ima just delete it
#help-19
timber spoke
swift lake
timber spoke
nvm
pulsar tiger
Can someone help me better understand the Monotonic Convergence Theem?
digital kestrel
you can usually show a sequence like that is bounded and monotone with induction
then to compute the limit we have $$\lim_{n\to\infty} x_{n+1} = \lim_{n\to \infty} 2 - \frac{1}{x_n}$$
clever fjordBOT
swashbuckler
digital kestrel
you do need to know (x_n) converges first to do that, which is the point of showing that it converges
digital kestrel
**swashbuckler**
but you can like, solve for $\lim_{n\to\infty} x_n$ here
clever fjordBOT
swashbuckler
bitter sleet
then to compute the limit we have $$\lim_{n\to\infty} x_{n+1} = \lim_{n\to \inft...
How does that help 👀
digital kestrel
it will be the root of a quadratic equation. if it has 2 roots you can hopefully rule one out
digital kestrel
How does that help 👀
do my last 2 messages explain that enough
pulsar tiger
Wait - I am confused. I don't think we have learned this approach
Is this some calculus approach here?
Or Real Analysis?
narrow crypt
Wait isn’t this the third uve posted this😭
bitter sleet
do my last 2 messages explain that enough
Ah i see now
digital kestrel
also $\lim_{n\to\infty} x_{n+1} = \lim_{n\to\infty} x_{n}$
clever fjordBOT
swashbuckler
pulsar tiger
If so, what resource can you point me to because I have no idea what is going on
digital kestrel
Wait - I am confused. I don't think we have learned this approach
what approach do you mean?
pulsar tiger
Wait isn’t this the third uve posted this😭
I am preparing for my exam. If I faill, I have to hang myself
So, stakes are high, my guy
pulsar tiger
what approach do you mean?
Everything you have told me here.
narrow crypt
Oh dear
digital kestrel
induction? the solving for the limit part?
pulsar tiger
Is this MCT?
pulsar tiger
induction? the solving for the limit part?
I am not familiar with either.
What is the induction part?
digital kestrel
i was saying you could use induction to show the sequence is bounded and monotone
which is what you need to show to apply the monotone convergence theorem
pulsar tiger
i was saying you could use induction to show the sequence is bounded and monoton...
How do I do that?
Induction proofs vary across the board.
digital kestrel
i cannot help right now as i need to go eat breakfast
bitter sleet
Induction proofs vary across the board.
Set a bound and do induction on the statement that the sequence is less than the bound
weary pelican
Like 2 - 1/3 for the Lower bound?
this will not be a satisfying lower bound
it will only work for x1 and x2
pulsar tiger
What is the sequence for this?
So, what does this start off as?
2- 1/3
That is x2
x3 = 2 - 1/(2 - 1/3)
Like this?
weary pelican
that would be x3 yes
anyways, to find the correct lower bound
Sometimes it's ok to look at "what happens if it does converge"
to extract the correct bounds
pulsar tiger
Would I be looking at the x_n+1 term for oconvergence?
weary pelican
**swashbuckler**
which is what was explained here
so if you call l = lim(x_n)
then find an equation that describes l
pulsar tiger
I am trying so hard
I am crying guys.
I can't do this
No matter how I try
I donn't want to kms but it is the only option if I fail
I can't do this
weary pelican
It's alright, you're doing well up until now
pulsar tiger
If I fail, I can't continue in this program
I just want to graduate and make my parents proud
I am too stupid
Spent 6 years in school only to not be able to pass this one class
Almost killing myself to learn this congtent but I can't
God wangts me to die
I should hang myself and not botherr no more people in this life
AW failrur like me doesn't derserv to liove
weary pelican
Take some time off for yourself, you're doing your best but working with this mentality is counterproductive
pulsar tiger
I can't take time off. I did that yesgteredya
Nothing helps
I can't grasp this no matter how hard I try
I honestly think I can't do this it is impossibe
weary pelican
We've been on this for only 7 mins
pulsar tiger
It is like a bunc hof static is gong on in my head
weary pelican
Let's go back to earlier
so what links the next term to the previous term is this statement given by the exercise
pulsar tiger
What links?
weary pelican
well if we have the previous term
we can use this equation to get the next
and so on
pulsar tiger
The previous term is xn = 3?
weary pelican
The previous term is xn = 3?
well that's the first term specifically, x1
every term in the sequence is the previous term to some next term
every xn has an xn+1
weary pelican
well yeah here the link is explicit
x_(n+1) can be computed
directly in terms of x_n
pulsar tiger
I am trying.
I am slow
weary pelican
It's ok, take your time
pulsar tiger
Is it bad that I don't know?
weary pelican
I've seen people three years in post-highschool not know what convergence means
pulsar tiger
I am a senior in University
weary pelican
and somehow they got good grades
pulsar tiger
This is my last class.
weary pelican
so I'm pretty confident in you if they can do it
pulsar tiger
If I can't pass, then everything was for nothing. There is no purpose to keep living
I can't tell what the lower bound it
I only know the upper bound
weary pelican
I can't tell what the lower bound it
everything in its due time
weary pelican
Can someone help me better understand the Monotonic Convergence Theem?
You had the correct idea that MCT was gonna be helpful
pulsar tiger
Wait, 1/xn = 1/x
weary pelican
But this problem is essentially hard and I don't blame you
weary pelican
Could it be that the Lower bound is 1?
alr so you stumbled on the correct lower bound, but I'd like to explain how you rigorously get to it
If you examine the first few terms, you get 3, then 2 - 1/3 = 5/3, etc...
this sequence seems to be decreasing at first glance
so if we were to do MCT
we need to show decreasing behavior, and a lower bound
and what better lower bound than what should be the limit
weary pelican
so what links the next term to the previous term is this statement given by the ...
so you take this statement
and you make "n go to infinity"
and you would get $l = 2 - \frac 1l$
clever fjordBOT
Raphaelisius Maximus MMIII
pulsar tiger
Is it because the lim(xn) = 0
weary pelican
Is it because the lim(xn) = 0
lim(xn) = l here
we're supposing it converges but we don't know the limit yet
pulsar tiger
Oh
weary pelican
so we call it l
and we solve
remember we're only doing this to get the most accurate lower bound
all of this preemptive talk we're doing is only something you'd put on your draft
weary pelican
**Raphaelisius Maximus MMIII**
solve for l
l = what should be the limit = best lower bound
weary pelican
here we have $l = 2 - \frac 1l$
clever fjordBOT
Raphaelisius Maximus MMIII
weary pelican
can you maybe rewrite this so there's no denominator?
sturdy cape
At this stage this is just algebra
weary pelican
to remove a denominator, you just multiply all around by the denominator
sturdy cape
**Raphaelisius Maximus MMIII**
That is to say, multiply all terms here by l
sturdy cape
yh
sturdy cape
l^2 - 2l - 1
(when you have this, you should still say =0 at the end)
pulsar tiger
This would give us a lower bound of 1
sturdy cape
l = 1 with two roots
Right, so you have your limit
weary pelican
alright so
Cool part, is that this wasn't just for finding a lower bound
Once you show that (x_n) converges
weary pelican
we're supposing it converges but we don't know the limit yet
well this isn't an assumption anymore
so you just copy paste this reasoning once you've shown convergence
and tada you have proof that the limit is 1
Let's leave this aside for now
we want to use MCT to prove convergence first
pulsar tiger
Just to recap, we had to find an upper bound and lower bound, and then we could establish convergence by observing the sequence to be decreasing?
weary pelican
Just to recap, we had to find an upper bound and lower bound, and then we could ...
you would actually need only one of those, depending on if your sequence is increasing or decreasing
If your sequence is increasing, you need to make sure it doesn't fly off to infinity
so you need to make sure there's a ceiling
an upper bound mathematically speaking
pulsar tiger
Oh, okay
What about convergence?
weary pelican
And if your sequence is decreasing, vice versa, you need a lower bound
bitter sleet
Oh, okay
What about convergence?
monotone convergence theorem
weary pelican
Oh, okay
What about convergence?
Well, once you have decreasing + lower bound
bitter sleet
Which u just showed
weary pelican
that's just the monotone convergence theorem that'll give it for you
the MCT gives you the convergence under certain conditions
you provide it with the conditions
it gives you back convergence
simple as that
pulsar tiger
MCT says that a sequence convergences then its sequence is bounded, right?
weary pelican
MCT says that a sequence convergences then its sequence is bounded, right?
you're forgetting what the M stands for
pulsar tiger
Monotone
weary pelican
MCT says that a sequence convergences then its sequence is bounded, right?
convergence -> bounded is a completely different thing
monotone + bounded (the correct way) -> convergence
pulsar tiger
Oh, its seq3uence has to be increasing or decreasing for it to be convergent, then it implies it is bounded?
weary pelican
monotone + bounded (the correct way) -> convergence
that's MCT
pulsar tiger
Oh, monotone and bounded
weary pelican
Oh, monotone and bounded
well, as we said, you only need one of upper or lower bounded
pulsar tiger
Oh, right.
weary pelican
Is it clearer for you? Do you want to go over something in more detail?
pulsar tiger
Since this problem was bounded above by 2, we knew that a lower bound must exist while recognizing this was decreasing
pulsar tiger
Is it clearer for you? Do you want to go over something in more detail?
Yeah, this makes sense.
What about the monotone of sbusequences?
weary pelican
Since this problem was bounded above by 2, we knew that a lower bound must exist...
Uh just to clarify
if a sequence is decreasing
then the first term is obviously bigger than all the others
and so that makes an upper bound
pulsar tiger
Oh, okay
weary pelican
what's not trivial is the lower bound
because it depends on how far below your sequence is going
If you want a monotone sequence to converge, you need to make sure it doesn't escape to +infinity or -infinity
so if it's heading down, don't let it escape to -infinity (decreasing needs lower bound)
if it's heading up, don't let it escape to +infinity (increasing needs upper bound)
Is that clearer?
pulsar tiger
Yeah
weary pelican
alright
So, quick recap
Our potential lower bound is 1
so now, we just have to prove it
We want to prove $\forall n\in \bN, x_n \geq 1$
clever fjordBOT
Raphaelisius Maximus MMIII
weary pelican
And how will we prove this?
pulsar tiger
We know that this is bounded and monotonically decreasing, so this should be convergent?
Earlier, we had lim(x_n) = L
Now L = 1, so we have lim(x_n) = 1
weary pelican
We know that this is bounded and monotonically decreasing, so this should be con...
So you're saying we "know" it is bounded and decreasing
but so far we haven't seen any proof of those two assertions
pulsar tiger
Yeah
weary pelican
alright
pulsar tiger
Wouldn't x_(n+1) = 2 - 1/x be enough when we take n to infinity
This would give us an upper bound of 2
WE know it is decreasing because no terms can be larger than 2
weary pelican
This would give us an upper bound of 2
x_(n+1) = 2 - 1/x_n
pulsar tiger
So, it is monotonically decreasing
weary pelican
there's multiple things wrong with what you did
weary pelican
Wouldn't x_(n+1) = 2 - 1/x be enough when we take n to infinity
mainly you're starting with assuming -1/x_n < 0?
so this implies you're assuming x_n > 0
not given
pulsar tiger
How would I do this then?
I am a suicidal stupid person
So, don't expect me to know things
weary pelican
**Raphaelisius Maximus MMIII**
we have this first to prove
pulsar tiger
**Raphaelisius Maximus MMIII**
I have no idea how to prove this.
weary pelican
I have no idea how to prove this.
this is a statement for all natural integers
pulsar tiger
I am so sorry. I should just leave this world.
I am wasting your time....
I don't know how to prove this for all natural numbers
I only know that those are natural numbers
weary pelican
Have you heard of proof by induction
narrow crypt
so we are trying to prove that the thing have lower bound?
pulsar tiger
I think we did that.
x - 1/x = x
This gave us x =2
*x = 1
This would be the lower bound
We need to use the MCT to prove convergence now
I am unsure how to do that
narrow crypt
what are the conditions
pulsar tiger
I know the two conditions are monotonically increasing or decreasing(IN this case, it is decreasing) and bounded(we did that)
I am unsure how to prove it inductively though
narrow crypt
we have established that it is decreasing already?
pulsar tiger
Oh, I just observed it to be decreasing
narrow crypt
..
pulsar tiger
I guess that is why I need help inductively showing it.
narrow crypt
lets see
lets recap,
[ x_{n+1} = 2- \frac{1}{x_n} ]
and ${x_0 = 3}$. Therefore,
${x_1 = 2 - 1/3 = 5/3 > 1}$. Now, let's see what we can do with ${x_2}$.
[ x_2 = 2 - \frac{1}{x_1}]
Since ${x_1}$ > 1,
[ \frac{1}{x_1} < 1.]
Therefore,
[ x_2 > 2 - 1 = 1]
hmm
weary pelican
this is 5/3 > 1
narrow crypt
oh ye
weary pelican
I'm back just for a small amount of time but you need induction to prove 1 is a lower bound for all x_n
narrow crypt
was wondering why its wrong
clever fjordBOT
k
weary pelican
ONCE you show lower bound is valid, you'll show decreasing
clever fjordBOT
k
narrow crypt
x_2 > 1 forces x_3 > 1 forces x_4 > 1 like domino and so on
this is the induction
x_n > 1 forces x_{n_1} to > 1
weary pelican
I need to show it is by the MCT
MCT will show you convergence, provided you fulfill its conditions first
narrow crypt
for all n
pulsar tiger
So, is that all then?
weary pelican
So, is that all then?
well you have to show the conditions before involving MCT
weary pelican
x_2 > 1 forces x_3 > 1 forces x_4 > 1 like domino and so on
so lower bounded is shown via this induction proof
pulsar tiger
Bounded + monotonically decreasing?
narrow crypt
Bounded + monotonically decreasing?
this is all for mct, yup
we're using induction to show the first one
narrow crypt
x_n > 1 forces x_{n_1} to > 1
this is the main hypothesis
can u use it to show us that 1 is the lower bound
narrow crypt
no like the proof like
Claim: ${\forall x \in \mathbb{N}(x_n > 1)}$.\
Base case: Consider ${x_0 = 3 > 1}$ and ${x_1 = 2 - 1/3 = 5/3 > 1}$.\
Inductive hypothesis: Assume ${x_n > 1}$ for ${n \geq 2}$. We will now show that ${x_{n+1} > 1}$ also.
weary pelican
1 < x_n < 2
don't worry about the upper bound at all for now
clever fjordBOT
k
pulsar tiger
no like the proof like
I am sorry. I wouldn't know how
narrow crypt
ok
so assume x_n > 1
and using our defn that x_(n+1) = 2 - 1/x_n
can we show that x_(n+1) > 1?
pulsar tiger
x_(n+1) = 2 - 1/x_n = 5/3 > 1
narrow crypt
where did the 5/4 come from
pulsar tiger
Oh, right.
weary pelican
so assume x_n > 1
all you know is this
can you prove x_(n+1) > 1 just with this information
pulsar tiger
I don't know....
This is what is going to get me killed.
No matter how hard you study, no matter how many times you review the definitions, if you aren't smart, I might as well kms
narrow crypt
calm down
narrow crypt
we all suck at smth
pulsar tiger
No, you don't
pulsar tiger
Are you failing it rn?
narrow crypt
we all have smth we're bad at it
tribal ingot
prayers up for you 🙏
pulsar tiger
That is nothing. Try 35/ 100
After studying meticulous and investing a lot of hours
tribal ingot
hey perfection only comes with practice
weary pelican
That is nothing. Try 35/ 100
My 4th year post-high had a lesson in Geometry, I failed hard, 5.2/20 on the final test
And hard study
there are hard subjects everywhere
tribal ingot
a winner is a loser who just tried one more time 
weary pelican
doesn't mean you should beat yourself over it
pulsar tiger
I will believe this if I can pass an exam
pulsar tiger
can you prove x_(n+1) > 1 just with this information
Can you give me some direction here? I can't think of anything
pulsar tiger
1/xn < 1
xn > 1
narrow crypt
could u elaborate ur derivation from our hypothesis x_(n) >1
weary pelican
1/xn < 1
xn > 1
1/xn < 1 is pretty good progress
narrow crypt
pulsar tiger
1/xn < 1 is pretty good progress
So, does that work?
narrow crypt
so we know how
x_n > 1
1/x_n < 1
intuitively
the function 1/x gets smaller and smaller as x increases
for x > 0
pulsar tiger
Oh, so we know that this monotonically decreases because 1/x_n < 1 and gets smaller as x increases?
pulsar tiger
Oh, so what is next?
pulsar tiger
Finish the IH?
narrow crypt
the lower bound proof can be used to prove the monotonic decrease? 👀
weary pelican
induction hypothesis
narrow crypt
induction hypothesis
ye
weary pelican
the lower bound proof can be used to prove the monotonic decrease? 👀
you'll see, yes
narrow crypt
you'll see, yes
x_(n+1) - x_n, right?
narrow crypt
What counts as proven?
u've shown clear derivation
from the hypothesis and how it leads to the conclusion
similar to logic
pulsar tiger
Guys - I am confused. Please guide me more
narrow crypt
alright
lets follow from this
Claim: ${\forall x \in \mathbb{N}(x_n > 1)}$.\
Base case: Consider ${x_0 = 3 > 1}$ and ${x_1 = 2 - 1/3 = 5/3 > 1}$.\
Inductive hypothesis: Assume ${x_n > 1}$ for ${n \geq 2}$. \
\
We will now show that ${x_{n+1} > 1}$ also.\
\
Firstly,
[ x_{n+1} = 2 - \frac{1}{x_n}]
by definition. By hypothesis, ${x_n > 1}$. Therefore, ${\frac{1}{x_n} < 1}$. Hence,
[ x_{n+1} > \text{(fill in here)} = 1.]
QED.
@pulsar tiger fill in the blank
weary pelican
<@802383993143164979> fill in the blank
wrong inequality side
narrow crypt
oopsie
clever fjordBOT
k
narrow crypt
what 😭
odd edgeBOT
@pulsar tiger Has your question been resolved?
narrow crypt
bro u good?
weary pelican
They can continue this once they've been unmuted or they finished their mute, I'll close this channel for now
.close
nova fractal
odd edgeBOT
nova fractal
is this correct
glass vault
where did 1/8 come from
nova fractal
I multiplied and divided by 8
desert jolt
mistake while transitioning from line 3 to line 4
desert jolt
also another calculation mistake in trnasitioning from second last line to last line
desert jolt
yes and also the other mistake i mentioned
uh also another mistake you did a wrong integration of sin function in both terms
last pawn
My question is this: There are 11 objects, 4 A's, 4 B's, 1 C, 1 D, 1 E. I want to figure out the probability that upon selecting 5 objects without replacement, that I select 1 C, 1 D, and atleast 1 A. The order here does not matter. I am having trouble knowing which permutation and combinatoric rules to even apply. What I have tried is introducing the event F which is the event that I select B or E. Then splitting it into 3 cases: (1A, 1C, 1D , 2F) or (2A, 1C, 1D, 1F) or (3A, 1C, 1D), and multiplying these by 5!/2! , 5!/2! and 5!/3! respectively. My final answer is 37/231 which I feel is on the low end but I am unsure. Any help would be appreciated!
nocturne brook
in counting the number of ways, the presence of C and D is forced, so we need only count the number of ways to choose 3 of the remaining 9 objects so that there's at least one A; this is (9 choose 3) - (5 choose 3). divide this by (11 choose 5) to obtain the actual probability
nocturne brook
My question is this: There are 11 objects, 4 A's, 4 B's, 1 C, 1 D, 1 E. I want t...
Sorry I didn't address your method, this is how I'm seeing it and I don't know if they're equivalent
last pawn
in counting the number of ways, the presence of C and D is forced, so we need on...
Thank you, that helps. I will attempt it again using that.
nocturne brook
We're taking advantage of the fact that picking C and D first is the same as picking them later, and in any order
I also treat the A's and such as distinct, but that's ok because they live in their own little collection, so when we don't want an A, we choose from the collection of all non-A's
last pawn
In this case, do we just treat C and D as being the product of their probabilities (1/11 x 1/10) and multiply that by the number of ways to choose 1 through 3 A's in a group of 3? I am struggling to wrap my head around this honestly
nocturne brook
hm, well my approach was to count all the designated good combinations and divide the count by the total number of combinations to get a probability
it felt cleaner than wrestling with probabilities the whole way through
last pawn
i see i see. ill try that
nocturne brook
My question is this: There are 11 objects, 4 A's, 4 B's, 1 C, 1 D, 1 E. I want t...
I think our answers are equivalent at least
so we're either both right or both wrong lol
last pawn
alright thank you, i am satisfied with 37/231 so i think i will close this. i appreciate the help
.close
odd edgeBOT
odd edgeBOT
Available help channel!
fresh notch
Hi, Im wondering if there’s a way to get three independent equations and one discrete solutions with these set of equations.
(V1-V2) = i1R1 + i2R2;
(V2-V3) = i3R3 - i2R2;
(V1-V3) = i1R1 + i3R3;
i1 = i2 + i3
I keep arriving at no discrete solutions, but i wonder if anyone can arrive at anything else. (The unknown values are R1,R2, and R3)
quasi sparrow
!show
odd edgeBOT
Show your work, and if possible, explain where you are stuck.
fresh notch
!show
But my linear algebra is very rusty so idk if im doing it right
My teacher insists that there’s a discrete solution
I cant find it
odd edgeBOT
@fresh notch Has your question been resolved?
fresh notch
<@&286206848099549185>
odd edgeBOT
@fresh notch Has your question been resolved?
fresh notch
Bruh i thought you were gonna help me
fresh notch
i have a issue
Go look at how to get help
fresh notch
<@&286206848099549185>
ivory grove
Hello, this is my first time in this server
!occupied
odd edgeBOT
Hello, this is my first time in this server
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).
ivory grove
welcome, btw
ivory grove
<@&286206848099549185>
I can't help u, sry
odd edgeBOT
@fresh notch Has your question been resolved?
odd edgeBOT
odd edgeBOT
Available help channel!
vale herald
guys if i want to do substitution method, do i have to rearrange the equation so that it becomes either y= or x=
nimble blaze
not really
you can rearrange to whatever you want that's convenient to substitute
though isolating a variable (and subbing that into the other equation) is the basic approach/ what you're usually taught when being introduced to this
odd edgeBOT
odd edgeBOT
Available help channel!
bright sorrel
Hey yall, I needed help with this pretty simple systems of equations problem (finding the one solution):
3x + y = 3
-3x -4y =6
So I gotta find the first equation in terms of y and then plug it into the second equation to find x right?
- -3x + 3 = y
- -3x -4(-3x+3)=6
- -3x + 12x -12=6
- 9x = -6
- x = -6/9
I’ve asked all my friends and we’ve all arrived at the same answer so far, but -6/9 isn’t one of the available answers for x in the practice test that I am taking. What did I do wrong?
late sinew
Step 4
late sinew
Hey yall, I needed help with this pretty simple systems of equations problem (fi...
How did you all do the same mistake
Did they even solve
lean yew
Hey yall, I needed help with this pretty simple systems of equations problem (fi...
you went too fast between steps 3 and 4 i presume
late sinew
Also even if u did get -6/9 u would've had to reduce it to simplest form
lean yew
your signs are wrong
wooden python
up to and including step 3 it is correct
lean yew
2 is substituting the rearranged eq 1 into eq 2, so yes
wooden python
it's the subsequent algebra and solving for x where you fumbled
bright sorrel
Step 3 is right?
wooden python
that's what i said, yes
wooden python
i literally said "up to and including step 3 is correct"; why ask me a second time :P
bright sorrel
So combining like terms, -3x + 12x = 9x right? And then you can
Ohhhhh
It’s negative 12
wooden python
separate the "collect like terms" step and the "handle the constants" step into two
bright sorrel
Is that it
wooden python
-3x+12x = 9x yes that part is correct
point is tho
you tried to kill two birds with one stone
you got one, but the other escaped your grasp
lean yew
more appropriate to say you got one, but the other bit you when you tried to loot its friend
lean yew
Is that it
try reworking step 4
lean yew
well done
bright sorrel
.close
odd edgeBOT
bright sorrel
.reopen
odd edgeBOT
✅
bright sorrel
Really quick for an equation like:
-8 = 4x - 12y
-10 = 5x + 15y
Am I allowed to multiply both equations in order to eliminate a variable?:
5 • -8 = 5 • 4x - 12y
-4 • -10 = -4 • 5x + 15y
Sorry for reopening
lean yew
multiply 1 to match the other
bright sorrel
How did you all do the same mistake
I have one friend, my other one lowkey doesn’t fw me
lean yew
you can multiply both, or you can multiply one by a fraction
mystic nova
It should be like this
5•(-8)=5•(4x-12y)
-4•(-10)=-4(5x+15y)
bright sorrel
It should be like this
5•(-8)=5•(4x-12y)
-4•(-10)=-4(5x+15y)
Thanks
But that isn’t allowed anyway when doing elimination I assume?
bright sorrel
you can multiply both, or you can multiply one by a **fraction**
Oh that’s tricky
For me at least
mystic nova
But that isn’t allowed anyway when doing elimination I assume?
Why isn't that allowed
bright sorrel
Why isn't that allowed
I dunno is it
bright sorrel
you can multiply both, or you can multiply one by a **fraction**
Oh! I can’t read
lean yew
For me at least
if that's the case, then pick one variable and do the LCM method with both equations i suppose
better safe than try something funky and make a careless mistake
odd edgeBOT
odd edgeBOT
Available help channel!
cerulean flare
$\text{Given } f(x) \text{ is continuous on } [0,1] \text{ and } \int_{0}^{1} f(x) ,dx = 0.\newline\text{ Prove that } \exists c \in (0,1) \text{ such that } f(c) = \int_{0}^{c} f(x) ,dx.$
cerulean flare
Gotta get used to some university maths or else im doomed
Anyways, i found this problem and it's very weird
Based on the problem, i think there must be an usage of IVT or Rolle's theorem
somewhere
desert jolt
Do you know lmvt
desert jolt
Do you know lmvt
Lagrange mean value theorem
cerulean flare
WAIT
clever fjordBOT
TargetVN
cerulean flare
Sorry for causing misunderstanding
desert jolt
Lagrange mean value theorem
.
cerulean flare
Yeah i do know that
faint knot
try seeing what happens if you mix up $\int_0^xf(t),dt$ and $\frac{f(1)-f(0)}{1-0}$ in some way
narrow crypt
just mvt is fine
clever fjordBOT
mtt
narrow crypt
that integral is antiderivative of f
faint knot
mvt and lmvt mean the same thing
faint knot
ooh Ive never seen that one before
cerulean flare
isnt cauchy mvt just lmvt but for 2 functions
cerulean flare
👀
desert jolt
i keep confusing cauchy's mvt and lmvt
Cmvt is an extension of lmvt anyways
cerulean flare
The what
cerulean flare
I mean it partially looks like a ODE
Oh nvm
I overthink
F'(c) = 0
wait what about F(c)
faint knot
So the problem is just proving F'(c) = F(c)?
personally, I would solve this by using MVT on some variant of f(x) that would turn this problem from dealing with integrals to dealing with derivatives
cerulean flare
Aha
I figured it out
g(x) = e^(-x) * F(x) is the function
g(0) = g(1) = 0 so I can do Rolle's theorem
g'(c) = e^(-c)[F'(c) - F(c)] = 0 => F'(c) = F(c)
Thx everyone
.close
odd edgeBOT
rigid orchid
odd edgeBOT
Available help channel!
sudden blaze
isnt a rank 2 tensor a 2 by 2 matrix?
odd edgeBOT
isnt a rank 2 tensor a 2 by 2 matrix?
@sudden blaze Has your question been resolved?
odd edgeBOT
odd edgeBOT
Available help channel!
next isle
how do you prove that some any function applied to every integer leads to a sort of unique solution that you want?
for example, how could i show that for some arbitrary polynomial function f(x): Z -> Z, how could I show that its bijective? I guess we can start off there idk what else to add
next isle
how do you prove that some any function applied to every integer leads to a sort...
^or show that it isnt bijective
fickle silo
can you even come up with any example of degree n >= 2 that is bijective?
odd edgeBOT
@next isle Has your question been resolved?
next isle
can you even come up with any example of degree n >= 2 that is bijective?
i dont know? the question can be n < 2
its just that
i want anything that works
i am stumpted from like anything geniunely
perhaps i should also say
partial polynomials
i think thats not the word for it
like
the same function according to some rule outputs different things
fickle silo
because thing is, with any polynomials of degree n >= 2, I guess you can show that by choosing K big enough, you can guarantee that it's monotone on the interval [K, inf), f(K) > f(y) for all y < K, and that the gaps between f(x) and f(x + 1) for x >= K are greater than 1, which gives you holes so it can't be bijective (maybe there is a simpler argument)
next isle
so what youre saying is like since polynomials just grow faster than +1 after the extremums you can guarantee that on the sides its definetely bijective for the first part
to the right of the rightmost extremum and left of the leftmost extremum
maybe i misunderstood
is K standard for some notation
next isle
because thing is, with any polynomials of degree n >= 2, I guess you can show th...
oh
okay so
not what i wanted
fickle silo
K is just some big integer
fickle silo
so what youre saying is like since polynomials just grow faster than +1 after th...
yeah I guarantee that it's definitely not bijective on its ends
next isle
that is lowkey easy to see
i see now
so it has to be n = 1
and obviously n= 1 and the coefficient of x^1 = 1 would work out
oh
no n=0 works out if like
i think
okay so lets just focus on
fickle silo
f(x) = 3
next isle
f(x) = mx + k
and still can be partial
depending on some rule
im using the word partial wrongly i think
cuz im like directly translating it from turkish
fickle silo
are you thinking of polynomials defined on a smaller interval?
next isle
like
i can say some things like
if x is a multiple of 4 then mx+k
else nx+p
also also $n,m \in \mathbb{Q}$
clever fjordBOT
Red.
next isle
even though its obvious i also want to just step on that also means they can be integers
fickle silo
so you could divide Z into different classes modulo 4 and apply a different polynomial on each class?
next isle
i mean perhaps with a certain ruleset you can get a specific f: A->B that n>=2 would work for but not the focus really just curious thought
next isle
so you could divide Z into different classes modulo 4 and apply a different poly...
i mean sure i think i get what that means
could you like slightly explain what that'd do
or show it
w an example
fickle silo
like I guess you could have 4 polynomials $f_0, f_1, f_2, f_3$
clever fjordBOT
rbit
fickle silo
and then $f(x) = f_i(x)$ if $x \equiv i$ mod $4$
clever fjordBOT
rbit
next isle
ohh i mean
from where i got my original question, this is related
i did this
but it didnt help a lot
it was just chaotic
fickle silo
that setup seems weird to me though
next isle
there was no way to prove it
cuz it was just pretty randomly moving around all functions
oh yeah wait
i should also say
since its unrelated to my original questions i got this from, i rather say that
$f(x) = mx + k$ and $m \in \mathbb{Q}, k$ is either 0, 1 or 2
clever fjordBOT
Red.
next isle
what I want to happen is kind of like
how do i find out how to do a proof against many rulesets
like for example the one i said where a ruleset can be
for 4 | x another thing
and everyting else other
it could also be like
for multiples of 3 but not 5 do something
for multiples of 5 but not 3 do something
for multiples of 3 and 5 do something
for everything else do something
and perhaps how do you tweak the outputs to work out
if you prove that the function isnt bijective
how do i change not the rulesets but the outputs
so that it is bijective
or prove that theres no bijective solution
fickle silo
I mean these rulesets can be pretty arbitrary, so not sure if there can be said anything, or I don't really understand what the goal here is
next isle
well
actually
i mean
i am 99% sure i proved that if the rulesets contain something where k isnt 0 for both and m is 2 distinct primes regardless of which power, it is never bijective
the reason why i said m is fractional is because
the problem arises with fractions
i mean the ruleset HAS TO BE such that no matter what, you never get non integer outputs for integer inputs
or else you just kinda say theres nothing to prove cuz it isnt even Z -> Z
i need literally anything that'd hint me at something that'll help me with fractions in a way
like a ruleset can be like
x/m when x is a multiple of m
otherwise do other things
until you get a multiple of m
sigh
ping me if someones here
odd edgeBOT
@next isle Has your question been resolved?
odd edgeBOT
@next isle Has your question been resolved?
next isle
bro nobody even SPOKE
sturdy cape
In general a function is bijective definitionally if it is both injective (f(x) = f(y) must imply x = y) and surjective (for any y there is some x such that y = f(x))
But essentially the thing that appears to be your question seems far too broad for any answer to be adequate...
next isle
But essentially the thing _that appears to be your question_ seems far too broad...
it doesnt really matter if i get just a hint at how to approach this
im not really looking for a direct answer
odd edgeBOT
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
next isle
i know its very broad
for any function with a specific ruleset and output, prove that its bijective or find a bijective output for the ruleset of your choice
the output can only be in the form f(x) = mx+k where k is either 0,1 or 2 and m is a rational number (in Q)
the ruleset specifically isnt like
that complex
rulesets that i look for is like
divisiblity
so like
f(x) = if x is divisible by 5, then (x/5) + 1, else 4x+2
for example
how could you approach finding answers
sleek moth
how do you prove that some any function applied to every integer leads to a sort...
If you'd like me to rephrase your question, I think a related interesting question is which polynomial functions with integer coefficients are surjective
next isle
to this type of things
sleek moth
I suspect that the identity polynomial is the only one
next isle
If you'd like me to rephrase your question, I think a related interesting questi...
hmm
which one is surjective cuz idk the english variants
i know bijective is both
and theres surjective and injective but idk which one is whihc
sturdy cape
What language do you know them in?
next isle
turkish
sleek moth
surjective means every output has its corresponding input
sturdy cape
("örten")
next isle
If you'd like me to rephrase your question, I think a related interesting questi...
is it asking surjective for f: Z -> Z
we can also do Z+ -> Z+
btw
i dont really care ab negatives
sturdy cape
(similarly "birebir" is injective or one-to-one)
next isle
(similarly "birebir" is injective or one-to-one)
👍
sleek moth
Well what you mean by "a polynomial function" depends on what your coefficients are, but if your coefficients are integers, then your corresponding polynomial function will be from Z to Z
next isle
well i rephrased it after the discussion
its not just polynomial
its specifically
mx + k
sleek moth
ah okay
next isle
cuz it kind of doesnt work for n >= 2
even if it did itd need like hundreds of rules
dont think thats manageable and
the original question IS just
mx+k
i just was curious if there could be n>=2
like the more generalized way to write this for specifically 2 rules is like
f(x) = {p | x, (x/p) + q;
p doesn't | x, mx+k}
$k,q \in {0,1,2}$
clever fjordBOT
Red.
next isle
$m \in \mathbb{Q}$
clever fjordBOT
Red.
next isle
m is often integer and even if its rational
it isnt something weird like
11/29
its often basic ratios like 2/3
and often also related to divisibility
🕙
sleek moth
well it's starting to sound like number theory and I fear number theory
next isle
$p \in \mathbb{Z}^+, m \in \mathbb{Q}^+; k,q \in {0,1,2}$
How do we prove that there exists or doesn't exist a certain ruleset for a function where $f: \mathbb{Z}^+ \rightarrow \mathbb{Z}^+$ is bijective for $f(x) =
\begin{cases}
p \mid x, & \frac{x}{p} +q \
p \nmid x, & mx+k
\end{cases}$
oh right
i gotta
one sec
clever fjordBOT
Red.
next isle
there
next isle
i mean we could also remove +q for convenience tbf
im unsure
would using any other like base-n help with finding out the answer cuz i feel like lower base and prime bases help with finding specific patterns that shows the answer better than base-10
just a thought i had asw
i tried for a long time base-3 where it looks like something is happening but geniunely i cant tell nothing cuz im so not used to base 3
or any other bases tbf
odd edgeBOT
@next isle Has your question been resolved?
faint knot
$p \in \mathbb{Z}^+, m \in \mathbb{Q}^+; k,q \in \{0,1,2\}$
How do we prove that...
use \mid and \nmid so that it formats properly
| is for absolute value
faint knot
also, is there a reason you have the text swapped for the cases
usually its function if condition
faint knot
you are literally the only person Ive seen to do that
usually its
[
f(x)=\begin{cases}\frac xp+q&\text{if }p\mid x\mx+k&\text{if }p\nmid x\end{cases}
]
clever fjordBOT
mtt
next isle
sorry
i do that
cuz like it makes more sense
if this rule applies, do this output
not
do this output, if this rule applies
faint knot
well a lot of things dont make sense conventionally
next isle
its fine
faint knot
thats not exactly an excuse
faint knot
too bad, you gotta do it the conventional way
next isle
❌
faint knot
sucks to suck bro
also, you know p | x could make more sense as x | p, so that it resembles x / p
but I dont see you doing that here
also @radiant ledge you gotta stop spamming that in the chat
faint knot
**Red.**
being more on topic, is there a specific reason k and q can only be 0, 1, or 2
you could have p = m = 1 and q = k = 3 and itd be just as bijective as for q = k = 2
@next isle oh more importantly, do Z^+ and Q^+ include 0
next isle
also, you know p | x could make more sense as x | p, so that it resembles x / p
interesting
next isle
being more on topic, is there a specific reason k and q can only be 0, 1, or 2
cuz
otherwise it doesnt work anyways
no 0
Z+ and Q+
are strictly positives
no 0
faint knot
bijective shows that every output is reached by exactly one of the cases of this piecewise function
for positive multiples of p, that accounts for some of the outputs
and for positive not-multiples of p, that accounts for the rest of the outputs
this is because:
if the outputs can be reached with x from one case and y from another, then f being injective would have x = y even though p | x but not y
also, every output requires an input, so at least one of the cases has to supply an input for the output
since p is an integer, we run into a problem
represent the positive multiples of p with pt for positive integer t
then f(pt) = (pt)/p + q = t + q
this would mean all integer outputs > q are handled by the top case
this leaves all the integer outputs < q to be handled, and only handled, by the bottom case
since the bottom case is also a line with rational coefficients, you can see we're running into a problem here
@next isle you could try considering this
if a set A is finite, a bijection from A to itself is called a "permutation"
that means you move/swap some of the elements of A around
for Z^+, that would mean a function that rearranges integers to other integers
an example would be having f(x) be x + 1 for odd x and x - 1 for even x
you can see all this does is swap 1 with 2, 3 with 4, 5 with 6, etc.
this f(x) is bijective and also has f(f(x)) = x
next isle
<@321693448115716096> you could try considering this
if a set A is finite, a bij...
this is interesting
faint knot
if you want one of the cases to be along the lines of "if p | x, then nx/p + q for positive n, p, q", then you have to consider that the kinds of inputs youre left having to put inside the output
a line isnt going to work, even if p | d
next isle
bijective shows that every output is reached by exactly one of the cases of this...
i dont really get the problem that well tbf but in the case of not divisible by p you cannot predict the multiples you get out of it except that none of the multiples will remain the same after mx+k
so you could get p
or you couldnt get p
its like unknownable
faint knot
...you know m can be written as a fraction of integers
m = n / d for coprime positive n, d
faint knot
multiplying m by d would get you an integer
next isle
but id consider that
in the most basic scenerio
youd want m to be an integer
just that keeping in mind it can be rational
faint knot
red you dont seem to be seeing the fundamental issue here
multiplying m by d gets you an integer
thats the alarm
faint knot
if you want one of the cases to be along the lines of "if p | x, then nx/p + q f...
you gotta have p | d or something to have a chance of this working
faint knot
otherwise having f(dt) for positive integer t can cause an issue
next isle
it was spesified before
faint knot
maybe you should have p be rational instead of an integer
next isle
if m is a fraction
then like
if it outputs not an integer
that function isnt
right
is that the issue
faint knot
you dont seem to know whats happening here
next isle
im very confused its 3 am
faint knot
maybe let me finish
next isle
okay
faint knot
you know a bijective function from Z^+ to Z^+ isnt supposed to output anything other than integers, right
otherwise itd be a bijective function from (some subset of Z^+) to Z^+
when you have a function f : A -> B, its implied that f(any member of A) must be a member of B
faint knot
<@321693448115716096> you could try considering this
if a set A is finite, a bij...
for bijective functions from Z^+ to Z^+, the best way I see them are as permutations
you rearrange the elements around
if its not a permutation, or if it cant seem to form a permutation, then it cant be bijective
next isle
how does it "form a permutation"
faint knot
thats the hard part - I have no idea and have to check on a case-by-case basis
but its enough to show that f is bijective to have that
faint knot
<@321693448115716096> you could try considering this
if a set A is finite, a bij...
take this example function here for instance
you can see that it maps evens to odds and odds to evens
and also, the function is invertible since f(f(x)) = x
so f is bijective, and so f is a permutation
(permutation implies bijective is easier to understand - permutations must place every element somewhere, so theyre surjective, and permutations cant place two elements in the same spot, so theyre injective)
next isle
ahh
so what we're looking for in the original function is to see if it forms a permutation
faint knot
thats just a conceptual way of thinking about it
next isle
uh huh
faint knot
however work-wise we're still talking about bijections
and for that, the proof isnt finished but f(x)'s case isnt looking too strong
personally Im using this permutation idea if I want to go about proving that f cannot be a bijection (unless you remove one of the cases)
next isle
i dont think it would make it much easier to say $m \in \mathbb{Z}^+$ no?
clever fjordBOT
Red.
faint knot
that makes it worse
next isle
really
faint knot
think about this
we had a previous problem when m was an element of Q^+
now would restricting m to also be an integer erase this problem?
next isle
well
faint knot
I was pointing to p | d since that was a line of reasoning I was using
we could require that
faint knot
alr so lets say p | d
this would mean that mpt, for some positive integer t, will eventually reach some integer of some sort
next isle
personally Im using this permutation idea if I want to go about proving that f c...
i mean the original question was, if you need to, add more cases and just try to realiably find out if something can be bijective and the minimum amount of cases must be 2 btw
faint knot
this permutation thing is the best I got
faint knot
<@321693448115716096> you could try considering this
if a set A is finite, a bij...
I mean you can group up like every n or so integers together and reverse them
so you can have this work for any number of positive integer cases
in the example, n was 2
faint knot
-1?
how?
you mean to say negative numbers arent allowed, even when subtracting x - 1?
faint knot
youre still sticking hard to f(x) arent you
faint knot
do you know why f doesnt work out?
next isle
k,q are 0,1,2
next isle
do you know why f doesnt work out?
you can write collatz with this function so
faint knot
collatz isnt bijective
next isle
i dont know why it doesnt work out
next isle
collatz isnt bijective
well theres some method to just
duplicate integers into 2 seperate sets
and just say
i allow every set to have multiple of each integer
its more unrelated though
its just that its like a weird function
faint knot
then it looks like you got lost trying to solve the collatz conjecture and ended up here
I was hoping you werent just trying for that
next isle
not really
this was a function i wrote after i read that
P = p1p2p3p4p5p6...pn
thingy
uhh
the proof for
infinitely many primes
one of them
the q = P+1 and no matter what
unique primes come out of it
made me curious
faint knot
is this weird function bijective?
next isle
+0,+1,+2 all work for this
next isle
is this weird function bijective?
well
now that i think about it a little better
i think i was trying to more similarly do it like collatz
like the function has some pattern
not necessarily
forcing one loop or infinity
just that
can you predict a pattern
for specific inputs
for m,k,q,p
and if you can
faint knot
a pattern in what way?
next isle
why or how
faint knot
iterating f()?
next isle
like something that isnt random or chaotic
you can like
yeah
iterating f()
and getting something
that is predictable
faint knot
you know the collatz conjecture is still unsolved
faint knot
you can always say you solved half of it if the other half is unsolved
next isle
yeah im cooked i know
faint knot
generalizing all the numbers isnt going to get us far since itll include the collatz conjecture
try this instead
I know there are examples that have solved this for functions of similar kinds
next isle
that'd be helpful
faint knot
they use particular tricks which I dont recognize
so unfortunately I cant point you there
there was something though I gotta find it
it was on mse?
<@&268886789983436800>
next isle
crazy
faint knot
next isle
oh the moment i load in the first thing i read was 2-adic
faint knot
nah thats just the valuation function
you know what it acts like:
v(odds) = 1
v(2 * odds) = 2
v(4 * odds) = 4
v(8 * odds) = 8
etc.
1 2 1 4 1 2 1 8 1 2 1 4 1 2 1 16 and so on
next isle
ohh
faint knot
it just has a fancy name since its related to p-adic numbers
faint knot
sort of counts how many zeros it has at the right
next isle
i learned like all of that in the last week so im unsure still but they lead me to places
most of the normal algebra looks like the algebera i did that hits a dead end
faint knot
yea, more likely than not people have already looked to attempt collatz with easier techniques
next isle
this T sounds like something interesting
next isle
this wasnt it, but maybe https://math.stackexchange.com/questions/4171307/collat...
which part of collatz is this about
faint knot
I havent really read it, let me look
next isle
my english isnt very good with these
new words i have to relearn
cuz i know these words in turkish and translator fools me sometimes
faint knot
the question is asking for some least integer which would cause collatz to diverge
next isle
diverge is go off to infinity?
faint knot
yep
faint knot
and seemed to reason that these kinds of integers, which are all odd, cannot be this least divergent integer
then again
maybe you gotta get reading to get up to date to the cutting edge on collatz
(I cant do this because I cant understand most of the words)
next isle
i know i can do this because i have a different perspective at something that i dont see anywhere
i started loving and studying math a lot cuz of this
faint knot
maybe its hidden in a paper you havent read yet
next isle
like i need this nudge
someone has to teach me something so immense
that when i see it i will know its the missing part from my proof
its like im looking for the arc reactor to power it
faint knot
maybe its hidden in a paper you havent read yet
next isle
maybe
faint knot
I saw something promising on mse but I cant find it again
next isle
itd be cool to see it
faint knot
it was able to essentially solve this for other forms that were easier to handle
next isle
you know what it acts like:
v(odds) = 1
v(2 * odds) = 2
v(4 * odds) = 4
v(8 * od...
could you explain this a little more
i think i gotta get the gist of it like better