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#help-42
calm coralBOT
Available help channel!
wide depot
I feel like I'm losing my fucking mind
calm coralBOT
I feel like I'm losing my fucking mind
wide depot
how do I evaluate this indefinite integral
granite dust
I feel like I'm losing my fucking mind
expand 
granite dust
u sub easier
wide depot
I've been trying to usub it and it's been giving me a headache
prime locust
pretty sure you gotta do parts
granite dust
no
it's called forced u sub
u = 5-x so x = 5-u
sub x for 5-u, sub 5-x for u
then u should get integral of -(5-u)u^9 du
then expand and evaluate
its a cool technique for integration bee
prime locust
don't mind me ig
granite dust
np
prime locust
i'm still in hs lol
dude high school calc is so light compared to the stuff I see in here
prime locust
yeah fr
granite dust
dude high school calc is so light compared to the stuff I see in here
hs math is nothing compared to undergrad math 😭
prime locust
I don't want to jump into calc 3 looking like a bozo
prime locust
calc bc does NOT cover 90% of the questions that this server asks
bc is so light compared to the stuff here it's genuinely baffling
calm coralBOT
viral isle
How to do 13 and 14 without a diagram only with calculus and derivative
calm coralBOT
Available help channel!
brittle sail
calm coralBOT
brittle sail
How to check it?
potent lotusBOT
@mellow harbor
brittle sail
n!/(3n+4)
mellow harbor
<@946379454056386560>
.
hexed grove
we want $$\frac{a_{n+1}}{a_n}$$
potent lotusBOT
Cycadellic
hexed grove
,rotate
potent lotusBOT
hexed grove
hold on
hexed grove
we want $$\frac{(n+1)!}{\prod\limits_{i=1}^{n+1} 3i+4}\cdot\frac{\prod\limits_{i=1}^n 3i+4}{n!}$$
potent lotusBOT
Cycadellic
blazing coyote
Trying to do this
calm coralBOT
Trying to do this
potent lotusBOT
What a wonderful world !
blazing coyote
so we have $(2 \cos (t) \sin(u) , 2\sin(t) \sin(u), 2\cos(u)); 0≤t≤2π; 0≤u≤π/4$
potent lotusBOT
What a wonderful world !
topaz raft
Wai
calm coralBOT
calm coralBOT
Available help channel!
frosty chasm
calm coralBOT
velvet osprey
frosty chasm
Ughh i tried doing the dot product of P1 and p2
velvet osprey
like... $\overrightarrow{OP_1} \cdot \overrightarrow{OP_2}$?
potent lotusBOT
Ann
velvet osprey
i don't see how that could help us
frosty chasm
Ya i was just trying random things i knew
velvet osprey
you understand that the problem asks you to find the distance from P to the line P1P2, yes?
frosty chasm
Ya
velvet osprey
do you know in general how to find the distance from a point to a line in 3D?
frosty chasm
Yes
velvet osprey
ok, how do you do it?
frosty chasm
Sqrt x^2+y^2+z^2
velvet osprey
no, point-to-line distance.
it is going to be a bit more complicated than just applying 3D pythagoras like you would for the distance between two points!
also missing parentheses
frosty chasm
No i haven't heard sth like that
velvet osprey
frosty chasm
Point to line distance
frosty chasm
https://youtu.be/5BDrdI3rdQY?feature=shared
Thanks
Do i use distance formula first to make a line first?
frosty chasm
https://youtu.be/5BDrdI3rdQY?feature=shared
There's a equation of the line in this vid
velvet osprey
ah i just realized...
this video's title is misleading
he's actually showing how to find point to plane distance
whoops
for a line it is going to be more complicated than the vid
so im gonna say ignore that and forgive me for giving you a misleading video
but you will still need the equation of line P1P2 for your purposes,
or at least the direction vector
frosty chasm
so im gonna say ignore that and forgive me for giving you a misleading video
It's ok
velvet osprey
the geometric idea is to drop a perpendicular from P onto P1P2
cedar rose
I think you can use scalar vector product to find projection of PP_1 on line P_1P_2 then you can just subtract this projection from PP_1
frosty chasm
I think you can use scalar vector product to find projection of PP_1 on line P_1...
Line p1 p2 from distance formula right??
cedar rose
P_1 P_2 from your problem
frosty chasm
But they are points
cedar rose
and? P_1P_2 is vector
frosty chasm
Yes
cedar rose
calm coralBOT
@frosty chasm Has your question been resolved?
calm coralBOT
@frosty chasm Has your question been resolved?
calm coralBOT
calm coralBOT
Available help channel!
craggy nebula
is there a reliable way to tell if an equation is a function without plotting it
say for example x^4 = y^2
craggy nebula
one way to go about doing it would be to input different values for x and see their corresponding y values
and imagine what the graph could be like
but is there a better and more reliable way
full quarry
a function is basically a relation
if u can relate x and y by any means u have a function ig
craggy nebula
if u can relate x and y by any means u have a function ig
yeah but x could have more than one corresponding value for y right, in that case it wouldn't be a function
cedar rose
is there a reliable way to tell if an equation is a function without plotting it...
if its function then there is not more than one y for each x, so you need to show that when you solve it as eq with unknown y you dont get 2 roots
craggy nebula
if its function then there is not more than one y for each x, so you need to sho...
thank you
@full quarry and thank you to you too
full quarry
if its function then there is not more than one y for each x, so you need to sho...
wait but isnt that just a type of function?
like isnt that one-one function?
cedar rose
wait but isnt that just a type of function?
he means that y(x) is function
cedar rose
like isnt that one-one function?
no we dont check that for each x there is unique y
craggy nebula
.close
violet grail
Someone please help me idk where to start this is practice for the AP exam
short salmon
wow, I can't believe precalculus is an AP now.
well, for the first question, perhaps think about what a composition of functions means
in fact, they sort of tell you with h(x)=g(f(x))
so plug in x=1 to get h(1)
short salmon
Is this calculator active sorry if I’m sounding stupid
probably not. it seems this is pretty doable without it
calm coralBOT
@violet grail Has your question been resolved?
violet grail
in fact, they sort of tell you with h(x)=g(f(x))
So wherever there is an X I put a 1
calm coralBOT
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Available help channel!
slender socket
why do they start with n = 2 not n=1 ?
calm coralBOT
why do they start with n = 2 not n=1 ?
velvet osprey
what is 1 - 1/1
slender socket
0
because I don't know if they want it because I can find an odd even rule that works if that isn't the first input for n
velvet osprey
idk what an "odd even rule" is
the reason the product starts from 2 and not 1 is to make it so the problem is not just trivial like "it's always 0 
"
slender socket
ok so what I noticed is for odd inputs the numerator was always increasing by 1 and the denomenator was just n and for evens I realizled it was n+1 for the numerator and the denomerator is 2n
slender socket
the reason the product starts from 2 and not 1 is to make it so the problem is n...
why would it always be 0 if it starts at 1?
would that just be the first input
velvet osprey
why would it always be 0 if it starts at 1?
i think we are either talking about two different things or you aren't understanding me
you are asking why the product is $\prod_{k=2}^n \paren{1-\frac{1}{k^2}}$ and not $\prod_{k=1}^n \paren{1-\frac{1}{k^2}}$, correct?
potent lotusBOT
Ann
calm coralBOT
@slender socket Has your question been resolved?
idle patio
$$\frac{k^2 - 1}{k^2}$$
potent lotusBOT
Klein Bottle
slender socket
For why we don't use 1 as an input?
slender socket
What do you think you can do from thid?
.
dusky flax
because I don't know if they want it because I can find an odd even rule that wo...
what exactly is an odd even rule
slender socket
what exactly is an odd even rule
For odds inputs the product equals (n-(n-1)/2)/n and for evens it is n+1/2n
velvet osprey
Yes that's the first part of my question
ok then the answer is: if the product were to start at k=1, then the first factor would end up as 0
and so ruin the entire product
velvet osprey
yes
slender socket
Ok thx
Also for the rule is there any way to combine it into one rule or must be 2 rules
serene talon
Also for the rule is there any way to combine it into one rule or must be 2 rule...
what rules are you trying to use
Oh I think I understand the question
serene talon
For odds inputs the product equals (n-(n-1)/2)/n and for evens it is n+1/2n
this seems wrong
even if you mean (n+1)/(2n)
steel cobalt
Yh
Does anyone here know how to do long division *
and yes
I can
I was literally just tryna help..? 😔
whole hinge
(modping addressed, carry on)
potent smelt
I was literally just tryna help..? 😔
You're good, I just misunderstood the problem.
my bad
slender socket
even if you mean (n+1)/(2n)
? Are you sure I was checking for values and it seems to be working
serene talon
? Are you sure I was checking for values and it seems to be working
for both even and odd, right?
slender socket
Yes
slender socket
Yes
serene talon
so, your queston was, can you combine both rules into one rule? and the answer is, there's only one rule
slender socket
So your saying that the odd even thing is wrong?
Or that it can be combined into one thing?
serene talon
You said that (n+1)/(2n) works for both even and odd n
I havent done the problem completely, but I thinkyou're right
so, you set up an induction proof.
Now, you asked if you can use n=1 as the starting point
You actually can, if you write it in product notation like this:
$$\prod_{k=2}^n \left({1-\frac{1}{k^2}}\right)=\frac{n+1}{2n}$$
slender socket
You said that (n+1)/(2n) works for both even and odd n
No one that only works for even
potent lotusBOT
gfauxpas
serene talon
No one that only works for even
let's try it
slender socket
Ye thats why I use the odd rule which is n-n-1/2 /n
slender socket
3-(3-1)/2 /3
slender socket
but this is of the form (n+1)/(2n)
How?
serene talon
(3+1)/(2*3) = 4/6=2/3
slender socket
Oh snap it does work
serene talon
now, about starting at 1
as people noted, the expression giving in the problem is zero if k starts at 1
This is true, and this ruins the formula. BUT
You can still start your induction at one if you'd like, for this reason
have you ever seen a sum like this?
$$\sum_{i=4}^1 6i$$
potent lotusBOT
gfauxpas
serene talon
where the bottom number is larger than the top?
actually i should ask first if you've seen the Σ notation for sums at all
slender socket
**gfauxpas**
No
slender socket
actually i should ask first if you've seen the Σ notation for sums at all
Yes
serene talon
So, when the bottom number is larger than the top, we say the sum is 0.
This is called the "empty sum" or "vacuous sum", the sum of no numbers.
Why 0? It's a convention that is consistent with everything you prove about sums
the reason is because "it just works" 
, not because it's a law of the universe
likewise, an expression like $\prod_{j=2}^1 j^2$ is defined to be 1. Why? because it makes formulas work. not because it's a law of the universe
potent lotusBOT
gfauxpas
serene talon
with that in mind, something like $\prod_{k=2}^n \left({1-\frac{1}{k^2}}\right)$ gives you, if you plug in n=0, 1 on the left side, and 1 on the right side
potent lotusBOT
gfauxpas
slender socket
What do you mean it makes the formula work?
serene talon
What do you mean it makes the formula work?
I mean like exactly this case. You have a rule about products, and you want to check if the product works for n=1 as well, if you define the product of no numbers at 1, the proof works
you familiar with the formula:
$1 + 2 + 3 + \cdots + n = \dfrac{n(n+1)}{2}$
?
potent lotusBOT
gfauxpas
serene talon
Probably
slender socket
Yes
serene talon
if you write:
$\sum_{k=1}^n k = \dfrac{n(n+1)}{2}$
then plugging in n=0 to both sides gives
$\sum_{k=1}^0 k = \frac{0(0+1)}{2} = 0$
potent lotusBOT
gfauxpas
serene talon
that's what I mean by "it makes formulas work"
slender socket
What about our formula ?
serene talon
empty product = 1
empty sum = 0
so if you put n=1 here you get 1 on the left and (1+1)/(2) = 2/2 = 1 on the right
But, this is only if you write it this way with Σ or Π
whole hinge
the reason is because "it just works" <:todd:436979598878834709> , not because i...
(I think "it just works" is a little misleading---it works because 0 is the identity under summation, which is why it makes sense to define the empty sum to be 0)
serene talon
And it's a good idea to write something like "Here we use that the vacuous product is defined to be 1"
serene talon
(I think "it just works" is a little misleading---it works because 0 is the iden...
Yeah I'm not sure how to say that without the student having background knowledge in abstract algebra, please explain to them if you can
slender socket
But, *this is only if you write it this way with Σ or Π*
Yes but for n=1 shouldn't it be 0
whole hinge
Just like how adding 0 to any number doesn't do anything to the number, adding an "empty sum" doesn't do anything, so the empty sum should be 0. That's one way you could think about it.
slender socket
What about for this formula .
serene talon
What about for this formula .
if "the product of no numbers" is 1
then "the sum from k=2 to k=n=1 of whatever" is one.
And the right hand for n=1 gives (1+1)/2 = 2/2 = 1
So you can prove this by induction by the base case being n=2 , or by n=1, but this would not be clear to the reader if you were using the notation of your homework assignment
slender socket
then "the sum from k=2 to k=n=1 of whatever" is one.
And the right hand for n=1 ...
Yes so isn't that the wrong answer
serene talon
1=1 is a true statement
it's just, in our example, not particularly informative or interesting
slender socket
Isn't the left side 0 or am I not understanding something
serene talon
Isn't the left side 0 or am I not understanding something
the SUM of no numbers is zero. the PRODUCT is zero
another way to think about similar to what @whole hinge said
for a sum, you want to be able to do things like
slender socket
So if we have our k start at 2 and we input 1 it will be 0
serene talon
So if we have our k start at 2 and we input 1 it will be 0
1, not zero
slender socket
the SUM of no numbers is zero. the PRODUCT is zero
?
serene talon
for a sum, you want to be able to do things like
$\sum_{i=1}^k 2i^2 + 0 = \sum_{i=0}^k 2i^2$
potent lotusBOT
gfauxpas
serene talon
you also want to do things like
$1 \cdot \prod_{i=1}^j 2i^2 = \prod_{i=0}^1 2i^2 \cdot \prod_{i=1}^j 2i^2$
potent lotusBOT
gfauxpas
serene talon
$=\prod_{i=0}^j 2i^2$
potent lotusBOT
gfauxpas
serene talon
$\sum_{i=1}^k 2i^2 + 0 = \sum_{i=0}^k 2i^2$
for this I meant
$\sum_{i=1}^k 2i^2 + 0 = \sum_{i=1}^k 2i^2 + \sum_{i=0}^1 2i^2$
potent lotusBOT
gfauxpas
serene talon
$=\sum_{i=0}^k 2i^2$
potent lotusBOT
gfauxpas
slender socket
One sec let me reread this I'm not following
serene talon
the point is we want "add no numbers to the number" to be the same as "do nothing to the number"
and "multiply no numbers by the number" to be the same as "do nothing to the number"
whole hinge
(Btw, I'm not sure if I understand how the topic of empty sums/products came up? Seems like they were just asking how to do the induction)
slender socket
the point is we want "add no numbers to the number" to be the same as "do nothin...
?
serene talon
(Btw, I'm not sure if I understand how the topic of empty sums/products came up?...
bigben asked if thebase case has to be n=2
serene talon
?
and "add no numbers" we want to be adding 0, and "multiply by no numbers" means "Multiply by 1", for it to leave the number the same
If this is too confusing there's nothing wrong with starting at n=2
sorry if I didnt explain it well
slender socket
And just clarifying this is when our bounds are not set right
serene talon
when the bounds aren't satisfied by anything, such as 2 < i < 2
3 <= i <= 1 better example
slender socket
Ok I'll just reread the conversation again
serene talon
at this point you can do the problem fully and just start at n=2 for the base case, this is for your own personal knowledge if you want to dig at "wait, I thought the base case has to be 1?" question
calm coralBOT
calm coralBOT
Available help channel!
edgy bluff
I need help with a arthemtic sequence question
calm coralBOT
I need help with a arthemtic sequence question
balmy bane
!da2a
calm coralBOT
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/
velvet osprey
,rotate 180
potent lotusBOT
edgy bluff
I can't figure out what to do with the median part of the question
balmy bane
(you had one too many rs in the command)
edgy bluff
(you had one too many `r`s in the command)
Oh
edgy bluff
Idk what the translation is exactly
balmy bane
!original
calm coralBOT
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
edgy bluff
Ye
velvet osprey
oh god ok i don't speak arabic at all
edgy bluff
See why I translated it
balmy bane
Lemme check...
ancient grotto
Call for reinforcements
balmy bane
Just underline the word you'd translated to "medians" rl quick?
plucky walrus
It could be average
edgy bluff
Just underline the word you'd translated to "medians" rl quick?
وسيطيها
velvet osprey
also i take it that's 29?
balmy bane
You've somehow pluralised it
velvet osprey
and not 2a
edgy bluff
and not `2a`
No
velvet osprey
but also what does "median" mean here lol
edgy bluff
You've somehow pluralised it
It's plural
edgy bluff
but also what does "median" mean here lol
No clue
sonic snow
but also what does "median" mean here lol
the middle most value
edgy bluff
That's the whole problem
sonic snow
for even its n/2 th and n/2+1 th term and for odd its (n+1)/2th term
slim meteor
wtf is this
plucky walrus
lol
balmy bane
It's Arabic ya [redacted]
balmy bane
BRO
balmy bane
Come on...
potent lotusBOT
edgy bluff
,rcw
potent lotusBOT
edgy bluff
,rcw
potent lotusBOT
edgy bluff
I got the first 2 equations
leaden thunder
you can just use the emojis after first rotate
slim meteor
My neck is about to snap.
edgy bluff
(use the emojis there)
Idk how
balmy bane
Just press them
leaden thunder
click
edgy bluff
So that's what those r forr
edgy bluff
so... where does 29 come into it though
It's supposed to be equation 3
Cause there are 3 variables
..
A/D/N
velvet osprey
ok well it looks like this is a language issue and we need an arabic speaker
because all i can say is that i don't understand what "median"/"medians" is supposed to signify in this context at all
edgy bluff
وسيط means the middle most number
balmy bane
Oh
balmy bane
IT's مجموع وسيطها
tranquil wasp
isnt there only one median
sonic snow
But there is no rule for the median in a sequence
if n is even, the n/2 th term and n/2 + 1 th term are median terms i believe
edgy bluff
if n is even, the n/2 th term and n/2 + 1 th term are median terms i believe
The question does tell us if it's even or odd
sonic snow
just assume its even
edgy bluff
I tried with median = n/2
plucky walrus
Q: An arithmetic sequence with a sum of its terms equal to 116 and the average equal to 29. Find the first term if the second term is 7.
I think we can let the first term be a, the second a + d, then a + 2d and so on.
balmy bane
"The sum of its medians"
Yeah, this isn't mathematical - that does say "sum of its median" - it should have probably been "ووسيطها 29" – meaning "and its middle term is 29."
edgy bluff
Yeah, this isn't mathematical - that _does_ say "sum of its median" - it should ...
If that's the case this is a easy question
I think
Let's assume your right
How do I go about solving it
potent lotusBOT
Removed the studying! role from you.
edgy bluff
The question has an error
sonic snow
'll get one only
well if you do the way i have interpretted the Qs:
||you'll get 2A+(N-1)D=29
=>N/2=116/29=>N=8||
edgy bluff
well if you do the way i have interpretted the Qs:
||you'll get 2A+(N-1)D=29
=>N...
Why
I saw someone do it like that
But aren't u suppose to multiply by n/2
sonic snow
But aren't u suppose to multiply by n/2
A+(N/2)D+A+(N/2-1)D=29
simplify and you'll get what i got
edgy bluff
U mean A×n/2?
sonic snow
U mean A×n/2?
uh no... , which part are you asking about?
edgy bluff
First part
sonic snow
ohh the sum
slim meteor
pika avatar
sonic snow
let me start from what i did slowly
edgy bluff
Okay
sonic snow
Is this clear?
edgy bluff
No
sonic snow
"Sum of median terms is 29"
edgy bluff
Oh u mean like that
calm coralBOT
calm coralBOT
Available help channel!
neon quest
hello
this is a very complicated question, so i'm not looking for anyone to solve it (but if you come up with a solution that's great), just to get some pointers and ideas on what to do next
it's one that i've invented myself, and i've been working on it these past few weeks, so it's not urgent at all
the question: find an explicit formula for the sum of x^a/b^x as x goes from 1 to infinity (condition because otherwise it diverges, although i haven't looked at b<0 at all yet)
i'll send my progress in a seperate message
neon quest
to explain my idea as to how to solve this, i'll show the method on a concrete example:
basically, i take away 1 from each of the fractions, then i take away 1 from each fraction except for 1/2 since that one is already 0, then 1 from each fraction except 1/2 and 1/4, etc.
this splits the sum into a bunch of geometric series, which i can compute, which gives the answer of 2
leaden thunder
should probably find what values of a that the series will even converge for
neon quest
should probably find what values of a that the series will even converge for
any, exponentials grow faster than polynomials
at least that was my line of thought
leaden thunder
oh if you've proven that then you can just explore in desmos
ratio test should be sufficient to prove it converges for any |b| > 1 and any a > 0
next step would probably to find a closed form for any positive integer a
you have it for a=1
neon quest
doing the same for x^2/2^x, i used the fact that every perfect square n can be written as the sum of the first n odd numbers
so 1=1, 4=1+3, 9=1+3+5, and so on
i extracted the 2 from each of the geometric series
then i added and subtracted 1 to each of the fractions
amazingly, the previous result shows up here, so i can simplify that one to 2 and get the result: 6
neon quest
you have it for a=1
i have it for a= 1,2,3 right now, and in theory i can get and integer a
let me write it out though
then i showed this
it's really obvious, because it's just x^n = 1 + 2^n - 1 + 3^n - 2^n + 4^n - 3^n and so on
and so all of the others except for x^n cancel out
applying that to the general case, i got this formula
which is really good, because the polynomial (x+1)^a - x^a is of the (a-1)th order
leaden thunder
missing a term on the left
leaden thunder
then i showed this
this is a telescoping sum
so there should be a first AND last term
neon quest
which is really good, because the polynomial (x+1)^a - x^a is of the (a-1)th ord...
i can plug in a=3, then in the sum on the right get some polynomial (in this case 3x^2+3x+1)
i can then split that large sum into a bunch of smaller sums, conveniently for which i already have the formulas
these are the results i have for now
neon quest
applying that to the general case, i got this formula
and, i want to somehow collapse this sum into a formula of its own
idk how though
and i haven't been able to write it with a := a+1 to try and get a recursive formula
leaden thunder
see if you can try to prove this
leaden thunder
i can plug in a=3, then in the sum on the right get some polynomial (in this cas...
the right side is a generalization of your polynomial here
but with z = 1/b
neon quest
see if you can try to prove this
woah where is this from?
it kind of resembles the nth sum of a geometric sequence
leaden thunder
the last two steps require induction on m
but how the formula was discovered i dunno
what you're solving for is effectively the definition of the special function called polylogarithm
leaden thunder
and the integer values of a have many interpretations
https://oeis.org/A000629
Number of necklaces of partitions of n+1 labeled beads
neon quest
2, 6, 26, 150
leaden thunder
don't know much combinatorics but it's interesting polylogarithm isn't on that webpage
neon quest
i recognize those numbers
it's when b=2 and a ranges from 1-4
what's a bead? (i googled it)
tysm for all of this, i would have liked to get a solution by myself but i don't think i have enough knowledge to do it
i'm a senior in high school and we have barely done any calculus, let alone something of this magnitude
my question is resolved
calm coralBOT
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lost hatch
A line segment parallel to the y-axis is placed in the limited area between the curves as shown in the figure below. Determine the maximum possible length of this line segment.
lost hatch
I can get the points where the functions meet, but no idea what after that
lost hatch
Mind explaining why the length is the difference?
leaden thunder
the distance between two points a and b is |a - b|
distance between two points is the length of the line segment
lost hatch
Mhh, right. Okay so derive y_1-y_2, find zeros and maximum?
hollow relic
calm coralBOT
hollow relic
ok so C obviously implies D
but i'm not sure if D implies C
i think it does because if we have M > 5, and f(x) > M
then f(x) > 5 > 0
actually no
f(x) = (1/|x|) + 5
f(x) holds for D but not C right
hollow relic
ye so for the f i defined, it holds for D but not C right
hollow relic
arbitrary
dapper sable
It doesn't say it's arbitrary.
hollow relic
it's not in the pic but it just says let a be an element of R
it's alright i got it now
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green cosmos
calm coralBOT
eternal shard
consider the level curves
green cosmos
u can exclude stupid part btw
eternal shard
how do they look like in the yz plane
green cosmos
they look like
one of those
cubic
functions
uhhh
the one after parabola
i forgot
what are they called
like the stupid cubic ones
eternal shard
yes
how would you describe a cubic function in the yz plane through the origin for example
a simple one
eternal shard
almost
y=z^3 would be like flipped
y=z^3 would be like x=y^3 but you want the y=x^3
version
green cosmos
,w graph y =z^3
potent lotusBOT
eternal shard
yes but notice the axis
eternal shard
welco
green cosmos
do u know
the difference
between adjective and noun
a noun is what u describe soemthing right
so i can say
the @calm coral is stupid
and like a adjective is a place or thing or something
a verb is
a verb is
uhhh
eternal shard
professional troller
eternal shard
verb is something something like closing this channel
calm coralBOT
eternal shard
describes what you do
calm coralBOT
✅
green cosmos
i js wanna make sure im doing this right
to sketch the graph
problem 13
so i set
each variable to 0
and then like graph the points
on their respectable axis
is that how the traces work
green cosmos
for traces you would set only one variable to 0
so
if i set x = 0
i would have 0 = y^2 + 4z^2
how would i get the points for the y and z
?
eternal shard
well are there even that many points that satisfy that
eternal shard
ya
green cosmos
ya
and to continue to graph i js
increase value of x
?
then like i get the larger points
eternal shard
for example, but me personally i would consider the curves at y=0 and z=0 and then draw a conclusion (hehe pun)
green cosmos
for example, but me personally i would consider the curves at y=0 and z=0 and th...
ok thx
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tawdry flame
Why do we need to check that ‘the operations are well defined’? Why does proving (c) make them well defined?
swift laurel
"well defined" means that if we put in the same input then we should get the same output no matter how we represent the inputs
so ordinary number addition is well defined because we have that if a = b and c = d, then a + c = b + d
short salmon
It's just another way of wording the proof of uniqueness of something
(if you've ever seen that word used before. e.g. existence and uniqueness theorems for solutions of a diffeq)
tawdry flame
When do i need to show that an operation is well defined?
swift laurel
an example of a function that isn't well defined is, a binary function # that takes two rational numbers and just adds their numerators, e.g. 2/3 # 6/7 = 8. this is not well defined because we can take two equivalent rational numbers and get different results, e.g. 4/6 # 6/7 = 10, even though 2/3 = 4/6
short salmon
uhhhhhh... wdym by "need to know"?
swift laurel
all binary operators (e.g. anything that we can call addition) should be well-defined. sometimes it's more obvious than other times
tawdry flame
I see
short salmon
a slightly more abstract example would be the pre-image of a non-injective function. For instance, trying to invert $f(x)=x^2$
potent lotusBOT
00100000
short salmon
where $x\in\R$
potent lotusBOT
00100000
short salmon
exactly
you can't invert the function, because the pre-image wouldn't be well-defined
a really common "structure" of these proofs is, assume your operation yields products x, y. show that it must be the case that x=y
tawdry flame
So ‘well defined’ is a property of functions in general?
short salmon
uhhhhh
well now I'm afraid of saying something inaccurate
so I will avoid answering that one
tawdry flame
No problem
I mean, the definition of a function f is a relation in which if x, y in dom f satisfy f(x)=f(y) then x=y
Is that what makes it well defined
short salmon
oh, I see what you're saying here
yeah, functions have well-defined outputs by definition
tawdry flame
Well-defined outputs?
short salmon
basically what you said
tawdry flame
Idk this feels like an abuse of notation
short salmon
¯_(ツ)_/¯
tawdry flame
‘Well defined’ seems to mean different things in different contexts
ashen sphinx
I mean, the definition of a function f is a relation in which if x, y in dom f s...
That's the contrapositive form of injectivity
Rather we need that x=y implies f(x)=f(y)
tawdry flame
I might have been confused
tawdry flame
Rather we need that x=y implies f(x)=f(y)
This looks right
ashen sphinx
If a function f satisfies that f(x)=f(y) implies x=y then f is injective by contraposition principle
ashen sphinx
So ‘well defined’ is a property of functions in general?
Yea sure
why not
in that case we would have + operator as a function
(not sure if it would act on sets or elements of sets , but it could act on sets)
tawdry flame
I think its more clear now, something is well defined iff x=y -> f(x)=f(y) for 1-ary functions
So in the example i provided
I need that s1=s1’, s2=s2’ -> f(s1,s2)=f(s1’,s2’)
Because (s1,s2)=(s1’,s2’) iff s1=s1’ and s2=s2’
ashen sphinx
Let $W_1 = (v_1 + W), W_2 = (v_2 + W)$ similarily write for $W_{1}^{'}, W_{2}^{'}$, then if $W_1 = W_{1}^{'}, W_2 = W_{2}^{'}$ , then $+(W_1,W_2) = +(W_{1}^{'},W_{2}^{'})$ is the statement of "well-defined" you would've proven if you proven c)
potent lotusBOT
Prelude to archbishop
short salmon
relevant mse: https://math.stackexchange.com/a/606928/1460038
tawdry flame
Yes tysm makes a lot of sense
tawdry flame
relevant mse: https://math.stackexchange.com/a/606928/1460038
Will have a look, ty
ashen sphinx
**Prelude to archbishop**
What is written before that is likely some equivalence relation on pairs of objects (W1,W2) , since the pair (W1,W2) = (W'1,W'2) by that equivalence relation and + takes the pair (W1,W2) to (W'1,W'2) , then the + operator is well-defined , because it maps equivalent pairs to an equivalent output. [ Note: this doesn't mean that + is injective , but it does hint at the possibility of it being thought of as a function in the standard sense , because it preserves that x=y implies f(x) = f(y)]
tawdry flame
What kind of equivalence relation?
ashen sphinx
(W1,W2) = (W'1,W'2) if W1 = W'1 and W2 = W'2
tawdry flame
Does that statement need any justification/
ashen sphinx
Why do we need to check that ‘the operations are well defined’? Why does proving...
literally the thing they state in c) after ",show that if (here they state"
ashen sphinx
Does that statement need any justification/
no
You assume that the pairs (W1,W2) = (W'1,W'2) are equivalent , you don't prove anything about that
You use that assumption to prove that + operator is well-defined in the sense that it takes the pair (W1,W2) to output x and the pair (W'1,W'2) that is equivalent to the pair (W1,W2) to the same output x
To prove that you of course want to understand first what the objects W1 = v1 + W are , and what + does to them, i.e how + is defined/thought of here
ashen sphinx
im a bit sleepy , but i suppose v + W is supposed to be like a set of elements {v + w, where w in W}
tawdry flame
Yes
ashen sphinx
addition is then gonna produce a set of elements {v1 + v2 + w: w in W}
tawdry flame
Yes
I think the result follows from b
I havent attempted it yet
Because v1+W=v’1+W implies v1-v1’ in W and similarly v2-v2’ in W
Since W is a vector space then
(v1+v2)-(v’1+v’2) in W
That shows coset addition if thats the right term to use is well defined
For scalar multiplication, we have a(v1-v’1) = av1 - av’1 in W since W is a space
I think thats done
ashen sphinx
well we will have +(W1,W2) --> (v1 + v2) + W , and +(W'1,W'2) --> (v'1 + v'2) + W , due to the properties of R-modules we have , v2 + (v1 + W) , and v'2 + (v'1 + W) , now we don't yet know whenever these are equal to each other from my understanding , but due to W1 = W'1 and W2 = W'2 we atleast know that v1 + W and (v'1 + W) are the same , not entirely sure how to get that v2 + (v + W) = v'2 + (v + W) tho ?
ashen sphinx
(v + W) need no longer be a subspace of V from what i understood
ashen sphinx
Are the field axioms a subset of what you call r-modules
i don't think we are doing any form of division, nor that we need scalars being commutative on multiplication for the + case, so without needing to worry about any issues related to higher generality we can prove statements about + in the more general context (unless simply scalars appear then we can substitute the word "R-module" with vector space ig lol)
i would say that pointwise form with v2 = v'2 is fine tho and faster way
cuz those elements were already fixed beforehand
so if they're equal the sets are kinda obviously equal
tawdry flame
Sounds kind of complicated
But that seems to also work
The online solution i found is same to mine but maybe not as rigorous to yours
ashen sphinx
you might want to prove this more rigorously if you want
actually hold up that might be slightly wrong
tawdry flame
Idk but im not used to proving things too rigorously
Such as im not sure how to prove things like a->b or c, and negation of implications
I’m in hs u see, lol
ashen sphinx
wait
wrong wording
i really didn't want to , but i guess i'm too skill issue/sleepy , so if i were to prove this right now, i would probably use basis
ashen sphinx
god damn
tawdry flame
I dont think they expect a proof as rigorous as your attempt
ashen sphinx
your proof works?
tawdry flame
The exercises only required the notion of a vector space and some s and n conditions for a subset of a vector space to be a subspace
It’s the same as an online source
I mean its the natural way to go cus it uses part (b)
ashen sphinx
Because v1+W=v’1+W implies v1-v1’ in W and similarly v2-v2’ in W
this seems fine so far
ashen sphinx
(v1+v2)-(v’1+v’2) in W
This no??
tawdry flame
We have (v1-v’1) in W, and (v2-v’2) in W, so their sum is in W since W is a space
Hence (v1-v’1)+(v2-v’2)=(v1+v2)-(v’1+v’2) in W
Using multiple vector space axioms
This means from b that
(v1+v2)+W=(v’1+v’2)+W
And using the addition defined we have (v1+W)+(v2+W)=(v’1+W)+(v’2+W)
As desired
Do you know any particular significance of this vector space S
ashen sphinx
(v1+v2)-(v’1+v’2) in W
alr nvm it works
tawdry flame
Do you know any particular significance of this vector space S
That which they denote the ‘quotient space V module W’
ashen sphinx
didnt look at how they were doing the minus stuff
tawdry flame
Yea it felt like a guided proof
I dont see any thing special about this V/W space though
ashen sphinx
That which they denote the ‘quotient space V module W’
I think it will be the equivalence classes under the quotient W of V , i.e if [x] is in V/W , then [x] := {all v in V s.t v - x in W}
ashen sphinx
I dont see any thing special about this V/W space though
It probably forms a vector space itself
tawdry flame
Yea no i kind of forgot everything about equivalence relations and classes I’ll find that out some other time
ashen sphinx
Yea no i kind of forgot everything about equivalence relations and classes I’ll ...
It important for algebra stuff
tawdry flame
Self studying undergraduate material is painful
ashen sphinx
in modern algebra you will divide algebras more than dividing x's lol
tawdry flame
I remember trying really hard to prove that equivalence classes form a partition or something lol
tawdry flame
in modern algebra you will divide algebras more than dividing x's lol
Dam
Yea i suppose linear algebra is a very small part of what it has to offer
ashen sphinx
I remember trying really hard to prove that equivalence classes form a partition...
Munkres might do
the set theory part
ashen sphinx
gn
tawdry flame
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calm coralBOT
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Available help channel!
vestal rampart
Did I perhaps do ts right?
calm coralBOT
Did I perhaps do ts right?
vestal rampart
I completely forgot how to do all this and my teacher is pissed!!!
nimble harbor
The second one is also wrong
vestal rampart
Fml.
potent lotusBOT
k
nimble harbor
So it is
${\frac{d}{dt}(9) - \frac{d}{dt} W^2(t) = 0 - \frac{d}{dt}(W^2(t))}$
Right?
potent lotusBOT
k
nimble harbor
Because it is W^2(t), the outer function is the squaring function so u need chain rule
Its a
${-2W(t) \frac{dW}{dt}}$
No?
potent lotusBOT
k
vestal rampart
Ok yes
potent lotusBOT
k
blazing coyote
calm coralBOT
blazing coyote
Is there any neat paramatrization here
velvet osprey
sure is
blazing coyote
because calculating the coss product using the usual paramtrization isn't going to be fun
velvet osprey
spherical coords no good?
remote mural
good morning miss ann
blazing coyote
spherical coords no good?
like (cos(u)cos(t), cos(u) sin(t), sin(t))
velvet osprey
this times 2
blazing coyote
something like that, is what you mean?
velvet osprey
but yes
blazing coyote
yea, but finding the cross product of their derivatives wrt t and u isn;t going to be fun. I suppose I could ask my prof if I can use it as a standard result?
swift laurel
with these parameterizions for common objects like cylinders, spheres, cones etc based on taking any two coordinates from a known coordinate system, you may as well write down the result of the cross product for future use because they come up a fair bit anyway
blazing coyote
I think the norm should just be 2r?
so I have $\iint_{S} (cos^2(u) cos^2(t) sin(t) + sin^2(u)cos^2(t) sin(t)) \cdot 2u dA$?
potent lotusBOT
What a wonderful world !
potent lotusBOT
What a wonderful world !
blazing coyote
we then have $x = (5- 3 \cos(t))$
potent lotusBOT
What a wonderful world !
calm coralBOT
@blazing coyote Has your question been resolved?
blazing coyote
so the region is paramatrized by (5-rcos(t),rcos(t), r sin(t)) 0≤r≤3; 0≤t≤2π
Is this fine so far?
<@&286206848099549185>
If this is right I can do it from here
.close
calm coralBOT
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sonic mason
can someone pls explain how the x^2-y^2 part came
calm coralBOT
can someone pls explain how the x^2-y^2 part came
naive meadow
can someone pls explain how the x^2-y^2 part came
$f(p,q)=\frac{q^2-p^2}{4}, f(y,x)=\frac{x^2-y^2}{4}$
potent lotusBOT
Buzzing Hornet
sonic mason
but how coz we jus took p =x-y
and q =x+y
im very new to this so could u pls break down what we're doing here
naive meadow
but how coz we jus took p =x-y
We've shown that f(p,q)=(q^2-p^2)/4 for all p,q
sonic mason
okay
naive meadow
So we just plug in y and x
sonic mason
😶
naive meadow
I'm not entirely understanding what the difficulty you're having is
We have a function f, we plug in y and x
sonic mason
like if u the question
in that we have it as f(x-y, x+y)
then we assume that x-y is p and x+y is q
so when we do take it as p and q
and we finally we a quen with that
how can we sutitute as x and y
cox we first assumed the calue of p to be x-y right
naive meadow
There isn't like some set x and y
You can think of the original problem statement as f(a+b,a-b)=ab instead if you want
sonic mason
okay
calm coralBOT
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tall moon
the larger circle's radius is R, the medium circle's radius is 2r, and the small circle's radius is r, if they are all tangent to eachother find R/r
tall moon
i eyeballed the drawing, dont expect it to be accurate
honestly i dont really have any ideas
alpine lynx
Here is an example.
calm coralBOT
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modern briar
hello! i was studying functions and i cant find a direct explanation
can even functions have like ^ odd
i dontt have the enough english for this, so im gonna just
what i mean is like can this be an even function if it has the 5x
modern briar
i made the function out of my ass btw its not relevant like solving or anything
velvet osprey
what's your native language
modern briar
turkish
polar bison
f(-x) = -f(x) is odd?
modern briar
yeah
modern briar
anyway no this function is not even
oh
polar bison
That's all u gotta check?
velvet osprey
this is a polynomial of even degree
but it is NOT an even function
velvet osprey
sure
modern briar
thank you :3
remote mural
calm coralBOT
versed sluice
No
remote mural
O then?
versed sluice
remote mural
Angle AOC is 131 ×2rifhht?
Angle at the centre is twice the angle at the circum?
versed sluice
Draw this
versed sluice
Angle at the centre is twice the angle at the circum?
That's for a chord
Angle at F would be 180 - B as it is a cyclic quadrilateral
remote mural
Oo ok
remote mural
Ok
versed sluice
Oh, u mean like AC is still the chord and B is the angle
I don't understand why that's wrong tho
Maybe they aren't supposed to be in the same side?
remote mural
Me not sure 🤷♀️
versed sluice
Okay
calm coralBOT
@remote mural Has your question been resolved?
jovial sedge
how is this a brilliant?
calm coralBOT
how is this a brilliant?
jovial sedge
.close
calm coralBOT
primal bear
😭
lean hull
calm coralBOT
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blazing coyote
Trying to figure out what exactly a boundary is
calm coralBOT
Trying to figure out what exactly a boundary is
blazing coyote
the best I can come up with, is that set of points for each every open ball centred at that point contains points both in the interior and exterior of it
ooh
icy galleon
a boundary is the intersection of the closure of a set and its complement?
but the two definitions are equivalent no?
blazing coyote
So my defn is right?
blazing coyote
So a sphere doesn't have a boundary as?
icy galleon
So a sphere doesn't have a boundary as?
wdym
blazing coyote
a sphere has a boundary
idts
icy galleon
closed ones, opened ones, you name it. spheres have boundaries. its kinda obvious...
blazing coyote
lyric zephyr
https://math.stackexchange.com/questions/4806976/why-does-a-sphere-not-have-a-bo...
oh
a sphere doesnt have a boundary in the sense that it never ends if ur existing on said surface
if u have a ball (a sphere that includes the inside), u could exist in that ball and the boundary of that ball would be the spherical surface of the ball
cuz thats where the ball meets the outside world
if u lived on a sphere (no inside, just the spherical surface, kinda like existing on the surface of the earth) there is no end
u could go on forever and ud just end up still on the sphere
like there is no "edge of the earth" à la flat earther
if u include the insides of the earth, then the earth's crust is the edge of the earth
also think about a square
if u lived inside a filled square the boundary would be the edges of the square, u can go no further than the edges of the square, if u came up onto one ud have to turn around
if u lived on the edges of the square only, there is no end, u can just go round and round
there is no place where u logically would have to stop and turn around cuz thats where the shape ends
in fact the boundary of the boundary of any shape is zero
calm coralBOT
@blazing coyote Has your question been resolved?
blazing coyote
u could go on forever and ud just end up still on the sphere
I don't get this
lyric zephyr
I don't get this
its like..
can u walk on the surface of the Earth until u get off the earth
a boundary is something u can walk (or swim idk) across to get outside of the thing u were in
if ur set was the earth, including the interior, u could start at the core and swim outward and boom u cross the crust of the earth and now ur in space
if ur set was just the surface of the earth
u cant do that
u cant walk to anywhere on the earth where earth meets space
blazing coyote
Yea, but if you centere a ball on the boundary of the sphere , you can go both inside the set ( the centre of the ball , and outside yeh set )
ooh
okay
I think I;m confusing S2 and the sphere as a subset of R^3
.close
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dreamy wyvern
can someone please explain this to me
calm coralBOT
can someone please explain this to me
dreamy wyvern
photos won’t send for me.
alpine lynx
Hmm...
dreamy wyvern
for
an airplane is flying at an airspeed of 500 km/h on a heading of 040 degrees, a 150 km/h wind is blowing from a bearing of 120 degrees. determine the ground velocity of airplane and direction of flight
so you get for the resultant vector, 496 km/h which i’m okay with
by using cosine law
then you use sin law to find theta, the direction
and you get 82.69
but then the textbook says
“the head of the original wind vector is 60 degrees off 180 degrees so the head of the translated wind vector is 60 degrees off 360 degrees”