what if codomain cardinality 5 and domain cardinality 4
#help-39
stoic imp
cinder flower
then there are no surjective functions
stoic imp
then there are no surjective functions
i see because then its not a function
cinder flower
stoic imp
we cant have two different output for same input
cinder flower
sure
stoic imp
ok
anyways by this analysis of yours we conclude that we are talking about bijective functions
cinder flower
yes, any surjective function here is bijective
stoic imp
ok about the conditions
modern talon
(as an aside, you have surjective function between two sets of the same size so it has to be injective)
cinder flower
one thing i wanted to mention is that if you look at outputs here, they look like a permutation of 1234. the permutation 1432. every bijection corresponds to a permutation in this way
so you can kinda just remove all the function language from the problem and just view it as counting permutations of 12345678 with some constraints
which i would personally find easier
stoic imp
how
cinder flower
how what?
stoic imp
4 x 4 x 6!
stoic imp
one thing i wanted to mention is that if you look at outputs here, they look lik...
the order matters
cinder flower
we want to count permutations of 12345678 for which
- the first two numbers sum to an odd number
- the 3rd number is less than the 4th
- the 4th number is less than the 5th
stoic imp
, w 446!
jolly parrotBOT
cinder flower
what is this a calculation of?
stoic imp
what is this a calculation of?
nothing, it is a brain fart
cinder flower
one thing i wanted to mention is that if you look at outputs here, they look lik...
by this correspondence here (well that was for 4 element sets but same idea)
stoic imp
we want to count permutations of 12345678 for which
1. the first two numbers su...
the problem is the consecutive numbers
cinder flower
the what
stoic imp
we want to count permutations of 12345678 for which
1. the first two numbers su...
the 2 and 3 conditions
cinder flower
are you saying it’s a problem as in you don’t know how to work with it, or that you don’t agree this is a correct reframing of the problem?
stoic imp
former
cinder flower
well the positions 3,4,5 need to be assigned three numbers. and once you choose the three that will go there, there is only one possible order to put them in
and for any three numbers, you can assign them there, in consecutive order
once you fix positions 1 and 2 and you have 6 numbers left, you can choose any 3 of them to go in positions 3,4,5
pearl pondBOT
@stoic imp Has your question been resolved?
ancient mauve
help me
ionic merlin
help me
Try drawing a free body diagram
And break all the forces in their x-y components
Show me ur work
pearl pondBOT
@ancient mauve Has your question been resolved?
ancient mauve
Try drawing a free body diagram
i know but do i look at it as a system or just one block
idle hedge
i know but do i look at it as a system or just one block
How does the system accelerate and how does the 25 kg block accelerate? That should tell you whether to consider the system or one block
ancient mauve
How does the system accelerate and how does the 25 kg block accelerate? That sho...
they accelerate the same
idle hedge
So, system or one block?
Hint: also consider external forces acting on the system VS the block
pearl pondBOT
@ancient mauve Has your question been resolved?
proud turret
How can I determine if this is an odd or even function when it's not passing through (0,0)?
feral sedge
Test some points using the definitions of even and odd
pearl pondBOT
@proud turret Has your question been resolved?
sharp smelt
pearl pondBOT
sharp smelt
I'm interested in how I'd find $E\left(X_1^2(1-X_2)^3\right)$
jolly parrotBOT
wai
sharp smelt
Like when you find the pdfs you find X_1 and X_2 are indepndent
but that's about it
Like what I've been able to come up with is
$E(X_1^2(1-X_2^3-3X_2+3X_2^2))$
jolly parrotBOT
wai
sharp smelt
the issue is then finding $E(X_2^3)$ quickly ( without explicitly finding the 3rd moment of $X_2^3$. And ofcourse this asumes $X_1^2$ and $X_2^3$ are indepndent, which I doubt
jolly parrotBOT
wai
modern talon
Like what I've been able to come up with is
on the right track
indep of x1 and x2 imply transformations are als indep
so their squared and cubed are indep
as for a quick calculation
ah
well
you just gotta calc it i think
sharp smelt
I just have to bite the bullet there I suppose
modern talon
indeed
modern talon
sorry twin
pearl pondBOT
sharp smelt
.reopen
pearl pondBOT
✅ Original question: #help-39 message
sharp smelt
sorry twin
hmm?
sharp smelt
.close
pearl pondBOT
pearl pondBOT
Available help channel!
stoic imp
pearl pondBOT
jolly parrotBOT
Renato
cursive wraith
what have you tried?
stoic imp
I dont know
wary bane
this is a counting problem yh?
one way of counting jumps out at me as maybe being faster / doable
stoic imp
care to elaborate?
wary bane
i was hoping you'd have at least some ideas 😭
|A| = |A n B| + |A n B'|
Because "not injective" is awkward to count on its own, I feel.
stoic imp
|A| = |A n B| + |A n B'|
I see
what does A represent
what does B represent
B is injective?
wary bane
|A| = |A n B| + |A n B'|
these are sets
this was meant to be my hint for how I think it might be faster to count, that you might try
wary bane
B is injective?
the functions with that domain/codomain which are injective, sure.
stoic imp
what is A
wary bane
**Renato**
you have 2 conditions here
if the universal set is all functions with that domain and codomain
one condition is the functions which satisfy f1 < f3 < f5
the other condition is the functions which satisfy not injective
i'm suggesting you let A be the functions which satisfy f1 < f3 < f5
B be the functions which satisfy injectivity
wary bane
|A| = |A n B| + |A n B'|
and then counting these 3 sets may turn out to be doable
cursive wraith
how do I count A and AnB
well, A is "functions f:{1,...,10}->{1,...,12} such that f(1) < f(3) < f(5)"
Let's start by choosing f(1), f(3) and f(5) for those
stoic imp
12 x 11 x 10
cursive wraith
not exactly
that would imply choosing f(1), f(3) different from f(1), and f(5) different from the other two
so they are 2 by 2 distinct, but that doesn't tell you if the inequality chain is verified
Start by choosing 3 distinct numbers, without saying which one is f(1), which one is f(3) and so on
How many ways are there to choose 3 distinct numbers among {1,...,12} (without ordering them)?
wary bane
i'd also point out if you're struggling to count these.
You can try a toy example like
f : {1, 2, 3} > {1, 2, 3, 4, 5}
and then have the condition be just
f1 < f2
for example
stoic imp
How many ways are there to choose 3 distinct numbers among {1,...,12} (without o...
12 x 11 x 10
cursive wraith
12 x 11 x 10
that's not it
choosing without ordering means choosing a subset of 3 elements
not a list
wary bane
if this was {1, 2, 3} and we were choosing 2 distinct numbers you would be answering 3 x 2
What are the 6 pairs you think we can choose here?
I reckon there are only 3.
3x2 = 6 would be with ordering.
(1, 2)
(2, 1)
(2, 3)
(3, 2)
(1, 3)
(3, 1)
stoic imp
3 x 3 x 2 x 1
cursive wraith
3 x 3 x 2 x 1
?
cursive wraith
3 x 3 x 2 x 1
what does each number mean here
stoic imp
well
for the triples
(3,2,1)
(3,1,2)
(2,1,3)
(2,3,1)
(1,2,3)
(1,3,2)
3 x 2 x 1 x 2
wary bane
there's miscomms
wary bane
How many ways are there to choose 3 distinct numbers among {1,...,12} (without o...
How many ways are there to choose 3 distinct numbers among {1,...,12} (without ordering them)?
there are 2 questions you could be answering here
this one
wary bane
if this was {1, 2, 3} and we were choosing 2 distinct numbers you would be answe...
or this one
stoic imp
ok
stoic imp
> How many ways are there to choose 3 distinct numbers among {1,...,12} (without...
12 x 11 x 10 x 2
wary bane
no
cursive wraith
12 x 11 x 10 x 2
where does the extra 2 come from now
wary bane
with order is counting the tuples of 3 elements (without any element repeated)
without order is counting the sets of 3 elements
cursive wraith
Your previous logic was:
"I choose the first number, then the second, then the third"
And that gave you 12 * 11 * 10
wary bane
(1, 9, 3) is distinct from (9, 1, 3)
{1, 9, 3} is the same as {9, 1, 3}
your first answer of 12 x 11 x 10 is counting the tuples of 3 elements (without any element repeated)
stoic imp
I need help
cursive wraith
That's what we're trying to do
stoic imp
Your previous logic was:
"I choose the first number, then the second, then the t...
what's wrong with that
cursive wraith
what's wrong with that
When I ask you to give me distinct numbers without ordering them
And you do this
"I choose the first number, then the second, then the third"
Can you understand that something is wrong with the way you're doing it?
no, it's worse
stoic imp
When I ask you to give me distinct numbers without ordering them
but you said distinct
cursive wraith
but you said distinct
I did
that's not my problem
When I ask you to give me distinct numbers without ordering them
This is my problem
I'm asking you to pick numbers without ordering them, and what you do is "pick the first, then the second, then the third"
you're ordering them when I'm asking you not to
thus overcounting
because for any triple {a,b,c}
there are 6 orderings possible (= 3!)
abc
acb
bac
bca
cab
cba
cursive wraith
6 x 12 x 11 x 10
you're now overcounting by a factor of 6 * 6
stoic imp
if 12 * 11 * 10 is overcounting by a factor of 6
how do you know
that 12 x 11 x 10 is overcounting by 6
cursive wraith
that 12 x 11 x 10 is overcounting by 6
because of what I just said
I want you to pick a subset {a,b,c} of 3 distinct numbers
cursive wraith
abc
acb
bac
bca
cab
cba
And you're actually counting that subset 6 times
because you decided to count them with ordering
when I asked them unordered
stoic imp
I see
this is what other helper was saying
in permutation you are counting tuples
in combinations you are counting sets
stoic imp
abc
acb
bac
bca
cab
cba
so if order doesn't matter I am counting the same set 6 times already
wary bane
it is for the reason of this confusion I avoid names like 'permutation' and 'combination', but rather refer to them as "order matters" or "order doesn't matter"
and then make sure I am not confusing myself with small toy examples of 3 to 4 elements
cursive wraith
if 12 * 11 * 10 is overcounting by a factor of 6
Well now we can go back to this
12 * 11 * 10 is counting 6 times the number of ways to choose {a,b,c} unordered from {1,...,12}
so the actual number of ways to choose {a,b,c} unordered is ...?
stoic imp
so the actual number of ways to choose {a,b,c} unordered is ...?
12 x 11 x 10 / 6
also please ping me
if replying
wary bane
also please ping me
to be precise, it is 12 x 11 x 10 / (3 x 2 x 1)
or even
12C3 = 12! / (9! x 3!)
wary bane
Let's start by choosing f(1), f(3) and f(5) for those
well, A is "functions f:{1,...,10}->{1,...,12} such that f(1) < f(3) < f(5)"
Let's start by choosing f(1), f(3) and f(5) for those
so that is the number of ways we can choose {f1, f3, f5}
and since order doesn't matter
we just simply put them in ascending order to get the inequality sign, as desired
does that make sense?
eg. we have {10, 5, 11}
that means we've chosen
f(1) = 5
f(3) = 10
f(5) = 11
the next step is given we fix f1, f3, f5, how many functions are there? So choose where the other numbers in the domain go
wary bane
if replying
^
(i might not be here later)
pearl pondBOT
@stoic imp Has your question been resolved?
cursive wraith
the next step is given we fix f1, f3, f5, how many functions are there? So choos...
@stoic imp we chose f(1), f(3), f(5) (and there are 12 * 11 * 10/6 ways to do so)
How many ways are there to choose the other images
(Number of ways to choose f(2)) = ...
(Number of ways to choose f(4)) = ...
etc
stoic imp
the next step is given we fix f1, f3, f5, how many functions are there? So choos...
7 numbers
from the 10 numbers from the domain
we are fixing 3
7 are left
wary bane
wait what
stoic imp
f is injective or not
wary bane
well, A is "functions f:{1,...,10}->{1,...,12} such that f(1) < f(3) < f(5)"
We were here. So f need not be injective
You can therefore map the other 7 numbers to any of the 12
stoic imp
yes
stoic imp
12 x 11 x 10 /6
stoic imp
please ping if you are replying
stoic imp
So in total, functions which satisfy just f1 < f3 < f5 would be?
12 x 11 x 10 /6 x 12^7
wary bane
ok
next.
The next step is to find the functions which satisfy f1 f3 f5 AND are injective. Which means choosing f1 f3 f5 the same way as before, but choosing the images of the other 7 numbers differently.
(since f1 < f3 < f5, the same method of choosing them as before will satisfy injectivity, thankfully)
stoic imp
|A| - |A n B| = |A n B'|
wary bane
|A| - |A n B| = |A n B'|
Yes this is now A n B
wary bane
12 x 11 x 10 /6 x 12^7
this is A
stoic imp
Yes this is now A n B
which one is which
wary bane
The next step is to find the functions which satisfy f1 f3 f5 AND are injective....
this is A n B
A = {functions with f1 < f3 < f5}
B = {injective functions}
A n B is therefore {injective functions with f1 < f3 < f5}
wary bane
which one is which
make sense?
stoic imp
ok
wary bane
|A| = 12 x 11 x 10 / 6 x 12^7
wary bane
(12 x 11 x 10 / 6) choose f1 f3 f5 order doesnt matter
aka choosing the set {f1 f3 f5} which must be 3 distinct elements
12^7 is choosing f2 f4 f6 f7 f8 f9 f10 (which need not be distinct)
=====
wary bane
The next step is to find the functions which satisfy f1 f3 f5 AND are injective....
So for A n B you can choose {f1 f3 f5} the same way as before
but u have to choose the others differently in order to satisfy injectivity
stoic imp
(12 x 11 x 10 / 6) choose f1 f3 f5 order doesnt matter
but we do care about the order or not
wary bane
but we do care about the order or not
do not care = doesnt matter
stoic imp
wary bane
{f1, f3, f5} is a set of 3 elements
it is only in this way we can count the triples which satisfy f1 < f3 < f5
If our set is {5, 2, 4} this uniquely determines what f1 f3 f5 must be
(aka theres a bijection between the assignment, make sense?)
stoic imp
there's a bijection because of the strict inequalities
wary bane
{5, 2, 4} corresponds uniquely to the assignment
f1 = 2
f3 = 4
f5 = 5
wary bane
it is the same assignment we d give to {2, 4, 5}
that is why we count with order not mattering
stoic imp
ok
stoic imp
12^7 is choosing f2 f4 f6 f7 f8 f9 f10 (which need not be distinct)
ok
I get A now...
what about AnB
wary bane
A n B is therefore {injective functions with f1 < f3 < f5}
Do you agree choosing f1 f3 f5 in the same way as we did before makes sense?
wary bane
then the challenge is to choose f2 f4 f6 f7 f8 f9 f10
choose these 7 so we have injectivity.
wary bane
explain 9! / 2
oh its 9! / 2! ?
sounds right, ill think about it for a bit
yeah i agree
stoic imp
(12 x 11 x 10 / 6) x ?
before the wildcard "?" is just addressing f(1) < f(2) < f(3)
then, if we need injective functions we already used 3 of the 12 spots, (12-3=9), (bijection because of strict inequalities)
furthermore, we got 9 spots left for the outputs, however our domain has only cardinality 10, and we used 3 spots, we have 7 spots more
9! / 2!
wary bane
So fully, i would write
$\frac{12!}{9!3!}\frac{9!}{2!}$
jolly parrotBOT
wary bane
And finally, i think u can address the original question, with this.
So the first keypoint was considering the common counting technique which is complements
You do this when you think the complement is maybe easier to count
(in this case, i feel it very much is)
stoic imp
|A| = 12 x 11 x 10 / 6 x 12^7
|AnB| = 12 x 11 x 10 / 6 x 9! / 2
|AnB^c| = 12 x 11 x 10 / 6 x 12^7 - 12 x 11 x 10 / 6 x 9! / 2
stoic imp
,w 12 * 11 * 10 / 6 * 12^7 - 12 * 11 * 10 / 6 * 9! / 2
jolly parrotBOT
stoic imp
or compute it, if u need to
i appreciate the help shuwi
counting is hard
,w 12 * 11 * 10 / 6 * 12^7 - 12 * 11 * 10 / 6 * 9! / 2 = 12^7 * 12 C 3 - 12 C 3 * 9!/2!
jolly parrotBOT
jolly parrotBOT
stoic imp
,w 12 * 11 * 10 / 6 * 12^7 - 12 * 11 * 10 / 6 * 9! / 2 = 12^7 * 220 - 220 * 9!/2!
jolly parrotBOT
pearl pondBOT
pearl pondBOT
Available help channel!
coral pecan
pearl pondBOT
coral pecan
Gng can someone like explain this to me step by step
I know it's like super basic stuff but I never understood column vectors , translations and all those
And like adding column vectors
ancient slate
Ok which one do you know understand
coral pecan
.reopen
pearl pondBOT
✅ Original question: #help-39 message
pearl pondBOT
@coral pecan Has your question been resolved?
ancient slate
both
ashen ivy
coral pecan
I was doing my 101 side quests
ancient slate
xD
coral pecan
listen carefully
Go ahead
ancient slate
AB is a vector and AC is one too
coral pecan
Wait can you give me like 2 minutes I put my noodles into the microwave
I'm so sorry
ancient slate
sure
coral pecan
I feel super bad but it's gonna get burnt
ancient slate
nah its np
coral pecan
@ancient slate
ancient slate
so AB and AC are vectors
coral pecan
Hmm
ancient slate
We get BC by adding BA and AC
coral pecan
That's why it's given as column vector yah?
ancient slate
BA is -AB
coral pecan
We get BC by adding BA and AC
I don't understand this part
ancient slate
mhm
ancient slate
ok wait
ancient slate
adding together 2 vectors are just putting one to the end of another
in this case
AB + BC = AC
coral pecan
adding together 2 vectors are just putting one to the end of another
But not this
coral pecan
Wait let me look at this for like 2 minutes maybe I'll get it
ancient slate
sure
coral pecan
Or any shape ?
ancient slate
idk what you mean
coral pecan
its one of the basics of vectors
Ya I'm just now learning basics of vector 🥲🥲🥲
ancient slate
coral pecan
idk what you mean
Like AB+BC=AC
ancient slate
same here AB + BC + CD = AD
coral pecan
same here AB + BC + CD = AD
Oh wait
Okk makes sense I think
Makes sense
Is it like the small sides added up is the big side or I'm I saying a bunch of bs
coral pecan
I feel like I'm one of those people that's really frustrating to teach maths to
ancient slate
I feel like I'm one of those people that's really frustrating to teach maths to
thats not true
like look there
coral pecan
I'm looking
coral pecan
I feel like I kinda got it for triangle but any other quad I don't think I'll do it properly
You get what I mean 😭😭
coral pecan
Ok
coral pecan
Yes ?
I know I think
(-2,-7 ) column vector is BA
And then AC IS 10,11
Column vector
ancient slate
then we can write them in so $-\binom{2}{7}+\binom{10}{11}$
jolly parrotBOT
Roy
ancient slate
that is $\binom{-2+10}{-7+11}$
jolly parrotBOT
Roy
coral pecan
Yes
ancient slate
so a is good
coral pecan
I want to learn how to use latex
coral pecan
so a is good
Hmm
ancient slate
lets do one thing at time
coral pecan
lets do one thing at time
Yes sorry I'll try to focus on one thing at a time
Continue
ancient slate
do you understand or not
coral pecan
Yes
ancient slate
ok now lets prove ABE is straight line
for ABE to be a straight line BE should be the multiple of AB
coral pecan
for ABE to be a straight line BE should be the multiple of AB
Yes
coral pecan
B+BE=E
Yes this makes sense
But why can I not do B -BE
Will that make E Like negative
ancient slate
we want to know BE because we know E
jolly parrotBOT
Roy
ancient slate
which is a multiple of $\binom{2}{7}$
jolly parrotBOT
Roy
ancient slate
no
coral pecan
No ?
ancient slate
$\binom{58}{203} = 29 * \binom{2}{7}$
jolly parrotBOT
Roy
coral pecan
Where did we get 29 from 😕
ancient slate
we just know
coral pecan
we just know
Wdym
ancient slate
because $\binom{58}{203} / \binom{2}{7} = 29$
jolly parrotBOT
Roy
coral pecan
Isn't this what I said tho
ancient slate
oh idk
ancient slate
i might have misunderstood
ancient slate
no because the question was to prove
coral pecan
What else should I do
ancient slate
we proved it
here is the full proof.
B + BE = E
BE = E - B = $\binom{58}{203}$
AB = $\binom{2}{7}$
$\binom{58}{203}$ = 29 * $\binom{2}{7}$ --> BE = 29*AB
pearl pondBOT
jolly parrotBOT
Roy
coral pecan
do you want to learn TEX?
Yes
But I'm in the middle of folding my clothes
And class
crystal dew
#latex-help would be a good place
ancient slate
xD
coral pecan
<#840667252793802752> would be a good place
Okk I'll check it out tomorrow
pearl pondBOT
Available help channel!
inland laurel
Let
$$
A=\begin{bmatrix}
1 & 0 & 0\
2 & 1 & 0\
3 & 2 & 1
\end{bmatrix}
$$
If $u_1$ and $u_2$ are column matrices such that
$$
A u_1=\begin{bmatrix}1\0\0\end{bmatrix}, \quad
A u_2=\begin{bmatrix}0\1\0\end{bmatrix},
$$
then find $u_1+u_2$.
jolly parrotBOT
BlackidoZΣ
inland laurel
a) mat [ 1 -1 -1]
b) mat [ -1 1 0]
c) [-1 1 -1]
d) [ - 1 - 1 0 ]
how to do dis
A(u1 + u2) = [ 1 1 0]
tropic saddle
well you could just check what A*(all of those options) is
fresh juniper
x²+y²-10x+16
can any one help to find the center and radius from this eqn
toxic lichen
x²+y²-10x+16
can any one help to find the center and radius from this eqn
!occupied
pearl pondBOT
x²+y²-10x+16
can any one help to find the center and radius from this eqn
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).
toxic lichen
A(u1 + u2) = [ 1 1 0]
you can find u1+u2 using this info
say by gaussian elimination
glass meadow
Only one option works for the first row anyway
inland laurel
Only one option works for the first row anyway
how do you check it
okay by options
glass meadow
A * [a b c] = [a x y]
The first row of A is 1 0 0, so whatever the first value of u_1 + u_2 is, it will be the first value of the result A*(u_1+u_2)
pearl pondBOT
pearl pondBOT
Available help channel!
thin badger
Anybody can help me with this math problem??
pearl pondBOT
Anybody can help me with this math problem??
toxic lichen
image uploading?
thin badger
image uploading?
I send it
pearl pondBOT
@thin badger Has your question been resolved?
hexed vortex
heya people so i dont understand how we are getting our delta... more precicely how are we getting our
constant to bound x^2+x+1
plush bramble
heya people so i dont understand how we are getting our delta... more precicely ...
which line specifically ?
"getting out delta" is vague
hexed vortex
getting the delta* misstyped it
hexed vortex
which line specifically ?
so it says let delta = min {1, eplsion/7}
how did we come to this conclusion?
plush bramble
they got delta by doing these steps and working backwards
a similar problem worked out here with more explanation:
https://www.milefoot.com/math/calculus/limits/DeltaEpsilonProofs03.htm
hexed vortex
a similar problem worked out here with more explanation:
https://www.milefoot.co...
even in that example where did the 5 spawn in from?
plush bramble
there's 2 examples which one are you reading
hexed vortex
there's 2 examples which one are you reading
second, non linear
plush bramble
because x is approaching 5
hexed vortex
ohhhhh okay okay
hexed vortex
a similar problem worked out here with more explanation:
https://www.milefoot.co...
dude thank you so so sos so much
this clears out a lot and i mean a lot of issues i was having
.close
pearl pondBOT
waxen basin
hey
pearl pondBOT
Available help channel!
south ice
Hello
south ice
I have a question
Regarding equations of planes
Say you have an equation of a plane such as 2x+3y-z6 = 9
You know that the normal vector would be <2,3,-6>
Now you’re left off with <x,y,z> as the other arbitrary vectors
So now all together we have <2,3,-6> dot <x,y,z> = 9??
But then it can’t be 9 if they must be orthogonal
It has to be 0
west sapphire
<x,y,z> isn't generally a vector in the plane, so <2,3,-6> doesn't have to be orthogonal to it
south ice
But the equation is equal to 9 which I’m assuming implies that the dot product of those two is equal to 9 which doesn’t entail orthogonality
west sapphire
but you can rewrite the plane equation in the form <2,3,-6> . <x-x0, y-y0, z-z0> = 0, where (x0, y0, z0) is a point in the plane
so <x-x0, y-y0, z-z0> is a vector parallel to the plane, and you get that it's orthogonal to the normal vector
south ice
Yes but it’s equal to 9 tho
west sapphire
what is equal to 9
west sapphire
yes?
west sapphire
but you can rewrite the plane equation in the form <2,3,-6> . <x-x0, y-y0, z-z0>...
see this
west sapphire
i don't think you have read anything i wrote above
south ice
i don't think you have read anything i wrote above
I did bro
I’m just wondering how we can make it equal to 9
west sapphire
what is "it"?
south ice
but you can rewrite the plane equation in the form <2,3,-6> . <x-x0, y-y0, z-z0>...
This
west sapphire
first find a point on the plane
south ice
If the equation is equal to 9, then that ⇒ they set the dot product equal to 9
Right ?
west sapphire
for example (6,1,1) is a point on the plane
thus <2,3,-6> . <6,1,1> = 9
so you can write:
<2,3,-6> . <x,y,z> = <2,3,-6> . <6,1,1>
or equivalently
<2,3,-6> . (<x,y,z> - <6,1,1>) = 0
or
<2,3,-6> . <x-6, y-1, z-1> = 0
the vector <x-6, y-1, z-1> is an arbitrary vector parallel to the plane
and it is orthogonal to <2,3,-6> as desired
south ice
<2,3,-6> . (<x,y,z> - <6,1,1>) = 0
So then
= <2,3,-6> dot <x-6,y-1,z-1> = 0
= 2x-12 + 3y-3 -6z +6 = 0
⇒ 2x + 3y -6z = 9
Ohhhh
west sapphire
yep!
south ice
So it’s all about where your origin is right
west sapphire
right exactly
<x,y,z> has its origin at <0,0,0> so it's not parallel to the plane (unless the plane passes through the origin)
whereas <x - x0, y - y0, z - z0> has its origin at <x0, y0, z0>, so if you choose the latter to be any point in the plane you're good
south ice
@west sapphire thanks for the help man
west sapphire
sure yw
south ice
.close
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lunar mist
im stuck with a task, it says create a graph like this
pearl pondBOT
im stuck with a task, it says create a graph like this
lunar mist
but it doesnt say what the graph is
how can i know what the function of this graph is?
versed mica
im stuck with a task, it says create a graph like this
🤔
can you show the full question
this looks like a right vs left riemann sum
versed mica
it never says you need to graph any particular function?
it just says any generic strictly decreasing function?
its a rough sketch
it doesn't ask to plot points or anything
lunar mist
it says create a figure
versed mica
ok and?
lunar mist
similar to the one in example 4
versed mica
in this example they did any strictly increasing one
so just draw one that strictly decreases
lunar mist
yes how do i do that
versed mica
why?
versed mica
well of course it needs to be a function
lunar mist
do you know a function that can work?
versed mica
you have to do this in geogebra?
lunar mist
yes
versed mica
you don't know any decreasing functions?
do you know any increasing functions?
if you do, then just negate it
lunar mist
i dont know any in my head
versed mica
how about something easy like an exponential
versed mica
why are you complicating it with other things
just do something of the form b^-x
,w plot 2^-x
jolly parrotBOT
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inland laurel
pearl pondBOT
toxic lichen
for part a you took the starting position of the projectile as (0,0) yes?
inland laurel
yes
toxic lichen
ok then the distance from the thrower is sqrt(x^2+y^2)
find d/dx (x^2+y^2) in terms of theta and try to figure out a condition to make it always positive
inland laurel
ok then the distance from the thrower is sqrt(x^2+y^2)
Is it fine to assume the parabolic path of projectile to be straight to say sqrt(x^2+y^2) by Pythagoras theorem
toxic lichen
we are not assuming anything to be straight
inland laurel
dr/dx = 2vcos(theta)*t
toxic lichen
"distance to the thrower" is the straight line distance from (0,0) to (x,y) only, trajectory does not matter
it does not mean distance traveled
toxic lichen
sure
inland laurel
dr/dx = 2vcos(theta) * t
this is positive when theta is positive and it is in the 4th quadrant
toxic lichen
dr/dx = 2vcos(theta) * t
this looks suspicious
can you show me what you got for r itself
inland laurel
r = x^2 + y^2
dr/dx = 2x
and from previous part x = vcos(theta) * t
toxic lichen
r = x^2 + y^2
dr/dx = 2x
and from previous part x = vcos(theta) * t
incorrect
dr/dx = 2x + 2y dy/dx
inland laurel
oops i forgot we are considering y in our function
toxic lichen
and what was y(x)?
toxic lichen
ok good. work out dy/dx for this
toxic lichen
dr/dx = 2x **+ 2y dy/dx**
then put into this
inland laurel
while differentiation of y(x) i am not able to identify which varible like x , theta is constant and which one is to be differentitate
toxic lichen
while differentiation of y(x) i am not able to identify which varible like x , t...
dy/dx
inland laurel
dy/dx = tan(theta) - gx/vcos^2(theta)
toxic lichen
why did v^2 become v?
anyway i guess you will have some kind of cubic with no constant term (if my foresight is correct)
so you can divide it by x then painstakingly work out the discriminant
maybe easier to do like y=Ax-Bx^2 and work out dr/dx in that way
inland laurel
dy/dx = theta(theta) - (g/(2v^2cos^2(theta)) (2x)
simplifying
dy/dx = tan(theta) - gx/[v^2 cos^2(theta]
pearl pondBOT
@inland laurel Has your question been resolved?
abstract rampart
pearl pondBOT
abstract rampart
,rotate
jolly parrotBOT
abstract rampart
How do I use integration by parts in this
abstract rampart
Write it as lnxx^-4
But how is the integral of lnx xlnx -x
I dk how to integrate lnx sorry
abstract rampart
The lnx and x^-4
Yeah ik we get x^-4 and lnx
quick venture
,w integrate lnx
abstract rampart
jolly parrotBOT
quick venture
That's the integrate
abstract rampart
That's the integrate
But how did we get that value: xlnx - x
Reverse chain rule states ∫u dx = ux - ∫x dx
If we take u=lnx
xlnx - (x²/2) should be the ans
Wait no
It's du
How do you ∫ x d(lnx)
toxic lichen
Write it as lnxx^-4
not with this lack of brackets please 😭 😭
summer imp
But how did we get that value: xlnx - x
You can compute the integral of ln(x) itself using integration by parts.
Now in this case however, you don’t need to compute it as it’s the wrong choice of u and dv.
You want to pick u and dv such that it overall gets simpler when you compute du/dx or int(dv), and in particular differentiating ln(x) gets simpler than integrating it.
abstract rampart
You can compute the integral of ln(x) itself using integration by parts.
Now i...
Why is the choice or u and dv wrong
Isn't that how we define the reverse chain rule
u=lnx and v=x
summer imp
I mean choosing u = 1/x^4 and dv =ln(x) dx is wrong (and would require you to compute the integral of dv, aka ln(x)
abstract rampart
I mean choosing u = 1/x^4 and dv =ln(x) dx is wrong (and would require you to co...
No but I integrated x^(-4) separately that was easy
What i don't get is when applying the reverse chain rule to lnx I can't get the desired integrated value
Instead I have to solve stuff like ∫x d(lnx)
summer imp
Instead I have to solve stuff like ∫x d(lnx)
If u = ln(x) you have du = dx/x
abstract rampart
If u = ln(x) you have du = dx/x
Oh
But still
How do you integrate x with respect to d(x/x)
jolly parrotBOT
Azyrashacorki
summer imp
Again I'm not totally following here because while this is how you would usually show that $\int \ln(x) dx = x\ln(x) - x + C$ and this is a useful thing to know how to do, your original problem doesn't require you to integrate $\ln(x)$ at any point.
jolly parrotBOT
Azyrashacorki
summer imp
And the fact that you mentioned you "integrated x^(-4) separately" and are now integrating ln(x) on its own makes me think something went wrong along the way, so it would be helpful to see what you've done so far working on the problem.
pearl pondBOT
@abstract rampart Has your question been resolved?
tender pulsar
you could've just... integrated 1/(x^4) and left lnx as is
pearl pondBOT
you could've just... integrated 1/(x^4) and left lnx as is
pearl pondBOT
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royal galleon
https://math.stackexchange.com/questions/1776315/intermediate-value-for-derivative-apostol-text. I am having trouble seeing how we can be sure that k will take every value from a to b. Wouldn't that only work is k = x everytime?
solar ember
Is your query about the asnwer posted to that question?
royal galleon
Is your query about the asnwer posted to that question?
No. It is about the photo in the question
worldly jewel
f' takes on every value from a to b because g takes on every value from a to b and g(x) = f'(k) for some k
sorry (f(b)-f(a))/(b-a) not b
royal galleon
f' takes on every value from a to b because g takes on every value from a to b a...
Yes but for f' to take it for every value in (a,b) don't we need it to work for all k in (a,b)
worldly jewel
wdym work for all k in (a, b)
the logic is like this
Suppose I give you some ξ ∈ (f'+(a), f'-(b)) and you wanna find some k such that f'(k) = ξ
Suppose for the first case that ξ lies between f'+(a) and (f(b)-f(a))/(b-a), then with g defined as in the picture you have g(x) = ξ for some x
Then, since g(x) = (f(x)-f(a))/(x-a), by the mean value theorem on the interval from a to x, you know (f(x)-f(a))/(x-a) is equal to f'(k) for some k lying between x and a
Hence, we have produced a k such that f'(k) = ξ as required.
solar ember
Yes but for f' to take it for every value in (a,b) don't we need it to work for ...
No, we don't need to show that every $k$ in $(a, b)$ gets used. We only need to show that for every value $c$ (where $c$ is a number between the endpoint derivatives), there exists at least one $k$ such that $f'(k) = c$.
jolly parrotBOT
Ajay
worldly jewel
The other case where ξ lies between (f(b)-f(a))/(b-a) and f'-(b) is similar.
royal galleon
One sec let me read all of this
royal galleon
Suppose I give you some ξ ∈ (f'+(a), f'-(b)) and you wanna find some k such that...
Why are you adding a and then subtracting b from f'?
jolly parrotBOT
royal galleon
Suppose for the first case that ξ lies between f'+(a) and (f(b)-f(a))/(b-a), th...
Just making sure this x is in (f'(a),g(b))
worldly jewel
no this x is in (a, b)
you know that g(a) = f'+(a), g(b) = (f(b)-f(a))/(b-a), and g is continuous
so you apply the regular intermediate value theorem to conclude the existence of some x in (a, b) such that g(x) = ξ
grave mirage
https://math.stackexchange.com/questions/1776315/intermediate-value-for-derivati...
I don't think that's written anywhere thi
royal galleon
So why are we doing the mean value theorem up to x and not just always from a to b
worldly jewel
because you want to equate this g(x) = (fx-fa)/(x-a) to some f'
royal galleon
But g(x) is defined from a to b?
worldly jewel
if you apply it from a to b you get some j such that f'(j) = (fb-fa)/(b-a), which is useless
worldly jewel
But g(x) is defined from a to b?
yes, but the goal is to show that this g(x) = (fx-fa)/(x-a) is equal to f'(k) for some k
royal galleon
if you apply it from a to b you get some j such that f'(j) = (f**b**-fa)/(**b**-...
Well doesn't that guarantee one value?
worldly jewel
well yes, it guarantees in particular that the point (fb-fa)/(b-a) = f'(j) for some j
but that's just one point in the whole interval
royal galleon
I'm still confused on how we make it that any point in the interval works
worldly jewel
Ok well backtrack a lil
The goal is to show that f'(k) = ξ for some k
We have shown that ξ = g(x) = (fx-fa)/(x-a) for some x
right?
royal galleon
We have shown that ξ = g(x) = (fx-fa)/(x-a) for some x
Well haven't we shown it for all x from a to b because on intermediate value theorem
worldly jewel
What do you mean?
for all x what
Do you understand why, ξ = g(x) = (fx-fa)/(x-a) for some x ?
royal galleon
Is it not because of the intermediate value theorem. If xi is in (f'(a), g(b))
Then intermediate value confirms that it is an out of g(x)
worldly jewel
Okay, so
we have
- g is continuous on [a, b]
- g(a) = f'+(a)
- g(b) = (fb-fa)/(b-a)
- and we assume that ξ is lying between f'+(a) and (fb-fa)/(b-a)
royal galleon
Yes
worldly jewel
Yes okay, so we apply the IVT to conclude...
royal galleon
That one of the ouputs of g(x) on that interval equals xi
worldly jewel
Yes that's right, but the way you word it is imprecise
"one of the ouputs of g(x) on that interval" is not a proper way to say it
royal galleon
Xi is guaranteed to be an output?
worldly jewel
You want to say, "ξ is attained by some x in (a, b)", or "g(x) = ξ for some x in (a, b)", etc
Always remember to bring your quantifiers with you
royal galleon
Understood
worldly jewel
Okay, so now we have this
- ξ = g(x) = (fx-fa)/(x-a) for some x
royal galleon
Yes
worldly jewel
So consider now the restriction of f to the interval [a, x]
Since it is differentiable, we can apply the mean value theorem to conclude...
Remember, the goal is to produce some k such that f'(k) = (fx-fa)/(x-a) = g(x) = ξ
royal galleon
Well we can produce that. Just by the mean value theorem. Since it guarantees us f'(k) = f(x) -f(a)/x-a for k in (a,x). I am just having trouble seeing how that concludes it. Since all we have done is shown that we can for one value k we can have f' = g(x)
worldly jewel
Okay so you know now that f'(k) = g(x) = ξ for some k in (a, x) ⊂ (a, b) right?
royal galleon
Ok so why is it proper subset and not just subset since x could be b
worldly jewel
⊂ can mean proper or non-proper subset fyi
it is weaker than < vs ≤
it's inconsistent but it's how most people use it 
royal galleon
I'm confused. I thought the one with the line under was more general???
worldly jewel
,tex some people use $\subset$ to mean $\subseteq$
jolly parrotBOT
royal galleon
What? Why? Isn't that just confusing.
worldly jewel
mhm
royal galleon
But that is just as good as if we applied mean value theorem on [a,b]
worldly jewel
No it is not just as good
Because you need f'(k) to equal this value ξ that we are given
royal galleon
I think my confusion stems from how can we treat f'(k) as f'
worldly jewel
There is no "treating f'(k) as f'"
Ok I think you may be confused with the high level structure of the proof
So let's say you have to prove some statement like this: prove that all values between 0 and 1 are attained by some function h
Here's the strategy: you can imagine an advisory (me), producing any number, ξ, lying between 0 and 1
You, the defender, have to describe a procedure, that produces a k, corresponding to this ξ, such that h(k) = ξ
Does that make sense?
royal galleon
So I always need to be able to find a k such that h(k) equal xi
worldly jewel
yes exactly
let's try this with h(x) = x²
I produce ξ = 1/4, can you give me a corresponding k?
royal galleon
K equal 1/2
worldly jewel
exactly
royal galleon
Or negative 1/2
worldly jewel
can you do this with any ξ ∈ [0, 1] that I give you? why or why not? can you describe a general procedure to produce a k, given an ξ?
royal galleon
So for x^2 given xi I sqrt it and find k
worldly jewel
yes exactly
so you have described a way, given any ξ ∈ [0, 1] from the advisory (me), for the defender (you) to produce a k so that h(k) = ξ
this is the general structure of this proof
It is very similar to ε-δ proofs that you have done in the past
royal galleon
We are saying that we can always find the ouputs of f' to equal xi
worldly jewel
mhm
royal galleon
Wait why are we restricting to [a,x]?
royal galleon
We want to show that the ouputs of the deirvate will also be xi given that our input is from a to x which is a subset of a to b but then doesn't a to b contain all subsets. Meaning that if we wanted f' to take the value xi given any value from a to b we should just use (a,b)
worldly jewel
No...
You want f'(k) to equal (fx-fa)/(x-a)
if you apply the MVT to f on [a, b], that would just give you some j for which f'(j) = (fb-fa)/(b-a), which is not what we needed
You may want to revise the statement of the mean value theorem
royal galleon
if you apply the MVT to f on [a, b], that would just give you some j for which f...
So this is because xi can equal any g(x) not just g(b)
And the mean value theorem tells us that the f'(k) that I am producing is the one that is equal to g(x) which is equal to xi
worldly jewel
So this is because xi can equal any g(x) not just g(b)
yes
worldly jewel
And the mean value theorem tells us that the f'(k) that I am producing is the on...
exactly
royal galleon
Ok so let me make sure I am understanding everything correctly. We have xi which is in (f'(a),f'(b)). Next we are looking at g(x) which is continuous on [a,b]. Now if xi is defined on (g(a),g(b)) then the ivt guarantees it. Meaning that there exists an x in (a,b) such that g(x) =xi and g(a)≤g(x)≤g(b). Next I want to show that given this xi I can produce a k where k in (a,x) such that f'(k) = xi. The mvt for f on [a,x] tells us that I can find a f'(k) = f(x)-f(a)/x-a =g(x) = xi. Since [a,x] is a subset of [a,b]. This means that given a xi defined by g(x) from a to b I can find you a corresponding k on one of the subitnervals that will give me the corresponding xi value. And since I can do that on [a,b] that means that f' takes on xi for some k on [a,b]
worldly jewel
yes this is essentially correct, except for a detail at the beginning
what we have proven is that any ξ between f'(a) and (fb-fa)/(b-a) is attained
but we want to show that any ξ between f'(a) and f'(b) are attained, so how do we remedy this?
the answer is to repeat the proof to show that any ξ between (fb-fa)/(b-a) is also attained
royal galleon
So go from g(b) to f'(b)
worldly jewel
since these two intervals (f'a to (fb-fa)/(b-a), and (fb-fa)/(b-a) to f'b) together cover the interval from f'a to f'b, we are done
royal galleon
I have one concern when I was writing it up I wrote that if xi is in the interval g(a) to g(b). That isnt guaranteed right?
Which is why we have to look at the other one too
worldly jewel
yes you basically prove 2 separate lemmas
worldly jewel
what we have proven is that any ξ between f'(a) and **(fb-fa)/(b-a)** is attaine...
this one is the one we have proven
pearl pondBOT
worldly jewel
Ok I see. Thx so much. Everything seems to be very clear now.
No worries. As an exercise you can try to write out the proof of the second lemma in detail
It is quite similar to the first
royal galleon
Will do. Thx again
pearl pondBOT
Available help channel!
warm bobcat
Did not understand this
pearl pondBOT
Did not understand this
warm bobcat
How is reflex POR 200?
glacial bluff
by the inscribed angle theorem and the size of PQR, which is subtended by the arc PR with reflex POR as its central angle.
indigo iris
Angle at the centre is double the angle at the circumference on the same arc, so ∠POR = 2 × 100° = 200° (this is a reflex angle).
The interior central angle in triangle POR is 360° − 200° = 160°.
OP = OR (radii of the circle), so triangle POR is isosceles.
Angles in triangle POR add to 180°, so 2∠OPR + 160° = 180°.
Solving gives ∠OPR = 10°.
stuck oracle
Angle at the centre is double the angle at the circumference on the same arc, so...
!nosols
pearl pondBOT
Angle at the centre is double the angle at the circumference on the same arc, so...
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.
indigo iris
oops
warm bobcat
So it doesn't have to be always like this?
glacial bluff
no.
pearl pondBOT
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Available help channel!
fading nexus
Could somebody just check these for me in the most ify on four
fresh solar
Could somebody just check these for me in the most ify on four
3 is incorrect and 4 is correct upto 4x=π\6+ 2πk
After that I couldn't understand your handwriting in 4
1 and 2 seems correct
pearl pondBOT
@fading nexus Has your question been resolved?
fading nexus
3 is incorrect and 4 is correct upto 4x=π\6+ 2πk
Wait where did I go wrong with 3
fading nexus
After that I couldn't understand your handwriting in 4
For my final answer in four it was x=pi/1.5+1/2pik , x=-pi/1.5+1/2pik
fresh solar
Wait where did I go wrong with 3
Sin7π/6 is -1/2 and cos inverse of -1/2 is 2π/3
fresh solar
For my final answer in four it was x=pi/1.5+1/2pik , x=-pi/1.5+1/2pik
Oh wait
Why is it 1.5?
It should be just 24
fading nexus
Sin7π/6 is -1/2 and cos inverse of -1/2 is 2π/3
Ohhh I had it drawn correctly but wrote it down wrong
fading nexus
ohhh
pearl pondBOT
@fading nexus Has your question been resolved?
plush bay
Q. A carton contains 25 light bulbs, eight of which are defective. What is the probability that if a sample of four bulbs is chosen, exactly two of them will be defective?
Here is my approach, I was hoping someone could tell me if this is correct or not.
I use the binomial approach here, total events are 4 and exactly 2 of those events must satisfy the condition that the bulb is defective
here is the formula:
$ \binom{4}{2} {\frac{8}{25}}^2 {\frac{17}{25}}^2 $
plush bay
ah how do I get the bot to mathify this
alright
$\binom{4}{2} {\frac{8}{25}}^2 {\frac{17}{25}}^2$
jolly parrotBOT
Nox
plush bay
here we go
the squared is on the entire fraction
$\binom{4}{2} (\frac{8}{25})^2 (\frac{17}{25})^2$
jolly parrotBOT
Nox
plush bay
### Q. A carton contains 25 light bulbs, eight of which are defective. What is t...
just wanted to know if this is the correct approach
plush bay
### Q. A carton contains 25 light bulbs, eight of which are defective. What is t...
okay so I gave it more thought and I figured I can't use the binomial distribution here since the probability changes when the first bulb is drawn
rough stream
Though the binomial distribution will be very close, there is a better choice
plush bay
$\binom{4}{2} \frac{8}{25} \frac{7}{25} \frac{17}{25} \frac{16}{25}$
jolly parrotBOT
Nox
plush bay
is this it?
plush bay
Though the binomial distribution will be very close, there is a better choice
mhm
oh wait no, the denominator
$\binom{4}{2} \frac{8}{25} \frac{7}{24} \frac{17}{23} \frac{16}{22}$
jolly parrotBOT
Nox
plush bay
this might be it
rough stream
That might actually be it.
However, there is a distribution for this, called the hypergeometric distribution
Where you draw from a population without replacement, and want to find n successes
pearl pondBOT
pearl pondBOT
Available help channel!
viscid turtle
3^x > -3
the answer is x∈R, right? an exponent base should not be negative or so i think
glass meadow
a^x > 0 for all real x and for all real a > 0
quick venture
For the highlighted bit why are the values different?shouldn't they be the same
dense jasper
For the highlighted bit why are the values different?shouldn't they be the same
$x+2$ vs $x-2$
jolly parrotBOT
Civil Service Pigeon
warped violet
For the highlighted bit why are the values different?shouldn't they be the same
Yes
Should be x+2
pearl pondBOT
@quick venture Has your question been resolved?
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