#help-33
marsh citrusBOT
tough wind
i have a series $S(x)=\sum_{n=0}^{+\infty} a_nx^n$ defined over $]-1,0[$ and i want to prove that for $x$ fixed in $]-1,0[$, S is bounded over $[x,0[$
elfin berryBOT
macilak
main idol
S(x) can't go off to infinity on [x, 0] since S(x) is defined on (-1, 0)
and you can use the triangle inequality repeatedly to say $|S(x)| \le \sum |a_n x^n|$
elfin berryBOT
riemann
tough wind
S(x) can't go off to infinity on [x, 0] since S(x) is defined on (-1, 0)
why can't it go to infinity when x approaches 0 ?
tough wind
and you can use the triangle inequality repeatedly to say $|S(x)| \le \sum |a_n ...
it doesn't lead to anything concluding, does it ?
main idol
tough wind
plug in small x into the series and see
i mean how do you prove it
main idol
**riemann**
this is a good start
tough wind
this is a good start
yeah but you can't say anything about the RHS
main idol
use |bc| = |b| * |c| for all real numbers b, c
tough wind
and then ?
main idol
try something for a couple mins before instantly asking for next steps
tough wind
i already tried triangular inequality
applying it the way you said doesn't give anything
marsh citrusBOT
@tough wind Has your question been resolved?
tough wind
**macilak**
<@&286206848099549185>
marsh citrusBOT
@tough wind Has your question been resolved?
marsh citrusBOT
@tough wind Has your question been resolved?
marsh citrusBOT
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fair gate
Yo
marsh citrusBOT
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frigid anchor
marsh citrusBOT
frigid anchor
i need help setting up the summation
if its 8 partitions, wouldnt delta(A) just be 1
and if you get all the points for the center of each unit square
and do x^2 * y
is it not just 42
wait i forgot to divide by 4
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small widget
Can you explain solving trigometic equations
marsh citrusBOT
Can you explain solving trigometic equations
small widget
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marsh citrusBOT
@small widget Has your question been resolved?
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wise oxide
How come first image they do 90/3 to find the area but the second image they just put it as 0,pi/2? cus second image doing 90/number in front of theta gives different answers
marsh citrusBOT
How come first image they do 90/3 to find the area but the second image they jus...
fervent rampart
because the solutions to sin theta = 0 and cos theta = 0 are different
wise oxide
because the solutions to sin theta = 0 and cos theta = 0 are different
Ok so when it’s cosine I always do 90/number in front of theta to get the bounds?
fervent rampart
that would give you one of the solutions to cos(n theta) = 0, yes
wise oxide
that would give you one of the solutions to cos(n theta) = 0, yes
So the other solution is always the negative of that one
?
fervent rampart
that's another solution, yes. there are several solutions
wise oxide
Bro what
So which am I supposed to use
So for cosine I can just do 90/number in front of n to get the upper bound and then times that by -1 to get the lower bound?
fervent rampart
look at the graph. each angle at which the loop goes through the origin is a different solution.
wise oxide
look at the graph. each angle at which the loop goes through the origin is a dif...
The simplest one is the one he’s doing in the vid?
fervent rampart
yes
wise oxide
Ok for cosine I can just do that for all the problems then?
Find the area of the one on the positive x axis and then times the answer by however how remaining petals there r to find the area of whole thing?
@fervent rampart ??
fervent rampart
sure
marsh citrusBOT
@wise oxide Has your question been resolved?
wise oxide
How do I find bounds
marsh citrusBOT
How do I find bounds
marsh citrusBOT
@wise oxide Has your question been resolved?
wise oxide
And solve for theta
Doesn’t that gimme a buncha solutions
marsh citrusBOT
@wise oxide Has your question been resolved?
meager badger
Doesn’t that gimme a buncha solutions
make sure it is within the period of sin(2theta)
wise oxide
Can I just do 180/n(upper bound) and 0(always lower bound) for sin function and for cosine functions I do 90/n for upper bound and times that by -1 for lower bound? Is that correct
@meager badger @lost smelt
meager badger
im not sure
lost smelt
(but make sure you're using radians when you actually evaluate the integral)
wise oxide
yeah
Ok for something like cos(theta) the bounds r pi/2(upper bound) and -pi/2(lower bound)?
elfin berryBOT
qianqian07
lost smelt
when you plug numbers in for theta, you need to make sure that it's in radians or you'll get the wrong answer
lost smelt
Ok for something like cos(theta) the bounds r pi/2(upper bound) and -pi/2(lower ...
yeah I think that sounds right
the derivation works the same way as sine
wise oxide
when you plug numbers in for theta, you need to make sure that it's in radians o...
I thought they were equal tho
Like pi/4 is 45 degree
lost smelt
but 2n*(45 degrees) doesn't really mean anything outside a trig function
lost smelt
**qianqian07**
the first term here has a theta outside a trig function
lost smelt
sin^2(n*theta)/2
that's what happens when you plug sin(n*theta) into the formula for polar integration
wise oxide
when you plug numbers in for theta, you need to make sure that it's in radians o...
Oh u mean for the bounds I need to change them to radians
Before I integrate the function?
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novel juniper
Let $y_1=6$ and $y_{n+1} = \frac{2y_n-6}{3}$
\
Use induction to show that $\forall n \in \N, y_n>-6$
marsh citrusBOT
Let $y_1=6$ and $y_{n+1} = \frac{2y_n-6}{3}$
\\
Use induction to show that $\for...
elfin berryBOT
ƒ(Why am. I here)=I don't Know
novel juniper
This is what I've done so fa
$y_1>-6
\
$y_n= \frac{2y_{n-1}-6}{3} geq -6$
\
$y_{n+1} = \frac{2y_n-6}{3} = \frac{2\frac{2y_{n-1}-6}{3} -6}{3}$
elfin berryBOT
ƒ(Why am. I here)=I don't Know
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novel juniper
[
y_1 > -6
]
[
y_n = \frac{2y_{n-1} - 6}{3} \geq -6
]
[
y_{n+1} = \frac{2y_n - 6}{3} = \frac{2 \frac{2y_{n-1} - 6}{3} - 6}{3}
]
novel juniper
what happened to your break
I realised I;m travelling soon, so I'll relax then
novel juniper
use induction hypothesis
how
elfin berryBOT
ƒ(Why am. I here)=I don't Know
devout mauve
literally just plug y_n>-6 in
novel juniper
Find a collection of sets such that $\bigcap_{i=1}^{n} B_i \neq \varnothing \forall n \in \N$, but $\bigcap_{i=1}^{\infty} B_i = \varnothing$
marsh citrusBOT
Find a collection of sets such that $\bigcap_{i=1}^{n} B_i \neq \varnothing \for...
novel juniper
I was thinking $B_i = (0 , \frac{1}{n})$
elfin berryBOT
ƒ(Why am. I here)=I don't Know
proud ice
Find a collection of sets such that $\bigcap_{i=1}^{n} B_i \neq \varnothing \for...
bad tex? Looks weird
every finite intersection is non-empty, but the total intersection is empty?
elfin berryBOT
ƒ(Why am. I here)=I don't Know
proud ice
**ƒ(Why am. I here)=I don't Know**
$\forall(n\in\bN)\left(\bigcap_{i=1}^n B_i\ne\emptyset\right)$
elfin berryBOT
SWR
proud ice
**ƒ(Why am. I here)=I don't Know**
this would work completely fine
novel juniper
I'm not sure how to show that though
tight furnace
Which part
novel juniper
I could use the archemedian property, but abbott hasn't intorduced that yet
proud ice
I'm not sure how to show that though
easy to show that each finite intersection is non-empty, I'm sure you know how to do that, yes?
novel juniper
easy to show that each finite intersection is non-empty, I'm sure you know how t...
yeah
proud ice
I could use the archemedian property, but abbott hasn't intorduced that yet
You will need that 
novel juniper
I mean I'm just asked to find an example, so I guess for now my intution is fine?
proud ice
yeah
tight furnace
Ye
novel juniper
Cool
tight furnace
You could let the nth set be {n, n+1, ...} too
that would dodge the Archimedean thing
proud ice
You could let the nth set be {n, n+1, ...} too
oh. That one took a second
hm wait.. 
Not seeing it
or am I?
hmm
that's a neat one
tight furnace
let the nth set be $\bN - {n, 2n, 3n, \ldots}$
elfin berryBOT
Dreyuk
tight furnace
tight furnace
ig ignore n=1
novel juniper
that works
tight furnace
All you need to show finite intersections are nonempty is infinitude of the primes
novel juniper
lol
proud ice
ezpz
Let $a_1=3$, let $a_{n+1}$ be the first twin prime (lesser value) that is greater than $a_n+2$, or $0$ if no such twin prime exists. Let $B_n=\bigcup_{i=1}^n a_i$. Would $\bigcap_{i=1}^{\infty} B_i$ be empty?
elfin berryBOT
SWR
novel juniper
I actually have another question
Prove $(\bigcup_{i=1}^{\infty} A_i)^C =. \bigcap_{i=1}^{\infty} A_i^C$
elfin berryBOT
ƒ(Why am. I here)=I don't Know
cedar stone
if e is in the first, than e is not in A_k for any k, s e in in A_k C for all k, so e is in the second
novel juniper
yes
cedar stone
we can do similar for e not in the first
proud ice
**ƒ(Why am. I here)=I don't Know**
A well-known property. Can even be extended to general unions and intersections
novel juniper
I think I just have to do standard double inclusion here?
proud ice
I think I just have to do standard double inclusion here?
maybe? I don't know it by that term
novel juniper
like prove each set is a subset of the other
proud ice
Suppose $x\in\bigcup_{i=1}^{\infty} A_i$, what does that mean in words?
elfin berryBOT
SWR
proud ice
like prove each set is a subset of the other
ah
novel juniper
$x \in A_i $for some $i \in \N$
elfin berryBOT
ƒ(Why am. I here)=I don't Know
proud ice
Yeah
This is actually definition of big union
$x\in\bigcup_{i=1}^{\infty} A_i\Leftrightarrow \exists(i\in\bN)(x\in A_i)$
elfin berryBOT
SWR
proud ice
So, what does it mean that $x\in\left(\bigcup_{i=1}^{\infty} A_i\right)^C$?
elfin berryBOT
SWR
proud ice
The complement would be the negation of this
mhm
novel juniper
$\forall( i \in \N)( x \notin A_i)$
elfin berryBOT
ƒ(Why am. I here)=I don't Know
proud ice
$\forall( i \in \N)( x \notin A_i)$
mhm
And what does $x\in A_i$ mean (in terms of set complements)?
elfin berryBOT
SWR
novel juniper
$x \notin A_i^C$
elfin berryBOT
ƒ(Why am. I here)=I don't Know
proud ice
**ƒ(Why am. I here)=I don't Know**
oops yeah
You are right
I meant to ask what $x\notin A_i$ meant, which of course would be $x\in A_i^C$
elfin berryBOT
SWR
proud ice
**ƒ(Why am. I here)=I don't Know**
But anyway, put that complement bit back in here, and you basically have the definition of big intersection (but with complements of your set family)
proud ice
If $x\in\bigcap_{i=1}^{\infty} \left(A_i^C\right)$, what does that mean?
elfin berryBOT
SWR
novel juniper
$x \in \bigcgup_{i=1}^{\infty} A_i$
elfin berryBOT
ƒ(Why am. I here)=I don't Know
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novel juniper
$x \notin \bigcup_{i=1}^{\infty} A_i$
proud ice
Is this what it means? If x is in the intersection of complements, then it is in the union of the original family?
elfin berryBOT
ƒ(Why am. I here)=I don't Know
proud ice
**ƒ(Why am. I here)=I don't Know**
Yeah this isn't wrong, but this is basically what you are trying to prove
Okay let me take a second here
proud ice
**ƒ(Why am. I here)=I don't Know**
You gave me this answer for big union.
novel juniper
I think just writing down the logical setences and comparing them should be enough proof
proud ice
**SWR**
To which I slightly refined for you here. We are definining arbitrary union using an existential qualifier. Arbitrary intersection is also defined with an existential qualifer. I'm basically asking for that definition (or it's equivalent in simple words)
proud ice
I think just writing down the logical setences and comparing them should be enou...
yes. That's basically where I was trying to get you. Just be sure your steps are sufficient
novel juniper
$x \in \bigcup_{i=1}^{\infty} \equiv \exists (i\in \N)(x \in A_i)$. The complement is the negation of this $x \in (\bigcup_{i=1}^{\infty} )^C \equiv \forall( i \in \N)(x \notin A_i)$
elfin berryBOT
ƒ(Why am. I here)=I don't Know
novel juniper
And $x \in (\bigcap_{i=1}^{n}A_i \equiv \forall ( i \in \N) (x \notin A_i)$ by defn
elfin berryBOT
ƒ(Why am. I here)=I don't Know
marsh citrusBOT
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jaunty garden
marsh citrusBOT
jaunty garden
If I had an integral like this why can’t I use u-sub by letting u = 1-x^2
rancid geode
then it becomes more complex, no?
rancid geode
${\int \frac{\sqrt{u}}{-2x} \dd u}$
elfin berryBOT
k
rancid geode
then u have to write x in terms of u to be integrable
jaunty garden
how did -2x end up in the denominator
vast magnet
-2x is not multipled right
rancid geode
du = -2x dx
$\dd u = -2x\dd x \implies -\frac{1}{2x}\dd u = \dd x$
elfin berryBOT
k
vast magnet
yeah because that's a different integral
rancid geode
its worth u subbing if u know the resulting term is more simplified
fickle shell
marsh citrusBOT
fickle shell
I feel like I can almost solve this
Im just missing one piece of information or something
The equation of the circle is (y)^2 + (x-2)^2 = 16
And I think we need to do something with |ax + by - c|/sqrt(a^2 + b^2)
The point on the circle where the tangent intersects (lets call it T)
I know the line through the midpoint of c and point T has an inverse slope to the line between T and P
I just dont know the coordinates of T so idk how I solve
inland brook
I just dont know the coordinates of T so idk how I solve
You need to use point slope form
fickle shell
You need to use point slope form
But I dont have the coordinates of T? Or the slope of the line
Im missing 3/5 unknowns
inland brook
Then substitute and solve for the missing
fickle shell
Then substitute and solve for the missing
But how do I solve? Thats the issue
Too many unknowns
inland brook
Hint: write all the variables in terms of one then substitute into point slope form substituting one of the x's and y's as P
fickle shell
In terms of one what sry?
inland brook
In terms of one what sry?
Like one variable x, y, z, etc
fickle shell
Hint: write all the variables in terms of one then substitute into point slope f...
What other equation do we use?
To get everything in terms of 1 variable
inland brook
To get everything in terms of 1 variable
Like isolate that variable
inland brook
Oh
fickle shell
I would love to get everything in terms of 1 variable
But I really dont see how
With the info we have
inland brook
Ok so you don't know how to find the slope?
inland brook
We know that the slope of the circle is equal to the slope of the line at the points we need
fickle shell
Wait what
inland brook
We know that the slope of the circle is equal to the slope of the line at the po...
Derivative
fickle shell
Wym slope of the circle
inland brook
Wym slope of the circle
You know how to find a tangent line for a for a point on a function?
inland brook
And you know that the derivative is the slope at the point where it intersects the circle think about it
fickle shell
And I think we need to do something with |ax + by - c|/sqrt(a^2 + b^2)
Are you sure were not supposed to use this equation 😭
inland brook
Why I don't see how
fickle shell
Ok nevermind I guess not
It was just that we couldve fit a lot of things in there
And it seemed related to this question
But ok
fickle shell
Ok
My issue is we dont have either of the red dots
The coordinates of them
If we did we would work out the slopes of everything
inland brook
but we can assume they are both part of the original equation
like the original circle equation had x and y lets assume (x1,y1) is the point of the intersection and swap them for x and y and do the same for the slope(derivative)
fickle shell
(y)^2 + (x-2)^2 = 16
inland brook
yes
inland brook
the same x1 and y1
inland brook
yes
inland brook
both
fickle shell
2(y1) + 2(x1) - 4
inland brook
where is dy/dx next to 2y
inland brook
what
inland brook
Forget about substituting just take the derivative of the circle equation
fickle shell
Forget about substituting just take the derivative of the circle equation
Why? What will that give us?
I dont understand anything were doing sry
inland brook
dy/dx didn't say dy/dx=m like in the picture
fickle shell
I need a break I will come back to this my brain is cooked
Thank you for your help to this point
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marsh citrusBOT
marsh citrusBOT
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ripe comet
the answer is -1, i have no idea how hmhm
marsh citrusBOT
the answer is -1, i have no idea how hmhm
formal fern
Is there a list of possible answers or something? It reads like a multiple choice question
proud ice
the answer is -1, i have no idea how hmhm
My first guess is to find local extrema
formal fern
no thats the function
ripe comet
red is answer
proud ice
I think as it goes to negative infinity
proud ice
(which rules out my local extrema idea)
ripe comet
I think as it goes to negative infinity
how is it not 1 for both goes to negative inf and positive inf
proud ice
i had them backwards
grizzled sonnet
sin(1/x) as x goes to negative infinity
1/x approaches 0 from the negative side, never reaching it. sin(0.1) > 0, sin(-0.1) < 0
so, this term is negative as x goes to negative infinity
and sqrt(x) is always positive, so positive * negative = negative
proud ice
Haven't looked at the oroblem much
grizzled sonnet
it cant be a positive number
ripe comet
sin(1/x) as x goes to negative infinity
1/x approaches 0 from the negative side...
oh i get it thxxx
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marsh citrusBOT
valid cape
miku
marsh citrusBOT
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drowsy bane
hi can you help me find if the following function is primitive recursive
marsh citrusBOT
hi can you help me find if the following function is primitive recursive
@drowsy bane Has your question been resolved?
marsh citrusBOT
marsh citrusBOT
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lyric kelp
here they gave the example that a polynomial with degree < n is not closure
lyric kelp
but the title is litreally degree < or equal to n?
devout mauve
the polynomials of degree =n do not form a subspace
the polynomials of degree <=n do form a subspace
marsh citrusBOT
@lyric kelp Has your question been resolved?
marsh citrusBOT
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radiant vine
LMFAOi have to tell this.
a 2008 student first grader in my high school, FAILED a recap (meaning from primary school) course of math. what a loser!
worn sluice
Its okay if somebody doesn't know maths
radiant vine
i see him rn
i can send u a pic
he’s so dumb probably
A RECAP
From primary school
ocean plaza
@supple wigeon @worn sluice stop responding. Spam account.
final parcel
marsh citrusBOT
final parcel
Is this right?
quaint elm
,w plot (x^2 + x - 3)^4, 12x + 25 from -3 to -1
elfin berryBOT
quaint elm
looks like it to me
lucid zenith
those are some weird plot lines
final parcel
Ok thanks
quaint elm
,w plot (x^2 + x - 3)^4, 12x + 25 from -2.01 to -1.99
elfin berryBOT
final parcel
Quick question
Is the interval on which a function is differentiable just the domain?
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marsh citrusBOT
marsh citrusBOT
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buoyant jetty
marsh citrusBOT
buoyant jetty
- Let $\mathbb{H} = {x \in \mathbb{R}^4 \mid kx_1 - 2x_3 - x_4 = 0}$ and $\mathbb{S} = \langle (1, 0, 1, 2), (0, 1, 1, -2) \rangle$. Find $k \in \mathbb{R}$ and a subspace $\mathbb{T}$ of $\mathbb{R}^4$ such that $\mathbb{S} \oplus \mathbb{T} = \mathbb{H}$ and $\mathbb{S}^\perp \cap \mathbb{T} \neq {0}$.
elfin berryBOT
why is math so hard anyways?
spark otter
3. Let $\mathbb{H} = \{x \in \mathbb{R}^4 \mid kx_1 - 2x_3 - x_4 = 0\}$ and $\ma...
!status
marsh citrusBOT
3. Let $\mathbb{H} = \{x \in \mathbb{R}^4 \mid kx_1 - 2x_3 - x_4 = 0\}$ and $\ma...
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
1
I was finding a basis for H
kx1 - 2x3 - x4 = 0
x4 = kx1 - 2x3
(x1,x2,x3,x4) = (x1,x2,x3,kx1-2x3)
(x1,x2,x3,kx1-2x3) = x1(1,0,0,k) + x2(0,1,0,0) + x3(0,0,1,-2)
H = <(1,0,0,k),(0,1,0,0),(0,0,1,-2)>
buoyant jetty
the dimension of T depends on k
because
I dont know if you know the dimension theorem of finite vector spaces
but basically is also called the grassman's formula (popularized by hermann grassman I think)
but basically
,, \textbf{This direct sum implies that} \ \mathbb{S} \oplus \mathbb{T} = \mathbb{H} \ \implies \dim(\mathbb{S}) + \dim(\mathbb{T}) - \dim(\mathbb{S} \cap \mathbb{T}) = \dim(\mathbb{S}) + \dim(\mathbb{T}) \ \textbf{because } \dim(\mathbb{S} \cap \mathbb{T}) = 0
elfin berryBOT
why is math so hard anyways?
buoyant jetty
but as I said dim(H) depends on the k parameter
I want to say dim(H) = 3 but that depends on the k
furthermore, S = <(1,0,1,2),(0,1,1,-2)> ==> dim(S) = 2 because one of the vectors is not a multiple of the other (because the second vector has a 0 in the first coordinate while the first vector not, for example
buoyant jetty
this is from the corollary here
let me find the orthogonal complement of S, that way we can see what is the dimension of S perp
,w Nullspace[{1,0,1,2},{0,1,1,-2}]
elfin berryBOT
buoyant jetty
a generic vector in S perp can be written as
(-x-2y, -x+2y, x,y) = x(-1,-1,1,0) + y(-2,2,0,1)
S perp = <(-1,-1,1,0),(-2,2,0,1)>
,, \mathbb{S}^{\perp} = \langle (-1,-1,1,0), (-2,2,0,1) \rangle
elfin berryBOT
why is math so hard anyways?
buoyant jetty
from the direct sum, we know SnT = {0} , they share only the zero vector
,, \mathbb{S} \oplus \mathbb{T} \implies \mathbb{S} \cap \mathbb{T} = {0}
elfin berryBOT
why is math so hard anyways?
buoyant jetty
other think to know is that since S has a basis with two vectors inside
the dimension of the orthogonal complement of S will be two aswell
since they are complementary subspaces
but basically the dimension of T has to be at least 1, since they share a nontrivial intersection with S perp
spark otter
I want to say dim(H) = 3 but that depends on the k
why?
spark otter
nope
you wrote yourself that H = <(1,0,0,k),(0,1,0,0),(0,0,1,-2)>
those three vectors are always gonna be linearly independent
if a(1,0,0,k) + b(0,1,0,0) + c(0,0,1,-2) = 0
then a = 0 (first coordinate)
b=0 (second coordinate)
c=0 (third coordinate)
dim(H) = 3 everytime
buoyant jetty
yeah I guess I agree
spark otter
yep
so we're looking for T spanned by a single vector
so if we write T = span(v)
what do we know about v?
buoyant jetty
v is contained in Sperp
spark otter
yep
spark otter
**why is math so hard anyways?**
so this can help us write v
buoyant jetty
S perp = <(-1,-1,1,0), (-2,2,0,1)>
v = a(-1,-1,1,0) + b(-2,2,0,1) with a,b nonzero scalars
spark otter
yep correct
now
before we find out a and b
there is a way to find out k first
remember what can be said "inclusion-wise" when we have A + B = C?
buoyant jetty
v must be contained in H, we can make it satisfy the equation of the plane in H
buoyant jetty
A + B = C ==> A is contained in C and B is contained in C
spark otter
A + B = C ==> A is contained in C and B is contained in C
yesss
so we're gonna use that v is in H in just a second
so we're gonna use that T is contained in H
but we haven't used that S is contained in H yet
buoyant jetty
first lets do S contained in H to find k
S = <(1,0,1,2),(0,1,1,-2)>
s in S with a,b nonzero scalars
s = a(1,0,1,2) + b(0,1,1-2)
s = (a,b,a+b,2a-2b)
now lets make this generic vector in S, satisfy the equation of the plane H
H : kx1 - 2x2 - x4 = 0
(x1,x2,x3,x4) = (a, b, a+b, 2a -2b)
k.a -2.b -(2a-2b) = 0
ak - 2b -2a +2b = 0
ak - 2a = 0
spark otter
it's 2x3 not 2x2
buoyant jetty
fuck my bad
H : kx1 -2x3 -x4 = 0
(x1,x2,x3,x4) = (a,b,a+b,2a-2b)
k.a - 2.(a+b) - (2a-2b) = 0
ak -2a -2b -2a + 2b = 0
ak -4a = 0
ak = 4a
spark otter
yes and that is true for all real inputs of a
buoyant jetty
my messages are not being sent
my internet is so ass
k = 4 !!! 
my internet is real bad, sorry if the messages are not arriving
idk why is so bad if I pay like 50 usd monthtly
like I am typing but they are not being sent
they are not being sent
buoyant jetty
H = <(1,0,0,4),(0,1,0,0),(0,0,1,-2)>
S perp = <(-1,-1,1,0),(-2,2,0,1)>
s' in S perp with a,b, nonzero scalars
s' = a(-1,-1,1,0) + b(-2,2,0,1)
s' = (-a-2b, -a + 2b, a, b)
H = {x in R4 | 4x1 - 2x3 -x4 = 0}
2(-a-2b) - 2(a) - (b) = 0
-2a -4b - 2a - b = 0
-4a-5b = 0
-4a = 5b
a = (-5/4) . b
s' = ((5/4)b -2b, (5/4)b + 2b, (-5/4) . b, b)
s' = ((-3/4)b , (13/4) . b, (-5/4) b, b)
,, \mathbb{S}^{\perp} \cap \mathbb{H} = \langle \left(\frac{-3}{4}, \frac{13}{4}, \frac{-5}{4}, 1 \right) \rangle \ \mathbb{S}^{\perp} \cap \mathbb{H} = \langle \left(-3, 13, -5, 4 \right) \rangle \ \mathbb{T} = \langle \left(-3, 13, -5, 4 \right) \rangle
elfin berryBOT
why is math so hard anyways?
buoyant jetty
- Let $\mathbb{H} = {x \in \mathbb{R}^4 \mid kx_1 - 2x_3 - x_4 = 0}$ and $\mathbb{S} = \langle (1, 0, 1, 2), (0, 1, 1, -2) \rangle$. Find $k \in \mathbb{R}$ and a subspace $\mathbb{T}$ of $\mathbb{R}^4$ such that $\mathbb{S} \oplus \mathbb{T} = \mathbb{H}$ and $\mathbb{S}^\perp \cap \mathbb{T} \neq {0}$.
elfin berryBOT
why is math so hard anyways?
buoyant jetty
thank you for the help, do you mind helping me check if this T satisfies S(+) T = H
aka if a basis for S + T generate H and SnT = {0}
S = <(1,0,1,2),(0,1,1,-2)>
T = <(-3,13,-5,4)>
S (+) T ==> S n T = {0} ==> <(1,0,1,2),(0,1,1-2)> n <(-3,13,-5,4)> = {0}
like one way I was thinking was to count the number of pivots in the column space, if its equal to 3, then this three vectors are linearly independent
spark otter
I'm checking that T is not in H
so I would recheck the computations before
4*(-3) - 2(-5) - 4 = -6, not 0
buoyant jetty
wait but why T is not in H
A + B = C implies B is contained in C
,, \mathbb{S} \oplus \mathbb{T} = \mathbb{H} \implies \mathbb{T} \subseteq \mathbb{H}
buoyant jetty
so I would recheck the computations before
so I messed up when finding T then or when finding k
elfin berryBOT
why is math so hard anyways?
buoyant jetty
where did I mess up?
marsh citrusBOT
@buoyant jetty Has your question been resolved?
marsh citrusBOT
@buoyant jetty Has your question been resolved?
marsh citrusBOT
@buoyant jetty Has your question been resolved?
spark otter
2(-a-2b) - 2(a) - (b) = 0
the problem is here, k = 4 and not 2
4x1 - 2x3 - x4 = 0
4(-a-2b)-2a - b = 0
-6a - 9b = 0
a = -3b/2 or 2a = -3b
2s' = (-2a-4b,-2a+4b,2a,2b) = (-b,7b,-3b,2b) = b(-1,7,-3,2)
and this matches
T = <(-1,7,-3,2)>
buoyant jetty
.solved
marsh citrusBOT
marsh citrusBOT
Available help channel!
craggy lake
Does anyone know how I can do this?
marsh citrusBOT
Does anyone know how I can do this?
main idol
have you learned complete the square
craggy lake
Oh I'm supposed to use the square formula?
main idol
no
i'm just answering this part
completing the square is a common method to solve those types of problems
craggy lake
Ok then how do I do this in a quick way? It took me about an hour during to finish this one question
elfin berryBOT
riemann
main idol
you're given a quadratic with b=8, c=9
just follow the right side with those values to find p and q
craggy lake
Please explain it to me like I know nothing about math 😭🙏
main idol
idk what you're confused by
craggy lake
Yeah I get that part but what do you mean follow the right side to fin the values ot p and q? Do I just 8x + (8/2)^2 + 9 - (8/2)^2?
main idol
compare this with
this
you should be able to identify p and q once you plug in b=8 and c=9
craggy lake
So wait
craggy lake
compare this with
I use this?
main idol
yes
craggy lake
Ok one moment 👍
marsh citrusBOT
@craggy lake Has your question been resolved?
storm flume
marsh citrusBOT
storm flume
ok so ive been kinda sorta guessing what i should be doing for everything else
but this time im stumped
im on question d
unreal oxide
I suppose it must be 8 + 2(¾)²
storm flume
wait why 3/4?
storm flume
OH +?!
OKAY WAIT CHAT THAT MIGHT MAKE SENSE
that would be 18/4 + 8 yes?
ok it make sense now!!
ty so much silversoldier!!
.close
marsh citrusBOT
marsh citrusBOT
Available help channel!
rocky nova
is there a fast way to calculate 10*(5^6) mod 13 on paper?
marsh citrusBOT
is there a fast way to calculate 10*(5^6) mod 13 on paper?
rocky nova
I know about square-multiply but that works best when I am given just a single int on the left side of the mod
5^6 on paper is not fun
rocky nova
sure
rocky nova
you're making jumps that I am not following
rocky nova
25 mod 13 = 12
static quarry
well 5^6 = (5^2)^3
and if you do that mod 13
you get:
(-1)^3
which is easy to compute
rocky nova
its -1 again
rocky nova
-10
rocky nova
how is -10 aka mod 13
static quarry
two numbers are equal mod 13 if they differ by an integer multiple of 13
so what is -10 also known as
when you work mod 13?
rocky nova
3
rocky nova
how come I can do the 5^6 mod 13 separately ?
static quarry
because they're easier to deal with
static quarry
ah you have the general fact that [xy] = [x][y] where [] indicates the equivalence class mod whatever
static quarry
let me try
sure try
rocky nova
@static quarry it works, or I got lucky, but i think it works
I still think you're way is better because it doesn't involve as many "big" numbers
@static quarry is all this written legal?
for the 2nd to last last, should I omit the "mod 13" ?
bingo bango bungo
buoyant jetty
marsh citrusBOT
elfin berryBOT
why is math so hard anyways?
buoyant jetty
this was cursed to me the day I got it on the exam
I think what happened was that the W subspace I found had no intersection with T somehow
or like S n H had no intersection
something extremely offsetting, if someone can help me out with this one 
marsh citrusBOT
@buoyant jetty Has your question been resolved?
marsh citrusBOT
@buoyant jetty Has your question been resolved?
marsh citrusBOT
@buoyant jetty Has your question been resolved?
wheat trench
Am I correct in thinking that if $\tan \theta = \frac{\sin \theta}{\cos \theta}$, then the length of the tangent segment is $\tan \theta$ multiplied by the length of the x-axis segment demarcated in pink?
marsh citrusBOT
Am I correct in thinking that if $\tan \theta = \frac{\sin \theta}{\cos \theta}$...
elfin berryBOT
Vulkanoid
marsh citrusBOT
@wheat trench Has your question been resolved?
wheat trench
.close
marsh citrusBOT
valid wing
is that y power n.?
marsh citrusBOT
@small stream Has your question been resolved?
marsh citrusBOT
@small stream Has your question been resolved?
quick nexus
Can you solve this plz
marsh citrusBOT
Can you solve this plz
novel juniper
!status
marsh citrusBOT
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
find maxima and minima
quick nexus
ye i know but how
quick nexus
i've already done fx -1/2 and fx + 1/2
.
valid wing
ye i know but how
differentiate it
quick nexus
How to prove that the first one (fx - 1/2) is < 0
And prove that the second (fx + 1/2) is > 0
wooden shore
Since xsinx satisfies -x≤xsinx≤x multiply (x^2+1) on all of them so we can just differentiate x/(x^2+1)
And find the maximum
atomic ruin
idk how to do
atomic ruin
<@&286206848099549185>
elfin berryBOT
amber birch
the other equation you need is say the normal vector is parallel to (2, 6 + t, 3 + t (lambda - 3)) - (7, 4, -1))
green point is free to move
so keep the other endpoint of the line fixed
amber birch
the other equation you need is say the normal vector is parallel to (2, 6 + t, 3...
once you find t, set the magnitude of (2, 6 + t, 3 + t (lambda - 3)) - (7, 4, -1)) equal to 3
you have another quadratic in lambda to solve
marsh citrusBOT
@atomic ruin Has your question been resolved?
atomic ruin
green point is free to move
im a little confused on this
amber birch
ah okay
atomic ruin
what is 2,6+t, 3+t(lambda-3)
amber birch
what is 2,6+t, 3+t(lambda-3)
ah it's the position vector of a point on line CD
I used your parametric equation for line CD
so (point on line CD) - (point (7, 4, -1) on line AB) = red direction vector in my drawing
atomic ruin
the other equation you need is say the normal vector is parallel to (2, 6 + t, 3...
wait why parallel not perpendicular
amber birch
wait why parallel not perpendicular
you've already found the normal vector right
if we were using perpendicular, we would use the direction vector of one of the lines is perpendicular to the red vector
atomic ruin
if we were using perpendicular, we would use the direction vector of one of the ...
arent both direction vectors perpendicular to my normal vector?
amber birch
arent both direction vectors perpendicular to my normal vector?
yes, that's the point
your normal vector should be parallel to the red line
for the shortest distance
atomic ruin
Oh
amber birch
since the red line will become a normal vector to both lines
atomic ruin
you are right
Oh
WAIT
so
line AB and line CD, we can express any point on them in the form of a+blambda with s and t right?
amber birch
,w (4, -1, 1) cross (0, 1, lambda - 3)
elfin berryBOT
atomic ruin
that should be my normal vector
amber birch
yeah and you did it correctly, I typed it in incorrectly before
amber birch
line AB and line CD, we can express any point on them in the form of a+blambda w...
yes!
atomic ruin
line AB and line CD, we can express any point on them in the form of a+blambda w...
but my normal vector has to be stretched or compressed to fit the line between those two ARBITRARY points
amber birch
you just need to make one point free to move
atomic ruin
you just need to make one point free to move
why only one?
amber birch
sorry I don't know the reasoning but somehow it works with only 1 point
like this
wait
atomic ruin
yeah
how would i do that? do u mind showing the steps
i briefly know this but not this method
tysm
atomic ruin
yes
amber birch
,w (2 - 7)/(2 - lambda) = (6 + t - 4)/(12 - 4 lambda) = (3 + t (lambda - 3) - (-1))/(4)
ah you need all 3 equations
atomic ruin
what hte hell is that 😭
elfin berryBOT
amber birch
what is going on
so like if (2, 3, 4 + a) and (4, 6, 10) are parallel let's say
then 2/4 = 3/6 = (4 + a)/10
atomic ruin
is that a line and a point
amber birch
ah I think you need to find the equation of the plane passing through say point A, and the plane passing through point C
amber birch
is that a line and a point
no, two lines
amber birch
ah I think you need to find the equation of the plane passing through say point ...
you need to do this instead
atomic ruin
why ddo we need a plane'
amber birch
why ddo we need a plane'
you need to use the fact that the distance between the line AB and the line CD is 3
but with your normal vector
you can make two planes, one plane contains line AB and one plane contains line CD
and they are parallel planes! cause you have used the normal vector, so both planes have the same normal vector ---> parallel
atomic ruin
this is so complicated 😭
amber birch
this is so complicated 😭
yeah I think you're missing knowledge
try some easier questions first and go over worked examples in your textbook or on YT
also consider joining the r/IGCSE Discord
despite the name, there's a really active AL further channel
atomic ruin
also consider joining the r/IGCSE Discord
shit thanks
i never knew this existed
but this is further maths
alr ill join
ty
marsh citrusBOT
marsh citrusBOT
Available help channel!
atomic ruin
x^3+2x^2+5=0
marsh citrusBOT
x^3+2x^2+5=0
atomic ruin
how do i use recursion to find (r1)^2+(r2)^(r3)^2?
spark otter
wdym recursion?
atomic ruin
you know what S1 is
yup
spark otter
x^3+2x^2+5=0
in this case
atomic ruin
idk how to find S2
spark otter
compare S2 and S1^2
atomic ruin
huh?
atomic ruin
yup
spark otter
S1 is the sum of roots
atomic ruin
yup
spark otter
what is (S1)^2
atomic ruin
im not sure sorry
spark otter
S1 = r1 + r2 + r3
so what is (S1)^2
compute it
wait do you know vieta's formula(s)?
atomic ruin
so what is (S1)^2
we trying to do (r1+r2+r3)^2 - something right
spark otter
yeah
atomic ruin
oh :(
atomic ruin
can i do a substitution
spark otter
on a degree 3
atomic ruin
y=x^2
atomic ruin
doesn't sound very "polynomial" to me
but
now we have
y^(3/2)+2y+5=0
y^(3/2) = -5-2y
square both sides
and you get a neat polynomial
spark otter
a worse polynomial than the one you started with
and still a cubic polynomial
so now instead of solving directly for x
you're now solving for x^2
in an equally complicated polynomial
spark otter
but i can just do
?
spark otter
I see what you did now
ok now last problem
"y^(3/2)" is very ambiguous
perhaps if you're trying to do the same thing but rigorously
say with x^3 + 2x^2 + 5 = 0
and do the sub "y = x^2" last, when it really works
atomic ruin
and do the sub "y = x^2" last, when it really works
wdym by this?
spark otter
because y^(3/2) doesn't make sense when x < 0
atomic ruin
thats true
spark otter
so the sub y = x^2 is at the very least, too soon
you can still do x^3 = -2x^2 - 5
and square
x^6 = (....)^2
and now the sub y = x^2 would work
marsh citrusBOT
marsh citrusBOT
Available help channel!
grim hemlock
How is that Green V/U line obtained from the definition V/U = {v + U | v element of V}
marsh citrusBOT
How is that Green V/U line obtained from the definition V/U = {v + U | v element...
@grim hemlock Has your question been resolved?
marsh citrusBOT
@grim hemlock Has your question been resolved?
marsh citrusBOT
@grim hemlock Has your question been resolved?
marsh citrusBOT
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grim hemlock
.close
acoustic plume
marsh citrusBOT
marsh citrusBOT
@acoustic plume Has your question been resolved?
marsh citrusBOT
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acoustic plume
<@&286206848099549185>
marsh citrusBOT
@acoustic plume Has your question been resolved?
eager flicker
I dont understand how to integrate this last term in the provided equation
main idol
wut
stray hound
yeah this is not very readable lol
main idol
what's given and what are you trying to integrate
eager flicker
y and c are both functions of z.
@main idol
I am all over the place here. Restarting: here is my entire question rewritten in mathcad for legibility.
main idol
what is dc?
eager flicker
i dropped the dz term as I separated it. It is 1*dc so after integration I get just the function, c.
i do not know if this is legal to do with the second term.
I've only done this before when it was two partial derivatives that were with respect to different variables multiplied together. (or with isolated partials like that dc term)
marsh citrusBOT
@eager flicker Has your question been resolved?
eager flicker
No
<@&286206848099549185>
For context: My intent is to derive an equation for dy/dz.
This is a species mass balance derived from the diffusion equation and the continuity equation
The screenshot I've already provided is a simplified version of my equation that zeroes in on exactly what i am stuck on.
This following screenshot is the work I've done so far on this equation.
Every variable is a function of z except for the following:
This is the context, but the only thing i care about is the question i ask in the following:
marsh citrusBOT
@eager flicker Has your question been resolved?
marsh citrusBOT
@eager flicker Has your question been resolved?
bold condor
What is 6⁶×5⁹÷⅔×%66=___
marsh citrusBOT
What is 6⁶×5⁹÷⅔×%66=___
smoky spire
use a calculator
regal coral
do you mean 66%?
brave bluff
use a calculator
Ding ding ding that's the right response
proud ice
What is 6⁶×5⁹÷⅔×%66=___
!redir
marsh citrusBOT
What is 6⁶×5⁹÷⅔×%66=___
This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.
proud ice
.close
marsh citrusBOT
jovial osprey
ngl thats overkill
regal coral
seemed on topic to me
proud ice
.reopen
jovial osprey
me too
marsh citrusBOT
✅
proud ice
ngl thats overkill
If you you guys believe this is legit, I'll keep it alive.
wary kite
troll
proud ice
I'll admit I acted hastily, but I see this play out too much
regal coral
well, it may be a troll
jovial osprey
they prob ran away now xD
proud ice
yeah I'm with knief, but if y'all want benefit of doubt, I respect that
still temple
hiii