patent belfry
doing ",w" before your message
#help-13
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quiet plover
what happened
drowsy sage
Sounds culty
teal meadow
mathematics drama
drowsy sage
There are only a certain number of rooms available.. do you need math help, or did you just want to chat?
teal meadow
do /kick
drowsy sage
Because if you just want to chat, feel free to head down to the general room
teal meadow
There are only a certain number of rooms available.. do you need math help, or d...
there's quite a few left its not a huge deal but yeah
try pinging instead
@ and then their name
itll autofill click the autofill
i think it worked
drowsy sage
Not really.
teal meadow
damnit hes too powerful
drowsy sage
No
I'm just active on the server.. hence the blue
You can't, but you can also ping moderators as a group if needed
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drowsy sage
If you want to kick people who are here to help, you're not going to make it very far
oh, feel free
broken bough
hello
daksh you are grounded sir
i have come to send you to your room
nah im jk
and no i dont bc you cant
every one of the 200,000 some people are in this channel btw daksh
ignore that part then
it was just a bit of pazazz
drowsy sage
"by the way"
broken bough
bro what am i experiencing
i need to sleep
ope
ignore me \
im just here for entertainment purposes
ok bye guys im no help so bye
drowsy sage
Well, you can't kick me, but you can right click my ugly pic and Block me... but I mean.. eventually someone is going to ping the <@&268886789983436800> and point out that you're not here for math help
slate lintel
i think you'd be better off in your own server where you are free to kick people
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craggy stratus
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craggy stratus
Show that if p > 1 then x^p - px + p >= 1 for all x that’s >= 0
What would the idea be to prove this?
I thought about derivating to see when the function grows/shrinks in value but i feel like that’s a rough approach
But if i do that can i derivate both sides in the difference equation?
so i get px^(p-1) - p >= 0
cedar kilnBOT
@craggy stratus Has your question been resolved?
sand cradle
What would the idea be to prove this?
A possible solution will most likely be with Bernoulli's inequality: [(1 + x)^p \geq 1 + xp.] But I haven't worked it out yet.
craggy stratus
I was thinking something like this
sand cradle
craggy stratus
But I’m not entirely sure on the bottom right part
I get confused because if i set x to 0 there 0^0+ isn’t 1..
But this is Also the derivative of the function..
But if I have x^1^+ - 1^(+)*x >= 0^-
Is there another way to show that the function will grow?
sand cradle
Is there another way to show that the function will grow?
I don't think it will always grow
Actually, I'm pretty sure it won't always grow
Think of p = 2
sand cradle
Yes
sand cradle
This is why it strongly hints towards this approach
craggy stratus
But surely we can approach this with derivatives and limits no?
sand cradle
You'd want to find the minimum of x^p - px + p, for x >= 0 and p > 1
craggy stratus
I don’t see How you manage to factor like that either
My brain is not braining with the (1+x)^p part
craggy stratus
You'd want to find the minimum of x^p - px + p, for x >= 0 and p > 1
Yeah so then we want the derivative, but should that be based on x or p..?
I think this is where i struggle since we have two variables😟
I feel like doing the limit for p > 1+ should work
sand cradle
Ok, this trivially follows by Bernoulli actually
craggy stratus
And then we get x^1+ - 1^+x >= 0-
And then i feel like i should be able to find out when that expression hits its lowest value somehow..
I thought about trying to factor out x but we can’t factor out stuff with exponents right?
sand cradle
[x^p - px + p \geq 1 ][\iff x^p \geq 1 + px - p ][\iff (1 + (x - 1))^p \geq 1 + p(x - 1)] which is true by Bernoulli's inequality $(1 + x)^p \geq 1 + px$.
Beautiful
craggy stratus
What happened in the LJS
sand cradle
Surely we can do this with Calculus too
craggy stratus
LHS
sand cradle
What happened in the LJS
I just rewrote x as 1 + (x - 1)
I can do that, 1 and -1 cancels out
craggy stratus
ok i think i see
sand cradle
Ok, now let's try to get a calculus proof
sand cradle
Let $f(x) = x^p - px + p$. Then $f'(x) = px^{p-1} - p = p(x^{p - 1} - 1)$. Setting that equal to $0$ gives us $x = 1$.
craggy stratus
So then it cancels before we apply the exponent
sand cradle
sand cradle
So then it cancels before we apply the exponent
Exactly
craggy stratus
That’s the part i struggled with, never saw this approach before, very nice
And yes this is the part i got to too.. and then it fell apart :D
sand cradle
Now $f''(x) = p(p-1)x^{p-2} > 0$ because $p > 1$ and $x > 0$. So there is a minimum at $x = 1$.
craggy stratus
So then we know that the min/max is at x=1 right?
sand cradle
sand cradle
So then we know that the min/max is at x=1 right?
Yeah, it's a minimum because the second derivative is always positive
craggy stratus
Hold up
How did you get to the conclusion that it’s positive again?
Because x can Also be 0
I think
sand cradle
Because x can Also be 0
Ah, then we can't generalize like I did
craggy stratus
But we can say that there’s a min/max at x=1 right?
sand cradle
But we can still say it's a maximum because [f''(x) = p(p-1)x^{p-2} ] and now $x = 1$ plugged in gives us $p(p-1)$
craggy stratus
And then we can check at x=0,5 and x=2 and compare
sand cradle
And we know p>1
sand cradle
And then we can check at x=0,5 and x=2 and compare
We can just use the second derivative and plug in x = 1
craggy stratus
Okok
sand cradle
We just can't say "the second derivative is always positive" because of x = 0
So we really have to plug in
craggy stratus
Sorry for dumb question but What does the second derivative ”tell us”?
Because first derivative is growth/shrinkage per x no? In that point
sand cradle
Concavity. A positive second derivative says the graph is concave up (left-curved), a negative one tells us it's concave down (right-curved)
,w plot x^3
wraith daggerBOT
sand cradle
x^3 is concave down for x < 0 (negative second derivative) and concave up for x > 0 (positive second derivative)
craggy stratus
Uhhh
But like x^3
Derivative is 3x^2
So then it tells is the speed at which it is growing
sand cradle
Yes
craggy stratus
6x would tell us What?
Another word for concavity plox
Because if we plug in x=-1
It’s -6x But it’s growing
sand cradle
If you imagine a car driving and f is the function describing the length of the path that it has passed, the first derivative is the speed and the second derivative the acceleration
wraith daggerBOT
craggy stratus
,w plot 6x
wraith daggerBOT
sand cradle
Because if we plug in x=-1
A negative second derivative tells us that the derivative is decreasing
craggy stratus
A negative second derivative tells us that the derivative is decreasing
Ok i think i might get it
sand cradle
Ok i think i might get it
Think of the car analogy
sand cradle
Yes
craggy stratus
F’’(x) = (p-1)px^p-2
And then we know that we have min/max at x=1
Plug it in to see what the acceleration is in that point
P(p-1)
which always is greater than 0
So if i understand it correctly this would mean there’s no chance that as x>=0 we would get a value which is less than 1 from the original function
sand cradle
f(1) will be the minimum value of the function (also called global minimum)
Be careful, not 1. 1 is just the x-value at which the minimum occurs
craggy stratus
So full Proof would be to say f’’(x) in the f’(x) = 0 is positive, ergo the inequality must be true
sand cradle
So full Proof would be to say f’’(x) in the f’(x) = 0 is positive, ergo the ineq...
Well we still have to see what f(1) is
sand cradle
Yeah
sand cradle
So we know f(x) >= 1
craggy stratus
Take care!!
sand cradle
Np
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crimson prairie
so you know how some aspects of series can be ignored when doing a test, like for the nth term test where you can ignore lower powers, what are all of the tests that I could apply that to
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grizzled hare
<@&286206848099549185> any1?
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@grizzled hare Has your question been resolved?
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@grizzled hare Has your question been resolved?
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@grizzled hare Has your question been resolved?
grizzled hare
<@&286206848099549185> please?
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uneven swallow
I really do not understand how I am getting this wrong. I don't know if I genuinely bad at math, but something does not add up.
unkempt totem
can you show your steps
livid hound
read the question carefully for what it actually wants
unique valley
i think you found angle theta, but they ask you to find cos(theta)
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weak spruce
What grade is trigonometry learned in? what's the earliest and latest grades countries teach students trig.
mighty shuttle
I learnt it in 10th
cedar kilnBOT
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weak spruce
I learnt it in 10th
what would be the earliest grade countries teach it?
vernal basin
I think 10th in the US
But there are accelerated programs
I learned it in 8th
But that is the earliest unit of trig
mossy pumice
LaGrange Multipliers: I think I have things right up to the system of equations. What is wrong here?
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knotty steeple
its physics but its a pretty simple question
cedar kilnBOT
its physics but its a pretty simple question
@knotty steeple Has your question been resolved?
uncut quarry
its physics but its a pretty simple question
so, did learn about kirchoff law?
knotty steeple
ye
uncut quarry
great, so u have two loops here, u get the voltage law of each one and then using the current law at the node, u can solve it
green crow
I'm having trouble visualizing this problem like drawing the picture
green crow
I attempted the problem yesterday and this is what I think the problem is telling me. Still have no clue what to do
winged umbra
yeah
green crow
Now how would I actually solve this problem
uncut quarry
Now how would I actually solve this problem
if its a frictionless table the the acculturation of the rocket would still the same up until it reaches the table edge and fall
green crow
Is acculturation another word for force
uncut quarry
so u need to know the velocity then first before it falls
winged umbra
yeah
uncut quarry
Is acculturation another word for force
f=ma
winged umbra
(force equals mass times acceleration)
green crow
To find velocity would I need a kinematic equation
uncut quarry
first of all what will be the acceleration here
green crow
Well to know that I would need to know the acceleration
uncut quarry
Well to know that I would need to know the acceleration
acceleration*
bad auto type
green crow
1.28 m/s^2
Did I make the right assumption of initial velocity being 0 because the problem says it is released from rest
uncut quarry
Did I make the right assumption of initial velocity being 0 because the problem ...
yes
now use motion equation
uncut quarry
yep
uncut quarry
alright, now do u know about projectile motion?
green crow
A little, not the best at it
I don't understand how projectile motion works here because usually it goes something like " A ball is thrown off a cliff a some initial velocity"
But didn't we say initial velocity is 0
uncut quarry
But didn't we say initial velocity is 0
yeah, that was befor the rocket falls of the table, but now when the rocket falls ut will fall with the initial velocity that we get
green crow
So the velocity I calculated is the new initial velocity?
uncut quarry
basically when the it was on the table, it only had a horizontal force, but now its in free falling the force of gravity is now in play
so we have two kind of motions, a horizontal one and a vertical one, and u would work each motion seperate from the other but they share the same time
green crow
So we only know acceleration in x is 1.28 and acceleration in y is -g
uncut quarry
yes, now what we need to know is the horizontal distance which is delta(x)
green crow
I'm stuck on what kinematic equation to use
I don't know what initial velocity is, final velocity or time
While falling
uncut quarry
alright, start witg what u have, what do u know about the horizontal motion? u only know the initial velocity, but u dont have time and u need to know the distance
then focus on the vertical motion, what do u about it?
like its initial velocity, and the distance
green crow
What was the initial velocity again
uncut quarry
What was the initial velocity again
of which motion
green crow
Horizontal
uncut quarry
well, what do u think when it began to fall vertically
green crow
I know for a fact final velocity is 0 because its eventually going to land on the ground
uncut quarry
did it have a vertical velocity befor when it was on the table?
green crow
did it have a vertical velocity befor when it was on the table?
I would say it doesn't
uncut quarry
I know for a fact final velocity is 0 because its eventually going to land on th...
nope, u cant assume that cuz it will hit the ground
and it will hit the ground with a speed
its not like it will slow its velocity
it will keep accelerating in the y until something stops it by force which is ground
uncut quarry
I would say it doesn't
yes, so initial y velocity is zero, right?
green crow
Yes
green crow
We don't know distance that's what I have to solve
uncut quarry
the vertical distance
green crow
We do know the vertical distance
uncut quarry
the vertical distance is like the height of the object
green crow
I thought the vertical distance is the height of the cliff
uncut quarry
I thought the vertical distance is the height of the cliff
yessss!!!
uncut quarry
We do know the vertical distance
oh wait , u said u do know
sorry my bad i thought u said dont know
whatever, so we have distance velocity and acceleration
whats left?
uncut quarry
whatever, so we have distance velocity and acceleration
or what can we do with these
green crow
I still don't know what the velocity is
uncut quarry
u know the initial y velocity
green crow
But I thought you said that before it falls of cliff
uncut quarry
the final values before it falls, is the initial values after it falls
logical it will not change its velocity when it starts to fall
uncut quarry
im talking about the moment when it begins to fall, at that moment it still will not a velocity in the y
after a smalle time, it will begin to accelerate and build speed in y
green crow
Is this like the rule constant velocity means zero acceleration but you reverse it
Based off your picture, Velocity in x must be a number and not 0 right?
uncut quarry
reverse it how?
uncut quarry
Based off your picture, Velocity in x must be a number and not 0 right?
the picture is about Vy
imagine if u drop a pencil from ur hand, it will start a Vy = 0
thats the same idea but here we have Vx equal a number
green crow
I would've said gravity is pulling the pencil downward
uncut quarry
I would've said gravity is pulling the pencil downward
thats acceleration
but that doesn't mean it has an initial velocity equal a number
cuz it clearly is at rest when u hold it, the moment u open ur hand and drop it, it began to build up speed from rest
green crow
If it builds up speed then how cna the speed be 0 in y
uncut quarry
If it builds up speed then how cna the speed be 0 in y
its initial value is zero
but after delta of time, it will have speed greater than zero
uncut quarry
If it builds up speed then how cna the speed be 0 in y
u are talking about the rocket here or the pencil?
green crow
In your picture is initial velocity in y 0 but once it reaches the cliff and falls the final velocity is 0 as well
uncut quarry
In your picture is initial velocity in y 0 but once it reaches the cliff and fal...
the final velocity is not zero, cuz if its it will hit the ground smoothly and softly and will stop by it self
but thats called a collision, where it hit the ground so hard
uncut quarry
but thats called a collision, where it hit the ground so hard
it will become different kind of motion
green crow
So we only know initial velocity in y is 0
uncut quarry
So we only know initial velocity in y is 0
yes,
green crow
This what I’m thinking
uncut quarry
This what I’m thinking
correct
from it u can get the time needed for the object to hit the ground
green crow
One problem I see is a_y=-g but we can’t have negative in square roots unless you want complex number
uncut quarry
One problem I see is a_y=-g but we can’t have negative in square roots unless yo...
also delta(y) is negative
cuz its toward the negative y
green crow
I got t=0.5714 seconds
uncut quarry
continue
green crow
I can solve for delta x but don't know if initial velocity in x is zero or not
uncut quarry
as i said before, final values befor it falls is the initial values when it began to fall
green crow
But I don't know final velocity in x either
uncut quarry
I got v_f ≈3.0776
dude then what is this!!!
green crow
How am I supposed to know it was in x direction
uncut quarry
cuz the object was moving in the x direction ._.
when the object is moving on the table, its moving in the x direction so it has a horizontal motion
uncut quarry
when the object is moving on the table, its moving in the x direction so it has ...
but it clearly dosent have a vertical motion in this time
green crow
I agree with that
uncut quarry
green crow
So I plugged in all the numbers and got the right answer
This was a really hard question
Don't think I could've done it myself
uncut quarry
Don't think I could've done it myself
and u have done it
the ideia is that when u work with projectile motion u work each of the vertical and horizontal motion seperately, where u will have some values known in one motion and what u need is in the other one
like we did, first we get time from vertical motion, then used it with horizontal motion
u just need to have a clear view of whats going
so next time i would suggest u first define what kind of motion have whether its just one type or different ones
then work each of them seperately and know that the final values of the first motion is the initial values of the second one
green crow
I did atleast know the kinematic equations and some basic assumptions
Sort of unrelated but I find my calculus 2 class easier than physics. This is the hardest subject I've come across ever
uncut quarry
Sort of unrelated but I find my calculus 2 class easier than physics. This is th...
yeah me two, with calculus and discrete math the rules is already known but with applied math u are restricted by the rules of physics
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mellow fog
hey i need help with this math algebra
cedar kilnBOT
hey i need help with this math algebra
mellow fog
i neeed to know what they did for this question's working out
how did they remove ln from both sides and then it becomes e^-115.38
i figured it out its allg
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vague sierra
questions 2. How can I find POI without finding the second derivative? it’s too complicated to find the second derivative so figured there should be a better way to find the POI
agile bramble
questions 2. How can I find POI without finding the second derivative? it’s too ...
nope, i think you really need to find the second derivative
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lime mortar
13-8+(14-6)=x
cedar kilnBOT
13-8+(14-6)=x
coral jewel
13-8+(14-6)=x
perform the calculations in the parentheses first
since theres nothing to complicate it, just do calculations from left to right
lime mortar
since theres nothing to complicate it, just do calculations from left to right
x=13-8+8 x=13-16 x=-3 is oke?
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lime mortar
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cedar kilnBOT
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fair vale
@cedar kiln
cedar kilnBOT
<@805217272087838750>
fair vale
<@&286206848099549185>
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fair vale
I am stuck in probabilty
@gleaming path
<@&286206848099549185>
<@&286206848099549185>
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fair vale
!done
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fair vale
.close
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ember lichen
How do I find the shaded area?
cedar kilnBOT
How do I find the shaded area?
slate lintel
it's a circle and you take out a square
low vine
Radius of circle is 1 then
slate lintel
2 cm is also the diagonal of the square
low vine
Radius of circle is 1 then
You can apply Pythagoras after that
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agile cradle
yo
pastel vault
if u want the primitive of this
Yeah, you have to use the chain rule
agile cradle
but for some reason its 2ln|2x-1| + c
agile cradle
ye
dapper cairn
its just u'/u
pastel vault
So given that integration and differentiation cancel each other out
dapper cairn
get 2 out
agile cradle
damn u type fast
pastel vault
You have to multiply by 1/2 so that 2 * 1/2 = 1
agile cradle
ooh shit ye
pastel vault
Yeah u-sub is another version of this, as you'll also get 1/2 if you do it that way
agile cradle
i thought it would be 4ln|2x-1| + c
so basically take this then multiply by the reverse integration thing
pastel vault
so basically take this then multiply by the reverse integration thing
Yep
agile cradle
ye right i forgot about that ty guys
agile cradle
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nova crypt
HELP ME FIND Q
cedar kilnBOT
HELP ME FIND Q
pastel vault
hi
sage field
hi
nova crypt
help pls
nova crypt
price index
sage field
wait a sec
nova crypt
jndex
pastel vault
I have a feeling they are proportional
nova crypt
i= q(current year)/q(base year)
pastel vault
So for example 150/165 = r/231
nova crypt
whats the formula for price index
its just index
nova crypt
yes
sage field
will that work?
nova crypt
whar
sage field
simplify it
nova crypt
111.429
sage field
let me check
nova crypt
but the answer q = 130
sage field
is it correct?
nova crypt
No
sage field
bruh
sage field
did you got p?
nova crypt
yes
nova crypt
thank
nova crypt
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novel aurora
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nocturne field
how can i prove that:
(MN)^-1 = N^-1*M^-1
cedar kilnBOT
how can i prove that:
(MN)^-1 = N^-1*M^-1
chrome delta
are those matrices?
nocturne field
yes
crimson delta
what does it mean to be the inverse of a matrix
nocturne field
um like 1/detM and then you flip a and d and multiply b and c by -1
buoyant latch
pastel vault
Here's a trick: start with $I = I$
Then $(MN)(MN)^{-1} = MN N^{-1} M^{-1}$
wraith daggerBOT
south
pastel vault
By associativity you can choose to do N N^(-1) first on the right
crimson delta
um like 1/detM and then you flip a and d and multiply b and c by -1
thats a specific case for 2x2 matrices. not what I mean
what is the definition of inverse matrix
pastel vault
um like 1/detM and then you flip a and d and multiply b and c by -1
Yes, actually this logic is nice
3B1B said something about this
Wait he didn't
chrome delta
chrome delta
Here's a trick: start with $I = I$
Then $(MN)(MN)^{-1} = MN N^{-1} M^{-1}$
i have this backwards lol
to justify why the order is reversed
pastel vault
i have this backwards lol
Yeah well the steps are reversible as long as M and N are not singular
It doesn't matter
chrome delta
(AB) (B^(-1) A^(-1)) = AI A^(-1) = I
pastel vault
Yeah that's also a proof
crimson delta
you also need the other way around
nocturne field
**south**
would i then write that :
I = M * N * N^-1 * M^-1
M^-1 * N^-1 = M ^ -1 * M * N^-1 * M * N^-1 * M^-1
and then simplify to :
(M * N)^-1 = N^-1 * M^-1
chrome delta
so just combine my thing with south's 
pastel vault
would i then write that :
I = M * N * N^-1 * M^-1
M^-1 * N^-1 = M ^ -1 * M * N...
Unfortunately your RHS is now $N^{-1} M N^{-1} M^{-1}$
wraith daggerBOT
south
nocturne field
ok ty i will try again
pastel vault
Nwnw
nocturne field
**south**
can i not just cancel out the MNs on both side to get to what is required in my proof? @pastel vault
because like they appear on both sides
pastel vault
can i not just cancel out the MNs on both side to get to what is required in my ...
You can't cancel in matrix multiplication
Dividing doesn't exist
You can only multiply both sides by the inverse
You can left multiply or right multiply, and you can't multiply something in between
nocturne field
say if i had like M * N * M ^-1 * N^-1 is that an identity
or does the matrix and its corresponding inverse have to be next to eachother
pastel vault
or does the matrix and its corresponding inverse have to be next to eachother
This
The order matters 100% in matrix multiplication
Cause AB is not equal to BA
nocturne field
yeah that makese sense ty
pastel vault
Nwnw
nocturne field
if you started with I = I could you then do:
(MN) * (MN)^-1 = I
M^-1 * N^ -1 * M * N * (MN)^-1 = I
(MN)^-1 = M^-1 * N^ -1
cedar kilnBOT
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frank bronze
cedar kilnBOT
frank bronze
i dont understand where this equation comes from, specifically the order of why its sin minus cos
void glen
this is the area bounded by the curves
when finding the exact area, we have to do the upper function minus the lower function or else the integral will evaluate to a negative number as we are subtracting a bigger number from a smaller number
and the upper function here is sine
cedar kilnBOT
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@frank bronze Has your question been resolved?
warm tangle
derivative is lim x->a f(x)-f(a)/x-a
If the limit exists why does somehow x-a cancels out, after lots of factoring and or rationalisation.
Eg a=1, f(x) = 1/sqrt(2x+2)
cedar kilnBOT
derivative is lim x->a f(x)-f(a)/x-a
If the limit exists why does somehow x-a ca...
@warm tangle Has your question been resolved?
dry latch
derivative is lim x->a f(x)-f(a)/x-a
If the limit exists why does somehow x-a ca...
the expression you've provided is the definition of the derivative of a function f(X)
The expression you've provided is the definition of the derivative of a function ( f(x) ) at a point ( a ), denoted as ( f'(a) ). It represents the rate of change of the function at that specific point.
[ f'(a) = \lim_{{x \to a}} \frac{f(x) - f(a)}{x - a} ]
The reason why the ( x - a ) factor seems to cancel out is because it represents the change in ( x ) relative to ( a ). When we take the limit as ( x ) approaches ( a ), we are essentially considering the rate of change of ( f(x) ) as ( x ) gets closer and closer to ( a ).
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garv
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warm tangle
My dude did you gpt this
cedar kilnBOT
My dude did you gpt this
main needle
warm tangle
I know 0/0 is weird
I just wanted to know why x-a always pops up in the numerator somehow so we can cancel it
Just like magic
sand cradle
I just wanted to know why x-a always pops up in the numerator somehow so we can ...
Well it needs to pop up or otherwise the limit is indeterminate
Then we'd have the numerator have some constant value and the denominator going to 0, since it has x - a and x approaches a
warm tangle
Then we'd have the numerator have some constant value and the denominator going ...
Fair but why though
I dont see how factoring and rationalisation pops out that
Is it a factor
sand cradle
Fair but why though
It's not always the case
e.g. take f(x) = e^x
There won't be an x - a in the numerator that pops up
eager wraith
VLAD
warm tangle
e.g. take f(x) = e^x
Isnt e special or sim
sand cradle
Isnt e special or sim
Yeah, the derivative of e^x is e^x
But e.g. also for f(x) = 2^x
There won't be an x - a that pops up in the numerator there
warm tangle
But e.g. also for f(x) = 2^x
So for all polynomials and rationals or sum?
Why does it pop up??
Whats happennin
main needle
well polynomials have a factor of (x-a) when they're 0 at x=0
might be annoying to see how this extends to 1/sqrt(...) though
warm tangle
So its beacuse 0 is a factor of 0?
main needle
idk what you mean
for the numerator f(x)-f(a), that's a thing that's 0 at x=a, and if it's a polynomial it would have an (x-a) factor
warm tangle
How?
main needle
it's from the fundamental theorem of algebra, that polynomials can be written as (x-root)(x-root)(x-root)...
warm tangle
I dont see how you use that
main needle
it says you can factor out x-a from the numerator
warm tangle
Would it mean we are factoring out x itself
main needle
just (polynomial that is 0 at x=a) can be factored as (x-a)(...)
and f(x)-f(a) does that because f(a)-f(a)=0
warm tangle
x-x=o , x=x?
I think i kinda see
How it somehow would happen indeed i may be wrong but i will not dwell on it thanks gotta go to next chapter thanks
warped cliff
Just making sure, this is true for all leminscates right?
cedar kilnBOT
Just making sure, this is true for all leminscates right?
@warped cliff Has your question been resolved?
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warped cliff
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abstract bane
I do not understand the step where it says that F( theta - epsilon) = 0 ??
abstract bane
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abstract bane
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late hemlock
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late hemlock
Can anyone confirm if I did it right?
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ashen palm
Hi, have a question about using the Wronskian to show linear independance.
ashen palm
I am wanting to show that the set of functions {x; cos2x; sin2x} is linearly independant on the real line
I find the determinant of the Wronskian to be equal to 8x
From the video/source I watched about this, I thought that if the determinant is zero at any point, then the function is linearly dependent
8(0) = 0
I assume I am wrong about that / misunderstood the source
Can someone clarify when the determinant has to be zero to make a function linearly dependant?
Does it need to be zero across all real numbers?
slim birch
I find the determinant of the Wronskian to be equal to 8x
wronskian implies linear independence only on the interval which w(cos2x, sin2x) != 0
ashen palm
wronskian implies linear independence only on the interval which w(cos2x, sin2x)...
so if I'm using the determinant of the wronskian for linear independence, I only care if the determinant is equal to zero when x != 0?
This is what my textbook states
does I not encompass 0?
slim birch
I find the determinant of the Wronskian to be equal to 8x
how did you do this?
ashen palm
this is the same problem being solved by my professor in the exact same way I did
excuse the hand writing, i struggle with it as much as you will
you end up with 8x(sin^2(2x)+cos^2(2x))+4 cos2x sin2x - 4cos2x sin2x
slim birch
ok cool
slim birch
does I not encompass 0?
in this case you would exclude 0 from your interval
linearly independent on {x in R | x != 0}
I believe that's a valid way to go about it
ashen palm
in this case you would exclude 0 from your interval
why would I do that
I understand that I can and that it would lead to that answer
I'm trying to understand why I'm excluding it
slim birch
because at x=0 you have x and sin2x being linearly dependent
note that sin2x and cos2x are always linearly independent for any interval however
ashen palm
because at x=0 you have x and sin2x being linearly dependent
is there a trivial way youre looking at x and sin2x and determining theyre dependant?
slim birch
wrt functions the only way i know of is to find the wronskian
luckily it was fairly painless for {x, sin2x} and by inspecting setting x=0 resulted in w=0
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broken sable
cedar kilnBOT
broken sable
precalc rotating functions into the (u,v) plane
i dont understand the step highlighted in red
warm crescent
they broke the parentheses and factorized per u^2, uv and v^2 terms
broken sable
they broke the parentheses and factorized per u^2, uv and v^2 terms
wait so youre saying distribute the 13 into the parantheses and the sixswrt3 and the 7
then combine all the like terms?
and then factor?
warm crescent
yes
broken sable
wait so its just a long grind?
oh ok wait i understand the step i initially didnt understand
warm crescent
yes but you can also do it almost instantly
broken sable
yes but you can also do it almost instantly
how?
broken sable
oh
warm crescent
broken sable
you just have to look at it a little differently right
ok yeah we chilling
thanks
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royal wasp
prove every number is either positive, negative, or zero
cedar kilnBOT
prove every number is either positive, negative, or zero
quartz knot
prove every number is either positive, negative, or zero
Use the Riemann hypothesis
royal wasp
can u elaborate?
dull oxide
Use the Riemann hypothesis
Do not troll in help channels
dull oxide
can u elaborate?
He's trolling. Ignore it. What class is this for? What information do you have available so far?
royal wasp
He's trolling. Ignore it. What class is this for? What information do you have a...
This might sound dumb but it isn't for a class.
I'm just wondering how I know every number is necessarily one of those 3
since I seem to assume it so much
dull oxide
I did this in set theory a long time ago, but I forgot the proof
royal wasp
does it require complicated proof?
isnt it as simple as saying something like:
every number is either 0 or not 0
if a number is not zero, it is either greater than zero or not greater than zero
former is positive, latter is negative
dull oxide
At the root, you need to prove that the reals are totally ordered
cedar kilnBOT
@royal wasp Has your question been resolved?
past frost
Please help me double differentiating the following equation. I know how to find the first derivative but I'm starting to get confused when I'm differentiating from that to find the second derivative. At the bottom of the second image is the first derivative.
Note: I was taught to use u, v, u', v' to differentiate
past frost
first derivative
cedar kilnBOT
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past frost
<@&286206848099549185>
cosmic steppe
languid zealot
Please help me double differentiating the following equation. I know how to find...
im in the progress of writing it out but my mouse is buggin rn
past frost
im in the progress of writing it out but my mouse is buggin rn
I don't mind if you do it on paper and take a photo as long as it's neat
languid zealot
I don't mind if you do it on paper and take a photo as long as it's neat
if you need to simplify further just make a common denominator
heres the product rule for the first term
also I did the work under the assumption that your first derivative is correct, i didnt actually take the time to check it
past frost
if you need to simplify further just make a common denominator
Thank you so much. You just saved my assignment
.close.
slate finch
Hello, can someone tell me how to do this, thankyou for any help
stiff vigil
Hello, can someone tell me how to do this, thankyou for any help
factorisation will definitely help for this question
unborn verge
can you try to factorize the numerator first
slate finch
ok thanks
inland ocean
$(a^2-b^2) = (a+b)(a-b)$
wraith daggerBOT
ColdTe²
worldly chasm
Not dumb, just missed one thing
slate finch
how could i miss that lol
worldly chasm
I miss obvious stuff all the time. 
slate finch
I miss obvious stuff all the time. <:cat_happycry:636407813144576010>
dont we all
anwyays thankyou
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cedar kilnBOT
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cerulean pine
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eager flame
binomial expression
cerulean pine
that doesnt explain it for me, sorry
eager flame
to find x's independent your teacher made k so you will get x from the expression
if you open it you will get many terms also consists x^2 or other
ı don't know how to teach it more obvius,sorry
cerulean pine
in the second line why did he do x^36-2k
eager flame
in the second line why did he do x^36-2k
independent term means it wont effected by any variable
so your teacher just set to variable 1
by making power of x 0
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distant snow
cedar kilnBOT
still holly
how do you even interpret this?
ebon thorn
YOLO
harsh zephyr
do you know what slope-intercept form is?
ebon thorn
It’s where the line touch the y or x ain’t it???
harsh zephyr
thats the axis intercepts
ebon thorn
Then where the slope touches the line
harsh zephyr
the slope intercept form is a form of the equation of a line
sage field
Then where the slope touches the line
it doesnt makes sense
harsh zephyr
y=mx+c
sage field
y = mx + c
harsh zephyr
so you just calculate the slope and then plug in a coordinate to find c
ebon thorn
What’s that
harsh zephyr
that works too
ebon thorn
What
sage field
y - y1 = (y1-y2)/(x1-x2) * (x -x1)
ebon thorn
Ohhh
harsh zephyr
shouldn't it be -6--6 on the top?
ebon thorn
Wait what
But the other 6 not negative
Oh wait
Thank you tho
Bro
So ain’t it 0/20
???
People???
harsh zephyr
yes
ebon thorn
Alr peace bud
harsh zephyr
same
ebon thorn
Peace to all
harsh zephyr
good luck with your work
ebon thorn
✌️☮️🪦🕊️
Thanks I need it
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Pls help me solve 8 questions under 20 min
upbeat nacelle
Can someone help me solve a permutation problem?
upbeat nacelle
ok
sage field
Can someone help me solve a permutation problem?
!help
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Can someone help me solve a permutation problem?
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harsh zephyr
whats the question
ebon thorn
I realize that I had to solve the problem
With slope intercept form
Like
Y=mx+b
But idk how to do that
harsh zephyr
what do you have so far
ebon thorn
Y
harsh zephyr
you have m don't you?
ebon thorn
No…
ebon thorn
Nah
I looked at the answer sheet to check
N it was wrong
Answer sheet says y=-6
So I’m tryna get that
harsh zephyr
oh okay
well you know y=0x+c
thats your current equation
so just substitute one of your points into that
and then solve for c
harsh zephyr
any one you like
harsh zephyr
no
ebon thorn
Damn…
harsh zephyr
one of the points on your line is (6,-6) correct?
harsh zephyr
therefor it must be a solution to your equation y=0x+c correct?
ebon thorn
The 6,-6
harsh zephyr
yes
ebon thorn
So do I just plug it in x??
wooden dagger
you both which country
ebon thorn
🇺🇸
wooden dagger
then why your name shubh
harsh zephyr
if you plus those numbers in
you get an equation with one variable, c
and then you can solve for it
ebon thorn
???
harsh zephyr
plus x=-6 and y=6 into your equation
cedar kilnBOT
lusty mantle
w
cedar kilnBOT
spark garden
Bro is not cooking 💀
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ionic finch
I want to do polynomial division without drawing the long division
ionic finch
ie essence out the idea in it and do it more algebraically
crimson delta
well there are other ways to write down polynomial division
like synthetic division
lapis maple
dusky coral
I want to do polynomial division without drawing the long division
uhh if the roots are rational, use rational root theorem
orchid perch
I want to do polynomial division without drawing the long division
i tend to do (factor)(polynomial of degree n-1) = (polynomial of degree n) and then expand out to create linear systems of equations
crimson delta
i tend to do (factor)(polynomial of degree n-1) = (polynomial of degree n) and t...
thats even worse than just doing normal polynomial division
orchid perch
hm
orchid perch
i tend to do (factor)(polynomial of degree n-1) = (polynomial of degree n) and t...
like (x-1)(ax^2+bx+c) = (whatever cubic youre dividing in this specific example case)
dusky coral
or maybe use vieta's to find a system of equations
orchid perch
mayb
dusky coral
youll get a system of three equations in three variables
dusky coral
oh its a quadratic, maybe try factorising it?
and see if any of the factors arfe common w f(x)
ionic finch
This is sort of what I did
where qi’s are the intermediate factors idk how to say
Does this shed idea of why long division of polynomial works?
essentially I’m doing this to make the standard way more sense
@crimson delta
crimson delta
I mean polynomial division is like long division for numbers
you are slowly subtracting big multiples of the thing you want to divide by
ionic finch
yeah, but long division of numbers kind of tweaks with place value system
ionic finch
right
ionic finch
here, how do i prove the existence of qi’s ?
crimson delta
well just compute them
ionic finch
I mean a proof
crimson delta
a way to compute them is a proof that they exist
polynomial division is an existence proof
ionic finch
with specific examples i can compute, but how to prove it exists
ionic finch
polynomial division is an existence proof
yeahh
crimson delta
well you have an algorithm. which always works
ionic finch
well you have an algorithm. which always works
How to prove correctness of this algorithm