Hmm, isnt that the same thing?
#help-41
pseudo crescent
honest dagger
Maybe? Imo its less work. But that's me
pseudo crescent
or do you mean lagrange interpolating polynom?
honest dagger
Idk the name. It's just what i do
honest dagger
That looks neat. Not what i do, but i think what i do is just that with extra work
pseudo crescent
That looks neat. Not what i do, but i think what i do is just that with extra wo...
how do you do it?
honest dagger
how do you do it?
I know P(x) will be my 7-degree polynomial with P(1)=1. So Q(x)=P(x)-1 will have Q(1)=0. So Q(x)=R(x)(x-1) for some sixth degree polynomial. I continue down until I get a line then solve for the two remaining points'
indigo cloud
I think that leads to newton interpolation
honest dagger
oh is that what I'm doing?
pseudo crescent
newton is same as lagrange?
indigo cloud
no, newton is different
honest dagger
I like to think that makes me as smart as Newton 
indigo cloud
google it. you can draw a funny table and compute a few values and then you are done
if you know the proof of the chinese remainder theorem, then lagrange polynomials are basically the polynomial version of that
as a fun fact
polynomial Interpolation is solving for p(x) = y_i mod (x-x_i)
indigo cloud
I always love it when two seemingly completely unrelated things are actually the same
this is my favourite example of that
limpid terrace
3 cos(2 x + 15 °) = sqrt(5)
Is it fine to just use the 15 degrees as is or is there like a special step
wide perch
its fine, just make sure you're consistent with degrees vs radians.
limpid terrace
Can wolfram be wrong? This looks quite weird to me
wide perch
If I had to guess it might be taking 15 as radians, not degrees. Try converting to radians first
limpid terrace
It's in degrees
vague thorn
Can wolfram be wrong? This looks quite weird to me
i mean, simplify the brackets
this seems okay to me
Wolfram converted 15 degrees to $\frac{\pi}{12}$ rad
grizzled pagodaBOT
@vague thorn
amber waspBOT
amber waspBOT
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deft vine
amber waspBOT
dull pike
What have u done in your first line?
deft vine
What have u done in your first line?
changing the formula to find the final temperature
if that's the formula that means I can't cancel out the volume?
vague thorn
why'd you multiply V2 on both sides and cancel V2s which are on the numerator
I suppose you meant to divide both sides by V2
vague thorn
well..
deft vine
vague thorn
$\frac{V_1}{T_1}=\frac{V_2}{T_2}$
$\implies\frac{1}{V_2}\times\frac{V_1}{T_1}=\frac{1}{\cancel{V_2}}\times\frac{\cancel{V_2}}{T_2}$
grizzled pagodaBOT
@vague thorn
vague thorn
yeah, yes
amber waspBOT
zinc goblet
.close
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tough mica
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cunning birch
what's your question?
tough mica
can someone explain with a simple example
I dont see how this property works, but is used heavily in my reader
I dont ... get it
@cunning birch please
cunning birch
Well
You have a transformation f
From V to W
And a way to interpret it is through a matrix representation
For example if you have bases Bv, Bw
M(Bv,Bw)(f) is a way to "compute" f
It reads a vector v of V in basis Bv, call X the coordinate vector in basis Bv
And it outputs f(v) written in the coordinates of basis Bw
Which would be obtained through matrix multiplication M(Bv,Bw)(f) * X
So to sum up, M(Bv,Bw)(f) computes f the following way: it takes as input the coordinates of your vector in basis Bv
And it outputs in coordinates of basis Bw
But Bv and Bw were arbitrary, I can replace those bases with other bases of V and W, and I would get another representation of f through those new bases
tough mica
So to sum up, M(Bv,Bw)(f) computes f the following way: it takes as input the co...
up until here what you said only explains left side of equality, besides, is an example going to be provided?
cunning birch
up until here what you said only explains left side of equality, besides, is a...
Yes, the question is: if I take two different representations of the same transformation f, say M(Bv,Bw)(f) and M(Bv',Bw')(f), how are they connected
Well first of all, one can freely switch from a basis B to a basis B' of the same vector space using "change of basis" matrices:
C_(B1,B2) is the matrix that takes as input a vector as written in basis B1
And outputs the same vector but written in basis B2
cunning birch
M(Bv,Bw)(f) is a representation of the transformation f
tough mica
that also changes basis?
cunning birch
Well sometimes change of basis is mandatory
Since V is possibly not the same vector space as W
Though there is a link
tough mica
examples should be provided because I am not following
cunning birch
C(B1,B2) = M(B1,B2)(Identity)
tough mica
that is true!! but idk why
cunning birch
Recall that M(Bv,Bw)(f) takes as input a vector v written in Bv
And outputs f(v) written in Bw
So
M(B1,B2)(Identity) takes as input a vector v written in B1
And outputs Identity(v) = v written in B2
cunning birch
C_(B1,B2) is the matrix that takes as input a vector as written in basis B1
Well
tough mica
Identity is a linear trasnformation here?
identity linear operator with domain and codomain wait
examples would be highly appreciated because reader straight jumps into proving the property
cunning birch
Ok
tough mica
cunning birch
Consider f:R²-> R³
f(x,y) = (x+y,y,0)
tough mica
domain is R2 codomain is R3
cunning birch
we consider the basis B = {(1,0),(1,1)} in R²
And the basis C = {(1,0,0),(0,1,0),(0,0,1)} in R³
Since f is a linear transformation
We can try to find M(B,C)(f)
tough mica
We can try to find M(B,C)(f)
i am trying
cunning birch
Ok
In the meantime I just write B = {b1,b2} and C = {c1,c2,c3}
To find M(B,C)(f), we have to recall that its 1st column is [f(b1)]_C
And second column is [f(b2)]_C
tough mica
Yeah but is it bad if I ask you to not use canonical basis for C
cunning birch
Sure, that makes the example harder for you but ok
tough mica
cunning birch
f(b1)_[(1,0,0)]?
cunning birch
[image]_basis
tough mica
Yeah
[f(b1)]_[C]
f(b1) = f(1,0)
f(x,y) = (x+y,y,0)
f: R2 -> R3
f(1,0)=(1,0,0)
[(1,0,0)]_[ C ] = ???
tough mica
(1,0,0) = 1.c1 + 0c2 + 0c3
cunning birch
[v]_C = (x,y,z)
tough mica
f(b2) = f(1,1)
Consider f:R²-> R³
f(x,y) = (x+y,y,0)
f(1,1) = (2,1,0)
[(2,1,0)]_[C] = (0,1,-1)
cunning birch
Yes
grizzled pagodaBOT
Renato Chavez
cunning birch
Yes
cunning birch
so
There are two ways to start from a vector v written in basis Bv'
and get f(v) written in basis Bw'
either you have M(Bv',Bw')(f) and you immediately get that
otherwise
if you know M(Bv,Bw)(f)
for some other bases Bv,Bw
so you start by
converting "v written in basis Bv' " into "v written in basis Bv "
then
use M(Bv,Bw)(f), since you now have v in basis Bv
to get f(v) in basis Bw
and finally
tough mica
an intuitive example can help to understand the right side of the equality
cunning birch
convert f(v) written in basis Bw, into f(w) written in basis Bw'
I literally told you intuitively how to understand the right hand side
tough mica
I am trying
cunning birch
start by converting the input from basis Bv' to basis Bv
then compute f(v) in basis Bw
then convert the output from basis Bw to basis Bw'
tough mica
start by converting the input from basis Bv' to basis Bv
What is basis Bv' and Bv in this case
cunning birch
f(b2) = f(1,1)
if you took this f again
considered B' = {(4,2),(1,1)}
and C' = {(1,4,4),(2,2,2),(0,1,-1)}
to find M(B',C')(f)
we could do the whole shenanigans we did for M(B,C)(f) earlier
or
cunning birch
we use this formula
tough mica
Let's use the formula but idk how is used how to use it
I can make drawings i have paper and pen btw
cunning birch
?
like
all we have to do
is find
the matrix that changes basis B' into basis B
and the matrix that changes C into basis C'
so
M(B',B)(identity)
and M(C,C')(identity)
tough mica
What's with this identity thing
tough mica
Domain and codomain?
cunning birch
depends with which bases you're working
if you're working with B and B'
then R^2 to R^2
if you're working with C and C'
R^3 to R^3
tough mica
id : V -> W in general but I'm this case?
cunning birch
no
id:V->V
or id:W->W
so
if you're working with the bases of R^2
id:R^2->R^2
tough mica
tough mica
But B is in R2 and C is in R3
cunning birch
when did I ever talk about M(B,C)(identity)
tough mica
M(B',B)(identity:R^2->R^2)
Okie
tough mica
when did I ever talk about M(B,C)(identity)
Sry i see now
M(B'B)(id)=C(B'B)
id : R2 -> R2
C' = {(1,4,4),(2,2,2),(0,1,-1)}
B' = {(4,2),(1,1)}
Let's find that matrix C(B'B)=M(B'B)(id : R2 -> R2)
B = {(1,0),(1,1)}
id(4,2)=(4,2)
[(4,2)]_(B) = (2,2)
tough mica
cunning birch
yes
that's the transformation you're working with on R^2->R^2
and you wanna find M(B',B)(I_R^2)
tough mica
id(1,1)=(1,1)
[(1,1)]_(B)=(0,1)
C' = {(1,4,4),(2,2,2),(0,1,-1)}
B' = {(4,2),(1,1)}
id(4,2)=(4,2)
[(4,2)]_(B) = (2,2)
,, C_{B'B} = \begin{pmatrix} 0 & 2 \ 1 & 2 \end{pmatrix}
grizzled pagodaBOT
Renato Chavez
tough mica
and now for $C_{BB'}$ we can invert this matrix (find the inverse)
grizzled pagodaBOT
Renato Chavez
tough mica
I was trying to use the formula as u said
tough mica
is C_BB' needed
unsure
,, M_{B'C'}(f) = C_{CC'} \cdot M_{BC}(f) \cdot C_{B'B} \ M_{B'C'}(f) = C_{CC'} \cdot \begin{pmatrix} 1 & 0 \ 0 & 1 \ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} 0 & 2 \ 1 & 2 \end{pmatrix}
tough mica
I am trying
grizzled pagodaBOT
Renato Chavez
tough mica
C' = {(1,4,4),(2,2,2),(0,1,-1)}
B' = {(4,2),(1,1)}
B = {(1,0),(1,1)}
C = {(1,0,0),(2,1,1),(0,0,1)}
Consider f:R²-> R³
f(x,y) = (x+y,y,0)
,, M_{B'C'}(f) = C_{CC'} \cdot M_{BC}(f) \cdot C_{B'B} \ M_{B'C'}(f) = C_{CC'} \cdot \begin{pmatrix} 1 & 0 \ 0 & 1 \ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} 0 & 2 \ 1 & 2 \end{pmatrix}
grizzled pagodaBOT
Renato Chavez
tough mica
I am not following
cunning birch
wdym
tough mica
im unsure wat u mean exactly
tough mica
maybe there is something dat i am missing
maybe i am not following, im not sure how this formula works, is hard
cunning birch
?
The formula is given
M(B',C')(f) = C_(CC') * M(B,C)(f) * C_(B'B)
there is C_(CC') in this formula
there is C_(B'B) in this formula
but there is no C_(BB')
so we don't need C_(BB')
tough mica
Okie
I see what you mean now
C' = {(1,4,4),(2,2,2),(0,1,-1)}
B' = {(4,2),(1,1)}
B = {(1,0),(1,1)}
C = {(1,0,0),(2,1,1),(0,0,1)}
@cunning birch
Like this?
I am trying
cunning birch
yeah
grizzled pagodaBOT
grizzled pagodaBOT
grizzled pagodaBOT
tough mica
@cunning birch
We need to check if the matrix multiplication yields the same as the left side of the equality
C' = {(1,4,4),(2,2,2),(0,1,-1)}
B' = {(4,2),(1,1)}
B = {(1,0),(1,1)}
C = {(1,0,0),(2,1,1),(0,0,1)}
Consider f:R²-> R³
f(x,y) = (x+y,y,0)
,w rref {{1,2,0,6},{4,2,1,2},{4,2,-1,0}}
grizzled pagodaBOT
cunning birch
if you want to check sure
tough mica
,w rref {{1,2,0,2},{4,2,1,1},{4,2,-1,0}}
grizzled pagodaBOT
tough mica
tough mica
if you want to check sure
I want
🤓
let me put all of this together in one equation
,, M_{B'C'}(f) = C_{CC'} \cdot M_{BC}(f) \cdot C_{B'B} \ M_{B'C'}(f) = C_{CC'} \cdot \begin{pmatrix} 1 & 0 \ 0 & 1 \ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} 0 & 2 \ 1 & 2 \end{pmatrix} \ \begin{pmatrix} \frac{-5}{3} & \frac{-1}{2} \ \ \frac{23}{6} & \frac{5}{4} \ \ 1 & \frac{1}{2} \end{pmatrix} = \begin{pmatrix} \frac{-1}{3} & \frac{-1}{3} & \frac{1}{6} \ \ \frac{2}{3} & \frac{7}{6} & \frac{-1}{12} \ \ 0 & 0 & \frac{-1}{2} \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \ 0 & 1 \ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} 0 & 2 \ 1 & 2 \end{pmatrix}
grizzled pagodaBOT
Renato Chavez
tough mica
@cunning birch
am I tripping here?
,w {{-1/3, -1/3, 1/6}, {2/3, 7/6, -1/12}, {0, 0, -1/2}} . {{1, 0}, {0, 1}, {0, -1}} . {{0, 2}, {1, 2}}
grizzled pagodaBOT
cunning birch
<@246787399991033858> rafi please
sorry I went away
why do you think there is a mistake?
oh
well let me compute those again
tough mica
tough mica
well let me compute those again
💖 
amber waspBOT
@tough mica Has your question been resolved?
tough mica
there is cerrainly a mistake, but im not sure where
otherwise RHS = LHS
<@&286206848099549185> @cunning birch please
😔
cunning birch
<@&286206848099549185> <@246787399991033858> please
Did you swap the columns?
tough mica
thank you for finding it that was it
,, M_{B'C'}(f) = C_{CC'} \cdot M_{BC}(f) \cdot C_{B'B} \ M_{B'C'}(f) = C_{CC'} \cdot \begin{pmatrix} 1 & 0 \ 0 & 1 \ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} 2 & 0 \ 2 & 1 \end{pmatrix} \ \begin{pmatrix} \frac{-5}{3} & \frac{-1}{2} \ \ \frac{23}{6} & \frac{5}{4} \ \ 1 & \frac{1}{2} \end{pmatrix} = \begin{pmatrix} \frac{-1}{3} & \frac{-1}{3} & \frac{1}{6} \ \ \frac{2}{3} & \frac{7}{6} & \frac{-1}{12} \ \ 0 & 0 & \frac{-1}{2} \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \ 0 & 1 \ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} 2 & 0 \ 2 & 1 \end{pmatrix}
grizzled pagodaBOT
Renato Chavez
tough mica
,w {{-1/3, -1/3, 1/6}, {2/3, 7/6, -1/12}, {0, 0, -1/2}} . {{1, 0}, {0, 1}, {0, -1}} . {{2,0}, {2,1}}
grizzled pagodaBOT
tough mica
,w {{-1/3, -1/3, 1/6}, {2/3, 7/6, -1/12}, {0, 0, -1/2}} . {{1, 0}, {0, 1}, {0, -...
it was clear explanation
I have been struggling with this topic for like ages
I feel a bit more confident after today sir
mystic lotus
@tough mica has your question been answered or do you still need help with something
tough mica
<@1187916384341078070> has your question been answered or do you still need help...
I was thanking him for his time
you can close the channel, but he wont see it
mystic lotus
if you ping him, he'll see it
tough mica
@cunning birch .solved
mystic lotus
.solved
tough mica
- Let ( B = {(1,1,0), (0,1,1), (0,1,0)} ) be a basis of (\mathbb{R}^3) and let ( f : \mathbb{R}^3 \to \mathbb{R}^2 ) be the linear transformation such that
[
M_{B,E}(f) =
\begin{pmatrix}
1 & -1 & 5 \
-1 & 5 & 1
\end{pmatrix}.
]
Let ( S = {(6,4,-1), (2,-1,-2)} ) and ( T = {x \in \mathbb{R}^2 \mid x_1 = 0} ). Find ( S \cap f^{-1}(T) ).
grizzled pagodaBOT
Renato Chavez
tough mica
f(1,1,0)=(1,-1)
(1,-1)(E)= (1,-1)
f(0,1,1)=(-1,5)
(-1,5)(E)=(-1,5)
f(0,1,0)=(5,1)
(5,1)_(E)=(5,1)
how do I find the preimage of the basis of T under f
let me find a basis of T first
x1 = 0
(0,x2) = x2(0,1)
T = <(0,1)>
how do I find f^{-1}(<(0,1)>)
there is a property of applying a linear transformation to a span
basically
,, f^{-1}(\langle (0,1) \rangle) = \langle f^{-1}(0,1) \rangle
grizzled pagodaBOT
Renato Chavez
tough mica
basically applying a linear transformation to a span let say
W = <w1,w2, w3>
f(W)= <f(w1),f(w2),f(w3)>
I need to find which vector under the linear transformation f spits out (0,1)
also W = <w1,w2,w3> = span{w1,w2,w3}
is just notation
f(1,1,0)=(1,-1)
f(0,1,1)=(-1,5)
f(0,1,0)=(5,1)
(0,1)=a(1,-1)+b(-1,5)+c(5,1)
(0,1)=(a-b+5c,-a+5b+c)
i) a-b+5c=0
ii) -a+5b+c=1
,w rref {{1,-1,5,0},{-1,5,1,1}}
please help
amber waspBOT
@tough mica Has your question been resolved?
tough mica
<@&286206848099549185>
amber waspBOT
@tough mica Has your question been resolved?
tough mica
<@&286206848099549185>
amber waspBOT
@tough mica Has your question been resolved?
tough mica
<@&286206848099549185>
rapid shore
**Renato Chavez**
Write down what $2 \times 3$ matrix $f$ is. Then finding $f^{-1}(T)$ is just a linear system and taking intersection with S is also a linear system
grizzled pagodaBOT
L
tough mica
wdym write down what 2x3 matrix f is
I am only given this info about f
f(1,1,0)=(1,-1)
f(0,1,1)=(-1,5)
f(0,1,0)=(5,1)
I am trying but I cant seep to make it work
@rapid shore 😭
rapid shore
**Renato Chavez**
you are given $M_{B, E}(f) = fB$, so $f = M_{B, E}(f)B^{-1}$.
grizzled pagodaBOT
L
rapid shore
I write $f$ for the matrix of $f$, i.e. $M_{E, E}(f)$
grizzled pagodaBOT
L
grizzled pagodaBOT
Renato Chavez
tough mica
B is a basis wtf
rapid shore
what does B denote?
B is the matrix whose columns are the vectors in B
yeah you can use a different notation if you want
tough mica
why multiplying M_(EE)(f) with B gives M_(BE)
rapid shore
use the definitions
tough mica
rapid shore
It follows easily from definitions that $M_{A, B}(f) = A^{-1}fB$. This is because both sides give the same result when applied to a standard basis vector $e_j$.
grizzled pagodaBOT
L
tough mica
I am trying
problem is A and B
f(1,1,0)=(1,-1)
f(0,1,1)=(-1,5)
f(0,1,0)=(5,1)
I need more handholding . . .
rapid shore
I need more handholding . . . <:cat_happycry:636407813144576010>
Sure lets go with what you have
rewrite it into a matrix equation for the matrix f
tough mica
,, M_{EE}(f) = \begin{pmatrix} 1 & 1 & 0 \ 0 & 1 & 1 \ 0 & 1 & 0 \end{pmatrix}
grizzled pagodaBOT
xbz
tough mica
I might be tripping but you mean like this?
rapid shore
I might be tripping but you mean like this?
no the dimension isn't even correct
tough mica
how can I fix it
rapid shore
how can I fix it
$f : \mathbb{R}^3 \to \mathbb{R}^2$ is a linear map, so there is a $3 \times 2$ matrix $F$ such that $f(x) = Fx$ for all $x \in \mathbb{R}^3$. Rewrite your 3 equations into an equation for $F$ and solve for $F$.
grizzled pagodaBOT
L
tough mica
,, M_{EE}(f) = \begin{pmatrix} 1 & -1 \ -1 & 5 \ 5 & 1 \end{pmatrix}
grizzled pagodaBOT
xbz
tough mica
I might be tripping, still
rapid shore
,, M_{EE}(f) = \begin{pmatrix} 1 & -1 \\ -1 & 5 \\ 5 & 1 \end{pmatrix}
where are you getting this? do you not know how to compute $M_{E, B}(f)B^{-1}$? It's just matrix multiplication, there is no way you can make a mistake here
grizzled pagodaBOT
L
rapid shore
it should be 2 x 3 since f maps a 3d space to a 2d space
my calculator gave that f is
amber waspBOT
@tough mica Has your question been resolved?
tough mica
mmm
tough mica
.solved
amber waspBOT
thick trench
Helo
amber waspBOT
Available help channel!
golden vortex
amber waspBOT
split sail
find dx/dt and dy/dt first
and then dy/dx is dy/dt divided by dx/dt
because of the chain rule
golden vortex
find dx/dt and dy/dt first
that's cos(t) and -sin(t) right?
split sail
then when you have dy/dx in terms of t, do your implicit differentiation I think
or repeat the same process
split sail
that's cos(t) and -sin(t) right?
nope
golden vortex
hmm
golden vortex
or repeat the same process
i think repeating the same process is easier
split sail
Maybe you can find d^2x/dt^2
and d^2y/dt^2 to find y’’
split sail
i think repeating the same process is easier
Yea
golden vortex
thanks i will get to that then
i was gonna close the channel but southlander, i will wait for you to finish typing
mint nacelle
i was gonna close the channel but southlander, i will wait for you to finish typ...
I was just brainstorming
you can go ahead and close the channel
amber waspBOT
mint nacelle
ah $\frac{d^2 y}{dx^2} = \frac{d}{dx} \frac{dy}{dx}$
$= \frac{1}{dx/dt} \frac{d}{dt} \left(\frac{dy}{dx} \right)$
grizzled pagodaBOT
southlander!
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Available help channel!
upper forge
How did they make Z1 the subject here
amber waspBOT
How did they make Z1 the subject here
rocky vapor
How did they make Z1 the subject here
barely readable , but if u dont understand u can ask
amber waspBOT
@upper forge Has your question been resolved?
upper forge
barely readable , but if u dont understand u can ask
Lol I actually understood
Thanks a lot
amber waspBOT
@upper forge Has your question been resolved?
amber waspBOT
amber waspBOT
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snow blade
b, 0, b^2, 0, b^3, 0, b^4...
what positive values of b makes the sequence convergent
hearty inlet
snow blade
💔
hearty inlet
when does b, b^2, b^3, ... converge
snow blade
okay so i was first thinking that b could be -1 because then it could be a telescoping series that keeps cancelling out and converges to 0
but it says b needs to be positive
so i thought b can be a fraction
less than 1
so it approaches 0? idk
hearty inlet
right but there is no series
verbal jewel
okay so i was first thinking that b could be -1 because then it could be a teles...
youre example isnt a series.
hearty inlet
b < 1
snow blade
would it be |b| < 1
cuz that's what i wrote
since itll need to be positive
but like
i fully guessed that
it just made sense but idk why its true
hearty inlet
since itll need to be positive
right so |b| = b
snow blade
is it just convergent becuase later it'll start approaching 0
or is thehre other reasoning
hearty inlet
is it just convergent becuase later it'll start approaching 0
yes
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dry crown
I have tried to study the motion of the point T, where T is the intersection of line L and circumference T.
Then I calculate PT in fuction of time and applied the pythagorean theorem to find OP in function of time. But can I write the angular velocity in function of time?
There's a bettere way to solve the problem?
Am I wrong?
Please help me
wind stirrup
thats some question you have there
Im just getting a feel for the problem
will the coordinates of P be (1,pi/4) after the Point of contact reaches 1,0?
@dry crown
if that is correct, when the line reaches (0,-1), the coordiates of P must be (pi/2, -1)
im not sure about it though
dry crown
wait, i'm calculating
dry crown
ahahahah
dry crown
it's a difficult problem, i found it on an italian admission test to Scuola Normale Superiore
dry crown
did you calculate though?
i've tried
wind stirrup
wow
dry crown
but according to the third equation i think it's a spiral
dry crown
im in 11th
wind stirrup
what
dry crown
but im italian, schools in Italy works different
dry crown
and I really like math and physics
wind stirrup
this is way too difficult for me to be honest
do me a favour and ping me when you/someone gets a solution
really interesting probelm
dry crown
ok bro
dry crown
I have tried to study the motion of the point T, where T is the intersection of ...
<@&286206848099549185>
wind stirrup
@dry crown is the distance of the point from the origin always equal to sqrt( 1 + (wt)^2)
oh you already figureed that out
I did it by rolling a circle on the line instead of other way round
anyway im out
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dry crown
.reopen
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✅
dry crown
I have tried to study the motion of the point T, where T is the intersection of ...
<@&286206848099549185>
native stag
I have tried to study the motion of the point T, where T is the intersection of ...
uh
isn't it just
wait whats t
native stag
what are you doing im super confused
is this circular motion
physics?
is this pure math
what is it
dry crown
oh, dear percy, you are discovering the fantastic world of math but with physics
officially, it's a question of math
dry crown
is this circular motion
yes, the motion of T is uniform circular motion
native stag
and what are you confused about?
dry crown
I don't know if I am right or wrong
native stag
a point on a circle is just $(r \cos \theta, r \sin \theta)$
grizzled pagodaBOT
Percy
dry crown
yes
native stag
you have $\theta$, you have $\omega$, what's the issue
grizzled pagodaBOT
Percy
native stag
.......yes
wind stirrup
P is not the POC of tangent
native stag
$\omega=\frac{d\theta}{dt}$
$\implies d\theta=\omega dt$
grizzled pagodaBOT
Percy
wind stirrup
torque is applied on the line which causes P to lift up off the circle
dry crown
he's saying that line L is rolling around the circle
native stag
I-
dry crown
and P is marked on the line, so when L rotates, P lift up off the circle
native stag
right i just dont understand the question ig
dry crown
it's asking basically the equations of motion in function of time (x(t) and y(t)) of the point P
now you understand?
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@dry crown Has your question been resolved?
dry crown
<@&286206848099549185>
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@dry crown Has your question been resolved?
compact wagon
amber waspBOT
weak zinc
[remember integrals are signed areas!]
compact wagon
huhh
so it would be negative right??
or since the -2 is negative then it would become positive/?
weak zinc
or since the -2 is negative then it would become positive/?
yep, and you should get the integral of f between 0 and 1 as negative 
compact wagon
.close
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brave zealot
question b
amber waspBOT
question b
outer hull
find the normal vector of pi 2
then calculate the angle between the pi 1 normal vector and pi 2 normal vector
that will also be the angle of between both planes
brave zealot
do i do dot product of the two vectors
outer hull
its included in the vector angle formula so sure
$\cos\alpha=\frac{\vec a\cdot\vec b}{\lvert\lvert\vec a\rvert\rvert\cdot\lvert\lvert\vec b\rvert\rvert}$
grizzled pagodaBOT
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stray eagle
Hi
stray eagle
So f(x) is concave up when f’(x)>0
Or when the slope is increasing
Oh yeah
Shit I forgot
So wouldn’t it be (-2,0) U (2,infinity)
plain coral
So wouldn’t it be (-2,0) U (2,infinity)
yes
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tough mica
\textbf{Exercise 18} Let ( R(x) ) be a function with 3 continuous derivatives at ( x = 0 ) such that
( R(0) = R'(0) = R''(0) = 0 ). Prove that
[
\frac{R(x)}{x^3} = \frac{R'''(c)}{3!}
]
for some ( c ) between ( 0 ) and ( x ). (\textit{Hint: use Cauchy's Theorem 3 times.})
grizzled pagodaBOT
xbz
tidal basin
\textbf{Exercise 18} Let \( R(x) \) be a function with 3 continuous derivatives ...
honestly I don't know what cauchy's theorem is but
$f(x) = cx^3$ so $f'''(x) = 3!c$
$\frac{cx^3}{x^3} = c$ and $\frac{3!c}{3!} = c$
grizzled pagodaBOT
Karma
tidal basin
can you elaborate what happened?
wym
i feel like im missing something lol this seems pretty trivial
tidal basin
derivatives are factorial by nature, eg x^3 derivative multiplies by 3, the next multiplies by 2 and so on.
If you keep derivating till you have no more x terms, your left with the initial exponent factorial multiplied by whatever constant might be attached
in the problem, the left side of the equivalence divides out the initial function, leaving you with the constant. the right side of the equivalence divides out the derivated factorial, leaving you with the cosntant.
tidal basin
persona 5 is a soundtrack that comes with a game, btw
idk if this is meant to be an insult, p5r is top 3 games oat and its not 3
just bc the ost is amazing doesnt mean the game isnt too
tough mica
problem is you are using f(x) = cx^3 and f'''(x) = 3!c wtf is dat
tidal basin
if f(x) is x^3 multiplied by a constant c (basically any scalar of x^3, it doesnt matter what the constant is the property still holds)
then the third derivative is 3!c
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full elk
The idea is to write $G(x) = x^3$, and then use Cauchy's MVT to show that you have some $c_1 \in (0,x)$ such that $\frac{R'(c_1)}{G'(c_1)} = \frac{R(x) - R(0)}{x^3 - 0^3} = \frac{R(x)}{x^3}$.
Then you apply it again on the LHS to get $c_2 \in (0,c1)$ that is equivalent to it and so on
If you do this 3 times you get the desired result.
grizzled pagodaBOT
Azyrashacorki
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@tough mica Has your question been resolved?
tough mica
The idea is to write $G(x) = x^3$, and then use Cauchy's MVT to show that you ha...
why can you put G(c1) at the bottom?
like I though x = 0
c is between 0 and x yeah
is just strange that x^3 can be at the denominator
since prob statement says x = 0
full elk
The problem statement says that we're trying to find some c between 0 and x, not that x = 0
tough mica
yeah fr srry
applying cauchy mvt once
,, \frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)}
grizzled pagodaBOT
xbz
tough mica
continuous [a,b]
differentiable (a,b)
b = x
and there exists an c in (0,x)
,, \frac{R(x) - R(0)}{x^3 - 0^3} = \frac{R'(c_1)}{3c_1^2} \ \frac{R(x)}{x^3} = \frac{R'(c_1)}{3c_1^2} \ \text{second time, for } R'(x) \text{ and } 3x^2 \text{ in the interval } \left[0,c1\right] \ \frac{R'(c_1) - R'(0)}{3c_1^2 - 0^2}
,, \frac{R(x) - R(0)}{x^3 - 0^3} = \frac{R'(c_1)}{3c_1^2} \ \frac{R(x)}{x^3} = \frac{R'(c_1)}{3c_1^2} \ \text{second time, for } R'(x) \text{ and } 3x^2 \text{ in the interval } \left[0,c1\right] \ \frac{R'(c_1) - R'(0)}{3c_1^2 - 0^2} = \frac{R''(c_2)}{6c_2} \ \frac{R'(c_1)}{3c_1^2} = \frac{R''(c_2)}{6c_2} \ \text{third time, for } R''(x) \text{ and } 6x \text{ in the interval } \left[0, c_2 \right] \ \frac{R''(c_2) - R''(0)}{6c_2 - 0} = \frac{R'''(c_3)}{6} \ \frac{R''(c_2)}{6c_2} = \frac{R'''(c_2)}{6}
grizzled pagodaBOT
xbz
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tropic dagger
If anyone understands how to use the SAS software, I am having trouble with this problem. Any help is appreciated. Use SAS programming to answer the following questions.
Submit related SAS codes and outputs. Explain the results in your own words to answer each problem.
Data set MPG presents the gasoline mileage performance for 32 automobiles. The variables in the
study are:
Dependent variable: Miles per gallon (y)
Independent variables:
x1: displacement (cubic in.)
x2: horsepower (ft-lb)
x3: torque (ft-lb)
x4: horsepower + torque (ft-lb)
x5: carburetor (barrels)
x6: weight (lbs.)
x7: type of transmission (1=automatic; 0=manual)
x8: no. of transmission speeds (3, 4, or 5)
-
Build a multiple linear regression model relating the response variable Miles per gallon to these
predictors. (Hint: x7 and x8 are qualitative variables).
Use SAS program to find the least square regression equation. -
Construct the ANOVA table. Test for significance of regression.
State the null and alternative hypotheses; report the F statistic and p-value.
What conclusions can you draw? (Use 𝛼= .05) -
Find the coefficient of determination, R^2 and Adjusted-R^2for the linear model you constructed.
Discuss your results -
Find a 95 % CI for the regression coefficient for x2.
tropic dagger
-
Fit a reduced regression model using x1-x4 and x7-x8 as the regressors. Give the least squares
equation for reduced model. -
Report the partial F statistic and p-value and make a conclusion whether or not to drop
both carburetor and weight have a significant relationship with the gasoline mileage. -
Find a 95 % CI for the regression coefficient for x3 in reduced model and compare it to
the one from full model. -
(Test for the qualitative variable) Test the hypothesis that no. of transmission speeds level
will have impact on Miles per gallon. Set up the null and alternative hypotheses;
Report the P- value and make your conclusion from SAS result. -
(Interaction effect) Test the hypothesis that the type of transmission
effect is the same for all three levels of no. of transmission speeds.
Set up appropriate model, hypotheses and report your results.
this is the data from the MPG file: y,x1,x2,x3,x4,x5,x6,x7,x8
18.9,350,165,260,425,4,3910,1,3
17,350,170,275,445,4,3860,1,3
20,250,105,185,290,1,3510,1,3
18.25,351,143,255,398,2,3890,1,3
20.07,225,95,170,265,1,3365,0,3
11.2,440,215,330,545,4,4215,1,3
22.12,231,110,175,285,2,3020,1,3
21.47,262,110,200,310,2,3180,1,3
34.7,89.7,70,81,151,2,1905,0,4
30.4,96.9,75,83,158,2,2320,0,5
16.5,350,155,250,405,4,3885,1,3
36.5,85.3,80,83,163,2,2009,0,4
21.5,171,109,146,255,2,2655,0,4
19.7,258,110,195,305,1,3375,1,3
20.3,140,83,109,192,2,2700,0,4
17.8,302,129,220,349,2,3890,1,3
14.39,500,190,360,550,4,5290,1,3
14.89,440,215,330,545,4,5185,1,3
17.8,350,155,250,405,4,3910,1,3
16.41,318,145,255,400,2,3660,1,3
23.54,231,110,175,285,2,3050,1,3
21.47,360,180,290,470,2,4250,1,3
16.59,400,185,295,480,4,3850,1,3
31.9,96.9,75,83,158,2,2275,0,5
29.4,140,86,90,176,2,2150,0,4
13.27,460,223,366,589,4,5430,1,3
23.9,133.6,96,120,216,2,2535,0,5
19.73,318,140,255,395,2,4370,1,3
13.9,351,148,243,391,2,4540,1,3
13.27,351,148,243,391,2,4715,1,3
13.77,360,195,295,490,4,4215,1,3
16.5,360,165,255,420,4,3660,1,3
pallid canopy
Yeaaaaa good luck getting help needing to read all that
tropic dagger
please i downloaded discord just to get help with this
brisk leaf
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@tropic dagger Has your question been resolved?
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@tropic dagger Has your question been resolved?
tardy dune
Can we speak of concepts like linearity when talking about the maps of operators $+$ and $\cdot$? Take the following definition of the $\cdot$ operator for vector spaces:
[
\cdot: (F\times V) \rightarrow V
]
where $F$ is the field over $V$. Linearity requires that for some transformation $T$ we have [
T(ax) = aT(x)
]
but can we really talk about "linearity" when the definition of linearity relies on the map of the operator we are considering to begin with? I hope my question is understandable to some extent...
grizzled pagodaBOT
Aero
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rough pawn
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thick shale
do you know this identity: $$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$
grizzled pagodaBOT
Pro_Hecker
thick shale
I think it will be easier if you go from the left
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woven venture
help
thick shale
argument being pi/4 implies that x and y coordinate are the same
thick shale
arguement = tan^-1(y/x)
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cursive wasp
I am looking for intuition as to why the inner product is greatest when the vectors are in the same direction and why it makes sense to define the length a vector as sqrt(<v,v>)
Specifically I would like an explanation as to how the axioms lead to this making sense. The dot product makes sense to me, but I am having difficulty accepting that the same concepts are true for the inner product.
indigo cloud
your first question is essentially answered by the cauchy schwarz inequality
$\langle u,v\rangle^2 \leq \langle u,u\rangle \langle v,v\rangle$
grizzled pagodaBOT
Denascite
indigo cloud
with equality if and only if u and v are linearly dependent, aka in the same direction
in other words, the inner product is bounded by some value and it can only achieve that value if u and v point in the same direction
aka it is greatest when they point in the same direction
cursive wasp
Fair enough thanks
stable leaf
Does cauchy schawrz apply for all inner products?
indigo cloud
yes
stable leaf
Thanks. Need to brush up my linear alg 😬
indigo cloud
as for the norm, not really sure about some intuitive argument. it turns out that if you define it that way then you actually satisfy the norm axioms (so thats good) and it generalizes what is the case for the dot product (so thats also good)
I think the point that it satisfies the norm axioms is quite a lot, those axioms capture several important things we think about "lengths"
stable leaf
Well, if the inner product is analogous to multiplication over the reals, then it makes some sense since mod x = sqrt(x*x) in the reals
cursive wasp
I think the point that it satisfies the norm axioms is quite a lot, those axioms...
Fair enough. I haven't heard of the norm axioms. Will check them out thanks
indigo cloud
norm(v) >= 0 with =0 only if v=0
norm(number*v) = abs(number)*norm(v)
norm(u+v) <= norm(u) + norm(v)
so lengths are positive
you can scale them
and triangle inequality
all very basic concepts that you would want from something you call "length"
stable leaf
The last one is just a special case of cauchy schwarz is it not?
indigo cloud
its proved using cauchy schwarz, yes
stable leaf
Also is inner product linear in the first or second arguments? I keep seeing conflicting answers in different books
indigo cloud
depends on the author
doesnt matter in the grand scheme of things
as long as you are consistent
stable leaf
Fair
cursive wasp
Thank you. Off topic, what is your opinion on the Von Neumann quote "Young man, in mathematics you don't understand things. You just get used to them."
To me it feels very important to understand the concepts.
indigo cloud
**Denascite**
if you have complex numbers then you of course need |<u,v>|^2 on the left here
it depends on the concept
stable leaf
Since anti symmetry really doesn't apply in the reals, its linear in both args in reals right
indigo cloud
yes
stable leaf
thank you so much💪
cursive wasp
Thank you for your responses
indigo cloud
I am not sure what concepts von neumann is talking about
stable leaf
While we are on the topic of linear algebra, what does a change of basis matrix with respect to a linear map mean? Does it mean its a matrix form of a map you give arg in Base A and get output i base B?
indigo cloud
it would be better if you open your own channel
I have to go but someone will surely help you
cursive wasp
.close
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cobalt vapor
hi
cobalt vapor
can anyone help me with this section right here, especially the e) one
For the pyramid S.ABCD, the bottom ABCD is the centroid parallelogram O. Call M and N the midpoints AB and CD respectively. Let Q be the point next to SB so that QS = 2QB. a) Proof: MN // (SBC) and MN // (SAD). b) Calling P the midpoint of SA. Prove that SB and SC are parallel to (MNP). c) Call K and L to be the focus of triangles ABC and SBC, respectively. Proof of KL // (SAC). d) Call I the intersection of the plane (OPQ) and the line AD. Calculate the ratio ID/IA e) Call (a) the plane containing DQ and parallel to the AC. Determine the intersection E and F of the plane (a) with the sides SA, SC, respectively. Calculate the area ratio of the DQF triangle to the DBC triangle.
sorry if the information isnt clear, english isnt my first language and i translated myself all the questions
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cobalt vapor
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split sail
Find the minima of 3x^4−20x^3−72x^2−168x+5000
neat wind
may as well use calculus
neat wind
then solve it using calculus and prove the result is correct without
placid wadi
im not aware of any methods this can be solved with that doesnt involve calculus at all
you need to at least know the nature of the curve
split sail
OK
inland pulsar
,w factorise 3x^4−20x^3−72x^2−168x+5000
grizzled pagodaBOT
inland pulsar
no roots hmm
neat wind
once you know where the minimum occurs it is not hard to prove that is where it is
but trying to find that location without using calculus is a death sentence (waste of time)
amber waspBOT
@split sail Has your question been resolved?
proven vapor
but trying to find that location without using calculus is a death sentence (was...
maybe there's something to do with a sum of squares
,w minimum value of 3x^4−20x^3−72x^2−168x+5000
grizzled pagodaBOT
proven vapor
valid cobalt
amber waspBOT
valid cobalt
i was thinking this is literally the definition for the limit form of the comparison test
but apparently there is a counter example
odd lynx
ok here's an example
odd lynx
but apparently there is a counter example
suppose a_k = 1 for all k
that would be convergent yes?
wait
nvm
shadow stump
you should note the conditions on the limit comparison test
amber waspBOT
@valid cobalt Has your question been resolved?
valid cobalt
no
blissful blade
...
valid cobalt
timid pulsar
think of one property to disprove it
valid cobalt
this is the conditions
timid pulsar
I think it's false right
valid cobalt
no sorry i meant no my answers wasnt solved
yep its meant to be wrong
its got to do with the fact the sequences can be negative
timid pulsar
ye what's the property for which it works
timid pulsar
oh was thinking about more of the monotony of the sequences
if there are not monotonous then we don't have it
valid cobalt
it can be any sequence we want
i mean we are finding a counterexample so anything is fine
provided its a real sequence
timid pulsar
Do you see one so ?
with not the monotony we want
like $(-1)^{n}/\sqrt(n)$ and $(-1)^{n}/(\sqrt(n) + (-1)^{n})$ for example seems to work cause we don't have the monotony
grizzled pagodaBOT
phoestaclies
valid cobalt
ok ill try that
timid pulsar
first one converges with leibniz criteria and the other not
You just want monotony at a certain rank for this property to work
valid cobalt
wait but both sequences here converge
timid pulsar
no don't think so
timid pulsar
ye let's do that
valid cobalt
oh i see ok
timid pulsar
what are the hypothesis in the ast ? maybe you should see your error
valid cobalt
sorry is the sqrt in the base meant to cover the entire denominator or just the n?
timid pulsar
just the n
valid cobalt
oh ok mb
ill try it now
sorry im struggling a bit to prove the second one diverges
i need to head off for a little bit but if you have any ideas ill give them a read in a bit
thanks
timid pulsar
ok I'll give you some hints but I don't know what's your property name (it will be a bit difficult :)) do you know notations like ~ , o(1/n) or O(1/n) for example, limited development ?
amber waspBOT
@valid cobalt Has your question been resolved?
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@valid cobalt Has your question been resolved?
timid pulsar
sry went eat, it will be complicated so
maybe need another example easier then
$log(1+ (-1)^{n}/\sqrt n)$ and $ (-1)^{n}/\sqrt n$
grizzled pagodaBOT
phoestaclies
timid pulsar
this should be good and maybe easier I don't know
you just need to prove that the serie of $log(1+ (-1)^{n}/\sqrt n)$ diverges
grizzled pagodaBOT
phoestaclies
exotic olive
so i had to find the foci and determine which is closest to the sun
exotic olive
i already found it
but i need help finding which is closest to sun
when i graph it just shows this
so what point would be closer
do i just say all four points?
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exotic olive
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exotic olive
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glad tiger
How am I meant to prove that height is proportional to volume for a cylinder filled with water?
glad tiger
h(t) is proportional to kV(t) where k is some constant of proportionality
split sail
pi r^2 h/ h = pi r^2 (constant)
glad tiger
why divide by h?
previous step was this, I finished it already
V(t) is quadratic
split sail
why divide by h?
wdym, it's by definition, we're taking a ratio
split sail
pi r^2 h/ h = pi r^2 (constant)
this is the proof
glad tiger
the height of a cylinder is proportional to the volume times a constant. If the volume is pi h r^2, then the derivative with respect to time (since r doesn't vary) is pi r^2 dh/dt
but
i dont think that's what im supposed to be doing
i cant just divide by h because h is a function
h(t) is proportional to V(t)
mellow hornet
I think you're overthinking this
glad tiger
im not sure what im supposed to be doing though is the problem
mellow hornet
you basically just explained that V(t) is h(t) times a constant
that's proportional automatically
grizzled pagodaBOT
Ari
glad tiger
oh and you can also see that when analyzing the changing in volume. Since the change in volume is some constant pi r^2 times the change in height, the volume itself is too proportional to the height
split sail
how did i explain that? how am I supposed to find the constant k
the constant is 1/ (pi r^2)
glad tiger
why 1 over?
oh because if h(t) = kV(t), then k = h(t)/V(t), and since V(t) = pi r^2 h(t), k = h(t)/(pi r^2 h(t)) = 1/pi r^2
split sail
umm it depends
amber waspBOT
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timber copper
Hello
I’ve got a bunch of these "true or false statements"
This one being e it’s obv false and I’ve managed to solve with n being a natural. I’m wondering if, though logical, it’s rigorously correct to say that the limit for natural n is the same as the limit for real n to prove the inequality
honest dagger
Hello
I’ve got a bunch of these "true or false statements"
This one being e it’...
They are, in fact, not the same
But, if the limit of reals is convergent, then the limit of naturals will converge to that same value
but in general, the two are not the same
timber copper
Ok
solar gust
SWR a tout dit
timber copper
Thanks
But would proving growth with n natural prove growth with any real deeming continuity?
Actually nvm this statement wouldn’t always be true.
timber copper
But, if the limit of reals is convergent, then the limit of naturals will conver...
Sorry to keep bothering, but wouldn’t the reciprocate statement be true?(if the natural converges in +inf then the real also does)
honest dagger
But would proving growth with n natural prove growth with any real deeming conti...
No
honest dagger
Sorry to keep bothering, but wouldn’t the reciprocate statement be true?(if the ...
No
If natural diverges to +inf, then the real will not converge, but it is not guaranteed to diverge to +inf either
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Construct the augmented matrix that corresponds to the following system of equations.
4(z−7)−4x+y=z
−3(7x−y)+4=5z
5x/2−(y−7z)=x
I am really struggling getting the third equation correct. My first thought is to multiple out the two on both sides. So then I am left with "5x - y - 7z = 2x" my initial instinct from there is to subtract 2x from 5x and that leaves me with "3x - y -7z = 0" but for some reason I keep getting the matix wrong. I've done this same structured problem about 5 times now and every single one has had a fraction in the first position. What am I doing wrong?
broken saffron
when you multiply by 2 on both sides you get -2y +14z, not -y-7z
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okay let me try taht
BINGO
i swear i did that a few times ago hm
but thank you !!
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