#help-36
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37, how are these equal?
But I can’t
I don’t get that they are equal
what do your derivatives give
I get 1/(x+1)^2 and 1/x+1
am i tripping or is that second one not right
try chain and exponent rules
indeed
We all good
Thanks
Wait in the second part how does it not contradict if
If they are equal the only variation in the functions can be a constant C
And I don’t see how a constant is the different between them
$-\frac{1}{x + 1} + \frac{x + 1}{x + 1} = \frac{x}{x + 1}$
south
Just add 1
But is that a constant
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I want to know how i could solve this
i couldnt even start
well do you know the kinematic equations
tht would be my teacher
ofc basic
yeah i do
so what would the equation for the position of the money bag be
uh idk i can't make out the values of acceleration and stuff if you know what i mean
well it’s a projectile
so it will have no horizontal acceleration but the vertical acceleration will be gravity
It is physics not maths
no way!
😖 i am a 10th grader i haven't learnt this
Yup it is 11th projectile
well you should know some projectile motion
If you are from India
can you solve it pls
!noans
The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.
The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.
bro ikt i am a helper too
ye helped some ppl already
alright mr "help" me
💀
Use range formula
so what is the first step
i asked you a question
Look it up in ur book
ok
And then use
Then divide displacement by vel to get time
yeah cos i didnt know how to find it by using just velocity
x=xo+vot+1/2at^2
it just makes it harder
lol how
i told you
i dont have time and acceleration
where?
gravity🤝🏻
like here
naught
you don’t need xo
in this problem
you could just do deltax
which is the same as x-xo
either will be zero
depending on how you define your coordinate system
alright nvm i am closing this channel
bruh
its totally out of my league
I need some sleep man
can you give me a step by step procedure how i can do this
i will learn the concept and solve it
you’re not even following what i’m telling you to do
so
you need to find the vertical position function
of the money bag
voy=vosintheta
a=-g
ok by using the kinematics
alright
you mean vy = sinx?
this would just change the sign for
no
then?
the initial velocity in the vertical direction is the initial velocity times the sine of theta
oh ok get it
👍🏻
and then we find the time and give the answer right?
well you need to know if superman gets there first
i think tht is easy
yea so how would you do it
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Hold up I’m just re-reading this and quite new to kinematics myself but shouldn’t this be final instead of initial for the first time you mention it
wdym
for the position function of the money bag?
it’s y=voyt+1/2 gt^2
Yes we’ll you typed initial velocity twice
yes
because
we’re taking the vertical position function
so we need the vertical component
of initial velocity
it has components in both the x and y direction
the x component is irrelevant to the vertical position
Yes I understand that
ok so what’s the problem
In your explanation
You say initial velocity = initial velocity times the sien of theta
no
i said initial velocity in the vertical direction
read
aka
vertical component
Oh I see
Yeah ur right I completely misinterpreted what your wrote
Ty
I understand at and it now
.close
Oh ok well then bye
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i might be conceptualising this wrongly
because like
what i would expect instead is
,, Wy_1,, y_2' = \mat[v]{ y_1 & y_2 \ y_1'' &y_2''} \0r\ne W'y_1,, y_2
Differentiate y1y2' - y1'y2?
right, so you get [\8{y_1'y_2' + y_1y_2''} - \8{y_1'' y_2 + y_1' y_2'} = y_1y_2'' - y_1'' y_2] right lmao. i didn't actually expect you'd need to actually differentiate the thing to retreieve the above. I thought it was like some weird substitution idk
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Yeah well like not rly any other way around it lol
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is it possible to tile a square with noncongruent right isosceles triangles?
yes
oh i meant a finite construction is also possible*
i referred to this website
ahh thats cool
is very interesting indeed 🤔
did u come up with this problem from ur own curiosity?
yea
oh that's fun
@blazing breach Has your question been resolved?
https://youtu.be/4HHUGnHcDQw?si=D_wStOJ78xOVJEg0&t=382
if you're curious about tiling stuff, i highly recommend to check out this story (starting at 6:20 in this video) about some fun and interesting tiling discoveries from last year : )
Quanta Magazine’s mathematics coverage in 2023 included landmark results in Ramsey theory and a remarkably simple aperiodic tile capped a year of mathematical delight and discovery.
Read about more math breakthroughs from this year at Quanta Magazine: https://www.quantamagazine.org/the-biggest-discoveries-in-math-in-2023-20231222/
00:05 Ramse...
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Can someone help me with these problems?
@heavy laurel Has your question been resolved?
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in this this system of equations:
$$ \left{\begin{matrix}
I. 2x+3y = a\
II. \ bx+12y = 100
\end{matrix}\right.$$
eitiel
find a and b given this set of equations have an infinite solutions
how do I start? infinite soulutions means I and II are equal to each other right? cause they are the same equation
Yeah
You might think of it as the second equation is a multiple of the first equation
Like here you can divide 12/3 to find the factor of multiplication between the two equations
Which is 4
Then by that you would be able to get a & b
@velvet saffron understandable?
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Just started learning about limits and continuity, found this to be slightly confusing:
for lim f(x) as x -> 3...
in the video it's said that the function is continuous.
but the function itself is not continuous correct?
only the part for where x approaches 3 the function is continuous.
please correct me where I'm wrong.
Because I did a quick search to understand what a continuous function is and what I got was a function that doesn't show breaks or disconnects/leaps.
which to me clearly isn't that.
if we look at the function near x=3, is there any holes or discontinuities?
Nope.
that would imply that at x=3 the function is continuous
but that doesn’t necessarily mean that it is continuous for all other numbers
Perfect that's what I thought as I said but had to confirm, sorry to waste your time. Thank you so much 🙏
You've helped me before as well, really appreciate your patience btw lol.
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im just a bit confused regarding the definition of homogeneity for ODEs. As far as I know there are three "definitions" (ill consider first order here)
1- if for y' + p(t)y = g(t), we have g(t) = 0
2- if for y' = f(t,y) we can have f(t,y) = f(y/t) = f(u)
3- if for y' = f(t,y) we have f(kt, ky) = k^n f(t,y)
@tranquil pine Has your question been resolved?
If 1 and 2 are precisely the same then f(t, y) = f(y/t) implies that p(t)y - g(t) = f(y/t) which would demand that p(t) = 1/t and g(t) = 0. But clearly we have more complicated homogeneous ODEs than just this one. So 1 and 2 are incompatible in general except in the one specific case.
Similarly 3 and 2 are incompatible.
I would imagine that these are describing 3 different senses of homogeneous, rather than 3 facets of one larger sense.
However, I couldn't swear to it. I do know that this was never brought up in my particular ODE class.
Weird. I can never get a good grip of what homogenous is supposed to mean in this case
with the different ways of mentioning it like here
Obviously they mentioned homogeneous itself many times, but I took the different usages to be just the homogeneous case of whatever particular solution case we were working on
It is interesting to think that they might be deeply related concepts, but I do not think they are
alright so i guess the question becomes what exactly do you consider the "homogenous case" of some solution
i mean surely there is some sort of relevance to the word "homogenous" maybe
Unfortunately, that's not something I'm equipped to answer, I took the engineering ODE class. 😦
I just learned homogeneous as the "nice and easy to solve" case lol
hmmm
seems like the term itself is totally arbitrary in that case then. Well, I will let this stay open for further input from others I suppose
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I do not know after I take three derivatives of YP, and the outcome is not same as the equation in red circle.
@dim brook Has your question been resolved?
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How do u solve q1 and 2
Do you know the period of sinx
yea 2pi/(1/2)
So it’s D
yes
I don’t rly understand q2
do u know which ones are definitly correct/definitly incorrect?
so CDE only has 1 correct or none, AB is the same
now A and B you can determine with the amplitude
and CDE u can determine by finding when sin(2pit/12-pi/2) is largest
For that do u have to also use the phase shift?
Or does phase shift only change the graph function
yea u need the phase shift because the time
it will move the graph pi/2 to the right
wait not pi/2
it will move it x amount right where 2pit/12 = pi/2
@shell panther Has your question been resolved?
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am I the only one whose brain can't seem to comprehend this problem?
"is it likely that the guaranteed condition is verified", I mean I'm fucking looking at test data that doesn't meet the condition
what am I supposed to do here even?
have you done hypothesis testing or is this just supposed to be a 95 percent confidence interval
this is part of the confidence interval section so I assume we use confidence interval
I already know how to conduct a hypothesis test and build a confidence interval
but what am I supposed to be testing against here anyway
i see
well it is asking for us to test whether or not the “guaranteed condition” is verified
it's not we can see it
or rather the alternative hypothesis would be that mu is less than 5
how does mu being less than 5 prove anything
well we don’t know that unless we do hypothesis testing
it does look like it doesn’t
but we can’t know for sure
if the "average" is less than 5, that doesn't gaurantee that some values are not higher than 5
hence the test will be usless
to show that most of them is less than 5
since we are doing it with 95 percent confidence
it will be even more accurate
where 95 percent of the samples would fall under the condition
but for the sake of what the question is already asking "is it likely that the guaranteed condition is verified?" isn't hte question already answered
supposing that the hypothesis test proves our alternative hypothesis
well that is more intuition
for stats we usually need statistical evidence to back it up
we can’t really just say, oh one value is greater so it must not
a sample can have its outliers
if there exists one value which does not meet the condition, then the condition is not verified
if there exists outliers then by definition the condition is not met
the condition is "it is gauranteed to be less than 5"
well the thing is, we are doing this with 95 percent confidence
not 100 percent
so there will be some samples that might not do something like that
but ig the question is a little extreme by saying that it is guaranteed
I don't think we're on the same page here
regardless, we have to do hypothesis testing or a confidence interval whether you like it or not
the very context of the problem doesn't need a hypothesis test
you don't disprove mathematical theorems with hypothesis testing
for "for all ...." theorems, you can disprove them with a single counter example
same with here
note how the question also asks if it is likely that the condition is verified, and not if the condition is verified
so just because there is one outlier that doesn’t satisfy the condition
doesn’t mean that it is unlikely that is satisfied
what you’re going at here, you should probably talk about with your teacher
they will most likely have a better way to counteract that than i do
whats up
sorry but I'm still not convinced
what you said up there is true that, yes the condition is not verified because there is one counterexample
but it isn’t asking if it is, rather if it is likely that it is true
when we look at a sample we assume all values are sampled from the population
non of them are "false"
whether they represent the bigger population is what we're testing
but in an existence proof, just one single value from a sample is enough
as in this case
again, that is verifying whether or not the condition IS true, not whether or not it is LIKELY to be true
I might be stupid but I don't see how there is a distinction between the two in this context
let’s say this, from the sample that we got there, there is one number above 5 which disproves the condition about the entire population
but this is just one sample out of the many samples we could’ve taken
and maybe a majority of these samples don’t have a number above 5
then that shows that we got a sample that has a low probability of occurring that disproves this condition
meaning that it is more likely that the condition is true than it isn’t
based on the sample we’ve taken
like if we took multiple samples, there might be a higher chance that i get a sample that doesn’t have a number above 5
which shows how it is more likely to prove the condition than it isn’t
doing it this way proves whether or not the condition is likely to be true
but doesn’t necessarily mean that the condition is true
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i am lost
the divisors is the prime factorization of n^2
$n = p_1^{k_1}p_2^{k_2}p_3^{k_3}....p_n^{k_n}$
Derivative
@tranquil pine Has your question been resolved?
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how do i use telescoping series for this
this is what i got, tell me if i got it good
$\lim_{n\to\infty} 1 + \dfrac{1}{2} - \dfrac{1}{nn!} - \dfrac{1}{n!(n+1)} = \dfrac{3}{2}$
Derivative
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urgent need to give back assigment
homework assigment
ah i used helpers tag for previous question i asked but then i asked another question and used again
i found the answer to my first question but to that question i did but wanted to make sure it is right
ye i know but scroll up i asked another question
ok so i broke the rules my bad
yes thanks for the explanation i wont do it again
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From personal experience is linear algebra or topology more fun(and or useful)? I don’t know which I want to take next year, I’m finishing up multi rn
linear algebra is like 50 times more useful and topology may/may not be more fun depending on how analysis heavy it is
Linear algebra is very useful
I'm wondering how you're getting to decide between Lin Alg and Topology though
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Suppose we roll 4 6 sided die. What is the expected largest integer to appear on any of the four die?
I can imagine enumerating and finding an answer, but that’s too tedious obviously. I wanted to used linearity of expectation, but I’m not sure how to apply it
It may be neither approach
I can’t think of any other way so maybe I am just supposed to count?
So to find E(2) we might say
WLOG let one die roll 2. then the other 3 have 2 options. We then need to multiple it by 4 for each one we fix, Which I feel like tells us
E(2) = 24/1296
wait that overcounts
Tbh I can’t think of any other way (there might be one) so I might just count all possibilities
We can count the probability of 1 with (1/6)^4
And the probability of 2 with (2/6)^4-(1/6)^4
hi. I'm new here. Why are there so many different help channels?
you can get your own help channel by going into one without a name next to the number and typing
I’m not confident I understand why
np
For the highest number to be 2, each number must be 2 or less
Which is (2/6)^4
But this over counts the possibility we got all 1s or less
So we subtract (1/6)^4
oh you’re right
Also am I just supposed to put the one with the highest expectation? Seems obvious but I just want to make sure
I’m not sure what you mean by that but you just add all the possibilities
the problem asks what is the expected largest integer that appears on any of the four dice
Can I be certain that evaluates to an integer?
No
But expected value doesn’t have to be an integer
If you wanted the integer that would be most likely to occur
It would be 6
My main worry is I’m not sure if my problem is a normal EV problem. It might be but it’s wording says “compute the expected value of the largest integer that appears on any of the four dice”
Expected value is just an average so it should just be normal
The expected value of a dice roll is 3.5 and that’s not an integer
So it shouldn’t be a problem
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@livid schooner Has your question been resolved?
You're using the form of the Lagrange error bound that has M
$|R_n (x)| < \frac{M}{(n + 1)!} |x - x_0|^{n + 1}$
south
So here M = 128/3
wait how did you get M
Now, if we want to find the error of the second degree term, what must n be?
pls continue
Yeah so this
Second degree = x^2
so the x part is 1/4 I assume?
Not quite
Remember -1/4 < x < 1/4 cause that's your radius of convergence
ok
so what will x1-x0 be?
cuz n=2 since it is f prime prime but Idk what x1-x0 be
It depends on the x you have within that interval
So what other variable must be 1/4 then
wiat so since x is between 1/4 and -1/4, so 0?
You mean x0 right
Yes
Right, so what's the maximum value that |x - x0| can be
There's no x1
x is just any number between -1/4 and 1/4
oh ok I get it
Yep and if you can figure that one out, you just need to figure out what n is, given you have the Taylor series to x^2
I'm not sure if you've learned this, but the maximum error of any Taylor series is just the next... (fill in the blank)
the next term in the order thing
like u expand the entire thing out
and the next one is error
Yes, it's the next term
So you know n then
yes
Mhm, the Taylor series is constructed in a way that it's the best polynomial approximation for a polynomial of degree n
So it's accurate to x^n
And so the next precision it may not be accurate to is x^(n + 1)
oh so that's why it's the error then
But many functions such as sin x = x - x^3/3! + x^5/5! or cos x = 1 - x^2/2! + x^4/4! have some coefficients equal to 0, so that's an upper bound for the max error, as in if you have sin x stopping at x^3 then the next term is x^5
Sometimes the max error may never be that big at all
I literally remember the sin one is 1-x^3/3! that it is all odd numbers
and cos is all even numbers
ok
Yeah that's actually where the names of even and odd functions come from
Even and odd powers of the Taylor series
I did not know that but damn your knowledge is impressive
Yeah it is haha
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The factorial is throwing me off
see if every next number in the sequence gets smaller by a percent? idk
@lyric quiver Has your question been resolved?
Do you know the definition of factorial
What techniques have you tried
I tried graphing / looking at the plots. Table graph, it looks like it goes to infinity, not to a specific number
I thought it diverges, but the answer key is almost stating that it converges? Not sure if I am reading that wrong or not though
Usually a factorial is more powerful than an exponent too
So to me, it's almost as if it's saying infinity / bigger infinity = goes to 0 ?
I asked for the definition
You might be able to get the answer through inspection, however I would use the ratio test
Ah, we have not learned about the ratio test yet
But this looks like... I could use the telescoping sum?
I am afraid I can't help you
That's ok, thank you for trying @silver maple
seems like a pretty perfect fit for ratio test
weird problem to assign without that
oh im stupid
nth term
Ah ok hmmm.. I'm trying to plug in the answers and see if I get any hits that way 💀
i mean
just consider the limit of the individual terms as n->\infty
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can someone tell me how to solve fractions equations
<@&286206848099549185>
Example?
There are some tips and tricks in fraction equations
For addition
I think it's proportion
You can use cross-multiplication on proportions.
oh ok
5/7 = x/35
what is this /
It means fraction
It can represent a fraction bar
oh ok
We cross-multiply to get 7x = 175
Then we can cancel out the 7 in the left side of the equation
by dividing
So it becomes 7x/7 = 175/7
x = 175/7
Therefore x = 25
can make in a pic so i can understand more
Wait for a bit
ok
Kinda messy, but it is what it is
As you can see here I’ve gotten rid of the fraction by multiplying evry term. Then just simply expand an solve, no fraction. (This is obviously more complicated because you can’t cross multiply)
So I always just get rid of the fractions then I solve
So what do you do to simplify what is the first step and the final step do you always divide ?
hmm
Yep, just cross-multiply, cancel out, divide, and you're done
Sorry if I got it more complicated 😂
rlly ?
Yeah!
no u didn't but i just want an easy method since tom is a test yn
If your not gonna come across the ones I did don’t worry bout it
so i'm going to solve a fraction and you tell me if i'm correct
the fraction is x/5 = 4/7
5x4=20
and then i divide 20 with 7
am i correct ?
or did i miss some steps ?
Yup ur correct
so what is the answer ?
oh
I saw the teacher looking for the lcm with fractions do i do the same thing to those type of fractions ?
and something about terms not sure
Ok
So teach was talking about adding and subtracting fractions which is far different
For that you need a common denominator
I’ll show you
ok
Sorry did you want that with variables or without
For multiplying and dividing fractions simply make multiply the numerators and multiply the denominators.
Sorry for division you do literally the same thing as multiplication but first you flip over one of the fractions then multiply them
Which ones? Like they equal something and you have to solve for x?
Oh
Your teacher just multiples everything by 15 because you can’t get rid of all the fraction in one go that’s all
If you need more help just ask I can’t help with much but I can help with this
idk how to solve those type of fractions
<@&286206848099549185>
Uhm
What is the problem
could u help with how to solve those type of fractions
6x=28 x=28/6
So it's easy
steps ?
Let me show you with a draw
Because my English is poor and cost to explain
ok
One moment please
take your time
Just multiply 6/7 by x and 4/x by 7 which is = 6x-28/7x=0
bring 4/x to the other side first
Goodbye
how about divison ?
thx
Step 1: bring 4/x to the LHS which is: 6/7 - 4/x=0
Step 2: u wanna combine them so u multiply 4/x by 7 and 6/7 by x which is: 6x-28/7x=0
Then u gotta get rid of the 7x
U multiply 7x on both sides
And since the RHS is 0, then 7x*0 =0
So 6x-28 = 0
Then bring 28 to the RHS and divide it by 6
what's the answer?
x = 14/3
no is 27/8
Huh
Did u find this answer or was it in the book?
Cuz I’m pretty sure it’s 14/3
my bad it's 45/2
Ohh u meant this question?
for which one
yea
bro i'm failing my test 💀
💀💀💀
i'm cooked
Do yk how to do this one?
that was the teache answer
today
Oh do u. Need help?
u can do it another way
I find it easier
why my here then lmao
🧍♂️
HAHAHHAHAA
okay okay
So like what u can do
I’ll draw
my test is on fraction and equations , 1 step equation 2 step equation .linear equations
bro..
Damnn
ya'll what do i doo
U can study during ur other classs
i can but it's gonna be useless since i don't know how to simplify equations or fractions equation
@brittle mesa what grade ur in
Wait a sec
Huh
my i......1.1.1.1.1.1.1.q is prob 5
Chillout💀
oh ur answering this question
3k/5 - 6 = k/3
What else😭😭😭
no
hahaha
hmm i'l try my best pls wish me gud luck on the test
@brittle mesa have you failed any courses ?
Nope
brooo ur smart
💀💀💀
so luck i'm pretty i could understand if u follow my words
so what is the first step like the first step to solve those kind of fraction
Ok so the first thing I did
broo u aren't a helper sorry i took so much of ur time
It’s ok lmfaooo
u don't have school ?
Holidays
ur in the us ?
I’m French
oh bonjour
…
lmao
.
Bring all the fractions together
So to do that
U need to put 3k/5 and k/3 to the same side
Which is what I did in the 2 first steps
use the bot pls
oh ok
Which one is it?
idk
Idk what else to do
Call a helper or sum💀
Way more qualified
My teaching skills mid
bro it's not u
i just lose braincells from learning math
@brittle mesa thank you
Hahahaha
It’s better if u ask ur teacher to explain in detail what u didn’t get
Face to face is more practical
too late
my friends are stupid
you know what idc what i get on the test but i rlly need to improve
cause my average is 33.38 %
alr later
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If a company makes x hundred A tires and y hundred B tires.
0=<x=<4
y = 40-10x / 5-x
The profit on grade A tires is twice that of the profit of B tires Find the number of each tires to be made to maximize profit
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Hello!
If I have the following
I'm well aware that I can do
ln( (1-x) / (x+3) ) = ln 1
e^ln(1-x/x+3) = e^ln(1)
(1-x) / (x+3) = 1
1-x = x+3
2x = -2
x = -1
however, say I didn't want to do the step where I put the ln terms together
what would that look like
would it be e^ln(1-x) - e^ln(x+3) = e^ln(1)
or would it be
e^ln(1-x) - ln(x+3) = e^ln(1)
e^(entire rhs) = 1
lhs i guess you mean
Yes
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Can anyone once again explain where and how multisets are used?
can you help me with question 4?
And can be used in regular notions of set
Some folk answered but ig I was busy and till I came
Channel vanished
Lol I am not eligible to answer ur question
You can ping helpers after 10 min ig
Channel occupied
im having exam tomorrow and im having hard time understanding them
can someone help me?
You should ask this in another channel
which channel?
<@&286206848099549185>
@tranquil pine , sorry, but I don't know so much about multisets
Good luck!
No issues bruv
Even I don't 🫂
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@tranquil pine Has your question been resolved?
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yo
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So I'm working on part 2, I understand it's supposed to be trivial, but I don't really understand.
jan Niku
I'm not understanding the next steps from here though, in the proofs that I'm reading
We do want to introduce a pointwise limit, right?
@mint orbit Has your question been resolved?
@mint orbit Has your question been resolved?
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i dont know how to work this discord server
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Go to the channel and read.
while youre here please help @neon slate 
i would be happy to but this is way out of my depth 😄
i dont know anything about analysis
@mint orbit Has your question been resolved?
@mint orbit Has your question been resolved?
Where does this f_m come from do you know
cauchyness
Oh it’s just some other function in the sequence
I’m not sure what $|f_k-f_m|\leq ||f_k-f_m||_\infty$ means
Frosst
It’s a function on the left and a real number on the right
How can I have inequalities between those 2 things
The infinity norm is a supremum
like if you have $||f|| = a$
jan Niku
then almost everywhere, $a\geq f$
jan Niku
or, $a$ is a supremum of $f$ except on some set of measure 0
Oh as a constant
jan Niku
sure
Ok
@mint orbit what are you struggling to get in this question
im not certain how to start
What about this
I think you’re just supposed to stack the Cauchy on the uniform convergence
Cauchy tells you you can squeeze that norm as tight as you wish
Then that tight bound is the same as picking any epsilon to show uniform convergence
thiis proof doesnt seem like a proof
i do not understand the one that i linked at all i just thought it might help bait someone to help me to be honest
i know the definition of cauchy
i just dont follow the rest of this proof
the proof i showed in that image seems to just jump from being cauchy to being convergent which is the primary meat of the proof and the part i dont understand
does that make sense
can u delete this and move to an open channel?
ok
Heyyy
where are you stuck? (if you are still stuck)
I guess I don't really know how to start, beyond "take some cauchy sequence"
I get the framework is like
#1 you need to find some candidate function that this cauchy sequence converges to
you'r following this?
#2 you need to show it does converge to that
i posted this to try to improve response rates
like "look the problem is easy"
i dont understand that proof at all
Convergence in sup norm is uniform convergence iirc
sorry its been a while since i took this
I was working with another person in my dept
he suggested that i like
so take cauchy to mean that we can find some k such that $||f_k - f_j|| < \epsilon$ given $k < j$
jan Niku
jan Niku
my analysis is not strong enough to understand the 4 or 5 jumps the author makes in that proof
Which jump specifically?
the proof just ends
it seems like it barely starts
I guess I need more details than this
like
i dont understand whats going on with the union
and then he just asserts that it converges uniformly but doesnt show it?
and then the proof just stops
the rest of this document is about p < infty
id like to try to track this out because i at least barely understand it
Ok lemme write out the details (or attept to rn)
by the way
does that step make sense
The limsup has to move closer to the fixed f_k
so you can use it as a candidate since its at least less than epsilon
but I'm not sure how to resolve this outside of some fixed k
Like the dream is that you just say
okay take $N_j$ to be the set where $|f_k - f_j| \geq \epsilon$ for some fixed $k,j$
jan Niku
then call $N = \bigcup _J N_j$ has $|N|=0$ and on $N^c$ you have that $|f_k - f_j| < \epsilon$
jan Niku
now it makes sense to propose the candidate $|f_k - \limsup _j f_j| < \epsilon$ because $\limsup _j f_j \leq f_j$ i think
jan Niku
and well the limsup will be within the epsilon ball
but its not clear to me how to resolve the epsilon here, which is fixed because of the fixed k
once we want to involve other k, presumably, to get convergence
I think i got it
iirc it depended on completedness of reals
but im just doublechcking if im a crank now LOL
yea my prof literally said
'key ingredient is completedness of R' ill send pic
its not enough
limsup isnt guaranteed to converge to f_k
it will only converge along j,k
the limsup might actually eternally be some non-negative distance away from f_k, we can't actually bound it without un-fixing k
oh i didnt do it this way oops ill check again tho
Ok so the only problem part here is
you want to pass to the reals at the end of the proof
All this Z_n,k nonsense is doing is carving out the bad set wher bad things happn
So you have:




