thx !
#help-13
stray crypt
old dirge
no problem
stray crypt
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cedar kilnBOT
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stray crypt
@old dirge i kinda have a problem since both p0 and p3 are equal to -4
stray crypt
there can be two global minimum ? @old dirge
stray crypt
well the global minimum is -4
oh, so can i say p(3)=p(0) =-4 >= p(x) for every x in the interval [0,4]
?
dense wraith
I have a question I am in 9th
silk gust
old dirge
oh, so can i say p(3)=p(0) =-4 >= p(x) for every x in the interval [0,4]
yeah ig
dusk mason
Can someone tell me all the things I need to relearn from alg and geo and what I need to MOST importantly learn for alg 2, I was the dumbest kid in all my math classes and I never really understood anything and Im rlly anxious cause Im a freshmen and people r just gonna see me as a dumb kid
fossil dawn
https://math.libretexts.org/Bookshelves
you can try using this as a guide to structure your relearning, imo
cedar kilnBOT
@dusk mason Has your question been resolved?
marsh flare
,tex Find the area of a loop of the curve $r = \alpha \sin 3\theta$
wraith daggerBOT
cat
marsh flare
im not very familiar with polar coordinates so i can't exactly start solving this. can someone help :(
proven summit
im not very familiar with polar coordinates so i can't exactly start solving thi...
sub 0 to r to get the upper and lower limits of the loop
karmic field
im not very familiar with polar coordinates so i can't exactly start solving thi...
Do you know the Identity for area with polar coordinates?
Can either prove it from x y coordinates or just take it from your course and use it
proven summit
this formula $\frac{1}{2}\int_{a}^{b} r^2 ,d\theta$
karmic field
Or prove it as a reimann sum
wraith daggerBOT
឵឵MxRgD
karmic field
Like think of it as instead of dividing into rectangles we divide into sectors
proven summit
sub 0 to r to get the upper and lower limits of the loop
after that you use those upper and lower limits for a and b in the formula shown
@marsh flare, are you here?
karmic field
**឵឵MxRgD**
1/2 r^2 dθ is formula of sector
And this is basically the sum of all of them from the starting till the last angle θ we are calculating over
cedar kilnBOT
@marsh flare Has your question been resolved?
marsh flare
And this is basically the sum of all of them from the starting till the last ang...
ahh i get it
cedar kilnBOT
ahh i get it
marsh flare
.close thank you @karmic field and @proven summit
cedar kilnBOT
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dusky furnace
Linear algebra question.
cedar kilnBOT
Linear algebra question.
dusky furnace
I've come to the following thought, and I want to confirm it:
If we have a vector space R^n, then any basis of that vector space, say B={e1,...,eN}, can be made into a change-of-basis matrix by using those basis vectors e1,...,eN as the columns of a matrix M. That matrix M converts vectors expressed in basis B into vectors expressed in the standard basis. Conversely, matrix M^-1 converts from vectors expressed in the standard basis, into ones expressed in basis B.
The crux of the question is that any basis of R^n can be used in this manner.
Is this correct?
void sand
that sounds right to me
paper edge
I've come to the following thought, and I want to confirm it:
If we have a vecto...
Yeah, that's the standard basis change procedure
Also note the given matrix will always be invertible, and thus its associated linear application will be bijective
Which is why basis changes "leave the space as is"
Note, however, that your procedure will transform vectors in the basis B to vectors in the canonical basis
The opposite procedure will require you to compute the inverse of M
dusky furnace
Cool. Thank you, both. This basis business is slowly starting to make sense.
🙂
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cedar kilnBOT
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livid vigil
cedar kilnBOT
livid vigil
I am confused what they mean by D with the arrow on top and how let get that equation for D,
I thought they mean a vector to the point D, but then I am not sure why the equation is true,
dapper raven
wherever the origin is
from which you calculate the vecotrs
vectors
if you add those fractions of B and C up together
you will get D
<@&268886789983436800>
livid vigil
Wouldn’t it be vector C minus vector B?
livid vigil
Wouldn’t it be vector C minus vector B?
Oh wait no,
dapper raven
in geogebra
i would give you the file but i don't know how to share it so it works
if you wanna interact with it you can write these out
livid vigil
(Pardon my childish language I don’t know the formal names for theses things)
If I walk along B, and I want to get point D, don’t I have to walk in the direction of BC? How can start at B, walk in the direction of C and get to D ?
dapper raven
that's one way of getting to D
imagine it this way
you want to walk to D
so you walk 9/13th of the way to B
and then from there you walk 4/13th of the way to C
flat mica
(Pardon my childish language I don’t know the formal names for theses things)
...
mathematicians talk this way
you're fine
dapper raven
in the sense that you take the walk from the origin
and the walks to B or C don't change after you start walking
dapper raven
add B' and C' together
first walk along B'
then walk along C'
you will get to D
livid vigil
Ahhhhh, this makes sense,
But how do you find the coefficients?
Why 9/13 and 4/13?
dapper raven
DB and DC are in a proportion of 4:9
livid vigil
Yes?
dapper raven
and so if you want the smallest whole number lengths of them
you will make DB 4
4cm
4 inch
4 unit
and DC 9
and together the length of BC will be 13
livid vigil
Okay?
livid vigil
How does the lengths of BC, BD, and DC tell me about length of vector B?
They are in different directions, the lengths should not be related?
dapper raven
they don't tell you anything about the length of the vector B
just about the relation between the vectors B C and D
in my drawing you can move the origin O around
so the length of B changes
but the relation between B and D doesn't
does that make sense
livid vigil
Kind of,
So I basically want to find the length of the red and blue vector,
But they are not in the say direction as BC, so I don’t understand why the length of BC is related to the lengths of the red and blue?
dapper raven
you can find the coefficients by finding D by going along B to C and vice versa
the red and blue are multiples of B and C
and so they are related to BC
as that is B - C or C - B depending on which direction you care
with the right combination of the red and blue ones
you can find a multiple of B - C
in this case it's 9/13 (B - C)
or 4/13 (B - C)
i can never remember which direction
do remember that you have B + k (B - C)
dapper raven
yes
cedar kilnBOT
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thorn fox
cedar kilnBOT
pastel vault
oh, if f(x) and g(x) have a root in common, alpha
then you know that f(alpha) = g(alpha), and the x^2 term drops out
but yeah the easier way would be to spot (alpha + beta) + (beta + gamma)
thorn fox
I got value of alpha beta gama in terms of lambda
Do I put value of one of the roots in eqn to get value of lembda?
Nvm
Solved it
Tq
.close
merry delta
I need help solving for the limit of this function. The exercise offers the solution but what i'm looking for help with is deriving that solution. This exercise comes from a section of a textbook covering the concept of "The Limit Laws" and I've provided a screenshot of the section's outcomes for context. I also included my work and the thinking behind the steps I took:
use direct substitution to obtain an undefined expression of the original function
$$
f'(x)=\lim_{ x \to -2^- } \frac{2x^2+7x-4}{x^2+x-2}
$$
$$
f'(x)=\lim_{ x \to -2^- } \frac{2(-2)^2+7(-2)-4}{(-2)^2+(-2)-2}
$$
$$
f'(x)=\lim_{ x \to -2^- } \frac{8-14-4}{4-4}
$$
$$
f'(x)=\lim_{ x \to -2^- } \frac{-10}{0}
$$
Then, simplify the function to help determine the limit.
factoring the numerator
$$
2x^2+7x-4
$$
$$
2x^2-x+8x-4
$$
$$
(2x^2-x)+(8x-4)
$$
$$
x(2x-1)+(2x-1)
$$
$$
(2x-1)(x-1)
$$
factoring the denominator
$$
x^2+x-2
$$
$$
x^2+2x-x-2
$$
$$
(x^2+2x)+(-x-2)
$$
$$
x(x+2)-(x+2)
$$
$$
(x+2)(x-1)
$$
simplifying the function
$$
\frac{(2x-1)(x-1)}{(x+2)(x-1)}
$$
$$
\frac{2x-1}{x+2}
$$
still an undefined expression
$$
f'(x)=\lim_{ x \to -2^- } \frac{2x-1}{x+2}
$$
$$
f'(x)=\lim_{ x \to -2^- } \frac{2(-2)-1}{(-2)+2}
$$
$$
f'(x)=\lim_{ x \to -2^- } \frac{5}{0}
$$
wraith daggerBOT
Tubberware
dire geode
I need help solving for the limit of this function. The exercise offers the solu...
do you know what $\lim_{ x\to -2^-}$ means?
wraith daggerBOT
riemann
dire geode
also don't use $f'(x)$ unless you actually have the derivative of a function. $f(x)$ is a function of $x$ and you have a limit that's either a number or limit that does not exist
wraith daggerBOT
riemann
merry delta
as x approaches -2 from the "bottom" right?
dire geode
'left' is a better word than 'bottom'
because the x axis is oriented like
...., -4, -3, -2, -1, 0, 1, 2, 3, 4, ....
merry delta
oh ok so my notation is alluding to the wrong thing, since we aren't looking for the derivative--we are looking for the limit of that function i shouldn't use $f'(x)$ but should use $f(x)$
wraith daggerBOT
Tubberware
sand cradle
There is no f defined anywhere
merry delta
so i'm trying to find what this function approaches as x approaches -2 from the left. and direct substitution doesn't work, even after simplifying. did i miss the point of simplifying in the way i did it? or did i do it wrong? or should i be doing something completely different that still satisfies the suggestion in the problem instructions "simplify the function to determine the limit"
sand cradle
so i'm trying to find what this function approaches as x approaches -2 from the ...
The idea when simplifying for limits is that you can cancel out an (x + 2) and it won't be undefined anymore after direct substitution
If you don't have an x + 2 factor in your factorization, you likely made a mistake somewhere
For example beginning from your fourth line factoring the numerator
merry delta
so this step contains a mistake?
$$
x(2x-1)+(2x-1)
$$
$$
(2x-1)(x-1)
$$
wraith daggerBOT
Tubberware
sand cradle
so this step contains a mistake?
$$
x(2x-1)+(2x-1)
$$
$$
(2x-1)(x-1)
$$
That too yes but also the step before it
merry delta
oooooh theres supposed to be an 8 pulled out
sand cradle
4 pulled out
merry delta
ohhh right right
i'm gonna go back and rework that real fast
or real careful
factoring the numerator (hopefully corrected)
$$
2x^2+7x-4
$$
$$
2x^2-x+8x-4
$$
$$
(2x^2-x)+(8x-4)
$$
$$
x(2x-1)+4(2x-1)
$$
$$
(2x-1)(x+4)
$$
wraith daggerBOT
Tubberware
sand cradle
Yep
merry delta
with that, this function looks like this now and I don't see anything i can do to eliminate the x+2 term:
simplifying the function
$$
\frac{(2x-1)(x+4)}{(x+2)(x-1)}
$$
sand cradle
Wait a sec let me check the calculations again
wraith daggerBOT
Tubberware
wraith daggerBOT
wraith daggerBOT
sand cradle
Yeah looks good
merry delta
ok cool so now i have a correct "simplified function" i suppose? the solution provided by my book says the limit is $-\inf$
sand cradle
(I just checked because a x + 2 term actually does not factor out)
sand cradle
ok cool so now i have a correct "simplified function" i suppose? the solution pr...
Yep, now you need to see what this approaches if you approach -2 from the left, which is easier in this factored form
$\dst \frac{(2x-1)(x+4)}{(x+2)(x-1)}$
sand cradle
**Kepe**
Surely as x goes to -2, the denominator will go to 0 and the numerator to some constant
You just want to know if this thing will be negative or positive (as you know it will go to either +infinity or -infinity (as it's of the form {-> (nonzero) Const.}/{-> 0}))
So now study each of the factors and determine their sign when very close to -2 from the left
merry delta
i'm just mulling over that one sec
flat mica
You just want to know if this thing will be negative or positive (as you know it...
for a formal exposition
merry delta
so we simplified it and determined it can't be evaluated with direct substitution, so my next thought would be that we now have something much easier to work with at least and we know that the denominator is going to be approaching 0 and the numerator will be approaching a constant --that for as close as I am getting to -2 from the left the faster this function is approaching infinity? would plugging in -2.001 be the right idea?
sand cradle
so we simplified it and determined it can't be evaluated with direct substitutio...
Yes, but you really just want to know the signs each of these factors will be having
If for example all of them are positive, you can conclude it's +infinity
So you don't actually need to plug anything in
Just in thought
For example, 2x - 1 will be somewhere around -5 when coming close to -2 from the left, so negative
What about x + 4?
merry delta
positive 2
merry delta
0
sand cradle
0
We care about the sign here though, we don't actually plug in -2 but come very close to it from the left
so will it be very small positively or very small negatively?
e.g. we go x = -2.1, x = -2.01, x = -2.001, ...
merry delta
I just want to take a step back real quick to see if I am following how to think of this. I get to a factored function, see i can't direct substitute and then i can tell this is approaching infinity of some kind because its undefined, and want to know whether its approaching positive or negative infinity so i do that by looking at the signage of the factors as we approach -2 from the left. plugging in -2 gives us a -10 in the numerator. and is that thorough enough to conclude its approaching negative infinity? or would i need to plugin some of the numbers on the left? e.g. we go x = -2.1, x = -2.01, x = -2.001.
oh i need to plug in from the left in the denominator to be sure about what we are approaching?
the numerator alone isn't enough to conclude what we're approaching
sand cradle
I just want to take a step back real quick to see if I am following how to think...
One should think in terms of 'approaching' for all terms by the way when considering limits. The numerator approaching -10 is not enough to conclude this is -infinity. Say the denominator approached 0 from the left too, then the two negative signs would cancel (for every x near -2) and it would give us +infinity
merry delta
so i could plug something to the left in the denominator and if thats positive, the we are approaching overall a negative infinity
sand cradle
yes
merry delta
if its negative then the numerator and the denominator both being negative means its overal a positive number and thus we're moving toward a postivie infinity?
sand cradle
Yes
merry delta
oooooooooooo
i seeeeeeeeeeeeee
thank you so much for working through that with me!!
sand cradle
np
merry delta
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cedar kilnBOT
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junior idol
Hello, can someone help me with this problem? I currently have that doing a move twice on any S_i turns it back into the original form (so, the set can be original or inverted, only two options) and that the order doesn't matter. However, I'm not sure what to do with this information. Thank you!
junior idol
There are 2^10 possible configurations for transforming the S_i's, but the TA told me that this is an overcount. However, I'm unable to figure out how much of an overcount it is.
junior idol
yes, but excluding (i, i)
violet flume
There are 2^10 possible configurations for transforming the S_i's, but the TA to...
yea this is my thought, but you'd need to invent a procedure to invert just some given j,k
or, show it's not possible
junior idol
yea this is my thought, but you'd need to invent a procedure to invert just some...
can you elaborate more on this
violet flume
if there are 2^10 configurations then any of them should be reachable
within 2^10 is any individual square as some color, and everything else the other
unless im miscounting too
so there must be some procedure to invert a single square, and nothing else
wait i am totally misunderstanding
silk gust
@junior idol isnt this the amsp combi hw from a couple of days ago 
?; either way i gave up on this problem 😭
junior idol
yea
junior idol
<@795432721869438996> isnt this the amsp combi hw from a couple of days ago <:KE...
how far did you get?
my solution from that day was SUPER bogus
silk gust
not that far, it was very incoherent which is why i didnt submit it
junior idol
oh
silk gust
i wanted to prove that the operation was commutative somehow
junior idol
it is
silk gust
but idk how to
junior idol
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cedar kilnBOT
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cinder umbra
i think the answer is 28 is it?
cedar kilnBOT
i think the answer is 28 is it?
silk gust
could you show your work? i got a different answer but my arithmetic is terrible
cinder umbra
ty
silk gust
were gonna just ignore that :)
cedar kilnBOT
silk gust
btw welcome to the mathcord!
surreal cave
btw welcome to the mathcord!
congrats on the blue name!! 
silk gust
tyty :)
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true sapphire
help
true sapphire
3rd part
stuck on it for months (1 day,jk)
only thing i got was, max time was 2.5 so total 2.5 and so, it was aove ground for 1 second so .5 below it will give time for h, so 2.5-.5 is 2 so 2 seconds
ik im wrong
buoyant latch
What have u tried
true sapphire
only thing i got was, max time was 2.5 so total 2.5 and so, it was aove ground f...
this
idk how people can do problems so fast
its like they have a secret
true sapphire
ok
true sapphire
im in pain
true sapphire
how do you know
what is the question
whaty should i focus on
it just said above h
buoyant latch
My particle is above some height for 1 second
buoyant latch
So if I draw a horizontal line and only look above it
I should see my particle for exactly 1 second
true sapphire
what do i do? teacher said to form suvat system but i cant
buoyant latch
Ya but you need to understand what you’re solving for first
true sapphire
are you saying i should make a graph
buoyant latch
A diagram is fine
buoyant latch
This is a diagram
true sapphire
yea
buoyant latch
If I asked, how long was the particle above 0m
That’s the same as finding how long it was in the air for
true sapphire
hm
buoyant latch
If I ask how long was the particle above 1m I’m just moving the floor to 1m
true sapphire
because math
buoyant latch
If I ask how long was the particle above 3m I’m moving the floor to 3m
buoyant latch
These are this diagram
If it’s any higher than 3m then I don’t start counting until it first goes above this height
true sapphire
so us aying for all those questions, use that diagram
buoyant latch
That’s this diagram
Not use that diagram
But draw a diagram for it
Reading the problem then turning it into a diagram is the big skill to learn here
true sapphire
hm so higher, than h,, so must be MAX point or near it and x could be uh 1?
buoyant latch
What?
true sapphire
so question said HIGHER so we can use the
buoyant latch
Do you agree that if I set h = 0 then the time is 2.5
The particle is above 0m for 2.5s
buoyant latch
How did you find 2.5s
buoyant latch
only thing i got was, max time was 2.5 so total 2.5 and so, it was aove ground f...
Here
buoyant latch
Yes but how did you do it
true sapphire
v is 0 coz max height (final velocity) and u is 25, and it is going against gravity so -9.8
0=25+-9.8(t) -> -25/-9.8
=t
uh
im done
true sapphire
yea
buoyant latch
can you show that as well
true sapphire
so dispalcement is 20 meters, so , s is 20, u is 25, v is v, a is -9.8, t is? i used s = s_0 + ut + \frac{1}{2} a t^2., and got t_1\approx0.99\text{ s},\quad
t_2\approx4.11\text{ s}, and noticed that two x values so subtarcted later from first
to get time
buoyant latch
where does it say displacement is 20m
buoyant latch
there should be 2 t values right
true sapphire
yes
true sapphire
yes
true sapphire
yes
true sapphire
why h-3
buoyant latch
well when we wanted to find above 23m we did 23-3 for 20
because the particle starts at 3m
true sapphire
yea
buoyant latch
we also have the additional constraint that the 2 roots of t here
need to be 1 unit apart
buoyant latch
this means the first time it goes above hm and the 2nd time when it's coming down
true sapphire
ig
buoyant latch
the 2 solutions for t will be the 2 timestamps those things happen
we need it to happen 1s apart
so try use that info on the quadratic
true sapphire
so u saying u just replaced displacement thing with h
buoyant latch
Yeah I want to find h as a function of t
true sapphire
huh
buoyant latch
Do you know about vieta’s formula
true sapphire
nope
buoyant latch
It’s otherwise known as the sums and products of roots of a quadratic
buoyant latch
It’s very useful in this case!
true sapphire
yay!
buoyant latch
The formula says that if I have a quadratic ax² + bx + c
true sapphire
yea
buoyant latch
Then the roots multiply to c/a and sum to -b/a
I think
Yes that’s right I just googled it
true sapphire
okay
buoyant latch
Now, we know that the 2 roots of t are 1 second apart
So clearly t₁ + t₂ = 2t₁ + 1
buoyant latch
h-3 = 25 t - 1/2 * 9.8 * t^2
If we look here, we see that our quadratic looks like -4.9t₂ + 25t + 3 - h
true sapphire
yea
buoyant latch
So we know that 2t₁ + 1 (the sum of the roots) = -b/a = 25/4.9
true sapphire
ye
buoyant latch
This immediately tells us what t₁ is
true sapphire
thank you
true sapphire
ill see this tomorrow, its late here
buoyant latch
Alright
cedar kilnBOT
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cinder umbra
can someone explain how to do this
cedar kilnBOT
can someone explain how to do this
cinder umbra
i get you can cut it in half and you can cut it again but i want to know how to get the answer
silk gust
do you understand how $\frac{3}{20}\cdot\frac{6}{6}$ numerically is still equivalent to $\frac{3}{20}$?
wraith daggerBOT
Mathemusician
cinder umbra
no
silk gust
ok uh
cinder umbra
im sorry
silk gust
what is $\frac{6}{6}$?
fossil dawn
have you learnt about LCM?
wraith daggerBOT
Mathemusician
fossil dawn
least/lowest common multiple?
fossil dawn
ok that's step 1
fossil dawn
isn't that one?
the realization that x/x for any number x = 1 is step 2
so then, to simplify something like 18/120
eh did i say LCM earlier? i mean GCD
sorry
cinder umbra
your fine
fossil dawn
aight good!
your job is then just to find the GCD of the top and bottom
so the GCD of 18 and 120 is?
silk gust
that's a common divisor, but not th greatest
fossil dawn
if you say it's 3, then dividing top and bottom by 3 gives you 6 and 40 respectively
there's still a common factor
cinder umbra
is there a khan video or something i can view to try to learn from it
fossil dawn
but if i may give you a hint, there's a fast way involving prime factorization
assuming you've learnt about that
fossil dawn
if you know what prime factorization is and can apply it properly, prime factorize both top and bottom
then just cancel out any number appearing in both sides
surreal cave
(\frac{18}{120}=\frac{2\cdot 3\cdot 3}{2\cdot3\cdot4\cdot5})
fossil dawn
i'll give you an example to get started
wraith daggerBOT
PajamaMamaLlama
cinder umbra
so we just take the common and use that to both sides
surreal cave
yep we take the common and remove them from top and bottom that's exactly right :)
surreal cave
do you think you could repeat that exericse for this fraction?
cinder umbra
yea
surreal cave
(oh 4 isn't prime 😅 that's my bad but exercise still reamins the same, prime factor it then cancel)
cinder umbra
if you see the thing can go in 3 time and 1/3 if you remove everything
but im guessing i don't really know but it seems easy to catch that one
surreal cave
if you see the thing can go in 3 time and 1/3 if you remove everything
could you elaborate a little more on what exactly you mean here? 
cinder umbra
and im really sorry
fossil dawn
if you can't explain in words, you can draw it out
surreal cave
and im really sorry
btw no need to be sorry when learning, backwards thinking thought
surreal cave
cinder umbra
but beside that i really have no idea what going on
can you give me one that isn't easy to catch that pattern?
silk gust
just by the way it looks... i feel like the thing can keep going smaller like 3/...
1/3 is the "smallest" though (as in, most reduced)
surreal cave
can you give me one that isn't easy to catch that pattern?
91/143
buoyant latch
i think that 79 is prime
cinder umbra
bruh
buoyant latch
fossil dawn
79 is prime. the only way you can cancel here is if the other number is a multiple of 79
surreal cave
sorry very late rn
buoyant latch
91 famously looks prime but isn't
cinder umbra
ok so far i got it's a odd number because 1 part
fossil dawn
did you accidentally hit enter too early
cedar kilnBOT
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Available help channel!
crimson sedge
convoluted question: if I have a Matrix A which is similar to a rotation matrix, then could I find A^n easily, like a diagonal matrix, or no?
upper abyss
If you can write that similarity out, then yes!
As (PRP')ⁿ = PRⁿP'
Where I'm using ' for inverse
crimson sedge
If you can write that similarity out, then yes!
does this property hold for all similar matrices... or only diagonalisable ones? im kinda confused
livid vigil
cedar kilnBOT
opal hinge
How do they get H = A + B + C ?
ever heard of euler line?
opal hinge
It's a line that pass through circumcenter, orthocenter and centroid
livid vigil
Ahh,
livid vigil
I see I see,
livid vigil
How do they get H = A + B + C ?
I still don’t see I how proves this?
(But I am also very tired lol)
opal hinge
OA+OB+OC=3OG+GA+GB+GC ( G is centroid
livid vigil
Where does the 3OG come from?
livid vigil
OOOHHHHH
livid vigil
OA+OB+OC=3OG+GA+GB+GC ( G is centroid
Okay this makes sense,
And 3OG + GA + GB + GC = H?
opal hinge
Why?
Just do some manipulation and you will see, it's a basic problem from vector chapter
Igtg
sorry
livid vigil
Okay,
I think I am very tired I will go to bed now and think more tomorrow lol,
.close
cedar kilnBOT
cedar kilnBOT
Available help channel!
shrewd blaze
I dont get it
cedar kilnBOT
I dont get it
slender ginkgo
!xy
cedar kilnBOT
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
slender ginkgo
Find vertex?
shrewd blaze
Domain and range
shrewd blaze
I understand it but what i dotn understand is how it became like -2{1/4}^2
Yup i know
Oh waitt
Ohhhh
I think i know what i missed
shrewd blaze
I dont get it
Um i also dont know how to find like the range and domain
Like input it
shrewd blaze
Then what to do?
proven summit
do you know how to find the maximum of the quadratic?
shrewd blaze
I dont get it
And also this
shrewd blaze
I dont get it
Is this correct?
proven summit
wait what nvm I was right, you do input 1/4 to find the maximum
proven summit
Y= -2x^2 + x - 2
I misread the quadratic as x^2 + x - 2
shrewd blaze
I also did
shrewd blaze
I dont get it
Is this correct?
proven summit
Yeah the working out is correct I believe
To find the range though, what do you know about the maximum point?
shrewd blaze
But then how do you put this on like
Wait ill send something
How you put it on like that form
The infinite or something
proven summit
it shouldn't be an equal sign btw, it should be ≤
shrewd blaze
Ohhh
proven summit
for the set notation when you wrote the range
shrewd blaze
How do you put it like that form
proven summit
How do you put it like that form
for set notation?
proven summit
I think you're talking about interval notation instead
shrewd blaze
Yesss
proven summit
the interval notation is correct
shrewd blaze
shrewd blaze
This thing
proven summit
it's just the equal sign that's wrong for the range for set notation
shrewd blaze
They said something about parabola up and down
proven summit
not sure what you mean
shrewd blaze
Like if down it would be like ( -infinite,N ). But if open it would be like ( N , infinite)
shrewd blaze
Like if down it would be like ( -infinite,_N_ ). But if open it would be like (...
Im not sure if thats the right way
How to know if like Less than or Greater than sign
proven summit
Like if down it would be like ( -infinite,_N_ ). But if open it would be like (...
for (-infinite,N) it means you can use all values up to N while (N, infinite) means you can use all values above N
proven summit
How to know if like Less than or Greater than sign
depends on the quadratic
shrewd blaze
depends on the quadratic
Howww
proven summit
it's in the form of ax^2 + bx + c
if the a is negative you are finding the maximum
if the a is posistive you are finding the minimum
shrewd blaze
A is the constant?
shrewd blaze
But if possitive its greater than
proven summit
But if possitive its greater than
it's greater than or equal, if a is posistive
shrewd blaze
Ohhhh
proven summit
it's also important that you include the "or equal too"
shrewd blaze
Is it like the same thing?
proven summit
Is it like the same thing?
wdym?
shrewd blaze
Like does it matter if its negative or positive
shrewd blaze
notice how different they are
This is what i mean by up and down parabola
proven summit
Like does it matter if its negative or positive
it matters yeah
proven summit
This is what i mean by up and down parabola
yeah just look at the 2 and -2
shrewd blaze
Does it change that?
proven summit
yes it changes that as well
nova snow
yeah just look at the 2 and -2
uncle riggidus ballus
proven summit
uncle riggidus ballus
shrewd blaze
yes it changes that as well
What will change?
proven summit
What will change?
it becomes [-17/8, ∞)
if it was a 2 instead from the quadratic we are working with
nova snow
if it was a 2 instead from the quadratic we are working with
infima ballus
proven summit
esentially yeah
proven summit
infima ballus
proven summit
esentially yeah
actually no that's wrong
shrewd blaze
it becomes [-17/8, ∞)
It added -2?
proven summit
there shouldn't be a minus sign on the infinity
for the up parabola
think of it as you approaching bigger values
shrewd blaze
Oh it is possitive
proven summit
it doesn't make sense for it to be negative
shrewd blaze
Oh yeaaaahhhh
proven summit
It added -2?
our a changed to 2
shrewd blaze
Does it apply for both range and domain?
proven summit
yes
well
generally the domain is for all real values of x
in a quadratic
it's the range that alters
shrewd blaze
So the domain is always like (-infinitely , infinity)
It does not change ever?
We will only put something on range?
Like value
proven summit
So the domain is always like (-infinitely , infinity)
[n, ∞) or (-∞, n] depending on what your minimum or maximum value of n is for the quadratic
also make sure to do ]
the ] adds equal as well
shrewd blaze
[n, ∞) or (-∞, n] depending on what your minimum or maximum value of n is for th...
I thought this is only used for range
You will only put it on range
proven summit
I thought this is only used for range
yeah only for the range
shrewd blaze
So domain depends on range?
shrewd blaze
I mean like
proven summit
you know what a domain is right?
domain is your input that you put into the function
shrewd blaze
If range is like (-Infinty, N]
Then domain is (-infinity, infinity)
proven summit
If range is like (-Infinty, N]
Then domain is (-infinity, infinity)
yes that is correct
if you wanted to do in interval notation
for the domain
good job 
proven summit
So it means Domain and range depends on the parabola
only the range depends on the quadratic
the domain for the quadratic is consistently going to be (-∞,∞)
shrewd blaze
What if the Parabola is like Up
Range would be [ N , infinity)
But domain is like (-infinity, infinity) still?
proven summit
What if the Parabola is like Up
Range would be [ N , infinity)
But domain is l...
yes but it would be (-∞,∞) just make sure you write [] and () correctly though
shrewd blaze
What if the Parabola is like Up
Range would be [ N , infinity)
But domain is l...
Theree
I would only use [] that if there is a value
Like number and not infinity
proven summit
you only use [] if it's a number that's also equal to it
hope everything is understandable to you now?
shrewd blaze
It’s just that earlier I didn’t listen to class so idk what to answer lol
proven summit
It’s just that earlier I didn’t listen to class so idk what to answer lol
Try to pay attention to class or watch videos on youtube
it can be confusing if you miss the info
shrewd blaze
Try to pay attention to class or watch videos on youtube
It is only once
Actually earlier i was trying to solve something that teacher send
Thats why i didnt listen
I tried to solve it on my own but wrong
proven summit
Ic
shrewd blaze
proven summit
yeah you showed this before
shrewd blaze
I tried to reverse engineer it
proven summit
you made a help channel in it
proven summit
Right
shrewd blaze
Thats why no use of phone
shrewd blaze
I thought i will just need to find the slope and the y intercept
Since i can just reverse engineer it
proven summit
not sure what you mean by reverse engineer it
shrewd blaze
I mean like
shrewd blaze
The table is not given only 3x - 1
proven summit
The table is not given only 3x - 1
you dind't really answer the question
shrewd blaze
I made the table thinking that if the y intercept is -1 then, x is 0 then y is negative 1
proven summit
you dind't really answer the question
you were asked to find it's domain and range
shrewd blaze
you dind't really answer the question
I thought the range would be, -7, -4 -1 2 5
proven summit
I thought the range would be, -7, -4 -1 2 5
No the range would be for all real values of y
shrewd blaze
And the domain is -2 -1 0 1 2
I thought that is the way
Since its kinda right in some sense
proven summit
you have to remember that for domain and range, we are looking for all the possible values
that can go in or out of the function
shrewd blaze
Thats how we do it last time like if the table is given
proven summit
Thats how we do it last time like if the table is given
I think that was for when they given you values of x and you had to workout the outputs of y
shrewd blaze
I dont think so
proven summit
hmm, the only thing I could see it is if they had a restricted domain
anyways you should watch a video to get more clarification
shrewd blaze
The table is given
So i kinda just reverse engineer it
But i guess i was wrong
Anyway i understand it completely thank to you thank youuu
proven summit
shrewd blaze
I want to perfect all math exam this upcoming exam
Anyway thanks
Thank you very much mxrgd if that is your name
proven summit
don't forget to close the help channel btw
proven summit
Thank you very much mxrgd if that is your name
it's not my name 
proven summit
Idk AHHAAHHA
anyways type in .close to close this channel
cedar kilnBOT
cedar kilnBOT
Available help channel!
lunar kettle
Hey so I need to do this without any tools
I've done integration by substitution
u = x^3+28 => du = 3x^2dx
int(1/u du) => F(u) = ln|u|
Now I need to determine the integral from -2 to -3 and give an answer with at least two decimal accuracy.
ln(-2) - ln(-3)
Now I don't know if there's a step I did wrong or a trick to knowing what natural log numbers are, but this seems impossible to do in the head for someone at the supposed start-of-uni level I'm at. Can anyone help me figure this out?
rugged tundra
Hey so I need to do this without any tools
I've done integration by substitution...
the limits change when you do a u-sub
lunar kettle
Oh? In what manner?
tranquil oracle
rugged tundra
at the end you may also use log rules to simplify the calculation
buoyant latch
You don’t even need u sub for this really
lunar kettle
You don’t even need u sub for this really
I'm advised to, as per the assignment, it's a practice in integrals
buoyant latch
Recognise that the integrand is in the form $\frac{f’(x)}{f(x)}$ and you immediately know the primitive to be $\ln|f(x)|$
wraith daggerBOT
frosst
supple grove
just plug u into $\ln(u)$
wraith daggerBOT
DaveyLovesSocks
supple grove
so that way you dont have to adjust any bounds
buoyant latch
Yeah but you don’t need to do the substitution with the bounds and stuff
lunar kettle
It's significantly more confusing when I'm being told to do several things at once, guys
I want a fundamental understanding of the task at hand, integration via substitution, I'm self taught and it's the first time I'm hearing substitution changes the limits, so that is what I'll be focusing on, but thank you nonetheless!
proven summit
You can watch a video that explains why we change them
rugged tundra
essentially all you have to do to change the limits is figure out what u is when x is that value
buoyant latch
I want a fundamental understanding of the task at hand, integration via substitu...
Ah I see
rugged tundra
you have a direct relationship between them
lunar kettle
I will go do that, but then it makes sense that I was so confused! I was missing an entire step. Thank you so much :)
buoyant latch
When you do integration it actually says $\int_{x=-3}^{x=-2}\frac{3x^2}{x^3 +28},dx$
wraith daggerBOT
frosst
buoyant latch
So when you do the substitution you need to change the bounds, because implicitly it’s saying to integrate from when x is something to x is something
lunar kettle
Yeah it makes a lot of sense! Honestly I'm just knee deep in a lot of studies so I'm a bit slower and faster to seek help, but thank you all for this, I'm very satisfied
What's the command to end help again?
buoyant latch
.close
cedar kilnBOT
lunar kettle
Perfect! thank you lmao
cedar kilnBOT
Available help channel!
dusky furnace
In some notes about linear algebra, I'm having trouble with a sentence fragment:
dusky furnace
I don't understand that logic jump. I don't understand M being 0 on the right side.
Why does a matrix M having nonzero eigenvalues means that there exist no vector that satisfy that equation? And, where does that equation come from?
royal finch
M isn't 0 it says that there is no non-zero vector v such that Mv equals the zero vector. It also doesn't say M has no eigenvalues. It says M doesn't have 0 as an eigenvalue.
dusky furnace
I agree about this part; I made a mistake in the language: "It also doesn't say M has no eigenvalues. It says M doesn't have 0 as an eigenvalue."
This would make sense to me Mv = 0. But, why the intermediate part Mv = 0v?
royal finch
That is the definition of an eigenvalue
dusky furnace
You're refering to the following, then?
Meaning that the v in that chunk is an eigenvector
and the 0 represents an would-be 0 eigenvalue?
cedar kilnBOT
@dusky furnace Has your question been resolved?
cedar kilnBOT
@dusky furnace Has your question been resolved?
dusky furnace
.close
odd elbow
I need help with my inverted pendulum math for my control systems. I need to double check my MATLAB math for converting my systems non linear equations to linear using the jacobian method for my Linear Quadratic Regulator controller.
dire geode
odd elbow
try <#326138793650421760>
I mean i could but its still math. the matlab just solves it for me i want to make sure my setup is right and that my equations are accurately simulating my system
dire geode
yea sure
odd elbow
this is the reference im using here "https://github.com/junggeehoon/Furuta-Pendulum/tree/main" and im using his matlab code to derive the matrices
mine is almost an identical setup, its a rotary inverted pendulum so im trying to see where maybe some discrpencies could be
dire geode
yea all the more reason to go to a specific channel for computing
odd elbow
alright
cedar kilnBOT
Available help channel!
cinder umbra
do i always need to add a 0 for money?
cedar kilnBOT
do i always need to add a 0 for money?
surreal cave
if the website you're using requires that then yes
cinder umbra
surreal cave
but mathematically $53.50 and $53.5 are equivalent
cinder umbra
so i wasn't wrong then
storm zealot
In solving decimals, you fill up the empty spaces at the end with 0's
cinder umbra
wait how do you know im doing decimals?
cinder umbra
but mathematically $53.50 and $53.5 are equivalent
also hi again
cedar kilnBOT
cedar kilnBOT
Available help channel!
cobalt holly
is this correct?
cedar kilnBOT
is this correct?
fossil dawn
show steps please
wicked mantle
is this correct?
Yesss
cedar kilnBOT
Available help channel!
ember heath
cedar kilnBOT
ember heath
ive gotten
tan(pi x ) = -pi x
how to tell what values satisfy
do i plot the graph?
cedar kilnBOT
@ember heath Has your question been resolved?
lavish vine
I guess the periodicity becomes the natural numbers so only 1 solution in each interval
ember heath
plot tan(t) = t for simplicity
but like from my graph im not able to tell properly if it intersects in some places or not
lavish vine
It has to intersect the graph at least once in each period
mighty shuttle
Note that $f(0)= f(1)$. Apply MVT
wraith daggerBOT
wai
mighty shuttle
Similarly f(n)=f(n+1)= ? ( this will give you your answer), or will atleast help you reduce the number of options
ember heath
Note that $f(0)= f(1)$. Apply MVT
i can apply rolles theorem right
mighty shuttle
i can apply rolles theorem right
sure
ember heath
so there will be atleast one f'(x) = 0 in n,n+1
mighty shuttle
and then find the period of the function which is?( well of the periodic part)
mighty shuttle
so there will be atleast one f'(x) = 0 in n,n+1
yes
but also note the period of sin(πx)
ember heath
and then find the period of the function which is?( well of the periodic part)
2?
pi x = 2pi?
ember heath
oh why not
because of that the values wont repeat right
and there wont be more than 1 value of x
ruling out D
ember heath
oh ok
dusk finch
MVT uses f to derive result about f' if that's what you are using
ember heath
MVT uses f to derive result about f' if that's what you are using
oh ok yeah
hi btw
long time
ember heath
so you want to examine the derivative
so how can i tell no. of pts where f'(x) = 0
mighty shuttle
I'm thinking
dusk finch
sketch is prolly easiest
mighty shuttle
You want to note the intervals on which it's increasing and decreasing I would think
ember heath
sketch is prolly easiest
sketch of what? tan (pi x) = -pi x?
dusk finch
xsin(pix)
ember heath
xsin(pix)
im not really good with these kinds of sketches.. where there are products involved
mighty shuttle
Alternatively note that as x>0 whether the function is increasing or decreasing depnds solely on sin(πx)
ember heath
shouldnt be for x>1?
mighty shuttle
Why
mighty shuttle
so what do you notice
ember heath
it is an even function?
dusk finch
Alternatively note that as x>0 whether the function is increasing or decreasing ...
By this, you mean that for x > 0, xsin(pix) is increasing iff sin(pix) is increasing?
mighty shuttle
By this, you mean that for x > 0, xsin(pix) is increasing iff sin(pix) is increa...
yes
dusk finch
that's not true unfortunately
mighty shuttle
it works for x>1, I think
dusk finch
that'd imply that the extrema of xsin(pix) are at the same x-coords as extrema of sin(pix)
which can easily be verified to be false
mighty shuttle
fair enough
ember heath
if we really wanted to avoid sketching
yes please 😔 or u can help me sketch
dusk finch
you could take the derivative explicitly and then apply sth like IVT instead of using rolles / MVT
ember heath
explicitly meaning?
edgy spade
aint this a pyq
dusk finch
product rule on xsin(pix)
edgy spade
ember heath
aint this a pyq
😔 dont judge me bro
ember heath
product rule on xsin(pix)
yeah thats what i did lol
tan pi x = - pi x
edgy spade
yeah thats what i did lol
tan pi x = - pi x
yeah
dusk finch
yeah thats what i did lol
tan pi x = - pi x
okay, cool
dusk finch
if you can graph that, it helps
dusk finch
and crucially, note that tanx grows faster than x, so make sure to draw it like that in your graph
ember heath
dusk finch
in reality, the tan should be stretched out vertically a bit more
pix is actually a tangent to it
but it doesnt matter that much
ember heath
im not able to draw the rest of the graph accurately
because of the pi x term
oh wait i do notice things tho
dusk finch
it doesnt need to be accurate
ember heath
oh yeah im getting B,C as correct
ember heath
it doesnt need to be accurate
yeah just understood that
well thank u
thanks @edgy spade
.close
cedar kilnBOT
ember heath
WE ARE PEASANTS HERE
edgy spade
WE ARE PEASANTS HERE
mighty shuttle
WE ARE PEASANTS HERE
???
mighty shuttle
np
cedar kilnBOT
Available help channel!
opal hinge
Find all integers $m \in [-20;20]$ satisfy $$m(1+mx^2)^2-x+1 \geq 0 \forall x \in \mathbb{R}.$$
wraith daggerBOT
Alexis_Fx
cedar kilnBOT
@opal hinge Has your question been resolved?
acoustic mica
**Alexis\_Fx**
for all integer m in set [1;20]
observ that negative m does not satisify because if we plug in x=1 we get m(1+m)^2>=0 which means m has to be positive
trivaly m=0 dose not satisfy
now we prove that for all integer m in set [1;20] satisfy that equation. first notice that the equation is true for all negative x because pos*pos+pos>=0 now we look at case where x>=1 observe that m(1+mx^2)^2>=(1+mx^2)>=x^2>=x-1 at last we look at case where 0<=x<1 , m(1+mx^2)^2-x+1>=m(1+mx^2)^2>=m>=1>=0
qtd
opal hinge
Cool exactly what I'm looking for, I did m(1+mx^2)^2-x+1> m(1+mx^2)-x+1 but that doesn't make any different i guess
Alright thanks:D
.close
cedar kilnBOT
cedar kilnBOT
Available help channel!
gloomy pewter
Does anyone know an iterative method for presenting tetration results with any index?
For explample:
F_tetration(a; b) = a^^b
lim (c → +∞) Iterative_approximate_F_tetration(a; b; c) = a^^b
gloomy pewter
Iterative_approximate_F_tetration — recurrent function
cedar kilnBOT
@gloomy pewter Has your question been resolved?
gloomy pewter
<@&286206848099549185>
cedar kilnBOT
@gloomy pewter Has your question been resolved?
gloomy pewter
For example:
F_tetration(2; 3) = 2^2^2 = 16
Iterative_approximate_F_tetration(2; 3; x) = a sequence of numbers where the larger the number X, the closer the result is to the tetration value, namely, it is approaching 16
tropic oxide
gloomy pewter
Now I will ask the question more precisely
Does anyone know an iterative method for presenting exponentiation results with any index of interativity?
For explample:
F_exponentiatio(a; b, c) = a^a^a^...("c" times)...^b
lim (d → +∞) Iterative_approximate_F_exponentiation(a; b; c, d) = a^a^a^...("c" times)...^b
surreal cave
[\left(\underbrace{a^{a^{a^{\dots}}}}_{c;\mathrm{times}}\right)^b]
is this what you're saying?
wraith daggerBOT
PajamaMamaLlama
flat mica
**PajamaMamaLlama**
@gloomy pewter this is what you mean?
pl.ease define "with any index of iterativity" it's not clear what you mean by that
abstract breach
I think OP wants a sequence that converges to a^^b
gloomy pewter
is this what you're saying? <:eeveethink:601029006418116608>
The number B is at the very top of the power tower
gloomy pewter
pl.ease define "with any index of iterativity" it's not clear what you mean by t...
The iterativity index is how many times a function is used in recursion
surreal cave
yes that's what I did, I just placed the parentheses there for ambiguity's sake
gloomy pewter
yes that's what I did, I just placed the parentheses there for ambiguity's sake
But this is incorrect from the point of view of notation, since it will look like a tower of C copies of number A, which is raised to the power B, and this is incorrect, since B is at the very top of the power tower
surreal cave
then I can simply remove the parentheses ({\underbrace{a^{a^{a^{\dots}}}}_{c;\mathrm{times}}}^b)
wraith daggerBOT
PajamaMamaLlama
gloomy pewter
I think OP wants a sequence that converges to a^^b
I need to understand what is the real and complex iterativity of exponentiation
Ordinary tetration is the application of an exponent with a certain base A with an iterativity that is equal to the number B. The exponent itself takes one as an argument if it is ordinary tetration. But if it is a nested power tower, then any other number can be used as an argument, and the height of the tower can also be any
a^^b = a^a^a^...("b" times)...^a^1
F(a;b;c) = exp_a^b(c)
flint cape
[he's found the LaTeX we might acc get somewhere with this]
But.. you know that tetration, like addition and multiplication, is a binary operator, right?
gloomy pewter
$$F(a; b; c) = {\underbrace{a^{a^{a^{\dots^{b}}}}}_{c\mathrm{times}}}$$
flint cape
I'm not sure what you're doing that necessarily requires it have 3 inputs...
gloomy pewter
$$F(a; b; c) = {\underbrace{a^{a^{a^{\dots^{b}}}}}_{\mathrm{c}}}$$
flint cape
$$F(a; b; c) = {\underbrace{a^{a^{a^{\dots^{b}}}}}_{c,\mathrm{times}}}$$
flint cape
**Waes (Wires)**
This?
gloomy pewter
Yes, but the curly bracket should only touch instances of the letter A and the dots
flint cape
Yes, but the curly bracket should only touch instances of the letter A and the d...
So (a^^c)^b?
oh wait that's not quite right
Why are you making such a function anyways?
gloomy pewter
$$F(a; b; c) = {\underbrace{a^{a^{a^{\dots}}}}_{\mathrm{c}}}^{^{^{b}}}$$
wraith daggerBOT
Pentalogue
