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#help-17
vocal sleetBOT
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restive basalt
Hi there, looking for some academic advise.
I'm currently doing a Masters' degree in Applied Mathematics. The primary reason I am doing this degree is to develop a better foundation in quantitative skills for a future career/PhD in Computer Science, with a focus on ML research. I already have a BSc. in Computer Science and a great deal of software development and applied ML experience (e.g. working with Pytorch, tensorflow, implementing papers), but not as much quantitative/theoretical skills.
I'd like some advise on what courses I should take for my upcoming semester
For my first semester, my options are:
- Real Analysis or Matrix Theory
For my second semester, my options are:
- A sequence of two courses on Advanced Differential Equations, or
- A sequence of two courses on Probability and Stochastic Process, or
- A sequence of two choices on Theory of Statistics.
What would you guys choose?
vocal sleetBOT
Hi there, looking for some academic advise.
I'm currently doing a Masters' degre...
restive basalt
My intuition is that I should do Matrix Theory, as I understand it to be more applicable to ML research, and then for the next semester do Probability & Stochastic Processes. What do you guys think?
Although, I personally find the curriculum for Real Analysis to be more interesting, since I already completed a class on Linear Algebra last semester, and the syllabus for Matrix Theory looks like a more proofs-based version of Linear Algebra.
flat whale
Hi there, looking for some academic advise.
I'm currently doing a Masters' degre...
#math-discussion or something. Help channels are for math problems not career advice
vocal sleetBOT
@restive basalt Has your question been resolved?
restive basalt
Thanks, will move the discussion there.
fiery elbow
kinda abstract, how to prove alpha=beta
vocal sleetBOT
kinda abstract, how to prove alpha=beta
fiery elbow
no visual proofs
flat whale
Can you use sum of angles to a straight line is 180deg
fervent wasp
kinda abstract, how to prove alpha=beta
theta+alpha = 180° because CD is a straight line.
theta+beta = 180° because AB is a straight line too
theta+alpha=theta+beta
Therefore, alpha = beta
fiery elbow
thanks
vocal sleetBOT
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raven bough
over here how in finding b^3 how is b^2/* b and b*b^2 coming the same ? It is given that the binary operation table is a cayley table , but from that i know that this structure is group and not a albelian group so how is it coming out to be commutative ?
vocal sleetBOT
over here how in finding b^3 how is b^2/* b and b*b^2 coming the same ? It is gi...
hidden gyro
this is an abelian group
raven bough
this is an abelian group
but how
unborn sundial
but how
based on the multiplication table. It is symmetric with respect to the main diagonal, so the group is an abelian one
raven bough
based on the multiplication table. It is symmetric with respect to the main diag...
but by defination isn't a cayley table just a group ?
unborn sundial
but by defination isn't a cayley table just a group ?
did not quite get the question. Yes, cayley table defines the group. In this case - an abelian group
raven bough
did not quite get the question. Yes, cayley table defines the group. In this cas...
if cayley table defines a group then how is this an albelian group ? like is the defination of cayley table extendable for albelian groups too ?
unborn sundial
sure, abelian groups - is a subset of the set of all groups. Cayley table can define any group, including the abelian ones
*any finite group, of course
raven bough
sure, abelian groups - is a subset of the set of all groups. Cayley table can de...
ah ok so there is possibility that a cayley table can be an albelian group .
ok got it thanks
.close
vocal sleetBOT
vocal sleetBOT
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vestal slate
Im new to writing proof and i wanted to know if this proof is correct or at least on right track
Show that for every a ∈ 𝐂 with a ≠ 0, there exists a unique b ∈ 𝐂 such
that ab = 1
for every a,b ∈ 𝐂 with a ≠ 0, let a=(c+di) b=(e+fi) where c,d,e,f ∈ R
ab= (c+di)(e+fi) = (ce - fd) + (cf + ed)i = 1
ce-fd=1
cf+ed=0
Fact that there exists a solution to these equations proves that there exists b ∈ 𝐂 such that ab=1
And fact that there is a unique answer to equations means that b is unique
vocal sleetBOT
**Im new to writing proof and i wanted to know if this proof is correct or at le...
jagged cargo
how do you know the last 2 lines are true?
vocal sleetBOT
@vestal slate Has your question been resolved?
vocal sleetBOT
vocal sleetBOT
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oak magnet
Hum, 2% + 2% != 4% according to my calculator
But if i do 2/100 + 2/100 = 4/100, am i missing something or its just some typo in calculator or even machine error ?
vocal sleetBOT
Hum, 2% + 2% != 4% according to my calculator
But if i do 2/100 + 2/100 = 4/100,...
fiery elbow
4/100 is 4%
fiery elbow
Hum, 2% + 2% != 4% according to my calculator
But if i do 2/100 + 2/100 = 4/100,...
is the ! a typo
stone gazelle
The calculator sees the second 2% as 2% of 2%
fiery elbow
like %50+%50=%75?
cedar burrow
Yeah it does 102% of 2%, like with investments
stone gazelle
Yes I believe so
vocal sleetBOT
vocal sleetBOT
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pearl pebble
90?
vocal sleetBOT
vocal sleetBOT
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vast shale
yo
pale perch
...
queen root
seems like a case of pythagoren theorem
vocal sleetBOT
@vast shale Has your question been resolved?
inner pumice
how would i go about doing these
viral copper
are we to assume that f is a quadratic in the first and cubic in the second
vocal sleetBOT
vocal sleetBOT
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spiral lintel
is AO*BO=CO*DO?
vocal sleetBOT
is AO\*BO=CO\*DO?
spiral lintel
That should be true according to property of a chord but that just looks very unintuitive to me
vast shale
aod and cob are similar
spiral lintel
aod and cob are similar
so AO*BO=CO*DO is true?
loud walrus
Yes
spiral lintel
Ok I get It now, thanks
silver tusk
is AO\*BO=CO\*DO?
yes
spiral lintel
.close
vocal sleetBOT
vocal sleetBOT
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undone jungle
hey im trying to understand the simplification here but im really struggling can you help me pls
vocal sleetBOT
hey im trying to understand the simplification here but im really struggling can...
@undone jungle Has your question been resolved?
sick turtle
hey im trying to understand the simplification here but im really struggling can...
Just simplify the stuff in the bracket
Has that answered your question?
vocal sleetBOT
@undone jungle Has your question been resolved?
mellow oyster
hey im trying to understand the simplification here but im really struggling can...
they've just added 8e^x * cos(2x) to both sides, then factored out the e^xcos2x to bring the +8 into the parentheses. the parentheses can also be simplified to -10c0 - 2d0 + 8 and 2c0 - 10d0, and since there are no e^xcos2x or e^xsin2x terms on the right side, these coefficients must be 0, which is where they get the final equations
plush fiber
Hello, I apologize if this is not the correct place to ask this, I dont need help with a math problem, but i have a question about the Dot Product and linear transformations
vocal sleetBOT
Hello, I apologize if this is not the correct place to ask this, I dont need hel...
plush fiber
if anyone is available to discuss it
vocal sleetBOT
No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/
plush fiber
From what i understand, if I have two matrices, A and B, which are both vectors. Their output C must be contained within the set of A or B (i think). I think that would mean that if A and B are both oriented so that the vector space of the cartesian plane can be reached as a combination of A and B, then the dot product would compress that total reachable vector space down onto the span of A. Would this be analagous to the perpsective of rotating the plane along a 3D axis orthogonal to the viewer so that that vector A appears to project downward in a straigh line onto the vector B
Sorry if this doesnt make total sense, im not super good at thinking like this yet
gaunt sparrow
What do you mean by “their output” when you talk about A and B?
gaunt sparrow
And you define their product as the usual vector component wise dot product?
Because C in this case is a scalar
Or are you considering A and B as matrices in the vector space of matrices
Like 2x2
plush fiber
ah i mean [a, b][c, d] only if you could imagine [c. d] as a 2x1 matrix and not a 1x2
so one of the matrices is oriented upright
also just for the sake of the question im asking im just assuming [a,b] is a unit vector or has been normalized already
Ok so just to understood, this whole question is related to projection. So the effect is projection A onto the span of B to form a 90 degree angle between the two
Or B onto A
Whichever one
gaunt sparrow
The dot product takes two vectors and spits out an element of the scalar field.
It doesn’t compress vectors, as it doesn’t even product a vector in the first place
plush fiber
i can show an example
gaunt sparrow
Yeah that might help a tad
plush fiber
just a moment ill make something in desmos real quick
The dot product is absolutely related to projection
im just trying to figure out why. I understand the cosine interpretation, but im not getting where the component representation ccomes from
If you scale one of the original vectors by the Dot Product, it creates a projection.
i just assumed that was a given but i should have been more clear i suppose
gaunt sparrow
Yes, the dot product gives a quantity that describes how much of vector A goes into vectors B’s direction provided B is unit
You use it in the computation of projections, so it’s normal that they appear to be related
subtle laurel
That's so easy bro
plush fiber
Sure that make complete sense. From that perspective, |A||B|cos(theta) makes total sense. But where does A.xB.x+A.yB.y comes from
that process seems totally arbitrary, other than a consequence of matrix multiplication
My intuition tells me it must come from some geometric relationship, but maybe there isnt one and im just reading too deep into this
A.xB.x+A.yB.y So im just wanting to know why this is equal to |A||B|cos(theta)
plush fiber
You use it in the computation of projections, so it’s normal that they appear to...
Are they actually not related at all?
plush fiber
A.xB.x+A.yB.y So im just wanting to know *why* this is equal to |A||B|cos(theta)
Also this wasnt really my original question, but this is really the source of where that question came from, since you answered it already
gaunt sparrow
They are related, sorry I made it sound as though it wasn’t the case.
The geometric interpretation is that it gives the length of the projection. There are probably better resources than I can explain as to
- Where the equivalence comes from
- An interpretation of the equivalence.
plush fiber
They are related, sorry I made it sound as though it wasn’t the case.
The geome...
No its ok, I do not always do a good job of explaining how i mean
plush fiber
They are related, sorry I made it sound as though it wasn’t the case.
The geome...
Ok, what im understanding is that the relationship results from trigonometric identies between the two cosine angles arising from the polar coordinate form which simplifies to the component form. I dont remember all of my trigonometric identities. EDIT: (correction) sin (a + b) = sin a cos b + cos a sin b.
So the component form must be what you get from simplifying the polar relationship in rectangular coordinates?
As the link says, the equations state the same thing, just that one is polar and one is rectangular
vocal sleetBOT
Available help channel!
simple kraken
vocal sleetBOT
simple kraken
I’m trying to make a math equation for my brother that the answer is where his present is but he is in a higher math level then me so I’m doing quadratics but differently then normal to try and make it challenging
pale perch
so whats the desired result
simple kraken
My goal is to create a quadratic equation where the second dif is 2 because is B on the code thing I made
I made it like 14,,30,,56 but idk if that’s a way I can make a problem for him to solve
14,~,30,~,56
pale perch
if i have ax^2+bx+c
the second diff is 2a
the first diff is 3a+b
the first term is a+b+c
my internet is not very good right now i warn
simple kraken
All good, so does that mean it’s possible solve 14,~,30,~,56 as the second diff being 2?
pale perch
is ~ meaning you have to find those terms?
simple kraken
Yep
vocal sleetBOT
@simple kraken Has your question been resolved?
hidden turret
im pretty sure this requirement is impossible but i cant prove it
simple kraken
im pretty sure this requirement is impossible but i cant prove it
That’s what I’m thinking but I think it’s possible through trial and error
mighty nacelle
14,30,56 is an arithmetic progression, so youre going to get a linear relationship
wait nvm
i cant do math lmao
dark kiln
you solve 14 30 56 and then you write x/2 instead of x
idk if it works i think it should
hm
well it doesn't
also you have 8,8,12... not 8,10,12...
yeah i changed to 12 20 30 42... now it works
vocal sleetBOT
@simple kraken Has your question been resolved?
half imp
no, there's only one quadratic going through those points like that and it has second diff 2.5
tacit birch
does anyone have an idea on how to calculate this?
vocal sleetBOT
does anyone have an idea on how to calculate this?
vocal sleetBOT
vocal sleetBOT
Available help channel!
vast shale
vocal sleetBOT
vast shale
vocal sleetBOT
@vast shale Has your question been resolved?
lunar latch
Hello,
I was solving a basic equation and I got stuck in factorising this
lilac pebble
first factor out 2x, it should make the rest of the process easier
undone jungle
Hello,
I was solving a basic equation and I got stuck in factorising this
خذ ال 2x عامل مشترك
lunar latch
خذ ال 2x عامل مشترك
يعطيك العافية
undone jungle
يعطيك العافية
الله يعافيك 
vocal sleetBOT
Available help channel!
vast shale
I need help putting this into LaTeX ;-;
vocal sleetBOT
I need help putting this into LaTeX ;-;
vast shale
I have no idea on how to put this into LaTeX, I know how to create Plots but idk how to customize it like so
tawny nacelle
I need help putting this into LaTeX ;-;
it might be better to ask in #latex-help
(spoiler: I also don't know how to put this in LaTeX 
)
but the people there might!
golden herald
I need help putting this into LaTeX ;-;
\begin{tikzpicture}
\draw[->] (-1, 0) -- (10, 0) node[right] {$x$};
\draw[->] (0, -1) -- (0, 7) node[above] {$f(x)$};
\foreach \x/\label in {1/$x_1$, 2/$x_2$, 3/$x_3$, 4/$x_4$, 5/$x_5$, 6/$x_6$, 7/$x_7$, 8/$x_8$, 9/$x_0$}
\draw[thick] (\x, 0.1) -- (\x, -0.1) node[below] {\label};
\draw[dotted] (9, 0) -- (9, 6);
\draw[->] (9, 6) -- (8.5, 6.5) node[left] {$f(x_0)$};
\draw[dotted] (0, 6) -- (-0.2, 6) node[left] {$x_0^2$};
\draw[fill] (9, 6) circle (2pt);
\end{tikzpicture}
twin meteorBOT
Aero
vast shale
```latex
\begin{tikzpicture}
\draw[->] (-1, 0) -- (10, 0) node[right] {$x$};
\d...
tysm!
tawny nacelle
Aero comes in to save the day
vast shale
it might be better to ask in <#840667252793802752>
ah sorry, I didnt know that channel exist
I was struggling so hard, I spent 10 mins searching for this online and couldnt find anyway ;-;
vast shale
.close
vocal sleetBOT
vocal sleetBOT
Available help channel!
obsidian vigil
Hey! Would anyone be able to check my work on this question?
vocal sleetBOT
Hey! Would anyone be able to check my work on this question?
obsidian vigil
answer is correct , but i just need someone to give me the greenlight on it
( soln kinda messy )
vocal sleetBOT
@obsidian vigil Has your question been resolved?
obsidian vigil
<@&286206848099549185>
lilac pebble
so far i got f(-z)=-f(z) and f(0)=0
obsidian vigil
i got it like cauchy's functional equation
obsidian vigil
so far i got f(-z)=-f(z) and f(0)=0
but yes i did get these
obsidian vigil
i got it like cauchy's functional equation
with x^2 and y^2, but then i js took it as f(p)+f(q)=f(p+q) and got it
obsidian vigil
so far i got f(-z)=-f(z) and f(0)=0
this is almost it now
f(x^2)=xf(x) plug in y=0
f(y^2)=yf(y) plug in x=0
since function is odd f(-y^2)=-f(y^2)=-yf(y)
f(x^2-y^2)=xf(x)-yf(y)=f(x^2)-f(y^2)
=>f(x^2)+f(-y^2)=f(x^2-y^2)
=>f(p)+f(q)=f(p+q)
=>f(p)=cp
this is just cauchys functiona equation so u cna get f(x)=cx
<@&286206848099549185> can somoene let me know if this is correct
lilac pebble
i think you only showed f(p)+f(q)=f(p+q) for p>=0, q<=0
obsidian vigil
if i interchange the variables i can prove the same but for p<=0 and q>=0 right
lilac pebble
yeah i think so
obsidian vigil
isnt that the same as proving for all real p and q
lilac pebble
what about p and q with the same sign?
obsidian vigil
either ways youl just get f(x^2)=cx^2 when x>=0, f(-x^2)=-cx^2 when x<=0
lilac pebble
apparently cauchy's functional equation has some pathological solutions but f(x^2)=xf(x) precludes those
obsidian vigil
apparently cauchy's functional equation has some pathological solutions but f(x^...
ah
obsidian vigil
apparently cauchy's functional equation has some pathological solutions but f(x^...
wait what
i thought f(x)=cx was the only solution that holds true for all reals
lilac pebble
wikipedia says otherwise
obsidian vigil
wikipedia says otherwise
not in the parts my puny brain can understand so im not gonna punch above my weight
obsidian vigil
ah
well, the question was part of a pset to prepare for a secondary school level contest so i dont think those solutions were in mind when framing the question
vocal sleetBOT
@obsidian vigil Has your question been resolved?
vocal sleetBOT
vocal sleetBOT
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analog nova
Hi, if lamda = v * f
vocal sleetBOT
Hi, if lamda = v * f
analog nova
How do i know the slope or the relation between v and f if its a straight or a curved line?
I know this is physics but I wanna understand from a mathematical perspective
thanks
modest dirge
You know their relationship from the formula you gave
analog nova
I know that v is related to 1/f
modest dirge
Its inverse since v = lambda / f
modest dirge
It’s not a line it’s a curve
analog nova
how do i know if it's curved
analog nova
It’s not a line it’s a curve
why?
modest dirge
Graph y = 1/x
analog nova
there are different formulas where it's a straight line
modest dirge
As x gets small y grows to infinity and as x grows big y goes to 0
analog nova
that also goes down
modest dirge
So it’s like a curve that goes in the first quadrant
analog nova
So it’s like a curve that goes in the first quadrant
When is it a straight line though
That goes down
analog nova
A straight line that goes up usually is the relation between a and b in a = b * z
modest dirge
Sorry you said up
Yes then z needs to be positive
But if z is negative the line goes down
Z represents the slope in that equation you gave
modest dirge
I can’t read Arabic
analog nova
First one in the left is a straight line that goes down
modest dirge
Yea and that has a negative z
analog nova
How
analog nova
Oh so like
modest dirge
And the third has z as positive
analog nova
If a=b*-z
modest dirge
It’s IS constant but the constant is negative
analog nova
then it's an inversed straight line
modest dirge
A constant is a number
And numbers can be negative
I’m saying that if the slope goes down then that means the slope is a negative number
The fourth graph in your picture is y=a/x
That’s the inverse one
It’s a curve see?
analog nova
So like a = b * c
a/b = c straight line that goes up
b/c = 1/a inversed curved line
analog nova
It’s a curve see?
What makes me go crazy that
the correct answer for it was the left one
the first one in the left
Which is lamda = v * f
v/f
Why not the one in the right
modest dirge
In that question the answer is a
The curved one is the relation between v and f
Or wait is it lambda = v/f
If that’s the case then it is a straight line since v=lambda * f
v=lambda * f is correct right?
analog nova
v=lambda * f is correct right?
yes
analog nova
The curved one is the relation between v and f
Could you give me an example for the curved one?
vocal sleetBOT
@analog nova Has your question been resolved?
crude bone
That‘d be the c
crude bone
speed of sound in solids and liquids doesn't significantly change with frequency, though the speed is higher than in air.
crude bone
The original answer is D
Weird,
analog nova
speed of sound in solids and liquids doesn't significantly change with frequency...
It didn't specify
crude bone
It didn't specify
Lemme think
vocal sleetBOT
@analog nova Has your question been resolved?
vocal sleetBOT
@analog nova Has your question been resolved?
vocal sleetBOT
vocal sleetBOT
Available help channel!
solar geyser
I have a problem that I got close to solving (I think) but I don't know how to finish it.
The problem is as follows:
There are 22 students that play 5 games of football. In each game, there are two teams of 11 students. If a team wins, every winning student gets 3 points, in case of a draw everybody gets 1 point. if youre on the loosing team you get 0 points.
Every student must be on the same team with all other students at least once.
Show that there are always five or more people with the same number of points.
My thoughts so far are that in case of a draw, everybody gets a point. This means that a draw "doesn't change anything". What I mean by that is if two people have the same number of points, they still will after a draw. This means that if two people have the same number of points, they must have won the same number of times.
Now there are five cases: no draw, 1 draw, 2 draws, ..., 5 draws.
In case 5, everybody has the same number of points.
Case 4: Only one game that is not a draw => 11 people win the same number of times
Case 3: In the first game (that's not a draw), 11 people have the same number of wins, in the "second" game there have to be 6 or more people in the same team that were in the same team previously at least once (can be shown with the pigeon hole theorem)
but then in case of one draw, two draws and no draw I don't know how to show that there are five people with the same number of wins
maybe you have an idea?
vocal sleetBOT
I have a problem that I got close to solving (I think) but I don't know how to f...
solar geyser
<@&286206848099549185>
vocal sleetBOT
@solar geyser Has your question been resolved?
solar geyser
<@&286206848099549185> some ideas at least?
verbal fiber
let me try
No Draw Case
There are always 11 students with the same number of points (either 0 or 3 points per game).
i got it by making a graph but idk if its correct
Two Draw Cases
There are always 6 students with the same number of points (either 2, 5, 8, or 11 points).
One Draw Case
There are always 11 students with the same number of points (either 1, 4, 7, 10, or 13 points).
idk how to explain but this is my answer :)
verbal fiber
yep i did
vocal sleetBOT
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viral python
vocal sleetBOT
viral python
Need help with this
tawny nacelle
as it turns out, our moderator @brittle patrol made a video explaining this!
https://www.youtube.com/watch?v=PXxvoXIql7M
nimble elm
.close
vocal sleetBOT
nimble elm
@viral python dont post memes here, read #rules and #❓how-to-get-help
tawny nacelle
hi Loch 
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Available help channel!
smoky cave
I am currently stuck on 19(b)(ii) I don’t know how to find p or q. I need to find the area of circumcircle of OAP first
vocal sleetBOT
I am currently stuck on 19(b)(ii) I don’t know how to find p or q. I need to fin...
finite owl
I am currently stuck on 19(b)(ii) I don’t know how to find p or q. I need to fin...
if you know that L is a tangent to the circle, that also means that there is a point in L that also belongs to the circle C
you can find p and similarly q
smoky cave
if you know that L is a tangent to the circle, that also means that there is a p...
So find the contact point first?
finite owl
yea you said you needed to find p or q right
and in this problem it looks helpful to know
smoky cave
yea you said you needed to find p or q right
I think need to find p and q first
smoky cave
Just past paper exam questions
finite owl
I am currently stuck on 19(b)(ii) I don’t know how to find p or q. I need to fin...
also the problem resolution is kinda weird you gotta show what you're doing for the other answers
they might deduct some points for that
smoky cave
also the problem resolution is kinda weird you gotta show what you're doing for ...
That is literally a typical question type in DSE 
finite owl
also the problem resolution is kinda weird you gotta show what you're doing for ...
mmm maybe reading it back na
well whatever
vocal sleetBOT
@smoky cave Has your question been resolved?
hexed umbra
is SSA and SAS the same
vocal sleetBOT
is SSA and SAS the same
hexed umbra
or different
cursive turret
different
hexed umbra
like this?
cursive turret
SSA instead of ASS.
kind light
ass
unborn beacon
vocal sleetBOT
unborn beacon
Proof
vocal sleetBOT
Available help channel!
versed pelican
quick question: how do I solve an expression that has X in it on a calculator?
it would make my life much quicker
vocal sleetBOT
quick question: how do I solve an expression that has X in it on a calculator?
i...
fiery elbow
put the x on one side and other stuff on other side of the equation
then insert the other stuff into the calculator
example
3x+7=2
subtract 7 from both sides
hard atlas
not every calculator can just solve equations
fiery elbow
3x=-5
then divide both sides by 3
x=-5/3
then input -5/3 into the calculator to find x
so isolate x
versed pelican
the main issue is that the question doesnt have =
hard atlas
then what do you mean with solve
versed pelican
english isnt really my first language but what im studying is (ensuring 2 functions are inverse to one another) and to do that i have to put one of the functions instead of X and to "solve" it
honestly idk how to phrase it srry
dim pumice
you mean simplify
versed pelican
yes
dim pumice
that's not something you should be doing with a calculator when learning about it
hard atlas
if you wanna cheat then wolframalpha is your best friend
versed pelican
I am not rlly learning about it
its a part of the lesson
and it would be much faster if i can solve it with a calculator
vocal sleetBOT
@versed pelican Has your question been resolved?
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@versed pelican Has your question been resolved?
stark berry
whats your question now?
versed pelican
how do I simplify an expression that has X in it with a calculator
stark berry
wolframalpha is a calculator
this is also a calculator
whats the context of this
fiery elbow
abacus is a calculator
ln(3x^2-5)=x^2-10 how do you solve this with a oldschool calculator
stark berry
you cannot
you can use a calculator for approximations using other methods
but you wont get an algebraic closed form solution
cosmic cloud
yes
versed pelican
kk thx
trim stratus
19b
fiery elbow
|PC|*|AP|=|BP|^2
if you’ve done a correctly
you can figure out the ratios of PC and PB
trim stratus
|PC|*|AP|=|BP|^2
but why
fiery elbow
you can figure out the ratios of PC and PB
3PC=2PB
fiery elbow
hmm
trim stratus
https://artofproblemsolving.com/wiki/index.php/Power_of_a_Point_Theorem
I don’t think I have learnt that
fiery elbow
let PC=2k and PB=3k
3k/6=(10+2k)/3k
from similarity
trim stratus
why PC=2k PB=3k
fiery elbow
9k^2=60+12k
3k^2-4k-20=0
(3k-10)(k+2)=0
3k=10
k=10/3
manic lance
how do i prove convergence
vocal sleetBOT
how do i prove convergence
finite owl
ratio test imo
should work
think its gonna diverge though intuitively
because n^n grows much faster than e^n and n!
or not it doesnt
finite owl
how do i prove convergence
vocal sleetBOT
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twin citrus
guys what is the general strategy to find range of functions like this one ?
oak magnet
find the range of the quotient inside and then apply sqrt to the positive part
so
the quotient range is R
twin citrus
so quotiend range is everything but 1
oak magnet
nah its everything
twin citrus
when its 1?
brittle reef
guys what is the general strategy to find range of functions like this one ?
i dont think it can be 1
oak magnet
but this is positive when
brittle reef
but i could definitely be wrong
brittle reef
i think the fact that the range of a function is equal to the domain of its inverse can be used as well
oak magnet
also working
brittle reef
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vale frigate
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vale frigate
can anyone explain this?
stark berry
whuch part do you not understand
stark berry
the first line?
vale frigate
no
"we can replace the y on both sides of this equation with dy/dx"
if we do that then it should be d/dx[d(dy/dx)/dx]
.close
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novel iris
Let $p$ and $q$ be prime numbers, where $2<p<q$, and let $M$ be the set of all positive integers $n$ s.t. $n^5$ is divisible by $p^3$ and $64q^{11}$. What is the smallest integer in $M$?
vocal sleetBOT
Let $p$ and $q$ be prime numbers, where $2<p<q$, and let $M$ be the set of all p...
twin meteorBOT
KySquared
novel iris
Idk how to solve this problem
My idea was to find a common multiple between 3,5, and 11 and then reduce from there
$n^{165}$ is div. by $p^{165}$ and $64q^{165}$\
$\implies \quad n^{165}$ is div. by $64(pq)^{165}$\
$\implies \quad n$ is div. by $64pq$,Thus n is div. by 64\
64pq gets reduced to 2pq and is what I got but I have no idea if thats right
twin meteorBOT
KySquared
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@novel iris Has your question been resolved?
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@novel iris Has your question been resolved?
novel iris
<@&286206848099549185>
brittle cipher
My idea was to find a common multiple between 3,5, and 11 and then reduce from t...
this will produce a number that's a bit too large. Since 2 < p < q you know they're relatively prime and can use the chinese remainder theorem to think of each prime power independently
try thinking about what power of p divides n first
novel iris
try thinking about what power of p divides n first
Uhmm
If p^3 divides n^5 then p^(3/5) divides n right?
brittle cipher
well p^(3/5) is an irrational number not an integer anymore
so it doesn't really make sense to talk about divisibility unless they're integers
brittle cipher
maybe start small one case at a time, suppose p doesn't divide n, then does p^3 divide n^5?
brittle cipher
yeah good
novel iris
Idk if that can be extended to all primes p and numbers n though
brittle cipher
it can, I think this is a good strategy to look at special cases too like this to get a feel for it
raising to power doesn't introduce new prime factors, so if n wasn't divisible by 3 before, raising it to a power won't introduce a 3 after
it'll just make more 7s in your example
novel iris
But i can say smth like p = 2… wait no, that was a restriction
brittle cipher
good catch
maybe even simpler is if p doesn't divide n, then p doesn't divide n^5
so we can't even have p^3 divide it either, since that's even more "p"
that make sense?
novel iris
Mhmm
novel iris
Yes
Since n=p
novel iris
Oh wait
n^5 would be divisible by p^3 but not the other way around
novel iris
brittle cipher
so if a divides b and b divides a, then a=b
maybe that's a bit too much of a diversion from what we were originally doing so we can get back on track now haha
but I think it helps to have that in mind as a sanity check I think
novel iris
Right lol
brittle cipher
so let's try to say this more like the problem did, if n=p then can we see that p^3 divides n^5?
don't substitute p=3 now
novel iris
No
brittle cipher
ah it does
if n=p then p^3 divides n^5 = p^5
maybe writing out the prime factorization helps to
p^3 = p*p*p and n^5 = p*p*p*p*p
novel iris
Wait give me 1 moment
brittle cipher
yup
brittle cipher
cool
so main thing we're focusing on now is looking at each prime factor of n individually
brittle cipher
yeah good
Can we extend that further, what primes do we know appear in the prime factorization of n?
novel iris
So we’re looking at the general case now correct?
brittle cipher
sort of, I'm really just looking for what primes appear at least once, not exactly how often
for instance, is p in the prime factorization of n?
is q? are there any other primes that are in n?
novel iris
for instance, is p in the prime factorization of n?
No because p doesn’t divide n
We’d have to look at n^5
In which case yes, p would appear 3 times
?
brittle cipher
it gets a little tricky but you're on the right track
for p^3 to divide n^5 we need p to divide n at least once
novel iris
Mhmm
brittle cipher
so that when we raise it to the 5th power it makes more ps in there that will be divisible by p^3
maybe we should go back to the case with p=3 I think that helped to see what's going on
when p=3 and n=3, we can see p^3 dividing n^5 would be the special case of 3^3 dividing 3^5
novel iris
Mhmm
brittle cipher
No because p doesn’t divide n
if p didin't divide n here, we couldn't have the exponent 5 make more of them for us, hopefully I'm not beating this to death too much here
novel iris
But if we’re raising n to the 5th power that would saying 3 dividing 3^5… ohhhhhh
brittle cipher
yesss
so to try to zoom out a bit so we're not stuck, the over all strategy is:
we checked no p dividing n earlier, that didn't work, then we checked a single p dividing n, and then we saw that worked
if it didn't we would try n=p^2 etc
but we don't really need to keep going now because we see that p^3 divides p^5 already
since for the problem we wanted the smallest n
also another way to see that is p^5/p^3 = p^2 or alternatively write it as p^5 = p^3 * p^2 to see that p^3 is a factor here
keep in mind n could have other prime factors, but they wouldn't really affect this simpler reasoning, for instance n=6 with p=3, the same reasoning works and the "2" in "6" there is just hanging out if that makes sense
I hope I'm not overloading you with too much here so if you need to breathe I understand haha
novel iris
Okay
In general we’d have to check if p^m divides n
From there we can make a conclusion about p^3 dividing n^5?
m is just some increment
brittle cipher
yeah you got it
now we'll do the same for the other primes, we're trying to minimize m basically and make sure we meet the conditions of the problem
novel iris
m = 0 can be considered as there being no p’s present to divide n so p^3 doesn’t divide n^5
when m= 1 there is one p present to divide n -> p^3 must divide n^5
And since we’re looking for the smallest p, we can just stop here
brittle cipher
perfect!
novel iris
brittle cipher
yeah keep cooking do you see if you can do the rest of the problem now for the other primes
it might require one extra trick we didn't talk about too much yet
let me know if you're getting stuck
novel iris
And since we’re looking for the smallest p, we can just stop here
Just realized this should be the smallest p^m not p
brittle cipher
oh good catch, I knew what you meant and didn't see it either haha
novel iris
For 64q^m dividing n
Are we looking at the entire number or are we only concerned with q?
Like there are no q’s present in the zero case, but 64 could still divide n
brittle cipher
For 64q^m dividing n
Are we looking at the entire number or are we only concerne...
good question, I would break it up into 2^6 and q^11 separately
q^11 since the q^m would be "inside" the n
and we'd do a similar case for 2^m as a factor of n
so we can treat 2 and q individually the same way we did for p
novel iris
I see
So instead of 64q^11 we consider 64 dividing n and q^11 dividing n
brittle cipher
yup exactly
and we know we can do that because 2 < q really tells us they're different primes
if they didn't give us that inequality at the start of the problem we'd be sorta stuck with thinking through more complicated things
novel iris
Which can further be expressed as 2^6 dividing n and q^11 dividing n
The first case is telling us that n must be even, but 2^6 can’t divide n^5 (if n = 2) so we have to look at 2^7?
Ahhhh screw it
I’ll just work it out instead of trying this in my head
(Taking n = 2) 2^1 does divide n -> but 2^6 can’t divide n^5 so keep going
(q= n = 5)q^1 divides n but q^11 dividing n^5 wouldn’t be an integer so we have to keep going
I think my logic is wrong since wouldn’t this mean we have to go up to 2^5 and q^5?
brittle cipher
I would suggest doing all one prime at a time instead, but this way works too
brittle cipher
I think my logic is wrong since wouldn’t this mean we have to go up to 2^5 and q...
let me look more closely at your work here
novel iris
As in take m to be prime?
brittle cipher
no I mean take n=2^m and look at n=2, n=4, n=8, etc... forget about q for now until you get it
novel iris
Ah
brittle cipher
so right now you are good 2^6 doesn't divide n=2 case 2^5, so now try n=4
novel iris
Yeah something seems wrong to me
For n = 2^m and 2^6 dividing n^5 to be true n would have to be 2^6
But the possible answers only go up to 2^5
@brittle cipher
novel iris
Yeah something seems wrong to me
For n = 2^m and 2^6 dividing n^5 to be true n w...
Since that would be 2^6 dividing into 2^30
brittle cipher
n=2^6 is too large
novel iris
2^6 dividing 2^(5m)
brittle cipher
so first case is plain n=2 and you saw 2^6 doesn't divide 2^5 so now go to n=2^2 does 2^6 divide (2^2)^5?
brittle cipher
2^6 dividing 2^(5m)
yes perfect
novel iris
Right
The only time this doesn’t cause a… DANG IT
brittle cipher
since you're good with looking at just the exponents then you know 5m >= 6 and m=1 is too small, but m=2 works
novel iris
I was dividing the powers in my head
brittle cipher
ah haha I see
novel iris
& on paper
brittle cipher
I see how it's easy to divide the exponents on accident when you need to subtract if I get you
novel iris
No i knew that, just a mistake on my end lol
We’re looking for what m 2^(5m)/2^6 >1, m = 2 makes this true
Similar case w/ q^(5m)/q^11>1, so 3
?
@brittle cipher
brittle cipher
Yeah that looks good to me
so now just a matter of combining all those and you should have your answer
novel iris
4pq^3
brittle cipher
Yeah
brittle cipher
But the possible answers only go up to 2^5
You said you had answer choices, I hope that's listed there!
novel iris
Yayyyyy
Yeah its there
brittle cipher
Awesome haha
brittle cipher
cool yeah I would recommend if you have more like this in the future post in #elementary-number-theory or #competition-math
you're welcome!
novel iris
I haven’t done much number theory but now im thinking of getting into it
I’m spending my break studying for the GRE
brittle cipher
the question channels like this are good too, but in those channels you might have better chance on letting it sit for a while and get a response eventually since I think fewer people know this sort of stuff
ah cool, yeah I didn't get interested in NT until later myself too
novel iris
the question channels like this are good too, but in those channels you might ha...
Will do
Have a great day
brittle cipher
yeah you too
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tired matrix
claim
tired matrix
what's a good online precalculus studying method?
silk osprey
read book, do exercises
tired matrix
i do not have a book to study from
silk osprey
openstax probably has a nice textbook for you
tired matrix
kk
silk osprey
look up openstax
steep crater
Or go to #book-recommendations
vocal sleetBOT
silk osprey
it is
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trim stratus
I wanna know how to do these two questions
vocal sleetBOT
I wanna know how to do these two questions
@trim stratus Has your question been resolved?
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lean cairn
can someone help me with quaternions?
vocal sleetBOT
can someone help me with quaternions?
lean cairn
in a) I have trouble expressing quaterinos
vocal sleetBOT
@lean cairn Has your question been resolved?
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@lean cairn Has your question been resolved?
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@lean cairn Has your question been resolved?
torpid narwhal
steep crater
!occupied
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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
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wide smelt
Hi, those two bracketed expressions, are they multiplied?
Oh they’re divided, sorry.
Ok, now we need to review something!
Now what is tan x?
Sin x / cos x
So that is almost exactly what we have here
But we need to find a good substitution.
Now we need to match to find α
Now √49 is 7.
So alpha is 7.
That means tan ( 7 ) = tan (√7x)
7=sqrt (7x)
49=7x
x=7.
steady oak
opaque geyser
Can someone tell me where im wrong?
vocal sleetBOT
Can someone tell me where im wrong?
opaque geyser
$$\lim_{x \to +\infty } (\frac{2x^4-2x-1}{2x^4-x})^{2x^{3}} = L \
ln(L) = \lim_{x \to +\infty } 2x^3 * ln(\frac{2x^4-2x-1}{2x^4-x}) = 2x^3 * ln(\frac{x^4(2-\frac{2}{x^3}-\frac{1}{x^4})}{x^4(2-\frac{1}{x^3})}) = 2x^3 * 0 = 0??? $$
$$\lim_{x \to +\infty } (\frac{2x^4-2x-1}{2x^4-x})^{2x^{3}} = L$$
twin meteorBOT
maxy
opaque geyser
$$ ln(L) = \lim_{x \to +\infty } 2x^3 * ln(\frac{2x^4-2x-1}{2x^4-x})$$
twin meteorBOT
maxy
opaque geyser
$$2x^3 * ln(\frac{x^4(2-\frac{2}{x^3}-\frac{1}{x^4})}{x^4(2-\frac{1}{x^3})}) = 2x^3 * 0 = 0???
$$
twin meteorBOT
maxy
tulip nebula
I thought it was 1^inf form at beginning
opaque geyser
ln (1) = 0
dense cedar
Oh I didn't see the power
opaque geyser
all of that must come -1 in order to be correct
tulip nebula
the term inside ln can be simplified into 1-((1/x^3+1/x^4)/(2-1/x^3)
tulip nebula
the term inside ln can be simplified into 1-((1/x^3+1/x^4)/(2-1/x^3)
dont know its of any use thou
opaque geyser
no lol
well chat gpt says that ln has a weird behaviour when you simplify inside of it
tulip nebula
I see
stark berry
what does that mean
tulip nebula
I have studied that if its 0* inf form then we have to manipulate it somehow into 0/0 or inf/inf form
So I was trying to see if I can manipulate term inside ln to make any use of it
opaque geyser
well i gave up, i think it's because the contribution of 1/x^n terms is not so irrelevant inside the logarithm
tulip nebula
Yeah, that approach may be wrong
opaque geyser
thanks anyways @tulip nebula
tulip nebula
thanks anyways <@561270415008333847>
Np
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quartz beacon
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tidal dock
can you send the original problem?
quartz beacon
sketch and find the area of region R the closed loop bounded by x=t^3-4t, y=t² (-2<=t<=2)
i get the area i just dont get the sketch
vocal sleetBOT
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buoyant talon
What formula would I use to generate something like this? It looks similar to a cube root function but I have no idea
vocal sleetBOT
What formula would I use to generate something like this? It looks similar to a ...
buoyant talon
heres something similar with the other side too
vocal sleetBOT
@buoyant talon Has your question been resolved?
kind light
xe^-x maybe? tweak some numbers naybe
cyan talon
What formula would I use to generate something like this? It looks similar to a ...
do you know what's supposed to happen if the slip ratio is super huge ?
(first of all can it be bigger than 1 or not?)
buoyant talon
it can be bigger than 1. I would assume the graph just goes on infinitely
cyan talon
does it decay all the way to 0, or does it plateau somewhere say 0.5 ?
buoyant talon
no clue lmao
I would imagine it plateaus probably
doing some more research, it looks like this is the pacejka tire model, so I think my question is answered
hazy escarp
i get the area i just dont get the sketch
@quartz beacon I think this is the graph
vocal sleetBOT
<@414787086802223115> I think this is the graph
hazy escarp
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grim ivy
lol
misty hare
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open epoch
delicate lynx
lol
little bluff
?!
hushed pewter
@delicate lynx did you have a question?
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minor rivet
can i hv help w question b pls ty
vocal sleetBOT
can i hv help w question b pls ty
stark berry
do you know the formula for circumference
wispy warren
circumference of what circle?
little bluff
circumference of what circle?
The cylindrical tank
minor rivet
do you know the formula for circumference
2pi r
stark berry
and whats r
smoky iron
Or pi * diameter
minor rivet
lmao
stark berry
i would reccomend drawing it out
minor rivet
idk how to deraw it 😭
stark berry
althoufhr intuitively i think yoy can guess the answer
vocal sleetBOT
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@minor rivet Has your question been resolved?
exotic ravine
can someone help me solve this
vocal sleetBOT
can someone help me solve this
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shell furnace
smn please help me with finding the sides of the triangle?
vocal sleetBOT
smn please help me with finding the sides of the triangle?
vocal sleetBOT
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vast shale
Yup
That is correct.
But you can use a triangle to solve this question
Whenver you have such quirky shapes
think about the fact that you have to find the area of NORMAL shapes
and then subtract normals shapes to find the area of the quirky shape
In this case, you can subtract the area of the 3 sectors of the circle from triangle ABC to find the black area
low haven
vocal sleetBOT
low haven
why this is true? i'm not able to see why
and alpha and beta are random arithmetic functions
hard atlas
the alpha(n) is a factor which you can move into the second sum
so then you are summing over all n<=x and m<=x/n
which is just the same as summing over all n,m with mn<=x
low haven
makes sense, because if n > x then the inner sum is going to be null
thank you
.close
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vast shale
hello. can someone help me to understand the mark scheme? im not sure why they divide it by 2
vocal sleetBOT
hello. can someone help me to understand the mark scheme? im not sure why they d...
vast shale
ok i understand why f'(x) < 0
vast shale
nah the answer is correct but im not sure why they are dividing it by 2
stark berry
oh
its the fraction
out of 0<x<2
so out of that interval
what interval inside 0<x<2 is the function decreasing
vast shale
what interval inside 0<x<2 is the function decreasing
?
stark berry
imagine a number line
0 to 2
out of that line
what section is the function decreasing
so its (10/11)/total
vast shale
so its (10/11)/total
hae ok. so between 1/11<x<1.
the function is decreasing. but wdym by that
i dont get it..
vast shale
so i need to find the fraction of the interval 0<x<2 for which f(x) is decreasing.
but we have 1/11 < x < 1
vast shale
how do u understand the last line
stark berry
the eaisiest way i can think of is
vast shale
ok so 1/11<x<1 is decreasing.. what does it mean by the fraction of the interval
vast shale
ok
vast shale
and find the ratio of their sizes
oh and how do we do that?
stark berry
(1-(1/11))/(2-0)
vocal sleetBOT
vast shale
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vocal sleetBOT
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vast shale
(1-(1/11))/(2-0)
ok so like if you imagine the interval 0<x<2 as a line of total length. and we are just looking for the proportion of the line where f(x) is decreasing. right?
vast shale
i finally understand the question
stark berry
strange question i must say
oblique juniper
vocal sleetBOT
fiery nimbus
=1/tan(x)
You can do it by using the identitty provided and then using tan(x)=sin/cos and then simplifying
oblique juniper
tan(90-x) = cot(x)
Tan(90-x) = cos(x) ?
grim lotus
no
oblique juniper
Tf is a cot
oblique juniper
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vocal sleetBOT
oblique juniper
Thanks for the help
grim lotus
💀
oblique juniper
What?
grim lotus
did you solve it?
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minor rivet
yo cansomeone help w part b
vocal sleetBOT
yo cansomeone help w part b
dense cedar
yo cansomeone help w part b
Any info about the range of sight ?
Circumference =
,calc 2pi20
twin meteorBOT
Result:
125.66370614359
The persons standing 60 m away
Ok so i got what they mean !!!
You see the persons 60 m away
We need tangents of length 60 m from the circle
dense cedar
minor rivet
oook
dense cedar
And again
AB = AE + R = 60+20 = 80
Also yk the tangents theorem
AD perpendicular to DB
So AD² + DB² = AE ²
This way we can find out the angle ABD
minor rivet
ooooooooh
minor rivet
js wondering can u do it without tangent theorem or is it like necessary
dense cedar
It's not necessary
But the concept that a tangent is perpendicular to the curve is necessary to say the triangle is right angled
minor rivet
oook
minor rivet
how do i fin d the length of ad ?
dense cedar
So just calc theta and solve the rest
minor rivet
So AD² + DB² = AE ²
wairt but wht ab this one
shell furnace
In this case, you can subtract the area of the 3 sectors of the circle from tria...
sorry for the late reply but that helps alot, thanks.
dense cedar
wairt but wht ab this one
Right angled triangle so you can use to find out the other side
It's not necessary