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# Accelerated Gossip via Stochastic Heavy Ball Method

In this paper we show how the stochastic heavy ball method (SHB) -- a popular method for solving stochastic convex and non-convex optimization problems --operates as a randomized gossip algorithm. In particular, we focus on two special cases of SHB: the Randomized Kaczmarz method with momentum and its block variant. Building upon a recent framework for the design and analysis of randomized gossip algorithms, [Loizou Richtarik, 2016] we interpret the distributed nature of the proposed methods. We present novel protocols for solving the average consensus problem where in each step all nodes of the network update their values but only a subset of them exchange their private values. Numerical experiments on popular wireless sensor networks showing the benefits of our protocols are also presented.

## Authors

• 16 publications
• 94 publications
• ### Provably Accelerated Randomized Gossip Algorithms

In this work we present novel provably accelerated gossip algorithms for...
10/31/2018 ∙ by Nicolas Loizou, et al. ∙ 0

• ### Randomized Progressive Hedging methods for Multi-stage Stochastic Programming

Progressive Hedging is a popular decomposition algorithm for solving mul...
09/25/2020 ∙ by Gilles Bareilles, et al. ∙ 0

• ### Revisiting Randomized Gossip Algorithms: General Framework, Convergence Rates and Novel Block and Accelerated Protocols

In this work we present a new framework for the analysis and design of r...
05/20/2019 ∙ by Nicolas Loizou, et al. ∙ 0

• ### Quickly Finding a Benign Region via Heavy Ball Momentum in Non-Convex Optimization

The Heavy Ball Method, proposed by Polyak over five decades ago, is a fi...
10/04/2020 ∙ by Jun-Kun Wang, et al. ∙ 0

• ### A Privacy Preserving Randomized Gossip Algorithm via Controlled Noise Insertion

In this work we present a randomized gossip algorithm for solving the av...
01/27/2019 ∙ by Filip Hanzely, et al. ∙ 0

• ### Momentum Accelerated Multigrid Methods

In this paper, we propose two momentum accelerated MG cycles. The main i...
06/30/2020 ∙ by Chunyan Niu, et al. ∙ 0

##### This week in AI

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## I Introduction

Average consensus is a fundamental problem in distributed computing and multi-agent systems. It comes up in many real world applications such as coordination of autonomous agents, estimation, rumour spreading in social networks, PageRank and distributed data fusion on ad-hoc networks and decentralized optimization. Due to its great importance there is much classical

[35, 7] and recent [38, 37, 4] work on the design of efficient algorithms/protocols for solving it.

One of the most attractive classes of protocols for solving the average consensus are gossip algorithms. The development and design of gossip algorithms was studied extensively in the last decade. The seminal 2006 paper of Boyd et al. [4] on randomized gossip algorithms motivated a fury of subsequent research and now gossip algorithms appear in many applications, including distributed data fusion in sensor networks [38], load balancing [6] and clock synchronization [11]. For a survey of selected relevant work prior to 2010, we refer the reader to the work of Dimakis et al. [8]. For more recent results on randomized gossip algorithms we suggest [40, 17, 28, 20, 24, 1]. See also [9, 2, 29, 14].

The main goal in the design of gossip protocols is for the computation and communication to be done as quickly and efficiently as possible. In this work, our focus is precisely this. We design randomized gossip protocols which converge to consensus fast.

### I-a The average consensus problem

In the average consensus (AC) problem we are given an undirected connected network with node set and edges . Each node “knows” a private value . The goal of AC is for every node to compute the average of these private values, , in a distributed fashion. That is, the exchange of information can only occur between connected nodes (neighbors).

### I-B Main Contributions

We present a new class of randomized gossip protocols where in each iteration all nodes of the network update their values but only a subset of them exchange their private information. Our protocols are based on recently proposed ideas for the acceleration of randomized Kaczmarz methods for solving consistent linear systems [22]

where the addition of a momentum term was shown to provide practical speedups over the vanilla Kaczmarz methods. Further, we explain the connection between gossip algorithms for solving the average consensus problem, Kaczmarz-type methods for solving consistent linear systems, and stochastic gradient descent and stochastic heavy ball methods for solving stochastic optimization problems. We show that essentially all these algorithms behave as gossip algorithms. Finally, we explain in detail the gossip nature of two recently proposed fast Kacmzarz-type methods: the randomized Kacmzarz with momentum (mRK), and its block variant, the randomized block Kaczmarz with momentum (mRBK). We present a detailed comparison of our proposed gossip protocols with existing popular randomized gossip protocols and through numerical experiments we show the benefits of our methods.

### I-C Structure of the paper

This work is organized as follows. Section II introduces the important technical preliminaries and the necessary background for understanding of our methods. A new connection between gossip algorithms, Kaczmarz methods for solving linear systems and stochastic gradient descent (SGD) for solving stochastic optimization problems is also described. In Section III the two new accelerated gossip protocols are presented. Details of their behaviour and performance are also explained. Numerical evaluation of the new gossip protocols is presented in Section IV. Finally, concluding remarks are given in Section V.

### I-D Notation

The following notational conventions are used in this paper. We write . Boldface upper-case letters denote matrices;

is the identity matrix. By

we denote the solution set of the linear system , where and . Throughout the paper, is the projection of onto (that is, is the solution of the best approximation problem; see equation (5)). An explicit formula for the projection of onto set is given by

 ΠL(x):=argminx′∈L∥x′−x∥=x−A⊤(AA⊤)†(Ax−b).

A matrix that often appears in our update rules is

 H:=S(S⊤AA⊤S)†S⊤, (1)

where

is a random matrix drawn in each step of the proposed methods from a given distribution

, and denotes the Moore-Penrose pseudoinverse. Note that is a random symmetric positive semi-definite matrix.

In the convergence analysis we use

to indicate the smallest nonzero eigenvalue, and

for the largest eigenvalue of matrix , where the expectation is taken over . Finally,

represents the vector with the private values of the

nodes of the network at the iteration while with we denote the value of node at the iteration.

## Ii Background-Technical Preliminaries

Our work is closely related to two recent papers. In [20], a new perspective on randomized gossip algorithms is presented. In particular, a new approach for the design and analysis of randomized gossip algorithms is proposed and it was shown how the Randomized Kaczmarz and Randomized Block Kaczmarz, popular methods for solving linear systems, work as gossip algorithms when applied to a special system encoding the underlying network. In [22], several classes of stochastic optimization algorithms enriched with heavy ball momentum were analyzed. Among the methods studied are: stochastic gradient descent, stochastic Newton, stochastic proximal point and stochastic dual subspace ascent.

In the rest of this section we present the main results of the above papers, highlighting several connections. These results will be later used for the development of the new randomized gossip protocols.

### Ii-a Kaczmarz Methods and Gossip Algorithms

Kaczmarz-type methods are very popular for solving linear systems with many equations. The (deterministic) Kaczmarz method for solving consistent linear systems was originally introduced by Kaczmarz in 1937 [15]. Despite the fact that a large volume of papers was written on the topic, the first provably linearly convergent variant of the Kaczmarz method—the randomized Kaczmarz Method (RK)—was developed more than 70 years later, by Strohmer and Vershynin [32]. This result sparked renewed interest in design of randomized methods for solving linear systems [25, 26, 10, 23, 39, 27, 31, 18]. More recently, Gower and Richtárik [12] provide a unified analysis for several randomized iterative methods for solving linear systems using a sketch-and-project framework. We adopt this framework in this paper.

In particular, the sketch-and-project algorithm [12] for solving the consistent linear system has the form

 xk+1 = xk−A⊤Sk(S⊤kAA⊤Sk)†S⊤k(Axk−b) (2) xk−A⊤Hk(Axk−b),

where in each iteration matrix is sampled afresh from an arbitrary distribution . In [12] it was shown that many popular algorithms for solving linear systems, including RK method and randomized coordinate descent method can be cast as special cases of the above update by choosing111In order to recover a randomized coordinate descent method, one also needs to perform projections with respect to a more general Euclidean norm. However, for simplicity, in this work we only consider the standard Euclidean norm. an appropriate distribution . The special cases that we are interested in are the randomized Kaczmarz (RK) and its block variant, the randomized block Kaczmarz (RBK).

Let be the unit coordinate vector in and let be column submatrix of the identity matrix with columns indexed by . Then RK and RBK methods can be obtained as special cases of the update rule (2) as follows:

• RK: Let , where

is chosen in each iteration independently, with probability

. In this setup the update rule (2) simplifies to

 xk+1=xk−Ai:xk−bi∥Ai:∥22A⊤i:. (3)
• RBK: Let , where is chosen in each iteration independently, with probability . In this setup the update rule (2) simplifies to

 xk+1=xk−A⊤C:(AC:A⊤C:)†(AC:xk−bC). (4)

In this paper we are interested in two particular extension of the above methods: the randomized Kaczmarz method with momentum (mRK) and its block variant, the randomized block Kaczmarz with momentum (mRBK), both proposed and analyzed in [22]. Before we describe these two algorithms, let us summarize the main connections between the Kaczmarz methods for solving linear systems and gossip algorithms, as presented in [20].

In [13, 30, 22], it was shown that even in the case of consistent linear systems with multiple solutions, Kaczmarz-type methods converge linearly to one particular solution: the projection of the initial iterate onto the solution set of the linear system. This naturally leads to the formulation of the best approximation problem:

 minx=(x1,…,xn)∈Rn12∥x−x0∥2subject toAx=b. (5)

Above, and is the standard Euclidean norm. By we denote the solution of (5).

In [20] it was shown how RK and RBK work as gossip algorithms when applied to a special linear system encoding the underlying network.

###### Definition II.1 ([20])

A linear system is called “average consensus (AC) system” when is equivalent to saying that for all .

Note that many linear systems satisfy the above definition. For example, we can choose and to be the incidence matrix of . In this case, the row of the system corresponding to edge directly encodes the constraint . A different choice is to pick and , where is the Laplacian of . Note that depending on what AC system is used, RK and RBK have different interpretations as gossip protocols.

From now on we work with the AC system described in the first example. Since , the general sketch-and-project update rule (2) simplifies to:

 xk+1=[I−A⊤HkA]xk. (6)

The convergence performance of RK and RBK for solving the best approximation problem (and as a result the average consensus problem) is described by the following theorem.

###### Theorem II.2 ([12, 13])

Let be the iterates produced by (2). Then where is the solution of (5), , and denotes the minimum nonzero eigenvalue of .

In [20], the behavior of both RK and RBK as gossip algorithms was described, and a comparison with the convergence results of existing randomized gossip protocols was made. In particular, it was shown that the most basic randomized gossip algorithm [4] (“randomly pick an edge and then replace the values stored at vertices and by their average”) is an instance of RK applied to the linear system , where the is the incidence matrix of . RBK can also be interpreted as a gossip algorithm:

###### Theorem II.3 ([20], RBK as a Gossip Algorithm)

Each iteration of RBK for solving works as follows: 1) Select a random set of edges , 2) Form subgraph of from the selected edges, 3) For each connected component of , replace node values with their average.

### Ii-B The Heavy Ball momentum

A detailed study of several (equivalent) stochastic reformulations of consistent linear systems was developed in [30]. This new viewpoint facilitated the development and analysis of relaxed variants (with relaxation parameter ) of the sketch-and-project update (2). In particular, one of the reformulations is the stochastic optimization problem

 minx∈Rnf(x):=ES∼D[fS(x)],where (7)
 fS(x):=12∥Ax−b∥2H=12(Ax−b)⊤H(Ax−b), (8)

and is the random symmetric positive semi-definite matrix defined in (1).

Under certain (weak) condition on , the set of minimizers of is identical to the set of the solutions of the linear system. In [30], problem (7) was solved via Stochastic Gradient Descent (SGD):

 xk+1=xk−ω∇fSk(xk), (9)

and a linear rate of convergence was proved despite the fact that is not necessarily strongly convex and that a fixed stepsize is used. Observe that the gradient of the stochastic function (8) is given by

 ∇fSk(x)A⊤Hk(Ax−b). (10)

and as a result, it is easy to see that for , the SGD update (9) reduces to the sketch-and-project update (2).

The recent works [21, 22] analyze momentum variants of SGD, with the goal to accelerate the convergence of the method for solving problem (7

). SGD with momentum—also known as the stochastic heavy ball method (SHB)—is a well known algorithm in the optimization literature for solving stochastic optimization problems, and it is extremely popular in areas such as deep learning

[33, 34, 16, 36]. However, even though SHB is used extensively in practice, its theoretical convergence behavior is not well understood. To the best of our knowledge, [21, 22] are the first that prove linear convergence of SHB in any setting.

The update rule of SHB for solving problem (7) is formally presented in the following algorithm:

Using the expression for the stochastic gradient (10), the update rule of SHB can be written more explicitly:

 xk+1=xk−ωA⊤Hk(Axk−b)+β(xk−xk−1). (11)

Using the same choice of distribution as in equation (3) and (4), we now obtain momentum variants of RK and RBK:

• RK with momentum (mRK):

 xk+1=xk−ωAi:xk−bi∥Ai:∥22A⊤i:+β(xk−xk−1)
• RBK with momentum (mRBK):

 xk+1=xk−ωA⊤C:(AC:A⊤C:)†(AC:xk−bC)+β(xk−xk−1)

In [22], two main theoretical results describing the behavior of SHB (and as a result also the special cases mRK and mRBK) were presented:

###### Theorem II.4 (Theorem 1, [22])

Choose . Let be the sequence of random iterates produced by SHB. Let (resp, ) be the smallest nonzero (resp. largest) eigenvalue of . Assume and and that the expressions and satisfy . Then

 E[∥xk−x∗∥2]≤qk(1+δ)∥x0−x∗∥2, (12)

and where and . Moreover, .

###### Theorem II.5 (Theorem 4, [22])

Let be the sequence of random iterates produced by SHB, started with , with relaxation parameter (stepsize) and momentum parameter . Let . Then there exists a constant such that for all we have

Using Theorem II.5 and by a proper combination of the stepsize and the momentum parameter , SHB enjoys an accelerated linear convergence rate in mean, [22].

###### Corollary 1

(i) If and , then the iteration complexity of SHB becomes: .
(ii) If and , then the iteration complexity of SHB becomes: .

## Iii Randomized Gossip protocols with momentum

Having presented SHB for solving the stochastic optimization problem (7) and describing its sketch-and-project nature (11), let us now describe its behavior as a randomized gossip protocol when applied to solving the AC system , where is the incidence matrix of the network.

Since , method (11) can be simplified to:

 (13)

In the rest of this section we focus on two special cases of (13): RK with momentum and RBK with momentum.

### Iii-a Randomized Kaczmarz Gossip with momentum

When RK is applied to solve an AC system , one recovers the famous pairwise gossip algorithm [4]. Algorithm 2 describes how the relaxed variant of randomized Kaczmarz with momentum behaves as a gossip algorithm. See also Figure (1) for a graphical illustration of the method.

###### Remark III.1

In the special case with (zero momentum) only the two nodes of edge update their values. In this case the two nodes do not update their values to their exact average but to a convex combination that depends on the stepsize . To obtain the pairwise gossip algorithm of [4], we should further choose .

Distributed Nature of the Algorithm: Here we highlight a few ways to implement mRK in a distributed fashion: Asynchronous pairwise broadcast gossip: In this protocol each node of the network

has a clock that ticks at the times of a rate 1 Poisson process. The inter-tick times are exponentially distributed, independent across nodes, and independent across time. This is equivalent to a global clock ticking at a rate

Poisson process which wakes up an edge of the network at random. In particular, in this implementation mRK works as follows: In the iteration (time slot) the clock of node ticks and node randomly contact one of its neighbors and simultaneously broadcast a signal to inform the nodes of the whole network that is updating (this signal does not contain any private information of node ). The two nodes share their information and update their private values following the update rule of Algorithm 2 while all the other nodes updating their values using their own information. In each iteration only one pair of nodes exchange their private values.

Synchronous pairwise gossip: In this protocol a single global clock is available to all nodes. The time is assumed to be slotted commonly across nodes and in each time slot only a pair of nodes of the network is randomly activated and exchange their information following the update rule of Algorithm 2. The remaining not activated nodes update their values using their own last two private values. Note that this implementation of mRK comes with the disadvantage that requires a central entity which choose the activate pair of nodes in each step.

Asynchronous pairwise gossip with common counter: The update rule of the nodes of the active pair in Algorithm 2 can be rewritten as follows:

 xk+1i=xki+β(xki−xk−1i)+ω2(xkj−xki)
 xk+1j=xkj+β(xkj−xk−1j)+ω2(xki−xkj)

In particular observe that in their update rule they have the expression which is precisely the update of all non activate nodes of the network. Thus if we assume that the nodes share a common counter that counts how many iterations take place and each node saves also the last iterate that it was activated then the algorithm can work in distributed fashion as follows:

Let us denote the number of total iterations (common counter) that becomes available to the activate nodes of each step as and let us define with the number of iterations between the current iterate and the last time that the node is picked (iteration ) then the update rule of the Algorithm 2 can be equivalently expressed as:

• Pick an edge at random following .

• The private values of the nodes are updated as follows:

 xk+1i=ik[xki+β(xki−xk−1i)]+ω2(xkj−xki)
 xk+1j=jk[xkj+β(xkj−xk−1j)]+ω2(xki−xkj)

; for any other node :

### Iii-B Connection with the accelerated gossip algorithm

In the randomized gossip literature there in one particular method closely related to our approach. It was first proposed in [5] and its analysis under strong conditions was presented in [17]. In this paper local memory is exploited by installing shift registers at each agent. In particular we are interested in the case of just two registers where the first stores the agent’s current value and the second the agent’s value before the latest update. The algorithm can be described as follows. Suppose that edge is chosen at time . Then,

• Node :

• Node :

• Any other node :

where . The method was analyzed in [17] under a strong assumption on the probabilities of choosing the pair of nodes that as the authors mentioned is unrealistic in practical scenarios, and for networks like the random geometric graphs. At this point we should highlight that the results presented in [22] hold for essentially any distribution and as a result such a problem cannot occur.

Note also that if we choose in the update rule of Algorithm 2, then our method is simplified to:

• Node :

• Node :

• Any other node :

In order to apply Theorem II.4, we need to assume that and which also means that . Thus for and momentum parameter it is easy to see that our approach is very similar to the shift-register algorithm. Both methods update the selected pair of nodes in the same way. However, in our case the other nodes of the network do not remain idle but instead also update their values using their own previous information.

Using the momentum matrix , the two algorithms above can be expressed as:

 xk+1=xk−ω2(xki−xkj)(ei−ej)+B(xk−xk−1). (14)

In particular, in our algorithm every element on the diagonal is equal to , while in [5] all values on the diagonal are zeros except for the two values .

###### Remark III.2

The shift register case and our algorithm can be seen as two limit cases of the update rule (14). In particular, the shift register method uses only two non-zero diagonal elements in , while our method has a full diagonal. We believe that further methods can be developed in the future by exploring the cases where more than two but not all elements of the diagonal matrix are non-zero. It might be possible to obtain better convergence if one carefully chooses these values based on the network topology. We leave this as an open problem for future research.

### Iii-C Randomized block Kaczmarz gossip with momentum

Recall that Theorem II.3 says how RBK (with no momentum and no relaxation) can be interpreted as a gossip algorithm. Now we use this result to explain how relaxed RBK with momentum works. Note that the update rule of RBK with momentum can be rewritten as follows:

 xk+1ω(I−A⊤HkA)xk+(1−ω)xk+β(xk−xk−1), (15)

where is the update rule of RBK (6).

Thus, in analogy to the simple RBK, in the step, a random set of edges is selected and connected components are formed as a result. This includes the connected components that belong to both sub-graph and also the singleton connected components (nodes outside the ). Let us define the set of the nodes that belong in the connected component at the step , such that and for any .

Using the update rule (15), Algorithm 3 shows how mRBK is updating the private values of the nodes of the network (see also Figure 2 for the graphical interpretation).

Note that in the update rule of mRBK the nodes that are not attached to a selected edge (do not belong in the sub-graph ) update their values via . By considering these nodes as singleton connected components their update rule is exactly the same with the nodes of sub-graph . This is easy to see as follows:

 xk+1ℓ = ω∑j∈Vkrxkj|Vkr|+(1−ω)xkℓ+β(xkℓ−xk−1ℓ) (17) = ωxkℓ+(1−ω)xkℓ+β(xkℓ−xk−1ℓ) = xkℓ+β(xkℓ−xk−1ℓ).
###### Remark III.3

In the special case that only one edge is selected in each iteration () the update rule of mRBK is simplified to the update rule of mRK. In this case the sub-graph is the pair of the two selected edges.

###### Remark III.4

In [20] it was shown that several existing gossip protocols for solving the average consensus problem are special cases of the simple RBK (Theorem II.3). For example two gossip algorithms that can be cast as special cases of the simple RBK are the path averaging proposed in [3] and the clique gossiping [19]. In path averaging, in each iteration a path of nodes is selected and its nodes update their values to their exact average (). In clique gossiping, the network is already divided into cliques and a through a random procedure a clique is activated and the nodes of it update their values to their exact average (). Since mRBK contains simple RBK as a special case for , we expect that these special protocols can also be accelerated with the addition of momentum parameter .

### Iii-D Mass preservation

One of the key properties of some of the most efficient randomized gossip algorithms is mass preservation. If a gossip algorithm has this property it means that the sum (and as a result the average) of the private values of the nodes remains fixed during the iterative procedure. That is, The original pairwise gossip algorithm proposed in [4] satisfied the mass preservation property, while exisiting accelerated gossip algorithms [5, 17] preserving a scaled sum.

In this section we show that the two proposed protocols presented above also have a mass preservation property. In particular, we prove mass preservation for the case of the block randomized gossip protocol (Algorithm 3) with momentum. This is sufficient since the Kaczmarz gossip with momentum (mRK) can be cast as special case.

###### Theorem III.1

Assume that . That is, the two registers of each node have the same initial value. Then for the Algorithms 2 and 3 we have for any and as a result, .

We prove the result for the more general Algorithm 3. Assume that in the step of the method connected components are formed. Let the set of the nodes of each connected component be so that and for any . Thus:

 ∑ni=1xk+1i=∑i∈Vk1xk+1i+⋯+∑i∈Vkqxk+1i (18)

Let us first focus, without loss of generality, on connected component and simplify the expression for the sum of its nodes: . By substituting this for all into the right hand side of (18) and from the fact that , we get Since , we have , and as a result for all .

## Iv Numerical Evaluation

We devote this section to experimentally evaluate the performance of the proposed gossip algorithms: mRK and mRBK. In particular we perform three experiments. In the first two we focus on the performance of the mRK, while in the last one on its block variant mRBK. In comparing the methods with their momentum variants we use the relative error measure where the starting vectors of values are taken to be always Gaussian vectors. For all of our experiments the horizontal axis represents the number of iterations. The networks used in the experiments are the cycle (ring graph), the 2-dimension grid and the randomized geometric graph (RGG) with radius . Code was written in Julia 0.6.3.

### Iv-a Impact of momentum parameter on mRK

Recall that in the simple pairwise gossip algorithm the two nodes that exchange information update their values to their exact average while all the other nodes remain idle. In our framework this method can be cast as special case of mRK when and . In this experiment we keep always the stepsize to be which means that the pair of the chosen nodes update their values to their exact average. We show that by choosing a suitable momentum parameter we can have faster convergence for all networks under study. See Figure 3 for more details.

### Iv-B Comparison with the Shift-Register

In this experiment we compare mRK with the shift register case when we choose the and in such a way in order to satisfy the connection establish in Section III-B. That is, we choose for any choice of . Observe that in all plots of Figure 4 our algorithm outperform the corresponding shift-register case.

### Iv-C Impact of momentum parameter on mRBK

In this experiment our goal is to show that the addition of momentum accelerates the RBK gossip algorithm proposed in [20]. Without loss of generality we choose the block size to be always equal to . That is the random matrix in the update rule of mRBK is always a column submatrix of the indetity matrix. Thus, in each iteration edges of the network are chosen to form the subgraph and the values of the nodes are updated according to Algorithm 3. Note that similar plots can be obtained for any choice of block size. We run all algorithms with fixed stepsize . It is obvious that by choosing a suitable momentum parameter we have faster convergence than when , for all networks under study. See Figure 5 for more details.

## V Conclusion and Future research

In this paper we present new accelerated randomized gossip algorithms using tools from numerical linear algebra and the area of randomized Kaczmarz methods for solving linear systems. In particular, using recently developed results on the stochastic reformulation of consistent linear systems we explain how stochastic heavy ball method for solving a specific quadratic stochastic optimization problem can be interpreted as gossip algorithm. To the best of our knowledge, it is the first time that such protocols are presented for average consensus problem. We believe that this work opens up many possible future venues for research. For example, using other Kaczmarz-type methods to solve particular linear systems we can obtain novel distributed protocols for average consensus. In addition, we believe that the gossip protocols presented in this work can be extended to the more general setting of distributed optimization where the goal is to minimize the average of convex functions in a distributed fashion.

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text:
# How is the normal ordering on the Natural Numbers defined in Zermelo set theory?

A fact that often gets mentioned in the elementary development of arithmetic in ZFC is that there are a bunch of different ways one could have defined the natural numbers. The most common alternative to the modern way of doing things is due to Zermelo, who sets $0=\emptyset$ and $n'=\{n\}$.

My question is how does one define the normal ordering on this collection using only the resources Zermelo allowed himself (i.e. The normal ZFC axioms, minus replacement, foundation, and a modified version of the axiom of infinity which states the existence of the above set rather than the first von Neumann ordinal)?

• I'd guess by induction: $n\leq n$ and if $n\leq m$, then $n\leq m'$. – Wojowu Apr 10 '16 at 14:39
• Or one would try to build the transitive hull of $x<\{x\}$ – Hagen von Eitzen Apr 10 '16 at 14:40
• Yeah, but how do you actually carry it out? Not sure how to do this without the normal ordinals or replacement... – I. Boon Apr 10 '16 at 14:49

$$a < b =_{\text{def} } \forall w \ [\forall x \ (a \in x \to x \in w) \land \forall x \forall y \ (x \in w \land x \in y \to y \in w) \to b \in w].$$


text:
# Misunderstood

Algebra Level 4

A teacher has given the following equation to his four best students, and then asked how many integer answers it had.

$\left [ \dfrac{12x+1}{x+12} \right ] = 10$

Al considered his teacher's notation to be a mere parenthesis, and quickly shouted his answer. Bob considered $$\left [ x \right ]$$ the floor function, and answered in time with his fellows Carl (which used the round function for his solution) and Dan (which preferred the ceiling function).

Which boy gave out the biggest answer?

×


text:
# Limit involving integration

Prove that the limit $$\lim_{x\to1}\frac{\int_{2}^{2x}\frac{u}{1+\ln(u/2)}du}{x-1}$$ exists and equals 4.

My thoughts involving this limit are that I will have to use L'hospital's rule at some point as the denominator will equal zero if I substitute 1 in for x, but I don't know how to handle the integral in the numerator.

• At first, I wasn't sure how to approach it through the FTC because I wasn't sure how to show the numerator was zero, but I just now see how the integral from 2 to 2 is simply zero. This is something for me to go off of here. – Peetrius Jul 24 '18 at 11:04
• I think the limit is not 1... – Szeto Jul 24 '18 at 12:22
• Thanks for the heads up. I fixed it. – Peetrius Jul 25 '18 at 1:12

Let $f(x):=\int_{2}^{2x}\frac{u}{1+\ln(u/2)}du$. Then
$$\lim_{x\to1}\frac{\int_{2}^{2x}\frac{u}{1+\ln(u/2)}du}{x-1}= \lim_{x\to1}\frac{f(x)-f(1)}{x-1}=f'(1).$$
• I think I see where this is going. So the structure of the limit at hand looks like the limit definition of the derivative, so I just need to take the derivative of $f(x)$ and then plug in 1, and I've evaluated the limit? – Peetrius Jul 24 '18 at 11:10
Since the integrand $f$ is continuous by mean value theorem the integral in numerator is equal to $2(x-1)f(c)$ for some $c$ between $2$ and $2x$. Thus the expression under limit is equal to $2f(c)$ and as $x\to 1$ we have $c\to 2$ so that by continuity of $f$ we have the desired limit as $2f(2)=4$.
Hint: $$\dfrac{d}{dx}\int_{2}^{2x}\frac{u}{1+\ln(u/2)}du=\dfrac{d}{dx}(2x) \left(\frac{2x}{1+\ln(2x/2)}\right) - \dfrac{d}{dx}(2) \left(\frac{2}{1+\ln(2/2)}\right)$$


text:
# Issue with Chapters Being Included in List of Figures

I was editing a table of contents earlier this week and, thanks to this forum, was able to get most of the formatting down. This post is still about the table of contents, but now it is in regards to a rather weird occurrence. After sending in my revised thesis, the reviewer noted that there should be double spacing throughout the entire table of contents. Also, the table of contents should follow the acknowledgments, then the list of tables, list of figures, and then the chapters. I am also using the UNT style file which I have slightly modified. In the MWE below, I was able to do most of this:

\documentclass[12pt, titlepage, openany]{amsbook}

\makeatletter
\renewcommand\tocchapter[3]{%
}
\makeatother

\makeatletter
\def\@tocline#1#2#3#4#5#6#7{\relax
\ifnum #1>\c@tocdepth % then omit
\else
\begingroup \hyphenpenalty\@M
\@ifempty{#4}{%
\@tempdima\csname r@tocindent\number#1\endcsname\relax
}{%
\@tempdima#4\relax
}%
\rightskip\@pnumwidth plus4em \parfillskip-\@pnumwidth
#5\leavevmode\hskip-\@tempdima #6\nobreak\relax
\ifnum#1<0\hfill\else\dotfill\fi\hbox to\@pnumwidth{\@tocpagenum{#7}}\par
\nobreak
\endgroup
\fi}
\makeatother

\makeatletter
\DeclareRobustCommand{\l@specialchapter}{%
\let\@tocpagenum\@gobble
\@tocline{-1}{8pt plus 1pt}{0pt}{}{}}
\makeatother

\usepackage{amsmath, setspace, fancyhdr, times, graphicx}
\usepackage{lipsum}

\begin{document}

\doublespacing

\tableofcontents

\begin{doublespace}
\listoftables
\end{doublespace}
\begin{doublespace}
\listoffigures
\end{doublespace}

\makeatletter
\let\protect\l@chapter\l@specialchapter}
\makeatother

\mainmatter

\begin{doublespace}
\chapter{..} \lipsum[1-20] %This is the chapter that gets placed in LoF
\chapter{..} \lipsum[1-20]
\chapter{..} \lipsum[1-20]
\chapter{..} \lipsum[1-20]
\chapter{..} \lipsum[1-20]
\end{doublespace}

\end{document}

Although I feel that I am almost done, there is a new issue of the first chapter being included in the List of Figures which is not good. I have tried to move around the location of the list of tables/figures, but then the word CHAPTER gets moved as well which is not acceptable. Also the doublespacing commands do not seem to be working on the table of contents.

Is there a way to fix this? Any help is deeply, and most sincerely, appreciated. Many thanks in advance.

UPDATE

So I have been trying really hard to correct this misplaced chapter issue and here is what I have noticed: If I use the \include{...} command to call the .tex files that I have written, then it seems that only the first one will be included in the list of figures, for some unknown reason (I use \include{chapter1}, ... , \include{chapter5}). I was trying to reverse engineer the problem starting from the MWE above which led me to notice this. I also tried to create an empty chapter, but to no avail. At this point, it seems like LaTeX is going haywire on me. If anyone has an idea or can come up with a solution, please let me know as this will help me finish this thesis that I am working on. Also, I would be glad to send a copy of these files so that one may further investigate/see what is going on. Thank you for your time.

• Is "Background Information" in chapter1.tex? Does it use \chapter*? You should really consider providing the community with something that doesn't require additional sources not available in your question. For example, instead of \include{chapter1}, use \chapter{..} \lipsum[1-20] (say). – Werner Jun 2 '16 at 18:32
• Yes, "Background Information" is in chapter1.tex, and it does not use the asterisk * after the command. I just edited the post according to your suggestion. – Jamil Villarreal Jun 2 '16 at 18:44
• ...no... not just a verbatim copy-and-paste of what I suggested. Instead, read up on how to create a minimal working example (MWE). We want to copy-and-paste-and-compile your entire code block without seeing an error, but also see exactly what you're seeing in the output. \lipsum[1-20] causes an error as you haven't included the lipsum package. – Werner Jun 2 '16 at 18:45
• ...please click the MWE link and read the resulting posts on TeX - LaTeX Meta. – Werner Jun 2 '16 at 18:45
• Ok. I apologize for carelessly having copy-and-paste that suggestion. I have included a working MWE that compiles without any errors. In that MWE, however, the table of contents does not include the first chapter in the list of figures. – Jamil Villarreal Jun 2 '16 at 19:16

A quick workaround would be to delete the following lines

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\makeatletter
\let\protect\l@chapter\l@specialchapter}
\makeatother

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

in your main tex file JamilVThesis.tex and add them as first lines in your file chapter1.tex:

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\makeatletter
\let\protect\l@chapter\l@specialchapter}

Now the text Chapter can be added to the TOC at the place where it belongs, in front of your chapters ...


text:
Thread: Show that gcd (a, m) is less than equal to gcd(a, mn) for any integers a,m and n.?

1. Show that gcd (a, m) is less than equal to gcd(a, mn) for any integers a,m and n.?

Show that gcd (a, m) $\displaystyle$\leq$$gcd(a, mn) for any integers a,m and n. 2. Originally Posted by Statsnoob2718 Show that gcd (a, m) \displaystyle \leq$$ gcd(a, mn) for any integers a,m and n.
Sometimes gcd(0,0) is defined as 0 which would provide an exception, but aside form that, show that gcd(a,m) | gcd(a,mn).

,

,

show that gcd(a,m) less than or equal to gcd (a,mn)

Click on a term to search for related topics.


text:
# Embree–Trefethen constant

In number theory, the Embree–Trefethen constant is a threshold value labelled β*.[1]

For a fixed positive number β, consider the recurrence relation

$x_{n+1}=x_n \pm \beta x_{n-1} \,$

where the sign in the sum is chosen at random for each n independently with equal probabilities for "+" and "−".

It can be proven that for any choice of β, the limit

$\sigma(\beta) = \lim_{n \to \infty} (|x_n|^{1/n}) \,$

exists almost surely. In informal words, the sequence behaves exponentially with probability one, and σ(β) can be interpreted as its almost sure rate of exponential growth.

We have

σ < 1 for 0 < β < β* = 0.70258 approximately,

so solutions to this recurrence decay exponentially as n→∞ with probability 1, and

σ > 1 for β* < β,

so they grow exponentially.

Regarding values of σ, we have:

The constant is named after applied mathematicians Mark Embree and Lloyd N. Trefethen.

## References

1. ^ Embree, M.; Trefethen, L. N. (1999). "Growth and decay of random Fibonacci sequences" (PDF). Proceedings of the Royal Society A: Mathematical, Physical and Engineering Sciences 455 (1987): 2471. Bibcode:1999RSPSA.455.2471T. doi:10.1098/rspa.1999.0412.


text:
+0

# help

+1
87
1

A number is called purple if it can be expressed in the form $$\dfrac{1}{2^a5^b}$$ for positive integers a > b. Find the sum of all purple numbers.

I'm not sure how to do this.

Jul 2, 2020

#1
+657
+2

Graphing the possible values for $$a>b$$ as positive integers. (Selected Points are the possible values)

_____________________________________________________________

a is y-axis

b is x-axis

By plugging in values of A and B...

Notice how if we set b = 1, we can get an infinite series as $$a$$ increases by increments of 1.

Notice how if we set b = 2, we can get the same thing except the starting term is different.

So the answer really is the sum of an infinite series of infinite series'.... very interesting!

_____________________________________________________________

Let us find the infinite series if b = 1 and $$a$$ increases to start off:

$$\frac{1}{20}+\frac{1}{40}+\frac{1}{80}...$$

_____________________________________________________________

The formula for an infinite series is

$$\frac{a_{(1)}}{1-r}$$

_____________________________________________________________

So the series for b = 1 is $$\frac{1}{10}$$.

By applying the same method, we know that the series for b = 2, b = 3, b = 4 is:

1/100

1/1000

1/10000

respectively...

So now we have to find the sum to THAT infinite series, which is:

$$\frac{\frac{1}{10}}{1-\frac{1}{10}} = \boxed{\text{math is fun}}$$

*an interesting thing is as you add the number of different variables, the higher dimensions the graph is. Interesting how we don't need to actually VISUALIZE higher dimensional graphs, we can just calculate the sum if you have the method down.

Jul 2, 2020
edited by AnExtremelyLongName  Jul 2, 2020


text:
mersenneforum.org Assignment Problems
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 2012-03-08, 00:14 #1 Prime95 P90 years forever!     Aug 2002 Yeehaw, FL 2×5×769 Posts Assignment Problems A manual user emailed me that these exponents are now registered to GPUto72: 50311747 (running, non-networked) 42576463 (running, networked) 45522661 (queued on networked CPU) 42578279 (ditto) 43094647 (running, non-networked) 43163753 (queued on networked CPU) 42578609 (running on networked CPU) The server should not have handed these out - so some investigation is required. In the meantime, can these exponents be quarantined and not handed out to any users (or if already handed out can you please stop testing until the user has a chance to complete his first time test)?
 2012-03-08, 00:39 #2 Dubslow Basketry That Evening!     "Bunslow the Bold" Jun 2011 40
 2012-03-08, 01:05 #3 kladner     "Kieren" Jul 2011 In My Own Galaxy! 2×3×1,693 Posts Whew! At first look I thought I had one of those, at 36.7% completion. A closer look showed that just the last three digits were the same (661).
 2012-03-08, 02:55 #4 Dubslow Basketry That Evening!     "Bunslow the Bold" Jun 2011 40
2012-03-08, 17:54   #5
chalsall
If I May

"Chris Halsall"
Sep 2002

22×7×359 Posts
George, PrimeNet has a problem.

Quote:
 Originally Posted by Prime95 A manual user emailed me that these exponents are now registered to GPUto72: 50311747 (running, non-networked) 42576463 (running, networked) 45522661 (queued on networked CPU) 42578279 (ditto) 43094647 (running, non-networked) 43163753 (queued on networked CPU) 42578609 (running on networked CPU) The server should not have handed these out - so some investigation is required. In the meantime, can these exponents be quarantined and not handed out to any users (or if already handed out can you please stop testing until the user has a chance to complete his first time test)?
I hope people don't think I'm ignoring this; I'm not. But I have "real" work to do as well, and it took a bit of time to drill down on these candidates against the knowledge I happen to have.

George... I used a new technique to aquire candidates, which I will PM you the details about shortly. It was, in my opinion, entirely "legal", and I used a "slaved" mprime and the PrimeNet API to do so. As in, it was entirely your code which did this.

However, I think four things are important...

1. Please note the reservation dates of the above candidates:

Code:
42576463 - 2008-11-24
42578279 - 2008-11-30
42578609 - 2008-12-08
43094647 - 2008-12-02
43163753 - 2008-12-02
45522661 - 2008-11-25
50311747 - 2008-10-28
1.1. Please note that no LL work had been done on them (or, at least, not reported).

2. The user in question is not questioning the "loss" of approximately 200 other candidates.

3. The user in question produces approximately two (2) LL results a month.

4. The user in question also has four "Trial Factoring" assignments from 2008-11-26, 1191 days ago, which have been independently completed by other workers.

4.1. To be explicit, the TF work was not coordinated by GPU72.

4.2. To be more explicit, the candidates I'm talking about are 51130021, 51163129, 51208343 and 51208343.

4.2.1. Don't you think it might make sense for PrimeNet to release these back into the Pool for LL work?

At the end of the day the GPU to 72 project's goal is to help find the next Mersenne Prime.

If finding bugs in PrimeNet is part of the process, I would argue that is simply a healthy example of the Scientific Method.

 2012-03-09, 00:56 #6 chalsall If I May     "Chris Halsall" Sep 2002 Barbados 274416 Posts So, as an experiment, I manually requested "Trial Factoring" work from PrimeNet shortly after midnight UTC. I was given two assignments below 45M to TF from 69 to 70. Note that these are below where "Spidy" is given assignments. And further note that these were assigned to "Anonymous" to LL on "2011-07-20". According to the historical data I have, these were both at approximately 43% complete of the LL test. George -- should I "hold off" doing any testing on these as well?
2012-03-09, 02:24   #7
Prime95
P90 years forever!

Aug 2002
Yeehaw, FL

2×5×769 Posts

Quote:
 Originally Posted by chalsall So, as an experiment, I manually requested "Trial Factoring" work from PrimeNet shortly after midnight UTC. I was given two assignments below 45M to TF from 69 to 70. Note that these are below where "Spidy" is given assignments. And further note that these were assigned to "Anonymous" to LL on "2011-07-20". According to the historical data I have, these were both at approximately 43% complete of the LL test. George -- should I "hold off" doing any testing on these as well?
I don't have strong feelings one way or the other on "holding off". I need to drill down and figure out why Primenet is releasing these as TF assignments.

To be clear, I'm not accusing anyone of doing anything underhanded. I just need to figure out how it happened and correct it. The user in question has been around a long time contributing with slow machines. He obviously becomes frustrated when "poached".

As to recycling seemingly abandoned TF assignments, we need to look into this. It is a separate issue.

 2012-03-10, 21:41 #8 Prime95 P90 years forever!     Aug 2002 Yeehaw, FL 2×5×769 Posts The user also reports these exponents were scooped: 42578197 42652487 42968837 43025869 43055297 43110383 If you have one of them, please don't LL it. Chalsall and I think we have figured out why this has happened so we can prevent this in the future.
 2012-03-10, 21:51 #9 Dubslow Basketry That Evening!     "Bunslow the Bold" Jun 2011 40
2012-03-10, 22:13   #10
Prime95
P90 years forever!

Aug 2002
Yeehaw, FL

2·5·769 Posts

Quote:
 Originally Posted by Dubslow This is perhaps a bold proposition...
I've just emailed the user explaining what happened. The fairest thing is to let the user reclaim as many of his exponents as possible. The fact that they are "milestone exponents" is irrelevant.

I've also suggested he join gpu72.com to get LL tests in the future. He enjoys testing smaller exponents (as do I) and this is the best way to get them.

2012-03-10, 22:15   #11
chalsall
If I May

"Chris Halsall"
Sep 2002

1005210 Posts

Quote:
 Originally Posted by Dubslow From the initial post here, it seems about half of the expos are not yet running. Seeing as these are milestone exponents, and we (you) are able to communicate with said user, would he be willing to not run those which aren't already going, and instead pass them out to forum goers here so that all those low expos can be run "in parallel" and completed as soon as possible?
I'm glad someone said it...

I have to say I agree. Why should someone be able to reserve (several hundred) candidates, and then sit on them for more than three years?

Please note the following data on these latest six candidates:

Code:
Candidate - Reserved Date - Estimated Completion
42578197 - 2008-12-02 - 2009-02-08
42652487 - 2008-11-24 - 2011-02-22
42968837 - 2008-12-07 - 2009-04-19
43025869 - 2008-11-24 - 2011-03-01
43055297 - 2008-11-24 - 2011-03-01
43110383 - 2008-11-24 - 2011-03-02
George... While I will abide by anything you tell me to do with regards to PrimeNet and this project, it has to be said that all of these would have been recycled long ago if PrimeNet followed the expiry rules you yourself have said are in place.

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# For all numbers x and y, the operation & is defined by x&y

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For all numbers x and y, the operation & is defined by x&y [#permalink]  27 Aug 2018, 09:32
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Question Stats:

80% (00:45) correct 20% (00:35) wrong based on 15 sessions
For all numbers x and y, the operation & is defined by $$x&y = \frac{x-y}{x-1}$$ . If $$(-3)&q=2$$, then q =

A. -11

B. -8

C. -5

D. 5

E. 11
[Reveal] Spoiler: OA

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Kudos [?]: 14 [0], given: 0

Re: For all numbers x and y, the operation & is defined by x&y [#permalink]  28 Aug 2018, 06:54
(-3-q)/(-3-1) = 2

solve this and you get q = 5
Re: For all numbers x and y, the operation & is defined by x&y   [#permalink] 28 Aug 2018, 06:54
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# For all numbers x and y, the operation & is defined by x&y

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text:
52, Divide Decimals by Whole Numbers – Page No. Multiples of 15: 15 = 5(7+9) As Antonio buys at least 100 of each type, so multiples of 18 are 18,36,54,72,90,108 as 108 is the least number and more than 100 and nearest to 100, so the least number of sheets he might buy= 108, Question 15. Question 15. Are you getting the free resources, updates, and special offers we send out every week in our teacher newsletter? of rows is 2800÷25= 112 rows. Amelia needs 24 grains of beads to make a bracelet. = 41.98+ 40.838, Explanation: 8.4 + (13.1 − 0.6)

Perform operations in parentheses. Answer: 4 packages of raisins and 3 packages of granola bars he should buy. Common factors are 1,5 Multiples of 6 are 6,12,18,24,30,36,42,48,54 Question 5. Explanation: Prime factors of 225 are 3×3×5×5 which is 3355 as the fourth digit of the code number is 2 more than the fourth digit, s0 5+2=7 and by replacing 7 in 3×3×5×5, then 3×3×5×7= 315. How many packets will there be for each sport? He cuts each one in half.

GCF is 9, e. 18,27 0.086-0.013= 0.073. Earnings per hour is $150. Answer: Maurice as 0.51 is greater than 0.482. Amoebas can range in size from 0.01 mm to 5 mm in length. Chapter 5 Extra Test for Grade 6 for either Retake or Extra Practice! Juanita purchased 5 CDs and total cost is$47.15, so cost of 1 CD is $47.15÷5=$9.43 Multiples of 24 are 24 How much remaining did mark has left?

It covers all the key concepts in the chapter including common denominators, estimating sums and differences using benchmarks, adding and subtracting fractions and mixed numbers with unlike denominators, patterns with, Go Math 4th Grade, Chapter 6 LCM is 36. So the LCM of 12 and 9. c. 9, 18

Multiples of 9 are 9,18,27,36,45,54 Question 4. 3.

= (16)×0.6 What is the greatest number of cards he can place in each packet? Explanation: Total no.of people attending a dinner party are 147 and 7 people can seat in each table, so 147÷7= 21 tables are needed for a dinner party. Go Math! 2.4375 Multiples of 5: 5,10,15. Explanation: Mr. Hernandez calculated in a wrong way. LCM is 10, Explanation:

Average cost of all CD’s i= (cost of all CD’s)÷(No.of CD’S) Multiples of 3: 53,6,9,12,15 There are 111 guests attending a party. Complete the table to show the average cost of the CDs each friend bought. = 1.14, Explanation: 54 + (9.2 − 1.413) Answer: 5 Sky jets would be needed and 50 empty seats would be there. She has 2 cups of walnuts. The combination for a lock is a 3-digit number. Question 10. Question 12. Question 6. Veggie burgers come in packages of 15, and buns come in packages of 6. As he have two $20 so he will receive$40-$29.08=$10.92. Explanation: As Kyle drives 429 miles on 33 gallons gas, so 429÷33= 13 miles he can drive on 1 gallon of gas. = 3×(7+4) Graph the number on the number line. Explanation: The place value of 5 in $69,500,000 is 500,000. Factors of 42: 1,2,3,6,7,14,21,42 Explanation: Each bag will hold only one type of muffin. Question 17. How would your answers to the previous problem have changed? Each bag needs to contain the same number of teaspoons, and each bag can contain only one spice. 4. Question 16. =4×(8+7) Common factors are 1,2 Multiples of 10: 10 Question 12. Use numbers and words to explain why or why not. How much did he pour into each glass? As a clue, she says the number is the least whole number that has three different prime factors. It covers all the key concepts in the chapter including common denominators, estimating sums and differences using benchmarks, adding and subtracting fractions and mixed numbers with … Every prize will have one kind of pin. Explanation: A factory produces 30,480 bolts in 12 hours. Factors of 15: 1,3,5,15 Question 3. Answer: Move the decimal point two places right. Multiples of 4 are 4,8,12 Four friends went shopping at a music store. Explanation: By distributive property (35+45) By distributive law 50+60 Joanie ran 0.25 of the distance during her lunch break. So the average annual growth for girls age 8 to age 14 is 1.96 inches. She only wants one color of mug on each shelf. Earnings per hour is 42.75÷4.5=$9.5 = 6.4 – 0.1376 So 30 pencils and 30 erasers are the least numbers we can buy without any leftover. Factors of 45: 1,3,5,9,15,45 A computer code is based on the prime factorization of 160. Question 6.

3, 9 Question 15. Answer: Dani’s secret number is 30 and the prime factorization is 2,3,5. Multiples of 12: 12,24,36,48,60. LCM is 108. Question 4. Mr. Yaw’s bookcase holds 20 nonfiction books and 15 fiction books. This assessment is aligned to the 5th grade chapter 6 Go Math test and curriculum. Ron divided 67.6 fluid ounces of orange juice evenly among 16 glasses. Factors of 31: 1,31 Chapter 10 Extra Test for Grade 6 for either Retake or Extra Practice! **FOLLOW ME for updates! Carrie; My Links = 21.3-(15.88) 6 for the 4th grade Go Math! Question 12. Multiples of 6: 6,2,18,24 Explanation: No.of meals are 64 and the total fee is $1,088. She spends$29.40. Factors of 35: 1,2,5,7,35 He has 18 teaspoons of cinnamon and 30 teaspoons of nutmeg. If he has 42 parakeets and 18 canaries, how many birds will he put in each cage? There are 16 sixth graders and 20 seventh graders in the Robotics Club. Lynn paid $4.75 for cereal,$8.96 for chicken, and $3.25 for soup. Question 4. of flowers, Let the no. Evaluate the expression using the order of operations. As beads are sold in packages of 20 Juanita needs 3 least number of packages to make necklaces with no beads leftover. Two different versions. On average, how many hours do they all volunteer per week? 2×2×12 Juan is reading a 312-page book for school. = 34.7-(16.97) Question 2. 60+12= (12×5)+(12×1) GCF is 6 Question 12. Explanation: The cost of pizza is$12.95 which was shared by 5 friends, so each person pays $12.95÷5=$2.59. If the monorail runs 18 hours each day, what is the average number of passengers riding each hour? 7,236 ÷ 9 = 804 Yes No, Question 6. Question 10. Paxtang Elementary 3530 Rutherford Street Harrisburg, PA 17111 • 717.561.1781 Quality to the Core Explanation: Length of one celled organism is 32 millimeters, as the photo was enlarged by a factor of 100, it’s actual length is 32×100= 3200 millimeters. Factors of 45: 1,3,5,9,15,45 = 4.92÷(0.8-(0.4)) The greatest number of groups she can have is 5. = 30.32. Answer: 30 pencils and 30 erasers are the least numbers we can buy without any leftover. For numbers 5a–5d, choose Yes or No to indicate whether each equation is correct. Factors of 150: 1,2,3,5,6,10,15,25,30,50,75,150 Explanation: = (88.9)÷7

Answer: Two consecutive prime numbers are 31 and 37 and PIN is 1147. Which company had the greater profit? Problem Solving + Applications – Page No. Audi A7 Competitors, All Rank 4 Essences Achievement, Nfl Teams In Alphabetical Order, Real Mccoy Outdoors, Yorba Linda Fault, Types Of Coffee Drinks Pdf, Similes In Disney Movies, Miniature Dollhouse Stores Near Me, Reddit Wallstreetbets Options, Queen Olga Of Greece Tattoo, Rdo Train Schedule, Bernie Parent Jr, Cube Surfer Online, Michael Keaton Partner, Dude Perfect Zoom, Thea 2: The Shattering Wiki, Can You Play 2v2 On Mortal Kombat 11, Kenny Loggins Ears, Someone Gets Hurt Lyrics, Nfl 2k5 Xbox One, Amadi Longest Name, Mallory Pugh Sister, Medina Rapper Phaedra, Monster Korean Drama Blind, Ryan Cartagena Illness, Narnia Fanfiction Lucy Injured, Dhakota Williams Net Worth, Roxane Bruneau Partition, Ricardo Martinelli Net Worth, Welcome To The Punch Ending Explained, Peace Talks Arc, James Crowder Faze Net Worth, Mongoose 24 Bmx, Why Kpop Is So Popular Essay, " />
52, Divide Decimals by Whole Numbers – Page No. Multiples of 15: 15 = 5(7+9) As Antonio buys at least 100 of each type, so multiples of 18 are 18,36,54,72,90,108 as 108 is the least number and more than 100 and nearest to 100, so the least number of sheets he might buy= 108, Question 15. Question 15. Are you getting the free resources, updates, and special offers we send out every week in our teacher newsletter? of rows is 2800÷25= 112 rows. Amelia needs 24 grains of beads to make a bracelet. = 41.98+ 40.838, Explanation: 8.4 + (13.1 − 0.6)

Perform operations in parentheses. Answer: 4 packages of raisins and 3 packages of granola bars he should buy. Common factors are 1,5 Multiples of 6 are 6,12,18,24,30,36,42,48,54 Question 5. Explanation: Prime factors of 225 are 3×3×5×5 which is 3355 as the fourth digit of the code number is 2 more than the fourth digit, s0 5+2=7 and by replacing 7 in 3×3×5×5, then 3×3×5×7= 315. How many packets will there be for each sport? He cuts each one in half.

GCF is 9, e. 18,27 0.086-0.013= 0.073. Earnings per hour is $150. Answer: Maurice as 0.51 is greater than 0.482. Amoebas can range in size from 0.01 mm to 5 mm in length. Chapter 5 Extra Test for Grade 6 for either Retake or Extra Practice! Juanita purchased 5 CDs and total cost is$47.15, so cost of 1 CD is $47.15÷5=$9.43 Multiples of 24 are 24 How much remaining did mark has left?

It covers all the key concepts in the chapter including common denominators, estimating sums and differences using benchmarks, adding and subtracting fractions and mixed numbers with unlike denominators, patterns with, Go Math 4th Grade, Chapter 6 LCM is 36. So the LCM of 12 and 9. c. 9, 18

Multiples of 9 are 9,18,27,36,45,54 Question 4. 3.

= (16)×0.6 What is the greatest number of cards he can place in each packet? Explanation: Total no.of people attending a dinner party are 147 and 7 people can seat in each table, so 147÷7= 21 tables are needed for a dinner party. Go Math! 2.4375 Multiples of 5: 5,10,15. Explanation: Mr. Hernandez calculated in a wrong way. LCM is 10, Explanation:

Average cost of all CD’s i= (cost of all CD’s)÷(No.of CD’S) Multiples of 3: 53,6,9,12,15 There are 111 guests attending a party. Complete the table to show the average cost of the CDs each friend bought. = 1.14, Explanation: 54 + (9.2 − 1.413) Answer: 5 Sky jets would be needed and 50 empty seats would be there. She has 2 cups of walnuts. The combination for a lock is a 3-digit number. Question 10. Question 12. Question 6. Veggie burgers come in packages of 15, and buns come in packages of 6. As he have two $20 so he will receive$40-$29.08=$10.92. Explanation: As Kyle drives 429 miles on 33 gallons gas, so 429÷33= 13 miles he can drive on 1 gallon of gas. = 3×(7+4) Graph the number on the number line. Explanation: The place value of 5 in $69,500,000 is 500,000. Factors of 42: 1,2,3,6,7,14,21,42 Explanation: Each bag will hold only one type of muffin. Question 17. How would your answers to the previous problem have changed? Each bag needs to contain the same number of teaspoons, and each bag can contain only one spice. 4. Question 16. =4×(8+7) Common factors are 1,2 Multiples of 10: 10 Question 12. Use numbers and words to explain why or why not. How much did he pour into each glass? As a clue, she says the number is the least whole number that has three different prime factors. It covers all the key concepts in the chapter including common denominators, estimating sums and differences using benchmarks, adding and subtracting fractions and mixed numbers with … Every prize will have one kind of pin. Explanation: A factory produces 30,480 bolts in 12 hours. Factors of 15: 1,3,5,15 Question 3. Answer: Move the decimal point two places right. Multiples of 4 are 4,8,12 Four friends went shopping at a music store. Explanation: By distributive property (35+45) By distributive law 50+60 Joanie ran 0.25 of the distance during her lunch break. So the average annual growth for girls age 8 to age 14 is 1.96 inches. She only wants one color of mug on each shelf. Earnings per hour is 42.75÷4.5=$9.5 = 6.4 – 0.1376 So 30 pencils and 30 erasers are the least numbers we can buy without any leftover. Factors of 45: 1,3,5,9,15,45 A computer code is based on the prime factorization of 160. Question 6.

3, 9 Question 15. Answer: Dani’s secret number is 30 and the prime factorization is 2,3,5. Multiples of 12: 12,24,36,48,60. LCM is 108. Question 4. Mr. Yaw’s bookcase holds 20 nonfiction books and 15 fiction books. This assessment is aligned to the 5th grade chapter 6 Go Math test and curriculum. Ron divided 67.6 fluid ounces of orange juice evenly among 16 glasses. Factors of 31: 1,31 Chapter 10 Extra Test for Grade 6 for either Retake or Extra Practice! **FOLLOW ME for updates! Carrie; My Links = 21.3-(15.88) 6 for the 4th grade Go Math! Question 12. Multiples of 6: 6,2,18,24 Explanation: No.of meals are 64 and the total fee is $1,088. She spends$29.40. Factors of 35: 1,2,5,7,35 He has 18 teaspoons of cinnamon and 30 teaspoons of nutmeg. If he has 42 parakeets and 18 canaries, how many birds will he put in each cage? There are 16 sixth graders and 20 seventh graders in the Robotics Club. Lynn paid $4.75 for cereal,$8.96 for chicken, and $3.25 for soup. Question 4. of flowers, Let the no. Evaluate the expression using the order of operations. As beads are sold in packages of 20 Juanita needs 3 least number of packages to make necklaces with no beads leftover. Two different versions. On average, how many hours do they all volunteer per week? 2×2×12 Juan is reading a 312-page book for school. = 34.7-(16.97) Question 2. 60+12= (12×5)+(12×1) GCF is 6 Question 12. Explanation: The cost of pizza is$12.95 which was shared by 5 friends, so each person pays $12.95÷5=$2.59. If the monorail runs 18 hours each day, what is the average number of passengers riding each hour? 7,236 ÷ 9 = 804 Yes No, Question 6. Question 10. Paxtang Elementary 3530 Rutherford Street Harrisburg, PA 17111 • 717.561.1781 Quality to the Core Explanation: Length of one celled organism is 32 millimeters, as the photo was enlarged by a factor of 100, it’s actual length is 32×100= 3200 millimeters. Factors of 45: 1,3,5,9,15,45 = 4.92÷(0.8-(0.4)) The greatest number of groups she can have is 5. = 30.32. Answer: 30 pencils and 30 erasers are the least numbers we can buy without any leftover. For numbers 5a–5d, choose Yes or No to indicate whether each equation is correct. Factors of 150: 1,2,3,5,6,10,15,25,30,50,75,150 Explanation: = (88.9)÷7

Answer: Two consecutive prime numbers are 31 and 37 and PIN is 1147. Which company had the greater profit? Problem Solving + Applications – Page No. Audi A7 Competitors, All Rank 4 Essences Achievement, Nfl Teams In Alphabetical Order, Real Mccoy Outdoors, Yorba Linda Fault, Types Of Coffee Drinks Pdf, Similes In Disney Movies, Miniature Dollhouse Stores Near Me, Reddit Wallstreetbets Options, Queen Olga Of Greece Tattoo, Rdo Train Schedule, Bernie Parent Jr, Cube Surfer Online, Michael Keaton Partner, Dude Perfect Zoom, Thea 2: The Shattering Wiki, Can You Play 2v2 On Mortal Kombat 11, Kenny Loggins Ears, Someone Gets Hurt Lyrics, Nfl 2k5 Xbox One, Amadi Longest Name, Mallory Pugh Sister, Medina Rapper Phaedra, Monster Korean Drama Blind, Ryan Cartagena Illness, Narnia Fanfiction Lucy Injured, Dhakota Williams Net Worth, Roxane Bruneau Partition, Ricardo Martinelli Net Worth, Welcome To The Punch Ending Explained, Peace Talks Arc, James Crowder Faze Net Worth, Mongoose 24 Bmx, Why Kpop Is So Popular Essay, ">

Sun block 1×$9.99=$9.99 Sammy has 3 pancakes. Divide. Explanation: Use numbers and words to explain your answer. Common Factors are 1,2,3,6,9,18 LCM is 54, Explanation: So he put 6 berries on each dessert and 6 strawberries and 7 blueberries in each type of fruit. No. So there will be 7 packets of baseball cards and 4 packets of football cards and each packet contains 3 cards. Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. 13, Problem Solving + Applications – Page No. How many groups of each type will there be? Cheese slices come in packages of 18. Factors of 16: 1,2,4,8,16 Multiples of 12: 12,24,36,48,60 Factors of 20 : __________, Explanation: Factors of 64:  1,2,4,8,16,32,64 13. What operation should you use to find the distance the zebra ran in 10 seconds? • Chapter 1 Performance Task • Answer Keys and ... Overview of Go Math! Each class has the greatest possible number of students. Answer: No. The least common multiple of 12 and 9 is _____ . Explanation: No.of cheese packages Gavin bought are 4 and each package weight is 1.08 kg. This packet can be used as an in class review and/or homework assignment before giving the chapter 6 Go Math! = 36.4+(9.2-(0.7))

Question 13. Jean needs $$\frac{1}{3}$$ cup of walnuts for each serving of salad she makes. Show your work. What is the greatest number of chairs that he can fit in each row? She wants to fill as many jars as possible. Factors of 54:  1,2,3,6,9,18,27,54 How many shelves does she need for each color? Multiples of 3: 3,6,92,15,18,21,24 If each bag of grapes has the same weight, how much does each bag weigh? N× 0.7= 3× 4.9 If there were 21 full flights and 64 reservations cancelled, which airplane made the flights? 2b. Explanation: (61.8 × 1.7) + 9.5 b. What is the greatest number of seedlings she can plant in each row? They are based more on the "on your own" sections from each lesson. Factors of 18: 1,2,3,6,9,18 LCM is 15, Explanation: What if Toby had decided to keep one baseball card for himself and sell the rest? Question 5.

52, Divide Decimals by Whole Numbers – Page No. Multiples of 15: 15 = 5(7+9) As Antonio buys at least 100 of each type, so multiples of 18 are 18,36,54,72,90,108 as 108 is the least number and more than 100 and nearest to 100, so the least number of sheets he might buy= 108, Question 15. Question 15. Are you getting the free resources, updates, and special offers we send out every week in our teacher newsletter? of rows is 2800÷25= 112 rows. Amelia needs 24 grains of beads to make a bracelet. = 41.98+ 40.838, Explanation: 8.4 + (13.1 − 0.6)

Perform operations in parentheses. Answer: 4 packages of raisins and 3 packages of granola bars he should buy. Common factors are 1,5 Multiples of 6 are 6,12,18,24,30,36,42,48,54 Question 5. Explanation: Prime factors of 225 are 3×3×5×5 which is 3355 as the fourth digit of the code number is 2 more than the fourth digit, s0 5+2=7 and by replacing 7 in 3×3×5×5, then 3×3×5×7= 315. How many packets will there be for each sport? He cuts each one in half.

GCF is 9, e. 18,27 0.086-0.013= 0.073. Earnings per hour is $150. Answer: Maurice as 0.51 is greater than 0.482. Amoebas can range in size from 0.01 mm to 5 mm in length. Chapter 5 Extra Test for Grade 6 for either Retake or Extra Practice! Juanita purchased 5 CDs and total cost is$47.15, so cost of 1 CD is $47.15÷5=$9.43 Multiples of 24 are 24 How much remaining did mark has left?

It covers all the key concepts in the chapter including common denominators, estimating sums and differences using benchmarks, adding and subtracting fractions and mixed numbers with unlike denominators, patterns with, Go Math 4th Grade, Chapter 6 LCM is 36. So the LCM of 12 and 9. c. 9, 18

Multiples of 9 are 9,18,27,36,45,54 Question 4. 3.

= (16)×0.6 What is the greatest number of cards he can place in each packet? Explanation: Total no.of people attending a dinner party are 147 and 7 people can seat in each table, so 147÷7= 21 tables are needed for a dinner party. Go Math! 2.4375 Multiples of 5: 5,10,15. Explanation: Mr. Hernandez calculated in a wrong way. LCM is 10, Explanation:

Average cost of all CD’s i= (cost of all CD’s)÷(No.of CD’S) Multiples of 3: 53,6,9,12,15 There are 111 guests attending a party. Complete the table to show the average cost of the CDs each friend bought. = 1.14, Explanation: 54 + (9.2 − 1.413) Answer: 5 Sky jets would be needed and 50 empty seats would be there. She has 2 cups of walnuts. The combination for a lock is a 3-digit number. Question 10. Question 12. Question 6. Veggie burgers come in packages of 15, and buns come in packages of 6. As he have two $20 so he will receive$40-$29.08=$10.92. Explanation: As Kyle drives 429 miles on 33 gallons gas, so 429÷33= 13 miles he can drive on 1 gallon of gas. = 3×(7+4) Graph the number on the number line. Explanation: The place value of 5 in $69,500,000 is 500,000. Factors of 42: 1,2,3,6,7,14,21,42 Explanation: Each bag will hold only one type of muffin. Question 17. How would your answers to the previous problem have changed? Each bag needs to contain the same number of teaspoons, and each bag can contain only one spice. 4. Question 16. =4×(8+7) Common factors are 1,2 Multiples of 10: 10 Question 12. Use numbers and words to explain why or why not. How much did he pour into each glass? As a clue, she says the number is the least whole number that has three different prime factors. It covers all the key concepts in the chapter including common denominators, estimating sums and differences using benchmarks, adding and subtracting fractions and mixed numbers with … Every prize will have one kind of pin. Explanation: A factory produces 30,480 bolts in 12 hours. Factors of 15: 1,3,5,15 Question 3. Answer: Move the decimal point two places right. Multiples of 4 are 4,8,12 Four friends went shopping at a music store. Explanation: By distributive property (35+45) By distributive law 50+60 Joanie ran 0.25 of the distance during her lunch break. So the average annual growth for girls age 8 to age 14 is 1.96 inches. She only wants one color of mug on each shelf. Earnings per hour is 42.75÷4.5=$9.5 = 6.4 – 0.1376 So 30 pencils and 30 erasers are the least numbers we can buy without any leftover. Factors of 45: 1,3,5,9,15,45 A computer code is based on the prime factorization of 160. Question 6.

3, 9 Question 15. Answer: Dani’s secret number is 30 and the prime factorization is 2,3,5. Multiples of 12: 12,24,36,48,60. LCM is 108. Question 4. Mr. Yaw’s bookcase holds 20 nonfiction books and 15 fiction books. This assessment is aligned to the 5th grade chapter 6 Go Math test and curriculum. Ron divided 67.6 fluid ounces of orange juice evenly among 16 glasses. Factors of 31: 1,31 Chapter 10 Extra Test for Grade 6 for either Retake or Extra Practice! **FOLLOW ME for updates! Carrie; My Links = 21.3-(15.88) 6 for the 4th grade Go Math! Question 12. Multiples of 6: 6,2,18,24 Explanation: No.of meals are 64 and the total fee is $1,088. She spends$29.40. Factors of 35: 1,2,5,7,35 He has 18 teaspoons of cinnamon and 30 teaspoons of nutmeg. If he has 42 parakeets and 18 canaries, how many birds will he put in each cage? There are 16 sixth graders and 20 seventh graders in the Robotics Club. Lynn paid $4.75 for cereal,$8.96 for chicken, and $3.25 for soup. Question 4. of flowers, Let the no. Evaluate the expression using the order of operations. As beads are sold in packages of 20 Juanita needs 3 least number of packages to make necklaces with no beads leftover. Two different versions. On average, how many hours do they all volunteer per week? 2×2×12 Juan is reading a 312-page book for school. = 34.7-(16.97) Question 2. 60+12= (12×5)+(12×1) GCF is 6 Question 12. Explanation: The cost of pizza is$12.95 which was shared by 5 friends, so each person pays $12.95÷5=$2.59. If the monorail runs 18 hours each day, what is the average number of passengers riding each hour? 7,236 ÷ 9 = 804 Yes No, Question 6. Question 10. Paxtang Elementary 3530 Rutherford Street Harrisburg, PA 17111 • 717.561.1781 Quality to the Core Explanation: Length of one celled organism is 32 millimeters, as the photo was enlarged by a factor of 100, it’s actual length is 32×100= 3200 millimeters. Factors of 45: 1,3,5,9,15,45 = 4.92÷(0.8-(0.4)) The greatest number of groups she can have is 5. = 30.32. Answer: 30 pencils and 30 erasers are the least numbers we can buy without any leftover. For numbers 5a–5d, choose Yes or No to indicate whether each equation is correct. Factors of 150: 1,2,3,5,6,10,15,25,30,50,75,150 Explanation: = (88.9)÷7

Answer: Two consecutive prime numbers are 31 and 37 and PIN is 1147. Which company had the greater profit? Problem Solving + Applications – Page No.


text:
# Utility Elasticity

## Framework

It is typically assumed that utility ($u$) as a function of income ($x$) takes this form: $$u = \left\{ \begin{array}{ll} \ln{(x)} & \epsilon = 1 \\ \frac{x^{1-\epsilon} - 1}{1 - \epsilon} & \epsilon \gt 0 \end{array} \right.$$ where $\epsilon$ is non-negative and represents how quickly additional income becomes useless.

Mathy people will note that the $\epsilon = 1$ case is the limit of the $\epsilon \gt 0$ case.

This formula doesn't have strong justification, but there is a discussion of it here Welfare weights.

Our task is to estimate $\epsilon$.

Cowell and Gardiner Welfare weights survey many methosd used to estimate $\epsilon$. Six years later, Evans took a stab at the same subject Evans. Though most of them are (in my opinion) ludicrously terrible estimation techniques, here are the methods they considered:

• Gevers et al; Glejser et al; Amiel et al - TODO
• We can assume utility is normally distributed and a function of only income and then solve for $\epsilon$ mathematically.
• We can assume governments tax everyone such that everyone's utility is decreased equally. Then, we can compute the $\epsilon$ that would imply this is true.
• We can ask people in surveys with hypothetical questions.
• Binswanger - TODO
• We can examine lifetime consumption patterns, which (if people are rational) should be largely determined by $\epsilon$.
• We can use the Fisher-Frisch-Fellner (FFF) model. This is the sole method I think is reasonable, so we'll discuss it more below.

There's also the popular method of plotting GDP on the x-axis and self-reported happiness or well-being on the y-axis.

## FFF Estimates

The FFF procedure assumes that utility functions are additively separatable between a particular good/service and all other goods and services. This means we can represent utility as $$u = f(c_0) + g(c_1)$$ where $f$ and $g$ are two utility functions, $c_0$ is how much of some particular good/service is consumed, and $c_1$ is how much of all other goods/services are consumed.

This assumption isn't widely accepted for any good/service. To quote Evans Evans:

Results for $e$ based on the FFF model have generally been disregarded in the literature, mainly on the grounds of the strong condition of additive separability (preference independence) that is required for the approach to be valid.

...

However, in the case of food (and some other broadly defined product groups), Fellner (1967), Selvanathan and Selvanathan (1993) and Evans and Sezer (2002) have all argued that preference independence is a plausible assumption. Furthermore, Selvanathan (1988) tested the want-independence assumption for broad aggregates, using OECD data, and found it to be empirically valid. [However,] both Cowell and Gardiner (1999) and Pearce and Ulph (1999) largely disregard empirical estimates of $e$ based on the model.

In any case, if we accept this is true for some particular good (e.g. food), then we can examine how the demand for food changes when incomes change and when prices change and deduce (with some calculus) what the implied value of $\epsilon$ is.

Evans then goes on to list some empirical estimates:

 Study Estimate of $\epsilon$ Nation Kula (2002) 1.64 India Evans and Sezer (2002) 1.6 UK Evans (2004a) 1.25 or 1.6 UK Evans (2004a) 1.3 or 1.8 France

## Historical Labor

Suppose, for sake of argument that $\epsilon = 1$. This implies utility is given by $$u = \ln{(x)}$$

Now suppose labor is additively separatable from consumption and people consume all their income. Now, utility is given by $$u = \ln{(w \cdot L)} - f(L)$$ where $w$ is the wage and $f$ is some function representing disutility from working.

If we assume people choose $L$ rationally, we can show (with a little calculus) that $$1 = L \cdot f'(L)$$

This implies that $L$ is independent of $w$, which means that we should expect people to work the same number of hours regardless of $w$.

Fortunately, we have a natural experiment for this. Over the last couple centuries, the average wage has increased dramatically. $\epsilon = 1$ predicts that labor hasn't changed.

Between 1948 and 2018, real GDP per hour worked increased 3.43-fold while total hours worked increased 2.57-fold Table 1.1.3 Table 1.1.5 Table 6.9B Table 6.9D.

However, during that same time, the US population increased 2.22-fold Historical National Population Estimates Population Clock, which means we really only saw a 16% increase in labor-per-capita despite a 243% increase in wages.

However, there have been a number of changes that bias this estimate:

1. Women entered the workforce. Their abor force participation increased from 32.7% to 57.1% between 1948 and 2018 Civilian Labor Force Participation Rate: Women.
2. The percent of the population of working age has increased. I don't have data for 1948 and 2018, but between 1955 and 2014, the proportion increased by 7.5% Working age population.
3. Transfers and benefits ("welfare") increased between 1948 and 2018 due to a variety of factors such as LBJ's The Great Society, Obamacare, the general rise of health care costs, and the increasing of life expectancy relative to a static retirement age.

Changes #1 and #2 bias our estimated change in $L$ as a result of changes in $w$ (16%) upwards, while change #3 biases it downwards. These changes (in addition to other cultural shifts) add significant uncertainty to our estimate. Altogether, I expect the effect of increasing wages on labor has been slightly negative.

Now suppose $\epsilon$ was large, say $\epsilon = 2$ (rather than 1), we can model utility as $$u = -\left(w \cdot L \right)^{-1} - L^\beta$$ where $\beta$ is some constant.

Again, using calculus and assuming rationality, we can prove $$1 = w \cdot L^{b+1}$$

Now, suppose this 3.43-fold increase in wages caused $L$ to decrease by 10%. This implies $b = 11.7$, which means shifting the working day from 8 to 9 hours would cause 3.75 times more disutility compared to the change from 7 to 8 hours. This seem ludicrously high to me.

This entire section on estimating $\epsilon$ from historical labor trends is rather speculative. However, it suggests to me that $\epsilon$ is probably closer to 1 than 2.

## Overall Estimate

While the FFF model suggests $\epsilon \approx 1.6$, my observations of the historical changes in labor demand as wages have increased suggest $\epsilon$ is probably closer to 1 than 2. For this reason, I'm going to adjust the meidan FFF estimates downward to settle on $\epsilon \approx 1.5$.

## Discount Rates

Using some basic calculus, expected value theory, and an assumption of little change in income inequality, we can prove the discount rate should be $$p + i + \epsilon \cdot r$$ where $p$ is the probability of social collapse, $i$ is inflation, and $r$ is the expected percent growth in incomes. We're going to ignore $p$ and $i$ and focus on $\epsilon$ and $r$ because

• $p$ is extremely difficult to nail down and also fairly irrelevent for discounting purposes. Even if the probability of humanity going extinct this century is 19% Todd, B., $p$ would be a mere 0.21%.
• $i$ typically doesn't matter unless you own bonds. If inflation is 5% higher, interest rates tend to be 5% higher and the stock market tends to grow 5% faster. In any case, you can compute expected inflation using TIPS spreads 10-Year Breakeven Inflation Rate.

Real GDP per capita has averaged 4.3% growth over the last 28 years GDP per capita, PPP (current international $), which suggests a global discount rate of around 6.5%. Are things different among the global poor? The incomes of people between the 5th percentile and 50th percentile increased by about 6.3% per year between 2003 and 2013. However, real global GDP increased 6.4% over the same time period, indicating that the discount rate is roughly the same among the global poor. Cowell, F. A., & Gardiner, K. (1999). Welfare weights (STICERD). Economics Research Paper, 20, 1999. Evans, D. J. (2005). The elasticity of marginal utility of consumption: estimates for 20 OECD countries. Fiscal studies, 26(2), 197-224. https://doi.org/10.1111/j.1475-5890.2005.00010.x. The World Bank. (n.d.). GDP per capita, PPP (current international$). Retrieved November 10, 2019, from https://data.worldbank.org/indicator/NY.GDP.PCAP.PP.CD. U.S. Bureau of Economic Analysis, “Table 1.1.3. Real Gross Domestic Product, Quantity Indexes” https://apps.bea.gov/iTable/iTable.cfm?reqid=19&step=2#reqid=19&step=2&isuri=1&1921=survey U.S. Bureau of Economic Analysis, “Table 1.1.5. Gross Domestic Product” https://apps.bea.gov/iTable/iTable.cfm?reqid=19&step=2#reqid=19&step=2&isuri=1&1921=survey U.S. Bureau of Economic Analysis, “Table 6.9B. Hours Worked by Full-Time and Part-Time Employees by Industry” https://apps.bea.gov/iTable/iTable.cfm?reqid=19&step=2#reqid=19&step=2&isuri=1&1921=survey U.S. Bureau of Economic Analysis, “Table 6.9D. Hours Worked by Full-Time and Part-Time Employees by Industry” https://apps.bea.gov/iTable/iTable.cfm?reqid=19&step=2#reqid=19&step=2&isuri=1&1921=survey Census Bureau, "Historical National Population Estimates: July 1, 1900 to July 1, 1999". https://www.census.gov/population/estimates/nation/popclockest.txt. Census Bureau, "Population Clock". https://www.census.gov/popclock/. U.S. Bureau of Labor Statistics, Civilian Labor Force Participation Rate: Women [LNS11300002], retrieved from FRED, Federal Reserve Bank of St. Louis; https://fred.stlouisfed.org/series/LNS11300002, November 11, 2019. Demography - Working age population - OECD Data. (n.d.). Retrieved from https://data.oecd.org/pop/working-age-population.htm. Todd, B. (2017, October 1). Why despite global progress, humanity is probably facing its most dangerous time ever. Retrieved November 11, 2019, from https://80000hours.org/articles/extinction-risk/#if-we-add-everything-together-whats-the-total-risk. Federal Reserve Bank of St. Louis, 10-Year Breakeven Inflation Rate [T10YIE], retrieved from FRED, Federal Reserve Bank of St. Louis; https://fred.stlouisfed.org/series/T10YIE, November 12, 2019.


text:
Mechanical Vibrations - 2 DOF System - Rod with 2 springs

• Engineering

Homework Statement:

Derive equations of motion in terms of the displacements $x_1$ and $x_2$

Relevant Equations:

F = ma
Hi,

So the question is to: derive the equations of motion for the following in terms of x1 and x2? The bar is assumed to be light and rigid.

(NB.
I know I posted another vibrations problem earlier in which I tried to use an energy approach to get to the equations of motion. However, we haven't been taught these, so I believe we are expected to reach these solutions by using force methods)

I am just wondering whether there is a general method when approaching these 2-DOF vibration problems? (e.g. introduce another variable for equations, specific places to take moments about, etc). I was wondering whether my method seems correct?

1) Resolve vertically downwards
I have decided to define an angle $\theta$ in the CCW direction from the point where the 2m mass is so that $\theta = \frac{x_1 - x_2}{3L}$

Also, I have identified that the COM is L away from the 2m mass. I have also made small angle approximations for the angles.

Therefore, we get the following when we resolve vertically downwards and get:
$$-k x_2 - k \left( x_2 + 2L \theta \right) = m \left( \ddot x_2 + L \ddot \theta \right)$$
rearranging and substituting in for $\ddot \theta$ leads to:
$$m \ddot x_1 + 2m \ddot x_2 + 2k x_1 + 4k x_2 = 0$$

2) Take moments about the centre of mass CCW direction
Because the bar is assumed to be rigid, I thought that the angle $\theta$ will be the same at the COM as it is at the right side

Taking moments, I get:
$$k x_2 L - 2mgL + mg(2L) - k \theta L^2 = I \ddot \theta$$
noting that: $I = m(2L)^2 + 2m(L)^2 = 6mL^2$ and rearranging we get:
$$6m \ddot x_1- 6m \ddot x_2 + 4k x_1 - 4k x_2 = 0$$

Combining these into matrix form, we get:
$$\begin{bmatrix} m & 2m \\ 6m & -6m \end{bmatrix} \begin{bmatrix} \ddot x_1 \\ \ddot x_2 \end{bmatrix} + \begin{bmatrix} 2k & 4k \\ 4k & -4k \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = 0$$

Does this equation look correct?

A final question on this: if I was to go on from here to find the natural frequencies/ etc., I would presume that it is okay to leave the matrices in their current form? I only ask that as the few examples I have seen in lectures only have 'M' matrices which are diagonal, but I presume that is reflective of the respective situation rather than a mathematical reason?

Thanks in advance for the help.

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