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## Supplemental Information for A Signal-Space Distance Measure for Nondispersive Optical Fiber''

This file contains the Supplemental Information cited in:

R. Rafie Borujeny and F. R. Kschischang, A Signal-Space Distance Measure for Nondispersive Optical Fibersubmitted to IEEE Transactions on Information Theory.

This file is a header file written in C containing the numerical values as well as the function that calculates the approximation of the adversarial distance.

Instructions:

This header file contains a routine that can be used to computed

the adversarial distance between two complex points according to

the formulation given in

@misc{borujeny2020signalspace,

title={A Signal-Space Distance Measure for Nondispersive Optical Fiber},

author={Reza Rafie Borujeny and Frank R. Kschischang},

year={2020},

eprint={2001.08663},

archivePrefix={arXiv},

primaryClass={cs.IT}

}

https://arxiv.org/abs/2001.08663

The distance calculations are performed based on the approximation

given in the above manuscript. For the normalized evolution equation

q'(z) = i * |q(z)| ^ 2 * q(z) + n(z), 0 < z < 1,

for two complex numbers z1 and z2, with polar coordinates

z1 = r1 * exp(i * t1),

z2 = r2 * exp(i * t2),

the approximation of the adversarial distance is given by

d(z1, z2) = (1 / 4) * (r1 - r2) ^ 2 +

a(r1, r2) * sin^2(( t2 - t1 - \phi(r1, r2) ) / 2).

The function a(r1, r2) is approximated using numerical methods,

as described in the reference manuscript, and the corresponding

values are given below in the array a_fit. The function distance

below calculates the adversarial distance between two points. For

use with the unnormalized equation

q'(z) = i * GAMMA * |q(z)| ^ 2 * q(z) + n(z), 0 < z < L,

you should change the assigned values to L and GAMMA in the

associated macros.

*/

/*

This file contains the following:

1- a macro named L, which represents the length of the

fiber in [km]

2- a macro named GAMMA, which represents the nonlinearity

coefficient in [1/W/km]

3- a macro INDEX(i,j), which converts 2-D index to 1-D

indexing to access the array a_fit (to be described later).

4- an array of doubles X, which contains the 31 grid points along the

x-axis. This grid points are the same along the y-axis as well. That

is, for any x in X, and any y in X, we have a corresponding

value a(x, y) in the array a_fit.

5- a double DX, which is equal to the increment between the grid points

in the x-axis, i.e., DX = X[1] - X[0]

6- a double DA, which is equal to the area of a surface area element,

i.e., DA = (X[1] - X[0]) * (X[1] - X[0])

7- a function distance, which takes two complex numbers in polar

coordinates and calculates the adversarial distance between them. This

function uses bi-linear approximation to estimate A(x,y) from the

existing numerical values given in a_fit.

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## 5G Campus Networks: Measurement Traces

This data set contains packet captures (PCAPs) of a 5G campus network.

The corresponding paper can be found at 5G Campus Networks: A First Measurement Study

Instructions:

The instructions are within the READMEs(.md/.pdf).

In addition the source code of this project is available on Github: https://github.com/justus-comnets/5g-campus-measurements

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1007 Views

## Massive User Transmission Enabled Combinatorial Based Non-Orthogonal Multiple Access Wireless Networks With Theoretical Analysis simulation

This is the MATLAB .fig file from the paper "Massive User Transmission Enabled Combinatorial Based Non-Orthogonal Multiple Access Wireless Networks With Theoretical Analysis"

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## Augmented Reality Streams for Cloud-Based Rendering

A promising technique to realize augmented reality on future light-weight glasses is to offload computationally extensive rendering tasks to the cloud. This however places considerable demands on the network as well as the air interface with respect to latency, reliability and throughput. For evaluation of these architectures and for traffic modelling, a dataset is provided, which contains realistic payloads of cloud-rendered augmented reality in form of video files.

Instructions:

Provided are the raw video files after rendering with a resolution of 7200x6360 pixels. For low-latency encoding libx264 ffmpeg version 4.2.4 was used with flags -preset ultrafast -tune zerolatency at a target bitrate of 8Mbit/s. The streaming is taking place with ffmpeg and the custom nut output muxer. The resulting packetized output is sent to a UDP port on localhost. The encoded video files as well as the captured traffic traces are provided.

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In this paper, we propose a novel resource management scheme that jointly allocates the transmitpower and computational resources in a centralized radio access network architecture. The networkcomprises a set of computing nodes to which the requested tasks of different users are offloaded. Theoptimization problem minimizes the energy consumption of task offloading while takes the end-to-end latency, i.e., the transmission, execution, and propagation latencies of each task, into account.

Instructions:

Notes on the simulation files:

1.   Number of single-antenna users, which is equal to the number of tasks

2.   Maximum acceptable latency of tasks

3.   Ratio of RAN latency to the maximum acceptable latency

5.   Data size of each task

After receiving the parameters, DTO.m executes the disjoint method and returns the outputs as in the following:

1.      Acceptance Ratio

3.   Propagation latency of all tasks

4.   Execution latency of all tasks

1.      Number of single-antenna users, which is equal to the number of tasks.

2.   Maximum acceptable latency of tasks

4.   Data size of each task.

After receiving the parameters, JTO.m executes the disjoint method and returns the outputs as in the following:

1.      Acceptance Ratio

3.   Propagation latency of all tasks

4.   Execution latency of all tasks.

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## In-browser and network traffic based web response time measurements

This repository contains the results of 30 public Internet browsing experiments, from a computer at the campus network of the Public University of Navarre, out of which 20 used plaintext HTTP browsing, while 10 used HTTPS. We present both the original data sources in the form of network packet traces and HAR waterfalls, as well as the processed results formatted as line-based text files.

Instructions:

Each experiment consisted of a Selenium-automated web browser (Google Chrome 80.0) visiting a set of predefined web sites, with all caching options disabled. Both network packet traffic traces and in-browser measurements were collected. The network measurements were collected using tcpdump running at the client, while in-browser measurements were collected through the HAR Export Trigger extension. We have uploaded both sets of files.

The sets of websites for HTTP and HTTPS experiments are different, as modern web sites usually support HTTPS but not HTTP. The HTTPS set was obtained by collecting the top 2000 web sites from the Alexa Top Ranking. The HTTP set is the subset of these top 2000 websites, those that supported plain-text HTTP. To extend the amount of measurements of plain HTTP traffic, each of these websites was crawled, following the embedded ‘http://’ links.

For each web resource requested by the browser, we computed the time elapsed between the HTTP request being sent and the response being fully received; this is referred to as the resource's response time. Each response time obtained, along with the URL for that resource, and the timestamp at which the request was made, is referred to as a sample. These samples are obtained from the browser measurements and from network traffic. For the HTTPS experiments, the network data was decrypted using the ephemeral per-session encryption keys generated by the web browser. The files containing these keys have also been uploaded.

A number of resources are requested more than once during each test, such as cascade style sheets or images. Although we deactivated the cache, the browser still sometimes reported some resources as requested with a false response time of zero, since the request is never issued to the server but obtained from a cache. Also, a small number of requests trigger an exception in the browser, which prevents data being collected at the client side, although the request and response are present in the network traffic. These behaviours complicate one-to-one comparisons between network and in-browser measurements because a different number of response times for a specific resource may be found in the network traffic and in the browser report. We exported to text files only the first response time seen for each resource with a unique URL. This filtering removes false measurements reported by the browser. In case this filtering is not desired, all the data can be obtained from the pcap and HAR files uploaded.

The dataset contains the original PCAP and HAR files, and also the post-processed files obtained from them. The raw data is contained in the raw_http.zip and raw_https.zip files, while the post-processed files are contained in the data.zip file. Inside the data.zip archive there are two directories, corresponding to the HTTP and HTTPS experiments respectively.

Both raw data archives contain files named X.pcap and subdirectories named X_har (with X being the name of each individual experiment), corresponding to the data gathered from network traces and in-browser measurements respectively. Inside each X_har directory, a .har file is stored for each visited site with the full download waterfall. Additionally, decryption keys for the HTTPS experiments are provided, under the name of X.key.

The data.zip archive contains three files for each experiment, amounting to a total of 60 and 30 files for HTTP and HTTPS respectively.

The three files describing each experiment contain line-based text data, and are named X_network_tresp.txt, X_browser_tresp.txt and X_conn_info.txt, with X being the name of each individual experiment. The first two files contain, on each line, space-separated fields describing a single request-response sample. X_network_tresp.txt contains the information gathered from network traces, while X_browser_tresp.txt was obtained from browser instrumentation. On the other hand, X_conn_info.txt contains, on each line, space-separated fields related to each TCP connection present during the experiment, obtained through network traces.

The connections in X_conn_info.txt and the samples in X_network_tresp.txt are associated through a unique connection ID field present in each line in both files. Note that this is a one-to-many relationship, meaning that a connection ID is associated to a single TCP stream (i.e. line in X_conn_info.txt), but one or more samples (i.e. lines in X_network_tresp.txt).

We describe below the line format for each file. This information is included as well in the "format.txt" file, located on the top level directory of the compressed archive.

X_conn_info.txt:

Connection ID

RTT (milliseconds)

Number of retransmissions

Number of sequence holes

Number of data packets, client to server

Number of data packets, server to client

X_network_tresp.txt:

Request timestamp (seconds)

Response time (seconds)

Requested URL

Response size (bytes)

Connection ID

X_browser_tresp.txt:

Request timestamp (seconds)

Response time (seconds)

Requested URL

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237 Views

## The Space-Integrated Ground Information Intelligent Network traffic data

In the simulation experiment, OPNET simulation software was used to simulate the Space-Integrated Ground Information Intelligent Network environment to obtain the data. A satellite network composed of six satellites was used to simulate the data flow formed by the ground communication traffic through this network. The simulated time was one week and the traffic data information of a node was randomly collected for analysis and learning.

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135 Views

## AI-Based Secure NOMA and Cognitive Radio enabled Green Communications: Channel State Information and Battery Value Uncertainties

In this paper, the security-aware robust resource allocation in energy harvesting cognitive radio networks is considered with cooperation between two transmitters while there are uncertainties in channel gains and battery energy value. To be specific, the primary access point harvests energy from the green resource and uses time switching protocol to send the energy and data towards the secondary access point (SAP).

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264 Views

## Perfect and Related Codes

This dataset is devoted to 1-perfect codes. Currently, it is mainly focused on the concatenated ternary perfect codes, but there is also an additional content, see (3+) below, (1-2) This dataset contains all inequivalent concatenated ternary 1-perfect codes of length 13. Additionally, it contains some components necessary to obtain such concatenated codes, namely, collections of disjoint ternary distance-3 Reed-Muller-like codes of length 9, see p.(1) below.

Instructions:

(1) Each text file in the dataset archived in H93.zip keeps collections of disjoint $(9,3^6,3)_3$ codes, coded in the following manner (the file names are coded with equivalence classes of the corresponding codes, from 0 to 3; some files are empty, for example, "n00123", because two codes from equivalence class 0, one code from class 1, one code from class 2, and one code from class 3 cannot be packed in a disjoint manner). Each line corresponds to one representative of such a collection. If the number of codes is M, then the line contains M records like "N T PPPPPPPPP". "N" denotes the number of a code in the list 7codes.py of seven permutably inequivalent codes (note that "permutably equivalent" is not the same as "equivalent"; only first 4 are inequivalent in the sense of $Aut(H(9,3))$, the last 3 are equivaleng to the code number 3). "T" is the number of a translation vector in the list 0: 000000000, 1: 120000000, 2: 102000000, 3: 100200000, 4: 100020000, 5: 100002000, 6: 100000200, 7: 100000020, 8: 100000002, and "PPPPPPPPP" is a coordinate permutation. To reconstruct the corresponding code, one should take the code number "N" from 7codes.py, apply the coordinate permutation "PPPPPPPPP" in the following manner: $$(x_0,x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8) \to (x_{P^{-1}(0)},x_{P^{-1}(1)},x_{P^{-1}(2)},x_{P^{-1}(3)},x_{P^{-1}(4)},x_{P^{-1}(5)},x_{P^{-1}(6)},x_{P^{-1}(7)},x_{P^{-1}(8)}),$$ and add the translation vector number "T" to all codewords.  After M records about the codes, each line contains: (ii) the record "-A", where A is the order of the automorphism group of the collection of codes; (iii) the record "+R", where 6+R is the rank of the collection, that is, the dimension of the union of the (non-translated) codes (the minimum rank is 6 and the maximum is 8=9-1 because of the all-parity check, so R is 0, 1, ro 2); (iv) the record "~" or "|", where "~" means that the collection can be continued to a collection of M+1 codes and "|" means that no $(9,3^6,3)_3$ code (satisfying the all-parity check) can be added to the complement (this can only happen when M is 6 or 9).

(2) The file "concat13" (arxived to concat13.xz; the original file is 1.7GiB) contains the list of all inequivalent concatenated ternary 1-perfect codes of length 13 in the following form. Each length-9 partition in the form described in p.(1) (the order of records is (i), (iii), (ii), and the last record contains generators of the group of permutations of 9 codes induced by the automorphism group of the partition) is followed by the list of concatenated codes obtained with this partitions. The records for each code are the following:

(i) "A" or "B" indicate the partition of $H(4,3)$ into 1-perfect codes.

(ii) The next 8 numbers show the permutation $p$ of the codes of the partition.

(iii) The record "+R" means that the rank of the code is 10+R (10, 11, or 12).

(iv) The record ":K" means that the dimension of the kernel of the code is K (the kernel is the set of all periods of the code, that is, the translations that send the code to itself).

(v) The record "/S" means that the code can be represented as a concatenated code in S different ways (partitions 9+4 of coordinates); for the rank-10 Hamming code this number is 13; for rank-12 codes it is 1; for rank-11 codes it is from 1 to 4.

(vi) The record "-A*K" means that the order of the automorphism group of the code is A*K, while A is the number of authomorphisms that keep the coordinate partition 9+4, i.e., the current concatenation structure (if (iv) is "/1", then K necessarily equals 1).

If the record (v) is "/1" or "/13", then there are no more records, because this guarantees that an equivalent code cannot be obtained by concatenation in another way (in particular, from a different partition).

If the record (v) is "/2", "/3", or "/4", then such codes were checked for isomorphism, and one more record is given:

(vii) This record contains a symbol "n" or "o" and some number. "n" means that the code is new, and it is not equivalent to any of the codes above, and the following number is its unique number (among the codes with symbol "n" in the record). "o" means that the code is not new and it is equivalent to one of the codes above, namely, the one with symbol "n" followed by the same number.

So (ATTENTION!!!), some lines of the database correspond to equivalent codes; to read only inequivalent codes, one should ignore the lines with symbol "o".

(3) The file all_hamming15.zip contains the database file (to reduce the size, the xz-compression was applied) with 64864800 check matrices of [15,11,3] different Hamming codes, the database file with the 232 inequivalent pairs of disjoint Hamming codes, the file readme.txt with instructions and some scripts.

(4) The file 3ary-perfect-fullrank.zip contains examples of ternary 1-perfect codes of length 13 (i.e.,  [13,59049,3]_3 codes), rank 13 (i.e., full-rank), and kernel dimension from 3 to 7. Each code is kept in a separate file where the codewords are listed in the ternary-vector form.

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## Control Accuracy Measurement of a Mixed Reality Application for Digital Twin based Crane Operation

The dataset corresponds the measurement that was implemented on the control accuracy of a mixed reality application for digital twin based crane.

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# R.D. Sharma solutions 10 Mathematics chapter 4 Triangles

## Chapter 4 - Triangles

#### Page 0

Fill in the blanks using correct word given in the brackets:−

All circles are __________. (congruent, similar)

Fill in the blanks using correct word given in the brackets:−

All squares are __________. (similar, congruent)

Fill in the blanks using correct word given in the brackets:−

All __________ triangles are similar. (isosceles, equilateral)

Two triangles are similar, if their corresponding angles are .......... (proportional, equal)

Two triangles are similar, if their corresponding sides are .......... (proportional, equal)

Fill in the blanks using correct word given in the brackets:−

Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)

Write the truth value (T/F) of each of the following statement

Any two similar figures are congruent.

Write the truth value (T/F) of each of the following statement

Any two congruent figures are similar.

Write the truth value (T/F) of each of the following statement

Two polygons are similar, if their corresponding sides are proportional.

Write the truth value (T/F) of each of the following statement

Two polygons are similar if their corresponding angles are proportional.

Write the truth value (T/F) of each of the following statement

Two triangles are similar if their corresponding sides are proportional.

Write the truth value (T/F) of each of the following statement

Two triangles are similar if their corresponding angles are proportional.

#### Page 0

In ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC

If AD = 6 cm, DB = 9 cm and AE = 8 cm, find AC.

In ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC

If "AD"/"DB"=3/4 and AC = 15 cm, find AE

In ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC

If "AD"/"DB"=2/3 and AC = 18 cm, find AE

In ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC

If AD = 4, AE = 8, DB = x – 4, and EC = 3x – 19, find x.

In ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC

If AD = 8cm, AB = 12 cm and AE = 12 cm, find CE.

In ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC

If AD = 4 cm, DB = 4.5 cm and AE = 8 cm, find AC.

In ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC

If AD = 2 cm, AB = 6 cm and AC = 9 cm, find AE.

In ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC

If "AD"/"BD"=4/5 and EC = 2.5 cm, find AE

In ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC

If AD = 8x − 7, DB = 5x − 3, AE = 4x − 3 and EC = (3x − 1), find the value of x.

In ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC

If AD = 4x − 3, AE = 8x – 7, BD = 3x – 1 and CE = 5x − 3, find the volume of x.

In ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC

If AD = 2.5 cm, BD = 3.0 cm and AE = 3.75 cm, find the length of AC.

In a ΔABC, D and E are points on the sides AB and AC respectively. For  the following case show that DE || BC

AB = 2cm, AD = 8cm, AE = 12 cm and AC = l8cm.

In a ΔABC, D and E are points on the sides AB and AC respectively. For  the following case show that DE || BC

AB = 5.6cm, AD = 1.4cm, AC= 7.2 cm and AE = 1.8 cm.

In a ΔABC, D and E are points on the sides AB and AC respectively. For  the following case show that DE || BC

AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm and AE = 2.8 cm.

In a ΔABC, D and E are points on the sides AB and AC respectively. For  the following case show that DE || BC

AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and EC = 5.5 cm.

In a ΔABC, P and Q are points on sides AB and AC respectively, such that PQ || BC. If AP = 2.4 cm, AQ = 2 cm, QC = 3 cm and BC = 6 cm, find AB and PQ.

In a ΔABC, D and E are points on AB and AC respectively such that DE || BC. If AD = 2.4cm, AE = 3.2 cm, DE = 2cm and BC = 5 cm, find BD and CE.

In below Fig., state if PQ || EF.

M and N are points on the sides PQ and PR respectively of a ΔPQR. For the following case, state whether MN || QR

PM = 4cm, QM = 4.5 cm, PN = 4 cm and NR = 4.5 cm

In three line segments OA, OB, and OC, points L, M, N respectively are so chosen that LM || AB and MN || BC but neither of L, M, N nor of A, B, C are collinear. Show that LN ||AC.

If D and E are points on sides AB and AC respectively of a ΔABC such that DE || BC and BD = CE. Prove that ΔABC is isosceles.

#### Page 0

In a ΔABC, AD is the bisector of ∠A, meeting side BC at D.

If BD = 2.5cm, AB = 5cm and AC = 4.2cm, find DC.

In a ΔABC, AD is the bisector of ∠A, meeting side BC at D.

If BD = 2cm, AB = 5cm and DC = 3cm, find AC.

In a ΔABC, AD is the bisector of ∠A, meeting side BC at D.

If AB = 3.5 cm, AC = 4.2 cm and DC = 2.8 cm, find BD.

In a ΔABC, AD is the bisector of ∠A, meeting side BC at D.

If AB = 10 cm, AC =14 cm and BC =6 cm, find BD and DC.

In a ΔABC, AD is the bisector of ∠A, meeting side BC at D.

If AC = 4.2 cm, DC = 6 cm and 10 cm, find AB

In a ΔABC, AD is the bisector of ∠A, meeting side BC at D.

If AB = 5.6 cm, AC = 6cm and DC = 3cm, find BC.

In a ΔABC, AD is the bisector of ∠A, meeting side BC at D.

If AD = 5.6 cm, BC = 6cm and BD = 3.2 cm, find AC.

In a ΔABC, AD is the bisector of ∠A, meeting side BC at D.

If AB = 10cm, AC = 6 cm and BC = 12 cm, find BD and DC.

In the following Figure, AE is the bisector of the exterior ∠CAD meeting BC produced in E. If AB = 10cm, AC = 6cm and BC = 12 cm, find CE.

In the following Figure, ΔABC is a triangle such that "AB"/"AC"="BD"/"DC", ∠B = 70°, ∠C = 50°. Find ∠BAD.

In ΔABC , if ∠1 = ∠2, prove that "AB"/"AC"="BD"/"DC"

D, E and F are the points on sides BC, CA and AB respectively of ΔABC such that AD bisects ∠A, BE bisects ∠B and CF bisects ∠C. If AB = 5 cm, BC = 8 cm and CA = 4 cm, determine AP, CE and BD.

In the following figure, check whether AD is the bisector of ∠A of ∆ABC in the following case:

AB = 5 cm, AC = 10 cm, BD = 1.5 cm and CD = 3.5 cm

In the following figure, check whether AD is the bisector of ∠A of ∆ABC in the following case:

AB = 4 cm, AC = 6 cm, BD = 1.6 cm and CD = 2.4 cm

In the following figure, check whether AD is the bisector of ∠A of ∆ABC in the following case:

AB = 8 cm, AC = 24 cm, BD = 6 cm and BC = 24 cm

In the following figure, check whether AD is the bisector of ∠A of ∆ABC in the following case:

AB = 6 cm, AC = 8 cm, BD = 1.5 cm and CD = 2 cm

In the following figure, check whether AD is the bisector of ∠A of ∆ABC in the following case:

AB =  5 cm, AC = 12 cm, BD = 2.5 cm and BC = 9 cm

#### Page 0

In below figure, If AB || CD, find the value of x.

In the below figure, If AB || CD, find the value of x.

In below figure, AB || CD. If OA = 3x – 19, OB = x – 4, OC = x – 3 and OD = 4, find x.

#### Page 0

In figure, ∆ACB ~ ∆APQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm, AP = 2.8 cm, find CA and AQ.

A vertical stick 10 cm long casts a shadow 8 cm long. At the same time a shadow 30 m long. Determine the height of the tower.

In the following figure, AB || QR. Find the length of PB.

In the following figure, XY || BC. Find the length of XY.

In a right angled triangle with sides a and b and hypotenuse c, the altitude drawn on the hypotenuse is x. Prove that ab = cx.

In the following Figure, ∠ABC = 90° and BD ⊥ AC. If BD = 8 cm and AD = 4 cm, find CD.

In the following Figure, ∠ABC = 90° and BD ⊥ AC. If AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm, find BC.

In the following Figure, DE || BC such that AE = (1/4) AC. If AB = 6 cm, find AD.

In the following figure, PA, QB and RC are each perpendicular to AC. Prove that 1/x+1/z+1/y

In below figure, ∠A = ∠CED, Prove that ΔCAB ~ ΔCED. Also, find the value of x.

The perimeters of two similar triangles are 25 cm and 15 cm respectively. If one side of first triangle is 9 cm, what is the corresponding side of the other triangle?

In ΔABC and ΔDEF, it is being given that: AB = 5 cm, BC = 4 cm and CA = 4.2 cm; DE=10cm, EF = 8 cm and FD = 8.4 cm. If AL ⊥ BC and DM ⊥ EF, find AL: DM.

D and E are the points on the sides AB and AC respectively of a ΔABC such that: AD = 8 cm, DB = 12 cm, AE = 6 cm and CE = 9 cm. Prove that BC = 5/2 DE.

D is the mid-point of side BC of a ΔABC. AD is bisected at the point E and BE produced cuts AC at the point X. Prove that BE : EX = 3 : 1

ABCD is a parallelogram and APQ is a straight line meeting BC at P and DC produced at Q. Prove that the rectangle obtained by BP and DQ is equal to the AB and BC.

In ΔABC, AL and CM are the perpendiculars from the vertices A and C to BC and AB respectively. If AL and CM intersect at O, prove that:

(i) ΔOMA and ΔOLC

(ii) "OA"/"OC"="OM"/"OL"

In Fig below we have AB || CD || EF. If AB = 6 cm, CD = x cm, EF = 10 cm, BD = 4 cm and DE = y cm, calculate the values of x and y.

ABCD is a quadrilateral in which AD = BC. If P, Q, R, S be the mid-points of AB, AC, CD and BD respectively, show that PQRS is a rhombus.

In Figure below, if AB ⊥ BC, DC ⊥ BC and DE ⊥ AC, Prove that Δ CED ~ ABC.

Given: AB ⊥ BC, DC ⊥ BC and DE ⊥ AC

To prove: ΔCED ~ ΔABC

Proof:

∠BAC + ∠BCA = 90°                  …(i) [By angle sum property]

And, ∠BCA + ∠ECD = 90°             …(ii) [DC ⊥ BC given]

Compare equation (i) and (ii)

∠BAC = ∠ECD                           …(iii)

In ΔCED and ΔABC

∠CED = ∠ABC                        [Each 90°]

∠ECD = ∠BAC                         [From (iii)]

Then, ΔCED ~ ΔABC               [By AA similarity]

In an isosceles ΔABC, the base AB is produced both the ways to P and Q such that AP × BQ = AC2. Prove that ΔAPC ~ ΔBCQ.

A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2m/sec. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using similarity criterion for two triangles, show that "OA"/"OC"="OB"/"OD"

In the following figure, ABC and AMP are two right triangles, right angled at B and M respectively, prove that:

ΔABC ~ ΔAMP

 (CA)/(PA) = (BC)/(MP)

A vertical stick of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

In below Figure, ΔABC is right angled at C and DE ⊥ AB. Prove that ΔABC ~ ΔADE and Hence find the lengths of AE and DE.

#### Page 0

If ∆ABC ~ ∆DEF such that area of ∆ABC is 16cm2 and the area of ∆DEF is 25cm2 and BC = 2.3 cm. Find the length of EF.

If ∆ABC is similar to ∆DEF such that ∆DEF = 64 cm2 , DE = 5.1 cm and area of ∆ABC = 9 cm2 . Determine the area of AB

Triangles ABC and DEF are similar If AC = 19cm and DF = 8 cm, find the ratio of the area of two triangles.

Triangles ABC and DEF are similar If area (ΔABC) = 36 cm2, area (ΔDEF) = 64 cm2 and DE = 6.2 cm, find AB.

Triangles ABC and DEF are similar If AB = 1.2 cm and DE = 1.4 cm, find the ratio of the areas of ΔABC and ΔDEF.

In figure below ΔACB ~ ΔAPQ. If BC = 10 cm, PQ = 5 cm, BA = 6.5 cm and AP = 2.8 cm,
find CA and AQ. Also, find the area (ΔACB): area (ΔAPQ)

The areas of two similar triangles are 81 cm2 and 49 cm2 respectively. Find the ratio of their corresponding heights. What is the ratio of their corresponding medians?

The areas of two similar triangles are 169 cm2 and 121 cm2 respectively. If the longest side of the larger triangle is 26 cm, find the longest side of the smaller triangle.

Two isosceles triangles have equal vertical angles and their areas are in the ratio 36 : 25. Find the ratio of their corresponding heights.

The areas of two similar triangles are 25 cm2 and 36 cm2 respectively. If the altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other.

The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas.

ABC is a triangle in which ∠A =90°, AN⊥ BC, BC = 12 cm and AC = 5cm. Find the ratio of the areas of ΔANC and ΔABC.

In Figure, DE || BC If DE = 4 cm, BC = 6 cm and Area (ΔADE) = 16 cm2, find the area of ΔABC.

In Figure, DE || BC If DE = 4cm, BC = 8 cm and Area (ΔADE) = 25 cm2, find the area of ΔABC.

In the given figure, DE || BC and DE : BC = 3 : 5. Calculate the ratio of the areas of ∆ADE and the trapezium BCED

In ΔABC, D and E are the mid-points of AB and AC respectively. Find the ratio of the areas of ΔADE and ΔABC

In Figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that (ar(ABC))/(ar(DBC)) = (AO)/(DO)

ABCD is a trapezium in which AB || CD. The diagonals AC and BD intersect at O. Prove that: (i) ΔAOB and ΔCOD (ii) If OA = 6 cm, OC = 8 cm,

Find:(a)  ("area"(triangleAOB))/("area"(triangleCOD))

(b)  ("area"(triangleAOD))/("area"(triangleCOD))

In ABC, P divides the side AB such that AP : PB = 1 : 2. Q is a point in AC such that PQ || BC. Find the ratio of the areas of ΔAPQ and trapezium BPQC.

The areas of two similar triangles are 100 cm2 and 49 cm2 respectively. If the altitude the bigger triangle is 5 cm, find the corresponding altitude of the other.

The areas of two similar triangles are 121 cm2 and 64 cm2 respectively. If the median of the first triangle is 12.1 cm, find the corresponding median of the other.

If ΔABC ~ ΔDEF such that AB = 5 cm, area (ΔABC) = 20 cm2 and area (ΔDEF) = 45 cm2, determine DE.

In ΔABC, PQ is a line segment intersecting AB at P and AC at Q such that PQ || BC and PQ divides ΔABC into two parts equal in area. Find "BP"/"AB"

The areas of two similar triangles ABC and PQR are in the ratio 9:16. If BC = 4.5 cm, find the length of QR.

ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 m, prove that area of ΔAPQ is one- sixteenth of the area of ABC.

If D is a point on the side AB of ΔABC such that AD : DB = 3.2 and E is a Point on BC such that DE || AC. Find the ratio of areas of ΔABC and ΔBDE.

If ΔABC and ΔBDE are equilateral triangles, where D is the mid-point of BC, find the ratio of areas of ΔABC and ΔBDE.

AD is an altitude of an equilateral triangle ABC. On AD as base, another equilateral triangle ADE is constructed. Prove that Area (ΔADE): Area (ΔABC) = 3: 4

#### Page 0

If the sides of a triangle are 3 cm, 4 cm, and 6 cm long, determine whether the triangle is a right-angled triangle.

If the sides of a triangle are 3 cm, 4 cm, and 6 cm long, determine whether the triangle is a right-angled triangle.

The sides of triangle is given below. Determine it is right triangle or not.

a = 7 cm, b = 24 cm and c = 25 cm

The sides of triangle is given below. Determine it is right triangle or not.

a = 7 cm, b = 24 cm and c = 25 cm

The sides of triangle is given below. Determine it is right triangle or not.

a = 9 cm, b = l6 cm and c = 18 cm

The sides of triangle is given below. Determine it is right triangle or not.

a = 9 cm, b = l6 cm and c = 18 cm

The sides of triangle is given below. Determine it is right triangle or not.

a = 1.6 cm, b = 3.8 cm and c = 4 cm

The sides of triangle is given below. Determine it is right triangle or not.

a = 1.6 cm, b = 3.8 cm and c = 4 cm

The sides of triangle is given below. Determine it is right triangle or not.

a = 8 cm, b = 10 cm and c = 6 cm

The sides of triangle is given below. Determine it is right triangle or not.

a = 8 cm, b = 10 cm and c = 6 cm

A man goes 15 metres due west and then 8 metres due north. How far is he from the starting point?

A man goes 15 metres due west and then 8 metres due north. How far is he from the starting point?

A ladder 17 m long reaches a window of a building 15 m above the ground. Find the distance of the foot of the ladder from the building.

A ladder 17 m long reaches a window of a building 15 m above the ground. Find the distance of the foot of the ladder from the building.

Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.

Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.

In an isosceles triangle ABC, AB = AC = 25 cm, BC = 14 cm. Calculate the altitude from A on BC.

In an isosceles triangle ABC, AB = AC = 25 cm, BC = 14 cm. Calculate the altitude from A on BC.

The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its tip reach?

The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its tip reach?

Two poles of height 9 m and 14 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.

Two poles of height 9 m and 14 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.

Using Pythagoras theorem determine the length of AD in terms of b and c shown in Figure.

Using Pythagoras theorem determine the length of AD in terms of b and c shown in Figure.

A triangle has sides 5 cm, 12 cm and 13 cm. Find the length to one decimal place, of the perpendicular from the opposite vertex to the side whose length is 13 cm.

A triangle has sides 5 cm, 12 cm and 13 cm. Find the length to one decimal place, of the perpendicular from the opposite vertex to the side whose length is 13 cm.

ABCD is a square. F is the mid-point of AB. BE is one third of BC. If the area of ΔFBE = 108 cm2, find the length of AC.

ABCD is a square. F is the mid-point of AB. BE is one third of BC. If the area of ΔFBE = 108 cm2, find the length of AC.

In an isosceles triangle ABC, if AB = AC = 13 cm and the altitude from A on BC is 5 cm, find BC.

In an isosceles triangle ABC, if AB = AC = 13 cm and the altitude from A on BC is 5 cm, find BC.

In a ΔABC, AB = BC = CA = 2a and AD ⊥ BC. Prove that

(i) AD = asqrt3

(ii) Area (ΔABC) = sqrt3 a2

In a ΔABC, AB = BC = CA = 2a and AD ⊥ BC. Prove that

(i) AD = asqrt3

(ii) Area (ΔABC) = sqrt3 a2

The lengths of the diagonals of a rhombus are 24 cm and 10 cm. Find each side of the rhombus.

The lengths of the diagonals of a rhombus are 24 cm and 10 cm. Find each side of the rhombus.

Each side of a rhombus is 10 cm. If one of its diagonals is 16 cm find the length of the other diagonal.

Each side of a rhombus is 10 cm. If one of its diagonals is 16 cm find the length of the other diagonal.

In an acute-angled triangle, express a median in terms of its sides.

In an acute-angled triangle, express a median in terms of its sides.

Calculate the height of an equilateral triangle each of whose sides measures 12 cm.

Calculate the height of an equilateral triangle each of whose sides measures 12 cm.

In right-angled triangle ABC in which ∠C = 90°, if D is the mid-point of BC, prove that AB2 = 4AD2 − 3AC2.

In right-angled triangle ABC in which ∠C = 90°, if D is the mid-point of BC, prove that AB2 = 4AD2 − 3AC2.

In Figure, D is the mid-point of side BC and AE ⊥ BC. If BC = a, AC = b, AB = c, ED
= x, AD = p and AE = h, prove that:

(i) b^2 = p^2 + ax + a^2/4

(ii) c^2 = p^2 - ax + a^2/4

(iii) b^2 + c^2 = 2p^2 + a^2/2

In Figure, D is the mid-point of side BC and AE ⊥ BC. If BC = a, AC = b, AB = c, ED
= x, AD = p and AE = h, prove that:

(i) b^2 = p^2 + ax + a^2/4

(ii) c^2 = p^2 - ax + a^2/4

(iii) b^2 + c^2 = 2p^2 + a^2/2

In the given figure, ∠B < 90° and segment AD ⊥ BC, show that

(i) b= h+ a+ x- 2ax

(ii) b2 = a2 + c2 - 2ax

In the given figure, ∠B < 90° and segment AD ⊥ BC, show that

(i) b= h+ a+ x- 2ax

(ii) b2 = a2 + c2 - 2ax

In ∆ABC, ∠A is obtuse, PB ⊥ AC and QC ⊥ AB. Prove that:

(i) AB ✕ AQ = AC ✕ AP

(ii) BC2 = (AC ✕ CP + AB ✕ BQ)

In ∆ABC, ∠A is obtuse, PB ⊥ AC and QC ⊥ AB. Prove that:

(i) AB ✕ AQ = AC ✕ AP

(ii) BC2 = (AC ✕ CP + AB ✕ BQ)

In a right ∆ABC right-angled at C, if D is the mid-point of BC, prove that BC2 = 4(AD2 − AC2).

In a right ∆ABC right-angled at C, if D is the mid-point of BC, prove that BC2 = 4(AD2 − AC2).

In a quadrilateral ABCD, ∠B = 90°, AD2 = AB2 + BC2 + CD2, prove that ∠ACD = 90°.

In a quadrilateral ABCD, ∠B = 90°, AD2 = AB2 + BC2 + CD2, prove that ∠ACD = 90°.

In an equilateral ΔABC, AD ⊥ BC, prove that AD2 = 3BD2.

In an equilateral ΔABC, AD ⊥ BC, prove that AD2 = 3BD2.

∆ABD is a right triangle right-angled at A and AC ⊥ BD. Show that

(i) AB2 = BC x BD

(ii) AC2 = BC x DC

(iii) AD2 = BD x CD

(iv) "AB"^2/"AC"^2="BD"/"DC"

∆ABD is a right triangle right-angled at A and AC ⊥ BD. Show that

(i) AB2 = BC x BD

(ii) AC2 = BC x DC

(iii) AD2 = BD x CD

(iv) "AB"^2/"AC"^2="BD"/"DC"

A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

An aeroplane leaves an airport and flies due north at a speed of 1000km/hr. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km/hr. How far apart will be the two planes after 1 hours?

An aeroplane leaves an airport and flies due north at a speed of 1000km/hr. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km/hr. How far apart will be the two planes after 1 hours?

Determine whether the triangle having sides (a − 1) cm, 2sqrta cm and (a + 1) cm is a right-angled
triangle.

Determine whether the triangle having sides (a − 1) cm, 2sqrta cm and (a + 1) cm is a right-angled
triangle.

#### Extra questions

In ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC

If AD = x, DB = x − 2, AE = x + 2 and EC = x − 1, find the value of x.

S

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text:
# Thermal efficiency

In thermodynamics, the thermal efficiency (${\displaystyle \eta _{\rm {th}}}$) is a dimensionless performance measure of a device that uses thermal energy, such as an internal combustion engine, a steam turbine or a steam engine, a boiler, furnace, or a refrigerator for example. For a heat engine, thermal efficiency is the fraction of the energy added by heat (primary energy) that is converted to net work output (secondary energy). In the case of a refrigeration or heat pump cycle, thermal efficiency is the ratio of net heat output for heating, or removal for cooling, to energy input (the coefficient of performance).

## Overview

Output (mechanical) energy is always lower than input energy

In general, energy conversion efficiency is the ratio between the useful output of a device and the input, in energy terms. For thermal efficiency, the input, ${\displaystyle Q_{\rm {in}}}$, to the device is heat, or the heat-content of a fuel that is consumed. The desired output is mechanical work, ${\displaystyle W_{\rm {out}}}$, or heat, ${\displaystyle Q_{\rm {out}}}$, or possibly both. Because the input heat normally has a real financial cost, a memorable, generic definition of thermal efficiency is[1]

${\displaystyle \eta _{\rm {th}}\equiv {\frac {\text{benefit}}{\text{cost}}}.}$

From the first law of thermodynamics, the energy output cannot exceed the input, and by the second law of thermodynamics it cannot be equal in a non-ideal process, so

${\displaystyle 0\leq \eta _{\rm {th}}<1}$

When expressed as a percentage, the thermal efficiency must be between 0% and 100%. Efficiency is typically less than 100% because there are inefficiencies such as friction and heat loss that convert the energy into alternative forms. For example, a typical gasoline automobile engine operates at around 25% efficiency, and a large coal-fuelled electrical generating plant peaks at about 46%, advances in Formula 1 motorsport regulations have pushed teams to develop highly efficient power units which peak around 45–50% thermal efficiency. The largest diesel engine in the world peaks at 51.7%. In a combined cycle plant, thermal efficiencies are approaching 60%.[2] Such a real-world value may be used as a figure of merit for the device.

For engines where a fuel is burned there are two types of thermal efficiency: indicated thermal efficiency and brake thermal efficiency.[3] This efficiency is only appropriate when comparing similar types or similar devices.

For other systems the specifics of the calculations of efficiency vary but the non dimensional input is still the same. Efficiency = Output energy / input energy

## Heat engines

Heat engines transform thermal energy, or heat, Qin into mechanical energy, or work, Wout. They cannot do this task perfectly, so some of the input heat energy is not converted into work, but is dissipated as waste heat Qout into the environment

${\displaystyle Q_{in}=W_{\rm {out}}+Q_{\rm {out}}\,}$

The thermal efficiency of a heat engine is the percentage of heat energy that is transformed into work. Thermal efficiency is defined as

${\displaystyle \eta _{\rm {th}}\equiv {\frac {W_{\rm {out}}}{Q_{\rm {in}}}}={\frac {{Q_{\rm {in}}}-Q_{\rm {out}}}{Q_{\rm {in}}}}=1-{\frac {Q_{\rm {out}}}{Q_{\rm {in}}}}}$

The efficiency of even the best heat engines is low; usually below 50% and often far below. So the energy lost to the environment by heat engines is a major waste of energy resources. Since a large fraction of the fuels produced worldwide go to powering heat engines, perhaps up to half of the useful energy produced worldwide is wasted in engine inefficiency, although modern cogeneration, combined cycle and energy recycling schemes are beginning to use this heat for other purposes. This inefficiency can be attributed to three causes. There is an overall theoretical limit to the efficiency of any heat engine due to temperature, called the Carnot efficiency. Second, specific types of engines have lower limits on their efficiency due to the inherent irreversibility of the engine cycle they use. Thirdly, the nonideal behavior of real engines, such as mechanical friction and losses in the combustion process causes further efficiency losses.

### Carnot efficiency

The second law of thermodynamics puts a fundamental limit on the thermal efficiency of all heat engines. Even an ideal, frictionless engine can't convert anywhere near 100% of its input heat into work. The limiting factors are the temperature at which the heat enters the engine, ${\displaystyle T_{\rm {H}}\,}$, and the temperature of the environment into which the engine exhausts its waste heat, ${\displaystyle T_{\rm {C}}\,}$, measured in an absolute scale, such as the Kelvin or Rankine scale. From Carnot's theorem, for any engine working between these two temperatures:[4]

${\displaystyle \eta _{\rm {th}}\leq 1-{\frac {T_{\rm {C}}}{T_{\rm {H}}}}\,}$

This limiting value is called the Carnot cycle efficiency because it is the efficiency of an unattainable, ideal, reversible engine cycle called the Carnot cycle. No device converting heat into mechanical energy, regardless of its construction, can exceed this efficiency.

Examples of ${\displaystyle T_{\rm {H}}\,}$ are the temperature of hot steam entering the turbine of a steam power plant, or the temperature at which the fuel burns in an internal combustion engine. ${\displaystyle T_{\rm {C}}\,}$ is usually the ambient temperature where the engine is located, or the temperature of a lake or river into which the waste heat is discharged. For example, if an automobile engine burns gasoline at a temperature of ${\displaystyle T_{\rm {H}}=816^{\circ }{\text{C}}=1500^{\circ }{\text{F}}=1089{\text{K}}\,}$ and the ambient temperature is ${\displaystyle T_{\rm {C}}=21^{\circ }{\text{C}}=70^{\circ }{\text{F}}=294{\text{K}}\,}$, then its maximum possible efficiency is:

${\displaystyle \eta _{\rm {th}}\leq \left(1-{\frac {294K}{1089K}}\right)100\%=73.0\%}$

It can be seen that since ${\displaystyle T_{\rm {C}}\,}$ is fixed by the environment, the only way for a designer to increase the Carnot efficiency of an engine is to increase ${\displaystyle T_{\rm {H}}\,}$, the temperature at which the heat is added to the engine. The efficiency of ordinary heat engines also generally increases with operating temperature, and advanced structural materials that allow engines to operate at higher temperatures is an active area of research.

Due to the other causes detailed below, practical engines have efficiencies far below the Carnot limit. For example, the average automobile engine is less than 35% efficient.

Carnot's theorem applies to thermodynamic cycles, where thermal energy is converted to mechanical work. Devices that convert a fuel's chemical energy directly into electrical work, such as fuel cells, can exceed the Carnot efficiency. [5][6]

### Engine cycle efficiency

The Carnot cycle is reversible and thus represents the upper limit on efficiency of an engine cycle. Practical engine cycles are irreversible and thus have inherently lower efficiency than the Carnot efficiency when operated between the same temperatures ${\displaystyle T_{\rm {H}}\,}$ and ${\displaystyle T_{\rm {C}}\,}$. One of the factors determining efficiency is how heat is added to the working fluid in the cycle, and how it is removed. The Carnot cycle achieves maximum efficiency because all the heat is added to the working fluid at the maximum temperature ${\displaystyle T_{\rm {H}}\,}$, and removed at the minimum temperature ${\displaystyle T_{\rm {C}}\,}$. In contrast, in an internal combustion engine, the temperature of the fuel-air mixture in the cylinder is nowhere near its peak temperature as the fuel starts to burn, and only reaches the peak temperature as all the fuel is consumed, so the average temperature at which heat is added is lower, reducing efficiency.

An important parameter in the efficiency of combustion engines is the specific heat ratio of the air-fuel mixture, γ. This varies somewhat with the fuel, but is generally close to the air value of 1.4. This standard value is usually used in the engine cycle equations below, and when this approximation is made the cycle is called an air-standard cycle.

• Otto cycle: automobiles The Otto cycle is the name for the cycle used in spark-ignition internal combustion engines such as gasoline and hydrogen fuelled automobile engines. Its theoretical efficiency depends on the compression ratio r of the engine and the specific heat ratio γ of the gas in the combustion chamber.[4]:558
${\displaystyle \eta _{\rm {th}}=1-{\frac {1}{r^{\gamma -1}}}\,}$
Thus, the efficiency increases with the compression ratio. However the compression ratio of Otto cycle engines is limited by the need to prevent the uncontrolled combustion known as knocking. Modern engines have compression ratios in the range 8 to 11, resulting in ideal cycle efficiencies of 56% to 61%.
• Diesel cycle: trucks and trains In the Diesel cycle used in diesel truck and train engines, the fuel is ignited by compression in the cylinder. The efficiency of the Diesel cycle is dependent on r and γ like the Otto cycle, and also by the cutoff ratio, rc, which is the ratio of the cylinder volume at the beginning and end of the combustion process:[4]
${\displaystyle \eta _{\rm {th}}=1-{\frac {r^{1-\gamma }(r_{\rm {c}}^{\gamma }-1)}{\gamma (r_{\rm {c}}-1)}}\,}$
The Diesel cycle is less efficient than the Otto cycle when using the same compression ratio. However, practical Diesel engines are 30% - 35% more efficient than gasoline engines.[7] This is because, since the fuel is not introduced to the combustion chamber until it is required for ignition, the compression ratio is not limited by the need to avoid knocking, so higher ratios are used than in spark ignition engines.
• Rankine cycle: steam power plants The Rankine cycle is the cycle used in steam turbine power plants. The overwhelming majority of the world's electric power is produced with this cycle. Since the cycle's working fluid, water, changes from liquid to vapor and back during the cycle, their efficiencies depend on the thermodynamic properties of water. The thermal efficiency of modern steam turbine plants with reheat cycles can reach 47%, and in combined cycle plants, in which a steam turbine is powered by exhaust heat from a gas turbine, it can approach 60%.[4]
• Brayton cycle: gas turbines and jet engines The Brayton cycle is the cycle used in gas turbines and jet engines. It consists of a compressor that increases pressure of the incoming air, then fuel is continuously added to the flow and burned, and the hot exhaust gasses are expanded in a turbine. The efficiency depends largely on the ratio of the pressure inside the combustion chamber p2 to the pressure outside p1[4]
${\displaystyle \eta _{\rm {th}}=1-{\bigg (}{\frac {p_{2}}{p_{1}}}{\bigg )}^{\frac {1-\gamma }{\gamma }}\,}$

### Other inefficiencies

One should not confuse thermal efficiency with other efficiencies that are used when discussing engines. The above efficiency formulas are based on simple idealized mathematical models of engines, with no friction and working fluids that obey simple thermodynamic rules called the ideal gas law. Real engines have many departures from ideal behavior that waste energy, reducing actual efficiencies below the theoretical values given above. Examples are:

• friction of moving parts
• inefficient combustion
• heat loss from the combustion chamber
• departure of the working fluid from the thermodynamic properties of an ideal gas
• aerodynamic drag of air moving through the engine
• energy used by auxiliary equipment like oil and water pumps.
• inefficient compressors and turbines
• imperfect valve timing

These factors may be accounted when analyzing thermodynamic cycles, however discussion of how to do so is outside the scope of this article.

## Energy conversion

For a device that converts energy from another form into thermal energy (such as an electric heater, boiler, or furnace), the thermal efficiency is

${\displaystyle \eta _{\rm {th}}\equiv {\frac {Q_{\rm {out}}}{Q_{\rm {in}}}}}$

where the ${\displaystyle Q}$ quantities are heat-equivalent values.

So, for a boiler that produces 210 kW (or 700,000 BTU/h) output for each 300 kW (or 1,000,000 BTU/h) heat-equivalent input, its thermal efficiency is 210/300 = 0.70, or 70%. This means that 30% of the energy is lost to the environment.

An electric resistance heater has a thermal efficiency close to 100%.[8] When comparing heating units, such as a highly efficient electric resistance heater to an 80% efficient natural gas-fuelled furnace, an economic analysis is needed to determine the most cost-effective choice.

### Effects of fuel heating value

The heating value of a fuel is the amount of heat released during an exothermic reaction (e.g., combustion) and is a characteristic of each substance. It is measured in units of energy per unit of the substance, usually mass, such as: kJ/kg, J/mol.

The heating value for fuels is expressed as the HHV, LHV, or GHV to distinguish treatment of the heat of phase changes:

• Higher heating value (HHV) is determined by bringing all the products of combustion back to the original pre-combustion temperature, and in particular condensing any vapor produced. This is the same as the thermodynamic heat of combustion.
• Lower heating value (LHV) (or net calorific value) is determined by subtracting the heat of vaporization of the water vapor from the higher heating value. The energy required to vaporize the water therefore is not realized as heat.
• Gross heating value accounts for water in the exhaust leaving as vapor, and includes liquid water in the fuel prior to combustion. This value is important for fuels like wood or coal, which will usually contain some amount of water prior to burning.

Which definition of heating value is being used significantly affects any quoted efficiency. Not stating whether an efficiency is HHV or LHV renders such numbers very misleading.

## Heat pumps and refrigerators

Heat pumps, refrigerators and air conditioners use work to move heat from a colder to a warmer place, so their function is the opposite of a heat engine. The work energy (Win) that is applied to them is converted into heat, and the sum of this energy and the heat energy that is moved from the cold reservoir (QC) is equal to the total heat energy added to the hot reservoir (QH)

${\displaystyle Q_{\rm {H}}=Q_{\rm {C}}+W_{\rm {in}}\,}$

Their efficiency is measured by a coefficient of performance (COP). Heat pumps are measured by the efficiency with which they add heat to the hot reservoir, COPheating; refrigerators and air conditioners by the efficiency with which they remove heat from the cold interior, COPcooling:

${\displaystyle \mathrm {COP} _{\mathrm {heating} }\equiv {\frac {Q_{\rm {H}}}{W_{\rm {in}}}}\,}$
${\displaystyle \mathrm {COP} _{\mathrm {cooling} }\equiv {\frac {Q_{\rm {C}}}{W_{\rm {in}}}}\,}$

The reason the term "coefficient of performance" is used instead of "efficiency" is that, since these devices are moving heat, not creating it, the amount of heat they move can be greater than the input work, so the COP can be greater than 1 (100%). Therefore, heat pumps can be a more efficient way of heating than simply converting the input work into heat, as in an electric heater or furnace.

Since they are heat engines, these devices are also limited by Carnot's theorem. The limiting value of the Carnot 'efficiency' for these processes, with the equality theoretically achievable only with an ideal 'reversible' cycle, is:

${\displaystyle \mathrm {COP} _{\mathrm {heating} }\leq {\frac {T_{\rm {H}}}{T_{\rm {H}}-T_{\rm {C}}}}=\mathrm {COP} _{\mathrm {heating,carnot} }}$
${\displaystyle \mathrm {COP} _{\mathrm {cooling} }\leq {\frac {T_{\rm {C}}}{T_{\rm {H}}-T_{\rm {C}}}}=\mathrm {COP} _{\mathrm {cooling,carnot} }}$

The same device used between the same temperatures is more efficient when considered as a heat pump than when considered as a refrigerator:

${\displaystyle \mathrm {COP} _{\mathrm {heating} }-\mathrm {COP} _{\mathrm {cooling} }=1\,}$

This is because when heating, the work used to run the device is converted to heat and adds to the desired effect, whereas if the desired effect is cooling the heat resulting from the input work is just an unwanted by-product. Sometimes, the term efficiency is used for the ratio of the achieved COP to the Carnot COP, which can not exceed 100%.[9]

## Energy efficiency

The 'thermal efficiency' is sometimes called the energy efficiency. In the United States, in everyday usage the SEER is the more common measure of energy efficiency for cooling devices, as well as for heat pumps when in their heating mode. For energy-conversion heating devices their peak steady-state thermal efficiency is often stated, e.g., 'this furnace is 90% efficient', but a more detailed measure of seasonal energy effectiveness is the annual fuel use efficiency (AFUE).[10]

### Heat exchangers

A counter flow heat exchanger is the most efficient type of heat exchanger in transferring heat energy from one circuit to the other. However, for a more complete picture of heat exchanger efficiency, exergetic considerations must be taken into account. Thermal efficiencies of an internal combustion engine are typically higher than that of external combustion engines.

## References

1. Fundamentals of Engineering Thermodynamics, by Howell and Buckius, McGraw-Hill, New York, 1987
2. GE Power’s H Series Turbine
3. The Internal Combustion Engine in Theory and Practice: Vol. 1 - 2nd Edition, Revised, MIT Press, 1985, Charles Fayette Taylor - Equation 1-4, page 9
4. Holman, Jack P. (1980). Thermodynamics. New York: McGraw-Hill. pp. 217. ISBN 0-07-029625-1.
5. Sharma, B. K. (1997). Electro Chemistry, 5th Ed. Krishna Prakashan Media. pp. E-213. ISBN 8185842965.
6. Winterbone, D.; Ali Turan (1996). Advanced Thermodynamics for Engineers. Butterworth-Heinemann. p. 345. ISBN 0080523366.
7. "Where does the energy go?". Advanced technologies and energy efficiency, Fuel Economy Guide. US Dept. of Energy. 2009. Retrieved 2009-12-02.
8. "Energy Saver - Department of Energy". www.energysavers.gov.
9. "Coefficient of Performance". Industrial Heat Pumps. Retrieved 2018-11-08.
10. HVAC Systems and Equipment volume of the ASHRAE Handbook, ASHRAE, Inc., Atlanta, GA, US, 2004


text:
# How to predict the Nikkei 225 at the end of 2022 in Excel

Here comes the new year, let’s do our annual forecast of the Nikkei 225 closing price.

The Figure 1 gives an overview of the stock price trends over the 72 years from 1950 to last year 2021. Please note that the vertical axis is logarithmic.

Figure 1

##### Nikkei 225 Index – 67 Year Historical Chart | MacroTrends

Step 1

Focusing on the daily rate of change of the stock price over the 72 years, classify the number of days that the stock price has risen and fallen, and the respective rate of change into the following four patterns.

Pattern A:  Number of days of rise > Number of days of fall,  Rate of rise > Rate of fall

Pattern B:  Number of days rising > Number of days falling,  Rate of rise < Rate of fall

Pattern C:  Number of days rising < Number of days falling,  Rate of rise > Rate of fall

Pattern D:  Number of days rising < Number of days falling,  Rate of rise < Rate of fall

Table 1 shows the changes over the past 72 years, categorized by the four patterns.

Table 1  Classification of expression patterns over the past 72 years

 Pattern Number of occurrences Average rate     of rise Average rate     of fall Average number of days of rise Average number of days of fall A 23 0.748% -0.680% 134 114 B 24 0.815% -0.886% 132 106 C 19 0.905% -0.843% 122 127 D 6 1.154% -1.195% 118 129 72

Step 2

Next, calculate the stock price to be reached at the end of 2022 for each pattern using the formula for stock price prediction.

Formula for predicting the stock price $x_{0}$ in $n$ days

$$x_{n}=x_{0} (1 + a)^p(1 – b)^q$$

$$(p + q = n, a, b > 0)$$.

$x_{0}$ is the initial value; if we are forecasting the closing price in 2022, we set the initial value to 28,791.71 yen, the value of the closing price in 2021. Let $a$ be the average rate of increase, $b$ the average rate of decrease, $p$ the number of days of increase, and $q$ the number of days of decrease.

For the derivation of the formula for stock price prediction, see “How accurate is the prediction of stock price one year from now? “ for more details.

Table 2 shows the result of the calculation using the data in Table 1. For example, there is a 40.9% probability that Pattern A will continue next year, in which case the closing price will be 35,910.72 yen (an increase of 24.7% over the previous year).

Table 2  Closing prices for 2022 predicted using the prediction formula

 Transition Pattern Number of Occurrences Probability of Occurrence Forecast of closing price  in 2022 Percentage change A → A 9 40.9% 35,910.72 24.7% A → B 6 27.3% 32,718.70 13.6% A → C 6 27.3% 29,268.90 1.7% A → D 1 4.5% 23,738.44 -17.6% 22 100.0%

Step 3

Finally, we check the above predictions against the probability distribution obtained by the other calculation method. While the stock price tries to increase exponentially, it always fluctuates up and down. The formula for this is

$$\frac{dx}{dt}=rx+vx\frac{dB(t)}{dt}$$

where where $x$ is the stock price, $r$ is the growth coefficient, $v$ is the fluctuation coefficient, and $B(t)$ is the daily random change. This equation represents the moment-to-moment changes in the stock price $x$. The accumulation of these changes is the trajectory of the stock price. The above equation is called a stochastic differential equation, and the calculation of the accumulation is called solving (integrating) the differential equation. The result is

$$x=x_{0}\ e^{(r-\frac{v^2}{2})t+vB(t)}$$

The equation tells us how much the stock price will be when time elapses by t. $x_{0}$ is the initial value. For more information on how to determine $r$ (growth coefficient), $v$ (fluctuation coefficient), and $B(t)$ (daily change), please see my April 21, 2020 blog, “When will the stock market recover from the Corona shock?” for more details. All calculations can be done in Excel.

Figure 4 shows the probabilistic distribution of the stock price when the initial value is the closing price in 2020. The orange line $µ$ is the center of the probability distribution, and the dashed lines show $1v, -1v, +2v, -2v…$. The larger the number, the lower the probability of falling into that range. Then, superimpose the four predictions above. We can see that they are all within the range of $pm 1v$ and there is no contradiction.

Figure 2  Probability distribution of Nikkei 225 and prediction of closing price at the end of 2022

The above is the forecast calculation. We will continue to check the forecast against the actual Nikkei 225 once a week this year!

The End


text:
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

era v0.3.1 is now available on CRAN:

install.packages("era")

I wrote about the basic concept of the package in my last post. It provides a representation of year-based time scales in R, built around two core classes: yr, a vector of years, and era, defining the parameters that relate those years to a moment in time. The idea is to make it easier to handle different era systems in a tidy analysis, and convert between them, as described in the introductory vignette:

library("tibble")
library("dplyr")

tribble(
~period,           ~start_ka,
"Late Holocene",   4.2,
"Mid Holocene",    8.326,
"Early Holocene",  11.7,
"Younger Dryas",   12.9,
"Bølling-Allerød", 14.7,
"Heinrich 1",      17.0
) %>%
mutate(start_ka = yr(start_ka, "ka")) %>%
mutate(start_bp = yr_transform(start_ka, era("BP")),
start_bce = yr_transform(start_ka, era("BCE")))
#> # A tibble: 6 x 4
#>   period          start_ka start_bp start_bce
#>
#> 1 Late Holocene     4.2 ka  4200 BP  2250 BCE
#> 2 Mid Holocene    8.326 ka  8326 BP  6376 BCE
#> 3 Early Holocene   11.7 ka 11700 BP  9750 BCE
#> 4 Younger Dryas    12.9 ka 12900 BP 10950 BCE
#> 5 Bølling-Allerød  14.7 ka 14700 BP 12750 BCE
#> 6 Heinrich 1         17 ka 17000 BP 15050 BCE

The major changes since the beta release (v0.2.0) flesh out the definition of an “era” so that a broader range of time scales and calendric year numbering systems can be represented as yr vectors:

• yr_transform() can now handle eras with different units.
• ‘Year units’ are now explicitly defined as the length of a year in days, with a new class era_year.
• Built-in definitions for lunar and solar hijra year numbering systems, and the Julian calendar, are now included (as examples of era systems that use different units).

These and other minor changes are described in full in the changelog. In the rest of this post, I want to explain the concepts behind this representation of years.

## Years as a real number time scale

era is concerned with year-based time scales; a measurement of the length of time elapsed between one instant and another, counted in years. This is the sense used by archaeologists when they say an object is about 5,000 years old, or geologists when they say the Holocene started 11.7 thousand years ago. It’s not quite the same as the everyday notion of a year, which is date (or part of a date), describing a span of time on a calendar.

To tie a set of numbers to a concrete measurement of time, we need to know:

• the epoch – the datum instant from which measurements are made, which could be a calendar epoch (e.g. the birth of Jesus in the Common Era) or some other conventionally-agreed point in time (e.g. the “Present” in radiometric dating);
• the year unit used;
• the direction in which years are counted, i.e. forwards or backwards from the epoch;
• the multiplicative scale used, when measurements are made in e.g. thousands of years rather than single years.

Together with two attributes used human-friendly printing (name and label), these four parameters comprise the era attribute of a yr vector in era. They also enable us to derive a generalised algorithm for transforming years between different era systems (yr_transform()).

Rather than something decomposable to a set of calendar dates (12 months, 365 days, etc.), yr vectors are treated as a real number scale with a year as its base unit. For example, it makes sense to think about both fractional years and negative years:

library("era")
frac_year <- function(x) {
as.numeric(format(x, "%Y")) + (as.numeric(format(x, "%j")) / 365.2425)
}

# It is now:
yr(frac_year(Sys.Date()), "CE")
#> # CE years :
#> [1] 2021.104
#> # Era: Common Era (CE): Gregorian years (365.2425 days), counted forwards from 0

# Julius Caesar was born:
yr_transform(yr(100, "BCE"), "CE")
#> # CE years :
#> [1] -100
#> # Era: Common Era (CE): Gregorian years (365.2425 days), counted forwards from 0

Unfortunately, a “year” is not a single or well-defined unit. era used the era_year class to represents different year units, defined as their average length in days. Although some units (e.g. the astronomical Julian year) can be precisely defined in this way, for most it is an approximation. For the most part we can get away with this because we are not concerned with exact dates, but it does have implications for the accuracy of some transformations, especially when they involve calendar eras.

## Time scales and calendars

In addition to “Before Present”-style eras used by palaeoscientists, era supports, and includes built-in definitions for, year numbering systems from contemporary and historic calendars. But since the package treats years as a time scale, not as dates, it can only represents the year numbering components of these calendars. If we were to implement the modern Iranian calendar as a Solar Hijri (SH) era, for example, we would only consider its:

• Epoch for year numbering – the day of Nowruz (the vernal equinox) in the year 622 CE. Since epochs are defined in era("CE"), we define this as approximately 622.2218, and subtract 1 because the numbering started at 1 SH.
• Year unit – the length of time between two Nowruz, which is close but not exactly the same as a Gregorian year. We define this as approximately 365.2424 days, based on astronomical data from the last few millennia.
• Direction and scale

This gives us:

sh <- era(label = "SH",
epoch = 622.2218 - 1,
name = "Solar Hijri",
unit = era_year("Nowruz", 365.2424),
scale = 1,
direction = 1)

These parameters should lead to a good approximation of the transformation between the SH era we defined and others, especially over longer time scales. But it isn’t precise, which we might see if we try to distinguish points close in time. For example, transforming the current beginning of this year in the Common Era (yr(2021.0, "CE")) into our new era gives us:

yr_transform(yr(2021, "CE"), sh)
#> # SH years :
#> [1] 1399.779
#> # Era: Solar Hijri (SH): Nowruz years (365.2424 days), counted forwards from 621.2218

In reality the 1 January 2021 was 12 Dey 1399. 1399 is a leap year, with 366 days, and 12 Dey is the 228th day of the year, so this is close to the value we would expect: 1399.787.

## Next release

For the next release of era, I plan to focus on the built-in era definitions, especially those derived from contemporary calendar systems. The development version already includes tweaks to the definitions of the Islamic calendars that have made the conversion from Gregorian years more accurate:

this_year()
#> # CE years :
#> [1] 2021
#> # Era: Common Era (CE): Gregorian years (365.2425 days), counted forwards from 0

this_year("AH")
#> # AH years :
#> [1] 1442
#> # Era: Anno Hegirae (AH): Islamic lunar years (354.36708 days), counted forwards from 621.539367680377

this_year("SH")
#> # SH years :
#> [1] 1399
#> # Era: Solar Hijri (SH): Nowruz years (365.2424 days), counted forwards from 621.221770467566

If you have any suggestions of eras to include, please do submit an issue on GitHub.


text:
Volume 307 - XVII International Workshop on Neutrino Telescopes (NEUTEL2017) - Session xiv - convener: alberto guglielmi – neutrino mass ordering
A Vision for Neutrino Particle Physics at the South Pole: from IceCube to PINGU
J.P. Athayde Marcondes de André*  on behalf of the IceCube-Gen2 collaboration
Full text: pdf
Pre-published on: March 28, 2018
Published on: April 05, 2018
Abstract
Located at the South Pole Station in Antarctica, the IceCube Neutrino Observatory is the world's largest neutrino telescope. In the clearest part of the ice sits a more densely instrumented section, DeepCore, that is able to measure neutrinos from 5 GeV to 80 GeV. Using DeepCore, neutrino oscillations can be observed via $\nu_{\mu}$ disappearance with precision comparable to that from accelerator experiments. With additional optical modules instrumenting the DeepCore volume, it is possible to further reduce the detector's energy threshold and improve the resolution of the detector at low energies. This allows measuring the $\nu_\tau$ appearance — which accompanies $\nu_{\mu}$ disappearance — at 10% precision or better, and determining the neutrino mass ordering. These are the key science goals of the proposed IceCube-Gen2 Phase 1 and PINGU, respectively.

Both current IceCube results on standard neutrino oscillations and sensitivities for the proposed Phase 1 and PINGU extensions of IceCube will be discussed in this talk.
DOI: https://doi.org/10.22323/1.307.0058
How to cite

Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete.

Open Access


text:
This data set is based on species in the North Sea (Blanchard et al.). It is a data.frame that contains all the necessary information to be used by the MizerParams() constructor. As there is no gear column, each species is assumed to be fished by a separate gear.

NS_species_params

## Format

A data frame with 12 rows and 7 columns. Each row is a species.

species

Name of the species

w_inf

The von Bertalanffy W_infinity parameter

w_mat

Size at maturity

beta

Size preference ratio

sigma

Width of the size-preference

R_max

Maximum reproduction rate

k_vb

The von Bertalanffy k parameter

Blanchard et al.

## Examples

if (FALSE) {
params <- MizerParams(NS_species_params)
sim = project(params)
plot(sim)
}


text:
You are currently browsing the monthly archive for July 2007.

I’ve just uploaded to the arXiv my lecture notes “Structure and randomness in combinatorics” for my tutorial at the upcoming FOCS 2007 conference in October. This tutorial covers similar ground as my ICM paper (or slides), or my first two Simons lectures, but focuses more on the “nuts-and-bolts” of how structure theorems actually work to separate objects into structured pieces and pseudorandom pieces, for various definitions of “structured” and “pseudorandom”. Given that the target audience consists of computer scientists, I have focused exclusively here on the combinatorial aspects of this dichotomy (applied for instance to functions on the Hamming cube) rather than, say, the ergodic theory aspects (which are covered in Bryna Kra‘s lecture notes from Montreal, or my notes from Montreal for that matter). While most of the known applications of these decompositions are number-theoretic (e.g. my theorem with Ben Green), the number theory aspects are not covered in detail in these notes. (For that, you can read Bernard Host’s Bourbaki article, Ben Green‘s Gauss-Dirichlet article or ICM article, or my Coates article.)

One of the oldest and most fundamental concepts in mathematics is the line. Depending on exactly what mathematical structures we want to study (algebraic, geometric, topological, order-theoretic, etc.), we model lines nowadays by a variety of standard mathematical objects, such as the real line ${\Bbb R}$, the complex line ${\Bbb C}$, the projective line $\Bbb{RP}^1$, the extended real line ${}[-\infty,+\infty]$, the affine line ${\Bbb A}^1$, the continuum $c$, the long line $L$, etc. We also have discrete versions of the line, such as the natural numbers ${\Bbb N}$, the integers ${\Bbb Z}$, and the ordinal $\omega$, as well as compact versions of the line, such as the unit interval ${}[0,1]$ or the unit circle ${\Bbb T} := {\Bbb R}/{\Bbb Z}$. Finally we have discrete and compact versions of the line, such as the cyclic groups ${\Bbb Z}/N{\Bbb Z}$ and the discrete intervals $\{1,\ldots,N\}$ and $\{0,\ldots,N-1\}$. By taking Cartesian products we then obtain higher-dimensional objects such as Euclidean space ${\Bbb R}^n$, the standard lattice ${\Bbb Z}^n$, the standard torus ${\Bbb T}^n = {\Bbb R}^n/{\Bbb Z}^n$, and so forth. These objects of course form the background on which a very large fraction of modern mathematics is set.

Broadly speaking, the line has three major families of structures on it:

1. Geometric structures, such as a metric or a measure, completeness, scales (coarse and fine), rigid motions (translations and reflection), similarities (dilation, affine maps), and differential structures (tangent bundle, etc.);
2. Algebraic structures, such group, ring, or field structures, and everything else that comes from those categories (e.g. subgroups, homomorphisms, involutions, etc.); and
3. One-dimensional structures, such as order, a length space structure (in particular, path-connectedness structure), a singleton generator, the Archimedean property, the ability to use mathematical induction (i.e. well-ordering), convexity, or the ability to disconnect the line by removing a single point.

Of course, these structures are inter-related, and it is an important phenomenon that a mathematical concept which appears to be native to one structure, can often be equivalently defined in terms of other structures. For instance, the absolute value $|n|$ of an integer $n$ can be defined geometrically as the distance from 0 to $n$, algebraically as the index of the subgroup $\langle n \rangle = n \cdot \Bbb Z$ of the integers ${\Bbb Z}$ generated by n, or one-dimensionally as the number of integers between 0 and $n$ (including 0, but excluding $n$). This equivalence of definitions becomes important when one wants to work in more general contexts in which one or more of the above structures is missing or otherwise weakened.

What I want to talk about today is an important toy model for the line (in any of its incarnations), in which the geometric and algebraic structures are enhanced (and become neatly nested and recursive), at the expense of the one-dimensional structure (which is largely destroyed). This model has many different names, depending on what field of mathematics one is working in and which structures one is interested in. In harmonic analysis it is called the dyadic model, the Walsh model, or the Cantor group model; in number theory and arithmetic geometry it is known as the function field model; in topology it is the Cantor space model; in probability it is the martingale model; in metric geometry it is the ultrametric, tree, or non-Archimedean model; in algebraic geometry it is the Puiseux series model; in additive combinatorics it is the bounded torsion or finite field model; in computer science and information theory it is the Hamming cube model; in representation theory it is the Kashiwara crystal model. Let me arbitrarily select one of these terms, and refer to all of these models as dyadic models for the line (or of objects derived from the line). While there is often no direct link between a dyadic model and a non-dyadic model, dyadic models serve as incredibly useful laboratories in which to gain insight and intuition for the “real-world” non-dyadic model, since one has much more powerful and elegant algebraic and geometric structure to play with in this setting (though the loss of one-dimensional structure can be a significant concern). Perhaps the most striking example of this is the three-line proof of the Riemann hypothesis in the function field model of the integers, which I will discuss a little later.

I’ve just uploaded to the arXiv the paper “The cubic nonlinear Schrödinger equation in two dimensions with radial data“, joint with Rowan Killip and Monica Visan, and submitted to the Annals of Mathematics. This is a sequel of sorts to my paper with Monica and Xiaoyi Zhang, in which we established global well-posedness and scattering for the defocusing mass-critical nonlinear Schrödinger equation (NLS) $iu_t + \Delta u = |u|^{4/d} u$ in three and higher dimensions $d \geq 3$ assuming spherically symmetric data. (This is another example of the recently active field of critical dispersive equations, in which both coarse and fine scales are (just barely) nonlinearly active, and propagate at different speeds, leading to significant technical difficulties.)

In this paper we obtain the same result for the defocusing two-dimensional mass-critical NLS $iu_t + \Delta u= |u|^2 u$, as well as in the focusing case $iu_t + \Delta u= -|u|^2 u$ under the additional assumption that the mass of the initial data is strictly less than the mass of the ground state. (When mass equals that of the ground state, there is an explicit example, built using the pseudoconformal transformation, which shows that solutions can blow up in finite time.) In fact we can show a slightly stronger statement: for spherically symmetric focusing solutions with arbitrary mass, we can show that the first singularity that forms concentrates at least as much mass as the ground state.

Last week, as mentioned previously, I attended a very inspiring interdisciplinary meeting at the Royal Society. It would be impossible for me to describe all 47 talks in detail; but I thought that I would take the time to discuss one representative talk, presented by Rosemary Grant FRS, on her research (together with several collaborators, including her husband), on the ongoing evolution (and speciation) of Darwin’s finches in the Galápagos islands; this was a particularly striking talk for me, as I had not been aware that such dramatic microevolutionary changes could be seen in real-time (Grant and her co-authors have tracked multiple generations of these finches for 35 years!). They have several interesting results; the talk that I am reproducing here is largely based on this article in Evolution.

I’ve added some new pages to this blog. Firstly, I’ve moved my old page on my books to a new page on this blog, so that each book now has its own page for publication information, errata, etc., as well as its own comments section for feedback (including the inevitable reports of future errata).

I also added a new page to my advice page on writing and submitting papers, on “taking advantage of the English language“.

Finally, I added some links on the sidebar of this blog to some of the more popular of my older posts and articles, and reorganised the categories slightly. In particular, my expository articles were moved out of the “short story” category (they didn’t fit particularly well with the other wordpress articles in this category) and into the “expository” category. (Now that there are a number of wordpress blogs in research mathematics, it might eventually be worth trying to establish some standards for use of mathematical tags, but this is hardly an urgent concern.)

[Update, July 18: Added the arXiv math categories, as well as a parent Mathematics category.]

This week I was in London, attending the New Fellows Seminar at the Royal Society. This was a fairly low-key event preceding the formal admissions ceremony; for instance, it is not publicised on their web site. The format was very interesting: they had each of the new Fellows of the Society give a brief (15 minute) presentation of their work in quick succession, in a manner which would be accessible to a diverse audience in the physical and life sciences. The result was a wonderful two-day seminar on the state of the art in many areas of physics, chemistry, engineering, biology, medicine, and mathematics. For instance, I learnt

• How the solar neutrino problem was resolved by the discovery that the neutrino had mass, which did not commute with flavour and hence caused neutrino oscillations, which have since been detected experimentally;
• Why modern aircraft (such as the Dreamliner and A380) are now assembled using (incredibly tough and waterproofed) adhesives instead of bolts or welds, and how adhesion has been enhanced by nanoparticles;
• How the bacterium Helicobacter pylori was recently demonstrated (by two Aussies :-) ) to be a major cause of peptic ulcers (though the exact mechanism is not fully understood), but has also been proposed (somewhat paradoxically) to also have a preventative effect against esophageal cancer (cf. the hygiene hypothesis);
• How recent advances in machine learning and image segmentation (including graph cut methods!) now allow computers to identify and track many general classes of objects (e.g. people, cars, animals) simultaneously in real-world images and video, though not quite in real-time yet;
• How large-scale structure maps of the universe (such as the 2dF Galaxy Redshift Survey) combine with measurements of the cosmic background radiation (e.g. from WMAP) to demonstrate the existence of both dark matter and dark energy (they have different impacts on the evolution of the curvature of the universe and on the current distribution of visible matter);
• … and 42 other topics like this. (One strongly recurrent theme in the life science talks was just how much recent genomic technologies, such as the genome projects of various key species, have accelerated (by several orders of magnitude!) the ability to identify the genes, proteins, and mechanisms that underlie any given biological function or disease. To paraphrase one speaker, a modern genomics lab could now produce the equivalent of one 1970s PhD thesis in the subject every minute.)

I’ve just uploaded to the ArXiV my paper “Norm convergence of multiple ergodic averages for commuting transformations“, submitted to
Ergodic Theory and Dynamical Systems. This paper settles in full generality the norm convergence problem for several commuting transformations. Specifically, if $(X, {\mathcal X},\mu)$ is a probability space and $T_1, \ldots, T_l: X \to X$ are commuting measure-preserving transformations, then for any bounded measurable functions $f_1, \ldots, f_l: X \to {\Bbb R}$, the multiple average

$\frac{1}{N} \sum_{n=0}^{N-1} \int_X T_1^n f_1 \ldots T_l^n f_l$ (1)

is convergent in the $L^2(X)$ norm topology (and thus also converges in probability). The corresponding question of pointwise almost everywhere convergence remains open (and quite difficult, in my opinion). My argument also does not readily establish a formula as to what this limit actually is (it really establishes that the sequence is Cauchy in $L^2$ rather than convergent).

The l=1 case of this theorem is the classical mean ergodic theorem (also known as the von Neumann ergodic theorem). The l=2 case was established by Conze and Lesigne. The higher l case was partially resolved by Frantzikinakis and Kra, under the additional hypotheses that all of the transformations $T_i$, as well as the quotients $T_i T_j^{-1}$, are ergodic. The special case $T_j=T^j$ was established by Host-Kra (with another proof given subsequently by Ziegler). Another relevant result is the Furstenberg-Katznelson theorem, which asserts among other things when $f_1=\ldots=f_l$ is non-negative and not identically zero, then the inner product of the expression (1) with f has a strictly positive limit inferior as $N \to \infty$. This latter result also implies Szemerédi’s theorem.

It is also known that the Furstenberg-Katznelson theorem can be proven by hypergraph methods, and in fact my paper also proceeds by a hypergraph-inspired approach, although the language of hypergraphs is not explicitly used in the body of the argument. (In contrast to the work of Host-Kra and Ziegler, no nilsystems appear in the proof.)

A few months ago, I was invited to contribute an article to Scholarpedia – a Wikipedia-like experiment (using essentially the same software, in fact) in which the articles are far fewer in number, but have specialists as the primary authors (and curators) and are peer-reviewed in a manner similar to submissions to a research journal. Specifically, I was invited (with Ben Green) to author the article on Szemerédi’s theorem. The article is now submitted (awaiting review) reviewed and accepted, and can be viewed on the Scholarpedia page for that theorem. Like Wikipedia, the page is open to edits or any other comments by any user (once they register an account and login); but the edits are moderated by the curators and primary authors, who thus remain responsible for the content.

Scholarpedia seems to be an interesting experiment, trying to blend the collaborative and dynamic strengths of the wiki system with the traditional and static strengths of the peer-review system. At any rate, any feedback on my article with Ben, either at the Scholarpedia page or here, would be welcome.

[Update, July 9: the article has been reviewed, modified, and accepted in just three days – a blindingly fast speed as far as peer review goes!]

This problem in compressed sensing is an example of a derandomisation problem: take an object which, currently, can only be constructed efficiently by a probabilistic method, and figure out a deterministic construction of comparable strength and practicality. (For a general comparison of probabilistic and deterministic algorithms, I can point you to these slides by Avi Wigderson).

I will define exactly what UUP matrices (the UUP stands for “uniform uncertainty principle“) are later in this post. For now, let us just say that they are a generalisation of (rectangular) orthogonal matrices, in which the columns are locally almost orthogonal rather than globally perfectly orthogonal. Because of this, it turns out that one can pack significantly more columns into a UUP matrix than an orthogonal matrix, while still capturing many of the desirable features of orthogonal matrices, such as stable and computable invertibility (as long as one restricts attention to sparse or compressible vectors). Thus UUP matrices can “squash” sparse vectors from high-dimensional space into a low-dimensional while still being able to reconstruct those vectors; this property underlies many of the recent results on compressed sensing today.

There are several constructions of UUP matrices known today (e.g. random normalised Gaussian matrices, random normalised Bernoulli matrices, or random normalised minors of a discrete Fourier transform matrix) but (if one wants the sparsity parameter to be large) they are all probabilistic in nature; in particular, these constructions are not 100% guaranteed to actually produce a UUP matrix, although in many cases the failure rate can be proven to be exponentially small in the size of the matrix. Furthermore, there is no fast (e.g. sub-exponential time) algorithm known to test whether any given matrix is UUP or not. The failure rate is small enough that this is not a problem for most applications (especially since many compressed sensing applications are for environments which are already expected to be noisy in many other ways), but is slightly dissatisfying from a theoretical point of view. One is thus interested in finding a deterministic construction which can locate UUP matrices in a reasonably rapid manner. (One could of course simply search through all matrices in a given class and test each one for the UUP property, but this is an exponential-time algorithm, and thus totally impractical for applications.) In analogy with error-correcting codes, it may be that algebraic or number-theoretic constructions may hold the most promise for such deterministic UUP matrices (possibly assuming some unproven conjectures on exponential sums); this has already been accomplished by de Vore for UUP matrices with small sparsity parameter.


text:
# Wondering about how to deal with a deflating balloon?

1. Feb 20, 2010

### jeebs

Hi,
Not sure if this is asked in the right section, but I am doing something which involves moving a small car, and I am trying to make predictions from an air resistance perspective. I'm trying to come up with a rough expression for the drag force on a balloon as it deflates. I've started with the general expression for the drag force $$F_d = 1/2\rho AC_dv^2$$.
To simplify things I'm assuming a spherical balloon, which has a radius rmax when it is fully inflated and a radius rmin when it is fully deflated, so it goes from having surface area A = pi(rmax)2 to pi(rmin)2. I realise the surface area of a sphere is 4pi.r2 but I need the area of the object presented to the oncoming air. I also realise that the velocity v is going to increase from zero to some maximum, but I'm ignoring that for now (perhaps unjustifiably, lol).

So assuming everything else (velocity v, air density rho and drag coefficient Cd) is kept constant, the balloon radius r determines the drag force Fd.

However, this is where I start to have problems, because I am not expecting r to decrease at a constant rate - my thinking is that, from the general spring force equation F=-kx, a balloon stretched (inflated) to the maximum is going to have a greater force trying to pull its molecules back to their original positions than if it is partially inflated. This means the rate at which air is forced out (ie. pressure decreases) falls with time.

I need to find out how r varies with time but all I can think of really is that its going to depend on the "stretchiness" of the balloon, ie. something analogous to the spring constant, k, and also the size of the hole that the air is escaping from. I'm not really sure where I should go from here, so if anyone has any ideas they think might be relevant I would love to hear them.

Thankyou.

Last edited: Feb 20, 2010


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Hyperbolic Functions - Maple Programming Help

Home : Support : Online Help : Math Apps : Trigonometry : MathApps/HyperbolicTrigonometricFunctions

Hyperbolic Functions

Main Concept

There are a total of six hyperbolic functions:

Summary of the Hyperbolic Function Properties

 Name Notation Equivalence Derivative Special properties Hyperbolic Sine sinh(x)  Hyperbolic Cosine cosh(x)  Hyperbolic Tangent tanh(x)  Hyperbolic Cosecant csch(x)  Hyperbolic Secant sech(x) Hyperbolic Cotangent coth(x)

Explore the properties of the Hyperbolic Functions.

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# Proving a language with equal occurences of ab, and cd is not a regular language using the Pumping Lemma

I am trying to show that $$A = \{w \in \{a,b,c,d\}^{*}|w \textrm{ has equal occurences of } ab \textrm{ and } cd\}$$ is not regular by using the Pumping Lemma.

My idea here was to use the string $$s = (ab)^{p}(cd)^{p}$$. Clearly, $$s \in A$$, and $$|s| = 2p \geq A$$. So the the three conditions of the pumping lemma hold.

Where I am stuck is picking the correct decomposition of $$s$$. My idea was to do the following:

We can decompose $$s$$ into $$xyz$$ such that:

• $$x = (ab)^{j}, j \geq 0$$
• $$y = (ab)^{k}, k \geq 1$$
• $$z = (ab)^{p-j-k}(cd)^{p}$$

Now consider if we "pumped" $$y$$. Let $$i = 2$$. Then $$xy^{2}x = (ab)^{j}(ab)^{2k}(ab)^{p-j-k}(cd)^{p} = (ab)^{p+k}(cd)^{p} \notin A$$. Therefore, we arrived at a contradiction and $$A$$ is not a regular language.

But my confusion here is that the $$xyz$$ I have chosen is not the only decomposition of $$s$$. For example, you could choose $$x = a$$, $$y = b$$, and $$z$$ is the rest of the string. Do I have to cover all possible cases? If so, this seems like a bad choice for $$s$$ -- any other ideas?

Another concern is that my chosen decomposition violates the condition $$|xy| \leq P$$ of the Pumping Lemma will not hold for $$p = 1$$.

Ok I will split the answer into two parts, one for each of your questions \ concerns \ confusions:

Q: Are we allowed to use the pumping lemma if we chose a specific decomposition, or do we need to show for all decompositions?

A: The pumping lemma states, $$\exists p.\forall w.|w|\ge p \rightarrow \exists x,y,z.w=xyz \land\text{ they are a valid decomposition }\land\forall k\in\mathbb{N}.xy^kz\in A$$.

and you are trying to show the negation of this statement. as you can see, there is an "exists" sign before the "x,y,z" and therefore in the negation there will be a "for all" sign.

Thus, you MUST go over ALL possible decompositions in order to use the pumping lemma in this way.

Q: This doesn't work for $$p=1$$ but works for every other value of $$p$$. is this a problem?

A: no! its absolutely fine.

Say, for example, you know how to prove for $$p=5$$ and you don't know how to do $$p=1$$. define $$w=(ab)^5(cd)^5$$, and you know that $$|w|\ge 5 \ge 1=p$$. but since $$|w|\ge 5$$ and you know how to solve for $$p=5$$, you know how to deal with $$w$$.

This means that its enough to prove for some value of $$p$$ and all other values BELOW it will be automatically proved too!

Also, you are in the right direction. This is a good choice of a word, but you will need to work a bit more to show that every decomposition cannot be always pumped.

I hope this helped you a bit :)

You need to consider all possible decompositions. There aren't so many cases to consider:

1. $$x \in (ab)^*$$, $$y \in (ab)^+$$.
2. $$x \in (ab)^*$$, $$y \in (ab)^*a$$.
3. $$x \in (ab)^*a$$, $$y \in (ba)^+$$.
4. $$x \in (ab)^*a$$, $$y \in (ba)^*b$$.

Go over them one by one.


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Autocovariance sequence (ACS)

1. Jul 14, 2008

thedean515

We have two definitions for the autocovariance of finite samples $y\left(t\right)$and
it is given as
$$\hat{r}\left(k\right)=\frac{1}{N-k}\sum_{t=K+1}^{N}y\left(t\right)y^{*}\left(t-k\right),\qquad0\le k\le N-1$$
and
$$\tilde{r}\left(k\right)=\frac{1}{N}\sum_{t=K+1}^{N}y\left(t\right)y^{*}\left(t-k\right),\qquad0\le k\le N-1$$

In addition we know that the autocovariance sequence for infinite
samples is
$$r\left(k\right)=E\left\{ y\left(t\right)y^{*}\left(t-k\right)\right\}$$
where $$E\left\{ \cdot\right\}$$is the expectation operator which
averages over the ensemble of realizations. Now I have been told that
$$E\left\{ \tilde{r}\left(k\right)\right\} =r\left(k\right)$$
and
$$E\left\{ \hat{r}\left(k\right)\right\} =\frac{N-\left|k\right|}{N}r\left(k\right)$$
but they would be the other way round, can we proof it?

2. Jul 23, 2008

thedean515

someone can help me