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In the history of statistics, economy and decision theory, the St. Petersburg paradox plays a key role. This lottery problem goes back a full three centuries to the mathematician Nicolas Bernoulli who first formulated the problem in 1713. Twenty five years later, in 1738, his nephew Daniel Bernoulli presented the problem to the Imperial Academy of Sciences in St. Petersburg. That presentation not only gave the paradox its name, it also created a lot of commotion amongst mathematicians.

And even today, following three centuries of discussion and debate, the problem continuous to intrigue scientists and lay people alike. Yet, the paradox got fully resolved in the middle of the previous century. I was reminded of this paradox when it came up here at Science 2.0 in the comment section of a recent blog post on math paradoxes. When I checked the Wikipedia page dedicated to this paradox, I was shocked to see that this article mentions several approaches that have been claimed to resolve the paradox, but relegates the real resolution to this paradox to a short remark.

Time for a blogpost.

Imperial Academy of Sciences, St. Petersburg

# St. Petersburg Lottery

The rules of the St. Petersburg lottery are straightforward: a fair coin is tossed repeatedly, until a tail appears, ending the game. The pot starts at $2 and is doubled each time a head appears. After the game ends, you win whatever is in the pot. So if the first coin toss results in a tail, you get$2. A head followed by a tail pays you $4. Head-head-tail yields$8, etc. Now the question is: how much are you willing to pay to enter this lottery? Statistical reasoning tells us to compute an ensemble average: investigate all potential outcomes and calculate the sum of the gains for each outcome weighted with the probability for that outcome. This weighted sum is known as the expected gain.

For the St. Petersburg lottery where one wins $2 with probability 1/2,$4 with probability 1/4, $8 with probability 1/8, etc. the expected dollar gain is (1/2) 2 + (1/4) 4 + (1/8) 8 + (1/16) 16 + ... = 1 + 1 + 1 + 1 + ... = infinite The divergence of the expectation value means that no finite amount suffices as a fair entry fee for this game. Yet, if you play the game, you are sure to win no more than a finite amount. Also in actual practice, participants subjected to this game are not willing to pay more than a small entry fees (typically less than$20).

What is happening here? Why is the expectation value for the St.Petersburg lottery such a poor representation of human actions in this lottery?

Interestingly, the resolution to the St. Petersburg paradox becomes particularly transparent when applying statistical physics insights.

# Steinhaus Sequence

Although I present the results in statistical physics language, the argument below follows the essence of Hugo Steinhaus' 1949 paper. It helps to investigate a simpler, deterministic game that shares its key characteristics with the St. Petersburg lottery.

Imagine an infinite sequence of powers of two. The numbers generated follow a deterministic rule, that is recursive in nature:

1) Start with an endless list of empty states and the number '1'
2) Double the number and, starting with the first empty state, fill every other empty state with the number
3) Repeat step 2) ad infinitum

Building a Steinhaus sequence

So we get a sequence that starts with the number '2', followed by a '4', then a '2', an '8', a '2', a '4', etc. If you would select at random one of he numbers generated, you would half of the time hit a '2', one quarter of the time a '4', and so on. Therefore, the expected value of numbers in the Steinhaus sequence is no different than that for the St. Petersburg lottery. More specifically, the Steinhaus expectation value diverges in exactly the same way as the St. Petersburg expectation value.

However, one may also ask what value we would obtain when averaging over subsequent numbers in the Steinhaus sequence. For the first two numbers, an average value of (2 + 4)/2 = 3 is observed. For the first four numbers we have (2 + 4 + 2 + 8)/4 = 4. Averaging over eight numbers we have (2 + 4 + 2 + 8 + 2 + 4 + 2 + 16)/8 = 5. We see a simple pattern developing. For 2n subsequent numbers we observe an average value of n + 2. This goes against the widespread intuition that the average gain per game should be independent of the number of games played. At least we expect this to be the case when the number of games played is large. Here, such is not the case.

Statistical physicists are familiar with this phenomenon. They refer to this peculiarity as 'non-ergodicity'. Ergodicity means that one can replace time averages with ensemble averages. Here, ergodicity does not hold as the time averages (the average values n+2 observed over 2n subsequent numbers) do not coincide with the ensemble average (the divergent expectation value). What is happening in any finite Steinhaus sequence, is that it visit large values contributing strongly to the average only logarithmically slowly, thereby causing a breakdown of the ergodic assumption.

# Ergodicity

Now back to the St. Petersburg lottery. Based on the learnings from the Steinhaus sequence, it is easy to observe that also a sequence of gains from a repeated St. Petersburg lottery shows a pronounced breakdown of the ergodicity assumption. It is straightforward to generate a long sequence of St. Petersburg lottery pay-offs. This can be done with a coin, or more efficiently with any spreadsheet program. As suggested by the Steinhaus exercise, we plot the average gain of the first 2n numbers for progressively larger values of n. This is done in below figure for n up to 15 (2n = 32,384) for two independent runs.

Although the average gains within each run fluctuate considerably, it is clear from the trends that the expected gains per game increases logarithmically with the number of games, just as we observed in the Steinhaus example. (To guide the eye, I have included in a fainter color the curve describing a pay-out per game of n+2 for 2n games.)

The conclusion is straightforward: only one type of averages are practically relevant in the St. Petersburg lottery. These are the the time averages. The St. Petersburg time averages are finite, but logarithmically dependent on the number of games played. As these time averages don't converge to a finite value in the long time limit, the ergodic hypothesis breaks down for St. Petersburg lottery results. Therefore, the ensemble average, aka the expectation value for the game, can be cast aside as a meaningless divergent quantity.

Math is simple if you're a physicist...


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# Topologically Transitive Action on a Metrizable Compact Topological Group

I'm trying to solve the following problem: Let $G$ be a metrizable compact topological group. Suppose for some $g_0 \in G$ the translation $L_{g_0}$ defined by $L_{g_0}g=g_0 g$ is topologically transitive. Then $G$ is an abelian group.

I know that since $L_{g_0}$ is topologically transitive, it is also minimal. I was hoping to show that the commutator of two elements is the identity and also wanted to incorporate the metrizability and compactness of the space somehow but didn't get too far.

Thanks

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1. Minimality means that every $L_{g_0}$-orbit is dense, in particular the orbit of the identity is dense.
2. The orbit of the identity is just the powers of $g_0$.
3. All powers of $g_0$ commute.
4. Multiplication is continuous and you can approximate any two group elements with powers of $g_0$.

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# The limit to continuous space and the danger of graph plots

Posted 1 year ago
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 In collaboration with Johan-Tobias Schäg, we were able to create a rule producing an ordinary 2D-grid (https://community.wolfram.com/groups/-/m/t/1946413). Wolfram himself also mentions grids several times (see for example: https://www.wolframphysics.org/technical-introduction/limiting-behavior-and-emergent-geometry/recognizable-geometry/). It is often assumed in several of the texts I read on this subject (the last link is an example) that in the limit of making grids finer and finer we obtain the well known 2-dimensional continuous space. This is not correct, however. In fact, it seems quite complicated to come up with a graph structure that resembles classical 2D-space even remotely (the same is true for higher dimensions as well).It is important to solve this problem since any model resembling physics should look like ordinary continuous space on large scales.In order to illustrate this, I will consider the simplest case known to us all; the regular 2D-grid (see the figure below). I have drawn a piece of a 2D-grid, as well as four points on the grid labeled A, B, C, and D. In the texts, I have read so far, there is no mention of how we think about distances in graphs. However, there are indirect hints. For example, when the subject of dimension and curvature is discussed, the distance $r$ from a point plays a crucial role. In these cases, the distance between two vertices is indirectly defined as the least number of (hyper)edges you would have to cross in order to get from one point to the other. This is a very logical definition, and I will keep using it here.Now keep in mind, that the graphs we draw do not illustrate these distances correctly. They are only representations of the graph. For example, take another look at the picture of the grid above. The distance between A and B is 1 (you could call one edge for the fundamental length unit, or you could simply omit units for simplicity) and the distance between A and D is 5. The distance between B and C is also 1. However, as the graph is drawn on this picture, the distance between A and C should be $\sqrt{2}$, while the true answer is that the distance is 2! In fact, you could never create graphs with non-integer distances. Okay, this is weird, but not problematic yet. After all, the graph should only look like ordinary space on large scales, and my example here is on the absolutely smallest scale possible. However, even on large scales, the answer is still the same. Imagine a huge grid with point A. Choose a direction and go $n$ steps in only that direction (left, right, up, and down are the possibilities on a grid). Now denote the point you end up at by B. There is no shorter path between the two points, which means that the distance between A and B is $n$. Now choose a new direction away from B, which is not the same (or opposite) as before. If you chose to go to the right away from A, you could choose to go upward now, for example. Go another $n$ steps up and label this point C. We are now in the same case as before, but by varying $n$ we can vary the scale of the setup as much as we want. What is the distance between A and C? One path to take is to go from A to B and then from B to C as above. In fact, there is no shorter path than this either! So the distance is $2n$. In the usual 2D-space, we expect the ratio between the distance from A til B (denoted $d(A,B)$) and from A til C to fulfill the following identity: $$\frac{d(A,C)}{d(A,B)}=\sqrt{2}$$ A non-integer value like this is no problem anymore (irrational numbers are, however, but never mind that for now). What is the case for our graph? Well, that is easily calculated: $$\frac{d(A,C)}{d(A,B)}=\frac{2n}{n}=2$$ The result is 2 no matter how huge $n$ is! You can make the grid as fine as you ever want it, but you will never get to the usual 2D-space. You could also say that the Pythagorean Theorem doesn't apply, which we know it has to in any physical space. The same problem arises for any regular grid.There are two main points up until now: 1) do not believe in the distances drawn on a graph (even simple, regular ones), and 2) just because you get an infinitely fine grid, you do not automatically get continuous space. All is not lost, however. Up until now, I have only mentioned regular grids. It is, however, not surprising if nature is not regular on the smallest of scales. In fact, I would be very surprised if that was really the case. Instead, a graph that appears 2-dimensional on large scales could very well look quite chaotic on the small scales. The vertices and edges could have very complicated and seemingly random connections (I have heard rumors of Wolfram mentioning this, but I am not sure). This could solve the problem, but the natural question then is this:How do we create a grid which obeys the Pythagorean Theorem on large scales?Other natural questions to ask are: How do we deal with angles on large scales? How and when are they defined? What happens to the irrational numbers that arise from the Pythagorean Theorem? Are they approximations? Can we create a large scale isotropic graph?
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Posted 1 year ago
 I think I've made a graph capable of approximate on large scale the euclidean space. The method to make it is literally "cheating" because is produced iterating a simple algorithm on the euclidean 2D space to produce a suitable mapping of the space dependant by a single parameter (an angle that is obtainable dividing by an integer $2\pi$, so in some sense a single positive integer) In the image (I apologize for GeoGebra, but I used the first thing I had available on the computer), I've used an angle of $15°$.Here you can find the project (I apologize again for having done the constructions manually, I'm thinking to write a real program that maybe compute some macroscopic distance and angle). If you change the partition angle (with $1°$ is visible) the partition cover finer and finer way the space. The use of a $\frac{2\pi}{n}$ angle is needed to obtain a FINITE graph from the computation. This with the fact that our universe is actually finite, is sufficient to say that at the end of the algorithm we have a finite number of nodes and edges. I'm going right now to make some suitable test at various scale.An important needed step of the algorithm is to remove the double-counting of some points, and in the case, the node already exists, connect the node for which the algorithm is being applied with the existing node instead of introducing a new one.To make sure the mapping preserves the circular symmetry from every point, the algorithm needs to be run for every node in the graph. The graph can be seen as completed when a successive application of the algorithm on any node, preserve the graph. I suppose this is possible from the finite space to map and from the $\frac{2\pi}{n}$ angle used for the partition.PS: I apologize if I have not explained myself clearly, I prepare some programs and as soon as possible I publish them to make the concept clearer and finally have numbers in hand.
Posted 1 year ago
 How do we create a grid which obeys the Pythagorean Theorem on large scales? That's also called Euclidean distance Can we create a large scale isotropic graph? Both of these questions were resolved in this recent post of mine: https://community.wolfram.com/groups/-/m/t/1945186
Posted 1 year ago
 Well i am not convinced yet. As there are many unclear things. Can you write a program (better a replacement rule based) which generates such a graph as you are thinking?Could you also proof algebraically that a left and up movement results in a length of sqrt(2) in the limit?
Posted 1 year ago
 I'm working on something similar, actually, I have some problem with the floating-point approximation in the really particular rule I'm using. Theoretically, the rule will converge for a portion of the euclidean space, but that isn't actually happening for approximation errors, so I'm rewriting all with a geometrical substitution rule (a little more complex to implement)The method I'm using (the circle mapping of the euclidean space) is intended to preserve some fundamental rotational symmetry. For the distance, I'm using the number of jumps, even if in the paper the part about Minkowskian distance is REALLY incomplete and inaccurate about the mapping of the coordinates on the spatial, causal and brachial space.
Posted 1 year ago
 What are you using floating point numbers for? You can just use my method instead. You can use it to build the graph by simple rules from which the Euclidean plane emerges: https://community.wolfram.com/groups/-/m/t/1945186If you want to build a 4D Minkowski space you can use my new method instead, where the rotational invariance emerges from Wigner rotations: https://community.wolfram.com/groups/-/m/t/1953906
Posted 1 year ago
 However, these (very beautiful) methods are not based on replacement rules. What I was experimenting with was a particular mapping of the Euclidean space through a specific substitution rule (which I am still studying) to preserve all (too much) of the Euclidean properties, even at an intermediate scale. For experimental working, I'm using a lot of rotation. Theoretically, with the right bounding condition the algorithm will need to converge, but the floating-point approximation breaks it all.Furthermore, your methods are for the spacetime causal graph (if I understand decently, but I still have many doubts about the various mappings of the Wolfram model), while here we were trying to build a purely Euclidean space, therefore working on the space hypergraph.I would also like to add that from the 4D Minkowsian method it is not evident, exactly as in the paper, how the mapping of the spacetime points on the whole lattice on which the Minkowskian distance is then defined, leaving room for an arbitrariness that leads to the complete breakage of any metric model based directly on a norm dependent on the coordinates and not on the topological characteristics of the graph.In the end, you can't have a metric if you don't first specify how to interpret the topology, including the coordinates of the space.But I may have understood cabbage for potatoes.
Posted 1 year ago
 I would also like to add that from the 4D Minkowsian method it is not evident, exactly as in the paper, how the mapping of the spacetime points on the whole lattice on which the Minkowskian distance is then defined, leaving room for an arbitrariness that leads to the complete breakage of any metric model based directly on a norm dependent on the coordinates and not on the topological characteristics of the graph. In the end, you can't have a metric if you don't first specify how to interpret the topology, including the coordinates of the space. I didn't read their paper but their findings on relativity seem to be complementary to mine. In my model each reference frame is a 16-cell honeycomb that provides you with coordinates.
Posted 1 year ago
 16-cell honeycomb that provides you with coordinates How? If there is an arbitrariness to obtain the coordinates, and it is not possible to obtain it directly from the topology, it is a huge problem.I believe that the use of lattice coordinates to define the Minkowskian norm is a huge conceptual error because it requires as a fact present here on page 13, that there is a valid graph embedding on a Euclidean space, and does not clearly demonstrate that it is guaranteed a priori.The solution can come free from the removal of the concept of microscopic coordinate and be based entirely on the topology and the metric of the graph, defining the metric using the typical distance function of a graph, i.e. the minimum jump number to connect two points. In this way there is a very important criterion for mapping coordinates, that is that they must respect the metric just introduced. In this way, an effective, various and universal coordinate system is obtained, because there is the certainty that it preserves the most important characteristic of the graph, it's metric, to which everything is conceptually traced.
Posted 1 year ago
 The solution can come free from the removal of the concept of microscopic coordinate and be based entirely on the topology and the metric of the graph, defining the metric using the typical distance function of a graph, i.e. the minimum jump number to connect two points. In this way there is a very important criterion for mapping coordinates, that is that they must respect the metric just introduced. In this way, an effective, various and universal coordinate system is obtained, because there is the certainty that it preserves the most important characteristic of the graph, it's metric, to which everything is conceptually traced. Yes or you can use the Lorentz distance which is the length of the longest path along the directed edges, which is what I mentioned at the end of my post. It's the path with the most jumps, that's what it's all about. From this you can derive the Euclidean distance which is the length of some shortest path.
Posted 1 year ago
 Questions for both your approaches (Leuenberger and Pasqua):Can you specify more clearly what edges and vertices your eventual graphs consist of? When you intertwine two grids, Leuenberger, what edges do you create? Or, Pasqua, what are the edges and vertices of your graph? I assume the vertices are the crossings, while the edges are the lines "doing the crossing"? It seems to me to become extremely complicated to determine distances between points, so can you elaborate on that?And mayby I am simply a bit slow, but how do you obtain Euclidean space in the limit? Remember that you cannot simply embed a graph in the plane and expect to "see" the correct distances.I assume we all use the basic graph metric, otherwise, can you specify what other metric you use and justify why?
Posted 1 year ago
 When you intertwine two grids, Leuenberger, what edges do you create? No edges are created, the grids are instead intertwined by unifying specified pairs of nodes into single nodes. But this method of construction is only for ease of explanation. In reality a grid would be born within an older grid and they would be joined from the start and they would grow simultaneosely together in time direction.
Posted 1 year ago
 Well i am not convinced yet. As there are many unclear things. Can you write a program (better a replacement rule based) which generates such a graph as you are thinking? Could you also proof algebraically that a left and up movement results in a length of sqrt(2) in the limit? I won't have time, You could try to program the many interweaving grids (not just A and B, but A,B,C,D,...) and then measure the long range graph distance between two points that are on grid A.
Posted 1 year ago
 This paper might be of relevance: https://arxiv.org/pdf/gr-qc/0311055.pdf
Posted 1 year ago
 If I understand clearly, this is about how to take a manifold and make it discrete such that Lorentz invariance is conserved.This is not what I was thinking though. Two problems: Time and space are fundamentally separated in the Wolfram model, and Lorentz invariance is guaranteed in "most" graphs automatically. The referenced text is interesting but maybe not that relevant. If I understand it correctly, the idea to make a manifold discrete is to pick points on a manifold discontinuously (an intuitive approach indeed). The metric between these points is obtained by the metric from the manifold. This would, for example, be equivalent to me requiring the diagonal distances in the grid to be $\sqrt{2}$. Especially the second problem was one of the main points of my post. Even when you can embed a graph, you simply cannot generally choose a metric like that. There is no way to know which manifold your graph would "fit into" (and they are certainly not unique).The entire idea is to build a graph from "bottom up" so to say, rather than already having a manifold and build a graph onto it. Also it should be mentioned (not only for this comment, but some of the others too), that even if you have a set of points you don't have a graph. You still need to define the edges as well.Having a graph with a manifold-induced metric is also kinda weird, since graphs should be thought of as abstract concepts which can (if desired) be drawn/embedded in an infinite different possible ways. Having a graph on a manifold "requires" you to embed it on that manifold in some sense.
Posted 1 year ago
 I think the point is if you draw a graph with each node having a random number of neighbors drawn from a Poisson process, this will eventually smooth out the metric so that you do not get the usual taxicab metric for a discrete cubic lattice.


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# Tag Info

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To elaborate on Dmitri Pavlov's comment, your questions have been answered in a recent paper by Yehonatan Sella. Sella give an example (Example 0.3) of a locally contractible space $X$ for which $\mathcal{S}^k(X) \xrightarrow{\mathrm{shf}_X} \tilde{\mathcal{S}}^k(X)$ is not surjective. Sella's space has $5$ points, and the argument that $\mathrm{shf}_X$ is ...

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I think the function you are looking for is $f(x)=\int_{\gamma_x} \alpha$ where $\gamma_x$ is a path from some chosen base point $x_0$ to $x$. But then for this to be well defined you have to say that all paths between $x$ and $x_0$ are homotopic, i.e. that your sphere is simply connected, so the whole argument gets a bit circular... P.S. Traying to write ...

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You used it when you wrote this equation: $$\frac{\ker \phi \oplus H}{\operatorname{im} \partial_1} = \frac{\ker \phi}{\operatorname{im} \partial_1} \oplus H$$ You're implicitly using the fact that $\operatorname{im} \partial_1 \subset \ker \phi \iff \phi \circ \partial_1 = 0$. Otherwise taking the quotient $\ker \phi / \operatorname{im} \partial_1$ doesn't ...

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In terms of the logical process of inducing maps, you can think in the following way: A continuous map $f : X\rightarrow Y$ induces chain maps $f_\# : C_k(X) \rightarrow C_k(Y)$ (for each $k$) which induces maps on homology $f_* : H_k(X) \rightarrow H_k(Y)$. It's important to note that $f_*$ exists only because $f_\#$ sends homologous cycles in $X$ to ...

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Consider the space $K$ given by the disjoint union of two circles. If $\phi$ is the map interchanging the two components (preserving the orientations), then we see that $K_{\phi}$ is a torus. However, $K\times I$ is the disjoint union of two tori, hence the Betti numbers are obviously different: $$(b_0(K_\phi),b_1(K_\phi),\ldots)=(1,2,1,0,\ldots)$$ ...

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If $f:M\to M$ is a diffeo, then there is a linear map $f^p:H^pM\to H^pM$ and then we can consider the spaces $$H^p(M)_f=H^p(M)/(1-f^p)H^p(M)$$ and $$H^p(M)^f=\ker(1-f^p).$$ One can then show that there are exact sequences $$0\to H^{p-1}(M)_f\to H^p(S_fM)\to H^p(M)^f\to 0$$ with $SM$ the mapping torus of $f$. One speedy way of getting this, by the way, is to ...

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Let me try to outline the proof, and you can fill in the details. We take an $n$-dimensional smooth quadric hypersurface $Q$ sitting in $\mathbf P^{n+1}$. By the Lefschetz hyperplane theorem, for $k \neq n$, the restriction maps $$H^k(\mathbf P^n, \mathbf Z) \rightarrow H^k(Q,\mathbf Z)$$ are isomorphisms, and so the cohomology groups of $Q$ in these ...

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I'm going to use singular cohomology. Recall that if $C_i(M)$ denotes the collection of singular $i$-chains, then for an abelian group $G$, $C^i(M; G) = \operatorname{Hom}(C_i(M), G)$ is the collection of singular $i$-cochains with values in $G$. Note, if $H$ is a subgroup of $G$, then $C^i(M; H) \subseteq C^i(M; G)$. There is a coboundary operator ...

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Note that the restriction of $p$ to $S\setminus\delta S$ is a homeomorphism onto $X$. So $X\cap Y$, being a subset of $X$, is homeomorphic to a subset of the open unit square. Which subset is that? Well, $p^{-1}(Y)$ is just a small neighborhood of $\delta S$, so $Y\cap X$ will look like a small open ring along the edge of the square, but not actually ...

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First, let us note that a map induces isomorphisms on cohomology with coefficients in all rings iff it induces isomorphisms on cohomology with coefficients in all abelian groups. Indeed, if $G$ is any abelian group, consider the ring $R=\mathbb{Z}\oplus G$, with multiplication defined by $(m,g)\cdot(n,h)=(mn,mh+ng)$. Then there is an isomorphism ...

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$\require{AMScd}$I'd rather think about the negation of your first condition. So call a space bad if there exists a decomposition $A \cup B$ into connected open sets with disconnected intersection. Call a space semi-bad if there exists a decomposition into open sets $A \cup B$ such that the induced map on relative homology $\tilde H_0(A \cap B) \to \tilde ... 2 Your De Rham cohomology algebra is$H^*(\mathbb P^3_\mathbb C)=\mathbb R[w]/(w^4)$, where$w=[\omega]$is the class of some closed$2$-form$\omega$. We then have$0\neq w^2=[\omega \wedge \omega]$, so that$\omega\wedge \omega$is not exact and a fortiori$\neq 0$. 2 Since$\alpha = (1,0)$and$\beta = (0,1)$generate$\Bbb{Z}^2$, let's see where your operator takes them. Let$T = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$. Then$T\alpha = (1,3)$and$T\beta = (2,4)$. Then for arbitrary,$a,b \in \Bbb{Z}$,$T(a \alpha + b \beta) = (a+2b, 3a+4b)$. As$a,b$independently range through all of the integers, you ... 2 The standard answer is to observe$S^1$is an Eilenberg-MacLane spaces, specifically that$S^1$is a model for$K(\Bbb Z, 1)$. (This relies on showing that all of the higher homotopy groups of$S^1$vanish.) In general, there is a natural isomorphism$H^n(X;G) \cong [X, K(G, n)]$. This is referred to as the representability of cohomology by ... 2$H_n(M,M-x;R)= H_n(M,M-x)\otimes R$. For each$r\in R$determines a covering space$M_r$of$M$consisting of points$\pm \mu_x\otimes r \in H_n(M,M-x;R)$where$\mu_x\in H_n(M,M-x)$is a generator. If$r=-r$i.e order$2$, then$M_r$is just a copy of$M$, otherwise it is isomorphic with the oriented$2-$sheeted cover$\bar{M}$. So observe that a ... 2 Let$M$be an orientable (connected, closed)$n$-manifold. Then classically$H_n(M;F) \cong F$. But moreover,$H_{n-1}(M;\mathbb{Z})$is torsion-free. This is Corollary 3.28 in Hatcher's book. The proof basically goes like this: if$H_{n-1}(M;\mathbb{Z})$weren't torsion free, by the UCT then$H_n(M; \mathbb{F}_p)$would be bigger than just$\mathbb{F}_p = ...

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There is a homotopy $H_t: C_r \to \Bbb R^2 - 0$ given by $H_t(z) = f(tz)$. At time $1$ this is just $f(z)$ and at time $0$ it is the constant map $f(0)$. Note that $f(tz) \neq 0$ by our assumption on $f(z)$. Therefore $f$ induces the same homomorphism on homology as the constant map does, and the latter clearly induces the zero homomorphism. Edit: When I ...

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Write $S^n$ as the union of two open sets $U= S^n-N$, $V=S^n-S$where $N,S$ are the north and south pole. Note that the intersection $U\cap V$ is connected if $n\geq 2$. $U,V$ are contracile (diffeomorphic to $R^,$, and on each of these sets the form $\alpha$ is exact and andmits primitive $f_U, f_V$. On the intersection $d(f_U-f_V)=0$ and by connexity ...

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It is hard to know what will give you an intuition without knowing what you want and what background you already have. If you are asking roughly what kind of geometric or topological information Cech cohomology reveals about a space, then answer is the same as for all cohomology theories: it reveals the global connectivity of spaces with relatively simple ...

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I'm confused. $H_1(\mathbb{R}P^3)=\mathbb{Z}/{2\mathbb{Z}}$, but is orientable (on edit, this answers your question 2). EDIT: Maybe it refers to the fact that $H^{m-1}(M)$ can only be torsion if the manifold $M$ is non-orientable. It is Corollary 3.28 in Hatcher.

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All of the homology groups of $\mathbb{S}^n$ are trivial, except of top and bottom one. The induced map $H_0(\mathbb{S}^n) \to H_0(\mathbb{R}\mathbb{P}^n)$ will always be isomorphism (this is very easy to calulate). The top homology group $H_n(\mathbb{R}\mathbb{P}^n)$ is either trivial or $\mathbb{Z}$, depending on whether $\mathbb{R}\mathbb{P}^n$ is ...

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No, having a CW complex structure is absolutely not enough to be a manifold. Being a CW complex is very easy, being a manifold is hard. Take the wedge sum of two circles $S^1 \vee S^1$ for example: it's a CW complex, but not a manifold. If a space $M$ is a compact manifold without boundary, then you can read its dimension using singular homology: it is the ...

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In applying the Mayer-Vietoris sequence, not only do you need to know how to evaluate homology groups of spaces, but you also need to evaluate induced homology homomorphisms of continuous maps. In this example, you have correctly evaluated the homology groups $H_1(K) \approx \mathbb{Z}$ and $H_1(K_0) \approx \mathbb{Z} \oplus \mathbb{Z}$. The additional ...

1

Let $\pi \colon M \to S^1$ denote the usual projection and let $C$ denote a non-zero homologous simple closed curve with no self intersections, i.e. an injective map $c$ from $S^1$ to $M$, which induces something non-zero on homology. We can isotop $c$ s.th. $\pi \circ c$ is a covering of $S^1$ (This is not that hard, you just make the map locally ...

1

As OP correctly remarked in the comments, my answer was wrong! And his proof was right (I guess). Well done, there's just a trivial mistake you've made: $H_2(Y)=0$ would imply that the map $H_2(Y)\to H_1(A\cap B)$ is the zero map. On the other hand, the image of this map is the kernel of the map $\phi$, which you correctly described; the kernel of this map ...

1

Recall that the $H_0$ of a path-connected space $X$ is isomorphic to $\mathbb{Z}$ via the augmentation map, which counts the sum of the coefficients of a given (class of a) $0$-chain, i.e., a formal sum of points of $X$. Therefore the generator $1 \in \mathbb{Z}$ corresponds to the class of any point $x \in X$; alternatively, if you pick $x,y \in X$, then ...

1

Why not directly prove it is isomorphic to the projective line? For example, changing variables to $(x_0+ix_1, x_0-ix_1,ix_2)$, this curve $M$ is same as the one given by $x_0x_1-x_2^2=0$ and you have a map $\mathbb{P}^1\to M$, by $(u,v)\mapsto (u^2, v^2, uv)$, which you can easily see to be an isomorphism.

1

This is the fundamental lemma in homological algebra (see here theorem $2.22$ page $36$ and the following corollary.) If you take the identity the thesis of the theorem says in particular that you obtain a chain homotopy

1

From Universal Coefficient Theorem we have the following natural s.e.s. for homology we have $$0 \to H_1(X;\mathbb{Z})\otimes \mathbb{Z}_2 \to H_1(X;\mathbb{Z}_2) \to \text{Tor}(H_0(X;\mathbb{Z});\mathbb{Z}_2)\to 0$$ since $H_0(X;\mathbb{Z})$ is $\mathbb{Z}$ the s.e.s becomes $$0 \to H_1(X;\mathbb{Z})\otimes \mathbb{Z}_2 \cong H_1(X;\mathbb{Z}_2) \to 0$$ ...

1

It is enough to show that $H_n(g):H_n(\Bbb RP^n)\to H_n(\Bbb RP^n)$ is multiplication by an odd integer because $2\deg(f)=\deg(p\circ f)=\deg(g\circ p)=2\deg(g)$. By the UCT, this is equivalent to showing that $$H_n(g;\Bbb Z/2):H_n(\Bbb RP^n;\Bbb Z/2)\to H_n(\Bbb RP^n; \Bbb Z/2)$$ is an isomorphism, which in turn is equivalent to the analogous statement for ...

Only top voted, non community-wiki answers of a minimum length are eligible


text:
Computing the Class Number H(-3)
 Theorem 1: The class number $H(-3) = 1$.
• Proof: Let $f(x, y)$ be a reduced binary quadratic form. Since $d = -3 < 0$, we must have that:
• So $a = 1$. But then $0 < b \leq a = c$ implies that $c = 1$ and $b = 1$, so $f(x, y) = x^2 + xy + y^2$ is a reduced binary quadratic form with discriminant $d = -3$. The inequality $-1 < b \leq a < c$ implies that $b = 0$ or $b = 1$. If $b = 0$ then $-3 = b^2 - 4ac = -4c$ can never be satisfied. If $b = 1$ then $-3 = b^2 - 4ac = 1 - 4c$ implies that $-4 = 4c$, so $c = 1$, which gives us $f(x, y)$.
• So the only reduced binary quadratic form with discriminant $d = -3$ is $f(x, y)$ and $H(-3) = 1$. $\blacksquare$


text:
## 07 August 2014

### Coursera Courses

In this post I review some of the Coursera courses that I completed.

Algorithms: Design and Analysis, Part 2. The course covers greedy, dynamic programming, and NP-completeness. The lectures were very clear, not too fast and not too slow. Caveat: I watched lectures only when I needed. My strategy was to try to solve the problems. If I had trouble, I looked at the slides. If the slides were cryptic, then I looked at the lectures. So, I ended up seeing very few lectures. I even learned an algorithm that I knew I should know but didn't: Karger's. I think it's my favorite example of a randomized algorithm.

Introduction to Genetics and Evolution. The course covers the basic math of genetics. I'll give two examples of things I remember now, more than one year after.

One idea is the Hardy-Weinberg equilibrium. Suppose each individual has one of the three genes $aa$, $aA$, and $AA$. I'll use the fancy word allele: it means one of $a$ and $A$. Reproduction works as follows: Randomly choose an allele from each of the two parents, and put them together. Hardy-Weinberg says that in an infinite population, the percent of each allele is constant. So, if you start with half $a$s and half $A$s, you'll remain with half $a$s and half $A$s.

The other idea is genetic drift. It says what happens when the population is not infinite. What happens is that you'll have random variations. Eventually, these lead to the extinction of one of the alleles.

Programming Languages. This course uses ML, Racket, and Ruby to illustrate basic principles of programming languages. As usual in this kind of courses, you write a few interpreters. The best time to take this course is after you've learned your second language. (I learned BASIC when I was 6; Pascal and C++ came when I was 14. I think I would've benefited most from this course when I was 15. So, if you are 15, then you should try it!)

The Modern World: Global History since 1760. What I remember most vividly about this course is rather strange. It is the soothing voice of the lecturer. Even when I was disagreeing with what he said, I felt compelled to keep listening.

From what I recall, the quizzes were boring: They simply tested a bunch of facts that happened to be mentioned in lectures. The videos, on the other hand, always told a nice story. These stories were clearly addressed to someone living in 2013. For example, the lectures not only said that the West was weak in 1760. They stressed this fact repeatedly, because a world in which the West is weak and insignificant is really difficult to imagine for the typical user of Coursera.

Natural Language Processing. For this course, like for the Algorithms one, I did problems first. I looked at the slides and watched the lectures only when I needed. I ended up looking at most slides and almost no lecture video. Which, I guess, means that the slides are self-sufficient. From what I recall, doing the homeworks involved a substantial effort.

The topics covered are things like probabilistic grammars, tagging problems (e.g., noun, verb, …), log-linear models, learning.

Discrete Optimization. Like for the other computer-related courses, I went directly for the homeworks. In addition, I delayed doing most homeworks for the last week. That turned out to be a huge mistake. I simply couldn't do the homeworks at the rate of one per day, as I planned. In fact, I thought at some point that I won't even get a passing grade. I did, but I also learned that Coursera sometimes has tough courses. So, what were the homeworks like? I recall: knapsack, graph coloring, euclidean traveling salesman. There was also some combination of bin packing and euclidean traveling salesman. More importantly, I recall some tricks that I learned while doing the homeworks. For example, for knapsack I used two tricks to get a good solution. The standard dynamic programming runs in time proportional to the total weight. This is troublesome because weight can be very large. In that case, use a trick from physics: pick your units of measurement. Strictly speaking, the unit of measure doesn't matter. But, when one works in kilometers, one usually doesn't keep track of millimeters. Thus, by ‘picking the unit’ I actually mean ‘ignore the less important bits’. When ignoring bits, you'll want to round up. This way, you never try to fit more than possible in the knapsack. The other trick is to repeat this process using the items that remained out, with better precision.

For travelling salesmen, I wrote some complicated solution that barely got medium points. It did two main things, as I recall. First, it used dynamic programming to do local optimizations. Second, it greedily improved the tour by using a set of moves that I noticed empirically. For an example of move, think of a crossing of edges. These two edges are far apart in the current best tour, so local optimizations won't notice the crossing. But, of course, you can look at all pairs of edges and check for intersections. Once you notice the interesection, it's clear what to do to shorten the tour.

A Brief History of Humankind. This course was like the other history course: trivial (useless?) quizzes, but fascinating lectures. Where the Modern History course would talk about personalities and their decisions, the Humankind History course would talk about the way of life of a typical person. In this sense, the two courses complement each-other nicely.

The most vivid image that this course left in my mind is that of ‘shared imaginary worlds’. Why do money have value? Because we believe that others believe that money have value. This kind of circular reasoning has a certain appeal for a computer scientist.

Linear and Integer Programming. This course was too easy for my taste. They do recommend a resource that looks intriguing: the convex optimization book by Stephen Boyd.

Automata. This course had two kinds of homework: multiple-choice quizzes and programming assignments. Unexpectedly, I enjoyed the quizzes more. Each quiz question tries to test whether you understood some concept. You get treated as a computer program: The quiz presents you with some (smallish) input data, you have to provide the right output. If you repeat the quiz, you get some other input data that you have to process. For one of the first quizzes, I managed to get a question wrong three times in a row. I was trying to do it quickly, and I ended up being very slow. The fourth time I moved slowly and checked my work. I think this phenomenon appears often in real life: People make fun of those that appear slow, but the slow ones often finish first.

Back to the course. Why did I not like the programming assignments? Because I was supposed to modify some Python scripts that were kinda crap.

To answer some questions, I had to look at the slides. These were extremely clear — I never needed to look at the video lectures.

Social and Economic Networks: Models and Analysis. I learned something new from almost every video lecture. The quizzes were rather easy, but only after I learned the vocabulary. It wasn't possible to grasp the definitions from the slides, but the videos were very clear. In other words, if you take this course and you don't already work in social networks, you'll have to watch the videos. Which is a good thing.

Here's an example of the kind of stuff I remember. Social networks are graphs. In various circumstances it makes sense to talk about the utility of a social network. A simple way to define this is as the sum of utilities of each vertex. For an example, think of the co-author graph. The vertices are authors, the edges represent an ongoing collaboration. One particular author benefits from the time given by its co-authors. In turn, this particular author offers his time to their co-authors. In the simplest model, the time is divided equally. Thus, the utility of a vertex $x$ is $1+\sum_{y \in N(x)} 1/|N(y)|$. (The $1$ comes from the time spent by $x$ on their own stuff, and $1/|N(y)|$ appears because $y$ divides its time equally between its neighbors $N(y)$, one of those neighbours being $x$.) Once you have such a simple model, you can ask questions like: Which configurations are pairwise stable (meaning that there is no local incentive to create or delete edges)? Which configurations correspond to global optima? For the answers, you'll have to take the course. (Well, you could find out the answers yourself, of course.)

Statistical Mechanics: Algorithms and Computations. This course is mostly about Markov chains. You can think of Markov chains as a way to sample complicated probability distributions, or as a way to estimate integrals. Come to think of it, the course is more about sampling and integrating than about Markov chains. For example, when you get to the quantum stuff, you see that computing path integrals is better done using Levy paths. These are a direct sampling technique, which works better than the Markov chain approach, in this case.

Also, these techniques can be used for optimization problems. One example is the traveling salesman problem, which makes an appearance in one of the reviews above. I was conceited, and expected that the program I wrote with more than one day of work would do better. After all, I looked at the Python program provided by the instructors and it seemed very similar, except: (1) it did no local optimization using dynamic programming, (2) the vocabulary of moves was a small subset of the one I accumulated. And yet, their implementation worked much better. Wat?? I had to take a look. It turns out there was only one difference: They sometimes allowed a move that made the tour worse. (They used simulated annealing, rather than greedy.) In other words, my solution was getting stuck in local optima. I'm not sure why I didn't think of this when I was doing the discrete optimization course. (It seems obvious with hindsight.) A good rule of thumb: Whenever you use greedy for an optimization problem, at least consider trying simulated annealing.

Starting Soon. Out of these courses, the ones that start soon are the following:


text:
# Performance Analysis of Millimeter Wave Wireless Power Transfer With Imperfect Beam Alignment

In this paper, the impact of imperfect beam alignment (IBA) on millimeter wave (mmWave) wireless power transfer (WPT) is investigated. We consider a mmWave WPT network, where the location of the energy transmitters follows a Poisson point process. Instead of the mostly used flat-top antenna model, we adopt the Gaussian antenna model suggested by the 3rd Generation Partnership Project (3GPP) for better accuracy. Two beam alignment error (BAE) models, i.e., truncated Gaussian and uniform models, are introduced to represent different BAE sources. We derive the probability density functions (PDFs) of the cascaded antenna gain with both BAE models and then provide the approximated PDFs for tractability. With the help of Fox's H function, the analytic expression for the energy coverage probability with nonlinear energy harvesting model is derived. Besides, we deduce a closed-form expression of the average harvested radio frequency (RF) energy. The simulation results verify our theoretical results and demonstrate the performance degradation incurred by BAE. It also shows that the flat-top antenna model cannot always provide accurate performance evaluation in the presence of BAE.

• 6 publications
• 173 publications
• 39 publications
• 11 publications
03/19/2020

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### Covert Millimeter-Wave Communication: Design Strategies and Performance Analysis

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08/02/2019

### Beam Codebook Design for 5G mmWave Terminals

A beam codebook of 5G millimeter wave (mmWave) for data communication co...
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### Beamspace Precoding and Beam Selection for Wideband Millimeter-Wave MIMO Relying on Lens Antenna Arrays

Millimeter-wave (mmWave) multiple-input multiple-out (MIMO) systems rely...
03/22/2019

### Hybrid Precoder and Combiner for Imperfect Beam Alignment in mmWave MIMO Systems

In this letter, we aim to design a robust hybrid precoder and combiner a...
03/29/2022

### Near-Field Hierarchical Beam Management for RIS-Enabled Millimeter Wave Multi-Antenna Systems

In this paper, we present a low overhead beam management approach for ne...

## I Introduction

In the foreseeable future, there would be huge amounts of low-power devices in wireless networks to perform information forwarding, data collecting, situation sensing [9]. These devices could be the nodes in Internet of Things (IoT), wireless sensor networks (WSNs) or Device-to-Device (D2D) systems. These low-power devices are usually powered by batteries and deployed in a broad area [14]. In order to prolong the network lifetime and maintain the sustainability of these nodes, far-field wireless power transfer (WPT) via radio frequency (RF) has been considered as a promising technology to energize massive battery-powered devices. The reason is that RF-WPT can provide a flexible and long-distance charging service, while being compatible with the existing wireless information networks [47, 23]. At present, RF-WPT has been successfully applied in various wireless networks [9, 14, 47, 23].

For addressing the demand of ever-increasing data transmission rates, millimeter wave (mmWave) frequencies are leveraged to fulfill the multi-Gigabit information transmission requirement [31]. Due to the small wavelength and blockage sensitivity of mmWave signals, strong directional antenna and small cell structure are suggested to enhance the energy and spectrum efficiency of mmWave communications [3]. As WPT also suffers from the severe power propagation loss and needs to avoid interference to existing wireless information networks, mmWave also benefits WPT. It has been proven by [21, 40] that mmWave WPT outperforms WPT with lower frequencies. In [50], lens array based mmWave WPT was proposed to charge multiple energy receivers. In [11], simultaneous wireless information and power transfer (SWIPT) was applied in the hybrid precoding mmWave system. Besides, the effect of rain attenuation on mmWave WPT was investigated by [20]. All that literature shows mmWave WPT is feasible.

To evaluate the system-level performance of mmWave WPT, stochastic geometry has been utilized to capture the effects of propagation loss and blockage, which dominate the received signal power [5, 15]. The energy coverage probability and average harvested energy of mmWave WPT were studied by [21], where the location of energy transmitters follows the Poisson point process (PPP). Furthermore, SWIPT was also introduced into stochastic mmWave networks [40, 38]. Both works verified that mmWave could improve the performance of SWIPT compared to lower frequency solutions. In [22], the energy coverage probability in the presence of human blockage was derived. Discretizing harvested energy into a finite number of power levels, the total coverage probability integrating information and energy transmission was derived by [51]. A beam-training based mmWave WPT scheme was proposed in [18], where the energy transmitter steers the energy beam along the direction in which it receives the strongest training signal. Considering the nonlinear behavior of energy harvesting, the coverage probability and average harvested energy were studied by [37]. In [41], the location of mmWave powered users was modeled as the Poisson cluster process and the energy and information coverage probabilities were derived.

Most of the aforementioned works, such as [21, 40, 38, 22, 51, 18, 37, 41], employ the flat-top antenna pattern for mathematical tractability. Nevertheless, it may incur great inaccuracy for evaluating system-level performance [13, 46]. Therefore, some literature considers more realistic antenna patterns in the performance analysis of mmWave WPT. In [16], the microstrip patch antenna and end-fire antenna models were used to analyze the coverage performance of SWIPT. [30] adopted the Fejér kernel model to represent the actual array antenna gain of energy transmitters. Both [16] and [30] assumed each energy transmitter is equipped with a directional antenna, while the antenna of energy receiver was assumed to be isotropic. In [39], both Tx and Rx were assumed to be equipped with directional antennas and the average harvested energy was derived.

Most of the existing works on mmWave communications/WPT have assumed that the beam direction of the associated Tx-Rx pair is perfectly aligned. However, due to direction estimation error and hardware imperfection, the beam alignment error (BAE) is always inevitable, and affects the received signal strength greatly

[42]. Therefore, it is necessary to investigate the performance of the mmWave transmission systems with imperfect beam alignment (IBA). By now, much effort has been dedicated to evaluating the performance of the mmWave wireless information networks in the IBA scenario, such as [8, 4, 34, 46, 32, 10]. Taking into account of the BAEs in the associated Tx-Rx pair and the Interfering Txs-Rx pairs, the information coverage probability with the flat-top antenna gain model was derived by [8]. With the same BAE setup, while, [4] adopted the cosine antenna and isotropic antenna patterns to model the antenna gains of the transmitters and receivers, respectively. In [34], the approximated ergodic capacity loss of the mmWave ad hoc network with both flat-top and Gaussian antenna gain models was derived. In [46], the authors employed the sinc and cosine antenna patterns to derive the information coverage probability, just considering the BAEs between the typical receiver and interfering transmitters. Alternatively, using the antenna gain model suggested by the 3rd Generation Partnership Project (3GPP), [32] investigated the impact of beam misalignment from interfering transmitters, and resorted to the curve-fitting method to derive the probability density function of the antenna gain with BAE. Furthermore, [10] employed both the 2D and 3D directional antenna gains and derived the information coverage probability with BAE incurred by the strongest interfering link.

To the best of our knowledge, the impact of BAE on mmWave WPT system has not been investigated and well understood to date. Additionally, in this paper, we consider the comprehensive IBA scenarios and the realistic 3GPP directional antenna model, which has not been well studied from the perspective of system-level performance evaluation in either mmWave communications or mmWave WPT. Specifically, there are two aspects of our system model to be highlighted. First, unlike [42, 8], we employ the 3GPP Gaussian gain model, which can approximate the realistic directional antenna pattern exactly, to explore the impact of BAE. The reason is that the flat-top model lacks the capability of depicting the roll-off effect of the mainlobe and thus is not suitable for accurately evaluating the impact of BAE on mmWave systems [34, 46, 32, 4]. Second, different from [34, 46, 32, 10, 4], we consider a more realistic IBA scenario, where all Txs and Rxs are equipped with directional antennas. Moreover, not only the BAE between the associated Tx-Rx pair but also the BAEs of the non-associated Txs to the typical Rx are simultaneously taken into consideration. Besides, the nonlinear energy harvesting model in the typical receiver is also assumed. With these models, we derive the analytic expression of the system-level performance of the mmWave WPT in the IBA scenario. For clarity, we summarize our main contribution as follows:

• We adopt the Gaussian antenna gain model suggested by the 3GPP to represent the realistic directional antenna pattern and also assume all transmitters and receivers are equipped with the same directional antenna. Considering the truncated Gaussian and uniform BAE distributions, we derive the exact probability density functions (PDFs) of the cascaded antenna gains of a Tx-Rx pair. Taking into account of the distribution characteristics of BAE and the strong directivity of mmWave antenna, we also provide the approximated PDFs for tractability.

• With the help of the Fox’s H function and its series expansion, we provide a novel solution to derive the analytic expression of the energy coverage probability of the mmWave WPT in the presence of BAE. Besides, the closed-form expression of average harvested RF energy is derived. We define the relative energy loss (REL) in order to further investigate the performance degradation incurred by BAE.

• Through Monte-Carlo simulations, we verify the derived energy coverage probability and average harvested RF energy. It is found that the widely used flat-top antenna gain model cannot always exactly evaluate the performance of the mmWave WPT system in the presence of BAE. We also conclude that BAE indeed lowers the energy coverage probability and the average harvested RF energy.

Notation: , , and represent the sets of all integers, real numbers and complex numbers, respectively.

is the expectation of random variable (r.v.)

and is the probability of event . For r.v. , and

stand for the probability density function (PDF) and cumulative distribution function (CDF) of

, respectively. For , is the modulus of . Given , means the Euclidean norm of

. For a complex vector

, and stand for transpose and conjugate transpose of , respectively. means the r.v.

follows Gamma distribution with PDF

, where is the Gamma function. is the Gaussian error function and is the Dirac delta function. is the natural logarithm of . For and , represents the binomial coefficient and equals to . For r.v. , is the Laplace transform of .

## Ii System Model

### Ii-a Network and Node Models

In this paper, we consider a mmWave wireless-powered network which consists of two types of nodes, i.e., Energy Transmitter (ETx) and Energy Receiver (ERx) (See the Fig.1 on the next page). The ETxs are connected to stable power sources, thus having the capability of emitting energy signals. While, the ERxs have to harvest energy to maintain their routine operation and information transmission. The location of all ETxs follows the homogeneous Poisson point process (HPPP) with intensity on the two-dimensional Euclidean plane. Suppose a saturated service scenario, where each ETx is assumed to serve one dedicated ERx in a WTP block [19, 12]. We study the performance of the typical ERx located at the origin. Let the ETx associated by the typical ERx be located at and .111Herein can be seen as the maximum allowable WPT distance for an ETx-ERx pair. Accordingly, in this work we investigate the worst performance of the considered mmWave WPT system [19, 12]. Then, due to Slivnyak’s Theorem [19], except the ETx at , the location of other ETxs still forms a PPP with the same intensity , denoted by . Without loss of generality, the WPT duration is assumed as unit time. It means in following context, the harvested energy and the harvested power have an equivalence.

### Ii-B Channel Model

For mmWave links, line-of-sight (LOS) and non-line-of-sight (NLOS) channels have sharply different propagation characteristics. Given a propagation distance , the path loss of a LOS mmWave link can be modeled as , while in the NLOS case, the path loss is . Here ( ) is the LOS (NLOS) path loss exponent and () represents the intercept of LOS (NLOS) link. In general, we have and [15, 3]. Moreover, we model the small-scale fading of each mmWave link as independent Nakagami fading with parameters and for LOS and NLOS scenarios, respectively. Thus, we denote and as the small-scale channel gains for LOS and NLOS links, respectively, and then express the LOS power gain as and NLOS power gain as .

The probability of a mmWave link experiencing LOS is modeled as a function with respect to the distance from Tx to Rx [31]. Several empirical and analytical blockage models had been reported in [31] and [3]. Considering tractability and accuracy, we employ the random shape theory model, by which the analytical LOS probability is coincident with the empirical expression of the 3GPP blockage model [3]. Hence, given the distance , the LOS and the NLOS probabilities can be separately expressed as and , where the blockage parameter is determined by the average number and average perimeter of buildings in the interested area. Similar to most literature, e.g., [Bai and Heath [5]][Comisso and Babich [10]], the correlation of LOS probabilities for different mmWave links is ignored. Consequently, the ETxs in can be divided into two independent sets, namely, LOS ETx set and NLOS ETx set. Both of which follow the inhomogeneous PPP (IPPP) with intensity and , denoted as and , respectively. As the ETx at has the distance to the typical ERx, the performance analysis in the LOS scenario is very similar to that in the NLOS scenario. Furthermore, considering the harvested energy plays a crucial role in maintaining the lifetime of wireless-powered nodes, we assume the ETx only selects the ERx experiencing LOS as its target receiver like [46, 12, 43].

### Ii-C Antenna Pattern

We assume that all ETxs and ERxs are equipped with the same directional antenna to perform mmWave beamforming. It is also assumed that all ETxs have the same transmit power . For the typical ERx, the received RF energy transmitted by the ETx at can be given by

 εx,i=Ptℓi(rx)hx,iG(ϕx)G(φx),i∈{L,N} (1)

where and stand for the antenna gains of ETx and ERx, respectively. and are the orientation angles relative to the boresights of ETx and ERx, respectively, and belong to the interval . Note that we ignore the energy harvested from the noise, as it is trivial compared with the received RF energy [37].

In order to theoretically evaluate the system-level performance, it is inevitable to depict the PDF of the cascaded antenna gain . For the uniform linear array (ULA), the Fejér kernel based sinc and cosine antenna patterns are employed to represent the directional antenna gain [46, 2, 30, 4]. Unfortunately, it is difficult to directly derive the exact analytic expression of the PDF of using the sinc or cosine antenna gain model. In [46, 30, 4], the receiver was assumed to be equipped with omni-directional antenna to avoid the cascaded directional antenna gain. Although [2] considered a cascaded antenna gain with the sinc antenna pattern, the nodes were not stochastically deployed.

Alternatively, the flat-top antenna gain model is mostly used in stochastic geometry coverage analysis for its mathematical tractability, e.g., [21, 38, 51, 40, 42, 8, 12, 18, 41, 22, 43, 37]. In the perfect beam alignment or slight BAE scenarios, the flat-top antenna gain model can provide the tractable theoretical expression of the system-level performance with acceptable accuracy. However, the flat-top antenna model is lack of the capability of depicting the roll-off effect of the mainlobe and incurs significantly inaccuracy for evaluating the system-level performance in the IBA scenario [46, 34, 35, 44, 32]. It is worth mentioning that in [13] a generalized flat-top model was proposed to approximate the practical antenna gain. Nevertheless, in the IBA case, it could incur relatively high complexity to derive the PDF of the cascaded antenna gain.

In [34, 35], a Gaussian antenna pattern, i.e.,

 G(θ)=(Gm−Gs)e−ηθ2+Gs,

where is the maximum mainlobe gain, is the sidelobe gain and is determined by the 3dB beamwidth, was used to represent the mmWave directional antenna gain. Nevertheless, to obtain the analytic PDF of the cascaded antenna gain , the side lobe was ignored in the analysis of [34, 35]. Moreover, the loss in ergodic capacity derived by [34, 35] only involves the truncated Gaussian BAE. Differently, in the non-stochastic mmWave networks [44, 49, 32] adopted the 3GPP Gaussian antenna model, which can depict the roll-off characteristic and match the measurement well. In this paper, therefore, we employ the 3GPP Gaussian antenna model, i.e.,

 G(θ)={Gme−ηθ2|θ|≤θ0,Gsθ0<|θ|≤π, (2)

where

 Gs=πΘ(θ0,θ3dB)+π−θ0,
 Θ(θ0,θ3dB)=∫θ00100.3(θ20−x2θ23dB)dx,

, is the main lobe (20dB) beamwidth, and is the half-power (3dB) beamwidth. Since , the continuity is ensured. According to the practical measurement reported by [36, 44], when , is approximately equal to . Then, with this empirical approximation, we further obtain , , and . For convenience, we introduce the normalized antenna gain , i.e.,

 ˜G(θ)={e−ηθ2|θ|≤θ0,gθ0<|θ|≤π, (3)

in which and . Obviously, with the empirical expression , the Gaussian antenna pattern is only determined by . We herein employ (3) to reduce the parameter number of the antenna radiation pattern. Besides, the cascaded normalized antenna gain is denoted by .

### Ii-D Imperfect Beam Alignment Models

According to (3), if we intend to maximize the harvested energy, we can let and . In practice, however, and are not necessarily equal to zero due to the direction estimation error and hardware imperfection [42, 8, 4].

For the associated ETx-ERx pair, as the BAE appears after the beam aligning procedure, and

are usually modeled as independent and identically distributed truncated Gaussian random variables with zero mean and standard deviation

[34, 8, 4]. Such that, the PDF of can be expressed as

 fϕx0(ψ)=e−ψ22σ2√2πσ2erf(π√2σ),    ψ∈[−π,π) (4)

which is named as the Gaussian BAE model. The standard deviation is usually used to indicate the variability of BAE. The larger is, the stronger statistical dispersion the BAE exhibits, which means the BAE becomes more severe. Observing (4), if , gradually converges to . It is corresponding to the fact that if the beam is perfectly aligned, the antenna gain equals to with probability 1.

For the non-associated ETx-ERx pairs, an ETx in or just aligns its beam with the boresight of its paired ERx. To facilitate understanding and representation, for the typical ERx, given the ETx at , , we can also treat and

as BAEs, which are usually modeled as independent uniform distribution over

[46, 8, 4], i.e.,

 fϕx(ψ)=12π,  ψ∈[−π,π), x∈Φi, i∈{N,L}. (5)

It is named as the uniform BAE model. Note that it is assumed that BAEs at ETxs and ERxs are independently distributed [34, 8, 42].

### Ii-E Energy Harvesting Model

As the small-scale gains of different mmWave links are independently distributed, the harvested RF power of the typical ERx can be written by

 εRF=ε0+εL+εN, (6)

where is the harvested power from the ETxs in , is the harvested power from the ETxs in , and is the harvested power from the ETx located at . Then, the harvested direct current (DC) power at the typical ERx is

 εDC=ζ(εRF). (7)

Note that is the RF-DC power conversion function. In practice, is a nonlinear function with respect to the input RF power [37, 30, 7]. Using the practical nonlinear energy harvesting model proposed in [7], we can write the harvested DC power as

 εDC=pm(1−exp(−paεRF))1+exp(−pa(εRF−pb)), (8)

where is the maximum DC power that can be harvested by the ERx and and are the constants determined by the rectifier circuit [7].

## Iii The PDFs of Antenna Gains

In this section, we derive the PDFs of the normalized antenna gains with the BAE following the truncated Gaussian or uniform distributions. Then, the PDFs of the cascaded antenna gains with two BAE models are derived. For tractability, we also provide the approximated PDFs of the cascaded antenna gains.

### Iii-a The PDFs of the Normalized Antenna Gains

###### Lemma 1.

If the PDF of the stochastic BAE is , the PDF of the normalized Gaussian antenna gain is given by

 f˜G(y)=1y√−ηlnyfψ(√−lnyη)+(1−P0)δ(y−g), (9)

where and .

###### Proof:

See Appendix A. ∎

Based on the Lemma 1, we have following two corollaries according to (4) and (5).

###### Corollary 1.

If the BAE

follows the truncated Gaussian distribution with zero mean and variance

over , the PDF of can be expressed as

 f˜G(y)=y12ησ2−1√2πησ2erf(π√2σ2)√−lny+(1−PG0)δ(y−g), (10)

where and .

Thus, for the associated ETx-ERx pair, the PDF of the normalized antenna gain of ERx or ETx is equal to (10).

###### Corollary 2.

If the BAE follows the uniform distribution over , the PDF of can be written by

 f˜G(y)=12πy√−ηlny+(1−PU0)δ(y−g), (11)

where and .

For the non-associated ETx-ERx pairs, the PDF of involved antenna gains is illustrated by the Corollary 2.

### Iii-B The PDFs of the Cascaded Antenna Gains

By (1), the cascaded antenna gain plays a crucial role in the system performance. Herein, we intend to derive the PDFs of with both BAE models respectively. To this end, we have following two theorems.

###### Theorem 1.

The PDF of the cascaded normalized antenna gain with truncated Gaussian BAE model can be written as (12) at the top of the next page.

See Appendix B ∎

###### Theorem 2.

The PDF of the cascaded normalized antenna gain with uniform BAE model can be written as (13) at the top of the next page.

###### Proof:

The proof of Theorem 2 is similar to that of Theorem 1, therefore, we omit the proof for clarity. ∎

### Iii-C The Approximated PDFs of the Cascaded Antenna Gains

Although the Theorem 1 and 2 show the exact PDFs of with the Gaussian and uniform BAE models respectively, the arctan functions in (12) and (13) make further analysis less tractable. Therefore, we provide two approximated PDFs, by considering the distribution characteristics of BAE models and the strong directivity of mmWave antenna.

#### Iii-C1 The Approximated PDF With the Gaussian BAE model

In the mmWave WPT system, the associated ETx-ERx pair is expected to employ the elaborately designed beam alignment algorithms, such as [48], to minimize and as much as possible. It is therefore reasonable to infer that and are less than in most cases. For instance, it is straightforwardly assumed by [28] that and lie in . With the same consideration, the authors in [34] and [35] ignored the cascaded antenna gain involving the sidelobe gain, because it has the relatively small value and happens in a very low probability in the Gaussian BAE scenario. Following these works, we also ignore sidelobe gain in the cascaded antenna gain, i.g., the component of (12) over . As a result, with the Gaussian BAE model can be approximately presented as

 f˜Ωx0(Ω)∼Ω12ησ2−12ησ2erf2(π√2σ2),   Ω∈[g,1]. (14)

When , we have .

#### Iii-C2 The Approximated PDF With the Uniform BAE model

Recalling the expression of , the strong directional antenna means is far less than . Hence, in the uniform BAE case, the event that equals to the product of two mainlobes occurs in an extremely small probability, i.e., . Furthermore, by (40) in the Appendix B, it can be inferred that the arctan term in (13) is generated by the product of two mainlobes. Therefore, it is reasonable to neglect the arctan term of (13) in the uniform BAE case. Then, we can approximate (13) by

 ⎧⎪ ⎪⎨⎪ ⎪⎩1−PU0πΩ√η√−lnΩg+δ(Ω−g2)(1−PU0)2,Ω∈[g2,g)14πηΩ,Ω∈[g,1] (15)

#### Iii-C3 Verification of the Approximated PDFs

To verify our approximations, we draw with Gaussian and uniform BAE models in Fig. 2 and Fig. 3, respectively. For the Gaussian BAE model, we draw the PDF of with as an example to verify our approximation.222, the approximated PDF of shows the similar accuracy with various . From the left subfigure of Fig. 2, we can see that with and , the approximated PDFs match the theoretical PDFs closely. As decreases, the PDF of tends to be a pulse-like function. We can infer that it asymptotically converges to as . While, in the right subfigure of Fig. 2, for , there is a slight difference between the results of (12) and (14). It is because the probability that the sidelobe appears in the cascaded antenna gain becomes larger when grows. Thus, with , the difference generated by our approximation seems extremely apparent. Accordingly, if , (14) is an appropriate approximation of (12). 333 Given , there is always for . While, given , we have for . Consequently, the probabilities of two mainlobes cascading with and , i.e., , are about 0.9946 and 0.9111, respectively. Apparently, for , , there is . That is why we here choose as the approximation condition. Note that it is weaker than the assumption that adopted by [28]. Moreover, from Fig. 3 we can see that the curves of (15) approach those of (13) extremely closely. Summarily, it is verified that both approximated PDFs can be used to analyze the system-level performance instead of the exact PDFs under our considered circumstance.

## Iv Energy Coverage Analysis

In this section, we focus on analyzing the energy coverage probability of the typical ERx. Energy coverage probability is defined as the probability that the harvested DC energy is larger than a pre-defined threshold, which is always the minimum required energy for information transmission or other operations.

In [33], the Meijer G-function was used to derive the analytic expression of information coverage probability of mmWave transmission. Theoretically speaking, we can also adopt this successful approach in our analysis. By [29], however, to obtain the analytic expression in the form of the Meijer G-function, the path loss exponents should be positive integers, which limits the application of the Meijer G-function based method. Alternatively, in this paper, without path loss exponent limitation, we provide an analytic expression of energy coverage probability with the help of Fox’s H function.444The Fox’s H function is a general function which can encompass almost all commonly used functions, e.g., Meijer G-function. Although the Fox’s H function is defined by an integral in a nonanalytic form, like the widely used Gamma function, Q-function, Hypergeometric function, Meijer G-function, etc., a look-up-table (LUP) storing the values of Fox’s H function can be generated via numerical methods. A Matlab program for evaluating the Fox’s H function was provided in [26]. More details about the Fox’s H function can be found in [24].

Letting be the DC energy threshold, we can write the energy coverage probability of the typical ERx as

 Pec=Pr(εDC>εth)=Pr(εRF>~εth), (16)

where is the equivalent RF energy threshold. In addition, according to , we let to observe the behavior of . If , there is . Note that , and are independent random variables. Following the widely adopted Gamma r.v. approximation [21, 37], i.e., using , , instead of , we can rewrite as

 Pec≈Eε0,εL,εN{Pr(μK<ε0+εL+εN~εth)}≈Eε0,εL,εN⎧⎨⎩[1−exp(−A(ε0+εL+εN)~εth)]K⎫⎬⎭,=K∑k=0(−1)k(Kk)Lε0(ak)LεL(ak)LεN(ak),k∈Z+ (17)

in which , and . Next, we derive the analytic expressions of , , and , respectively.

### Iv-a The Analytic Expression of Lε0(ak)

As for , and are independent random variables, we have

 Lε0(ak)=E˜Ωx0,hx0,L{e−akPtG2mℓL(r0)˜Ωx0hx0,L}=E˜Ωx0⎧⎨⎩(1+akPtG2mCLr−αL0˜Ωx0mL)−mL⎫⎬⎭≈∫1g(1+akPtG2mCLr−αL0˜Ωx0mL)−mL˜Ω12ησ2−1x02ησ2erf2(π√2σ2)d˜Ωx0(a)=F(1)−F(g)erf2(π√2σ2) (18)

where , , , and is the Gauss hypergeometric function [29]. Note that in we resort to the following equation [1],

 ∫10b(1+zt)−atb−1dt=2F1(a,b;1+b;−z),∀z≥−1,a>0,b>0,z,a,b∈R.

When , i.e., the perfect beam alignment scenario, we can easily obtain

 Lε0(ak)=(1+akPtG2mCLr−αL0mL)−mL. (19)

### Iv-B The Analytic Expression of LεL(ak)

Define , . As follows the IPPP with intensity , the 2-tuple forms a marked IPPP (MIPPP). Due to the probability generating functional (PGFL) of MIPPP [19], we have

 LεL(ak)=EΦL,ωx,L⎧⎨⎩∏x∈ΦLe−akPtG2mCLr−αLxωx,L⎫⎬⎭=e−2πλt∫∞0(1−EωL{e−akPtG2mCLr−αLωL})e−βrrdr(b)=e−2πλt(1β2−EωL{∫∞0e−βr−akPtG2mCLr−αLωLrdr}). (20)

In , due to the Fubini’s theorem [6], we exchange the order of integral and expectation operations. Due to [Mathai and Haubold [25], (1.9.5)], we have

 ∫∞0e−βr−akPtG2mCLr−αLωLrdr=1αLβ2H2,00,2[β(akPtG2mCLωL)1αL∣∣(2,1)(0,1αL)]. (21)

Note that is the Fox’s H function and is defined by [24, 25]. Consequently, there is

 e−2πλt⎛⎝1β2−1αLβ2EωL⎧⎨⎩H2,00,2⎡⎣β(akPtG2mCLωL)1αL∣∣∣(2,1)(0,1αL)⎤⎦⎫⎬⎭⎞⎠. (22)

Before solving the expectation of with respect to , we introduce the Lemma 2 as follows.

###### Lemma 2.

For , there is

 Hm,np,q[xy∣∣(a1,A1)⋯(ap,Ap)(b1,B1)⋯(bq,Bq)]=xb1B1∞∑t=0(1−x1B1)tt!Hm,np,q[y∣∣(a1,A1)⋯(ap,Ap)(t+b1,B1)(b2,B2)⋯(bq,Bq)].
###### Proof:

See [Mathai et al. [24], (1.88)]. ∎

By the Lemma 2, we can further attain

In , we apply Binomial theorem. Hence, there is

 LεL(ak)≈exp⎧⎪ ⎪⎨⎪ ⎪⎩−2πλt⎡⎢ ⎢⎣1β2−1αLβ2∞∑t=0t∑q=0(−1)q(tq)χL,q+2αLt!H2,00,2[β(akPtG2mCL)1αL∣∣∣(t+2,1)(0,1αL)]]}, (23)

where , .

###### Proposition 1.

If the PDF of follows (15), , there is

 E{˜Ωz}=1−PU0π√ηgz√πerf(√−zlng)√z+1−gz4πηz+(1−PU0)2g2z, (24)
###### Proof:

By (15), it is straightforward to obtain Proposition 1. The detailed proof is omitted to save space. ∎

###### Proposition 2.

If , , , there is

 E{hz}=Γ(m+z)Γ(m)mz, (25)
###### Proof:

By the PDF of , it is conveniently to prove Proposition 2. ∎

Therefore, substituting Proposition 1 and 2 into (23), we can obtain the analytic expression of .

### Iv-C The Analytic Expression of LεN(ak)

Similarly to (20), defining , we have

 (26)

Next, we derive the expressions of and separately.

Firstly, after some manipulations, it is easy to know

 L1(ak)=12(akPtG2mCN)2αNχN,2αNΓ(1−2αN), (27)

where , , which can be also written in the analytic form due to Proposition 1 and 2. Following the derivations from (21) to (23), similarly, we can obtain

 L2(ak)≈1β2−1αNβ2∞∑t=0t∑q=0(−1)q(tq)χN,q+2αNt!×H2,00,2[β(akPtG2mCN)1αN∣∣∣(t+2,1)(0,1αN)]. (28)

Consequently, the analytic expression of is also derived. Therefore, we can get the analytic expression of with , , and .

It is worth noting that for applying Proposition 1 and 2, we need to let . Clearly, it always holds for the piratical condition . So, we put no limitation on the path loss exponents.

## V Average Harvested Energy

Although the derived can be used to evaluate the energy coverage performance of mmWave WPT, it may not provide explicit and direct insight into the effect of BAE. Thus, in this section, the average harvested energy is addressed to further investigate the effect of BAE.

By [21, 37], the average harvested DC energy can be expressed as

 εavg=εminPec(εmin)+∫∞εminPec(ε)


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text:
# Solving a functional equation in complex analysis

Is there any simple (general) way to find the complex solution of a functional-equation?

For example, it is given $f(z) = f\big(\frac{z^2+4}{3}\big)$, and the question is to find all the functions $f(z)$ witch satisfy the given condition.

In an another topic I have seen a interesting solution, where another equation ($f(2z) = f^2 (z)$) was solved by differencing, defining a new function, and using the identity theorem. This solution is elegant, but I think that it doesn't help in the case above, and in many other cases. So, what is "the usual way" (if it exists) of solving this?

$$f(z) = f\big(\frac{z^2+4}{3}\big)$$

I am happy to assume $f$ is an entire function, so that you can differentiate freely.

• If you differentiate $f$ you are assuming that $f$ is analytic. This is a very strong condition that you have not included in your problem. – ajotatxe Jul 12 '16 at 17:30
• Of course, this condition isn't given in my problem (but I think that it was in the similar topic here), so I probably would lose some of the solutions by taking that $f$ is a holomorphic function. (I tried to derive looking for some solutions, I understand that it is not necessary the whole solutions set) – Icarus 369 Jul 12 '16 at 17:44
• A typical functional equation of this sort will have no nonconstant analytic solutions, so it makes a huge difference whether you require that. – Eric Wofsey Jul 12 '16 at 18:13
• Assume that $f$ is an entire function (holomorphic over the whole complex plane), so the differentiation is possible and allowed. – Icarus 369 Jul 12 '16 at 18:24

Let me assume you are interested in a functional equation of the form $f(z)=f(g(z))$ where $g(z)$ is some entire function and you want solutions $f(z)$ which are entire. In this case, for most choices of $g(z)$, there will be no nonconstant solutions. Indeed, suppose that there is some $a\in\mathbb{C}$ such that $g(a)=a$. If $|g'(a)|<1$, then if you start with a point $z_0$ near $a$ and iterate $g$, you will get closer and closer to $a$. Since $f$ must take the same value at all these points, we find that $f$ takes the same value at a set of points that accumulates at $a$, so by the identity theorem $f$ must be constant. On the other hand, if $|g'(a)|>1$, then $g$ has a holomorphic inverse $g^{-1}$ near $a$ such that $f(z)=f(g^{-1}(z))$ for all $z$ near $a$ and $|(g^{-1})'(a)|<1$, and we can apply the argument above with $g^{-1}$ to again find that $f$ must be constant.
So if $g$ has a fixed point $a$ at which $|g'(a)|\neq 1$, then any entire solution to $f(z)=f(g(z))$ must be constant. In particular, for instance, it is easy to verify that this condition holds for $g(z)=\frac{z^2+4}{3}$. If $g$ has a fixed point $a$ with $|g'(a)|=1$, I'm not sure whether you can always find a grand orbit of $g$ that accumulates at $a$; probably this is known and someone who knows complex dynamics better than me could answer that question.
• Thank you, your answer was very helpful - most of the equations that I had to solve are of the type $f(z) = f(g(z))$. I still have only one question, what to do if there is a situation $f(h(z)) \pm f(z)= g(z)$? For example, if we have $f(z) - f(z^2) = z$, what is the usual aproach? I tried something (again with derivinig), but my logic looks circular, so I would be grateful for an advice. – Icarus 369 Jul 13 '16 at 7:28
• If $h(0)=0$, one technique will probably often be useful is looking at the Taylor series of $f$. You can write each side of the equation as a power series, and then equate them term by term, which will give you a bunch of equations relating the Taylor coefficients of $f$. If it is particularly simple to write down the Taylor series of $g(z)$ and $h(z)$ (as in your example), these equations will be easy to understand. – Eric Wofsey Jul 13 '16 at 7:35


text:
1- In each building identify and name three architectural *functional elements

1- In each building identify and name three architectural *functional elements that exist within it or define it.2- Analyze if these are functioning well or not and why (for each building) ?3- Analyze by comparison, which of the two buildings and their specific 3 functions  is more successful and why  and how is the less successful one so.*note*~ functional elements can be the same, but not a must.*Please be sure to write minimum two *paragraph long answer (per question) with intelligent reasoning to your answer.*A paragraph should be 100 to 200 words long, or be no more than five or six sentences. A good paragraph should not only be measured in characters, words, or sentences.The true measure of your paragraphs should be ideas.


text: You can simply replace this command with brew install --universal libyaml and you should be able to continue, but I think you will run into more issues shortly after. You can continue down this path with Fuerte, or backup to Electric, up to you.