of course, it will give different areas for different functions
#help-23
opal fractal
quasi bison
AI slop video
oak cobalt
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marsh wraith
h : S→Nx{f} ∪ Nx{g} with S to ( f(i),f) and with T to ( g(i),g) how to prove h is bijection right here
peak estuary
what are f,g,S,T?
peak estuary
yes that certainly doesnt work
and is also not what you wrote
consider finding h:SuT -> Z
and I hope S and T are disjoint, otherwise you have to do some extra bookkeeping
safe radishBOT
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@marsh wraith Has your question been resolved?
warm warren
In the context of this question
what if $i\in S$ and $i\in T$? You've not considered that while defining $h(i)$
flat frigateBOT
Annie Maqionde
marsh wraith
**Annie Maqionde**
h is not bijective
warm warren
h is not bijective
Apologies if I read it wrong, but in the pic you've sent, you clearly stated that $h$ is a bijection.
flat frigateBOT
Annie Maqionde
warm warren
And eitherways, no function can possibly map a single preimage to two different images.
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vale cypress
hi, i want to know how to sketch the argand diagram from arguments?
I always keep on getting it wrong, and Im getting really frustrated
marsh wraith
hi, i want to know how to sketch the argand diagram from arguments?
I always ke...
Do you know how we get arg(z) from z1
vale cypress
not rlly ://
marsh wraith
not rlly ://
Wait a minute I just send to you pic how to do that
split kayak
Argand basically is just a plot over the Complex Plane
Now, you have to regions and have to find their intersection
split kayak
yea
Now, do you agree that |z - (0+0i)| <= 1 is just the same?
And if so, what do you think happens when you have |z - (1+0i)| <= 1
split kayak
almost
vale cypress
*0,-1 centre , 1 radii
split kayak
Well, |z-z_1| <= is just a circle shifted to have its center on z_1
split kayak
Remember that
a+bi is the analog to (x,y)
and re^arg(z)i is analog to (r,theta).
vale cypress
is it just a line of -1 slope that passes thru z1?
oh wait what? then what should it look like ://
split kayak
Its a sort of "pizza slice" of the whole domain
I suppose you had trig at some point already
If i tell you
0 < theta < pi
You should know thats the upperhalf of the (x,y) plane
split kayak
not quite sadly
You have to consider that the argument is shifted by z_1 being in there.
split kayak
sort of, yep.
vale cypress
but then how would you draw that out
thats the most confusing part where i don't get
split kayak
you can think that (z-z1) makes the new center of the plane be z1
If so, try to find how 0 < arg(z) <3pi/4 would look like and then shift accordinglt
vale cypress
Ohhhh
so basically
its like this slanted line to z1
then
flat?
cuz the argument starts on positive x-axis. and with z1 being the origin, it just goes flat after reaching it?
vale cypress
ahh
gotcha
just a quick lil second question
for this question
do i just solve for the sector area then - C1's area under curve between P and 0?
split kayak
prob theres a better method but i cant think of it rn
vale cypress
and better question
is the area under C1 curve the area between OP and the intersection between C1 and the initial line
cuz I'm kinda confused with that part a lil, don't really get which area is the area under curve
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spark lynx
My mean is 106.31, which is accurate
52.83
66.51
75.43
78.1
82.18
83
85.09
86.48
89.04
89.24
90.29
90.29
90.4
96.41
98.62
98.73
100.34
103.37
105.45
107.95
109.18
117.06
117.8
120.77
122.49
124.6
124.8
130.47
131.49
133.18
133.36
137.72
140.92
141.22
166.19
Why is my standard deviation calculation so absurdly off
It should be around 24-30
I think intuitively
oh wait lmao nvm this is because thats variance
I'm stupid I need to still calculate standard deviation
.close
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ocean vine
(3)6 plz help
safe radishBOT
(3)6 plz help
normal moss
(3)6 plz help
What's (3)6
normal moss
!xy
safe radishBOT
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.
ocean vine
3x6 is the problem
normal moss
Otherwise I assume it's 3*6
plucky elk
so any help>
have you used a calculator before
ocean vine
my teacher said i couldnt
plucky elk
oh what grade does your teacher teach?
ocean vine
5th
plucky elk
oh 5th grade, are you 11 years old?
mint plume
ocean vine
6
normal moss
I really don't see the appeal in doing this
ocean vine
now can yoou teach me dirivtive algebra eginomics?
plucky elk
<@&268886789983436800> troll
normal moss
thanks
vivid stream
@ocean vine Just so we've made it clear: do not shitpost in help channels please. If you want to shitpost, go to #chill
I'm closing this channel now.
.close
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storm rock
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storm rock
Could someone help me figure this out
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@storm rock Has your question been resolved?
flat frigateBOT
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upper kraken
It is actually half the area you delimited
And since you want to integrate y=r^2+1 for 0< r < 1 for a rotating point, by the polar change of variable, you get the integral for theta from 0 to 2pi of the integral for r from 0 to 1 of (r^2+1)r dr dtheta
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rich jay
I don't understand why he added 45 under 45 and 0 with 45 below it, this is "Solving quadratic equations using square roots" and he said this is a "way" to do it, does that mean I can still do it the regular way? Which is having the first number of the equation be the fraction of A and B? while ignoring C (equals 0)?
safe radishBOT
I don't understand why he added 45 under 45 and 0 with 45 below it, this is "Sol...
rich jay
Sorry if my explanation is overcomplicated, I'm not very good with english math so...
austere goblet
this way works and is quite faster iff b = 0 (that is, you don't have a bx term)
of course, all of your regular tricks like quadratic formula still work
rich jay
this way works and is quite faster iff b = 0 (that is, you don't have a bx term)
Why would he do it this way though? He basically only got rid of 0 right??
austere goblet
because doing it this way is the fastest given the conditions
also, the question explicitly wanted the person to sqrt both sides
severe pond
Why would he do it this way though? He basically only got rid of 0 right??
no
its just isolating the variable term
your "way" of doing this isn't really a "way" at all
austere goblet
oh hi knief
severe pond
its just using the conclusion of the actual method used here from your memorization
severe pond
oh hi knief
hello hanako
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ancient onyx
help
hollow dock
burnt mantle
try splitting the denominator and notice how cosx is bounded
ancient onyx
both
ancient onyx
yes
severe pond
so lim x-> inf is just asking for the limiting behavior of the function as x gets arbitrarily large
does the function blow up to infinity or does it approach some finite value or does the function have no limiting behavior because it oscillates or something
,w plot e^(-x)
flat frigateBOT
severe pond
here we say lim x-> inf e^(-x) = 0 because regardless of how close to 0 you want the function to be, we can always find some point after which the function is within that distance from 0
and by "find some point after which" i mean there is some x value so that for all x values to the right of it...
in the usual limit we speak of x values within some distance from say a for lim x -> a. for lim x-> inf we consider all of the x values to the right of whatever point at which the function satisfies the desired inequality |f(x) - L| < eps
safe radishBOT
@ancient onyx Has your question been resolved?
covert sphinx
could i get help with these questions?
safe radishBOT
could i get help with these questions?
@covert sphinx Has your question been resolved?
covert sphinx
could i get help with these questions?
<@&286206848099549185> 
spice canyon
all?
spice canyon
im working on 17 so i guess I’d like help with #18 first
in 18. A box with a side of 20 is given. And a cylinder inside of it has a diameter and height of 20.
then
They're saying that the amount of volume between the volume of cylinder and cube is equal to 1717. Aka. They filled the difference of the two volumes with water amd it ca,e out to be 1717
so first let's calculate the area then we will go on from there.
volume*
volume of the box = 20×20×20 = 8000
Volume of cylinder = (pi)r²h . h is 20 and r is diameter/2 which is 10, so.
10²×20×pi = 2000pi.
Now. 8000-2000pi is the difference in volume
And what else was the difference in volume?
spice canyon
im working on 17 so i guess I’d like help with #18 first
Ivor can you do it from here? Try to understand. If you don't then tell me I'll try to help.
covert sphinx
Ivor can you do it from here? Try to understand. If you don't then tell me I'll ...
ok!
so the difference is 6000 cm3?
spice canyon
could i get help with these questions?
For 19. The upper part seems a bit tricky, I'm sure you can find the volume,e of the cuboid with the lengths of 30×8×6.
For the upper curved part. You see that the entire thing has a length of 20. Height of 10 width of 8cm (it shaes the same with as the other figure) but. There's a big portion cut from the middle of it , it's also given to be in the shape of a semicircle. Since it's a semicircle in 2d If you add a width to a semi circle then it becomes half of the part of a cylinder similar to how extending a circle and giving it height makes it a cylinder
spice canyon
so the difference is 6000 cm3?
No no the difference is given.
You need to calculate the value of pi
spice canyon
For 19. The upper part seems a bit tricky, I'm sure you can find the volume,e of...
So from here. Calculate the volume of the half-cylinder thing. And subtract it from the volume,e of the above cuboid.
spice canyon
how do i do that? <:catthink:526983414226747393>
Isn't 8000-2000pi the difference in volume?
And didn't the question say that they filled the remaining space of the cube with a water of volume 1717?
What is the remaining space equal to?
covert sphinx
spice canyon
So now
covert sphinx
i get it now
spice canyon
Nice
Now you treat pi like a variable. Then calculate for pi
Upto four digits the question says
spice canyon
So from here. Calculate the volume of the half-cylinder thing. And subtract it f...
Here's the thing for 19. Try to think from here and if isn't clear then ask
covert sphinx
Now you treat pi like a variable. Then calculate for pi
wait dumb question but isn’t pi a fixed irrational number?
so like
spice canyon
wait dumb question but isn’t pi a fixed irrational number?
It is but you wanna treat it l8ke a variable if you wanna calculate for it.
spice canyon
How would you say that 3.1415926 = 22/7
😭
Also I'm sure that there are like a billon approximations for pi.
covert sphinx
yeah
i think a) 3.142
wait but how do i get 3 significant figures
@spice canyon got 1783 cm3 for 19
covert sphinx
wait but how do i get 3 significant figures
this was my confusion
cerulean berry
wait but how do i get 3 significant figures
just remove the last number??
3.14 is 3 sf
covert sphinx
just remove the last number??
what was the point of them making that a question then 😭
cerulean berry
no idea
covert sphinx
im crine
cerulean berry
they just wanted to see ur rounding skills
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wise cloud
Why is distance formula usually written as 2-1 like x2-x1 but not x1-x2 while the answer is the same
safe radishBOT
Why is distance formula usually written as 2-1 like x2-x1 but not x1-x2 while th...
wise cloud
or there is no explanation for that
naive dragon
Why is distance formula usually written as 2-1 like x2-x1 but not x1-x2 while th...
x2-x1, y2-y1 refers to a vector with tail (x1,y1) and the tip (x2,y2) that's why
Not really important tho
shut barn
I am trying to do 3.1.1
I don't really understand how to plot bifurcation diagrams in general and I am not even sure about my vector fields as well.
As for my work,
I have used the quadratic formula from high school to find that we have $r^2 - 4 = 0 \text{ or } |r| \geq 2$
To find where the saddle note happens, I need to look at where $f(x, r) = 0$, differentiating it with respect to $x$ I get $f_x = r + 2x$
With this I identified the 5 different scenarios
- 2 fixed points if $r > 2$ or $r < 2$, (stable -> unstable)
- 1 fixed point at $x = -1$ or $x = 1$ and $r = 2$ (half stable)
- No fixed points at $ - 2 < r < 2$
So I might have an idea of how to sketch vector fields for different r, but I really don't know how to sketch the bifurcation diagram. How do I do it? Where do I start?
flat frigateBOT
Kazuchin
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@shut barn Has your question been resolved?
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@shut barn Has your question been resolved?
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@shut barn Has your question been resolved?
wild cape
@shut barn to get the bifurcation diagram we plot f(x,r)=0 where r is the horizontal axis and x is the vertical. in this case we can get an easy plot by solving r=...
safe radishBOT
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shut barn
Thank you!
fleet valve
This is such a small thing that’s stumping me, I broke down the equation. Typically when it comes down to solving a radical like this it’s only one numbers.
Like if it was only -5 to the square root of 25 it’d -25
If it was 20 (the product of 25-5) it’s be 100.
But that’s not the answer. Could I get explained how to do this with a practice problem?
safe radishBOT
This is such a small thing that’s stumping me, I broke down the equation. Typica...
fleet valve
I also just don’t think I understand how to solve for radicals, as that is the radical sign I believe. When solving for radicals I’m Square rooting the number on the outside right..? Hence why 5 is the square root of 25
quasi bison
This is such a small thing that’s stumping me, I broke down the equation. Typica...
$25 - 5 \times \sqrt{25}$. the order of operations is not changed here and doesn't require any special rules to be put in place.
flat frigateBOT
Ann
quasi bison
but also the "outside number" is not the "answer" to the "square root".
it is simply a number you multiply to the result of the rooting operation.
for example, $6 \sqrt{16}$ isn't the statement ``the square root of 16 is 6'', and in fact it isn't any sort of statement at all;
rather it is an instruction saying: \ ``take the square root of 16, then multiply 6 with the result.''
flat frigateBOT
Ann
flat frigateBOT
Ann
quasi bison
though a pair of brackets wrapped around a root isn't really necessary. because the root symbol (in proper handwriting) covers all of the stuff inside of the root, it acts as its own bracket.
fleet valve
I’m reading this, don’t take my silence as absence. I’m just writing notes
split kayak
though a pair of brackets wrapped around a root isn't really necessary. because ...
the obscure upperline brackets
fleet valve
I do have a question, why is “5” being put in parentheses and not “-5”?
flat frigateBOT
Ann
quasi bison
same shit though
fleet valve
Oh and I can see how that’d simplify into 25-(5 sqrt 25
Okay
OKAY!
No further questions thank you so much
.done
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covert rain
Why is the maximum probability 1, why can't it be 5?
safe radishBOT
Why is the maximum probability 1, why can't it be 5?
upbeat swan
by definition
probability means
how much out of 100%
so
how much from 1
so naturally 1 is the highest you can go
probability 5 would mean
500%
covert rain
But why is the sample space a subset of ?
upbeat swan
what?
upbeat swan
It’s not that the sample space is a subset of something else
It’s that every event is a subset of the sample space
covert rain
But the sample space Is a subset of the sample space ?
covert rain
But then what changes between a subset that has the same elements as the set and the set?
upbeat swan
its the same exact set then
covert rain
So why is it a subset?
upbeat swan
subset can also mean equal to the set
quasi bison
So why is it a subset?
"subset" is like ≤
the symbol is even ⊆ as a visual reminder of that
but also you want to be able to speak of events that are guaranteed to happen
bc that is what Ω means as an event
covert rain
Why is it needed =?
quasi bison
but also you want to be able to speak of events that are guaranteed to happen
upbeat swan
Why is it needed =?
because a set obviously contains itself and you need to account for this case
quasi bison
Why i Need to account for this case
thing become quite inconvenient otherwise
do you want to see exactly where and how it blows up in your face? or no
upbeat swan
Why i Need to account for this case
this is the same as asking
why do we use ≤
and not < all the time
quasi bison
ok let's consider a very basic kind of probability space
rolling a dice
nothing fancy just standard shit
EXCEPT we disallow Ω as an event.
i.e. for us any proper subset of {1,2,3,4,5,6} will be considered an event, but {1,2,3,4,5,6} is not.
do you follow so far, yes or no.
covert rain
Yes
quasi bison
ok
now, what you'd want is that if A and B are events, then A ∪ B should also be an event, right?
after all, it's just those two events with a logical or between them.
covert rain
By what property is AUB an event again?
quasi bison
it's literally an axiom of probability spaces.
(if you reject this also then i have nothing else to say to you. as in that case you're definitely rejecting too much to be doing anything resembling probability theory)
covert rain
I have seen that given n events with union operations/other operations other events can be constructed, but this does not say if they are always so.
quasi bison
unless stated otherwise, it is implied to always be so.
rejecting that means dealing with edge cases and weirdness and having to bloat every single statement with "as long as it isn't the entire space" or whatever.
desert juniper
wait, hold up. Why is omega not allowed as an event? isnt it the sure event with probability 1?
upbeat swan
I have seen that given n events with union operations/other operations other eve...
i'm preety sure it's always the case
by definition
covert rain
But then isn't it enough to say event=subset?
upbeat swan
no
an event is always a subset of the sample space, but not every subset has to be an event.
quasi bison
wait, hold up. Why is omega not allowed as an event? isnt it the sure event with...
goofy joe wanted to see what happens if we disallow it.
but then we stopped at him calling into question why the union of events was an event.
covert rain
I fixed
quasi bison
But what if for example A event -> notA event?
in a general probability space, the sigma-algebra answers the question of what subsets of Ω are considered events.
covert rain
I don't understand now why it's not enough to say that event=subset
upbeat swan
for example
Ω={1,2,3,4}
and
F={∅, Ω, {1,2}, {3,4}} (the event set)
now we look at the subset {1}. It is a subset of Ω, but {1}∈/F. So {1} is not an event in this probability space.
quasi bison
I don't understand now why it's not enough to say that event=subset
you can declare all subsets of Ω to be events. that will make a valid probability space.
there are just other subcollections that can also act as events but which are not the full 2^Ω.
covert rain
Ω={1,2,3,4}
and
F={∅, Ω, {1,2}, {3,4}} (the event set)
now we look at the subs...
Omega finite -> all can be events
upbeat swan
if omega is finite then you CAN take all subsets as events. You dont HAVE TO take all subsets as events
upbeat swan
yes
gray mulch
Why is the maximum probability 1, why can't it be 5?
You could invent a system where total likelihood is 5, but what for?
Consider this - if you have a box of 5 balls, 3 blue and 2 red, the probability of pulling out a blue is 3/5, a red is 2/5. You can't pull out 25 balls if there are only 5
Thus the probability as defined wouldn't make sense
If you are guaranteed to pull out a red or a blue ball, the probability is 5/5 = 1 < 5
So you are not guaranteed
Which is a contradiction
Again, only in this system of axioms
Which are called Kolmogorov's axioms.
One can invent a new set of axioms where this would make sense, but it just makes the math unnecessarily complicated
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covert rain
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earnest mirage
Hello, could someone check if this proof looks good please?
\begin{Theorem}
Assume $A$, $B$ and $C$ are sets.
If $A \cup B = A \cup C$ and $A \cap B = A \cap C$, then $B = C$.
\end{Theorem}
\begin{proof}
Let $A$, $B$ and $C$ be sets.
Either $A = \emptyset$ or $A \neq \emptyset$.
Suppose $A = \emptyset$.
In which case $A \cap B = A \cap C = \emptyset$ holds.
Also, $A \cup B = B$ and $A \cup C = C$, and because $A \cup B = A \cup C$, we have that $B = C$.\
Now, suppose $A \neq \emptyset$.
If $B$ and $C$ are both empty, then $B = C = \emptyset$, which is what we wanted to prove.
Suppose now that one of $B$ or $C$ is empty and the other non-empty, or both non-empty.
Without loss of generality, suppose $B = \emptyset$ and $C \neq \emptyset$.
Then $A \cup B = A$.
And since $A \cup B = A \cup C$, we have that $A = A \cup C$.
On the other hand, $A\cap B = \emptyset = A \cap C$.
Because $A$ is non-empty, $A = A \cup C$, and $A \cap C = \emptyset$, it must mean that that $C = \emptyset$.
Thus, $B=C$.\
Lastly, suppose $B \neq \emptyset$ and $C \neq \emptyset$.
Let $x \in B$.
Now, either $x \in A$ or $x \notin A$.
If $x \in A$, then $x \in A \cap B$.
Since $A \cap B = A \cap C$, then $x \in A \cap C$, which means that $x \in C$.
Since $x$ was chose arbitrarily, it holds for any $x \in B$.
Therefore, $x \in B$ implies $x \in C$, , so, by the definition of the subset, $B \subseteq C$.
On the other hand, suppose $x \notin A$.
Then $A \cup B$ holds because $x \in B$.
And since $A \cup B = A \cup C$, and $x \notin A$, it must be that $x \in C$.
In either case, $x \in B$ implies $x \in C$, so, by the definition of the subset, $B \subseteq C$.
The rest follow from a similar case with $x \in C$, and with that we have that $C \subseteq B$.
We shown that $B \subseteq C$ and $C \subseteq B$, thus $B = C$.
Therefore, in any case, if $A \cup B = A \cup C$ and $A \cap B = A \cap C$, then $B = C$.
\end{proof}
flat frigateBOT
Mor Bras
earnest mirage
\begin{Definition}[Subset of a set]
Suppose $A$ and $B$ are sets. If every element in $A$ is also an element of $B$, then $A$ is a \emph{subset} of $B$, which is denoted $A \subseteq B$.
\end{Definition}
\begin{Definition}[Union of sets]
The \emph{union} of sets $A$ and $B$ is the set $A \cup B = { x : x \in A \text{ or } x \in B}$.
\end{Definition}
\begin{Definition}[Intersection of sets]
The \emph{intersection} of sets $A$ and $B$ is the set $A \cap B = { x : x \in A \text{ and } x \in B}$.
\end{Definition}
flat frigateBOT
Mor Bras
tardy mango
This is unnecessarily long because of your casework
earnest mirage
For sure
tardy mango
Let $x \in B$.
\begin{itemize}
\item If $x \in A$, then $x \in A \cap B$. Since $A \cap B=A \cap C$, we have that $x \in A \cap C$, and so $x \in C$.
\item If $x \notin A$, we still have that $x \in A \cup B$ since $x \in B$. Since $A \cup B=A \cup C$, it follows that $x \in A \cup C$. But because $x \notin A$, it follows that $x \in C$.
\end{itemize}
In both cases, $x \in C$. So, $B \subseteq C$. But by symmetry, we can show that $C \subseteq B$, which allows us to conclude that $B=C$.
flat frigateBOT
Civil Service Pigeon
tardy mango
This is all you had to write
earnest mirage
I wanted to be as comprehensive as possible, covering all cases without exception
Just using the definitions
cloud hound
I guess I'll also demonstrate the contradiction idea I suggested before, as another example of a shorter proof
tardy mango
I wanted to be as comprehensive as possible, covering all cases without exceptio...
I don't see what generality you lose or gain from this but ok
earnest mirage
If B is empty, then there is no x in B
A, B, and C are sets, either they are empty or not
I tried working with subsets, but I couldn't find a way
cloud hound
\begin{theorem}
Assume $A$, $B$ and $C$ are sets.
If $A \cup B = A \cup C$ and $A \cap B = A \cap C$, then $B = C$.
\end{theorem}
\begin{proof}
Suppose $B \neq C$.
Then there exists an element in one of the sets that is not in the other.
WLOG, let $x \in B$ but $x \notin C$.
\begin{case}
If $x \in A$, then $x \in A \cap B$ but $x \notin A \cap C$ and by Definition 3, so we have $A \cap B \neq A \cap C$.
\end{case}
\begin{case}
If $x \notin A$, then $x \in A \cup B$ but $x \notin A \cup C$ by Definition 2, so we have $A \cup B \neq A \cup C$.
\end{case}
Combining these two cases, if $A \cap B = A \cap C$ \textit{and} $A \cup B = A \cup C$, then $B = C$.
\end{proof}
Coolempire93
cloud hound
And here we don't need to check the empty case because we have equality in the first place if both are empty
earnest mirage
```latex
\begin{theorem}
Assume $A$, $B$ and $C$ are sets.
If $A \cup B = A \cu...
Thank you for your proof. At the moment, I'm working with direct proof about sets, later I'll be learning about the contrapositive and contradictions.
cloud hound
Ah makes sense
earnest mirage
That is why I'd like to know if the proof looks good
cloud hound
I guess the way to do it directly in less steps would be (and then after I finish this one I'll look at yours)
earnest mirage
I appreciate your comments on proofs, even if it's not clear to me because I'm learning how to write proofs
earnest mirage
Let $x \in B$.
\begin{itemize}
\item If $x \in A$, then $x \in A \cap B$. Since...
So, thanks for your comment!
cloud hound
I guess the way to do it directly in less steps would be (and then after I finis...
Actually the way you did it is exactly what I was about to do
Your proofs looks good
Only change I would make would be to polish some wording
\begin{proof}
Let $A$, $B$ and $C$ be sets.
Then either $A$ is empty or $A$ is nonempty.
First, suppose that $A$ is empty.
If so, then $A \cap B = A \cap C = \emptyset$ holds.
Also, $A \cup B = B$ and $A \cup C = C$, so because $A \cup B = A \cup C$, we have that $B = C$.
Now, suppose that $A$ is nonempty.
If $B$ and $C$ are both empty, then $B = C = \emptyset$, which is what we wanted to prove.
Suppose then that one of $B$ or $C$ is nonempty.
\begin{case}
Without loss of generality, let $B = \emptyset$ and $C \neq \emptyset$.
Then $A \cup B = A$, and since $A \cup B = A \cup C$, we have that $A = A \cup C$.
On the other hand, $A\cap B = \emptyset = A \cap C$.
Because $A$ is non-empty, $A = A \cup C$, and $A \cap C = \emptyset$, we must have that $C = \emptyset$.
Hence, $B=C$.
\end{case}
\begin{case}
Lastly, suppose $B \neq \emptyset$ and $C \neq \emptyset$.
Let $x \in B$.
Now, either $x \in A$ or $x \notin A$.
If $x \in A$, then $x \in A \cap B$.
Since $A \cap B = A \cap C$, then $x \in A \cap C$, which means that $x \in C$.
On the other hand, suppose $x \notin A$.
Then $A \cup B$ holds because $x \in B$.
And since $A \cup B = A \cup C$, and $x \notin A$, it must be that $x \in C$.
In either case, $x \in B$ yields $x \in C$, so, by the definition of the subset, $B \subseteq C$.
By the same argument, we may show that $C \subseteq B$.
Hence, as $B \subseteq C$ and $C \subseteq B$, we have $B = C$.
\end{case}
Therefore, in any case, if $A \cup B = A \cup C$ and $A \cap B = A \cap C$, then $B = C$.
\end{proof}
Coolempire93
cloud hound
There's more I would change if it were mine but I tried to stick mostly to structural changes to lower the length
But in general words are preferred over symbols where the meaning is clear, for the sake of the reader and clarity
safe radishBOT
@earnest mirage Has your question been resolved?
cloud hound
Ah actually case 1 is impossible
You begin by assumine B =/= C
And derived a contradiction at that
Because you said C nonempty and showed that C empty
Making B /= C impossible, i.e. B = C
It's a simple fix
\begin{proof}
Let $A$, $B$ and $C$ be sets.
Then either $A$ is empty or $A$ is nonempty.
First, suppose that $A$ is empty.
If so, then $A \cap B = A \cap C = \emptyset$ holds.
Also, $A \cup B = B$ and $A \cup C = C$, so because $A \cup B = A \cup C$, we have that $B = C$.
Now, suppose that $A$ is nonempty.
If $B$ and $C$ are both empty, then $B = C = \emptyset$, which is what we wanted to prove.
Suppose then that one of $B$ or $C$ is nonempty.
\begin{case}
Without loss of generality, let $B = \emptyset$.
Then $A \cup B = A$, and since $A \cup B = A \cup C$, we have that $A = A \cup C$.
On the other hand, $A\cap B = \emptyset = A \cap C$.
Because $A$ is non-empty, $A = A \cup C$, and $A \cap C = \emptyset$, we must have that $C = \emptyset$.
Hence, $B=C$.
\end{case}
\begin{case}
Lastly, suppose $B \neq \emptyset$ and $C \neq \emptyset$.
Let $x \in B$.
Now, either $x \in A$ or $x \notin A$.
If $x \in A$, then $x \in A \cap B$.
Since $A \cap B = A \cap C$, then $x \in A \cap C$, which means that $x \in C$.
On the other hand, suppose $x \notin A$.
Then $A \cup B$ holds because $x \in B$.
And since $A \cup B = A \cup C$, and $x \notin A$, it must be that $x \in C$.
In either case, $x \in B$ yields $x \in C$, so, by the definition of the subset, $B \subseteq C$.
By the same argument, we may show that $C \subseteq B$.
Hence, as $B \subseteq C$ and $C \subseteq B$, we have $B = C$.
\end{case}
Therefore, in any case, if $A \cup B = A \cup C$ and $A \cap B = A \cap C$, then $B = C$.
\end{proof}
Coolempire93
cloud hound
All I did was remove where it said 'and C =/= emptyset'
We don't need to assume it (because we never use it)
Instead, we derive that C = emptyset = B
earnest mirage
Yes, from B empty we get C empty, so B = C
cloud hound
Actually we can extend the reasoning further to simplify
Rather than talking about 'both nonempty' or 'both empty'
Just consider 1 of the sets
You said either A is empty or A is nonempty
Now either B is empty or B is nonempty
In the first case we derive that B = C
In the second case we have case 2
Because the reasoning of case 1 could be applied to C, we do not need to check when C is empty, and can immediately jump to both nonempty
(further trying b nonempty and C empty leads to the same contradiction that I removed above)
\begin{proof}
Let $A$, $B$ and $C$ be sets.
Then either $A$ is empty or $A$ is nonempty.
First, suppose that $A$ is empty.
If so, then $A \cap B = A \cap C = \emptyset$ holds.
Also, $A \cup B = B$ and $A \cup C = C$, so because $A \cup B = A \cup C$, we have that $B = C$.
Now, suppose that $A$ is nonempty.
WLOG, consider set $B$. $B$ must either be empty or nonempty.
\begin{case}
Suppose $B$ is empty.
Then $A \cup B = A$, and since $A \cup B = A \cup C$, we have that $A = A \cup C$.
On the other hand, $A\cap B = \emptyset = A \cap C$.
Because $A$ is non-empty, $A = A \cup C$, and $A \cap C = \emptyset$, we must have that $C = \emptyset$.
Hence, $B=C$.
\end{case}
\begin{case}
Now suppose $B$ is nonempty. Because $B$ was chosen without loss of generality, we may also suppose that $C$ is nonempty.
Let $x \in B$.
Now, either $x \in A$ or $x \notin A$.
If $x \in A$, then $x \in A \cap B$.
Since $A \cap B = A \cap C$, then $x \in A \cap C$, which means that $x \in C$.
On the other hand, suppose $x \notin A$.
Then $A \cup B$ holds because $x \in B$.
And since $A \cup B = A \cup C$, and $x \notin A$, it must be that $x \in C$.
In either case, $x \in B$ yields $x \in C$, so, by the definition of the subset, $B \subseteq C$.
By the same argument, we may show that $C \subseteq B$.
Hence, as $B \subseteq C$ and $C \subseteq B$, we have $B = C$.
\end{case}
Therefore, in any case, if $A \cup B = A \cup C$ and $A \cap B = A \cap C$, then $B = C$.
\end{proof}
Coolempire93
earnest mirage
How did you set your Texit?
I see that you used the triple ticks to format LaTeX
Well, you can DM me if you like to
cloud hound
,texconfig autotex_level
flat frigateBOT
cloud hound
The triple ticks is what's called a codeblock
earnest mirage
Yes, I forgot the name of it, I used with my notes in markdown
earnest mirage
Actually we can extend the reasoning further to simplify
Thank you very much for your help!
.close
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cloud hound
No problem! 🙂
earnest mirage
\begin{proof}
Let $A$, $B$ and $C$ be sets.
\begin{case}
Suppose $B$ is empty.
\end{case}
\begin{case}
Now suppose $B$ is nonempty.
\end{case}
\end{proof}
Mor Bras
Compile Error! Click the 
reaction for more information.
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faint glade
Why do we not pay attention to the -a here? $\frac{a^2}{-a^4}$
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Why do we not pay attention to the -a here? $\frac{a^2}{-a^4}$
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Vortac
faint glade
Like, the result when simplifying is $\frac{1}{a2}$ right?
flat frigateBOT
Vortac
median vigil
that is incorrect
proud tree
**Vortac**
think of it as $- \left (\frac{a^2}{a^4}\right)$
faint glade
ahh okay
flat frigateBOT
Green
faint glade
and $\frac{(-a)^2}{a^4} $ is $\frac{1}{a^2}$ because $(-a)^2$ makes a positive?
flat frigateBOT
Vortac
median vigil
no
median vigil
you can't just move the - inside the square like that, you are changing the order of operations
faint glade
Its a separate question from the first one
proud tree
oh
faint glade
my bad for not clarifying
proud tree
then yes
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lime frigate
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proud tree
!15m
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rigid blaze
!15min
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quick trellis
<@&286206848099549185>
????
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compact plover
label every angle. you'll find some similar triangles
use properties of similar triangles
lime frigate
ok let me try that
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shut gust
Please Help I Got It Its System Is That
00000000.00110111
00000000= 0
00110111= .37
0011 .3
0111 .7
But I am Not Getting Correct Answer.Even Adding ^ i Couldnt UnderStand Pls Help
shut gust
@vagrant jolt
onyx tiger
<@397068881362944011>
shut gust
hey
onyx tiger
Hi
shut gust
can you help me
onyx tiger
No, idk binary
onyx tiger
The ping is @ Helpers
onyx tiger
No space
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btw, a friendly reminder to only use the ping once after 15 minutes have passed since the start of your session, so please don't spam, just in case.
onyx tiger
btw, a friendly reminder to only use the ping **once** after 15 minutes have pas...
-# you got ghost ping because i deleted
austere goblet
last warning: please do NOT DM me for math help. I have pinged you once to tell you the same thing the first time you tried that. the next attempt will be met with a modping and a block.
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shut gust
.reopen
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✅ Original question: #help-23 message
@shut gust Has your question been resolved?
shut hound
Please Help I Got It Its System Is That
00000000.00110111
00000000= 0
00110...
Can you show your work converting 0.37 and 0.94 to binary?
shut gust
shut hound
Hint: $(0.3)_{10} = (0.01\overline{0011})_2$
flat frigateBOT
@shut hound
shut gust
The Answer Should
shut hound
Of the question, sure
shut gust
But couldnt Gte That
shut hound
I mean yea, I see that
shut gust
Can You Send Me Solved Version Of That Question Just Once So Can Solve ALl Other My Self This will Help ME A Lot
shut hound
!nosols
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As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
shut hound
!noans
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shut gust
But
shut hound
I'm telling you though
shut gust
Ok
shut hound
the issue is you did your conversion wrong
shut hound
<@946379454056386560>
This is a secondary hint
shut gust
OK Gimem THat
shut hound
$(0.3)_{10} = (0.01\overline{0011})$\\
$(0.37)_{10} = (0.3)_{10} + (0.07)_{10}$\\
$\implies (0.37)_{10} > (0.3)_{10}$
@shut hound
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shut gust
.solved
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edgy isle
it's about 6.26 1
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it's about 6.26 1
edgy isle
take $X=(0,1)\cup {2}\subseteq\mathbb{R}$ as a metric space, then $C=(0,1)$ and $D={2}$ are both non-empty, closed and bounded subsets.
But then if $r=2$ the latter property is obviously satisfied, but then isn't $h(C,D)=2$? Meaning that the $\Longleftarrow$ implication is wrong?
flat frigateBOT
rbit
edgy isle
closed in X
tawdry plover
o oop
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obsidian oracle
yeah it seems you need C and D to be compact for 1) to be correct
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steel roost
Hi,
I need to find zeros of a mimo control system, which is equivalent to finding s where P(s) loses rank.
For the case where D is a diagonal matrix, or same inputs as outputs, P(s) can be solved as an eigenvalue problem.
But what approach can I take if D isnt? How can I find the s where P(s) loses rank, while P isnt a square matrix
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Hi,
I need to find zeros of a mimo control system, which is equivalent to findin...
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steel roost
.reopen
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✅ Original question: #help-23 message
hard crest
i really don't remember that much of this part of linear algebra, but my brain is telling me the QR decomposition would be useful here
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steel roost
I found an approach online that uses the first n columns of matrices U and V to bring P into a square form and then solve the eigenvalue problem.
Do you think that could be a valid approach? And how can I interpret this multiplication: U_n* P V_n.
A non square matrix can ever only get a rank that equals the smallest dimension right? Is it maybe a projection into a space of all actual internal dimensions of P, where we can then look for values that reduce the rank of P?
hard crest
A non square matrix can ever only get a rank that equals the smallest dimension right?
correct, an m x n matrix has a maximum rank of min(m, n)
oh singular value decomp. yeah sure
steel roost
i know that the rank is equal to the number of non zero singular values, but its still very hard to imagine what to do exactly
since i still need to find a way to solve for s
hard crest
yeah i'm not sure that actually helps you :/
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steel roost
thanks for the help 😅
slender steeple
hi, how do you make a confidence interval for the population variance in excel
slender steeple
<@&286206848099549185>
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slender steeple
https://support.microsoft.com/en-us/office/confidence-t-function-e8eca395-6c3a-4...
that's for standard error
im looking for the confidence interval of the population variance
which uses the chi square distribution
vale vessel
that's for standard error
a normal chi squared test is CHISQ.TEST(actual_range,expected_range)
but i'm not 100% i'm sure i'm understanding what you're asking, so i'll let someone else take it (am slightly scatterbrained)
slender steeple
not using it for independence tests on hypotheses, sorry
i think
its time to ping helpers
<@&286206848099549185>
vale vessel
i also imagine you could just hardcode a formula from the chi squared variance formula, since i'm not sure excel has a built in function for the thing you are attempting to calculate
slender steeple
i see
is there a function for gettng the left-tailed value of a chisq dist?
im only seeing a right tailed value
vale vessel
unsure
you may just have to hardcode the formula
you're looking for this formula, right?
vale vessel
ye i mean your best bet is probably to just hardcode whatever formula you are looking for
as sucky as that is
vale vessel
the good news is that ${\chi}_{1-a/2, N-1}^2$ is constant
flat frigateBOT
Mirror
slender steeple
in excel ofc
vale vessel
but how does one get the value of that
CHISQ.INV.RT(probability, deg_freedom) seems helpful but not 100% sure
gray mulch
i have to actually type shit now
Here's a video of someone setting this up in excel https://youtu.be/OhC5-TCNyNg?si=HYNMKYIDKFIMS3t1
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pearl furnace
how to add and subtract like terms
vale vessel
how to add and subtract like terms
!occupied
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how to add and subtract like terms
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slender steeple
Here's a video of someone setting this up in excel https://youtu.be/OhC5-TCNyNg?...
thanks
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burnt nymph
Hey guys how to approach identity based q
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Hey guys how to approach identity based q
quasi bison
general questions like this are better in #study-discussion, but also what do you mean by "identity based q"?
are you talking about questions that go "prove such and such trig identity"?
burnt nymph
are you talking about questions that go "_prove_ such and such trig identity"?
No like a³+b³,a^n-b^n,(a+b+c)²
. close
quasi bison
it's .close without a space
burnt nymph
.close
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twin warren
do i just overtake this now…. how do i get a fresh help page
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do i just overtake this now…. how do i get a fresh help page
quiet plume
It claims it automatically when you write something in a free channel.
What do you need help with?
twin warren
How can I graph a dual cone with 2 real axes and one imaginary axis?
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river sundial
Hello. Can someone look over my work please I am sure I did the steps right but I’m lost at the end
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Hello. Can someone look over my work please I am sure I did the steps right but ...
river sundial
lone void
i dont think it needed to be this lengthy
severe pond
u = x^2 + 16 maybe
lone void
x^3 sqrt(x^2+16) can be written as (x^2)(x)sqrt(x^2+16) and u can take x^2+16=t
river sundial
so was the answer wrong? or is it just a format issue when i do the final step when i take the limits
bc honestly its a lengthy trig sub problem yes but i dont see any mistakes tbh
severe pond
we are suggesting an easier way to do this
trig sub is ugly
i certainly didn’t bother reading through 20 lines of work
severe pond
yes but on my exam this is what the professor looks for
why?
i don’t think you understand our approach
river sundial
make sure that there is no cheating
ima just close this and figure it out ig thanks anyway
river sundial
ur advice was just do it better im telling u this is practice ill get it
thank you
.close
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severe pond
$\int_0^4 x^3 \sqrt{x^2 + 16} \dd{x} = \int_{16}^{32} \frac{u - 16}{2}\sqrt{u} \dd{u}$
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knief
quasi jewel
so like i wanna calculate the amount of light years it takes to drain out a femboy
yall can answer that or nah
💔
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bitter timber
so i kind of visualized a "trend line", you can sort of see that the function goes down by 3 ish after 4 blocks to the right, so that suggested a "mean" of y = -3/4 x
then the oscillations come from adding a sine component (not cos bc we have 0 at x = 0), so I just played with the sine coefficient until the oscillations looked comparable
neon vapor
you can kinda compare peaks and troughs
it barely increases from trough to peak and then decreases by a lot from trough to peak
neon vapor
no scale here, so its a bit of a malformed question
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glass edge
The total of 4 values is 50. Another value is added and now the mean is 20. What was the 5th value
safe radishBOT
The total of 4 values is 50. Another value is added and now the mean is 20. What...
severe pond
glass edge
nothing
severe pond
brilliant
well
can you translate the first sentence into an equation
call the first four values x_1, x_2, x_3, x_4
glass edge
a+b+c+d=50
severe pond
or that
glass edge
(a+b+c+d+e)/5=20
glass edge
e=100(a+b+c+d)
severe pond
🤔
glass edge
no
severe pond
you are forgetting something
glass edge
100/a+b+c+d
flat frigateBOT
knief
severe pond
what happens after multiplying by 5
glass edge
$(a+b+c+d+e)/5=20\a+b+c+d+e=100\e=100-a+b+c+d$
flat frigateBOT
Tan
severe pond
parentheses
glass edge
$e=100-(a+b+c+d)$
flat frigateBOT
Tan
severe pond
good
severe pond
well done
glass edge
thanks
but
i have one more question
Intervals Frequency
10-19 4
20-29 7
30-39 4
40-49 6
50-59 2
Find median and mean
mean is easy
its the multiplication of interval by frequency
divided by the sum of frequency
but median.?
idk
glass edge
could you help me tho
viscid monolith
Whoops. can't let my thoughts slip
glass edge
i think im wrong
glass edge
statistics is sooo long ago..
u saying its outdated?
viscid monolith
No. I meant its so long ago since i took it
glass edge
oh
viscid monolith
Heres an example
Age groups :
1 - 10 : 15 people
11 - 20 : 30 people
21 - 30 : 7 people
31 - 40 : 2 people
If i were to put it into words
Age groups :
Kids ; 15
teens ; 30
young adults ; 7
not-so-young adults ; 2
If i find where exactly it hits the middle, that is
15 + 30 + 7 + 2 = 54
54/2 = 27
The median lies on the 27th position
So i would count from 0 to 27 starting from the first ordered list
The 27th position would be the teen age group.
glass edge
understood
viscid monolith
Uh but this is for even ya
glass edge
but then what the heck is the formula (n+1)/2 for??!
viscid monolith
If there are odd terms, you would get
yes u'd get that formula
Okay, heres an example
there are 5 males 6 females
total = 11 right?
glass edge
yes
glass edge
So is the median females or males?
uhm
viscid monolith
It cant be half male and half female right?
glass edge
female
viscid monolith
So we take (5+6 +1)/2 where 5+6 is your n terms.
glass edge
because 5.5 is approximately 6
viscid monolith
but then what the heck is the formula (n+1)/2 for??!
Follow this
Then you'd get the 6th position
hence the median would be the female at 6th position
glass edge
how do you have two constants for one n term
viscid monolith
n is the total number of terms in that data set
glass edge
n is the total number of terms in that data set
i see
glass edge
and an alternate route you could do is add the frequency and divide it by 2 to get the position?
okay
viscid monolith
Uh yes.
viscid monolith
In your case yes, the summation of the frequency would give you n number of terms
viscid monolith
Hold on. thats not how it works
glass edge
In your case yes, the summation of the frequency would give you n number of term...
4+7+4+6+2 isnt equal to 5 tho?
viscid monolith
Nono,
Okay let me try to draw it out
in this case, the median height is 168 where the third kid lies
viscid monolith
So
161, 167, 168, 168, 172. -> Number of terms : 5
Since there are 5 terms, it is odd.
The position at which the median occurs is 6/2 = 3rd position
viscid monolith
but you did the exact thing though
I force it to be an even number
glass edge
what
viscid monolith
This is the definition of median
*The median is the middle value in a dataset when numbers are arranged in order, dividing the data into two equal halves; *
glass edge
right
viscid monolith
Two equal halves
viscid monolith
if you take the sum of ur frequency what do you get?
glass edge
if you take the sum of ur frequency what do you get?
23
viscid monolith
Intervals Frequency
10-19 4
20-29 7
30-39 4
40-49 6
50-59 2
Here's your uh original question
glass edge
Intervals Frequency
10-19 4
20-29 7
30-39 4
...
there're 4 numbers in the 10-19 interval
viscid monolith
Can i say
a = 10-19
b = 20-29
c = 30-39
d = 40-49
e = 50-59?
glass edge
1
viscid monolith
Based on frequency
glass edge
7
viscid monolith
aaaa,bbbbbbb,cccc,dddddd,ee
In this case let me write it this way
Do we see there is total of 23 terms?
glass edge
yes
viscid monolith
Okay, i'll be showing you WHY we are using (23 + 1)/2
Okay, 23/2 is around 11.5 right?
glass edge
yes
glass edge
yes
glass edge
no
viscid monolith
okay
glass edge
its b+c/2?
glass edge
24/2
viscid monolith
24/2 = 12
glass edge
=13
viscid monolith
12!!
glass edge
=12
viscid monolith
AHHAHA
aaaa,bbbbbbb,cccc,dddddd,ee
We see that we isolated it at "c"
Can you count on either sides, how many terms are there?
glass edge
11
viscid monolith
Are they equal?
glass edge
yes
viscid monolith
Then c that lies in the middle
glass edge
but what if the number was even
viscid monolith
Then it makes ur life even simpler
glass edge
then you take the average of the two middle numbers
viscid monolith
Lets use the same case
aaaa,bbbbbb,cccc,dddddd,ee
I removed 1 b, you should get 22 terms
glass edge
okay
glass edge
yes
glass edge
i got 11?
viscid monolith
Nope yes, youre right
viscid monolith
So if we think about it right,
a,b,c,d,e,f
For this even set
The middle lies in between
c and d
glass edge
yes
viscid monolith
The following only works if its numerical values
for example
161, 162, 167, 172, 174, 178
The middle value is between 167 and 172
Then you will have to take (167+172)/2
This only works if your set is a numeric value and there are even terms
viscid monolith
no, the middle value lies in between 167 and 172
glass edge
for example
161, 162, 167, 172, 174, 178
but there's an odd set of numbers?
viscid monolith
oh i thought i had 6 terms
glass edge
,w median of 161, 162, 167, 172, 174
flat frigateBOT
viscid monolith
ok my example is supposed to have 6 terms
glass edge
,w (167+172)/2
flat frigateBOT
glass edge
ok my example is supposed to have 6 terms
oh
viscid monolith
Whatever it is this is a very simple representation of what im trying to say
If you have odd terms
Within the set, there is a certain term where the number of terms on the right side is equals to the number of terms on the left side
Eg.
1,2,3,4,5,6,7
glass edge
understood
viscid monolith
The median is 4 and there are 3 terms on each side.
glass edge
but what about intervals
viscid monolith
but what about intervals
What about the intervals?
glass edge
are you supposed to take the average of intervals?