i thought no correct answers but that was marked wrong
#help-28
torn jolt
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i thought no correct answers but that was marked wrong
torn jolt
i don’t need to be walked through it, can someone tell me if i’m wrong or the system is?
shrewd hamlet
torn jolt
ok in that case idk what i’m doing wrong. bc i checked everything twice AND with a calculator and got the same answers every time
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slate night
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gritty rose
for 15)
You can only use partial fractions when the power of the bottom is greater than the power of the top polynomial
5 > 3
slate night
You can only use partial fractions when the power of the bottom is greater than ...
ah ok
slate night
You can only use partial fractions when the power of the bottom is greater than ...
if u can, could you explain why?
gritty rose
Your resulting sum gets the wrong power in the numerator
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gritty rose
You can only use partial fractions when the power of the bottom is greater than ...
Greater than or equal to
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slim sleet
no clue what this means
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no clue what this means
static bramble
,rccw
glossy valveBOT
sleek apex
go to sleep then
sleek apex
wake up early
serene hazel
won’t have time
you can only finish this by 5 mins instead of playing overwatch for 3 hours
I'd agree with arctrices
wake up early
slim sleet
my overwatch is running in the background on my of
pc
i’m not even on my pc lol
i decided to skip the question
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split otter
part b?
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solid yew
Is there a literature that uses exactly “m%n=y” in a set-builder notation like this ➡️ {x ∈ Z | x % 2 = 0}
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muted vigil
just wanted help to see if my plan would work: going to do a triple integral, and I found the equation of the plane as y=6-2z-2x
muted vigil
balmy rose
the equation looks good
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restive hawk
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restive hawk
can anyone help me on this
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lunar quarry
hi
lunar quarry
and it is colored with 3 colors
how to prove that there must be four separate square in the same color that forms a rectangle
the process i have now is that
by Pigeonholing there must be 4 same color in the 1 row(length of 10)
in the rest 12 rows
we can only consider the line following the first 4 same colors square
and every row of four square must have one color repeated
so either that color is same as the color of the first row or not
in the second case
if the color is not the one in the first row, then we will have two choice of color
and by C(4,2), we will have six choice of position
which is totally 12 possibile case
but we only have 12 row now
so that won't be able to mach the Pigeonhole?
<@&286206848099549185> need some help pls
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round plank
b33 4.5
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cold fern
can someone help
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can someone help
neon fulcrum
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cold fern
Would that be my equation
or would i have to draw out a line like this
opps sorry i was looking on the wrong equatioon
cold fern
yes sorry
It’s wrong
Did it this way also and it’s wrong
i guess ill just figure it out tomorroe
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bitter beacon
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buoyant pewter
how to do this
you can here, for your own exercise, calculate the nth partial sum of this series, and then show that this sum tends to infinity, so the series will diverge based on the definition
buoyant pewter
how to do this
$\lim_{n \to \infty } \sum_{k=1}^{n}\ln\left( 1+\frac{1}{k} \right)=+\infty $
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stable nest
bitter beacon
i don't know how to find n th partial sum but i rewrote the ln as ln(n+1)-ln(n) and concluded that since both diverges then ln(n+1)-ln(n) also diverges
bitter beacon
no you cant
why?
buoyant pewter
such trick only works for convergent servies
bitter beacon
hmm calculus is intuition based science
buoyant pewter
listen
that is basic theorem
you are not allowed to substract divergent series
$S_{n}=\sum_{k=1}^{n}ln\left( 1+\frac{1}{k} \right)$
glossy valveBOT
Joanna Angel
buoyant pewter
that is n-th partial sum
of yoru series
and you should evaluate its defintion
we call it as investigaitng series on defintion
on the ba of defintion
i show you one lione as a hint:
$S_{n}=\sum_{k=1}^{n}ln\left( 1+\frac{1}{k} \right)=\sum_{k=1}^{n}\left[ ln\left( k+1 \right)-lnk \right]$
glossy valveBOT
Joanna Angel
bitter beacon
limit as n goes to inf?
buoyant pewter
yes ofc
buoyant pewter
no
it is positive
look
$S_{n}=\sum_{k=1}^{n}ln\left( 1+\frac{1}{k} \right)=\sum_{k=1}^{n}\left[ ln\left( k+1 \right)-lnk \right]=\\\left[\left( ln2-ln1 \right)+\left( ln3-ln2 \right)+\left( ln4-ln3 \right)+_{\cdots }+\left( lnn-ln\left( n-1 \right)+\left( ln\left( n+1 \right)-lnn \right) \right) \right]=\\ln\left( n+1 \right)\to +\infty $
today bot has issues lol
$=\sum_{k=1}^{n}\left[ ln\left( k+1 \right)-lnk \right]=\\\left[\left( ln2-ln1 \right)+\left( ln3-ln2 \right)+\left( ln4-ln3 \right)+...+\left( lnn-ln\left( n-1 \right)+\left( ln\left( n+1 \right)-lnn \right) \right) \right]=\\ln\left( n+1 \right)\to +\infty $
$=\sum_{k=1}^{n}\left[ ln\left( k+1 \right)-lnk \right]$
glossy valveBOT
Joanna Angel
buoyant pewter
$\left[\left( ln2-ln1 \right)+\left( ln3-ln2 \right)+\left( ln4-ln3 \right)+...+\left( lnn-ln\left( n-1 \right)+\left( ln\left( n+1 \right)-lnn \right) \right) \right]$
glossy valveBOT
Joanna Angel
bitter beacon
ok
buoyant pewter
it is the elementary solution
bitter beacon
so to be clear:
if two a and b series diverges we can not say anything about a+b or a-b or a/b or a*b ?
buoyant pewter
right
buoyant pewter
if both are convergenet then you can make any arithemtic operations on them
you may add them
multiplicaiton is litle bit diffeertn
since
product serieses
requires
specicif formual
fo cauchy
but additiona and subtraction and taking constant out fo series are ok
bitter beacon
ok and it's not relevant but i have this question
if limit of two functions f and g dnt than what can we say about f-g f+g f*g f/g
dne
buoyant pewter
if they are finite
then all your operations are alowed
but
if they are infinite or
bitter beacon
if they are finite
funite means limit exists?
bitter beacon
if they are infinite or
in this case?
buoyant pewter
finite
infinite is problem
in such ase
lim f(x) = + infi, lim g(x) = +iinf, and you can t use
lim f(x) - g(x)
there are such 7 symbols
i write them
bitter beacon
lim f(x) - g(x)
so we can't use means it might be finite or dne?
buoyant pewter
$\left[ \frac{0}{0} \right],\left[ \frac{\infty }{\infty } \right],\left[ 0\cdot \infty \right],\left[ +\infty -\infty \right],\left[ 1^{\infty } \right],\left[ \infty ^{0} \right],\left[ 0^{0} \right]$
glossy valveBOT
Joanna Angel
buoyant pewter
if such situaion appears
you can't tell result at once
you need to use variours methods
to obtain result
buoyant pewter
yw 🙂
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digital plover
how would i find this?
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how would i find this?
glacial pasture
find the perpendicular bisector of the line joining them, thats the general method
digital plover
i'm supposed to use the distance formula but i dk how to apply it
the mid point is (0, 1) so i dont get why the answer is only b and not a
glacial pasture
x=0 is a vertical line, points on it can easily be closer to one than the other
as for using the distance formula, one momento
$$\sqrt{(x-0)^2+(y+2)^2}=\sqrt{(x-0)^2+(y-4)^2}$$
$$x^2+(y+2)^2=x^2+(y-4)^2$$
$$y^2+4y+4=y^2-8y+16$$
$$12y=12$$
$$y=1$$
glossy valveBOT
AℤØ
glacial pasture
is how that would work
glacial pasture
just subtracted x^2 from each side
glacial pasture
nw
digital plover
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boreal zealot
Can someone help me understand why if Ax = 0 has more than one solution then A is not invertible?
torn jolt
well if it had a non-trivial solution that would imply that the columns aren't linearly independent
rare dock
if A is invertible then Ax = 0 must have one solution, x = A^(-1)*0
boreal zealot
I guess a better question would be why is a matrix not invertible if one of its columns aren't independent
rare dock
just gonna add that's a more complicated fact than needed for your original question
boreal zealot
I know it's because the determinant would be 0 but we still haven't reached determinants so I guess that's why I still don't understand(?)
rare dock
determinants don't really demystify that anyway
torn jolt
[
Ax = 0 \implies A^{-1}Ax = A^{-1}0 \implies x = 0
]
glossy valveBOT
torn jolt
this is the textbook proof usually if that's what you are wondering
supposing that A is invertible
idle kelp
You should probably just think about what it means for something to be an inverse. If you have many different vectors going to 0, starting from 0 you can't tell what the original input is.
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rancid flame
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rancid flame
sorry for the quality. i cant understand where the 30-(p+6) comes from and the reasoning behind it
i get it that p is greater or equals to 10 or p is greater or equals to 24
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rancid flame
anyone? 🙂
open igloo
sorry for the quality. i cant understand where the 30-(p+6) comes from and the r...
There's a property
$^nC_r = ^{n}C_{n-r}$
glossy valveBOT
Lorentz
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last locust
in convoluting these two functions, does it matter which of the two use as the unit step function and which of them will be the other one>
last locust
Like can I just arbitarily choose g(t-τ) to ge te^-t or f(t)?
last locust
ok thanks
but actually, if I chose the step function to be the exponential function, the solution would look different. But im guessing it would still be the same as if I chose f(t) to be the step function
because taking the integral would yield f(0)-f(t) if I chose f(t) to be the unit step unction, and it would be something like t*e^-(t-tau) or sth if I chose the exponential one to be the step function
by step function I mean g(t-tau)
kindred grove
because taking the integral would yield f(0)-f(t) if I chose f(t) to be the unit...
not even sure what you're saying here
last locust
According to the convolution property we're rewriting one of the functions (lets say f(t)) to f(tau), and the other one (lets say g(t)) to g(t-tau) right?
and then taking the integral of them from 0 to t with respect to tau
kindred grove
ok
last locust
Im asking if what I am saying is correct?
last locust
I just don't see where your f(0)-f(t) comes from for example
before i get to that, let me ask you
last locust
According to the convolution property we're rewriting one of the functions (lets...
so If i want to convolute sin(t) and cos(t), I can either do the integral from 0 to t of sin(tau)(cos(t-tau)dtau or I can do cos(tau)sin(t-tau)dtau and it would be the same?
kindred grove
yes
last locust
ok thats it then, so if i have the integral from 0 to t of f(t-tau), don't I have f(t-t)-f(t-0) which is f(0)-f(t)?
kindred grove
why do you think you can split your integral like that
$\int f(x)g(x) \dd{x} \neq \left(\int f(x)\dd{x} \right)\left(\int g(x)\dd{x} \right)$ in general
glossy valveBOT
aPlatypus
kindred grove
they're completely not
take f(x)=g(x)=x
even for ultra simple functions it's already wrong
last locust
hm
Okay so what I am getting from this is that i cant split the integral when two functions are multiplied
kindred grove
yes
last locust
so this cannot be further simplitfied
kindred grove
if you don't know more things about f, there's not much you can do
hazy hollow
Can someone please help me verify the answer to this equation
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placid needle
can anyone guide me with steps on how to solve this?
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can anyone guide me with steps on how to solve this?
hot herald
start with a pythagorean trig identity
dusky locust
1+tan^2v = ?
open igloo
Ye
placid needle
and then what's next?
hot herald
recognise/recall a common trig integral/derivtive
open igloo
$\int 4{\sec^2}x$ dx?
glossy valveBOT
Lorentz
placid needle
4 tanx + c?
hot herald
yeh
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@torn jolt Has your question been resolved?
keen harness
anyone know to solive this lmao
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anyone know to solive this lmao
keen harness
help
quaint prawn
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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.
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spiral vigil
are you trying to find z?
keen harness
spiral vigil
what happened to the original question
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limber flicker
yoyoyo
limber flicker
I have completely forgotten how to test if certain things cancel due to symmetries, how would I go about doing that ?
nvm
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languid cradle
What is the difference between tangential speed and angular speed
viral jasper
one depends on the radius, the other does not
umbral dome
imagine you and your friend are running around a pair of circular tracks, one inside the other. you and your friend start and finish at the same time. that means that you and your friend cover the same number of full cycles in the same time, meaning you have the same angular speed. (sometimes we split a full cycle into angles but it's the same idea)
however, you're on the outside track, which is longer, so in order to have the same angular speed, you have to run a longer distance in the same time, so you have a larger tangential speed (which is the same classic meters per second speed you're used to)
umbral dome
Is this like radius per second
angular speed is usually measured in cycles per second, or we can split one full cycle into 360 degrees or 2π radians, and also use degrees per second or radians per second
where one cycle is getting all the way back to where you started, a full circle
languid cradle
imagine you and your friend are running around a pair of circular tracks, one in...
Awsomness
And thank you
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worthy star
pls help
marsh rose
torn jolt
Please open another channel
torn jolt
What are the coordinates of the vertex of the parabola?
worthy star
(4,2)
torn jolt
And is it an upwards or a downwards parabola?
worthy star
opens up
torn jolt
Oh yes I see now that you already have drawn those points
worthy star
Yeah 👍
my friend tried the equation below it but the y int came out to be 10
in vertex for@
torn jolt
So the equation is indeed in the form y=a(x-4)^2+2
Now you have to determine a right?
And you know that the point (0, 7) lies on the parabola
worthy star
Now you have to determine a right?
Yeah
worthy star
And you know that the point (0, 7) lies on the parabola
ohhh plug that into y and x?
torn jolt
Exactly
worthy star
kk let me try this
i can’t solve this cause school just ended 💀
but how do i close this channel
torn jolt
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iron sand
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iron sand
How to solve this limit without lhospital
I did this but I must have done something wrong because I got stuck
(╯°□°)╯︵ ┻━┻
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iron sand
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iron sand
<@&286206848099549185>
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lament garnet
ummmmmm
this is rather confusing for multiple reasons to me
first the main reason is that why couldn't we just say that the entire limit is 1
because 1^x = 1
sin(0) is 0
cos(0) is 1
but there is one more reason for confusion in my opinion
lament garnet
I did this but I must have done something wrong because I got stuck
ummmm that second step you did
third equality
lament garnet
I did this but I must have done something wrong because I got stuck
$ln(x) \neq x - 1$
glossy valveBOT
Bishop
lament garnet
$ln(x) \leq x - 1$
glossy valveBOT
Bishop
lament garnet
I'm unsure of the viability of this method when it comes to combination of functions like f(x)/g(x) , but i will assume it works
iron sand
$ln(x) \neq x - 1$
i used this lim x->0 ln(x+1)/x=1
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lament garnet
i used this lim x->0 ln(x+1)/x=1
again i don't get what you're trying to achieve
iron sand
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iron sand
i am trying to get rid of ln
lament garnet
why go down the route of some strange trig manipulations that dont even give much considering that you have cos(sin(x^4)) chained functions
but you have cos(sin(x^4)) - 1 times 1/cos^3(x^4) - 1
oh nvm
wait but why even do all that in the first place
lament garnet
is that the entire limit exists:
i agree that we cant assure that the limit of say 3pi/ln(cos^3(x^4)) exists
but we assume that the entire limit exists
that limit will be equal to some c
so multiply it by the limit of ln(cos^3(x^4))
this limit exists i believe
doing this we get cd = 0
iron sand
but we assume that the entire limit exists
well the entire limit is 0/0 isnt it?
so we need to modify it
lament garnet
but if lim f(x) = c
then limit exists
?????????????????????????
so if lim f(x) is 0/0 then it doesn't exist
thus no modifying f(x) will make lim f(x) exist
so why are you assuming that lim f(x) is 0/0
when it might not be
it just appears to be if you plug in x = 0
am i right?
iron sand
it just appears to be if you plug in x = 0
ah yes i just did this
lament garnet
yea but that doesnt imply that lim f(x) is 0/0
its just that the method is not working in that particular case
lament garnet
ah yes i just did this
but as i said
we assume that lim f(x) exists
then use product rule of limits
we choose our function to be ln(cos^3(x^4)) whose limit is 0
oh wait
idle kelp
This conversation is going all over the place and I don't really want to read what has been already been said.
idle kelp
Your first step is fine to rewrite the limit using e^ln(cos(sin(x^4))) etc
idle kelp
After that use the small angle approximation on sin(x^4) to replace it with x^4 and then use the small angle approximation on cos(x^4) and cos^3(x^4)) to get cancellation. That pretty much completes the problem.
lament garnet
or alternatively
lament garnet
I did this but I must have done something wrong because I got stuck
on third equality the one with cos(sin(x^4)) -1 / cos^3x^4 - 1 you multiply by sin(cos(x))/sin(cos(x)) aka by 1 because then sin(cos(x)) cos^3(x^4) - sin(cos(x)) will go to sin(cos(x)) and on the top you will have 0 so the entire limit goes to 0
you will have 0 because
cos(sin(0)) is 1
sin(cos(x)) * 1 = sin(cos(x))
$sin(cos(x))( cos(sin(x^4)) - 1 ) \to 0$
glossy valveBOT
Bishop
lament garnet
you are multiplying by 1 which lim exists
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idle kelp
You are writing nonsense. Near 0 $\cos(\theta) \approx 1 - \frac{\theta^2}{2} + \mathcal{O}(\theta^4)$ and $\sin(\theta)\approx \theta + \mathcal{O}(\theta^3)$. Hence
$\lim_{x\rightarrow 0} \frac{\cos(\sin(x^4))-1}{\cos^3(x^4)-1} = \lim_{x\rightarrow 0} \frac{\cos(x^4)-1}{(1-\frac{x^8}{2})^{3}-1} = \lim_{x\rightarrow 0} \frac{-\frac{x^8}{2}}{(1-\frac{x^8}{2})^{3}-1} = \lim_{x\rightarrow 0} \frac{-\frac{x^8}{2}}{-\frac{x^{24}}{8} + \frac{3x^{16}}{4} - \frac{3x^8}{2}} = 1/3.$ It is certainly not 0.
glossy valveBOT
JessicaK
void marsh
this is ohysics but can someone help
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this is ohysics but can someone help
primal condor
So vector sum of forces? Does it not work out correctly?
lament garnet
So vector sum of forces? Does it not work out correctly?
person C is cutting the tree down
but they didn't say what kind of force that applies
there is a force of action of his tool acting on the tree probably?
primal condor
So, I guess that's just colour to the problem.
If there is no way to include an aspect, it may not be intended to be included.
The numbers seem to be bearings. There is also no mention of the height, eg the inclination of the ropes.
So, a well-worded problem? Probably not.
lament garnet
You are writing nonsense. Near 0 $\cos(\theta) \approx 1 - \frac{\theta^2}{2} +...
so anyway i don't understand what any of that even is xD But i can provide more clarity for my reasoning
we assume it exist , so it equals to c
i should say I obtain*
but you get the idea
so the question is
maybe about the notation
im tired so i treated cos^3(x^4) as cos(x^4)cos(x^4)cos(x^4)
not sure if im supposed to treat it as cos(cos(cos...
in that case its most likely different
lament garnet
im tired so i treated cos^3(x^4) as cos(x^4)cos(x^4)cos(x^4)
thats how i treated it*
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lament garnet
You are writing nonsense. Near 0 $\cos(\theta) \approx 1 - \frac{\theta^2}{2} +...
Oh my nvm what i wrote earlier
You are obviously right
lament garnet
https://cdn.discordapp.com/attachments/839950625269088276/1182113929858662480/42...
this step is where i can't plug values because i get 0 in the denominator
void marsh
So, a well-worded problem? Probably not.
How do i solve it then cuz im@so confused
lament garnet
How do i solve it then cuz im@so confused
Doesn't a force of 80 + 90 = 170N work out?
void marsh
Doesn't a force of 80 + 90 = 170N work out?
what abt the angles tho wouldnt they affect it
lament garnet
what abt the angles tho wouldnt they affect it
Say 45* angle
Away from what exactly?
lament garnet
what abt the angles tho wouldnt they affect it
Maybe it's meant to be like
45 angles away in an anti clockwise manner
Rather than clockwise
It's hard to imagine clockwise to me
because then the person would be pulling through the tree imo
Well in this 2d scenario they would pull anti clockwise
lament garnet
what abt the angles tho wouldnt they affect it
I would assume that
That its anti clockwise only for drawing/imagination purposes
But mathematically it wouldnt matter
It could be a unit vector in either direction since its imaginary 2d scenario
But here is the thing
lament garnet
what abt the angles tho wouldnt they affect it
The magnitude of the resultant is basically what is meant by the net force right
Sry im very sleepy now so it takes a while
lament garnet
what abt the angles tho wouldnt they affect it
but i think i found out a satisfying answer
void marsh
Sry im very sleepy now so it takes a while
All good
void marsh
but i think i found out a satisfying answer
Whatd you get?
lament garnet
Whatd you get?
So its kinda hard to explain in my condition now but uhh
Basically
You need some normalized vectors
Unit vectors
So like of magnitude that is unital
To use them to set up an coordinate system
To be able to describe your two vectors
One inclined at an angle 45* degrees away from the axis in clockwise or anticlockwise direction doesnt matter
and the second onr inclined at 345* angle
Having their coordinates you can then find the coordinates of the resultant vector by adding the coordinates of the respective vectors up
this lets you find the magnitude of the resultant
If that is too bad idea then just figure out something else , but the main idea stays that you need to get the magnitude of the resultant vector
That magnitude is then what i believe would be a satisfying answer here if we include angles , because thats the influence of direction on vectors
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torn jolt
Can someone show me equation?
subtle prairie
if that is a parallelogram then u can set the parallel sides equal to each other and then use a system of equations
but thats only if the top and bottom are equal and the left line and right line are equal
so you would get y = 4x - 1 and y + 3 = 5x - 1
from there u could move the 3 over to isolate y and set both equations equal to each other and try to solve for x
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torn jolt
How do I simplify $\frac{\frac{3}{2+h}-\frac{3}{2}}{h}$
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How do I simplify $\frac{\frac{3}{2+h}-\frac{3}{2}}{h}$
glossy valveBOT
Matt
torn jolt
i dont think you can subtract the fractions on top because the denominator isnt the same so im not sure what to do
buoyant pewter
expand the given fraction by an expression that eliminates the "tiny" denominators
that means
multiply ,nominator and denomiantor by the same expresison like:
2(2 + h)
torn jolt
2(2 + h)
so multiply 2+h by 2 and 2 by 2+h?
so itll look smth like $\frac{3}{2(2+h)}-\frac{3}{(2+h)2}$
glossy valveBOT
Matt
torn jolt
Ok
buoyant pewter
Ok
$\frac{\frac{3}{2+h}-\frac{3}{2}}{h}=\frac{6-3\left( 2+h \right)}{2h\left( h+2 \right)}$
glossy valveBOT
Joanna Angel
buoyant pewter
by 2(2 +h)
generally memorise this:
$\frac{\frac{a}{b}-\frac{c}{d}}{h}=\frac{ad-bc}{h\cdot bd}$
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Joanna Angel
buoyant pewter
that is elemtary maths
buoyant pewter
yw
dim crystal
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dim crystal
needing some help on where to start
primal condor
What's "correlation coeficient"?
white geyser
the angle between vectors is sometimes called that.
dim crystal
basically where direct of the variables are going correct?
primal condor
the angle between vectors is sometimes called that.
Ok, I'll leave it to you to explain.
I know. Suggestion: Ask the OP what they know, so you know what level to speak at?
dim crystal
ooop
primal condor
Well highly correlated would mean the values of x and y agree well, so the scatter is narrow.
white geyser
so for a correlation coefficient of 1, we would get x = y more or less.
primal condor
Either positive of negative, of course.
white geyser
-1 would be x = -y
dim crystal
right
white geyser
we don't have to be exact here
which one of these scatterplots seems to follow a linear relation in the positive direction the most?
dim crystal
the last one
white geyser
good, so that one would definitely have the highest positive correlation coefficient, which is?
dim crystal
0.76?
white geyser
yes
dim crystal
that one makes sense now
rare dock
another kuromi pfp
dim crystal
i was thinking the last one but i thought i was confused
white geyser
now what about the opposite.? y = -x
dim crystal
another kuromi pfp
urs is cuter 🥹
rare dock
urs is cuter 🥹
aw haha i like the evil look in yours
dim crystal
now what about the opposite.? y = -x
the 3rd option for sure
white geyser
good job, so what do you think the coefficient for that one is?
dim crystal
hmm..
rare dock
hello riemann
dim crystal
that’s where i have difficult time figuring out the numbers that correlate
white geyser
alright well that blue line would represent coefficient of -1
so whats our closest option to -1?
my god i cant type today
dim crystal
-0.20
white geyser
think about the number line. closest to -1
white geyser
so the third graph is definitely -0.94
dim crystal
great that also made sense
white geyser
the last two are not so easy to recognize. but if we look at the second one a bit, do we see any pattern?
dim crystal
they’re kinda formed in the middle
white geyser
the second graph?
white geyser
dim crystal
yes
dim crystal
right
dim crystal
y=-x?
white geyser
right, they resemble y=-x, so the correlation is also negative
dim crystal
so would it be the same concept of finding something closer to -1
white geyser
and if we look at the first one real quick, we definitely don't see much correlation at all
so we know the first one will be close to 0, either negative or positive, and the second will definitely be negative
so what are our options?
-0.2, and -0.53 right?
dim crystal
right
white geyser
so which one do you think the second plot is
dim crystal
i think -0.94
white geyser
we already used that value
white geyser
why did you choose -0.20 over -0.53 for the second graph?
white geyser
good
does what we did kind of make sense?
we are looking for a general pattern in the points
white geyser
dim crystal
i will be saving that😂👍🏼
white geyser
dim crystal
need help on where to start
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static bramble
you start and end by computing, there aren't really steps here
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odd canyon
I need help with this
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odd canyon
Idk what im doing
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gilded tapir
so in a club there are 320 seats they make lines in each line there are same amount of seats the club owner added 4 seats to each line and he also added 1 more line after that there where 420 seats in total
gilded tapir
I need to find how mani line there are now
With a equation
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cursive dock
so in a club there are 320 seats they make lines in each line there are same am...
let the number of seats in a row be x and start with y rows. then you have xy=320.
what changes to get =420?
proven jay
why is the complex logarithm not analytic in the negative real axis ?
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tough marsh
Definitely doesn't need L'hôpital
Try to make the division go inside the square root.
After simplification, you can just plug that in directly.
buoyant pewter
$\lim_{x \to -\infty }\frac{\sqrt{x^{2}+1}}{x}=\lim_{x \to -\infty }\frac{\sqrt{x^{2}}\cdot \sqrt{1+\frac{1}{x^{2}}}}{x}=\\\lim_{x \to -\infty }\frac{\left| x \right|\cdot \sqrt{1+\frac{1}{x^{2}}}}{x}=\lim_{x \to -\infty }\frac{-x\sqrt{1+\frac{1}{x^{2}}}}{x}=\\=\lim_{x \to -\infty }\left( -\sqrt{1+\frac{1}{x^{2}}} \right)=-1$
glossy valveBOT
Joanna Angel
buoyant pewter
the difference would be if you compute limnit as x -> + infinity, then it wud be 1
tough marsh
Well, inf/inf is an indeterminate form.
buoyant pewter
Remeber that: $\forall _{x\in \mathbb{R}}\text{ }\sqrt{x^{2}}=\left| x \right|$
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Joanna Angel
tough marsh
I mean, you can just use:
$\lim_{x \to -\infty }\frac{\sqrt{x^{2}+1}}{x}=\lim_{x \to \infty }\frac{\sqrt{x^{2}+1}}{-x}=-\lim_{x \to \infty }\sqrt{\frac{x^{2}+1}{x^2}}$
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Xwtek
buoyant pewter
<@1035542975503138836>?
it is also important to write :$\frac{1}{x^{2}}$
glossy valveBOT
Joanna Angel
buoyant pewter
inside the radical form
pseudo cape
inside the radical form
But in what I wrote, we wouldn't have that at all
buoyant pewter
yes
lament garnet
Definitely doesn't need L'hôpital
oh right its trivial mb, eitherway works so doesn't matter much
pseudo cape
You shouldn't write your third and fourth line
Just cancel
You get x
That goes to infinity
buoyant pewter
It wouldn't be necessary either for how I did it
the teachers at schools want to see very elementary method, hence i showed it in a such way
pseudo cape
the teachers at schools want to see very elementary method, hence i showed it in...
Alright, yep
lament garnet
just sqrt(x^-2) = x^-1 thus it yields sqrt((x^2 + 1)/x^2) which is clearly sqrt(x^-2 + 1)
pseudo cape
just sqrt(x^-2) = x^-1 thus it yields sqrt((x^2 + 1)/x^2) which is clearly sqrt(...
Where do we have a sqrt(x^-2)?
lament garnet
x^-1 is outside
pseudo cape
Ah
smoky wing
\begin{align*}\rainbow*{
\lt x{-\8} \df{\s{x^2+1}}x &= -\lt x{-\8} \s{\df{x^2+1}{x^2}}\\
&= - \s{ \lt x{-\8} \df{x^2+1}{x^2} } \\
&= -\s{ \lt x{-\8} \df{1 + \f1{x^2}}{1} }
}
\end{align*}
amazing
tough marsh
I feel like this question is given before the student learns L'Hôpital, though. So, I think we should avoid it first. People don't learn derivative before they mastered limits.
lament garnet
I feel like this question is given before the student learns L'Hôpital, though....
Alright if you say so
pseudo cape
sqrt(x^-2) = (x^-2)^(1/2)
[\lim_{x \to -\infty} \frac{\sqrt{x^2 + 1}}{x} = \lim_{x \to -\infty} \frac{\sqrt{x^2 + 1}}{-\sqrt{x^2}} = \lim_{x \to -\infty} -\sqrt{\frac{x^2 + 1}{x^2}}] is what you probably mean
Right?
Similar to what Joanna had
lament garnet
\[\lim_{x \to -\infty} \frac{\sqrt{x^2 + 1}}{x} = \lim_{x \to -\infty} \frac{\sq...
yes
pseudo cape
Yep. But we need -sqrt(x^-2) in that case
pseudo cape
Yes
lament garnet
sqrt doesnt matter (if insides converge)
pseudo cape
I mean this step [\lim_{x \to -\infty} \frac{\sqrt{x^2 + 1}}{x} = \lim_{x \to -\infty} \frac{\sqrt{x^2 + 1}}{-\sqrt{x^2}}]
x goes to negative infinity
sqrt(x^2) = |x| goes to positive infinity
So we need **-**sqrt(x^2)
lament garnet
ok
smoky wing
im back
pseudo cape
That gives us -1 at the end, yep
smoky wing
\rainbow*{\begin{align*}
\lt x{-\8} \df{\s{x^2+1}}x &= -\lt x{-\8} \s{\df{x^2+1}{x^2}}\\
&= - \s{ \lt x{-\8} \df{x^2+1}{x^2} } \\
&= -\s{ \lt x{-\8} \df{1 + \f1{x^2}}{1} }
\end{align*}}
Pure
pseudo cape
```tex
\rainbow*{\begin{align*}
\lt x{-\8} \df{\s{x^2+1}}x &= -\lt x{-\8} \s{\df...
Yeah exactly, this is basically
pseudo cape
this
smoky wing
buoyant pewter
**Pure**
it is correct, but it requires the assumption of continuity of the radical function, to make a step to put limes inside the radical, if continuity has not been shown then no, but if so it is also great
smoky wing
of course
tough marsh
I mean, you should have been given that already.
smoky wing
Yes, I'd assume bob420 has already seen\proved continuity of sqrt(x)
somber eagle
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somber eagle
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somber eagle
where did the 1- come from?
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torn jolt
where did the 1- come from?
It looks like they just did long division or something similar to get that
viscid imp
It looks like they just did long division or something similar to get that
long division feels overkill here
rewrite the numerator as (u^2+2)-2
and it falls out immediately
torn jolt
or that
viscid imp
my favourite trick when dealing with algebraic fractions hehehe
torn jolt
leave it to me to swat another fly using a sledgehammer
somber eagle
ahhh ok
damn how am i gonna see this in an exam
cant i do integration by parts here
viscid imp
damn how am i gonna see this in an exam
if you practice it for a bit, adding and subtracting stuff in the numerator to get an integer and a simpler fraction becomes very standard
somber eagle
alright
viscid imp
gonna be honest ive never actually done a single algebraic long division
i always just bash it with this thing, or compare coefficients
viscid imp
:))
lament garnet
where did the 1- come from?
$\frac{2}{u^2 + 2} - 1 = \frac{2}{u^2 + 2} - \frac{u^2 + 2}{u^2 + 2} = \frac{2 - u^2 + 2}{u^2 + 2}$ (motivation ??? You have two in the denominator , 2/(u^2 + 2) is trivial integration, because of 2 cancelling out )
glossy valveBOT
Bishop
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somber eagle
thanks alot!!
somber eagle
no prob man
lament garnet
But you get the idea
pearl trout
compact briar
or diagonal
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wispy mauve
can someone tell me what the formal name for this method is?
I cant find any source explaining it
essentially we take a trigonometric series and convert it to euler's form and sum it from there
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warm grotto
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warm grotto
How do I set up this integral?
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1. I don't know where to begin.
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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.
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1
torn jolt
this is multivar?
warm grotto
yes
torn jolt
I haven’t worked with multivar in a while but it looks like you will want to use polar coordinates
for instance
how can I modify z = 4 - x^2 - y^2
warm grotto
you're sure you have to convert it?
torn jolt
you don’t have to, but it makes the integral easier
Some good stuff to know
$x^2 + y^2 = r^2 \newline x = r\cos{\theta}\newline y = r\sin{\theta}$
glossy valveBOT
pulse
torn jolt
you could also write that as z = 4 - r^2
torn jolt
now try that with the y = sqrt(4 - x^2)
you’ll notice a catch with it if you do it right
warm grotto
r = -2, 2
torn jolt
nice
warm grotto
so does that give me the bounds of r?
-2 to 2?
no, I don't think that makes sense
scratch that lol
torn jolt
that 2 should be your upper r bound if I’m not mistaken (i’m flipping through my old notes rn lol)
warm grotto
ok
sacred zodiac
You'll get r goes from 0 to 2 (you can't have a negative r!)
torn jolt
with these i used to try to visualize the shape itself
and whether you are rotating a full circle or just half a circle or something else
sacred zodiac
what is the area in the xy plane bound by y = 0 and y = √(4-x^2)?
You're totally right its the top half circle
y = √(4-x^2) is already just the top half circle, and y = 0 is the x axis
So that gives you your theta bounds!
torn jolt
^
torn jolt
Lol all g
torn jolt
y = √(4-x^2) is already just the top half circle, and y = 0 is the x axis
@warm grotto does this part make sense
torn jolt
just to make sure youre on the right page which bounds do u have for theta?
torn jolt
nahnah
ok so
i’ll draw this out
When you graph the y equation, notice how we are covering only the upper half of the circle
warm grotto
oh right
torn jolt
if we traverse half of the circle in polar, we are traveling from 0 to pi
if we travel the full circle it’s 0 to 2pi
warm grotto
that's right, sorry, I don't know what I was thinking
torn jolt
all g
torn jolt
almost there
ancient sandal
Hey is this application of integrals jus wondering?
warm grotto
which is 2 pi
torn jolt
dont forget z bound
ancient sandal
I finished integrals today and am starting application of integrals.
warm grotto
lmao
torn jolt
REAL
warm grotto
right??
torn jolt
cal 2 worse ngl
warm grotto
true
ancient sandal
cal 2 worse ngl
I heard the horror stories.
torn jolt
I got a B in 2
ancient sandal
I hear that Calc 2 is jus integral calculus pretty much?
torn jolt
Series :(
ancient sandal
Yk, I thought when I learnt abt how to calculate the region between two points of a function we are integrating that I'd be ecstatic or sum shit, but naw not rlly. I had way more fun with derivatives ngl.
ancient sandal
Lmao easy ass math for u smart lads
mild horizon
lol
torn jolt
but yea last thing you missed is the z bounds, and you need to account for the jacobian
mild horizon
Discrete math is kinda destroying me rn too lol
Oh nah 💀
warm grotto
but yea last thing you missed is the z bounds, and you need to account for the j...
how do I go about that?
torn jolt
how do I go about that?
the z bound goes on the inside because r is in it
mild horizon
It's really easy to learn if you have someone who can explain it well
torn jolt
for the jacobian, just put r inside the integral (you will learn how the math behind the jacobian at the very end of the class)
warm grotto
what is the z bound again?
torn jolt
you want everything below the surface, so it goes 0 to 4 - r^2
mild horizon
@viral jasper For example, you can't have a bad teacher teach anything to anyone because they can't explain it properly, but an exceptional teacher will teach them in a heartbeat
torn jolt
<@374795910032523266> For example, you can't have a bad teacher teach anything t...
no fr tho
this messed me up in cal 2 and kinda in linear algebra but i still somehow have an A
mild horizon
Rip
torn jolt
Like
I learn to do the HW
the professor grades so lightly
I can just kinda do it right and get a 95 and I hate it LOL
mild horizon
my mom just made a new email, guess the name 💀
torn jolt
mild horizon
nope
torn jolt
specifically hotmail
mild horizon
taxpro1@proton.mail 💀
torn jolt
PROTON
mild horizon
Yes
warm grotto
you want everything below the surface, so it goes 0 to 4 - r^2
ohh ok
so this?
sorry I disappeared for a minute, was making tea
torn jolt
Looks good to me
torn jolt
torn jolt
Rip
do u think i can somehow learn all the conceptual stuff in 3 days before my exam LOL
mild horizon
do u think i can somehow learn all the conceptual stuff in 3 days before my exam...
NO
torn jolt
i figured
eh as long as i can compute everything i’ll get an A somehow
that’s happened all semester
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frank grove
uhm
frank grove
I don
don't know how to solve this
Like count the lowest number.
@torn jolt
I know that this is a propyne
3,3-dichloroprop-1-yne???
Is this correct idk
.close
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