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1:05 AM
Anyone here?
 
 
2 hours later…
3:27 AM
Is there a reason why $\bmod 4$ shows up so often in number theory? What's special about 4?
 
3:48 AM
0
Q: A Goldbach "Goldfish": symmetries of a graph representing addition of certain of the primes.

Shine On You Crazy DiamondDo you see the goldfish? It's a top view of a goldfish, because that's one direction in which a fish would look symmetric. Anyway, you construct it by adding two edges to certain even numbers, namely those that aren't double a prime $2p$. If they are of that form, then connect an edge from t...

@Ultradark
Dudes, I made a goldfish by adding a few small primes
Animals are our gods
@user76284 4 = 2 + 2 is the smallest even number that is the sum of two primes.
Ramanujan was friends with all the primes, I'm only friends with the primes up to about 103. Lol
Yes, that's a goldfish, for sure.
:D
 
4:18 AM
Hells yea, I'm getting mad upvotes by the community on that one
:>
 
 
1 hour later…
5:34 AM
@user76284 I'm guessing because of quadratic reciprocity, but don't take my word for it.
 
 
1 hour later…
6:57 AM
Aye yai yai
The comments...
fucking christ
People just get conditioned by life, and don't treat every encounter as something that could be different than before.
 
7:40 AM
Morning all
 
 
3 hours later…
11:00 AM
Does anyone knows an algorithm that relates a bit of mathematics, namely permutations: python-forum.io/…
 
 
3 hours later…
1:34 PM
Question for myself: Let $A$ by a square matrix. If we may partition this matrix as $A=[A_1|A_2]$ where $A_1,A_2$ both have full rank, must $A$ itself have full rank?
I guess that's equivalent to: Suppose I have vectors $\{u_k\}_{k=1}^n,\{v_k\}_{k=1}^m \subseteq \mathbb{R}^{n+m}$. If both sets of vectors are linearly independent, must their union be linearly independent?
The first way I wrote that made me think $A$ would have to have full rank, but that rewriting of it makes me dubious...
ok, yeah: Consider the case where $n=m$ and $A_1=A_2$ are full rank. Then $A$ will have rank $n=m<n+m$.
In my case of interest I further have $A_1^\top A_2=0$. This would rule out the previous example since then $A_1^\top A_2=A_1^\top A_1=0$ and therefore $A_1=A_2=0$
in that case I'd have $A^\top A = \begin{bmatrix} A_1^\top A_1 & 0 \\ 0 & A_2^\top A_2\end{bmatrix} = A_1^\top A_1 \oplus A_2^\top A_2$
Then the rank of the direct sum is the sum of the ranks, so $\text{rank}(A^\top A)=\text{rank}(A_1^\top A_1)+\text{rank}(A_2^\top A_2)=n+m$. But $\text{rank}(A^\top A)=\text{rank}(A)$, so $A$ is indeed full rank.
 
2:07 PM
@Semiclassical "full rank" is "spanning" not "linearly independent" right
and your submatrices generally cant be full rank right
 
full rank means rank = min(col dim, row dim)
since that's as large as the rank can possibly be for a rectangular matrix
 
oh
 
2:46 PM
Hi, I'm a bit confused about how to end this problem.
 
your formula for the profit is incorrect, ab + (1-a)(1-b) - a(1-b) is the correct formula
multiply it out to get a(3b-2) +1-b, now its clear how you must choose a (depending on b) in order to get a good profit
 
Hmm, why is my profit formula incorrect
 
for example rock beats scissors, so if a is the prob you play rock and b the prob the enemy plays scissors, you will get a term contributing + ab in your profit, you have a - ab term
 
oh I see, I mismatched. I guess lets just assume then that $b$ is probability my opponent plays paper
 
What makes this problem interesting, I imagine, is that presumably your opponent is trying to maximize their profit as well
 
3:00 PM
Right, but since they can't play rock we have an advantage. However, essentially to minimize their loss the $b = \frac{1}{3}$ is saying that they should play paper one-third of the time and play scissors two-thirds of the time
From this though, I am not sure what we should play or what the value of $a$ should be
 
Actually, it's a little unclear as I look at it. The problem only specifies that -you- gain/lose a dollar if you win/lose, not that said dollar goes to your opponent
 
I apologize, assume that your dollar goes to opponent and vice versa
 
in order for the problem to make much sense it seems like you'd explicitly need your profit to be their loss
ah, good
 
edited for typos and made problem clearer: imgur.com/a/V2MjaS1
 
i imagine you also should have $(\partial P/\partial a)=...$
For my taste, though, what I'd suggest is plotting your profit as a function of $a$ for various $b$
(In that vein, note what kind of function $P(a,b)$ is with respect to $a$ with $b$ held fixed.)
 
3:08 PM
How do you prove that $A^T A \ge 0$ when $A$ is $n \times n$ and $n \in \mathbb{N}$?
and if $A^TA = 0$ if and only if $A = 0$
 
Try writing out the case of $n=2$ explicitly.
 
@Semiclassical If I just do the partial derivative for b or $\frac{\delta b}{\delta P}$ I get that $a= \frac{1}{3}$
 
actually, wait
Do you mean $A^\top A\geq 0$ element-by-element or in the sense of being positive semi-definite?
The former doesn't actually seem true to me.
 
positive-definite means exactly what?
 
There's a few definitions. One is that $M$ is positive semi-definite when $v^\top M v\geq 0$ for any vector $v$.
 
3:10 PM
@Tuki symmetric matrix that has positive eigenvalues is usually most common def
 
okey
Also isnt $A^TA$ symmetric?
 
$\Bbb Q(\zeta_{p^2})/\Bbb Q$ has a subextension of degree $p$ over $\Bbb Q$ by Galois theory (since its Galois group is $\Bbb Z/p \times \Bbb Z/p-1$): why is it unique?
 
Yeah, that's another. I'm fond of the one I gave because the link to the name is clearer (the function $Q(v)=v^\top M v$ is nonnegative for all possible $v$, so it's definitely not negative)
@Tuki yeah.
but, again, this all depends on what exactly you mean by $A^\top A\geq 0$
fixed
For the importance, consider the matrix $A=\begin{pmatrix} 1 & -2 \\ 0 & 1\end{pmatrix}$
then $$A^\top A=\begin{pmatrix} 1 & 0 \\ -2 & 1 \end{pmatrix} \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & -2 \\ -2 & 5\end{pmatrix}$$
 
I have to prove that $\langle A , B \rangle = \text{tr}(A^TB)$ for any $A,B \in \mathbb{R}^{n \times n}$ where $n \in \mathbb{N}$
 
Oh hey, frobenius inner product
as in, you need to prove that that's an inner product?
 
3:16 PM
So I have the set of axioms I need to show
the last one being the one I mentioned
yes
 
Wouldn't it be $\text{tr}(A^\top A)\geq 0$ then?
 
Yes I think but I need proof for this
 
The proof for tr(A^T A)>=0 is pretty easy: Just write out the definition of matrix multiplication and take the trace.
 
for the latter part I have $\text{tr}(A^TA) = \langle A , B \rangle = 0 \iff A = 0$
 
Any slick proof that $\int z^n \bar{z}^{m}$ over unit disk is always zero for $m \neq n$?
 
3:18 PM
I'm really not seeing why that necessitates showing $A^\top A\geq 0$
@genescuba my suggestion is that you avoid thinking about calculus here.
Suppose your opponent picks a particular b. Then what kind of function is P(a)?
 
Well I mean the definition has two parts: $$ \text{tr}(A^TA) \ge 0; \text{ and if } \text{tr}(A^TA) = 0 \iff A = 0$$
 
Here's a suggestion: Write out $\text{tr}(A^\top A)$ explicitly for the case of n=2
 
Sure I'll try that one
Not quite sure why I find proofs always this hard
 
For a start, is there some "symmetry" reason why $\int z^n=0$ over unit disk? for odd $n$ I see the symmetry $z \mapsto -z$
Ok and in general I can multiply $z \mapsto \zeta z$
Where $\zeta^n=-1$
 
@Emolga is that a path or a area integra
 
3:27 PM
area
 
then your substitution $z\mapsto \zeta z$ should work
 
Yes but I really want the former integral,
$\int z^n \bar z^m = 0$
 
$\overline z^n z^m = |z|^n z^{m-n}$
do the same trick
 
Mmm
Oh right!!
thanks
 
So on diagonal matrix product always produces squared terms when I have $A^TA$
anything squared $\ge 0$
 
3:33 PM
Right.
For a fuller proof, you can write out the definition of matrix multiplication: $C=AB\leftrightarrow C_{jk}=\sum_l A_{jl}B_{lk}$
What does that become in the case of $A^\top A$?
It's also a good exercise to write $\text{tr}(A^\top B)$ in terms of matrix elements
 
I've shown the $\text{tr}(AB)=\text{tr}(BA)$ by writing the matrix product by using two sums.
Then simply manipulating the summation order and concluding that they are the same
 
ok. It's still worth writing out tr(A^T B) explicitly in matrix elements.
 
If $f \in L^2$ and $\int f = t$ on $\{x^2+y^2 \leq t\}$ for every $t < 1$, how do I justify $\int f = 1$ on $t=1$?
 
Something-something completeness?
 
Then I have another problem which I find really difficult for some unknown reason
-1
Q: Proof that $\vec{v} \in U \iff \text{proj}_U(\vec{v})=\vec{v}$

TukiProblem Let $V$ be finite-dimensional inner product space and $U$ it's subspace. Let's assume that $\vec{v} \in V$. Show that $\vec{v} \in U \iff \text{proj}_U(\vec{v}) = \vec{v}$ Attempt to show Let's assume subspace $U$ has orthogonal basis ($\vec{u}_1,\vec{u}_2,\vec{u}_3,\dots \vec{u}_n$) ...

My approach should cover $"\implies"$ but doesn't cover the other way around
 
3:41 PM
@tuki btw, the proof that $A^\top A$ is positive semi-definite is trivial for the definition I gave: For any vector $v$, one has $v^\top (A^\top A)v=(Av)^\top Av = \|Av\|^2$ which is certainly nonnegative
so $A^\top A\geq 0$ is immediate from that definition of "matrix $\geq 0$"
 
@Emolga Let $\chi_r$ be the characteristc function on the ball of radius $≤r$, then you have $\langle \chi_r, f\rangle =r$. Check that $\chi_r\to\chi_1$ as $r\to 1$ in $L^2$ norm, then $r=\langle \chi_r, f\rangle \to \langle \chi_1, f\rangle$ and the right hand side is $1$
 
$||.||$ as in $\sqrt{a^2+b^2+c^2+\dots}$?
 
This last convergence
 
Right. $v=(v_k)_{k=1}^n\implies \|v\|^2=\sum_k v_k^2$
 
$⟨χ_r,f⟩→⟨χ_1,f⟩$
 
3:43 PM
sum of squared elements, right.
 
follows from $\chi_r\to\chi_1$ in $L^2$ norm
 
to link this to your problem, if $\{e_k\}$ is an orthonormal basis then the trace can be written as $\text{tr}(A^\top A)=\sum_{k} e_k^\top A^\top A e_k$
and since $v^\top A^\top A v\geq 0$ for all $v$, that trace is a sum of nonnegative quantities
 
So automatic from continuity of inner product? Surprisingly no dominated convergence needed then
 
right
 
@Emolga you can use dominated convergence if you want
 
3:46 PM
so certainly nonnegative itself. of course, you can also just write $\text{tr}(A^\top A)=\sum_k \|A e_k\|^2$
 
(@s.harp But your approach avoids it which is cool)
 
but, to connect back to what I was saying: one has $(A^\top B)_{jk} =\sum_l (A^\top)_{jl}B_{lk}=\sum_l A_{lj}B_{lk}$
 
@Emolga i think the statement $f, g\in L^2 \implies \overline f\cdot g\in L^1$ is dominated convergence (+ Cauchy-Schwarz for $\Bbb C$), which is necessary for the inner product to be defined so that is whats going on the background
 
and then $\text{tr}(A^\top B)=\sum_{k} (A^\top B)_{jj}=\sum_{j,l} A_{lj}B_{lj}$
which means you take the two matrices, multiply them element-wise, and then add everything up
equivalently, you could imagine flattening the matrices A,B into vectors
in which case summing over j,l is nothing more than the dot product of those vectors
 
Oh, brushed under the rug by the very existence of an inner product on $L^2$.
nice
 
3:51 PM
So there's a bunch of different way to think about the Frobenius inner product
the linear algebra problem I've been banging my head against is this one: math.stackexchange.com/q/3331782/137524
 
Yes quite fascinating indeed
 
I think the clever thing to do is form the partitioned matrices $A=[\alpha|\alpha_\bot]$ and $B=[\beta|\beta_\top]$, in which case $A,B$ are full-rank square matrices.
Can't quite see to the end of it tho, which is annoying
 
$\begin{bmatrix}\alpha \\ \alpha_\perp\end{bmatrix}\begin{bmatrix}\beta & \beta_\perp\end{bmatrix} = \begin{bmatrix} \alpha\beta & \alpha\beta_\perp \\ \alpha_{\perp}\beta &\alpha_\perp\beta_\perp\end{bmatrix}$
 
yeah, I got that far
I think the trick is to form $AA'BB'$ in two different ways
Working that out now
 
you can use the "usual" formula to invert the "$2\times 2$" matrices
 
4:04 PM
doesn't assume the submatrices commute?
 
true
in particular it needs the blocks to be of equal size
 
4:15 PM
hmm...
$$A'A = \begin{pmatrix}\alpha' \\ \alpha'_\bot\end{pmatrix} \begin{pmatrix}\alpha & \alpha_\bot\end{pmatrix} = \begin{pmatrix}\alpha' \alpha & 0 \\ 0 & \alpha'_\bot \alpha_\bot \end{pmatrix}$$
$$AB'=\begin{pmatrix}\alpha & \alpha_\bot\end{pmatrix} \begin{pmatrix}\beta' \\ \beta'_\bot\end{pmatrix}=\alpha \beta'+\alpha_\bot \beta'_\bot$$
$$A'AB'= \begin{pmatrix}\alpha' \alpha & 0 \\ 0 & \alpha'_\bot \alpha_\bot \end{pmatrix} B' = A'(\alpha \beta'+\alpha_\top\beta'_\top)$$
That seems awfully close
 
4:54 PM
@s.harp think I've got a proof if I assume the matrices $(\alpha,\beta_\bot)$ and $(\beta,\alpha_\bot)$ are nonsingular
But I can't tell if that's stronger than the premises provided.
Suppose $(\alpha,\beta_\bot)$ is singular. Then $(\alpha,\beta_\bot)(\alpha,\beta_\bot)'=\alpha\alpha'+\beta_\bot\beta_\bot'$ would have to be singular. But $\alpha$ and $\beta_\top$ are both assumed to be full-rank.
 
5:30 PM
@s.harp found an example where $\alpha'_\bot \beta'_\bot=\beta'\alpha=0$ and so the required inverses don't exist
So the premises stated aren't enough to guarantee the formula makes sense.
 
@Semiclassical i had gone running sorry
 
my brain is a bit too fried to keep track of matrices right now :)
 
lol
it's a simple example: $\alpha=\beta_\bot=e_1$, $\beta=\alpha_\bot=e_2$
All four vectors trivially have full rank, and $\alpha'\alpha_\bot=\beta'\beta_\bot=0$
But $\alpha'_\bot \beta_\bot = \beta'\alpha=0$ as well, so the inverses don't exist
 
the dimensions dont work with the conditions tho
oh wait they do
or they dont
 
5:39 PM
N=2, R=1, rank = R=N-R = 1 for all four
 
ok you are right XD I thought $N=1$
 
lol
note that the example given is precisely one where $(\alpha,\beta_\bot)$ and $(\beta,\alpha_\bot)$ are singular
If they're non-singular, then one has the following:
$$I
=\begin{pmatrix} \alpha & \beta_\bot \end{pmatrix} \begin{pmatrix} \alpha & \beta_\bot \end{pmatrix}^{-1} \begin{pmatrix} \beta' \\ \alpha_\bot' \end{pmatrix}^{-1}\begin{pmatrix} \beta' \\ \alpha_\bot' \end{pmatrix}
= \begin{pmatrix} \alpha & \beta_\bot \end{pmatrix}\left[\begin{pmatrix} \beta' \\ \alpha_\bot' \end{pmatrix} \begin{pmatrix} \alpha & \beta_\bot \end{pmatrix} \right]^{-1}\begin{pmatrix} \beta' \\ \alpha_\bot' \end{pmatrix}$$
blah, that didn't work
hng
There we go
But under the assumptions given the term in brackets is just $\begin{pmatrix} \beta' \alpha & 0 \\ 0 & \alpha'_\bot \beta_\bot\end{pmatrix}$
at which point one multiplies the formula out to get the desired result
 
6:02 PM
posted an updated answer + counterexample
 
 
2 hours later…
7:36 PM
Is @Semiclassical still here?
 
depends on what your question is :P
 
Well I thought if you could help with this
0
Q: Proof that $\vec{v} \in U \iff \text{proj}_U(\vec{v})=\vec{v}$

TukiProblem Let $V$ be finite-dimensional inner product space and $U$ it's subspace. Let's assume that $\vec{v} \in V$. Show that $\vec{v} \in U \iff \text{proj}_U(\vec{v}) = \vec{v}$ Attempt to show Let's assume subspace $U$ has orthogonal basis ($\vec{u}_1,\vec{u}_2,\vec{u}_3,\dots \vec{u}_n$) ...

 
Seems like there's a bunch of answers there already.
 
Well I dont understand the answer V.Ch gave me, Kitter catter explains something I know already which doesn't help much
 
I take V.Ch's suggestion to be an alternative method of solution, rather than a patch on yours
 
7:40 PM
I think hes approach to this problem is different
I would like to patch mine so I could understand this
I've tried method using the "error vector" but I don't think it works
 
following their line: Given a subspace U, you decompose $V=U\oplus U^\perp$
 
what exactly is this symbol?
 
$U^\top$ is the orthogonal complement of $U$
 
@Semiclassical you mean $\oplus$
 
yes $U^{\perp}$
 
7:43 PM
that is, it's the set of all vectors in $V$ which are perpendicular to vectors in $U$
oops fixed
 
yes this one @s.harp
 
as an example, take 3D euclidean space
 
if $V, W$ are two subspaces of a vector space $X$, do you know what $V+W$ means?
 
@s.harp No I don't
those subspaces combined is my guess
but havent seen this before
 
$V+W= \{ v+w \mid v\in V, w\in W\}$ :)
 
7:45 PM
oh okey
 
if your subspace is a line (that is, the 1-dimensional subspace of vectors parallel to some v) then the orthogonal complement is the plane which is perpendicular to that line (that is, the 2-dimensional subspace of vectors perpendicular to that v)
 
\oplus means exactly what?
 
$V\oplus W$ has a more special meaning, but if $V$ and $W$ are both subspaces of the same vector space $X$ then $V\oplus W$ only makes sense if $V\cap W=\{0\}$, in such a case $V\oplus W = V+W$
 
^ "line through the origin / plane through the origin" in the above
 
@Semiclassical I think I understand what it means that something is in orthogonal complement, the definition of this was introduced in lecture yes.
Also done few practical problems with this
 
7:47 PM
so basically it means that you can write $v=u + u^\perp$ where $u\in U$ and $u^\perp$ is perpendicular to $u$.
 
@Tuki there is a point about orthogonal complements relating the symbol $\oplus$ that semiclassical was trying to make
 
oh ok
 
Main thing now is to ask what happens if you apply $P_U$ to that $v$--or, more generally, to the entire vector space $V$
do you know what $P_U(v)$ gives?
 
$P_U(v)$? I dont exactly know what $P_U$ means perpendicular to U?
or projection?
 
it's just $P_U(v)=\text{proj}_U(v)$
 
7:51 PM
okey
 
(that's the notation which V.Ch used)
 
It should give $P_U(v)=v$ since $v$ is already at U?
 
Did I say $v$ was in $U$?
 
nope
well the result should be vector in $U$
if we just assume $v \in V$
 
Right. But if $v=u+u_\perp$, then the answer is more specific
 
7:53 PM
is a transcendental divided by a transcendental always another transcendental
 
e/e=1
@tuki if nothing else, try plugging $v=u+u_\perp$ into your formula
 
$u$ as in orthogonal basis vector for $U$?
 
bleh, i keep swapping between ^perp and _perp
$u$ as any vector in $U$
 
so it's possible that a transcendental raised to a transcendental divided by a transcendental is equal to an integer?
 
well then I would have just $P_U(v) = u + u_\perp$
 
7:56 PM
Nope.
What do you know about $u_\perp$?
 
that it's perpendicular to u?
and $u \in U$
 
Not just $u$. It's perpendicular to every vector in $U$.
 
${T_1}^{\frac{T_2}{T_3}}$
 
yes except itself
 
What?
 
7:57 PM
@Ultradark take a look at $e^{\pi}/e^{\pi}$ (although its not know whether $e^{\pi}$ is transcendental
 
Well if you say "every" it would include also $u$
 
$u_\perp$ is perpendicular to every vector in $U$. But $u_\perp$ is not itself in $U$
 
oh yes sorry
 
So, what vectors do you know are in $U$?
besides your chosen $u$ itself, of course
 
there isnt anything else?
 
7:59 PM
@s.harp do you think it's possible that $e^{\frac{T^2}{T}}$ is an integer?
 
@Ultradark $T=\ln 2$
 
where $T$ is fixed transcendental number
 
and the projected(v) is in U
 
@tuki try looking at what you wrote in your post
 
well besides $P_U(v),(u_1,u_2,\dots,u_n) \in U$ I don't find anything else?
 
8:03 PM
No, you don't find anythin else. But you didn't include the $n$ basis vectors before.
All you said was "there isn't anything else."
So. If the $n$ basis vectors are in $U$, what can we conclude about $u_\perp$?
 
its perpendicular to all of the basis vectors?
 
Right. So what is $P_U(u_\perp)$?
Look at your formula before you answer.
 
@s.harp $e^{\frac{T}{\ln(10)}}=10$ I looked at your example. I'm wondering if there's some $T$ that makes this true
 
Yes. $T=\ln 10$.
 
@Semiclassical ln(10)^2
 
8:07 PM
no
 
woops, yes
 
okay good
 
zero vector?
 
well if i write the projection I have $\langle u^\perp, u_n \rangle$
if they are perpendicular this is 0?
 
8:08 PM
Correct. So $P_U(u_\perp)=0$.
Moreover, if you compute $P_U(v)=P_U(u+u^\perp)$, then you'll end up with $\langle u+u^\perp,u_n\rangle$ in your formula
So what happens then?
 
then you can write it as $\langle u, u_n \rangle + \langle u^{\perp}, u_n \rangle$
 
Right. And so?
 
the term with $u^\perp$ goes to 0 and only $\langle u, u_n \rangle$ stays
but isnt $\langle u, u_n \rangle$ also zero hmm?
 
Why would it be? $u$ is a vector in $U$
 
Oh sorry i was thinking they are orthogonal but they are not indeed
 
8:13 PM
$u^\perp$ wasn't, but $u$ still is
right.
in fact, those values of $\langle u,u_n\rangle$ should be pretty familiar
 
well similiar to $\langle v, u_i \rangle = c_1 \langle u_1, u_i \rangle \dots $ just without the constant
 
Let's try this differently. If $u$ is a vector in $U$, what can you say about it? (You already have this in your question.)
 
well it can be expressed as linear combination by some basis in $U$?
 
yes. write that out.
 
Heya @Mathein :)
 
8:22 PM
well $u = c_1 u_1 + c_2 u_2 + c_3u_3 \dots c_n u_n$ where $n \in \mathbb{N}$ and $c \in \mathbb{R}$ and where $(u_1,u_2,u_3,\dots,u_n)$ is orthogonal basis for U
 
Sure. And you already know $v=u_\perp+u$, so $v=u_\perp+\sum_{k=1}^n c_k u_k$
So now: compute $P_U(v)$ and report what you get.
 
8:44 PM
$$ P_U(v) = \frac{ \langle \sum_{k=1}^{n} c_k u_k, u_1 \rangle }{ \langle u_1, u_1 \rangle }u_1 + \frac{ \langle \sum_{k=1}^{n} c_k u_k, u_2 \rangle }{ \langle u_2, u_2 \rangle }u_2 + \dots + \frac{ \langle \sum_{k=1}^{n} c_k u_k, u_n \rangle }{ \langle u_n, u_n \rangle }u_n = \frac{c_n \langle u_n,u_n \rangle}{\langle u_n, u_n \rangle}u_n $$
The last "=" doesn't hold i think
this is just getting hard to write
I mean by the last "=" that the sum can be written as shown
then we have just that = $c_n u_n$
at the end
 
@ÍgjøgnumMeg hey
 
-Hi @Mathei @ÍgjøgnumMeg
 
@Semiclassical Any thoughts on that?
 
Buonasera @Alessandro
 
How are you?
 
8:54 PM
well, I'd simplify that to $\langle \sum_{k=1}^n c_k u_k, u_1\rangle = \sum_{k=1}^n c_k \langle u_k,u\rangle$
 
Sto bene, grazie. Sto scrivendo la mia tesi. E tu?
 
plus you can write that entire summation as $$P_U(v)=\sum_{j=1}^n \sum_{k=1}^n c_k \frac{\langle u_k,u_j\rangle}{\langle u_j,u_j\rangle}$$
 
yes
 
So, what do you know about $\langle u_k,u_j\rangle$?
 
Scusa per parlare italiano. Sono contento di tutte le occasioni di esercitarmi
 
8:58 PM
not quite sure
 
In other words: What do you know about the angle between the basis vectors?
 
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