#help-49
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well what does fermat's christmas theorem say about 3 mod 4 primes?
if p | a^2 + b^2 then
nothing that i know of
oh ok wait lemme pull up the statement
oh ok maybe it's not bundled together with the statement of fermat's christmas theorem
but yeah a common thing is that if p | a^2 + b^2, then p | a and p | b
this isn't too hard to prove
but basically use this to conclude that ugly numbers are those whose prime factorisation does not contain a 1 mod 4 prime
this completely characterises ugly numbers and finishes both a and b
btw do you mind if we move to dms to not worry about the channel closing? our timezones are so horribly mismatched theres only really a 2 hour overlap between us ralking :(
yh sure
ok so basically since k isnt divisible by a 1 mod 4 then neither does a and b
as in try to use this
let's say $k = 2^t p_1 p_2 \cdots p_l$ where all the $p_i$s are 3 mod 4 primes
LY
then we have $2k^2 = a^2 + b^2$
LY
so $p_t$ divides $a$ and $b$
LY
can you carry on like this to finish?
(these aren't necessarily distinct primes)
ok nw!
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How do I go about this
$\frac{df^{-1}(a)}{dx}=\frac{1}{f'(f^{-1}(a))}$. also, you're given that $f(4)=6$, so $f^{-1}(6)=4$, and $f'(4)=\frac{1}{5}$
fish
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nyone know the answer to this?
if P(t) is a matrix of continuous functions, then the wronskians of any two fundemental sets of solutions of x' = P(t) x are exactly the same. True or False
i got true but gpt is telling me its false
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!status
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
1
well all primes past 2 are odd, so if there are any triplets like that, they would all be odd numbers. so write down some potential candidates.
for example: 7, 9, 11
what's wrong with that triplet? 9 isn't prime because it's divisible by 3
okay now try 11, 13, 15
notice a pattern starting to emerge?
No
15 is not prime
what number seems to always mess the sequence up?
in each of these triplets there is a number that isn’t prime because it’s a multiple of what?
3
suppose you have such a prime triplet, then the first number of the triplet must be either 1 or 2 mod 3
what do you know about one of the other two numbers after it mod 3?
@tidal turret do you know what mod 3 means
it’s the remainder after dividing by 3
so a number that is 1 mod 3 is a number of the form 3n + 1, where n is an integer
your first number of your prime triplet, assuming there is one, has to be either 3n + 1 or 3n + 2
if it was 3n + 0 or 3n + 3 it would be divisible by 3
contradicting the fact that it’s prime
Ye
then each number after that is 2 more than the previous number
so suppose your first number is 3n + 1
the next number is 3n + 1 + 2
which can’t be prime because …
Div3 remainder0
right
then suppose your first number is 3n + 2
the next number is 3n + 4, nothing wrong so far
the third number in the triplet would be 3n + 6
Id there any proof
is there any proof of what?
Only triplet 357
well yeah it’s the proof i just gave you lol
suppose there’s a prime triplet with this construction, then you’ll always find that one of the numbers is divisible by 3 and therefore not prime
the one exception is the triplet (3,5,7) in which case yes the first number is divisible by 3, but is clearly prime
3 is the only multiple of 3 that is prime
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the red 2y at the bottom doesnt exist in the solution im wondering why?
how do we get rid of it?
You should have du/dy = 2y
u is a function of y so chain rule
Which implies du = 2y dy instead
So that gets replaced: you dropped the dy
In the integral
so the 2y just gets replaced by du/dy?
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how do you solve the integral of 1/sqrt(x)?
Power rule for integrals
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If x²+y²+z²=20
and x+y+z=0
Is there a solution?
yes
as long as when they are added without the sqrt they = 0
like you could have x=1 y=-3 and z=2 and when they are squared they will equal a pos number
yes. one is a sphere with the radius of 2sqrt5, while one is a plane cutting through said sphere
so according to what fungus said, then any solution is where they will intersect eachother
im not too sure whether this is abstract thought or you are doing 3 dimensions, but either work
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maybe not really a math question but do you guys learn math better with lecture notes or actual books?

I only use notes that I make by myself from books tbh
memory from explanation 👍
how do you refresh your memory then?
aahh then yeah
i tend to actually internalize the concepts better during the lecture if i don't take notes
i can't say the same to everyone else
My advice would be to, just write formulas and create a formulae list, and look to proofs from books whenever you are unable to recall them.
if i understand the concept it will remain engrained in my brain
idk i have something like a logical memory
something i made up
that stops working once you get to incredibly dumb stuff to memorise
probably
I just remember how to derive them instead 🥲
exactly
to this day I still don't remember what's the derivative of cot(x) 💀
it depends, i learned calculus really well from lectures and i learned geometry almost solely from books, but i think you kind of need both
What are math notes
lecture notes eventually just become books so i don’t distinguish them too much
facts
I only make sst notes!!
literally all my notes = the book but keeping on the necessary and non trival stuff
here's an example of a set of lecture notes my professor decided to use in one of my topics classes, this set of lecture notes is what we used but it really reads more like a book with exercises and everything https://arxiv.org/pdf/1011.1690
so is there really gonna be that much of a difference later on? not really
i think the best thing that you can do is go talk to people about what you've learned, and then it won't matter whether you got it from lecture notes or from a book
what kind of subject are you trying to learn? we also have #book-recommendations if you are trying to choose something to read
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✅
Aren’t they too 😭exhaustive most of the time. Like imagine going over 500 pages
I have time
I'm only doing maths
and one other subject evry sem
Well, 2
but yeah
I have time
just work 9-10 hours a day( assuming you only have 3 hours of classes a day)
Ohh i see, interesting. So you’ve lost been using lecture notes most of the time and then if necessary you find exercises from relevant textbooks?
or is this only relevant when you’re at a moderately high level of mathematical maturity
reading math books isn't like reading a novel
you are supposed to spend months on it
Yes uh but well eventually it’ll take longer than just a course on it? Idk unless i’m doing smthing wrong
For example, i’m guessing you guys are aware of the mainstream calc book: stewart
u have like 1000-2000 pages on that book
it’ll probably take ages to go over the entire thing

I also don't see how are lecture notes less time expensive than books. Either they cover a thinner topic, which could be solved by only reading the relevant topic in the book or they are written more densely, which makes you spend more time on understanding the material. (A book would arguably be a better option in this case, because if someone explains it to you it's easier than thinking about it on your own)
yeah, stewart is bad, like very bad
get a better book imo
I used it for one sem, I hated it
took me two months to do stewart, thomas is better though
dont take everything by the number of pages
the content matters alot
but most books are still pretty big right? Idk for me it takes over 5-6 months to go over a book
This is longer than me just working through an entire course on the subject/topic
who knows if the actual content is in the first 100 pages and the remaining 900 are acknowledgements
depends on the book
Lol
for example stewart also covers multivariable calculus
some books will take 6 months, It took me 4 months just to do a third of LADR
yeah see so that's what i mean
not only single variable
courses are often longer, often miss some topics and also don't bring as much clarity as a book does, having a reference book with a course is always good
which is like 2-3 courses in uni
I don't have an year to spend on LADR for example if i need to get it over with in 1 semester or less
😭
i'd rather read 500 pages than attend all lectures of a course
no don't universities offer lectures notes online
Just do the relvant topics from ladr
like you can google "lecture notes ...."
depends on the uni
so i was curious if those are better
more concise and maybe use the books for the problems?
But also that raises the question
LADR + strang + david c lay is doable in 4-5 months
on what problems should i be doing
strang took ages for me just for the first few chapters
including dual spaces?
maybe cuz i was doing all the problems
are you doing all the problems
yes
and strang's book doesn't have repetitive problems like stewart's for example
so i thought it'd be instructive to do all of them?
honestly you should decide yourself
but like okay we all have time constraints so i'm looking for a practical alternative
Hmm, maybe just do the first 5 problems from each exercise in LADR
you should find what you prefer and choose it
i can't invest years of my life, for example, just learning linear algebra when i need it solely for a foundation for another overarching goal
depending on how deep you want to go
maybe that makes sense?
my bad i didnt consider that i already had a first course in LA before starting LADR so it was quicker for me to finish all three books
LADR is a 2 sem book if you're a beginner IMO
it makes sense but at the same time there is no method that leads to this
whether using a book or lectures or whatever
you wont take years in learning linear algebra
how many LA courses does your university offer?
is that axler?
i'm not in uni
yes
okay so my idea was strang -> axler
I'd prefer axler+ strang as axler avoids dets
strang also avoids dets
it's generally two so you can use strang for the first course and start LADR towards the end of the course
dets are really important
so okay how many problems do you guys do in general? Or rather how do you choose which problems to work on for a given book?
I used axler against my prof's reccomendation for the first course 
Around 6 a day if the topic is new
like I'm currently doing RA
based
hardly get 3-4 problems done a day from abbott
does LADR have any prereqs? u need to know proof writing?
Yes
so wouldn't the book take ages?
😭
even in your case
I'm doing it a semseter early
so I have time
I practically have a year before I take RA
well, 8 months
what all did you take so far?
I've taken calc 1, LA 1 and proof writing
yes
technically not even integration is covered
and epsilon-delta limits
cuz in the US they call it calc 2
yeah fair
I tried to use spivak for calc 1, but that didn't go well 
that's rough
i was halfway through the strang book
and half way through multivariable calculus
but then i dropped all of it and just decided to do a bit of comp math so i'm revising precalc and stuff
and i constantly get discouraged by how long a book takes
😭 i guess it's normal
whoops
serge lang's LA
anyway so for this
what do you do that allows you to maybe work thru all that in 4-5 months?
u don't do all the problems?
yeah i skip problems that look doable or similar to the ones i have done before, problem overlap over all three books is huge of course since all three books have the same primary focus that is linear algebra
definitely
that sounds scary
okay well i just gotta go with the flow i guess
thanks guys
i really haven't used strang tbh, only used it for transformation, but im sure it's a really good beginners book judging from the way he teaches
It's a bit verbose
But if you do all of his questions u have a good handle on that topic
generally
i see
langs LA is a bit too much... well FIS is a good one
yeah i think i agree with that, haven't read FIS till now
FIS is a big book ,I'm currently using H&K's book
@gusty falcon Has your question been resolved?
ideally one would ask for help before you’ve spent 8 hours on a single linear algebra problem
😭
what did you do?
You'd work through books/lecture notes?
yes, working through books is good
being unable to do exercises is usually a sign that you didn’t understand the material
so they’re good litmus tests
i have spent a long time on some problems but in things like LADR (or LADW if you prefer) you really should be pairing material from the book and exercises, and maybe a lecture series, and ask for help when you need it or feel like you don’t understand a problem or are stuck on an exercise
again the exercises in LADR aren’t as scary as some would make it out to seem
I see
Can you do calculus and linear algebra in tandem?
Like together?
Or do both courses need some kinda proof writing experience

you can
what kind of calculus do you wanna do
like intro real analysis?
or just like straight up calculus
midway between the two
ladr will have proofs in it for the record
not intro real analysis cuz i've heard you need proof and a lot of other things to get started
you’ll pick up on how to prove things as you go
but also not high school calculus because i know most of the algorithms lol
and hs calc is just mostly algorithms
you don’t need so much extra stuff, knowing what a proof by contradiction (and a few other things) is and a willingness to spend time internalizing definitions is probably enough
eh
i’ve gone back and forth on my opinions on it
everyone i asked said it was a bad idea
maybe it depends on what book do you take for RA honestly
and at the very least try to get a solid mastery over calc (so i thought of just revising calc again)
abbott?
I think Abbott is pretty friendly to go into even if you don't have any proof experience
or is there a lecture series i can follow 😭
oh so no prereqs?
there is no harm in shrimply trying, and if you really dislike it or feel stuck or frustrated try to gain a better foundation as needed. if you don’t like that idea then you can try an intro proofs style thing or philosophical logic (<- underrated!) style thing
omggg philosophy again 😭
yeah or you can read abbott and see how you like it
have you done any philosophy?
no actually
philosophy just reminded me of that probability question we both worked on
😭
as long as you don't take Rudin as your first intro to RA it's not that bad
it’s good to know some logic, the idea of fallacies, and deduction using formal logical rules, and some history behind it (personal opinion, i think it should be required instead of an intro proof class, though maybe some people would get mad at me for saying this)
no i don't really want to jump into actual RA cuz i guess i can't handle that yet
but at the same time i don't want a calc book either
smth in between
the only in between one i know is spivak
but ... i guess that's very hard in itself 😭
Apostol has some good proof based clac book
you could try reading a non-proof based linear algebra book
non proof-based*
sorry
its a two volume long book and the second volume contains some multi var calc stuff +LA
i always put the hyphen in the wrong place
Strang?
Yeah strang has a bit of proof
which were great
Oh
I see
Is that the "in between"
also is spivak the "In between" too?
Yes
yes to both questions?
there are some books which are more friendly intro to RA but their difficulty can be considered as in between
yes
that’s intro to RA?
i mean spivak
or is it a bit lower level than that
don’t really want an intro to RA book cuz no proof writing experience but yeah
It's a great book
that's fair then again some people have their first intro to proofs by RA or LA
what makes spivak hard?

yeah i think i’ll work through that 
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A triangle ( \triangle ABC ) is inscribed in a circle with a radius of 4. The length of side ( AB ) is 6. Side ( BC ) has a length of ( 4\sqrt{3} ) and is the longest side of the triangle.
Calculate the length of side ( AC ) of triangle ( \triangle ABC ). Why can't I use the law of cosines right away?
john3515
although i think it would only help if Δ was 0
@obtuse wing Has your question been resolved?
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Do you know stars and bars?
@paper whale Has your question been resolved?
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What did I do wrong?
Why would it be wrong ?
,w dy/dx = (x+1)/y
Well wolfram agree also
yo when ur done can u pls help
You're welcome
.close
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Bro
sorry
You opened a channel
i didn’t see the channel
Its ok
.close
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just a question I thought of: If an event happens 10 times every day (at the same times) what is the average time a person would have to wait for the next event? is it just 24/10 hours? or could it be more complicated cause a person could have to wait for the next day for the next event
it would be more complicited
imagine all 10 events happened right after each other
then the expected wait time would be 24/2
as in basically the same time
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@tidal turret Has your question been resolved?
@tidal turret Has your question been resolved?
can you translate?
is it find all vectors such that the image of these vectors is 1,3,2?
and find a first?
Let $B = {(-1, 2, 1), (1, 1, 0), (1, 0, 0)}$ be a basis of $\mathbb{R}^3$, and let $f : \mathbb{R}^3 \to \mathbb{R}^3$ be the linear transformation such that
[
M_{BE}(f) =
\begin{pmatrix}
a & 2 & 1 \
1 & 2 & 1 \
2 & 2 & 1
\end{pmatrix}.
]
Find $a \in \mathbb{R}$ such that $f(1, 2, -1) = (1, 3, 2)$, and for the value of $a$ obtained, find all vectors $\mathbf{v} \in \mathbb{R}^3$ such that $f(\mathbf{v}) = (1, 3, 2)$.
938c2cc0dcc05f2b68c4287040cfcf71
Here is the translation sir
Ye
Exactly, sir
,w inverse {{-1,1,1},{2,1,0},{1,0,0}}
,w {{a,2,1},{1,2,1},{2,2,1}} * {{0,0,1},{0,1,-2},{1,-1,3}} * {{1},{2},{-1}}
Okay yeah so a = 3
Ye
Look, if i multiply M_BE(f) with M_EB(id) i get M_EE(f) and then i multiply by (1,2,-1)
You see what i mean
Yes yes I understand
M_BE(f) is the matrix representation for f in the basis B
so for the normal basis, first multiply by the inverse of the basis B to get your vector in terms of B's basis vectors
then multiply by M_BE
as your final transformation
so now you need to solve this matrix equation
And what is M_BE(id)?? That's the vectors of B placed as columns in a matrix!!!
Problem what is M_EB(id)?? That's the inverse of M_BE(Id)
,w {{3,2,1},{1,2,1},{2,2,1}} * {{0,0,1},{0,1,-2},{1,-1,3}} * {{x},{y},{z}}
$$ \begin{pmatrix}
3 & 2 & 1
1 & 2 & 1
2 & 2 & 1
\end{pmatrix} \cdot
\begin{pmatrix}
0 & 0 & 1
0 & 1 & -2
1 &-1& 3
\end{pmatrix} v =
\begin{pmatrix}
1
3
2
\end{pmatrix} $$
vehnil
$$ \begin{pmatrix}
3 & 2 & 1 \
1 & 2 & 1 \
2 & 2 & 1
\end{pmatrix} \cdot
\begin{pmatrix}
0 & 0 & 1 \
0 & 1 & -2 \
1 &-1& 3
\end{pmatrix} v =
\begin{pmatrix}
1 \
3 \
2
\end{pmatrix} $$
938c2cc0dcc05f2b68c4287040cfcf71
ah do they just not show up in text
weird
this is what you need to solve right?
Ye
v is a generic vector in R3
,, \begin{cases} x + y + 2z = 1 \ x + y = 3 \ x + y + z = 2 \end{cases}
okay yeah
938c2cc0dcc05f2b68c4287040cfcf71
best bet is to multiply the two matrices together
and then do rref with an augmented matrix to find the entries of v
and seems like you already have the product of the mtrices together
,w rref {{1,1,2,1},{1,1,0,3},{1,1,1,2}}
set either x_1 or x_2 as the free variable and solve
we know that x_3 = -1
x_1 + x_2 = 3
- x1 + x2 = 3
- x3 = -1
(x1,x2,x3)=(3-x2, x2, -1) = x2(-1,1,0)+(3,0,-1)
perfect
but you probably want to parametrize with a different variable like t
(x1,x2,x3) = (-1,1,0)t +(3,0,-1) for t \in R
its good work!!
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where did i go wrong?
(shut up i did not just write A^2C^2 LMAO)
@viral dagger Has your question been resolved?
Uhh, how did you get AC*BC = 72?
36 is the area of ABD, not ABC.
Yea, I kinda expected that youd get negative since you got the hypotenus AB as 6.
Like, AB has to be longer than 12
i mean is it not
no?
Due to the area being 36, theres a hard limit that BC<6
since AD gotta be larger than 12
Since AB is the hypotenuse, it has to be larger than 24
and that skews the ratio of AB/BC
Lets say AD = 18. So, we have BC = 4, and the angle bisector thm makes AB = 6, which is way less than AC.
So, AD gotta be even longer. Lets say AD = 36. That makes BC = 2, and angle bisector thm makes AB = 6 (still)
So no matter how long you make AD, the hypotenus is fixed to be 6, its not gonna grow any longer
Maybe, you can try to find a lowest possible area for ABD such that the problem has a solution
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Hi! $2^n$ plays a fundamental role in discrete calculus, similar to $e^x$ in infinitesimal calculus, it is it's own derivative.\
When solving linear homogeneous \textit{difference} equations do we use $2^{rn}$ instead of $e^{rx}$?
woomy
i.e, are the solutions to linear homogeneous difference equations of the form:
$$C_1 2^{r_1 n} + C_2 2^{r_2 n } +... + C_k 2^{r_k n}$$
Where $r_k$ are the roots of the characteristic equation for the DE
woomy
@gusty birch Has your question been resolved?
<@&286206848099549185>
@gusty birch Has your question been resolved?
@gusty birch Has your question been resolved?
Specifically I need help solving this:
I know the answer and if my above assumption is correct then its pretty much solved.
(of course the solution being the Binet's Formula and the sequence in question being the fibonacci sequence)
assuming deltas mean differentiation.
the characteristic equation you look for is r^2 + r - 1 = 0
the two roots of this will give you the solution such that f = c_1e^r_1x + c_2e^r_2x
which leads to c_1cos(x) + c_2sin(x)
with the initial conditions you will find the coefficients c_1 and c_2 and you are done
The deltas are differences not derivatives
Which are the discrete version of derivatives
correction cos(r_1x) and sin(r_2x)
if the function is continuous and differentiable you can take its laplace transform that gives the characteristic equation
Thats what my question is about, if i can treat differences the same as derivatives when solving ODEs
all I know is this
no, it's still e^rx, right?
wait no i'm dumb
My thought process for it becoming 2^rx is that since 2^x is its own difference, the reasoning we go through to build the solution for linear homogeneous ordinary differential equations should be the same, but instead of derivatives we use differences
ok ok ok
so if you have uhhhh
ok so let's take the fibonacci sequence
$f_n = f_{n-1} + f_{n-2}$
Kaisheng21
then that's m^2 = m + 1, so m^2 - m - 1 = 0
let m_1 and m_2 be the roots
then the general solution turns out to be $Am_1^n + Bm_2^n$ where $A$ and $B$ are arbitrary constants
Kaisheng21
oh so i dont exponentiate anything?
yeah it just works out
no i mean, the roots get exponentiated but they arent in the exponents
sure
unlike for ODEs
Kaisheng21
yea makes sense but its just weird that it kinda just works out lol
wait so
And now i just use the initial conditions to find out C1 and C2
yes
this seems different though
cause i have the first and second difference, you have different terms of the sequence
?
f_n-1 is the n-1th term of the sequence no?
can't you just turn the differences into the different terms of the sequence?
it'll end up as some recurrence relation in the end
okay yes sorry i was typing to write down my though process
yess thankyouuu!!
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The measures of angles $A$ and $B$ are both positive, integer numbers of degrees. The measure of angle $A$ is a multiple of the measure of angle $B$, and angles $A$ and $B$ are complementary angles. How many measures are possible for angle $A$?
938c2cc0dcc05f2b68c4287040cfcf71
what have you tried
don't know where to start
can you help me
write what's in the question as equations
A = 180-B
@tidal turret Has your question been resolved?
A= k(B -180)
@tidal turret Has your question been resolved?
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(K+1)B=180
how
A+B=180
<@&286206848099549185>
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Have the student work THROUGH the problem and guide them ALONG the wae
from here use the fact that k is an integer
king, can you guide me through
it might benifit you to write that k = (180/b) - 1
since k is an integer, the right hand side must also be an integer
@tidal turret Has your question been resolved?
why? how did you arrived to that
i divided both sides by b and then subtracted 1
from here
oh ye
so?
b has to be a factor of 180? or what
just make sure to subtract 1 because b =180 gives k = 0
,w divisors of 180
18 divisors
wdym
B(k+1)=180
(k+1)=180/B
k = 180/B - 1
k = (180/b) - 1
so 17 divisors?
yeah
answer is 17 then
i meant that literally plugging in b = 180 would give A = 0 which ig works but probably not counted in the actual problem
yes
measures are positive integers
oh ok
ohh noo
uhm
oh i see now
a and b complementary means a + b = 90
not 180
you can do the excact same steps with that now
whoops mb i should have noticed before
, w divisors 90
k = (90/b) - 1
great
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How would I solve this question:
ok
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from part a to part b, im just wondering how they relate to each other for c
cause these are two different graphs, normally when doing trapezoidal dont you compare the same graph to its exact value?
also im wondering why im differentiating y=xlnx
i dont see what relevance is has to b
nvm i understand it know, i hate this question
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Happy new year!
For c
Idempotent mean M^2 =M
So
M^2 - M =0
Which gives us M= 0 matrix
And M= identity matrix
matrix algebra is not quite the same as real number algebra in this respect
when you solve polynomials like that, you are implicitly using the fact that if ab = 0, then a = 0 or b = 0. but that's not true of matrices
Yeah
They gave an example on that
But I thought the matrix is similar to the real number algebra in case of polynomial solving
:p
Found an example
1 1
0 0
well it gives you some solutions (since it's still true that any matrix * 0 = 0) but not necessarily all of them
Is there a way to find an example without trial and error?
solve a bunch of equations
like $\begin{bmatrix} a&b\c&d \end{bmatrix} \cdot \begin{bmatrix} a&b\c&d \end{bmatrix} = \begin{bmatrix} a&b\c&d \end{bmatrix}$
math_rocks
That's gonna take a lot of effort 
so solve $\begin{bmatrix} a^2+bc& ab+ bd \ ca+dc & cb + d^2 \end{bmatrix}= \begin{bmatrix} a&b\c&d \end{bmatrix}$
math_rocks
you can also exploit the fact that $det(A) = 0$ or $1$
math_rocks
Ah
a notable example of idempotent matrices are projections (which you will probably learn later)
Det polynomials follow real number algebra
Okay
Thanks cloud,wai
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✅
Can someone check it please
Not sure about h
It is true for a 2x2 matrix
But how to prove it for an nxn matrix?
there is a property connecting transpose and inverse
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✅
Are the rest of the right ?
the last one is sus
Lop
Lol
$(A+B)^{-1}= A^{-1} + B^{-1}$
And since A inverse and B inverse exist ,A+B whole inverse also exists?
where did you get that formula?
it's not in the book for a reason
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Could someone explain the proof of the bolzaon-weistrass theorm
specifically this part
Let $(a_n)$ be a bounded seqeunce so that there exists $M>0$ satisfying $\abs{a_n} \leq M , \forall n \in \N$ bisect the closed interval $-[M,M]$ into two half intervals at 0. Now it must be that atleast one of thse intervals constains an infinite number of terms in the sequence $(a_n)$
why is this
math_rocks
OK
$-M \leq a_n \leq M$
math_rocks
If you got it why are you still here?
for all $n \in \N$
math_rocks
Be nice lol
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why is it wrong
to calculate no of toss needed for n on a dice like this
you forgot $- \left ( \frac{5}{6} \right )^n}$ on the left
Slayer05
Compile Error! Click the
reaction for more information.
(You may edit your message to recompile.)
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like n would be the solution to that equation?
Oh i see what you're trying to do
the RHS is the probability that you get a given number after at most n tosses
yea
well, is that probability ever gonna be 1?
You could e.g. keep tossing 1s and never toss a single 6
no matter how many times you toss the dice, you cant be certain that the given number appeared
the expression on RHS will only approach 1
so how do i recitfy the approach
wdym?
What exactly are you trying to calculate?
You could e.g. modify your question to be
"After how many tosses can I be x% certain that the number appeared on the dice"
or "How many tosses will I need on average to get the given number"
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"Determine the value d such that the plane
$x+y+z=d$ tangents hyperbolid $x^2+y^2-z^2=1$"
Merineth 🇸🇪
Could someone help me with this one ?
When doing exercises such as these, should i be thinking in 2d to make it easier for me to understand what to do?
So it would be looking something like this? Where the red line is the plane and the blue circle is the paraboloid
yes
And to my understanding, to solve this exercise it has something to do with gradients, normalvalues and them being parallel?
So we start with getting the gradient for the paraboloid like you mentioned
$\grad f(x,y,z) = (2x,2y,-2z)$
Merineth 🇸🇪
cool
So the gradient of the paraboloid gives vectors in the entirety of the space such as the purple arrows, and they point in the direction of the fastest rate of change. And someone said it's always away from the paraboloiid.
But we are only interested in the green arrow, right=
So we need to find the corresponding x,y,z values that match the intersection between the plane and the paraboloid, right?
Are my concepts correct so far? Like in my understanding?
I'd instead say, consider a point (a,b,c) on the paraboloid, The equation of the plane is then $z - \sqrt{a^2+b^2-1} = 2a(x-a)+2b(y-b)$
math_rocks
I think so, yes
I don't understand how you got this
That seems to be wrong
Considering that isn't mentioned in the answer sheet
given a curve, do you remember how to find the tangent plane?
I'm using that method
My understanding here is that i'm trying to find a vector that is paralell to the normal vector of the plane
$\grad f || (1,1,1)$
Merineth 🇸🇪
sure, that works too
Merineth 🇸🇪
The book says it's like this
I don't understand why the k?
Could this be the reason?
That when we solve for x,y and z giving us :
$x = 2/k\y=2/k\z = -2/k$
Merineth 🇸🇪
That it could be any of the green dashed lines but we are only interested in the blue dashed line?
yes, only in the blue dashed line
To be honest, I'm not comfortable with this method, best wait for someone else
This is pretty straightforward if you’ve aware of how to express a tangent to a surface
In your case the surface is described by x^2 + y^2 - z^2 = 1
You pretty much already said it here
That’s essentially what I was gonna say
But phrased like this
Let F(x,y,z) = x^2 + y^2 - z^2
Then F(x,y,z) = 1 describes ur surface
And gradF(a,b,c) • (x-a, y-b, z-c) = 0
Is the tangent at the point (a,b,c)
Let F(x,y,z) = x^2 + y^2 - z^2
^So this describes the paraboloid but it doesn't have a fixed location in space
Then F(x,y,z) = 1 describes ur surface
This describes both the paraboloid in space and also it's location?
Would that be right?
Yea sorta, think of it in a dimension lower, when you have say f(x,y) = x^2 + y^2 then we can describe certain curves through the level curves of f
The same thing goes for our F but instead F = 1 is a specific level surface
I honestly can't remember too much from Linear Algebra. But if i took the dot product between two vectors and they = 0, that meant that they are paralell to each other?
That they are orthogonal to eachother
Yeah
Okok so we take the vector from the gradient
dot it with the normal vector of the plane
and should be = 0
It’s alot easier to visualize in a lower dimension imo, but yeah basically that
I'm not entirely sure i understand this tho
Since our F can only be graphed in 4D it can be quite tricky to make that jump
Well let’s think about it in a dimension lower
Say we had f(x,y) = x^2 + y^2
And we were looking at the curve f = 1
ok
Now what does grad f represent with respect to its level curves?
Well the gradient of that function would represent every point in space with a vector which points towards the fastest rate of change?
I still have no idea what level curves are
Someone tried explaining it to me yesterday for lik 6 hours
Are you Swedish?
Har du hört talas om nivå kurvor?
Nej, allt måste man lära sig själv.
Det enda jag vet är att om vi har en karta och "kapar" en figur horisontellt då är linjerna som skapar figuren nivå kurvorna?
Japp
f = 1 t.ex är ju snittet som görs med planet z = 1 och ytan som f utgör i grafen
Det är lite så jag försöker tänka när jag arbetade med denna figuren
Ja, så vi kapar "kartan" när z = 1
Yes
Detta kallas då för en nivå kurva till f
Speciellt så säger då gradienten av f något mer
Att för varje punkt på nivå kurvan så är gradienten vid denna punkt vinkelrät/ortogonal mot kurvan
Kan det kännas rimligt?
Ja det känns nog okej.
För då menar du dom blå streckade pilarna, tex?
om då den blå circkeln var våra nivåkurvor till x^2+y^2
Precis
Okej det låter 100% rimligt
Härligt, då om vi tittar tillbaka till detta
På motsvarande sätt här kan vi säga att
grad f(a,b) • (x-a, y-b) = 0 är elevationen för tangenten till kurvan vid punkten (a,b)
punkten a,b. Är det punkten där dom tangerar?
Om vi t.ex använder samma f som innan och tittar på kurvan f = 1,
så kan vi t.ex kolla på punkten (0,1), detta är en punkt på kurvan.
Tangenten vid den punkt till kurvan blir då
(2*0, 2*1) • (x - 0, y - 1) = 0
Dvs.
2(y-1) = 0,
Dvs.
y = 1, vilket är ju rimligt!
Mm men jag förstår fortfarande inte varför du tar x-a och y-b i den andra vektorn
Vi kan tänka det som att vi förskjuter koordinat systemet så att vårt origo är punkten (a,b)
Eftersom vektorer ju alltid börjar från origo
Um det visste jag inte..
Det blir lite för mycket nya koncept
Jag lär mig främst genom visuellt, jag kan inte bara anta att saker och ting börjar vid origo därav (x-a, y-b) därav osv...
Jag förstår att den gröna pilen är den vi söker. Den är parallell med planets normalvektor. Så jag måste ju börja med att hitta den vektorn
Okej låt oss säga så här, tänkt att vi var intresserade av att hitta tangenten till en kurva vid (0,0)
Låt oss säga att kurvan faktiskt går igenom här
Nej?
Va jo?
Ganska säker på att någon förklarade samma sak igår
Om den har mer än 1 point of intersection
då går ju planet igenom hyperboloiden vilket betyder att den inte är tangent emd den
Jo visst det är ett sätt att göra det på, men det kan väldigt lätt bli knepigt
Jo kanske det men det är väl den enda sättet jag förstår
Det var därför jag försökte förklara det på annat här
Ingen stoppar dig ifrån att försöka på detta dock
Då det verkar vara väldigt rimligt för dig
ja men problermet är att jag har suttit med denna uppgfiten i 15 h
och ingen kan förklara på ett sätt som är förståeligt
alla ska försöka med sitt egna sätt
Så hur vill du ha det? Lära dig eller sitta och sucka?
Vad menar du sitta och sucka?
Vi försöker vårt bästa
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.CLOSE
1.- Let $S = \langle (1, -1, -1, 0), (-1, 1, 2, 1) \rangle$, $T = { x \in \mathbb{R}^4 \mid x_1 + x_2 - 2x_3 + x_4 = 0 }$, and
$W = { x \in \mathbb{R}^4 \mid x_1 - 3x_3 + x_4 = 0 }$.
Find, if possible, a scalar $a \in \mathbb{R}$ and a linear transformation $f : \mathbb{R}^4 \to \mathbb{R}^4$ such that
$f(1, 0, 0,-1) = (a, 1, 1, -3)$ and $f(S) \oplus f(T) = W$.
938c2cc0dcc05f2b68c4287040cfcf71
start by finding a basis of T
x1 = -x2+2x3-x4
(x1,x2,x3,x4)=(-x2+2x3-x4,x2,x3,x4)
x2(-1,1,0,0)+x3(2,0,1,0)+x4(-1,0,0,1)
T=<(-1,1,0,0),(2,0,1,0),(-1,0,0,1)>
x1= 3x3 -x4
(x1,x2,x3,x4)=(3x3-x4,x2,x3,x4)
x3(3,0,1,0)+x2(-1,1,0,0)+x4(-1,0,0,1)
W=<(3,0,1,0),(-1,1,0,0),(-1,0,0,1)>
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@tidal turret Has your question been resolved?
@tidal turret Has your question been resolved?
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I am curious now 
It was a random roblox gif xd
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Thx i needed to know for some reason 