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frosty plume
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frosty plume
Task 8: give a regular tetrahedron ABCD w/ edges legth=2, let M and N be the midpoints of AC and CD respectively, these statements are true or false? Explain?
vivid yoke
Task 8: give a regular tetrahedron ABCD w/ edges legth=2, let M and N be the mid...
Which part you're stuck with
frosty plume
<:pandapopcorn:616252654175256579> Which part you're stuck with
I think d), I did abc
How to prove if its t or f?
How vector mb x vector mn =1/2?
vivid yoke
can you find the angle between MN and MB?
vivid yoke
use dot product
vivid yoke
nvm
vivid yoke
Mb.mn=|mb||mn|cosbmn right?
yeah
vivid yoke
sài cosine law trong tam giác BMN ấy, BN , MN, BM tính đc cả rr
frosty plume
Ô wtf ng việt a
vivid yoke
Ô wtf ng việt a
đâu cần bất ngờ v
vivid yoke
sài cosine law trong tam giác BMN ấy, BN , MN, BM tính đc cả rr
tính đc cos góc BMN rồi sài tích vô hướng thôi
frosty plume
Ng việt thì ns luôn đi cứ ns ta bất đồng ngôn ngữ
vivid yoke
Ng việt thì ns luôn đi cứ ns ta bất đồng ngôn ngữ
Giả sử mk bận ko giúp đc, thì người khác vào thế vẫn hiểu nãy h làm được đến đâu rr
Nếu nói tiếng việt thì chỉ có ng việt bt
frosty plume
ng ở đây giải đc cả toán VMO cơ
Bt là siêu r xem xem giúp có nhiệt tình k th
Dù j cx pk lm hết có chỗ nào khó quá ms lên đây nhờ
vivid yoke
Bt là siêu r xem xem giúp có nhiệt tình k th
cs đó nhma lựa giờ hỏi
Tầm tối tối vs sáng hẳn là hay có ng giúp
Tầm 7-9h sáng/ tối
Đêm cx hay có ng giúp
vivid yoke
Bt là siêu r xem xem giúp có nhiệt tình k th
Có ng nhiệt tình cực còn lên gg dịch, dịch tiếng anh sang t.v để nhắn cho đứa ko giỏi t.a cơ kk
frosty plume
Toàn bên mỹ đk
vivid yoke
Toàn bên mỹ đk
Bên mỹ nhiều, nhma chủ yếu ng Ấn
frosty plume
Thế cos bmn=60 nhỉ
vivid yoke
Thế cos bmn=60 nhỉ
BMN ko phải 60 độ đâu
frosty plume
Bên mỹ nhiều, nhma chủ yếu ng Ấn
Ấn thì super giỏi toán r
frosty plume
BMN ko phải 60 độ đâu
45..
frosty plume
BMN ko phải 60 độ đâu
Thế bro tính đc độ dài mb mn nn?
vivid yoke
Thế bro tính đc độ dài mb mn nn?
um
vivid yoke
Lm s để bt góc bmn=?°
định lý cos
frosty plume
Kpk lập phg đâu mà dễ
vivid yoke
đầy j cách lm
midnight plankBOT
@frosty plume Has your question been resolved?
frosty plume
đầy j cách lm
Ok câu đấy đúng
midnight plankBOT
Ok câu đấy đúng
midnight plankBOT
frosty plume
Đang mở 1 kênh đs
T còn 3 bài nx bro giúp t k
midnight plankBOT
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vivid yoke
T còn 3 bài nx bro giúp t k
gửi bài lên đây đi
frosty plume
vivid yoke
okay cần giúp câu nào
golden crescent
câu nào ông ơi
midnight plankBOT
vivid yoke
vivid yoke
ko DMs giúp toán đâu, có j thì lên đây thôi
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polar nexus
Need help
mystic condor
how would you find DAE if you were given numbers
rare maple
Such kinds of problems are just make equations and hope for the best
surreal moon
what does C-E-D mean?
rare maple
It means collinear i think
midnight plankBOT
Need help
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
tbh, just write all angles in terms of <C and <B, and you'll get it
surreal moon
(remember: bisector)
polar nexus
1/2(<C - <B)
surreal moon
what is <CAB in terms of <C and <B
midnight plankBOT
@polar nexus Has your question been resolved?
polar nexus
what is <CAB in terms of <C and <B
<CAB?
surreal moon
angle CAB
steep cape
YOO I also needed help with this
midnight plankBOT
@polar nexus Has your question been resolved?
rare maple
@polar nexus what they mean by this is that we need to eliminate angle A from all the places we can
For example in a triangle A+B+C = 180
So A=180-B-C
You can do this for every single angle and get your answer
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viral dagger
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viral dagger
whats the reasoning for the fact that you can just multiply (n-10)(n-11)/2 by 2, cant the same reasoning be used to multiply 90 by 2 aswell?
gusty portal
No I don't think you can multiply 90 by 2
The games would be winners - winners (n-10)(n-11)/2
losers-losers -90
And losers winners ,which is also (n-10)(n-11)/2 , I don't understand why it's equal to winners winners but 90 will definitely not be multiplied by 2
Oh also it's already multiplied by 2
It's multiplied by 2 because they earned half their points against the 10 lowest players
viral dagger
It's multiplied by 2 because they earned half their points against the 10 lowest...
cant you reverse that logic and say that the losers earnt the other half of that points from losers-winners
the reason losers winners is (n-10)(n-11)/2 is cause that the total points of what the winners got from losers-winners is the same as the total points of what the winners got from winners-winners right? but cant we use the same concept to get the fact that the total points of the losers from losers-winners is the same as the total points of the losers from losers-losers
oh wait uhh
gusty portal
cant you reverse that logic and say that the losers earnt the other half of that...
Winners aren't the 10 lowest
gusty portal
the reason losers winners is (n-10)(n-11)/2 is cause that the total points of wh...
Yeah I got why it's equal now
viral dagger
OHHH wait i just realized it lmaoo they accounted for that
gusty portal
OHHH wait i just realized it lmaoo they accounted for that
Yeah 45x2
viral dagger
yeahhh sorry
gusty portal
All good
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fervent burrow
midnight plankBOT
fervent burrow
Why is my answer wrong
fallen galleon
6sqrt(25) is not 150
hard meteor
fallen galleon
i think you just completely ignored the square root was there
fallen galleon
Why is my answer wrong
.
midnight plankBOT
frosty plume
ko DMs giúp toán đâu, có j thì lên đây thôi
frosty plume
<:sadthink:899608203736780821>
Thế t lm bn vs bro đc k
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manic bison
how many ordered pairs (a,b,c) exist, where all are distinct and a + c = b , all of them should be from the set 1 to 999
midnight plankBOT
how many ordered pairs (a,b,c) exist, where all are distinct and a + c = b , all...
rain wasp
how many ordered pairs (a,b,c) exist, where all are distinct and a + c = b , all...
my first thought is to fix a and choose c such that a + c <= 999
so for a = 1 you have (1, 3, 2), (1, 4, 3), ..., (1, 999, 998), which is 997 options
increment a and count the number of options, see if you notice a pattern
manic bison
increment a and count the number of options, see if you notice a pattern
so c will be from 1 to 999-a?
rain wasp
just to be sure, by "all are distinct" are you referring to each ordered set is distinct, or a b c are distinct?
i was assuming the latter
tawdry nymph
OOH I think of something like this
You just need to findd first that a and c are distinct cause b would automatically be distinct since like 1 + 2 = not 1 or 2 obviously and I did this and found a pattern
1 + 2
1 + 3
1 + 4
...
1 + 998
2 + 3
2 + 4
2 + 5
...
2 + 997
3 + 4
3 + 5
3 + 6
...
3 + 996
Obviously we need to remove both 2 + 1 and 2 + 2 for set of 2 and 3 + 1 , 3 +2, and 3 + 3
Cause for 2 + 1, 3 + 1 and 3 + 2, they have all mentioned in the previous list while obviously you cannot have two numbers of with the same number as they are not distinct anymore
And in order to get the total number of pairs, just look at the pattern and use arithmetic summation to find the total for each set
Maybe, for this one, you can just use the arithmetic summation and replace n with x so that we can have a generalized formula not only for 999, but for numbers higher than 999
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thin relic
fresh sparrow
what is [ ]? floor?
frozen lake
ok so ig the [ ] are just used as normal brackets
fresh sparrow
there are only two pairs of a and b that satisfy this
frozen lake
then the key idea to solve this eq is to let all 3 be equal to k
then solve for k using whatever equation pops up
write a and b completely in k
fresh sparrow
that's one way, another way is to notice that
1-a = -b => b = a-1
so ab = a(a-1) = 20-a
solve for a
frozen lake
ah thats smart
but it doesnt happen in many such 3-equalities
best strategy to solve it every time is to make all of them equal to k
then write each variable in terms of k
OP u there?
shell wigeon
"best strategy" is highly subjective
frozen lake
indeed
shell wigeon
You already have an system of two equations, no need to introduce a third variable
Anyway that's assuming the brackets are parentheses, which is unlikely given the problem is asking for a least upper bound
frozen lake
there are only two pairs of a and b that satisfy this
actually only one pair
a and b are positive remember?
frozen lake
Anyway that's assuming the brackets are parentheses, which is unlikely given the...
hmm thats tru too
fresh sparrow
a and b are positive remember?
ah woops
shell wigeon
It's almost certain the brackets stand for floor, but at this point we may never know 
fresh sparrow
there are no solutions if it's floor 💀
thin relic
Do we get more values if the brackets were gif functions
thin relic
there are no solutions if it's floor 💀
Rlly
fresh sparrow
seems like so
thin relic
OMG
shell wigeon
there are no solutions if it's floor 💀
? There sure are
thin relic
there are only two pairs of a and b that satisfy this
They are?
And how did you get these
fresh sparrow
or wait
fresh sparrow
? There sure are
yes mb
shell wigeon
You're basically saying that you don't know what your question is...
thin relic
You're basically saying that you don't know what your question is...
I got the question like this
thin relic
You're basically saying that you don't know what your question is...
And yea that is the biggest problem of this problem
I didn't make this problem
shell wigeon
Where did you get it from then?
thin relic
*Where* did you get it from then?
Practice paper
Online
shell wigeon
Link?
thin relic
Link?
It is an online course on an app
So can't rlly send the link
But I have the solution and I don't understand what is going on
Should I give it?
shell wigeon
Why didn't you mention that at the very start?
Your actual question wasn't "how to solve this", it was "can you help me understand this solution", wasn't it?
thin relic
Why didn't you mention that at the very start?
Yes
Because I thought it is not good to post solution before question gets discussed
shell wigeon
You're the one asking for help, not us
thin relic
This is it
Sorry for wasting your time
thin relic
You're the one asking for help, not us
Yea but I think there not much one can do to help someone understand a solution
shell wigeon
So yes, the brackets do mean floor
thin relic
Btw now can you help?
thin relic
So yes, the brackets do mean floor
Where is it ?
shell wigeon
It's not explicitly mentioned in the solution, I'm just deducing that from reading the solution
It's the only thing that makes sense in this context
Which part of the solution do you need help understanding?
thin relic
Which part of the solution do you need help understanding?
The middle part
When they start using inequalities
And what is with N
How is it 15
shell wigeon
Do you understand why N can't be 14?
thin relic
Do you understand why N can't be 14?
Yeaa
shell wigeon
Well, N is obviously an integer, so if you want to minimize N, you just take the first possible value, N = 15
shell wigeon
It's asking for a least upper bound of a+b, so we want a and b to be as big as possible
shell wigeon
Well a and b are relatively close because of the first equality
thin relic
Yup
shell wigeon
So yes they both must be relatively large, which drives N to be small
thin relic
Oh ok
shell wigeon
Once you know N=14 isn't possible because of [ab], N=15 is the obvious solution
shell wigeon
Cool 
fervent burrow
midnight plankBOT
fresh sparrow
how do i do part b
two lines are parallel if they have the same gradient (aka same "m")
steel crest
how do i do part b
i recommend first using the formula
a line with slope m going through the point (x₁,y₁) has the formula:
y-y₁=m(x-x₁)
and then solve for y
and simplify into y=mx+c
steel crest
y-y₁=m(x-x₁)
this is called the "point-slope formula" because you use it when you have the slope and a point
delicate frigate
Graph theory
midnight plankBOT
Graph theory
delicate frigate
i numbered them so its easy
independent are set of those nodes who are not adjacent
thats what i know
no idea how to approach
runic hamlet
just try to find some independent sets
delicate frigate
1,8
2,7
3,8
6,7
4,7
5,8
1,5
2,4
2,6
2,5....
wait those sets should have only 2 elements ?
or more is fine ?
wait
i think i mma group them
{1,5,8,3}
{2,4,7,6}
thats it?
answer is 4?
am i right?
@runic hamlet
sorry for the ping
runic hamlet
why is there no set with size 5
delicate frigate
why is there no set with size 5
because there isnt, idk why
but i cant visbly see
any set with 5 elements
who are non adjacent
runic hamlet
I dont know how well you have to argue that there isnt such a set of size 5
one argument might go as follows: if you have a set of size 5, you need to take at least 3 of the middle six nodes. but its easy from that observation to go through all options
delicate frigate
one argument might go as follows: if you have a set of size 5, you need to take ...
and if we choose two random non adjacent nodes from the middle six, there does not exist any third node which is non adjacent to other two chosen nodes
ANYWAYS, the question does not ask why the cardinality is 4(4 is correct btw)
i have another to ask, may i ?
what does it mean by 'coloring'
is it what i used to do when i was three?
runic hamlet
it should be explained somewhere in your notes
it means choosing colors for the vertices so that adjacent vertices have different colors
may sometimes also mean choosing colors for the edges
snow dawn
What is your method of solving this or better still, what is the issue with it?
!status
midnight plankBOT
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
it asks for the smallest number of colors you need
delicate frigate
it means choosing colors for the vertices so that adjacent vertices have differe...
ah
okay
Yes it should be in my notes, my bad here
will check it
should not be a problem i think
thanks for the help
.close
midnight plankBOT
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solemn barn
.Ask
polar nexus
This looks easy tbh
midnight plankBOT
This looks easy tbh
solemn barn
might be easy but how do i solve tht
polar nexus
Have u tried anything yet?
polar nexus
?
scenic wyvern
note that your name is on the channel, not count fabio
solemn barn
oh lol
scenic wyvern
for reference, OP (count fabio in this case), when taking a new channel, don't start your message with a .
this suppresses opening a new channel
otherwise, please proceed
wintry sand
Try letting a=log(x), b =log(y)
solemn barn
Try letting a=log(x), b =log(y)
dosnt help not with log3x and 4y
polar nexus
note that your name is on the channel, not count fabio
Oh yea
midnight plankBOT
@polar nexus Has your question been resolved?
scenic wyvern
or just close the thing
maybe just open another channel
solemn barn
maybe just open another channel
did it
fallen rock
midnight plankBOT
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To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.
midnight plankBOT
@fallen rock Has your question been resolved?
midnight plankBOT
molten bay
Is this question wrong?
shell wigeon
It's oddly phrased but e^x isn't the only function that satisfies f(x+y) = f(x)f(y)
last slate
Is this question wrong?
e^kx works too
and its the set of functions you're supposed to look at in this problem
molten bay
It seems wrong to me
last slate
Can you explain why?
lyric charm
f(x+y) = f(x)f(y) is satisfied by functions other than e^x too.
in fact the functions that satisfy it are those of the form f(x) = a^x for real constant a > 0
molten bay
how can I find value of derivatives
lyric charm
$f(x) = a^x$ so $f'(x)$ is?
grand pondBOT
Ann
lyric charm
$a^x \ln(a)$
grand pondBOT
Ann
molten bay
Did I say differently?
lyric charm
yes you did
molten bay
I edited
lyric charm
"a^x log a, base x"
molten bay
Is this fine?
lyric charm
the log is base e not x
tired osprey
lol
molten bay
😅
Hang on i will be back
a^x ln(a) fine
thanks for little things
f'(5)=a^5 ln(5)
last slate
in fact the functions that satisfy it are those of the form f(x) = a^x for real ...
The world if this was actually true
molten bay
10 ln(5)
lyric charm
The world if this was actually true
oh sorry do you have a counterexample to this
last slate
oh sorry do you have a counterexample to this
Look up Cauchy's functional equation
lyric charm
maybe one that's also differentiable everywhere, given the question talks about derivatives?
molten bay
i gave the hint from my world
last slate
maybe one that's also _differentiable everywhere_, given the question talks abou...
Yes differentiable then gives your solution
last slate
Look up Cauchy's functional equation
Then take e^f where f is a shitty sol that works for the Cauchy
molten bay
I'm asking what will be the value of log(5)?
slender walrus
log(5) = log(5)
slender walrus
why do you still think there is a problem with the question
molten bay
I was thinking there is not D
shell wigeon
Because he hasn't properly solved it
slender walrus
why?
molten bay
What?
shell wigeon
f'(5)=a^5 ln(5)
How do you get this?
molten bay
If you guys were okay with D
Then it could have been easy
f(x+y)=f(5)=a^5
f'(x+y)=f'(5)=a^5log(5)
given value of a^5 is 10
So 10 log(5)
shell wigeon
f'(5)=a^5 ln(5)
I did, it's this one
shell wigeon
But feel free to leave, I'm not the one asking for help
molten bay
I got it
shell wigeon
Yes
molten bay
f(2+3)=f(5)=10=a^2×a^3=a^5
slender walrus
derivative was fine
plugging in x=5
was faulty
molten bay
a^5 ln(a)
slender walrus
check that again
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sinful lynx
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sinful lynx
I don't understand the 2nd step
wont it be (a^2+2ab+b^2) - a^2-2ab+b^2) / ab
what happens after that??
mortal mirage
difference of squares :)
sinful lynx
ion get it
mortal mirage
wont it be (a^2+2ab+b^2) - a^2-2ab+b^2) / ab
yes, but this can be simplified
to that
mortal mirage
ion get it
$m^2-n^2=(m+n)(m-n)$
grand pondBOT
Mathemusician
mortal mirage
have you seen this before?
sinful lynx
yess
mortal mirage
so do you understand how they got the second line?
mortal mirage
(a^2+2ab+b^2) - a^2-2ab+b^2) / ab
a^2 - a^2 + b^2 - b^2 +2ab +2ab/ ab?
you dont even need to do this
sinful lynx
you dont even need to do this
well it doesn't make sense does it
cedar mason
admittedly just expanding is far cleaner
mortal mirage
(a^2+2ab+b^2) - a^2-2ab+b^2) / ab
a^2 - a^2 + b^2 - b^2 +2ab +2ab/ ab?
but yes this is correct
sinful lynx
but yes this is correct
huh
mortal mirage
you can simplify it a bit more though
a^2 - a^2 + b^2 - b^2 +2ab +2ab is the numerator, right?
some things cancel out :p
sinful lynx
a^2 - a^2 + b^2 - b^2 +2ab +2ab is the numerator, right?
yes
do both square b's and a's get cancelled?
mortal mirage
yea
cedar mason
$(a+b)^2-(a-b)^2 = a^2+2ab+b^2 - (a^2 -2ab +b^2) \
= a^2-a^2 + 2 ab + 2ab + b^2 -b^2 = 4ab$
grand pondBOT
Percy
sinful lynx
how did they get what they got
mortal mirage
well (2a)(2b) is 4ab as well, no?
cedar mason
and you can do that mentally so much cleaner than the diff of squares
sinful lynx
well (2a)(2b) is 4ab as well, no?
OH
sinful lynx
and you can do that mentally so much cleaner than the diff of squares
I don't get how they got to their step though
I mean, they skipped a lot of simplifcation right
cedar mason
theyre using difference of squares
$(a+b)^2 - (a-b)^2 = (a+b+a-b)(a+b-a+b) = (2a)(2b)$
grand pondBOT
Percy
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tidal turret
midnight plankBOT
tidal turret
need help with v)
sudden yacht
TRANSLATION?
golden crescent
crop
tidal turret
TRANSLATION?
f,g are bijective, prove fog is bijective
golden crescent
💔
lavish venture
f,g are bijective, prove fog is bijective
didnt you prove iii and iv
tidal turret
didnt you prove iii and iv
yes, need help with v)
lavish venture
i don’t think you understand though
bijective means injective and surjective
so there’s nothing to prove
tidal turret
i don’t think you understand though
wdym?
lavish venture
didnt you prove iii and iv
if you’ve proved these
tidal turret
oh, right
I just need to do them one after another in one proof
which is a skill in itself
lavish venture
well
not really
just say "by iii fog is injective" and "by iv fog is surjective" so fog is bijective
no need to redo the proof
tidal turret
ok
exotic stratus
no need to redo the proof
It's not the same thing though, if you consider that the questions are statements in isolation
lavish venture
It's not the same thing though, if you consider that the questions are statement...
no it definitely is
just because a function has some other properties as well is irrelevant
exotic stratus
no it definitely is
Well yes, if it were the same f, then sure, but I don't think that's guaranteed
lavish venture
it has nothing to do with the proof of it being injective and surjective
lavish venture
Well yes, if it were the same f, then sure, but I don't think that's guaranteed
f is arbitrary wdym
there is no "same f"
exotic stratus
Yes that is what I mean
lavish venture
the proof applies universally to all functions that are injective/surjective
this is the point of making them arbitrary and having a general statement
you don’t prove examples
exotic stratus
Oh you mean something like, since (f is bijective => f is injective, g is bijective => g is injective) => fog is injective and then (f is bijective => f is surjective, g is bijective => g is surjective) => fog is surjective, since fog is injective and surgective <=> fog is bijective, qed
tidal turret
tldr?
exotic stratus
By definition of bijection, iii, iv, fog is bijective
tidal turret
oh, so at the end you agree with kniefy
exotic stratus
Yea
lavish venture
tldr?
since f, g are bijective they are both injective and surjective so iii and iv together imply that the composition is injective and surjective (hence bijective)
lavish venture
recall your definitions
tidal turret
well I proved them, but, idk
lavish venture
wdym?
tidal turret
Like I need to practice more surjective and bijective proofs, because
lavish venture
i mean just remember that bijective means injective and surjective
tidal turret
i am bad at proof writing
sudden yacht
well I proved them, but, idk
Definitions are not to be proven
exotic stratus
I didnt even notived iii and iv implies v
It's sometimes not immediately obvious, so don't beat yourself up about it
tidal turret
i mean just remember that bijective means injective and surjective
i know that, I definitely remember this, is just that I left v) for the end and I thought it was unrelated with the previous exercises, but as you guys said, we can re use the proofs from 3 and 4
lavish venture
usually if an exercise has multiple parts then the parts are related in some way
scenic wyvern
some multipart exercises actually force you to reuse what you've proven or worked out in earlier parts
tidal turret
this was adviced to me from a classmate, the other day, and I keep re proving the wheel
scenic wyvern
look out especially for the word "Hence,"
lavish venture
hence show that
scenic wyvern
(not sure what the Spanish equivalent is)
lavish venture
hénce
tidal turret
we dont want to make it offtopic dudes, I will be closing
lavish venture
🤣
tidal turret
i appreciate it hard
lavish venture
they’ll take my helpful away for spam
lavish venture
thank you sir
tidal turret
.solved
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quaint shoal
Does this seem to be correct?
midnight plankBOT
Does this seem to be correct?
quaint shoal
haven’t done natural log in a while
lavish venture
sure
lavish venture
the process where I did the ln is right too?
what is ln^x
split warren
Does this seem to be correct?
YES IT IS
u only use the definition for x> 0 there which is ln(x)
quaint shoal
Notation is questionable
yea im starting to remember some stuff now
it should be e^x=1 instead right?
gaunt nimbus
Yes
quaint shoal
that’s proper notation I believe?
split warren
u only use the definition for x> 0 there which is ln(x)
also ln(x) is continuos so it should be fine
gaunt nimbus
Yup
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zealous schooner
How do you prove that the interchange of the order of integration is valid here?
midnight plankBOT
How do you prove that the interchange of the order of integration is valid here?
zealous schooner
In general, when the region isn't rectangular, how would you justify this sort of interchange? The only theorems I know of which deal with this are the Fubini and Tonelli theorems, both of which apply to rectangular regions only
Not to mention that the integral is not absolutely convergent so we can't use that anyway
mortal falcon
i think this might be more straightforward, if we write the improper integral as a limit of a bound, then we're just interchanging on a bounded, connected domain which is fine, and then applying the limit will give us the improper bound on the outside again
unless you're asking abt generally interchanging on bounded domains
zealous schooner
i think this might be more straightforward, if we write the improper integral as...
You mean $$\int_0^\infty\int_{x-1}^{x+1} f(x,t)\dd{t}\dd{x} = \lim_{b\to\infty}\int_0^b\int_{x-1}^{x+1}f(x,t)\dd{x}\dd{t} = ...$$
grand pondBOT
kheer257
mortal falcon
yeah
carmine sigil
Lebegesue integrability, no? For Fubini's theorem?
zealous schooner
Lebegesue integrability, no? For Fubini's theorem?
Doesn't fubini's theorem only work for rectangular surfaces
carmine sigil
oh, my bad, I didn't notice the bounds were dependent
jaunty ivy
Fubini's works if the absolute integrability condition holds
zealous schooner
That's the problem
carmine sigil
oh, of course, that's pretty slick
carmine sigil
yup
zealous schooner
That still doesn’t solve the problem though because we don’t have absolute integrability
zealous schooner
But then how do we justify taking the limit back inside?
mortal falcon
the limit is always outside im pretty sure
if you interchange in terms of b, i think b stays outside
im sying i think bc im too lazy to do it out
but based on the steps u showed, infinity is still on the outside
so the limit never had to go "inside" anything
zealous schooner
You’re right
The region will be a trapezium
And b stays in the outside
Okay that works
zealous schooner
unless you're asking abt generally interchanging on bounded domains
Is there something else you can do in general?
The indicator function thing was pretty cool
mortal falcon
i wanna say you can interchange pretty generally on bounded domains? idk, sort of from a compactness-y perspective, but idk how to formalize this intuition, and that obviously already fails when ur integral is improper on a bounded domain
hmm i think i need to read more abt this
zealous schooner
Well the domain being infinite isn't a big deal when the function is lebesgue integrable over the whole thing right?
mortal falcon
yeah that makes sense
zealous schooner
I must confess that I've never studied measure theory, so most of these terms mean very little to me
mortal falcon
i guess in some sense.... i feel like there's some tradeoff between boundedness of the integrand and boundedness of the region
but that's just intuitive nonsense thats probably not real lol
zealous schooner
I must confess that I've never studied measure theory, so most of these terms me...
sigma-finite just means that you can cover the whole space with countably many sets with finite measure right?
mortal falcon
hopefully i understand this better after this semester
zealous schooner
i guess in some sense.... i feel like there's some tradeoff between boundedness ...
well unbounded functions can't converge surely
mortal falcon
er like not on generically unbounded domains right
mortal falcon
i guess i mean moreso that a bounded domain might imply some structure
where an unbounded function behaves nicely
but thats totally unfounded and i dont wanna keep talking abt stuff i have no substance for 
mortal falcon
sigma-finite just means that you can cover the whole space with countably many s...
yeah
zealous schooner
hmm
mortal falcon
well unbounded functions can't converge surely
wait an unbounded function totally can converge on an unbounded domain, idk how interesting that is but like just take any piecewise function of two 1/x^p functions with appropriate convergences
idk what the point of that is though lol or if thats relevant at all to this
im also okay with just... ignoring unbounded functions right now
i think for bounded integrable functions on bounded regions, you totally can perform interchanges, and that's just from the indicator argument i think
and that applies here anyway for any finite b
zealous schooner
wait an unbounded function totally can converge on an unbounded domain, idk how ...
Well then the whole function is bounded
Each piece is bounded in the relevant domain
zealous schooner
i think for bounded integrable functions on bounded regions, you totally can per...
Yeah
But sinx/x isn’t lebesgue integrable
mortal falcon
f(x) defined from R+ to R
f(x) = 1/x^(1/2) from 0 to 1
f(x) = 1/x^2 from 1 to inf
f isnt bounded here
zealous schooner
Which is why we needed to do the b thing
zealous schooner
f(x) defined from R+ to R
f(x) = 1/x^(1/2) from 0 to 1
f(x) = 1/x^2 from 1 to i...
Ahh true lol
mortal falcon
yeah i just mean i guess my intuition for why we could interchange once we wrote the improper integral as a limit was just bc we had a bounded function on a bounded region
mortal falcon
unless you're asking abt generally interchanging on bounded domains
that was where i was coming from with this, i was just not grasping the bit that my brain was assuming lol
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sturdy prism
f(n)=n squared +n + 2 , n∈N.
Show that f n( ) is always even.
midnight plankBOT
f(n)=n squared +n + 2 , n∈N.
Show that f n( ) is always even.
jaunty ivy
f(n)=n squared +n + 2 , n∈N.
Show that f n( ) is always even.
what do you notice about n(n+1) + 2
try with some examples like n=1, n=2, n=3
raw egret
he meant, try putting n=1,2,3.. and see the number you're getting
what type of number are you getting, even or odd?
sturdy prism
even
raw egret
how do u pove its even tho
as dj said, n^2+n+2 can be written as n(n+1) + 2 and this is will always give even integer for n belongs to N why? because
take any n, say your n is odd, then your n+1 has to be even
and vice verse, if your n is even itself, your n+1 would be odd
raw egret
now, multiplying any even integer with odd results in even
raw egret
because it'll have a factor of 2, which makes it even, and you're adding 2 to it, which is again an even integer
sturdy prism
it says
raw egret
even + even = even
sturdy prism
raw egret
yeah, its the same thing as we discussed
raw egret
uh no, its more like
they just called n(n+1) as '2m' because n(n+1) is always even and you can write any even integer in the form '2m/2p' whatever variable
because 2m will always result in an even integer
they did that so its easier to understand the next step
sturdy prism
yes
when i proof
do i need to state
that
like
if n+1 = even therefore n will be odd and vice versa
and afterwards state
that
both equals even
because
even x odd =even
raw egret
either that or you can do as they said
they said the same thing, just in more short manner
that product of 2 consecutive numbers will be even because one of them will be even, other will be odd,and their multiplication results in even
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exotic ridge
im confused about why the question is asking to evaluate with x=1 and x=2, when I simplify the expression, I get
midnight plankBOT
im confused about why the question is asking to evaluate with x=1 and x=2, when ...
exotic ridge
$(3x-4)(x-2)$
grand pondBOT
UltraSonicSpeed
cursive zenith
I think they just mean that they want you to plug in x=1 and x=2 for the unsimplified equation and the simplified equation and check that you get the same result
exotic ridge
oh
seems a bit strange to me
its probably relevant to the main question since this is just a warm up for the main question
cursive zenith
what is the main question?
sullen bolt
would my goal here to factor this into some (y+x-constant)(x²+y²+...constant)?
sullen bolt
translation: given x, y are real numbers which correspond with:
find y-x
peak walrus
I would say probably (x-y-constant)(x^2+y^2+constant)
dawn dagger
Maybe even (x-y+a)(x²+bx+cxy+dy+y²) and compare terms
midnight plankBOT
@sullen bolt Has your question been resolved?
sullen bolt
Maybe even (x-y+a)(x²+bx+cxy+dy+y²) and compare terms
seems unlikely
dawn dagger
What does that even mean
dawn dagger
If you manage to factor it, you can simply apply the zero product theorem.
sullen bolt
multiplied the expression by -1, seems much easier to express in terms of (y-x) now
keen saddle
I would say probably (x-y-constant)(x^2+y^2+constant)
You should try this, if it doesn't work
keen saddle
Maybe even (x-y+a)(x²+bx+cxy+dy+y²) and compare terms
Try this
Because the latter one might be super long
So better to check for the easier case first
dawn dagger
I looked it up and it is lengthy
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@sullen bolt Has your question been resolved?
sullen bolt
.close
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raven zinc
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proud abyss
why does this not work? i am trying to print the second largest element in an array #include <iostream> using namespace std; int main() { int n, max1, max2; cout << "enter the number of elements in your array: "; cin >> n; int a[n]; cout << "enter elements of the array: "; for(int i = 0; i < n; i++) { cin >> a[i]; } max1 = a[0]; max2 = a[0]; for(int i = 0; i < n; i++) { if(a[i] > max1) { max1 = a[i]; } if(a[i] > max1 && max2 != max1) { max2 = a[i]; } } cout << "the second largest element is: " << max2; } ⠀
midnight plankBOT
why does this not work? i am trying to print the second largest element in an ar...
shell wigeon
If you encounter a new maximum, you first set max1 and then a[i] > max1 is obviously false
So max2 is never set
proud abyss
{
max2 = a[i];
}``` this block?This block is never executed, yes
proud abyss
so i create another for loop for it?
proud abyss
right
see
what i am trying to think of is
i first find the largest element in the array
and then iterate through the whole array again and find a number which is larger than all elements but the maximum one
if that makes sense
rain wasp
you can do this in one pass btw
proud abyss
definitely, but i am just trying to think my way through it
shell wigeon
What you're thinking of makes sense but you can indeed do it in a single loop
proud abyss
i get that. but i would rather first try to work on what comes to my mind first, and then work on the most optimal solution
rain wasp
you have three ways to do this: one pass solution, two passes solution, or the meme solution
try not to overthink, it's much simpler than you think
rain wasp
Think of the maximums like a queue
again, this will help
proud abyss
i do not know how to proceed with that information
shell wigeon
Try an example manually
proud abyss
okay, but, if i were to do it in one pass, what approach would i be thinking of?
proud abyss
honestly, the way i thought earlier. i would, again, just search for the highest element and then look for the second highest one. i am sorry if i am sounding slow right now, it's just that i cannot think of another (easier) way to solve it.
shell wigeon
It's ok, first time for everything
Let's do it step by step
First I see a 3, so I take that as my current max
Then I see a 4, it's bigger, so this is my new current max
If the list stopped right there, what would be the second max?
proud abyss
3
shell wigeon
Right, so I take that as my current second max
Next, a 2. Well, it's smaller than both the max and the second max, so nothing to do
Then, a 5. What do I do now?
proud abyss
i set it as my current max and check if it is greater than my current second max?
ah wait
it already would be
so i update both of them simultaneously, rather than making another loop for it?
shell wigeon
Exactly
Whenever you encounter a new maximum, the previous maximum becomes the new second maximum
proud abyss
ah so i swap max2 with max1 if i find a new max
right
thank you so much! for being patient with me, too
shell wigeon
Not "swap", more like "overwrite" or just "set"
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woeful turret
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woeful turret
not sure what to do with this
usually my go to move is to use series expansions
but i cant do that for the arctan x
lyric charm
can't like forbidden or can't like unable?
woeful turret
unable
lyric charm
$\arctan'(x) = \frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 + \dots$
grand pondBOT
Ann
woeful turret
like i dont know the series expansion
lyric charm
integrate this term-by-term and get the series for arctan
woeful turret
oh
lyric charm
C=0
woeful turret
so it does have an easy expansion
lyric charm
also suggest rewriting the log thing as $\frac{\log(1+x)-\log(1-x)}{2}$ if you havent done so already & recognize in it the odd component of $\log(1+x)$
grand pondBOT
Ann
woeful turret
what about the other inverse trignometric functions?
lyric charm
$\arcsin'(x) = (1-x^2)^{-1/2}$
grand pondBOT
Ann
woeful turret
oh and arccos?
vivid remnant
this can be done via binomial expansion
real. bionomial makes things easier
lyric charm
oh and arccos?
pi/2 - arcsin
woeful turret
oh right
alright i think ill just remember these expansions
cant go through the pain of deriving them
thanks a lot
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graceful ferry
given permutation $\alpha = ( 1, 11, 5, 7, 2)(4, 10, 13)(8, 9, 6) \in S_{13}$ let $\beta \in S_{13}$ such that it is an odd permutation, it doesnt have a fixed point, and $\beta^2=\alpha$ find $\beta$
midnight plankBOT
given permutation $\alpha = ( 1, 11, 5, 7, 2)(4, 10, 13)(8, 9, 6) \in S_{13}$ le...
graceful ferry
How do i do this
dreamy lichen
separate it with commas pls
graceful ferry
mb
visual tiger
given permutation $\alpha = ( 1, 11, 5, 7, 2)(4, 10, 13)(8, 9, 6) \in S_{13}$ le...
what have you tried?
graceful ferry
I know how to get beta from the given that beta^2=alpha but i dont know how to change the fixed points 3 and 12 while keeping it an odd permutation
visual tiger
12 and 13*?
grand pondBOT
prograce
visual tiger
also, know that 5cycle * 3cycle * 3cycle is an even permutation
graceful ferry
3 and 12, typo in permutation
visual tiger
what is your current beta?
graceful ferry
also, know that 5cycle * 3cycle * 3cycle is an even permutation
Oh,,, wait then something is wrong wait
Nvm i fixed it,yes i mean change it from even to odd
beta I found is currently (7, 2, 1, 11,5)(10, 13, 4)(9, 6, 8)
how do i work with the fixed points/
dreamy lichen
Nvm i fixed it,yes i mean change it from even to odd
beta I found is currently ...
What are your current fixed points?
graceful ferry
3 and 12
dreamy lichen
so you need to somehow unfix them, without changing beta^2
visual tiger
uhm your beta is actually bad
try computing even the first cycle squared
what you wrote is actually just alpha itself
(1 2 3) = (2 3 1) = (3 1 2)
graceful ferry
Ok im lost then I thought u just move them around
I did that with another permutation
visual tiger
I did that with another permutation
... then I'm scared that might be wrong too
moving them around all to the right by 1 will not change the permutation
because literally every number in a cycle is sent to the one on its right
so if you move them all around equally
graceful ferry
I had alpha=(1 4 7)(25 3 9) and beta=(1 7 4)(2 5 3 9) while beta^5=alpha in S_9
visual tiger
you don't change the number that's on the right of another
this one is fine
because (2 5 3 9)^4 = id
turbid plinth
How can I maximize cos a + cos b + cos c if a+b+c =π
visual tiger
How can I maximize cos a + cos b + cos c if a+b+c =π
!occupied
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How can I maximize cos a + cos b + cos c if a+b+c =π
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
visual tiger
I had alpha=(1 4 7)(25 3 9) and beta=(1 7 4)(2 5 3 9) while beta^5=alpha in S_9
but you didn't "move everything around" like you did with the first exercise
graceful ferry
because (2 5 3 9)^4 = id
Yea since 5=1(mod4) since its a 4 cycle it doesnt change,
graceful ferry
In the first exercise 2=2(mod 3) and 2=2(mod 5)
visual tiger
(1 4 7) became (1 7 4), not something like (7 1 4)
graceful ferry
(1 4 7) became (1 7 4), not something like (7 1 4)
Oh wait i have no idea how this happened
I thought I moved it around two times
visual tiger
I thought I moved it around two times
that's not.. how cycles work xd
graceful ferry
Yes I read what you wrote lol
visual tiger
if you move every number in a cycle around equally, you don't change the number that's on the right of another
visual tiger
Yes I read what you wrote lol
ok
graceful ferry
How do I do it then?
visual tiger
so the cool property of cycles of odd length is you can compute their 'square roots' by computing their powers
say my cycle is named $\sigma$
graceful ferry
OHH
grand pondBOT
Raphaelisius Maximus MMIII
visual tiger
and I want some $\sigma'$ such that $(\sigma')^2 = \sigma$
grand pondBOT
Raphaelisius Maximus MMIII
visual tiger
just take $\sigma'$ as a power of $\sigma$, say $\sigma^k$
grand pondBOT
Raphaelisius Maximus MMIII
visual tiger
we want $\sigma^{2k} = \sigma$
grand pondBOT
Raphaelisius Maximus MMIII
graceful ferry
Ok
visual tiger
$2k - 1 \equiv 0 \mod m$, m is the length of the cycle
grand pondBOT
Raphaelisius Maximus MMIII
pulsar zephyr
$3k-2\equiv 0 \mod a$, a is the length of the cycle
grand pondBOT
Curiosity
visual tiger
$3k-2\equiv 0 \mod a$, a is the length of the cycle
?
visual tiger
beta=(1 5 2 11 7)(4 13 10)(8 6 9)(3 12)
eeeeeh we're not quite there for the 5cycle
what power of ( 1, 11, 5, 7, 2) did you compute?
graceful ferry
Oh i did 2,shouldve been 3 right?
visual tiger
uhhh
I hope you didn't compute (1 5 2 11 7)^3 accidentally
that's not the cycle that alpha has
graceful ferry
Yea
visual tiger
alright
change "beta is odd" to "beta is even" in your question statement
and here you go
midnight plankBOT
dreamy lichen
change "beta is odd" to "beta is even" in your question statement
Wait is that even possible
visual tiger
Wait is that even possible
yes. HINT (prograce do not unspoiler if you want a go): ||A 6cycle squared gives two 3cycles||
dreamy lichen
oh, just realized
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tiny shale
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dreamy lichen
I'd try to rewrite that fraction to only have constant in the numerator
that'll make everything more convenient
lyric charm
finding the range of a function is something there's unfortunately not really any general procedure for
dreamy lichen
I'd try to rewrite that fraction to only have constant in the numerator
for a hint on how to do this:
You'll want that 6x to somehow cancel with 2x + 3. For that, it needs to be a multiple of it. Try adding some constant to 6x, which will make it a multiple of 2x + 3
tiny shale
dreamy lichen
this didnt help much, you still have x in the numerator
tiny shale
Factor out 6x?
dreamy lichen
that will just move 6x outside
dreamy lichen
for a hint on how to do this:
You'll want that 6x to somehow cancel with 2x + 3....
try answering this
6x + what is a multiple of 2x + 3?
grand pondBOT
MathIsAlwaysRight
dreamy lichen
**MathIsAlwaysRight**
now you can split up that fraction
$\frac{6x}{2x+3}=\frac{6x+9}{2x+3}-\frac{9}{2x+3}$
like this
grand pondBOT
MathIsAlwaysRight
dreamy lichen
and in the first fraction, it cancels (because we chose 6x + 9 to be a multiple of 2x + 3)
dreamy lichen
why 0 <= x <= 3
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✅
dreamy lichen
Because 3 is max
3 isnt max
9/(2x+3) can be negative
and subtracting negative will result in positive
but do you know how does a graph of 9 / (2x + 3) look like?
tiny shale