easitest way you can
|| Let t=x-3 hence t^2 + |t| -11=0
If t=a is a root so is t=-a
So their sum would be
t_1+t_2=x_1 - 3+ x_2 - 3= a- a=0
So x_1 + x_2 =6 ( x_1 and x_2 are roots )
We will count how many pair of roots with sum 6
Now look at t^2 + t - 11=0
Using Vieta we know that t_1 • t_2= -11 hence one negative and one positive, but in t^2 + |t| -11 =0 only positive works
So there are only one solution for t hence 2 solutions for x with mean x_1 + x_2 = 6 as stated ||
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@old kettle Has your question been resolved?
keen hare
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keen hare
i feel like im missing something here
final saddleBOT
@keen hare Has your question been resolved?
vital crag
How did you get this partial fraction decomposition
,w partial fraction decomposition [(s^2 + 9)((s-4)^2 + 9)]^(-1)
soft zealotBOT
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keen hare
and im back
final saddleBOT
and im back
loud sundial
(In the future, please make the first message you send the actual question since that's what the bot pins. This saves us the effort of having to change the pin manually.)
loud sundial
missing parentheses to indicate multiplication
keen hare
(In the future, please make the first message you send the actual question since...
noted
this was the theorom used in the book so i think i followed it well
keen hare
missing parentheses to indicate multiplication
will do
loud sundial
this was the theorom used in the book so i think i followed it well
Yes that's fine
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dawn whale
does anyone know if cubics have absolute maximums?
final saddleBOT
does anyone know if cubics have absolute maximums?
drowsy epoch
depends on your cubic
dawn whale
what type of cubics have them
drowsy epoch
well if you have a closed interval then the cubic will always attain an absolute max
blissful meadow
does anyone know if cubics have absolute maximums?
If there’s no specified interval, then cubics on R (or any odd degree polynomial for that matter) always fail to have absolute extrema
drowsy epoch
-# I could imagine on open intervals it might still be possible to construct it in a way to make the local max to a global max
brazen breach
-# I could imagine on open intervals it might still be possible to construct it ...
idt it's possible
blissful meadow
-# I could imagine on open intervals it might still be possible to construct it ...
Yes you could imagine taking a small interval around one of the peaks or valleys and you would get a global extrema in that interval
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@dawn whale Has your question been resolved?
chrome ermine
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chrome ermine
21
timber leaf
What have you done
slow coral
i tried usub
for all of them?
chrome ermine
im doing only 21
river steppe
self learning analysis for the first time, which opens with some set theory. where have i gone wrong here?
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self learning analysis for the first time, which opens with some set theory. whe...
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solemn cloud
@scarlet sequoia @scarlet sequoia
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<@246787399991033858> <@246787399991033858>
plucky rover
drowsy epoch
twice a charme
silk meadow
Hello! I need some assistance with some math.
final saddleBOT
To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.
vital crag
<@246787399991033858> <@246787399991033858>
!noping
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<@246787399991033858> <@246787399991033858>
Please do not ping individual helpers unprompted.
@solemn cloud Has your question been resolved?
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cosmic summit
do you know how to prove that two distinct lines intersect at least once?
wet thicket
uh
you know the funniest part about all this
ive spent genuine hours on this question and i solve it like as soon as i put up a post 😔
.close
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chilly raven
mine
chilly raven
chilly raven
I found the roots
lofty notch
B
lofty notch
I want how
ask chatgpt
spring cypress
The correct answer is B.
chilly raven
stop giving the answers
lofty notch
ok
rancid leaf
i hate you hamza
lofty notch
i hate you hamza
ty
lofty notch
hhhh
chilly raven
help
brazen breach
multiply (x-2)(x-3) to both sides and consider cases(in which (x-2)(x-3) is positive or negative so you can consider swapping the intequality sign)
and expand
don't find just the roots
spring cypress
you can do that easier
rancid leaf
you can do that easier
chatgpt
final saddleBOT
B
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
lofty goblet
try making the RHS 0 first
jagged flare
what the hell is this channel
chilly raven
try making the RHS 0 first
ya done
is it not clear enough, you are not allowed to hand out the answer
chilly raven
I found the roots
lofty goblet
what did you get
spring cypress
subtract 1 from both sides and simplify to get zero on one side so you can consider the signs of the numerator and denominator at once
chilly raven
(-5+-√5)/2
lofty notch
<:catcutethink:827414434808725515> is it not clear enough, you are not allowed ...
ok
ty
chilly raven
(-5+-√5)/2
and 3,2
spring cypress
what are your roots @chilly raven where are you fro
jagged flare
first of all, do not give out direct answers. as stated in the rules, helpers should help the student work through the problem, and as helpers you should guide them. and second of all, please do not use ai!
on a less serious note ai is fucking stupid at aritmethic. use wolfram alpha
chilly raven
first of all, do not give out direct answers. as stated in the rules, helpers sh...
yea
rancid leaf
on a less serious note ai is fucking stupid at aritmethic. use wolfram alpha
chatgpt
chilly raven
bro y done these people leave
jagged flare
@rancid leaf please dont troll in the help channels :(
rancid leaf
<@764891391640731690> please dont troll in the help channels :(
oops
chilly raven
<@764891391640731690> please dont troll in the help channels :(
imma redo the question once again
jagged flare
if you really want to shitpost go to general or something
chilly raven
if you really want to shitpost go to general or something
I got my mistake
I took the -1 to the numerator directly
.close
final saddleBOT
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candid pulsar
um what
candid pulsar
question 35
i know the answer should be a vector but expanding the determinant doesn't seem to help
candid pulsar
yes
south dirge
-# are you indian by any chance
candid pulsar
yes
candid pulsar
oh ok then
bronze viper
those who know
I dont
south dirge
Balaji publication?
candid pulsar
yeah that
south dirge
Nahi ruko idts this is simplifying using row and column opertaions
bronze viper
black book
Can you specify? @candid pulsar
candid pulsar
that's it's common name
south dirge
Can you specify? <@702550180154834984>
Its a popular book for toigher question for jee students
candid pulsar
advanced problems in mathematics by balaji publications
bronze viper
Its a popular book for toigher question for jee students
ah got it
south dirge
It deals questions on all concepts asked in jee adv
final saddleBOT
candid pulsar
backtracked from the options
candid pulsar
thanks
old quarry
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lime crest
Let $n \geq 5$ be an integer. Prove that the following are equivalent: $\$ $(i)$ Both of $n,n+1$ aren't prime. $\$ $(ii)$ The nearest integer to $\frac{(n-1)!}{n(n+1)}$ is even. $\$ (note: if $k$ is an integer, the nearest integer to $k+1/2$ is $k+1$
final saddleBOT
Let $n \geq 5$ be an integer. Prove that the following are equivalent: $\\$ $(i)...
soft zealotBOT
Copter
lime crest
i dont know how to handle (ii) but im pretty sure if n, n+1 arent prime then (n-1)!/n(n+1) is an integer
<@&286206848099549185>
tired marlin
ew factorial
valid reaction
karmic glen
so for ii) ->i)
this is probably done by contradiction
- n prime n+1 composite
- n composite n+1 prime
halcyon ether
i dont know how to handle (ii) but im pretty sure if n, n+1 arent prime then (n-...
have you shown it's true?
halcyon ether
there sure are a lots of theorems
halcyon ether
i dont know how to handle (ii) but im pretty sure if n, n+1 arent prime then (n-...
brief idea, if n can be express as k*p^i with p is a prime then there is a number less than n which is k*p^{i-1} and (k-1)*p^{i-1} in (n-1)! the only case it's invalid is when n is a prime itself
also gcd(n,n+1)=1 iirc
sharp yew
Reminder:if done with channel please .close
final saddleBOT
@lime crest Has your question been resolved?
lime crest
have u learnt Wilson theorem
yea
lime crest
this is probably done by contradiction
yeah but how would i deal with "closest integer to"..
halcyon ether
yeah but how would i deal with "closest integer to"..
can you like, make it more clear the way the round number to the nearest integer
oh wait nvm it's the standard way
lime crest
**Copter**
this is all thats given
halcyon ether
maybe showing if (n-1)!/(n(n+1)) = p/q ( gcd(p,q) = 1) then p is always a multiple of 2, that would probably lead to smth
lime crest
i think its known that floor(...) is even so maybe if i do the ceil case to imply one of n,n+1 prime
karmic glen
**Copter**
maybe this is useful $\frac{(n-1)!}{n(n+1)} = \frac{(n-1)!}{n} - \frac{(n-1)!}{n+1}$
soft zealotBOT
k
karmic glen
if n is prime (n-1)! = -1 (mod n) by Wilson, so (n-1)! = kn - 1 for some k
in this case n+1 is composite
so this (n-1)!/(n+1) is an integer
lime crest
so this (n-1)!/(n+1) is an integer
yeah i see that
then we bound (n-1)!/n?
if n >2 then the closest integer to it is k - (n-1)!/(n+1)
karmic glen
since $(n-1)! = kn - 1$ for some $k$ we have $\frac{kn - 1}{n} - \frac{(n-1)!}{n+1} = k - \frac{1}{n} - B$
soft zealotBOT
k
karmic glen
if n >2 then the closest integer to it is k - (n-1)!/(n+1)
oh yes sorry I got a little lost
thats exactly what you said
hmm
maybe we can figure out the parity of k and B
lime crest
well B is probably even cuz the power of two in n+1 wouldnt cancel n-1!
lime crest
by wilsons n! = -1 mod n+1 and the inverse of n mod n+1 is -1 so (n-1)! = 1 mod n+1
karmic glen
<@&268886789983436800>
lime crest
oh right mb
karmic glen
yup
lime crest
..?
karmic glen
one second
karmic glen
n+1>2 we have the nearest integer must be (n-1)!/n +k
what about (n-1)!/n?
lime crest
probably even, but how would i prove that
soft zealotBOT
k
lime crest
:c
lime crest
do you know legendre's formula?
but wouldnt that depend on n
= floor(n-1/2) + floor(n-1/4) + ...
karmic glen
but wouldnt that depend on n
it does yes
lime crest
hmm
karmic glen
you could do cases like n odd n power of two n = m * 2^k
karmic glen
yeah
lime crest
what about the other
karmic glen
i dont know how to handle (ii) but im pretty sure if n, n+1 arent prime then (n-...
have you shown that (ii) is an integer?
lime crest
nah
lime crest
then the => direction is finished
wait, didnt we prove the contrapositive of <=
loud sundial
**Copter**
lime crest
yeye i got it now
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woven ledge
$g:\mathbb{N}\to \mathbb{N}$, $(g(n)+m)(g(m)+n)$ is a perfect square $\forall m,n \in \mathbb{N}$. Find all g
final saddleBOT
$g:\mathbb{N}\to \mathbb{N}$, $(g(n)+m)(g(m)+n)$ is a perfect square $\forall m...
soft zealotBOT
CherryMan
tender pollen
positive or nonnegative
woven ledge
not including 0
woven ledge
f(x) = g(x)-x
(f(n)+m+n)(f(m)+m+n) = k^2
= (m+n)^2 +f(n)f(m) + (m+n)(f(n)+f(m))
(m+n)^2 <= (f(n)+m+n)(f(m)+m+n) <= ((f(m)+f(n))/2 + m+n)^2
if we keep m+n constant
tender pollen
f(x) = g(x)-x
i would guess ||induction on c=f(1), prove f(n)=f(n+1)?||
(f(n)+2n+1)(f(n+1)+2n+1)
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willow dirge
Hey, when doing a top down view, if I have a 50x50ft square and my scale is 1 foot = 12 pixels I have a 600px by 600px square. When I tilt the camera by 45 degrees to get the angle for the scene, that square should still be 600px wide, but how many pixels in the other direction (depth)? I believe it should be 600px * sin(45degrees) = 424px correct?
final saddleBOT
Hey, when doing a top down view, if I have a 50x50ft square and my scale is 1 fo...
fathom meteor
What is depth referring to here? Is it the horizontal axis after the square is rotated?
willow dirge
Looking directly downwards on a 50x50ft square.
The square is not rotated, the camera perspective is.
If width is X, depth is Y.
acoustic vale
are we assuming the bottom of the square is aligned with the camera?
fathom meteor
Oh wait you don't mean rotation in the plane parallel to the square?
willow dirge
acoustic vale
this may go faster if you could draw a representation
willow dirge
Looking straight down on a square. The square is 50ftx50ft.
acoustic vale
ok
willow dirge
If I rotate my camera 45 degrees and place it in the center of the frame once more, how deep does it appear to be at that angle?
Added other object for reference.
fathom meteor
Oh it is pitch got it
fathom meteor
Are we assuming like isometric perspective? Because if not I think it would depend on the distance of the camera from the square
fathom meteor
Yes.
Oh, can't help you then sorry but someone else should be able to help you faster now that you've elaborated
willow dirge
Ah, okay.
No worries, thank you for the help in clarification!
Okay to restate then:
When doing a top down view, if I have a 50x50ft square and my scale is 1 foot = 12 pixels I have a 600px by 600px square. When I pitch the camera upwards by 45 degrees to get the angle for the scene, that square should still be 600px wide, but how many pixels does it appear to be in the other direction (depth)? I believe it should be 600px * sin(45degrees) = 424px correct?
final saddleBOT
@willow dirge Has your question been resolved?
acoustic vale
theres not enough information to give an answer, you'll need to figure out the axis of rotation
the hypothetical is underconstrained
poor wifi ough
that was sent half an hour ago lemme catch up rq
@willow dirge ill write out an equation just give be a sec
willow dirge
Ah no worries.
The axis of rotation?
I will say, I kind of confirmed my math on that.
acoustic vale
like where is it rotating
yeah that was supposed to send half an hour ago so i hadnt read the other parts
willow dirge
All good. The object doesn't move. It's just the camera.
But here.
50x50 square straight down.
acoustic vale
holy stars my brain is lagging right now
acoustic vale
that should be correct
willow dirge
0.355 * 12 = 4.25.
So yeah, that's my proof. 😄
To get the depth of any object at an angle of 45 degrees, multiple the actual depth by sin(45).
That's not math.
acoustic vale
it took me a second to process the question cause i was thinking in art form 😭😭
dang he deleted it
yeah that should be correct tho
willow dirge
😄 No worries at all. Sometimes just helps to talk it out with someone.
Thanks!
Have a great rest of your day!
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gentle zephyr
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gentle zephyr
\textbf{Exercise 5} Encode the following numbers in base 2, using the precision and representation form indicated in each case. Compare the results.
\begin{itemize}
\item $0_{10} \longrightarrow$ using 8 bits, sign-magnitude notation and two's complement notation.
\item $-1_{10} \longrightarrow$ using 8 and 16 bits, in both cases two's complement and sign-magnitude notation.
\item $255_{10} \longrightarrow$ using 8 bits unsigned notation and 16 bits two's complement notation.
\item $-128_{10} \longrightarrow$ using 8 and 16 bits, in both cases two's complement notation.
\item $128_{10} \longrightarrow$ using 8 bits unsigned notation and 16 bits two's complement notation.
\end{itemize}
soft zealotBOT
Renato
gentle zephyr
shouldnt be too hard
final saddleBOT
@gentle zephyr Has your question been resolved?
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@gentle zephyr Has your question been resolved?
old kettle
hi guys
old kettle
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signal vector
<@&268886789983436800> advertisement (scam, most likely)
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scenic notch
say i were to take the complement of a set {a in E | a has condition b}, would that be the set of {a is still in E | a has condition not b}?
final saddleBOT
say i were to take the complement of a set {a in E | a has condition b}, would t...
barren hound
if you want to talk about the complement of a set, you need a universe to consider
scenic notch
I guess my question is does, a in E act as a universe?
barren hound
not inherently
e.g. "the complement of the even integers" could be the odd integers, the odd integers + rational numbers, the odd integers + real numbers, or could even include complex or fancier numbers
depending on what you've identified as the universal set
scenic notch
I see
Mainly, I want to find the complement of the set ${a | (a^m \in I, \forall m \in \mathbb{N}) \vee
(\exists n\in \mathbb{N}-{1} \ s.t \ \forall s\in \mathbb{N} \wedge 1 \leq i \leq n-1, a^{n(s-1)+i}
\in I \wedge a^{ns} \in R)}$ for the universe of real numbers
barren hound
guh
scenic notch
am I just tripping
barren hound
there's so much
soft zealotBOT
greaterthan.333
scenic notch
where I is the irrationals and R is the rationals
I think I'm cooking myself because part of the statement relates to a series and I tried to "simplify" it
${a | (a^m \in I, \forall m \in \mathbb{N}) \vee
(\exists n\in \mathbb{N}-{1} \ s.t \ \forall s\in \mathbb{N}, {a^{n(s-1)+1}, a^{n(s-1)+2}$ $,\ldots, a^{n(s-1)+n-1}}
\subseteq I \wedge a^{ns} \in R)}$
makes more sense, maybe
soft zealotBOT
greaterthan.333
final saddleBOT
@scenic notch Has your question been resolved?
scenic notch
.close
unkempt gazelle
Can anyone help me with some questions
final saddleBOT
Can anyone help me with some questions
unkempt gazelle
Trying to solve it for 25 mins
verbal juniper
send em
proud igloo
!da2a
final saddleBOT
Asking the actual question right away is more likely to get responses.
Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.
final saddleBOT
@unkempt gazelle Has your question been resolved?
cold cloud
Can anyone help me with some questions
What's the question
brittle drift
why can't i do 5xy-8x=-28-3y
final saddleBOT
why can't i do 5xy-8x=-28-3y
south dirge
why can't i do 5xy-8x=-28-3y
Are you proposing you statement as a step or as the final answer?
south dirge
like instead of what they do in line 3, what if i do that
Okay sure go ahead!
You would still get the same answer
south dirge
Nope you would get the same answer!
final saddleBOT
final saddleBOT
Available help channel!
slate python
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
soft zealotBOT
slate python
ok let me try that
median abyss
Be sure to take the mod of the square root on the rhs
Beta - Alpha is a positive number, since beta > Alpha
If you dont, youll only get one value of lambda
final saddleBOT
@slate python Has your question been resolved?
final saddleBOT
@slate python Has your question been resolved?
bold turtle
night raft
twas a brilliant question
heady portal
Indeed
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Available help channel!
pine sand
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@pine sand Has your question been resolved?
strange pelican
ihave seen this before
final saddleBOT
pine sand
.reopen
final saddleBOT
✅ Original question: #help-36 message
pine sand
ihave seen this before
Where?
neat wolf
how do I start this problem? my mind tells me about menelaus theorem?
i think not
this can be solved with middle school math
final saddleBOT
@pine sand Has your question been resolved?
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Available help channel!
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polar agate
<@&268886789983436800>
stable dew
final saddleBOT
stable dew
,,rcw
woeful haven
which q
soft zealotBOT
Trollstar
woeful haven
idk either
woeful haven
wait ima check
stable dew
it's denominator
barren pebble
,rccw
soft zealotBOT
barren pebble
Send it again bro
barren pebble
,rccw
soft zealotBOT
stable dew
patter still repeate after 3 terms
pattern
f100 is f1 not 2(f1)
it's my question...
barren pebble
<@&268886789983436800>
stable dew
thanks
barren pebble
it's my question...
Yep, nvm
woeful haven
yo
stable dew
every three term repeats
yeah
woeful haven
-_-
stable dew
woeful haven
is the answer 5/3?
woeful haven
the 2/3 and 2/3 are positive and -1/2
stable dew
the 2/3 and 2/3 are positive and -1/2
huh
stable dew
f1 = (x-1)/x
I'm getting f1=x-1
oh nvm
i see da x now
man these type of questions got too much calculation
strange pelican
<@&268886789983436800> how you doin
opaque ember
doin good eatin lunch
strange pelican
do you need help
final saddleBOT
@stable dew Has your question been resolved?
gentle zephyr
final saddleBOT
soft zealotBOT
Renato
valid carbon
is that a ratio
leaden moon
i think that's gcd
valid carbon
ah?
gentle zephyr
greatest common divisor
lofty goblet
the gcd of the two divides any linear combination
gentle zephyr
the gcd of the two divides any linear combination
this^
d = gcd(x, y) => d | x and d | y
=> d | ax + by
lofty goblet
try considering the combination ||2(3n+2^(n+1)) - 3(2n-2^n)||
gentle zephyr
yeah then the 2n and 3n cancels
we get 2^(n+2) - 3.2^n
but personally speaking there is a better linear combination I think, no?
say d | a(3n + 2^{n+1}) + b(2n - 2^n)
a = 1 and b = 2 is better
after that you get that: d | 7n
lofty goblet
ah thats good
gentle zephyr
now what
lofty goblet
try to show that any divisor d of 3n+2^(n+1) and 2n-2^n must be coprime to n
gentle zephyr
what
lofty goblet
you have d | 7n
if d has a common factor with n, then d could be something larger like 7n or any divisor of it
so we show gcd(d,n)=1 to restrict d to 1 or 7
gentle zephyr
if d has a common factor with n, then d could be something larger like 7n or any...
what?
what do you mean by this?
can you rephrase it differently
lofty goblet
7n is multiple of d, an example is if d is 21 and n is 15
but d has a factor of 3 that comes from n=15
since d can share prime factors with n, d can be larger than 7
we need to show that d CANNOT share any factors with n
gentle zephyr
I am not understanding
7n is a multiple of d, sure
but the rest that you said is a little hard to understand
proud igloo
final saddleBOT
@gentle zephyr Has your question been resolved?
dim matrix
but the rest that you said is a little hard to understand
we know that d divides 7n. this means d is a factor of the product 7 * n.
so, we have two possibilities: it only divides the '7' (d = 1 or 7), or it divides the n (examples: n = 30, d could be any of the divisors of 30: 2,3,5,6,10,15...)
Kiwie is trying to prove that d cannot be one of these dividers. if we show that gcd(d,n)=1, we prove that they are coprimes (so d is none of the divisors). so we are just left with d being divisor of 7.
to prove gcd(d,n)=1, we can assume that there exists a prime p which p|d, p|n, and we need to get to a contradiction in the original expressions 3n+2^(n+1) and 2n-2^n
dim matrix
try to show that any divisor d of 3n+2^(n+1) and 2n-2^n must be coprime to n
this part
gentle zephyr
to prove gcd(d,n)=1, we can assume that there exists a prime p which p|d, p|n, a...
care to expand on that, I understand now what you guys mean
dim matrix
do you see that if p divides d, then p must divide these 2 expressions?
3n+2^(n+1) and 2n-2^n
gentle zephyr
p | d , d | x , d | y => p | x , p | y?
gentle zephyr
3n+2^(n+1) and 2n-2^n
how does that work
dim matrix
p | d , d | x , d | y => p | x , p | y?
it works exactly like that
but instead of x,y, we are using n, as in our proof
gentle zephyr
you mean divisibility is transitive
dim matrix
"we can assume that there exists a prime p which p|d, p|n"
gentle zephyr
but instead of x,y, we are using n, as in our proof
x = 3n+2^(n+1) and y = 2n-2^n
dim matrix
3n+2^(n+1) and 2n-2^n
coming back to this:
but if p also divides n, we can remove the terms 3n,2n
so it leaves p hanging to divide 2^n+1 and -2^n
do you know how to end in the contradiction from here?
gentle zephyr
coming back to this:
but if p also divides n, we can remove the terms 3n,2n
how does that work
assume p | 3n + 2^(n+1)
assume p | 2n - 2^n
then what?
gentle zephyr
do you know how to end in the contradiction from here?
no?
dim matrix
then what?
this p | 3n + 2^(n+1), p | 2n-2^n is the same as saying p | d, since d is the linear combination of these two.
the method for showing that gcd(d,n)=1 (of course there are many, this is just the most common) is assuming the inverse, that there exists a prime where p|d (that is, p | 3n + 2^(n+1), p | 2n-2^n) AND p|n, and show that it is impossible (proof by contradiction)
because if it was true that p|d and p|n, gcd(d,n) wouldnt be 1, it would be p
remember, we are doing as the second step of the answer. the first step you already did: you found that d | 7n
dim matrix
we know that d divides 7n. this means d is a factor of the product 7 * n.
so, w...
the proof is to do this restriction
let's start the proof (i'll do until the part i discussed instead of step by step because i gotta go)
Assume that gcd(d,n) = 1. Suppose there exists a prime p of which p divides d and p divides n.
If p divides d, and d = a(3n+2^(n+1)) + b(2n-2^n). by the transitive property of divisibility, p divides 3n+2^(n+1) and p divides 2n-2^n. (#)
but if p also divides n, p consequently divides any kn, k being a integer. so, p divides 3n and p divides 2n.
as a result of the transitive property (again), p must divide 2^(n+1) and p must divide 2^n to satisfy (#).
now, to finish: what is the only prime number that satisfies this condition?
cold cloud
Ywah
dim matrix
this weird notation is gcd(3n+2^n+1,2n-2^n)
dim matrix
yes
final saddleBOT
@gentle zephyr Has your question been resolved?
obsidian mountain
quicker way to determine $\sum_{n=0}^{9}\sum_{m=0}^{9}\binom{n}{m}\bmod 2$
soft zealotBOT
zeta theta beta eta
obsidian mountain
what I used was nCr(n, m) = 0 if m > n and nCr(k, k) = 1, then I brute forced the remaining 45 terms
but im looking for a method which doesn't involve brute forcing
deep condor
<@&268886789983436800>
obsidian mountain
[
\begin{array}{c|cccccccccc}
n\backslash m & 0&1&2&3&4&5&6&7&8&9 \ \hline
0&1&0&0&0&0&0&0&0&0&0\
1&1&1&0&0&0&0&0&0&0&0\
2&1&0&1&0&0&0&0&0&0&0\
3&1&1&1&1&0&0&0&0&0&0\
4&1&0&0&0&1&0&0&0&0&0\
5&1&1&0&0&1&1&0&0&0&0\
6&1&0&1&0&1&0&1&0&0&0\
7&1&1&1&1&1&1&1&1&0&0\
8&1&0&0&0&0&0&0&0&1&0\
9&1&1&0&0&0&0&0&0&1&1
\end{array}
]
soft zealotBOT
zeta theta beta eta
obsidian mountain
maybe there is some pattern?
thin sequoia
looks like sierpinski traingle to me, 1 + 1 = 0, 1 + 0 = 0 + 1 = 1, 0 + 0 = 0 kinda parity thing
a generating function of 2 variables might work here
final saddleBOT
@obsidian mountain Has your question been resolved?
obsidian mountain
looks like sierpinski traingle to me, 1 + 1 = 0, 1 + 0 = 0 + 1 = 1, 0 + 0 = 0 ...
youre right
its pascals triangle mod 2
didn't even notice that beforehand
is there a way to calculate the sum quickly
with the normal pascals triangle the sum of the nth row from n = 0 is 2^n
thin sequoia
obsidian mountain
but when you sum these these are all powers of 2
maybe there is some connection
karmic glen
**zeta theta beta eta**
we can change inner sum to be up to n
thin sequoia
^ ye cuz basically half the table is just zeroes
karmic glen
mod 2
obsidian mountain
the mod 2 is inside the summation
karmic glen
are you taking something else mod 2
obsidian mountain
sorry for not including the brackets
karmic glen
this might be useful
https://en.wikipedia.org/wiki/Lucas's_theorem
final saddleBOT
@obsidian mountain Has your question been resolved?
obsidian mountain
maybe it is, but i don't see why it's useful for the moment
obsidian mountain
this might be useful
https://en.wikipedia.org/wiki/Lucas%27s_theorem
why do you think this might be useful
karmic glen
for mod 2 this theorem says that n C m is the product of m_i C n_i (mod 2), where m_i and n_i are binary representations of m and n respectively. From here this product is 1 only if for every bit n_i >= m_i
karmic glen
**zeta theta beta eta**
from this you can perhaps show that the number of 1's in each row is 2^(number of 1 bits in the representation of the number on the left)
karmic glen
for mod 2 this theorem says that n C m is the product of m_i C n_i (mod 2), wher...
take some n we are asking how many m there are such that n C m = 1. For every bit that is 1 in n there are two choices for the bit m_i (from this theorem we must have n_i >= m_i), since n_i >= m_i for all i then surely n >= m so any m is fine and there are two choices iff n_i = 1
soo this is the same as summing 2^(number of bits of n) of n = 1,..,9
@obsidian mountain
karmic glen
this reduces the time complexity from n^2 to nlogn, cool
final saddleBOT
@obsidian mountain Has your question been resolved?
candid pulsar
how would i try finding the maxima of a function like\\
$f(x) = \frac{\sqrt{2}x + \sqrt{x^2-1}}{x^2+1}$
final saddleBOT
how would i try finding the maxima of a function like\\\\
$f(x) = \frac{\sqrt{2}...
soft zealotBOT
candid pulsar
equating f'(x) = 0 seems hard
$(x^2+1)\left(\sqrt{2} + \frac{x}{\sqrt{x^2-1}}\right) = 2x(\sqrt{2}x + \sqrt{x^2-1})$
soft zealotBOT
lofty goblet
i would try a substitution maybe
drowsy epoch
$(x^2+1)\left(\sqrt{2} + \frac{x}{\sqrt{x^2-1}}\right) = 2x(\sqrt{2}x + \sqrt{x^...
Is that f'?
candid pulsar
the equation you get on trying to solve for f'(x) = 0
drowsy epoch
,w D[(sqrt(2)x+sqrt(x^2-1))/(x^2+1),x]
soft zealotBOT
lofty goblet
i would try a substitution maybe
||x=secθ|| and then it makes f'(x)=0 easier
candid pulsar
||x=secθ|| and then it makes f'(x)=0 easier
ok this simplifies to $\sqrt{2}t^3 + 3t^2 - 2 = 0$, where t = $\sin(\theta)$
soft zealotBOT
lofty goblet
final saddleBOT
little ermine
Can I get help
fathom meteor
Can I get help
You can wait for this channel to get moved to Math Help (Available) or send your question in a channel that's already in that category
final saddleBOT
Available help channel!
shell tide
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coral zenith
!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
3
coral zenith
okay show your working or tell how you did it?
coral zenith
you are supposed to find out the weight on the surface so r here should be the radius of the planet and not the orbital distance
coral zenith
ill plug it in and run it by you give me 2 mins
sure
coral zenith
I think tht should be correct
final saddleBOT
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hexed sedge
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hexed sedge
what do i need to show
ocean yoke
yea he right
rugged merlin
as far as i understand
Hi big F
hexed sedge
so i multiply and it needs to result into identity matrix?
sturdy cypress
yes
final saddleBOT
final saddleBOT
Available help channel!
warm python
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rocky tusk
which direction?
warm python
I've established earlier a function is integrable iff $\forall \varepsilon>0, \exists P_{\varepsilon}$ such that $U(f,P_{\varepsilon})-L(f,P_{\varepsilon})< \varepsilon$. Letting $\varepsilon =\frac {1}{n}$ we get a function is integrable iff $\forall \frac{1}{n}>0 \exists P_{\frac{1}{n}}$ such that $U(f,P_{\frac{1}{n}})-L(f,P_{\frac{1}{n}})< \frac{1}{n}$.
As there exists a bijection b/w $\N$ and $\left{ \frac{1}{n} \bar n \in \N \right}$. We can say $U(f,P_{n})-L(f,P_{n})< \frac{1}{n}$.
We're done
warm python
which direction?
I think both can be done at once?
rocky tusk
i guess yea if you just manipulate that condition
soft zealotBOT
Wai
junior token
this is Riemann integrable, right?
muted prairie
it's probably not for general n
warm python
well, in that case I just need to show one of the limits exist
after which I can use algebra of limits
does that sound about right
warm python
oh yeah, this is the backwards direction I think
How does this not do both at once
junior token
How does this not do both at once
yeah it does
weird how they gave you this exercise if you could just prove that result beforehand
warm python
though the 2nd part is a bit more interestng
warm python
the second step isnt clean. not all eps can be written as 1/n
for every eps 1,2,\dots N, we can. And then by the arch property for every eps there's a 1/n less then it
desert mantle
better
warm python
Could I have a hint for the 2nd part
oh ,f is integrable so $U(f)$ exists. Then we can quite easily start off with an abritrary partition and just consider refinements of the same
cool
soft zealotBOT
Wai
tropic mirage
The cafe has a late night where they open at 9am and close at 8pm twelve minutes after opening, number of cats decrease to a minimum of 5 cats, before increasing to maximum of 21 cats when breakfast served at 9:30am
The number of customers at the cafe reaches its first peak of 24 customer at 9.12am. THe number of customer peaks every 42 minutes , with a minimum of 4 customer
find the time intervals, in the first three hours from opening, where there are 12 or more cats 12 or more customers
The cafe is considered stressful for the cats when there are 12 or more cats and 12 or more customer in the main lounge. Generalise the times for when the cat cafe is stressful for the cats
final saddleBOT
The cafe has a late night where they open at 9am and close at 8pm twelve minutes...
tropic mirage
how can i find the general solution n integer
timber leaf
The cafe has a late night where they open at 9am and close at 8pm twelve minutes...
what have you done
@tropic mirage if you finish with your other channel please close it
lofty goblet
maybe treat the number of cats and customers as periodic functions
with period of 42min
final saddleBOT
@tropic mirage Has your question been resolved?
tropic mirage
maybe treat the number of cats and customers as periodic functions
they dif
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warm python
I don't get how to do (c)
do I determine if (x^2) is maximal
wel it is so Z[x]/(x^2) is a field
the question though of what characteristic remains
grim nebula
x^2 = 0 in that ring
warm python
x^2 = 0 in that ring
yes
grim nebula
how can it be a field
warm python
oh
grim nebula
it's not even a domain
kindred mortar
Z[x^2] isn't maximal
can you think of an ideal containing (x^2) ?
warm python
can you think of an ideal containing (x^2) ?
other than (x) no
then how do I describe the ring structure?
do I just say not a field and move on?
grim nebula
you can list out the elements of Z[x]/(x^2) and state explicitly the multiplication
kindred mortar
other than (x) no
well (x) is a proper ideal of Z[x]
warm python
${(kx^{2n+1})/(x^2) \mid k \in \Z} \cup { (x^2) }$
soft zealotBOT
Wai
lost crown
Yoi
left trail
Can someone explain how the graph is below the axis for 0 to 1
final saddleBOT
Can someone explain how the graph is below the axis for 0 to 1
silver needle
log is negative for values 0 to 1.
left trail
But the graph is above the x axis the whole time
silver needle
that is the graph of 1/x
the left graph shows 1/x, of which the area (or integral) is ln(x).
You do not count values 0 to 1 in the integral, because 1/x diverges as you approach 0
left trail
Ok so how does L(1)=0 show that it is below the axis though
silver needle
$\int \frac{1}{x},dx = \ln|x| + C$
soft zealotBOT
prfn
drowsy epoch
1/x > 0 => log is incr, so there exists a C where it intersects x-axis and for x<C log is negative
strange pelican
<@&268886789983436800>
silver needle
1/x > 0 => log is incr, so there exists a C where it intersects x-axis and for x...
^
since the integral diverges, it cannot be evaluated as it approaches 0
left trail
1/x > 0 => log is incr, so there exists a C where it intersects x-axis and for x...
Ok so I see so our log function is an increasing function it intersects the x axis at 1 so under 1 it will be negative
drowsy epoch
Yes
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silver needle
in TNB frames, is there a physical meaning of the unit binormal vector and torsion?
final saddleBOT
in TNB frames, is there a physical meaning of the unit binormal vector and torsi...
silver needle
past that the binormal vector is the perpendicular vector to the normal and tangent vectors
blissful meadow
in TNB frames, is there a physical meaning of the unit binormal vector and torsi...
B tells you the direction in which the curve is escaping its osculating plane (the plane spanned by T and N), and the torsion tells you how fast it's escaping that plane.
So a curve which sits vaguely within a plane around a point has low torsion (with 0 torsion if it's completely in the plane, i.e. planar), and a curve which deviates drastically from its osculating plane at a point has high torsion.
final saddleBOT
@silver needle Has your question been resolved?
silver needle
something like that
crystal lion
im kinda confused on how the mark scheme got the solutions of 1/2 ln2 and -1/2 ln2
final saddleBOT
im kinda confused on how the mark scheme got the solutions of 1/2 ln2 and -1/2 l...
blissful meadow
$\ln(\sqrt{2})) = \frac{1}{2} \ln(2)$
soft zealotBOT
Azyrashacorki
crystal lion
oh yeah i forgot u could do that lol