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inland patio
Let $f:\mathbb R^n\times\mathbb R^m\to\mathbb R^m$ be $C^1$ in an open set containing the point $(a,b)$. Let $M$ be the $m\times m$ matrix $$(D_{n+j}f^i(a,b)),\quad 1\leq i,j\leq m.$$ Define $F:\mathbb R^n\times\mathbb R^m\to\mathbb R^n\times\mathbb R^m$ by $F(x,y)=(x,f(x,y))$ and let $F'(a,b)$ be the Jacobian matrix at $(a,b)$. Maybe it is obvious, but I don't see why $$\det F'(a,b)=\det M?$$
midnight plankBOT
Let $f:\mathbb R^n\times\mathbb R^m\to\mathbb R^m$ be $C^1$ in an open set conta...
grand pondBOT
Philip
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last slate
What are conjugates and in that matter why real numbers don't/can't have conjugates
midnight plankBOT
What are conjugates and in that matter why real numbers don't/can't have conjuga...
slender walrus
who says real numbers can't have conjugates
obtuse basin
"Conjugate" refers to many things (e.g., complex conjugate, conjugates of expressions with square roots, conjugate by an element in a group, etc.), which one are you mentioning?
last slate
"Conjugate" refers to many things (e.g., complex conjugate, conjugates of expres...
Complex conjugates ig
last slate
who says real numbers can't have conjugates
I was taught conjugate of real number is real number itself
And conjugate of complex number is negative of that complex number
obtuse basin
The latter is not necessarily true, you were told lies
The complex conjugate of a complex number has the same real part and the same complex part but with sign changed
Anyway it's still unclear what you meant by real numbers not having conjugates
last slate
The complex conjugate of a complex number has the same real part and the same co...
Exactly
last slate
The complex conjugate of a complex number has the same real part and the same co...
So conjugate is 4 + 5i is 4 - 5i
Why sigh of 4 isn't changed?
And again brings me back to question.. What is conjugate?
slender walrus
a + b and a - b
are conjugates of each other
you can view
4 + 5i as 5i + 4
and 5i - 4 would be a valid conjugate (though not what would usually be considered as the complex conjugate)
last slate
`a + b` and `a - b`
are conjugates of each other
A single number cannot have conjugate?
Like i Or 5i Or something like that
slender walrus
for complex conjugates we consider
a + bi and a - bi
i = 0 + i
so its complex conjugate would be -i
whoops forgot the -
last slate
whoops forgot the -
But
i = i + 0
So i - 0
So why can't it be i?
slender walrus
for complex conjugates we consider
a + bi and a - bi
viewing i as i + 0
i is technically a conjugate (though not what would usually be considered as the complex conjugate)
last slate
> for complex conjugates we consider
> a + bi and a - bi
Any specific reason for this m
This?*
slender walrus
just the way someone defined it i guess
last slate
just the way someone defined it i guess
Real
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round wharf
why did it simplify to that ?
midnight plankBOT
why did it simplify to that ?
kind zealot
cos(a + b) = cos(a)cos(b) - sin(a)sin(b)
kind zealot
"Trigonometric identities" should give you a few good ones
tardy wyvern
Trignometric products ?
round wharf
"Trigonometric identities" should give you a few good ones
Alright, thank you
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tardy wyvern
I have a fair idea i need to find a realtion b/w area and the sides and angle
but idk what the relation could bne
be
i need to dervaite wrt time and then subts
but idk what the relation is
could someone help me with finding the relation? and if possible explain how?
sharp coral
we can use the trigonometric formula for the area of a triangle. if $a$ and $b$ are sides of a triangle and $\theta$ is the angle between them, then the area is [ A = \frac 12 ab \sin \theta ]
grand pondBOT
cloud
tardy wyvern
we can use the trigonometric formula for the area of a triangle. if $a$ and $b$ ...
oh thanx got it
c.lose
.close
midnight plankBOT
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autumn folio
is my math correct here?
midnight plankBOT
is my math correct here?
twilit field
!original
midnight plankBOT
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.
autumn folio
given that at a certain time s1 is cut off, calculate s2.
autumn folio
twilit field
The forces acting on $M_2$ are $T_2$ and it's weight , right?
autumn folio
yes
grand pondBOT
ƒ(Why am. I here)=I don't know
twilit field
whereas the forces acting on $M_1$ are $M_1's weight , T_2 and T_1$ I think
grand pondBOT
ƒ(Why am. I here)=I don't know
twilit field
ah
in that case the only forces acting on it at that instant would be $T_1 and M_1g$ IMO
grand pondBOT
ƒ(Why am. I here)=I don't know
grand pondBOT
ƒ(Why am. I here)=I don't know
autumn folio
so its what i wrote
grand pondBOT
ƒ(Why am. I here)=I don't know
autumn folio
btw the y axis is up
twilit field
and $M_1 g -T_2=M_1a$
grand pondBOT
ƒ(Why am. I here)=I don't know
twilit field
add these now
autumn folio
I think your answer is right
but my answer says that when s1 is cutted off, the system would have acceleration upwards, how does that make sense?
twilit field
in that case, I have no idea
have you tried googling this?
or ask in the physics server, or wait for someone else
ping helpers 15m after you post your problem
autumn folio
its been 30, ill ping
given that at a certain time s1 is cut off, calculate s2.
according to my calculations(y axis is upwards) the system would have acceleration upwards, whats wrong? <@&286206848099549185>
frail willow
can someone help me with my math
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twilit field
have a doubt here, why is the only way to express 0 as 0+0+0.....
midnight plankBOT
have a doubt here, why is the only way to express 0 as 0+0+0.....
jagged saffron
whats your definition of direct sum
twilit field
I mean I understand why it is using the other defn
but it doesn't make much sense using this defn
jagged saffron
so by that very definition there is only one way to write 0 in the direct sum
jagged saffron
what do you mean 1-1?
twilit field
I mean the vector v - v
jagged saffron
what is v?
midnight plankBOT
last slate
@twilit field r u in college now
twilit field
<@796939110815629322> r u in college now
starting college in 2 months
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brisk lantern
could someone explains to me why I can do the following operation using complex exponentials?
midnight plankBOT
could someone explains to me why I can do the following operation using complex ...
brisk lantern
cant*
brisk lantern
I mean the two equations don't end up equaling each other
brisk lantern
is there a reason you think you can't?
First line and last line
lunar ocean
yeah those are just two unrelated identities involving the trig functions, there doesn't seem to be any conflict between them
the bottom identity is essentially $\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)$
grand pondBOT
bee [it/its]
lunar ocean
because $e^{ix}e^{iy} = e^{i(x+y)}$
grand pondBOT
bee [it/its]
lunar ocean
and so in fact you can derive the top identity, because $\cos(x-y) = \cos(x)\cos(y) + \sin(x)\sin(y)$, using the fact that $\cos$ is even and $\sin$ is odd
grand pondBOT
bee [it/its]
lunar ocean
so summing them together you get $2\cos(x)\cos(y)$, and then the $\frac12$ cancels that $2$ and you get $\cos(x)\cos(y)$
grand pondBOT
bee [it/its]
brisk lantern
oh okay I see I might need a second to digest that hahahahaa
brisk lantern
**bee [it/its]**
Perfect I just worked it out. Makes sense now. Thank you very much
midnight plankBOT
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brisk lantern
and so in fact you can derive the top identity, because $\cos(x-y) = \cos(x)\cos...
Hey so I actually ended up plotting the functions and ended up with the following:
where the blue line plot is: $\cos(x)cos(y)$
grand pondBOT
yoga545
brisk lantern
and the orange line is $Re{exp(jx)exp(jy)$
grand pondBOT
yoga545
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
lunar ocean
wow that's a somewhat ridiculous scale to plot it on
but yeah it makes sense that it would look like that
brisk lantern
but yeah it makes sense that it would look like that
oh okay I thought they should end up looking like the same plot at the end if i not mistaken
I thought instead of carrying out the formula using cos(x)*cos(y) I could just use Re{exp(jx)*exp(jy)} and it would result in the same answer
midnight plankBOT
brisk lantern
.reopen
midnight plankBOT
✅
brisk lantern
hmmm is there a way to use complex exponentials to carry out cos(x)*cos(y)
wary trail
if you know sin(x)sin(y) lol
lunar ocean
Re(e^ix) * Re(e^iy)? lol
brisk lantern
oh okay I see I'd have to convert back into cos
theres actually no way to do it in the complex exponential realm
well I guess I found a way....
just convert cos(x) and cos(y) to their respective 1/2(e^jz+e^-jz) and carry out the multiplication...
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last slate
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last slate
we just started learning about capacitance
there is an irregular cavity in a spherical conductor with +q
why did it induce exactly -q charge?
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last slate
<@&286206848099549185>
vital sinew
You might get better responses in the physics discord
sharp coral
since charges are free to move in a conductor, they will always move in such a fashion that the electric field is 0 on the inside of the conductor (otherwise, the charge would keep moving)
note that if the electric field inside the solid part is 0, then the electric flux across any gaussian surface enclosing the cavity but inside the conductor must be 0
gauss's law dictates that this flux is proportional to the net charge enclosed by the surface, so the net charge of the cavity and its surface must be 0
last slate
gauss's law dictates that this flux is proportional to the net charge enclosed b...
So our argument is based on off a arbitrary Gaussian surface like the one I marked in blue
sharp coral
yes
vital sinew
He's asking what your definition of the sine function is
Well, this definition doesn't really work if we want the argument to be greater than or equal to 90 degrees
One definition of sine is via the unit circle
If we rotate around the unit circle by theta degrees, the x coordinate of the point will be sin(theta)
So for example, sin(120) will be sqrt3/2
vital sinew
So for example, sin(120) will be sqrt3/2
Now, we can see that sin(theta)=sin(180-theta) because the circle is symmetric
I think this is bad pedagogy
Not to say that it's wrong, but I think a better approach is to just show them the definition that lets you prove it
vital sinew
If we rotate around the unit circle by theta degrees, the x coordinate of the po...
Sorry, I meant the y value here
Yes, the sine at the degrees is the second coordinate
Not to say that it's wrong
Because 180-theta will be the point at theta reflected across the y axis, and reflection across the y axis will keep the y coordinate the same
The x coordinate is the cosine value, the y coordinate is the sine value
But they're also just the coordinates of the points on the circle (the circle is the solutions to the equation x^2+y^2=1)
What do you mean by different values?
It will always show the values of sin and cosine
Ok so, the unit circle is a circle centered at the origin (the point (0,0)) that has radius 1
We're going to define our point at 0 degrees to be the point (1,0), and the point at theta degrees will be the point we get when we rotate that point around the circle by theta degrees
The x coordinate of the point at theta degrees will be cosine, the y coordinate is sine
This is how we define sine and cosine (though notice that for theta<90, this will actually get us the same value as the right triangle definition, because we have a right triangle with hypotenuse 1 and legs of length [x coordinate] and [y coordinate])
Yes
This is how we define sine and cosine for theta>90
Define in the sense of an English word
That's just what we say the function is
Yeah
The unit circle is how we know what the value of sine and cosine are if the angle is greater than 90
Well, how would you find sin(91) using the right triangle definition?
Right, but that doesn't work for sin(91), does it?
Because we can't have a triangle that has an angle of 91 and an angle of 90
You also can't do it for 90, because we can't have a triangle that has two 90 degree angles
If we did, then one of the angles would be 0 degrees (which clearly doesn't work)
Yes, this is possible
Remember, the angles in a triangle have to add up to 180, and they all have to be positive
But with our new definition using the unit circle, we can find sin for values that can't be part of a triangle (like 270 or 180)
Yes
Well, how would you find sin(180) without the unit circle definition
Well you can use the right triangle defn to find sin(89.5) but yeah, for anything 90 degrees or more you have to use the circle
Sorry?
Yes, the circle definition is just the unit circle
Yes, you need the unit circle definition for sin(theta) to even have a meaning for theta>90
Sure, I guess you could say that
How do you think the calculator knows
Well, the real answer is "calculus"
But the calculus works because we define it using the unit circle
vital sinew
Well, the real answer is "calculus"
(unless it's 60,45,or 30)
Though you can use the unit circle to just visually figure out things like sin(120)=sin(60) or sin(90)=0
Basically
last slate
Weird question but what is the largest possible shape for a set perimeter
Ex. If I wanted the largest possible area, but the perimeter would not exceed 50 feet, what would said shape be?
Psure it's a circle but maybe I'm wrong
midnight plankBOT
Weird question but what is the largest possible shape for a set perimeter
Ex. I...
surreal moon
Weird question but what is the largest possible shape for a set perimeter
Ex. I...
Pretty sure it's a circle. And I think square is the smallest. Or I could have those backwards
You can prove this in differential geometry, but that was ages ago for me
dull yoke
i think its the other way around
wait
lets say we have 10 m of cord: r = 5/pi -> area = 25/pi m^2 (CIRCLE)
square: 10m of cord means each side is 2.5m, and thus area = 6.25 m^2
surreal moon
If square has perimeter 1, then s=1/4 and area is 1/16. If circle has circumference 1, then r=1/2pi and A=1/4pi
1/4pi>1/16 since pi<4
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lusty cave
midnight plankBOT
lusty cave
Can someone please check my work here
And then I also have one more question to check
dull yoke
looks right
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last slate
Hi! I'm sorry to ask about this here, I can't find a trigonometry-only discussion. I don't understand the reflections of the sine graph.
midnight plankBOT
Hi! I'm sorry to ask about this here, I can't find a trigonometry-only discussio...
last slate
I'm trying to reason it doing drawing but I don't think they are of any value
Why does this happen?
If the angle is negative it means we're moving clockwise. So, both the x and y values are decreasing.
Why is the graph of -sen(x) the same as sen(-x)?
Why does -sen(x) exist?
OMG I GOT IT
.close
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ocean plaza
Given a group representation $\rho: G \to GL(V)$, over a $K$-vector field $V$, we can define a natural $K(G)$-module structure on $V$. In the definition of an absolutely irreducible representation, they use the notion of extending the field of the representation, but I don't get how that is defined
midnight plankBOT
Given a group representation $\rho: G \to GL(V)$, over a $K$-vector field $V$, w...
grand pondBOT
Eduude
midnight plankBOT
ocean plaza
how is this defined?
midnight plankBOT
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last slate
Looking for a pdf that can explain me everything I need to know from "find the area drown" in highschool.
midnight plankBOT
Looking for a pdf that can explain me everything I need to know from "find the a...
@last slate Has your question been resolved?
slender cobalt
A parabola and a line closing up an area
..
what is blud talking about
are you talking about finding the area between 2 curves?
you gotta take the integral of the absoloute value of the difference of the two functions.
in practice you'd integrate the top function - the bottom function
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chrome vessel
On the second line it should be $\sum_{n=0}^\infty \sum_{i=1}^n p(1-p)^i$ (edit: nvm)
wait nvm, my mistake
do you know the sum of geometric series?
so what would $\sum_{n=1}^\infty (1-p)^n$ be?
grand pondBOT
qwertytrewq
chrome vessel
if you want to include 0 sure. What would $\sum_{n=0}^\infty (1-p)^n$ be?
grand pondBOT
qwertytrewq
grand pondBOT
qwertytrewq
chrome vessel
Hint: what is $(1-p)^i\sum_{n=0}^\infty (1-p)^n$
grand pondBOT
qwertytrewq
grand pondBOT
qwertytrewq
chrome vessel
notice that its very very similar to $\sum_{n=i}^\infty (1-p)^n$
grand pondBOT
qwertytrewq
chrome vessel
do you know where I am going at?
can you show that
$$\sum_{m=0}^\infty (1-p)^{m+i}=\sum_{n=i}^\infty (1-p)^n$$
chrome vessel
yes that the idea, but you get why they are equal right?
its reindexing
Note that you can write
chrome vessel
its reindexing
Note that you can write
Note that you can write
$$(1-p)^i+(1-p)^{i+1}+\cdots$$
as
$$\sum_{n=i}^\infty (1-p)^n$$
or
$$\sum_{m=0}^\infty (1-p)^{m+i}$$
grand pondBOT
qwertytrewq
chrome vessel
yeah!
chrome vessel
**l'agit**
Now you can simplify the sum
aren't we all?
before it all clicks
yeah i think its right
grand pondBOT
l'agit
chrome vessel
yeah the $\frac{ (1-p)^{i}}{p}$ you computed
btw u can edit your original message and texit will recompile
grand pondBOT
qwertytrewq
chrome vessel
btw i made a typo, (1-p)^i should be on the numerator
and from here you will see it will be another geometric series
no worries
edgy schooner
If $|z_1| = |z_2| = |z_3| =1$ and $z_1 + z_2 + z_2 = 0$ then, what is the area of the triangle whose vertices are $z_1, z_2$ and $z_3$
midnight plankBOT
If $|z_1| = |z_2| = |z_3| =1$ and $z_1 + z_2 + z_2 = 0$ then, what is the area o...
grand pondBOT
Pro_Hecker
edgy schooner
I don't know how to approach this problem
I think these form an equilateral triangle, but i don't see how
last slate
**Pro\_Hecker**
Omega 😏
Cube root of unity
@edgy schooner
BTW did you solve the previous question?
last slate
What curve was it?
last slate
what should i do, how did you get to cube root?
From the given conditions,you can come to the conlusion that cube roots of unity satisfy z
edgy schooner
did you just assume it or it is only case possible?
edgy schooner
Hmm,I got a parabola by solving it by your method and a line by another
how?
which parabola?
last slate
did you just assume it or it is only case possible?
Modulus of z is 1 and z1+z2+z3=0
And the fi4st thing that came to my mind was w
last slate
did you just assume it or it is only case possible?
That is not the only case
edgy schooner
if z1+z2+z3/2 =0 so , centroid lies on origin
chrome vessel
If $|z_1| = |z_2| = |z_3| =1$ and $z_1 + z_2 + z_2 = 0$ then, what is the area o...
note that solution is up to rotation so we can fix z_1=1
edgy schooner
if z1+z2+z3/2 =0 so , centroid lies on origin
so how can i say it is equilateral triangle
chrome vessel
now we have |z_2|=|z_3|=1 and they sum to -1
edgy schooner
ok
chrome vessel
now we have |z_2|=|z_3|=1 and they sum to -1
HINT: write z_2=a+ci, then z_2+z_3=-1 gives you z_3=-1-a-ci
try to use the equalities to figure out what a and c are
edgy schooner
why is z_2 + z_3 = -1?
chrome vessel
why is z_2 + z_3 = -1?
because we assumed z_1 is 1 (note that is z_1,z_2,z_3 is a solution, we can rotate the 3 vectors by a certain degree and it will still be a solution)
so we can rotate z_1 so the it is 1
edgy schooner
ok
edgy schooner
yes
chrome vessel
HINT: write z_2=a+ci, then z_2+z_3=-1 gives you z_3=-1-a-ci
👍 now we reduced it to two variables, and the simultaneous equation is not hard to solve
edgy schooner
there are 3 variable z_3, a and c
chrome vessel
there are 3 variable z_3, a and c
z_2+z_3=-1 gives us z_3 in terms of a and c
note that z_2+z_3=-1 comes from z_1+z_2+z_3=0
edgy schooner
$\sqrt{(-1-a)^2 + (c)^2} = \sqrt{a^2+c^2} =1$
grand pondBOT
Pro_Hecker
grand pondBOT
qwertytrewq
chrome vessel
now it becomes clear that they are indeed the roots of unity
grand pondBOT
Pro_Hecker
edgy schooner
oh
last slate
Oh
chrome vessel
OH
last slate
Lol
chrome vessel
yup!
edgy schooner
Answer is $\frac{3\sqrt{3}}{4}$
grand pondBOT
Pro_Hecker
edgy schooner
thanks for the help
chrome vessel
Answer is $\frac{3\sqrt{3}}{4}$
lemme double check
last slate
You can just directly find the height and the base length
chrome vessel
probably right
chrome vessel
You can just directly find the height and the base length
that works too but all roads leads to 3\sqrt{3}/4
edgy schooner
.close
midnight plankBOT
chrome vessel
Answer is $\frac{3\sqrt{3}}{4}$
oh yeah btw since we rotated z_1 to 1, you'd have to justify that the area is invariant through rotation. However, this is not that hard to show.
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last slate
FUNCTIONS
anyone please help me on how to solve this question cause i legit have no idea how to solve it
midnight plankBOT
# FUNCTIONS
anyone please help me on how to solve this question cause i legit h...
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brisk lantern
Is there a way to simplify the following to just cos and sin without the j?
midnight plankBOT
Is there a way to simplify the following to just cos and sin without the j?
brisk lantern
In this case they are just constants
oh okay I see
I thought the i would impoart a pi/2 phase shift
for the sinusoidal terms somehow
midnight plankBOT
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surreal moon
I thought the i would impoart a pi/2 phase shift
You can reduce it to just one trig function, yes
The exact phase shift depends on A and B though
gray pumice
let X be an evenly distributed random variable on {1,2,3,4,...,n}.
(I know the expected value and variance.)
Approximate P{X>=3n/4} using Chebychev's formula.
midnight plankBOT
let X be an evenly distributed random variable on {1,2,3,4,...,n}.
(I know the e...
gray pumice
The problem I'm having is that I don't see how
$P(|X-\mu|\geq k) \leq \frac{\sigma^2}{k^2}$
grand pondBOT
Bob Goldham
gray pumice
can be used to approximate $P(X\geq \frac{3}{4}n)$
grand pondBOT
Bob Goldham
grand pondBOT
Bob Goldham
gray pumice
$\sigma^2 = \frac{3(n+1)^2}{4}+\frac{(n+1)(2n+1)}{6}$
grand pondBOT
Bob Goldham
gritty hatch
Q
gritty hatch
Q1: When factoring polynomials, aren't you able to factor and multiply in any way you want so long as it works? Why specefically binomials multiplied together?
junior flower
what?
dusky flint
Q1: When factoring polynomials, aren't you able to factor and multiply in any wa...
well its technically easier to find roots
gritty hatch
A: Polynomials, when factored, typically are broken into binomials multiplied together.
dusky flint
A: Polynomials, when factored, typically are broken into binomials multiplied to...
he is asking their benefits
junior flower
A: Polynomials, when factored, typically are broken into binomials multiplied to...
for quadratics, what else could they be broken down into?
dusky flint
for quadratics, what else could they be broken down into?
well they could be made into squares
dusk furnace
Usually "factor a polynomial" means breaking it into as small factors as possible -- and if you can get linear factors, there'll be nothing smaller to get.
dusky flint
for example (x+c)^2 + d = 0
junior flower
that’s not factored
dusk furnace
That can be a valid rewriting, but it wouldn't be described as "factoring".
gritty hatch
for quadratics, what else could they be broken down into?
A: e.g., if they simply tried multiplying them raw, or any method not utilizing binomial multiplication?
dusk furnace
Uhm. If you have two binomials, any method that results in their product is "binomial multiplication".
dusky flint
that’s not factored
i was explaining what other forms they can be broken into
dusk furnace
That term describes the outcome, not a particular method.
dusky flint
please read above linking text
dusky flint
A: e.g., if they simply tried multiplying them raw, or any method not utilizing ...
like completing squares etc.
gritty hatch
✅
Q: What if someone tried multiplying them raw after breaking up the pieces? Like constantly re-organizing the problem to multiply the correct parts. Or would that still somehow end up in binomials?
dusk furnace
It's not very clear from your description what the someone is trying to achieve.
At worst they would be wasting time doing something that won't help them reach their goal.
At best they would reach it.
gritty hatch
A: they wish to solve the quadratics in the way described within this video. https://youtu.be/GuwofNeT9ok
dusk furnace
Okay, so you've followed that process and gotten something like (x+5)(x+2), and your question is what would happen if you actually carry out that multiplication? You should be getting the original quadratic back; otherwise something has gone wrong along the way.
gritty hatch
Q: What principles or processes would one use to determine that was a possible answer?
I: It appears to complex to easily determine such a result.
olive matrix
nope it's pretty simple, 15 - 2 = 13
gritty hatch
Q: They also appear to be attempting to solve for what input value as X would generate a result of zero from the function/operations done unto it.
Guess: Probably for behavior modeling purposes or something.
gritty hatch
nope it's pretty simple, 15 - 2 = 13
Q: Is there no process? Do you simply look at it and try different options with your knowledge of the operations that typically occure and find a fitting result?
dusk furnace
Judging from the thumbnail, the video (which I don't care to watch) seems to present essentially a guess-and-check procedure that will sometimes work, but also sometimes doesn't produce a result.
A more reliable and systematic method would be to use the quadratic formula to find the roots of ax² + bx + c = 0. If we call those roots p and q, then we can factor ax² + bx + c = a(x-p)(x-q).
olive matrix
that video is roughly the method that I use, and then fall back to quadratic formula if that doesn't work
valid rover
^^^
olive matrix
but i've probably factored.... at least hundreds, maybe thousands of quadratic expressions
so for a lot of them I can look at it and instantly get the answer
gritty hatch
Q: Would you sometimes add rather than subtract in that formula in the quantity you wrote?
Q: So this is simply the fast estimation/guess-based method?```
olive matrix
I wouldn't characterize it as estimation but it does involve guessing to an extent
What's with the pictures
gritty hatch
What's with the pictures
uhm....
I am thinking really hard, and adjusting as I go.
olive matrix
okay. I don't understand this question:
Q: Would you sometimes add rather than subtract in that formula in the quantity you wrote?
what formula?
gritty hatch
olive matrix
it will always be subtracting p and q
gritty hatch
Does X-P represent positive or negative inputs?
olive matrix
however, if p is for example -3, then the expression will involve (x - -3) which is equivalent to (x + 3)
gritty hatch
Thank you.
R: Can you write, in summation, what each part of the formula does in a goal-oriented simplified manner?
olive matrix
no
gritty hatch
Ok.
gritty hatch
olive matrix
open desmos, type in y = a(x - p)(x - q) and then when it asks if you want to add sliders say yes to all
then you can move a, p, and q around
and see how the function changes
gritty hatch
gritty hatch
olive matrix
move each slider in turn. what happens?
hard umbra
type in (p, 0), (q, 0) as well and check the label box
gritty hatch
P and Q move the base, (the whole function graphed) whereas A determines how much is added or subtracted as the line goes on (I assume this because of the ever-growing curve.)
junior flower
hi open up my e-girl eyes and snoseph
silver falcon
how is this possible
midnight plankBOT
how is this possible
flat veldt
how is this possible
Show original question
silver falcon
Show original question
soz for low quality
full working out
<@&286206848099549185>
please and thankyou
midnight plankBOT
@silver falcon Has your question been resolved?
dusky goblet
full working out
Bro is studying on how to make a rocket to get to moon
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flat veldt
full working out
midnight plankBOT
https://www.indeed.com/career-advice/career-development/how-to-calculate-z-score
flat veldt
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zenith garden
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zenith garden
i got 87.70, but i checked ith https://www.gauthmath.com/solution/1-If-continuous-random-variable-X-is-normally-distributed-with-mu-8-7-and-sigma--1699973383528453 and they say otherwise,, who is right?
broken stone
I get 8.69%. 14 is above the mean and we're asking for greater than 14, so it definitely can't be more than 50%. Do you use tables or a calculator?
zenith garden
tables
zenith garden
i got 87.70, but i checked ith https://www.gauthmath.com/solution/1-If-continuou...
these guys say 12.30 though..
broken stone
these guys say 12.30 though..
Well, the z-score for this is $z=\frac{14-8.7}{3.9}=1.36$. When I look up 1.36 in a z-table, I get $P(X<14)=0.9131$, so to get the desired answer, we do $P(X>14)=1-P(X<14)=1-0.9131=0.0869=8.69%$
grand pondBOT
DonDoesMath
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wild glen
Could someone help work me through this double integral problem? I am getting a completely wrong answer (44/3 instead of 2/3), and I am wondering if I maybe did not set up my integrals correctly... I have narrowed the issue down to the 88/3x^3 final integral, which should actually be 4/3, but I have no idea where I went wrong.
midnight plankBOT
Could someone help work me through this double integral problem? I am getting a ...
wild glen
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wild glen
<@&286206848099549185>
wild glen
<@&286206848099549185> anybody? I'm really stuck
crisp oxide
Where does that 40x2 come from?
wild glen
Where does that 40x2 come from?
5/2x times 16x
When I distribute, everything cancels except for the x^2 terms
crisp oxide
Yeah that’s fine, the problem is in the integral setups somewhere
I suspect there’s an issue with your outer bounds
wild glen
Like the 2x+4 and 2x-4?
wild glen
I suspect there’s an issue with your outer bounds
Or the x upper bounds, +-2
crisp oxide
<@&286206848099549185> anybody? I'm really stuck
Well, I guess I’m confused too
crisp oxide
Oh oh oh the answer is 2/3, yeah then the x bounds should be -0.5 to 0.5
wild glen
Sorry I think what I said was confusing
wild glen
Oh oh oh the answer is 2/3, yeah then the x bounds should be -0.5 to 0.5
well its 2/3rds after dividing by area
Which I am pretty confident is 32
Relatively confident
Though that depended on the x bounds
But how do I figure out the initial bounds? Because at first I was thinking I set y=0 to figure out x, but then I realized if y is negative, x can be greater than 2 and if y is positive x can be less than -2...
Oh but there are the two inequalities
Ok yeah so -2<=x<=2 is definitely right for x
crisp oxide
Which I am pretty confident is 32
Are you sure?
wild glen
Are you sure?
Well that would change if the bounds changed because the calculation was based on the bounds
Wait so actually
It is supposed to just be the first octant
So then x would be 0 to 2 and y would be 0 to 4
?
So then is y still 2x+4?
Wait never mind
That was a different question
crisp oxide
I mean, 1/2bh for top and bottom: 1/2(4)(4)2=16
wild glen
It is a trapezoid though not a triangle?
crisp oxide
Two triangles top and bottom
wild glen
Oh true
What could be wrong with my A(R), then?
And 16 doesn't get me any closer to the answer because I still have 88/3
My y bound must be wrong
nova scaffold
Looks like an algebra problem
wild glen
Looks like an algebra problem
Well I mean the area problem is but the main problem isn't
crisp oxide
And 16 doesn't get me any closer to the answer because I still have 88/3
That’s an algebra error somewhere:
wild glen
That’s an algebra error somewhere:
😬 Ok but even that divided by 16 is still 8/3 not 2/33...
crisp oxide
So if we know the area should be 16, we should be able to figure out the bounds on A(r) like so:
And using this we get
Which is still slightly off
wild glen
Wait so actually
And how can it be 1, if x clearly goes to -2 and 2?
What calculator are you using?
wild glen
Yes, to satisfy both inequalities it does
Ok so it looks like it doesn't matter what a is but b has to be 1?
nova scaffold
I was wrong it's a bound error that looks like a sign error. Your lower y bound is -2x-4 I believe.
nova scaffold
Your integral should end up x^4+8/3x^3 and x^4 cancels due to symmetry
wild glen
I am going to do it on paper again and see if it works
wild glen
Your integral should end up x^4+8/3x^3 and x^4 cancels due to symmetry
😭 I still get A(R) = 32 though
nova scaffold
That's right, it's the area of the parallelogram.
rhombus I forgot my shapes for a second
wild glen
That's right, it's the area of the parallelogram.
How? If each quadrant is a traingle, 2*4/2 is 4 in each quadrant
totals 16
Area of a rhombus is d1*d2/2, it is 8 from top corner to bottom and 4 from left corner to right, also 16
nova scaffold
My brain might be dying.
crisp oxide
My brain might be dying.
I’m with ya
wild glen
Me too
crisp oxide
Well I figured out the area thing I think
wild glen
Well I figured out the area thing I think
Like that it is 16, or why the double integral is saying that it is 32?
wild glen
Oh that does make sense
crisp oxide
Yeah
So it’s an integral bounds issue, but we’re still confused 🙃
But we’re better informed now!
wild glen
So the problem is that it is taking into account that extra area
I somehow need to bound it on all four sides, right?
crisp oxide
And that explains why my trick was giving the same area, but wrong bounds
wild glen
Oh yeah
So |2x+y|=4 and |2x-y|=4 are the outer bounds
Does that make y=|2x-4| true or not necessarily
So basically I need to divide in half
That's it
But I can't just divide in half
Right?
wild glen
Like I keep getting double what the answer should be
Because the area between the bounds is double
crisp oxide
Like I keep getting double what the answer should be
Because you’re double counting the outer areas
crisp oxide
But if you didn’t have linear boundaries it wouldn’t be exactly double
crisp oxide
But there's no way to have both x and y bounded by equations?
Unless you parameterize or something
You could decompose it, but that feels unnecessary
nova scaffold
You can decompose it. I believe since y=2x+4 and y=-2x-4 are reflections of each other across the x axis you can justify just multiplying one side of the integral by 2 giving you the proper area so long as the bounds of x are from the vertex to the mid point of the rhombus. or something along those lines
wild glen
You can decompose it. I believe since y=2x+4 and y=-2x-4 are reflections of each...
I'm not sure I understand
You mean making x go from 0 to 2?
crisp oxide
I'm not sure I understand
You break it up for each easier region, eg the four quadrants in this case
wild glen
Yes, that makes sense
So in the case of the first integral, though, would it not also take the area of the space bounded by blue, purple, and black?
It seems like it works but I am just trying to understand why
Oh wait I see, nothing below the blue line
Thank you both, I never would have come up with that on my own
nova scaffold
I think you can do it in two. $$\int_{-2}^{0}\int_{-2x-4}^{2x+4}dydx + \int_{0}^{2}\int_{2x-4}^{-2x+4}dydx$$
grand pondBOT
Critzzzy
wild glen
That makes sense too, yeah
nova scaffold
But like I said, I think an argument of symmetry can be applied since the lines are parallel and symmetric, at least for the area part. The function 5xy+x^2 might need to be done in parts since I can't justify it off the top of my head.
wild glen
Oh yeah, but if f(x,y)=5xy+x^2 f(-1,1) dne f(1,1)
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jovial citrus
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jovial citrus
how is OD = OA + BC?
to find a particular vector let’s say AB, you basically find another vector that travels to AB (iygwim) right?
how does OA to BC land at OD?
ik OA + AD would
but why OA + BC
is it related to them being parallel? but how
midnight plankBOT
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jovial citrus
nobskibidi
last slate
is it related to them being parallel? but how
Both the vectors are parallel and have the same magnitude thus BC=AD
@jovial citrus
jovial citrus
Both the vectors are parallel and have the same magnitude thus BC=AD
Is AB = DC too?
last slate
✅
jovial citrus
Ohhh
last slate
Yes,ig
jovial citrus
Okay
jovial citrus
last slate
Everything is correct
jovial citrus
then why is it different from the answer key which says OD is (3,3,1)?
last slate
Hmm,idk
last slate
Oh it's the same question !?
@jovial citrus
The figure it wrong
Dumb me,didn't see it
The parallelogram is ABCD ,not ABDC
Get it?
jovial citrus
may I ask what’s that letter on the right form A?
below A is D right?
and C is at the other corner
last slate
Xd it's nothing
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@jovial citrus Has your question been resolved?
jovial citrus
another question
For Q3, May I ask how did they know that squareroot 10 = magnitude of r?
like, isn’t R supposed to be connecting to a different point from the position vectors?
looking at this illustration
position vector of R is connecting to somewhere else on the line
it’s not connecting to P
it definitely feels like I’m wrong somewhere about this
but I can’t exactly pinpoint where I went wrong in terms of my knowledge about r
last slate
For Q3, May I ask how did they know that squareroot 10 = magnitude of r?
It's not magnitude of r
BTW r is a point
Position vector*
It's the distance of a point from origin
Since origin is anyways 0 they ignored it
@jovial citrus
jovial citrus
BTW r is a point
is r any point on the vector?
jovial citrus
R is any point on the line
Okay
Let’s say we got this graph
Does this mean r = OP?
last slate
Does this mean r = OP?
I didn't get you
jovial citrus
I didn't get you
last slate
Gtg
jovial citrus
Oh okay
jovial citrus
like
last slate
Ping someone else or you can wait
jovial citrus
you see the curve there’s point p
and r also ends at point P
Does that mean OP = r
jovial citrus
Ping someone else or you can wait
okai
last slate
you see the curve there’s point p
there is no curve in the above question
midnight plankBOT
@jovial citrus Has your question been resolved?
twilit field
$\int \cos\left(2\theta\right)\ln\left(\frac{\left(\cos\left(\theta\right)+\sin\left(\theta\right)\right)}{\cos\left(\theta\right)-\sin\left(\theta\right)}\right)d\left(\theta\right)$
midnight plankBOT
$\int \cos\left(2\theta\right)\ln\left(\frac{\left(\cos\left(\theta\right)+\sin...
grand pondBOT
ƒ(Why am. I here)=I don't know
twilit field
I was thinking that $tan(\theta + \frac{\pi}{4})$ may help
grand pondBOT
ƒ(Why am. I here)=I don't know
twilit field
,w differentiate ln(tan(\theta + \frac{\pi}{4}))
grand pondBOT
twilit field
close
hard shard
i feel like conjugate would help big time
twilit field
**ƒ(Why am. I here)=I don't know**
from here?
gloomy cedar
Just chainrule if you want the dumb version
twilit field
**ƒ(Why am. I here)=I don't know**
this is the OG problem
gloomy cedar
Mb i only saw the wolfram thing
cerulean narwhal
hey
twilit field
👋
gloomy cedar
I'm on a mobile I doubt I can try things
cerulean narwhal
split cos 2 x = cos^2x - sin^2x
gloomy cedar
Ohh
midnight plankBOT
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near raven
so I have this question and its solution, but i still dont quite understand whats going on in red box i made
midnight plankBOT
so I have this question and its solution, but i still dont quite understand what...
near raven
also i am aware its in dutch but i put the translations where necessary
like i can kind of guess why they would input that the period is pi, but is it like a testing process or is there some other way they can deduce the period is pi ?
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craggy yew
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craggy yew
what is the result
olive matrix
the thing that they're trying to to prove
craggy yew
the thing that they're trying to to prove
why is a=c and b=d not valid if b and d and squares of rationals
runic hamlet
you can for example have 3+sqrt(4)=2+sqrt(9)
craggy yew
ohhh
craggy yew
you can for example have 3+sqrt(4)=2+sqrt(9)
couldnt that also count as square of integers?
craggy yew
then why are rational numbers specified
runic hamlet
you can have 3+sqrt(9/4) = 4 + sqrt(1/4)
but I didnt think that was a good example to make
cause its slightly harder to see
craggy yew
you can have 3+sqrt(9/4) = 4 + sqrt(1/4)
so is it valid for integers, whole numbers and natural numbers aswell?
runic hamlet
if something is valid for rationals then that includes integers, whole numbers and natural numbers
because those are all rationals
craggy yew
oh okay
craggy yew
because those are all rationals
could u give an example of the same with real numbers?
runic hamlet
it will feel constructed
0+sqrt(2) = sqrt(2) + sqrt(0)
where on the right sqrt(2) is c and sqrt(0) is sqrt(d)
midnight plankBOT
@craggy yew Has your question been resolved?
craggy yew
0+sqrt(2) = sqrt(2) + sqrt(0)
but a is not equal to c here
craggy yew
you can have 3+sqrt(9/4) = 4 + sqrt(1/4)
but it does not hold for rational numbers aswell
runic hamlet
ok we may be talking about different things
rational numbers have the caveat that a and c are only allowed to be different if b and d are squares of rational numbers
real numbers dont have that kind of restriction
or well, technically they do but thats cause all positive real numbers are squares of real numbers
silver falcon
Bro is studying on how to make a rocket to get to moon
huh
midnight plankBOT
@craggy yew Has your question been resolved?
craggy yew
oh okay
craggy yew
ok we may be talking about different things
(another question)
why is sqrt a - sqrt b whole square equal to sqrt c - sqrt d whole square
<@&286206848099549185>
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craggy yew
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last slate
1st question
midnight plankBOT
1st question
last slate
black book I see
yeah gonna start solving from chp 1 again
sad part is i did it long ago
couldnt do now for some reason bruh
i want full solution @zealous schooner
sinx lies from -1/4 ,1/4
the one with log base root2 got no problem
but with the bigger one i simply did greater than 0
ah shit
i got this thingg
its was easy
now that i have sinx range [-1/4,1/4] i can square it, multiply by 16 add 1 and so on to finally convert it into the question
got it
thanks
.close
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twilit field
I don't get this, could someone walk me though this
midnight plankBOT
I don't get this, could someone walk me though this
prime hornet
what is the definition of linear independence?
Axler is just using it to show that these vectors are LI
twilit field
prime hornet
right.
so call the vectors v1, v2, v3
Axler is writing a1v1 + a2v2 + a3v3 = 0, and then showing that a1, a2, and a3 must be 0
ergo, v1, v2, v3 are LI
prime hornet
twilit field
the last bit still confuses me though
prime hornet
in general, to show that a set of vectors is LI, you should always write a linear combination and set to 0
twilit field
the $a_1=a_2..=0$
grand pondBOT
ƒ(Why am. I here)=I don't know
prime hornet
are you asking how he deduced that, or what he means by that?
twilit field
how he deduced that
hard umbra
twilit field
they are all 0, right
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twilit field
could someone explain this
midnight plankBOT
could someone explain this
twilit field
I'm unable to understand why $a_0=a_1....=0$
grand pondBOT
ƒ(Why am. I here)=I don't know
hard umbra
a polynomial is 0 if and only if all of its coefficients are 0
fading palm
the polnomials are the same when they have the same coeffiecnts at x^i
twilit field
a polynomial is 0 if and only if all of its coefficients are 0
how does that imply linear independence
hard umbra
by definition of linear independence
midnight plankBOT
twilit field
the polnomials are the same when they have the same coeffiecnts at x^i
thanks
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pulsar sun
I want to make a program in which given 3 points coordinate in 2D before and after a rotation (that i don't know, it's random) I can deduce the rotation. But my teacher said that sometimes the rotation doesn't exists with 3 points or more but I can't see why. Can someone explainto me ?
midnight plankBOT
I want to make a program in which given 3 points coordinate in 2D before and aft...
surreal moon
I want to make a program in which given 3 points coordinate in 2D before and aft...
Depends on other constraints. Please refine your question more
pulsar sun
what does this means, what are the contraints ?
we rotate around the origin and we are in 2D
we conserve the norm
@surreal moon
pulsar sun
Then you really only need one point
what this means ?
surreal moon
Suppose your point is $(x, y)$, then its angle from the x axis is $\atan(y, x)$. If you're given its rotated point $(x', y')$, then its angle would be $\atan(y', x')$. So the angle of rotation is just their difference.
grand pondBOT
SWR
pulsar sun
ye but if I have 3 points or more it's my original question, I already made it for 2 points
surreal moon
ye but if I have 3 points or more it's my original question, I already made it f...
describe what you mean, because it's not making sense.
Give me an example
pulsar sun
ok so imagine I have three point (x,y,z) and also (x',y',z') that are the points after a rotation from an angle lets say theta that we want to deduce. Does the rotation exists, I mean I think yes because we already know there was a rotation but my teacher say maybe not in some cases. Or does he wanted to say if I have (x,y,z) and (x',y',z') that conserve the norm from the origin and distance between them does there exist a rotation between (x,y,z) and (x',y',z') ? Is it clear ?
strong lava
you assume the rotation is about the origin?
pulsar sun
ye rotation is around the origin
strong lava
Then you only need one point to find what it has to be
Then look at whether the other 2's movement corresponds, if so there is a rotation, if not there isn't
surreal moon
ok so imagine I have three point (x,y,z) and also (x',y',z') that are the points...
By (x, y, z), do you mean a point in 3 dimensions?
pulsar sun
yes but why after 3 points or more the rotation can be impossible ?
pulsar sun
By (x, y, z), do you mean a point in 3 dimensions?
no i mean 3 points in a plane sorry for the confusion
strong lava
For instance if (1,0) goes to (0,1) you know it's 90° ccw
So if you also see (0,1) -> (1,0) it's impossible
pulsar sun
like coordinate of x can be (x1,y1)
surreal moon
like coordinate of x can be (x1,y1)
call the points p1, p2, and p3 to avoid confusion
pulsar sun
For instance if (1,0) goes to (0,1) you know it's 90° ccw
So if you also see (0,...
? (0,1) to (1,0) impossible ?
pulsar sun
no
strong lava
Well try to visualize things
pulsar sun
but we have a rotation of -90 no ?
strong lava
For instance if (1,0) goes to (0,1) you know it's 90° ccw
So if you also see (0,...
which is contradictory with the other point which imposes that it be a +90° rotation
pulsar sun
but we actually have a rotation between this 2 points ? I dont understand you
I don't want the rotation to be the same
strong lava
ok maybe me giving two purposes to the same points was a bad idea
strong lava
I don't want the rotation to be the same
what
pulsar sun
like I just want to know if there exists a rotation
strong lava
aren't you trying to find a rotation R such that R(x) = x', R(y) = y', R(z) = z'
pulsar sun
yes
strong lava
For instance if (1,0) goes to (0,1) you know it's 90° ccw
So if you also see (0,...
x = (1, 0), y = (0, 1)
x' = (0, 1), y' = (1, 0)
to send x to x' you need 90° ccw
to send y to y' you need 90° cw
not compatible
pulsar sun
sry what cw and ccw means
strong lava
clockwise and counterclockwise
pulsar sun
ty
strong lava
hence the commands ,rcw and ,rccw for images (r for rotate)
pulsar sun
but (x,y) and (x',y') are the same point
strong lava
ok maybe me giving two purposes to the same points was a bad idea
yeah but just rotate y, y' by 180° and the point still stands
if that confuses you
pulsar sun
so there is no rotation between them
strong lava
y = (0, -1) and y' = (-1, 0)
strong lava
so there is no rotation between them
yeah, each can be rotated to the other, but not by the same rotation in each case
pulsar sun
and if the points are not the same, cause this case we just say there is no rotation because our points are the same
strong lava
y = (0, -1) and y' = (-1, 0)
no
pulsar sun
i mean x'= y and y'=x
strong lava
y = (0, -1) and y' = (-1, 0)
Not anymore
strong lava
Pick any 3rd you want
pulsar sun
Pick any 3rd you want
and it will not work