2 mins max
#help-39
toxic lichen
wheat finch
AHHH
wait
can you teach me how to do just the first fraction
because i dont understand how to do it if theres multiple fraction in 1 fraction
next ore
pearl pondBOT
next ore
i have no idea what to do here
slow grove
note x^4 + x^3 + x^2 + x = (x^5 - 1) / (x - 1) - 1
thats not quite right
next ore
is it
slow grove
correct now but can be written simpler as (x^5 - x)/(x - 1)
next ore
and also we are given a function with degree 4
acoustic path
i did not know that
this is geometric sum
next ore
this is geometric sum
ohh i see
acoustic path
correct now but can be written simpler as (x^5 - x)/(x - 1)
how it is written does not matter much since the idea is you clear the x - 1 anyway
slow grove
true
acoustic path
spotting the geometric sum is just the most motivated way to do this
next ore
but we got an extra constant too
wait can we do
f(x)=(x^4+x^3+x^2+x-1) Q(x) + g(x) ?
if we put a b and c in this first term becomes 0
and we get f(a)=g(a)
Q(x) is some random function we get when we divide f(x) with x^4+x^3+x^2+x-1 and g(x) is just the remainder
acoustic path
wait can we do
f(x)=(x^4+x^3+x^2+x-1) Q(x) + g(x) ?
this is essentially the idea
next ore
this is essentially the idea
yes but idk how i can be sure that the remainder will be nothing else but g(x)
acoustic path
by finding Q
acoustic path
i would say yes
the problem presupposes g(1) has a unique value and certainly you have found a valid choice of g
as for whether or not this is actually true, 🤷♂️, but this is almost guaranteed the intended way of solving this problem
pearl pondBOT
Available help channel!
outer hare
Would the dimention of this not be n+1?
pearl pondBOT
Would the dimention of this not be n+1?
outer hare
my book says the basis is ${x^{n-1},...,1}$ and i dont understand why it stops at n-1
jolly parrotBOT
BOSS
sharp vigil
it defines it as the degree being "less than" n
wet osprey
Less than n
outer hare
time for bed
wet osprey
Not the
outer hare
Not the
yeah sorry
@wet osprey is it to think think of a basis as a genirator for the vector space?
wet osprey
It pretty much is yeah
sharp vigil
it's a bit weird though because usually P_n denotes vector spaces of degree at most n
outer hare
it's a bit weird though because usually P_n denotes vector spaces of degree at m...
yeah I did the problem like this
didnt even notice the less then
@sharp vigil
Would the middle be enough to show subspaces
or do i need ot show it out more mathimatically
this is just practice btw not turning it in for anthing i just want feedback
sorry if its messy asf lmao
sharp vigil
i would be a little careful of calling the zero vector "identity" since that makes me think of the identity function
wet osprey
Can you describe them
outer hare
1 acts when scaler multiplication is there
wet osprey
Woah
outer hare
so like it surves as an identity for scaler multiplication
wet osprey
Be very careful now
wet osprey
You’re claiming that we can multiply polynomials together in this vector space
outer hare
no scaler multiplication
wet osprey
You called the scalar multiplication identity “the 1 polynomial”
wet osprey
What actually is the zero vector in F[x]?
sharp vigil
the (scalar) multiplicative identity is a scalar whereas the additive identity is a vector
outer hare
in abstract algebra terms, I know its an abelian group under scaler product
by definition
is the opperation just vector addition and scaler multiplication?
if so they kinda have different identities no?
like a ring maybe idk
wet osprey
They do
outer hare
(this might be very wrong im just trying to understand)
wet osprey
But the scalar multiplicative identity comes from the multiplicative identity of the underlying field
The additive identity is something in the vector space V itself
outer hare
ah ok
wet osprey
Here the zero vector is the polynomial p, where p sends every input to 0
outer hare
ok so for subspaces
what exactly is the third condition, which identity am i checking for?
wet osprey
ie. p(x) = 0 for all x is a polynomial that lives in F[x]
sharp vigil
for subspaces we require them to be nonempty
wet osprey
This guy is the zero vector in F[x]
wet osprey
ie. p(x) = 0 for all x is a polynomial that lives in F[x]
No this one
sharp vigil
the zero vector is a very convenient way of checking nonemptyness, because if it's nonempty but doesn't contain 0 then it's not closed
outer hare
ie. p(x) = 0 for all x is a polynomial that lives in F[x] ok
okie wait
can i just say
0 vector and 1 vector
or are there better vocab
wet osprey
What the heck is a 1 vector
outer hare
shflskajdg
wet osprey
The 0 vector makes sense sure
hazy pilot
(1,1,1,1,1,1)
outer hare
thought i was cooking
sharp vigil
0 vector is a common term for additive identity
wet osprey
(1,1,1,1,1,1)
Now you need to specify a basis
wet osprey
scalers have 1
Yes the • from the field has a multiplicative identity
sharp vigil
yeah the mutliplicative identity for scalar mult is just the multiplicative identity for the field (this is one of the axioms)
wet osprey
And you use the same identity for the multiplication between scalars and vectors in the vector space
outer hare
yeah
so waht im understanding is
the 1 comes from scalers and is not nesisiarly present in vectors
like ie there is not nessisarely an element $v \in V$ where \alpha v = \alpha
jolly parrotBOT
BOSS
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
wet osprey
the 1 comes from scalers and is not nesisiarly present in vectors
Yes, for F[x], the “1 vector” or “1 polynomial” that you called is actually just the 1 in F
sharp vigil
well there is no such operation (a priori) as multiplying two vectors together to get a vector back
wet osprey
Yes, for F[x], the “1 vector” or “1 polynomial” that you called is actually just...
Which is, obviously not a vector, it lives in the field F
outer hare
Which is, obviously not a vector, it lives in the field F
i mean the thing is
F[x] is also an integral domin
wet osprey
**BOSS**
Compile Error! Click the <:errors:504266556054962196> reaction for more...
I don’t know what 𝛂v = 𝛂 means
outer hare
so in this case
sharp vigil
alpha is a scaler
well there is also no such operation (inherently) that multiplies a scalar and a vector to get a scalar
wet osprey
What operation is between the 𝛂 and v
outer hare
well there is also no such operation (inherently) that multiplies a scalar and a...
ur right
fuck
kinda cool tho
wet osprey
Cos scalar multiplication on V (a vector space over F) has the form • : F x V -> V
outer hare
ok yeah
that makes more sense
i knew that too i just got a little too lost in discribing 1
however in this case
We do have a multiplicative identity because F[x] is an integral domain
however
that does not mean anything for vector spaces
sharp vigil
you can multiply polynomials in this case yes, it just doesn't have anything to do with the vector space structure
outer hare
ok cool
now in terms of abstract algebra
when they say vector spaces are an abelian group
what exactly is the opperation?
wet osprey
Oh dear how did we get to vector fields
sharp vigil
vector addition
outer hare
ok cool, and then scaler multiplcation just like
an added requirment for vector spaces?
sharp vigil
yes, that's the added structure that makes them more than just abelian groups
outer hare
Ok thats cool
alr tysm
I was just trying to figure out how to properly think of it in terms of abstract algebra
yall goated fr for dealing with my stupidity
outer hare
tysm lwky feel a lot better lmao
alr im gonna try and finish this so i can understand maps between vector spaces
hopefully
wet osprey
At least you recognise what I mean when I say your words don’t make sense
outer hare
tbf i didnt do lin alg yet and for some reason this just randomly popped up at the end of my abs alg class
i didnt know it would be required
but ive always wanted to learn it properly so
ill take the oppertunity
wet osprey
LA is really sick
outer hare
yeah lwky ive been suffering because i havent taken it fully yet
i hadd a lin alg / diff eq class
where it was basicly all diff eq
and the next one is super proof based so ive been a little scared
wet osprey
One of my professor said that if you have a math problem and you can convert it into LA then you’ve pretty much solved it
outer hare
i took real analysis before lin alg lmao
outer hare
One of my professor said that if you have a math problem and you can convert it ...
yee i took a class on optimization and its insane how much u can do
computationally
span and stuff and understanding the proper proffs behind it is smth i need to do tho
wet osprey
Shouldn’t be too hard if you can do abstract algebra
outer hare
ill finish these exercises and come back if i got questions
im gonna take it next sem
lets see how it goes
we use axler so its fire
tropic saddle
One of my professor said that if you have a math problem and you can convert it ...
"mathematicians can do two things. linear algebra and reducing things to linear algebra"
sharp vigil
that's why they spend like 80% of ode course on linear, because those are the ones we can do (and the linear algebra behind it is actually pretty cool when they bother to teach it)
pearl pondBOT
@outer hare Has your question been resolved?
outer hare
that's why they spend like 80% of ode course on linear, because those are the on...
mine was
the exact opposite lmao
sadly
outer hare
is this talking about linear transformations?
pearl pondBOT
is this talking about linear transformations?
outer hare
obsidian 💪💪
yessirr
outer hare
yes it's about linear transformations
ok bet
i was trying to understand my class notes lmao
.close
outer hare
what would be a basis for this subspace?
pearl pondBOT
what would be a basis for this subspace?
outer hare
Here is my proof for it being a subspace:
for some $x,y \in \mathbb{R^3}$ lets define x_n as elements of x and y_n as elements of y
jolly parrotBOT
BOSS
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
calm wing
x_n in dollar signs
calm wing
💀💀
jolly parrotBOT
BOSS
outer hare
huh
calm wing
^3 outside
outer hare
yee but there is a 3 inside which looks cooler
calm wing
it's trying to find mathbb of "^3"
jolly parrotBOT
BOSS
jolly parrotBOT
artemetra
outer hare
fr
calm wing
mathjax works slightly differently
outer hare
ah
bro why is my ipad not lettting me copy paste
everything i have is broken
smh
ok so
@calm wing clearly this is a subspace
what would the basis be?
i dont think its just 3,-2,1
because then you can get like
3,-4,1
outer hare
wait why does that work
calm wing
any two linearly independent vectors span a plane
outer hare
ngl im learning this through abs alg so idk much bout the physical spaces of lin alg
calm wing
ngl im learning this through abs alg so idk much bout the physical spaces of lin...
yeah i went the opposite route
outer hare
not any 2 tho they have to be like lenth n right?
outer hare
yeah i went the opposite route
much more logical lmfao
calm wing
no
calm wing
not any 2 tho they have to be like lenth n right?
any length works
(non zero obviously)
outer hare
doesnt the basis always have to be a certain dimention
calm wing
wdym
outer hare
calm wing
dimension can refer to s vouple of things here
calm wing
yes
calm wing
that equation is a plane equation
outer hare
oh
outer hare
bruh
lmfao
wait wdym plane equation why cant this be like
3 different vectors in x,y,z
calm wing
when you learn about rank nullity and stuff about kernels and images you'll see why this has to be 2d
outer hare
lmfaoooooooooooo im doing a problem with kernals right after
to hopefully understand homomorphisms more
calm wing
3 different vectors in x,y,z
if they are all linearly independent, two of them span a plane, and the third one lies outside of the plane
outer hare
lwky im learning this right after rings
so thats kinda cool
from waht im understanding
groups -> rings -> vector spaces -> fields
calm wing
interesting approach
outer hare
in terms of struction
outer hare
interesting approach
just what my book does
calm wing
which book
outer hare
calm wing
no i meant the name 😭
outer hare
lmfaoo
check dms @calm wing
oh also
thats the next question lmao
so i guess ill find out why its in 2d space now
.close
pearl pondBOT
pearl pondBOT
Available help channel!
vestal pelican
in this limit we separate 3 from 3/n because of $$
\lim (a \cdot f(x)) = a \cdot \lim (f(x))
$$ rule
jolly parrotBOT
Rony
vestal pelican
but in this limit we dont, why
compact ridge
but in this limit we dont, why
that goes to $\frac{3}{1 + 2 \cdot 0} = 3$
jolly parrotBOT
south
vestal pelican
so we can separate it
compact ridge
wdym by separate?
compact ridge
the 2/sqrt n part
yes
compact ridge
well the really important idea is that you can sub in 0 directly
you definitely need to check whether it's $\lim_{n \to \infty}$ or $\lim_{n \to 0}$
jolly parrotBOT
south
vestal pelican
yeah i guess idk why im overcomplicating this
compact ridge
well the really important idea is that you can sub in 0 directly
it's based on one of the algebraic limit theorems
pearl pondBOT
compact ridge
yeah i guess idk why im overcomplicating this
find y first
also this isn't your channel
make a new channel or else this one will close suddenly
compact ridge
I’m kinda new
vestal pelican
find y first
im finding if serie is convergent or divergent
compact ridge
im finding if serie is convergent or divergent
sorry, I pinged you accidentally
vestal pelican
oh ok
pearl pondBOT
Available help channel!
viscid minnow
Hi, I'm working on a tweak for a tabletop game, and my question is about PROBABILITY, which I know very little about, hopefully it's not too much game-language:
The game rules are currently system A:
**A) ** To hit someone with a blade, a dice roll (D20) is made to meet or exceed a target number, such as 15+. Here we know the chance of success without any modifiers on the roll is 25% (5/20). To poison someone in addition to hitting them, you make a second roll vs a second target number. Example. On a hit with a sword vs the enemys Armor Class (AC), roll another D20 vs their Constitution (CON) to see if they're Poisoned. Two rolls.
My idea was to change it to save time during play by using system B:
B) Instead of prompting another roll, we use the first result against both target numbers (sequentially). So in the example above, we would first roll to meet or exceed the AC (which would be a hit that deals damage), and then we check if the result also meets or surpasses the CON which would result in target being Poisoned. One roll.
My question is, how does the probability change by using B instead of A? Is it more or less likely for the extra effect to occur with the change, and by how much?
I appreciate any help, THX.
pearl pondBOT
Hi, I'm working on a tweak for a tabletop game, and my question is about **PROBA...
@viscid minnow Has your question been resolved?
viscid minnow
<@&286206848099549185>
viscid minnow
idk how many times im alowed to pong <@&286206848099549185>
steady crescent
idk how many times im alowed to pong <@&286206848099549185>
So, to clarify you want to ask if the probability of rolling greater than 15 on one die is the same as two die?
The answer is no for that.
Probability with one die is 1/4 and with two dice is (1/4)^2 = 1/16
Three dice would be (1/4)^3, etc
Assuming fair dice ofc
viscid minnow
Thanks for helping out! So I guess I'm curious as to how I approach this when moving forward. It's like onion layers right? We're not checking the second layer unless we meet the first. What if the second layer is higher? or Lower? Does it change anything?
like if we need to meet 15 for the first dice, and the second ~~score ~~target is 16, or vice versa?
steady crescent
Sorry, I'm not really following
viscid minnow
No worries I'll try to clarify
So in your first message you assumed the target number was 15 for both targets, right? What happens when we roll one dice to meet 15, which is 1/4, and only if that succeds, we roll an additional dice to see for an extra effect. Only this time the target is higher or lower, let's say the target is 10 for the second effect. Would that be 1/4 * 1/2?
steady crescent
Ah okay, I'm following
So of you are looking for the probability that both die succeed, the answer is yes it would be 1/4*1/2
If you're looking for the probability that the second succeeds given the first succeeds it would just be 1/2
steady crescent
So of you are looking for the probability that both die succeed, the answer is y...
Assuming it's independent, which seems pretty fair
viscid minnow
What do you mean by independent? In this case theres like a barrier, we dont check for poison unless the blade passes through the armor. So we never roll the second die unless the first one succeeds. I assume that means its not independent, and it would be 1/4*1/2?
steady crescent
I mean, in think case ita a bit tricky to understand. But it basically means that one dice roll doesn't affect the other. Say for example, you roll the second die every time no matter if the first succeeds and just disreguard the result. The result of the first die doesn't affect the result of the second.
I hope that makes sense. In this case, it's a pretty reasonable assumption
viscid minnow
Ah yeah that makes sense, at least if the first has a higher difficulty than the latter. If the first one needs to meet 10, and the second 15, it’s different no?
steady crescent
Ah yeah that makes sense, at least if the first has a higher difficulty than the...
Different how?
Generally the probabilities are arbitrary, you could swap them out for symbols like p and q.
The methods stay the same as far as I can think
viscid minnow
I imagined that hitting 15 is easier than first having to hit 10, and then 15. Since if we roll 10 times, we first remove 50% of those that fail, and only from the 5 that succeed, we can get 1/4 that hits 15… wait i think you helped me solve the original question.
So if i change the ruling from this double roll, to just one roll that we check against both, its just:
vs 15/15 = 1/4 because its the same
vs 15/10 = 1/4 because anytime the first succeeds, the second automatically succeeds, which it doesnt in the original ruling.
vs 10/15 = here im confused still. We roll once, we have a 1/2 chance to hit 10, and a 1/4th to hit 15
Is it still arbitrary for the single roll (thats checked vs two targets) if the first target is lower than the second?
I think i get how the double roll is calculated now at least
Let me know if i lost you i can try reformulating
steady crescent
I mean, to be honest I can't fully grasp what you're trying to achieve, I thought I answered it before, but apparently not. I'd recommend learning or doing a bit of research on probability at the very least to get some of the terminology and notation down so you can be more precise in what you're asking (at least from a methematical perspective). I think then the answer should follow quite naturally. If you still have trouble after reformulating your question you can always ask again, but for now I think you just need to learn a little bit more which will help someone try to understand exactly what you're after.
Sorry, I hope that didn't come off as rude. I just genuinely think it would be helpful.
viscid minnow
I mean, to be honest I can't fully grasp what you're trying to achieve, I though...
No worries, I will definitely try to learn some more fundamentals, I thought I had the basics down, and of course I have the game-terminology spinning around in my head at the same time as well so that probably doesn't help heh. In purely practical terms, I wanted to know if and by how much I would be changing the balance of the game by removing that extra step of rolling an additional die for bonus effects, and instead just using a single roll to succeed on both to save time.
I'll try to revise the basics and see if I grasp this whole issue better afterwards
You've definitely been helpful though, you did answer my question, I might have had more than one question in my head is all. Thank you for you time!
forest sigil
pearl pondBOT
forest sigil
How can to find value of A
late sun
You have the value for B and C, you need to use those to find A. You know that this holds for all x, so you can plug any value of x in that you like (that isn’t -1 or 1/2) and solve
Say, x = 0
late sun
That looks right to me
You can test it by trying it with different x values in an exam, or if you wanna check it quick pop it in desmos
past perch
im a bit perplexed do you not do partial fraction decomposition by expanding and comparing coefficients?
i havent actually done it but from what i've seen
forest sigil
im a bit perplexed do you not do partial fraction decomposition by expanding and...
In book it is showing comparing coefficient of x^2 but I don't know how to dp it
past perch
you expand the rhs
forest sigil
Yes
past perch
expand this
forest sigil
How
late sun
Yeh you can do it like that and solve the system that comes out, it’s personal preference
forest sigil
expand this
Expanding this will make it more complex
past perch
Expanding this will make it more complex
eh it's just a 3 variable system of simultaneous equations
thanks though!
past perch
easy check, plug your values of A B and C
forest sigil
easy check, plug your values of A B and C
I did but the answer is wrong
past perch
,w (-1/9)(x+1)(2x-1) + (4/3)(2x-1) + (2/9)(x+1)^2
jolly parrotBOT
forest sigil
It is wrong
past perch
can you elaborate? i think you mean
that they fractions are cleaner
you can write it as $-\frac{1}{9\left(x+1\right)}$
in essence it's the same thing
forest sigil
The answer is log(2x-1/x+1 ) -4/3(1/x+1)
past perch
i assume you're talking about the answer of the integral
forest sigil
Yes
past perch
Yes
you have to integrate your partial fractions now
your PARTIAL FRACTIONS are correct
forest sigil
Okk
past perch
The answer is log(2x-1/x+1 ) -4/3(1/x+1)
are you sure?
i think you're missing something here
you'll see when you integrate
pearl pondBOT
pearl pondBOT
Available help channel!
lone nimbus
If I'm finding the curve of intersection between z = sqrt(5-x^2-y^2) and z = 1, I get x^2+y^2=4, is this a cylinder or a circle?, because usually in 3D x^2+y^2=4 is a cylinder (by itself)
wet osprey
well you did say curve of intersection
a cylinder isn't so much a curve as it is a surface
so probably a circle
since you have also the equation z = 1
lone nimbus
by itself x^2+y^2=4, what is it?
wet osprey
that's a cylinder yes
lone nimbus
so when we put the two functions equal to each other, and get x^2+y^2=4, since it's by itself, wouldn't it be a cylinder? (before adding the z = 1 component)
wet osprey
which 2 functions are we talking about
lone nimbus
so the two surfaces i listed
lone nimbus
z = sqrt(5-x^2-y^2) and z = 1
wet osprey
saying z = sqrt(5-x^2-y^2) and z = 1 is different to saying x^2+y^2=4
the second gives no restriction on the value of z
lone nimbus
when we put them equal to each other, and get x^2+y^2=4, wouldn't that be a cylinder unless you attach the z = 1?
wet osprey
it would yes
wet osprey
saying `z = sqrt(5-x^2-y^2) and z = 1` is different to saying `x^2+y^2=4`
again, this is different
pearl pondBOT
@lone nimbus Has your question been resolved?
midnight haven
If I'm finding the curve of intersection between z = sqrt(5-x^2-y^2) and z = 1, ...
You're essentially finding a level curve of the function z = f(x,y) = sqrt(5-x²-y²)
A level curve is a curve (not a surface) obtained when you compute f(x,y) = c for some constant c
You get x²+y²=4, which is a circle. Consider this as a slice when you cut the given surface horizontally at z=1. It's the locus of all points (x,y,z) lying on z=f(x,y) such that z=1, so it's not a cylinder
lone nimbus
So the equation x^2+y^2=4 can represent a circle or a cylinder, and here it's a circle
midnight haven
Indeed. It's context dependent
lone nimbus
is this similar to the 2D case? for example, if we have two lines that intersect, we have x = a being the x-coordinate of our point, here x^2+y^2=4 is the curve?
im trying to find the similarity
midnight haven
lone nimbus
what would you call x^2+y^2=9? x was the x-coordinate of your point, here x^2+y^2=9 is the what?
midnight haven
I would implicitly assume it's a circle unless mentioned that it's a cylinder
lone nimbus
I know it's a circle, but what would u call it? the curve of intersection?
midnight haven
I assume you mean that what if you took horizontal slices of the circle right?
Like in this case
You would get a level set
pearl pondBOT
midnight haven
For example, for x² +y² = 9, we have y = f(x) = sqrt(9-x²) (consider the upper semicircle for now)
For some f(x) = c, you would get a level set, for example f(x) = 0 gives {(-3,0),(3,0)}
Have a great day
pearl pondBOT
Available help channel!
dusky basin
I dont know what to do or how to find G.
pearl pondBOT
I dont know what to do or how to find G.
grim fractal
So the key is these angles are determined by the arcs they form
Notice how they both form (minor) arc EH
toxic lichen
I dont know what to do or how to find G.
are you familiar with circle theorems?
perhaps something called the inscribed angle theorem?
pearl pondBOT
@dusky basin Has your question been resolved?
dusky basin
perhaps something called the _inscribed angle theorem?_
ive heard of inscribed angles but no the inscribed angle theorem
toxic lichen
an inscribed angle equals half the measure of its subtended arc
what is this theorem called in your class
dusky basin
Im not sure we have a name for it. I know it's something we've gone over though.
pastel umbra
Yh, that last bullet point
A corollary (small follow-up) of that is that two inscribed angles from the same chord, on the same side of the chord, are equal
dusky basin
But I've already failed that question though. Could ya'll help me with this one
pastel umbra
Taking H to be a centre, what do you know about HG and HF?
dusky basin
Wait if the triangle is 180 degrees and I take that point and then subtract 79 from 180.
180-79=101 n since theres 2 points missing i divide it by 2 in order to find the missing points?
dusky basin
Central Angle is 2 times more right?
inland ivy
Let $f$ be an analytic function which has a Taylor series [ f(y)=\sum_{n=0}^\infty \frac{f^{(n)}(x)}{n!}(y-x)^n ] for $y\in \Omega$ where $\Omega$ is some neighbourhood of $x$. Replacing $y$ with $x+y$ so that $x+y\in\Omega$, [ f(x+y)=\sum_{n=0}^\infty \frac{f^{(n)}(x)}{n!}y^n=e^{y\dv{x}}f(x) ] where the exponential is opened as a formal power series: [ e^{y\dv{x}}:=\sum_{n=0}^\infty \frac{\left(y\dv{x}\right)^n}{n!}=\sum_{n=0}^\infty \frac{y^n}{n!}\dv{^n}{x^n} ] Thus, [ f(x+y)-f(x)=\left(e^{y\dv{x}}-1\right)f(x) ] and by inputting $y=1$ we get the equation [ \boxed{\Delta=e^{\dv{x}}-1} ] where $\Delta$ is the forward difference operator, given by [ (\Delta f)(x)=f(x+1)-f(x). ] Abusing notation, we can now write [ \boxed{\dv{x}=\log(1+\Delta)=\sum_{n=1}^\infty (-1)^{n+1} \frac{\Delta^n}{n}} ]
jolly parrotBOT
kheer257
inland ivy
what is this and why does it work
how do you even talk of the convergence of series like this
tropic saddle
define a relevant norm on the space
inland ivy
define a relevant norm on the space
what would that look like in this context?
tropic saddle
dunno. I would need to remember much more from functional analysis than I do
inland ivy
there's also more cursed stuff
where you take the e^{y d/dx} - 1 operator to the other side and then invoke the taylor series for x/(e^x-1) to get the Euler-Maclaurin formula
tropic saddle
you only need that norm(d/dx) is a thing
and d/dx is a linear function on some function space
so you can get your operator norm
but there is a bit of a problem here as C^infty isnt a normed space iirc
so you have to fix that
inland ivy
hmm
tropic saddle
but frankly open a book on functional analysis
this needs a few weeks of background
pearl pondBOT
@inland ivy Has your question been resolved?
inland ivy
Taking the "inverse" of the $e^{\dv{x}}-1$ operator on both sides, we get [ \left(e^{\dv{x}}-1\right)^{-1}\Delta=1\implies \dv{x}\left(e^{\dv{x}}-1\right)^{-1}\Delta=\dv{x}. ] Note that the exponential generating function for the Bernoulli Numbers is given by [ \frac{t}{e^t-1} = \sum_{k=0}^\infty B_k\frac{t^k}{k!}, ] so once again abusing notation, [ \dv{x}\left(e^{\dv{x}}-1\right)^{-1}=\sum_{k=0}^\infty\frac{B_k}{k!}\dv{^k}{x^k}. ] Thus, [ \left(\dv{x}\left(e^{\dv{x}}-1\right)^{-1}\Delta f\right)(x)=\sum_{k=0}^\infty \frac{B_k}{k!}\big(f^{(k)}(x+1)-f^{(k)}(x)\big)=f'(x). ] Replacing $f'(x)$ with $f(x)$ and treating $f^{(-1)}(x)$ as $\int f \dd{x}$, along with $B_0=1$ yields [ \int_x^{x+1} f(t)\dd{t} + \sum_{k=1}^\infty \frac{B_k}{k!}\big(f^{(k-1)}(x+1)-f^{(k-1)}(x)\big)=f(x). ] Summing from $x=m$ to $x=n-1$, where $m$ and $n$ are integers, $$\begin{gathered} \sum_{x=m}^{n-1} \left(\int_x^{x+1} f(t) \dd{t} + \sum_{k=1}^\infty \frac{B_k}{k!}\big(f^{(k-1)}(x+1)-f^{(k-1)}(x)\big)\right) = \sum_{x=m}^{n-1} f(x) \ \implies \int_m^n f(t) \dd{t} + \sum_{k=1}^\infty \frac{B_k}{k!}\big(f^{(k-1)}(n)-f^{(k-1)}(m)\big)=\sum_{x=m}^{n-1} f(x) \end{gathered}$$ Truncating the infinite series at $k=p$ we get the Euler-Maclaurin Formula: [ \boxed{\sum_{k=m}^{n-1} f(k) - \int_m^n f(t) \dd{t} = \sum_{k=1}^p \frac{B_k}{k!} \big(f^{(k-1)}(n)-f^{(k-1)}(m)\big) + R_p} ] where the remainder term $R_p$ can usually be made small for large enough $p$.
jolly parrotBOT
kheer257
pearl pondBOT
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cobalt oak
I am trying to define a set S that contains all elements less than and coprime to n using set builder notation. I currently have written that S={n|gcd(n,x)=1, x e Z, x<n}, but the way I am using x here I don't think is correct. How could I write this set?
toxic lichen
yeah you need x at the beginning not n
$S = {x : x \in \bZ, 1 \leq x < n, \gcd(x,n)=1}$ (presumably you don't want all the negatives?)
jolly parrotBOT
Ann
cobalt oak
yeah I don't want negatives, nice catch
does the colon operate similar to a such that here?
I am very unfamiliar with set builder notation
tropic saddle
: and | are the same thing in that context
cobalt oak
ahh ok, thank you two very much :)
toxic lichen
if you've got the divides symbol in there, which is a vertical bar, it's better to use a colon
for clarity
otherwise it's up to you
tropic saddle
generally vertical bars appear much more often
: is a symbol thats basically never used
cobalt oak
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pearl pondBOT
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worn basalt
is anyone familiar with spss software?
pearl pondBOT
is anyone familiar with spss software?
hazy pilot
Please don't ask ur question in other help channels :)
worn basalt
oh ok i'm sorry I don't know how this works, I just need help really bad
radiant terrace
I'm personally not but it might help people to give more info on the sw
worn basalt
whats the sw 
radiant terrace
Short for software
pearl pondBOT
@worn basalt Has your question been resolved?
worn basalt
ohh, its a statistics software
peak kayak
anyone in y11?
pearl pondBOT
anyone in y11?
plush bramble
ruby cargo
!da2a
pearl pondBOT
No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/
peak kayak
.close
pearl pondBOT
pearl pondBOT
Available help channel!
warm wraith
I'm stuck at this question for 2 days and I can't solve it....
pearl pondBOT
I'm stuck at this question for 2 days and I can't solve it....
@warm wraith Has your question been resolved?
signal atlas
I'm stuck at this question for 2 days and I can't solve it....
Have you tried simplifying the left side too?
warm wraith
yes, I already tried both side but it didn't work
sinful quest
i would start by using cot^2 = csc^2 - 1 on the right side and simplifying
sinful quest
yes, I already tried both side but it didn't work
you should be able to turn the right hand side into the left hand side
or the other way around
pearl pondBOT
@warm wraith Has your question been resolved?
sinful quest
yes, I already tried both side but it didn't work
have you tried using cot^2 = csc^2 - 1
sinful quest
i tried converting everything to sins and cosines and it got messy pretty quickly, instead just try to make the right side look like the left, so once you use the formula I gave earlier just distribute the sin^2 C into the brackets and see how you can make the two sides more similar
what is something you could do to get the right looking more like the left
pearl pondBOT
@warm wraith Has your question been resolved?
forest sigil
pearl pondBOT
forest sigil
Someone explain
forest sigil
differentiate to see
What will it be
forest sigil
Arctan??
versed mica
,w derivative arctan(x)
jolly parrotBOT
versed mica
know your derivatives sir
forest sigil
Tan inverse
versed mica
arctan
forest sigil
We call it tan inverse
versed mica
nah
forest sigil
@versed mica
versed mica
🤔
versed mica
you need to factor x^2-1 first before partial fractions
$x^2 + 1 = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}$
jolly parrotBOT
knief
forest sigil
Oh a²-b² = (a+b)(a-b)
light helm
didn't you just ask this
versed mica
lmao
pearl pondBOT
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Available help channel!
dense ledge
pearl pondBOT
dense ledge
prove that BS=CS=IS
autumn fossil
it should suffice to prove that IBS and ICS are isosceles
dense ledge
it should suffice to prove that IBS and ICS are isosceles
thats kinda the question but i dont know how to say it
autumn fossil
I'd try some angle chasing
give those angles names and fill in as much angles as yo ucan
dense ledge
give those angles names and fill in as much angles as yo ucan
if i have IBS=ACB , does that mean A,I,S is on the same line ?
weary ledge
Is that not given information?
dapper kraken
if i have IBS=ACB , does that mean A,I,S is on the same line ?
i dont see why it would?
also isnt A,I,S colinear by construction
dense ledge
Is that not given information?
the only thing that was given is BI,IC,IS is equal
dapper kraken
the only thing that was given is BI,IC,IS is equal
hm?
arent you asked to figure out the fact that BI,IC,IS are equal?
dense ledge
also isnt A,I,S colinear by construction
i mean is true, but not given
dapper kraken
!xy
pearl pondBOT
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.
dense ledge
arent you asked to figure out the fact that BI,IC,IS are equal?
...
i just think that i should try a different solution
but nvm
my teacher gonna solve it now
so
.close
pearl pondBOT
dapper kraken
ts pmo 🥀
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Available help channel!
timid jetty
hello
timid jetty
how do i solve for x and y if 2 expressions are given
tropic saddle
depends on the expressions
timid jetty
but theyre not equal to 0
oak ivy
so what
timid jetty
i mean its not given
tropic saddle
but generally you can solve for one of the variables in one of the equations and then plug that into the other equation
oak ivy
ah
timid jetty
and
tropic saddle
!original
pearl pondBOT
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.
autumn fossil
autumn fossil
Those 2 expressions dont provide any information about x or y whatsoever
if you're still stuck on that
so "solving for x and y" has no meaning
timid jetty
- Find the value(s) of x and y in the following, if they are consistent -
i. 4x-2y=3 and 3x-y=12
ii. 3x+5y-3 and 8x-3y+6
iii. 17y=3x and 2y=6
i copy pasted the question
autumn fossil
17. Find the value(s) of x and y in the following, if they are consistent -
i. 4...
okay now this is a meaningful question
timid jetty
ii subpart
tropic saddle
are you sure these arent typos?
tropic saddle
yes
oak ivy
They both have to be equal to 0 I think
timid jetty
but im not sure...
oak ivy
cause a1/a2 neq b1/b2
timid jetty
yes
oak ivy
they intersect and the way to find them is equating both I think
oak ivy
inconsistent?
consistent
timid jetty
consistent
oak ivy
yes
oak ivy
You know the answer?
oak ivy
Yea ask your teacher
pearl pondBOT
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Available help channel!
midnight reef
Consider the series $\sum_{n=1}^{\infty} \frac{n}{(n+1)!}$. Find a formula for the partial sums $S_n$
midnight reef
Any hint pls?
jolly parrotBOT
Halex
glass root
Try to find a recurrence relation for n/(n+1)!
pearl pondBOT
@midnight reef Has your question been resolved?
midnight reef
Try to find a recurrence relation for n/(n+1)!
I got $S_n = 1 - \frac{1}{(n+1)!}$, is this true?
jolly parrotBOT
Halex
midnight reef
By telescope sum, right?
midnight reef
I got the same by just doing telescope sum
midnight reef
.close
pearl pondBOT
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normal sigil
Hi guys.
normal sigil
I know this seems simple, but here's what I wrote:
Since cos(pi/2) = 0 we go to -sin(pi/2) which is -1 therefore this is the first term and the next term will be the remainder ie
$\cos{(x)} = (-1)(x - \frac{\pi}{2}) + \left(\frac{-\cos{(c)}}{2!} (x - \frac{\pi}{2})^2 \right)$
jolly parrotBOT
Bete Puttigieg 🐢
normal sigil
where c is between pi/2 - x and pi/2.
However, the solution appears to have used the next nonzero derivative of cos evaluated at pi/2, so it used the third term as the remainder bit.
Is this common/what you are supposed to do?
pearl pondBOT
@normal sigil Has your question been resolved?
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sharp smelt
Find all values of $x$ for which $2^x<8$
pearl pondBOT
Find all values of $x$ for which $2^x<8$
jolly parrotBOT
What a wonderful world !
sharp smelt
This is from spivak, chapter 1, so I'll have to prove this , can't use logs
( I think?)
I think we can't use logs
It's not explicitly mentioned, but I suppose we can't.
sharp smelt
I'll do this axiomatically
ruby cargo
**What a wonderful world !**
I think you should just be good to use $2^a < 2^b \iff a < b$
jolly parrotBOT
@ruby cargo
ruby cargo
$\forall a, b \in \bR$ ofc
jolly parrotBOT
@ruby cargo
sharp smelt
We start by stating the following conventions:
$2^{x} = \underbrace{2 \cdot 2 \cdots \cdot 2}{\text{x times}}$,
$2^{-x} = \underbrace{\frac{1}{2} \cdot \frac{1}{2} \cdots \cdot \frac{1}{2}}{\text{x times}}$
$2^{x} \cdot \frac{1}{2} < 8 \cdot \frac{1}{2}$
if $a < b$ and $c$ is positive, then $ac < bc$
$(2^{x} \cdot \frac{1}{2}) \cdot \frac{1}{2} < 8 \cdot (\frac{1}{2} \cdot \frac{1}{2})$
if $a < b$ and $c$ is positive,then $ac < bc$
$(2^{x} \cdot \frac{1}{2}) \cdot \frac{1}{2} \cdot \frac{1}{2} < 8 \cdot (\frac{1}{2} \cdot \frac{1}{2}) \cdot \frac{1}{2}$
if $a < b$ and $c$ is positive, then $ac < bc$
We then have $2^{x - 3} < 1$
hmm
ruby cargo
Is this just for integers?
sharp smelt
Is this just for integers?
for now, yea
jolly parrotBOT
What a wonderful world !
ruby cargo
You can also say, $2^a < 2^b \implies \exists c \in \bN \text{ s.t. } 2^a \cdot 2^c = 2^b \implies 2^{a + c} = 2^b \implies a + c = b \implies a > b$
jolly parrotBOT
@ruby cargo
ruby cargo
Yea
midnight haven
And that includes logs to prove I think
ruby cargo
$2^a \cdot 2^b = 2^{a + b}$ in particular
jolly parrotBOT
@ruby cargo
midnight haven
Actually if a and b are natural then logs aren't necessary
ruby cargo
And that includes logs to prove I think
Not really, because we're just comparing inputs by the virtue of the fact that $2^x$ is a bijection
jolly parrotBOT
@ruby cargo
ruby cargo
You can do this for any bijection, without requiring the inverse
sharp smelt
This definition will only give you x ∈ ℕ
yea, I know
ruby cargo
This definition will only give you x ∈ ℕ
Yea that's what they need rn
sharp smelt
Find all values of $x$ for which $2^x<8$
what do I do now is the question
full pine
help
full pine
i do not know how to continue solving this problem
pearl pondBOT
@full pine Has your question been resolved?
little grove
pearl pondBOT
little grove
mental vine
well you gotta do the derivative of y=-ln(x+2) right?
little grove
yeah i got -1/x+2
i don't understand why the max value of c is is a f(0) the intercept of the actual graph
what if there is another point on the graph where the tangent line intercepts the y axis at a higher point than the yint of the graph?
mental vine
hmm I am not sure\
I mean
y=mx+b, thus y=-x/(x+2)+b
y=-ln(x+2)
-ln(x+2)=-x/(x+2)+b
b=x/(x+2)-ln(x+2)
b=1-(2/(x+2)+ln(x+2))
so we need to find when 2/(x+2)+ln(x+2) is minimized
maybe take another derivative of that?
this should solve it, but it is not very efficient, anyone got other ideas?
little grove
mm that's a bit much this is a one mark question
long yarrow
what if there is another point on the graph where the tangent line intercepts th...
it's not possible to get a higher value
if it's higher then the line wouldn't be a tangent line
foggy scarab
bruh
yall fr
the graph is downwards
so minimum c is when x is =
0
c ranges from -infinity to 0
fossil jewel
0 isnt even one of the options dawg
foggy scarab
x=0 bro
that means -ln(0+2)
=-ln2
which is the answer ;-;
this is the most common sensical question
idk if that word even exists
but u get it
foggy scarab
so we need to find when 2/(x+2)+ln(x+2) is minimized
this is the best answer
but visualization is op
just see how tangents move
pearl pondBOT
@little grove Has your question been resolved?
little grove
idk why its not clicking for me
little grove
so minimum c is when x is =
do you mean maximum?
i think i'll come back to this question another time because right now it's not making sense idk why it looks pretty straight forward
queen gazelle
bro is not slick 🥀
little grove
bro is not slick 🥀
i wasn't trying to be
queen gazelle
i wasn't trying to be
didnt they solve in methods
little grove
didnt they solve in methods
sometimes one explaination just clicks for me rather than another
but idk i'm not getting it
queen gazelle
lemme get my cas
little grove
lemme get my cas
im going to close the channel cause i need to do other work, but if you want you can send in dms so i can look at it later
terse tartan
i do not know how to continue solving this problem
I might be able to help.
pearl pondBOT
@little grove Has your question been resolved?
little grove
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outer hare
Why is $\mathbb{Q}(\sqrt{-2} + \sqrt{-3})$ a feild extention of $\mathbb{Q}(\sqrt{-2})$
jolly parrotBOT
BOSS
outer hare
Im new to extention fields
My book explained it through linear combinations, which i did not understand
tropic saddle
didnt we just talk about this?
the word linear combination doesnt matter
fields are closed under addition and multiplication
toxic lichen
Why is $\mathbb{Q}(\sqrt{-2} + \sqrt{-3})$ a feild extention of $\mathbb{Q}(\sqr...
what is under the radicals there
negative numbers or positive?
tropic saddle
if you subtract sqrt2+sqrt3 from sqrt3-sqrt2 you get what?
outer hare
didnt we just talk about this?
I was told to go back and learn vector spaces more and i did
outer hare
negative numbers or positive?
oh this was a mistake it should be positive
outer hare
if you subtract sqrt2+sqrt3 from sqrt3-sqrt2 you get what?
-2\sqrt2?
outer hare
which is therefore an element of the field
why
tropic saddle
fields are closed under addition and multiplication
.
tropic saddle
sqrt3-sqrt2=1/(sqrt2+sqrt3)
tropic saddle
and thereforr also in the field
outer hare
sorry
yeah ok so it has multiplicative inverces
so we showed that
-2\sqrt2 is in the field
cool
tropic saddle
and if you divide that by -2 you get?
.
tropic saddle
-2 is in Q
outer hare
yeah just realized
we get sqrt 2
so that means
Ok let me undertand this fully
$\mathbb{Q}(\sqrt2)$ is the smallest field with $\mathbb{Q}$ and $\sqrt2$
jolly parrotBOT
BOSS
outer hare
right?
tropic saddle
yes
outer hare
Then $\mathbb{Q}(\sqrt2 + \sqrt3)$ is the smallest field with $\mathbb{Q}$ and $\sqrt2+\sqrt3$
jolly parrotBOT
BOSS
outer hare
We want to show that this is an extention field
so we have to show that it also cotains $\sqrt2$ as $\mathbb{Q}$ is given
jolly parrotBOT
BOSS
tropic saddle
yea
jolly parrotBOT
BOSS
outer hare
we know it has the multiplicative inverce $\sqrt3 - \sqrt2$
jolly parrotBOT
BOSS
outer hare
then $\sqrt2+\sqrt3 - (\sqrt3 - \sqrt2) = 2\sqrt2$
jolly parrotBOT
BOSS
outer hare
idk where the negitive came from last time
tropic saddle
cause I said it the other way around
outer hare
anyway, 2 is in the rationals so we can just devide by two and get \sqrt2 meaning it contains it and thus contains F
outer hare
cause I said it the other way around
ah any reason
tropic saddle
ah any reason
no
outer hare
Ok
ok wait so
with the same logic
can i not just show $\sqrt a + \sqrt b \in \mathbb{Q}(\sqrt a)(\sqrt b)$
jolly parrotBOT
BOSS
tropic saddle
yes
outer hare
@tropic saddle idk why i was so scared of this
thanks for going through it slowly lmao
tropic saddle
for the record, writing it in one equation we have 1/2(sqrt2+sqrt3)-1/2(sqrt3-sqrt2)=sqrt2
outer hare
ok so in the other direction I just have to show $\sqrt a$ and $\sqrt b$ are in $\mathbb{Q}(\sqrt a + \sqrt b)$$
jolly parrotBOT
BOSS
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
tropic saddle
thats a linear combination of sqrt2+sqrt3 and sqrt3-sqrt2
outer hare
wait is it 1/2 or 1/the entire first part
tropic saddle
1/2
outer hare
oh yeah ur jsut reqriting what we did
outer hare
thats a linear combination of sqrt2+sqrt3 and sqrt3-sqrt2
why does that work
tropic saddle
otherwise I would have written 1/(2(...))
outer hare
are they generaters for the set or something?
tropic saddle
wdym work
outer hare
or just the way its explained
like why is it important to explain it that way thater then just showing how we got it
yk
tropic saddle
thats just something called a linear combination
outer hare
No i get that
outer hare
the author just never used lin alg verbage befroe this
ok
until now
we used it for like
ideals with multiple genirators
which is why i asked
something like this
tropic saddle
its always the same idea, yes
outer hare
tropic saddle
in linear algebra speak, they are two linearly independent vectors over Q
tropic saddle
and all field extensions are vector spaces over the subfield
outer hare
$\mathbb{Q}(\sqrt2)$ are scalers
jolly parrotBOT
BOSS
outer hare
over the vector space $\mathbb{Q}(\sqrt2+\sqrt3)$
jolly parrotBOT
BOSS
outer hare
not enough for a basis
ok so they were just showing
tropic saddle
ok yes over that space
outer hare
taht u can write it as a combination of two factors
I learned lin alg just for this btw
so
its like whatever my book covered
tropic saddle
you should learn lin algebra anyway
tropic saddle
its the single most important class in math