eg for the first line segment
#help-28
sacred sparrow
meaning if we want to integrate over its tangential line, we integrate over V starting with V(0,0) and ending at V(x1, 0)
it's irrelevant how many lines you have, since you can dissect them into individual lines
sacred sparrow
meaning if we want to integrate over its tangential line, we integrate over V st...
And we can express this integral as
Integral[u=0->x1]V(u,0)du
since this integral will start at V(0,0)
and end at V(x1,0)
now the same principle for the second line segment
we start at (x1, 0)
and end at (x1, x2)
untold vessel
right. and so we have V(x_1,x_2) = (x_1*x_2,x_1)
sacred sparrow
meaning we need an integral variable that goes from 0 to x2 for the second component
sacred sparrow
right. and so we have V(x_1,x_2) = (x_1*x_2,x_1)
Yes you can insert right after
untold vessel
why do they insert for u in the thing?
sacred sparrow
therefore Integral[u=0->x2]V(x1,u)du
sacred sparrow
why do they insert for u in the thing?
do you mean why they need u?
because you want to integrate over the line
and an integral "iterates" from a starting value to an end value
and they called that iterator u
sacred sparrow
it's like if I want you to integrate f(x) = x² from 1 to 3
here you already have a variable, x which you "iterate" with
Integral[x=1->3]x²dx
untold vessel
how come its seperated into two parts, called V_1 and V_2?
sacred sparrow
how come its seperated into two parts, called V_1 and V_2?
I think both just refer to V
since they didn't define V1, V2 before
untold vessel
they refer to the components (x1*x2, x1)
sacred sparrow
Integral[u=0->x1]V1(u,0)du = Integral[u=0->x1]0du
sacred sparrow
how does this relate geometrically
well they're the two lines you regard
untold vessel
okay! thanks. How come we don't use the tangential line method where we multiply inside the integral with the jacobian?
sacred sparrow
okay! thanks. How come we don't use the tangential line method where we multiply...
The splitting is simpler here
Or rather, using the Jacobian works as well, but you'd choose the segmentation method here instead
If possible you usually tend to substitutions instead of the Jacobian
untold vessel
The splitting is simpler here
Alright, I used sympy to calculate it and got x1*x2
How do I determine if V is a gradient vector field?
.close
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sacred sparrow
How do I determine if V is a gradient vector field?
It's not since it would need to be the gradient of some scalar field
untold vessel
It's not since it would need to be the gradient of some scalar field
how do I make sure of that?
Actually, I want to show that the line integral is not path independent @sacred sparrow
sacred sparrow
how do I make sure of that?
The partial derivatives aren't equal
V1 derived by y
And V2 derived by x
Are not equal
You can also lookup the definition of gradient vector field properties
sacred sparrow
Actually, I want to show that the line integral is not path independent <@100018...
Well for two line segments it doesn't matter which path you take
The only relevant aspect is that the end points remain the same
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fervent lava
My answers are:
Domain: [-1, infinity)
Range: [-2, 2) U (2, 4]
however the answers say the range is just: [-2, 4]
How is this possible if there is an asymptote?
torn jolt
I think it may mean that the asymptote is after x=4
and there isn't a hole at x=0
so the range is continuous
fervent lava
But they are extending the asymptote throughout all x values so how can it cross it?
sacred sparrow
How is this possible if there is an asymptote?
at x=0 the function still has the value y=2
hot herald
its a misconception that asympotes can't be crossed
sacred sparrow
the asymptote can be crossed, you just get arbitrarily close to it as x tends towards inf. / -inf.
hot herald
horizontal asymptotes describe end behaviour,
that has little relevance to whether that value can be attained at an earlier value
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minor shuttle
minor shuttle
<@&286206848099549185>
neat aspen
what have you tried
minor shuttle
what have you tried
uhh
well i drew the diagram
but like idrk how to do the problem itself
glossy valveBOT
Pratham_Shetty
minor shuttle
sorry
how do u know theyre perpendicular
oop alr
i havent another question
how to do b
minor shuttle
.....?
slate violet
how to do b
Since you only need the ratios between sides for trig ratios
You can let BC = 2, AC = 1
So you can use the sine rule more easily
Also 14a) should be sin A or sin C
sin C cause 14b) asks you for sin A
wispy schooner
the problem: Let $\psi\in C^{1}(\mathbb{R})$ such that $x[\psi(x)]^{2}\to0$ as $|x|\to\infty$ and $\int_{\mathbb{R}}|\psi(x)|^{2},dx=1$. Show that $$1\leq4\int_{\mathbb{R}}x^{2}|\psi(x)|^{2},dx\int_{\mathbb{R}}|\psi'(x)|^{2},dx$$
what I've done so far: $||\psi(x)||_{L^{2}}=1$. and then I'm stuck and can't figure out what to do.
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Etaoin Shrdlu
wispy schooner
my guess is that I need to use the fact that L^2 is a hilbert space somehow and turn the integrals on that one side into some statement about inner products
maybe use parseval or plancharel somewhere
or maybe cauchy-schwarz, but I don't understand why $\int_{\mathbb{R}}x|\psi(x)||\psi'(x)|,dx=\frac{1}{2}$ which is what I would need for that
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Etaoin Shrdlu
wispy schooner
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plucky terrace
Not sure how to solve this
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Not sure how to solve this
plucky terrace
I looked it up tbh, but it’s still wrong and I’d like to understand it
hot herald
where's arcsin coming from
plucky terrace
Ignore the answer I googled it
hot herald
perhaps this would be more recognisable written as
$$f(x) = 4 - \frac{3}{1+x^2}$$
glossy valveBOT
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plucky terrace
So the 3 isn’t simplified with the (1+x^2)?
hot herald
wdym
no. the power of -1 thats applied to the (1+x^2) takes priority from order of operations
3 is being multiplied to that
hot herald
antidervitive of 4 is pretty straight forward
there's an integral identity that can be applied to the fraction
(or if you aren't allowed to use that, go through the trig sub starting with x = tan(t))
plucky terrace
Ah, I see so in the answer I looked up, they must’ve mistaken square root of 1-x2 as the regular
Which explains why they got arcsin
hot herald
what exactly were you looking up / reading
plucky terrace
Ok for the satisfying part
I came up with 4x-3arctan(x) + C
satisfying f(1) = -7
Would be
4(1)-3arctan(1) + C = -7
4-3arctan + C = -7
-7-4 is -11
-11 = -3arctan + C
11/3arctan = C?
@hot herald
hot herald
no
arctan(1) ** isn't** the product of some expression arctan and 1
arctan is a function
arctan(1) is the expression representing the value when the number 1 is plugged into that function
plucky terrace
Ok?
plucky terrace
Yes, is arctan supposed to represent another trig value?
plucky terrace
Yeah?
hot herald
from the relationship,
arctan(1) is the value between -pi/2 and pi/2 that when tan is applied, will give 1
tan(what value between -pi/2 and pi/2?) = 1
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upbeat haven
Hey, I can't wrap my head around projecting a function onto another function.
This is the result (green) I'm getting from projecting 1 onto x^2
This might be wrong tho
upbeat haven
I guess what I'm asking is maybe an intuitive or visual explanation on what is actually going on. I feel like it's way easier to understand how vectors project but this, no clue.
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silver crest
need help
silver crest
help pleaseeeee
sage stream
need help
Binomial distribution
E(X) = np
Var(X) = np(1-p)
And find the square root of variance to get your standard deviation
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silver crest
Binomial distribution
so i only need to calculate sd and that it
sage stream
so i only need to calculate sd and that it
No you need the expected value and the s.d
silver crest
the expected value is the mean right?
sage stream
You expect a certain amount of people +/- the margin of error to develop the disease
sage stream
the expected value is the mean right?
Yes
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sage needle
Hi im trying to write these linearised navier stokes equations in a compact tensor form:
continuity:
$\frac{\partial\rho\prime}{\partial t}+\frac{\partial}{\partial x}\left(\bar{\rho}u\prime+\rho\prime\bar{U}\right)+\frac{\partial}{\partial y}\left(\bar{\rho}v\prime+\rho\prime\bar{V}\right)+\frac{\partial}{\partial x}\left(\bar{\rho}w\prime+\rho\prime\bar{W}\right)=0$
x-momentum
$\bar{\rho}\left(\frac{\partial u\prime}{\partial t}+\bar{U}\frac{\partial u\prime}{\partial x}+\bar{V}\frac{\partial u\prime}{\partial y}+\bar{W}\frac{\partial u\prime}{\partial z}+u\prime\frac{\partial\bar{U}}{\partial x}+v\prime\frac{\partial\bar{U}}{\partial y}+w\prime\frac{\partial\bar{U}}{\partial z}\right)+\rho\prime\left(\bar{U}\frac{\partial\bar{U}}{\partial x}+\bar{V}\frac{\partial\bar{U}}{\partial y}+\bar{W}\frac{\partial\bar{U}}{\partial z}\right)=-\frac{\partial p\prime}{\partial x}+\frac{1}{R}\left{\frac{\partial}{\partial x}\left[\bar{\mu}\left(\varpi_2\frac{\partial u\prime}{\partial x}+\varpi_0\frac{\partial v\prime}{\partial y}+\varpi_0\frac{\partial w\prime}{\partial z}\right)+\mu\prime\left(\varpi_2\frac{\partial\bar{U}}{\partial x}+\varpi_0\frac{\partial\bar{V}}{\partial y}+\varpi_0\frac{\partial\bar{W}}{\partial z}\right)\right]+\frac{\partial}{\partial y}\left[\bar{\mu}\left(\frac{\partial u\prime}{\partial y}+\frac{\partial v\prime}{\partial x}\right)+\mu\prime\left(\frac{\partial\bar{U}}{\partial y}+\frac{\partial\bar{V}}{\partial x}\right)\right]+\frac{\partial}{\partial z}\left[\bar{\mu}\left(\frac{\partial w\prime}{\partial x}+\frac{\partial u\prime}{\partial z}\right)+\mu\prime\left(\frac{\partial\bar{W}}{\partial x}+\frac{\partial\bar{U}}{\partial z}\right)\right]\right}$
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sage needle
y-momentum:
$\bar{\rho}\left(\frac{\partial v\prime}{\partial t}+\bar{U}\frac{\partial v\prime}{\partial x}+\bar{V}\frac{\partial v\prime}{\partial y}+\bar{W}\frac{\partial v\prime}{\partial z}+u\prime\frac{\partial\bar{V}}{\partial x}+v\prime\frac{\partial\bar{V}}{\partial y}+w\prime\frac{\partial\bar{V}}{\partial z}\right)+\rho\prime\left(\bar{U}\frac{\partial\bar{V}}{\partial x}+\bar{V}\frac{\partial\bar{V}}{\partial y}+\bar{W}\frac{\partial\bar{V}}{\partial z}\right)=-\frac{\partial p\prime}{\partial y}+\frac{1}{R}\left{\frac{\partial}{\partial x}\left[\bar{\mu}\left(\frac{\partial u\prime}{\partial y}+\frac{\partial v\prime}{\partial x}\right)+\mu\prime\left(\frac{\partial\bar{U}}{\partial y}+\frac{\partial\bar{V}}{\partial x}\right)\right]+\frac{\partial}{\partial y}\left[\bar{\mu}\left(\varpi_0\frac{\partial u\prime}{\partial x}+\varpi_2\frac{\partial v\prime}{\partial y}+\varpi_0\frac{\partial w\prime}{\partial z}\right)+\mu\prime\left(\varpi_0\frac{\partial\bar{U}}{\partial x}+\varpi_2\frac{\partial\bar{V}}{\partial y}+\varpi_0\frac{\partial\bar{W}}{\partial z}\right)\right]+\frac{\partial}{\partial z}\left[\bar{\mu}\left(\frac{\partial v\prime}{\partial z}+\frac{\partial w\prime}{\partial y}\right)+\mu\prime\left(\frac{\partial\bar{V}}{\partial z}+\frac{\partial\bar{W}}{\partial y}\right)\right]\right}$
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sage needle
z-momentum:
$\bar{\rho}\left(\frac{\partial w\prime}{\partial t}+\bar{U}\frac{\partial w\prime}{\partial x}+\bar{V}\frac{\partial w\prime}{\partial y}+\bar{W}\frac{\partial w\prime}{\partial z}+u\prime\frac{\partial\bar{W}}{\partial x}+v\prime\frac{\partial\bar{W}}{\partial y}+w\prime\frac{\partial\bar{W}}{\partial z}\right)+\rho\prime\left(\bar{U}\frac{\partial\bar{W}}{\partial x}+\bar{V}\frac{\partial\bar{W}}{\partial y}+\bar{W}\frac{\partial\bar{W}}{\partial z}\right)=-\frac{\partial p\prime}{\partial y}+\frac{1}{R}\left{\frac{\partial}{\partial x}\left[\bar{\mu}\left(\frac{\partial w\prime}{\partial x}+\frac{\partial u\prime}{\partial z}\right)+\mu\prime\left(\frac{\partial\bar{W}}{\partial x}+\frac{\partial\bar{U}}{\partial z}\right)\right]+\frac{\partial}{\partial y}\left[\bar{\mu}\left(\frac{\partial v\prime}{\partial z}+\frac{\partial w\prime}{\partial y}\right)+\mu\prime\left(\frac{\partial\bar{V}}{\partial z}+\frac{\partial\bar{W}}{\partial y}\right)\right]+\frac{\partial}{\partial z}\left[\bar{\mu}\left(\varpi_0\frac{\partial u\prime}{\partial x}+\varpi_0\frac{\partial v\prime}{\partial y}+\varpi_2\frac{\partial w\prime}{\partial z}\right)+\mu\prime\left(\varpi_0\frac{\partial\bar{U}}{\partial x}+\varpi_0\frac{\partial\bar{V}}{\partial y}+\varpi_2\frac{\partial\bar{W}}{\partial z}\right)\right]\right}$
glossy valveBOT
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sage needle
Energy:
in which $\varpi_j=j+\sfrac{\lambda}{\mu}$, {\lambda} is the stokes hypothesis = -2/3{\mu}
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sage needle
i got it for continuity:
$\frac{\partial\rho'}{\partial t}+\frac{\partial\left({\bar{\rho}u}_i'+\rho'\bar{u_i}\right)}{\partial x_j}=0$
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hot idol
So, I've got conveyors and each of these conveyors can be seperated with a conveyor splitter, which have 4 connections; three for output and one for input. Contrary to this, there are also conveyor mergers which have 4 connections; one for output.
And, these conveyors can divide/merge a certain number with minimum 2 and maximum 3.
Considering that we have 10 iron rods, which all have to be seperated into groups of two, with a maximum divisions amount of 3 and minimum 2.
Write down each step in order to divide 10 iron rods into 2 pieces each. Mind that iron rods are a positive integer, in which the conveyor splitters split in order (incase if there is a piece in addition) first; right, second; left, third; middle.
hot idol
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torn gust
So, I've got conveyors and each of these conveyors can be seperated with a conve...
the goal isn't that clear. do you want to turn 10 into five groups of 2 using splitter/mergers? reminds me of this game https://steamcommunity.com/sharedfiles/filedetails/?id=2962055149
hot idol
the goal isn't that clear. do you want to turn 10 into five groups of 2 using s...
no. it's satisfactory
hot idol
the goal isn't that clear. do you want to turn 10 into five groups of 2 using s...
yes i do
the goal is the thing you wrote
torn gust
I'd probably split in half for 5-5 then split each in 3 for 2-2-1 with 2-2-1 and combine the 1's
hot idol
I'd probably split in half for 5-5 then split each in 3 for 2-2-1 with 2-2-1 and...
You cannot
The order changes everytime when a new
group of iron rods come in
torn gust
oh so using order isn't really useful ok
torn gust
like does this picture cycle between 4-3-3, 3-4-3, and 3-3-4?
torn gust
hmmm maybe it's even impossible without smarter conveyor parts, since splitting 10 into 2's or 3's can't become a division by 5
torn gust
https://forums.factorio.com/viewtopic.php?t=6806 has the idea of feeding some output back up into the input to get more kinds of fractions 
hot idol
<https://forums.factorio.com/viewtopic.php?t=6806> has the idea of feeding some ...
I think we find the constant of the ratio then divide by the denominator, then merge by the numerator
Is that right
Nevermind. It cannot be true since the ratio constant cannot contain numbers other than 2 and 3
while it is forced to contain 5
torn gust
like if you split an input into 6 and fed one line back in that'd be a 1/5 splitter
might not work depending on the situation though
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modest pumice
Is this true?
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sharp vine
sinθ - 1 is cosθ neither
modest pumice
Welp I've been lied to
sharp vine
recall sin^2θ + cos^2θ = 1
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cloud seal
I'm practicing a test question and I'm unable to navigate through this, can anyone help me please?
amber isle
I'm practicing a test question and I'm unable to navigate through this, can anyo...
apply kirchoffs loop law
on the circuit
thick hedge
derive it using KCL and KVL
it's a second order ODE
so if you don't know second order ODEs you may have some trouble
cloud seal
Alright lemme try that out, thanks
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flint kraken
can somebody tell me where i am going wrong here? I know the correct answer but for the live of me i cannot see the correct step:
flint kraken
it is about finding the fourier transform of sine if you know the transform of cosine
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flint kraken
so i figured this but the step in red is surely illegal the way i wrote it
<@&286206848099549185> anyone up for it? I am sure it is a really small detail in the setup i am missing,,,
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native matrix
Root of x^2-3x+m+2=0 is x1 x2 and x1(1-x2)+x2=3m Find m
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Root of x^2-3x+m+2=0 is x1 x2 and x1(1-x2)+x2=3m Find m
native matrix
Is the question
I expanded the bottom one and saw x1+x2 and -x1x2
Used vieta on the top to replace those and found -5/4 but that isnt in the answers
Where did i go wrong
plain fox
Used vieta on the top to replace those and found -5/4 but that isnt in the answe...
product = m+2
so - (x1x2) = - (m+2) = -m -2
plain fox
yeah sum = -b/a
native matrix
Thought it depended on the degree so wrote it as one - from the maşn one
supple sail
is m = -2?
native matrix
is m = -2?
Not in the answers
supple sail
Not in the answers
holdup
whats the original ques?
native matrix
Root of x^2-3x+m+2=0 is x1 x2 and x1(1-x2)+x2=3m Find m
This
supple sail
x1 and x2 are roots
or x1 times x2 is the root?
native matrix
(-1)^n an-1/an
plain fox
x1 and x2 are roots of the first quadratic
native matrix
Thought it was like that and since the degree was 2 put 2 on the beginning
$(-1)^n an-1/an$
glossy valveBOT
Hiya
plain fox
whats an?
supple sail
is m = 1/4
native matrix
Coef
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native matrix
Of the highest degree guy
supple sail
is m = 1/4
i got this
i think im correct
heres how i did it
native matrix
Should be
What i dont understand is
Why is there 2 negatives at the beginning when n is 2
(-1)^2 should be positive no?
It does if it has another negative it becomes -5/4
Found the link ill watch the video about vieta soon since theres quite a bit i dont understand btw what u typing you were typing for a solid 7 mins
supple sail
Let x1 and x2 be the roots
of the given equation
then by vieta we can say
x1 + x2 = 3
and x1x2 = m+2
now x2 = 3-x1
putting in second eqn
x1(3-x1) = m+2
3x1-x1^2 = m +2
rearranging
**x1^2 -3x1 = -2 -m **
now onto the second given equation
x1(1-x2)+x2=3m
put x2 = 3-x1
x1(x1-2) + 3-x1 = 3m
x1^2 -2x1 +3 -x1 = 3m
x1^2 -3x1 +3 = 3m
**x1^2 -3x1 = 3m-3 **
comparing the bold equations
3m-3 = -2-m
4m = 1
**m = 1/4 **
supple sail
Found the link ill watch the video about vieta soon since theres quite a bit i d...
been 4 mins bro 😭
native matrix
Let x1 and x2 be the roots
of the given equation
then by vieta we can say
x1...
Lul close enough also can i solve it using vieta without this?
hollow mica
hollow mica
native matrix
Rule for the -
hollow mica
supple sail
Lul close enough also can i solve it using vieta without this?
idts it can solved with vieta without this approach
you gotta form equations
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tribal viper
…
tribal viper
How what when where why
fallow oasis
proven mango
if you search this up, apparently the arm span and a person’s height is a one to one ratio
tribal viper
Oh
proven mango
never knew that
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meager sedge
How can we prove combinatorially using bijections that (n choose k) = (n-1 choose k) + (n-1 choose k-1)? I was given a hint to "map a set A as a subset of [n] of size k to A if n isn't in A and to A\{n} if n is in A."
fast peak
is that the exact wording of the hint?
meager sedge
is that the exact wording of the hint?
Yeah.
meager sedge
[n] = {1,2,...,n}
fast peak
what they mean is the following: for any subset A of [n], obviously either n in A or not
meager sedge
so i think i understand the intuition so that's why we add the combinatorics
The biggest thing is how do I make a bijection
Instead of just explaining it
in fact it says bijections plural which is even more confusing for me
fast peak
so, we consider the two sets $N = {A \subseteq [n]: n\in A, |A|=k}$ and $M={A\subseteq [n]: n\notin A, |A|=k}$
yes?
meager sedge
n is arbitrary right? We can choose any rando element in reality
But we choose n cause the hint says?
glossy valveBOT
Denascite
fast peak
you mean any element from [n] and then make the distinction of whether thats in A or not?
yeah thats arbitrary
choosing n works quite well tho
meager sedge
Ok so I think I agree with your sets
fast peak
in the next step, we consider the sets $S={B\subseteq [n-1]: |B|=k}$ and $T={B\subseteq [n-1]: |B|=k-1}$
glossy valveBOT
Denascite
meager sedge
Sure
fast peak
how many elements do S and T have?
fast peak
no
meager sedge
Oh this looks weird is it like sets of all possible subsets
So then it's probably some choose formula
|S|=(n-1 choose k) and |T|=(n-1 choose k-1)?
meager sedge
S ={all sets B that are subsets of {1,2,...,n-1} such that the size of B is k}=all possible choices of k elements in a set of size n-1=(n-1 choose k)
T={all sets B subsets of {1,2,...,n-1} such that the size of B is k-1}=all possible choices of k-1 elements in a set of size n-1=(n-1 choose k-1)
fast peak
How can we prove combinatorially **using bijections** that (n choose k) = (n-1 c...
and the hint basically already says how these bijections should look like
after that you can conclude that (n choose k) = |M|+|N| = |S|+|T| and are done
I have to go
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meager sedge
So basically we know that $N = {A \subseteq [n]: n\in A, |A|=k}$ and $M={A\subseteq [n]: n\notin A, |A|=k}$ added together is n choose k by intuition, but $|N| = T={B\subseteq [n-1]: |B|=k-1}$ and $|M| = S={B\subseteq [n-1]: |B|=k}$
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meager sedge
.close
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pliant heron
Can anyone verify if this is correct or not
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leaden ermine
Can anyone verify if this is correct or not
You sure that n can be any integer?
pliant heron
You sure that n can be any integer?
Nope that is what I was unsure about
leaden ermine
z³ = 1 would have 3 solutions fundamentally
the thing is if those 3 fundamental solutions satisfy z^3/4 = i
leaden ermine
wdym
pliant heron
for n
leaden ermine
oh yea
but since there are only 3 solutions
for example take
n = {0,1,2}
at least one n is a solution
for n = 0 i would get 1 and 1^3/4 = i is not the case
so you remain with n = 1 or n = 2
leaden ermine
you can take 3 arbitrary but subsequent n
pliant heron
Ahh okie
leaden ermine
because the equation z^n = w has n solutions that are equidistant by 2π/n units (given n is a natural number)
pliant heron
i don't know what equidistant means unfortunately
like equal everywhere, as in a circle?
leaden ermine
it just means the solutions are equally far distributed
pliant heron
ohh okay
leaden ermine
like if you were to cut a cake into pieces but equal pieces
pliant heron
k got it : )
pliant heron
for n = 0 i would get 1 and 1^3/4 = i is not the case
what did you plug 0 into? The equation I found?
pliant heron
mhm
pliant heron
but since there are only 3 solutions
how do you know this?
leaden ermine
because the equation z^n = w has n solutions that are equidistant by 2π/n units ...
.
leaden ermine
so what you got
pliant heron
so what you got
n = 1, 2, 3?
pliant heron
Well my calculator was in the wrong mode
haha
Should be in radians. I had it in degrees
leaden ermine
yes
glossy valveBOT
𝔸dωn𝓲²s
leaden ermine
So since I wrote it in exp form (it's faster)
which of these satisfies your intial equation?
glossy valveBOT
𝔸dωn𝓲²s
leaden ermine
I would suggest to try the exp form, since you can tell it instantly
glossy valveBOT
𝔸dωn𝓲²s
leaden ermine
Which of these result to i?
leaden ermine
yes
Now I was doint ^(3/4) on both sides
And using De Moivre I wrote the result
like n = 3/4
leaden ermine
**𝔸dωn𝓲²s**
Hence I multiply theta with 3/4 in the argument
It's a shame that you haven't learnt exponential form when you were introduced to the complex world x)
pliant heron
But i foudn that r = 1^1/3 so that is just one?
pliant heron
so why do we still have to multiply in 3/4
leaden ermine
because I did ^(3/4) on both sides
leaden ermine
**𝔸dωn𝓲²s**
.
pliant heron
Oh wait I was doing it separately
Like plugging into my equation first and then putting it into the given form to check if it gives i
Did you just do it all together?
leaden ermine
yea
leaden ermine
what are you doing actually?
leaden ermine
**𝔸dωn𝓲²s**
You really dont need a calculator for this tho
pliant heron
what are you doing actually?
pliant heron
k i'll do it all at once
leaden ermine
**𝔸dωn𝓲²s**
Why don't you do this instead to rather check which solution satifies the intial equation?
pliant heron
n =1
pliant heron
yup
can see why i don't need a calculator now haha
is this just a matter of guess and check?
Is there some other way that gives a definitive answer?
leaden ermine
no not really guess
the thing is we transformed the intial equation into such form z^n = w
but it can be that not all solutions are part of the intial equation
so we would have to check
leaden ermine
**𝔸dωn𝓲²s**
here for n = 2 you can tell that it is not satisfying it because you get -1
not i
pliant heron
yeahh
There is one thing I am confused about in a similar example my prof did
would you be able to clarify it
I will send a pic
leaden ermine
i hope so.
pliant heron
She did one similar to this with a real number.
My questions are :
- How do we know we want to find all such theta is between 0 and 2pi
- Why did the interval she did do 0 <= (2n +1) / 15 <= 2
leaden ermine
Yes ok lemme take a deep look
pliant heron
Thanks!!
leaden ermine
She did one similar to this with a real number.
My questions are :
1. How do w...
- How do we know we want to find all such theta is between 0 and 2π
If I understand correctly, you are asking* why that is the case, well we expect n solutions within a cycle/period, hence 15 solutions between 0 and 2π because otherwise we wouldn't have 15 solutions if they were not within one period.
- Why did the interval she did do 0 ≤ (2n +1) / 15 ≤ 2
If you take a closer look at 2(cos(π) + isin(π)) that is the same as 2[cos(π +2πn) + isin(π+2πn)] = 2[cos(π(2n+1)) + isin(π(1+2n))]
pliant heron
Alright thank you!
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neon eagle
can someone confirm if i did this right or wrong
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can someone confirm if i did this right or wrong
lilac flume
uh do u have to find the derivative w.r.t x?
neon eagle
sorry for not giving context
they told us to differentiate, i just put (theta^2) as one term and sin(2theta) as the other
neon eagle
uh do u have to find the derivative w.r.t x?
im soz idk what w.r.t.x means
slate violet
i mean i think i did it right i kind of just wanted confirmation
You're differentiating with respect to theta right
slate violet
If so, that's correct
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tacit sierra
Can someone help me with linear relations 😭
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Can someone help me with linear relations 😭
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@tacit sierra Has your question been resolved?
tacit sierra
<@&286206848099549185>
ruby wolf
<@&286206848099549185>
Hii
tacit sierra
Hello
tacit sierra
TY
tacit sierra
ok
ruby wolf
hmmm
draw me a coordinate plane in the example box
pls
it should look something like this
tacit sierra
Alright one sec
ruby wolf
np!
tacit sierra
Just draw any coordinate?
ruby wolf
hmmm
i accidentally skipped hte first page
we'll circle back to that
anyways
lets draw a coordinate
(2,3)
:D
and show me what you get
ok?
tacit sierra
MB
ruby wolf
lol
tacit sierra
😭
tacit sierra
No shot ima fit that in there
ruby wolf
okk how about uhm
night lotus
Sorry for the horrendous writing 💀
use a ruler by the way if you got one. Should make your life easier 😄
tacit sierra
use a ruler by the way if you got one. Should make your life easier 😄
Im so un prepared i dont even have an eraser 😭
ruby wolf
Im so un prepared i dont even have an eraser 😭
oki soo get those supplies ready next time
night lotus
Im so un prepared i dont even have an eraser 😭
yeah just get them later
ruby wolf
mhm
tacit sierra
Thanks
ruby wolf
for now cross it out
tacit sierra
Alr
ruby wolf
kk
so uhm
i wish you could like live stream this mannn
lol its so much harder
ok uhm
ruby wolf
in a private chat technically yes
tacit sierra
Yea
ruby wolf
if ur willing to go there i can help you but i wont be able to talk
tacit sierra
Alg
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zenith haven
Hello how we can determine the radius of convergence ?
I have reached | (1/3) (2x-1) | <1
but final ans is R=3/2
stiff musk
informally, the (-1)^n, the n in the denominator, don't affect the radius of convergence
blissful silo
does ; make a sentence a compound or complex sentence?
zenith haven
informally, the (-1)^n, the n in the denominator, don't affect the radius of con...
Am sorry but Can you explain more plz ?
zenith haven
does ; make a sentence a compound or complex sentence?
Not sure how this math related it combines 2 sentences
blissful silo
Not sure how this math related it combines 2 sentences
so compound or complex?
stiff musk
Am sorry but Can you explain more plz ?
so you got as far as $$\left|\frac{2x-1}{3}\right| < 1$$
glossy valveBOT
Bungo
zenith haven
so compound or complex?
dude come private and gimme the sentence
zenith haven
so you got as far as $$\left|\frac{2x-1}{3}\right| < 1$$
yep correct
stiff musk
just multiply both sides by 3
blissful silo
dude come private and gimme the sentence
I have a headache; I need to go home
stiff musk
$|2x-1| < 3$
glossy valveBOT
Bungo
stiff musk
and now see what range of x satisfies this
zenith haven
@stiff musk
okay so now +1-3 <2x<3+1
then -2<2x<4
then -1<x<2
stiff musk
yes
all good so far
so now what's the midpoint of the interval (-1,2)
or if you don't care about that, just find the radius as half the length of that interval
zenith haven
( 2- (-1) )/2= 3/2
so basically the radius of conversion is half the distance of the range I got right ?
stiff musk
half the length of that interval yea
3/2 is correct
but it's not +/-3/2 around zero
the series is centered at the midpoint of (-1,2)
which is 1/2
zenith haven
which is 1/2
oh so we take absolute value of the points we have ?
and then combined /2
right ?
stiff musk
no it's just their sum divided by 2
-1 + 2 is 1
divide by 2 is 1/2
that's where the series is centered
blissful silo
huh?
zenith haven
that's where the series is centered
okey what next after we got the center 1/2 ?
stiff musk
okey what next after we got the center 1/2 ?
depends on what you want to know
blissful silo
thanks
stiff musk
you now have the radius of convergence as well as the center
and those are usually the things you care about
zenith haven
depends on what you want to know
I mean if we have the center which is half the interval we multiply by 2 to get the radius of conv or what
I did not get how we can obtain radius from the center
Sorry to bother either way
stiff musk
no you can't get one from the other
but you can get both from the interval
(-1, 2)
the radius is half the length
and the center is the midpoint of the interval
zenith haven
and the center is the midpoint of the interval
gotyou.. so basically for radius we get the length
2- (-1) = 3
and then half the length 3/2 am I right ?
stiff musk
yep correct
zenith haven
Dude... You can't imagine how did you help me now
Thanks a bundle !!
zenith haven
.close
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blissful silo
check dm plz
ok
𝗡𝗼𝘄 𝗽𝗹𝗮𝘆𝗶𝗻𝗴:
"Surround Sound - 21 Savage"
01:09 ━━━━●───── 02:17
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torn jolt
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torn jolt
Am I correct?
lilac flume
is tg tangent?
torn jolt
yes
lilac flume
$tan(-\theta)=-tan(\theta)$
glossy valveBOT
swerriee
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torn jolt
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torn jolt
Whats going on?
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granite torrent
.close
i assume u figured it out but the idea is when u take the roots of cosx = 0 and 1 - 2cosx = 0 you have to consider all possibilities of theta in the range (-pi, pi]
though the "book answer" u write down there should be ± pi/3
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formal cloak
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formal cloak
how do i even get started on the problem
i tried to find the tangent line but you’d get two different lines depending on where you start
<@&286206848099549185>
someone please its 3 am 😭
been on this problem for half an hour now
anyone?
i’ve tried everything i could think of
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@formal cloak Has your question been resolved?
torn jolt
i tried to find the tangent line but you’d get two different lines depending on ...
have you considered the bounds it gives you
formal cloak
yes
torn jolt
draw a diagram of the problem
formal cloak
i’ve tried finding tangents with those bounds
i’ve deduced it to be optimization
tried watching videos on it
nothing remotely close to this
also tried graphing the tangents based on the answers
none of them helped
wait
those are wrong
i guess i can just graph out each one and find the integral and get the answer or something
but thats inefficient
and prob inaccurate as well
whatever its 4 am
thanks for the great and wonderful help
torn jolt
Try finding a general equation for the lines of the tangent
Then use that to find the points it intercepts on the x/y axis for any given value
then find the formula for the area of the triangle based on those given points
formal cloak
that’s inefficient
it’ll take up the 15 min i have to take the quiz to do that
i know theres a way to do it via derivatives and optimization
i just don’t know since there’s no definitive points
and if there was a general formula
it’d be y-(12-t^2)=-2x(x-t)
torn jolt
well do it now
while u arent doing the test
And learn how to solve it
and u might find some shortcuts to@get the answer straight away
formal cloak
i’ve been trying to learn for the past hour
nothing in my notes
nothing in videos
all i know its optimization
and the best bet i have is to hope to dear god it isn’t on the quiz or the ap exam
how the heck do you even optimize something without atleast a given point
if i do dy/dx i get 0
but i know 2 is the answer
what i’ve got this down to is that I need to find the x coordinate of the point of intersection between the curve and the hypotenuse of the triangle
but to do that i have to find the smallest area
wait
would it be tangent to the point closest to the origin
is that where it’d minimized
nope
doesn’t work
<@&286206848099549185> can someone seriously please help
been on this problem for ages
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torn jolt
i assume u figured it out but the idea is when u take the roots of cosx = 0 and ...
But, what happents with the second answear.
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raven flax
How can a limit be the same for both an open and closed value?
thick hedge
wdym?
raven flax
for e.g the limit of x as x approaches 3 for f(x) = x is 3 (3 is included). But the same limit for x(x-3/x-3) is also 3 (3 is not included). Isn't a limit saying a value which can't be obtained like x is constantly going towards 3 but never reaching it?
neat aspen
what does x * (x-3) / (x-3) look like when it is not 3? it exactly like and has the same behavior as x
torn jolt
for e.g the limit of x as x approaches 3 for f(x) = x is 3 (3 is included). But ...
Right, $\ds\6fx = x\8{\4{x-3}{x-3}}$ indeed has a discontinuity at $x = 3$ and $\6f3$ is undefined. But, like you said, a limit approaches the value and since this is a removable discontinuity, we cab still conclude that $\ds\lim_{x \to 3} \6fx = 3$
glossy valveBOT
raven flax
but if a limit approaches a vlaue but never hits it, how is the first funciton also equal 3
neat aspen
the limit is: what hsould happen to this function when it goes to 3?
you can apply the same thought, for f(x) = x, then the limit "never hits 3" but what should happen as f(x) gets extremely close to 3?
raven flax
but it does hit 3
neat aspen
yes but that is a nice coincidence
raven flax
oh ok so it has nothing to do with if the value x approaches is included or not?
neat aspen
correct
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rich verge
17
rich verge
where would i start
barren tundra
where is the question
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@rich verge Has your question been resolved?
nova sigil
Label each number by E or O depending on if it's even or odd then look at how many ways you can rearrange Es and Os in a circle to produce the above
Then you will have 2 different circles that can be filled in different orders.
dapper sigil
16x6
Just pair an even with an odd, multiple each pair by 2 because you can interhcnage them I.e (2,3) can be written as (3,2) so we need to take into account these. So we'll multiply 2x2x2x2 to get 32 and finally I think here we take permutation of 4-1 because we made 4 sets, we do the - 1 because it's a circle.
32x(4-1)!/2= 16x6
Divide by 2 because arrangements in circle get repeated
But I make mistakes always in math so I'm not certain. Although I'm sure the question can be solved this way
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worthy nest
How would i determine that $H_{0}^n and H_{-1}^n$ for all n>=1 are smooth submanifolds of $\mathbb{R}^{n+1}$ without using the regular value theorem.
glossy valveBOT
PiuPiu
torn jolt
Can anyone help me out with this one
worthy nest
@torn jolt can you go into a different chat xd
i started this one just before you started writing
torn jolt
Its okay man ill leave
worthy nest
How would i determine that $H_{1}^n and H_{-1}^n$ for all n>=1 are smooth submanifolds of $\mathbb{R}^{n+1}$ without using the regular value theorem.
glossy valveBOT
PiuPiu
worthy nest
<@&286206848099549185>
worthy nest
**PiuPiu**
- this has to be H_1 and not H_0
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wicked cliff
why i can use 5
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why i can use 5
wicked cliff
You can use 5 instead of 3/5?
wicked cliff
this wrong?
wicked cliff
that'd be undefined
.
torn jolt
this wrong?
i mean yeah arccos(5) is simply not defined
wicked cliff
but you can cancel? cos(arcos(x)) = x?
wicked cliff
i mean yeah arccos(5) is simply not defined
or cos(arcos(x))
white smelt
technically, cos(arccos(x)) = x if you use the extended definition of arccos into the set of complex numbers
but that isnt required here since 3/5 is in the domain of arccos
torn jolt
but you can cancel? cos(arcos(x)) = x?
its a composite function. So with cos(arccos(x)) you compute the arccos(x) while taking into account its domain, and then the range of values of arccos(x) will be the domain of cos(x). But this doesn't work if x is not a part of the domain of arccos(x) which is [-1, 1] to begin with
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hidden ingot
a stationery shop orders a batch of pens from a factory. The factory could make a batch of pens in 20 days using 12 machines. Due to a fault only 9 machines were used for the first 8 days. All 12 machines were used from day 9 onwards. Work out the total number of days taken to make the batch of pens.
slate violet
a stationery shop orders a batch of pens from a factory. The factory could mak...
You can think of this using the unit of day-machines
The batch of pens needs 240 day-machines of work
So can you set up a linear equation from this?
hazy dome
Oh
hazy dome
<#1018700472988749904>
Is it possible for you to solve, I just posted it in 45
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simple falcon
So I don't know how to continue with this question as I can't figure out how to move the y to the other side with the dy
slate violet
So I don't know how to continue with this question as I can't figure out how to ...
It's not separable
Divide both sides by x and rearrange to get y' + p(x) y = q(x)
Then do you know how to continue?
slate violet
Wait what? Sorry I don't know
Have you covered how to do first order linear DEs?
Okay you need to go study this
simple falcon
Thank you ;-; I will go study that
slate violet
No worries
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runic spruce
Any errors?
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Any errors?
quaint prawn
Any errors?
yes there is a mistake in the first line
what is the derivative of (1-delta)L^(gamma) with respect to L?
also, don't use X for multiplication, use *
quaint prawn
yes
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cloud snow
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cloud snow
i need help understanding how to solve (a) (ii)
what proccess is needed to solve for it
cloud snow
nvm i found out how to do (a) (ii). I need help with b. How do I prove something is independant?
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mint aurora
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mint aurora
I am having trouble on step 4
I am supposed to do INV norm
and for this type I have been doing 4500/SQ 1000 for my standard deviation
but when I try to do that to get the 80th percentile I get a error
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outer badger
honestly just need a refresher in stats and excel
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honestly just need a refresher in stats and excel
outer badger
Im trying to solve for these questions and I already calculated the z scores for 380 500 and 420.5 but I have no clue what to do with them
I also know I need to use =NORM.S.DIST for 7 and =NORM.DIST for 8 but idk how to use the functions
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outer badger
no
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outer badger
i think i did most of them
idk how to do the ones with the empirical rule and the single value
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covert wing
$2\cos(2x)+3\sin(x) = \frac{5}{2}$
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$2\cos(2x)+3\sin(x) = \frac{5}{2}$
glossy valveBOT
Reclipse
last pewter
To me, it seems clear that cos(2x) and sin(x) must both be 1/2 at least for one of the solutions
so id start there
using the unit circle
stiff musk
you can use the trig identity for cos(2x)
torn jolt
**Reclipse**
Try expanding Cos2x and making a quadratic in terms of Sinx
covert wing
Try expanding Cos2x and making a quadratic in terms of Sinx
Can you explain more in terms of how please :(
stiff musk
you know cos(2x) = cos^2(x) - sin^2(x) i assume?
and then you can write cos^2(x) in terms of sin
or maybe you have a formula for cos(2x) involving only sin memorized
torn jolt
Can you explain more in terms of how please :(
Are you familiar with the formula of Cos(A+B)?
covert wing
Are you familiar with the formula of Cos(A+B)?
I haven't done trig in a long time and I forgot most of trig (a+-b) functions
glossy valveBOT
who
torn jolt
for A-B, simply substitute B with -B
covert wing
Yup I remember these now appreciates
torn jolt
since we require Cos2x here, substitute B by A
covert wing
Ok so I wrote the equation in the end as $8\sin^2(x) + 3\sin(x) - 9 = 0$ and after this I can just substitute $\sin(x)$ with Y for eg. and then it's a quadratic
glossy valveBOT
Reclipse
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abstract anchor
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abstract anchor
Exterior Inequality Theorem
How do I find the x?
What equation is used here???
I know all exterior = 360
I know all interior = 180
(x+20) > 34
<@&286206848099549185>
ripe horizon
abstract anchor
Yes
ripe horizon
what do you think is the value of this angle
torn jolt
90?
ripe horizon
they form a linear pair right
abstract anchor
180 - (x+20)
ripe horizon
so they must equal 180
ripe horizon
180 - (x+20)
correct
abstract anchor
knowing remote exterior angles are congruent
What
All remote exterior angles are congruent
abstract anchor
Is this right
Is this right
Remote1 = 180 - (180 - (x+20) + 34)
Remote2 = 180 - (x+20)
Finding x
180 = 180 - (180 - (x+20) + 34) + 180 - (x+20)
-74 = -2x
37 = x
Bro
Why does it need me to compute that long
Aint there a shorter formula
Is that even right
Guys
gritty patrol
if its an inequality
and you know that the theorem is x+20>34
then x > 14
so any angle larger than 14 degrees would be a solution for x
abstract anchor
Is it possible to compute the exact value of x with those given values?
Or just the range?
gritty patrol
here is what i tried until i realised that it is an inequality
abstract anchor
Aww man