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Distributed Compression for the Uplink of a Backhaul-Constrained Coordinated Cellular Network

# Distributed Compression for the Uplink of a Backhaul-Constrained Coordinated Cellular Network

Aitor del Coso and Sébastien Simoens Aitor del Coso is with the Centre Tecnològic de Telecomunicacions de Catalunya (CTTC), Av. Canal Olímpic S/N, 08860, Castelldefels, Spain. E-mail: aitor.delcoso@cttc.esSébastien Simoens is with Motorola Labs PARIS, Parc Les Algorithmes, 91193, Saint-Aubin, France. E-mail: simoens@motorola.com. This work was partially supported by the internship program of Motorola Labs, Paris. Also, it was partially funded by the European Comission under projects COOPCOM (IST-033533) and NEWCOM++ (IST-216715).
###### Abstract

We consider a backhaul-constrained coordinated cellular network. That is, a single-frequency network with multi-antenna base stations (BSs) that cooperate in order to decode the users’ data, and that are linked by means of a common lossless backhaul, of limited capacity . To implement receive cooperation, we propose distributed compression: BSs, upon receiving their signals, compress them using a multi-source lossy compression code. Then, they send the compressed vectors to a central BS, which performs users’ decoding. Distributed Wyner-Ziv coding is proposed to be used, and is optimally designed in this work. The first part of the paper is devoted to a network with a unique multi-antenna user, that transmits a predefined Gaussian space-time codeword. For such a scenario, the compression codebooks at the BSs are optimized, considering the user’s achievable rate as the performance metric. In particular, for the optimum codebook distribution is derived in closed form, while for an iterative algorithm is devised. The second part of the contribution focusses on the multi-user scenario. For it, the achievable rate region is obtained by means of the optimum compression codebooks for sum-rate and weighted sum-rate, respectively.

EDICS: WIN-INFO, MSP-CAPC, MSP-MULT, WIN-CONT.

## I Introduction

Inter-cell interference is one of the most limiting factors of current cellular networks. It can be partially, but not totally, mitigated resorting to frequency-division multiplexing, sectorized antennas and fractional frequency reuse [1]. However, a more spectrally efficient solution has been recently proposed: coordinated cellular networks [2]. They consist of single-frequency networks with base stations (BSs) cooperating in order to transmit to and receive from the mobile terminals. Beamforming mechanisms are thus deployed in the downlink, as well as coherent detection in the uplink, to drastically augment the system capacity[3, 4]. Hereafter, we only focus on the uplink channel.

Preliminary studies on the uplink performance of coordinated networks consider all BSs connected via a lossless backhaul with unlimited capacity [5][6]. Accordingly, the capacity region of the network equals that of a MIMO multi-access channel, with a supra-receiver containing all the antennas of all cooperative BSs [7]. Such an assumption seems optimistic in short-mid term, as operators are currently worried about the costs of upgrading their backhaul to support e.g., HSPA traffic load. To deal with a realistic backhaul constraint, two approaches have been proposed: i) distributed decoding[8, 9], consisting on a demodulating scheme distributely carried out among BSs, based on local decisions and belief propagation. Decoding delay appears to be its main problem. ii) Quantization [10], where BSs quantize their observations and forward them to decoding unit. Its main limitation relies on its inability to take profit of signal correlation between antennas/BSs; thus, introduces redundancy into the backhaul.

This paper considers a new approach for the network: distributed compression. The cooperative BSs, upon receiving their signals, distributely compress them using a multi-source lossy compression code [11]. Then, via the lossless backhaul, they transmit the compressed signals to the central unit (also a BS); which decompresses them using its own received signal as side information, and finally uses them to estimate the users’ messages. Distributed compression has been already proposed for coordinated networks in [12, 13, 14]. However, in those works, authors consider single-antenna BSs with ergodic fading. We extend the analysis here to the multiple-antenna case with time-invariant fading.

The compression of signals with side information at the decoder is introduced by Wyner and Ziv in [15, 16]. They show that side information at the encoder is useless (i.e., the rate-distortion tradeoff remains unchanged) to compress a single, Gaussian, source when it is available at the decoder [16, Section 3]. Unfortunately, when considering multiple (correlated) signals, independently compressed at different BSs, and to be recovered at a central unit with side information, such a statement can not be claimed. Indeed, this is an open problem, for which it is not even clear when source-channel separation applies [17]. To the best of authors knowledge, the scheme that performs best (in a rate-distortion sense) for this problem is Distributed Wyner-Ziv (D-WZ) compression [18]. Such a compression is the direct extension of Berger-Tung coding to the decoding side information case [19, 20]. In turn, Berger-Tung compression can be thought as the lossy counterpart of the Slepian-Wolf lossless coding [21]. D-WZ coding is thus the compresssion scheme proposed to be used, and is detailed in the sequel.

Summary of Contributions. This paper considers a single-frequency network with multi-antenna BSs. The first base station, denoted , is the central unit and centralizes the users’ decoding. The rest, , are cooperative BSs, which distributely compress their received signals using a D-WZ code, and independently transmit them to via the common backhaul of aggregate capacity . In the network, time-invariant, frequency-flat channels are assumed, as well as transmit and receive channel state information (CSI) at the users and BSs, respectively.

The first part of the paper is devoted to a network with a single user, equipped with multiple antennas. It aims at deriving the optimum compression codebooks at the BSs, for which the user’s transmission rate is maximized. Our contributions are the following:

• First, Sec. II revisits Wyner-Ziv coding [16, Section 3] and Distributed Wyner-Ziv coding [19], and adapts them to our compression scenario.

• For the single user transmitting a given Gaussian codeword, Sec. III proves that the optimum compression codebooks at the BSs are Gaussian distributed. Accordingly, the compression step is modelled by means of Gaussian ”compression” noise, added by the BSs on their observations before retransmitting them to the central unit.

• Considering a unique cooperative BS in the network (i.e., ), Sec. IV derives in closed form the optimum ”compression” noise for which the user’s rate is maximized. We also show that conditional Karhunen-Loève transform plus independent Wyner-Ziv coding of scalar streams is optimal.

• The compression design is extended in Sec. V to arbitrary BSs. The optimum ”compression” noises (i.e., the optimum codebook distributions) are obtained by means of an iterative algorithm, constructed using dual decomposition theory and a non-linear block coordinate approach [22, 23]. Due to the non-convexity of the noises optimization, only local convergence is proven.

The second part of the paper extends the analysis to a network where multiple users transmit simultaneously. For it, the achievable rate region is described resorting to the weighted sum-rate optimization:

• First, the sum-rate of the network is derived in Sec. VI, adapting previous results a single-user. Later, the weighted sum-rate, and its associated optimum compression ”noises”, are obtained by means of an iterative algorithm, constructed using dual decomposition and Gradient Projection [23].

Notation. denotes expectation. , and stand for the transpose of , conjugate transpose of and complex conjugate of , respectively. . denotes mutual information, entropy. The derivative of a scalar function with respect to a complex matrix is defined as in [24], i.e., . In such a way, e.g., . Moreover, we compactly write , and . A sequence of vectors is compactly denoted by . Furthermore, to define block-diagonal matrices, we state , with square matrices. stands for convex hull. Finally, the covariance of random vector conditioned on random vector is denoted by and computed

## Ii Compression of Vector Sources

The aim of compression within coordinated networks is to make the decoder extract the more mutual information from the reconstructed signals. Known rate-distortion results apply to this goal as follows.

### Ii-a Single-Source Compression with Decoder Side Information

Consider Fig. 1 with . Let be a zero-mean, temporally memoryless, Gaussian vector to be compressed at . Assume that it is the observation of the signal transmitted by user , i.e., . compresses the signal and sends it to , which makes use of its side information to decompress it. Finally, once reconstructed the signal into vector , the decoder uses it to estimate the message transmitted by the user. Wyner’s results [16] apply to this problem as follows.

###### Definition 1 (Single-source Compression Code)

A compression code with side information at the decoder is defined by two mappings, and and three spaces and , where

 fn:Yn1→{1,⋯,2nρ} gn:{1,⋯,2nρ}×Yn0→^Yn1.
###### Proposition 1 (Wyner-Ziv Coding [16])

Let the random vector with conditional probability satisfy the Markov chain , and let and be jointly Gaussian. Then, considering a sequence of compression codes with side information at the decoder:

 1nI(Xns;Yn0,gn(Yn0,fn(Yn1)))=I(Xs;Y0,^Y1) (1)

as if:

• the compression rate satisfies

 I(Y1;^Y1|Y0)≤ρ, (2)
• the compression codebook consists of random sequences drawn i.i.d. from , where ,

• the encoding outputs the bin-index of codewords that are jointly typical with the source sequence . In turn, outputs the codeword that, belonging to the bin selected by the encoder, is jointly typical with .

{proof}

The proposition is proven in [16, Lemma 5] using joint typicality arguments.

### Ii-B Multiple-Source Compression with Decoder Side Information

Consider Fig. 1. Let be zero-mean, temporally memoryless, Gaussian vectors to be compressed independently at , respectively. Assume that they are the observations at the BSs of the signal transmitted by user , i.e., . The compressed vectors are sent to , which decompresses them using its side information and uses them to estimate the user’s message. Notice that the architecture in Fig. 1 imposes source-channel separation at the compression step, which is not shown to be optimal. However, it includes the coding scheme with best known performance: Distributed Wyner-Ziv coding [18]. It applies to the setup as follows.

###### Definition 2 (Multiple-source Compression Code)

A compression code with side information at the decoder is defined by mappings, , , and , and spaces , and , where

 fin:Yni→{1,⋯,2nρi},  i=1,⋯,N gn:{1,⋯,2nρ1}×⋯×{1,⋯,2nρN}×Yn0→^Yn1×⋯×^YnN.
###### Proposition 2 (Distributed Wyner-Ziv Coding [18])

Let the random vectors , , have conditional probability and satisfy the Markov chain . Let and , be jointly Gaussian. Then, considering a sequence of compression codes with side information at the decoder:

 1nI(Xns;Yn0,gn(Yn0,f1n(Yn1),⋯,fNn(YnN)))=I(Xs;Y0,^Y1:N) (3)

as if:

• the compression rates satisfy

 I(YG;^YG|Y0,^YcG)≤∑i∈Gρi∀G⊆{1,⋯,N}, (4)
• each compression codebook , consists of random sequences drawn i.i.d. from , where .

• for every , the encoding outputs the bin-index of codewords that are jointly typical with the source sequence . In turn, outputs the codewords , that, belonging to the bins selected by the encoders, are all jointly typical with .

{proof}

The proposition is proven for discrete sources and discrete side information in [18, Theorem 2]. Also, the extension to the Gaussian case is conjectured therein. The conjecture can be proven by noting that D-WZ coding is equivalent to Berger-Tung coding with side information at the decoder[19]. In turn, Berger-Tung coding can be implemented through time-sharing of successive Wyner-Ziv compressions [20], for which introducing side information at the decoder reduces the compression rate as in (4). Due to space limitations, we limit the proof to this sketch.

Now, we can present the coordinated cellular network with D-WZ coding.

## Iii System Model

Let a single source , equipped with antennas, transmit data to base stations , each one equipped with antennas. The BSs, as in typical 3G networks, are connected (through radio network controllers) to a common lossless backhaul of aggregate capacity , and is selected to be the decoding unit. This user-to-BSs assignment is assumed to be given by upper layers and out of the scope of the paper111The derivation of the optimum set of BSs to decode the user is out of the scope of our study. We refer the reader to e.g, [6] for assignment algorithms and selection criteria..

The source transmits a message mapped onto a zero-mean, Gaussian codeword , drawn i.i.d. from random vector and not subject to optimization. The transmitted signal, affected by time-invariant, memory-less fading, is received at the BSs under additive noise:

 Yni=Hs,i⋅Xns+Zni,  i=0,⋯,N (5)

where is the MIMO channel matrix between user and , and is AWGN. Channel coefficients are known at both the BSs and at the user, while has centralized knowledge of all the channels within the network.

### Iii-a Problem Statement

Base stations , upon receiving their signals, distributely compress them using a D-WZ compression code. Later, they transmit the compressed vectors to , which recovers them and uses them to decode. Considering so, the user’s message can be reliably decoded iif [12, Theorem 1]:

 Rs ≤ limn→∞1nI(Xns;Yn0,gn(Yn0,f1n(Yn1),⋯,fNn(YnN))) = I(Xs;Y0,^Y1:N).

Second equality follows from (3) in Prop. 2. However, equality only holds for compression rates satisfying the set of constraints (4). As mentioned, in the backhaul there is only an aggregate rate constraint , i.e., , . Therefore, the set of constraints (4) can be all re-stated as:

 I(YG;^YG|Y0,^YcG)≤R∀G⊆{1,⋯,N}. (7)

Furthermore, from the Markov chain in Prop. 2, the following inequality holds

 I(YG;^YG|Y0,^YcG)≤I(Y1:N;^Y1:N|Y0)∀G⊆{1,⋯,N}. (8)

Therefore, forcing the constraint to hold makes all constraints in (7) to hold too. Accordingly, the maximum transmission rate of user is obtained from optimization:

 C = max∏Ni=1p(^Yi|Yi) I(Xs;Y0,^Y1:N) s.t.  I(Y1:N;^Y1:N|Y0)≤R,
###### Theorem 1

Let . Optimization (III-A) is solved for Gaussian conditional distributions . Thus, the compressed vectors can be modelled as , where is independent, Gaussian, ”compression” noise at . That is,

 C=maxΦ1,⋯,ΦN⪰0 logdet(I+Qσ2rH†s,0Hs,0+QN∑n=1H†s,n(σ2rI+Φn)−1Hs,n) (10) s.t.  logdet(I+diag(Φ−11,⋯,Φ−1N)RY1:N|Y0)≤R.

where the conditional covariance follows (54).

{proof}

See Appendix B for the proof.

###### Remark 1

The maximization above is not concave in standard form: although the feasible set is convex, the objective function is not concave on .

### Iii-B Useful Upper Bounds

Prior to solving (10), we present two upper bounds on it.

###### Upper Bound 1

The achievable rate in (10) is upper bounded by

 (11)
###### Upper Bound 2

The achievable rate in (10) satisfies

 C≤I(Xs;Y0)+R=logdet(I+1σ2rHs,0QH†s,0)+R. (12)
{proof}

See Appendix C for the proof.

###### Remark 2

Notice that, independently of the number of BSs, the achievable rate is bounded above by the capacity with plus the backhaul rate.

## Iv The Two-Base Stations Case

We first solve (10) for . As mentioned, the objective function, which has to be maximized, is convex on . In order to make it concave, we change the variables , so that

 C = maxA1⪰0 logdet(I+Qσ2rH†s,0Hs,0+QH†s,1(A1σ2r+I)−1A1Hs,1) s.t.  logdet(I+A1RY1|Y0)≤R.

The objective has turned into concave. However, the constraint now does not define a convex feasible set. Therefore, Karush-Kuhn-Tucker (KKT) conditions become necessary222Notice that all feasible points are regular. but not sufficient for optimality. To solve the problem, we need to resort to the general sufficiency condition [23, Proposition 3.3.4]: first, we derive a matrix for which the KKT conditions hold. Later, we demonstrate that the selected matrix also satisfies the general sufficiency condition, thus becoming the optimal solution. The optimum compression noise is finally recovered as . This result is presented in Theorem 2:

###### Theorem 2

Let and the conditional covariance (see Appendix A-A):

 RY1|Y0=Hs,1(I+Qσ2rH†s,0Hs,0)−1QH†s,1+σ2rI, (14)

with eigen-decomposition . The optimum ”compression” noise at is , with

 ηj=[1λ(1σ2r−1sj)−1σ2r]+, (15)

and is such that .

{proof}

See Appendix D for the proof

### Iv-a Practical Implementation

The optimum compression in Theorem 2 can be carried out using a practical Transform Coding (TC) approach. With TC, first transforms its received vector using an invertible linear function and then separately compresses the resulting scalar streams [25]. We show that the conditional Karhunen-Loève transform (CKLT) is an optimal linear transformation [26]. First, let recall that multiplying a vector by a matrix does not change the mutual information [27], i.e., and . From Theorem 2, the optimum compressed vector satisfies , with and . Therefore, the following compressed vectors are also optimal

 ^Y1=U†Y1+U†Z∗c, (16)

where vector is referred to as the CKLT of vector . Notice now that is diagonal. Therefore, the elements of the compressed vector are conditionally uncorrelated given . Likewise, so are the elements of vector . Due to this uncorrelation, each element of vector can be compressed, without loss of optimality, independently of the compression of the others elements, at a compression rate , [16]. From Theorem 2 we validate that . This demonstrates that CKLT plus independent coding of streams is optimal, not only for minimizing distortion as shown in [26], but also for maximizing the achievable rate of coordinated networks.

## V The Multiple-Base Stations Case

Consider now assisted by cooperative BSs. The achievable rate follows (10) where, as previously, the objective function is not concave over , . To make it concave, we change the variables: , , so that:

 C=maxA1,⋯,AN⪰0 logdet(I+Qσ2rH†s,0Hs,0+QN∑n=1H†s,n(Anσ2r+I)−1AnHs,n) (17) s.t.  logdet(I+diag(A1,⋯,AN)RY1:N|Y0)≤R.

Again, the feasible set does not define a convex set. Our strategy to solve the optimization is the following: first, we show that the duality gap for the problem is zero. Later, we propose an iterative algorithm that solves the dual problem, thus solving the primal too. An interesting property of the dual problem is that the coupling constraint in (17) is decoupled [23, Chapter 5].

### V-a The dual problem

Let the Lagrangian of (17) be defined on , and as:

 L(A1,⋯,AN,λ) = logdet(I+Qσ2rH†s,0Hs,0+QN∑n=1H†s,n(Anσ2r+I)−1AnHs,n) (18) −λ⋅(logdet(I+diag(A1,⋯,AN)RY1:N|Y0)−R).

The dual function for follows [22, Section 5.1]:

 g(λ)=maxA1,⋯,AN⪰0L(A1,⋯,AN,λ). (19)

The solution of the dual problem is then obtained from

 C′=minλ≥0g(λ). (20)
###### Lemma 1

The duality gap for optimization (17) is zero, i.e., the primal problem (17) and the dual problem (20) have the same solution.

{proof}

The duality gap for problems of the form of (17), and satisfying the time-sharing property, is zero [28, Theorem 1]. Time-sharing property is defined as follows: let be the solution of (17) for backhaul rates , respectively. Consider for some . Then, the property is satisfied if and only if , . That is, if the solution of (17) is concave with respect to the backhaul rate . It is well known that time-sharing of compressions cannot decrease the resulting distortion [27, Lemma 13.4.1], neither improve the mutual information obtained from the reconstructed vectors. Hence, the property holds for (17), and the duality gap is zero.

We then solve the dual problem in order to obtain the solution of the primal. First, consider maximization (19). As expected, the maximization can not be solved in closed form. However, as the feasible set (i.e., ) is the cartesian product of convex sets, then a block coordinate ascent algorithm333Also known as Non-Linear Gauss-Seidel Algorithm [29, Section II-C]. can be used to search for the maximum [23, Section 2.7]. The algorithm iteratively optimizes the function with respect to one while keeping the others fixed. It has been previously used to e.g., solve the sum-rate problem of MIMO multiple access channels with individual and sum-power constraint [30][31]. We define it for our problem as:

 At+1n=argmaxAn⪰0L(At+11,⋯,At+1n−1,An,Atn+1,⋯,AtN,λ), (21)

where is the iteration index. As shown in Theorem 3, the maximization (21) is uniquely attained.

###### Theorem 3

Let the optimization and the conditional covariance matrix (See Appendix A-A)

 RYn|Y0,^Ycn=Hs,n⎛⎝I+Q⎛⎝1σ2rH†s,0Hs,0+∑p≠nH†s,p(Apσ2rI+I)−1ApHs,p⎞⎠⎞⎠−1QH†s,n+σ2rI (22)

with eigen-decomposition . The optimization is uniquely attained at , where

 ηj=[1λ(1σ2r−1sj)−1σ2r]+,  j=1,⋯,Nn. (23)
{proof}

See Appendix E-A for the proof.

Function is continuously differentiable, and the maximization (21) is uniquely attained. Hence, the limit point of the sequence is proven to converge to a local maximum [23, Proposition 2.7.1]. To demonstrate convergence to the global maximum, it is necessary to show that the mapping is a block contraction444See [32, Section 3.1.2] for the definition of block-contraction. for some [32, Proposition 3.10]. Unfortunately, we were not able to demonstrate the contraction property on the Lagrangian, although simulation results suggest global convergence of our algorithm always.

Once obtained through the Gauss-Seidel Algorithm555Assume hereafter that the algorithm has converged to the global maximum of ., it remains to minimize it on . First, recall that is a convex function, defined as the pointwise maximum of a family of affine functions [22]. Hence, to minimize it, we may use a subgradient approach as e.g., that proposed by Yu in [31]. The subgradient search consists on following search direction such that

 g(λ′)−g(λ)λ′−λ≥h∀λ′. (24)

Such a search is proven to converge to the global minimum for diminishing step-size rules [29, Section II-B]. Considering the definition of , the following satisfies (24):

 h=R−logdet(I+diag(A1,⋯,AN)RY1:N|Y0). (25)

Therefore, it is used to search for the optimum as:

 increase λ if  h≤0  or  decrease λ if  h≥0. (26)

Consider now as the initial value of the Lagrange multiplier. For such a multiplier, the optimum solution of (19) is and the subgradient (25) is (See Appendix E-B). Hence, following (26), the optimum value of is strictly lower than one. Algorithm 1 takes all this into account in order to solve the dual problem, hence solving the primal too. As mentioned, we can only claim convergence of the algorithm to a local maximum.

### V-B Practical Implementation

In the network, Distributed Wyner-Ziv compression can be practically implemented using a simple Successive Wyner-Ziv (S-WZ) approach [20][33, Theorem 3]. To describe it, let us recall that the optimum compression noises are obtained from Algorithm 1, and let be a given permutation on . For such a permutation, the S-WZ coding is defined as follows:

• Parallel Compression: compresses its received vector using a single-source Wyner-Ziv code with decoder side information (following Proposition 1), at a compression rate

 ρπ(1) = (27) = logdet(I+(Φ∗π(1))−1RYπ(1)|Y0).

The conditional covariance is calculated in (53). In parallel, , compresses its signal using a single-source Wyner-Ziv code with decoder side information , at a rate

 ρπ(n) = (28) = logdet(I+(Φ∗π(n))−1RYπ(1)|Y0,^Yπ(1:n−1)).

In this case, the conditional covariance can be calculated from (56).

• Successive Decompression: first recovers the codeword using side information ; later, it successively recovers codewords , , using as side information.

It is easy to check the optimality of the S-WZ coding:

 N∑n=1ρπ(n) = N∑n=1I(Yπ(n);^Yπ(n)|Y0,^Yπ(1:n−1)) = N∑n=1I(Y1:N;^Yπ(n)|Y0,^Yπ(1:n−1)) = I(Y1:N;^Y1:N|Y0) = R.

Second equality comes from the Markov chain in Proposition 2, and third from the chain rule for mutual information; The fourth follows from the fact that satisfy the constraint (10) with equality. Unfortunately, transform coding is not (generally) optimum for S-WZ with , since the eigenvectors of , and those of does necessarily match.

## Vi The Multiple User Scenario

In previous sections, we considered a single user within the network. To complement the analysis, we study hereafter multiple (i.e., two) senders transmitting simultaneously. The users, and , transmit two independent messages , , mapped onto codewords , , respectively. Codewords are drawn i.i.d. from random vectors , and are not subject to optimization. Hence, now, the BSs receive:

 Yni=2∑u=1Hu,iXnu+Zni,  i=0,⋯,N. (30)

Here, is the MIMO channel between user and , and . As previously, signals at are distributely compressed using a D-WZ code, and later sent to , which centralizes decoding. Using standard arguments, the set of transmission rates , at which messages , can be reliably decoded is [27][14]:

 (31)

The union in (31) is explained by the fact that compression codebooks might be arbitrary chosen at the BSs. Notice that the boundary points of the region can be achieved using superposition coding (SC) at the users, successive interference cancellation (SIC) at the , and (optionally) time-sharing (TS). Furthermore, as for the single-user case, the optimum conditional distributions , at the boundary of the region can be proven to be Gaussian666Recall that , . We omit the proof due to space limitations.. Therefore, the union in (31) can be restricted to compressed vectors of the form , where . That is:

 C=coh⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝⋃Φ1,⋯,ΦN∈c(R)⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩R1≤logdet(I+Q1σ2rH†1,0H1,0+Q1∑Nn=1H†1,n(σ2rI+Φn)−1H1,n)R2≤logdet(I+Q2σ2rH†2,0H2,0+Q2∑Nn=1H†2,n(σ2rI+Φn)−1H2,n)R1+R2≤logdet(I+Qσ2rH†s,0Hs,0+Q∑Nn=1H†s,n(σ2rI+Φn)−1Hs,n)⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠ (32)

Where , and , for . Covariance is calculated in Appendix A-B. To evaluate such a region, we resort to the weighted sum-rate (WSR) optimization [34, Sec. III-C]. That is, we express

 C={(R1,R2):αR1+(1−α)R2≤R(α),∀α∈[0,1]}, (33)

with the maximum WSR, given weights and


text:
# When Do I Send Emails?

Work and the holidays have been distracting me from blogging and fun side projects for a couple of months, so I'm easing back into it with a really quick and easy Gmail data-wrangling post.

When I first started playing around with my Gmail data, I mentioned that I wanted to get some of the stats that Xobni used to provide before they were swallowed by the Yahoo! black hole. A couple of the simpler stats to compile are "what days of the week do I send most of my email?" and "what time of day do I send the most emails?".

In order to do any time-based analysis on my emails, I'm going to need the dates and times they were sent. So I've taken the PowerShell script from a while back and made a slight modification; in the $props hash I'm adding a field called SentDate: $props = @{
Id = $mimeMessage.MessageId To =$mimeMessage.To.ToString()
From = $mimeMessage.From.ToString() FromEmail =$fromEmail
Subject = $mimeMessage.Subject Body =$bodyText
Format = $actualFormat SentDate =$mimeMessage.Date.ToUniversalTime()
}

MimeKit provides the sent date of the email as a DateTimeOffset; to keep things consistent, I'm converting everything to UTC at this stage.

From there, I import the data into pandas as per usual and filter it down to just the emails sent by me:

import pandas as pd
import numpy as np
import humanfriendly
import matplotlib.pyplot as plt
plt.style.use('ggplot')

df = pd.read_csv('../times.csv', header = 0)

fromMe = df.query('FromEmail == "[my email]"')

It turns out that indexing your data by date/time in pandas is pretty easy; you just create a DateTimeIndex:

temp = pd.DatetimeIndex(fromMe['SentDate'], tz = 'UTC').tz_convert('America/Denver')

Here I'm telling pandas to create an index by the SentDate field, and that the field is already in UTC. Then I'm converting all of those dates and times to my local timezone so that the data makes sense from my local perspective. This mostly works, because I mostly live in the Mountain timezone. Some of my data will be a little skewed because of emails sent while traveling and a few months where I lived in the Eastern time zone, but not so much that I care. In a later post I might look at how this data changes over time, which is more interesting (I might even be able to identify when and where I was traveling based on that data).

But for now, let's break down the data in temp and shove it back into the original dataset:

fromMe['DayOfWeek'] = temp.dayofweek
fromMe['Hour'] = temp.hour
fromMe['Year'] = temp.year

Now for each email from me, I've got a column that tells me what hour of the day and day of the week it was sent. From there, aggregating it and charting it are a snap:

# Number of emails sent by day of week
sentDayOfWeek = fromMe.groupby(['DayOfWeek']).agg({'Id' : 'count'})
sentDayOfWeek['Id'].plot(kind='bar', figsize=(6, 6), title='Emails Sent By Day Of Week')
plt.show()

# Number of emails sent by hour of day
sentHourOfDay = fromMe.groupby(['Hour']).agg({'Id' : 'count'})
sentHourOfDay['Id'].plot(kind='bar', figsize=(6, 6), title='Emails Sent By Hour Of Day')
plt.show()

The data is about what I'd expect; more emails on Monday than any other day (0 == Monday on this chart) and the majority of emails sent during the workday (with a dip around lunch).

Aggregating by year provides a bit of mystery, though:

sentYear = fromMe.groupby(['Year']).agg({'Id' : 'count'})
sentYear['Id'].plot(kind='bar', figsize=(6, 6), title='Emails Sent By Year')
plt.show()
The numbers vary quite a bit more than I'd expect. 2004 makes sense; I only started using Gmail in July of that year. And the next couple of years shows me using Gmail more and more over my old Lycos account. The spike in 2011 also seems reasonable, as that's when I stopped working at an office with an Exchange server, so my day-to-day email load shifted. But the dips in 2012 and 2015? No idea. I'll have to dig further into those.
Comment

# Simple Photo De-duplication with PowerShell

After several failed attempts and false starts over the course of the last decade or so, this summer I vowed to finally get my wife's photo library and my own properly merged. For various reasons, we both had large libraries of photos from various events and vacations which had a lot of overlap, but not 100% overlap.

This meant I needed to examine two large collections of photos and copy all of the non-duplicate photos from one to the other. And I couldn't rely on file names or paths at all, because my wife is good about renaming and organizing her photos, while I am not. So I whipped up a PowerShell script to gather the SHA-1 hashes of every file; by comparing them all, I could find and ignore the duplicates. It's not perfect - it'll only find pictures which are exact duplicates. If my collection has the original and my wife's collection just has the "red-eye reduction" version, we'll end up with both in the final collection. But it considerably reduced the amount of de-duplication work we have to do by hand.

Here's the function which actually gathers all the photo data for a folder (and all its child folders):

function Get-PhotoData {
param([string]$path)$results = @()

$files = Get-ChildItem$path -Recurse -Filter *.jpg

$total = ($files | measure).Count

$files | % {$i=1} {
Write-Host "Processing $_ ($i of $total)"$props = @{
Name = $_.Name Path =$_.FullName
Size = $_.Length Hash = (Get-FileHash$_.FullName -Algorithm SHA1).Hash
}

$results += (New-Object PSObject -Property$props)
$i++ }$results
}

I ran that function against my wife's photo collection (which was effectively our master collection) and my own:

$master = Get-PhotoData -path$masterPath
$toMerge = Get-PhotoData -path$toMergePath

In theory, I could then have used Compare-Object to figure out which items in my collection were safe to delete (i.e., they already existed in my wife's collection):

$safeToDelete = Compare-Object -IncludeEqual -ExcludeDifferent -ReferenceObject$toMerge -DifferenceObject $master -Property Hash -PassThru | Select-Object -ExpandProperty Path  This would give me a list of paths to photos in my collection which had a matching SHA1 hash to a photo already in my wife's collection. Or, I could find the list of items in my collection which were missing from her collection: $toMove = Compare-Object -ReferenceObject $master -DifferenceObject$toMerge -Property Hash -PassThru | ? { $_.SideIndicator -eq '=>' } | Select-Object -ExpandProperty Path  Moving each item in that collection would then be easy: $toMove | % {$i=1} { Write-Host "Moving$_ ($i of$total)"
Move-Item $_ "[destination path]"$i++
}

For small enough collections, this works great. But if you've got a large enough photo collection you might start running into performance problems with Compare-Object. If that's the case, with a little extra effort and a little bit of Python you can figure out your $safeToDelete list much faster. First, we dump our photo data to a couple of files: $master | Select Name, Path, Size, Hash | Export-Csv -NoTypeInformation "master.csv"
$toMerge | Select Name, Path, Size, Hash | Export-Csv -NoTypeInformation "toMerge.csv"  Now we throw together a quick Python program using pandas to read in the two data sets, merge them into a single data set by matching on the file size and hash, and dump the output to another file: import pandas as pd # Read in our email data file master = pd.read_csv('../master.csv', header = 0) toMerge = pd.read_csv('../toMerge.csv', header = 0) both = pd.merge(toMerge, master, on=['Size', 'Hash']) both.to_csv('../safeToDelete.csv')  The new .csv file will have columns 'Path_x' and 'Path_y'; since we had toMerge as the first parameter to merge, Path_x is a list of all the files in that collection which can be deleted. More Python-savvy folks than me can probably handle the deletion straight from the Python script, but I just did it with PowerShell: $toDelete = (Import-Csv .\safeToDelete.csv).Path_x

$total = ($toMove | measure).Count

$toDelete | % {$i=1} {
Write-Host "Deleting $_ ($i of $total)" Remove-Item$_
$i++ }  Of course, don't go running any of this code or deleting any files until you've backed your folders up somewhere safe; if you make any mistakes (or any of my code is totally broken), you'll want a safety net in place. Comment ### E.Z. Hart # How Long Would It Take To Read All My Email? This is part of a series on mining your own Gmail data. For this post I want to tackle a fun question: how long would it take to read all of my email if that's all I did, 24/7? It's one of those questions that should interest anyone who's concerned about information overload or is looking to pare down their information consumption: "Just how much of my time is theoretically committed to my inbox?" First, the obvious: nobody actually does this. No one actually reads every email they receive from start to finish (as anyone who's dealt with email in a corporate environment knows all too well). Most of us have filters (both electronic and mental) set up to glean the info we need and skip the rest. And I'll bet that a lot of email is written without any expectation that the whole thing will be read; the author may be well aware that different portions of the email are relevant to different recipients, or that the email itself will only be interesting to a subset of the mailing list (e.g., many marketing emails). So it's not the one super-relevant data point that should make people completely re-think their information consumption habits or anything like that. But it is fun to think about, and as one data point among many others it might prove interesting or useful. On to the fun part - actually coming up with a number! Like most people, I'm getting new emails all the time. So technically I should be taking into account all the new emails I receive while I'm still reading through my old ones. But that's hard, so I'm not going to bother. Instead, I'm just going to assume I've stopped getting emails at all while I'm reading. Which means that getting a basic number is easy - I just have to count all the words in all my emails, divide that by the number of words per minute I read, and I've got the number of minutes it would take to read everything. The first thing I need to do is go back to my PowerShell script and pull in the body of each email. This is where we hit the first snag - HTML emails. For doing word counts, I really don't want to look at HTML emails, because there's a ton of junk in there which a human won't be reading. Luckily, most email clients which send HTML emails also include a text version; in those cases, we'll just extract that text portion of the email and ignore the HTML. Unfortunately, this isn't always the case; when there's not a text version available, we'll just have to get the HTML and figure out how to deal with it later. As usual, MimeKit will be doing most of the work. This version of the script is pretty similar to our previous ones, except that we have to loop through the possible body formats for each message to figure out which formats are available. We always check for the 'Text' format first, because that's the one we really want. If that's not available, we run through the others until we find one that works. The relevant changes are the hash of the possible formats, which we use for iteration and for tracking the number of emails of each type: $formats = @{
[MimeKit.Text.TextFormat]::Text = 0;
[MimeKit.Text.TextFormat]::Flowed = 0;
[MimeKit.Text.TextFormat]::Html = 0;
[MimeKit.Text.TextFormat]::Enriched = 0;
[MimeKit.Text.TextFormat]::RichText = 0;
[MimeKit.Text.TextFormat]::CompressedRichText = 0
}

And the section where we determine what the actual format is and store it:

    $bodyText =$null
$actualFormat =$null

# Run through all the enumeration values
# The pipe through sort ensures that we check them in the enum order,
# which is great because we prefer text over flowed over HTML, etc.
$formats.Keys | sort | % { # try each Format until we find one that works if($actualFormat -eq $null) { # Try to get the body in the current format$bodyText = $mimeMessage.GetTextBody($_)
if($bodyText) {$actualFormat = $_ } } } if($actualFormat -eq $null) {$unknownFormat += 1;
$actualFormat = "Unknown" } else {$formats[\$actualFormat] += 1;
}

You can find the full script here.

A couple of notes:

1. This isn't perfect; sometimes MimeKit can't really figure out what the format is. For example, I have some Skype notification emails which MimeKit thinks are HTML only, but are in fact text. I'm not sure why MimeKit gets confused (probably incorrect headers in the original emails), but out of about 43,000 emails only a couple dozen seem to have issues, so I'm not going to worry about it.
2. In all of my emails, the only two formats returned were Text and HTML. This might have something to do with what Gmail supports; I've seen some posts that suggest Gmail doesn't support Flowed, though those may be outdated. In any case, I'm only really dealing with Text and HTML in my word counts.

Once we've got the data, we can load it up in pandas and do some counting. Doing a naive count of the words in the plain text emails is trivial; we just define a method that uses Python's split method with None as the delimiter argument, and then look at the length of the returned list. Here's what textWordCount looks like:

def textWordCount(text):
if not(isinstance(text, str)):
return 0

return len(text.split(None))

But the HTML emails are problematic because most of the content is markup that the user will never actually read. So we need to strip all that markup out and just count the words in the text portions of the HTML. To do that, we create another method which parses the HTML email content using the amazing Beautiful Soup library, strips away the style, script, head, and title parts, and extracts the text from what's left using get_text(). Once we've got the actual human-readable text, we can run it through our usual word counting method:

def htmlWordCount(text):
if not(isinstance(text, str)):
return 0

soup = bsoup(text, 'html.parser')

if soup is None:
return 0

stripped = soup.get_text(" ", strip=True)

[s.extract() for s in soup(['style', 'script', 'head', 'title'])]

stripped = soup.get_text(" ", strip=True)

return textWordCount(stripped)

I took a couple of online tests to get an idea of how fast I read and came up with 350 words per minute. With that bit of data, we can now add some more columns to our data and figure out the total time to read all the emails:

def wordCount(row):

if(row['Format'] == 'Html'):
return htmlWordCount(row['Body'])

return textWordCount(row['Body'])

averageWordsPerMinute = 350

# Count the words in each message body
emails['WordCount'] = emails.apply(wordCount, axis=1)
emails['MinutesToRead'] = emails['WordCount'] / averageWordsPerMinute

# Get total number of minutes required to read all these emails
totalMinutes = emails['MinutesToRead'].sum()

# And convert that to a more human-readable timespan
timeToRead = humanfriendly.format_timespan(totalMinutes * 60)

The full script is here, if you're playing at home.

Running that against all of my Gmail gives me:

>>> timeToRead
'2 weeks, 6 days and 18 hours'

So if I sat down and read at my fastest speed 24/7 for three weeks straight with no breaks, no sleep, and never slowing down, I could finish reading every word of every email I've ever received in my Gmail account. If I only read them 8 hours a day, it'd take me about 9 weeks to finish.

That's actually less than I expected, though "two whole months of your life spent just reading your email" is a still a bit sobering.

Sobering enough that I'm not going to try to compute this for my other four email accounts, anyway.

Comment

# Mining Your Gmail Data - Part 6

First off, let's take a look at the second question that came up at the end of the last post: ignoring the Media Type (the 'application/', 'video/', etc.) from the MIME type.

That turns out to be pretty easy - the script from last time already collected that data, because MimeKit already made it available. We just need to adjust our pandas script to group on 'MediaSubtype' instead of 'MimeType':

types = notFromMe.groupby(['MediaSubtype'])

That cleaned things up a lot. But we still have the second question from the last post: what's behind octet-stream?

Application/octet-stream is basically the generic binary file option; most likely the original client which uploaded the file didn't specify the type. But we can make an educated guess about the type based on the file name extension, where we have it. So we'll write a quick function which takes a row of data and, if the Media Subtype is 'octet-stream', returns the file name extension from the FileName column:

import os.path

...

def filetype(row):
if not(isinstance(row['ContentTypeName'], str)):
return ''
if row['MediaSubtype'] == 'octet-stream':
return os.path.splitext(row['ContentTypeName'])[1]
return row['MediaSubtype']

We can run that function against our data and put the results in a new column which we'll call 'FileType':

notFromMe['FileType'] = notFromMe.apply(lambda row: filetype(row), axis = 1)

Now, instead of grouping by MediaSubtype, we just group by FileType. This isn't perfect - some of our data is getting discarded because there's not enough info between Media Subtype and FileName to figure out what kind of attachment it is. But the data is mostly good, and gives us a much more useful chart:

I'm also running this chart with a threshold of 0.02 for the 'other' section, to clean up the less-frequent file types. The whole script can be found here.

So, if I'm looking to downsize my Gmail backup, I should probably concentrate on JPEGs, videos (wmv and mpeg), and PDFs.

Comment

format: plain


text:
Contest Duration: - (local time) (210 minutes) Back to Home
Official

## D - Shopping Editorial by rng58_admin

Let’s consider the game as follows. Initially, each player is given $$K$$ cards, numbered $$1$$ through $$K$$. In each turn, the player chooses one of the cards in his hand at random, and if the card is not a joker (described later), he wins the game, and leaves the game. At this point, if the card numbered $$x$$ is just played, all other cards numbered $$x$$ (in other players’ hands) will be changed to jokers. The game ends when $$K$$ players leave.

What happens after each player takes exactly one turn? For each player, one of the following happens:

• Wins and leaves the game.
• There are $$K - 1$$ cards in his hand. Exactly $$J - 1$$ of them are jokers, where $$J$$ is the number of people who have already left the game. (Since he didn’t win the game in the first round, the card he played must be a joker).

A key observation is that, after each remaining player takes the same number of turns (call it $$t$$), all remaining player has the same number of jokers in hands (that is, $$J - t$$, where $$J$$ is the number of people who have already left the game.)

Thus, we can define the following DP state:

$$dp[n][a][b][i]$$: $$N$$ people are remaining, each player has $$a$$ cards, $$b$$ of them are jokers. What is the probability that the $$i$$-th player wins?

To get a recurrence formula to compute $$dp[n][a][b][i]$$, iterate over two parameters $$l, r$$ - the number of people who leave the game in the next turn to the left / right of the $$i$$-th player. We get an $$O(N^6)$$ solution with a very small constant.

posted:
last update:


text:
# If $X,Y,Z$ are independent random variables having identical density functions

If $X,Y,Z$ are independent random variables having identical density functions $f(x) = e^{-x}$ , $0 < x < \infty$ , derive the joint probability density of the vector $(U,V,W)$ , where

\begin{align} U &= X + Y \\ V &= X + Z \\ W &= Y + Z \end{align}

I'm not sure on how to do this but, if were to find the joint probability density... I would do moment generate function for each $U,$ $V$ and $W$ and then multiple those.

• If you knew that $U,$ $V$ and $W$ were already independent, then you would be able to use the method you want, but they don't seem independent at first glance (and they aren't, look at my solution below). – Will M. Dec 18 '16 at 0:36

One idea is to use the change of variables theorem. Consider the random vector $(X,Y,Z)$ and notice that: $$f_{(X,Y,Z)}(x,y,z) = e^{-(x+y+z)} \mathbf{1}_{\mathbb{R}_+^3}(x,y,z).$$ Define the transformation $$T:(x,y,z) \mapsto (x + y, x + z, y + z)$$ which is linear and possesses the inverse $$S:(u,v,w) \mapsto \left(\dfrac{u+v-w}{2},\dfrac{u - v + w}{2}, \dfrac{-u+v+w}{2} \right).$$ By the change of variables theorem, the random vector $(U,V,W) = T(X,Y,Z)$ has density $$f_{(U,V,W)}(u,v,w) = f_{(X,Y,Z)}(S(u,v,w)) |\det \mathbf{D}S(u,v,w)|.$$ Now, $\det \mathbf{D}S(u,v,w) = \det S = -\dfrac{1}{2},$ so \begin{align} f_{(U,V,W)}(u,v,w) &=\dfrac{1}{2} f_{(X,Y,Z)}\left(\dfrac{u+v-w}{2},\dfrac{u - v + w}{2}, \dfrac{-u+v+w}{2} \right) \\ &= \dfrac{1}{2} e^{-\frac{u+v+w}{2}} \mathbf{1}_{\mathbb{R}_+^3} \left(\dfrac{u+v-w}{2},\dfrac{u - v + w}{2}, \dfrac{-u+v+w}{2} \right) \\ &= \dfrac{1}{2} e^{-\frac{u+v+w}{2}} \mathbf{1}_{\mathrm{G}}(u,v,w), \end{align}
where $\mathrm{G}$ is the set of $(u,v,w)$ such that the three relations $u + v \geq w,$ $u + w \geq v$ and $v + w \geq u$ hold. (Naturally, these three relations imply $u,v,w \geq 0$).


text:
# Does anyone take the Wightman axioms seriously?

+ 4 like - 0 dislike
1363 views

Does anyone take the Wightman axioms seriously? Mainly with respect to quantum gravity or gauge theores, abelian or non-abelian? Anyone doing any research on axiomatization of QFTs in some way?

This post imported from StackExchange Physics at 2014-04-11 15:21 (UCT), posted by SE-user user33923
question with a caveat for those thinking there is a "solid mathematical background" for what we do in physics... lol...

This post imported from StackExchange Physics at 2014-04-11 15:21 (UCT), posted by SE-user user33923
You give me a hard time figuring out if you are seriously interested in or just making fun of and sniping at theoretical physics ...

This post imported from StackExchange Physics at 2014-04-11 15:21 (UCT), posted by SE-user Dilaton
I am seriously interested and sniping at theoretical physicists (while being one...) now, seriously, these aspects are important and I worked today some hours on this. There is apparently no axiomatization of QFT and the Wightman axioms appear very interesting if one starts quitting them one by one in the attempt to solve BH-"paradoxes"... so, if someone knows something, please answer! And stop assuming I am just spending time here for nothing!

This post imported from StackExchange Physics at 2014-04-11 15:21 (UCT), posted by SE-user user33923
More on Wightman axioms: physics.stackexchange.com/search?q=is%3Aq+Wightman+axioms

This post imported from StackExchange Physics at 2014-04-11 15:21 (UCT), posted by SE-user Qmechanic
Ok, I see ;-). Maybe you could edit what you told me in the comment unto the question, as you presently have 3 close votes hanging around your neck here ... :-/. I voted to leave open, as I think this is interesting too.

This post imported from StackExchange Physics at 2014-04-11 15:21 (UCT), posted by SE-user Dilaton
@user33923 Maybe you should update your question accordingly. I think to say if they are taken seriously is a little bit harsh. Maybe inspirehep.net/record/1094230 is an interesting read.

This post imported from StackExchange Physics at 2014-04-11 15:21 (UCT), posted by SE-user ungerade
at least a good starting point, thanks!

This post imported from StackExchange Physics at 2014-04-11 15:21 (UCT), posted by SE-user user33923

+ 11 like - 0 dislike

The question sounds like this, for a classical physicist:

Does anyone take Lagrangian mechanics seriously?

Wightman axioms describe qft how it should be within a certain paradigm, assuming that some fundamental difficulties affecting perturbative qft can be solved in some way (but without suggesting any solution). It just presents the final theory, I mean, that including interactions, as it should be in that view. There is no guaratee however that it is a correct and complete picture of the world. In particular because, the mentioned difficulties could be a clue, and probably are, of new physics like string theory or other structures relevant at very high energy or very small scales. Moreover the description of gauge theories within Wigthman formulation is by no means straigthforward. Nevertheless this approach stands as a mathematically solid framework where proofs of physically fundamental statements of qft have been rigorously built up. I mean, for instance, spin statistic theorem, cpt theorem and so on. However, it does not mean that these results would not arise from other formulations based on different physics. I think that Wightman axioms can be viewed as Lagrangian mechanics with respect to "real" classical physics. Lagrangian formulation is a model where some important relationships between crucial notions can be analysed, I think of the interplay between conserved quantities and symmetries for instance. On the other hand, it is however clear that Lagrangian formulation is too physically naive, since for instance it does not properly consider forces due to friction that reveal the existence of another level of reality (I mean thermodynamics and microscopic physics... It assumes that physical objects are pictured by differential geometry disregarding the discrete microphysical structures...). The core of Garding Wightman Streater‘s formulation has produced other formulations of qft that insist on the notion of local field. A textbook on those ideas is Haag's one. These ideas have been implemented to develop qft in curved spacetime, with application to black hole physics in particular and, recently, to cosmology. I belong to that community of mathematical physicists. The uv renormalized procedure has ben completely reformulated in curved spacetime into a generally covariant framework without assuming the existence of a preferred vacuum in view of the absence of Poincare' symmetry.

This post imported from StackExchange Physics at 2014-04-11 15:21 (UCT), posted by SE-user V. Moretti
answered Feb 5, 2014 by (2,075 points)

This post imported from StackExchange Physics at 2014-04-11 15:21 (UCT), posted by SE-user Zoltan Zimboras
(of course, the rest is important but...) First two lines, to me, answer the question:)

This post imported from StackExchange Physics at 2014-04-11 15:21 (UCT), posted by SE-user c.p.
I also like this answer, although I don't mark it as "accepted answer"... I will comment on this a bit later...

This post imported from StackExchange Physics at 2014-04-11 15:21 (UCT), posted by SE-user user33923
Ok, first: Lagrangeans or Hamiltonians are choices. You can use lots of other functions with the same effects. From this point of view one should not take them extremely seriously either. Next, a set of axioms may or may not be unique and in some cases you can replace some axioms with some theorems and derive the axioms as theorems from them. So, no, you should not take axioms extremely "serious" either.

This post imported from StackExchange Physics at 2014-04-11 15:21 (UCT), posted by SE-user user33923
Actually, there are non Lagrangian or Hamiltonian in W. axioms. Yes it is true that "a set of axioms may or may not be unique and in some cases you can replace some axioms with some theorems and derive the axioms as theorems from them", but it is true for every theoretical construction in theoretical physics. That is the way of life: We should not take axioms for anything extremely "serious". It is already a miracle that something so technically complicated like renormalization in QFT or GR, or QM in general give some physically sensible result.

This post imported from StackExchange Physics at 2014-04-11 15:21 (UCT), posted by SE-user V. Moretti
I know Valter, these are muddy waters. There is a lot more I thought about this, but I don't know if it's worth it to share here...

This post imported from StackExchange Physics at 2014-04-11 15:21 (UCT), posted by SE-user Pedro Lauridsen Ribeiro
I would be interested to hear other opinions, as muddy as they might be

This post imported from StackExchange Physics at 2014-04-11 15:21 (UCT), posted by SE-user user33923
+ 3 like - 0 dislike

The Wightman axioms are nearly irrelevant for those physicists who are satisfied with the nonrigorous level of arguments typical for most of theoretical physics. But they cannot be ignored by anyone doing mathematical physics, i.e., physics on a mathematically rigorous level, as they are universally believed to capture the minimal properties observable fields must have.

Note that the Wightman axioms capture only observable fields, hence cannot be applied directly to gauge fields (for which only the associated curvature fields are observable).

There is a fairly dedicated group of people from all over the world working on Algebraic and Axiomatic Quantum Field Theory, primarily in the framework of local quantum physics, closely related to the Wightman setting. See

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text:
# How do you multiply mixed rational numbers like a decimal and a fraction?

Apr 1, 2018

See explanation below.

#### Explanation:

To multiply two rational numbers without a calculator, it is easiest to express them both as fractions, then multiply their numerators and denominators.

For example, if you have

1.6xx5/9=?

You can express $1.6$ as $\frac{8}{5}$, so it becomes

$\frac{8}{\cancel{5}} \times \frac{\cancel{5}}{9} = \frac{8}{9} = 0.88888 \ldots$

How did I know that $1.6 = \frac{8}{5}$?

Try to see that 1.6 can be expressed as:

$\textcolor{red}{1.6}$

$= 1 + \frac{6}{10}$

$= \frac{10}{10} + \frac{6}{10}$

$= \frac{16}{10}$

$= \frac{\cancel{2} \times 8}{\cancel{2} \times 5}$

$= \textcolor{red}{\frac{8}{5}}$

Similarly, if we had a really weird number like 1.234, we can do the same thing:

$\textcolor{red}{1.234}$

$= 1 + \frac{2}{10} + \frac{3}{100} + \frac{4}{1000}$

$= \frac{1000}{1000} + \frac{200}{1000} + \frac{30}{1000} + \frac{4}{1000}$

$= \frac{1234}{1000}$

$= \frac{\cancel{2} \times 617}{\cancel{2} \times 500}$

$= \textcolor{red}{\frac{617}{500}}$

Using this method, ANY rational number, by definition, can be expressed as a fraction.

Hope this helps!


text:
# Applying a Two-site 'Gate' to an MPS

A very common operation with matrix product states (MPS) is multiplication by a two-site operator or 'gate' which modifies the MPS. This procedure can be carried out in an efficient, controlled way which is adaptive in the MPS bond dimension.

Say we have an operator $G^{s'_3 s'_4}_{s_3 s_4}$ which is our 'gate' and which acts on physical sites 3 and 4 of our MPS, as in the following diagram:

To apply this gate in a controlled manner, first 'gauge' the MPS such that either site 3 or 4 is the 'orthogonality center'.

MPS psi; //psi is an object of type MPS
//... prepare psi, for example by initializing it to a particular state
psi.position(3); //shift the gauge position to site 3

Next, contract the gate tensor G with the MPS tensors for sites 3 and 4

auto wf = psi(3)*psi(4);
wf *= G;
wf.noPrime();

Finally, use the singular value decomposition (SVD) to factorize the resulting tensor, multiplying the singular values into either U or V. Assign these two tensors back into the MPS to update it.

auto [U,S,V] = svd(wf,inds(psi(3)),{"Cutoff=",1E-8});
psi.set(3,U);
psi.set(4,S*V);

Note that in the SVD above, we set a truncation error cutoff to truncate the smallest singular values and control the size of the resulting MPS. Other cutoff values can be used, depending on the desired accuracy, as well as limits on the maximum bond dimension ("MaxDim").

Complete code example:

//assume we have an MPS 'psi' and a gate 'G'
//G is an ITensor with indices s3, s4, s3', s4'
//where s3 and s4 are the third and fourth physical
//indices of the MPS

psi.position(3);

auto wf = psi(3)*psi(4);
wf *= G;
wf.noPrime();

auto [U,S,V] = svd(wf,inds(psi(3)),{"Cutoff=",1E-8});
psi.set(3,U);
psi.set(4,S*V);

Back to Formulas
Back to Main


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# Internal Rate of Return (IRR)

## What Is Internal Rate of Return (IRR)?

The internal rate of return (IRR) is a metric used in financial analysis to estimate the profitability of potential investments. IRR is a discount rate that makes the net present value (NPV) of all cash flows equal to zero in a discounted cash flow analysis.

IRR calculations rely on the same formula as NPV does. Keep in mind that IRR is not the actual dollar value of the project. It is the annual return that makes the NPV equal to zero.

Generally speaking, the higher an internal rate of return, the more desirable an investment is to undertake. IRR is uniform for investments of varying types and, as such, can be used to rank multiple prospective investments or projects on a relatively even basis. In general, when comparing investment options with other similar characteristics, the investment with the highest IRR probably would be considered the best.

### Key Takeaways

• The internal rate of return (IRR) is the annual rate of growth that an investment is expected to generate.
• IRR is calculated using the same concept as net present value (NPV), except it sets the NPV equal to zero.
• IRR is ideal for analyzing capital budgeting projects to understand and compare potential rates of annual return over time.
1:30

## Formula and Calculation for IRR

The formula and calculation used to determine this figure are as follows:

\begin{aligned} &\text{0}=\text{NPV}=\sum_{t=1}^{T}\frac{C_t}{\left(1+IRR\right)^t}-C_0\\ &\textbf{where:}\\ &C_t=\text{Net cash inflow during the period t}\\ &C_0=\text{Total initial investment costs}\\ &IRR=\text{The internal rate of return}\\ &t=\text{The number of time periods}\\ \end{aligned}

### How to Calculate IRR

1. Using the formula, one would set NPV equal to zero and solve for the discount rate, which is the IRR.
2. The initial investment is always negative because it represents an outflow.
3. Each subsequent cash flow could be positive or negative, depending on the estimates of what the project delivers or requires as a capital injection in the future.
4. However, because of the nature of the formula, IRR cannot be easily calculated analytically and instead must be calculated iteratively through trial and error or by using software programmed to calculate IRR (e.g., using Excel).

### How to Calculate IRR in Excel

Using the IRR function in Excel makes calculating the IRR easy. Excel does all the necessary work for you, arriving at the discount rate you are seeking to find. All you need to do is combine your cash flows, including the initial outlay as well as subsequent inflows, with the IRR function. The IRR function can be found by clicking on the Formulas Insert (fx) icon.

Here is a simple example of an IRR analysis with cash flows that are known and annually periodic (one year apart). Assume a company is assessing the profitability of Project X. Project X requires $250,000 in funding and is expected to generate$100,000 in after-tax cash flows the first year and grow by $50,000 for each of the next four years. In this case, the IRR is 56.72%, which is quite high. Excel also offers two other functions that can be used in IRR calculations: the XIRR, and the MIRR. XIRR is used when the cash flow model does not exactly have annual periodic cash flows. The MIRR is a rate-of-return measure that includes the integration of cost of capital and the risk-free rate. 4:20 #### How to Calculate IRR in Excel ## Understanding IRR The ultimate goal of IRR is to identify the rate of discount, which makes the present value of the sum of annual nominal cash inflows equal to the initial net cash outlay for the investment. Several methods can be used when seeking to identify an expected return, but IRR is often ideal for analyzing the potential return of a new project that a company is considering undertaking. Think of IRR as the rate of growth that an investment is expected to generate annually. Thus, it can be most similar to a compound annual growth rate (CAGR). In reality, an investment will usually not have the same rate of return each year. Usually, the actual rate of return that a given investment ends up generating will differ from its estimated IRR. ## What Is IRR Used For? In capital planning, one popular scenario for IRR is comparing the profitability of establishing new operations with that of expanding existing operations. For example, an energy company may use IRR in deciding whether to open a new power plant or to renovate and expand an existing power plant. While both projects could add value to the company, it is likely that one will be the more logical decision as prescribed by IRR. Note that because IRR does not account for changing discount rates, it’s often not adequate for longer-term projects with discount rates that are expected to vary. IRR is also useful for corporations in evaluating stock buyback programs. Clearly, if a company allocates substantial funding to repurchasing its shares, then the analysis must show that the company’s own stock is a better investment—that is, has a higher IRR—than any other use of the funds, such as creating new outlets or acquiring other companies. Individuals can also use IRR when making financial decisions—for instance, when evaluating different insurance policies using their premiums and death benefits. The general consensus is that policies that have the same premiums and a high IRR are much more desirable. Note that life insurance has a very high IRR in the early years of policy—often more than 1,000%. It then decreases over time. This IRR is very high during the early days of the policy because if you made only one monthly premium payment and then suddenly died, your beneficiaries would still get a lump sum benefit. Another common use of IRR is in analyzing investment returns. In most cases, the advertised return will assume that any interest payments or cash dividends are reinvested back into the investment. What if you don’t want to reinvest dividends but need them as income when paid? And if dividends are not assumed to be reinvested, are they paid out, or are they left in cash? What is the assumed return on the cash? IRR and other assumptions are particularly important on instruments like annuities, where the cash flows can become complex. Finally, IRR is a calculation used for an investment’s money-weighted rate of return (MWRR). The MWRR helps determine the rate of return needed to start with the initial investment amount factoring in all of the changes to cash flows during the investment period, including sales proceeds. ## Using IRR with WACC Most IRR analyses will be done in conjunction with a view of a company’s weighted average cost of capital (WACC) and NPV calculations. IRR is typically a relatively high value, which allows it to arrive at an NPV of zero. Most companies will require an IRR calculation to be above the WACC. WACC is a measure of a firm’s cost of capital in which each category of capital is proportionately weighted. All sources of capital, including common stock, preferred stock, bonds, and any other long-term debt, are included in a WACC calculation. In theory, any project with an IRR greater than its cost of capital should be profitable. In planning investment projects, firms will often establish a required rate of return (RRR) to determine the minimum acceptable return percentage that the investment in question must earn to be worthwhile. The RRR will be higher than the WACC. Any project with an IRR that exceeds the RRR will likely be deemed profitable, although companies will not necessarily pursue a project on this basis alone. Rather, they will likely pursue projects with the highest difference between IRR and RRR, as these likely will be the most profitable. IRR may also be compared against prevailing rates of return in the securities market. If a firm can’t find any projects with IRR greater than the returns that can be generated in the financial markets, then it may simply choose to invest money in the market. Market returns can also be a factor in setting an RRR. Analyses will also typically involve NPV calculations at different assumed discount rates. ## IRR vs. Compound Annual Growth Rate The CAGR measures the annual return on an investment over a period of time. The IRR is also an annual rate of return. However, the CAGR typically uses only a beginning and ending value to provide an estimated annual rate of return. IRR differs in that it involves multiple periodic cash flows—reflecting that cash inflows and outflows often constantly occur when it comes to investments. Another distinction is that CAGR is simple enough that it can be calculated easily. ## IRR vs. Return on Investment (ROI) Companies and analysts may also look at the return on investment (ROI) when making capital budgeting decisions. ROI tells an investor about the total growth, start to finish, of the investment. It is not an annual rate of return. IRR tells the investor what the annual growth rate is. The two numbers normally would be the same over the course of one year but won’t be the same for longer periods of time. ROI is the percentage increase or decrease of an investment from beginning to end. It is calculated by taking the difference between the current or expected future value and the original beginning value, divided by the original value and multiplied by 100. ROI figures can be calculated for nearly any activity into which an investment has been made and an outcome can be measured. However, ROI is not necessarily the most helpful for lengthy time frames. It also has limitations in capital budgeting, where the focus is often on periodic cash flows and returns. ## Limitations of the IRR IRR is generally most ideal for use in analyzing capital budgeting projects. It can be misconstrued or misinterpreted if used outside of appropriate scenarios. In the case of positive cash flows followed by negative ones and then by positive ones, the IRR may have multiple values. Moreover, if all cash flows have the same sign (i.e., the project never turns a profit), then no discount rate will produce a zero NPV. Within its realm of uses, IRR is a very popular metric for estimating a project’s annual return. However, it is not necessarily intended to be used alone. IRR is typically a relatively high value, which allows it to arrive at an NPV of zero. The IRR itself is only a single estimated figure that provides an annual return value based on estimates. Since estimates in IRR and NPV can differ drastically from actual results, most analysts will choose to combine IRR analysis with scenario analysis. Scenarios can show different possible NPVs based on varying assumptions. As mentioned, most companies do not rely on IRR and NPV analyses alone. These calculations are usually also studied in conjunction with a company’s WACC and an RRR, which provides for further consideration. Companies usually compare IRR analysis to other tradeoffs. If another project has a similar IRR with less up-front capital or simpler extraneous considerations, then a simpler investment may be chosen despite IRRs. In some cases, issues can also arise when using IRR to compare projects of different lengths. For example, a project of short duration may have a high IRR, making it appear to be an excellent investment. Conversely, a longer project may have a low IRR, earning returns slowly and steadily. The ROI metric can provide some more clarity in these cases, although some managers may not want to wait out the longer time frame. ## Investing Based on IRR The internal rate of return rule is a guideline for evaluating whether to proceed with a project or investment. The IRR rule states that if the IRR on a project or investment is greater than the minimum RRR—typically the cost of capital, then the project or investment can be pursued. Conversely, if the IRR on a project or investment is lower than the cost of capital, then the best course of action may be to reject it. Overall, while there are some limitations to IRR, it is an industry-standard for analyzing capital budgeting projects. ## IRR Example Assume a company is reviewing two projects. Management must decide whether to move forward with one, both, or neither. Its cost of capital is 10%. The cash flow patterns for each are as follows: Project A • Initial Outlay =$5,000
• Year one = $1,700 • Year two =$1,900
• Year three = $1,600 • Year four =$1,500
• Year five = $700 Project B • Initial Outlay =$2,000
• Year one = $400 • Year two =$700
• Year three = $500 • Year four =$400
• Year five = $300 The company must calculate the IRR for each project. Initial outlay (period = 0) will be negative. Solving for IRR is an iterative process using the following equation:$0 = Σ CFt ÷ (1 + IRR)t

where:

• CF = net cash flow
• IRR = internal rate of return
• t = period (from 0 to last period)

-or-

$0 = (initial outlay * −1) + CF1 ÷ (1 + IRR)1 + CF2 ÷ (1 + IRR)2 + ... + CFX ÷ (1 + IRR)X Using the above examples, the company can calculate IRR for each project as: ### IRR Project A:$0 = (−$5,000) +$1,700 ÷ (1 + IRR)1 + $1,900 ÷ (1 + IRR)2 +$1,600 ÷ (1 + IRR)3 + $1,500 ÷ (1 + IRR)4 +$700 ÷ (1 + IRR)5

IRR Project A = 16.61 %

### IRR Project B:

$0 = (−$2,000) + $400 ÷ (1 + IRR)1 +$700 ÷ (1 + IRR)2 + $500 ÷ (1 + IRR)3 +$400 ÷ (1 + IRR)4 + \$300 ÷ (1 + IRR)5

IRR Project B = 5.23 %

Given that the company’s cost of capital is 10%, management should proceed with Project A and reject Project B.

## What does internal rate of return mean?

The internal rate of return (IRR) is a financial metric used to assess the attractiveness of a particular investment opportunity. When you calculate the IRR for an investment, you are effectively estimating the rate of return of that investment after accounting for all of its projected cash flows together with the time value of money. When selecting among several alternative investments, the investor would then select the investment with the highest IRR, provided it is above the investor’s minimum threshold. The main drawback of IRR is that it is heavily reliant on projections of future cash flows, which are notoriously difficult to predict.

## Is IRR the same as ROI?

Although IRR is sometimes referred to informally as a project’s “return on investment,” it is different from the way most people use that phrase. Often, when people refer to ROI, they are simply referring to the percentage return generated from an investment in a given year or across a stretch of time. But that type of ROI does not capture the same nuances as IRR, and for that reason, IRR is generally preferred by investment professionals.

Another advantage of IRR is that its definition is mathematically precise, whereas the term ROI can mean different things depending on the context or the speaker.

## What is a good internal rate of return?

Whether an IRR is good or bad will depend on the cost of capital and the opportunity cost of the investor. For instance, a real estate investor might pursue a project with a 25% IRR if comparable alternative real estate investments offer a return of, say, 20% or lower. However, this comparison assumes that the riskiness and effort involved in making these difficult investments are roughly the same. If the investor can obtain a slightly lower IRR from a project that is considerably less risky or time-consuming, then they might happily accept that lower-IRR project. In general, though, a higher IRR is better than a lower one, all else being equal.

### Article Sources

Investopedia requires writers to use primary sources to support their work. These include white papers, government data, original reporting, and interviews with industry experts. We also reference original research from other reputable publishers where appropriate. You can learn more about the standards we follow in producing accurate, unbiased content in our editorial policy.
1. Microsoft. “IRR function.” Accessed Sept. 15, 2020.

2. Microsoft. “Insert Function.” Accessed Sept. 15, 2020.

3. Microsoft. “XIRR function.” Accessed Sept. 15, 2020.

4. Microsoft. “MIRR function.” Accessed Sept. 15, 2020.


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# Multiple Polinomial Regression

1. Jun 17, 2011

### acmilanhn

Hi friends,
I need to get the surface equation like function f(x,y) that passes through 66 points.
I attached the file xls where is the points, and a pdf file where is the graphics of the points.

Can somebody help say me that I have to do for solve it?

#### Attached Files:

File size:
172.2 KB
Views:
73
• ###### Puntos x, y, z.xls
File size:
23 KB
Views:
72
2. Jun 17, 2011

### TheoMcCloskey

acmilanhn,

Does the approximation need to truely go through each of the data points (that is, approximate exactly at the data points) or is a more general approximation acceptable (one where the predicted values approximates the data at the data points)?

The former would generally involve an interpolation approximation while the latter could be treated more simply by, say, Least Squares.

3. Jun 17, 2011

### acmilanhn

I think my friend that using least squares can work, with r2=0.999999. But the same time, I need to graph the surface therefore I have to get the equation of the surface...

4. Jun 18, 2011

### TheoMcCloskey

You can perform a multilinear regression (least squares) using a design matrix consisting of two dimensional monomials.

For example, a third degree (bi-cubic) approximation can be generated if you consider as basis functions the following:

$$[1,x,x^2,x^3, y,yx,yx^2, y^2, y^2x, y^3]$$

Construct a design matrix with columns computed with your data using this basis. Then regress. You'll obtain the various coefficents for each term. Examine the degree of fit.

Your prediction formula would then be:

$$\hat{z}(x,y) = g_0(x) + y \, \left(g_1(x) + y \, \left(g_2(x) + g_3(x) \, y \right) \right)$$
with
$$g_0(x)= c_0+x \, (c_1 + x \, (c_2 + c_3 \, x ))$$
$$g_1(x)= c_4+x \, (c_5 + c_6 \, x)$$
$$g_2(x)= c_7+c_8 \, x$$
$$g_3(x)= c_9$$

A higher degree of fit requires a higher order polynomial. But this can have disadvantages as the number of terms can easily grow large. Also, you have to be concerned about the number of data and order of fit as well as the possibilities of numerical ill-condition.

However, this approach is not difficult. In fact, it can easily be done in Excel. For example, see the enclosed Excel file where I took your file and performed this analysis.

#### Attached Files:

• ###### Puntos x, y, z (TMcC).xls
File size:
87 KB
Views:
71
5. Jun 18, 2011

### acmilanhn

excelent my friend.... how to use Excel for solve it?

6. Jun 19, 2011

### TheoMcCloskey

acmilanhn,

There are a couple ways to perform this. You can use the Regression tool from the Analysis Toolpak (which is what I did) or you could just as easily use the LINEST Excel (array) function. The Regression tool will give you a static solution while the LINEST function will give you a dynamic solution (one that changes when inputs change).

Either one will do the job. Look them up in Excel help.


text:
## Calculus 10th Edition

a: $dy/dt=-5/8$ b: $dx/dt=3/2$
$xy=4$: given $x(dy/dt)+y(dx/dt)$: chain rule $dy/dt=(-y/x)dx/dt$ $dx/dt=(-x/y)dy/dt$ a: $x=8$, $y=1/2$, $dx/dt=10$ $dy/dt = -((1/2)/8)\times(10)=-5/8$ b: $x=1$, $y-4$, $dy/dt=-6$ $dx/dt= -((1/4)\times(-6)=3/2$


text:
# If inertial mass were equal to for example half of gravitational mass why would things not still fall at same rate?

I have read that it is the equality of inertial and gravitational mass that causes all objects to fall at the same rate irrespective of size. But if this were not true, what effect would we see? Would inertial mass being half g-mass mean an object twice as massive would accelerate twice as fast? If so, why?

EDIT: Here is where it is asserted that the equality is responsible for all objects falling at same rate:

Why do two bodies of different masses fall at the same rate (in the absence of air resistance)?

• There is always a constant of proportionality, like G in gravity. So, if you introduce another constant like "2" it won't make a difference. It would just imply that the force equations have a different constant. – Peter R Aug 8 '16 at 19:12
• So the equivalence of g-mass and e-mass is not responsible for all objects falling at same rate? – Jeff Aug 8 '16 at 19:13
• Imagine a baloon. The difference would be that it would work even in vacuum. – peterh Aug 8 '16 at 19:13
• What do you mean by 'equality of inertial and gravitational mass'? – obliv Aug 8 '16 at 19:14
• Indeed, when physicists refer to gravitational mass and inertial mass being the same, they mean there is a constant of proportionality relating them. For them to be not equal, it would mean more massive objects have less inertia, or have a non-linear relationship between gravity and inertia – Jim Aug 8 '16 at 19:14

Say you have Earth's mass as $M$ and your object's mass and distance from Earth is $m$ and $r$

Using Newtonian gravity we have the Force experiences by the object to be $F_g = \frac{GMm}{r^2}$ where $G$ is a constant.

Now the force experienced by earth due to the object's mass is $F_{g2} = \frac{GMm}{r^2}$

Acceleration of both objects, $F_1 = ma_1 \to \frac{GMm}{r^2} = ma_1$ so $a_1 = \frac{GM}{r^2}$

$F_2 = \frac{GMm}{r^2} = Ma_2 \to a_2 = \frac{Gm}{r^2}$

The acceleration that the earth experiences is negligible in comparison. The acceleration the object experiences is only due to Earth's mass. (Note that for considerably large objects the acceleration due to the object's mass would make it seem like the object is accelerating quicker towards Earth than other objects because Earth is moving towards the object fairly quickly too)

If inertial mass $F = ma$ was different from the gravitational mass you would just have a different acceleration for each object BUT every object would still be accelerated at the same magnitude.

ex: $m_g = \frac{m_i}{2}$ then $F_g = \frac{GMm_g}{r^2} = m_ia$ so that $a = \frac{GM}{2r^2}$ and you'd have halved acceleration.

• Hey @Obliv; saw your profile brought to my notice by SirC. I know lately, quite has changed in the chat; obe has left; 0celot is banned and there is not the same warm atmosphere used to be and lately, I'm too busy to attend the chat- if this is the reason behind your latest profile update, then I would be really sorry; we cheered your presence during the days we chatted; some great users like G.Bergaron do come in the chat along with old guys like ACM, JR. Anyway, Bernardo, obe, I, SirC are available in Skype; do contact us and contact me here when you require. All the best to your future. – user36790 Dec 13 '16 at 13:52

Assume that Force by each if $$W = \frac{GM_gm_g}{R^2}$$

Where $$m_g$$ and $$M_g$$ are the gravitational masses of both body under test and earth respectively Suppose that $$m_i$$ is the body inertial mass Assuming $$m_i=km_g$$

Where k is not 1 but a proportionality constant

Newton 2nd law:

$$W = \frac{GM_gm_g}{R^2}=k m_g a$$

Where $$a$$ is the acceleration

==>

$$a=\frac{G M_g m_g}{k R^2}$$ ==> independent of the both values of $$m_i$$, and $$m_g$$, but different than the case when:

k=1

The acceleration is different in its absolute value but independent of the inertial/gravitational mass of the falling body

• Please use MathJax for writing equations. – exp ikx May 30 at 6:57


text:
Home

 Solve any problem (step by step solutions) Input table (Matrix, Statistics)
Mode :
SolutionHelp
Solution will be displayed step by step (In 2 parts)
Solution
Problem: Backward, 2x^3-4x+1, [2,4], x=2.1, N=4 [ Calculator, Method and examples ]

Solution:
Your problem -> Backward, 2x^3-4x+1, [2,4], x=2.1, N=4

Equation is f(x) = 2x^3-4x+1.

h = (b-a)/N

h = (4 - 2) / 4 = 0.5

The value of table for X and Y

 X Y 2 2.5 3 3.5 4 9 22.25 43 72.75 113

Newton's backward differentiation method to find solution
Newton's backward differentiation table is as follows.
 X Y(X) gradY grad^2Y grad^3Y grad^4Y 2 9 13.25 2.5 22.25 7.5 20.75 1.5 3 43 9 0 29.75 1.5 3.5 72.75 10.5 40.25 4 113

The value of x at you want to find the f(x) : x = 2.1

Solution provided by AtoZmath.com
Any wrong solution, solution improvement, feedback then Submit Here
Want to know about AtoZmath.com and me


text:
# Volatility models using Rugarch

I have estimated sGARCH, EGARCH and TGARCH, which some for particular models are significant. For others, the alpha remain insignificant using various innovations such as the skewed variants of the normal or student $t$ distributions.

I am tempted to rely on the models that give significant estimates and then ignore the ones that do not. For instance, if the EGARCH-std is not significant in terms of alpha and the EGARCH-sstd is significant, I prefer to focus on the later for further analysis.

Please let me know if I am on the right path.

-
What are you trying to do? If you're doing an empirical study and you decide to go with the one model that generates statistical significance then this sounds like data snooping bias. – Jase Dec 29 '12 at 8:26

The answer depends on what you will use the models for. For example, if you care about prediction, you should use different metrics (model fit, prediction accuracy on hold-out samples, etc) to determine what works best.

-


text:
## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$B$
The coin always comes up with tails, thus the probability of any event containing heads will be $0$ and the probability of an event containing only tails will be $1$. Thus the answer is B.


text:
### Enforcing Semantic Integrity on Untrusted Clients in Networked Virtual Environments

Somesh Jha, Stefan Katzenbeisser, Christian Schallhart, Helmut Veith, and Stephen Chenney

##### Abstract

During the last years, large-scale simulations of realistic physical environments which support the interaction of multiple participants over the Internet have become increasingly available and economically viable, most notably in the computer gaming industry. Such systems, commonly called networked virtual environments (NVEs), are usually based on a client-server architecture where for performance reasons and bandwidth restrictions, the simulation is partially delegated to the clients. This inevitable architectural choice renders the simulation vulnerable to attacks against the semantic integrity of the simulation: malicious clients may attempt to compromise the physical and logical rules governing the simulation, or to alter the causality of events a posteriori. In this paper, we initiate the systematic study of semantic integrity in NVEs from a security point of view. We argue that naive policies to enforce semantic integrity involve intolerable network load, and are therefore not practically feasible. We present a new provably secure semantic integrity protocol based on cryptographic primitives which enables the server system to audit the local computations of the clients on demand. Our approach facilitates low network and CPU load, incurs reasonable engineering overhead, and maximally decouples the auditing process from the soft real time constraints of the simulation.

Available format(s)
Category
Applications
Publication info
Published elsewhere. An extended abstract of this paper appears at the IEEE Symposium on Security and Privacy, 2007
Keywords
protocolsaudit trails
Contact author(s)
skatzenbeisser @ acm org
History
Short URL
https://ia.cr/2007/056

CC BY

BibTeX

@misc{cryptoeprint:2007/056,
author = {Somesh Jha and Stefan Katzenbeisser and Christian Schallhart and Helmut Veith and Stephen Chenney},
title = {Enforcing Semantic Integrity on Untrusted Clients in Networked Virtual Environments},
howpublished = {Cryptology ePrint Archive, Paper 2007/056},
year = {2007},
note = {\url{https://eprint.iacr.org/2007/056}},
url = {https://eprint.iacr.org/2007/056}
}

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