So u will get 2 x and 2y
#help-26
rigid cloak
yes
shell orbit
mx -1y + c
shell orbit
We can also get c using the point of A
rigid cloak
Yes good enough
I said that only
U satisfy for c
Now only m
Okok
Good enough
Yea we done here
shell orbit
For x and y we use the midpoint of c2 (8,5)
shell orbit
oh?
shell orbit
(0,5) and what?
shell orbit
Yes
shell orbit
I was going to use (ax + by + c)/sqrt(a^2 + b^2) = length of radius of circle c2
shell orbit
yeye
rigid cloak
b=-1
shell orbit
Yep
shell orbit
ah oops
shell orbit
6
rigid cloak
Okok
shell orbit
c2 is (y - 5)^2 + (x - 8)^2 = 36
shell orbit
Is it wrong
shell orbit
U did not read it i mean
read which?
shell orbit
I dont understand
shell orbit
(y-5)^2+(x-14)(x-2)=0
this is c2?
rigid cloak
Yes
rigid cloak
c2 is (y - 5)^2 + (x - 8)^2 = 36
But yes it ultimately forms this only
I saw it now
My bad
Carry on
@shell orbit
Find
shell orbit
Oh my god
rigid cloak
Why am I getting +- √(3/29)
shell orbit
🤦♂️
shell orbit
c1 is (y + 1)^2 + (x + 0)^2 = 16
shell orbit
Sry hahaha
shell orbit
Im still calculating 1 moment
rigid cloak
Okok
shell orbit
Yes me too
rigid cloak
Okok
Good
Now just put m in
(y-y1)=m(x-x1)
Where x1 and y1 are coordinates of one point on line that is A
shell orbit
(y-5)=-sqrt(5)/2(x-0)
rigid cloak
What
shell orbit
I did what you said
shell orbit
Ah you can do them together?
shell orbit
Ah oke
topaz sinewBOT
topaz sinewBOT
Available help channel!
tiny finch
The question of this task is if P and R lie on the line that A B make?
My question is why we chose the direction vector s to point up, aka so that s = AB vector, why didn't we chose so that its direction is downwards aka. so that s = BA vector?
topaz sinewBOT
The question of this task is if P and R lie on the line that A B make?
My questi...
@tiny finch Has your question been resolved?
neon iron
Find the equation of the line L
topaz sinewBOT
Find the equation of the line L
neon iron
I also tried forming a triangle with 2R as an hypotenuse and R as one cathetus, but I shouldn't hvae done this, because I am assuming things and it gives the wrong answer if I also apply it to the line above
<@&286206848099549185>
topaz sinewBOT
@neon iron Has your question been resolved?
delicate dragon
is that a tangent?
ruby tree
I also tried forming a triangle with 2R as an hypotenuse and R as one cathetus, ...
You'll need to compute R some way or another
Do you know the equation for a circle center on (p,q) with radius r?
neon iron
Do you know the equation for a circle center on (p,q) with radius r?
Well, I know it, but from school and I'm in a pre-university academy. I am catching up on some topics and I haven't learned that equation yet, so I shouldn't use it
ruby tree
Well I have a solution that doesn't involve that but it very much feels like cheating
neon iron
mm why cheating?
ruby tree
Like it's most likely not how you're supposed to solve this
neon iron
like using more advanced math?
ruby tree
I'm not sure how you're supposed to solve this without knowing how to represent a circle as an equation and without using my "cheaty" method
Not particularly advanced, no
Do you know which answer is correct already?
neon iron
Do you know which answer is correct already?
yes
ruby tree
Tell me
neon iron
C
ruby tree
Ok
So let me show you my odd method but again it's not how you're supposed to do this
Also just to be sure, we know that the small arc is a semi-circle and the big arc is a quarter of a circle, right?
neon iron
So let me show you my odd method but again it's not how you're supposed to do th...
mm ok
ruby tree
Alright so we only have 5 options, let's rewrite them as y = ...
- A) y = (x-175)/7
- B) y = (-x+135)/7
- C) y = (-x+175)/7
- D) y = 7x-150
- E) y = -7x+135
Do you agree with this?
ruby tree
Alright and we know that the line L crosses the big arc at x=0 and also somewhere else, but since it's a quarter of a circle, its slope is only slightly downward
ruby tree
Right
So B and C are our only options
Now which one is correct depends on that radius R
The line crosses the y axis at (0, 2R), correct?
neon iron
yes
ruby tree
If the answer is B, then 2R = 135/7 which is approximately 19.3
If the answer is C, then 2R = 25
neon iron
mm right
ruby tree
Ok now we know that 2R must be greater than the distance between P and the origin
neon iron
and how do you do it w the equation of a circle
?
perhaps I am expected to discover it by myself
ruby tree
You can see that since P is on the green circle, if you draw the purple circle, it will intersect the x axis at a value less that the green circle
neon iron
yes
ruby tree
It's always going to be the case until P is at the very edge and the purple and blue circles coincide
So no matter where P is on the green circle, the distance between P and the origin is less than the diameter of the green circle
ruby tree
Now it turns out that this distance between P and the origin is sqrt(16^2+12^2) = 20
And that's more than the 135/7 from option B
We know that 2R is more than that, so it has to be option C, 25
neon iron
you're right
and how do you do it w the equation of a circle?
perhaps I am expected to discover it by myself
ruby tree
First step is to find the green circle, with center (R,0), so its equation is (x-R)^2 + y^2 - P^2 = 0
It passes through the origin, so (0-R)^2 + 0^2 - P^2 = 0, meaning P = R
It also passes through P, so (16-R)^2 + 12^2 - R^2 = 0
That gives you 2R = 25
Then the blue circle is easy, center (0,0) and radius 2R
Then the black line through points P and (2R,0)
Then the intersections between that black line and the blue circle (that's the orange lines)
It gives you x=7 or x=25
Plugging x=7 back into the line equation gives you y=24
So finally you have the line L going through (0,25) and (7,24)
The corresponding equation is answer C
ruby tree
Note that instead of finding an equation for the black line, you can also say that P is the midpoint between the two intersections between that line and the big arc, because there is an isosceles triangle where P is on the perpendicular bisector
So if one intersection is at x=25, the other is at x = 16 - (25-16) = 7
neon iron
mm right, is also the midpoint
ruby tree
Seems a bit more complicated
neon iron
yeah
ruby tree
Anyway, I still don't know how you're supposed to do this without the equation of a circle
neon iron
me neither
sometimes that happens
I have exercises that I didn't have the information needed to solve it
well, thx
ruby tree
neon iron
.close
topaz sinewBOT
topaz sinewBOT
Available help channel!
lean jackal
Hello, I would like to ask this linear algebra question, related to affine spaces and barycentric coordinates/combination. I am given this proof from a textbook, but I don't understand it
topaz sinewBOT
Hello, I would like to ask this linear algebra question, related to affine space...
lean jackal
The main thing is, I dont understand the part where a+l is just switched back to a
Like I know that a+l because of operation in an affine space, it would still be inside of A, and I know that the sum of the constants yi from the field K add up to 1 (it's from the definition of barycentric)
shadow salmon
notice that you can rewrite the sum, and you have a term left which is just zero, the one they mention
(and they factor)
lean jackal
yes, i see that [ 1-sum(yi) ] * l = 0
because the sum of yi's are just 1
but im still confused tho
like how do i manipulate the LHS of the equation
shadow salmon
sum yi(ai -a -l) = sum yi(ai-a) - l sum yi
and then factor out l with the l you have on the outside
lean jackal
how do you factor out the l?
isnt the l a vector?
shadow salmon
and the rest are scalars?
lean jackal
wait so
shadow salmon
ok so you have a + l + sum yi(ai -a - l) = a + l + sum yi(ai-a) - l sum yi on the LHS, right?
lean jackal
i mean i understand summation properties but its just that vectors and scalars are just messing with my mind right now
im so sorry
pseudo jetty
,, a + l + \sum_{i = 0}^s y_i (a_i - a \2r{- l}) = a + l + \sum_{i = 0}^s y_i(\2r{-l}) + \sum_{i = 0}^s y_i(a_i - a)
thorny flameBOT
lean jackal
ah so you just do basic separation and factorization and since yi and (ai-a) contains index i you cant put them out
shadow salmon
i mean i understand summation properties but its just that vectors and scalars a...
what part is messing with you?
lean jackal
and l here is just independent so i can just put it outside in some sense
pseudo jetty
i guess you could say it like that
lean jackal
ah i see, after this may i have a hint on what i should do?
pseudo jetty
wdym after this
shadow salmon
thats it
shadow salmon
i would recommend refreshing yourself with algebra if that's what is the confusing part here?
lean jackal
pseudo jetty
,, y_i(a_i - a \2r{-l}) = y_i(a_i - a) + y_i(\2r{-l})
thorny flameBOT
lean jackal
im so sorry like i just realized i asked a silly question
i understand what you mean now
shadow salmon
no need to apologize, after all this is a help channel
pseudo jetty
please ask any question you'd like
lean jackal
really thanks for all your help, backoseph and aslan
well
this is related to the above but if i may ask
pseudo jetty
,, \sum_{i = 0}^s y_i(\2r{-l}) = \2r{-l} \sum_{i = 0}^s y_i = \2r{-l}
thorny flameBOT
lean jackal
it's still the same concept but now the sum of the xi's are equal to 0
pseudo jetty
yeah
lean jackal
ah okay, i will try this then
i got back to where it started
wait mb i tried again
so can this say that sum xi(ai-a) doesnt depend on a?
because it now contains only xi and ai?
.close
topaz sinewBOT
pseudo jetty
so can this say that sum xi(ai-a) doesnt depend on a?
because it now contains on...
do you know that you can take linear combos of the a_i?
they come from an affine space, so surely you can only add them after you subtract off the a
lean jackal
wait so you mean i did wrong steps here?
i dont understand
a_i comes from the affine space A because they're just points or 'coordinates' from the set A right?
but im confused to the meaning of taking the linear combos
.reopne
.reopen
topaz sinewBOT
✅
lean jackal
allg
pseudo jetty
a_i comes from the affine space A because they're just points or 'coordinates' f...
the difference of two affine points give you a point in a linear space
at which point you can use the vector space operations like addition
you cannot add two points in affine space
lean jackal
wait difference between two affine points say a is in A and b is in A
then a-b wouldnt it be a vector? or am i wrong here
pseudo jetty
yes a - b would be a vector in a linear space
so then it makes sense to take linear combos
but for instance the sum of a_is themselves does not make sense
you have to take a sum of (a_i - a)
lean jackal
yes, i think i understand, because say a1 + a2 + a3 in an affine space doesnt make sense
because usually you would add a vector to a point, like say a is in the affine A
and v is from the associated vector space V
lean jackal
you have to take a sum of (a_i - a)
but how do you do this? taking the sum of ai-a?
like do it one by one? or something else?
pseudo jetty
i'm just giving an example but since a_i and a come from A, their difference is in V
so taking sums is just as you'd do in V
lean jackal
so i dont need to do it mathematically, but i just need to say that
ai-a basically just represents a vector from the associated vector space V from the affine space (A,V)?
or am i still wrong here?
pseudo jetty
i'm not sure what you mean by do it mathematically
a_i - a is in V
you can just add them by the vector space axioms
lean jackal
wait mb i tried again
no i mean like are these steps wrong? because i did it just like the example before?
i understand the concept you just taught me, but im confused how its related to this calculation
pseudo jetty
i got back to where it started
this is the correct calculation
pseudo jetty
wait mb i tried again
this afterwards does not make sense
you can't separate the a_i from the -a in the sum
the fact that you even have a_i - a is what's allowing you take the sum in the first place
lean jackal
ahh i see, separating it would be wrong because the subtraction results to a vector, and x_i times a does not make sense as well?
pseudo jetty
yeah
lean jackal
by the way a point cant be multiplied by another point?
lean jackal
i got back to where it started
okay so after obtaining the result, which comes back to the beginning, how should i say/conclude?
pseudo jetty
and nothing else
pseudo jetty
okay so after obtaining the result, which comes back to the beginning, how shoul...
essentially the choice of a doesn't matter
lean jackal
like ai-a is just a vector from V and xi(ai-a) is just linear combos of vector inV?
pseudo jetty
because you can replace it with another a' = a + l and you get the same result
pseudo jetty
like ai-a is just a vector from V and xi(ai-a) is just linear combos of vector i...
yep
lean jackal
OHHH
lean jackal
because you can replace it with another a' = a + l and you get the same result
my brain just understood this
pseudo jetty
lean jackal
so in some sense it's saying that the a is not 'unique'?
like you can add a with another vector l
and still get the same result?
pseudo jetty
well that's kind of a way to phrase it
lean jackal
or am i saying it wrong
alright i mustve said some mistake, but i get the meaning now
pseudo jetty
i'd rather say that the expression does not depend on the choice of a
lean jackal
alright i get it
pseudo jetty
lean jackal
and a+l still lies on the affine A because it's the basic binary operation that defines the affine space in the beginning?
l being a vector in V (and a being a point in the affine A)
pseudo jetty
if you have another a' in A, then the difference a' - a is an element of V
lean jackal
ahh i see, so it's better to phrase it that way
pseudo jetty
lean jackal
you just explained something i could not understand for a whole day thanks a lot brother
pseudo jetty
lean jackal
idk if its normal to be stuck on something trivial like this for a day,
pseudo jetty
nah it's fine
lean jackal
👍
pseudo jetty
the abstractness can be confusing at first if you don't have good concrete examples
lean jackal
that is very true
alright brother, im not gonna waste more of your time if you want to help others
pseudo jetty
i got a class now so i gotta dash anyway
lean jackal
but thanks a lot and have a nice day/evening
pseudo jetty
topaz sinewBOT
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Available help channel!
blazing scarab
hi, how would I go about factoring this to find all real solutions?
blazing scarab
I have tried putting the -2 on the other side and factoring out x^2, but i got stuck there
like this: $x^2(x^2-3)=-2$
thorny flameBOT
KXLI
topaz sinewBOT
@blazing scarab Has your question been resolved?
blazing scarab
.close
olive lichen
$\int_C\frac{-e^{-i\omega}}{\omega^2+2\gamma \omega i-\omega_0^2}d \omega$\
Where $\omega \in \mathbb{C}$ and $C$ is the semicircular arc in the upper half plane with a radius -> $\infty$
topaz sinewBOT
$\int_C\frac{-e^{-i\omega}}{\omega^2+2\gamma \omega i-\omega_0^2}d \omega$\\
Whe...
thorny flameBOT
KySquared
olive lichen
I’ve already shown it equals 0 for t<0. But the way I did it felt really slow
Can anyone help with speeding up the proof or give tips on how I could get to the conclusion faster?
I went about it by brute force and expanded everything
I’m trying now to instead use polar representation but I’m kinda stuck
$\int_C\frac{-e^{-i\omega t}}{\omega^2+2\gamma \omega i-\omega_0^2}d \omega \longrightarrow \int_{0}^{\pi}\lim_{R \to \infty}\frac{-e^{-iRe^{i\theta}t}}{(Re^{i\theta})^2+2\gamma Re^{i\theta}i-\omega_0^2} iRe^{i\theta} d\theta$
thorny flameBOT
KySquared
cinder sequoia
i mean as R tends to infinity, the numerator is a decaying exponential in R and the denominator behaves linearly
assuming you use the R from the differential to reduce the quadratic denominator to a linear one
try using jordan’s lemma if you want, specifically since your contour is the upper half semicircle and since your integrand is multiplied by a complex exponential
olive lichen
Thats the problem
I’m not allowed to directly use jordan’s lemma
cinder sequoia
oh
there are common ways to bound the absolute value using inequalities
like triangle and reverse triangle
that’s your best bet without jordan’s
olive lichen
My idea was to change the complex number in the exponential from polar to rectangular
Then expand from there
cinder sequoia
generally leaving in polar is best here
but you might be able to bound the exponential in the numerator by expanding using euler’s formula
olive lichen
But my concern is that $e^{-iwt}=e^{yt}e^{-itx}$
thorny flameBOT
KySquared
olive lichen
Which kinda still depends on R by the polar definition
Would I still be able to pull it out the limit regardless?
I’m in the upper half hemisphere so $y\geq 0$ so it theoretically works out. But idk if selectively choosing what is in polar/what’s in cartesian is allowed
thorny flameBOT
KySquared
cinder sequoia
so the absolute value of the integrand is <= |R| * |e^(-iRte^(iθ))| / |denominator|
uh i’m just tryna check if this works
then |e^(-iRte^(iθ))| = |e^(-iRt(cos(θ) + isin(θ)))| = |e^(-iRtcos(θ))| * |e^(Rtsin(θ))| <= |e^(Rtsin(θ))|
that make sense?
@olive lichen
olive lichen
I follow
cinder sequoia
and we’re concerned about when t < 0?
as θ varies from 0 to π, sin(θ) is always positive, so tsin(θ) is always negative, so Rtsin(θ) is some negative number as R tends to infinity
which means the exponent is always negative and gets increasingly more negative, so |e^(Rtsin(θ))| <= |e^0| = 1
following?
@olive lichen sorry for pinging i just wanna make sure you see it
olive lichen
Ahh okay
cinder sequoia
makes sense?
olive lichen
Yeah it makes sense. I went a similar route but I haven’t considered using the triangle inequality
Pretty much I noticed the entire function was being multiplied by $e^{yt} \geq 0$ for $y \in C$ but that would cause everything to diverge thus to make it converge t must be <0
thorny flameBOT
KySquared
cinder sequoia
yeah that's pretty much necessary
cinder sequoia
then |e^(-iRte^(iθ))| = |e^(-iRt(cos(θ) + isin(θ)))| = |e^(-iRtcos(θ))| * |e^(Rt...
so then this exponential has absolute value <= 1
cinder sequoia
so the absolute value of the integrand is <= |R| * |e^(-iRte^(iθ))| / |denominat...
so absolute value of integral is <= |R| / |denominator|
which is |R / (R^2 * complex exponential + R * stuff * complex exponential - constant)|
you could technically be more rigorous here and find a way to upper bound the denominator using reverse triangle inequality
but at this point it's clear that letting R tend to infinity causes that upper bound to vanish to 0
showing that your integral vanishes on the contour
olive lichen
then |e^(-iRte^(iθ))| = |e^(-iRt(cos(θ) + isin(θ)))| = |e^(-iRtcos(θ))| * |e^(Rt...
For this, wouldn’t $\abs{e^{-iR\cos{(\theta)}t}}$ just be 1?
thorny flameBOT
KySquared
olive lichen
@cinder sequoia
cinder sequoia
yes, so actually that last sign should be an equality, not an inequality
most complex analysis i’ve seen doesn’t care too much about the specifics tbh they’re quite hand wavey with the inequalities
olive lichen
Gotcha
But ah okay
Yeah this is basically what I did except without abs
I ended up with $e^{yt}$(fraction1 - fraction2$i$)
thorny flameBOT
KySquared
neat sierra
can somebody check my solution
topaz sinewBOT
can somebody check my solution
neat sierra
Rewrite into a quadratic in terms of x:
4x^2 + (4y-16)x + (2y^2-4y+23)
To look at the minimum value for x, we can use the shortcut h=-b/2a
this gives -y/2 + 2 as the minimum for x
subbing into x and simplifying
we get y^2 + 4y + 7
and the minimum of this is 3
I did work out all the steps
i didnt want to say them all as it’d be too long, just listed the crucial steps
(the correct answer is 3, just want to know if i got it by chance or is the logic right)
rigid cloak
Okok
rigid cloak
!noans
topaz sinewBOT
The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.
rigid cloak
!status
topaz sinewBOT
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
Which step are u on
neat sierra
i dont want to be rude but i clearly am in stage 4
i just want to see if the logic is right
as i ended up getting the right answer
thorny flameBOT
rigid cloak
@neat sierra
neat sierra
yea
neat sierra
its minimize the expression
and i didnt minimize x or u
x or y
i put the expression in terms of y
and then minimzed it entirey
are you a troll
,w min (4x^2+2y^2+4xy-16x-4y+23)
thorny flameBOT
neat sierra
minima
does my logic look right to you
oppais a troll literally just looked through his chat logs
raven sparrow
The idea is good.
You're minimizing the function for a fixed y, yielding an expression for the minimizing x in terms of that fixed y.
Then you plug it back into the original expression to minimize the whole value in terms of y.
In general, if you've learned that in the context of the class, you can always use calculus, but again it's a good approach, although it might not work for general functions (here it's "easy" since we can find the optimal x right away, which might not be the case for higher degree expressions, for instance.)
neat sierra
The idea is good.
You're minimizing the function for a fixed y, yielding an exp...
So the logic is there?
Also i cant use calculus
I would if i could
raven sparrow
Yeah I figured you were restricted in some way.
neat sierra
Because its a contest problem for high school math and im not allowed to use calculus
neat sierra
Yeah I figured you were restricted in some way.
Thanks so much
raven sparrow
But yeah it's logically sound
neat sierra
.close
topaz sinewBOT
topaz sinewBOT
Available help channel!
mild shale
I’m trying to find the angle between vectors and can’t figure it out when I put it into my ti 84 it comes out with a domain error
topaz sinewBOT
I’m trying to find the angle between vectors and can’t figure it out when I put ...
@mild shale Has your question been resolved?
mild shale
<@&286206848099549185>
sudden temple
what's the matter?
mild shale
I cant solve this vector
sudden temple
I’m trying to find the angle between vectors and can’t figure it out when I put ...
C is the resultant of A and B right?
mild shale
yes
rigid cloak
Use dot product
topaz sinewBOT
@mild shale Has your question been resolved?
brazen dawn
topaz sinewBOT
barren zodiac
split it into a couple shapes that you can calculate the area of, calculate and add them together
clear juniper
Hi
barren zodiac
I don't see a smart way/shortcut to do this honestly, yeah takes a bit of patience but should be pretty easy though
clear juniper
is the problem finsihed here?
neon iron
yea, just cut the figure into different parts and label them starting from A
that is what i did in primary school though...
brazen dawn
so that's right
brazen dawn
yea, just cut the figure into different parts and label them starting from A
Kk
ty
so is it 6 times 6 19 times 7 28 times 14 and 19 times 4
barren zodiac
yea just add them all up
brazen dawn
i did is wrong
barren zodiac
what'd you get?
clear juniper
Hey
barren zodiac
usually they all can be solved the same way, split them into simple shapes and add their area
barren zodiac
i'm done with this
u can close this with .close
long cave
!status
topaz sinewBOT
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
7
topaz sinewBOT
@brazen dawn Has your question been resolved?
inland plover
topaz sinewBOT
inland plover
Yo guys how did that long equation simplify to that?
summer dock
imagine a simpler case where
,, \frac{a}{b^{\frac{1}{2}}} - \frac{c}{4b^{\frac{3}{2}}}
imagine doing this subtraction
the first step would be to make the bases the same
thorny flameBOT
[code{RED}]
junior folio
Not that hard, multiply the bases together.
Then multiply the top by the opposite bottom
summer dock
Not that hard, multiply the bases together.
not necessary
topaz sinewBOT
@inland plover Has your question been resolved?
topaz sinewBOT
topaz sinewBOT
Available help channel!
proper dagger
topaz sinewBOT
proper dagger
What is wrong with my working out for this question?
sudden temple
that's a geometric sequence not Arithmetic
proper dagger
that's a geometric sequence not Arithmetic
I see, what steps can I do then?
sudden temple
I see, what steps can I do then?
follow the same procedure except use the formula of geometric sequenxe
proper dagger
follow the same procedure except use the formula of geometric sequenxe
$T_n=ar^{n-1}$?
thorny flameBOT
Jshy
thorny flameBOT
77²
proper dagger
and also remember the fact that common ratio = $\frac{a_{n+1}}{a_n}$
hmm I dont believe i have learnt this
sudden temple
hmm I dont believe i have learnt this
it is just the next term over previous term
proper dagger
I see, in this case its the 3rd and 5th terms though
sudden temple
I see, in this case its the 3rd and 5th terms though
yup
and two variables
a and r
proper dagger
yup
slightly lost, do I need to figure out r first before i can determine a using the formula?
sudden temple
I see, in this case its the 3rd and 5th terms though
see from here, you can make two equations in terms of a and r, right?
sudden temple
yup
topaz sinewBOT
@proper dagger Has your question been resolved?
proper dagger
cheers, thank you for the help that makes sense
fierce sierra
So for H, its basically asking for functions such that the second derivative = 8281 * the original function. So that means the exponential, sinh and cosh are solutions. So does that mean H is the span of (exp(-91x), exp(91x), sinh(91x), cosh(91x))? And as H is closed under addition and scalar multiplication, it means its a subspace of S?
topaz sinewBOT
So for H, its basically asking for functions such that the second derivative = 8...
@fierce sierra Has your question been resolved?
neon iron
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neon iron
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sour bloom
what is the intuition behind this
if the points were uniformly spaced, what do you expect the answer to be?
sour bloom
the amount of points in an area should be proportional to the area
the measure of the area
do you get that idea?
sour bloom
what intuition are you trying to gain
sour bloom
what do you want to say about random behaviour
for random walks, a main idea is probably distance from origin
neon iron
i just want an intuition to apply to better understand probability
brazen inlet
Can someone please explain the fluid force centroid section
topaz sinewBOT
Can someone please explain the fluid force centroid section
brazen inlet
I get the integral for fluid force, and I get centroids, but when it tries to combine the two I’m not following
Specifically “moment about surface level line…”
Nvm I get it now
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sacred kettle
Let's say I have a convex function f : R^n -> R.
If f((x1, x2, ..., xn)) < f((x'1, x'2, ..., x'n)), then the minimum is necessarily attained for a xopt on the side of x1 (meaning if x'1 > x1 then the interval [x'1;+inf) is useless and x'1 < x1 then the interval (-inf; x'1] is useless), etc .. ?
topaz sinewBOT
Let's say I have a convex function f : R^n -> R.
If `f((x1, x2, ..., xn)) < f((x...
topaz sinewBOT
@sacred kettle Has your question been resolved?
sacred kettle
no
carmine marten
not every convex function has a minimum? so you can't say anything whether a minimum is necessarily attained without at least the condition of a minimum existing
sacred kettle
mmh good point, what if the convex function is unimodal on top of it then?
carmine marten
Consider $f(x,y)=(x+y)^2+y^2$. Then $f(-1,1) < f(-0.5,1)$ so if I understand correctly you're claiming that the optimum solution will have first coordinate not in $[-0.5,\infty)$? this is false, since the minimum is at $(0,0)$
thorny flameBOT
Edward II
trail swan
Would the graph of D be correct? Thats the one I picked
topaz sinewBOT
Would the graph of D be correct? Thats the one I picked
ruby tree
Yes
topaz sinewBOT
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tiny finch
Why did we get +-pi/6?
topaz sinewBOT
Why did we get +-pi/6?
sudden temple
!occupied
topaz sinewBOT
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).
mint crescent
Why did we get +-pi/6?
The ref angle for cos(x)=(sqrt 3)/2 is pi/6, but cosine is also positive in the fourth quadrant, so we consider -pi/6 (the angle in the fourth quadrant with a reference angle of pi/6) as well
tiny finch
The ref angle for cos(x)=(sqrt 3)/2 is pi/6, but cosine is also positive in the ...
I only understood that cos is positive in the 4. quadrant, but the rest I didn't. Could you give me a simpler explanation?
Why do we consider -pi/6 if it is negative?
mint crescent
(the angle in the fourth quadrant with a reference angle of pi/6) as well
Note that 2pi - pi/6 is coterminal with -pi/6
tiny finch
> (the angle in the fourth quadrant with a reference angle of pi/6) as well
What does coterminal mean?
mint crescent
tiny finch
I'm sorry but I still don't understand... I don't get why -pi/6 is in the positive quadrant aka the 4. quadrant of cos.
tiny finch
I understand this part but I don't understand how it fits to answer my question
topaz sinewBOT
@tiny finch Has your question been resolved?
tiny finch
Why did we get +-pi/6?
<@&286206848099549185>
ivory moon
I'm sorry but I still don't understand... I don't get why -pi/6 is in the positi...
More help?
tiny finch
Please🥺
ivory moon
There is+ - pi/6 because in the first and fourth quadrant, cos is positive
tiny finch
but -pi/6 is negative
ivory moon
Here, think it as Cos being the X values, and sin as thr y values.
ivory moon
but -pi/6 is negative
The angle is, but x is positive, y is negative
So sin-pi/6 would be negative
In th fourth quadrant, it’s expressed as (x,-y)(x, y are natural numbers)
So x is pos and y is neg
Dm me if you have questions
tiny finch
Dm me if you have questions
Can I add you as a friend?
ivory moon
Sure
tiny finch
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topaz sinewBOT
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proper lotus
Probability
topaz sinewBOT
Probability
proper lotus
Random variable and varience
opal vault
Probability
!status
topaz sinewBOT
Probability
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
Like how did they put a whole square there
opal vault
Like how did they put a whole square there
it's how you compute variance
Var(aX+b) = a^2Var(X)
if you can't see why, go back to the definition E(X^2) - E(X)^2 (Koenig-Huygens formula)
proper lotus
The thing is this isn't tought in our school and its also not in our syllabus so that is why i dont knew the full info
Thanks
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tepid mauve
topaz sinewBOT
tepid mauve
ok so
i'll show my work process and wjere i got stuck
2log5(x)=log5(225)
log5(x^2)=log5(225)
log5(225)-log5(x^2)=0
log5(225/x^2) = 0
dense lily
Log of negative values are ...
whole spade
theres a better way
tepid mauve
Log of negative values are ...
undefined
safe pewter
Isin't it x^2 = 225?
tepid mauve
raise to the power of 5
oh
2log5(x)=log5(225)
log5(x^2)=log5(225)
(log5(x^2))^5=(log5(225))^5
x^2=225
x= -15, 15
x=15, as you cannot take the log of negative nums
is this right?
whole spade
i guess so
dense lily
2log5(x)=log5(225)
log5(x^2)=log5(225)
(log5(x^2))^5=(log5(225))^5
x^2=225
x= -1...
Yes
dense lily
If you only need real values of 2logx
tepid mauve
yes?
tepid mauve
yeah this is alg 2 i don't think they're looking for complex solutions
ty for all the help
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topaz sinewBOT
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fickle fable
prove or disprove:
if a_n as a positive seq approaching zero, so there is a sub seq a_n_k such that the sum from 1 to infty of (a_n_k) is convergent
I think that this statement is false and a counter example is the harmonic series,
however I did not know how to prove that sum(1/n) does not have any subseq that converges
frigid scroll
i believe that counter example does not work. for example, choose a_n_k = 1/k^2, which converges
fickle fable
i believe that counter example does not work. for example, choose a_n_k = 1/k^2,...
oooh 😂😂😂😂😂
how did i not notice
frigid scroll
lol yeah
cunning snow
Can some one help me in the 18th problem
topaz sinewBOT
Can some one help me in the 18th problem
@cunning snow Has your question been resolved?
cunning snow
<@&286206848099549185>
topaz sinewBOT
@cunning snow Has your question been resolved?
worn mica
had to guess on this one
topaz sinewBOT
had to guess on this one
worn mica
don't know what to do here
snow nimbus
don't know what to do here
Multiply (ax+b)(2x+d) out
worn mica
2ax^2 + 2bx + cax + bc
snow nimbus
Great
worn mica
i set 2ax^2 to 6x^2 and got a = 3 but idk if thats gonna help
snow nimbus
Let's compare the coefficient without an x. What do you get?
worn mica
bc = 54
worn mica
yes i see now, since c and b multiply to get 54, 54 divided by any of those numbers will get us a integer
snow nimbus
Just 54/b = c. And c is an integer by definition
worn mica
yep thanks
snow nimbus
You're welcome
worn mica
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pseudo sonnet
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sour carbon
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sour carbon
I’m pretty sure Grogg must travel to an even amount of stations
and I think I have to find the amount of ways for 2,4,6,8, and 10
keen raptor
I’m pretty sure Grogg must travel to an even amount of stations
This is not true
sour carbon
Oh
or how about the amount of lines we use?
they have to be even
So you can use 2,4 or 6lines
And then find the cases for each?
keen raptor
Also not true
keen raptor
all I can think is try to simplify the diagram. Cut off all the loose edges and maybe find a way to redraw
sour carbon
So you can use 2,4 or 6lines
why would this not be true?
keen raptor
Hmm
Are you thinking we can't continue along the same line after visiting a station?
Otherwise it's not true because it's easy to find examples where it's false
keen raptor
Ok I see what you mean
"using" a line meaning actually going on it, not just passing through it
Or well
sour carbon
Yeah sorry for th confusion
keen raptor
I was thinking of the line segments
Np
So yes you need to use an even number of lines, though not necessarily an even number of line segments
sour carbon
How about for 4 and 6 lines
topaz sinewBOT
@sour carbon Has your question been resolved?
keen raptor
How many choices for the first line?
topaz sinewBOT
@sour carbon Has your question been resolved?
left moth
topaz sinewBOT
topaz sinewBOT
@left moth Has your question been resolved?
tiny finch
One small question. A point in linear algebra is never actually a point but a vector from the coordinate 0 of the point A, correct or incorrect, if incorrect then why?
topaz sinewBOT
One small question. A point in linear algebra is never actually a point but a ve...
@tiny finch Has your question been resolved?
tiny finch
<@&286206848099549185>
craggy haven
it's more like you can think about them both ways
sometimes one is more useful than the other
dense crescent
It's kinda hard to give example ig
But u will understand when to use it
Instincts
tiny finch
how do I visualize a task then
dense crescent
?
tiny finch
sorry, not the best question. How do you visualize a task when solving it?
dense crescent
For geometry it is best to draw the diagram
tiny finch
like for example here in step 4, how the heck do you remember to do that?
It's not even a pattern so that I can solve multiple geometry tasks it's just intuition that I don't know how to build, please help me find a way
tiny finch
Find a point in which the line crosses through the plane. Whats the angle between p (line) and pi (plane)?
tiny finch
I tried to solve it by myself but there is no way I would remember to do the 4. step on my own so I just gave up
tiny finch
Here u have equation of line and plane right
yes
dense crescent
yes
So here
U convert the line into parametric
And u get a parametric point
U then substitute it into the plane
Which gives the value of t
U put t in parametric
U get point
tiny finch
I get that, but how do you build intuition to solve this without looking at the answer?
dense crescent
That is the process u generally use
For finding intersection
Basically
U got 2 equations
But 3 unknowns
So u try to convert it to have less unknowns
How do u do that?
U compare stuff in line
To get parametric
In one unknown
Then u sub that in plane to get the values of unknown
tiny finch
Ye I kind of get it
dense crescent
Nice
tiny finch
Is there a way to solving geometry tasks in general?
tiny finch
Way of thinking or something of help, some system, visualisation...?
dense crescent
U fine that then u have many ways to solve a question
dense crescent
Way of thinking or something of help, some system, visualisation...?
Basically draw stuff
Try manipulation by drawing lines and all
At start ull face some issues
But later on ull get the idea on ur own
tiny finch
but I drew this and still didnt know what to do after the 3. step
I guess I would get half points for finding the angle where p and pi intersect but I'm not satisfied with that
I'm guessing practice is the only thing that will help me here?
thank you for the help, I have to go to sleep. Good night
dense crescent
Ok so
U cannot find the angle directly
Remember from plane equation
U can get normal vector?
tiny finch
nono I get the problem I understand it
tiny finch
Im just talking in general
topaz sinewBOT
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Available help channel!
lapis siren
how many five letter passwords can be made from the letters of the word ESPRESSO?
topaz sinewBOT
how many five letter passwords can be made from the letters of the word ESPRESSO...
lapis siren
i got 820, just want to check if its right cos my answers say 760
This is my messy working out
I just characterised the letters into triple, double and single letters then found out the different cases that can be made
Am I over counting in any case?
stuck hearth
5! for single
2 * 5C2 * 4 * 3 * 2 for 1 pair
5C2 * 3C2 * 3 for 2 pair
5C3 * 4 * 3 for 1 triple
5C3 for 1 triple and 1 pair
i think
lapis siren
hm that also adds to 820
stuck hearth
ye
topaz sinewBOT
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barren radish
help explain pls
topaz sinewBOT
help explain pls
barren radish
inscribed angles
barren radish
I could be wrong
what grade you in
barren radish
ok so im pretty sure this is solvedd by using inscribed angles
barren radish
we dont know y or x
ivory sorrel
yeah, I know
barren radish
we only know the arc lengths
ivory sorrel
is there any more information given?
steady escarp
I don't think there's enuf info too
ivory sorrel
We don't know any lengths(that could help us)
steady escarp
help explain pls
If that line were moved up towards the red line then the angles x and y will be different
So there's no unique answer unless something else it known
neon iron
howd u find z
crimson kayak
What do those 2 values 66 and 96 define
Which angle are they referring to ?
livid flame
What do those 2 values 66 and 96 define
Which angle are they referring to ?
Exactly the question I was gonna ask
pseudo sonnet
Is this the complete original problem? @barren radish
I feel as though we must be missing information
tardy lantern
What do those 2 values 66 and 96 define
Which angle are they referring to ?
The outside of the circle
barren radish
Is this the complete original problem? <@902333706474238014>
yes
pseudo sonnet
Huh.
bitter hemlock
this is just inscribed angle theorem
pseudo sonnet
oh, no wonder I couldn't think of a method lol
crimson kayak
this is just inscribed angle theorem
There's no angle at the center ....
pseudo sonnet
my Euclidean geometry is extremely limited outside of angle chasing 
bitter hemlock
There's no angle at the center ....
an angle with the vertex at the center of the circle (red) will have 66 degrees (same angle measure as the arc it's describing), an angle with the vertex somewhere on the circle itself will have half the measure of the arc it's describing (green), it's a bit clearer if the green is on either side of the center, but it ultimately doesn't matter
crimson kayak
an angle with the vertex at the center of the circle (red) will have 66 degrees ...
Okay the 66 degrees is the angle of the arc
bitter hemlock
so it's a bit clearer in the picture of the theorem i pasted (angle O is double angle C), but no matter how you move A, B, C it holds true
yes
crimson kayak
Then the question is solved
bitter hemlock
the answer is 33
sharp dew
Hello
cinder sequoia
no
sharp dew
Reverse chain rule?
hazy pumice
Do I need chain rule for this
No, u-sub will do
sharp dew
I hate u-sub
cinder sequoia
Reverse chain rule
technically yes since there's a composition but not necessary since the derivative is just 1
sharp dew
Is there any intuitive alternative
hazy pumice
I hate u-sub
U sub is the easiest trick in the toolbox
sudden temple
Is there any intuitive alternative
$\int{f'(ax+b)}dx=\frac{f(ax+b)}{a}+K$
it is just a generalized thing derived from u-sub
but works only for linears
sharp dew
Am I right
sudden temple
Am I right
uh no
cinder sequoia
what's the extra stuff you multiplied by
sharp dew
what's the extra stuff you multiplied by
the integral of x+1
sharp dew
the integral of x+1
the "indefinite" integral I suppose
sharp dew
it is just a generalized thing derived from u-sub
This can be only solved by U-sub I suppose?
sudden temple
This can be only solved by U-sub I suppose?
yeah
cinder sequoia
you don't multiply by the integral
ivory moon
Am I right
I think the second part shouldn’t be there