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12:04 AM
Is this proof correct? Let $D$ be a domain. By the theorem of existence of harmonic conjugates, we get that every harmonic function on $D$ has a harmonic conjugate.
In other words, for every harmonic function $u$ on $D$, there exists a harmonic function $v$ on $D$ such that $f=u+i\,v$ is analytic on $D$. Since $f$ is analytic on $D$, so $f$ is infinitely (complex) differentiable. It follows that $u$ and $v$ are infinitely differentiable. In particular, this show that $u$ has partial derivatives of all orders.
 
@LeakyNun Functors o-0
 
12:22 AM
@user330477 no. You have just asserted the conclusion without making an argument.
 
@MikeMiller Essentially this is what people like Leaky Nun are telling me and saying telling me it is the basic idea.
 
12:48 AM
@Semiclassical well, I've done just that. I showed that: $$\sum_{k=0}^n a_k\sum_{j=0}^m b_j = \sum_{l=0}^{n+m}\sum_{q=0}^{l} {a_{l-q}\cdot b_{q}\cdot x^{q}} - 2X$$ where 2X are the outliers that don't fit into LHS and then I showed that for a special case of $$a_k = \binom{n}{k}$$ $$k\gt n$$ which by definition makes $$2X=0$$ and completes the proof
 
I am lost working on a series of questions about continuous functions. (i.e.)
f(x) = ax + b, x greater_than_or_equal 1;
f(x) = x^2, x < 1

Find all values of a,b such that f(x) is continuous
Find all values of a, b such that f'(x) is continuous
Can anyone help me on the right approach to questions such as this?
 
1:08 AM
Denote $f(z)$ the analytic continuation of $\sum_{n=0}^\infty\chi(n)z^n$, where $\chi$ denotes Dirichlet character $\mod k(k>1)$. Show that $f(z)=\chi(-1)f(1/z)$. Can anyone help me? I don't know where to begin.
 
@Michael.P cool.
 
2:08 AM
Hi, can anyone tell me if I have any deleted questions? I remember asking a question but can't find it now...
 
@stewbasic I'm kind of new here so I don't know how to answer your question, but while you're here can you help me with a basic function composition problem? I'm not sure what my teacher is asking me to do.
 
Sure... I guess there are more questions than answers here :P
 
Thank you so much! I'll just type it in really quick...
Write the equation of the function $g(x)$ if $g(x) = f(x+2)+4$ and $f(x) = x^3-7$.
I don't know what the $f(x+2)$ part means.
 
@stewbasic Unfortunately this isn't something users have access to: you can only search for your own deleted questions, and even then only if you have >10k rep on the site.
You can view them, but only if you have a link.
 
@Startec Since the function is definitely continuous for x!=1, it's continuous exactly when the limits from left and right at x=1 agree. So you should set lim_{x\to1^-}f(x)=lim_{x\to1^+}f(x) to get an equation involving a and b
@MikeMiller I don't have 10k rep :/. I read here that a moderator might be able to help:
https://math.meta.stackexchange.com/questions/13948/is-there-any-way-to-see-my-deleted-questions-or-answers
@CaptainAmerica16 $f(x+2)$ is what you get from $f(x)$ by replacing $x$ with $x+2$. In this case $f(x+2)=(x+2)^3-7$.
 
2:15 AM
@stewbasic Wow, I can tell I'm getting sleepy. Thanks for the help, lol.
 
Yep, I know. Whence why it is unfortunate. :)
 
Ok, well thanks for the information anyway :)
 
 
2 hours later…
4:08 AM
$lim_{x\to1^-}f(x)=lim_{x\to1^+}f(x)$
 
4:28 AM
tell that to |x|/x
 
sP_
If e^-x = z, is x = -ln(z)?
 
@sP_ yep
 
sP_
(y)
 
4:53 AM
@sP_ I think $x=-Ln(z)$ (or $x=-\ln z+2k\pi i).
 
Zee
I need to fix my sleep schedule , slept at six woke at six pm
Should I sleep at 3 pm and then 6 pm the next day and then 9 pm the day after?
Or get drunk and sleep in 2 hours , 3 am
 
Probably you can try to force yourself be awake until the time you planned to sleep.
 
Zee
I can , idk what is best though
I have drugs to keep me awake or put me to sleep
 
In my experience it's easier to roll a sleep schedule forward than it is back. When I get out of whack like that, that's what I do.
 
Zee
That’s my experience too , sleep is such a hassle man
 
5:02 AM
Btw If u want to sleep early, just woke up early and play phone games.
 
Zee
Actually, I used to do that! Had a tv with call of duty in my bed room , but then I ended up playing it before sleep and that was a disaster lol
Am just gonna get drunk and sleep at 3 am
Pray for me
 
5:51 AM
How do we use wolfram alpha to find volume of a region
For example, volume of region bounded by sphere and cylinder whose equation is known. I need region bounded by that, its volume.
 
Try converting it to an integral.
 
That is a problem, Also I want to solve it on my own, the problem is of finding volume of intersection of three perpendicular pipes
I had hard time visualising, but finally cut the region into 48 parts
And arrived at an answer, I need to check whether its correct, and if its incorrect, ill look for error
 
I don't know whether W|A can do this, but I know Mathematica can.
 
I have mathematica available at college
Actual question was finding volume of region {$x^2+y^2 < 1, y^2+z^2 < 1, x^2+z^2 < 1$}
 
Try this command: Integrate[1, {x, y} [Element] ...]
Sorry I must be offline now. cya
 
5:58 AM
Alright ill look for this command!
I tried this command:
Integrate[1, {x, y, z} E {x^2 + y^2 < 1, y^2 + z^2 < 1, z^2 + x^2 < 1}]
Instead of E, there I wrote the belong to set symbol
It says invalid integration limits!
 
Zee
6:31 AM
Why use wolfram when you can use math ?
Why eat Burger King when you can eat your mother’s cooking ?
 
6:47 AM
Why troll when you can say things seriously?
 
Ok, my turn
Why live when you finally die?
 
@Zee I have tried using maths to get to an answer which is most certainly wrong, I just found my answer is around 12, but answer should already be less than 8 certainly
 
@MikeMiller lol, it's complicated but is there any alternative?
Given a real polynomial equation $P$ of degree $n$, e.g. $a_nx^n+a_{n-1}x^{n-1}+\dots+a_0=0$, and it has a root $\alpha_i$ with multiplicity $k$, why $\alpha_i$ will satisfy $P,P',\dots,P^{'(k-1)}$?
0
Q: Why a root $\alpha_i$ with multiplicity $m_i$ will satisfy up to $(m_i-1)$-th derivatives of polynomial of degree $n$?

Isana YashiroI'm reading a proof of linear recurrence relation with constant coefficients of order $k$, which gives the formula of $a_n$ when some roots of its characteristic equation has multiplicity $\ge2$ . In one step it says: \begin{align*} &\textrm{Since }\alpha_i\textrm{ satisfy}\\ &C_n\alpha^n+C_{n-1...

 
7:10 AM
I have an excercise from class equation. Rule out as many of the following as possible as class equation for group of order $10$ (i) $1+1+1+2+5$ (ii) $1+2+2+5$ (iii) $1+2+3+4$ (iv) $1+1+2+2+2+2$ , (iii) can't be a class equation. Is there any other that can't be class eqn ?
 
the centre is a subgroup
so (i) is impossible
(ii) should be D5
(iv) is impossible because the centre would have 2 elemenets, so G/Z(G) is cyclic, so G is abelian to start with, so it would be all 1
 
reasoning for (i) is not clear :-(
Can you expln it?
 
the centre is the number of $1$s
(why?)
 
@LeakyNun oh, you mean cardinality of centre is number of $1$s, as orbit of elements from centre under action of conjugation is itself only.
right?
 
right
 
7:24 AM
@LeakyNun also nt getting why G/Z(G) is cyclic!
is it because |G/Z(G)|=5 ?
@LeakyNun got it , thanks :-)
 
ok
 
if anyone could help, I've added a bounty:
2
Q: Combinatorial meaning of $L_m=S_m-{m\choose m-1}S_{m+1}+{m+1\choose m-1}S_{m+2}\mp\dots+(-1)^{n-m}{n-1\choose m-1}S_n$

Isana YashiroI want to understand the meaning behind the coefficients of the following formula, $L_m$: Let $U$ be a finite set, and there are $n$ properties defined on it: $a_1,a_2,\dots,a_n,$ and let $S_m$ denote the sum of all cardinalities of $m$-intersection(s) of $A_{1\le i\le n}, $ where $A_i=\{x\in U|...

 
Zee
7:39 AM
@LeakyNun oh come on , I was serious , not trolling
BTW guys , if you didn’t watch Akira , you need to
 
hi the HNQ list showed me this question
8
Q: 9-4+1 does not equal to 4?

ac1002I ran across the following math problem where there is arithmetic involved: $$9-4+1$$ Where supposedly the answer is 6? I entered into my computer and calculator and got the same result so I realized there was something strange in this math problem because it does not follow the PEMDAS pattern. ...

the OP seems to be confused about why the answer isn't 4, but by what logic is the answer supposed to be 4 here?
the question and answers seems to be well received but I don't get why this is even a question
may be I'm not understanding something can some regulars of this site please shed some light onto this?
 
> Your confusion lies in thinking that Addition supersedes Subtraction. The actual precedence set by "PEMDAS" is

Parenthetical terms first
Exponents
Any multiplication OR division (equal precedence!)
Any addition OR subtraction (equal precedence!)
This is why I suggested you think of it as PE(MD)(AS).
It is one of those examples of small things that one must get it right
and we all tend to overlook when we get very advanced into maths
(Not to mention a good mathematics education example)
 
8:00 AM
I see so the op is looking at it as 9-(4+1)
or at least that's the only way I can think of why the answer could be thought to be 4
 
yup
Misconceptions about PE(MD)(AS) is very common to the point many joke problems are based on it
 
Zee
High school math is more difficult than college math major math , I said it and will repeat
 
I wonder if the extension is known as PTE(MD)(AS)
Also I think it should be
P(ER)(MD)(AS)
R=Roots
or maybe even:
PF(ER)(MD)(AS)
F=any function that is not listed here
In mathematics and computer programming, the order of operations (or operator precedence) is a collection of rules that reflect conventions about which procedures to perform first in order to evaluate a given mathematical expression. For example, in mathematics and most computer languages, multiplication is granted a higher precedence than addition, and it has been this way since the introduction of modern algebraic notation. Thus, the expression 2 + 3 × 4 is interpreted to have the value 2 + (3 × 4) = 14, not (2 + 3) × 4 = 20. With the introduction of exponents in the 16th and 17th centuries,...
Now...
$$\min(💥(\text{PE(MD)(AS)})) = ?$$
Actually, the description of this operator needs some update:
 
8:16 AM
so, we're doing natural deduction at uni. The lecturer presented 8 "base" inference rules, and then a bunch more rules for convenience that he said we could prove using the 8 original ones
we weren't asked to do it, but I'm curious, so I tried doing a few
some were mostly trivial, but I'm having serious issues with others
first of all, here are the 8 rules: pastebin.com/fpkgaVR0
and these are some additional rules: pastebin.com/Pua3gBNX
// easiest one was probably
¬E1
¬¬A ⊦ A
1. ¬¬A      by data
2. ¬A → ¬¬A
  1. ¬A     assume
  2. ¬¬A    by 1
3. ¬A → ¬A
  1. ¬A     assume
  2. ¬A     by 3.1
4. A        by ¬E 2,3
 

  Logic

This room is meant for discussion about logic, including found...
 
oh, thanks
 
21820 is an expertise in logic
 
what's 21820/
oh
 
Does anyone know the number of ways to arrange 8 identical blue balls and 4 identical white balls in a circle such that no 2 white balls are adjacent to each other?
 
Zee
8:37 AM
@PrashinJeevaganth it’s the number of ways you can put identical 8 balls in 4 boxes such that each box will contain atleast 1 ball
Or equivalently , the number of ways you can put 4 identical balls in 4 boxes
 
A very simple question: If you know some very useful trick for solving some math problem, would you prefer sharing all of them to your friend and/or people?
Answer this question with format: (yes/no/other, your rep.)
I have a conjecture that those prefer sharing will have high rep.
 
Counterexample
(yes,1033)
 
8:53 AM
@Zee Does it matter whether the boxes are distinct?
 
Zee
@PrashinJeevaganth yes
This is the intuitive explanation
Each white ball needs to have a blue ball after it otherwise you would half to adjacent white balls
Hence if you have 4 blue balls and 4 white balls you get
White blue white blue white blue white blue
Now you add 4 blue balls
You can place 4 balls between the first two white balls or maybe between the last two white balls or maybe in the middle or whatever way fancy you
This is equivalent to how many ways you can put 4 identical balls into 4 different boxes
 
Actually ...
This was an extension question from an easier problem of how many ways are there to arrange 8 boys and 4 girls in a circle such that no 2 girls are seated next to each other
@Zee I made the box last time in terms of the boys and my question became the number of ways to put 4 balls into 8 boxes such that each box is at most 1
Somehow I tried applying your method of more balls into smaller number of box and I'm sort of undercounting
 
9:15 AM
if you glue a blue ball to the right of each white ball the problem becomes equivalent to the problem of arranging 4 blue balls and 4 whiteblue doubleballs in a circle
though depending on if you count rotations as different or not, the number might be different
 
@mercio Isn't it a default assumption for any 2 circles with similar rotational view to be considered identical and not contribute to additional count?
 
it should be, yes
 
@mercio wait are you talking about the first problem of identical blue balls or the girls and boys problem?
 
isn't it the same problem ?
 
Zee
I can’t really see anything wrong with my intuition , you got four slots and you gotta arrange 4 balls in them , (note some of them may be empty)
 
9:23 AM
@Zee did you start computing for your intuition?
If ur method A arrives at the same answer as my method B, then it means I wasn't good at calculating with method A
@mercio it feels like it isn't the same problem
because 2 girls cannot be identical
but 2 balls could be identical
 
ah
 
what's more if you glued them in a fixed orientation
it still doesn't force a distinction factor
 
it does
 
Zee
Lol two girls cannot be identical
 
what if they are octuplets ?
 
9:25 AM
well let's not talk about special cases ... the question will specifically say if they are twins or something ... but it's not said so they are different problems
 
Zee
@PrashinJeevaganth did you get 35?
 
for the 8 blue and 4 white problem?
 
Zee
Y
Ya
 
hmm I can't do the 8 blue 4 white problem because I'm caught with the confusion of distinction
How about the 8 boys and 4 girls problem such that no 2 girls are seated together
 
Zee
that’s the same problem
 
9:28 AM
o..o''''
 
I got 8467200
no
it isn't the same problem ...
 
Zee
Stop being a specist
Humans and balls are created equal
 
is that a joke ...
 
I have 10 for the blue and white balls problem
 
Zee
Ya :p
 
9:30 AM
even twins have distinct names ...
 
Zee
@mercio that’s not comforting
How did you come up with that?
 
counting
 
Zee
Four unlabeled balls into four labeled boxes?
What’s the formula for that?
That’s my answer
 
I counted the ways to put 4 blue and 4 whiteblue in a line, that's 70
then for all of the arrangement but 6, they are in classes of 8 when you place them in a circle
so that's (70-2-4)/8
and then you add the two backs
one is in a class of 2 and on is in a class of 4
70 = 8+8+8+8+8+8+8+8+4+2
so that's 10 ways
if i am not mistaken those are
ababababbbbb
abababbabbbb
abababbbabbb
abababbbbabb
ababbababbbb
ababbabbabbb
ababbabbbabb
ababbbababbb
ababbbabbabb
abbabbabbabb
if you count rotations as different then you get instead (3/2) * 70 = 105
 
Zee
I can’t do you justice so I will stop here
 
9:39 AM
if you count rotations as different and distinguishable balls that should be 105 * 4! * 8! = a lot
 
Zee
Kinda drunk so good night
 
good night
if the balls are distinguishable and rotations are not different that should be 105 * 4! * 8! / 12
= still a lot
 
@mercio ah great at least I am on the same page as you for the distinct case
 
yeah, still a lot = 8467200
 
1
Q: Limit of Coefficient Sequence and Radius of Convergence

user193319 Show that if the power series $\sum_{n=0}^\infty a_n x^n$ has radius of convergence $R$ and if $\lim_{n \to \infty} |a_{n+1}/a_n|$ exists, then the value of this limit is $R$ I think there might be a typo in the problem: I think the conclusion should be that $1/R$ is the limit of the sequenc...

 
9:51 AM
@Zee is it advisable to do this kind of "distinct identity" problems just by enumeration?
Can't really count them in a intuitive way for myself
I definitely believe the answer is small, so 10 feels believable
 
the difficulty is that some arrangements are invariant when rotated by a fraction of a whole turn
abbabbabbabb and ababbbababbb
so you have to find them and be careful about them when you go about dividing 70 by 8
 
lol I don't even understand your working that is the problem ... I think since the problem was about inserting 4 identical white balls into identical boxes in a circle
I will jsut do it one by one and try to cancel repetitions in subcases
feels very rigourous
it can't be a simple divide and conquer and multiplication principle since everything depends on another
 
putting 4 identical balls in 4 distinguishable boxes in a line is easy, it's the 70
 
I know that
haha
well it's about getting the 70 to divide by 8 ...
 
well when you place them in a circle, you have to regroup the 70 solutions into groups that give the same solution on the circle
most of the time, those groups have 8 elements
so that's why i say "dividing by 8"
there is just a handful of exceptions
 
9:59 AM
"Handful"
haha
1 is bad enough
 
if you had 7 blues it would be easier since 7 is a prime number
 
5 whites and 9 blues?
wait what does this have to do with 13?
 
I messed up a little
(oh putting 4 identical balls in 4 distinguished boxes is 35)
(but then it's about dividing it by 4 cuz you don't have exactly the same things as me in your classes)
 
wait ... why is it 35?
are the box in a circle?
 
no
 
10:07 AM
or does it matter?
ok
 
it does
 
oh gosh
I enumerated to 10 ways
its a crazy process
 
great !
 
Well it would be crazy to enumerate for some x distinct A and y distinct B ... gotta find a proper way to count it
 
11:06 AM
@mercio do you happen to know how many ways to pick 3 identical points in a circle?
I got 42, want a person to check with
 
If $f \in C^{(\infty)}(a,b)$ and $c \in (a,b)$, show that if there is an $r>0$ with $(c-r,c+r) \subseteq (a,b)$ such that $\frac{f^{(n)}(c)}{n!}(x-c)^n \to 0$ when $|x-c|<r$, then $f$ can be represented as a power series in an interval about $c$.
I want to show that $f(x) = \lim_{N \to \infty} \sum_{n=0}^N \frac{f^{(n)}(c)}{n!}(x-c)^n$, but I just don't see how to do it...
I'm searching my book for the right theorem, but none of them seem relevant...
 
11:52 AM
That looks like taylor's theorem
as what they are asking for is to prove that if the nth term vanishes, then a taylor series exists for f in that interval
In calculus, Taylor's theorem gives an approximation of a k-times differentiable function around a given point by a k-th order Taylor polynomial. For analytic functions the Taylor polynomials at a given point are finite-order truncations of its Taylor series, which completely determines the function in some neighborhood of the point. The exact content of "Taylor's theorem" is not universally agreed upon. Indeed, there are several versions of it applicable in different situations, and some of them contain explicit estimates on the approximation error of the function by its Taylor polynomial. Taylor...
 
12:22 PM
@PrashinJeevaganth what
 
@mercio sorry I meant pick 3 identical points out of 15 identical points in a circle
 
@user193319 $\exp(-1/x^2)$ says hi
 
Conjecture: Many pathological examples to continuity theorems always contain function of the following form:
$\frac{1}{f(x)}$ for some $x$ such that $f(x)=0$
I wonder if there is a name to classes of functions like $x^n \sin \frac{1}{x^m}$, $e^{\frac{1}{x^n}}$ etc.
More symbolically, what is:
$$\text{Name}(💥(\text{Pathological counterexample functions to continuity theorems})[\text{Exclude category theory and above}])$$
 
@PrashinJeevaganth I get 31
 
Hmm?
Mind showing your working?
 
12:29 PM
NB: The evaluation of the following:
$$💥(\text{Pathological counterexample functions to continuity theorems})$$
Does not fit in the chat
 
how many ways are there to pick 3 points out of 15 points on a line ?
 
no in a circle
 
yes but i'm asking for a line
 
@mercio Ah, so $e^{-1/x^2}$ is a counterexample?
 
I think so
 
12:31 PM
oh you're asking me?
 
yes
 
If I am more free, will explore why that is a counterexample
 
extended with $f(0) = 0$
 
Oh, I see. Let me work out the details. Thanks!
 
@mercio I think it's 13?
 
12:33 PM
Could it explain to me why the $Area=-2\int_{D}H^{2}dA$ decreases when we doformace the surface in the direction of $H\overrightarrow{n}$, where $D\subset
\mathbb{R}
^{2}$?
 
13 is awfully low
 
oh wait it should be more than the circle
lol I forgot to count a node
@mercio 91?
 
not at all
 
XD ok I don't know what went wrong with my working
 
 
4 hours later…
4:55 PM
hi , for $A,B \in M_n(\Bbb C)$ , can someone help me understand why $AB^* = \overline{BA^*}$ ? $A^* = adj(A)$
 
hi, why is $PGL_2(\Bbb R)$ a scheme?
I get how to see $GL_2(\Bbb R)$ as a scheme
$Spec(\Bbb{R}[w,x,y,z,q]/((wz-xy)q-1))$
But I don't know how to deal with wanting an action to identify scalar matrices
I imagine the answer is: It's $PGL_2(\Bbb R)/\Bbb G_m(R)$ and then I have to interpret that
 
$PGL_2(\Bbb R) = \{[w:x:y:z] \in \Bbb RP^3 \mid wz-xy \ne 0\}$
so it's an open subset of $\Bbb RP^3$
 
Oh I see my likely mistake. It's not an affine scheme is it
$PGL_2(\Bbb R)=\{[w:x:y:z:q]\in\Bbb RP^4\mid (wz-xy)q=1\}$ closed in $\Bbb RP^4$
Or does one need to say $V((wz-xy)q-a^3)$ from $\Bbb{R}[w,x,y,z,q,a]$?
And then it's the scheme $Proj(\Bbb{R}[w,x,y,z,q,a]/((wz-xy)q-a^3)))$ perhaps?
Maybe I'll ask on main
 
5:15 PM
i actually need thlpo prove $tr(AB^*) = tr(BA^*)$ , not what i wrote before. someone can help?
 
@Liad $A=(a_{ij})_{i,j=0}^n$ and $B=(b_{ij})_{i,j=0}^n$, do you know what the diagonal entries of $AB^*$ are?
 
$tr(AB^*) = \sum_{i=1}^n \sum_{k=1}^n a_{i,k} (-1)^{k+i}B_{k,i}$ where $B_{k,i}$ is the k-i th minor
@user469689 am i right?
 
@Liad forget about it being $A$ and $B^*$, just prove it for any $A,B$
It holds in that generality
 
@user469689 zero locus of a non-homogeneous polynomial doesn't make sense in projective space
 
Oh the star switches, I wasn't reading it right
 
5:24 PM
ah. ok
 
@loch I fixed that in the next line by homogenising I guess
 
and over $\Bbb C$ we need $\overline{tr(BA^*)}$
 
anyway as leaky said it's a proper open subset of projective space (in particular it's quasi-projective but not projective), so you won't be able to exhibit it as vanishing locus of polynomials in projective space..
the problem with what you wrote there is that $a$ could be $0$
 
You're right
 
but anyway that's ok - after all you want to see that it's locally an affine variety - which is still true!
 
5:51 PM
Hello @LeakyNun ?
 
Hey everyone!
 
@Daminark gwarn
 
hey @Daminark
Hi @ÍgjøgnumMeg
 
hey @Mathein
 
6:15 PM
Hey @I
 
6:28 PM
let $k:[0,1] \to [0\infty)$ be measurable and let $X$ be all the function (module equality a.e) that $\int_0^1 |(fx)|^2 k(x) dx \lt \infty$. i want to find $f \ne 0$ s.t $\int_0^1 |(f(x))|^2 k(x)dx =0$, someone can help?
i guess f should have some relation with $k(x)$.. but i can't find it.
 
maybe start with an example, e.g. $k(x)=1/x$?
oh, needs to be valid on the endpoints
 
i thought maybe $f(x) = \dfrac{g(x)}{\sqrt(k(x)}$ with $g\ne 0$ and have integral (squared) zero?
 
I guess $k(x)=x$ is the most obvious boring example. In which case, I can't see how you'd ever have $\int_0^1 x |f(x)|^2 \,dx=0$
I mean, you can take $f=0$ of course. And $f=0$ a.e. would presumably work as well
but in the latter case you've got something equivalent to $f=0$ in $X$
 
$f\ne 0$ ..
 
well, yes. i'm just saying that the trivial case, and variations thereof, are the only thing which spring to mind
 
6:34 PM
$f $ doesn't need to be continuous
 
Eh. Not sure that's enough, if you want some $f$ which isn't equal to 0 a.e.
 
that's how the question stated :/
the next one is to find a condition on $k(x)$ s.t $(f,g) = \int_0^1 f(x) \overline g(x) k(x)dx$ would be inner product
so the $f$ we are looking for must have a relation to $k$ somehow..
 
well, start by finding examples for simple $k$'s I guess
e.g. $k(x)=1,$ $k(x)=x$
 
@Liad show us a screenshot of the original problem?
 
it's not in English..
 
6:39 PM
the smallest circumventing circle around a triangle is only dependant of the largest side of the triangle, just realized that, now, how can i claim a nobelprize? jk
 
$X$ is as i said. $(f,g)$ as i wrote..
domain of the functions is $[0,1]$
 
i'm not good with analysis counterexamples, tbh. but I find it hard to see how you could have $\int_0^1 |f(x)|^2\,dx=0$ without having $f=0$ a.e.
(which is what you'd need for the case $k(x)=1$)
 
f doesn't need to be continuous
 
@Liad it doesn't matter, just send us
 
tbf, I do mostly have this in mind: en.wikipedia.org/wiki/…
 
6:41 PM
hmm ok O_o , one sec.
 
which as stated is indeed for continuous functions
but they do state a version of it for discontinuous functions.
 
@LeakyNun fine ?
 
$f$ has codmain $\mathbb{C}$ in there. that probably doesn't matter, though, since one still would have $|f(x)|\geq 0$
 
nope
$f$ has domain $[0,1]$
 
codomain
 
6:45 PM
ah. misread.. ^^
 
@Semiclassical I am really surprised that fact has a name
 
@Liad What language is that?
 
@MikeMiller so was I tbh
 
Hebrew
 
it's a handy name tho
 
6:47 PM
When I clicked on 'the fundamental lemma of calculus of variations' I was expecting something more robust, I suppose
 
yeah
there's comparatively little discussion of the variational context
@MikeMiller there's an interesting question there: What does it mean for a lemma/theorem to be 'fundamental'?
 
hehe, someone won a fields medal for proving a fundamental lemma
 
lol, really
 
@MikeMiller i saw him smoking yesterday!
 
is there a known bijective function from a unitsquare to a unitcircle? if that even makes sense. A function that maps every coordinate from the square to a coordinate from the circle, but bijective.
 
do you mean the square as a whole, or just the boundary?
 
@Semiclassical the square and circle as a whole
 
i didnt realize langlands is still alive
 
a lot of "fundamental" whatevers are like mega easy tbf
fundamental theorem of riemannian geo is like really stupid
 
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