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3100 | https://www.sans.org/security-resources/glossary-of-terms | Glossary of Cyber Security Terms
Become your company’s cyber security thesaurus. Find the definition of the most commonly used cyber security terms in our glossary below.
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A-C
Access Control
Access Control ensures that resources are only granted to those users who are entitled to them.
Access Control List (ACL)
A mechanism that implements access control for a system resource by listing the identities of the system entities that are permitted to access the resource.
Access Control Service
A security service that provides protection of system resources against unauthorized access. The two basic mechanisms for implementing this service are ACLs and tickets.
Access Management Access
Management is the maintenance of access information which consists of four tasks: account administration, maintenance, monitoring, and revocation.
Access Matrix
An Access Matrix uses rows to represent subjects and columns to represent objects with privileges listed in each cell.
Account Harvesting
Account Harvesting is the process of collecting all the legitimate account names on a system.
ACK Piggybacking
ACK piggybacking is the practice of sending an ACK inside another packet going to the same destination.
Active Content
Program code embedded in the contents of a web page. When the page is accessed by a web browser, the embedded code is automatically downloaded and executed on the user's workstation. Ex. Java, ActiveX (MS)
Activity Monitors
Activity monitors aim to prevent virus infection by monitoring for malicious activity on a system, and blocking that activity when possible.
Address Resolution Protocol (ARP)
Address Resolution Protocol (ARP) is a protocol for mapping an Internet Protocol address to a physical machine address that is recognized in the local network. A table, usually called the ARP cache, is used to maintain a correlation between each MAC address and its corresponding IP address. ARP provides the protocol rules for making this correlation and providing address conversion in both directions.
Advanced Encryption Standard (AES)
An encryption standard being developed by NIST. Intended to specify an unclassified, publicly-disclosed, symmetric encryption algorithm.
Algorithm
A finite set of step-by-step instructions for a problem-solving or computation procedure, especially one that can be implemented by a computer.
Applet
Java programs; an application program that uses the client's web browser to provide a user interface.
ARPANET
Advanced Research Projects Agency Network, a pioneer packet-switched network that was built in the early 1970s under contract to the US Government, led to the development of today's Internet, and was decommissioned in June 1990.
Asymmetric Cryptography
Public-key cryptography; A modern branch of cryptography in which the algorithms employ a pair of keys (a public key and a private key) and use a different component of the pair for different steps of the algorithm.
Asymmetric Warfare
Asymmetric warfare is the fact that a small investment, properly leveraged, can yield incredible results.
Auditing
Auditing is the information gathering and analysis of assets to ensure such things as policy compliance and security from vulnerabilities.
Authentication
Authentication is the process of confirming the correctness of the claimed identity.
Authenticity
Authenticity is the validity and conformance of the original information.
Authorization
Authorization is the approval, permission, or empowerment for someone or something to do something.
Autonomous System
One network or series of networks that are all under one administrative control. An autonomous system is also sometimes referred to as a routing domain. An autonomous system is assigned a globally unique number, sometimes called an Autonomous System Number (ASN).
Availability
Availability is the need to ensure that the business purpose of the system can be met and that it is accessible to those who need to use it.
Backdoor
A backdoor is a tool installed after a compromise to give an attacker easier access to the compromised system around any security mechanisms that are in place.
Bandwidth
Commonly used to mean the capacity of a communication channel to pass data through the channel in a given amount of time. Usually expressed in bits per second.
Banner
A banner is the information that is displayed to a remote user trying to connect to a service. This may include version information, system information, or a warning about authorized use.
Basic Authentication
Basic Authentication is the simplest web-based authentication scheme that works by sending the username and password with each request.
Bastion Host
A bastion host has been hardened in anticipation of vulnerabilities that have not been discovered yet.
BIND
BIND stands for Berkeley Internet Name Domain and is an implementation of DNS. DNS is used for domain name to IP address resolution.
Biometrics
Biometrics use physical characteristics of the users to determine access.
Bit
The smallest unit of information storage; a contraction of the term "binary digit;" one of two symbolsN"0" (zero) and "1" (one) - that are used to represent binary numbers.
Block Cipher
A block cipher encrypts one block of data at a time.
Blue Team
The people who perform defensive cybersecurity tasks, including placing and configuring firewalls, implementing patching programs, enforcing strong authentication, ensuring physical security measures are adequate and a long list of similar undertakings.
Boot Record Infector
A boot record infector is a piece of malware that inserts malicious code into the boot sector of a disk.
Border Gateway Protocol (BGP)
An inter-autonomous system routing protocol. BGP is used to exchange routing information for the Internet and is the protocol used between Internet service providers (ISP).
Botnet
A botnet is a large number of compromised computers that are used to create and send spam or viruses or flood a network with messages as a denial of service attack.
Bridge
A product that connects a local area network (LAN) to another local area network that uses the same protocol (for example, Ethernet or token ring).
British Standard 7799
A standard code of practice and provides guidance on how to secure an information system. It includes the management framework, objectives, and control requirements for information security management systems.
Broadcast
To simultaneously send the same message to multiple recipients. One host to all hosts on network.
Broadcast Address
An address used to broadcast a datagram to all hosts on a given network using UDP or ICMP protocol.
Browser
A client computer program that can retrieve and display information from servers on the World Wide Web.
Brute Force
A cryptanalysis technique or other kind of attack method involving an exhaustive procedure that tries all possibilities, one-by-one.
Buffer Overflow
A buffer overflow occurs when a program or process tries to store more data in a buffer (temporary data storage area) than it was intended to hold. Since buffers are created to contain a finite amount of data, the extra information - which has to go somewhere - can overflow into adjacent buffers, corrupting or overwriting the valid data held in them.
Business Continuity Plan (BCP)
A Business Continuity Plan is the plan for emergency response, backup operations, and post-disaster recovery steps that will ensure the availability of critical resources and facilitate the continuity of operations in an emergency situation.
Business Impact Analysis (BIA)
A Business Impact Analysis determines what levels of impact to a system are tolerable.
Byte
A fundamental unit of computer storage; the smallest addressable unit in a computer's architecture. Usually holds one character of information and usually means eight bits.
Cache
Pronounced cash, a special high-speed storage mechanism. It can be either a reserved section of main memory or an independent high-speed storage device. Two types of caching are commonly used in personal computers: memory caching and disk caching.
Cache Cramming
Cache Cramming is the technique of tricking a browser to run cached Java code from the local disk, instead of the internet zone, so it runs with less restrictive permissions.
Cache Poisoning
Malicious or misleading data from a remote name server is saved [cached] by another name server. Typically used with DNS cache poisoning attacks.
Call Admission Control (CAC)
The inspection and control all inbound and outbound voice network activity by a voice firewall based on user-defined policies.
Cell
A cell is a unit of data transmitted over an ATM network.
Certificate-Based Authentication
Certificate-Based Authentication is the use of SSL and certificates to authenticate and encrypt HTTP traffic.
CGI
Common Gateway Interface. This mechanism is used by HTTP servers (web servers) to pass parameters to executable scripts in order to generate responses dynamically.
Chain of Custody
Chain of Custody is the important application of the Federal rules of evidence and its handling.
Challenge-Handshake Authentication Protocol (CHAP)
The Challenge-Handshake Authentication Protocol uses a challenge/response authentication mechanism where the response varies every challenge to prevent replay attacks.
Checksum
A value that is computed by a function that is dependent on the contents of a data object and is stored or transmitted together with the object, for the purpose of detecting changes in the data.
Cipher
A cryptographic algorithm for encryption and decryption.
Ciphertext
Ciphertext is the encrypted form of the message being sent.
Circuit Switched Network
A circuit switched network is where a single continuous physical circuit connected two endpoints where the route was immutable once set up.
Client
A system entity that requests and uses a service provided by another system entity, called a "server." In some cases, the server may itself be a client of some other server.
Cloud Computing
Utilization of remote servers in the data-center of a cloud provider to store, manage, and process your data instead of using local computer systems.
Cold/Warm/Hot Disaster Recovery Site
Hot site. It contains fully redundant hardware and software, with telecommunications, telephone and utility connectivity to continue all primary site operations. Failover occurs within minutes or hours, following a disaster. Daily data synchronization usually occurs between the primary and hot site, resulting in minimum or no data loss. Offsite data backup tapes might be obtained and delivered to the hot site to help restore operations. Backup tapes should be regularly tested to detect data corruption, malicious code and environmental damage. A hot site is the most expensive option. Warm site. It contains partially redundant hardware and software, with telecommunications, telephone and utility connectivity to continue some, but not all primary site operations. Failover occurs within hours or days, following a disaster. Daily or weekly data synchronization usually occurs between the primary and warm site, resulting in minimum data loss. Offsite data backup tapes must be obtained and delivered to the warm site to restore operations. A warm site is the second most expensive option. Cold site. Hardware is ordered, shipped and installed, and software is loaded. Basic telecommunications, telephone and utility connectivity might need turning on to continue some, but not all primary site operations. Relocation occurs within weeks or longer, depending on hardware arrival time, following a disaster. No data synchronization occurs between the primary and cold site, and could result in significant data loss. Offsite data backup tapes must be obtained and delivered to the cold site to restore operations. A cold site is the least expensive option.
Collision
A collision occurs when multiple systems transmit simultaneously on the same wire.
Competitive Intelligence
Competitive Intelligence is espionage using legal, or at least not obviously illegal, means.
Computer Emergency Response Team (CERT)
An organization that studies computer and network INFOSEC in order to provide incident response services to victims of attacks, publish alerts concerning vulnerabilities and threats, and offer other information to help improve computer and network security.
Computer Network
A collection of host computers together with the sub-network or inter-network through which they can exchange data.
Confidentiality
Confidentiality is the need to ensure that information is disclosed only to those who are authorized to view it.
Configuration Management
Establish a known baseline condition and manage it.
Cookie
Data exchanged between an HTTP server and a browser (a client of the server) to store state information on the client side and retrieve it later for server use. An HTTP server, when sending data to a client, may send along a cookie, which the client retains after the HTTP connection closes. A server can use this mechanism to maintain persistent client-side state information for HTTP-based applications, retrieving the state information in later connections.
Corruption
A threat action that undesirably alters system operation by adversely modifying system functions or data.
Cost Benefit Analysis
A cost benefit analysis compares the cost of implementing countermeasures with the value of the reduced risk.
Countermeasure
Reactive methods used to prevent an exploit from successfully occurring once a threat has been detected. Intrusion Prevention Systems (IPS) commonly employ countermeasures to prevent intruders form gaining further access to a computer network. Other counter measures are patches, access control lists and malware filters.
Covert Channels
Covert Channels are the means by which information can be communicated between two parties in a covert fashion using normal system operations. For example by changing the amount of hard drive space that is available on a file server can be used to communicate information.
Crimeware
A type of malware used by cyber criminals. The malware is designed to enable the cyber criminal to make money off of the infected system (such as harvesting key strokes, using the infected systems to launch Denial of Service Attacks, etc.).
Cron
Cron is a Unix application that runs jobs for users and administrators at scheduled times of the day.
Crossover Cable
A crossover cable reverses the pairs of cables at the other end and can be used to connect devices directly together.
Cryptanalysis
The mathematical science that deals with analysis of a cryptographic system in order to gain knowledge needed to break or circumvent the protection that the system is designed to provide. In other words, convert the cipher text to plaintext without knowing the key.
Cryptographic Algorithm or Hash
An algorithm that employs the science of cryptography, including encryption algorithms, cryptographic hash algorithms, digital signature algorithms, and key agreement algorithms.
Cut-Through
Cut-Through is a method of switching where only the header of a packet is read before it is forwarded to its destination.
Cyber-Attack
A cyber- attack is any unauthorized attempt to access, disrupt, steal, or damage computer systems, networks, or data. Learn more.
Cybersecurity Risk Assessment
A cybersecurity risk assessment is the systematic process of identifying, analyzing, and evaluating potential threats, vulnerabilities, and impacts to an organization’s digital assets. Learn more.
Cyclic Redundancy Check (CRC)
Sometimes called "cyclic redundancy code." A type of checksum algorithm that is not a cryptographic hash but is used to implement data integrity service where accidental changes to data are expected.
D-F
Daemon
A program which is often started at the time the system boots and runs continuously without intervention from any of the users on the system. The daemon program forwards the requests to other programs (or processes) as appropriate. The term daemon is a Unix term, though many other operating systems provide support for daemons, though they're sometimes called other names. Windows, for example, refers to daemons and System Agents and services.
Data Aggregation
Data Aggregation is the ability to get a more complete picture of the information by analyzing several different types of records at once.
Data Breach
A data breach is a security incident in which sensitive, protected, or confidential information is accessed, stolen, or disclosed without authorization. Learn more.
Data Custodian
A Data Custodian is the entity currently using or manipulating the data, and therefore, temporarily taking responsibility for the data.
Data Encryption Standard (DES)
A widely-used method of data encryption using a private (secret) key. There are 72,000,000,000,000,000 (72 quadrillion) or more possible encryption keys that can be used. For each given message, the key is chosen at random from among this enormous number of keys. Like other private key cryptographic methods, both the sender and the receiver must know and use the same private key.
Data Mining
Data Mining is a technique used to analyze existing information, usually with the intention of pursuing new avenues to pursue business.
Data Owner
A Data Owner is the entity having responsibility and authority for the data.
Data Warehousing
Data Warehousing is the consolidation of several previously independent databases into one location.
Datagram
Request for Comment 1594 says, "a self-contained, independent entity of data carrying sufficient information to be routed from the source to the destination computer without reliance on earlier exchanges between this source and destination computer and the transporting network." The term has been generally replaced by the term packet. Datagrams or packets are the message units that the Internet Protocol deals with and that the Internet transports. A datagram or packet needs to be self-contained without reliance on earlier exchanges because there is no connection of fixed duration between the two communicating points as there is, for example, in most voice telephone conversations. (This kind of protocol is referred to as connectionless.)
Day Zero
The "Day Zero" or "Zero Day" is the day a new vulnerability is made known. In some cases, a "zero day" exploit is referred to an exploit for which no patch is available yet. ("day one"-> day at which the patch is made available).
Decapsulation
Decapsulation is the process of stripping off one layer's headers and passing the rest of the packet up to the next higher layer on the protocol stack.
Decryption
Decryption is the process of transforming an encrypted message into its original plaintext.
Defacement
Defacement is the method of modifying the content of a website in such a way that it becomes "vandalized" or embarrassing to the website owner.
Defense In-Depth
Defense In-Depth is the approach of using multiple layers of security to guard against failure of a single security component.
Demilitarized Zone (DMZ)
In computer security, in general a demilitarized zone (DMZ) or perimeter network is a network area (a subnetwork) that sits between an organization's internal network and an external network, usually the Internet. DMZ's help to enable the layered security model in that they provide subnetwork segmentation based on security requirements or policy. DMZ's provide either a transit mechanism from a secure source to an insecure destination or from an insecure source to a more secure destination. In some cases, a screened subnet which is used for servers accessible from the outside is referred to as a DMZ.
Denial of Service
The prevention of authorized access to a system resource or the delaying of system operations and functions.
Dictionary Attack
An attack that tries all of the phrases or words in a dictionary, trying to crack a password or key. A dictionary attack uses a predefined list of words compared to a brute force attack that tries all possible combinations.
Diffie-Hellman
A key agreement algorithm published in 1976 by Whitfield Diffie and Martin Hellman. Diffie-Hellman does key establishment, not encryption. However, the key that it produces may be used for encryption, for further key management operations, or for any other cryptography.
Digest Authentication
Digest Authentication allows a web client to compute MD5 hashes of the password to prove it has the password.
Digital Certificate
A digital certificate is an electronic "credit card" that establishes your credentials when doing business or other transactions on the Web. It is issued by a certification authority. It contains your name, a serial number, expiration dates, a copy of the certificate holder's public key (used for encrypting messages and digital signatures), and the digital signature of the certificate-issuing authority so that a recipient can verify that the certificate is real.
Digital Envelope
A digital envelope is an encrypted message with the encrypted session key.
Digital Signature
A digital signature is a hash of a message that uniquely identifies the sender of the message and proves the message hasn't changed since transmission.
Digital Signature Algorithm (DSA)
An asymmetric cryptographic algorithm that produces a digital signature in the form of a pair of large numbers. The signature is computed using rules and parameters such that the identity of the signer and the integrity of the signed data can be verified.
Digital Signature Standard (DSS)
The US Government standard that specifies the Digital Signature Algorithm (DSA), which involves asymmetric cryptography.
Disassembly
The process of taking a binary program and deriving the source code from it.
Disaster Recovery Plan (DRP)
A Disaster Recovery Plan is the process of recovery of IT systems in the event of a disruption or disaster.
Discretionary Access Control (DAC)
Discretionary Access Control consists of something the user can manage, such as a document password.
Disruption
A circumstance or event that interrupts or prevents the correct operation of system services and functions.
Distance Vector
Distance vectors measure the cost of routes to determine the best route to all known networks.
Distributed Scans
Distributed Scans are scans that use multiple source addresses to gather information.
Domain
A sphere of knowledge, or a collection of facts about some program entities or a number of network points or addresses, identified by a name. On the Internet, a domain consists of a set of network addresses. In the Internet's domain name system, a domain is a name with which name server records are associated that describe sub-domains or host. In Windows NT and Windows 2000, a domain is a set of network resources (applications, printers, and so forth) for a group of users. The user need only to log in to the domain to gain access to the resources, which may be located on a number of different servers in the network.
Domain Hijacking
Domain hijacking is an attack by which an attacker takes over a domain by first blocking access to the domain's DNS server and then putting his own server up in its place.
Domain Name
A domain name locates an organization or other entity on the Internet. For example, the domain name "www.sans.org" locates an Internet address for "sans.org" at Internet point 199.0.0.2 and a particular host server named "www". The "org" part of the domain name reflects the purpose of the organization or entity (in this example, "organization") and is called the top-level domain name. The "sans" part of the domain name defines the organization or entity and together with the top-level is called the second-level domain name.
Domain Name System (DNS)
The domain name system (DNS) is the way that Internet domain names are located and translated into Internet Protocol addresses. A domain name is a meaningful and easy-to-remember "handle" for an Internet address.
Due Care
Due care ensures that a minimal level of protection is in place in accordance with the best practice in the industry.
Due Diligence
Due diligence is the requirement that organizations must develop and deploy a protection plan to prevent fraud, abuse, and additional deploy a means to detect them if they occur.
DumpSec
DumpSec is a security tool that dumps a variety of information about a system's users, file system, registry, permissions, password policy, and services.
Dumpster Diving
Dumpster Diving is obtaining passwords and corporate directories by searching through discarded media.
Dynamic Link Library
A collection of small programs, any of which can be called when needed by a larger program that is running in the computer. The small program that lets the larger program communicate with a specific device such as a printer or scanner is often packaged as a DLL program (usually referred to as a DLL file).
Dynamic Routing Protocol
Allows network devices to learn routes. Ex. RIP, EIGRP Dynamic routing occurs when routers talk to adjacent routers, informing each other of what networks each router is currently connected to. The routers must communicate using a routing protocol, of which there are many to choose from. The process on the router that is running the routing protocol, communicating with its neighbor routers, is usually called a routing daemon. The routing daemon updates the kernel's routing table with information it receives from neighbor routers.
Eavesdropping
Eavesdropping is simply listening to a private conversation which may reveal information which can provide access to a facility or network.
Echo Reply
An echo reply is the response a machine that has received an echo request sends over ICMP.
Echo Request
An echo request is an ICMP message sent to a machine to determine if it is online and how long traffic takes to get to it.
Egress Filtering
Filtering outbound traffic.
Emanations Analysis
Gaining direct knowledge of communicated data by monitoring and resolving a signal that is emitted by a system and that contains the data but is not intended to communicate the data.
Encapsulation
The inclusion of one data structure within another structure so that the first data structure is hidden for the time being.
Encryption
Cryptographic transformation of data (called "plaintext") into a form (called "cipher text") that conceals the data's original meaning to prevent it from being known or used.
Ephemeral Port
Also called a transient port or a temporary port. Usually is on the client side. It is set up when a client application wants to connect to a server and is destroyed when the client application terminates. It has a number chosen at random that is greater than 1023.
Escrow Passwords
Escrow Passwords are passwords that are written down and stored in a secure location (like a safe) that are used by emergency personnel when privileged personnel are unavailable.
Ethernet
The most widely-installed LAN technology. Specified in a standard, IEEE 802.3, an Ethernet LAN typically uses coaxial cable or special grades of twisted pair wires. Devices are connected to the cable and compete for access using a CSMA/CD protocol.
Event
An event is an observable occurrence in a system or network.
Exponential Backoff Algorithm
An exponential backoff algorithm is used to adjust TCP timeout values on the fly so that network devices don't continue to timeout sending data over saturated links.
Exposure
A threat action whereby sensitive data is directly released to an unauthorized entity.
Extended ACLs (Cisco)
Extended ACLs are a more powerful form of Standard ACLs on Cisco routers. They can make filtering decisions based on IP addresses (source or destination), Ports (source or destination), protocols, and whether a session is established.
Extensible Authentication Protocol (EAP)
A framework that supports multiple, optional authentication mechanisms for PPP, including clear-text passwords, challenge-response, and arbitrary dialog sequences.
Exterior Gateway Protocol (EGP)
A protocol which distributes routing information to the routers which connect autonomous systems.
False Rejects
False Rejects are when an authentication system fails to recognize a valid user.
Fast File System
The first major revision to the Unix file system, providing faster read access and faster (delayed, asynchronous) write access through a disk cache and better file system layout on disk. It uses inodes (pointers) and data blocks.
Fast Flux
Protection method used by botnets consisting of a continuous and fast change of the DNS records for a domain name through different IP addresses.
Fault Line Attacks
Fault Line Attacks use weaknesses between interfaces of systems to exploit gaps in coverage.
File Transfer Protocol (FTP)
A TCP/IP protocol specifying the transfer of text or binary files across the network.
Filter
A filter is used to specify which packets will or will not be used. It can be used in sniffers to determine which packets get displayed, or by firewalls to determine which packets get blocked.
Filtering Router
An inter-network router that selectively prevents the passage of data packets according to a security policy. A filtering router may be used as a firewall or part of a firewall. A router usually receives a packet from a network and decides where to forward it on a second network. A filtering router does the same, but first decides whether the packet should be forwarded at all, according to some security policy. The policy is implemented by rules (packet filters) loaded into the router.
Finger
A protocol to lookup user information on a given host. A Unix program that takes an e-mail address as input and returns information about the user who owns that e-mail address. On some systems, finger only reports whether the user is currently logged on. Other systems return additional information, such as the user's full name, address, and telephone number. Of course, the user must first enter this information into the system. Many e-mail programs now have a finger utility built into them.
Fingerprinting
Sending strange packets to a system in order to gauge how it responds to determine the operating system.
Firewall
A logical or physical discontinuity in a network to prevent unauthorized access to data or resources.
Flooding
An attack that attempts to cause a failure in (especially, in the security of) a computer system or other data processing entity by providing more input than the entity can process properly.
Forest
A forest is a set of Active Directory domains that replicate their databases with each other.
Fork Bomb
A Fork Bomb works by using the fork() call to create a new process which is a copy of the original. By doing this repeatedly, all available processes on the machine can be taken up.
Form-Based Authentication
Form-Based Authentication uses forms on a webpage to ask a user to input username and password information.
Forward Lookup
Forward lookup uses an Internet domain name to find an IP address
Forward Proxy
Forward Proxies are designed to be the server through which all requests are made.
Fragment Offset
The fragment offset field tells the sender where a particular fragment falls in relation to other fragments in the original larger packet.
Fragment Overlap Attack
A TCP/IP Fragmentation Attack that is possible because IP allows packets to be broken down into fragments for more efficient transport across various media. The TCP packet (and its header) are carried in the IP packet. In this attack the second fragment contains incorrect offset. When packet is reconstructed, the port number will be overwritten.
Fragmentation
The process of storing a data file in several "chunks" or fragments rather than in a single contiguous sequence of bits in one place on the storage medium.
Frames
Data that is transmitted between network points as a unit complete with addressing and necessary protocol control information. A frame is usually transmitted serial bit by bit and contains a header field and a trailer field that "frame" the data. (Some control frames contain no data.)
Full Duplex
A type of duplex communications channel which carries data in both directions at once. Refers to the transmission of data in two directions simultaneously. Communications in which both sender and receiver can send at the same time.
Fully-Qualified Domain Name
A Fully-Qualified Domain Name is a server name with a hostname followed by the full domain name.
Fuzzing
The use of special regression testing tools to generate out-of-spec input for an application in order to find security vulnerabilities. Also see "regression testing".
G-I
Gateway
A network point that acts as an entrance to another network.
gethostbyaddr
The gethostbyaddr DNS query is when the address of a machine is known and the name is needed.
gethostbyname
The gethostbyname DNS quest is when the name of a machine is known and the address is needed.
GNU
GNU is a Unix-like operating system that comes with source code that can be copied, modified, and redistributed. The GNU project was started in 1983 by Richard Stallman and others, who formed the Free Software Foundation.
Gnutella
An Internet file sharing utility. Gnutella acts as a server for sharing files while simultaneously acting as a client that searches for and downloads files from other users.
Hardening
Hardening is the process of identifying and fixing vulnerabilities on a system.
Hash Function
An algorithm that computes a value based on a data object thereby mapping the data object to a smaller data object.
Hash Functions
(cryptographic) hash functions are used to generate a one way "check sum" for a larger text, which is not trivially reversed. The result of this hash function can be used to validate if a larger file has been altered, without having to compare the larger files to each other. Frequently used hash functions are MD5 and SHA1.
Header
A header is the extra information in a packet that is needed for the protocol stack to process the packet.
Hijack Attack
A form of active wiretapping in which the attacker seizes control of a previously established communication association.
Honey Client
see Honeymonkey.
Honey pot
Programs that simulate one or more network services that you designate on your computer's ports. An attacker assumes you're running vulnerable services that can be used to break into the machine. A honey pot can be used to log access attempts to those ports including the attacker's keystrokes. This could give you advanced warning of a more concerted attack.
Honeymonkey
Automated system simulating a user browsing websites. The system is typically configured to detect web sites which exploit vulnerabilities in the browser. Also known as Honey Client.
Hops
A hop is each exchange with a gateway a packet takes on its way to the destination.
Host
Any computer that has full two-way access to other computers on the Internet. Or a computer with a web server that serves the pages for one or more Web sites.
Host-Based ID
Host-based intrusion detection systems use information from the operating system audit records to watch all operations occurring on the host that the intrusion detection software has been installed upon. These operations are then compared with a pre-defined security policy. This analysis of the audit trail imposes potentially significant overhead requirements on the system because of the increased amount of processing power which must be utilized by the intrusion detection system. Depending on the size of the audit trail and the processing ability of the system, the review of audit data could result in the loss of a real-time analysis capability.
HTTP Proxy
An HTTP Proxy is a server that acts as a middleman in the communication between HTTP clients and servers.
HTTPS
When used in the first part of a URL (the part that precedes the colon and specifies an access scheme or protocol), this term specifies the use of HTTP enhanced by a security mechanism, which is usually SSL.
Hub
A hub is a network device that operates by repeating data that it receives on one port to all the other ports. As a result, data transmitted by one host is retransmitted to all other hosts on the hub.
Hybrid Attack
A Hybrid Attack builds on the dictionary attack method by adding numerals and symbols to dictionary words.
Hybrid Encryption
An application of cryptography that combines two or more encryption algorithms, particularly a combination of symmetric and asymmetric encryption.
Hyperlink
In hypertext or hypermedia, an information object (such as a word, a phrase, or an image; usually highlighted by color or underscoring) that points (indicates how to connect) to related information that is located elsewhere and can be retrieved by activating the link.
Hypertext Markup Language (HTML)
The set of markup symbols or codes inserted in a file intended for display on a World Wide Web browser page.
Hypertext Transfer Protocol (HTTP)
The protocol in the Internet Protocol (IP) family used to transport hypertext documents across an internet.
Identity
Identity is whom someone or what something is, for example, the name by which something is known.
Incident
An incident as an adverse network event in an information system or network or the threat of the occurrence of such an event.
Incident Handling
Incident Handling is an action plan for dealing with intrusions, cyber-theft, denial of service, fire, floods, and other security-related events. It is comprised of a six step process: Preparation, Identification, Containment, Eradication, Recovery, and Lessons Learned.
Incident Response
Incident response is the structured process of identifying, managing, and mitigating the effects of cybersecurity incidents to minimize damage, recover operations, and prevent future occurrences. Learn more.
Incremental Backups
Incremental backups only backup the files that have been modified since the last backup. If dump levels are used, incremental backups only backup files changed since last backup of a lower dump level.
Inetd (xinetd)
Inetd (or Internet Daemon) is an application that controls smaller internet services like telnet, ftp, and POP.
Inference Attack
Inference Attacks rely on the user to make logical connections between seemingly unrelated pieces of information.
Information Security
Information security (InfoSec) refers to the practice of protecting information from unauthorized access, disclosure, alteration, destruction, or disruption. Learn more.
Information Warfare
Information Warfare is the competition between offensive and defensive players over information resources.
Ingress Filtering
Ingress Filtering is filtering inbound traffic.
Input Validation Attacks
Input Validations Attacks are where an attacker intentionally sends unusual input in the hopes of confusing an application.
Integrity
Integrity is the need to ensure that information has not been changed accidentally or deliberately, and that it is accurate and complete.
Integrity Star Property
In Integrity Star Property a user cannot read data of a lower integrity level then their own.
Internet
A term to describe connecting multiple separate networks together.
Internet Control Message Protocol (ICMP)
An Internet Standard protocol that is used to report error conditions during IP datagram processing and to exchange other information concerning the state of the IP network.
Internet Engineering Task Force (IETF)
The body that defines standard Internet operating protocols such as TCP/IP. The IETF is supervised by the Internet Society Internet Architecture Board (IAB). IETF members are drawn from the Internet Society's individual and organization membership.
Internet Message Access Protocol (IMAP)
A protocol that defines how a client should fetch mail from and return mail to a mail server. IMAP is intended as a replacement for or extension to the Post Office Protocol (POP). It is defined in RFC 1203 (v3) and RFC 2060 (v4).
Internet Protocol (IP)
The method or protocol by which data is sent from one computer to another on the Internet.
Internet Protocol Security (IPsec)
A developing standard for security at the network or packet processing layer of network communication.
Internet Standard
A specification, approved by the IESG and published as an RFC, that is stable and well-understood, is technically competent, has multiple, independent, and interoperable implementations with substantial operational experience, enjoys significant public support, and is recognizably useful in some or all parts of the Internet.
Interrupt
An Interrupt is a signal that informs the OS that something has occurred.
Intranet
A computer network, especially one based on Internet technology, that an organization uses for its own internal, and usually private, purposes and that is closed to outsiders.
Intrusion Detection
A security management system for computers and networks. An IDS gathers and analyzes information from various areas within a computer or a network to identify possible security breaches, which include both intrusions (attacks from outside the organization) and misuse (attacks from within the organization).
IP Address
A computer's inter-network address that is assigned for use by the Internet Protocol and other protocols. An IP version 4 address is written as a series of four 8-bit numbers separated by periods.
IP Flood
A denial of service attack that sends a host more echo request ("ping") packets than the protocol implementation can handle.
IP Forwarding
IP forwarding is an Operating System option that allows a host to act as a router. A system that has more than 1 network interface card must have IP forwarding turned on in order for the system to be able to act as a router.
IP Spoofing
The technique of supplying a false IP address.
ISO
International Organization for Standardization, a voluntary, non-treaty, non-government organization, established in 1947, with voting members that are designated standards bodies of participating nations and non-voting observer organizations.
Issue-Specific Policy
An Issue-Specific Policy is intended to address specific needs within an organization, such as a password policy.
ITU-T
International Telecommunications Union, Telecommunication Standardization Sector (formerly "CCITT"), a United Nations treaty organization that is composed mainly of postal, telephone, and telegraph authorities of the member countries and that publishes standards called "Recommendations."
J-L
Jitter
Jitter or Noise is the modification of fields in a database while preserving the aggregate characteristics of that make the database useful in the first place.
Jump Bag
A Jump Bag is a container that has all the items necessary to respond to an incident inside to help mitigate the effects of delayed reactions.
Kerberos
A system developed at the Massachusetts Institute of Technology that depends on passwords and symmetric cryptography (DES) to implement ticket-based, peer entity authentication service and access control service distributed in a client-server network environment.
Kernel
The essential center of a computer operating system, the core that provides basic services for all other parts of the operating system. A synonym is nucleus. A kernel can be contrasted with a shell, the outermost part of an operating system that interacts with user commands. Kernel and shell are terms used more frequently in Unix and some other operating systems than in IBM mainframe systems.
Lattice Techniques
Lattice Techniques use security designations to determine access to information.
Layer 2 Forwarding Protocol (L2F)
An Internet protocol (originally developed by Cisco Corporation) that uses tunneling of PPP over IP to create a virtual extension of a dial-up link across a network, initiated by the dial-up server and transparent to the dial-up user.
Layer 2 Tunneling Protocol (L2TP)
An extension of the Point-to-Point Tunneling Protocol used by an Internet service provider to enable the operation of a virtual private network over the Internet.
Least Privilege
Least Privilege is the principle of allowing users or applications the least amount of permissions necessary to perform their intended function.
Legion
Software to detect unprotected shares.
Lightweight Directory Access Protocol (LDAP)
A software protocol for enabling anyone to locate organizations, individuals, and other resources such as files and devices in a network, whether on the public Internet or on a corporate Intranet.
Link State
With link state, routes maintain information about all routers and router-to-router links within a geographic area, and creates a table of best routes with that information.
List Based Access Control
List Based Access Control associates a list of users and their privileges with each object.
Loadable Kernel Modules (LKM)
Loadable Kernel Modules allow for the adding of additional functionality directly into the kernel while the system is running.
Log Clipping
Log clipping is the selective removal of log entries from a system log to hide a compromise.
Logic bombs
Logic bombs are programs or snippets of code that execute when a certain predefined event occurs. Logic bombs may also be set to go off on a certain date or when a specified set of circumstances occurs.
Logic Gate
A logic gate is an elementary building block of a digital circuit. Most logic gates have two inputs and one output. As digital circuits can only understand binary, inputs and outputs can assume only one of two states, 0 or 1.
Loopback Address
The loopback address (127.0.0.1) is a pseudo IP address that always refer back to the local host and are never sent out onto a network.
M-0
MAC Address
A physical address; a numeric value that uniquely identifies that network device from every other device on the planet.
Malicious Code
Software (e.g., Trojan horse) that appears to perform a useful or desirable function, but actually gains unauthorized access to system resources or tricks a user into executing other malicious logic.
Malware
A generic term for a number of different types of malicious code.
Mandatory Access Control (MAC)
Mandatory Access Control controls is where the system controls access to resources based on classification levels assigned to both the objects and the users. These controls cannot be changed by anyone.
Man-in-the-Middle Attack (MitM)
A man-in-the-middle attack is a type of cyber attack in which the attacker secretly intercepts and relays messages between two parties who believe they are communicating directly with each other. The attack is a type of eavesdropping in which the attacker intercepts and then controls the entire conversation.
Masquerade Attack
A type of attack in which one system entity illegitimately poses as (assumes the identity of) another entity.
md5
A one way cryptographic hash function. Also see "hash functions" and "sha1"
Measures of Effectiveness (MOE)
Measures of Effectiveness is a probability model based on engineering concepts that allows one to approximate the impact a give action will have on an environment. In Information warfare it is the ability to attack or defend within an Internet environment.
Monoculture
Monoculture is the case where a large number of users run the same software, and are vulnerable to the same attacks.
Morris Worm
A worm program written by Robert T. Morris, Jr. that flooded the ARPANET in November, 1988, causing problems for thousands of hosts.
Multi-Cast
Broadcasting from one host to a given set of hosts.
Multi-Homed
You are "multi-homed" if your network is directly connected to two or more ISP's.
Multiplexing
To combine multiple signals from possibly disparate sources, in order to transmit them over a single path.
National Institute of Standards and Technology (NIST)
National Institute of Standards and Technology, a unit of the US Commerce Department. Formerly known as the National Bureau of Standards, NIST promotes and maintains measurement standards. It also has active programs for encouraging and assisting industry and science to develop and use these standards.
Natural Disaster
Any "act of God" (e.g., fire, flood, earthquake, lightning, or wind) that disables a system component.
Netmask
32-bit number indicating the range of IP addresses residing on a single IP network/subnet/supernet. This specification displays network masks as hexadecimal numbers. For example, the network mask for a class C IP network is displayed as 0xffffff00. Such a mask is often displayed elsewhere in the literature as 255.255.255.0.
Network Address Translation
The translation of an Internet Protocol address used within one network to a different IP address known within another network. One network is designated the inside network and the other is the outside.
Network Mapping
To compile an electronic inventory of the systems and the services on your network.
Network Taps
Network taps are hardware devices that hook directly onto the network cable and send a copy of the traffic that passes through it to one or more other networked devices.
Network-Based IDS
A network-based IDS system monitors the traffic on its network segment as a data source. This is generally accomplished by placing the network interface card in promiscuous mode to capture all network traffic that crosses its network segment. Network traffic on other segments, and traffic on other means of communication (like phone lines) can't be monitored. Network-based IDS involves looking at the packets on the network as they pass by some sensor. The sensor can only see the packets that happen to be carried on the network segment it's attached to. Packets are considered to be of interest if they match a signature.Network-based intrusion detection passively monitors network activity for indications of attacks. Network monitoring offers several advantages over traditional host-based intrusion detection systems. Because many intrusions occur over networks at some point, and because networks are increasingly becoming the targets of attack, these techniques are an excellent method of detecting many attacks which may be missed by host-based intrusion detection mechanisms.
Non-Printable Character
A character that doesn't have a corresponding character letter to its corresponding ASCII code. Examples would be the Linefeed, which is ASCII character code 10 decimal, the Carriage Return, which is 13 decimal, or the bell sound, which is decimal 7. On a PC, you can often add non-printable characters by holding down the Alt key, and typing in the decimal value (i.e., Alt-007 gets you a bell). There are other character encoding schemes, but ASCII is the most prevalent.
Non-Repudiation
Non-repudiation is the ability for a system to prove that a specific user and only that specific user sent a message and that it hasn't been modified.
Null Session
Known as Anonymous Logon, it is a way of letting an anonymous user retrieve information such as user names and shares over the network or connect without authentication. It is used by applications such as explorer.exe to enumerate shares on remote servers.
Octet
A sequence of eight bits. An octet is an eight-bit byte.
One-Way Encryption
Irreversible transformation of plaintext to cipher text, such that the plaintext cannot be recovered from the cipher text by other than exhaustive procedures even if the cryptographic key is known.
One-Way Function
A (mathematical) function, f, which is easy to compute the output based on a given input. However given only the output value it is impossible (except for a brute force attack) to figure out what the input value is.
Open Shortest Path First (OSPF)
Open Shortest Path First is a link state routing algorithm used in interior gateway routing. Routers maintain a database of all routers in the autonomous system with links between the routers, link costs, and link states (up and down).
OSI
OSI (Open Systems Interconnection) is a standard description or "reference model" for how messages should be transmitted between any two points in a telecommunication network. Its purpose is to guide product implementers so that their products will consistently work with other products. The reference model defines seven layers of functions that take place at each end of a communication. Although OSI is not always strictly adhered to in terms of keeping related functions together in a well-defined layer, many if not most products involved in telecommunication make an attempt to describe themselves in relation to the OSI model. It is also valuable as a single reference view of communication that furnishes everyone a common ground for education and discussion.
OSI layers
The main idea in OSI is that the process of communication between two end points in a telecommunication network can be divided into layers, with each layer adding its own set of special, related functions. Each communicating user or program is at a computer equipped with these seven layers of function. So, in a given message between users, there will be a flow of data through each layer at one end down through the layers in that computer and, at the other end, when the message arrives, another flow of data up through the layers in the receiving computer and ultimately to the end user or program. The actual programming and hardware that furnishes these seven layers of function is usually a combination of the computer operating system, applications (such as your Web browser), TCP/IP or alternative transport and network protocols, and the software and hardware that enable you to put a signal on one of the lines attached to your computer. OSI divides telecommunication into seven layers. The layers are in two groups. The upper four layers are used whenever a message passes from or to a user. The lower three layers (up to the network layer) are used when any message passes through the host computer or router. Messages intended for this computer pass to the upper layers. Messages destined for some other host are not passed up to the upper layers but are forwarded to another host. The seven layers are: Layer 7: The application layer...This is the layer at which communication partners are identified, quality of service is identified, user authentication and privacy are considered, and any constraints on data syntax are identified. (This layer is not the application itself, although some applications may perform application layer functions.) Layer 6: The presentation layer...This is a layer, usually part of an operating system, that converts incoming and outgoing data from one presentation format to another (for example, from a text stream into a popup window with the newly arrived text). Sometimes called the syntax layer. Layer 5: The session layer...This layer sets up, coordinates, and terminates conversations, exchanges, and dialogs between the applications at each end. It deals with session and connection coordination. Layer 4: The transport layer...This layer manages the end-to-end control (for example, determining whether all packets have arrived) and error-checking. It ensures complete data transfer. Layer 3: The network layer...This layer handles the routing of the data (sending it in the right direction to the right destination on outgoing transmissions and receiving incoming transmissions at the packet level). The network layer does routing and forwarding. Layer 2: The data-link layer...This layer provides synchronization for the physical level and does bit-stuffing for strings of 1's in excess of 5. It furnishes transmission protocol knowledge and management. Layer 1: The physical layer...This layer conveys the bit stream through the network at the electrical and mechanical level. It provides the hardware means of sending and receiving data on a carrier.
Overload
Hindrance of system operation by placing excess burden on the performance capabilities of a system component.
P-S
Packet
A piece of a message transmitted over a packet-switching network. One of the key features of a packet is that it contains the destination address in addition to the data. In IP networks, packets are often called datagrams.
Packet Switched Network
A packet switched network is where individual packets each follow their own paths through the network from one endpoint to another.
Partitions
Major divisions of the total physical hard disk space.
Password Authentication Protocol (PAP)
Password Authentication Protocol is a simple, weak authentication mechanism where a user enters the password and it is then sent across the network, usually in the clear.
Password Cracking
Password cracking is the process of attempting to guess passwords, given the password file information.
Password Sniffing
Passive wiretapping, usually on a local area network, to gain knowledge of passwords.
Patch
A patch is a small update released by a software manufacturer to fix bugs in existing programs.
Patching
Patching is the process of updating software to a different version.
Payload
Payload is the actual application data a packet contains.
Penetration
Gaining unauthorized logical access to sensitive data by circumventing a system's protections.
Penetration Testing
Penetration testing is used to test the external perimeter security of a network or facility. Learn more.
Permutation
Permutation keeps the same letters but changes the position within a text to scramble the message.
Personal Firewalls
Personal firewalls are those firewalls that are installed and run on individual PCs.
Pharming
This is a more sophisticated form of MITM attack. A user’s session is redirected to a masquerading website. This can be achieved by corrupting a DNS server on the Internet and pointing a URL to the masquerading website’s IP. Almost all users use a URL like www.worldbank.com instead of the real IP (192.86.99.140) of the website. Changing the pointers on a DNS server, the URL can be redirected to send traffic to the IP of the pseudo website. At the pseudo website, transactions can be mimicked and information like login credentials can be gathered. With this the attacker can access the real www.worldbank.com site and conduct transactions using the credentials of a valid user on that website.
Phishing
The use of e-mails that appear to originate from a trusted source to trick a user into entering valid credentials at a fake website. Typically the e-mail and the web site looks like they are part of a bank the user is doing business with.
Ping of Death
An attack that sends an improperly large ICMP echo request packet (a "ping") with the intent of overflowing the input buffers of the destination machine and causing it to crash.
Ping Scan
A ping scan looks for machines that are responding to ICMP Echo Requests.
Ping Sweep
An attack that sends ICMP echo requests ("pings") to a range of IP addresses, with the goal of finding hosts that can be probed for vulnerabilities.
Plaintext
Ordinary readable text before being encrypted into ciphertext or after being decrypted.
Point-to-Point Protocol (PPP)
A protocol for communication between two computers using a serial interface, typically a personal computer connected by phone line to a server. It packages your computer's TCP/IP packets and forwards them to the server where they can actually be put on the Internet.
Point-to-Point Tunneling Protocol (PPTP)
A protocol (set of communication rules) that allows corporations to extend their own corporate network through private "tunnels" over the public Internet.
Poison Reverse
Split horizon with poisoned reverse (more simply, poison reverse) does include such routes in updates, but sets their metrics to infinity. In effect, advertising the fact that there routes are not reachable.
Polyinstantiation
Polyinstantiation is the ability of a database to maintain multiple records with the same key. It is used to prevent inference attacks.
Polymorphism
Polymorphism is the process by which malicious software changes its underlying code to avoid detection.
Port
A port is nothing more than an integer that uniquely identifies an endpoint of a communication stream. Only one process per machine can listen on the same port number.
Port Scan
A port scan is a series of messages sent by someone attempting to break into a computer to learn which computer network services, each associated with a "well-known" port number, the computer provides. Port scanning, a favorite approach of computer cracker, gives the assailant an idea where to probe for weaknesses. Essentially, a port scan consists of sending a message to each port, one at a time. The kind of response received indicates whether the port is used and can therefore be probed for weakness.
Possession
Possession is the holding, control, and ability to use information.
Post Office Protocol, Version 3 (POP3)
An Internet Standard protocol by which a client workstation can dynamically access a mailbox on a server host to retrieve mail messages that the server has received and is holding for the client.
Practical Extraction and Reporting Language (Perl)
A script programming language that is similar in syntax to the C language and that includes a number of popular Unix facilities such as sed, awk, and tr.
Preamble
A preamble is a signal used in network communications to synchronize the transmission timing between two or more systems. Proper timing ensures that all systems are interpreting the start of the information transfer correctly. A preamble defines a specific series of transmission pulses that is understood by communicating systems to mean "someone is about to transmit data". This ensures that systems receiving the information correctly interpret when the data transmission starts. The actual pulses used as a preamble vary depending on the network communication technology in use.
Pretty Good Privacy (PGP)TM
Trademark of Network Associates, Inc., referring to a computer program (and related protocols) that uses cryptography to provide data security for electronic mail and other applications on the Internet.
Private Addressing
IANA has set aside three address ranges for use by private or non-Internet connected networks. This is referred to as Private Address Space and is defined in RFC 1918. The reserved address blocks are: 10.0.0.0 to 10.255.255.255 (10/8 prefix) 172.16.0.0 to 172.31.255.255 (172.16/12 prefix) 192.168.0.0 to 192.168.255.255 (192.168/16 prefix)
Program Infector
A program infector is a piece of malware that attaches itself to existing program files.
Program Policy
A program policy is a high-level policy that sets the overall tone of an organization's security approach.
Promiscuous Mode
When a machine reads all packets off the network, regardless of who they are addressed to. This is used by network administrators to diagnose network problems, but also by unsavory characters who are trying to eavesdrop on network traffic (which might contain passwords or other information).
Proprietary Information
Proprietary information is that information unique to a company and its ability to compete, such as customer lists, technical data, product costs, and trade secrets.
Protocol
A formal specification for communicating; an IP address the special set of rules that end points in a telecommunication connection use when they communicate. Protocols exist at several levels in a telecommunication connection.
Protocol Stacks (OSI)
A set of network protocol layers that work together.
Proxy Server
A server that acts as an intermediary between a workstation user and the Internet so that the enterprise can ensure security, administrative control, and caching service. A proxy server is associated with or part of a gateway server that separates the enterprise network from the outside network and a firewall server that protects the enterprise network from outside intrusion.
Public Key
The publicly-disclosed component of a pair of cryptographic keys used for asymmetric cryptography.
Public Key Encryption
The popular synonym for "asymmetric cryptography".
Public Key Infrastructure (PKI)
A PKI (public key infrastructure) enables users of a basically unsecured public network such as the Internet to securely and privately exchange data and money through the use of a public and a private cryptographic key pair that is obtained and shared through a trusted authority. The public key infrastructure provides for a digital certificate that can identify an individual or an organization and directory services that can store and, when necessary, revoke the certificates.
Public-Key Forward Secrecy (PFS)
For a key agreement protocol based on asymmetric cryptography, the property that ensures that a session key derived from a set of long-term public and private keys will not be compromised if one of the private keys is compromised in the future.
QAZ
A network worm.
Race Condition
A race condition exploits the small window of time between a security control being applied and when the service is used.
Radiation Monitoring
Radiation monitoring is the process of receiving images, data, or audio from an unprotected source by listening to radiation signals.
Ransomware
A type of malware that is a form of extortion. It works by encrypting a victim's hard drive denying them access to key files. The victim must then pay a ransom to decrypt the files and gain access to them again.
Reconnaissance
Reconnaissance is the phase of an attack where an attackers finds new systems, maps out networks, and probes for specific, exploitable vulnerabilities.
Reflexive ACLs (Cisco)
Reflexive ACLs for Cisco routers are a step towards making the router act like a stateful firewall. The router will make filtering decisions based on whether connections are a part of established traffic or not.
Registry
The Registry in Windows operating systems in the central set of settings and information required to run the Windows computer.
regression analysis
The use of scripted tests which are used to test software for all possible input is should expect. Typically developers will create a set of regression tests that are executed before a new version of a software is released. Also see "fuzzing".
Request for Comment (RFC)
A series of notes about the Internet, started in 1969 (when the Internet was the ARPANET). An Internet Document can be submitted to the IETF by anyone, but the IETF decides if the document becomes an RFC. Eventually, if it gains enough interest, it may evolve into an Internet standard.
Resource Exhaustion
Resource exhaustion attacks involve tying up finite resources on a system, making them unavailable to others.
Response
A response is information sent that is responding to some stimulus.
Reverse Address Resolution Protocol (RARP)
RARP (Reverse Address Resolution Protocol) is a protocol by which a physical machine in a local area network can request to learn its IP address from a gateway server's Address Resolution Protocol table or cache. A network administrator creates a table in a local area network's gateway router that maps the physical machine (or Media Access Control - MAC address) addresses to corresponding Internet Protocol addresses. When a new machine is set up, its RARP client program requests from the RARP server on the router to be sent its IP address. Assuming that an entry has been set up in the router table, the RARP server will return the IP address to the machine which can store it for future use.
Reverse Engineering
Acquiring sensitive data by disassembling and analyzing the design of a system component.
Reverse Lookup
Find out the hostname that corresponds to a particular IP address. Reverse lookup uses an IP (Internet Protocol) address to find a domain name.
Reverse Proxy
Reverse proxies take public HTTP requests and pass them to back-end webservers to send the content to it, so the proxy can then send the content to the end-user.
Risk
Risk is the product of the level of threat with the level of vulnerability. It establishes the likelihood of a successful attack.
Risk Assessment
A Risk Assessment is the process by which risks are identified and the impact of those risks determined.
Risk Averse
Avoiding risk even if this leads to the loss of opportunity. For example, using a (more expensive) phone call vs. sending an e-mail in order to avoid risks associated with e-mail may be considered "Risk Averse"
Rivest-Shamir-Adleman (RSA)
An algorithm for asymmetric cryptography, invented in 1977 by Ron Rivest, Adi Shamir, and Leonard Adleman.
Role Based Access Control
Role based access control assigns users to roles based on their organizational functions and determines authorization based on those roles.
Root
Root is the name of the administrator account in Unix systems.
Rootkit
A collection of tools (programs) that a hacker uses to mask intrusion and obtain administrator-level access to a computer or computer network.
Router
Routers interconnect logical networks by forwarding information to other networks based upon IP addresses.
Routing Information Protocol (RIP)
Routing Information Protocol is a distance vector protocol used for interior gateway routing which uses hop count as the sole metric of a path's cost.
Routing Loop
A routing loop is where two or more poorly configured routers repeatedly exchange the same packet over and over.
RPC Scans
RPC scans determine which RPC services are running on a machine.
Rule Set Based Access Control (RSBAC)
Rule Set Based Access Control targets actions based on rules for entities operating on objects.
S/Key
A security mechanism that uses a cryptographic hash function to generate a sequence of 64-bit, one-time passwords for remote user login. The client generates a one-time password by applying the MD4 cryptographic hash function multiple times to the user's secret key. For each successive authentication of the user, the number of hash applications is reduced by one.
Safety
Safety is the need to ensure that the people involved with the company, including employees, customers, and visitors, are protected from harm.
SANS
SysAdmin, Audit, Network, Security
Scavenging
Searching through data residue in a system to gain unauthorized knowledge of sensitive data.
Secure Electronic Transactions (SET)
Secure Electronic Transactions is a protocol developed for credit card transactions in which all parties (customers, merchant, and bank) are authenticated using digital signatures, encryption protects the message and provides integrity, and provides end-to-end security for credit card transactions online.
Secure Shell (SSH)
A program to log into another computer over a network, to execute commands in a remote machine, and to move files from one machine to another.
Secure Sockets Layer (SSL)
A protocol developed by Netscape for transmitting private documents via the Internet. SSL works by using a public key to encrypt data that's transferred over the SSL connection.
Security Policy
A set of rules and practices that specify or regulate how a system or organization provides security services to protect sensitive and critical system resources.
Segment
Segment is another name for TCP packets.
Sensitive Information
Sensitive information, as defined by the federal government, is any unclassified information that, if compromised, could adversely affect the national interest or conduct of federal initiatives.
Separation of Duties
Separation of duties is the principle of splitting privileges among multiple individuals or systems.
Server
A system entity that provides a service in response to requests from other system entities called clients.
Session
A session is a virtual connection between two hosts by which network traffic is passed.
Session Hijacking
Take over a session that someone else has established.
Session Key
In the context of symmetric encryption, a key that is temporary or is used for a relatively short period of time. Usually, a session key is used for a defined period of communication between two computers, such as for the duration of a single connection or transaction set, or the key is used in an application that protects relatively large amounts of data and, therefore, needs to be re-keyed frequently.
SHA1
A one way cryptographic hash function. Also see "MD5"
Shadow Password Files
A system file in which encryption user password are stored so that they aren't available to people who try to break into the system.
Share
A share is a resource made public on a machine, such as a directory (file share) or printer (printer share).
Shell
A Unix term for the interactive user interface with an operating system. The shell is the layer of programming that understands and executes the commands a user enters. In some systems, the shell is called a command interpreter. A shell usually implies an interface with a command syntax (think of the DOS operating system and its "C:>" prompts and user commands such as "dir" and "edit").
Signals Analysis
Gaining indirect knowledge of communicated data by monitoring and analyzing a signal that is emitted by a system and that contains the data but is not intended to communicate the data.
Signature
A Signature is a distinct pattern in network traffic that can be identified to a specific tool or exploit.
Simple Integrity Property
In Simple Integrity Property a user cannot write data to a higher integrity level than their own.
Simple Network Management Protocol (SNMP)
The protocol governing network management and the monitoring of network devices and their functions. A set of protocols for managing complex networks.
Simple Security Property
In Simple Security Property a user cannot read data of a higher classification than their own.
Smartcard
A smartcard is an electronic badge that includes a magnetic strip or chip that can record and replay a set key.
Smishing
Smishing is a combination of the terms "SMS" and "phishing." It is similar to phishing, but refers to fraudulent messages sent over SMS (text messaging) rather than email.
Smurf
The Smurf attack works by spoofing the target address and sending a ping to the broadcast address for a remote network, which results in a large amount of ping replies being sent to the target.
Sniffer
A sniffer is a tool that monitors network traffic as it received in a network interface.
Sniffing
A synonym for "passive wiretapping."
Social Engineering
A euphemism for non-technical or low-technology means - such as lies, impersonation, tricks, bribes, blackmail, and threats - used to attack information systems.
Socket
The socket tells a host's IP stack where to plug in a data stream so that it connects to the right application.
Socket Pair
A way to uniquely specify a connection, i.e., source IP address, source port, destination IP address, destination port.
SOCKS
A protocol that a proxy server can use to accept requests from client users in a company's network so that it can forward them across the Internet. SOCKS uses sockets to represent and keep track of individual connections. The client side of SOCKS is built into certain Web browsers and the server side can be added to a proxy server.
Software
Computer programs (which are stored in and executed by computer hardware) and associated data (which also is stored in the hardware) that may be dynamically written or modified during execution.
Source Port
The port that a host uses to connect to a server. It is usually a number greater than or equal to 1024. It is randomly generated and is different each time a connection is made.
Spam
Electronic junk mail or junk newsgroup postings.
Spanning Port
Configures the switch to behave like a hub for a specific port.
Split Horizon
Split horizon is a algorithm for avoiding problems caused by including routes in updates sent to the gateway from which they were learned.
Split Key
A cryptographic key that is divided into two or more separate data items that individually convey no knowledge of the whole key that results from combining the items.
Spoof
Attempt by an unauthorized entity to gain access to a system by posing as an authorized user.
SQL Injection
SQL injection is a type of input validation attack specific to database-driven applications where SQL code is inserted into application queries to manipulate the database.
Stack Mashing
Stack mashing is the technique of using a buffer overflow to trick a computer into executing arbitrary code.
Standard ACLs (Cisco)
Standard ACLs on Cisco routers make packet filtering decisions based on Source IP address only.
Star Property
In Star Property, a user cannot write data to a lower classification level without logging in at that lower classification level.
State Machine
A system that moves through a series of progressive conditions.
Stateful Inspection
Also referred to as dynamic packet filtering. Stateful inspection is a firewall architecture that works at the network layer. Unlike static packet filtering, which examines a packet based on the information in its header, stateful inspection examines not just the header information but also the contents of the packet up through the application layer in order to determine more about the packet than just information about its source and destination.
Static Host Tables
Static host tables are text files that contain hostname and address mapping.
Static Routing
Static routing means that routing table entries contain information that does not change.
Stealthing
Stealthing is a term that refers to approaches used by malicious code to conceal its presence on the infected system.
Steganalysis
Steganalysis is the process of detecting and defeating the use of steganography.
Steganography
Methods of hiding the existence of a message or other data. This is different than cryptography, which hides the meaning of a message but does not hide the message itself. An example of a steganographic method is "invisible" ink.
Stimulus
Stimulus is network traffic that initiates a connection or solicits a response.
Store-and-Forward
Store-and-Forward is a method of switching where the entire packet is read by a switch to determine if it is intact before forwarding it.
Straight-Through Cable
A straight-through cable is where the pins on one side of the connector are wired to the same pins on the other end. It is used for interconnecting nodes on the network.
Stream Cipher
A stream cipher works by encryption a message a single bit, byte, or computer word at a time.
Strong Star Property
In Strong Star Property, a user cannot write data to higher or lower classifications levels than their own.
Sub Network
A separately identifiable part of a larger network that typically represents a certain limited number of host computers, the hosts in a building or geographic area, or the hosts on an individual local area network.
Subnet Mask
A subnet mask (or number) is used to determine the number of bits used for the subnet and host portions of the address. The mask is a 32-bit value that uses one-bits for the network and subnet portions and zero-bits for the host portion.
Switch
A switch is a networking device that keeps track of MAC addresses attached to each of its ports so that data is only transmitted on the ports that are the intended recipient of the data.
Switched Network
A communications network, such as the public switched telephone network, in which any user may be connected to any other user through the use of message, circuit, or packet switching and control devices. Any network providing switched communications service.
Symbolic Links
Special files which point at another file.
Symmetric Cryptography
A branch of cryptography involving algorithms that use the same key for two different steps of the algorithm (such as encryption and decryption, or signature creation and signature verification). Symmetric cryptography is sometimes called "secret-key cryptography" (versus public-key cryptography) because the entities that share the key.
Symmetric Key
A cryptographic key that is used in a symmetric cryptographic algorithm.
SYN Flood
A denial of service attack that sends a host more TCP SYN packets (request to synchronize sequence numbers, used when opening a connection) than the protocol implementation can handle.
Synchronization
Synchronization is the signal made up of a distinctive pattern of bits that network hardware looks for to signal that start of a frame.
Syslog
Syslog is the system logging facility for Unix systems.
System Security Officer (SSO)
A person responsible for enforcement or administration of the security policy that applies to the system.
System-Specific Policy
A System-specific policy is a policy written for a specific system or device.
T-Z
T1, T3
A digital circuit using TDM (Time-Division Multiplexing).
Tamper
To deliberately alter a system's logic, data, or control information to cause the system to perform unauthorized functions or services.
TCP Fingerprinting
TCP fingerprinting is the user of odd packet header combinations to determine a remote operating system.
TCP Full Open Scan
TCP Full Open scans check each port by performing a full three-way handshake on each port to determine if it was open.
TCP Half Open Scan
TCP Half Open scans work by performing the first half of a three-way handshake to determine if a port is open.
TCP Wrapper
A software package which can be used to restrict access to certain network services based on the source of the connection; a simple tool to monitor and control incoming network traffic.
TCP/IP
A synonym for "Internet Protocol Suite;" in which the Transmission Control Protocol and the Internet Protocol are important parts. TCP/IP is the basic communication language or protocol of the Internet. It can also be used as a communications protocol in a private network (either an Intranet or an Extranet).
TCPDump
TCPDump is a freeware protocol analyzer for Unix that can monitor network traffic on a wire.
TELNET
A TCP-based, application-layer, Internet Standard protocol for remote login from one host to another.
Threat
A potential for violation of security, which exists when there is a circumstance, capability, action, or event that could breach security and cause harm.
Threat Assessment
A threat assessment is the identification of types of threats that an organization might be exposed to.
Threat Model
A threat model is used to describe a given threat and the harm it could to do a system if it has a vulnerability.
Threat Vector
The method a threat uses to get to the target.
Time to Live
A value in an Internet Protocol packet that tells a network router whether or not the packet has been in the network too long and should be discarded.
Tiny Fragment Attack
With many IP implementations it is possible to impose an unusually small fragment size on outgoing packets. If the fragment size is made small enough to force some of a TCP packet's TCP header fields into the second fragment, filter rules that specify patterns for those fields will not match. If the filtering implementation does not enforce a minimum fragment size, a disallowed packet might be passed because it didn't hit a match in the filter. STD 5, RFC 791 states: Every Internet module must be able to forward a datagram of 68 octets without further fragmentation. This is because an Internet header may be up to 60 octets, and the minimum fragment is 8 octets.
Token Ring
A token ring network is a local area network in which all computers are connected in a ring or star topology and a binary digit or token-passing scheme is used in order to prevent the collision of data between two computers that want to send messages at the same time.
Token-Based Access Control
Token based access control associates a list of objects and their privileges with each user. (The opposite of list based.)
Token-Based Devices
A token-based device is triggered by the time of day, so every minute the password changes, requiring the user to have the token with them when they log in.
Topology
The geometric arrangement of a computer system. Common topologies include a bus, star, and ring. The specific physical, i.e., real, or logical, i.e., virtual, arrangement of the elements of a network. Note 1: Two networks have the same topology if the connection configuration is the same, although the networks may differ in physical interconnections, distances between nodes, transmission rates, and/or signal types. Note 2: The common types of network topology are illustrated
Traceroute (tracert.exe)
Traceroute is a tool the maps the route a packet takes from the local machine to a remote destination.
Transmission Control Protocol (TCP)
A set of rules (protocol) used along with the Internet Protocol to send data in the form of message units between computers over the Internet. While IP takes care of handling the actual delivery of the data, TCP takes care of keeping track of the individual units of data (called packets) that a message is divided into for efficient routing through the Internet. Whereas the IP protocol deals only with packets, TCP enables two hosts to establish a connection and exchange streams of data. TCP guarantees delivery of data and also guarantees that packets will be delivered in the same order in which they were sent.
Transport Layer Security (TLS)
A protocol that ensures privacy between communicating applications and their users on the Internet. When a server and client communicate, TLS ensures that no third party may eavesdrop or tamper with any message. TLS is the successor to the Secure Sockets Layer.
Triple DES
A block cipher, based on DES, that transforms each 64-bit plaintext block by applying the Data Encryption Algorithm three successive times, using either two or three different keys, for an effective key length of 112 or 168 bits.
Triple-Wrapped
S/MIME usage: data that has been signed with a digital signature, and then encrypted, and then signed again.
Trojan Horse
A computer program that appears to have a useful function, but also has a hidden and potentially malicious function that evades security mechanisms, sometimes by exploiting legitimate authorizations of a system entity that invokes the program.
Trunking
Trunking is connecting switched together so that they can share VLAN information between them.
Trust
Trust determine which permissions and what actions other systems or users can perform on remote machines.
Trusted Ports
Trusted ports are ports below number 1024 usually allowed to be opened by the root user.
Tunnel
A communication channel created in a computer network by encapsulating a communication protocol's data packets in (on top of) a second protocol that normally would be carried above, or at the same layer as, the first one. Most often, a tunnel is a logical point-to-point link - i.e., an OSI layer 2 connection - created by encapsulating the layer 2 protocol in a transport protocol (such as TCP), in a network or inter-network layer protocol (such as IP), or in another link layer protocol. Tunneling can move data between computers that use a protocol not supported by the network connecting them.
UDP Scan
UDP scans perform scans to determine which UDP ports are open.
Unicast
Broadcasting from host to host.
Uniform Resource Identifier (URI)
The generic term for all types of names and addresses that refer to objects on the World Wide Web.
Uniform Resource Locator (URL)
The global address of documents and other resources on the World Wide Web. The first part of the address indicates what protocol to use, and the second part specifies the IP address or the domain name where the resource is located. For example, .
Unix
A popular multi-user, multitasking operating system developed at Bell Labs in the early 1970s. Created by just a handful of programmers, Unix was designed to be a small, flexible system used exclusively by programmers.
Unprotected Share
In Windows terminology, a "share" is a mechanism that allows a user to connect to file systems and printers on other systems. An "unprotected share" is one that allows anyone to connect to it.
User
A person, organization entity, or automated process that accesses a system, whether authorized to do so or not.
User Contingency Plan
User contingency plan is the alternative methods of continuing business operations if IT systems are unavailable.
User Datagram Protocol (UDP)
A communications protocol that, like TCP, runs on top of IP networks. Unlike TCP/IP, UDP/IP provides very few error recovery services, offering instead a direct way to send and receive datagrams over an IP network. It's used primarily for broadcasting messages over a network. UDP uses the Internet Protocol to get a datagram from one computer to another but does not divide a message into packets (datagrams) and reassemble it at the other end. Specifically, UDP doesn't provide sequencing of the packets that the data arrives in.
Virtual Private Network (VPN)
A restricted-use, logical (i.e., artificial or simulated) computer network that is constructed from the system resources of a relatively public, physical (i.e., real) network (such as the Internet), often by using encryption (located at hosts or gateways), and often by tunneling links of the virtual network across the real network. For example, if a corporation has LANs at several different sites, each connected to the Internet by a firewall, the corporation could create a VPN by (a) using encrypted tunnels to connect from firewall to firewall across the Internet and (b) not allowing any other traffic through the firewalls. A VPN is generally less expensive to build and operate than a dedicated real network, because the virtual network shares the cost of system resources with other users of the real network.
Virus
A hidden, self-replicating section of computer software, usually malicious logic, that propagates by infecting - i.e., inserting a copy of itself into and becoming part of - another program. A virus cannot run by itself; it requires that its host program be run to make the virus active.
Vishing (voice or VoIP phishing)
Vishing refers to phishing attacks that involve the use of voice calls, using either conventional phone systems or Voice over Internet Procotol (VoIP) systems.
Voice Firewall
A physical discontinuity in a voice network that monitors, alerts and controls inbound and outbound voice network activity based on user-defined call admission control (CAC) policies, voice application layer security threats or unauthorized service use violations.
Voice Intrusion Prevention System (IPS)
Voice IPS is a security management system for voice networks which monitors voice traffic for multiple calling patterns or attack/abuse signatures to proactively detect and prevent toll fraud, Denial of Service, telecom attacks, service abuse, and other anomalous activity.
War Chalking
War chalking is marking areas, usually on sidewalks with chalk, that receive wireless signals that can be accessed.
War Dialer
A computer program that automatically dials a series of telephone numbers to find lines connected to computer systems, and catalogs those numbers so that a cracker can try to break into the systems.
War Dialing
War dialing is a simple means of trying to identify modems in a telephone exchange that may be susceptible to compromise in an attempt to circumvent perimeter security.
War Driving
War driving is the process of traveling around looking for wireless access point signals that can be used to get network access.
Web of Trust
A web of trust is the trust that naturally evolves as a user starts to trust other's signatures, and the signatures that they trust.
Web Server
A software process that runs on a host computer connected to the Internet to respond to HTTP requests for documents from client web browsers.
WHOIS
An IP for finding information about resources on networks.
Windowing
A windowing system is a system for sharing a computer's graphical display presentation resources among multiple applications at the same time. In a computer that has a graphical user interface (GUI), you may want to use a number of applications at the same time (this is called task). Using a separate window for each application, you can interact with each application and go from one application to another without having to reinitiate it. Having different information or activities in multiple windows may also make it easier for you to do your work. A windowing system uses a window manager to keep track of where each window is located on the display screen and its size and status. A windowing system doesn't just manage the windows but also other forms of graphical user interface entities.
Windump
Windump is a freeware tool for Windows that is a protocol analyzer that can monitor network traffic on a wire.
Wired Equivalent Privacy (WEP)
A security protocol for wireless local area networks defined in the standard IEEE 802.11b.
Wireless Application Protocol
A specification for a set of communication protocols to standardize the way that wireless devices, such as cellular telephones and radio transceivers, can be used for Internet access, including e-mail, the World Wide Web, newsgroups, and Internet Relay Chat.
Wiretapping
Monitoring and recording data that is flowing between two points in a communication system.
World Wide Web ("the Web", WWW, W3)
The global, hypermedia-based collection of information and services that is available on Internet servers and is accessed by browsers using Hypertext Transfer Protocol and other information retrieval mechanisms.
Worm
A computer program that can run independently, can propagate a complete working version of itself onto other hosts on a network, and may consume computer resources destructively.
Zero Day
The "Day Zero" or "Zero Day" is the day a new vulnerability is made known. In some cases, a "zero day" exploit is referred to an exploit for which no patch is available yet. ("day one" - day at which the patch is made available).
Zero-Day Attack
A zero-day (or zero-hour or day zero) attack or threat is a computer threat that tries to exploit computer application vulnerabilities that are unknown to others or undisclosed to the software developer. Zero-day exploits (actual code that can use a security hole to carry out an attack) are used or shared by attackers before the software developer knows about the vulnerability.
Zero-Day Exploit
A zero-day exploit refers to a cyberattack that takes advantage of a software, hardware, or firmware vulnerability that is unknown to the vendor or the public. Learn more.
Zombies
A zombie computer (often shortened as zombie) is a computer connected to the Internet that has been compromised by a hacker, a computer virus, or a trojan horse. Generally, a compromised machine is only one of many in a botnet, and will be used to perform malicious tasks of one sort or another under remote direction. Most owners of zombie computers are unaware that their system is being used in this way. Because the owner tends to be unaware, these computers are metaphorically compared to zombies.
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3-way handshake
Machine A sends a packet with a SYN flag set to Machine B. B acknowledges A's SYN with a SYN/ACK. A acknowledges B's SYN/ACK with an ACK. |
3101 | https://web.mit.edu/fluids-modules/www/highspeed_flows/ver2/bl_Chap2.pdf | 1 I-campus project School-wide Program on Fluid Mechanics Modules on High Reynolds Number Flows K. P. Burr, T. R. Akylas & C. C. Mei CHAPTER TWO TWO-DIMENSIONAL LAMINAR BOUNDARY LAYERS 1 Introduction.
When a viscous fluid flows along a fixed impermeable wall, or past the rigid surface of an immersed body, an essential condition is that the velocity at any point on the wall or other fixed surface is zero. The extent to which this condition modifies the general character of the flow depends upon the value of the viscosity. If the body is of streamlined shape and if the viscosity is small without being negligible, the modifying effect appears to be confined within narrow regions adjacent to the solid surfaces; these are called boundary layers. Within such layers the fluid velocity changes rapidly from zero to its main-stream value, and this may imply a steep gradient of shearing stress; as a consequence, not all the viscous terms in the equation of motion will be negligible, even though the viscosity, which they contain as a factor, is itself very small.
A more precise criterion for the existence of a well-defined laminar boundary layer is that the Reynolds number should be large, though not so large as to imply a breakdown of the laminar flow.
2 Boundary Layer Governing Equations.
In developing a mathematical theory of boundary layers, the first step is to show the existence, as the Reynolds number R tends to infinity, or the kinematic viscosity ν tends to zero, of a limiting form of the equations of motion, different from that obtained by putting ν = 0 in the first place. A solution of these limiting equations may then reasonably be expected to describe approximately the flow in a laminar boundary layer 2 for which R is large but not infinite. This is the basis of the classical theory of laminar boundary layers.
The full equation of motion for for a two-dimensional flow are: ∂u ∂t + u∂u ∂x + v∂u ∂y = −1 ρ ∂p ∂x + ν ∂2u ∂x2 + ∂2u ∂y2 , (2.1) ∂v ∂t + u∂v ∂x + v∂v ∂y = −1 ρ ∂p ∂y + ν ∂2v ∂x2 + ∂2v ∂y2 , (2.2) ∂u ∂x + ∂v ∂y = 0, (2.3) where the x and y variables are, respectively, the horizontal and vertical coordinates, u and v are, respectively, the horizontal and vertical fluid velocities and p is the fluid pressure. A wall is located in the plane y = 0. We consider non-dimensional variables x′ = x L, (2.4) y′ =y δ , (2.5) u′ = u U , (2.6) v′ = v U L δ , (2.7) p′ = p ρU 2, (2.8) t′ =tU L , (2.9) where L is the horizontal length scale, δ is the boundary layer thickness at x = L, which is unknown. We will obtain an estimate for it in terms of the Reynolds number R. U is the flow velocity, which is aligned in the x-direction parallel to the solid boundary. The non-dimensional form of the governing equations is: ∂u′ ∂t′ + u′∂u′ ∂x′ + v′∂u′ ∂y′ = −∂p′ ∂x′ + ν UL ∂2u′ ∂(x′)2 + ν UL L2 δ2 ∂2u′ ∂(y′)2, (2.10) ∂v′ ∂t′ + u′ ∂v′ ∂x′ + v′∂v′ ∂y′ = − L δ 2 ∂p′ ∂y′ + ν UL ∂2v′ ∂(x′)2 + ν UL L δ 2 ∂2v′ ∂(y′)2, (2.11) ∂u′ ∂x′ + ∂v′ ∂y′ = 0, (2.12) 3 where the Reynolds number for this problem is R = UL ν .
(2.13) Inside the boundary layer, viscous forces balance inertia and pressure gradient forces.
In other words, inertia and viscous forces are of the same order, so ν UL L δ 2 = O(1) ⇒δ = O(R−1/2L).
(2.14) Now we drop the primes from the non-dimensional governing equations and with equa-tion (2.14) we have ∂u ∂t + u∂u ∂x + v∂u ∂y = −∂p ∂x + 1 R ∂2u ∂x2 + ∂2u ∂y2 , (2.15) 1 R ∂v ∂t + u∂v ∂x + v∂v ∂y = −∂p ∂y + 1 R2 ∂2v ∂x2 + ∂2v ∂y2 , (2.16) ∂u ∂x + ∂v ∂y = 0.
(2.17) In the limit R →∞, the equations above reduce to: ∂u ∂t + u∂u ∂x + v∂u ∂y = −∂p ∂x + ∂2u ∂y2 , (2.18) −∂p ∂y = 0, (2.19) ∂u ∂x + ∂v ∂y = 0.
(2.20) Notice that according to equation (2.19), the pressure is constant across the boundary layer. In terms of dimensional variables, the system of equations above assume the form: ∂u ∂t + u∂u ∂x + v∂u ∂y = −1 ρ ∂p ∂x + ν ∂2u ∂y2 , (2.21) −1 ρ ∂p ∂y = 0, (2.22) ∂u ∂x + ∂v ∂y = 0.
(2.23) 4 To solve the system of equations above we need to specify initial and boundary conditions.
3 Steady State Laminar Boundary Layer on a Flat Plate.
We consider a flat plate at y = 0 with a stream with constant speed U parallel to the plate. We are interested in the steady state solution. We are not interested in how the flow outside the boundary layer reached the speed U. In this case, we need only to consider boundary conditions and the equation (2.18) simplifies since ∂u ∂t = 0. At the plate surface there is no flow across it, which implies that v = 0 at y = 0.
(3.24) Due to the viscosity we have the no slip condition at the plate. In other words, u = 0 at y = 0.
(3.25) At infinity (outside the boundary layer), away from the plate, we have that u →U as y →∞.
(3.26) For the flow along a flat plate parallel to the stream velocity U, we assume no pressure gradient, so the momentum equation in the x direction for steady motion in the boundary layer is u∂u ∂x + v∂u ∂y = ν ∂2u ∂y2 (3.27) and the appropriate boundary conditions are: 5 At y = 0, x > 0;u = v = 0, (3.28) At y →∞, all x;u = U, (3.29) At x = 0;u = U.
(3.30) These conditions demand an infinite gradient in speed at the leading edge x = y = 0, which implies a singularity in the mathematical solution there. However, the assump-tions implicit in the boundary layer approximation break down for the region of slow flow around the leading edge. The solution given by the boundary layer approximation is not valid at the leading edge. It is valid downstream of the point x = 0. We would like to reduce the boundary layer equation (3.27) to an equation with a single dependent variable. We consider the stream function ψ related to the velocities u and v according to the equations u =∂ψ ∂y , (3.31) v = −∂ψ ∂x .
(3.32) If we substitute equations (3.31) and (3.32) into the equation (3.27), we obtain a partial differential equation for ψ, which is given by ∂ψ ∂y ∂2ψ ∂x∂y −∂ψ ∂x ∂2ψ ∂y2 = ν ∂3ψ ∂y3 , (3.33) and the boundary conditions (3.28) and (3.29) assume the form At y = 0, x > 0;∂ψ ∂y = ∂ψ ∂x = 0, (3.34) At y →∞, all x;∂ψ ∂y = U.
(3.35) The boundary value problem admits a similarity solution. We would like to reduce the partial differential equation (3.33) to an ordinary differential equation. We would like to find a change of variables which allows us to perform the reduction mentioned above.
6 3.1 Similarity Solution.
We look for a one-parameter transformation of variables y, x and ψ under which the equations for the boundary value problem for ψ are invariant. A particularly useful transformation is y =λay′, (3.36) x =λbx′, (3.37) ψ =λcψ′.
(3.38) Requiring the invariance of the equation (3.33) with respect to the transformation (3.36) to (3.38) gives ∂ψ′ ∂y′ ∂2ψ′ ∂x′∂y′ −∂ψ′ ∂x′ ∂2ψ′ ∂(y′)2 = λ−c−a+bν ∂3ψ′ ∂(y′)3, (3.39) and performing the same for the boundary conditions gives λc−a∂ψ′ ∂y′ = 0 for y′ = 0 and x′ > 0, (3.40) λc−b∂ψ′ ∂x′ = 0 for y′ = 0 and x′ > 0.
(3.41) The requirement of invariance in the equations (3.39) to (3.41) leads to the algebraic relations −c −a + b = 0 c −a = 0 b = a/2 c = b c = a/2, (3.42) which suggest that we take the change of variables ψ = UBx1/2f(η) with η = A y x1/2.
(3.43) 7 If we substitute the change of variables above into equation (3.33), we obtain an ordinary differential equation for f(η), given by −1 2UBf(η)f ′′(η) = Aνf ′′′(η), (3.44) and we chose B = U −1/2√ 2ν (3.45) so that η = U 2νx 1/2 y, (3.46) such that ψ = (2νUx)1/2f(η).
(3.47) The change of variables given by equations (3.46) and (3.47) is analog to the change of variables used in the Rayleigh problem discussed in Capter 1, where instead of x/U we have time t. The analogy is that the disturbance due to the plate spreads out into the stream at the rate given by the unsteady problem (Rayleigh problem), but at the same time it is swept downstream with the fluid. The Rayleigh problem in Chapter 1 can be used to give an approximate solution to the problem here. The expression for the flow velocity u for the Rayleigh problem can be used to estimate the downstream velocity relative to the plate by identifying t with x/U.
This is equivalent to the following approximations in the momentum equation u∂u ∂x + v∂u ∂y = ν ∂2u ∂x2 + ∂2u ∂y2 .
Firstly, the convection terms on the left are replaced by the approximation U∂u/∂x, and, secondly, the term ν∂2u/∂x2 is neglected in the viscous terms on the right. In 8 this way, we obtain the diffusion equation for u with t replaced by x/U. The boundary layer approximation retains the convection terms in full and makes only the second sim-plification. The Rayleigh approximation obviously overestimates the convection effects; hence its prediction of the boundary layer thickness will be too small and the value of the shear stress to great.
The equation that f(η) has to satisfy is ff ′′ + f ′′′ = 0 (3.48) with boundary conditions: f ′(η) = 0 at η = 0 (3.49) f(η) = 0 at η = 0 (3.50) f ′(η) = 1 at η →∞ (3.51) This is a boundary value problem for the function f(η) which has no closed form solution, so we need to solve it numerically. Solving boundary value problems numerically is not an easy task. We would like to reduce this boundary value problem to an initial value problem. For the equation (3.48) this is possible. If F(η) is any solution of equation (3.48), also f = CF(Cη) (3.52) is a solution, with C an arbitrary constant of homology. Then, lim η→∞f ′(η) = C2 lim η→∞F ′(Cη) = C2 lim η→∞F ′(η).
(3.53) Since f ′(η) = 1 for η →∞, we have that C = lim η→∞F ′(η) −1/2 (3.54) 9 We know from equation (3.52) that f ′′(0) = C3F ′′(0), and if we specify F ′′(0) = 1, we have from equation (3.54) that f ′′(0) = lim η→∞F ′(η) −3/2 .
(3.55) To obtain the value f ′′(0), we need to evaluate numerically the initial value problem for F(η), given by equation (3.48) with initial conditions F(0) = F ′(0) = 0 and F ′′(0) = 1, for a large value of η to obtain F ′(η) as η →∞and f ′′(0) (see equation (3.55). Therefore, the initial value problem for f(η) equivalent to the boundary value problem given by equations (3.48) to (3.51) is given by the ordinary differential equation (3.48) plus the initial conditions f(0) =0, (3.56) f ′(0) =0, (3.57) f ′′(0) = lim η→∞F ′(η) −3/2 .
(3.58) Below we illustrate the result of the numerical integration of the initial value problem given by equation (3.48) and initial conditions (3.56) to (3.58) in the figures 1. We obtained numerically that f ′′(0) ≈0.46960.
10 Figure 1: Functions f(η), f ′(η) and f ′′(η). The horizontal axis represents the range of values of η considered, and in the vertical axis we have the values of the functions f(η), f ′(η) and f ′′(η).
Once we have computed numerically f(η) and its derivatives up to second order, we can obtain the velocity components (u, v) at any point (x, y) of the flow domain, according to the equations 11 u =∂ψ ∂y = Uf ′(η), (3.59) v = −∂ψ ∂x = − Uν 2x 1/2 f(η) −1 2Uyf ′(η).
(3.60) 3.1.1 Vorticity.
The vorticity of the flow in Cartesian coordinates in term of dimensional variables is given by ω = ∂u ∂y −∂v ∂x.
(3.61) If we non-dimensionalize the vorticity, according to ω = U L ω′, (3.62) the non-dimensional form of the equation (3.61) is U L ω′ = U δ ∂u′ ∂y′ −Uδ L2 ∂v′ ∂x′, (3.63) and since δ ∼O(R−1/2L), we have that ω′ = R1/2∂u′ ∂y′ −R−1/2 ∂v′ ∂x′, (3.64) and in the limit R →∞we have ω′ ∼R1/2∂u′ ∂y′ .
(3.65) In the context of the boundary-layer approximation, the vorticity in terms of dimensional variables is given by ω = µ∂u ∂y .
(3.66) 12 We can write the vorticity for the Blasius boundary layer similarity solution by placing equation (3.47) for ψ into the equation (3.66), which gives ω = µ∂2ψ ∂y2 = µ U 3 2νx 1/2 f ′′(η) (3.67) 3.1.2 Stress.
The normal components of the stress perpendicular and parallel to the flat plate ex-pressed non-dimensionally are τxx ρU 2 = − p ρU 2 + 2 ν U 2 ∂u ∂x | {z } O(1/R) , (3.68) τyy ρU 2 = − p ρU 2 + 2 ν U 2 ∂v ∂y | {z } O(1/R) .
(3.69) Therefore, in the limit R →∞, we have that τxx = τyy = −p (3.70) The shearing stress over surfaces parallel to the wall is τxy = µ ∂u ∂y + ∂v ∂x , (3.71) which is approximated in the same way as the vorticity, as follows.
ρU 2τ ′ xy = ρν U δ |{z} U L R1/2 ∂u′ ∂y′ + Uδ L2 |{z} U L R−1/2 ∂v′ ∂y′ = ρU 2R−1/2∂u′ ∂y′ + ρU 2R−3/2 ∂v′ ∂x′ (3.72) In the limit R →∞, the shear stress is given by 13 τ ′ xy ∼R−1/2∂u′ ∂y′ (3.73) or in terms of dimensional variables τxy = µ∂u ∂y .
(3.74) For the Blasius laminar boundary layer similarity solution given by equation (3.47), the shear stress τxy is given by τxy ∼µ∂2ψ ∂y2 = µ U 3 2νx 1/2 f ′′(η), (3.75) 4 Boundary-layer Thickness, Skin friction, and En-ergy dissipation.
According to equation (2.22), the pressure across the boundary layer is constant in the boundary-layer approximation, and its value at any point is therefore determined by the corresponding main-stream conditions. If U(x, t) now denotes the main stream velocity, so that −1 ρ dp dx = ∂U ∂t + U ∂U ∂x .
(4.76) Elimination of the pressure from equation (2.21) gives in terms of dimensional variables the boundary layer momentum equation ∂u ∂t + u∂u ∂x + v∂u ∂y = ∂U ∂t + U ∂U ∂x + ν ∂2u ∂y2 , (4.77) and from equation (2.23), we have the mass conservation equation ∂u ∂x + ∂v ∂y = 0.
(4.78) 14 In most physical problems the solutions of the boundary layer equations (4.77) and (4.78) are such that the velocity component u attains its main-stream value U only asymptotically as R1/2y/L →∞. The thickness of the layer is therefore indefinite, as there is always some departure from the asymptotic value at any finite distance y from the surface. In practice the approach to the limit is rapid and a point is soon reached beyond which the influence of viscosity is imperceptible. It would therefore be possible to regard the boundary layer thickness as a distance δ from the surface beyond which u/Y > 0.99, for example, but this is not sufficiently precise (since ∂u/∂y is small there) for experimental work, and is not of theoretical significance.
The scale of the boundary layer thickness can, however, be specified adequately by certain lengths capable of precise definition, both for experimental measurement and for theoretical study. These measures of boundary layer thickness are defined as follows: • Displacement thickness δ1: δ1 = Z ∞ 0 1 −u U dy (4.79) • Momentum thickness δ2: δ2 = Z ∞ 0 u U 1 −u U dy (4.80) • Energy thickness δ3: δ3 = Z ∞ 0 u U 1 −u2 U 2 dy (4.81) The upper limit of integration is taken as infinity owing to the asymptotic approach of u/U to 1, but in practice the upper limit is the point beyond which the integrand is negligible.
Uδ1 is the diminution, due to the boundary layer, of the volume flux across a normal to the surface; the streamlines of the outer flow are thus displaced away from the surface 15 through a distance δ1. Similarly, ρU 2δ2 is the flux of the defect of momentum, and 1 2ρU 3δ3 is the flux of defect of kinetic energy.
Two other quantities related to these boundary layer thickness are the skin friction τω and the dissipation integral D. The skin friction is defined as the shearing stress exerted by the fluid on the surface over which it flows, and is therefore the value of τxy at y = 0, which by (3.74) is τω = µ ∂u ∂y y=0 (4.82) in terms of dimensional variables. The rate at which energy is dissipated by the action of viscosity has been shown to be µ ∂u ∂y 2 per unit time per unit volume, and D is the integral of this across the layer: D = Z ∞ 0 µ ∂u ∂y 2 dy.
(4.83) Consequently, D is the total dissipation in a cylinder of small cross-section with axis normal to the layer per unit time per unit area of cross-section.
4.1 Quantities for the Blasius Boundary Layer Solution.
For the Blasius similarity solution for a two-dimensional boundary layer given by equa-tion (3.47), we can compute the the quantities defined above: • Displacement thickness δ1: δ1 = 2νx U 1/2 Z ∞ 0 (1 −f ′(η))dη = 2νx U 1/2 lim η→∞(η −f(η)) (4.84) • Momentum thickness δ2: 16 δ2 = 2νx U 1/2 Z ∞ 0 f ′(η)(1 −f ′(η))dη = 2νx U 1/2 f(η)|∞ 0 −ff ′|∞ 0 + Z ∞ 0 ff ′′dη and taking into account the boundary conditions (3.49) to (3.51), we obtain that δ2 = 2νx U 1/2 Z ∞ 0 ff ′′dη (4.85) • Energy thickness δ3: δ3 = 2νx U 1/2 Z ∞ 0 f ′(η)(1 −f ′(η)2)dη = 2νx U 1/2 f(η)|∞ 0 −f(f ′)2|∞ 0 + 2 Z ∞ 0 ff ′f ′′dη and taking into account the boundary conditions (3.49) to (3.51), we obtain that δ3 = 2νx U 1/2 2 Z ∞ 0 ff ′f ′′dη (4.86) • Skin friction τω: τomega = µ U 3 2νx 1/2 f ′′(η)|η=0, and according to the initial condition f ′′(0) = 1, we have that τomega = µ U 3 2νx 1/2 (4.87) • Dissipation integral D: D = µ U 3 2νx Z ∞ 0 f ′′(eta)2dη (4.88) 17 5 Momentum and Energy Equations.
The skin friction and dissipation are connected with the boundary-layer thickness by two equations which represent the balance of momentum and energy within a small section of the boundary layer.
5.1 Momentum Integral We can write equation (4.77) in the form −ν ∂2u ∂y2 = ∂ ∂t(u −U) + U ∂U ∂x −u∂u ∂x −v∂u ∂y , (5.89) and if we multiply equation (4.78) by (u −U) we obtain (U −u)∂u ∂x + (U −u)∂v ∂y = 0.
(5.90) If we add these two equations, we obtain −ν ∂2u ∂y2 = ∂ ∂t(u −U) + ∂ ∂x(Uu −u2) + (U −u)∂U ∂x + ∂ ∂x(vU −vu).
(5.91) Next, we integrate with respect to y from 0 to ∞. This yields −ν ∂u ∂y |y=0 = ∂ ∂t Z ∞ 0 (u −U)dy + ∂ ∂x Z ∞ 0 (Uu −u2)dy + ∂U ∂x Z ∞ 0 (U −u)dy + (vU)|y=0, (5.92) since ∂u/∂y and v(U −u) tend to zero as y →∞. If there is no suction at the body surface, vy=0 = 0. We assume that there is no suction at the body surface, and by taking into account equations (4.79) to (4.82), we can write equation (5.92) in the form τω ρ = ν ∂u ∂y |y=0 = ∂ ∂t(Uδ1) + ∂ ∂x(U 2δ2) + U ∂U ∂x δ1 (5.93) 18 5.2 Energy Integral.
We multiply equation (4.77) and (4.78) respectively by 2u and (U 2 −u2), so we have (4.77) →−2uν ∂2u ∂y2 = 2u ∂ ∂t(U −u) + 2uU ∂U ∂x −2u2∂u ∂x −2vu∂u ∂y , (5.94) (4.77) →0 = (U 2 −u2)∂u ∂x + (U 2 −u2)∂v ∂y, (5.95) and by adding them we obtain 2ν ∂u ∂y 2 −2ν ∂ ∂y u∂u ∂y = ∂ ∂t(Uu −u2) + U 2 ∂ ∂t 1 −u U + ∂ ∂x(U 2u −u3) + ∂ ∂y(vU 2 −vu2) (5.96) If we integrate the equation above with respect to y from 0 to ∞, and if we take into account equations (4.79) to (4.83), we obtain 2D ρ = ∂ ∂t(U 2δ2) + U 2 ∂ ∂tδ1 + ∂ ∂x(U 3δ3), (5.97) since v(U 2−u2) and ∂u/∂y both tend to zero as y tends to infinite and the term (vU 2)y=0 is zero because it is assumed that there is no suction on the plate. The energy integral may also be regarded as an equation for the “kinetic energy defect” 1 2ρ(U 2 −u2) per unit volume, namely ∂ ∂t Z ∞ 0 1 2ρ(U 2 −u2)dy + ∂ ∂x Z ∞ 0 1 2ρ(U 2 −u2)udy = D + ρ∂U ∂t Z ∞ 0 (U −u)dy (5.98) 6 Approximate Method Based on the Momentum Equation: Pohlhausen’s Method.
One of the earliest and, until recently, most widely used approximate methods for the solution of the boundary layer equation is that developed by Pohlhausen. This method is based on the momentum equation of K´ arm´ an, which is obtained by integrating the 19 boundary layer equation (4.77) across the layer, as shown in section 5.1. In the case of steady flow over an impermeable surface, the momentum equation (5.93) reduces to τω ρU 2 = d dxδ2 + 2δ2 + δ1 U dU dx (6.99) where τω = µ(∂u/∂y)|y=0 is the skin friction, δ1 = R ∞ 0 (1 −u/U)dy is the displacement thickness, and δ2 = R ∞ 0 (u/U)(1 −u/U)dy is the momentum thickness of the boundary layer. The boundary conditions for the boundary layer equations (4.77) and (4.78) are u = v = 0 at y = 0, (6.100) u →U(x) as y →∞.
(6.101) From the boundary layer equations (4.77) and (4.78) and their derivatives with respect to y, a set of conditions on u can be derived with the aid of the boundary conditions (6.100) and (6.101). These conditions are From (6.100) →u = 0 at y = 0, (6.102) From (4.77) and (6.100) →∂2u ∂y2 = −U ν dU dx at y = 0, (6.103) From ∂ ∂y(4.77), (4.78) and (6.100) →∂3u ∂y3 = 0 at y = 0, (6.104) From ∂2 ∂y2(4.77), ∂ ∂y(4.78) and (6.100) →∂4u ∂y4 = 1 ν ∂u ∂y ∂2u ∂y∂x at y = 0, (6.105) . . . . . . , and at y →∞we have u →U ⇒∂u ∂y →0 ⇒. . . ⇒∂(n)u ∂y(n) →0 ⇒. . .
(6.106) In the Pohlhausen’s method, and similar approximate methods, a form for the velocity profile u(x, y) is sought which satisfies the momentum equation (6.99) and some of the boundary condition (6.102) to (6.106). It is hoped that this form will approximate to 20 the exact profile, which satisfies all the conditions (6.102) to (6.106) as well as (6.99).
The form assumed is u U = f(η) with η = y δ(x), (6.107) where δ(x) is the effective total thickness of the boundary layer. The function f(η) may also depend on x through certain coefficients which are chosen so as to satisfy some of the conditions (6.102) to (6.106). Altough strictly the conditions at infinity are only approached asymptotically, it is assumed that these conditions can be transfered from infinity to y = δ without appreciable error. Thus equations (6.102) to (6.106) become f(0) = 0 (6.108) f ′′(0) = −Λ = δ2 ν dU dx (6.109) f ′′′(0) = 0 (6.110) f (iv)(0) = δ3 ν f ′ d dx U f ′ δ |η=0 (6.111) . . . . . .
f(1) = 1 and f ′(1) = f ′′(1) = . . . = f (n)(1) = . . . = 0 (6.112) For the assumed velocity profile (6.107), we have that τω ρU 2 = ν δU f ′(0), (6.113) δ1 = δ Z 1 0 (1 −f)dη, (6.114) δ2 = δ Z 1 0 f(1 −f)dη, (6.115) If the form assumed for f(η) involves m unknown coefficients, these can be specified using m of the boundary conditions (6.108) to (6.112) and the remaining unknown δ can be determined from the momentum equation (6.99). Also, if only the conditions (6.108), (6.109) and (6.110) are considered, τωδ/µU, δ1/δ and δ2/delta are functions of 21 Λ alone. In this case, substitution of relations (6.113) to (6.115) into (6.99) leads to an equation of the form ν Uδf ′(0) = d dx δ Z 1 0 f(1 −f)dη + 2δ U dU dx Z 1 0 f(1 −f)dη + δ U dU dx Z 1 0 (1 −f)dη ν Uδf ′(0) = dδ dx Z 1 0 f(1 −f)dη + δ U dU dx 2 Z 1 0 f(1 −f)dη + Z 1 0 (1 −f)dη f ′(0) U = δ ν dδ dx dU dx R 1 0 f(1 −f)dη dU dx + 1 2 δ2 ν d2U dx2 R 1 0 f(1 −f)dη dU dx −1 2 δ2 ν d2U dx2 R 1 0 f(1 −f)dη dU dx + 1 U δ2 ν dU dx 2 Z 1 0 f(1 −f)dη + Z 1 0 (1 −f)dη (6.116) Since Λ = (δ2/ν)dU/dx, we can write the equation above in the form dΛ dx = 1 U dU dx ( f ′(0) R 1 0 f(1 −f)dη −Λ " 2 − R 1 0 (1 −f)dη R 1 0 f(1 −f)dη #) + d2U dx2 dU dx Λ 2 (6.117) or simply dΛ dx = 1 U dU dx g(Λ) + d2U dx2 dU dx h(Λ) (6.118) where g(Λ) = ( f ′(0) R 1 0 f(1 −f)dη −Λ " 2 − R 1 0 (1 −f)dη R 1 0 f(1 −f)dη #) (6.119) h(Λ) =Λ/2 (6.120) Pohlhausen used a family of velocity profiles given by the quartic polynomial u/U = f(η) = 2η −2η3 + η4 + 1 6Λη(1 −η)3 (6.121) chosen to satisfy the boundary conditions f(0) = 0, f ′′(0) = −Λ, f(1) = 1, f ′(1) = f ′′(1) = 0 22 With the chosen velocity profile (6.121), we can evaluate the functions g(Λ) and h(Λ), and then integrate numerically equation (6.118) to obtain Λ as a function of x and through equation δ2 = νΛ/(dU/dx). By the substitution of Λ = Λ(x) and δ = δ(x) in the assumed velocity profile f(η), we obtain the velocity profile for any x and y.
7 Stagnation Point flow.
For an ideal fluid the flow against an infinite flat plate in the plane y = 0 is given by u =Ux, (7.122) v = −Uy, (7.123) where U is a constant.
When viscosity is included, it still must be true that u is proportional to x, for small x and for all y.Thus, for small x, at least, we may take u = kxF(y) (7.124) The governing equations for steady flow in terms of dimensional variables are given by equations (2.1) to (2.3). If we substitute equation (7.124) into the continuity equation (2.3), we obtain ∂v ∂y = −kF(y) (7.125) This suggest that we take u =kxf ′(Ay), (7.126) v = −Bf(Ay).
(7.127) Now, we substitute equations (7.126) and (7.127) into the governing equations (2.1) to (2.3). We obtain from 23 (2.1) →k2xf ′(Ay)2 −Bf(Ay)kxf ′′(Ay)A = −1 ρ ∂p ∂x + νkxA2f ′′′(Ay), (7.128) (2.2) →B2f(Ay)Af ′(Ay) = −1 ρ ∂p ∂y −νBf ′′(Ay)A2, (7.129) (2.3) →kf ′(Ay) −BAf ′(Ay) = 0.
(7.130) This last equation implies that BA = k, so we can write equations (7.128) and (7.130) in the form (7.128) →k2x(f ′)2 −k2xf ′′f = −1 ρ ∂p ∂x + νkxA2f ′′′ (7.131) (7.129) →Bkff ′ = −1 ρ ∂p ∂y −νkAf ′′ (7.132) We can solve equation (7.131) for the pressure derivative with respect to x to obtain −1 ρ ∂p ∂x = k2x[(f ′)2 −ff ′′ −ν kA2f ′′′] (7.133) Since the the term between brackets in the right side of the equation above is not a function of x, we set (f ′)2 −ff ′′ −ν kA2f ′′′ = 1, (7.134) which implies, according to equation (7.133), that −1 ρ ∂p ∂x = k2x →−1 ρp = 1 2k2x2 + C(y) (7.135) If we substitute equation (7.135) into equation (7.132), we have the relation Bkff ′ = dC dy −νkAf ′′, (7.136) and if we integrate this equation, we obtain that 24 C(y) = 1 2B k Af 2 + νkf ′ −C0.
(7.137) Next, we substitute the expression for C(y) above into the equation (7.135) for the pressure, which gives −1 ρp = 1 2k2x2 + 1 2B k Af 2 + νkf ′ −C0.
(7.138) To simplify equation (7.134), we chose the coefficient of f ′′′ equal to one, which implies that ν A2 k = 1 →A = r k ν , (7.139) and since BA = k, we have that B = √ νk.
(7.140) With equations (7.138) and (7.139), we can write the equation (7.138) as follows p0 −p ρ = 1 2k2x2 + 1 2νkf 2 + νkf ′ (7.141) with C0 = p0/ρ and p0 is the pressure as y →∞. The function f(η), where η = p k/νy satisfies the ordinary differential equation (f ′)2 −ff ′′ −f ′′′ −1 = 0 (7.142) with boundary conditions: • No slip condition at the plate surface.
u = 0 at y = 0 →f ′(η) = 0 at η = 0, (7.143) 25 • No flux across the plate: v = 0 at y = 0 →f(η) = 0 at η = 0, (7.144) • At y →∞, we have u = kx, which implies that f ′(η) = 1 as η →∞ (7.145) In summary, the function f(η) is the solution of the boundary value problem given by the equations (7.142) to (7.145), which has no closed form solution. The equation ordinary differential equation (7.142) is non-linear and has to be solved numerically together with the boundary conditions (7.143) to (7.145). Figure 2 below illustarte the result of the numerical evaluation of the boundary value problem given by equations (7.142) to (7.145) for η in the range 0 < η < 10.
26 Figure 2: Functions f(η), f ′(η) and f ′′(η). The horizontal axis represents the range of values of η considered, and in the vertical axis we have the values of the functions f(η), f ′(η) and f ′′(η).
Once we know the values of f(η), f ′(η) and f ′′(η), we can obtain the velocities u and v given, respectivelly, by equations (7.126) and (7.127) with B and A given, respectively, by equations (7.139) and (7.140), the pressure field is given by equation (7.141).
27 8 Two-dimensional Laminar Jet.
We consider a two-dimensional jet as illustrated in the figure below. x is the horizontal coordinate and y is the vertical coordinate. u and v are, respectively, the horizontal and vertical fluid velocities. The jet in the direction of the x axis generates a flow where the fluid velocity along the y axis tends to zero. We assume that the boundary layer approximation is valid and the governing equation for the fluid motion are equations (2.21) to (2.23), but with ∂u/∂t. The pressure does not vary in the y direction according to equation (2.22), so the pressure is constant across the boundary layer and its gradient is given by the pressure gradient outside the boundary layer. For this problem there is no pressure gradient, so the governing equations for the fluid motion are u∂u ∂x + v∂u ∂y = ν ∂2u ∂y2 , (8.146) ∂u ∂x + ∂v ∂y = 0 (8.147) Next, we integrate equation (8.146) with respect to the y variable from −∞to +∞, which gives Z ∞ −∞ (u∂u ∂x + v∂u ∂y )dy = ν Z ∞ −∞ ∂2u ∂y2 dy (8.148) We can write v∂u ∂y = ∂ ∂y(vu) −u∂v ∂y, and if we multiply the continuity equation (8.147) by u, we can write the equation above as v∂u ∂y = ∂ ∂y(vu) −u∂u ∂x, (8.149) Next, we substitute equation (8.149) into equation (8.148), we perform the integration and we use the boundary condition 28 u →0 as y →±∞.
(8.150) Equation (8.148) simplifies to ∂ ∂x Z +∞ −∞ u2dy = 0 (8.151) Thus, the momentum flux is constant in x. In other words, ρ Z ∞ −∞ u2dy = M (8.152) We call δ(x) the order of the magnitude of the boundary layer thickness at position x.
Since the orifice is very small we have δ(0) →0. At the orifice the equation (8.152) can be written as ρu(0)2δ(0) = M, (8.153) which implies that u(0) = δ(0)−1/2 (8.154) The Mass flux at the orifice is ρu(0)δ(0) ∼δ(0)1/2 →0, (8.155) hence unimportant. A jet is the result of a momentum source, not a volume source.
Next, we are going to solve the boundary layer equations for the jet. We introduce the stream function ψ related to the velocities u and v according to the equations u =∂ψ ∂y (8.156) v = −∂ψ ∂x (8.157) 29 In terms of the stream function, the x momentum equation (8.146) assume the form ∂ψ ∂y ∂2ψ ∂y∂x −∂ψ ∂x ∂2ψ ∂y2 = ν ∂3ψ ∂y3 (8.158) with the boundary conditions ψ →0 as y →±∞ (8.159) and ρ Z ∞ −∞ ∂ψ ∂y 2 dy = M (8.160) We are going to solve the boundary value problem given by equations (8.158) to (8.160) by looking for a similarity solution. We look for a one-parameter transformation of variables y, x and ψ under which the equations of the boundary value problem mentioned above are invariant. A particular useful transformation is x = λax′, y = λby′ and ψ = λcψ′ (8.161) The requirement of invariance of the boundary value problem (8.158) to (8.160) under such transformation implies that c + b −a =0 (8.162) c −b =0 (8.163) and no information is gained from equation (8.160). From the equations (8.162) and (8.163) we obtain c = a/3 and b = 2a/3 (8.164) This suggest that we take 30 ψ Bx1/3 = f(η) with η = Cy x2/3, (8.165) where the coefficients B and C are chosen to simplify the appearance of the final equa-tion. We substitute (8.165) into equation (8.158), which gives the ordinary differential equation 3f ′′′ + (f ′)2 + ff ′′ = 0, (8.166) and for ψ and η we have ψ = Mνx ρ 1/3 f(η) with η = M ρν2x2 1/3 (8.167) the boundary conditions become f ′(±∞) →0, f(0) = f ′′(0) = 0(symmetry) (8.168) and 1 = Z ∞ −∞ |f ′(η)|2dη (8.169) Now we integrate equation (8.166) once with respect to η, which gives 3f ′′ + ff ′ = constant = 0, and integrating again 3f ′ + 1 2f 2 = c2 (8.170) We now write f = √ 2F and η = 3 √ 2ζ. then equation (8.170) assumes the form 31 dF dζ + F 2 = c2 → dF/c 1 −(F/c)2 = cdζ (8.171) which can be integrated in closed form, so we have cζ = tanh−1(F/c) (8.172) since F(0) = 0. Thus f = √ 2F = √ 2c tanh cη 3 √ 2 .
(8.173) We substitute the equation above for f in the boundary condition (8.169), which gives 1 = c3√ 2 3 Z ∞ −∞ sech4(cζ)d(cζ) = 4 √ 2c3 9 →c3 = 9 4 √ 2 (8.174) Finally, we write χ = M 48ρν2 1/3 y x2/3 (8.175) The final solution has the form ψ = 9Mνx 2ρ 1/3 tanh(χ) (8.176) and for the velocities we have u = ∂ψ ∂y = 3M 2 32ρ2νx 1/3 sech2(χ) (8.177) v = −∂ψ ∂x = Mν 6ρx2 1/3 (2χsech2χ −tanh χ) (8.178) Next, we discuss the implications of the results obtained above. The jet width can be defined by χ = ±χ0 such that u →0. Then from equation (8.177), we realize that 32 the jet width is proportional to x2/3. The centerline velocity, let say umax, according to equation (8.177) is proportional to x−1/3. As χ →±∞, according to equation (8.178), v →∓(Mν/6ρx2)1/3. This implies a that there is entrainment from the jet eddges as show in the figure 4. If we define the Reynolds number as R = umaxδ/nu, equation (8.177) implies that R ∝x1/3. To illustrate the streamlines for the flow generated by the jet, we present figure 3. The velocity field is illustrated in figure 3.
Figure 3: Streamlines obtained from equation (8.176) with M = 1000kg/sec2, ν = 0.01m2/sec and ρ = 1000kg/m3.
33 Figure 4: Velocity field obtained from equations (8.177) and (8.178) with M = 1000kg/sec2, ν = 0.01m2/sec and ρ = 1000kg/m3. |
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Qual è la dimensione normale di un asciugamano da bagno?
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Ultimo aggiornamento:27 agosto 2025
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Qual è la dimensione normale di un asciugamano da bagno?
Non riuscite a scegliere la dimensione giusta dell'asciugamano per il vostro marchio? Ordinare dimensioni sbagliate può frustrare i clienti e danneggiare la vostra reputazione. Questa guida chiarisce le dimensioni standard degli asciugamani.
Le dimensioni di un normale asciugamano da bagno sono in genere comprese tra 27 e 30 pollici di larghezza e 52 e 58 pollici di lunghezza (circa 70 x 130 cm - 75 x 145 cm). Queste dimensioni sono pensate per essere abbastanza grandi da asciugare comodamente un adulto medio dopo la doccia.
Conoscere le dimensioni degli asciugamani non è solo un dettaglio tecnico. Le dimensioni di un asciugamano influiscono direttamente sul modo in cui il cliente vive il prodotto, dalla sensazione di lusso alla semplice funzione quotidiana. Scegliere le dimensioni giuste è il primo passo fondamentale per sviluppare un asciugamano che le persone ameranno, useranno e ricompreranno. Vediamo quali sono le domande più comuni che mi vengono poste dai proprietari di marchi e dai responsabili degli acquisti per essere sicuri di fare la scelta giusta.
Indice dei contenutinascondersi
1 Che dimensioni ha un asciugamano normale in cm?
2 Che dimensioni hanno gli asciugamani da bagno utilizzati dagli hotel?
3 Che cos'è un asciugamano a grandezza naturale?
4 Qual è la lunghezza media di un asciugamano?
5 Conclusione
Che dimensioni ha un asciugamano normale in cm?
Pensate in pollici mentre il vostro produttore parla in centimetri? Questa semplice discrepanza può causare grossi problemi e costosi errori nell'ordine. Ecco le dimensioni necessarie.
Un normale asciugamano da bagno in centimetri misura solitamente tra i 70 e i 75 cm di larghezza e tra i 130 e i 145 cm di lunghezza. Queste misure metriche sono lo standard globale per la produzione tessile e garantiscono la coerenza tra fornitori e regioni diverse.
Quando ci si approvvigiona di asciugamani, soprattutto per un marchio globale, è fondamentale parlare la stessa lingua del produttore. Nei miei anni di lavoro presso TowelTrend, ho visto la confusione tra pollici e centimetri causare ritardi e campioni errati. Confermiamo sempre le specifiche in entrambe le unità per evitare qualsiasi problema. Queste misure standard non sono casuali. Si basano sulle larghezze standard dei telai di tessitura, il che aiuta a massimizzare l'efficienza del tessuto e a ridurre gli sprechi. È grazie a questa efficienza che possiamo offrire prezzi competitivi, anche per ordini di quantità inferiori a 500 pezzi. Per gli acquirenti, la comprensione delle dimensioni comuni è fondamentale per fare un acquisto intelligente.
Dimensioni comuni degli asciugamani (imperiali e metriche)
| Tipo di asciugamano | Dimensioni in pollici (larghezza x lunghezza) | Dimensioni in centimetri (larghezza x lunghezza) | Uso primario |
--- --- |
| Asciugamano | Da 12" x 12" a 13" x 13" | Da 30 x 30 cm a 33 x 33 cm | Lavaggio di viso e corpo |
| Asciugamano | Da 16" x 28" a 18" x 30". | Da 40 x 70 cm a 45 x 75 cm | Asciugare le mani |
| Asciugamano da bagno | Da 27" x 52" a 30" x 58". | Da 70 x 130 cm a 75 x 145 cm | Asciugare il corpo dopo la doccia |
| Foglio da bagno | Da 35" x 60" a 40" x 70". | Da 90 x 150 cm a 100 x 180 cm | Impacco su tutto il corpo, lusso |
Che dimensioni hanno gli asciugamani da bagno utilizzati dagli hotel?
Volete dare ai vostri clienti la sensazione di un hotel lussuoso e soffice? Gli asciugamani standard per la vendita al dettaglio potrebbero non essere sufficienti per creare un'esperienza di alto livello. Scoprite le dimensioni preferite dagli hotel.
Gli hotel utilizzano in genere asciugamani da bagno di dimensioni maggiori, spesso più simili a un lenzuolo da bagno, che misurano circa 30 x 58 pollici (75 x 145 cm) o addirittura 35 x 60 pollici (90 x 150 cm). Queste dimensioni maggiori trasmettono un senso di lusso e offrono una migliore copertura agli ospiti.
La scelta di un asciugamano più grande nel settore dell'ospitalità è molto intenzionale. Non è solo una questione di dimensioni, ma di valore percepito e di comfort. Un asciugamano più grande e più pesante dà una sensazione di qualità superiore e di benessere, che eleva l'esperienza dell'ospite. Ricordo di aver lavorato con un nuovo cliente di un boutique hotel che stava decidendo tra un asciugamano standard di 70×140 cm e una versione più lussuosa di 80×160 cm. Abbiamo prodotto campioni di entrambi. Dopo aver provato la differenza, hanno scelto il formato più grande. In seguito mi hanno detto che era uno dei dettagli più apprezzati nelle recensioni degli ospiti. Oltre al lusso, questi asciugamani sono anche costruiti per durare nel tempo, per resistere a continui lavaggi industriali ad alta temperatura: ecco perché un GSM (grammi per metro quadrato) elevato è importante quanto le dimensioni.
Dimensioni degli asciugamani d'albergo: Lusso e praticità
| Tipo di asciugamano | Dimensione tipica (cm) | Peso tipico (GSM) | Caratteristiche principali |
--- --- |
| Asciugamano da bagno standard per hotel | 70 x 140 cm | 500-600 GSM | Durevole, assorbente, economico |
| Asciugamano da bagno per hotel di lusso | 80 x 150 cm | 600-750 GSM | Più grande, più pesante, sensazione di alta gamma |
| Asciugamano da piscina per hotel | 90 x 180 cm | 550-650 GSM | Extra-lunghi per il relax, spesso a righe |
Che cos'è un asciugamano a grandezza naturale?
Avete mai sentito il termine "asciugamano full-size" e vi siete chiesti cosa significasse? Ordinare basandosi su termini di marketing vaghi è un rischio che non dovreste correre. Definiamolo chiaramente.
Un "asciugamano intero" è semplicemente un altro nome per un asciugamano da bagno standard. È l'asciugamano multiuso più comune, di solito misura circa 27 x 52 pollici (70 x 130 cm), progettato per asciugarsi completamente dopo il bagno o la doccia.
Il linguaggio del marketing può talvolta creare confusione nel mondo della produzione. Termini come "full-size", "dinner-plate" o "queen-size" possono avere significati diversi per persone diverse. Nel settore degli asciugamani, "full-size" si riferisce quasi sempre all'asciugamano da bagno standard di cui abbiamo parlato. Si distingue dall'asciugamano per le mani, che è più piccolo, e dal telo da bagno, che è significativamente più grande. Quando un nuovo cliente richiede un "asciugamano a grandezza naturale", il mio primo passo è sempre quello di chiarire le dimensioni esatte di cui ha bisogno in centimetri o pollici. Questa semplice verifica è una parte fondamentale del nostro processo di TowelTrend. Elimina le supposizioni e garantisce che il prodotto finale corrisponda perfettamente alla visione del cliente, evitando i tempi e le spese di una nuova produzione.
Decodificare la terminologia degli asciugamani
| Termine comune | Conosciuto anche come | Dimensione tipica (cm) | Il migliore per |
--- --- |
| Asciugamano a grandezza naturale | Asciugamano da bagno standard | 70 x 130 cm | Asciugatura quotidiana e multiuso |
| Foglio da bagno | Asciugamano da bagno di lusso | 90 x 150 cm | Copertura extra e sensazione di benessere |
| Asciugamano | Asciugamano per ospiti | 40 x 70 cm | Asciugare le mani nei bagni |
Qual è la lunghezza media di un asciugamano?
Avete bisogno di una risposta rapida e semplice per le lunghezze standard degli asciugamani? Tirando a indovinare si può ottenere un prodotto troppo corto per essere funzionale o troppo lungo per essere pratico.
La lunghezza media di un asciugamano da bagno standard è compresa tra 130 cm e 145 cm (circa 52-58 pollici). Questa lunghezza è stata appositamente studiata per essere abbastanza lunga da avvolgere comodamente la vita o il busto di un adulto medio dopo la doccia.
Sebbene l'asciugamano da bagno sia il punto di riferimento più comune, la "lunghezza media" dipende in realtà dalla funzione cui l'asciugamano è destinato. Ogni tipo di asciugamano ha una lunghezza ottimizzata per il suo uso specifico. Ad esempio, un asciugamano per le mani è più corto perché serve solo per asciugare le mani. Un asciugamano da spiaggia è extra-lungo per potersi sdraiare comodamente senza toccare la sabbia. Pensare all'uso finale è fondamentale quando si sviluppa un nuovo prodotto di asciugamani. Una volta ho lavorato con un marchio di fitness che voleva creare un asciugamano per la palestra. La lunghezza dell'asciugamano da bagno standard era troppo corta e larga per essere drappeggiata sulle panche. Abbiamo collaborato per produrre un asciugamano personalizzato, più lungo e più stretto. È stato un grande successo perché ha risolto un problema reale per i loro clienti.
Lunghezza media tra i vari tipi di asciugamani
Asciugamano: La lunghezza media è di circa 70 cm (28 pollici). Perfetto per essere appeso vicino al lavandino.
Asciugamano da bagno: La lunghezza media è di circa 135 cm (54 pollici). Ideale per asciugare il corpo.
Foglio da bagno: La lunghezza media è di circa 150-180 cm (60-70 pollici). Offre un avvolgimento completo e lussuoso.
Asciugamano da spiaggia: La lunghezza media è di circa 180 cm (70 pollici). Ideato per il relax.
Conclusione
Scegliere la dimensione giusta dell'asciugamano significa far coincidere le dimensioni con le esigenze dell'utente e con la promessa del marchio. Chiarire le specifiche esatte con il produttore assicura un prodotto di successo fin dall'inizio.
Messaggi correlati:
Quali sono le dimensioni standard degli asciugamani da conoscere?Quali sono le misure standard degli asciugamani da bagno da conoscere?Cosa definisce il cotone a fibra lunga?Come trovare il telo da bagno perfetto?
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3103 | https://mathoverflow.net/questions/20549/basis-for-modular-forms-of-half-integral-weight | Stack Exchange Network
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Basis for modular forms of half-integral weight
Given a character $\chi$ and $k$ odd how can one compute a basis for the space of modular forms $M_\frac{k}{2}(\Gamma_0(4),\chi)$. By compute a basis I mean, finding the beginning of the Fourier expansions. I am looking for computer programs, which can do that for me.
I have heard of the package SAGE, which seems to do the job for integral weight modular forms. There is even the function but the examples all have q-expansions starting with q, so I guess this is not really a basis for the space of all modular forms but only cusp forms.
MAGMA does not seem to include this functionality, either.
So, are there any packages which can do this? Since I have not found a package, I have some doubts that there is really an algorithm working in general. If there is no algorithm known to handle this, what methods are available in order to compute a basis "by hand"?
Thanks.
4 Answers 4
Edit: Here's a rather silly method that should work if SAGE is just giving you cusp forms: $\Gamma_0(4)$ has a single normalized cusp form of weight 6, given by $\eta(2\tau)^{12} = q - 12q^3 + 54q^5 - \dots$, so take your basis of cusp forms of weight $k/2 + 6$, and divide each element by this form to get a basis of the space of modular forms of weight $k/2$.
Edit in response to Buzzard: Thanks for pointing out that I should make this argument. Here is a proof that the cusp form has minimal vanishing at all cusps. $\Gamma_0(4)$ is conjugate to $\Gamma(2)$ by $\tau \mapsto 2\tau$, so it suffices to check that $\Delta(\tau)$, the square of $\eta(\tau)^{12}$, vanishes to twice the minimum order at each cusp of $\Gamma(2)$. The quotient $\Gamma(1)/\Gamma(2) \cong S_3$ acts transitively on the cusps of $X(2)$ with stabilizers of order 2, so the quotient map to $X(1)$ has ramification degree 2 at each cusp. $\Delta(\tau)$ is invariant under the weight 12 action of $\Gamma(1)$, and $\Delta(\tau)$ has minimal vanishing at infinity on $X(1)$.
Old answer: If you have a cusp form of weight $k/2$ for $\Gamma_0(4)$ (e.g., given to you by SAGE), you can multiply it by the modular function $\frac{\eta(\tau)^8}{\eta(4\tau)^8} = q^{-1} - 8 + 20q - 62q^3 + 216q^5 - \dots$ to get a modular form of the same weight, that is nonvanishing at one of the three cusps and vanishing at the other two. If you want a form that is nonzero at one of the other cusps, you can multiply by $\frac{\eta(4\tau)^8}{\eta(\tau)^8}$ (has a pole at zero) or by $\frac{\eta(\tau)^{16}\eta(4\tau)^8}{\eta(2\tau)^{24}}$ (pole at $1/2$). [Constant term $-8$ added Sept. 23, in response to an email correction from Michael Somos.]
"MAGMA does not seem to include this functionality, either."
Basis(HalfIntegralWeightForms(DirichletGroup(4).1^2,11/2));
[
1 - 88q^3 - 330q^4 - 4224q^7 - 7524q^8 - 30600q^11 + O(q^12),
q + 4q^3 + 56q^4 + 132q^5 + 224q^6 + 512q^7 + 912q^8 + 1525q^9 +
2752q^10 + 4044q^11 + O(q^12),
q^2 + 6q^3 + 20q^4 + 56q^5 + 130q^6 + 256q^7 + 472q^8 + 800q^9 +
1266q^10 + 1970q^11 + O(q^12)
]
Basis(HalfIntegralWeightForms(DirichletGroup(112).1^2,3/2));
[
1 + 2q^16 + 2q^28 + O(q^30),
q - q^21 + 2q^29 + O(q^30),
...
]
Here is a standard approach: One has the Jacobi $\theta$-function
$\sum_{n = -\infty}^{\infty} e^{2 \pi i n^2 \tau}$, which is weight $1/2$
on $\Gamma_1(4)$. Thus multiplication by $\theta$ induces an embedding
$M_{k+\frac{1}{2}}(\Gamma_1(4N)) \hookrightarrow M_{k+1}(\Gamma_1(4 N)),$
for any integer $N$.
It is not too hard to determine the image: given an element $f$ in
$M_{k+1}(\Gamma_1(4 N))$, one must determine if $f/\theta$ is holomorphic
in the upper half-plane, and at the cusps. This is just a question
of $f$ having zeroes at the location of the zeroes of $\theta$. One can
use this condition to compute the dimension of the image, and with more
effort one should be able to find an actual basis of the image (although
I have never tried to implement this latter step myself, and I don't know
how hard it is in practice).
$$\theta(\tau) = \prod (1-q^{2k})(1+q^{2k-1})^2,$$
You might also try Eichler and Zagier's book on the theory of Jacobi forms. For example, they show how to compute half-integer weight mf's of weight k+1/2 and level N from Jacobi forms when k is odd.
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3104 | https://www.reddit.com/r/maths/comments/1cv8q0g/inequalities_with_logarithms/ | inequalities with logarithms : r/maths
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•1 yr. ago
justafleecehoodie
inequalities with logarithms
Help: 16 - 18 (A-level)
when i solve the inequality by taking log to the base of 0.6 on both sides, i dont understand why my inequality should flip and how itll flip. when the solution set solved it using log to the base of 10, log to the base of 10 of 0.6 is negative so the inequality sign flips.
when i asked my teacher about it, he mentioned something on negative logs or fractional logs im not entirely sure, but ive never heard of either of those terms before (whichever one it was) and im confused.
maybe i should stick to log to the base 10, but what if im taking log to the base ten of a bigger number that results in me not knowing what to do with the inequality again. i would really appreciate some help on it :D
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3105 | https://www.electronics-tutorials.ws/resistor/resistivity.html | Published Time: 2017-04-16T08:52:15+00:00
Resistivity and Electrical Conductivity
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Home/Resistors/Resistivity
Resistivity
Resistivity of materials is the resistance to the flow of an electric current with some materials resisting the current flow more than others
Ohms Law states that when a voltage (V) source is applied between two points in a circuit, an electrical current (I) will flow between them encouraged by the presence of the potential difference between these two points. The amount of electrical current which flows is restricted by the amount of resistance (R) present. In other words, the voltage encourages the current to flow (the movement of charge), but it is resistance that discourages it.
We always measure electrical resistance in Ohms, where Ohms is denoted by the Greek letter Omega, Ω. So for example: 50Ω, 10kΩ or 4.7MΩ, etc. Conductors (e.g. wires and cables) generally have very low values of resistance (less than 0.1Ω), so for circuit analysis calculations we can assume that wires have zero resistance and neglect them from our calculations.
Insulators (e.g. plastic or air) on the other hand generally have very high values of resistance (greater than 50MΩ), therefore we can ignore them also for circuit analysis as their value is far too high.
But the electrical resistance between two points can depend on many factors such as the conductors length, its cross-sectional area, the temperature, as well as the actual material from which it is made. For example, let’s assume we have a piece of wire (a conductor) that has a length L, a cross-sectional area A and a resistance R as shown.
A Single Conductor
The electrical resistance, R of this simple conductor is a function of its length, L and the conductors area, A. Ohms law tells us that for a given resistance R, the current flowing through the conductor is proportional to the applied voltage as I = V/R. Now suppose we connect two identical conductors together in a series combination as shown.
Doubling the Length of a Conductor
Here by connecting the two conductors together in a series combination, that is end to end, we have effectively doubled the total length of the conductor (2L), while the cross-sectional area, A remains exactly the same as before. But as well as doubling the length, we have also doubled the total resistance of the conductor, giving 2R as: 1R + 1R = 2R.
Therefore we can see that the resistance of the conductor is proportional to its length, that is: R∝L. In other words, we would expect the electrical resistance of a conductor (or wire) to be proportionally greater the longer it is.
Note also that by doubling the length and therefore the resistance of the conductor (2R), to force the same current, i to flow through the conductor as before, we need to double (increase) the applied voltage as now I = (2V)/(2R). Next suppose we connect the two identical conductors together in parallel combination as shown.
Doubling the Area of a Conductor
Here by connecting the two conductors together in a parallel combination, we have effectively doubled the total area giving 2A, while the conductors length, L remains the same as the original single conductor. But as well as doubling the area, by connecting the two conductors together in parallel we have effectively halved the total resistance of the conductor, giving 1/2R as now each half of the current flows through each conductor branch.
Thus the resistance of the conductor is inversely proportional to its area, that is: R 1/∝A, or R∝1/A. In other words, we would expect the electrical resistance of a conductor (or wire) to be proportionally less the greater is its cross-sectional area.
Also by doubling the area and therefore halving the total resistance of the conductor branch (1/2R), for the same current, i to flow through the parallel conductor branch as before we only need half (decrease) the applied voltage as now I = (1/2V)/(1/2R).
So hopefully we can see that the resistance of a conductor is directly proportional to the length (L) of the conductor, that is: R∝L, and inversely proportional to its area (A), R∝1/A. Thus we can correctly say that resistance is:
Proportionality of Resistance
But as well as length and conductor area, we would also expect the electrical resistance of the conductor to depend upon the actual material from which it is made, because different conductive materials, copper, silver, aluminium, etc all have different physical and electrical properties.
Thus we can convert the proportionality sign (∝) of the above equation into an equals sign simply by adding a “proportional constant” into the above equation giving:
Electrical Resistivity Equation
Where: R is the resistance in ohms (Ω), L is the length in metres (m), A is the area in square metres (m 2), and where the proportional constant ρ (the Greek letter “rho”) is known as Resistivity.
Electrical Resistivity
The electrical resistivity of a particular conductor material is a measure of how strongly the material opposes the flow of electric current through it. This resistivity factor, sometimes called its “specific electrical resistance”, enables the resistance of different types of conductors to be compared to one another at a specified temperature according to their physical properties without regards to their lengths or cross-sectional areas. Thus the higher the resistivity value of ρ the more resistance and vice versa.
For example, the resistivity of a good conductor such as copper is on the order of 1.72 x 10-8 ohm metre (or 17.2 nΩm), whereas the resistivity of a poor conductor (insulator) such as air can be well over 1.5 x 10 14 or 150 trillion Ωm.
Materials such as copper and aluminium are known for their low levels of resistivity thus allowing electrical current to easily flow through them making these materials ideal for making electrical wires and cables. Silver and gold have much low resistivity values, but for obvious reasons are more expensive to turn into electrical wires.
Then the factors which affect the resistance (R) of a conductor in ohms can be listed as:
The resistivity (ρ) of the material from which the conductor is made.
The total length (L) of the conductor.
The cross-sectional area (A) of the conductor.
The temperature of the conductor.
Resistivity Example No1
Calculate the total DC resistance of a 100 metre roll of 2.5mm 2 copper wire if the resistivity of copper at 20 o C is 1.72 x 10-8 Ω metre.
Data given: resistivity of copper at 20 o C is 1.72 x 10-8, coil length L = 100m, the cross-sectional area of the conductor is 2.5mm 2 which is equivalent to a cross-sectional area of: A = 2.5 x 10-6 metres 2.
That is 688 milli-ohms or 0.688 Ohms.
We said previously that resistivity is the electrical resistance per unit length and per unit of conductor cross-sectional area thus showing that resistivity, ρ has the dimensions of ohms metre, or Ωm as it is commonly written. Thus for a particular material at a specified temperature its electrical resistivity is given as:
Electrical Resistivity, Rho
Electrical Conductivity
While both the electrical resistance (R) and resistivity (or specific resistance) ρ, are a function of the physical nature of the material being used, and of its physical shape and size expressed by its length (L), and its sectional area (A), Conductivity, or specific conductance relates to the ease at which electric current con flow through a material.
Conductance (G) is the reciprocal of resistance (1/R) with the unit of conductance being the siemens (S) and is given the upside down ohms symbol mho, ℧. Thus when a conductor has a conductance of 1 siemens (1S) it has a resistance is 1 ohm (1Ω). So if its resistance is doubled, the conductance halves, and vice-versa as: siemens = 1/ohms, or ohms = 1/siemens.
While a conductors resistance gives the amount of opposition it offers to the flow of electric current, the conductance of a conductor indicates the ease by which it allows electric current to flow. So metals such as copper, aluminium or silver have very large values of conductance meaning that they are good conductors.
Conductivity, σ (Greek letter sigma), is the reciprocal of the resistivity. That is 1/ρ and is measured in siemens per metre (S/m). Since electrical conductivity σ=1/ρ, the previous expression for electrical resistance, R can be rewritten as:
Electrical Resistance as a Function of Conductivity
Then we can say that conductivity is the efficiency by which a conductor passes an electric current or signal without resistive loss. Therefore a material or conductor that has a high conductivity will have a low resistivity, and vice versa, since 1 siemens (S) equals 1Ω-1. So copper which is a good conductor of electric current, has a conductivity of 58.14 x 10 6 siemens per metre.
Resistivity Example No2
A 20 metre length of cable has a cross-sectional area of 1mm 2 and a resistance of 5 ohms. Calculate the conductivity of the cable.
Data given: DC resistance, R = 5 ohms, cable length, L = 20m, and the cross-sectional area of the conductor is 1mm 2 giving an area of: A = 1 x 10-6 metres 2.
That is 4 mega-siemens per metre length.
Resistivity Summary
We have seen in this tutorial about resistivity, that resistivity is the property of a material or conductor that indicates how well the material conducts electrical current. We have also seen that the electrical resistance (R) of a conductor depends not only on the material from which the conductor is made from, copper, silver, aluminium, etc. but also on its physical dimensions.
The resistance of a conductor is directly proportional to its length (L) as R∝L. Thus doubling its length will double its resistance, while halving its length would halve its resistance. Also the resistance of a conductor is inversely proportional to its cross-sectional area (A) as R∝1/A. Thus doubling its cross-sectional area would halve its resistance, while halving its cross-sectional area would double its resistance.
We have also learnt that the resistivity (symbol: ρ) of the conductor (or material) relates to the physical property from which it is made and varies from material to material. For example, the resistivity of copper is generally given as: 1.72 x 10-8 Ωm. The resistivity of a particular material is measured in units of Ohm-Metres (Ωm) which is also affected by temperature.
Depending upon the electrical resistivity value of a particular material, it can be classified as being either a “conductor”, an “insulator” or a “semiconductor”. Note that semiconductors are materials where its conductivity is dependent upon the impurities added to the material.
Resistivity is also important in power distribution systems as the effectiveness of the earth grounding system for an electrical power and distribution system greatly depends on the resistivity of the earth and soil material at the location of the system ground.
Conduction is the name given to the movement of free electrons in the form of an electric current. Conductivity, σ is the reciprocal of the resistivity. That is 1/ρ and has the unit of siemens per metre, S/m. Conductivity ranges from zero (for a perfect insulator) to infinity (for a perfect conductor). Thus a super conductor has infinite conductance and virtually zero ohmic resistance.
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Ravi Dhakad Nice content Posted on November 08th 2024 | 7:55 pm Reply
ADITYA Really very informative thank you!! Posted on July 25th 2024 | 11:38 am Reply
Shifau Love it Posted on May 16th 2024 | 5:37 am Reply
thatonedude i think this was quite informative but i have to say who made 1+1=2 why cant it equal 3 Posted on May 02nd 2024 | 12:00 pm Reply
leslie pangilinan I’m so glad that this page help me with my problem Posted on March 20th 2024 | 4:54 am Reply
Okolie mary It is very good app
Thanks Posted on February 01st 2024 | 6:09 pm Reply
amppowergy very useful article Posted on December 10th 2023 | 8:47 pm Reply
Varney H. Gulu - Gayflor I am a new member in this group Posted on October 02nd 2023 | 10:47 am Reply
Sasha Thank you so much for the help! Posted on August 27th 2023 | 8:28 am Reply
Sasha Thank you so much for the explanation about the equation related to resistance and length and the equation related to resistance and the cross-sectional area! It truly helped me! Thank you! Posted on August 27th 2023 | 8:20 am Reply
Sanjay Required Conversation chart Posted on June 27th 2023 | 11:00 am Reply
Mubiru David kato Good Posted on March 28th 2023 | 1:28 am Reply
Leila Ehab if the resistivity of a wire = 0.5 ohm.m , then the product of the resisitivity of the material of this wire by it’s electric conductivity equals ? Posted on February 26th 2023 | 2:36 pm Reply
Samuel Appreciate Posted on February 13th 2023 | 6:45 am Reply
Sathish Kumar B I have gained a lot of knowledge via this Tutorials
Thank you! Posted on January 18th 2023 | 7:32 am Reply
Deepak pal we enjoy reading Posted on January 10th 2023 | 11:39 am Reply
Alex In the equation you have put 10^-6 and it should be 10^-3 as milli prefix
10^-6 is micro μ of the cross sectional 2.5mm^2 Posted on December 16th 2022 | 8:41 pm Reply
Wayne Storr In the tutorial example, the cross-sectional area of the conductor was stated as being 2.5 mm 2. As there are one million square millimeters within one square meter (1000 x 1000). Then 2.5mm 2 is equal to 2.5/1,000,000 = 0.0000025m 2 or 2.5 x 10-6 meters 2. Thus, the tutorial is correct as given. Posted on December 17th 2022 | 8:53 am Reply
AMON Am interested in posting on thia site ams am learning alot . Posted on September 29th 2022 | 7:20 am Reply
Dennis murunga Good Posted on September 25th 2022 | 3:44 pm Reply
Shelby Bruno no Posted on September 04th 2022 | 9:24 pm Reply
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Read more Tutorials inResistors
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3. Resistors in Series
4. Resistors in Parallel
5. Resistors in Series and Parallel
6. Potential Difference
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8. Resistors in AC Circuits
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3106 | https://www.diy.org/article/isosceles_trapezoid | ##### DIY TV
Facts for Kids
An isosceles trapezoid is a four-sided figure with one pair of parallel sides and the non-parallel sides being equal in length.
Construction And Visualization
Comparison With Other Trapezoids
Isosceles Trapezoids In Geometry
Definition Of Isosceles Trapezoid
Properties Of Isosceles Trapezoids
Fun Facts About Isosceles Trapezoids
Formulas Related To Isosceles Trapezoids
Real World Applications Of Isosceles Trapezoids
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Did you know?
🔺 An isosceles trapezoid has one pair of parallel sides.
📏 The non-parallel sides of an isosceles trapezoid are of equal length.
🛠️ The angles at each end of the parallel sides are equal in an isosceles trapezoid.
🌐 The area of an isosceles trapezoid can be calculated using the formula: Area = 1/2 × (Base1 + Base2) × Height.
📐 Isosceles trapezoids are a type of quadrilateral.
🔄 The diagonals of an isosceles trapezoid are congruent.
🏗️ An isosceles trapezoid can be symmetric with respect to its median.
🏅 The median of an isosceles trapezoid is the average of the lengths of the two bases.
🌟 The vertices opposite to each other are not necessarily equal in an isosceles trapezoid.
⚖️ Isosceles trapezoids have rotational symmetry of 180 degrees.
Introduction
An isosceles trapezoid is a special shape in geometry! 🟠It has four sides, two of which are parallel, and the other two are equal in length. Picture a bridge with slants! The word "isosceles" comes from the Greek words for "equal legs," describing those two sides. Isosceles trapezoids can be found in buildings, art, and even furniture! It’s important to know about this shape because it helps us understand more complex shapes and designs. Plus, learning about them can turn you into a geometry whiz! 😃
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Construction And Visualization
You can easily draw an isosceles trapezoid! 🎨Start with a horizontal line at the bottom for the lower base. Then, find the center point, measure equal lengths upwards for the legs, and draw parallel lines for the upper base. Use a ruler to ensure the parallel lines are equal. You can bring your creation to life with colors and patterns! 🌈To visualize this shape in real life, try crafting models using LEGO or paper. Building and creating shapes can make learning geometry fun! 📚
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Comparison With Other Trapezoids
Isosceles trapezoids are different from other trapezoids! 🤔Regular trapezoids can have non-equal legs, making it less symmetrical. For example, think about a “skewed” trapezoid where legs aren’t the same length; it doesn’t have the same cool properties! Also, in an isosceles trapezoid, the diagonals are equal in length, while in other trapezoids, they may not be. Understanding these differences will help you recognize the unique beauty of isosceles trapezoids compared to their trapezoid cousins! 📏
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Isosceles Trapezoids In Geometry
In geometry class, isosceles trapezoids are a key topic! 📘They belong to the larger family of trapezoids, which have one pair of parallel sides. In a geometry lesson, you’ll learn to identify their properties and understand how they differ from other shapes! You can even investigate their angles and congruency (matching shapes). Students often use tools like protractors and rulers to measure and draw them. Drawing isosceles trapezoids can help make geometry fun and engaging, bringing math to life! 🎨
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Definition Of Isosceles Trapezoid
An isosceles trapezoid has two parallel sides called bases and two equal sides called legs. 🤔The parallel sides can be of different lengths, like a tall drink glass next to a smaller skinny glass—both flat on the top! The top parallel side is called the "upper base", and the bottom is the "lower base." Their unusual symmetry makes them special and important when studying shapes. So next time you see a trapezoid, think about whether it’s isosceles or not! 🏗️
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Properties Of Isosceles Trapezoids
Isosceles trapezoids have some neat properties! First, the angles next to each base are equal. 🟡So, if one angle is 60 degrees, the other angle next to it will also be 60 degrees. This makes them symmetrical! The lengths of the legs are also equal, which means if you cut them in half, both sides will match perfectly! Additionally, the diagonals (the lines that connect opposite corners) are also equal in length! All these properties help us recognize and create isosceles trapezoids easily. 📏
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Fun Facts About Isosceles Trapezoids
Did you know that an isosceles trapezoid is sometimes called a "trapezium" in countries like the UK? 🇬🇧 It’s true! Also, ancient Greek mathematician Euclid studied trapezoids over 2,000 years ago! 🕰️ The tallest building in the world, the Burj Khalifa in Dubai, has parts shaped like isosceles trapezoids! Isn't that neat? You can even find them in games and puzzles. 🧩Explore and spot isosceles trapezoids in your daily life, and you might find them everywhere!
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Formulas Related To Isosceles Trapezoids
When dealing with isosceles trapezoids, there are helpful formulas! 🧮To find the area (the space inside), you can use the formula: Area = 1/2 × (Base 1 + Base 2) × Height. The bases are the parallel sides, and the height is the vertical distance between them. Also, if you want to find the perimeter (the distance around), the formula is: Perimeter = Base 1 + Base 2 + 2 × Leg Length. These formulas can help you in math problems or when measuring real shapes! 📐
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Real-world Applications Of Isosceles Trapezoids
Isosceles trapezoids are found all around us! 🏢For example, the rooftops of some buildings are shaped like isosceles trapezoids to help with rain runoff! You might also see them in furniture designs, like tables and lamps, making them pretty and functional. Even in art, this shape can create attractive patterns! 🚪In nature, some leaves and flowers may have this shape too. Understanding isosceles trapezoids can connect you to the world around you in fascinating ways! 🌼
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Isosceles TrapezoidQuiz
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3107 | https://sites.pitt.edu/~mgahagan/SandDquestions.pdf | Supply and demand practice questions Hint: draw a graph to illustrate each problem in the space provided.
Simple shifts: 1. Incomes increase. In a graph of the market for bus rides (an inferior good) we would expect: a. The demand curve to shift to the left b. The demand curve to shift to the right.
c. The supply curve to shift upwards.
d. The supply curve to shift downwards.
e. Neither the supply nor the demand curve shifts.
2.As a result of the increase in income, we should expect to see that price will – and quantity will -- in the new equilibrium in the market for bus rides.
a. increase – increase e. increase – be uncertain b. increase – decrease f. be uncertain - increase c. decrease – increase g. decrease – be uncertain d. decrease – decrease h. be uncertain – decrease 3. Incomes increase. In the market for iPods (a normal good) , we would expect: [same options as question 1] 4. Incomes increase. In the market for iPods, we would expect price will – and quantity will – in the new equilibrium.
[same options as question 2] Simple shifts (continued) 5. The price of computer memory chips increases. In the market for computers, we would expect to see: a. The demand curve to shift to the left b. The demand curve to shift to the right.
c. The supply curve to shift upwards.
d. The supply curve to shift downwards.
e. Neither the supply nor the demand curve shifts.
6. The price of computer memory chips increases. In the market for computers, we would expect price will – and quantity will – in the new equilibrium.
a. increase – increase e. increase – be uncertain b. increase – decrease f. be uncertain - increase c. decrease – increase g. decrease – be uncertain d. decrease – decrease h. be uncertain – decrease 7. New techniques in the production of LCD screens make it possible to produce them at lower marginal cost. In the market for televisions using LCD screens, we would expect to see: [same options as question 5] 8. New techniques in the production of LCD screens make it possible to produce them at lower marginal cost. In the new equilibrium in the market for televisions using LCD screens, price will – and quantity will --.
[same options as question 6] Double shifts (again, draw graphs to help your intuition) 1. Wages of bus drivers increase. At the same time, incomes of consumers generally increase. In the market for bus rides, we should expect to see curves shift. The supply curve will --- and the demand curve will ---.
a. shift up – shift to the left e. shift up – remain unchanged b. shift up – shift to the right f. shift down – remain unchanged c. shift down – shift to the left g. remain unchanged – shift left d. shift down – shift to the right h. remain unchanged – shift right 2. As a result of the simultaneous increase in the wages of bus drivers and of consumer incomes, we would expect that price will -- and the quantity of bus rides will --- in the new equilibrium.
a. increase – increase e. increase – be uncertain b. increase – decrease f. decrease – be uncertain c. decrease – increase g. be uncertain -- increase d. decrease – decrease h. be uncertain – decrease.
3. The price of gasoline falls and consumer incomes generally increase. In the market for bus rides, we should expect to see a curve or curves shift. The supply curve will – and the demand curve will --.
[same options as question 1] 4. As a result of the simultaneous fall in gasoline prices and the rise in consumer incomes, we should expect to see in the new equilibrium that price will – and the quantity of bus rides will --.
[same options as question 2] Double shifts (continued) 5. The price of tea increases. At the same time, robots are developed which prove to lower the cost of production of coffee . In the market for coffee, we should expect to see curves shift. The supply curve will --- and the demand curve will ---.
a. shift up – shift to the left e. shift up – remain unchanged b. shift up – shift to the right f. shift down – remain unchanged c. shift down – shift to the left g. remain unchanged – shift left d. shift down – shift to the right h. remain unchanged – shift right 6. As a result of the simultaneous increase in the price of tea and of the drop in the cost of production of coffee, we would expect that price of coffee will -- and the quantity of coffee will --- in the new equilibrium.
a. increase – increase e. increase – be uncertain b. increase – decrease f. decrease – be uncertain c. decrease – increase g. be uncertain -- increase d. decrease – decrease h. be uncertain – decrease.
7. Wages of workers who pick coffee rise. At the same time, the price of half-and-half (a complement for coffee) decreases. In the market for coffee, we should expect to see curves shift. The supply curve will --- and the demand curve will ---.
[same options as question 5] 8. As a result of the simultaneous rise in the wages of coffee workers and fall in the price of half-and-half, we would expect that price of coffee will -- and the quantity of coffee will --- in the new equilibrium.
[same options as question 6] Answers to supply and demand multiple choice questions: Simple shifts: Quest ions 1-2 (income increase). Since we are looking at an inferior good (bus rides), the quantity demanded will decline at any given price (Richer consumers will buy a car and not ride the bus as often).
Graphically, the demand curve shifts to the left (option a in question 1) and we expect the price and quantity to both drop (option d in question 2).
Questions 3-4. Incomes increase and we are looking at a normal good. Richer consumers throw away their old Walkman and buy an iPod. At any given price, the quantity of iPods demanded will increase.
Graphically, the demand curve shifts to the right (option b in question 3) and we expect price and quantity to both increase (option a in question 4).
Questions 5-6. Increased price of computer memory chips raises the cost of production of computers. The supply curve shifts up (option c), indicating that computer producers want to pass the price increase on to consumers. We expect price to increase but quantity to decrease. (option b). Note: -- a movement upward on the graph is a decrease in supply.
-- when a supply curve shifts, price and quantity move in opposite directions.
Questions 7-8. Lower marginal cost means producers are willing to accept lower prices for the television sets. The supply curve for televisions shifts down (an increase in supply), and we expect price to decrease and quantity to increase. Options D and C are correct.
Double shifts: Questions 1-2. The wage increase for bus drivers increases the cost of production, and the supply curve will shift up and to the left (a decrease in supply). Higher incomes reduce the demand for bus rides. Option A of qu. 1 describes the curve shifts; both changes tend to reduce quantity, but pull in opposite directions on the price.
Option H of question 2 indicates that price is uncertain and quantity definitely decreases.
Questions 3-4. Falling gas prices reduce the cost of production of bus rides and shift supply down and to the right. Consumer incomes rise, reducing the demand for bus rides. Option C of question 3 describes the shifts.
Price definitely falls; quantity change is uncertain. Option F of question 4 is the appropriate one.
Questions 5-6. When the price of tea (a substitute for coffee) increases, the demand for coffee will increase as some tea drinkers switch to coffee. The demand curve shifts rightwards. At the same time, robotic coffee pickers lower the cost of production of coffee, so the supply curve shifts down and to the right. Option D describes the curves shifting. Both changes tend to increase the quantity of coffee; however the increase in demand tends to pull price up, and the robots tend to pull price down. We are uncertain whether price will be higher or lower in the new equilibrium, but we know quantity will be higher. Option G is correct for question 6.
Questions 7-8. Higher wages for coffee pickers increase the cost of production, so supply shifts up and to the left on the graph. A fall in the price of a complement for coffee will lead some to demand more coffee at the same coffee price. Option B of question 7 describes the shifts; the result is a definitely higher price (producers want it as a result of higher input prices; consumers are willing to pay it as a result of the fall in half-and-half prices) but an uncertain quantity (producers want to reduce the quantity supplied at any given price; consumers want more at any given price). Option E is appropriate for question 8. |
3108 | https://economics.yale.edu/sites/default/files/world_population_1800_1938_yale_seminar.pdf | 1 Giovanni Federico and Antonio Tena-Junguito [New York University Abu Dhabi and Universidad Carlos III, Madrid] How many people on earth? World population 1800 1938 1 The number of people is one of the most basic information about any society but we hardly know it until the early 20th century. The easily accessible sources refer to few advanced countries and thus widely used data-bases build series for other countries with interpolations. In this paper we fill this gap by re-estimating series of population for all existing polities from 1800 to 1938 using first-hand sources and the country-specific literature. We are thus able to date, albeit tentatively, the start of the demographic transition in data-scarce peripheral countries and to measure the impact of major demographic crises such as the Tai’ping civil war, World War One and the Spanish flu. JEL keywords I10, J11, J13 [Address for contact Giovanni.Federico@nyu.edu] {provisional text, comments invited] 1. Introduction: why it is important The number of people is one of the most basic information about any society and it is essential for debates on three issues are now attracting much interest among economic historians and economists such as the demographic transition, the role of market integration in reducing the impact of agricultural production failures and the effect of pandemics. The conventional wisdom argues that market integration and economic growth reduced the size of demographic shocks in the 19th and early 20th century. The former reduced impact of crop failures (e.g. O’ Grada and Chevet 2002, Burgess and Donaldson 2011), while economic growth improved nutrition and health conditions. It made it possible to invest in provision of clean waters and sanitation, which greatly helped to fight waterborne diseases, including cholera (Haines 2000, Chapman 2019, Chaudhary and Lindert 2021). Spanish flu was an exception for its aerial transmission, which caused a world-wide impact and a huge death toll (Beach et al 2021). The demographic transition has long been studied by demographers and economic historians (Chesnais 1992, Kirby 1996, Lee 2003, Alter and Clark 2010, Davenport and Saito 2021, Perrin forthcoming) but it has enjoyed a revival of interest among economist (Guinnaine 2011), because it features prominently in unified growth theory models. Scholars have explored the patterns of transition (Delventhal et al 2021) and analyzed the causes of decline in fertility (cf e.g. Murtin 2013) and the relations between long-term life expectancy and GDP change (cf. e.g. Cervellati and Sunde 2011) These works are based on very partial and often defective evidence. Before 1950 series of population and above all vital statistics are available only for a handful of advanced countries. For instance Murtin (2013) use a for a balanced sample of 16 advanced countries from 1870 to 2000. The book by Chesnais (1992) on demographic transition covers the whole world, but its information on the periphery is absent or very 1 We thank J.C. Bassino, D. Chilosi, J. Fourie, P. van Der Eng, A. Graziosi, A. Markevitch, L.Maravall-Buckwalter, C. O’Grada, M.Saleh, M.Voutilainen and the participants to the Groningen Growth and Development Center 25th anniversary conference, Yale Economic history seminar, Maddison project conference (Utrecht) EHES conference (Groningen) for comments and advice on data. 2 sketchy and he advocates using population trends as a proxy. The deaths from Spanish flu are estimated with a mix of actual data, guesstimates by experts and econometric estimate of excess deaths. Thus, the number of covered countries and the results differ rather widely, from 17-24 million (Spreeunwenberg et al 2018) to 35-44 million deaths (Athukorala and Athukorala 2020), with unsubstantiated guesses up to 100 million2. Also world-wide long-term trends are poorly measured. Historical demographers have made substantial efforts to produce country-specific series but their results have seldom trickled down in world-wide data-bases. The most commonly used data-sets are based on partial and outdated scholarship (McEvedy-Jones 1978, Maddison 1995) or, as Gapminder, they rely on hardly transparent interpolations from these same historical data-sets. Yet, these data are being widely used in econometric analysis (Guinnaine 2021). This paper aims at filling this gap. We estimate series of population at current borders for all polities from 1800 to 1938, and then we sum them to get the continent and world total(s). When possible, we use official statistics, but in most cases we rely on the work by demographic historians who have painstakingly collected and, when necessary corrected, official data and any available information. We fill gaps with interpolations or extrapolations, taking into account whenever possible the population movement in neighboring or otherwise comparable polities. Needless to say, our figures are of widely different quality, ranging from the almost perfect for Scandinavian countries to the mere guesswork for Sub-Saharian Africa. The amount of guesswork is the greater the less capable the administration was, and thus as a rule earlier data are less reliable than more recent ones. Most of the figures of the first half of the 19th century are highly conjectural and there are almost no data on the 18th century. The next section surveys the estimates of world population, Section Three discusses in general our sources and Section Four sketches out our strategy of estimation, with details by polity in Appendix I. Section Five outlines long-term population trends results and compare them with previous estimates. In a nutshell, world population has been growing with underlying faster rate, as more and more countries entered the demographic transition but the overall growth was substantially slowed down by two major exogenous shocks (plus minor local ones), the Tai’Ping civil war in China in the 1850s-early 1860s and the late 1910s world crisis (the combination of World War One, Spanish flu and Russian civil war and famine). Section Six tests the conventional wisdom about the decline in volatility and analyses the frequency and severity of demographic crises. Section Seven deals with the demographic transition, focusing on the periphery. Section Eight explores the robustness of our series. We estimate the likely margins of errors of the aggregate series, on the basis of our assessment of the quality of individual polity ones and we discuss the biases for the analysis of volatility from linear interpolation and for the dating of the demographic transition from the omission of migration flows (with a much expanded data-set on natural increase of population). on the conclusions about population volatility and demographic transition. Section Nine concludes with some ideas for further research. 2) What do we (pretend to) know 2 This figure, which is often quoted in the historical literature, is a speculative guess by Johnson and Mueller (2002). The sum of their country estimates yield a total of 33-43 million, but they increase it to 50 million, and add that even this figure ‘may be substantially lower than the real toll, perhaps as much as 100% understated’ (p.115).. 3 In 1661, the Italian Jesuit Riccioli put forward the first known estimate of world population at ‘less than one billion’ (Korenjak 2018) and since then many scholars have followed his example. The American demographer Willcox (1940) lists 68 benchmark estimates of world population plus an almost yearly series from 1851 to 1927 in the various editions of Hubner’s Geographical Atlas. Willcox was one the pioneers of modern work on the issue, jointly with Carr-Saunders (1936), and these two authors have been highly influential in all the following literature (Caldwell and Schindlmayr 2002). A possibly incomplete list includes about thirty estimates for world population in the 19th and 20th centuries since the 1930s but quite a few of them simply reproduce figures from some other authors (see for an analysis of these derivations Appendix ?). Thus, in Table 1 we report only the ‘original’ estimates, which feature relevant changes in figures by continent and (hence) for the world population. 4 Table 1 Estimates of world population 1800 1820 1850 1870 1890 1900 1910 1913 1920 1930 1935 1940 1950 Hubner 1055a 1300 1514 1551 1568 1652 1712 2029b Willcox (1940) 919 1091 1571 1995 Carr-Saunders (1936) 906 1171 1608 2057c Swaroop (1951) 906 1171 2378d Bennett (1954) 919 1163 1555 2368d Clark (1967) 890 1668 McEvedy and Jones (1978) 900 1200 1625 2500 Biraben (1979) 954 1241 1633 2520 Klein Goldewijk and Battjes (1995) 1532 1638 1767 1914 2084 2281 2511 League of Nations/UN 1834 2008 2216 2532 UN 1999 980 1260 1650 1750 1860 2070 2300 2520 Maddison (2010) 1042 1276 1563 1792 1863 2299 2528 HYDE 3.1 990 1263 1654 1777 1912 2092 2307 2545 quoted by Willcox (1940); a 1951;b 1929; c 1933; d 1949 5 All estimates tell a similar story, with an accelerating growth throughout the period, but differences between them are substantial, especially in the first half of the 19th century and for poor countries. The worldwide rates of growth for the 19th century range from 0.51% (Klein Goldewijk et al 2010) to 0.63% (Swaroop 1951) – a difference by almost a quarter. The highest rate for Africa, 0.50% (Klein Goldewijk et al 2010) is 3.4 time larger than minimum one, 0.11% (Clark 1967). The maximum/minimum ratio is 1.67 for Asia, 1.43 for America (presumably as sum of a very low difference North and a large one in the South) and only 1.20 for Europe. In the first half of the 20th century, the differences are decidedly small for the world (rates between 0.80% and 0.85%) but still substantial for some continents, especially Africa3. Only five estimates report yearly data by polity, and not all of them quote their sources in any detail. The Statistical Yearbook of the League of Nations, the forerunner of the current UN population data (United Nations), published figures for 1913 and series from 1922 to 1938, supplementing the available official data with crude estimates for missing countries. McEvedy and Jones (1978) plot graphs for some major countries and macro-areas since 1 AD, adding figures for benchmark years (in the period of interest 1800, 1850 and 1900). The Maddison project website reports the original estimates by A. Maddison of world population in the 1990s as a by-product of his well-known reconstruction of historical GDP series. He collected the (then) available data and integrated them with benchmark guess-estimates for many other polities. His data base features a growing number of population series by polity (Figure 1 blue bars, left axis) from about 20 before 1870 to 60 on the eve of World War Two, which accounted for three quarters of world population according to Maddison’s own estimate (red line, right axis) for most of the period, with an increase to about 85% in the 1930s4 3 The ratio between maximum and minimum figures range from 1.14 for Asia to 1.83 for Africa – in the latter case between 0.68% (Swaroop 1951) and 1.25% (Mc Evedy Jones 1978). Our computation excludes the two estimates by Willcox (1940) and Carr-Saunders (1936), which refer to 1935. Including them the difference among rates of growth would be much larger. 4 Here we use Maddison’s original data-set, as detailed in Maddison 1995 and 2010. The most recent version of the Maddison project website reports the country data for three benchmarks only (1820, 1870 and 1913). The number of series is bound to differ between Maddison’s and our data-base because the former is at current and the latter at 1995 borders and because Maddison reports aggregate data for groups of small polities (e.g. 21 Caribbean). We compute Maddison’s yearly series of world population (the denominator of the ratio in Figure 1) by interpolating linearly his figures for 1820, 1870, 1913 and 1940. 6 -- Figure 1 Maddison’s world population data-base Last but not least, two on-line data-bases publish data by polity: the environmental data-base HYDE 3.1 (Klein Goldewijk et al 2010) reports data at ten-year intervals since 8000 years BC for 237 polities and the Gapminder ( accessed July 2021) yearly series for 197 polities from 1800 onwards 5. The origin of these data is mysterious: data-bases quote general sources such as Maddison, without any detail on their methods of estimation 6. 5 Other sources, including statistical yearbooks of European states (e.g. Annuaire Statistique Francaise Statistiches Jahrbuch, Board of Trade ). and other contemporary statistical collections (e.g. the Statesman’s yearbook) collect but do not sum them in a world total. Etemad (2007 Appendix D) provides figures for all colonies in 1760, 1830, 1880, 1913 and 1938 without detailing his sources. The CLIO-Infra project (Van Vleuten and Kok (2014), as reported in its website ( accessed Sept 5 2021), lists several standard sources (Maddison 1995 and 2010, McEvedy-Jones 1978, Mitchell 2007, Kuczynski 1948-1949 and so on) without any further detail on how these have been used or how gaps have been filled. 6 The HYDE 3.1 writes in the source file ’Historical population numbers of McEvedy and Jones (1978), Livi-Bacci (2007), and Maddison (2003), Denevan (1992) form the basis of our national historical population estimates. Supplemented with the sub-national population numbers of Populstat (Lahmeyer, 2004, pers. comm.; who provides data for several time periods varying per country), time series were constructed for each province or state of every country of the world’. An earlier version of the same HYDE data-base (Klein Goldewijk et al 1995) used a transparent, albeit highly questionable, statistical interpolation method. They assume a logistic curve between the earliest available data from Mitchell’s International Historical statistics and the start of United Nations population series in 1950, with some adjustments. Gapminder v6 states in its explicative note ( - accessed 5 Sept 2021) ‘We use Maddison population data improved by CLIO INFRA in April 2015 and Gapminder v3 documented in greater detail by Mattias Lindgren. The main source of v3 was Angus Maddison’s data which is maintained and improved by CLIO Infra Project. The updated Maddison data by CLIO INFRA were based on the following improvements: i. Whenever estimates by Maddison were available, his figures are being followed in favor of estimates by Gapminder; ii. For Africa, estimates by Frankema and Jerven (2014) for the period 1850-1960 have been added to the existing database; – For Latin America, estimates by Abad and Van Zanden (2014) for the period 1500-1940 have been added’. In all fairness, it is to be added that Gapmider does not claim any historical accuracy. In their 0 0.2 0.4 0.6 0.8 1 1.2 0 10 20 30 40 50 60 70 80 90 100 1820 1825 1830 1835 1840 1845 1850 1855 1860 1865 1870 1875 1880 1885 1890 1895 1900 1905 1910 1915 1920 1925 1930 1935 1940 Number Ratio 7 3) Our sources Knowing the number of their subjects, as potential taxpayers and/or soldiers, has always been a major concern for rulers and governments since the early antiquity. The first counts of population may date back 4000 years BC Babylonia and all great empire, including Rome and China, followed suit. The first ‘modern’ censuses, with detailed information for all individuals, were taken in the early 19th century (Shryock et al 1971, Thorvaldsen 2018). In 1855-1864 only 24 sovereign countries had taken a census, and the number rose to 49 in 1925-1934 (Shryock et al 1971 tab 2.1), plus 18 colonies (Kuczynski 1937). The number jumped to 150 in 1955-1964, after the strong prodding by the United Nations. By definition, censuses are snapshots: a full reconstruction of movements of population needs yearly data, which can be kept only by registering the number of birth, deaths and migrations. In Europe, the Catholic local clergy started to collect these data in the Middle Ages (the earliest surviving French register dates back to 1303). In the 16th century the compilation of these registers was made compulsory by several states (e.g. England in 1538, France in 1539) and by the Catholic church with the Council of Trento (1563/1614). Following the pioneering work by Henry, historical demographers have used parish registers to produce series of local population. Wrigley and Schofield (1989) have used 404 such registers to produce their monumental Population history of England and they have been imitated by scholars for other countries, such as Northern Italy (Galloway 1994) and Germany (Pfister and Fertig 2010). Unfortunately, there is no guarantee that the surviving registers yield a representative sample of parishes (Spagnoli 1977). In the 18th century some European states started to collect parish data and in the early 19th century they set up centralized population registers (Wilke 2004, Poulain and Herm 2013). There is no need to dwell further on the historical evolution of demographic sources. It is however important to stress that in the 19th and early 20th century the ideal combination of modern censuses and accurate centralized population registers can be found in few, mostly advanced, countries. For instance, China had a state registration system with no fiscal purposes, the bao-jia system, since 1741 (Ho 1959:36-50 and 68-73), but not a proper census until 1953. Demographic sources are problematic also in some advanced countries. A national register of birth and death was set up in ten American states only as late as 1900, and was extended to the whole country only in 1933 (Haines 2000). Furthermore, the censuses of the 19th and early 20th century undercounted population, up to 10-15% according to some local tests (Coale and Zelnick 1963, Coale and Rives 1973, Shryock et al 1971, Parkinson 1991, Steckel 1991 Thorvaldsen 2018 pp.98-). Unfortunately, so far, no scholar has put forward a corrected series for the whole period and the Population Section of the Millennial version of the Historical Statistics US (2006) mostly reproduces the official data. The shortcomings of censuses is, unsurprisingly, more serious for poor countries and colonies, especially for large mainland territories, while counting people was easier in small islands7. Much depended on the website they state that ‘our data is more consistent over time and space than most other sources, because we dare to fill all the gaps in the sources. We dare this because our purpose is to show people the big picture, and they won’t understand it if its full of holes’ ( accessed February 20 2022) 7 We highlight this point by using, in the description of sources (Appendix I), the words count, enumeration or ‘census’ between brackets rather than census 8 size and capabilities of the civil service, and especially of its native staff. In many cases, such as the Dutch East Indies (Boomgaard-Gooszen 1991), the accuracy of official sources has been improving over time thanks to learning by doing and to greater resources. In India the geographical coverage of British-organized censuses was progressively extended to native states (Davis 1968). These improvements are a mixed blessing as they might bias upwards the rate of growth of population if they reduce the gap between counted and real population. The problem of reliability is deemed most serious for Sub-Saharian Africa, where most ‘censuses’ were a mix of information from village chiefs and guesswork by colonial administrators (Kuczynski 1937, 1948-1953). Table 2 tests this claim by comparing the total population with the estimates for the same polities/years by Frankema and Jerven (2014) 8. Table 2 The reliability of colonial censuses Number British French Italian Total Total 24 12 6 42 ratio <0.9 14 7 3 24 ratio>1.05 4 3 1 8 Average ratio total 86.1 87.7 95.8 96.0 ratio <1 77.5 80.9 67.8 77.2 ratio>1 112.0 140.4 127.5 122.6 The results are somewhat better than expected: only about a half of censuses are seriously undervalued (ratio census/estimate less than 0.9), while a fifth are overvalued (ratios over 1.05). The (unweighted) average ratio implies a very low undervaluation. Unfortunately, these reassuring results do not help much our work, as almost all these censuses were taken in the 1920s and 1930s: before World War One official sources for Sub-Saharian Africa are extremely scarce. Historical demographers have integrated the missing or defective official data with (variants of ) three methods, the use of ‘experts’ estimates, the backward projection and the forward projection i) ‘Experts’ were (usually Western) people, explorers, missionaries or diplomats, who put forward population estimates in books, travel accounts, memories, official dispatches and the like. These figures have been widely used, but are to be handled with caution. Most ‘experts’ had personal experience of limited areas only and applied their knowledge to the whole territories of interest. E.g. the early navigators estimated the population of Oceania islands by looking at coasts, or at tracts of them, and making wild inferences on the population of the (unobserved) interior. As a result, the population figures varied a lot: for instance, the members of the Cook expedition estimated the population of Hawaii to range between 240 and 400K (Nordyke 1989 13). Furthermore, in some contested areas, ‘experts’ were politically motivated: according to Karpat (1985 p.4), the estimates for the Balkans ‘strongly reflect the political biases 8 The French and Italian data refer to ‘censuses’ as reported respectively from Statistical Yearbook France and Statistical Yearbook Italy while for British colonies we rely on the list by Kuczynski (1948- 1953). We include only colony-wide censuses, dropping area-specific enumerations (e.g. for the capital city and its surroundings), approximate counts or guesstimates. 9 of the writers or of their informants and worst of all, in some of them the statistics were blatantly manipulated or falsified outright’. ii) The backward projection method have been widely used for Africa, following the pioneering work by Manning (2010, 2014). The starting point are population censuses for 1950, which have been endorsed by the United Nations, in spite of some doubts about their reliability 9. He extrapolated them backwards to 1850 with area-specific coefficients, which he estimated by adjusting upwards or downwards the decadal rate of growth of Indian population (Davis 1968) according to the evolution of each African macro-area (e.g. the intensity of slave exports) 10. Frankema and Jerven (2014) have used also rates of change in Indonesia and the Philippines, as land-abundant areas closer to the (West African) factor endowment, and have modified the adjustment coefficients to take into account demographic shocks. Last but not least, in still unfinished work, Manning and Nickleach (2014) have further refined the method for the period 1650-1890, taking into account the age and composition of the population and also the slave flows, both outside and inside the continent. iii) the forward projection method has been used to estimate the population of Pacific islands at contact, the time of first arrival of Europeans (Kirch and Rallu 2007). The method is highly speculative, as it needs hypotheses about the time of first settlement of each island, the number of first settlers and the rate of growth of the native population, which depends on the natural increase, on migration flows on and the frequency and impact of demographic shocks. Thus, whenever possible, the authors supplement their extrapolations with archeological evidence on the number of sites or on agricultural land use, with additional hypotheses, respectively, on the rates of occupation rates of dwelling and of the extraction of surplus by the élites. However, also the archeological evidence must be used with caution, as sites might have been occupied at different moments in time (e.g. within a slash-and burn agricultural system) 4) The construction of our data base The whole data-base features a total of 174 series, of different length, from 1800 to 1938 for a total of 21815 observations. We detail the sources and methods of estimation in Appendix I, while here we discuss only the general criteria First and foremost, in order to get a world total, we cover all polities, even if there are no solid data but only guesstimates or nothing at all. Second, unlike Maddison and Gapminder, as a rule, our series refer to polities at current borders. This solution is historically more sensible and avoids the adjustment to different borders, which would have needed additional information and/or guesses, increasing the risk of errors, without any obvious advantage. Thus, the list of polities differs in time following changes in the political map. Just to take the most extreme example, our data-base includes a series for Austria-Hungary until the dissolution of the empire in 1918 and then separate series for the successor states, Austria, Hungary, 9 Frankema and Jerven (2014) discuss at length the shortcomings of African post-war censuses and reckon that they undervalued population by about 8% (221.7 million vs 240 million). Also the Latin American post-war censuses were undervalued – cf. the discussion by Yanez et al (2012). They correct the pre-war censuses, as far back in time as 1817 for Cuba, with a polity-specific constant coefficient (on average about 6%). The assumption of a constant bias contrasts with the results of an earlier comparison between censuses and other demographic sources by Collver (1965). 10 In the first application of his method for the West African countries, Manning (1988) used as baseline the data from pre-war colonial censuses and assumed steady rates of population growth 0.5-1%. 10 Poland, Czechoslovakia and the Kingdom of Serbs, Croats and Slovenians (later Yugoslavia), while Tyrol and Dalmatia are included in the Italian population. However, we apply this rule with some flexibility. We extend series for African colonies to 1800, as there are no information about pre-colonial polities, we estimate series for Italy and Germany before their unification, in 1861 and 1870 and we keep separate Belgium and Netherlands during their short-lived political union in 1815-1830. We estimate separately series for Ottoman territories in Europe and in Asia, in order to get consistent continent-wise series, while we treat formally Ottoman territories in North Africa (e.g. Algeria and Tunisia before French conquest) as separate polities. Of course, we have taken as much care as possible to avoid omission and double counting of territories. In theory, we are aiming at measuring present population at mid-year. We prefer present (de facto) rather than resident (de jure) population, as most historical (and present-day) censuses (Thorvaldsen 2018 p.156), because it minimizes the distortions from domestic and international migrations 11. Our series include all present people, including military personnel, and, most notably, natives. The issue is very relevant for some colonies and especially in countries of Western Settlement, which often treated native population separately in official sources, as in Australia (Smith 1980) or omitted them altogether, as the United States before 1890. Their omission would severely bias the results, as the share of natives on total population collapsed for the joint effect of the decrease in their absolute number and of immigration of white settlers (e.g. natives were over 90% of the Australian population until 1825, a third in 1850 and a mere 2% on the eve of World War One). The series for white settlers only were bound to underestimate total population at any specific moment in time and to overestimate its growth. We have tried to follow as much as possible these criteria but this has not been possible in very many cases. For each polity, we have chosen the series or, when not available, the benchmark year figures which seem more solid. Whenever possible, we use country specific sources or population and area specific data-sets such as Rothenbacher 2002 for Western Europe and 2013 for Central and Eastern Europe and Bulmer-Thomas (2012) for the Caribbean. We rely on general purpose data-base by Mitchell (2007) and Maddison (2010) only as a last resort 12. When data are insufficient or dubious, we have taken into account trends in comparable polities and also the 1950 population of the polity. The United Nations (2019) report figures for all polities in the world, as we assume them to be accurate, although many of them are estimates. We have filled gaps between figures for benchmark years with linear interpolation while we have estimated any missing years at the end or, much more frequently, at the beginning of the period by taking into account trends in the polity (or in in neighbouring ones) in adjacent years 13 When possible, we have corrected the 11 If the local population registers are not updated, the resident population would be greater (smaller) than the present in countries or regions of emigration (immigration). The distortion was quite large in Russia, where former peasants resisted cancellation from the registers of village population as this entailed the loss of their right to access to common land. This effect caused official series to be overstated by about 5% of the population - i.e. by about 8.5 million (Markevich and Harrison 2011 Appendix tab A8). 12 Actually the most comprehensive source on historical population used to be the data-base POPULSTAT by J.Lahmeyer, which however seems to be off-line as of March 2021. It reported figures for total population at current borders, without attempting to fill gaps nor, a fortiori, to compute world-wide totals. The references list included a large number of contemporary reference works (e.g. Statesman Yearbooks or the Almanach Gotha) and world atlases, while the author seemed unaware of historical data-base such as Mitchell and Maddison (2010) and of the country-specific historical literature. Thus, we do not use it 13 In quite a few cases we have estimated population changes in the 1930s by taking into account the implicit rate of growth from 1938 to 1950 and in the 1950s (United Nations 2019). These latter can be considered, given the pattern of demographic transition, an upper bound of rates in the pre-war period. 11 linear interpolation with data on the impact on population of demographic crises, such as famines or epidemics in any specific year of the period. In particular, we have used the information on deaths from the Spanish flu from country-specific sources and from the recent paper by Athukorala and Athukorala (2020)14. In contrast, we cannot correct linear interpolations with yearly data on international migrations because these latter are usually available for countries with good population registers and thus annual population series. Last but not least, when possible we have adjusted end-year annual data to the mid-year by averaging two consecutive years. Africa, the Arabian peninsula and Oceania need some further comments. We have estimated separately with polity-specific sources the series for the six North African countries, for eight small islands and for South Africa after 1910. There are no sufficient sources for as many as 31 polities – almost the whole Sub-Saharian Africa and thus we have had to rely on the backward extrapolations by Manning and Nickleach (2014) and Frankema and Jerven (2014) 15. Table 3 compares their estimates for macro-areas in three benchmark year 16. Table 3 Estimate of African population, 1800-1950 (millions) Northern Northeast Central West East Southern Total Manning et al. 2014 1800 24.9 21.1 18.7 47 31.8 7.8 151.3 Manning et al. 2014 1850 27 20.8 16.1 46.2 30.9 8.3 149.3 Frankema et al.2014 1850 12.8 28.5 20.8 25.9 22.1 4.1 114.1 Manning et al 2014 1890 28.2 20.7 15.6 46.4 29.8 8.7 149.4 Frankema et al. 2014 1890 19.4 28.2 24.4 31.8 24.4 6.2 134.4 Frankema et al.2014 1950 44.3 31.1 31.3 64.0 34.1 17.1 221.8 The differences in rates of change for Sub-Saharian Africa are not so large from 1890 to 1950 (0.66% for Manning-Nickleach 2014, 0.83% for Frankema and Jerven 2014), when both use a similar approach, but 14 As a rule, we allocate the excess deaths to 1918 and/or 1919 with information of timing of the epidemic and we estimate population change until 1917 extrapolating previous rates of growth and then we interpolate linearly to the next available figure 15 The polities are Angola, Botswana (Bechuanaland), British East Africa, Cameroon, Congo, Eritrea, French Equatorial Africa, French Somalia, French West Africa, Gambia, Ghana (Gold Coast), Guinea Bissau, Lesotho (Basutoland), Liberia, Madagascar, Malawi (Nyasaland), Mozambique, Namibia (German South West Africa), Nigeria, Rwanda and Burundi, Sierra Leone, Spanish Guinea, Somalia (jointly Italian and British), Sudan, Swaziland, Tanganika (German East Africa), Togo (German West Africa), Western Sahara, Zambia (Northern Rhodesia) and Zimbabwe (Southern Rhodesia). 16 In this table, following Manning and NIckleach 2014 Map 1.1, West Africa includes French West Africa (present- day Mauritania, Senegal, Guinee, Ivory Coast, Dahomey, Niger, Upper Volta, Mali), Togo, Liberia Guinea Bissau, Gambia, Nigeria, Ghana (Gold Coast) and Sierra Leone, Central Africa includes, Cameroon French Equatorial Africa (present-day Central African Republic, Chad, Gabon, Congo-Brazzaville), Equatorial Guinea, Angola, Belgian Congo, Malawi (Nyasaland) and Zambia (Northern Rhodesia), Northern Africa includes Morocco, Western (Spanish) Sahara, Algeria, Tunisia, Lybia and Egypt, North-East Africa Somalia, Djibouti, Eritrea, Somalia and Sudan, East Africa includes British East Africa (Kenya and Uganda), Mozambique, Madagascar, Tanganyka, Rwanda Burundi, Southern Africa includes South Africa, Namibia (German South-West Africa), Botswana (Bechuanaland), Lesotho (Basutoland) and Zimbabwe (Southern Rhodesia). 12 wide in the previous forty years. Ultimately, we prefer to use the Frankema-Jerven series from 1850 onwards. 17 The Manning-Nickleach series (2014) are still provisional and refer to macro-areas which do not coincide with colonial polities. Their very sophisticated modelling strategy needs a lot of assumptions about unknown parameters, such as life expectancy, rates of survival of slaves after capture, division of surviving captives (as slaves) between Africa and overseas markets etc. Furthermore, Manning and Nickleach (2014) in their main projection assume that, after the end of the slave exports to the Americas and until 1890, the number of captured people in each sub-region remained as high as at the peak of transatlantic trade 18. We use the Manning-Nickleach (2014) rates of change by macro-region to extrapolate backward to 1800 the population by polity in 1850. It is the only available estimate, and anyway the figures for slave export are based on the Eltis data-base on transatlantic voyages. Official sources for polities in the Arabian peninsula are totally lacking (except the British colony of Aden) and estimates are scarce, partial and/or wide off the mark. Thus, we use a ‘Manning-type’ approach: we extrapolate backwards the 1950 UN figures by hypothesizing that their population grew by 0.3% yearly from 1800 to 1870, 0.6% from 1870 to 1914 and by 1.3% yearly from 1914 onwards (roughly the rates of neighbouring countries) The case of Oceania is somewhat different. The colonial censuses for the 20th century are deemed reasonably accurate and indeed in most cases the results tally well with the United Nations data after 1950. In contrast, trends in the 19th century are very controversial. Traditionally, historical demographers have relied on estimates from ‘experts’ and in the second half of the 19th century on enumerations by missionaries. These latter were fairly well aware of the size of their flock and of potential converts and indeed the results of their enumerations are often confirmed by later censuses. In contrast, the early estimates by ‘experts’ often differed quite widely. The historical demographers have given more weight to figures closer to the late 19th century ones, discarding some early estimates as implausibly high and thus they have downplayed the decline in population after the contact (McArthur 1967, Campbell 2006). On the other hand, the forward projection method yields much higher figures, similar or even higher than the present-day population. This implies a large collapse in population, which was caused by the ‘fatal impact’ with European diseases. The most controversial case is population of the Hawaii. The consensus among historical demographers settled for a figure in the middle range of the quoted estimates by members of the Cook expedition, even if Schmitt (1972) argued strongly for a low figure. In contrast, Stannard (1989), put forward a range of 0.8-1 million at contact in 1778 (double the population in 1940) hypothesizing the arrival of about 100 people in the first century AD and a 0.52% yearly rate of population growth. Later work has used so-called Dye-Tomori model, which relies on changes in the charcoal quantities in archeological sites to build an index of population movements (Kirch 2007). Extrapolating backward the earliest reliable figures yields a population at contact close to the lower bound of the historical demographers’ estimated range. In our estimation, unless we have additional polity-specific evidence (e.g. on major epidemics or on civil wars), we have adopted a conservative view. We have assumed that population were stable before the 17 We have re-computed the Frankema-Jerven (2014) series for Ethiopia, because the website by mistake reports a constant population. We do not discuss separately these series in Appendix one, unless, as for Congo or South Africa, there are alternative modern sources. 18 This assumption would underestimate the population of the slave-exporting areas if, as equally likely, total exports had instead declined. The bias in estimates of the African population would be equal to the difference between actual and hypothesized captures for slaves exported outside the continent. In contrast, the enslavement of people in other Africa areas would affect mostly the polity-specific series, while total African population would differ only by the additional deaths in capture and by the difference in birth rates between freed and captive population. Frankema and Jerven (2014) do not quote the slave trade after 1850 and we assume that they take it into account in the adjustment of reference rates. 13 start of substantial interactions with Europeans (which in most cases began later than the first contact) and that it declined after it at different rates, not exceeding 2% per year, until the earliest reliable figure. 5) The results: world population in the long run Overall, the world population increased from just above 1 billion in 1800 to 2.2 billion, which corresponds to a log rate of 0.54%. A visual inspection (Figure 2, left axis) suggests an almost linear increase but the joint TS/DS model by Razzaque et al (2007) returns a not significant rate. Indeed, the yearly changes (even when smoothed with a 11 year moving average as in Figure 2, right axis) show two massive decelerations, in the mid-19th century, and in the late 1910s. Figure 2 World population and its yearly change 1800-1938 A Bai-Perron (2003) test singles out five breaks in 1824, 1846, 1866, 1886 and 1919 and Table 4 reports (left-hand column) the rates of change for corresponding intervals, plus (right-hand column) additional medium term rates Table 4 Rates of change, world population, 1800-1938 Short term ° Medium-term 1800-1823 0.32 1800-1850 °° 0.50 1824-1846 0.47 1847-1866 0.20 1850-1864° 0.20 1867-1886 0.62 1865-1913°° 0.78 1887-1919 0.77 1913-1920° 0.29 1920-1938 1.17 1920-1938° 1.17 ° log linear; °° Razzaque et al (2007) 0 500000 1000000 1500000 2000000 2500000 1800 1806 1812 1818 1824 1830 1836 1842 1848 1854 1860 1866 1872 1878 1884 1890 1896 1902 1908 1914 1920 1926 1932 1938 -0.002 0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 Changes (11 year moving average Population 14 World population grew in the first half of the 19th century roughly as much as in the 18th century, if one believes in the available estimates 19. The growth accelerated in the 19th and 20th century, to culminate in the second half of the 1960s with rates around or above 2% per year. The time pattern of growth differed substantially between continents, as Figure 3 shows Figure 3 The distribution of world population by continent, 1800-1938 In relative terms, the African and Asian population declined, the European (here including the whole of Russia) increased and the American and Oceanian boomed. The changes were already massive by 1870 and were mostly over on the eve of World War One. Trends slowed down or even reversed in interwar years, but total changes were small in comparison with pre-war ones. The African share of world population fell by a fifth in the first half of the 19th century, from 11.3% to 8.9%, slid by a further percentage point to 7.8% in 1913 and in 1938 was roughly at the level of the mid 1900s. The Asian share declined over the whole period by 13 points, from 67% to 55% and a third of the change was cumulated in the 1850s and 1860s. Likewise, two thirds of the relative rise of Europe from 18% in 1800 to 27% in 1913 were cumulated before 1870. The European share fell during the war by one point and continued to decline afterwards. In contrast, the American share rose almost as fast from 1870 to 1913 (from 6.5% to 10.4%) than in the first seventy years of the century (from 2.5% to 6.5%) and rose further to 11.9% in 1918. 19 The rates for the 18th century hover around 0.4% per year, ranging from 0.33% for Clark (1967) to 0.46% for Maddison (until 1820) and to 0.48% according to HYDE 3.1 (Klein Goldewijk et al 2010). This latter suggests a slowdown in growth from 0.57% in 1700-1750 to 0.39% in 1750-1800, but it is an exception. All other estimates report higher rates in the second half of the century - e.g. McEvedy Jones (1978) 0.33% in 1700-1750 and 0.45% in 1750-1800. 0% 10% 20% 30% 40% 50% 60% 70% 80% 90% 100% 1800 1805 1810 1815 1820 1825 1830 1835 1840 1845 1850 1855 1860 1865 1870 1875 1880 1885 1890 1895 1900 1905 1910 1915 1920 1925 1930 1935 Asia Europe Africa Americas Oceania 15 We will return to the causes of these change in the next sections. Before that, we need to ask to what extent our results modify the conventional wisdom about long term trends. We start by comparing population at the relevant benchmark years with the main alternative estimate, by Maddison (Figure 4) 20 Figure 4 World population, 1820-1940: a comparison with Maddison The differences in long-run rates of growth to 1940 (0.63% the present estimate, 0.66% Maddison) are small and this is a strong evidence of Maddison’s unique historical skills 21 On the other hand, there are non negligible divergences in the time profile. Our estimate is 3.4% higher than Maddison in 1820, very similar in 1913 and lower by 1% in 1920 22. Thus, Maddison’s data overestimates by some percentage points the cumulated growth from 1820 to 1913 (by 66% vs 72%) and over the 1910s crisis (almost 4% vs 2.5%). The differences between our estimates and other series are somewhat greater, at least in the 19th century. In Figure 5 we compare our results with two widely quotes estimates by McEvedy and Jones (1978) and HYDE 3.1 (Klein et al 2010) and with an average of other major independent ones23 20 For the purpose of this comparison, we compute the population in 1940 by extrapolating our estimate of population in 1938 with the growth rate for the previous five years (1.3% for Asia and Africa, 1.2% for Americas and world population, 1.1% for Europe and Oceania). 21 One may add that the RMSE for the comparable yearly series, which however refer mostly to advanced countries with good sources, is quite low – around 10%. 22 The population in 1940 is computed, for comparative purposes only, by extrapolating the rates of growth by continent in 1934-1938 – and thus it might be overstated as it does not include an adjustment for war-time losses. 23 The average is computed on seven sources common to all three benchmarks, Willcox (1940), Carr-Saunders (1936), Swaroop (1951), Bennett (1954), Durand (1967), Biraben (1979) and United Nations (1999). We add Clark (1967) in 1800 and 1900 and Hubner Geographical Atlas (as quoted by Willcox 1940) n 1851 and 1900. We do not add a bar for the late 1930s because the estimates refer to different benchmark years. 0 500 1000 1500 2000 2500 1820 1870 1900 1913 1920 1940 Maddison (2010) Federico Tena 16 Figure 5 World population, 1800-1900: a comparison with other estimates Our estimate is higher than all others in 1800 (up to 10% higher) and it grows more slowly.24 As expected, differences by continent are greater (Table 5), and they are clearly related to the abundance of sources 25. They are fairly small for Europe (without Russia), and quite large for Oceania and Africa. These latter would have been larger if we had used the Manning-Nickleach (2014) estimates also for the period 1850-1890 (cf. Table 2). 24 Our estimate implies an annual rate of growth 0.47% 1800-1900 vs 0.51% for HYDE 3.1, 0.59% according to McEvedy and Jones (1978) and 0.55% for the average, with a maximum of 0.63% according to Durand (1967). 25 In the columns for the average of estimates, the cells ‘Europe’ (excluding Russia) and ‘Russia’ are empty because they treat differently Russia, sometimes reporting separate figures and sometime including in Europe. 0 200 400 600 800 1000 1200 1400 1600 1800 1800 1850 1900 McEvedy and Jones (1978) HYDE 3.1 Average Federico Tena 17 Table 5 Comparison with other estimates, ratios by continent McEvedy-Jones (1978) HYDE 3.1 Average ‘core’ Maddison 1800 1850 1900 1800 1850 1900 1800 1850 1900 1820 1900 1940 Africa 1.63 1.39 1.22 1.33 1.09 0.95 1.16 1.11 1.03 1.53 1.22 0.98 America 1.01 0.97 1.01 0.88 0.95 0.99 0.89 0.95 0.93 1.01 1.00 1.00 Asia 1.10 1.04 0.95 1.00 0.99 0.96 1.11 1.08 0.98 1.01 1.03 1.01 Europe 1.01 0.99 0.99 0.98 0.97 0.96 0.96 0.94 1.04 Russia 0.89 0.94 1.08 0.88 1.09 1.11 0.90 1.08 0.88 Oceania 1.23 1.11 0.97 3.61 1.95 1.25 1.41 1.33 1.10 6.27 1.45 1.33 18 Finally, Figure 6 compares the present estimate with the yearly series in the Gapminder website, computing also the RMSE to measure the polity-specific differences 26 Figure 6 Comparison with Gapminder The Gapminder series, as other estimates, overvalues the growth of population relative to the present one, in the ‘long’ 19th century but the difference is not large. In contrast, the short term movements are quite different (the correlation between the cyclical components from Hodrick-Prescott filters is a mere 0.48) and the RMSE is very high (44% on average). The RMSE is to some extent boosted by an imperfect adjustment of present-day borders to historical ones, but some large differences (e.g. for India) are difficult to explain 27 The League of Nations is a somewhat different case. Its estimate is very similar to the present one in 1913, with a modest 17% RMSE and again in 1925 (the present estimate is about 1% higher and the RMSE is down to 10%), but diverges sharply in the 1930s. In 1938 our estimate is 5% higher and the RMSE is 42%. Most of this divergence depends on a different assessment of trends in China. The League of Nations puts forward an ‘approximate estimate’ with stable population at 450 millions for the whole interwar period, while according to our estimate (and consistently with the first post war census in 1953) the Chinese population rose from 454 in 1925 to 530 millions in 1938. These comparisons show that the so far available series capture long run trends in world population, given the unavoidable margin of error of all estimates, including the present one (cf. Section eight). In contrast, yearly series are either missing, as in most scholarly works, with the partial exception of Maddison, or of 26 We compute the RMSE on a set of comparable polities to avoid bias results upwards. Anyway, the omitted polities account for a very small share of total (less than 5%) for both Gapminder and the League of Nations 27 The sum of Pakistan, India, Bangladesh and Myanmar according to Gapminder exceeds the population of British India according to censuses, as re-worked by Davis (1968) and endorsed by Maddison 1995 and Sivasubramonian 2000 by 17%, throughout the period. Such a divergence suggests some major difference in geographical coverage but it is impossible to find it with the available information on Gapminder sources. 0 0.2 0.4 0.6 0.8 1 1.2 1800 1805 1810 1815 1820 1825 1830 1835 1840 1845 1850 1855 1860 1865 1870 1875 1880 1885 1890 1895 1900 1905 1910 1915 1920 1925 1930 1935 Ratio FT/Gapminder RMSE 19 questionable reliability for lack of information (for Gapminder). In the next Section we will use our newly compiled polity data for historical analysis. 6) The results: volatility and demographic crises As a first step, we test the conventional wisdom about volatility, which we compute as the residuals from a Hodrick-Prescott filter and, as a robustness check, also with the normalized five-year difference as advocated by Hamilton for yearly series (2018). We normalize the residuals with the trend component and we plot them in Figure 7 for continents and world. Figure 7 Volatility by continent 20 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 1800 1825 1850 1875 1900 1925 Africa -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 1800 1825 1850 1875 1900 1925 America -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 1800 1825 1850 1875 1900 1925 Asia -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 1800 1825 1850 1875 1900 1925 Europe -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 1800 1825 1850 1875 1900 1925 Oceania -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 1800 1825 1850 1875 1900 1925 World The results do not tally with the conventional wisdom. There is not a clear downward trend and volatility does not appear inversely related to level of development. It is high in Oceania but very low in Africa, and higher Europe than Americas and Asia. These results might be affected by the use of linear interpolations which by definition reduce volatility (cf Section Eight). This is surely the case of low volatility of African series, as almost all series for Sub-Saharian polities are obtained as five-years periods with constant rates - s (with a jump between them). Furthermore, a decline in share of series obtained with linear interpolation might balance the decline in volatility. We test this latter hypothesis by adding add the share as in our regression to estimate the rate of change in volatility. 28 The results (Table 6) confirm the first impression 28 The dependent variables are computed as the logs of square root(s) of the ratios of squared numerator and denominator (e.g. the residuals and the trend component of the Hodrick-Prescott filters). 21 Table 6 Rate of change in volatility of population series, 1800-1938 HP Hamilton 1800-1938 With share 1800-1938 Africa 3.17 3.66 3.21 America 2.20 1.70 -0.09 Asia 0.78 0.71 1.18 Europe 3.23 3.77 0.13 Oceania 1.28 1.25 1.90 World 1.10 1.00 0.80 There is no consistent sign of declining volatility with either method, with or without the share of interpolated observations as control. If any, the variance increased, but results, with the exception of Africa, are not consistent enough to draw any firm conclusion. Volatility depends on demographic crises. In theory, one could define a crisis any observation sufficiently below trend (e.g. by two SD lower), but results would be biased by the use of linear interpolations. We thus adopt a more conservative definition of crisis as absolute decline in population, for whatever reason (epidemics, famines, wars) 29. Out of a total of 21423 yearly changes by polity, population increased in 17461, remained constant in 979 and declined in 2983 i.e. about a sixth. The contrast with the 5.3% share in 1951-1991, before the start of mass emigration which reduced population of several Eastern European countries. The distribution in time of these crises (Fig 8) shows a sharp drop in the second half of the 19th century, and then a rise which peaked, predictably, in 1919 with the Spanish flu. Figure 8 Number of crises and share of world population 29 We adjust only for declines in population caused by territorial losses, as in France in 1871 and Germany and Russia in 1919. We assume that in that year, population changed as much as in the average of the two neighbouring years (e.g. 1870 and 1872). We need not to adjust for the dissolution of Austria-Hungary and of the Ottoman empire because the series end in 1919. 22 However most affected policies were small and indeed the effects on world population, as measured by the share of population in affected policies (Fig 9), are smaller. In the first half of the century, crises affected mostly Africa and Oceania, around mid century Asia (or, more precisely China) and in after 1913 Europe, while the thirty years before World War One, the heyday of first globalization, crises were very rare outside Africa, which were hit by the rinderpest and European conquest 30 Figure 9 Share of population affected by demographic crises, by period/continent 30 According to the estimates by Frankema and Jerven (2014), Africa accounted for almost two thirds of all crises in 1883-1913 (340 out of 576). In those years, population declined in 13-15 polities in SubSaharian Africa (out of 41) with between a third and a quarter of the population. 0 0.1 0.2 0.3 0.4 0.5 0.6 0 10 20 30 40 50 60 1802 1807 1812 1817 1822 1827 1832 1837 1842 1847 1852 1857 1862 1867 1872 1877 1882 1887 1892 1897 1902 1907 1912 1917 1922 1927 1932 1937 number crises share world population 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Africa Americas Asia Europe Oceania World 1802-1852 1853-1882 1883-1913 1913-1938 1802-1938 23 Our definition, as Table 7 shows, captures two very different categories of ‘crises’, short and severe shocks and long run, slower, downward trends 31 Table 7 Duration and severity of demographic crises Number crises Years of crises Yearly decrease Average cumulated decrease One year 56 56 -3.11 -3.11 Two years 51 102 -1.14 -3.85 Three years 18 54 -2.53 -7.38 Over three years 157 2983 -0.56 -10.68 Total 282 3195 -1.53 -5.83 The worst demographic shocks plagued the tiny island of Cabo Verde, which, in spite of its name, was subject to disastrous droughts. There were five such crises, and in the worst one, in 1830-1832, population collapsed from 76K to 47K – i.e. by 45% . The Irish famine and related diseases claimed 1-1.5 million lives, but the total population of the island fell, because of emigration, from the peak of 8.3 million in 1845 to 6.3 in 1852 – i.e. by a quarter (Mitchell 1988). As a consequence, the total population of the United Kingdom declined, in spite of the fast rise elsewhere in the country, from 1846 to 1851 by about 3%. The devastating famine of North China 1877-1878 claimed between 9 and 13 million lives (Aird 1968 p.265): in the most affected province, Shanxi, ca 5.5 – i.e. about 15% of population (and just for a comparison over a third of the population of UK in those years). The long run downward trends accounted for over 90% of ‘crises’ in our definition. The worst case is the Danish Virgin island where, according to the estimates by Bulmer Thomas (2012) population declined in 91 years out of 139 – i.e. in all years but the initial ones (1800-1835), the final ones (1930-1938) and a short period of stagnation from 1846 to 1850. However, downwards trends were longer and more frequent in Sub-Saharian Africa and Oceania: the former accounted for more than half of observations in long run trends (1565 out of 2983) but the latter had proportionally more, 41% vs 28%. The exact timing, speed and duration of these long crises is uncertain as the rates for these continents are, at best, informed guesses, but the evidence for these long-term crises is overwhelming. In Africa the population declined because of 31 There is no good independent source on demographic shocks. The EMDAT data-base ( deal with natural disasters only since 1900 and it seems to underestimate the number of deaths. It lists 387 events, but in only eight cases the death toll exceeds one million people 24 the slave exports (Manning and Nicleach 2014), in other continents because of European infectious diseases which natives had little or no immunity against. By 1800, the collapse of native population had already happened in South America and in Mexico, was on-going in North America and had just started in Oceania. The extent of population decline depended on the type of diseases and on the population density. Some diseases were more lethal than others and transmission depended on intensity of contacts. The wide spaces of North America reduced contacts between infected native groups (Thornton 2000) and the length of trips from Europe might have initially slowed down transmission of some diseases (Bushnell 1993). Yet, our source suggests a 62% decline of native population of North America 1800 to 1890 (Thornton and Marsh-Thornton 1981). In Oceania the decline of the native population was as large as in the Americas and there was very little rebound before World War Two. According to the best estimates, the total aboriginal population of Australia fell from 541K 1800 to 71K 1936 (i.e. by 87%). Yet the impact of these demographic catastrophes on world population was small and sometimes negligible. If the native population had been growing at 0.5% yearly since 1800 rather than declining, the population of Oceania would have been 2.3 million higher in 1938 and the world total only 0.1% greater. 32 In the extreme and somewhat implausible case that the 1800 native population were ten times the historical minimum, as hypothesized by Stannard (1989) for Hawaii, the population Oceania would have been 8 millions (vs 3) but world population would have been only 0.5% higher. Even major human tragedies barely affected long term trends. The African rinderpest killed about 0.5% of world population, the North Chine famine 0.6-0.9% world population. Indeed, the world population declined in only eight years out of 139, in 1826, 1862-1866 and in 1918-19. The first one is the consequence of a sudden fall in Chinese population, from 392 to 385 million, which however does not correspond to any known catastrophe and thus might well be spurious. The 1862-1866 decline it coincided with the bloodiest phase of the Tai’ping civil war. In the fourteen years from the 1852 peak, when the war had just started, to 1871, when the last Tai’ping army was defeated, population collapsed from 439 to 359 million (i.e. by almost a fifth), and about half of these losses were cumulated in 1862-186633. The 1917/19 decline was determined by three different shocks, the Great War, the Russian post revolution famine and civil war and the Spanish flu. Table 8 compares the actual population changes for selected countries and European belligerents with a crude estimate of excess mortality34. Table 8 Population changes 1917-1919 32 For Australia and New Zealand we obtain the counterfactual series as sum of natives and white population with data from Historical Statistics Australia and Historical Statistics New Zealand. For Hawaii we rely on data from Schmitt (1968 tab 16 and 26). We estimate the counterfactual series for full-blood natives only, while we include ‘part Hawaiians in the series for ‘others’ (mostly Japanese immigrants), assuming there were 100 ‘others’ in 1800. For other islands the sources do not distinguish between natives and immigrants, who anyway were much fewer than in Hawaii or, a fortiori, Australia and New Zealand. Thus, we implicitly assume that all population was native-born and we select for the counterfactual the period of absolute decline -i.e. 1800-1885 for Polynesia (a total decline of 50%), in Micronesia 1830-1913 (37%) in 1820-1901 Melanesia, without Papua New Guinea (-29%). We extrapolate the population in the final year of the period with the actual growth rate. 33 The excess deaths, extrapolating the rate of the late 1840s (0.4% year) could be up to 110 million. The figure is higher than standard estimates of war losses as it includes also the losses from other revolts in the 1860s and deaths from famine and diseases. It is however close to recent Chinese estimates (Platt 2012 p.308) 34 Cf. for Indonesia and India Appendix I. For the other countries we compute excess mortality as the difference between actual and counterfactual population in 1918 and 1919. We estimate this latter by extrapolating the actual population in 1917 (or 1918), thus inclusive of previous war losses and deaths from the first wave of the flu, with the ‘normal’ rate of change, as proxied by the 1909-1913 rate. For Russia we use the series at interwar borders from Markevich and Harrison 2011 tab A7-A9. 25 Actual change Excess mortality 1917 to 1918 1918-1919 1917 to 1918 1918-1919 China 1405 (+0.32%) 760 (+0.17%) -1147 (0.26%) -1800 (-0.41%) Indonesia -1072 (-1.97%) 523(+0.98%) -1690 (-3.10%) -82 (-0.15%) India -8966 (-2.79%) 1460 (+0.46%) -8965(-2.77%) -3031 (-0.93%) Russia -774 (-0.45%) -1756 (-1.21%) -3520 (-2.07%) -4075 (-2.79%) WWI belligerents -2247 (-0.78%) -4866 (-1.71%) World -6218 (-0.34%) -2644 (-0.14%) -22519 (-1.22%) -18899 (-1.01%) excluding Russia, USA and dominions Our estimate implies that the flu caused about 33 million deaths, but figure could be somewhat underestimated 35. These data are not available for most countries and thus they have been substituted by rough guesses. Thus, the number of covered countries and the results differ rather widely, from 17-24 million (Spreeunwenberg et al 2018) to 35-44 million deaths (Athukorala and Athukorala 2020), with unsubstantiated guesses up to 100 million36. About 10-12% of world population has been estimated with linear interpolation between benchmarks or, as in the case of Africa, with constant rate of change over the period 1915-192037. Furthermore, the figure of 3 million deaths in China, where anyway the flu might have been more benign than elsewhere (Chen and Leung 2007), is lower than current estimates of the losses, varying between 4 and 9.5 million deaths (Johnson and Mueller 2002, Athukorala and Athukorala 2020 Barro et al 2020). On the other hand, it seems highly unlikely that the true total losses reached 40 million. This conclusion implies that the Tai’ping civil war, or more in general the mid-19th century crisis, was the largest demographic shock in the whole period. In absolute terms is comparable to, if not higher than, World War Two, when world population was double a century earlier. The impact of these two world-wide crises can be measured with a simple back of the envelope counterfactual exercise. We compute no-crisis population series assuming that the Chinese population had been growing from 1851 to 1871 as fast as in 1830-1850 and that the world population had been growing 1914-1919 as fast as in 1890-1914 38. The comparison of the three counterfactual estimates (Figure 12) shows that the Tai’ping war affected world population much more than the later 1910s crisis: world population in 1938 would have been 2429 million, rather than 2254 (or 7.7% greater) without the civil war 35 This figure is obtained as residual deducting from the total excess death (40.8 million) the non pandemic deaths in Russia (4.8 million) and WWI belligerents (about 3 million). These latter are computed by deducting the available estimates of pandemic deaths (Athukorala and Athukorala 2020) from the total losses. This latter source does not cover all countries (missing Bulgaria, Romania, Serbia) and thus we use the upper bound of the implicit range. Total war losses, including diseased soldiers, were about 1.8 millions in Russia and 6.7 millions in other European countries (Becker 1999 p.80). 36 This figure, which is often quoted in the historical literature, is a speculative guess by Johnson and Mueller (2002). The sum of their country estimates yields a total of 33-43 million, but they increase it to 50 million, and add that even this figure ‘may be substantially lower than the real toll, perhaps as much as 100% understated’ (p.115). See for other estimates of total losses Athukorala and Athukorala 2020 tab 1. 37 Johnson and Mueller (2002) hypothesize a total of 2.4 million deaths in Africa, but, given the state of sources, this figure is pure guesswork. 38 We extrapolate the post crisis trend with the actual rate, which could exceed the steady state one to compensate for the losses. However, there is no much evidence of such an acceleration. The rates for China were marginally higher in the mid-1870s than in the 1840s but the increase was short lived. As said, the worldwide rate of growth rose gradually from 1921 to the eve of World war rather than jumping after the end of crisis and then returning to the normal. 26 and 2137 without the late 1910s crisis (2.8% greater). Without both major crises, the population in 1938 would have been about 244 million greater (10.8%): if one factor in also the ‘minor crises’ the total losses might well exceed 300 million. Figure 10 Counterfactual, no-crisis series of world population 7) The results: dating the demographic transition The textbook model of demographic transition (Kirby 1996, Lee 2003) implies an inverted U-pattern of the rate of natural increase of population. It is nil or very low before the transition, rises in the first stage for the reduction of mortality while fertility remain high and eventually return to very low (or zero) levels in the second stage, when fertility converge to mortality levels. This pattern would cause a permanent upward shift between two (almost) constant population levels. As all textbook models, also this one does not necessarily correspond to historical patterns, as Figure 11 shows, with two examples of major European countries, with suitably long demographic series. Figure 11 The demographic transition in history (rates per thousand) a) United Kingdom, 1541-1980 0 500000 1000000 1500000 2000000 2500000 3000000 1800 1806 1812 1818 1824 1830 1836 1842 1848 1854 1860 1866 1872 1878 1884 1890 1896 1902 1908 1914 1920 1926 1932 1938 Baseline No Tai'ping war No World War One No Tai'ping and WWI 27 Sources: Mitchell 1988 Data in five years averages for data constraint Birth and death rates per thousands; population index 1800-1804=100 Population Henry and Blayo 1975 tab 22 for 1740-1800, our estimate 1800-1938, Annuaire Statistique 1945-1949, United Nations 1950 ff; birth and death rates; Henry and Blayo 1975 tab 22 1740-1830, Annuarie Statistique 1913 1830-1850, Rothenbacher 2002 1850-1945, Annuaire Statistique 1957 1945-50, United Nations 1950-2015 Even if movements of birth and death rates do not conform to the model, the transition did cause an increase in population. Indeed, population growth would be a perfect proxy for the transition if net migrations were zero. This is by definition true for world-wide population and thus our results (Section five) over the whole period implies that transition had started in the 19th century and it has been accelerating in the first half of the 20th century. In the early 1950s, the rates were already higher than the pre-war peak (1.7% vs. 1.3% in 1929-1933) and grew to 2% in 1965-1970, to decline to 1.1% in 2015-2020 (United Nation 0 100 200 300 400 500 600 700 800 900 1000 1547 1563 1579 1595 1611 1627 1643 1659 1675 1691 1707 1723 1739 1755 1771 1787 1803 1819 1835 1851 1867 1883 1899 1915 1931 1947 1963 1979 birth death population 0 50 100 150 200 250 300 350 400 450 1740-1744 1750-1754 1760-1764 1770-1774 1781-1784 1790-1794 1800-1804 1810-1814 1820-1824 1830-1834 1840-1844 1850-1854 1860-1864 1870-1874 1880-1884 1890-1894 1900-1904 1910-1914 1920-1924 1930-1934 1940-1944 1950-1955 1960-1965 1970-1975 1980-1985 1990-1995 2000-2005 2010-2015 birth death population 28 2019). The world-wide increase is however consistent with a wide range of possible patterns by country (or area). Indeed the comprehensive, if not necessarily accurate, post-war United Nations data confirm the different timing of transition. The rates of natural increase have been consistently very high in Africa, up to a maximum of 2.9% in the early 1980s, initially high but declining in Asia and Latin America since the 1960s, fairly low (in the region of 1.5%) and declining in North America and in Oceania, initially low (ca 1%) and falling fast down to stagnation or decrease in Europe. One would thus infer that by 1950, the demographic transition was in the final stage in Europe, North America and Oceania, while it was in the ascending stage in Asia, Latin America in Africa. Unfortunately, the available pre-war data on rates of natural increase (Section Eight) are insufficient to pursue further this analysis. The coverage is partial even in the interwar years (about a half of the countries and a third of observations) and it is very limited in the 19th century. A first look at rates of total population growth by polity highlights a major difference with post war trends. Rates over 2% were the norm after 1950 but fairly rare before 1938 . The 2% threshold, which was the norm at peaks after WWII, was very rarely attained: yearly increase of 11-year averages exceed 2% only in 11% of cases (2215 out of 19718) 39. Most of these cases are inflated by immigration, and not only in the usual suspects but also, for shorter periods, in countries, such as Malaysia (Kim 1998), Hawaii (Nordyke 1989) and possibly Mauritius (Lutz and WiIls 1994). We explore in more detail the pattern by grouping polities in nineteen areas, taking into account location, level of development and migration balances40. We then plot five year moving averages of population growth in these areas, ranking them (roughly) according to long-term growth rates and adding a 50 per thousand line to mark the world long term average rate (Figure 12) Figure 12 Rates of growth in population growth by area, five year moving averages a) demographic crises: Pacific islands and Sub-Saharian Africa 39 As a rule, we omit from computation of 11-year averages the initial and final years. Therefore, most series run from 1806 to 1933 and other are shorter – e.g. the series for Poland (original series from 1919 to 1938) refers to 1924-1933. 40 We divide America between North, immigration areas (Brazil, Argentina and Uruguay), Oceania between Immigration (Australia and New Zealand) and the rest (Micronesia, Melanesia and Polynesia), Africa between North and Sub-Saharian, Asia between Middle East, China, India, South East Asia (Indonesia etc), East Asia (Japan and Korea). Europe between United Kingdom, other North Western (France, Scandinavia etc.), Central (Germany, Austria-Hungary and successor states), Southern (Italy, Spain, Portugal), Eastern (Russia, Poland after 1918) and Balkans. For the sake of readability, in the series for Balkans and Central Europe we omit all averages including the year 1919, to avoid the spurious effect of boundary changes between countries belonging to different areas. In that year, the Southern provinces of the former Habsburg empire (included Central Europe) was transferred to Yugoslavia (included in the Balkans). 29 b) traditional societies: Asia c) Traditional societies: Mediterranean -10 -5 0 5 10 15 20 1803 1808 1813 1818 1823 1828 1833 1838 1843 1848 1853 1858 1863 1868 1873 1878 1883 1888 1893 1898 1903 1908 1913 1918 1923 1928 1933 Other Pacific islands Sub-SaharianAfrica World average -30 -25 -20 -15 -10 -5 0 5 10 15 20 25 1803 1808 1813 1818 1823 1828 1833 1838 1843 1848 1853 1858 1863 1868 1873 1878 1883 1888 1893 1898 1903 1908 1913 1918 1923 1928 1933 China India East Asia (Korea and Japan) South-East Asia World average 30 d) advanced Europe e) traditional societies: Eastern Europe -10 -5 0 5 10 15 20 1803 1808 1813 1818 1823 1828 1833 1838 1843 1848 1853 1858 1863 1868 1873 1878 1883 1888 1893 1898 1903 1908 1913 1918 1923 1928 1933 Middle East North Africa Mediterranean World average -10 -5 0 5 10 15 20 1803 1808 1813 1818 1823 1828 1833 1838 1843 1848 1853 1858 1863 1868 1873 1878 1883 1888 1893 1898 1903 1908 1913 1918 1923 1928 1933 United Kingdom NW Europe Central World average 31 f) traditional societies: South America g) immigration countries -25 -20 -15 -10 -5 0 5 10 15 20 25 30 1803 1808 1813 1818 1823 1828 1833 1838 1843 1848 1853 1858 1863 1868 1873 1878 1883 1888 1893 1898 1903 1908 1913 1918 1923 1928 1933 Balkans Balkans Eastern Europe Russia World average World average 0 5 10 15 20 25 1803 1808 1813 1818 1823 1828 1833 1838 1843 1848 1853 1858 1863 1868 1873 1878 1883 1888 1893 1898 1903 1908 1913 1918 1923 1928 1933 Other Latin America Caribbean World average 32 A visual look suggests six stylized facts i) as expected, population grew very fast in immigration countries, with the notable exception of Australia and New Zealand in the first half of the 19th century, for the collapse of native populations. But the growth was not only from immigration: the rates of natural increase were very high, at the top of the world scale at least before World War I. In the United states, the (estimated) rate remained above 2% until the mid 1850s and declined below 1% (well above the world average) only during the Great Depression ii) the rates of growth in advanced European countries were rather low in the first decades of the 19th century (with the partial exception of pre-famine United Kingdom) and declined further in the period, with some major fluctuations in Germany. These trends suggest that these countries were starting the final stage of demographic transition 41. iii) in North Africa, East and South-East Asia the pattern resembles the textbook model of the transition. The population stagnated until the 1860s or 1870s and then it started to grow at accelerating rates. The rates exceeded 1-1.2% in the 1930s. and then grew further after the war, with maxima well above 2%. In none of these areas, the transition can be considered complete as rates of total (and natural) increase still exceeded 1% in 2015-2020. iv) In Pacific island and Sub-Saharian Africa population had been stagnant or declining in the early 19th century and thus the fast growth of interwar years included a substantial component of rebound. On the 41 In the 1930s, when migrations were negligible, the population growth was 0.4% per year in North-Western Europe (including France) and United Kingdom and 0.6% in Central Europe. In the 1950s and 1960s, the natural increase rose somewhat in Northern Europe (United Kingdom and Scandinavia), remained almost stable in Central Europe (France and Germany) but population rose faster thanks to immigration – up to 1% per year. -60 -40 -20 0 20 40 60 80 100 120 1803 1808 1813 1818 1823 1828 1833 1838 1843 1848 1853 1858 1863 1868 1873 1878 1883 1888 1893 1898 1903 1908 1913 1918 1923 1928 1933 Aus+NZ Immigration Latin America (Arg+Bra+URU) USA+Canada World average 33 eve of World War One, population was 15% higher than in 1800 in SubSaharian Africa and 10% lower in Pacific islands. Given the poor quality of the data, it is difficult to interpret the pre-war trends as clear evidence of demographic transition. v) in contrast, there is no evidence of accelerating population growth until the very of the period in the two most populous countries, China and India. Indian rates remained extremely low (less than 0.2%) until WWI, increased to about 1% in the late 1920s and to less than 1.5% in the late 1930s. The growth accelerated after WWII, peaking in the late 1970s-early 1980s. The Chinese population grew very slowly after the shock of the mid 19th century, with a short-lived acceleration in the mid 1930s. As in India, rates increased after the war, but the growth was modest for the combined effect of the disasters of the Great Leap Forward and of the of the one-child policy. vi) last but not least, in quite a few cases, movements which can be interpreted as a beginning of the transition were interrupted by exogenous crises and had to re-start. The most relevant example is Russia/Soviet Union. The rate of population growth, after a first, temporary increase in the 1820s and a modest retrenchment in the 1840s-1850s, remained quite high (1.5% to 2%) from the 1870s to WWI. Total population more than doubled from 1850 to the eve of WWI. The growth resumed at quite brisk pace in the 1920s and then after World War Two, with high but declining birth rates (Rothenbacher 2012 tab. SU 4A). It is tempting to suggest that this decline could have started earlier without the disasters of wars and famine. In the Balkans, World War One caused a sudden stop to population growth, which then resumed at slightly higher rates (about 1.2-1.3%) in the 1920s and 1930s and continued after the war 8) Robustness checks: shall we trust these figures? 8.1 Our interpretation in two previous Sections would be the more convincing the more reliable the series are. Thus, in the first part of the Section we will assess the quality of the polity series and estimate the effect of errors following Feinstein and Thomas (2001). We assume a symmetric band as we have already taken into account in the estimation any evidence of asymmetric errors (e.g. of undervaluation of official statistics). Then we deal with two specific sources of bias, the extensive use of linear interpolation and the migratory flows, which affect respectively our analysis of the demographic crises and our use of changes in total population as proxy for natural increase. Linear interpolations make timing of demographic transition less precise and, above all it biases downward volatility and thus downplays the impact of crises. Net emigration (immigration) causes total population growth to underestimate (overestimate) the natural increase and thus to bias downward (upward) the extent of the first stage of transition. 8.2 We assess the reliability of the series by classifying separately each year/polity observation, following the pioneering work by Durand (1977). He distinguished four classes, from A (good quality censuses or well kept yearly population records) to D (pure conjectures) and reckons that class A figures account for 48% of the estimate of world population in 1900 by Clark (1967), B for 14% B and C for 38%. We expand his classification to five classes, A ‘excellent’: fully trustworthy estimates, based on ‘modern censuses’ and/or complete population registers. B ‘good’: interpolation between modern censuses and partial or not fully reliable censuses 34 C ‘fair’: interpolations between not fully reliable censuses or estimates by scholars with solid evidence D ‘poor’: interpolations of C or estimates by scholars with weak evidence E ‘conjectures’: all other estimates, including extrapolations for periods before the earliest available estimate Our sources use the word census for a widely different range of population-counting exercises, but many of them do not fully meet the modern standards (and thus in Appendix I are labelled as ‘count’ or ‘enumeration’). A census can be defined ‘modern’ if was taken on a given date, possibly the same throughout the whole territory, counted all individuals (not just males and/or potential taxpayers), listed them separately with additional information (e.g. age, sex, occupation) and had data collected and elaborated by trained professional clerk rather than by local administrators or tribal chiefs (Baffour et al 2013). We define complete state-organized population registers of births and deaths for the whole population. On the other side of the reliability range, we reserve E for the most tentative figures. For instance, we classify as D the estimates by Frankema and Jerven (2014) for population in Sub-Saharian Africa 1850-1938 and as E our extrapolations to 1800. Some polity series are obtained as sum of differently reliable data – e.g. typically good ones for white settlers and guesstimates for natives. In these cases, we classify results with a crude weighting. The unweighted shares of different categories (Figure ? a) paint a quite dismal picture. For most of the period, the number of polities with A-class (‘excellent’) estimates was very small and, is spite of some growth since the 1880s, on the eve of World War Two, they accounted for a mere 15% of all polities. The class includes only European countries until the 1870s, when the Japanese (and Korean) statistics improve substantially 42. The number of E (‘conjectures’) fell drastically from two thirds to a mere tenth, but jointly with ‘poor’ estimates (D) they still accounted for two fifths of all observations in the late 1930s. However, most of these poor-quality series relative to small polities. The three most populous non-European countries, China, India and Russia, had a majority of C and D years, few Es (India) and also some Bs (Russia). Thus, the population-weighted shares (Figure ?) are less depressing: they show big spikes in census years and some improvement over time, but much less clear than for the trade statistics (Federico and Tena 2016). The A-class (‘excellent’) figures account for about 5% of the total in the early 19th century and grows to about a quarter on the eve of World War Two. Also the share of second best data (B or ‘good’) has been increasing – so that the two top categories accounted for almost a half of observation in 1931 (census year) and for a third in the late 1930s. On the other hand the progress was not steady: for instance the share of population in countries with C data (‘fair’) was quite high in the 1830s and 1840s, thanks to the good working of Chinese registration system, but collapsed after the outbreak of the Tai’ping war. Figure 13 Distribution of world population by quality of data: 42 The quality of data in countries of Western Settlement is not consistent with their high income. We have already quoted the shortcoming of American statistics (Section Three). The Australian and New Zealand data are uncertain until the early 20th century for the large but poorly counted native population. The Canadian censuses are fairly good after 1851 and very good in the 20th century but the official sources estimate intercensal population with linear interpolation. 35 a) unweighted b) population-weighted How much do the potential errors affect the estimates? Clearly, a major mistake in a polity series might jeopardize the interpretation of the demographic history of that polity but its effect on aggregate series depend on the size of the polity and on the possible compensating effects of opposite-sign mistakes in other series. By definition, it is impossible to detect mistakes in polity series without additional evidence – i.e. further research. It is however possible to compute the aggregate margin of error of estimate (Feinstein and Thomas 2002). We first attach a margin of error to each observation according to our assessment of its quality - 2.5% for A estimates (± 1.25% around the ‘true’ value), 7.5% for Bs (±3.75%), 0% 10% 20% 30% 40% 50% 60% 70% 80% 90% 100% 1800 1805 1810 1815 1820 1825 1830 1835 1840 1845 1850 1855 1860 1865 1870 1875 1880 1885 1890 1895 1900 1905 1910 1915 1920 1925 1930 1935 Excellent Good Fair Poor Conjectures 0% 10% 20% 30% 40% 50% 60% 70% 80% 90% 100% 1800 1805 1810 1815 1820 1825 1830 1835 1840 1845 1850 1855 1860 1865 1870 1875 1880 1885 1890 1895 1900 1905 1910 1915 1920 1925 1930 1935 Excellent Good Fair Poor Conjectures 36 17.5% Cs (±8.75%) 32.5% for Ds (±16.25%) and over 40% for Es (with band ±25%). Then we compute the standard error as the sum of variance, under the assumption that errors are independent. Almost all our population series are based on polity-specific sources, whose errors are likely to be independent. The exception might be the series for Sub-Saharian Africa, which are computed with similar rates by macro-area. However, errors will be common only if the true rates of change not only differed from the common one, but also equal across countries. Furthermore, the aggregate bias is small, as Sub-Saharian Africa accounted for only about 5% of world population. The anecdotal evidence on the organization population statistics would suggest that the margins of error were lower for advanced countries and declining over time. This hypothesis is only partially confirmed by the movements in (normalized) standard deviation (Figure 14) Figure 14 Standard deviation of estimates as share of population The world data show a modest improvement in time (from 18.7% to 12.4%) and, over the whole period a decidedly lower error for Europe (6.5%) than for other continents, including the Americas. Changes in Europe are heavily affected by the quality of Russian statistics: the 1897 census anchored firmly the registration data for about twenty year, but the quality deteriorated with the outbreak of the war (as in other belligerent countries) and hardly improved in interwar years, as population became a politically contentious issue in Soviet Union. The fairly high level of error for Americas and its increase before the war reflects the combination of (relatively) poor quality of United States statistics and of the relative rise of its population, from a quarter of the continent in the early 19th century to half after civil war. Last but not least, the spectacular decline in error for Oceania, from 31% to 6.5%, is a consequence of the change in composition of population – whites rising from 0.2% to 75%. In spite of its low share on world population, the fall in error in Oceania accounts for half the total world decline. 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 Africa America Asia Europe Oceania World 1800-1820 1821-1870 1871-1913 1914-1938 1800-1938 37 Given the interval of confidence, the potential error depends on the size of population of each area (Figure 15) Figure 15 Errors in estimates (thousands) a) Africa b) Americas c) Asia 0 50000 100000 150000 200000 250000 1800 1806 1812 1818 1824 1830 1836 1842 1848 1854 1860 1866 1872 1878 1884 1890 1896 1902 1908 1914 1920 1926 1932 1938 Minimum Baseline Maximum 0 50000 100000 150000 200000 250000 300000 1800 1806 1812 1818 1824 1830 1836 1842 1848 1854 1860 1866 1872 1878 1884 1890 1896 1902 1908 1914 1920 1926 1932 1938 Minimum Baseline Maximum 38 d) Europe e) Oceania 0 200000 400000 600000 800000 1000000 1200000 1400000 1600000 1800 1806 1812 1818 1824 1830 1836 1842 1848 1854 1860 1866 1872 1878 1884 1890 1896 1902 1908 1914 1920 1926 1932 1938 Minimum Baseline Maximum 0 100000 200000 300000 400000 500000 600000 700000 1800 1806 1812 1818 1824 1830 1836 1842 1848 1854 1860 1866 1872 1878 1884 1890 1896 1902 1908 1914 1920 1926 1932 1938 Minimum Baseline Maximum 39 f) World The figures highlight some continent-specific features, such as the long-term reduction of errors in Oceania, the bulge in Europe in the second half of the 1910s and 1920s and the systematic shrinking of the band in the Americas in census years. World-wide, the modest reduction in confidence interval is compensated by the increase in total population, so that the corresponding margin of error of the aggregate estimate fluctuates around 200 million people, with a minimum of 141 million in 1841, a census year, and a maximum of 265 in 1917. It is possible to compute, given the yearly band, the greatest possible error in long-term trends as the differences between the lower (upper) bound in 1800 and the upper (lower) bound in 1938. They are substantial, but do not change the historical narrative. Our baseline estimate yields a 123.7 % increase in world population, from 1008 to 2254 million, while the alternatives range from a minimum of 94.1% (1103 to 2142) to a maximum of 159.5% (from 912 to 2367 million). 0 2000 4000 6000 8000 10000 12000 14000 1800 1805 1810 1815 1820 1825 1830 1835 1840 1845 1850 1855 1860 1865 1870 1875 1880 1885 1890 1895 1900 1905 1910 1915 1920 1925 1930 1935 Minimum Baseline Maximum 0 500000 1000000 1500000 2000000 2500000 1800 1806 1812 1818 1824 1830 1836 1842 1848 1854 1860 1866 1872 1878 1884 1890 1896 1902 1908 1914 1920 1926 1932 1938 Min Baseline Max 40 8.3 We have been forced to estimate population with linear interpolation in about half of observations (9157 out of 21509 testable cases) 43. Figure 16 Number of polities with linearly interpolate population The distribution in time shows the expected (downward) spikes in the census years, which often correspond to change from one rate of interpolation to another, and a sizeable decline in the first decades of the 19th century. Since the 1870s, the yearly number of observations in non-census years remains stable around 60 – i.e. about a half. Most of these observations refers to African or Middle Eastern polities, with small population. Indeed, the share of world population estimated with linear interpolation is much lower (Figure 17) Figure 17 Share of population estimate with linear Interpolations 43 We compute this figure as sum of all consecutive years with equal change, including cases of no change, which usually are outcome of assumption as well. We lose the two initial observations of each series and therefore the number of cases is lower the number of population estimates. 0 20 40 60 80 100 120 1802 1807 1812 1817 1822 1827 1832 1837 1842 1847 1852 1857 1862 1867 1872 1877 1882 1887 1892 1897 1902 1907 1912 1917 1922 1927 1932 1937 41 The share is substantially lower across the whole period (18% vs 43%) and much lower from 1821 to 1902 (10% vs 44% for number of polities). It increases again later, almost exclusively because the series for China shows same rate of growth for two or more years, without necessarily being linearly interpolated. The data by continent (Figure 18) confirm the key role of Asia in determining the world changes Figure 18 Linear interpolations as share of population, by continent 8.4 Figure 19 shows the extent of potential bias from the omission of migrations by comparing total and natural population growth in the largest immigration and largest immigration country in the age of mass migrations (Hatton and Williamson 1998) 0 10 20 30 40 50 60 1802 1807 1812 1817 1822 1827 1832 1837 1842 1847 1852 1857 1862 1867 1872 1877 1882 1887 1892 1897 1902 1907 1912 1917 1922 1927 1932 1937 Share population Share population withotu China 0 10 20 30 40 50 60 70 80 Africa Americas Asia Europe Oceania World 1802-1821 1822-1902 1903-1938 1800-1938 42 Figure 19 Natural and total population increase, per thousands, five years averages a) United States b) Italy Sources: United States population Federico Tena data-base natural increase United States estimated as a residual, deducting the net immigration (Historical statistics series Aa13) Italy ISTAT on line table 2.3 0 5 10 15 20 25 30 35 40 1822 1826 1830 1834 1838 1842 1846 1850 1854 1858 1862 1866 1870 1874 1878 1882 1886 1890 1894 1898 1902 1906 1910 total Natural 0 2 4 6 8 10 12 14 1864 1866 1868 1870 1872 1874 1876 1878 1880 1882 1884 1886 1888 1890 1892 1894 1896 1898 1900 1902 1904 1906 1908 1910 total Natural 43 As expected our series of total population underestimates the natural increase (and thus the extent of the first stage of the demographic transition) in Italy and overestimates it in the United States. The biases are substantial: at the peak the difference is half percentage point increase in Italy (population 5.8 vs natural 10.5 in 1896-1900) and one a half in the United States (20.4 natural vs 35.5 total in 1850-1854). However, they are not sufficiently large to change the overall picture of the on-going transition in both countries, with high but declining rates of natural increase in the United States and fairly high rates in Italy We extend the analysis to other countries by building a data-base of rates of natural increase with two different methods. When possible, we get directly data on birth and death rates from sources such as Mitchell’ s International Historical statistics and Rothenbacher (2002 and 2013). When these latter are not available, we compute rates as total growth less net immigration (Ferenczi Willcox 1929). Overall, we have been able to muster long series for most advanced countries (but also for some less developed ones, such as Chile since 1850) and snapshots for quite a few other ones for a total of 81 polities 44. We plot differences total and natural increase of population in 1908-1913, the peak of mass migrations and 1920-1924 and 1934-38, the years of most extensive country coverage (Figure 20) Figure 20 Average difference between total to natural increase a) 1908-1913 (58 polities)45 b) 1920-1924 (68 polities) 44 We have been able to collect 3286 rates from direct sources and estimate an additional 749 with from the indirect computations. They account for 9% or potential observations in 1801-1869, 24% 1870-1913 and 34% 1914-1938 45 For the sake of readability of the figure, we omit Hong-Kong (-100.5) Basutoland (+ 369.4) and Swaziland (+73.5). These two latter are rather suspicious, as the estimate of total population refers to the whole Southern Africa, disregarding the specific situation of these two enclaves in South Africa. -30 -25 -20 -15 -10 -5 0 5 10 15 20 Hawaii Canada New Zeland Trinidad & Tobago… Uruguay Chile Ceylon (Sri Lanka) Seychelles Paraguay Malawi Romania Switzerland Bermuda Netherlands Iceland South Africa Italy Austria-Hungary Barbados 44 c) 1934-1938 (71 polities) There are some cases of quite large differences, but their share on world population is low (Table 9) Table 9 Aggregate biases from migration Shares on covered % covered Mild (± 2‰) Substantial (± 5‰) Large (± 10‰) 1908-1913 49.6 77.3 22.6 3.1 1920-1924 51.4 79.7 24.5 13.8 1934-1938 57.2 60.3 17.6 0.0 -30 -25 -20 -15 -10 -5 0 5 10 15 20 Hong-Kong Greece Denmark New Zeland Cyprus France South Africa Philippines Hungary Trinidad & Tobago… Paraguay Serbia/Yugoslavia Jamaica Iceland Netherlands Poland Romania Lithuania Seychelles Belgium Egypt Germany Cabo Verde -10 -5 0 5 10 15 20 Venezuela Chile Argentina Greece Barbados Czechoslowakia Serbia/Yugoslavia Finland Denmark France New Zeland Belgium Iceland Netherlands Costa Rica New Foundland Ceylon (Sri Lanka) Cyprus Panama Guatemala 45 9) Conclusions This work has confirmed the broad outline from Maddison (and others). The world population did grow in the long run and the growth accelerated in the early 20th century, as an harbinger of the post war population boom. The lowest absolute yearly increase (1.05% in 2019-2020) was exceeded only in the second half of the 1920s and in the 1930s, with an average rate around 1.1%. The yearly data, albeit imperfect, offer two relevant additional results. First, world was still plagued by crises – both short shocks and longer trends… Some of these crises claimed high proportion of the affected areas. Especially devastating the (admittedly poorly measured) decline at collapse and slave trade. However, most crises had limited impact on the growth of world population. There were two exceptions, the late 1910s one, with World War One, Spanish flu and Russian revolution and the Tai’ping war. Both reduced significantly the growth rate, but the impact of the latter was much stronger. The pre-WWII world thus differed substantially from the post-WWII world, when local crises were rare and less severe and world population kept on growing.. Second we have got some insights on the transition. By 1938, the transition completed or very advanced in high income countries, while in the periphery the record seems mixed. In some cases, transition was just starting, in others it had begun earlier but the process protracted in a slow/motion timing. To same extent the pattern reflected exogenous events. Last but not least, we have to repeat that our series are imperfect. It seem unlikely that they could be significantly improved by the discovery of new official sources – historical demographers have done quite a good job. In some cases, the series might be improved by systematic work on micro-data, such as parish registers. Where not available, possible to use archeological sources – but it seems very unlikely that we will 46 References Aird John S. 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China the West and the epic story of the Taiping civil war, New York: A.Knopf Poulain, Michel and Anne Herm (2013) ‘Central Population Registers as a Source of Demographic Statistics in Europe’ Population 68, pp. 183 212 51 Razzaque, Mohammad, Philip Osafa-Kwaako and Roman Grynberg (2007) ‘The problems of commodity dependence’ In Roman Grynberg and Samantha Newton (eds) Commodity prices and development Oxford University Press Oxford pp. 7-16 Rothenbacher Franz (2002) The European population Palgrave Basingstoke Rothenbacher Franz (2012) The Central and East European Population Since 1850 Basingstoke: Palgrave Macmillan Schmitt Robert (1972) ‘New estimates of the pre-censal population on Hawaii’ Journal of the Polynesian society 80 pp. 237-243 Shryock, Henry, Jacob Siegel and associates (1971) The methods and materials of demography Washington US Department of commerce Smith, Leonard Robert (1980) The Aboriginal Population of Australia s, Canberra: Australian National University Pres Spreeunwenberg, Peter, Madelon Kroneman and John Paget (2018) ‘Reassessing the global mortality burden of the 1918 influenza pandemic’ American journal of epidemiology 187 n.2 pp.2561-2567 Steckel Richard (1991) ‘The Quality of Census Data for Historical Inquiry: A Research Agenda’ Social science history 15 pp.579-599 Stannard David E. (1989) Before the horror Social science research institute Honolulu: University of Hawai’i Stern, D.I. and Kaufmann, R.K., Estimates of global anthropogenic sulphate emissions 1860 - 1993. Center for Energy and Environmental Studies, Boston University, Working paper series 9602. 1996.Chemosphere 33 Swaroop S (1951) ‘Growth of the population in the world’ WHO Epidemiological and vital statistics report IV n.4 pp.162-169 Thornton, Russel (2000) ‘Population history of native North Americans’ in Haines Michael and Richard H. Steckel (eds.) A population history of North America Cambridge: Cambridge University Press pp.9-50 Thornton Russell and Joan Marsh-Thornton (1981) ‘Estimating Prehistoric American Indian Population Size For United States Area: Implications of the Nineteenth Century Population Decline and Nadir’ American Journal of Physical Anthropology 55 pp. 47-53 (1981) Schmitt Robert (1972) ‘New estimates of the pre-censal population on Hawaii’ Journal of the Polynesian society 80 pp. 237-243 Smith, Leonard Robert (1980), The Aboriginal Population of Australia, Australian National University Press, Canberra 52 Shryock, Henry, Jacob Siegel and associates (1971) The methods and materials of demography Washington: US Department of commerce Spagnoli, Paul G.(1977) ‘Population History from Parish Monographs: The Problem of Local Demographic Variations’ Journal of Interdisciplinary History, 7 pp. 427-452 Spreeunwenberg, Peter, Madelon Kroneman and John Paget (2018) ‘Reassessing the global mortality burden of the 1918 influenza pandemic American journal of epidemiology 187 pp.2561-2567 Statisches Jahrbuch Deutsche Reich 1938 Swaroop S. (1951) ‘Growth of the population in the world’ WHO Epidemiological and vital statistics report IV n.4 (April) pp.162-169 Thorvaldsen Gunnar (2018) Censuses and Census Takers A Global History Abingdon: Routledge Tuve, G.L., Energy, environment, population, and food. A Wiley-Interscience Publication. John Wiley & Sons, New York, 264 pp. 1976. United Nations 1973 The determinants and consequences of population trends Population studies, 50 p.10 United Nations (1999) The world at six billions New York: United nations United Nations (2019) Population Division Department of Economic and Social Affairs World Population Prospects: The 2019 Revision POP/DB/WPP/Rev.2019/POP/F01-1 (accessed August 20 2021) Van der Vleuten. Lotte and Jan Kok (2014) ‘Demographic trends since 1820’ in Jan Luiten Van Zanden et al (ed) How was life? Global well-being since 1820, Paris: OECD pp.37-55 Wilke Jurgen (2004) ‘From parish register to the ‘‘historical table’’: The Prussian population statistics in the 17th and 18th centuries’, History of the Family, 9 pp. 63–79 Willcox Walter (1940) Studies in American demography Ithaca NY Cornell University Press Woytinsky, W.S. and Woytinsky, E.S., World population and production, The Lord Baltimore Press, New York. 1953. Wrigley Edward Anthony and Roger Schofield (1989) The population history of England 1541-1871: a reconstruction. Cambridge: Cambridge University Press Yáñez César, Rodrigo Rivero, Marc Badia-Mirò & Anna Carreras Marín (2012). La Población de los países latinoamericanos desde el siglo XX hasta 2008. Ensayo de Historia Cuantitativa. GHIP Working Paper n 17 |
3109 | https://www.zhihu.com/question/634726921 | 截距如何影响线性回归模型的预测能力？ - 知乎
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截距是线性回归模型中的常数项，它代表了模型在自变量为 0 时预测的因变量值。截距会影响线性回归模型的预测能力，具体表现在以下几个方面：
影响模型的预测偏差：截距越大，模型的预测偏差越大。这是因为截距代表了模型在自变量为 0 时预测的因变量值，而现实世界中，自变量为 0 的可能性很小。因此，截距越大，模型的预测值就会越偏离实际值。
影响模型的预测线的形状：截距会影响模型的预测线的形状。截距越大，模型的预测线就会越靠近 y 轴。
影响模型的预测精度：截距会影响模型的预测精度。截距越大，模型的预测精度越低。这是因为截距越大，模型的预测偏差越大，从而导致预测精度下降。
因此，在选择线性回归模型时，需要考虑截距的影响。如果截距过大，可以通过调整模型的参数来降低截距。
具体来说，可以通过以下方法降低截距：
增加自变量的数量：增加自变量的数量可以帮助模型更好地拟合数据，从而降低截距。
使用正则化技术：正则化技术可以帮助模型降低模型复杂度，从而降低截距。
调整模型的参数：可以通过调整模型的参数，例如权重系数和截距，来降低截距。
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截距表示当自变量的取值为零时，因变量的预测值。在回归模型中，截距的存在可以使模型具有更好的拟合性能，其主要作用是调整预测结果的基准值。
发布于 2023-12-19 17:57
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PQ is the chord passing through centre of the ellipse a2x2+b2y2=1. If the square of its length is the H.ML of the squares of length of major and minor axes, then slope of line may be equal to
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| Question Text | PQ is the chord passing through centre of the ellipse a2x2+b2y2=1. If the square of its length is the H.ML of the squares of length of major and minor axes, then slope of line may be equal to |
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Houston firefighter who rescued three others from apartment fire speaks
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Houston firefighter rescued 3 from apartment fire
The Houston firefighter who ran into harms way to save the lives of three of his fellow firefighter spoke exclusively with FOX 26's Rashi Vats about his story.
The Brief
HOUSTON - The Houston firefighter credited with heading straight into harms way and saving the lives of three fellow firefighters in an apartment fire is speaking to us exclusively about that fateful day.
Houston firefighter Angel Gaitan, 32, is an Army veteran who commutes all the way from San Antonio to Houston to work.
Who is Houston firefighter Angel Gaitan?
What we know:
Gaitan quit the fire department after working there for one and a half years.
"My car was breaking down almost every time I came. Sometimes I had to Uber to work from wherever I was broken down, and it just became stressful. I was spending more money than I needed to just to get to work," said Gaitan.
He was rehired last month in December after learning about the new "Reignite Your Passion Campaign" to help with the firefighter shortage.
FULL: Firefighter speaks about saving fellow crew
The Houston firefighter credited with heading straight into harms way and saving the lives of three fellow firefighters in an apartment fire speaks exclusively about that fateful day.
3 firefighters injured in apartment fire
The backstory:
Around 5:30 a.m. on Jan. 7, firefighters responded to a report of an apartment fire with people trapped on Jarmese Street near Coffee Street.
Officials say Station 46 crews conducted a primary search while others worked to control the fire.
RELATED: 3 Houston firefighters injured while battling fire at apartment complex
At some point, Senior Captain Edward Escamilla, Firefighter Darren Jones, Firefighter Jonathan Guzman became trapped.
Firefighter Gaitan's rescue
What we know:
Thirty minutes before his shift was over, Gaitan got the call. "Things happened. The building flashed. That’s when the mayday was dropped," said Gaitan.
"There was fire all around us, it was behind us. You shouldn’t let the fire get behind you, but our concern then was getting our guys out," he said. "I could see his gear burned off, as well as Jones. You could see skin."
Gaitan was able to quickly get the firefighters out.
"I feel like everything I did in my life led me to that day. It was just kind of hard to grasp it a little bit."
Gaitan's response:
Gaitan says while the commute to and from work is long, but a small price to pay.
Get news, weather and so much more on the new FOX LOCAL app
"It was an honor to be there that day. Not just there, but here. Even if that day didn’t happen. Just showing up to work here every day is just amazing. I feel like a kid again. Being here is just awesome." Over in San Antonio are his wife and 2-year-old son, and they are expecting their second baby girl any day!
What you can do:
As far as the firefighters- all three are making great progress. To find out how you can help the firefighters recover, visit this website.
The Source: Anchor Rashi Vats spoke one-on-one with firefighter Angel Gaitan.
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3112 | https://www.storyofmathematics.com/range-of-square-root-function/ | Range of Square Root Function - Understanding its Limits and Behavior
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Range of Square Root Function – Understanding its Limits and Behavior
Range of Square Root Function – Understanding its Limits and Behavior
JUMP TO TOPIC [show]
Exploring the Range of a Square Root Function
Determining Range
Real-World Examples
Conclusion
The range of a square root function like f(x)=x plays a crucial role in understanding how this function behaves. For any given real number ( x ), the square root function returns a value that is the square root of ( x ).
The domain of this function, which is the set of all possible real number inputs, is limited to x≥0 since the square root of a negative number is not a real number.
Consequently, the range — the set of possible outputs or function values — consists of real number values starting from 0 and extending to positive infinity, denoted as [0,+∞).
As I continue to explore function characteristics, keeping in mind the connection between the domain and range is essential.
This relationship ensures that mappings from inputs to outputs are well-defined and that the function operates within the constraints of real numbers. Stay tuned as I dive deeper into the implications of this foundational concept in the realm of mathematics.
Exploring the Range of a Square Root Function
In this section, we’ll understand how to pinpoint the range of a square root function and observe its application in real-world examples.
Determining Range
The range of a square root function represents the set of all possible output values. For the standard square root function, f(x)=x, the domain comprises all non-negative real numbers, expressed as x≥0 or [0,+∞) in interval notation.
Consequently, since the square root of a number is always non-negative, the range is also all non-negative real numbers, or [0,+∞) in interval notation.
Square root functions are increasing functions with their graphs resembling one-half of a sideways parabola with the vertex at the origin (0, 0).
This is because square root functions are one-to-one, meaning each element in the domain corresponds to a unique element in the range.
Real-World Examples
The concept of square root functions is pervasive in real-world examples, often without us even realizing it. Take, for instance, the time it takes an object to fall from a certain height.
If we ignore air resistance, the time is proportional to the square root of the distance fallen due to gravity. Here, the distance is a radicand, which inputs into our square root function, while the time is the function’s output, falling within the function’s range.
Another example can be found in architecture, particularly in determining the ratio of heights and widths, which sometimes involves square roots to maintain symmetry and balance.
The aesthetically pleasing ratio often called the golden ratio, involves the square root of 5. Here, the design elements are governed by a ratio that is influenced by a square root function, showcasing the relevance of the function’s range in practical applications.
Conclusion
In this exploration of the range of the square root function, I’ve discussed the fundamental aspects of the function and its behavior.
The range is essentially determined by the nature of the square root operation. A square root, by definition, does not produce negative results; therefore, the function
f(x)=x
can only yield non-negative values. This is reflected in the range, which consists of all real numbers ( y ) greater than or equal to zero, formally written as:
[0,+∞)
Understanding the domain and range of functionslike the square root is not only important for academic purposes but also for real-world applications where these concepts play a crucial role.
As I unpacked the characteristics of this function, remember that its graph is a curve starting at the origin (0, 0) and extends indefinitely in the( x)-positive direction, while remaining above the x-axis.
Reflecting on mathematical concepts like this reinforces how mathematical functions have distinct properties defining their overall behavior.
Familiarity with these function properties enables me to handle a variety of mathematical problems and practical scenarios with competence and confidence.
Posted byWilliam SmithFebruary 1, 2024Posted inBlog
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3113 | https://en.wikipedia.org/wiki/Brine | Jump to content
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Contents
(Top)
1 In nature
2 Uses
2.1 Iodine and bromine mining
2.2 Lithium and magnesium mining
2.3 Chlorine production
2.4 Refrigerating fluid
2.5 Water softening and purification
2.6 Culinary
2.7 De-icing
2.8 Quenching
3 Desalination
3.1 Characteristics
3.2 Dissolved chemicals
3.3 Heavy metals
3.4 Discharge
3.5 Marine environment
3.6 Mitigation measures
3.7 Regulation
4 Wastewater
5 Composition and purification
6 See also
7 References
Brine
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From Wikipedia, the free encyclopedia
Concentrated solution of salt in water
For other uses, see Brine (disambiguation).
| |
| Part of a series on |
| Water salinity |
| Salinity levels |
| Fresh water (< 0.05%)Brackish water (0.05–3%)Saline water (3–5%)Brine (> 5% up to 26%–28% max) |
| Bodies of water |
| Seawater Salt lake Hypersaline lake Salt pan Brine pool Bodies by salinity |
| v t e |
Brine (or briny water) is a high-concentration solution of salt (typically sodium chloride or calcium chloride) in water. In diverse contexts, brine may refer to the salt solutions ranging from about 3.5% (a typical concentration of seawater, on the lower end of that of solutions used for brining foods) up to about 26% (a typical saturated solution, depending on temperature). Brine forms naturally due to evaporation of ground saline water but it is also generated in the mining of sodium chloride. Brine is used for food processing and cooking (pickling and brining), for de-icing of roads and other structures, and in a number of technological processes. It is also a by-product of many industrial processes, such as desalination, so it requires wastewater treatment for proper disposal or further utilization (fresh water recovery).
In nature
[edit]
Main article: Saline water
Brines are produced in multiple ways in nature. Modification of seawater via evaporation results in the concentration of salts in the residual fluid, a characteristic geologic deposit called an evaporite is formed as different dissolved ions reach the saturation states of minerals, typically gypsum and halite. Dissolution of such salt deposits into water can produce brines as well. As seawater freezes, dissolved ions tend to remain in solution resulting in a fluid termed a cryogenic brine. At the time of formation, these cryogenic brines are by definition cooler than the freezing temperature of seawater and can produce a feature called a brinicle where cool brines descend, freezing the surrounding seawater.
The brine cropping out at the surface as saltwater springs are known as "licks" or "salines". The contents of dissolved solids in groundwater vary highly from one location to another on Earth, both in terms of specific constituents (e.g. halite, anhydrite, carbonates, gypsum, fluoride-salts, organic halides, and sulfate-salts) and regarding the concentration level. Using one of several classification of groundwater based on total dissolved solids (TDS), brine is water containing more than 100,000 mg/L TDS. Brine is commonly produced during well completion operations, particularly after the hydraulic fracturing of a well.
Uses
[edit]
Iodine and bromine mining
[edit]
Iodine, essential for human health, is obtained on a commercial scale from iodide-rich brines. The purification begins by converting iodide to hydroiodic acid, which is then oxidized to iodine using chlorine. The iodine is then separated by evaporation or adsorption. Bromine is also obtained from brines. Akin to the production of iodine, the process exploits the easy oxidation of bromide into bromine, again using chlorine as the oxidant. The product bromine can be selectively collected by exploiting its volatility.
Lithium and magnesium mining
[edit]
Major deposits of lithium are in the form of brines. Magnesium is also produced in part from waste brine from various sources, such as potash production. Crude magnesium oxides and chlorides mixtures are converted into magnesium metal by electrolysis.
Chlorine production
[edit]
Main article: Chlorine production
Elemental chlorine can be produced by electrolysis of brine (NaCl solution). This process also produces sodium hydroxide (NaOH) and hydrogen gas (H2). The reaction equations are as follows:
Cathode: 2 H+ + 2 e− → H2 ↑
Anode: 2 Cl− → Cl2 ↑ + 2 e−
Overall process: 2 NaCl + 2 H2O → Cl2 + H2 + 2 NaOH
Refrigerating fluid
[edit]
Brine (primarily cheap brines based on calcium chloride and sodium chloride) is used as a secondary fluid in large refrigeration installations to transport thermal energy. It is used because the addition of salt to water lowers the freezing temperature of the solution, significantly enhancing its heat transport efficiency at low cost. The lowest freezing point obtainable for NaCl brine (called its eutectic point) is −21.1 °C (−6.0 °F) at the concentration of 23.3% NaCl by weight.
Because of their corrosive properties, salt-based brines have been replaced by organic liquids such as ethylene glycol.
Sodium chloride brine spray is used on some fishing vessels to freeze fish. The brine temperature is generally −5 °F (−21 °C). Air blast freezing temperatures are −31 °F (−35 °C) or lower. Given the higher temperature of brine, the system efficiency over air blast freezing can be higher. High-value fish usually are frozen at much lower temperatures, below the practical temperature limit for brine.
Water softening and purification
[edit]
Brine is used for regeneration of ion-exchange resins. After treatment, ion-exchange resin beads saturated with calcium and magnesium ions from the treated water, are regenerated by soaking in brine containing 6–12% NaCl. The sodium ions from brine replace the calcium and magnesium ions on the beads.
Culinary
[edit]
Main article: Brining
Brine is a common agent in food processing and cooking. Brining is used to preserve or season the food. Brining can be applied to vegetables, cheeses, fruit and some fish in a process known as pickling. Meat and fish are typically steeped in brine for shorter periods of time, as a form of marination, enhancing its tenderness and flavor, or to enhance shelf period.
De-icing
[edit]
In lower temperatures, a brine solution can be used to de-ice or reduce freezing temperatures on roads.
Quenching
[edit]
Quenching is a heat-treatment process when forging metals such as steel. A brine solution, along with oil and other substances, is commonly used to harden steel. When brine is used, there is an enhanced uniformity of the cooling process and heat transfer.
Desalination
[edit]
The desalination process consists of the separation of salts from an aqueous solution to obtain fresh water from a source of seawater or brackish water; and in turn, a discharge is generated, commonly called brine.
Characteristics
[edit]
The characteristics of the discharge depend on different factors, such as the desalination technology used, salinity and quality of the water used, environmental and oceanographic characteristics, desalination process carried out, among others. The discharge of desalination plants by seawater reverse osmosis (SWRO), are mainly characterized by presenting a salinity concentration that can, in the worst case, double the salinity of the seawater used, and unlike of thermal desalination plants, have practically the same temperature and dissolved oxygen as the seawater used.
Dissolved chemicals
[edit]
The discharge could contain trace chemical products used during the industrial treatments applies, such as antiscalants, coagulants, flocculants which are discarded together with the discharge, and which could affect the physical-chemical quality of the effluent. However, these are practically consumed during the process and the concentrations in the discharge are very low, which are practically diluted during the discharge, without affecting marine ecosystems.
Heavy metals
[edit]
The materials used in SWRO plants are dominated by non-metallic components and stainless steels, since lower operating temperatures allow the construction of desalination plants with more corrosion-resistant coatings. Therefore, the concentration values of heavy metals in the discharge of SWRO plants are much lower than the acute toxicity levels to generate environmental impacts on marine ecosystems.
Discharge
[edit]
The discharge is generally dumped back into the sea, through an underwater outfall or coastal release, due to its lower energy and economic cost compared to other discharge methods. Due to its increase in salinity, the discharge has a greater density compared to the surrounding seawater. Therefore, when the discharge reaches the sea, it can form a saline plume that can tends to follow the bathymetric line of the bottom until it is completely diluted. The distribution of the salt plume may depend on different factors, such as the production capacity of the plant, the discharge method, the oceanographic and environmental conditions of the discharge point, among others.
Marine environment
[edit]
Brine discharge might lead to an increase in salinity above certain threshold levels that has the potential to affect benthic communities, especially those more sensitive to osmotic pressure, finally having an effect on their abundance and diversity.
However, if appropriate mitigation measures are applied, the potential environmental impacts of discharges from SWRO plants can be correctly minimized. Some examples can be found in countries such as Spain, Israel, Chile or Australia, in which the mitigation measures adopted reduce the area affected by the discharge, guaranteeing a sustainable development of the desalination process without significant impacts on marine ecosystems. When noticeable effects have been detected on the environment surrounding discharge areas, it generally corresponds to old desalination plants in which the correct mitigation measures were not implemented. Some examples can be found in Spain, Australia or Chile, where it has been shown that saline plumes do not exceed values of 5% with respect to the natural salinity of the sea in a radius less than 100 m from the point of discharge when proper measures are adopted.
Mitigation measures
[edit]
The mitigation measures that are typically employed to prevent negatively impacting sensitive marine environments are listed below:
A well-designed discharge mechanism, employing the use of efficient diffusers or pre-dilution of discharges with seawater
An environmental evaluation study, which assesses the correct location of the discharge point, considering geomorphological and oceanographic variables, such as currents, bathymetry, and type of bottom, which favor a rapid mixing process of the discharges;
The implementation of an adequate environmental surveillance program, which guarantees the correct operation of the desalination plants during their operational phase, allowing an accurate and early diagnostics of potential environmental threats
Regulation
[edit]
Currently, in many countries, such as Spain, Israel, Chile and Australia, the development of a rigorous environmental impact assessment process is required, both for the construction and operational phases. During its development, the most important legal management tools are established within the local environmental regulation, to prevent and adopt mitigation measures that guarantee the sustainable development of desalination projects. This includes a series of administrative tools and periodic environmental monitoring, to adopt preventive, corrective and further monitoring measures of the state of the surrounding marine environment.
Under the context of this environmental assessment process, numerous countries require compliance with an Environmental Monitoring Program (PVA), in order to evaluate the effectiveness of the preventive and corrective measures established during the environmental assessment process, and thus guarantee the operation of desalination plants without producing significant environmental impacts. The PVAs establishes a series of mandatory requirements that are mainly related to the monitoring of discharge, using a series of measurements and characterizations based on physical-chemical and biological information. In addition, the PVAs could also include different requirements related to monitoring the effects of seawater intake and those that may potentially be related to effects on the terrestrial environment.
Wastewater
[edit]
Main article: Industrial wastewater treatment § Brine treatment
Brine is a byproduct of many industrial processes, such as desalination, power plant cooling towers, produced water from oil and natural gas extraction, acid mine or acid rock drainage, reverse osmosis reject, chlor-alkali wastewater treatment, pulp and paper mill effluent, and waste streams from food and beverage processing. Along with diluted salts, it can contain residues of pretreatment and cleaning chemicals, their reaction byproducts and heavy metals due to corrosion.
Wastewater brine can pose a significant environmental hazard, both due to corrosive and sediment-forming effects of salts and toxicity of other chemicals diluted in it.
Unpolluted brine from desalination plants and cooling towers can be returned to the ocean. From the desalination process, reject brine is produced, which proposes potential damages to the marine life and habitats. To limit the environmental impact, it can be diluted with another stream of water, such as the outfall of a wastewater treatment or power plant. Since brine is heavier than seawater and would accumulate on the ocean bottom, it requires methods to ensure proper diffusion, such as installing underwater diffusers in the sewerage. Other methods include drying in evaporation ponds, injecting to deep wells, and storing and reusing the brine for irrigation, de-icing or dust control purposes.
Technologies for treatment of polluted brine include: membrane filtration processes, such as reverse osmosis and forward osmosis; ion exchange processes such as electrodialysis or weak acid cation exchange; or evaporation processes, such as thermal brine concentrators and crystallizers employing mechanical vapour recompression and steam. New methods for membrane brine concentration, employing osmotically assisted reverse osmosis and related processes, are beginning to gain ground as part of zero liquid discharge systems (ZLD).
Composition and purification
[edit]
Brine consists of concentrated solution of Na+ and Cl− ions. Other cations found in various brines include K+, Mg2+, Ca2+, and Sr2+. The latter three are problematic because they form scale and they react with soaps. Aside from chloride, brines sometimes contain Br− and I− and, most problematically, sulfate SO2−4. Purification steps often include the addition of calcium oxide to precipitate solid magnesium hydroxide together with gypsum (CaSO4), which can be removed by filtration. Further purification is achieved by fractional crystallization. The resulting purified salt is called evaporated salt or vacuum salt.
See also
[edit]
Brine mining – Extracting materials from saltwater
Brinicle – Sea ice formation
Brine pools – Anoxic pockets of high salinity on the ocean bottom
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[edit]
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^ 5. A. Giwa, V. Dufour, F. Al Marzooqi, M. Al Kaabi, S.W. Hasan, "Brine management methods: Recent innovations and current status", Desalination, Volume 407, 2017, pp. 1–23
^ "Reverse Osmosis Desalination: Brine disposal". Lenntech. Retrieved 18 July 2017.
^ "Novel Technology for Concentration of Brine Using Membrane-Based System" (PDF). Water Today. Retrieved 31 August 2019.
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CBSE Math
The Pythagorean Theorem
The Pythagorean Theorem
Concept Summary:
The Pythagorean Theorem defines the relationship between the three sides of a right-angled triangle, stating that the square of the hypotenuse c is equal to the sum of the squares of the other two sides a and b: c 2=a 2+b 2.
The converse of the Pythagorean Theorem states that if c 2=a 2+b 2,then the triangle is a right triangle.
Pythagorean triples are sets of three whole numbers that satisfy the Pythagorean Theorem, such as (3, 4, 5). Multiples of Pythagorean triples are also Pythagorean triples.
If a 2+b 2>c 2,the triangle is acute.
If a 2+b 2<c 2,the triangle is obtuse.
Ask your own question
Which of the following sets are Pythagorean triples? (8,15,17) (1, √3, 2) (9,12,16) (8,11,14) (20,21,29) (30,40,50)
Write the correct answer in the gap. 25+6= ____
Triangle ABC has side lengths 10, 24, and 26. Do the side lengths form a Pythagorean triple? Explain.
Graph each pair of points. Then, find the distance between the points (4,2) and (-3,4).
A popular school is located at the coordinates (3, 5) where x and y are measured in miles. Children who live in houses that are 4 or more miles from the school are allowed to ride the bus. Suppose children live in House A, House B, House C, and House D. Determine which of the houses are located 4 or more miles from the school. House C is located at the coordinates (− 4, − 2). How far is the house from the school in miles?
A popular school is located at the coordinates (3, 5) where x and y are measured in miles. Children who live in houses that are 4 or more miles from the school are allowed to ride the bus. Suppose children live in House A, House B, House C, and House D. Determine which of the houses are located 4 or more miles from the school. House D is located at the coordinates (5, 8). How far is the house from the school in miles?
Two football players are located at points [A] and [B] in a rectangular football field as shown. Point [A] is located yards [(yd)] from the east edge and [25 , yd] from the south edge; Point [B] is located [12 , yd] from the east edge and [0 , yd] from the south edge. What is the distance, in yards, between the two players? (Round your answer to the nearest tenth of a yard.)
The students' houses in grade 10 Lapu-Lapu are mapped on a coordinate grid with the origin being at the school which is located at (3,4). Angela's house is at (5,5), Kent's is at (2,3), Johan's is at (-4,-7), and Michael's is at (8,2). Who among the four students is nearest to the school?
Imagine the triangle described by the points ( − 5 , 3 ) , ( x , 3 ) , and ( x , y ) where x > − 5 and y > 3 . Write an expression for the length of the horizontal leg in terms of x . Write an expression for the length of the vertical leg in terms of y . Write an expression for the length of the hypotenuse in terms of x and y .
What is the correct formulation of the distance formula for the coordinates (−32,−61) and (−21,−33)?
What is the distance between point A(3a,2a) and point D(a,6a)?
What is the distance between point A (3a, 2a) and point D (a, -6a)?
What is the distance between point A(3a,2a) and point D(-a,6a)?
What is the distance between point A(3a,2a) and point D(-a,-6a)?
Rosa is looking at a map of the park that is laid out on a coordinate plane. Rosa is at (1, –1). The shelter house is at (–2, –4) and the fossil exhibit is at (3, 2). Each unit on the map represents 100 feet. How much closer is Rosa to the fossil exhibit than to the shelter house? Round to the nearest foot.
Square the following numbers. Options: a. 4 b. 6 c. 13 d. 10
Find the distance between the point (9.4,2.6) and (9.4,-9).
Triangle RST is located at the ordered pairs R (-8,1), S (3,4) and T (1,-2). Calculate the length and slope of each side of Triangle RST. Use your work from the previous questions to justify whether or not triangle RST is a right triangle.
What is the distance between points A(-4,1) and B(-1,1) in the polygon with vertices A(-4,1), B(-1,1), C(-1,-3), and D(-4,-3)?
Given two points (0,3) and (-3,-3), can you find the distance between them and their midpoint?
Find the distance between the following pairs of point: (5,1) and (5,-4).
Draw the triangle described by the points ( 5 , 3 ) , ( 5 , 7 ) , and ( - 4 , 7 ). What is the length of the horizontal leg? What is the length of the vertical leg? What is the length of the hypotenuse?
Find the distance between the following pairs of points: (4, 6) and (-3, 7).
Find the distance between the pair of points (9/2,1/7) and (1/2,8/7). If necessary, express the answer in simplified radical form and then round to two decimal places.
Find the distance between the points (-6,-10) and (9,-10).
Find the distance between the following pairs of points: (-2,-4), (3,-2).
Is the distance between the points (0, 6) and (2, 3) equal to 5 √73 or √13? True/False
Find the distance between the following pairs of points: (-6,2) and (-8,2).
What is the length of the shortest leg of a right triangle whose vertices are at (-4,2), (5,5), and (5,2)?
Find the distance between the pair of points. If necessary, express answers in simplified radical form and then round to two decimal places. (0,0) and (-5,12)
Find the distance between the pair of points (-4,-6) and (4,-5). If necessary, express answers in simplified radical form and then round to two decimal places.
Find the distance between the points (-3, 2) and (4, -5). Is it equal to √10, 7√2, or 2√2? Fill in the blank.
Triangle STU has vertices at S( – 5,5), T(5, – 7), and U( – 7, – 6). Is triangle STU an isosceles right triangle? True/False.
The sides of a triangle have lengths 10, 18, and 20. What kind of triangle is it? Is it acute, obtuse or right?
Using the distance formula, d = √((x2 - x1)2 + (y2 - y1)2), what is the distance between point (-2, 2) and point (2, -3) rounded to the nearest tenth?
The sides of a triangle have lengths 43, 39, and 9. This triangle is an obtuse triangle. True/False
Find the distance between the pair of points (-4,-4) and (4,-3). If necessary, express the answer in simplified radical form and then round it to two decimal places. Is this true or false?
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3115 | https://www.math.unipd.it/~monti/tesi/Bosio.pdf | Department of Pure and Applied Mathematics Bachelor’s Degree in Mathematics Maximal Function and Lebesgue Differentiation Theorem Supervisor: Prof. Monti Roberto Candidate: Bosio Niccolò Academic year 2020/2021 Quote: Any good idea can be stated in fifty words or less.
(Stanislaw M. Ulam, "Adventures of a Matematician") Contents Introduction 5 1 The study of Lebesgue Differentiation Theorem on Rn with Lebesgue measure 7 2 The study of Lebesgue Differentiation Theorem on doubling metric spaces 19 3 The study of Lebesgue Differentiation Theorem on Rn with Radon measures 23 Conclusion 31 Bibliography 33 3 Introduction This thesis develops around Lebesgue’s Differentiation Theorem, progres-sively trying to generalize its statement.
The first chapter will focus on Rn with the lebesgue measure, here the ’Max-imal function’ will be introduced and its properties will be studied.
The differentiation theorem, in fact, will descend as a corollary from the theorem of the Maximal function.
Subsequently we will treat Vitali’s cover lemma, necessary for previous proofs, and the last part of the first chapter will end with some application of the theorem, in particular the Marcinkiewicz integral will be introduced and some properties of this function will be studied.
At this point it is natural to wonder if the Rn environment is necessary to prove the previous theorems; in the second chapter we will try to answer this question by showing that in some metric spaces equipped with regular Borel measures, it is possible to show the same theorems as the first chapter using a cover lemma very similar to vitali’s one.
Finally, in the last chapter, the study of the differentiation theorem in Rn with measures other than that of Lebesgue will be deepened. We will show that it is possible to generalize the previous concepts but not for free, in fact it is necessary to narrow yourself to metric balls and not to general families of regular sets as in the first two cases.
The latter section will close with a study of the cover lemma used to show the third variant of the theorem: the Besicovitch cover lemma.
5 Chapter 1 The study of Lebesgue Differentiation Theorem on Rn with Lebesgue measure In this chapter we are going to study the properties of the maximal func-tion and the applications on Differentiation Theorem. We will analyze the behaviour of this function in Rn with the Lebesgue measure.
Let’s start with some definitions: Definition 1.1. We define the maximal function as: M(f)(x) = sup r>0 1 m(Br(x)) Z B |f(y)| dy.
(1.1) It is to be noticed that nothing excludes the possibility that M(f)(x) is infinite for any given x.
Definition 1.2. Let be λ(α) = m(x | |g(x)| > α). This is called the distri-bution function of g with g a measurable function.
Obviously we are interested in studying the properties of this function when is defined on a non bounded set. This function describes the relative largeness of the function g: it is very useful for many applications and for the proof of the principal theorem of this chapter.
Observation 1.3. It’s useful to notice that Z Rn |g(y)|p dy = p Z ∞ 0 αp−1 λ(α) dα.
(1.2) We will use this equality during the proof of Theorem 1.5.
7 8CHAPTER 1. THE STUDY OF LEBESGUE DIFFERENTIATION THEOREM ON RN WITH Proof.
Z Rn |g(y)|p dy = Z Rn Z |g(y)|p 0 1 dt !
dy = Z Rn Z ∞ 0 χ0,|g(y)p| dt dy = = Z ∞ 0 Z Rn χ{y∈Rns.a.|g(y)|p>t} dy dt = Z ∞ 0 λ(t 1 p )dt.
Now we use this replacement: t 1 p = α, dt = pαp−1dα˙ The result is: p Z ∞ 0 αp−1 λ(α) dα.
Observation 1.4. If g ∈L∞we can claim that ∥g∥∞= inf{α | λ(α) = 0} and Z Rn |g(y)| dy ≥αλ(α).
(1.3) Proof. Te first equality comes from the definition of ∥g∥∞. Let’s prove the 1.3 inequality: Z Rn |g(y)| dy ≥ Z {x | g(x)>α} |g(y)|dy ≥ Z {x | g(x)>α} α dy = α λ(α) After these small observations we can introduce the first important the-orem of this chapter: Theorem 1.5. Let f : Rn − →R and A a constant depending on p and n a) If f ∈Lp(Rn); 1 ≤p ≤∞we have that: M(f) < ∞a.e.
b) If f ∈L1(Rn) we have that forall α > 0 : m{y | M(f)(x) > α} ≤A α Z Rn |f(x)| dx.
c) If f ∈Lp(Rn); 1 < p ≤∞we have that: M(f) ∈Lp(Rn) and ∥M(f)∥p ≤A ∥f∥p.
9 Observation 1.6. In c), when p=1 and f ̸= 0, we can’t claim that M(f) ∈ L1(Rn): We can observe that M(f)(x) = sup r>0 1 m(Br(x) Z B |f(y)| dy ≥ C |x|n ; ∀|x| ≥1 where C is a constant depending on ∥f∥1 and |x|n comes from the measure of the n-dimensional ball. Now if we try to computing ∥M(f)∥1: ∥M(f)∥1 ≥ Z Rn C |x|n dnx and passing by the n-dimensional spherical coordinates we found: Z ∞ 0 K∗ rn rn−1 dr = +∞ so we have to ask stronger conditions to f than the integrability to be M(f) integrable.
()=K is a constant that comes from the change of variables rule and C, the prior constant.
Observation 1.7. The result obtained in b) is the best possible estimator: if we consider f = δ0 we have: M(f)(x) ≥ 1 m(B|x|(x)) Z B δ0(y) dy = 1 m(B) = 1 C |x|n .
In this case λ(α) = m{x | 1 |x|n > C α} = m(B 1 C α )(0) = V ol(B1(0)) 1 Cnαn = 1 Cn−1αn .
Now we put n=1 and we have concluded.
Now we can prove the Theorem 1.5.
Proof. b) We define Eα = {x | M(f)(x) > α} so forall x ∈Eα exists Br(x) such that: Z B |f(y)| dy ≥ Z B α dx ≥α m(Br(x)).
This implies: m(B) ≤∥f∥1 α .
Thanks to the Vitali’s 5 covering lemma we can extract some disjointed balls such that ∞ X k=0 mBk ≥Cm(Eα).
10CHAPTER 1. THE STUDY OF LEBESGUE DIFFERENTIATION THEOREM ON RN WIT So ∥f∥1 ≥ Z ∪Bk |f(y)| dy ≥α ∞ X k=0 m(Bk) ≥α C m(Eα) Let A = 1 C then we find ∥f∥1 A α ≥m(Eα) = λ(α).
Now let’s prove a) and c): If f ∈L∞then sup r>0 1 m(Br(x)) Z B |f(y)| dy ≤∥f∥∞ m(B) m(B) = ∥f∥∞ and so ∥M(f)∥∞= ∥f∥∞.
Now we consider p ∈(1; ∞) : Let be f1(x) = ( f(x) if |f(x)| > α 2 0 otherwise It’s obvious that |f(x)| ≤|f1(x)| + α 2 ; ∀x ∈R and M(f)(x) ≤M(f1 + α)(x) = M(f)(x) + α.
That implies: {x | M(f)(x) > α} ⊂{x | M(f1)(x) + α 2 > α} = {x | M(f1)(x) > α 2 } Now we can use point b): m(Eα) = m{x | M(f)(x) > α} ≤2 A ∥f1∥∞ α which mean m(Eα) ≤2 A α Z {x | |f(x)|> α 2 } |f(x)| dx.
Let g = M(f) and λ the distribution function of g: Z Rn |g(x)|p dx = p Z ∞ 0 αp−1 λ(α) dα 11 (Observation 1.3).
So ∥g∥p p =p Z ∞ 0 αp−1 m(Eα) dα ≤p Z ∞ 0 αp−1 2 A α Z |f|> α 2 |f(x)| dx !
dα = =p Z ∞ 0 αp−1 2 A α Z Rn |f(x)|χ|f|> α 2 (x) dx dα = ∗ =p 2 A Z Rn |f(x)| Z |f|> α 2 αp−2 dα !
dx = p 2 A Z Rn |f(x)|(2|f(x)|)p−1 p −1 dx = =2p A p p −1 Z Rn |f(x)|p dx = Ap ∥f∥p.
=In this passage we used Tonelli-Fubini.
So we managed to prove that: ∥M(f)∥p p ≤A ∥f∥p p.
There is a very important corollary of this theorem called the Lebesgue differentiation Theorem: Corollary 1.8. If f ∈Lp(Rn), 1 ≤p ≤∞; or if f is locally integrable then: lim r→0 1 m(Br)(x) Z B f(y)dy = f(x) (1.4) Proof. Let define fr(x) = 1 m(Br(x)) Z B f(y) dy; r > 0.
It’s easy to prove that: r →0 ⇒∥fr −f∥p →0.
We know that C0 c is dense in Lp so we can write f = f1 + f2 with f1 ∈C0 c and ∥f2∥p < ϵ, forall ϵ ∈R.
Define ∆(y) = ∥f(x −y) −f(x)∥p and writing ∆(y) with f = f1 + f2 we find ∆≤∆1 + ∆2.
Notice that ∆1 →0 for y →0 as an immediate consequence of the uniform convergence (f1 is continuous with a compact support) and ∆2 ≤2ϵ so 12CHAPTER 1. THE STUDY OF LEBESGUE DIFFERENTIATION THEOREM ON RN WIT ∆(y) →0 for y →0.
Let be φr(y) = 1 m(Br(x)) χB(y) and notice that Z Rn φr(y) dy = 1.
Now we have fr(x −y) = Z B f(x −y) 1 m(B)dy = Z Rn f(x −y)χB(y) m(B) dy = = Z Rn f(x −y)φr(y) dy = (f ∗φr)(x).
Clearly f ∗φr −f = Z Rn[f(x −y) −f(x)]φr(y) dy, because Z Rn φr(x) dx = 1.
So ∥fr −f∥p =∥f ∗φr −f∥p ≤ Z Rn ∆(y)|φr(y)| dy = = Z Rn ∆(ry)|φ1(y)| dy →0 for r →0 where in the first inequality we used Minkowsky and in the last limit we used the Lebesgue dominated convergence theorem.
What remains to be seen is that limr→0 fr(x) exists almost everywhere: for this purpose we denote forall g ∈L1 and x ∈Rn, Ωg(x) = | lim sup r→0 gr(x) −lim inf r→0 gr(x)|.
(We reduce the consideration to the case p=1).
If g is continuous with compact support, then gr→ →g so Ωg = 0.
If g ∈L1 we can use b) of Theorem 1.5: m{x| 2M(g)(x) > ϵ} ≤2 A ϵ ∥g∥1.
Clearly Ωg ≤2M(g), thus m{x| Ωg(x) > ϵ} ≤2 A ϵ ∥g∥1.
13 Figure 1.1: this is the limit case where diam(Bi) = 1 2 diam(Bj) and they are tangent Finally we can write g as g1 + g2 with g1 ∈C1 c and g2 such that ∥g2∥1 < ϵ2 so m{x| Ωg(x) > ϵ} ≤2 A ϵ ϵ2 →0 for ϵ →0.
Now we must study the Vitali’s 5 covering Lemma that we used for the proof of the Theorem 1.5 and we will explain why the constant A of the 1.5 Theorem is 2( 5n p p−1) 1 p .
Lemma 1.9. Let E be a measurable subset of Rn which is covered by the union of a family of balls {Bi}, of bounded diameter. Then from this family we can select a disjointed subsequence, B1, ..., Bn, ... (finite or infinite) such that: P n m(Bn) ≥Cm(E).
Proof. Let’s take B1; ...; Bk such that diam(Bk+1) ≥1 2{sup(d(Bj) | Bj is disjointed with B1; ...; Bk)}.We call {1; ...; k} = K.
This sequence could be finite or infinite: if it’s finite then it’s impossible to find Bk+1 disjointed with B1; ...; Bk so we have to show that if we define B∗ i = 5Bi forall i ∈{1; ...; k} we find S i≤k B∗ i ⊃E.
Let’s show that forall j ∈N Bj ⊂∪i≤kB∗ i : We know that forall Bj exists i ≤k such that diam(Bi) ≥1 2 diam(Bj) and Bi ∩Bj ̸= ∅, otherwise we would have taken Bj in K.
So 5Bi ⊃Bj.
The same proof works for an infinite set of index K with P k m(Bk) < ∞; if it’s equal to ∞the proof is trivial.
14CHAPTER 1. THE STUDY OF LEBESGUE DIFFERENTIATION THEOREM ON RN WIT Figure 1.2: we multiplied the radius of the small ball by 5.
Now a legitimate question is: what if we try to apply the theorem 1.5 with a different family of sets? We will see that it works if the family has a properties: Definition 1.10. A family of sets F is regular if forall S ∈F exists Br(0) such that S ⊂Br(0) and m(S) ≥C m(B).
Definition 1.11. We can define a different maximal function: MF(f(x)) = sup S∈F 1 m(s) Z S |f(x −y) dy.
(1.5) Observation 1.12. MF(f(x)) ≤C−1M(f(x)) so MF satisfies the same conclusions of Theorem 1.5.
In particular, if f is locally integrable we have: lim S∈F; m(S)→0 1 m(S) Z S f(x −y) dy = f(x) (1.6) Proof. We know that exists Br(0) such that B ⊃S and C m(B) ≤m(S) so sup S∈F 1 m(S) Z S |f(x −y)| dy ≤sup S∈F 1 C m(B) Z S |f(x −y)| dy ≤ ≤ 1 C m(B) Z B |f(y)| dy ≤C−1 M(f(x)).
Observation 1.13. The equation 1.4 is true almost everywhere but the exeptional set where 1.4 is not valid depends on F. Our goal is to find the exeptional set depending on f; we consider the relation: lim r→0 1 m(Br(0)) Z B |f(y) −c| dy = |f(x) −c| (1.7) 15 It’s valid almost everywhere except on a set Ec such that m(Ec) = 0.
Now forall c ∈Q be E = S c∈Q Ec : obviously m(E) = 0 and forall x / ∈E the relation 1.5 works.
Definition 1.14. An element x ∈E is called "point of density" of E if: lim r→0 m(E ∩Br(0)) m(Br(0)) = 1 (1.8) Observation 1.15. If we apply (1.5) to x "point of density" of E with f = χE we find: lim r→0 1 m(Br(0)) Z B χE(y) dy = χE(x) = 1 = lim r→0 m(E ∩Br(0)) m(Br(0)) (1.9) Proposizione 1.16. For almost every x ∈E the limit (1.5) holds then almost every x ∈E is a point of density.
From now we will consider E = E and it’s not restrictive because m is a regular measure.
δ(x; F) will be the distance from x to F closed set. Obviously δ(x; F) = 0 ⇐ ⇒x ∈F; ∀x ∈F, δ(x + y; F) ≤|y|; ∀ϵ > 0, ∃η | |y| < η such that δ(x + y; F) < ϵ|y|.
Proposizione 1.17. Let F a closed set. For almost every x ∈F δ(x + y; F) = o(|y|).
Proof. Let x ∈F be point of density.
Obviously B(x + y; ϵ |y|) ⊂B(x; |y| + ϵ |y|).
We claim that exists z ∈F such that z ∈B(x + y; ϵ|y|) if |y| is ’small’.
Otherwise: m(F ∩B(x; |y| + ϵ |y|)) m(B(x; |y| + ϵ |y|)) ≤m(B(x; |y| + ϵ |y|)) −m(B(x + y; ϵ |y|)) m(B(x; |y| + ϵ |y|)) ≤1− ϵ 1 + ϵ n ̸= 1 so x is not a point of density. ⊥ Thus exists z ∈F such that z ∈B(x + y; ϵ |y|) forall ϵ small, therefore: δ(x + y; F) ≤δ(x + y; z) ≤ϵ |y|.
The last thing that we present in this chapter is the Marcinkiewicz inte-gral: I(x) = Z {|y|≤1} δ(x + y; F) |y|n+1 dy (1.10) 16CHAPTER 1. THE STUDY OF LEBESGUE DIFFERENTIATION THEOREM ON RN WIT Figure 1.3 Theorem 1.18. Let I(x) be the Marcinkiewicz integral and F a closed set in Rn.
a) If x ∈F c then I(x) = ∞.
b) For almost every x ∈F : I(x) < ∞ Proof. a) Obviously exists c > 0 such that δ(x + y) > c; this implies I(x) > c Z Rn 1 |y|n+1 dy = ∞ b) this result is a simple consequence of the following lemma.
Lemma 1.19. Let F be a closed set whose complement has finite measure.
Let I∗(x) = Z Rn δ(x + y) |y|n+1 dy.
(1.11) Then I∗(x) < ∞for almost every x ∈F. Moreover: Z F I∗(x) dx ≤c m(F c).
(1.12) Proof. Clearly is sufficient to prove (1.10) since the integrand is positive: Z F I∗(x) dx = Z F Z Rn δ(x + y) |y|n+1 dy dx = Z F Z Rn δ(y) |x −y|n+1 dy dx = = Z F Z F c δ(y) |x −y|n+1 dy dx = Z F c δ(y)( Z F dx |x −y|n+1 ) dy.
Let’s consider the integral on F: y ∈F c implies |x −y| ≥δ(y) thus: Z F I∗(x) dx ≤ Z |x|≥δ(y) dx |x|n+1 = Z ∞ δ(y) c rn−1 rn+1 dr = c δ(y).
17 So Z F I∗(x) dx ≤ Z F c c δ(y) δ(y)−1 dy = c m(F c).
Now we can finish the proof of the theorem 1.18: Fm = F ∪Bm(0)c.
Fm is closed and (F ∪Bc m)c ⊂Bm.
We can apply the lemma 1.19 to Fm with δm the distance from Fm.
Forall F exists m > 0 such that δ(x+y) = δm(x+y) if |y| ≤1 and x ∈Bm−2.
For the lemma 1.19: I(x) < ∞for almost every x ∈Bm−2 ∩F.
Chapter 2 The study of Lebesgue Differentiation Theorem on doubling metric spaces In this chapter we will redefine the structures as in the previous chapter but generalizing the definition of Maximal Function on some particular metric spaces: doubling metric spaces. Let’s start defining the properties of the measure that we need for this chapter: Definition 2.1. Let µ be a outer measure on (X, d) metric space.
It’s called a Borel regular measure if: 1) For every B Borel set and for every A ⊂X we have: (Borel condition) µ(A) = µ(A ∩B) + µ(A ∩Bc).
2) For every set A ⊂X there exists an open set B ⊃A such that: (regular condition from above) µ(A) = µ(B).
3) For every set A ⊂X there exists a compact set K ⊂A such that: (regular condition from below) µ(A) = µ(K).
Definition 2.2. Let (X, d, µ) be a metric measure space, it is said to be doubling if there exists a constant C > 0 such that: 0 < µ(B2r(x)) ≤C µ(Br(x)) ∀x ∈X, r ∈R.
(2.1) Looking at the properties of the Lebesgue measure that we used for the Vitali’s covering lemma we notice the importance of the previous property of m: we said that if we multiply the diameter of the selected balls for 5 we found 19 20CHAPTER 2. THE STUDY OF LEBESGUE DIFFERENTIATION THEOREM ON DOUBL that every balls is contained in this new set and this implied the thesis: P n m(Bn) ≥Cm(E).
It’s looks like very trivial that m(B5r(x)) = C m(Br(x)) but if we change measure and we chose µ as a non positive measure we could even find µ(B5r(x)) < µ(Br(x)). So if finding C such that 0 < µ(B2r(x)) ≤C µ(Br(x)) is impossible, the proof of the Vitali’s covering theorem falls apart.
Now we can define the maximal function in (X, µ) doubling space with µ regular and Borel measure.
Definition 2.3. Let f ∈Lp(X), we define the maximal function of f: M(f)(x) = sup δ>0 1 µ(Bδ(x) Z Bδ(x) |f| dµ.
(2.2) From now we will use the following notation: sup δ>0 1 µ(Bδ(x) Z Bδ(x) |f| dµ = sup δ>0 Z \ B |f| dµ.
Observation 2.4. If X = Rn and M(f) is lower semi-continuous than M(f) is measurable but this claim is not true with a general metric space.
For this reason we define the following "decentralized" maximal function: ˜ M(f)(x) = sup B |x∈B Z \ B |f| dµ.
(2.3) We observe that ˜ M(f) ≥M(f) and ˜ M is lower semi-continuous, in fact {x| ˜ M(f)(x) > α} is open forall α ∈R.
Proof. Let’s prove that {x| ˜ M(f)(x) > α} is open: if x ∈{ ˜ M(f) > α} then ˜ M(f)(x) > α so we know that exists B such that R B |f| dµ > α now clearly if y ∈B than ˜ M(f)(y) > α thus B ⊂{ ˜ M(f) > α}.
Now we are ready to state the Maximal Function Theorem: Theorem 2.5. Let (X, d) be a doubling metric spaces and µ a regular Borel measure on X.
If p > 1 and f ∈Lp(X) then ˜ M(f) ∈Lp(X) and exists a constant Cp depending on p such that: ∥˜ M(f)∥p ≤Cp(X)∥f∥p.
(2.4) If p = 1 then: µ({ ˜ M(f) > α}) ≤Cp α ∥f∥1.
21 Before proving this theorem we need a new covering lemma, similar to the Vitali’s one but for compact sets.
Lemma 2.6. Let (X, d, µ) be a doubling metric space. Let K ⊂X a compact set and let {Bi | i ∈I} a finite cover of X. Then there exists a subcover { ¯ Bj | j ∈J ⊂I} such that: 1) ¯ Bj ∩¯ Bi = ∅for i ̸= j 2) K ⊂S 3 ¯ Bj Proof. Let be ¯ B1 ∈{Bi}i∈I such that diam( ¯ Bi) = max{diam(Bi)|i ∈I}.
Let be ¯ B2 such that ¯ B1 ∩¯ B2 = ∅and diam( ¯ B2) = max{diam(Bi)|i ∈I ̸= 1}.
(Notice that exists because the cover is finite.) With this method we define a subset of I.
Now it’s easy to see that if we consider A = Sp i=1 ¯ Bi every Bj, j ∈I is contained in A.
Now we can prove Theorem 2.5: Proof. 1) Let’s start with p = 1. We have to prove that if f ∈L1(X) : µ({ ˜ M(f) > α}) ≤C α ∥f∥1.
Let K be a subset of { ˜ M(f) > α}, thus: ∥f∥1 = Z X |f| dµ ≥ Z K |f| dµ ≥µ(K) α C.
The last inequality comes from lemma 2.6 and the regularity of µ: if x ∈K then ˜ M(f)(x) > α and so exists Bx such that: 1 µ(Bx) Z Bx |f| dµ > α = ⇒µ(Bx) < 1 α Z Bx |f| dµ.
So if we take {Bi|i ∈I} a cover of K and we extract { ¯ Bj|j ∈J ⊂I} with the properties of lemma 2.6: µ(K) ≤ X j∈J µ(3 ¯ Bj) ≤3 X j∈J µ( ¯ Bj) ≤3 α X j∈J Z ¯ Bj |f| dµ = 3 α Z ∪¯ Bj |f| dµ ≤C α ∥f∥1.
(2.5) Thanks to the regularity of µ we can conclude.
2) Now let’s consider p > 1 : We define f1 : X − →R such that: f1 = ( f(x) if |f(x)| > α 0 otherwise.
Clearly: |f| ≤|f1| + α; ∀x ∈X.
22CHAPTER 2. THE STUDY OF LEBESGUE DIFFERENTIATION THEOREM ON DOUBL Thanks to the monotony of ˜ M we can claim that ˜ M(f) ≤˜ M(f1) + ˜ M(α) = ˜ M(f1) + α This implies: { ˜ M(f) > 2α} = { ˜ M(f) −α > α} ⊂{ ˜ M(f1) > α}, so µ{ ˜ M(f) > 2α} ≤µ{ ˜ M(f1) > α} ≤(∗)C α Z X |f1| dµ = C α Z |f|>α |f| dµ ()= this equality comes from the series of inequalities 2.5.
Now we can prove 2.4: ∥˜ M(f)∥p p = Z X | ˜ M(f)|p dµ = p Z ∞ 0 αp−1 µ{x| ˜ M(f)(x) > α} dα ≤p Z ∞ 0 αp−1 C 2α Z |f|>α |f| dµ !
dα = p C 2 Z ∞ 0 αp−2 Z X |f|χ|f|>α dµ dα = p C 2 Z X |f| Z α<|f| αp−2 dα !
dµ = p C 2(p −1)∥f∥p p.
Chapter 3 The study of Lebesgue Differentiation Theorem on Rn with Radon measures In this third chapter we are going to study the same theorem but with a different type of measures: the Radon measures. Let’s start with a definition: Definition 3.1. A Radon measure defined on the σ-algebra of Borel sets is a regular measure that is finite on all of the compact sets of X.
Our goal is to study the differentiation of Radon measures and to do that we must define some functions: Definition 3.2. Let µ, ν be Radon measures on Rn.
For each point x ∈Rn, define: Dµν(x) ( lim supr→0 ν(B(x,r)) µ(B(x,r)); µ(B) > 0 ∀r > 0 +∞otherwise Dµν(x) = ( lim infr→0 ν(B(x,r)) µ(B(x,r)); µ(B) > 0∀r > 0 +∞otherwise Definition 3.3. If Dµν(x) = Dµν(x) < ∞we say that ν is differentiable with respect to µ at x and we write: Dµν(x) = Dµν(x) = Dµν(x) We will call Dµν the density of ν with respect to µ.
We want to study two things about the density function: 1) when exists, 2) when ν can be recovered by integrating Dµν. To reach this goal we must start with an important lemma: 23 24CHAPTER 3. THE STUDY OF LEBESGUE DIFFERENTIATION THEOREM ON RN WIT Lemma 3.4. Fix 0 < α < ∞: A ⊂{x ∈Rn | Dµν(x) ≤α} = ⇒ν(A) ≤αµ(A) (3.1) B ⊂{x ∈Rn | Dµν(x) ≥α} = ⇒ν(A) ≥αµ(A) (3.2) Proof. We can assume µ(Rn) and ν(Rn) finite, since we could otherwise consider these two measures restricted to compact subsets of Rn.
Fix ϵ > 0 and U open subset of {x ∈Rn | Dµν(x) ≤α} such as A ⊂U.
Let’s define F = {B | B = B(a, r), a ∈A, B ⊂U, ν(B) ≤(a + ϵ)µ(B)}.
We notice that inf{r | B(a, r) ∈F} = 0 ∀a ∈A so we can use a corollary of the Besicovitch covering theorem that assures us that exists a countable collection G of disjoint balls in F such that ν(A − [ B∈G B) = 0 Then: ν(A) ≤ X B∈G ν(B) ≤(α + ϵ) X B∈G µ(B) ≤(α + ϵ)µ(U).
We can conclude thanks to the regularity of ν and µ.
Now we can enunciate a theorem that answer to the first question we did after the definition 3.3 about the density function: Theorem 3.5. Let µ and ν be Radon measures on Rn. Than Dµν exists and is finite µ-almost everywhere. Furthermore, Dµν is measurable.
Proof. As before we can assume µ(Rn) and ν(Rn) < ∞.
1)Dµν exists and is finite µ a.e.: Let I = {x | Dµν(x) = ∞}, and for all 0 < a < b let R(a, b) = {x | Dµν(x) < a < b < Dµν(x) < ∞}.
Now we can apply the lemma 3.4 observing that for each α > 0, I ⊂ {x | Dµν(x) ≥α}. So we can claim that µ(I) ≤1 αν(I).
Sending α →∞we conclude that µ(I) = 0.
Now we must show that Dµν exists µ a.e.: Using lemma 3.4 we can claim the following inequalities: ν(R(a, b)) ≤aµ(R(a, b)) and ν(R(a, b)) ≥bµ(R(a, b)) but b > a = ⇒µ(R(a, b)) = 0.
Now we can write: {x | Dµν(x) < Dµν(x) < ∞} = [ 0 0: fr(x) = ( ν(B(x,r)) µ(B(x,r)) if µ(B) > 0 +∞ if µ(B) = 0 is µ-measurable because is quotient of measurable functions. But Dµν = lim r→0 fr = lim k→∞f 1 k µ −a.e.
and so Dµν is µ-measurable.
This part was indispensable to define the absolutely continuity and the mutually singularity of measures, definitions that are mandatory to enunciate the differentiation theorem for Radon measures. In the following pages we are going to define such things and at the end we will be ready for the Lebesgue differentiation theorem.
Definition 3.6. The measure ν is absolutely continuous with respect to µ provided µ(A) = 0 = ⇒ν(A) = 0 ∀A ⊂Rn.
Written: ν ≪µ.
26CHAPTER 3. THE STUDY OF LEBESGUE DIFFERENTIATION THEOREM ON RN WIT Definition 3.7. The measures ν and µ are mutually singular if there exists a Borel subset B such that: µ(Rn −B) = ν(B) = 0.
Written: ν ⊥µ.
Theorem 3.8. Let ν and µ be Radon measures on Rn, with ν ≪µ. Then ν(A) = Z A Dµν dµ (3.4) for all µ-measurable sets A ⊂Rn.
Proof. Let A be µ-measurable. Then there exists a Borel set B with A ⊂B, and µ(B −A) = 0. Thus ν(B −A) = 0 and so A is ν-measurable. This prove that each µ-measurable set is also ν-measurable.
Let’s define: Z = {x ∈Rn | Dµν(x) = 0} and I = {x ∈Rn | Dµν(x) = ∞}; Thanks to the theorem 3.5 I and Z are µ-(thus ν) measurable sets and by the same theorem µ(I) = ν(I) = 0. Also lemma 3.4 implies ν(Z) ≤αµ(Z) for all α > 0; thus ν(Z) = 0. So ν(Z) = 0 = Z Z Dµν dµ and ν(I) = 0 = Z I Dµν dµ.
This works because we are integrating a measurable and well defined func-tion (the set of point where Dµν is infinite has measure equal to zero) on a zero-measure set hence the integral is obviously zero.
Now fix 1 < t < ∞and define for each integer m Am = A ∩{x ∈Rn | tm ≤Dµν(x) < tm+1}.
Then Am is µ-(and ν) measurable (as it’s intersection on measurable sets) and: A − ∞ [ m=−∞ Am ⊂Z ∪I ∪{x | Dµν(x) ̸= Dµν(x)}.
27 This implies: µ(A − ∞ [ m=−∞ Am) = ν(A − ∞ [ m=−∞ Am) = 0 and consequently: ν(A) = ∞ X m=−∞ ν(Am) ≤ X m tm+1µ(Am) (thanks to the lemma 3.4) ∀t ∈R = t X m tmµ(Am) ≤t X m Z Am Dµν dµ = t Z A Dµν dµ Similarly: ν(A) = X m ν(Am) ≥ X m tmµ(Am) = 1 t X m tm+1µ(Am) ≥1 t Z A Dµν dµ In conclusion: 1 t Z A Dµν dµ ≤ν(A) ≤t Z A Dµν dµ Sending t →1 we prove the theorem.
The last theorem we need to reach our goal: proving the Lebesgue dif-ferentiation theorem, is the Lebesgue differentiation theorem.
Theorem 3.9. Let µ and ν be Radon measures on Rn.
Then ν = νac + νs where νac, νs are Radon measures on Rn and νac ≪µ, νs ⊥µ Furthermore Dµν = Dµνac, Dµνs = 0 and consequently we have: ν(A) = Z A Dµν dµ + νs(A).
for each Borel set A ⊂Rn.
Proof. We assume ν(Rn), µ(Rn) < ∞.
Define Z = {A ⊂Rn | µ(Rn −A) = 0} and we chose Bk ∈Z s.a.
ν(Bk) ≤inf A∈Z ν(A) + 1 k 28CHAPTER 3. THE STUDY OF LEBESGUE DIFFERENTIATION THEOREM ON RN WIT Now we call B = \ k Bk so we have: µ(Rn −B) ≤ X k µ(Rn −Bk) = 0 and ν(B) ≤inf A∈Z ν(A).
(3.5) Define νac = ν| B, νs = ν| Bc It’s easy to verify that νac ≪µ and νs ⊥µ: if we take A ⊂B we have µ(A) = 0, and we suppose ν(A) > 0 = ⇒B −A ∈ Z and ν(B −A) < ν(B) that is a contradiction to the equation 3.5.
On the other and we can see: µ(Rn −B) = 0 thus νs ⊥µ. Finally, fix α > 0 and set C = {x ∈B | Dµνs(x) ≥α} According to lemma 3.4 αµ(C) ≤νs(C) = 0 and therefore Dµνs = 0 a.e.
This implies Dµνac = Dµν, µ a.e.
Now we can start with the Lebesgue differentiation theorem for Radon measures, let’s begin with some notation: Definition 3.10. Like in the previous chapter we will denote Z \ E f dµ = 1 µ(E) Z E f dµ Theorem 3.11. Let µ be a Radon measure on Rn and f ∈L1 loc(Rn, µ).
Then lim r→0 Z \ B(x,r) f dµ = f(x) for µ a.e. x ∈Rn.
29 Proof. For Borel B ⊂Rn, define ν±(B) = Z B f± dµ and for A ⊂Rn ν±(A) = inf{ν±(B) | A ⊂B, B Borel}.
Now ν± are Radon measures and so we can apply the theorem 3.8.
Consequently lim r→0 Z \ B(x,r) f dµ = lim r→0 Z \ B f+ −f−dµ = lim r→0 1 µ(B)(ν+(B) −ν−(B)) (by theorem 3.8) = Dµν+(x) −Dµν−(x) = f+(x) −f−(x) = f(x), µ −a.e.
Conclusion 31 Bibliography Elias M. Stein, "Singular Integrals and Differentiability Properties of Func-tions", PRINCETON UNIVERSITY PRESS, NEW JERSEY, 1970. Pages 3-20.
Lawrence C. Evans, Ronald F. Gariepy, "Measure Theory and Fine Proper-ties of Functions", CRC PRESS, 1992. Pages 26-49.
33 |
3116 | https://file.scirp.org/Html/12-6302161_47178.htm | Creative Education Vol.5 No.11(2014), Article
ID:47178,6
pages
DOI:10.4236/ce.2014.511110
Solving Equations with Parameters
Bat-Sheva Ilany1, Dina Hassidov2
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1Beit-Berl College, Kfar Saba, Israel
2Western Galilee College, Akko, Israel
Email: bat77i@gmail.com, hasidov@netvision.net.il
Copyright © 2014 by authors and Scientific Research Publishing Inc.
This work is licensed under the Creative Commons Attribution International License (CC BY).
Received 23 April 2014; revised 16 May 2014; accepted 6 June 2014
ABSTRACT
This study examines how high school pupils and student teachers tackle equations that are presented in various ways. Some presentations are non-standard, for example having to express “a” (usually perceived as a parameter) in terms of “x” (usually perceived as a variable). We investigate how subjects solve those equations as well as the differences between pupils and student teachers. Our findings suggest that equations containing parameters are more difficult to solve than equations without parameters. The difficulties are associated with the letters that should be expressed and in the arrangement of the equation. The results of this study may expand the knowledge of teachers on the subject of equations with parameters and on the specific difficulties encountered by students during the solution process.
Keywords:Equations, Parameters, High School Pupils, Student Teachers, Mathematics Education, Mathematics Knowledge, Cognition and Mathematics
Theoretical Framework
“Problems with parameters test the solver’s knowledge of mathematics deeply and detect his weak points” (Sedivy, 1976). Sedivy also claims that pupils encounter more difficulties with equations with parameters than with numerical coefficient equations. Algebraic equations containing parameters represent larger classes of equations and more general forms of quantitative relationships compared to equations without parameters.
Davis & Henkin (1978) wrote on the importance of quadratic equations in general and with parametric coefficients in particular. They describe how they interpret understanding of quadratic equations. They provide a long list of skills as well as the mathematical knowledge which are required in order to understand and solve quadratic equations. Davis and Henkin argue that understanding and solving equations with parameters improve students’ understanding of equations in general.
Skemp (1987) argues that learning rules without reasons allows the pupil to function only within a narrow framework, and to deal merely with standard problems. However, this approach limits the students’ understanding and they would not be able to cope with more complex tasks. Understanding enables the pupil to handle non-standard problems which cannot be solved by mechanical application of formulas. Fischbein and Muzicant (2002) indicate that pupils learned primarily in procedural manner and that hence they often fail to distinguish between conceptual and procedural knowledge. Many pupils find it difficult to recognize that a letter represents a number and do not know how to work with symbolic values (Kieran, 1992, 2014).
It is widely agreed that it is important for teachers to recognize how their students perceive the mathematical topics, in particular the specific reasons for making errors (Almog & Ilany, 2012).
In elementary courses students get used to x (and later on y and z) being variables, and letters such as a, b, c being constants or parameters. However, in more advanced courses they come across questions where the letters are used differently, and they find it very confusing. For example, integration problems where x is fixed and y is the variable. Or volume questions where a, b, c are used as variables.
Pupils solve at school countless exercises of equations in which the letter x represents a variable that should be expressed in terms of the letters a, b, c etc… In this study, subjects were given a variety of equations with non-standard presentation, as mentioned above.
Purpose
The study objectives are to examine and identify the difficulties of high school pupils and pre-service student teachers in solving equations with parameters. In addition, to find out if there are significant differences between pupils and student teachers in regards to solving such equations. Which procedures are used and which difficulties are encountered in solving equations with parameters.
Research Questions
1) How do the subjects solve equations with parameters where they have to express various letters in terms of other letters and parameters? In particular when the equations are not arranged in a standard manner.
2) Are there differences between pupils and student teachers in regards to the way they solve such equations. And if so, what are the differences.
Examples of equations investigated in the article (the numbering of the questions follows that of the questionnaire):
• Question 16: Find x (in terms of a) in (fairly standard equation, with variable x and parameter a).
• Question 1: Find c (in terms of x) in (less standard as it is required to express c in terms of x).
• Question 9: Find b (in terms of x) in (less standard as it is enquired to rearrange the equation “properly” and then to express b in terms of x).
• Question 21: The following equation is linear in x, though might mistakenly appear to be a quadratic equation in terms of a:.
We refer to the necessary knowledge needed for solving simple equations as a procedural knowledge. Equations with non-standard presentations call for more flexible approach.
Method
The study sample consisted of 115 mathematics student teachers in their third and fourth year of studies, and 133 twelfth grade pupils of high mathematics level (5 unit mathematics stream) in four secondary schools.
Instruments: A questionnaire was designed for the study and administered to both groups. Part of the questionnaire consists of six questions which include equations with parameters. This study reports the findings of two questions of first degree equations (questions (5) and (21)) and three questions of second degree equations (questions 16, 1 and 9). The questions are listed in Table 1 and Table3
Procedure: To understand the subjects’ solution process, five pupils and six student teachers were interviewed. The interview questions were designed after the questionnaire was filled out and analysed. The purpose was to clarify and prod further the findings of the questionnaire. In addition, open observations were made in order to examine closely the subjects’ responses.
Table 1. Questions 5 and 21, analysis of the results (first degree equations with parameters).
Note: (P = Pupils, S.T. = Student Teachers). The subjects expressed x correctly but some of them failed to specify valid domain and then to simplify the expression. See Table 2 for details.
Data Analysis: Quantitative analysis was carried out via descriptive statistics (tables that list percentage of success), c2 tests and t-tests. Qualitative analysis was carried out by observations and interviews as mentioned above. The results were then analysed according to the emerging criteria.
Results
The results of the first degree equations5 and 21 are presented in Table1
Table 1 shows that a large percentage of high school pupils (22%) did not answer question 21. A very small percentage of the subjects who expressed correctly x in terms of a also specified the correct domain of the solution. Some of the subjects coped only with standard questions which required straightforward approach. Overall, only 36% of pupils and 73% of the student teachers gave a complete solution to question 21.
In question 5, subjects were asked to express the letter min terms of x. m is a letter that is usually perceived as a parameter (Ilany, 1997, 1998). In this question we have a first degree equation for which the solution process should be simple, but only 67% of pupils gave the correct answer and 29% erred. We posit that the mistakes stem from the confusion associated with non-standard use of parameters and variables (Examples are given below). For the student teachers the situation was better, 88% solved correctly and 10% made errors associated with the non-standard presentation.
In question 21, the equation is of first degree in x and x had to be expressed in terms of a. This presentation of the equation was associated with various mistakes in the solution process of the subjects as x is usually perceived as a variable (Ilany, 1997). It is worth noting that equation 21 is actually linear in x, as x is the variable of degree one. However, due to the a2 in the equation, some students were confused and perceived the equation as quadratic.
In order to solve the above equation it is required to open parenthesis, gather similar expressions and rearrange the equation. There is a restriction on a (a ≠ −1), and several algebraic operations are required to solve the equation. Numerous mistakes were made by the subjects who attempted to solve this question, especially the pupils. Only 36% of the pupils expressed x in terms of a correctly. That is, they found that:
However, most of the students who expressed x correctly failed to specify the valid domain and to simplify the resulting equation (see Table 2).
For the student teachers, the situation was better: 73% of them answered question 21 correctly. An analysis of the errors revealed that 41% of pupils compared to 14% of student-teachers made errors associated with the “non-standard” letters (see Table 1).
We chose to explore three second degree equations in this study. In each equation it is required to express a given letter by another one. Errors were divided into two types: mistakes in calculation and errors resulting from confusion associated with the non-familiar usage of letters: ignoring letters, finding the wrong letter and circular expression of letters. (Examples will be cited in Table 3: questions 16, 1, and 9).
According to Table 3, high percentage of pupils (16%) did not answer question 9. There was a significant difference between student teachers and high school pupils in all of the three questions. The three equations have a similar structure (Table 3). The difference between them is that in question 16 (5x2 + 8ax + 4a2 = 0), x has to be found, a letter perceived as a variable according to Ilany (1997, 1998), while in the other equations the
Table 2 . Question 21—Subjects who expressed x correctly, (some of them did not complete the process).
Note: (P=Pupils, S.T. = Student Teachers).
Table 3. Questions 16, 1, and 9, analysis of the results.
Note: (P = Pupils, S.T. = Student Teachers).
letters that should be found are b or c. These letters are usually perceived as parameters, hence they caused confusion.
The results show that more students answered question 16 correctly compared to questions 1 and 9 (see Table 3), which addresses Research Question 1. (That is, that solvers tend to get confused when required to express a, b, c or m (normally function as parameters) in terms of x (which usually stands for a variable).
A significant difference was found between student teachers and pupils in two respects. To begin with, the student teachers scored better on all questions. In addition, a smaller proportion of student teachers seems to have been confused by the change and rearrangement of variables (70% of pupils answered question 16 correctly versus 28% for question 9; the corresponding figures for student teachers are 88% vs. 43%).
Table 3 presents three questions of similar structure. The first question (16) is fairly standard (Expressing x in terms of other parameters).
In the second question (1), we have a non-standard use of letters (expressing x in terms of c).
The third question also involves a non-standard use of variables, as b has to be expressed in terms of x. Moreover, you have to rearrange the equation in order to solve it.
Comparing the results of questions 9 and 1 highlights the significance of the arrangement of the equation. In question 9 the equation () resembles the equation in question 1 (), but is not arranged in the usual order, that is, higher exponents written first. Here we received the lowest results (only 28% of pupils and 43% of student teachers solved the equation correctly). The reasons for these results are attributed to the difficulty in arranging the equation and finding the non-standard letter b in terms of x.
Question 1 was also discussed in the interview. The interviewer emphasized that b or c has to be expressed in terms of x. In spite of this fact most respondents expressed x in terms of b or c. One interviewee that made that mistake said: ”Even though you do not want x, you want b, I automatically find x, because this is a quadratic equation”. She made an attempt to express b saying: “I cannot find b, I know that there are two solutions because D > 0, but I cannot find it, I think that it is impossible”. At the end of the interview, we went back to the question. This time we rearranged the equation as:, and the interviewee expressed b in terms of x without any difficulty. In other words, it was difficult for the student to even start the solution process when faced with non-standard arrangement and letters, although she had no technical difficulty solving such equations once rearranged properly. The same student teacher managed to rearrange and successfully solve an “untidy” equation where x had to be expressed in terms of b, that is, standard use of letters. After the interview she said:”From now on I will ask my pupils to express different letters, not necessarily in the usual order”.
It is interesting to note that rearranging the equation in question 9 is one way to solve it. Another way is to interchange letters, that is, to swap x and b, solve for x, then swap the letters again. None of the subjects applied this method.
Examples of errors resulting from using non-standard letters in the above questions:
In question 1 the correct answer is:. In this case it was difficult for some participants to accept an “unfinished solution”, even though this expression cannot be simplified further, except that the domain ought to be specified.
Consequently, some subjects who solved the question correctly were not satisfied with their answer and crossed it. However, we did consider the crossed solutions to be correct.
A related mistake was expressing x instead of c (2% of student teachers).
“You cannot find c as it is a parameter” (5% of student teachers).
“c is a parameter and therefore can be any number.”
An interesting error in question 16 was getting the correct answers then omitting “a” from those answers. That is, the student wrote:
A similar mistake was mixing up “x” and “a”:
Examples of errors in question 9: Finding x instead of b (11% of pupils, 5% of student teachers).
Expressing b in terms of b. (6% of pupils, 10% of student teachers):
Similarly, in question 21, 4% of the high school students expressed x in terms of x, for example:
Examples of mistakes that took place while attempting to solve:
“Impossible to express x on its own.”
“Don’t know” (32% of students, 5% of student teachers).
Turn x into x2 (6% of students, 2% of student teachers).
Expressed a instead of x (3% of students).
Examples of mistakes that took place while attempting to solve question 5: Find m in:
Found x (3% of students).
“Impossible” (10% of students, 7% of student teachers).
Turned the equation into a quadratic equation for m (3% of students).
Discussion
The results of this study indicate that for equations with parameters, a large percentage of student teachers, and even larger proportion of pupils, erred due to confusion associated with non-standard use of letters, (for example, having to express a in terms of x). Being faced with equations where the parameters and variables were not arranged in a standard manner made it even more difficult for the subjects to solve those equations.
In general, equations containing parameters are more difficult to solve than equations without parameters. Subjects tend to solve equations mechanically, and hence they make numerous mistakes when they come across non-standard questions which call for deeper understanding.
Key factors that are often problematic for students are the type of the letter that has to be expressed, the arrangement of the equation to be solved, the power of the variable, the power of the parameter and the type of the algebraic techniques required.
Implications for teaching how to solve equations with parameters The findings of this study, in particular the large number of mistakes, raise questions in regards to the efficacy of the teaching of this topic. Hence, in order to improve students’ understanding of equations with parameters, it is recommended to improve students’ understanding concerning the use of letters in such equations. For example, by careful reading of what is required in the question, by adopting a more flexible approach in regards to which letters can be expressed in terms of other letters, by making sure that the expression is simplified properly, and by being aware of the differences and similarities between equations with or without parameters.
Bat-Sheva Ilany, Dina Hassidov We recommend including in the school mathematics curriculum more equations with various parameters and variables, as well as equations which are not presented in the usual order. Doing so will encourage students to think more deeply about the topic and to gain deeper understanding.
Almog, N., & Ilany, B.-S. (2012). Absolute Value Inequalities: High School Students’ Solutions and Misconceptions. Educational Studies in Mathematics, 81, 347-364.
Davis, R. B., & Henkin, L. (1978). Inadequately Tested Aspects of Mathematics Learning—Testing, Teaching and Learning. Report of a Conference on Research on Testing, National Institute of Education, 49-63.
Fischbein, E., & Muzicant, B. (2002). Richard Skemp and His Conception of Relational and Instrumental Understanding: Open Sentences and Open Phrases. In D. Tall, & M. Thomas (Eds.), Intelligence, Learning and Understanding in Mathematics: A Tribute to Richard Skemp (pp. 49-77). Flaxton, QLD: Post Pressed Publishers.
Ilany, B. (1998). The Elusive Parameter. In The 23rd Annual Meeting of the International Group for the Psychology of Mathematics Education (Vol. 4, p. 265). Stellenbosch: University of Stellenbosch.
Ilany, B. (1997). The Concepts of Variable and Parameter of Pre-Service Teachers and High School Students. Doctoral Dissertation, Tel Aviv: Tel Aviv University. (in Hebrew)
Kieran, C. (1992). The Learning and Teaching of School Algebra. In D. A. Grouws (Ed.), Handbook of Research on Mathematics Teaching and Learning (pp. 390-419).
Kieran, C. (2014). Algebra Teaching and Learning. Encyclopedia of Mathematics Education (pp. 27-32). Springer: Springer Reference.
Sedivy, J. (1976). A Note on the Role of Parameters in Mathematics Teaching. Educational Studies in Mathematics, 7, 121-126.
Skemp, R. (1987). The Psychology of Learning Mathematics. Hillsdale, NJ: Lawrence Erlbaum Associates. |
3117 | https://www.mdpi.com/1422-0067/23/22/14103 | Identification of the Karyopherin Superfamily in Maize and Its Functional Cues in Plant Development
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Table of Contents
Abstract
Introduction
A General View of the Karyopherin Superfamily
Importin α Family in Maize and Arabidopsis
Importin β Family in Maize and Arabidopsis
Functional Cues of Karyopherins in Hormone Signaling and Plant Development
Conclusions and Perspectives
Supplementary Materials
Author Contributions
Funding
Institutional Review Board Statement
Informed Consent Statement
Data Availability Statement
Conflicts of Interest
References
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Open Access Review
Identification of the Karyopherin Superfamily in Maize and Its Functional Cues in Plant Development
by
Lu Jin
Lu Jin
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1,
Guobin Zhang
Guobin Zhang
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2,
Guixiao Yang
Guixiao Yang
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1 and
Jiaqiang Dong
Jiaqiang Dong
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1,
1
The Key Laboratory of Plant Development and Environmental Adaptation Biology, Ministry of Education, School of Life Sciences, Shandong University, Qingdao 266237, China
2
College of Agronomy, Shandong Agricultural University, Taian 271018, China
Author to whom correspondence should be addressed.
Int. J. Mol. Sci.2022, 23(22), 14103;
Submission received: 5 October 2022 / Revised: 6 November 2022 / Accepted: 13 November 2022 / Published: 15 November 2022
(This article belongs to the Special Issue Regulatory Mechanisms of Auxin in Plant Growth and Development)
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Abstract
Appropriate nucleo-cytoplasmic partitioning of proteins is a vital regulatory mechanism in phytohormone signaling and plant development. However, how this is achieved remains incompletely understood. The Karyopherin (KAP) superfamily is critical for separating the biological processes in the nucleus from those in the cytoplasm. The KAP superfamily is divided into Importin α (IMPα) and Importin β (IMPβ) families and includes the core components in mediating nucleocytoplasmic transport. Recent reports suggest the KAPs play crucial regulatory roles in Arabidopsis development and stress response by regulating the nucleo-cytoplasmic transport of members in hormone signaling. However, the KAP members and their associated molecular mechanisms are still poorly understood in maize. Therefore, we first identified seven IMPα and twenty-seven IMPβ genes in the maize genome and described their evolution traits and the recognition rules for substrates with nuclear localization signals (NLSs) or nuclear export signals (NESs) in plants. Next, we searched for the protein interaction partners of the ZmKAPs and selected the ones with Arabidopsis orthologs functioning in auxin biosynthesis, transport, and signaling to predict their potential function. Finally, we found that several ZmKAPs share similar expression patterns with their interacting proteins, implying their function in root development. Overall, this article focuses on the Karyopherin superfamily in maize and starts with this entry point by systematically comprehending the KAP-mediated nucleo-cytoplasmic transport process in plants, and then predicts the function of the ZmKAPs during maize development, with a perspective on a closely associated regulatory mechanism between the nucleo-cytoplasmic transport and the phytohormone network.
Keywords:
Karyopherin; Importin α; Importin β; nucleo-cytoplasmic transport; maize; NLS; NES; phytohormone signaling; auxin; root development
1. Introduction
Eukaryotic cells establish separate functional spaces for transcription and translation in the nucleus and cytoplasm. The nuclear pores and nuclear pore complexes (NPCs) across the nuclear envelope link two cellular compartments for high-efficiency molecular exchange channels [1,2]. Disordered phenylalanine- and glycine-rich nucleoporins (FG-Nups) are distributed in the center of NPCs, serving as a bidirectional permeability gate to restrict arbitrary translocation of macromolecules [3,4]. Ions, metabolites, and signal-independent small molecules diffuse freely through the NPCs; macromolecules such as proteins, RNAs, and some complexes more than ~5 nm or ~40 kDa in size are usually signal-dependent active transport-mediated by a range of nuclear transport receptors (NTRs) [5,6]. An evolutionarily conserved superfamily of soluble receptors is primarily responsible for the nucleo-cytoplasmic transport (NCT) of the macromolecules, known as Karyopherins (KAPs) or Importins (IMPs) [7,8]. The KAPs play central roles in substrate screening and transport via recognition of the specific short-peptide signals displayed on cargos, referred to as nuclear localization signals (NLSs) or nuclear export signals (NESs) [9,10].
Appropriate nucleo-cytoplasmic partitioning of specific proteins is the critical intracellular step for executing downstream physiological functions [11,12]. However, how the intracellular distribution of nuclear proteins is regulated remains incompletely understood. KAPs may act as upstream regulators of the functional components for gene regulation, chromatin modulation, and signal transduction [13,14,15]. Some published reports have demonstrated the pivotal roles of several KAP members in plant growth, reproduction, immunity, stress response, and epigenetic regulation (Table S1). However, the role of only one member of the KAP superfamily in maize has been revealed: its role in mediating the nuclear accumulation of Opaque2 (O2) to promote zein biosynthesis in kernel development [16,17]. In contrast, the other members of the KAP superfamily and their functions in maize are still unknown.
Therefore, this review first identifies seven IMPα and twenty-seven IMPβ genes in the maize genome, then starts with this entry point to review the evolution traits of the KAP superfamily, the KAP-mediated nucleo-cytoplasmic transport pathway, and the recognition rules for substrates with nuclear localization signals (NLSs) or nuclear export signals (NESs) in plants. Furthermore, we spotlight the regulatory roles of nucleo-cytoplasmic transport in phytohormone signaling and execution. Next, we searched for the protein interaction partners of the ZmKAPs and selected the ones with Arabidopsis orthologs functioning in auxin biosynthesis, transport, and signaling to predict their potential function. Lastly, several ZmKAPs were observed to share similar expression patterns with their interacting proteins, implying their potential functions in root development.
2. A General View of the Karyopherin Superfamily
The Karyopherin superfamily is categorized into Importin α (IMPα) and Importin β (IMPβ) based on structural and functional features . Genome-wide identification of the IMPα or IMPβ families in Saccharomyces cerevisiae, Danio rerio, Homo sapiens, Mus musculus, Arabidopsis thaliana, and Solanum tuberosum has been successively reported [18,19,20,21]. The IMPαs serve as a protein adaptor between cargo and IMPβ1 in the classical nuclear import pathway in yeast and mammals, and most IMPβs can independently mediate nuclear–cytoplasmic transport [22,23]. Generally, IMPβs are divided into importins and exportins, while a few IMPβs perform a dual role in both nuclear import and export, such as ScKAP142/ScKAP122, HsXPO4/HsXPO7, and HsIPO13 in yeast and humans [24,25,26,27,28]. However, these bidirectional receptors are demonstrably undetermined in plants. In addition, the function of some KAPs remains poorly understood in plants (Supplementary Materials Table S1).
2.1. Evolution of the Karyopherins
Based on validated members of the IMPα and IMPβ families from yeast, humans, and Arabidopsis, each protein sequence was used as a query to perform BLASTP searches against the blue-green algae (Nostoc), green algae (Chlamydomonas reinhardtii), bryophyte (Marchantia polymorpha), pteridophyte (Selaginella moellendorffii), gymnosperm (Thuja plicata), angiosperm (Amborella trichopoda), and maize genomes (Figure 1). For this analysis, the KAPs are an ancient gene superfamily existing in all eukaryotes. In blue-green algae, a few sequences referred to as HEAT (Huntingtin, elongation factor 3 (EF3) 1, protein phosphatase 2A (PP2A) 2, and the yeast PI3-kinase TOR1) repeat domain-containing proteins share a low similarity with IMPα and IMPβ, which may suggest the evolutionary source of their unique properties. In eukaryotes, the KAP superfamily is highly conserved from single-celled to multicellular organisms. Among plant species, the PLANTKAP clade in the IMPβ family is unique to embryophyte plants. Analogously, there is also an embryophyte plant-specific group in the IMPα family, and we named this clade PLANTα (Figure 1).
Figure 1. Phylogenetic tree of the Karyopherin superfamily. Based on validated members of the IMPα and IMPβ families from yeast, humans, and Arabidopsis, each protein sequence was used as a query to perform BLASTP searches in Phytozome v13 ( (accessed on 4 October 2022)), NCBI ( (accessed on 4 October 2022)), and MaizeGDB ( (accessed on 4 October 2022)), remove the non-representative splicing forms of the same gene locus, and confirm sequences of non-redundant candidates by phylogenetic analysis with the homologous series of the other species. Saccharomyces cerevisiae (Sc), Homo sapiens (Hs), Chlamydomonas reinhardtii (Cre), Marchantia polymorpha (Mapoly), Selaginella moellendorffii (Smo.), Thuja plicata (Thupl.), Amborella trichopoda (AmTr.), Arabidopsis thaliana (At), Zea mays (Zm); ZmIMPα proteins in blue font and ZmIMPβ in red.
Results ultimately identified seven IMPαs and twenty-seven IMPβs in maize, named based on their subfamily affiliation (Table 1). In comparison to Amborella trichopoda and Arabidopsis, the members of maize KAPs undergo family expansions, especially in the IMPβ family. The lineage of maize experienced a tetraploidy period combined with two genomes, the Maize1 and Maize2, accompanied by whole genome duplication (WGD) [29,30]. As shown in Supplementary Materials Table S2, sixteen of thirty-four KAP genes experienced duplication and retained elements from ancient tetraploid maize genomes. There are fourteen KAP genes that may undergo uneven gene loss after WDG. Among the thirty-four KAP genes, eighteen genes come from the Maize1 subgenome and twelve genes are from the Maize2 subgenome. In addition, four maize KAP genes may be dispersed as duplicate genes.
Table 1. List of putative Karyopherin gene family members in Zea mays.
2.1.1. Importin α
IMPαs in animals include three subfamilies designated α1, α2, and α3 . Group α1, found in all eukaryotes, is believed to be the earliest progenitor of IMPαs and gave birth to the other two groups, which function in development and differentiation for the evolution of metazoan animals [31,32]. Eight of nine IMPαs in Arabidopsis belong to subfamily α1, and the remaining one is a non-conventional isoform . Replication events based on group α1 are distinct between animals and plants, which may have taken unique evolutionary paths to bring forth particular clades. ZmIMPα1-5 and AtIMPA1-8 are orthologous to ScSRP1 and HsKAPNA1/5/6, belonging to clade α1. ZmIMPα7 is the ortholog of AtIMPA9 as the non-conventional isoform. This specific group is also present in other species, except for Chlamydomonas reinhardtii. Therefore, we named PLANTα as an embryophyte plant-specific group of the IMPαs. In addition, ZmIMPα6 failed to classify into any group, and it appeared to be another gene duplication.
2.1.2. Importin β
IMPβs are a large conserved family in which the number of members varies slightly across eukaryotes, and can be divided into fifteen subfamilies (Figure 1). The ZmIMPβs lack the XPO6 subfamily and have a PLANTKAP group without a noticeable difference from other eukaryotic plants. The distribution pattern of IMPβ subclasses may be established before the evolutionary expansion of eukaryotes, accompanied by continuous selective pressure leading to a secondary loss of the IMPβ orthologs . The lack of the XPO6 subfamily in Arabidopsis is likely to be a representative loss event, and an analogous situation is available in yeast (XPO4/6/7) . PLANTKAP is a paralogous expansion cluster identified in embryophyte-specific land plants . It indicates the fifteen IMPβ subfamilies that are conserved in eukaryotes but at the same time accompanied by ortholog expansion or paralog secondary loss. A report shows decreased IMPβ subfamilies during the evolution of the potato genome, but increased homologous genes within the IMB1 and IMB3 subfamilies in Solanum tuberosum . Analogous duplication events might have observably promoted the expansion of the composition of ZmIMPβ members, especially in the IMB1, XPOT, XPO2, TNPO3, and PLANTKAP subfamilies compared to HsIMPβs and AtIMPβs.
2.2. The Karyopherin-Mediated NCT Pathway
2.2.1. The Classical Nuclear Protein Import Cycle in Yeast and Mammals
The classical nuclear protein import cycle in which IMPα and IMPβ1 cooperate has been well characterized in yeast and mammals . It includes three steps: (I) In the cytoplasm, cargos with classical NLS (cNLS) are recognized by the IMPαs, linking with the IMPβ1 to form an IMPα/β1 heterodimer localized to the nuclear envelope . Then, the IMPβ1 directly interacts with the FG-Nups to facilitate transport of the cargo–IMPα–IMPβ1 complex across the NPCs . (II) Once the imported complex reaches the nucleus, a conformation change triggered by high-affinity RanGTP binding to the IMPβ1 results in the primary dissociation of IMPβ1 from the IMPα-cargo . This irreversible dissociation also influences the conformational change in IMPα itself and accelerates the release of cargo from the IMPα . (III) Lastly, the empty IMPα is recycled by exportin CAS back to the cytoplasm in preparation for the next round of nuclear import .
2.2.2. The IMPα- and IMPβ-Mediated Nuclear Transport Pathway in Plants
Although the classical transport cycle has yet to be confirmed in plants, several reports have shown a conservative mechanism of the IMPα/β-mediated nuclear protein import pathway. A bimolecular fluorescence complementation (BiFC) assay shows the interaction between AtKPNB1 and four AtIMPAs (AtIMPA1, AtIMPA2, AtIMPA4, and AtIMPA6) . The exportin AtXPO2/AtCAS can be specifically bound to AtIMPA1, AtIMPA2, AtIMPA3, and AtIMPA4 in yeast two-hybrid (Y2H) analysis . The AtIMPA2 interacts with the N-terminal region of AtXPO5/AtHASTY to mediate its nuclear shuttling from the cytoplasm to the nucleus . Additionally, the vitro nuclear import assay demonstrates that rice IMPα1 can form a complex with mouse IMPβ1 and cNLS cargo . Interestingly, another report shows that AtIMPα can mediate the nuclear accumulation of NLS cargo independent of IMPβ . It implies that IMPα may not only act as a protein adaptor but also possibly independently mediate a unique nuclear import pathway in plants.
2.2.3. The IMPβ-Dependent Nuclear Translocating Pathway
In eukaryotic cells, the IMPβ family dominates the nuclear translocation transport of most proteins and RNAs . These cargos, with distinctive signals, can directly interact with importins or exportins to constitute multiple non-classical transport pathways [47,48]. These parallel pathways share a similar mechanism to the classical nuclear import cycle in their multivalent interaction with the FG-Nups and directional regulation by the Ran (Ras-like nuclear protein) system, as well as their functional redundancy in the transportation of the same cargos [49,50,51]. The IMPαs and the IMPβs are probably evolutionarily related proteins defined by two helical secondary structures, Armadillo-like (Arm) and the HEAT repeats, which provide interaction scaffolds for multiple protein ligands [52,53]. That might lead to differences in protein conformation flexibility between IMPαs and IMPβs that impact their affinities for specific cargos.
3. Importin α Family in Maize and Arabidopsis
3.1. Protein Domain Distribution and Gene Expression Profiles of the ZmIMPαs
Three conserved domains, an N-terminal importin-β-binding (IBB) domain followed by a consecutive ARM repeat region and an atypical ARM repeat at the C-terminal, are predicted by the NCBI and Pfam database of both the AtIMPA and the ZmIMPα proteins (Figure 2A). These conserved domains form a highly similar protein structure in most IMPαs, suggesting their comparable biological function.
Figure 2. IMPORTINα family in maize and Arabidopsis. (A) Schematic view of the domains conserved between AtIMPA and ZmIMPα proteins according to Pfam Database ( (accessed on 4 October 2022)) and CCD Tools ( (accessed on 4 October 2022)); (B) Heat map of the expression pattern of ZmIMPα genes, with the expression value calculated by log2 (FPKM). SAM: shoot apical meristem, NU: nucellus, em: embryo, en: endosperm, HAP: Hours after Pollination, DAP: Day after Pollination; (C) Signatures of the Importin β binding (IBB) domain of the ZmIMPα1 protein predicted by AlphaFold Protein Structure Database ( (accessed on 4 October 2022)); multiple amino acid sequences of the IBB domain aligned using CLUSTALW, three conserved motifs highlighted in red and rectangle boxes.
The flexible IBB domain is the central zone in recruiting IMPβ1 . In the AtIMPαs, the IBB domain appears to be absent in AtIMPA8 and AtIMPA9. In AtIMPA8 this is due to a reduction in the partial sequence at the N-terminal, while AtIMPA9 seems to have an undefined region. Compared to its homolog in the PLANTα group, ZmIMPα7 contains the IBB domain at the N-terminal. The sequence alignment suggests that the vacant N-terminal of AtIMPA9 may have a similar function to the IBB domain (Figure 2C).
The ARM array and atypical ARM are responsible for cargo loading and CAS binding [55,56]. In the ZmIMPαs, ZmIMPα5 appears to be an incomplete gene copy with a closer kinship to ZmIMPα4. The lack of the multi-ARM repeats region and the atypical ARM may result in nuclear transport function deficiency. Additionally, members in the PLANTα group display a reduced ARM array, which may lead to differences with other isoforms in substrate recognition.
The RNA-seq-based B73 gene expression data from twenty-one tissues at different growth stages were selected and analyzed . As shown in Figure 2B, the ZmIMPα genes display a constitutive expression pattern in various organs. ZmIMPα4 (Zm00001d009850) shows high expression in the endosperm (en), which may relate to its role in the transcriptional regulation of storage proteins [16,17]. The expression levels of ZmIMPα1/2/3/4 are noticeably higher than those of the other isoforms in the IMPα group. A report shows that AtIMPA9 is highly expressed in the leaves during pathogen infection . The expression profile may imply their potential functional redundancy or differentiation in response to specific ambient cues (Supplementary Materials Figure S1).
3.2. Multifunctionality of the IBB Domain
The IBB domain is a critical molecular connector between IMPα and IMPβ, and it is also an ingenious regulator for the activity of IMPα itself. The crystal structure of IMPα in mammalians shows the IBB domain containing an internal NLS that binds to its NLS-binding site and functions as an autoinhibited regulator . The auto-inhibitory action can be displaced by IMPβ1 binding to fulfill its affinity switch to cargos . The alkaline amino acid 54 KRR 56 (Lys-Arg-Arg) in the IBB domain of ScSRP1 acts as an auto-inhibitory NLS sequence [60,61]. The mutation of 54 KRR 56 does not impact the interaction with IMPβ, but it will lead to the failure of cargo to be released in the nucleus . Additionally, the other two conserved alkaline amino acids in the IBB domain, 33 RXXR 36 and 44 RXXXR 48 (X for any residue), are likely to significantly affect the binding activity of IMPβ1 . This shows flexible switching roles of the IBB domain in auto-inhibition, interaction with IMPβ, and cargo release.
The protein conformation of the ZmIMPαs displays a similar structure except in ZmIMPα5. As represented in ZmIMPα1, the IBB domain folds back to occupy the NLS-binding surface (Figure 2C). That auto-inhibited state can be switched from closed to open by cooperative binding of the NLS cargo and IMPβ to the IMPα . In the putative IBB domain sequence, three clusters of alkaline amino acids in the AtIMPAs and ZmIMPαs show subtle distinctions or variations (Figure 2C). The first two clusters in the PLANTα groups show distinct features, such as (Q/N) RRR and KERRE. The RRRR cluster is conservative in other IMPαs such as ScSRP1 and HsKPNA1. The RKXKR motif is the primary pattern in group α1 except for ZmIMPα1/2 (RKSRR), suggesting that amino acid R (arginine) at both ends is likely the most conserved residue. The terminal residue of the last cluster is random, e.g., KRX.
Moreover, a recent study in mice shows that a DNA-binding region can be identified in the IBB domain of KPNA2 and characterized to overlap the conserved alkaline amino acid region . This suggests that the IBB domain may act as a common interacting domain for multiple binding partners involved in the functional switching of the transport and non-transport pathways of IMPαs .
3.3. The ARM Repeat and Classical NLS Recognition
A series of ARM repeats in IMPαs is mainly responsible for cargo loading and releasing by cooperating with the IBB domain . The consecutive stacking ARM repeats generate a superhelical structure and the inner concave surface of the protein provides NLS-binding grooves for the cargos, which include the major and minor binding pockets for recognizing positively charged amino acid clusters in NLSs [66,67]. NLSs with short and regular amino acid clusters generally divide into classical (cNLS) and non-classical NLSs (ncNLS) based on residue composition . The monopartite (MP) and bipartite (BP) motifs are two common types of cNLSs mainly recognized by IMPαs . In addition, the LSD1-type zinc finger motifs possibly act as NLSs bound to the IMPα . That indicates that more potential signals are yet to be discovered and interpreted.
The first identified cNLS in the simian virus 40 (SV40) large T antigen, composed of seven amino acids, was PKKKRKV (Pro-Lys-Lys-Lys-Arg-Lys-Val), identified as an MP-cNLS bound to the major site of the IMPαs . There are five classes of MP-cNLS motifs with a distinctive preference for the major or minor binding sites of IMPαs differently in yeast, plant, and mammals [72,73]. As shown in Table 2, the Class I type MP-cNLS seems to be the most common, while others exhibit flexible variation [74,75,76,77]. AtIMPA1/2/3 can recognize Class I/II/V NLS-containing proteins . The NLS of PIP5K2 is analogous to the Class III consensus motifs and is recognized by AtIMPA6/9 .
Table 2. Classification of NLSs and NESs recognized by KAPs in plants.
As shown in Figure 2C, the autoinhibitory sequences in the IBB domain (KRR and RRRR) may act as a BP-cNLS, folding back to occupy the major and minor sites when the IMPα is in an empty state to prevent futile nuclear translocation of unloaded import complexes . In rice, OsIMPAα1 may show binding activity to variable motifs on different proteins, suggesting a mutual co-recognition mechanism in BP-cNLS [80,81]. Additionally, there is more than one NLS displayed on cargo; for example, AtMINIYO has two NLSs that may promote its accumulation in the nucleus .
4. Importin β Family in Maize and Arabidopsis
4.1. The Characteristic Domains of IMPβ Proteins
Compared to the high similarity among the IMPα proteins, the IMPβs may represent a more flexible transport receptor family containing various functional domains (Figure 3A). The increased numbers of homologous genes in the IMB1, IMB2, IMB3, XPO2, XPOT, TNPO3, and PLANTKAP subfamilies form a larger family than the AtIMPβs. The conserved domains stay the same in importin and exportin subfamilies between maize and Arabidopsis, implying that members of each group hold potential functional resemblances. As shown in Figure 3B, most ZmKAPβ genes display a constitutive expression pattern suggesting their indispensable roles in maize growth and development. The homologous genes appear to have differential expression levels in each subfamily, potentially indicating neo- and sub-functionalization of these proteins.
Figure 3. IMPORTINβ family in maize and Arabidopsis. (A) Schematic representation of the conserved domains between AtIMPβ and ZmIMPβ proteins according to Pfam Database and CCD Tools; (B) Heat map of the expression profile of ZmIMPβ genes in different tissues, with the expression value calculated via log2 (FPKM). SAM: shoot apical meristem, NU: nucellus, em: embryo, en: endosperm, HAP: Hours after Pollination, DAP: Day after Pollination.
4.1.1. Importin
Four ZmIMB1s with high protein similarity are classed into the IMB1 subfamily. ZmIMB1a and ZmIMB1b appear to be the closest orthologs to AtKPNB1, while ZmIMB1c and ZmIMB1d show higher kinship to the other two ARM repeat superfamily proteins, At3G08943 and At3G08947. The importins contained in the IMB1/2/3/4/5, IPO8, KA120, PLANTKAP and TNPO3 subfamilies independently mediate nuclear import. In Arabidopsis, AtKPNB1 and AtSAD2 have shown prominent functions in responses to various abiotic stresses [41,87]. AtTRN1, AtKETCH1, and AtSAD2 have demonstrated different roles in microRNA biogenesis and activity regulation [88,89,90]. Both AtKA120 and AtMOS14 act as modifiers of Suppressor of npr1-1, constitutive (SNC1) to affect plant immunity response [91,92]. In yeast and mammals, KAP122 and IPO13 may act as bidirectional receptors [26,28]. The protein domain of the TNPO3 subfamily shares high similarity with the exportins in Arabidopsis and maize, which may imply their function in nuclear export, and still needs further verification in plants.
4.1.2. Exportin
The exportins exhibit unique domains in each group and remain highly consistent in Arabidopsis and maize. The XPO1 domain is a common feature among the XPO1, XPOT, XPO5, and TNPO3 subfamilies. In the XPO1 subfamily, the CRM1_C domain may contribute to the transition from an extended to a compact conformation in NES–cargo binding [93,94]. A report suggests that the CRM1_C domain in AtXPO1 functions to facilitate virus infection in the nuclear export of viral replicase . Members in the XPO2 subfamily have two related domains, CSE1 and CAS/CSE1, which appear to form a flexible conformation that changes upon cargo binding . XPOT and XPO5 are primarily involved in the nuclear export of multiple RNAs to the cytoplasm . The EXPORTIN-T and EXPORTIN-5 domains are likely to provide the binding pocket for various RNAs [98,99]. Remarkably, however, the link between the protein conformation of the IMBβs and their distinctive cargos is still an open question. This may be inseparable from the function of these conserved domains and still needs further exploration and verification.
4.2. The Function of the IBN_N Domain and Ran System
The Importin-beta N-terminal domain (IBN_N) is a typical structural feature at the N-terminal of most IMPβs (Figure 3A). It seems to play a role in cooperation with the Ran system. Several reports show that the residues at the N-terminals of KPNB1, TPNO1, and CSE1 provide the first interactive interface for the Ran protein [37,38,100,101]. The crystal structure of XPO4 in mammals has revealed four distinct Ran-interaction sites, and the N-terminal is in charge of the first Ran-binding site . In Arabidopsis, the Ran interacts with the amino terminus in AtHASTY, AtTRN1, and AtMOS14 [83,91,103]. In addition, the IBN_N domain of AtXPO1 appears to support the binding activity of virus protein to impact mosaic virus replication .
The IMPβs bound to RanGTP are the direct target regulated by the Ran system [23,38]. In the nucleus, RanGTP binds to IMPβ1 to facilitate the disassembly of the IMPα- β1 cargo . In the cytoplasm, Ran-binding protein 1 (RanBP1) and RanBP2 cooperate with RanGTPase-activating protein 1 (RanGAP1) to hydrolyze RanGTP to RanGDP for releasing IMPβ1 [104,105]. The gradient distribution of the RanGTP/GDP in the nucleus and cytoplasm ensures the proper direction of the nucleocytoplasmic traffic . Therefore, the RanGTP/GDP transformation, the KAP-mediated cargo transport, and the restriction of the NPC complex constitute a multiple-layer control for NCT .
4.3. The Non-Classical NLS and NES Recognized by Importin β
4.3.1. The Non-Classical NLS
Unlike arginine or lysine residue-enriched cNLSs, only a few ncNLSs or other types of NLSs are structurally characterized and recognized according to their particular IMPβs [68,107]. The PY (proline-tyrosine) motif is a distinguishing feature of the ncNLS that interacts with members of the IMB2 and IMB4 subfamilies [82,108,109]. The PY-NLS has loose sequence motifs in a disordered structure and its overall basic charge is irregular and variable among different cargos [107,110]. The M9 domain of human heterogeneous nuclear ribonucleoprotein A1 (hnRNP A1) with a typical PY-NLS interacts with HsTNPO1 . In Arabidopsis, two small RNA-binding proteins, AtGRP7 and AtGRP8, contain an M9-like domain to interact with the ortholog AtTRN1 . The difference in several amino acid residues between the M9 and M9-like domains suggests a discrepancy in the PY-NLS between plants and animals (Table 2). Additionally, the PY motifs seem to function not just in nucleocytoplasmic shuttling. AtIMB4 interacts with the PY motifs in FRA1 kinesin to inhibit its motility and protect protein stabilization in the cytoplasm . Additionally, there are two other types of NLSs, recognized by their designated IMPβs in yeast and human. ScKAP121 and HsIPO5 can bind to a specific IK (isoleucine-lysine-rich)-NLS with a consensus motif K-V/I-X-K-X1–2-K/H/R . HsTNPO3 can mediate the cellular trafficking of SR proteins (serine/arginine-rich proteins) through interaction with the RS (arginine–serine) repeat domain . However, these two analogous NLS are still unknown in plants.
4.3.2. NES
NES is a leucine-rich peptide signal in the nuclear export process, primarily recognized by the exportin XPO1/CRM1 . A set of ten consensus sequence patterns apply to the NES family in animals and plants [114,115]. As shown in Table 2, the NES motifs of zinc finger transcription factor OXS2 members show high conservation in Arabidopsis, rice, and maize [84,85]. NES and NLS may coexist in transcription factors such as AtFHY1 and OsWRKY62, suggesting a dynamic nucleocytoplasmic distribution of the nuclear proteins in plant developmental and environmental responses [74,80]. A similar situation also presents itself in plant virus proteins that may facilitate the virus’s replication cycle in plant host cells . Generally, these identified NLSs or NESs are linear targeting signals for IMPαs or IMPβs. In addition, the folded domains in some cargos are likely to bind to IMPβ as well, and that may be related to the particular conformation of the IMPβs [111,116]. However, for other exportins, the more extensive identification signals are still an outstanding problem requiring further elucidation of the potential interaction mechanism.
5. Functional Cues of Karyopherins in Hormone Signaling and Plant Development
5.1. The Roles of Arabidopsis KAPs in Hormone Signaling
Phytohormones are important in regulating transcriptional networks in plant growth and environmental adaption . Recently, some encouraging progress has been made in understanding the regulatory roles of the Arabidopsis KAPs in plant hormone pathways, and a schematic illustration is shown in Figure 4. Of note, this motivates a stepwise progression towards new insight into the more regulatory components in the phytohormone network.
Figure 4. A schematic illustration of Arabidopsis KAP-mediated nucleo-cytoplasmic transport in hormone signaling for plant development. (A) Cytokinin regulates cell division by promoting nuclear shuttling of transcription factor MYB3R4, mediated by AtIMPA3 and AtIMPA6, in the shoot apical meristem (SAM). (B) AtIMB4 mediates the nuclear partitioning of GRF-INTERACTING FACTOR1 (GIF1)/ANGUSTIFOLIA3 and JANUS, which antagonistically regulate PLETHORA1 (PLT1) transcription. (C) AtSAD2 and AtKPNB1 act as negative regulators in abscisic acid (ABA) signaling. The atsad2 mutant displays an ABA hypersensitivity response during seed germination and seedling growth. AtKPNB1 is involved in controlling ABA-induced stomatal closure under drought conditions. (D) AtXPO1A mediates the nuclear export of a WD40 repeat-containing protein, XIW1 (XPO1-interacting WD40 protein 1), which maintains the stability of ABA INSENSITIVE 5 (ABI5) in the nucleus. The schematic illustration was drawn with BIORENDER ( (accessed on 4 October 2022)).
5.1.1. AtIMB4 and PLT1-Mediated Root Development
PLETHORA (PLT) family members encoding AP2 class transcription factors depend on auxin response . Auxin-induced PLTs form a gradient to control the location of the stem cell region and root meristem size. . AtIMB4 is a positive regulator in root meristem size . It is involved in transcriptional regulation for the PLT1 gene by mediating the nuclear accumulation of two antagonistic cargos, JANUS and GIF1 .
5.1.2. AtIMPA3/6 and Cytokinin-Activated Cell Division in Shoot Apical Meristem
Myb-domain protein 3R4 (MYB3R4) transcription factor is highly expressed in the shoot apical meristem and enriched in the dividing cells to activate the expression of the cell cycle genes during mitosis . Generally, MYB3R4 is mainly localized in the cytoplasm, and AtIMPA3 acting together with AtIMPA6 mediates its rapid nuclear accumulation triggered by cytokinin at the G2/M transition .
5.1.3. AtIMPβs and ABA Signaling in Response to Abiotic Stress
There are three IMPβs, AtSAD2, AtKPNB1, and AtXPO1A, shown to be involved in ABA signaling in responses to abiotic stress. AtSAD2 is initially found in the abscisic acid (ABA) hypersensitivity response during seed germination and seedling growth as a negative regulator of ABA sensitivity, suggesting its potential function in ABA signaling . AtKPNB1 also acts as a negative regulator at early steps in ABA signaling, and it might play an essential role in controlling ABA-induced stomatal closure under drought conditions [41,123]. Conversely, AtXPO1A mediates the nuclear export of a WD40 repeat-containing protein XPO1-interacting WD40 protein 1 (XIW1) . In the nucleus, XIW1 interacts with the key transcription factor ABA INSENSITIVE 5 (ABI5) in the ABA signaling pathway to maintain its stability and further positively regulate the ABA response .
5.2. The Predicted Interacting Protein of the ZmKAPs Involved in the Auxin Pathway
Reports in Arabidopsis suggest that the Karyopherin-mediated nucleocytoplasmic shuttling of signal molecules is the critical link to the hormone signal transduction chain [120,121,124]. However, more signal elements remain to be discovered for obtaining a better understanding of the role played by KAPs in the phytohormone network, especially for corn growth and development. Therefore, to understand the functional cues of the ZmKAPs, we explored the putative interacting proteins using plant.MAP and STRING database [125,126]. As the function of most proteins in maize is not yet studied, we selected their orthologs in Arabidopsis involving auxin biosynthesis, transport, and signaling to discuss their potential functionality links (Table 3).
Table 3. Predicted interacting protein of the ZmKAPs.
5.2.1. Auxin Biosynthesis
The tryptophan (TRP)-dependent/indole-3-pyruvic acid (IPyA) pathway in two-step auxin biosynthesis has been well characterized to finely tune the local auxin synthesis in response to various internal development cues and external stimuli . AtIMPA1/2/3 play redundant roles in the nuclear import of LHP1 and are necessary for flowering regulation . In auxin biosynthesis, LHP1 links SUPERMAN (SUP) and polycomb repressive complex 2 (PRC2) to repress the expression of YUC1 and YUC4 genes and fine-tune local auxin signaling in the floral meristem . However, another report shows that LHP1 is a positive regulator for YUC genes in leaves, suggesting its complicated roles in auxin biosynthesis in different tissues or at different developmental stages . Chromatin remodeling factors CHR11 and CHR17 and Arabidopsis DEAH-box splicing factor PRP16 are predicted to be the downstream targets for IMPα1/2/3/4 and IPO8 (Table 3). CHR11 and CHR17 form a complex with AGAMOUS (AG) at the proximal region of the YUC4 promoter to control its chromatin accessibility for transcription regulation in the floral meristem . The expression of the YUC4 gene is regulated via alternative splicing to generate two splice variants with tissue-specific distributions . The mutation of PRP16 disturbs the expression trait of YUC4 transcript variants in seedlings and cauline leaves, as well as the expression of several other genes involving auxin biosynthesis .
5.2.2. Auxin Transport
Intercellular directional auxin transport depends on PIN-FORMED (PIN) auxin efflux transporters . IMPα1/2/3/4 and IMB1 seem to be responsible for the nuclear import of NAP1-related protein NRP1/2 and nuclear RNA polymerase II subunit NRPB2, which may influence the expression and location of PIN proteins (Table 3). Histone chaperones NRP1 and NRP2 are recruited at the PIN1 locus for local chromatin modulation and coordinate with the Arabidopsis chromatin-remodeling factor INOSITOL AUXOTROPHY 80 (AtINO80) to control the size of meristem inflorescence . NRBP2 is the second-largest subunit of RNA pol II required in mRNA and non-coding RNA biosynthesis . The root tips of the nrpb2-3 mutant display strongly decreased expression and positioning of the PIN1/2/3 proteins, which may change local auxin levels, resulting in WUSCHEL-RELATED HOMEOBOX 5 (WOX5) ectopic expression in the root apical meristem (RAM) . In addition, PRP16 seems to regulate the expression of most PIN genes in flowers or seedlings and influences the proper subcellular localization of PIN1 in roots as well .
5.2.3. Auxin Signaling
The SKP1/CULLIN1/F-BOX(SCF)-type E3 ubiquitin ligase complex is critical for auxin perception and signaling in the nucleus . The F-box proteins TRANSPORT INHIBITOR RESPONSE 1/AUXIN SIGNALING F-BOX (TIR1/AFB), as auxin receptors, mediate the degradation of Auxin/Indole-3-Acetic Acid (AUX/IAA) transcriptional repressors via 26S proteasome (26SP) to release AUXIN RESPONSE FACTOR (ARF) transcription factors, leading to transcriptional reprogramming .
HDA6 is a negative regulator of gene expression, and AtXPO1A functions as an anti-silencing factor by mediating the nucleocytoplasmic partitioning of HDA6 . HDA6 and HDA9 may act synergistically in the auxin signaling pathway to regulate valve cell elongation, and they exhibit functional redundancy in the expression of the ARF4 gene in silique valves . The ortholog HDA108 (Zm00001d050139) is essential for maize development, and the mutant exhibits defects in fertility due to altered ear and tassel growth and microgametogenesis in the anthers .
IMPα, IMB1, IMB3 and XPO1 appear to interact with HSP90, CAND1, CSN4, and UBP14 proteins, which may be involved in the regulation of the SCF complex (Table 3). HSP90 acts as a chaperone of TIR1 to facilitate its nuclear localization and positively regulates its auxin receptor function in the nucleus [142,143]. Increased temperature promotes HSP90-mediated rapid nuclear accumulation of TIR1, suggesting its role in integration between temperature and auxin signaling . CAND1 is likely to function in the assembly and disassembly cycles of the SCF complex through its interactions with CULLIN1 (CUL1) to regulate SCF TIR1 activity . The COP9 signalosome (CSN), composed of eight subunits (CSN1-8), is a conserved nuclear protein complex required for the dynamic modification of cullin . The csn mutant exhibits impaired auxin responses, which may be related to SCF TIR1/AFBs-mediated protein degradation . CSN4 is involved in the control of adventitious root (AR) formation and modulates the activity of CUL1 by affecting de-neddylation for CUL1-NEDD8 . UPB14 acts on the turnover of cellular proteins via 26SP-mediated degradation and is likely to function with TIR1, ARF7, and AUX1 in auxin signaling . A reduction in UPB14 activity results in delayed lateral root primordium (LRP) initiation and impaired lateral root growth, which may be related to the stabilization of the AUX/IAA repressor proteins in the mutant [149,150].
IMB3, IMB4, IPO8 and XPOT are predicted to be potential interaction factors for tRNA-specific methyltransferase TRM4B (Table 3). TRM4B mediates posttranscriptional methylation of RNA cytosine residues to 5-methylcytosine (m5C), including tRNAs, mRNAs, and noncoding RNAs . It promotes the m5C modification of SHORT HYPOCOTYL 2 (SHY2) and INDOLEACETIC ACID-INDUCED PROTEIN 16 (IAA16) mRNA and plays a positive role in mRNA stability in root development . Chromatin remodeling protein PKL and WD40-containing protein PRL1 may serve as the interaction targets of the IMPαs and IMB1s (Table 3). PKL interacts with RETINOBLASTOMA-RELATED 1 (RBR1) to serve as a transcriptional repressor of LATERAL ORGAN BOUNDARIES-DOMAIN 16 (LBD16), which functions in the symmetric division of lateral root (LR) founder cells [153,154]. The suppression of the PKL–RBR1 complex may be relieved from the LBD16 promoter by the IAA14/ARF7/ARF19 signaling pathway to facilitate LR formation . PRL1 encodes a nuclear WD40 protein that has a pleiotropic effect on sugar and several hormone responses and is necessary for the activity of the root stem cell niche and maintenance of the meristem size [155,156]. PRL1 has cell- and tissue-specific expression traits in RAM during primary root growth and appears to configure WOX5 expression in the quiescent center (QC) to act as an upstream regulator of the PLT1/PLT2 dependent pathway .
Additionally, IMPα1/2/3/4, IMB3, and IMB4 appear to interact with another WD40 protein, PCN, and a regulatory component of 26SP. The PCN gene encodes a nuclear WD40 protein that may integrate auxin signaling into the organization and maintenance of apical meristems . It appears to coordinate with BODENLOS (BDL) and TOPLESS (TPL) to mediate the repression of MONOPTEROS (MP) genes and other targets in the auxin signaling pathway . The regulatory particle AAA-ATPase 5a (RPT5a) is a 26SP subunit that possibly facilitates substrate recognition and unfolding [158,159]. In the rpt5a mutant, drastically aberrant auxin and cytokinin responses in roots suggest a role of RPT5a in adjusting the auxin/cytokinin signaling balance to maintain RAM morphology under high boron stress .
5.3. Expression Profiles of ZmKAPs and Corresponding Interaction Partners in Root Development
Several orthologs of interaction partners have shown regulatory roles in root development. To gain additional insight into the potential function of ZmKAPs and correlated interacting partners, we searched for their detailed gene expression patterns in roots through RNA-seq based B73 gene atlas data . Figure 5 shows that seven candidates have similar temporal–spatial expression profiles to those of their putative interacting ZmKAP genes.
Figure 5. Gene expression profiles of ZmKAPs and interacting partners in the root. PR: Primary Root, MZ: Meristem Zone, EZ: Elongation Zone, DZ, Differentiation Zone, CP: Cortical Parenchyma, SR: Seminal Roots, Z1: Zone 1(root tips region), Zone 2 (from the end of Z1 to the point of root hair or lateral root initiation), Zone 3 (lower half of differentiation zone); Zone 4 (upper half of differentiation zone), CR: Crown Roots, BR: Brace Roots, DAS: Day After Sowing, V: Vegetative.
In Arabidopsis, the NRP1/2 double mutant displays a smaller meristem and shorter root than the wild type . Zm00001d050874/ZmNAP1 and Zm00001d016935/ZmNFA104 are orthologs of AtNRP1 and -2 that show high transcription levels in the primary roots and the root tip region. The expression level of ZmIMPα3/4 is the same as that of ZmNAP1, and that of ZmIMPα1/2 is the same as that of ZmNFA104. In maize, the Zm00001d020898/ZmHSP4 gene has upregulated expression induced by heat stress . The Arabidopsis HSP90 affects temperature-mediated root and hypocotyl growth through modulating the auxin response . ZmHSP4 shows high expression levels in primary roots and crown roots, and ZmIMPα1/2 may present co-expression patterns with ZmHSP4 during crown root development. OsCAND1 is a regulator of the G2/M transition for meristem cells involved in the emergence of crown root primordia . In maize, the ortholog of CAND1, Zm00001d053813, exhibits the same expression pattern as ZmIMPα1/2 in root development.
Analogously, Zm00001d008743, Zm00001d020810, Zm00001d013330, Zm00001d033912, and Zm00001d030554 have high expression levels in the primary roots and the root tip region, which may be closely related to the root meristem zone. Zm00001d020810 appears to interact with more than one ZmKAP, while ZmIMB4 and ZmIPO8a exhibit a more similar transcriptional trend to the UPB14 ortholog in maize. ZmIMB3b seems to have the same expression profile in roots as the other three interaction partners. In addition, Zm00001d030554 is the ortholog of the nucleolus localization protein APUM23, and the mutation of APUM23 displays reduced and mislocalized auxin maxima within the root tips, suggesting its potential role in auxin homeostasis maintenance .
6. Conclusions and Perspectives
The KAP-mediated nucleo-cytoplasmic transport of biomacromolecules is the core link in organizing genome activities and triggering downstream cell behaviors. The KAP superfamily and their regulatory mechanisms are highly conserved among eukaryotes and display critical roles in various intracellular biological processes with indispensability in plant growth and development (Supplementary Materials Table S1). However, the KAP superfamily in corn has yet to be studied. Hence, identifying the ZmKaps is essential for understanding new genetic regulatory mechanisms in maize biology. The comparable sub-familial distribution and functional features between maize and Arabidopsis suggest their potential similarity in biological functions and cargo recognition mechanisms (Figure 1). Meanwhile, the expanded number of members in the ZmIMB1, ZmIMB2, ZmIMB3, ZmPLANTKAP, ZmXOP2, ZmTNPO3 and ZmXPOT subfamilies may link to the more complex cellular activities in the physiological environment (Figure 3). The proper nucleo-cytoplasmic partitioning of nuclear proteins is a vital mechanism in the plant signaling pathway, including the members of various hormone signal transduction chains . In searching for the interaction partners of ZmKAPs, we obtained some function cues of ZmKAPs in the auxin pathway (Table 3). Although these cues are enlightening, these potential actors still need to be further explored and investigated in maize.
Considering some transient protein–protein interactions in cells is likely far beyond what the database describes; more interaction partners of ZmKAPs and dynamic transport mechanisms remain to be uncovered in the hormone signal transduction chain. Additionally, how to transport some low-stability proteins or cargos lacking nuclear localization signal motifs remains to be illustrated. For example, the F-BOX protein TIR1 lacks an NLS, and HSP90 serves as its chaperone to function in the folding of the nascent protein and promote its nuclear localization . That is probably one of the nucleo-cytoplasmic transport modes, whereas the vast majority of the regulatory networks of phytohormone-related specific transcription factors remain yet unknown. In other respects, KAPs exhibit multifunction beyond the transport receptors in maintaining protein stability, epigenetic regulation, and miRNA processing and movement [43,82,165,166]. That will contribute to a deep understanding of the functional characteristics in the ZmKAP superfamily. In the future, based on the use of the predicted KAP information to build up a mutant library via reverse genetics techniques such as the CRISPR/CAS9 system, these are all meaningful subjects that warrant additional exploration in maize growth and development.
Supplementary Materials
The supporting information can be downloaded at: References [167,168,169,170,171,172,173,174,175,176,177,178,179] are cited in the supplementary materials.
Author Contributions
Writing and editing, L.J.; formal analysis, G.Z.; visualization, G.Y.; supervision and project administration, J.D. All authors have read and agreed to the published version of the manuscript.
Funding
This research was supported by the National Natural Science Foundation of China (Grant number: 32171921) and the Natural Science Foundation of Shandong Province, China (Grant number: ZR2021MC013).
Institutional Review Board Statement
Not applicable.
Informed Consent Statement
Not applicable.
Data Availability Statement
Data sharing not applicable.
Conflicts of Interest
The authors declare no conflict of interest.
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Figure 1. Phylogenetic tree of the Karyopherin superfamily. Based on validated members of the IMPα and IMPβ families from yeast, humans, and Arabidopsis, each protein sequence was used as a query to perform BLASTP searches in Phytozome v13 ( (accessed on 4 October 2022)), NCBI ( (accessed on 4 October 2022)), and MaizeGDB ( (accessed on 4 October 2022)), remove the non-representative splicing forms of the same gene locus, and confirm sequences of non-redundant candidates by phylogenetic analysis with the homologous series of the other species. Saccharomyces cerevisiae (Sc), Homo sapiens (Hs), Chlamydomonas reinhardtii (Cre), Marchantia polymorpha (Mapoly), Selaginella moellendorffii (Smo.), Thuja plicata (Thupl.), Amborella trichopoda (AmTr.), Arabidopsis thaliana (At), Zea mays (Zm); ZmIMPα proteins in blue font and ZmIMPβ in red.
Figure 2. IMPORTINα family in maize and Arabidopsis. (A) Schematic view of the domains conserved between AtIMPA and ZmIMPα proteins according to Pfam Database ( (accessed on 4 October 2022)) and CCD Tools ( (accessed on 4 October 2022)); (B) Heat map of the expression pattern of ZmIMPα genes, with the expression value calculated by log2 (FPKM). SAM: shoot apical meristem, NU: nucellus, em: embryo, en: endosperm, HAP: Hours after Pollination, DAP: Day after Pollination; (C) Signatures of the Importin β binding (IBB) domain of the ZmIMPα1 protein predicted by AlphaFold Protein Structure Database ( (accessed on 4 October 2022)); multiple amino acid sequences of the IBB domain aligned using CLUSTALW, three conserved motifs highlighted in red and rectangle boxes.
Figure 3. IMPORTINβ family in maize and Arabidopsis. (A) Schematic representation of the conserved domains between AtIMPβ and ZmIMPβ proteins according to Pfam Database and CCD Tools; (B) Heat map of the expression profile of ZmIMPβ genes in different tissues, with the expression value calculated via log2 (FPKM). SAM: shoot apical meristem, NU: nucellus, em: embryo, en: endosperm, HAP: Hours after Pollination, DAP: Day after Pollination.
Figure 4. A schematic illustration of Arabidopsis KAP-mediated nucleo-cytoplasmic transport in hormone signaling for plant development. (A) Cytokinin regulates cell division by promoting nuclear shuttling of transcription factor MYB3R4, mediated by AtIMPA3 and AtIMPA6, in the shoot apical meristem (SAM). (B) AtIMB4 mediates the nuclear partitioning of GRF-INTERACTING FACTOR1 (GIF1)/ANGUSTIFOLIA3 and JANUS, which antagonistically regulate PLETHORA1 (PLT1) transcription. (C) AtSAD2 and AtKPNB1 act as negative regulators in abscisic acid (ABA) signaling. The atsad2 mutant displays an ABA hypersensitivity response during seed germination and seedling growth. AtKPNB1 is involved in controlling ABA-induced stomatal closure under drought conditions. (D) AtXPO1A mediates the nuclear export of a WD40 repeat-containing protein, XIW1 (XPO1-interacting WD40 protein 1), which maintains the stability of ABA INSENSITIVE 5 (ABI5) in the nucleus. The schematic illustration was drawn with BIORENDER ( (accessed on 4 October 2022)).
Figure 5. Gene expression profiles of ZmKAPs and interacting partners in the root. PR: Primary Root, MZ: Meristem Zone, EZ: Elongation Zone, DZ, Differentiation Zone, CP: Cortical Parenchyma, SR: Seminal Roots, Z1: Zone 1(root tips region), Zone 2 (from the end of Z1 to the point of root hair or lateral root initiation), Zone 3 (lower half of differentiation zone); Zone 4 (upper half of differentiation zone), CR: Crown Roots, BR: Brace Roots, DAS: Day After Sowing, V: Vegetative.
Table 1. List of putative Karyopherin gene family members in Zea mays.
| | Gene Name a | Locus ID b | Chromosomal Location c | Transcript ID | Putative Proteins d |
:---: :---: :---: |
| Chr | Chr_start | Chr_end | Length (aa) | MW (kDa) | Subcellular Location |
| IMPα | ZmIMPα1 | Zm00001d008345 | 8 | 5938159 | 5944491 (−) | T001 | 527 | 57.85 | Nucleus/Cytoplasm |
| ZmIMPα2 | Zm00001d040274 | 3 | 35350411 | 35356233 (+) | T001 | 529 | 57.95 | Nucleus/Cytoplasm |
| ZmIMPα3 | Zm00001d037606 | 6 | 131468248 | 85071126 (−) | T001 | 528 | 58.13 | Nucleus/Cytoplasm |
| ZmIMPα4 | Zm00001d009850 | 8 | 85065908 | 131476305 (−) | T005 | 529 | 58.20 | Nucleus/Cytoplasm |
| ZmIMPα5 | Zm00001d040153 | 3 | 29316628 | 29318539 (+) | T004 | 183 | 20.42 | Nucleus/Cytoplasm |
| ZmIMPα6 | Zm00001d022536 | 7 | 179671127 | 179674969 (+) | T008 | 568 | 61.71 | Nucleus |
| ZmIMPα7 | Zm00001d008640 | 8 | 15537598 | 15544131 (+) | T002 | 526 | 56.54 | Nucleus/Cytoplasm |
| IMPβ | ZmIMB1a | Zm00001d030694 | 1 | 153742904 | 153749377 (+) | T002 | 1074 | 116.51 | Nucleus/Cytoplasm |
| ZmIMB1b | Zm00001d041556 | 3 | 127112005 | 127118515 (−) | T002 | 987 | 107.94 | Nucleus/Cytoplasm |
| ZmIMB1c | Zm00001d038021 | 6 | 145393970 | 145398983 (−) | T001 | 879 | 96.77 | Nucleus/Cytoplasm |
| ZmIMB1d | Zm00001d010512 | 8 | 118588081 | 118591573 (−) | T001 | 876 | 96.15 | Nucleus/Cytoplasm |
| ZmIMB2a | Zm00001d002936 | 2 | 27303853 | 27322287 (+) | T010 | 891 | 98.80 | Nucleus/Cytoplasm |
| ZmIMB2b | Zm00001d026696 | 10 | 150180000 | 150204009 (+) | T005 | 890 | 98.86 | Nucleus/Cytoplasm |
| ZmIMB3a | Zm00001d021893 | 7 | 165532355 | 165542662 (−) | T002 | 1126 | 123.26 | Nucleus/Cytoplasm |
| ZmIMB3b | Zm00001d033632 | 1 | 269308497 | 269321829 (+) | T008 | 1132 | 123.78 | Nucleus/Cytoplasm |
| ZmIMB4 | Zm00001d028511 | 1 | 37580598 | 37594480 (−) | T008 | 1047 | 114.94 | Nucleus/Cytoplasm |
| ZmIMB5 | Zm00001d045725 | 9 | 35215354 | 35239296 (−) | T001 | 1028 | 113.35 | Nucleus envelope/Cytosol |
| ZmIPO8a | Zm00001d050526 | 4 | 96801958 | 96828810 (+) | T008 | 1145 | 128.01 | Nucleus envelope/Cytosol |
| ZmIPO8b | Zm00001d016479 | 5 | 164246013 | 164266562 (+) | T001 | 1036 | 131.22 | Nucleus envelope/Cytosol |
| ZmKA120 | Zm00001d007225 | 2 | 225128676 | 225140701 (+) | T019 | 1115 | 116.49 | Nucleus/Cytoplasm |
| ZmXPO1a | Zm00001d012815 | 5 | 776419 | 787466 (+) | T022 | 1151 | 132.30 | Nucleus envelope/Cytosol |
| ZmXPO1b | Zm00001d034914 | 1 | 305341236 | 305352529 (−) | T037 | 1122 | 128.54 | Nucleus envelope/Cytosol |
| ZmXPO2a | Zm00001d033764 | 1 | 272997605 | 273005246 (+) | T002 | 981 | 108.32 | Nucleus/Cytoplasm |
| ZmXPO2b | Zm00001d013417 | 5 | 10817793 | 10829003 (+) | T005 | 982 | 108.52 | Nucleus/Cytoplasm |
| ZmXPOTa | Zm00001d022125 | 7 | 170837895 | 170846340 (+) | T002 | 978 | 107.96 | Nucleus/Cytoplasm |
| ZmXPOTb | Zm00001d006845 | 2 | 217770387 | 217778785 (+) | T005 | 1024 | 113.22 | Nucleus/Cytoplasm |
| ZmXPO4 | Zm00001d032704 | 1 | 235324863 | 235346931 (−) | T036 | 1165 | 129.84 | Nucleus/Cytoplasm |
| ZmXPO5 | Zm00001d009270 | 8 | 49685540 | 49721525 (+) | T001 | 1175 | 130.20 | Nucleus/Cytoplasm |
| ZmXPO7 | Zm00001d037100 | 6 | 112267718 | 112290870 (+) | T051 | 1067 | 121.03 | Nucleus/Cytoplasm |
| ZmTNPO3a | Zm00001d052632 | 4 | 195421971 | 195454136 (+) | T005 | 1038 | 114.09 | Cytoplasm |
| ZmTNPO3b | Zm00001d014033 | 5 | 29806869 | 29825510 (−) | T001 | 564 | 62.27 | Cytoplasm |
| ZmTNPO3c | Zm00001d032699 | 1 | 235073263 | 235106928 (−) | T030 | 981 | 109.64 | Cytoplasm |
| ZmPLANTKAPa | Zm00001d048628 | 4 | 1742938 | 1750365 (+) | T001 | 1092 | 120.63 | Nucleus envelope/Cytosol |
| ZmPLANTKAPb | Zm00001d019335 | 7 | 28400881 | 28407780 (−) | T004 | 655 | 73.54 | Nucleus envelope/Cytosol |
a Name refers to systematic designation among members of the Karyopherin family applied to Zea mays based on homology against Arabidopsis thaliana and Homo sapiens; b Gene accession number in maizeGDB (MAIZE GENETICS AND GENOMICS DATABASE); c Chromosomal location of the ZmIMPα and ZmIMPβ genes based on the Zm-B73-REFERENCE-GRAMENE (V4.0); d Basic physicochemical properties of the putative ZmIMPα and ZmIMPβ proteins, and subcellular location predicted by UniProt ( (accessed on 4 October 2022)).
Table 2. Classification of NLSs and NESs recognized by KAPs in plants.
| Type | Consensus Motifs | Cargo | Sequence | NTR | Source |
:---: :---: :---: |
| MP-cNLS | Class I— KR (K/R) R or K (K/R) RK | AtFHY1/AtFHL | 40KKRK | AtIMPA1 | Arabidopsis |
| AtPARP2 | 48KRKR | AtIMPA2 | Arabidopsis |
| AtLHP1 | 173 RKRKRK | AtIMPA1/2/3 | Arabidopsis |
| AtMINIYO | 253 KLKKRRK | AtIMPA4 | Arabidopsis |
| Class II— (P/R) XXKR (^DE) (K/R) | AtVRN1 | 173 PTPTPKIPKKRGRKKKNADPE | AtIMPA1/2/3 | Arabidopsis |
| Class III— KRX (W/F/Y) XXAF | AtPIP5K2 | 239 ATRKRSSVDSGAG SLTGEKIFPRIC | AtIMPA6/9 | Arabidopsis |
| Class IV— (R/P) XXKR (K/R) (^DE) | – | – | – | – |
| Class V— LGKR (K/R) (W/F/Y) | VQ-protein | 92 LGLGKRKRGPGVSGGKQTKRRSR | AtIMPA1/2/3 | Arabidopsis |
| BP-cNLS | Class VI— KRX10–12K(KR) (KR) or KRX10–12K(KR) X (K/R) | AtMINIYO | 1401RKR–1414RYKK, | AtIMPA4 | Arabidopsis |
| OsWRKY62/OsWRKY76 | 8RK–36KKK | OsIMPα1 | Oryza Sativa |
| OsCOP1 | 294RKKR–312KRR | OsIMPα1b | Oryza sativa |
| ZmOpaque2 | 230RKRK–241RRSRYRK | OsIMPα1b ZmIMPα4 | Oryza sativa , Zea mays |
| PY-NLS | (basic/hydrophobic) Xn— (R/H/K) (X)2–5 PY | AtFRA1 | 311KKRK–320PY | AtIMB4 | Arabidopsis |
| M9-like domain | AtGRP7 | 112 SGGGG SYGGGGGRREGGGGYSG | AtTRN1 | Arabidopsis |
| Other NLS | Zinc finger motifs | PsLSD1 | 7CNGCRNMLLYPRGATNVCCALC– 46CGGCRTLLMYTRGATSVRCSCC– 84CANCRTTLMYPYGAPSVKCAVC | AtIMPA1 | Pisum sativa |
| NES | Φ-X2–3-Φ-X2–3-Φ-X-Φ | OXS2 | 699LEAWIEQMQL/LGALLEQMQL Arabidopsis [84,85] |
| AtFHY1 | 54LLPL | AtXPO1 | Arabidopsis |
| OsWRKY62 | 308VDQIPHIPV | AtXPO1 | Oryza Sativa |
| CMV 2b | 79L-85L-87L | AtXPO1 | Mosaic Virus |
cNLS: classical nuclear locational signals. MP: monopartite, BP: bipartite, PY: Proline-Tyrosine, NES: nuclear export signals, NTR: nuclear transport receptor, X: any amino acid, ^D/E: any amino acid except Asp or Glu, Φ: for Leu/Val/Ile/Phe/Met. FHY1: FAR-RED elongated hypocotyl 1, FHL: FHY1-like, PARP: poly (ADP-Ribose) polymerase, LHP1: like heterochromatin protein 1, VRN1: vernalization1, PIP5K2: phosphatidylinositol 4-phosphate 5-kinase 2, VQ-protein: VQ motif-containing protein, COP1: photomorphogenic 1, FRA1: fragile fiber 1, GRP7: glycine-rich RNA-binding protein, OXS2: oxidative stress 2, CMV 2b: cucumber mosaic virus 2b.
Table 3. Predicted interacting protein of the ZmKAPs.
| NTR | Putative Interactor in Maize | Interactive Score | Ortholog of the Putative Interactor in Arabidopsis |
:---: :---: |
| Name | Gene ID |
| ZmIMPα1/2/3/4 (P, S) | Zm00001d009312 | P-0.208, S-0.582 | CHR11 | AT3G06400 |
| Zm00001d040831 | CHR17 | AT5G18620 |
| ZmIMPα1/2/3/4 (S) | Zm00001d014449 | S-0.781 | LHP1 | AT5G17690 |
| ZmIMPα1/2/3/4 (P) | Zm00001d050874 | P-0.242 | NRP1 | AT5G17690 |
| Zm00001d016935 | NRP2 | AT1G74560 |
| ZmIMB1c/d (P) | Zm00001d033218 | P-0.333 | NRPB2 | AT4G21710 |
| Zm00001d013683 |
| ZmIMPα1/2 (P) | Zm00001d020898 | P-0.631 | HSP90.2 | AT5G56030 |
| Zm00001d031332 |
| ZmIMPα1/2 (P), ZmXPO1 (S) | Zm00001d053813 | P-0.208, S-0.582 | CAND1 | AT2G02560 |
| ZmIMPα1/2/3/4 (P), ZmIMB1 (P) | Zm00001d028143 | P-0.243, P-0.363 | CSN4 | AT5G42970 |
| ZmIMPα1/2 (P), ZmIMB3 (P) | Zm00001d008743 | P-0.299, P-0.255 | UBP14 | AT3G20630 |
| ZmIMPα1/2/3/4 (P), ZmIMB1c/d (P) | Zm00001d045109 | P-0.299, P-0.303 | PKL | AT2G25170 |
| ZmIMPα4 (S) | Zm00001d033309 | S-0.421 | PRL1 | AT4G15900 |
| ZmIMB3, ZmIMB4 (S), ZmIPO8 (S), ZmXPOT (S) | Zm00001d020810 | S-0.716, S-0.640, S-0.505 | TRM4B | AT2G22400 |
| ZmIMPα1/2/3/4 (P), ZmXPO1/5 (S) | Zm00001d050139 | P-0.270, S-0.805 | HDA6 | AT5G63110 |
| ZmIMB3, ZmIMB4 (S) | Zm00001d013330 | S-0.639 | PCN | AT4G07410 |
| Zm00001d033912 |
| ZmIPO8 (S) | Zm00001d006459 | S-0.655 | PRP16 | AT5G13010 |
| ZmIMB3, ZmIMB4 (S) | Zm00001d030554 | S-0.489 | APUM23 | AT1G72320 |
| ZmIMPα1/2/3/4 (P) | Zm00001d037481 | P-0.231 | RPT5A | AT3G05530 |
| Zm00001d018409 |
(S) for Data analysis from STRING ( (accessed on 4 October 2022)), (P) for data analysis from plant.MAP ( (accessed on 4 October 2022)), (P, S) for Data from both STRING and plant.MAP databases. NTR: Nuclear transport receptor, CHR11/17: CHROMATIN REMODELING 11/17, LHP1: LIKE HETEROCHROMATIN PROTEIN 1, NRP1/2: NAP1-RELATED PROTEIN 1/2, NRPB1/2: Nuclear RNA polymerase II (RNA Pol II) subunit 2, HSP90: HEAT SHOCK PROTEIN 90, CAND1: Cullin-Associated and Neddylation-Dissociated, CSN4: CONSTITUTIVE PHOTOMORPHOGENIC9 (COP9) signalosome subunit 4, UBP14: UBIQUITIN-SPECIFIC PROTEASE14, PKL: PICKLE, PRL1: Pleiotropic Regulatory Locus 1, TRM4B: tRNA-specific methyltransferase 4B, HDA6: Histone deacetylase 6, PCN: POPCORN, PRP16: pre-mRNA-processing factor 16, APUM23: Arabidopsis Pumilio 23, RPT5a: Regulatory particle AAA-ATPase 5a.
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Jin, L.; Zhang, G.; Yang, G.; Dong, J. Identification of the Karyopherin Superfamily in Maize and Its Functional Cues in Plant Development. Int. J. Mol. Sci.2022, 23, 14103.
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Jin L, Zhang G, Yang G, Dong J. Identification of the Karyopherin Superfamily in Maize and Its Functional Cues in Plant Development. International Journal of Molecular Sciences. 2022; 23(22):14103.
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Jin, Lu, Guobin Zhang, Guixiao Yang, and Jiaqiang Dong. 2022. "Identification of the Karyopherin Superfamily in Maize and Its Functional Cues in Plant Development" International Journal of Molecular Sciences 23, no. 22: 14103.
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Jin, L., Zhang, G., Yang, G., & Dong, J. (2022). Identification of the Karyopherin Superfamily in Maize and Its Functional Cues in Plant Development. International Journal of Molecular Sciences, 23(22), 14103.
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Jin, L.; Zhang, G.; Yang, G.; Dong, J. Identification of the Karyopherin Superfamily in Maize and Its Functional Cues in Plant Development. Int. J. Mol. Sci.2022, 23, 14103.
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Jin L, Zhang G, Yang G, Dong J. Identification of the Karyopherin Superfamily in Maize and Its Functional Cues in Plant Development. International Journal of Molecular Sciences. 2022; 23(22):14103.
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Jin, Lu, Guobin Zhang, Guixiao Yang, and Jiaqiang Dong. 2022. "Identification of the Karyopherin Superfamily in Maize and Its Functional Cues in Plant Development" International Journal of Molecular Sciences 23, no. 22: 14103.
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Jin, L., Zhang, G., Yang, G., & Dong, J. (2022). Identification of the Karyopherin Superfamily in Maize and Its Functional Cues in Plant Development. International Journal of Molecular Sciences, 23(22), 14103.
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Facial Swelling As It Relates To TMJ Issues
by Michael Karegeannes | Aug 6, 2018 | Symptoms, Treatment | 7 comments
I recently received two inquiries in one week requesting explanations for facial swelling and as it might relate to TMJ issues. One request was from a visitor of my TreatingTMJ.com website, suffering some facial swelling after dry needling to their masseter muscle and the other was from another PT colleague who has a young teen presenting with intermittent facial swelling, so I took this as a sign that maybe I need to put a blog together to highlight some of the more common causes, but not all-inclusive, causes of facial swelling.
Pain originating in the salivary glands is typical of inflammatory, infectious, traumatic, or neoplastic origin. Common salivary gland disorders that are accompanied by pain include sialoadenitis, sialolithiasis, epidemic parotitis, and tumors. Usually, diagnosis of salivary gland pain is not difficult, due to the accompanying signs and symptoms, such as pain occurring on eating, or swelling, firmness, or tenderness of the affected gland. I will describe below a few of these.
Sialadenitis
Is an infection of the salivary glands. It is usually caused by a virus or bacteria. The parotid (in front of the ear) and submandibular (under the chin) glands are most commonly affected. Sialadenitis may be associated with pain, tenderness, redness, and gradual, localized swelling of the affected area.
Sialolithiasis
Is the medical term for the formation of salivary duct stones in a patient’s salivary glands. The main function area of the salivary glands is to produce saliva in the mouth, which aids in our everyday lives by making chewing, swallowing, talking, and even eating possible. The formation of salivary gland stones can occur as a result of an infection, virus, and in certain cases, chemicals in our saliva that can become crystallized and block the salivary ducts.
Salivary Duct Stones
Can occur in all salivary glands, but are most commonly found in the submandibular glands, which are located in the back of the mouth and on both sides of the jaw. Stones in the parotid glands are also possible but are much rarer.
Epidemic Parotitis (mumps)
Mumps is an acute, contagious, viral disease that causes painful enlargement of the salivary or parotid glands. Unvaccinated children between the ages of 2 and 12 are most commonly infected, but the infection can occur in other age groups. The mumps are caused by a virus which is spread from person-to-person by respiratory droplets or direct contact with articles that have been contaminated with infected saliva. Avoiding this contact would subsequently greatly reduce the chance for the infection to take hold.
Tumors
We will hold off on that for now and proceed with 2 more thoughts:
Sjogrens syndrome
Sjogren’s (SHOW-grins) syndrome is a disorder of your immune system identified by its two most common symptoms — dry eyes and a dry mouth.
The condition often accompanies other immune system disorders, such as rheumatoid arthritis and lupus. In Sjogren’s syndrome, the mucous membranes and moisture-secreting glands of your eyes and mouth are usually affected first — resulting in decreased tears and saliva.
Although you can develop Sjogren’s syndrome at any age, most people are older than 40 at the time of diagnosis. The condition is much more common in women. Treatment focuses on relieving symptoms.
Symptoms
The two main symptoms of Sjogren’s syndrome are:
Dry eyes. Your eyes might burn, itch or feel gritty — as if there’s sand in them.
Dry mouth. Your mouth might feel like it’s full of cotton, making it difficult to swallow or speak.
Some people with Sjogren’s syndrome also have one or more of the following:
Joint pain, swelling, and stiffness
Swollen salivary glands — particularly the set located behind your jaw and in front of your ears
Skin rashes or dry skin
Vaginal dryness
Persistent dry cough
Prolonged fatigue
Dental Abscess with Facial Cellulitis
A dental abscess is an infection at the base of a tooth. It means a pocket of pus has formed at the tip of a tooth root in your jaw bone. If the infection isn’t treated, it can appear as a swelling on the gum near the tooth. More serious infections spread to the face. This causes your face to swell (cellulitis). This is a very serious condition. Once the swelling begins, it can spread quickly.
A dental abscess usually starts with a crack or cavity in a tooth. The pain is often made worse by drinking hot or cold beverages or biting on hard foods. The pain may spread from the tooth to your ear or the area of your jaw on the same side.
I hope you have found this blog helpful and informative. I have found over the years as you continue to treat more TMJ/TMD patients, you will inevitably run into those cases that are more medical in nature and require your referral expertise to the team of specialists you have built relations with.
All the best!
Mike
Author
Recent Posts
Michael Karegeannes
Owner / PT at Freedom Physical Therapy Services, S.C.
Michael Karegeannes, PT, MHSc, LAT, MTC,CFC, CCTT, CMTPT is the owner of Freedom Physical Therapy Services in WI and is one of the few physical therapists in the United States recognized as a Certified Cervical and Temporomandibular Therapist with the AAOP and PTBCTT.
Latest posts by Michael Karegeannes (see all)
Bruxism vs Bracing - July 28, 2025
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7 Comments
Adam on June 9, 2020 at 6:24 AM Any follow up on PT colleague who has a young teen presenting with intermittent facial swelling?
Reply
2. Lisa Hill on March 8, 2021 at 3:46 AM I have the problem of the swelling on my face. My face gets swelled after using these cautions my face gets better. Thanks for the information.
Reply
Robyn on January 29, 2022 at 12:06 PM I have been having facial swelling after dry needing on the masseter muscle on the left side. Noone will do anything
Reply
Susan on March 22, 2021 at 4:50 PM Where on the face does swelling occur with tmj? I have swelling and my tooth hurts but nothing is found with the tooth. I also have tmj but the swelling is very low off from my chin. Is that tmj swelling?
Reply
4. Melissa on April 30, 2021 at 7:36 PM I am really freaking out. I have swelling on the edge of my face and my massage therapist think its a swollen salivary gland, but at the same time I have TMJ symptoms such as ringing, facial pain, clicking on right and deviation to right when I open my mouth. My lower and upper jaw does not align anymore. I dont know what I have anymore.
Reply
5. Laura Callihoo on May 1, 2021 at 3:10 PM My face swells and when it does it pops like a pimple inside my mouth. It has done this for 30 years and I even popped it in front of a doctor and almost hit her with the pus but nothing has ever been done about it.
Reply
John on June 25, 2021 at 5:46 PM I have been diagnosed with tmd with bilateral dislocation of the cartilage. I haven’t had any corrective surgery but I still have all the classic freys syndrome symptoms. Is it possible my tmd has caused this.
Reply
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17: Electrochemistry
Unit 4: Equilibrium in Chemical Reactions
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Fri, 14 Aug 2020 21:36:10 GMT
17.3: Concentration Effects and the Nernst Equation
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3. General Chemistry
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7. 17.3: Concentration Effects and the Nernst Equation
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17.3: Concentration Effects and the Nernst Equation
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17.2: The Gibbs Free Energy and Cell Voltage
17.4: Batteries and Fuel Cells
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Table of contents
1. Learning Objectives
2. The Effect of Concentration on Cell Potential: The Nernst Equation
1. The Power of the Nernst Equation
2. Example (\PageIndex{1})
1. Strategy:
2. Solution
3. Exercise \(\PageIndex{1}\)/Unit_4:_Equilibrium_in_Chemical_Reactions/17:_Electrochemistry/17.3:_Concentration_Effects_and_the_Nernst_Equation#Exercise_.5C(.5CPageIndex.7B1.7D.5C))
Concentration Cells
Example (\PageIndex{2})
Strategy:
Solution
Exercise (\PageIndex{2})
Using Cell Potentials to Measure Solubility Products
Example (\PageIndex{3}): Solubility of lead(II) sulfate
Strategy:
Solution
Exercise (\PageIndex{3})
Using Cell Potentials to Measure Concentrations
Example (\PageIndex{4}): Measuring pH
Strategy:
Solution
Exercise (\PageIndex{4})
Summary
Learning Objectives
Relate cell potentials to Gibbs energy changes
Use the Nernst equation to determine cell potentials at nonstandard conditions
Perform calculations that involve converting between cell potentials, free energy changes, and equilibrium constants
The Nernst Equation enables the determination of cell potential under non-standard conditions. It relates the measured cell potential to thereaction quotient and allows the accurate determination of equilibrium constants (including solubility constants).
The Effect of Concentration on Cell Potential: The Nernst Equation
Recall that the actual free-energy change for a reaction under nonstandard conditions, (\Delta{G}), is given as follows:
[\Delta{G} = \Delta{G°} + RT \ln Q \label{Eq1} ]
We also know that (ΔG = −nFE_{cell}) (under non-standard conditions) and (ΔG^o = −nFE^o_{cell}) (under standard conditions). Substituting these expressions into Equation (\ref{Eq1}), we obtain
[−nFE_{cell} = −nFE^o_{cell} + RT \ln Q \label{Eq2} ]
Dividing both sides of this equation by (−nF),
[E_\textrm{cell}=E^\circ_\textrm{cell}-\left(\dfrac{RT}{nF}\right)\ln Q \label{Eq3} ]
Equation (\ref{Eq3}) is called the Nernst equation, after the German physicist and chemist Walter Nernst (1864–1941), who first derived it. The Nernst equation is arguably the most important relationship in electrochemistry. When a redox reaction is at equilibrium ((ΔG = 0)), then Equation (\ref{Eq3}) reduces to Equation (\ref{Eq31}) and (\ref{Eq32}) because (Q = K), and there is no net transfer of electrons (i.e., E cell = 0).
[E_\textrm{cell}=E^\circ_\textrm{cell}-\left(\dfrac{RT}{nF}\right)\ln K = 0 \label{Eq31} ]
since
[E^\circ_\textrm{cell}= \left(\dfrac{RT}{nF}\right)\ln K \label{Eq32} ]
Substituting the values of the constants into Equation (\ref{Eq3}) with (T = 298\, K) and converting to base-10 logarithms give the relationship of the actual cell potential (E cell), the standard cell potential (E°cell), and the reactant and product concentrations at room temperature (contained in (Q)):
[E_{\textrm{cell}}=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \label{Eq4} ]
The Power of the Nernst Equation
The Nernst Equation ((\ref{Eq3})) can be used to determine the value of E cell, and thus the direction of spontaneous reaction, for any redox reaction under any conditions.
Equation (\ref{Eq4}) allows us to calculate the potential associated with any electrochemical cell at 298 K for any combination of reactant and product concentrations under any conditions. We can therefore determine the spontaneous direction of any redox reaction under any conditions, as long as we have tabulated values for the relevant standard electrode potentials. Notice in Equation (\ref{Eq4}) that the cell potential changes by 0.0591/n V for each 10-fold change in the value of (Q) because log 10 = 1.
Example (\PageIndex{1})
The following reaction proceeds spontaneously under standard conditions because E°cell > 0 (which means that ΔG° < 0):
[\ce{2Ce^{4+}(aq) + 2Cl^{–}(aq) -> 2Ce^{3+}(aq) + Cl2(g)}\;\; E^°_{cell} = 0.25\, V \nonumber ]
Calculate (E_{cell}) for this reaction under the following nonstandard conditions and determine whether it will occur spontaneously: [Ce 4+] = 0.013 M, [Ce 3+] = 0.60 M, [Cl−] = 0.0030 M, (P_\mathrm{Cl_2}) = 1.0 atm, and T = 25°C.
Given: balanced redox reaction, standard cell potential, and nonstandard conditions
Asked for: cell potential
Strategy:
Determine the number of electrons transferred during the redox process. Then use the Nernst equation to find the cell potential under the nonstandard conditions.
Solution
We can use the information given and the Nernst equation to calculate E cell. Moreover, because the temperature is 25°C (298 K), we can use Equation (\ref{Eq4}) instead of Equation (\ref{Eq3}). The overall reaction involves the net transfer of two electrons:
[2Ce^{4+}{(aq)} + 2e^− \rightarrow 2Ce^{3+}{(aq)}\nonumber ]
[2Cl^−{(aq)} \rightarrow Cl{2(g)} + 2e^−\nonumber ]
so n = 2. Substituting the concentrations given in the problem, the partial pressure of Cl 2, and the value of E°cell into Equation (\ref{Eq4}),
[\begin{align}E_\textrm{cell} & =E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \
& =\textrm{0.25 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{[\mathrm{Ce^{3+}}]^2P_\mathrm{Cl_2}}{[\mathrm{Ce^{4+}}]^2[\mathrm{Cl^-}]^2}\right) \
& =\textrm{0.25 V}-[(\textrm{0.0296 V})(8.37)]=\textrm{0.00 V}\end{align} \nonumber ]
Thus the reaction will not occur spontaneously under these conditions (because E = 0 V and ΔG = 0). The composition specified is that of an equilibrium mixture
Exercise (\PageIndex{1})
Molecular oxygen will not oxidize (MnO_2) to permanganate via the reaction
[\ce{4MnO2(s) + 3O2(g) + 4OH^{−} (aq) -> 4MnO^{−}4(aq) + 2H2O(l)} \;\;\; E°_{cell} = −0.20\; V\nonumber ]
Calculate (E_{cell}) for the reaction under the following nonstandard conditions and decide whether the reaction will occur spontaneously: pH 10, (P_\mathrm{O_2})= 0.20 atm, [MNO 4−] = 1.0 × 10−4 M, and T = 25°C.
Answer
E cell = −0.22 V; the reaction will not occur spontaneously.
Applying the Nernst equation to a simple electrochemical cell such as the Zn/Cu cell allows us to see how the cell voltage varies as the reaction progresses and the concentrations of the dissolved ions change. Recall that the overall reaction for this cell is as follows:
[Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)\;\;\;E°cell = 1.10 V \label{Eq5} ]
The reaction quotient is therefore (Q = [Zn^{2+}]/[Cu^{2+}]). Suppose that the cell initially contains 1.0 M Cu 2+ and 1.0 × 10−6 M Zn 2+. The initial voltage measured when the cell is connected can then be calculated from Equation (\ref{Eq4}):
[\begin{align}E_\textrm{cell} & =E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log\dfrac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]}\
& =\textrm{1.10 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{1.0\times10^{-6}}{1.0}\right)=\textrm{1.28 V}\end{align} \label{Eq6} ]
Thus the initial voltage is greater than E° because (Q<1). As the reaction proceeds, [Zn 2+] in the anode compartment increases as the zinc electrode dissolves, while [Cu 2+] in the cathode compartment decreases as metallic copper is deposited on the electrode. During this process, the ratio Q = [Zn 2+]/[Cu 2+] steadily increases, and the cell voltage therefore steadily decreases. Eventually, [Zn 2+] = [Cu 2+], so Q = 1 and E cell = E°cell. Beyond this point, [Zn 2+] will continue to increase in the anode compartment, and [Cu 2+] will continue to decrease in the cathode compartment. Thus the value of Q will increase further, leading to a further decrease in E cell. When the concentrations in the two compartments are the opposite of the initial concentrations (i.e., 1.0 M Zn 2+ and 1.0 × 10−6 M Cu 2+), Q = 1.0 × 10 6, and the cell potential will be reduced to 0.92 V.
Figure (\PageIndex{1}): The Variation of E cell with Log Q for a Zn/Cu Cell. Initially, log Q < 0, and the voltage of the cell is greater than E°cell. As the reaction progresses, log Q increases, and E cell decreases. When [Zn 2+] = [Cu 2+], log Q = 0 and E cell = E°cell = 1.10 V. As long as the electrical circuit remains intact, the reaction will continue, and log Q will increase until Q = K and the cell voltage reaches zero. At this point, the system will have reached equilibrium.
The variation of E cell with (\log{Q}) over this range is linear with a slope of −0.0591/n, as illustrated in Figure (\PageIndex{1}). As the reaction proceeds still further, (Q) continues to increase, and E cell continues to decrease. If neither of the electrodes dissolves completely, thereby breaking the electrical circuit, the cell voltage will eventually reach zero. This is the situation that occurs when a battery is “dead.” The value of (Q) when E cell = 0 is calculated as follows:
[\begin{align}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q=0 \
E^\circ &=\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \
\log Q &=\dfrac{E^\circ n}{\textrm{0.0591 V}}=\dfrac{(\textrm{1.10 V})(2)}{\textrm{0.0591 V}}=37.23 \
Q &=10^{37.23}=1.7\times10^{37}\end{align} \label{Eq7} ]
Recall that at equilibrium, (Q = K). Thus the equilibrium constant for the reaction of Zn metal with Cu 2+ to give Cu metal and Zn 2+ is 1.7 × 10 37 at 25°C.
The Nernst Equation: The Nernst Equation (opens in new window) [youtu.be]
Concentration Cells
A voltage can also be generated by constructing an electrochemical cell in which each compartment contains the same redox active solution but at different concentrations. The voltage is produced as the concentrations equilibrate. Suppose, for example, we have a cell with 0.010 M AgNO 3 in one compartment and 1.0 M AgNO 3 in the other. The cell diagram and corresponding half-reactions are as follows:
[\ce{Ag(s)\,|\,Ag^{+}}(aq, 0.010 \;M)\,||\,\ce{Ag^{+}}(aq, 1.0 \;M)\,|\,\ce{Ag(s)} \label{Eq8} ]
cathode:
[\ce{Ag^{+}} (aq, 1.0\; M) + \ce{e^{−}} \rightarrow \ce{Ag(s)} \label{Eq9} ]
anode:
[\ce{Ag(s)} \rightarrow \ce{Ag^{+}}(aq, 0.010\; M) + \ce{e^{−}} \label{Eq10} ]
Overall
[\ce{Ag^{+}}(aq, 1.0 \;M) \rightarrow \ce{Ag^{+}}(aq, 0.010\; M) \label{Eq11} ]
As the reaction progresses, the concentration of (Ag^+) will increase in the left (oxidation) compartment as the silver electrode dissolves, while the (Ag^+) concentration in the right (reduction) compartment decreases as the electrode in that compartment gains mass. The total mass of (Ag(s)) in the cell will remain constant, however. We can calculate the potential of the cell using the Nernst equation, inserting 0 for E°cell because E°cathode = −E°anode:
[\begin{align} E_\textrm{cell}&=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \[4pt] &=0-\left(\dfrac{\textrm{0.0591 V}}{1}\right)\log\left(\dfrac{0.010}{1.0}\right) \[4pt] &=\textrm{0.12 V} \end{align} \nonumber ]
An electrochemical cell of this type, in which the anode and cathode compartments are identical except for the concentration of a reactant, is called a concentration cell. As the reaction proceeds, the difference between the concentrations of Ag+ in the two compartments will decrease, as will E cell. Finally, when the concentration of Ag+ is the same in both compartments, equilibrium will have been reached, and the measured potential difference between the two compartments will be zero (E cell = 0).
Example (\PageIndex{2})
Calculate the voltage in a galvanic cell that contains a manganese electrode immersed in a 2.0 M solution of MnCl 2 as the cathode, and a manganese electrode immersed in a 5.2 × 10−2 M solution of MnSO 4 as the anode (T = 25°C).
Given: galvanic cell, identities of the electrodes, and solution concentrations
Asked for: voltage
Strategy:
Write the overall reaction that occurs in the cell.
Determine the number of electrons transferred. Substitute this value into the Nernst equation to calculate the voltage.
Solution
A This is a concentration cell, in which the electrode compartments contain the same redox active substance but at different concentrations. The anions (Cl− and SO 4 2−) do not participate in the reaction, so their identity is not important. The overall reaction is as follows:
[\ce{ Mn^{2+}}(aq, 2.0\, M) \rightarrow \ce{Mn^{2+}} (aq, 5.2 \times 10^{−2}\, M)\nonumber ]
B For the reduction of Mn 2+(aq) to Mn(s), n = 2. We substitute this value and the given Mn 2+ concentrations into Equation (\ref{Eq4}):
[ \begin{align} E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \[4pt] &=\textrm{0 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{5.2\times10^{-2}}{2.0}\right) \[4pt] &=\textrm{0.047 V}\end{align} \nonumber ]
Thus manganese will dissolve from the electrode in the compartment that contains the more dilute solution and will be deposited on the electrode in the compartment that contains the more concentrated solution.
Exercise (\PageIndex{2})
Suppose we construct a galvanic cell by placing two identical platinum electrodes in two beakers that are connected by a salt bridge. One beaker contains 1.0 M HCl, and the other a 0.010 M solution of Na 2 SO 4 at pH 7.00. Both cells are in contact with the atmosphere, with (P_\mathrm{O_2}) = 0.20 atm. If the relevant electrochemical reaction in both compartments is the four-electron reduction of oxygen to water:
[\ce{O2(g) + 4H^{+}(aq) + 4e^{−} \rightarrow 2H2O(l)} \nonumber ]
What will be the potential when the circuit is closed?
Answer
0.41 V
Using Cell Potentials to Measure Solubility Products
Because voltages are relatively easy to measure accurately using a voltmeter, electrochemical methods provide a convenient way to determine the concentrations of very dilute solutions and the solubility products ((K_{sp})) of sparingly soluble substances. As you learned previously, solubility products can be very small, with values of less than or equal to 10−30. Equilibrium constants of this magnitude are virtually impossible to measure accurately by direct methods, so we must use alternative methods that are more sensitive, such as electrochemical methods.
Figure (\PageIndex{1}): A Galvanic ("Concentration") Cell for Measuring the Solubility Product of AgCl. One compartment contains a silver wire inserted into a 1.0 M solution of Ag+, and the other compartment contains a silver wire inserted into a 1.0 M Cl− solution saturated with AgCl. The potential due to the difference in [Ag+] between the two cells can be used to determine (K_{sp}). (CC BY-NC-SA; Anonymous by request)
To understand how an electrochemical cell is used to measure a solubility product, consider the cell shown in Figure (\PageIndex{1}), which is designed to measure the solubility product of silver chloride:
[K_{sp} = [\ce{Ag^{+}}][\ce{Cl^{−}}]. \nonumber ]
In one compartment, the cell contains a silver wire inserted into a 1.0 M solution of Ag+; the other compartment contains a silver wire inserted into a 1.0 M Cl− solution saturated with AgCl. In this system, the Ag+ ion concentration in the first compartment equals K sp. We can see this by dividing both sides of the equation for K sp by [Cl−] and substituting:
[\begin{align}[\ce{Ag^{+}}] &= \dfrac{K_{sp}}{[\ce{Cl^{−}}]} \[4pt] &= \dfrac{K_{sp}}{1.0} = K_{sp}. \end{align} \nonumber ]
The overall cell reaction is as follows:
Ag+(aq, concentrated) → Ag+(aq, dilute)
Thus the voltage of the concentration cell due to the difference in [Ag+] between the two cells is as follows:
[\begin{align} E_\textrm{cell} &=\textrm{0 V}-\left(\dfrac{\textrm{0.0591 V}}{1}\right)\log\left(\dfrac{[\mathrm{Ag^+}]\textrm{dilute}}{[\mathrm{Ag^+}]\textrm{concentrated}}\right) \nonumber \[4pt] &= -\textrm{0.0591 V } \log\left(\dfrac{K_{\textrm{sp}}}{1.0}\right) \nonumber \[4pt] &=-\textrm{0.0591 V }\log K_{\textrm{sp}} \label{Eq122} \end{align} ]
By closing the circuit, we can measure the potential caused by the difference in [Ag+] in the two cells. In this case, the experimentally measured voltage of the concentration cell at 25°C is 0.580 V. Solving Equation (\ref{Eq122}) for (K_{sp}),
[\begin{align}\log K_\textrm{sp} & =\dfrac{-E_\textrm{cell}}{\textrm{0.0591 V}}=\dfrac{-\textrm{0.580 V}}{\textrm{0.0591 V}}=-9.81 \[4pt]
K_\textrm{sp} & =1.5\times10^{-10}\end{align} \nonumber ]
Thus a single potential measurement can provide the information we need to determine the value of the solubility product of a sparingly soluble salt.
Example (\PageIndex{3}): Solubility of lead(II) sulfate
To measure the solubility product of lead(II) sulfate (PbSO 4) at 25°C, you construct a galvanic cell like the one shown in Figure (\PageIndex{1}), which contains a 1.0 M solution of a very soluble Pb 2+ salt [lead(II) acetate trihydrate] in one compartment that is connected by a salt bridge to a 1.0 M solution of Na 2 SO 4 saturated with PbSO 4 in the other. You then insert a Pb electrode into each compartment and close the circuit. Your voltmeter shows a voltage of 230 mV. What is K sp for PbSO 4? Report your answer to two significant figures.
Given: galvanic cell, solution concentrations, electrodes, and voltage
Asked for: K sp
Strategy:
From the information given, write the equation for K sp. Express this equation in terms of the concentration of Pb 2+.
Determine the number of electrons transferred in the electrochemical reaction. Substitute the appropriate values into Equation (\ref{Eq12}) and solve for K sp.
Solution
A You have constructed a concentration cell, with one compartment containing a 1.0 M solution of (\ce{Pb^{2+}}) and the other containing a dilute solution of Pb 2+ in 1.0 M Na 2 SO 4. As for any concentration cell, the voltage between the two compartments can be calculated using the Nernst equation. The first step is to relate the concentration of Pb 2+ in the dilute solution to K sp:
[\begin{align}[\mathrm{Pb^{2+}}][\mathrm{SO_4^{2-}}] & =K_\textrm{sp} \
[\mathrm{Pb^{2+}}] &=\dfrac{K_\textrm{sp}}{[\mathrm{SO_4^{2-}}]}=\dfrac{K_\textrm{sp}}{\textrm{1.0 M}}=K_\textrm{sp}\end{align} \nonumber ]
B The reduction of Pb 2+ to Pb is a two-electron process and proceeds according to the following reaction:
Pb 2+(aq, concentrated) → Pb 2+(aq, dilute)
so
[\begin{align}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{0.0591}{n}\right)\log Q \
\textrm{0.230 V} & =\textrm{0 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{[\mathrm{Pb^{2+}}]\textrm{dilute}}{[\mathrm{Pb^{2+}}]\textrm{concentrated}}\right)=-\textrm{0.0296 V}\log\left(\dfrac{K_\textrm{sp}}{1.0}\right) \
-7.77 & =\log K_\textrm{sp} \
1.7\times10^{-8} & =K_\textrm{sp}\end{align} \nonumber ]
Exercise (\PageIndex{3})
A concentration cell similar to the one described in Example (\PageIndex{3}) contains a 1.0 M solution of lanthanum nitrate [La(NO 3)3] in one compartment and a 1.0 M solution of sodium fluoride saturated with LaF 3 in the other. A metallic La strip is inserted into each compartment, and the circuit is closed. The measured potential is 0.32 V. What is the K sp for LaF 3? Report your answer to two significant figures.
Answer
5.7 × 10−17
Using Cell Potentials to Measure Concentrations
Another use for the Nernst equation is to calculate the concentration of a species given a measured potential and the concentrations of all the other species. We saw an example of this in Example (\PageIndex{3}), in which the experimental conditions were defined in such a way that the concentration of the metal ion was equal to K sp. Potential measurements can be used to obtain the concentrations of dissolved species under other conditions as well, which explains the widespread use of electrochemical cells in many analytical devices. Perhaps the most common application is in the determination of [H+] using a pH meter, as illustrated below.
Example (\PageIndex{4}): Measuring pH
Suppose a galvanic cell is constructed with a standard Zn/Zn 2+ couple in one compartment and a modified hydrogen electrode in the second compartment. The pressure of hydrogen gas is 1.0 atm, but [H+] in the second compartment is unknown. The cell diagram is as follows:
[\ce{Zn(s)}|\ce{Zn^{2+}}(aq, 1.0\, M) || \ce{H^{+}} (aq, ?\, M)| \ce{H2} (g, 1.0\, atm)| Pt(s) \nonumber ]
What is the pH of the solution in the second compartment if the measured potential in the cell is 0.26 V at 25°C?
Given: galvanic cell, cell diagram, and cell potential
Asked for: pH of the solution
Strategy:
Write the overall cell reaction.
Substitute appropriate values into the Nernst equation and solve for −log[H+] to obtain the pH.
Solution
A Under standard conditions, the overall reaction that occurs is the reduction of protons by zinc to give H 2 (note that Zn lies below H 2 in Table P2):
Zn(s) + 2H 2+(aq) → Zn 2+(aq) + H 2(g) E°=0.76 V
B By substituting the given values into the simplified Nernst equation (Equation (\ref{Eq4})), we can calculate [H+] under nonstandard conditions:
[\begin{align}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}n\right)\log\left(\dfrac{[\mathrm{Zn^{2+}}]P_\mathrm{H_2}}{[\mathrm{H^+}]^2}\right) \
\textrm{0.26 V} &=\textrm{0.76 V}-\left(\dfrac{\textrm{0.0591 V}}2\right)\log\left(\dfrac{(1.0)(1.0)}{[\mathrm{H^+}]^2}\right) \
16.9 &=\log\left(\dfrac{1}{[\mathrm{H^+}]^2}\right)=\log[\mathrm{H^+}]^{-2}=(-2)\log[\mathrm{H^+}] \
8.46 &=-\log[\mathrm{H^+}] \
8.5 &=\mathrm{pH}\end{align} \nonumber ]
Thus the potential of a galvanic cell can be used to measure the pH of a solution.
Exercise (\PageIndex{4})
Suppose you work for an environmental laboratory and you want to use an electrochemical method to measure the concentration of Pb 2+ in groundwater. You construct a galvanic cell using a standard oxygen electrode in one compartment (E°cathode = 1.23 V). The other compartment contains a strip of lead in a sample of groundwater to which you have added sufficient acetic acid, a weak organic acid, to ensure electrical conductivity. The cell diagram is as follows:
[Pb_{(s)} ∣Pb^{2+}(aq, ? M)∥H^+(aq), 1.0 M∣O_2(g, 1.0 atm)∣Pt_{(s)}\nonumber ]
When the circuit is closed, the cell has a measured potential of 1.62 V. Use Table P2 to determine the concentration of Pb 2+ in the groundwater.
Answer
(1.2 \times 10^{−9}\; M)
Summary
The Nernst equation can be used to determine the direction of spontaneous reaction for any redox reaction in aqueous solution. The Nernst equation allows us to determine the spontaneous direction of any redox reaction under any reaction conditions from values of the relevant standard electrode potentials. Concentration cells consist of anode and cathode compartments that are identical except for the concentrations of the reactant. Because ΔG = 0 at equilibrium, the measured potential of a concentration cell is zero at equilibrium (the concentrations are equal). A galvanic cell can also be used to measure the solubility product of a sparingly soluble substance and calculate the concentration of a species given a measured potential and the concentrations of all the other species.
17.3: Concentration Effects and the Nernst Equation is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.
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3120 | https://www.kiragoldner.com/teaching/DS320/spring22/L21.pdf | DS 320 Algorithms for Data Science Lecture #21 Spring 2022 Prof. Kira Goldner Zero-Sum Games and the Minimax Theorem Consider the game Rock-Paper-Scissors, where as usual, paper covers rock, scissors cuts paper, and rock breaks scissors (that is: the former beats the latter in the comparison). In a face-off, the winner earns +1 and the loser earns -1. If two of the same type face each other, then there is a tie, and both earn 0.
The matrix below shows the game of Rock-Paper Scissors depicted as a zero-sum-game.
Sup-pose that brothers Ron and Charlie Weasley are facing off. Each brother must choose a strategy.
In the language of the payoffmatrix below, Ron is the row player, and he must choose a row to play as his strategy. Similarly, Charlie is the column player and he just choose which column to play. If Ron chooses row i and Charlie chooses column j, then the payoffto Ron will be aij, and the payoffto Charlie will be −aij, hence the term “zero-sum.” Thus, the row and column players prefer bigger and smaller numbers, respectively.
Rock Paper Scissors Rock 0 -1 1 Paper 1 0 -1 Scissors -1 1 0 Order of Turns • Typically, RPS is played by both players simultaneously choosing their strategies.
• But what if I made you go first? That’s obviously unfair—whatever you do, I can respond with the winning move.
• Now what if I only forced you to commit to a probability distribution over rock, paper, and scissors? (Then I respond choosing a strategy, and then nature flips coins on your behalf.) You can protect yourself by randomizing uniformly among the three options—then, no matter what I do, I’m equally likely to win, lose, or tie.
The minimax theorem states that, in general games of “pure competition,” a player moving first can always protect herself by randomizing appropriately.
The Minimax Theorem Notation: • m×n payoffmatrix A—aij is the row player’s payofffor outcome (i, j) when row player plays strategy i and column player plays strategy j • mixed row strategy x (a distribution over rows) • mixed column strategy y (a distribution over columns) Expected payoffof the row player: m X i=1 n X j=1 Pr[outcome (i, j)] aij = m X i=1 n X j=1 Pr[row i chosen] | {z } =xi Pr[column j chosen] | {z } =yj aij = xT Ay The minimax theorem is the amazing statement that turn order doesn’t matter.
Theorem 1 (Minimax Theorem). For every two-player zero-sum game A, max x min y xT Ay = min y max x xT Ay .
(1) On the left, the row player goes first, choosing a strategy to maximize their payoffand protect against the fact that the column player goes second and adapts to their strategy. The right is the opposite situation. The value of the game (value that both sides will equal) is 0 in this case: the first player will play randomly and the second will respond arbitrarily.
From LP Duality to Minimax This is not the original or only argument, but we will now derive Theorem 1 from LP duality arguments. The first step is to formalize the problem of computing the best strategy for the player forced to go first.
Two issues: (1) the nested min/max, and (2) the quadratic (nonlinear) character of xT Ay in the decision variables x, y.
Observation 2. The second player never needs to randomize. If the row player goes first and chooses any distribution x, the column player can then simply compute the expected payoff(with respect to x) of each column and choose the best.
In math, we have argued that max x min y xT Ay = max x n min j=1 xT Aej (2) = max x n min j=1 m X i=1 aijxi !
(3) where ej is the jth standard basis vector, corresponding to the column player deterministically choosing column j.
We’ve solved one of our problems by getting rid of y. But there is still the nested max/min.
Specifically, we introduce a decision variable v, intended to be equal to (2), and max v subject to v − m X i=1 aijxi ≤0 for all j = 1, . . . , n m X i=1 xi = 1 x1, . . . , xm ≥0 and v ∈R.
Note that this is a linear program. Rewriting the constraints in the form v ≤ m X i=1 aijxi for all j = 1, . . . , n makes it clear that they force v to be at most minn j=1 Pm i=1 aijxi.
If (v∗, x∗) is an optimal solution, then v∗= minn j=1 Pm i=1 aijxi. By feasibility, v∗cannot be larger than minn j=1 Pm i=1 aijx∗ i . If it were strictly less, then we can increase v∗slightly without destroying feasibility, yielding a better feasible solution (contradicting optimality).
Since the linear program explicitly maximizes v over all distributions x, its optimal objective function value is v∗= max x n min j=1 xT Aej = max x min y xT Ay (4) Now, we do the same thing for the column player, where the column player moves first: min w subject to w − n X j=1 aijyj ≥0 for all i = 1, . . . , m n X j=1 yj = 1 y1, . . . , yn ≥0 and w ∈R.
At an optimal solution (w∗, y∗), y∗is the optimal strategy for the column player (when going first, assuming optimal play by the row player) and w∗= min y m max i=1 eT i Ay = min y max x xT Ay (5) These two linear programs are duals! For example, the one unrestricted variable (v or w) corre-sponds to the one equality constraint in the other linear program (Pn j=1 yj = 1 or Pm i=1 xi = 1, respectively). The n x variables correspond to the remaining dual constraints, and the m y vari-ables correspond to the remaining primal constraints. Then strong duality implies that v∗= w∗; in light of (4) and (5), the minimax theorem follows directly.
Online Learning and the Multiplicative Weights Algorithm Think back to when we learned about caching or job scheduling. We always assumed that we knew everything that was coming in advance and could make decisions about the future. What if we couldn’t see the future? This is called an online setting, not like the internet, but as if the input is waiting on line.
An Online Problem 1. The input arrives “one piece at a time.” 2. An algorithm makes an irrevocable decision each time it receives a new piece of the input.
Now, for an Online Decision-Making Problem, we should consider the event when you have a bunch of experts advising you on the stocks or the weather, and you have to choose one to trust each day.
Or, equivalently, a bunch of actions you could take. Each day (or time step), you get to see how right or wrong the experts are—they are assigned some reward (or loss) by an adversary. Your goal is to come up with a strategy of how to choose experts as time goes on such that, after you choose your strategy for each successive time step, the adversary assigns rewards, and you get the best rewards (or minimal losses) possible. The adversary knows your (possibly randomized) strategy, but does not see the result of the randomness until after assignment rewards.
Online Decision-Making At each time step t = 1, 2, . . . , T : a decision-maker picks a probability distribution pt over her experts or actions A an adversary picks a reward vector rt : A →[−1, 1] an action at is chosen according to the distribution pt, and the decision-maker receives reward rt(at) the decision-maker learns rt, the entire reward vector The input arrives “one piece at a time.” What should we compare to?
Thus far, we’ve been trying to achieve optimal solutions, or comparing to optimal solutions assuming we know full information about the future and what is optimal. Does that still make sense?
Example 1 (Comparing to the Best Action Sequence). Suppose your set of experts (or actions) is A = {1, 2}. Each day t, the adversary chooses the reward vector rt as follows: if the algorithm chooses a distribution pt for which the probability on action 1 is at least 1 2 , then rt is set to the vector (−1, 1). Otherwise, the adversary sets rt equal to (1, −1).
This adversary forces the expected reward of the algorithm to be nonpositive, while ensuring that the reward of the best action sequence in hindsight is T. Thus, the algorithm’s approximation is x/T where x ≤0—no approximation at all.
Example 1 tells us that we should not be trying to compare to the Best Action Sequence—this is too strong of a goal. Instead, we compare it to the reward incurred by the best fixed action in hindsight. In words, we change our benchmark from T X i=1 N max i=1 rt i to N max i=1 T X i=1 rt i.
Definition 1 (Regret). Fix reward vectors r1, . . . , rT . The regret of the action sequence a1, . . . , aT is N max i=1 T X t=1 rt i | {z } best fixed action − T X t=1 rt(at) | {z } our algorithm .
(6) We’d like an online decision-making algorithm that achieves low regret, as close to 0 as possible (and negative regret would be even better). Notice that the worst-possible regret in 2T (since rewards lie in [−1, 1]). We think of regret Ω(T) as an epic fail for an algorithm. What is the justification for the benchmark of the best fixed action in hindsight? First, simple and natural learning algorithms can compete with this benchmark. Second, achieving this is non-trivial: as the following examples make clear, some ingenuity is required. Third, competing with this benchmark is already sufficient to obtain many interesting applications (see end of this lecture and all of next lecture).
The Multiplicative Weights Algorithm No-Regret Algorithm Design Principles 1. Past performance of actions should guide which action is chosen at each time step, with the probability of choosing an action increasing in its cumulative reward. (Recall that we need a randomized algorithm to have any chance.) 2. The probability of choosing a poorly performing action should decrease at an exponential rate.
The first principle is essential for obtaining regret sublinear in T, and the second for optimal regret bounds.
The MW algorithm maintains a weight, intuitively a “credibility,” for each action. At each time step the algorithm chooses an action with probability proportional to its current weight. The weight of each action evolves over time according to the action’s past performance. |
3121 | https://medlineplus.gov/genetics/condition/rheumatoid-arthritis/ | Rheumatoid arthritis: MedlinePlus Genetics
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Rheumatoid arthritis
Description
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Rheumatoid arthritis is a disease that causes chronic abnormal inflammation, primarily affecting the joints. The most common signs and symptoms are pain, swelling, and stiffness of the joints. Small joints in the hands and feet are involved most often, although larger joints (such as the shoulders, hips, and knees) may become involved later in the disease. Joints are typically affected in a symmetrical pattern; for example, if joints in the hand are affected, both hands tend to be involved. People with rheumatoid arthritis often report that their joint pain and stiffness is worse when getting out of bed in the morning or after a long rest.
Rheumatoid arthritis can also cause inflammation of other tissues and organs, including the eyes, lungs, and blood vessels. Additional signs and symptoms of the condition can include a loss of energy, a low fever, weight loss, and a shortage of red blood cells (anemia). Some affected individuals develop rheumatoid nodules, which are firm lumps of noncancerous tissue that can grow under the skin and elsewhere in the body.
The signs and symptoms of rheumatoid arthritis usually appear in mid- to late adulthood. Many affected people have episodes of symptoms (flares) followed by periods with no symptoms (remissions) for the rest of their lives. In severe cases, affected individuals have continuous health problems related to the disease for many years. The abnormal inflammation can lead to severe joint damage, which limits movement and can cause significant disability.
Frequency
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Rheumatoid arthritis affects about 1.3 million adults in the United States. Worldwide, it is estimated to occur in up to 1 percent of the population. The disease is two to three times more common in women than in men, which may be related to hormonal factors.
Causes
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Rheumatoid arthritis probably results from a combination of genetic and environmental factors, many of which are unknown.
Rheumatoid arthritis is classified as an autoimmune disorder, one of a large group of conditions that occur when the immune system attacks the body's own tissues and organs. In people with rheumatoid arthritis, the immune system triggers abnormal inflammation in the membrane that lines the joints (the synovium). When the synovium is inflamed, it causes pain, swelling, and stiffness of the joint. In severe cases, the inflammation also affects the bone, cartilage, and other tissues within the joint, causing more serious damage. Abnormal immune reactions also underlie the features of rheumatoid arthritis affecting other parts of the body.
Variations in dozens of genes have been studied as risk factors for rheumatoid arthritis. Most of these genes are known or suspected to be involved in immune system function. The most significant genetic risk factors for rheumatoid arthritis are variations in human leukocyte antigen (HLA) genes, especially the HLA-DRB1 gene. The proteins produced from HLA genes help the immune system distinguish the body's own proteins from proteins made by foreign invaders (such as viruses and bacteria). Changes in other genes appear to have a smaller impact on a person's overall risk of developing the condition.
Other, nongenetic factors are also believed to play a role in rheumatoid arthritis. These factors may trigger the condition in people who are at risk, although the mechanism is unclear. Potential triggers include changes in sex hormones (particularly in women), occupational exposure to certain kinds of dust or fibers, and viral or bacterial infections. Long-term smoking is a well-established risk factor for developing rheumatoid arthritis; it is also associated with more severe signs and symptoms in people who have the disease.
Learn more about the genes associated with Rheumatoid arthritis
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HLA-B
HLA-DPB1
HLA-DRB1
IRF5
PTPN22
RBPJ
RUNX1
STAT4
Additional Information from NCBI Gene:
AFF3
ARID5B
BLK
C5
CCL21
CCR6
CD2
CD28
CD40
CD5
CD58
CTLA4
FCGR2A
FCGR2B
GATA3
IKZF3
IL2
IL21
IL2RA
IL2RB
IL6R
IL6ST
IRAK1
IRF8
KIF5A
NFKBIL1
PADI4
PIP4K2C
POU3F1
PRDM1
PRKCQ
PTPRC
PXK
RASGRP1
RCAN1
REL
SPRED2
TAGAP
TLE3
TNFAIP3
TNFRSF14
TRAF1
TRAF6
TYK2
Inheritance
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The inheritance pattern of rheumatoid arthritis is unclear because many genetic and environmental factors appear to be involved. However, having a close relative with rheumatoid arthritis likely increases a person's risk of developing the condition.
Other Names for This Condition
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Arthritis, rheumatoid
RA
Additional Information & Resources
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Genetic Testing Information
Genetic Testing Registry: Rheumatoid arthritis
Patient Support and Advocacy Resources
National Organization for Rare Disorders (NORD)
Clinical Trials
ClinicalTrials.gov
Catalog of Genes and Diseases from OMIM
RHEUMATOID ARTHRITIS; RA
Scientific Articles on PubMed
PubMed
References
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Carmona L, Cross M, Williams B, Lassere M, March L. Rheumatoid arthritis. Best Pract Res Clin Rheumatol. 2010 Dec;24(6):733-45. doi: 10.1016/j.berh.2010.10.001. Citation on PubMed
Diogo D, Kurreeman F, Stahl EA, Liao KP, Gupta N, Greenberg JD, Rivas MA, Hickey B, Flannick J, Thomson B, Guiducci C, Ripke S, Adzhubey I, Barton A, Kremer JM, Alfredsson L; Consortium of Rheumatology Researchers of North America; Rheumatoid Arthritis Consortium International; Sunyaev S, Martin J, Zhernakova A, Bowes J, Eyre S, Siminovitch KA, Gregersen PK, Worthington J, Klareskog L, Padyukov L, Raychaudhuri S, Plenge RM. Rare, low-frequency, and common variants in the protein-coding sequence of biological candidate genes from GWASs contribute to risk of rheumatoid arthritis. Am J Hum Genet. 2013 Jan 10;92(1):15-27. doi: 10.1016/j.ajhg.2012.11.012. Epub 2012 Dec 20. Citation on PubMed or Free article on PubMed Central
Eyre S, Bowes J, Diogo D, Lee A, Barton A, Martin P, Zhernakova A, Stahl E, Viatte S, McAllister K, Amos CI, Padyukov L, Toes RE, Huizinga TW, Wijmenga C, Trynka G, Franke L, Westra HJ, Alfredsson L, Hu X, Sandor C, de Bakker PI, Davila S, Khor CC, Heng KK, Andrews R, Edkins S, Hunt SE, Langford C, Symmons D; Biologics in Rheumatoid Arthritis Genetics and Genomics Study Syndicate; Wellcome Trust Case Control Consortium; Concannon P, Onengut-Gumuscu S, Rich SS, Deloukas P, Gonzalez-Gay MA, Rodriguez-Rodriguez L, Arlsetig L, Martin J, Rantapaa-Dahlqvist S, Plenge RM, Raychaudhuri S, Klareskog L, Gregersen PK, Worthington J. High-density genetic mapping identifies new susceptibility loci for rheumatoid arthritis. Nat Genet. 2012 Dec;44(12):1336-40. doi: 10.1038/ng.2462. Epub 2012 Nov 11. Citation on PubMed or Free article on PubMed Central
Raychaudhuri S, Sandor C, Stahl EA, Freudenberg J, Lee HS, Jia X, Alfredsson L, Padyukov L, Klareskog L, Worthington J, Siminovitch KA, Bae SC, Plenge RM, Gregersen PK, de Bakker PI. Five amino acids in three HLA proteins explain most of the association between MHC and seropositive rheumatoid arthritis. Nat Genet. 2012 Jan 29;44(3):291-6. doi: 10.1038/ng.1076. Citation on PubMed or Free article on PubMed Central
Stahl EA, Raychaudhuri S, Remmers EF, Xie G, Eyre S, Thomson BP, Li Y, Kurreeman FA, Zhernakova A, Hinks A, Guiducci C, Chen R, Alfredsson L, Amos CI, Ardlie KG; BIRAC Consortium; Barton A, Bowes J, Brouwer E, Burtt NP, Catanese JJ, Coblyn J, Coenen MJ, Costenbader KH, Criswell LA, Crusius JB, Cui J, de Bakker PI, De Jager PL, Ding B, Emery P, Flynn E, Harrison P, Hocking LJ, Huizinga TW, Kastner DL, Ke X, Lee AT, Liu X, Martin P, Morgan AW, Padyukov L, Posthumus MD, Radstake TR, Reid DM, Seielstad M, Seldin MF, Shadick NA, Steer S, Tak PP, Thomson W, van der Helm-van Mil AH, van der Horst-Bruinsma IE, van der Schoot CE, van Riel PL, Weinblatt ME, Wilson AG, Wolbink GJ, Wordsworth BP; YEAR Consortium; Wijmenga C, Karlson EW, Toes RE, de Vries N, Begovich AB, Worthington J, Siminovitch KA, Gregersen PK, Klareskog L, Plenge RM. Genome-wide association study meta-analysis identifies seven new rheumatoid arthritis risk loci. Nat Genet. 2010 Jun;42(6):508-14. doi: 10.1038/ng.582. Epub 2010 May 9. Citation on PubMed or Free article on PubMed Central
Viatte S, Plant D, Raychaudhuri S. Genetics and epigenetics of rheumatoid arthritis. Nat Rev Rheumatol. 2013 Mar;9(3):141-53. doi: 10.1038/nrrheum.2012.237. Epub 2013 Feb 5. Citation on PubMed or Free article on PubMed Central
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SAT Math
Course: SAT Math>Unit 2
Lesson 4: Graphs of linear equations and functions: foundations
Graphs of linear equations and functions | Lesson
Graphing linear equations — Basic example
Graphing linear equations — Harder example
Graphs of linear equations and functions: foundations
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Graphs of linear equations and functions: foundations
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Graphing linear equations — Basic example
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Watch Sal work through a basic Graphing linear equations problem.
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mazia 2 years ago Posted 2 years ago. Direct link to mazia's post “Anyone here please answer...” more Anyone here please answer am i going to pass sat this December 😭
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L. E. 2 years ago Posted 2 years ago. Direct link to L. E.'s post “Yes! You will pass the SA...” more Yes! You will pass the SAT because it's impossible to fail it. The SAT is not graded on a pass/fail basis, it simply shows what you know and what you don't. Even if you get every single answer wrong, your score will be a 400.
If you mean whether you'll get a competitive score for your college admissions, that depends. What are your practice SAT scores? Are you consistently practicing to improve that score? How well do you feel like you're doing in the different reading-writing/math sections of the SAT?
Those are questions only you can answer, but yes, even if your score is lower than you'd like, it's impossible to fail the SAT.
Hope this helped, and good luck!
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Hamid a year ago Posted a year ago. Direct link to Hamid's post “Khan Academy is like a sc...” more Khan Academy is like a school run by one teacher. That's crazy lol
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Lyric a year ago Posted a year ago. Direct link to Lyric's post “There are lots of teacher...” more There are lots of teachers here on Khan Academy. Sal just teaches (probably) 75% of them as he is great at Math, Biology, Chemistry, Physics, Cosmology and... almost everything.
Well, we have to keep in mind that he is an MIT Grad and holds an MBA from Harvard.
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Brainy Red A! a year ago Posted a year ago. Direct link to Brainy Red A!'s post “Thanks, Annelise! Very ki...” more Thanks, Annelise! Very kind of you.
R u a photog?
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Shravan Illickkal Suresh 2 years ago Posted 2 years ago. Direct link to Shravan Illickkal Suresh's post “Hey sal how you doin?” more Hey
sal how you doin?
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smithanthonyenyinnaya 2 years ago Posted 2 years ago. Direct link to smithanthonyenyinnaya's post “Please I don't this graph...” more Please I don't this graph and xan someone Please explain it more
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Eva 2 years ago Posted 2 years ago. Direct link to Eva's post “you should first find 2 p...” more you should first find 2 points that the line touches where it is obvious what the coordinates are (so where it touches the point of a square?) here the two "obvious" ones are where it meets with x and y. The x one is (0,6) and the y one is (0, 1). On the graph, you can also see b (where the line crosses y) and that is gonna be 1. Now using these numbers you can start to fill out the equation we have been using y = xm + b. b is 1 so y = xm + 1. Now to find m(the slope) you have to take the two points we made for ourselves. after subtracting the right numbers we will get -1/6 (if you don't know which numbers to subtract you should rewatch the videos) now we know y = x-1/6 + 1. and now simply reorder the equation and tada!
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Alfonso Martinez 2 years ago Posted 2 years ago. Direct link to Alfonso Martinez's post “gogeta is better than veg...” more gogeta is better than vegito
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Daria Cosciug 2 years ago Posted 2 years ago. Direct link to Daria Cosciug's post “Isn't option 4 correct to...” more Isn't option 4 correct too? Because for x=0 it's 6y=-6 which is equal to y=1.
Am I wrong?
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muhammadahmadkhan2812 2 years ago Posted 2 years ago. Direct link to muhammadahmadkhan2812's post “yeah. but what about Y=0?...” more yeah. but what about Y=0? the answer would come -6. and so wrong. X should be 6 when y=0. not -6. Hope that helps. :)
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efidavis 2 years ago Posted 2 years ago. Direct link to efidavis's post “I dont like math no more” more I dont like math no more
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137375 2 years ago Posted 2 years ago. Direct link to 137375's post “great!” more great!
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codermonloader135 2 years ago Posted 2 years ago. Direct link to codermonloader135's post “bout wat?” more bout wat?
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122744 2 months ago Posted 2 months ago. Direct link to 122744's post “not a single train of tho...” more not a single train of thought in this guy teaching this. when you do mx+b he just -- what are you even solving for? are you just trying to match it up to one of the solutions you pick? What?
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aryana 2 months ago Posted 2 months ago. Direct link to aryana's post “hey! i'm 4 days late so i...” more hey! i'm 4 days late so i'm sure you already figured this out, but i just wanna reply to help anyone else confused.
i agree that the way he explains it is kind of confusing, the way i learned it is doing rise/run for the slope. basically, if you see the linear line going through an exact "clean" point (like in the video, the line was going through (0,1)) then you need to count how many units to the left/right it is moving until it makes it to the next exact point, (6,0).
seeing that it moved over by 6, we know the denominator of our slope is 6, and since it moved down by 1, we know the numerator of our slope is -1. therefore our slope is -1/6.
for the y intercept (b), just look at where the line is intercepting the y axis!
(like i said, i'm pretty sure you already know this-- but i'm putting it out there for anyone confused :D)
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Video transcript
[Instructor] A line is graphed in the xy-plane as shown. Which of the following equations represents the line? They give us a bunch of equations here and so there is several ways we can tackle it. When we look at this, I could see there's two interesting points here, there's the point when x is, let me just write this and actually I'm gonna write a little lower so we can look at it the same time that I look at the equation choices. So we see that when x is a zero, y is one. So that is the y intercept we could say. And then when we could see when x is six, y is zero. When x is six, y is zero, so a very kind of basic way of approaching this is see well, when x is a zero, y needs to be equal to one. When x is zero we get six y is equal to one, well then y is gonna be equal to one sixth, rule that one out. When x is equal to zero, y needs to be equal to one. If x is zero then six y equals six. Yeah, y is going to be equal to one. Now when y is zero, x needs to be equal to six. So if y is 0, this goes away and x is equal to six. So we're done, this is our choice. Now there's other ways that we could do it. We could write it first in slope intercept form and then convert to this form right over here. So let's do it that way as well. We could say that the equation of this line is gonna be y, if I write it in y equals mx plus b form where m is the slope and b is the y intercept. We already know that b is equal to one. So we already know that's one, and what's the slope? Well slope is our change in y for given change in x and we see when our change in x is positive six, when our change in x is positive six, our change in y is negative one, so our slope is we decrease in y by one when we increase in x by six is negative 1/6. So the equation of the line y is equal to negative 1/6 x, this is the slope plus one. And then we could convert to the forms that we have here, so lets see, we could add 1/6 x to both sides and you're gonna get one over six x plus y is equal to one, and that's not quite what we have here. All the coefficients on x are just one. So we can multiply both sides of this times six, and we would get x plus six y is equal to six. Which is exactly the choice that we picked.
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3123 | https://allen.in/dn/qna/646665147 | 1/(i-1) + 1/(i+1) is a. purely imaginary b. purely rational number. c. purely integer d. negative integer
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3125 | https://courses.physics.illinois.edu/phys211/su2012/Text/ch10.pdf | Center of Mass A) Overview This unit expands our study of mechanics from single particles to systems of particles. We will introduce the very important concept of the center of mass of a system of particles and determine the center of mass for both discrete and continuous mass distributions. We will use Newton’s second law to obtain the equations of motion for the center of mass of a system of particles. We will also obtain a version of the work-kinetic energy theorem, called the center of mass equation, that can be applied to a system of particles, B) Systems of Particles and the Center of Mass So far we have only considered the motion of simple objects. We have intentionally not considered the motion of, for example, an object composed of two different sized balls connected to the ends of a rod In the next few units we will develop the tools to understand the motion of more complicated systems of objects such as these. We will discover that their behavior can be understood by applying what we already know, and we will see that the equations describing their motion are remarkably similar to those we have already developed. We will start by introducing a new concept which will play a key role in what follows, namely that of the center of mass. Quite simply put, the center of mass of an object is just the average location of the mass that makes up the object. For a simple symmetric object like a ball or box of uniform density we will see that the center of mass is just at the center of the object. For less simple shapes we will have to perform a calculation to determine the location of the center of mass. The procedure we will adopt for finding the average position of the all of the mass contained in some system of objects will be to simply take a mass-weighted average of the positions of the individual parts. Namely, we will define the location of the center of mass of a system of particles to be equal to the sum of the positions of the individual particles with each one weighted by its own fraction of the total mass of the system as shown for a system of discrete masses in Figure 10.1. In the next section we will determine the center of mass for a two particle system. This concrete example will illustrate the general procedure and hopefully make clear why this definition makes sense. C) Center of Mass for a Two-Body System We’ll start by considering an object made of only two point particles, labeled 1 and 2. We will assume that we know the masses of the two particles as well as their locations along the x axis as shown in Figure 10.2. Since the particles lie along the x-axis, the calculation of the position of their center of mass is straightforward. 2 1 2 2 1 1 m m r m r m RCM + + ≡ ⇒ 2 1 2 2 1 1 m m x m x m xCM + + ≡ If the masses are equal, we see that the center of mass is located at the average value of x1 and x2. This position is halfway between them, which certainly makes sense. If, on the other hand, one mass is twice as big as the other, it will count for twice as much in the average, which means the center of mass will be closer to the heavier particle than the lighter one, which also seems reasonable. So far we have considered just the one-dimensional case in which both particles lie along the x axis. This procedure can be easily extended to more than one dimension, though, using vector addition. We start by writing the expression for the location of the center of mass in terms of the masses of the two particles and the vectors that locate each of the two particles. 2 1 2 2 1 1 m m r m r m RCM + + ≡ Figure 10.1 The definition of the center of mass for a system of three discrete masses.
Figure 10.2 Two masses located along the x-axis. We can rewrite this formula so that the vector locating the center of mass is equal to the sum of two vectors, the displacement vector of one particle and another vector that is proportional to the difference in the displacement vectors of the two particles. ( ) 1 2 2 1 2 1 2 1 2 2 1 1 r r m m m r m m r m r m RCM − + + = + + ≡ We can think of this equation as a map that tells us how to get to the center of mass: We first go to one of the objects and then we go a fraction of the way to the other object, where this fraction is determined by the masses, as shown in Figure 10.3. If the second object has the same mass as the first, we go half way. If the second object is heavier than the first, we go more than half way, and if the second is lighter than the first we go less than halfway. The beauty of this approach is that we can see that the location of the center of mass does not depend on our choice of the origin or the orientation of our coordinate system; the center of mass always lies at the same fixed point along the line connecting the two objects. We have just demonstrated an important result, namely, that the center of mass is a property of the system itself; it does not depend on the way we choose to look at the system. Indeed, we will show in a later unit that the center of mass of a rigid system is the same as its balance point! D) Center of Mass for Systems of More than Two Particles We can extend our definition of the center of mass for systems containing more than two particles by simply summing up the mass-weighted displacement vectors for each particle. i i i total CM r m M R ∑ ≡ 1 It is usually easier to break this vector equation into components and evaluate each component separately. Just to make sure we know how this works, let’s do an example involving eight equal mass particles located on the corners of a cube, as shown in Figure 10.4. Figure 10.3 The location of the center of mass of a two body system is located along the vector difference (r2 – r1) of the two displacements.. To find the x-coordinate of the center of mass we need to sum the x-coordinates of each of the eight particles weighted by the ratio of the mass of each particle to the total mass of the system. In this example the particles all have the same mass and their x-coordinates are either zero or L, so that the sum is easy to evaluate, and we find that the x-coordinate of the center of mass is just equal to ½ L. 2 4 8 1 1 L mL m x m M X i i i ttoal CM = = ∑ ≡ We will get the same results for both the y and z coordinates, so that we see that the center of mass is at the center of the box, as expected. Now suppose the cube was a solid, made up of millions of atoms rather than just eight particles on the corners. We suspect the answer would be the same, that the center of mass is still in the middle, but how can we prove this conjecture when actually performing the sum over millions of atoms seems difficult, if not impossible? Once again calculus comes to the rescue! Rather than calculating the product of position and mass for individual particles, we just integrate the position vector over all infinitesimal mass elements dm contained in the cube. We will do this calculation in the next section. E) Center of Mass for Continuous Mass Distribution Before we actually evaluate any integral, let’s make sure we understand exactly how we adjust our definition of the center of mass when we are dealing with a continuous mass distribution. The big idea is that we have to replace a discrete sum by a continuous sum, an integral. In the discrete sum we evaluated the product of the mass of each part of the system and its position and then added them up. In the continuous sum we are doing the same thing, the only difference now is that are dividing a continuous object up into an infinite number of tiny volume elements each having a mass dm. ∫ ≡ dm r M R total CM 1 Since here we are integrating over the volume of a cube, a 3 dimensional object, the integral itself must be evaluated in all three dimensions, x, y and z. To evaluate the mass element dm, we take the product of the volume of the element and the mass density (the mass per unit volume ρ) of the cube. dxdydz dV dm ρ ρ = = Our job now is to evaluate this triple integral. The first step is to break our equation for the center of mass vector Rcm into x, y and z components. We will start with the x-equation and first determine the limits of integration. In each direction, the cube is Figure 10.4 A system of eight equal mass particles located at the corners of a cube of side L. located between the origin and a distance L from the origin. If we assume the mass density ρ is a constant, then it can be taken outside of the integral. ∫ ∫ ∫ = L L L total CM dz dy xdx M X 0 0 0 1 ρ The resulting three-dimensional integral is equal to the product of three one-dimensional integrals, each of which is evaluated separately. Requiring the product of the mass density and the volume of the cube to be equal to the total mass of the cube, we obtain the expected result, that the c-coordinate of the center of mass of the cube is just equal to ½ L. ( )( )( ) ( ) L L M L L L L M X total total CM 2 1 2 1 3 2 2 1 1 = = = ρ ρ The y and z component calculations are absolutely identical to the x-component calculation, giving us the expected result, that the center of mass of the box is at its center! F) Center of Mass of a System of Objects We now know how to find the center of mass of a collection of point particles as well as that of a continuous solid object. What happens when we want to find the center of mass of a collection of solid objects? Figure 10.5 shows two objects, labeled a and b. By definition, the center of mass of the system is found by integrating the position vector over all of the mass in the system. Since the system is made of two objects, the total integral is just the sum of two separate integrals, one for each object. ∫ ∫ + = a b total CM dm r dm r M R 1 If we multiply and divide each of these integrals by the mass of the object, we certainly haven't changed anything, but we can now see that the numerator just becomes the mass of each object times the position of its center of mass. ( ) b CM b a CM a total b b b a a a total CM R M R M M M dm r M M dm r M M R , , 1 1 + = ∫ + ∫ = Therefore, we have just arrived at a simple procedure for finding the center of mass of a system of solid objects. Namely, we just treat each object as a point particle with all of its mass located at its center of mass! That’s all there is to it! Figure 10.5 The center of mass of the two objects can be calculated simply by treating each object as a point particle having a mass equal to the total mass of the object G) Dynamics of the Center of Mass To this point we have defined the concept of the center of mass and have shown how to find it for any system of objects. With this knowledge in hand, we can finally do some physics. We will start with our definition for the center of mass of a system of objects and take the derivative of this expression with respect to time. dt r d m M dt R d i i i total CM ∑ = 1 The left hand side of the equation becomes the velocity of the center of mass, and the numerator on the right hand side becomes the sum of the mass times velocity for each object in the system. We have already defined the product of the mass and velocity of an object as its momentum. Therefore, the numerator on the right hand side is just equal to the total momentum of the system. i i i total CM v m M V ∑ = 1 We will now take another derivative with respect to time. The left hand side of the equation becomes the acceleration of the center of mass, and the numerator on the right hand side becomes the sum of the mass times acceleration for each object in the system. dt v d m M dt V d i i i total CM ∑ = 1 The product of the mass and acceleration of an object is just equal to the total force on that object. total i i net i i i total CM M F a m M A ∑ = ∑ = , 1 Now this sum of the total forces acting on all the objects in the system could get unwieldy if the number of objects gets large. The really good news, though, is that we can simplify this sum significantly by realizing that any forces that act between two objects that are both in the system will cancel in this sum Newton’s third law requires that all such forces always come in pairs of equal magnitude and opposite direction! ∑ = ∑ + ∑ ∑ = ≠ i i external j i ij i i external i i net F F F F , , , Hence the sum of all forces acting on all objects in the system just reduces to the sum of all forces acting on the system from the outside, or the total external force. total External Net total i i external CM M F M F A , , = ∑ = We have finally arrived at something that looks exactly like Newton’s second law, but rather than applying just to point particles as before, this expression relates the total external force on the whole system to the acceleration of the center of mass of the system! Consequently, we can say that no matter how complicated a system of objects may be, the center of mass of the system behaves in the same simple way that a point particle does. Indeed, in the next section, we will use this equation to obtain a generalization of the work-kinetic energy theorem for systems of particles. H) Center of Mass Equation In section E of unit 7, we integrated Newton’s second law to obtain the work-kinetic energy theorem for point particles: namely, that the change in kinetic energy of a particle is equal to the work done on that particle by the net force. net W K = ∆ We will now extend this result to systems of particles. We start from our result from the last section: that the acceleration of the center of mass of a system of particles is equal to the total external force acting on a system of particles divided by the total mass of the system. total External Net total i i external CM M F M F A , , = ∑ = This equation looks exactly like Newton’s second law for a point particle that has the total mass of the system (Mtotal) and is located at the center of mass of the system. Consequently, we can perform exactly the same derivation we made in unit 7 to obtain the work-kinetic energy theorem for a point particle. The result here is an equation, often called the center of mass equation, that looks exactly like the work-kinetic energy theorem we derived for point particles. ( ) ∫ • = ∆ CM External Net CM CM total d F A V M ℓ , 2 2 1 The only differences lie in the subscripts. These subscripts are important, however. The change in kinetic energy term is calculated as if the system were a particle of mass Mtotal and were moving with the velocity of the center of mass. We define the right hand side of the equation as the “macroscopic work done by the net force”. This macroscopic work is calculated as if all forces were acting on this particle located at the center of mass. We can now see why we did not need to worry about the microscopic work done by kinetic friction when we were calculating the motion of the box skidding to a stop in the last unit. If we consider the box to be a system of particles, we see that the change in kinetic energy of the box is exactly equal to the macroscopic work done by the net force, the kinetic friction force. The microscopic work done by the kinetic friction force at the interface of the surfaces of the box and the floor determines the additional thermal energy in the box and the floor, but does not determine the motion of the center of mass of the box. I) Example: the Astronaut and the Wrench We will end with a simple example to illustrate some of the concepts we have developed in this unit. Imagine you are an astronaut far out in space. You have just finished fixing a space telescope using a big wrench whose mass is one tenth as big as yours. You realize you have no way to get back to your spaceship which is 20 meters away from you, so you throw the wrench as hard as you can in a direction away from the spaceship which causes you to move in the opposite direction, toward the spaceship. When you finally reach the space ship, how far away are you from the wrench? The key concept needed to answer this question is that the acceleration of the center of mass of a system will be zero if the external force on the system is zero. In this case, we define the system to be you and the wrench, and the center of mass of the system is initially at rest a distance of 20 meters from your spaceship. Since there are no external forces acting on the system and the center of mass is initially at rest, the location of the center of mass of the system can never change! If we choose the initial location of the center of mass to be at x = 0, the center of mass will always be at x = 0. The location of the center of mass of the system is determined from its definition. 0 = + + = wrench astronaut wrench wrench astronaut astronaut CM M M x M x M X Multiplying both sides by the total mass, we obtain our result that the product of the position and mass of the wrench is always equal to minus the product of your position and mass. astronaut astronaut wrench wrench x M x M − = Since the mass of the wrench is 1/10 of your mass, the wrench will always be ten times as far away from the center of mass as you are, and it will always be on the opposite side of the center of mass from you. astronaut wrench astronaut wrench x M M x − = When you are at the spaceship, 20 meters to the left of the center of mass, the wrench will be 200 meters to the right of the center of mass, which is 220 meters from the spaceship |
3126 | https://pmc.ncbi.nlm.nih.gov/articles/PMC3265327/ | Postprandial lipoprotein metabolism; VLDL vs chylomicrons - PMC
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Clin Chim Acta
. Author manuscript; available in PMC: 2012 Jul 15.
Published in final edited form as: Clin Chim Acta. 2011 Apr 19;412(15-16):1306–1318. doi: 10.1016/j.cca.2011.04.018
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Postprandial lipoprotein metabolism; VLDL vs chylomicrons
Katsuyuki Nakajima
Katsuyuki Nakajima
1 School of Health Sciences, Faculty of Medicine, Gunma University, Maebashi, Gunma, Japan
2 Otsuka Pharmaceuticals Co., Ltd, Tokushima, Japan
4 Department of Lipidology and Division of Cardiology, Kanazawa University Graduate School of Medical Science, Kanazawa, Japan
5 Department of Molecular Biosciences, School of Veterinary Medicine and Department of Nutrition, University of California, Davis, CA, USA
9 Nutrition Clinic, Kagawa Nutrition University, Tokyo, Japan
Find articles by Katsuyuki Nakajima
1,2,4,5,9, Takamitsu Nakano
Takamitsu Nakano
1 School of Health Sciences, Faculty of Medicine, Gunma University, Maebashi, Gunma, Japan
2 Otsuka Pharmaceuticals Co., Ltd, Tokushima, Japan
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1,2, Yoshiharu Tokita
Yoshiharu Tokita
1 School of Health Sciences, Faculty of Medicine, Gunma University, Maebashi, Gunma, Japan
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1, Takeaki Nagamine
Takeaki Nagamine
1 School of Health Sciences, Faculty of Medicine, Gunma University, Maebashi, Gunma, Japan
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1, Akihiro Inazu
Akihiro Inazu
3 Department of Laboratory Sciences, Kanazawa University Graduate School of Medical Science, Kanazawa, Japan
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3, Junji Kobayashi
Junji Kobayashi
4 Department of Lipidology and Division of Cardiology, Kanazawa University Graduate School of Medical Science, Kanazawa, Japan
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4, Hiroshi Mabuchi
Hiroshi Mabuchi
4 Department of Lipidology and Division of Cardiology, Kanazawa University Graduate School of Medical Science, Kanazawa, Japan
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4, Kimber L Stanhope
Kimber L Stanhope
5 Department of Molecular Biosciences, School of Veterinary Medicine and Department of Nutrition, University of California, Davis, CA, USA
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5, Peter J Havel
Peter J Havel
5 Department of Molecular Biosciences, School of Veterinary Medicine and Department of Nutrition, University of California, Davis, CA, USA
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5, Mitsuyo Okazaki
Mitsuyo Okazaki
6 Skylight Biotech Inc,. Akita, Japan
7 Department of Vascular Medicine and Geriatrics, Graduate School of Medicine, Tokyo Medical and Dental University, Tokyo, Japan
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6,7, Masumi Ai
Masumi Ai
8 Department of Life Sciences and Bioethics, Graduate School of Medicine, Tokyo Medical and Dental University, Tokyo, Japan
9 Nutrition Clinic, Kagawa Nutrition University, Tokyo, Japan
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8,9, Akira Tanaka
Akira Tanaka
7 Department of Vascular Medicine and Geriatrics, Graduate School of Medicine, Tokyo Medical and Dental University, Tokyo, Japan
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1 School of Health Sciences, Faculty of Medicine, Gunma University, Maebashi, Gunma, Japan
2 Otsuka Pharmaceuticals Co., Ltd, Tokushima, Japan
3 Department of Laboratory Sciences, Kanazawa University Graduate School of Medical Science, Kanazawa, Japan
4 Department of Lipidology and Division of Cardiology, Kanazawa University Graduate School of Medical Science, Kanazawa, Japan
5 Department of Molecular Biosciences, School of Veterinary Medicine and Department of Nutrition, University of California, Davis, CA, USA
6 Skylight Biotech Inc,. Akita, Japan
7 Department of Vascular Medicine and Geriatrics, Graduate School of Medicine, Tokyo Medical and Dental University, Tokyo, Japan
8 Department of Life Sciences and Bioethics, Graduate School of Medicine, Tokyo Medical and Dental University, Tokyo, Japan
9 Nutrition Clinic, Kagawa Nutrition University, Tokyo, Japan
✉
Correspondence to: Dr. Katsuyuki Nakajima, Ph.D, School of Health Sciences, Faculty of Medicine, Gunma University, 371-8511, 3-39-22, Showa-machi, Maebashi, Gunma, Japan 371-8511, Phone; +81-90-1124-7298, nakajimak05@ybb.ne.jp
Issue date 2011 Jul 15.
© 2011 Elsevier B.V. All rights reserved.
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PMCID: PMC3265327 NIHMSID: NIHMS348748 PMID: 21531214
The publisher's version of this article is available at Clin Chim Acta
Abstract
Since Zilversmit first proposed postprandial lipemia as the most common risk of cardiovascular disease, chylomicrons (CM) and CM remnants have been thought to be the major lipoproteins which are increased in the postprandial hyperlipidemia. However, it has been shown over the last two decades that the major increase in the postprandial lipoproteins after food intake occurs in the very low density lipoprotein (VLDL) remnants (apoB100 particles), not CM or CM remnants (apoB48 particles). This finding was obtained using the following three analytical methods; isolation of remnant-like lipoprotein particles (RLP) with specific antibodies, separation and detection of lipoprotein subclasses by gel permeation HPLC and determination of apoB48 in fractionated lipoproteins by a specific ELISA. The amount of the apoB48 particles in the postprandial RLP is significantly less than the apoB100 particles, and the particle sizes of apoB48 and apoB100 in RLP are very similar when analyzed by HPLC. Moreover, CM or CM remnants having a large amount of TG were not found in the postprandial RLP. Therefore, the major portion of the TG which is increased in the postprandial state is composed of VLDL remnants, which have been recognized as a significant risk for cardiovascular disease.
Keywords: Postprandial remnant lipoproteins, Chylomicron remnants, Very low density lipoprotein (VLDL) remnants, apoB-48, apoB-100, Remnant-like lipoprotein particles (RLP), RLP-triglyceride (RLP-TG), RLP-cholesterol (RLP-C)
1. Introduction
Plasma triglycerides (TG) are known to be a surrogate for TG-rich lipoproteins (TRL) and are present in chylomicrons (CM), very low density lipoproteins (VLDL) and their remnants. TRL and their remnants are significantly increased in the postprandial plasma and are known to predict the risk of coronary heart disease (CHD) (1, 2), independent of the total cholesterol, LDL or HDL cholesterol level. Recently, non-fasting TG levels have come to be known as a significant risk indicator for CHD events (3–5). Zilversmit (6) first proposed that the postprandial CM is the most common risk factor for atherogenesis in persons who do not have familial hyperlipoproteinemia. The hypothesis of postprandial CM and CM remnants came to be widely accepted as a major cause of common atherogenesis, because it was well established that CM are significantly increased in the intestine after food intake and a large amount of CM flow into the blood stream through the thoracic duct. Therefore, CM and CM remnants have been thought to be the major lipoproteins in the postprandial hyperlipidemia until recently.
Furthermore, Type I and V chylomicronemia, which are associated with severe lipemia, are often confused with common alimentary lipemia. However, severe lipemia is is most commonly associated with a deficiency of lipoprotein lipase (LPL) or apoC II, and significantly elevated CM and CM remnants are detected in the fasting plasma of these cases (7). Type III hyperlipidemia is a kind of postprandial genetic defect which results in significantly elevated apoB48 in beta-VLDL and is associated with the low affinity of apoE2/2 for the receptors which clear remnant lipoproteins (8–11). Alimentary lipemia is often reflected by the increased turbidity of fat emulsions, not CM, in plasma in terms of lipid concentrations. Therefore, the severity of lipemia in the postprandial plasma often weakly correlates with the plasma triglyceride concentration.
Nevertheless, many of the current textbooks on alimentary lipemia or postprandial hyperlipidemia still indicate CM and/or CM remnants as the major lipoproteins which are increased in plasma after food intake. In addition, approximately 80% of the postprandial increase of triglycerides has been considered to be accounted for by the apo B-48 containing lipoproteins until recently (12). This is based on the fact that the CM and/or CM remnants are the major triglyceride-carrier in the postprandial state, with each particle carrying a very large number of triglyceride molecules. Therefore, large quantities of triglycerides are thought to be transported by a very small number of CM or CM remnant particles. As the CM particle size has not been determined, it has remained unclear whether the major TRL in the postprandial plasma was CM-derived or not.
To investigate the characteristics of postprandial lipoproteins, we developed a new immunoseparation method which enables the direct isolation of remnant lipoproteins as remnant-like lipoprotein particles (RLP) from the postprandial plasma and examination of the concentration and particle size of CM and VLDL remnants (13, 14). This method has the capacity to isolate both CM and VLDL remnants from the plasma simultaneously as RLP, which fulfill most of the biochemical characteristics of CM and VLDL remnant lipoproteins.
In this manuscript, we have demonstrated that the major remnant lipoproteins associated with postprandial hyperlipidemia are in fact not CM remnants, but VLDL. To confirm this finding, three new analytical methods developed in Japan during the last two decades were employed; the isolation of remnant-like lipoprotein particles (RLP) using specific antibodies (13, 14), separation and detection of lipoprotein subclasses by a gel permeation high-performance liquid chromatographic (HPLC) system (15) and determination of apoB48 in fractionated lipoproteins by a highly specific ELISA (16 ).
2. Metabolism of CM, VLDL and their remnants in plasma
Figure 1 shows the metabolic pathway of CM and VLDL. CM is secreted by the intestine after fat consumption. CM particles contain apo B-48 as a structural protein, which in humans is formed exclusively in the intestine after tissue-specific editing of the apo B-100 mRNA (17, 18). It appears that apo B-48 containing particles are continuously secreted from the enterocyte, and at times of excessive triglyceride availability, lipid droplets fuse with nascent lipoprotein particles, resulting in the secretion of enormous chylomicrons [19, 20]. Once the CM particle reaches the plasma compartment, apoA-I dissociates very rapidly (21) and acquires apo Cs, in particular apo C-II, to enable efficient unloading of its massive triglyceride content after binding to the lipoprotein lipase (LPL) which is bound to the endothelium . High density lipoproteins (HDL) are a major reservoir for the apo Cs and apo E, but in conditions with low HDL concentrations (found most often in hypertriglyceridemic subjects). CM may receive apo Cs and apo E from resident VLDL particles. The half-life of CM triglycerides in healthy subjects is very short, approximately 5 min . The half-life of CM particles has been very difficult to estimate due to the difficulty of obtaining adequate labeling of CM. The CM particle half-life is certainly longer than for CM triglycerides and seems to be quite heterogenous. Certain pools of CM remnants have a very long residence time, at least as long as similar-sized VLDL particles [24, 25].
Figure 1.
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After fat intake, the intestine secretes chylomicrons (CM), the triglycerides of which are lipolyzed by lipoprotein lipase (LPL). The LPL reaction constitutes the initial process in theformation of triglyceride-rich lipoprotein (TRL) remnants. The VLDL secretion process is partly regulated by the rate of FFA influx to the liver. VLDL triglycerides are lipolyzed by endothelial-bound lipoprotein lipase and VLDL remnant particles are formed. The final TRL remnant composition is modulated by the cholesterol ester transfer protein (CETP) reaction with HDL, hepatic lipase (HL), and the exchange of soluble apolipoproteins such as C-I, C-II, C-III and E. (A) The great majority of the remnants are removed from plasma by receptor-mediated processes and the principal receptors are the LDL receptor and the LDL-receptor-related protein (LRP). It is probable that the CM remnants use both of these routes, whereas the VLDL remnants are more likely to use only the LDL receptor.
Furthermore, a major proportion of the CM remnants leave the plasma compartment quite rapidly while still quite large, i.e., 75 nm in diameter . There is competition for lipolysis: CM and VLDL mix in the blood and the two TRL species compete for the same lipolytic pathway [25, 26]. It has been shown that endogenous TRL accumulate in human plasma after fat intake and the mechanism behind this phenomenon is explained by the delayed lipolysis of the apo B-100 TRL particles due to competition with CM for the sites of LPL action . Similarly, endogenous TRL accumulate in rat plasma due to competition with a CM-like triglyceride emulsion for the common lipolytic pathway . The increase in the number of TRL apo B-100 particles is actually far greater than that of the apo B-48 containing lipoproteins in the postprandial state . Of note, the accumulation of large TRL apo B-100 particles seems to be a particular sign in hypertriglyceridaemic patients with CAD compared with healthy hypertriglyceridaemic subjects, suggesting a link between the accumulation of large VLDL and the development of atherosclerosis .
VLDL particles are secreted continuously from the liver (Figure 1). In contrast to CM and their remnants, they are characterized by their apo B-100 content. The secretion of VLDL is under complex regulation, as the larger and more triglyceride-rich VLDL species are under strict insulin control in a dual sense. First, a number of more or less insulin-sensitive mechanisms regulate the availability of triglycerides for VLDL production. The free fatty acids (FFA) which are generated by lipolysis in adipose tissue through the action of hormone-sensitive lipase provide a major source for hepatic VLDL secretion. Insulin stimulates the endothelial expression of lipoprotein lipase (LPL), the key enzyme in TRL metabolism, in a post-transcriptional manner [30, 31]. Hepatic uptake of poorly lipolyzed VLDL or CM remnant particles may also contribute to the hepatocellular triglyceride availability. Similarly, reduced uptake of FFA in adipose and muscle tissues after LPL-mediated lipolysis of CM and VLDL shunts FFA to the liver . Finally, the liver has the capacity of de novo synthesis of triglycerides and VLDL.
In contrast, the metabolic pathway of VLDL by hepatic triglyceride lipase (HTGL) seems to be still controversial because of the difficulties of measurements. HTGL has been reported to metabolize comparatively small remnant lipoproteins, although to a lesser extent than LPL. However, our recent studies have shown no correlation between HTGL activity and plasma TG, RLP-C, RLP-TG or the small dense LDL-C concentration in humans (33), although we did find a strong inverse correlation between LPL activity and these parameters in both the fasting and postprandial state (Table 1). Previous studies proposed the concept of HTGL role to the remnant metabolism seemed to be mainly based on the animal studies using anti-HTGL antibodies in monkeys and rats and found the accumulation of remnant lipoproteins in plasma after the HTGL specific antibody treatment (34, 35). As it is well known that small dense LDL (sd LDL) is positively correlated with TG and remnant lipoproteins in plasma, these data support the concept that remnant lipoproteins are the precursor of sd LDL and are metabolized in the same pathway by LPL (33). From these data, HTGL does not seem to play a significant role in the metabolic pathway of remnant lipoproteins, in contrast to previous reports (34–38), but instead, plays a definitive role in HDL metabolism in humans.
Table 1.
Single linear regression analysis of plasma lipids, lipoproteins, lipases and ANGPTL3 in 20 volunteers
| | |
:---: |
| | TG | RLP-C | RLP-TG | sdLDL-C | HDL-C | LPL | HTGL | ANGPTL3 |
| Fasting |
| TG | | | | | | |
| RLP-C | 0.72 | | | | | |
| RLP-TG | 0.88 | 0.91 | | | | |
| sdLDL-C | 0.56 | 0.28 | 0.43 | | | |
| HDL-C | −0.37 | −0.19 | −0.23 | −0.36 | | |
| LPL | −0.43 | −0.27 | −0.25 | −0.22 | 0.33 | |
| HTGL | −0.06 | −0.01 | 0.02 | −0.10 | −0.49 | 0.08 |
| ANGPTL3 | 0.15 | 0.20 | 0.16 | −0.12 | 0.36 | −0.13 | −0.50
| Postprandial |
| TG | | | | | | |
| RLP-C | 0.86 | | | | | |
| RLP-TG | 0.95 | 0.86 | | | | |
| sdLDL-C | 0.44 | 0.36 | 0.33 | | | |
| HDL-C | −0.42 | −0.32 | −0.42 | −0.30 | | |
| LPL | −0.38 | −0.40 | −0.42 | −0.21 | 0.35 | |
| HTGL | 0.09 | −0.05 | 0.24 | −0.20 | −0.47 | −0.19 |
| ANGPTL3 | 0.00 | −0.22 | −0.04 | −0.12 | 0.24 | −0.27 | −0.25
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p<0.05,
p<0.01,
P<0.001
3. Biochemical characteristics of CM and VLDL remnants
TRL remnants are formed in the circulation when apoB-48 containing CM of intestinal origin or apoB-100 containing VLDL of hepatic origin are converted by lipoprotein lipase (and to a lesser extent by hepatic lipase according to the description commonly given in the literature) into smaller and denser particles (36–38). Compared with their nascent precursors, TRL remnants are depleted of triglyceride, phospholipid, apoCs (and apoA-I and apoA-IV in the case of CM) and are enriched in cholesteryl esters and apoE (40, 41). They can thus be identified, separated, or quantified in plasma on the basis of their density, charge, size, specific lipid components, apolipoprotein composition and apolipoprotein immunospecificity (42). Each of these approaches has provided useful information about the structure and function of remnant lipoproteins, and has helped to establish the role of TRL remnants in the pathogenesis of atherosclerosis. Accurate measurement and characterization of plasma remnant lipoproteins, however, has proven to be difficult for the following reasons: (1) despite their reduced size and triglyceride content, they are difficult to differentiate from their triglyceride-rich precursors; (2) due to their rapid plasma catabolism, they exist in plasma at relatively low concentrations; and (3) since remnants are at different stages of catabolism, they are markedly heterogeneous in size and composition. TRL is known to become progressively smaller, denser and less negatively charged as they are converted into TRL remnants. They gradually lose triglycerides and, in relative terms, become enriched with cholesteryl esters. They also reduce their complement of C apolipoproteins (apoC-I, apoC-II, and apoC-III), which are replaced by apoE. At any given time, there is a continuous spectrum of different-sized remnants in the blood. Some of these particles are of intestinal origin. They contain apoB-48 and are present in greater concentrations after a fat-rich meal. The majority, however in both the fed and fasted state, contain apoB-100 and are derived from the liver. Depending on the extent to which they have been lipolyzed, the different species of TRL contain different proportions of triglyceride and cholesterol and may or may not contain apoCs or apoE. Remnant lipoproteins are thus structurally and compositionally diverse, which has made it necessary to develop a variety of specific biochemical techniques for the detection, quantification and characterization of these lipoproteins. In the light of such difficulties, we endeavored to find a new approach to isolate remnant lipoproteins. An approach which could separate variety of remnant lipoproteins from the normal apoB100 (nascent VLDL and LDL) carrying lipoproteins. A specific anti-apoB-100 antibody which does not recognize α-helix structure of apoB-51 region was developed and used for the isolation of the apoE-rich VLDL remnants (Figure 2) (14).
Figure 2.
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The amino acid sequence of the epitope region of the anti-apoB-100 antibody (JI-H) is homologous to an amphipathic helical region of apoE, which suggests that apoE can compete for binding of the antibody with the B-51 amphipathic helical epitope (2270–2321) ( Ref 14). (Nakajima et al. J Clin Ligand Assay 1996;19:177–83.)
4. Daily rhythm of plasma cholesterol, TG and remnant lipoproteins and the changes in the lipoprotein levels after a fat load
The plasma triglyceride concentration fluctuates throughout the day in response to the ingestion of meals. Even if measured after a 10- to 12-hour overnight fast (as is normal clinical practice), triglyceride levels vary considerably more than LDL and HDL cholesterol levels.
As non-fasting TG levels are now known to be a significant risk for CHD events (3–5), the analysis of postprandial lipoproteins, rather than the fasting state, has come to be recognized as more important. We reported that non-fasting TG correlated more strongly with remnant lipoproteins than fasting TG . The correlations between postprandial TG and remnant lipoprotein concentrations were significantly more robust when compared with fasting TG vs remnant lipoprotein concentrations. In particular, the increase of postprandial RLP-TG from fasting RLP-TG contributed to approximately 80% of the increase of postprandial total TG from total fasting TG (Table 2). The greater predictive value of non-fasting TG levels associated with cardiovascular events is directly correlated with the increased levels of remnant lipoproteins in the postprandial state.
Table 2.
Lipids, lipoproteins, RLP-TG/RLP-C ratio and RLP-TG/total TG ratio after oral fat load in 18 male volunteers
| | 0 hr | 2 hr | 4 hr | 6 hr |
:---: :---:
| Median | 75%tile | Median | 75%tile | Median | 75%tile | Median | 75%tile |
| TC (mg/dl) | 200 | 217 | 196 | 219 | 193 | 210 | 200 | 212 |
| TG (mg/dl) | 85 | 104 | 140 | 169 | 162 | 206 | 125 | 139 |
| HDL-C(mg/dl) | 62 | 74 | 62 | 73 | 60 | 71 | 61 | 75 |
| RLP-C(mg/dl) | 4.8 | 6.1 | 6.6 | 7.7 | 7.1 | 9.4 | 6.1 | 8 |
| RLP-TG(mg/dl) | 12 | 16 | 63 | 77 | 64 | 101 | 39 | 58 |
| RLP-TG/RLP-C | 2.5 | 2.7 | 8.7 | 10.1 | 9.3 | 13 | 5.8 | 8.7 |
| RLP-TG/TG | 0.13 | 0.15 | 0.45 | 0.47 | 0.42 | 0.56 | 0.34 | 0.41 |
| |
| ◿ TG (mg/dl) - | 56 | 75 | 74 | 102 | 35 | 63 |
| ◿ RLP-TG (mg/dl) - | 51 | 68 | 51 | 85 | 29 | 41 |
| ◿ RLP-TG/◿ TG (ratio) - | 0.82 | 0.94 | 0.8 | 0.91 | 0.82 | 0.94 |
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0 hr vs 2 hr, 4 hr, and 6 hr by U-test.
; p<0.05,
; p<0.01
◿ TG; postprandial minus fasting (0 hi) TG, ◿ RLP-TG: postprandial minus fasting (0 hr) RLP-TG
As shown by Stanhope et al. (44) in Figure 3 in their fructose treatment study, plasma TG levels are significantly increased during the day associated with food intake. It was only in the early morning that TG levels in all cases returned to the basal levels. When fructose was administered, a significant increase of TG after food intake compared with the regular meal was observed. Plasma cholesterol levels did not significantly change during the day. Figure 3 shows that blood samples were collected every hour during the 24h day. Between blood samplings, each subject consumed a standardized meal (9 am, 1pm and 6 pm) containing 55% of the energy as carbohydrate, 30% as fat and 15% as protein. The energy content of the meals was based on each subject’s energy requirement as determined by the Mifflin equation (45). Figure 3 shows that the TG levels in generally healthy volunteer plasma were the highest at 2 AM in the very early morning, indicating that postprandial conditions continue even past midnight during in the course of a day, except in the early morning. Remnant lipoprotein levels increased significantly for most of the day except the early morning, a similar profile TG. These increases may depend on the kind of foods. The typical carbohydrate rich Japanese meal did not increase the levels of TG and remnants during a given day compared with a fat-rich meal such as in the typical Western diet (44, 46), as shown by Ai et al. (47) and Sekihara et al. (48).
Figure 3.
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Twenty four-hour circulating plasma TG concentrations in subjects before and after 2, 8 and 10 weeks of consuming fructose-sweetened beverages (n = 17). The plasma TG levels were found to be significantly increased during the day associated with food intake in a fructose treatment study. Only in the early morning did the TG levels in all cases returned to the basal levels, and were highest in the middle of the night. Blood samples were collected every hour for 24hs. Between blood samplings, each subject consumed a standardized breakfast (9:00 h), lunch (13:00 h) and dinner (18:00 h) containing 55% of the meal energy as carbohydrate, 30% fat and 15% protein. (Ref. 43) (Stanhope et al. J Clin Invest. 2009; 119:1322–34)
Other postprandial studies have been conducted by oral fat load test. One typical study performed in our laboratory was carried out in 6 male and 6 female (postmenopausal) Japanese volunteers aged 39–60 (mean 52 years) who were generally healthy with no apparent disease (49). Among these 12 volunteers, there were three cases of mild hyperlipidemia (Type IIb). All participants performed an oral fat tolerance test (OFTT) as previously reported (50). Briefly, after a 12 h fast, the subjects ingested 17g fat /m 2 body surface area (OFTT cream, Jomo foods, Takasaki. Gunma). The test meal (OFTT cream) had a water content of 56.9%, while lipids accounted for 32.9%, protein for 2.5%, carbohydrates for 7.4% and minerals for 0.3%. The fat was 64.3% saturated, 29.3% monounsaturated and 3.5% polyunsaturated. Blood samples were drawn before and 2, 4 and 6 h after an oral fat load. Plasma apo B-48 significantly increased and correlated with the TG levels in postprandial plasma, however, apoB (more than 98% apoB-100) did not (Figure 4). These results suggest that CM or CM remnants carrying a large amount of TG are the major component of the increase in the postprandial remnant lipoproteins. However, the apoB concentration was far greater than apoB-48 in the plasma, as shown in Figure 4. The ApoB-100 in LDL decreased during a fat load, which resulted in there being no change in apoB despite the increase of apoB100 in the postprandial RLP.
Figure 4.
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Postprandial changes in the plasma TG, apoB-48 and apoB concentrations before and after an oral fat load. TG displayed a close correlation with apoB-48, but not with apoB (Ref.48) (Nakano et al., Ann Clin Biochem. 2011; 48: 57–64)
The origin of the cholesterol increase in the postprandial TRL remnants has been investigated by several researchers. Both a single fatty meal and long-term diet have been reported to exhibit a cholesterol accumulation in the TRL fraction less than expected, arguing for a rather limited role of dietary cholesterol in determining the TRL remnant cholesterol level [51, 52]. In fact, the calculated proportion of dietary cholesterol present in the CM fraction was very low, i.e. only one out of 99 cholesterol molecules originated directly from the meal . Obviously, the dietary cholesterol is diluted by the cholesterol undergoing enterohepatic recirculation, but the delayed and incomplete absorption of cholesterol (compared to triglycerides) also argues that cholesterol and triglycerides are not incorporated at similar rates into CM particles. It is likely that the majority of the cholesterol is absorbed at a later stage, further down in the gastrointestinal tract, where the abundance of triglycerides is reduced. Therefore, in the postprandial state the accumulation of cholesterol in CM remnant particles is limited in comparison with the cholesterol-enrichment of VLDL remnants. The major source of the cholesterol in the TRL remnants comes from HDL as a result of CETP activity, which will be described later in this manuscript.
5. Isolation of remnant-like lipoprotein particles (RLP) using specific antibodies and the diagnostic characteristics of the RLP-cholesterol (RLP-C) assay
An assay system based on the recognition of TRL remnants according to their apolipoprotein content and immune-specificity has been developed that provides a quantitative and clinically applicable approach to the measurement of plasma remnant lipoproteins (13, 14). This is the first method to separate remnant lipoproteins from TRL using specific antibodies so as to isolate the remnant fraction under moderate conditions. In this assay, RLP is separated from plasma in unbound fraction by immunoaffinity chromatography with a gel containing an anti–apoA-I antibody and a specific anti-apoB-100 monoclonal antibody (JI-H) (JIMRO II, Otsuka, Tokyo). The former antibody recognizes all HDL and any newly synthesized CM containing apoA-I, whereas the latter antibody recognizes all apoB-100 containing lipoproteins, except for certain particles enriched in apoE. Anti-apoB-100 antibody JI-H recognizes the B51 region of apoB100 and CM has no epitope in this region. Therefore, CM lacking apoA-I, which is defined as the CM remnant (21), is not recognized by the gel and all of the apoB-48 particles in the plasma are isolated in the unbound RLP fraction. The reason the anti–apoB-100 antibody does not recognize the apoE-enriched RLP is not entirely clear, although the amino acid sequence of the epitope region of the apoB-100 antibody is homologous to an amphipathic helical region of apoE, which suggests that apoE should be able to compete for binding of the antibody to its epitope, located between 2270–2320 amino acid from N-terminal on apoB-100 (14) (Figure 2). The amphipathic helical peptide (2293–2301) of the chemically synthesized antibody epitope reacted with the anti apoB-100 antibody (JI-H) and exhibited potent reverse cholesterol transport activity as apoA-I, apo E or HDL. Of further note, this peptide showed an inhibitory effect on CETP activity as well (unpublished data).
HDL, LDL, large CM and the majority of VLDL are thus retained by the gel. The unbound RLP are made up of remnant VLDL containing apoB-100 and CM containing apoB-48, which are routinely measured in terms of cholesterol, although they can also be quantified in terms of triglycerides or specific apolipoproteins (ie, apoB, apoC-III, or apoE) (54, 55). The plasma concentration of RLP-C has been shown to be significantly correlated with the plasma concentration of total TG, VLDL-TG and VLDL-C. It has not been significantly correlated with LDL cholesterol or LDL apoB (56, 57). The physical and chemical properties of lipoproteins which are not recognized by the apoB-100 monoclonal antibody JI-H, subsequently isolated by ultracentrifugation at a density 1.006 g/mL, have been described (58). These lipoproteins contained more molecules of apoE and cholesteryl esters than those that were bound, consistent with them being remnant-like lipoproteins. They had slow pre-b electrophoretic mobility compared with the bound VLDL fraction and ranged in size from 25 to 80 nm. Other lipoproteins, however, may be present when the JI-H monoclonal antibody (together with an anti–apoA-I antibody) is used to isolate RLP by immunoaffinity chromatography from total plasma in the absence of ultracentrifugation (13, 14). HPLC analysis of RLP fractions isolated in this way from normolipidemic and diabetic subjects (14), and fast protein liquid chromatographic analysis of RLP from type III and type IV patients (55), have revealed considerable size heterogeneity in RLP, with particles ranging in size from VLDL to LDL. The relative amount of lipid and apolipoprotein in RLP can also vary considerably from one individual to another. Hypertriglyceridemic patients have more triglyceride and apoC-III, and less apoE, relative to the apoB in RLP, than do normolipidemic subjects (54, 55). Hypertriglyceridemic patients invariably have elevated levels of RLP-C, and the clinical usefulness of this assay depends on the studies which show that RLP-C concentration predicts the presence of coronary or carotid atherosclerosis independently of the plasma triglyceride level (59, 60).
The median concentration of RLP-C is 5.9 mg/dL in 35- to 54-year-old American men and 4.6 mg/dL in similarly-aged women (57). RLP-C is higher in older versus younger subjects,(13, 56), men versus women (56,57), postmenopausal versus premenopausal women(56), the fed versus the fasted state (50, 61), individuals with diabetes (62), patients with familial dysbetalipoproteinemia, (13, 55, 63, 64), hemodialysis patients (48,65) and patients with coronary artery restenosis after angioplasty (66). It has been demonstrated that the RLP-C concentration is significantly higher in patients with CAD than in control subjects (13, 56, 57, 67–69). The potential atherogenicity of RLP-C is supported by the observation that RLP can promote lipid accumulation in mouse peritoneal macrophages (70), stimulate whole-blood platelet aggregation (71, 72) and impair endothelium-dependent vasorelaxation (73). The physiological and pathophysiological aspects of RLP have been investigated extensively by many researchers using isolated RLP fraction in various diseases (74), but relatively little is known about the biochemical composition of RLP or the extent to which this composition varies from one individual to another.
6. Fractionation and analysis of RLP with a gel permeation HPLC system
Gel permeation HPLC is a method for quantifying lipoproteins by particle size (15, 75). Online post column detection of lipid components in lipoproteins particles after separation by size provides quantitative lipoprotein size distribution from whole serum or the RLP fraction. The following combined techniques were used: 1) TSKgel LipopropakXL columns which can separate a wide range of lipoprotein particle sizes from CM to HDL (76), 2) an on-line enzymatic reaction for TC and TG in the separated effluents from the column (77) and 3) 20 component peak analysis using a Gaussian curve fitting technique(78). These techniques can provide CM-cholesterol (CM-C), VLDL-cholesterol (VLDL-C), LDL-C, HDL-C, and total cholesterol (TC) and their triglycerides (CM-TG, VLDL-TG, LDL-TG, HDL-TG and total TG) together with their VLDL, LDL and HDL subclasses. This HPLC method has satisfactory performance on sensitivity, reproducibility and accuracy compared to the reference methods (76). The LipoSearch HPLC analytical service for the TC and TG concentration in the major subclasses of lipoproteins was conducted at Skylight Biotech, Inc (Akita, Japan.:
This HPLC system is appropriate for the profiling of TG rich lipoproteins and postprandial samples because of its capacity to separate CM and VLDL. The high sensitivity of this method enables an examination of the particle size distribution in the RLP fractions separated by immunoseparation gels, which are very low in concentration compared with LDL and HDL in plasma. Moreover, the dual detection system of TG and TC provides useful information for determining the heterogeneity of the RLP particles separated by immunoaffinity gels. Using 100μL of supernatant in immnoaffinity gel suspension solution after 2 h incubation (5μL serum or plasma with 300μL immunioaffinity gel solution), the RLP-C and RLP-TG profiles can be directly observed with an on-line dual detection method using gel permeation HPLC (77). Figure 5 shows a typical profile of the postprandial RLP monitored by TC and TG reagents in normolipidemic and hyperlipidemic cases after a fat load (49). The particle sizes are shown in a range of VLDL to LDL before (0 h) and after (2, 4, 6h) an oral fat load. A small peak at the void retention time was detected at 2h and 4h, in each case as large TG-rich lipoproteins. The clearance of the VLDL fraction was significantly delayed in a hyperlipidemic subject compared to a normolipidemic subject.
Figure 5.
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Typical profiles of RLP-TG and RLP-C in the postprandial plasma of a normolipidemic (TG<150 mg/dL) and a hyperlipidemic (TG>150 mg) subject after oral fat load. The RLP fraction was monitored by TC and TG with an HPLC system. After 5μL serum was incubated with 50μL of immnoaffinity gel (300 μL of gel suspension solution) for 2h as the RLP-C assay, 100μL of the supernatant of the unbound fraction were applied to the HPLC system. RLP was always detected with a large VLDL particle size (thin line; RLP-C, thick line; RLP-TG) and a small peak in the void fraction comprised of the CM size. The clearance of RLP-C and RLP-TG was delayed in the hyperlipidemic case, while clearance in the normolipidemic case returned to the basal level in 6h ( Ref. 97) (Nakano et al., Clin Chim Acta 2011; 412: 71–78)
Another approach is to isolate a large amount of RLP from the plasma for further HPLC fractionation. For example, 0.2ml aliquots of plasma were applied to columns containing 2ml of immunoaffinity mixed gel. The plasma samples were incubated with the immunoaffinity mixed gel at room temperature for 30min. Lipoproteins unbound from the gel (containing primarily CM and VLDL remnants) were eluted with 3.5 ml of 10mM phosphate buffered saline (pH 7.2). The unbound fraction was concentrated with an Amicon Ultra filter (Millipore, USA) for HPLC fractionation and analysis. The concentrated RLP fraction was fractionated by HPLC and 0.35 ml of aliquot in each tube was collected for the determination of TC, TG, apoB-48 and apoB-100 (49).
7. Determination of serum apoB-48 with ELISA
The characteristics and development of the apo B-48 ELISA assay using monoclonal antibodies have been reported by the two groups in Japan[16, 79, 80]. The ELISA for apo B-48 for this study was obtained from Shibayagi (Shibukawa, Gunma) (16). The assay uses a monoclonal antibody raised against a C terminal decapeptide of the apo B-48 protein and was calibrated using recombinant apo B-48 antigen (81). The monoclonal antibody has no cross-reactivity with apo B-100, as verified by ELISA and Western blotting, with more than 90% recovery of apo B-48, and the assay has within- and between run coefficients of variation of 4.8% and 5.4%, respectively. The assay was tested in healthy fasting Japanese volunteers with mean reported values of 0.460 ± 0.15 mg/dL (range 0.27–0.81 mg/dL). In healthy volunteers tested at 0, 2, 4 and 6 h after a 40g fat load, serum apo B-48 and TG increased approximately 2-fold, with a similar peak time of 3 to 4 h [16, 80]. The apoB48 ELISA determined the apoB48 protein in the large TG-rich lipoprotein particles in severe lipemic plasma by pre-treatment with detergent (0.1% Triton X-100). The assay had a sensitivity of 0.25 ng/mL; and a linear dynamic range at 2.5 to 40 ng/mL. Interference with the assay values was noted in the hemoglobin values of at least 106 mg/dL and bilirubin values of at least 10 mg/dL. To further validate this assay, whole plasma and the TRL fraction (after subjecting plasma to ultracentrifugation at its own density of 1.006 g/mL for 18 hours) was studied (82). TRL samples were subjected to gel electrophoresis for the separation of apo B-100 and apo B-48, followed by gel scanning, showed that the mean TRL apo B-48 value was 1.01 mg/dL (SD, 0.43 mg/dL) using ELISA and 0.50 mg/dL (SD, 0.20 mg/dL) using gel scanning. The correlation coefficient or r value between the 2 methods was 0.82 (P <001) (82), highly correlated. Therefore, although the 2 methods correlate well, and the gel method yielded TRL apo B-48 concentrations that were approximately 50% lower than the ELISA values, indicating that a correction factor of 2.02 would need to be applied to prior kinetic studies (12) in which the TRL apo B-48 levels had been assessed by gel scanning. Furthermore, Nakano et al. (83) determined the concentration of apoB-48 and apoB-100 carrying lipoprotein particles extracted from human aortic atherosclerotic plaques in sudden death cases. The plasma apo-B48 level was shown to be significantly elevated in type III cases, as previously reported (55, 80). Therefore, this method was applicable to the determination of the apoB48 concentration in the fractionated RLP using an HPLC system.
8. Postprandial remnant metabolism in CETP deficiency
Plasma cholesteryl ester transfer protein (CETP) mediates the CE/TG exchange from HDL to TG-rich lipoproteins which forms remnant lipoproteins (Figure 1). Fielding et al. (84) reported that CETP activity was significantly increased in the postprandial state almost in parallel with the increase of plasma TG. The RLP-C and RLP-TG levels were increased along with elevated CETP activity in patients with nephrotic range proteinuria (85). Therefore, CETP activity and remnant formation are evidently closely associated.
CETP deficiency results in a low LDL/high HDL phenotype including apoE-rich large HDL. Large HDL may provide cholesteryl ester and apoE to CM/VLDL during lipolysis in the postprandial state, accelerating remnant lipoprotein formation and uptake in the liver. To investigate the role of CETP in postprandial lipoprotein metabolism, the lipid levels of plasma RLP were determined in one homozygous and three heterozygous CETP deficiency cases and controls with an apoE3/3 phenotype by Inazu et al. (86). After oral fat-load, the area under the curve (AUC) of the TG, RLP-TG and apoB48 levels was remarkably decreased in heterozygous CETP deficiency as compared to the controls (Figure 6). Similarly, the homozygote had a significantly low AUC for the TG, RLP-TG and apoB-48 levels.
Figure 6.
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The plasma TG, RLP-C, RLP-TG and ApoB-48 concentrations after a fat-load. The bars indicate the S.E.M and each time point is shown for the homo-, heterozygotes and controls. The open circles indicate controls; half-shaded circles indicate heterozygotes; closed circles indicate data from a CETP-deficient homozygote as a reference. CETP deficiency results in a significantly reduced TRL remnant formation in the postprandial state. (Ref. 85)(Inazu et al. Atherosclersois 2008; 196: 953-57).
HPLC analysis in the homozygote showed that the increased RLP-C was not due to the conventional VLDL size RLP, but to those of large HDL size. In heterozygotes, a bimodal distribution of the RLP-C level was found for the conventional VLDL and large HDL sizes. Subjects with CETP deficiency appeared to have low levels of TG response and diminished remnant lipoprotein formation after a fat-load.
Ai et al. (87) also found a unique profile of serum postprandial RLP-C and RLP-TG during a fat load in a male CETP deficiency patient, aged 40, whose serum HDL-C, apo-AI and TG levels were abnormally elevated. From the genetic analysis, this case was a compound heterozygote with s known mutation of Intron 14 G(+1) > A and R268X. The former mutation is common, but the latter had not been reported previously in the Japanese population. This is the reason the CETP level was less than the detectable limit. Because of this CETP deficiency, HDL is unable to exchange its cholesteryl ester with the TG of other lipoprotein particles, including RLP (86–88). In this case, it is not clear whether the CETP deficiency was the cause of the profile of the postprandial RLP-C and RLP-TG. A similar CETP deficient case with abnormally elevated TG was previously reported by Ritsch et al. (89) and precise genetic analysis of CETP was performed to find the cause of the dissociation between the cholesteryl ester and TG transportation in plasma. However, they could not find any specific cause in that case, and she was completely CETP deficient as well as being an apo-E2 carrier. The serum levels of RLP-C and RLP-TG as well as total TG usually increase and decrease in parallel after an oral fat load in normal individuals. Also, both RLP-C and RLP-TG in CETP deficient cases are usually reduced, as reported by Inazu et al. (86). However, in this case subject, the serum RLP-C level was highly elevated in the fasting state and did not increase after a fat load, but rather decreased, while the RLP-TG and total TG levels significantly increased after a fat load, with a delayed peak time compared with normal control subjects. This phenomenon indicated that CETP and HDL played an important role in the formation of RLP-C ( but not RLP-TG), as has been reported for postprandial lipid metabolism in homo- and heterozygous CETP deficiency cases (86) and the in vitro formation of RLP with CETP deficiency by Okamoto et al. (88). However, interestingly, the RLP-TG and total TG levels in this case subject increased significantly at 240 mins after a fat load, like those in common hyperlipidemic cases. The trend of the case subject was similar to that of individuals treated with estrogen, whose serum RLP-C level is reduced, but RLP-TG level increases, after the treatment (90, 91). This means that the major metabolic pathways of RLP-C and RLP-TG in the postprandial state are controlled independently, although the RLP particle itself is of the same structure as other lipoproteins i.e. comprised of TC, TG, phospholipids and apolipoproteins, and of isolated by the same immunoseparation method. In the case subject, found an extremely elevated plasma ANGPTL3 level, which was discovered as an inhibitory modulator of LPL and HTGL in mice (92). However, it was recently reported that ANGPTL3 associates more strongly with EL or HTGL, which controls HDL-C metabolism, but not with TG or remnants in humans (33, 93). As the case subject Ai et al. reported (87) showed nearly normal LPL and HTGL activity in post-heparin plasma, ANGPTL3 was shown not to affect RLP-TG levels associated with LPL and HTGL activities (33). However, the lack of CETP together with enhanced EL or HTGL inhibition by elevated ANGPTL3 may have affected a significantly increase of the HDL-C level, especially apo-E-rich HDL in this case.
Another interesting dissociation between RLP-C and RLP-TG in the postprandial plasma was observed in one of the heterozygous CETP-deficient subjects (CC-2) after a fat load (87). The serum RLP-C and RLP-TG levels reportedly increase and decrease in parallel after an oral fat load in most of the study cases (46, 95, 96). However, one heterozygous CETP-deficient subject with increased RLP-C and RLP-TG (VLDL size remnants) after a fat load exhibited RLP-TG which started to separately decrease at 240 mins and RLP-C in 360 mins. This dissociation may be associated with the magnitude of CETP deficiency, in which case CC-2 still enhances the formation of VLDL size RLP-C and RLP-TG, but is apparently insufficient to complete the normal pathway between CETP and LPL (97). These cases may be associated with some genetic disorder of either CETP or its activity, resulting in the dissociated clearance of RLP-C and RLP-TG.
As LPL metabolizes TRL and CETP enhances the formation of both CM and VLDL remnants, the balance of LPL and CETP activity may determine the major components of the postprandial remnant lipoproteins. Further, CETP deficiency itself may not be atherogenic, whereas together with elevated RLP may it be atherogenic and pose a risk for CHD. These cases may help to clarify the controversy whether CETP deficiency is atherogenic or not.
9. RLP-TG as a marker for the analysis of postprandial remnant lipoproteins
RLP-C has been considered as a risk factor for cardiovascular disease for the last two decades. Most of the studies in the literature have reported the RLP-C concentration in the fasting state (74). The lack of sensitivity of the TG measurement in the RLP fraction made us delay an intended investigation of RLP-TG. Recently, we have established a satisfactory assay of RLP-TG which enabled us to determine the concentration of RLP-TG reference range in the Japanese population (98).
The simultaneous measurement of TG and cholesterol in RLP resulted in a RLP-TG/RLP-C ratio which reflects the RLP particle size, as shown by the HPLC profile reported by Okazaki et al. (99). The RLP-TG/RLP-C ratios showed the the variety of the RLP particle sizes in various lipid disorders and under different physiological conditions; For example, Type III cases exhibited a significantly lower RLP-TG/RLP-C ratio which indicated the presence of a higher cholesterol content in RLP (mainly IDL), while the RLP-TG/RLP-C ratio was increased significantly in the postprandial state, indicating a significantly increased TG content in RLP (mainly large VLDL), as shown in Figure 5. Therefore, the RLP-TG/RLP-C ratio predicted the particle size of RLP in a manner comparable with the HPLC profile.
Although RLP-C also increased after an oral fat load, the changes in the RLP-C/total TG ratio in the postprandial state were not significantly different from the ratio in the fasting state. This is because the percentage of RLP-C in the total TG (RLP-C/total TG ratio) was approximately only 5% in the fasting and 4% in the postprandial state. However, a RLP-C/total TG ratio above 10% in the fasting state is now commonly used for the detection of Type III hyperlipidemia (63, 64, 100). In contrast, the RLP-TG/total TG ratio was approximately 10–15 % in the fasting state, and the ratio increased significantly to more than 40% in the postprandial state. Therefore, we investigated the postprandial changes in the RLP-TG/total TG ratio after an oral fat load (49, 98). The postprandial RLP-TG/total TG ratio increased more than 3 fold in 2 h and RLP-TG made up almost half of the total TG (Table 2). In terms of total TG, we have also studied the TRL-TG ratio with RLP-TG, which is the major fraction of the total TG as VLDL (d<1.006). The RLP-TG/TRL-TG ratio changed significantly more than the RLP-TG/total TG ratio.
The particle size predicted by the RLP-TG/RLP-C ratio reflects a significant time-dependent increase of TG in RLP particles in the postprandial state. RLP-TG increased 5.3 fold in 4h, while RLP-C increased 1.5 fold, as shown in Table 2. Therefore, we have found that RLP-TG is a better maker than RLP-C for a direct comparison with total TG in the postprandial state. The RLP-TG/total TG ratio may reflect LPL activity, because LPL activity was shown to be inversely correlated with the concentration of RLP and TG (33). If a subject has a fasting RLP-TG/total TG ratio above 0.13 (95% percentile in TG less than 150 mg/dL ) (98) or an RLP-TG above 20 mg/dL in the fasting state, it may be a subject who is still in the postprandial state and this reflects the delayed remnant lipoprotein metabolism because of disturbed LPL activities. A higher RLP-TG/total TG ratio may be associated with an increased risk for CHD (29, 59, 60, 101).
10. Characteristics of the RLP isolated from postprandial plasma
As reported by Karpe et al. (29), CM is more susceptible to lipoprotein lipase (LPL) than is VLDL. Therefore, the greater susceptibility to LPL of CM rather than VLDL may more readily result in the formation and accumulation of VLDL remnants than CM remnants in the postprandial state. Therefore, the accumulation of large RLP particles in 4h after an oral fat load may be due to the delayed metabolism of VLDL by LPL. Schneeman et al. (102) also reported the postprandial responses (after fat load) of apoB-48 and apoB-100 were highly correlated with those of TRL triglycerides. Although the increase in apoB-48 represented a 3.5-fold difference in concentration as compared with a 1.6-fold increase in apoB-100, apoB-100 accounted for approximately 80% of the increase in lipoprotein particles in TRL. The increase in the number of apoB-100 particles in RLP (VLDL remnants) is actually far greater than that of the apoB-48 particles (CM remnants) in the postprandial state (102,103).
The RLP isolated from plasma in healthy subjects after an oral fat load using an immunoaffinity mixed gel was significantly increased (49). Figure 5 shows the typical profile of RLP particle size in the range of VLDL to LDL at 0, 2, 4 and 6h after an oral fat load, when monitored by TC and TG with HPLC in a normolipidemic and hyperlipidemic subject, respectively. CM or CM remnants (apoB48) could be detected at void volume (a retention time of 15 min). Figure 7 shows a typical RLP profile isolated from the plasma of a Type IIb hyperlipidemic subject 4 h after an oral fat load. RLP was fractionated by HPLC equipped with gel permeation columns (TSK Lipopropak XL, Toso, Tokyo) (77) and 0.35 ml of aliquot each was collected for the determination of TC, TG, apoB-48 and apoB-100 (Figure 7). The peak of the TC and TG retention time in RLP fractionated by HPLC was observed almost at the same particle size as the apoB-100 peak, but was not the same as the apoB48 peak, as shown in Figure 7. Furthermore, the apoB-48 particle size was similar or smaller than that of apoB-100. These results clearly demonstrate that postprandial RLP do not have any apoB-48 particles carrying a large amount of TG in the void fraction, which may thus be categorized as nascent CM (76, 77). The scale (perpendicular axis) of the apoB-100 (left) and apoB-48 (right) concentration in Figure 7 is 4 fold different, but there are similar areas under the curve. Therefore, the significantly higher concentrations of apoB-100 than apoB-48 in postprandial RLP of similar particle size were found in both normolipidemic and hyperlipidemic cases (49). These results show that approximately 80% of the RLP in the postprandial state is composed of large VLDL with apoB100. The increase of small VLDL such as intermediate density lipoproteins (IDL, Sf 12–20) in the postprandial state has been reported to not be increased in the postprandial state (12, 104), even though IDL has often been defined as a typical remnant lipoprotein.
Figure 7.
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A typical profile of postprandial RLP in a hyperlipidemic subject. RLP was isolated and fractionated by HPLC from the postprandial plasma of a hyperlipidemic subject (fasting level: TC; 238, TG; 196, HDL-C; 53, LDL-C; 111, RLP-C; 7.0, RLP-TG; 69, apoB; 92, apoB48; 0.86 mg/dL) (in) 4 h after an oral fat load. The RLP fractionated by HPLC was monitored for total cholesterol (TC) and triglycerides (TG) (top) as well as for apoB-48 and apoB-100 (bottom). The scale (perpendicular axis) of the apoB-100 (left) concentration is 4 fold higher than that of apoB-48 (right), but displays a similar area under the curve. The major particles in RLP were of VLDL size as monitored by apoB-100, and the comparatively smaller size of CM as monitored by apoB-48 ( Ref. 48) (Nakano et al., Ann Clin Biochem. 2011; 48: 57–64.)
11. Postprandial remnant hyperlipoproteinemia in sudden cardiac death (SCD)
Clinical studies have shown that elevated plasma TG levels greatly increase the risk of sudden cardiac death. Results from the Paris Prospective Study (105) and The Apolipoprotein Related Mortality Risk Study (AMORIS) in Sweden (106) as well as the coronary heart disease mortality in a 24-year follow-up study in China (107) demonstrated that increased TG was a strong risk factor for fatal myocardial infarction. However, plasma TG levels vary often over even a short period of time. Therefore, it has been difficult to identify the relationship between clinical events and elevated TG in the long term prospective studies until recently (3–5).
If the lipid and lipoprotein levels in postmortem plasma correctly reflect the antemortem levels, these data would provide the same values with the results obtained from the prospective studies, which are difficult to obtain in that they require long-term observation to evaluate. The plasma levels of lipids and lipoproteins in sudden death cases may reflect the condition of the subject at the moment of fatal clinical events followed by certain inevitable postmortem alterations, but nevertheless still usefully reflect the physiological conditions when the fatal events had occurred. Therefore, we analyzed postmortem plasma under well-controlled conditions to clarify the cause or risk of sudden cardiac death. Plasma RLP-C and RLP-TG levels vary greatly within a short time do the TG levels, unlike other stable lipoprotein markers such as HDL-C and LDL-C. Hence, the cross-sectional study of RLPs at the moment of sudden death may be superior to a prospective study of RLP in determining the potential risk for CHD (108). During the investigations of sudden death cases, we found that the postmortem alterations of lipoproteins in plasma were unexpectedly slight (109) compared with the proteins or other bio-markers. Moreover, these plasma lipoprotein levels were very similar to those determined in living patients, based on the clinical studies in our laboratory.
More than two thirds of the SCD cases observed in our studies, including Pokkuri death syndrome (PDS; sudden cardiac death cases without coronary atherosclerosis), showed stomach full, indicating a strong association with postprandial remnant hyperlipoproteinemia. Significant remnant hyperlipoproteinemia was observed in the plasma of SCD cases compared with the control death cases (69, 103, 110–114). These data suggest that the increased RLP in SCD cases may be mainly composed of CM remnants. However, unexpectedly, we found no significant differences in the apoB-48 levels in plasma or in terms of RLP apoB48, but found a significant increase of RLP aoB100 levels in SCD compared to the control cases (103). The RLP apoB100 levels were significantly increased in the SCD cases in the postprandial state (when RLP-C and RLP-TG were significantly increased), however, neither the plasma apoB48 nor RLP apoB48 level was significantly increased. These results strongly suggested that the major subset of RLP associated with fatal clinical events with the stomach full was the apoB-100 carrying particles, not the apoB-48 particles.
The absolute amount of apoB-100 in RLP is much greater (approximately 7 fold) than that of apoB-48 in RLP. Also the particle size of apoB-48 and apoB-100 was very similar, as shown in Figure 7 (114). Furthermore, we often found SCD cases who had consumed alcohol on a full stomach. It is known that alcohol increases fatty acids in the liver and enhances VLDL production while inhibiting LPL activity (115). Alcohol intake with a fatty meal is well known to greatly increase TG in the postprandial state. The intake of alcohol together with a fatty meal may thus enhance the production of apoB100 carrying VLDL in the liver and increase VLDL remnants by inhibiting LPL activity, resulting in an increase in the potential risk of coronary artery in SCD cases (116).
12. Conclusion
Three new analytical methods were used to investigate the characteristics of postprandial lipoprotein metabolism. These were a method of isolating the remnant lipoproteins as RLP using specific antibodies, gel permeation HPLC and apoB48 ELISA. The major portion of the TG which increased in the postprandial state was shown to be TRL remnant lipoproteins. Although Zilversmit proposed CM or CM remnants as the major lipoproteins which increased in the postprandial lipemia, we have shown that the major portion of the postprandial lipoproteins which increased after food intake was comprised of VLDL remnants. The amount of apoB48 particles was shown to be much lower than apoB100 particles in the postprandial RLP, and the particle sizes of apoB48 in RLP were similar with those of apoB100 or smaller when analyzed by gel permeation HPLC system. We did not find CM or CM remnants having a large amount of TG in postprandial RLP. Therefore, the major part of postprandial TRL remnants is composed of VLDL remnants (approximately 80% or more), not CM remnants. It was also found that TG versus RLP-TG concentrations in the postprandial state correlated significantly higher than in the fasting state. Furthermore, it was demonstrated that approximately 80 % of the increased TG in the postprandial state consisted of the TG in remnant lipoproteins. Taken together, we propose the following equation for the estimated concentration of the increased remnant lipoproteins in the postprandial plasma; (Postprandial TG - Fasting TG) x 0.8 = RLP-TG, to reflect the increase in the postprandial plasma as being mainly composed of VLDL remnants.
Acknowledgments
The authors wish to thank Dr. Richard Havel, University of California San Francisco, and Dr. Ernst Schaefer, Tufts University Boston for their long term collaboration on remnant lipoprotein research. Also we greatly thank Drs. Sanae Takeichi, Takeichi Medical Research Laboratory, and Masaki Q Fujita, Keio University, for their research collaboration on sudden cardiac death and remnant lipoproteins.
Footnotes
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Abstract
1. Introduction
2. Metabolism of CM, VLDL and their remnants in plasma
3. Biochemical characteristics of CM and VLDL remnants
4. Daily rhythm of plasma cholesterol, TG and remnant lipoproteins and the changes in the lipoprotein levels after a fat load
5. Isolation of remnant-like lipoprotein particles (RLP) using specific antibodies and the diagnostic characteristics of the RLP-cholesterol (RLP-C) assay
6. Fractionation and analysis of RLP with a gel permeation HPLC system
7. Determination of serum apoB-48 with ELISA
8. Postprandial remnant metabolism in CETP deficiency
9. RLP-TG as a marker for the analysis of postprandial remnant lipoproteins
10. Characteristics of the RLP isolated from postprandial plasma
11. Postprandial remnant hyperlipoproteinemia in sudden cardiac death (SCD)
12. Conclusion
Acknowledgments
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References
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3127 | https://www.droracle.ai/articles/267200/stemi-management-guidelines | What are the management guidelines for ST-Elevation Myocardial Infarction (STEMI)?
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What are the management guidelines for ST-Elevation Myocardial Infarction (STEMI)?
Medical Advisory BoardAll articles are reviewed for accuracy by our Medical Advisory Board
Educational purpose only • Exercise caution as content is pending human review
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Last updated: August 17, 2025 • View editorial policy
Management Guidelines for ST-Elevation Myocardial Infarction (STEMI)
Primary percutaneous coronary intervention (PCI) is the preferred reperfusion strategy for all STEMI patients presenting within 12 hours of symptom onset, and should be performed as rapidly as possible with a target door-to-balloon time of less than 90 minutes.1, 2
Initial Assessment and Management
Immediate Interventions
Administer aspirin 162-325 mg (chewable or non-enteric coated) immediately upon STEMI diagnosis 1, 2
Administer a P2Y12 inhibitor loading dose as early as possible:
Clopidogrel 600 mg, OR
Prasugrel 60 mg (contraindicated in patients with history of stroke/TIA or weight <60 kg), OR
Ticagrelor 180 mg 1, 2, 3
Initiate anticoagulation therapy:
Unfractionated heparin: 70-100 U/kg IV bolus if no planned GP IIb/IIIa inhibitor; 50-70 U/kg if planned GP IIb/IIIa inhibitor 2
Bivalirudin may be preferred in patients at high risk of bleeding 1, 2
Administer supplemental oxygen if oxygen saturation <90% 4
Reperfusion Strategy Decision
Primary PCI (Class I, Level A) for:
Patients presenting within 12 hours of symptom onset 1
Patients with contraindications to fibrinolytic therapy regardless of time delay 1
Patients with cardiogenic shock or acute severe heart failure regardless of time delay 1
Fibrinolytic therapy if:
Primary PCI cannot be performed within 120 minutes of first medical contact 2
Patient presents within 12 hours of symptom onset 1
PCI after 12 hours is reasonable if:
Clinical and/or ECG evidence of ongoing ischemia between 12-24 hours after symptom onset (Class IIa, Level B) 1
Procedural Considerations for Primary PCI
Stent Selection
Placement of a stent (bare-metal or drug-eluting) is recommended (Class I, Level A) 1
Use bare-metal stents in patients with:
High bleeding risk
Inability to comply with 1 year of dual antiplatelet therapy
Anticipated invasive or surgical procedures within the next year 1
Avoid drug-eluting stents in patients unable to tolerate or comply with prolonged DAPT (Class III: Harm) 1
Procedural Cautions
Do not perform PCI of a non-infarct artery during primary PCI in hemodynamically stable patients (Class III: Harm) 1
Do not use fondaparinux as the sole anticoagulant for primary PCI due to risk of catheter thrombosis (Class III: Harm) 1, 2
Post-PCI Antithrombotic Therapy
Antiplatelet Therapy
Continue aspirin 81 mg daily indefinitely 1, 2
Continue P2Y12 inhibitor for 1 year after stent placement:
Clopidogrel 75 mg daily, OR
Prasugrel 10 mg daily (5 mg if weight <60 kg), OR
Ticagrelor 90 mg twice daily 1, 2, 3
Anticoagulation
Continue anticoagulation for a minimum of 48 hours and preferably for the duration of hospitalization (up to 8 days) 2
Management of Complications
Cardiogenic Shock
Perform immediate angiography and PCI when indicated 1
Initiate intra-aortic balloon counterpulsation if shock is not quickly reversed with pharmacological therapy 4
Consider early revascularization (PCI or CABG) for patients <75 years with shock developing within 36 hours of MI 4
Use intra-arterial monitoring 4
Consider pulmonary artery catheter monitoring 4
Pulmonary Congestion
Administer oxygen to maintain saturation >90% 4
Give morphine sulfate 4
Initiate ACE inhibitors within 24 hours (start with low dose, e.g., captopril 1-6.25 mg) unless systolic BP <100 mmHg 4
Consider intra-aortic balloon pump for refractory pulmonary congestion 4
Post-Cardiac Arrest
Initiate therapeutic hypothermia as soon as possible in comatose patients with STEMI and out-of-hospital cardiac arrest 1, 2
Perform immediate angiography and PCI when indicated 1
Secondary Prevention
Pharmacological Therapy
Initiate high-intensity statin therapy as early as possible 2
Start beta-blockers within a few days if no contraindications 5
Initiate ACE inhibitors within 24 hours for patients with:
Anterior infarction
Pulmonary congestion
LVEF <40% 5, 2
Consider aldosterone blockade for patients with LVEF ≤40% and heart failure or diabetes 2
Treat hypertension to target <140/90 mmHg (<130/80 mmHg for patients with diabetes or chronic kidney disease) 5
Initiate hypoglycemic therapy to achieve HbA1c <7% in diabetic patients 5
Anticoagulation for Specific Indications
Warfarin (INR 2.0-3.0) for patients with:
Persistent or paroxysmal atrial fibrillation
LV thrombus (for at least 3 months)
LV dysfunction with extensive regional wall-motion abnormalities 5
Lifestyle Modifications
Encourage regular exercise (minimum 30 minutes daily or at least 3-4 times per week) 5
Implement cardiac rehabilitation programs 5, 2
Promote smoking cessation with appropriate support 2
Follow-up Assessment
Perform echocardiography during hospitalization to assess:
Left and right ventricular function
Mechanical complications
Left ventricular thrombus 2
Common Pitfalls to Avoid
Delaying primary PCI beyond recommended timeframes
Using prasugrel in patients with history of stroke/TIA or weight <60 kg 3
Discontinuing dual antiplatelet therapy prematurely, especially in the first few weeks after acute coronary syndrome 3
Administering beta-blockers or calcium channel blockers to patients with frank cardiac failure 4
Using short-acting dihydropyridine calcium channel blockers for hypertension 5
References
1
Guideline
Guideline Directed Topic Overview
Dr.Oracle Medical Advisory Board & Editors, 2025
2
Guideline
Acute Management of ST-Elevation Myocardial Infarction (STEMI)
Praxis Medical Insights: Practical Summaries of Clinical Guidelines, 2025
3
Drug
Official FDA Drug Label For prasugrel (PO)
FDA, 2025
4
Guideline
Guideline Directed Topic Overview
Dr.Oracle Medical Advisory Board & Editors, 2025
5
Guideline
Guideline Directed Topic Overview
Dr.Oracle Medical Advisory Board & Editors, 2025
Related Questions
What is the management of ST-Elevation Myocardial Infarction (STEMI)?What is the management approach for a patient with anterolateral ST elevation myocardial infarction (MI)?What is the management of an inferior ST-Elevation Myocardial Infarction (STEMI) on an electrocardiogram (ECG)?What is the recommended treatment for Non-ST-Elevation Myocardial Infarction (NSTEMI)?What is the management approach for ST-Elevation Myocardial Infarction (STEMI) as a hospitalist?Why can't potassium be given with dextrose (glucose) saline in hypokalemic periodic paralysis?What is the recommended dose of phosphate enema for pediatric patients?How does thyrotoxicosis exacerbate hypokalemic periodic paralysis?When should anticoagulant therapy be initiated in patients with ST-Elevation Myocardial Infarction (STEMI)?What is the management for exposure to Methicillin-resistant Staphylococcus aureus (MRSA)?
Professional Medical Disclaimer
This information is intended for healthcare professionals. Any medical decision-making should rely on clinical judgment and independently verified information. The content provided herein does not replace professional discretion and should be considered supplementary to established clinical guidelines. Healthcare providers should verify all information against primary literature and current practice standards before application in patient care. Dr.Oracle assumes no liability for clinical decisions based on this content.
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3128 | https://www.doubtnut.com/qna/98743788 | The condition that xn+1−xn+1 shall be divisible by x2−x+1 is that
n=6k+1
n=6k−1
n=3k+1
none of these
The correct Answer is:a
To determine the condition under which xn+1−xn+1 is divisible by x2−x+1, we can follow these steps: Step 1: Factor the divisor The polynomial x2−x+1 can be factored using its roots. The roots are given by: ω=e2πi/3andω2=e−2πi/3 Thus, we can express x2−x+1 as: x2−x+1=(x−ω)(x−ω2) Step 2: Apply the divisibility condition For xn+1−xn+1 to be divisible by x2−x+1, it must equal zero when x=ω and x=ω2. Therefore, we need to check: 1. f(ω)=ωn+1−ωn+1=0 2. f(ω2)=(ω2)n+1−(ω2)n+1=0 Step 3: Evaluate f(ω) Calculating f(ω): f(ω)=ωn+1−ωn+1 Factoring out ωn: f(ω)=ωn(ω−1)+1 Setting this equal to zero gives: ωn(ω−1)+1=0⟹ωn(ω−1)=−1 Step 4: Evaluate f(ω2) Calculating f(ω2): f(ω2)=(ω2)n+1−(ω2)n+1 Factoring out (ω2)n: f(ω2)=(ω2)n(ω2−1)+1 Setting this equal to zero gives: (ω2)n(ω2−1)+1=0⟹(ω2)n(ω2−1)=−1 Step 5: Analyze the conditions From both evaluations, we have two conditions: 1. ωn(ω−1)=−1 2. (ω2)n(ω2−1)=−1 Step 6: Determine values of n To satisfy these equations, we observe the periodic nature of powers of ω: - ω3=1 - Thus, ωn will repeat every 3 values. We find that: - For n≡0mod3: ωn=1 - For n≡1mod3: ωn=ω - For n≡2mod3: ωn=ω2 By substituting these into the conditions, we find that n must satisfy: - n≡1mod3 Conclusion Thus, the condition that xn+1−xn+1 is divisible by x2−x+1 is: n≡1mod3
To determine the condition under which xn+1−xn+1 is divisible by x2−x+1, we can follow these steps:
Step 1: Factor the divisor
The polynomial x2−x+1 can be factored using its roots. The roots are given by:
ω=e2πi/3andω2=e−2πi/3
Thus, we can express x2−x+1 as:
x2−x+1=(x−ω)(x−ω2)
Step 2: Apply the divisibility condition
For xn+1−xn+1 to be divisible by x2−x+1, it must equal zero when x=ω and x=ω2. Therefore, we need to check:
1. f(ω)=ωn+1−ωn+1=0
2. f(ω2)=(ω2)n+1−(ω2)n+1=0
Step 3: Evaluate f(ω)
Calculating f(ω):
f(ω)=ωn+1−ωn+1
Factoring out ωn:
f(ω)=ωn(ω−1)+1
Setting this equal to zero gives:
ωn(ω−1)+1=0⟹ωn(ω−1)=−1
Step 4: Evaluate f(ω2)
Calculating f(ω2):
f(ω2)=(ω2)n+1−(ω2)n+1
Factoring out (ω2)n:
f(ω2)=(ω2)n(ω2−1)+1
Setting this equal to zero gives:
(ω2)n(ω2−1)+1=0⟹(ω2)n(ω2−1)=−1
Step 5: Analyze the conditions
From both evaluations, we have two conditions:
1. ωn(ω−1)=−1
2. (ω2)n(ω2−1)=−1
Step 6: Determine values of n
To satisfy these equations, we observe the periodic nature of powers of ω:
- ω3=1
- Thus, ωn will repeat every 3 values.
We find that:
- For n≡0mod3: ωn=1
- For n≡1mod3: ωn=ω
- For n≡2mod3: ωn=ω2
By substituting these into the conditions, we find that n must satisfy:
- n≡1mod3
Conclusion
Thus, the condition that xn+1−xn+1 is divisible by x2−x+1 is:
n≡1mod3
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The term multiplicative inverse is generally used to represent the reciprocal of a given number. A number when multiplied with its multiplicative inverse gives one as a product. The term inverse can refer to different operations, elements, functions across different fields of study in mathematics. We have Inverse Functions, additive inverses and multiplicative inverse where they can take upon different meanings depending upon our domain of study; real numbers, polynomial functions, trigonometric functions, matrices, vectors and so on in mathematics .
Through this article, we will learn how to find the multiplicative inverse of integers, fractions, complex numbers and more with solved examples.
Multiplicative Inverse
The multiplicative inverse of a particular number is the number that when multiplied by the original number gives 1 as the outcome. The concept is commonly used in the simplification of mathematical expressions.
Mathematically, for a number, say P, the multiplicative inverse is represented by 1/P or (P^{-1}).
Multiplicative inverses of numbers are also referred to as reciprocals. Any given number when multiplied by its multiplicative inverse results in one. Let us learn the multiplicative inverse of integers, fractions, mixed fractions, complex numbers, zero and matrices.
Multiplicative Inverse of Integers
Integers in the number system is defined as the collection of positive and negative ‘counting numbers’, as well as zero, that does not have a fractional component(…-3, -2, -1, 0, 1, 2, 3…). The multiplicative inverse of integers is the reciprocal of the integer itself.
For example, for an integer, say, 12, if we multiply it by 1/12, the outcome will be 1. Thus, the multiplicative inverse of 12 is 1/12. Just like positive integers, for a negative integer, say -21 the multiplicative inverse is -1/21.
We can also say that the multiplicative inverse of a positive integer is always positive and that of a negative integer is always negative.
Multiplicative Inverse of a Fraction
A simple fraction in maths consists of a numerator and denominator and is represented in a p/q format where (q\neq 0[/layex]
The multiplicative inverse of a unit fraction(a type of fraction where the numerator is equal to 1) is represented by 1/x is x. As the product of 1/x and x is 1. Similarly, the multiplicative inverse of a fraction that is of the form; p/q is q/p. As the multiplication of p/q × q/p = 1. For example, for the fraction 3/8, the multiplicative inverse is 8/3.
Multiplicative Inverse of a Mixed Fraction
A type of fraction where there is a whole number and a fractional part is said to be a mixed fraction. To determine the multiplicative inverse of a mixed type fraction, we transform the mixed fraction into an improper fraction, i.e., a fraction whose numerator is greater than its denominator and then take its reciprocal.
For example for a mixed fraction;()2\frac{4}{5}), we convert it to an improper fraction; (\frac{5}{14}). Then, the multiplicative inverse is;(\frac{5}{14}). The multiplicative inverse of a mixed fraction is always a proper fraction, i.e., a fraction whose numerator is smaller than its denominator.
Also, read about here.
Multiplicative Inverse of Complex Numbers
A complex number is a mix of real and imaginary numbers, the representation of a complex number is z=x + iy. The ‘x’ implies the real part and is denoted by Re(z), and ‘iy’ signifies the imaginary part that is denoted by Im(z).
The multiplicative inverse of a complex number denoted by Z is 1/Z. Also, the Multiplication of Complex Numbers of Z and1/Z is 1.
Multiplicative Inverse of 0
According to the definition, the multiplication of a number and its multiplicative inverse is one. When it comes to zero, its product with any number is zero only. Also, the inverse of zero(0) is written as 1/0 but its value is not defined. Therefore, the multiplicative inverse of 0 does not exist.
Multiplicative Inverse of Matrices
A matrix is a rectangular array of ‘mxn’ numbers arranged in m horizontal rows and n vertical columns. Unique multiplicative inverse exists only for square matrices,i.e., matrices who have the same number of rows and columns. The multiplicative inverse of a square matrix of order nxn is another matrix of order nxn such that their product gives an identity matrix of order nxn. If (A^{-1}) is the multiplicative inverse of matrix A, then (AA^{-1}= A^{-1}A = I).
An identity matrix I has 1’s in its principal diagonal and the rest of the elements are zero. The formula for the inverse of a matrix A is;
(A^{-1}=\frac{1}{\left|A\right|}.\ adj\ A)
Know more about Proper Fractions here.
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Steps to Find Multiplicative Inverse
Now that we know, what is the multiplicative inverse of a number and how it is denoted for different types of numbers. Let us understand the steps to find the multiplicative inverse for integers, fractions, complex numbers and matrices.
Steps to find multiplicative inverse of rational numbers
Let us find the multiplicative inverse of 2/3.
Step 1: The first step is to divide the given number by 1. The first step is to divide it by 1, which will result in 1/(2/3) = 3/2.
Step 2: Now express the number in the form of a fraction. If our number is p, then its reciprocal fraction is 1/p. This will result in 1/(2/3) = 3/2.
Step 3: For an expression that contains multiple terms, simplify the second step to obtain the answer. Therefore, the reciprocal of 2/3 is 3/2.
Learn about linear equations in one variables and linear equations in two variables
Steps to find multiplicative inverse of mixed fraction
Step 1: Transform the given mixed fraction into an improper fraction first.
If the mixed fraction is=(2\frac{1}{3}), its equivalent improper fraction=(\frac{7}{3})
Step 2: Now take the reciprocal of the improper fraction and this is the required multiplicative inverse.
The reciprocal of (\frac{7}{3}) is (\frac{3}{7}).
Steps to find multiplicative inverse of complex numbers
Step 1: For a complex number x + iy, take the reciprocal i.e. 1/(x+iy).
Step 2: To simplify the fraction 1/(x+iy), multiply and divide the fraction by its conjugate i.e. (x-iy).
Step 3: Simplifying the fraction will give the required inverse.
Learn more about the different Operations of Complex Numbers here.
Steps to find multiplicative inverse of matrices
Any non-singular matrix(A=\left[a_{ij}\right]) of order n is said to have an inverse if there is another non-singular square matrix B of order n, such that;(A ⋅ B = B ⋅ A = I_n).Here, (I_n) is the identity matrix of order n.
To determine the inverse of a matrix, it must be a square matrix, i.e the no of rows and numbers of columns must be the same ( m=n ).
The determinant of the given matrix should not be equal to zero, i.e. ( \left|A\right|\ne0). If the determinant is not zero we find the adjoint of the matrix.
Once we have the adjoint of the given matrix, the inverse of a square matrix A of order n is calculated using the formula: (A^{-1}=\frac{1}{\left|A\right|}.\ adj\ A).
We can even obtain the inverse of a matrix, by using either elementary transformation(elementary row or elementary column operations) approach as well.
Check out our inverse of matrix article to find the detailed steps.
Modular Multiplicative Inverse
The modular multiplicative inverse of an integer say ‘q’ is another integer ‘x’ that is congruent to 1 concerning the modulus m.
Mathematically expressed as;
qx ≡ 1 (mod m).
Where,
q is an integer
m is an integer
q & m are relatively prime
Also, the modular multiplicative inverse of an integer q exists if and only if q and m are relatively prime. This implies that gcd(q, m) = 1.
Also, read about Inverse Trigonometric Functions here.
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Solved Examples of Multiplicative Inverse
Now that we are well aware of what a multiplicative inverse is and the step to find the same for different numbers, let us head towards some solved examples for more practice.
Solved Example 1: Find the multiplicative inverse of the following and verify the answer as well.
A) -6/13
B)(3\frac{2}{5})
C)-43
Solution: There are three different types of numbers for which we need to find the multiplicative inverse, let us start with option A. A) -6/13 According to the steps the reciprocal of -6/13 is its multiplicative inverse. That is (\frac{-13}{6})
For the verification, the multiplication of the two should be equal to one; (-\frac{6}{13}\times\frac{-13}{6}=1). Thus the answer is verified.
B)(3\frac{2}{5})
This given fraction is a mixed fraction, first convert it into an improper fraction and then take the reciprocal.
For (3\frac{2}{5}), the improper fraction is (\frac{17}{5}), the reciprocal for the same is (\frac{5}{17}).
For the verification, the multiplication of the number and its reciprocal should be equal to one; (\frac{17}{5}\times\frac{5}{17}=1).
C)-43
-43 is a integer, its multiplicative inverse (-\frac{1}{43}). Also, (-43\times\left(-\frac{1}{43}\right)=1). Hence the result is verified.
Learn the various concepts of Like Fractions here.
Solved Example 2: Find the multiplicative inverse of (2+i\sqrt{5}).
Solution: Given, (2+i\sqrt{5}) is a complex number.
To start with take the reciprocal of (2+i\sqrt{5}), i.e. is equal to (\frac{1}{2+i\sqrt{5}}).
Now simplify the fraction using the concept of conjugate.
(\frac{1}{2+i\sqrt{5}}\frac{2-i\sqrt{5}}{2-i\sqrt{5}})
Using the identity (\left(a+b\right)\left(a-b\right)=a^2-b^2), simplify the expression.
(\frac{2-i\sqrt{5}}{4+5})
Thus the multiplicative inverse of (2+i\sqrt{5})=(\frac{2-i\sqrt{5}}{9}).
Real in detail about the Real Numbers here.
Solved Example 3: What does the multiplicative inverse property say?
Solution: The multiplicative inverse property states that the product of a given number and its multiplicative inverse is equal to one(1).
That is (a. a^{-1} = 1)
Solved Example 4: Check if the given statement is true or false.
A) The multiplicative inverse of 1 is consistently one.
B) The multiplicative inverse of 0 is not defined.
C) The multiplication of a number with its multiplicative inverse is equal to one.
Solution: Let us check the statement one by one. A) The multiplicative inverse of 1 is consistently one. Statement one is true as the reciprocal of 1 is always one, and so is its multiplicative inverse. B) The multiplicative inverse of 0 is not defined. Statement B is also true, as the 1/0 is not defined and hence the multiplicative inverse of 0 is not defined. C) The multiplication of a number with its multiplicative inverse is equal to one. Statement C is the most important property for the verification of multiplicative inverse, and hence it is true as well.
We hope that the above article is helpful for your understanding and exam preparations. Stay tuned to the Testbook App for more updates on related topics from Mathematics, and various such subjects. Also, reach out to the test series available to examine your knowledge regarding several exams.
| |
| If you are checking Multiplicative Inverse article, also check the related maths articles in the table below: |
| Markup | Consecutive integers |
| Factors of 54 | Factors of 32 |
| Factors of 24 | Factors of 21 |
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Multiplicative Inverse FAQs
What is the multiplicative inverse meaning?
The reciprocal of the given number obtained is such that when it is multiplied by the actual number the resultant is equal to one.
How to find the multiplicative inverse?
To obtain the multiplicative inverse, take the reciprocal of the number, that is 1/p is the multiplicative inverse for a number p
What is the multiplicative inverse of 1?
The multiplicative inverse of 1 is 1 itself, as the reciprocal or inverse of 1 is equal to 1.
The multiplicative inverse of a negative rational number is?
It Is known to us that the product of any rational number with its multiplicative inverse is 1 equal to one. Thus the multiplicative inverse of a negative rational number should be negative, such that the product is equal to 1.
What is the multiplicative inverse of − 4?
.The multiplicative inverse of − 4 is -1/4.
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General Rheumatology
Behçet’s Disease
History and Epidemiology
Behçet’s disease — also known as Behçet’s syndrome, Behçet’s vasculitis, Adamantiades-Behçet’s syndrome, and Silk Road disease — is a chronic inflammatory disorder affecting multiple systems throughout the body. Symptoms of Behçet’s disease may have been described as early as the 5th century by Hippocrates1 and were reported by Adamantiades, another Greek physician, in 1931.2
However, Behçet’s disease was named after Turkish dermatologist and scientist Hulusi Behçet, the first to recognize the key triad of symptoms contributing to the disease in 1937.3 Behçet’s disease is most prevalent throughout the Middle East and East Asia (370 patients per 100,000 in Turkey, 80 per 100,000 in Iran, and 14 per 100,000 in China) but is rarer in the United States (5.2 per 100,000), Europe (0.64 per 100,00 in the United Kingdom), and Africa (7.6 per 100,000 in Egypt).4
Behçet’s Disease Diagnosis & Presentation
Behçet’s disease causes inflammation in blood vessels throughout the body, with veins more commonly affected than arteries.4 The International Study Group (ISG) for Behçet disease criteria are the most widely used.5 To meet these criteria, patients much have recurrent oral ulcers (at least three times over a 12-month period), as well as two of the following symptoms:
Recurrent genital ulcers
Eye lesions (eg, retinal vasculitis)
Skin lesions (eg, erythema nodosum)
A positive pathergy test
Despite their widespread usage, the ISG criteria are not without limitations. Eighty percent of patients used to formulate the ISG criteria were from the Middle East, where gastrointestinal symptoms and involvement is less common compared to in other parts of the world. Furthermore, these criteria do not consider the prevalence of symptoms.6 Additional diagnostic criteria for Behçet disease have been developed,7 but the new criteria have lower specificity than the ISG criteria.8
Additional symptoms of Behçet’s disease include pericarditis, arthritis (especially in the lower limbs), abdominal pain, hemoptysis, and pleuritis.9,10 Neurological involvement — dubbed neuro-Behçet disease — is less frequent compared to other symptoms but is considered a more devastating manifestation. Three quarters of patients with neurological involvement display parenchymal nervous system lesions and may present with hemiparesis, dysarthria, ataxia, cranial neuropathies, and severe headache.11,12
Behçet’s Disease Quick Facts
Behçet’s Disease Quick Facts: Infographic Embed Code
Infographic by: Rheumatology AdvisorRheumatology Advisor – Behcet’s Disease
Diagnostic Workup
There is no single laboratory test used to diagnose Behçet’s disease. The diagnostic workup typically begins by taking a detailed history, completing a systematic examination and workup — including serology and laboratory tests (blood and urine tests, skin biopsies) — and referring to the ISG diagnostic criteria specified above (i.e., recurrent genital ulcers, eye/skin lesions, and a positive pathergy test.13
Almost all patients with Behçet’s disease (97% to 99%) present with multiple, painful, and recurrent oral ulcers affecting the soft palate, hard palate, tongue, lips, tonsils, buccal mucosa, and gingiva. Genital lesions are slightly less common (seen in more than 80% of patients), occurring on the scrotum in males and on the vulva or vagina in females.14
Imaging can assist in determining the extent of organ involvement in Behçet’s disease:
X-rays or arthrocentesis for arthritis
CT (including angiography) for bleeding, thrombosis, and aneurysms or vessel involvement in the chest and abdomen
MRI for neurological involvement
Echocardiography for cardiac involvement
Ultrasonography for differentiation of lesions and biopsies
Retinal imaging for peripheral lesions
Optical coherence tomography to confirm macular edema.6
Gastrointestinal involvement can be determined using fecal calprotectin levels.6 A lumbar puncture can evaluate if meningitis is present.
Behçet’s Disease Differential Diagnosis
The diagnosis of Behçet’s disease can be challenging due to the lack of specific or characteristic laboratory findings. Consequently, laboratory tests and investigations are undertaken to rule out other conditions. The symptoms of Behçet’s disease can appear similar other conditions, including:
Inflammatory bowel disease (IBD)
Seronegative arthritis
Systemic lupus erythematosus (SLE)
Herpetic infections23
Behçet’s Disease Management
The management of Behçet’s disease aims to control symptoms and prevent organ damage by decreasing inflammation and/or suppressing the immune system.17,18 Consequently, it is difficult to define a standard treatment for Behçet’s disease, as the management approach will be dictated by symptom presence and severity, as well as other patient and provider factors.6 However, most patients with Behçet’s disease display an undulating pattern of disease activity, where they improve over time as their symptoms become milder.19
Table 1. Range of management approaches used in Behçet disease.6,15
| | |
--- |
| Symptom | Treatment(s) |
| Ulcers and arthritis | Colchicine (effective for arthritis, conflicting evidence for oral ulcers)20 Mouthwash containing local anesthetic (for oral ulcers) NSAIDS Apremilast Immunosuppressive or biological agents (eg, azathioprine, thalidomide, adalimumab, IFNα, and etanercept) Monoclonal antibodies (ustekinumab, secukinumab) TNF inhibitors (certolizumab, golimumab) Topical corticosteroids |
| Thrombotic complications | Rest and limb elevation Compression stockings Topical antibiotics TNF inhibitors |
| Arterial and nervous system involvement | Methylprednisolone pulses prior to oral prednisolone and immunosuppressives Cyclophosphamide TNF inhibitors |
| Ocular disease | Immunosuppressive or biological agents Infliximab, adalimumab Dexamethasone implants Aspirin Retinal laser photocoagulation Vitrectomy Topical (cortico)steroids (eg, fluocinolone acetonide) |
| Gastrointestinal involvement | Mesalazine Corticosteroids Immunomodulators (azathioprine, 6- mercaptopurine) TNF inhibitors (infliximab, adalimumab) |
Monitoring Behçet’s Disease Side Effects
The prevention of post-thrombolytic complications in Behçet’s disease is controversial. If patients develop an arterial aneurysm, the bleeding can be fatal.6 It is not uncommon for patients to rapidly develop cataracts following local steroid use for ocular symptoms, but the treatment remains highly effective. The larger risk associated with steroid use is the development of secondary glaucoma. Close follow-ups are required.6
Adalimumab, while shown to be an effective treatment, is associated with infections and injection site reactions.21 Other drugs, such as IFNα, can lead to a reduced quality of life due to the frequency of adverse events associated with these medications.22
Frequently Asked Patient Questions
What is Behçet’s disease?
Behçet’s disease is a chronic inflammatory disorder affecting blood vessels in multiple body systems, with veins more commonly affected than arteries.
What are the symptoms of Behçet’s disease and how do they aid in the diagnostic process?
To meet the diagnostic criteria for Behçet’s disease, patients must have recurrent oral ulcers and two of the following symptoms: recurrent genital ulcers, eye lesions, skin lesions, and a positive pathergy test. Additional symptoms include pericarditis, arthritis, abdominal pain, hemoptysis, and pleuritis. Neurological involvement is less frequent but considered a more devastating manifestation.
How is Behçet’s disease treated?
Pharmacotherapy used in the management of Behçet’s disease is multi-faceted. Treatment is used based on symptoms and includes colchicine, NSAIDs, immuno-suppressants, biological agents, corticosteroids, and TNF inhibitors. Additional treatment options for arterial and nervous system involvement of Behçet’s disease include methylprednisolone pulses before oral prednisolone and cyclophosphamide. Likewise, ocular disease may be treated with a variety of treatments, including infliximab, adalimumab, dexamethasone implants, aspirin, retinal laser photocoagulation, and topical corticosteroids. Finally, gastrointestinal involvement may be treated with mesalazine, and immunomodulators.
Updated February 16, 2024
References
Feigenbaum A. Description of Behçet’s Syndrome in the Hippocratic Third Book of Endemic Diseases. Br J Ophthalmol. 1956;40(6):355–357. doi: 10.1136/bjo.40.6.355
Adamantiades B. Sur un case d’iritis á hypopyon récidivant. Ann Oculist (Paris). 1931;168:271–278.
Behçet H. Über rezidiverende, apthöse durch ein Virus verursachte Geschwüre am Mund, am uge und an der Genitalen. Dermatol Wochenschr. 1937;105:1152–1157.
Yazici H, Seyahi E, Hatemi G, Yazici Y. Behçet syndrome: a contemporary view. Nat Rev Rheumatol. 2018;14(2): 107–119. doi: 10.1038/nrrheum.2017.208
International Study Group for Behçet’s Disease. Criteria for diagnosis of Behçet’s disease. Lancet. 1990;335(8697):1078-1080.
Yazici Y, Hatemi G, Bodaghi B, et al. Behçet syndrome. Nat Rev Dis Primers. 2021;7(1):67. doi: 10.1038/s41572-021-00301-1
International Team for the Revision of the International Criteria for Behçet’s Disease (ITR-ICBD). The International Criteria for Behçet’s Disease (ICBD): a collaborative study of 27 countries on the sensitivity and specificity of the new criteria. J Eur Acad Dermatol Venereol. 2014;28(3):338-347. doi: 10.1111/jdv.12107
Blake T, Pickup L, Carruthers D, et al. Birmingham Behçet’s service: classification of disease and application of the 2014 International Criteria for Behçet’s Disease (ICBD) to a UK cohort. BMC Musculoskelet Disord. 2017;18(1):101. doi: 10.1186/s12891-017-1463-y
Bolster MB. MKSAP 15 Medical Knowledge Self-assessment Program: Rheumatology. Philadelphia: American College of Physicians; 2009.
Hatemi G, Seyahi E, Fresko I, Hamuryudan V. Behçet’s syndrome: a critical digest of the recent literature. Clin Exp Rheumatol. 2012;33(6 Suppl 94):S3-14.
Al-Araji A, Kidd DP. Neuro-Behçet’s disease: epidemiology, clinical characteristics, and management. Lancet Neurol. 2009;8(2): 192–204. doi:10.1016/S1474-4422(09)70015-8
Uygunoğlu U. & Siva A. Nervous system involvement in Behçet’s syndrome. Curr Opin Rheumatol. 2019;31(1):32-39. doi: 10.1097/BOR.0000000000000562
Davatchi F. Diagnosis/Classification Criteria for Behcet’s Disease. Patholog Res Int. 2012;2012:607921. doi: 10.1155/2012/607921
Adil A, Goyal A, Quint JM. Behcet Disease. StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2022.
15. Behçet’s Syndrome. Rare Disease Database. Published 2018. Accessed 22 July 2022.
Kiafar M, Faezi ST, Kasaeian A, et al. Diagnosis of Behçet’s disease: clinical characteristics, diagnostic criteria, and differential diagnoses. BMC Rheumatol. 2021;5(1):2. doi: 10.1186/s41927-020-00172-1
Shahneh F, Mohammadian M, Babaloo Z, Baradaran B. New approaches in immunotherapy of behçet disease. Adv Pharm Bull. 2013;3(1):9-11. doi: 10.5681/apb.2013.002
Alpsoy E, Akman A. Behçet’s disease: an algorithmic approach to its treatment. Arch Dermatol Res. 2009;301(10):693-702. doi: 10.1007/s00403-009-0990-2
Kural- Seyahi E, Fresko I, Seyahi N, et al. The long-term mortality and morbidity of Behçet syndrome: a 2-decade outcome survey of 387 patients followed at a dedicated center. Medicine. 2003;82(1):60-76. doi: 10.1097/00005792-200301000-00006
Leccese P, Ozguler Y, Christensen R, et al. Management of skin, mucosa and joint involvement of Behçet’s syndrome: A systematic review for update of the EULAR recommendations for the management of Behçet’s syndrome. Semin Arthritis Rheum. 2019;48(4):752-762. doi: 10.1016/j.semarthrit.2018.05.008
Suzuki Y, Hagiwara T, Kobayashi M, et al. Long-term safety and effectiveness of adalimumab in 462 patients with intestinal Behçet’s disease: results from a large real-world observational study. Intest Res. 2021;19(3):301-312. doi: 10.5217/ir.2020.00013
Ozguler Y, Leccese P, Christensen R, et al. Management of major organ involvement of Behçet’s syndrome: a systematic review for update of the EULAR recommendations. Rheumatology. 2018;57(12):2200-2212. doi: 10.1093/rheumatology/key242
Adil A, Goyal A, Quint JM. Behcet Disease. In: StatPearls. NCBI Bookshelf version. StatPearls Publishing; 2022. Accessed August 2, 2022.
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How can you find the minimum positive values of x and y in ax = by + c, when a, b, c, x, y are all integers?
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Peter v. Elbing
Algebra is favorite. Calculus is necessary. · Author has 714 answers and 938.2K answer views
·8y
Originally Answered: How can you find the minimum values of x and y in ax = by + c, when a, b, c, x, y are all integers? ·
If a,b a,b are not relative prime, c c has to be divisible by gcd(a,b)gcd(a,b). Otherwise the equation has no solution. But I don’t understand, what you mean by “minimum values of x and y”. x x or y y might be negative with no limit. So you have to define the notion “minimum values of x and y” first.
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Jackpots that "must drop" before a certain value or time. For example: Red Tiger provider often has jackpots that must trigger before a specific time of day (like before 11:00 PM). If it's already 10:59 PM, you can say with certainty that it's about to drop.
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Most answers to this question boil down to one thing: you can't.
And this is indeed true for the vast majority of slot machines. Jackpots are completely random and don't depend on timing, so predicting the moment of a win is impossible.
But, as always, there are exceptions. Let's break it down.
Cases where you can predict a jackpot
Jackpots that "must drop" before a certain value or time. For example: Red Tiger provider often has jackpots that must trigger before a specific time of day (like before 11:00 PM). If it's already 10:59 PM, you can say with certainty that it's about to drop.
Relax Gaming has must-drop before X amount jackpots. For instance, on May 8, 2025, I saw a jackpot that had to drop before 3,000,000 EUR, and it had already reached 2,985,300 EUR. Within an hour, it actually triggered at 2,986,133 EUR.
In such moments, you can almost guarantee that "now is the time." Of course, the odds remain random, but the probability of payouts increases as you approach the limit, and players worldwide start playing more actively.
I even made some bets myself just in case. It likely had positive expected value.
What about regular progressive jackpots?
For standard progressive jackpots (like NetEnt's: Mega Joker, Divine Fortune, Vegas Night Life, and others), you can't predict the exact moment of winning.
But you can calculate the break-even point for players.
When a jackpot reaches a certain amount (for example, 20,000 EUR for Mega Joker), the game becomes theoretically break-even, and with each bet, the expected value (EV) increases.
That's when so-called jackpot hunters appear, who will play until it drops. That's why you can say that after this threshold, the jackpot is about to trigger.
I've done this myself, so I know how it works. Here's what it looks like on a graph.
After the "threshold" value, the jackpot grows disproportionately fast thanks to players with an advantage.
Bottom line
In 99% of cases, you won't be able to determine when a slot will trigger a jackpot.
But for must-drop jackpots (by time or amount), you can say with near certainty that if there are a couple of minutes or a couple hundred euros left, now is a very good time.
For some regular progressive jackpots, you can only track when they become mathematically profitable, meaning hunters will soon knock them out.
So yes, in rare cases you can tell when a slot is about to pay out a jackpot.
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Mehdi Khayeche
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Benedito Freire
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Alexey Godin
, Ph.D. Mathematics & Economics, Moscow State University (1998) · Author has 75 answers and 1.4M answer views
·Updated 5y
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If (a−b)(b−c)(c−a)=a+b+c(a−b)(b−c)(c−a)=a+b+c and a,b,c a,b,c are positive integers, what is the minimum possible value of a+b+c a+b+c?
First of all, let’s notice that by the pigeonhole principle, two of a,b,c a,b,c have the same parity, implying that one of a−b a−b, b−c b−c, c−a c−a is even, hence LHS is even and so is a+b+c a+b+c.
Let x=a−b x=a−b , y=b−c y=b−c.
Notice that c−a=−(a−b)−(b−c)=−x−y c−a=−(a−b)−(b−c)=−x−y so :
(a−b)(b−c)(c−a)=x y(−x−y)=−x 2 y−y 2 x(1)(1)(a−b)(b−c)(c−a)=x y(−x−y)=−x 2 y−y 2 x
and:a+b+c=(a−b)+2(b−c)+3 c=x+2 y+3 c(2)(2)and:a+b+c=(a−b)+2(b−c)+3 c=x+2 y+3 c
Let’s rewrite our equation and work modulo 3 3 :
−x 2 y−y 2 x=x+2 y+3 c≡x−y(mod 3)(3)(3)−x 2 y−y 2 x=x+2 y+3 c≡x−y(mod 3)
If x≢0(mod 3)x≢0(mod 3) then x 2≡1(mod 3)x 2≡1(mod 3). So:
−x 2 y−y 2 x≡−y−y 2 x(mod 3)−x 2 y−y 2 x≡−y−y 2 x(mod 3) . Combining this with (3)(3) gives:
[math]-y^2x \equiv x \pmod 3 \Rightarrow y^2 \equiv -1 [/math]
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First of all, let’s notice that by the pigeonhole principle, two of a,b,c a,b,c have the same parity, implying that one of a−b a−b, b−c b−c, c−a c−a is even, hence LHS is even and so is a+b+c a+b+c.
Let x=a−b x=a−b , y=b−c y=b−c.
Notice that c−a=−(a−b)−(b−c)=−x−y c−a=−(a−b)−(b−c)=−x−y so :
(a−b)(b−c)(c−a)=x y(−x−y)=−x 2 y−y 2 x(1)(1)(a−b)(b−c)(c−a)=x y(−x−y)=−x 2 y−y 2 x
and:a+b+c=(a−b)+2(b−c)+3 c=x+2 y+3 c(2)(2)and:a+b+c=(a−b)+2(b−c)+3 c=x+2 y+3 c
Let’s rewrite our equation and work modulo 3 3 :
−x 2 y−y 2 x=x+2 y+3 c≡x−y(mod 3)(3)(3)−x 2 y−y 2 x=x+2 y+3 c≡x−y(mod 3)
If x≢0(mod 3)x≢0(mod 3) then x 2≡1(mod 3)x 2≡1(mod 3). So:
−x 2 y−y 2 x≡−y−y 2 x(mod 3)−x 2 y−y 2 x≡−y−y 2 x(mod 3) . Combining this with (3)(3) gives:
−y 2 x≡x(mod 3)⇒y 2≡−1(mod 3)−y 2 x≡x(mod 3)⇒y 2≡−1(mod 3) which can never happen.
It follows that 3 3 divides x x, so (3)(3) can be rewritten as 0≡−y(mod 3)0≡−y(mod 3)
i.e. 3 3 also divides y y, hence 3 3 divides −x−y−x−y implying that 3⋅3⋅3 3⋅3⋅3 divides x y(−x−y)x y(−x−y) and we already know it’s even, so the minimum value we’re looking for is at least 2⋅27=54 2⋅27=54 but this bound can be achieved by setting a=21,a=21,b=15,c=18 b=15,c=18 since:
(21–15)(15–18)(18–21)=6⋅(−3)⋅(−3)=54=21+15+18(21–15)(15–18)(18–21)=6⋅(−3)⋅(−3)=54=21+15+18
So we’re done. ■◼
EDIT:
I wanted to shed some light on how I came up with the solution.
First of all, whenever a question asks for a min (resp. max) of some expression, the strategy is usually to find a lower (resp. upper) bound and then giving an example that shows the bound can be achieved.
Then I like to observe the equations given and try to extract as much information as I can following the KISS principle. For example:
We know a+b+c>0 a+b+c>0, so (a−b)(b−c)(c−a)>0(a−b)(b−c)(c−a)>0 , this tells us two things:
The integers must be distinct, which gives a (naive) lower bound 1+2+3=6 1+2+3=6. Next thing you should do is try to show that 6 6 can be achieved, but you’ll quickly realize that it’s not possible.
a−b,b−c,c−a a−b,b−c,c−a cannot be all positive, but their product is. This implies that one of them is positive and two are negative. I can assume that a a is the largest of the three, so a−b>0 a−b>0 ,c−a<0 c−a<0 hence b−c<0 b−c<0 i.e. a>c>b a>c>b.
(Side note : Here’s another cool thing we can notice : (a−b)(b−c)(c−a)(a−b)(b−c)(c−a) does not change if I add the same constant to the variables, while a+b+c a+b+c would change by a multiple of 3 3, this reduces the problem from an equality to a congruence modulo 3 3.)
When dealing with integers, one of the most common ways to prove a lower bound is through divisibility. The principle is fairly straightforward : if n,m>0 n,m>0 and n n divides m m then n≤m n≤m:
Checking parity is the first thing I did, since 2 2 is the smallest prime number. This proves that a+b+c a+b+c must be even.
What comes after 2 2? That’s right, 3 3. We can’t apply the pigeonhole principle in this case, but here’s what we can say: if two of a,b,c a,b,c have the same remainder modulo 3 3, then one of a−b a−b,b−c b−c,c−a c−a is a multiple of 3 3, so 3 3 divides a+b+c a+b+c. “But what if they all have different remainders..” you say? No problem: that would imply that their remainders are 0,1,2 0,1,2 so a+b+c≡0+1+2≡0(mod 3)a+b+c≡0+1+2≡0(mod 3) ! So in all cases a+b+c a+b+c is a multiple of 3 3.
Do we check divisibility by 5 5 next? You could, but it will lead you no where for the following reason : With 5 5 possible remainders and only 3 3 variables, there’s just too much degrees of “freedom” for the remainders, and we’d probably end up with 20 20 ish possible combinations. No thanks.
So far we proved that our number is divisible by 6 6. 6 6 doesn’t work, so you would try 12,18 12,18 but you will be disappointed. At this point you should ask yourself; do I need to extract more information on a+b+c a+b+c? And how would I go about that?
Remember that:
a+b+c=(a−b)(b−c)(c−a)=(a−b)(c−b)(a−c)a+b+c=(a−b)(b−c)(c−a)=(a−b)(c−b)(a−c)
I rewrote it this way so the three factors are positive which makes things easier. Then I noticed that a−b a−b is the sum of the other two factors. So if I let x=c−b x=c−b and y=a−c y=a−c, then x+y=a−b x+y=a−b and:
a+b+c=x y(x+y)(1)(1)a+b+c=x y(x+y)
Interesting, so the number we’re looking for can be written in that form (i.e. a product of two numbers and their sum). This helps us in reducing the possible values of a,b,c a,b,c when trying to find potential solutions. For example:
When checking 12 12, the only possible combination of divisors that can satisfiy (1)(1) is 1⋅3⋅4 1⋅3⋅4
When checking 18 18, you won’t be able to find solutions since 18 can be written as 6⋅3⋅1 6⋅3⋅1, 9⋅2⋅1 9⋅2⋅1, 3⋅3⋅2 3⋅3⋅2 , 18⋅1⋅1 18⋅1⋅1. But none of those possibilities satisfy (1)(1)
I kept trying this until I got to 30 30, before I realized I needed more information.
As I mentioned earlier, there’s no point in trying to find prime factors >3>3 , so this leaves us with powers of 2 2 and powers of 3 3. Powers of 2 2 won’t get you anywhere (I tried), so you proceed to working modulo 3 3, which is what I did and got that a+b+c a+b+c must be a multiple of 27 27. Combining this with the fact that it has to be even, we obtain a lower bound of 54 54.
Then I magically came up with with the values 21,18,15 21,18,15 b̶e̶c̶a̶u̶s̶e̶ ̶m̶y̶ ̶b̶r̶a̶i̶n̶ ̶i̶s̶ ̶a̶ ̶h̶u̶m̶a̶n̶ ̶c̶a̶l̶c̶u̶l̶a̶ ̶ … I can’t even say that with a straight face. Again, no wizardry involved, this is how I got the solutions:
I want to find x,y x,y that satisfy x y(x+y)=54 x y(x+y)=54. A quick check would tell you that the only combination of factors that satisfy that are 3⋅3⋅6 3⋅3⋅6. Let’s solve for a a,b b, and c c:
3=x=c−b⇒c=b+3 3=x=c−b⇒c=b+3
3=y=a−c⇒a=c+3=b+6 3=y=a−c⇒a=c+3=b+6
What is b b though? Well, we have a third equation, remember our goal is to prove that 54 54 is feasible, so we need :
a+b+c=54⇒(b+6)+b+(b+3)=54⇒b=54–9 3=15 a+b+c=54⇒(b+6)+b+(b+3)=54⇒b=54–9 3=15
and it follows that c=18 c=18 and a=21 a=21. Bam! Done.
This process might seem time-consuming (and it is, at first) but once you work on dozens/hundreds of problems, these things will become reflexes. So my advice to you is : Practice. Practice. Practice.
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Vijay Mankar
HoD (Electronics) at Government Polytechnic Nagpur · Author has 6.9K answers and 11.1M answer views
·1y
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If a b c−a b−b c−c a+a+b+c=2008 a b c−a b−b c−c a+a+b+c=2008 for positive integers a,b,a,b, and c,c, then how do I find the minimum value of a+b+c a+b+c?
a b c−a b−b c−c a+a+b+c=2008 a b c−a b−b c−c a+a+b+c=2008
Subtracting 1 1 from both sides
a b c−a b−b c−c a+a+b+c−1=2008−1=2007 a b c−a b−b c−c a+a+b+c−1=2008−1=2007
a b(c−1)−b(c−1)−a(c−1)+(c−1)=2007 a b(c−1)−b(c−1)−a(c−1)+(c−1)=2007
(c−1)[a b−b−a+1]=2007(c−1)[a b−b−a+1]=2007
(a−1)(b−1)(c−1)=2007=3 2×223(a−1)(b−1)(c−1)=2007=3 2×223
Being cyclic in nature let's assume WLOG a≤b≤c a≤b≤c
1×1×2007⟹(a,b,c)=(2,2,2008)1×1×2007⟹(a,b,c)=(2,2,2008)
1×3×(3×223=669)⟹(a,b,c)=(2,4,670)1×3×(3×223=669)⟹(a,b,c)=(2,4,670)
1×9×223⟹(a,b,c)=(2,10,224)1×9×223⟹(a,b,c)=(2,10,224)
3×3×223⟹(a,b,c)=(4,4,224)3×3×223⟹(a,b,c)=(4,4,224)
(a+b+c)m i n=232(a+b+c)m i n=232
When (a,b,c)=(4,4,224)(a,b,c)=(4,4,224)
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Ragu Rajagopalan
Passionate Maths solver ;Reviving knowledge after 3 decades · Author has 10K answers and 7.5M answer views
·1y
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If a b c−a b−b c−c a+a+b+c=2008 a b c−a b−b c−c a+a+b+c=2008 for positive integers a,b,a,b, and c,c, then how do I find the minimum value of a+b+c a+b+c?
a b c−a b−b c−c a+a+b+c=2008 a b c−a b−b c−c a+a+b+c=2008
⟹a b(c−1)−a(c−1)−b(c−1)+(c−1)=2007⟹a b(c−1)−a(c−1)−b(c−1)+(c−1)=2007
⟹(c−1)⋅[a b−a−b+1]=2007⟹(c−1)⋅[a b−a−b+1]=2007
⟹(a−1)⋅(b−1)⋅(c−1)=2007⟹(a−1)⋅(b−1)⋅(c−1)=2007
Factors of 2007 are :(1,3,9,223,669,2007)Factors of 2007 are :(1,3,9,223,669,2007)
Possible values of (a-1), (b-1) and (c-1):Possible values of (a-1), (b-1) and (c-1):
[Math Processing Error]\begin{matrix}-----&-----&-----&-------&\(a-1)&(b-1)&(c-1)&(a+b+c)&\-----&-----&-----&-------&\1&1&2007&2012&\1&3&669&676&\1&9&223&236&\3&3&223&232&\-----&-----&-----&-------&\
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a b c−a b−b c−c a+a+b+c=2008 a b c−a b−b c−c a+a+b+c=2008
⟹a b(c−1)−a(c−1)−b(c−1)+(c−1)=2007⟹a b(c−1)−a(c−1)−b(c−1)+(c−1)=2007
⟹(c−1)⋅[a b−a−b+1]=2007⟹(c−1)⋅[a b−a−b+1]=2007
⟹(a−1)⋅(b−1)⋅(c−1)=2007⟹(a−1)⋅(b−1)⋅(c−1)=2007
Factors of 2007 are :(1,3,9,223,669,2007)Factors of 2007 are :(1,3,9,223,669,2007)
Possible values of (a-1), (b-1) and (c-1):Possible values of (a-1), (b-1) and (c-1):
−−−−−−−−−−−−−−−−−−−−−−(a−1)(b−1)(c−1)(a+b+c)−−−−−−−−−−−−−−−−−−−−−−1 1 2007 2012 1 3 669 676 1 9 223 236 3 3 223 232−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−(a−1)(b−1)(c−1)(a+b+c)−−−−−−−−−−−−−−−−−−−−−−1 1 2007 2012 1 3 669 676 1 9 223 236 3 3 223 232−−−−−−−−−−−−−−−−−−−−−−
∴(a+b+c)m i n=232 when(a,b,c)=(4,4,224)∴(a+b+c)m i n=232 when(a,b,c)=(4,4,224)
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In the equation (x 2+y)(x+y 2)=(x+y)3(x 2+y)(x+y 2)=(x+y)3, x x and y y are positive integers. What is the greatest possible value of x+y x+y?
How do you prove that there are no positive integer solutions for (x-y) (x+y) = y(x-y)?
If x and y are positive integers such 13x+4y=100, then what is the value of x+y?
If x and y are positive integers and x-y=5, what is the least possible value of x+y?
What is the value of ZZ if X and Y are two positive integers that satisfy XY = ZZ? What is the value of X+Y+ZZ in this case?
Gerald Bieniek
Lives in Corpus Christi, TX (1990–present) · Author has 5K answers and 1.7M answer views
·1y
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What is the minimum value of a/(b+c) +b/(a+c) +c/(a+b) when a, b, and c are positive?
So, we are assuming that all three values are at least 1, and we are looking for the smallest value possible.
We start with the obvious choices, a = 1, b =1, c = 1
1/(1+1) + 1/(1+1) + 1/(1+1) = ?
1/2 + 1/2 + 1/2 = 3/2 = 1.5
And we’ll find that if a = b = c, that’s always the answer because:
a/2a + a/2a + a/2a
1/2 + 1/2 + 1/2 = 3/2 = 1.5
So, what if 1 number was larger?
2/(1+1) + 1/(2+1) + 1/(1+2) = ?
2/2 + 1/3 + 1/3 =
1 + 1/3 + 1/3 = 1.666
And if two were larger, yet equal?
2/(2+1) + 2/(1+2) + 1/(2+2)
2/3 + 2/3 + 1/4 = ?
8/12 + 8/12 + 3/12 = 19/12 = 1.5833
And if all three numbers are different?
1/(2+3) + 2/(
Continue Reading
So, we are assuming that all three values are at least 1, and we are looking for the smallest value possible.
We start with the obvious choices, a = 1, b =1, c = 1
1/(1+1) + 1/(1+1) + 1/(1+1) = ?
1/2 + 1/2 + 1/2 = 3/2 = 1.5
And we’ll find that if a = b = c, that’s always the answer because:
a/2a + a/2a + a/2a
1/2 + 1/2 + 1/2 = 3/2 = 1.5
So, what if 1 number was larger?
2/(1+1) + 1/(2+1) + 1/(1+2) = ?
2/2 + 1/3 + 1/3 =
1 + 1/3 + 1/3 = 1.666
And if two were larger, yet equal?
2/(2+1) + 2/(1+2) + 1/(2+2)
2/3 + 2/3 + 1/4 = ?
8/12 + 8/12 + 3/12 = 19/12 = 1.5833
And if all three numbers are different?
1/(2+3) + 2/(1+3) + 3/(1+2)
1/5 + 2/4 + 3/3
14/20 + 1 = 1.70
So, the smallest option is the simplest
a = b = c
a/(b+c) + b/(a+c) + c/(a+b) = 1.5
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Shambhu Bhat
Retired professor in engineering ;Very fond of mathematics · Author has 6.6K answers and 4.9M answer views
·Apr 16
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What is the maximum value of xy if x and y are integers and x + y = a?
x+y=a x+y=a
⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩a is even say 2 b 4 x y=(x+y)2−(x−y)2(x−y)2 is minimum at 0 when x=y=b 4 x y m a x=(x+y)2−0=a 2 x y m a x=a 2 4{a is even say 2 b 4 x y=(x+y)2−(x−y)2(x−y)2 is minimum at 0 when x=y=b 4 x y m a x=(x+y)2−0=a 2 x y m a x=a 2 4
⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩a is odd say 2 b+1 4 x y=(x+y)2−(x−y)2 minimum value of(x−y)2=1 when x=b+1;y=b x y m a x=a 2−1 4{a is odd say 2 b+1 4 x y=(x+y)2−(x−y)2 minimum value of(x−y)2=1 when x=b+1;y=b x y m a x=a 2−1 4
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Kwok Choy Yue
B.Sc in Mathematics, The Chinese University of Hong Kong (Graduated 1978) · Author has 1.3K answers and 1.4M answer views
·11mo
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How do you solve y=ax²+bx+c goes (-1,_9), (-2,-7) and (3,5) using the point to find the values of a, b and c?
() suppose the points are (-1,-9), (-2,-7) and (3,5).
Continue Reading
() suppose the points are (-1,-9), (-2,-7) and (3,5).
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Asaf Aharoni
Software Engineer at Google (company) (2017–present) · Upvoted by
Surjeet Kaushik
, PhD Mathematics, Indian Institute of Technology, Hyderabad (2018) · Author has 231 answers and 235.2K answer views
·5y
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The positive integers a,b and c are all different. None of them is a square but ab,bc and ca are all squares.What will be the minimum value of a+b+c?
Any positive integer can be expressed as r⋅s 2 r⋅s 2 for some positive integers r,s r,s where r r is square-free. That representation is unique (r r comprises all primes that appear with an odd power in the number). Since a⋅b a⋅b is square, it follows that r(a)=r(b)r(a)=r(b). So we have some positive integers r,s a,s b,s c r,s a,s b,s c such that a=r⋅s 2 a,b=r⋅s 2 b,c=r⋅s 2 c a=r⋅s a 2,b=r⋅s b 2,c=r⋅s c 2 and r is square-free. Since a,b,c are different it follows that s a,s b,s c s a,s b,s c are different. Let's choose the minimal values for each of r,s a,s b,s c r,s a,s b,s c: r r cannot be below 2 (if we choose 1 then a,b,c a,b,c will be squares) and the minimal value
Continue Reading
Any positive integer can be expressed as r⋅s 2 r⋅s 2 for some positive integers r,s r,s where r r is square-free. That representation is unique (r r comprises all primes that appear with an odd power in the number). Since a⋅b a⋅b is square, it follows that r(a)=r(b)r(a)=r(b). So we have some positive integers r,s a,s b,s c r,s a,s b,s c such that a=r⋅s 2 a,b=r⋅s 2 b,c=r⋅s 2 c a=r⋅s a 2,b=r⋅s b 2,c=r⋅s c 2 and r is square-free. Since a,b,c are different it follows that s a,s b,s c s a,s b,s c are different. Let's choose the minimal values for each of r,s a,s b,s c r,s a,s b,s c: r r cannot be below 2 (if we choose 1 then a,b,c a,b,c will be squares) and the minimal values for the s's are 1,2,3 1,2,3. So the minimal values for a,b,c a,b,c is 2⋅1 2,2⋅2 2,2⋅3 2=2,8,18 2⋅1 2,2⋅2 2,2⋅3 2=2,8,18 and their sum is 28.
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Kamal Jain
know a bit of elementary math. · Author has 293 answers and 523.5K answer views
·5y
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If (a−b)(b−c)(c−a)=a+b+c(a−b)(b−c)(c−a)=a+b+c and a,b,c a,b,c are positive integers, what is the minimum possible value of a+b+c a+b+c?
If (a−b)(b−c)(c−a)=a+b+c(a−b)(b−c)(c−a)=a+b+c(a−b)(b−c)(c−a)=a+b+c and a,b,c a,b,c a,b,c are positive integers, what is the minimum possible value of a+b+c a+b+c a+b+c?
Let us do modulo 3 arithmetic, and denote the remainders by -1, 0, and 1.
If all 3 numbers have different remainders, LHS is not divisible by 3 while RHS is. This is contradiction.
So at least two of them have the same remainders, then LHS is divisible by 3, therefore RHS is also divisible by 3. That means that the third number has the same remainder too. That means all three numbers have the same remainders. That means each of the term on LHS is divisible by 3. Th
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If (a−b)(b−c)(c−a)=a+b+c(a−b)(b−c)(c−a)=a+b+c(a−b)(b−c)(c−a)=a+b+c and a,b,c a,b,c a,b,c are positive integers, what is the minimum possible value of a+b+c a+b+c a+b+c?
Let us do modulo 3 arithmetic, and denote the remainders by -1, 0, and 1.
If all 3 numbers have different remainders, LHS is not divisible by 3 while RHS is. This is contradiction.
So at least two of them have the same remainders, then LHS is divisible by 3, therefore RHS is also divisible by 3. That means that the third number has the same remainder too. That means all three numbers have the same remainders. That means each of the term on LHS is divisible by 3. That means the product is divisible by 27. Note that RHS is not zero, so LHS is not allowed to be zero.
If you do the same logic modulo 2, two of the three numbers have same parity, which shows that LHS is even.
The smallest such possibility is 54.
Let us see is we can find that. The only way to get 54 on LHS is 6x3x3. Do not worry about the sign of the product, since you can rotate a,b,c values among themselves to make LHS positive. Use any method of solving linear equations, gives you a solution of 15, 18, 21. QED
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Kwok Choy Yue
B.Sc in Mathematics, The Chinese University of Hong Kong (Graduated 1978) · Author has 1.3K answers and 1.4M answer views
·2y
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Given that x and y are positive integers and x+xy+y=2022, what are all the possible values of x+y?
Some algebra:
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Some algebra:
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Max Gretinski
Studied Mathematics · Author has 6.3K answers and 2.3M answer views
·Apr 15
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What is the maximum value of xy if x and y are integers and x + y = a?
Let y = a - x, where a is a fixed real number.
Define f(x) = x(a - x) = ax - x 2 x 2
Notice that this is the product of x and y.
Then f is differentiable with f’(x) = a - 2x.
Since the graph of y = f(x) is a parabola opening downward, the point where f’(x) = 0 — the vertex of the parabola — is the maximum.
a - 2x = 0
x = a 2 a 2
For this value of x, y = a - x = a 2 a 2.
The maximum occurs when x and y are equal. That maximum value is
f(a 2 a 2) =a 2 a 2 a 2 a 2= a 2 4 a 2 4
Since x and y must be integers, a is also an integer.
If a is even, we have the situation above.
If a is odd, th
Continue Reading
Let y = a - x, where a is a fixed real number.
Define f(x) = x(a - x) = ax - x 2 x 2
Notice that this is the product of x and y.
Then f is differentiable with f’(x) = a - 2x.
Since the graph of y = f(x) is a parabola opening downward, the point where f’(x) = 0 — the vertex of the parabola — is the maximum.
a - 2x = 0
x = a 2 a 2
For this value of x, y = a - x = a 2 a 2.
The maximum occurs when x and y are equal. That maximum value is
f(a 2 a 2) =a 2 a 2 a 2 a 2= a 2 4 a 2 4
Since x and y must be integers, a is also an integer.
If a is even, we have the situation above.
If a is odd, then either x is odd and y is even, or x is even and y is odd. This is essentially the same situation. The maximum value will be the greatest integer less than a 2 4 a 2 4.
Example:
Let a = 7.
Then a 2 4=49 4 a 2 4=49 4
The greatest integer less than (or equal to) 49 4 49 4 is 48 4=12 48 4=12 In this case, x = 3 and y = 4 (or vice-versa).
Note: since a is odd, a 2 a 2 is 1 more than a multiple of 4. Therefore, we may refine our solution to indicate that the maximum value is a 2−1 4 a 2−1 4.
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Robert Paxson
BSME in Mechanical Engineering, Lehigh University (Graduated 1983) · Author has 3.9K answers and 4M answer views
·1y
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The graph of y=ax^2+bx+c has a minimum at (-7/4, -169/8) and assess through (-5, zero). What are the values of a, b, and c?
y=a x 2+b x+c y=a x 2+b x+c
y′=2 a x+b y′=2 a x+b
At the minimum, y′=0 y′=0:
2 a(−7 4)+b=0 2 a(−7 4)+b=0
b=7 2 a b=7 2 a
such that the equation becomes:
y=a x 2+7 2 a x+c y=a x 2+7 2 a x+c
At (−7 4,−169 8)(−7 4,−169 8), we get:
−169 8=a(−7 4)2+7 2 a(−7 4)+c−169 8=a(−7 4)2+7 2 a(−7 4)+c
16 c=49 a−338 16 c=49 a−338
At (−5,0)(−5,0), we get:
0=a(−5)2+7 2 a(−5)+c 0=a(−5)2+7 2 a(−5)+c
c=−15 2 a c=−15 2 a
From above, we found:
16 c=49 a−338 16 c=49 a−338
16(−15 2 a)=49 a−338 16(−15 2 a)=49 a−338
a=2 a=2
b=7 2(2)=7 b=7 2(2)=7
c=−15 2(2)=−15 c=−15 2(2)=−15
The required equation is:
y=2 x 2+7 x−15 y=2 x 2+7 x−15
A plot looks like this:
Continue Reading
y=a x 2+b x+c y=a x 2+b x+c
y′=2 a x+b y′=2 a x+b
At the minimum, y′=0 y′=0:
2 a(−7 4)+b=0 2 a(−7 4)+b=0
b=7 2 a b=7 2 a
such that the equation becomes:
y=a x 2+7 2 a x+c y=a x 2+7 2 a x+c
At (−7 4,−169 8)(−7 4,−169 8), we get:
−169 8=a(−7 4)2+7 2 a(−7 4)+c−169 8=a(−7 4)2+7 2 a(−7 4)+c
16 c=49 a−338 16 c=49 a−338
At (−5,0)(−5,0), we get:
0=a(−5)2+7 2 a(−5)+c 0=a(−5)2+7 2 a(−5)+c
c=−15 2 a c=−15 2 a
From above, we found:
16 c=49 a−338 16 c=49 a−338
16(−15 2 a)=49 a−338 16(−15 2 a)=49 a−338
a=2 a=2
b=7 2(2)=7 b=7 2(2)=7
c=−15 2(2)=−15 c=−15 2(2)=−15
The required equation is:
y=2 x 2+7 x−15 y=2 x 2+7 x−15
A plot looks like this:
Upvote ·
9 6
Vijay Mankar
HoD (Electronics) at Government Polytechnic Nagpur · Author has 6.9K answers and 11.1M answer views
·Updated 2y
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Given that x and y are positive integers and x+xy+y=2022, what are all the possible values of x+y?
x+x y+y=2022 x+x y+y=2022
x+x y+y+1=2023 x+x y+y+1=2023
(x+1)(y+1)=2023=7×17 2(x+1)(y+1)=2023=7×17 2
As x,y∈Z+x,y∈Z+
So, if (x+1)=7,(y+1)=17 2⟹(x,y)=(6,288)(x+1)=7,(y+1)=17 2⟹(x,y)=(6,288)
if (x+1)=17,(y+1)=7×17⟹(x,y)=(16,118)(x+1)=17,(y+1)=7×17⟹(x,y)=(16,118)
if (x+1)=17 2,(y+1)=7⟹(x,y)=(288,6)(x+1)=17 2,(y+1)=7⟹(x,y)=(288,6)
if (x+1)=7×17,(y+1)=17⟹(x,y)=(118,16)(x+1)=7×17,(y+1)=17⟹(x,y)=(118,16)
if (x+1)=1,(y+1)=2023⟹(x,y)=(0,2022)(x+1)=1,(y+1)=2023⟹(x,y)=(0,2022)
if (x+1)=2023,(y+1)=1⟹(x,y)=(2022,0)(x+1)=2023,(y+1)=1⟹(x,y)=(2022,0)
As pointed by Girija Warrier I am removing above solution, x=0 OR y=0,x=0 OR y=0,as it is mentioned x,y,x,y, as positive integers.
So,
(x+y)={294,134}(x+y)={294,134}
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BENZOYL CHLORIDE (Benzoic acid chloride )
BENZOYL CHLORIDE; Benzenecarbonyl chloride; Benzoic acid chloride; alpha-Chlorobenzaldehyde; CAS #: 98-88-4; UN #: 1736;EC Number: 202-710-8
Preferred IUPAC name: Benzoyl chloride;Synonyms: Benzoic acid chloride.Product Information: CAS number: 98-88-4; EC index number: 607-012-00-0; EC number: 202-710-8;Formula: C₇H₅ClO; Chemical formula:C₆H₅COCl
;Molar Mass: 140.57 g/mol; HS Code: 2916 32 00
α-chlorobenzaldehyde; benzenecarbonyl chloride;benzoic acid chloride
The most important application of benzoyl chloride is the production of benzoyl peroxide, that is used as initiator in the polymer industry.
It is also used as precursor for benzophenone, in the pharmaceutical industry and as precursor of agro-chemicals.
Applications & Features:
Production of Benzoyl Peroxide (catalyst for the polymer industry)
Precursor of Benzophenone and several other UV-stabilizers
Precursor of Agrochemicals and Pharmaceutical applications
Dyes and pigments
Benzoyl chloride, also known as benzenecarbonyl chloride, is an organochlorine compound with the formula C6H5COCl. It is a colourless, fuming liquid with an irritating odour.
It is mainly useful for the production of peroxides but is generally useful in other areas such as in the preparation of dyes, perfumes, pharmaceuticals, and resins.
Definition: An acyl chloride consisting of benzene in which a hydrogen is replaced by an acyl chloride group. It is an important chemical intermediate for the manufacture of other chemicals, dyes, perfumes, herbicides and pharmaceuticals.
CAS No.98-88-4; Chemical Name:Benzoyl chloride; Synonyms: BzCl;Basic Red 1;BENZOXALONE;Benzoylchlorid;BENZOYL CHLORIDE;BenzoylChlorideGr;chloruredebenzoyle;-Chlorobenzaldehyde;Benzoyl Chloride >BENZOYL CHLORIDE, ACS
Stability:Stable. Combustible. Incompatible with strong oxidizing agents, water, alcohols, strong bases. Reacts violently with DMSO and vigorously with alkalies.
Application:Benzoyl chloride is an organochlorine useful for the production of peroxides
Benzoyl chloride appears as a colorless fuming liquid with a pungent odor. Flash point 162°F. Lachrymator, irritating to skin and eyes. Corrosive to metals and tissue. Density 10.2 lb / gal. Used in medicine and in the manufacture of other chemicals.
Preparation: Benzoyl chloride is produced from benzotrichloride using either water or benzoic acid.
As with other acyl chlorides, it can be generated from the parent acid and other chlorinating agents phosphorus pentachloride, thionyl chloride, or oxalyl chloride.
It was first prepared by treatment of benzaldehyde with chlorine.
An early method for production of benzoyl chloride involved chlorination of benzyl alcohol.
Reactions: It reacts with water to produce hydrochloric acid and benzoic acid.
Benzoyl chloride is a typical acyl chloride. It reacts with alcohols to give the corresponding esters. Similarly, it reacts with amines to give the amide.
It undergoes the Friedel-Crafts acylation with aromatic compounds to give the corresponding benzophenones and related derivatives.
With carbanions, it serves again as a source of "PhCO+"
Benzoyl peroxide, a common reagent in polymer chemistry, is produced industrially by treating benzoyl chloride with hydrogen peroxide and sodium hydroxide.
Benzoyl chloride appears as a colorless fuming liquid with a pungent odor. Flash point 162°F. Lachrymator, irritating to skin and eyes. Corrosive to metals and tissue.
Density 10.2 lb / gal. Used in medicine and in the manufacture of other chemicals.
Benzoyl chloride Chemical Properties,Uses,Production
Physical and Chemical Properties
Its pure product is a colorless and transparent flammable liquid, which is smoking exposed to air in the air. In Industry, it is slightly pale yellow, with a strong pungent odor. Its steam has a strong stimulating effect for eye mucous membranes, skin and respiratory tract, by stimulating the mucous membranes and eyes tear. Benzoyl chloride Melting point is-1.0 ℃, boiling point is 197.2 ℃, and the relative density is 1.212 (20 ℃), while a flash point is 72 ℃, and refractive index (n20) is 1.554. It is soluble in the ether, chloroform, benzene and carbon disulfide. It can gradually decomposed in water or ethanol, ammonia, which generates benzoic acid, generating benzamide, ethyl benzoate and hydrogen chloride. In the laboratory, it can be obtained by distillation of benzoic acid and phosphorus pentachloride under anhydrous conditions. Industrial production process can be obtained by the use of thionyl chloride benzaldehyde. Benzoyl chloride is an important intermediate for preparing dyes, perfumes, organic peroxides, resins and drugs. It is also used in photography and artificial tannin production, which was formerly used as an irritant gas in chemical warfare.
Application
Used for dye intermediates, initiator, UV absorbers, rubber additives, medicine etc.
Benzoyl chloride is intermediate of herbicide metamitron, and insecticide propargite, benzene hydrazine or intermediate food.
Benzoyl chloride is used for organic synthesis, dye and pharmaceutical raw material, manufacturing initiator benzoyl peroxide, t-butyl peroxybenzoate, pesticides and herbicides. Benzoyl chloride is an important benzoyl and benzyl reagent. Most of benzoyl chloride is used in the production of benzoyl peroxide, and secondly for the production of benzophenone, benzyl benzoate, benzyl cellulose. Benzoyl peroxide catalyzes polymerization initiator for the monomer plastic, polyester, epoxy, acrylic resin production, self-curing agent, which is a glass fiber material, fluorine rubber, silicone crosslinking agents, oil refined, bleached flour, fiber decolorizing.Further reaction with the acid chloride can also produce acid anhydride, and benzoic acid anhydride is the main purpose for acylation agents, which can also be used as a bleaching agent and flux of a component, as well as it can also be used for the preparation of benzoyl peroxide over.
Chemical Properties: Transparent, colorless liquid; pungent odor; vapor causes tears. Soluble in ether and carbon disulfide; decomposes in water. Combustible.
Chemical Properties: Benzoyl chloride is a colorless to slight brown liquid with a strong, penetrating odor.
Uses: Benzoyl Chloride is used in the manufacturing of dye intermediates.
Uses: For acylation, i.e., introduction of the benzoyl group into alcohols, phenols, and amines (Schotten-Baumann reaction); in the manufacture of benzoyl peroxide and of dye intermediates. In organic analysis for making benzoyl derivatives for identification purposes.
Definition: A liquid acyl chloride used as a benzoylating agent.
Production Methods: Benzoyl chloride can be prepared from benzoic acid by reaction with PCl5 or SOCl2, from benzaldehyde by treatment with POCl3 or SO2 Cl2, from benzotrichloride by partial hydrolysis in the presence of H2SO4 or FeCl3, from benzal chloride by treatment with oxygen in a radical source, and from several other miscellaneous reactions. Benzoyl chloride can be reduced to benzaldehyde, oxidized to benzoyl peroxide, chlorinated to chlorobenzoyl chloride and sulfonated to m-sulfobenzoic acid. It will undergo various reactions with organic reagents. For example, it will add across an unsaturated (alkene or alkyne) bond in the presence of a catalyst to give the phenylchloroketone:
Synthesis Reference(s)
General Description: A colorless fuming liquid with a pungent odor. Flash point 162°F. Lachrymator, irritating to skin and eyes. Corrosive to metals and tissue. Density 10.2 lb / gal. Used in medicine and in the manufacture of other chemicals.
Reactivity Profile: Benzoyl chloride reacts violently with protic solvents such as alcohols, with amines and amides (for example dimethylformamide [Bretherick 1979 p. 6] ) and with inorganic bases. Causes the violent decomposition of dimethyl sulfoxide [Chem. Eng. News 35(9): 87 1957]. May react vigorously or explosively if mixed with diisopropyl ether or other ethers in the presence of trace amounts of metal salts [J. Haz. Mat., 1981, 4, 291]. Friedel-Crafts acylation of naphthalene using Benzoyl chloride, catalyzed by AlCl3, must be conducted above the melting point of the mixture, or the reaction may be violent [Clar, E. et al., Tetrahedron, 1974, 30, 3296].
Hazard: Highly toxic. Strong irritant to skin, eyes, and mucous membranes, and via ingestion, inhala- tion. Upper respiratory tract irritant. Probable carcinogen.
Health Hazard
INHALATION: may irritate eyes, nose and throat. INGESTION: causes acute discomfort. SKIN: causes irritation and burning.
Chemical Reactivity: Reactivity with Water Slow reaction with water to produce hydrochloric acid fumes. The reaction is more rapid with steam; Reactivity with Common Materials: Slow corrosion of metals but no immediate danger; Stability During Transport: Not pertinent; Neutralizing Agents for Acids and Caustics: Soda ash and water, lime; Polymerization: Does not occur; Inhibitor of Polymerization: Not pertinent.
Safety Profile: Confirmed carcinogen with experimental tumorigenic data by skin contact.
Physicochemical Information
Boiling point: 197.2 °C (1013 hPa)
Density: 1.21 g/cm3 (20 °C)
Explosion limit: 2.5 - 27 %(V)
Flash point: 93 °C
Ignition temperature: 600 °C
Melting Point: -0.6 °C
pH value: 2 (1 g/l, H₂O, 20 °C)
Vapor pressure: 0.5 hPa (20 °C)
Refractive Index: 1.5537 (20 °C, 589 nm)
Solubility(20 °C): (decomposition)
BENZOYL CHLORIDE
98-88-4
Benzenecarbonyl chloride
Benzoic acid, chloride
Benzoylchloride
alpha-Chlorobenzaldehyde
benzoic acid chloride
benzoylchlorid
Benzaldehyde, alpha-chloro-
HSDB 383
EINECS 202-710-8
UN1736
BRN 0471389
Benzoyl chloride, 99%, pure
Benzoyl chloride, 98+%, ACS reagent
Benzoyl chloride, ReagentPlus(R), >=99%
benzoyl chlorid
benzoyl choride
bezoyl chloride
benzoic chloride
BzCl
benzoyl chloride-
Benzoyl chloride [UN1736] [Corrosive]
.alpha.-Chlorobenzaldehyde
Benzaldehyde, |A-chloro-
Benzoyl chloride, ACS reagent, 99%
UN 1736
CAS-98-88-4
Benzoyl chloride
ALPHA-CHLOROBENZALDEHYDE
BENZALDEHYDE, ALPHA-CHLORO-
BENZENECARBONYL CHLORIDE
BENZOIC ACID CHLORIDE
BENZOYL CHLORIDE
DIBENZOYL CHLORIDE {BENZOYL CHLORIDE}
BENZOYL CHLORIDE
98-88-4
Benzenecarbonyl chloride
Benzoic acid, chloride
Benzoylchloride
alpha-Chlorobenzaldehyde
benzoic acid chloride
benzoylchlorid
Benzaldehyde, alpha-chloro-
EINECS 202-710-8
UN1736
BRN 0471389
Benzaldehyde, .alpha.-chloro-
benzoyl chlorid
benzoyl choride
bezoyl chloride
benzoic chloride
BzCl
benzoyl chloride-
PhCOCl
Bz-Cl
Benzoyl chloride [UN1736] [Corrosive]
.alpha.-Chlorobenzaldehyde
Benzaldehyde, |A-chloro-
247-558-3 [EINECS]
471389
98-88-4 [RN]
a-Chlorobenzaldehyde
Benzaldehyde, α-chloro-
benzoic acid chloride
Benzoic acid, chloride
Benzoyl chloride [ACD/Index Name] [ACD/IUPAC Name] [Wiki]
Benzoyl chloride [UN1736] [Corrosive]
Benzoylchlorid [German] [ACD/IUPAC Name]
Chlorure de benzoyle [French] [ACD/IUPAC Name]
DM6600000
VTY8706W36
100-09-4 [RN]
2719-27-9 [RN]
4-09-00-00721 (Beilstein Handbook Reference) [Beilstein]
43019-90-5 [RN]
52947-05-4 [RN]
59748-37-7 [RN]
ANISIC ACID
Benzaldehyde, α-chloro-
Benzenecarbonyl chloride
BENZOYL CHLORIDE-(RING-13C6)
Benzoyl Chloride, ACS reagent
BENZOYL CHLORIDE|BENZOYL CHLORIDE
benzoylchloride
Benzoyl-d5 Chloride
Cyclohexanecarbonyl chloride [ACD/Index Name] [ACD/IUPAC Name]
EINECS 202-710-8
Hexahydrobenzoyl chloride
InChI=1/C7H5ClO/c8-7(9)6-4-2-1-3-5-6/h1-5
MFCD01865658 [MDL number]
O-CHLOROFORMYLBENZENE
PS-10801
UNII:VTY8706W36
UNII-VTY8706W36
α-Chlorobenzaldehyde
α-Chlorobenzaldehyde
苯甲酰氯 [Chinese]
benzoilklorür (tr)
benzoil klorür (tr)
benzoil klorid (tr)
benzoilklorid (tr)
benzoilklorit (tr)
benzoil klorit (tr)
bensoüülkloriid (et)
bentsyylikloridi (fi)
benzoil klorid (sl)
benzoil-klorid (hr)
benzoil-klorid (hu)
benzoilchloridas (lt)
benzoile cloruro (it)
benzoilhlorīds (lv)
benzoylchlorid (cs)
benzoylchlorid (da)
Benzoylchlorid (de)
benzoylchlorid (sk)
benzoylchloride (nl)
benzoylklorid (no)
benzoylklorid (sv)
chlorek benzoilu (pl)
chlorek kwasu benzoesowego (pl)
chlorure de benzoyle (fr)
cloreto de benzoílo (pt)
cloruro de benzoílo (es)
cloruro di benzoile (it)
clorură de benzoil (ro)
klorur tal-benżojl (mt)
βενζοϋλοχλωρίδιο (el)
бензоил хлорид (bg)
CAS names: Benzoyl chloride
IUPAC names
Benzoic acid chloride
BENZOYL CHLORIDE
Benzoyl Chloride
Benzoyl chloride
benzoyl chloride
Benzoylchlorid
Trade names
BENZOESAEURECHLORID
BENZOLCARBONYLCHLORID
BENZOYL CHLORIDE
Benzoyl chloride
BENZOYLCHLORID
PHENYLCARBONYLCHLORID
BENZOYL CHLORIDE (BENZOİL KLORÜR;)
CAS NUMBER: 98-88-4
EC NUMBER: 202-710-8
SYNONYMS: BENZOYL CHLORIDE, 98+%, ACS REAGENT; AK4020300005; Benzoyl Chloride; C7H5ClO; 98-88-4; Benzoyl chloride; ACS reagent, 99%; MFCD00000653; PASDCCFISLVPSO-UHFFFAOYSA-N; BENZOYL CHLORIDE; 98-88-4; Benzenecarbonyl chloride; Benzoic acid; chloride; Benzoylchloride; alpha-Chlorobenzaldehyde; benzoic acid chloride; benzoylchlorid; Benzaldehyde; alpha-chloro-a-Chlorobenzaldehyde; CCRIS 802; HSDB 383; EINECS 202-710-8; UN1736; BRN 0471389; Benzaldehyde; .alpha.-chloro- CHEBI:82275; PASDCCFISLVPSO-UHFFFAOYSA-N ; SBB059783 ; Benzoyl chloride ReagentPlus(R) >=99% ; benzoyl choride; bezoyl chloride ; UNII-VTY8706W36 ; benzoic chloride ; BzCl ; benzoyl chloride-PhCOCl; Bz-Cl ; PubChem22045 ; Benzoyl chloride [UN1736][Corrosive] ; AC1L1OPD ;.alpha.-Chlorobenzaldehyde ;Benzaldehyde; |A-chloro- DSSTox_CID_6631 ACMC-20aj01; SCHEMBL1241 ;BENZOIC ACID ; CHLORIDE ; DSSTox_RID_78168 ; DSSTox_GSID_26631 ; BENZOYL CHLORIDE ACS ; 4-09-00-00721 ; KSC486Q7P ; AC1Q3G68 ; CHEMBL2260719 ; DTXSID9026631 ; CTK3I6877 ; KS-00000UUU ; Benzoyl chloride 99% 250g ; Benzoyl chloride AR >=99% ; Benzoyl chloride LR >=99% ; MolPort-001-768-889 ; OTAVA-BB 1051706 ; LABOTEST-BB LTBB000456 ; VTY8706W36 ; CS-B1785 ; ZINC2041164 ; Tox21_200431 ; ANW-75551 ; MFCD00000653 ; STL264120 ; Benzoyl chloride, ACS reagent 99% ; AKOS000121308 ; AS00010 ; MCULE-3627399529 ; RP20639 ; TRA0028457 ; TRA0031506 ; UN 1736 ; CAS-98-88-4 ; Benzoyl chloride purum >=99% (GC) ; Benzoyl chloride ReagentPlus(R) 99% ; NCGC00248610-01 ; NCGC00257985-01 ; Benzoyl chloride [UN1736] [Corrosive] ; Benzoyl chloride p.a. 98-100.5% ; CJ-32526 ; LS-42590 ; OR034273 ; OR382473 ; SC-76545 ; DB002645 ; TC-164296 ; TR-038640 ; B0105 ; FT-0622741 ; ST51046056 ; Benzoyl chloride SAJ first grade >=98.0% ; C19168 ; 11273-EP2270015A1 ; 11273-EP2270114A1 ; 11273-EP2272841A1 ; 11273-EP2272972A1 ; 11273-EP2272973A1 ; 11273-EP2275407A1 ; 11273-EP2275412A1 ; 11273-EP2277848A1 ; 11273-EP2277872A1 ; 11273-EP2277878A1 ; 11273-EP2277880A1 ; 11273-EP2279750A1 ; 11273-EP2281813A1 ; 11273-EP2281815A1 ; 11273-EP2284157A1 ; 11273-EP2286811A1 ; 11273-EP2287159A1 ; 11273-EP2287167A1 ; 11273-EP2292595A1 ; 11273-EP2292610A1 ; 11273-EP2292621A1 ; 11273-EP2295415A1 ; 11273-EP2295429A1 ; 11273-EP2295437A1 ; 11273-EP2298767A1 ; 11273-EP2298775A1 ; 11273-EP2305629A1 ; 11273-EP2305658A1 ; 11273-EP2305687A1 ; 11273-EP2308838A1 ;11273-EP2308858A1 ; 11273-EP2308861A1 ; 11273-EP2308865A1 ; 11273-EP2308877A1 ; 11273-EP2311814A1 ; 11273-EP2311816A1 ; 11273-EP2311817A1 ; 11273-EP2311824A1 ; 11273-EP2311830A1 ; 11273-EP2311840A1 ; 11273-EP2314575A1 ; 11273-EP2314578A1 ; 11273-EP2314587A1 ; 11273-EP2314593A1; 11273-EP2316827A1; 11273-EP2371831A1; 30500-EP2298734A2 ; 30500-EP2308873A1 ; 30500-EP2311835A1 ; 109242-EP2281861A2 ; 109242-EP2295422A2 ; 109242-EP2298769A1 ; I01-4397 ; F2190-0038 ; InChI=1/C7H5ClO/c8-7(9)6-4-2-1-3-5-6/h1-5 ; InChI=1S/C7H5ClO/c8-7(9)6-4-2-1-3-5-6/h1-5H ; PASDCCFISLVPSO-UHFFFAOYSA-N ; benzaldehyde ; a-chloro- ; benzenecarbonyl chloride ; benzoic acid chloride ; benzoic acid ; chloride ; benzoylchloride ; a-chlorobenzaldehyde ; BENZOİL ; KULORİT ; BENZOİLKULORİT ; benzoil ; benzoıl ; chlorite ; benzoılchlorite ; benzoylchloride ; PASDCCFISLVPSO-UHFFFAOYSA-N ; Benzoic acid chloride ; 98-88-4 ; Benzenecarbonyl Chloride ; alpha-Chlorobenzaldehyde ; Benzoic Acid Chloride ; PASDCCFISLVPSO-UHFFFAOYSA-N ; a-Chlorobenzaldehyde Benzaldehyde ?-chloro-benzoic acid chloride ; Benzoic acid, chloride ; Benzoyl chloride ; Benzoylchlorid ; Chlorure de benzoyle ; 100-09-4 ; 2719-27-9 ; 4-09-00-00721 ; 52947-05-4 ; 59748-37-7 ; ANISIC ACID ; Benzaldehyde, ?-chloro- ; Benzenecarbonyl chloride ; BENZOYL CHLORIDE-(RING-13C6) ; benzoylchloride ; Cyclohexanecarbonyl chloride ; EINECS 202-710-8 ; Hexahydrobenzoyl chloride ; InChI=1/C7H5ClO/c8-7(9)6-4-2-1-3-5-6/h1-5 ; MFCD01865658 ; O-CHLOROFORMYLBENZENE ; UNII:VTY8706W36 ; UNII-VTY8706W36 ; ?-Chlorobenzaldehyde ; ?-Chlorobenzaldehyde ; ???? ; C7H5ClO ; MFCD00000653 ; benzaldehyde, a-chloro- ; benzenecarbonyl chloride ; benzoic acid chloride ; benzoic acid, chloride ; benzoylchloride ; a-chlorobenzaldehyde ; p-Chloromethyl benzoyl chloride ; BENZOYL CHLORIDE ; Hydrolysis of Benzoyl Chloride ; PASDCCFISLVPSO-UHFFFAOYSA-N ; 3-(Chlorosulfonyl)benzoyl chloride 98% ; Benzoic acid ; chloride ; benzaldehyde ; alpha-chloro ; Benzoyl Chloride ; cloruro de benzoilo ; 2,4,6-Tris(prop-2-yl)benzoyl chloride ; 2716-5-12; C16H23ClO; Benzoyl Chloride (O) Pure Benzoyl Chloride ; Alpha-Chlorobenzaldehyde ; Benzenecarbonyl Chloride ; Benzoic Acid Chloride ; BENZOYL CHLORIDE ; BENZOİL KLORÜR ; Benzenecarbonyl chloride ; Benzoic acid chloride ; alpha-Chlorobenzaldehyde; 2,3 Dichloro Benzoyl Chloride ; 2,3,4,5,6-PENTAFLUORO-BENZOYL CHLORIDE ; Cloruro benzoilico o cloruro de benzoilo; 4-(Chloromethyl)benzoyl chloride ?97% ;Benzenecarbonyl chloride ; Alpha-chlorobenzaldehyde ; Benzoic acid; BENZOYIL; BENZOYIL KLORİT; BENZOYİLCLORİT; BENZOYILCLORIT; BENZOILKLORID; benzoyıl; benzoyıl klorıt; benzoyilclorit; benzoyılclorıt; benzoılklorıd; benzoil; benzoyıl; benzoyil; chloride; chlorite; kulorit; kulorid; klorür; klorur; benzoyılkulorur; benzoil; benzoilklorür; benzoyiklorür; kulorür; benzoyl chloride; Benzoik asit klorid, Benzenecarbonyl chloride, Benzoic acid, Chloride benzoylchlorid, alpha-Chlorobenzaldehyde, a-Chlorobenzaldehyde, Benzaldehyde, alpha-chloro-CCRIS 802, Benzaldehyde, .alpha.-chloro-, HSDB 383, CHEBI:82275, benzoyl-chloride,benzoyl chloride, Benzoyl chlorite, Acılıopd, Chlorobenzaldehyde, Benzoıc acıd chlorıde, Benzoil klorit, Benzoil Klorid, Benzil Klorit, Benzil Kılorid; BENZOYL CHLORIDE; Benzenecarbonyl chloride; Benzoic acid, chloride; benzoylchlorid; alpha-Chlorobenzaldehyde; a-Chlorobenzaldehyde; Benzaldehyde, alpha-chloro-; Benzaldehyde, .alpha.-chloro-; HSDB 383; CHEBI:82275; PASDCCFISLVPSO-UHFFFAOYSA-N; EINECS 202-710-8; benzoyl-chloride; benzoyl chloride-; PubChem22045; Benzoyl chloride [UN1736] [Corrosive]; ; Benzoyl chloride [UN1736] [Corrosive]; C7H5ClO / C6H5COCl; BENZOİL KLORÜR; benzoil klorür; benzoil klorit; benzoil kloride; benzoil chloride; 247-558-3 [EINECS]; 98-88-4 [RN]; a-Chlorobenzaldehyde, Benzaldehyde, ?-chloro-; benzoic acid chloride; Benzoic acid, chloride; Benzoyl chloride [ACD/IUPAC Name] [Wiki]; Benzoyl chloride [UN1736] [Corrosive]; Benzoylchlorid [German] [ACD/IUPAC Name]; Chlorure de benzoyle [French] [ACD/IUPAC Name]; ALPHA-CHLOROBENZALDEHYDE; BENZALDEHYDE, ALPHA-CHLORO-; BENZENECARBONYL CHLORIDE; BENZOIC ACID CHLORIDE; BENZOYL CHLORIDE; DIBENZOYL CHLORIDE {BENZOYL CHLORIDE}; benzoil klorit; benzoil klorür; benzoil chlorür; benzoil chloride; benzoil klorite;
General Description
Benzoyl Chloride;
A colorless fuming liquid with a pungent odor. Flash point 162°F. Lachrymator, irritating to skin and eyes. Corrosive to metals and tissue. Density 10.2 lb / gal. Used in medicine and in the manufacture of other chemicals.
50 grams of dry benzoic acid are treated in a 500 ml flask, with 90 grams of finely pulverized phosphorus pentachloride. The mixture is mixed well, upon which, after a short time, the reaction takes place with an energetic evolution of hydrochloric acid, and the reaction mass becomes liquid. During the reaction a lot of heat is released. After standing a short time, the completely liquid mixture is twice fractionated by collecting fraction which boils at ~ 200° C, yielding 90 % of benzoyl chloride.
Benzoyl chloride, also known as benzenecarbonyl chloride, is an organochlorine compound with the formula C6H5COCl. It is a colourless, fuming liquid with an irritating odour. It is mainly useful for the production of peroxides but is generally useful in other areas such as in the preparation of dyes, perfumes, pharmaceuticals, and resins.
Benzoyl chloride is a colorless fuming liquid with a pungent odor. Flash point 162°F. Lachrymator, irritating to skin and eyes. Corrosive to metals and tissue. Density 10.2 lb / gal. Used in medicine and in the manufacture of other chemicals.
Physical and Chemical Properties
Benzoyl Chloride;
Its pure product is a colorless and transparent flammable liquid, which is smoking exposed to air in the air. In Industry, it is slightly pale yellow, with a strong pungent odor. Its steam has a strong stimulating effect for eye mucous membranes, skin and respiratory tract, by stimulating the mucous membranes and eyes tear. Benzoyl chloride Melting point is -1.0 ?, boiling point is 197.2 ?, and the relative density is 1.212 (20 ?), while a flash point is 72 ?, and refractive index (n20) is 1.554. It is soluble in the ether, chloroform, benzene and carbon disulfide. It can gradually decomposed in water or ethanol, ammonia, which generates benzoic acid, generating benzamide, ethyl benzoate and hydrogen chloride. In the laboratory, it can be obtained by distillation of benzoic acid and phosphorus pentachloride under anhydrous conditions. Industrial production process can be obtained by the use of thionyl chloride benzaldehyde. Benzoyl chloride is an important intermediate for preparing dyes, perfumes, organic peroxides, resins and drugs. It is also used in photography and artificial tannin production, which was formerly used as an irritant gas in chemical warfare.
Used for dye intermediates, initiator, UV absorbers, rubber additives, medicine etc. Benzoyl chloride is intermediate of herbicide metamitron, and insecticide propargite, benzene hydrazine or intermediate food. Benzoyl chloride is used for organic synthesis, dye and pharmaceutical raw material, manufacturing initiator benzoyl peroxide, t-butyl peroxybenzoate, pesticides and herbicides. In pesticides, it is a new insecticide, which is inducible isoxazole parathion (Isoxathion, Karphos) intermediate. Benzoyl chloride is an important benzoyl and benzyl reagent. Most of benzoyl chloride is used in the production of benzoyl peroxide, and secondly for the production of benzophenone, benzyl benzoate, benzyl cellulose. Benzoyl peroxide catalyzes polymerization initiator for the monomer plastic, polyester, epoxy, acrylic resin production, self-curing agent, which is a glass fiber material, fluorine rubber, silicone crosslinking agents, oil refined, bleached flour, fiber decolorizing Wait. Domestic original benzoyl chloride manufacturing enterprises are more than 20. Some of the manufacturers also produce acid chloride, and the production capacity is 10,000t. However, according to the 2003 survey, the profit is too low, because of the use of small polluting production line, while the use of polluting route is controlled by the government restrictions, and a further raw material price increases. Therefore most of the manufacturers stop the production. Further reaction with the acid chloride can also produce acid anhydride, and benzoic acid anhydride is the main purpose for acylation agents, which can also be used as a bleaching agent and flux of a component, as well as it can also be used for the preparation of benzoyl peroxide over.
Reagents for the analysis, but also for spices, organic synthesis.
APPLICATIONS
Benzoyl Chloride;
Acyl is a radical formed from an organic acid by removal of a hydroxyl group. The general formula of acyl compound is RCO-. Acyl halide is one of a large group of organic substances containing the halocarbonyl group, have the general formula RCO·X, where X is a halogen atom (fluorine, chlorine, bromine, iodine, and astatine) and R may be aliphatic, alicyclic, aromatic, and H etc. In substitutive chemical nomenclature, their names are formed by adding '-oyl' as a suffix to the name of the parent compound; ethanoyl chloride, CH3COCl, is an example. The terms acyl and aroyl halides refer to aliphatic or aromatic derivatives, respectively. Acyl halides are made by replacing the -OH group in carboxylic acids by halogen using halogenating agents. They react readily with water, alcohols, and amines and are widely used in organic synthetic process whereby the acyl group is incorporated into the target molecules by substitution of addition-elimination sequence called acylation reaction. Acylation reaction involves substitution by an electron donor (nucleophile) at the electrophilic carbonyl group (C=O). Common nucleophiles in the acylation reaction are aliphatic and aromatic alcohols, both of which give rise to esters and amines (RNH2) which give amides. The carboxylic acid (X = OH) itself can function as an acylating agent when it is protonated by a strong acid catalyst as in the direct esterification of an alcohol. Two common acylation agents, with the general formula RCOX, are acid halides (X = halogen atom) and anhydrides (X = OCOR). Schotten-Baumann reaction is an acylation reaction that uses an acid chloride in the presence of dilute alkali to acylate the hydroxyl and amino group of organic compounds. There are also other acylating agents. Benzoyl Chloride belongs to acyl halides. Acyl halides are involved in acetylation process which introduce an acetyl group (CH3CO-) into compounds. Benzoyl Chloride decomposes violently by heating or on exposure to moist air or water. It reacts violently with strong oxidants, metals (especially iron), alkali and earth alkali metals, bases and wide range of organic substances such as amines, dimethyl sulfoxide and alcohols. The reactions cause fire and explosion hazard. It is used to introduce benzenecarbonyl groups into compounds. Typical reactions undergone by benzoyl chloride are the Schotten-Baumman reaction (the benzoylation of compounds containing a hydrogen), and the Friedel-Crafts reactions (preparation of substituted benzophenones). It is used in manufacturing peroxides such as a benzoyl peroxide and t-butyl perbenzoate. It is also used in the synthesis of benzophenone and its derivatives used in manufacturing pesticides, pharmaceuticals, perfume fixative, polymerization catalyst, benzolating agents, and dyestuffs.
Genel açıklama
Benzoil klorür;
Duygusal kokusu olan renksiz bir köpürme sıvısı. Parlama noktası 162 ° F. Lachrymator, cildi ve gözleri tahriş eder. Metallere ve dokulara karşı koroziftir. Yoğunluk 10,2 lb / gal. Tıpta ve diğer kimyasalların üretiminde kullanılır.
50 gram kuru benzoik asit 500 ml'lik bir şişe içinde ince öğütülmüş fosfor pentaklorür ile 90 gram muamele edilir. Karışım iyi karıştırılır, kısa bir süre sonra reaksiyon hidroklorik asitin enerjik bir şekilde evrimi ile gerçekleşir ve reaksiyon kütlesi sıvı hale gelir. Reaksiyon esnasında çok fazla ısı açığa çıkar. Kısa bir süre bekledikten sonra, tamamen sıvı olan karışım iki kat daha bölünerek% 200 benzoil klorür verecek şekilde ~ 200 ° C'de kaynayan fraksiyon toplar.
Benzoikarbonil klorür olarak da bilinen benzoil klorür, C6H5COCl formülüne sahip bir organochlorin bileşiğidir. Bu, tahriş edici bir koku veren, renksiz, köpüren bir sıvıdır. Aslında peroksitlerin üretimi için yararlıdır, ancak genellikle boyalar, parfümler, farmasötikler ve reçinelerin hazırlanması gibi diğer alanlarda da yararlıdır.
Benzoil klorür, keskin kokusu olan renksiz bir köpürme sıvısıdır. Parlama noktası 162 ° F. Lachrymator, cildi ve gözleri tahriş eder. Metallere ve dokulara karşı koroziftir. Yoğunluk 10,2 lb / gal. Tıpta ve diğer kimyasalların üretiminde kullanılır.
Fiziksel ve kimyasal özellikler
Benzoil klorür;
Saf ürünü, havada havaya maruz bırakılan, renksiz ve şeffaf yanıcı bir sıvıdır. Endüstride, hafif soluk sarı, güçlü bir keskin kokusu vardır. Buhar, göz mukoza zarları, cilt ve solunum yolu için, mukoza zarlarını ve göz yırtıklarını uyararak güçlü bir uyarıcı etkiye sahiptir. Benzoil klorür Erime noktası -1.0 ?, kaynama noktası 197.2 ° C ve nispi yoğunluk 1.212 (20 ° C) iken, parlama noktası 72 ° C, refraktif indeks (n20) ise 1.554'dür. Eter, kloroform, benzen ve karbon disülfide çözünür. Su veya etanol, benzoik asit üreten amonyak, benzamit, etil benzoat ve hidrojen klorür üretirken giderek ayrışabilir. Laboratuarda, susuz koşullar altında benzoik asit ve fosfor pentaklorid damıtımı ile elde edilebilir. Endüstriyel üretim süreci tiyonil klorür benzaldehid kullanılarak elde edilebilir. Benzoil klorür boyalar, parfümler, organik peroksitler, reçineler ve ilaçlar hazırlamak için önemli bir ara maddedir. Ayrıca kimyasal savaşta tahriş edici bir gaz olarak kullanılan fotoğraf ve suni tanen üretiminde de kullanılır.
Boya ara maddeleri, başlatıcı, UV emici, kauçuk katkı maddeleri, ilaç vb için kullanılır. Benzoil klorür herbisit metamitron ve insektisid propargit, benzen hidrazin veya ara gıdanın ara ürünüdür. Benzoil klorür, organik sentez, boya ve farmasötik hammadde, başlatıcı benzoil peroksit, t-butil peroksibenzoat, böcek öldürücüler ve herbisitler üretmek için kullanılır. Pestisitlerde, yeni bir insektisittir ve bu indüksiyon izoksazol paratiyonudur (izoaksiyon, Karphos) ara madde. Benzoil klorür, önemli bir benzoil ve benzil reaktifidir. Benzoil klorürün birçoğu benzoil peroksit üretiminde, ikinci olarak benzofenon, benzil benzoat, benzil selüloz üretimi için kullanılır. Benzoil peroksit, bir cam elyaf malzemesi, florür kauçuğu, silikon çapraz bağlama ajanları, yağ rafine edilmiş, ağartılmış un, elyaf renk giderme Bekleme monomer plastik, polyester, epoksi, akrilik reçine üretimi, kendini sertleştirici ajan için polimerizasyon başlatıcı katalize eder. Yurtiçi orijinal benzoil klorür üretim işletmelerinin 20'den fazla. Üreticilerin bazıları da asit klorür üretmek ve üretim kapasitesi 10.000t. Bununla birlikte, 2003 araştırmasına göre, kirleten rota kullanımının hükümet kısıtlamaları ile kontrol edildiği ve küçük bir kirleten üretim hattının kullanılması nedeniyle karın çok düşük olduğu ve bir başka hammadde fiyatının arttığı belirtildi. Bu nedenle imalatçıların çoğu üretimi durdurmaktadır. Asit klorid ile reaksiyona girmek ayrıca asit anhidrit üretebilir ve benzoik asit anhidrid, bir ağartma maddesi ve bir bileşen akışı olarak da kullanılabilen asilasyon ajanları için asıl amaçtır, ayrıca hazırlama için de kullanılabilir Benzoil peroksit üzerinden.
Analiz için reaktifler, aynı zamanda baharatlar, organik sentez için kullanılır.
UYGULAMALAR
Benzoil klorür;
Asil, bir hidroksil grubunun çıkarılmasıyla bir organik asitten oluşan bir radikaldir. Açil bileşiğinin genel formülü RCO- dir. Asil halid, halokarbonil grubu içeren, X'in halojen atomu (florin, klorin, bromin, iyot ve astatin) olduğu genel formül RCO · X'e sahip büyük bir organik madde grubundan biridir ve R, alifatik, alisiklik, aromatik ve H vb. içerir. Yedek kimyasal adlandırmada, adları, ana bileşiğe bir sonek olarak '-oil' eklenerek oluşturulur; etanoil klorür, CH3COCl, bir örnektir. Açil ve aroil halidleri terimleri sırasıyla alifatik veya aromatik türevlere karşılık gelir. Asil halidler, halojenleme ajanları kullanılarak karboksilik asitlerdeki -OH grubunun halojen ile değiştirilmesi ile hazırlanır. Su, alkoller ve aminler ile kolayca reaksiyona girer ve organik sentetik proseste yaygın olarak kullanılırlar ki, açil grubu asilasyon reaksiyonu olarak adlandırılan ekleme-eliminasyon sekansının yerini alarak hedef moleküllere dahil edilir. Asilasyon reaksiyonu, elektrofilik karbonil grubundaki (C = O) bir elektron vericisi (nükleofil) ile ikame edilmesini içerir. Asilasyon reaksiyonundaki yaygın nükleofiller alifatik ve aromatik alkoldür ve her ikisi de amitler üreten aminler (RNH2) ve esterleri oluştururlar. Karboksilik asit (X = OH) kendisi, bir alkolün doğrudan esterifikasyonu gibi kuvvetli bir asit katalizörü ile protonlandığında bir asile edici madde olarak işlev görebilir. RCOX genel formülü ile iki ortak asilasyon ajanı asit halidler (X = halojen atomu) ve anhidritlerdir (X = OCOR). Schotten-Baumann reaksiyonu, organik bileşiklerin hidroksil ve amino grubunu asile etmek için seyreltik alkali varlığında asit klorür kullanan bir asilasyon reaksiyonudur. Ayrıca başka asilasyon ajanları da vardır. Benzoil Klorid asil halojenürlere aittir. Asil halidler, bileşiklere bir asetil grubu (CH3CO-) katan asetilasyon prosesine katılırlar. Benzoil Klorid, ısıtma yoluyla veya nemli havaya veya suya maruz bırakıldığında şiddetli biçimde parçalanır. Güçlü oksidanlar, metaller (özellikle demir), alkali ve toprak alkali metalleri, bazlar ve aminler, dimetil sülfoksit ve alkoller gibi çeşitli organik maddeler ile şiddetle tepki verir. Tepkiler yangın ve patlama tehlikesine neden olur. Benzenkarbonil gruplarını bileşiklere dahil etmek için kullanılır. Benzoil klorür tarafından geçirilen tipik reaksiyonlar, Schotten-Baumman reaksiyonu (bir hidrojen içeren bileşiklerin benzoilasyonu) ve Friedel-Crafts reaksiyonları (ikame edilmiş benzofenonların hazırlanması) 'dur. Bir benzoil peroksit ve t-bütil perbenzoat gibi peroksitlerin imalatında kullanılır. Aynı zamanda böcek ilacı, ilaç, parfüm fiksatif, polimerizasyon katalizörü, benzolasyon maddeleri ve boyarmaddeleri imalinde kullanılan benzofenon ve türevlerinin sentezinde de kullanılır.
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3133 | https://courses.lumenlearning.com/mathforliberalartscorequisite/chapter/simplifying-and-evaluating-expressions-with-integers/ | Simplifying and Evaluating Expressions With Integers That Use Addition | Mathematics for the Liberal Arts Corequisite
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Mathematics for the Liberal Arts Corequisite
Module 4: Graph Theory
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Simplifying and Evaluating Expressions With Integers That Use Addition
Learning Outcomes
Add and subtract integers
Simplify variable expressions for a given value
Evaluate variable expressions with integers
Now that you have modeled adding small positive and negative integers, you can visualize the model in your mind to simplify expressions with any integers.
For example, if you want to add 37+(−53)37+(−53), you don’t have to count out 37 37 blue counters and 53 53 red counters.
Picture 37 37 blue counters with 53 53 red counters lined up underneath. Since there would be more negative counters than positive counters, the sum would be negative. Because 53−37=16 53−37=16, there are 16 16 more negative counters.
37+(−53)=−16 37+(−53)=−16
Let’s try another one. We’ll add −74+(−27)−74+(−27). Imagine 74 74 red counters and 27 27 more red counters, so we have 101 101 red counters all together. This means the sum is -101.-101.
−74+(−27)=−101−74+(−27)=−101
Look again at the results of −74−(27)−74−(27).
Addition of Positive and Negative Integers5+3 5+3−5+(−3)−5+(−3)
both positive, sum positive both negative, sum negative
When the signs are the same, the counters would be all the same color, so add them.
−5+3−5+35+(−3)5+(−3)
different signs, more negatives different signs, more positives
Sum negative sum positive
When the signs are different, some counters would make neutral pairs; subtract to see how many are left.
Exercises
Simplify:
19+(−47)19+(−47)
−32+40−32+40
Solution:
Since the signs are different, we subtract 19 19 from 47 47. The answer will be negative because there are more negatives than positives.
19+(−47)−28 19+(−47)−28
The signs are different so we subtract 32 32 from 40 40. The answer will be positive because there are more positives than negatives
−32+40 8−32+40 8
Tip: Think of positive numbers as money you have and negative numbers as money you owe. This will help you determine if your answer is positive or negative. (-4) + 7 would be owing $4 and having $7, once you settle up, you still have $3. So the answer would be positive 3.
Another example is (-3) + (-5). This means you owe $3 and you owe $5, so you owe $8, which would be represented by -8.
If you have 6 + (-10) and we think in terms of money, you have $6 but you owe $10. Once you settle up, you still owe $4. This gives you an answer of -4.
try it
example
Simplify: −14+(−36)−14+(−36)
Show Solution
Solution:
Since the signs are the same, we add. The answer will be negative because there are only negatives.
−14+(−36)−50−14+(−36)−50
try it
The techniques we have used up to now extend to more complicated expressions. Remember to follow the order of operations.
example
Simplify: −5+3(−2+7)−5+3(−2+7)
Show Solution
Solution:
−5+3(−2+7)−5+3(−2+7)
Simplify inside the parentheses.−5+3(5)−5+3(5)
Multiply.−5+15−5+15
Add left to right.10 10
try it
Watch the following video to see another example of how to simplify an expression that contains integer addition and multiplication.
Evaluate Variable Expressions with Integers
Remember that to evaluate an expression means to substitute a number for the variable in the expression. Now we can use negative numbers as well as positive numbers when evaluating expressions. In our first example we will evaluate a simple variable expression for a negative value.
example
Evaluate x+7 when x+7 when
x=−2 x=−2
x=−11 x=−11
Show Solution
Solution:
Evaluate x+7 x+7 when x=−2 x=−2
Substitute −2−2 for x x.−2+7−2+7
Simplify.5 5
Evaluate x+7 x+7 when x=−11 x=−11
x+7 x+7
Substitute −11−11 for x x.−11+7−11+7
Simplify.−4−4
Now you can try a similar problem.
try it
In the next example, we are give two expressions,n+1 n+1, and −n+1−n+1. We will evaluate both for a negative number. This practice will help you learn how to keep track of multiple negative signs in one expression.
example
When n=−5 n=−5, evaluate
n+1 n+1
−n+1−n+1
Show Solution
Solution:
Evaluate n+1 n+1 when n=−5 n=−5
n+1 n+1
Substitute −5−5 for n.−5+1−5+1
Simplify.−4−4
Evaluate −n+1−n+1 when n=−5 n=−5
−n+1−n+1
Substitute −5−5 for n.−(−5)+1−(−5)+1
Simplify.5+1 5+1
Add.6 6
Now you can try a similar problem.
try it
Next we’ll evaluate an expression with two variables, where one of the variables is assigned a negative value.
example
Evaluate 3 a+b 3 a+b when a=12 a=12 and b=−30 b=−30.
Show Solution
Solution:
3 a+b 3 a+b
Substitute 12 12 for a and −30−30 for b.3(12)+(−30)3(12)+(−30)
Multiply.36+(−30)36+(−30)
Add.6 6
Now you can try a a similar problem.
try it
In the next example, the expression has an exponent as well as parentheses. It is important to remember the order of operations, you will need to simplify inside the parentheses first, then apply the exponent to the result.
example
Evaluate (x+y)2(x+y)2 when x=−18 x=−18 and y=24 y=24.
Show Solution
Solution:
This expression has two variables. Substitute −18−18 for x x and 24 24 for y y.
(x+y)2(x+y)2
Substitute −18−18 for x x and 24 24 for y y.(−18+24)2(−18+24)2
Add inside the parentheses.(6)2(6)2
Simplify 36 36
Now you can try a similar problem.
try it
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CC licensed content, Original
Revision and Adaptation. Provided by: Lumen Learning. License: CC BY: Attribution
CC licensed content, Shared previously
Ex: Simplify an Expression With Integers Using the Order of Operations. Authored by: James Sousa (Mathispower4u.com). Located at: License: CC BY: Attribution
Question ID: 145013, 145014, 145018, 145022, 145023, 145024, 145025. Authored by: Alyson Day. License: CC BY: Attribution. License Terms: IMathAS Community License CC-BY + GPL
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3134 | https://www.youtube.com/watch?v=tNIpbHBKQG8 | Kinetic Friction Skidding to a Stop
Timothy Palladino
3980 subscribers
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Posted: 18 May 2011
From www.PhysicsAccordingtoPalladino.org
Calculating the distance required to stop a car while skidding.
22 comments
Transcript:
One of the things that you're forced to memorize in driver's education courses is that it's better to pump your brakes instead of locking up your brakes if you need to come to a stop relatively quickly. So, in this session, we're going to figure out why it is better to pump your brakes than to jam on your brakes or lock them up. So, if you imagine that you have some car and this will be the car and you are initially traveling in this direction and let's call this the positive x direction. And so let's write a coordinate system here. So this will be our coordinate system where you have motion in the positive x direction and in the upward direction is the positive y direction. So if this is our velocity vector then the car needs to come to a rest or slow down and the acceleration will be in the opposite direction of the velocity. So notice that the velocity vector is in the opposite direction of the acceleration vector and so the object's going to slow down. Now, in order for this object to slow down, there needs to be a net force in the direction opposite to the direction that this object is traveling. So, notice another fundamental concept that the net force is in the same direction as the acceleration. So, let's explore how to solve this problem. Now, the first thing that we need to do is we need to draw a free body diagram representing the forces acting on this object. So, in this case, here's our object which we'll represent with a dot. Now, this object has mass. Therefore, it has weight. The other force acting on this car is going to be the normal force. The ground has to push the car in the upward direction. Now, what I want to do is draw something that's technically not part of the free body diagram, but is very important to understanding this problem, and that's the velocity vector. This car is initially moving in this direction. So, now we have to figure out the direction in which the net force is acting. And we sort of already did. We said the net force was acting in the opposite direction that the car was moving and in this case this is going to be the frictional force or the kinetic frictional force. This is going to be the force responsible for opposing the forward motion of this car. So this is our free body diagram representing all the forces acting on this object. So what we need to do now is apply Newton's second law. Now, Newton's second law says that if you add up all the forces, and in this case, I'm going to write it in the vector form because we need to look at the forces acting in two directions, and that's going to equal the mass of the object times the acceleration of the object. Now, the forces are acting on this in two different directions. So, we can write this as two separate equations, which say that if you add up the forces in the y direction, it's going to equal the mass of the object times the acceleration of the object in the y direction. And if you add up the forces in the x direction, it's going to equal the mass of the object times the acceleration of the object in the x direction. So let's first look at the forces acting on this car in the y direction. So in this case, we have two forces acting in the y direction. Now in the y direction, which I'll write it again like this as the sum of the forces in the y direction is going to equal the mass of the object times the acceleration of the object in the y direction. Now this car is sitting on the ground. It's not being accelerated in the upward direction. And so the acceleration of this object is going to be zero. It's not moving in the upward direction. The velocity is not changing in that direction. So when you add up all the forces on this object, they're going to equal zero. And in this case, we already said that there's two forces acting on this object. There's the normal force and there's the weight force. And those two forces add up to be zero. Now to find the magnitude of the normal force, which is what we need in order to determine the frictional force, you need to take the weight force and move it to the other side. So to do that, you add the weight to both sides. And when you do that, what you should see is that the normal force equals the weight force in this case. Now, there are instances in which the normal force does not equal the weight force, but in this case, we're assuming we're driving along a completely horizontal surface or flat surface. So this is the first relationship that we need to know. Now the second relationship that we need need to know is the sum of the forces in the x direction. So when you add up the forces in the x direction, it's going to equal the mass of the object times the acceleration of the object in the x direction. Now if you go back to the free body diagram, there's only one force acting in the x direction. And if you go back to our coordinate system, this was thex direction. And so this frictional force is acting in the negativex direction. It's in the direction opposite to the direction of motion. And so when you add this using Newton's second law, what you should see is that you get minus F subscript K, the force due to kinetic friction is going to equal the mass of the object time the acceleration of the object in the X direction. You know this object is accelerating in the X direction because it has to slow down otherwise it would continue its state of motion. Now you know a relationship for the kinetic frictional force and the kinetic frictional force I'll write it off to the side here. You know that the force due to kinetic friction is going to equal the coefficient of kinetic friction times the value of the normal force. And in this case we already figured out a relationship to the normal force. We said the normal force is equal to the weight of the car. And you should also know that weight equals mass time the gravitational acceleration. So what you can now do is rewrite this as the force due to kinetic friction equals mu k. And what you're going to do is you're going to substitute in place of this normal force which you also see here the weight force which in this case equals the mass of the object time the gravitational acceleration. So this is just going to equal mass time the gravitational acceleration. And this is one special case. If you were not on a horizontal surface, the normal force would not equal this. So now what we're going to do is we're going to take this value right here for the frictional force in this case the kinetic frictional force and we're going to plug it in to this term right here. And so when you do that what you get is negative and I'm going to do this in parenthesis so that you see the substitution that I'm making. mu k the coefficient of kinetic friction time the mass of the object time the gravitational acceleration of the object is going to equal the mass of the object times the acceleration of the object in the x direction. Now notice a few things. Notice on this side you have a mass term. On this side you also have a mass term. So what you can do now is cancel those two terms out. And what you should see is that you get minus mu k the coefficient of kinetic friction times the gravitational acceleration is going to equal the acceleration in the x direction. Now the really cool thing about this is that the rate at which you slow down does not depend on the mass of the car. It only only depends on the gravitational acceleration and the kinetic friction between the cars tires and the road. And we've assumed that we haven't modified the car at all because if we did some modifications, we could change the value of the normal force. All right. So now that we've found the acceleration of the car, that is the rate at which this car is going to slow down, we can figure out how far this car will travel while it's slowing down. So if you remember back from the initial part of the problem, the car's initial velocity was 40 mph. And this is a quick conversion that you should be able to do now to convert miles hour to meters/s. And what you should see is when you do that, and I'll do it out really quickly, is 1 mile is 1.61 km and 1 kilometer is 1,000 m. And you know, 1 hour has 3,600 seconds. And when you do that all out, what you should get is a initial velocity of about 17.9 m/s. That's going to be your initial velocity. Notice that this is also a kinematic problem besides just a force problem. Now, this car is going to come to a stop. So, its final velocity should be 0 m/s. Now, one of the things that Newton's second law did was allow us to figure out the acceleration of this object. And in this case, we figured out it was negative mu k the coefficient of kinetic friction times the gravitational acceleration of this object. Now in this case we said the coefficient of kinetic friction was 0.85. And remember that's a unitless number. And the acceleration due to gravity on earth is 9.8 m/s squared. And when you do this multiplication out, you should get 8.33. And notice I forgot a negative sign. So let's make sure we explicitly put that back in. m/s squared. And this negative sign is telling us that this object's going to be slowing down because the acceleration is in the opposite direction of the velocity. And so this object's going to be slowing down at a rate of 8.33 m/s every single second. Now, the thing that we're looking for is how far this car will travel during during the period of time in which it's slowing down. So that's our question mark. That's what we're looking for. So now you can go to your kinematic equation that says the final velocity squar equals the initial velocity squared plus 2 the acceleration times the distance it travels while it's slowing down given this initial velocity and final velocity. Now you know the you know the final velocity is zero because this car comes to a stop and so 0 equ= v initial^ 2 + 2 the acceleration time the distance it travels. And so what you need to do next is you need to take this term and move it over to the other side. To do that, you need to subtract it from both sides. So minus V initial^ 2 minus V initial squared from both sides. What you do to one side, you have to do to the other side of an equation. And what you should see is that this works out to be v initial^ 2 = 2 the acceleration time the distance this object travels. And so since we're looking for delta x, what you need to do to both sides of the equation now is divide by 2 the acceleration. And when you do that, you should notice that this term cancels out with that term. And you get delta x or the distance this object travels is going to be v initial squar / 2 the acceleration. Now the initial velocity, so minus the initial velocity, which was 17.9 m/s square the entire term. And now you're going to divide it by 2 the acceleration, which worked out to be 8.33 m/s squared. And when you do that out, you get 19.2 m. So this car travels a distance of 192 m while it's slowing down and skidding to a stop. The other thing to notice is that it's a positive 19.2. It's not a negative 19.2. If you go back to our coordinate system in the very beginning, so let's go back to our initial coordinate system which is right here. We said that we were traveling in the positive x direction. So of course the distance that we traveled during this period of time where the car is coming to a stop is going to be a positive number. All right. In the next video, we'll take a look at what happens when the car pumps its brakes as opposed to lock up its brakes. |
3135 | https://www.youtube.com/watch?v=gP4XW-8wU9s | Translations on the Coordinate Plane
MooMooMath and Science
583000 subscribers
23 likes
Description
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Posted: 9 Aug 2022
Geometry translations explained.. Learn how to complete translation on a coordinate plane.
A transformation is an operation that maps a preimage or the original figure onto a new figure called the image. This is the figure with prime marks.
A translation is when you slide a figure from one position to another without turning the preimage.
Math Standard
MGSE8.G.1 Recognize/describe a reflection,translation, and rotation.
English
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Transcript:
[Music] let's take a look at translations a transformation is an operation that Maps a pre-image or the original figure onto a new figure called the image this is the figure with the prime marks or an apostrophe a translation is when you slide a figure from one position to another without turning the pre-image so if we look here this one is our pre-image because there are no Prime marks or apostrophes and here is our new image here are some reminders that you can look at it moves horizontally left if it's negative and right if it's positive or if you add and then y up and down so vertically if it's positive or you add and then negative or subtract it goes down so let's look the pre-image is translated or slid to the left five units so we go 1 2 3 4 5 that shows five units and it went up three units 1 2 3 to get to that new point so the notation XY becomes x - 5 and y + 3 to get to our new image with our new points so let's look at an example it says graph triangle jkl with vertices -34 1 3 and -41 and then we're going to graph the translated image using this information it'll tell us how to do our translation so first we're going to graph each of these points -3 4 be sure to label them 1 3 and -4 1 so here's our pre-image and we want to use this information to figure out how to translate so x + 2 means I'm going to go to the right two and Y - 5 means I go down five so for our pre-image points we have -3 4 1 3 and4 1 in our new image we're going to get our new points so we take our x value to the right or add two so our x value here is going to be 1 here it will be three and here it will be -2 for this one we subtract so we subtract or we move down five so 4 - 5 is -1 3 - 5 is -2 and 1 - 5 is -4 and then we plot these new points with with our Prime marks or our apostrophe so -11 [Music] |
3136 | https://fahrenheittocelsius.org/373-15-kelvin-to-celsius | 373.15 Kelvin to Celsius ▷ What is 373.15 K in °C?
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F to C ConversionsWelcome to F to C, our Fahrenheit to Celsius category with posts covering temperature conversions from °F to °C. Each article explains a specific temperature conversion ranging from absolute zero to 1000 degrees Fahrenheit, and even beyond that. Every post comes with the formula for a changing a specific temperature in °F to °C, in addition we show you how to conduct the math. What’s more, each article also has a converter and a feedback form. We then change your °F to the units Delisle, Rankine, Newton, Kelvin, Réaumur and Rømer. As there are quite a number of pages with posts explaining the conversion of x degrees Fahrenheit to Celsius, the recommended way to locate a certain F to C transformation is by means of the search form located at the bottom or in the sidebar. The results page contains all posts pertinent to your °F to °C query.
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C to F ConversionsWelcome to C to F, our Celsius to Fahrenheit category covering temperature conversions from °C to °F. Our posts contain a specific temperature conversion from absolute zero to 1000 degrees Celsius, and beyond, to degrees in Fahrenheit. Every article comes with the formula for a changing a particular temperature in °C to °F, in addition to details on how to conduct the math. Moreover, each posts has a converter as well as a comment form to ask questions or to say something. We also change your °C to the units Newton, Kelvin, Réaumur, Rømer, Delisle and Rankine. Here are quite a number of pages with posts explaining the conversion of x degrees Celsius to Fahrenheit. Thus, the best way to locate a certain C to F transformation is using the search form in the sidebar or at the bottom. In the result page, you can find all articles deemed relevant to your °C to °F query.
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F to C ConversionsWelcome to F to C, our Fahrenheit to Celsius category with posts covering temperature conversions from °F to °C. Each article explains a specific temperature conversion ranging from absolute zero to 1000 degrees Fahrenheit, and even beyond that. Every post comes with the formula for a changing a specific temperature in °F to °C, in addition we show you how to conduct the math. What’s more, each article also has a converter and a feedback form. We then change your °F to the units Delisle, Rankine, Newton, Kelvin, Réaumur and Rømer. As there are quite a number of pages with posts explaining the conversion of x degrees Fahrenheit to Celsius, the recommended way to locate a certain F to C transformation is by means of the search form located at the bottom or in the sidebar. The results page contains all posts pertinent to your °F to °C query.
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Home » K to C » 373.15 Kelvin to Celsius
373.15 Kelvin to Celsius
by Mark
Table of Contents
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Temperature Converter
373.15 Kelvin to Celsius Formula
Convert 373.15 Kelvin to Celsius
373.15 K to °C
373.15 Kelvin to Degrees Celsius
Conclusion
Quick Conversion Table
Welcome to 373.15 Kelvin to Celsius.
This article is about changing the measurement for temperature from 373.15 K to °C.
Here you can find everything about converting 373.15 Kelvin to degrees Celsius, including the formula and useful information regarding the temperature units.
For the temperature conversion of 373.15 Kelvin to centigrades you can also use our calculator: 🙂
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If you have been looking for 373.15 Kelvin in Celsius, then you are right here, too. 🙂
Read on to learn all about 373.15 Kelvin to °C.
Kelvin has no degrees, so we either write 373.15 Kelvin or use the symbol K to denote the temperature: 373.15 K.
In contrast, 373.15 degrees Celsius is synonym for 373.15 Celsius and for 373.15 centigrades; using the unit symbol we write: 373.15 °C.
373.15 Kelvin to Celsius Formula
The 373.15 Kelvin to Celsius formula is: [°C] = [373.15] – 273.15. Thus, we get:
373.15 K to °C = 100 Celsius
Note that the the unit of measurement Kelvin has no degrees, so we either write 373.15 Kelvin or use the symbol K to denote the temperature: 373.15 K.
In contrast, 373.15 degrees Celsius is synonym for 373.15 Celsius and for 373.15 centigrades; using the unit symbol we write: 373.15 °C.
Here you can change 100 Celsius to Kelvin.
Convert 373.15 Kelvin to Celsius
To convert 373.15 Kelvin to Celsius simply subtract 273.15 from the temperature in K: 100 °C = 373.15 – 273.15.
If you like you can also use our converter above.
Enter the temperature in Kelvin, e.g. 373.15, in the first field using a decimal point if applicable.
Moving the bottom input spinner you may also calculate 373.15 degrees Celsius to Kelvin.
Besides 373.15 K to C, other temperature conversions on our site include:
378.15 kelvin to celsius
379.15 kelvin to celsius
380.15 kelvin to celsius
373.15 K to °C
How much is 373.15 Kelvin in Celsius?
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There, in the search form, you can find many temperature conversions including, but not limited to, 373.15 K to C.
For example, enter 373.15 K in °C.
373.15 Kelvin to Degrees Celsius
As stated before, 373.15 Kelvin to degrees Celsius is the same as 373.15 Kelvin to centigrades.
Remember that the equivalence of 373.15 K to C is exactly 273.15 less than in degrees Celsius.
Also, keep in mind that 373.15 degrees kelvin is a misconception. It’s 373.15 Kelvin without degrees.
If you have been searching for 373.15 K in Celsius, or 373.15 K to Celsius, then you have all the answers, too.
The same applies if you typed convert 373.15 K to C in your preferred search engine.
For further information about the scales and units of measurement related to 373.15 Kelvin to degrees C read our article Kelvin to Celsius which you can find in the header menu.
Conclusion
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Quick Conversion Table
| Kelvin | Celsius |
--- |
| 372.65 K | 99.5 °C |
| 372.66 K | 99.51 °C |
| 372.67 K | 99.52 °C |
| 372.68 K | 99.53 °C |
| 372.69 K | 99.54 °C |
| 372.7 K | 99.55 °C |
| 372.71 K | 99.56 °C |
| 372.72 K | 99.57 °C |
| 372.73 K | 99.58 °C |
| 372.74 K | 99.59 °C |
| 372.75 K | 99.6 °C |
| 372.76 K | 99.61 °C |
| 372.77 K | 99.62 °C |
| 372.78 K | 99.63 °C |
| 372.79 K | 99.64 °C |
| 372.8 K | 99.65 °C |
| 372.81 K | 99.66 °C |
| 372.82 K | 99.67 °C |
| 372.83 K | 99.68 °C |
| 372.84 K | 99.69 °C |
| 372.85 K | 99.7 °C |
| 372.86 K | 99.71 °C |
| 372.87 K | 99.72 °C |
| 372.88 K | 99.73 °C |
| 372.89 K | 99.74 °C |
| 372.9 K | 99.75 °C |
| 372.91 K | 99.76 °C |
| 372.92 K | 99.77 °C |
| 372.93 K | 99.78 °C |
| 372.94 K | 99.79 °C |
| 372.95 K | 99.8 °C |
| 372.96 K | 99.81 °C |
| 372.97 K | 99.82 °C |
| 372.98 K | 99.83 °C |
| 372.99 K | 99.84 °C |
| 373 K | 99.85 °C |
| 373.01 K | 99.86 °C |
| 373.02 K | 99.87 °C |
| 373.03 K | 99.88 °C |
| 373.04 K | 99.89 °C |
| 373.05 K | 99.9 °C |
| 373.06 K | 99.91 °C |
| 373.07 K | 99.92 °C |
| 373.08 K | 99.93 °C |
| 373.09 K | 99.94 °C |
| 373.1 K | 99.95 °C |
| 373.11 K | 99.96 °C |
| 373.12 K | 99.97 °C |
| 373.13 K | 99.98 °C |
| 373.14 K | 99.99 °C |
| 373.15 K | 100 °C |
| 373.16 K | 100.01 °C |
| 373.17 K | 100.02 °C |
| 373.18 K | 100.03 °C |
| 373.19 K | 100.04 °C |
| 373.2 K | 100.05 °C |
| 373.21 K | 100.06 °C |
| 373.22 K | 100.07 °C |
| 373.23 K | 100.08 °C |
| 373.24 K | 100.09 °C |
| 373.25 K | 100.1 °C |
| 373.26 K | 100.11 °C |
| 373.27 K | 100.12 °C |
| 373.28 K | 100.13 °C |
| 373.29 K | 100.14 °C |
| 373.3 K | 100.15 °C |
| 373.31 K | 100.16 °C |
| 373.32 K | 100.17 °C |
| 373.33 K | 100.18 °C |
| 373.34 K | 100.19 °C |
| 373.35 K | 100.2 °C |
| 373.36 K | 100.21 °C |
| 373.37 K | 100.22 °C |
| 373.38 K | 100.23 °C |
| 373.39 K | 100.24 °C |
| 373.4 K | 100.25 °C |
| 373.41 K | 100.26 °C |
| 373.42 K | 100.27 °C |
| 373.43 K | 100.28 °C |
| 373.44 K | 100.29 °C |
| 373.45 K | 100.3 °C |
| 373.46 K | 100.31 °C |
| 373.47 K | 100.32 °C |
| 373.48 K | 100.33 °C |
| 373.49 K | 100.34 °C |
| 373.5 K | 100.35 °C |
| 373.51 K | 100.36 °C |
| 373.52 K | 100.37 °C |
| 373.53 K | 100.38 °C |
| 373.54 K | 100.39 °C |
| 373.55 K | 100.4 °C |
| 373.56 K | 100.41 °C |
| 373.57 K | 100.42 °C |
| 373.58 K | 100.43 °C |
| 373.59 K | 100.44 °C |
| 373.6 K | 100.45 °C |
| 373.61 K | 100.46 °C |
| 373.62 K | 100.47 °C |
| 373.63 K | 100.48 °C |
| 373.64 K | 100.49 °C |
| 373.65 K | 100.5 °C |
– Article written by Mark, last updated on November 29th, 2023
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3137 | https://www.symbolab.com/popular-trigonometry/trigonometry-17100 | sin(x-pi/4)=0
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Solution
sin(x−4 π)=0
Solution
x=2 πn+4 π,x=π+2 πn+4 π
+1
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x=4 5∘+36 0∘n,x=22 5∘+36 0∘n
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sin(x−4 π)=0
General solutions for sin(x−4 π)=0
sin(x) periodicity table with 2 πn cycle:
x 0 6 π4 π3 π2 π3 2 π4 3 π6 5 πsin(x)0 2 12 22 31 2 32 22 1x π 6 7 π4 5 π3 4 π2 3 π3 5 π4 7 π6 11 πsin(x)0−2 1−2 2−2 3−1−2 3−2 2−2 1
x−4 π=0+2 πn,x−4 π=π+2 πn
x−4 π=0+2 πn,x−4 π=π+2 πn
Solve x−4 π=0+2 πn:x=2 πn+4 π
x−4 π=0+2 πn
0+2 πn=2 πn x−4 π=2 πn
Move 4 πto the right side
x−4 π=2 πn
Add 4 π to both sides x−4 π+4 π=2 πn+4 π
Simplify x=2 πn+4 π
x=2 πn+4 π
Solve x−4 π=π+2 πn:x=π+2 πn+4 π
x−4 π=π+2 πn
Move 4 πto the right side
x−4 π=π+2 πn
Add 4 π to both sides x−4 π+4 π=π+2 πn+4 π
Simplify x=π+2 πn+4 π
x=π+2 πn+4 π
x=2 πn+4 π,x=π+2 πn+4 π
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3138 | https://pmc.ncbi.nlm.nih.gov/articles/PMC3104041/ | Structural and dynamic properties of linker histone H1 binding to DNA - PMC
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Biomicrofluidics
. 2011 May 4;5(2):024104. doi: 10.1063/1.3587096
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Structural and dynamic properties of linker histone H1 binding to DNA
Rolf Dootz
Rolf Dootz
1 Max Planck Institute for Dynamics and Self-Organization, Bunsenstraße 10, 37073 Göttingen, Germany
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1, Adriana C Toma
Adriana C Toma
2 Department of Chemistry, University of Basel, Klingelbergstrasse 80, 4056 Basel, Switzerland
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2, Thomas Pfohl
Thomas Pfohl
1 Max Planck Institute for Dynamics and Self-Organization, Bunsenstraße 10, 37073 Göttingen, Germany
2 Department of Chemistry, University of Basel, Klingelbergstrasse 80, 4056 Basel, Switzerland
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1,2,a)
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1 Max Planck Institute for Dynamics and Self-Organization, Bunsenstraße 10, 37073 Göttingen, Germany
2 Department of Chemistry, University of Basel, Klingelbergstrasse 80, 4056 Basel, Switzerland
a)
Author to whom correspondence should be addressed. Electronic mail: thomas.pfohl@unibas.ch.
Received 2011 Jan 7; Accepted 2011 Apr 15; Collection date 2011 Jun.
Copyright © 2011 American Institute of Physics
PMC Copyright notice
PMCID: PMC3104041 PMID: 21629560
Abstract
Found in all eukaryotic cells, linker histones H1 are known to bind to and rearrange nucleosomal linker DNA. In vitro, the fundamental nature of H1∕DNA interactions has attracted wide interest among research communities—from biologists to physicists. Hence, H1∕DNA binding processes and structural and dynamical information about these self-assemblies are of broad importance. Targeting a quantitative understanding of H1 induced DNA compaction mechanisms, our strategy is based on using small-angle x-ray microdiffraction in combination with microfluidics. The usage of microfluidic hydrodynamic focusing devices facilitates a microscale control of these self-assembly processes, which cannot be achieved using conventional bulk setups. In addition, the method enables time-resolved access to structure formation in situ, in particular, to transient intermediate states. The observed time dependent structure evolution shows that the H1∕DNA interaction can be described as a two-step process: an initial unspecific binding of H1 to DNA is followed by a rearrangement of molecules within the formed assemblies. The second step is most likely induced by interactions between the DNA and the H1’s charged side chains. This leads to an increase in lattice spacing within the DNA∕protein assembly and induces a decrease in the correlation length of the mesophases, probably due to a local bending of the DNA.
INTRODUCTION
The linker histone family is a heterogeneous family of highly tissue-specific, basic proteins, which exhibit significant variations in sequence.1, 2 However, most eukaryotic linker histones H1 share a similar, tripartite structure consisting of a globular domain flanked by two lysine-rich tails, a shorter amino-terminal domain (N-tail) and a longer carboxyl-terminal one (C-tail).3 Linker histones H1 are known to attach close to the entry and exit sites of linker DNA on the nucleosome core, bringing together two linker DNA segments.4, 5, 6 The globular domain with a diameter of 2.9 nm is the only domain that is folded in solution exhibiting three α-helices.7 Its most noticeable structural features are the two DNA binding sites situated on opposite sides of the molecule,8 which, with the help of the C-terminal domain, render H1-chromatin binding highly dynamic.9
Despite positioning along the nucleosome core particle, it is not the globular domain but rather the highly positively charged C-tail that imparts to linker histones their unique ability to bind to linker DNA through nonspecific electrostatic interactions.6, 10, 11In vivo, the absence of C-tails leads to greatly reduced chromatin binding.12 Binding of linker histones to the linker DNA facilitates the shift of the chromatin structure toward more condensed, higher order forms.13 Although a chromatin fiber lacking linker histones is able to fold to a certain extent,14 there is abundant evidence that the highly ordered chromatin compaction of the 30 nm fiber is only attained in the presence of linker histones.5, 13, 15In vivo studies on H1 depleted chicken cells have shown an altered chromatin structured as well as chromosomal aberrations and increased DNA damage during replication, suggesting that H1 was involved in transcription regulation.16 Linker histones help select a specific folding state from among the set of compact states reached in its absence by contributing to the free energy of chromatin folding.17 This suggests that linker histones are of central importance in genome organization and regulation.
Since the linker histone’s position on the nucleosome is still a matter of debate, understanding more closely its interaction with DNA upon condensing is therefore essential for understanding the role H1 plays in gene regulation and chromatin structure. It has been argued that H1∕DNA assemblies are an excellent model system for studying aspects of the interaction of H1 with chromatin. H1∕DNA interaction in vitro10, 18, 19, 20 was reported to be dependent on the ionic strength,21, 22, 23, 24 suggesting an interaction governed by electrostatics. At high salt concentrations, the screening of interactions was so significant that assembling was no longer reported,25 denoting a rather weak binding in comparison to protamine-DNA assemblies that can still be formed even at monovalent salt concentrations as high as 1.3 M.26 Thus, reinforcing the belief that H1 is a highly mobile chromatin component compared to the germinal chromatin, which is made mechanically stable and transcriptionally inactive by strong protamine binding. In a more general context, the H1∕DNA system can be regarded as a model for nonspecific DNA-protein interactions. Herein, using the innovative junction between x-ray scattering and microfluidics, we probed linker histone∕DNA interaction dynamics and structure formation in real-time. A microfluidic channel with a cross geometry provides means to control mixing of H1 and DNA uncovering a two-step assembly process. The nonspecific binding of linker histones to DNA leads to the formation of primary assemblies with columnar structures governed by electrostatic interactions. As the binding reaction evolves and the concentration of H1 histones increases, the columnar structure rearranges to less organized assemblies with smaller correlation lengths. We attribute the latter observation to an additional bending of DNA molecules induced by the interaction of the linker histones tails with DNA.
EXPERIMENTAL SECTION
Materials
Calf thymus linker histone H1 (isolated lysine-rich fraction27) and lyophilized highly polymerized calf-thymus DNA were purchased from Sigma-Aldrich GmbH, Taufkirchen, Germany. Both components were solubilized in 18.2 MΩ cm water (Millipore GmbH, Schwalbach, Germany) to final concentrations of c H1=10 mg ml−1 and c DNA=2.5 mg ml−1, respectively. At physiological p H conditions, H1 possesses a molecular weight of M w=21.5 kDa and 55 positive charges,22 whereas DNA molecules carry two negative charges per base pair.
Microfluidic devices
X-ray compatible Kapton-Steel-Kapton microfluidic devices have been fabricated as described elsewhere.28 Briefly, the microchannel structure is spark eroded in a thin stainless steel plate, resulting in a microchannel structure, which is open on both sides. The thickness of the plate defines the height of the microchannels. Adhesive Kapton foils coated with a poly(siloxane) layer (thickness of 53 μm, Dr. Müller GmbH, Allingen, Germany) are placed on both sides of the steel plate, such that the foils seal the device and serve as x-ray transparent windows to the microchannel. Four holes are punched into the bottom Kapton foil, fitting the channel ends of the steel plate and serving as inlets of the microfluidic device. Using a thin double sided sticking tape with cavities at the inlet positions and the measuring area, the microfluidic device is mounted on a poly(methylmethacrylate) (PMMA) slab assisting the connection to the fluid pumping system. The center region of the PMMA support is milled out to provide an undisturbed pathway for the x-ray beam. The connection to the pumping system is established by Teflon tubing implemented into male nuts (ProLiquid, Überlingen, Germany) that are screwed in the sockets of the PMMA support. The channels we used had a width of 100–150 μm and a depth ranging from 200 to 300 μm.
Microfocused small-angle x-ray measurements
Small-angle x-ray scattering experiments were performed at beamline ID10b at the European Synchrotron Radiation Facility in Grenoble, France. Two-dimensional (2D) scattering patterns were collected using a charge-coupled device (CCD) detector with a fluorescent screen. Beryllium compound refractive lenses focused the x-ray beam of 8 keV (λ=0.155 nm) down to a diameter of 20 μm. The microchannel device was loaded on an x-y stage to probe specific positions using the microfocused x-ray beam. The positional accuracy of absolute coordinates in the microdevice was on the order of the beam size. All CCD images were taken at ambient temperature with exposure times of 30 s∕position and azimuthally averaged to produce one-dimensional intensity profiles I(q). Using a Lorentzian fit, the peak positions and Δ q in I(q) are determined. Bulk experiments were performed in quartz capillaries, and a very broad peak has been detected for all measured N∕P ratios (Fig. 1). Local concentrations are translated into assembly compositions given in terms of the relative charge ratio N∕P. N is the total number of positive amine charges of H1 and P is the total number of negatively charged DNA phosphate groups.
Figure 1.
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Small angle x-ray scattering profiles of H1/DNA bulk assemblies at different N∕P charge ratios.
Finite element modeling
FEMLAB software (Comsol, Inc., Burlington, MA) was used to perform finite element modeling simulations of conditions within the microchannel device. The incompressible Navier–Stokes equation was solved in two dimensions using about 20 000 elements to obtain a solution for the diffusive mixing in the microflow. The velocity fields (and the corresponding strain rates per position) and concentration profiles were subsequently calculated. The viscosity of the formed H1∕DNA assemblies’ η complex and the diffusion constant D H1 were used to match the experimentally recorded shape of the hydrodynamically focused DNA stream and of the formed H1∕DNA aggregates. All other parameters, such as channel geometry, flow rates, and the viscosity of the DNA solution, are known. For each flow velocity, two finite element simulations are performed. In order to accurately determine relevant fit parameters, physical conditions in the microchannel device are first simulated with high precision [i.e., high number of finite elements) for a close-up region around the confluence area [x=−200–500 μm, Fig. 3a]. Independent of the flow velocity and throughout all simulations, the viscosity of H1∕DNA aggregates was fitted to η complex≈3×10 3⋅η water=2.7 Pa s. This result is on the same order of magnitude as the results known from other polymer hydrogels.29 A diffusion constant of D H1≈2×10−10 m 2 s−1 was found, which is close to the result one obtained from the Stokes–Einstein relation (r globular domain≈1.5 nm and D SE≈1.7×10−10 m 2 s−1) under purely aqueous conditions. To describe physical conditions at positions further down the reaction channel, a second simulation extending over the whole length of the device [x=−200–12 000 μm, Fig. 3b] has been performed using fit parameter values determined in the first set of simulations.
Figure 3.
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Real-time monitoring of linker histone H1 induced DNA compaction in a hydrodynamic focusing device. (a) Simulation results (top) and birefringence data (bottom) close to the confluence region are contrasted (u DNA=600 μm s−1). The product of the assembly reaction appears in the diffusion cone of side and main stream components due to its highly increased viscosity. (b) Simulated H1 concentration profile of the whole device (x=−200–12000 μm). Dependence of the N∕P ratio on the position along x (c) and the time t (d) in the middle of the outlet channel (y=0).
RESULTS AND DISCUSSION
Hydrodynamic focusing device
Aside from advantages, such as reduced sample volumes and the possibility of high throughput and parallel operations, microfluidic techniques are particularly useful for the investigation of biomaterials. More importantly, using microfluidics and x rays significantly reduces radiation induced damage of the sample—an important experimental consideration given the damage x rays cause to protein solutions.30 The hydrodynamic focusing device used here consists of two perpendicular microchannels in the form of a cross with three inlets and one outlet. A semidiluted aqueous DNA solution is injected into the reaction channel and hydrodynamically focused by two side streams of aqueous H1 solutions. Figure 2 gives a schematic representation of the experimental setup.
Figure 2.
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Schematic representation of the experimental x-ray setup. (I) Syringe pumps with aqueous solutions of DNA and H1. (II) Kapton microfluidic device. (III) Microfocused x-ray beam. (IV) X-ray diffraction pattern.
Laminar flow conditions due to microscale channel dimensions force mixing to occur purely by diffusion. Following the confluence of the microchannels, H1 molecules diffuse into the DNA stream establishing stable concentration gradients. Mixing and concentration distributions in the reaction channel can be adjusted by controlling the main and side channel velocities. Flow velocities are chosen such that concentration gradients of reactants extend along the measurable length of the device. It follows that the composition of the formed assemblies varies at every accessible point along the reaction channel. This allows access to the dynamics of H1∕DNA structure formation in situ, in particular, to transient intermediate states. Thus, it is essential to first quantify concentrations and therefore the composition of the aggregates. This is achieved by a detailed comparison of microscopic images and finite element simulations of flow conditions in the microfluidic device.
Quantifying local conditions in microfluidic devices
Polarized light microscopy has been used widely to study DNA assemblies, which are known to be highly birefringent.31, 32 The birefringence signal is increased upon self-assembly of DNA aggregates under conditions of alignment and elongation, which are imposed by using microfluidic devices. Concurrently, combining microfluidics with polarized light microscopy provides a fast and easy access to direct imaging of H1 induced DNA compaction. Data are acquired in the reaction channel (main channel) at three different flow velocities: u DNA=60, 150, and 600 μm s−1. The flow velocities in the side channels are varied such that a flow velocity ratio u H1∕u DNA=1 is maintained.
Finite element simulations of the flow profiles inside the microchannels are performed and compared to experiments in order to elucidate the experimental parameters. In Fig. 3a, simulation data are shown for u DNA=600 μm s−1 and contrasted to the corresponding experimental results. Owing to the symmetry of the microfluidic device in two dimensions, it is sufficient to simulate half of the device. Following the intersection of the two fluid flows, H1 molecules diffuse into the DNA stream and the H1∕DNA interactions can be observed along the reaction channel. The optically birefringent pattern reflects that DNA chains in the assemblies are orientationally ordered due to the superimposed flow. For direct comparison with the experimental results, the modeled velocity profile white arrows in Fig. 3a in the hydrodynamic focusing device are overlaid to the recorded birefringence image [Fig. 3a (bottom)]. The strong increase in local viscosity connected to the compaction reaction can be exploited to visualize H1∕DNA assemblies in the simulations. Insignificant deviations of simulation results from the experimentally recorded shape are observed in the crossing area, where the center stream is slightly expanding into the side channels. These deviations result from the fact that simulations are performed in two dimensions, whereas the experimental system is affected by additional walls at the top and the bottom of the device. Apart from this detail, experiments and simulations show good agreement. The result of the corresponding simulations over the whole range of the device is given in Fig. 3b.
From simulations, local experimental parameters, such as flow velocities and concentrations, can be obtained at each position. In Fig. 3c, N∕P ratios are plotted for three different flow velocities as a function of the position x along the center of the outlet channel (y=0). Flow velocities result in final charge ratios at the furthest measurable point of the device (x≈12 mm) of N∕P<2.4, 3.3, and 5.1 for u DNA=60, 150, and 600 μm s−1, respectively. Local flow velocities can be used to translate positional changes along each streamline into corresponding reaction time coordinates t. Figure 3d shows the t dependence of N∕P represented as well for three different flow velocities. For larger t, N∕P ratios unite into a single curve. Deviations at the initial time states reflect differences in the flow velocity depending strain rate, .33 Thus, in Fig. 3d it is clearly emphasized that the composition of the formed assemblies (for three different flow velocities) only depends on the charge ratio N∕P and not on the elapsed time. Once the information concerning the local concentrations and hence assemblies’ compositions is at hand, it is possible to analyze H1∕DNA structure formation in detail.
Small angle x-ray diffraction of H1∕DNA assemblies
Spatially resolved small-angle x-ray (micro)diffraction is employed as the principle method of analysis to access relevant molecular length scales for the study of biomolecular interactions. Data are obtained at different x-positions along the main channel for all three different flow velocities. In Fig. 4, characteristic 2D diffraction images are shown. The observed alignment is due to the elongational flow at the confluence of the solution streams. Structural information can be obtained by analyzing the radial integrated intensity profiles. The diffraction intensity is plotted as a function of the scattering vector q, which is inversely proportional to the periodic distance between reticular planes d.34 Since DNA has a significantly higher electron density than H1 proteins, the DNA self-assembly promotes or leads to the formation of mesophases that dominate the scattering profile. At positions close to the middle of the cross channels, the diffraction intensity curve is composed of a broad peak with a shoulder in the low q value region. The best decomposition of this broad peak is successfully made by fitting two Lorentzian functions, yielding peak positions q 1 and q 2. In Fig. 4, this is shown for the scattered intensity at x=100 μm.
Figure 4.
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Representative 2D x-ray diffraction images (right) obtained at u DNA=150 μm s−1 in the middle of the outlet channel (y=0) at different positions x and the extracted, radially averaged q-dependence of scattering intensities (left). I(q) plots are offset for clarity.
In Fig. 5a, the dependence of the peak positions q 1 and q 2 on the channel position x is given for the data set recorded at u DNA=150 μm s−1. To elucidate their dependence on assembly composition, it is reasonable to plot quantities of interest versus N∕P obtained from simulations. This is shown for q 1 and q 2 in Fig. 5b.
Figure 5.
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Dependence of peak positions q 1 and q 2 on the position x along the outlet channel (a) and on the N∕P ratio (b) shown exemplarily for the data set recorded at a flow velocity of u DNA=150 μm s−1.
Plotting quantities extracted from x-ray diffraction data obtained at different flow velocities against N∕P allows for the superposition of all data onto a single plot showing their dependence on composition. In Fig. 6a, this is demonstrated for peak positions q 1 (lower curve) and q 2 (upper curve) measured along the streamline at the center of the reaction channel (y=0). The three data sets obtained at u DNA=60, 150, and 600 μm s−1 show excellent agreement with deviations between different data sets of less than 0.01 nm−1. Local N∕P ratios are highly dependent on the diffusion of H1 molecules. Accordingly, the fact that the data obtained at different flow velocities are in good agreement and collapse onto single curves establishes the validity of the experimental method and the high degree of consistency between experiments and simulations.
Figure 6.
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Dependence of the peak position (a), the lattice spacing (b), the intensity ratio of the two peaks (c), and the correlation length given in terms of lattice spacing (for H1 and dendrimers) (d) on the local N∕P ratio.
At low N∕P ratios, peak positions of q 1=1.76 nm−1 and q 2=1.90 nm−1 are observed. With increasing N∕P, q 1 and q 2 are simultaneously shifted toward lower q values with minima at N∕P≈0.2 of q 1=1.73 nm−1 and q 2=1.88 nm−1, respectively. As follows, the peak position q 1 increases monotonically, moving toward higher q values, whereas q 2 levels off at q 2=1.90 nm−1. Eventually, for all the studied velocities, for N∕P>1.8 (corresponding to x>2200 μm) in Fig. 6a, the peak at q 2 disappears, leaving a single peak at q 1. Associated with the disappearance of the peak at q 2, the remaining peak q 1 shows a q max position at 1.78 nm−1. Increasing H1 concentration even further (i.e., increasing the charge ratio N∕P) results in smaller q values yielding q 1=1.72 nm−1 at the furthermost observable position x=12000 μm along the reaction channel [N∕P≈3.3, Fig. 6a] for a flow velocity of 60 μm s−1.
Structure of H1∕DNA mesophases
Owing to the absence of higher order peaks, the detailed structure of the formed H1∕DNA mesophases cannot be ruled out. This is mainly due to the fact that spatial constrains of the beamline limited the observable q-range to q<2.67 nm−1. Furthermore, correlation lengths of the formed H1∕DNA aggregates are on the order of 10–70 nm (described in Sec. 3E). For systems with such a reduced long-range ordering, scattering peaks of a structure with square symmetry are expected at a position , which is well situated in the accessible q-range and should be therefore observable. Furthermore, minima of the form factor of the globular domain, which could account for the absence of extra information about a hexagonal or a square phase, are situated at 1.55 and 2.66 nm−1 and are therefore not expected to be of influence. Accordingly, from the absence of peaks at this position, the structure of the mesophase exhibiting the peak at q 1 can be ruled out to be most likely a hexagonal one.35, 36 Unfortunately, it is not possible to narrow down the structure of the mesophase exhibiting the peak at q 2. Moreover, it was proven experimentally that the globular domain of H1 (d=2.9 nm) primarily defines distances between neighboring DNA strands;37, 38 thus, one can imagine a simplified version of linker histones by approximating them with lower generation dendrimers of similar size and charge (e.g., PAMAM G2). Recently, a systematic structural study has been conducted, which revealed that PAMAM2∕DNA assemble most favorably into hexagonal lattices, their results indicating as well a B conformation maintaining upon interaction.39 Assuming hexagonal ordering, the lattice spacing d can be calculated according to the following relation: . In Fig. 6b, lattice spacing’s d 1 and d 2—corresponding to q 1 and q 2—are given and their dependence on H1∕DNA assembly composition is shown in terms of the charge ratio N∕P. The observed lattice spacings are in the range of 3.8–4.2 nm.
The particular three domain structure of the linker histone induces a complex interaction with DNA leading to the formation of unique structures. Although unique, the DNA molecules within these structures are organized in a columnar phase as it was shown in earlier studies using polarizing microscopy for investigating DNA dense phases induced by other polycations.40 Previous work using electron microscopy to visualize H1∕DNA complexes on an electron microscopy grid revealed that H1 molecules are sandwiched between two DNA helices forming a tram-track-like pattern8 with a diameter of 3.8 nm,21 agreeing with interhelical values d 2 plotted in Fig. 6b. However, interhelical distances within complexes formed with poly-L-lysine (PLL) at a N∕P ratio of 1.7 and no added salt were found close to 2.7 nm,34 thus below our experimental values. Although the inner organization of the helices with the linear PLL is hexagonal, the higher lattice values we found are most certainly due to the structured globular domain of the linker histones. In order to enable comparability with results in literature, it is useful to translate N∕P into mass fraction w∕w of H1 to DNA. The x-ray data presented in Fig. 6b exhibit a maximal lattice spacing of both coexisting phases at N∕P≈0.2 (w∕w≈0.2). This is in line with the fact that for linear DNA molecules and low-salt conditions, small amounts of H1, w∕w≈0.15, produce complete incorporation of all DNA molecules into extremely large aggregates.18
Dynamics of H1∕DNA assembly structure formation
Figure 4 shows the existence of two overlapping diffraction peaks. The evolution of these peaks at different positions along the reaction channel is an evidence of a two-step process. At first, H1 binds electrostatically to DNA and the formed assemblies give rise to the diffraction peak at q 2. A closer study of this diffraction peak reveals columnar packing with DNA helices linked together by H1 molecules. The second step of the process is most probably due to the successive rearrangements of molecules in the formed assemblies, which result in a structure yielding the peak at q 1. Infrared (IR) spectroscopy studies have shown changes in the structure of the C-terminal domain of H1 upon DNA binding.41 In our experiments, this conformational transition can be monitored in terms of the intensity ratio I 2∕I 1 of the two Bragg reflections [Fig. 6c]. A coexistence regime occurs over a relatively wide range of N∕P ratios and is characterized by an overlapping of the two relatively sharp peaks at q 1 and q 2. With increasing N∕P, I 2∕I 1 is gradually reduced, reaching zero at N∕P≈1.8. This is the assembly composition at which the peak at q 2 is completely lost. The scattering is further characterized by a single peak at q 1 that both shifts to lower q values and broadens in q with a further increase of N∕P.
N∕P≈1.8 corresponds to 15 base pairs (bp) of DNA per H1 molecule. This result is in excellent agreement with sedimentation titration binding data that reported a binding density of one H1 molecule per 10–13 bp (N∕P≈2.5–1.9) (Ref. 42)—a value relatively independent of the salt concentration in the range of c salt=14–350 mM.22, 42 Moreover, our result is also consistent with nuclease digestion studies of chromatin, which have shown that each linker histone protects approximately 10 bp from each end of the chromatosomal DNA.15, 23, 43 Furthermore, considering the case of DNA∕PAMAM2 assemblies, each dendrimer has been estimated to cover seven phosphate groups.39
The experimental setup we used did not allow for a complete distinction between reaction time and composition dependent effects. In Fig. 3d, the dependence of N∕P on the reaction time t is given for all three flow velocities. For flow velocities of u=150 and 600 μm s−1, N∕P≈1.8 is reached at t≈2.5 s. For u=60 μm s−1, this assembly composition is only reached at t≈4.1 s. These deviations are due to several factors, such as the complex interaction of the flow fields influenced by the local viscosities and diffusion. However, although the reaction time is almost doubled for u=60 μm s−1, a vanishing of the peak at q 2 is only observed for t≥4.1 s. This indicates that the compaction mechanism of H1 and DNA is rather diffusion limited and conforms to the observations of similar sized dendrimer∕DNA interactions (article in preparation).
Domain size of H1∕DNA assemblies
Based on the combination of microfluidics technology with x-ray microdiffraction, we demonstrate that the interactions of H1 with DNA follow a two-step dynamic process. The efficiency of DNA compaction by H1 is influenced by multiple factors, including flow velocities, diffusion, and viscosity.
In addition to peak positions, average domain sizes of H1∕DNA assemblies can be determined from the full width at half maximum Δ q of the reflections at q 1 and q 2. The domain size corresponds to a typical correlation length L C=2π∕Δ q. To ensure comparability, it is useful to analyze q∕Δ q, which corresponds to the correlation lengths given in terms of the lattice spacing, q∕Δ q=L C∕d. The N∕P-dependence of q 1∕Δ q 1 and q 2∕Δ q 2 is shown in Fig. 6d. A clear dependence on the flow velocity of both q 1∕Δ q 1 and q 2∕Δ q 2 is evident. At low N∕P ratios in the coexistence region, q 2∕Δ q 2 is significantly higher than q 1∕Δ q 1, exhibiting values of 15–23 (L C2≈48–73 nm) and 5–9 (L C1≈19–33 nm), respectively. With increasing N∕P, q 2∕Δ q 2 shows a strong decrease starting around the charge neutral point characterized by N∕P=1. For all three flow velocities, maximal values q 1∕Δ q 1=6, 7, and 9, respectively, are found at N∕P≈1.8 when the feature at q 2 is disappeared. This finding is consistent with the evolution of the ratio of intensity I 2∕I 1 shown in Fig. 6c. Parallel to the observed shift in peak position q 1 to smaller q values with further increasing H1 concentration, domain sizes decrease to about 4 d at the furthest recorded assembly composition (N∕P≈3.3).
The ratio q∕Δ q allows for a comparison of H1∕DNA domain sizes to the values obtained in dendrimer∕DNA assemblies. Compared to linker histones, PPI generation 4 and PAMAM generation 3 dendrimers have a similar size and charge. These DNA assemblies have similar lattice spacing (d PPI4=3.1–3.6 nm, d PAMAM3=3.8–4.3 nm, and d H1=3.8–4.2 nm), although the process of structure evolution is not similar (article in preparation). At comparable strain rates and charge ratios (N∕P≈1), PPI 4∕DNA and PAMAM 3∕DNA assemblies exhibit domain sizes of approximately 23 d and 27 d, respectively. The range of these values is added to Fig. 6d (delimited by the blue lines), being comparatively close to those of q 2∕Δ q 2 and differing significantly from q 1∕Δ q 1.
The x-ray diffraction patterns show that at low charge ratios two mesophases coexist within the H1∕DNA assemblies. Since both phases experience identical experimental conditions (N∕P, strain rate, etc.), the differences in correlation lengths of L C2≈2.5–3⋅L C1 suggest that the transition from an ordered mesophase (L C2) corresponding to the columnar organization of helices immediately after binding H1 to a less organized structure (L C1) is mediated by the rearrangement of the histone tails. For a clearer illustration of the two-step model, we propose a schematic representation that is shown in Fig. 7. As presented in Fig. 6d, the correlation length of domains with extended tails (L C2) is rapidly reduced after N∕P≈1.1 (28 bp of DNA per each H1 molecule), implying that the tails might distort the order of the columnar phase and bend the DNA. Several studies have previously suggested that chromatosomal linker DNA is bent by the C-terminal domain of H1 forming a stemlike structure.44, 45, 46, 47 This structure has also been implicated in the formation of the 30 nm chromatin fiber.48 Even though similar condensed DNA structures have been observed with linker histone H1 and polyamines,49 the effect on the compaction process seems to be H1 specific since a coexistence of two dense phases is present from the beginning of the diffusive mixing. In order to validate the two-step binding model proposed in this study, future experiments should be performed in physiological conditions using truncated H1 molecules lacking the N and C charged terminal domains. Moreover, assemblies of DNA and H1 have been proven to be more compact at physiological salt (0.2 M) rather than at diluted salt (0.05 M),50 Therefore, we envisage complementary experiments in order to investigate the structure evolution at variable ionic conditions.
Figure 7.
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Schematic representation of the H1/DNA interaction mechanism. In a first step, H1 molecules bind unspecifically to DNA with extended tails. In a second step, H1 tails fold upon interaction with DNA, distorting and bending thereby the DNA structure.
CONCLUSIONS
Based on the combination of microfluidics technology with x-ray microdiffraction, we demonstrate that the interactions of H1 with DNA follow a two-step dynamic process. The efficiency of DNA compaction by H1 is influenced by multiple factors, including flow velocities, diffusion, and viscosity. We have seen that by superimposing external strain on biomaterials, it is possible to significantly improve the correlation length of the formed assemblies, underling the advantage of using flow for self-assembling materials compared to the bulk mixing of the components. We show that the first binding step is primarily due to electrostatic interactions between the DNA and the linker histone. Our results suggest that the organization of this phase is columnar hexagonal and is composed of domains with a long correlation length. Furthermore, due to the most certain rearrangement of histone tails within this dense phase, a loss of ordering is observed. Thus, domains with shorter correlation lengths are revealed, which might be attributed to an additional bending of the DNA.
A potential direction for future studies involves monitoring of nucleosome core particle (NCP) arrays∕H1 assembly dynamic formation and structure evolution in microflow. The NCP∕H1 assemblies in microflow may shed light on the 30 nm fibber formation in real-time. Our approach may be generalized and used to access additional relevant biophysical problems of chromatin compaction and decompaction in a single microfluidic experiment.
ACKNOWLEDGMENTS
We gratefully acknowledge fruitful discussions with Heather Evans, Stephan Herminghaus, Sarah Köster, Alexander Otten, Bernd Struth, Oleg Konovalov, Anatoly Snigirev, Dagmar Steinhauser, and Sravanti Uppaluri. We thank Udo Krafft for excellent technical assistance. We acknowledge the ESRF for provision of synchrotron radiation facilities at beamline ID10b. This work was supported by the Deutsche Forschungsgemeinschaft [within the Emmy-Noether-Program (PF 375∕2)] and Sonderforschungsbereich [755 “Nanoscale photonic imaging”].
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3139 | https://www.schoolmykids.com/learn/periodic-table/cl-chlorine | Home
Periodic Table
Cl - Chlorine
Chlorine (Cl)- Atomic, Physical & Chemical Properties, Uses, and Periodic Table Trends
ChlorineElement Information, Facts, Properties, Trends, Uses, Comparison with other elements
Element 17 of Periodic table is Chlorine with atomic number 17, atomic weight 35.453. Chlorine, symbol Cl, has a Base Centered Orthorhombic structure and Yellow color. Chlorine is a Halogens element. It is part of group 17 (fluorine family). Discover everything about Chlorine Facts, Physical Properties, Chemical Properties, Electronic configuration, Atomic and Crystal Structure.
Chlorine (Cl) - Comprehensive Element Profile, Properties & Uses
In this comprehensive guide, you'll learn about Chlorine's unique chemical and physical properties, trends in the periodic table, isotopes, and its historical significance. We'll also cover its abundance, crystal structure, electron configuration, and health & safety guidelines. Explore how Chlorine compares with other elements and discover its many uses.
Chlorine is a chemical element with symbol Cl and atomic number 17. It also has a relative atomic mass of 35.5. Chlorine is in the halogen group (17) and is the second lightest halogen following fluorine.
Chlorine belongs to group 17 of the periodic table having trivial name halogens.You can also downloadPrintable Periodic Table of Elements Flashcards for Chlorine in a PDF format.
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Chlorine Facts
Read key information and facts about element Chlorine
| | |
--- |
| Name | Chlorine |
| Atomic Number | 17 |
| Atomic Symbol | Cl |
| Atomic Weight | 35.453 |
| Phase | Gas(Diatomic Gas) |
| Color | Yellow |
| Appearance | pale yellow-green gas |
| Classification | Halogens |
| Natural Occurance | Primordial |
| Group in Periodic Table | 17 |
| Group Name | fluorine family |
| Period in Periodic Table | period 3 |
| Block in Periodic Table | p-block |
| Electronic Configuration | [Ne] 3s2 3p5 |
| Electronic Shell Structure (Electrons per shell) | 2, 8, 7 |
| Melting Point | 171.6 K |
| Boiling Point | 239.11 K |
| CAS Number | CAS7782-50-5 |
Neighborhood Elements
17
Cl
| | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
| 1 | | Atomic # Electronic Shell # Atomic Weight HGas HgLiquid CSolid | | | | --- | Metals | Metalloids | NonMetals | | Alkali metals | Alkali earth metals | Lanthanoids | Transition metals | Post-transition metals | Other nonmetals | Halogens | Nobel gas | | Actinoids | | 2 He Helium 4.003 |
| 2 | 3 Li Lithium 6.941 | 4 Be Beryllium 9.012 | 5 B Boron 10.811 | 6 C Carbon 12.011 | 7 N Nitrogen 14.007 | 8 O Oxygen 15.999 | 9 F Fluorine 18.998 | 10 Ne Neon 20.180 |
| 3 | 11 Na Sodium 22.990 | 12 Mg Magnesium 24.305 | 13 Al Aluminium 26.982 | 14 Si Silicon 28.085 | 15 P Phosphorus 30.974 | 16 S Sulfur 32.065 | 17 Cl Chlorine 35.453 | 18 Ar Argon 39.948 |
| 4 | 19 K Potassium 39.098 | 20 Ca Calcium 40.078 | 21 Sc Scandium 44.956 | 22 Ti Titanium 47.867 | 23 V Vanadium 50.941 | 24 Cr Chromium 51.996 | 25 Mn Manganese 54.938 | 26 Fe Iron 55.845 | 27 Co Cobalt 58.933 | 28 Ni Nickel 58.693 | 29 Cu Copper 63.546 | 30 Zn Zinc 65.409 | 31 Ga Gallium 69.723 | 32 Ge Germanium 72.640 | 33 As Arsenic 74.922 | 34 Se Selenium 78.960 | 35 Br Bromine 79.904 | 36 Kr Krypton 83.798 |
| 5 | 37 Rb Rubidium 85.468 | 38 Sr Strontium 87.620 | 39 Y Yttrium 88.906 | 40 Zr Zirconium 91.224 | 41 Nb Niobium 92.906 | 42 Mo Molybdenum 95.940 | 43 Tc Technetium 98 | 44 Ru Ruthenium 101.070 | 45 Rh Rhodium 102.906 | 46 Pd Palladium 106.420 | 47 Ag Silver 107.868 | 48 Cd Cadmium 112.411 | 49 In Indium 114.818 | 50 Sn Tin 118.710 | 51 Sb Antimony 121.760 | 52 Te Tellurium 127.600 | 53 I Iodine 126.904 | 54 Xe Xenon 131.293 |
| 6 | 55 Cs Cesium 132.905 | 56 Ba Barium 137.327 | 57- 71La- Lu Lanthanides | 72 Hf Hafnium 178.490 | 73 Ta Tantalum 180.948 | 74 W Tungsten 183.840 | 75 Re Rhenium 186.207 | 76 Os Osmium 190.230 | 77 Ir Iridium 192.217 | 78 Pt Platinum 195.078 | 79 Au Gold 196.967 | 80 Hg Mercury 200.590 | 81 Tl Thallium 204.383 | 82 Pb Lead 207.200 | 83 Bi Bismuth 208.980 | 84 Po Polonium 209 | 85 At Astatine 210 | 86 Rn Radon 222 |
| 7 | 87 Fr Francium 223 | 88 Ra Radium 226 | 89- 103Ac- Lr Actinides | 104 Rf Rutherfordium 261 | 105 Db Dubnium 262 | 106 Sg Seaborgium 266 | 107 Bh Bohrium 264 | 108 Hs Hassium 269 | 109 Mt Meitnerium 268 | 110 Ds Darmstadtium 281 | 111 Rg Roentgenium 272 | 112 Cn Copernicium 285 | 113 Nh Nihonium 284 | 114 Fl Flerovium 289 | 115 Mc Moscovium 288 | 116 Lv Livermorium 292 | 117 Ts Tennessine 294 | 118 Og Oganesson 294 |
| Lanthanides | 57 La Lanthanum 138.905 | 58 Ce Cerium 140.116 | 59 Pr Praseodymium 140.908 | 60 Nd Neodymium 144.240 | 61 Pm Promethium 145 | 62 Sm Samarium 150.360 | 63 Eu Europium 151.964 | 64 Gd Gadolinium 157.250 | 65 Tb Terbium 158.925 | 66 Dy Dysprosium 162.500 | 67 Ho Holmium 164.930 | 68 Er Erbium 167.259 | 69 Tm Thulium 168.934 | 70 Yb Ytterbium 173.040 | 71 Lu Lutetium 174.967 |
| Actinides | 89 Ac Actinium 227 | 90 Th Thorium 232.038 | 91 Pa Protactinium 231.036 | 92 U Uranium 238.029 | 93 Np Neptunium 237 | 94 Pu Plutonium 244 | 95 Am Americium 243 | 96 Cm Curium 247 | 97 Bk Berkelium 247 | 98 Cf Californium 251 | 99 Es Einsteinium 252 | 100 Fm Fermium 257 | 101 Md Mendelevium 258 | 102 No Nobelium 259 | 103 Lr Lawrencium 262 |
Explore our interactive periodic table
Download Periodic Table Flash Cards - Chlorine Element
Table of Content
History
Abundance
Physical Properties
Thermal Properties
Crystal Structure
Atomic & Orbital Properties
Chemical Properties
Isotopes
Databases
Health and Safety Parameters and Guidelines Compare Chlorine with other elements
How to Locate Chlorine on Periodic Table
Periodic table is arranged by atomic number, number of protons in the nucleus which is same as number of electrons. The atomic number increases from left to right. Periodic table starts at top left ( Atomic number 1) and ends at bottom right (atomic number 118). Therefore you can directly look for atomic number 17to findChlorineon periodic table.
Another way to read periodic table and locate an element is by using group number (column) and period number (row). To locate Chlorineon periodic table look for cross section of group 17and period3in the modern periodic table.
Explore Interactive Periodic Table to Understand and Learn Cool Trends
Chlorine History
The element Chlorinewas discovered byW. Scheelein year1774in Sweden. Chlorinewas first isolated byW. Scheelein1774. Chlorinederived its name from the Greek word chloros, meaning 'greenish yellow'.
| | |
--- |
| Discovered By | W. Scheele |
| Discovery Date | 1774 in Sweden |
| First Isolation | 1774 |
| Isolated by | W. Scheele |
Obtained it from hydrochloric acid, but thought it was an oxide. Only in 1808 did Humphry Davy recognize it as an element.
Download printable flash card for Chlorine periodic table PDF
Chlorine Uses
Chlorine is used in water treatment and as an antiseptic. Large amounts of chlorine are also used to produce paper, plastics, solvents, and textiles
Water Treatment: Chlorine is used in water treatment to disinfect drinking water and swimming pools.
Bleaching Agent: Chlorine is used as a bleaching agent in the textile, paper, and laundry industries.
Chemical Manufacturing: Chlorine is used in the production of various chemicals, including plastics like PVC (polyvinyl chloride).
Chlorine Presence: Abundance in Nature and Around Us
The table below shows the abundance of Chlorinein Universe, Sun, Meteorites, Earth's Crust, Oceans and Human Body.
| | ppb by weight (1ppb =10^-7 %) | ppb by atoms (1ppb =10^-7 %) |
---
| Abundance in Universe | 1000 | 40 |
| Abundance in Sun | 8000 | 300 |
| Abundance in Meteorites | 380000 | 160000 |
| Abundance in Earth's Crust | 170000 | 100000 |
| Abundance in Oceans | 19870000 | 3470000 |
| Abundance in Humans | 1200000 | 210000 |
Crystal Structure of Chlorine
The solid state structure of ChlorineisBase Centered Orthorhombic.
The Crystal structure can be described in terms of its unit Cell. The unit Cells repeats itself in three dimensional space to form the structure.
Unit Cell Parameters
The unit cell is represented in terms of its lattice parameters, which are the lengths of the cell edges Lattice Constants (a, b and c)
| a | b | c |
---
| 622.35 pm | 445.61 pm | 817.85 pm |
and the angles between them Lattice Angles (alpha, beta and gamma).
| alpha | beta | gamma |
---
| π/2 | π/2 | π/2 |
The positions of the atoms inside the unit cell are described by the set of atomic positions ( xi, yi, zi) measured from a reference lattice point.
The symmetry properties of the crystal are described by the concept of space groups. All possible symmetric arrangements of particles in three-dimensional space are described by the 230 space groups (219 distinct types, or 230 if chiral copies are considered distinct.
| | |
--- |
| Space Group Name | Cmca |
| Space Group Number | 64 |
| Crystal Structure | Base Centered Orthorhombic |
| Number of atoms per unit cell | 2 |
The number of atoms per unit cell in a simple cubic, face-centered cubic and body-centred cubic are 1,4,2 respectively.
Chlorine Atomic and Orbital Properties
Chlorineatoms have 17electrons and the electronic shell structure is [2, 8, 7] with Atomic Term Symbol (Quantum Numbers)2P3/2.
| | |
--- |
| Atomic Number | 17 |
| Number of Electrons (with no charge) | 17 |
| Number of Protons | 17 |
| Mass Number | 35 |
| Number of Neutrons | 18 |
| Shell structure (Electrons per energy level) | 2, 8, 7 |
| Electron Configuration | [Ne] 3s2 3p5 |
| Valence Electrons | 3s2 3p5 |
| Valence (Valency) | 5 |
| Main Oxidation States | -1, 1, 3, 5, 7 |
| Oxidation States | -1, 1, 2, 3, 4, 5, 6, 7 |
| Atomic Term Symbol (Quantum Numbers) | 2P3/2 |
Bohr Atomic Model of Chlorine - Electrons per energy level
| n | s | p | d | f |
--- ---
Ground State Electronic Configuration of Chlorine - neutral Chlorine atom
Abbreviated electronic configuration of Chlorine
The ground state abbreviated electronic configuration of Neutral Chlorine atom is [Ne] 3s2 3p5. The portion of Chlorine configuration that is equivalent to the noble gas of the preceding period, is abbreviated as [Ne]. For atoms with many electrons, this notation can become lengthy and so an abbreviated notation is used. This is important as it is the Valence electrons 3s2 3p5, electrons in the outermost shell that determine the chemical properties of the element.
Unabbreviated electronic configuration of neutral Chlorine
Complete ground state electronic configuration for the Chlorine atom, Unabbreviated electronic configuration
1s2 2s2 2p6 3s2 3p5
Electrons are filled in atomic orbitals as per the order determined by the Aufbau principle, Pauli Exclusion Principle and Hund’s Rule.
As per the Aufbau principle the electrons will occupy the orbitals having lower energies before occupying higher energy orbitals. According to this principle, electrons are filled in the following order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p…
The Pauli exclusion principle states that a maximum of two electrons, each having opposite spins, can fit in an orbital.
Hund's rule states that every orbital in a given subshell is singly occupied by electrons before a second electron is filled in an orbital.
Atomic Structure of Chlorine
Chlorine atomic radius is 79 pm, while it's covalent radius is 99 pm.
| | |
--- |
| Atomic Radius Calculated | 79 pm(0.79 Å) |
| Atomic Radius Empirical | 100 pm (1 Å) |
| Atomic Volume | 22.4129 cm3/mol |
| Covalent Radius | 99 pm (0.99 Å) |
| Van der Waals Radius | 175 pm |
| Neutron Cross Section | 35.3 |
| Neutron Mass Absorption | 0.033 |
Spectral Lines of Chlorine - Atomic Spectrum of Chlorine
A spectral line is a dark or bright line in an otherwise uniform and continuous spectrum, resulting from an excess or deficiency of photons in a narrow frequency range, compared with the nearby frequencies. Spectral lines are often used to identify atoms and molecules.
Spectral lines are the result of interaction between a quantum system and a single photon. A spectral line may be observed either as an emission line or an absorption line.
Spectral lines are highly atom-specific, and can be used to identify the chemical composition of any medium. Several elements, including helium, thallium, and caesium, were discovered by spectroscopic means. They are widely used to determine the physical conditions of stars and other celestial bodies that cannot be analyzed by other means.
Emission spectrum of Chlorine
Absorption spectrum of Chlorine
Chlorine Chemical Properties: Chlorine Ionization Energies and electron affinity
The electron affinity of Chlorine is 349 kJ/mol.
| | |
--- |
| Valence | 5 |
| Electronegativity | 3.16 |
| ElectronAffinity | 349 kJ/mol |
Ionization Energy of Chlorine
Ionization energy is the amount of energy required to remove an electron from an atom or molecule.in chemistry, this energy is expresed in kilocalories per mole (kcal/mol) or kilojoules per mole (kJ/mol).
Refer to table below for Ionization energies of Chlorine
Here are the ionization energies of Chlorine (Cl) both in electron volts (eV) and in kilojoules per mole (kJ/mol).
| Ionization energy number | Enthalpy in kJ/mol | Energy (eV) |
---
| 1st | 1251.2 | 12.968 |
| 2nd | 2298 | 23.817 |
| 3rd | 3822 | 39.612 |
| 4th | 5158.6 | 53.465 |
| 5th | 6542 | 67.803 |
| 6th | 9362 | 97.031 |
| 7th | 11018 | 114.194 |
| 8th | 33604 | 348.282 |
| 9th | 38600 | 400.062 |
| 10th | 43961 | 455.625 |
| 11th | 51068 | 529.284 |
| 12th | 57119 | 591.999 |
| 13th | 63363 | 656.713 |
| 14th | 72341 | 749.764 |
| 15th | 78095 | 809.400 |
| 16th | 352994 | 3658.538 |
| 17th | 380760 | 3946.313 |
The conversion from kJ/mol to eV is done using the formula: Energy (kJ/mol) = Energy (eV) x 96.485 Energy (kJ/mol)=Energy (eV)x96.485 where 1 eV = 96.485 kJ/mol.
1 electronvolt (eV) is equal to 96.485 kilojoules per mole (kJ/mol)
Chlorine Physical Properties
Refer to below table for Chlorine Physical Properties
| | |
--- |
| Density | 0.003214 g/cm3 |
| Molar Volume | 22.4129 cm3/mol |
Elastic Properties
| | |
--- |
| Young Modulus
| Shear Modulus
| Bulk Modulus | 1.1 GPa |
| Poisson Ratio
Hardness of Chlorine - Tests to Measure of Hardness of Element
| | |
--- |
| Mohs Hardness
| Vickers Hardness
| Brinell Hardness
Chlorine Electrical Properties
Electrical resistivity measures element's electrical resistance or how strongly it resists electric current.The SI unit of electrical resistivity is the ohm-metre (Ω⋅m). While Electrical conductivity is the reciprocal of electrical resistivity. It represents a element's ability to conduct electric current. The SI unit of electrical conductivity is siemens per metre (S/m).
Chlorine is a Insulator. Refer to table below for the Electrical properties of Chlorine
| | |
--- |
| Electrical conductors | Insulator |
| Electrical Conductivity | 0.01 S/m |
| Resistivity | 100 m Ω |
| Superconducting Point
ChlorineHeat and Conduction Properties
| | |
--- |
| Thermal Conductivity | 0.0089 W/(m K) |
| Thermal Expansion
ChlorineMagnetic Properties
| | |
--- |
| Magnetic Type | Diamagnetic |
| Curie Point
| Mass Magnetic Susceptibility | -7.2e-9 m3/kg |
| Molar Magnetic Susceptibility | -5.11e-10 m3/mol |
| Volume Magnetic Susceptibility | -2.31e-8 |
Optical Properties of Chlorine
| | |
--- |
| Refractive Index | 1.000773 |
Acoustic Properties of Chlorine
| | |
--- |
| Speed of Sound | 206 m/s |
ChlorineThermal Properties - Enthalpies and thermodynamics
Refer to table below for Thermal properties of Chlorine
| | |
--- |
| Melting Point | 171.6 K(-101.55 °C, -150.790 °F) |
| Boiling Point | 239.11 K(-34.04 °C, -29.272 °F) |
| Critical Temperature | 416.9 K |
| Superconducting Point
Enthalpies of Chlorine
| | |
--- |
| Heat of Fusion | 3.2 kJ/mol |
| Heat of Vaporization | 10.2 kJ/mol |
| Heat of Combustion
ChlorineIsotopes - Nuclear Properties of Chlorine
Chlorinehas 24isotopes, with between28and51nucleons. Chlorinehas 2stable naturally occuring isotopes.
Isotopes of Chlorine - Naturally occurring stable Isotopes: 35Cl, 37Cl.
| Isotope | Z | N | Isotope Mass | % Abundance | T half | Decay Mode |
--- --- ---
| 28Cl | 17 | 11 | 28 | Synthetic |
| 29Cl | 17 | 12 | 29 | Synthetic |
| 30Cl | 17 | 13 | 30 | Synthetic |
| 31Cl | 17 | 14 | 31 | Synthetic |
| 32Cl | 17 | 15 | 32 | Synthetic |
| 33Cl | 17 | 16 | 33 | Synthetic |
| 34Cl | 17 | 17 | 34 | Synthetic |
| 35Cl | 17 | 18 | 35 | 75.78% | Stable |
| 36Cl | 17 | 19 | 36 | Synthetic |
| 37Cl | 17 | 20 | 37 | 24.22% | Stable | N/A |
| 38Cl | 17 | 21 | 38 | Synthetic |
| 39Cl | 17 | 22 | 39 | Synthetic |
| 40Cl | 17 | 23 | 40 | Synthetic |
| 41Cl | 17 | 24 | 41 | Synthetic |
| 42Cl | 17 | 25 | 42 | Synthetic |
| 43Cl | 17 | 26 | 43 | Synthetic |
| 44Cl | 17 | 27 | 44 | Synthetic |
| 45Cl | 17 | 28 | 45 | Synthetic |
| 46Cl | 17 | 29 | 46 | Synthetic |
| 47Cl | 17 | 30 | 47 | Synthetic |
| 48Cl | 17 | 31 | 48 | Synthetic |
| 49Cl | 17 | 32 | 49 | Synthetic |
| 50Cl | 17 | 33 | 50 | Synthetic |
| 51Cl | 17 | 34 | 51 | Synthetic |
Regulatory and Health - Health and Safety Parameters and Guidelines
The United States Department of Transportation (DOT) identifies hazard class of all dangerous elements/goods/commodities either by its class (or division) number or name. The DOT has divided these materials into nine different categories, known as Hazard Classes.
| | |
--- |
| DOT Numbers | 1017 |
| DOT Hazard Class
NFPA 704 is a Standard System for the Identification of the Hazards of Materials for Emergency Response. NFPA is a standard maintained by the US based National Fire Protection Association.
The health (blue), flammability (red), and reactivity (yellow) rating all use a numbering scale ranging from 0 to 4. A value of zero means that the element poses no hazard; a rating of four indicates extreme danger.
| | | |
---
| NFPA Fire Rating | N/A | N/A |
| NFPA Health Rating | N/A | N/A |
| NFPA Reactivity Rating | N/A | N/A |
| NFPA Hazards | N/A |
| | |
--- |
| Autoignition Point
| Flashpoint
Database Search
List of unique identifiers to search the element in various chemical registry databases
| Database | Identifier number |
--- |
| CAS Number - Chemical Abstracts Service (CAS) | CAS7782-50-5 |
| RTECS Number | RTECSFO2100000 |
| CID Number | CID24526 |
| Gmelin Number
| NSC Number
Compare Chlorinewith other elements
Compare Chlorinewith Group 17, Period3and Halogenselements of the periodic table.
Compare Chlorinewith all Group 17 elements
Compare Chlorine with AstatineCompare Chlorine with BromineCompare Chlorine with FluorineCompare Chlorine with IodineCompare Chlorine with Tennessine
Compare Chlorinewith all Period 3 elements
Compare Chlorine with ArgonCompare Chlorine with AluminiumCompare Chlorine with PhosphorusCompare Chlorine with SulfurCompare Chlorine with MagnesiumCompare Chlorine with SiliconCompare Chlorine with Sodium
Compare Chlorinewith all Halogens elements
Chlorine vs Astatine ComparisonChlorine vs Bromine ComparisonChlorine vs Fluorine ComparisonChlorine vs Iodine ComparisonChlorine vs Tennessine Comparison
Explore our interactive periodic tablePeriodic Table Element Comparison
Frequently Asked Questions (FAQ)
Find the answers to the most frequently asked questions about Chlorine
1. What is the electronic configuration of Chlorine?
The electronic configuration of Chlorine is 1s2 2s2 2p6 3s2 3p5.
2. What is the abbreviated electronic configuration of Chlorine?
The abbreviated electronic configuration of Chlorine is [Ne] 3s2 3p5. To form abbreviated notation of electronic configuration, the completely filled subshells are replaced by the noble gas of the preceding period in square brackets.
3. What is the symbol of Chlorine?
Symbol of Chlorine is Cl. Chlorine is a chemical element with symbol Cl and atomic number 17.
4. What is the position of Chlorine in the Periodic Table?
Chlorine is a chemical element with the symbol Cl and atomic number 17. Chlorine is the 17 element on the periodic table. It is located in group 17 and period 3 in the modern periodic table.
5. How do you find Chlorine on the periodic table?
Chlorine is the 17 element on the periodic table. Chlorine is located in group 17 and period 3 in the modern periodic table.
6. What is the atomic number of Chlorine?
The atomic number of Chlorine is 17.
7. What is the color of Chlorine?
Chlorine is of Yellow color.
8. Who discovered Chlorine?
The element Chlorine was discovered by W. Scheele in year 1774 in Sweden. Chlorine was first isolated by W. Scheele in 1774.
9. How many valence electrons does a Chlorine atom have?
Chlorine has 5 valence electrons. Chlorine has 17 electrons out of which 5 valence electrons are present in the 3s2 3p5 outer orbitals of atom.
10. What is the melting Point of Chlorine?
Melting Point of Chlorine is 171.6 K.
11. What is the density of Chlorine?
Density of Chlorine is 0.003214 g/cm3.
12. What is Chlorine used for?
Chlorine is used for: Water Treatment: Chlorine is used in water treatment to disinfect drinking water and swimming pools. Bleaching Agent: Chlorine is used as a bleaching agent in the textile, paper, and laundry industries. Chemical Manufacturing: Chlorine is used in the production of various chemicals, including plastics like PVC (polyvinyl chloride).
13. What is the boiling Point of Chlorine?
Boiling Point of Chlorine is 239.11 K.
14. What is the melting Point of Chlorine in Kelvin?
Melting Point of Chlorine in Kelvin is 171.6 K.
15. What is the boiling Point of Chlorine in Kelvin?
Boiling Point of Chlorine in Kelvin is 239.11 K.
16. What is the melting Point of Chlorine in Celsius?
Melting Point of Chlorine in Celsius is -101.55 °C.
17. What is the boiling Point of Chlorine in Celsius?
Boiling Point of Chlorine in Celsius is -34.04 °C.
18. What is the melting Point of Chlorine in Fahrenheit?
Melting Point of Chlorine in Fahrenheit is -150.79 °F.
19. What is the boiling Point of Chlorine in Fahrenheit?
Boiling Point of Chlorine in Fahrenheit is -29.27 °F.
20. What is the electronic configuration of Chlorine 17?
The electronic configuration of Chlorine will be 1s2 2s2 2p6 3s2 3p5.
21. How do you write the electron configuration for Chlorine?
The electronic configuration of Chlorine will be 1s2 2s2 2p6 3s2 3p5.
22. How to locate Chlorine on the periodic table?
On the periodic table, elements are listed in order of increasing atomic number. Chlorine is the 17 element on the periodic table. Chlorine is located in group 17 and period 3 in the modern periodic table. |
3140 | https://chemistry.stackexchange.com/questions/143726/in-metal-complex-electron-counting-is-the-neutral-or-ionic-formalism-more-corre | coordination compounds - In metal complex electron counting, is the neutral or ionic formalism more correct? - Chemistry Stack Exchange
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In metal complex electron counting, is the neutral or ionic formalism more correct?
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I have been practicing finding total valence electron counts by both methods, and a few points have arisen that have confused me.
1) Which method gives the true d-electron count?
From what I understand, the ionic method assumes that both electrons from a bond go to the more electronegative atom, and the bond is 100% ionic. This lines up with formal oxidation state which uses the same premise. As a result, d-electrons are usually calculated as completely lost from the metal, giving a positive formal oxidation state.
In the neutral method, the d-electron count doesn't change as all calculations assume that M-L bonds are cleaved such that both species are neutral, and metal oxidation state stays at 0.
This results in 2 different values for a d-electron count, and I was wondering if the value actually means anything. I guess it doesn't matter in the end as the same number of electrons are put into the overall MO scheme, but I was wondering if being asked the d-electron count has any point to it. This can be seen in the below example from here:
2) If the neutral formalism is used, where does the electron deficiency in positively charged complexes come from?
In the ionic method, we assume everything is charged and just take electrons out of the metal to account for the overall charge (even this seems strange, why do we take electrons from the metal when we could take them from anywhere else to give the same MO scheme?). But in the neutral formalism, if everything is assumed neutral, where are we supposed take the electrons from?
3) If the electrons don't have to be taken from the metal, what is the point of assigning metal oxidation state?
I always hear statements like "d8 metals such as Pt(II) commonly form square planar complexes", but both the assignment of d8 and Pt(II) differ when using a different formalism. So is the use of the ionic method and oxidation of the metal specifically simply a convention that allows us to compare complexes quickly by their total electron count?
It seems to me like the correct answer is somewhere in between the 2 methods (between 100% ionic and 100% covalent), and the use of one or the other is arbitrary, but any corrections would be appreciated.
Also, if oxidation state is arbitrary, is the argument that "metals with lower ionisation energies reach higher oxidation states so can form more bonds" the same as saying "metals that have lower ionisation energies generally have larger radii so can fit more covalent bonds around them"?
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edited Dec 11, 2020 at 18:20
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1 You have good questions, and I don't have the time to address all in detail, but the gist is that the electron counting method is arbitrary (that's why they give the same answer, and you should be worried if they don't). Those merely tell you the total number of electrons in the MO diagram, they are not meant to reflect some physical reality about their distribution. Oxidation state and d-electron count aren't arbitrary. Those tell you how many of those electrons are in the metal d-orbitals (or t2g/eg orbitals, etc. for different geometry). That has a great influence on chemical behaviour.orthocresol –orthocresol 2020-12-11 18:33:39 +00:00 Commented Dec 11, 2020 at 18:33
Well, technically the concepts of oxidation state and d-electron count can be stretched to a point where they don't make much sense (e.g. when there is heavy delocalisation), but as long as you don't go there...orthocresol –orthocresol 2020-12-11 18:36:32 +00:00 Commented Dec 11, 2020 at 18:36
@orthocresol Looking at an MO scheme, it doesn't seem to matter where you decide to take the electrons from, it will still give the same number of electrons in t2g/eg orbitals. Does this mean using the formal oxidation state of the metal and taking electrons from the metal to account for charge gives you a not so arbitrary quantification of eg/t2g electron count that we just CALL a d-electron count, ignoring the ligand character?Jabbamanga –Jabbamanga 2020-12-11 18:44:44 +00:00 Commented Dec 11, 2020 at 18:44
1 The eg/t2g orbitals are mainly metal d-orbitals (you can see that from an MO scheme), so yes, the oxidation state or "d-electron count" pretty much ignore the ligand contribution to these orbitals. Of course, these are not 100% metal d-orbital character, which is why the "truth" is somewhere in between. That starts to really break down when... well, I already mentioned it. :-)orthocresol –orthocresol 2020-12-12 01:05:28 +00:00 Commented Dec 12, 2020 at 1:05
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I think, the answer to number 1 will remove the need to answer the subsequent questions:
1) Which method gives the true d-electron count?
Neither!
You cannot, and I want to stress that, derive the oxidation state of a metal in a complex solely by looking at its ligands. You can make educated guesses and in some cases only one oxidation state is plausible but there are many cases—most notably those denoted {F e N O}X n{F e N O}X n where such an assingment is meaningless to impossible. Without consulting spectroscopic data, it is absolutely impossible to tell whether a {F e N O}X 7{F e N O}X 7 compound is [F e+I I(N O X∙)][F e+I I(N O X∙)] or a [F e+I I I(N O X−)][F e+I I I(N O X−)]. In fact, entire papers have been published arguing one way or the other.
This essentially renders your follow-up questions entirely moot, but I do wish to take the time to point out that the metal’s oxidation state along with the corresponding oxidation state of the ligands is theoretically important information that can predict the electrochemical behaviour of a complex. The only problem is that it cannot always be neatly predicted as shown above. It is worth noting, though, that certain complexes have been assigned metal oxidation states beyond reasonable doubts (e.g. [C u(N H X 3)X 4(H X 2 O)X 2][C u(N H X 3)X 4(H X 2 O)X 2]), and in such cases the predictions of reactivity are absolutely adequate.
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answered Dec 16, 2020 at 15:02
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3141 | https://math.stackexchange.com/questions/375270/size-of-largest-prime-factor | number theory - Size of largest prime factor - Mathematics Stack Exchange
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Size of largest prime factor
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It is well known and easy to prove that the smallest prime factor of an integer n n is at most equal to n−−√n. What can be said about the largest prime factor of n n, denoted by P 1(n)P 1(n)? In particular:
What is the probability that P 1(n)>n−−√P 1(n)>n ?
More generally, what is the expected value of the size of P 1(n)P 1(n), measured by log P 1(n)log n logP 1(n)logn ?
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edited Apr 28, 2013 at 20:48
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Size of the largest prime of p p is p p itself. Where p p is a prime.This is obvious, though.Inceptio –Inceptio 2013-04-28 14:50:57 +00:00 Commented Apr 28, 2013 at 14:50
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Take the negation and this is a very well-known question: what is the probability that all prime factors of n n are ≤n−−√≤n? The answer is known quite generally: for any real u≥1 u≥1, the probability that the prime factors of n n are ≤n 1/u≤n 1/u is given by the Dickman–de Bruijn rho function, defined by a delay-differential equation. For u=2 u=2 we have ρ(u)=1−log 2 ρ(u)=1−log2, as in Ross Millikan's answer, but there is a very easy calculation that gives this particular case:
{n≤x:P 1(n)>n−−√}=∑p#{n≤x,nx√⌊x/p⌋=x log 2+O(x/log x),#{n≤x:P 1(n)>n}=∑p#{n≤x,nx⌊x/p⌋=x log2+O(x/logx),
where the main term comes from Mertens' theorem on ∑p 1/p∑p 1/p and the error terms can be deduced from the Prime Number Theorem (or Chebyshev's upper bound on π(x)π(x)).
Here, by convention, p p is assumed to only take prime values. The reason this is so simple is that no n n here can have more than one prime factor >n−−√>n.
The answer to your second question is known as the Golomb-Dickman constant. Wikipedia gives it as about 0.62433 0.62433, but I doubt anything is known about its rationality, say.
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edited Apr 28, 2013 at 15:29
answered Apr 28, 2013 at 15:15
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Very nice, thanks.lhf –lhf 2013-04-29 00:58:40 +00:00 Commented Apr 29, 2013 at 0:58
Is there something you could suggest to further read about it?, it feels trivial but I get lost as to why it looks like ∑p≤x√(p−1)+∑p>x√⌊x/p⌋∑p≤x(p−1)+∑p>x⌊x/p⌋ and either how Mertens/PNT give the estimates, thanks.Dabed –Dabed 2020-03-14 16:57:32 +00:00 Commented Mar 14, 2020 at 16:57
1 @DanielD. If p≤x−−√p≤x then the nx−−√p>x then the n≤x n≤x condition dominates and the summand counts multiples of p p up to x x.Erick Wong –Erick Wong 2020-03-14 17:12:20 +00:00 Commented Mar 14, 2020 at 17:12
Thanks again that was useful Dabed –Dabed 2020-03-22 00:02:10 +00:00 Commented Mar 22, 2020 at 0:02
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In Hans Riesel, Prime Numbers and Computer Methods for Factorization, he gives a few approaches to largest and second largest prime factor. On pages 157-158, he gives a heuristic for a "typical" factorization, that suggests the largest gives
log P 1/log n≈1−1/e≈0.6321,logP 1/logn≈1−1/e≈0.6321,
log P 2/log n≈(1−1/e)/e≈0.2325.logP 2/logn≈(1−1/e)/e≈0.2325.
On page 161 he mentions that Knuth and Trabb-Pardo get 0.624,0.210 0.624,0.210 with a more rigorous argument. This is 1976, Theoretical Computer Science, volume 3, pages 321-348. Analysis of a Simple Factorization Algorithm. So I would say you want to get a copy of Knuth and Trabb-Pardo, which is reproduced, with later comments, in KNUTH
He then presents the Erdos-Kac theorem on pages 158-159, finally giving probability distribution curves for the three largest prime factors on page 163. These graphs would be what I call "cumulative distribution functions," being the integral of the "probability distribution function." These are also taken from Knuth and Trabb -Pardo. Let me make a jpeg.
KNOTE: The table on page 163 of ρ 1(α)ρ 1(α) agrees exactly with the table of ρ(u)ρ(u) in Erick's link on the Dickman-de Bruijn function. So, I think you have a winner.
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edited Apr 28, 2013 at 20:34
answered Apr 28, 2013 at 19:23
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I should have looked in Riesel. I had tried Pomerance but with not much luck. Thanks.lhf –lhf 2013-04-29 00:59:17 +00:00 Commented Apr 29, 2013 at 0:59
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In Mathworld it states that the probability that P 1(n)>n−−√P 1(n)>n is log 2 log2. The first few rough numbers are given in OEIS A064052 but there are no references.
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answered Apr 28, 2013 at 14:49
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Ah, I missed that when I searched. Thanks.lhf –lhf 2013-04-29 00:52:10 +00:00 Commented Apr 29, 2013 at 0:52
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Posting my first Math-StackExchange answer...
According to this section of Wikipedia, due to Dixon's theorem, the probability of largest prime factor of n n to be less than n 1/m n 1/m is approximately m−m m−m for any real m≥1 m≥1.
So probability of largest prime factor to be less than n−−√=n 1/2 n=n 1/2 is approximately 2−2=1/4=0.25 2−2=1/4=0.25. To be less than n−−√3=n 1/3 n 3=n 1/3 is approximately 3−3=1/27≈0.037 3−3=1/27≈0.037.
I don't know the details of this theorem, I just found this quotation in Wikipedia and thought it might be useful for you. Also I don't know how approximate is this formula.
I tried to check this formula experimentally and wrote Python code for that (using Pollard-Rho and Fermat algorithms). Don't know if according to rules it is allowed to post code here on Math-StackExchange, so providing just links:
You can see (run) my code in action here (and here is a copy of my code just in case if first link is broken, second link is not runnable).
Results for 10K checked 64-bit numbers here:
Checked nums: 10242
Expected: 1.0: 1.00000, 1.5: 0.54433, 2.0: 0.25000, 2.5: 0.10119, 3.0: 0.03704, 3.5: 0.01247, 4.0: 0.00391, 4.5: 0.00115, 5.0: 0.00032, 5.5: 0.00008
Actual: 1.0: 1.00000, 1.5: 0.51541, 2.0: 0.23203, 2.5: 0.09008, 3.0: 0.03202, 3.5: 0.01021, 4.0: 0.00321, 4.5: 0.00105, 5.0: 0.00038, 5.5: 0.00005
Here are pairs of m (from formula above) and probability. So Expected (by formula above) is close to Actual (experimental with factoring of 64-bit numbers), especially close for larger m. Maybe formula is more precise for larger than 64-bit numbers that I checked or for more amount of tested numbers.
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3142 | https://artofproblemsolving.com/wiki/index.php/Radical_axis?srsltid=AfmBOorrR_tcm2Hjjy2SglCiKDUM-WJAn7MH6ddORRKB1Qx1AmmM3P59 | Art of Problem Solving
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Radical axis
Contents
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1 Introduction
2 Definitions
3 Results
4 Proofs
5 Exercises
6 Problems
6.1 Simple
6.2 Intermediate
6.3 Olympiad
7 See Also
Introduction
The theory of radical axis is a priceless geometric tool that can solve formidable geometric problems fairly readily. Problems involving it can be found on many major math olympiad competitions, including the prestigious USAMO. Therefore, any aspiring math olympian should peruse this material carefully, as it may contain the keys to one's future success.
Not all theorems will be fully proven in this text. The objective of this document is to introduce you to some key concepts, and then give you a chance to derive some of the beautiful results on your own. In that way, you will understand and retain the information in here much more solidly. Finally, your newfound knowledge will be tested on a few challenging problems that are exemplary examples on how radical axis theory can be used and why it pertains to that situation. I hope after you read this text, you will become a better math student, armed with another tool to solve difficult problems. But, anyway, good luck. i
Definitions
The power of point with respect to circle (with radius and center ), which shall thereafter be dubbed , is defined to equal .
Note that the power of a point is negative if the point is inside the circle.
The radical axis of two non-concentric circles is defined as the locus of the points such that the power of with respect to and are equal. In other words, if are the center and radius of , then a point is on the radical axis if and only if (i.e., the radical axis is the line that one gets when you subtract the equations of two circles).
Results
Theorem 1: (Power of a Point) If a line drawn through point P intersects circle at points A and B, then .
Theorem 2: (Radical Axis Theorem)
a. The radical axis is a line perpendicular to the line connecting the circles' centers (line ).
b. If the two circles intersect at two common points, their radical axis is the line through these two points.
c. If they intersect at one point, their radical axis is the common internal tangent.
d. If the circles do not intersect, and if one does not fully contain the other, their radical axis is the perpendicular to through point A, the unique point on such that .
Theorem 3: (Radical Axis Concurrence Theorem) The three pairwise radical axes of three circles concur at a point, called the radical center.
Theorem 4: (Radical Centre of Intersecting Circles) (EGMO Theorem 2.9) Let and be two circles with centers and . Select two points A and B on and C and D on . Then the following are equivalent
lie on a circle with center not on line .
Lines and intersect on the radical axis of and .
Theorem 5: (EGMO Lemma 2.11) Let ABC be a triangle and consider a point in its interior. Suppose that is tangent to and , ray bisects
Proofs
Theorem 1 is trivial Power of a Point, and thus is left to the reader as an exercise. (Hint: Draw a line through P and the center.)
Theorem 2 shall be proved here. Assume the circles are and with centers and and radii and , respectively. (It may be a good idea for you to draw some circles here.)
First, we tackle part (b). Suppose the circles intersect at points and and point P lies on . Then by Theorem 1 the powers of P with respect to both circles are equal to , and hence by transitive . Thus, if point P lies on , then the powers of P with respect to both circles are equal.
Now, we prove the inverse of the statement just proved; because the inverse is equivalent to the converse, the if and only if would then be proven. Suppose that P does not lie on . In particular, line does not intersect X. Then intersects circles and a second time at distinct points and , respectively. (If is tangent to , for example, we adopt the convention that ; similar conventions hold for . Power of a Point still holds in this case. Also, notice that and cannot both equal , as cannot be tangent to both circles.) Because is not equal to , so does not equal , and thus by Theorem 1 is not congruent to , as desired. This completes part (b).
For the remaining parts, we employ a lemma:
Lemma 1: Let be a point in the plane, and let be the foot of the perpendicular from to . Then .
The proof of the lemma is an easy application of the Pythagorean Theorem and will again be left to the reader as an exercise.
Lemma 2: There is an unique point P on line such that .
Proof: First show that P lies between and via proof by contradiction, by using a bit of inequality theory and the fact that . Then, use the fact that (a constant) to prove the lemma.
Lemma 1 shows that every point on the plane can be equivalently mapped to a line on . Lemma 2 shows that only one point in this mapping satisfies the given condition. Combining these two lemmas shows that the radical axis is a line perpendicular to , completing part (a).
Parts (c) and (d) will be left to the reader as an exercise. (Also, try proving part (b) solely using the lemmas.)
Now, try to prove Theorem 3 on your own! (Hint: Let P be the intersection of two of the radical axes.)The proof of this theorem along with the proof of Theorem 4are given in EGMO as Example 2.7 and Theorem 2.9.
Theorem 5 Let intersect at at and let , . Clearly, is a radical axis of , . We see thatas desired.
Exercises
If you haven't already done so, prove the theorems and lemmas outlined in the proofs section. Note: No solutions will be provided to the following problems(laziness). If you are stuck, ask on the forum.
Problem 1. Two circles and intersect at and . Point is located on such that and ~~15~~. If the radius of is , find the radius of . Note: An error in this problem previously rendered it unsolvable.
Problem 2. Solve 2009 USAMO Problem 1. If you already know how to solve it.
Problem 3. Two circles P and Q with radii 1 and 2, respectively, intersect at X and Y. Circle P is to the left of circle Q. Prove that point A is to the left of if and only if .
Problem 4. Solve 2012 USAJMO Problem 1.
Problem 5. Does Theorem 2 apply to circles in which one is contained inside the other? How about internally tangent circles? Concentric circles?
Problem 6. Construct the radical axis of two circles. What happens if one circle encloses the other?
Problem 7. Solve 1995 IMO Problem 1 in two different ways. Compare your solutions with the solutions provided.
Problems
Simple
Let triangle points and be given. Denote
Prove that
Proof
WLOG, the order of the points is as shown on diagram.
Intermediate
2021 AIME Problem 13. Circles and with radii and , respectively, intersect at distinct points and . A third circle is externally tangent to both and . Suppose line intersects at two points and such that the measure of minor arc is . Find the distance between the centers of and .
Olympiad
2014 USAMO Problem 5. Let be a triangle with orthocenter and let be the second intersection of the circumcircle of triangle with the internal bisector of the angle . Let be the circumcenter of triangle and the orthocenter of triangle . Prove that the length of segment is equal to the circumradius of triangle .
2017 USAMO Problem 3. Let be a scalene triangle with circumcircle and incenter Ray meets at and again at the circle with diameter cuts again at Lines and meet at and is the midpoint of The circumcircles of and intersect at points and Prove that passes through the midpoint of either or
2012 IMO Problem 5. Let be a triangle with , and let be the foot of the altitude from . Let be a point in the interior of the segment . Let be the point on the segment such that . Similarly, let be the point on the segment such that . Let . Prove that .
See Also
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Choose the adjective petulant to describe a person or behavior that is irritable in a childish way.
The adjective petulant is a disapproving term used to describe a bad-tempered child, an adult behaving like an angry child or behavior of this type. It's one thing to be angry or annoyed but if someone is petulant, they're acting in an unreasonable or unjustified manner. Petulant came to English in the late 16th century from the Latin petulantem, "forward, insolent" but was not recorded to mean "childishly irritable" until the late 1700s.
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The Scarlet Letter
Nathaniel Hawthorne
After having a child out of wedlock, Hester Prynne is shunned by her Puritan community and forced to wear a scarlet "A" on her clothing — but Hester is not the only one who has transgressed. This classic novel explores guilt, sin, and hypocrisy.
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3144 | https://artofproblemsolving.com/wiki/index.php/Circular_Inversion?srsltid=AfmBOoojaOydvDqzj1dyFhg1PuHxB-qn-9be_b_8-ZCKKTfE3aSb3lWG | Art of Problem Solving
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Circular Inversion
Circular Inversion, sometimes called Geometric Inversion or simply Inversion, is a transformation where point in the Cartesian plane is transformed based on a circle with radius and center such that , where is the transformed point on the ray extending from through .
Note that , when inverted, transforms back to . All points outside of are transformed inside , and vice versa. Points on transform to themselves, meaning . Finally, the transformation of is debated on its existence. Some call the transformation the ideal point, which is infinitely far away and in every direction. Others claim that this point does not have an inverse.
Geometric Inversion technically refers to many different types of inversions, however, if Geometric Inversion is used without clarification, Circular Inversion is usually assumed.
Circular Inversion can be a very useful tool in solving problems involving many tangent circles and/or lines.
Contents
[hide]
1 Basics of Circular Inversion
1.1 Inversion of a Circle intersecting O
1.2 Inversion of a Circle not intersecting O
1.3 General Formula for the Radius of a Circle in Terms of the Radius of its Inverse Circle
1.4 Problems
1.4.1 Problem 1
Basics of Circular Inversion
Inversion of a Circle intersecting O
The first thing that we must learn about inversion is what happens when a circle which intersects the center of the inversion, , is inverted. Let us have circle , with diameter . is chosen arbitrarily on circle . Points and represent the inversions of and , respectively. is the radius of . We seek to show that circle inverts to a line perpendicular to through .
By the definition of inversion, we have and .
We can combine the two equations to get . Rewriting this gives:
Also, since is a diameter of circle , must be right.
Now, we consider and . They share an angle - , and we know that Therefore, we have SAS similarity. Therefore, must be right. From there, it follows that all points on circle will be inverted onto the line perpendicular to at .
Therefore, the inversion of circle becomes a line.
Note that, if circle extends beyond , the argument still holds. All one needs to do is shuffle things around.
Inversion of a Circle not intersecting O
Now, we study the inversion of a circle not intersecting the center of inversion. Let us have circle not intersecting , the center of , the circle which we invert around. The points where intersect circle are points and , respectively. Point is arbitrary and on circle . We invert points , , and , producing , , and , respectively. We draw and . Because is a diameter, must be right. We wish to show that circle inverts to another circle.
The definition of inversion tells us that . From here, we obtain that and By SAS symmetry (exploiting ), the ratios tell us that:
Therefore, we have and . Note that , which must equal . Therefore, . But . Therefore, . As this holds for any , all points on circle will invert to a point on a circle with diameter .
General Formula for the Radius of a Circle in Terms of the Radius of its Inverse Circle
This is how circular inversion is useful in the first place - we find the radius of an inverted circle to find the radius of the original circle. Let the original circle be and the inverted circle be , with radii of and , respectively. The radius of the circle of inversion is . We draw the tangent line of circle intersecting O. We know that this is also a tangent line to circle from the result from part 2 - the tangent line, by definition, intersects circle at exactly one point, and for every intersection point, part 2 says that there will be another intersection point. Therefore, the tangent line to circle intersects circle at exactly one point, necessitating this line to be a tangent line. Call the intersections and , respectively. We have .
We have and = . We can write an equation for by dividing:
From the definition of inversion, we have . Subsituting yields:
From Power of a Point, we know that , which equals . Subsistuting gives , and solving for gives:
Alternately,
Problems
Problem 1
In the figure below, semicircles with centers at and and with radii 2 and 1, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter . The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered at is drawn externally tangent to the two smaller semicircles and internally tangent to the largest semicircle. What is the radius of the circle centered at ?
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3145 | https://www.cancerresearchuk.org/about-cancer/hodgkin-lymphoma/stages | Stages of Hodgkin Lymphoma | Cancer Research UK
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Home About cancer Hodgkin lymphoma Stages of Hodgkin lymphoma
Stages of Hodgkin lymphoma
The stages of Hodgkin lymphoma tell you about the number and places in the body affected by lymphoma.
Knowing the stage of Hodgkin lymphoma helps your doctor decide what treatment you need.
Doctors use the Lugano classification system to stage Hodgkin lymphoma. There are 4 stages, from stage 1 to 4. Doctors can also simplify the staging into early, intermediate or advanced stage.
How doctors work out your stage
Doctors look at whether your lymphoma is on:
the same side of the diaphragm or
on both sides of the diaphragm
They also look at whether it is inside or outside the lymphatic system .
They will measure the size of the lymphoma. They do this by doing various tests, such as a CT or PET scan. Your doctor will also take your symptoms into consideration.
The diaphragm
The diaphragm (pronounced dia-fram) is the big breathing muscle that separates the chest from the tummy (abdominal) area.
Doctors use the diaphragm as a guide because it is about halfway down the body.
Inside or outside of the lymphatic system
Doctors look at whether your lymphoma is affecting your lymph nodes and organs inside the lymphatic system. These are called lymphatic sites. If it is affecting areas outside the lymphatic system they are called extranodal sites.
Lymphatic sites
Lymphatic sites include a group of lymph nodes or an organ in the lymphatic system, such as the:
thymus
spleen
tonsils
Extranodal sites
Extranodal sites are sometimes called extralymphatic sites. They are outside the lymphatic system and include the:
lungs
liver
blood
bone marrow
kidneys
brain
Your doctor may use the letter E (for extranodal) after the stage number if you have lymphoma outside the lymphatic system. For example stage 1E. Your doctor or nurse can explain what this means for you.
B symptoms
Your doctor will add the letter B to your stage (for example, stage 1B) if you have any of the following symptoms:
heavy sweating at night
high temperatures that come and go, often at night
unexplained weight loss, more than a tenth (10%) of your body weight in the last 6 months
If you don't have any of these symptoms your doctor will add the letter A to your stage (for example, stage 2A).
People with B symptoms may need more treatment than those without them.
Bulky disease
This means you have either:
lymphoma that is 10cm or more
lymphoma in the centre of your chest (mediastinum) which is a third of the width of your chest or bigger
Stage 1
This means that you have one of the following:
lymphoma in a single lymph node or one group of lymph nodes, or an organ of the lymphatic system (such as the thymus)
lymphoma in an extranodal site (this is called stage 1E)
Below is an example of stage 1 lymphoma.
Stage 2
This means one of the following:
your lymphoma is in two or more groups of lymph nodes
your lymphoma is in an extranodal site and one or more groups of lymph nodes (this is called stage 2E)
In both cases, the 2 sites of lymphoma are on the same side of the diaphragm.
Below is an example of stage 2.
Stage 3
This means that you have lymphoma on both sides of the diaphragm.
Below is an example of stage 3. The lymphoma is in lymph nodes above the diaphragm and the spleen.
Stage 4
Stage 4 means one of the following:
lymphoma is in the lymph nodes and an extranodal site
lymphoma is in more than one extranodal site, for example the liver, bones or lungs
Below is an example of stage 4 lymphoma.
Hodgkin lymphoma stages made simple
Your doctor may describe your lymphoma as being early, intermediate or advanced stage.
They will consider risk factors when deciding what stage your Hodgkin lymphoma is. The risk factors are:
having bulky disease in the centre of your chest (mediastinum)
being 50 years or older
if your red blood cells (erythrocytes) are sticking together (erythrocyte sedimentation rate or ESR
having lots of nodal areas with lymphoma
Early stage
Early (limited) stage generally means you have stage 1 or stage 2 Hodgkin lymphoma with no risk factors. It is also called early stage favourable.
Intermediate stage
Intermediate stage usually means you have stage 1 or 2 with one or more risk factor. Your doctor might call this stage early stage unfavourable Hodgkin lymphoma.
Advanced stage
Advanced stage generally means you have stage 3 or stage 4 Hodgkin lymphoma. However stage 2 with B symptoms and bulky disease or extranodal sites is usually treated as advanced stage (stage 2BX or 2BE).
Hodgkin lymphoma treatment
Treatment for stage 1 and 2 Hodgkin lymphoma is usually 2 to 4 cycles of chemotherapy. You might also have radiotherapy.
Treatment for stage 3 and 4 Hodgkin lymphoma is usually between 4 to 6 cycles of chemotherapy. You might also have:
steroids
radiotherapy
targeted therapy
Treatment for nodular lymphocyte-predominant Hodgkin lymphoma (NLPHL) is mostly the same as classical Hodgkin lymphoma. However if you have stage 1A NLPHL with no risk factors you might have radiotherapy on it’s own. You will usually have a PET-CT scan after some treatment so your doctor can see if your lymphoma has changed.
Read more about the treatments for Hodgkin lymphoma
Hodgkin lymphoma that comes back
Hodgkin lymphoma that has come back after it has been treated and gone away is called recurrent or relapsed disease.
Your doctors will not stage it in the same way as when you were first diagnosed. You can still have treatment for relapsed Hodgkin lymphoma and this will often work well. Your treatment may include:
chemotherapy, usually a high dose
a stem cell transplant
targeted therapy
immunotherapy
radiotherapy
For some people NLPHL can change (transform) into non Hodgkin lymphoma. If your doctor thinks your NLPHL has come back, you will have another biopsy and they will check if it has transformed. You will have different treatment if your lymphoma has transformed into non Hodgkin lymphoma.
References
AJCC Cancer Staging Manual (8th edition)
American Joint Committee on Cancer
Springer, 2017
Guideline for the first-line management of Classical Hodgkin Lymphoma — A British Society for Haematology guideline
George A. Follows and others
British journal of haematology, 2022. Volume 197, issue 5.
Essential Haematology (8th edition)
V Hoffbrand, P Moss, J Pettit
Wiley, 2019
Hodgkin lymphoma: ESMO Clinical Practice Guidelines for diagnosis, treatment and follow-up
D. A. Eichenauer and others
Annals of Oncology 29 (Supplement 4): 19–29, 2018
Recommendations for Initial Evaluation, Staging, and Response Assessment of Hodgkin and Non-Hodgkin Lymphoma: The Lugano Classification
Bruce D. Cheson, and others
Journal of Clinical Oncology, 2014. Volume 32, number 27
Last reviewed:
18 Jun 2024
Next review due:
18 Jun 2027
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Types of Hodgkin lymphoma
Treatment for Hodgkin lymphoma
Survival for Hodgkin lymphoma
Living with Hodgkin lymphoma
Symptoms of Hodgkin lymphoma
About Hodgkin lymphoma
Mental health and cancer
We know that it is common to struggle with your mental health when you have cancer or care for someone with cancer.
We have information to help
Related links
Types of Hodgkin lymphoma
The two types of Hodgkin lymphoma are classical Hodgkin lymphoma and nodular lymphocyte predominant Hodgkin lymphoma (NLPHL). There are also subtypes. Knowing the type helps your doctor choose the right treatment.
Treatment for Hodgkin lymphoma
Hodgkin lymphoma treatment is usually very successful and most people are cured. The treatment you have depends on the type and stage of your Hodgkin lymphoma, as well as your general health. You might have more than one treatment.
Survival for Hodgkin lymphoma
Survival for Hodgkin lymphoma is generally good, particularly if you are diagnosed early. However survival depends on many factors and no one can tell you exactly how long you will live. Find out more about survival.
Living with Hodgkin lymphoma
Get practical and emotional support to help you cope with a diagnosis of Hodgkin lymphoma, and life during and after treatment.
Symptoms of Hodgkin lymphoma
The most common symptom of Hodgkin lymphoma is swollen lymph nodes. Check the possible symptoms of Hodgkin lymphoma and when to see your doctor.
About Hodgkin lymphoma
Hodgkin lymphoma is a cancer of a type of white blood cell called lymphocytes. Find out more about the tests, treatments and support available if you have Hodgkin lymphoma.
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6.2.5 Jensen's Inequality
Remember that variance of every random variable $X$ is a positive value, i.e., \begin{align}%\label{} Var(X)=EX^2-(EX)^2 \geq 0. \end{align} Thus, \begin{align}%\label{} EX^2 \geq (EX)^2. \end{align} If we define $g(x)=x^2$, we can write the above inequality as \begin{align}%\label{} E[g(X)] \geq g(E[X]). \end{align} The function $g(x)=x^2$ is an example of convex function. Jensen's inequality states that, for any convex function $g$, we have $E[g(X)] \geq g(E[X])$. So what is a convex function? Figure 6.2 depicts a convex function. A function is convex if, when you pick any two points on the graph of the function and draw a line segment between the two points, the entire segment lies above the graph. On the other hand, if the line segment always lies below the graph, the function is said to be concave. In other words, $g(x)$ is convex if and only if $-g(x)$ is concave. We can state the definition for convex and concave functions in the following way:
Definition Consider a function $g: I \rightarrow \mathbb{R}$, where $I$ is an interval in $\mathbb{R}$. We say that $g$ is a convex function if, for any two points $x$ and $y$ in $I$ and any $\alpha \in [0,1]$, we have \begin{align}%\label{} g(\alpha x+ (1-\alpha)y) \leq \alpha g(x)+ (1-\alpha)g(y). \end{align} We say that $g$ is concave if \begin{align}%\label{} g(\alpha x+ (1-\alpha)y) \geq \alpha g(x)+ (1-\alpha)g(y). \end{align}
Note that in the above definition the term $\alpha x+ (1-\alpha)y$ is the weighted average of $x$ and $y$. Also, $\alpha g(x)+ (1-\alpha)g(y)$ is the weighted average of $g(x)$ and $g(y)$. More generally, for a convex function $g: I \rightarrow \mathbb{R}$, and $x_1$, $x_2$,...,$x_n$ in $I$ and nonnegative real numbers $\alpha_i$ such that $\alpha_1+\alpha_2+...+\alpha_n=1$, we have \begin{align}\label{eq:conv} g(\alpha_1 x_1+\alpha_2 x_2+...+\alpha_n x_n) \leq \alpha_1 g(x_1)+ \alpha_2 g(x_2)+...+\alpha_n g(x_n) &\qquad(6.4) \end{align} If $n=2$, the above statement is the definition of convex functions. You can extend it to higher values of $n$ by induction.
Now, consider a discrete random variable $X$ with $n$ possible values $x_1$, $x_2$,...,$x_n$. In Equation 6.4, we can choose $\alpha_i=P(X=x_i)=P_X(x_i)$. Then, the left-hand side of 6.4 becomes $g(EX)$ and the right-hand side becomes $E[g(X)]$ (by LOTUS). So we can prove the Jensen's inequality in this case. Using limiting arguments, this result can be extended to other types of random variables.
Jensen's Inequality:
If $g(x)$ is a convex function on $R_X$, and $E[g(X)]$ and $g(E[X])$ are finite, then \begin{align}%\label{} E[g(X)] \geq g(E[X]). \end{align}
To use Jensen's inequality, we need to determine if a function $g$ is convex. A useful method is the second derivative.
A twice-differentiable function $g: I \rightarrow \mathbb{R}$ is convex if and only if $g''(x) \geq 0$ for all $x \in I$.
For example, if $g(x)=x^2$, then $g''(x) =2 \geq 0$, thus $g(x)=x^2$ is convex over $\mathbb{R}$.
Example
Let $X$ be a positive random variable. Compare $E[X^a]$ with $(E[X])^{a}$ for all values of $ a\in \mathbb{R}$.
Solution
First note
| | |
--- |
| $E[X^a]=1=(E[X])^{a},$ | $\textrm{ if }a=0$, |
| $E[X^a]=EX=(E[X])^{a},$ | $\textrm{ if }a=1.$ |
So let's assume $a\neq 0,1$. Letting $g(x)=x^a$, we have \begin{align}%\label{} g''(x)=a(a-1)x^{a-2}. \end{align} On $(0, \infty)$, we can say $g''(x)$ is positive, if $a<0$ or $a>1$. It is negative, if $01$. It is concave, if $0| | |
--- |
| $E[X^a] \geq (E[X])^{a},$ | $\textrm{ if }a<0 \textrm{ or }a>1$, |
| $E[X^a] \leq (E[X])^{a},$ | $\textrm{ if }0 |
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【抽样调查】分层随机抽样
第2部分：分层随机抽样
第2部分：分层随机抽样
概述
简单估计量
简单估计量的性质
无偏性
方差
总值的相关推论
比例的相关推论
比率估计量
比率估计量的性质
期望与均方误差
分别比估计和联合比估计的比较
回归估计量
分别回归估计
联合回归估计
样本量分配
比例分配
最优分配与Neyman分配
总样本量的确定
指定方差上限
给定总费用
概述
分层随机抽样的思路：当(N,n)都较大，总体单元之间的差异也较大时，简单随机抽样会出现高成本、低精度情形，解决方法是将总体划分为若干个子总体、减少总体单元之间的差异。假设在各个子总体内已经满足实施简单随机抽样的条件，则可以在各个子总体内独立地进行简单随机抽样，再将各个子总体参数的估计值进行加权，得到总体参数的估计。
层：如果一个包含(N)个单位的总体可以分成不重不漏的(L)个子总体，即每个单元必定属于且仅属于一个子总体，则这样的子总体称为层。有(N_1+\cdots+N_L=N)。
分层抽样：在每一层中独立进行抽样，总的样本由各层样本组成，总体参数又按照各层样本参数的汇总作出估计。有(n_1+\cdots+n_L=n)。
分层随机抽样：每层的样本，都独立地按照简单随机抽样进行，这样的分层抽样称为分层随机抽样。
(h)：层。从而(N_h)代表第(h)层的单位总数，(n_h)代表第(h)层的样本数。
(i)：层内单位号。从而(Y_{hi})代表第(h)层第(i)个总体单元，(y_{hi})代表第(h)层第(i)个样本单元。
(W_h)：层权，即(W_h=\dfrac{N_h}{N})。
(f_h)：层内抽样比，即(f_h=\dfrac{n_h}{N_h})。
(\bar Y_h,Y_h,S_h^2)：层内总体参数（均值、总值与方差）。
(\bar y_h,y_h,s_h^2)：层内样本参数（样本均值、样本总值与样本方差）。
简单估计量
分层抽样首先根据各层的样本，计算出各层均值(\bar Y_h)的适当估计值(\hat {\bar Y}_h)，然后再使用总体层权加权平均，得到总体均值(\bar Y)的估计，即
[\hat {\bar Y}_{st}=\sum_{h=1}^{L}W_h\hat{\bar Y}_{h}=\frac{1}{N}\sum_{h=1}^{L}N_h\hat {\bar Y_h}. ]
对于分层随机抽样，每一层的(\hat{\bar Y}_h)就是(h)层的样本均值(\bar y_h)，即
[\hat{\bar Y}_{st}=\sum_{h=1}^{L}W_h\bar y_h=\frac{1}{N}\sum_{h=1}^{L}N_h\bar y_h. ]
注意这里的线性形式。
简单估计量的性质
无偏性
定理：对于分层随机抽样，(\hat{\bar Y}_{st})是(\bar Y)的无偏估计。
先证明关于总体均值的这个性质：总体均值等于各层均值关于层权的加权平均。
[\bar Y=\frac{1}{N}\sum_{i=1}^{N}Y_i=\frac{1}{N}\sum_{h=1}^{L}\sum_{i=1}^{N_h}Y_{hi}=\frac{1}{N}\sum_{h=1}^{L}N_h\bar Y_h=\sum_{h=1}^{L}W_h\bar Y_h. ]
由期望的线性运算性质，有
[\mathbb{E}\left(\sum_{h=1}^{L}W_h\hat{\bar Y}_h \right)=\sum_{h=1}^{L}W_h\mathbb{E}(\bar y_h)=\sum_{h=1}^{L}W_h\bar Y_h=\bar Y. ]
方差
先回顾一个结果：对于简单随机抽样，(\bar y)的方差是
[\mathbb{D}(\bar y)=\frac{1-f}{n}S^2. ]
定理：对于分层抽样，有
[\mathbb{D}(\hat{\bar Y}_{st})=\sum_{h=1}^{L} W_h^2\mathbb{D}(\hat{\bar Y}_{h}) ]
特别对分层随机抽样，记(\bar y_{st})为简单估计量，有
[\mathbb{D}(\bar y_{st})=\sum_{h=1}^{L}W_h^2\frac{1-f_h}{n_h}S_h^2=\sum_{k=1}^{L}\left(\frac{1}{n_h}-\frac{1}{N_h} \right)W_h^2S_h^2=\sum_{h=1}^{L}\frac{W_h^2S_h^2}{n_h}-\sum_{h=1}^{L}\frac{W_hS_h^2}{N}. ]
这里
[S_h^2=\frac{1}{N_h-1}\sum_{i=1}^{N_h}(Y_{hi}-\bar Y_h)^2. ]
第一个等号，因为是简单随机抽样，使用(\dfrac{1-f_h}{n_h}S_h^2)直接替代(\mathbb{D}(\hat{\bar Y}_{h}))得到。
第二个等号，显然
[\frac{1-f_h}{n_h}=\frac{N_h-n_h}{n_hN_h}=\frac{1}{n_h}-\frac{1}{N_h}. ]
第三个等号，只需注意到
[\sum_{h=1}^{L}\frac{W_h^2S_h^2}{N_h}=\sum_{h=1}^{L}\frac{N_h}{N}\frac{W_hS_h^2}{N_h}=\sum_{h=1}^{L}\frac{W_hS_h^2}{N}. ]
在简单随机抽样中，我们有：(\mathbb{E}(s^2)=S^2)，因此对(\mathbb{D}(\bar y_{st}))可以有下述的无偏估计。
定理：对于分层随机抽样，(\mathbb{D}(\bar y_{st}))的无偏估计量为
[\hat {\mathbb{D}}(\bar y_{st})=v(\bar y_{st})=\sum_{h=1}^{L}W_h^2\frac{1-f_h}{n_h}s_h^2=\sum_{h=1}^{L}\left(\frac{1}{n_h}-\frac{1}{N_h} \right)W_h^2s_h^2=\sum_{k=1}^{L}\frac{W_h^2s_h^2}{n_h}-\sum_{k=1}^{L}\frac{W_hs_h^2}{N}. ]
这里(s_h^2)是第(h)层样本的样本方差。
[s_h^2=\frac{1}{n_h-1}\sum_{i=1}^{n_h}(y_{hi}-\bar y_h)^2. ]
这里只是直接用(s_h^2)替代了(S_h^2)，应注意(n_h\ne 0)，否则无法计算(s_h^2)。
由此，能够给出总体均值的(1-\alpha)置信区间：
[\left[\bar y_{st}- z_{\alpha/2}\sqrt{v(\bar y_{st})},\quad \bar y_{st}+ z_{\alpha/2}\sqrt{v(\bar y_{st})}\right] ]
总值的相关推论
推论：对于分层随机抽样，总体总量(Y)的简单估计量(\hat Y_{st}=N\bar y_{st})有如下性质：
(\mathbb{E}(\hat Y_{st})=Y)。
(\mathbb{D}(\hat Y_{st})=\displaystyle\sum_{h=1}^{L}N_h(N_h-n_h)\dfrac{S_h^2}{n_h})。
(v(\hat Y_{st})=\displaystyle\sum_{h=1}^{L}N_h(N_h-n_h)\dfrac{s_h^2}{n_h})。
直接运用期望、方差的线性运算公式即可，注意到(W_h=\dfrac{N_h}{N})，有
[\mathbb{E}(\hat Y_{st})=N\mathbb{E}(\bar y_{st})=N\bar Y=Y,\ \begin{aligned} \mathbb{D}(\hat Y_{st})&=N^2\mathbb{D}(\bar y_{st})\ &=N^2\sum_{h=1}^{L}W_h^2\frac{1-f_h}{n_h}S_h^2\ &=\sum_{h=1}^{L}N_h^2\frac{N_h-n_h}{n_hN_h}S_h^2\ &=\sum_{h=1}^{L}N_h(N_h-n_h)\frac{S_h^2}{n_h}. \end{aligned} ]
对(v(\hat Y_{st}))，直接以(s_h^2)替代(\mathbb{D}(\hat Y_{st}))中的(S_h^2)即可。
比例的相关推论
比例是特殊的均值，只是此时的变量全部是(0-1)变量。在此先回顾一下关于(0-1)变量的相关结果，记(p=\bar y_{0-1})。
[\mathbb{E}(p)=P,\ S^2=\frac{1}{N-1}NP(1-P),\ s^2=\frac{1}{n-1}np(1-p),\ \mathbb{D}(p)=\frac{1-f}{n}\frac{N}{N-1}P(1-P),\ v(p)=\frac{1-f}{n-1}p(1-p). ]
推论：对于分层随机抽样，总体比例(P)的简单估计量(p_{st}=\displaystyle\sum_{h=1}^{L}W_hp_h)有如下性质：
(\mathbb{E}(p_{st})=P)。
(\mathbb{D}(p_{st})=\dfrac{1}{N^2}\displaystyle\sum_{h=1}^{L}N_h^2\dfrac{N_h-n_h}{N_h-1}\dfrac{P_hQ_h}{n_h})。记(Q_h=1-P_h)。
(v(p_{st})=\dfrac{1}{N^2}\displaystyle\sum_{h=1}^{L}\dfrac{N_h(N_h-n_h)}{n_h-1}p_hq_h)。记(q_h=1-p_h)。
如果(N_h)很大，则(N_h\approx N_h-1)，于是
[\mathbb{D}(p_{st})\approx\frac{1}{N^2}\sum_{h=1}^{L}N_h(N_h-n_h)\frac{P_hQ_h}{n_h}=\sum_{h=1}^{L}W_h^2\frac{1-f}{n_h} P_hQ_h. ]
期望是显然的，注意到(S_h^2)可以被(\dfrac{N_hP_hQ_h}{N_h-1})所替代，于是
[\begin{aligned} \mathbb{D}(p_{st})=\sum_{h=1}^{L}W_h^2\dfrac{1-f_h}{n_h}S_h^2=\sum_{h=1}^{L}W_h^2\frac{1-f_h}{n_h}\frac{N_hP_hQ_h}{N_h-1}=\frac{1}{N^2}\sum_{h=1}^{L}N_h^2\frac{N_h-n_h}{N_h-1}\frac{P_hQ_h}{n_h}. \end{aligned} ]
对(v(p_{st}))，就用(\dfrac{n_hp_hq_h}{n_h-1})替代(\mathbb{D}(p_{st}))中的(\dfrac{N_hP_hQ_h}{N_h-1})，从而
[v(p_{st})=\frac{1}{N^2}\sum_{h=1}^{L}N_h^2\frac{N_h-n_h}{N_h-1}\frac{P_hQ_h}{n_h}\frac{\frac{n_hp_hq_h}{n_h-1}}{\frac{N_hP_hQ_h}{N_h-1}}=\frac{1}{N^2}\sum_{h=1}^{L}\frac{n_h(N_h-n_h)p_hq_h}{n_h-1}. ]
对于总体中具有指定特征的单元总数(A)，由于(A=NP)，故对应的有(a_{st}=Np_{st})。
比率估计量
比率估计量与分层随机抽样有两种结合方式，所得的估计量称为分别比估计(separate ratio estimator)和联合比估计(combined ratio estimator)。
分别比估计：对每一层样本分别考虑比估计量，然后对各层的比估计量进行加权平均，即先比后加权。
联合比估计：对比率的分子和分母分别加权计算出总体均值或总体总量的分层估计量，然后用对应的分层估计量来构造比估计，即先加权后比。
分别比估计：
[\bar y_{RS}=\sum_{h=1}^{L}W_h\bar y_{Rh}=\sum_{h=1}^{L}W_h\frac{\bar y_h}{\bar x_h}\bar X_h,\ \hat Y_{RS}=N\bar y_{RS}=\sum_{h=1}^{L}N_h\bar y_{Rh}=\sum_{h=1}^{L}\frac{\bar y_h}{\bar x_h}X_h=\sum_{h=1}^{L}\hat Y_{Rh}. ]
联合比估计：
[\hat R_c\xlongequal{def}\frac{\bar y_{st}}{\bar x_{st}}, \ \bar y_{RC}=\frac{\bar y_{st}}{\bar x_{st}}\bar X=\hat R_c\bar X,\ \hat Y_{RC}=N\bar y_{RC}=N\frac{\bar y_{st}}{\bar x_{st}}\bar X=\frac{\bar y_{st}}{\bar x_{st}}X=\hat R_cX. ]
比率估计量的性质
期望与均方误差
先回顾简单随机抽样里，关于比估计量期望、方差的性质。
[\mathbb{E}(r)\approx R,\ \mathbb{D}(r)\approx\frac{1}{\bar X^2}\frac{1-f}{n}(S_y^2-2RS_{yx}+R^2S_x^2).\ \mathbb{E}(\bar y_{R})\approx R\bar X=\bar Y,\ \mathbb{D}(\bar y_{R})\approx \frac{1-f}{n}(S_y^2-2RS_{yx}+R^2S_x^2). ]
定理：对于分层随机抽样的分别比估计，若各层的样本量(n_h)都比较大，则有
[\mathbb{E}(\bar y_{RS})\approx \bar Y,\ \mathrm{MSE}(\bar y_{RS})\approx\mathbb{D}(\bar y_{RS})\approx\sum_{h=1}^{L}\frac{W_h^2(1-f_h)}{n_h}(S^2_{yh}+R_h^2S_{xh}^2-2R_h\rho_hS_{yh}S_{xh}). ]
这里(f_h)是第(h)层的抽样比，(S_{xh}^2,S_{yh}^2,\rho_h)分别是第(h)层指标(X,Y)的方差以及它们的相关系数，
[R_h=\frac{\bar Y_h}{\bar X_h}=\frac{Y_h}{X_h},\quad \rho_h=\frac{S_{xyh}}{S_{xh}S_{yh}}. ]
估计均方误差时，通常分别用(s_{xh}^2,s_{yh}^2)作为(S_{xh}^2,S_{yh}^2)的估计，用(s_{xyh})作为(S_{xyh})的估计，用(\hat R_{h})作为(R_h)的估计。
只要运用分层比估计的线性性质即可，有
[\mathbb{E}(\bar y_{RS})=\sum_{h=1}^{L}W_h\mathbb{E}(\bar y_{Rh})\approx\sum_{h=1}^{L}W_h\bar Y_h=\bar Y,\ \mathrm{MSE}(\bar y_{RS})\approx\mathbb{D}(\bar y_{RS})=\sum_{h=1}^{L}W_h^2\mathbb{D}(\bar y_{RS})\approx\sum_{h=1}^{L}W_h^2\frac{1-f_h}{n_h}(S_{yh}^2-2R_hS_{xyh}+R_h^2S_{xh}^2). ]
定理：对于分层随机抽样的联合比估计，若总样本量(n)较大，则有
[\mathbb{E}(\bar y_{RC})\approx \bar Y,\ \mathrm{MSE}(\bar y_{RC})\approx \mathbb{D}(\bar y_{RC})\approx \sum_{h=1}^{L}\frac{W_h^2(1-f_h)}{n_h}(S_{yh}^2+R^2S_{xh}^2-2R\rho_hS_{yh}S_{xh}). ]
注意均方误差处，(RC)与(RS)的主要区别在于(R)与(R_h)。对(R)的估计，一般使用(\hat R_c=\dfrac{\bar y_{st}}{\bar x_{st}})。
联合比估计的证明与分层比估计略有不同，注意(n)较大时(\bar x_{st}\approx \bar X)。
[\bar y_{RC}-\bar Y=\frac{\bar y_{st}}{\bar x_{st}}\bar X-R\bar X=\frac{\bar X}{\bar x_{st}}(\bar y_{st}-R\bar x_{st})\approx \bar y_{st}-R\bar x_{st}. \ \mathbb{E}(\bar y_{RC}-\bar Y)\approx\mathbb{E}(\bar y_{st}-R\bar x_{st})=\bar Y-R\bar X=0,\ \Downarrow \ \mathbb{E}(\bar y_{RC})\approx \bar Y. ]
令(G_{hi}=Y_{hi}-RX_{hi})，则(\bar G_h=\bar Y_h-R\bar X_h)；令(\bar g_{st}=\bar y_{st}-R\bar x_{st})，则(\mathbb{E}(\bar g_{st})=0)。因此
[\mathbb{D}(\bar y_{RC})\approx \mathbb{E}(\bar y_{RC}-\bar Y)^2\approx \mathbb{E}(\bar y_{st}-R\bar x_{st})^2=\mathbb{E}(\bar g_{st}^2)=\mathbb{D}(\bar g_{st}). ]
于是
[\begin{aligned} \mathbb{D}(\bar y_{RC})\approx \mathbb{D}(\bar g_{st})& =\sum_{h=1}^{L}W_h^2\frac{1-f_h}{n_h}S_{gh}^2\ &=\sum_{h-1}^{L}\frac{W_h^2(1-f_h)}{n_h}\left[\frac{1}{N_h-1}\sum_{i=1}^{N_h}(G_{hi}-\bar G)^2 \right]\ &=\sum_{h-1}^{L}\frac{W_h^2(1-f_h)}{n_h}\left[\frac{1}{N_h-1}\sum_{i=1}^{N_h}[(Y_{hi}-\bar Y)-R(X_{hi}-\bar X)]^2 \right]\ &=\sum_{h-1}^{L}\frac{W_h^2(1-f_h)}{n_h}(S_{yh}^2-2RS_{xyh}+R^2S_{xh}^2). \end{aligned} ]
分别比估计和联合比估计的比较
如果(R=R_h)，即每一层的总体比值都严格等于整个总体的比值，或者(\rho_h=\dfrac{R+R_h}{2}\dfrac{S_{xh}}{S_{yh}})时，分别比估计的精度与联合比估计精度是一样的。
当各层的(n_h)都比较大时，采用分别比估计更有效。当某些层的样本量(n_h)不够大时，采用联合比估计更有效。
回归估计量
回归估计量中，先回归后加权的称为分别回归估计(separate regression estimator)，先加权后回归的称为联合回归估计(combined regression estimator)，这与比率估计量类似。
分别回归估计
分别回归估计为
[\bar y_{lrs}=\sum_{h=1}^{L}W_h\bar y_{lrh}=\sum_{h=1}^{L}W_h[\bar y_h+\beta_h(\bar X_h-\bar x_h)],\ \hat Y_{lrs}=\sum_{h=1}^{L}N_h\bar y_{lrs}=\sum_{h=1}^{L}N_h[\bar y_h+\beta_h(\bar X_h-\bar x_h)]. ]
这里(\beta_h)是各层的回归系数，分层回归估计时各层的回归系数可以不相同，这种情况下分别回归估计量更合适。
回顾简单随机抽样的回归估计(\hat y_{lr}=\bar Y+\beta(\bar X-\bar x))，回归估计的显然是参数的无偏估计，且方差为
[\mathbb{D}(\bar y_{lr})=\frac{1-f}{n}(S_y^2-2\beta S_{yx}+\beta^2S_{x}^2). ]
而分别回归估计回归估计的简单加权平均，故自然地成立以下定理。
定理：当各层的回归系数(\beta_h)是事先给定的常数时，
[\mathbb{E}(\bar y_{lrs})=\bar Y,\ \mathbb{D}(\bar y_{lrs})=\sum_{h=1}^{L}\frac{W_h^2 (1-f_h)}{n_h}(S_{yh}^2+\beta_h^2S_{xh}^2-2\beta_hS_{xyh}). ]
且为使方差最小，应取
[\beta_h=\frac{S_{xyh}}{S_{xh}^2}=B_h, ]
此时对应的有
[\min \mathbb{D}(\bar y_{lrs})=\sum_{h=1}^{L}\frac{W_h^2(1-f_h)}{n_h}S_{yh}^2(1-\rho_h^2). ]
如果(\beta_h)不能事先设定，则取(\beta_h)为(B_h)的最小二乘估计(b_h)，即样本回归系数：
[b_h=\frac{\mathrm{cov}(x,y)}{s_x^2}=\frac{\displaystyle\sum_{i=1}^{n_h}(y_{hi}-\bar y)(x_{hi}-\bar x)}{\displaystyle\sum_{i=1}^{n_h}(x_{hi}-\bar x_h)^2}. ]
联合回归估计
联合回归估计，即先对(\bar Y)和(\bar X)作分层估计，进而构造总体均值的联合回归估计(\bar y_{lrc})。
[\bar y_{lrc}=\bar y_{st}+\beta(\bar X-\bar x_{st}). ]
定理：如(\beta)是设定的常数，则
[\mathbb{E}(\bar y_{lrc})=\bar Y,\ \mathbb{D}(\bar y_{lyc})=\sum_{h=1}^{L}\frac{W_h^2(1-f_h)}{n_h}(S_{yh}^2+\beta^2S_{xh}^2-2\beta S_{xyh}). ]
由于比率估计是回归估计的一种特例，故证明步骤也类似，即令(G_{hi}=Y_{hi}+\beta(\bar X-X_{hi}))，于是(\bar G_{h}=\bar Y_h+\beta(\bar X-\bar X_h))。同时对所抽取的样本，(g_{hi}=y_{hi}+\beta(\bar X-x_{hi}))，从而(\bar g_{st}=\bar y_{st}+\beta(\bar X-x_{st})=\bar y_{lrc})。
从而
[\begin{aligned} \mathbb{D}(\bar g_{st})&=\sum_{h=1}^{L}W_h^2\frac{1-f_h}{n_h}S_{gh}^2\ &=\sum_{h=1}^{L}W_h^2\frac{1-f_h}{n_h}\frac{1}{N_h-1}\sum_{i=1}^{N_h}[(Y_{hi}-\bar Y)-\beta( X_{hi}-\bar X)]^2\ &=\sum_{h=1}^{L}W_h^2\frac{1-f_h}{n_h}(S_{yh}^2+\beta^2S_{xh}^2-2\beta S_{xyh}). \end{aligned} ]
利用最小二乘法，可以给出(\mathbb{D}(\bar g_{st}))取最小值时，(\beta)的最小二乘解(B_c)为
[B_c=\frac{\displaystyle\sum_{h=1}^{L}\left(\frac{W_h^2(1-f_h)S_{xyh}}{n_h} \right)}{\displaystyle\sum_{h=1}^{L}\left(\frac{W_h^2(1-f_h)S_{xh}^2}{n_h} \right)}. ]
如果(\beta)未知，则取(B_c)的样本估计(b_c)来替代。
样本量分配
比例分配
比例分配(proportional allocation)指
[\frac{n_h}{N_h}=\frac{n}{N},\quad f_h=f,\quad h=1,2,\cdots,L. ]
自加权：若总体总量的一个无偏估计量可以表示成样本基本单元的变量值总值(\hat Y)（或均值(\bar y)）的一个常数倍，即(\hat Y=ky)或(\hat {\bar Y}=k\bar y)，则称这种估计量为自加权（等加权）的。
[\bar y_{prop}=\sum_{h=1}^{L}W_h\bar y_h=\sum_{h=1}^{L}\frac{n_h}{n}\left(\frac{1}{n_h}\sum_{i=1}^{n_h}y_{hi} \right)=\frac{1}{n}\sum_{h=1}^{L}\sum_{i=1}^{n_h}y_{hi}=\bar y, \ \hat Y_{prop}=N\bar y_{prop}=\frac{N}{n}y=\frac{1}{f}y. ]
最优分配与Neyman分配
最优分配(optimum allocation)指：在分层随机抽样中，对于给定的费用，使得估计量的方差(\mathbb{D}(\bar y_{st}))达到最小；或者给定的估计量方差(V)，使总费用达到最小的各层样本量分配。
总费用函数：以线性函数为例，可表示为
[C_T=c_0+\sum_{h=1}^{L}c_hn_h. ]
由于
[\mathbb{D}(\bar y_{st})=\sum_{h=1}^{L}\frac{W_h^2S_h^2}{n_h}-\sum_{h=1}^{L}\frac{W_hS_h^2}{N}, ]
所以构造目标函数为
[C'V'=(C_T-c_0)\left(V+\sum_{h=1}^{L}\frac{W_hS_h^2}{N} \right), ]
对此目标函数求极小值，得到下面的定理。
定理：对于分层随机抽样，若费用函数为线性的，目标函数如下：
[C'V'=(C_T-c_0)\left(V+\sum_{h=1}^{L}\frac{W_hS_h^2}{N} \right), ]
则最优分配为
[f_h=\frac{n_h}{n}=\frac{W_hS_h/\sqrt{c_h}}{\displaystyle\sum_{h=1}^{L}\dfrac{W_hS_h}{\sqrt{c_h}}}. ]
特别当各层的单位抽样费用相等，即(c_n=c)时，有
[f=\frac{n_h}{n}=\frac{W_hS_h}{\displaystyle\sum_{h=1}^{L}W_hS_h}=\frac{N_hS_h}{\displaystyle\sum_{h=1}^{L}N_hS_h}. ]
此为Neyman分配。
可将目标函数改写为
[C'V'=\left(\sum_{h=1}^{L}c_hn_h \right)\left(\sum_{h=1}^{L}\frac{W_h^2S_h^2}{n_h} \right)=\left[\sum_{h=1}^{L}(\sqrt{c_hn_h})^2 \right]\left[\sum_{h=1}^{L}\left(\frac{W_hS_h}{\sqrt{n_h}} \right)^2 \right], ]
由Cauchy-Schwarz不等式，((\sum a_h^2)(\sum b_h^2)\ge \sum(a_hb_h)^2)，当且仅当(a_h/b_h=K)为一常数时等号成立，故
[C'V'\ge \left(\sum_{h=1}^{L}\sqrt{c_h}W_hS_h \right)^2, ]
[\frac{\sqrt{c_h}n_h}{W_hS_h}=K ]
为常数时成立，故(n_h=\dfrac{KW_hS_h}{\sqrt{c_n}})。
总样本量的确定
指定方差上限
如果指定了方差上限(V)，则
[V=\sum_{h=1}^{L}\frac{W_h^2S_h^2}{n_h}-\sum_{h=1}^{L}\frac{W_hS_h^2}{N}. ]
对于确定的样本量分配：(n_h=nw_h)，则
[V=\frac{1}{n}\sum_{h=1}^{L}\frac{W_h^2S_h^2}{w_h}-\sum_{h=1}^{L}\frac{W_h^2S_h^2}{N},\ n=\frac{\displaystyle\sum_{h=1}^{L}\frac{W_h^2S_h^2}{w_h}}{V+\dfrac{1}{N}\displaystyle\sum_{h=1}^{L}W_h^2S_h^2}. ]
对于比例分配：(n_h=nW_h)，则
[V=\sum_{h=1}^{L}\frac{W_hS_h^2}{n}-\sum_{h=1}^{L}\frac{W_hS_h^2}{N},\ n=\frac{\displaystyle\sum_{h=1}^{L}W_hS_h^2}{V+\dfrac{1}{N}\displaystyle\sum_{h=1}^{L}W_hS_h^2}\xlongequal{n_0=\frac{1}{V}\sum\limits_{h=1}^{L} W_hS_h^2}\frac{n_0}{1+n_0/N}. ]
对于内曼分配：(w_h=\dfrac{W_hS_h}{\sum_{h=1}^{L}W_hS_h})，有
[n=\frac{\left(\displaystyle\sum_{h=1}^{L}W_hS_h \right)^2}{V+\dfrac{1}{N}\displaystyle\sum_{h=1}^{L}W_hS_h^2}. ]
如果给定的是绝对误差限，也可以转化为指定(V)的情况，此时
[\mathbb{P}(|\bar y_{st}-\bar Y|\le d)=1-\alpha\Leftrightarrow \mathbb{P}\left(\left|\frac{\bar y_{st}-\bar Y}{\sqrt{\mathbb{D}(\bar y_{st})}} \right|\le \frac{d}{\sqrt{\mathbb{D}(\bar y_{st})}} \right)=1-\alpha,\ \mathbb{D}(\bar y_{st})=\frac{d^2}{(z_{\alpha/2})^2}. ]
如给定的是相对误差限，则结合(d=r\bar Y)，还需要对(\bar Y)进行估计。
给定总费用
[C_T=c_0+\sum_{h=1}^{L}c_hn_h, ]
[n_h=\frac{kW_hS_h}{\sqrt{c_h}}, ]
于是
[C_T-c_0=\sum_{h=1}^{L}c_h\left(\frac{kW_hS_h}{\sqrt{c_h}} \right)=k\sum_{h=1}^{L}\sqrt{c_h}W_hS_h,\ k=\frac{C_T-c_0}{\sum_{h=1}^{L}\sqrt{c_h}W_hS_h},\quad n_h=\frac{(C_T-c_0)W_hS_h/\sqrt{c_h}}{\sum_{h=1}^{L}\sqrt{c_n}W_hS_h},\ n=\sum_{h=1}^{L}n_h=(C_T-c_0)\frac{\displaystyle\sum_{h=1}^{L}\frac{W_hS_h}{\sqrt{c_n}}}{\displaystyle\sum_{h=1}^{L}\sqrt{c_n}W_hS_h}. ]
posted @ 2021-04-12 18:51 江景景景页 阅读(3711) 评论(1) 收藏 举报
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Does cosine similarity imply collinearity even for Hilbert spaces?
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If two vectors are cosine similar, that is the angle between them is 0, then the vectors are collinear. For Non-zero vectors, then one vector is the scalar multiple of the other.
Does this still apply to general Hilbert spaces? If not, then in which function spaces and what inner products can we conclude that one function is a scalar multiple of the other if they're cosine similar?
edit: i.e are functions f f and g g scalar multiples of each other when
⟨f,g⟩∥f∥∥g∥=1⟨f,g⟩‖f‖‖g‖=1
linear-algebra
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edited Feb 2, 2024 at 2:07
Rain ZhaoRain Zhao
asked Feb 2, 2024 at 1:36
Rain ZhaoRain Zhao
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1 do you mean when you have equality in the cauchy schwarz inequality?Mr. Cooperman –Mr. Cooperman 2024-02-02 01:48:08 +00:00 Commented Feb 2, 2024 at 1:48
@Timkinsella I mean if ⟨f,g⟩∥f∥∥g∥=1⟨f,g⟩‖f‖‖g‖=1 Rain Zhao –Rain Zhao 2024-02-02 02:04:39 +00:00 Commented Feb 2, 2024 at 2:04
1 google the phrase "equality in the cauchy schwarz inequality" and stuff should come up Mr. Cooperman –Mr. Cooperman 2024-02-02 02:08:24 +00:00 Commented Feb 2, 2024 at 2:08
@Timkinsella thank you!Rain Zhao –Rain Zhao 2024-02-02 12:07:04 +00:00 Commented Feb 2, 2024 at 12:07
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Yes, it is true that f f and g g are necessarily multiples of each other if
∣∣∣⟨f,g⟩∥f∥∥g∥∣∣∣=1.|⟨f,g⟩‖f‖‖g‖|=1.
This result is sometimes referred to as the "equality case" of the Cauchy-Bunyakovsky-Schwarz inequality.
More generally, the CBS inequality states that |⟨f,g⟩|≤∥f∥∥g∥|⟨f,g⟩|≤‖f‖‖g‖, which is equivalent to saying that the cosine similarity always has absolute value less than or equal to 1 1. The result regarding the equality case is that |⟨f,g⟩|=∥f∥∥g∥|⟨f,g⟩|=‖f‖‖g‖ if and only if f f and g g are scalar multiples of each other.
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answered Feb 2, 2024 at 2:13
Ben GrossmannBen Grossmann
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Thank you so much!Rain Zhao –Rain Zhao 2024-02-02 12:08:14 +00:00 Commented Feb 2, 2024 at 12:08
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3150 | https://dhananjayparkar.wordpress.com/portfolio/dimension-analysis-questions-and-answers/ | Dhananjay Parkar
Physics has made our life easy
Dimension Analysis Questions and Answers
Q1: Define the term ‘Dimension’
Answer: The term ‘dimension’ is used to refer to the physical nature of a quantity and the type of unit used to specify it. Mathematically dimensions of a physical quantity are the powers to which the fundamental quantities must be raised.
e.g. Dimension of velocity = Displacement / time = [L]/[T] = [M0][L1][T-1]
Q2: What are dimensional constants?
Answer: Constants which possess dimensions are called dimensional constants. E.g. Planck’ Constant.
Q3: What are dimensional variables?
Answer: Those physical quantities which possess dimensions but do not have a fixed value are called dimensional variables. E.g. Displacement, Force, velocity etc.
Q4: What are dimensionless quantities?
Answer: Physical quantities which do not possess dimensions are called dimensionless quantities. E.g. Angle, specific gravity, strain. In general, physical quantity which is a ratio of two quantities of same dimension will be dimensionless.
Q5: Define the principle of homogeneity of dimensions. On What principle is it based?
Answer: The principle of homogeneity of dimensions states that an equation is dimensionally correct if the dimensions of the various terms on either side of the equation are the same.
This principle is based on the fact that two quantities of the same dimension only can be added up, and the resulting quantity also possess the the same dimension.
i.e. In equation X + Y = Z is valid if the dimensions of X, Y and Z are same.
Q6: Who introduced Dimension Analysis
Answer: Fourier (Joseph Fourier – French Mathematician)
Q7: List the basic dimensions.
Answer:
Q8: What are the uses (applications) of dimensional analysis?
Answer: The applications of dimensional analysis are:
To convert a physical quantity from one system of units to another.
To check the dimensional correctness of a given equation.establish a relationship between different physical quantities in an equation.
Q9 (NCERT): A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion:
(a) y = a sin 2π t/T
(b) y = a sin vt
(c) y = (a/T) sin t/a
(d) y = (a 2) (sin 2πt / T + cos 2πt / T )
(a = maximum displacement of the particle, v = speed of the particle. T = time-period
of motion). Rule out the wrong formulas on dimensional grounds.
Answer:
Given,
Dimension of a = displacement = [M0L1T0]
Dimension of v (speed) = distance/time = [M0L1T-1]
Dimension of t or T (time period) = [M0L0T1]
Trigonometric function sine is a ratio, hence it must be dimensionless.
(a) y = a sin 2π t/T (correct ✓ )
Dimensions of RHS = [L1] sin([T].[T-1] ) = [M0L1T0] = LHS (eqation is correct).
(b) y = a sin vt (wrong ✗)
RHS = [L1] sin([LT-1] [T1]) = [L1] sin([L]) = wrong, since trigonometric function must be dimension less.
(c) y = (a/T) sin t/a (wrong ✗)
RHS = [L1] sin([T].[L-1] ) = [L1] sin([TL-1] ) = wrong, sine function must be dimensionless.
(d) y = (a 2) (sin 2πt / T + cos 2πt / T ) (correct ✓ )
RHS = [L1] ( sin([T].[T-1] + cos([T].[T-1] ) = [L1] ( sin(M0L1T0) + cos(M0L1T0) )
= [L1] = RHS = equation is dimensionally correct.
Q10(NCERT): A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ mo of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes
Answer:
Dimension of m (mass) = [M1L0T0]
Dimension of m0 (mass) = [M1L0T0]
Dimension of v (velocity) = [M0L1T-1]
∴ Dimension of v2 = [M0L2T-2]
Dimension of c (velocity) = [M0L1T-1]
Applying principle of homogeneity of dimensions, [LHS] = [RHS] = [M1L0T0]
⇒ The equation (1- v2)½ must be dimension less, which is possible if we have the expressions as:
(1 – v2/c2) The equation after placing ‘c’
Q11: Check the following equation for calculating displacement is dimensionally correct or not
(a) x = x0 + ut + (1/2) at2
where, x is displacement at given time t
xo is the displacement at t = 0
u is the velocity at t = 0
a represents the acceleration.
(b) P = (ρgh)½
where P is the pressure,
ρ is the density
g is gravitational acceleration
h is the height.
Answer:
(a) x = x0 + ut + (1/2) at2
Applying principle of homogeneity, all the sub-expressions of the equation must have the same dimension and be equal to [LHS]
Dimension of x = [M0L1T0]
Dimensions of sub-expressions of [RHS] must be [M0L1T0]
⇒ Dimension of x0 (displacement) = [M0L1T0] = [LHS]
Dimension of ut = velocity x time = [M0L1T-1][M0L0T1] = [M0L1T0] = [LHS]
Dimension of at2 = acceleration x (time)2 = [M0L1T-2][M0L0T-2] = [M0L1T0] = [LHS]
∴ The equation is dimensionally correct.
(b) P = (ρgh)½
Dimensions of LHS i.e. Pressure [P] = [M1L-1T-2]
Dimensions of ρ = mass/volume = [M1L-3T0]
Dimensions of g (acceleration) = [M0L1T-2]
Dimensions of h (height) = [M0L1T0]
Dimensions of RHS = [(ρgh)½] = ([M1L-3T0]. [M0L1T-2].[M0L1T0])½ = ([M1L-1T-2])½
= [M½L-½T-1] ≠ [LHS]
Q.12 (NCERT): A man walking briskly in rain with speed v must slant his umbrella forward making an angle θ with the vertical. A student derives the following relation between θ and v : tan θ = v and checks that the relation has a correct limit: as v → 0, θ →0, as expected. (We are assuming there is no strong wind and that the rain falls vertically for a stationary man). Do you think this relation can be correct ? If not, guess the correct relation.
Answer: Given, v = tanθ
Dimensions of LHS = [v] = [M0L1T-1]
Dimension of RHS = [tanθ] = [M0L0T0] (trigonometric ratios are dimensionless)
Since [LHS] ≠ [RHS]. Equation is dimensionally incorrect.
To make the equation dimensionally correct, LHS should also be dimension less. It may be possible if consider speed of rainfall (Vr) and the equation will become:
tan θ = v/Vr
Q.13: Hooke’s law states that the force, F, in a spring extended by a length x is given by F = −kx.
According to Newton’s second law F = ma, where m is the mass and a is the acceleration.
Calculate the dimension of the spring constant k.
Answer: Given, F = -kx
⇒ k = – F/x
F = ma, the dimensions of force is:
[F] = ma = [M1L0T0].[M0L1T-2] = [M1L1T-2]
Therefore, dimension of spring constant (k) is:
[k] = [F]/[x] = [M1L1T-2].[M0L-1T0] = [M1L0T-2] or [MT-2] ….. (answer)
Q.14: Compute the dimensional formula of electrical resistance (R).
Answer: According to Ohm’s law
V = IR or R = V/I
Since Work done = QV where Q is the charge
⇒ R = W/QI = W/I2t (I = Q/t)
Dimensions of Work [W] = [M1L2T-2]
∴Dimension of R = [R] = [M1L2T-2][A-2T-1] = [M1L2T-3A-2] … (answer)
Q.15: A calorie is a unit of heat or energy and it equals about 4.2 J where 1J = 1 kg m2 s–2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time is γ s. Show that a calorie has a magnitude 4.2 α–1 β–2 γ2 in terms of the new units.
Answer: Considering the unit conversion formula,
n1U1 = n1U2
n1[M1aL1bT1c] = n2[M2aL2bT2c]
Given here, 1 Cal = 4.2 J = 4.2 kg m2 s–2.
n1 = 4.2, M1 = 1kg, L1 = 1m, T1 = 1 sec
and
n2 = ?, M2 = α kg, L2 = βm, T2 = γ sec
The dimensional formula of energy is = [M1L2T-2]
⇒ a = 1, b =1 and c = -2 Putting these values in above equation,
n2= n1[M1/M2]a[L1/L2]b[T1/T2]c
= n1[M1/M2]1[L1/L2]2[T1/T2]-2
= 4.2[1Kg/α kg]1[1m/βm]2[1sec/γ sec]-2 = 4.2 α–1 β–2 γ2 … (answer)
Q.16: The kinetic energy K of a rotating body depends on its moment of inertia I and its angular speed ω. Considering the relation to be K = kIaωb where k is dimensionless constant.
Find a and b. Moment of Inertia of a spehere about its diameter is (2/5)Mr2
Answer:
Dimensions of Kinetic energy K = [M1L2T-2]
Dimensions of Moment of Inertia (I) = [ (2/5)Mr2] = [ML2T0]
Dimensions of angular speed ω = [θ/t] = [M0L0T-1]
Applying principle of homogeneity in dimensions in the equation K = kIaωb
[M1L2T-2] = k ( [ML2T0])a([M0L0T-1])b
[M1L2T-2] = k [MaL2aT-b]
⇒ a = 1 and b = 2
⇒ K = kIω2 … (answer)
Q.17: What are the limitations of Dimensional Analysis?
Answer: Limitations of Dimensional Analysis are:
It cannot determine value of dimensionless constants.
We cannot use this method to equations involving exponential and trigonometric functions.
It cannot be applied to an equation involving more than three physical quantities.
It is a too not a solution i.e. It can check only if the equation is dimensionally correct or not. But cannot say the equation is absolutely correct.
Q.18: Convert 1 Newton into dyne using method of dimensions.
Answer: Dimensions of Force = [M1L1T-2]
Considering dimensional unit conversion formula i.e. n1[M1aL1bT1c] = n2[M2aL2bT2c]
⇒ a = 1, b = 1 and c = -2
In SI system, M1 = 1kg, L1 = 1m and T1 = 1s
In cgs system, M2 = 1g, L2 = 1cm and T2 = 1s
Putting the values in the conversion formula,
n2 = n1(1Kg/1g)1.(1m/1cm)1(1s/1s)-2 = 1.(103/1g)(102cm) = 105dyne …(answer)
Q.19: The centripetal force (F) acting on a particle (moving uniformly in a circle) depends on the mass (m) of the particle, its velocity (v) and radius (r) of the circle. Derive dimensionally formula for force (F).
Answer: Given, F ∝ ma.vb.rc
∴ F = kma.vb.rc (where k is constant)
Putting dimensions of each quantity in the equation,
[M1L1T-2] = [M1L0T0]a. [M0L1T-1]b. [M0L1T0]c = [MaLb+cT+cT-b]
⇒ a =1, b +c = 1, -b = -2
⇒ a= 1, b = 2, c = -1
∴ F = km1.v2.r-1 = kmv2/r
Q.20: If the velocity of light c, gravitational constant G and planks constant h be chosen as fundamental units, find the value of a gram, a cm and a sec in term of new unit of mass, length and time respectively.
(Take c = 3 x 1010 cm/sec, G = 6.67 x 108 dyn cm2/gram2 and h = 6.6 x 10-27 erg sec)
Answer: Given,
c = 3 x 1010 cm/sec
G = 6.67 x 108 dyn cm2/gm2
h = 6.6 x 10-27 erg sec
Putting respective dimensions,
Dimension formula for c = [M0L1T-1] = 3 x 1010 cm/sec …. (I)
Dimensions of G = [M-1L3T-2] = 6.67 x 108 dyn cm2/gm2 …(II)
Dimensions of h = [M1L2T-1] = 6.6 x 10-27 erg sec …(III)
(Note: Applying newton’s law of gravitation, you can find dimensions of G i.e. G = Fr2/(mM)
Similarly, Planck’s Constant (h) = Energy / frequency)
To get M, multiply eqn-I and III and divide by eqn.-II,
⇒ [M0L1T-1].[M1L2T-1].[M1L-3T2]
= ( 3 x 1010 cm/sec).( 6.6 x 10-27 erg sec)/ 6.67 x 108 dyn cm2/gm2
⇒[M2] = 2.968 x 10-9
⇒[M] = 0.5448 x 10-4 gm
or 1gm = [M]/0.5448 x 10-4 = 1.835 x 10-4 unit of mass
To obtain length [L], eqn.-II x eqn-III / cube of eqn.-I i.e.
[M-1L3T-2].[M1L2T-1].[M0L-3T3]
= (6.67 x 108 dyn cm2/gm2 ).( 6.6 x 10-27erg sec)/(3 x 1010 cm/sec)3
⇒ [L2] = 1.6304 x 10-65 cm2
⇒ [L] = 0.4038 x 10-32 cm
or 1cm = [L]/ 0.4038 x 10-32 = 2.47 x 10-32unit of length
In eqn-I, [M0L1T-1] = 3 x 1010 cm/sec
⇒ [T] = [L] ÷ 3 x 1010cm/s
⇒ [T] = 0.4038 x 10-32 cm ÷ 3 x 1010cm/s = 0.1345 x 10-42 s
or 1s = [T]/0.1345 x 10-42 s = 7.42 x 1042 unit of time
Q 21: A student while doing an experiment finds that the velocity of an object varies with time and it can be expressed as equation:
v = Xt2 + Yt +Z .
If units of v and t are expressed in terms of SI units, determine the units of constants X, Y and Z in the given equation.
Answer: Given, v = Xt2 + Yt +Z
Dimensions of velocity v = [M0L1T-1]
Applying applying principle of homogeneity in dimensions, terms must have same dimension.
[v] = [Xt2] + [Yt] + [Z]
∴ [v] = [Xt2]
⇒ [X] = [v] /[t2] = [M0L1T-1] / [M0L0T2] = [M0L1T-3] ….(i)
Similarly, [v] = [Yt]
⇒ [Y] = [v] / [t] = [M0L1T-1]/ [M0L0T-1] = [M0L1T-2] …(ii)
Similarly, [v]= [Z]
[Z] = [M0L1T-1] …(iii)
⇒ Unit of X = m-s-3
⇒ Unit of Y = m-s-2
⇒ Unit of Z = m-s-1
Q.22: Express Capacitance in terms of dimensions of fundamental quantities i.e. Mass (M), Length(L), Time(T) and Ampere(A)
Answer: Capacitance(C) is defined as the ability of a electric body to store electric charge.
∴ Capacitance (C) = Total Charge(q) / potential difference between two plates (V)
= Coulomb/ Volt
∵ Volt = Work done (W)/ Charge(q) = Joule/Coulomb
⇒ Capacitance (C) = Charge(q)2/ Work(W)
∵ Charge (q) = Current (I) × Time(t)
Dimension of [q] = [AT] ———– (I)
Dimension of Work = Force × distance = [MLT-2][L] = [ML2T-2] ——— (II)
Putting values of I and II,
[C] = ([AT])2/ [ML2T-2] = [M-1L-2T2+2A2] = [M-1L-2T4A2]
Physical Quantities having the same dimensional formula:
a. impulse and momentum.
b. force, thrust.
c. work, energy, torque, moment of force, energy
d. angular momentum, Planck’s constant, rotational impulse
e. force constant, surface tension, surface energy.
f. stress, pressure, modulus of elasticity.
g. angular velocity, frequency, velocity gradient
h. latent heat, gravitational potential.
i. thermal capacity, entropy, universal gas constant and Boltzmann’s constant.
j. power, luminous flux.
Q.23: If Force (F), velocity (V) and acceleration (A) are taken as the fundamental units instead of mass, length and time, express pressure and impulse in terms of F, V and A.
Answer: We know that Force = mass ✕ acceleration
⇒ mass = FA-1
and length = velocity ✕ time = velocity ✕ velocity ÷ acceleration = V2A-1
and time = VA-1
∵ Pressure = Force ÷ Area = F ÷ (V2A-1)2 = FV-4A2
Impulse = Force ✕ time = FVA-1
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Dhananjay Parkar M.Sc.(Physics) B.Ed., DCGC( RIE Bhopal ) Worked at Ratlam Ujjain Indore Bikaner Sri Ganganagar( Raj.) Vasco ( Goa ) Nazira Haflong( Assam) and Agartala. Presently working as PGT (Physics) atKendriya Vidyalaya No.1 Devlali, Nashik (Maharashtra)
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3151 | https://artofproblemsolving.com/downloads/printable_post_collections/72602?srsltid=AfmBOopMzQIEAHNyoKiNqy-VFnE0G24qBynnkJrZ2doZAqzraDsWOvzY | AoPS Community 100 Geometry Solutions
Solutions for these problems:
www.artofproblemsolving.com/community/c72602 by CaptainFlint, hotstuffFTW
1 We let the x, y, and z be the radii of circles A, B, and C respectively. We then have the following systems of equations:
x − y = 6
x − z = 5
y + z = 9 .
Adding the first two equations we get 2x − (y + z) = 11 . Because y + z = 9 , we have
2x − 9 = 11 2x = 20
x = 10 .
2 Let ∠CAD = x. We know that ∠ADC = x and ∠ACD = 180 − 2x. Let ∠DAB = y. We know that ∠ADB = 180 −x, so ∠ABD = 180 −(180 −x+y) = x−y. Because ∠CAB −∠ABC = 30 ◦,we know that (x + y) − (x − y) = 30 = ⇒ y = 15 ◦ .
3A BCDEFxx
2222−x
Let the point of tangency be F . By the BC = F C = 2 and AE =
EF = x. Thus DE = 2 − x. The on 4CDE yields
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1AoPS Community 100 Geometry Solutions
DE 2 + CD 2 = CE 2
(2 − x)2 + 2 2 = (2 + x)2
x2 − 4x + 8 = x2 + 4 x + 4
x = 12
Hence CE = F C + x = 52 ⇒ 52 .
4A BCDM
663
It is given that ∠AM D ∼= ∠CM D . Since ∠AM D and ∠CDM are alternate interior angles and
AB ‖ DC , ∠AM D ∼= ∠CDM −→ ∠CM D ∼= ∠CDM . Use the Base Angle Theorem to show
DC ∼= M C . We know that ABCD is a rectangle, so it follows that M C = 6 . We notice that
4BM C is a 30 − 60 − 90 triangle, and ∠BM C = 30 ◦. If we let x be the measure of ∠AM D,
then
2x + 30 = 180 2x = 150
x = 75
5 Drawing the square and examining the given lengths,
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2AoPS Community 100 Geometry Solutions A BCDEF
x
3
x
3
xx
2x
32x
3
30 30 30
you find that the three segments cut the square into three equal horizontal sections. Therefore, (x being the side length), √x2 + ( x/ 3) 2 = 30 , or x2 + ( x/ 3) 2 = 900 . Solving for x, we get
x = 9 √10 , and x2 = 810 . Hence, the area of the square is 810 .
6AB CDE
Since ∠DEC = ∠DBC = 90 , quadrilateral DEBC is cyclic, from which the result follows due to same inscribed arcs.
7 By the pigeonhole principle, there must be 3 points with lengths a, a, and 2a between them. Since a + a ¡= 2a, these 3 points cannot form a triangle. Instead, the three points must be collinear. Let them be A, B, and C such that AB = BC = a and AC = 2 a. Let the last point be D. D is a distance of a from two of the points among A, B, and C. Because the four points are distinct, D must form an equilateral triangle with A and B or with B and C. Without loss of generality, let AD = BD = AB = a. Let the foot of the perpendicular from D to AB be K.Because AD = BD , AK = BK = 12 (AB ) = a
2
. By the Pythagorean theorem, DK = a√32 . Thus,
b2 = DC 2 = DK 2 + KC 2 = 3a2
4
9a2
4
= 3 a2. Therefore, ba = √3.
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3AoPS Community 100 Geometry Solutions
8CA BLMN
Since quadrilateral ABLM is cyclic, ∠M LB = 180 − ∠A = 90 ◦ = ∠M LC . Thus ∠CM L = 90 −
∠C. We also have ∠C = ∠AN M since ALCN is cyclic. Then since ∠M AN = 180 − ∠A = 90 ◦,
we get
∠AM N = 90 − ∠AN M = 90 − ∠C = ∠CM L.
Since the vertical angles are congruent, L, M, N are collinear.
9XAB CIYDE
Let D and E be the feet of the altitudes from I to BC and AB respectively. Since IX = IA and
ID = IE we have that 4IDX ∼= 4IEA by HL congruence (both are clearly right triangles). It is also easy to show that 4IDX ∼= 4IDY. So, AE = s − a = XD = 12 XY, and thus
XY = b + c − a = 1400 + 1800 − 2014 = 1186 .
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4AoPS Community 100 Geometry Solutions
10 Notice that, since ADBE is cyclic, we want to show that it is an isosceles trapezoid. Thus, it suffices to prove that DB
_ = AE
_. However, we have AB
_ = AC
_ and AD
_ = CE
_ (since AD = CE ). Thus, we have DB
_ = AB
_ − AD
_ = AC
_ − CE
_ = AE
_. QED
11 Let P be the shapes perimeter and A its area. Note that if we dilate the shape by a factor of
k, its perimeter becomes kP and its are becomes k2A. Thus, if we want our dilated shape to equiable, or kP = k2A, we should dilate the shape by a factor of k = PA .
12 By simple angle chasing, we notice that triangle AEB is similar to EF C . Let the side length of the square be a, EC = x and thus BE = a − x. Because of similarity, it is
a
4 = x
3 ⇔ x = 34 a.
That yields BE = 14 a. Using the Pythagorean Theorem in triangle AEB the result is
a2 + ( 14 a) = 16 ⇔ 17 16 a2 = 16 ⇔ a2 = 16 2
17
which is the area we searched.
13 ∠ABN = ∠AM N since they are subtended from the same chord. Since ∠M XN = ∠M Y N =90 ◦, it follows that M , X, Y , and N are concyclic. Then, since M and X form angles subtended from the same chord, ∠Y XN = ∠Y M N = ∠AM N = ∠ABN . Because BN is an extension of XN , and because ∠Y XN = ∠ABN , it follows that AB ‖ XY .
14 Extend AE, DF and BE, CF to their points of intersection. Since 4ABE ∼= 4CDF and are both 5 − 12 − 13 right triangles, we can come to the conclusion that the two new triangles are also congruent to these two (use ASA, as we know all the sides are 13 and the angles are mostly complementary). Thus, we create a square with sides 5 + 12 = 17 .
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5AoPS Community 100 Geometry Solutions A BCDEFGH
EF is the diagonal of the square, with length 17 √2; the answer is EF 2 = (17 √2) 2 = 578 .
15 Let O be 4ABC s circumcenter. Note that ∠BF E = 12 ∠BOE = 14 ∠BOC = 12 ∠A. Similarly,
∠AEF = 12 ∠B and ∠DF B = 12 ∠C. Since ∠DF E + ∠AEF = ∠DF B + ∠BF E + ∠AEF =
12
(∠A + ∠B + ∠C) = 90 ◦, it follows that DF ⊥ AE .
16 We position face ABC on the bottom. Since [4ABD ] = 12 = 12 · AB · hABD , we find that hABD = 8 . The height of ABD forms a
30 − 60 − 90 with the height of the tetrahedron, so h = 12 (8) = 4 . Thus, the volume of the tetrahedron is 13 Bh = 13 15 · 4 = 20 .
17
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6AoPS Community 100 Geometry Solutions P1
P2
P3
P4
A3
A1
O
Since the quadrilateral is orthodiagonal, we can write the statement we want to show as D2 =
P1P 22 + P3P 24 . Let A1 and A3 be the antipodes of P1 and P3, respectively, and let O be the center of the circle. The claim is that reflecting P1P2P3P4 over the diameter of the circle that is parallel to P1P3 results in the new quadrilateral A3P4A1P2. We can prove this by noting that
P1OP 3 ∼= A3OA 1, so P1 goes to A3 and P3 goes to A1 (and obviously P2 and P4 switch places). Now, after reflecting, we get that A1P2 = P3P4, so now we want to show that D2 = P1P 22 +A1P 22
which is true by the Pythagorean Theorem on 4P1P2A1.
18 Notice that the midpoint of AB , the centers of the two circles, and the center of the sphere form a rectangle. Thus, if we know the sides of the rectangle we can compute the distance from the center of the sphere to the midpoint of AB . Then, since this line is perpendicular to
AB , we can then compute the radius of the sphere. First, we find the distances from the centers of the circles to midpoint of AB . They are √54 2 − 21 2
and √66 2 − 21 2. Thus, the distance from the center of the sphere to the midpoint of AB is √54 2 − 21 2 + 66 2 − 21 2. Finally, we apply pythagorean theorem once more to get the radius of the sphere to be √54 2 − 21 2 + 66 2 − 21 2 + 21 2. Squaring, we get that R2 = 6831 .
19 Extend AB and CD to meet at X. We see that ∠AXD is a right angle, so XM and XN are medians of triangle XBC and XAD , respectively. We can also see that X, M, N are collinear. Hence, XN = AN and XM = BM , so we easily find that M N = XN − XM = 1004 − 500 = 504 .
20
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7AoPS Community 100 Geometry Solutions A BCDEF
First remark that E and F are symmetric with respect to the altitude from C in 4ABC , so
EF ‖ AB . Since CF BD is a cyclic quadrilateral, we have
∠CF D = ∠CBD = ∠CAB = ∠CF E.
Hence D, E, F are collinear as desired.
21 Let the center of the circle with radius 1 be A. Let the circle shaded grey have center B and let the circle shaded black that is adjacent to that grey circle be C. Let r be the radius that we want to find. Clearly, BC = 2 r, AB = 4 − r and AC = r + 1 . We can also see that, from symmetry, ∠BAC = 60 ◦. From Law of Cosines, we have
4r2 = ( r + 1) 2 + (4 − r)2 − 2( r + 1)(4 − r) cos 60 ◦)
This reduces down to r2 + 9 r − 13 , and plugging into the quadratic formula:
r = −9 + √133 2
We arrive at −9 + 133 + 2 = 126 .
22 Let D′ be a point on the same side of AB as D such that the circumcircle of 4ABC is tan-gent to BD ′. Then ∠BAC = ∠CBD ′ so B, D, D ′ are collinear. Hence BD is (also) tangent, as desired.
23 Clearly, we see that CQM B and AN P C are cyclic quadrilaterals, implying that ∠CM B =
∠CQB and ∠CN A = ∠CP A . Therefore, we can deduce that ∠M CN = 180 ◦ − (∠CM N +
∠M N C ) = 180 ◦ − (90 ◦ − A/ 2 + 90 ◦ − B/ 2) = 45 ◦.
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8AoPS Community 100 Geometry Solutions
24 Note that since AM CN is cyclic, we have ∠AN M = ∠ACB and ∠AM N = ∠ACD = ∠CAB
so 4M AN ∼ 4 ABC by AA similarity.
25 (a) We angle chase. Since AH is perpendicular to BC , ∠BAH = 90 − ∠B. Now consider the circumcircle of ABC . We have ∠AOC = 2 ∠B (since it inscribes the same arc but goes through the center). Since triangle OAC is isosceles ( OA = OC ), we have ∠OAC = (180 − 2∠B)/2 = 90 − ∠B.(b) Notice that ∠HAO = |∠A−∠BAH −∠OAC |. Substituting in our values from part (a), we get
∠HAO = |∠A−(90 −B)−(90 −B)| = |∠A+2 ∠B−180 | = |(∠A+∠B+∠C)+( ∠B−∠C)−180 | =
|∠B − ∠C|. QED
26 We know ∠AP B = ∠CP D , so ∠AP B + ∠BP C = ∠CP D + ∠BP C , or ∠AP C = ∠BP D . We also know AP P C = P B P D , so by SAS similarity 4P AC ∼ 4 P BD .
27 Note that ∠OY X = ∠OXY , so ∠Y ZO = ∠OZX . Let ∠Y ZO = α, then by Law of Cosines on triangles Y ZO and OZX we get 121 + 49 − 2 · 11 · 7 cos α = 121 + 169 − 2 · 11 · 13 cos α, which simplifies to cos α = 10 11 . Plugging this back in we have OY 2 = 170 − 140 = 30 , so OY = √30 .
28 Let ∠AKF = ∠F KB = α and ∠AM H = ∠HM D = β. Also let F E ∩ HG = X. Then because
∠KAF = ∠DCB , we have ∠M F X = α + C, so ∠M XF = 180 − (α + β + C). However we also know that 2α + C + D = 180 and that 2β + C + B = 180 , so adding these we see that
2α+2 β +2 C +( B +D) = 360 . But we also know that B +D = 180 because quadrilateral ABCD
is cyclic, so we get that α + β + C = 90 . Which means ∠M XF = 180 − 90 = 90 . From this we conclude that triangles EM X and F M X are congruent, and triangles HKX and GKX are congruent. Thus EX = XF , HX = XG , and HG ⊥ F E , so EGF H is a rhombus.
29 Analitic 5 77
ABCMPH
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9AoPS Community 100 Geometry Solutions
We will proceed by coordinates. Let A = (0 , 0) , C = (14 , 0) . Since a 13-14-15 is composed of a 5-12-13 and 9-12-15, we have B = (5 , 12) . Let H be the foot of the altitude from B to AC.
We also have H = (5 , 0) and M = (7 , 0) . Now from BH = 12 , HM = 2 , we get BM = √148 .Also since P M is a median to the hypotenuse of 4AP C, we have M P = M C = 7 . Now the coordinates of P are just a weighted average of those of B and M :
P = 7√148 B +
√148 −7√148
M =
(
5(7) + 7( √148 − 7)
√148 , 12(7)
√148
)
=
(
7√37 − 7
√37 , 42
√37
)
Now, keeping a steady hand, using the distance formula gives AP = 7
√
2 − 2√37 , P C = 7
√
2 + 2√37
(see, it all worked out nicely in the end), and computing AP ·P C
2
gives [AP C ] = 294 √37 so p+q +r =294 + 2(37) = 368 .
29 Synthetic With some simple calculations we see that AH = 5 , HM = 2 , and M C = M P = 7 . Also
BH = 12 . So by the Pythagorean theorem we have BM = 2 √37 . Let the foot of the altitude from P to AC be D. Then M P M B = P D BH . Or 72√37 = P D
12
. Solving we have P D = 42 √37 Thus the area of 4P AC is 12 · 14 · 42 √37 = 294 √37 = 294 √37 37 . Therefore p + q + r = 294 + 37 + 37 = 368 .
30 OABCPDEME′
D′
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10 AoPS Community 100 Geometry Solutions
Let O be the circumcentre of 4ABC and M be the midpoint of BC . Then its clear that P M ⊥
BC . Since ∠BDP = 90 ◦ = 190 −∠P M B , quadrilateral P M BD is cyclic. So we have ∠M DP =
∠M BP . Also ∠ADM = 90 ◦ − ∠M DP = 90 ◦ − ∠M DP . As BP is tangent to the circumcircle of 4ABC we must have ∠M BP = BC
_
2 = ∠A. Therefore ∠ADM = 90 ◦ − ∠M DP = 90 ◦ − ∠A.Let D′ = DM ∩ AE then ADD ′ is right. So DD ′ is the D-altitude of 4ADE . Similarly EE ′ is the E-altitude of 4ADE which means that M is the orthocentre of 4ADE .
31 Note that HH ABH C and HH ACH B are cyclic. Then ∠HC HAA = ∠HC BH B = 90 − ∠BAC and ∠AH AHB = ∠ACH C = 90 − ∠BAC .Therefore, H lies on the angle bisector of ∠HC HAHB .The same argument can be repeated for the angles HC HB HA and HAHC HB to show that H
is the incenter.
32 Let Y and Z be the intersections of AC with the circle, w.l.o.g. CY < CZ . Then clearly CY =97 − 86 = 11 and CZ = 97 + 86 = 183 . Now, CB · CX = CY · CZ = 11 · 183 = 3 · 11 · 61 .If BX and CX have integer lengths, this also holds for BC and since BC ≥ CX ≥ 11 (the latter holds by the Triangle Inequality) the only possible values are CX = 11 , V C = 183 and
CX = 33 , BC = 61 . But in the first case, ACX would be degenerated which implies X ∈ AC ,only possible if X = C but this contradicts AX = 86 6 = 97 = AC . Hence CX = 33 and
BC = 61 is the only possible solution.
33 Let ∠BAC = A, ∠ABC = B, and ∠BCA = C. We know that ∠BOC = 2 A, so the measure of ̂
BN C = 4 A. Then A = 4A−PQ
_
2
, which means PQ
_ = 2 A, and ∠P N Q = A. Because the center of circle τ also lies on the perpendicular bisector of BC , we know that ∠N OB = ∠N OC = A.Thus ∠N QC = ∠N OC = A, so ∠N QA = 180 − A, and ∠N P A = 180 − A. From this we conclude AP N Q is a parallelogram.
34
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11 AoPS Community 100 Geometry Solutions
Drawing the center of the square to each of the 8 vertices we see that we are looking for the area of 4 sectors of a circle with central angle 45 ◦, 4 right triangles with length and width equal to the half the side of the square, and 4 tiny external right triangles. The sectors form a semicircle with radius
√22
, so the area of that is π
4
. The 4 big right triangles have a total area of 4 · 12 · 12
2
= 12 . Then the small external triangles are isosceles right triangles, and have height
√2−12
, so their area is
4 ·
( √2−12
)2
2 = 3 − 2√22
So the total area is π
4 + 2 − √2 .
35 Take O the midpoint of the diameter, M, O, P, S are concyclic, so ∠M P S = ∠M OS ; since ST
has constant length, we are done.
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12 |
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Pharmacokinetics
This chapter will be most useful after having a basic understanding of the material in Chapter 2, Pharmacokinetics in Goodman & Gilman’s The Pharmacological Basis of Therapeutics, 12th Edition. This chapter also draws from content in Chapter 5, Membrane Transporters and Drug Response; Chapter 6, Drug Metabolism; and parts of Chapter 7, Pharmacogenetics. The drugs presented in this chapter are used to illustrate general pharmacokinetic principles. The pharmacokinetic, pharmacodynamic, and therapeutic uses of drugs described in Chapter 2 are discussed in more detail in subsequent chapters. Neither a Mechanisms of Action Table nor a Clinical Summary Table is included in this chapter because this information is provided in subsequent chapters.
In addition to the material presented here, the 12th Edition contains:
• A review in Chapter 2 of cell membranes and their physicochemical properties that affect the movement of drugs to their site of action
• A discussion in Chapter 2 of different routes of drug administration and their relative advantages and disadvantages
• A detailed discussion in Chapter 2 of renal drug excretion, biliary and fecal excretion, and excretion by other routes
• A comprehensive discussion of design and optimization of dosage regimens in Chapter 2, including examples of dosing calculations that take into consideration bioavailability, clearance, and distribution
• Appendix II Design Optimization of Dosage Regimens: Pharmacokinetic Data in Goodman & Gilman’s The Pharmacological Basis of Therapeutics, 12th Edition provides a summary of basic pharmacokinetic data for a number of drugs that are in common clinical use, as well as information that is useful in individualizing dosing in a given patient
• Chapter 5 contains a detailed review of membrane transporters, including their role in absorption, distribution, and clearance of xenobiotics, and their role in adverse drug responses
• Table 5-1 Regulation of Transporter Expression by Nuclear Receptors provides detailed information about drug regulation of transporter expression mediated through specific nuclear receptors
• A description in Chapter 5 of the role of efflux transporters in the blood-brain barrier (BBB) and blood-cerebrospinal fluid (CSF) barrier
• Chapter 6 provides a comprehensive discussion of drug metabolism enzymes and illustrations that show many of the key chemical reactions catalyzed by these enzymes
CHARACTERISTICS OF A DRUG THAT PREDICT ITS MOVEMENT AND AVAILABILITY AT ITS SITES OF ACTION
• Molecular size and structural features
• Degree of ionization
• Relative lipid solubility of its ionized and nonionized forms
• Its binding to serum and tissue proteins
LEARNING OBJECTIVES
Understand key concepts and terms that determine drug pharmacokinetics, including absorption, distribution, metabolism, and excretion.
Understand the mechanisms by which drugs cross membranes and the physicochemical factors that influence this transfer.
Be able to predict the ionization state of a drug that is a weak acid or base knowing the drug’s pKa and the pH of the fluid.
Describe the role of membrane transporters in drug absorption, distribution, and clearance.
Know the major mechanisms by which drugs are metabolized and excreted from the body.
Understand the major pharmacokinetic mechanisms that can result in drug interactions.
Know the pharmacokinetic mechanisms that give rise to interpatient variability in drug response and toxicity.
Understand the clinical pharmacokinetic principles used to calculate dosing required to achieve steady-state drug concentrations in plasma.
Understand how route of administration can affect the bioavailability and clearance of drugs.
PASSIVE TRANSPORT VERSUS ACTIVE TRANSPORT ACROSS CELLULAR BARRIERS
• Passive transport is the dominant transport mechanism in the disposition of most drugs (Figure 2-1).
FIGURE 2-1 The variety of ways drugs move across cellular barriers in their passage throughout the body. Transmembrane movement of drug generally is limited to unbound drug; thus drug-protein complexes constitute an inactive reservoir of drug. Unbound drugs cross membranes either by passive processes or by mechanisms involving the active participation of components of the membrane. In passive transfer, the drug molecule usually penetrates by diffusion along a concentration gradient by virtue of its solubility in the lipid bilayer. Such transfer is directly proportional to the magnitude of the concentration gradient across the membrane, to the lipid-water partition coefficient of the drug, and to the membrane surface area exposed to the drug. The greater the partition coefficient, the higher is the concentration of drug in the membrane and the faster is its diffusion. After a steady-state is attained, the concentration of the unbound drug is the same on both sides of the membrane if the drug is a nonelectrolyte. For ionic compounds, the steady-state concentrations depend on the electrochemical gradient for the ion and on differences in pH across the membrane, which will influence the state of ionization of the molecule disparately on either side of the membrane and can effectively trap drug on 1 side of the membrane. Carrier-mediated active transport is characterized by a direct requirement for energy.
▶ In passive diffusion, the drug molecule usually penetrates cellular barriers by diffusion along a concentration gradient by virtue of its solubility in the lipid bilayer.
▶ Paracellular transport through intercellular gaps occurs across capillary endothelium and is important in filtration across the glomerulus in the kidney but is limited in some tissues such as the CNS where capillaries have tight intercellular junctions.
▶ Facilitated diffusion describes a carrier-mediated transport process in which there is no input of energy, and therefore enhanced movement of the involved substance is down a chemical gradient.
• Active transport is characterized by a direct requirement for energy, movement against an electrochemical gradient, saturability, selectivity, and competitive inhibition by cotransported compounds.
CASE 2-1
In Case 1-1, a gentleman is seeking an allergy medicine at his local pharmacy to relieve his hay fever symptoms. One of the nonprescription drugs he finds contains diphenhydramine, which can cause drowsiness or other CNS effects as discussed in part d of Case 1-1.
a. How is diphenhydramine able to cause CNS effects?
First-generation antihistamines such as diphenhydramine can cross the BBB and cause CNS depression such as somnolence. Diphenhydramine and the other first-generation antihistamines contain a tertiary amino group linked by a 2- or 3-atom chain to 2 aromatic substituents (see Figure 32-3 in Goodman & Gilman’s The Pharmacological Basis of Therapeutics, 12th Edition). The only ionizable group in diphenhydramine and these other agents is the tertiary amino group, which has a pKa ~9.0. Thus, at physiological pH (pH 7.4), the drug is largely unionized (see Figure 2-2) and very lipophilic and can easily diffuse through lipid membranes, including the BBB.
FIGURE 2-2 Influence of pH on the distribution of a weak acid between plasma and gastric juice separated by a lipid barrier. A. The dissociation of a weak acid, pKa = 4.4. B. Dissociation of the weak acid in plasma (pH 7.4) and gastric acid (pH 1.4). The uncharged form, HA, equilibrates across the membrane. Blue numbers in brackets show relative concentrations of HA and A–. In this example, the ratio of nonionized to ionized drug in plasma is 1:1000; in gastric juice, the ratio is 1:0.001, as given in brackets. The total concentration ratio between the plasma and the gastric juice therefore would be 1000:1 if such a system came to a steady-state. For a weak base with a pKa of 4.4 (eg, chlordiazepoxide), the ratio would be reversed, as would the thick horizontal arrows, which indicate the predominant species at each pH. Accordingly, at steady state, an acidic drug will accumulate on the more basic side of the membrane and a basic drug on the more acidic side.
WEAK ELECTROLYTES AND THE INFLUENCE OF PH
• Many drugs are weak acids or bases that are present in solution as both the nonionized and ionized species.
▶ Nonionized molecules are usually more lipid soluble and can diffuse readily across the cell membrane.
▶ Ionized molecules usually are less able to penetrate the lipid membrane because of their low lipid solubility.
• The transmembrane distribution of a weak electrolyte is influenced by its pKa and the pH gradient across the membrane.
• The ratio of nonionized to ionized drug at a given pH is readily calculated from the Henderson-Hasselbalch equation (Equation 2-1), where pKa is the drug’s acid dissociation constant.
• Figure 2-2 illustrates the partitioning of a weak acid (pKa = 4.4) between plasma (pH = 7.4) and gastric juice (pH = 1.4).
• At steady state, an acidic drug will accumulate on the more basic side of the membrane and a basic drug on the more acidic side, a phenomenon known as ion trapping.
• In the kidney tubules where a lipid soluble (uncharged) drug can be reabsorbed by passive diffusion, excretion of the drug can be promoted by altering the pH of the urine to favor the ionized state (A– or BH+).
▶ Alkaline urine favors excretion of weak acids; for example, elevation of urine pH with sodium bicarbonate will increase the excretion of weak acids such as aspirin (pKa ~3.5) and urate (pKa ~5.8).
▶ Acid urine favors excretion of weak bases.
b. Are there other antihistamines that this patient could take that do not cause drowsiness?
The second-generation antihistamines, such as loratadine (see Chapter 21), do not cause drowsiness. These second-generation histamine H1 antagonists are ionized at physiological pH and as a consequence do not readily cross the BBB. For example, the pKa of loratadine is ~4.3, and is thus largely ionized at pH 7.4 and not able to diffuse readily through lipid membranes.
DRUG ABSORPTION, BIOAVAILABILITY, AND ROUTES OF ADMINISTRATION
• Absorption is the movement of a drug from its site of administration into the central compartment (Figure 2-3).
FIGURE 2-3 The interrelationship of the absorption, distribution, binding, metabolism, and excretion of a drug and its concentration at its sites of action. Possible distribution and binding of metabolites in relation to their potential actions at receptors are not depicted.
• Bioavailability is the fractional extent to which a dose of drug reaches its site of action or a biological fluid from which the drug has access to its site of action.
• A fraction of the administered and absorbed dose of drug will be inactivated or diverted in the intestine and liver before it can reach the general circulation and be distributed to its sites of action.
▶ If the metabolic or excretory capacity of the liver and the intestine for the drug is large, bioavailability will be reduced substantially (first-pass effect).
• The route of drug administration can affect the time course of drug effects, extent of absorption, bioavailability, first-pass effect, variability of drug effects, and adverse effects (see Table 2-1).
TABLE 2-1 Some Characteristics of Common Routes of Drug Administration
• Knowledge of drugs that undergo significant metabolism or require active transport across the intestinal and hepatic membranes can be used to predict some pharmacokinetic interactions because drugs may compete for metabolism and transport.
DISTRIBUTION OF DRUGS
• Distribution refers to the movement of drug into interstitial and intracellular fluids following absorption or systemic administration (see Figure 2-3).
• Distribution reflects a number of physiological factors and the particular physicochemical properties of the individual drug.
• Physiological factors that determine the rate of delivery and potential amount of drug distributed into tissue include:
▶ Cardiac output
▶ Regional blood flow
▶ Capillary permeability
▶ Tissue volume
• During the initial distribution phase, the liver, kidney, brain, and other well-perfused organs receive most of the drug.
• The second distribution phase to muscle, most viscera, skin, and fat is slower and may require minutes to several hours before the concentration of drug in tissue is in equilibrium with that in blood.
▶ The second phase involves a far larger fraction of body mass (eg, muscle) than does the initial phase and generally accounts for most of the extravascularly distributed drug.
• With exceptions such as the brain, diffusion of drug into the interstitial fluid occurs rapidly because of the highly permeable nature of the capillary endothelial membrane.
• Tissue distribution is determined by the partitioning of drug between blood and the particular tissue.
▶ Lipid solubility and transmembrane pH gradients are important determinants of tissue uptake for drugs that are either weak acids or bases; however, ion trapping associated with transmembrane pH gradients is generally not large because the pH difference between tissue and blood (~7.0 vs 7.4) is small.
▶ The most important determinant of blood-tissue partitioning is the relative binding of drug to plasma proteins and tissue macromolecules that limits the concentration of free drug.
c. Besides the CNS, where else in the body might the net charge of a drug be affected by pH?
The partitioning of a weak acid (pKa = 4.4) between plasma (pH = 7.4) and gastric juice (pH = 1.4) is depicted in Figure 2-2. Assume that the gastric mucosal membrane behaves as a simple lipid barrier with a high electrical resistance that is permeable only to the lipid-soluble, nonionized form of the acid. The ratio of nonionized to ionized drug at each pH is readily calculated from the Henderson-Hasselbalch equation (see Equation 2-1). In the example of Figure 2-2, the ratio of nonionized to ionized drug in plasma is 1:1000; in gastric juice, the ratio is 1:0.001, as given in brackets in Figure 2-2. The total concentration ratio between the plasma and the gastric juice therefore would be 1000:1 if such a system came to a steady-state. For a weak base with a pKa of 4.4 (eg, chlordiazepoxide), the ratio would be reversed, as would the thick horizontal arrows in Figure 2-2, which indicate the predominant species at each pH. Accordingly, at steady state, an acidic drug will accumulate on the more basic side of the membrane and a basic drug on the more acidic side.
The effects of net charge are observable elsewhere in the body, such as in the kidney tubules. Urine pH can vary over a ride range, from 4.5 to 8. As urine pH drops (as [H+] increases), weak acids (A–) and weak bases (B) will exist to a greater extent in their protonated forms (HA and BH+); the reverse is true as pH rises, where A– and B will be favored. In the kidney tubules where a lipid-soluble (uncharged) drug can be reabsorbed by passive diffusion, excretion of the drug can be promoted by altering the pH of the urine to favor the ionized state (A– or BH+). Thus, alkaline urine favors excretion of weak acids; acid urine favors excretion of weak bases. Elevation of urine pH (by giving sodium bicarbonate) will promote urinary excretion of weak acids such as aspirin (pKa~3.5) and urate (pKa~5.8). This principle of ion trapping is an important process in drug distribution.
The establishment of concentration gradients of weak electrolytes across membranes with a pH gradient is a physical process and does not require an active electrolyte transport system. All that is necessary is a membrane preferentially permeable to one form of the weak electrolyte and a pH gradient across the membrane. The establishment of the pH gradient, however, is an active process.
CASE 2-2
A 68-year-old woman has symptoms of angina when she climbs stairs or engages in strenuous activity. She is prescribed nitroglycerin (glyceryl trinitrate, see Chapter 16) to take prophylactically before she engages in any activity that might cause angina symptoms.
a. The patient’s pharmacist instructs her that she is to place the nitroglycerin tablet under her tongue a couple of minutes before strenuous activity to prevent angina. Why is this route of administration used for this drug?
Sublingual administration of nitroglycerin results in rapid onset of action (peak concentrations within 4 minutes of administration) and avoids the first-pass effect. Absorption from the oral mucosa has special significance for certain drugs despite the fact that the surface area available is small. Venous drainage from the mouth is to the superior vena cava, bypassing the portal circulation and thereby protecting the drug from rapid intestinal and hepatic first-pass metabolism. Nitroglycerin is effective when retained sublingually because it is nonionic and has very high lipid solubility. Thus, the drug is absorbed very rapidly. Nitroglycerin also is very potent; absorption of a relatively small amount produces the therapeutic effect (“unloading” of the heart; see Chapter 16). Nitroglycerin that is swallowed undergoes nearly complete first-pass metabolism as the result of enzymatic denitration in the liver. Thus, its bioavailability is only ~0.01 (ie, ~1%) when swallowed.
b. To prevent first-pass effects, what other routes of administration might be effective for this agent?
Because nitroglycerin has very high lipid solubility, it can be absorbed through the skin and administered via controlled-release transdermal patches. This route of administration is effective for patients requiring chronic administration of nitroglycerin to prevent symptoms of angina, but can result in the development of tolerance unless the patch is removed for 8-12 hours each day (see Case 16-1).
PLASMA PROTEIN BINDING OF DRUGS
• Many drugs circulate in the bloodstream bound to plasma proteins.
▶ Albumin is a major carrier for acidic drugs.
▶ α1-Acid glycoprotein binds basic drugs.
▶ Certain drugs may bind to proteins that function as specific hormone carrier proteins, such as the binding of estrogen or testosterone to sex hormone–binding globulin or the binding of thyroid hormone to thyroxin-binding globulin.
• Plasma protein binding is a nonlinear, saturable process.
• Binding of a drug to plasma proteins limits its concentration in tissues and at its site of action because only unbound (free) drug is in equilibrium across membranes (see Figure 2-3).
• Appendix II in Goodman & Gilman’s The Pharmacological Basis of Therapeutics, 12th Edition provides plasma protein binding percentages for a number of commonly used drugs.
• The extent of plasma protein binding also may be affected by disease-related factors.
▶ Changes in protein binding due to disease states and drug–drug interactions are clinically relevant mainly for a small subset of the so-called high-clearance drugs of narrow therapeutic index that are administered intravenously.
• Drug excretion by the kidneys, transport, and metabolism can be limited by binding to plasma protein.
CASE 2-3
A 59-year-old man who was diagnosed with coronary artery disease undergoes angioplasty and receives a coronary stent. To prevent platelet thrombosis of the stent, he is prescribed a standard dosing regimen of clopidogrel, a drug that inhibits platelet aggregation (see Chapter 19). After several weeks of clopidogrel therapy, his platelet function is measured and it is determined that the dose of clopidogrel is too low to effectively inhibit platelet aggregation.
a. What kind of drug is clopidogrel and how might this explain the poor response to therapy with this agent?
Clopidogrel is a prodrug that requires bioactivation, primarily by CYP2C19. There is wide interindividual variability in the capacity of clopidogrel to inhibit platelet aggregation, and some patients are designated resistant to the antiplatelet effects of the drug. This variability reflects, at least in part, genetic polymorphisms in the CYPs involved in the metabolic activation of clopidogrel, most importantly CYP2C19. Clopidogrel-treated patients with the loss-of-function CYP2C192 allele exhibit reduced platelet inhibition compared with those with the wild-type CYP2C191 allele and experience a higher rate of cardiovascular events. Even patients with the reduced-function CYP2C193, 4, or 5 alleles may derive less benefit from clopidogrel than those with the full-function CYP2C191 allele. Although CYP3A4 also contributes to the metabolic activation of clopidogrel, polymorphisms in this enzyme do not appear to influence clopidogrel responsiveness.
TISSUE BINDING OF DRUGS
• Many drugs accumulate in tissues at higher concentrations than those in the extracellular fluids and blood.
• Tissue accumulation may be a result of active transport or, more commonly, binding to cellular constituents such as proteins, phospholipids, or nuclear proteins.
• Tissue binding is generally reversible and can serve as a reservoir that prolongs drug action in that same tissue or at a distant site reached through the circulation.
▶ Tissue binding and accumulation also can produce local toxicity.
• Many lipid-soluble drugs are stored by physical solution in the neutral fat; hence, fat may serve as a reservoir for lipid-soluble drugs.
• Some drugs, divalent metal ion chelating agents, and heavy metals may accumulate in bone by adsorption onto the bone crystal surface and eventual incorporation into the crystal lattice.
DRUG MEMBRANE TRANSPORTERS
• Transporters are membrane proteins that control the influx (uptake) of essential nutrients and ions, and the efflux of cellular waste, environmental toxins, drugs, and other xenobiotics.
• The functions of membrane transporters may be facilitated (equilibrative, not requiring energy) or active (requiring energy); see Figure 2-1.
• The transport of drugs is primarily mediated by 2 major superfamilies, ABC (ATP-binding cassette) and SLC (solute carrier) transporters.
▶ Most ABC proteins are primary active transporters, which rely on ATP hydrolysis to actively pump their substrates across membranes (Table 2-2).
TABLE 2-2 ABC Transporters Involved in Drug Absorption, Distribution, and Excretion Processes
■ There are 49 known genes for ABC protein which include P-glycoprotein (P-gp, encoded by ABCB1, also termed MDR1) and the cystic fibrosis transmembrane regulator (CFTR, encoded by ABCC7).
▶ The SLC superfamily includes genes that encode facilitated transporters and ion-coupled secondary active transporters (Table 2-3).
TABLE 2-3 Families in the Human Solute Carrier Superfamily
■ Approximately 315 SLC transporters have been identified in the human genome, many of which serve as drug targets or in drug absorption and disposition, including the serotonin (5-HT) and dopamine transporters (SERT, encoded by SLC6A4; DAT, encoded by SLC6A3).
• Uptake and efflux transporters determine the plasma and tissue concentrations of endogenous compounds and xenobiotics, and thereby can influence the systemic or site-specific toxicity of drugs.
• Genetic mutations and polymorphisms in drug membrane transporters can lead to significant variability in individual patient drug disposition and response (for examples, see Table 2-4).
TABLE 2-4 Examples of Genetic Polymorphisms Influencing Drug Response
DISTRIBUTION OF DRUGS INTO THE CNS
• Distribution of drugs into the CNS from the blood is unique because:
▶ Brain capillary endothelial cells have continuous tight junctions, thus drug penetration into the brain depends on transcellular rather than paracellular transport.
▶ The BBB, which consists of brain capillary endothelial cells and pericapillary glial cells, have unique properties that limit drug transport.
▶ At the choroid plexus, a blood-CSF barrier is present with epithelial cells that are joined by tight junctions, rather than endothelial cells.
• The more lipophilic a drug, the more likely it is to cross the BBB (see Case 2-1).
• Drugs may penetrate into the CNS by specific uptake transporters normally involved in the transport of nutrients and endogenous compounds from blood into the brain and CSF.
• The functional BBB and blood-CSF barrier involve efflux transporters that are capable of removing a large number of chemically diverse drugs from the CNS (see Figure 2-4, and Tables 2-2 and 2-3).
FIGURE 2-4 Transepithelial or transendothelial flux. Transepithelial or transendothelial flux of drugs requires distinct transporters at the 2 surfaces of the epithelial or endothelial barriers. These are depicted diagrammatically for transport across the small intestine (absorption), the kidney and liver (elimination), and the brain capillaries that comprise the BBB. Asymmetrical transport across a monolayer of polarized cells, such as the epithelial and endothelial cells of brain capillaries, is called vectorial transport. Vectorial transport is important in the efficient transfer of solutes across epithelial or endothelial barriers. From the viewpoint of drug absorption and disposition, vectorial transport plays a major role in hepatobiliary and urinary excretion of drugs from the blood to the lumen and in the intestinal absorption of drugs. In addition, efflux of drugs from the brain via brain endothelial cells and brain choroid plexus epithelial cells involves vectorial transport. The ABC transporters mediate only unidirectional efflux, whereas SLC transporters mediate either drug uptake or efflux.
▶ MDR1 (P-gp) and the organic anion–transporting polypeptide (OATP) are 2 of the more notable efflux transporters expressed in the BBB and blood-CSF barrier (BCSFB).
▶ Efflux transporters are expressed in brain capillary endothelial cells and the choroid plexus.
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3153 | https://math.stackexchange.com/questions/2186743/understanding-rolles-theorem | Stack Exchange Network
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Understanding Rolle's theorem.
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I think this is a very basic question but somehow I am unable to understand the answer.
Find the number of zeroes of $f(x) = x^3 + x + 1$.
Answer in the book :
$f^\prime (x) = 3x^2 + 1$ since $3x^2 + 1 \ge 0$ for all $x \in \mathbb{R}$ therefore $f^\prime (x) \ge 1 $ for all $x \in \mathbb{R}$
Therefore by Rolle's theorem $f(x)$ has at most one zero in its domain.
Now $f(-1) = -1$ and $f(0) = 1$ therefore by Intermediate value theorem $f(x)$ has at least one zero in the interval $[-1, 0]$.
Thus $f(x)$ has one zero.
Rolle's theorem : A function which is continuous $[a,b]$ and differentiable on $(a,b)$ such that $f(a) = f(b) = 0$, then there exist at least a point $c \in [a,b]$ for which $f^\prime(c) = 0$
In the first part of the proof I am unable to understand how Rolle's theorem is applied because we don't know any $a,b$ where $f(a) = f(b) = 0$ .
How can conclude that $f(x)$ has at most one zero ?
calculus
real-analysis
derivatives
continuity
rolles-theorem
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asked Mar 14, 2017 at 19:21
user312097user312097
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$\begingroup$ If $f$ had (at least) two zeros then Rolle's theorem would imply that $f'$ has a zero. But your initial calculation shows that $f'$ has no zero. $\endgroup$
Martin R
– Martin R
2017-03-14 19:23:56 +00:00
Commented Mar 14, 2017 at 19:23
$\begingroup$ When proving an implication of the form $P => Q$ you don't need to first prove that $P$ is true. Hence it is irrelevant that you are not able to find $a$ and $b$ with $f(a) = f(b) = 0$. $\endgroup$
John Coleman
– John Coleman
2017-03-14 19:26:08 +00:00
Commented Mar 14, 2017 at 19:26
$\begingroup$ @JohnColeman can you state P and Q in context of my question ? $\endgroup$
user312097
– user312097
2017-03-14 19:28:26 +00:00
Commented Mar 14, 2017 at 19:28
$\begingroup$ $P$ is $f$ has at least two roots. $Q$ is some contradiction (involving $f'$ having a zero when it is known to be nonzero). You are trying to prove that $P$ is false by contradiction, hence the fact that you can't prove it true is irrelevant. Of course you can't prove it true, because it isn't. $\endgroup$
John Coleman
– John Coleman
2017-03-14 19:31:39 +00:00
Commented Mar 14, 2017 at 19:31
1
$\begingroup$ You need to be able to engage in hypothetical reasoning. If $f$ had at least two roots, say $a$ and $b$, then something nonsensical follows. I can reason like "If the Loch Ness Monster exists then it lives in Scotland" without worrying about the fact that I lack good photographs of said beastie. $\endgroup$
John Coleman
– John Coleman
2017-03-14 19:36:03 +00:00
Commented Mar 14, 2017 at 19:36
| Show 10 more comments
2 Answers 2
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Suppose the function has two zeros, then by Rolle, $f'(x)$ must be zero for some real number $c$ strictly between the two roots. This is impossible, since $f'(x) > 0$ for all real $x.$ So, the function has at most one zero.To show it has a zero, note that $f(-1) < 0$ and $f(0)>0,$ and appeal to the intermediate value theorem.
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edited Mar 14, 2017 at 22:50
DanielWainfleet
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answered Mar 14, 2017 at 20:34
Chris LearyChris Leary
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2
$\begingroup$ I already got the answer. Thanks anyway. $\endgroup$
user312097
– user312097
2017-03-14 20:42:38 +00:00
Commented Mar 14, 2017 at 20:42
$\begingroup$ My edit was to change $f'(x)\geq 0$ to $f'(x)>0$. Likely a typo. $\endgroup$
DanielWainfleet
– DanielWainfleet
2017-03-14 22:51:29 +00:00
Commented Mar 14, 2017 at 22:51
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0
$\begingroup$
The derivative is always positive, so the function is monotonically increasing, hence it will have one root at most if the minimum value is less than 0.
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answered Mar 14, 2017 at 19:25
Abdullah al allahAbdullah al allah
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$\begingroup$ I guess your answer is correct but it does not help me as it does not solve my concerns. $\endgroup$
user312097
– user312097
2017-03-14 19:43:57 +00:00
Commented Mar 14, 2017 at 19:43
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3154 | https://achievethecore.org/coherence-map/3/14/134/134 | Operations And Algebraic Thinking
Represent And Solve Problems Involving Multiplication And Division.
Major Cluster
3.OA.A.1
Interpret products of whole numbers), e.g., interpret 5 × 7 as the total number of objects in 5 groups of 7 objects each. For example, describe a context in which a total number of objects can be expressed as 5 × 7.
Operations And Algebraic Thinking
Represent And Solve Problems Involving Multiplication And Division.
Major Cluster
3.OA.A.2
Interpret whole-number quotients of whole numbers), e.g., interpret 56 ÷ 8 as the number of objects in each share when 56 objects are partitioned equally into 8 shares, or as a number of shares when 56 objects are partitioned into equal shares of 8 objects each. For example, describe a context in which a number of shares or a number of groups can be expressed as 56 ÷ 8.
Operations And Algebraic Thinking
Represent And Solve Problems Involving Multiplication And Division.
Major Cluster
3.OA.A.4
Determine the unknown whole number) in a multiplication or division equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations 8 × ? = 48, 5 = ? ÷ 3, 6 × 6 = ?.
Operations And Algebraic Thinking
Understand Properties Of Multiplication And The Relationship Between Multiplication And Division.
Major Cluster
3.OA.B.6
Understand division as an unknown-factor problem. For example, find 32 ÷ 8 by finding the number that makes 32 when multiplied by 8.
Operations And Algebraic Thinking
Represent And Solve Problems Involving Multiplication And Division.
Major Cluster
3.OA.A.3
Use multiplication and division within 100) to solve word problems in situations involving equal groups, arrays, and measurement quantities, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem.
See Glossary, Table 2
Example Task
3.OA Gifts from Grandma, Variation 1
Provided by Illustrative Mathematics
Task
Juanita spent $9 on each of her 6 grandchildren at the fair. How much money did she spend?
Nita bought some games for her grandchildren for $8 each. If she spent a total of $48, how many games did Nita buy?
Helen spent an equal amount of money on each of her 7 grandchildren at the fair. If she spent a total of $42, how much did each grandchild get?
Solutions
Solution: Tape diagram
This task needs a tape diagram solution; one is under development.
Solution: Writing multiplication equations for division problems
Sandra spent 6 groups of $9, which is dollars all together.
Since the number of games represent the number of groups, but we don’t know how many games she bought, this is a "How many groups?" division problem. We can represent it asorSo Nita must have bought 6 games.
Here we know how many grandchildren there are (so we know the number of groups), but we don’t know how much money each one gets (the number of dollars in each group). So this is a "How many in each group?" division problem. We can represent it asorSo Helen must have given each grandchild $6.
Download Example Task
Progressions
Relating Equal Group situations to Arrays, and indicating rows or columns within arrays, can help students see that a corner object in an array (or a corner square in an area model) is not double counted: at a given time, it is counted as part of a row or as a part of a column but not both.
Problems in terms of “rows” and “columns,” e.g., “The apples in the grocery window are in 3 rows and 6 columns,” are difficult because of the distinction between the number of things in a row and the number of rows. There are 3 rows but the number of columns (6) tells how many are in each row. There are 6 columns but the number of rows (3) tells how many are in each column. Students do need to be able to use and understand these words, but this understanding can grow over time while students also learn and use the language in the other multiplication and division situations.
Please reference page 24 in the Progression document
Download Progressions PDF
Tasks
Water Balloons Classroom Supplies Two Interpretations of Division
Lessons
Multiplication and the Meaning of the Factors
Assessments
Foundations of Multiplication and Division Mini-Assessment
Assessment Item
Smarter Balanced Assessment Item Illustrating 3.OA.A.3
Focus
Focus in Grade 3
Operations And Algebraic Thinking
Solve Problems Involving The Four Operations, And Identify And Explain Patterns In Arithmetic.
Major Cluster
3.OA.D.8
Solve two-step word problems using the four operations. Represent these problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding.
This standard is limited to problems posed with whole numbers and having wholenumber answers; students should know how to perform operations in the conventional order when there are no parentheses to specify a particular order (Order of Operations).
Operations And Algebraic Thinking
Use The Four Operations With Whole Numbers To Solve Problems.
Major Cluster
4.OA.A.1
Interpret a multiplication equation as a comparison, e.g., interpret 35 = 5 × 7 as a statement that 35 is 5 times as many as 7 and 7 times as many as 5. Represent verbal statements of multiplicative comparisons as multiplication equations.
Operations And Algebraic Thinking
Use The Four Operations With Whole Numbers To Solve Problems.
Major Cluster
4.OA.A.2
Multiply or divide to solve word problems involving multiplicative comparison, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem, distinguishing multiplicative comparison from additive comparison.
See Glossary, Table 2
Number And Operations-Fractions
Build Fractions From Unit Fractions By Applying And Extending Previous Understandings Of Operations On Whole Numbers.
Major Cluster
4.NF.B.4
Apply and extend previous understandings of multiplication to multiply a fraction) by a whole number).
Grade 4 expectations in this domain are limited to fractions with denominators 2, 3, 4, 5, 6, 8, 10, 12, and 100.
Measurement And Data
Solve Problems Involving Measurement And Conversion Of Measurements From A Larger Unit To A Smaller Unit.
Supporting Cluster
4.MD.A.3
Apply the area and perimeter formulas for rectangles in real world and mathematical problems. For example, find the width of a rectangular room given the area of the flooring and the length, by viewing the area formula as a multiplication equation with an unknown factor.
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3155 | https://courses.csail.mit.edu/6.854/18/Scribe/s27-sweepline/s27-sweepline.html | # Lecture 1 - Sweep Line Algorithms
[\text{ 11/8/2004 }] [\text{6.854 - Advanced Algorithms}] [\text{Professor David Karger}] [\text{Aidan Downes}]
Introduction
Sweep line algorithms are used in solving planar problems. The basic outline of a sweep line algorithm is as follows:
Sweep a line across problem plane.
As the line sweeps across the plane, events of interest occur, keep track of these events.
Deal with events that occur at the line leaving a solved problem behind.
Convex Hull
Given a set of points in a plane, the smallest convex polygon that encloses all of the points in the set is the convex hull of the set. A polygon is convex if any line segment connecting two points in the polygon is entirely closed by the polygon. A convex hull can be represented by it's vertices in cyclic order.
Lower Bounds for Convex Hull Algorithms
The lower bound for Convex Hull Algorithms is (\Omega (n \log n)). This can be proved by reducing the sorting problem to an instance of the convex hull problem. One possible reduction is shown below:
Let (x_1, x_2, ..., x_n) be the input
Let (p_i = (x_i, x_i^2))
Find the convex hull for all (p_i)'s.
Return the (x) value for each point (remember a convex hull is represented by its vertices in cyclic order).
Let (T(n)) be the time to compute the convex hull. Then, the total time for sorting using the above algorithm is (O(n + T(n))). Since the lower bound for sorting (using the comparison model) is (\Omega (n \log n)), then (T(n) = \Omega (n \log n)) 1.
Sweep Line algorithm for Convex Hulls
This algorithm finds the upper envelop of a convex hull. A similar algorithm can be used to find the lower envelop. The algorithm uses two data structures. The first data structure stores the points that make up the hull of points seen so far. The second data structure stores the points not seen yet in sorted order by x-coordinate. A simple list can be used for the first data structure while a heap is well suited for the purpose of the second data structure.
The events of interest for this sweep line algorithm is the arrival of a new point as the sweep line travels from left to right. As each event occurs, the algorithm updates the current convex hull. There are two cases to consider.
Case 1, a "right" turn : The existing convex hull can be extended to include the new point without violating the convexity. In this case, the new point is directly added to the Convex Hull. Case 2, a "left" turn: Adding this new point to the Convex Hull violates its convexity. In this case, we add the new point to the Convex Hull but we need to do some surgery on the Convex Hull as following: Perform a convexity check on the predecessor of the new point. If the convexity check fails, delete the point from the envelope. Now the new point has a new predecessor. Repeat the convexity check and deletion on the predecessor of the new point until the convexity check is satisfied.
The full algorithm is as follows:
insert point into heap ordered by x-coordinate
Delete-Min from heap
Run Convexity Check
if not convex delete elements from hull backwards till convex check passes
add new point to hull
go to line 2
Runtime:
Line 1 runs in (O(n \log n)). Line 2 runs (n) times with (O(\log n)) each time and (O(n \log n)) in total. The convexity check in line 3 is (O(1)) and runs (n) times. To analyze line 4, note that an element can only be deleted once. For each element deleted, constant work is performed, and therefore (O(n)) work is performed in total. Therefore the total running time is (O(n \log n)).
Segment Intersections
Segment intersection algorithms returns the coordinates of all pairwise intersections given a set of line segments as input. The naive algorithm tests every segment pair for intersection and runs in (O(n^2)). The sweep line algorithm solves the problem in (O((n + k) \lg n)) time. If (k) is not huge (less than (n^2)) then the sweep line algorithm is a significant improvement over the naive algorithm.
The sweep line algorithm we will use to solve the segment intersection sweeps a line from the top to the bottom of the plane, reporting intersections as they are encountered. As the sweep line moves across the plane it forms intersections with line segments in the plane. When a line segments intersects the sweep line, we say it is active.
Note that in order for two segments to intersect they must be adjacent to each other at some point on the sweep line. This allows us to only check for intersections between adjacent pairs of segments. The algorithm uses two different data structures. One data structure stores all active segments ordered by intersection point with the sweep line. The other data structure is a priority queue which stores events ordered by the distance from the sweep line.
The events of interest in this sweep line algorithm are:
the sweep line encounters(intersects) a new segment. The line segment becomes active.
an active segment no longer intersects the sweep line. The segment is no longer active.
two active segments intersect each other
Case 1: Insert the line segment into the sweep line status data structure. Then check if the line segment intersects any of its new neighbors. If there is an intersection add the intersection to the event queue.
Case 2: Remove the line segment from the sweep line status data structure. The former neighbors of the disposed line segment are now adjacent to each other. Check if the line segments intersects. If they do, add the intersection to the event queue.
Case 3: When there is an intersection event, report the intersection and swap the involved line segments positions in the sweep line status data structure. Each of the swapped line segments have one new neighbor. Check for intersections between the swapped line segments and its new neighbors. Add any intersections to the event queue.
The full algorithm is as follows:
Add segment activation and deactivation events to event queue
Dequeue an event from the event queue
If the event is an activation event then add the line segment to the sweep line status data structure and check for intersections.
If the event is a deactivation event, remove the line from the sweep line status data structure and check for intersections.
If the event is a intersection event, report the intersection, swap the lines in the sweep line status data structure and check for intersections.
Go to line 1.
Runtime:
Line 1 runs in (O(n)) time. The number of times that the loop (lines 2-6) runs is equal to the number of events. There are (k) swaps, (n) arrivals and (n) departures. Therefore there are (O(n+k)) events and the loop runs (O(n+k)) times. If we use a binary tree to represent the sweep line status and a heap to represent the priority queue, each iteration of the loop takes (O(\log n)) time. Therefore, the total runtime is (O((n+k)\log n)).
Voronoi Diagrams
Given a set of points a plane, we want to be able to answer nearest neighbor queries. More specifically, given any point in the plane we want to return the closet point in the given set of points. One idea for solving the problem is to divide the plane regions according to closest point. This reduces the problem to determining the region of the query point.
Voronoi diagrams are the partitioning of a plane with (n) points into (n) convex polygons such that each polygon contains exactly one point and every point in a given polygon is closer to its central point than to any other. Voronoi diagrams can be used to solve the nearest neighbor problem.
Size of voronoi diagrams
The size of a voronoi diagrams is (O(n)).
Proof: Add a point at infinity such that all edges (including infinite edges) have two endpoints. Euler's formula states that (n_v - n_e + n_f = 2). Not that the number of faces, (n_f), in the voronoi diagram is equal to (n). Also note that every voronoi vertex has degree (3) but every edge has two end points. Therefore, we have
| | | |
---
| (2 n_e ) | (\geq ) | ( 3(n_v + 1)\nonumber ) |
| (2(n + n_v - 2) ) | (\geq) | ( 3(n_v + 1)\nonumber ) |
| (n_v ) | (\leq ) | ( 2n - 7 ) |
Since the number of voronoi vertices is linear in the number of input points, the size of the voronoi diagram is (O(n)).
Sweep line algorithm for constructing Voronoi Diagram
In this algorithm we sweep a line from the top to the bottom of the plane and maintain a beachfront of parabola, points equidistant for the sweep line and seen sites (input points). Our algorithm maintains the following invariant: The voronoi diagram is complete and unchanging behind the beachfront.
As the sweep lines moves, the parabola forming the beach front also descend. [(x-x_f)^2 + (y-y_f)^2 = (y-t)] is the equation for a parabola where ((x_f, y_f)) is coordinate of the focus point and (t) is the parameter for the sweep line. Fixing (x) and solving for (\frac{dy}{dt}) produces
| | |
--- |
| (2(y - y_f) \frac{dy}{dt} ) | (= 2(y-t)(\frac{dy}{dt} - 1)) |
| (\frac{dy}{dt}) | (= -\frac{y-t}{t-y_f}) |
Thus the higher the focus of the parabola the slower the parabola descends.
The events of interest in our Voronoi Diagram are:
Site Events. This occurs when the sweep line crosses a new site. When this occurs we create a new parabola for the site. Initially the parabola is a vertical line but it widens to normal parabola as the sweep line continues moving. The new parabola creates two new intersections of parabola in the beach front. We insert these intersections into ordered list of parabolas. (Parabolas are represented by the intersection points).
Movement Events. These events occur as the sweep line moves across the plane, changing the beach front parabolas. There are two cases to consider.
An inner parabola crosses an outer parabola. This does not happen. Consider the moment when the inner parabola becomes adjacent with the outer parabola. The inner parabola has a higher focus than the outer parabola so it descends slower. Therefore the inner parabola does not cross the outer parabola.
A parabola hits the intersection of two other parabolas. All three sites are equidistance from the intersection. The middle parabola cannot absorb the intersection because its focus is too high and thus it descends slower. Therefore the middle parabola disappears from the beach front. This is also known as the circle event.
In recap, site events create parabolas and circle events destroy parabolas. The full algorithm is as follows:
Add all site events to the event queue.
While queue is not empty:
Dequeue next event.
If its site event add parabola to the beachfront and compute future circle events with neighbors.
If its a circle event then remove parabola from beachfront and check for future circle events.
Runtime: Each event requires (O(\log n)) time to process, mostly updating data structures for the event queue and parabolas. And there are (O(n)) events as each site is responsible for at most one site event and one circle event. Therefore, the total runtime is therefore (O(n \log n)).
Note that the argument presented here only works for convex hall algorithms that are based on a comparison model and also it relies on the requirement that the points of convex hull should be reported "in order". However, there are stronger results showing that even deciding which points are on the hull takes (\Omega(n \log n)) even in a more general "algebraic decision tree" model that allows arbitrary algebraic comparisons.© |
3156 | https://www.pnw.edu/wp-content/uploads/2020/03/Lecture-Notes-11-5.pdf | 192 Chapter 11. Probability and Calculus (LECTURE NOTES 11) Chapter 11 Probability and Calculus Determination of probability for a continuous random variable involves integrating a probability density function for this random variable.
We first look at general probability density functions and their expected value and variance, then look at special probability density functions; specifically, the uniform, exponential and normal density functions and their expected values and variances.
11.1 Continuous Probability Models The (cumulative) distribution function for random variable X is F(x) = P (X ≤x) , −∞< x < ∞, and has properties • limx→−∞F(x) = 0, • limx→∞F(x) = 1, • if x1 < x2, then F(x1) ≤F(x2); that is, F is nondecreasing.
Also, limn→∞F(xn) = F(x); that is, F must be right continuous (which determines where the solid and empty endpoints are on the graph of a distribution function). A random variable X with distribution function F(x) is continuous if F(x) is continuous, and the first derivative of F(x) exists and continuous except, possibly, at a finite number of points in a finite interval. The (probability) density function, f(x), is given by f(x) = dF(x) dy = F ′(x), and so, also, F(x) = Z x −∞f(t) dt Properties of the density function for a continuous random variable are • f(x) ≥0, for all x, −∞< x < ∞, 193 194 Chapter 11. Probability and Calculus (LECTURE NOTES 11) • R ∞ −∞f(x) dx = 1.
And, P(a ≤X ≤b) = P(X ≤b) −P(X ≤a) = F(b) −F(a) = Z b a f(x) dy.
Exercise 11.1 (Continuous Probability Models) 1. Discrete function: number of heads when flipping a coin twice.
Let the number of heads flipped in two flips of a coin be a random variable X.
There are four possible cases, TT (X = 0), HT or TH (X = 1) and HH (X = 2) and so, assuming each of the four cases are equally likely, then the probability density function is x (number of heads) 0 1 2 P (X = x) 1 4 2 4 1 4 or, P (X = 0) = 0.25, P (X = 1) = 0.50 and P (X = 2) = 0.25. 0 1 2 0.25 0.50 0.75 1 discrete density f(y) distribution F(y) 0 1 2 0.25 0.50 0.75 1 Figure 11.1: Density and distribution function: flipping a coin twice (a) F(0) = P (X ≤0) = P (X = 0) = (choose one) (i) 0 (ii) 0.25 (iii) 0.75 (iv) 1.
(b) F(1) = P (X ≤1) = P (X = 0) + P (X = 1) = (choose one) (i) 0 (ii) 0.25 (iii) 0.75 (iv) 1.
(c) F(2) = P (X ≤2) = P (X = 0) + P (X = 1) + P (X = 2) = (choose one) (i) 0 (ii) 0.25 (iii) 0.75 (iv) 1.
Section 1. Continuous Probability Models (LECTURE NOTES 11) 195 (d) Also, if x < 0, F(x) = P (X ≤x) = (choose one) (i) 0 (ii) 0.25 (iii) 0.75 (iv) 1.
(e) And, if x ≥2, F(x) = P (X ≤x) = (choose one) (i) 0 (ii) 0.25 (iii) 0.75 (iv) 1.
(f) So, in this case, F(x) = 0, x < 0 0.25, 0 ≤x < 1 0.75, 1 ≤x < 2 1, x ≥2.
(i) True (ii) False (g) The discontinuous “step function” graph of this F(x) is given in the figure. Notice that F(x) is right continuous, which is indicated by the solid and empty endpoints on the graph of this distribution function.
(i) True (ii) False (h) P (X < 1) = P (X = 0) = 0.25 = (choose one) (i) F (0) (ii) F (1) (iii) F (2) (iv) F (3).
(i) P (X < 2) = P (X = 0) + P (X = 1) = 0.75 = (choose one) (i) F (0) (ii) F (1) (iii) F (2) (iv) F (3).
(j) P (X > 1) = 1 −P (X ≤1) = 1 −F(1) = 1 −0.75 = (choose one) (i) 0 (ii) 0.25 (iii) 0.75 (iv) 1.
2. Another discrete distribution function.
Let random variable X have distribution (not density), F(x) = 0, x < 0 1 3, 0 ≤x < 1 1 2, 1 ≤x < 2 1, x ≥2.
(a) F(1) = P (X ≤1) = P (X = 0) + P (X = 1) = (choose one) (i) 0 (ii) 1 6 (iii) 1 3 (iv) 1 2.
196 Chapter 11. Probability and Calculus (LECTURE NOTES 11) (b) P (X < 2) = P (X = 0) + P (X = 1) = (choose one) (i) 0 (ii) 1 6 (iii) 1 3 (iv) 1 2.
(c) P (X ≤1.5) = P (X = 0) + P (X = 1) = (choose one) (i) 0 (ii) 1 6 (iii) 1 3 (iv) 1 2.
(d) p(1) = P (X = 1) = P (X ≤1) −P (X ≤0) = F(1) −F(0) = (choose one) (i) 0 (ii) 1 6 (iii) 1 3 (iv) 1 2.
(e) p(2) = P (X = 2) = P (X ≤2) −P (X ≤1) = F(2) −F(1) = (choose one) (i) 0 (ii) 1 6 (iii) 1 3 (iv) 1 2.
(f) Random variable X is discrete, not continuous, because the associated F(x) is a discontinuous (“step”, in this case) function.
(i) True (ii) False (g) Notice that (1) limx→−∞F(x) = 0, (2) limx→∞F(x) = 1, (3) if x1 < x2, then F(x1) ≤F(x2); that is, F is nondecreasing.
(i) True (ii) False 3. Continuous distribution function: waiting time.
Let the time waiting in line, in minutes, be described by the random variable X which has the following probability density (not distribution), f(x) = 0, x < 2, 1 2, 2 ≤x < 4, 0, x ≥4.
(a) The chance of waiting at most x = 1 minute is F(1) = P (X ≤1) = Z 1 −∞f(x) dx = Z 1 −∞0 dx = (choose one) (i) 0 (ii) 0.1 (iii) 0.5 (iv) 1.
Section 1. Continuous Probability Models (LECTURE NOTES 11) 197 0 1 2 3 0.25 0.50 0.75 1 4 continuous density f(y) 0 1 2 3 0.25 0.50 0.75 1 4 distribution F(y) Figure 11.2: Distribution function: waiting time (b) The chance of waiting any time less than 2 minutes, x < 2, F(x) = P (X ≤x) = Z x −∞0 dx = (i) 0 (ii) 0.1 (iii) 0.5 (iv) 1.
(c) The chance of waiting at most x = 3 minutes is F(3) = P (X ≤3) = Z 3 −∞f(x) dx = Z 2 −∞0 dy + Z 3 2 1 2 dx = 0 + x 2 x=3 x=2 = 3 2 −2 2 = (choose one) (i) 0 (ii) 0.1 (iii) 0.5 (iv) 1.
(d) The chance of waiting at most x = 5 minutes is F(5) = P (X ≤5) = Z 5 −∞f(x) dx = Z 2 −∞0 dy + Z 4 2 1 2 dy + Z 5 4 0 dx = 0 + x 2 x=4 x=2 + 0 = 4 2 −2 2 = (choose one) (i) 0 (ii) 0.1 (iii) 0.5 (iv) 1.
Integrals involving zero will not always be explicitly stated as they are here; instead of R 5 −∞f(x) dy = R 2 −∞0 dy + R 4 2 1 2 dy + R 5 4 0 dy, integral R 5 −∞f(x) dy = R 4 2 1 2 dy will be used instead.
198 Chapter 11. Probability and Calculus (LECTURE NOTES 11) (e) The chance of waiting any time more than 4 minutes, x ≥4, F(x) = P (X ≤x) = Z x −∞f(x) dy = Z 4 2 1 2 dx = (i) 0 (ii) 0.1 (iii) 0.5 (iv) 1.
(f) In general, the distribution function is F(x) = 0, x < 2, x 2 −1, 2 ≤x < 4, 1, x ≥4.
Random variable X is continuous because, as shown in the figure, even though the density, f(x), is a discontinuous function, the associated distribution, F(x), is a continuous function in this case.
(i) True (ii) False (g) F(3) = 1 2 is both the (positive) area under density f(x) from −∞up to 3 and also the point on distribution F(x) at x = 3.
(i) True (ii) False (h) Since X is a continuous random variable P (X < 3) = Z 3 2 1 2 dt = x 2 x=3 x=2 = 3 2 −2 2 = 1 2 = (i) P (X > 3) (ii) P (X < 4) (iii) P (X ≤3) (iv) P (X ≤4).
and, consequently, the chance of waiting exactly x = 3 minutes is zero, P (X = 3) = Z 3 3 f(x) dx = 0.
(i) Since P (2.5 < X < 3) = Z 3 2.5 1 2 dy = x 2 x=3 x=2.5 = 3 2 −2.5 2 = 0.5 2 = 1 4 = (choose one or more) (i) P (2.5 ≤X < 3) (ii) P (2.5 ≤X ≤3) (iii) P (2.5 < X < 3) (iv) P (2.5 < X ≤3).
because the chance of waiting exactly x minutes is zero, P (X = x) = Z x x f(t) dt = 0.
Section 1. Continuous Probability Models (LECTURE NOTES 11) 199 (j) (i) True (ii) False Function f(x) is a probability density here because both f(x) ≥0 and also R f(x) dx = 1 on interval [2, 4).
4. Continuous distribution function: triangle.
Let random variable X have the following probability density, f(x) = 0, x < 2, 1 6x, 2 ≤x < 4, 0, x ≥4.
In other words, f(x) = x 6 on [2, 4), equivalently [2, 4], and zero elsewhere.
0 1 2 3 0.25 0.50 0.75 1 4 density f(y) 0 1 2 3 0.25 0.50 0.75 1 4 distribution F(y) Figure 11.3: Density and distribution function: triangle (a) (i) True (ii) False The distribution function is F(x) = R x −∞0 dt = 0, x < 2, R 2 −∞0 dt + R x 2 t 6 dt = 0 + t2 12 it=x t=2 = x2 12 −4 12, 2 ≤x < 4, R 2 −∞0 dt + R t=4 t=2 t 6 dt + R ∞ 4 0 dt = 0 + x2 12 it=4 t=2 + 0 = 1, x ≥4.
In other words, F(x) = x2 12 − 4 12 on [2, 4) or equivalently [2, 4] and zero elsewhere. Both density and distribution are given in the figure.
(b) F(1) = (choose one) (i) 0 (ii) 5 12 (iii) 7 12 (iv) 1.
(c) F(2) = 22 12 −4 12 = (choose one) (i) 0 (ii) 5 12 (iii) 9 12 (iv) 1.
(d) F(3) = 32 12 −4 12 = (choose one) (i) 0 (ii) 5 12 (iii) 9 12 (iv) 1.
200 Chapter 11. Probability and Calculus (LECTURE NOTES 11) (e) F(5) = (choose one) (i) 0 (ii) 5 12 (iii) 9 12 (iv) 1.
(f) Also, P(1 < X < 3) = P(X < 3) −P(X < 1) = P(X ≤3) −P(X ≤1) = F(3) −F(1) = 32 12 −4 12 !
−0 = (choose one) (i) 0 (ii) 5 12 (iii) 9 12 (iv) 1.
(g) P(2.5 < X < 3.5) = F(3.5) −F(2.5) = 3.52 12 −4 12 − 2.52 12 −4 12 = (choose one) (i) 0 (ii) 0.25 (iii) 0.50 (iv) 1.
(h) P(X > 3.5) = 1 −P(X ≤3.5) = 1 −F(3.5) = 1 − 3.52 12 −4 12 = (choose one) (i) 0.3125 (ii) 0.4425 (iii) 0.7650 (iv) 1.
(i) (i) True (ii) False Function f(x) is a probability density here because both f(x) ≥0 and also R f(x) dx = 1 on interval [2, 4].
5. Continuous distribution function: triangle with unknown k.
Let random variable X have the probability density f(x) = kx on the interval [1, 5] and zero elsewhere. What is k? Find P(2.5 < X < 3.5).
(a) The distribution function is F(x) = Z x 1 kt dt = kt2 2 #t=x t=1 = kx2 2 −k 2 = k(x2 −1) 2 (i) True (ii) False (b) What is k? Since 1 ≤x < 5 and the total probability must equal 1, F(5) = k(52 −1) 2 = 24k 2 = 1, then k = (choose one) (i) 1 11 (ii) 1 12 (iii) 1 13 (iv) 1 14.
Section 1. Continuous Probability Models (LECTURE NOTES 11) 201 (c) In other words, f(x) = kx = 1 12x, and F(x) = k(x2 −1) 2 = 1 12 × x2 −1 2 = (x2 −1) 24 on [1, 5] and zero elsewhere.
(i) True (ii) False (d) P(2.5 < X < 3.5) = F(3.5) −F(2.5) = 3.52−1 24 − 2.52−1 24 = (choose one) (i) 0 (ii) 0.25 (iii) 0.50 (iv) 1.
6. Another continuous probability function with unknown k.
Find k such that f(x) = kx2 is a probability density function where x is in the interval [1, 5]. Then find P(X ≥2).
(a) Since the distribution function is F(x) = Z x 1 kt2 dt = kt3 3 #t=x t=1 = kx3 3 −k 3 = k(x3 −1) 3 and since 1 ≤x < 5 and the total probability must equal 1, F(5) = k(53 −1) 3 = 124k 3 = 1, then k = (i) 3 124 (ii) 2 124 (iii) 1 124 (b) And so F(x) = k(x3 −1) 3 = 3 124 × x3 −1 3 = (i) 3 124(x3 −1) (ii) 2 124(x3 −1) (iii) 1 124(x3 −1) (c) And also f(x) = kx2 = 3 124x2 = (i) 3 124x2 (ii) 2 124x2 (iii) 1 124x2 (d) And so P(X ≥2) = 1 −P(X < 2) = 1 −F(2) = 1 − 1 124((2)3 −1) = (i) 116 124 (ii) 117 124 (iii) 118 124 202 Chapter 11. Probability and Calculus (LECTURE NOTES 11) 7. Probability density function and improper integration Consider function f(x) = 3e−3x defined on [0, ∞). Find F(x) and P(X ≥2). Show R ∞ 0 f(x) dx = 1.
(a) By substituting u = −3x, then du = −3 du and so F(x) = Z 3e−3x dx = − Z e−3x(−3) dx = − Z eu du = −eu = (i) −e−3x (ii) e−3x (iii) −3e−3x then F(x) = Z x 0 3e−3t dt = Z x 0 3e−3t dt = h −e−3tit=x t=0 = −e−3x −(−e−k(0)) = (i) −e−3x + 1 (ii) e−3x + 1 (iii) 1 −e−3x (b) And so P(X ≥2) = 1 −P(X < 2) = 1 −F(2) = 1 − 1 −e−3(2) ≈ (i) 0.002 (ii) 0.003 (iii) 0.004 (c) (i) True (ii) False Z ∞ 0 3e−3x dx = lim b→∞ Z b 0 3e−3x dx = lim b→∞ h −e−3xix=b x=0 = lim b→∞ h −e−3b −(−ex(0)) i = 1 or, equivalently, lim b→∞F(b) = lim b→∞1 −e−3b = 1 11.2 Expected Value and Variance of Continuous Random Variables The expected value, E(X), of a discrete random variable, X is given by E(X) = X x xP(X = x).
The expected value, also called the mean, µ, E(X) = µ, is, roughly, a weighted average of X.
The expected value, E(X), of a continuous random variable, X is given by E(X) = Z ∞ −∞xf(x) dx.
Section 2. Expected Value and Variance of Continuous Random Variables (LECTURE NOTES 11)203 The variance, Var(X), is Var(X) = σ2 = E[(X −µ)2] = E(X2) −[E(X)]2 = E(X2) −µ2 with associated standard deviation, σ = √ σ2, in both discrete and continuous cases.
The median of X is the m that satisfies P(X ≤m) ≥1 2 and P(X ≥m) ≥1 2.
Exercise 11.2 (Expected Values for Continuous Random Variables) 1. Discrete function: number of heads when flipping a coin twice.
Let the number of heads flipped in two flips of a coin be a random variable X with probability density function x (number of heads) 0 1 2 P (X = x) 1 4 2 4 1 4 (a) Expected number of flips of the coin µ = E(X) = X x xP(X = x) = 0 × 1 4 + 1 × 2 4 + 2 × 1 4 = (i) 1 2 (ii) 3 4 (iii) 1 (iv) 3 2.
(b) E (X2).
E X2 = X x x2P(X = x) = 02 × 1 4 + 12 × 2 4 + 22 × 1 4 = (i) 1 2 (ii) 3 4 (iii) 1 (iv) 3 2.
(c) Variance in number of flips of coin.
σ2 = Var(X) = E X2 −µ2 = 3 2 −(1)2 = (i) 1 2 (ii) 3 4 (iii) 1 (iv) 3 2.
(d) Standard deviation in number of flips of coin.
σ = √ σ2 = s 1 2 ≈ (i) 0.57 (ii) 0.61 (iii) 0.67 (iv) 0.71.
204 Chapter 11. Probability and Calculus (LECTURE NOTES 11) (e) Determine P(µ −σ < X < µ + σ) P(µ−σ < X < µ+σ) ≈P (1 −0.71 < X < 1 + 0.71) ≈P (0.29 < X < 1.71) = (i) 1 2 (ii) 3 4 (iii) 1 (iv) 3 2.
Look at the table above: how much probability is between 0.29 and 1.71?
(f) Median. Since P(X ≤1) = 3 4 > 1 2 and P(X ≥1) = 3 4 > 1 2 the median is m = (i) 1 2 (ii) 3 4 (iii) 1 (iv) 3 2.
2. Discrete function: number of heads when flipping an unfair coin twice.
Let the number of heads flipped in two flips of a coin be a random variable X with probability density function x (number of heads) 0 1 2 P (X = x) 1 2 1 4 1 4 (a) Expected number of flips of the coin µ = E(X) = X x xP(X = x) = 0 × 1 2 + 1 × 1 4 + 2 × 1 4 = (i) 11 16 (ii) 3 4 (iii) 1 (iv) 5 4.
(b) E (X2).
E X2 = X x x2P(X = x) = 02 × 1 2 + 12 × 1 4 + 22 × 1 4 = (i) 11 16 (ii) 3 4 (iii) 1 (iv) 5 4.
(c) Variance in number of flips of coin.
σ2 = Var(X) = E X2 −µ2 = 5 4 − 3 4 2 = (i) 11 16 (ii) 3 4 (iii) 1 (iv) 5 4.
Section 2. Expected Value and Variance of Continuous Random Variables (LECTURE NOTES 11)205 (d) Standard deviation in number of flips of coin.
σ = √ σ2 = s 11 16 ≈ (i) 0.77 (ii) 0.81 (iii) 0.83 (iv) 0.91.
(e) Determine P(µ −σ < X < µ + σ) P(µ−σ < X < µ+σ) ≈P 3 4 −0.83 < X < 3 4 + 0.83 ≈P (−0.08 < X < 1.58) = (i) 1 2 (ii) 3 4 (iii) 1 (iv) 3 2.
Look at the table above: how much probability is between -0.08 and 1.58?
(f) Median. Since P(X ≤1) = 3 4 > 1 2 and P(X ≥1) = 1 2 ≥1 2 the median is m = (i) 1 2 (ii) 3 4 (iii) 1 (iv) 3 2.
3. Continuous distribution function: Time and cost of cell phone.
Let the amount of time spent (in minutes) on a cell phone call be represented by random variable X with the following probability density, f(x) = ( 1 6x, 2 ≤x ≤4, 0, elsewhere.
(a) Expected value. The expected amount of time on the call is, µ = E(X) = Z ∞ −∞xf(x) dx = Z 4 2 x 1 6x dx = Z 4 2 x2 6 dx = x3 18 #4 2 = 43 18−23 18 = (i) 23 9 (ii) 28 9 (iii) 31 9 (iv) 35 9 .
(b) E (X2).
E X2 = Z ∞ −∞x2f(x) dx = Z 4 2 x2 1 6x dx = Z 4 2 x3 6 dx = x4 24 #4 2 = 44 24−24 24 = (i) 9 (ii) 10 (iii) 11 (iv) 12.
206 Chapter 11. Probability and Calculus (LECTURE NOTES 11) (c) Variance. Variance in the amount of time on the call is, σ2 = Var(X) = E X2 −µ2 = 10 − 28 9 2 = (i) 23 81 (ii) 26 81 (iii) 31 81 (iv) 35 81.
(d) Standard deviation. Standard deviation in time spent on the call is, σ = √ σ2 = s 26 81 ≈ (i) 0.57 (ii) 0.61 (iii) 0.67 (iv) 0.73.
(e) Determine probability time falls within one standard deviation of mean.
P(µ −σ < X < µ + σ) ≈ P 28 9 −0.57 < X < 28 9 + 0.57 ≈ P (2.54 < X < 3.68) = Z 3.68 2.54 x 6 dx = x2 12 #3.68 2.54 = (3.68)2 12 −(2.54)2 12 ≈ (i) 0.49 (ii) 0.59 (iii) 0.69 (iv) 0.79.
(f) Median. Since the distribution function is F(x) = Z x 2 t 6 dt = t2 12 #t=x t=2 = x2 12 −22 12 = x2 −4 12 then median m occurs when F(m) = P(X ≤m) = m2 −4 12 = 1 2 so m2 −4 12 = 1 2 m2 −4 = 6 m2 = 10 m = √ 10 ≈ (i) 1.16 (ii) 2.16 (iii) 3.16 Section 2. Expected Value and Variance of Continuous Random Variables (LECTURE NOTES 11)207 4. Another continuous distribution example.
Let random variable X have the following probability density, f(x) = x, 0 < x < 1, 2 −x, 1 ≤x < 2, 0, elsewhere.
(a) Expected value.
µ = E(X) = Z ∞ −∞xf(x) dx = Z 1 0 x (x) dx + Z 2 1 x (2 −x) dx = Z 1 0 x2 dx + Z 2 1 2x −x2 dx = "x3 3 #1 0 + "2x2 2 −x3 3 #2 1 = 13 3 −03 3 !
+ 22 −23 3 !
− 12 −13 3 !
= (i) 1 (ii) 2 (iii) 3 (iv) 4.
(b) E (X2).
E X2 = Z 1 0 x2 (x) dx + Z 2 1 x2 (2 −x) dx = Z 1 0 x3 dx + Z 2 1 2x2 −x3 dx = "x4 4 #1 0 + "2x3 3 −x4 4 #2 1 = 14 4 −04 4 !
+ 2(2)3 3 −24 4 !
− 2(1)3 3 −14 4 !
= (i) 4 6 (ii) 5 6 (iii) 6 6 (iv) 7 6.
(c) Variance.
σ2 = Var(X) = E X2 −µ2 = 7 6 −12 = (i) 1 3 (ii) 1 4 (iii) 1 5 (iv) 1 6.
(d) Standard deviation.
σ = √ σ2 = s 1 6 ≈ 208 Chapter 11. Probability and Calculus (LECTURE NOTES 11) (i) 0.27 (ii) 0.31 (iii) 0.41 (iv) 0.53.
(e) Determine P(µ −σ < X < µ + σ).
P(µ −σ < X < µ + σ) ≈ P (1 −0.41 < X < 1 + 0.41) ≈ P (0.59 < X < 1.41) = Z 1 0.59 x dx + Z 1.41 1 (2 −x) dx = "x2 2 #1 0.59 + " 2x −x2 2 #1.41 1 = 12 2 −0.592 2 !
+ 2(1.41) −1.412 2 !
− 2(1) −12 2 !
≈ (i) 0.35 (ii) 0.45 (iii) 0.55 (iv) 0.65.
(f) Median. Since the distribution function is F(x) = Z x 0 t dt = t2 2 #t=x t=0 = x2 2 −02 2 = x2 2 then median m occurs when F(m) = P(X ≤m) = m2 2 = 1 2 so m = (i) 1 (ii) 1.5 (iii) 2 11.3 Special Probability Density Functions Three special probability density functions are discussed: uniform, exponential and normal. The continuous uniform distribution of random variable X, defined on the interval [a, b], has density f(x) = ( 1 b−a, a ≤x ≤b, 0, elsewhere, distribution function, F(x) = 0 x ≤a, x−a b−a a < x < b, 1 x ≥b, Section 3. Special Probability Density Functions (LECTURE NOTES 11) 209 and its expected value (mean), variance and standard deviation are, µ = E(X) = a + b 2 , σ2 = Var(X) = (b −a)2 12 , σ = q Var(X).
The continuous exponential random variable X has density f(x) = ( ae−ax, 0 ≤x < ∞, 0, elsewhere, distribution function, F(x) = ( 0 x < 0, 1 −e−ax 0 ≤x < ∞, and its expected value (mean), variance and standard deviation are, µ = E(X) = 1 a, σ2 = V (Y ) = 1 a2, σ = 1 a.
The continuous normal distribution of random variable X, defined on the interval (−∞, ∞), has density with parameters µ and σ, f(x) = 1 σ √ 2πe−(1/2)[(x−µ)/σ]2 and its expected value (mean), variance and standard deviation are, E(X) = µ, Var(X) = σ2, σ = q Var(X).
A normal random variable, X, may be transformed to a standard normal, Z, f(z) = 1 √ 2πe−z2/2, where µ = 0 and σ = 1 using the following equation, Z = X −µ σ .
The distribution of this density does not have a closed–form expression and so must be solved using numerical integration methods. We will be using the TI-84+, rather than tables, to obtain approximate numerical answers.
Exercise 11.3 (Special Probability Density Functions) 1. Potatoes. An automated process fills one bag after another with Idaho potatoes.
Although each filled bag should weigh 50 pounds, in fact, because of the differing shapes and weights of each potato, each bag weighs anywhere from 49 pounds to 51 pounds, with the following uniform density: f(x) = ( 0.5, 49 < x ≤51, 0, elsewhere.
210 Chapter 11. Probability and Calculus (LECTURE NOTES 11) (a) Since a = 49 and b = 51, the distribution is F(x) = 0 x < 49, x−49 51−49 49 ≤x < 51, 1 x ≥52, and so graphs of density and distribution are given in the figure.
(i) True (ii) False 0 49 50 51 density f(y) 0 49 50 51 distribution F(y) 0.25 0.50 0.75 1 0.25 0.50 0.75 1 Figure 11.4: Distribution function: continuous uniform (b) The chance the bags weight between 49.5 and 51 pounds is P(49.5 < X < 51) = F(51) −F(49.5) = 51 −49 51 −49 −49.5 −49 51 −49 = (i) 0.25 (ii) 0.50 (iii) 0.75 (iv) 1.
Notice P (49.5 < X < 51) = 51−49.5 2 = 0.75 (c) Probability question.
P(X > 49.5) = 1 −P(X ≤49.5) = 1 −F(49.5) = 1 −49.5 −49 51 −49 = (i) 0.25 (ii) 0.50 (iii) 0.75 (iv) 1.
(d) Probability question.
P(X ≥49.5) = 1 −P(X < 49.5) = 1 −F(49.5) = 1 −49.5 −49 51 −49 = (i) 0.25 (ii) 0.50 (iii) 0.75 (iv) 1.
(e) Mean. What is the mean weight of a bag of potatoes?
µ = E(X) = a + b 2 = 49 + 51 2 = (i) 49 (ii) 50 (iii) 51 (iv) 52.
Section 3. Special Probability Density Functions (LECTURE NOTES 11) 211 (f) What is the standard deviation in the weight of a bag of potatoes?
σ = s (b −a)2 12 = s (51 −49)2 12 = (i) 0.44 (ii) 0.51 (iii) 0.55 (iv) 0.58.
2. Another example of a uniform distribution.
f(x) = ( k, −4 < x ≤8, 0, elsewhere, where k is an unknown constant.
0 -4 0 8 density f(y) k 1 area = 1 Figure 11.5: Distribution function: continuous uniform (a) What is k?
Since the uniform has nonzero probability defined in the range −4 ≤x < 8 with length 8 −(−4) = 12, and rectangular area under any uniform must be equal to 1, k = (i) 1 11 (ii) 1 12 (iii) 1 13 (iv) 1 14.
In other words, f(x) = ( 1 12, −4 < x ≤8, 0, elsewhere.
(b) Since a = −4 and b = 8, the distribution is F(x) = x −(−4) 8 −(−4) = (i) x−4 8−4 (ii) x+4 8−4 (iii) x−4 8+4 (iv) x+4 8+4.
212 Chapter 11. Probability and Calculus (LECTURE NOTES 11) (c) Probability.
P(−7 < X < 0) = F(0) −F(−7) = 0 + 4 8 + 4 −0 = (i) 1 12 (ii) 2 12 (iii) 3 12 (iv) 4 12.
Notice P (−7 < X < 0) = 0−(−4) 12 (d) Probability.
P(1 < X < 10) = F(10) −F(1) = 1 −1 + 4 8 + 4 = (i) 7 12 (ii) 8 12 (iii) 9 12 (iv) 10 12.
(e) Mean.
µ = E(X) = a + b 2 = −4 + 8 2 = (i) 1 (ii) 2 (iii) 3 (iv) 4.
(f) Standard deviation.
σ = s (b −a)2 12 = s (8 −(−4))2 12 ≈ (i) 1.93 (ii) 2.69 (iii) 3.46 (iv) 4.33.
(g) Probability between mean and 1 standard deviation above mean P(µ < X < µ + σ) ≈ P (2 < X < 2 + 3.46) P (2 < X < 5.46) = F(5.46) −F(2) = 5.46 −(−4) 8 + 4 −2 −(−4) 8 + 4 = (i) 0.29 (ii) 0.35 (iii) 0.45 (iv) 0.51.
Notice P (2 < X < 5.46) = 5.46−1 12 = 0.28 3. Exponential: waiting time for emails. Density is f(x) = ( ae−ax, 0 ≤x < ∞, 0, elsewhere, Section 3. Special Probability Density Functions (LECTURE NOTES 11) 213 and corresponding distribution function is F(x) = ( 0 x < 0, 1 −e−ax 0 ≤x < ∞.
(a) If a = 1 2, chance of waiting at most 1.1 minutes is P(X ≤1.1) = F(1.1) = 1 −e−1 2 (1.1) ≈ (i) 0.32 (ii) 0.42 (iii) 0.45 (iv) 0.48.
(b) If a = 3, P(X ≤1.1) = F(1.1) = 1 −e−3(1.1) ≈ (i) 0.32 (ii) 0.42 (iii) 0.75 (iv) 0.96.
(c) If a = 5, P(X < 1.1) = F(1.1) = 1 −e−5(1.1) ≈ (i) 0.312 (ii) 0.432 (iii) 0.785 (iv) 0.996.
(d) If a = 3, the chance of waiting at least 0.54 minutes P(X > 0.54) = 1 −F(0.54) = 1 − 1 −e−(3)(0.54) = e−(3)(0.54) ≈ (i) 0.20 (ii) 0.22 (iii) 0.29 (iv) 0.34.
(e) If a = 3, chance of waiting between 1.13 minutes and 1.62 minutes, P(1.13 < X < 1.62) = F(1.62) −F(1.13) = 1 −e−(3)(1.62) − 1 −e−(3)(1.13) = e−(3)(1.13) −e−(3)(1.62) ≈ (i) 0.014 (ii) 0.026 (iii) 0.034 (iv) 0.054.
(f) Expectation and Variance.
For a = 3, µ = 1 a = (i) 1 2 (ii) 1 3 (iii) 1 4 (iv) 1 5.
For a = 1 3, µ = 1 a = (i) 2 (ii) 3 (iii) 4 (iv) 5.
For a = 3, σ = 1 a = (i) 1 3 (ii) 1 5 (iii) 1 7 (iv) 1 9.
For a = 1 3, σ = 1 a = (i) 2 (ii) 3 (iii) 4 (iv) 5.
214 Chapter 11. Probability and Calculus (LECTURE NOTES 11) (g) Probability between mean and 1 standard deviation below mean If a = 3, P(µ −σ < X < µ) = P 1 3 −1 3 < X < 1 3 = P 0 < X < 1 3 = F(0) −F 1 3 = e−(3)(0) −e−(3) 1 3 ≈ (i) 0.33 (ii) 0.43 (iii) 0.53 (iv) 0.63.
4. Nonstandard normal: IQ scores.
It has been found that IQ scores, Y , can be distributed by a normal distribution.
Densities of IQ scores for 16 year olds, Y1, and 20 year olds, Y2, are given by f(y1) = 1 16 √ 2πe−(1/2)[(y−100)/16]2, f(y2) = 1 20 √ 2πe−(1/2)[(y−120)/20]2.
A graph of these two densities is given in the figure.
f(x) f(x) x 20 year old IQs 16 year old IQs µ = 100 µ = 120 σ = 20 σ = 16 Figure 11.6: Normal distributions: IQ scores (a) Mean IQ score for 20 year olds is µ = (choose one) (i) 100 (ii) 120 (iii) 124 (iv) 136.
(b) Average (or mean) IQ scores for 16 year olds is µ = (choose one) (i) 100 (ii) 120 (iii) 124 (iv) 136.
(c) Standard deviation in IQ scores for 20 year olds σ = (choose one) (i) 16 (ii) 20 (iii) 24 (iv) 36.
Section 3. Special Probability Density Functions (LECTURE NOTES 11) 215 (d) Standard deviation in IQ scores for 16 year olds is σ = (choose one) (i) 16 (ii) 20 (iii) 24 (iv) 36.
(e) Normal density for 20 year old IQ scores is (choose one) (i) broader than normal density for 16 year old IQ scores.
(ii) as wide as normal density for 16 year old IQ scores.
(iii) narrower than normal density for 16 year old IQ scores.
(f) Normal density for the 20 year old IQ scores is (choose one) (i) shorter than normal density for 16 year old IQ scores.
(ii) as tall as normal density for 16 year old IQ scores.
(iii) taller than normal density for 16 year old IQ scores.
(g) Total area (probability) under normal density for 20’s IQ scores is (i) smaller than area under normal density for 16’s IQ scores.
(ii) the same as area under normal density for 16’s IQ scores.
(iii) larger than area under normal density for 16’s IQ scores.
(h) Number of different normal densities: (choose one) (i) one (ii) two (iii) three (iv) infinity.
5. Percentages: IQ scores.
Densities of IQ scores for 16 year olds, Y1, and 20 year olds, Y2, are given by f(y1) = 1 16 √ 2πe−(1/2)[(y−100)/16]2, f(y2) = 1 20 √ 2πe−(1/2)[(y−120)/20]2.
(a) For the sixteen year old normal distribution, where µ = 100 and σ = 16, P(Y1 < 84) = Z 84 −∞ 1 16 √ 2πe−(1/2)[(y−100)/16]2 dy1 ≈ (choose one) (i) −0.1587 (ii) 0.1587 (iii) 0.3587 (iv) 0.8413.
(2nd DISTR 2:normalcdf(−2nd EE 99, 84, 100, 16) Notice the normalcdf function has four arguments: normalcdf( low, high, µ, σ).
In this case, the “low” number is “−2nd EE 99” and approximates negative infinity. The “high” number is 84. Finally, this is a nonstandard normal, where the µ and σ are 100 and 16, respectively..
(b) P(Y1 < 100) = (choose one) (i) 0.4413 (ii) 0.5000 (iii) 0.6587 (iv) 0.8413.
(2nd DISTR 2:normalcdf(−2nd EE 99, 100, 100, 16).) (c) P(84 < Y1 < 100) = (choose one) (i) 0.3413 (ii) 0.4901 (iii) 0.5587 (iv) 0.7413.
(2nd DISTR 2:normalcdf(84, 100, 100, 16).) 216 Chapter 11. Probability and Calculus (LECTURE NOTES 11) (d) For the twenty year old normal distribution, where µ = 120 and σ = 20, P(84 < Y2 < 100) = (choose one) (i) 0.0413 (ii) 0.1227 (iii) 0.3597 (iv) 0.5413.
(2nd DISTR 2:normalcdf(84, 100, 120, 20).) (e) Consider the following table of probabilities and possible values of proba-bilities. Use the figure.
f(x) x 100 84 (a) N(100,16 ) P(X > 84) = ?
f(x) f(x) f(x) x 100 96 120 (b) (d) P(96 < X < 120) = ?
x 120 96 N(120,20 ) P(96 < X < 120) = ?
(c) x 120 84 N(120,20 ) P(X > 84) = ?
N(100,16 ) 2 2 2 2 Figure 11.7: Normal probabilities: IQ scores Column I Column II (a) P(Y1 > 84) ≈ (a) 0.4931 (2nd DISTR 2:normalcdf(84, 2nd EE 99, 100, 16)) (b) P(96 < Y1 < 120) ≈ (b) 0.9641 (2nd DISTR 2:normalcdf(96, 120, 100, 16)) (c) P(Y2 > 84) ≈ (c) 0.8413 (2nd DISTR 2:normalcdf(84, 2nd EE 99, 120, 20)) (d) P(96 < Y2 < 120) ≈ (d) 0.3849 (2nd DISTR 2:normalcdf(96, 120, 120, 20)) Using your calculator and the figure above, match the four items in column I with the items in column II.
Column I (a) (b) (c) (d) Column II 6. Standard normal.
Normal densities of IQ scores for 16 year olds, Y1, and 20 year olds, Y2, are Section 3. Special Probability Density Functions (LECTURE NOTES 11) 217 given by f(y1) = 1 16 √ 2πe−(1/2)[(y−100)/16]2, f(y2) = 1 20 √ 2πe−(1/2)[(y−120)/20]2.
Both densities may be transformed to a standard normal with µ = 0 and σ = 1 using the following equation, Z = Y −µ σ .
60 80 100 120 140 160 180 X, nonstandard -3 -2 -1 0 1 2 3 Z, standard 52 68 84 100 116 132 148 X, nonstandard -3 -2 -1 0 1 2 3 Z, standard 16 year olds mean 100, SD 16 20 year olds mean 120, SD 20 110 Figure 11.8: Standard normal and (nonstandard) normal (a) Since µ = 100 and σ = 16, a 16 year old who has an IQ of 132 is z = 132−100 16 = (choose one) (i) 0 (ii) 1 (iii) 2 (iv) 3 standard deviations above the mean IQ, µ = 100.
(b) A 16 year old who has an IQ of 84 is z = 84−100 16 = (choose one) (i) −2 (ii) −1 (iii) 0 (iv) 1 standard deviations below the mean IQ, µ = 100.
(c) Since µ = 120 and σ = 20, a 20 year old who has an IQ of 180 is z = 180−120 20 = (choose one) (i) 0 (ii) 1 (iii) 2 (iv) 3 standard deviations above the mean IQ, µ = 120.
(d) A 20 year old who has an IQ of 100 is z = 100−120 20 = (choose one) (i) −3 (ii) −2 (iii) −1 (iv) 0 standard deviations below the mean IQ, µ = 120.
218 Chapter 11. Probability and Calculus (LECTURE NOTES 11) (e) Although both the 20 year old and 16 year old scored the same, 110, on an IQ test, the 16 year old is clearly brighter relative to his/her age group than is the 20 year old relative his/her age group because z1 = 110 −100 16 = 0.625 > z2 = 110 −120 20 = −0.5.
(i) True (ii) False (f) The probability a 20 year old has an IQ greater than 90 is P(Y2 > 90) = P Z2 > 90 −120 20 = P (Z2 > −1.5) = (i) 0.93 (ii) 0.95 (iii) 0.97 (iv) 0.99.
(nonstandard (x): 2nd DISTR 2:normalcdf(90, 2nd EE 99, 120, 20) or standard (z): 2nd DISTR 2:normalcdf( 90−120 20 , 2nd EE 99, 0, 1).) (g) The probability a 20 year old has an IQ between 125 and 135 is P(125 < Y2 < 135) = P 125 −120 20 < Z2 < 135 −120 20 = P (0.25 < Z2 < 0.75) = (i) 0.13 (ii) 0.17 (iii) 0.27 (iv) 0.31.
(nonstandard (x): 2nd DISTR 2:normalcdf(125, 135, 120, 20) or standard (z): 2nd DISTR 2:normalcdf 125−120 20 , 135−120 20 , 0, 1 .) (h) If a normal random variable Y with mean µ and standard deviation σ can be transformed to a standard one Z with mean µ = 0 and standard deviation σ = 1 using Z = Y −µ σ , then Z can be transformed to Y using Y = µ + σZ.
(i) True (ii) False (i) A 16 year old who has an IQ which is three (3) standards above the mean IQ has an IQ of y1 = 100 + 3(16) = (choose one) (i) 116 (ii) 125 (iii) 132 (iv) 148.
(j) A 20 year old who has an IQ which is two (2) standards below the mean IQ has an IQ of y2 = 120 −2(20) = (choose one) (i) 60 (ii) 80 (iii) 100 (iv) 110.
(k) A 20 year old who has an IQ which is 1.5 standards below the mean IQ has an IQ of y2 = 120 −1.5(20) = (choose one) (i) 60 (ii) 80 (iii) 90 (iv) 95.
Section 3. Special Probability Density Functions (LECTURE NOTES 11) 219 (l) The probability a 20 year old has an IQ greater than one (1) standard deviation above the mean is P(Z2 > 1) = P (Y2 > 120 + 1(20)) = P (Y2 > 140) = (choose one) (i) 0.11 (ii) 0.13 (iii) 0.16 (iv) 0.18.
(standard (z): 2nd DISTR 2:normalcdf(1, 2nd EE 99, 0, 1) or nonstandard (x): 2nd DISTR 2:normalcdf(120 + 1(20), 2nd EE 99, 120, 20).) |
3157 | https://math.stackexchange.com/questions/104603/geometric-series-and-generating-functions | geometric series and generating functions - Mathematics Stack Exchange
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geometric series and generating functions
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For x>1 x>1, the geometric series ∑i=0 n x i∑i=0 n x i is equal to x n+1−1 x−1 x n+1−1 x−1.
By getting the limit,
lim n→∞∑i=0 n x i=lim n→∞x n+1−1 x−1 lim n→∞∑i=0 n x i=lim n→∞x n+1−1 x−1
=∞=∞
However, the generating function ∑n≥0 x n∑n≥0 x n is equal to 1 1−x 1 1−x since
∑n≥0 x n−x∑n≥0 x n=(1+x+x 2+x 3+…)−(x+x 2+x 3+…)=1∑n≥0 x n−x∑n≥0 x n=(1+x+x 2+x 3+…)−(x+x 2+x 3+…)=1
⇒(1−x)∑n≥0 x n=1⇒(1−x)∑n≥0 x n=1
⇒∑n≥0 x n=1 1−x⇒∑n≥0 x n=1 1−x
Obviously, 1 1−x 1 1−x is a finite number.
What is the underlying concept that makes them yield different results?
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asked Feb 1, 2012 at 14:09
edgaredgar
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1 A "generating function" need not be a function in the usual sense.André Nicolas –André Nicolas 2012-02-01 14:49:55 +00:00 Commented Feb 1, 2012 at 14:49
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Although this is not the intended interpretation, one could also make sense of the identity
∑n≥0 x n=(1−x)−1(†)(†)∑n≥0 x n=(1−x)−1
formally; i.e., in the ring of formal power series CC. The user Bill Dubuque explains the idea in several posts in this site; see, e.g., his answer to the post Product of two power series.
The advantage in interpreting (†)(†) formally is that since x x is an indeterminate rather than a number, there are no longer any convergence issues. In this sense, the series ∑n≥0 x n∑n≥0 x n is the same as the finite thing(1−x)−1(1−x)−1.
The flip side of this is that one can no longer plug in values into the series: it does not make sense to talk about the sum of the series at x=1 2 x=1 2. Thus the statement "the series converges for |x|<1|x|<1 and diverges otherwise" is meaningless under this interpretation.
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edited Apr 13, 2017 at 12:21
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answered Feb 1, 2012 at 16:46
SrivatsanSrivatsan
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2 (I am fully aware that there is a chance the OP may not understand the answer. This post is meant for all readers, not just the OP.)Srivatsan –Srivatsan 2012-02-01 16:47:31 +00:00 Commented Feb 1, 2012 at 16:47
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They're not different. The generating function is only defined for when |x|<1|x|<1. Similarly, the geometric series only converges for |x|<1|x|<1. And in that case, lim x n=0 lim x n=0 and you see that the two expressions are the same.
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answered Feb 1, 2012 at 14:18
davidlowryduda♦davidlowryduda
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The given derivation of the fact that ∑n≥0 x n=1 1−x∑n≥0 x n=1 1−x, doesn't address convergence of the intermediate series at all. You can't ignore this issue in a derivation and then hope to plug in numbers to the result.
The derivation given shows that the given identity is valid in the context of formal power series, which are important in the study of generating functions.
As @mixedmath correctly states, you may only substitute numbers into this identity if |x|<1|x|<1.
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answered Feb 1, 2012 at 14:33
Austin RochfordAustin Rochford
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3: Derivatives
Calculus (OpenStax)
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3.1: Defining the Derivative
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3.1: Defining the Derivative
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3.0: Prelude to Derivatives
3.1E: Exercises for Section 3.1
Page ID 2490
Gilbert Strang & Edwin “Jed” Herman
OpenStax
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Learning Objectives
2. Tangent Lines
1. Definition: Difference Quotient
2. Definition: Tangent Line
3. Example 3.1.1: Finding a Tangent Line
1. Solution
4. Example 3.1.2: The Slope of a Tangent Line Revisited/03:_Derivatives/3.01:_Defining_the_Derivative#Example_.5C(.5CPageIndex.7B2.7D.5C):_The_Slope_of_a_Tangent_Line_Revisited)
1. Solution/03:_Derivatives/3.01:_Defining_the_Derivative#Solution_2)
5. Example 3.1.3: Finding the Equation of a Tangent Line/03:_Derivatives/3.01:_Defining_the_Derivative#Example_.5C(.5CPageIndex.7B3.7D.5C):_Finding_the_Equation_of_a_Tangent_Line)
1. Solution/03:_Derivatives/3.01:_Defining_the_Derivative#Solution_3)
6. Exercise 3.1.1/03:_Derivatives/3.01:_Defining_the_Derivative#Exercise_.5C(.5CPageIndex.7B1.7D.5C))
The Derivative of a Function at a Point
Definition: Derivative
Example 3.1.4: Estimating a Derivative
Solution
Exercise 3.1.2
Example 3.1.6: Finding a Derivative
Solution
Example 3.1.7: Revisiting the Derivative
Solution
Exercise 3.1.4
Velocities and Rates of Change
Example 3.1.8: Estimating Velocity
Solution
Exercise 3.1.5
Definition: Instantaneous Rate of Change
Example 3.1.9: Chapter Opener: Estimating Rate of Change of Velocity
Example 3.1.10: Rate of Change of Temperature
Solution
Example 3.1.11: Rate of Change of Profit
Solution
Exercise 3.1.6
Key Concepts
Key Equations
Glossary
Learning Objectives
Recognize the meaning of the tangent to a curve at a point.
Calculate the slope of a tangent line.
Identify the derivative as the limit of a difference quotient.
Calculate the derivative of a given function at a point.
Describe the velocity as a rate of change.
Explain the difference between average velocity and instantaneous velocity.
Estimate the derivative from a table of values.
Now that we have both a conceptual understanding of a limit and the practical ability to compute limits, we have established the foundation for our study of calculus, the branch of mathematics in which we compute derivatives and integrals. Most mathematicians and historians agree that calculus was developed independently by the Englishman Isaac Newton (1643–1727) and the German Gottfried Leibniz (1646–1716), whose images appear in Figure 3.1.1. When we credit Newton and Leibniz with developing calculus, we are really referring to the fact that Newton and Leibniz were the first to understand the relationship between the derivative and the integral. Both mathematicians benefited from the work of predecessors, such as Barrow, Fermat, and Cavalieri. The initial relationship between the two mathematicians appears to have been amicable; however, in later years a bitter controversy erupted over whose work took precedence. Although it seems likely that Newton did, indeed, arrive at the ideas behind calculus first, we are indebted to Leibniz for the notation that we commonly use today.
Figure 3.1.1: Newton and Leibniz are credited with developing calculus independently.
Tangent Lines
We begin our study of calculus by revisiting the notion of secant lines and tangent lines. Recall that we used the slope of a secant line to a function at a point (a,f(a)) to estimate the rate of change, or the rate at which one variable changes in relation to another variable. We can obtain the slope of the secant by choosing a value of x near a and drawing a line through the points (a,f(a)) and (x,f(x)), as shown in Figure 3.1.2. The slope of this line is given by an equation in the form of a difference quotient:
m s e c=f(x)−f(a)x−a
We can also calculate the slope of a secant line to a function at a value a by using this equation and replacing x with a+h, where h is a value close to a. We can then calculate the slope of the line through the points (a,f(a)) and (a+h,f(a+h)). In this case, we find the secant line has a slope given by the following difference quotient with increment h:
m s e c=f(a+h)−f(a)a+h−a=f(a+h)−f(a)h
Definition: Difference Quotient
Let f be a function defined on an interval I containing a. If x≠a is in I, then
Q=f(x)−f(a)x−a
is a difference quotient.
Also, if h≠0 is chosen so that a+h is in I, then
Q=f(a+h)−f(a)h
is a difference quotient with increment h.
These two expressions for calculating the slope of a secant line are illustrated in Figure 3.1.2. We will see that each of these two methods for finding the slope of a secant line is of value. Depending on the setting, we can choose one or the other. The primary consideration in our choice usually depends on ease of calculation.
Figure 3.1.2: We can calculate the slope of a secant line in either of two ways.
In Figure 3.1.3a we see that, as the values of x approach a, the slopes of the secant lines provide better estimates of the rate of change of the function at a. Furthermore, the secant lines themselves approach the tangent line to the function at a, which represents the limit of the secant lines. Similarly, Figure 3.1.3b shows that as the values of h get closer to 0, the secant lines also approach the tangent line. The slope of the tangent line at a is the rate of change of the function at a, as shown in Figure 3.1.3c.
Figure 3.1.3: The secant lines approach the tangent line (shown in green) as the second point approaches the first.
In Figure 3.1.4 we show the graph of f(x)=x and its tangent line at (1,1) in a series of tighter intervals about x=1. As the intervals become narrower, the graph of the function and its tangent line appear to coincide, making the values on the tangent line a good approximation to the values of the function for choices of x close to 1. In fact, the graph of f(x) itself appears to be locally linear in the immediate vicinity of x=1.
Figure 3.1.4: For values of x close to 1, the graph of f(x)=x and its tangent line appear to coincide.
Formally we may define the tangent line to the graph of a function as follows.
Definition: Tangent Line
Let f(x) be a function defined in an open interval containing a. The tangent line to f(x) at a is the line passing through the point (a,f(a)) having slope
(3.1.1)m t a n=lim x→af(x)−f(a)x−a
provided this limit exists.
Equivalently, we may define the tangent line to f(x) at a to be the line passing through the point (a,f(a)) having slope
(3.1.2)m t a n=lim h→0f(a+h)−f(a)h
provided this limit exists.
Just as we have used two different expressions to define the slope of a secant line, we use two different forms to define the slope of the tangent line. In this text we use both forms of the definition. As before, the choice of definition will depend on the setting. Now that we have formally defined a tangent line to a function at a point, we can use this definition to find equations of tangent lines.
Example 3.1.1: Finding a Tangent Line
Find the equation of the line tangent to the graph of f(x)=x 2 at x=3.
Solution
First find the slope of the tangent line. In this example, use Equation 3.1.1.
\displaystyle \begin{align} m_{tan}&=\lim_{x→3}\frac{f(x)−f(3)}{x−3} && \text{Apply the definition.}\[4pt] &=\lim_{x→3}\frac{x^2−9}{x−3} && \text{Substitute }f(x)=x^2\text{ and }f(3)=9\[4pt] &=\lim_{x→3}\frac{(x−3)(x+3)}{x−3} =\lim_{x→3}(x+3)=6 && \text{Factor the numerator to evaluate the limit.}\end{align}
Next, find a point on the tangent line. Since the line is tangent to the graph of f(x) at x=3, it passes through the point (3,f(3)). We have f(3)=9, so the tangent line passes through the point (3,9).
Using the point-slope equation of the line with the slope m=6 and the point (3,9), we obtain the line y−9=6(x−3). Simplifying, we have y=6x−9. The graph of f(x)=x 2 and its tangent line at 3 are shown in Figure 3.1.5.
Figure 3.1.5: The tangent line to f(x) at x=3.
Example 3.1.2: The Slope of a Tangent Line Revisited
Use Equation 3.1.2 to find the slope of the line tangent to the graph of f(x)=x 2 at x=3.
Solution
The steps are very similar to Example 3.1.1. See Equation 3.1.2 for the definition.
m t a n=lim h→0f(3+h)−f(3)h Apply the definition.=lim h→0(3+h)2−9 h Substitute f(3+h)=(3+h)2 and f(3)=9=lim h→09+6h+h 2−9 h Expand and simplify to evaluate the limit.=lim h→0h(6+h)h=lim h→0(6+h)=6
We obtained the same value for the slope of the tangent line by using the other definition, demonstrating that the formulas can be interchanged.
Example 3.1.3: Finding the Equation of a Tangent Line
Find the equation of the line tangent to the graph of f(x)=1/x at x=2.
Solution
We can use Equation 3.1.1, but as we have seen, the results are the same if we use Equation 3.1.2.
\displaystyle \begin{align} m_{tan}&=\lim_{x→2}\frac{f(x)−f(2)}{x−2} && \text{Apply the definition.}\[4pt] &=\lim_{x→2}\frac{\frac{1}{x}−\frac{1}{2}}{x−2} && \text{Substitute }f(x)=\frac{1}{x}\text{ and }f(2)=\frac{1}{2}\[4pt] &=\lim_{x→2}\frac{\frac{1}{x}−\frac{1}{2}}{x−2}⋅\frac{2x}{2x} && \text{Multiply numerator and denominator by }2x\text{ to simplify fractions.}\[4pt] &=\lim_{x→2}\frac{(2−x)}{(x−2)(2x)} && \text{Simplify.}\[4pt] &=\lim_{x→2}\frac{−1}{2x} && \text{Simplify using }\frac{2−x}{x−2}=−1,\text{ for }x≠2.\[4pt] &=−\frac{1}{4} && \text{Evaluate the limit.}\end{align}
We now know that the slope of the tangent line is −1 4. To find the equation of the tangent line, we also need a point on the line. We know that f(2)=1 2. Since the tangent line passes through the point (2,1 2) we can use the point-slope equation of a line to find the equation of the tangent line. Thus the tangent line has the equation y=−1 4x+1. The graphs of f(x)=1 x and y=−1 4x+1 are shown in Figure 3.1.6.
Figure 3.1.6:The line is tangent to f(x) at x=2.
Exercise 3.1.1
Find the slope of the line tangent to the graph of f(x)=x at x=4.
Hint
Use either Equation 3.1.1 or Equation 3.1.2. Multiply the numerator and the denominator by a conjugate.
Answer
1 4
The Derivative of a Function at a Point
The type of limit we compute in order to find the slope of the line tangent to a function at a point occurs in many applications across many disciplines. These applications include velocity and acceleration in physics, marginal profit functions in business, and growth rates in biology. This limit occurs so frequently that we give this value a special name: the derivative. The process of finding a derivative is called differentiation.
Definition: Derivative
Let f(x) be a function defined in an open interval containing a. The derivativeof the function f(x) at a, denoted by f′(a), is defined by
(3.1.3)f′(a)=lim x→af(x)−f(a)x−a
provided this limit exists.
Alternatively, we may also define the derivative of f(x) at a as
(3.1.4)f′(a)=lim h→0f(a+h)−f(a)h.
Example 3.1.4: Estimating a Derivative
For f(x)=x 2, use a table to estimate f′(3) using Equation 3.1.3.
Solution
Create a table using values of x just below 3 and just above 3.
| x | x 2−9 x−3 |
--- |
| 2.9 | 5.9 |
| 2.99 | 5.99 |
| 2.999 | 5.999 |
| 3.001 | 6.001 |
| 3.01 | 6.01 |
| 3.1 | 6.1 |
After examining the table, we see that a good estimate is f′(3)=6.
Exercise 3.1.2
For f(x)=x 2, use a table to estimate f′(3) using Equation 3.1.4.
Hint
Evaluate (x+h)2−x 2 h at h=−0.1,−0.01,−0.001,0.001,0.01,0.1
Answer
6
Example 3.1.6: Finding a Derivative
For f(x)=3x 2−4x+1, find f′(2) by using Equation 3.1.3.
Solution
Substitute the given function and value directly into the equation.
\displaystyle \begin{align} f′(x)&=\lim_{x→2}\frac{f(x)−f(2)}{x−2} && \text{Apply the definition.}\[4pt] &=\lim_{x→2}\frac{(3x^2−4x+1)−5}{x−2} && \text{Substitute }f(x)=3x^2−4x+1\text{ and }f(2)=5.\[4pt] &=\lim_{x→2}\frac{(x−2)(3x+2)}{x−2} && \text{Simplify and factor the numerator.}\[4pt] &=\lim_{x→2}(3x+2) && \text{Cancel the common factor.}\[4pt] &=8 && \text{Evaluate the limit.}\end{align}
Example 3.1.7: Revisiting the Derivative
For f(x)=3x 2−4x+1, find f′(2) by using Equation 3.1.4.
Solution
Using this equation, we can substitute two values of the function into the equation, and we should get the same value as in Example 3.1.6.
\displaystyle \begin{align} f′(2)&=\lim_{h→0}\frac{f(2+h)−f(2)}{h} && \text{Apply the definition.}\[4pt] &=\lim_{h→0}\frac{(3(2+h)^2−4(2+h)+1)−5}{h} && \text{Substitute }f(2)=5\text{ and }f(2+h)=3(2+h)^2−4(2+h)+1.\[4pt] &=\lim_{h→0}\frac{3(4+4h+h^2)-8-4h+1-5}{h} && \text{Expand the numerator.}\[4pt] &=\lim_{h→0}\frac{12+12h+3h^2-12-4h}{h} && \text{Distribute and begin simplifying the numerator.}\[4pt] &=\lim_{h→0}\frac{3h^2+8h}{h} && \text{Finish simplifying the numerator.}\[4pt] &=\lim_{h→0}\frac{h(3h+8)}{h} && \text{Factor the numerator.}\[4pt] &=\lim_{h→0}(3h+8) && \text{Cancel the common factor.}\[4pt] &=8 && \text{Evaluate the limit.} \end{align}
The results are the same whether we use Equation 3.1.3 or Equation 3.1.4.
Exercise 3.1.4
For f(x)=x 2+3x+2, find f′(1).
Hint
Use either Equation 3.1.3, Equation 3.1.4, or try both.
Answer
f′(1)=5
Velocities and Rates of Change
Now that we can evaluate a derivative, we can use it in velocity applications. Recall that if s(t) is the position of an object moving along a coordinate axis, the average velocity of the object over a time interval [a,t] if t>a or [t,a] if t<a is given by the difference quotient
(3.1.5)v ave=s(t)−s(a)t−a.
As the values of t approach a, the values of v ave approach the value we call the instantaneous velocity at a. That is, instantaneous velocity at a, denoted v(a), is given by
(3.1.6)v(a)=s′(a)=lim t→as(t)−s(a)t−a.
To better understand the relationship between average velocity and instantaneous velocity, see Figure 3.1.7. In this figure, the slope of the tangent line (shown in red) is the instantaneous velocity of the object at time t=a whose position at time t is given by the function s(t). The slope of the secant line (shown in green) is the average velocity of the object over the time interval [a,t].
Figure 3.1.7: The slope of the secant line is the average velocity over the interval [a,t]. The slope of the tangent line is the instantaneous velocity.
We can use Equation 3.1.6 to calculate the instantaneous velocity, or we can estimate the velocity of a moving object by using a table of values. We can then confirm the estimate by using Equation 3.1.5.
Example 3.1.8: Estimating Velocity
A lead weight on a spring is oscillating up and down. Its position at time t with respect to a fixed horizontal line is given by s(t)=sint (Figure 3.1.8). Use a table of values to estimate v(0). Check the estimate by using Equation 3.1.6.
Figure 3.1.8: A lead weight suspended from a spring in vertical oscillatory motion.
Solution
We can estimate the instantaneous velocity at t=0 by computing a table of average velocities using values of t approaching 0, as shown in Table 3.1.2.
Table 3.1.2: Average velocities using values of t approaching 0| t | sint−sin0 t−0=sint t |
--- |
| −0.1 | 0.998334166 |
| −0.01 | 0.9999833333 |
| −0.001 | 0.999999833 |
| 0.001 | 0.999999833 |
| 0.01 | 0.9999833333 |
| 0.1 | 0.998334166 |
From the table we see that the average velocity over the time interval [−0.1,0] is 0.998334166, the average velocity over the time interval [−0.01,0] is 0.9999833333, and so forth. Using this table of values, it appears that a good estimate is v(0)=1.
By using Equation 3.1.6, we can see that
v(0)=s′(0)=lim t→0sint−sin0 t−0=lim t→0sint t=1.
Thus, in fact, v(0)=1.
Exercise 3.1.5
A rock is dropped from a height of 64 feet. Its height above ground at time t seconds later is given by s(t)=−16t 2+64,0≤t≤2. Find its instantaneous velocity 1 second after it is dropped, using Equation 3.1.6.
Hint
v(t)=s′(t). Follow the earlier examples of the derivative using Equation 3.1.6.
Answer
−32 ft/s
As we have seen throughout this section, the slope of a tangent line to a function and instantaneous velocity are related concepts. Each is calculated by computing a derivative and each measures the instantaneous rate of change of a function, or the rate of change of a function at any point along the function.
Definition: Instantaneous Rate of Change
Theinstantaneous rate of changeof a function f(x) at a value a is its derivative f′(a).
Example 3.1.9: Chapter Opener: Estimating Rate of Change of Velocity
Reaching a top speed of 270.49 mph, the Hennessey Venom GT is one of the fastest cars in the world. In tests it went from 0 to 60 mph in 3.05 seconds, from 0 to 100 mph in 5.88 seconds, from 0 to 200 mph in 14.51 seconds, and from 0 to 229.9 mph in 19.96 seconds. Use this data to draw a conclusion about the rate of change of velocity (that is, its acceleration) as it approaches 229.9 mph. Does the rate at which the car is accelerating appear to be increasing, decreasing, or constant?
Figure 3.1.9: (credit: modification of work by Codex41, Flickr)
Solution: First observe that 60 mph = 88 ft/s, 100 mph ≈146.67 ft/s, 200 mph ≈293.33 ft/s, and 229.9 mph ≈337.19 ft/s. We can summarize the information in a table.
Table 3.1.3: v(t) at different values of t| t | v(t) |
--- |
| 0 | 0 |
| 3.05 | 88 |
| 5.88 | 147.67 |
| 14.51 | 293.33 |
| 19.96 | 337.19 |
Now compute the average acceleration of the car in feet per second on intervals of the form [t,19.96] as t approaches 19.96, as shown in the following table.
Average acceleration| t | v(t)−v(19.96)t−19.96=v(t)−337.19 t−19.96 |
--- |
| 0.0 | 16.89 |
| 3.05 | 14.74 |
| 5.88 | 13.46 |
| 14.51 | 8.05 |
The rate at which the car is accelerating is decreasing as its velocity approaches 229.9 mph (337.19 ft/s).
Example 3.1.10: Rate of Change of Temperature
A homeowner sets the thermostat so that the temperature in the house begins to drop from 70°F at 9 p.m., reaches a low of 60° during the night, and rises back to 70° by 7 a.m. the next morning. Suppose that the temperature in the house is given by T(t)=0.4t 2−4t+70 for 0≤t≤10, where t is the number of hours past 9 p.m. Find the instantaneous rate of change of the temperature at midnight.
Solution
Since midnight is 3 hours past 9 p.m., we want to compute T′(3). Refer to Equation 3.1.3.
\displaystyle \begin{align} T′(3)&=\lim_{t→3}\frac{T(t)−T(3)}{t−3} && \text{Apply the definition.}\[4pt] &=\lim_{t→3}\frac{0.4t^2−4t+70−61.6}{t−3} && \text{Substitute }T(t)=0.4t^2−4t+70\text{ and }T(3)=61.6.\[4pt] &=\lim_{t→3}\frac{0.4t^2−4t+8.4}{t−3} && \text{Simplify.}\[4pt] &=\lim_{t→3}\frac{0.4(t−3)(t−7)}{t−3}\[4pt] &=\lim_{t→3}0.4(t−7) && \text{Cancel.}\[4pt] &=−1.6 && \text{Evaluate the limit.} \end{align}
The instantaneous rate of change of the temperature at midnight is −1.6°F per hour.
Example 3.1.11: Rate of Change of Profit
A toy company can sell x electronic gaming systems at a price of p=−0.01x+400 dollars per gaming system. The cost of manufacturing x systems is given by C(x)=100x+10,000 dollars. Find the rate of change of profit when 10,000 games are produced. Should the toy company increase or decrease production?
Solution
The profit P(x) earned by producing x gaming systems is R(x)−C(x), where R(x) is the revenue obtained from the sale of x games. Since the company can sell x games at p=−0.01x+400 per game,
R(x)=xp=x(−0.01x+400)=−0.01x 2+400x.
Consequently,
P(x)=−0.01x 2+300x−10,000.
Therefore, evaluating the rate of change of profit gives
\displaystyle \begin{align} P′(10000)&=\lim_{x→10000}\frac{P(x)−P(10000)}{x−10000}\[4pt] &=\lim_{x→10000}\frac{−0.01x^2+300x−10000−1990000}{x−10000}\[4pt] &=\lim_{x→10000}\frac{−0.01x^2+300x−2000000}{x−10000}\[4pt] &=100 \end{align}.
Since the rate of change of profit P′(10,000)>0 and P(10,000)>0, the company should increase production.
Exercise 3.1.6
A coffee shop determines that the daily profit on scones obtained by charging s dollars per scone is P(s)=−20s 2+150s−10. The coffee shop currently charges $3.25 per scone. Find P′(3.25), the rate of change of profit when the price is $3.25 and decide whether or not the coffee shop should consider raising or lowering its prices on scones.
Hint
Use Example 3.1.11 for a guide.
Answer
P′(3.25)=20>0; raise prices
Key Concepts
The slope of the tangent line to a curve measures the instantaneous rate of change of a curve. We can calculate it by finding the limit of the difference quotient or the difference quotient with increment h.
The derivative of a function f(x) at a value a is found using either of the definitions for the slope of the tangent line.
Velocity is the rate of change of position. As such, the velocity v(t) at time t is the derivative of the position s(t) at time t.
Average velocity is given by v ave=s(t)−s(a)t−a. Instantaneous velocity is given by v(a)=s′(a)=lim t→as(t)−s(a)t−a.
We may estimate a derivative by using a table of values.
Key Equations
Difference quotient
Q=f(x)−f(a)x−a
Difference quotient with increment h
Q=f(a+h)−f(a)a+h−a=f(a+h)−f(a)h
Slope of tangent line
m t a n=lim x→af(x)−f(a)x−a
m t a n=lim h→0f(a+h)−f(a)h
Derivative of f(x) at a
f′(a)=lim x→af(x)−f(a)x−a
f′(a)=lim h→0f(a+h)−f(a)h
Average velocity
v ave=s(t)−s(a)t−a
Instantaneous velocity
v(a)=s′(a)=lim t→as(t)−s(a)t−a
Glossary
derivativethe slope of the tangent line to a function at a point, calculated by taking the limit of the difference quotient, is the derivativedifference quotient
of a function f(x) at a is given by
f(a+h)−f(a)h or f(x)−f(a)x−a
differentiationthe process of taking a derivativeinstantaneous rate of changethe rate of change of a function at any point along the function a, also called f′(a), or the derivative of the function at a
This page titled 3.1: Defining the Derivative is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Gilbert Strang & Edwin “Jed” Herman (OpenStax) via source content that was edited to the style and standards of the LibreTexts platform.
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3.0: Prelude to Derivatives
3.1E: Exercises for Section 3.1
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Exponential Decay
Quantum and Nuclear
Episode 516: Exponential and logarithmic equations
Lesson for 16-19
Activity time 145 minutes
Level Advanced
Students may find this mathematical section difficult. It is worth pointing out that they have already covered the basic ideas of radioactive decay in the earlier episodes.
Lesson Summary
Discussion: The exponential decay equation (15 minutes)
Student question: An example using the equation (20 minutes)
Discussion: The logarithmic form of the equation (15 minutes)
Worked example: Using the log equation (20 minutes)
Student questions: Practice calculations (60 minutes)
Discussion (optional): Using Lilley’s formula (15 minutes)
Discussion: The exponential decay equation
Explain that the equation
N=N 0×e-λ t
can be used to generate an exponential decay graph. Work through a numerical example, perhaps related to the dice-throwing analogue (N 0=100 ; λ=1 6). Make sure that your students know how to use the e x key on their calculators.
Emphasise that similar equations apply to activity
A=A 0×e-λ t
and count rate
C=C 0×e-λ t
(Not all the radiation emitted in all directions by a source will collected by a detector lined up in one direction from a source)
Student question: An example using the equation
Set students the task of drawing a graph for a lab source, e.g. Co-60 ( λ=0.132 y-1 , C 0=200 counts s-1 ). They should first calculate and tabulate values of C at intervals of 1 year, and then draw a graph. From the graph, deduce half-life. Does this agree with the value from
T ½ =ln(2)λ
or
T ½ =0.693 λ
Discussion: The logarithmic form of the equation
Point out that a straight line graph is usually more useful than a curve, particularly when dealing with experimental data. Introduce the equation ln N=ln (N 0)−λ t. Emphasise that this embodies the same relationship as the exponential equation. Use a sketch graph to how its relationship to the straight line equation
y=m x+c
( intercept=ln(N 0) , gradient=−l).
Worked examples: Using the log equation
Start with some experimental data (e.g. from the decay of protactinium), draw up a table of ln (count rate) against time. Draw the log graph and deduce l (and hence half-life). The experimental scatter should be obvious on the graph, and hence the value of a straight line graph can be pointed out.
You will need to ensure that your students can find natural logs using their calculators.
Student questions: Practice calculations
Your students should now be able to handle a range of questions involving both e x and ln functions. It is valuable to link them to some of the applications of radioactive materials (e.g. dating of rocks or ancient artefacts, diagnosis and treatment in medicine, etc).
Episode 516-1: Decay in theory and practice (Word, 42 KB)
Episode 516-2: Radioactive decay with exponentials (Word, 77 KB)
Episode 516-3: Radioactive decay used as a clock (Word, 48 KB)
Radio carbon dating
Episode 516-4: Two important dating techniques (Word, 29 KB)
Discussion (optional): Using Lilley’s formula
Some students may benefit from a simpler approach to the mathematics of radioactive decay, using Lilley's formula
fraction= ½ n.
When first introduced at pre-16 level, radioactivity calculations are limited to integral number of half lives. After 1, 2, 3, …, half-lives, 1/2, 1/4, 1/8, … remains. The pattern here is that after n half lives, a fraction= ½ n remains to decay.
This formula works for non integral values of n ; i.e. it also gives the fraction remaining yet to decay after any non-whole number of half-lives (e.g. 2.4, or 3.794). To use this formula, a little skill with a calculator is all that is required.
For example: The T ½ of 14 6 C is 5730 years. What fraction of a sample of 14 6 C remains after 10 000 years? Answer:
fraction remaining= ½ n
The number of half lives,
n=10 000 5730
n=1.745
Thus the
fraction remaining= ½ 1.745
And using the y x button on a calculator gives fraction=0.298.
If students know how to take logarithms (or can lean the log version of the formula), they can solve other problems:
How many years will it take for 99% of 60 27 Co to decay if its half life is 5.23 yr?
The fraction remaining is 1%, so
fraction=0.01.
Hence
0.01= ½ n
Taking logs of both sides gives
ln(0.01)=n×ln( ½ )
giving
n=6.64 half lives,
and so the
number of years=6.64×5.23 years
number of years=34.7 years.
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3160 | https://phys.libretexts.org/Bookshelves/Electricity_and_Magnetism/Essential_Graduate_Physics_-_Classical_Electrodynamics_(Likharev)/09%3A_Special_Relativity/9.03%3A_4-vectors_Momentum_Mass_and_Energy | Skip to main content
9.3: 4-vectors, Momentum, Mass, and Energy
Last updated
: Mar 5, 2022
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9.2: Relativistic Kinematic Effects
9.4: More on 4-vectors and 4-tensors
Page ID
: 57037
Konstantin K. Likharev
Stony Brook University
( \newcommand{\kernel}{\mathrm{null}\,})
Before proceeding to the relativistic dynamics, let us discuss the mathematical formalism that makes all the calculations more compact – and more beautiful. We have already seen that the three spatial coordinates {x,y,z} {x,y,z} and the product ct ct are Lorentz-transformed similarly – see Eqs. (18)-(19) again. So it is natural to consider them as components of a single four-component vector (or, for short, 4-vector),
{x0,x1,x2,x3}≡{ct,r},
{x0,x1,x2,x3}≡{ct,r},(9.48)
with components
x0≡ct,x1≡x,x2≡y,x3≡z.Space-time 4-vector
x0≡ct,x1≡x,x2≡y,x3≡z.Space-time 4-vector(9.49)
According to Eqs. (19), its components are Lorentz-transformed as
xj=3∑j′=0Ljj′x′j′,Lorentz transform: 4-form
xj=∑j′=03Ljj′x′j′,Lorentz transform: 4-form(9.50)
where Ljj′ Ljj′ are the elements of the following 4×4 Lorentz transform matrix
(γβγ00βγγ0000100001).Lorentz transform matrix
Since such 4-vectors are a new notion for this course and will be used for many more purposes than just the space-time transform, we need to discuss the general mathematical rules they obey. Indeed, as was already mentioned in Sec. 8.9, the usual (three-component) vector is not just any ordered set (string) of three scalars {Ax,Ay,Az}; if we want it to represent a reference-frame-independent physical reality, the vector’s components have to obey certain rules at the transfer from one reference frame to another. In particular, in the non-relativistic limit the vector’s norm (its magnitude squared),
A2=A2x+A2y+A2z,
should be invariant with respect to the transfer between different reference frames. However, a naïve extension of this approach to 4-vectors would not work, because, according to the calculations of Sec. 1, the Lorentz transform keeps intact the combinations of the type (7), with one sign negative, rather than the sum of all components squared. Hence for the 4-vectors, all the rules of the game have to be reviewed and adjusted – or rather redefined from the very beginning.
An arbitrary 4-vector is a string of 4 scalars,23
General 4-vector{A0,A1,A2,A3},
whose components Aj, as measured in the systems 0 and 0’ shown in Fig. 1, obey the Lorentz transform relations similar to Eq. (50):
Lorentz transform: general 4-vectorAj=3∑j′=0Ljj′A′j′.
As we have already seen on the example of the space-time 4-vector (48), this means in particular that
Lorentz invarianceA20−3∑j=1A2j=(A′0)2−3∑j=1(A′j)2.
This is the so-called Lorentz invariance condition for the 4-vector’s norm. (The difference between this relation and Eq. (52), pertaining to the Euclidian geometry, is the reason why the Minkowski space is called pseudo-Euclidian.) It is also straightforward to use Eqs. (51) and (54) to check that an evident generalization of the norm, the scalar product of two arbitrary 4-vectors,
Scalar 4-productA0B0−3∑j=1AjBj,
is also Lorentz-invariant.
Now consider the 4-vector corresponding to a small interval between two close world events:
{dx0,dx1,dx2,dx3}={cdt,dr};
its norm,
Interval(ds)2≡dx20−3∑j=1dx2j=c2(dt)2−(dr)2,
is of course also Lorentz-invariant. Since the speed of any particle (or signal) cannot be larger than c, for any pair of world events that are in a causal relation with each other, (dr)2 cannot be larger than (cdt)2, i.e. such time-like interval (ds)2 cannot be negative. The 4D surface separating such intervals from space-like intervals (ds)2<0 is called the light cone (Fig. 9).
Now let us assume that two close world events happen with the same particle that moves with velocity u. Then in the frame moving with the particle (v = u), on the right-hand side of Eq. (58) the last term equals zero, while the involved time is the proper one, so that
ds=cdτ,
where dτ is the proper time interval. But according to Eq. (21), this means that we can write
dτ=dtγ,
where dt is the time interval in an arbitrary (besides being inertial) reference frame, while
β≡uc and γ≡1(1−β2)1/2=1(1−u2/c2)1/2
are the parameters (17) corresponding to the particle’s velocity ( u) in that frame, so that ds=cdt/γ.24
Let us use Eq. (60) to explore whether a 4-vector may be formed using the spatial components of the particle’s velocity
u={dxdt,dydt,dzdt}.
Here we have a slight problem: as Eqs. (22) show, these components do not obey the Lorentz transform. However, let us use dτ≡dt/γ, the proper time interval of the particle, to form the following string:
{dx0dτ,dx1dτ,dx2dτ,dx3dτ}≡γ{c,dxdt,dydt,dzdt}≡γ{c,u}.4-velocity
As it follows from the comparison of the middle form of this expression with Eq. (48), since the time-space vector obeys the Lorentz transform, and τ is Lorentz invariant, the string (63) is a legitimate 4-vector; it is called the 4-velocity of the particle.
Now we are well equipped to proceed to relativistic dynamics. Let us start with such basic notions as the momentum p and the energy E-so far, for a free particle.25 Perhaps the most elegant way to “derive” (or rather guess26) the expressions for p and E as functions of the particle’s velocity u, is based on analytical mechanics. Due to the conservation of v, the trajectory of a free particle in the 4D Minkowski space {ct,r} is always a straight line. Hence, from the Hamilton principle,27 we may expect its action S, between points 1 and 2, to be a linear function of the space-time interval (59):
Free particle: actionS=α∫21ds≡αc∫21dτ≡αc∫t2t1dtγ,
where α is some constant. On the other hand, in analytical mechanics, the action is defined as
S≡∫t2t1Ldt,
where L is particle’s Lagrangian function.28 Comparing these two expressions, we get
L=αcγ≡αc(1−u2c2)1/2.
In the non-relativistic limit (u<<c), this function tends to
L≈αc(1−u22c2)=αc−αu22c.
To correspond to the Newtonian mechanics,29 the last (velocity-dependent) term should equal mu2/2. From here we find α=−mc, so that, finally,
Free particle: Lagrangian functionL=−mc2(1−u2c2)1/2≡−mc2γ.
Now we can find the Cartesian components pj of the particle’s momentum as the generalized momenta corresponding to the corresponding components rj(j=1,2,3) of the 3D radius-vector r:30
pj=∂L∂˙rj≡∂L∂uj=−mc2∂∂uj(1−u21+u22+u23c2)1/2=muj(1−u2/c2)1/2≡mγuj.
Thus for the 3D vector of momentum, we can write the result in the same form as in non-relativistic mechanics,
p=mγu≡Mu,Relativistic momentum
using the reference-frame-dependent scalar M (called the relativistic mass) defined as
M≡mγ=m(1−u2/c2)1/2≥m,Relativistic mass
m being the non-relativistic mass of the particle. (More often, m is called the rest mass, because in the reference frame in that the particle rests, Eq. (71) yields M=m.)
Next, let us return to analytical mechanics to calculate the particle’s energy E (which for a free particle coincides with its Hamiltonian function H):31
E=H=3∑j=1pjuj−L=p⋅u−L=mu2(1−u2/c2)1/2+mc2(1−u2c2)1/2≡mc2(1−u2/c2)1/2.E=Mc2
which expresses the relation between the free particle’s mass and its energy.32 In the non-relativistic limit, it reduces to
E=mc2(1−u2/c2)1/2≈mc2(1+u22c2)=mc2+mu22,
the first term mc2 being called the rest energy of a particle.
Now let us consider the following string of 4 scalars:
{Ec,p1,p2,p3}≡{Ec,p}.4-vector of energy-momentum
Using Eqs. (70) and (73) to represent this expression as
{Ec,p}=mγ{c,u},
and comparing the result with Eq. (63), we immediately see that, since m is a Lorentz-invariant constant, this string is a legitimate 4-vector of energy-momentum. As a result, its norm,
(Ec)2−p2,
is Lorentz-invariant, and in particular, has to be equal to the norm in the particle-bound frame. But in that frame, p=0, and, according to Eq. (73), E=mc2, and so that the norm is just
(Ec)2=(mc2c)2≡(mc)2,
so that in an arbitrary frame
(Ec)2−p2=(mc)2.
This very important relation33 between the relativistic energy and momentum (valid for free particles only!) is usually represented in the form34
Free particle: energyE2=(mc2)2+(pc)2.
According to Eq. (70), in the so-called ultra-relativistic limit u→c, p tends to infinity, while mc2 stays constant so that pc/mc2→∞. As follows from Eq. (78), in this limit E≈pc. Though the above discussion was for particles with finite m, the 4-vector formalism allows us to consider compact objects with zero rest mass as ultra-relativistic particles for which the above energy-to-moment relation,
E=pc, for m=0,
is exact. Quantum electrodynamics35 tells us that under certain conditions, the electromagnetic field quanta (photons) may be also considered as such massless particles with momentum p=ℏk. Plugging (the modulus of) the last relation into Eq. (78), for the photon’s energy we get E=pc=ℏkc=ℏω. Please note again that according to Eq. (73), the relativistic mass of a photon is not equal to zero: M=E/c2=ℏω/c2, so that the term “massless particle” has a limited meaning: m=0. For example, the relativistic mass of an optical phonon is of the order of 10−36 kg. On the human scale, this is not too much, but still a noticeable (approximately one-millionth) part of the rest mass me of an electron.
The fundamental relations (70) and (73) have been repeatedly verified in numerous particle collision experiments, in which the total energy and momentum of a system of particles are conserved – at the same conditions as in non-relativistic dynamics. (For the momentum, this is the absence of external forces, and for the energy, the elasticity of particle interactions – in other words, the absence of alternative channels of energy escape.) Of course, generally only the total energy of the system is conserved, including the potential energy of particle interactions. However, at typical high-energy particle collisions, the potential energy vanishes so rapidly with the distance between them that we can
use the momentum and energy conservation laws using Eq. (73).
As an example, let us calculate the minimum energy Emin of a proton (pa), necessary for the well-known high-energy reaction that generates a new proton-antiproton pair, pa+pb→p+p+p+¯p, provided that before the collision, the proton pb had been at rest in the lab frame. This minimum corresponds to the vanishing relative velocity of the reaction products, i.e. their motion with virtually the same velocity (ufin), as seen from the lab frame – see Fig. 10.
Due to the momentum conservation, this velocity should have the same direction as the initial velocity (umin) of proton pa. This is why two scalar equations: for energy conservation,
mc2(1−u2min/c2)1/2+mc2=4mc2(1−u2fin/c2)1/2,
and for momentum conservation,
mu(1−u2min/c2)1/2+0=4mufin(1−u2fin/c2)1/2,
are sufficient to find both umin and ufin. After a conceptually simple but technically somewhat tedious solution of this system of two nonlinear equations, we get
umin=4√37c,ufin=√32c.
Finally, we can use Eq. (73) to calculate the required energy; the result is Emin=7mc2. (Note that at this threshold, only 2mc2 of the kinetic energy Tmin=Emin−mc2=6mc2 of the initial moving particle, go into the “useful” proton-antiproton pair production.) The proton’s rest mass, mp≈1.67×10−27 kg, corresponds to mpc2≈1.502×10−10 J≈0.938GeV, so that Emin≈6.57GeV.
The second, more intelligent way to solve the same problem is to use the center-of-mass (c.o.m.) reference frame that, in relativity, is defined as the frame in that the total momentum of the system vanishes.36 In this frame, at E=Emin, the velocity and momenta of all reaction products are vanishing, while the velocities of protons pa and pb before the collision are equal and opposite, with an initially unknown magnitude u′. Hence the energy conservation law becomes
2mc2(1−u′2/c2)1/2=4mc2,
readily giving u′=(√3/2)c. (This is of course the same result as Eq. (81) gives for ufin .) Now we can use the fact that the velocity of the proton pa in the c.o.m. frame is (−u′), to find its lab-frame speed, using the velocity transform (25):
umin=2u′1+u′2/c2.
With the above result for u′, this relation gives the same result as the first method, umin=(4√3/7)c, but in a simpler way.
Reference
23 Such vectors are said to reside in so-called 4D Minkowski spaces – called after Hermann Minkowski who was the first one to recast (in 1907) the special relativity relations in a form in which the spatial coordinates and time (or rather ct) are treated on an equal footing.
24 I have opted against using special indices (e.g., βu,γu) to distinguish Eqs. (17) and (61) here and below, in a hope that the suitable velocity (of either a reference frame or a particle) will be always clear from the context.
25 I am sorry for using, just as in Sec. 6.3, the same traditional notation (p) for the particle’s momentum as had been used earlier for the electric dipole moment. However, since the latter notion will be virtually unused in the balance of this course, this may hardly lead to confusion.
26 Indeed, such a derivation uses additional assumptions, however natural (such as the Lorentz-invariance of S), i.e. it can hardly be considered as a real proof of the final results, so that they require experimental confirmation. Fortunately, such confirmations have been numerous – see below.
27 See, e.g., CM Sec. 10.3.
28 See, e.g., CM Sec. 2.1.
29 See, e.g., CM Eq. (2.19b).
30 See, e.g., CM Sec. 2.3, in particular Eq. (2.31).
31 See, e.g., CM Eq. (2.32).
32 Let me hope that the reader understands that all the layman talk about the “mass to energy conversion” is only valid in a very limited sense of the word. While the Einstein relation (73) does allow the conversion of “massive” particles (with m≠0) into particles with m=0, such as photons, each of the latter particles also has a non-zero relativistic mass M, and simultaneously the energy E related to this M by Eq. (73).
33 Please note one more simple and useful relation following from Eqs. (70) and (73): p=(E/c2)u.
34 It may be tempting to interpret this relation as the perpendicular-vector-like addition of the rest energy mc2 and the “kinetic energy” pc, but from the point of view of the total energy conservation (see below), a better definition of the kinetic energy is T(u)≡E(u)−E(0).
35 It is briefly reviewed in QM Chapter 9.
36 Note that according to this definition, the c.o.m.’s radius-vector is R=ΣkMkrk/ΣkMk≡Σkγkmkrk/Σkγkmk, i.e. is generally different from the well-known non-relativistic expression R=Σkmkrk/Σkmk.
9.2: Relativistic Kinematic Effects
9.4: More on 4-vectors and 4-tensors |
3161 | https://www.youtube.com/watch?v=eLwraf_A80U | Nuclear Binding Energy Per Nucleon & Mass Defect Problems - Nuclear Chemistry
The Organic Chemistry Tutor
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This nuclear chemistry video tutorial explains how to calculate the nuclear binding energy per nucleon for an isotope as well as the mass defect. The mass defect is the difference between the mass of the nucleus and the mass of the nucleons that make up the nucleus such as the protons and neutrons. The change in energy is equal to the mass defect times the square of the speed of light. This video explains how to convert joules into MeV or mega electron volts. This video contains plenty of examples and practice problems.
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Transcript:
in this video we're going to talk about how to calculate the mass defect of an isotope and also how to calculate the nuclear binding energy per nucleon so let's start with this problem what is the mass defect of carbon-12 the mass defect is the difference in the mass of the nucleus and the mass of the individual particles that make up the nucleus so let's talk about carbon 12. carbon 12 has a mass of 12 and an atomic number six the number of protons is equal to the atomic number so carbon 12 has six protons the difference between the mass number and the atomic number is the number of neutrons so 12 minus six is six now for an atom the number of protons and electrons are the same an atom is electrically neutral so an atom of carbon 12 has six electrons ions have different numbers of protons and electrons now let's calculate the mass of the nucleus separately the mass of the nucleus is the difference between the mass of the carbon atom and the mass of the electrons in that atom so we said that carbon has six electrons now let's calculate the mass of the carbon atom in kilograms let me do this at the bottom here now the atomic mass of carbon is 12 amu 12 atomic mass units and you need to know that one atomic mass unit is 1.66054 times 10 to the negative 27 kilograms and you can look that up in the internet it's not hard to find just type in amu to kilograms in google or something and it should come up now let's go ahead and multiply those two numbers so the mass of carbon in kilograms is 1.99 265 times 10 to the negative 26 kilograms and then -6 now the mass of the electron in kilograms is 9.11 times 10 to the negative 31 kilograms all right so let's go ahead and plug this in so the mass of the carbon nucleus is 1.99210 times 10 to the negative 26 kilograms so notice that the mass of the electron is relatively insignificant if you look at the difference between these two numbers they're almost the same 1.99265 versus 1.99210 so let's say if you decide to neglect this value your answer shouldn't change too much but if you want to get the exact answer subtract the mass of the carbon atom by the number of electrons present so now that we have that let's get the other stuff the mass of the particles so we said that carbon-12 has six protons and six neutrons so the mass of the particles is going to be six mp six protons and six mn for neutrons so we have the mass of a proton here it is so that's a 1.67262 times 10 to the negative 27 a lot of numbers to write down and then we have the mass of a neutron which is a little bit heavier than a proton but it's almost the same all right so let's figure out what that's going to be so if i typed in everything correctly this is what i got 2.00853 times 10 to the negative 26 kilograms so hopefully you got that too so now we can calculate the mass defect so the mass of the nucleus that's 1.9921 times 10 to the negative 26 kilograms and then the mass of the individual particles that make up the nucleus is this number and so let's see what this is going to be so this is going to be negative 1.643 times 10 to the negative 28 kilograms so that's the difference between the mass of the carbon nucleus and the mass of the individual subatomic particles that make up the nucleus so that's the mass defect of carbon 12. now let's calculate the nuclear binding energy per nucleon so when you use this equation delta e is equal to delta m c squared so the mass defect is this number c represents the speed of light which is 3 times 10 to the 8 meters per second and then we have to square that number so the change in energy is negative 1.47 times 10 to the negative 11 joules per individual nucleus now you might be wondering why is this answer negative now what this means is that if you take the six protons and the six neutrons and put them together to form the carbon nucleus energy will be released since mass is lost so what really happens is the mass is converted to energy so as those six protons and six neutrons come together to form the nucleus that change in mass that difference in mass was converted into the energy that was released by that process so that's why it's negative now the reverse is positive let's say if you have a carbon nucleus it's going to take this much energy to break the nucleus into its individual particles so the sine depends on what you're trying to do if you're trying to bring the particles together to make up a nucleus then energy will be released delta e will be negative if you're trying to put in energy to break up the nucleus into its individual particles then delta e is positive now to calculate the binding energy the energy that we need to break the nucleus that's going to be a positive value so we need to calculate the binding energy per nucleon now you need to know what a nucleon is what do you think a nucleon is a nucleon is basically a particle in the nucleus so protons and neutrons are considered to be nucleons not electrons so carbon-12 has six protons and six neutrons so six plus six is twelve so we could say that it has 12 nucleons so to calculate the binding energy per nucleon the nuclear binding energy it's going to be this value well let me convert it so let's start with this number this is the energy per nucleus and one carbon-12 nucleus has 12 nucleons and so the unit nucleus cancels and so now we have the joules per nucleon and so this comes out to be 1.232 times 10 to the negative 12 joules per nucleon so that's the binding energy per nucleon now sometimes you may need to report your answer in mega electron volts per nucleon if you need to do that you can use this conversion one mega electron volt is 1.6 times 10 to the negative 13 joules so the unit joules will cancel and let's go ahead and divide these two numbers so you should get 7.7 mega electron volt per nucleon so that is the nuclear binding energy of carbon-12 that's all you need to do number two calculate the change in energy in joules and mega electron volts if 5 moles of nitrogen 14 nuclei were formed from protons and neutrons so if you want to try this problem feel free to pause the video and work on it so let's begin so we said that the mass defect is the difference between the mass of the nucleus and the mass of the individual particles that make up the nucleus so in this case we're combining protons and neutrons to make up the nitrogen nuclei and so energy is going to be released when that happens anytime a bond is formed between two atoms or two particles energy is released so we should expect a negative energy value for our answer now let's calculate the mass of the nucleus first so it's going to be the mass of the nitrogen atom minus the seven electrons that are in the nitrogen cloud now keep in mind nitrogen 14 has an atomic number of seven so it has seven protons seven neutrons and also seven electrons the number of neutrons is going to be 14 minus seven and for an atom the protons and electrons are the same if the atom is electrically neutral which by definition it is now let's calculate the mass of the nitrogen atom so we're given the mass and atomic mass units so that's 14.0031 amu now based on the last problem we know that one atomic mass unit is one point six six zero five four times ten to the negative twenty seven kilograms so that's not going to change and so the mass of the nitrogen atom is 2.32527 times 10 to the negative 26 kilograms and then the mass of the electron is 9.11 times 10 to the minus 31 kilograms so let's go ahead and plug that in so you should get 2.32463 times 10 to the negative 26 kilograms so now we need to calculate the mass of the subatomic particles in the nucleus and so we said that nitrogen 14 has seven protons and seven neutrons so the mass of a proton it's 1.67262 times 10 to the negative 27 kilograms and then the mass of a neutron is 1.67493 times 10 to the negative 27 kilograms so you should get this number two point three four three two eight five times ten to the negative twenty six so now let's go ahead and use this equation so the mass defect is the difference between these two numbers so if we subtract 2.32463 times 10 to the negative 26 by 2.34 two eight five to the same power you should get negative one point eight six five five times ten to the negative twenty eight kilograms so that is the mass defect in this problem that's the change in mass so when the seven protons and the seven neutrons get together to form the nitrogen nucleus this is how much mass is lost and so that mass is changed into energy so now let's calculate how much energy is released during the process so this is going to be negative 1.8655 times 10 to the negative 28 kilograms multiplied by the square of the speed of light and so the change in energy is negative 1.679 times 10 to the negative 11 joules and this is per nucleus so that's how much energy is released if only one nucleus is formed what about if five moles of nitrogen nuclei were formed how much energy would be released let's start with five moles now one mole is equal to avogadro's number 6.022 times 10 to the 23 and we're dealing with the nitrogen nuclei and the energy change for just one nucleus is 1.679 times 10 to the negative 11 joules and so we can see the unit moles will cancel and the nucleus will cancel as well so this will give us the energy change in joules if five moles of nitrogen nuclei were formed from protons and neutrons and so this is equal to 5.055 times 10 to the 13 joules and so that's the answer in joules now if you want the answer in mega electron volts simply convert it divided by 1.6 times 10 to the 13 joules okay that's supposed to be a j so if you're wondering why the sudden change in the screen that's just me editing this video because i made it before but i noticed that i missed the negative sign one mega electron volt is not positive 13 but it's 1.6 times 10 to the negative 13. and so i had to correct that mistake and so the answer that you should get it's going to be 3.16 times 10 to the 26 mega electron volts so one little negative sign can change your answer drastically so that's the answer that you should have and so that's it for this video thanks for watching and have a good day you |
3162 | https://bcan.org/intravesical-therapy/ | Skip to content
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Intravesical Therapy
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What is intravesical therapy?
Intravesical therapy is a treatment that puts medicine directly into the bladder. This way, the medicine goes straight to the cancer cells without affecting the rest of the body. It is helpful for treating tumors that are hard to remove, like CIS, or high-risk non-muscle invasive bladder cancer (NMIBC), which is more likely to return after surgery. The treatment is given through a small tube called a catheter, and patients hold the medicine in their bladder for a certain time before it is drained out.
Intravesical Therapy for NMIBC (non-muscle invasive bladder cancer) includes:
BCG (Bacille Calmette-Guerin) Immunotherapy
BCG is a type of immunotherapy that uses weakened bacteria to activate the immune system against bladder cancer cells. It’s administered directly into the bladder for patients with high-grade, non-invasive bladder cancers, like CIS or T1 tumors that have not spread into the muscle. BCG is usually given in an initial six-week cycle, and doctors may recommend additional maintenance doses to help reduce the risk of cancer coming back. BCG has been effective for many patients, though it can cause side effects like bladder irritation, fever, or mild bleeding.
What about the current BCG shortage?
ImmunityBio is working to address the shortage and ongoing patient need of Bacillus Calmette-Guérin (BCG) by developing a new form of BCG called recombinant mycobacterium BCG (or rBCG), and making it available in the United States both through clinical trials and an Expanded Access program for patients that do not qualify for clinical trials. You can learn more about it here.
Adstiladrin (Nadofaragene Firadenovec-vncg)
Adstiladrin is a gene therapy approved for patients with carcinoma in situ (CIS) who did not respond to BCG. Delivered directly into the bladder, Adstiladrin introduces a gene called interferon alpha-2b, which activates immune cells to target and kill cancer cells. This local approach minimizes side effects outside of the bladder, although some patients may experience bladder irritation. For some, Adstiladrin can be effective in reducing or eliminating cancer that remains after BCG treatment
Anktiva (Nogapendekin Alfa Inbakicept-pmln)
Anktiva is an immunotherapy that enhances the immune response against bladder cancer by activating a protein called interleukin-15. It’s approved for patients with CIS who didn’t respond to BCG and is used together with BCG to boost its effectiveness. Delivered through a catheter into the bladder, Anktiva can help reduce cancer in some patients, though results vary. Side effects are usually limited to the bladder, including irritation.
BCG, Adstiladrin, and Anktiva all work to stimulate the immune system to fight cancer cells in the bladder. This is different from other immunotherapies like checkpoint inhibitors, which block pathways cancer cells use to hide from the immune system. Each therapy has its own side effects and benefits, so it is important to talk with your doctor about which might work best for your type of cancer and treatment plan.
Other Intravesical Therapies
Chemotherapy (chemo) drugs can be put right into the bladder through a catheter. These drugs kill actively growing cancer cells. Intravesical chemotherapy is most often used when intravesical immunotherapy with BCG doesn’t work. Giving chemo right into the bladder instead of injecting it into the bloodstream means these medicines that can kill cancer, usually do not reach and effect other parts of the body. This helps people avoid many of the side common effects that a person who receives systemic chemotherapy might experience.
Intravesical chemotherapy is a treatment for bladder cancer that also puts special medicine directly into the bladder using a thin tube called a catheter. This type of chemotherapy is usually used when another treatment called BCG does not work. By delivering the medicine straight to the bladder, it targets the cancer cells without affecting other parts of the body. This helps to reduce many of the side effects that people might have if they received chemotherapy through their bloodstream.
Gemcitabine
Gemcitabine is a medicine that can help stop bladder tumors from coming back after surgery (TURBT). It is usually well tolerated, meaning most people don’t have serious side effects. Some people may have minor problems, like needing to urinate more often. When taken with another medicine known as oral alkalization (a method to help balance acidity), it causes very few side effects like nausea, vomiting, hair loss, or low blood counts.
Mitomycin C
Mitomycin Cis another medicine used after TURBT that can help reduce the chance of bladder tumors returning by up to 50%. One good thing about Mitomycin C is that it does not easily enter the bloodstream, which makes it less risky than chemotherapy given through an IV. Some side effects can include pain when urinating or irritation of the bladder lining, which can feel like a urinary tract infection. These side effects are temporary and will go away once the treatment stops.
Gemcitabine with Docetaxel
Gemcitabine with Docetaxeltreatment combines gemcitabine with another medicine called docetaxel, and both are given directly into the bladder in a series of visits to the doctor’s office after TURBT. The side effects are similar to those of gemcitabine alone, and during the visit, one drug is given, drained out of the bladder, and then the second drug is administered.
Vicinium
Vicinium is a treatment for patients whose cancer has returned after BCG therapy. It targets a specific part of the cancer cells in the bladder. About half of the patients may have side effects, but most of these are not serious. Vicinium requires a more frequent treatment schedule, starting with two times a week for six weeks, then weekly for six weeks, and then every other week for two years. It is important to note that Vicinium is not yet approved by the U.S. Food and Drug Administration.
Valrubicin
Valrubicin may also be given to some patients, especially if their carcinoma in situ (CIS) bladder cancer has not responded to BCG treatment and they cannot have surgery right away to remove the bladder.
Questions for Your Doctor
Discuss with your doctor which intravesical therapy might suit your situation, how long treatment might last, and what side effects to monitor.
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Angle bisector
This is an AoPSWiki Word of the Week for June 6-12
Contents
[hide]
1 Angle Bisector
2 Features of Angle Bisectors
3 Proofs
3.1 Angle Bisector Locus Theorem
3.2 Angle Bisector Theorem
3.3 Internal and External Angle Bisectors Are Perpendicular
4 See also
Angle Bisector
For an angle, the (internal) angle bisector of is the line from such that the angle between this line and is congruent to the angle between this line and :
An angle also has an external angle bisector, which bisects external angle :
The external angle is defined by and the two angle bisectors are perpendicular to each other.
Features of Angle Bisectors
The distances from a point on an angle bisector to both of its sides are equal.
The angle bisectors are the locus of points which are equidistant from the two sides of the angle.
A reflection about either angle bisector maps the two sides of the angle to each other.
In a triangle, the Angle Bisector Theorem gives the ratio in which the angle bisector cuts the opposite side.
In a triangle, the internal angle bisectors (which are cevians) all intersect at the incenter of the triangle. The internal angle bisector of one angle and the external angle bisectors of the other two angles all intersect at an excenter of the triangle.
A bisector of an angle can be constructed using a compass and straightedge.
Triangle with incenterI, incircle (blue), angle bisectors (orange), and external angle bisectors (green)
Proofs
Angle Bisector Locus Theorem
Theorem: A point lies on the internal angle bisector of if and only if it is equidistant from the sides and .
Proof:
Let point lie on the angle bisector of . Drop perpendiculars from to lines and , meeting them at points and respectively.
Since and both triangles and are right triangles with a shared hypotenuse and equal angles, they are congruent by AAS. Hence,
Conversely, if is equidistant from and , then , and lies on the angle bisector of .
Angle Bisector Theorem
Theorem: In triangle , if the internal angle bisector of intersects at point , then
Proof:
Let , and draw the angle bisector .
Construct a line through parallel to , and let it intersect the extension of at point .
Since (by alternate interior angles), triangles and are similar by AA.
Therefore,
Internal and External Angle Bisectors Are Perpendicular
Theorem: The internal and external angle bisectors of any angle are perpendicular.
Proof:
Let the internal and external angle bisectors divide and its supplement into two equal parts.
The internal angle is , and the external angle is .
So the internal bisector makes an angle of with one side, and the external bisector makes an angle of with the same side. Adding them: so the two bisectors are perpendicular.
See also
Cevian
Geometry
Stewart's Theorem
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3164 | https://web.mit.edu/2.14/www/Handouts/PoleZero.pdf | MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING 2.14 Analysis and Design of Feedback Control Systems Understanding Poles and Zeros 1 System Poles and Zeros The transfer function provides a basis for determining important system response characteristics without solving the complete differential equation. As defined, the transfer function is a rational function in the complex variable s = σ + jω, that is H(s) = bmsm + bm−1sm−1 + . . . + b1s + b0 ansn + an−1sn−1 + . . . + a1s + a0 (1) It is often convenient to factor the polynomials in the numerator and denominator, and to write the transfer function in terms of those factors: H(s) = N(s) D(s) = K (s −z1)(s −z2) . . . (s −zm−1)(s −zm) (s −p1)(s −p2) . . . (s −pn−1)(s −pn) , (2) where the numerator and denominator polynomials, N(s) and D(s), have real coefficients defined by the system’s differential equation and K = bm/an. As written in Eq. (2) the zi’s are the roots of the equation N(s) = 0, (3) and are defined to be the system zeros, and the pi’s are the roots of the equation D(s) = 0, (4) and are defined to be the system poles. In Eq. (2) the factors in the numerator and denominator are written so that when s = zi the numerator N(s) = 0 and the transfer function vanishes, that is lim s→zi H(s) = 0.
and similarly when s = pi the denominator polynomial D(s) = 0 and the value of the transfer function becomes unbounded, lim s→pi H(s) = ∞.
All of the coefficients of polynomials N(s) and D(s) are real, therefore the poles and zeros must be either purely real, or appear in complex conjugate pairs. In general for the poles, either pi = σi, or else pi, pi+1 = σi±jωi. The existence of a single complex pole without a corresponding conjugate pole would generate complex coefficients in the polynomial D(s). Similarly, the system zeros are either real or appear in complex conjugate pairs.
1 Example A linear system is described by the differential equation d2y dt2 + 5dy dt + 6y = 2du dt + 1.
Find the system poles and zeros.
Solution: From the differential equation the transfer function is H(s) = 2s + 1 s2 + 5s + 6.
(5) which may be written in factored form H(s) = 1 2 s + 1/2 (s + 3)(s + 2) = 1 2 s −(−1/2) (s −(−3))(s −(−2)).
(6) The system therefore has a single real zero at s = −1/2, and a pair of real poles at s = −3 and s = −2.
The poles and zeros are properties of the transfer function, and therefore of the differential equation describing the input-output system dynamics. Together with the gain constant K they completely characterize the differential equation, and provide a complete description of the system.
Example A system has a pair of complex conjugate poles p1, p2 = −1 ± j2, a single real zero z1 = −4, and a gain factor K = 3. Find the differential equation representing the system.
Solution: The transfer function is H(s) = K s −z (s −p1)(s −p2) = 3 s −(−4) (s −(−1 + j2))(s −(−1 −j2)) = 3 (s + 4) s2 + 2s + 5 (7) and the differential equation is d2y dt2 + 2dy dt + 5y = 3du dt + 12u (8) 2 Figure 1: The pole-zero plot for a typical third-order system with one real pole and a complex conjugate pole pair, and a single real zero.
1.1 The Pole-Zero Plot A system is characterized by its poles and zeros in the sense that they allow reconstruction of the input/output differential equation. In general, the poles and zeros of a transfer function may be complex, and the system dynamics may be represented graphically by plotting their locations on the complex s-plane, whose axes represent the real and imaginary parts of the complex variable s.
Such plots are known as pole-zero plots. It is usual to mark a zero location by a circle (◦) and a pole location a cross (×). The location of the poles and zeros provide qualitative insights into the response characteristics of a system. Many computer programs are available to determine the poles and zeros of a system from either the transfer function or the system state equations . Figure 1 is an example of a pole-zero plot for a third-order system with a single real zero, a real pole and a complex conjugate pole pair, that is; H(s) = (3s + 6) (s3 + 3s2 + 7s + 5) = 3 (s −(−2)) (s −(−1))(s −(−1 −2j))(s −(−1 + 2j)) 1.2 System Poles and the Homogeneous Response Because the transfer function completely represents a system differential equation, its poles and zeros effectively define the system response.
In particular the system poles directly define the components in the homogeneous response. The unforced response of a linear SISO system to a set of initial conditions is yh(t) = n i=1 Cieλit (9) where the constants Ci are determined from the given set of initial conditions and the exponents λi are the roots of the characteristic equation or the system eigenvalues. The characteristic equation is D(s) = sn + an−1sn−1 + . . . + a0 = 0, (10) and its roots are the system poles, that is λi = pi, leading to the following important relationship: 3 Figure 2: The specification of the form of components of the homogeneous response from the system pole locations on the pole-zero plot.
The transfer function poles are the roots of the characteristic equation, and also the eigenvalues of the system A matrix.
The homogeneous response may therefore be written yh(t) = n i=1 Ciepit.
(11) The location of the poles in the s-plane therefore define the n components in the homogeneous response as described below: 1. A real pole pi = −σ in the left-half of the s-plane defines an exponentially decaying component , Ce−σt, in the homogeneous response.
The rate of the decay is determined by the pole location; poles far from the origin in the left-half plane correspond to components that decay rapidly, while poles near the origin correspond to slowly decaying components.
2. A pole at the origin pi = 0 defines a component that is constant in amplitude and defined by the initial conditions.
3. A real pole in the right-half plane corresponds to an exponentially increasing component Ceσt in the homogeneous response; thus defining the system to be unstable.
4. A complex conjugate pole pair σ ± jω in the left-half of the s-plane combine to generate a response component that is a decaying sinusoid of the form Ae−σt sin (ωt + φ) where A and φ are determined by the initial conditions. The rate of decay is specified by σ; the frequency of oscillation is determined by ω.
5. An imaginary pole pair, that is a pole pair lying on the imaginary axis, ±jω generates an oscillatory component with a constant amplitude determined by the initial conditions.
4 6. A complex pole pair in the right half plane generates an exponentially increasing component.
These results are summarized in Fig. 2.
Example Comment on the expected form of the response of a system with a pole-zero plot shown in Fig. 3 to an arbitrary set of initial conditions.
Figure 3: Pole-zero plot of a fourth-order system with two real and two complex conjugate poles.
Solution: The system has four poles and no zeros. The two real poles correspond to decaying exponential terms C1e−3t and C2e−0.1t, and the complex conjugate pole pair introduce an oscillatory component Ae−t sin (2t + φ), so that the total homogeneous response is yh(t) = C1e−3t + C2e−0.1t + Ae−t sin (2t + φ) (12) Although the relative strengths of these components in any given situation is determined by the set of initial conditions, the following general observations may be made: 1. The term e−3t, with a time-constant τ of 0.33 seconds, decays rapidly and is significant only for approximately 4τ or 1.33seconds.
2. The response has an oscillatory component Ae−t sin(2t + φ) defined by the com-plex conjugate pair, and exhibits some overshoot. The oscillation will decay in approximately four seconds because of the e−t damping term.
3. The term e−0.1t, with a time-constant τ = 10 seconds, persists for approximately 40 seconds. It is therefore the dominant long term response component in the overall homogeneous response.
5 Figure 4: Definition of the parameters ωn and ζ for an underdamped, second-order system from the complex conjugate pole locations.
The pole locations of the classical second-order homogeneous system d2y dt2 + 2ζωn dy dt + ω2 ny = 0, (13) described in Section 9.3 are given by p1, p2 = −ζωn ± ωn ζ2 −1.
(14) If ζ ≥1, corresponding to an overdamped system, the two poles are real and lie in the left-half plane. For an underdamped system, 0 ≤ζ < 1, the poles form a complex conjugate pair, p1, p2 = −ζωn ± jωn 1 −ζ2 (15) and are located in the left-half plane, as shown in Fig. 4. From this figure it can be seen that the poles lie at a distance ωn from the origin, and at an angle ± cos−1(ζ) from the negative real axis.
The poles for an underdamped second-order system therefore lie on a semi-circle with a radius defined by ωn, at an angle defined by the value of the damping ratio ζ.
1.3 System Stability The stability of a linear system may be determined directly from its transfer function. An nth order linear system is asymptotically stable only if all of the components in the homogeneous response from a finite set of initial conditions decay to zero as time increases, or lim t→∞ n i=1 Ciepit = 0.
(16) where the pi are the system poles. In a stable system all components of the homogeneous response must decay to zero as time increases. If any pole has a positive real part there is a component in the output that increases without bound, causing the system to be unstable.
6 In order for a linear system to be stable, all of its poles must have negative real parts, that is they must all lie within the left-half of the s-plane. An “unstable” pole, lying in the right half of the s-plane, generates a component in the system homogeneous response that increases without bound from any finite initial conditions. A system having one or more poles lying on the imaginary axis of the s-plane has non-decaying oscillatory components in its homogeneous response, and is defined to be marginally stable.
2 Geometric Evaluation of the Transfer Function The transfer function may be evaluated for any value of s = σ + jω, and in general, when s is complex the function H(s) itself is complex. It is common to express the complex value of the transfer function in polar form as a magnitude and an angle: H(s) = |H(s)| ejφ(s), (17) with a magnitude |H(s)| and an angle φ(s) given by |H(s)| = ℜ{H(s)}2 + ℑ{H(s)}2, (18) φ(s) = tan−1 ℑ{H(s)} ℜ{H(s)} (19) where ℜ{} is the real operator, and ℑ{} is the imaginary operator. If the numerator and denomi-nator polynomials are factored into terms (s −pi) and (s −zi) as in Eq. (2), H(s) = K (s −z1)(s −z2) . . . (s −zm−1)(s −zm) (s −p1)(s −p2) . . . (s −pn−1)(s −pn) (20) each of the factors in the numerator and denominator is a complex quantity, and may be interpreted as a vector in the s-plane, originating from the point zi or pi and directed to the point s at which the function is to be evaluated. Each of these vectors may be written in polar form in terms of a magnitude and an angle, for example for a pole pi = σi +ωi, the magnitude and angle of the vector to the point s = σ + ω are |s −pi| = (σ −σi)2 + (ω −ωi)2, (21) ̸ (s −pi) = tan−1 ω −ωi σ −σi (22) as shown in Fig. 5a. Because the magnitude of the product of two complex quantities is the product of the individual magnitudes, and the angle of the product is the sum of the component angles (Appendix B), the magnitude and angle of the complete transfer function may then be written |H(s)| = K m i=1 |(s −zi)| n i=1 |(s −pi)| (23) ̸ H(s) = m i=1 ̸ (s −zi) − n i=1 ̸ (s −pi).
(24) The magnitude of each of the component vectors in the numerator and denominator is the distance of the point s from the pole or zero on the s-plane. Therefore if the vector from the pole pi to the point s on a pole-zero plot has a length qi and an angle θi from the horizontal, and the vector from 7 Figure 5: (a) Definition of s-plane geometric relationships in polar form, (b) Geometric evaluation of the transfer function from the pole-zero plot.
the zero zi to the point s has a length ri and an angle φi, as shown in Fig. 5b, the value of the transfer function at the point s is |H(s)| = K r1 . . . rm q1 . . . qn (25) ̸ H(s) = (φ1 + . . . + φm) −(θ1 + . . . + θn) (26) The transfer function at any value of s may therefore be determined geometrically from the pole-zero plot, except for the overall “gain” factor K. The magnitude of the transfer function is proportional to the product of the geometric distances on the s-plane from each zero to the point s divided by the product of the distances from each pole to the point. The angle of the transfer function is the sum of the angles of the vectors associated with the zeros minus the sum of the angles of the vectors associated with the poles.
Example A second-order system has a pair of complex conjugate poles a s = −2±j3 and a single zero at the origin of the s-plane. Find the transfer function and use the pole-zero plot to evaluate the transfer function at s = 0 + j5.
Solution: From the problem description H(s) = K s (s −(−2 + j3))(s −(−2 −j3)) = K s s2 + 4s + 13 (27) The pole-zero plot is shown in Fig. 6. From the figure the transfer function is |H(s)| = K (0 −5)2 (0 −(−2))2 + (5 −3)2 (0 −(−2))2 + (5 −(−3))2 = K 5 4 √ 34 (28) 8 Figure 6: The pole-zero plot for a second order system with a zero at the origin.
and ̸ H(s) = tan−1(5/0) −tan−1(2/2) −tan−1(8/2) = −310 (29) 3 Frequency Response and the Pole-Zero Plot The frequency response may be written in terms of the system poles and zeros by substituting jω for s directly into the factored form of the transfer function: H(jω) = K (jω −z1)(jω −z2) . . . (jω −zm−1)(jω −zm) (jω −p1)(jω −p2) . . . (jω −pn−1)(jω −pn) .
(30) Because the frequency response is the transfer function evaluated on the imaginary axis of the s-plane, that is when s = jω, the graphical method for evaluating the transfer function described above may be applied directly to the frequency response. Each of the vectors from the n system poles to a test point s = jω has a magnitude and an angle: |jω −pi| = σ2 i + (ω −ωi)2, (31) ̸ (s −pi) = tan−1 ω −ωi −σi , (32) as shown in Fig. 7a, with similar expressions for the vectors from the m zeros. The magnitude and phase angle of the complete frequency response may then be written in terms of the magnitudes and angles of these component vectors |H(jω)| = K m i=1 |(jω −zi)| n i=1 |(jω −pi)| (33) ̸ H(jω) = m i=1 ̸ (jω −zi) − n i=1 ̸ (jω −pi).
(34) 9 Figure 7: Definition of the vector quantities used in defining the frequency response function from the pole-zero plot. In (a) the vector from a pole (or zero) is defined, in (b) the vectors from all poles and zeros in a typical system are shown.
As defined above, if the vector from the pole pi to the point s = jω has length qi and an angle θi from the horizontal, and the vector from the zero zi to the point jω has a length ri and an angle φi, as shown in Fig. 7b, the value of the frequency response at the point jω is |H(jω)| = K r1 . . . rm q1 . . . qn (35) ̸ H(jω) = (φ1 + . . . + φm) −(θ1 + . . . + θn) (36) The graphical method can be very useful for deriving a qualitative picture of a system frequency response. For example, consider the sinusoidal response of a first-order system with a pole on the real axis at s = −1/τ as shown in Fig. 8a, and its Bode plots in Fig. 8b. Even though the gain constant K cannot be determined from the pole-zero plot, the following observations may be made directly by noting the behavior of the magnitude and angle of the vector from the pole to the imaginary axis as the input frequency is varied: 1. At low frequencies the gain approaches a finite value, and the phase angle has a small but finite lag.
2. As the input frequency is increased the gain decreases (because the length of the vector increases), and the phase lag also increases (the angle of the vector becomes larger).
3. At very high input frequencies the gain approaches zero, and the phase angle approaches π/2.
As a second example consider a second-order system, with the damping ratio chosen so that the pair of complex conjugate poles are located close to the imaginary axis as shown in Fig. 9a. In this case there are a pair of vectors connecting the two poles to the imaginary axis, and the following conclusions may be drawn by noting how the lengths and angles of the vectors change as the test frequency moves up the imaginary axis: 1. At low frequencies there is a finite (but undetermined) gain and a small but finite phase lag associated with the system.
2. As the input frequency is increased and the test point on the imaginary axis approaches the pole, one of the vectors (associated with the pole in the second quadrant) decreases in length 10 Figure 8: The pole-zero plot of a first-order system and its frequency response functions.
and at some point reaches a minimum. There is an increase in the value of the magnitude function over a range of frequencies close to the pole.
3. At very high frequencies, the lengths of both vectors tend to infinity, and the magnitude of the frequency response tends to zero, while the phase approaches an angle of π radians because the angle of each vector approaches π/2.
The following generalizations may be made about the sinusoidal frequency response of a linear system, based upon the geometric interpretation of the pole-zero plot: 1. If a system has an excess of poles over the number of zeros the magnitude of the frequency response tends to zero as the frequency becomes large. Similarly, if a system has an excess of zeros the gain increases without bound as the frequency of the input increases. This cannot happen in physical energetic systems because it implies an infinite power gain through the system.
2. If a system has a pair of complex conjugate poles close to the imaginary axis, the magnitude of the frequency response has a “peak”, or resonance at frequencies in the proximity of the pole. If the pole pair lies directly upon the imaginary axis, the system exhibits an infinite gain at that frequency.
3. If a system has a pair of complex conjugate zeros close to the imaginary axis, the frequency response has a “dip” or “notch” in its magnitude function at frequencies in the vicinity of the zero. Should the pair of zeros lie directly upon the imaginary axis, the response is identically zero at the frequency of the zero, and the system does not respond at all to sinusoidal excitation at that frequency.
4. A pole at the origin of the s-plane (corresponding to a pure integration term in the transfer function) implies an infinite gain at zero frequency.
5. Similarly a zero at the origin of the s-plane (corresponding to a pure differentiation) implies a zero gain for the system at zero frequency.
11 Figure 9: The pole-zero plot for a second-order system and its its frequency response functions.
3.1 A Simple Method for constructing the Magnitude Bode Plot directly from the Pole-Zero Plot The pole-zero plot of a system contains sufficient information to define the frequency response except for an arbitrary gain constant. It is often sufficient to know the shape of the magnitude Bode plot without knowing the absolute gain. The method described here allows the magnitude plot to be sketched by inspection, without drawing the individual component curves. The method is based on the fact that the overall magnitude curve undergoes a change in slope at each break frequency.
The first step is to identify the break frequencies, either by factoring the transfer function or directly from the pole-zero plot. Consider a typical pole-zero plot of a linear system as shown in Fig.
10a. The break frequencies for the four first and second-order blocks are all at a frequency equal to the radial distance of the poles or zeros from the origin of the s-plane, that is ωb = √ σ2 + ω2.
Therefore all break frequencies may be found by taking a compass and drawing an arc from each pole or zero to the positive imaginary axis. These break frequencies may be transferred directly to the logarithmic frequency axis of the Bode plot.
Because all low frequency asymptotes are horizontal lines with a gain of 0dB, a pole or zero does not contribute to the magnitude Bode plot below its break frequency. Each pole or zero contributes a change in the slope of the asymptotic plot of ±20 dB/decade above its break frequency. A complex conjugate pole or zero pair defines two coincident breaks of ±20 dB/decade (one from each member of the pair), giving a total change in the slope of ±40 dB/decade. Therefore, at any frequency ω, the slope of the asymptotic magnitude function depends only on the number of break points at frequencies less than ω, or to the left on the Bode plot. If there are Z breakpoints due to zeros to the left, and P breakpoints due to poles, the slope of the curve at that frequency is 20 × (Z −P) dB/decade.
Any poles or zeros at the origin cannot be plotted on the Bode plot, because they are effectively to the left of all finite break frequencies. However, they define the initial slope. If an arbitrary starting frequency and an assumed gain (for example 0dB) at that frequency are chosen, the shape of the magnitude plot may be easily constructed by noting the initial slope, and constructing the 12 Figure 10: Construction of the magnitude Bode plot from the pole-zero diagram: (a) shows a typical third-order system, and the definition of the break frequencies, (b) shows the Bode plot based on changes in slope at the break frequencies curve from straight line segments that change in slope by units of ±20 dB/decade at the breakpoints.
The arbitrary choice of the reference gain results in a vertical displacement of the curve.
Figure 10b shows the straight line magnitude plot for the system shown in Fig. 10a constructed using this method. A frequency range of 0.01 to 100 radians/sec was arbitrarily selected, and a gain of 0dB at 0.01 radians/sec was assigned as the reference level. The break frequencies at 0, 0.1, 1.414, and 5 radians/sec were transferred to the frequency axis from the pole-zero plot. The value of N at any frequency is Z −P, where Z is the number of zeros to the left, and P is the number of poles to the left. The curve was simply drawn by assigning the value of the slope in each of the frequency intervals and drawing connected lines.
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Algebra 1 Course - Unit 3 - Solve Single Step and Multi Step Equations
06 - Solve One-Step Equations with Multiplication and Division, Part 1
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3166 | https://www.cureffi.org/2019/06/05/using-genetic-data-to-estimate-disease-prevalence/ | CureFFI.org
Using genetic data to estimate disease prevalence
introduction
Traditionally, people have estimated how common a genetic disease is by observing the disease itself. With the advent of large databases of population genetic variation, such as gnomAD, there arises a new possibility: estimating the prevalence of a disease based on the frequency of a causal allele in the population. Is this approach legitimate? What are the caveats? Since reviewing a paper that used this approach last year [Gao 2018], I’ve spent some time pondering this issue. In this blog post I will consider the why and how of disease prevalence estimation, the opportunities and potential issues with using genetic data for this purpose, and some potential guidelines for how to do it right.
how to measure how common a disease is
There are a few different ways to measure how common a disease is. Prevalence, where it has a precise definition, is the number of people symptomatic with a disease at any given point in time; I’ve also heard this called “point prevalence”. (Informally, the word “prevalence” can also just refer to how common a disease is, by any measure.) Incidence is the number of new cases (or sometimes new births), per unit time, usually annually. Lifetime risk is the probability of any random person (from birth) eventually developing the disease. The term “genetic prevalence” seems to be used to refer to the proportion of people in the population who have a causal genotype, and assuming full penetrance, this should be equivalent to lifetime risk.
The below diagram illustrates conceptually the different things these three terms are measuring:
These different measures may be appropriate for different purposes, and many people, even scientists, get confused between them, so it’s important to be clear about what one is estimating and how one arrived at the estimate. At the same time, all of these measures are inter-related, and if you have an estimate of disease duration and age of onset you can derive one from the other. So although I’ve used the term “prevalence” in the title of the post, what’s of interest here, broadly speaking, is simply any metric of how common a disease is.
why this is important
It’s very important to know how common a disease is, for several reasons. Prevalence is one input into the financial models that industry uses to evaluate an investment in R&D to make a new drug, so knowing how common your disease is can affect whether anyone bothers to try to cure it. People’s idea of how common a disease is can affect how they interpret data around them. For example, the misperception that prion disease only affects one in a million people makes things that are actually expected — like occasionally observing a married couple who both develop the disease — seem like crazy coincidences, and drives people to search for crazy explanations like horizontal transmission of prion infection between people. And many diseases, especially rare ones, are under-diagnosed, and if you have a good measure of disease prevalence or incidence that is orthogonal to just counting diagnoses, then you have an idea of how far you have to go in improving diagnosis.
Despite this importance, there are a great many diseases for which we really don’t know how common they are. There are diseases for which there are simply no published estimates, or only outdated estimates. A disease can appear more common the harder you look [Klug 2013], so for instance, as prion surveillance has gotten better over the past couple of decades, prion disease has appeared to be more and more common. And there are those where published estimates disagree wildly with one another, and it’s not clear why. For example, one meta-analysis found that estimates of Huntington’s disease prevalence in people of European ancestry were 10 times higher than in Asian populations [Pringsheim 2012] — does that reflect real differences in biology, differences in diagnosis, or just differences in study design? Considering how important it is, I think this topic is under-studied: there should be more efforts to estimate how common diseases are.
traditional approaches to estimating how common a disease is
It’s worth taking a moment to consider the methods by which people have traditionally estimated how common a disease is, and what different biases and caveats come with each.
The traditional methods for measuring prevalence per se all revolve around counting people with the actual disease. If you’re interested in knowing how many people you could recruit for research or trials, your best resource may be a good registry of patients with the disease. But except in rare cases like cystic fibrosis (the CF Foundation Patient Registry is purported to be darned-near exhaustive) a registry will typically ascertain only a fraction of the true disease prevalence — it’s heavily self-selected for the most motivated patients who want to volunteer for research. In the Huntington’s disease meta-analysis cited above, most prevalence studies reviewed medical records or genetic testing records, others mailed out surveys to doctors or to patients themselves, or even went door-to-door [Pringsheim 2012]. One study aimed to exhaustively ascertain Huntington’s patients in British Columbia by combining every possible approach to tracking them down [Fisher & Hayden 2014]. While some approaches are more thorough than others, the goal of all these methods is to find and count sick people. Remember that prevalence is the number of people sick at one moment in time. If one group of people (say, the patients in one country, or with one disease subtype) live longer with a disease, either due to a different progression rate, better medical care, or simply being kept alive longer on life support, then that group of people will appear more “prevalent” even if their incidence (rate of new cases) is the same as other groups. Example: prion disease patients in Japan survive on average three times longer than patients in Europe [Nagoshi 2011]. The authors speculate in the Discussion that the difference is largely physician reluctance to take patients off of life support. If so, does the longer duration mean prion disease is more “prevalent” in Japan in any meaningful sense? I would argue no — and yet if you count sick patients, you may get an answer three times higher than you would get in Europe.
There are also methods for measuring incidence. The Huntington’s meta-analysis noted above [Pringsheim 2012] includes incidence studies as well, and most are basically the same approaches as the prevalence studies — surveys, medical records review, and so on — but asking the question of how many new cases arise per year instead of asking how many patients are around today. Prion disease studies may count diagnoses made per year at national surveillance centers [Klug 2013] or count the death certificates listing prion disease as cause of death per year [Holman 2010]. ALS studies may count cases referred to national or regional registries [Logroscino 2010]. As above, the big challenge here is that people still only count if they are diagnosed correctly. It’s difficult to know if differences between countries or populations are due to true biological differences or differences in the healthcare system.
Finally, there are also ways to directly estimate lifetime risk. Indeed, for fatal diseases, review of death records can be one such method — divide deaths due to the disease of interest by total deaths in some time period, and you have the lifetime risk in the general population. This has been done for prion disease in the U.S. and U.K.
All the different measures and different approaches listed above share one key source of bias: they depend on correct diagnosis. Neither physicians nor patients nor medical records nor death certificates can tell you that someone should count towards your total if they were never daignosed correctly. So if your hope is to find out whether the condition you’re studying is under-diagnosed, none of these approaches can tell you the answer.
genetics to estimate how common a disease is
Enter a completely different approach: instead of counting disease diagnoses, let’s count disease-causing alleles in people’s DNA. This approach isn’t new. Not long after CFTR ΔF508 was identified as the predominant cause of cystic fibrosis [Kerem 1989], studies began screening for this and other mutant CFTR alleles in the general population try to determine the carrier frequency [Abeliovich 1992]. For years prior, people had inferred what the carrier frequency must be based on the prevalence of cystic fibrosis among live births; now suddenly they had the ability to directly measure the carrier frequency, and ask whether the predicted prevalence matched observation.
That worked for cystic fibrosis in the 1990s only because cystic fibrosis is so darn common: ΔF508 alone has an allele frequency of 1.3% in people of European ancestry. Most genetic diseases are far, far rarer. Even for Huntington’s disease, which is far from being the rarest genetic disease out there, the earliest study I’m aware of that screened for mutant HTT alleles in the general population was only a few years ago, and even then, the sample size was still small enough to leave fairly large error bars around the true genetic prevalence — they found that 3 out of 7,315 people had high-penetrance alleles [Kay 2016].
For many diseases, then, only in the past few years with ExAC and gnomAD have we finally arrived at the sample size needed to query genetic prevalence. For yet other diseases, we are still not there yet, but we may be soon. Therefore I expect that in the coming years we’ll see a lot more studies using gnomAD or other genetic datasets to estimate how common a disease is. Which makes now a good time to consider how to do this right.
the risk of circularity and interpretation
Your first question may be, isn’t this approach circular? Allele frequency is one factor used in deciding whether a genetic variant is pathogenic [Richards 2015], and disease prevalence is one factor used in deciding what allele frequency cutoff is reasonable [Whiffin & Minikel 2017]. So isn’t it circular to then add up the allele frequencies of pathogenic variants in order to estimate disease prevalence?
I would argue that circularity is a risk that one should be aware of, and much of this post will address what that awareness should entail, but I will argue that the specter of circularity does not as a blanket rule preclude this approach entirely. That’s because, done right, assertions of pathogenicity should be based primarily on strong lines of evidence that are orthogonal to allele frequency. Such lines of evidence will often include Mendelian segregation or de novo status, and depending on your disease and your gene, might also include things like functional annotation and behavior in in vitro or cell culture systems. As I’ve argued in my guide to using allele frequency, a variant’s frequency in the population should be just a filter to throw out the most obvious false positives or low-penetrance variants, and often, you’ll find that the variants thrown out by this approach were those that didn’t actually have strong evidence for pathogenicity to begin with [Minikel 2016].
issues to consider
penetrance and prevalence
All that being said, allele frequency information is only informative about how common a disease is, to the extent that you are confident about the penetrance of all the genetic variants being included.
To see why this is so, first consider the simplest case: all cases of a disease are attributable to a single genetic variant, and that variant always causes the disease. Here, the lifetime risk in the general population is equal to the frequency of the disease-causing genotype. We’ll call these variables P(D) for probability of developing disease, and P(G) for probability of disease-causing genotype. For allele frequency p, this means that for a recessive disease, P(D) = P(G) = p2. For a dominant disease, P(D) = P(G) = 2p(1-p) ≈ 2p, or, if heterozygotes are fertile and homozygotes viable, P(D) = P(G) = p2 + 2p(1-p) ≈ 2p.
Next suppose that the variant is not fully penetrant, but instead has penetrance between 0 and 1 denoted by P(D|G), meaning, probability of developing disease given the causal genotype. Now, instead of just P(D) = P(G), we have P(D) = P(G) × P(D|G).
Finally suppose that not all cases of the disease are attributable to just a single variant either. Now we have to also consider the proportion of disease cases explained by the variant, denoted P(G|D), for the probability of having this particular genotype given the disease state. Our equation now becomes P(D) = P(G) × P(D|G) / P(G|D).
So we find ourselves looking at Bayes’ theorem! Or, simply a rearranged version of the formula we used to calculate penetrance in [Minikel 2016], introduced in this post. Or, if you plug in 2p or p2 for P(G), and divide P(G|D) into separate terms for genetic and allelic heterogeneity, then you have a rearranged version of the formula we used for allele frequency filtering described in [Whiffin & Minikel 2017]. In other words, there is nothing new here. We have just performed some simple algebra to isolate the variable of interest. In [Minikel 2016] we isolated penetrance, P(D|G), in [Whiffin & Minikel 2017] we isolated genotype frequency, P(G), and now in this blog post we’re talking about isolating P(D). If you know three of the variables, you can estimate the fourth.
But there’s the rub — to solve the equation, you can only have one unknown. In practice, there’s usually some uncertainty around one of the input variables, so instead of determining with certainty the output variable, you can only put some bounds on it.
For instance: prion disease is well-enough surveilled that we are reasonably confident about lifetime risk in the general population. You could convince me that prion disease is under-diagnosed by half, but you couldn’t convince me it is under-diagnosed by 30-fold, which is what it would have taken to explain the frequency of supposedly disease-causing variants in the population. Therefore, given case and population control allele frequency information, we could estimate penetrance [Minikel 2016] — only roughly and on a log scale, but that was enough to change clinical practice. For allele frequency filtering, we left the uncertainty up to the user, inviting people to set conservative limits on what values they would believe for disease prevalence and proportion of cases attributable to a variant, and a minimum penetrance that they cared about, and then computing a loose upper bound on what allele frequency was plausible [Whiffin & Minikel 2017].
When we think about estimating how common a disease is, we’re not too worried about the proportion of cases attributable to any one particular variant, P(G|D), because we are going to sum across a number of variants. If the disease has been well-enough studied for the most common causes to have already been identified, then collectively the variants we include in our analysis will explain most cases. Mathematically, this is equivalent to saying P(D) = Σ P(G) × P(D|G) / P(G|D), and ΣP(G|D) ≈ 1. There’s an “if” there, of course, and I’ll return later to the question of which variants are being included. But assuming we’ve included enough of the right variants, we can eliminate the P(G|D) term and simply say that P(D) = Σ P(G) × P(D|G), or in other words, lifetime risk of our disease in the general population is the sum, over all causal variants, of the frequency of the disease-causing genotype times the penetrance. What the advent of large population control genetic datasets has done, then, is simply to give us a reasonable empirical estimate of allele frequency and thus P(G), so that now we are down to two unknowns: P(D) and P(D|G). Commonness and penetrance. To the extent that you know one, you can estimate the other.
And this is not a case where one can just hand-wave and say, “well, the few best-studied variants in my disease appear to be highly penetrant, so I’ll assume they all are.” Because when we say penetrance or P(D|G), we’re not only talking about whether a variant confers 99% versus only 80% disease risk — what really kills your analysis is the variants where penetrance is zero, or close to it, because the variant is not confidently associated to disease in the first place. Such variants can be a majority of reported variants in a gene — for instance, for PRNP we found that only 27 out of 70 reportedly disease-causing variants had evidence for Mendelian segregation or a de novo as basis for an assertion of Mendelian disease risk [Minikel 2018]. But worse still, it’s very likely that any variants you include that are not truly causal, or confer only very low risk, will have a much, much higher allele frequency than the truly causal variants in your dataset. To see why, we turn to the allele frequency spectrum.
allele frequency spectrum
Most human genetic variants are extremely rare, and we are continuing to realize just how much this is the case. In the early days of sequencing small datasets, on the order of 100 people, it was found that about 90% of variants are common (>1% allele frequency), but the larger the dataset, the more the frequency distribution skews rare, such that in gnomAD the majority of variants are singletons, seen just once in 141,456 people. One intuition for why it should be so is that, based on DNA mutation rates [Samocha 2014, Lek 2016], we estimate that of all possible SNPs, most that are compatible with life probably exist in some human somewhere in a heterozygous state — that’s three possible mutations per base, at 3 billion base pairs in the haploid genome, so on the order of 9 billion unique variants in almost 8 billion people. But we still have not seen most of those variants, so they must be very rare globally (though a handful may be common in some population we just haven’t yet sampled).
So most variants in gnomAD are singletons, corresponding to a nominal allele frequency of 3 × 10-6 and a true allele freuency likely lower, and most variants on Earth are rarer still. And if most variants in general are rare, pathogenic variants skew yet rarer still, because they are under pressure from purifying selection. Thus, even if false positives were to account for only a small fraction of unique variants (and in fact they may account for a large fraction), the fact that they are not under purifying selection means they are likely to be enriched for variants that, even though still below some pre-specified allele frequency cutoff, could be many times more common than your true causal variants. Thus, there exists a risk that a small number of false positives could account for the vast majority of cumulative allele frequency and for a dramatic inflation in any prevalence estimate you derive.
As an example, say you were applying this approach to prion disease (and for simplicity of example, say you’re using global and not population-specific allele frequencies in gnomAD). You might assume 1 in 5,000 risk in the population, most common variant causes 5% of cases, plug in numbers and settle on a maximum tolerated allele count of 6 globally in gnomAD v2, corresponding to an allele frequency of about 1 in 50,000, for variants of at least 50% penetrance. That would filter out the most common variants that are obviously either benign or very, very low-risk, such as P39L, E196A, and R208H — so far so good. But it would still leave in, for example, S97N, V203I, and Q212P, which collectively have an allele count of 10, and none of which have evidence for Mendelian segregation. Compare this to E200K, the most common cause of genetic prion disease and the only variant with evidence for Mendelian segregation with prion disease that does appear in gnomAD — it has an allele count of just 1. This illustrates that even with a fairly aggressive allele frequency filter, those variants that may not be truly associated to disease, or that may be much less penetrant than the assumption you put into the allele frequency cutoff calculation, can very quickly outnumber confidently disease-associated alleles by 10 to 1.
In other words, the allele frequency spectrum means that we need to be very careful about what variants are included in the analysis — so let’s turn to that topic next.
which variants to include
It’s not trivial to identify which genetic variants to include in an analysis to estimate disease commonness. That’s because fact that a variant has been reported as pathogenic — in ClinVar or in a Google Scholar search or in any other database I’ve seen — is neither very sensitive nor very specific for true pathogenic variants.
Reportedly pathogenic variants are a place to start, but unfortunately, you need to curate them. ClinVar has things like different levels of review and different numbers of submitters and assertion criteria provided or not provided, and to be sure, these are helpful. But so far I have not seen many genes for which the curation captured in ClinVar is so deep that a hard filter on just these existing annotations would be adequate. The best approach is to go back to primary scientific literature (and, if your gene has any variants asserted in ClinVar but never published, the original assertion criteria provided). First consider the potential criteria for pathogenicity outlined by the American College of Medical Genetics in [Richards 2015] and think about which criteria make sense biologically for your disease and your gene. Once you’ve decided what you’re looking for, set up a spreadsheet and go through every variant that’s been published. For prion disease, we decided the only credible criteria for assertions of high risk were Mendelian segregation with three affected individuals in one family, or de novo variants, and we found 27 of 70 variants had one of these [Minikel 2018]. You might also visit the allele frequency app to decide on a maximum credible allele frequency and then check this against gnomAD, but probably this will merely serve to increase your confidence that you were right to throw out many variants based on your literature curation. If you find that your filter is actually throwing out variants that did have good evidence for assertions of pathogenicity, then you have to consider whether A) the criteria you settled on make sense for your gene, B) any of your assumptions were wrong, or C) you are in fact discovering evidence that your disease may be more common than you thought.
The next question is which, if any, variants to include that have not previously been reported as pathogenic. It is virtually certain that there are pathogenic variants in your gene out there in the world that have not yet been published or have not yet even been seen in a case, and (particularly for recessive diseases) it’s reasonably likely that some of these may appear in gnomAD. But how to sift these out from the heap of benign variants in gnomAD? The answer is that it’s hard. In some genes where loss-of-function is a well-established disease mechanism, it may be appropriate to include loss-of-function variants seen in gnomAD in your calculation, with caveats. First, these should be nonsense, frameshift, and essential splice site only, and rated as high-confidence LoF. Second, you should curate the variants, especially any with higher allele frequency than the others — look at screenshots of the raw reads to make sure the variant call is real, look at the pext track in gnomAD to make sure the affected exon is actually expressed in your disease-relevant tissue [Cummings 2019], look at UCSC genome browser to make sure you believe the annotation as loss-of-function is real, and investigate any suspiciously non-random positional distributions of LoF variants across the coding sequence [Minikel 2019]. This is all because again, as explained above, even if just one or a few false positives were to sneak in, they might have a far higher allele frequency than genuine LoF variants.
One thing you should never do is to include missense variants never reported as pathogenic on the basis that they are predicted to be deleterious. Unfortunately, as of 2019, our ability to interpret missense variants is just not good enough to use them in this way. There are various tools for predicting which missense variants are deleterious — SIFT, PolyPhen, and CADD are the best-known. These tools do enrich for pathogenic variants in the aggregate — for just one line of evidence, see Figure 2E from [Lek 2016], which shows that the missense variants that PolyPhen and CADD predict to be worst have higher MAPS scores, meaning they are more often singletons, held at low allele frequency by purifying selection, than the variants predicted to be benign. But because of the allele frequency spectrum issue discussed earlier, enriching for pathogenicity in the aggregate is not good enough — if even a few benign variants sneak in, they will dominate the analysis. This is especially true for genes that are not missense-constrained — Figure 3F of [Lek 2016] shows that for the least missense-constrained genes, even PolyPhen “probably damaging” or CADD “high” missense variants are on average barely above synonymous variants in terms of their MAPS scores.
interpreting allele frequency
Next, you need to consider the interpetation of the allele frequency data themselves. Here, the issues to consider are very different for dominant versus recessive diseases.
For dominant diseases, to the extent that you assume the variants in question are penetrant, you are assuming that the people in gnomAD with those variants actually have, or will develop, the disease. Note that, per the gnomAD FAQ:
We have made every effort to exclude individuals with severe pediatric diseases from the gnomAD data set, and certainly do not expect our data set to be enriched for such individuals, but we typically cannot rule out the possibility that some of our participants do actually suffer from your disease of interest.
Thus, if your dominant disease is severe, and pediatric onset, then gnomAD allele frequency is not a sound basis for estimating how common your disease is, because gnomAD should be quite depleted for individuals with the disease. If your dominant disease is adult onset, then you could be in business, but with caveats. The median age of people ascertained in gnomAD is something like 55, so if you’re talking about a disease that is fatal at age 50, you should expect gnomAD to be significantly depleted for causal variants. Conversely, if your disease presents with a phenotype specifically ascertained in any of the cohorts included in gnomAD, then causal variants could be enriched. These sorts of caveats will differ for every dataset — 23andMe, UKBB, and so on — but in general it is a good idea to think about who is in the dataset and whether allele frequencies for your disease can be accurately estimated.
For recessive diseases, the issues involved are different, because your calculation will be based overwhelmingly on heterozygous carriers rather than people with the disease-causing genotype. Thus, questions of who is in the dataset and whether your disease phenotype is enriched or depleted do not come into play. But, questions of population structure start to matter more. You’ll presumably be adding up the cumulative allele frequency of all causal variants and then squaring it to arrive at an estimate of homozygote and compound heterozygote frequency. In that act of squaring, there is an opportunity to get pretty different answers depending on how you assume people choose mates. You might have one variant much more common in a particular population, such that the cumulative allele frequency is 0.02% overall in gnomAD, but 0.01% in Europeans but 0.05% in East Asians. Squaring 0.02% may not be meaningful, instead, the conclusion may be that you expect there to be just a handful of affected individuals of European ancestry across North America and Europe, but a few hundred affected individuals in China. There is no doubt that one needs to think about this in terms of populations, but where to stop — is East Asian good enough? Do you need to know if the carriers of your variant are indeed Chinese? And from which province? Or what if the population where a pathogenic variant is most common has a high rate of consanguinity — is p2 still a good estimate of homozygote frequency, or do you need to account for autozygosity? On these issues, all I can say is you have to consider these issues and do the best you can with available data, concluding what you can conclude and caveating what you can’t.
Another situation that arises, though, is where the population in which pathogenic alleles are most frequent is a population where the disease has never been well-studied, and the variant in question has only been seen in one or two cases. Here, you have to redouble your efforts to curate and be dead certain whether you believe the evidence for pathogenicity of that variant. It’s possible that you’ve just discovered that your disease is way more common in South Asia than anyone ever guessed, and now that you know where to look for more patients, within a few years you’ll be part of an awesome international consortium gathering valuable new natural history data and preparing for clinical trials. But it’s also possible that the one variant driving that allele frequency in South Asia is just not highly penetrant, or perhaps, for example, is only penetrant when found in trans to an even worse allele, whereas it does not cause disease in a homozygous state. In such cases, unless the evidence for the one variant is rock solid, I’d interpret the results to mean “this would be very interesting if true, and therefore motivates studies to see if this variant is indeed penetrant and whether this disease is indeed common in population X”.
which variants account for most cases
Earlier, I noted that the formula relating penetrance and prevalence also contains a term for the proportion of cases caused by a given variant, P(G|D). I made the simplifying assumption that, summed over all the variants in consideration, this is close to 1. I’ll now explain that assumption a bit more. Under “which variants to include”, above, I argued for pretty strict filtering based on literature evidence for pathogenicity, and for not including much or anything in the way of “predicted” pathogenic variants, so it is certain that not all disease-causing variants will be included, so that ΣP(G|D) does not quite get to 1. Why am I OK with this?
Most importantly, at the end of the day, you’re going to have wide confidence intervals around your estimate of disease commonness anyway, so if you only include variants explaining 90% or 80% or even 60% of cases, you’re in decent shape — that’s a much better outcome than including a bunch of “predicted” pathogenic variants that are actually benign, and ending up with an estimate that is inflated by 10-fold.
Moreover, it’s often the case that just a few pathogenic variants explain a large fraction of cases, and these are usually the variants that get discovered first when a disease is characterized — the long tail of variants discovered later often does not add a ton to cumulative allele frequency. In prion disease, just three variants — E200K, P102L, and D178N, all discovered in 1989 - 1992, explain 85% of cases with a high-risk variant [Minikel 2018]. Or to raise a more recent example, by far the most common LoF allele in NGLY1, R401X, with allele frequency an order of magnitude higher than any other, was seen in the very first patient identified [Need 2012, Enns 2014]. Why is it often the case that one or a few variants explain a large fraction of cases? One reason is that different DNA mutations arise at very different rates [Samocha 2014, Lek 2016]. CpG transitions (C→T where the adjacent base is G) occur spontaneously 10 times more often than other transitions (such as T→C), and 100 times more often than transversions (such as T→A). There are other PRNP mutations that are just as terrible but far less likely to arise spontaneously, whereas E200K, P102L, and D178N are all CpG transitions that have recurred many times in human history. Another reason in recessive diseases is simply drift: if pathogenic variants are rare, then pathogenic genotypes are squaredly rare, and so the variants in question are not under very intense natural selection, so one such variant may happen to get a bit more common than the others. A third reason is population bottlenecks, which allowed a few pathogenic variants to rise to unusually high frequency in populations like Finns and Ashkenazi Jews.
I could only think of one example where a variant explaining a significant fraction of cases was discovered late — COL6A1 was identified as a disease gene over 20 years ago (see OMIM #120220) but a single variant explaining 25% of unsolved cases (a smaller fraction of total cases) was discovered only in 2017 [Cummings 2017]. That’s because it was a deep intronic variant with a novel mechanism of creating a pseudoexon, and it was identified only by RNA-seq.
There could certainly be more instances like COL6A1, but I’m still more comfortable underestimating disease commonness by a bit than taking the risk of over-inclusion — especially since a variant like that one in COL6A1 would never have turned up as “predicted” pathogenic anyway. But certainly, the potential that common, undiscovered variants are out there is a caveat that might need to be made for some disease genes if you are trying to use allele frequency to estimate commonness.
As an aside, a related question that sometimes arises is whether and how allele frequency data can be used to figure out which pathogenic variants account for most cases of a disease. Suppose, for instance, that you’re Vertex Pharmaceuticals 20 years ago, just starting to find hit compounds that can rescue the phenotype of gating mutations in CFTR, and you want to know which compounds are most advanceable clinically based on which ones rescue variants with enough patients to run a trial. Here, what you really want is a registry or a case series, but often you don’t have that. A review of published cases in the literature is helpful, though it could be biased in either direction — maybe more common mutations are over-published because there is a large enough N to write a paper about, or maybe rarer mutations are published because of the novelty factor. You might then turn to allele frequency data from gnomAD or elsewhere, and you may be able to get an answer, but all the same caveats from this post will apply. For diseases too rare to even yet get a reasonable allele frequency estimate from gnomAD, your best guess may simply come down to mutation rates — all else being equal, if there are two known pathogenic variants and one is a CpG while the other is a transversion, the CpG is the one you’re more likely to see again. (Some indels such as short tandem repeats can also have rather high mutation rates [Ballantyne 2010]).
conclusions
Overall, I believe that estimation of disease prevalence (or incidence, or lifetime risk) is an important topic and, for many diseases, we still know too little. Allele frequency information from population databases such as gnomAD provides a new opportunity to estimate how common a disease is, and does not result in too much circular reasoning provided that pathogenicity assertions are based on good evidence orthogonal to allele frequency. But there are a lot of caveats to consider, hence, the following call-out box summarizing suggestions I’ve made above:
| |
| DO examine the primary literature evidence for pathogenicity, and be strict about what you include DO decide what criteria make sense for pathogenicity assertions for your gene/disease DON'T take assertions of pathogenicity in ClinVar or in the literature at face value DON'T assume that all variants that pass an allele frequency cutoff are pathogenic DO include high-confidence LoF variants that you have curated, if LoF is a well-established disease mechanism DON'T include missense variants never reported as pathogenic, even if they are "predicted" to be deleterious DO consider how ascertainment and population stratification affect your estimates DON'T assume that that your penetrance assumptions are necessarily correct when calculating prevalence — consider scenarios and different possible interpretations |
A few suggested guidelines for using allele frequency information to estimate how common a disease is.
I would love to see a consensus from the genetics community over the coming years on whether and how this sort of study can be done right. I welcome comments on whether this all sounds reasonable, and what other guidelines or considerations you would suggest.
About Eric Vallabh Minikel
Eric Vallabh Minikel is on a lifelong quest to prevent prion disease. He is a scientist based at the Broad Institute of MIT and Harvard.
Follow @cureffi |
3167 | https://www.khanacademy.org/math/get-ready-for-7th-grade/xa46d6dd638f86863:get-ready-for-rates-proportional-relationships/xa46d6dd638f86863:equivalent-ratios/e/equivalent-ratio-word-problems--basic- | Equivalent ratios with equal groups (practice) | Khan Academy
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Get ready for 7th grade math
Course: Get ready for 7th grade math>Unit 3
Lesson 1: Equivalent ratios
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Equivalent ratios
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Equivalent ratios with equal groups
PA.Math: CC.2.1.6.D.1
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Problem
The following diagram describes the volume of blue and red paint in a mixture.
Blue volume Red volume
What is the ratio of blue paint to total paint in the mixture?
Choose 1 answer:
Choose 1 answer:
(Choice A) 3:8 A 3:8
(Choice B) 5:8 B 5:8
(Choice C) 5:3 C 5:3
(Choice D) 8:3 D 8:3
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3168 | https://byjus.com/jee/geometric-progression-solved-problems/ | A geometric progression is a sequence where the succeeding term is ‘r’ times the preceding term. If a paper is folded four or five times, calculating the height of the stack of the paper after being folded four to five times is an example of geometric progression. In this article, we come across geometric progression solved problems along with the definition and properties of common ratio.
Geometric Progression Definition
A sequence in which the first term is non-zero and the ratio of each term to its previous term remains constant. The ratio is represented by the letter “r” and is called the common ratio.
The properties of a geometric progression are listed below:
Finite and infinite geometric progression
The geometric progression with a finite number of terms is a finite geometric progression, whereas a GP with an infinite number of terms is an infinite geometric progression.
Important Results
The formulae related to a geometric progression are given below:
| | | |
---
| Representation of GP | a, ar, ar², ar³, ……… arn-1, arn | a = first term r = common ratio n = the last term in GP and, Sn = sum to n terms of GP |
| The common ratio of GP | (\begin{array}{l}r=\frac{a_{n}}{a_{n-1}}\end{array} ) |
| The nth term of GP | an = arn-1 |
| Sum to n terms of a GP | Sn = a (1−rn)/(1−r ) or Sn = a(rn -1)/(r-1) |
or
Sn = a(rn -1)/(r-1)
To Determine Whether the Given Sequence Is a GP or Not
If the common ratio between the terms is a constant, then it is said to be a geometric sequence.
For instance, check whether the sequence 8, 16, 32, 64, 128 . . . . . is a GP.
Solution:
Given
(\begin{array}{l}a_{1}=8\ a_{2}=16\ a_{3}=32\ a_{4}=64\ a_{5}=128\ r=\frac{a_{2}}{a_{1}}=\frac{16}{8}=2\ r=\frac{a_{4}}{a_{3}}=\frac{64}{32}=2\\end{array} )
Since the common ratio is a constant, it is a GP.
Properties of Common Ratio [r]
Related articles
Geometric Progression
Geometric Mean
Solve Problems on Geometric Progression for IIT JEE
Example 1: If the 4th, 7th and 10th terms of a G.P. be a, b, and c, respectively, then what is the relation between a, b, and c?
Solution:
Let the first term of G.P. = A and common ratio = r
We know that the nth term of G.P. = Arn−1
Now t4 = a = Ar3, t7 = b = Ar6 and t10= c = Ar9
b2 = (Ar6)2
= A2r12
ac = (Ar3) (Ar9) = A2r12
So, b2 = ac is the relation between a, b and c.
Example 2: If a, b, c are pth, qth and rth terms of a G.P., then what is (c / b)p (b / a)r (a / c)q equal to?
Solution:
a = ARp−1, b = ARq−1, c = ARr−1
(c / b)p (b / a)r (a / c)q = [(ARr−1 / ARq−1) ]p [(ARq−1 / ARp−1)]r [(ARp−1/ ARr−1)]q
= R(r−q)p+(q−p)r+(p−r)q
= R0
= 1
Example 3: If the 6th term of a G.P. is 32, and its 8th term is 128, then find the common ratio of the G.P.
Solution:
T6 = 32 and T8 = 128
⇒ ar5= 32 ….. (i) and
ar7= 128 …..(ii)
Dividing (ii) by (i), we have
r2=4
r = 2
Example 4: Find the sum of the series 6 + 66 + 666 + ……….upto n terms.
Solution:
Given series 6 + 66 + 666 + ……….upto n terms
= [6 / 9] (9 + 99 + 999 + ….. upto n terms)
= 2/3
= [2 / 3] ((10 + 102 + 103 + ………. + upto n terms) −n)
= [2 / 3] [10 (10n−1) / 10−1] −n)
= [2 / 3] [10 (10n−1) / 9] −n)
= [2/3] [10n+1 – 10 – 9n]/9
= 2 (10n+1− 9n − 10) / 27
Example 5: The sum of a few terms of any ratio series is 728. If the common ratio is 3 and the last term is 486, then what will be the first term of the series?
Solution:
nth term of series = arn−1 = a(3)n−1 = 486 …. (i)
Sum of n terms of series is Sn= a (3n−1) / [3−1] = 728 ….(ii)(∵r >1)
From (i), a (3n / 3) = 486 or [a 3n] = 3 × 486 = 1458
From (ii), [a 3n] − a = 728 2 or [a 3n] − a = 1456
1458 − a = 1456
or a = 2
Example 6: If three geometric means be inserted between 2 and 32, then find the third geometric mean.
Solution:
The GP will be 2, g1 , g2, g3, 32 where a = 2, ar = g1, ar2 = g2, ar3 = g3 and ar4 = 32
Now, 2 × r4 = 32
⇒ r4 = 16 = (2)4
⇒ r = 2
Then, the third geometric mean = ar3 = 2 × 23 = 16
Example 7: An increasing GP is formed by three positive numbers. The new numbers form an AP if the middle term of the geometric progression is doubled. Find the common ratio of the GP.
Solution:
Let a, ar, ar2 be in GP.
If the middle term of the GP is doubled, then the new numbers are in AP.
⇒ a, 2ar, ar2 are in AP.
⇒ 4ar = a + ar2
⇒ r2 – 4r + 1 = 0
⇒ r = 2 ± √3
Since the series is an increasing GP, r = 2 + √3.
Example 8: Find the 9th and nth terms of a G.P 4, 16, 64 …
Solution:
Here, a = 4
r = 16/4 = 4
a9 = arn-1 = 4 × 49-1 = 4 × 48 = 49
Also an = arn-1 = 4(4)n-1 = 4n.
Example 9: Find the sum of the first 5 terms of the G.P 2, 8, 32, …
Solution:
Here, a = 2
r = 8/2= 4
n = 5
Sn = a(rn – 1)/(r – 1)
= 2(45 – 1)/(4 – 1)
= 2(45 – 1)/3
= 682
Hence, the required sum is 682.
Example 10:The geometric mean of two numbers is 6, and their arithmetic mean is 6.5. The numbers are
Solution:
Let a and b be the numbers.
GM = 6
√(ab) = 6
ab = 36
b = 36/a …(i)
AM = 6.5
(a + b)/2 = 6.5
(a + b) = 13 ..(ii)
Substitute b in (ii)
(a + 36/a) = 13
a2 – 13a + 36 = 0
(a – 9)(a – 4) = 0
So a = 9 or a = 4
Hence, the numbers are 9 and 4.
Geometric Progression and Its Properties
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3169 | https://www.youtube.com/watch?v=dsF7SpMHiDs | Definite Integral of Piecewise Function
Jonathan Walters
6020 subscribers
135 likes
Description
22541 views
Posted: 17 Sep 2020
Here we tackle a definite integral of a piecewise function.
The key to evaluating a definite integral for a piecewise function is to break up the bound of integration everywhere there is a new piece in the function.
Then just use the appropriate piece of the function over that sub interval.
Thanks for watching!
If you have any questions, just post them down in the comments!
-dr. dub
10 comments
Transcript:
hey guys what's up today we're looking at a definite integral of a piecewise function and this piecewise function is actually a straight line for x between negative two and zero so that's like this straight line you know over here and then from 0 to 2 it's this parabola opening down that actually intersects the x-axis right here at 2. so this is like 2 right here and then over here this is negative 2. and we want to find the area under this curve that's what the definite integral of a positive function can be represented as it's a positive area under the curve so we want to find the solution we want to basically do the integral so the integral from negative 2 to two of f of x depends on where we are so if our if our bounds are between negative two and zero we need to use the part of the function that is two so this integral from negative 2 to 2 f of x dx can be broken up intermediately as the integral from negative 2 to 0 f of x dx plus the integral from 0 to 2 f of x dx and we basically do this because we want to break it up depending on where our pieces are so our pieces change at x equals 0 so we're going to use that as our intermediate point to break up this integral and so this integral is going to become the integral from negative 2 to 0. well what is f equal to on this interval well it's equal to 2 so 2 dx plus integral from 0 to 2 4 minus x squared dx so depending on where you are it determines what piece of the function you use so basically this first integral is representing this first area a1 we'll call it so a1 for area 1 is the first integral so a1 is this first integral a2 is the second integral it's going to be the second area over here so this area a2 is basically the part that's under the parabola so now all we have to do is just do the integration using the fundamental theorem of calculus part two so basically just find an antiderivative so two x from negative two to zero plus four x minus x cubed over 3 from 0 to 2. so we're finding anti-derivatives here and then plugging in the bounds so this first one is going to become 2 times 0 minus negative 2 plus 4 times 2 minus 1 third times two cubed minus zero plus zero so just plugging in those bounds all right so this becomes two times two just four plus and then here we have eight minus and then two cubed is eight so that's eight thirds so four plus eight minus eight thirds that's 12 minus eight thirds which is 36 minus eight thirds or 36 over three minus eight over three which is 28 over 3 and that's our answer so that would be the area under the curve so that's how we integrate a piecewise function we just break it break the integral up into the bounds that have different pieces and then use those pieces over those sub intervals you |
3170 | https://static.bigideasmath.com/protected/content/pe/hs/sections/alg2_pe_03_04.pdf | Section 3.4 Using the Quadratic Formula 121 Using the Quadratic Formula Deriving the Quadratic Formula Work with a partner. Analyze and describe what is done in each step in the development of the Quadratic Formula. Step Justifi cation ax 2 + bx + c = 0 ax 2 + bx = −c x 2 + b — a x = − c — a x 2 + b — a x + ( b — 2a ) 2 = − c — a + ( b — 2a ) 2 x 2 + b — a x + ( b — 2a ) 2 = − 4ac — 4a2 + b2 — 4a2 ( x + b — 2a ) 2 = b2 − 4ac — 4a2 x + b — 2a = ± √ — b2 − 4ac — 4a2 x = − b — 2a ± √— b2 − 4ac — 2 ∣ a ∣ x = −b ± √— b2 − 4ac —— 2a Using the Quadratic Formula Work with a partner. Use the Quadratic Formula to solve each equation.
a. x 2 − 4x + 3 = 0 b. x 2 − 2x + 2 = 0 c. x 2 + 2x − 3 = 0 d. x 2 + 4x + 4 = 0 e. x 2 − 6x + 10 = 0 f. x 2 + 4x + 6 = 0 Communicate Your Answer Communicate Your Answer 3. How can you derive a general formula for solving a quadratic equation? 4. Summarize the following methods you have learned for solving quadratic equations: graphing, using square roots, factoring, completing the square, and using the Quadratic Formula.
REASONING ABSTRACTLY To be profi cient in math, you need to create a coherent representation of the problem at hand. Essential Question Essential Question How can you derive a general formula for solving a quadratic equation?
3.4 The result is the Quadratic Formula.
122 Chapter 3 Quadratic Equations and Complex Numbers Lesson 3.4 Solving an Equation with Two Real Solutions Solve x 2 + 3x = 5 using the Quadratic Formula.
SOLUTION x 2 + 3x = 5 Write original equation.
x 2 + 3x −5 = 0 Write in standard form.
x = −b ± √— b2 − 4ac —— 2a Quadratic Formula x = −3 ± √—— 32 − 4(1)(−5) —— 2(1) Substitute 1 for a, 3 for b, and −5 for c.
x = −3 ± √ — 29 — 2 Simplify. So, the solutions are x = −3 + √ — 29 — 2 ≈ 1.19 and x = −3 − √ — 29 — 2 ≈ −4.19.
COMMON ERROR Remember to write the quadratic equation in standard form before applying the Quadratic Formula.
What You Will Learn What You Will Learn Solve quadratic equations using the Quadratic Formula. Analyze the discriminant to determine the number and type of solutions. Solve real-life problems.
Solving Equations Using the Quadratic Formula Previously, you solved quadratic equations by completing the square. In the Exploration, you developed a formula that gives the solutions of any quadratic equation by completing the square once for the general equation ax 2 + bx + c = 0. The formula for the solutions is called the Quadratic Formula.
Quadratic Formula, p. 122 discriminant, p. 124 Core Vocabulary Core Vocabulary Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Solve the equation using the Quadratic Formula. 1. x 2 − 6x + 4 = 0 2. 2x 2 + 4 = −7x 3. 5x 2 = x + 8 Check Graph y = x 2 + 3x − 5. The x-intercepts are about −4.19 and about 1.19. ✓ Core Core Concept Concept The Quadratic Formula Let a, b, and c be real numbers such that a ≠ 0. The solutions of the quadratic equation ax 2 + bx + c = 0 are x = −b ± √— b2 − 4ac —— 2a .
5 −10 −7 10 Zero X=1.1925824 Y=0 Section 3.4 Using the Quadratic Formula 123 Solving an Equation with One Real Solution Solve 25x 2 − 8x = 12x − 4 using the Quadratic Formula.
SOLUTION 25x 2 − 8x = 12x − 4 Write original equation. 25x 2 − 20x + 4 = 0 Write in standard form.
x = −(−20) ± √—— (−20)2 − 4(25)(4) ——— 2(25) a = 25, b = −20, c = 4 x = 20 ± √ — 0 — 50 Simplify.
x = 2 — 5 Simplify. So, the solution is x = 2 — 5 . You can check this by graphing y = 25x 2 − 20x + 4. The only x-intercept is 2 — 5 . Solving an Equation with Imaginary Solutions Solve −x 2 + 4x = 13 using the Quadratic Formula.
SOLUTION −x 2 + 4x = 13 Write original equation. −x 2 + 4x − 13 = 0 Write in standard form. x = −4 ± √—— 42 − 4(−1)(−13) ——— 2(−1) a = −1, b = 4, c = −13 x = −4 ± √— −36 —— −2 Simplify. x = −4 ± 6i — −2 Write in terms of i. x = 2 ± 3i Simplify. The solutions are x = 2 + 3i and x = 2 − 3i.
ANOTHER WAY You can also use factoring to solve 25x 2 − 20x + 4 = 0 because the left side factors as (5x − 2)2.
COMMON ERROR Remember to divide the real part and the imaginary part by −2 when simplifying.
Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Solve the equation using the Quadratic Formula. 4. x 2 + 41 = −8x 5. −9x 2 = 30x + 25 6. 5x − 7x 2 = 3x + 4 Check Graph y = −x 2 + 4x − 13. There are no x-intercepts. So, the original equation has no real solutions. The algebraic check for one of the imaginary solutions is shown.
−(2 + 3i )2 + 4(2 + 3i ) = ? 13 5 − 12i + 8 + 12i = ? 13 13 = 13 ✓ Check 1.25 −1 −0.5 4 Zero X=.4 Y=0 12 −50 −8 10 124 Chapter 3 Quadratic Equations and Complex Numbers Analyzing the Discriminant Find the discriminant of the quadratic equation and describe the number and type of solutions of the equation.
a. x 2 − 6x + 10 = 0 b. x 2 − 6x + 9 = 0 c. x 2 − 6x + 8 = 0 SOLUTION Equation Discriminant Solution(s) ax 2 + bx + c = 0 b2 − 4ac x = −b ± √— b2 − 4ac —— 2a a. x 2 − 6x + 10 = 0 (−6)2 − 4(1)(10) = −4 Two imaginary: 3 ± i b. x 2 − 6x + 9 = 0 (−6)2 − 4(1)(9) = 0 One real: 3 c. x 2 − 6x + 8 = 0 (−6)2 − 4(1)(8) = 4 Two real: 2, 4 Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Find the discriminant of the quadratic equation and describe the number and type of solutions of the equation. 7. 4x 2 + 8x + 4 = 0 8. 1 — 2 x 2 + x − 1 = 0 9. 5x 2 = 8x − 13 10. 7x 2 − 3x = 6 11. 4x 2 + 6x = −9 12. −5x2 + 1 = 6 − 10x Analyzing the Discriminant In the Quadratic Formula, the expression b 2 − 4ac is called the discriminant of the associated equation ax 2 + bx + c = 0. x = −b ± √— b2 − 4ac —— 2a discriminant You can analyze the discriminant of a quadratic equation to determine the number and type of solutions of the equation.
Core Core Concept Concept Analyzing the Discriminant of ax 2 + bx + c = 0 Value of discriminant b2 − 4ac > 0 b2 − 4ac = 0 b2 − 4ac < 0 Number and type of solutions Two real solutions One real solution Two imaginary solutions Graph of y = ax2 + bx + c x y Two x-intercepts x y One x-intercept x y No x-intercept Section 3.4 Using the Quadratic Formula 125 Writing an Equation Find a possible pair of integer values for a and c so that the equation ax2 − 4x + c = 0 has one real solution. Then write the equation.
SOLUTION In order for the equation to have one real solution, the discriminant must equal 0. b2 − 4ac = 0 Write the discriminant. (−4)2 − 4ac = 0 Substitute −4 for b. 16 − 4ac = 0 Evaluate the power. −4ac = −16 Subtract 16 from each side. ac = 4 Divide each side by −4.
Because ac = 4, choose two integers whose product is 4, such as a = 1 and c = 4. So, one possible equation is x2 − 4x + 4 = 0.
ANOTHER WAY Another possible equation in Example 5 is 4x2 − 4x + 1 = 0. You can obtain this equation by letting a = 4 and c = 1.
Methods for Solving Quadratic Equations Method When to Use Graphing Use when approximate solutions are adequate. Using square roots Use when solving an equation that can be written in the form u2 = d, where u is an algebraic expression.
Factoring Use when a quadratic equation can be factored easily.
Completing the square Can be used for any quadratic equation ax 2 + bx + c = 0 but is simplest to apply when a = 1 and b is an even number.
Quadratic Formula Can be used for any quadratic equation.
Concept Summary Concept Summary Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com 13. Find a possible pair of integer values for a and c so that the equation ax2 + 3x + c = 0 has two real solutions. Then write the equation. The table shows fi ve methods for solving quadratic equations. For a given equation, it may be more effi cient to use one method instead of another. Suggestions about when to use each method are shown below.
Check Graph y = x2 − 4x + 4. The only x-intercept is 2. You can also check by factoring. x2 − 4x + 4 = 0 (x − 2)2 = 0 x = 2 ✓ 7 −2 −3 8 Zero X=2 Y=0 126 Chapter 3 Quadratic Equations and Complex Numbers Solving Real-Life Problems The function h = −16t 2 + h0 is used to model the height of a dropped object. For an object that is launched or thrown, an extra term v0t must be added to the model to account for the object’s initial vertical velocity v0 (in feet per second). Recall that h is the height (in feet), t is the time in motion (in seconds), and h0 is the initial height (in feet).
h = −16t 2 + h0 Object is dropped.
h = −16t 2 + v0t + h0 Object is launched or thrown.
As shown below, the value of v0 can be positive, negative, or zero depending on whether the object is launched upward, downward, or parallel to the ground. Modeling a Launched Object A juggler tosses a ball into the air. The ball leaves the juggler’s hand 4 feet above the ground and has an initial vertical velocity of 30 feet per second. The juggler catches the ball when it falls back to a height of 3 feet. How long is the ball in the air?
SOLUTION Because the ball is thrown, use the model h = −16t 2 + v0t + h0. To fi nd how long the ball is in the air, solve for t when h = 3. h = −16t2 + v0t + h0 Write the height model. 3 = −16t2 + 30t + 4 Substitute 3 for h, 30 for v0, and 4 for h0. 0 = −16t2 + 30t + 1 Write in standard form.
This equation is not factorable, and completing the square would result in fractions. So, use the Quadratic Formula to solve the equation. t = −30 ± √—— 302 − 4(−16)(1) ——— 2(−16) a = −16, b = 30, c = 1 t = −30 ± √— 964 —— −32 Simplify. t ≈ −0.033 or t ≈ 1.9 Use a calculator. Reject the negative solution, −0.033, because the ball’s time in the air cannot be negative. So, the ball is in the air for about 1.9 seconds.
Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com 14. WHAT IF? The ball leaves the juggler’s hand with an initial vertical velocity of 40 feet per second. How long is the ball in the air? V0 > 0 V0 < 0 V0 = 0 Section 3.4 Using the Quadratic Formula 127 1. COMPLETE THE SENTENCE When a, b, and c are real numbers such that a ≠ 0, the solutions of the quadratic equation ax2 + bx + c = 0 are x = __. 2. COMPLETE THE SENTENCE You can use the ____ of a quadratic equation to determine the number and type of solutions of the equation. 3. WRITING Describe the number and type of solutions when the value of the discriminant is negative. 4. WRITING Which two methods can you use to solve any quadratic equation? Explain when you might prefer to use one method over the other.
Exercises 3.4 Vocabulary and Core Concept Check Vocabulary and Core Concept Check In Exercises 5–18, solve the equation using the Quadratic Formula. Use a graphing calculator to check your solution(s). (See Examples 1, 2, and 3.) 5. x2 − 4x + 3 = 0 6. 3x2 + 6x + 3 = 0 7. x2 + 6x + 15 = 0 8. 6x2 − 2x + 1 = 0 9. x2 − 14x = −49 10. 2x2 + 4x = 30 11. 3x2 + 5 = −2x 12. −3x = 2x2 − 4 13. −10x = −25 − x2 14. −5x2 − 6 = −4x 15. −4x2 + 3x = −5 16. x2 + 121 = −22x 17. −z2 = −12z + 6 18. −7w + 6 = −4w2 In Exercises 19–26, fi nd the discriminant of the quadratic equation and describe the number and type of solutions of the equation. (See Example 4.) 19. x2 + 12x + 36 = 0 20. x2 − x + 6 = 0 21. 4n2 − 4n − 24 = 0 22. −x2 + 2x + 12 = 0 23. 4x2 = 5x − 10 24. −18p = p2 + 81 25. 24x = −48 − 3x2 26. −2x2 − 6 = x 27. USING EQUATIONS What are the complex solutions of the equation 2x2 − 16x + 50 = 0? ○ A 4 + 3i, 4 − 3i ○ B 4 + 12i, 4 − 12i ○ C 16 + 3i, 16 − 3i ○ D 16 + 12i, 16 − 12i 28. USING EQUATIONS Determine the number and type of solutions to the equation x2 + 7x = −11.
○ A two real solutions ○ B one real solution ○ C two imaginary solutions ○ D one imaginary solution ANALYZING EQUATIONS In Exercises 29–32, use the discriminant to match each quadratic equation with the correct graph of the related function. Explain your reasoning.
29. x2 − 6x + 25 = 0 30. 2x2 − 20x + 50 = 0 31. 3x2 + 6x − 9 = 0 32. 5x2 − 10x − 35 = 0 A. x y 2 4 −4 −8 B. x y 20 −40 8 −4 C. x y 20 10 8 4 −4 D. x y 15 25 35 5 10 6 2 −2 Monitoring Progress and Modeling with Mathematics Monitoring Progress and Modeling with Mathematics Dynamic Solutions available at BigIdeasMath.com 128 Chapter 3 Quadratic Equations and Complex Numbers ERROR ANALYSIS In Exercises 33 and 34, describe and correct the error in solving the equation.
33. x2 + 10x + 74 = 0 x = −10 ± √—— 102 − 4(1)(74) ——— 2(1) = −10 ± √— −196 —— 2 = −10 ± 14 — 2 = −12 or 2 ✗ 34. x2 + 6x + 8 = 2 x = −6 ± √—— 62 − 4(1)(8) —— 2(1) = −6 ± √ — 4 — 2 = −6 ± 2 — 2 = −2 or −4 ✗ OPEN-ENDED In Exercises 35–40, fi nd a possible pair of integer values for a and c so that the quadratic equation has the given solution(s). Then write the equation. (See Example 5.) 35. ax2 + 4x + c = 0; two imaginary solutions 36. ax2 + 6x + c = 0; two real solutions 37. ax2 − 8x + c = 0; two real solutions 38. ax2 − 6x + c = 0; one real solution 39. ax2 + 10x = c; one real solution 40. −4x + c = −ax2; two imaginary solutions USING STRUCTURE In Exercises 41–46, use the Quadratic Formula to write a quadratic equation that has the given solutions. 41. x = −8 ± √— −176 —— −10 42. x = 15 ± √— −215 —— 22 43. x = −4 ± √— −124 —— −14 44. x = −9 ± √— 137 — 4 45. x = −4 ± 2 — 6 46. x = 2 ± 4 — −2 COMPARING METHODS In Exercises 47–58, solve the quadratic equation using the Quadratic Formula. Then solve the equation using another method. Which method do you prefer? Explain. 47. 3x2 − 21 = 3 48. 5x2 + 38 = 3 49. 2x2 − 54 = 12x 50. x2 = 3x + 15 51. x2 − 7x + 12 = 0 52. x2 + 8x − 13 = 0 53. 5x2 − 50x = −135 54. 8x2 + 4x + 5 = 0 55. −3 = 4x2 + 9x 56. −31x + 56 = −x2 57. x2 = 1 − x 58. 9x2 + 36x + 72 = 0 MATHEMATICAL CONNECTIONS In Exercises 59 and 60, fi nd the value for x. 59. Area of the rectangle = 24 m2 (2x − 9) m (x + 2) m 60. Area of the triangle = 8 ft2 (x + 1) ft (3x − 7) ft 61. MODELING WITH MATHEMATICS A lacrosse player throws a ball in the air from an initial height of 7 feet. The ball has an initial vertical velocity of 90 feet per second. Another player catches the ball when it is 3 feet above the ground. How long is the ball in the air? (See Example 6.) 62. NUMBER SENSE Suppose the quadratic equation ax2 + 5x + c = 0 has one real solution. Is it possible for a and c to be integers? rational numbers? Explain your reasoning. Then describe the possible values of a and c. Section 3.4 Using the Quadratic Formula 129 63. MODELING WITH MATHEMATICS In a volleyball game, a player on one team spikes the ball over the net when the ball is 10 feet above the court. The spike drives the ball downward with an initial vertical velocity of 55 feet per second. How much time does the opposing team have to return the ball before it touches the court?
64. MODELING WITH MATHEMATICS An archer is shooting at targets. The height of the arrow is 5 feet above the ground. Due to safety rules, the archer must aim the arrow parallel to the ground.
5 ft 3 ft a. How long does it take for the arrow to hit a target that is 3 feet above the ground? b. What method did you use to solve the quadratic equation? Explain. 65. PROBLEM SOLVING A rocketry club is launching model rockets. The launching pad is 30 feet above the ground. Your model rocket has an initial vertical velocity of 105 feet per second. Your friend’s model rocket has an initial vertical velocity of 100 feet per second. a. Use a graphing calculator to graph the equations of both model rockets. Compare the paths. b. After how many seconds is your rocket 119 feet above the ground? Explain the reasonableness of your answer(s). 66. PROBLEM SOLVING The number A of tablet computers sold (in millions) can be modeled by the function A = 4.5t 2 + 43.5t + 17, where t represents the year after 2010. a. In what year did the tablet computer sales reach 65 million? b. Find the average rate of change from 2010 to 2012 and interpret the meaning in the context of the situation. c. Do you think this model will be accurate after a new, innovative computer is developed? Explain.
67. MODELING WITH MATHEMATICS A gannet is a bird that feeds on fi sh by diving into the water. A gannet spots a fi sh on the surface of the water and dives 100 feet to catch it. The bird plunges toward the water with an initial vertical velocity of −88 feet per second. a. How much time does the fi sh have to swim away? b. Another gannet spots the same fi sh, and it is only 84 feet above the water and has an initial vertical velocity of −70 feet per second. Which bird will reach the fi sh fi rst? Justify your answer. 68. USING TOOLS You are asked to fi nd a possible pair of integer values for a and c so that the equation ax2 − 3x + c = 0 has two real solutions. When you solve the inequality for the discriminant, you obtain ac < 2.25. So, you choose the values a = 2 and c = 1. Your graphing calculator displays the graph of your equation in a standard viewing window. Is your solution correct? Explain.
10 −10 −10 10 69. PROBLEM SOLVING Your family has a rectangular pool that measures 18 feet by 9 feet. Your family wants to put a deck around the pool but is not sure how wide to make the deck. Determine how wide the deck should be when the total area of the pool and deck is 400 square feet. What is the width of the deck?
x x x x x x x x 18 ft 9 ft 130 Chapter 3 Quadratic Equations and Complex Numbers 70. HOW DO YOU SEE IT? The graph of a quadratic function y = ax2 + bx + c is shown. Determine whether each discriminant of ax2 + bx + c = 0 is positive, negative, or zero. Then state the number and type of solutions for each graph. Explain your reasoning. a. x y b. x y c. x y 71. CRITICAL THINKING Solve each absolute value equation. a. ∣ x2 – 3x – 14 ∣ = 4 b. x2 = ∣ x ∣ + 6 72. MAKING AN ARGUMENT The class is asked to solve the equation 4x2 + 14x + 11 = 0. You decide to solve the equation by completing the square. Your friend decides to use the Quadratic Formula. Whose method is more effi cient? Explain your reasoning. 73. ABSTRACT REASONING For a quadratic equation ax2 + bx + c = 0 with two real solutions, show that the mean of the solutions is − b — 2a . How is this fact related to the symmetry of the graph of y = ax2 + bx + c?
74. THOUGHT PROVOKING Describe a real-life story that could be modeled by h = −16t 2 + v0t + h0. Write the height model for your story and determine how long your object is in the air.
75. REASONING Show there is no quadratic equation ax2 + bx + c = 0 such that a, b, and c are real numbers and 3i and −2i are solutions.
76. MODELING WITH MATHEMATICS The Stratosphere Tower in Las Vegas is 921 feet tall and has a “needle” at its top that extends even higher into the air. A thrill ride called Big Shot catapults riders 160 feet up the needle and then lets them fall back to the launching pad.
a. The height h (in feet) of a rider on the Big Shot can be modeled by h = −16t2 + v0 t + 921, where t is the elapsed time (in seconds) after launch and v0 is the initial vertical velocity (in feet per second). Find v0 using the fact that the maximum value of h is 921 + 160 = 1081 feet.
b. A brochure for the Big Shot states that the ride up the needle takes 2 seconds. Compare this time to the time given by the model h = −16t2 + v0t + 921, where v0 is the value you found in part (a). Discuss the accuracy of the model.
Maintaining Mathematical Proficiency Maintaining Mathematical Proficiency Solve the system of linear equations by graphing. (Skills Review Handbook) 77. −x + 2y = 6 78. y = 2x − 1 x + 4y = 24 y = x + 1 79. 3x + y = 4 80. y = −x + 2 6x + 2y = −4 −5x + 5y = 10 Graph the quadratic equation. Label the vertex and axis of symmetry. (Section 2.2) 81. y = −x2 + 2x + 1 82. y = 2x2 − x + 3 83. y = 0.5x2 + 2x + 5 84. y = −3x2 − 2 Reviewing what you learned in previous grades and lessons |
3171 | http://clubztutoring.com/ed-resources/math/cosecant-definitions-examples-6-7-6-5-2/ | Cosecant: Definitions and Examples - Club Z! Tutoring
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Home / Educational Resources / Math Resources / Cosecant: Definitions and Examples
Cosecant, often abbreviated as csc, is a trigonometric function that is the reciprocal of the sine function. It is one of the six trigonometric functions, along with sine, cosine, tangent, cotangent, and secant, that are used to relate the angles and sides of a right-angled triangle.
The cosecant function is defined as the ratio of the length of the hypotenuse of a right-angled triangle to the length of its opposite side. In mathematical terms, if we have a right-angled triangle with an angle ?, the cosecant of that angle is given by:
csc(?) = hypotenuse / opposite
where hypotenuse is the longest side of the triangle, and opposite is the side opposite to the angle ?.
The cosecant function has many applications in mathematics, science, and engineering. It is used in the calculation of various physical phenomena, such as the amplitude and frequency of waves, the velocity and acceleration of objects in motion, and the behavior of electrical circuits.
Properties of Cosecant Function
Like the other trigonometric functions, the cosecant function has certain properties that are important to understand. Here are some of the key properties of the cosecant function:
Periodicity: The cosecant function is periodic with a period of 2?. This means that the cosecant function repeats itself every 2? radians or 360 degrees. In other words, if we add or subtract a multiple of 2? to the angle ?, the value of the cosecant function remains the same.
Range: The range of the cosecant function is the set of all real numbers except for 0. This is because the cosecant function is undefined at angles where the opposite side is equal to 0, which corresponds to the vertical asymptotes of the function.
Symmetry: The cosecant function is an odd function, which means that csc(-?) = -csc(?). This property is a consequence of the reciprocal relationship between the sine and cosecant functions.
Relationship with Other Trigonometric Functions: The cosecant function is related to the other trigonometric functions by various identities, such as:
csc(?) = 1/sin(?) sin(?) = 1/csc(?)
These identities allow us to express the values of one trigonometric function in terms of another, which can be useful in simplifying expressions and solving problems.
Applications of Cosecant Function
The cosecant function has a wide range of applications in various fields, including mathematics, physics, engineering, and computer science. Here are some examples of how the cosecant function is used:
Wave Phenomena: The cosecant function is used in the analysis of wave phenomena, such as sound waves, light waves, and electromagnetic waves. The amplitude and frequency of a wave can be expressed in terms of the cosecant function, which allows us to model and understand the behavior of waves.
Electrical Circuits: The cosecant function is used in the analysis of electrical circuits, particularly in the calculation of impedance and admittance. These concepts are important in the design and analysis of electronic devices, such as amplifiers, filters, and oscillators.
Motion and Mechanics: The cosecant function is used in the analysis of motion and mechanics, particularly in the calculation of velocity, acceleration, and force. These concepts are important in the study of physics and engineering, and are used in the design and analysis of machines and structures.
Introduction The trigonometric functions are essential tools in mathematics and have a wide range of applications in fields such as engineering, physics, and computer graphics. One of these functions is the cosecant, which is also known as the reciprocal of the sine function. In this article, we will explore the cosecant function and its properties, including its definition, its graph, and its various applications.
Definition of Cosecant The cosecant function is defined as the reciprocal of the sine function. It is denoted by csc(x) and can be written as:
csc(x) = 1 / sin(x)
Where x is an angle in radians.
The cosecant function is undefined at values of x where sin(x) = 0, which corresponds to the vertical asymptotes of the cosecant graph.
Properties of Cosecant Function
Periodicity: The cosecant function has a period of 2?. This means that the function repeats itself after every 2? radians.
Range: The range of the cosecant function is (-?, -1] ? [1, ?).
Asymptotes: The cosecant function has vertical asymptotes at x = k?, where k is an integer.
Symmetry: The cosecant function is an odd function, which means that csc(-x) = -csc(x).
Zeroes: The cosecant function has zeroes at x = k?, where k is an integer, except for the cases where k is zero.
Examples of Cosecant Function
Find the value of csc(?/3)
Solution: csc(?/3) = 1/sin(?/3) = 1/(?3/2) = 2/?3
Find the values of x where csc(x) = -2
Solution: csc(x) = -2 1/sin(x) = -2 sin(x) = -1/2 x = 7?/6 + 2k? or 11?/6 + 2k?, where k is an integer.
Find the period of the cosecant function.
Solution: The period of the cosecant function is 2?, which means that the function repeats itself after every 2? radians.
Find the vertical asymptotes of the cosecant function.
Solution: The vertical asymptotes of the cosecant function occur where the sine function equals zero. This happens at x = k?, where k is an integer.
Find the range of the cosecant function.
Solution: The range of the cosecant function is (-?, -1] ? [1, ?).
Applications of Cosecant Function
Electrical Engineering: The cosecant function is used in electrical engineering to calculate the impedance of a circuit. The impedance is the ratio of the voltage applied to a circuit to the current that flows through it. The impedance is given by the formula:
Z = V/I = R + jX = R + j/(?C)
Where R is the resistance of the circuit, X is the reactance of the circuit, ? is the frequency of the voltage applied to the circuit, and C is the capacitance of the circuit. The reactance is given by:
X = 1/(?C) = 1/(2?fC)
Quiz
Q1. What is the cosecant function? A1. The cosecant function (csc) is a trigonometric function that represents the reciprocal of the sine function. It is defined as csc(x) = 1/sin(x).
Q2. What is the domain of the cosecant function? A2. The domain of the cosecant function is all real numbers except for the values of x where sin(x) = 0. In other words, the domain is the set of all x such that x ? n?, where n is an integer.
Q3. What is the range of the cosecant function? A3. The range of the cosecant function is the set of all real numbers except for values between -1 and 1 (inclusive).
Q4. What is the period of the cosecant function? A4. The period of the cosecant function is 2?, which means that the function repeats itself every 2? units.
Q5. What are the even and odd properties of the cosecant function? A5. The cosecant function is an odd function, which means that csc(-x) = -csc(x). It has no even property, which means that csc(-x) ? csc(x).
Q6. What is the graph of the cosecant function? A6. The graph of the cosecant function is a continuous curve that oscillates between positive and negative infinity as it approaches the asymptotes at x = n?, where n is an integer.
Q7. What are the asymptotes of the cosecant function? A7. The asymptotes of the cosecant function are the lines x = n?, where n is an integer. These lines represent the values of x for which sin(x) = 0, and therefore, the cosecant function is undefined.
Q8. What is the reciprocal identity of the cosecant function? A8. The reciprocal identity of the cosecant function is sin(x) = 1/csc(x). This identity states that the sine of an angle is equal to the reciprocal of its cosecant.
Q9. What is the quotient identity of the cosecant function? A9. The quotient identity of the cosecant function is cot(x) = cos(x)/sin(x) = 1/(sin(x)/cos(x)) = 1/tan(x) = csc(x)/sec(x). This identity relates the cosecant function to the other trigonometric functions.
Q10. What is the inverse cosecant function? A10. The inverse cosecant function (csc^-1) is the inverse of the cosecant function. It is defined as csc^-1(y) = sin^-1(1/y), where y is a number between -1 and 1 (exclusive).
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Cosecant:
Illustration
Definition
The cosecant csc z is the function defined by csc z | congruent | 1/(sin z) | = | (2i)/(e^(i z) - e^(-i z)), where sin z is the sine. The cosecant is implemented in the Wolfram Language as Csc[z].
Related terms
Flint Hills series | inverse cosecant | secant | sine
Related Wolfram Language symbol
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3172 | https://www.geneseo.edu/~heap/courses/333/equivalence_relation.pdf | 1 Equivalence Relation Definition 1. An equivalence relation is a relationship on a set, generally denoted by “∼”, that is reflexive, symmetric, and transitive for everything in the set.
1. (Reflexivity) a ∼a, 2. (Symmetry) if a ∼b then b ∼a, 3. (Transitivity) if a ∼b and b ∼c then a ∼c.
Equivalence relations are often used to group together objects that are similar, or “equiv-alent”, in some sense.
2 Examples Example: The relation “is equal to”, denoted “=”, is an equivalence relation on the set of real numbers since for any x, y, z ∈R: 1. (Reflexivity) x = x, 2. (Symmetry) if x = y then y = x, 3. (Transitivity) if x = y and y = z then x = z.
All of these are true.
Example: Let “≃” denote the relation on the set of symmetric matrices (recall symmetric means A = At) defined as follows. A ≃B if A = Bt. This is an equivalence relation since for any symmetric matrices A, B, C: 1. (Reflexivity) A ≃A since A = At.
2. (Symmetry) If A ≃B then A = Bt, but then B = (Bt)t = (A)t = At. And this implies B ≃A.
3. (Transitivity) If A ≃B and B ≃C then A = Bt and B = Ct. Of course, since these are symmetric matrices, we know B = Bt. Thus A = Bt = B = Ct, and therefore A ≃C.
(Of course this is actually a “stupid” equivalence relation since it is the same as the “is equal to” equivalence relation on the set of symmetric matrices. Do you see why?) Non-example: The relation “is less than or equal to”, denoted “≤”, is NOT an equivalence relation on the set of real numbers. For any x, y, z ∈R, “≤” is reflexive and transitive but NOT necessarily symmetric.
1. (Reflexivity) Of course x ≤x is true since x = x.
2. (Symmetry) If x ≤y then it is not necessarily true that y ≤x. For example, 5 ≤7, but 7 ≰5.
3. (Transitivity) If x ≤y and y ≤z then x ≤z since x ≤y ≤z.
1 |
3173 | https://www.wikiwand.com/en/articles/Capital_accumulation | Capital accumulation
Dynamic that motivates pursuit of profit, central tenet of capitalism From Wikipedia, the free encyclopedia
Capital accumulation is the dynamic that motivates the pursuit of profit, involving the investment of money or any financial asset with the goal of increasing the initial monetary value of said asset as a financial return whether in the form of profit, rent, interest, royalties or capital gains. The goal of accumulation of capital is to create new fixed capital and working capital, broaden and modernize the existing ones, grow the material basis of social-cultural activities, as well as constituting the necessary resource for reserve and insurance. The process of capital accumulation forms the basis of capitalism, and is one of the defining characteristics of a capitalist economic system.
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Definition
In economics and accounting, capital accumulation is often equated with investment of profit income or savings, especially in real capital goods. The concentration and centralisation of capital are two of the results of such accumulation (see below).
Capital accumulation refers ordinarily to:
and by extension to:
Both non-financial and financial capital accumulation is usually needed for economic growth, since additional production usually requires additional funds to enlarge the scale of production. Smarter and more productive organization of production can also increase production without increased capital. Capital can be created without increased investment by inventions or improved organization that increase productivity, discoveries of new assets (oil, gold, minerals, etc.), the sale of property, etc.
In modern macroeconomics and econometrics the term capital formation is often used in preference to "accumulation", though the United Nations Conference on Trade and Development (UNCTAD) refers nowadays to "accumulation". The term is occasionally used in national accounts.
Measurement of accumulation
Accumulation can be measured as the monetary value of investments, the amount of income that is reinvested, or as the change in the value of assets owned (the increase in the value of the capital stock). Using company balance sheets, tax data and direct surveys as a basis, government statisticians estimate total investments and assets for the purpose of national accounts, national balance of payments and flow of funds statistics. Usually, the reserve banks and the Treasury provide interpretations and analysis of this data. Standard indicators include capital formation, gross fixed capital formation, fixed capital, household asset wealth, and foreign direct investment.
Demand-led growth models
In macroeconomics, following the Harrod–Domar model, the savings ratio (
s
{\displaystyle s}
) and the capital coefficient (
k
{\displaystyle k}
) are regarded as critical factors for accumulation and growth, assuming that all saving is used to finance fixed investment. The rate of growth of the real stock of fixed capital (
K
{\displaystyle K}
) is:
where
Y
{\displaystyle Y}
is the real national income. If the capital-output ratio or capital coefficient (
k
K
Y
{\displaystyle k={K \over Y}}
) is constant, the rate of growth of
Y
{\displaystyle Y}
is equal to the rate of growth of
K
{\displaystyle K}
. This is determined by
s
{\displaystyle s}
(the ratio of net fixed investment or saving to
Y
{\displaystyle Y}
) and
k
{\displaystyle k}
.
A country might, for example, save and invest 12% of its national income, and then if the capital coefficient is 4:1 (i.e. $4 billion must be invested to increase the national income by 1 billion) the rate of growth of the national income might be 3% annually. However, as Keynesian economics points out, savings do not automatically mean investment (as liquid funds may be hoarded for example). Investment may also not be investment in fixed capital (see above).
Assuming that the turnover of total production capital invested remains constant, the proportion of total investment which just maintains the stock of total capital, rather than enlarging it, will typically increase as the total stock increases. The growth rate of incomes and net new investments must then also increase, in order to accelerate the growth of the capital stock. Simply put, the bigger capital grows, the more capital it takes to keep it growing and the more markets must expand.
The Harrodian model has a problem of unstable static equilibrium, since if the growth rate is not equal to the Harrodian warranted rate, the production will tend to extreme points (infinite or zero production). The Neo-Kaleckians models do not suffer from the Harrodian instability but fails to deliver a convergence dynamic of the effective capacity utilization to the planned capacity utilization. For its turn, the model of the Sraffian Supermultiplier grants a static stable equilibrium and a convergence to the planned capacity utilization. The Sraffian Supermultiplier model diverges from the Harrodian model since it takes the investment as induced and not as autonomous. The autonomous components in this model are the Autonomous Non-Capacity Creating Expenditures, such as exports, credit led consumption and public spending. The growth rate of these expenditures determines the long run rate of capital accumulation and product growth.
Marxist concept
Karl Marx borrowed the idea of capital accumulation or the concentration of capital from early socialist writers such as Charles Fourier, Louis Blanc, Victor Considerant, and Constantin Pecqueur. In Marx's critique of political economy, capital accumulation is the operation whereby profits are reinvested into the economy, increasing the total quantity of capital. Capital was understood by Marx to be expanding value, that is, in other terms, as a sum of capital, usually expressed in money, that is transformed through human labor into a larger value and extracted as profits. Here, capital is defined essentially as economic or commercial asset value that is used by capitalists to obtain additional value (surplus-value). This requires property relations which enable objects of value to be appropriated and owned, and trading rights to be established. Marx argued that capital has the tendency for concentration and centralization in the hands of richest capitalists
According to Marxism during periods of stagnation in capitalism, the accumulation process is increasingly oriented towards investment on military and security forces, real estate, financial speculation, and luxury consumption. In that case, income from value-adding production will decline in favour of interest, rent and tax income, with as a corollary an increase in the level of permanent unemployment. Capital accumulation of the means of production in Marxist thought leads to the formation of the bourgeoisie.
"Accumulation of capital" sometimes also refers in Marxist writings to the reproduction of capitalist social relations (institutions) on a larger scale over time, i.e., the expansion of the size of the proletariat and of the wealth owned by the bourgeoisie. In the first volume of Das Kapital Marx had illustrated this idea with reference to Edward Gibbon Wakefield's theory of colonisation, and further refers to the "fetishism of capital" reaching its highest point with interest-bearing capital, because of how capital appeared to grow almost of its own accord.
The Marxist analysis of capital accumulation and the development of capitalism identifies systemic issues with the process that arise with expansion of the productive forces. A crisis of overaccumulation of capital occurs when the rate of profit is greater than the rate of new profitable investment outlets in the economy, arising from increasing productivity from a rising organic composition of capital (higher capital input to labor input ratio). This depresses the wage bill, leading to stagnant wages and high rates of unemployment for the working class while excess profits search for new profitable investment opportunities. Marx believed that this cyclical process would be the fundamental cause for the dissolution of capitalism and its replacement by socialism, which would operate according to a different economic dynamic. Anarchists hold that the state always maintains a form of capital accumulation to the elite, even in self proclaimed socialist states and that for true equality the state should be abolished.
Effects
The effects of wealth accumulation results in increased savings for the individual. If economic growth is shared unevenly between different groups of the population wealth inequality emerges. Extreme wealth inequality can result in oligarchy, in which super rich individuals hold most power and money in society.
See also
References
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Home » Fractions » Simplify Fractions
Simplify Fractions
Easy How-To w/ 8 Step-by-Step Examples!
// Last Updated: November 8, 2020 - Watch Video //
How to simplify fractions?
Jenn, Founder Calcworkshop®, 15+ Years Experience (Licensed & Certified Teacher)
Great question!
And in today’s math lesson, you’re going learn two different methods to do just that.
Overview
The word simplify means to make something easier to do or understand.
So, reducing or simplifying fractions means we make the fraction as simple as possible.
We do this by dividing the numerator and the denominator by the largest number that can divide into both numbers exactly. In other words, we divide the top and bottom by the biggest number they have in common.
Note that the numerator and denominator are called “terms” of a fraction. So, if we simplify a fraction, we reduce the fraction to the simplest terms.
For example, notice the three rectangles below where the left-hand side of each rectangle is shaded and the fraction that results represents the shaded part out of the whole rectangle.
Three Equivalent Fractions — Visual
All three images represent the same part to whole, which indicates that each resulting fraction is equivalent to the other. But what is important to recognize is that while they are equal, only one fraction is in simplest terms — 1/2.
And just as this example indicates, our goal is to transform a fraction by creating an equivalent fraction whose terms no longer have any common factors as noted by Lumen Learning.
So, how do we reduce fractions?
There are two methods:
Guess and Check
Greatest Common Factor (GCF)
Guess And Check Method
The guess and check method is when we choose a number we know divides evenly into both the numerator and denominator, but we aren’t sure if it’s the largest. So, we may need to continue to reduce the fraction further in necessary.
Worked Example #1
Let’s look at a specific problem.
Suppose we want the simplified fraction of 24/36.
Using the “guess and check method,” we may notice that 24 and 36 are both divisible by 3.
And after reducsing both terms and get 8/12.
Finding Equivalent Fractions — Example
But we quickly notice that 8 and 12 still have a factor in common.
Namely, they are both divisible by 4. This means we need to simplify further.
Notice that this new fraction of 2/3 is fully simplified because neither the numerator nor the denominator has any factors left in common.
Simplifying Fraction — Example
GCF Method
Whereas the greatest common factor (GCF) method will always give us the largest number for which we should divide.
Worked Example #2
Now, let’s work the same example using the GCF method.
First, let’s use factor trees to find the prime factorization of both the numerator and denominator.
And then we identify the GCF from the prime factorization. Remember, when we find the GCF from a list of prime factors, we choose the fewest of what is common.
GCF Using Prime Factorization
This means that when reducing the fraction 24/36, we should divide both terms by 12, as this is the GCF for both terms.
Reduce Fractions To Lowest Term
Worksheet (PDF) —Hands on Practice
Put that pencil to paper in these easy to follow worksheets — test your knowledge!
Simplifying Fractions — Practice Problems
Simplifying Fractions —Step-by-Step Solutions
Final Thoughts
Both methods are perfectly acceptable, and it comes down to personal preference as to which technique you wish to employ.
It is apparent that if your “guess” is also the GCF, you will simplify your fraction fast, as the GCF will always yield the lowest possible reduction.
Throughout this lesson, we will look at numerous examples of how to reduce fractions to simplest form as well as some applications problems where we will first create a fraction and then reduce it to the lowest terms.
Let’s jump right in!
Video Tutorial — Full Lesson w/ Detailed Examples
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Rich 04-27-25 - PA, United States
I am a teacher who had not done anything with Calculus for over 10 years. This was a great resource for me to review some content that I did not remember. The videos and examples were so well explained that I used some of those concepts and examples when I presented the topics to my students. I would strongly recommend this to students who are looking for additional help or teachers that need to brush up on specific topics.
Chris 04-27-25 - NV, United States
Jen is the best.I struggled with the provided material given to me at ASU and resorted to YouTube videos and google.I caught wind of Jen and her website from a few classmates and I've been using her since.I have now completed calculus for Engineers: 1,2,3; DiffEq, and Linear Algebra and Jen helped me from calc2 forward. Subscribing to her was just 'another tool in my toolbox' and definitely added more confidence when doing my homework.I highly recommend.
Keshav 04-24-25 - United Kingdom
I loved it and it helped me get a better understanding of the topics. The ample amount of examples was really helpful to make sure that I understand what I am doing and how to solve various types of problems
Syniah N.04-24-25 - CA, United States
Jen is an exceptional math teacher and a rare find. In a university setting where many instructors prioritize research over teaching, Jen stands out for her ability to explain complex topics in a clear, approachable way.Before discovering this resource, I often felt lost in math classes where instructors wouldn't simplify formal equations for beginners. Jen changed that completely. Thanks to her, I passed Calculus II with an A.I'm truly grateful and highly recommend this resource to anyone struggling with math
Stephanie B.09-23-25 - FL, United States
As a homeschool mom to a math savvy teenager, I needed a strong resource for a subject that strikes fear in the hearts of most people - Calculus. Jenn's videos were exactly what I was looking for: thorough explanations, notes, and tons of example problems. Not only that, this is one of the few formats where you can actually see the person doing the teaching and she uses a real whiteboard, so taking notes feels like you are in a classroom. I can confidently say my daughter has a rock solid foundation after learning through these videos. On the occasion where she had a question I could not begin to answer, I was delighted to learn we could email Jenn and receive a prompt reply within a detailed explanation. I don't know any other instructional platform with such exceptional customer service. We cannot recommend Calcworkshop enough.
Gray A.09-23-25 - TN, United States
I had no idea what was happening in my calculus class before I started using CalcWorkshop. Within 1 lesson I learned more than my professor taught in 3 weeks. CalcWorkshop works.
Ali M.09-15-25 - NM, United States
It was amazing from start to finish of my usage of it. Great teaching style, relatable and great practice problems and setup. Genuinely such a great tool. Step-by-step instructions and effortless after a couple problems.
Autobrian20 09-14-25 - CA, United States
All videos are very clear and informative, so easy to learn complex concepts. Hardest thing was figuring out what chapters to learn alongside what I'm learning in class but that depends on your class. Wish I could just take these as my classes. Would not have made it without these videos.
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I have been so super confident ever since I encountered the Calcworksop. This platform was so convenient for me it enabled me to study effectively. Jenn is an absolute guru, I love her so much. She gets it! I do feel I come to tears with the amount of progress I've made with calculus. If anyone ever has any struggle with calculus, Calcworkshop will rescue you. And in a brief amount of time will you start to make progress. This website is a miracle TRUST ME. Calculus 1,2 and 3 can all be courses that you can get great grades out of from this website. And there are more courses you can do well in for mathematics. Again this website saved my life! Jenn Forever!
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Math is hard and math can be boring, so I really appreciate Jenn's enthusiasm in ALL of her videos. She explains concepts so well, points out the common mistakes, and goes through many examples. I am retired and learning math, which I didn't take in my adult years, so it's been really hard to learn on my own. But I want to learn physics and I will need calculus for that. I love how Jenn has organized her courses to focus on what will be needed for calculus and then the calculus courses, as well as linear algebra and differential equations.
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Excellent explanations on every topic of precalculus and calculus. Great examples. Easy to understand. Thank you!
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Literally taught me calc because my actual teacher didn't. Jen explains in a way that is simple and just makes sense. My calc teacher in high school taught super vaguely and just expected us to understand how to make connections right away. This website actually saved me from failing and helped a lot with studying for the AP test!!
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I used CalcWorkshop because I needed help with college calculus I required for entrance to a grad program I am pursuing. The instruction provided through my university was vague at best. As a full-time professional, wife, and mother of 5, it was difficult for me to go in for the tutoring sessions offered by my university. I love how Jenn provided the repetition I needed to solidify the processes in my brain and reviewed the algebra tricks while going through the comprehensive variety of problems in detail. I believe that anything can be learned if enough time is spent trying. The one-stop-shop in CalcWorkshop saved me the frustration of searching YouTube for examples and helped me make more efficient use of my time. In the course evaluation for my university, I mentioned that I used CalcWorkshop for support. I am likely to re-subscribe for myself and for my children. Thanks, again.
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This has been the best resource I have used in my whole academic career. I was struggling in my courses and this helped me go from barely passing my college classes to getting As. I highly recommend this to everyone. The calc workshop is a game changer and is just absolutely amazing. Thank you so much!
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I am very grateful to have found this site and above all I am grateful for Jen's work. It is notorious the passion for teaching and doing good to the community that continues with the hope and dream of dedicating themselves to mathematics despite the fact that the educational system or teachers that we may encounter along the way are not the most cooperative. As a personal experience, I had a terrible year academically in terms of mathematics, actually no one in my class understood the teacher, I thought that this was no longer my thing, but this online course made me regain confidence in me and knowing that from now on I will have a tool that can accompany me if I have any doubt is a great relief.
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Being in my thirties and returning to school to pursue a degree in Mechanical Engineering and hadn't taken a math class in over 25 years, Calc Workshop was the only reason I survived pre calc/trig, calc 1, and calc 2. Jenn breaks down every step of every type of problem into a logical and easy to remember process. This subscription has been worth every penny and I have recommended it to everyone I know taking math. I will be re-subscribing for calc 3 and differential equations next year. Thank you Jenn for such an amazing tool and helping me achieve success in these challenging courses, I couldn't have done it without you!
Deathbyintegrals 05-09-25 - United States
Overall it served my intended purpose which was to get me through double, triple, and line integrals. I like how the site is organized and I absolutely appreciate the clear and concise video lessons. Should I need a math refresher or assistance in a future course I will absolutely be back.
Dillan 05-06-25 - UT, United States
Her videos are very well made, you can tell she put a lot of work into how she introduces and walks you through all the material. I used her videos to help me with calculus 3. I just canceled my subscription because I wasn't in need of the information anymore and 30$ is a lot to pay if your not gonna watch the videos. I will probably re subscribe for my future math classes. Her videos are super good!Her website is well organized and easy to get around. I did not use any of her other features on her website I just used her posted videos from the website and I found that plenty sufficient for me to learn the material. Overall, I felt like 30$ a month was a little expensive but this is such a great product that I would 100% subscribe again!
Amanda M.05-02-25 - AZ, United States
Calcworkshop is an outstanding resource for anyone looking to strengthen their understanding of calculus. Jen's clear and approachable teaching style made even the most complex topics feel manageable. As a math teacher, I found her step-by-step explanations incredibly helpful in breaking down difficult concepts for my own students. The scaffolding she provides is thoughtfully designed, building confidence and deepening comprehension at every stage. Whether you're looking for a refresher or a fresh perspective on solving problems, I highly recommend Calcworkshop as a go-to tool for mastering calculus.
A Reviewer 04-28-25 - United States
It was everything u needed to know simplifed in one video. Love it.
Rich 04-27-25 - PA, United States
I am a teacher who had not done anything with Calculus for over 10 years. This was a great resource for me to review some content that I did not remember. The videos and examples were so well explained that I used some of those concepts and examples when I presented the topics to my students. I would strongly recommend this to students who are looking for additional help or teachers that need to brush up on specific topics.
Chris 04-27-25 - NV, United States
Jen is the best.I struggled with the provided material given to me at ASU and resorted to YouTube videos and google.I caught wind of Jen and her website from a few classmates and I've been using her since.I have now completed calculus for Engineers: 1,2,3; DiffEq, and Linear Algebra and Jen helped me from calc2 forward. Subscribing to her was just 'another tool in my toolbox' and definitely added more confidence when doing my homework.I highly recommend.
Keshav 04-24-25 - United Kingdom
I loved it and it helped me get a better understanding of the topics. The ample amount of examples was really helpful to make sure that I understand what I am doing and how to solve various types of problems
Syniah N.04-24-25 - CA, United States
Jen is an exceptional math teacher and a rare find. In a university setting where many instructors prioritize research over teaching, Jen stands out for her ability to explain complex topics in a clear, approachable way.Before discovering this resource, I often felt lost in math classes where instructors wouldn't simplify formal equations for beginners. Jen changed that completely. Thanks to her, I passed Calculus II with an A.I'm truly grateful and highly recommend this resource to anyone struggling with math
Stephanie B.09-23-25 - FL, United States
As a homeschool mom to a math savvy teenager, I needed a strong resource for a subject that strikes fear in the hearts of most people - Calculus. Jenn's videos were exactly what I was looking for: thorough explanations, notes, and tons of example problems. Not only that, this is one of the few formats where you can actually see the person doing the teaching and she uses a real whiteboard, so taking notes feels like you are in a classroom. I can confidently say my daughter has a rock solid foundation after learning through these videos. On the occasion where she had a question I could not begin to answer, I was delighted to learn we could email Jenn and receive a prompt reply within a detailed explanation. I don't know any other instructional platform with such exceptional customer service. We cannot recommend Calcworkshop enough.
Gray A.09-23-25 - TN, United States
I had no idea what was happening in my calculus class before I started using CalcWorkshop. Within 1 lesson I learned more than my professor taught in 3 weeks. CalcWorkshop works.
Ali M.09-15-25 - NM, United States
It was amazing from start to finish of my usage of it. Great teaching style, relatable and great practice problems and setup. Genuinely such a great tool. Step-by-step instructions and effortless after a couple problems.
Autobrian20 09-14-25 - CA, United States
All videos are very clear and informative, so easy to learn complex concepts. Hardest thing was figuring out what chapters to learn alongside what I'm learning in class but that depends on your class. Wish I could just take these as my classes. Would not have made it without these videos.
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3175 | https://study.com/skill/practice/calculating-torque-questions.html | Calculating Torque Practice | Physics Practice Problems | Study.com
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Calculating Torque
AP Physics 1 Skills Practice
A 5.0 N force is applied at a distance r = 1.5 m from the axis of rotation {eq}O {/eq}, as shown below. The angle between the force and the distance vector, {eq}\theta=25^\circ{} {/eq}. Calculate the torque, and determine its sign and direction.
Answers:
-3.2 N m, directed toward the screen
-7.5 N m, directed toward the screen
7.5 N m, directed out of the screen
3.2 N m, directed out of the screen
When a 42 N force is applied at a distance r = 27 m from the axis of rotation {eq}O {/eq}, a torque will be generated (see the image below). If the angle between the force and the distance vector, {eq}\theta=15^\circ{} {/eq}, calculate the torque, and determine its sign and direction.
Answers:
-1100 N m, directed out of the screen
-290 N m, directed toward the screen
290 N m, directed out of the screen
1100 N m, directed toward the screen
We have a 14 N force applied at a distance r = 0.33 m from the axis of rotation {eq}O {/eq} (see the image below). Find the torque, and determine its sign an direction if the angle between the force and the distance vector, {eq}\theta=78^\circ{} {/eq}.
Answers:
4.6 N m, directed out of the screen
-4.5 N m, directed toward the screen
4.5 N m, directed toward the screen
-4.6 N m, directed out of the screen
A force of magnitude 8.0 N (shown in the image below) is applied at a distance r = 0.75 m from the axis of rotation {eq}O {/eq}. The angle between the force and the distance vector, {eq}\theta=35^\circ{} {/eq}. Calculate the torque, and determine its sign and direction.
Answers:
-6.0 N m, directed toward the page
-3.4 N m, directed toward the page
3.4 N m, directed out of the page
6.0 N m, directed out of the page
John applies a 6.6 N force at a distance r = 0.35 m from the axis of rotation {eq}O {/eq}, as shown in the image. The angle between the force and the distance vector, {eq}\theta=65^\circ{} {/eq}. Calculate the torque, and determine its sign and direction.
Answers:
-2.1 N m, directed toward the page
-2.3 N m, directed out of the page
2.1 N m, directed out of the page
2.3 N m, directed toward the page
An engineer applies a 12 N force at a distance r = 0.83 m from the axis of rotation {eq}O {/eq}, as shown below. The angle between the force and the distance vector, {eq}\theta=38^\circ{} {/eq}. Calculate the torque, and determine its sign and direction.
Answers:
6.1 N m, directed out of the screen
-10 N m, directed out of the screen
-6.1 N m, directed toward the screen
10 N m, directed toward the screen
A 9.5 N perpendicular force is applied at a distance r = 1.3 m from the axis of rotation {eq}O {/eq} (the angle between the force and the distance vector, {eq}\theta=90^\circ{} {/eq}). Calculate the torque, and determine its sign and direction.
Answers:
-12 N m, directed toward the screen
12 N m, directed toward the screen
-12 N m, directed out of the screen
12 N m, directed out of the screen
Sofia applies a 2.4 N force at a distance r = 0.68 m from the axis of rotation {eq}O {/eq}, as shown in the image. The angle {eq}\theta=55^\circ{} {/eq}. Calculate the torque, and determine its sign and direction.
Answers:
-1.6 N m, directed toward the screen
1.3 N m, directed out of the screen
-1.3 N m, directed toward the screen
1.6 N m, directed out of the screen
A robot arm applies a 230 N force at a distance r = 1.1 m from the axis of rotation {eq}O {/eq}, as shown in the image. The angle between the force and the distance vector, {eq}\theta=8.5^\circ{} {/eq}. Calculate the torque, and determine its sign and direction.
Answers:
250 N m, directed out of the screen
37 N m, directed out of the screen
-250 N m, directed toward the screen
-37 N m, directed toward the screen
When a 44 N force is applied at a distance r = 22 m from the axis of rotation {eq}O {/eq}, as shown in the image, a torque is created. The angle between the force and the distance vector, {eq}\theta=30^\circ{} {/eq}. Calculate the torque, and determine its sign and direction.
Answers:
-480 N m, directed out of the screen
-480 N m, directed toward the screen
970 N m, directed out of the screen
-970 N m, directed toward the screen
Assume a 4.1 N force is applied at a distance r = 2.6 m from the axis of rotation {eq}O {/eq}, as shown below. The angle between the force and the distance vector, {eq}\theta=33^\circ{} {/eq}. Calculate the torque, and determine its sign and direction.
Answers:
5.8 N m, directed out of the screen
11 N m, directed toward the screen
-11 N m, directed out of the screen
-5.8 N m, directed toward the screen
A 6.2 N force is applied at a distance r = 1.4 m from the axis of rotation {eq}O {/eq}, and it is in the same direction as the distance vector, as shown below. Calculate the torque, and determine its sign and direction.
Answers:
-8.7 N m, directed out of the screen
-8.7 N m, directed toward the screen
8.7 N m, directed toward the screen
0 N m, no direction
You have a 17 N force applied at a distance r = 0.89 m from the axis of rotation {eq}O {/eq} (see the image below). Find the torque, and determine its sign and direction if the angle between the force and the distance vector, {eq}\theta=43^\circ{} {/eq}.
Answers:
10 N m, directed out of the screen
-10 N m, directed toward the screen
15 N m, directed toward the screen
-15 N m, directed out of the screen
Let's assume a force of magnitude 4.5 N, shown in the image below, is applied at a distance r = 0.25 m from the axis of rotation {eq}O {/eq}. The angle between the force and the distance vector, {eq}\theta=52^\circ{} {/eq}. Calculate the torque, and determine its sign and direction.
Answers:
-1.1 N m, directed out of the screen
0.89 N m, directed out of the screen
-0.89 N m, directed toward the screen
1.1 N m, directed toward the screen
Karen applies a 3.7 N force at a distance r = 0.22 m from the axis of rotation {eq}O {/eq}, as shown in the image. The angle {eq}\theta=28^\circ{} {/eq}. Calculate the torque, and determine its sign and direction.
Answers:
0.38 N m, directed out of the screen
-0.81 N m, directed toward the screen
-0.38 N m, directed toward the screen
0.81 N m, directed out of the screen
A 62 N force is applied at a distance r = 12 m from the axis of rotation {eq}O {/eq}, as shown below. The angle between the distance vector and positive x axis is {eq}20^\circ{} {/eq} while the angle between the force vector and x axis is {eq}70^\circ{} {/eq}. Calculate the torque, and determine its sign and direction.
Answers:
570 N m, directed out of the screen
250 N m, directed out of the screen
-570 N m, directed toward the screen
-250 N m, directed toward the screen
A 25 N force is applied at a distance r = 14 m from the axis of rotation {eq}O {/eq}, as shown below. The distance vector is directed along the y axis, and the force vector is parallel to x axis. Calculate the torque, and determine its sign and direction.
Answers:
350 N m, directed toward the screen
-350 N m, directed out of the screen
350 N m, directed out of the screen
-350 N m, directed toward the screen
We have a 38 N force applied at a distance r = 16 m from the axis of rotation {eq}O {/eq}, as shown below. The angle between the distance vector and positive x axis is {eq}120^\circ{} {/eq} while the force vector is parallel to x axis (see the figure). Calculate the torque, and determine its sign and direction.
Answers:
530 N m, directed out of the screen
-530 N m, directed toward the screen
-300 N m, directed out of the screen
300 N m, directed toward the screen
Consider a 60 N force applied at a distance r = 3.3 m from the axis of rotation {eq}O {/eq}, as shown below. The distance vector is directed along the x axis in the negative direction, and the force vector is parallel to y axis in the negative y direction (see the figure). Calculate the torque, and determine its sign and direction.
Answers:
200 N m, directed out of the screen
200 N m, directed toward the screen
-200 N m, directed toward the screen
-200 N m, directed out of the screen
An engineer draws a scheme wherein a 9.2 N force is applied at a distance r = 0.18 m from the axis of rotation {eq}O {/eq}, as shown below. The distance vector is directed along the positive x axis, while the force vector is parallel to y axis and is pointing in the negative y direction (see the figure). Calculate the torque, and determine its sign and direction.
Answers:
-1.7 N m, directed out of the screen
1.7 N m, directed toward the screen
-1.7 N m, directed toward the screen
1.7 N m, directed out of the screen
You are trying to close a door by applying a 2.7 N force at a distance 0.75 m from the axis of rotation {eq}O {/eq}, as shown below. The angle between the force and the door frame, {eq}\theta=40^\circ{} {/eq}. Calculate the torque, and determine its sign and direction.
Answers:
-2.0 N m, directed out of the screen
1.3 N m, directed out of the screen
-1.3 N m, directed toward the screen
2.0 N m, directed toward the screen
David pulls a rod holding it 0.32 m from the axis of rotation {eq}O {/eq} and applying a 12 N force, as shown in the image below. The angle between the force and the rod, {eq}\theta=35^\circ{} {/eq}. Calculate the torque, and determine its sign and direction.
Answers:
-3.8 N m, directed toward the screen
-2.2 N m, directed toward the screen
2.2 N m, directed out of the screen
3.8 N m, directed out of the screen
A rod can rotate around an axis of rotation {eq}O {/eq}, and a mass is hanging 0.47 m from the axis of rotation, which applies an 8.5 N force, as shown in the image below. The angle between the force and the rod, {eq}\theta=35^\circ{} {/eq}. Calculate the torque, and determine its sign and direction.
Answers:
4.0 N m, directed out of the screen
2.3 N m, directed out of the screen
-2.3 N m, directed toward the screen
-4.0 N m, directed toward the screen
A board stands vertically, and it can flip around the axis of rotation {eq}O {/eq} as shown below. You throw an object, and it hits the board at a location 1.3 m from the axis of rotation. The object exerts a force of 38 N, and the angle between the force and the board, {eq}\theta=63^\circ{} {/eq}. Calculate the torque, and determine its sign and direction.
Answers:
-44 N m, directed toward the screen
49 N m, directed out of the screen
44 N m, directed out of the screen
-49 N m, directed toward the screen
A chamber door can rotate around the axis of rotation {eq}O {/eq} as shown below. A force of 350 N is applied 0.025 m from the axis of rotation at an angle {eq}\theta=45^\circ{} {/eq}. Calculate the torque, and determine its sign and direction.
Answers:
-8.8 N m, directed toward the screen
8.8 N m, directed out of the screen
-6.2 N m, directed toward the screen
6.2 N m, directed out of the screen
Frank applies an 18 N force at a distance r = 0.56 m from the axis of rotation {eq}O {/eq}, as shown below. The angle between the force and the distance vector, {eq}\theta=14^\circ{} {/eq}. Calculate the torque, and determine its sign and direction.
Answers:
2.4 N m, directed out of the screen
-10 N m, directed toward the screen
10 N m, directed out of the screen
-2.4 N m, directed toward the screen
Laura sets an experiment wherein she applies a 7.7 N force at a distance r = 0.15 m from the axis of rotation {eq}O {/eq}, as shown in the image. The angle {eq}\theta=68^\circ{} {/eq}. Calculate the torque, and determine its sign and direction.
Answers:
-1.1 N m, directed toward the screen
1.2 N m, directed out of the screen
-1.2 N m, directed toward the screen
1.1 N m, directed out of the screen
An equipment is set up, so a 550 N force can be applied at a distance r = 1.7 m from the axis of rotation {eq}O {/eq}, as shown in the image. The angle between the force and the distance vector, {eq}\theta=85^\circ{} {/eq}. Calculate the torque, and determine its sign and direction.
Answers:
940 N m, directed toward the screen
-940 N m, directed out of the screen
-930 N m, directed toward the screen
930 N m, directed toward the screen
Albert designs an apparatus wherein he applies a 0.75 N force at a distance r = 2.9 m from the axis of rotation {eq}O {/eq}, as shown in the image. The angle {eq}\theta=33^\circ{} {/eq}. Calculate the torque, and determine its sign and direction.
Answers:
-1.2 N m, directed toward the screen
1.2 N m, directed out of the screen
-2.2 N m, directed toward the screen
2.2 N m, directed out of the screen
In a device design, a 0.12 N force is applied at a distance r = 0.26 m from the axis of rotation {eq}O {/eq}, as shown in the image. As a result, a torque will be created. The angle between the force and the distance vector, {eq}\theta=62^\circ{} {/eq}. Calculate the torque, and determine its sign and direction.
Answers:
0.028 N m, directed out of the screen
0.031 N m, directed out of the screen
-0.031 N m, directed toward the screen
-0.028 N m, directed toward the screen
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3176 | https://mathoverflow.net/questions/370471/are-there-any-identities-for-alternating-binomial-sums-of-the-form-sum-k-0 | Stack Exchange Network
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Are there any identities for alternating binomial sums of the form $\sum_{k=0}^{n} (-1)^{k}k^{p}{n \choose k} $?
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In equations (20) - (25) of Mathworld's article on binomial sums, identities are given for sums of the form $$\sum_{k=0}^{n} k^{p}{n \choose k}, $$ with $p \in \mathbb{Z}_{\geq 0}$. I wonder whether identities also exist for the alternating counterparts: $$\sum_{k=0}^{n} (-1)^{k}k^{p}{n \choose k} .$$ Furthermore, I'm interested in results for the same sum that is cut off, i.e. when the summands go from $k=0$ to some $D.
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edited Aug 30, 2020 at 19:18
Max Lonysa MullerMax Lonysa Muller
asked Aug 30, 2020 at 18:58
Max Lonysa MullerMax Lonysa Muller
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A rewrite of formula (10) on MathWorld (replacing the summation index $k-i\mapsto i$) gives the desired formula: $$\sum_{k=0}^{n} (-1)^{k}k^{p}{n \choose k} =(-1)^n n! S_2(p,n),$$ where $S_2(p,n)$ is the Stirling number of the second kind (the number of ways of partitioning a set of $p$ elements into $n$ non-empty subsets). It is remarkable that the alternating sum equals zero for $p.
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edited Aug 30, 2020 at 20:23
answered Aug 30, 2020 at 19:28
Carlo BeenakkerCarlo Beenakker
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$\begingroup$ It's not so remarkable that the alternating sum is zero for $pp$ is 0. $\endgroup$
Richard Stanley
– Richard Stanley
2020-08-30 21:48:28 +00:00
Commented Aug 30, 2020 at 21:48
$\begingroup$ I think letting the LHS sum run to infinity allows it to be defined for fractional $n$ and still maintain the same behavior on the integers. An optimist might hope that this suggests a definition of an analytic continuation of the Stirling numbers $\endgroup$
Sidharth Ghoshal
– Sidharth Ghoshal
2024-06-05 01:23:07 +00:00
Commented Jun 5, 2024 at 1:23
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For the cutoff version:
We can get a subtraction-free formula for the cutoff version, which should be sufficient to get asymptotics, by the same idea that gives a simple bijective proof of the identity that Carlo Beenakker mentioned. That is:
$k^p$ counts maps from a $p$-element set $[p]$ to a $k$-element set
Thus $\binom{n}{k} k^p$ counts pairs of a $k$-element subset $S$ of an $n$-elements set $[n]$ with a map from $[p]$ to $S$. In other words, it counts maps $f$ from $[p]$ to $[n]$ together with a $k$-element subset $S$ of $[n]$ containing the image of $f$.
So $\sum_{k=0}^d (-1)^k \binom{n}{k} k^p$ is the sum over maps $f: [p] \to n$ of the sum over subsets $S$ of $[n]$, containing the image of $f$, of size at most $k$, of $(-1)^{|S|}$. We may assume the image of $f$ has size $\leq d < n $ and thus that there is some element $e$ not in the image of $f$. We can cancel each subset with $e\notin S$ with the $S \cup {e}$, as these have opposite signs. The only subsets that fail to cancel are those that have size exactly $d$ and do not contain $e$, of which there are $\binom{n - | \operatorname{Im}(f) | -1}{ d - |\operatorname{Im}(f)| } $.
With $S_2(p,j)$ again the Stirling numbers of the second kind, the number of maps from $[p]$ to $[n]$ with image of size $j$ is $ \frac{n!}{ (n-j)!} S_2(p,j) $, so the sum is
$$ (-1)^d \sum_{j=0}^d S_2(p,j) \frac{n!}{(n-j)!} \binom{ n-j-1}{d-j} $$
$$= (-1)^d \frac{n!}{ (n-1-d)!} \sum_{j=0}^d S_2(p,j) \frac{1}{(n-j)} \frac{1}{(d-j)!} $$
(If $d=n$ then all subsets cancel and so only the terms with $| \operatorname{Im} f| =n$ remain, so we just obtain the count of surjections from $[p]$ to $[n]$, as in Carlo Beenakker's answer.)
Alternately, a formula-based proof:
we have $$ k^p = \sum_{j=0}^k S_2( p,j) \frac{k!}{ (k-j)!} $$ ( a standard identity.) so
$$\sum_{k=0}^d (-1)^k k^p {n \choose k} = \sum_{j=0}^d \sum_{k=j}^d (-1)^k S_2( p,j) \frac{k!}{(k-j)!} {n \choose k} $$ and $$\frac{k!}{(k-j)!}{n\choose k} = \frac{k! n!}{ (k-j)! k! (n-k)! } = \frac{n!}{ (k-j)! (n-k)!} = \frac{n!}{(n-j)!} \binom{n-j}{k-j} $$ so $$ \sum_{k=0}^d (-1)^k k^p {n \choose k} = \sum_{j=0}^d \sum_{k=j}^d (-1)^k S_2( p,j) \frac{n!}{(n-j)!} \binom{n-j}{k-j}$$ $$ = \sum_{j=0}^d (-1)^d S_2( p,j) \frac{n!}{(n-j)!} \binom{n-j-1}{d-j} = (-1)^d \frac{n!}{ (n-1-d)!} \sum_{j=0}^d S_2(p,j) \frac{1}{(n-j)} \frac{1}{(d-j)!} $$
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edited Aug 30, 2020 at 21:12
answered Aug 30, 2020 at 20:58
Will SawinWill Sawin
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$\begingroup$ Thank you! It is at times like these that I wish I could accept two answers. $\endgroup$
Max Lonysa Muller
– Max Lonysa Muller
2020-08-31 11:12:14 +00:00
Commented Aug 31, 2020 at 11:12
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Up to the factor $(-1)^n$, the uncut sum is $$s_{p,n}:=\sum_{k=0}^n(-1)^{n-k}\, k^p\,\binom nk.$$ As noted in the comment by Richard Stanley, $$s_{p,n}=(\Delta^n f_p)(0),$$ where $f_p(x):=x^p$ and $(\Delta f)(x):=f(x+1)-f(x)$. Here and in what follows, $x$ denotes any real number.
It is easy to check by induction on $n$ that for any smooth enough function $f$ we have $$(\Delta^n f)(x)=Ef^{(n)}(x+S_n),$$ where $f^{(n)}$ is the $n$th derivative of $f$, $S_n:=U_1+\cdots+U_n$, and $U_1,\dots,U_n$ are independent random variables uniformly distributed on the interval $[0,1]$. So, $$s_{p,n}=n!\binom pn ES_n^{p-n} \tag{1}$$ for $p=0,1,\dots$ and $n=0,1,\dots$. In particular, it follows that $s_{p,n}=0$ for $n=p+1,p+2,\dots$, as noted in the answer by Carlo Beenakker.
In fact, (1) holds for all real $p\ge n$ (and $n=0,1,\dots$), and then, obviously, $$0
If $p-n\ge1$, then, in view of Jensen's inequality, the lower bound $0$ on $s_{p,n}$ in (2) can be greatly improved, to $$b_{p,n}:=n!\binom pn \Big(\frac n2\Big)^{p-n}.$$
Moreover, by the law of large numbers, $S_n/n\to1/2$ in probability (say). Also, $0\le S_n/n\le1$. So, by dominated convergence, from (1) we immediately get the following asymptotics: if $n\to\infty$ and $p-n\to a$ for some real $a>0$, then $$s_{p,n}\sim b_{p,n}.$$
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edited Aug 31, 2020 at 3:18
answered Aug 31, 2020 at 0:02
Iosif PinelisIosif Pinelis
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You can find plenty of documentation on Gould's site. Maybe it could be useful. The link is Interesting files are Vol.1.PDF to Vol. 8.PDF.
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edited Aug 30, 2020 at 20:14
answered Aug 30, 2020 at 20:05
Luciano PetrilloLuciano Petrillo
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Number Theory =============
Partitioning the set of positive integers into n subsets
by Puzzled417, Sep 16, 2016, 3:59 PM
Does there exist an integer that satisfies the following condition?
The set of positive integers can be partitioned into nonempty subsets such that an arbitrary sum of integers, one taken from each of any of the subsets, lies in the remaining subset.
(Source: 1995 IMO Shortlist/N7)
Solution
We prove that such a number does not exist. We can't have since then the sets would be equal. Now suppose and let and be numbers in . Let be any element in which may be the same as or . Assume, for sake of contradiction, that and belong to different subsets. If one of them, say , also belongs to while the other belongs to some other subset, say , choose in for . Then is in , so that is in . On the other hand, is in , so that is in , a contradiction. The only other case is where neither nor belongs to , say is in and is in . Then is in while is in , a contradiction. Thus, and must belong to the same subset.
For choose in and let , where . Then is also in . We may assume that is in . Since is in the same subset as , which is , it follows that contains all the even numbers. If is even, then is an even number in , a contradiction. Now suppose is odd. Then is in and hence odd. Thus, must be even, so that consists of precisely the even numbers. Then is the same subset as . Thus, all odd numbers greater than are in the same subset, say . But then also, a contradiction.
This post has been edited 1 time. Last edited by Puzzled417, Sep 16, 2016, 8:18 PM
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Why can't all chemist/IUPAC agree to make equilibrium constant as dimensionless?
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I mean the gold book defines two terms as the equilibrium constant - the standard/thermodynamic equilibrium constant and generic equilibrium constant.
To me, only the thermodynamic one seems correct cause it is derived from thermodynamic factors or activities and Gibbs free energy. The other one just seems like a senseless ratio/pedagogical tool to teach this concept of "Law of Mass Action" which itself doesn't have any thermodynamic basis but only a kinetic one.
So, is the second one useful into some other parts of chemistry, if not why can't we do away with it.
Or is it because kinetic equilibrium without thermodynamic equilibrium is a serious thing.
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edited Apr 2, 2022 at 11:13
Ian Bush
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asked Apr 2, 2022 at 3:24
Deepak AryaDeepak Arya
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Why do you think the law of mass action cannot be derived from thermodynamics factors ?Maurice –Maurice 2022-04-02 08:28:01 +00:00 Commented Apr 2, 2022 at 8:28
There are cases where the TD K looks like a senseless pedagogical tool looking good in textbooks, while not always applicable in real life.Poutnik –Poutnik 2022-04-02 10:14:13 +00:00 Commented Apr 2, 2022 at 10:14
It is both important when teaching chemistry and also in "real life" to understand that from kinetics we can get equilibrium constants. I have seen how using kinetics data for gold chloride complexes interconverting that we can calculate the stability constants for complexes such as AuCl4-.Nuclear Chemist –Nuclear Chemist 2022-04-02 10:47:20 +00:00 Commented Apr 2, 2022 at 10:47
1 if you ever use Δ G=−R T ln(K e q)Δ G=−R T ln(K e q) then K e q K e q has to be dimensionless, which means that all equilibrium constants have to be dimensionless. It is a historical/convenient thing that units are still given, and we implicitly assume that the K K are divided by 1 unit to make them dimensionless. The Gold book definition gives them as dimensionless.porphyrin –porphyrin 2022-04-03 07:04:49 +00:00 Commented Apr 3, 2022 at 7:04
@porphyrin what you are saying is just the end results where we have already defined relation between chemical potential and activities (which are dimensionless), this just strengthen the point that we should do away generic equilibrium constant. I am looking for some strong counter arguments. One good one is by Karsten Theis below.Deepak Arya –Deepak Arya 2022-04-03 17:13:32 +00:00 Commented Apr 3, 2022 at 17:13
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The three K K's of biochemistry
It is common to use K K with a unit in biochemistry. There are three main uses, and the unit customarily has the dimension of a concentration (e.g. m o l L−1 m o l L−1).
K d K d: Dissociation constant, typically of a ligand bound to a biomolecule. You would say "the ligand has nanomolar affinity" to express that the dissociation constant is about 1⋅10−9 1⋅10−9. This means that at nanomolar concentration, there would be a significant fraction of ligand-bound biomolecule (with the assumption that the ligand is always at higher concentration than the biomolecule).
K i K i: Inhibition constant, essentially the dissociation constant of the inhibitor binding to an enzyme. When the inhibitor concentration is equal to the inhibition constant, more than 50% is bound.
K m K m: Michaelis–Menten constant, a combination of three rate constants. This describes a steady state of an enzymatic reaction. When the substrate concentration is equal to K m K m, the enzymatic rate is half-maximal (compared to saturating concentrations of substrate).
In all three cases, you assume a bimolecular association event with no cooperativity.
Here is a snapshot from the Table 1. Recommended Symbols and their Units [1, p. 290]:
Symbol K i K i A K i c k i j K i u K m K m A K s K s A Meaning inhibition constant (inhibition type unspecified)inhibition constant for A competitive inhibition constant rate constant for step from E i to E j uncompetitive inhibition constant Michaelis constant (or Michaelis concentration)Michaelis constant for A substrate dissociation constant value of K s for substrate A Customary unit m o l d m−3 m o l d m−3 m o l d m−3 as k m o l d m−3 m o l d m−3 m o l d m−3 m o l d m−3 m o l d m−3 Symbol Meaning Customary unit K i inhibition constant (inhibition type unspecified)m o l d m−3 K i A inhibition constant for A m o l d m−3 K i c competitive inhibition constant m o l d m−3 k i j rate constant for step from E i to E j as k K i u uncompetitive inhibition constant m o l d m−3 K m Michaelis constant (or Michaelis concentration)m o l d m−3 K m A Michaelis constant for A m o l d m−3 K s substrate dissociation constant m o l d m−3 K s A value of K s for substrate A m o l d m−3
Relation to thermodynamic equilibrium constant
The relation is loose. For example, even if these are pH-dependent, you would not include the hydronium concentration in the scheme but instead quote the values at a specific pH. You would also combine distinct protonation states in a single constant (for examples for ATP).
Time to wash your hands
As a physical chemist, you might feel sullied by the muddy definitions above. However, they are part of the daily language in biochemical labs and in the pharmaceutical industry. This would be the "A" of IUPAC.
Logarithms and units
The argument and the result of a logarithm do not have units, but measurements do. Somewhere before taking a logarithm, you have to have a step that gets rid of the units. Here are some examples:
p H=−logH X+(1)p H=−log[H X+]
Δ G∘=−R T ln(K)(2)(2)Δ G∘=−R T ln(K)
ln k 2 k 1=E A R(1 T 1−1 T 2)(3)(3)lnk 2 k 1=E A R(1 T 1−1 T 2)
Textbooks often are not explicit about what happens to the units. In (1), you have to divide the concentration by the standard state (unless you use activities instead of concentration). In (2), you "dropped" the units when calculating the equilibrium constant. In (3), you get away with rate constants that have units because they cancel out. It would be a fine convention to have equilibrium constants with units, and then divide them by Q∘Q∘, the reaction quotient for the standard state. So whether K K is defined as dimensionless or with units is not directly linked to thermodynamic or kinetic definition, it is a convention.
Kinetics vs thermodynamics
K m K m by its nature is defined through kinetics, as is K i K i. For determining K d K d, however, you typically would do an equilibrium measurement. Combined with the (first-order) rate constant of dissociation, k o f f k o f f, you can calculate the (second-order) rate constant of association. For non-covalent complexes, this will be fast, near diffusion controlled, in many instances. So there is a connection between kinetics and thermodynamics, but it is not directly related to the choice whether K K has units or is dimensionless.
Reference
Nomenclature Committee of the International Union of Biochemistry (NC-IUB). Symbolism and Terminology in Enzyme Kinetics (Recommendations 1981). Eur J Biochem1982, 128 (2–3), 281–291. DOI: 10.1111/j.1432-1033.1982.tb06963.x.
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edited Apr 3, 2022 at 13:39
answered Apr 2, 2022 at 12:53
Karsten♦Karsten
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1 Other fields have chemistry have similar things, for example Henry's Law constants (K H K H) are generally reported with units because there is no single standard definition.Andrew –Andrew 2022-04-02 19:57:07 +00:00 Commented Apr 2, 2022 at 19:57
2 This answer seems to me like just a series of facts without an explantion of the chemistry. One should aim to understand chemistry rather than merely memorise it. I would be more happy if the answer showed the relationship between kinetics and the thermodynamic idea of an equilibrium.Nuclear Chemist –Nuclear Chemist 2022-04-03 06:12:37 +00:00 Commented Apr 3, 2022 at 6:12
1 @NuclearChemist I gave it a shot. However, I can not be responsible for your happiness.Karsten –Karsten♦ 2022-04-03 13:42:02 +00:00 Commented Apr 3, 2022 at 13:42
1 @KarstenTheis to me the logarithm argument doesn't make sense. In fact it will point to flaws in our thermodynamic theory if some variable came under logarithm and all we have to say is look logarithm are unitless , so let's divide by something to make our variable unitless. That's just a bad theory/argument. It is unitless cause activities are unitless and activities are unitless cause they are defined w.r.t standard states.Deepak Arya –Deepak Arya 2022-04-03 17:32:27 +00:00 Commented Apr 3, 2022 at 17:32
1 It is sloppy use of math. DeltaG = -RTlnKeq + RTlnQ = -RT [lnKeq -lnQ] = -RTln[Keq/Q]. The units of Keq and Q are the same so the expression is unitless. At equilibrium Q=Keq so deltaG = 0. Under standard conditions All activities are defined as 1. so deltaG = deltaG = _RTln[Keq/1] still unitless.jimchmst –jimchmst 2022-04-05 05:29:01 +00:00 Commented Apr 5, 2022 at 5:29
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3179 | https://www.youtube.com/watch?v=SMr6tDC0-EM | If A+B+C=pi and if cos3A+cos3B+cos3C=1 then show that one angle must be 120^@.
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If A+B+C=pi and if cos3A+cos3B+cos3C=1 then show that one angle must be 120^@.
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कि आज डाउनलोड करें डॉक्टर पर होगा आपके सभी मैच केमिस्ट्री फिजिक्स बायोलॉजी डोंट कर सपा या बसपा में पशुओं की फोटो खींचो उसे करो करो और तुरंत वीडियो सलूशन पाव डाउनलोड Now की जगह क्वेश्चन में हमें का योग बना इसे प्लस प्लस रिकॉर्ड उपाय इनको 3S प्लस कौस रवि प्लस कौस इक्वल टू वन शामिल है फ्रंट करना है शो करना है तुम्हें गर्ल मस्ट 1520 ई एच ए करने के लिए क्या करेंगे सबसे पहले हम क्या करते हैं प्लस बी प्लस सी कोड लिखे है 158 प्लस को भी प्लस सीट बंटवारे के भाई दोनों साइड छिपा कर व डिफरेंट क्रीम बहुत सेट किए हैं पथरी ऐड कि पृथ्वी एक रस्सी 1835 ई कौन सी मेरा किया जाएगा मेरा तृषा जाएगा फ्री पाई - कि 383 बी कि यह मारा गया अब उसके बाद करना है उसके बाद हमें CR यहां पर अब मैं क्या करता हूं कॉस्ट लेता हूं को स्कैंडल ऑथर्स मेरे पास आएगा कौन सी इक्वल टू कैश थ्री फाइव माइनर मेरे प्रिय कि पृथ्वी हुआ था अब कॉस्ट 5 मिनट या होता है को थी पर - इट मेरे पास है - का सेट अप यह क्या आएगा माइनस कॉस ए कि चाहिए ए प्लस टीवी मैं एक उल्टा कौन सी कि अभी को अतिथि कि यह जो वैल्यू परपस है यह मैं किसी क्वेश्चन मुक्त का यह क्वेश्चन है इस क्वेश्चन एंड पुट करते हैं लिए कोई क्वेश्चन का तिवारी को 303 भी अलग और इसी को खोज चाहिए कि प्रकाश 3b कैंडी क्रश कोर्स c-equator का सेवन तो हमें क्या करना है गौर से की गई वाली पोस्ट करनी है तो करते हैं कलियां देखो को स्तरीय ए प्लस कौन-कौन सी ईद - कोर्स case-2 ए समय को से प्लस बी का होल क्यूब प्लस बी प्लस खोज यह कि को स्थाई जो से प्लस बी का फॉर्मूला मेरे पास बना कौशिक को सभी को का कॉस्पी को स्टीव भिंडी - मैं साइन इन साइन भी यहां क्या बनेगा साइंस लैब साइंस पेपर कि इक्वल टू वन लुट अब देखो मैच बैक ओपन करता हूं ब्रेड को ओपन करने से पहले को क्या बना को स्त्री ऐड प्लस को कोसती भी - को अतिशीघ्र कौन-कौन सी बी ए प्लस कि साहित्य के साथ टीवी हुआ था लुट इक्वल टू के वध सी इतने पार्ट को मैं क्या लिख लेता हूं देखो इतना पार्ट है कि इतने पार्ट को इतनी पार्टी का ले सकता हूं मैं वन माइनस वन माइनस कौस थीटा माइनस कॉस हुई दो मैं वन माइनस कॉस चीफ ए प्लस कि साइंसेज़ ए साइंटिफिक व्हाट दोएस वन सेवन कैंसर लाइट बिल हमारे पास क्या बचा है कि संयंत्र इस एंट्री भी कि इक्वल टू वन - स्वास्थ्य मैं वन माइनस कॉस 3b कि हमारे पास क्या बचेगा विटामिंस अट त्रिवेंद्रम ए साइंटिफिक लिखेंगे साइन थ्री यह कि बाय वन माइनस कौस स्क्वायर इक्वल टू गो मैं वन माइनस कॉस चीफ मैं कौन हूं के साथ बेबी आप देखोगे ना लुट इसको में 40 बताऊं साईं टू इन 272 कौन-कौन वन माइनस कॉस जगह बैक टू साइंस टो इटावा टू से हमें कली हुआ क्लॉस कैसे तबीयत थोड़ी ठीक है बनेगा हाउ टू थे कॉस्केट टैटू सीट अमेरिका 382 इक्वल टू सेमी दर टू पॉस कर थे रिवर टू सिगरेट अपऑन इसको मिलता हूं साइन टू हां जी बिट्टू भी थे इंडियन साइंटिस्ट यह देखो थे सेंट्रल टू साइंस सेंटर फॉर स्टेटस ब्लुटूथ साइन कि सिगरेट टू कि को स्ट्रीम कि वे टू इक्वल टू टू कैश फिर copy2 कि यह फोन में काया हाउ टू को स्ट्रैट 372 हुआ था कौए द टू साइन MP3 बाय टू कौन-कौन सी बाय टू यह देखो इसको सिक्वेंस लौट इसको से इसको कैन सिलेक्ट दूसरे टू एंड ठोसे टू यहां क्या बचा क्या हो जाएगा साइंस व 21 ए साइंटिफिक टू बी कि इक्वल टू कौर थे रिवर 24312 भी तो अब क्या बना मेरे पास देखो पॉसिबल टू हुए अ क्वेश्चन पति बैक टू माइनर साइन 382 ए साइन आफ बटु इक्वल टू केसेस मेरा को से कोसमी सनशाइन भी एक फॉर्म लाखों से लक्ष्मी का कोर्स 372 ए प्लस बी प्लस टू इक्वल टू कि है वापस आएगा कोर्स फाइबर टू अब को सिक्वेंस लैस शिकायत 382 प्लस थ्री ल तू काया कि फाइबर 2अक्टूवर कैंसल 2153 काज जगत को प्लस Music 125 बट दोएस थ्री कि फाइबर थ्री लेता गया छह और हमें क्या दिया वे प्लस बी प्लस सी इक्वल टू के 130 सीटें ही जीत पाएगी 130 - 6 इक्वल 21 हमारे पास 10 करना था हमने कर दिया है ए क्लासिक शेड्यूल से लेकर नहीं आया है कि मेन और एडवांस तिलक टर्मिनस ज्यादा प्रेम का भरोसा आज डू व्यवस्था कीजिए अपने डाउट 804 सुपर क |
3180 | https://www.teacher.org/lesson-plan/addition-and-subtraction/ | Representing Addition and Subtraction Lesson Plan | Teacher.org
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Home » Lesson Plans » » Representing Addition and Subtraction
Representing Addition and Subtraction
Brittany Zae
Teacher
This lesson is designed to help students represent addition and subtraction with objects, fingers, mental images, drawings, sounds (e.g., claps), acting out situations, verbal explanations, expressions, or equations.
Grade Level: K - 1st
Subject:
Length of Time: About 45 Minutes
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Common Core Alignment
CCSS: MATH.CONTENT.K.OA.A.1 - Represent addition and subtraction with objects, fingers, mental images, drawings 1, sounds (e.g., claps), acting out situations, verbal explanations, expressions, or equations.
Objectives & Outcomes
Students will be able to express addition and subtraction equations in multiple different ways, by adding and subtracting physical items through multiple activities, discussions, and observations.
Materials Needed
Animal cards (cards with images of animals on each)
Chart paper for KWL
Laptop/projector
Internet access to view
Procedure
Opening to Lesson
Watch video to review animals (stop at certain points to identify terms and check for understanding)
Then, play ‘flash review’ where you quickly flash ocean animal cards and vocabulary cards and have students respond with what it is as you put them on the board.
Actively listen as students answer guided questions relating to the video as well as tell you the correct name of the picture shown.
Also try describing things and see if they can think of the correct term.
Body of Lesson
Direct Teaching:
Directing students to the sorted cards on the board, ask students to count the sea animals, the maps, the blue animals, etc. As you count five animals, write the equation in numbers below. Explain to the students that you don’t just have to write this equation; you can show with your fingers or blocks or even just say it, etc.
Guided Practice 1
Next, review addition and subtraction using the students. Ask them to stand up, as you have each student select a card, then ask a select amount sit down. “If you’re a dolphin, sit down… how many are left?” Write the equation on the board and repeat. (Reverse the activity and write an equation and see if the students can show it. i.e. 5-2=3, ask what animal is there five of? Then have those animals stand up and if they can show the equation with their body)
Guided Practice 2
Students will then be split into two teams to play charades. The game will consist of a bowl of equations where they will have to act out their equation using nonverbal cues. The teams will alternate until all the children have been chosen and the equations have been answered. (Provide additional opportunities for practices as needed, and observe)
Independent Practice
Students will sit back at their desks and complete their worksheet of three equations from the game: i.e. 1. Draw: 1+1 = 2 using dolphins. 2. Write three – two = one, using numbers… etc. (Use proximity to walk around the room, provide time for reflection and assessment, and review final written assignment)
Closing
Reference the completed KWL chart
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Assessment & Evaluation
Through guided questions, close monitoring and informal observation; the teacher will be able to assess student’s ability and understanding of the subject being introduced. Through class discussions as well as eliciting responses when finalizing the KWL chart, the teacher will be able to evaluate students.
Modification & Differentiation
When students are counting from the mystery bag, challenge higher-level students by asking them to count on their own, or even specifically how many blue animals they have, etc. Accommodate for lower level students by monitoring closely to provide extra assistance as needed. Allow struggling students to verbalize what their worksheet may say, or show you the equation using their hands. A number of examples could also be shown on the board or given as a supplementary resource. Students could also be paired in groups in order to help student collaborate and understand the objectives better. Sporcle.com could also be available for any students that finish early.
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3181 | https://www.purplemath.com/modules/syseqgen5.htm | What are advanced systems of non-linear equations? | Purplemath
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Solving More-Advanced Systems of Non-Linear Equations
Intro ConceptsGraphing ConceptsSolving Simple SystemsSolving Intermediate SystemsSolving Advanced SystemsSolving w/ Quadratic Formula
Purplemath
The basic process for solving more complicated systems of non-linear equations remains the same as for the previous systems; namely, solve one of the equations for one of the variables, plug that information into the other equation, and solve the resulting one-variable equation.
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In "real life", naturally, you can expect advanced systems of non-linear equations to generally not be algebraically solvable. (Instead, you would have to use technology — a graphing calculator, a dedicated numerical package, etc.)
Thankfully, though, you can safely expect that the systems they give you in your algebra and calculus classes will be solvable. It just might take some time... and plenty of scratch-paper.
Solve the system of nonlinear equations:
y = x 2
x 2 + (y − 2)2 = 4
From the form of the equations, I (should) know that this system contains a parabola and a circle. Assuming that the two curves touch or cross, there might be one, two, three, or four intersection points. I'll have to do the algebra to find out.
According to the graph, I see that there should be three solutions to this system:
The solution at the origin is pretty clear (though I still need to prove that the curves do in fact touch at the origin), but the other two intersection points' coordinates may be messy.
From the first equation, I think I'll plug in "y" in place of "x 2" in the second equation, and solve:
x 2 + (y − 2)2 = 4
y + (y − 2)2 = 4
y + (y 2 − 4 y + 4) = 4
y 2 − 3 y = 0
y (y − 3) = 0
y = 0,y = 3
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Okay; I've got two values for y. Now I need to find the corresponding x-values. I'll find them by plugging the y-values back into the first equation.
When y = 0, the corresponding x-value is:
y = x 2
0 = x 2
0 = x
(This is the solution at the origin that I'd been expecting.)
When y = 3, the corresponding x-value is:
y=x 2 3=x 2±3∣=x\small{ \begin{aligned} y &= x^2 \ 3 &= x^2 \ \pm \sqrt{3\;\vphantom{|}} &= x \end{aligned} }y 3±3∣=x 2=x 2=x
Then my solutions are the following three points:
(−3∣,3),\small{ \boldsymbol{ \color{purple}{ \left(-\sqrt{3\;\vphantom{|}},\; 3\right), }}}(−3∣,3),
(0,0),\small{ \boldsymbol{ \color{purple}{ (0,\;0), }}}(0,0),
(3∣,3)\small{ \boldsymbol{ \color{purple}{ \left(\sqrt{3\;\vphantom{|}},\; 3\right) }}}(3∣,3)
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Numerical analysis
numerical
Solve the following system:
3 x 2 + 2 y 2 = 35
4 x 2 − 3 y 2 = 24
Yikes! Okay...
I can rearrange the first equation to get:
x 2(35 3)+y 2(35 2)=1\small{ \dfrac{x^2}{(\normalsize{\frac{35}{3}})} + \dfrac{y^2}{(\normalsize{\frac{35}{2}})} = 1 }(3 35)x 2+(2 35)y 2=1
This tells me that the first equation is an ellipse. However, rather than graphing this using ellipse formulas, I can also solve to get a "plus-minus" expression that I can graph as two equations:
y=±35−3 x 2 2\small{ y = \pm \sqrt{\dfrac{35 - 3x^2}{2}\;} }y=±2 35−3 x 2
(To graph this, I would plug this into my graphing calculator as two graphs, one graph for the "plus" part of the ellipse, and another for the "minus" part.)
The second equation rearranges as:
x 2 6−y 2 8=1\small{ \dfrac{x^2}{6} - \dfrac{y^2}{8} = 1 }6 x 2−8 y 2=1
...which is an hyperbola. The second equation also solves (for use in my graphing calculator) as:
y=±(4 3)x 2−8\small{ y = \pm \sqrt{\left(\normalsize{\frac{4}{3}}\right)x^2 - 8\;} }y=±(3 4)x 2−8
Whatever format I end up using (the ellipse and the hyperbola center-vertex forms, or the "plus-minus" for-calculator forms), this system graphs as:
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From what I can see, there appear to be four solutions to this system. To find these solutions algebraically, I will choose to solve the second equation for x 2 (rather than just x), and plug the resulting expression into the first equation, which I will then solve for y.
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(It's okay that I "only" solve for x 2, because neither equation has a linear x-term. There is no need, in this particular case, to do any more solving.)
4 x 2−3 y 2=24\small{ 4x^2 - 3y^2 = 24 }4 x 2−3 y 2=24
4 x 2=3 y 2+24\small{ 4x^2 = 3y^2 + 24 }4 x 2=3 y 2+24
x 2=(3 4)y 2+6\small{ x^2 = \left(\normalsize{\frac{3}{4}}\right)y^2 + 6 }x 2=(4 3)y 2+6
Then, subsituting into the first equation for the x 2, I get:
3 x 2+2 y 2=35\small{ 3x^2 + 2y^2 = 35 }3 x 2+2 y 2=35
3(3 4 y 2+6)+2 y 2=35\small{ 3\left(\normalsize{\frac{3}{4}}y^2 + 6\right) + 2y^2 = 35 }3(4 3y 2+6)+2 y 2=35
9 4 y 2+18+2 y 2=35\small{ \normalsize{\frac{9}{4}}y^2 + 18 + 2y^2 = 35 }4 9y 2+18+2 y 2=35
9 y 2+72+8 y 2=140\small{ 9y^2 + 72 + 8y^2 = 140 }9 y 2+72+8 y 2=140
17 y 2=68\small{ 17y^2 = 68 }17 y 2=68
y 2=4\small{ y^2 = 4 }y 2=4
y=±2\small{ y = \pm 2 }y=±2
So I've found two values for y. I'll plug each of these values into the second equation, and solve for the corresponding values of x.
When y = −2, the corresponding value of x is:
x 2=(3 4)y 2+6\small{ x^2 = \left(\normalsize{\frac{3}{4}}\right)y^2 + 6 }x 2=(4 3)y 2+6
=(3 4)(−2)2+6\small{ = \left(\normalsize{\frac{3}{4}}\right)(-2)^2 + 6 }=(4 3)(−2)2+6
=(3 4)(4)+6\small{ = \left(\normalsize{\frac{3}{4}}\right)(4) + 6 }=(4 3)(4)+6
=3+6=9\small{ = 3 + 6 = 9 }=3+6=9
x=±3\small{ x = \pm 3 }x=±3
Okay; so I got two values of x for when y=−2. I suspect I'll get two values for when y=+2, as well.
When y = 2, the corresponding value of x is:
x 2=(3 4)y 2+6\small{ x^2 = \left(\normalsize{\frac{3}{4}}\right)y^2 + 6 }x 2=(4 3)y 2+6
=(3 4)(+2)2+6\small{ = \left(\normalsize{\frac{3}{4}}\right)(+2)^2 + 6 }=(4 3)(+2)2+6
=(3 4)(4)+6\small{ = \left(\normalsize{\frac{3}{4}}\right)(4) + 6 }=(4 3)(4)+6
=3+6=9\small{ = 3 + 6 = 9 }=3+6=9
x=±3\small{ x = \pm 3 }x=±3
Then my solution consists of the following four points:
(−3, −2), (−3,2), (3,−2), (3,2)
Note: In this one specific sort of case, the points can also be written as (±3,± 2), since all of the "plus-minus" combinations are included. This is not, in general, true; you are not, in general, able to combine the coordinates like this.
To keep yourself on the safe side — that is, to avoid accidentally combining coordinates that don't actually go together — it's probably a good idea always to list out the various solution points, like I did at the end of the last exercise above.
Also, I could have solved this last system maybe quicker if I'd tried the addition method. If I'd multiplied the first equation by 3 and the second equation by 2, I'd have gotten:
1 9 x 2+6 y 2=105 1 8 x 2−6 y 2=1 48‾17 x 2+6 y 2=153\small{ \begin{array}{r} \phantom{1}9x^2 + 6y^2 = 105 \[0.5em] \underline{ \phantom{1}8x^2 - 6y^2 = \phantom{1}48} \[0.5em] 17x^2 \phantom{+ 6y^2} = 153 \end{array} }1 9 x 2+6 y 2=105 1 8 x 2−6 y 2=1 4817 x 2+6 y 2=153
Then x 2 = 9, so x=±3. Plugging these values back into one of the original equations would then return the corresponding y-values.
If you happen to notice that the addition method will work to get you started, then don't be afraid to use this other method. There is no rule that says that you absolutely must use substitution the whole way through.
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3182 | https://justinmath.com/solving-differential-equations-by-substitution/ | Solving Differential Equations by Substitution
by Justin Skycak (@justinskycak) on March 03, 2019
Non-separable differential equations can be sometimes converted into separable differential equations by way of substitution.
This post is part of the book Justin Math: Calculus. Suggested citation: Skycak, J. (2019). Solving Differential Equations by Substitution. In Justin Math: Calculus.
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Sometimes, non-separable differential equations can be converted into separable differential equations by way of substitution.
For example, $y’+y=x$ is a non-separable differential equation as-is. However, we can make a variable substitution $u=x-y$ to turn it into a separable differential equation. Differentiating both sides of $u=x-y$ with respect to $x$, and interpreting $y$ as a function of $x$, we have $u’=1-y’$, so $y’=1-u’$. Substituting, the equation becomes separable and thus solvable in terms of $u$.
Lastly, to find what $y$ is, we can solve for $y$ in our original substitution $u=x-y$.
Choosing the Right Substitution
In general, to determine what substitution we need to perform, it is helpful to rearrange the equation until we see a group of terms whose derivative also appears in the equation.
After rearranging the above equation, we see that $u=x^2+y^2$ is a good substitution. We rewrite the equation in terms of $u$, solve it, and then solve for $y$ in terms of $x$.
We don’t always have to use addition in our substitutions. In the equation below, for example, we require the substitution $u=xy$.
Exercises
Use substitution to solve the following differential equations. (You can view the solution by clicking on the problem.)
This post is part of the book Justin Math: Calculus. Suggested citation: Skycak, J. (2019). Solving Differential Equations by Substitution. In Justin Math: Calculus.
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Tags: Calculus, Differential Equations |
3183 | https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-2-new/ab-3-4/v/derivative-inverse-sine | Use of cookies
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3184 | https://home.ubalt.edu/ntsbarsh/business-stat/otherapplets/comcount.htm | Combinatorial Mathematics: How to Count Without Counting
: This site is a part of the JavaScript E-labs learning objects for decision making. Other JavaScript in this series are categorized under different areas of applications in the MENU section on this page.
Professor Hossein Arsham
The following is a collection of JavaScript for computing permutations and combinations counting with or without repetitions.
Many disciplines and sciences require the answer to the question: How Many? In finite probability theory we need to know how many outcomes there would be for a particular event, and we need to know the total number of outcomes in the sample space.
Combinatorics, also referred to as Combinatorial Mathematics, is the field of mathematics concerned with problems of selection, arrangement, and operation within a finite or discrete system. Its objective is: How to count without counting. Therefore, One of the basic problems of combinatorics is to determine the number of possible configurations of objects of a given type.
You may ask, why combinatorics? If a sample spaces contains a finite set of outcomes, determining the probability of an event often is a counting problem. But often the numbers are just too large to count in the 1, 2, 3, 4 ordinary ways.
A Fundamental Result: If an operation consists of two steps, of which the first can be done in n1ways and for each of these the second can be done in n2 ways, then the entire operation can be done in a total of n1× n2 ways.
This simple rule can be generalized as follow: If an operation consists of k steps, of which the first can be done in n1 ways and for each of these the second step can be done in n2 ways, for each of these the third step can be done in n3 ways and so forth, then the whole operation can be done in n1 × n2 × n3 × n4 ×.. × nk ways.
Numerical Example: A quality control inspector wishes to select one part for inspection from each of four different bins containing 4, 3, 5 and 4 parts respectively. The total number of ways that the parts can be selected is 4×3×5×4 or 240 ways.
Factorial Notation: the notation n! (read as, n factorial) means by definition the product:
n! = (n)(n-1)(n-2)(n-3)...(3)(2)(1).
Notice that by convention, 0! = 1, (i.e., 0! º 1) . For example, 6! = 6×5×4×3×2×1 = 720
Permutations versus Combination: A permutation is an arrangement of objects from a set of objects. That is, the objects are chosen from a particular set and listed in a particular order. A combination is a selection of objects from a set of objects, that is objects are chosen from a particular set and listed, but the order in which the objects are listed is immaterial.
The number of ways of lining up k objects at a time from n distinct objects is denoted by n P k, and by the preceding we have:
n P k = (n)(n-1)(n-2)(n-3)......(n-k+1)
Therfore, The number of permutations of n distinct objects taken k at a time can be written as:
n P k = n! / (n - k) !
Combinations: There are many problems in which we are interested in determining the number of ways in which k objects can be selected from n distinct objects without regard to the order in which they are selected. Such selections are called combinations or k-sets. It may help to think of combinations as a committee. The key here is without regard for order.
The number of combinations of k objects from a set with n objects is n C k. For example, the combinations of {1,2,3,4} taken k=2 at a time are {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, for a total of 6 = 4! / [(2!)(4-2) !] subsets.
The general formula is:
n C k = n! / [k! (n-k) !].
Permutation with Repetitions: How many different letter arrangements can be formed using the letters P E P P E R?
In general, there are multinomial coefficients:
n! / (n1! n2! n3! ... nr!)
different permutations of n objects, of which n1 are alike, n2, are alike, n3 are alike,..... nr are alike. Therefore, the answer is 6! /(3! 2! 1!) = 60 possible arrangements of the letters P E P P E R.
#### MENU:
1. Permutation of n objects in a group of size k
2. Permutation of n objects in a group of size k, repetitions allowe
3. Combination of n objects in a group of size k
4. Combination of n objects in a group of size k, repetitions allowed
Enter positive integer values for both n and k, and then click on the Calculate.
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| Permutation of n objects in a group of size k, k £ n |
| Permutation of n objects in a group of size k, repetitions allowed |
| Combination of n objects in a group of size k, k £ n |
| Combination of n objects in a group of size k, repetitions allowed |
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3185 | https://ntrs.nasa.gov/api/citations/19670020584/downloads/19670020584.pdf | TECHNICAL REPORT 66-1 I J. P. LaSALLE I APRIL, 1968 r ' AN INVARIANCE PRINCIPLE IN THE THEORY OF STABILITY ff 653 July 66 CENTER FOR DYNAMICAL SYSTEMS (CODE) AN INVARIANCE PRINCIPLE I N TKE THEORY O F STABILITY J. P. LaSalle ' Center f o r Dynamical Systems I Brown University 1 . Introduction. The purpose of t h i s paper i s t o give a unified presenta- tion of Liapunov's theory of stability that includes the classical Liapunov theorems on s t a b i l i t y and instability as w e l l as t h e i r more recent extensions. beginnings some time ago. It was, however, the use made of t h i s idea by Yoshizawa i n [ l ] i n his study of nonautonomous differential equations and by Hale i n i n h i s study of autonomous functional differential equations that caused the author t o return t o t h i s subject and t o adopt the general approach and point of view of t h i s paper. by ordinary differential equations which demonstrate the essential nature of a Liapunov function and which may be useful i n applications. Of greater importance, however, is the possibility, as already in- dicated by Hale's results for functional differential equations, The idea being exploited here had i t s This produces some new results for dynamical systems defined This research was supported i n part by the Nationalperonautics and Space Administration under- Grant No. NGR-40-002-015.hd under--Contract No, Ms8-11264,' i n part by the Uniced States A i r Force through the A i r Force Office of Scientific Research underAGrant No and i n part by the United States Army Research OfficejDurham, under Contract N o DA-31-124-ARO-D-270. -,- 1 A F - A F O S R - ~ ~ ~ ~ ~ ~ , / i 2 that these ideas can be extended t o more general classes of dynam- i c a l systems. It i s hoped, for instance, t h a t it may be possible t o do t h i s for some special types of dynamicalsystems defined by p a r t i a l differential equations. In section 2 we present some basic results for ordinary differential equations. theorem for nonautonomous systems and i s a modified version of Yoshizawa's Theorem 6 i n [l]. A simple example shows t h a t the conclusion of t h i s theorem i s the best possible. However, when- ever the l i m i t s e t s of solutions are known t o have an invariance property then sharper results can be obtained. This "invariance principle" explains the t i t l e of t h i s paper. It had its origin for autonomous and periodic systems i n - [SI, although w e present here improved versions of those results. lished an invariance property for almost periodic systems and ob- tains thereby a similar s t a b i l i t y theorem for almost periodic systems. Since l i t t l e attention has been paid t o theorems which make possible estimates of regions of attraction (regions of asymp- t o t i c stability) f o r nonautonomous systems results of t h i s type are included. Section 3 is devoted t o a brief discussion of some of Hale's recent results for autonomous functional differential equations. Theorem 1 is a fundamental s t a b i l i t y Miller i n has estab- 2. Ordinary differential equations. Consider the system 3 n+ 1 where x i s an n-vector, f i s a continuous function on R t o R and s a t i s f i e s any one of the conditions guaranteeing unique- 'ness of solutions. For each x i n Rn w e define 1x1 = (xl + ... + xn) d(x,E) = Min [ Ix-yl : y i n E). n 2 2 3 , and for E a closed s e t i n Rn w e define , Since we do not wish t o confine our- selves t o bounded solutions, we introduce the point at define d(x,m) = 1xl-l . Thus when we write E = E U[m), w e shall mean d(x,E) = Min{d(x,E), d(x,m)), I f x ( t ) i s a solution of (l), w e say that x ( t ) approaches E as t - + m if d(x(t),E) 4 9 as t --$a. If we can find such a s e t E, w e have obtained in- formation about the asymptotic behavior of x ( t ) as t + 0 0 . The best that w e could hope t o do i s t o find the smallest closed s e t m and R t h a t x ( t ) approaches as t -00. This set R i s called the positive limit s e t of x ( t ) and the points p i n R are called the positive limit points of x ( t ) . I n exactly the same way one -- defines x ( t ) + E as t --f -= , negative l i m i t sets, and negative limit points. This i s exactly G. D. Birkhoff's concept of l i m i t sets. A point p i s a positive l i m i t point of x ( t ) i f and only if there i s a sequence of times tn approaching 00 as n -+a and such that x(tn) + p as n 4 o . In the above it may be that the maximal interval of definition of x ( t ) i s [ 3 , ~ ) . This causes no difficulty since i n the results t o be presented here w e need only with respect t o time t replace mby T. W e usually ignore 4 t h i s possibility and speak as though our solutions are def'irred on [o,w) or (-m,m) . i n Let V(t,x) be a C fanction Oil [O,m) x R t o R, and l e t G be m y s e t i n Rn . W e s h a l l say that V i s a Liapunov function on G for equation ( 1 ) i f V(t,x) 2 0 ana. V(t,x) 5 -W(x) 5 0 for a l l t > 0 and a l l x i n G where W i s continuous on Rii t o R a d W e define (c i s the closure of G) - E = (x, W(x) = 0, x i n G ) . The following result i s then a modified but closely re- lated version of Yoshizawa's Theorem 6 i n [ l ] . THEOREM 1. If V i s a Liapunov function on G for equation ( l ) , then each solution x ( t ) of (1) t h a t remains i n G for a l l t > to 2 0 approaches E = E U (a] as t + c o , provided one of the following conditions i s satisfied: (i) For each p i n there i s a neighborhood 1\s of p such t h a t If(t,x)l i s bounded for a l l t > 0 and a i l x i n N. (ti) W i s C1 and i s bounded from above o r below along each solution which remains i n G f o r all t > t o 2 0 . 1 I t a I I I I I 1 I I I 1 c I I I I P ! 1 1 I I I I I If E i s bounded, then for t > t 2 0 either 0 each solution approaches E Thus t h i s theorem explains 5 of (1) that remains i n G or 03 as t + - . precisely the nature of the information given by a Liapunov function. relative t o a s e t G defines a s e t E which under the conditions of the theorem contains (locates) a l l the positive l i m i t sets of solutioils which f o r positive time remain i n applying the result i s t o find "good" Liapunov functions. instance, the zero function V = 0 i s a Liapunov function for tiie whole space Rn information since E = R . It i s t r i v i a l but useful f o r appli- cations t o note that if V, and V , are Liapunov functions on G, A Liapunov function G. The problem i n For aiid condition (ii) i s satisfied but gives no in- n then V = V + V2 1 If E i s smaller Liapunov function I L i s a l s o a Liapunov function and E = E ll E2 . 1 t h m either El or E2 , -then V i s a "better" than either E or E2 and i s always at l e a s t as 1 "good" as either of the two. Condition (5) of Theorem 1 i s essentially the one used by Yoshizawa. i s satisfied and condition (i) i s not. W e now look at a simple example where condition (ii) The example also shows t h a t the conclusion of the theorem is the best possible. Consider 2 + p(t)? + x = 0 where p ( t ) 2 6 > 0 . Define 2 V = x 2 2 + y , 2 2 where y = 2 . Then V = -p(t)y 6 - 6y and V i s a Liapunov 2 2 function on R2 N o w W = 6y and = 2 6 ~ = - 2 S ( x y + p(t)) ) -26xy. Since all solutions are evidently bounded for all t > 0, 6 condition (ii) is satisfied. Here E i s the x-axis (y = 0) and f o r each solution x ( t ) , y ( t ) = 2(t) + 5 as t + . Noting t h a t the equation x ( t ) = 1 t e-t , w e see t h a t t h i s i s the best possible r e s u l t with- out further restrictions on p . t x + (2 + e )S + x = 0 has a solution I n order t o use Theorem 1 there must be some means of determinLng which SolEtiOiis remain i n G . The following corollary, which i s an obvious consequence of Theorem 1, gives one way of doing this and also provides for nonautonomous systems a method f o r estimating regions of attraction. Corollary 1. Assume t h a t there exist continuous functions u(x) and v(x) on R t o R such t h a t U(X) 5 V(t,x) 5 V(X) for all t 2 0 . 4 - of Qrl containing G+ . If V i s a Liapunov f’Jnction on G for (1) and the conditions of Theorem 1 ( 1 ) starting i n G + at any time t 2 o remains i n G f o r all t > to and approaches E as t + w e If G i s bounded and Eo = r G C G+ , then Eo i s an attractor and G i s i n its region of attraction. n + + Define Q7 = {x ; U(X) < v} and l e t G be a component Let G denote the component of Q = {x ; v(x) < 71 1 1 are satisfied, then each solution of 0 + I n general w e know t h a t i f x ( t ) i s a solution of (l)--in fact, i f x ( t > is any continuous function on R t o R”-- then i t s positive limit s e t i s closed and connected. If x ( t ) i s bounded, then i t s positive limit s e t i s compact. There are, how- 7 1 I I I I I I I I 1 I I 1 ever, special classes of differential equations where the limit s e t s of solutions have an additional invariance property which makes possible a refinement o t ' Theore= 1. m e first of these are the autonomous systems a = f(x) (3) The limit sets of solutions of (3) are invariant sets. If x ( t ) i s defined on [O,m) and i f p i s a positive l i m i t point of x ( t ) , then the points on the solution through v a l of definition are positive l i m i t points of x ( t ) . If x ( t ) i s boEded for t > 0 , then it i s defined un [O,mj, i t s positive limit s e t R i s compact, noneinpty and solutions through points p of R are defined on ( -a , .
.
) (ioe., R i s invariant). If the maximal domain of definition of x ( t ) for t > 0 i s f i n i t e , then x ( t ) has no f i n i t e positive l i m i t points: t h a t is, i f the maxirrlal ijnierval of definilion of x ( t ) f o r L > 0 i s [O,p), then x ( t ) + a as t -+p . As w e have said before, w e w i l l always speak as though our solutions are defined on and it should be remembered t h a t f i n i t e escape time i s always a possibility unless there is, as f o r example i n Corollary 2 below, some condition t h a t rules it out. In Corollary 3 below, the solutions might well go t o infinitjr iii f i n i t e time. p on i t s maximal inter- (-m,m) The invariance property of the l i m i t sets of solutions of autonomous systems Let V be a C 1 function on R t o R . If G i s any arbitrary ( 3 ) now ena5les us t o refine Theorem 1 . n 8 set i n Rn , w e say t h a t V i s a Liapunov function on G for equation (3) if f l = (grad V ) . f does not change sign on G . Define E = ( x ; i ( x ) = 3 , x ir, G 1 , where G i s ihe closure of G . Let M be the largest invariant s e t i n E . M w i l l be a closed set. The fundamental s t a b i l i t y theorem for autonomous systems i s then the following: - - THEORFM 2. I f V i s a Liapunov function on G for ( 3 ) , then each solution x ( t ) of (3) that remains i n G for a l l t > 0 (t < 0) approaches M = M U (m) as t -+ m (t -m). If M i s bounded, then either x ( t ) - + M or x ( t ) + m as t - + m (t -m) . This one theorem contains a l l of the usual Liapunov l i k e theorems on s t a b i l i t y and instability of autonomous systems. Here however, there are no conditions of definiteness for V or V , and it i s often possible t o obtain stabilitjr information about a system w i t i i these more general types of Liapunov functions. The f i r s t corollary below i s a s t a b i l i t y result which for applications has been quite useful aid the second i l l u s t r a t e s how one obtains infomatton on instability. Cetaev's instability theorem i s similarly an immediate consequence of Theorem 2 (see section 3). v COROLLARY 2. Let G be a component of Q = ( x ; V ( x ) < 7 ) . Ass-me tliai G i s bon-ided, V 6 0 on G , and M = M n G c G . Then M i s an attractor and G i s i n i t s region of attraction. I f , i n addition, V i s constant on the boundary of Mo , then 7 0 - 0 1 1 1 I I I 1 I I I 1 I I 1 I 1 1 I I 9 Mo i s a stable attractor. Note that if Mo consists of a single point p , then p i s asymptotically stable and G provides an estimate of i t s region of asymptotic stability. C O R O L L A R Y 3. Assume that relative t o (3) that V > 0 on G and on the boundary of G that V = 0 . Then each solution of (3) starting i n G approaches m as t -+ m (or possibly i n f i n i t e time). There are also some special classes of nonautonomous systems where the l i m i t sets of solutions have an invariance property. The simplest of these are periodic systems (see [ 3 ] ) . 2 = f ( t , x ) , f ( t + T,x) = f ( t ) for a l l t and x . (4) Eere i n order t o avoid introducing the concept of a periodic approach of a solution of (4) t o a s e t and the concept of a periodic l i m i t point l e t us confine ourselves t o solutions of (4) which are bounded for t > 0 . Let R be the positive l i m i t s e t of such a solution x(t), and l e t p be a point i n R . Then there i s a solution of (4) starting a t p which remains i n R for a l l t i n (-w,oo) ; that is, i f one s t a r t s at p at the proper time the solution remains i n R for a l l time. Tnis i s the sense now i n which R i s an invariant set. Let V(t,x) be C1 on R x Rn and periodic i n t of period T . For an arbitrary s e t G of Rn we say that V i s a Liapunov function - on G for x ( t ) 1 0 for the geriodic system (4) if V does not change sign for a l l t and a l l x i n G . Define E = { (t,x); V(t,x) = 0, x i n 3 } and l e t M b e the union of a l l solutions x ( t ) of (4) with the property that ( t , x ( t ) ) i s i n E for a l l t . M could be called "the largest invariant s e t relative t o E". following version of Theorem 2 for periodic systems: One then obtains the THEOFEM 3. I f V i s a Liapunov function on G for the periodic system ( 4 ) , then each solution of (4) that i s bounded and remains In G for a l l t > 0 (t < 0) approaches M as t -+ 00 ( t + - 0 ~ ) . I n Miller showed that the limit sets of solutions of almost periodic systems have a similar invariance property and from t h i s he obtains a result quite l i k e Theorem 3 for almost periodic systems. systems a whole chain of theorems on s t a b i l i t y and i n s t a b i l i t y quite similar t o that for autonomous systems. For example, one has This then yields f o r periodic and almost periodic + COKOLLARY 4. l e t G be a component of Q . Let G be the compofient of Q = { x; V(t,x) < 1 1 f o r some t i n [O,T] } containing i s bounded, V 5 0 for a l l t and a l l x i n G , and i f M = M fl G C G+, then Mo i s an attractor and G i s i n its region of attractioc. If V(t,x) = q ( t ) for all t and a l l x on the boundary of Mo , then Mo i s a stable attractor. Let Qrl = { x; V(t,x) < 7, all t i n [O,T] } , and + + 1 1 G + . If G 0 7 + Our l a s t example of an invariance principle for ordinary 11 differential equations is that due to Yoshizawa in [ l ] for "asymp- totically autonomous" systems. It is a consequence of Theorem 1 and results by Markus and Opial (see [ l ] for references) on the limit sets of such systems. A system of the form is said to be asymptotically autonomous if ( i ) g ( t , x ) -+9 as t - + w uniformly for x in an arbitrary compact set of Rn , (ii) J I h ( t , q ( t ) ) l dt < w Yor all cp bounded and continuous on [O,w) to Rn . The combined results of Markus and Opial then W 0 state that the positive limit sets of solutions of (3) are in- variant sets of k = F ( x ) . Using this, Yoshizawa then improved Theorem 1 for asymptotically autonomous systems. It turns out to be useful, as we shall illustrate in a moment on the simplest possible example, in studying systems ( 1 ) which are not necessarily asymptotically autonomous to state the theorem in the following manner: THEOREM 4 . known that a solution x ( t ) o f ( 1 ) remains in G for t > 0 If, in addition to the conditions of Theorem 1 , it is and is also a solution of an asymptotically autonomous system (?), then x ( t ) approaches M = M U {OD) as t + w , where M is the largest invariant set of 2 = F ( x ) in E . It can happen that the system ( 1 ) is itself asymptotically autonomous in which case the above theorem can be applied. However, ..- A i d as the following example illustrates, the original system may not i t s e l f be asymptotically autonomous but it s t i l l may be possible t o construct for each solution of (1) an asymptotically autonomous system (7) which it also satisfies. Consider again the example ? = y (6) y = - x - P ( t ) Y , o < s ~ p ( t ) ~ m for all t > 0 N o w w e have the additional assumption t h a t i s bounded from above. Let (x(t), y ( t ) ) be any solution of (6). As was argued previously below Theorem 1, a l l solutions are bounded and y(t) + 0 as t + Q) . N o w (E(t), y(t)) satisfies k = y , y = p ( t ) - -x - p ( t ) y ( t ) , and t h i s system i s asymptotically autonomous t o (") E i s the x-axis and the largest invariant s e t of () i n E i s the origin. 2 = y , $ = -x . With the same Liapunov function as before, Thus for (6) the origin i s asymptotically stable i n the large 3. Autonomous functional differential equation. Difference differential equations of the form 2(t) = f ( t , x ( t ) , x ( t - r ) ) 9 r > O (7) have been studied almost as long as ordinary differential equations and these as well as other types of systems are of the general form 13 where x i s i r i Rn and x i s the function defined on [-r,O] by x ( 2 ) = x(t+.r), -r S 2 6 0. Thus xt i s the function that describes the past history of the system on the interval and i n order t o consider it as an element i n the space C of contizuous fmc-tions a l l defined on the same interval [ - r , O ] , x , i s taken t o be the function whose graph i s the translation of the graph of x on the interval [ t - r , t ] t o the interval [-r,O] . Since such equations have had a long history it seems surprising that; it i s only within the l a s t 1 0 years or so t h a t the geometric theory of ordinary differential equations has been successfully carried over t o functional differential equations. has demonstrated the effectiveness of a geometric approach i n ex- tending the classical Liapunov theory, including the converse theorems, t o functional differential equations. aspects of their theory which have yielded t o t h i s geometric approach can be found i n the paper [ g ] by Hale. t o present Hale's extension i n [ 2 ] of the results of Section 2 of t h i s paper t o autonomous functional differential equations t t [t-r,t] b Krasovskii An account of other What w e wish t o do here is K = f ( x ) . t (9) It i s t h i s extension t h a t has had so far the greatest success i n studying s t a b i l i t y properties of the solutions of systems (9), and it i s possible that t h i s may lead t o a similar theory for special classes of systems defined by p a r t i a l differential equations. With r 4 3 the space C i s the space of continuous 14 functions cp on [ - r , O ] t o Rn with l{cpll = m a x (Icp(~)l ; -r 5 T s 01. Convergence i n c i s uniform conver- gence on [ - r , O ] . A function x defined on [ - r , m ) t o Rn t o said t o be a solution of (9) satisfying - the i n i t i a l condition cp a t time t = 0 i f there i s an a > 0 such that %(t) = f ( for a l l t i n [O,a) and x = cp . Remember x = cp means X ( T ) = cp(z), -r 5 T 5 0. A t t = 0, k i s the right hand deriv- ative. The existence uniqueness theorems are quite similar t o those for ordinary differential equations. I f f i s locally Lipschitzian on C, then f o r each cp i n C there i s one and only one solution of (9) and the solution depends continuously on cp . The solution can also be extended i n C for t > 0 as long as it remains bounded. As i n Section 2, w e w i l l always speak as though solutions are defined on [ - r , m ) . The space C i s now the s t a t e space of (9) and through each point cp of C there i s the motion o r flow x starting a t cp defined by the solution x ( t ) of (9) satisfying a t time t = 0 the i n i t i a l condition c p ; x 0 5 t 0. One of the differences here i s that i n C closed and bounded sets are not always compact. Another i s that t although w e have uniqueness of solutions i n the future two motions starting from different i n i t i a l conditions can come together i n f i n i t e time t > 0; after this they coincide for t 2 t . (The motions define semi-groups and not necessarily groups. ) 0 0 Hale i n has, however, shown that the positive l i m i t s e t s R of bounded motions x are nonempty, compact, connected, invariant sets i n C . Invariance here i s i n the sense that, i f x i s a motion starting at a point of R, then there i s an exten- sion onto (-w,-r] such that x ( t ) i s a solution of (9) for a l l t t t i n (-m,w) and x remains i n R for a l l t . With t h i s result he i s then able t o obtain a result which i s similar t o t Corollary 1 of Section 2. For cp E C l e t x (cp) denote the motion defined by (9) t starting a t cp . For V a continuous function on C t o R define V and QR by 1 - +(cp) = l i m 7 rv(xT(cp))-v(cp)l. T - + and THEOREM 3. If V i s a Liapunov function on G for (9) and x i s a trajectory of (9) which remains i n G and i s bounded for t > 0, then x + M as t + 00 . t t 1 6 Hale has also given the following more useful version of t h i s result. COROLLARY 5 . Define Q = (~p; V(cp) < v] and l e t G be Q or a component of Q . Assume that V i s a Liapunov function on G for (9) and that either (i) G i s bounded or (iii) [ c p ( O ) [ i s bounded for cp i n G . Then each trajectory starting i n G approaches M as t + 00 . v v v The following i s an extension of Fetaevls instability theorem. i n , which should have stated "V(cp) > 0 on U when cp # 0 and V ( 0 ) = 0 ' ' and at the end 'I... intersect the boundary of C . . . ' I . This i s clear from h i s proof and i s necessary since he wanted t o generalize the usual statment of Cetaev's theorem t o in- clude the possibility that the equilibrium point be inside well as on i t s boundary. This i s a somewhat simplified version of Hale's Theorem 4 Y Y U as COROLLARY 6. Let p e C be an equilibrium point of (9) contained i n the closure of an open s e t U and l e t N be a neighborhood of p . Assume t h a t (i) V i s a Liapunov function on G = U f' N, (ii) M n G i s either the empty s e t or p, (5%:) V(cp) < 1 1 on G when cp # p9 and (iv) V(p) = 7 and V(cp) = 7 on t h a t part of the boundary of G inside N. Then p i s unstable. I n fact, if No i s a 5ourided neighborhood of p properly contained i n N then each trajectory starting at a point of Go = G fl No other than p leaves No i n f i n i t e time. I 1 I 1 1 I I I I I 1 1 1 I 17 Proof. trajectory starting inside either leave Conditions (i) and (iv) imply that it cannot reach o r approach that part of the boundary of Go inside No nor can it approach p as t + m . N o w (ii) states that there are no points of M on t h a t part of the boundary of No inside G . Hence each such trajectory must leave No i n f i n i t e time. Since p i s either i n the interior or on the boundary of contains such trajectories, and p is therefore unstable. B y the conditions of the corollary and Theorem 6 each - a t a point other than p must Go approach i t s boundary or approach p . GO, G, each neighborhood of p In it was shown t h a t the equilibrium point cp = 0 of R ( t ) = a x 3 (t) + bx 3 (t-r) was unstable if a > 0 and Ibl < I a1 . Using the same Liapunov function and Theorem 6 w e can show a b i t more. With 4a -r 4 t 4a t-r V(X,) = - - (t) + $ .
I xb(e)ae and which i s nonpositive when (negative definite with re- spect t o cp(0) and cp(-r)); that is, V i s 8 Liapimov function c,n C and E = (cp; cp(0) = cp(-r) = O } . Therefore M i s simply the n u l l function cp = 0 . I f ' a > 0, the region G = { c p ; V(cp) < 0) I bl < 1 a 1 1 8 i s nonempty, and no trajectory starting i n G can have cp = 0 as a positive l i m i t point nor can it leave G . Hence by Theorem 5 each trajectory starting i n G must be unbounded. Since cp = 3 i s a boundary point of G, it is unsk;i'!)l.?. It i s also easily seen t h a t if a < 0 and lbl < 1 ai, then cp = 0 i s asymptotically stable i n the large. I n Hale has also extended t h i s theory f o r systems with i n f i n i t e lag of significant examples of the applications of t h i s theory. (r = m), and i n that same paper gives a number I I 1 1 1 I I I I 1 1 1 1 I I 1 1 1 I I I I I I I I I I I I I I 1 1 I I 1 I Ret'erences [l] Yoshizawa, T., Asymptotic behavior of solutions of a system of differential equations, Contrib. t o Diff. Eq., 1(1963), 371-387 Hale, J., Sufficient conditions for stability and instability of autonomous functional differential equations, J. of Diff. Eq., 1(1963), 432-482. LaSalle, J., Some extensions of Liapunov's second method, IRE Trans. on Circuit Theory, CT-7 ( 1 $ 0 ) , 320-527. LaSalle, J., Asymptotic stability criteria, Proc. of Symposia i n Applied Mathematics, Vol. 13, Hydrodynamic Instability, Amer. Math. SOC., Providence, R.I., 1962, 299-307. LaSslle, J., and Lefschetz, So, Stability by Liapunov's Direct Method with Applications, Academic Press, N e w York, 1961. Miller, R,, O n almost periodic differential equations, B u l . Amer. Math. SOC o , 70 (l964), 792-793 Miller, R., Asymptotic behavior of nonlinear delay-differential equatLms, J, of D i f f . Eq., 3(1965), 293-305. Krasovskii, N.N., Stability of Motion, S t a n Y m i University Press, 1963 (translation of 1939 Russian Edition). Hale, J., "Geometric theory of functional differential equations", Proc. of An International Symposium on Differential Equations and Dynamical Systems, University of Puerto Rico,Mayaguez, P. R., Dec. 1963, Academic Press, N e w York ( t o appear). |
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Tools & Reference>Nephrology
Azotemia Workup
Updated: Mar 09, 2023
Author: Moro O Salifu, MD, MPH, MBA, MACP; Chief Editor: Vecihi Batuman, MD, FASN more...;)
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Sections
Azotemia
Sections Azotemia
Overview
Background
Pathophysiology
Etiology
Epidemiology
Prognosis
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Presentation
History
Physical Examination
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DDx
Workup
Laboratory Studies
Ultrasonography
Computed Tomography and Magnetic Resonance Imaging
Abdominal Radiography, Pyelography, and Angiography
Radionuclide Studies
Kidney Biopsy
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Treatment
Pharmacologic and Supportive Therapy
Surgical Relief of Obstruction
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Medication
Medication Summary
Diuretics, Other
Volume Expanders
Corticosteroids
Alpha/Beta Adrenergic Agonists
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Media Gallery;)
References;)
Workup
Laboratory Studies
For the initial evaluation, obtain a complete blood count (CBC), a biochemical profile, urinalysis, and urine electrolyte concentrations. In addition to establishing the presence of systemic disease, these tests may reveal clues to the origin of the azotemia. Diagnostic indices are commonly used to differentiate prerenal azotemia from intrarenal or postrenal azotemia (see in the image below).
Diagnostic indices in azotemia. Although such indices are helpful, it is not necessary to perform all these tests on every patient. Comparison should always be made with patients' baseline values to identify trends consistent with increase or decrease in effective circulating volume. Use of some of these indices may be limited in certain clinical conditions, such as anemia (hematocrit), hypocalcemia (serum calcium), decreased muscle mass (serum creatinine), liver disease (blood urea nitrogen [BUN], total protein, and albumin), poor nutritional state (BUN, total protein, and albumin), and use of diuretics (urine sodium). Fractional excretion of urea and fractional excretion of trace lithium appear to be superior for assessing prerenal status in patients on diuretics.
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Prerenal azotemia
In prerenal azotemia, hemoconcentration results in elevation of the hematocrit and total protein/albumin, calcium, bicarbonate, and uric acid levels from baseline values. Urinary findings include the following:
Oliguria (urine volume < 500 mL/day) or anuria (< 100 mL/day)
High specific gravity (> 1.015)
Normal sediment
Low sodium (< 20 mEq/L; fractional excretion of sodium [FENa] < 1%)
When volume depletion is predominant, exaggerated proximal tubular reabsorption results in azotemia, hypernatremia, and elevated levels of calcium, uric acid, and bicarbonate, whereas hemoconcentration results in elevation of total protein, albumin, and hematocrit levels from baselines. Patients with hypoperfusion due to decreased cardiac output or effective arterial volume is present, patients exhibit edema, hyponatremia, and hypoalbuminemia. Hematocrit and calcium, uric acid, and bicarbonate levels vary widely. These patients often are critically ill.
The FENa has traditionally been used to differentiate prerenal azotemia from ATN. An FENa below 1% suggests a prerenal cause (eg, volume depletion), whereas an FENa above 2% suggests acute tubular necrosis (ATN). Because the FENa is based on the fact that sodium reabsorption is enhanced in the setting of volume depletion, active use of diuretics may elevate the FENa even when volume depletion is present, yielding misleading values.
Alternatives to the FENa in this setting include the fractional excretion of urea or urea nitrogen (FEUrea) and the fractional excretion of uric acid (FEUA); excretion of urea and uric acid excretion is not influenced by diuretics. An FEUrea below 35% or an FEUA below 9-10 % suggests a prerenal etiology of acute kidney injury (AKI), whereas an FEUrea above 50% or an FEUA above 10-12 % suggests acute tubular necrosis (ATN).
Intrarenal azotemia
On blood studies, findings that may suggest intrarenal azotemia include the following:
Anemia
Thrombocytopenia
Hypocalcemia
High–anion gap metabolic acidosis
Plasma BUN–creatinine ratio < 20
On urine studies, findings that may suggest intrarenal azotemia include the following:
Low specific gravity (< 1.015)
Active sediment (see Pathophysiology)
High sodium (> 40 mEq/L; FENa > 5%)
Low osmolality
In patients with long-standing chronic kidney disease (CKD), renal ultrasonography usually shows small, contracted kidneys. However, normal-sized or large kidneys may be seen in CKD from some causes, such as HIV nephropathy, diabetes, and renal amyloidosis. The renal sonogram usually is diagnostic for patients with polycystic kidney disease. In patients with active urinary sediment, progressive azotemia, proteinuria, or normal-sized kidneys on ultrasonography, a kidney biopsy should be considered. Consultation with a nephrologist is imperative in all such patients.
Postrenal azotemia
Urinary indices in postrenal azotemia due to complete bilateral obstruction are usually nondiagnostic. The prima facie finding here is anuria, occasionally accompanied by hypertension. Urine output still may be present if overflow (in bladder outlet obstruction) or partial ureteral obstruction is present.
A Foley catheter should be inserted as part of the initial evaluation to rule out obstruction below the bladder outlet. Unilateral ureteral obstruction rarely leads to azotemia; it occurs acutely (as a result of obstruction from calculi, papillary necrosis, or hematoma), producing renal colic, or may be chronic and asymptomatic, producing hydronephrosis.
Bilateral partial obstruction may be associated with azotemia in the presence of normal urine output. When patients are subjected to maneuvers that increase urinary flow (eg, diuretic renography or perfusion pressure flow studies), they may exhibit an increase in size or pressure of the collecting system or experience pain.
In addition to azotemia, polyuria due to loss of concentrating ability and type 1 renal tubular acidosis, with hyperkalemia, hypercalcemia from a metastatic pelvic tumor, and elevated prostate-specific antigen (PSA) levels, may be clues to postrenal azotemia. Hydronephrosis in the absence of hydroureter may be seen in early (< 3 days) obstruction, retroperitoneal process, or partial obstruction.
Renal ultrasonography (see below) is the test of choice for ruling out obstructive uropathy. If the renal sonogram is equivocal, a furosemide (Lasix) washout scan (see Radionuclide Studies) should be performed.
Next: Ultrasonography
Ultrasonography
Renal ultrasonography is the most commonly used renal imaging study because of its ease of use and broad applicability for the following purposes :
Determination of kidney size and echogenicity, which is important when considering kidney biopsy; small echogenic kidneys (< 9 cm) may suggest scarring from advanced renal disease, whereas normal or large kidneys with smooth contours may indicate a potentially reversible process
Differentiation of cystic lesions from solid lesions
Diagnosis of urinary tract obstruction (for which it is the test of choice)
Detection of kidney stones
Doppler renal ultrasonography can be used to evaluate renal vascular flow (eg, for identification of renal vein thrombosis, renal infarction, or renal artery stenosis).
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Next: Ultrasonography
Computed Tomography and Magnetic Resonance Imaging
Computed tomography (CT) is complementary to ultrasonography, especially when the diagnosis is uncertain. Contrast nephrotoxicity should be weighed against the benefits. CT is used for the following purposes:
Differentiation of neoplastic lesions from simple cysts (in most cases)
Radiologic diagnosis of renal stone disease, including radiolucent stones
Evaluation and staging of renal cell carcinoma
Diagnosis of renal vein thrombosis
Diagnosis of polycystic kidney disease; it is more sensitive than ultrasonography for this task, particularly in younger patients
Knipp et al describe successful use of a technique for computed tomographic angiography (CTA) of the abdomen and pelvis in azotemic patients that uses a reduced iodinated contrast volume and low kilovolt (peak) [80-kV(p)] with iterative reconstruction. Their retrospective study in 103 patients with end-stage renal disease found that this technique allows for satisfactory abdominal/pelvic CTA with a 50% reduction in contrast volume and a 43% mean radiation dose reduction, compared with a standard 120-kV(p) CTA protocol.
Magnetic resonance imaging (MRI) or magnetic resonance angiography (MRA) is used only when CT and ultrasonography are nondiagnostic. These modalities are standard for diagnosis of renal vein thrombosis and are also used in the evaluation of renal cell carcinoma and renal artery stenosis or vasculitis.
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Abdominal Radiography, Pyelography, and Angiography
If symptoms suggest nephrolithiasis, a plain film of the abdomen is performed to screen for presence of a radiopaque stone. Calcium-containing, struvite, and cystine stones can be identified, but radiolucent ones, such as uric acid stones, will be missed.
Intravenous pyelography (IVP) can provide detailed information concerning calyceal anatomy and the size and shape of the kidney. It is extremely useful for detecting renal stones. IVP is the preferred technique for evaluation and diagnosis of certain structural disorders (eg, chronic pyelonephritis, medullary sponge kidney, and papillary necrosis). It can provide data on the degree of obstruction. The risk of contrast nephrotoxicity should be weighed against the benefits of making a diagnosis that will not change management.
Retrograde or anterograde pyelography is of limited usefulness now that renal ultrasonography is more widely available. It may be used in patients with a high index of suspicion for hydronephrosis in whom sonograms appear normal, such as those with retroperitoneal fibrosis.
Renal arteriography is used in polyarteritis nodosa and renal artery stenosis to demonstrate multiple aneurysms or stenoses. Because of the availability of procedures that do not require contrast material (eg, ultrasonography, MRI, and MRA) and thus do not carry a risk of contrast nephrotoxicity, this test is less commonly used than it once was.
Renal venography is the standard for diagnosis of renal vein thrombosis. However, it poses a risk of contrast nephrotoxicity.
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Next: Ultrasonography
Radionuclide Studies
Technetium-99m dimercaptosuccinic acid (99mTc DMSA) is heavily distributed within the renal parenchyma at first pass and so is best for detecting renal parenchymal scarring.
Technetium diethylenetriamine pentaacetic acid (99mTc DTPA) is heavily filtered at first pass and therefore is best for qualitative assessment of kidney function (filtration and excretion). Because it is heavily filtered, it is most sensitive in detecting urine leaks after kidney transplantation. For the same reason, it is also used concomitantly with a furosemide washout scan (see below) for assessing functional obstruction of the collecting system.
Mercaptoacetyltriglycine (MAG3) is evenly distributed at first pass in the kidney and so is best for qualitative assessment of perfusion, filtration, and excretion. It is the preferred test for assessing those 3 aspects of function after kidney transplantation. It can be used with furosemide to detect urine leaks or functional obstruction, though 99mTc DTPA scanning remains the test of choice for these conditions. Voiding cystourethrography can be performed with a radionuclide study to detect vesicoureteral reflux.
In a furosemide washout scan, the renal scan usually is performed first. Then, if needed, the furosemide washout is done after the radionuclide has accumulated in the collecting system. Furosemide is used as a part of the renogram to separate nonobstructive hydronephrosis from obstructive hydronephrosis. If there is no obstruction, furosemide-induced flow containing little or no radionuclide will fill the collecting system, washing out radionuclide-containing urine. If obstruction is present, the radionuclide is not washed out as quickly.
The half-life or clearance of the radioisotope is plotted on a curve. A half-life shorter than 10 minutes is considered normal, one longer than 20 minutes is considered obstruction, and one of 10 to 20 minutes is subject to further interpretation.
Conditions that can make it difficult to interpret the furosemide washout curve include a megaureter or pelvis that accepts a large bolus of urine and poor kidney function. In patients with a megaureter, it can be difficult to determine when the renal pelvis is full, and in patients with kidney disease, the onset of furosemide action may be delayed. To overcome the problem of poor kidney function or relative hypovolemia if a patient has been fasting, the patient should be well hydrated with intravenous (IV) fluids before the study.
The test also is operator dependent, in that the furosemide should be administered at a time when the renal pelvis is believed to be full. A full bladder also delays washout of isotope. Therefore, the patient’s bladder must be catheterized before the study can be performed.
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Kidney Biopsy
When glomerulonephritis, vasculitis, and (occasionally) interstitial nephritis are suspected, kidney biopsy is indicated to establish the correct diagnosis and guide therapy. The following are common indications for kidney biopsy:
Isolated glomerular proteinuria or hematuria
Nephrotic syndrome
Acute nephritic syndrome
Unexplained acute or subacute kidney injury
Percutaneous kidney biopsy is associated with potential complications. Severe bleeding causing hypotension occurs in 1-2% of patients. Bleeding necessitating transfusion occurs in about 0.1-0.3% of patients. Bleeding complications can be minimized by performing pre-procedure coagulation studies: bleeding time, prothrombin time (PT), activated partial thromboplastin time (aPTT), and platelet count.
Nonsteroidal anti-inflammatory drugs (NSAIDs) should be stopped at least 1 week before a scheduled elective biopsy. Patients on warfarin should be started on heparin at least 3 days before kidney biopsy. Patients who are taking heparin for other reasons should stop the drug for at least 1 day.
Contraindications for percutaneous kidney biopsy include the following:
Uncorrectable bleeding diathesis
Small kidneys
Severe hypertension
Multiple bilateral cysts or renal tumor
Hydronephrosis
Active renal or perirenal infection
Uncooperative patient
Percutaneous biopsy may be performed in selected patients with a solitary kidney because of the generally low risk of bleeding. Open biopsy may be performed if a percutaneous attempt is either unsuccessful or contraindicated and if the benefits of diagnosis outweigh the risks. When percutaneous biopsy is contraindicated but a diagnosis is necessary, a transvenous transjugular renal core biopsy can be performed. With this approach, bleeding occurs intravascularly, thereby reducing the risk of hematoma.
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Treatment & Management
References
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Fenske W, Stork S, Koschker AC, et al. Value of fractional uric acid excretion in differential diagnosis of hyponatremic patients on diuretics. J Clin Endocrinol Metab. 2008 Aug. 93(8):2991-7. [QxMD MEDLINE Link].
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Graph shows relation of glomerular filtration rate (GFR) to steady-state serum creatinine and blood urea nitrogen (BUN) levels. In early renal disease, substantial decline in GFR may lead to only slight elevation in serum creatinine. Elevation in serum creatinine is apparent only when GFR falls to about 70 mL/min.
Diagnostic indices in azotemia. Although such indices are helpful, it is not necessary to perform all these tests on every patient. Comparison should always be made with patients' baseline values to identify trends consistent with increase or decrease in effective circulating volume. Use of some of these indices may be limited in certain clinical conditions, such as anemia (hematocrit), hypocalcemia (serum calcium), decreased muscle mass (serum creatinine), liver disease (blood urea nitrogen [BUN], total protein, and albumin), poor nutritional state (BUN, total protein, and albumin), and use of diuretics (urine sodium). Fractional excretion of urea and fractional excretion of trace lithium appear to be superior for assessing prerenal status in patients on diuretics.
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Moro O Salifu, MD, MPH, MBA, MACP Associate Professor, Department of Internal Medicine, Chief, Division of Nephrology, Director of Nephrology Fellowship Program and Transplant Nephrology, State University of New York Downstate Medical Center
Moro O Salifu, MD, MPH, MBA, MACP is a member of the following medical societies: American College of Physicians-American Society of Internal Medicine, American Medical Association, American Society for Artificial Internal Organs, American Society of Diagnostic and Interventional Nephrology, American Society of Nephrology, American Society of Transplantation, National Kidney Foundation
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Subodh J Saggi, MD, MPH, FACP, FASN Professor of Medicine (with Tenure), Fellowship Program Director (Nephrology and Nephrology-Critical Care), Medical Director, Pancreas Transplant Program, State University of New York Downstate College of Medicine
Subodh J Saggi, MD, MPH, FACP, FASN is a member of the following medical societies: American College of Physicians, American Society of Nephrology, American Society of Transplantation
Disclosure: Received research grant from: Vertex Pharmaceuticals; Aurinia Pharmaceuticals, Hi-Bio Pharmaceuticals.I did not receive any direct research payments from the entities mentioned below, but our Research Foundation at SUNY Downstate Health Sciences University did and all research funds given for research went to research coordinator (other personal support), for their salary support and for operationalizing the research., IRB payments, etc. Such funds were accounted for direct conduct of research and approved by Ethics Department at Albany, SUNY.
Sonalika Agarwal, MBBS Attending Physician, Department of Internal Medicine (Nephrology/Transplant Nephrology), Kings County Hospital
Sonalika Agarwal, MBBS is a member of the following medical societies: American College of Physicians, American Medical Association, American Society of Nephrology, American Society of Transplantation, Delhi Nephrology Society, National Kidney Foundation
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Vecihi Batuman, MD, FASN Professor of Medicine, Section of Nephrology-Hypertension, Deming Department of Medicine, Tulane University School of Medicine
Vecihi Batuman, MD, FASN is a member of the following medical societies: American College of Physicians, American Society of Hypertension, American Society of Nephrology, Southern Society for Clinical Investigation
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Onyekachi Ifudu, MD, MBBS Director of Inpatient Dialysis Services, Associate Professor, Department of Internal Medicine, State University of New York Health Science Center at Brooklyn
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George R Aronoff, MD Director, Professor, Departments of Internal Medicine and Pharmacology, Section of Nephrology, Kidney Disease Program, University of Louisville School of Medicine
George R Aronoff, MD is a member of the following medical societies: American Federation for Medical Research, American Society of Nephrology, Kentucky Medical Association, and National Kidney Foundation
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Patterns of genomic differentiation between two Lake Victoria cichlid species, Haplochromis pyrrhocephalus and H. sp. ‘macula’
Shohei Takuno1na1,
Ryutaro Miyagi2,3,
Jun-ichi Onami4,
Shiho Takahashi-Kariyazono1,
Akie Sato5,
Herbert Tichy6,
Masato Nikaido7,
Mitsuto Aibara2,
Shinji Mizoiri2,
Hillary D. J. Mrosso8,
Semvua I. Mzighani2,8,
Norihiro Okada2,9,10 &
…
Yohey Terai
ORCID: orcid.org/0000-0003-3353-34201,2na1
BMC Evolutionary Biology
volume 19, Article number: 68 (2019)
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Abstract
Background
The molecular basis of the incipient stage of speciation is still poorly understood. Cichlid fish species in Lake Victoria are a prime example of recent speciation events and a suitable system to study the adaptation and reproductive isolation of species.
Results
Here, we report the pattern of genomic differentiation between two Lake Victoria cichlid species collected in sympatry, Haplochromis pyrrhocephalus and H. sp. ‘macula,’ based on the pooled genome sequences of 20 individuals of each species. Despite their ecological differences, population genomics analyses demonstrate that the two species are very close to a single panmictic population due to extensive gene flow. However, we identified 21 highly differentiated short genomic regions with fixed nucleotide differences. At least 15 of these regions contained genes with predicted roles in adaptation and reproductive isolation, such as visual adaptation, circadian clock, developmental processes, adaptation to hypoxia, and sexual selection. The nonsynonymous fixed differences in one of these genes, LWS, were reported as substitutions causing shift in absorption spectra of LWS pigments. Fixed differences were found in the promoter regions of four other differentially expressed genes, indicating that these substitutions may alter gene expression levels.
Conclusions
These diverged short genomic regions may have contributed to the differentiation of two ecologically different species. Moreover, the origins of adaptive variants within the differentiated regions predate the geological formation of Lake Victoria; thus Lake Victoria cichlid species diversified via selection on standing genetic variation.
Background
The molecular basis of the incipient stage of speciation is of great interest in genetics, ecology, and evolutionary biology . Cichlid species in Lake Victoria are a suitable model system to study this stage of speciation . Lake Victoria harbors more than 500 endemic cichlid species [3, 4]. They are thought to have experienced an explosive adaptive radiation during a very short evolutionary period because Lake Victoria dried up at the end of the Pleistocene and was refilled only 15,000 years ago [5, 6]. Indeed, levels of genetic differentiation among species are low, and the species share a large number of nucleotide polymorphisms including differentiated variants shown by studies using a small number of genetic markers, restriction site associated DNA (RAD) data, and whole genome sequencing data [7,8,9,10,11,12,13,14]. Nevertheless, fixed genetic differences between species are expected at loci responsible for adaptive traits and, as a consequence, for speciation. One of the best examples is the long wavelength-sensitive opsin gene (LWS), which exhibits a high level of genetic differentiation with fixed genetic differences among Lake Victoria cichlid species [15,16,17,18,19]. As expected, LWS alleles are variable among species adapted to different light environments created by different turbidities and different depths and are responsible for speciation by sensory drive [15, 17]. Such variation of LWS alleles among species would have originated from the admixture of two divergent lineages . The other gene with fixed genetic differences is the rod opsin gene (RH1) for scotopic vision. RH1 alleles are differentiated among species from different turbidities and depths, and adapted to their ambient light environments .
The joint effect of gene flow and divergent selection shapes the pattern of genomic differentiation between an incipient species pair. At the very beginning of this stage, a small part of genes could be involved in reproductive isolation and/or local adaptation . In the latter case, divergent selection acts on these genes, where one allele is advantageous in a species and the other allele is advantageous in its counterpart because separated populations adapt to different niches/environments. Gene flow actually occurs around the target sites of divergent selection but offspring of migrants with the non-adaptive allele are immediately selected out from the species, and as a consequence, the effective migration rate is decreased. On the other hand, gene flow is allowed in other genomic regions and suppresses differentiation, leading to the heterogeneity of genetic differentiation [21,22,23,24].
Indeed, the lines of empirical evidence of speciation with gene flow have been recently increased, in plants , insects [26, 27], and cichlid species [7, 8, 28, 29]. Despite gene flow, highly differentiated genomic regions between species exist in the genome, and these regions bear genes related to local adaptation such as pigmentation and visual perception in crows , beak shape in Darwin’s finches , and RH1 for scotopic vision in cichlids . Also, fixed nucleotide differences have been observed in such short genomic regions that emerge only when divergent selection effectively acts [28, 30,31,32,33,34].
Recently, this pattern of genomic differentiation has been reported between two Lake Victoria species, Pundamillia nyererei and P. pundamillia that both live in rocky habitats [7, 8]. On the other hand, many Lake Victoria cichlid species are distributed along the bottom where there is a soft sandy–muddy substrate, and the pattern of genomic differentiation between species from a sandy–muddy bottom has not been analyzed. The differentiated genomic regions between closely related species from a sandy–muddy bottom may contain candidate genes related to adaptation to micro-habitats, and these candidate genes provide an opportunity to study the incipient evolutionary process in a soft-bottom, benthic ecosystem.
In this study, we focused on two cichlid species from Mwaburugu to reveal the pattern of genomic differentiation between closely related species living in a sandy–muddy habitat. Mwaburugu is a consistently shallow area (2–3 m) with a sandy–muddy bottom, located in the eastern region of Speke Gulf (Fig. 1a). The two species exhibit morphological and behavioral differences (Fig. 1a). H. pyrrhocephalus inhabits the middle layer (mainly 7–13 m in Mwanza gulf, south part of Lake Victoria) , and H. sp. ‘macula’ is a demersal species [36, 37]. In Mwaburugu, however, the two species distribute in sympatry at a 1- to 3-m depth (Fig. 1a). H. sp. ‘macula’ is a phytoplankton eater, while H. pyrrhocephalus is a zooplanktivore . Males of these two species exhibit different nuptial colorations , but the distribution of the hue index values largely overlap and are different from other species in Mwaburugu . In cichlids, color perception is important for mate choice [17, 39,40,41]. Therefore, we expected that the reproductive isolation of these two species may be incomplete, and as a consequence, the genomic differentiation between them may be low. Indeed, we found that most of their genome did not show significant differentiation between the two species.
Results
Summary of population genetic statistics
We performed population genomic analyses in H. pyrrhocephalus and H. sp. ‘macula’ from the same locality in Lake Victoria (Fig. 1a). We sampled 20 individuals (10 males and 10 females) for each species, extracted DNA, and constructed Pool-seq libraries (Methods). After mapping the paired-end short reads to the reference genome sequence, the final coverage was ~ 34× in both species. We extracted 4,967,102 and 5,109,870 sites with 80–200× coverage for H. pyrrhocephalus and H. sp. ‘macula,’ respectively (Additional file 2: Text S1). We first inferred the site frequency spectrum (SFS) for each species . We estimated 329,613 (6.64%) and 322,956 (6.32%) polymorphic sites in H. pyrrhocephalus and H. sp. ‘macula,’ respectively. The folded SFSs were very similar between the two species (Additional file 1: Figure S1A), and almost 70% of segregating sites were inferred to be singletons in each species. Nucleotide diversity was 0.00717 and 0.00678 for the two species. Tajima’s D values were highly negative (less than − 2 in both species), indicating an excess of rare alleles in each species. This observation suggests population expansion accordingly, and we inferred demographic parameters based on the SFSs using δa/δi (Additional file 2: Text S1). As expected, we detected rapid population expansion after the colonization of Lake Victoria; the current effective population size was on the order of 106 (Additional file 2: Text S1).
Note that the power and accuracy of a Pool-seq analysis would be low for allele frequency estimation, especially for alleles at lower frequency . Nevertheless, nucleotide diversity and Tajima’s D values are consistent with previous estimates based on the Sanger method [12, 48]. We further note that we utilized sites with a much higher coverage than the genomic average, and high coverage can indicate repetitive regions. After excluding sites within repetitive regions, we obtained nearly the same statistics for each species (Additional file 2: Text S1).
Two Lake Victoria cichlid species would be close to a single panmictic population
We measured the level of population differentiation between the two species. We estimated the allele frequency at every site given the estimated rate of sequence errors (Materials and Methods, Additional file 2: Text S1). We used 7,516,409 polymorphic sites, at which the coverage was ≥40 for both species. When the minor allele frequency in the total dataset was higher than 0.05, > 80% of sites exhibited shared polymorphisms, as observed in Lake Victoria cichlids [7,8,9,10,11,12,13,14, 48], suggesting that the species are closely related. We applied FSTstatistics , which is suitable to quantitatively assess the level of differentiation between such closely related species or populations. This statistic is calculated by.
(1)
where TW and TB represent the coalescent time of samples from the same population and that of samples from different populations, respectively; μis the mutation rate per generation, and πW and πB represent the average nucleotide diversity within each population and the average pairwise nucleotide divergence between the two populations, respectively. To measure the level of population differentiation, TW was used as a control. When two sets of samples are from the same population, TW and TB are expected to be equal and FST is close to 0.
We calculated FST for each of the 7,516,409 segregating sites. Because the accuracy of allele frequency estimates is low when coverage is low , FST is expected to become relatively high at such sites. As expected, we found a slight, but detectable, negative correlation between FST and coverage (blue plot in Fig. 1b). Nevertheless, average FST values were around zero even in alleles with low coverage (< 0.0043; Fig. 1b), indicating almost no population differentiation.
We examined whether we could treat the two species as a single panmictic population. In theory, when the migration rate is high enough (i.e., when the population migration rate, 4 Nm, is higher than 10, where N is the effective population size and m is the migration rate per gamete per generation), the pattern of nucleotide polymorphisms in samples from two populations is expected to be similar to that in samples from a single population . To test this, a standard coalescent simulation is not sufficient because we used Pool-seq data. Thus, to generate null distributions of FST values given our coverage, we simulated both the coalescent process under panmixia and the Pool-seq process with the inferred population expansion (Additional file 2: Text S1). The distributions of simulated FST values exhibited a similar tendency (orange plot in Fig. 1b), though the observed distributions were slightly skewed toward negative values. When a single segregating site was analyzed, FST statistics became negative when allele frequencies in the two sample sets were very similar. This indicates a slight excess of the proportion of segregating sites with very similar allele frequencies between the two species. The cause of this excess is not known, but the observed and expected distributions were largely overlapping. Thus, we reasoned that the two species might be close to a single panmictic population (see also Fig. 2a).
We have three caveats in this section. First, as in the previous section, we excluded sites in repetitive regions, and we obtained essentially the same results as in Fig. 1b (Additional file 2: Text S1). The second caveat is that a Pool-seq analysis is less powerful for detecting population differentiation, especially when minor allele frequency is low . However, in our case we could detect population differentiation when FST was ≥0.01, and when the sample size was set to be 40 in each population (Additional file 2: Text S1). Furthermore, we excluded singletons and obtained essentially the same results as in Fig. 1b. Finally, the females of H. sp. ‘macula’ are morphologically similar to other Haplochromis or Enterochromis species, and it is possible that these female specimens included misidentification of species. To avoid the misclassification of species, we performed the same analysis only in male individuals. Despite the reduction in sample size, we obtained essentially same result (Additional file 3: Figure S2).
Identification of highly differentiated genomic regions
Despite minimal genomic differentiation, H. pyrrhocephalus and H. sp. ‘macula’ exhibit differences in morphology and habitat . Thus, we performed a genome scan to search for candidate genes that are subject to divergent selection using sites with 20–200× coverage in both species (Additional file 2: Text S1). We used 10-kb window size with 2-kb increments for the scan, and calculated πW, πB, and FST for each window. The spatial distributions of πW and πB are almost the same across the genome, indicating that TW and TB are almost the same and FST values are close to zero (see two typical genomic regions in Fig. 2a) due to the panmixia of the two species (Fig. 1b). On the other hand, the LWS gene, a prime example of target genes under divergent selection [15, 17, 18], exhibited a remarkable pattern. The spatial patterns of πW, πB, and FST around LWS are shown in Fig. 2b. We observed that πB values are significantly higher than πW, and as a consequence, there is a clear peak of FST in the 14-kb region that includes the LWS gene. Despite the low accuracy of Pool-seq, these statistics are very close to the estimates reported in a previous study, which determined the sequences by the Sanger method . Outside the 14-kb region, πW and πB values are almost equal as in Fig. 2a. Furthermore, we found seven fixed nucleotide differences between H. pyrrhocephalus and H. sp. ‘macula’ within the peak (green triangles in Fig. 2b) that indicate strong signatures of divergent selection under the pressure of extensive migration.
To screen candidate genes under divergent selection, we initially filtered windows with the top 0.1% of FST values (FST > 0.372). We performed a neutrality test to see if such high FST values are observed without natural selection. We simulated a coalescent process and Pool-seq process with 20× coverage to maximize the variance of a null distribution (Additional file 2: Text S1), and obtained a very small P-value (P < 10− 5; false discovery rate < 0.014). We further screened such windows that exhibited a clear peak of FST values as in LWS genes (Fig. 2b) and fixed nucleotide differences within the peaks. In total, we detected 21 highly differentiated regions (14–28 kb). Hereafter, we focused on these 21 short differentiated regions (DRs), and three examples of DRs are shown in Fig. 2c-e and the others in Additional file 4: Figure S3. We searched for genes in the DRs by BLASTN and explored their biological roles in terms of adaptation and speciation. Nineteen out of 21 DRs included 1–3 genes (Table 1), and 28 total genes were found in the DRs.
As mentioned above, allele frequency estimates in regions of low coverage are not accurate, especially for alleles with low frequencies . Thus, we repeated the analysis, discarding segregating sites with minor allele frequency ≤ 0.05, and identified the same set of 21 regions. Furthermore, we verified the fixed nucleotide differences by Sanger sequencing for 5 of 21 loci (we selected a subset of 5 DRs due to limited amounts of DNA samples) (Additional file 1: Figure S1C). To avoid the misclassification of species, we repeated the analysis using only male samples as in the previous section. We obtained essentially the same result (Additional file 3: Figure S2B) as in the LWS genes (Fig. 2b) and in other DRs.
The pattern of polymorphisms around the target site of divergent selection
The hitchhiking effect of divergent selection under the pressure of migration is more limited than that of positive selection (i.e., selective sweep). To show this, we performed a population genetic simulation under a simplified model (Methods). We assumed an isolation with migration model with two populations (populations 1 and 2), into which we incorporated mutation, recombination, migration, divergent selection, and random genetic drift. The locus I is subject to divergent selection, and has two alleles, A and a. The A allele is advantageous in population 1, and the a allele is advantageous in population 2. At the start of simulations, locus I is monomorphic for the A allele, and a linked neutral locus (locus II) is under mutation-drift equilibrium. At the time T = 0, we introduced the a allele in population 2 with the initial frequency of 1/2 N and ran simulations for 8 N generations. The beneficial alleles (allele A in population 1 and allele a in population 2) are expected to immediately reach an equilibrium frequency, , when population size is infinite:
(2)
where m is migration rate per generation, and s is the selection coefficient. As long as s is much higher than m (e.g., s/m > > 5), is very close to 1. We ran simulations 100,000 times each with a variety of pairs of m and s, and calculated expected πW, πB, and FST at several time points. We confirmed that the frequencies of the beneficial alleles in both populations quickly reached in finite populations, and the patterns of polymorphisms were qualitatively consistent among the pairs of m and s.
We show the result with 4 Nm = 50, and 4Ns = 400 in greater detail in Fig. 3a. First, the a allele quickly reaches in population 2 which causes a strong reduction of πW in long genomic regions such as in the case of a selective sweep event [51, 52]. However, the signature of selective sweep is quickly eliminated by the effects of migration and recombination, and a short differentiated region appears with fixed nucleotide differences as shown in Fig. 2b-e and Additional file 3: Figure S3. This situation is very similar to the process of neofunctionalization of duplicated genes (an analog to divergent selection in our model) under the pressure of interlocus gene conversion (analogs to migration and recombination), where a short differentiated peak between duplicates appears around the target site of neofunctionalization . We further simulated the situation in which the a allele is derived from standing genetic variation because it has been proposed that adaptive variants are derived from standing variation (; see the section “Lake Victoria cichlid species diversified via selection on standing genetic variation” below). We ran simulations with s = 0 until the frequency of the a allele reached 20%, and then divergent selection started to act. The result is shown in Fig. 3b, and we found that the shrinkage of the differentiated region is faster than that in Fig. 3a.
A high false positive rate for the FST-outlier approach has been pointed out previously, and using absolute nucleotide divergence (or πB) is recommended [54, 55]. FST can be increased without divergent selection when πW is reduced as in Eq. (1), for example, by a classical selective sweep or by background selection. When one focuses on highly differentiated species, we may not be able to ignore this flaw of the FST-outlier approach (i.e., when TB is much older than TW; FST > 0.1 as in ). However, a high false positive rate is expected when using πB in low-differentiated, young species pairs with extensive migration such as Haplochromis species (Figs. 1b and 2). This is because high πB regions occasionally appear due to the large variance of coalescent time in the ancestral population. If such regions have evolved in a neutral manner, πW values are expected to be equal to πB due to migration, and FST is close to zero. Thus, employing FST should perform better in our case.
In our DRs, the possibility of a classical selective sweep without divergent selection is unlikely. As shown in Fig. 3, we observe the sweep-like strong reduction of πW that actually increases FST. However, without divergent selection, such reduction of πW is quickly eliminated. Both divergent selection and migration are required to maintain short differentiated regions as observed in Fig. 2b-e and Additional file 3: Figure S3. Background selection is also unlikely. The effect of background selection is remarkable on regions with low recombination rates. Purifying selection purges deleterious mutations, and also linked neutral variants together. As a consequence, πW is slightly reduced in relatively long genomic regions , and FST values become high. However, the lengths of DRs are very short (14–28 kb; Fig. 2b–e; Additional file 3: Figure S3). Furthermore, the inferred population sizes in Haplochromis species are fairly large (Additional file 2: Text S1), and therefore the population recombination rate (4Nr, where N is effective population size and r is recombination rate per generation) is so high that recombination effectively reduces the effect of background selection. More importantly, in both cases, we would not expect to observe fixed nucleotide differences under the pressure of migration without divergent selection, however, we did (Fig. 2b-e, Additional file 3: Figure S3).
The roles of fixed differences in DRs
Among the genes in DRs (Table 1), six nonsynonymous fixed differences were located in the coding region of LWS (Additional file 1: Figure S1C). These nonsynonymous substitutions make 8 and 9 nm shifts in the absorption spectra of LWS photo-pigments, with 11-cis retinal (A1-) and 11-cis 3- dehydroretinal (A2-derived retinal), respectively , suggesting that the fixed differences are responsible for the functional difference of LWS pigments between two species.
In contrast, four fixed differences in another opsin, melanopsin A, were not located in the coding region, but are in the upstream region of the gene (Fig. 4a), raising the possibility of differential expression. We quantified the expression level of this gene in the eyes of six individuals each from H. pyrrhocephalus and H. sp. ‘macula’ by real-time qPCR. We detected a significant difference in expression between species (Fig. 4a), indicating the possibility that these substitutions may cause the expression difference of melanopsin A in these two species.
The color pattern of cichlids is one of the most variable traits among species in Lake Victoria; therefore, we performed a pooled RNA-seq analysis to screen for the candidates of differentially expressed genes in the anterior part of the lateral skin of these two species. In total, 172 contigs showed differential expression between the two species (P < 0.05). These included three genes in DRs, P450 aromatase, UDP-N-acetylglucosamine transporter, and aloxe3. The fixed differences in these genes were also located in their upstream regions (Fig. 4b–d). Therefore, we quantified the expression levels of these genes in the anterior part of the lateral skin of ten individuals each from both species. In all three genes, the expression levels in the anterior part of the lateral skin were completely different between the two species (Fig. 4b–d, P < 0.001). These results also indicate the possibility of differential expression of genes caused by the fixed differences.
Lake Victoria cichlid species diversified via selection on standing genetic variation
We inferred the origin of the putative adaptive variants (i.e., fixed differences) in cichlid species. We determined the sequences (~ 1 kbp), including the fixed differences, within 16 DRs from Lakes Victoria, Malawi, Tanganyika, and riverine haprochromis species. We constructed phylogenetic trees based on our sequences and orthologous sequences in other cichlid genomes . No tree for any DR showed monophyly of Lake Victoria species (Fig. 5a and Additional file 5: Figure S4A), suggesting that these alleles in DRs arose as presumably neutral variants in the ancestral population of the two species. By contrast, five trees showed monophyly of the Lake Victoria superflock, including species from Lake Victoria and surrounding rivers [9, 16] (Additional file 5: Figure S4F), suggesting that the adaptive variants arose in the common ancestor of this monophyletic clade (riverine origin: Fig. 5b and Additional file 5: Figure S4B). The origin of variants in these five DRs are coincident with the scenario that two divergent lineages admixed in the ancestor of the Lake Victoria superflock . Astatotilapia burtoni and species from Lake Victoria, surrounding rivers, Lake Malawi and tribe Tropheini (Lake Tanganyika) were mixed in a monophyletic clade in the other 11 trees (Additional file 4: Figure S4E), suggesting that the mutations accumulated in the common ancestor of this lineage [Fig. 5c and Additional file 4: Figure S4C, modern haplochromines origin ]. Hence, the origin of the adaptive variants can be traced back to some point in the ancestral lineage of modern haplochromines.
Discussion
In this study, we focused on H. pyrrhocephalus and H. sp. ‘macula,’ which are distributed in sympatry in Mwaburugu (Fig. 1a) where the bottom is a sandy–muddy substrate with no rocks, woody debris, or other structures.
In the genome scan, we detected 21 DRs (14–28 kb) including 28 genes. LWS, located in DR5 (Fig. 2b), is involved in adaptation to different light environments and contributes to speciation by sensory drive [15, 17], suggesting that DRs contain genes responsible for cichlid adaptation and reproductive isolation. DR2 and DR12 contain the vision-related genes ventral anterior homeobox 2 (vax2) and melanopsin A, respectively (Fig. 2c and d; Table 1). The vax2 gene is involved in the regulation of retinal development and melanopsin A is involved in the photic regulation of the circadian clock . H. pyrrhocephalus and other pelagic zooplanktivorous species (e.g. H. laparogramma, H. heusinkveldi) migrate toward the surface in the evening and stay during the night to forage for zooplankton . H. pyrrhocephalus has very large double cone photoreceptor cells in their retinas for high sensitivity to light . The genes involved in retinal development and circadian clock regulation may contribute to the specific features of H. pyrrhocephalus. Although we did not analyze other pelagic zooplanktivorous species, DR2 and DR12 might be shared among ecologically similar species with H. pyrrhocephalus. Other DRs contained several genes involved in developmental processes, such as the development of the brain (DR6), epidermis (DRs 1, 19, and 20), and fin muscles (DR9) (Table 1, references in Additional file 6: Table S1). The importance of brain activity for sociality and reproduction was reported in cichlids . Both the epidermis structure and fin muscles may be related to species-specific characteristics; the epidermis directly interacts with the external environment and fin muscles affect fish mobility. P450 aromatase (DR11) is responsible for estrogen synthesis and plays a regulatory role in sex determination, gametogenesis, central nervous system development, and reproductive behavior , which are important traits for sexual selection. Interestingly, Atlantic cod populations are also differentiated at this gene . Chronic hypoxia has been observed in Lake Victoria . Cichlid species adapt to hypoxia by multiple strategies . Aryl hydrocarbon receptor nuclear translocator gene (DR16) is involved in physiological adaptation to hypoxia and the G-protein coupled receptor 4 (DR3) regulates breathing by CO2 stimulation . Lake Victoria cichlid species may experience hypoxia in deep water, heavily vegetated shallow shoreline habitats, and dense algal blooming in open water . Since H. pyrrhocephalus, except in Mwaburugu, inhabits deeper water (inferred to be hypoxic) than H. sp. ‘macula’, these genes may be involved in adaptation to different oxygen concentrations. Intestinal mucin (DR8) functions as a host-specific determinant affecting the gut microbiota composition, which is important for digestion . Intestinal mucin may be involved in the digestion of species-specific food such as phytoplankton and zooplankton in H. sp. ‘macula’ and H. pyrrhocephalus, respectively. Mex3a (DR21) is associated with brain aging in the short-lived fish Nothobranchius furzeri, a model of aging studies , and may be related to interspecific differences in aging in Lake Victoria cichlids. In total, we detected genes with predicted roles in adaptation and speciation within at least 15 out of 21 DRs (Table 1), suggesting that the DRs contain genes responsible for adaptation and reproductive isolation. Two DRs (DR10 and DR13) did not contain genes, but might contain regulatory regions of genes (e.g., DR10 was located 30 kb upstream of tbx3).
How are the observed fixed differences related to functional differences in genes in DRs? The nonsynonymous fixed differences in LWS cause 7-nm shifts in the absorption spectra of LWS pigments . The fixed differences in four genes (melanopsin A, P450 aromatase, UDP-N-acetylglucosamine transporter, and aloxe3) were located in the upstream regions of the genes and are associated with significant differences in expression between the two species. P450 aromatase expression in the anterior part of lateral skin might partly explain the difference in regulation of color pattern formation between the two species (Fig. 1a); this gene plays a regulatory role in various sexual traits . The different expression levels of these four genes suggest that the fixed differences affect gene expression levels (Fig. 4). Hence, the fixed differences observed in DRs are expected to affect protein functions or gene expression. These results support the hypothesis that the divergence of Lake Victoria cichlid species is explained by differentiation in short genomic regions containing genes responsible for adaptation. Each of the short genomic regions may be responsible for the adaptive traits, and the combination of these traits including LWS adaptation may lead to reproductive isolation between ecologically different species.
It has been argued that standing genetic variation is important for the recent radiation of cichlid lineages based on genome-wide shared polymorphisms among different cichlid lineages [13, 14, 69], many of which are likely selectively neutral. Recently, Meier et al. reported that the ancient admixture event between distantly related lineages (Congolese and Upper Nile) would increase genetic variation that would have contributed to the morphological diversity and adaptive traits in the Lake Victoria region superflock , suggesting the importance of standing genetic variation. However, the origins of adaptive variants are not fully elucidated. We examined whether or not the origin of adaptive variants in our DRs are the same age as the two divergent lineages in Meier et al. 2017 by tracing back to the origins of the putative adaptive variants (i.e., fixed differences in DRs) in cichlid species. The phylogenetic trees based on the five DR sequences were in agreement with this prediction. Furthermore, the origins of the other 11 DRs were older than the common ancestor of the Lake Victoria region superflock. These estimations suggest that the adaptive variants in DRs can be traced back to some point in the ancestral lineage of modern haplochromines .
In this study, we demonstrate extensive gene flow between ecologically differentiated species, 21 DRs, and an ancient origin of the adaptive variants responsible for species divergence. These findings provide new insight into a long-standing question: why do cichlids represent the most successful radiation in Lake Victoria during a very short evolutionary period? The ancient origin of adaptive variants within DRs provides an important clue. These mutations tended to accumulate after the split of the modern haplochromine lineage, supporting the idea that the radiation of Lake Victoria species occurred via selection on standing genetic variation . It may also be explained by extensive gene flow, which enables species to share functional sequences that may have promoted adaptation to various environmental conditions within Lake Victoria. Indeed, the same LWS allele that is adaptive to local light environments has been observed in multiple species [15, 18, 19]. In this study, we focused on two species in Lake Victoria. Additional genomic sequences of multiple individuals from additional species pairs with extensive gene flow will allow us to paint a more comprehensive picture of the role of gene flow in Lake Victoria cichlid radiation.
Methods
Sample information
Two Lake Victoria cichlid species were used, i.e., Haplochromis pyrrhocephalus Witte and Witte-Maas (1987) and H. sp. ‘macula.’ These species are widely distributed in Lake Victoria [3, 71] and inhabit Mwaburugu at the east end of Speke Gulf (Fig. 1a). All specimens were collected in sympatry by netting (1.5-m height) at a 1- to 3-m depth in Mwaburugu. All fish were collected by M.A. and S.M. in 2004–2006. The identification of all specimens was verified by M.A. and S.M.
Species identification: Haplochromis pyrrhocephalus is one of the most common pelagic-sublittoral species in the eastern Speke gulf. The species is recognized to be in the Yssichromis group because of the slender body (body depth 27.5–31.1% of standard length in the original description and 26.2–30.3% in our measurement, see material and methods in ). This species is distinguished from all other Haplochromines by a combination of the slender body and male nuptial coloration: 1) orange to red coloration on the head, unpaired fins, and egg dummies, and 2) absence of a lateral band. We collected 14 species from Mwaburugu (Fig. 1a), southeastern Speke gulf, where no slender-bodied species were found except this species. Therefore, we identified slender-bodied females within the range described above as H. pyrrhocephalus.
Haplochromis sp. 'macula' was described by the Haplochromis Ecology Survey Team (HEST) in Leiden University and subsequently re-described by Seehausen with male nuptial coloration. Male of the species is relatively easy to identify because of the bright red coloration on the head, anterior body, and dorsal fin membrane, and yellow to green coloration on the posterior body and caudal peduncle. Among all species that we collected in Mwaburugu, H. sp. 'macula' is morphologically different from the other species by the combination of the following traits: 1) dorsal head profile is straight or weakly moderately curved (vs. moderately curved in the other species); 2) oral teeth in outer jaw are weakly compressed (vs. cylindrical to weakly compressed); 3) flange of main cusp of oral teeth in outer jaw are relatively prominent (vs. without or weak flange); and 4) arrangement of anterior teeth in outer jaw is relatively dense, with posterior end of the tooth slightly overlapping to the anterior end of the neighboring tooth (vs. not overlapping). In this study, we chose females which possessed all of these characteristics as H. sp. 'macula'.
Additional genetic information: all specimens used in the present study were subjected to the species identification procedure described above prior to the analysis of opsin genes . Among all species that we collected in Mwaburugu, one LWS gene allele Py and H was exclusively fixed in H. pyrrhocephalus and H. sp. 'macula', respectively. In particular, the Py allele was only found in H. pyrrhocephalus. Thus, these two species possess a genetic biomarker specific to species. Note, however, that we did not identify these species by using genetic information.
Pooled genomic DNA sequencing (Pool-seq) and mapping
Genomic DNAs were extracted from caudal or pectoral fins of wild-caught individuals using the DNeasy Blood & Tissue Kit (Qiagen, Hilden, Germany). All tissues were dissected and kept in 100% ethanol until use. Equal amounts of DNAs (500 ng) extracted from 10 males and 10 females each from H. pyrrhocephalus and H. sp. ‘macula’ were pooled. In each species, DNAs from males and females were pooled separately. Libraries were constructed using the TruSeq DNA LT Sample Prep Kit (Illumina, San Diego, CA, USA) and the sequences were determined (paired-end 100 bp) using the Illumina HiSeq2000 platform. The paired-end short reads were mapped to the reference genome sequence of Pundamilia nyererei using Bowtie 2 specifying “--score-min L,0,-0.2.”
The number of high-quality alleles was counted at every site using Samtools with the arguments “-C50 -q20 -Q30” [ver. 0.1.19-44,428 cd; ]. If an indel was called, the site was filtered out, including the regions 9 bp upstream and downstream of the site. Sites with ≥3 nucleotides were removed. Sites with coverage of ≥20× were retained. Finally, if PV4 information was available, a site was filtered out if the P-value for strand bias or tail distance bias was less than 10− 4 or if the P-value for baseQ bias was less than 10− 100 according to the default settings in VCFtools .
Population genomic analyses
Pool-seq data for each species were initially analyzed separately. Mono- and bi-allelic sites with coverage of 80–200× were screened for each species to calculate the site frequency spectrum (SFS) by applying the EM algorithm developed by Boitard et al. (Additional file 2: Text S1). The demographic history of each species was inferred from the SFS using δa/δi (Additional file 2: Text S1).
Population differentiation between the two cichlid species was analyzed. Sites with coverage of 20–200× were extracted for both species, and the allele frequency was estimated at every site by applying Eq. (1) in :
(3)
where Yi is the number of derived alleles at the ith genomic position in n sampled chromosomes, Zi is the observed reads at the ith position with coverage ri, Zi,j (1 ≤ j ≤ ri) is an indicator variable equal to 1 if the jth read has the derived mutation, and 0 otherwise, and ε is the error rate of sequencing (Additional file 2: Text S1). Because we did not know the ancestral state of alleles, we used Pf(Zi|Yi) = 1/2P(Zi|Yi) + 1/2P(1-Zi|Yi) to fold the SFS. We estimated the SFS as the probability is maximized. We also used the Eq. (3) to estimate an allele frequency at every site (Additional file 2: Text S1). A genome scan was performed to identify highly differentiated genomic regions. We performed a sliding window analysis in 10-kb windows with 5-kb increments after discarding windows, in which < 50% of sites were covered by ≥20× reads. We calculated the FST value in each window and screened windows with the top 0.1% of FST values. A neutrality test was performed using the ms software . A coalescent simulation of 80 chromosomes was performed given the inferred population expansion (Additional file 2: Text S1). The length of the simulated region was set to 10 kb, which was the same as the window size used for the genome scan. The 80 chromosomes were randomly divided into subsamples of 40 to simulate a panmictic population. Then, Pool-seq data were simulated, the allele frequency was estimated for each site, and FST was calculated. These processes was repeated 100,000 times. To maximize the variance of the null distribution of FST, we assumed no recombination and set coverage to 20× for both species.
The effect of divergent selection under the pressure of migration
We assumed an isolation with migration model with two populations (populations 1 and 2) with population size, N, into which we incorporated mutation, recombination, migration, divergent selection, and random genetic drift. We consider a two-locus biallelic model. The locus I has A and a alleles and is the target of divergent selection, where the fitness values of A and a in population 1 are 1 and 1–s, and those in population 2 are 1–s and 1, and s is the selection coefficient. We assume that divergent selection acts in an additive manner. The locus II has B and b alleles with no phenotypic effect. Symmetric mutation occurs only in locus II at rate μ per generation to measure the hitchhiking effect of divergent selection. Recombination between the two loci is incorporated at rate r per generation. Symmetric migration occurs at the rate m per gamete per generation between the populations. Let the frequencies of A-B, A-b, a-B, and a-b in population 1 be x1, x2, x3, and x4, respectively. As such, let those in population 2 be y1, y2, y3, and y4, respectively. The expectations of these frequencies in the next generation can be given by the following recursion equations:
(3a)
(3b)
(3c)
(3d)
(3e)
(3f)
(3g)
(3i)
where Dx = x1 x4 – x2 x3 and Dy = y1 y4 – y2 y3.
We simulated the pattern of polymorphisms around the target site of divergent selection. We fixed N = 1000, and 4Nμ = 0.01. We used a wide range of 4Nr to be 0.1~200. We assumed that the locus I is monomorphic for A in both populations, and performed a pre-run until the DNA polymorphism in locus II reached a mutation-drift equilibrium using the Eq. (3a, b, c, d, e, f, g, h, i). At time T = 0, we introduced a in population 2 with the initial frequency, 1/2 N, and ran the simulation for 8 N generations. We ran simulations for 100,000 cycles each with the variety of pairs of m and s, and calculated expected πW, πB, and FST in locus II at several time points.
RNA-seq and assembly
To screen the candidates of differentially expressed genes between the two species, a pooled RNA-seq analysis was performed. Total RNAs were extracted from the anterior part of the lateral skins of five males each from H. pyrrhocephalus and H. sp. ‘macula.’ Equal amounts of total RNAs (1 μg) were pooled, libraries were constructed using the TruSeq RNA Library Preparation Kit (Illumina), and the sequences were determined (paired-end 100 bp) using the Illumina HiSeq2000 platform. De novo assembly of paired-end short reads (7.7 Gbp) of H. sp. ‘macula’ was performed using the CLC genomic workbench ( with automatic word size. The short reads from both species (H. pyrrhocephalus, 6.3 Gbp; H. sp. ‘macula,’ 7.7 Gbp) were mapped to the assembled sequences (50,240 contigs) and the expression levels of sequences were compared between species using the CLC genomic workbench. The sequences with different expression between species (t-test with Bonferroni correction, P < 0.05) were differentially expressed candidate genes. In total, 50,240 contigs were tested and 172 (0.3%) showed differential expression. For gene identification, the differentially expressed contig sequences were subjected to a BLASTN search against the NCBI non-redundant nucleotide sequences database ( The differentially expressed contigs that were found in DRs were selected as candidate genes for differential expression between the two species.
Real-time qPCR
The expression of the candidate genes for differential expression screened by pooled RNA-seq were further analyzed by real-time qPCR (qPCR) between laboratory-reared individuals of H. sp. ‘macula’ and H. pyrrhocephalus. Fishes were 9–12 months old and were kept at 25 °C under commercial fluorescent lights with a 12 h light-dark cycle. To sample eye tissues, six individuals each from both species were euthanized under anesthesia using ethyl 4-aminobenzoate at 10 h after the light was turned on, and right eyes were enucleated. The eyes were immediately placed on ice in RNAlater (Ambion, Austin, TX, USA) and the cornea and lens were removed. The remaining eye samples were stored in fresh RNAlater at ˗80 °C until further use. To sample skin tissues, 10 individuals (five males and five females) each from both species were euthanized as described above at 5 h after the light was turned on. The euthanized fishes were immediately placed on ice in RNAlater and subsequent dissection was performed in this solution. A square area of the anterior part of the lateral skin was dissected. After muscle attached to the dissected skin was removed, the skin was cut into 2–5 mm2 pieces. The skin pieces were stored in fresh RNAlater at ˗80 °C until further use.
Total RNA was extracted from the eye and skin samples using TRIzol RNA Isolation Reagent (Thermo Fisher Scientific, Waltham, MA, USA) according to the manufacturer’s instructions and quantified using a NanoDrop 2000c spectrophotometer (Thermo Fisher Scientific). First-strand cDNA was reverse-transcribed from 500 ng of the eye total RNA or 1 μg of the skin total RNA using a PrimeScript RT Reagent Kit with gDNA Eraser (TaKaRa). The eye cDNA samples were diluted 25-fold in PCR-grade water for the amplification of melanopsin A. The skin cDNA samples were diluted 1.5- or 20-fold in PCR-grade water for the amplification of aloxe3 and UDP-N-acetylglucosamine transporter or for P450, respectively. Target genes and an internal control gene (GAPDH) were amplified from the cDNA samples in a 25-μl total volume of PCR solution containing 12.5 ml of SYBR Premix Ex Taq II (TaKaRa), 3 ml of the diluted cDNA samples, and 10 pmol each of the following forward and reverse primers: melanopsin A: 5′ − TGGAGCTTTCATCGATGGCTACAAC− 3′ and 5′ − GATGCCTACAGCAAGGATGACAACAC− 3′; GAPDH: 5′ − GCCCACGCAAACATCATTC− 3′ and 5′ − GTCAGATCCACCACTGACACATC− 3′; aloxe3: 5′ − GAAGCTGCAAGGTGACAGGACTATTG− 3′ and 5′ − TGAGATGGTCAAGTTCGTCACCATG− 3′; P450: 5′ − GAGAAATCTGAACGCAGACTGCAAAC− 3′ and 5′ − GGACAGCAGTGACTTCTGATGCTCTATC− 3′; UDP-N-acetylglucosamine transporter: 5′ − AGCGAGGACAGGACCATCAAGAG− 3′ and 5′ − GAGACACGTATTTTAGCCTGGAGGAAAG− 3′. PCRs were performed using the Thermal Cycler Dice Real Time System II (TaKaRa) with the following conditions: 95 °C for 30 s, followed by 40 cycles of 95 °C for 5 s and 60 °C for 30 s. Correction of the PCR efficiency for each primer set was performed using a standard curve drawn from the dilution series of the cDNA samples. GAPDH was used as an internal control. Each sample was measured at least two times for technical replicates.
DR sequence determination and phylogenetic tree construction
To confirm the fixed differences in differentiated regions (DRs), sets of primers were designed for four DRs (DR11, DR12, DR17, and DR19) to amplify regions including fixed differences. The primer sequences are listed below. The primers for DR5 were reported previously [15, 16]. Five DRs were amplified by PCR with the following conditions: a denaturation step for 3 min at 94 °C followed by 30 cycles of denaturation for 1 min at 94 °C, annealing for 30 s at 55 °C, and extension for 30 s at 72 °C. PCR products were purified and the sequences were determined using the Applied Biosystems Automated 3130xl Sequencer (Applied Biosystems, Waltham, MA, USA).
To construct phylogenetic trees, the sequences of DRs (~ 1 kb) and orthologous sequences from genome sequence data were used. Sets of primers were designed for 16 DRs (the remaining DRs failed to amplify), and these were amplified by PCR with the following reaction conditions: a denaturation step for 3 min at 94 °C followed by 30 cycles of denaturation for 1 min at 94 °C, annealing for 30 s at 55 °C, and extension for 1.5 min at 72 °C. The primer sequences are listed below. PCR products were cloned into the T-Vector pMD20 vector (Takara, Shiga, Japan) and the sequences were determined using the Applied Biosystems Automated 3130xl Sequencer. The genomic DNAs used as templates for amplification were the Lake Victoria species H. pyrrhocephalus, H. sp. ‘macula,’ and H. piceatus; riverine species H. sp. ‘katonga’, H. sp. ‘kitilda-rukwa’, and H. sp. ‘muzu’ (see Additional file 4: Figure S4F for localities); Lake Malawi species Labidochromis caeruleus, Melanochromis auratus, Labeotropheus trewavasae, Pseudotropheus lombardoi, and Dimidiochromis strigatus; and Lake Tanganyika species Tropheus moorii, T. duboisi, T. brichardi, Simochromis pleurospilus, Petrochromis macrognathus, Cyprichromis coloratus, Ectodus descampsi, Perissodus eccentricus, and Neolamprologus tretocephalus. The consensus sequences of the genomes of H. pyrrhocephalus and H. sp. ‘macula’ were constructed from the mapping results of the paired-end short reads to the reference genome sequence of P. nyererei [14, 76] using the CLC genomic workbench. Orthologous sequences of DR sequences were obtained by BLASTN searches and from genome sequence data for Pundamilia nyererei, Metriaclima zebra, Astatotilapia burtoni, Neolamprologus brichardi, and Oreochromis niloticus , and consensus sequences of H. pyrrhocephalus and H. sp. ‘macula.’ In the case of DR5, upstream (2 kbp) and downstream (2 kbp) sequences of LWS were determined following methods described in previous studies [15, 18] and using sequences from previous studies [15, 17, 18] deposited in databases. Each of the DR sequences was aligned and subjected to a phylogenetic analysis using the maximum likelihood method with 1000 bootstrap replications in MEGA ver. 6 .
Gene ontology analysis
The sequences of DRs (14–28 kbp) were used as queries for BLASTN searches against the NCBI nucleotide database ( The sequences of genes in DRs were subjected to a gene ontology analysis using DAVID and Blast2GO . The details of the Gene Ontology Analysis and primer sequences can be found in Supplemental Material online.
Abbreviations
SFS:
: site frequency spectrum
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Acknowledgments
We thank S. Wright and J. Gojobori for providing helpful comments on the early version of our manuscript, and the Tanzania Fisheries Research Institute (TAFIRI) for help with sampling.
Funding
This work was supported by KAKENHI Grant Number JP23570269 and JP26440209 to Y.T., and an internal SOKENDAI grant to Y.T.
Availability of data and materials
The nucleotide sequences were deposited in GenBank under accession numbers LC129373–LC129499 and in the DDBJ Sequenced Read Archive under accession numbers DRX051884–DRX051889.
Author information
Author notes
Shohei Takuno and Yohey Terai contributed equally to this work.
Authors and Affiliations
Department of Evolutionary Studies of Biosystems, SOKENDAI (The Graduate University for Advanced Studies), Shonan Village, Hayama, Kanagawa, 240-0193, Japan
Shohei Takuno, Shiho Takahashi-Kariyazono & Yohey Terai
2. Graduate School of Bioscience and Biotechnology, Tokyo Institute of Technology, 4259 Nagatsuta-cho, Midori-ku, Yokohama, 226-8501, Japan
Ryutaro Miyagi, Mitsuto Aibara, Shinji Mizoiri, Semvua I. Mzighani, Norihiro Okada & Yohey Terai
3. Department of Biological sciences, Tokyo Metropolitan University, 1-1 Minamiosawa, Hachioji, Tokyo, 197-0397, Japan
Ryutaro Miyagi
4. JST (Japan Science and Technology Agency), NBDC (National Bioscience Database Center), 5-3, Yonbancho, Chiyoda-ku, Tokyo, 102-0081, Japan
Jun-ichi Onami
5. Department of Anatomy and Cytohistology, School of Dental Medicine, Tsurumi University, 2-1-3 Tsurumi, Tsurumi-ku, Yokohama, 230-8501, Japan
Akie Sato
6. Max-Planck-Institut für Biologie, Abteilung Immungenetik, Corrensstrasse 42, D-72076, Tübingen, Germany
Herbert Tichy
7. School of Life Science and Technology, Department of Life Science and Technology, Tokyo Institute of Technology (Tokyo Tech), 2-12-1, Ookayama, Meguro ward, Tokyo, Japan
Masato Nikaido
8. Tanzania Fisheries Research Institute (TAFIRI), Mwanza, Tanzania
Hillary D. J. Mrosso & Semvua I. Mzighani
9. Department of Life Sciences, National Cheng Kung University, 701, Tainan, Taiwan
Norihiro Okada
10. Foundation for Advancement of International Science (FAIS), Tsukuba, Japan
Norihiro Okada
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Shohei Takuno
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Contributions
ST performed the next-generation sequence data analysis, population genomic analysis, determination of DRs and long DRs, and manuscript writing. RM developed the research concept, and performed DNA and RNA extraction, quantitative PCR analysis, the identification of genes in DRs, and manuscript writing. JO performed gene ontology analyses for genes in DRs and long DRs. STK determined DR sequences and performed phylogenetic analyses. AS managed riverine Haplochromis species samples and analyzed DR sequences. HT performed sampling and identification of riverine Haplochromis species. MN provided helpful discussion. MA performed sampling and identification of Lake Victoria species. SM performed sampling and identification of Lake Victoria species. HDJM performed sampling of Lake Victoria species. SM was involved in management and sampling of Lake Victoria species. NO arranged the sampling of Lake Victoria species. YT developed the research concept, planned the research, and performed DNA and RNA extraction, mapping of short reads, identification of genes in DRs and long DRs, determination of DR sequences for phylogenetic analyses, and manuscript writing. All authors read and approved the final manuscript.
Corresponding authors
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Norihiro Okada or Yohey Terai.
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Additional files
Additional file 1:
Figure S1. (A) Site frequency spectrum of Lake Victoria cichlids. The white and black bars represent H. pyrrhocephalus and H. sp. ‘macula,’ respectively. (B) Demographic model of Lake Victoria cichlids. See the section Demographic Model and Parameter Estimation for details. (C) Nucleotide and indel frequencies within five DRs. We amplified and determined the sequences, including fixed differences, of four DRs from 20 individuals each of H. sp. ‘macula’ and H. pyrrhocephalus. Positions indicate the positions from the first nucleotides of the determined sequences. The frequencies of nucleotides in the coding region of LWS were verified in a previous study (14). (PDF 155 kb)
Additional file 2
Text S1. Supporting Text. (DOCX 572 kb)
Additional file 3
Figure S2. Population genomic analyses using only male individuals. (A) Site frequency spectrum in H. pyrrhocephalus (white bars; 126,843 SNPs) and H. sp. ‘macula’ (black bars; 146,456 SNPs). Nucleotide diversity and Tajima’s D values were almost same as those in all samples. (B) Average FST values (±1 SD) against coverage for Pool-seq data as in Fig. 1b (5,662,990 SNPs). The blue and orange dots represent observed and simulated values under panmixia. (C) The spatial patterns of average nucleotide diversity within each species (πW; pink), average pairwise nucleotide divergence between species (πB; blue), and FST (green) in and around LWS gene as in Fig. 2b. The green arrows represent fixed nucleotide differences between species. (PDF 279 kb)
Additional file 4:
Figure S3. The spatial patterns of average nucleotide diversity within species (πW; pink), average pairwise nucleotide divergence between species (πB; blue), and FST (green) in and around DRs. The green arrows represent fixed nucleotide differences between species. (PDF 1356 kb)
Additional file 5:
Figure S4. The origins of mutations in DRs. Three phylogenetic trees represent the accumulation of mutations in the common ancestral species of (A) Lake Victoria species, (B) Lake Victoria and riverine Haplochromis species, and (C) tribe Tropheini in Lake Tanganyika, Lakes Malawi, Victoria, and riverine Haplochromis species. The tree topologies constructed from sequences of each DR were consistent with (D) “Riverine origin” or (E) “Modern haplochromine origin.” Scale bars indicate the number of substitutions per site. (F) The LWS sequences were determined from three riverine species: H. sp. ‘katonga’ from Katonga, H. sp. ‘kitilda-rukwa’ from Kitilda-Rukwa, and H. sp. ‘muzu’ from Muzu. (PDF 554 kb)
Additional file 6:
Table S1. Genes in DRs. (PDF 149 kb)
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Takuno, S., Miyagi, R., Onami, Ji. et al. Patterns of genomic differentiation between two Lake Victoria cichlid species, Haplochromis pyrrhocephalus and H. sp. ‘macula’.
BMC Evol Biol 19, 68 (2019).
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Keywords
Cichlids
Population genomics
Adaptation
Speciation
Genomic islands of speciation |
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RE57R07: Brokerage Relationships
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RE57R07: Brokerage Relationships
A Description of Brokerage Relationships
HISTORY OF BROKERAGE RELATIONSHIPS
Fundamentals of agency are unchanged for hundreds of years, but application of agency law and other brokerage relationships are constantly changing.
History of Agency: English common law evolved from master-servant relationship. Law of agency is a common-law concept.
Respondeat Superior "Let the master answer. This maxim means that a master is liable in certain cases for the wrongful acts of his servant, and a principal for those of his agent." Black's Law Dictionary
Review history of agency, disclosure form in Massachusetts and its evolution into the 2005 revision. This section applies to real estate brokers or salespeople engaged in the purchase or sale of land with a building intended for use as a one to four unit residential dwelling or to the purchase or sale of land on which a building is intended to be constructed for use as a one or two unit residential dwelling.
Define basic terms in real estate brokerage relationships (use examples of each).
AGENT - "One authorized to represent and to act on behalf of another person (called the principal)." The Language of Real Estate (the principal i.e. client (buyer/seller) or real estate broker (broker of record).
CLIENT - (also known as a Principal) A person who is represented by an agent. Client relationship = fiduciary relationship.
CUSTOMER - A buyer or seller who is unrepresented by the real estate licensee
DESIGNATED AGENT Individual Agency - "A designated agent is that agent designated to be the agent for the buyer or seller to the exclusion of all other agents in the brokerage (real estate office or firm). Another salesperson in the firm could be designated the agent of the other party without thereby creating a dual agency for the individual agents." The Language of Real Estate
DISCLOSED DUAL - Simultaneously representing both the buyer and
AGENT - the seller with the informed consent of both
FACILITATOR - Otherwise known as a transactional broker/salesperson or non-agent. The Facilitator works to complete the transaction. Although bound by license law and MGL Ch. 93A, they do not have a fiduciary relationship with the seller or the buyer. They do not represent either party in the transaction. The Facilitator must disclose all known material defects that exist. Failure to do so could result in a Chapter 93A violation. Their duties consist of accounting and any other Facilitator duties undertaken.
PRINCIPAL - (also known as a Client) The primary party in a transaction, e.g. the buyer or the seller, the party who hires an agent to represent them
SUBAGENT - An agent of a person who is already acting as agent for a principal (agent of an agent)
AGENT vs. BROKER - While an individual may act in both capacities, and often does, the terms have different meanings.
Use of "Massachusetts Mandatory Licensee-Consumer Disclosure"
When it must be presented:
No matter what relationship is established with a buyer or a seller, the disclosure is to be provided by the licensee immediately when all three of the following circumstances exist.
At the first personal meeting
To discuss a specific property
With a prospective buyer or seller
Explain how to complete the mandatory disclosure form:
Seller's agent, or buyer's agent, or facilitator
Traditional agency vs. designated agency
Who can sign the form? Who must sign the form?
If a consumer declines to sign the form, there is a check-off box.
Broker/salesperson is still required to complete the licensee section of the form and provide their license number.
Consumer still must receive a copy of the form they will not sign.
Broker must retain mandatory disclosure form for three years.
Open House exception (254 CMR 3.00(13)(a)(3)) - at an open house the Real Estate licensee must conspicuously post and/or provide with other written materials any relationship so that the attendees can understand the relationship they may have with the licensee conducting the open house.
Consent to dual or designated agency is a separate step.
OTHER TERMS AND DEFINITIONS
FIDUCIARY
"A relationship that implies a position of trust or confidence wherein one person is usually entrusted to hold or manage property or money for another. The term fiduciary describes the faithful relationship owed by an attorney to a client or by a broker (and salesperson) to a principal. The fiduciary owes complete allegiance to the client." The Language of Real Estate
INFORMED CONSENT
"Consent to a certain act that is given after a full and fair disclosure of all facts needed to make a conscientious choice."
The Language of Real Estate
VICARIOUS LIABILITY
A principal may be responsible for the actions of their agent.
"Liability created not because of a person's actions, but because of the relationship between the liable person and other parties. For example, a real estate broker is vicariously liable for the acts of his or her salespeople while acting on behalf of the broker even if the broker did nothing to cause the liability. (See respondaet superior)". - The Language of Real Estate
Duties that may be required of licensees Referred to as "OLD CAR"
"OLD CAR"
Obedience Agent must carry out all lawful instructions of client.
Loyalty Agent must act in best interest of client.
Disclosure Agent must disclose all information relevant to client.
Confidentiality Duty to keep confidential client's information or discussion. Duty survives termination of agency relationship. Duty does not apply to legally required disclosures such as known physical hazardous conditions of property.
Accountability Agent must protect and account for all money, documents, or other personal property given to her by the client.
Reasonable Care & Due Diligence
Agent must act competently, capable of performing duties within scope of license requirements.
Types of Brokerage Relationships in Massachusetts and the Duties Required for each
(NOTE: Review of duties for agents and facilitator should be brief. Stand-alone continuing education courses are offered for seller and buyer agency and facilitation to provide further detailed reference on the responsibilities of these roles).
SELLER AGENCY - Real estate agent represents seller as the client and treats buyer as a customer. Agent's duties to Seller client include "OLD CAR".
SELLER SUBAGENCY - When seller client expressly authorizes his/her agent to use other agents to market seller's property. Subagent's duties to seller include "OLD CAR". Specific legal requirements to offer subagency to cooperating brokers are: Written informed consent must be obtained from seller client to offer subagency, the consent must state the following: That the broker may cooperate with another broker who is then a subagent of the seller. Further, that vicarious liability is the potential for a seller to be held liable for a misrepresentation or an act or omission of the subagent and that the seller authorizes the broker or salesperson to offer subagency.
BUYER AGENCY - Real estate agent represents buyer as the client in real estate transaction. Agent's duties to buyer include "OLD CAR":
BUYER SUBAGENCY (not commonly used) - Only when buyer client expressly authorizes his broker to use other agents to assist buyer in locating property for purchase. Subagent's duties to buyer include "OLD CAR". Specific legal requirements to offer subagency to cooperating brokers are: Written consent must be obtained from buyer client to offer subagency, the consent must state the following: That the broker may cooperate with another broker who is then a subagent of the buyer. Further, that vicarious liability is the potential for a buyer to be held liable for a misrepresentation or an act or omission of the subagent and that the buyer authorizes the broker or salesperson to offer subagency.
FACILITATOR - A facilitator assists the seller and/or buyer in reaching an agreement, but does not represent either the seller or buyer in the transaction. A facilitator is NOT an agent. Although bound by license law and MGL Ch. 93A, they do not have a fiduciary relationship with the seller or the buyer. They do not represent either party in the transaction. The Facilitator must disclose all known material defects that exist. Failure to do so could result in a Chapter 93A violation. This relationship must be disclosed at the first personal meeting to discuss a specific property. Facilitator's duties include only the "AR" of "OLD CAR" (a/k/a transactional broker/salesperson, non-agent and contract broker/salesperson).
DUAL AGENCY - A real estate licensee who represents both the seller and buyer in the same transaction is a disclosed dual agent with written informed consent of both the buyer and the seller (see below). Fiduciary duties are owed to both buyer and seller. Undisclosed dual agency is illegal. A dual agent assists the seller and buyer in a transaction but shall be neutral with regard to any conflicting interest of the seller and buyer. Consent may be obtained in a listing agreement, buyer agency agreement, or on a stand-alone consent form. If consent if obtained prior to the identification of a transaction then a "Notice of Dual Agency" must be provided prior to the execution of the offer to purchase. Sample form for consent is available atwww.mass.gov/dpl/boards/re. Duties of a disclosed dual agent include only the "CAR" of "OLD CAR".
Written consent form must state: A real estate broker or salesperson may act as a dual agent who represents both prospective buyer and seller with their informed written consent. A dual agent is authorized to assist the buyer and seller in a transaction, but shall be neutral with regard to any conflicting interest of the buyer and seller. Consequently, a dual agent will not have the ability to satisfy fully the duties of loyalty, full disclosure, reasonable care and obedience to lawful instructions, but shall still owe the duty of confidentiality of material information and the duty to account for funds.
Buyers and sellers should understand that material information received from either client that is confidential may not be disclosed by a dual agent except if disclosure:
is expressly authorized;
is required by law;
is intended to prevent illegal conduct; or
is necessary to prosecute a claim against a person represented or to defend a claim against the broker or salesperson.
This duty of confidentiality shall continue after termination of the brokerage relationship.
2. Notice form must provide:
"When consent to dual agency has been given by a seller or prospective purchaser in advance of the identification of a potential transaction, written notice of dual agency must also be given by the broker or salesperson to the seller and prospective purchaser after a transaction has been identified stating that the broker is dual agent with regard to the transaction. Written notice of dual agency shall satisfy section 13(b) herein and such written notice shall be given prior to the seller and prospective purchaser entering into a written agreement for the purchase or sale of residential property." Section (13) (b) (2) of 254CMR 2.00
3. Consent forms must be signed by both the buyer and seller AND the broker or salesperson; however, signatures may appear on counterparts (separate forms).
4. Consent forms must be kept for a minimum of three (3) years.
5. Process for dual agency should always be:
1. disclosure
2. consent
3. notice (if applicable)
ALTERNATIVE BUSINESS MODELS
TRADITIONAL AGENCY - When a licensee enters into an agency relationship with either a buyer or a seller, all licensees in the same brokerage firm automatically are agents of that buyer or seller. In the event of an in-house transaction where the brokerage firm represents both the buyer and seller, a dual agency might occur.
SINGLE AGENCY - The licensee or firm only represents either buyers or sellers but never both and will never practice dual agency.
DESIGNATED AGENCY - With written consent a real estate agent can be appointed by another real estate agent (the appointing agent) to represent either the buyer or seller provided the buyer or seller expressly agree to such designation. (Appointments are done within the same firm.) The real estate agent once so designated is then the agent for either the buyer or seller, but never both. The consumer who is represented becomes the agent's client. The designated agent must put their client's interests first and be an advocate for the best price and terms for their client. In situations where the appointing agent appoints both a designated seller's agent and a designated buyer's agent for the same transaction, the appointing agent becomes a "dual agent (see above for description)." Consent may be obtained in a listing agreement, buyer agency agreement, or in a stand-alone consent form. If consent is obtained prior to the identification of a transaction, then a "Notice of Designated Agency" must be provided prior to the execution of the offer to purchase.
Sample form for consent is available online atwww.mass.gov/dpl/boards/re. Duties include "OLD CAR".
If you are a seller, you are advised that:
The designated seller's agent will represent the seller and will owe the seller the duties of loyalty, full disclosure, confidentiality, to account for funds, reasonable care and obedience to lawful instruction;
All other licensees affiliated with the appointing broker will not represent the seller nor will they owe the other duties specified in paragraph (a) to that seller, and may potentially represent the buyer;
If designated agents affiliated with the same broker represent the seller and buyer in a transaction, the appointing broker shall be a dual agent and neutral as to any conflicting interests of the seller and buyer, but will continue to owe the seller and buyer the duties of confidentiality of material information and to account for funds.
If you are a buyer you are advised that:
The designated buyer's agent will represent the buyer and will owe the buyer the duties of loyalty, full disclosure, confidentiality, accountability for funds, reasonable care and obedience to lawful instruction;
All other licensees affiliated with the appointing broker will not represent the buyer nor will they have the other duties specified in paragraph (a) to that buyer, and potentially may represent the seller;
If designated agents affiliated with the same broker represent the seller and buyer in a transaction, the appointing broker shall be a dual agent and neutral as to any conflicting interests of the seller and buyer, but will continue to owe the seller and buyer the duties of confidentiality of material information and to account for funds.
POLICY/LEGAL ISSUES FOR DESIGNATED AGENCY PRACTICE
All firms practicing designated agency should adopt an office policy regarding their practices to ensure the protection of confidential information of clients.
Designated agency must start with correct disclosure referenced above.
All agents of the firm should be disclosing their affiliation with a firm that practices designated agency (refer to above usage of agency disclosure form). Agents should not practice single agency and designated agency -
CONFLICT OF INTEREST. (USE EXAMPLES!!!)
The designated real estate broker or salesperson exclusively represents the seller or buyer and is responsible for the performance of any duties owed to the seller or buyer. The designated broker or salesperson may not share known or acquired information with any other real estate agent or person that would harm the seller's or buyer's interest in the real estate transaction, except for known material defects in real property. The designated broker or salesperson shall have an affirmative obligation to disclose known material defects in real property.
The appointment by a broker or salesperson of another affiliated broker or salesperson to represent a seller or buyer shall not limit the liability or responsibility of the appointing agent for any breach of duty by the designated broker or salesperson. The appointment of the broker or salesperson to represent the seller or buyer shall extend only to those brokers or salespersons as appointed by the appointing broker or salesperson and consented to by the seller or buyer.
CONCLUSION Legal requirements regarding disclosure should be emphasized.
Important note: Setting compensation among competitors is a violation of Anti-Trust Laws.
The Brokerage Relationships course will provide an overview to the types of relationships available in real estate brokerage in Massachusetts. This Continuing Education module is not intended to be a comprehensive study on the practice of agency. All brokers and salespeople are encouraged to take a full module on agency in conjunction with this course to further enhance their knowledge and professional practice.
Required Handouts
MASSACHUSETTS MANDATORY LICENSEE - CONSUMER RELATIONSHIP DISCLOSURE
"Massachusetts Consent for Dual Agency" form
"Massachusetts Consent for Designated Agency" form
Suggested References
The Language of Real Estate by John Reilly, 6 th edition
254 CMR 3.0MGL C 112 § 87AAA 3/4
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3189 | https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/Elementary_Number_Theory_(Clark)/01%3A_Chapters/1.13%3A_The_Functions__and | d
n
n
d∣n
d=pf11pf22⋯pfrr
i
0≤fi≤ei.
fi
ei+1
{0,1,2,…,ei}
(e1+1)(e2+1)⋯(er+1)
f1,f2,…,fr
1.13.1
1.13.2
n=pe11pe22⋯perr
r
r=1
n=pe11
1.13.2
1≤r≤k
r=k+1
n=pe11⋯pekkpek+1k+1
p1,…,pk,pk+1
ei≥0
a=pe11⋯pekk
b=pek+1k+1
gcd(a,b)=1
1.13.1
σ(n)=σ(a)σ(b)
σ(a)=(pe1+11−1p1−1)⋯(pek+1k−1pk−1)
1.13.2
σ(b)=pek+1+1k+1−1pk+1−1
σ(n)=(pe1+11−1p1−1)⋯(pek+1+1k+1−1pk+1−1).
r=k+1
r≥1
a
b
1
d∣ab⇔d=d1d2
d1∣a
d2∣b
ab
a
b
1,a1,…,as
a
1,b1,…,bt
b
σ(a)σ(b)=1+a1+a2+…b+as,=1+b1+b2+…b+bt.
n=ab
1,b1,b2,…c,bt,a1⋅1,a1⋅b1,a1⋅b2,…c,a1⋅bt,a2⋅1,a2⋅b1,a2⋅b2,…c,a2⋅bt,⋮as⋅1,as⋅b1,as⋅b2,…c,as⋅bt.
gcd(a,b)=1
aibj=akbℓ
ai=ak
bj=bℓ
1+b1+…b+bt=σ(b)a11+a1b1+…b+a1bt=a1σ(b)⋮as⋅1+asb1+…b+asbt=asσ(b).
σ(n)=σ(b)+a1σ(b)+a2σ(b)+…b+a3σ(b)=(1+a1+a2+…b+as)σ(b)=σ(a)σ(b).
p
pk
1,p,p2,…,pk
σ(pk)=1+p+p2+⋯+pk=pk+1−1p−1,
a
b
gcd(a,b)=1
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1.13: The Functions σ and τ
Last updated
: Aug 17, 2021
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1.12: Fermat Primes and Mersenne Primes
1.14: Perfect Numbers and Mersenne Primes
Page ID
: 82295
( \newcommand{\kernel}{\mathrm{null}\,})
Definition 1.13.11.13.1
For n>0n>0 define: τ(n)=the number of positive divisors of n,σ(n)=the sum of the positive divisors of n.
τ(n)σ(n)=the number of positive divisors of n,=the sum of the positive divisors of n.
Example 1.13.11.13.1
12=3⋅2212=3⋅22 has positive divisors 1,2,3,4,6,12.
1,2,3,4,6,12.
Hence τ(12)=6
τ(12)=6
and σ(12)=1+2+3+4+6+12=28.
σ(12)=1+2+3+4+6+12=28.
Definition 1.13.21.13.2: Proper Divisor
A positive divisor dd of nn is said to be a proper divisor of nn if d<nd<n. We denote the sum of all proper divisors of nn by σ∗(n)σ∗(n).
Note that if n≥2n≥2 then σ∗(n)=σ(n)−n.
σ∗(n)=σ(n)−n.
Example 1.13.21.13.2
σ∗(12)=16σ∗(12)=16.
Definition 1.13.31.13.3: Perfect
n>1n>1 is perfect if σ∗(n)=nσ∗(n)=n.
Example 1.13.31.13.3
The proper divisors of 66 are 11, 22 and 33. So σ∗(6)=6σ∗(6)=6. Therefore 66 is perfect.
Exercise 1.13.11.13.1
Prove that 2828 is perfect.
The next theorem shows a simple way to compute σ(n)σ(n) and τ(n)τ(n) from the prime factorization of nn.
Theorem1.13.11.13.1
Let n=pe11pe22⋯perr,r≥1,
n=pe11pe22⋯perr,r≥1,
where p1<p2<⋯<prp1<p2<⋯<pr are primes and ei≥0ei≥0 for each i∈{1,2,…,r}i∈{1,2,…,r}. Then
τ(n)=(e1+1)(e2+1)⋯(er+1)τ(n)=(e1+1)(e2+1)⋯(er+1)
σ(n)=(pe1+11−1p1−1)(pe2+12−1p2−1)⋯(per+1r−1pr−1)σ(n)=(pe1+11−1p1−1)(pe2+12−1p2−1)⋯(per+1r−1pr−1).
Proof
: Proof of (1) From the Fundamental Theorem of Arithmetic every positive factor d of n will have its prime factors coming from those of n. Hence d∣n iff d=pf11pf22⋯pfrr where for each i: 0≤fi≤ei. That is, for each fi we can choose a value in the set of ei+1 numbers {0,1,2,…,ei}. So, in all, there are (e1+1)(e2+1)⋯(er+1) choices for the exponents f1,f2,…,fr. So (1) holds.
Proof of (2) We first establish two lemmas, Lemma 1.13.1 "13: The Functions σ and τ") and Lemma 1.13.2 "13: The Functions σ and τ").
Let n=pe11pe22⋯perr. Our proof is by induction on r. If r=1, n=pe11 and the result follows from Lemma 1.13.2 "13: The Functions σ and τ"). Suppose the result is true when 1≤r≤k. Consider now the case r=k+1. That is, let n=pe11⋯pekkpek+1k+1 where the primes p1,…,pk,pk+1 are distinct and ei≥0. Let a=pe11⋯pekk, b=pek+1k+1. Clearly gcd(a,b)=1. So by Lemma 1.13.1 "13: The Functions σ and τ") we have σ(n)=σ(a)σ(b). By the induction hypothesis σ(a)=(pe1+11−1p1−1)⋯(pek+1k−1pk−1) and by Lemma 1.13.2 "13: The Functions σ and τ") σ(b)=pek+1+1k+1−1pk+1−1 and it follows that σ(n)=(pe1+11−1p1−1)⋯(pek+1+1k+1−1pk+1−1). So the result holds for r=k+1. By PMI it holds for r≥1.
Before proving this let’s look at an example. Take n=72=8⋅9=23⋅32n=72=8⋅9=23⋅32. The theorem says τ(72)=(3+1)(2+1)=12σ(72)=(24−12−1)(33−13−1)=15⋅13=195.
τ(72)σ(72)=(3+1)(2+1)=12=(24−12−1)(33−13−1)=15⋅13=195.
Lemma1.13.11.13.1
Let n=abn=ab where a>0a>0, b>0b>0 and gcd(a,b)=1gcd(a,b)=1. Then σ(n)=σ(a)σ(b)σ(n)=σ(a)σ(b).
Proof
: Since a and b have only 1 as a common factor, using the Fundamental Theorem of Arithmetic it is easy to see that d∣ab⇔d=d1d2 where d1∣a and d2∣b. That is, the divisors of ab are products of the divisors of a and the divisors of b. Let 1,a1,…,as denote the divisors of a and let 1,b1,…,bt denote the divisors of b. Then σ(a)=1+a1+a2+…b+as,σ(b)=1+b1+b2+…b+bt. The divisors of n=ab can be listed as follows 1,b1,b2,…c,bt,a1⋅1,a1⋅b1,a1⋅b2,…c,a1⋅bt,a2⋅1,a2⋅b1,a2⋅b2,…c,a2⋅bt,⋮as⋅1,as⋅b1,as⋅b2,…c,as⋅bt. It is important to note that since gcd(a,b)=1, aibj=akbℓ implies that ai=ak and bj=bℓ. That is there are no repetitions in the above array.
If we sum each row we get 1+b1+…b+bt=σ(b)a11+a1b1+…b+a1bt=a1σ(b)⋮as⋅1+asb1+…b+asbt=asσ(b). By adding these partial sums together we get σ(n)=σ(b)+a1σ(b)+a2σ(b)+…b+a3σ(b)=(1+a1+a2+…b+as)σ(b)=σ(a)σ(b). This proves the lemma.
Lemma1.13.21.13.2
If pp is a prime and k≥0k≥0 we have σ(pk)=pk+1−1p−1.
σ(pk)=pk+1−1p−1.
Proof
: Since p is prime, the divisors of pk are 1,p,p2,…,pk. Hence σ(pk)=1+p+p2+⋯+pk=pk+1−1p−1, as desired.
Exercise1.13.21.13.2
Find σ(n)σ(n) and τ(n)τ(n) for the following values of nn.
n=900n=900
n=496n=496
n=32n=32
n=128n=128
n=1024n=1024
Exercise 1.13.31.13.3
Determine which (if any) of the numbers in Exercise 1.13.21.13.2 are perfect.
Exercise 1.13.41.13.4
Does Lemma 1.13.11.13.1 hold if we replace σσ by σ∗σ∗?
Answer
: The answer is no, but find explicit numbers a and b such that the result fails yet gcd(a,b)=1.
1.12: Fermat Primes and Mersenne Primes
1.14: Perfect Numbers and Mersenne Primes |
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3191 | https://www.doubtnut.com/qna/642565750 | In a family of 3 children, the probability of having at least one boy is 7 8 (b) 1 8 (c) 5 8 (d) 3 4
To find the probability of having at least one boy in a family of three children, we can follow these steps: Step 1: Determine the Sample Space The first step is to identify all the possible combinations of boys (B) and girls (G) in a family of three children. Each child can either be a boy or a girl, leading to the following combinations: 1. BBB (3 boys) 2. BBG (2 boys, 1 girl) 3. BGB (2 boys, 1 girl) 4. BGG (1 boy, 2 girls) 5. GBB (2 boys, 1 girl) 6. GBG (1 boy, 2 girls) 7. GGB (1 boy, 2 girls) 8. GGG (3 girls) Thus, the total number of outcomes is 8. Step 2: Identify Favorable Outcomes Next, we need to count the outcomes that satisfy the condition of having at least one boy. The outcomes that have at least one boy are: 1. BBB 2. BBG 3. BGB 4. BGG 5. GBB 6. GBG 7. GGB Counting these, we find there are 7 outcomes that have at least one boy. Step 3: Calculate the Probability The probability of an event is calculated using the formula: Probability=Number of Favorable OutcomesTotal Number of Outcomes In this case, the number of favorable outcomes is 7, and the total number of outcomes is 8. Therefore, the probability of having at least one boy is: Probability=78 Conclusion Thus, the probability of having at least one boy in a family of three children is 78. ---
To find the probability of having at least one boy in a family of three children, we can follow these steps:
Step 1: Determine the Sample Space
The first step is to identify all the possible combinations of boys (B) and girls (G) in a family of three children. Each child can either be a boy or a girl, leading to the following combinations:
BBB (3 boys)
BBG (2 boys, 1 girl)
BGB (2 boys, 1 girl)
BGG (1 boy, 2 girls)
GBB (2 boys, 1 girl)
GBG (1 boy, 2 girls)
GGB (1 boy, 2 girls)
GGG (3 girls)
Thus, the total number of outcomes is 8.
Step 2: Identify Favorable Outcomes
Next, we need to count the outcomes that satisfy the condition of having at least one boy. The outcomes that have at least one boy are:
BBB
BBG
BGB
BGG
GBB
GBG
GGB
Counting these, we find there are 7 outcomes that have at least one boy.
Step 3: Calculate the Probability
The probability of an event is calculated using the formula:
Probability=Number of Favorable OutcomesTotal Number of Outcomes
In this case, the number of favorable outcomes is 7, and the total number of outcomes is 8. Therefore, the probability of having at least one boy is:
Probability=78
Conclusion
Thus, the probability of having at least one boy in a family of three children is 78.
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3192 | https://static.bigideasmath.com/protected/content/pe/fl2/msfl2_6pe_03.pdf | Algebraic Expressions and Properties 3.1 Algebraic Expressions 3.1 Algebraic Expressions 3.2 Writing Expressions 3.2 Writing Expressions 3.3 Properties of Addition and Multiplication 3.3 Properties of Addition and Multiplication 3.4 The Distributive Property 3.4 The Distributive Property 5 6 4 3 2 1 10 12 8 6 4 2 15 18 12 9 6 3 20 24 16 12 8 4 25 30 20 15 10 5 30 36 24 18 12 6 1 2 3 4 5 6 5 6 4 3 2 1 5 6 4 3 2 1 10 12 8 6 4 2 15 18 12 9 6 3 20 24 16 12 8 4 25 30 20 15 10 5 30 36 24 18 12 6 1 2 3 4 5 6 5 6 4 3 2 1 “Did you know that 5 3 6 5 6 3 5, but 5 4 6 Þ 6 4 5?” “Only certain operations like addition and multiplication preserve equality when you switch the numbers around.” “Descartes, evaluate this expression when x 5 2 to determine the number of cat treats you are going to eat today.” “Remember that you evaluate an algebraic expression by substituting the value of x into the expression.” 3 Interpreting Numerical Expressions (MACC.5.OA.1.2) Example 1 Write a sentence interpreting the expression 3 × (19,762 + 418). 3 × (19,762 + 418) is 3 times as large as 19,762 + 418. Example 2 Write a sentence interpreting the expression (316 + 43,449) + 5. (316 + 43,449) + 5 is 5 more than 316 + 43,449. Example 3 Write a sentence interpreting the expression (20,008 − 752) ÷ 2. (20,008 − 752) ÷ 2 is half as large as 20,008 − 752. Write a sentence interpreting the expression.
1. 3 × (372 + 20,967) 2. 2 × (432 + 346,322) 3. 4 × (6722 + 4086) 4. (115 + 36,372) + 6 5. (392 + 75,325) + 78 6. (352 + 46,795) + 100 7. (30,929 + 425) ÷ 2 8. (58,742 − 721) ÷ 2 9. (96,792 + 564) ÷ 3 (MACC.5.OA.1.1, MACC.6.EE.1.1) Example 4 Simplify 42 ÷ 2 + 3(9 − 5). First: Parentheses 42 ÷ 2 + 3(9 − 5) = 42 ÷ 2 + 3 ⋅ 4 Second: Exponents = 16 ÷ 2 + 3 ⋅ 4 Third: Multiplication and Division (from left to right) = 8 + 12 Fourth: Addition and Subtraction (from left to right) = 20 Simplify the expression. 10. 32 + 5(4 − 2) 11. 3 + 4 ÷ 2 12. 10 ÷ 5 ⋅ 3 13. 4(33 − 8) ÷ 2 14. 3 ⋅ 6 − 4 ÷ 2 15. 12 + 7 ⋅ 3 − 24 x 0 1 2 4 4 + x 5 6 “Great! You’re up to x = 2. Let’s keep going.” What You Learned Before 110 Chapter 3 Algebraic Expressions and Properties Algebraic Expressions 3.1 Work with a partner. a. You babysit for 3 hours. You receive $12. What is your hourly wage? ● Write the problem. Underline the important numbers and units you need to solve the problem. ● Read the problem carefully a second time. Circle the key word for the question. ● Write each important number or word, with its units, on a piece of paper. Write +, −, ×, ÷, and = on fi ve other pieces of paper. ● Arrange the pieces of paper to answer the key word question, “What is your hourly wage?” ● Evaluate the expression that represents the hourly wage. hourly wage = ÷ Write. = Evaluate. So, your hourly wage is $ per hour. b. How can you use your hourly wage to fi nd how much you will receive for any number of hours worked?
ACTIVITY: Reading and Re-Reading 1 How can you write and evaluate an expression that represents a real-life problem?
You babysit for 3 hours. You receive $12.
What is your hourly wage?
hourly wage ($ per hour) Algebraic Expressions In this lesson, you will ● use order of operations to evaluate algebraic expressions.
● solve real-life problems.
Learning Standard MACC.6.EE.1.2c COMMON CORE Section 3.1 Algebraic Expressions 111 Use what you learned about evaluating expressions to complete Exercises 4 –7 on page 115.
Work with a partner. Use the strategy shown in Activity 1 to write an expression for each problem. After you have written the expression, evaluate it using mental math or some other method. a. You wash cars for 2 hours. You receive $6. How much do you earn per hour?
b. You have $60. You buy a pair of jeans and a shirt. The pair of jeans costs $27. You come home with $15. How much did you spend on the shirt? c. For lunch, you buy 5 sandwiches that cost $3 each. How much do you spend? d. You are running a 4500-foot race. How much farther do you have to go after running 2000 feet? e. A young rattlesnake grows at a rate of about 20 centimeters per year. How much does a young rattlesnake grow in 2 years? ACTIVITY: Reading and Re-Reading 2 3. IN YOUR OWN WORDS How can you write and evaluate an expression that represents a real-life problem? Give one example with addition, one with subtraction, one with multiplication, and one with division.
Make Sense of Quantities What are the units in the problem? How does this help you write an expression?
Math Practice 112 Chapter 3 Algebraic Expressions and Properties Lesson 3.1 Write each expression using exponents. a. d ⋅ d ⋅ d ⋅ d Because d is used as a factor 4 times, its exponent is 4. So, d ⋅ d ⋅ d ⋅ d = d 4. b. 1.5 ⋅ h ⋅ h ⋅ h Because h is used as a factor 3 times, its exponent is 3. So, 1.5 ⋅ h ⋅ h ⋅ h = 1.5h3.
Identify the terms, coeffi cients, and constants in each expression. a. 5x + 13 b. 2z2 + y + 3 5x + 13 2z2 + y + 3 Terms: 5x, 13 Terms: 2z2, 1y, 3 Coeffi cient: 5 Coeffi cients: 2, 1 Constant: 13 Constant: 3 Identify the terms, coeffi cients, and constants in the expression. 1. 12 + 10c 2. 15 + 3w + 1 — 2 3. z2 + 9z EXAMPLE Writing Algebraic Expressions Using Exponents 2 EXAMPLE Identifying Parts of an Algebraic Expression 1 Exercises 8 –13 Key Vocabulary algebraic expression, p. 112 terms, p. 112 variable, p. 112 coeffi cient, p. 112 constant, p. 112 An algebraic expression is an expression that may contain numbers, operations, and one or more symbols. Parts of an algebraic expression are called terms. 5p + 4 Lesson Tutorials The numerical factor of a term that contains a variable is a coeffi cient.
A term without a variable is called a constant.
A symbol that represents one or more numbers is called a variable.
Study Tip A variable by itself has a coeffi cient of 1. So, the term y in Example 1(b) has a coeffi cient of 1. Section 3.1 Algebraic Expressions 113 Study Tip You can write the product of 4 and n in several ways. 4 ⋅ n 4n 4(n) a. Evaluate k + 10 when k = 25. k + 10 = 25 + 10 Substitute 25 for k. = 35 Add 25 and 10. b. Evaluate 4 ⋅ n when n = 12. 4 ⋅ n = 4 ⋅ 12 Substitute 12 for n. = 48 Multiply 4 and 12. 6. Evaluate 24 + c when c = 9. 7. Evaluate d − 17 when d = 30.
EXAMPLE Evaluating Algebraic Expressions 3 Evaluate a ÷ b when a = 16 and b = 2 — 3 . a ÷ b = 16 ÷ 2 — 3 Substitute 16 for a and 2 — 3 for b. = 16 ⋅ 3 — 2 Multiply by the reciprocal of 2 — 3 , which is 3 — 2 . = 24 Multiply.
Evaluate the expression when p = 24 and q = 8. 8. p ÷ q 9. q + p 10. p − q 11. pq EXAMPLE Evaluating an Expression with Two Variables 4 Exercises 33 –36 Exercises 25 –32 Exercises 16 –21 Write the expression using exponents.
4. j ⋅ j ⋅ j ⋅ j ⋅ j ⋅ j 5. 9 ⋅ k ⋅ k ⋅ k ⋅ k ⋅ k To evaluate an algebraic expression, substitute a number for each variable. Then use the order of operations to fi nd the value of the numerical expression.
114 Chapter 3 Algebraic Expressions and Properties a. Evaluate 3x − 14 when x = 5. 3x − 14 = 3(5) − 14 Substitute 5 for x. = 15 − 14 Using order of operations, multiply 3 and 5. = 1 Subtract 14 from 15. b. Evaluate z 2 + 8.5 when z = 2. z 2 + 8.5 = 22 + 8.5 Substitute 2 for z. = 4 + 8.5 Using order of operations, evaluate 22. = 12.5 Add 4 and 8.5.
Evaluate the expression when y = 6.
12. 5y + 1 13. 30 − 24 ÷ y 14. y 2 − 7 15. 1.5 + y 2 EXAMPLE Evaluating Expressions with Two Operations 5 You are saving money to buy a skateboard. You begin with $45 and you save $3 each week. The expression 45 + 3w gives the amount of money you save after w weeks. a. How much will you have after 4 weeks, 10 weeks, and 20 weeks? b. After 20 weeks, can you buy the skateboard? Explain. a. b. After 20 weeks, you have $105. So, you cannot buy the $125 skateboard. 16. WHAT IF? In Example 6, the expression for how much money you have after w weeks is 45 + 4w. Can you buy the skateboard after 20 weeks? Explain.
EXAMPLE Real-Life Application 6 Number of Weeks, w 45 + 3w Amount Saved 4 45 + 3(4) 45 + 12 = $57 10 45 + 3(10) 45 + 30 = $75 20 45 + 3(20) 45 + 60 = $105 Exercises 43–51 Substitute the given number of weeks for w. Section 3.1 Algebraic Expressions 115 9+(-6)=3 3+(-3)= 4+(-9)= 9+(-1)= Exercises 3.1 1. WHICH ONE DOESN’T BELONG? Which expression does not belong with the other three? Explain your reasoning. 2x + 1 5w ⋅ c 3(4) + 5 y ÷ z 2. NUMBER SENSE Which step in the order of operations is fi rst? second? third? fourth? Add or subtract from left to right.
Multiply or divide from left to right.
Evaluate terms with exponents.
Perform operations in parentheses. 3. NUMBER SENSE Will the value of the expression 20 − x increase, decrease, or stay the same as x increases? Explain.
Write and evaluate an expression for the problem. 4. You receive $8 for raking leaves for 2 hours. What is your hourly wage? 5. Music lessons cost $20 per week. How much do 6 weeks of lessons cost? 6. The scores on your fi rst two history tests were 82 and 95. By how many points did you improve on your second test? 7. You buy a hat for $12 and give the cashier a $20 bill. How much change do you receive?
Identify the terms, coeffi cients, and constants in the expression. 8. 7h + 3 9. g + 12 + 9g 10. 5c2 + 7d 11. 2m2 + 15 + 2p2 12. 6 + n2 + 1 — 2 d 13. 8x + x2 — 3 Terms: 2, x2, y Coeffi cient: 2 Constant: none ✗ 14. ERROR ANALYSIS Describe and correct the error in identifying the terms, coeffi cients, and constants in the algebraic expression 2x2y. 15. PERIMETER You can use the expression 2ℓ + 2w to fi nd the perimeter of a rectangle where ℓ is the length and w is the width. a. Identify the terms, coeffi cients, and constants in the expression. b. Interpret the coeffi cients of the terms.
1 Help with Homework w 116 Chapter 3 Algebraic Expressions and Properties 2 4 3 Write each expression using exponents. 16. b ⋅ b ⋅ b 17. g ⋅ g ⋅ g ⋅ g ⋅ g 18. 8 ⋅ w ⋅ w ⋅ w ⋅ w 19. 5.2 ⋅ y ⋅ y ⋅ y 20. a ⋅ a ⋅ c ⋅ c 21. 2.1 ⋅ x ⋅ z ⋅ z ⋅ z ⋅ z 3 ⋅ n ⋅ n ⋅ n ⋅ n = 4n3 ✗ 22. ERROR ANALYSIS Describe and correct the error in writing the product using exponents. 23. AREA Write an expression using exponents that 5d represents the area of the square.
How many were going to St. Ives?
Kits, cats, sacks, wives Each cat had seven kits Each sack had seven cats Each wife had seven sacks I met a man with seven wives As I was going to St. Ives 24. ST. IVES Suppose the man in the St. Ives poem has x wives, each wife has x sacks, each sack has x cats, and each cat has x kits. Write an expression using exponents that represents the total number of kits, cats, sacks, and wives going to St. Ives. ALGEBRA Evaluate the expression when a = 3, b = 2, and c = 12. 25. 6 + a 26. b ⋅ 5 27. c − 1 28. 27 ÷ a 29. 12 − b 30. c + 5 31. 2a 32. c ÷ 6 33. a + b 34. c − a 35. c — a 36. b ⋅ c 37. ERROR ANALYSIS Describe and correct the error in evaluating the expression when m = 8. 38. LAWNS You earn 15n dollars for mowing n lawns. How much do you earn for mowing one lawn? seven lawns? 39. PLANT After m months, the height of a plant is 10 + 3m millimeters. How tall is the plant after eight months? three years?
Copy and complete the table. 40. x 3 6 9 x ⋅ 8 41. x 2 4 8 64 ÷ x 42. FALLING OBJECT An object falls 16t 2 feet in t seconds. You drop a rock from a bridge that is 75 feet above the water. Will the rock hit the water in 2 seconds? Explain. 5m + 3 = 5 ⋅ 8 + 3 = 5 ⋅ 11 = 55 ✗ Section 3.1 Algebraic Expressions 117 ALGEBRA Evaluate the expression when a = 10, b = 9, and c = 4.
43. 2a + 3 44. 4c − 7.8 45. a — 4 + 1 — 3 46. 24 — b + 8 47. c 2 + 6 48. a 2 − 18 49. a + 9c 50. bc + 12.3 51. 3a + 2b − 6c 52. MOVIES You rent x new releases and y standard rentals. Which expression tells you how much money you will need? 3x + 4y 4x + 3y 7(x + y) 53. WATER PARK You fl oat 2000 feet along a “Lazy River” water ride. The ride takes less than 10 minutes. Give two examples of possible times and speeds. Illustrate the water ride with a drawing.
54. SCIENCE CENTER The expression 20a + 13c is the cost (in dollars) for a adults and c students to enter a science center.
a. How much does it cost for an adult? a student? Explain your reasoning.
b. Find the total cost for 4 adults and 24 students.
c. You fi nd the cost for a group. Then the numbers of adults and students in the group both double. Does the cost double? Explain your answer using an example.
d. In part (b), the number of adults is cut in half, but the number of students doubles. Is the cost the same? Explain your answer.
x in. 55. The volume of the cube is equal to four times the area of one of its faces.
What is the volume of the cube?
Find the value of the power. (Section 1.2) 56. 35 57. 83 58. 74 59. 28 60. MULTIPLE CHOICE Which numbers have a least common multiple of 24? (Section 1.6) ○ A 4, 6 ○ B 2, 22 ○ C 3, 8 ○ D 6, 12 5 John Smith Bob Newman Joe Holyman Based on a true story New Releases $4 Standard Rentals $3 118 Chapter 3 Algebraic Expressions and Properties Writing Expressions 3.2 How can you write an expression that represents an unknown quantity?
Work with a partner. You use a $20 bill to buy lunch at a café. You order a sandwich from the menu board shown. a. Complete the table. In the last column, write a numerical expression for the amount of change received. b. REPEATED REASONING Write an expression for the amount of change you receive when you order any sandwich from the menu board. c. Compare the expression you wrote in part (b) with the expressions in the last column of the table in part (a). d. The café offers several side dishes, each at the same price. You order a chicken salad sandwich and two side dishes. Write an expression for the total amount of money you spend. Explain how you wrote your expression. e. The expression 20 − 4.65s represents the amount of change one customer receives after ordering from the menu board. Explain what each part of the expression represents. Do you know what the customer ordered? Explain your reasoning.
ACTIVITY: Ordering Lunch 1 Sandwich Price (dollars) Change Received (dollars) Reuben BLT Egg salad Roast beef prices include tax Algebraic Expressions In this lesson, you will ● use variables to represent numbers in algebraic expressions.
● write algebraic expressions.
Learning Standard MACC.6.EE.1.2a COMMON CORE Section 3.2 Writing Expressions 119 Work with a partner. a. Complete the table. b. Here is a word problem that uses one of the expressions in the table.
You arrive at the café 8 minutes sooner than your friend. Your friend arrives at 6:42 P.M. When did you arrive? Which expression from the table can you use to solve the problem? c. Write a problem that uses a different expression from the table.
ACTIVITY: Words That Imply Addition or Subtraction 2 Use what you learned about writing expressions to complete Exercises 9 –12 on page 122.
Variable Phrase Expression n 4 more than a number m the difference of a number and 3 x the sum of a number and 8 p 10 less than a number n 7 units farther away t 8 minutes sooner w 12 minutes later y a number increased by 9 Work with a partner. Match each phrase with an expression.
the product of a number and 3 n ÷ 3 the quotient of 3 and a number 4p 4 times a number n ⋅ 3 a number divided by 3 2m twice a number 3 ÷ n ACTIVITY: Words That Imply Multiplication or Division 3 4. IN YOUR OWN WORDS How can you write an expression that represents an unknown quantity? Give examples to support your explanation.
Use Expressions How do the key words in the phrase help you write the given relationship as an expression?
Math Practice 120 Chapter 3 Algebraic Expressions and Properties Lesson 3.2 Some words imply math operations.
Operation Addition Subtraction Multiplication Division Key Words and Phrases added to plus sum of more than increased by total of and subtracted from minus difference of less than decreased by fewer than take away multiplied by times product of twice of divided by quotient of Write the phrase as an expression. a. 8 fewer than 21 21 − 8 The phrase fewer than means subtraction. b. the product of 30 and 9 30 × 9, or 30 ⋅ 9 The phrase product of means multiplication.
EXAMPLE Writing Numerical Expressions 1 When writing expressions involving subtraction or division, order is important. For example, the quotient of a number x and 2 means x ÷ 2, not 2 ÷ x.
Common Error Write the phrase as an expression. a. 14 more than a number x x + 14 The phrase more than means addition. b. a number y minus 75 y − 75 The word minus means subtraction. c. the quotient of 3 and a number z 3 ÷ z, or 3 — z The phrase quotient of means division.
Write the phrase as an expression. 1. the sum of 18 and 35 2. 6 times 50 3. 25 less than a number b 4. a number x divided by 4 5. the total of a number t and 11 6. 100 decreased by a number k EXAMPLE Writing Algebraic Expressions 2 Exercises 3–18 Lesson Tutorials Section 3.2 Writing Expressions 121 The length of Interstate 90 from the West Coast to the East Coast is 153.5 miles more than 2 times the length of Interstate 15 from southern California to northern Montana. Let m be the length of Interstate 15. Which expression can you use to represent the length of Interstate 90? ○ A 2m + 153.5 ○ B 2m − 153.5 ○ C 153.5 − 2m ○ D 153.5m + 2 2m + 153.5 The correct answer is ○ A .
EXAMPLE Writing an Algebraic Expression 3 The word times means multiplication. So, multiply 2 and m.
The phrase more than means addition. So, add 2m and 153.5.
You plant a cypress tree that is 10 inches tall. Each year, its height increases by 15 inches. a. Make a table that shows the height of the tree for 4 years. Then write an expression for the height after t years. b. What is the height after 9 years? a. The height is increasing, so add 15 each year as shown in the table. So, the height after year t is 10 + 15t. b. Evaluate 10 + 15t when t = 9. 10 + 15t = 10 + 15(9) = 145 After 9 years, the height of the tree is 145 inches. 7. Your friend has 5 more than twice as many game tokens as your sister. Let t be the number of game tokens your sister has. Write an expression for the number of game tokens your friend has. 8. WHAT IF? In Example 4, what is the height of the cypress tree after 16 years?
EXAMPLE Real-Life Application 4 Year, t Height (inches) 0 10 1 10 + 15(1) = 25 2 10 + 15(2) = 40 3 10 + 15(3) = 55 4 10 + 15(4) = 70 When t is 0, the height is 10 inches.
You can see that an expression is 10 + 15t.
Exercises 27–30 10 in.
Study Tip Sometimes, like in Example 3, a variable represents a single value. Other times, like in Example 4, a variable can represent more than one value.
122 Chapter 3 Algebraic Expressions and Properties 9+(-6)=3 3+(-3)= 4+(-9)= 9+(-1)= Exercises 3.2 1. DIFFERENT WORDS, SAME QUESTION Which is different? Write “both” expressions.
12 more than x x increased by 12 x take away 12 the sum of x and 12 2. REASONING You pay 0.25p dollars to print p photos. What does the coeffi cient represent?
Write the phrase as an expression. 3. 5 less than 8 4. the product of 3 and 12 5. 28 divided by 7 6. the total of 6 and 10 7. 3 fewer than 18 8. 17 added to 15 9. 13 subtracted from a number x 10. 5 times a number d 11. the quotient of 18 and a number a 12. the difference of a number s and 6 13. 7 increased by a number w 14. a number b squared 15. the sum of a number y and 4 16. the difference of 12 and a number x 17. twice a number z 18. a number t cubed ERROR ANALYSIS Describe and correct the error in writing the phrase as an expression. 19. the quotient of 8 and a number y 20. 16 decreased by a number x y — 8 ✗ x − 16 ✗ 21. DINNER Five friends share the cost of a dinner equally. a. Write an expression for the cost per person. b. Make up a total cost and test your expression. Is the result reasonable? 22. TV SHOW A television show has 19 episodes per season. a. Copy and complete the table. Seasons 1 2 3 4 5 Episodes b. Write an expression for the number of episodes in n seasons.
Give two ways to write the expression as a phrase. 23. n + 6 24. 4w 25. 15 − b 26. 14 − 3z 1 2 Help with Homework Section 3.2 Writing Expressions 123 Write the phrase as an expression. Then evaluate when x = 5 and y = 20.
27. 3 less than the quotient of a 28. the sum of a number x and 4, number y and 4 all divided by 3 29. 6 more than the product of 8 30. the quotient of 40 and the and a number x difference of a number y and 16 31. MODELING It costs $3 to bowl a game and $2 for shoe rental.
a. Make a table for the cost of up to 5 games.
b. Write an expression for the cost of g games.
c. Use your expression to fi nd the cost of 8 games.
32. PUZZLE Florida has 8 less than 5 times the number of counties in Arizona. Georgia has 25 more than twice the number of counties in Florida.
a. Write an expression for the number of counties in Florida.
b. Write an expression for the number of counties in Georgia.
c. Arizona has 15 counties. How many do Florida and Georgia have?
33. PATTERNS There are 140 people in a Singing Competition 1 2 3 4 5 150 100 125 75 50 25 0 Round Contestants after each round 110 95 80 65 125 singing competition. The graph shows the results for the fi rst fi ve rounds.
a. Write an expression for the number of people after each round.
b. How many people compete in the ninth round? Explain your reasoning.
34. NUMBER SENSE The difference between two numbers is 8. The lesser number is a. Write an expression for the greater number.
35. One number is four times another. The greater number is x. Write an expression for the lesser number.
Evaluate the expression. (Skills Review Handbook) 36. 8 + (22 + 15) 37. (13 + 9) + 37 38. (13 × 6) × 5 39. 4 × (7 × 5) 40. MULTIPLE CHOICE A grocery store is making fruit baskets using 144 apples, 108 oranges, and 90 pears. Each basket will be identical. What is the greatest number of fruit baskets the store can make using all the fruit? (Section 1.5) ○ A 6 ○ B 9 ○ C 16 ○ D 18 3 4 124 Chapter 3 Algebraic Expressions and Properties 3 Study Help Make information wheels to help you study these topics. 1. evaluating algebraic expressions 2. writing algebraic expressions After you complete this chapter, make information wheels for the following topics. 3. Commutative Properties of Addition and Multiplication 4. Associative Properties of Addition and Multiplication 5. Addition Property of Zero 6. Multiplication Properties of Zero and One 7. Distributive Property 8. factoring expressions You can use an information wheel to organize information about a topic. Here is an example of an information wheel for identifying parts of an algebraic expression.
“My information wheel for Fluffy has matching adjectives and nouns.” In the algebraic expression 5p + 4, 5p and 4 are the terms.
In the algebraic expression 5p + 4, p is a variable.
In the algebraic expression 5p + 4, 5 is a coefficient.
In the algebraic expression 5p + 4, 4 is a constant.
Identifying parts of an algebraic expression Graphic Organizer 1.1–1.3 3.1–3.2 Quiz Sections 3.1–3.2 Quiz 125 Identify the terms, coeffi cients, and constants of the expression. (Section 3.1) 1. 6q + 1 2. 3r 2 + 4r + 8 Write the expression using exponents. (Section 3.1) 3. s ⋅ s ⋅ s ⋅ s 4. 2 ⋅ t ⋅ t ⋅ t ⋅ t ⋅ t Evaluate the expression when a = 8 and b = 2. (Section 3.1) 5. a + 5 6. ab 7. a2 − 6 Copy and complete the table. (Section 3.1) 8. x x + 6 1 2 3 9. x 3x − 5 3 6 9 Write the phrase as an expression. (Section 3.2) 10. the sum of 28 and 35 11. a number x divided by 2 12. the product of a number m and 23 13. 10 less than a number a 14. COUPON The expression p − 15 is the amount Coupon Good for $15 off any purchase of $75 or more you pay after using the coupon on a purchase of p dollars. How much do you pay for a purchase of $83? (Section 3.1) 15. AMUSEMENT PARK The expression 15a + 12c is the cost (in dollars) of admission at an amusement park for a adults and c children. Find the total cost for 5 adults and 10 children. (Section 3.1) 16. MOVING TRUCK To rent a moving truck for the day, it costs $33 plus $1 for each mile driven. (Section 3.2) a. Write an expression for the cost to rent the truck. b. You drive the truck 300 miles. How much do you pay?
Quiz Quiz Progress Check 126 Chapter 3 Algebraic Expressions and Properties Properties of Addition and Multiplication 3.3 Does the order in which you perform an operation matter?
Work with a partner. Place each statement in the correct oval. a. Fasten 5 shirt buttons. b. Put on a shirt and tie. c. Fill and seal an envelope. d. Floss your teeth. e. Put on your shoes. f. Chew and swallow. Order Matters Order Doesn’t Matter Think of three math problems using the four operations where order matters and three where order doesn’t matter.
ACTIVITY: Does Order Matter?
1 When you commute the positions you switch their positions.
of two stuffed animals on a shelf, Commute Work with a partner. a. Which of the following are true? 3 + 5 = ? 5 + 3 3 − 5 = ? 5 − 3 9 × 3 = ? 3 × 9 9 ÷ 3 = ? 3 ÷ 9 b. The true equations show the Commutative Properties of Addition and Multiplication. Why do you think they are called commutative? ACTIVITY: Commutative Properties 2 Equivalent Expressions In this lesson, you will ● use properties of operations to generate equivalent expressions.
Learning Standards MACC.6.EE.1.3 MACC.6.EE.1.4 COMMON CORE Section 3.3 Properties of Addition and Multiplication 127 Use what you learned about the properties of addition and multiplication to complete Exercises 5 – 8 on page 130.
You have two best friends. Sometimes And sometimes you associate you associate with one of them. with the other. Associate Work with a partner. a. Which of the following are true? 8 + (3 + 1) = ? (8 + 3) + 1 8 − (3 − 1) = ? (8 − 3) − 1 12 × (6 × 2) = ? (12 × 6) × 2 12 ÷ (6 ÷ 2) = ? (12 ÷ 6) ÷ 2 b. The true equations show the Associative Properties of Addition and Multiplication. Why do you think they are called associative? ACTIVITY: Associative Properties 3 4. IN YOUR OWN WORDS Does the order in which you perform an operation matter? Give examples to support your explanation. 5. MENTAL MATH Explain how you can add the sum in your head. 11 + 7 + 12 + 13 + 8 + 9 6. SECRET CODE The creatures on a distant planet use the symbols ■, ◆, ★, and ●for the four operations. a. Use the codes to decide which symbol represents addition and which symbol represents multiplication. Explain your reasoning. 3 ● 4 = 4 ● 3 3 ★4 = 4 ★3 2 ● (5 ● 3) = (2 ● 5) ● 3 2 ★(5 ★3) = (2 ★5) ★3 0 ● 4 = 0 0 ★4 = 4 b. Make up your own symbols for addition and multiplication. Write codes using your symbols. Trade codes with a classmate. Decide which symbol represents addition and which symbol represents multiplication.
Use Counterexamples What do the false equations tell you about the Associative Properties?
Math Practice 128 Chapter 3 Algebraic Expressions and Properties Lesson 3.3 Commutative Properties Words Changing the order of addends or factors does not change the sum or product.
Numbers 5 + 8 = 8 + 5 Algebra a + b = b + a 5 ⋅ 8 = 8 ⋅ 5 a ⋅ b = b ⋅ a Associative Properties Words Changing the grouping of addends or factors does not change the sum or product.
Numbers (7 + 4) + 2 = 7 + (4 + 2) (7 ⋅ 4) ⋅ 2 = 7 ⋅ (4 ⋅ 2) Algebra (a + b) + c = a + (b + c) (a ⋅ b) ⋅ c = a ⋅ (b ⋅ c) Key Vocabulary equivalent expressions, p. 128 Expressions with the same value, like 12 + 7 and 7 + 12, are equivalent expressions. You can use the Commutative and Associative Properties to write equivalent expressions. a. Simplify the expression 7 + (12 + x). 7 + (12 + x) = (7 + 12) + x Associative Property of Addition = 19 + x Add 7 and 12. b. Simplify the expression (6.1 + x) + 8.4. (6.1 + x) + 8.4 = (x + 6.1) + 8.4 Commutative Property of Addition = x + (6.1 + 8.4) Associative Property of Addition = x + 14.5 Add 6.1 and 8.4. c. Simplify the expression 5(11y). 5(11y) = (5 ⋅ 11)y Associative Property of Multiplication = 55y Multiply 5 and 11.
Simplify the expression. Explain each step. 1. 10 + (a + 9) 2. ( c + 2 — 3 ) + 1 — 2 3. 5(4n) EXAMPLE Using Properties to Write Equivalent Expressions 1 Lesson Tutorials Exercises 5 – 8 Study Tip One way to check whether expressions are equivalent is to evaluate each expression for any value of the variable. In Example 1(a), use x = 2. 7 + (12 + x) = 19 + x 7 + (12 + 2) = ? 19 + 2 21 = 21 ✓ Section 3.3 Properties of Addition and Multiplication 129 Addition Property of Zero Words The sum of any number and 0 is that number.
Numbers 7 + 0 = 7 Algebra a + 0 = a Multiplication Properties of Zero and One Words The product of any number and 0 is 0. The product of any number and 1 is that number.
Numbers 9 ⋅ 0 = 0 Algebra a ⋅ 0 = 0 4 ⋅ 1 = 4 a ⋅ 1 = a a. Simplify the expression 9 ⋅ 0 ⋅ p. 9 ⋅ 0 ⋅ p = (9 ⋅ 0) ⋅ p Associative Property of Multiplication = 0 ⋅ p = 0 Multiplication Property of Zero b. Simplify the expression 4.5 ⋅ r ⋅ 1. 4.5 ⋅ r ⋅ 1 = 4.5 ⋅ (r ⋅ 1) Associative Property of Multiplication = 4.5 ⋅ r Multiplication Property of One = 4.5r EXAMPLE Using Properties to Write Equivalent Expressions 2 You and six friends are on the team, so use the expression 7x, not 6x, to represent the cost of the T-shirts.
Common Error You and six friends play on a basketball team. A sponsor paid $100 for the league fee, x dollars for each player’s T-shirt, and $68.25 for trophies. Write an expression for the total amount the sponsor paid.
Add the league fee, the cost of the T-shirts, and the cost of the trophies. 100 + 7x + 68.25 = 7x + 100 + 68.25 Commutative Property of Addition = 7x + 168.25 Add 100 and 68.25. An expression for the total amount is 7x + 168.25.
Simplify the expression. Explain each step. 4. 12 ⋅ b ⋅ 0 5. 1 ⋅ m ⋅ 24 6. (t + 15) + 0 7. WHAT IF? In Example 3, your sponsor paid $54.75 for trophies. Write an expression for the total amount the sponsor paid.
EXAMPLE Real-Life Application 3 Exercises 9 – 23 130 Chapter 3 Algebraic Expressions and Properties Exercises 3.3 9+(-6)=3 3+(-3)= 4+(-9)= 9+(-1)= Tell which property the statement illustrates. 5. 5 ⋅ p = p ⋅ 5 6. 2 + (12 + r) = (2 + 12) + r 7. 4 ⋅ (x ⋅ 10) = (4 ⋅ x) ⋅ 10 8. x + 7.5 = 7.5 + x 9. (c + 2) + 0 = c + 2 10. a ⋅ 1 = a 11. ERROR ANALYSIS Describe and correct the (7 + x) + 3 = (x + 7) + 3 Associative Property of Addition ✗ error in stating the property that the statement illustrates.
Simplify the expression. Explain each step. 12. 6 + (5 + x) 13. (14 + y) + 3 14. 6(2b) 15. 7(9w) 16. 3.2 + (x + 5.1) 17. (0 + a) + 8 18. 9 ⋅ c ⋅ 4 19. (18.6 ⋅ d ) ⋅ 1 20. ( 3k + 4 1 — 5 ) + 8 3 — 5 21. (2.4 + 4n) + 9 22. (3s) ⋅ 8 23. z ⋅ 0 ⋅ 12 24. GEOMETRY The expression 12 + x + 4 represents the perimeter of a triangle. Simplify the expression. 25. SCOUT COOKIES A case of Scout cookies has 10 cartons. A carton has 12 boxes. The amount you earn on a whole case is 10(12x) dollars. a. What does x represent? b. Simplify the expression. 1. NUMBER SENSE Write an example of a sum of fractions. Show that the Commutative Property of Addition is true for the sum. 2. OPEN-ENDED Write an algebraic expression that can be simplifi ed using the Associative Property of Addition. 3. OPEN-ENDED Write an algebraic expression that can be simplifi ed using the Associative Property of Multiplication and the Multiplication Property of One. 4. WHICH ONE DOESN’T BELONG? Which statement does not belong with the other three? Explain your reasoning.
7 + (x + 4) = 7 + (4 + x) 9 + (7 + w) = (9 + 7) + w (3 + b) + 2 = (b + 3) + 2 (4 + n) + 6 = (n + 4) + 6 1 2 Help with Homework Section 3.3 Properties of Addition and Multiplication 131 26. STRUCTURE The volume of the rectangular prism is 12.5 ⋅ x ⋅ 1. a. Simplify the expression. 12.5 1 x b. Match x = 0.25, 12.5, and 144 with the object. Explain. A. siding for a house B. ruler C. square fl oor tile Write the phrase as an expression. Then simplify the expression. 27. 7 plus the sum of a number x and 5 28. the product of 8 and a number y multiplied by 9 Copy and complete the statement using the specifi ed property. 29. 30. 31. 32. 33. 34. HATS You and a friend sell hats at a fair booth. You sell 16 hats on the fi rst shift and 21 hats on the third shift. Your friend sells x hats on the second shift. a. Write an expression for the number of hats sold. b. The expression 37(14) + 10x represents the amount that you both earned. How can you tell that your friend was selling the hats for a discounted price? c. You earned more money than your friend. What can you say about the value of x?
Evaluate the expression. (Section 1.3) 35. 7(10 + 4) 36. 12(10 − 1) 37. 6(5 + 10) 38. 8(30 − 5) Find the prime factorization of the number. (Section 1.4) 39. 37 40. 144 41. 147 42. 205 43. MULTIPLE CHOICE A bag has 16 blue, 20 red, and 24 green marbles. What fraction of the marbles in the bag are blue? (Skills Review Handbook) ○ A 1 — 5 ○ B 4 — 15 ○ C 4 — 11 ○ D 11 — 15 Property Statement Associative Property of Multiplication 7(2y) = Commutative Property of Multiplication 13.2 ⋅ (x ⋅ 1) = Associative Property of Addition 17 + (6 + 2x) = Addition Property of Zero 2 + (c + 0) = Multiplication Property of One 1 ⋅ w ⋅ 16 = 132 Chapter 3 Algebraic Expressions and Properties The Distributive Property 3.4 How do you use mental math to multiply two numbers?
Work with a partner. a. MODELING Draw two rectangles of the same width but with different lengths on a piece of grid paper. Label the dimensions. b. Write an expression for the total area of the rectangles. ( × ) + ( × ) c. Rearrange the rectangles by aligning the shortest sides to form one rectangle. Label the dimensions. Write an expression for the area. × ( + ) d. Can the expressions from parts (b) and (c) be set equal to each other? Explain. e. REPEATED REASONING Repeat this activity using different rectangles. Explain how this illustrates the Distributive Property. Write a rule for the Distributive Property.
ACTIVITY: Modeling a Property 1 When you distribute something you give that thing to each to each person in a group, person in the group. à Distribute à Equivalent Expressions In this lesson, you will ● use the Distributive Property to fi nd products.
● use the Distributive Property to simplify algebraic expressions.
Learning Standards MACC.6.NS.2.4 MACC.6.EE.1.3 MACC.6.EE.1.4 COMMON CORE Section 3.4 The Distributive Property 133 Use what you learned about the Distributive Property to complete Exercises 5–8 on page 137.
Work with a partner. Use the Distributive Property and mental math to fi nd the product. a. Sample: 6 × 23 6 × 23 = 6 × (20 + 3) Write 23 as the sum of 20 and 3. = (6 × 20) + (6 × 3) Distribute the 6 over the sum. = 120 + 18 Find the products. = 138 Add. So, 6 × 23 = 138. b. 5 × 17 c. 8 × 26 d. 20 × 19 e. 40 × 29 f. 25 × 39 g. 15 × 47 ACTIVITY: Using Mental Math 3 4. Compare the methods in Activities 2 and 3. 5. IN YOUR OWN WORDS How do you use mental math to multiply two numbers? Give examples to support your explanation. Work with a partner. Use the method shown to fi nd the product. a. Sample: 23 × 6 23 × 6 120 Multiply 20 and 6. + 18 Multiply 3 and 6. 138 Add. So, 23 × 6 = 138. b. 33 × 7 c. 47 × 9 d. 28 × 5 e. 17 × 4 ACTIVITY: Using Mental Math 2 23 is 20 + 3.
Find Entry Points How can you rewrite the larger number as the sum of two numbers so that you can use mental math?
Math Practice 134 Chapter 3 Algebraic Expressions and Properties Lesson 3.4 Distributive Property Words To multiply a sum or difference by a number, multiply each number in the sum or difference by the number outside the parentheses. Then evaluate.
Numbers 3(7 + 2) = 3 × 7 + 3 × 2 Algebra a(b + c) = ab + ac 3(7 − 2) = 3 × 7 − 3 × 2 a(b − c) = ab − ac Use the Distributive Property and mental math to fi nd 8 × 53. 8 × 53 = 8(50 + 3) Write 53 as 50 + 3. = 8(50) + 8(3) Distributive Property = 400 + 24 Multiply. = 424 Add.
EXAMPLE Using Mental Math 1 Lesson Tutorials Use the Distributive Property to fi nd 1 — 2 × 2 3 — 4 . 1 — 2 × 2 3 — 4 = 1 — 2 × ( 2 + 3 — 4 ) = ( 1 — 2 × 2 ) + ( 1 — 2 × 3 — 4 ) Distributive Property = 1 + 3 — 8 Multiply. = 1 3 — 8 Add.
Use the Distributive Property to fi nd the product. 1. 5 × 41 2. 9 × 19 3. 6(37) 4. 2 — 3 × 1 1 — 2 5. 1 — 4 × 4 1 — 5 6. 2 — 7 × 3 3 — 4 EXAMPLE Using the Distributive Property 2 Rewrite 2 3 — 4 as the sum 2 + 3 — 4 .
Exercises 5 –16 Key Vocabulary like terms, p. 136 Section 3.4 The Distributive Property 135 Use the Distributive Property to simplify the expression. a. 4(n + 5) 4(n + 5) = 4(n) + 4(5) Distributive Property = 4n + 20 Multiply. b. 12(2y − 3) 12(2y − 3) = 12(2y) − 12(3) Distributive Property = 24y − 36 Multiply. c. 9(6 + x + 2) 9(6 + x + 2) = 9(6) + 9(x) + 9(2) Distributive Property = 54 + 9x + 18 Multiply. = 9x + 54 + 18 Commutative Property of Addition = 9x + 72 Add 54 and 18.
Use the Distributive Property to simplify the expression. 7. 7(a + 2) 8. 3(d − 11) 9. 7(2 + 6 − 4d) EXAMPLE Simplifying Algebraic Expressions 3 José is x years old. His brother, Felipe, is 2 years older than José. Their aunt, Maria, is three times as old as Felipe. Write and simplify an expression that represents Maria’s age in years.
Name Description Expression José He is x years old.
x Felipe He is 2 years older than José. So, add 2 to x.
x + 2 Maria She is three times as old as Felipe. So, multiply 3 and (x + 2).
3(x + 2) 3(x + 2) = 3(x) + 3(2) Distributive Property = 3x + 6 Multiply. Maria’s age in years is represented by the expression 3x + 6.
EXAMPLE Real-Life Application 4 Exercises 17 –32 Study Tip You can use the Distributive Property when there are more than two terms in the sum or difference.
136 Chapter 3 Algebraic Expressions and Properties 10. Alexis is x years old. Her sister, Gloria, is 7 years older than Alexis. Their grandfather is fi ve times as old as Gloria. Write and simplify an expression that represents their grandfather’s age in years.
In an algebraic expression, like terms are terms that have the same variables raised to the same exponents. Constant terms are also like terms. 5x + 19 + 2x + 2 Use the Distributive Property to combine like terms.
Like terms Like terms Simplify each expression. a. 3x + 9 + 2x − 5 3x + 9 + 2x − 5 = 3x + 2x + 9 − 5 Commutative Property of Addition = (3 + 2)x + 9 − 5 Distributive Property = 5x + 4 Simplify. b. y + y + y y + y + y = 1y + 1y + 1y Multiplication Property of One = (1 + 1 + 1)y Distributive Property = 3y Add coeffi cients. c. 7z + 2(z − 5y) 7z + 2(z − 5y) = 7z + 2(z) − 2(5y) Distributive Property = 7z + 2z − 10y Multiply. = (7 + 2)z − 10y Distributive Property = 9z − 10y Add coeffi cients.
Simplify the expression. 11. 8 + 3z − z 12. 3(b + 5) + b + 2 EXAMPLE Combining Like Terms 5 Exercises 39 –53 Section 3.4 The Distributive Property 137 Exercises 3.4 1. WRITING One meaning of the word distribute is “to give something to each member of a group.” How can this help you remember the Distributive Property? 2. OPEN-ENDED Write an algebraic expression in which you use the Distributive Property and then the Associative Property of Addition to simplify. 3. WHICH ONE DOESN’T BELONG? Which expression does not belong with the other three? Explain your reasoning.
2(x + 2) 5(x − 8) 4 + (x ⋅ 4) 8(9 − x) 4. Identify the like terms in the expression 8x + 1 + 7x + 4.
9+(-6)=3 3+(-3)= 4+(-9)= 9+(-1)= Use the Distributive Property and mental math to fi nd the product. 5. 3 × 21 6. 9 × 76 7. 12(43) 8. 5(88) 9. 18 × 52 10. 8 × 27 11. 8(63) 12. 7(28) Use the Distributive Property to fi nd the product. 13. 1 — 4 × 2 2 — 7 14. 5 — 6 × 2 2 — 5 15. 5 — 9 × 4 1 — 2 16. 2 — 15 × 5 5 — 8 Use the Distributive Property to simplify the expression. 17. 3(x + 4) 18. 10(b − 6) 19. 6(s − 9) 20. 7(8 + y) 21. 8(12 + a) 22. 9(2n + 1) 23. 12(6 − k) 24. 18(5 − 3w) 25. 9(3 + c + 4) 26. 7(8 + x + 2) 27. 8(5g + 5 − 2) 28. 6(10 + z + 3) 29. 4(x + y) 30. 25(x − y) 31. 7(p + q + 9) 32. 13(n + 4 + 7m) 33. ERROR ANALYSIS Describe and correct the error in rewriting the expression. 34. ART MUSEUM A class of 30 students visits an art museum PRICES Museum Child (under 5) Student Regular Senior Free $8 $12 $10 Free $x $4 $3 Exhibit and a special exhibit while there. a. Use the Distributive Property to write and simplify an expression for the cost. b. Estimate a reasonable value for x. Explain. c. Use your estimate for x to evaluate the original expression and the simplifi ed expression in part (a). Are the values the same? 6(y + 8) = 6y + 8 ✗ 1 2 3 Help with Homework 138 Chapter 3 Algebraic Expressions and Properties 35. FITNESS Each day, you run on a treadmill for r minutes and lift weights for 15 minutes. Which expressions can you use to fi nd how many minutes of exercise you do in 5 days? Explain your reasoning.
5(r + 15) 5r + 5 ⋅ 15 5r + 15 r (5 + 15) 36. SPEED A cheetah can run 103 feet per second. A zebra can run x feet per second. Use the Distributive Property to write and simplify an expression for how much farther the cheetah can run in 10 seconds.
UNIFORMS Your baseball team has 16 players. Use the Distributive Property to write and simplify an expression for the total cost of buying the items shown for all the players.
37. Pants: $10 Belt: $x and or or 38. Jersey: $12 Socks: $4 or and and or Hat: $x Simplify the expression. 39. 6(x + 4) + 1 40. 5 + 8(3 + x) 41. 7(8 + 4k) + 12 42. x + 3 + 5x 43. 7y + 6 − 1 + 12y 44. w + w + 5w 45. 4d + 9 − d − 8 46. n + 3(n − 1) 47. 2v + 8v − 5v 48. 5(z + 4) + 5(2 − z) 49. 2.7(w − 5.2) 50. 2 — 3 y + 1 — 6 y + y 51. 3 — 4 ( z + 2 — 5 ) + 2z 52. 7(x + y) − 7x 53. 4x + 9y + 3(x + y) 54. ERROR ANALYSIS Describe and correct 8x − 2x + 5x = 8x − 7x = (8 − 7)x = x ✗ the error in simplifying the expression.
ALGEBRA Find the value of x that makes the expressions equivalent.
55. 4(x − 5); 32 − 20 56. 2(x + 9); 30 + 18 57. 7(8 − x); 56 − 21 58. REASONING Simplify the expressions and compare. What do you notice? Explain. 4(x + 6) (x + 6) + (x + 6) + (x + 6) + (x + 6) 5 Section 3.4 The Distributive Property 139 GEOMETRY Write and simplify expressions for the area and perimeter of the rectangle.
59. x à 8 8 60. 5.5 á x 12 61. 9 7 x 5 62. FUNDRAISER An art club sells 42 large candles and Price: $10 Cost: $x Profit â Price Ź Cost Price: $5 Cost: $y 56 small candles.
a. Use the Distributive Property to write and simplify an expression for the profi t.
b. A large candle costs $5, and a small candle costs $3. What is the club’s profi t?
63. REASONING Evaluate each expression by (1) using the Distributive Property and (2) evaluating inside the parentheses fi rst. Which method do you prefer? Is your preference the same for both expressions? Explain your reasoning.
a. 2(3.22 − 0.12) b. 12 ( 1 — 2 + 2 — 3 ) 64. REASONING Write and simplify an expression for the difference between the perimeters of the rectangle and the hexagon. Interpret your answer. 2x 2x á 7 x 2x x á 8 x á 6 x 2x 65. Add one set of parentheses to the expression 7 ⋅ x + 3 + 8 ⋅ x + 3 ⋅ x + 8 − 9 so that it is equivalent to 2(9x + 10).
Evaluate the expression. (Section 2.4, Section 2.5, and Section 2.6) 66. 4.871 + 7.4 − 1.63 67. 25.06 − 0.049 + 8.995 68. 15.3 ⋅ 9.1 − 4.017 69. 29.24 ÷ 3.4 ⋅ 0.045 70. MULTIPLE CHOICE What is the GCF of 48, 80, and 96? (Section 1.5) ○ A 12 ○ B 16 ○ C 24 ○ D 480 140 Chapter 3 Algebraic Expressions and Properties Lesson Tutorials Factoring Expressions Extension 3.4 Factoring an Expression Words Writing a numerical expression or algebraic expression as a product of factors is called factoring the expression. You can use the Distributive Property to factor expressions.
Numbers 3 ⋅ 7 + 3 ⋅ 2 = 3(7 + 2) Algebra ab + ac = a(b + c) 3 ⋅ 7 − 3 ⋅ 2 = 3(7 − 2) ab − ac = a(b − c) Which expression is not equivalent to 16x + 24? ○ A 2(8x + 12) ○ B 4(4x + 6) ○ C 6(3x + 4) ○ D (2x + 3)8 Each choice is a product of two factors in which one is a whole number and the other is the sum of two terms. For an expression to be equivalent to 16x + 24, its whole number factor must be a common factor of 16 and 24. Factors of 16: 1 , 2 , 4 , 8 , 16 Circle the common factors. Factors of 24: 1 , 2 , 3, 4 , 6, 8 , 12, 24 The common factors of 16 and 24 are 1, 2, 4, and 8. Because 6 is not a common factor of 16 and 24, Choice C cannot be equivalent to 16x + 24.
Check: 6(3x + 4) = 6(3x) + 6(4) = 18x + 24 ≠ 16x + 24 ✗ So, the correct answer is ○ C .
EXAMPLE Identifying Equivalent Expressions 2 Key Vocabulary factoring an expression, p. 140 Study Tip When you factor an expression, you can factor out any common factor.
Factor 20 − 12 using the GCF. Find the GCF of 20 and 12 by listing their factors. Factors of 20: 1 , 2 , 4 , 5, 10, 20 Circle the common factors. Factors of 12: 1 , 2 , 3, 4 , 6, 12 The GCF of 20 and 12 is 4.
Write each term of the expression as a product of the GCF and the remaining factor. Then use the Distributive Property to factor the expression. 20 − 12 = 4(5) − 4(3) Rewrite using GCF. = 4(5 − 3) Distributive Property EXAMPLE Factoring a Numerical Expression 1 Equivalent Expressions In this extension, you will ● use the Distributive Property to produce equivalent expressions.
Learning Standards MACC.6.NS.2.4 MACC.6.EE.1.3 MACC.6.EE.1.4 COMMON CORE Extension 3.4 Factoring Expressions 141 You receive a discount on each book you buy for your electronic reader. The original price of each book is x dollars. You buy 5 books for a total of (5x − 15) dollars. Factor the expression. What can you conclude about the discount?
Find the GCF of 5x and 15 by writing their prime factorizations. 5x = 5 ⋅ x Circle the common prime factor. 15 = 5 ⋅ 3 So, the GCF of 5x and 15 is 5. Use the GCF to factor the expression. 5x − 15 = 5(x) − 5(3) Rewrite using GCF. = 5(x − 3) Distributive Property The factor 5 represents the number of books purchased. The factor (x − 3) represents the price of each book. This factor is a difference of two terms, showing that the price x of each book is decreased by $3. So, the factored expression shows a $3 discount for every book you buy. The original expression shows a total savings of $15.
EXAMPLE Factoring an Algebraic Expression 3 Factor the expression using the GCF.
1. 7 + 14 2. 44 − 11 3. 18 − 12 4. 70 + 95 5. 60 − 36 6. 100 − 80 7. 84 + 28 8. 48 + 80 9. 2x + 10 10. 15x + 6 11. 26x − 13 12. 50x − 60 13. 36x + 9 14. 14x − 98 15. 10x − 25y 16. 24y + 88x 17. REASONING The whole numbers a and b are divisible by c. Is a + b divisible by c ? Is b − a divisible by c ? Explain your reasoning. 18. OPEN-ENDED Write fi ve expressions that are equivalent to 8x + 16. 19. GEOMETRY The area of the parallelogram is 4 ft (4x + 16) square feet. Write an expression for the base. 20. STRUCTURE You buy 37 concert tickets for $8 each, and then sell all 37 tickets for $11 each. The work below shows two ways you can determine your profi t. Describe each solution method. Which do you prefer? Explain your reasoning. profi t = 37(11) − (37)8 = 407 − 296 = $111 profi t = 37(11) − (37)8 = 37(11 − 8) = 37(3) = $111 Quiz 3.3–3.4 142 Chapter 3 Algebraic Expressions and Properties Tell which property the statement illustrates. (Section 3.3) 1. 3.5 ⋅ z = z ⋅ 3.5 2. 14 + (35 + w) = (14 + 35) + w Simplify the expression. Explain each step. (Section 3.3) 3. 3.2 + (b + 5.7) 4. 6 ⋅ (10 ⋅ k) Use the Distributive Property and mental math to fi nd the product. (Section 3.4) 5. 6 × 49 6. 7 × 86 Use the Distributive Property to simplify the expression. (Section 3.4) 7. 5(x − 8) 8. 7(y + 3) Simplify the expression. (Section 3.4) 9. 6q + 2 + 3q + 5 10. 4r + 3(r − 2) Factor the expression using the GCF. (Section 3.4) 11. 12 + 21 12. 16x − 36 13. GEOMETRY The expression 18 + 7 + (18 + 2x) + 7 represents the perimeter of the trapezoid. Simplify the expression. (Section 3.3) 14. MOVIES You and four of your friends go to a movie and each buy popcorn. (Section 3.4) a. Use the Distributive Property to write an expression for the total cost to buy movie tickets and popcorn. Simplify the expression. b. Choose a reasonable value for x. Evaluate the expression. 15. GEOMETRY The length of a rectangle is 16 inches, and its area is (32x + 48) square inches. Factor the expression for the area. Write an expression for the width. (Section 3.4) Progress Check 18 à 2x 18 7 7 Movie Tickets Snacks Movie Tickets Snacks Student $8 Adult $10 Candy $3 Popcorn $x Chapter Review 143 3.2 Writing Expressions (pp. 118–123) Write the phrase as an expression. a. a number z decreased by 18 z − 18 The phrase decreased by means subtraction. b. the sum of 7 and the product of a number x and 12 7 + 12x The phrase sum of means addition. The phrase product of means multiplication.
3 Chapter Review Review Key Vocabulary Review Examples and Exercises algebraic expression, p. 112 terms, p. 112 variable, p. 112 coeffi cient, p. 112 constant, p. 112 equivalent expressions, p. 128 like terms, p. 136 factoring an expression, p. 140 Vocabulary Help a. Evaluate a ÷ b when a = 48 and b = 8. a ÷ b = 48 ÷ 8 Substitute 48 for a and 8 for b. = 6 Divide 48 by 8. b. Evaluate y2 − 14 when y = 5. y 2 − 14 = 52 − 14 Substitute 5 for y. = 25 − 14 Using order of operations, evaluate 52. = 11 Subtract 14 from 25.
Evaluate the expression when x = 20 and y = 4. 1. x ÷ 5 2. y + x 3. 8y − x 4. GAMING In a video game, you score p game points and b triple bonus points. An expression for your score is p + 3b. What is your score when you earn 245 game points and 20 triple bonus points?
3.1 3.1 Algebraic Expressions (pp. 110–117) i s 144 Chapter 3 Algebraic Expressions and Properties 3.3 3.3 Properties of Addition and Multiplication (pp. 126–131) a. Simplify the expression (x + 18) + 4. (x + 18) + 4 = x + (18 + 4) Associative Property of Addition = x + 22 Add 18 and 4. b. Simplify the expression (5.2 + a) + 0. (5.2 + a) + 0 = 5.2 + (a + 0) Associative Property of Addition = 5.2 + a Addition Property of Zero c. Simplify the expression 36 ⋅ r ⋅ 1. 36 ⋅ r ⋅ 1 = 36 ⋅ (r ⋅ 1) Associative Property of Multiplication = 36 ⋅ r Multiplication Property of One = 36r Simplify the expression. Explain each step. 10. 10 + (2 + y) 11. (21 + b) + 1 12. 3(7x) 13. 1(3.2w) 14. 5.3 + (w + 1.2) 15. (0 + t) + 9 16. GEOMETRY The expression 7 + 3x + 4 represents the perimeter of the triangle. Simplify the expression.
3x 7 4 Write the phrase as an expression. 5. 11 fewer than a number b 6. the product of a number d and 32 7. 18 added to a number n 8. a number t decreased by 17 9. BASKETBALL Your basketball team scored 4 fewer than twice as many points as the other team. a. Write an expression for the number of points your team scored. b. The other team scored 24 points. How many points did your team score? Chapter Review 145 3.4 3.4 The Distributive Property (pp. 132–141) a. Use the Distributive Property to simplify 3(n + 9). 3(n + 9) = 3(n) + 3(9) Distributive Property = 3n + 27 Multiply. b. Simplify 5x + 7 + 3x − 2. 5x + 7 + 3x − 2 = 5x + 3x + 7− 2 Commutative Property of Addition = (5 + 3)x + 7 − 2 Distributive Property = 8x + 5 Simplify. c. Factor 14x − 49 using the GCF. Find the GCF of 14x and 49 by writing their prime factorizations. 14x = 2 ⋅ 7 ⋅ x Circle the common prime factor. 49 = 7 ⋅ 7 So, the GCF of 14x and 49 is 7. Use the GCF to factor the expression. 14x − 49 = 7(2x) − 7(7) Rewrite using GCF. = 7(2x − 7) Distributive Property Use the Distributive Property to fi nd the product. 17. 3 — 4 × 2 1 — 3 18. 4 — 7 × 4 5 — 8 19. 1 — 5 × 5 10 — 11 Use the Distributive Property to simplify the expression. 20. 2(x + 12) 21. 11(b − 3) 22. 8(s − 1) 23. 6(6 + y) 24. 25(z − 4) 25. 35(w − 2) 26. HAIRCUT A family of four goes to a salon for haircuts. The cost of each haircut is $13. Use the Distributive Property and mental math to fi nd the product 4 × 13 for the total cost.
Simplify the expression. 27. 5(n + 3) + 4n 28. t + 2 + 6t 29. 3z + 4 + 5z − 9 Factor the expression using the GCF. 30. 15 + 35 31. 36x − 28 32. 16x + 56y 146 Chapter 3 Algebraic Expressions and Properties Evaluate the expression when a = 6 and b = 8. 1. 4 + a 2. a − 6 3. ab Write the phrase as an expression. 4. twice a number x 5. 25 more than 50 6. 40 divided by 5 Simplify the expression. Explain each step. 7. 3.1 + (8.6 + m) 8. (10 ⋅ n) ⋅ 7 9. 3(15w) Use the Distributive Property to simplify the expression. 10. 4(x + 8) 11. 12( y − 5) Simplify the expression. 12. 4(q + 2) − 6 13. 3(2 + 5r) + 11 14. s + 3s + 4s 15. 4t − 2 − 2t + 7 Factor the expression using the GCF. 16. 18 + 24 17. 40 − 16 18. 15x + 20 19. 32x − 40y 20. SOCCER GAME Playing time is added at the end of a soccer game to make up for stoppages. An expression for the length of a 90-minute soccer game with x minutes of stoppage time is 90 + x. How long is a game with 4 minutes of stoppage time? 21. GEOMETRY The expression 15 ⋅ x ⋅ 6 represents the volume of a rectangular prism with a length of 15, a width of x, and a height of 6. Simplify the expression. 22. PARTY FAVORS You make party favors for an event. You tie 9 inches of ribbon around each party favor. Write an expression for the amount of ribbon you need for n party favors. The ribbon costs $3 for each yard. Write an expression for the total cost of the ribbon.
1 Chapter Test 3 Test Practice 1. The student council is organizing a school fair. Council members are making signs to show the prices for admission and for each game a person can play. Let x represent the number of games. Which expression can you use to determine the total amount, in dollars, a person pays for admission and playing x games? (MACC.6.EE.1.2a) A. 2.25 C. 2 + 0.25x B. 2.25x D. 2x + 0.25 2. Which property does the equation below represent? (MACC.6.EE.1.3) 17 ⋅ 44 + 17 ⋅ 56 = 17 ⋅ 100 F. Distributive Property H. Associative Property of Multiplication G. Multiplication Property of One I. Commutative Property of Multiplication 3. At a used book store, you can purchase two types of books. You can use the expression 3h + 2p to fi nd the total cost for h hardcover books and p paperback books. What is the total cost, in dollars, for 6 hardcover books and 4 paperback books? (MACC.6.EE.1.2c) 4. What is the value of 9.6 × 12.643? (MACC.6.NS.2.3) A. 12.13728 C. 1213.728 B. 121.3728 D. 12,137.28 Standards Assessment 147 Standards Assessment 3 Test-Taking Strategy After Answering Easy Questions, Relax “After answering easy questions, relax and try the harder ones. For this, 3(4) = 12. So, it is B.” SCHOOL FAIR SCHOOL FAIR Admission Price per game $2.00 $0.25 Hardcover Books - $3 Paperback Books - $2 3 148 Chapter 3 Algebraic Expressions and Properties 5. What is the value of 4.391 + 5.954? (MACC.6.NS.2.3) F. 9.12145 H. 9.345 G. 9.245 I. 10.345 6. Which number pair has a greatest common factor of 6? (MACC.6.NS.2.4) A. 18, 54 C. 30, 60 B. 30, 42 D. 36, 60 7. Properties of Addition and Multiplication are used to simplify an expression. 36 ⋅ 23 + 33 ⋅ 64 = 36 ⋅ 23 + 64 ⋅ 33 = 36 ⋅ 23 + 64 ⋅ (23 + 10) = 36 ⋅ 23 + 64 ⋅ 23 + 64 ⋅ 10 = x ⋅ 23 + 64 ⋅ 10 = 2300 + 640 = 2940 What number belongs in place of the x? (MACC.6.EE.1.3) 8. Which property was used to simplify the expression? (MACC.6.EE.1.3) (47 × 125) × 8 = 47 × (125 × 8) = 47 × 1000 = 47,000 F. Distributive Property G. Multiplication Property of One H. Associative Property of Multiplication I. Commutative Property of Multiplication 9. What is the value of the expression below when a = 5, b = 7, and c = 6? (MACC.6.EE.1.2c) 9b − 4a + 2c A. 29 C. 55 B. 31 D. 78 Standards Assessment 149 10. Which equation correctly demonstrates the Distributive Property? (MACC.6.EE.1.4) F. a(b + c) = ab + c G. a(b + c) = ab + ac H. a + (b + c) = (a + b) + (a + c) I. a + (b + c) = (a + b) ⋅ (a + c) 11. Which expression is equivalent to 3 3 — 5 ÷ 6 1 — 2 ? (MACC.6.NS.1.1) A. 5 — 18 × 13 — 2 C. 9 — 5 ÷ 6 — 2 B. 18 — 5 × 2 — 13 D. 18 — 5 ÷ 2 — 13 12. Which number pair does not have a least common multiple of 24? (MACC.6.NS.2.4) F. 2, 12 H. 6, 8 G. 3, 8 I. 12, 24 13. Use the Properties of Multiplication to simplify the expression in an effi cient way. Show your work and explain how you used the Properties of Multiplication. (MACC.6.EE.1.3) (25 × 18) × 4 14. You evaluated an expression using x = 6 and y = 9. You correctly got an answer of 105. Which expression did you evaluate? (MACC.6.EE.1.2c) A. 3x + 6y C. 6x + 9y B. 5x + 10y D. 10x + 5y 15. Which number is equivalent to the expression below? (MACC.6.EE.1.1) 2 × 12 − 8 ÷ 22 F. 2 H. 8 G. 4 I. 22 |
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Properties of Rational Numbers
The properties of rational numbers help us to distinguish them from the other types of numbers. Rational numbers consist of integers, whole numbers, and natural numbers. They can be represented in the form of a fraction p/q or as terminating decimal numbers, or as non-terminating but repeating decimal numbers. The properties of rational numbers include the associative property, the commutative property, the distributive property, and the closure property. Let us read about all the properties of rational numbers on this page.
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| 1. | What are the Properties of Rational Numbers? |
| 2. | Closure Property of Rational Numbers |
| 3. | Commutative Property of Rational Numbers |
| 4. | Associative Property of Rational Numbers |
| 5. | Distributive Property of Rational Numbers |
| 6. | Additive Property of Rational Numbers |
| 7. | Multiplicative Property of Rational Numbers |
| 8. | FAQs on Properties of Rational Numbers |
What are the Properties of Rational Numbers?
When numbers can be expressed in the form of p/q, then they are considered to be rational numbers, here both p and q are integers and q ≠ 0. There are six properties of rational numbers, which are listed below:
Closure Property
Commutative Property
Associative Property
Distributive Property
Multiplicative Property
Additive Property
Let us explore these properties on the four arithmetic operations (Addition, subtraction, multiplication, and division) in Mathematics.
Closure Property of Rational Numbers
The closure property of rational numbers states that when any two rational numbers are added, subtracted, or multiplied, the result of all three cases will also be a rational number. Let us read about how the closure property of rational numbers works on all the basic arithmetic operations. We will understand this property on each operation using various examples.
Let us take two rational numbers 1/3 and 1/4, and perform basic arithmetic operations on them.
For Addition: 1/3 + 1/4 = (4 + 3)/12 = 7/12. Here, the result is 7/12, which is a rational number. We say that rational numbers are closed under addition. That is, for any two rational numbers a and b, (a + b) is also a rational number.
For Subtraction: 1/3 - 1/4 = (4 - 3)/12 = 1/12. Here, the result is 1/12, which is a rational number. We say that rational numbers are closed under subtraction. That is, for any two rational numbers a and b, (a - b) is also a rational number.
For Multiplication: 1/3 × 1/4 = 1/12. Here, the result is 1/12, which is a rational number. We say that rational numbers are closed under multiplication. That is, for any two rational numbers a and b, (a × b) is also a rational number.
For Division: 1/3 ÷ 1/4 = 4/3. Here, the result is 4/3, which is a rational number. But we find that for any rational number a, a ÷ 0 is not defined. So rational numbers are not closed under division. However, if we exclude zero then the collection of all other rational numbers are closed under division.
Commutative Property of Rational Numbers
The commutative property of rational numbers states that when any two rational numbers are added or multiplied in any order it does not change the result. But in the case of subtraction and division if the order of the numbers is changed then the result will also change. We will understand this property on each operation using various illustrations.
Let us again take two rational numbers 1/3 and 1/4, and perform basic arithmetic operations on them.
For Addition: 1/3 + 1/4 = 1/4 + 1/3 = 7/12. We say that addition is commutative for rational numbers. That is, for any two rational numbers a and b, a + b = b + a.
For Subtraction: 1/3 - 1/4 ≠ 1/4 - 1/3 = 1/12 ≠ -1/12. We can see that subtraction is not commutative for rational numbers. That is, for any two rational numbers a and b, a - b ≠ b - a.
For Multiplication: 1/3 × 1/4 = 1/4 × 1/3 = 1/12. We can see that multiplication is commutative for rational numbers. This means, a × b = b × a for any two rational numbers a and b.
For Division: 1/3 ÷ 1/4 ≠ 1/4 ÷ 1/3 because 4/3 ≠ 3/4. We can see that the expressions on both sides are not equal. This means, a ÷ b ≠ b ÷ a for any two rational numbers a and b. So division is not commutative for rational numbers.
Associative Property of Rational Numbers
The associative property of rational numbers states that when any three rational numbers are added or multiplied the result remains the same irrespective of the way numbers are grouped. However, in the case of subtraction and division if the order of the numbers is changed then the result will also change. We will understand this property on each operation using various illustrations.
For Addition: For any three rational numbers, the associative property for addition is expressed as A, B, and C, (A + B) + C = A + (B + C). For example, (1/3 + 1/4) + 1/2 = 1/4 + (1/3 + 1/2) = 13/12. We say that addition is associative for rational numbers.
For Subtraction: For any three rational numbers, the associative property for subtraction is expressed as A, B, and C, (A - B) - C ≠ A - (B - C). For example, (1/3 - 1/4) - 1/2 ≠ 1/3 - (1/4 - 1/2). We can see that subtraction is not associative for rational numbers.
For Multiplication: For any three rational numbers, the associative property for multiplication is expressed as A, B, and C, (A × B) × C = A × (B × C). For example, (1/3 × 1/4) × 1/2 = 1/4 × (1/3 × 1/2) = 1/24 = 1/24. We can see that multiplication is associative for rational numbers.
For Division: For any three rational numbers, the associative property for division is expressed as A, B, and C, (A ÷ B) ÷ C ≠ A ÷ (B ÷ C). For example, (1/3 ÷ 1/4) ÷ 1/2 ≠ 1/4 ÷ (1/3 ÷ 1/2) because 8/3 ≠ 3/8. We can see that the expressions on both sides are not equal. So division is not associative for rational numbers.
Distributive Property of Rational Numbers
The distributive property of rational numbers states that if any expression with three rational numbers A, B, and C is given in form A (B + C), then it can be solved as A × (B + C) = AB + AC. This applies to subtraction also which means A (B - C) = AB - AC. This means operand A is distributed between the other two operands, i.e., B and C. This property is also known as the distributive property of multiplication over addition or subtraction. Let us learn how the distributive property of rational numbers works. We will understand this property using the illustration given below.
Example: Solve 1/2(1/6 + 1/5)
Solution: The given expression is of the form A (B + C) = A × (B + C) = AB + AC
1/2(1/6 + 1/5) = (1/2 × 1/6) + (1/2 × 1/5) = 11/60
Let us solve the same expression with subtraction.
Example: Solve 1/2(1/6 - 1/5)
Solution: The given expression is of the form A (B - C) = A × (B - C) = AB - AC
1/2(1/6 - 1/5) = (1/2 × 1/6) - (1/2 × 1/5) = -1/60
Additive Property of Rational Numbers
There are two basic additive properties of rational numbers, the additive identity property and the additive inverse property. For any rational number a/b, where b ≠ 0 these two properties are illustrated below.
Let us understand the additive identity property and the additive inverse property with the help of examples.
Additive Identity Property
The additive identity property of rational numbers states that the sum of any rational number (a/b) and zero is the rational number itself. Suppose a/b is any rational number, then a/b + 0 = 0 + a/b = a/b. Here, 0 is the additive identity for rational numbers. Let us understand this with an example:
3/7 + 0 = 0 + 3/7 = 3/7
Additive Inverse Property
The additive inverse property of rational numbers states that if a/b is a rational number, then there exists a rational number (-a/b) such that, a/b + (-a/b) = (-a/b) + a/b = 0.
For example, the additive inverse of 3/7 is (-3/7).
(3/7) + (-3/7) = (-3/7) + 3/7 = 0.
Multiplicative Property of Rational Numbers
There are two basic multiplicative properties of rational numbers, the multiplicative identity property, and the multiplicative inverse property. Let us understand these properties with examples.
Multiplicative Identity Property
The additive identity property of rational numbers states that the product of any rational number and 1 is the rational number itself. Here, 1 is the multiplicative identity for rational numbers. If a/b is any rational number, then a/b × 1 = 1 × a/b = a/b. For example: 5/3 × 1 = 1 × 5/3 = 5/3.
Multiplicative Inverse Property
The multiplicative inverse property of rational numbers states that for every rational number a/b, b ≠ 0, there exists a rational number b/a such that a/b × b/a = 1. In this case, a rational number b/a is the multiplicative inverse of a rational number a/b. For example, the multiplicative inverse of 7/3 is 3/7. (7/3 × 3/7 = 1).
Note: Every rational number multiplied with 0 gives 0. If a/b is any rational number, then a/b × 0 = 0 × a/b = 0. For example, 7/2 × 0 = 0 × 7/2 = 0.
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Examples on Properties of Rational Numbers
Example 1: Fill in the blanks using the properties of rational numbers.
a) 2/3 + 1/6 = __ + 2/3
b) 21 × 23 × 32 = 32 × __ × 23
Solution: Using the commutative property of rational numbers, we can fill in the blanks.
a) 2/3 + 1/6 = 1/6 + 2/3
b) 21 × 23 × 32 = 32 × 21 × 23
2. Example 2: Solve 7/2(1/6 + 1/4) by using the distributive property of rational numbers.
Solution:
Using the distributive property of rational numbers let us write the given expression in the form A (B + C) = A × (B + C) = AB + AC
= 7/2(1/6 + 1/4)
= 7/2 × (1/6 + 1/4)
= (7/2 × 1/6) + (7/2 × 1/4)
= (7/12) + (7/8)
= 35/24
3. Example 3: If 8/3 × (7/6 × 5/4) = 35/9, then find the product of (8/3 × 7/6) × 5/4.
Solution:
The associative property of rational numbers says that for any three rational numbers (A, B, and C) expression can be expressed as (A × B) × C = A × (B × C)
Given = 8/3 × (7/6 × 5/4) = 35/9
Using the associative property of rational numbers, we can conclude that (8/3 × 7/6) × 5/4 is also equal to 35/9.
To verify this, first, let us solve the terms inside the brackets.
(8/3 × 7/6) × 5/4
= 56/18 × 5/4
= 35/9
Hence, 8/3 × (7/6 × 5/4) = (8/3 × 7/6) × 5/4 = 35/9.
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FAQs on Properties of Rational Numbers
What are the Six Important Properties of Rational Numbers?
The six major properties of rational numbers are listed below:
Closure Property
Commutative Property
Associative Property
Distributive Property
Multiplicative Property
Additive Property
What is the Distributive Property of Rational Numbers?
The distributive property states, if p, q, and r are three rational numbers, then the relation between the three is given as, p × (q + r) = (p × q) + (p × r). For example, 1/3(1/2 + 1/5) = (1/3 × 1/2) + (1/3 × 1/5) = 7/30. This property is also known as the distributivity of multiplication over addition. This property is also applicable to subtraction which says p × (q - r) = (p × q) - (p × r). For example, 1/3(1/2 - 1/5) = (1/3 × 1/2) - (1/3 × 1/5) = 1/10.
The Commutative Property of Rational Numbers is Applicable on Which Two Operations?
The commutative property of rational numbers is applicable for addition and multiplication. Example, for addition 1/6 + 1/4 = 1/4 + 1/6 = 5/12, and for multiplication 1/3 × 1/7 = 1/7 × 1/3 = 1/21.
What are the Two Multiplicative Properties of Rational Numbers?
The two basic multiplicative properties of rational numbers are the multiplicative identity property and the multiplicative inverse property. Let us understand the two with examples.
Multiplicative identity for rational numbers is expressed as, p/q × 1 = 1 × p/q = p/q. For example: 5/4 × 1 = 1 × 5/4 = 5/4.
Multiplicative Inverse for rational numbers is expressed as p/q × q/p = 1 such that p/q is the multiplicative inverse of q/p. For example, the multiplicative inverse of 7/4 is 4/7. (7/4 × 4/7 = 1).
What is the Difference Between the Associative Property and the Commutative Property of Rational Numbers?
The commutative property of rational numbers says that A + B = B + A (here, A and B are rational numbers in a form of p/q), and on the other hand, the associative property of rational numbers states that (A + B) + C = A + (B + C) (here, A, B, and C are rational numbers in a form of p/q).
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3194 | https://simple.wikipedia.org/wiki/Blackbody_radiation | Blackbody radiation - Simple English Wikipedia, the free encyclopedia
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The curve of blackbody radiation for highest (blue), intermediate, and less-high temperatures (red)
Blackbody radiation is radiation produced by heated objects, particularly from a blackbody. A blackbody is an object that absorbs all radiation (visible light, infrared light, ultraviolet light, etc.) that falls on it. This also means that it will also radiate at all frequencies that heat energy produces in it.
Everything glows, depending on its temperature. Hotter things glow more in shorter wavelengths. Cooler things don't glow so much, especially in short wavelengths. Partly the radiation wavelength depends on what the material is, and partly it doesn't. The part that only depends on temperature, and not on composition, is called blackbody radiation.
Quantum wave
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This part of the story of radiation was first explained by James Clerk Maxwell via wave theory but the predicted and actual intensity vs. frequency curves did not go together right. At higher frequencies classical physics predicted that more and more energy would be radiated from the body until the energy became infinite. This broke the first law of thermodynamics which is a fundamental part of all physics. This was called the ultraviolet catastrophe.
When it was realised that classical physics did not work for blackbody radiation, the German physicist Max Planck explained their relationship by saying that there are individual things (he did not try to guess what kind of things) that vibrate, each at its frequency. Each wave of each frequency has its special energy level. A single x-ray is very high photon energy and can go right through the human body.
A single wave or photon of infrared light is very low energy, cannot go through the human body, and can only warm it. Planck's good thinking was to realize that to get a single wave at the x-ray frequency, it was necessary to have a big enough package of energy (or "quantum") to make such a strong wave. So if a blackbody took in a single wave at the x-ray frequency, then it could give off an x-ray at some later time. But if the blackbody only took in infra-red light it would not matter how much of it was absorbed.
It could only give off infrared light and could not give off even ordinary red light, much less any higher energy light such as ultraviolet light or x-ray radiation. Planck said that the total energy given off by a blackbody at any particular frequency is equal to the number of the "vibrating things" (see above), n, that was vibrating at a given frequency, f times a special constant, h, that turns frequency units into energy units. The equation is:
E = n h f
The constant he made, h, is called the Planck constant.
The idea that a unit of light at a given frequency always has the same energy, the idea that there is a quantum of energy for each unit of light at a given frequency, became the doorway into quantum mechanics, so the idea of a blackbody is something that is basic to modern physics. It shows up in discussions of a wide variety of physics topics having to do with energies and frequencies.
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3195 | https://groupprops.subwiki.org/wiki/Projective_special_linear_group:PSL(2,Z) | Projective special linear group:PSL(2,Z) - Groupprops
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This article is about a particular group, i.e., a group unique upto isomorphism. View specific information (such as linear representation theory, subgroup structure) about this group
View a complete list of particular groups (this is a very huge list!)[SHOW MORE])
VIEW FACTS ABOUT THIS GROUP: All factsGroup property satisfactionsSubgroup property satisfactionsGroup property dissatisfactionsSubgroup property satisfactionsVIEW DEFINITIONS USING THIS AS EXAMPLE: All terms |Group properties |Subgroup properties|VIEW FACTS USING THIS AS EXAMPLE: All facts |Group property non-implications |Subgroup property non-implications |Group metaproperty dissatisfactionsSubgroup metaproperty dissatisfactions
Definition
The group P S L(2,Z){\displaystyle PSL(2,\mathbb {Z} )}, also sometimes called the modular group, is defined in the following equivalent ways:
It is the projective special linear group of degree two over the ring of integers. In other words, it is the quotient of the special linear group:SL(2,Z) by the subgroup ±I{\displaystyle \pm I}.
It is the inner automorphism group of braid group:B3, i.e., the quotient of B 3{\displaystyle B_{3}} by its center.
It is the free product of cyclic group:Z2 and cyclic group:Z3.
Definition by presentation
The group can be defined by the following presentation, where 1{\displaystyle 1} denotes the identity element:
As a projective special linear group: ⟨S,T∣S 2=1,(S T)3=1⟩{\displaystyle \langle S,T\mid S^{2}=1,(ST)^{3}=1\rangle } where S{\displaystyle S} is the image of the matrix (0−1 1 0){\displaystyle {\begin{pmatrix}0&-1\1&0\\end{pmatrix}}} and T{\displaystyle T} is the image of the matrix (1 1 0 1){\displaystyle {\begin{pmatrix}1&1\0&1\\end{pmatrix}}}.
As a matrix group: ⟨x,y∣x 2=y 3=1⟩{\displaystyle \langle x,y\mid x^{2}=y^{3}=1\rangle }
| Function | Value | Explanation |
---
| order | infinite (countable) | As P S L(2,Z){\displaystyle PSL(2,\mathbb {Z} )}: The group is the quotient of the countably infinite group S L(2,Z){\displaystyle SL(2,\mathbb {Z} )} by its center, which is a subgroup of order two. As a free product: any free product of nontrivial finite groups is countably infinite. |
| exponent | infinite | As P S L(2,Z){\displaystyle PSL(2,\mathbb {Z} )}: The group is the quotient of S L(2,Z){\displaystyle SL(2,\mathbb {Z} )}, which has infinite exponent, by a finite subgroup. More explicitly, the image of (1 1 0 1){\displaystyle {\begin{pmatrix}1&1\0&1\\end{pmatrix}}} has infinite order. |
| minimum size of generating set | 2 | As P S L(2,Z){\displaystyle PSL(2,\mathbb {Z} )}: Follows from S L(2,Z){\displaystyle SL(2,\mathbb {Z} )} being 2-generated. As inner automorphism group of braid group B 3{\displaystyle B_{3}}: Follows from B 3{\displaystyle B_{3}} being 2-generated. As free product of cyclic group:Z2 and cyclic group:Z3: Follows because each of the free factors is cyclic, so we get a generating set of size 2. NOTE: In all interpretations, we can rule out the possibility of a generating set of size 1 because cyclic implies abelian and the group is non-abelian. |
| subgroup rank | infinite (countable) | Use that S L(2,Z){\displaystyle SL(2,\mathbb {Z} )} has infinite subgroup rank |
Group properties
| Property | Satisfied? | Explanation | Corollary properties satisfied/dissatisfied |
--- --- |
| 2-generated group | Yes | See explanation for minimum size of generating set above | satisfies: finitely generated group, countable group |
| Noetherian group | No | See explanation for subgroup rank above | |
| finitely presented group | Yes | Any of the definitions (P S L{\displaystyle PSL}, free product) gives a finite presentation | |
| solvable group | No | contains subgroup isomorphic to free group:F2 -- see Sanov subgroup in SL(2,Z) is free of rank two | dissatisfies: nilpotent group, abelian group |
| group satisfying no nontrivial identity | Yes | contains subgroup isomorphic to free group:F2 -- see Sanov subgroup in SL(2,Z) is free of rank two and note that the image in P S L(2,Z){\displaystyle PSL(2,\mathbb {Z} )} of the Sanov subgroup is isomorphic to it. | |
| SQ-universal group | Yes | contains subgroup isomorphic to free group:F2 -- see Sanov subgroup in SL(2,Z) is free of rank two and note that the image in P S L(2,Z){\displaystyle PSL(2,\mathbb {Z} )} of the Sanov subgroup is isomorphic to it. | |
| residually finite group | Yes | The kernels of the homomorphisms P S L(2,Z)→P S L(2,Z/n Z){\displaystyle PSL(2,\mathbb {Z} )\to PSL(2,\mathbb {Z} /n\mathbb {Z} )} for natural numbers n{\displaystyle n} are normal subgroups of finite index and their intersection is trivial. | satisfies: finitely generated residually finite group |
| Hopfian group | Yes | Follows from finitely generated and residually finite implies Hopfian | satisfies: finitely generated Hopfian group |
GAP implementation
| Description | Functions used |
--- |
| FreeProduct(CyclicGroup(2),CyclicGroup(3)) | FreeProduct, CyclicGroup |
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3196 | https://www.sciencedirect.com/science/article/pii/S0022314X15002139 | Bits of 3n in binary, Wieferich primes and a conjecture of Erdős - ScienceDirect
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Journal of Number Theory
Volume 158, January 2016, Pages 268-280
Bits of 3 n in binary, Wieferich primes and a conjecture of Erdős
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Let p and q be distinct primes. We show that digits of the base q expansions of p n are equidistributed on average (averaging over n). More precisely, for fixed m, we first prove a result for the first m q-adic bits of p n (averaging over n), then taking the large m limit we show equidistribution. A non-averaged version of this result would imply a conjecture of Erdős which states that there are only finitely many n such that the base 3 expansion of 2 n omits a 2. We prove our results by proving a nonexistence theorem for “higher Wieferich primes”.
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Keywords
Wieferich primes
Elementary number theory
Erdős
Units
1. Introduction
In [Erd79] Erdős conjectured that there are only finitely many powers of 2 whose ternary expansion omits a 2. We will refer to this conjecture as “Erdős' Conjecture”.
Progress towards this conjecture has been in the form of upper bounds on the function N(X)=#{n≤X:(2 n)3 omits a 2}, which, according to Erdős' conjecture, should be constant on an infinite interval. We explain the notation: for a prime q and a number N we will let (N)q denote the base q expansion of N. We view a base q expansion as a string of numbers from the set {0,1,…,q−1}. For example (3)2=11. The best known bound on N(X) is due to Narkiewicz [Nar80] who showed N(X)≤1.62 X α 0, where α 0=log 3(2)≈0.630. We refer the reader to [Lag09] for readable proofs and Narkiewicz type bounds for certain dynamical generalizations of this problem. See in particular [Lag09, Theorem 1.4, Proof on p. 20 of arxiv version]. See [Lag09, Conjecture E] for a refinement of Erdős' conjecture.
For p and q distinct primes, the present paper studies the structure of (p n)q as q→∞. Computer experimentation has led the authors to believe that base q digits of (p n)q are equidistributed as n→∞. We will now formalize this statement: for a∈{0,…,q−1} let d n(a) be the number of a's appearing in (p n)q.
Conjecture 1
For all p and q distinct primes and every a∈{0,…,q−1},(1.1)lim n→∞d n(a)n log q(p)=1 q.
Remark 2
The equidistribution statement (1.1) in the case p=2 and q=3 implies Erdős' conjecture. To see this, one argues by contrapositive: Suppose Erdős' conjecture was false. This says 0 is a limit point of the sequence {d n(2)}n≥0. This implies equation (1.1) is false.
The observation of Conjecture 1 and its implication of the Erdős' conjecture appears to be absent in the literature. When p=3 and q=2, Conjecture 1 says that the 0's and 1's appearing (3 n)2 are equidistributed as n→∞. Table 1 contains the first several members of the sequence {(3 n)2}n≥0.
Table 1. The first few values of d n(a) for p=3 and q=2.
| n | 3 n | (3 n)2 | d n(0) | d n(1) |
--- ---
| 0 | 1 | 1 | 0 | 1 |
| 1 | 3 | 11 | 0 | 2 |
| 2 | 9 | 1001 | 2 | 2 |
| 3 | 27 | 11011 | 1 | 4 |
| 4 | 81 | 1010001 | 4 | 3 |
For p=2 and q=3 the graph of {d n(2)log 3(2 n)}n≥1 is provided in Fig. 1.
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Fig. 1. The proportion of 2's in (2 n)3.
The present paper proves an averaged version Conjecture 1. Before stating our result we fix some notation. Fix distinct primes p and q, a natural number m and a∈{0,…,q−1}. If m≤⌈log q(p n)⌉ defines d n,m(a) and d n.m′(a) to be the numbers of a's in the first m digits and remaining digits of (p n)q respectively,1 so that d n(a)=d n,m(a)+d n,m′(a). Note that d n,m(a)=d n(a) when m=⌈log q(p n)⌉, the number of base q digits in (p n)q. If m>⌈log q(p n)⌉ then we will let d n,m(a)=d n(a) and d n,m′(a)=0. Concretely, if we write p n=a 0+a 1 q+⋯+a N q N, where N=⌊log q(p n)⌋ and a i∈{0,1,…,q−1} for 0≤i≤N then d n,m(a)=#{i:0≤i<m,a i=a}. In this paper we prove the following result.
Theorem 3
Let p and q be distinct primes and let a∈{0,…,q−1}.
(1)For every m≥1 we have(1.2)lim N→∞1 N∑n=1 N d n,m(a)m=1 h m∑n=1 h m d n,m(a)m where h m=#H m and H m=〈p〉⊂(Z/q m)×.
(2)In view of(1.2), the average proportion of a's in the first m digits is A m(a):=1 h m∑n=1 h m d n,m(a)m . In the limit we have lim m→∞A m(a)=1/q.
The proof of Theorem 3 uses a theorem about (Z/q m)× and makes contact with the theory of so-called Wieferich primes. We introduce some terminology to explain the main lemma used to prove Theorem 3.
Definition 4
A prime q is called (classical) Wieferich if one of the following equivalent conditions holds:
(1)2 q−1≡1 modulo q 2.
(2)The multiplicative order of 2 in (Z/q 2)× is q−1.
(3)The multiplicative group generated by 2 modulo q is isomorphic to the multiplicative group generated by 2 modulo q 2.
Such primes were first investigated in [Wie09] in connection to Fermat's Last Theorem. There he proved that if x q+y q=z q is a Fermat triple then q is Wieferich. It is an open problem whether there exist infinitely many Wieferich primes (even assuming the ABC conjecture). The infinitude of non-Wieferich primes is implied by the ABC conjecture [Sil88] and the distribution of Wieferich primes known relative to the Crandall–Dilcher–Pomerance conjecture which is a version of the Lang–Trotter conjectures. A review of these facts can be found in [Lan90, p. 42]. More details on the derivation of the asymptotic can be found in [Kat14, Section 2]. We refer the reader to [CDP97] for details on numerical searches for Wieferich primes.
We generalize the notion of a Wieferich prime for the purposes of our discussion.
Definition 5
Let p and q be distinct primes. Let's call a prime q p-Wieferich at r if the multiplicative group generated by p modulo q r is isomorphic to the group generated by p modulo q r+1.
In this notation classical Wieferich primes are simply 2-Wieferich primes at 2. Note that Table 1 for example shows that 2 is 3-Wieferich at 3 since the third column of digits is all zeros.
We can now state our main lemma which we used to prove Theorem 3.
Theorem 6
Let p and q be distinct primes.#{n:q is p-Wieferich at n}<∞.
This theorem appears in the body as Theorem 12. The proof depends on a modest generalization of the standard structure theorem for q-adic unit groups Z q× for which the authors could not find an adequate reference and hence have provided. In particular we show that for m sufficiently large the groups generated by p modulo q m contain subquotients of the form (1+q s Z)/(1+q m Z)≅Z/q m−s with s<m (see Theorem 12).
2. p-adic units
Let q be a prime. To describe the structure of (Z/q r)× it is sufficient and convenient to describe the units of Z q=lim←Z/q m, the q-adic integers. This section aims to make standard theorems in elementary number theory explicit for the purpose of later use.
Theorem 7
(See[Ser73, Chapter 1].)
(1)The units of Z q factor as a direct sum (written multiplicatively here): Z q×=T⋅U , with T={x∈Z q;x q=x}U=1+q Z q.
(2)
We have the following isomorphisms:
(a)T≅(Z/q)×,for all q ;
(b)U≅{(Z q,+),q≠2 Z/2⊕Z q,q=2 .
Comments on Theorem 7 part (1) and part (2a)
The group T is commonly referred to as the Teichmuller elements of Z q. The isomorphism T≅Z/q is given by the so-called Teichmuller map τ:(Z/q)×→Z q× as defined by τ(x¯)=lim n→∞x q n, where x∈Z q and we let x¯ denote its residue class in Z/q. One can extend this map to all of Z q and we note that τ(x) only depends on the residue class of x modulo q (it is a standard fact that is well-defined). One notes that reduction modulo q and τ are inverse operations which tell us that the exact sequence(2.1)1→1+q Z q→(Z q)×→(Z/q)×→1 splits. This splitting proves part (1). □
Examining the proof, we observe that the direct sum decomposition Z q×=T⋅U in Theorem 7 part (1) can be made explicit.
Corollary 8
The factorization of Z q×=T⋅U inTheorem 7can be made explicit. For x∈Z q×we have x=τ(x)(1+q a(x)),a(x)=(x/τ(x)−1)/q.∈1+q Z q.
The proof of part (2b) of Theorem 7 when q≠2 amounts to showing that (1+q Z q)/(1+q n Z q) is cyclic. The strategy is to pick some α∈1+q Z q and show {α q i}i=0 n−1 are distinct modulo q n. This will follow from the contraction property of the q th power map below (Lemma 11).
Our observation is that one can apply the same trick to smaller balls around the identity of G m(Z q), i.e. to α∈1+q r Z q. The goal of the rest of this section is to prove the following strengthening of (2b) of Theorem 7.
Theorem 9
Suppose one of the following holds:
(1)q>2 , s≥1 and r>s .
(2)q=2 , s≥2 and r>s .
Then for all α∈(1+q s Z q)∖(1+q s+1 Z q)we have〈α‾〉=(1+q s Z)/(1+q r Z)≤(Z/q r)×.The group generated by α‾has order q s−r .
Remark 10
The isomorphism in part (2b) of Theorem 7 is the case s=1 of Theorem 9. Explicitly, the isomorphism in part (2b) of Theorem 7 is given by x↦α x.
Lemma 11
Contraction property of q th power map
Let q≥2 or s≥2 . α∈(1+q s Z q)∖(1+q s+1 Z q)⟹α q∈(1+q s+1 Z q)∖(1+q s+2 Z q).
In this lemma we view U=1+q Z q as the unit ball around the identity of the multiplicative group G m(Z q)=Z q×. Observing that we may decompose U into annuli,U=1+q Z q=∐s≥1(1+q s Z q)∖(1+q s+1 Z q), the theorem says that the map x↦x q contracts each annulus in this decomposition to the neighboring annulus one level closer to the identity.
Proof of Lemma 11
For any a s∈Z q∖0 one can verify the formulas(2.2)(1+a s q s)q=1+a s+1 q s+1,(2.3)a s+1:=(1+a s q s)q−1 q s+1=∑j=1 q 1 q(q j)a s j q s(j−1)∈Z q. If a s∈Z q× then by examining (2.3) modulo q we see that a s+1∈Z q× under the hypothesis that q≥2 or s≥2.
In the case that q=2 we have a s+1=a s(1+a s 2 s−1) using formula (2.3) and difference of squares. Here it is necessary to have s≥2 as 1+a s may be congruent to 0 modulo 2. □
Proof of Theorem 9
Suppose that r>s≥2 and q is any prime. Let a 1∈Z q× and define α=1+q s a 1. For every t>0 we have(2.4)(α)q t=(1+q a 1)q t∈(1+q s+t Z q)∖(1+q s+t+1 Z q) by the contraction property (Lemma 11). Consider the reduction of (2.4) modulo q r. Observe that(α)q r−s≡1 mod q r and α,α q,…,α q r−s are distinct.
Since #(1+q s Z)/(1+q r Z)=#Z/q r−s=q r−s and α q r−s−1≠1 modulo q r, α must have order q r−s as an element of (Z/q r)× and hence generate all of (1+q s Z)/(1+q r Z). □
3. Proof of Theorem 6
Using the results we proved in the section on p-adic units (Section 2), we are now able to prove Theorem 6.
Theorem 12 No higher Wieferich primes
Let p and q be distinct primes. Let H r be the cyclic group generated by p modulo q r . There exists some s (depending on p and q) such that for all r>s sufficiently large, the group H r contains a subquotient isomorphic to the cyclic group(1+q s Z)/(1+q r Z)≅(Z/q)r−s .
In particular if K r denotes the kernel of the natural quotient map H r→H r−1 then for all r>s the kernel K r is nontrivial (which means q is not p-Wieferich at r).
Proof
We would like the show K r is nontrivial. Observe the following reduction. Let U r denote the reduction of U modulo q r. By the factorization of Z q×=T⋅U (Theorem 7) it suffices to show that the kernel of U r∩H r→U r−1∩H r−1 is nontrivial.
Since p n is not torsion in Z q we have 1≠α:=p/τ(p)∈(1+q t Z q)∖(1+q t+1 Z q). For some t depending on p and q (cf. Corollary 8).
We claim that some power of α is congruent to 1 modulo q 2.
Case q≠2:Raising α to the power q will achieve this by the contraction lemma (Lemma 11).
Case q=2:If t>1 we are ok. Suppose now that t=1. Write α=1+2 a. Suppose n=0 mod 4 and n>3. We will show that (1+2 a)n∈(1+4 Z 2). In this situation(n j)(2 a)j=0 mod 4 for j≥3. We now have α n=1+(n 1)2 a+(n 2)(2 a)2 mod 4. Since(n 1)2 a+(n 2)(2 a)2=2 n(a+(n−1)a 2)=0 mod 4 we can see that α n=(1+2 a)n∈1+4 Z 2. (It suffices to take n=4.)
This shows the claim.
We can now suppose there exists some power of α, which we will call β which is a member of (1+q s Z q)∖(1+q s+1 Z q) for some positive s.
We have 〈β‾〉=(1+q s Z)/(1+q r Z) for all r>s by Theorem 9. Hence for all r>s, we have#(1+q s Z q)/(1+q r Z q)=q s−r, by Theorem 9 the surjective map(1+q s Z q)/(1+q r Z q)→(1+q s Z q)/(1+q r−1 Z q) has nontrivial kernel of size Z/q. This proves that K r is nontrivial for every r>s>2. □
4. Proof of Theorem 3
In what follows it will be convenient to think of elements in Z/q n or Z q=lim←Z/q n in decimal form. For a sequence of elements a i∈{0,…,q−1} we use the notation(4.1)(a n…a 2 a 1 a 0)q:=a 0+a 1 q+a 2 q 2+⋯+a n q n. Again, the digits of (a n…a 2 a 1 a 0)q are ordered with a 0 being the first digit and a n being the last digit.
Lemma 13
Fix p and q distinct primes. Let H m be the multiplicative group generated by p in(Z/q m)×.
(1)For every m the first m digits in the sequence(p n)q are periodic in n. The period is the order of the subgroup H m .
(2)lim N→∞1 N∑n=1 N d n,m(a)m=1#H m∑i=1#H m d n,m(a)
Proof
The first m digits of a number a∈N can be determined by a mod q m. Since a∈(Z/q m)×, the group of units, there exists some number h m such that p h m≡1 mod q m.
Part (2) follows from part (1). □
Remark 14
In the statement of Theorem 3, we used the notation A m(a)=lim N→∞1 N∑n=1 N d n,m(a)m. This will appear again later.
Let p and q be distinct primes and let H m be the group generated by p in (Z/q m)×. Define K m=ker(H m→H m−1). Since ker((Z/q m)×→(Z/q m−1)×)={1+q m−1 a mod q m:a∈Z/q}≅Z/q. We know K m is either isomorphic to Z/q or 1. This means that #K m=1 or #K m=q. As sets we have the following description of K m:(4.2)K m={{00⋯01 q,10⋯01 q,…,(q−1)0⋯01 q},#K m=q{00⋯01 q},#K m=1.
We will show that one can determine inductively the total number of bits equal to a in the sequence {p n mod q m} given by the behavior of K m. The following notation will become useful:h m:=#H m,k m:=#K m,t m(a):=∑n=1 h m d n,m(a),for a∈{0,…,q−1}. Observe that t m(a) is just the total number of a's appearing in the sequence {(p n mod q m)q}n=1 h m. Also observe that we also have the equality A m(a)=t m(a)/m h m. Here A m(a) was defined in part (2) of Theorem 3 to be the average number of a's in the first m bits of p n as n→∞.
The following lemma says we can determine distribution digits in H m from the distribution of digits in H m−1.
Lemma 15
In the case k m=1 we have t m(a)=t m−1(a),h m=h m−1.In the case k m=q we have t m(a)=q t m−1(a)+h m−1,h m=q h m−1.
Proof
If h∈H m−1 defines h˜∈H m to be the lift of h where we append a zero on the left, i.e. (h˜)q=0(h)q as strings so that h=π m,m−1(h˜) with π m,m−1 the projection of H m onto H m−1. Observe that we have the partition(4.3)H m=⋃h∈H m−1 h˜K m=⋃h∈H m−1 π m,m−1−1(h).
•If #K m=1 then every element of H m (viewed as a string) is an element of H m−1 (as string) just with an extra 0 appended to the end. This proves h m=h m−1. The equality t m(a)=t m−1(a) is a trivial consequence of this.
•Suppose #K m=q. The equality h m=q h m−1 is trivial. We now work on showing t m(a)=q t m−1(a)+h m−1. Let (b m−2…b 1 1)q∈H m−1 and (c 0…01)q=c q m−1+1∈K m (recall equation (4.1) for the meaning of this notation). We have(0 b m−2…b 1 b 0)q⋅(c 0…01)q=(b m−1 b m−2…b 1 b 0)q b m−1=c⋅b 0 mod q. We know that the b 0's are equidistributed over {1,…,q−1} in H m−1 and that c∈{0,…,q−1} uniquely determines the element of K m. For each element of (b m−2…b 1 b 0)q∈H m−2, consider the preimage π m,m−1−1((b m−2…b 0)q)={((c b 0)b m−2…b 1 b 0)q:c∈Z/q}. Since (Z/q)× is a group{a:a∈(Z/q)×}={a⋅b 0:a∈(Z/q)×} this implies that π m,m−1−1((b m−2…b 1 b 0)q)={(b m−1 b m−2…b 0)q:b m−1∈Z/q} which follows from (4.3).
□
We now derive some formulas for A m(a). The main idea of this proof is that k m=q pulls digits of p n toward equidistribution and k m=1 pulls the distribution of the bits of p n toward having more zeros. In the situation where k m=q the “new bit” is completely equidistributed. Note in particular that for all m we have 0≤A m(a)≤1 from which it is easy to see that k m=q “pushes” A(a,m) towards equidistribution 1/q.
Lemma 16
(1)For m>2 we have A m(a)=(1−1 m)A m−1(a)+1 q(q−1)m(k m−1).
(2)Define k¯m=k m−1 for m≥2 and define k¯1=q . For all a∈{1,…,q−1}we have(4.4)A m(a)=1 q(q−1)k¯1+k¯2+⋯+k¯m m.
Proof
We analyze the formula by cases:
k m=1:(the density of a's in H m is strictly decreasing). We have H m≅H m−1 and that elements of H m−1 give an element of H m by just tacking a zero at the end. We have t m(a)=t m−1(a)h m=h m−1 which implies A m(a)=t m(a)m h m=m−1 m⋅t m−1(a)(m−1)h m−1=(1−1 m)A m−1(a).
k m=q:(density of a's will approach the equilibrium). We have t m=q t m−1+h m−1 h m=q h m−1 which gives A m(a)=t m m h m=q t m−1+h m−1 m h m=(1−1 m)A m−1(a)+1 q m.
We now solve the recurrence relation to give the formula in part (2). This proof is by induction. Fix some a∈{1,…,q−1}. Note that A 1(a)=1/(q−1) since p generates the unit group mod q which has (q−1) elements, so the base case is trivial. We now do the inductive step and suppose the formula holds for m and prove it for m+1.A m+1(a)=m m+1 A m(a)+k¯m+1 q(q−1)(m+1)=1 q(q−1)(m+1)[k¯1+k¯2+⋯+k¯m]+k¯m+1 q(q−1)(m+1)=1 q(q−1)k¯1+k¯2+⋯+k¯m+1 m+1, which proves our result. □
Remark 17
Note that (1) shows that A m(a)=1 h m∑n=1 h m d n,m(a)m only depends on whether a is zero or nonzero. This follows from A 1(a)=1/(q−1) for all a∈{1,…,q−1} as p generates (Z/q)× together with the recurrence.
Supposing A m(a) was completely independent of a we would have q A m(a)=∑a=0 q−1 A m(a)=1 which implies A(m)=1/q. This would give an easy proof of our result.
We have now related the distribution of bits to the condition about “Wieferich primes”.
Lemma 18
With the notation as above we have lim m→∞A m(a)=1/q⇔1=lim m→∞1 m(q−1)∑j=1 m k¯j.
Proof
Follows directly from Lemma 16 part (2) and the definition of A m(a). □
We now prove that lim n→∞1 n(q−1)∑j=1 n k¯j=1. To do this we need to study the multiplicative group generated by p modulo q r.
Theorem 19
With the notation as above and k¯j=#K j−1 we have lim n→∞1 n(q−1)∑j=1 n k¯j=1.
In particular this implies lim m→∞A m(a)=1/q for all a∈{0,…,q−1}.
Proof
Since K j≤ker(Z/q j→Z/q j−1)≅Z/q it can only have order q or 1. By Theorem 12, K j is nontrivial for all but a finite number of j and hence k¯j must be equal to q for all but a finite number of j.
The second part follows from Lemma 18. □
Acknowledgments
We would like to thank Cris Moore, Robert Lemke Oliver and Carl Pomerance for helpful comments and suggestions. The first author visited MSRI during the preparation of this manuscript. Research at MSRI is supported in part by NSF grant DMS-0441170.
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Journal of Number Theory, Volume 164, 2016, pp. 166-178 Jonas Kibelbek, …, Hao Yuan View PDF
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3197 | https://www.sciencing.com/distance-vs-displacement-whats-the-difference-why-it-matters-w-diagram-13720227/ | Science
Physics
Distance Vs Displacement: What's The Difference & Why It Matters (W/ Diagram)
By
Kevin Beck
Updated
Physics, at its core, is about describing the motion of objects through space in terms of their position, velocity and acceleration as a function of time.
As centuries progressed and humans expanded the power of observational tools at their disposal, this pursuit of learning exactly what objects are doing in physical space and when has grown to include extremely small objects, such as atoms and even their components, with the whole field of quantum physics, or quantum mechanics, arising as a result.
Still, the first things any physics student learns are the basic laws and equations of Newtonian mechanics. Thus usually starts with one-dimensional motion and moves on to motion in two dimensions (up-down and side-to-side) such as projectile motion, introducing Earth's unique gravitational acceleration of 9.8 meters per second per second (m/s2).
Once you have become skilled at using these in concert in your study of motion and the nature of classical mechanics, you will have developed a better appreciation for differences that seem trivial at first glance but are actually anything but trivial, such as the difference between distance and displacement.
Distance vs. Displacement
Distance vs. Displacement
Distance and displacement are commonly confused terms in physics that are important to get correct. Distance is a scalar quantity, the total distance traveled by an object; displacement is a vector quantity, the shortest path in a straight line between the starting position and final position.
Dana Chen | Sciencing
The difference between a vector quantity and a scalar quantity is that vector quantities include information about direction; scalar quantities are simply numbers. "Half-arrows" above a variable indicates that it is a vector quantity. The expression for the total displacement r of a particle in an x, y-coordinate plane, in vector notation, is:
Here, i and j are "unit vectors" in the x- and y-direction respectively; these are used to draw the components of a given vector quantity that points in a direction other than an axis, and their own magnitude is 1 by convention.
Calculating Distance vs. Calculating Displacement
Calculating Distance vs. Calculating Displacement
Anything that moves in relation to a fixed reference frame is covering distance. A person pacing back and forth at 2 m/s waiting for a bus to arrive and continually returning to the same spot has a speed of 2 m/s but a velocity of 0. How is this possible?
Physicists use the initial and final position to calculate displacement of an object, which is just the shortest path from its initial position a to its final position b even if the object did not take this direct, straight-line path to get there. Displacement mathematically assumes the form d = xf – xi, or horizontal displacement is equal to final position minus initial position).
Why the Distinction Matters
Why the Distinction Matters
Distance traveled is needed to calculate average speed (i.e., total distance over a period of time). Both distance and speed are scalar quantities, so they are naturally found together. Displacement is needed to find the final position of an object; it tells not only the distance from the starting position, but also the net direction of travel.
Because displacement is a vector quantity, it, not distance, must be used to find average velocity, another vector quantity. Average velocity is the total displacement of an object over a period of time. If you ride your bicycle around an oval for an hour and cover 20 miles, your average speed is 20 mi/hr, but your average velocity is zero because of the lack of displacement from your starting position.
On a similar note, if road signs included "VELOCITY LIMIT" instead of "SPEED LIMIT" varieties, it would be a lot easier to get out of a speeding ticket. All you'd have to do is make sure you pulled over in the same spot the officer first spotted you, and you could argue that, the distance of your trip aside, your displacement is clearly zero, rendering your velocity zero by definition. (Okay, maybe not such a good idea for various reasons!)
Distance and Displacement: Examples
Distance and Displacement: Examples
Consider the following scenarios:
A car drives three blocks north and four blocks east. The
total distance the object travels is 4 + 3 = 7 blocks. But the total displacement is the shortest distance from where the car begins and ends its trip, which is a diagonal line, the hypotenuse of a right triangle with legs 3 and 4. From the Pythagorean theorem, 32 + 42 = 25, so the length of the hypotenuse is the square root of this value, which is 5. The displacement vector points from initial position to final position.
A person walks north from their house 100 meters to the
park, and then returns home before continuing 20 meters south to check the mail. A FitBit or GPS watch would indicate a total distance walked of 100 m + 100 m + 20 m = 220 m. But if the starting point is the house situated at the origin (the point 0, 0 on a coordinate plane) and the final position is the mailbox, which is at (0, −20), the person ends up only 20 meters away from where they began, making the total displacement −20 m. The negative sign is important because a frame of reference was chosen to situate the park in the positive direction on the x-axis. It could have been arranged the opposite way, in which case the person's displacement would be + 20 m instead of −20 m.
An athlete runs 10 km on a standard 400-meter track before
breakfast (25 laps). What is the total distance they traveled? (10 kilometers.) What is the total displacement? (0 m, though reminding the runner of this after the race may be unwise!)
Position, Time and Other Variables of Motion
Position, Time and Other Variables of Motion
Specifying an object's position in space is a starting point for countless physics problems. For the most part, beginning and intermediate exercises use one-dimensional (x only) or two-dimensional (x and y) systems to keep the problems from being overly difficult, but the principles extend to three-dimensional space as well.
A particle moving in two-dimensional space can be assigned x- and y-coordinates for its position, its rate of change of position (velocity v) and its rate of change of velocity (acceleration a). Time, of course, is labeled t.
Newton's Laws of Motion
Newton's Laws of Motion
Much of classical physics relies on the equations describing motion derived by the great scientist and at mathematician Isaac Newton. Newton's laws of motion are to physics what DNA is to genetics: They contain most of the story and are essential to it.
Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless acted on by an external force. Newton's second law is perhaps the least well recognized of the three by the general public because it cannot be easily reduced to a simple phrase, and instead asserts that net force equals the product of mass and acceleration:
The third law states that every action (i.e., force) in nature has an equal and opposite reaction.
The position of an object at constant velocity is represented by a linear relationship:
where x0 is the displacement at time t=0.
The Importance of Reference Frames
The Importance of Reference Frames
This takes on greater importance in advanced physics, but it's important to emphasize that when physicists declare that something is "in motion," they mean with respect to a coordinate system or other reference frame that is fixed with respect to the variables in the problem. For example, it's fair to say that if a road's speed limit is 100 km/hr, it is implies that the Earth itself, though clearly not stationary in absolute terms, is treated as such in context.
Albert Einstein is best known for his theory of relativity, and his special relativity idea was one of the most groundbreaking in the history of modern thought. Without incorporating reference frames into his work, Einstein would not have been able to adapt Newton's equations in the early 20th century to suit relativistic particles, which deal with very high speeds and low masses.
References
Physics LibreTexts: Motion in Two Dimensions
Georgia State University: HyperPhysics: Position and Displacement
Indiana University – Purdue University Indianapolis: Scalars and Vectors
NASA: Newton's Laws of Motion
University of Rochester: Albert Einstein and the Theory of Relativity
Cite This Article
MLA
Beck, Kevin. "Distance Vs Displacement: What's The Difference & Why It Matters (W/ Diagram)" sciencing.com, 28 December 2020.
APA
Beck, Kevin. (2020, December 28). Distance Vs Displacement: What's The Difference & Why It Matters (W/ Diagram). sciencing.com. Retrieved from
Chicago
Beck, Kevin. Distance Vs Displacement: What's The Difference & Why It Matters (W/ Diagram) last modified August 30, 2022. |
3198 | https://www.liebertpub.com/doi/10.1089/andro.2021.0020 | Back to Top
Hyperandrogenism in Women with Polycystic Ovarian Syndrome: Pathophysiology and Controversies
Authors: Sarah A. Kanbour and Adrian S. Dobs adobs@jhu.eduAuthors Info & Affiliations
Publication: Androgens: Clinical Research and Therapeutics
Volume 3, Issue Number 1
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Abstract
Polycystic ovary syndrome (PCOS) is an essential differential diagnosis of hyperandrogenism to consider in women, with most women presenting with irregular menses, infertility, and hirsutism. The diagnosis is challenging because it is a heterogenous disorder with significant variations in its associated features. Classic PCOS (hyperandrogenism plus ovulatory dysfunction) affects about 10% of reproductive-age women. Hyperandrogenism (clinical, biochemical, or both) is a hallmark of the syndrome. Approximately 60–76% of PCOS women are hirsute and hyperandrogenemia is observed in 75–90% of these women. Androgen assays are not always reliable at the lower levels detected in women with PCOS. This article will attempt to review the pathophysiology and controversies of hyperandrogenism in PCOS women, its clinical and biochemical presentations, and management.
Introduction
The diagnosis of polycystic ovary syndrome (PCOS) requires two of the following abnormalities: clinical or biochemical hyperandrogenism, ovulatory dysfunction, and polycystic ovarian morphology as well as the exclusion of mimicking conditions such as Cushing syndrome (1 mg dexamethasone suppression test) especially when hypogonadotropic hypogonadism is present, nonclassical congenital adrenal hyperplasia (early follicular-phase early-morning plasma level of 17-hydroxy progesterone), hyperprolactinemia (prolactin level), and thyroid dysfunction (thyroid stimulating hormone and free thyroxine). First described by Stein and Leventhal in 1935, this condition has been called “polycystic ovary disease,” with many today proposing that “polycystic ovary syndrome” is best used.
This article will attempt to review the pathophysiology and controversies of hyperandrogenism in PCOS women, its clinical and biochemical presentation, and management.
Normal Physiology of Androgens in Women
Androgens have a role in female health and contribute to bone density, muscle mass, and female sexual function. Androgens include dehydroepiandrosterone sulfate (DHEA-S), dehydroepiandrosterone (DHEA), androstenedione, testosterone, and dihydrotestosterone. Testosterone and dihydrotestosterone are the most biologically active.1 Androgen biosynthesis occurs in the ovaries (25%), adrenal gland (25%), and peripheral tissues (50%) such as the liver, skin, and fat.2 Table 1 summarized the source and estimated normal concentrations of androgen in women, which may vary with the laboratory.
Table 1. Androgens in Women
| Androgens | Reference range | | Source | | | Comments |
--- --- ---
| Ovary (25%) | Adrenal (25%) | Periphery (50%) |
| DHEA-S | 18–29 years | 44–332 μg/dL (1.19–9.00 μmol/L) | None | 100% | None | Modulated by ACTH. Influenced by prolactin, IGF-1, and estrogen levels |
| 30–39 years | 31–228 μg/dL (0.84–6.78 μmol/L) |
| 40–49 years | 18–244 μg/dL (0.49–6.61 μmol/L) |
| DHEA | 100–1000 ng/dL (3–35 nmol/L) | | 20% | 50% | 30% (from DHEA-S) | |
| Androstenedione | 30–200 ng/dL (1.05–6.98 nmol/L) | | 50% | 50% | None | Circadian and menstrual cycle variation. Suppressed by corticosteroid |
| Testosterone | Total: 8–60 ng/dL (0.3–2.1 nmol/L) Free: 0.3–1.9 ng/dL (0.01-0.07 nmol/L) | | 50% | 25% | 25% (from androstenedione) | Circadian and menstrual cycle variation. Most biologically active |
| Dihydrotestosterone | 2 ng/dL | | None | None | 100% (from testosterone) | Most biologically active |
Table adapted from Burger.1
ACTH, adrenocorticotropic hormone; DHEA, dehydroepiandrosterone; DHEA-S, dehydroepiandrosterone sulfate; IGF-1, insulin-like growth factor 1.
Numerous studies suggest that androgen insufficiency diminishes women's sexual desire and energy,3 although it is not conclusive that treatment with testosterone increases libido.4 The differential diagnosis of hyperandrogenism is generally divided by adrenal (Cushing syndrome and congenital adrenal hyperplasia) versus ovarian (PCOS, hyperthecosis, and carcinoma). Whereas mild hyperandrogenism is characterized by hirsutism, acne, and male-pattern baldness, severe androgen excess results in virilization (voice deepening and clitoromegaly), suggesting the possibility of ovarian hyperthecosis or androgen-secreting tumor. Most guidelines recommend further evaluation if the DHEA-S concentration is >700 μg/dL (>19.0 μmol/L)5 or serum total testosterone concentration exceeds 150 ng/dL (>5.2 nmol/L).6
Ovarian hyperthecosis has sometimes been referred to as an extreme form of PCOS. It is a non-neoplastic condition of androgen excess classically described by the presentation of significant and progressive hirsutism or virilization due to ovarian interstitial cells differentiation into active luteinized theca cells capable of producing androstenedione and testosterone. Total testosterone concentration is usually >150 ng/dL and almost all women have obesity and hyperinsulinemia. It is more common in postmenopausal women. The diagnosis is confirmed histologically.7
Diagnosis of PCOS
There is no specific test to diagnose PCOS, rather three sets of criteria have been developed for PCOS diagnosis. Each set involves different combinations of hyperandrogenism, ovulatory dysfunction, and polycystic ovarian morphology8 (Table 2). As defined by the National Institutes of Health diagnostic criteria (hyperandrogenism plus ovulatory dysfunction), “classic” PCOS affects 6–10% of reproductive-age women. The prevalence rate is almost twice as high under the broader Androgen Excess and PCOS society criteria (hyperandrogenism plus ovulatory dysfunction or polycystic ovarian morphology).9
Table 2. Diagnostic Criteria for Polycystic Ovary Syndrome
| Variable | National Institutes of Health | Rotterdam | Androgen excess and PCOS society |
--- --- |
| Hyperandrogenisma | + | ≥2 of 3 criteria | + |
| Ovulatory dysfunctionb | + | ≥1 of 2 criteria |
| Polycystic ovarian morphological featuresc | − |
Table derived from McCartney et al.8
a
Clinical (hirsutism assessed by the Ferriman–Gallwey scale, acne, and male-pattern hair loss) or biochemical hyperandrogenism (total and/or free testosterone, androstenedione, and/or DHEA-S level above the upper 95th percentile of 98 healthy non-hirsute eumenorrheic women).
b
Menses are interval <21 days or >35 days or eumenorrhea with progesterone <3 to 4 ng/mL a week before anticipated menses (days 21 or 22).
c
Polycystic ovarian morphological features are defined as 12 or more antral follicles (2–9 mm in diameter) in either ovary, an ovarian volume that is >10 mL in one or both ovary.
PCOS, polycystic ovary syndrome.
Because of the overlap between normal pubertal physiology changes and PCOS clinical features, the diagnosis should not be made within 2 years of menarche.10 In practice, the diagnosis of PCOS is usually a challenge because it is a heterogenous disorder with significant variations in its associated features.
Clinical and Biochemical Hyperandrogenism
Hyperandrogenemia is the biochemical hallmark of PCOS. High levels of androgens are detected in 75–90% of PCOS patients with oligomenorrhea,11,12 and their concentrations often increase with the severity of the phenotype.13 Hyperandrogenemia is defined as a free/total testosterone, androstenedione, and/or DHEA-S level above the upper 95th percentile of 98 healthy non-hirsute eumenorrheic women.14 Testosterone assays (free/total testosterone) have low specificity and sensitivity at the lower levels detected in PCOS women. When measured, total testosterone level is preferably drawn early morning, during the early follicular phase of the menstrual cycle.
Mass spectrometry–based assays of total testosterone may be more precise compared with other clinical assays.15 Calculated free testosterone or free androgen index (FAI) [(total testosterone × 100)/sex hormone–binding globulin (SHBG)] further increases the diagnostic accuracy. FAI ≥6 is considered abnormal. Androstenedione appears to be a more sensitive test to assess for androgen excess compared with testosterone levels.16 In a study of 86 PCOS women fulfilling the Rotterdam criteria, elevated serum testosterone level was found in 65%, whereas serum androstenedione concentrations were above the reference range in 88%.17 Although the ovaries are the main source of hyperandrogenemia in PCOS,17 elevated levels of adrenal DHEA-S are seen in 25–35% of these women.11,12
Clinical manifestations of hyperandrogenemia are hirsutism, acne, and androgenic alopecia. Hirsutism is a more specific feature of hyperandrogenism than acne and alopecia. Hirsutism is defined as excessive terminal hair that appears in a male pattern in women. The Ferriman–Gallwey (FG) scale quantitates the extent of hair growth in different androgen-sensitive sites. A score of 8 or more on the FG scale indicates that a woman is hirsute.18 Ethnic variations add to the challenge in the interpretation of the FG score. A score of 8–15 indicates mild hirsutism, and a score >15 indicates moderate or severe hirsutism. Among PCOS patients, ∼60–76% are hirsute,19 with an average score of 8 ± 5.20 Hirsutism results from the interaction between androgen levels and the sensitivity of the hair follicle to androgen.18
A free testosterone level that is at least twice the upper limit of normal will result in some degree of hirsutism.21 However, the severity of hirsutism does not correlate well with the androgen levels.18 Insulin has direct effect on hair follicle growth and the severity of hirsutism may be a result of differences in the degree of hyperinsulinism. Lower levels of hyperinsulinism are associated with lower androgen levels and less severe hirsutism, whereas higher levels of hyperinsulinism favor the development of hirsutism and worsening hyperandrogenemia.22
Androgens also stimulate sebocytes and follicular keratinocytes leading to their hyperplasia and the impaction of the pilosebaceous unit that characterize acne.18 Androgenic alopecia is a male-pattern hair loss involving the anterior, mid, or temporal scalp and/or the vertex of the scalp and is the result of circulating androgens that suppress hair growth on the scalp.19
Pathophysiology: Hyperandrogenemia and Hyperinsulinism
The identification of the primary underlying pathophysiology of PCOS is still unclear. Clinicians and investigators have debated whether the primary event is ovarian, pituitary, or hypothalamic in origin, or related to obesity and its hyperinsulinism (Fig. 1).
PCOS is characterized by an increased frequency of gonadotropin-releasing hormone (GnRH) pulsatility that selectively increases luteinizing hormone (LH) secretion.23 LH stimulates multiple steroidogenic enzymes in the theca cells of the ovary, leading to theca cell hyperplasia and increased testosterone production.24 Because of relative follicle-stimulating hormone (FSH) deficiency, testosterone is not completely aromatized and degraded by the granulosa cells.
The increased levels of testosterone feeds back on the hypothalamus, decreasing the ability of estradiol and progesterone to slow down GnRH pulse frequency.25 The ovaries do not appear to be the primary abnormality in PCOS since they have the ability of responding promptly to changes in gonadotropin secretion; ovulation occurs in response to the surge in secretion of FSH stimulated by clomiphene citrate. Furthermore, weight reduction in obese patients reduces estrone and insulin levels, normalizes gonadotropin secretion, and regulates menstrual cycles in women with PCOS.26,27
The PCOS pathophysiology may originate centrally at the pituitary level and is explained by the preferentially inhibitory action of estrogen on FSH release coupled with a relative insensitivity of FSH release. This is supported by the observation that among postmenopausal women, PCOS women had lower FSH levels and SHBG and higher FAI compared with controls matched for weight, BMI and waist-to-hip ratio. The two groups did not differ in regard to LH level, DHEA-S, androstenedione, total testosterone, estradiol, and estrone.28,29 Furthermore, in the largest most comprehensive association study of PCOS genotype, mutations in the region of the FSH-beta gene were identified that positively correlate with an elevation in circulating LH concentration.30 LH release does not appear to be the major defect in PCOS as there is evidence of an intact positive feedback mechanism of estrogen on LH release and surge.31
Hyperinsulinemia also plays a major role in patients with PCOS and it is exacerbated by hyperandrogenism-related visceral fat accumulation.25 Hyperinsulinemia stimulates the hypothalamic–pituitary–adrenal axis both centrally32 and peripherally,32,33 leading to increased adrenal androgen production. DHEA-S levels are associated with insulin resistance in women with PCOS.34 Insulin and testosterone decrease hepatic production of SHBG, elevating the levels of free testosterone.35 Testosterone, in return, decreases the sensitivity of the feedback effects of estradiol and progesterone on the hypothalamus and pituitary.19
Insulin resistance is the reduced ability of insulin to mediate its actions on glucose metabolism resulting in an increased amount of insulin. Insulin sensitivity is measured with the following techniques, arranged according to their accuracy: fasting glucose, homeostasis model assessment (HOMA), oral glucose tolerance test (OGTT), frequently sampled intravenous glucose tolerance test (FSIVGTT), and euglycemic hyperinsulinemic clamp.25 The prevalence of insulin resistance in PCOS women was 53% for the modified FSIVGTT,36 23–35% for impaired glucose tolerance, and 4–10% for type 2 diabetes mellitus.25 The increase in insulin responses during OGTT was two times higher than age- and weight-comparable reproductively normal control women.25
Management of Hyperandrogenism
Hirsutism can be managed with direct hair removal methods and with medical therapies against androgen production and action. Mechanical hair removal (shaving, plucking, and waxing) is recommended for mild hirsutism and if not effective, pharmacological therapy (combined oral contraceptive pills and antiandrogens) is added.5
Combined oral contraceptives pills are the hallmark treatment of PCOS to reduce hyperandrogenism through different mechanisms. They suppress LH and, therefore, ovarian androgen secretion, and stimulate the production of SHBG from the liver, which increases androgen binding and reduces serum-free androgen concentrations.37 Progestins also increase testosterone clearance.38 All combined oral contraceptive pills appear to be equally effective for hirsutism,5 reducing FG scores by 7 points5 with the hirsutism scores almost reaching that of control group of non-hirsute women.39 The changes in the hair thickness become visible after at least 6 months of therapy. Antiandrogen therapy can be added if suboptimal response after 6 months8 (Table 3). There is also evidence that oral contraceptives decrease the severity of acne in women with PCOS.41
Table 3. Hirsutism in Women with Polycystic Ovary Syndrome (Average Ferriman–Gallwey Scores of 8 ± 5)
| Severity of hirsutism | Management of hirsutism | Response to management |
---
| Normal: FGS <8 Mild: 8–15 Moderate—severe: >15 | Mechanical hair removal | Effective for mild hirsutism |
| OCP | ↓ FGS by 7 |
| Antiandrogens (combined with OCP) • Spironolactone (up to 100 mg/d) • Finasteride (up to 5 mg/d) | ↓ FGS by 4 |
| Weight loss | Observed with 35% weight loss |
Table derived from Swiglo et al.40
FGS, Ferriman–Gallwey scores; OCP, combined oral contraceptive.
Antiandrogen therapy includes spironolactone (aldosterone-antagonist), finasteride (5α-reductase inhibitor), and flutamide (androgen receptor antagonist). Daily spironolactone 100 mg, finasteride 2.5–5 mg, and flutamide 500 mg improve hirsutism scores compared with placebo, reducing FG scores by 2–5 points and by 4 points on average and by 15–40% within 6 months after the start of therapy.40 There was no significant difference in effectiveness among the three antiandrogens, although spironolactone is often prescribed because of its relative favorable safety profile.
Flutamide is no longer a recommend option because of hepatotoxicity risk.5 The maximal effectiveness of spironolactone occurs at about 3 months and continues at 12 months after initiating treatment.42 Spironolactone may mildly reduce circulating androgen levels, which may further contribute to the progressive regression of hirsutism.42 There is a danger that a male fetus could be feminized in women taking antiandrogens and, therefore, a reliable contraception must be used.
As discussed earlier, lower insulin levels are related to lower androgen levels and less severe hirsutism in women with PCOS.22 Insulin-lowering drugs and insulin sensitizers such as metformin and thiazolidinediones are not more effective than placebo for hirsutism treatment,5 despite lowering serum testosterone levels by ∼20–25% in women with the PCOS.8 However, subgroup analyses showed that insulin sensitizers may be more effective than placebo in overweight and obese women treated for >6 months.43
Weight loss also lowers insulin levels. Loss of at least 5% of body weight in overweight and obese patients with PCOS lowers androgen levels (total and free testosterone, FAI, and DHEA-S), increases SHBG, improves insulin sensitivity (OGTT, HOMA, fasting glucose, and insulin levels), and decreases hirsutism.44 These findings are particularly pronounced after excessive weight loss, such as in the context of bariatric surgery.45,46 In one study, hirsutism had resolved in 29% of women with obesity after 35% weight loss by gastric bypass with a mean follow-up of ∼4 years.47 There are no randomized trials and no long-term data on hyperandrogenism and metabolic outcomes with weight loss.
Fertility and PCOS
PCOS is one of the leading causes of infertility among women of reproductive age. Ovulatory dysfunction is characterized by chronic anovulation in the presence of normal FSH and estradiol concentrations. This is in contrast to conditions such as hypercortisolism and hyperprolactinemia that suppress the hypothalamic-pituitary-ovarian signaling and lead to secondary amenorrhea characterized by estrogen deficiency. As previously discussed, the relative FSH deficiency increases testosterone production. In turn, testosterone decreases the normal feedback effects of estradiol and progesterone on GnRH pulse frequency, preventing LH surge and ovulation. Hyperinsulinemia increases adrenal androgen production, further exacerbating the situation.
Considering the prevalence of obesity and hyperinsulinism, it seems reasonable to assume that weight reduction in women with PCOS would provide benefit. Indeed, modest weight loss of 5–10% has been associated with resumption of ovulation.48 Higher the weight loss has been associated with better menstrual cycle regulation and fertility. Lifestyle interventions results in a mean weight loss of 5–8%. Glucagon-like peptide-1 receptor agonist is now emerging as a therapeutic option for obese women with PCOS. Liraglutide is the most studied option in this group and is associated with 5–6% weight loss. Most studies report improved menstrual patterns with Liraglutide. There is also evidence of increased spontaneous pregnancies and improved in vitro fertilization pregnancy rates.49
Semaglutide has not been studied in PCOS, but it is an attractive option as it has more impact on weight loss (up to 15%).50 Bariatric surgery is an alternative strategy for weight loss in women with PCOS. Based upon several small observational studies, all obese women who lost weight after Roux-en-Y gastric bypass had restored menstrual cycles at approximately 3–4 months after the operation and an increased ability to conceive within 2 years of surgery.45,51,52 Ovulation induction is expensive and not very effective. The live birth rate among women who received clomiphene and aromatase inhibitor letrozole was about 20–30%.8 Current guidelines recommend against the routine use of metformin in obese women with PCOS for infertility, because although it increases ovulatory rates and pregnancy rates, it does not improve live birth rates.53
Controversies
PCOS women are at increased risk for metabolic syndrome, as evidenced by a higher prevalence of insulin resistance, type 2 diabetes mellitus, dyslipidemia, and hypertension. This risk is above what is observed in simple obesity.54 The hyperandrogenic phenotype, in particular, appears to have the highest risk of metabolic syndrome. One putative mechanism is that androgens excess increases adipose tissue lipid accumulation, leading to insulin resistance. Hyperinsulinemia then further exacerbates androgen generation.55 Interestingly, women with isolated hyperandrogenemia have normal insulin sensitivity in comparison with PCOS patients.56,57 Furthermore, gender-affirming testosterone therapy does not decrease insulin sensitivity and may even be associated with an improvement.58
Screening for cardiometabolic risk factors is recommended and includes measurement of the weight, body mass index (BMI), waist circumference, and blood pressure at each visit, and a lipid panel every 2 years (unless if there is significant weight gain).8
PCOS consequences on bone mineral density is controversial, partially because of the wide spectrum of clinical presentations. Although hyperandrogenemia and hyperinsulinemia have anabolic effects on bones, the increased inflammation, decreased vitamin D levels osteoprotegerin, aromatase, and growth hormone, as well as the amenorrhea have negative impact on bone formation and reabsorption.59 Women with PCOS with ovulatory dysfunction often have acyclic production of 17b-estradiol, with concentrations similar to that of the follicular phase; this is significantly lower than the average 17b-estradiol concentration observed in women with normal menstrual cycle. This stable and suboptimal level of 17b-estradiol negatively affects bone density.60
At physiological levels, insulin has a positive impact on the bone; it increases the proliferation of osteoblast and decreases the impact of parathyroid hormone. However, hyperinsulinemia decreases osteoprotegerin and insulin-like growth factors, which may lead to increased bone reabsorption.
Androgens' effect on bones are also complex. Androgens can stimulate osteoblastic cell proliferation but may also indirectly increase inflammatory mediators (interleukin-1β and tumor necrosis factor-α), inhibiting osteoblast differentiation. A recent meta-analysis showed that women with PCOS and BMI <27 kg/m2 have decreased levels of the bone formation marker osteocalcin and reduced spinal and femur bone density compared with controls. These findings were not seen in women with PCOS and BMI ≥27 kg/m2.59
Although androgen deficiency is associated with reduced sexual drive, its impact on women with PCOS is limited. In a recent meta-analysis of 21 observational studies, female sexual dysfunction (FSD) was more prevalent in women with PCOS than without (odd ratio of 1.39). FSD is defined as persistent or recurrent problems with sexual response, desire, orgasm, or pain, which can have adverse effects on quality of life and interpersonal relationship. In the subscale analysis, however, women with PCOS scored lower for pain. There was no significant difference between the two groups regarding the scoring for desire, lubrication, arousal, orgasm, or satisfaction. The exact mechanism is not clear, but was speculated to be related to low self-esteem and emotional distress that might impair sexual function and interpersonal relationships.
In an observational study of 72 women with PCOS and androgen excess treated with oral contraceptive pills, there was significant improvement in dyspareunia, orgasm, and satisfaction at 6 months. More research is required to confirm these findings.61
Another area of controversy involves the overlap of autism with PCOS and other conditions associated with hyperandrogenemia. A recent large systematic review from low-quality independent cohorts found mixed support for the association between androgen levels in mothers and autistic traits and language ability.62 Conversely, elevated prenatal estrogens may be as a risk factor for autism.63 Estrone levels is increased in women with PCOS because of extraglandular aromatization of increased circulating androstenedione levels.64
Conclusions
PCOS is an important differential diagnosis of hyperandrogenism to consider in women, with most women presenting with irregular menses, infertility, and hirsutism. The diagnosis should be based on clinical signs (hirsutism, etc.) and/or hyperandrogenemia. Treatment options should be based on targeted treatment, for example, electrolysis and shaving, suppression of testosterone production and/or testosterone receptor blockade, and weight loss. Limited available data indicate that mortality among PCOS patients occurs at a similar rate as in the general population, although there may be decreased longevity related to increased risk of cardiovascular disease. Control of obesity is particularly important for women with PCOS.65
Abbreviations Used
ACTH
: adrenocorticotropic hormone
BMI
: body mass index
DHEA
: dehydroepiandrosterone
DHEA-S
: dehydroepiandrosterone sulfate
FAI
: free androgen index
FG
: Ferriman–Gallwey
FGS
: Ferriman–Gallwey scores
FSD
: female sexual dysfunction
FSH
: follicle-stimulating hormone
FSIVGTT
: frequently sampled intravenous glucose tolerance test
GnRH
: gonadotropin-releasing hormone
HOMA
: homeostasis model assessment
IGF-1
: insulin-like growth factor 1
LH
: luteinizing hormone
OCP
: combined oral contraceptive
OGTT
: oral glucose tolerance test
PCOS
: polycystic ovary syndrome
SHBG
: sex hormone–binding globulin
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Information & Authors
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Published In
Androgens: Clinical Research and Therapeutics
Volume 3 • Issue Number 1 • 2022
Pages: 22 - 30
Copyright
© Sarah A. Kanbour and Adrian S. Dobs 2022; Published by Mary Ann Liebert, Inc.
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This Open Access article is distributed under the terms of the Creative Commons License [CC-BY] ( which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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Published in print: 2022
Published online: 3 March 2022
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Topics
Body mass index
Diagnostic techniques
Endocrine and metabolic disorders
Gonadal disorders
Hyperandrogenism
Medicine, Surgery & Diagnosis
Physical examination
Polycystic ovarian syndrome
Authors
Affiliations
Sarah A. Kanbour
Division of Endocrinology, Diabetes and Metabolism, Department of Medicine, The Johns Hopkins University School of Medicine, Baltimore, Maryland, USA.
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Adrian S. Dobs adobs@jhu.edu
Division of Endocrinology, Diabetes and Metabolism, Department of Medicine, The Johns Hopkins University School of Medicine, Baltimore, Maryland, USA.
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Notes
Address correspondence to: Adrian S. Dobs, MD, MHS, Division of Endocrinology, Diabetes and Metabolism, Department of Medicine, The Johns Hopkins University School of Medicine, 1830 E Monument Street, Suite 333, Baltimore, MD 21287, USA, adobs@jhu.edu
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Authors' Contributions
S.A.K. wrote the article. A.S.D. provided critical feedback and helped shape it.
Author Disclosure Statement
S.A.K. and A.S.D. have no conflicts of interest.
Funding Information
This research did not receive any specific grant from funding agencies in the public, commercial, or not-for-profit sectors.
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Figures
FIG. 1. PCOS pathophysiology. DHEA-S, dehydroepiandrosterone sulfate; FSH, follicle-stimulating hormone; GnRH, gonadotropin-releasing hormone; LH, luteinizing hormone; PCOS, polycystic ovary syndrome; SHBG, sex hormone–binding globulin.
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Tables
Table 1. Androgens in Women
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Table 2. Diagnostic Criteria for Polycystic Ovary Syndrome
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Table 3. Hirsutism in Women with Polycystic Ovary Syndrome (Average Ferriman–Gallwey Scores of 8 ± 5)
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References
References
Burger HG. Androgen production in women. Fertil Steril. 2002;77(Suppl. 4):S3–S5.
Crossref
PubMed
Google Scholar
a [...] are the most biologically active.
b [...] Table adapted from Burger.
Longcope C. Adrenal and gonadal androgen secretion in normal females. Clin Endocrinol Metab. 1986;15(2):213–228.
Go to Citation
Crossref
PubMed
Google Scholar
Bolour S, Braunstein G. Testosterone therapy in women: A review. Int J Impot Res. 2005;17(5):399–408.
Go to Citation
Crossref
PubMed
Google Scholar
Jayasena CN, Alkaabi FM, Liebers CS, Handley T, Franks S, Dhillo WS. A systematic review of randomized controlled trials investigating the efficacy and safety of testosterone therapy for female sexual dysfunction in postmenopausal women. Clin Endocrinol (Oxf). 2019;90(3):391–414.
Go to Citation
Crossref
PubMed
Google Scholar
Martin KA, Anderson RR, Chang RJ, et al. Evaluation and treatment of hirsutism in premenopausal women: An endocrine society clinical practice guideline. J Clin Endocrinol Metab. 2018;103(4):1233–1257.
Crossref
PubMed
Google Scholar
a [...] concentration is >700 μg/dL (>19.0 μmol/L)
b [...] pills and antiandrogens) is added.
c [...] to be equally effective for hirsutism,
d [...] reducing FG scores by 7 points
e [...] option because of hepatotoxicity risk.
f [...] than placebo for hirsutism treatment,
O'Driscoll JB, Mamtora H, Higginson J, Pollock A, Kane J, Anderson DC. A prospective study of the prevalence of clear-cut endocrine disorders and polycystic ovaries in 350 patients presenting with hirsutism or androgenic alopecia. Clin Endocrinol (Oxf). 1994;41(2):231–236.
Go to Citation
Crossref
PubMed
Google Scholar
Markopoulos MC, Kassi E, Alexandraki KI, Mastorakos G, Kaltsas G. Hyperandrogenism after menopause. Eur J Endocrinol. 2015;172(2):R79–R91.
Go to Citation
Crossref
PubMed
Google Scholar
McCartney CR, Marshall JC. CLINICAL PRACTICE. Polycystic ovary syndrome. N Engl J Med. 2016;375(1):54–64.
Crossref
PubMed
Google Scholar
a [...] and polycystic ovarian morphology
b [...] Table derived from McCartney et al.
c [...] added if suboptimal response after 6 months
d [...] levels by ∼20–25% in women with the PCOS.
e [...] inhibitor letrozole was about 20–30%.
f [...] if there is significant weight gain).
Dumesic DA, Oberfield SE, Stener-Victorin E, Marshall JC, Laven JS, Legro RS. Scientific statement on the diagnostic criteria, epidemiology, pathophysiology, and molecular genetics of polycystic ovary syndrome. Endocr Rev. 2015;36(5):487–525.
Go to Citation
Crossref
PubMed
Google Scholar
Hoeger KM, Dokras A, Piltonen T. Update on PCOS: Consequences, challenges, and guiding treatment. J Clin Endocrinol Metab. 2021;106(3):e1071–e1083.
Go to Citation
Crossref
PubMed
Google Scholar
Azziz R, Carmina E, Dewailly D, et al. Task force on the phenotype of the polycystic ovary syndrome of the androgen excess and PCOS society. The Androgen Excess and PCOS Society criteria for the polycystic ovary syndrome: The complete task force report. Fertil Steril. 2009;91(2):456–488.
Google Scholar
a [...] of PCOS patients with oligomenorrhea,
b [...] DHEA-S are seen in 25–35% of these women.
Huang A, Brennan K, Azziz R. Prevalence of hyperandrogenemia in the polycystic ovary syndrome diagnosed by the National Institutes of Health 1990 criteria. Fertil Steril. 2010;93(6):1938–1941.
Crossref
PubMed
Google Scholar
a [...] of PCOS patients with oligomenorrhea,
b [...] DHEA-S are seen in 25–35% of these women.
Pasquali R, Zanotti L, Fanelli F, et al. Defining hyperandrogenism in women with polycystic ovary syndrome: A challenging perspective. J Clin Endocrinol Metab. 2016;101(5):2013–2022.
Go to Citation
Crossref
PubMed
Google Scholar
Knochenhauer ES, Key TJ, Kahsar-Miller M, Waggoner W, Boots LR, Azziz R. Prevalence of the polycystic ovary syndrome in unselected black and white women of the southeastern United States: A prospective study. J Clin Endocrinol Metab. 1998;83(9):3078–3082.
Go to Citation
PubMed
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Topics
Body mass index
Diagnostic techniques
Endocrine and metabolic disorders
Gonadal disorders
Hyperandrogenism
Medicine, Surgery & Diagnosis
Physical examination
Polycystic ovarian syndrome
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3199 | https://www1.udel.edu/chem/sametz/101Fall09/Ch8temp.pdf | Chapter 8 Chemical Bonding I: Basic Concepts Copyright McGraw-Hill 2009 1 Copyright McGraw-Hill 2009 2 Copyright McGraw-Hill 2009 8.1 Lewis Dot Symbols • Valence electrons determine an element’s chemistry.
• Lewis dot symbols represent the valence electrons of an atom as dots arranged around the atomic symbol.
• Most useful for main-group elements Copyright McGraw-Hill 2009 3 Copyright McGraw-Hill 2009 Lewis Dot Symbols of the Main Group Elements Copyright McGraw-Hill 2009 4 Copyright McGraw-Hill 2009 Write Lewis dot symbols for the following: (a) N (b) S2 (c) K+ Copyright McGraw-Hill 2009 5 Copyright McGraw-Hill 2009 Write Lewis dot symbols for the following: (a) N (b) S2 (c) K+ K+ N• •• • • •• •• S •• •• 2 Copyright McGraw-Hill 2009 6 Copyright McGraw-Hill 2009 8.2 Ionic Bonding Na• Cl• •• •• •• + Na+ Cl •• •• •• •• + IE1 + EA1 = 496 kJ/mol 349 kJ/mol = 147 kJ/mol f H = – 410.9 kJ/mol m.p. = 801oC • Ionic bond: electrostatic force that holds oppositely charge particles together • Formed between cations and anions • Example Copyright McGraw-Hill 2009 7 Microscopic View of NaCl Formation Copyright McGraw-Hill 2009 8 Copyright McGraw-Hill 2009 NaCl(s) Na+(g) + Cl(g) Hlattice = +788 kJ/mol Because they are defined as an amount of energy, lattice energies are always positive.
+ --------+ + + + + + + • Lattice energy = the energy required to completely separate one mole of a solid ionic compound into gaseous ions Copyright McGraw-Hill 2009 9 Copyright McGraw-Hill 2009 Q = amount of charge d = distance of separation d Q1 Q2 • Coulombic attraction: 2 2 1 d Q Q F • Lattice energy (like a coulombic force) depends on • Magnitude of charges • Distance between the charges Copyright McGraw-Hill 2009 10 Lattice energies of alkali metal iodides Copyright McGraw-Hill 2009 11 Copyright McGraw-Hill 2009 The ionic radii sums for LiF and MgO are 2.01 and 2.06 Å, respectively, yet their lattice energies are 1030 and 3795 kJ/mol. Why is the lattice energy of MgO nearly four times that of LiF?
Copyright McGraw-Hill 2009 12 Copyright McGraw-Hill 2009 • Born-Haber cycle: A method to determine lattice energies Copyright McGraw-Hill 2009 13 Copyright McGraw-Hill 2009 • Born-Haber cycle for CaO Ca(s) + (1/2)O2(g) CaO(s) Ca(g) #1 #1 Heat of sublimation = Hf[Ca(g)] = +178 kJ/mol Ca2+(g) #2 #2 1st & 2nd ionization energies = I1(Ca) + I2(Ca) = +1734.5 kJ/mol O(g) #3 #3 (1/2) Bond enthalpy = (1/2) D(O=O) = Hf[O(g)] = +247.5 kJ/mol O2(g) #4 #4 1st & 2nd electron affinities = EA1(O) + EA2(O) = +603 kJ/mol + #5 #5 (Lattice Energy) = Hlattice[CaO(s)] = (the unknown) #6 #6 Standard enthalpy of formation = Hf[CaO(s)] = 635 kJ/mol +178 +1734.5 +247.5 +603 Hlatt = 635 Hlattice = +3398 kJ/mol Copyright McGraw-Hill 2009 14 Copyright McGraw-Hill 2009 8.3 Covalent Bonding • Atoms share electrons to form covalent bonds.
• In forming the bond the atoms achieve a more stable electron configuration.
•H H• + •• H H H–H or Copyright McGraw-Hill 2009 15 Copyright McGraw-Hill 2009 • Octet: Eight is a “magic” number of electrons.
• Octet Rule: Atoms will gain, lose, or share electrons to acquire eight valence electrons Na• Cl• •• •• •• + Na+ Cl •• •• •• •• + Examples: H• •O• •• •• H• O •• •• •• •• H H + + Copyright McGraw-Hill 2009 16 Copyright McGraw-Hill 2009 •Lewis Structures •H H• + •• H H H–H Cl• •• •• •• + Cl• •• •• •• Cl •• •• •• •• Cl •• •• •• Cl •• •• •• Cl •• •• •• – Shared electrons Bonds Non-bonding valence electrons Lone pairs Copyright McGraw-Hill 2009 17 Copyright McGraw-Hill 2009 • Multiple Bonds - The number of shared electron pairs is the number of bonds.
Cl •• •• •• •• Cl •• •• •• Cl •• •• •• Cl •• •• •• – Single Bond O •• •• C •• •• •• •• O •• •• O •• •• O •• •• =C= Double Bond •• •• N •• •• •• N N •• •• N Triple Bond Copyright McGraw-Hill 2009 18 Copyright McGraw-Hill 2009 • Bond strength and bond length bond strength single < double < triple bond length single > double > triple N–N N=N NN Bond Strength 163 kJ/mol 418 kJ/mol 941 kJ/mol Bond Length 1.47 Å 1.24 Å 1.10 Å Copyright McGraw-Hill 2009 19 8.4 Electronegativity and Polarity • Nonpolar covalent bond = electrons are shared equally by two bonded atoms • Polar covalent bond = electrons are shared unequally by two bonded atoms Copyright McGraw-Hill 2009 20 red high electron density green intermediate electron density blue low electron density • Electron density distributions + -H – F alternate representations H – F Copyright McGraw-Hill 2009 21 Copyright McGraw-Hill 2009 • Electronegativity: ability of an atom to draw shared electrons to itself.
- More electronegative elements attract electrons more strongly.
• relative scale • related to IE and EA • unitless • smallest electronegativity: Cs 0.7 • largest electronegativity: F 4.0 Copyright McGraw-Hill 2009 22 Copyright McGraw-Hill 2009 Electronegativity: The Pauling Scale Copyright McGraw-Hill 2009 23 Copyright McGraw-Hill 2009 Variation in Electronegativity with Atomic Number Copyright McGraw-Hill 2009 24 Copyright McGraw-Hill 2009 • Polar and nonpolar bonds 2.1 - 2.1 = 0.0 4.0 - 2.1 = 1.9 4.0 - 0.9 = 3.1 nonpolar covalent polar covalent ionic > 2.0 is ionic Copyright McGraw-Hill 2009 25 Copyright McGraw-Hill 2009 • Dipole moments and partial charges - Polar bonds often result in polar molecules.
- A polar molecule possesses a dipole.
- dipole moment () = the quantitative measure of a dipole = Qr r +Q – Q + -H – F SI unit: coulomb•meter (C•m) common unit: debye (D) 1D = 3.34 1030 C•m HF 1.82 D HCl 1.08 D HBr 0.82 D HI 0.44 D Copyright McGraw-Hill 2009 26 Copyright McGraw-Hill 2009 8.5 Drawing Lewis Structures 1) Draw skeletal structure with the central atom being the least electronegative element.
2) Sum the valence electrons. Add 1 electron for each negative charge and subtract 1 electron for each positive charge.
3) Subtract 2 electrons for each bond in the skeletal structure.
4) Complete electron octets for atoms bonded to the central atom except for hydrogen.
5) Place extra electrons on the central atom.
6) Add multiple bonds if atoms lack an octet.
Copyright McGraw-Hill 2009 27 Copyright McGraw-Hill 2009 What is the Lewis structure of NO3 ?
1) Draw skeletal structure with central atom being the least electronegative.
O O – N – O – 4) Complete electron octets for atoms bonded to the central atom except for hydrogen.
:O: :O – N –O: – : : : : : 18e – 6) Add multiple bonds if atoms lack an octet.
:O: :O – N = O: – : : : : 24e 5) Place extra electrons on the central atom.
2) Sum valence electrons. Add 1 for each negative charge and subtract 1 for each positive charge.
NO3(1 5) + (3 6) + 1 = 24 valence e 24e 3) Subtract 2 for each bond in the skeletal structure.
6 e Copyright McGraw-Hill 2009 28 Copyright McGraw-Hill 2009 Copyright McGraw-Hill 2009 29 Copyright McGraw-Hill 2009 8.6 Lewis Structures and Formal Charge • The electron surplus or deficit, relative to the free atom, that is assigned to an atom in a Lewis structure.
Formal charges are not “real” charges.
H: orig. valence e= 1 non-bonding e= 0 1/2 bonding e= 1 formal charge = 0 O: orig. valence e= 6 non-bonding e= 4 1/2 bonding e= 2 formal charge = 0 Example: H2O = H:O:H : : Total valence electrons Formal Charge = Total non-bonding electrons Total bonding electrons 11 2 Copyright McGraw-Hill 2009 30 Copyright McGraw-Hill 2009 Example: Formal charges on the atoms in ozone 6 4 1 2 4 0 6 2 1 2 6 1 6 6 1 2 2 1 O O O O O O Copyright McGraw-Hill 2009 31 Copyright McGraw-Hill 2009 Formal charge guidelines −A Lewis structure with no formal charges is generally better than one with formal charges.
−Small formal charges are generally better than large formal charges.
−Negative formal charges should be on the more electronegative atom(s).
Example: Answer: or ?
H C O H C O H H H C O H •• •• + C O H H •• •• Copyright McGraw-Hill 2009 32 Copyright McGraw-Hill 2009 Identify the best structure for the isocyanate ion below: (a) :C = N = O::: – :C N – O::: – (c) (b) :C – N O::: – 2 +1 3 +1 +1 +1 1 1 0 Copyright McGraw-Hill 2009 33 Copyright McGraw-Hill 2009 33 Copyright McGraw-Hill 2009 Identify the best structure for the isocyanate ion below: (a) :C = N = O::: – :C N – O::: – (c) (b) :C – N O::: – 2 +1 3 +1 +1 +1 1 1 0 Copyright McGraw-Hill 2009 34 Copyright McGraw-Hill 2009 8.7 Resonance Two resonance structures, their average or the resonance hybrid, best describes the nitrite ion.
:O – N = O: : : : : – :O = N – O: : : : : – Solution: The double-headed arrow indicates resonance.
:O – N = O: : : : : – These two bonds are known to be identical.
• Resonance structures are used when two or more equally valid Lewis structures can be written.
Example: NO2 Copyright McGraw-Hill 2009 35 Copyright McGraw-Hill 2009 Benzene: C6H6 Additional Examples Carbonate: CO3 2 or Copyright McGraw-Hill 2009 36 Copyright McGraw-Hill 2009 8.8 Exceptions to the Octet Rule • Exceptions to the octet rule fall into three categories: −Molecules with an incomplete octet −Molecules with an odd number of electrons −Molecules with an expanded octet Copyright McGraw-Hill 2009 37 Copyright McGraw-Hill 2009 • Incomplete Octets Example: BF3 (boron trifluoride) BF3 (1 3) + (3 7) = 24 val. e :F: :F – B = F: – : : : : +1 -1 −Common with Be, B and Al compounds, but they often dimerize or polymerize.
Example: Cl Cl Cl Be Be Be Be Cl Cl Cl–––––––––––––––– :F: :F – B – F: – : : : : : no octet Copyright McGraw-Hill 2009 38 Copyright McGraw-Hill 2009 • Odd Numbers of Electrons Example: NO (nitrogen monoxide or nitric oxide) NO (1 5) + (1 6) = 11 valence e Example: NO2 (nitrogen dioxide) NO2 (1 5) + (2 6) = 17 val. e :N=O:: .
:N=O:: .
Are these both equally good? :O=N –O:::: .
:O–N=O:::: .
:O=N –O:::: .
:O–N=O:::: .
Are these all equally good? better 0 0 1 +1 0 0 0 0 0 0 0 +1 1 1 +1 0 best Copyright McGraw-Hill 2009 39 Copyright McGraw-Hill 2009 • Expanded Octet −Elements of the 3rd period and beyond have d-orbitals that allow more than 8 valence electrons.
XeF2 = :F – Xe – F: : : : : ::: (Xe has 10 valence electrons) 22 valence e (S has 12 valence electrons ) F F S F F–––– :F: – –:::::: :F::::::::: SF6 = 48 valence e Copyright McGraw-Hill 2009 40 Copyright McGraw-Hill 2009 8.9 Bond Enthalpy • Bond enthalpy is the energy associated with breaking a particular bond in one mole of gaseous molecules.
HCl(g) H(g) + Cl(g) Ho = 431.9 kJ Cl2(g) Cl(g) + Cl(g) Ho = 243.4 kJ O2(g) O(g) + O(g) Ho = 495.0 kJ N2(g) N(g) + N(g) Ho = 945.4 kJ single bonds double bond triple bond −For diatomic molecules these are accurately measured quantities. −Bond enthalpy is one measure of molecular stability. −Symbol: Ho Copyright McGraw-Hill 2009 41 Copyright McGraw-Hill 2009 −Bond enthalpies for polyatomic molecules depend upon the bond’s environment. −Average bond enthalpies are used for polyatomic molecules.
• Provide only estimates H = 435 kJ H H – C – H H – – H H – C H – – + H H = 410 kJ 6% less + H H H – C H – – H – C – H H – – H H – C H – – H – C H – – Copyright McGraw-Hill 2009 42 Copyright McGraw-Hill 2009 • Prediction of bond enthalpy reactants atoms products BE(r) BE(p) enthalpy Ho = BE(reactants) BE(products) Copyright McGraw-Hill 2009 43 Copyright McGraw-Hill 2009 Example: Calculate the enthalpy of reaction for CH4(g) + Br2(g) CH3Br(g) + HBr(g) Solution: Consider ONLY bonds broken or formed.
H H – C – H H – – Br – Br H – Br + + H H – C – Br H – – Hrxn = [BE(C–H) + BE(Br–Br)] – [BE(C–Br) + BE(H–Br)] = [(413) + (193)] – [(276) + (366)] = – 36 kJ/mol Copyright McGraw-Hill 2009 44 Copyright McGraw-Hill 2009 Copyright McGraw-Hill 2009 45 Key Points • Lewis dot symbols • Ionic bonding • Lattice energy • Born-Haber cycle • Covalent bonding • Octet rule • Lewis structures • Bond order • Bond polarity Copyright McGraw-Hill 2009 46 Key Points • Electronegativity • Dipole moment • Drawing lewis structures • Formal charge • Resonance structures • Incomplete octets • Odd numbers of electrons • Expanded octets • Bond enthalpy |
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