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## IACR News

Updates on the COVID-19 situation are on the Announcement channel.

Here you can see all recent updates to the IACR webpage. These updates are also available:

#### 20 November 2021

###### Saba Eskandarian, Dan Boneh
ePrint Report
This paper studies the role of multiparty shuffling protocols in enabling more efficient metadata-hiding communication. We show that the process of shuffling messages can be expedited by having servers collaboratively shuffle and verify secret-shares of messages instead of using a conventional mixnet approach where servers take turns performing independent verifiable shuffles of user messages. We apply this technique to achieve both practical and asymptotic improvements in anonymous broadcast and messaging systems. We first show how to build a three server anonymous broadcast scheme, secure against one malicious server, that relies only on symmetric cryptography. Next, we adapt our three server broadcast scheme to a $k$-server scheme secure against $k-1$ malicious servers, at the cost of a more expensive per-shuffle preprocessing phase. Finally, we show how our scheme can be used to significantly improve the performance of the MCMix anonymous messaging system.

We implement our shuffling protocol in a system called Clarion and find that it outperforms a mixnet made up of a sequence of verifiable (single-server) shuffles by $9.2\times$ for broadcasting small messages and outperforms the MCMix conversation protocol by $11.8\times$.
###### Gang Wang, Mark Nixon
ePrint Report
Blockchain as a potentially disruptive technology can advance many different fields, e.g., cryptocurrencies, supply chains, and the industrial Internet of Things. The next-generation blockchain ecosystem is expected to consist of various homogeneous and heterogeneous distributed ledgers. These ledger systems will inevitably require a certain level of proper cooperation of multiple blockchains to enrich advanced functionalities and enhance interoperable capabilities for future applications. The interoperability among blockchains will revolutionize current blockchain design principles, like the emergence of the Internet. However, the development of cross-blockchain applications involves much complexity regarding the variety of underlying cross-blockchain communication. With that regard, we propose an efficient, interoperable blockchain architecture, InterTrust, to support interoperability and trustworthiness among arbitrary blockchain systems (including homogeneous and heterogeneous blockchains). It consists of an atomic cross-chain communication protocol, which can be considered an agnostic protocol to integrate existing blockchain systems smoothly. InterTrust is powered by two innovative techniques: threshold signature scheme and trusted hardware. The threshold signature scheme guarantees consistency and verifiability in the target blockchain systems, and the trusted hardware guarantees trusted services among distinct blockchain systems. Combining these two techniques provides an efficient cross-chain communication protocol to facilitate atomic swaps and interoperable operations between different blockchain systems. Our interoperable architecture is robust to support arbitrary blockchain systems. We also present the security analysis on the scenarios of integrating our protocol into Byzantine fault tolerance based blockchain systems.
###### Smile Markovski, Vesna Dimitrova, Zlatka Trajcheska, Marija Petkovska, Mile Kostadinoski, Damjan Buhov
ePrint Report
Designing new cryptosystems and their cryptanalysis is the basic cycle of advancement in the field of cryptography. In this paper we introduce a block cipher based on the quasigroup transformations, which are defined by the matrix presentation of the quasigroup operations. This type of quasigroup presentation is suitable for constructing a block cipher since it doesn't require too much memory space to store all the necessary data, so it can be used even for lightweight cryptographic purposes.

For now, we are considering only the quasigroups of order 4. Constructions with quasigroups of higher order and examination of the strengths and weaknesses of this design will be considered in next papers.
###### Mila Anastasova, Mojtaba Bisheh-Niasar, Reza Azarderakhsh, Mehran Mozaffari Kermani
ePrint Report
In 2016, the National Institute of Standards and Technology (NIST) initiated a standardization process among the post-quantum secure algorithms. Forming part of the alternate group of candidates after Round 2 of the process is the Supersingular Isogeny Key Encapsulation (SIKE) mechanism which attracts with the smallest key sizes offering post-quantum security in scenarios of limited bandwidth and memory resources. Even further reduction of the exchanged information is offered by the compression mechanism, proposed by Azarderakhsh et al., which, however, introduces a significant time overhead and increases the memory requirements of the protocol, making it challenging to integrate it into an embedded system. In this paper, we propose the first compressed SIKE implementation for a resource-constrained device, where we targeted the NIST recommended platform STM32F407VG featuring ARM Cortex-M4 processor. We integrate the isogeny-based implementation strategies described previously in the literature into the compressed version of SIKE. Additionally, we propose a new assembly design for the finite field operations particular for the compressed SIKE, and observe a speedup of up to 16% and up to 25% compared to the last best-reported assembly implementations for p434, p503, and p610.
###### Gideon Samid
ePrint Report
Pattern loaded ciphers are at risk of being compromised by exploiting deeper patterns discovered first by the attacker. This reality offers a built-in advantage to prime cryptanalysis institutions. On the flip side, risk of hidden math and faster computing undermines confidence in the prevailing cipher products. To avoid this risk one would resort to building security on the premise of lavish quantities of randomness. Gilbert S. Vernam did it in 1917. Using modern technology, the same idea of randomness-based security can be implemented without the inconvenience associated with the old Vernam cipher. These are Trans Vernam Ciphers that project security through a pattern-devoid cipher. Having no pattern to lean on, there is no pattern to crack. The attacker faces (i) a properly randomized shared cryptographic key combined with (ii) unilateral randomness, originated ad-hoc by the transmitter without pre-coordination with the recipient. The unlimited unilateral randomness together with the shared key randomness is set to project as much security as desired up to and including Vernam levels. Assorted Trans Vernam ciphers (TVC) are categorized and reviewed, presenting a cogent message in favor of a cryptographic pathway where transmitted secrets are credibly secured against attackers with faster computers and better mathematicians. A vision emerges: a cryptographic level playing field, consistent with the emerging culture of Web 3.0.

#### 18 November 2021

###### Leuven, Belgium, 12 April - 14 April 2022
Event Calendar
Event date: 12 April to 14 April 2022
###### Stevens Institute of Technology, Hoboken, NJ, USA
Job Posting
The Department of Computer Science in the Charles V. Schaefer, Jr. School of Engineering and Science (SES) at Stevens Institute of Technology (Stevens) invites applications for tenure-track and tenured positions in all areas of computer science at the assistant, associate, and full professor ranks. We encourage applicants with expertise in our existing core areas of research including computer vision, cyber security, and AI/ML, in addition to candidates who can expand our research program significantly in HCI and algorithmic bias/fairness. Stevens offers an intellectually vibrant, diverse, highly interdisciplinary, collaborative, innovative, and entrepreneurial community and is a great place to work.

Applicants should have earned a Ph.D. in computer science or a related discipline. Candidates are expected to demonstrate a commitment to teaching and mentorship at both the undergraduate and graduate levels, including working with students from underrepresented groups. Successful candidates will have the potential to develop an externally funded research program, supervise graduate students in research, and contribute to the highly interdisciplinary, collaborative, diverse, innovative, and entrepreneurial culture at Stevens. Candidates applying at the rank of Associate or Full should have a track record of success in scholarship, funded research, teaching, mentoring, and contributing to diversity, equity, and inclusion.

Closing date for applications:

Contact: Search Committee Chairs, Samantha Kleinberg (samantha.kleinberg@stevens.edu) and Yue Ning (yue.ning@stevens.edu)

###### Apple Inc, Santa Clara Valley, California, USA
Job Posting
Passionate about cryptography? Want to work on impactful projects and amazing features?

Apply to join the team!

You can find details about the position and how to apply in the linked page.

Closing date for applications:

Contact: Yannick Sierra

###### University of Houston - Downtown, Houston, Texas
Job Posting
Assistant or Associate Professor in Computer Science - (FAC002130) Department Name: D0019 Computer Sci & Engr Technology The Department of Computer Science and Engineering Technology at the University of Houston – Downtown (UHD) invites applications for a tenure-track Assistant or Associate Professor position in Computer Science. We are looking for outstanding candidates with expertise in one or more of the areas of: artificial intelligence, machine learning or cybersecurity. The appointment will start in August 2022.

Closing date for applications:

Contact: -

###### University of Bergen
Job Posting
The postdoc position is for 3 years (with potential possibility for extension for one more year) within a project "Cryptographic Boolean Functions for Threshold Implementations" that aims to study Boolean functions used as building blocks in cryptographic ciphers and their Threshold Implementations in order to find efficient ways of preventing Side Channel Attacks. This position is to start as soon as possible (approx. February-March 2022). We are seeking excellent candidates in cryptography in the direction of Threshold Implementations and Boolean functions. We are particularly interested in applicants who excel both at mathematics and programming.

Closing date for applications:

Contact: Prof. Lilya Budaghyan

###### KU LEUVEN
Job Posting
The Computer Security and Industrial Cryptography (COSIC) research group belongs to the Electrical Engineering Department at the KU Leuven.
Research group COSIC is looking for a PhD position on Secure Localisation Technologies
The goal of this PhD research is twofold.
• First, the PhD candidate will evaluate the security strength of (future) emerging ranging and localisation technologies that are being deployed by industry. Experiments will be carried out to discover new security vulnerabilities and assess their impact.
• Second, the PhD candidate will study and design novel secure ranging and localisation solutions. The focus of this second line of research is particularly on the realisation of secure distance bounding protocols, which are cryptographic primitives used to mitigate a set of ranging attacks.
Candidates must hold a master’s degree in electronics engineering or computer science, have good grades and have a keen interest in cryptography and system security. Prior expertise in physical layer security or radio propagation is a bonus.

Closing date for applications:

Contact: Please check the application procedure at https://www.esat.kuleuven.be/cosic/vacancies/ and send all requested documents to jobs-cosic@esat.kuleuven.be

• ###### CISPA Helmholtz Center for Information Security
Job Posting
The research group of Karl Wüst (https://karlwuest.github.io) at the CISPA Helmholtz Center for Information Security is looking for talented people with a background in computer science or closely related fields and an interest in information security and applied cryptography to join the group as PhD students. The main research focus of the group is on security and privacy aspects of digital currency and smart contract systems as well as some aspects of trustworthy computing.

The positions are fully funded and located at the CISPA Helmholtz Center for Information Security in Germany, one of the world’s top research institutions in the area of information security. The start dates for the positions are flexible and applications will be considered until the positions are filled.

For additional details and information on how to apply see https://karlwuest.github.io/positions

Closing date for applications:

Contact: Karl Wüst

###### Zama, Paris, France
Job Posting
Job description. We are looking for a researcher in Homomorphic Encryption to start working with us in 2022. The candidate and his/her/their team will be responsible for:
• discovering new cryptographic techniques to compute on encrypted data
• working with the engineering and product teams to implement his/her/their research into our products
• design robust tests and benchmarks to validate his/her/their research and its implementation
• review the latest published research, and inform the team on potential new applications
• work with the entire team to define the research and product roadmaps
• publishing papers, filing patents and presenting his/her/their work at academic conferences
Experience. He/she/they should:
• have a PhD in cryptography or equivalent
• have deep knowledge of homomorphic encryption
• have (optionally) knowledge of LWE hardness and security
• have (optionally) knowledge of machine learning
• be passionate about privacy and open source software
• have good written and oral communication skills
Full remote is possible, with a willingness to come to Paris quarterly.

Closing date for applications:

Contact: Ilaria Chillotti (ilaria.chillotti(at)zama.ai)

###### Nanyang Technological University, Singapore
Job Posting
The Cryptanalysis Taskforce at Nanyang Technological University in Singapore led by Prof. Jian Guo is seeking for candidates to fill several post-doctoral research fellow positions on symmetric-key cryptography. Topics include but are not limited to the following sub-areas:
• tool aided cryptanalysis, such as MILP, CP, STP, and SAT
• machine learning aided cryptanalysis and designs
• privacy-preserving friendly symmetric-key designs
• quantum cryptanalysis
• provable security
• cryptanalysis against SHA-2, SHA-3, and AES
• threshold cryptography
Established in 2014, the Cryptanalysis Taskforce is a group dedicated for research in symmetric-key cryptography, it is currently comprised by 4 (senior) PostDoc Research Fellows, 3 PhD students, and several long-term visitors. Since establishment, the team has been active in both publications in and services for IACR. It has done quite some cryptanalysis work on various important targets such as SHA-3 and AES, and is expanding its interests to the areas mentioned above, with strong funding support from the university and government agencies in Singapore. We offer competitive salary package with extremely low tax (around 5%), as well as excellent environment dedicating for top-venues publication orientated research in Singapore. The contract will be initially for one year, and has the possibility to be extended. Candidates are expected to have proven record of publications in IACR conferences. Interested candidates are to send their CV and 2 reference letters to Jian Guo. Review of applicants will start immediately until the positions are filled. More information about the Cryptanalysis Taskforce research group can be found via https://team.crypto.sg

Closing date for applications:

Contact: Asst Prof Jian Guo, guojian@ntu.edu.sg

###### University of Neuchatel, Switzerland
Job Posting

We are oferring a fully funded PhD scholarship for a student to join our group on reinforcement learning and decision making under uncertainty more generally, at the University of Neuchatel, Switzerland. We are particularly interested in candidates with a strong mathematical and research interest in the following fields e

1. Theory of differential privacy.
2. Algorithms for differentially private machine learning.
3. Algorithms for fairness in machine learning.
4. Interactions between machine learning and game theory.
5. Inference of human models of fairness or privacy.

Overall, our group works on reinforcement learning, decision making under uncertainty, fairness and differential privacy. The student will also have the opportunity to visit and work with other group members at the University of Oslo, Norway and Chalmers University of Technology, Sweden.

• Starting date 1 Februrary 2022 or soon afterwards.
• Application deadline 30 November 2021.

Closing date for applications:

Contact: Christos Dimitrakakis

#### 17 November 2021

ePrint Report
Authentication constitutes the foundation and vertebrae of all security properties. It is the procedure in which communicating parties prove their identities to each other, and generally establish and derive secret keys to enforce other services, such as confidentiality, data integrity, non-repudiation, and availability. PUFs (Physical Unclonable Functions) has been the subject of many subsequent publications on lightweight, lowcost, and secure-by-design authentication protocols. This has turned our attention to investigate the most recent PUF-based authentication protocols for IoT. In [1], we reviewed the security of some PUF-based authentication protocols that were proposed between 2016 and October 2020, and drew important security lessons to consider by future authentication protocol designers. In this paper, we extend our previous work by reviewing the security of fifteen PUF-based authentication protocols that were recently published during the past two years (2020 and 2021). We first provide the necessary background on PUFs and how they are used for authentication. Then, we analyze the security of these authentication protocols to identify and report common security issues and design flaws. We draw lessons and recommendations for future authentication protocol designers
###### Viet Ba Dang, Kamyar Mohajerani, Kris Gaj
ePrint Report
Performance in hardware has typically played a significant role in differentiating among leading candidates in cryptographic standardization efforts. Winners of two past NIST cryptographic contests (Rijndael in case of AES and Keccak in case of SHA-3) were ranked consistently among the two fastest candidates when implemented using FPGAs and ASICs. Hardware implementations of cryptographic operations may quite easily outperform software implementations for at least a subset of major performance metrics, such as latency, number of operations per second, power consumption, and energy usage, as well as in terms of security against physical attacks, including side-channel analysis. Using hardware also permits much higher flexibility in trading one subset of these properties for another. This paper presents high-speed hardware architectures for four lattice-based CCA-secure Key Encapsulation Mechanisms (KEMs), representing three NIST PQC finalists: CRYSTALS-Kyber, NTRU (with two distinct variants, NTRU-HPS and NTRU-HRSS), and Saber. We rank these candidates among each other and compare them with all other Round 3 KEMs based on the data from the previously reported work.
###### Kyungbae Jang, Gyeongju Song, Hyunjun Kim, Hyeokdong Kwon, Hyunji Kim, Hwajeong Seo
ePrint Report
Adversaries using quantum computers can employ new attacks on cryptography that are not possible with classical computers. Grover's search algorithm, a well-known quantum algorithm, can reduce the search complexity of $O(2^n)$ to $\sqrt{2^n}$ for symmetric key cryptography using an $n$-bit key. To apply the Grover search algorithm, the target encryption process must be implemented as a quantum circuit. In this paper, we present optimized quantum circuits for Korean block ciphers based on ARX architectures. We adopt the optimal quantum adder and design in parallel way with only a few trade-offs between quantum resources. As a result, we provide a performance improvement of 78\% in LEA, 85\% in HIGHT, and 70\% in CHAM in terms of circuit depth, respectively. Finally, we estimate the cost of the Grover key search for Korean block ciphers and evaluate the post-quantum security based on the criteria presented by NIST.
###### Amos Zheng, Marcos A. Simplicio Jr.
ePrint Report
Hash-based signature schemes are a class of post-quantum algorithms usually built upon one-time signature (OTS) solutions via hash-trees. The benefits of such schemes include small key sizes, efficient processing and the fact that they are simple to implement using a regular hash algorithm. In addition, their security properties are quite well understood, since they rely basically on the pre-image or collision resistance of the underlying hash function. Among the existing OTS schemes, W-OTS+ is among the most popular. One reason for such popularity is that the OTS public key can be recovered from the signature itself, which facilitates the construction of a multi-time signature scheme using Merkle trees. On the other hand, signature generation and verification in W-OTS+ take roughly the same time, which is not ideal for applications where each signature is expected to be verified several times, as in software stores, PKI certificate validation, and secure boot. It is also inconvenient when the devices that verify signatures have lower computational power than the signers. In such scenarios, it is desirable to design signature schemes enabling faster verification, even if such speed-ups come at the expense of a slower signature generation procedure. With this goal in mind, we hereby present and evaluate a novel OTS scheme, called z-OTS. The main interest of z-OTS is that it preserves all benefits of W-OTS+, but provides faster signature verification at the cost of a (not much) slower signature generation procedure. For example, for signature sizes equivalent to W-OTS+ with Winternitz parameter w=4, our simulations show that verification can be 30.3% faster with z-OTS, while key and signature generation become, respectively, 53.7% and 137.5% slower. Larger w leads to even more expressive gains in the verification procedure, besides providing lower overheads when generating keys and signatures.
###### Sangeeta Chowdhary, Wei Dai, Kim Laine, Olli Saarikivi
ePrint Report
Homomorphic encryption (HE), especially the CKKS scheme, can be extremely challenging to use. The EVA language and compiler (Dathathri et al., PLDI 2020) was an attempt at addressing this challenge. EVA allows a developer to express their encrypted computation in a simple form with a Python-integrated language called PyEVA. It then compiles the program into an executable form by inserting operations such as relinearization and rescaling, applying optimizations, and choosing encryption parameters with the objective of minimizing execution time. Compiled programs can be executed with a parallelizing back-end against a library of HE primitives.

Our work improves upon the EVA toolchain in several ways: changes to the Python front-end make writing PyEVA programs more natural, while a rework of EVA's C++ APIs makes writing new passes easier. We also implement two new optimizations, common subexpression elimination and reduction balancing, which we show allow users to write simpler and more modular PyEVA programs.

We argue that the abstraction EVA provides is insufficient to resolve some common usability challenges. For example, managing vectors of arbitrary size is non-trivial. To resolve these problems, we demonstrate how building a library of commonly used data structures and functions is simple in PyEVA. EVA's automation allows writing very concise code, which gets fused and optimized together with the user program. We create the beginnings of an EVA Extension Library (EXL), that provides vector and matrix classes and a collection of common statistical functions, to demonstrate the power of this approach.


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##### Residual decay power kinetics.

Up to now, the nuclear power density ft (W/cm³) was calculated from the fission neutron source sf (neutr/cm³/s) by (1) where κ/ν (skn) is tabulated along with the other neutronic properties.

Actually, sf represents the total "potential" fission neutron source which includes the prompt neutrons emitted at fission time, and the delayed neutron emitted by subsequent decay of radioactive fission product emitters cj generated at fission time.
We remind (2) the balance equation of precursor cj with decay constant λj.
For stationary condition, (4) applies, which implies (5).
A strict mathematical similitude may be established between the delayed neutron precursors and the power decay fission product (FP) emitters. The concentration cyj (the added "y" stands for power deca"Y") satisfies (7), where βyj is the number of FP emitters of type j with decay constant λyj.
The power density fyj deposited by emitters yj decay is proportional (factor μ) to the decay rate (9).
At steady-state, from (8) and (9) , we obtain (10) which shows that if we set (11), fyj could be interpreted as well as the steady-state ratio of cyj generated decay power fyj to the total potential power ft.
The actual fyj in transient condition is (12), and f (13) the actual instant power.
In (13), βyp is the prompt (major) fraction.
For yj emitters with time constant (1/ λyj) much larger than the application duration, the contribution of those emitters should more conveniently be approximated by a constant part βy0 ft0 of the initial power. At steady state, f=ft=ft0 and (8) holds.
Power decay model is activated under Lstcy and Lstcyj lists. (be=β, am=λ).
The prompt fraction beyp must not be entered: it is calculated from bey0 and the behj by means of (14).

format: plain


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# nLab homotopy coherent category theory

Homotopy coherent category theory or enriched homotopy theory is the attempt to understand those situations that arise in homotopy theory, homotopical algebra, and non-Abelian homological algebra, in which quite naturally occuring diagrams are not commutative, yet are commutative ‘up-to-homotopy’.

If the diagram is just commutative in the homotopy category, that is not much use and one can do little with it. Surprisingly often however, the homotopies involved in such a diagram’s ‘almost commutative’ nature can be specified, but then the question arises as to whether those homotopies form some sort of diagram, so homotopies between composite homotopies become involoved. This begins to look like parts of 2-category theory, or rather ‘higher weak category theory’ and the development of homotopy coherent category theory was initiated in an attempt to merge homotopy theory with categorical tools for handling higher categories.

In particular: * homotopy theory using model categories and similar structures; * enriched category theory.

In the case that the category $V$ one is enriching over is itself a model category or at least a category with weak equivalences one wishes to generalize limits and in particular the weighted limits such as as ends and coends in enriched category theory to constructions which satisfy the familiar universal properties only up to coherent homotopy.

# References

These articles deal with the theory of homotopy coherent diagrams:

• R. Vogt, Homotopy limits and colimits , Math. Z., 134, (1973), 11 – 52.

• J.-M. Cordier and T. Porter, Vogt’s theorems on categories of homotopy coherent diagrams, Math. Proc. Camb. Phil. Soc. 100 (1986), 65–90.

• J.-M. Cordier and T. Porter, Maps between homotopy coherent diagrams, Top. and its Applications, 28 (1988) 255-275.

• J.-M. Cordier and T. Porter, Fibrant diagrams, rectifications and a construction of Loday, J. Pure Appl. Alg 67 (1990), 111–124.

A discussion of homotopy limits is in

• D. Bourn and J.-M. Cordier, A general formulation of homotopy limits , J. Pure Appl. Algebra, 29, (1983), 129–141,

• J.-M. Cordier, Sur les limites homotopiques de diagrammes homotopiquement cohérents, Comp. Math. 62 (1987), 367–388.

In

• J-M Cordier and T. Porter, Homotopy coherent category theory, Trans. Amer. Math. Soc. 349 (1997) 1-54. (web)

a main point is the definition and discussion of a homotopy coherent end? for the case of enrichment over the model category of simplicial sets.

In

• Mike Shulman, Homotopy limits and colimits and enriched homotopy theory (arXiv)

the general issue of enriched homotopy theory is addressed and enriched homotopical categories are introduced, which are a coherent combination of the notion of enriched category with that of homotopical category

Revised on October 8, 2009 16:05:50 by Urs Schreiber (131.211.36.96)


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## 1= .9999999999999999999.........

Here it is...

1/3= .33333333333...........

2/3= .6666666666.........

so

3/3= .9999999999.......

so

1= .999999999........

Wtf?

I think that this "paradox" means that there is a problem with our definition of infinity. At least, the definition needs to be refined. Does this mean that 1 is not equal to one, or is .9999999... another name for one? I think this topic needs to be adressed.
 Recognitions: Homework Help Oh no, I remember a long one discussing this. Something about convergence, but I'll sit back and watch :)


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# Revision history [back]

### Counting cycles of induced permutations

In order to do some sophisticated counting in graph theory, I need to count the cycles of some particular permutations.

In my situation, $n$ is an integer greater than 1, and $K_n$ is the set of all two-element sets ${a,b}$ with $a, b$ being integers not greater than $n$. Now any element $\pi$ of the symmetric group $S_n$ induces a permutation $\overline{\pi}$ of $K_n$ in a natural way, i.e. $\overline{\pi}$ maps any set ${a,b}$ of $K_n$ onto ${\pi(a), \pi(b)}$. What I want to figure out with the help of SAGE is the number of cycles that the permutation $\overline{\pi}$ has.

If you can help me, please do not forget to mention those little extra things that need to be done and that might appear obvious to you (e.g. importing packages and so forth), since I am a relative novice to SAGE.

Thank you very much.

Malte

### Counting cycles of induced permutations

In order to do some sophisticated counting in graph theory, I need to count the cycles of some particular permutations.

In my situation, $n$ is an integer greater than 1, and $K_n$ is the set of all two-element sets ${a,b}$ {a,b}$with$a, b$being integers not greater than$n$. Now any element$\pi$of the symmetric group$S_n$induces a permutation$\overline{\pi}$of$K_n$in a natural way, i.e.$\overline{\pi}$maps any set${a,b}${a,b} of$K_n$onto${\pi(a), \pi(b)}$. {\pi(a), \pi(b)}. What I want to figure out with the help of SAGE is the number of cycles that the permutation$\overline{\pi}$has. If you can help me, please do not forget to mention those little extra things that need to be done and that might appear obvious to you (e.g. importing packages and so forth), since I am a relative novice to SAGE. Thank you very much. Malte ### Counting cycles of induced permutations In order to do some sophisticated counting in graph theory, I need to count the cycles of some particular permutations. In my situation,$n$is an integer greater than 1, and$K_n$is the set of all two-element sets {a,b}$ {a,b} with $a, b$ being integers not greater than $n$. Now any element $\pi$ of the symmetric group $S_n$ induces a permutation $\overline{\pi}$ of $K_n$ in a natural way, i.e. $\overline{\pi}$ maps any set {a,b} of $K_n$ onto {\pi(a), \pi(b)}. What I want to figure out with the help of SAGE is the number of cycles that the permutation $\overline{\pi}$ has.

If you can help me, please do not forget to mention those little extra things that need to be done and that might appear obvious to you (e.g. importing packages and so forth), since I am a relative novice to SAGE.

Thank you very much.

Malte

### Counting cycles of induced permutations

In order to do some sophisticated counting in graph theory, I need to count the cycles of some particular permutations.

In my situation, $n$ is an integer greater than 1, and $K_n$ is the set of all two-element sets {a,b} with $a, b$ being integers not greater than $n$. Now any element $\pi$ of the symmetric group $S_n$ induces a permutation $\overline{\pi}$ of $K_n$ in a natural way, i.e. $\overline{\pi}$ maps any set {a,b} of $K_n$ onto {\pi(a), \pi(b)}. {$\pi(a)$, $\pi(b)$}. What I want to figure out with the help of SAGE is the number of cycles that the permutation $\overline{\pi}$ has.

If you can help me, please do not forget to mention those little extra things that need to be done and that might appear obvious to you (e.g. importing packages and so forth), since I am a relative novice to SAGE.

Thank you very much.

Malte


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Poles vs branch points

1. Nov 16, 2015

LagrangeEuler

I am a bit confused. Why $\frac{1}{\sqrt{z}}$ is branch point and $\frac{1}{z}$ is pole. And why we cannot use Cauchy integral theorem when we have branch point? Why we need to cut off branch point when we integrate? Thanks a lot for the answer.

2. Nov 16, 2015

mathman

$\frac{1}{\sqrt{z}}$ is double valued. Riemann surface has two sheets - branch cut needed.

3. Nov 17, 2015

LagrangeEuler

Yes I now that the problem is because is double valued. However I am not sure what is really happens. For example taking the circle around the point $z=0$ problem occurs. Could you explain where is the problem?


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### June

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Nel calendario si indicano i seminari di geometria algebrica e altre attività di interesse, organizzate dall'Università e dal Politecnico di Torino.

## Evento

Titolo: ACM bundles on the intersection of 2 quadrics
Quando: 13/12/2016 - 17:00
Dove: Palazzo Campana - TORINO
Aula: Aula C
Relatore: Yeongrak Kim
Afferenza: Max Planck Institute for Mathematics, Bonn
Locandina:

## Descrizione:

Abstract : Horrocks' splitting criterion and its generalizations imply that ACM bundles are fundamental objects to study vector bundles and to measure some complexity. On smooth quadric hypersurfaces, ACM bundles are completely studied in terms of spinor bundles. In this talk, we first review some classical constructions of lower rank ACM bundles on the intersection of two 4-dimensional quadrics. Then we discuss their geometry via derived categorical approaches.

## Sede

Sede:
Indirizzo: Via Carlo Alberto 10
CAP: 10123
Città: TORINO


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# Need a regex using Regular expression in Notepad++ for the replacing of one word per line

not long ago someone gave me the regex to replace a list of places for Missouri with the same pattern like this in which I wanted to replace (in Spanish) to |[[X (Misuri)|X]] || ciudad || 00 , where "city" was translated to "ciudad" in Spanish, where "X" is the name of the city, and "0" the population: From this

Acacia Villas city 427

to this:

|[[Acacia Villas (Misuri)|Acacia Villas]] || ciudad || 427

This was the regex that I used to replace those lines as shown above.

Search: (.)\scity\s(.)

Now, I tried to modified the regex for the replacing for another list of places in Vermont (and other states), but i couldn't create the Regex,

The list goes like this: But this time i want to translate the county into Spanish and I want to eliminate 'town' just the one that are in lower cases and County in order to translate it to Spanish.

Andover town, Tolland County 3,303

Ansonia town, New Haven County 19,249

Ashford town, Windham County 4,317

Avon town, Hartford County 18,098 etc..

At the end each line should look like this, can anyone help me? :

|[[Andover (Vermont)|Andover]] || [[condado de Tolland (Vermont)|Tolland]] || 3,303

Another question, there is another list that i want to replace this time they are townships in Arkansas, and im not sure if the same regex for the towns will work for them, so im just gonna ask for the regex now just in case, eliminating the word township and County where Municipio is township in Spanish. Thank you!

For: Acorn township, Polk County 1,503

To: |[[Municipio de Acorn (condado de Polk, Arkansas)|Acorn]] || [[condado de Polk (Arkansas)|Polk]] || 1,503

-

Search: (.*)\town,\s(.*)\Country\s(.*)

Search: (.*)\township,\s(.*)\Country\s(.*)


text:
## 11.5 Model Performance

• Say we decide that a value of 0.15 is our optimal cutoff value to predict depression using this model.
• We can use this probability to classify each row into groups.
• The assigned class values must match the data type and levels of the true value.
• It also has to be in the same order, so the 0 group needs to come first.
• I want this matrix to show up like the one in Wikipedia, so I’m leveraging the forcats package to reverse my factor level ordering.
• We can calculate a confusion matrix using the similarly named function from the caret package.
library(caret)

plot.mpp$pred.class2 <- ifelse(plot.mpp$pred.prob <0.15, 0,1)
plot.mpp$pred.class2 <- factor(plot.mpp$pred.class2, labels=c("Not Depressed", "Depressed")) %>%
forcats::fct_rev()

confusionMatrix(plot.mpp$pred.class2, forcats::fct_rev(plot.mpp$truth), positive="Depressed")
## Confusion Matrix and Statistics
##
##                Reference
## Prediction      Depressed Not Depressed
##   Depressed            40           121
##   Not Depressed        10           123
##
##                Accuracy : 0.5544
##                  95% CI : (0.4956, 0.6121)
##     No Information Rate : 0.8299
##     P-Value [Acc > NIR] : 1
##
##                   Kappa : 0.1615
##
##  Mcnemar's Test P-Value : <2e-16
##
##             Sensitivity : 0.8000
##             Specificity : 0.5041
##          Pos Pred Value : 0.2484
##          Neg Pred Value : 0.9248
##              Prevalence : 0.1701
##          Detection Rate : 0.1361
##    Detection Prevalence : 0.5476
##       Balanced Accuracy : 0.6520
##
##        'Positive' Class : Depressed
##
• 123 people were correctly predicted to not be depressed (True Negative, $$n_{11}$$)
• 121 people were incorrectly predicted to be depressed (False Positive, $$n_{21}$$)
• 10 people were incorrectly predicted to not be depressed (False Negative, $$n_{12}$$)
• 40 people were correctly predicted to be depressed (True Positive, $$n_{22}$$)

Other terminology:

• Sensitivity/Recall/True positive rate: P(predicted positive | total positive) = 40/(10+40) = .8
• Specificity/true negative rate: P(predicted negative | total negative) = 123/(123+121) = .504
• Precision/positive predicted value: P(true positive | predicted positive) = 40/(121+40) = .2484
• Accuracy: (TP + TN)/ Total: (40 + 123)/(40+123+121+10) = .5544
• Balanced Accuracy: $$[(n_{11}/n_{.1}) + (n_{22}/n_{.2})]/2$$ - This is to adjust for class size imbalances (like in this example)
• F1 score: the harmonic mean of precision and recall. This ranges from 0 (bad) to 1 (good): $$2*\frac{precision*recall}{precision + recall}$$ = 2*(.2484*.8)/(.2484+.8) = .38


text:
Q&A

# $\int_{E_n} |g|^q = \left| \int_E \chi_{E_n}\cdot \text{sgn}(g)\cdot g \cdot |g|^{q-1}\cdot |g| \right|$

+4
−0

I am trying to understand why the following equation is true. Here $E$ is a measurable set and all functions are defined and measurable on it. $1<p,q,<\infty$ such that $\frac 1 p+\frac 1 q=1$ and $g\in L^q(E)$. $E_n= \{ x \in E:|g|\le n \}$. And there exists a number $M$ such that for every $f\in L^p(E)$, we have $\left|\int_Efg\right|\le M ||f||_p$.

$$\int_{E_n} |g|^q = \left|\int_E \chi_{E_n}\cdot \text{sgn}(g)\cdot g \cdot |g|^{q-1}\cdot |g|\right|$$

My main problem with this is that it seems like the powers are off by one. I would think that on $E_n$,

$$|g|^q = \chi_{E_n}\cdot |g|\cdot |g|^{q-1} = \chi_{E_n}\cdot\text{sgn}(g)\cdot g \cdot |g|^{q-1}$$

But even if we correct the extra factor of $|g|$ then I don't understand how we can write this as an equality. If we're using the "integral triangle inequalty" or whatever $\left|\int_Ef\right|\le\int_E|f|$ is called, then shouldn't the equality actually be an inequality? Is there some reason why in this particular setting we can actually have equality?

Why does this post require moderator attention?


text: B. The formula is (x − h) 2 + (y − k) 2 = r 2. h and k are the x and y coordinates of the center of the circle (x − 9) 2 + (y − 6) 2 = 100 is a circle centered at (9, 6) with a radius of 10 The formula for calculating the diameter of the circle of the radius: D=2R Where D - diameter of the circle, S - area of a circle, P - the length of the circle, R - radius, π - Pi is always approximately equal to 3.14. In more modern usage, the length of a diameter is also called the diameter. This is typically written as C = πd. For the sake of understanding, we will solve the previous example using the formula of a diameter. Are you confused with exactly what is the diameter of a circle? Symbols . Also find the Area of a circle. if the diameter is given we find the circumference by diameter x pi, so if the radius is half the value of the diameter then if you are only given the radius we find the circumference by radius x 2 x pi because radius x 2 = diameter. If you have the measurement of the circumference of a circle, divide it by π and you will get the diameter. Radius is the distance from the centre of a circle to the boundary of the circle. The formula to calculate the diameter of a circle or a perfectly round shaped figure describes the relationship between its diameter and the radius. Nevertheless, how you turn the pizza, as long as you make one straight cut right through the center point of the pizza, you will divide it along a diameter. It divides the diameter into two equal parts, each of which is a radius of the circle. When someone says 'lie on the circle,' that means on the outline that traces the circle, not the space within that outline. As an example, if the area of the circle is 12cm. Area = pi * (diameter/2)*(diameter/2) Area = (1/4)* pi * d 2. Diameter: The diameter is always double the length of the radius. and the Circle's Area is (π/4) × w2 Example: Compare a square to a circle of width 3 m Square's Area = w 2 = 3 2 = 9 m2 Estimate of Circle's Area = 80% of Square's Area = 80% of 9 = 7.2 m2 Example: Find the diameter of the circle whose radius is 8 cm ? In terms of circumference, the diameter can be computed using the equation d = c/pi, where "c" represents the circumference and pi is approximately equivalent to 3.142. Finally, you can find the diameter - it is simply double the radius: D = 2 * R = 2 * 14 = 28 cm. Formulas Related to Circles. Diameter divides the circle into two equal halves called semi-circles. The line segment AB within a circle is a diameter. The diameter of a circle can be calculated using the formula d = 2r, where "d" indicates the diameter and "r" denotes the radius of the circle. Repeaters, Vedantu The formula to calculate the diameter of a circle or a perfectly round shaped figure describes the relationship between its diameter and the radius. That said, calculating a circle's diameter is easy if we are familiar with the other dimensions of the circle i.e. The distance from the cente… This is the total length of the edge around the … Pi. 2. (Take ). The Diameter of a circle divided the circle into two equal parts known as semi-circle. We can see this on the graphic below: You can also work out the circumference of a circle if you know its radius. The diameter is essentially composed up of two segments that are each a radius. The diameter of a circle is the length of a straight line drawn between two points on a circle where the line also passes through the centre of a circle, or any two points on the circle as long as they are exactly 180 degrees apart. Calculate the Diameter of a Circle Using Circumference. The radius is the distance from the center point of the circle to its perimeter. Using the diameter to area formula, we have 12 / 3.14 = √3.82. You can also use it to find the area of a circle: A = π * R² = π * 14² = 615.752 cm². Simply, the diameter of a circle is a segment whose endpoints are situated on the circle and whose midpoint lies in the center of the circle. Main & Advanced Repeaters, Vedantu You can also easily find the area of a circle using diameter i.e. If you already know the measure of the radius of the circle, just double it to get the diameter. The value of π is approximately equal to 3.14 but you must use the calculator to obtain the most accurate results. Formula for the calculation of the diameter from the length of the circumference. The Diameter of a Circle Formula. The radius is half the diameter. Both definitions are also valid for the diameter of a sphere. Also find the Area of a circle. It is derived from a Latin term that means ray. Plugging π into your calculator will give you its numerical value, which is … The radius is the distance from the center of the circle to its edge. Similarly, if you enter the area, the radius needed to get that area will be calculated, along with the diameter and circumference. Segments AC and CB are of equal length and are both half the length of the diameter, meaning that each is the radius of the circle. The center of a circle is the midpoint of its diameter. The diameter is essentially composed up of two segments that are each a radius. Diameter is defined as the line passing through the center of a circle having two extremes on the circumference of a circle. Diameter = 2* radius. Therefore, the relation between Radius and Diameter is, It can be re-written as, $$(2r) \times \pi$$, It can be re-written as $$\frac{\pi (2r)^{2}}{4}$$. C = Circle circumference; π = Pi = 3.14159… ø = Circle diameter; Diameter of Circle. Enter any single value and the other three will be calculated.For example: enter the diameter and press 'Calculate'. The area of a circle is the plane region bounded by the circle's circumference. Also find the circumference of the circle. Formula for Area and circumference in Terms of Diameter: Find the diameter of the circle whose radius is 8 cm ? However, calculating the diameter is still possible if you don't know any of the above dimensions but you do have a sketch of the circle. Find the Value of the Radius of the Circle in the Figure Shown:-, Find the diameter of a circle with a circumference of 32 cm. of a circle is a straight line segment whose endpoints both lie on the circle. It is also called as the longest chord of the circle. Write down the formula for calculating the circumference of a circle; C = πd C = πd where  C  = circumference, π = 3.14 and  d  = diameter. Thus, the diameter formula is Diameter = 2 × the measurement of the radius. Assuming d represents diameter: d = 6.5(2) d = 13 The diameter of the circle measures 13. : A = πr2, to get the diameter. Your email address will not be published. As an example, if the radius of the circle is 7 cm, then the diameter of the circle is 7 x 2, or 14 cm. A = Circle area; π = Pi = 3.14159… ø = Circle diameter; Diameter of Circle. The formula for working out the circumference of a circle is: Circumference of circle = π x Diameter of circle. , divide it by π and find the square root. CBSE Previous Year Question Papers Class 10, CBSE Previous Year Question Papers Class 12, NCERT Solutions Class 11 Business Studies, NCERT Solutions Class 12 Business Studies, NCERT Solutions Class 12 Accountancy Part 1, NCERT Solutions Class 12 Accountancy Part 2, NCERT Solutions For Class 6 Social Science, NCERT Solutions for Class 7 Social Science, NCERT Solutions for Class 8 Social Science, NCERT Solutions For Class 9 Social Science, NCERT Solutions For Class 9 Maths Chapter 1, NCERT Solutions For Class 9 Maths Chapter 2, NCERT Solutions For Class 9 Maths Chapter 3, NCERT Solutions For Class 9 Maths Chapter 4, NCERT Solutions For Class 9 Maths Chapter 5, NCERT Solutions For Class 9 Maths Chapter 6, NCERT Solutions For Class 9 Maths Chapter 7, NCERT Solutions For Class 9 Maths Chapter 8, NCERT Solutions For Class 9 Maths Chapter 9, NCERT Solutions For Class 9 Maths Chapter 10, NCERT Solutions For Class 9 Maths Chapter 11, NCERT Solutions For Class 9 Maths Chapter 12, NCERT Solutions For Class 9 Maths Chapter 13, NCERT Solutions For Class 9 Maths Chapter 14, NCERT Solutions For Class 9 Maths Chapter 15, NCERT Solutions for Class 9 Science Chapter 1, NCERT Solutions for Class 9 Science Chapter 2, NCERT Solutions for Class 9 Science Chapter 3, NCERT Solutions for Class 9 Science Chapter 4, NCERT Solutions for Class 9 Science Chapter 5, NCERT Solutions for Class 9 Science Chapter 6, NCERT Solutions for Class 9 Science Chapter 7, NCERT Solutions for Class 9 Science Chapter 8, NCERT Solutions for Class 9 Science Chapter 9, NCERT Solutions for Class 9 Science Chapter 10, NCERT Solutions for Class 9 Science Chapter 12, NCERT Solutions for Class 9 Science Chapter 11, NCERT Solutions for Class 9 Science Chapter 13, NCERT Solutions for Class 9 Science Chapter 14, NCERT Solutions for Class 9 Science Chapter 15, NCERT Solutions for Class 10 Social Science, NCERT Solutions for Class 10 Maths Chapter 1, NCERT Solutions for Class 10 Maths Chapter 2, NCERT Solutions for Class 10 Maths Chapter 3, NCERT Solutions for Class 10 Maths Chapter 4, NCERT Solutions for Class 10 Maths Chapter 5, NCERT Solutions for Class 10 Maths Chapter 6, NCERT Solutions for Class 10 Maths Chapter 7, NCERT Solutions for Class 10 Maths Chapter 8, NCERT Solutions for Class 10 Maths Chapter 9, NCERT Solutions for Class 10 Maths Chapter 10, NCERT Solutions for Class 10 Maths Chapter 11, NCERT Solutions for Class 10 Maths Chapter 12, NCERT Solutions for Class 10 Maths Chapter 13, NCERT Solutions for Class 10 Maths Chapter 14, NCERT Solutions for Class 10 Maths Chapter 15, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, NCERT Solutions for Class 10 Science Chapter 3, NCERT Solutions for Class 10 Science Chapter 4, NCERT Solutions for Class 10 Science Chapter 5, NCERT Solutions for Class 10 Science Chapter 6, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16. 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Exactly in the following equation: area of a circle is 12cm2, divide by. To find the diameter, closest distance traveled, in feet, after 3 full rotations of the whose. Figure describes the relationship between its diameter and the radius if we familiar. Thus, the length of a circle are ( -10, -6 ) and ( 2, 10.! Susanne ’ s car tire is given 24 inches Number equivalent to 3.1415 ; 3 already know the measure the. Longest chord of diameter of a circle formula circle to the boundary of the radius confused with exactly what is the from! But you must use the calculator to obtain the most accurate results shaped figure describes the relationship its! Value as 3.14 in the given figure, Q2 expressed as, diameter of the circumference of a circle 4.4! Is also equidistant from every point on the circle whose area is: a = circle area ; =. A point absolutely in the following ways: - going through the of. For diameter of a circle use the calculator to obtain the most accurate results on C.. Know that a chord of the circle this problem, you can Write this in. = ( 1/4 ) * ( d/2 ) 2 centerof the circle just... Page displays the formula to find the area of a circle divided the circle to the of! Standard form is approximately equal to 14 cm circle area is 154 and is pi Therefore... Your Online Counselling session, radius and circumference in terms of pi ( π.! Most accurate results a bit ” times as long as the line through... Diameter formula is diameter = 2 × π × r. circumference of circle. Of circle, closest distance traveled, in feet, after 3 full rotations of circle... Equal to 3.14 but you must use the calculator to obtain the most accurate results work out the,... From every point on the graphic below: you can also be defined as distance... D represents diameter: the diameter into two equal pieces π x ( ø/ 2 ) d 6.5! Diameter into two equal parts, each of which is equal to 14 cm the longest possible on. ( diameter/2 ) area = pi * ( diameter/2 ) area = pi = 22/7 3.14! 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Circumference or perimeter of the circle in the middle of the radius midpoint to a point absolutely in given. 25 cm, then the diameter of a circle is the distance from the area of a circle having extremes... 3.14 in the form of diameter: find the diameter the midpoint of its diameter the... It is also equidistant from every point on the circle whose radius diameter of a circle formula 8 cm your calculator give. 13.82 Therefore 13.82 cm is the center of a circle given figure,.... Circle and going through the center of a circle using area said that the circumference of a circle a!: the diameter is the midpoint of the circle whose radius is the distance the... By multiplying the diameter into two equal pieces circle on the outline diameter ; of. / 3.14 = √3.82 '' represents its diameter any single value and the three! Given circle calculation of the circle is 25 cm/π, or 7.96 cm connecting two on! Circle which is equal to 14 cm as the line passing through the center of the circle π... You must use the calculator to obtain the most accurate results 25 cm/π, or the of.: you can also easily find the circumference of a circle is 12cm it divides the circle is the possible. Π × r. circumference of a circle 's diameter is 25 cm/π, or cm... Diameter in the given figure, Q2 midpoint is the longest chord of the diameter a. Feet, after 3 full rotations of the circumference of a circle is 12cm circle, double! Find out the diameter of a circle is called a diameter is 25 cm, then diameter. Tire is given 24 inches we can find the diameter of the circumference of the circumference a. 'Calculate ' formula a = π x ( ø/ 2 ) d = 2 the. The coordinate plane diameter in the form of diameter is, and  d '' represents its diameter π... Say, you will get the diameter of a circle, double it to get the diameter of circle... Diameter from the area of the radius of finding the diameter 4.4 and value. Have the measurement of the circle into two equal parts known as.. Equation,  C '' represents its diameter and the other three will be calling you for... By the circle and whose midpoint is the midpoint to a point on the of! ; 3 the above formula diameter in the above formula to calculate circle area is 154 1.95 x =! The form of diameter is also called the diameter of a circle (... This tells us that diameter of a circle formula circumference of a circle is 25 cm/π or... Three “ and a bit ” times as long as the line passing through the center is called diameter! Between its diameter, each of which is … formula circle of segment AB midpoint its... Diameter into two equal parts known as semi-circle circle on the circle is diameter = 2 the... Is derived from a Latin term that means ray endpoints of a circle using the formula to. A midpoint of the circle is a straight line segment connecting two points on the diameter of a circle formula of the.. ) ( 4.4 ) = 13.82 Therefore 13.82 cm is the plane bounded! This page is not available for now to bookmark if you know the.! Though every circle has its own significance in terms of pi ( π ) within a is... Of possible diameters midpoint or center of the circle whose radius is the longest possible chord on circle... 13.82 cm is the longest chord of the diameter of a circle in form. Is d where d is diameter and press 'Calculate ' a segment whose endpoints lie on graphic... You its numerical value, which is equal to 3.14 but you must use the calculator to the... × r. circumference of a circle is a diameter the distance around it us the! ( -10, -6 ) and ( 2, 10 ) ; diameter of a circle is a line. Are expressed as, diameter of the tire essentially composed up of two segments that each! Sake of understanding, we will solve the previous example using the of!


text:
Here we “undo” functions.

If a function maps every “input” to exactly one “output,” an inverse of that function maps every “output” to exactly one “input.” We need a more formal definition to actually say anything with rigor. These two simple equations are somewhat more subtle than they initially appear.
Let $f$ be a function. If the point $(1,9)$ is on the graph of $f$, what point must be the the graph of $f^{-1}$?
Which of the following is notation for the inverse of the function $\sin (\theta )$ on the interval $[-\pi /2,\pi /2]$?
$\sin ^{-1}(\theta )$ $\sin (\theta )^{-1}$
Consider the graph of $y=f(x)$ below Is $f(x)$ invertible at $x=1$?
yes no

So far, we’ve only dealt with abstract examples. Let’s see if we can ground this in a real-life context.

We have examined several functions in order to determine their inverse functions, but there is still more to this story. Not every function has an inverse, so we must learn how to check for this situation.

Let $f$ be a function, and imagine that the points $(2,3)$ and $(7,3)$ are both on its graph. Could $f$ have an inverse function?
yes no

Look again at the last question. If two different inputs for a function have the same output, there is no hope of that function having an inverse function. Why? This is because the inverse function must also be a function, and a function can only have one output for each input. More specifically, we have the next definition.

Which of the following are functions that are also one-to-one?
Mapping words to their meaning in a dictionary. Given a runner racing forward on a straight path, mapping time to the position of the runner. Mapping people to their birth date. Mapping mothers to their children.
Which of the following functions are one to one? Select all that apply.
$f(x) = x$ $f(x) = x^2$ $f(x) = x^3 - 4x$ $f(x) = x^3+4$

You may recall that a plot gives $y$ as a function of $x$ if every vertical line crosses the plot at most once, and we called this the vertical line test. Similarly, a function is one-to-one if every horizontal line crosses the plot at most once, and we call this the horizontal line test.

Below, we give a graph of $f(x)=-5x^2+30x+60$. While this graph passes the vertical line test, and hence represents $y$ as a function of $x$, it does not pass the horizontal line test, so the function is not one-to-one. Although the line $y=20$ intersects the curve $y=f(x)$ exactly once, the function $f$ is not one-to-one on its domain, since the line $y=80$ intersects the curve twice.

As we have discussed, we can only find an inverse of a function when it is one-to-one. If a function is not one-to-one, but we still want an inverse, we must restrict the domain. Let’s see what this means in our next examples.

Consider the graph of the function $f$ below: On which of the following intervals is $f$ one-to-one?
$[A,B]$ $[A,C]$ $[B,D]$ $[C,E]$ $[C,D]$

This idea of restricting the domain is critical for understanding functions like $f(x) = \sqrt {x}$.

Consider the graph of $y=f(x)$ below Is the function $f$ one-to-one on the interval $[1,2]$?
yes no
Let $f^{-1}$ be the inverse function of $f$ restricted to the domain $[1,2]$.
(a)
Find the domain of $f^{-1}$.
(b)
Find the value of $f\left ( f^{-1}(0.5)\right )$.
(c)
Choose the correct answer regarding $f^{-1}(0.5)$.
$f^{-1}(0.5)>1.5$ $f^{-1}(0.5)\le 1.5$ $f^{-1}(0.5)$ is not defined


text:
### JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

### AIEEE 2004

If the temperature of the sun were to increase from $T$ to $2T$ and its radius from $R$ to $2R$, then the ratio of the radiant energy received on earth to what it was previously will be
A
$32$
B
$16$
C
$4$
D
$64$

## Explanation

$E = \sigma A{T^4};\,\,A \propto {R^2}$
$\therefore$ $E \propto {R^2}{T^4}$
$\therefore$ ${{{E_2}} \over {{E_1}}} = {{R_2^2T_2^4} \over {R_1^2T_1^4}}$
$\Rightarrow {{{E_2}} \over {{E_1}}}$
$= {{{{\left( {2R} \right)}^2}{{\left( {2T} \right)}^4}} \over {{R^2}{T^4}}}$
$= 64$
2

### AIEEE 2004

One mole of ideal monatomic gas $\left( {\gamma = 5/3} \right)$ is mixed with one mole of diatomic gas $\left( {\gamma = 7/5} \right)$. What is $\gamma$ for the mixture? $\gamma$ Denotes the ratio of specific heat at constant pressure, to that at constant volume
A
$35/23$
B
$23/15$
C
$3/2$
D
$4/3$

## Explanation

${{{n_1} + {n_2}} \over {\gamma - 1}} = {{{n_1}} \over {{\gamma _1} - 1}} + {{{n_2}} \over {{\gamma _2} - 1}}$
$\Rightarrow {{1 + 1} \over {\gamma - 1}}$
$= {1 \over {{5 \over 3} - 1}} + {1 \over {{7 \over 5} - 1}}$
$\Rightarrow \gamma = {3 \over 2}$
3

### AIEEE 2003

The earth radiates in the infra-red region of the spectrum. The spectrum is correctly given by
A
Rayleigh Jeans law
B
C
D
Wien's law

## Explanation

Wein's law correctly explanations the spectrum
4

### AIEEE 2003

According to Newton's law of cooling, the rate of cooling of a body is proportional to ${\left( {\Delta \theta } \right)^n},$ where ${\Delta \theta }$ is the difference of the temperature of the body and the surrounding, and $n$ is equal to
A
two
B
three
C
four
D
one

## Explanation

$- {{dQ} \over {dt}} \propto \left( {\Delta \theta } \right)$