#help-4
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deft condor
hello
deft condor
it seems correct
my proof is right?
charred burrow
yes
near edge
I try to solve this exercise, but it seems the statement is false, A and B are not in bijection. Do you know what is the correct statement for this exercise ?
vale dockBOT
I try to solve this exercise, but it seems the statement is false, A and B are n...
@near edge Has your question been resolved?
near edge
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rigid whale
not sure if you need the help anymore but i'd say that you should reflect each walk at it's level "lambda", flipping all of the steps before that time gives a bijection between "walks that ever reach lambda by n" and "walks that end >= lambda without reaching lambda early", and by symmetry i think this should show that
ofc the "walk" in this case is basically a sequence of partial sums (where S_0 = 0, S_1 = epsilon_1, S_2 = epsilon_1 + epsilon_2, S_n = epsilon_1 + epsilon_2 + ... + epsilon_n), where each epsilon_i is an independent step taking value +1 or -1 with equal probability
so basically a path on an integer line that moves one unit up or down at each step
i mean i haven't tried to do the problem by hand and this is my best attempt based on looks but i think by symmetry it should be corrected?
it's a bit complex and abstract but i hope you get what i mean
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indigo temple
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indigo temple
Is this even a valid question? Since an inverse of a function is defined only if it's bijective
glass kelp
any restriction?
green urchin
It's not inverse, rather it's the preimage
green urchin
What set will give you this set of image
indigo temple
How do you know that, does the notation tell you anything if it's a preimage or an inverse?
green urchin
How do you know that, does the notation tell you anything if it's a preimage or ...
It's generally understood that f inverse (D) where D is a set gives us the preimage
The question is
I forgot whether every point in D should be traversed by the function
Lemme check abott rq
gleaming kindle
How do you know that, does the notation tell you anything if it's a preimage or ...
its given in def p3.30
green urchin
@indigo temple hope that helps
indigo temple
So, does every point need to be traversed?
indigo temple
its given in def p3.30
I see
indigo temple
So, does every point need to be traversed?
I guess no right? Since they never mention anything about it?
zinc spear
I guess no right? Since they never mention anything about it?
if for some real number imput x, the output x^2 is in the interval [-1,2] then x is an element of f^-1([-1,2]). This preimage set « collects » all the x in the domain of f which satisfies this condition.
indigo temple
But [-1,0) has no preimage
zinc spear
yeah the preimage can be the empty set, for instance here no x in the domain will be sent in [-1,0) because x^2 >= 0., but some will land in [0,2]
indigo temple
Oh okay, i was confused for a while since function is a nonempty relation but preimage is a set not a function
Tyty
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stark wedge
ie what if it's the queen of spades
so handle these cases separately
either the top card is the queen of spades (1/52), or it's some other spade (12/52)
in each case, how many queens remain among the other 51 cards?
wraith heart
_word_ gives italics word
rocky lotusBOT
qoo
Hello. I got this problem for homework last week from a summer program, but I'm really confused on how to start because the cube ABCDEFG isn't fixed. The TA suggested to used $a^2 + b^2 + c^2 = 13^$ in some way, but I'm not sure how that connects to this problem. Can someone give me a hint on how to start this problem
```Compilation error:```! Missing { inserted.
<to be read again>
$
l.49 ... suggested to used $a^2 + b^2 + c^2 = 13^$
in some way, but I'm not ...
A left brace was mandatory here, so I've put one in.
You might want to delete and/or insert some corrections
so that I will find a matching right brace soon.
(If you're confused by all this, try typing `I}' now.)```
compact knot
A product is supplied to a store by four companies. The ratio of these companies to the product is
1 : 2 : 3 : 4, and the probabilities that the product is unusable are for each company in the following order: 0.5, 0.4, 0.3
and 0.2. A product is selected at random in a store and it is found to be unusable. What is the probability that the selected product is from a third company?
Is my answer correct?
compact knot
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finite olive
1/x + 2/y = 1/6
How many pairs of positive integers satisfy the equation?
finite olive
wraith heart
multiply both sides by xy
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@finite olive Has your question been resolved?
finite olive
multiply both sides by xy
6y + 12x = xy
finite olive
Hit and trial ??
finite olive
keep going
Not able to think after that
wraith heart
6y + 12x = xy
try factoring xy - 6y - 12x + c = (x - a)(y - b) and find c = ab
finite olive
try factoring xy - 6y - 12x + c = (x - a)(y - b) and find c = ab
I did it
It’s 12
ab = 12
finite olive
4 and 3
wraith heart
when you expand (x-4)*(y-3) you don't get -6y - 12x
finite olive
I mean to say ab = 72
wraith heart
so what's a and b?
finite olive
(x-6)(y-12)= 72
wraith heart
right
finite olive
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fickle rose
<@&268886789983436800>
fickle rose
what eve is this? 
chrome beacon
what the heck haha
wooden frigate
gone
fickle rose
RIP-By- 🤑 HY45DW3 @shadow moss
shadow moss
RIP-by-? 😂 CODE @ everyone
wooden frigate
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summer parrot
The question at the top is the question btw, but my answer which is at the bottom is wrong according to my work book, supposedly the exponent on the A is 3, not 2. I’m unsure what I did wrong
neat igloo
what's the circle up here
summer parrot
Zero
river shale
final adder
a^0, then?
summer parrot
Yes
normal hollow
a⁰ is not a
summer parrot
What?
summer parrot
Right okay yea I see what I did now, I had multiplied the outer exponent and I put it as one for some reason
Thanks
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orchid mango
Math nerds pls help me out! I’m working through ellipse geometry and came across this:
Is the distance from the co-vertex to the focus of an ellipse always equal to the distance from the center to a vertex?
If yes, is there any formal proof or reference (book, paper, website) that supports this claim? I want to be sure it's not just a coincidence from specific examples. Thanks!
vale dockBOT
Math nerds pls help me out! I’m working through ellipse geometry and came across...
@orchid mango Has your question been resolved?
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@orchid mango Has your question been resolved?
unique tapir
Math nerds pls help me out! I’m working through ellipse geometry and came across...
yes!
do you know the definition of an ellipse?
orchid mango
well I do know it is a curved shaped that looks like an oval. May you explain?
unique tapir
it is defined as the set of all points such that the sum of the distances to the 2 foci is constant
given some constant value
(you can think of it as a generalization of a circle. a circle is the set of all points such that the distance to the center is constant. introducing two foci allows it to have a flattened shape.)
orchid mango
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lament flicker
how should i try to solve this
vale dockBOT
how should i try to solve this
lament flicker
what i know is that
combinatoricly 2010 c 4020 is 0
hence we only need to count
2011 c 2 - 2011 c 4 + 2011 c 6 ...... 2011 c 2010
stark wedge
well, we only need to consider n going up to 1005 yes
lament flicker
2011 c 2 + 2011 c 4 + .... 2011 c 2010 = 2^2010 - 1
viral idol
2011 c 2 + 2011 c 4 + .... 2011 c 2010 = 2^2010 - 1
but there's the $(-1)^{n + 1}$
rocky lotusBOT
1 divided by 0 equals Infinity
viral idol
so if $n$ is odd, then it should be positive, but otherwise, you should put a minus instead
rocky lotusBOT
1 divided by 0 equals Infinity
lament flicker
**1 divided by 0 equals Infinity**
i dont see where it is wrong ?
lament flicker
but there's the $(-1)^{n + 1}$
no its diffrent
viral idol
2011 c 2 + 2011 c 4 + .... 2011 c 2010 = 2^2010 - 1
you put all pluses in here
lament flicker
you put all pluses in here
ik
stark wedge
let your sum be $S$; then you can recast it as $$S = \sum_{k=0}^{2011} \binom{2011}{k} c_k$$ where the sequence $(c_k)$ is $(0, 0, 1, 0, -1, 0, 1, \dots)$
lament flicker
2011 c 2 - 2011 c 4 + 2011 c 6 ...... 2011 c 2010
i need to count this
stark wedge
consider also $f(x) = \sum_{k=0}^{2011} \binom{2011}{k} x^k = (1+x)^{2011}$
rocky lotusBOT
Ann
viral idol
i need to count this
oh alr
rocky lotusBOT
Ann
stark wedge
in fact i will go a tiny bit more abstract and ask this: for which sequences c_k can you easily calculate that sum as the value of f at a point
stark wedge
i try using imaginary but still i couldnt get the answer
can you show your attempt
lament flicker
can you show your attempt
alr
stark wedge
complex numbers are a good route here
lament flicker
so what i get is
(-1+i)^2011
i think i could add it with (-1-i)^2011
to eliminate all the odd term
stark wedge
thats a good idea but needs some refinement
lament flicker
thats a good idea but needs some refinement
yeah thats right
stark wedge
so the key is to consider sums of this form: $$\sum_{k=0}^{2011} \binom{2011}{k} c_k$$ for various sequences $(c_k)$
rocky lotusBOT
Ann
stark wedge
f(0) corresponds to the sequence (1, 0, 0, 0, 0, ...)
f(i) corresponds to the sequence (1, i, -1, -i, 1, i, ...)
f(-i) corresponds to the sequence (1, -i, -1, i, 1, -i, ...)
so [f(i)+f(-i)] gives (2, 0, -2, 0, 2, ...)
so you can get to the sequence (0, 0, 1, 0, -1, 0, 1, ...) in this fashion
lament flicker
so [f(i)+f(-i)] gives (2, 0, -2, 0, 2, ...)
just asking what does the function of f mean here
is it like
i^n
lament flicker
so [f(i)+f(-i)] gives (2, 0, -2, 0, 2, ...)
and is it the same as ((-1+i)^2011 +(-1-i)^2011)/2
stark wedge
$f(x) = \sum_{k=0}^{2011} \binom{2011}{k} x^k = (1+x)^{2011}$ but it'll come in later
rocky lotusBOT
Ann
stark wedge
for now i just want you to think in terms of values of f and defer any evaluation until we've cooked up our target sum
lament flicker
for now i just want you to think in terms of values of f and _defer_ any evaluat...
alr
lament flicker
f(i) corresponds to the sequence (1, i, -1, -i, 1, i, ...)
still what does f(x) here mean ?
lament flicker
see here
yeah alr
we choose i because it could give us 1 and -1 with diffrence of 1 term right ?
since 1 , i, -1
for 2, 3 ,4
stark wedge
[f(i)+f(-i)] gives (2, 0, -2, 0, 2, ...)
so multiplying that by -1/2
-[f(i) + f(-i)]/2 gives (-1, 0, 1, 0, -1, ...)
which is almost what we want
we just need to add f(0) to it
rocky lotusBOT
Ann
compact orchid
hey guys.. im new here
lament flicker
**Ann**
why do i get 1- 2^1005 i
(1+i)^2011 = (1+i)^1005 (1+i) right
so we get (2i)^1005 (1+i)
for the other one we get
(-2i)^1005 (1-i)
lament flicker
$S = 1 - \frac12 ((1+i)^{2011} + (1-i)^{2011}) = 1 - \Re( (1+i)^{2011})$
adding them togetherwe get 1-1/2(2^2006 * i )
hence we get 1-2^2005 i
i think i make a mistake here
since i should get 1 - 2^2005
stark wedge
2006 
stark wedge
how did $2^{2006}i$ happen though \thonk
rocky lotusBOT
Ann
stark wedge
i think there was some funny business on your part
(1+i)^2011 = (2i)^1005 (1+i) = 2^1005 * i^1005 * (1+i)
lament flicker
(1+i)^2011 = (2i)^1005 (1+i) = **2^1005** * i^1005 * (1+i)
yep
stark wedge
i^1005 = i
lament flicker
yeah
stark wedge
i(1+i) = -1+i
lament flicker
i think there was some funny business on your part
i suppose
stark wedge
so the real part comes out as -2^1005
lament flicker
so the real part comes out as -2^**1005**
i was calculating the lhs one
stark wedge
you were making your own life a bit too difficult
lament flicker
lament flicker
you were making your own life a bit too difficult
yeah....
i want to make sure 🙏
i think i make mistake in my calculation
oh yeah i found my mistake
lament flicker
you were making your own life a bit too difficult
i want to ask something if i need to apply 2 times to change 1 into -1
in sense of 1, i, -1
where i is 90 degree from 1 in complex plane
So there is an imaginary number 60 degrees from 1, such that it needs to be used 3 times to change 1 to -1
?
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analog herald
\text{In triangle } ABC, \text{ we are given the conditions: } \angle ABD + \angle BCE = \angle ACE + \angle CBD \quad \text{and} \quad BD = CE. \text{ We need to prove that: } AB = AC.
How can i solve it? thanks
analog herald
$\text{In triangle } ABC, \text{ we are given the conditions: } \angle ABD + \angle BCE = \angle ACE + \angle CBD \quad \text{and} \quad BD = CE. \text{ We need to prove that: } AB = AC.$
rocky lotusBOT
hecker
rough talon
$\text{In triangle } ABC, \text{ we are given the conditions: } \angle ABD + \angle BCE = \angle ACE \angle CBD \quad \text{and} \quad BD = CE. \text{ We need to prove that: } AB = AC$
How can i solve it? thanks
rocky lotusBOT
ihave<skissue>
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@analog herald Has your question been resolved?
north walrus
i think you need to prove that $\triangle ACE\cong\triangle ABD$
rocky lotusBOT
Allen
north walrus
BD=CE (1 side)
sleek nebula
\text{In triangle } ABC, \text{ we are given the conditions: } \angle ABD + \ang...
Wait this is simo question-
Oh youre him okay mb
north walrus
um but AB and AC doesn't look equal
sleek nebula
he said that this is a generalisation so perhaps utilise similar techniques?
@analog herald the rv guy said
rmb the congruence triangle flip
sleek nebula
um but AB and AC doesn't look equal
the diagram is horrible
vale dockBOT
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marsh forge
got it
apologies for sending the solution but i have to go soon, the main gist is to show BEC = BDC
(is this even correct lol)
@analog herald
analog herald
<@1398538692569600062>
thanks, ill look at it
vale dockBOT
thanks, ill look at it
analog herald
.reopen
river shale
already opened
sleek nebula
Indeed
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analog herald
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lament flicker
using ord i found the first 2 solution which is (2,2) and (2,4)
lament flicker
and p=n is trivial
i need to prove that there will be no other solution
but how should i try to prove it
oh i can use ord to get
n^2p = 1 mod p^n +1
then 2p| p^n (fermat little theorem)
since n<=p we can just count manually
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marsh forge
A point $D$ is chosen inside an acute-angled triangle $ABC$ with $AB > AC$ so that $\angle BAD = \angle DAC$. A point $E$ is constructed on the segment $AC$ so that $\angle ADE = \angle DCB$. Similarly, a point $F$ is constructed on the segment $AB$ so that $\angle ADF = \angle DBC$. A point $X$ is chosen on the line $AC$ so that $CX = BX$. Let $O_1,O_2$ be circumcenters of the triangles $ADC$ and $DXE$. Prove that the lines $BC,EF,O_1O_2$ are concurrent.
rocky lotusBOT
Copter
marsh forge
status is 1
well one thing i observe is that the circumcircles of DEF and DCB are (maybe) tangent to eachother? how do i prove that
marsh forge
**Copter**
also uhh equivalent to proving the radical axis of circles BCEF, DEF, BDC intersect at that point
opal pendant
where do people find this shit bro 😭
marsh forge
straight from the depths of hell
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marsh forge
<@&286206848099549185>
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dull heart
Wtf is this shit 😭 frm where you found this
warped willow
this is scary
vale dockBOT
@marsh forge Has your question been resolved?
modest kettle
but it looks like AC > AB here...
marsh forge
oops
well ignore that 😭
i mean theyre still concurrent so i thought my diagram was right
midnight pier
where do people find this shit bro 😭
lmao
midnight pier
copter archives
Whats that?
slate folio
well one thing i observe is that the circumcircles of DEF and DCB are (maybe) ta...
tf they're actually tangent
marsh forge
yea
also let T be the intersection of EF and BC, inverting from T with radius TD probably gives something
slate folio
it's easy to prove but I'm not sure if it helps
marsh forge
but i have to go💔
marsh forge
it's easy to prove but I'm not sure if it helps
radical axis maybe? we find some other concyclic
slate folio
?
ye I pretty much used radical axis
since all lines in (DBC) and (DFE) have known properties
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dusty crescent
@wicked zenith
where you said v is p1 - p2, do you mean my original angle between two points, and the p1 and p2 are the two points?
vale dockBOT
<@211475256689426432>
where you said v is p1 - p2, do you mean my original angl...
@dusty crescent Has your question been resolved?
wicked zenith
<@211475256689426432>
where you said v is p1 - p2, do you mean my original angl...
yes, p1 and p2 are your original points, and v = p1 - p2 is the vector pointing from p2 to p1
dusty crescent
yes, p1 and p2 are your original points, and `v = p1 - p2` is the vector pointin...
ok so how does z become a whole number
if v is a vector
wicked zenith
z isn't a number, it's a vector
dusty crescent
ok now im very confused.
wicked zenith
x, y, z are vectors, and (x,y,z) is the matrix whose action is to rotate points in the direction of v
dusty crescent
ok so how do i get from those to a single 3 value vector of either degrees or radians
dusty crescent
i have two parts
i wanna draw a line thru them.
so i have a center point between the two parts, two points, and i want to find what angle i need to spawn the a new part at in order to draw that line
i want the angle i feed this function to be in the format of a 3 value vector in either degrees or radians, that represents X, Y, and Z rotations, or pitch, yaw, and roll
wicked zenith
you want pitch, yaw, and roll angles
pitch = arcsin(-x_z)
yaw = arctan(x_y / x_x
roll = arctan(y_z / z_z)
where x, y, and z are calculated as earlier
dusty crescent
funk yeah
wicked zenith
these will be in whatever units your arctrig functions spit out
What game engine or programming language are you using. There should be a LookingTo function which will do all this math for you.
dusty crescent
amber junco
i need some help with aleks math is anyone willing to hop in a call to help
hollow sentinel
aleks?
also we can just help you here
amber junco
its like an online site my school uses for our math classes
hollow sentinel
then you can ask whatever questions you have and we can help you
vale dockBOT
@amber junco Has your question been resolved?
flint phoenix
i need some help with aleks math is anyone willing to hop in a call to help
Sure, can you post it here?
smoky quarry
can i ask for help too even if its not a math question but math related
modern rune
can i ask for help too even if its not a math question but math related
what's the question
smoky quarry
i've loved math since my whole highschool and i wanna pursue actuarial but i feel like im too dumb for it, as if im not capable for it. But being an actuary has so many perks aside from salary, they're in demand and the unemployment rate is almost 1%
im still debating if i should take it
frozen ledge
i've loved math since my whole highschool and i wanna pursue actuarial but i fee...
is the unemployment rate that low?
flint phoenix
i've loved math since my whole highschool and i wanna pursue actuarial but i fee...
Dun plz
Stay away from math, as far as possible
smoky quarry
but i like math......
frozen ledge
i've loved math since my whole highschool and i wanna pursue actuarial but i fee...
if you enjoy it then go for it
smoky quarry
i mean i dont even like other subjects to the point when my last gf broke up with me i js ignored it and did math
then when i cant understand it i'd cry for math
frozen ledge
i mean i dont even like other subjects to the point when my last gf broke up wit...
typical
smoky quarry
HAHAHAHHAHA I AM ASIAN
frozen ledge
lol
frozen ledge
i've loved math since my whole highschool and i wanna pursue actuarial but i fee...
in all likelihood you are not too dumb btw
smoky quarry
ok. i will do actuary and be rich and have a pet orangutan at my home
frozen ledge
ok. i will do actuary and be rich and have a pet orangutan at my home
word
flint phoenix
ok. i will do actuary and be rich and have a pet orangutan at my home
It’s hard to foresee whether you’ll be rich tho
smoky quarry
oh dang
flint phoenix
Like there exists homeless cs graduates as well
frozen ledge
define rich
smoky quarry
Like there exists homeless cs graduates as well
SHHHH
frozen ledge
there are unemployed actuaries in this server
smoky quarry
define rich
can buy an orangutan
frozen ledge
👍
smoky quarry
there are unemployed actuaries in this server
is that serious...
flint phoenix
smoky quarry
there are unemployed actuaries in this server
is. that. fr.
frozen ledge
is. that. fr.
yep
flint phoenix
is. that. fr.
Aside from actuaries, do you consider any other options?
smoky quarry
stats is hella boring i cant lie
oh really
frozen ledge
at least for me
smoky quarry
what do you do
frozen ledge
i like pure math
steep mauve
What is the significance of Complex stationary points and what are their properties? I am especially concerened about how are they visually represented on complex function graphs such as vector fields and domain coloring graphs?
vale dockBOT
What is the significance of Complex stationary points and what are their propert...
@steep mauve Has your question been resolved?
fair wing
wut
fair wing
should i send here?
jovial edge
This channel is already closed, send your problem in other channel
vale dockBOT
Available help channel!
midnight pier
i saw this problem on internet and i was trying out i did abc + acb - 35 = 360 but i aint getting the right answer idk where i messed up
glass kelp
How did u get that eq in the first place
rigid trench
the measure of arc bc is not 35 right
glass kelp
Note that the sum of opposite angles of a cyclic quadrilateral is 180 deg
rigid trench
Note that the sum of opposite angles of a cyclic quadrilateral is 180 deg
💀 what? we need cyclic quad?
midnight pier
isnt abc + acb - 35 = 360 right?
glass kelp
How did u solve this then
glass kelp
isnt abc + acb - 35 = 360 right?
Why tho
rigid trench
i saw this problem on internet and i was trying out i did abc + acb - 35 = 360 b...
abc + acb - 35 = 360 -> abc + acb ||- 70|| = 360
small crystal
how do i sovle this? my brain legit stopped i cannot do a percentage problem 😭
tidal swift
focus on this
then replace all mentions of z with the function statement
then solve
tidal swift
please show
frozen ledge
how do i sovle this? my brain legit stopped i cannot do a percentage problem 😭
it’s not pz(w)
frozen ledge
how do i sovle this? my brain legit stopped i cannot do a percentage problem 😭
if z(w) decreases by p% for each increase by 1 in w this translates to z(w + 1) = z(w)(1 - p/100)
yes
small crystal
but why aint my desmos working
frozen ledge
well it would be 1 - p/100 = z(w + 1)/z(w) so p = 100(1 - z(w + 1)/z(w))
i’m not sure how you got what you have
small crystal
i’m not sure how you got what you have
wdym
frozen ledge
here
small crystal
i tried to do [z(w+1)+z(w)/z(w)] = p%
frozen ledge
if z(w) decreases by p% for each increase by 1 in w this translates to z(w + 1) ...
^
small crystal
well it would be 1 - p/100 = z(w + 1)/z(w) so p = 100(1 - z(w + 1)/z(w))
i did ts in desmos but its giving nothing 😭
frozen ledge
$p = 100\left(1 - \frac{(0.819)^{2(x + 1)}}{(0.819)^{2x}}\right)$
rocky lotusBOT
knief
small crystal
why would you need desmos for this?
graphing calculator might be faster
rocky lotusBOT
Result:
32.9239
put f(x)-f(x+1) on the numerator
small crystal
i did ts
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covert goblet
how might you find the shortest distance between the line through the points (1,3,1) and (1,5,-1) and the line through the points (0,2,1) and (1,2,-3)?
noble anvil
how might you find the shortest distance between the line through the points (1,...
Can you write the equation of both the lines for me
noble anvil
Correct
Now, since they aren't parallel, they can be either skew or intersecting
To find the shortest distance between skew lines, there's a straight forward formula
Let me pull it up hold on
Ok, so the shortest distance between skew lines is $|(\vec b - \vec a) \cdot \hat{(\vec p X \vec q)}|$
Where $\cdot$ is dot product and X is cross product
rocky lotusBOT
noble anvil
And the lines are of the form r1=a+lambda p and r2= b + mu q
So can you apply that here?
covert goblet
my syllabus doesnt teach cross product
covert goblet
ye i forgot how ur supposed to do it
noble anvil
Are you sure you weren't taught cross product?
This is how it would look in the magnitude form:
$|\vec a \cross \vec b| = |a||b|\sin(\theta)$
rocky lotusBOT
covert goblet
nah we arent allowed
noble anvil
idk then, I'll have to think a bit, if someone else has the answer feel free to chime in
jovial edge
Maybe we can prove it without cross product? Is it allowed?
covert goblet
yes
noble anvil
For reference, with cross product it would be
Ok, so the shortest distance between skew lines is $|(\vec b - \vec a) \cdot \hat{(\vec p X \vec q)}|$
rocky lotusBOT
Compile Error! Click the 
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noble anvil
Have you been taught determinants @covert goblet
covert goblet
im pretty sure ur supposed to use the length of the perpendicular segment that connects them
covert goblet
Have you been taught determinants <@446221660631072778>
nope
i think its supposed to be:
Create an arbitrary vector perpendicular to both lines
Find the vector that connects the 2 position vectors of both lines (you can choose)
Project one onto the other (the booklet should say which I forgot) and find the magnitude
i dont really get it tho
noble anvil
Ok let's try something
(this is how the formula i wrote is derived btw)
covert goblet
are we finding b-a
noble anvil
No, we're finding d
covert goblet
ok
noble anvil
Do you understand the diagram?
covert goblet
yes
noble anvil
Ok, so can you find sin theta
covert goblet
sintheta = d/(b-a)
covert goblet
liek thru trig manipulation or from the graph
covert goblet
costheta = (b-a)^2-d^2 / b-a
noble anvil
why upon d
covert goblet
mb
covert goblet
no t getting anything
noble anvil
show
noble anvil
y is ur pfp different on mobile lol
nitro
covert goblet
ok
noble anvil
we'll get a quartic convertible to quadratic to solve
covert goblet
grr why am i getting d^2 = -2
noble anvil
show
covert goblet
noble anvil
Ok let's try something
look at diagram
covert goblet
its position vectors minused right
so the magnitude wouldnt matter which one u did first
covert goblet
why tho if r1 = 1,3,1 + lambda(0,2,-2), r2 = 0,2,1 + mu(1,0,-4)
noble anvil
also this is wrong
noble anvil
u said alpha square is 2 but put 4 for alpha square in alpha^2 - d^2
noble anvil
can u show so we can be sure no mistake this time
covert goblet
noble anvil
ah ok, so theta cant be pi/2
because that would mean the triangle has two right angles
covert goblet
Ur right
covert goblet
Ok
noble anvil
this is rather tedious to have to do every time, so you should just remember the end result we got $\frac{d^2 + ((\alpha)^2-d^2)^2}{(\alpha)^2} = 1$
rocky lotusBOT
covert goblet
Thx
noble anvil
yw
covert goblet
gpt got 5sqrt2/6 u tho
covert goblet
Ok
covert goblet
damn
i think what im meant to do is
find l1 and l2
use direction vectors d1 and d2 and dot product with the line joining the position vectors
so p2p1 dot d1 =0, p2p1 dot d2 = 0,
solve simmultaneously for lambda and mu
plug lambda or mu into their respective equation
find the magnitude of the line
actually maybe dont neeed that arbitrary part
noble anvil
ok then whats p1p2
noble anvil
ok we'll do this
covert goblet
apparently this works even when they arent parallel on the 2d plane
noble anvil
ok firstly find the posn vector
vale dockBOT
@covert goblet Has your question been resolved?
rocky lotusBOT
noble anvil
Now we need to find it's projection on $\vec d$
rocky lotusBOT
noble anvil
Now $\vec d$ we can assume as (x,y,z)
rocky lotusBOT
noble anvil
Then take dot product with each line's parallel vec
So $\vec d \cdot \vec p = 0$ and $\vec d \cdot \vec q = 0$
rocky lotusBOT
covert goblet
yep
noble anvil
Do till here first
covert goblet
and set z = 1 for simplicity
covert goblet
but its a direction vector so ur taking some scalar multiple of it right
noble anvil
Do till here first
covert goblet
2y-2z=0, x-4z=0
the way i did it was setting z as 1 and solving for x and y, then just subbing z as 1
noble anvil
r1 = 1,3,1 + lambda(0,2,-2), r2 = 0,2,1 + mu(1,0,-4)
covert goblet
if u have d paralel = 4,1,1 and u project the line p1p2 which is 1,1,0 onto it you get 5/18 (4,1,1)
took the magnitude of that and got 5sqrt2/6
noble anvil
ye is ok to assume z
so u get d vec as 4,1,1
then taking projection would be $\frac{\vec{p1p2}\cdot\vec{d}}{|\vec {d}|}$
rocky lotusBOT
covert goblet
thats the magnitude of projection right
noble anvil
ye
vale dockBOT
noble anvil
.reopen
vale dockBOT
✅
covert goblet
bot is baiting lol
ok so i did that and got 5sqrt2 / 6
maybe check the cross product formula again
or maybe that one is for some other application
noble anvil
took the magnitude of that and got 5sqrt2/6
yeah this is right through the projection thing
i ll check with cross prod
covert goblet
okk
river shale
noble anvil
ye is correct
last time mi did bit of mistake while checking, forgor square root
noble anvil
ye is correct
@covert goblet free now 
noble anvil
:)
vale dockBOT
vale dockBOT
Available help channel!
golden cloak
Number 5: X is a set. Y is a subset of X. To show: S(Y) is a Subgroup of S(X). With S(X) is the group of all Isomorphisms from X to X. I don’t have a clue at all bc I started Group theory a few days ago
bronze moss
Number 5: X is a set. Y is a subset of X. To show: S(Y) is a Subgroup of S(X). W...
should specify: the group operation is composition.
try and verify all the axioms
its the set of all bijective maps, where the operation is composition
golden cloak
Ok, so as an example e should be in S(Y). But what even is e in this context. Would it be like a function that when it’s composed with another function gives back the other function?
bronze moss
Ok, so as an example e should be in S(Y). But what even is e in this context. Wo...
they should specify what S(Y) is as a subset of S(X): S(Y) is the set of isomorphisms from X to X that fixes X\Y (so effectively you are only changing Y).
alpine lance
hello
golden cloak
they should specify what S(Y) is as a subset of S(X): S(Y) is the set of isomorp...
I thought S(Y) are just the Isomorphisms from Y to Y. This also confused me bc what would the output be when I plug in an element from X that isn’t in Y
bronze moss
I thought S(Y) are just the Isomorphisms from Y to Y. This also confused me bc w...
S(Y) in this case is not even a subset of S(X) how could it be a subgroup😂
alpine lance
anyone can help ?
bronze moss
but in my definition, my S(Y) is isomorphic to your S(Y)
golden cloak
I really don’t understand it, idk
ebon glade
the question doesnt say that it is a subgroup
"S(Y) can be interpreted as a subgroup of S(X) in a canonical way"
would be a slightly better translation
bronze moss
oh that makes more sense
golden cloak
Oh sorry
golden cloak
Ok
bronze moss
so you can interpret an element of S(Y) as an element of S(X)
golden cloak
Makes sense
Maybe map every element from X that isn’t in Y to e? But that doesn’t feel so natural
bronze moss
Maybe map every element from X that isn’t in Y to e? But that doesn’t feel so na...
there is no e in X though, X is an arbitrary set
ebon glade
lets take some more explicit sets
bronze moss
and this makes the map fail to be an bijection of sets
ebon glade
lets say Y=N
and X=R
so you have bijections N->N
and you want to extend them to bijections R->R
golden cloak
and this makes the map fail to be an bijection of sets
Oh yes
ebon glade
in a "natural" way
ebon glade
so what could you do with elements R\N
golden cloak
I mean couldn’t I just do the same to these elements?
bronze moss
And note that you want to extend it in such a way where group operation is preserved: for f,g\in S(Y), (the extension of f)*(the extension of g)=(extension of fg)
ebon glade
the same meaning what?
golden cloak
Just applying the soma function bc they are the natural numbers but “can do more”, I can’t explain
Yeah…
ebon glade
you cant just plug 2.69164... into a function N->N
bronze moss
No
you want to map the identity in S(N) to the identity in S(R), maybe that'd be a clue
golden cloak
What is the identity? f(x)=x?
Ok
So I map f(x)=x for the natural numbers to g(x)=x for the reals?
But that’s just maping to the same function with another domain
golden cloak
Aren’t there functions that aren’t feasible for this?
bronze moss
But that’s just maping to the same function with another domain
maybe not think of them as some universal functions where domain can vary
golden cloak
Ok?
bronze moss
f is a map that maps 0->0, 1->1,...
how is the domain extended to R?
ie. what are we doing to R\N?
golden cloak
g maps x -> x for all elements in the reals
bronze moss
g maps x -> x for all elements in the reals
specifically, we are extending f, by making it identity on R\N
bronze moss
And we can do this for all functions over the natural numbers?
try a couple test it out
hollow merlin
ebon glade
its important that you dont think about functions as formulas
golden cloak
Okay
golden cloak
try a couple test it out
I won’t find any, but I’ll have to provide proof, don’t I?
bronze moss
but if in doubt u can always try a couple to convince yourself
bronze moss
So now I have to proof the axioms?
you have to prove that this correspondence really correspond an element of S(Y) to S(X)
you have to show that this correspondence is a group isomorphism between S(Y) and a subgroup of S(X).
bronze moss
isomorphism is basically what they mean as "interpret as a subgroup of S(X)": they mean a isomorphic copy of S(Y) lies in S(X) naturally
golden cloak
isomorphism is basically what they mean as "interpret as a subgroup of S(X)": th...
That’s making a lot more sense
bronze moss
I thought that I had to proof that S(Y) is a subgroup of S(X)…
denascite corrected that, I was also confused because in ur statement S(Y) is not a subset
golden cloak
denascite corrected that, I was also confused because in ur statement S(Y) is no...
Yeah, I’ll translate word for word from now on…
But how do I construct that Isomorphism?
bronze moss
But how do I construct that Isomorphism?
you have to extend an isomorphism of Y->Y to an isomorphism of X->X. This in turn gives you a map from S(Y)->S(X)
intuitively, we want to fix this Y->Y map inside of X->X. And make sense of where X\Y maps
golden cloak
Mhm, but even that. I can’t just change the range of the Isomorphism from Y to X
Okay
bronze moss
Mhm, but even that. I can’t just change the range of the Isomorphism from Y to X
you have to be more explicit
oh wait
yes u r right, but you are missing where X\Y maps
because we want an isomorphism of X
not just a map from Y to X
the map from Y->X is not a isomophism (it is not surjective)
golden cloak
But how should I do that? A set isn’t always ordered, is it?
golden cloak
the map from Y->X is not a isomophism (it is not surjective)
Yes of course
bronze moss
But how should I do that? A set isn’t always ordered, is it?
well you are only missing where X\Y maps. A set could be pretty random but there is one particular map that is pretty standard
golden cloak
well you are only missing where X\Y maps. A set could be pretty random but there...
Just map them to themselves?
bronze moss
Just map them to themselves?
try it, see if it makes an isomorphism from Y->Y to an isomorphism from X->X
lets say Y={1,2,3,4,5} and X=N for example
so it is easier to construct
golden cloak
Yes, I think that should work bc there can’t be any “overlap”
bronze moss
Yes, I think that should work bc there can’t be any “overlap”
cool, this extension also feels natural. Maybe you can show that this is an isomorphism between S(Y) and a subgroup of S(X)
golden cloak
So I have to show that I(ab) = I(a)*I(b)?
bronze moss
So I have to show that I(ab) = I(a)*I(b)?
and identity to identity
ok nvm
yeah
i forgot for group homo its implied
golden cloak
Ok
This is like the second time I wanna proof this… I don’t now. Should I just do an straight forward proof?
Ok, I have to think
So when I have I(ab). Then ab = a(b) which would just be a permutation of the elements of Y. And I(a)I(b) = I(a) of I(b). Now we have the function g(a) from X to X that maps every element that is also an element of Y to a(Y) and every element from X\Y to itself.
Now I(a) maps a to g(a). And thus the permutation of the elements from Y that “went through” I(a) of I(b) got maped to the same permutation as a(b) or just to themselves.
Idk, in my head it makes sense but I can’t express it… hope you at least get a little bit of what I’m trying to say…
bronze moss
So when I have I(ab). Then ab = a(b) which would just be a permutation of the el...
what does is g and g(a)?
a,b are functions from Y->Y right?
golden cloak
Yes
And g takes a and extends it by maping all of the elements that aren’t in Y to themselves
bronze moss
And g takes a and extends it by maping all of the elements that aren’t in Y to t...
but thats precisely what I does
golden cloak
Ok, so what was your question?
bronze moss
aha ok, lemme read it again
bronze moss
So when I have I(ab). Then ab = a(b) which would just be a permutation of the el...
u should really say I maps a to g(a). I(a) here is just an element of S(X).
you should prove it more rigorously by showing that, for each $x\in X$, $I(ab)(x)=(I(a)I(b))(x)$, because if their mapping of every $x\in X$ is the same, then they are the same function.
golden cloak
No, I meant I(a) to be the Isomorphism from S(Y) to S(X)
rocky lotusBOT
qwertytrewq
golden cloak
Ok
bronze moss
No, I meant I(a) to be the Isomorphism from S(Y) to S(X)
wait i don't get it, what is I(b) then
a, b are functions from S(Y). why is I(a) a function from S(Y) to S(X)?
golden cloak
Just the same Isomorphism but with another function as it’s input
bronze moss
Just the same Isomorphism but with another function as it’s input
so by definition I(a) is an element of S(X)?
since I is the mapping from S(Y) to S(X)?
golden cloak
I think so? I(a) should map a to g(a)
bronze moss
I think so? I(a) should map a to g(a)
anothing thing to point out: g is same as I. so you didn't need to introduce g
I is mapping every a to the extension of a to all of X
golden cloak
But g is the extension
bronze moss
But g is the extension
but you also defined I to be the extension
that is. I(a) is the extension of a
golden cloak
Then I wrote it down wrong. I should be the Isomorphism that maps S(Y) to S(X) and g is just a function that maps X to X
golden cloak
Then I wrote it down wrong. I should be the Isomorphism that maps S(Y) to S(X) a...
I is not an Isomorphism
Sorry
bronze moss
Then I wrote it down wrong. I should be the Isomorphism that maps S(Y) to S(X) a...
g is a function that maps X to X. then what is a?
a is not in X so whats g(a)?
golden cloak
a is a function that maps Y to Y
bronze moss
so what does g(a) mean when g is a function from X->X and a is a function from Y->Y
rocky lotusBOT
qwertytrewq
golden cloak
Yes
bronze moss
fair. but you could also just use $I(a)$ as is, because it is equal to $g_a$. But if you prefer $g_a$ sure.
rocky lotusBOT
qwertytrewq
golden cloak
Oh yeah, that makes sense. Idk, I wanted to write it out carefully but I suppose that didn’t work…
bronze moss
Oh yeah, that makes sense. Idk, I wanted to write it out carefully but I suppose...
Let me help you get started.
Let $I$ be the function from $S(Y)\to S(X)$, such that for all $f\in S(Y)$, and for all $x\in X$
$$(I(f))(x)=\begin{cases} f(x)& x\in Y\ x &\text{otherwise}\end{cases}$$
show that $I$ is well defined (it really sends an element in $S(Y)$ to one in $S(X)$). And show that $I(fg)=I(f)I(g)$ by showing that for all $x\in X$, $(I(fg))(x)=(I(f)I(g))(x)$.
golden cloak
Ok, I’ll try
rocky lotusBOT
qwertytrewq
golden cloak
So I would just do two cases: $x \in Y$ and $x \in X \setminus Y$
bronze moss
So I would just do two cases: $x \in Y$ and $x \in X \setminus Y$
btw \setminus is the command
golden cloak
Ty
rocky lotusBOT
Icephoenix
golden cloak
I think the case $x \in X \setminus Y$ is pretty straightforward
rocky lotusBOT
Icephoenix
golden cloak
And when x is in Y then g(x) is also an element of Y. So I(fg) = I(f)I(g)
bronze moss
I think the case $x \in X \setminus Y$ is pretty straightforward
still, write it down just to be sure
bronze moss
you don't need us to check it?
golden cloak
So bc x is not in Y I(g)(x) is also not in Y bc the output is just x. Then I(f)(x) is also x. When I have I(fg)(x) the output is also x bc x is still not in Y
Ok
Oh my good, this book is just full of these kind of problems…
.close
wet tundra
vale dockBOT
wet tundra
how do i do bi
I have this
do i use simultaneous equation
Gulp
I think I went wrong somewhere
glass kelp
pardon?
inner mirage
vale dockBOT
@wet tundra Has your question been resolved?
north scarab
what is the note saying?
vale dockBOT
what is the note saying?
tidal swift
what is cos 90?
stiff lily
what happens to the equation when A = 90°?
north scarab
a^2=b^2+c^2-2bccos(90degrees)
glass kelp
hi hell
tidal swift
and what is cos 90?
north scarab
and what is cos 90?
0 I assume
glass kelp
👍
stiff lily
it is 0, but you shouldn't assume
north scarab
well if i put in calc, it will give me 0
glass kelp
well if i put in calc, it will give me 0
so it is 0 😭
tidal swift
it is 0
north scarab
yes
tidal swift
don't have to assume
stiff lily
and with that, what will your equation simplify to?
north scarab
a^2=b^2+c^2
tidal swift
and what would that be?
north scarab
pytahogras thereoum!
north scarab
but why is it A specefically
stiff lily
the way they represented the cosine rule used A as the angle
tidal swift
you can change it around if you want
glass kelp
do u remember the shi with ${\theta}$ and ${\varphi}$
rocky lotusBOT
k
tidal swift
b^2 = c^2 + a^2 - 2ac cos B
it's the same thing
i remember telling you something about two sides, then the angle in between those two sides
tidal swift
it's more or less the same here
on the RHS, you have the sum of squares of two sides
north scarab
so the side we are trying to find has to be opposite the cos angle
tidal swift
the LHS side is opposite the cos angle, yes
north scarab
I had a question, do you review your old notes, for example algebra basics
because I heard if you don't use your old notes, you will eventually lose them
or it's not needed for people that are already good in maths
tidal swift
ever heard of the second exposure effect?
north scarab
I think so
tidal swift
further topics build on what you already know
so you will naturally review previous topics as you move onto further ones
not all previous topics see the same treatment depending on field though, which is why sometimes we forget about the simplest stuff
north scarab
well, do you review your algebra basics because I know u are advanced already
tidal swift
yes. i read Lial et al. a bit every day
arguably not as much as i used to since i'm now studying calc, linalg and abstract alg
north scarab
whats lial
tidal swift
but i still do
north scarab
arguably not as much as i used to since i'm now studying calc, linalg and abstra...
i see
tidal swift
yeah. why not
north scarab
true
tidal swift
sometimes if you get stuck on the harder topics, a refresher on the easy ones can give you fresh perspectives and avoid tunnel-visioning
turbid ridge
is this channel empty ?
turbid ridge
okay my bad
north scarab
we're literally talking here sir
lmao
tidal swift
because then you will just lose more time if u skip reviewing
not just that
you get stuck using the same method that has been proven NOT to work
north scarab
understandable
@tidal swift we are referring to A
in the
right angle position right
tidal swift
this A
those lines are probably just to help you understand the cosine law and where it comes from
north scarab
ah alr
light saddle
vale dockBOT
vale dockBOT
Available help channel!
balmy wagon
For the x^2 term, isn't the coefficient supposed to be 27/64? or 27/32 before distribution of the 1/2?
i got the same steps as the mark scheme, even the ((-1/2)(-3/2)((3/2)^2))/2 bit of the expansion but i keep getting 27/32 (pre distribution) and 27/64 as the final coefficient for the x^2 term.
Everything else looks the same as my working out, so idk if i made a mistake somewhere or if the mark scheme is wrong
balmy wagon
istg, if i just made another computation error and used up a help channel for it 💔
stone aspen
istg, if i just made another computation error and used up a help channel for it...
it is 27/64 unless im stupid
wraith heart
For the x^2 term, isn't the coefficient supposed to be 27/64? or 27/32 before di...
Is this mark scheme or your work
balmy wagon
Is this mark scheme or your work
mark scheme
stone aspen
(not mutually exclusive events)
stone aspen
Is this mark scheme or your work
im guessing it's the ms
balmy wagon
it is 27/64 unless im stupid
i've done the same calculation like 5 different times in different formats and got the SAME ANSWER, istg if i'm STILL missing smth
,w ((-1/2)(-3/2)((3/2)^2))/2
rocky lotusBOT
balmy wagon
ok, the markscheme is just wrong then, 27/32 is correct (pre distribution)
thanks sam and riemann
.close
vale dockBOT
vale dockBOT
Available help channel!
analog schooner
I was thinking about this proof of the product rule (https://youtu.be/YG15m2VwSjA) which uses area/ volume / the 4-D counterpart of volume / etc. for 2, 3, 4,... functions multiplied together
Each time there is some area / volume / whatever that is ignored because it is df * dg * dh *..., so it is negligible.
However, if we use this for a series of infinitely many functions multiplied together, the term that we would have to ignore would be a product of infinitely many infinitesimals. So it would be of the 0 * infinity indeterminate form.
My question is- how can we neglect this "volume" now, when it may not be an infinitesimal
analog schooner
Yes
rocky lotusBOT
Dhruv
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mrx
analog schooner
\hspace