well, we can solve for $\lim_{x\rightarrow2^+}$ and $\lim_{x\rightarrow2^-}$ separately
#help-49
shy rampart
last slate
and now we just solve for LHL and RHL
grand pondBOT
nameless individual
last slate
rihgt?
shy rampart
yep'
zealous schooner
Yes
zealous schooner
The very definition of continuity of a function involves limits
last slate
limit wont exist if a function is discontinuous at that point?
shy rampart
no
last slate
we haven't been taught continuity yet
zealous schooner
A function $f(x)$ is said to be continuous at $x=a$ iff. $$\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)$$
shy rampart
*iff
if a function is discontinuous, it can be because $\lim$ doesn't exist, but it can also be because $f$ doesn't exist or the two doesn't agree
zealous schooner
The limit must exist and that limit must be equal to the value of the function at the point
shy rampart
its the exact opposite: function is discontinuous if (not necessarily iff) limit DNE
zealous schooner
You can have cases where the limit exists but it doesn’t agree with the value at the point, which is called a removable discontinuity
If the limit doesn’t exist, but both the RHL and LHL exist then it’s a jump discontinuity
shy rampart
You can have cases where the limit exists but it doesn’t agree with the value at...
(or if $f$ is not defined at the point)
grand pondBOT
nameless individual
zealous schooner
And if either of the limits diverge to +-infinity it’s called an infinite discontinuity
zealous schooner
(or if $f$ is not defined at the point)
Yes this too
shy rampart
and otherwise it's called a wierd discontinuity
zealous schooner
🦇
last slate
If the limit doesn’t exist, but both the RHL and LHL exist then it’s a jump disc...
can this be called as jump discontinuity?
zealous schooner
can this be called as jump discontinuity?
Yes that’s exactly it
Wait no
This is a removable discontinuity
The left and right limits exist
And they’re equal
But they don’t match the value of the function
last slate
If the limit doesn’t exist, but both the RHL and LHL exist then it’s a jump disc...
u messed up here ig then
shy rampart
its a removable discontinuity, as in you can remove it by creating a new function $F$ defined to be
$f(x)$ iff $x$ is not the point of discontinuity
$\lim f(x)$ otherwise
grand pondBOT
nameless individual
zealous schooner
u messed up here ig then
Nope
When the left and right limits exist but they are not equal then it’s a jump discontinuity
But the case you made there the limits were equal
last slate
oh ok so if u lift the left horizontal line a lil bit
its gonna be a jump discontinuity?
shy rampart
But the case you made there the limits were equal
actually I think the RHS limit is one pixel higher
last slate
😭
shy rampart
so it is a jump discontinuity
last slate
cmon bro 😹
zealous schooner
its gonna be a jump discontinuity?
Uhh, if you lift only one side then yes.
shy rampart
he said "the left"
zealous schooner
Ah yeah
Then yes
Basically if the function is approaching two different values depending on which side you come from then it’s a jump discontinuity
If there’s just a single outlier point and the function seems to approach the same point from both sides then it’s a removable discontinuity
shy rampart
unless you life the LHS by one pixel, in which case it will align
last slate
so in a 🥜 🐚 we changed fractional part to integer part because fractional part function is discontinuous?
zealous schooner
And if approaching from one of the sides the function blows up to +infinity or something then it’s an infinite discontinuity
zealous schooner
so in a 🥜 🐚 we changed fractional part to integer part because fractional part...
Both of them are discontinuous
At exactly the same points
But it’s easier to work with GIF function
Typically
last slate
oh ok
shy rampart
And if approaching from one of the sides the function blows up to +infinity or s...
and then there's $\sin\frac1x$
grand pondBOT
nameless individual
last slate
my teacher told me to always find out lhl and rhl whenever theres a []
or {}
or if the function is changing its nature
zealous schooner
and then there's $\sin\frac1x$
Disgusting
zealous schooner
my teacher told me to always find out lhl and rhl whenever theres a []
Yes exactly
Whenever there’s some discontinuity find both limits separately
shy rampart
or if the function is changing its nature
usually, yes
shy rampart
Disgusting
that's what is called a wierd discontinuity
last slate
its acutally "weird" and not "wierd" 🤓
zealous schooner
that's what is called a **wierd discontinuity**
That’s not the actual name, is it??
shy rampart
probably not
shy rampart
its acutally "weird" and not "wierd" 🤓
that's weerd
last slate
lmfao
shy rampart
its acutally "weird" and not "wierd" 🤓
its actually "actually" and not "acutally"
last slate
btw was this necessary?
shy rampart
btw was this necessary?
you can postpone it, but you cant avoid it
shy rampart
cant we just find both the limits using the first one
and how would you do that?
zealous schooner
cant we just find both the limits using the first one
You can yeah
At the end you’ll do the substitution anyway
This just makes it easier
zealous schooner
Uhhh, well…
last slate
is this wrong?
last slate
That’s not how you *really* calculate limits
then how do we do that? let x=a-h and then apply lim h tends to 0+
like this?
for lhl
zealous schooner
then how do we do that? let x=a-h and then apply lim h tends to 0+
Yeah
That’s the proper way to do it
last slate
yeah
zealous schooner
here a is 2
zealous schooner
Sure, but just know that that doesn’t necessarily give you the value of the limit
last slate
yeah i will use the proper method then
shy rampart
it really depends on the rigorous your course is
last slate
its pretty rigorous ig
zealous schooner
True
If it’s really rigorous prove the limit using epsilon delta
Ugh autocorrect is so bad
last slate
uh idk what that is 😭
shy rampart
the $(\epsilon,\delta)$ definition of limits
grand pondBOT
nameless individual
last slate
idk and i dont think i should really be knowing about it rn
shy rampart
dats da dafanition of not rigorous lol
last slate
we've just started calc
imma close now
thanks kheerii
i think ive seen u on
chem server
thanks nameless
shy rampart
if you want it
https://mathworld.wolfram.com/Epsilon-DeltaDefinition.html
shy rampart
thanks nameless
ur welcome
zealous schooner
chem server
Ooh yeah I was wondering where I recognise you from
shy rampart
dont stress over it though, this web page gives an extremely formal presentation
zealous schooner
You answered a question or something of mine?
last slate
yea ig i dont really remember tho
last slate
dont stress over it though, this web page gives an **extremely** formal presenta...
oh ok
ill check it out
shy rampart
as in, ridikulusly formal
midnight plankBOT
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hollow pike
Does { x| x is a real number and greater than 5}
Translate to { you see this element/variable/or whatever it possibly could be that we are just gonna use X as the symbol for whatever it could be wall block (or a coma/fullstop) The X is a real number and greater than 5}
midnight plankBOT
Does { x| x is a real number and greater than 5}
Translate to { you see this el...
sharp coral
the {} denotes a set (essentially a huge collection of numbers), which is the set of all real numbers larger than 5 (where x just denotes any element [number] in the set)
hollow pike
Just a heads up, I dont understand english math that well so sorry
sharp coral
if you replaced x with y, or any other letter, you would get the same set
hollow pike
the {} denotes a set (essentially a huge collection of numbers), which is the se...
So basically when you put {} its already a set?
Or is it just the outershell of it
hollow pike
if you replaced x with y, or any other letter, you would get the same set
I understand that part 👍
sharp coral
anything inside {} is generally a set, { | } is set-builder notation, where the first part is a variable representing any element in the set and the second part is the condition that it needs to satisfy in order to belong to the set
hollow pike
Does { x| x is a real number and greater than 5}
Translate to { you see this el...
So is my interpretation correct?
hollow pike
anything inside {} is generally a set, { | } is set-builder notation, where the...
Is the | = such that?
midnight plankBOT
@hollow pike Has your question been resolved?
sharp coral
in words, yes
placid siren
midnight plankBOT
placid siren
where did the 3 come from?
carmine sigil
15 cancels with the 5 and produces 3, 6 cancels with 2 and produces the 3.
carmine sigil
not quite
carmine sigil
we have (6/5) * (-15/2), this is the same as (6 * -15)/(5 * 2) = 6/2 * (-15/5)
carmine sigil
[
\frac{6}{5} \times \frac{-15}{2} = \frac{6 \times -15}{5 \times 2} = \frac{6}{2} \times \frac{-15}{5}
]
grand pondBOT
OmnipotentEntity
carmine sigil
maybe this is a little bit more clear.
midnight plankBOT
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stable mirage
whas up here
midnight plankBOT
whas up here
versed apex
whas up here
It asked for the domain of f, not f'
stable mirage
OH
versed apex
cruel of them to ask it that way ngl
stable mirage
its so silly
versed apex
agreed
stable mirage
.close
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zenith garden
I think this one is C, am I right?
midnight plankBOT
I think this one is C, am I right?
zenith garden
because of 5 sqrt of x
sage olive
yes
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iron whale
im part of the way through a 3-variable system (im required to use subsitution, no matrices) and I got y=-18/11 where when I checked this solution with wolfram alpha it said y=-2. Where did I go wrong?
midnight plankBOT
im part of the way through a 3-variable system (im required to use subsitution, ...
surreal moon
im part of the way through a 3-variable system (im required to use subsitution, ...
here
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viscid girder
hello, to integrate sqrt (1+(1 / x ^ (2/3))), i am being told to set u to x^2/3 and simply to sqrt (1/u), but when I do this I am left with an x variable in the integral, was I given wrong instructions?
midnight plankBOT
hello, to integrate sqrt (1+(1 / x ^ (2/3))), i am being told to set u to x^2/3 ...
@viscid girder Has your question been resolved?
pearl idol
hello, to integrate sqrt (1+(1 / x ^ (2/3))), i am being told to set u to x^2/3 ...
lemme just make sure I understand the question, you're trying to integrate
[\int\sqrt{1 + \frac{1}{x^{2/3}}},\dd x?]
grand pondBOT
biased_estimatERIC
viscid girder
i set du = 2 / (3x^1/3) *dx and dx = du / 2 / (3x^1/3)
i then simplified the integral to sqrt 1/u * du, but when I plugged in du I was left with x in the integral
midnight plankBOT
@viscid girder Has your question been resolved?
twilit field
can I have this verified ?
midnight plankBOT
can I have this verified ?
subtle blaze
how did you get the span of w_1, w_2, ..., w_m is what you got
that's not the definition of span
you have a habit of asserting things without explanation
twilit field
I meant that set
subtle blaze
well how do you know that's true
pearl idol
is this better?
you might want to practice your set notation a little bit
It should be something like
[\operatorname{span}(v_1, \dots, v_m) = {a_1v_1 + \dots + a_m v_m \mid a_1,\dots,a_m \in F}]
grand pondBOT
biased_estimatERIC
pearl idol
your idea is correct, it's just somewhat confusingly written
here's how I'd go about proving that span(v1, ..., vm) is contained within span(w1, ..., wm)
pearl idol
no worries
actually I'll prove the other direction, span(w1, ..., wm) is contained within span(v1, ..., vm)
that one is a bit more straightforward
twilit field
isn't the process identical for both?
pearl idol
If I wanted to spell out a lot of the details, I might write something like this:
Let (v \in \operatorname{span}(w_1, \dots, w_m)). Then, since
[\operatorname{span}(w_1, \dots, w_m) = {b_1w_1 + \dots + b_m w_m \mid b_1,\dots,b_m \in F},]
we know that there exist (b_1, \dots, b_m \in F) such that
[v = b_1w_1 + b_2w_2 + \dots + b_mw_m.]
We can rewrite this as
[\begin{split}v &= b_1v_1 + b_2(v_1+v_2) + \dots + b_m(v_1+\dots+v_m)\&= (b_1+\dots+b_m)v_1 + (b_2+\dots+b_m)v_2 + \dots + b_mv_m.\end{split}]
Therefore, since
[\operatorname{span}(v_1, \dots, v_m) = {a_1v_1 + \dots + a_m v_m \mid a_1,\dots,a_m \in F},]
we see by setting (a_k = \sum_{i=k}^m b_i) that (v \in \operatorname{span}(v_1, \dots, v_m)).
This argument is a bit overly verbose, but maybe it's good for you to be somewhat verbose when you're beginning out so that you can check every detail is correct
twilit field
hmm,thanks a lot!
pearl idol
isn't the process identical for both?
How would you prove that span(v1, ..., vm) is contained within span(w1, ..., wm)?
You can use this argument as a template
Remember:
- to prove a set A is equal to set B, you need to prove that A is a subset of B and B is a subset of A
- to prove a set A is a subset of B, you need to take an arbitrary element x of A and show that x is also in B
These proof techniques are very important
hard umbra
This argument is a bit overly verbose, but maybe it's good for you to be somewha...
i don't like your wording 
pearl idol
snow you should rewrite it then 
hard umbra
Let (v \in \operatorname{span}(w_1, \dots, w_m)). Then, since
[\operatorname{span}(w_1, \dots, w_m) = {b_1w_1 + \dots + b_m w_m \mid b_1,\dots,b_m \in F},]
we would like to show that there exist (b_1, \dots, b_m \in F) such that
[v = b_1w_1 + b_2w_2 + \dots + b_mw_m.]
We can rewrite the above equation as
[\begin{split}v &= b_1v_1 + b_2(v_1+v_2) + \dots + b_m(v_1+\dots+v_m)\&= (b_1+\dots+b_m)v_1 + (b_2+\dots+b_m)v_2 + \dots + b_mv_m.\end{split}]
Therefore, since
[\operatorname{span}(v_1, \dots, v_m) = {a_1v_1 + \dots + a_m v_m \mid a_1,\dots,a_m \in F},]
we see by setting (a_k = \sum_{i=k}^m b_i) that (v \in \operatorname{span}(v_1, \dots, v_m)).
grand pondBOT
pearl idol
Let \(v \in \operatorname{span}(w_1, \dots, w_m)\). Then, since
\[\operatorname{...
wait "we would like to show" is not quite right, is it? that's part of the hypothesis
hard umbra
oh right it is
twilit field
$span(v_1 \dots v_m)={a_1v_1 , a_2v_2 \dots a_mv_m| a_i\in F}$
$span(w_1 \dots w_m)={b_1w_1,b_2w_2 \dots |b_i \in F}$
\
we know that $w_i=v_1+v_2 \dots v_i$
so we have $span(w_1 \dots w_m)={b_1v+1+b_2(v_1+v_2) \dots| b_i \in F}$
pearl idol
i read it backwards
yeah I changed my mind about which direction to prove at some point so you're not alone lol
pearl idol
ah yeah my bad
hard umbra
i think thats why i thought you were proving the other way
pearl idol
fixed ty
grand pondBOT
ƒ(Why am. I here)=I don't Know
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(You may edit your message to recompile.)
pearl idol
@twilit field I had a typo on the last line switching a and b in the original sorry, here's the corrected version ^
twilit field
Thanks
grand pondBOT
biased_estimatERIC
pearl idol
(also realized I forgot a word, thanks snow)
twilit field
I'd probably say we can similarly find $b_i$ in terms of $a_k$
grand pondBOT
ƒ(Why am. I here)=I don't Know
pearl idol
how?
grand pondBOT
ƒ(Why am. I here)=I don't Know
pearl idol
Yeah, to be extra rigorous, can you prove that that choice works? (By induction on m)
grand pondBOT
ƒ(Why am. I here)=I don't Know
twilit field
is that the aim?
subtle blaze
dont you have to minus the other ones as well
subtle blaze
**ƒ(Why am. I here)=I don't Know**
check b_3
grand pondBOT
ƒ(Why am. I here)=I don't Know
twilit field
so we have $a_3-a_2+a_1-a_1$
grand pondBOT
ƒ(Why am. I here)=I don't Know
grand pondBOT
ƒ(Why am. I here)=I don't Know
twilit field
so I essentially have to show $b_n=a_n-a_{n-1}$
grand pondBOT
ƒ(Why am. I here)=I don't Know
midnight plankBOT
@twilit field Has your question been resolved?
twilit field
I think I'll think a bit about the induction step
thanks for the help everyone !
.close
midnight plankBOT
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wide marsh
how do this
midnight plankBOT
how do this
wide marsh
the answer looks sus as hell
wide marsh
nothing yet
i was incredibly stupified by wolfram alpha
but would ibp be appropriate
mighty hamlet
You need to show your own workings first before we can start guiding you
wide marsh
aaand is ibp ok
astral garnet
i mean i dont see why u couldnt
wooden knoll
This is a homework task (not graded) - helps us practice word problems for exam
Context: Manufacturing and Packaging
Manufacturers have various demands for the packagings of their products; one of these is to keep costs to a minimum. It is possible for packages of different dimensions to have equal volume but larger or smaller surface areas.
Task
Select two existing packages utilised for common household products. One has to be a simple prism (rectangular prism or cylinder) and the other one has to be more complex (tapered sides or multiple prisms).
Using a calculus approach, redesign their packages so that their volumes stay the same but their surface areas are reduced which makes production costs lower
midnight plankBOT
This is a homework task (not graded) - helps us practice word problems for exam
...
undone ivy
hey can any one help me tomorrow at 8:30 with some Hw it’s an online class pls i would do it but im out of time and my class finishes tm.
wooden knoll
hey can any one help me tomorrow at 8:30 with some Hw it’s an online class pls i...
this is occupied
sorry
go to #help-21｜아리스킨충1
or somewhere else
midnight plankBOT
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wooden knoll
.close
midnight plankBOT
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prime lintel
midnight plankBOT
prime lintel
i don''t know any direction to take this in
astral garnet
unless theres some faster way i dont see, i'd suggest letting z=a+ib or something similar and solve for a and b
by substituting*
grim vector
try passing by trig form
Let z + 2 = cos θ + i sin θ
1 /(z + 2) = cos θ − i sin θ
==> (z + 1) / (z + 2) = 1 − 1 / (z + 2) = 1 − ( cos θ − i sin θ )
= ( 1 − cos θ ) + i sin θ
sage helm
,w imaginary part of (a + 1 + ib)/(a + 2 + ib)
grand pondBOT
sage helm
stupud
astral garnet
lol
grim vector
Im ( (z + 1) / (z + 2) ) = sin θ , sin θ = 1 /5
cos θ = ± √(1 − 1 /25) = ± 2√6 / 5
So real part of this is |cos θ| = 2√6 / 5
grim vector
it gets really really long
can't make it less long for this question
prime lintel
can't make it less long for this question
hmm i see
thanks then i will try it again and see if it works out
.close
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craggy yew
how did underlined eq 1 give us underlined eq 2
midnight plankBOT
how did underlined eq 1 give us underlined eq 2
craggy yew
i do not understand how change of base was used
sage helm
yeah what did they do
lmao
Oh wait
it doesn't go from 1 to 2 as you labeled
You need to show us more context
midnight plankBOT
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craggy yew
.close
midnight plankBOT
maiden tree
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twin ridge
I'm doing this inverse matrix problem and I'm not sure how I got this one wrong.
midnight plankBOT
I'm doing this inverse matrix problem and I'm not sure how I got this one wrong.
twin ridge
I found that det(A)=27, then did $\frac{1}{det(A)}\begin{bmatrix} 5 & 2\ 9 & 9\ \end{bmatrix}$
last slate
what is the |A|
grand pondBOT
Narutoes
twin ridge
what is the |A|
the wat
grim vector
**Narutoes**
ok and ?
last slate
**Narutoes**
A^-1 is correct
grim vector
How is this helpful at all?
last slate
do matrix multiplication to find x
twin ridge
It doesn't want me to actually solve it, they just want me to show x as a product of A^-1 and b
twin ridge
You got got I think
Wouldn't be the first time. It's some software that the teacher tried to program himself.
shy rampart
HAHAHHAHAHHAHHAHAHIUGFUIQWDGWIHHAUHAHHHHAHHHAHAH
stupid teacher
please ditch him/her
twin ridge
I can't, I need this class for my major lol
shy rampart
(ok im exagerating)
shy rampart
I can't, I need this class for my major lol
LOL
twin ridge
The teacher is actually pretty good, this software just isn't.
shy rampart
is it possible to send me the website?
shy rampart
The teacher is actually pretty good, this software just isn't.
i said i was exagerating
twin ridge
It's through WeBWorK, but the teacher is trying to do some weird fancy programming in the background to tailor to our specific university.
shy rampart
It's through WeBWorK, but the teacher is trying to do some weird fancy programmi...
so i cant access it? sad
twin ridge
I think I get this error at least twice per 10 problem assignment.
shy rampart
whats the warning message?
twin ridge
twin ridge
That's an example of one of the errors. I haven't kept a log of all of em, but it may be that none of them are his.
He said he wasn't able to replicate it, so sucks to suck.
shy rampart
LMA
twilit field
can someone verify this?
midnight plankBOT
can someone verify this?
runic hamlet
why does that mean a+b=0 and a-b=0. where are you using that you are working over R
twilit field
ok, for this to be zero, both the real part and imagnary parts have to be zero
so a+ai+b-bi=0
so a+b=0
and a-b=0
runic hamlet
why does that not work if a,b are allowed to be in C
twilit field
then we get a quadratic with two roots
if we're working over the field R a and b \in R
no?
runic hamlet
where are you getting a quadratic from
runic hamlet
what I mean is: if we think of C as a vector space over C, then 1+i and 1-i are linearly dependent. so somewhere your proof must fail. where
subtle blaze
Dena you always seem to ask really good questions
runic hamlet
I try
twilit field
That's the next question in the book xD. OK, give me a minute to figure this out
so $a=(e+fi), b=(h+ji)$
grand pondBOT
ƒ(Why am. I here)=I don't Know
hard shard
are you familiar with how complex numbers multiply?
twilit field
yes
subtle blaze
why does that mean a+b=0 and a-b=0. where are you using that you are working ove...
This is the question you want to answer
subtle blaze
what I mean is: if we think of C as a vector space over C, then 1+i and 1-i are ...
This is only an extra question to further elaborate why the previous question is good
twilit field
why does that mean a+b=0 and a-b=0. where are you using that you are working ove...
if we're working over R all it means is that a and b are real
right?
runic hamlet
if you havent said anything like "because a,b are real it follows that ..." then you havent used that a,b are real
runic hamlet
rest is fine
twilit field
thanks!
runic hamlet
formatting could be better
twilit field
I'll work on it
runic hamlet
yeah its a long term goal
twilit field
.close
midnight plankBOT
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twilit field
I'm kind of lost here
midnight plankBOT
I'm kind of lost here
merry pewter
i think the easiest way is to assume that it is linearly dependent and show that the original list is also linearly dependent
merry pewter
what do you mean
grand pondBOT
ƒ(Why am. I here)=I don't Know
merry pewter
ok so what are you asking?
merry pewter
err ok? i don't really know what you are trying to say
tough spindle
Suppose $a\left(v_1 - v_2 \right) + b\left(v_2 - v_3 \right) + c\left(v_3 - v_4 \right) + d\left(v_4 \right) = 0$ for some coefficients $a, b, c, d \in \mathbb{R}$. Simplifying, we get $av_1 + \left(b - a \right)v_2 + \left(c - b \right)v_3 + \left(d - c \right)v_4 = 0$. By linear independence of $v_1, v_2, v_3, v_4$, all the coefficients are equal to zero, so we have $0 = a, 0 = b - a = b - 0 = b, 0 = c - b = c - 0 = c, 0 = d - c = d - 0 = d$, which proves linear independence.
grand pondBOT
PowerUp
merry pewter
yeah
merry pewter
!nosols
midnight plankBOT
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
twilit field
err ok? i don't really know what you are trying to say
what I was trying to say is setting a=e, f-e=c, g-f=c and h-g=d
should be enough to complete the proof
right?
thanks a lot everyone!
.close
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wide marsh
can anyone help me with this integral? ive tried ibp but im not sure if itll lead anywhere cause the resulting integral seems pretty difficult
midnight plankBOT
can anyone help me with this integral? ive tried ibp but im not sure if itll lea...
wide marsh
like should i change what i make u and dv? or is that integral actually not that hard
midnight plankBOT
@wide marsh Has your question been resolved?
wide marsh
<@&286206848099549185>
midnight plankBOT
@wide marsh Has your question been resolved?
wide marsh
pls help lol
stark sable
can anyone help me with this integral? ive tried ibp but im not sure if itll lea...
i think if you do ibp with the e^x separate it might eventually lead to itself and you do the 2int trick
but its a pain, im still working out if that works
wide marsh
what is the 2int trick
stark sable
if youre doing ibp and the integral eventuall repeats itself so
og integral= some stuff - og integral
you can combine the two original integrals onto one side and just says its the stuff/2
wide marsh
ahhh i see
i thought that stuff only worked with trig integrals using ibp
well in any case could you please let me know if you find anything
stark sable
i think its more common to see it with the trig ones
ye lol ill try to find if it does
wide marsh
ok tysm!
stark sable
okay im stumped, might wanna ask your prof about it
theres no ibp, trig sub, pfd, any of the usual calc things to try ig
steady stag
someone please help with question 21
midnight plankBOT
someone please help with question 21
fathom knoll
what does m angle mean? is it just what the angle is?
so angle YWP is -1 + 4x
and angle YWX is 7x+2
ok cool
so well you know that WP is an angle bisector
so in particular angle 1 = angle 2 = (angle YWX)/2
steady stag
im not sure
steady stag
i think so yes
steady stag
sorry i was replying to m angle
fathom knoll
so you know that angle 1 = (angle YWX)/2
steady stag
yes yes
fathom knoll
ok
fathom knoll
sure feel free to dm if u dont get help quickly here
midnight plankBOT
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Available help channel!
somber gazelle
here is the problem
midnight plankBOT
here is the problem
somber gazelle
Here is my work (excuse the crudeness please)
my approach was just to draw it all out and get as many angles as i could
after using sine and cosine rule a billion times im pretty sure i got the right answer of around 7250
but i entered it in and its saying its wrong
im assuming its expecting integer answers so maybe my approach and/or answer is just completely wrong but idk
midnight plankBOT
@somber gazelle Has your question been resolved?
somber gazelle
<@&286206848099549185>
midnight plankBOT
@somber gazelle Has your question been resolved?
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@somber gazelle Has your question been resolved?
somber gazelle
<@&286206848099549185>
somber gazelle
my brothers in christ the question aint even that hard i just needa know if i did everything right 😭
anyone please
midnight plankBOT
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somber gazelle
nah im pretty sure im invisible atp
sage path
what do u want help w
somber gazelle
show me how to approach the problem
midnight plankBOT
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mossy cargo
midnight plankBOT
mossy cargo
I don't understand how two fermat's number are coprime 00:16
then I don't understand how for each of the fermat's number there should be different primes
small jasper
I don't understand how two fermat's number are coprime 00:16
Hint: Show that $$F_n=2+\prod^{n-1}_{k=1} F_k$$
grand pondBOT
Civil Service Pigeon
mossy cargo
**Civil Service Pigeon**
he used this
wait a minute I haven't seen what he did with it
I'm back
he did recursive relation
to prove that one is the only divisor for Fk and Fn
i) I didn't understand what recursive relation is
ii) I don't understand why he used (n ≥ 1) in the relationship
iii) I don't understand why should k < n
iv) why should both sides be divisible and why should the reminder be a natural number rather than decimals
v) how does this proof, "to prove that one is the only divisor for Fk and Fn" really prove two fermat's number are coprime
vi) then how for each of the fermat's number there should be different primes
midnight plankBOT
@mossy cargo Has your question been resolved?
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@mossy cargo Has your question been resolved?
midnight plankBOT
@mossy cargo Has your question been resolved?
small jasper
i) I didn't understand what recursive relation is
ii) I don't understand why he...
i. Look this up
ii. Because if n<1, then you get the product from 0 to a negative number, which doesn’t make sense
iii. Same logic as ii
iv. Because otherwise one side will have a factor the other doesn’t, and hence they can’t be equal. Also, the remainder is always defined to be an integer, so yeah.
v. Recall the definition of “coprime”
vi. No idea what you’re asking here
lucid vapor
Ive been stuck on these for hours, what are the legs?
midnight plankBOT
Ive been stuck on these for hours, what are the legs?
fleet idol
just use sin, cos, tan
last slate
Ive been stuck on these for hours, what are the legs?
How far have you gotten?
midnight plankBOT
@lucid vapor Has your question been resolved?
twilit field
#1260937723452330155 message can I have a hint ?
midnight plankBOT
https://discord.com/channels/268882317391429632/1260937723452330155/126102671018...
@twilit field Has your question been resolved?
Available help channel!
summer ravine
is the slope of a linear function graph same for every reference?
midnight plankBOT
is the slope of a linear function graph same for every reference?
summer ravine
for example: y = 2x+3
hollow dome
yes, because it is linear
summer ravine
slope can be calculated by (2x+3)/x as y = 2x+3, right?
hollow dome
if y = ax+b then slope is a
midnight plankBOT
summer ravine
.reopen
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✅
summer ravine
so if we construct a triangle to find the slope, the perpendicualr will be 2x? or 2x+3 as y = 2x+3?
the base x.
oh i get it. the perpendicular will be 2x+3-3 because it forms a trapezium.
.close
lapis beacon
midnight plankBOT
lapis beacon
Do i just find s'(t) then plug in 4 for t
hearty rune
that would be the velocity
hearty rune
acceleration would be the time derivative of velocity
last slate
you might have to just plug it in the function and then use an acceleration formula but idk
on s(4)
broken stone
You'd want to take the derivative twice to get acceleration, then plug in t=4, since you're starting with the function for position
lapis beacon
acceleration would be the time derivative of velocity
what does that mean
lapis beacon
You'd want to take the derivative twice to get acceleration, then plug in t=4, s...
so get the second derivative?
hearty rune
v(t)=s'(t)
a(t)=v'(t)=s''(t)
lapis beacon
what does getting the second derivative give you
midnight plankBOT
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Available help channel!
wraith oriole
Should it be T/0.01
rotund plume
okay so i'm trying to develop a game about simple 2D engeniring with gears and stuff.
so i'm currently making an way to auto align the gears i already have the position bein auto aligned, that works by just bringing the gear you are placing to the closest point of the circunference of the gear that is already placed
now i want to be able to auto align the rotation of the gear.
To do that i u have the piece code at the second image, it just sets the rotation of the sprite to the angle of the dirction vector from the mouse to the center of the gear that is already placed, multiply it by a factor in this case 3.5 + an offset.
i don't know if i can ask help here if i dont have any idea of how to come up with a formula to this but, i wanted to be able to determine the factor to multiply based on the number of teeth on the gear that is being placed in relation to the gear that is already placed so when it rotates it always aligns the teeth
the gears are static untill you play the start button so i don't have to calculate it while the gears are spinning
sorry for the wall of text and thank you if you read all of this
if you could send me an article about something like this i would also greatly apreciate it.
midnight plankBOT
okay so i'm trying to develop a game about simple 2D engeniring with gears and s...
main current
Does the original sprite always have a tooth directly facing upward?
Or: What can we use to identify the teeth location of the original sprite?
rotund plume
Does the original sprite always have a tooth directly facing upward?
no but i can make them
rotund plume
Or: What can we use to identify the teeth location of the original sprite?
what do u mean?
midnight plankBOT
@rotund plume Has your question been resolved?
midnight plankBOT
@rotund plume Has your question been resolved?
midnight kindle
midnight plankBOT
slender walrus
they're taking shortcuts and didn't notate it properly
its supposed to be
(coefficient of numerator)/(coefficient of denominator) = 7/3
so you don't care about the x
twin ridge
Consider the matrix transformation $T:\mathbb{R}^2\to\mathbb{R}^2$ that assigns to a vector x the closest vector to a projection. Find the matrix that defines the projection of any vector x onto the line $y=x$ in the Cartesian plane.
midnight plankBOT
Consider the matrix transformation $T:\mathbb{R}^2\to\mathbb{R}^2$ that assigns ...
grand pondBOT
Narutoes
twin ridge
I'm still working on this one. Just trying to get an actual answer.
rose trout
How do you usually find a matrix for a transformation?
twin ridge
Iirc, the matrix of a transformation is determined by how it affects the standard matrix.
So [1 0] and [0 1]
rose trout
Yeah. The columns of the transformation matrix is made of the transformed basis vectors
So (1,0) and (0,1). You want to find where those are being sent.
twin ridge
When I tried [1 0] and [0 1] I think I got [sqrt(2)/2 sqrt(2)/2] for both. Pretend these are all column vectors; I'm lazy.
rose trout
Indeed.
twin ridge
Because they make a 45 90 triangle with that unit vector
So $e_1\to \frac{e_1\sqrt{2}}{2}+\frac{e_2\sqrt{2}}{2}$ and $e_2\to \frac{e_1\sqrt{2}}{2} + \frac{e_2\sqrt{2}}{2}$?
grand pondBOT
Narutoes
rose trout
Yes that's correct, now you can put their coordinates as columns of your transformation matrix.
twin ridge
Wait
Is the transformation matrix actually just $\begin{bmatrix} \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\ \end{bmatrix}$?
grand pondBOT
Narutoes
twin ridge
I got that in class when I initially got the problem but it didn't quite look right in my head.
rose trout
Yes that's right.
twin ridge
Bruh.
I got this in 5 minutes of doing it in class but my teacher told me to look at it again. Was he yanking my chain?
I'm a little confused though.
rose trout
One sec
twin ridge
I tried this with an x of [1 1], since that x is already on y=x
That gives me a projection of [sqrt(2) sqrt(2)], but shouldn't it just give me back [1 1]?
sage olive
why isn't the projection matrix
0.5 0.5
0.5 0.5
?
rose trout
neil will know better I think I'm tired hahaha
twin ridge
I was told the transformation matrix is [ T(e1) T(e2) ]
twin ridge
why isn't the projection matrix
0.5 0.5
0.5 0.5
?
I also considered this as well.
I wasn't sure tbh.
sage olive
I was told the transformation matrix is [ T(e1) T(e2) ]
i know that this method should work, just transforming the basis vectors to get the columns
rose trout
So I guess the projections were wrong then?
twin ridge
Entirely possible.
sage olive
but this should also work using the least squares approach
twin ridge
gross
sage olive
you're orthogonally projecting a vector in R^2 onto the subspace spanned by the vector (1,1)
twin ridge
Right.
agile elk
Does anyone know how to find A
sage olive
we know that A(A^TA)^-1A^T is the projection matrix, where A has the basis vectors of your subspace as the columns
sage olive
Does anyone know how to find A
!occupied
midnight plankBOT
Does anyone know how to find A
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
sage olive
we know that A(A^TA)^-1A^T is the projection matrix, where A has the basis vecto...
so i don't see why A(A^TA)^-1A^T isn't your answer where A = (1, 1)
twin ridge
Am I supposed to use A transpose here?
sage olive
Am I supposed to use A transpose here?
where?
twin ridge
Fair enough.
sage olive
so if A is the column vector (1,1)
then A^TA is (2), so its inverse is 1/2
so you have (1,1) (1/2) (1 1)
uh that becomes
(1/2, 1/2) (1 1) = (1/2 1/2, 1/2 1/2)
so using this approach i'm getting that your projection matrix is the 2x2 matrix with each entry 1/2
but that seems wrong?
twin ridge
The reason I considered all 0.5s is that if I take the point on the other side of my original x, then the projection would be the midpoint.
So if I have [1 0], the reflection of that would be [0 1], then the midpoint is the projection
sage olive
we know that (1,0) and (0,1) should map to (sqrt2, sqrt2)
twin ridge
Right.
And the same logic should apply to every vector
Since the reflected vector creates a line with a slope of -1.
So I know that every single vector I reflect will be orthogonal to y=x
The midpoint of the two tails, that is.
sage olive
we know that (1,0) and (0,1) should map to (sqrt2, sqrt2)
okay then my projection matrix filled with 1/2s can't be right since it maps (1,0) and (0,1) to (1/2, 1/2)
strange that that approach didn't work though
twin ridge
Right?
sage olive
why isn't the correct matrix the 2x2 matrix filled with sqrt2?
i feel like this is it
sage olive
we know that (1,0) and (0,1) should map to (sqrt2, sqrt2)
based off this
twin ridge
Take the vector [1 1]. Shouldn't it just map back to [1 1] since it's already projected onto y=x?
twin ridge
Weird.
rose trout
Yeah the projections are 1/2 everywhere, not sqrt(2)/2.
$proj_{(1,1)} (1,0) = \frac{1}{2} (1,1)$
grand pondBOT
Azyrashacorki
twin ridge
@misty gorge you seemed sad you missed the last convo. Want a stab at it?
sage olive
not that it's reliable but this tries normalizing the basis vector first
(1,1) works fine then but what about the standard basis vectors?
twin ridge
That looks like a bunch .5's to me
rose trout
The issue is when you computed T(e_1) in the first place you divided by the norm of (1,1), aka sqrt(2), and not its square, this gives the 1/2
Granted I did the same thing
misty gorge
sorry how much progress has been made exactly?
misty gorge
it's u dot vhat times vhat, where vhat is the unit vector pointing in a 45 degree angle from the positive x axis
then the matrix is recoverable from that
sage olive
the reason we kept thinking the standard basis vectors should be getting mapped to (sqrt2, sqrt2) is because we were assuming the closest vector on y = x would've been the projection straight up so we were drawing a 45-45 triangle
rose trout
it's u dot vhat times vhat, where vhat is the unit vector pointing in a 45 degre...
I read vhat as what with a german accent at first
sage olive
the reason we kept thinking the standard basis vectors should be getting mapped ...
but we need to be orthogonal to y = x, not the x axis
for the smallest distance to the line
twin ridge
I did the orthogonal drawing, I just did the numbers wrong.
My brain thought the length of the vector was the coordinates.
sage olive
so it's agreed that the standard basis vectors map to (1/2, 1/2)?
twin ridge
Yee.
misty gorge
so let vhat be (1/sqrt(2), 1/sqrt(2)), then you can take x = (a,b), and then you get (a/sqrt(2) + b/sqrt(2)) (1/sqrt(2),1/sqrt(2)) = ((a+b)/2,(a+b)/2)
sage olive
great, that was our problem then
twin ridge
Wait
sage olive
that should not have been that hard 💀
misty gorge
so let vhat be (1/sqrt(2), 1/sqrt(2)), then you can take x = (a,b), and then you...
so the matrix is $\begin{pmatrix} 1/2 & 1/2 \ 1/2 & 1/2 \end{pmatrix}$
grand pondBOT
smaytrix group
twin ridge
Ok yee
Now I just gotta do the other version.
The other version looks like pain.
Now I have to do a composition of projections. Rotate then shrink.
misty gorge
what are you rotating by
twin ridge
The angle would be that curved red arrow
Which I can apparently get using sines and cosines.
misty gorge
okay so what is the transformation specifically
twin ridge
The first one is rotate the vector until it is in line with y=x
misty gorge
okay, then what
twin ridge
Then shrink it until the tail gets to the intersection point of the altitude.
misty gorge
okay, and that's it?
twin ridge
Yeah.
It's just a little strange for me since I don't have a ton of experience with rotation matrices.
misty gorge
this should turn out to be the same matrix
twin ridge
I was given to rotate a specific amount of degrees, I want something like $\begin{bmatrix} cos\theta & -sin\theta\ sin\theta & cos\theta\ \end{bmatrix}$ or something like that.
twin ridge
And then multiply by a scalar to shrink it.
misty gorge
if you have a specific amount of degrees, then yes you do $\begin{bmatrix}\cos(x) & \sin(x) \ -\sin(x) & \cos(x)\end{bmatrix}$, where x is the angle
twin ridge
That's the actual matrix given in class.
twin ridge
How do I set it to rotate towards the y=x though?
twin ridge
Sometimes I have to rotate clockwise and sometimes counter
twin ridge
Rotations aren't a linear transformation?
midnight plankBOT
@twin ridge Has your question been resolved?
midnight plankBOT
midnight plankBOT
Available help channel!
trim oasis
We're given that f(x) = sin(ωx + φ), where ω > 0 and |φ| ≤ π/2.
The smallest positive period of f(x) is T.
r(x) is the derivative of f(x).
g(x) = f(x) + r(x)
α(x) is an odd function.
The maximum value of α(x) is √5.
We need to find f(T/8). (edited)
midnight plankBOT
We're given that f(x) = sin(ωx + φ), where ω > 0 and |φ| ≤ π/2.
The smallest pos...
@trim oasis Has your question been resolved?
drifting root
hi, can someone help me with what vector bc is?
midnight plankBOT
hi, can someone help me with what vector bc is?
@drifting root Has your question been resolved?
lusty pike
How should I start part d
midnight plankBOT
How should I start part d
lusty pike
Please can you just tell me the name of the technique needed not work me through it 🙂
The answer to b is 2 - 3i
fringe onyx
wdym the answer to b is 2 - 3i
it is asking for a diagram
@lusty pike
do you need help with a?
"dividing complex numbers"
midnight plankBOT
@lusty pike Has your question been resolved?
craggy yew
midnight plankBOT
craggy yew
the answer is 130
<@&286206848099549185> what is the solution to get this answer
pseudo owl
base change formula
Like log_4(b)=log_2(b)/2
log_2(a)=2log_4(a)
@craggy yew
You here?
Put this values in the first equation
I've got b^2-a^2=128
Now hit and trial
craggy yew
Put this values in the first equation
i did that before and subtracted the 2 equations 2 get b^2-a^2=128
pseudo owl
i did that before and subtracted the 2 equations 2 get b^2-a^2=128
Got the same
that is
pseudo owl
(b+a)(b-a)=128
Look at the factors of 128
128 1
64 2
and son on
Try all cases and find a,b
Its a diophantine eqn.
craggy yew
Try all cases and find a,b
no other way?
pseudo owl
no other way?
Yeah...
craggy yew
128 1
64 2
how will they help tho
pseudo owl
It is interesting to note that a,b are not neccessarily integers
craggy yew
?
pseudo owl
That is one of the possible approach
craggy yew
can u give an example by using the factors of 128?
pseudo owl
Yeah
(b-a)(b+a)=128*1
Say b-a=128
b+a=1
Which gives b=129/2
Which is a real number
a=63.5
ab=4095.75
craggy yew
(b-a)(b+a)=128*1
Say b-a=128
b+a=1
why
craggy yew
(im gonna ask alot of dumb questions cuz im new to problem solving)
craggy yew
(b-a)(b+a)=128*1
Say b-a=128
b+a=1
how did we get b-a=128
pseudo owl
how did we get b-a=128
Coz we have rewritten 128= 128*1
Now we compare both the equations
We can interchange the values of a,b
craggy yew
wdym
tough spindle
a and b may not be integers though
The question is not saying that they need to be integers, so trying to put b - a = 128 does not work.
pseudo owl
The question is not saying that they need to be integers, so trying to put b - a...
we do get non integral values of a,b satisfying that
see non integers can also add up to integers
Like 2.5+1.5=4
Its all about long case work tbf
tough spindle
Yeah but I doubt we need to make that assumption to find a and b.
pseudo owl
What else can we do?
grand pondBOT
L'empereur
tough spindle
Put $x = \log_2(a)$ and $y = \log_4(b)$. Since $b^2 - a^2 = 128$, replace $a$ and $b$ to get $2^{4y} - 2^{2x} = 128 = 2^7$. Now divide by $2^7$ on both sides to get $2^{4y - 7} - 2^{2x - 7} = 1$. Now what we could do is show that (assuming $2^{4y - 7}$ and $2^{2x - 7} $ are integers) that the only way powers of two could subtract to get an odd number is to have 2x - 7 = 0.
grand pondBOT
PowerUp
craggy yew
$b^2-a^2=(b+a)(b-a)$
how we get b-a=128 by doin that
craggy yew
Put $x = \log_2(a)$ and $y = \log_4(b)$. Since $b^2 - a^2 = 128$, replace $a$ an...
nice
pseudo owl
how we get b-a=128 by doin that
Leave it for now
Its how you solve diophantine eqns, but the process used by powerup is more simpler
I'll suggest you to focus on that.
craggy yew
Its how you solve diophantine eqns, but the process used by powerup is more simp...
can u still tell me about it (just in case i see a similar question in a future problem and cant find other methods)
i gtg now
powerup what do u think will be the next steps
tough spindle
Since $x = \frac{7}{2}$, we can find $a$ and $b$ by substituting $x$ into the equations above.
grand pondBOT
PowerUp
pseudo owl
can u still tell me about it (just in case i see a similar question in a future ...
Yes, sure.
I'll take an easy example
Say ab=10
Now we know that 10= 5x2
So we write ab=5x2
Now we compare both sides, can you see that a=5 and b=2 satisfy the eqn?
tough spindle
So after finding x, just calculate $a$ and $b$ (we have $x = \log_2(a), y = \log_4(b), b^2 - a^2 = 128$).
grand pondBOT
PowerUp
pseudo owl
You will get y=2
midnight plankBOT
@craggy yew Has your question been resolved?
wraith oriole
Why isn’t it T/0.01
midnight plankBOT
Why isn’t it T/0.01
latent wolf
To find the solutions to the inequality ( b^2 - 6400 > 0 ), we can follow these steps:
-
Rewrite the inequality: ( b^2 - 6400 > 0 ).
-
Factor the inequality: ( b^2 - 6400 = (b - 80)(b + 80) ).
-
Set each factor to zero to find the critical points:
[
b - 80 = 0 \quad \text{or} \quad b + 80 = 0
]
This gives ( b = 80 ) and ( b = -80 ). -
Determine the intervals to test: The critical points divide the number line into three intervals:
[
(-\infty, -80), \quad (-80, 80), \quad (80, \infty)
] -
Test a value in each interval to see if the inequality ( (b - 80)(b + 80) > 0 ) is satisfied:
- For ( b \in (-\infty, -80) ): Choose ( b = -100 )
[
(-100 - 80)(-100 + 80) = (-180)(-20) > 0
]
This interval satisfies the inequality. - For ( b \in (-80, 80) ): Choose ( b = 0 )
[
(0 - 80)(0 + 80) = (-80)(80) < 0
]
This interval does not satisfy the inequality. - For ( b \in (80, \infty) ): Choose ( b = 100 )
[
(100 - 80)(100 + 80) = (20)(180) > 0
]
This interval satisfies the inequality.
- For ( b \in (-\infty, -80) ): Choose ( b = -100 )
-
Combine the intervals where the inequality is satisfied:
[
b \in (-\infty, -80) \cup (80, \infty)
]
So, the solutions to the inequality ( b^2 - 6400 > 0 ) are ( b \in (-\infty, -80) \cup (80, \infty) ).
grand pondBOT
Shaf
wraith oriole
To find the solutions to the inequality \( b^2 - 6400 > 0 \), we can follow thes...
Bro what
Why isn’t it T/0.01
latent wolf
Why isn’t it T/0.01
look
multiply both sides by 100
100T = P - 40000
P = 100T+40000
there
latent wolf
Why isn’t it T/0.01
are u practicing for the sat tooo
midnight plankBOT
@wraith oriole Has your question been resolved?
bleak pier
midnight plankBOT
wraith oriole
are u practicing for the sat tooo
Yes
midnight plankBOT
@bleak pier Has your question been resolved?
midnight plankBOT
@bleak pier Has your question been resolved?
formal igloo
i need help
midnight plankBOT
i need help
potent veldt
formal igloo
lone sorrel
which questions do u need help with?
formal igloo
all
dapper cairn
😭😭💀
formal igloo
<@587889029920456714> lock in bro
mb
formal igloo
do you know the formula for the slope of a line?
rise over run
formal igloo
idek gang
abstract glen
!status
midnight plankBOT
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
1
abstract glen
have you learnt this yet
midnight plankBOT
@formal igloo Has your question been resolved?
formal igloo
help
dense harness
whats the problem
midnight plankBOT
formal igloo
.reopen
chrome wren
this integral can be calculated in 2 parts right?
midnight plankBOT
this integral can be calculated in 2 parts right?
@chrome wren Has your question been resolved?
fringe onyx
this integral can be calculated in 2 parts right?
Yes
bitter wharf
I think I'm going psycho with this
midnight plankBOT
I think I'm going psycho with this
fringe onyx
How are you doing it?