uhhh
#help-13
hot crag
crimson sedge
,w expand a^2+(a+2)^2+(a+4)^2+1
lyric narwhal
Oh this should be easy then
hot crag
and the middle term is always divisible by 12
so all that's left is 3a^2 + 21
now, what form are squares of odd integers always in
crimson sedge
either 3k or 3k+1
lyric narwhal
either 3k or 3k+1
Yes, but what about mod 4
crimson sedge
Oh
hot crag
odd perfect squares but are a specific type
they can always be represented as 8n + 1
try it
the proof is quite fun too
albeit simple
crimson sedge
Oh yeah n triangular
lyric narwhal
Oh yeah n triangular
Yess
hot crag
what do you get
crimson sedge
24n + 24
crimson sedge
Damn, I over complicated it 
crimson sedge
Thanks 👍
hot crag
it's just by chance that i decided to expand it and noticed this
np
anything else?
if not
cedar kilnBOT
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cedar kilnBOT
hot crag
have a good day
cedar kilnBOT
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wispy brook
Need help on this number 3 question
cedar kilnBOT
Need help on this number 3 question
idle tusk
,rotate
wraith daggerBOT
brave copper
yea i cant even read that
idle tusk
there's an english translation
karmic cairn
yea i cant even read that
theres english underneath it
brave copper
ah
wispy brook
its mixed up, mb
idle tusk
Need help on this number 3 question
which part do you need help with
wispy brook
a1 and a2
cedar kilnBOT
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
1
idle tusk
well
for both of them, you just have to do the transformations on just point V
let's try a1
idle tusk
1
X is a reflection along the line x=4. can you show where V ends up after X?
hint: this is where the line x=4 is
wispy brook
wait, i am taking the pic now
austere pawn
What a reflection is, is a transformation along a line where any point doesn't change its y-value and the x-value is the same distance but in the opposite direction of the line
In this case*
cedar kilnBOT
@wispy brook Has your question been resolved?
cedar kilnBOT
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Available help channel!
potent flower
1+6C5+7C5+8C5+...+49C5 = ..........
cedar kilnBOT
1+6C5+7C5+8C5+...+49C5 = ..........
potent flower
how to start with this
crimson sedge
whats the question
crimson sedge
the original question
potent flower
the original question
that's the original question
crimson sedge
wheres it from
upper laurel
youll need to post a screenshot or a picture of the question instead of typing it out
crimson sedge
i dont think any work sheet/book/site will only say that as its really ambiguous
potent flower
wheres it from
It's from my textbook but it's in Arabic 🥸
spiral stag
What does C mean?
crimson sedge
What does C mean?
combinatorics
grand shale
1+6C5+7C5+8C5+...+49C5 = ..........
50 c 6
spiral stag
ah
grand shale
hockey stick identity
potent flower
hockey stick identity
whatttttttttttttt?
grand shale
look it up
potent flower
what is kck + k+1 c k?
IDK
grand shale
try using algebra
vestal quarry
Find Range of f(θ)=2cos
2
θ−6sinθ+1
potent flower
Find Range of f(θ)=2cos
2
θ−6sinθ+1
go to an available channel
potent flower
try using algebra
Is this identity useful : nCm+nC(m-1)=(n+1)Cm ?
grand shale
Mhm
potent flower
like splitting every term
grand shale
now extend it
potent flower
now extend it
how
potent flower
what is kck + k+1 c k?
k+2??
potent flower
try it out
It will work for only the first two terms
it will become 7+ 7C5 + .......
grand shale
kck + k+1 c k + k+2 c k = k+1 c k+1 + k+1 c k + k+2 c k= k+2 c k+1 + k+2c k = k+3 c k+1
potent flower
kck + k+1 c k + k+2 c k = k+1 c k+1 + k+1 c k + k+2 c k= k+2 c k+1 + k+2c k = k+...
whyyyyy
I don't understand anything
ah now I get it
grand shale
Alr
grand shale
its not
potent flower
k+1 c k+1 + k+1 c k + k+2 c k= k+2 c k+1 + k+1 c k
grand shale
read ir properly
potent flower
read ir properly
k+1 c k+1 + k+1 c k = k+2 c k+1
that I know
then the last term is k+2ck
and the other term on the rhs is k+1ck
ha?
cedar kilnBOT
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Available help channel!
zealous mauve
Hi
zealous mauve
I need help finding determinatn of this matrix
with #9
Here are my steps
3 -6 9
0 3 4
0 1 5
3 -6 9
0 3 4
0 0 -11
then determinatn should be -11 * 3 * 3 which is -99
but book answer is 33
plucky owl
3 -6 9
0 3 4
0 0 -11
I believe it had to do with getting this third row here. You multiplied the previous third row by -3 which affected it by a factor of -3
mental trail
you multiplied row 3 by -3 didn't you?
and you didn't multiply the determinant by 1/(-3) in consequence
zealous mauve
I believe it had to do with getting this third row here. You multiplied the prev...
but doesn't adding one row to antoher by a multiple not change anything?
plucky owl
But you had to multiple the third row by -3
mental trail
but doesn't adding one row to antoher by a multiple not change anything?
but you didn't do R3 <- R3 + ...R2
you did R3 <- -3R3 + ...R2
mental trail
,rotate
wraith daggerBOT
mental trail
no it's not wrong. the theorem is correct in the sense that
zealous mauve
or did I misunderstand
mental trail
it doesn't say "when a multiple of one row A is added to a multiple of another"
it says "when a multiple of one row A is added to another"
zealous mauve
ah
cedar kilnBOT
@zealous mauve Has your question been resolved?
cedar kilnBOT
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Available help channel!
frail yoke
can someone explain the answer, I guesssed it but I have no idea
long arrow
well, do you recognize what kind of transformation does f(x-2) - 1 describe?
frail yoke
Shift up 1 I think and shift 2 right
long arrow
2nd part is correct, the 1st one is not
frail yoke
So down?
long arrow
-4 + 2 = -2
-5 - 1 = -6
frail yoke
Thank you!! I thought it was something else
And shift up would have been +# outside () right?
long arrow
yeah
glass spruce
prove that the two sequences converge to the same limit:
cedar kilnBOT
prove that the two sequences converge to the same limit:
south tundra
Note that the sequence of products (p_n defined by p_n = a_nb_n) is constant, this will simplify the sequences
glass spruce
i have no idea where to start on this
south tundra
Note that the sequence of products (p_n defined by p_n = a_nb_n) is constant, th...
.
Also, are you sure this is everything that's given?
glass spruce
yeah but then what
glass spruce
Also, are you sure this is everything that's given?
yes
south tundra
Anyway, you get that a_{n+1} = 1/2 (a_n + p/a_n) where p is the constant that I mentioned
Note p > 0
glass spruce
isnt p_n = a_1(b_1)?
south tundra
Using monotone convergence theorem, you can show a_n converges and find it's limit using the recursive formula
south tundra
isnt p_n = a_1(b_1)?
Yes
glass spruce
Using monotone convergence theorem, you can show a_n converges and find it's lim...
whats the theorem and what formula?
south tundra
The theorem states that if a sequence is monotone decreasing (resp. increasing) and bounded below (resp. bounded above), then it converges
glass spruce
ah
south tundra
Anyway, you get that a_{n+1} = 1/2 (a_n + p/a_n) where p is the constant that I ...
And I referred to this formula
twilit escarp
The theorem states that if a sequence is monotone decreasing (resp. increasing) ...
Does the adjacent sequences theorem would be efficient there ?
south tundra
Does the adjacent sequences theorem would be efficient there ?
Can't find the theorem in google, what does it state?
twilit escarp
Can't find the theorem in google, what does it state?
I translate it litteraly, it says that if theyre two sequence with on decreasing, one increasing, and the difference converges towards zero then they have the same limit
twilit escarp
0<bn<bn+1<an+1<an maybe should lead to something
Like this
daring kelp
Alternatively $0<=a_{n+1}-b_{n+1}=\frac{a_n-b_n}{2}\cdot\frac{a_n-b_n}{a_n+b_n}<\frac{a_n-b_n}{2}$. So the differences between the terms of the sequences is strictly decreasing and it's bounded below by 0, so the differences converge to 0 meaning they have the same limit.
wraith daggerBOT
Daniel
south tundra
Hmm, a_{n+1} - p/a_{n+1} is gonna be (a_n + p/a_n)/2 - 2p/(a_n + p/a_n)
Maybe could work but showing convergence to zero of the difference will be cumbersome
twilit escarp
Alternatively $0<=a_{n+1}-b_{n+1}=\frac{a_n-b_n}{2}\cdot\frac{a_n-b_n}{a_n+b_n}<...
Yeah this is the one i was thinking ty
glass spruce
how did you know that the sequences are monotone?
south tundra
**Daniel**
I don't think this shows convergence to 0
south tundra
Let s_{n+1} = 1 + 1/s_n, we have 0 < s_n < s_{n+1} but not s_n -> 0 whenever s_1 > 0
daring kelp
No surely it does
If we take the $i$'th term then we have $a_i-b_i<\frac{a_1-b_1}{2^i}$
wraith daggerBOT
Daniel
daring kelp
RHS can obviously be made arbitrarily small by taking i big enough
south tundra
That's better
daring kelp
**Daniel**
this is off-by-one but it doesn't matter
glass spruce
Anyway, you get that a_{n+1} = 1/2 (a_n + p/a_n) where p is the constant that I ...
how did you get this?
south tundra
Note that the sequence of products (p_n defined by p_n = a_nb_n) is constant, th...
According to this, b_n = p/a_n
south tundra
Solve the inequality a_{n+1} <= a_n, make sure each step is an equivalence so that writing them from bottom up will result in the proof of being monotone decreasing
But actually before that you need to know the lower bound in this case
glass spruce
But actually before that you need to know the lower bound in this case
of a_n?
0 ig?
south tundra
a_n > b_n is irrelevant to the solution I'm presenting
south tundra
a_n > p_n/a_n
Right, so far so good
glass spruce
how can i know the infimum?
south tundra
But you can't show that yet
glass spruce
Right, so far so good
but this is only true if its descending
it doesnt prove that its descending
south tundra
how can i know the infimum?
Okay so a useful way would be to assume that the sequence converges, find its limit, which is the infimum, and only then show that the limit converges via the theorem
glass spruce
but wouldnt its limit be the infimum only if its montone?
south tundra
In this case yes
glass spruce
so id need to prove its montone first
south tundra
You don't need to prove that what you found is the infimum when you are proving the theorem
south tundra
so id need to prove its montone first
Not necessarily
glass spruce
You don't need to prove that what you found is the infimum when you are proving ...
why do i need the infimum then?
south tundra
Again, assume the sequence to converge, you will find out the infimum and then show that this is the lower bound without the assumption of convergence
And then monotonocity follows
south tundra
why do i need the infimum then?
For this sequence it turns out you need the infimum as your chosen lower bound to show that it's monotone decreasing
glass spruce
right but how did you know that the sequence is monotone?
intuitively i mean
like how did you think of that?
south tundra
Hmm, a natural way would be to just guess that by looking at the first terms, then discover that you need a_n >= sqrt(p) for that to happen, so you try to see if you can prove sqrt(p) to be a lower bound first
My personal answer is that I've seen this family of sequences already in an exercise which contained the hints that I am suggesting
south tundra
Again, you don't need to prove that it's the infimum, for now think of this as guessing
glass spruce
guessing what?
south tundra
Either way, show that sqrt(p) is a lower bound and then that the sequence is monotone decreasing
glass spruce
what does power bound mean?
south tundra
guessing what?
That the limit indeed converges and sqrt(p) is the infimum
south tundra
what does power bound mean?
Nothing, I made a typo there, meant lower bound
glass spruce
ok and how do i show that its a lower bound?
south tundra
It can be done by induction
glass spruce
is there something shorter? lol
south tundra
Hmm the base case is not so obvious though
Actually
a_1 > b_1 can be used for the base case
glass spruce
oh yeah
actually why are we proving that a_n is monotone?
the question wants us to prove that they converge to the same limit first, then to find that limit
south tundra
To apply monotone convergence theorem
And then find the limit using the recursive formula
Once the limit of a_n is known, so will be the limit of b_n
glass spruce
then wed have to do the same thing for b_n no?
glass spruce
Once the limit of a_n is known, so will be the limit of b_n
why?
glass spruce
oh
south tundra
So you will have b_n -> p/(limit of a_n)
glass spruce
but then why does it want us to find the limit later?
i think im supposed to solve this differently
glass spruce
So you will have b_n -> p/(limit of a_n)
yeah
south tundra
There was a different solution proposed above which doesn't involve actually finding the limits
south tundra
Right
glass spruce
so i need to prove that p/a_n > sqrt(p)
south tundra
One thing I realized just now is that induction isn't necessarily, AM-GM suffices
glass spruce
what's that
south tundra
so i need to prove that p/a_n > sqrt(p)
Are you sure the sign of inequality should be that way? Show your work
south tundra
what's that
It states that the arithmetic mean of nonnegative real numbers is greater than or equal to their geometric mean
In the case of just two numbers, it says (x + y)/2 >= sqrt(xy) when x, y >= 0
glass spruce
glass spruce
In the case of just two numbers, it says (x + y)/2 >= sqrt(xy) when x, y >= 0
oh
right
south tundra
The last step doesn't seem valid, a_n >= sqrt(p) and a_n + p/a_n >= 2sqrt(p) does not imply sqrt(p) + p/a_n >= sqrt(p)
You could rather use the fact that p/a_n <= p/sqrt(p) = sqrt(p)
glass spruce
The last step doesn't seem valid, a_n >= sqrt(p) and a_n + p/a_n >= 2sqrt(p) doe...
yeah its not a step, its what i need to prove in order to prove the above
thats what i meant by the question mark
south tundra
I see
glass spruce
You could rather use the fact that p/a_n <= p/sqrt(p) = sqrt(p)
can you use the latex renderer?
south tundra
Anyway, this shows that sqrt(p) is a lower bound
wraith daggerBOT
sibber
south tundra
can you use the latex renderer?
[ a_n \ge \sqrt{p} \implies \frac1{a_n} \le \frac1{\sqrt{p}} \implies \frac p{a_n} \le \frac p{\sqrt{p}} = \sqrt{p} ]
wraith daggerBOT
A Lonely Bean
glass spruce
done i need to prove that its bigger?
south tundra
What's bigger than what?
glass spruce
p/a_n > sqrt(p)
south tundra
No, that's not even true
south tundra
p/a_n <= sqrt(p) and then a_{n+1} >= sqrt(p) follows
glass spruce
how does it follow
i need to prove that a_n + p/a_n >= 2sqrt(p)
and what i have is that p/a_n <= sqrt(p)
and a_n > sqrt(p)
south tundra
Hmm, you are right, I messed up somewhere
Weirdly enough I remember proving this by induction somehow
south tundra
Either way we can forget about induction and use am-gm
glass spruce
ill just use the average inequaloty
yeah
so now i know that a_n+1 has a lowerbound
how do i prove its monotone
south tundra
Solve the inequality a_{n+1} <= a_n, make sure each step is an equivalence so th...
.
You already did this and ended up with a_n >= sqrt(p), which has just been proven
glass spruce
oh right
south tundra
So we've shown the sequence to be monotone decreasing and bounded below
This means that it converges
The limit can be found by letting n -> infinity in the equation a_{n+1} = (a_n + p/a_n)/2 (you might want to name the limit with a letter)
south tundra
not that a_n >= sqrt(p)
Yes, this was assumed
But the principle induction tells us that what we've shown is enough to conclude a_n >= sqrt(p) for all n
glass spruce
oh its still induction
glass spruce
The limit can be found by letting n -> infinity in the equation a_{n+1} = (a_n +...
ok so now what
south tundra
Choose a letter to name the limit of a_n
glass spruce
t
south tundra
You have a_n -> t and a_{n+1} -> t, so t = (t + p/t)/2
Solve for t
Keep in mind that t is positive
glass spruce
is there a theorem that states that shifting sequences doesnt change the limit?
south tundra
Pretty sure there's no conventional name for it, it is too obvious for that
glass spruce
lol yeah but ive seen obvious things that do have named theorems
anyway
does limit arithmetic allow me to do lim bn = p/(lim an) ?
glass spruce
thanks a lot!
south tundra
Turns out a1 > b1 is not necessary for the problem
a2 >= b2 holds regardless of how a1 and b1 compare
glass spruce
why is that
south tundra
(a1 + b1)/2 >= 2a1b1/(a1 + b1)
(a1 + b1)^2 >= 4a1b1
a1^2 + 2a1b1 + b1^2 >= 4a1b1
a1^2 - 2a1b1 + b1^2 >= 0
(a1 - b1)^2 >= 0
Each step is an equivalence (as long as a1 + b1 is nonzero, which holds in this case)
And the last inequality is true for all a1, b1 in R
glass spruce
oh yeah
south tundra
Actually we didn't really make use of a1 > b1 anyway so nvm
glass spruce
we did in the induction base
south tundra
We ditched induction
glass spruce
isnt that how we proved a_n >= sqrt(p)?
glass spruce
yeah and we assumed that its true for a_n
south tundra
Which makes sense
south tundra
yeah and we assumed that its true for a_n
No
south tundra
Anyway, is everything clear?
cedar kilnBOT
cedar kilnBOT
Available help channel!
vernal field
what does this mean
cedar kilnBOT
what does this mean
vernal field
what does the E[X_i | X_1, ..., X_i-1] = X_i-1 thing mean? it's really confusing lol
thanks
astral bay
i'm guessing it's like
if $X_1, X_2, \dots, X_{i-1}$ have ``already happened'' (if we know what their value are)
wraith daggerBOT
bee [it/its]
astral bay
then the expected value of $X_i$ is the same as whatever the value of $X_{i-1}$ was
wraith daggerBOT
bee [it/its]
astral bay
...maybe
worn meteor
.help
cedar kilnBOT
Commands:
- clopen:
.close,.reopen - consensus:
.poll - factoids:
.tag - help:
.help - version:
.version
Type .help <command name> for more info on a command.
astral bay
does that make any sense? it's 2 am i'm not necessarily particularly coherent
worn meteor
worn meteor
Pls slove this
flint plinth
the basic points being: if you have the entire "history" x_1 through x_n, and you want to know the expected value of X_(n+1) given this history:
(1) only X_n is relevant
(2) and the expected value given the history is X_n itself, meaning your best guess as to what happens next is what happened most recently
cedar kilnBOT
@vernal field Has your question been resolved?
dull briar
5
dull briar
crimson sedge
How far have you gotten so far?
crimson sedge
Unless there is something I don't know about the question, I believe this is solvable
dull briar
so help me to solv it
crimson sedge
Show me what u did :p
crimson sedge
youre here to do it
Not what the servers for lmao
crimson sedge
To help u do it and ideally learn, we not allowed to give out answers u gotta work for em
dull briar
ok
so
i multiply 3 in first bracket so i have (3w-3\2)
then i multiply (3w-3\2) to (w+1\4)
and here where awaerthing goes wrong
resolver says
that
it should be
3w>2 - 3\4w - 3\2 w -3\8= 2w>2-3\4w
but i have
3w>2 + 3\4w - 3\2 w -3\8= 2w>2-3\4w same but here is positive
so when i pass -3\4w through ecual i cant delete them
do you understand,
???
crimson sedge
I believe the resolver- whatever that is lol- is right
dull briar
why i get his positive
3w>2 + 3\4w why is it positive ??? it should be negative but i get it positive
crimson sedge
then i multiply (3w-3\2) to (w+1\4)
3w * w = 3w^2
-(3/2) * w = -3w/2
3w * 1/4 = 3w/4
-3/2 * 1/4 = -3/8
Add that together and you get the resolvers answer
You likely swapped the signs of the 2nd and 3rd step, or at least that makes the most sense
Idk if this helps but personally I'd do the 3 last in the multiplying, that way there's less variables- it's closer to squaring a binomial
That's just a preference though
But I mean you did the hard part of the math tbh 😄 just a sign issue
crimson sedge
You're wrong I'm right 😎
So uh I 2nd guessed myself and you were right lol, you can see this because I did exactly what you did and got your same answer and just simplified it too far then misread something lmao mb, but yeah you are right that should be where you are
In that case idrk what resolver is doing 🔥
crimson sedge
3w * w = 3w^2
-(3/2) * w = -3w/2
3w * 1/4 = 3w/4
-3/2 * 1/4 = -3/8
Add that toge...
^
Damn I feel goofy now ngl 😩
dull briar
can i get another person please because there is a lot of slang a lot of words but not the exact reason why i am wrong
crimson sedge
No I'm saying you're right
dull briar
again
the answers are
and i cant get them
do you get my confusion write now?
@crimson sedge ????
sombody pelase help me
why should i wait
???
moderators,
crimson sedge
I thought you wanted someone else but sure I can
Your answer was correct I just made an error
The program thing whatever is wrong
crimson sedge
Your answer was correct I just made an error
Well I was right I just misread something, the steps I provided are all true and same as yours
crimson sedge
Well for starters, where did the (8w^2)/16 come from in the last 2 you wrote?
crimson sedge
Yeah no reason for that- in fact the terms with just w are the easiest cause they perfectly combine to 0
dull briar
no they not
crimson sedge
-3w/2 +2(3w/4)
dull briar
they are opsitive
both
thats the problem i cant get zero
i should sume them
they are both positive
crimson sedge
-3w/2 +2(3w/4)
Right it was 3/2 that was negative but still
When combined they make 3/2, and that cancels the negative 3/2 we already have
dull briar
3\4w +3\4w= 2 3\2w
crimson sedge
Did you mean to put the 2 before 3/2?
crimson sedge
Just checking not being degrading lol, though it would be 3/2: 3/4 + 3/4 = 6/4 = 3/2
crimson sedge
3w^2+ 3\4w+ 3\4w- 3\2w- 3\8 -2w^2=0
If you were working with the other 3/2 here, remember it is negative and thus cancels out
dull briar
If you were working with the other 3/2 here, remember it is negative and thus ca...
no they not
they cant they have 2 before them
becaouse we have 2 w
crimson sedge
I do not get what you mean, why would they not combine? We still have more to do but we would reach 3w^2 - 2w^2 - 3/8 = 0 after doing that
dull briar
crimson sedge
Where do you get 2(3/2) though? That is 4x the value of 3/4
crimson sedge
Just checking not being degrading lol, though it would be 3/2: 3/4 + 3/4 = 6/4 =...
As I said here 3/4 + 3/4 would get us 3/2, where did you get the 2nd one?
dull briar
w
w+w=2w
what the heck
w is connected to 3\4
we cant just conbine them without w
it should be negative
crimson sedge
we cant just conbine them without w
Oh I think I see what you are saying now, to verify you're saying
3/4 + 3/4 = 3/2
and
w + w = 2
so together they make 2(3w/2)
Am I right about your thoughts or no?
dull briar
here second exponent is negative
i have the same but second exponent postive
if i had it negative
but i dont have it negative
so i have trouble why?
@crimson sedge can you answer that question
crimson sedge
Variable not exponent (just so you don't confuse someone :p) and you have the same thing they just simplified to a further point which I'm trying to get you to do but you won't follow where I'm trying to take you, but I'll answer you anyway, if there's something stopping you from fully reading my messages let me know and we can try to work something out
On the left side you put
3x^2 + 3w/4 - 3x/2 - 3/8
They put
3x^2 - 3w/4 - 3/8
If we combine like terms in the work you were doing, you would get
3x^2 - 3w/4 - 3/8
Which is what I was explaining how to do
You have the same thing they just combined like terms
dull briar
dull briar
Variable not exponent (just so you don't confuse someone :p) and you have the sa...
oooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo
i seeeeeeeeeeeeeeeeeeeeeeeeeeee
they are soooooooooooo freeky
okey
i have
they are not correct
crimson sedge
Well what you have isn't correct
crimson sedge
Oh I think I see what you are saying now, to verify you're saying
3/4 + 3/4 = 3/...
Thats why I was asking this to try to understand where your error comes from
crimson sedge
But you didnt answer me still lol
dull briar
i am so confused that i cant answer yo
crimson sedge
So you said that you should multiply 3/2 by 2, because they both have a w, right? Therefore 3w/4 + 3w/4 = 2(3w/2)
Or am I wrong?
crimson sedge
yes]
Alrighty that helps, cause now I know its an issue with your fundamental algebra
I'll reference what you said before
w + w = 2w
That makes sense, but what about this?
2w + 2w = 8w
You can probably see there is only 4 w's, but with your logic there should be 8 since the w's double it, which is false, when adding variables you simply add their coefficients, so in that case the coefficients are both 2, so we simply add 2
2w + 2w = 4w
Which makes sense! So what about the first problem which your proposed? Remember that if a variable doesn't have a written coefficient, the coefficient is actually 1, so that is the same as
1w + 1w = 2w
dull briar
why,
crimson sedge
So same there, 3x/4 + 3x/4, you add the coefficients, 3/4 + 3/4 = 3/2, so that equals 3x/2
crimson sedge
but i get +
How so? We both agreed earlier on an equation
crimson sedge
Variable not exponent (just so you don't confuse someone :p) and you have the sa...
Here, so why do you get + now?
dull briar
i will jump in the window right now
dull briar
Alrighty that helps, cause now I know its an issue with your fundamental algebra...
who sad that 2w +2w= 8w
crimson sedge
who sad that 2w +2w= 8w
Your logic, if you had to multiply by 2 because both had w then you would get 8
That's essentially what you were doing to get 2 (3/2)
You added them together, but multiplying by 2 because of the variable is wrong, you should have only added the coefficients
crimson sedge
So you said that you should multiply 3/2 by 2, because they both have a w, right...
It was just an example to show why this is wrong :p
crimson sedge
Variable not exponent (just so you don't confuse someone :p) and you have the sa...
We had - though
Idk why you have + now
You even called it freeky
dull briar
you cant amagine in what codition i am now
i dont understand anything
AAAAAAAAAAA
I GOT IT I GOT IT
I GOT IYTTTTT
T
TTT
T
TT
crimson sedge
Ooh?
dull briar
YEEEEEES
BUT I STILL
dont understand
why
3\4x + 3\4x =3/2x and not to 2 3\2x
is that becaouse they are one type
and x is not envolved in action and just like "familiraty sign"
crimson sedge
Alrighty that helps, cause now I know its an issue with your fundamental algebra...
That's what I was talking about here, you might need to revise some of your basic algebra
Yeah because they're like terms it's easier
Say 3/4 = y
yx + yx = 2yx
2y = 2(3/4) = 3/2
If that helps any too
Just touch on some of your fundamentals
dull briar
not it makes m even more confused
crimson sedge
Thats fine lol youll figure it out
dull briar
.close
cedar kilnBOT
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acoustic quartz
cedar kilnBOT
acoustic quartz
yea
acoustic quartz
wdym
acoustic quartz
how did u figure out which function to compare it to so quickly tho
formal sierra
like basically for comparison
acoustic quartz
cause i tend to have trouble with finding which function i should use to compare
formal sierra
you remove all things that would be unesecary
i cant spell
so here 1 is very small right?
acoustic quartz
So stuff with more weight basically?
acoustic quartz
so the x in the numerator i can just make that 1 right
cause that has no weight in comparison to e^x?
or
acoustic quartz
wdym cancels out with the bottom?
formal sierra
x/x^3
acoustic quartz
x^-2
formal sierra
yes
acoustic quartz
but x^-2 in comparison to e^-x is nothing right so i can remove that?
formal sierra
theres not really like a perfect thing to compare it to
yeah you can remove it but it shouldnt matter
as long as you can know it will converge
do you know p series test?
acoustic quartz
oh
formal sierra
1/x does not converge
acoustic quartz
oh right
formal sierra
dont think about it too hard
if it looks like it converges in hindsight
it probadly does
and any exponents at bottom is almost automatically a converge
formal sierra
np
glacial jetty
I need to make sure my math is mathing
If I price something at 9.99 and they take 12% I will get 8.79
mighty shuttle
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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
rotund fox
I need to make sure my math is mathing
If I price something at 9.99 and they tak...
yeah i guess
mighty shuttle
I need to make sure my math is mathing
If I price something at 9.99 and they tak...
yeah
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potent ledge
Anyone know how to do this?
cedar kilnBOT
Anyone know how to do this?
potent ledge
Idk how to start
lone schooner
Anyone know how to do this?
what do you have to do find theta?
clear umbra
Idk how to start
tan = sin / cos
potent ledge
tan = sin / cos
I did that but then there’s two unknowns
potent ledge
what do you have to do find theta?
Use the trig identities
lone schooner
Use the trig identities
what's the end goal?
clear umbra
well after using it you should have 4sinθcosθ = sinθ right?
lone schooner
what's the end goal?
to prove it ? to find theta?
potent ledge
what's the end goal?
Ur supposed to find theta
potent ledge
potent ledge
well after using it you should have 4sinθcosθ = sinθ right?
I have this
clear umbra
which is = 0
potent ledge
Yes
clear umbra
but for a fraction to equal 0 the numerator has to be 0
so you have 4sinθcosθ - sinθ = 0
potent ledge
I see
clear umbra
then what should you do after
clear umbra
yep
potent ledge
Ohh I see tysm!
lone schooner
@potent ledge see for θ = 0,pi,2pi,...,npi it's satisfying
now go for values of θ in which sinθ and tanθ are not 0 in those values you can cut out sinθ and write it as 4 = secθ then
θ = sec^-1 4
this will save you the identities
thorn robin
How many pairs of solutions (m, n) exist for the equation n = √(100 − m^2)where both
m and n are integers?
apprently the answer is 7
however what i did was n^2 = 100-m^2
n^2+m^2 = 100
n, m can be
6,8
-6,8
6,-8
-6,-8
8,6
8,-6
-8,6
-8,-6
0,10
0,-10
-10,0
10,0
so there should be 12 solutions?
humble karma
They probably don't count pairs of the form (a,b) and (b,a) to be distinct.
thorn robin
oh ok, but if that's the case shouldn't there be 6 solutions?
sacred grail
(n, m)
- 6,8
- 6,-8
- 8,6
- 8,-6
- 0,10
- 0,-10
- 10,0
thorn robin
(n, m)
1. 6,8
2. 6,-8
3. 8,6
4. 8,-6
5. 0,10
6. 0,-10
7. 10,0
yes but what about
-6,-8 and -8,-6
sacred grail
they are not solutions
cedar kilnBOT
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crimson sedge
evaluate the limit $\lim_{x \to 0} \frac{\sin{x}}{1 - \cos{x}}$
wraith daggerBOT
Emploice Muswashans
crimson sedge
Shouldn't this evaluate to undefined
formal prism
Use L hopital
livid hound
why do you think this is undefined?
crimson sedge
we have a little rule to evaluate the limits that are indeterminate 🤗
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swift fossil
using series how can i find the fraction that represents 1.24123123123123... i am having trouble finding out what n is in 123(1/10)^(n-1) in the series
digital cliff
you sure your series expression is correct?
swift fossil
I dont think it is
crimson sedge
yeah seems like it to me as well
digital cliff
indeed it is not
swift fossil
but i know it has to be solved using geom series because that is what we learned
crimson sedge
can you send the entire question?
digital cliff
you can write it as 0.01+1.23123123...
the second bit is 1.23(1/?)^(n-1), what would ? be?
swift fossil
question 29
digital cliff
what is ?
swift fossil
i was able to do 23 with ease but the next one assigned is 29 which i am having trouble with
crimson sedge
I am not quite sure how we would solve this with geometric series
but I know another method
swift fossil
go ahead maybe its something we learned that i am not recalling
crimson sedge
let's say we have $x = 1.24 \bar{123}$
wraith daggerBOT
Emploice Muswashans
swift fossil
thats fine
crimson sedge
I'll jut write it in my notebook and send
swift fossil
ok thanks
digital cliff
you can write it as 0.01+1.23123123...
the second bit is 1.23(1/?)^(n-1), what w...
the way it wants is here, if you figure what the ? should be then just apply the sum to infinity formula
swift fossil
the way it wants is here, if you figure what the ? should be then just apply the...
let me try to write in my notebook and see
swift fossil
i think i have a thing for ya
yes
crimson sedge
yeah exactly
that's what I was doing
split of the 1.24 from the repeating decimals
lament iris
yep
swift fossil
the way it wants is here, if you figure what the ? should be then just apply the...
what is the ? im confused is it a multiple of 10?
crimson sedge
if you have 0.123123..... its pretty easy to convert that
digital cliff
what is the ? im confused is it a multiple of 10?
it is a multiple of 10 yes
swift fossil
1000
swift fossil
no but there are 2 decimal places and then 3
swift fossil
let me try in my notebook
crimson sedge
cedar kilnBOT
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rigid hazel
hey could i get help with putting this in the final form it wants in this question, i am on e and have done the rest of the questions
frozen spindle
,rotate
wraith daggerBOT
frozen spindle
you're good, just use the stirling approximation and take the limit
rigid hazel
whats the stirling approximation
and how do i get m^k in my numberator and not j^k
@frozen spindle
im still a bit confused
.close
cedar kilnBOT
molten compass
whats the stirling approximation
lmao i was just about to learn it
frozen spindle
I just meant the formula they gave you
(it's really stirling in disguise, I didn't realize they didn't tell you that)
rigid hazel
thats ok
frozen spindle
are you allowed to use the fact that e^x = (1+x/N)^N
if not, then you can derive that first
rigid hazel
i did not know that, how would you derive it?
does it get rid of my j^k in the numberator as in the formula i need an m^k
molten compass
isn't that just a dummy variable thing
rigid hazel
im only a first year physics student aha sorry must seem i dont know a lot
frozen spindle
e = lim (1+1/N)^N [N \to \infty], replace N with N/x and we get
e = lim (1+x/N)^(N/x) [N/x \to \infty], which entails by some limit rule that
e = (lim (1+x/N)^(N) [N/x \to \infty])^(1\x), if x is fixed then we can just take the limit and get
e = (lim (1+x/N)^(N) [N \to \infty])^(1\x), raising both sides to x we get
e^x = lim (1+x/N)^(N) [N \to \infty]
there might be a shorter way to do that, I'm going off of memory here
rigid hazel
cool
frozen spindle
but the point is that with this in hand, we do the little binomial expansion trick here
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tribal sonnet
|L(M)|
M is a turing machine
what does this means?
length of strings in L is less than 3?
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|L(M)|
M is a turing machine
what does this means?
length of strings in L is les...
@tribal sonnet Has your question been resolved?
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pastel grove
in real life..
vague valve
In a randomly shuffled 40-card deck, you draw 4 cards at the top to be your starting cards. You have the option to use a mulligan, where you discard a starting card in your hand and redraw another card. Mulligans do not need to be done all at once.
Discarded cards do not go back into the deck, and redrawn cards are drawn from the top of the remaining deck.
The probability of getting a specific hand with mulligans has been determined (#help-13 message), but the likelihood of specific hands that have duplicate cards hasn't been calculated yet.
The deck can consist of 40 identical cards, 1 of 40 different cards, or any combination in between. What’s the chance of drawing a specific hand?
fading jewel
"What is the chance you get any specific hand" So you just want 1/total?
I don't understand the language of this problem
What is a "specific hand" and does the composition of your deck have anything to do with it?
solid juniper
this question is too general honestly
fair geyser
agreed
fading jewel
I think so as well, and the language is weird
solid juniper
describe a more specific situation you want to find the probability of
dusky peak
Do i understand it correctly that we dont even know the starting deck?
fading jewel
"The deck can contain 40 of the same card, 1 of 40 different cards, and everything in between, but for most cases it will be 4x of 12 unique cards."
What does that even mean
Yeah we don't know
And then "most cases it will be 4x of 12 unique cards"
How often is "most cases"
How often do other cases occur?
It isn't specific enough to be solvable
fair geyser
you can only solve a specific problem, so it only make sense to give a specific problem, but they gave the specific example anyway
just ignore the rest
vague valve
hm
fading jewel
Lmao
vague valve
sorry i went to do something
fading jewel
hm
What is a "specific hand" and what do you mean by "The deck can contain 40 of the same card, 1 of 40 different cards, and everything in between, but for most cases it will be 4x of 12 unique cards."
That second statement is too general
vague valve
sorry about that
fair geyser
What is a "specific hand" and what do you mean by "The deck can contain 40 of th...
ignore that
vague valve
wdym
fading jewel
ignore that
I want to solve his problem
fair geyser
7 aces, 11 jacks, 3 queens, 16 kings, 3 10s. what is the chance that you'll able to get an ace, jack, and queen in your starting hand, you can redraw any of the 4 initial cards
vague valve
yeah
fading jewel
In a deck of 40 correct?
vague valve
yes
it would be nice to have an equation for any deck composition but for a start this makes sense
fading jewel
What is the chance you redraw AT LEAST 1?
vague valve
nah you have to have all 3 in your starting hand
fading jewel
Wait are we not drawing 4 cards?
vague valve
we draw 4 cards, and we can redraw any of the drawn cards 1 time each
fading jewel
Ok wait I think I get it you have 4 slots for 3 cards
vague valve
we want to have king ace jack in our starter hand, it doesnt matter if its KAJK, KAJJ, KAJQ, it just needs king ace jack
fading jewel
Probability of the initial draw having all three cards is
(7c1 * 11c1 * 3c1 * 19c1)/(40c4)
Do we agree?
solid juniper
solid juniper
i do not agree with this
you are overcounting in the numerator
because the 37c1 part includes kings aces jacks ace jack queen oops edited, 2 different problems stated but ig we're doing this one
fading jewel
because the 37c1 part includes ~~kings aces jacks~~ ace jack queen oops edited, ...
So 19c1?
40 - 11 - 7 - 3
Wait the last one could be another ace king or jack
We don't care right?
It just has to include at least 1 of each
solid juniper
not 100% clear but yea i'd assume that's allowed
so still not right if that's the case
solid juniper
well we care about not double counting things
fading jewel
Sorry if that sounds dumb, havent had to do combinatorics in a long time
solid juniper
Then why would it not be 37?
call the aces A1, A2, ..., the jacks J1, J2 etc
do you see that A1 J1 Q1 A2 and A2 J1 Q1 A1 will be counted as different hands?
or in other words, A1 J1 Q1 A2 gets counted twice
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vague valve
or in other words, A1 J1 Q1 A2 gets counted twice
do we or do we not want to count them twice
solid juniper
depends how you count them ig. in que's way, you don't want those to be counted as different
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muted bear
card bartering advanced
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@vague valve Has your question been resolved?
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@vague valve Has your question been resolved?
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@vague valve Has your question been resolved?
unborn cradle
Probability of the initial draw having all three cards is
(7c1 * 11c1 * 3c1 * ...
We need inclusion-exclusion
unborn cradle
yeah
we want (40 choose 4) - (combos without ace) U (combos without jack) U (combos without queen)
Let
A = set of combos without ace
J = set of combos without jack
Q = set of combos without Q
then |AUJUQ| = |A| + |J| + |Q| - |A∩J| - |A∩Q| - |J∩Q| + |A∩J∩Q|
and we want (40 choose 4) - |AUJUQ|
that's without mulligans, with mulligans it's a bit more complicated.
We probably need 3 cases:
drawing 0 of the cards we want, mulligan all 4 cards
drawing 1 of the cards we want, mulligan 3 cards
drawing 2 of the cards we want, mulligan 2 cards
vague valve
yeah that would make sense
cedar kilnBOT
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unborn cradle
Drawing 0 of the cards we want is just |A∩J∩Q|
unborn cradle
For drawing 1 of the cards we want there are 3 possibilities. Let's start with drawing Ace but not Jack or Queen
that's |J∩Q∩(A^c)| where A^c is the complement of A
If you think about it, these are all the cases where we don't draw Jack or Queen - all the cases where we don't draw Ace, Jack, or Queen.
|J∩Q∩(A^c)| = |J∩Q| - |A∩J∩Q|
so if we just draw ace we get:
|J∩Q| - |A∩J∩Q|
if we just draw jack we get:
|A∩Q| - |A∩J∩Q|
if we just draw queen we get:
|A∩J| - |A∩J∩Q|
now the only remaining case is drawing two of the cards we want
Let's say we draw Ace and Jack but not Queen
we have
|(A^c)∩(J^c)∩Q|
|Q| = |((A^c)∩(J^c))∩Q| + |(AUJ)∩Q|
so rearranging we get
|(A^c)∩(J^c)∩Q| = |Q| - |(AUJ)∩Q|
then |(AUJ)∩Q| = |A∩Q| + |J∩Q| - |A∩J∩Q|
so
|(A^c)∩(J^c)∩Q| = |Q| - |A∩Q| - |J∩Q| + |A∩J∩Q|
So drawing Ace and Jack but not Queen is:
|Q| - |A∩Q| - |J∩Q| + |A∩J∩Q|
drawing Ace and Queen but not Jack is:
|J| - |A∩J| - |J∩Q| + |A∩J∩Q|
drawing Jack and Queen but not Ace is:
|A| - |A∩J| - |A∩Q| + |A∩J∩Q|
then we can use a similar process to calculate the probability of redrawing what we want after the mulligan
unborn cradle
@vague valve does that make sense so far?
vague valve
<@308779860346798090> does that make sense so far?
well the ∩ stuff is a bit over my head but vaguely yes
unborn cradle
oh it means intersection
are you familiar with unions and intersections?
The idea is we want to transform all the unions to intersections because the size of the intersections are easy to calculate
|A| = ((40 - #aces) choose 4)
|A∩J| = ((40 - #aces - #jacks) choose 4)
|A∩J∩Q| = ((40 - #aces - #jacks - #queens) choose 4)
for example
vague valve
that makes sense
noble granite
If your question has been answered, please type .close to allow other people to use the help channels.
vague valve
it hasnt
unborn cradle
which part do you still need help with?
vague valve
well i haven't seen what the chance is for the example i gave
i was hoping for an equation that i could plug numbers into
unborn cradle
ok I can show you how to calculate it for that example
vague valve
alr
unborn cradle
maybe in like 45 minutes I'm working rn
vague valve
take your time, you've helped a bunch
unborn cradle
ok so this is going to be a really long calculation
we want
chance of drawing all 3 + chance of drawing 0 * chance of successful mulligan + chance of drawing 1 * chance of successful mulligan + chance of drawing 2 * chance of successful mulligan
chance of drawing all 3 = ((40 choose 4) - |AUJUQ|)/(40 choose 4)
Let's start with calculating the size of all the intersections
|A| = ((40 - 7) choose 4) = 40920
|J| = ((40 - 11) choose 4) = 23751
|Q| = ((40 - 3) choose 4) = 66045
|A∩J| = ((40 - 7 - 11) choose 4) = 7315
|A∩Q| = ((40 - 7 - 3) choose 4) = 27405
|J∩Q| = ((40 - 11 - 3) choose 4) = 14950
|A∩J∩Q| = ((40 - 7 - 11 - 3) choose 4) = 3876
40 choose 4 = 91390
so
|AUJUQ| = |A| + |J| + |Q| - |A∩J| - |A∩Q| - |J∩Q| + |A∩J∩Q|
= 40920 + 23751 + 66045 - 7315 - 27405 - 14950 + 3876
= 84922
chance of drawing all 3 = ((40 choose 4) - |AUJUQ|)/(40 choose 4)
= (91390 - 84922)/91390 = 6468/91390 = 3234/45695 ~ 7.077%
chance of drawing 0 is just |A∩J∩Q|/(40 choose 4) = 3876/91390 = 102/2405 ~ 4.241%
then chance of a successful mulligan after drawing zero is just the 3234/45695 ~ 7.077% we found previously
so we have 102/2405 * 3234/45695 = 329868/109896475 ~ 0.3002%
next we do chance of drawing 1 * chance of successful mulligan
which is actually 3 separate cases
Just Ace = |J∩Q| - |A∩J∩Q| = 14950 - 3876 = 11074
Just Jack = |A∩Q| - |A∩J∩Q| = 27405 - 3876 = 23529
Just Queen = |A∩J| - |A∩J∩Q| = 7315 - 3876 = 3439
Then the chances are
11074/91390 = 5537/45695 ~ 12.12%
23529/91390 = ~ 25.75%
3439/91390 = 181/4810 ~ 3.763%
respectively
now we're going to mulligan 3 cards, so we'll divide by 39 choose 3 instead of 40 choose 4
and we have to adjust the calculations for drawing what we want accordingly as well
so say we draw just Ace then mulligan 3
We want ((39 choose 3) - |J* U Q*|)/(39 choose 3)
J* and Q* are modified J and Q since we're only drawing 3 and the pool of cards is different
we can use inclusion exclusion again
|J* U Q*| = |J*| + |Q*| - |J*∩Q*|
|J*| = (39 - 11) choose 3 = 3276
|Q*| = (39 - 3) choose 3 = 7140
|J*∩Q*| = (39 - 11 - 3) choose 3 = 2300
so |J* U Q*| = |J*| + |Q*| - |J*∩Q*| = 3276 + 7140 - 2300 = 8116
also (39 choose 3) = 9139
so ((39 choose 3) - |J* U Q*|)/(39 choose 3)
= (9139 - 8116)/9139 = 1023/9139 ~ 11.19%
Now we multiply that by the chance of drawing just Ace:
5537/45695 * 1023/9139 = 5664351/417606605 ~ 1.356%
now we have to do the same thing again for just Jack and just Queen lolol
are you getting the general idea now at least?
I can continue if needed
vague valve
yeah i think so lol, its just very complicated
honestly i knew it was complex but i didn't think it would be this much, but it does make sense
the example i gave isn't very applicable to what im using it for but i was thinking you could figure out the steps for it and then just apply it to the other situations
unborn cradle
There might be a way to simplify it, I'm not sure
the mulligans are really what makes it complicated
It would probably we worth coding a simulation that you can just plug the numbers into
calculator, not simulation
though a simulation might give a good approximation
vague valve
calculator, not simulation
you made one for the meter calculations, although im assuming this one would take quite a bit more time
unborn cradle
not neccessarily, is it always a 3 card combo?
vague valve
nah, it could be 1 2 3 or 4
although 4/1 seem easier to calculate
for the majority of cases it'll be a 2/3 card combo, and 4 of them in the deck per card. theres always 40 cards in the deck though
cedar kilnBOT
@vague valve Has your question been resolved?
vague valve
@unborn cradle are you here?