@last slate Has your question been resolved?
#help-49
midnight plankBOT
tacit jungle
y=-3/4x+11/4
midnight plankBOT
y=-3/4x+11/4
midnight plankBOT
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paper owl
i set my u to cos2x+5sin2x and dv to e^-4x (i used tabular method)
when should i stop? im confused
midnight plankBOT
i set my u to cos2x+5sin2x and dv to e^-4x (i used tabular method)
when should ...
small jasper
tabular doesn't rlly work on $e^{-at} \sin bt$ integrals
grand pondBOT
Civil Service Pigeon
small jasper
well at least the traditional start to finish
You should note that you eventually get a $e^{-at} \sin bt$ back again barring some constant
grand pondBOT
Civil Service Pigeon
small jasper
take up to that section and write out the equation you'd get from doing regular integration by parts
tidal turret
midnight plankBOT
tidal turret
how does this work?
tawdry laurel
uh can you help us translate
tidal turret
tawdry laurel
wat
tidal turret
are you helping or not?
shell wigeon
How does what work?
tidal turret
How does what work?
the whole observation
shell wigeon
... which part? There are 5 points there
tidal turret
i mean
the whole part of like
expressing the number of subsets as a combination
so like, from the beginning
shell wigeon
You can make a subset by choosing some number of elements from a set
If the set has n elements total, you make a k-element subset by choosing k from n; the number of ways to do that is known as "n choose k"
"n choose k" is written nCk or C(n, k) or vertically like the binomial coefficient
$\binom{n}{k}$
grand pondBOT
Nel
shell wigeon
Anything else?
shell wigeon
Do you understand how it's n in the previous point?
tidal turret
oh, so it's like n subsets of length n-1
shell wigeon
Yeah
Choosing n-1 elements from n total is equivalent to choosing just 1 and then taking the "complement"
shell wigeon
It's the same idea though
midnight plover
The equality is easy to demonstrate but the rest i dont understand spanish
shell wigeon
Take S = {1, 2, 3, 4, 5}
There are 5C2 = 10 ways to make a 2-element subset ({1,2}, {1,3}, ..., {4,5})
For every 2-element subset, there is a corresponding 3-element subset, for example {2,5} corresponds to {1,3,4}
So there are just as many ways to make a 3-element subset; hence 5C2 = 5C3
tidal turret
The equality is easy to demonstrate but the rest i dont understand spanish
i literally provided a translation
tidal turret
So there are just as many ways to make a 3-element subset; hence 5C2 = 5C3
you mean that the number of all the subsets of S with cardinality 3 is the same as the number of subsets pf S with cardinality 2?
shell wigeon
Yeah
tidal turret
how does that work?
shell wigeon
I literally just said how
For every 2-element subset, there is a corresponding 3-element subset
There is a bijection between the set of 2-element subsets and the set of 3-element subsets
tidal turret
Hopefully there is no duplicates
tidal turret
> For every 2-element subset, there is a corresponding 3-element subset
i mean this argument is kinda abstract to just fully believe this
tidal turret
it's true but i can't think of a counterexample when n choose k is not n choose n-k
shell wigeon
... because there is no counterexample
mortal falcon
why dont you try proving for yourself that the complement operation is a bijection on sets
it's not difficult to show injectivity and surjectivity
tidal turret
can i get some help with the proof
mortal falcon
let's say that for a set S of n elements, we consider the set of subsets of S of size k as S1, and the set of subsets of S of size n-k as S2. now look at the function f: S1 -> S2 defined as
f(A) = A^c
Remember again that the elements of S1 and S2 are subsets of S.
for instance, if S = {1, 2, 3, 4, 5} and k = 2 then
f({1, 3}) = {2, 4, 5}
we want to show f is a bijection between S1 and S2, so our goal is to show that
-
it sends every subset of size 2 to a unique subset of size 3, and
-
every subset of size 3 is mapped to by at least 1 subset of size 2.
these are both pretty straightforward from properties of complements
if thats maybe not illustrative, you can also think about the individual elements
for 1), we need to show that for any two subsets A and B of size 2,
if f(A) = f(B), then A = B
in other words, if A^c = B^c, then A = B
do you see why this is true?
midnight plover
let's say that for a set S of n elements, we consider the set of subsets of S of...
Ye thats clear
mortal falcon
im more concerned if renato understands it
shell wigeon
For any finite set S of n elements, you can build a k-element subset A where k <= n. The n-k elements that are not in A form the complement, S\A. The complement of that, S\(S\A), is just A again. So the complement operation is invertible, and so it's a bijection.
tidal turret
let's say that for a set S of n elements, we consider the set of subsets of S of...
wait a minute
mortal falcon
does Nel's answer make sense? its more concise than my answer
tidal turret
for 1), we need to show that for any two subsets A and B of size 2,
if f(A) = f...
well
tidal turret
For any finite set S of n elements, you can build a k-element subset A where k <...
how is it A
S-(SnA^c)
Sn(SnA^c)^c
Sn( S^c u A)
(SnS^c)u(SnA)
mortal falcon
S\A is equivalent to A^c here bc A is a subset of S
just think of (A^c)^c
what does that equal?
tidal turret
S\A is equivalent to A^c here bc A is a subset of S
yeah S is the universe i think
tidal turret
just think of (A^c)^c
A
mortal falcon
right, so the ^c operation is invertible, its inverse is itself
for any set A, (A^c)^c = A
shell wigeon
(SnS^c)u(SnA)
(yeah, and that's SnA = A)
tidal turret
(yeah, and that's SnA = A)
yeah exactly
shell wigeon
So, that's it, the complement operation is a bijection, so there are as many subsets as complements
In particular, as many k-element subsets as complements of k-element subsets, and these have n-k elements
tidal turret
i mean we showed injectivity
shell wigeon
?
mortal falcon
uh yeah same here
are you thinking of left-invertibility?
invertibility is a bit stronger
tidal turret
invertibility implies left and right inverse exists?
shell wigeon
An inverse function is both a left-inverse and a right-inverse
The existence of an inverse function is equivalent to the function being bijective
tidal turret
so i didn't understood, the inverse function of the complement is the complement or what?
shell wigeon
Yeah, the complement operation is its own inverse
tidal turret
involution time
shell wigeon
..sure
mortal falcon
yes its an involution
tidal turret
So there are just as many ways to make a 3-element subset; hence 5C2 = 5C3
i mean now i think i understand this a bit more
wat abt this
shell wigeon
Maybe try listing subsets of {1,2,3,4,5} and their complements if you're not completely convinced yet
shell wigeon
This just says that any n-element set has 2^n subsets
mortal falcon
what's your question? the blurb above explains what's going on
which part are you confused by
tidal turret
i don't understand why he sums every counting of the subsets of length 1, length 2 to length n
mortal falcon
so our goal is to count the total number of subsets, right
shell wigeon
Every subset of an n-element set has between 0 and n elements
nCk gives you the number of k-element subsets
tidal turret
oh, i think is saying the number of subsets of An is 2^n, and that is equal to the sum of the counting of subsets of length 1 to n
shell wigeon
The sum of all those nCk for k between 0 and n gives you the total number of subsets
0 to n
tidal turret
yeah
though is a bit handwavey how he got that the cardinality of the pset of An is 2^n
tidal turret
so our goal is to count the total number of subsets, right
should i just assume the cardinality of the powerset of a set with n elements is 2^n?
there must be a logical explanation
shell wigeon
Surely that was previously shown
It's not exactly hard to see why anyway: to make any subset of S, you either include or exclude each element of S
That's 2 choices per element of S, so 2*2*...*2, n times = 2^n
tidal turret
i see
i appreciate the help really i cannot emphasize this enough
i will be closing, thanks
.close
midnight plankBOT
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Available help channel!
worldly pine
Hi,
The 'average' topic is going very difficult for me. What should I do?
midnight plankBOT
Hi,
The 'average' topic is going very difficult for me. What should I do?
worldly pine
After every 2 or 3 questions, something new concept appears and I have to learn something new all the time.
I have completed only 60 questions on average, and things have heated up.
placid ridge
After every 2 or 3 questions, something new concept appears and I have to learn ...
I mean that seems like a good thing
strange ruin
What do you mean by the average topic?
worldly pine
Topic name - Average
strange ruin
As in you're self learning a whole range of topics?
worldly pine
As in you're self learning a whole range of topics?
No,
actually from a teacher
placid ridge
Probably faster if you can give an exercise
strange ruin
Yeah, can we see a question that you've donw
worldly pine
Yeah, can we see a question that you've donw
Do I show you an example of the question?
strange ruin
Yes please
lyric charm
yeah let's see the actual questions that confused you
that way we know what to explain or clear up
worldly pine
I generally mean..'the smart approach the teacher gives is kinda confusing...I get the basic but he tell us that we should go for smart approach
OK wait
lyric charm
yeah we want to see what exactly this "smart approach" is also
worldly pine
This is basic
And this is the different method
Do you want more questions?
The second one is the deviation method
placid ridge
And this is the different method
Theyre the same methods, just that both sides is divided by a number in this one (or at least i think, could just be it being simplified)
edgy aurora
what grade are you in?
worldly pine
Theyre the same methods, just that both sides is divided by a number in this one...
He told us different concept for the later one
placid ridge
Oh actually i think i get it
76n +72×15 becomes
76n+ 76×15-4×15
=76(n+15) -4×15
So on the left hand side we get 76 +(-4×15)/(n+15)
edgy aurora
i think it is 45
edgy aurora
He told us different concept for the later one
what grade are you in
worldly pine
what grade are you in
Undergraduate
midnight plankBOT
@worldly pine Has your question been resolved?
twilit field
midnight plankBOT
twilit field
$P(A \cap B)= P(A)P(B)$
grand pondBOT
wai
twilit field
$P((A \cup B)^c) = 1- P(A \cup B) = 1- [ P(A)+ P(B)-P(A) P(B)] = (1- P(A)) + P(B)(1-P(B)) = (1-P(A))(1-P(B))$
and we're done
grand pondBOT
wai
twilit field
That's it I suppose?
muted panther
What if you show instead that A and B^c are independent?
Then, you can use that property a second time to show that A^c and B^c
are independent
And I feel there is an algebra mistake on the right of the third equality
runic hamlet
you managed to write a chain of equalities without using A^C and B^C anywhere
you know, the events you are supposed to show something for
muted panther
you managed to write a chain of equalities without using A^C and B^C anywhere
On the left, with Morgan
twilit field
you managed to write a chain of equalities without using A^C and B^C anywhere
I-P(A)= P(Ac)
runic hamlet
On the left, with Morgan
well yes I know. but its not in the proof
twilit field
And similarly for p(B)
runic hamlet
well write it down
muted panther
well yes I know. but its not in the proof
I agree with you for that!
runic hamlet
the chain should start with P(A^c cap B^c) and end with P(A^c)P(B^c)
muted panther
But anyway, you made an algebra mistake near the end.
Also, it is sufficient to show that if E and F are two independent events, then E and F^c are too. This implies that E^c and F^c are independent too, by the intermediate property.
slow thorn
That's it I suppose?
yeah i guess...
i mean you could extend it a bit at the start and the end
with P(A^c cap B^c) = ... = P(A^c)P(B^c)
muted panther
yeah i guess...
There are also algebra mistakes in the chain
slow thorn
oh yeah
slow thorn
$P((A \cup B)^c) = 1- P(A \cup B) = 1- [ P(A)+ P(B)-P(A) P(B)] = (1- P(A)) + P(B...
the (1-P(A)) - P(B)(1-P(A)) part i think
but yeah sth like tht
midnight plankBOT
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Available help channel!
scarlet lodge
midnight plankBOT
scarlet lodge
so i did part a... and I was abt to start part b
umm so for my base... i assume id set list L to be an empty list
bold peak
Yes
bold peak
Yes
scarlet lodge
and then reverse that again
by using the defintioon again
alr but the IS is where i find difficulty
@bold peak
bold peak
Well you have it true for a list of length n
A list of length n + 1 is just a list of length L with one element added to the end
So you can use (1)
scarlet lodge
so (x:L)?
scarlet lodge
well ok how bou tmy inductive hypothesis
can u read over that
Assume that for some list , the following holds: rev(rev(L)) = L
tha seems right to me
bold peak
Assume that for some list , the following holds: rev(rev(L)) = L
Not exactly
Inductive hypothesis should be, for all lists of length n, rev(rev(L)) = L
Also please tag me when you reply so I can get to it faster
scarlet lodge
alr so wrong heading
scarlet lodge
Also please tag me when you reply so I can get to it faster
sorry that felt rude to do 😭
@bold peak
bold peak
No worries no worries
scarlet lodge
and then for my inductive step
we need to show that
rev(rev(x : L)) = x : L
@bold peak
scarlet lodge
wdym
bold peak
The question
scarlet lodge
so we can use rev ( x : L) = rev L + x
bold peak
Part 1
bold peak
Also you should probably state somewhere that Rev(x) = x
Yes
You write rev L + x as rev L : x
scarlet lodge
yes
which would be the reverse of the definition
correct?
so we get
rev (rev(L) + [x])
and then use the property from part a?
@bold peak
bold peak
Yes
scarlet lodge
God I hate their notation
sorry my professor has his own textbook
and he disagrees with like most common mathematicians 😭
scarlet lodge
oh ok i think ik what to do
bold peak
In left cancellation he cancels K but in right cancellation he cancels M
Lmk if you get stuck anywhere
scarlet lodge
do u do
rev ( K ++ m ) = rev (L++M)
so rev K + rev M = rev L + rev M
@bold peak
bold peak
Yes
bold peak
so rev K + rev M = rev L + rev M
Other way around no
scarlet lodge
yea yea
bold peak
The order reverses
scarlet lodge
revM + rev K
=
Rev M + rev L
ao then substract rev M
and rev K = rev L ?
@bold peak
scarlet lodge
oh aply rev rev
on both sides
so rev rev(K) = K
and rev rev (L) = L
so K = L
?
@bold peak
scarlet lodge
thank u
bold peak
Anytime!
bold peak
No worries
bold peak
scarlet lodge
sooo heres hwat i was thinking 😭
for max (m,n )
thats the same as
n + (m monus n)
?
bold peak
I thought you misspelt minus
scarlet lodge
for example
bold peak
Quotient sorta thing?
scarlet lodge
ummm sorta just removes all non natural numbers
bold peak
I'll Google it brb
scarlet lodge
thank u so much
bold peak
Oh difference if it's non negative and zero o/w
scarlet lodge
yea
bold peak
Yes
scarlet lodge
good job 😭
thing is i was using an ai as well
and it gave me a whole different answer
bold peak
Well don't use an AI firstly
Secondly, yours is correct and you can verify it using cases
What happens when m > n for example
And what happens when n > m
scarlet lodge
yea yea
scarlet lodge
Well don't use an AI firstly
this doesnt look correct at all 😭
bold peak
That's just wrong
scarlet lodge
Well don't use an AI firstly
thats why im asking u fr 😊
bold peak
That literally just simplifies to 2m - n
Wait no that's a monus in between
God damn it
It's correct but overcomplicated
scarlet lodge
so both work yes
bold peak
Ye
scarlet lodge
so how do i write out my work for this
bold peak
It just says define
Do you have to prove
If you have to, then you just do it by cases
scarlet lodge
and how do i do that....
scarlet lodge
yea
scarlet lodge
ummm
bold peak
Take your time
scarlet lodge
it gives u the smae thing no?
scarlet lodge
thats the issue 😭
hwo would u do it
scarlet lodge
m - n
scarlet lodge
m
scarlet lodge
0
bold peak
And n + 0 = ?
scarlet lodge
n
bold peak
So it works
bold peak
In that case m monus n = m - n
scarlet lodge
Just substituting so we can work with familiar operations
sorry sorry
just confirming thats all
bold peak
No worries no worries
scarlet lodge
but my thing is can i just start the problem with m monus n + n
or do i have to prove till that pooint
bold peak
Also hope you don't mind me metamorphosising in the middle of the conversation
bold peak
but my thing is can i just start the problem with m monus n + n
It really depends on your prof
scarlet lodge
bcs it states this
bold peak
The wording of the question suggests that you just have to state the forms
It doesn't say anything about verification
scarlet lodge
so just define them
yes?
so find a general form for max
and a general form for min
scarlet lodge
so min....
ummm maybe
well whats the form for min
i get monus is max essentially
but
scarlet lodge
but it doesnt work for that
cuz then u get n -m +m
so u get just n
which woudl be the max not the min
scarlet lodge
u get max regardless 😭
bold peak
(n monus m) + n
If n > m you get n - m + n
Which is 2n - m
Lol
As a hint
Since it's a min
You'll need to use regular subtraction somewhere
scarlet lodge
would u need monus twice?
bold peak
Well let's test it
scarlet lodge
so m - max (m-n, 0)
scarlet lodge
yes
so u get m - (m-n)
so m -m + n
which is n
and if n > m and we want m
u get 0
and m - 0 is m
bold peak
Yeah
scarlet lodge
alr alr i have it on a piece of paper
scarlet lodge
so part b....
bold peak
Yup
scarlet lodge
its very obvious
uz one has to be the max and the other has to be the min
so itd naturally solve out to
m + n 😭
can i just add the two fucntions we made
so
(m monus n) + n + m -(m monus n)
i mean naturally we could cancel m monus n 😭
@bold peak
scarlet lodge
or we could verify both cases?
m > n
u get
m -n + n + m -(m-n)
so m -n +n +m -m +n
bold peak
Ehhh you don't need to
scarlet lodge
so it cancels out to m + N
bold peak
It just works directly
scarlet lodge
umm cuz i dont want to hold u
im abt to shwoer.... and i dont want u waiting on me
but i do have more problems and youve been very very helpful
@bold peak
bold peak
Ah no worries no worries
It's quite late for me so I'll prolly sleep
But feel free to add and dm me
bold peak
Interesting to say the least
scarlet lodge
prof cooked on this problem set 😭
bold peak
I do not know if I see the pattern
Lmao
This is smth I'll prolly have to sit down and write out lol
scarlet lodge
can we not just do trial and error
its going
it should be
m= 0
n = 2
for row 5 right
?
but yea i dont see a pattern
@bold peak
bold peak
Yeah me neither
bold peak
We are just doing trial and ertor
I don't see a way but I'm also too sleepy lol
Is this due soon
bold peak
Well I hope someone else can help you cuz I'm at 12% brain power rn
scarlet lodge
alra rl
gn gn
ttyl hopefully
i lowk hate this class
its so easy but they make it so hard for no reason 😭
@bold peak
scarlet lodge
mhmm 😭
bold peak
Or a responsive prof
bold peak
ttyl hopefully
I mean feel free to dm me lol
scarlet lodge
he does comp sci and math
bold peak
Plus we'll probably run into each other in help channels
scarlet lodge
smart ass guy but
bold peak
Fair enough I guess
scarlet lodge
annoying ass class 😭
scarlet lodge
tysm
scarlet lodge
What?
scarlet lodge
any thoughts
@steel crest any ideas on this
scarlet lodge
Umm idt so no
steel crest
well I dont know the pattern, did you keep on going for more rows looking for a pattern
steel crest
the implication is you need to do more rows to see the pattern
grand pondBOT
gfauxpas
steel crest
the first term is a triangular number, and im sure thats not a coincidence. you familiar with triangular numbers?
scarlet lodge
Yea
steel crest
`T₁ = 1
∘
T₂ = 3
∘
∘ ∘
T₃ = 6
∘
∘ ∘
∘ ∘ ∘
T₄ = 10
∘
∘ ∘
∘ ∘ ∘
∘ ∘ ∘ ∘
T₅ = 15
∘
∘ ∘
∘ ∘ ∘
∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘`
scarlet lodge
Yea
steel crest
this hints to a way to pair two natural numbers to one natural number , implied is that it's a bijective way
this makes sense because
let's make an array
scarlet lodge
this hints to a way to pair two natural numbers to one natural number , implied ...
Could u explain this
steel crest
im about to
(0,0) (0,1) (0,2) (0,3) … (1,0) (1,1) (1,2) (1,3) … (2,0) (2,1) (2,2) (2,3) … (3,0) (3,1) (3,2) (3,3) … ⋮ ⋮ ⋮ ⋮ ⋱
it's saying there's a bijection between N and NxN
scarlet lodge
What’s bijextion mean
steel crest
I can draw NxN as an array that goes infinitely down and to the right
means an invertible function
injective and surjective, if youve seen those terms
one-to-one and onto
scarlet lodge
Alr alr
steel crest
so one way that's natural to make sure you go through each pair of numbers once and only once
is to thread a needle through them, so to speak
the first pass will be through (0,0)
the second one will be (0,1)->(1,0)
the third will be (0,2)->(1,1)->(2,0)
fourth will be (0,3)->(1,2)->(2,1)->(3,0)
scarlet lodge
So it appears to be going through the diagonals
steel crest
and thats why theyre triangular numbers
scarlet lodge
Ohhhh
steel crest
because the upper left corner is a triangle
scarlet lodge
That actually makes. A lot of sense😭
steel crest
right?
and then you have the +n term
that's where you are on the last diagonal, you havent necessarily gone through the whole diagonal yet
anyway
not sure if this helps you find the pattern, but, it probably does!@
twilit field
midnight plankBOT
twilit field
$y_{n+1}-y_n= \frac{1}{2n+2}+ \frac{1}{2n+1} - \frac{1}{n+1}$
which is negative
so the seqeunce is decreasing
grand pondBOT
wai
twilit field
Now I have to prove it's unbounded
I'd like to proceed without proving it's not cauchy
Oh, easy enough
Let $L$ be an infimum of y_n
grand pondBOT
wai
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
eternal pawn
so the seqeunce is decreasing
are you sure?
eternal pawn
hint: ||integrals||
twilit field
I was instead planning to show that it doesn't have a supremum
eternal pawn
that will be difficult
twilit field
idts
lemme try
let $y_n$ have a supremum, $L$. $y_n$ , then converges to $L$ by MCT. So $\forall \varepsilon>0 , \exists N \in \N: n≥N \implies \abs{y_n-L}< \varepsilon$.
\ So $-\varepsilon+L≤y_n≤L + \varepsilon$.
grand pondBOT
wai
twilit field
uh, can I do anything with this
twilit field
why do you think it diverges?
each term is essentially the harmonic sum to 2n terms
eternal pawn
it's not really the entire harmonic sum though
twilit field
I'm aware
eternal pawn
||so that you don't waste your time:|| ||maybe it doesn't diverge...||
twilit field
y_n coverges...?
eternal pawn
yep
eternal pawn
you can always do it with integrals, if you're familiar with them, just to check it converges
grand pondBOT
The current time for math_rocks is 12:50 AM (IST) on Mon, 06/10/2025.
runic hamlet
just bound each term then
.
and: ||how many terms are there||
@twilit field so you see it tomorrow
midnight plankBOT
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Available help channel!
woven forum
hey
woven forum
I have this function defined
and they want f to be continuous at x=4
so its rather pretty simple, lim(5x²-7/x) = lim(ax) while both approach x->4
and find a
now theyre asking if f is
- differentiable
- continuously differentiable
- not differentiable
on R+
thats where Im stuck
tall bronze
Yoo
woven forum
1st one means derivative of f'(R+) exists, 2nd one means if f'(R+) is continuous, and 3rd one means if derivative on f(R+) doesnt exist
right?
tall bronze
Let me check
You mentioned three conditions about the derivative of f at the point x=4
tall bronze
What do you mean no?
midnight plankBOT
@woven forum Has your question been resolved?
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sorry its a broader question
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night gyro
this guy took the 3 to the other side why didnt he multiply the whole thing by 3 he only multiplied the 9
midnight plankBOT
this guy took the 3 to the other side why didnt he multiply the whole thing by 3...
fallen sparrow
this guy took the 3 to the other side why didnt he multiply the whole thing by 3...
im not sure what youre saying, but he did multiply both sides by 3
fallen sparrow
the 3 on which your arrow starts and the 3 on which it ends is the same 3
its multiplied on both sides
night gyro
this guy took the 3 to the other side why didnt he multiply the whole thing by 3...
now look at the guys solution he only multiplies the 9
9*3 which is 27
fallen sparrow
yes which is correct
night gyro
yh im confused on tht as i think whole thing should be mulitiplied
like here
underlined
fallen sparrow
so that 4 gets cancelled by the 4 in RHS
divide the equation by 4
night gyro
so i cant do this?
fallen sparrow
you can
but theres no point
write the equation again
wait what
bruh
why did you multiply 3 to 4 and another 3 to 9?
@night gyro
why are you distributing 3 on 4 as well as 9
its simple multiplication
(4 * 9) * 3 = 4 * 9 * 3 and not (4 * 3) * (9 * 3)
fallen sparrow
he did though
fallen sparrow
(4 * 9) * 3 = 4 * 9 * 3 and not (4 * 3) * (9 * 3)
do you understand this
fallen sparrow
he did though
just not in the way you think he did
night gyro
fallen sparrow
it should have multiplied the 4x squared aswell
no?
night gyro
ur multiplieng the whole thing
fallen sparrow
ur multiplieng the whole thing
my guy
fallen sparrow
(4 * 9) * 3 = 4 * 9 * 3 and not (4 * 3) * (9 * 3)
did you even read this
night gyro
i get it
(2x) squared had bracket and 9x does not
thats why he only multiplied by 3
oblique ridge
It doesn't have to do with brackets
12x^2 times 9x would also have been right for the rhs
night gyro
why wouldnt 12 x squared and 27x be wrong?
oblique ridge
because that would be multiplying by 3 twice, which is the same as multiplying by 9
you can multiply every part if they're added together, that's what you're remembering
but if they're multiplied together you only multiply one part
night gyro
but if they're multiplied together you only multiply one part
is that right
oblique ridge
Yup, that's okay to do
midnight plankBOT
oblique ridge
lol, ty
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night gyro
how to do this
midnight plankBOT
how to do this
night gyro
i was thinking 5+6+7 and 7+9+8 and make both parts the same
gusty falcon
i was thinking 5+6+7 and 7+9+8 and make both parts the same
yes
night gyro
then
night gyro
really?
gusty falcon
yes
start: 5 : 6 : 7 : 18
end: 7 : 9 : 8 : 24
when writing it as part:part:part:whole
finally u want the wholes to match
night gyro
this wht i get
night gyro
lcm?
gusty falcon
yes
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white moss
i dont understand $\pi: V \mapsto V/U$ and $\tilde{T}: V/(\ker T) \mapsto W$
grand pondBOT
Altanis
white moss
i understand that the quotient map takes a vector v and gives a translate of U via v + U
but i dont understand what tilde T(v + ker T) = Tv means
i dont understand what v + ker T is
like i know it's just v added to all the elements of the kernel
white moss
but i dont understand what tilde T(v + ker T) = Tv means
but i dont understand the use of this or like i dont understand what this is supposed to mean conceptually
toxic nova
(Presumably) what T tilde here is the natural bijection you get from V/ker(T) -> im(T) if T: V->W
The first isomorphism theorem for vector spaces states that if T: V -> W this induces an isomorphism T': V/ker(T) -> im(T)
white moss
sorry gimme 5min
toxic nova
If youre thinking about it in terms of dimensions (at least for finite dim), dim(V/U) = dim(V) - dim(U), so dim(V/ker(T)) = dim(V) - dim(Ker(T))= dim(Range(T)) by rank nullity theorem therefore theyre isomorphic
jaunty ivy
intuitively, u can think of this as T being a map that splits V into either elements that get "killed" (sent to 0) or elements that dont. this is precisely the rank nullity u develop once u have established dim V = dim W implies (in both directions) the existence of some isomorphism between these spaces
white moss
intuitively, u can think of this as T being a map that splits V into either elem...
yeah i understand how rank nullity shows this, it says the domain comprises vectors that map to an image and vectors that map to nothing
jaunty ivy
it is the other way around, this shows rank nullity
white moss
oh
i learned rank nullity before this so
anyways i was looking back at translates and it seems that a vector space can be expressed as a bunch of disjoint translates
and you can say two vectors are under some form of equivalence if they belong to the same translate that forms the vector space
so the quotient map just takes a vector and maps it to its appropriate translate
white moss
If youre thinking about it in terms of dimensions (at least for finite dim), dim...
i understand this already
i just dont understand like
idk
i don't understand the intuition behind T'(v + ker T) = T(v)
why does it exist
white moss
i don't understand the intuition behind T'(v + ker T) = T(v)
it feels like some form of reversal of the quotient map
like if the quotient map was pi: W -> V/nullT (T: V -> W)
then tilde T would be the inverse
but the quotient map makes sense to me, it just associates a vector with its appropriate translate
jaunty ivy
by linearity, u can do T'(v) + T'(u), u in ker T and get that since T' is just a restriction on T, it must act the same way as T for all the common domain elements. T'(u) = 0 then
white moss
yeah so i understand that T(u + v) for u being any vector and v being an element of ker T equals Tu
how does that relate to tilde T
is tilde T a generalization of T(u + v) for v being any vector in ker T?
like i dont see why you would want to map a translate of v (specifically, the translate over ker T) to Tv
hard umbra
T maps every element of the translate v + kerT to Tv
white moss
T maps every element of the translate v + kerT to Tv
yeah i understand that
white moss
now you try to define T on V/kerT
hmm i see
so it is just a generalization of this idea
hard umbra
the elements of V/kerT are v + kerT
white moss
why is it called an induced map?
hard umbra
well it's induced by T
toxic nova
and it is the only reasonable definition
As mentioned here if you have a T you only have one reasonable definition of the map T'
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compact crow
For the directional gradient with respect to vector u, how do you find the components? Mainly is it always just u = (1/sqrt(2),1/sqrt(2)?
midnight plankBOT
For the directional gradient with respect to vector u, how do you find the compo...
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twilit field
During assembly a product is equipped with 5 control switches . each of which has a probability of 0.04 of being defective. What's the probability that 2 defective switches are encountered before 5 non-defective ones.
midnight plankBOT
During assembly a product is equipped with 5 control switches . each of which ha...
placid ridge
I thought theres only 5 switches?
cedar bone
During assembly a product is equipped with 5 control switches . each of which ha...
like on a single product?
twilit field
I'm confused as to why
a product is equipped with 5 control switches
Matters
i'll send a photo of the question
4.78
twilit field
like on a single product?
I have no idea
cedar bone
I have no idea
my best guess is that you continously see products with 5 switches
whats the chance you see one with 5 functional switches before you see one with at least 2 defective switches
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half idol
Z/nz is isomorphic to zn for all n?
midnight plankBOT
Z/nz is isomorphic to zn for all n?
half idol
Or n has to be prime
spiral rock
for all n yes
tbh you could consider these the same objects
they have the same construction pretty much
half idol
What if n is prime
Actually i am confusing with some results or i am mixing please clarify both of these
spiral rock
this has nothing to do with n being prime or not
if n is prime both are finite fields
but it doesn't change much for your original question
half idol
Or field, ring?
spiral rock
if n is prime both are finite fields
.
half idol
0,1,2,3
2×2
Non zero elements making 0 element right
So integral domain fails
Hence not field
Am I right?
half idol
I forget how do we find zero divisors
I guess list divisors
Then remove primes?
1,2,4
1 and 4?
spiral rock
we don'f find them
zero divisors would be elements of your ring that "divide 0" i.e. some non-zero multiple of them is 0
like in this case 2 * 2 = 0
an integral domain has only 0 as a zero-divisor
half idol
Do we not check units?
runic hamlet
check in what sense. you have to show that every nonzero element is a unit
which is equivalent in the finite setting to showing that every nonzero element is not a zero divisor
although you dont usually do it that way
spiral rock
,rotate
grand pondBOT
spiral rock
yes
runic hamlet
well I mean its isomorphic to other groups, sure
half idol
U(8)~ z2×z2×z2
runic hamlet
its a finite abelian group so you can always write it as a product of cyclic groups
spiral rock
U(8) is of order phi(8)=4
runic hamlet
U(n) is cyclic iff n=1,2,4 or p^r or 2p^r where p^r is an odd prime power
spiral rock
its not
3 * 3 = 1
so 3 is of order 2 in U(8)
and also 5*5=1
and also 7*7=1
well basically U(8) is Z2xZ2
runic hamlet
I cant help you if you cant read
runic hamlet
if I meant that I would have written it
half idol
half idol
well basically U(8) is Z2xZ2
True. Klien geoup
I am sorry it is my English reading problem and understanding not yours
Sorry denascite
.close
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heady spire
hi
pearl hull
just post your question
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half idol
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half idol
How do I start it?
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hearty vine
My question is regarding functions
Theres a well known result that any function can be decomposed as a surjection followed by an injection
And our teacher asked us to think about whether the reverse is true, that any function can be decomposed as an injection followed by a surjection
I tried looking online about this but only founda things about the more well known result
I think i have a proof for it but id like to check if its actually true
The proof goes smth like this:
f:A->B, lets have h1:A->C and h2:C->B where |C| = |A|+|B|
h1 will be chosed to be some injection from A to C which we know exists due to the cardinalities
h2 will work like this: for every element b in B, if it has a preimage a in A, then send h1(a) to b, and if it doesnt then send any of the unpaired elements left in C to b, so in other words take elements from C\Im(h1), which we know we can do due to the cardinalities, and this means h2 will be a surjection and h2(h1(x)) = f(x) for all x in A
I think it works but im not quite sure
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My question is regarding functions
Theres a well known result that any function ...
lyric charm
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worthy kite
oh oops wrong channel sowwy