idk if you dipped too
#help-39
pearl pondBOT
toxic lichen
idk if you dipped too
yes i did dip
pearl pondBOT
✅
toxic lichen
nope busy sorry
pearl pondBOT
pearl pondBOT
Available help channel!
lost zealot
Hi, I've never learnt trigonometry before and I tried to search youtube vids about it but I can't get a very clear understanding :') I know this much about it, but I'm not sure how to find seperate the hypotenuse of the smaller triangle and the larger triangle to find the height of the flag pole
compact ridge
Hi, I've never learnt trigonometry before and I tried to search youtube vids abo...
if you apply right-angled trig with 400 paces = 1000 ft and the 22 degree angle
you will get the height of the flagpole
lost zealot
if you apply right-angled trig with 400 paces = 1000 ft and the 22 degree angle
What's the formula for it?
compact ridge
What's the formula for it?
h is opposite the 22 degree angle
and 1000ft is adjacent to the 22 degree angle (it's not the hypotenuse = longest side!)
so SOHCAHTOA, you would use tan
tan(22) = h/1000
1000 tan(22) = h
lost zealot
1000 tan(22) = h
404ft?
compact ridge
,w 1000 tan(22 deg)
jolly parrotBOT
lost zealot
Oh nuts 😭
lost zealot
yeah that's the problem with this damn exercise
Thanks for the help 🙏
compact ridge
they shouldn't have the 22 degree angle there
then you'd just know to use similar triangles which gives you another answer, definitely less than the 330 ft
compact ridge
Thanks for the help 🙏
np!
lost zealot
then you'd just know to use similar triangles which gives you another answer, de...
Yeah it's confusing... the height of the flagpole is taller than the building
compact ridge
Yeah it's confusing... the height of the flagpole is taller than the building
sighs textbook makers are motivated by marketing and profit...
not by the actual maths or even common sense
lost zealot
😭
compact ridge
university textbooks or those aimed outside the traditional education system are a million times better
they still have errors but nowhere near this bad..
lost zealot
they still have errors but nowhere near this bad..
well at least i have a better idea of how to solve questions like these now 🫠
thanks again
.close
pearl pondBOT
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midnight haven
pearl pondBOT
midnight haven
hello so like
it's a transition matrix
though the trick is finding "last week's" shares
and we have the current situation's shares
and i don't know how to go back in time with matrices..?
like do i just use gauss-jordan
with an augmented matrix
of the current situation and P?
or rather
can I call my current X_0, X_1 instead
yeah idk
,w 2 * (0.25, 0.5, 0.5, 0.45) - (0.5, 0.5, 0.25, 0.375)
jolly parrotBOT
midnight haven
i mean..
i got the right answers after the rref but why..? <@&286206848099549185>
nvm i got it
lol
im stupid ig
.close
pearl pondBOT
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remote iris
Any idea how to change this function (formula and blue graph) to be more like red graph?
toxic lichen
for what purpose
remote iris
Oh jeez, that's a long story.
toxic lichen
like you could write down a great many equations that give you this sort of behavior with x-ints at 0, 0.6 and 1
remote iris
Sure.
Hmm, I suppose you're asking about the constraints I have for the function?
toxic lichen
the constraints, and/or the context that the function lives in from which one could derive constraints.
remote iris
The 0.6 isn't even really the important part. Though 0 and 1 are.
If I were to start at the beginning, then the problem was that I wanted to make a dataset that had values averaging about 0.6 to be closer to having values averaging 0.5, while also making the ends (values near 0 and 1) be more populated.
But I also can't push any values out of the 0 to 1 range.
So I guess I wanted to make the distribution of the values more uniform in that range.
toxic lichen
a dataset that had values averaging about 0.6 to be closer to having values averaging 0.5
i cant parse this
remote iris
This is the data I have.
toxic lichen
what am i looking at
remote iris
x has the value of the items
y has the quantity of items that have that value
So, you can see it's not only leaning a bit towards the right side, but that theres a lot more values near the middle.
After adding the initially listed function I get... (give me a sec to make the chart)
So, this distribution is a bit more uniform, but it still has more values at the right side.
pearl pondBOT
@remote iris Has your question been resolved?
pearl pondBOT
@remote iris Has your question been resolved?
remote iris
I've been looking at more functions, like
Found out another restriction, which is that the derivative of this adjusting function must stay in the range -1 to 1, because otherwise items in the dataset could switch places.
So that's as far as I can push this cubic function before it breaks.
But oddly enough, the maximum of the derivative doesn't want to shift much from 0.5.
I've almost found looking at the graph of the derivative more useful.
Like this is pretty nice. (But still centered around 0.5, which I don't like.)
Intuitively, it seems to me like the areas where the derivative is negative, are the areas where the data gets "squeezed closer together", and the areas where it's positive, are the areas the data gets "spread further apart".
So maybe something like the red graph would be better. But I'm still not sure how to get that.
remote iris
Oh hey, that looks good...
But then the integral of that...
meager leaf
So maybe something like the red graph would be better. But I'm still not sure ho...
um you could use a mixture of gaussians?
create three guassians centered at 0 0.6 and 1 and then add them up?
remote iris
Oh hey, that looks good...
But then the integral of that...
This one wouldn't work anyways. The values are not 0 at 0 and 1.
remote iris
um you could use a mixture of gaussians?
create three guassians centered at 0 0....
Not sure what you mean.
Don't Guassians never reach 0?
And if I added 3 of them, I'd get 3 lumps?
meager leaf
And if I added 3 of them, I'd get 3 lumps?
yeh
meager leaf
Not sure what you mean.
Don't Guassians never reach 0?
oh you also want them to be zero at certain points?
remote iris
Oh, uh... not for the derivative, I guess.
But I need to actually get the original function, not the derivative.
And I'm not sure if taking the antiderivative of a gaussian is... a nice thing to try.
meager leaf
can you tell me again what your original goal was?
cuz from what i understood you just want to create a dataset that takes high values near 0 0.6 and 1
i might have misunderstood that
remote iris
So, at a high level, I want to make the distribution of data in my dataset more uniform.
remote iris
This is the data I have.
I'd like for this to be closer to some sort of mostly flat bell curve centered around 0.5
remote iris
Any idea how to change this function (formula and blue graph) to be more like re...
And I think that a function that looks kiiind of like this could do it.
meager leaf
hmmm so is it like
you already have a bunch of data and you want to fit a function to it?
or do you want to manipulate your data itself?
remote iris
The latter.
meager leaf
ohh you wanna change your data
interesting
remote iris
So, if I just add or remove some value from each of the data points, I can get that data to have a different distribution.
But I don't want the data leaving the 0-1 range, and I don't want any data points switching places.
So that means that the function that adjusts it must be 0 at points 0 and 1.
And the derivative of that function must be between -1 and 1.
The derivative should also be negative at points 0 and 1 because that's where I have the least data points.
And ideally positive somewhere around 0.6, because that's where I have the most data points.
remote iris
.close
eager jewel
pearl pondBOT
eager jewel
my friend sent me this question
glacial sequoia
is it f(x) + f(y) or f(x + f(y))
eager jewel
is it f(x) + f(y) or f(x + f(y))
thats exactly what i was unsure about...i think its f(x+f(y)) cause if it was the other then f(x) would cancel
so can someone help
might be multi correct btw
rustic tendon
Take everything to one side
Since it need to be independent of y and x set every coefficient to 0
eager jewel
Since it need to be independent of y and x set every coefficient to 0
OHH thats smart
wait so was the key here noticing that x is in the equation which meant it was linear?
eager jewel
I guess you could sub f(x) = ax + b and try to find the coefficients
cause ive learnt this method but i struggle to find when to apply it
rustic tendon
then you can match the degrees of both sides
the right side is degree n
left side is degree n^2
n^2 = n ==> n = 0 or 1
fiery lance
yo
toxic fractal
!show
pearl pondBOT
Show your work, and if possible, explain where you are stuck.
toxic fractal
mm
thing is, those are not the equations you wanna use
(at least not yet)
you know that the mass A starts at an initial position of 2.5m and no velocity
and you know that $x=x_0+v_0t+at^2$
jolly parrotBOT
LordFelix
toxic fractal
where x is the position, x_0 the initial position, v_0 initial velocity, a acceleration, and t time
you're given x, x_0, v_0 and t
fiery lance
you're given x, x_0, v_0 and t
yeah sure
but why doesn't my method work aswell?
what's the difference
because there isn't tension yet?
toxic fractal
because you're missing two variables. You're missing m and a
you're also missing T, so that's 3 variables you dont know with a single equation
proven pawn
fiery lance
yo
fiery lance
why do we do
T - k = 2.5a
1.5g - T = 1.5a
wouldn't this mean that particle Q would be going down as the tension is less than 1.5g?
but it going down is impossible
as P is heavier
it would just hang there
pearl pondBOT
@fiery lance Has your question been resolved?
eager jewel
as P is heavier
That's not how it works
It is about how much acceleration is getting created
P is not contributing the force it's the friction
If we had a 1000kg block on smooth table it will still fall
sage walrus
sup
sage walrus
I've got 2 straight lines:
0.5y = -x + 8
3y = -x + 18
I need to find the intersection point. how can I do that?
ancient locust
The intersections points are when the equations are equal
blissful cloak
isolate y on both sides, these are the same y's, the equations in terms of x, equate them, solve it, then plug
ancient locust
One way would be to rearrange each equation into y= and then solve for x
sage walrus
what I did is:
-x + 8 = -x + 18
ancient locust
Yeah
ancient locust
Yeah
sage walrus
like, do I also change it according to the process while trying to even out the y
okay lemme try
sage walrus
Yeah
what I got is:
-2x + 16 = -3x + 54
is that correct?
no it isn't, because then x = 38, that doesn't make sense and it's also the wrong answer
@ancient locust help please
ancient locust
Hey
ancient locust
0.5y =-x + 8 therefore multiply by 6 and get 3y= -6x+48 then equating -6x+48 = -x +18
sage walrus
0.5y =-x + 8 therefore multiply by 6 and get 3y= -6x+48 then equating -6x+48 = -...
but why 6 ?
ancient locust
x=5
sage walrus
in the answers page, x is 6 and y is 4
ancient locust
0.5y =-x + 8 therefore multiply by 6 and get 3y= -6x+48 then equating -6x+48 = -...
Starting from here
You will get the correct answer
-5x=-30
Do you understand why we are making them the same y
sage walrus
so we get
-6x + 48 = -x + 18
finally got it right 😭
now we gotta solve for y
how do you do that again?
ancient locust
I’m not gonna say
sage walrus
I'll try
ancient locust
You got this
sage walrus
I got y=-4?
past flax
can someone help me
ancient locust
Wrong chat bro
past flax
which one
ancient locust
Go above and there is available chats
Hrmmm
There is no mistake in ur equation
I’d have another look at ur working
pearl pondBOT
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waxen oar
hey ! can anyone help me with proving that this function is integrable
jovial kiln
just show it is riemann integrable since it only has a single discontinuity at x=0
waxen oar
how would i find some partition that shows that tho ?
vital crescent
how would i find some partition that shows that tho ?
usually, continuous functions are nice, and one can prove that in the case where the function is continuous, it is Riemann integrable.
It turns out Riemann integrability carries over to finite jump discontinuities.
And you would realize that the only thing that is not "nice" about this function is happening at 0, where there is a jump. To deal with this, for a partition P, you notice that the "damage" that can be done by the "not so nice point" 0 is limited by the part of the partition that contains 0. And definition of Riemann integrability lets us control the size! THis means that I can make the not nice part, in both U(f,P), and L(f,P) to create an error less than epsilon. This combines to a total error of less than 2epsilon created by this jump discontinuity. I'm sure, from here you can figure out how to make the "nice parts" (the parts where the function is continuous), to have a total error of less than epsilon (which gives you a total error of 3epsilon)
waxen oar
usually, continuous functions are nice, and one can prove that in the case where...
ahhhh , i gotcha. so i need to take some partitions that are basically very close to 0 but not 0
which can be controlled by epsilon
vital crescent
ahhhh , i gotcha. so i need to take some partitions that are basically very clos...
yeah and suppose that 0 is contained in the bit of the partition [a,b] where we control the inteval to, say, length at most epsilon. Then the error created by U(f.P), and L(f,P), on that particular interval, is like at most $(b-a)\times 1-(b-a)\times 0=(b-a)<\epsilon$
jolly parrotBOT
qwertytrewq
vital crescent
you can alao give a very "rough" bound by $(b-a)\max(f)-(b-a)\min(f)$
jolly parrotBOT
qwertytrewq
vital crescent
since continuous functions with finitely many jump discontinuity are always bounded, max(f)-min(f) can just be treated as a constant (and we pick our epsilon around that constant)
This should give you intuition for how to prove that functions with $n$ jump discontinuities are Riemann integrable: if we use partition $P$ where each bit in the partition is bounded by $\epsilon$ length, the error that the not so nice bits can create is at most $\epsilon n (\max(f)-\min(f))$ (try and see why). And for the rest (the continuous bits), you can bound normally.
jolly parrotBOT
qwertytrewq
vital crescent
also by jump discontinuity I mean that the function should behave well in each jump: For example I don't consider
$$\begin{cases} \frac{1}{x} & 0<x\leq 1\ -1& -1\leq x\leq 0\end{cases}$$
to be a jump discontinuity (formally, the limit from the left and from the right at each discontinuity should both be finite)
jolly parrotBOT
qwertytrewq
pearl pondBOT
@waxen oar Has your question been resolved?
fresh remnant
pearl pondBOT
fresh remnant
so why is that the semicircle is assumed to be the top half and not the bottom half?
plush bramble
"underneath the x-axis"
fresh remnant
this difference effectively changes the area of the rectangle that is being subtracted upon the semicircle
fresh remnant
"underneath the x-axis"
but couldnt the semicircle be oriented both ways?
west sapphire
yea this is very poorly worded
fresh remnant
it would still be under the x-axis
west sapphire
both semicircles would qualify
fresh remnant
but the right answer assumed the top half of the semi circle
so the rectangle has an area of 72
sq units
but if it were oriented the other way it would be more
yielding an incorrect answer
oh you know what
its all multiple choice
and they all incorporate 72
so it has to be the top half
.close
steady wolf
did i do b and c wrong
pearl pondBOT
did i do b and c wrong
sharp vigil
the angle between vectors is defined as the angle they make when lined up tail-to-tail
which makes it supplementary to the angle you marked on the diagram
steady wolf
the angle between vectors is defined as the angle they make when lined up tail-t...
huh
im confused
can you draw it
sharp vigil
,tikz
{[thick, -Latex]
\draw (0,0) coordinate (O) -- node[above left]{$\vec a$} (4,2) coordinate(A);
\draw (A) -- node[above right]{$\vec b$} ++(2,-2) coordinate(B);
\draw (B) -- node[below]{$\vec c$} (O);
\draw (A) -- node[above left]{$\vec a$} ++(4,2) coordinate(A2);
}
\draw pic[draw, angle eccentricity=2, "$\theta$"]{angle=O--A--B}
pic[draw, angle eccentricity=2, "$\alpha$"]{angle=B--A--A2};
jolly parrotBOT
cloud
sharp vigil
the angle you were finding is theta but the angle you need to find is alpha
sharp vigil
we definite the angle between vectors as the angle between them if they share the same starting point
robust sentinel
,tikz
{[thick, -Latex]
\draw (0,0) coordinate (O) -- node[above left]{$\vec a$} ...
i wish to be a latex wizard like u someday
steady wolf
we definite the angle between vectors as the angle between them if they share th...
OH WAIT
"between" so like the angle between the 2 tips?
sharp vigil
yes
steady wolf
thank youuuuuuuuuuuuuuuuuu
steady wolf
i wish to be a latex wizard like u someday
fr very impressive
junior loom
thank youuuuuuuuuuuuuuuuuu
there is a way to define it for vectors regardless of position which may not help you here but is good to know
the first definition is what was said here
the second is not dependant on position
pearl pondBOT
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Available help channel!
outer hare
If $\alpha$ is the zero for sone $p(x) \in F[x]$ do we have an isomorphism from $F[\alpha] \cong F[x]/ \langle p(x) \rangle$
jolly parrotBOT
BOSS
pearl pondBOT
@outer hare Has your question been resolved?
safe prairie
theres 4 variables here, how am i supposed to see the bijection?
safe prairie
i always see bijection as being able to be represented by a monotonically increasing graph
wet osprey
What’s the example on
safe prairie
thats all
wet osprey
It’s sort of a known thing to go from x, y to u, v by doing u = x+y, v = x-y
You can rewrite this in matrix form and get a linear transformation
In fact the determinant of this matrix ends up being J
safe prairie
they require bijection for j to work or am i reading it wrong
lets say i have
u = xy^2
v = x+2y
how should i see bijection?
wet osprey
You wouldn’t pick that
That looks ass
And non linear
But you can just look at J for this transformation
And see if it’s non zero for all x, y
pearl pondBOT
@safe prairie Has your question been resolved?
safe prairie
But you can just look at J for this transformation
so write out j no matter what, i can just invalidate it after checking anyway?
pearl pondBOT
pearl pondBOT
Available help channel!
sharp smelt
Let $w,x,y,z$ be elements of a group $G$
\begin{enumerate}
\item solve for $y$ given $xyz^{-1}w=1$
\item suppose $xyz=1$. Does it follow that $yzx=1$ Does it follow that $yxz=1$
\end{enumerate}
Answers:
\begin{enumerate}
\item $yz^{-1}w=x^{-1} \implies y=x^{-1}w^{-1}z$
\item $xyz=1 \implies yz=x^{-1} \implies yzx=1
\xyz=1 \implies y=x^{-1}z^{-1} \implies yxz=1$
\end{enumerate}
jolly parrotBOT
What a wonderful world !
vital estuary
?
sharp smelt
As in multiplying from the left by x for instnace
vital estuary
oh ive always known it as left/right multiplying
sharp smelt
Yea, probably should add it
vital estuary
yeah, just to be clear
past jay
pearl pondBOT
pearl pondBOT
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Available help channel!
mortal spade
prove that if a quadratic has more than 2 zeros then a=b=c=0
mortal spade
my proof was that we can express any polynomial as a(x-m)(x-n)(x-o)...... since the zeros are more than 3
we have the degree as the number of 0s which is greater than 2
the coefficient of the degree term is a
since the given polynomial is quadratic
we have a=0
is this correct
acoustic path
if you’re allowed to use the fundamental theorem of algebra, then sure this is trivial
acoustic path
you can just say that a polynomial with more than 2 zeroes factors as [the way you wrote it]
ig your proof doesn’t fully use the fta, i’m sort of objecting to the way it is written (since “we can express any polynomial as a(x - m)(x - n)(x - 0)…” is a statement that suggests you’re using fta, but you don’t need it)
mortal spade
i meant that it is expressable in this form
mortal spade
thanks for the help
pearl pondBOT
Available help channel!
sharp smelt
Let $S$ be a subgroup of the additive group $Z^{+}$ .Either $S$ is the trivial subgroup ${0}$, or else it has the form $\Z a$ where $a$ is the smallest positive integer in $S$
jolly parrotBOT
What a wonderful world !
sharp smelt
I don't get the second part
tropic saddle
wdym not get it
pearl pondBOT
pearl pondBOT
Available help channel!
ancient sapphire
I’m not really too confident in my answer, can someone look over it for me?
pearl pondBOT
pearl pondBOT
Available help channel!
hushed dove
pearl pondBOT
hushed dove
for b if you do 25,000 + (n-1)1250= 44,000
you get n=16.2
so i thought it would be 17 complete years
but the markscheme is showing 16 years
but isn't n=16 still less than 44,000?
shadow spindle
so i thought it would be 17 complete years
Is 16.2 a complete year?
shadow spindle
Oh you’re studying for your IB test?
hushed dove
Oh you’re studying for your IB test?
hey yeah lol
hushed dove
Is 16.2 a complete year?
no
hushed dove
but if you do 25,000+(16-1)1250
it is less than 44,000
and the question is asking when it will be greater right?
shadow spindle
it is less than 44,000
Well it does not have to be exact since it’s asking for complete years not the exact amount of time being passed
hushed dove
but the question is asking for when it will be greater
shadow spindle
Oh I see what you confused about one sec
shadow spindle
but the question ***is*** asking for when it will be greater
I see now she earns it at the start of each year
shadow spindle
but the question ***is*** asking for when it will be greater
Not the end of each year
shadow spindle
what did I do wrong?
Looks like you did it right
hushed dove
i hardly doubt it since this is the ib markscheme
shadow spindle
i hardly doubt it since this is the ib markscheme
True the mark scheme is weird
shadow spindle
i hardly doubt it since this is the ib markscheme
Sl Analysis?
hushed dove
HL AI
hushed dove
lol
shadow spindle
lol
Yeah your mistake is that the 5 percent you get initially
Not after the first year
shadow spindle
it says it starts the subsequent year though?
Yeah it’s written weird
hushed dove
if its subsequent year then how can he get interest in the first year
the markscheme also does (n-1) and not just n so it seems they also want you to think it only gets deposited in the next year
shadow spindle
if its subsequent year then how can he get interest in the first year
Yeah it makes no sense but they do like to confuse you
shadow spindle
?
There might be something deeper but I don’t see it
Someone else proably will do better
I remeber when I took it most of it was trick questions
hushed dove
ait ill see if anyone can explain
hushed dove
I remeber when I took it most of it was trick questions
damn
shadow spindle
math pp3 is my last test
Yeah it’s pretty interesting that banks give you immediate interest
shadow spindle
then im free from this horrible program
I see they do give you a hint in the trick questions
Notice that they specifically said January 1st
And then they said at the start of each subsequently year
They specifically chose January 1st on purpose
They don’t usually add unnecessary info on the IB test
hushed dove
but how does this mean they get interest in the first year?
shadow spindle
but how does this mean they get interest in the first year?
The person who wrote it is bad at English
hushed dove
😭
shadow spindle
😭
Subsequent does mean after some time passes but they purposely included January 1st as the date she invested the money
hushed dove
i mean that is probs true
shadow spindle
i mean that is probs true
And it’s more of a math test then an English test
hushed dove
ill just see if anyone else can give another explanation
shadow spindle
ill just see if anyone else can give another explanation
I mean maybe but I don’t think there is
hushed dove
yeah
shadow spindle
yeah
It’s safe to assume for the questions for subjects that are before HL math there are trick questions
shadow spindle
yeah
Also remeber the HL test is an international test not strictly for English speakers
So it’s possible the person who wrote it is not an English speaker
shadow spindle
if thats okay
True but no one is showing up sadly
hushed dove
its fine ill just leave this channel open for now
shadow spindle
its fine ill just leave this channel open for now
But yeah like I said notice how they specifically stated January 1st as the day of investment
hushed dove
But yeah like I said notice how they specifically stated January 1st as the day ...
yeah im aware
im just looking for a second opinion if its okay
shadow spindle
yeah im aware
Also after 15 minutes you can @ the helpers
hushed dove
i understand your argument that it could be an english error
oh okay ty
<@&286206848099549185>
hushed dove
hi helpers this is the original message
pearl pondBOT
@hushed dove Has your question been resolved?
hushed dove
bruh
flint parrot
$(\int_{2}^{4} \sqrt{x^2-6x+9} ,dx)$
pearl pondBOT
$(\int_{2}^{4} \sqrt{x^2-6x+9} ,dx)$
jolly parrotBOT
Yousef
pure rapids
$(\int_{2}^{4} \sqrt{x^2-6x+9} ,dx)$
whats special about x^2-6x+9
flint parrot
how do i handle absolute values in integration?
like it becomes (x-3)^2 and then the absolute value of x-3
like what's next
equal it to 0 ?
pure rapids
oh wait
flint parrot
sorry?
wet osprey
how do i handle absolute values in integration?
Look at your bounds
Does the absolute value matter between 2 and 4
flint parrot
like between 4 and 3, 3 and 2, 2 and -3 ?
pure rapids
sry
pure rapids
Does the absolute value matter between 2 and 4
It kinda does tho
x-3 is positive
after 3 but negative before 3
wet osprey
Then split your integral into 2
pure rapids
Hmm
blissful cloak
$(\int_{2}^{4} \sqrt{x^2-6x+9} ,dx)$
(\int_{2}^{4}\sqrt{x^2-6x+9}dx=\int_{2}^{4}\sqrt{(x-3)^2}dx)
(=\int_{2}^{4}|x-3|dx=\int_{2}^{3}|x-3|dx+\int_{3}^{4}|x-3|dx)
jolly parrotBOT
PajamaMamaLlama
pearl pondBOT
@flint parrot Has your question been resolved?
pearl pondBOT
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Available help channel!
brisk horizon
pearl pondBOT
brisk horizon
Need a quick clarification on the Maclaurin Series.
where (0) is, that is what we plug into the x value of the derivative f^(n)?
So for example.
f(x) = e^x
f^(n) = e^x, and this does not change no matter how many derivatives we take of e^x
f^(n) * (0) = e^(0) = 1
toxic lichen
f^(n) * (0) is wildly wrong
that (0) isn't a magical number by which you multiply f^(n)
the (0) means you evaluate the function at x=0
no more and no less
brisk horizon
Sweet
toxic lichen
f(0) does not mean the function f multiplied by some mystery object called (0)
brisk horizon
Yepyep.
So in the case of f^(n)=e^x, x=0, and so we have e^0 = 1
toxic lichen
yes
hybrid gate
What I mean is just to solve it without any calculators
past perch
take liberty in drawing this line
hybrid gate
take liberty in drawing this line
is there more to it
past perch
properties of angles between parallel lines
hybrid gate
So do I just add them and then minus by 180
hybrid gate
So they're interior angles
past perch
So do I just add them and then minus by 180
think of it as x = y + z
do you have a way to find y and z separately
given what i've told you
hybrid gate
are y and z the numbers given
past perch
precisely
past perch
yup
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hybrid gate
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cedar scarab
Can someone explain Lie Algebra to me? I missed that day of class and apparently it's a heavy topic for our final.
cedar scarab
.close
pearl pondBOT
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stoic imp
pearl pondBOT
stoic imp
Being $$\mathcal{S} = { x \in \mathbb{R}^3 \mid x_1 + 2x_2 + x_3 = 0}$$ and $$\mathcal{B} = {(-1,2,2); (-1,2,1); (1,-1,0)}$$ basis of $\mathbb{R}^3$. $$\$$ Find all $\mathbf{v} \in \mathcal{S}$ that have same coordinates in basis $\mathcal{B}$ and in the canonical basis
jolly parrotBOT
938c2cc0dcc05f2b68c4287040cfcf71
karmic fern
Suppose we write v in standard coordinates; also suppose we let B^-1 to be the change-of-basis matrix that converts standard coordinates to coordinates in B.
Then, you're asked: for which vectors does B^-1 v = v, and I think you can figure out how to solve that
karmic fern
you may not have covered it by name, but the concepts are basic linear algebra
stoic imp
we shouldnt use that machinery mate
karmic fern
k
tropic saddle
it boils down to writing down the same linear system of equations
stoic imp
(x,y,z) = a(-1,2,2) + b(-1,2,1) + c(1,-1,0)
(x,y,z) = (a,b,c)
i) x = -a -b +c
ii) y = 2a + 2b -c
iii) z = 2a + b
iv) x = a
v) y = b
vi) z = c
,, \begin{cases} x = -x -y + z \ y = 2x + 2y - z \ z = 2x + y \end{cases} \ \begin{cases} 0 = -2x - y + z \ 0 = 2x + y -z \ 0 = 2x + y - z\end{cases} \ 0 = x + 2y + z
,w nullspace {{-2,-1,1},{2,1,-1},{2,1,-1},{1,2,1}}
stoic imp
all v € S that have same coordinates in B and same coordinates in standard basis are the scalar multiples of this vector (1,-1,1)
so the solution of this exercise can be given as a line that passes through the origin
L : X = k(1,-1,1) + (0,0,0)
.close
brisk pike
how am i supposed to do this
pearl pondBOT
how am i supposed to do this
brisk pike
(question coming please wait!)
Clearly A and C are wrong
but how do I decide between B and D
I know oxygen will have a high second IE because it has to break its stable half-filled valence orbital
but how do I know its the highest?
and I cant find a way that helps me see why D is wrong
brisk pike
where did a chemistry question come from in a mathematics server
because yes
sharp smelt
For D studying p-block elements may help
brisk pike
For D studying p-block elements may help
that chapter is in 12th grade no
@sharp smelt
<@&286206848099549185>
brisk pike
ty
past perch
for D how can the ionic radius of Li+ be more than Mg2+, Mg has a whole extra shell or am i stupid
am i reading this qs wrong
gimme a sec
brisk pike
for D how can the ionic radius of Li+ be more than Mg2+, Mg has a whole extra sh...
check data apparently thats true
past perch
is it something to do with the charge then hmm
past perch
past perch
Li+ rad is 76 and Mg2+ is 65 apparently
brisk pike
how does this make sense
past perch
pm
brisk pike
Li+ rad is 76 and Mg2+ is 65 apparently
showed 79 pm to me smh
brisk pike
na+ > li+ > mg2+ >al3+ > be 2+
like half of them are iso and the other half are iso with others
past perch
hmm yeah
brisk pike
iso meaning isoelectronic
brisk pike
and how tf am i supposed to derive that
common sense would say that the concentrated charge of mg2+ protons should overpower the shielding by s and p electrons
so zeff of mg 2+ > li+
but no way thats always true
this is like coming up with a logical argument for arguments sake
sharp smelt
this is like coming up with a logical argument for arguments sake
Welcome to inorganic chem
sharp smelt
there's a reason people really don't like chemistry
ooh, you're listening to oppenhimer
nice
pearl pondBOT
sharp smelt
there's a reason people really don't like chemistry
You're green now, you have access to a new channel :D
brisk pike
.reopen
pearl pondBOT
✅
brisk pike
@sharp smelt help me with chem again pelase 🙂
so
why would fluorines second electron affinity be ~0
wouldnt it be like crazy high?? because F- doesnt want electrons
to add another electorn u'd need alot of energy
pearl pondBOT
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hoary meadow
how do i compute the laurent series of 1/(z-w)
pearl pondBOT
how do i compute the laurent series of 1/(z-w)
@hoary meadow Has your question been resolved?
pearl pondBOT
flint wyvern
am i right with B here?
pearl pondBOT
am i right with B here?
flint wyvern
was gonna go with D but B makes more sense to me
spare lark
Your radius can be negative ?
misty adder
it can
blissful cloak
Your radius can be negative ?
yeah some textbooks allow for r<0
(-r,theta+pi)=(r,theta)
flint wyvern
doesnt it just mean the point is in the opposite direction from the angle
misty adder
i mean i just think of negative radius as it starting from that value
flint wyvern
okay got it
spare lark
doesnt it just mean the point is in the opposite direction from the angle
Why not just playing with the angle then 😭
blissful cloak
Why not just playing with the angle then 😭
¯_(ツ)_/¯
flint wyvern
so yeah ill go with B
blissful cloak
recall to convert for cartesian to polar: (r=\sqrt{x^2+y^2},\theta=\text{atan2}\left(y,x\right))
misty adder
ah yes that thing i was about to type
jolly parrotBOT
PajamaMamaLlama
pearl pondBOT
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outer hare
pearl pondBOT
outer hare
Would it just be Q?
rough stream
And Q(√3) but I believe those are the only two
plucky python
Any more formal explanation
well i mean consider a subfield Q(sqrt(3)) >= L >= Q
outer hare
Ok
plucky python
use tower law to think about the possible degrees of [L:Q]
plucky python
if F is a field of characteristic 0
then it must contain a copy of Z inside
the smallest field which contains Z is Q
i mean that isn't too hard to show
plucky python
so that means every char 0 field F must contain a copy of Q
outer hare
use tower law to think about the possible degrees of [L:Q]
Towers law lets you stack fields right
outer hare
so that means every char 0 field F must contain a copy of Q
Ok
plucky python
Towers law lets you stack fields right
i mean not entirely sure what you mean by stack fields but yeah it lets you calculate degrees of field extensions
outer hare
I thought towers law is
plucky python
so that means every char 0 field F must contain a copy of Q
(btw Q is called the prime subfield here)
outer hare
[E:F]=[E:K][K:F]
plucky python
[E:F]=[E:K][K:F]
yh
outer hare
So yeah for every field containing Q Q is the smallest field
So if there is another one it would also contain Q
plucky python
So if there is another one it would also contain Q
yh
outer hare
use tower law to think about the possible degrees of [L:Q]
Ok so
For degrees
It’s always just natural numbers
We know the degree of Q(sqrtn) = F is 2 right
For all n
Right
So the degree has to be less than 2
The only other possible degree is 1 but that happens when
Two fields are the same
Right
@plucky python
Ok I understand what ur trying to say
2 is prime thus it has no other factors
outer hare
[E:F]=[E:K][K:F]
So if there was another sub field it would have to meet this criteria
But the only other factor is 1 which gives us back Q
outer hare
Ok wait
@plucky python does that logic work for like
for example here
Because 2 is prime there are no other subfields, so can i assert that L = K(\sqrtd)
plucky python
Because 2 is prime there are no other subfields, so can i assert that L = K(\sqr...
nah doesn't work like that
as in yeah you know that there are no subfields other than K or L
but why does that mean L = K(sqrt(d))
(quick sanity check: this same argument works for [L:K] = 3, but it's definitely not true that L = K(cuberoot(d)) for some d in K)
(also another sanity check/mild spoiler: ||this actually isn't true if K has characteristic 2||)
outer hare
nah doesn't work like that
fuck
outer hare
(also another sanity check/mild spoiler: ||this actually isn't true if K has cha...
we didnt really havily do characterisitcs ngl
hmm
wait
what does it mean for [L:K]=2
does that not meal L is a simple extention of K with some \sqrtn not in K?
also the wording of the question makes me feel like i can just let L = K[\sqrt d] and prove rest of the claims from there
We know that $[L:K] = 2$. Firstly notice that 2 is prime, thus the only subgroups of $L$ are $K$ and itself.
Letting $L = K[\sqrt d]$ we know that $[L:K] = 2$ as the minimal polynomial is $(x^2-d)$
jolly parrotBOT
BOSS
outer hare
We know that $[L:K] = 2$. Firstly notice that 2 is prime, thus the only subgroups of $L$ are $K$ and itself.
Letting $L = K[\sqrt d]$ we know that $[L:K] = 2$ as the minimal polynomial is $(x^2-d)$
Now assume $d = c^2$ for some $c \in k$. then $L = K[\sqrt d] = K[c]$ and then as C is already in K we simply get $[K(c):K] = 1$ which is a contradiction.
jolly parrotBOT
BOSS
outer hare
@plucky python ok i think i made it work
some mistakes but
idk why this would not make sense
sorry had it type it out lmao
plucky python
your going about it the wrong way
why "letting L = K(sqrt(d))"
the point of the question is to prove that starting from the fact that [L:K] = 2
a field is not K shoved with something whose nth power is in K
a field is any arbitrary set equipped with some operations that make it nice
in fact, if i take K to be some disgusting field like R(X, Y), it doesn't scream out to me why it's obviously true that L = K(sqrt(d))
outer hare
@plucky python is every prime field iso to Q?
plucky python
does that not meal L is a simple extention of K with some \sqrtn not in K?
that's not how it works
L could be any field
outer hare
your going about it the wrong way
my bad im trying to figure out
ok
gmm
hmm*
how would i approach this then
plucky python
it turns out there's a theorem that says that you can always take L to be slightly nice
but a priori we don't know that yet
outer hare
it turns out there's a theorem that says that you can always take L to be slight...
any way to expand on this
plucky python
how would i approach this then
so L is a dimension 2 K vector space right?
so if i take any alpha in L but not in K, L = K(alpha)
a priori alpha need not square to something in K
but maybe we can modify our generating alpha to be something nicer
plucky python
any way to expand on this
primitive element theorem
outer hare
Ok wait for the notation
plucky python
<@573875411365265478> is every prime field iso to Q?
so the prime subfield of a field is just the field it has to contain
outer hare
does $[L:K] = 2$ mean ll elements in L are of degree 2
jolly parrotBOT
BOSS
outer hare
over K
plucky python
if K is characteristic 0 then it's Q, if it's characteristic p then it's F_p
but anyway if u don't really do characteristics then whatever
outer hare
ill note it down
plucky python
does $[L:K] = 2$ mean ll elements in L are of degree 2
if you've shown a statement like that in class/in a previous example sheet, fine
but if not, why does an element of L but not in K have degree 2?
anyway it's getting late for me and i'm kinda tired now, i think my hints are starting to get worse lol
you can wait here in case some other helper picks up from where i left u off
if not generally the advanced mathematics section is more active for uni problems, try posting to #groups-rings-fields
midnight haven
pearl pondBOT
midnight haven
im not getting what is this trying to say
p => Q and Q => P is P iff Q or P <=> Q ?
sharp vigil
P iff Q, alternatively denoted as P <=> Q, is a notational shorthand for (P => Q) AND (Q => P)
midnight haven
i get that the implication goes both side so both has to be false or true for it to be true
but what they meant by p => Q and Q => P is P
is P?
midnight haven
OH
OHHHHHHHHHH
they meant it is as P iff Q
oh shit
oh yeah lol
and that can also be represented as
P <==> Q
alright i see that makes sense
thank you cloud
.close
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midnight haven
P => Q , are the statements P and Q need to be propositions
pearl pondBOT
P => Q , are the statements P and Q need to be propositions
midnight haven
or can they be predicates too?
sharp vigil
they can be either
sharp vigil
yes, that is a true statement
sharp vigil
yes, both of them are predicates
midnight haven
Cool!
does statements consists of just
predicates and propositions
or there is more?
sharp vigil
those are the basic ones that you will encounter in this sort of class
midnight haven
oh i see so they do go much more in higher
classes
alright thank you once again!
.close
pearl pondBOT
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cedar scarab
This is one of the review questions on my final. I don't even know where to begin with this.
cedar scarab
The red interval is the answer, but I don't know how they got that at all.
None of the homework problems ever had us solve where we had an x bar, a population, AND a standard deviation. Is one of those a red herring?
pearl pondBOT
@cedar scarab Has your question been resolved?
junior loom
None of the homework problems ever had us solve where we had an x bar, a populat...
when you know the population variance, which you usually don't,
you can improve on the student t distrbution and use the z distirbution instead
cedar scarab
The greek thingy is population std?
junior loom
population variance
junior loom
yes
cedar scarab
you can improve on the student t distrbution and use the z distirbution instead
How would this work then? I have a list of formulas for the exam and I don't see any that use the variance
junior loom
you take the square root of varince to get standard deviation
do you have a formula for the Z-distribution confidence interval on your sheet?
cedar scarab
I do not have any confidence interval formulas on here. Dang that sucks.
junior loom
well I mean
do you have formulas for difference of means?
can you upload the sheet, is it digital?
cedar scarab
It's digital, but it doesn't have that one as far as I can see.
cedar scarab
cedar scarab
Heard.
junior loom
so you see it uses the variance actually
not the standard deviation
only, well, there's a wrinkle here
the true difference, what on the sheet is mu1-mu2
is unknown
cedar scarab
$z=\frac{(x_1-x_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}$
jolly parrotBOT
MJames
cedar scarab
Are we talking about this one?
junior loom
yeah
cedar scarab
I'm blind af how did I miss that
junior loom
except here
you dont know u1-u2
so you're trying to estimate it
dont solve for z, solve for (u1-u2) instead
cedar scarab
When you say u, do you mean the mu thing?
junior loom
yes Im not wearing my glasses and thought it was a u
cedar scarab
No worries, just wanted to make sure we're on the same page.
junior loom
call $\Delta \mu = \mu_1 - \mu_2$ to make bookkeeping easier
jolly parrotBOT
gfauxpas
cedar scarab
So it would be $(x_1-x_2)-(\mu_1-\mu_2)=z\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$
jolly parrotBOT
MJames
junior loom
and solve for Delta mu on the LHS on its own (or the RHS of course doesnt matter)
almost
cedar scarab
So $\Delta\mu=(x_1-x_2)\pm z\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$?
jolly parrotBOT
MJames
junior loom
yes!
cedar scarab
Ok so plugging stuff in would be $\delta\mu = (22.3-24.1)\pm z\sqrt{\frac{5.04}{32}+\frac{10.15}{43}}$?
jolly parrotBOT
MJames
cedar scarab
z comes from the 98, so I gotta find that number
cedar scarab
Would it be 2.33?
junior loom
so you're really looking for P(|Z|<z) = 0.98
but that's the same as P(Z<z)=0.99 , right?
cedar scarab
Yeah, that's what I looked for
junior loom
,w InverseCDF[NormalDistribution[0,1], 0.99]
jolly parrotBOT
junior loom
there you go
cedar scarab
How the heck do I get a number that precise without a table?
cedar scarab
I don't get that calculator on the test though
junior loom
do you get ANY calculator on the test
cedar scarab
Yeah, a graphing calc
cedar scarab
TI-Nspire
junior loom
googing it, i dont know that one
cedar scarab
It's the standard for the high school district here so the college got them too for the education majors.
junior loom
cedar scarab
I don't think she wants us to do that, since she never mentioned it. I'm assuming she wants us to go off table.
I think I do recall her actually saying go off table for test, but Google it for homework.
junior loom
well you either need a table or a calculator
it's not like a function y = x^2 that you can just solve by hand
cedar scarab
We get all the tables.
cedar scarab
I got 2.33, since it corresponded to 0.9901
This entire exam is just "Do you know which formula to use" so I'm stressing a lil
junior loom
yeah all you can get from this is that it's in between 2.32 and 2.33, closer to 2.33
cedar scarab
Last exam I skirted by with an 87 by just trying to match the variables I had values for with equations, but it didn't work for all of them.
cedar scarab
To be fair, I was stressing super hard about my Modern Algebra final that I took today, then I got a 92.
junior loom
what country are you in
cedar scarab
USA
junior loom
is this a standardized test?
cedar scarab
It's a university exam
junior loom
oh.
some tests allow students to choose a table OR a calculator
cedar scarab
So no. The teacher gives everyone different versions.
junior loom
but you have to use waht you're allowed to use
cedar scarab
I get both.
junior loom
okay
anything else?
good luck
maybe write the z* for 0.98, 0.99, 0.9, 0.95 circled already
since those are the usual confidence choices
cedar scarab
Yeah I just wanna run a couple of these by you so I think I'm in the correct direction
junior loom
I actually gotta go, but, lots of comkpetent people on this serveer
cedar scarab
junior loom
cedar scarab
Nw. Thanks
junior loom
okay one more look
cedar scarab
I'm seeing a "population mean" a sample mean, and a sample std.
junior loom
so which test
cedar scarab
It says assume it is normally distributed, so I would use a z?
So $z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$?
jolly parrotBOT
MJames
junior loom
assume it's normally distributed gets around the problem of "am I justified in using a normal approximation? do I have the hypotheses of the central limit theorem"
it's just saying, don't worry about that
cedar scarab
Normal distribution = z score, no?
junior loom
normal distribution allows you to use z score only if you know the population variance
junior loom
if you dont have the population variance but you're assuming normal, you use the the t-distribution which is meant for that
jolly parrotBOT
MJames
junior loom
that's to compare two means
cedar scarab
Oh right