Let R be a relation over set A, then: Prove that IA∘R=R=R∘IA

IA= Identity Relation
∘= composition symbol
I don’t know where to start proving this question

complexity theory – Proof that a relation is in FP

How we can prove that the relation: $R= left{0,1right}^*times left{0,1right}^* in FP$
I understand that we need to find a polytime algorithm to decide whether $(x,y) in R$ since $(x,y)in R= left{0,1right}^*times left{0,1right}^*$
How can we find this? And this is enough to prove that $R in FP$?

probability – Ways of solving the following recurrence relation system.

Consider the following system of linear recurrence relations.

$$begin{aligned} p_n &= a cdot p_{n-1} – c_{n-1}\ c_{n-1} &= p_{n-1} – b cdot p_{n-2} + c_{n-2}end{aligned}$$

with $p_0 = 1$ and $c_1 = 1$.

I’ve tried to represent $p_n$ as a finite linear combination of $p_{k}, k < n$, but this doesn’t work for me. Maybe there is any chance to represent the final solution?

Any ideas?

recurrence relation – Two dimensional recursive function in $Olog n$ time complexity

It is well known that a recursive sequence or $1$-d sequence can be solved in $O log n$ time given that it has the form

$$a_n=sum_{k=1}^{n} C_ka_{n-k}$$

where $C_k$ is a constant. Examples would include polynomials like $n^2$ or $n^3$, exponentials like $2^n$, and Fibonacci Numbers defined by $a_n=a_{n-1}+a_{n-2}, a_0=0, a_1=1$. Factorials would not be included for example, because they are defined by $a_n=na_{n-1}$, and $C_k=n$ is not constant.

Let $a_{n,k}$ be a two dimensional sequence defined by

$$a_{n,k} = sum_{i=0,j=0, (ine n text{ }∧text{ }jne k)}^{n,k}C_{i,j}a_{i,j}$$

where $C_{i,j}$ is constant.

Is it possible to compute $a_{n,k}$ in logarithmic time (a.k.a $O log n$) or better?

I know one case where this is possible, namely if $a_{n,k}$ are coefficients of polynomials in a sequence ($1$-d recursive sequence), such that the degrees of each term differ by $1$. In this case, the diagonals of the sequence $a_{n,n+t}$ for some constant $t$, are $1$-d constant recursive functions, whose terms can be computed in $Olog n$ time. This doesn’t use the initial relation in terms of two indices though as expected.

However, in certain cases, the diagonals do not form $1$-d constant recursive functions, namely when the degree difference of consecutive polynomial terms is more than $1$.

algorithm analysis – Finding a tight bound for a recurrence relation

algorithm analysis – Finding a tight bound for a recurrence relation – Computer Science Stack Exchange

Column vectors of a matrix and its relation to the span of a Subspace S

I am bit stuck in the understanding of how to identify if the column vectors of a matrix span a Subspace S for instance? Example: If there is a projection matrix that projects a vector onto a Subspace S, how can we conclude that the column vectors of the Projection matrix spans the Subspace S?

focal length to aperture relation while zooming – is my russian lens lying to me?

I have this old soviet/russian “Granit” 80-200mm lens, connected to my Canon 550d DSLR through M42 adapter. I also have “modern” kit 18-55mm lens.

What I don’t get is how aperture numbers work on these two, they don’t seem to mean the same thing, so clearly I’m missing something.

Aperture number, e.g f/4.5 (wide open in Granit lens) should mean that opening diameter is 4.5 times smaller than focal length. So for 80mm I should get about 17mm opening, which sounds reasonable. But when I zoom that lens in, the same f/4.5 should give me wider and wider opening, which is not the case, opening stays exactly the same as it was at 80mm. It doesn’t even seem to be physically able to give me this 44mm (f/4.5 at 200mm) opening, aperture housing just isn’t big enough.

So by my understanding, after zooming from 80mm to 160mm, I should be at f/9? which should give me much darker photo, given the same ISO and time. But this is not the case, I can zoom in and out freely without correcting anything and still get correct exposure.
Also, I don’t think that this “silent” change of aperture number would be acceptable when this lens was brand new and all the photos were taken on film, using manual separate light meters.

On my modern lens (18-55mm, f/3.5-5.6), when I zoom it while holding depth of field preview button, I can see and hear electronically controlled aperture opening and closing a little, even when set to like f/9 which is available through entire focal range. So keeping f/9 requires change in opening size.

So what gives? Is this russian lens just lying to me (but why correct exposure then?), or isn’t f number related to focal length at all, or at least is not directly related? Why does Canon lens change opening diameter, but Granit doesnt?

Also it seems that aperture housing is moving back and forth when zooming my modern lens (but its hard to tell by looking through lens), and I’m rather sure it doesn’t move in Granit, since it’s mechanically controlled and just seems to stay in place – this part I don’t get at all, but since I’m not sure it really moves, maybe lets just skip that.

Lore of relation between warlock and wild magic

I have a question about wild-magic areas in DnD.
I’m playing D&D 5e but my question is more lore related. If you want I’ll accept an dead-magic answer 🙂

Are wild-magic zones affecting warlock?

As I interpret the lore wild-magic is basically wound in Weave, in magic threads(or net). So it shouldn’t affect divine magic as it’s a “different type” of magic. The problem is that warlock also has some entity from whom he receive magic. We can say it’s type of divine magic, can’t we? Warlock is praying to someone or something and effect something is happening, pretty much like cleric, it’s a favour. A highier beeing should be able to cast a spell without any difficulty therefore warlock should too.

Is there any in-lore reason that wild-magic affects or not warlock or is it a gm decision? If so, give me the reasons.

time complexity – solving Recurrence relation f T(n)=T(n/2)+log log(n)

T(n)=T(n/2)+log log(n)--------1
T(n)=T(n/4)+log log(n/2)+log log(n)-------------2
T(n)=T(n/8)+log log(n/4)+log log(n/2)+log log(n)-----------3
T(n)=T(n/2k)+log log(n/k)+....+log log(n/2)+log log(n)
T(n)=1+log log(n/k)+....+log log(n/2)+log log(n)
 n/2k=1 => n=2k
 T(n)=1+log log(2K/k)+....+log log(n/2)+log log(n)
 T(n)=1+log log(2)+....+log log(n/2)+log log(n)
 T(n)=1+log (log(2)*....*log(n/2)+log(n))

I ARRIVED AT THIS POINT BUT I DIDN’T KNOW HOW TO CONTENU ANY HELP PLEASE

ag.algebraic geometry – Relation between flag varieties of Langlands dual groups

Let $G$ be a reductive algebraic group over $mathbb C$ and let $G^vee$ be its Langlands dual group. What are the precise relationships between the complete flag variety $X$ of $G$ and the flag variety $X^vee$ of $G^vee$? Are there interesting morphisms of varieties that connect them?

Apologize if the question is vague. I am new to Langlands dual groups, but recently discovered that some properties of modules of $mathfrak g$ can be related to Schbert cell structures on the dual side. It will also be appreciated if anyone can point out references on relationships between geometries of $X$ and $X^vee$.

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