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12:11 AM
Can anyone solve this inverse Z-transform problem? $\frac{1 - 2z^{-1} + z^-2}{1 - z^-2}
$\frac{1 - 2z^{-1} + z^-2}{1 - z^-2}$
12:25 AM
@Lewis I thought inverse Z-transform isn't well-defined since $x(n) = u[n]$ and $x(n) = -u[-n-1]$ give the same Z-transform?
@LeakyNun yeah you have to specify the ROC but this problem doesn't have one
then maybe you should use $\frac{1}{1-z^2} = 1 + z^2 + z^4 + \cdots$
which has a ROC of 1
@LeakyNun Hmm ok
I mean Z-transform is not injective so "inverse Z-transform" doesn't make sense
yeah there's also a case where ROC is the complement of a disc
Wiki assumes a causal ROC for inverse Z-transform
@Lewis what do you think
Well, the answer is x(n) = {1, -2, 2, -2, 2 ...}
I assume it was just long division but maybe I messed it up
12:37 AM
then that's anti-causal?
2 hours later…
2:54 AM
any help on this?
Q: Exactly one of the three points $L,M,N$ lies on the triangle $ABC$, where $AL, BM, CN$ are proper Cevian lines

BAYMAX $AL, BM, CN$ are proper Cevian lines and are concurrent at an ideal point. To prove - Exactly one of the three points $L,M,N$ lies on the triangle $ABC$. I was thinking that from Ceval's theorem we have, $\frac{BL}{LA}.\frac{BN}{NC}.\frac{BM}{MC} =1$. And $P$ is the only ideal point(I am t...

2 hours later…
4:44 AM
> Orthogonal Polynomials: Monic orthogonal polynomials, three term recurrence relationship, best approximation by polynomials in inner product induced norms, Chebyshev and Legendre polynomials. Fourier series.
Polynomial Interpolation: Lagrange, Newton and canonical forms of interpolating polynomial, existence and uniqueness, divided differences, error analysis, Runge phenomenon.
Mini-Max Approximation: Best approximation by polynomials in the uniform
sense, Chebyshev equioscillation theorem, optimal interpolation points zeros
this course is very algorithmic
1 hour later…
6:08 AM
Please provide an example of field extension that is not algebraic field extension.
6:36 AM
Can $F(\alpha)=\{a_0+\alpha a_1+\ldots+\alpha^{n-1} a_{n-1}:a_0,\ldots,a_{n-1}\in F\}$ hold if $\alpha$ is transcendental over $F$, for some $n$?
7:13 AM
Well $1/\alpha\in F(\alpha)$ @Silent
and if $1/\alpha=p(\alpha)$ for some polynomial $p$ then $\alpha$ is a root of $xp(x)-1$
which can't happen if $\alpha$ is transcendental
@AkivaWeinberger Why did you assume $1/\alpha=p(\alpha)$?
Isn't that the right-hand side of your equation?
$a_0+\alpha a_1+\dotsb+\alpha^{n-1}a_{n-1}$ is $p(\alpha)$ with $p(x)=a_0+\dotsb+a_{n-1}x^{n-1}$
(Also: did you mean $F(\alpha)$ or $F[\alpha]$)
(In the latter case you'd want to use $\alpha^n$ instead of $1/\alpha$ I guess)
oh! thank you. I thought that you were talking about some irreducible polynomial that one does in the beginning of field theory classes, while proving Kronecker Theorem, that $F[x]/(p(x))$ is field ext of F
Sorry for misunderstanding.
@AkivaWeinberger I meant smallest field containing $F$ and $\alpha$.
So that contains $1/\alpha$
If $\alpha$ is transcendental then $F[\alpha]$ and $F(\alpha)$ are infinite-dimensional vector spaces over $F$
i see
7:29 AM
Q: Digamma equation identification

Akiva WeinbergerI was messing around with the digamma function the other day, and I discovered this identity: $$\psi\left(\frac ab\right)=\sum_{\substack{\large\rho^b=1\\\large\rho\ne1}}(\rho^a-1)\ln(1-\bar\rho)-\gamma$$ when $0<\dfrac ab\le1$. It's unusual in that it sums over the $b$-eth roots of unity (which ...

@AkivaWeinberger But doesn't that mean that there is some linear combination of basis vectors that gives $1/\alpha$? While your argument above shows that no finite combination gives $1/\alpha$
I completely forgot about this
I have no idea how I did that
(re: digamma thing)
@Silent $\{\alpha^n:n=0,1,2,\dots\}$ isn't a basis for $F(\alpha)$
You'd need $\{\alpha^n:n=\dots,-2,-1,0,1,2,\dots\}$
7:31 AM
Oh wait
That's not even enough actually
because we need $1/(\alpha+1)$ in it as well
so I dunno what would be an easy basis
If I had to guess: $\alpha^n$ for nonnegative $n$, plus $\alpha^n/(p(\alpha)^m)$ with $n<\deg(p^m)$ for all irreducible $p$s
'cause you can do partial fractions
Hm. Can you write $\dfrac x{(x+1)^2}$ as a sum of "simpler" rational functions? You can't, right?
So, $a$ is nilpotent if there exists $n\in\mathbb{N}$ such that $a^n=0$. This is in regards to the Ring's multiplication operation, correct?
@AkivaWeinberger hint: partial fractions (i.e. Chinese Remainder Theorem)
@Rithaniel yes
Whereas, like, $\dfrac1{x^2-1}=\frac12\dfrac1{x-1}-\frac12\dfrac1{x+1}$, and $\dfrac x{x^2-1}=\frac12\dfrac1{x-1}+\frac12\dfrac1{x+1}$
Alright, danke
@AkivaWeinberger oops I just saw you said "partial fractions"
7:37 AM
@LeakyNun Not if the denominator is a power of an irreducible, I'm pretty sure
what we said aren't mutually exclusive
What does $F[\alpha]$ mean by the way? I think it means set of all elements which we get by plugging in $\alpha$ in each polynomial.
Smallest ring containing $F$ and $\alpha$
$\Bbb R[x]$ is the polynomial ring
So, if $\alpha$ is algebraic then $F[\alpha]=F(\alpha)$?
(Prove this!)
How do you know that $\dfrac1{\sqrt[3]2+\sqrt[3]3}\in\Bbb R[\sqrt[3]2+\sqrt[3]3]$?
Hint: vector spaces
Hm actually that might not be the best example
How do you know that $\dfrac1{\sqrt[3]2+1}\in\Bbb R[\sqrt[3]2]$?
Or, in general, how do you know that $\dfrac1{\alpha+1}\in\Bbb R[\alpha]$ if $\alpha$ is algebraic?
7:47 AM
I have a second small hint with more words if you want
@AkivaWeinberger Shall i take cube of $\dfrac1{\sqrt[3]2+1}$ to begi?
That would give you $\dfrac1{2+3\cdot2^{2/3}+3\cdot2^{1/3}+1}$ or $\dfrac1{3\sqrt[3]4+3\sqrt[3]2+3}$ if I'm not mistaken
@AkivaWeinberger this compulsory course I'm taking is as algorithmic as possible ._. help
And you want to escape the class in O(ln n) time
What's the course?
7:52 AM
"numerical analysis"
they spent like 9 lectures (out of 30) talking about polynomial interpolation
it's probably actually interesting
but like I need to spend time to learn this stuff lol
All I know about that is Lagrange interpolation
yes, they spent 9 lectures talking about it
Why would you need 9 lectures for it? You can describe it in a page
(interpolation is unique btw, there's only one polynomial of degree n to interpolate n+1 points, so every interpolation is "Lagrange interpolation")
Proof: vector spaces
7:55 AM
@AkivaWeinberger corollary: a polynomial of degree n has at most n roots? :P
@AkivaWeinberger they talk about how to compute the coefficients and how much the error is
@LeakyNun Oh, that's another proof
@AkivaWeinberger I can't think of any other way!
@Silent Hint: give me a basis for $\Bbb R[\sqrt[3]2]$
What does multiplication by $\sqrt[3]2+1$ do to that vector space
@LeakyNun This sounds like it could be interesting but it's not the sort of thing I know a lot about
one basis is $\{1,\sqrt[3]2\}$
No it's not
You also need $\sqrt[3]4$
8:01 AM
@AkivaWeinberger btw there is an alternative basis of the polynomials of degree <= n given n+1 complex numbers $z_0, \cdots, z_n$: $\prod_{k \ne j} \frac{z - z_j}{z_k - z_j}$
that's Lagrange interpolation
@AkivaWeinberger how did you figure that out? By multiplying by $\sqrt[3]2+1$? Also, how do we know there is no other element in basis?
@AkivaWeinberger replace $\Bbb R$ with $\Bbb Q$ everywhere
Oh crap you're right @LeakyNun
$\Bbb R[\sqrt[3]2]$ is just $\Bbb R$
Right so $\Bbb Q[\sqrt[3]2]$
but the deal is
let $\alpha$ be $\sqrt[3]2$. We know that $\alpha^3-2=0$, yeah?
8:05 AM
If $\alpha^2$ were a combination of $1$ and $\alpha$, then we'd have $\alpha^2=A+B\alpha$
which would mean $\alpha^2-B\alpha-A=0$
which would mean that $\alpha$ is the root of a quadratic polynomial
but $\sqrt[3]2$ isn't the root of any quadratic polynomial
So we get $\{1,\alpha,\alpha^2\}$ as a basis
or $\{1,\sqrt[3]2,\sqrt[3]4\}$
Well, we still need to show that those generate everything in $\Bbb Q[\sqrt[3]2]$
$\alpha^4\in\Bbb Q[\alpha]$, right?
When $\alpha=\sqrt[3]2$, what's $\alpha^4$?
It's $\alpha^3\cdot\alpha=2\sqrt[3]2$
8:09 AM
i was typing :)
Aw, sorry
so $\alpha^4$ is in the span of $\{1,\alpha,\alpha^2\}$
We can show inductively that $\alpha^n$ is in the span of $\{1,\alpha,\alpha^2\}$ for all $n$
and we know that the $\alpha^n$ span $\Bbb Q[\alpha]$
8:11 AM
thank you!
OK so now we have our vector space and we have a basis for it. $\Bbb Q[\sqrt[3]2]=\{a+b\sqrt[3]2+c\sqrt[3]4:a,b,c\in\Bbb Q\}$
What does multiplication by $\sqrt[3]2+1$ do to our basis vectors?
it gives elements like $(a+c)+\sqrt[3]2(a+b)+\sqrt[3]4(b+c)$
any idea what X^reg means? for a topological space X
8:19 AM
$a+2c$ at the start
because $\sqrt[3]4\cdot\sqrt[3]2=2$ @Silent
We can make it a matrix if we want to
0&1&1\end{bmatrix}\begin{bmatrix}a\\b\\c\end{bmatrix} = \begin{bmatrix}a+2c\\a+b\\b+c\end{bmatrix}$$
That's one way to think about it
Let's go with this way first
Do you know how to compute the inverse of a matrix?
Well what would we want to solve
in order to write $\dfrac1{\sqrt[3]2+1}$ as $a+b\sqrt[3]2+\sqrt[3]4$
This is the same as solving $1=(\sqrt[3]2+1)(a+b\sqrt[3]2+\sqrt[3]4)$
8:25 AM
Which means it's the same as solving $1=(a+2c)+(a+b)\sqrt[3]2+(b+c)\sqrt[3]4$
Which means it's the same as solving
0&1&1\end{bmatrix}\begin{bmatrix}a\\b\\c\end{bmatrix} = \begin{bmatrix}1\\0\\0\end{bmatrix}$$
Why 1 in place of a+2c only?
I mean, can we take e.g., (1/3,1/3,1/3)?
I need to go for a bit
ok ok. good bye
8:31 AM
6 mins ago, by Akiva Weinberger
Which means it's the same as solving $1=(a+2c)+(a+b)\sqrt[3]2+(b+c)\sqrt[3]4$
Think of the LHS as $1+0\sqrt[3]2+0\sqrt[3]4$
You want $a+2c=1$, $~a+b=0$, and $b+c=0$
@AkivaWeinberger so sorry. so solution is $(a,b,c)=(1/3,-1/3,1/3)$.
So, I'm looking for a proof of a fairly basic fact. That $1+(-1)^nx^{n+1}=(1+x)\sum_{i=0}^n(-1)^ix^i$. I'm blanking on what exactly to search for, though.
@AkivaWeinberger, so sorry that i wasted so much of your time.
@Silent Not at all
So $\dfrac1{\sqrt[3]2+1}={}?$
8:59 AM
(I have found the proof I was looking for. It's weird, I feel that I must have somehow skipped a class at some point where polynomial division was explored and verified.)
@Rithaniel $\begin{matrix}1&-x&+x^2&-x^3&\\&x&-x^2&+x^3&-x^4\end{matrix}$
Ah, that's a pretty simple way to show it.
Okay, yeah, I knew it was a basic fact.
I'm gonna keep that arrangement of elements in my head for future reference.
This is a geometric progression @Rithaniel
Yeah, related to the geometric series.
9:24 AM
well there's a better way to do it using Cayley Hamilton (@AkivaWeinberger @Silent)
He needs to answer $\dfrac1{\sqrt[3]2+1}={}?$ first
@AkivaWeinberger Let $p$ interpolate $(z_0, f_0), \cdots, (z_n, f_n)$ and $q$ interpolate $(z_0, f_0), \cdots, (z_{n-1}, f_{n-1})$
then $p-q$ is zero at $z_0, \cdots, z_{n-1}$, so $p-q = C (z-z_0) (z-z_1) \cdots (z-z_{n-1})$
evaluate at $z_n$ to give $f_n - q(z_n) = C \prod (z_n - z_k)$
so $C = \dfrac{f_n - q(z_n)}{\prod (z_n - z_k)}$
which is also the $z^n$-coefficient of $p$
and is called the divided difference, notated $f[z_0, \cdots, z_n]$
it has a nice formula: $f[z_0, \cdots, z_n] = \dfrac{f[z_0, \cdots, z_{n-1}] - f[z_1, \cdots, z_n]}{z_0 - z_n}$
which should remind you of the first difference method
10:16 AM
@AkivaWeinberger It is $\frac{1}{3}(1+\sqrt[3]2+\sqrt[3]4)$
Sorry! $\frac{1}{3}(1-\sqrt[3]2+\sqrt[3]4)$. Can't edit now
10:32 AM
errrm if $M$ is an f.g. free $\Bbb Z$-module, why is it also f.g. as an $\mathcal{O}_K$-module ? (ring of integers of number field $K$) I'm pretty sure this is obvious but I can't remember why! Is it literally just because $\Bbb Z \subset \mathcal{O}_K$
@LeakyNun oh. i think i spotted in artin algebra
@ÍgjøgnumMeg yes
fair one
10:47 AM
@Silent Ya
Arright so I mentioned there were two ways to see this
This is the more computational way
The other is an existence proof
Remember that our original question was just showing that $\dfrac1{\sqrt[3]2+1}$ is in $\Bbb Q[\sqrt[3]2]$
So the deal is that $\Bbb Q[\sqrt[3]2]$ is a finite-dimensional vector space (specifically, three-dimensional)
The important thing is, any injective linear map from a finite-dimensional vector space to itself is also a surjection.
(and thus a bijection)
Now, $x\mapsto(\sqrt[3]2+1)x$ is a linear map from $\Bbb Q[\sqrt[3]2]$ to itself
and it's injective
Therefore, it's surjective
Therefore $1$ is in its image
which means there's some $x\in\Bbb Q[\sqrt[3]2]$ such that $(\sqrt[3]2+1)x=1$
Hi. Am I wrong about this solution?
10:52 AM
meaning that $\dfrac1{\sqrt[3]2+1}\in\Bbb Q[\sqrt[3]2]$
Isn't that nice?
And the nice thing is that it generalizes
If $\alpha$ is algebraic, then $\Bbb Q[\alpha]$ is finite-dimensional (why? what's its dimension?)
and so for any $A\in\Bbb Q[\alpha]$, we have that $x\mapsto Ax$ is an injective linear map from $\Bbb Q[\alpha]$ to itself
which means it's surjective
which means there's an $x\in\Bbb Q[\alpha]$ solving $Ax=1$
which means $1/A\in\Bbb Q[\alpha]$
(Why doesn't this work when $\alpha$ is transcendental? For example, $1/\pi\notin\Bbb Q[\pi]$. Where does this proof go wrong in that case?)
its really nice. thinking about your last question
@LeylaAlkan If it's symmetric, wouldn't $t_{[321]}$ be zero because they'd all cancel out?
@AkivaWeinberger $S_4/V_4 = ?$
ok $S_4/V_4 = S_3$
@AkivaWeinberger yes
@LeakyNun 'cause you can write everything in $S_4/V_4$ as $[(1xyz)]$
11:01 AM
ah because $V_4$ is transitive
it's quite hard to imagine a surjection $S_4 \to S_3$
Or, rather, $\left[\begin{pmatrix}1&x\\y&z\end{pmatrix}\right]$
is there a geometric meaning to all this
because I view $V_4$ as symmetries of a rectangle
(acting on the corners of the rectangle)
Isn't $(3,0)$ tensor symmetric? @AkivaWeinberger
Question: What does the degree of an algebraic number refer to again?
11:03 AM
@LeylaAlkan I don't actually know much about tensors
@Rithaniel Degree of its minimal polynomial
@Rithaniel the degree of its minimal.... ok
(the smallest polynomial of which it is a root)
Hah, snipe
@LeakyNun I also kinda sniped your "geometric meaning" question with the rectangle thing
but it's clear with $S_4$ being the symmetries of a cube as well
@AkivaWeinberger In that case it is not finite dimensional as you told me earlier today. Hence injective does not imply surjective.
Ya @Silent
Color the faces of a cube in three colors so that opposite sides get colored the same @LeakyNun
11:04 AM
Symmetries of the cube permute the colors
@AkivaWeinberger nice
Thank you sooo much for all this help akiva
And the nontrivial elements of $V_4$ corresponds to 180 degree rotations about some axis @LeakyNun
@Silent No problem
@LeakyNun Do you know how to do a Rubik's cube?
So, $[\mathbb{Q}[\alpha ]:\mathbb{Q}]=n$ where $n$ is the degree of $\alpha$. Is this correct?
There's a simpler version where you only allow yourself to make 180 turns of the faces
and try to solve it with those as well
Simpler 'cause there's far fewer combinations it can end up in
Every face can have at most two colors on it
@Rithaniel Yeah
11:06 AM
Also, here's a puzzle on the Rubik's cube that feels like it should be simpler than the full thing but isn't
Only allow yourself to turn the top and right faces
Mix it up with those and solve it with those
There's this 2x2x3 section of the cube that stays constant when you do it like that
and it feels like it should be easier, but it isn't
in my experience at least
11:48 AM
@AkivaWeinberger yeah
@AkivaWeinberger why would Tchebyshev polynomials be related to minimizing $\sup \prod |(x-x_j)|$
Those are $\cos(n\arccos x)$, yeah?
given $n$ find $x_0, \cdots, x_n \in [-1, 1]$ minimizing $\displaystyle \sup_{-1 \le x \le 1} \prod_{j=0}^n |x - x_j|$
Or something similar
Hm so the log derivative is $\sum\frac1{x-x_j}$, yeah?
'cause the derivative of $\ln|x|$ is $\frac1x$, not $\frac1{|x|}$
11:57 AM
so we're looking at the.. harmonic mean of $x_j$?
maybe not
No, it's not translationally invariant
I dunno if that's the right term but it would make sense
anyway so $\sum \frac1{x^\ast-x_j} = 0$
(or $x^\ast = \pm 1$)
So you gave a hint with the Chebyshev there
Problem is I don't know what to do with that
What's $(x-\rho)(x-\bar\rho)$ where $\rho$ is a root of unity
brb digging up notes for Legendre transform
Chebyshev of first or second kind?
12:07 PM
Question: Are there any finite rings with a right identity but no left identity?
Probably first kind if you’re doing minimax
@Semiclassical you probably know how to do this :P
13 mins ago, by Leaky Nun
given $n$ find $x_0, \cdots, x_n \in [-1, 1]$ minimizing $\displaystyle \sup_{-1 \le x \le 1} \prod_{j=0}^n |x - x_j|$
It presumably hinges on Chebyshev polynomials as orthogonal
If it were the sum of the squares we'd want $x$ to be at their mean I think
I don’t remember what the inner product for T_n is tho
12:08 PM
same as L^2
@Rithaniel Are there any infinite rings like that?
I guess one should be able to deduce it from $\int_0^\pi \cos(m t)\cos(n t)\,dt$ for non-negative integers n,m
how does orthogonality help anyway?
@AkivaWeinberger there must be a 3b1b-style answer to this question lol
good question
unfortunately I am not 3b1b
12:12 PM
@LeakyNun The gold standard
Well I hope so, because one of my assignments asks for an example of a ring with a right identity but no left identity.
My initial impulse is to look at the log of the objective function
Is $\{a,b\}$ with $a^2=a$, $b^2=b$, $ab=a$, and $ba=b$ associative?
In this case, both $a$ and $b$ are right identities
Looking at the comments on this question: math.stackexchange.com/q/2710542/137524
And then we can make our ring $\{0,a,b,a+b\}$ (characteristic $2$) @Rithaniel
That's finite
12:18 PM
Seems like the idea of looking for rings with more than one right identity is correct
Indeed, I saw some stuff about a unique right identity automatically being a left identity.
Ah that's cool
If $e$ is a unique right identity then $b(ea-a)=bea-ba=0$
Wait no
@LeakyNun my brain is not awake yet but I’ll note that you can instead consider the objective function to be a product of squares
so $ea-a+e$ is another right identity so $ea-a+e=e$ so $ea=a$ so $e$ is a left identity
This is just the proof from the link
I think that's how it goes, yeah. The basic of ring theory are wild.
12:23 PM
I thought I could simplify it
I could not
The x which maximizes that (for a particular choice of xi’s) will also maximize the original product
At which point I can maybe imagine how Chebyshev will enter? Not sure until I have a chance to write something down tho
Multiplication table
Something along the lines of expressing $\prod_i (x-x_i)$ as a sum of Chebyshev polys
@Semiclassical wat
(Too lazy to figure out how to typeset that)
Thing on the left times thing on the top
Dunno if that's the usual convention but whatever
Notice that $a+b$ is an annihilator
12:27 PM
Hrm. Not sure what I said actually helps tho
If you have two right identities then their difference must be an annihilator
@AkivaWeinberger just tell him it's the monoid ring
In fact I think that "unique right identity" is equivalent to "no right annihilators"
@Semiclassical how on earth is prod (x-x_i) a sum of Tchebyshev polynomials...
and then $ea-a$ is a right annihilator
12:28 PM
also they're only orthogonal in L^2 not L^infty
which means $ea-a=0$ so $ea=a$
because it’s a polynomial
And that's a better proof that a unique right identity is a left identity
Add a concept but lose the annoying $ea-a+e$ thing
@Semiclassical Can we say anything about $\sup (f+g)$ given $\sup f$ and $\sup g$
I doubt it
Well $\sup(f+g)\le\sup f+\sup g$ isn't it
But that's the only thing we can say
12:33 PM
The issue with my line of thinking is not “can you expand $\prod_{i=1}^n (x-x_i)$ in Chebyshev polynomials.” That’s trivial, since you can express the monomials 1,x,x^2,...x^n in terms of Chebyshevs
The problem is why it’s worth doing in the first place :/
@Semiclassical so what's your experience with minimax?
Are the Chebyshev polynomials monic?
Very thin tbh
I’m aware of it but have not dealt with it
12:36 PM
@AkivaWeinberger no, the leading coefficient is n
No, that’s not right either is it
How does $\int fg$ relate to $\sup f$ and $\sup g$
For n>0, the leading coefficient is 2^(n-1)
$\frac12\int_{-1}^1fg~dx\le\sup|f|\sup|g|$, yeah?
I don’t think that’s going to help much here
I guess my thought right now would be to take a few examples of small n
12:42 PM
No I think this is good
When is that an equality
Oh never mind
Also we need some way to keep the zeroes in $[-1,1]$
Lol, one simple way
'cause otherwise we could just do $x^2-c$ for higher and higher $c$ or something and make $\sup_{[-1,1]}$ go to minus infinity
Let $x=\cos t$ and $x_i=\cos t_i$ @AkivaWeinberger
@Semiclassical And there's in fact a funny combinatorial interpretation for the coefficient matrix when you do that.
For expressing the monomials that is.
Sounds right
12:53 PM
The Chebishev polynomial U_n is defined by U_n(cos theta) = sin((n+1)theta) / sin(theta).
That’s chebyshev of the second kind tho
Yes, that's how the funny path counting stuff works well.
Also, it’s the quotient of those not the product
oh yeah
Now if x and y are positive integers, then let c_{y+x, y-x} be the number of paths from (0,0) to (x,y) where each edge of the path does a step of either (0,1) or (1,0), and the path never goes to a point (x',y') above the diagonal where x'<y'.
And if k and l have different parity then c_{k,l}=0.

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