happy new ear
#help-28
maiden vapor
rare dock
i should have read the proof so i could participate
quartz flare
Yeah
I didn't really know how ot say that
maiden vapor
anyways... Bubble sort on k+1 elements does extra comparison involving a_{k+1} in every outer pass and swaps involving elements from the first k positions and the (k+1)-th position can occur early and change the configuration of the first k positions in ways that are not the same as the n=k run.
quartz flare
๐ค
Oh
Well yes not the whole algorithm
But it's the same comparisons that would occur in bubble sort with the first k
And then an extra comparison with the k+1th
Every outer loop
quartz flare
anyways... Bubble sort on k+1 elements does extra comparison involving a_{k+1} *...
Yes this is true
I just shut my eyes and pretended not to worry about that
maiden vapor
no
quartz flare
I think I can try the invariant proof idea
maiden vapor
yes, that's the standard way
quartz flare
Although I do wonder how hard it would be to prove things for algorithms that are really random and nasty
maiden vapor
Well yes not the whole algorithm
the point is your wording is incorrect and the inference does not follow.
take [2,1,3,0] with k = 3 and you will get a counter example to the inference in your proof
quartz flare
Yes that's why I noticed and then hear-no-evil-see-no-evil'd ๐
Basically intending to abuse the indexing
maiden vapor
quartz flare
Since the book is wishywashy with it
maiden vapor
well, go ahead with the invariant proof
quartz flare
Am doing ๐
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@quartz flare Has your question been resolved?
quartz flare
Okay so I started attempting to write a lemma
\begin{lemma}[The Bubble-Up Property]
After the $i$th iteration of the outer \textbf{for} loop of Algorithm 4, the last $i$ elements are in their final positions.
\end{lemma}
Specifically
glossy valveBOT
Coolempire2026
quartz flare
But it turns out to just be the whole proof
So it can barely fit in one message ๐
But I'm this close to done
\begin{proof}
We intend to show that after the $i$th iteration of the outer \textbf{for} loop, the last $i$ elements are in increasing order, and in their final positions.
We go by double induction on $i$ and $j$.
\begin{case}[Base Case]
We seek to show that the largest value ends in the $n$th position of the sequence after the $i = 1$st iteration.
If $j = 1$, then $a_1$ and $a_2$ are compared in the inner loop, such that if $a_1 > a_2$, then $a_1$ and $a_2$ interchange positions, and if not (i.e., $a_1 \leq a_2$), they remain in their positions. In either case, the greater ends in position $a_2$.
If $n = 2$, both \textbf{for} loops terminate here, and the $1$st iteration of the outer \textbf{for} loop yields the last element in its final position (since the array is returned as-is).
Otherwise, the loop continues:
Suppose that for $j = k$, the largest value among the first $k$ elements ends up at the $k+1$th position.
Consider that when the loop continues to $j := k+1$, $a_{k+1}$ and $a_{k+2}$ are compared, so that the greater of the two ends up in position $a_{k+2}$ as described earlier. Hence, when $j = k+1$, the largest value among the first $k+1$ elements ends up in the $k+2$nd position.
We may continue this induction up to $j := n - i = n - 1$, where the inner \textbf{for} loop terminates, so that the largest value in the sequence ends up in position $a_n$.
\end{case}
\begin{case}[Inductive Step]
Now suppose that after the $k$th iteration of the outer \textbf{for} loop, the last $k$ elements are in their final positions, and consider the $i := k+1$th iteration.
Relying on the induction from the first half of this proof, after the $n-i$th iteration of the inner \textbf{for} loop, the greatest value among the first $n-i$ elements of the sequence ends in the $n-i+1$th position after the $i$th iteration of the outer \textbf{for} loop.
Because $i$ increases in each iteration,
\end{case}
\end{proof}
Coolempire2026
quartz flare
well, go ahead with the invariant proof
Okay I'm out of space but hopefully you can see my intention
at the end all that was left to show is
- because i increases each iteration and j only goes up to n - i, the top n - i + 2 elements are unchanged after every iteration
- the n - i + 1th element isn't changed because it greater than the first n - i elements
- therefore they are increasing and in final order
Okay but because this was written originally for the lemma I think it can be improved now that I realize that it's just the whole proof together
Let me see if I can tune it
Now that I know what's missing/what I want as well
maiden vapor
I shall wait for the tuned version then
quartz flare
\begin{lemma}[The Bubble-Up Property]
After each iteration of the outer \textbf{for} loop, the greatest value among the first $n - i + 1$ elements ends in position $a_{n - i + 1}$.
\end{lemma}
\begin{proof}
We go by induction on $j$, showing that after the $j$th iteration of the inner \textbf{for} loop, the greatest value among the first $j+1$ elements ends in position $a_{j+1}$.
When $j := 1$, $a_1$ and $a_2$ are compared in the inner loop, such that if $a_1 > a_2$, then $a_1$ and $a_2$ interchange positions, and if not (i.e., $a_1 \leq a_2$), they remain in their positions. In either case, the greater ends in position $a_2 = a_{j+1}$.
Now, suppose that after the $j := k$th iteration of the inner \textbf{for} loop, the largest value among the first $k+1$ elements ends up at the $k+1$th position.
Consider that when the loop continues to $j := k+1$, $a_{k+1}$ and $a_{k+2}$ are compared, so that the greater of the two ends up in position $a_{k+2}$ as described earlier. Hence, after the $j := k+1$th iteration, the largest value among the first $k+2$ elements ends up in position $a_{k+2} = a_{j+1}$.
We may continue this induction up to $j := n - i$, where the inner \textbf{for} loop terminates, so that the largest value among the first $n-i+1$ elements ends up in position $a_{n-i+1}$.
\end{proof}
Coolempire2026
quartz flare
I believe this is the Lemma I originally wanted
Continuing now for the actual proof
quartz flare
\begin{lemma}[The Bubble-Up Property]
After each iteration of the outer \textbf{for} loop, the greatest value among the first $n - i + 1$ elements ends in position $a_{n - i + 1}$.
\end{lemma}\vspace{1em}
\begin{proof}[Revised Proof of the Correctness of Algorithm 4]
We intend to show that after the $i$th iteration of the outer \textbf{for} loop, the last $i$ elements end in increasing order.
First note that the largest value among the whole sequence ends in the $n$th position of the sequence after the $i = 1$st iteration by Lemma 1, because $a_{n-i+1} = a_n$. Therefore, $a_n \geq a_\ell$ for all $\ell \leq n$.
Now suppose that for $i := k$, we have the last $k$ elements of the sequence in increasing order. After the $i := k+1$th iteration of the outer \textbf{for} loop, the largest value among the first $n - k$ elements ends in position $a_{n-k}$ by Lemma 1. And because $a_{n-k+1}$ is the greatest value among the first $n - k + 1$ elements, we must have $a_{n-k} \leq a_{n-k+1}$. Therefore, because the last $k$ elements of the sequence are in increasing order, we have $a_{n-k} \leq a_{n-k+1} \leq \cdots \leq a_n$, i.e., the last $n-k+1$ elements of the seuqnece are in increasing order.
We may induce to $i := n-1$, which yields the last $n-1$ elements in increasing order. Finally, we note that after the the $i := n-1$th iteration of the outer \textbf{for} loop, the greatest value among the first $n - i + 1 = 2$ elements of the sequence ends in position $a_2$ by Lemma 1, so we have $a_1 \leq a_2$. Since the latter $n-1$ elements of the sequence are in increasing order, we must end with $a_1 \leq a_2 \leq \cdots a_n$ after the $n-1$th (final) iteration of the outer \textbf{for} loop, and thus Algorithm 4 yields the given sequence in increasing order.
\end{proof}
Coolempire2026
quartz flare
Okay done now!
maiden vapor
looks good
small fixes
ends in position a_{n-i+1}
a_{n-i+1} is a value, not a position.
so either ends up as the element a_{n-i+1} or ends in position n-i+1
quartz flare
You'll probably be horrified at how many times they're used interchangeably across both ๐
maiden vapor
And because a_{n-k+1} is the greatest value among the first n-k+1 elementsโฆ
This is not true by itself unless you also say it was placed there in the previous pass and never moved again (which bubble sort guarantees because the inner loop doesnโt reach that far). So say
By the induction hypothesis, positions n-k+1,\dots,n are already fixed and are not touched during pass k+1.
maiden vapor
You'll probably be horrified at how many times they're used interchangeably acro...
one needs not be sloppy just to accommodate for such sloppiness
quartz flare
> And because a_{n-k+1} is the greatest value among the first n-k+1 elementsโฆ
T...
Ah yes originally it had the part about things being in their final position
Then I was like surely it's implicit somewhere in here
maiden vapor
and lastly if you already prove after i=n-1, the last n-1 elements are sorted, the first element must be smallest automatically
mhm
quartz flare
This is really good
maiden vapor
quartz flare
I don't know when I'll get to insertion sort since I leave for the math conference day after tomorrow but now I have an understanding of how to approach
maiden vapor
quartz flare
I would ask if there's any sessions you'd be interested in picures from but you probably know all the subjects already ๐
In any case thank you ๐
๐ if only I could use you as my personal evaluator
Well I'll start hearing your comments in my head soon enough anyway
quartz flare
I'll be writing a proof and hear "don't be sloppy just to allow sloppiness"
maiden vapor
I would ask if there's any sessions you'd be interested in picures from but you ...
well, if there's anything interesting to share, do so
maiden vapor
I'll be writing a proof and hear "don't be sloppy just to allow sloppiness"
amazing
quartz flare
well, if there's anything interesting to share, do so
idek what's interesting for you ๐ hlounge will probably see at least a few things though
Time for bed 
Thanks again ๐
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maiden vapor
Time for bed <:sleep:911870909516754964>
gightoolemplud
quartz flare
booly
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tame cradle
i need some inequality help
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i need some inequality help
tame cradle
i know s<<N
as s<log base 2 N
now we know b=N/l we have k<s/2 i need to prove b>k now if we assume k>B then we have s/2>N/l or l>2N/s we know s>2 and odd
how can we prove this is always true
or this is always false'
dusky hedge
You want to prove what
hidden harbor
!original
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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.
tame cradle
now how can we say this is never true thus our assumption is false
i have been asked to prove or disprove b>K
and given these conditions
s<<N
s<log base 2 N
hidden harbor
so you should have the original with you
tame cradle
b=N/l
k<s/2
one sec lemme show it
now i am trying by contradiction
by assuming
b<k
or k>b
hidden harbor
that's not the original
and you can choose the variables so that it holds or not at your choosing
hidden harbor
because it's not complete
tame cradle
cant i do anything??
tame cradle
he is saying this is
hidden harbor
so are you, but that doesnt make it true
i can also say that i'm riding a pink unicorn in the moon
tame cradle
what can i do he is insisting
hidden harbor
you can tell him that he sucks and that he should find the original
tame cradle
he is saying just prove this lmao
tame cradle
you can tell him that he sucks and that he should find the original
and get my friend lost over a help channel nvm
lmao
i will say i was not able to
hidden harbor
i mean, your friend is trolling you. That doesnt mean you can troll us
tame cradle
no problem its not that important
hidden harbor
so either tell him he sucks, or troll him back
tame cradle
how can i troll him?
dry arch
felix is right, i have a counter example
tame cradle
guide
hidden harbor
it's your friend
tame cradle
do you have any such pic
dry arch
s=3
n=100
k=0
l=-1
b=-100
tame cradle
i dont know man
hidden harbor
i knew l negative was gonna appear
tame cradle
what are these even
dry arch
all the givens are satisfied
hidden harbor
literally the first thing i thought for fucking shit up
dry arch
don't escalate, just look at the facts
tame cradle
one sec lemme insist on asking maybe he gives
imma lie i am asking my prof
he has sent me this now
lmao man he is such a sucker
he sent me that instead of this
its a statement based question
;-;
dry arch
it says prove or disprove
hidden harbor
n being a positive integer. First line already was not included on what you said
tame cradle
now'
tame cradle
n being a positive integer. First line already was not included on what you said
i am also seeing it now
hidden harbor
and reading the rest, you had left out pretty much everything important
tame cradle
for first
tame cradle
and reading the rest, you had left out pretty much everything important
he has sent me this just now when i said imma ask my prof
hidden harbor
like, you literally left out ALL the context
tame cradle
okok i get it
but at the last he sent it or i would have said i cant and went away
:>
hidden harbor
well, i did tell you i could guarantee it wasnt the original
tame cradle
can we even solve this
?
he is somewhat a math freak guy randomly sends me such typa shit
i am a 2nd chem e major so i just come to the server lmao
;-;
hidden harbor
well, 2nd line is redundant, because you get that from the definition of n being square-free
tame cradle
well, 2nd line is redundant, because you get that from the definition of n being...
idk
ok yes i saw it now
man i can still only think of contradicting
like can we try b<k
??
like disprove
hidden harbor
hm, still looks incomplete, what is a?
tame cradle
idk lemme ask him
one sec
man he is saying leave the question he disproved it
i guess this is it
thanks still for helping
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zenith grail
How to find whether P= โ32 - โ24 and Q= โ50 - โ48 is bigger or smaller than P. And please specify the method
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How to find whether P= โ32 - โ24 and Q= โ50 - โ48 is bigger or smaller than P. A...
charred hawk
How to find whether P= โ32 - โ24 and Q= โ50 - โ48 is bigger or smaller than P. A...
P= 0.76 , Q= 0.14 (rounded answers). Assuming these are square roots?
Plug into a calculator root 32 minus root 24 to get 0.76
craggy tapir
computing numerically is probably not the intended approach
craggy tapir
How to find whether P= โ32 - โ24 and Q= โ50 - โ48 is bigger or smaller than P. A...
my first intuition would be trying to factorize it
charred hawk
If numerical is not the ideal approach, the two can be deduced by perfect squares as a method
thick hedge
How to find whether P= โ32 - โ24 and Q= โ50 - โ48 is bigger or smaller than P. A...
โ32= 4โ2 for instance, something like that for every number
craggy tapir
i might have an approach, but i don't guarantee it's the most efficient one
the idea is to compute P - Q and see its sign
spiral vigil
my first thought was to compare P^2 and Q^2
zenith grail
my first thought was to compare P^2 and Q^2
I think it had something to do with that also
craggy tapir
that also works, since P and Q are both positive
outer lotus
How to find whether P= โ32 - โ24 and Q= โ50 - โ48 is bigger or smaller than P. A...
Use rationalization
outer lotus
Use rationalization
โa - โb = (a-b)/(โa+โb)
It's easier to compare because of the plus sign in the denominator
outer lotus
This way proving P is bigger is very straightforward
It has a bigger numerator and a smaller denominator
chrome elbow
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chrome elbow
can anyone help me with this please
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What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
i got the equation of the circle and the straight line but idk what to do after
bright bronze
circle doesnt intersect with the line means that the distance between the center and the line is larger than the radius
bright bronze
yes
chrome elbow
i got -px-3 + root 83
bright bronze
its better if you show your work
chrome elbow
i cant take a pic but i can write it out
bright bronze
if you dont know latex, instead of writing math steps, just describe the steps
chrome elbow
line - y=3x-9 so (3,-9) circle - (x+px)^2+(y-2)^2=24+(px)^2 so center is (-px,2)
bright bronze
what does that even mean?
bright bronze
Step 1: Find the center of the circle
Step 2: Find the distance between the center and the line
Can you tell me the answers to these steps?
bright bronze
line - y=3x-9 so (3,-9) circle - (x+px)^2+(y-2)^2=24+(px)^2 so center is (-px,...
thats not the center
chrome elbow
what does that even mean?
mb i sent by accidnet
chrome elbow
thats not the center
oh
bright bronze
well, theres still a bit of a problem with the y coordinate
onyx glen
may i suggest an alternative route
chrome elbow
is it positive
chrome elbow
may i suggest an alternative route
yea sure
onyx glen
treat the line and circle as a pair of simultaneous equations. rewrite the line as y=3x-9 and substitute into the circle equation
get a quadratic in x
declare its discriminant to be <0
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unique saffron
How can i show that if you have a R-module M and B is a subset of M and for all R modules N and for all images \alpha: B->N there exists a unique module morphism \theta:M->N for which \theta(v)=\alpha(v) for all v in B that then M is a free module. this is kind off the invere of the universal property of free modules but i dont know how to show this
zenith kernel
How can i show that if you have a R-module M and B is a subset of M and for all ...
Thatโs exactly the definition of free module, M is the free module generated by B. So nothing to show, unless you have a different definition then I show you why they are equivalent
unique saffron
But shouldn't i show explicitly that M is then generated by B?
zenith kernel
So M is generated by B and B is R-independent you want to show?
maiden vapor
seems so
zenith kernel
For any b, consider f_b: B->R, mapping b to 1 and c !=b to 0
unique saffron
For any b, consider f_b: B->R, mapping b to 1 and c !=b to 0
is that how i would construct the basis ?
zenith kernel
I mean there are two steps
First, let N=ฮฃRb: b in B
Consider B->M/N mapping every element to 0, this will show you M=N
Second my $f_{b}$ will let you construct $\alpha=\oplus f_{b}: B \to \oplus R_{b}$ where $R_{b}=R$ for any $b \in B$
glossy valveBOT
Cogwheels of the mind
zenith kernel
This will let you prove itโs independent
onyx glen
Second my $f_{b}$ will let you construct $\alpha=\oplus f_{b}: B \to \oplus R_{b...
\bigoplus btw
zenith kernel
$\bigoplus$
glossy valveBOT
Cogwheels of the mind
zenith kernel
I see
unique saffron
and generative would come out of the fact that \theta is fully know by alpha ?
zenith kernel
First, let $N=\sum_{b \in B}Rb$.
Consider $B \to M/N$ mapping every element to $0$, this will show you $M=N$.
Second my $f_{b}$ will let you construct $\alpha=\oplus f_{b}: B \to \bigoplus_{b \in B} R_{b}$ where $R_{b}=R$ for any $b \in B$ will let you prove $B$ is $R$-independent.
torn fossil
help
zenith kernel
and generative would come out of the fact that \theta is fully know by alpha ?
B generates M will be shown once you find out that in first step, there are two homomorohisms M->M/N. The canonical projection and the 0 homomorphism, that can extend this constant 0 mapping B->M/N
B being independent is shown just by existence of an extension in second step
unique saffron
Is there an easier way to prove this because im not so advanced on modules
glossy valveBOT
Cogwheels of the mind
zenith kernel
Np
unique saffron
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random turtle
Trigonometry: Trying to find where I went wrong. Problem is to find angle A. I used law of cosines to find my first side "a", which is kind of close? Then I tried to find one of the angles "B" by using law of sines, but it's not even close to 4.76ยฐ ๐ตโ๐ซ My work is on the lined sheet of paper.
random turtle
slate violet
**Trigonometry**: Trying to find where I went wrong. Problem is to find angle A....
seems like you forgot how to do right-angled trig
random turtle
I'm trying to find the right steps to take. By first identifying what type of triangle this is. I know it is a SAS type of triangle. The next step is to follow the formulas.
slate violet
I'm trying to find the right steps to take. By first identifying what type of tr...
I'm telling you that you do not need the law of sines or cosines
because it's a right-angled triangle
so you can just use right-angled trig
queen flame
so you need this?
slate violet
so you need this?
the other angle
random turtle
so you need this?
Angle BCA
random turtle
do you know arctan?
Nope
queen flame
hmm..
slate violet
okay, do you know that tan = opp/adj?
random turtle
okay, do you know that tan = opp/adj?
Yeah Soh Cah Toa
slate violet
nice
okay, so for angle BAC
the side opposite to that is the 6
the side adjacent to that is the 0.5 (it's not AC cause that's the longest side, the hypotenuse)
random turtle
Okay ๐ค
slate violet
okay, do you know that tan = opp/adj?
okay, so what's tan(BCA)?
random turtle
tan = 6/0.5
random turtle
I don't
slate violet
I don't
it's just how you find the angle given tan(angle)
tan(BCA) = 6/0.5
apply arctan on both sides, arctan(tan BCA) = arctan(6/0.5)
BCA = arctan(6/0.5)
then use your calculator to find the value
queen flame
hmm.. i thought you wrote sin inverse so you would be familiar with it, (worth a shot)
glossy valveBOT
south
glossy valveBOT
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zealous stirrup
Okay thanks guys! I'll revisit this.
Learn trigonometric ratios
zealous stirrup
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hoary ember
given $5\cos(x)+12\sin(y)=15$, find the maximum of $5\sin(x)+12\cos(y)$
glossy valveBOT
ihave<skissue>
hoary ember
yea status 1 idfk
zenith kernel
sin(x+ฯ/2)=cos(x)
slate violet
you get nice things if you square both equations (5 sin x + 12 cos y = k for some k)
zenith kernel
Forget what I said, I follow south
twin wolf
you get nice things if you square both equations (5 sin x + 12 cos y = k for som...
south w the clutch
hoary ember
hold on chat gimme a few mins
slate violet
south w the clutch
I've seen this a million times
this is a nice twist on the problem since usually k is given
twin wolf
I've seen this a million times
this is my first time
hoary ember
$$225=25\cos^2(x)+144\sin^2(y)=25(1-\cos^2(x))+144(1-\sin^2(y))+120\cos(x)\sin(y)=169-(25\sin^2(x)+144\cos^2(y))+120\cos(x)\sin(y)$$
$$25\sin^2(x)+144\cos^2(y)+120\cos(x)\sin(y)=-56$$
$$25\sin^2(x)+144\cos^2(y)+120\sin(x)\cos(y)=k^2$$
oh wait
glossy valveBOT
ihave<skissue>
hoary ember
$$k^2=120\sin(x)\cos(y)-120\cos(x)\sin(y)-56$$
glossy valveBOT
ihave<skissue>
slate violet
right and now the trig identity
tall onyx
Seeing ihaveskissue asking about a problem that's not olympiad type is kinda weird
to me at least
hoary ember
Seeing ihaveskissue asking about a problem that's not olympiad type is kinda wei...
no this is oly
tall onyx
no this is oly
wut
hoary ember
not an "advanced" one tho
slate violet
exactly you'd think they'd have learnt this already
okay yeah I think I also did learn this in oly too
hoary ember
right and now the trig identity
uhh which?
slate violet
there are some textbooks which cover this
hoary ember
oh wait product to sum?
glossy valveBOT
ihave<skissue>
hoary ember
120-56=64
k=8
tall onyx
okay yeah I think I also did learn this in oly too
You studied for oly math?
slate violet
<:kongouderp:483988392099446794> You studied for oly math?
only for regionals then I didn't bother
I didn't like comp maths anyways
like I got to AMC 12 level pretty much
hoary ember
ok tbh the thought of squaring crossed my mind but i immediately gave up after seeing that it became sin cos and cos sin as i dont think ive seen this manip (actually i think ive seen this once)
slate violet
that's good you've seen it
hoary ember
yeah i should keep that in mind
slate violet
I mean your brain is full of comp maths that I don't know
so everyone has different strengths
if you're preparing for calculus that trig identity should become automatic
hoary ember
ngl i feel like im weaker in comp than proof based oly
tall onyx
I didn't like comp maths anyways
Agree, I kinda hate comp math
slate violet
product-to-sum is nice (and also gets used in comp maths a lot, heh) but you only really need to use it in say, Fourier series integration
or there are certain complex number trig questions too
slate violet
ngl i feel like im weaker in comp than proof based oly
oh even after doing RA BMO1 still feels impossible
it feels like they're cheating cause the proof is easy once you've seen it
this kind of shit
hoary ember
oh even after doing RA BMO1 still feels impossible
yea im struggling abit with bmo too
hoary ember
this kind of shit
hm wait i feel like ive seen this
oh i helped someone with that problem (and failed)
okay guys thank you!
.solved
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light moth
is there way to motivate algebraic manipulation its mostly just guesswork
slate violet
is there way to motivate algebraic manipulation its mostly just guesswork
the more you practice, the more useful ideas you have in your toolbox
like, you really have to practice doing questions
you can write out just the approach, how you would do it step by step, instead of solving every question in full
then your brain figures out the patterns on its own and we call that 'intuition'
languid osprey
Hello
languid osprey
.reopen
stone matrix
Uhh
onyx glen
.reopen
not how that command works.
but welcome to the server.
do you have a math question to ask right now?
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radiant roost
how do u solve this kinda equation
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how do u solve this kinda equation
grave elm
you can either multiply by the denominator, expand everything and get a quadratic equation or subtract 20 from both sides and subtract it from the fraction
grave elm
there has to be a better way than regular expansion?
maybe a substitution would reduce the pain
actually no
if it was 0.4 - x it could, but not with +
restive geyser
Mult. the denominator, and strategically multiply both sides by 100
grave elm
only by 10, the other 10 is from the 20
restive geyser
oh, even better then, yh
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arctic bough
how do u solve this kinda equation
tbh thats an easy expansion so u could just do that
else idk anything else ngl
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shell pawn
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shell pawn
At last question, some people did it with recurrence and it worked
But some guys proved that the question was wrong
They convinced the teacher and everyone got a 2+
shell pawn
The confusing part is that we changed xโฐ
I hope that people give me there opinion on this
glossy valveBOT
mtt
left notch
Hoi
shell pawn
nvm this is $f(x)=\frac x{1+x\ln x}$
Yep
shell pawn
The last question 3)b)
spice knot
since you didnt translate this, Im taking a guess
in 3a) the variable alpha is defined as the unique positive solution to x ln x = 1
problem is alpha > 1
so in 3b) it wouldnt make sense to have x0 = alpha
also x ln x < 0 for 0 < x < 1
Im going to assume this alpha is ignored and we use some general variable a where 0 < a < 1
another problem is the yn = e^(1/xn) at the bottom
this would mean y0 = e^(1/x0), even though earlier y0 = x0
so that means x0 = y0 = e^(1/x0)
solving for x0,
x0 = e^(1/x0)
ln x0 = 1/x0
x0 ln x0 = 1
x0 = alpha
so again the question thinks x0 is alpha which we cant do
I think what we gotta do to fix this is ignore the 0 < x0 < 1 definition
since 0 < f(x) < 1 for all x > 0, that means 1) and 2) are still true
if you ignore 0 < x0 < 1, the question can be solved
you can prove using induction that yn is always the same thing as e^(1/xn)
and then use that to show what yn converges to
@shell pawn ok you here? I figured this out
shell pawn
since you didnt translate this, Im taking a guess
Sorry for not translating
Montrer que= probe that
I think other words are similar to English
spice knot
I guessed them, theyre alr
shell pawn
another problem is the yn = e^(1/xn) at the bottom
this would mean y0 = e^(1/x0)...
That what we all did
shell pawn
you can prove using induction that yn is always the same thing as e^(1/xn)
We all used induction
They said yes it works but it's somehow wrong
I didn't see clearly there prove
I'll try to get it
Are you interested to see it later ? ( I'll translate)
spice knot
ok
to me the only issue here is that the definition in III says that 0 < x0 < 1
alpha > 1 though
if you ignore 0 < x0 < 1, the question works correctly nvm you also have to change 2) too
it also works correctly in 2) that any sequence xn will converge to 1 for all x0 > 0
so you erase 0 < x0 < 1 in III then change 2) to account for this and youre good
spice knot
I didn't see clearly there prove
I'll try to get it
Are you interested to see i...
show the proof they have
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torn jolt
how would one go about solving this?
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how would one go about solving this?
torn jolt
/what topic or smth smth i could study to learn how to solve this
silk galleon
/what topic or smth smth i could study to learn how to solve this
similarity rules, scale factor
silk galleon
how would one go about solving this?
the triangle with 3 and P is similar to the one with 23 and H
silk galleon
since the value of p is not known thats basically the best u can get
silk galleon
itโs because itโs being timsed by
right
silk galleon
uh you really cant do much else
torn jolt
so how do we determine h
silk galleon
okay then
if you have P
you just find 23(whatever P is)/3
and then add the units (meters)
torn jolt
lowkey havenโt calculated p
silk galleon
do you need help with doing 3a also
torn jolt
yes please and thank you
torn jolt
i see
silk galleon
you have one angle of the triangle
can you figure out anything else about the triangle?
silk galleon
you got it
torn jolt
sohcatoa?
silk galleon
yeah
so we can figure out the height of the triangle or the hypotenuse, but we're trying to find the height
so the hypotenuse doesn't really help with that
torn jolt
wait let me process
silk galleon
sounds good
silk galleon
its called theta btw
torn jolt
wth am i missing
silk galleon
opposite is only part of p
torn jolt
right
torn jolt
right
silk galleon
if you wanna write it in terms of p then opposite=p-1.6
silk galleon
its just this part of p
start by finding the length of this line
you were getting it with the sohcahtoa
silk galleon
sure
torn jolt
tan (37) ?
silk galleon
uhhh
torn jolt
wait
silk galleon
yeah something with that
silk galleon
you got it
bleak osprey
Yep
silk galleon
yep
silk galleon
,w tan 37 deg=(p-1.6)/4
glossy valveBOT
silk galleon
4.6 yeah
torn jolt
and then add the height of the lampy thing
silk galleon
and then add the height of the lampy thing
uh you already did that
silk galleon
right
torn jolt
3h is equal to 23p
23p = 23x4.6
23x4.6=105.8
105.8 divided by 3
is 35.3
h = 35.3
silk galleon
,calc 23(4.6)/3
glossy valveBOT
Result:
35.266666666667
right
torn jolt
can i just ask one more question cus tbh i kinda guessed this stepped
how did we know that 3h = 23p part
rather let me clarify
alright
mightโve cheated and answered it but
small height / big height = small distance / big distance?
bleak osprey
I think it's proportion bro
torn jolt
thank you for your help @silk galleon i appreciate it
silk galleon
small height / big height = small distance / big distance?
right, if theres one triangle inside another
torn jolt
right, if theres one triangle inside another
of course
silk galleon
again, its called similar triangles
torn jolt
gotcha
right i think the reason i struggled with 3a is because i lowkey forget how sohcotoa goes
ik amateur mistake
torn jolt
tbh my maths teacher makes me feel like a bit of an idiot for that
but then again ur much further along in ur maths career than me
torn jolt
i was gonna say i totally understand and i hate sohcahtoa
damn really
torn jolt
i mean
to be fair where do we encounter that may right triangles
many
ahhh i hate the IB
silk galleon
literally the whole reason we do trig
is for sailors to figure out which angle to sail at
none of us are doing that
torn jolt
true
so many questions in my paper are about sohcatoa
and just ultimately triangles in general
i also equally hate linear graphs
silk galleon
the next classes you take will use maybe 1/10 of what you learned about trig
its pointless
torn jolt
ahhh okay mb the role confused me
silk galleon
oh i just have that so i can talk in the advanced math channels
torn jolt
high school = uk secondary right
torn jolt
oh i just have that so i can talk in the advanced math channels
gotcha
torn jolt
oh youre in the uk
yes but iโm an ib student
torn jolt
ahh
hs includes what we would call sixth form aswell?
like sixth formers do a levels which are the exams u apply to university with
silk galleon
probably not no
i mean those exams are replaced by the act and the sat
which are basically totally independent of grade school
torn jolt
thatโs confusing
silk galleon
its MONEY is what it is
torn jolt
yeahhh i feel so cooked for exams in like two weeks
silk galleon
oh god
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north linden
I am having trouble solving this simultaneous differential equation, I think itโs because Iโve messed up diagonalisation. Could someone take a look at my working please?
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I am having trouble solving this simultaneous differential equation, I think itโ...
@north linden Has your question been resolved?
lone osprey
I am having trouble solving this simultaneous differential equation, I think itโ...
What math is this?
Linear?
north linden
What math is this?
Linear algebra with differential equations
cedar briar
can't you do it using substitution?
north linden
Yes
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north linden
I skipped it earlier cause it could be done with substitution but I want to practice diagonalisation and doing it with the general method
.reopen
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lone osprey
somber pike
I am having trouble solving this simultaneous differential equation, I think itโ...
The P^-1 is not correct
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manic bramble
I have this passage in a book about calculus. It seems to have a wrong trig identity in it. Can someone confirm if this is the case?
manic bramble
The derivative of cot x is -csc^2 x, and the trig identity for sin^2 x is also 1/csc^2, not 1/sec^2 x .
robust slate
,w differentiate cot x
glossy valveBOT
robust slate
yeah they fucked up
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robust slate
thanks for the confirm.
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hot kernel
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onyx glen
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4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
main vault
You assume its geometric first
onyx glen
k^2+k+1=0 is impossible when k is real
hallow quail
That's x^2-x-1=0
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elder hinge
examine he conttinuitty of the function. i have a horrible prof and i need help solving this
woeful moth
examine he conttinuitty of the function. i have a horrible prof and i need help ...
Alright, so since you need to check continuity: Do you know what the definition is?
elder hinge
so you need to find the limit for the left and right side?
onyx glen
would also help if you knew some practical things about continuity, such as that polynomial functions are continuous
so that if your function was just x^2, or just x, or just x^3+1, and not this piecewise stuff, then it would be continuous everywhere
woeful moth
so you need to find the limit for the left and right side?
yes, for which points in this case? And taking into consideration what Ann said so you don't do more work than necessary
onyx glen
is this accessible to you y/n
elder hinge
is this accessible to you y/n
no
elder hinge
yes, for which points in this case? And taking into consideration what Ann said ...
0 and 1
onyx glen
bruh
so then you have to do bureaucracy to prove that the function is continuous in (-โ, 0) and in (0, 1) and in (1, +โ) too?
do you have to use epsilon bullshit or no
woeful moth
0 and 1
thats correct, but a prerequisite would be that you know that polynomials are always continuous as Ann said. If not, you would have to prove all other points as well
elder hinge
do you have to use epsilon bullshit or no
unfortunately yes
woeful moth
alright, you take over ๐ I don't know that
elder hinge
queen gull
<@&268886789983436800>
elder hinge
but can you please explain how i am supposed to find fx 1 and 0
i just substtitutte them for x right
elder hinge
examine he conttinuitty of the function. i have a horrible prof and i need help ...
bcc for fx 1 it would be 1 1 and 2, and for 0 it would be 0 0 and 1
and im pretty sure this isnt what im supposed to do at the start
bc that would be too easy to show its discontinuous
chagpt is telling me to do this but i dont really understand what the 3 rows of the function mean
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onyx glen
chagpt is telling me to do this but i dont really understand what the 3 rows of ...
!noai
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chagpt is telling me to do this but i dont really understand what the 3 rows of ...
Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).
onyx glen
however it sounds like there's a lot you're missing so i can at least give you the info on what "the 3 rows of the function" mean
this is called a piecewise function and it is kind of like an if-statement in a programming language
def f(x):
if x <= 0:
return x**2
if 0 < x < 1:
return x
if x >= 1:
return x**3 + 1
@elder hinge if you are familiar with programming at all, here is python code that describes your function
hardy wolf
In another perspective, this piecewise function is consists of three different functions, each with their own domain.
Usually the domains are "connected"
For example, the following is the graph of g(x) where
g(x) = {
xยฒ + 1 if x < 1,
e^-x if x >= 1
It consists of two separate functions (graphs), and where they cut off depend on their domain
Note that domain is the "x < 1" and "x >= 1"
hardy wolf
For example, the following is the graph of g(x) where
g(x) = {
xยฒ + 1 if x < 1,
...
Another way to think of this is: When x < 1, the graph will be xยฒ + 1. When x >= 1, the graph will be e^-x
hoary ember
how many solutions are there to
$$\sqrt{x+52-14\sqrt{x+3}}+\sqrt{x+28-10\sqrt{x+3}}=2$$
glossy valveBOT
ihave<skissue>
hoary ember
by guessing i got that x=22 and 33 works
onyx glen
substitute u := sqrt(x+3)
zenith kernel
Let t=sqrt(x+3)
onyx glen
sniped
scarlet pike
onyx glen
actually yeah i'll let you figure out how that plays out @hoary ember
it'll be spoilers to say anything more
hoary ember
that actually does something?
onyx glen
sure does
hoary ember
$\sqrt{u+49+\sqrt{u}}+\sqrt{u+25+\sqrt{u}}=2$
glossy valveBOT
ihave<skissue>
hoary ember
oop wait im sily
onyx glen
also actually no that's wrong
onyx glen
$\sqrt{a^2} = |a| \neq a$
glossy valveBOT
Ann
onyx glen
also u itself and not sqrt(u)
onyx glen
hoary ember
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hoary ember
Thank you ann!
austere orbit
Can someone proof that sin 15 is equal to (โ6 - โ2)/ 4
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Can someone proof that sin 15 is equal to (โ6 - โ2)/ 4
@austere orbit Has your question been resolved?
ionic anvil
use sine subtraction formula, $\sin(A-B)=\sin A\cos B-\cos A\sin B$
mystic oar
Helll
mystic oar
hello
Perpendiculars BE and CF are dropped onto AD from the points B and C of a semicircle with diameter AD. The straight lines AB and DC intersect in P, and the segments EC and BF intersect in Q. Prove that PQ perpendicular AD
this is combined figure
This is it broken down
Now, can any good man/woman help me confer is angle ABE = angle DCF,
I did this question using coordinate geometry and it's done
im trying to use similarities.
@help
<@&286206848099549185>
vapid shell
<@&286206848099549185>
Yes sir
mystic oar
Now, can any good man/woman help me confer is angle ABE = angle DCF,
Please help me here sir
vapid shell
Ok let me try
mystic oar
Thankyou
cobalt summit
what if y9u try to prove PQ||EF ?
mystic oar
I asked 2 AI, they say the angles are equal.
cobalt summit
CF*
mystic oar
how would that work, please elaborate
mystic oar
I solved this using cordinate geometry
But there I used a online tool
so that's out of questions in exams
vapid shell
I solved this using cordinate geometry
Yes bro there are 2 angle equal
zealous stirrup
Hemisphere?
mystic oar
just a semicircle, i fear
cobalt summit
similar triangles? ฮCFD ~ ฮPOD (O is the point of intersection with AD)
if you prove that then it should also work.
ehh.
how would you even.
mystic oar
howd we prove them similar?
mystic oar
how would you even.
exactly this is kinda weird
primal cipher
<@&286206848099549185>
Whot
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tight cloak
A rectangular sheet of paper ABCD has sides ab as a and bc 2a. The side bcd paper is folded alongside diagonal bd such that it is perpendicular to plane of the unfolded sheet of paper. The position of B becomes B'. Find AB'
delicate torrent
!status
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1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
?
tight cloak
2
turbid badge
what have you tried
tight cloak
I am struggling to write the x and y coordinates, I know the z coordinate
Nvm figured it out
.close
stone matrix
!done
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night musk
for number 8 what is there to proof? He says show that they are equal but then says that the average are sometimes equal. The case being where a = 1 and b=0. Every other case M_h and M_f cannot be eqaul
queen gull
you need to show Mh=Mf for every a != 0 and b
night musk
you need to show Mh=Mf for every a != 0 and b
Yes I understand but it seems like there are only equal if a=1 and b=0
queen gull
rememeber you have g in the definition of Mf
night musk
I tried this out with two functions on desmos and these averages were not equal unless it was that case
queen gull
rememeber you have g in the definition of Mf
it makes them equal
night musk
rememeber you have g in the definition of Mf
Wdym? G is just inverse of f
queen gull
exactly
night musk
We have nothing about f
We could get it in terms of h(x) and a and b but I didn't see how that could help
queen gull
try it
fast peak
I tried this out with two functions on desmos and these averages were not equal ...
show what you did
night musk
show what you did
fast peak
thats not how you define inverse functions
the graph is for some reason correct
but its the wrong function
try g(4)
its 16, not 2
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fast peak
of course
proven lynx
can anyone help here
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can anyone help here
grave elm
proven lynx
exactly tysm
so im confused by the "or"
So A customer orders online or an event trip by phone
idk how to do this
ping me pls if anyone helps
help
dusky locust
Or is a union and is a intersection
proven lynx
Or is a union and is a intersection
So: "So,Ko,Mo,Eo,Et" ?
dusky locust
I have no idea what these things mean
delicate sail
anyone help
proven lynx
I have no idea what these things mean
Sporting events Online, Konzerts online โฆ
buoyant ravine
anyone help
please post your question in an open help channel please, such as #help-25 . Thanks :)
proven lynx
i need help anyone ๐ญ
dusky locust
i need help anyone ๐ญ
There is a language barrier
buoyant ravine
with the b)
it appears that part of b) is cut off? at least according to the translation 
proven lynx
it appears that part of b) is cut off? at least according to the translation <:c...
oh
it says and calculate the probability of event A
For that i have to just add
the stuff
but im confused about what stuff includes event A
buoyant ravine
alright well i'm assuming that they want you to use something akin to a venn diagram
proven lynx
alright well i'm assuming that they want you to use something akin to a venn dia...
i got tbat
i did that already
my bad
i did a tree diagram
buoyant ravine
can you send a picture?
proven lynx
yes
proven lynx
can you send a picture?
i didnt calculate the other things
But so we have theseevnts
buoyant ravine
ok cool
i'm assuming that when they say "set notation" they mean that they want the union/intersection of some sets?
proven lynx
ok cool
i'm assuming that when they say "set notation" they mean that they want ...
exactly
ur so smart
im gonna cry
but the problem is
idk what they mean by or
like
it doesnt make sense do i just add them all or
buoyant ravine
you're smart too! 
so consider two sets with some overlap
you have one set is the set of people who ordered online, and another set is one in which they ordered by phone
proven lynx
you have one set is the set of people who ordered online, and another set is one...
yes
buoyant ravine
ok before we proceed dyk the notation of $\cup$
glossy valveBOT
Francium
glossy valveBOT
Francium
proven lynx
**Francium**
yes
its like
Itโs everything together but the numbers they share appear once
buoyant ravine
ok, good, so what's the union of A={1,2,3} and B={3,5}
buoyant ravine
alrighty so that's essentially what or does in set theory :)
buoyant ravine
A or B = $A\cup B$ for two sets A and B
glossy valveBOT
Francium
buoyant ravine
usually this is used for situations in real life rather than pure math paper stuff haha
proven lynx
Now for the actual exercise like
A customer orders online or an event trip by phone
So: "So,Ko,Mo,Eo,Et"
but
yeah i dont rly understand
buoyant ravine
well from my understanding you essentially only have 6 sets
those who book sporting events, concerts, musicals, event trips, by phone, and online
(S, K, M, E, T, O)
some of these sets have overlap, denoted by your So, Ko, etc.
proven lynx
some of these sets have overlap, denoted by your So, Ko, etc.
Wdym overlap
buoyant ravine
Wdym overlap
overlap=intersection
proven lynx
Yeah do
but
Sporting events online, concerts online, musicals online, event trips online, or event trips telephonically
How do u notate the or in this case with actual numbers
do i just mark Event trips telphonically too
buoyant ravine
well for sporting events online, you can say something like $S\cap O$
glossy valveBOT
Francium
proven lynx
wait ill show u how i notate it