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# Audit Log Overflow Due To Log Rotation
This book, which provides comprehensive coverage of the ever-changing field of SSL/TLS and Web PKI, is intended for IT security professionals, system administrators, and developers, with the main focus on getting things done. Troubleshooting. I see quite a few audit organizations that include a Web-based explanation to their clients how the audit process works. This is an advantage over the per-deployment logging configuration as the redeploy is not required for logging changes to take affect. On the Server Maintenance page, under Log Files, click Generate Log File Snapshot. log_truncate_on_rotation (boolean) When logging_collector is enabled, this parameter will cause PostgreSQL to truncate (overwrite), rather than append to, any existing log file of the same name. By default the log file snapshot is saved to: C:\ProgramData\Tableau\Tableau Server\data\tabsvc\files\log-archives. 39, which can be obtained in the tools/adb39 under the IDE directory. ” One of the main advantages of using agents is the ability to easily get the agent configured to monitor logs of any type no matter where those logs live on your file system. But relay output is not reliable and gives random switching sometimes. -Summary: Integer overflow lead to heap overflow +Summary: Heap Overflow due to integer overflows [2016-06-20 08:54 UTC] nguyenvuhoang199321 at gmail dot com Another code block that lead to integer overflow when check encrypt data is not a block cipher. How do you define the name for the localhost access log file to something other than false2015-11-11? How to create the access log files with name localhost_access_log. Orchestrate your PKI infrastructure to ensure that your keys and certificates are always valid, trusted and compliant. The Overflow is the official quarterly publication of Irrigation Australia in WA region. rotation synonyms, rotation pronunciation, rotation translation, English dictionary definition of rotation. However, truncation will occur only when a new file is being opened due to time-based rotation, not during server startup or size-based rotation. 28, urgency: MEDIUM. It got me thinking: Shouldn't someone who has a. ES 1 to 5 cover the integrity, objectivity and independence for auditors and apply in the audit of financial statements. That library is a thin wrapper above different logging frameworks. (i) in case of an auditor (whether an inpidual or audit firm), the period for which the inpidual or the firm has held office as auditor prior to the commencement of the Act shall be taken into account for calculating the period of five consecutive years or ten consecutive years,. Internal controls include the policies and procedures that financial institutions •establish to reduce risks and ensure they meet operating, reporting, and compliance objectives. Unable to connect to the vCenter Server as services are not started. old Use /dev/null for clean file. Learn, teach, and study with Course Hero. Computational Science Stack Exchange is a question and answer site for scientists using computers to solve scientific problems. The log method is responsible to handle log event according for the mentioned equation. Now I wondered how it could be that our clocks measure 24 hours for one day instead of only 23 hours, 56 minutes, and 4. Moreover, since it is proportional to $\omega$ (more exactly, for running perpendicular to the rotation axis, it's $2\omega v$), it only falls off with the square root of the radius, so I guess that is the determining factor to decide which radius is. By default, Tableau Server log file archives are gathered in a zip file called logs. This changelog does not contain all updates and fixes to the Tomcat connectors (yet). log Now, until you stop the script, all input and output in the Terminal will be stored in screen. 33-8220, "Standards Related to Listed Company Audit Committees") require audit committees to approve all audit services provided to the company, whether provided by the principal auditor or other firms. It got me thinking: Shouldn't someone who has a. One per rotation (core and non-core) submitted with the Supervisor’s Report at the end of each rotation and due by 31 May/31 October. x? We are not able to find how to configure http connector logging. Sometimes this is useful after adding new entries to logrotate, or if old log files have been removed by hand, as the new files will be created, and logging will con- tinue correctly. For the ACC Network to succeed, it needs unique content (not overflow) New, 32 comments At a time when cord-cutting is hurting every network, how does the ACC create something compelling on cable?. Running /etc/cron. What's the purpose of key-rotation? Does it have any effect on the probability of keys being breached in the first place? Does it refer to avoiding access after a breach to all past data, all future. Setting log files to mode 600 should. Audit quality in a mandatory audit firm rotation setting In order to improve Vietnamese auditors' knowledge of audit quality, independent audit and factors influencing audit quality, auditing firms need to provide more professional training for their staff. Upon reaching this size, the audit log will be rotated. - The purpose of this paper is to introduce the problem of the lack of auditor independence in the Egyptian context, how it might affect the audit quality, through assessing reasons behind the voluntary switching of auditors, whether this switch is in the side of improving audit quality or not and the suggestion of the mandatory auditor rotation as a solution to such a problem. Audit logs should be protected by implementing strict access control. They perform a decent job to collect events on running systems but they need to deploy extra piece of software on the target operating systems. This log template is useful for planning ahead or creating a record of the day's events. The auditd daemon writes "Audit daemon is low on disk space for logging" in messages log file when the log destination filesystem does not have enough free space. In the Audit Log Archive/Purge dialog box, click Continue to archive and purge the logs. D&D 1st edition, D&D 5e and Dark Heresy gaming sessions are going great and now it is time to introduce a new gaming system into the. Due to the rotation you get a Coriolis force. The entire audit of an issuer must be conducted by independent. old Optional: Use Gzip on copy of log. Of course, now that I've rebooted the server the SQL Server (MSSQLSERVER) service won't start, and I'm pretty sure it's because the event trace can't audit. One problematic scenario is when your /var/log/messages file fills up due to logging misconfiguration. Question 12. How to add the suffix log to the filename in JBoss EAP 6. /var/log/audit/audit. n+1, and the last log file being deleted) the current log file is truncated without disturbing the writing process. In this environment, it would be suggested that /var/log/audit be on its own partition. Using a logging profile also allows for runtime changes to the configuration. A set of log files is unique to a storage group. The story dates back to 2016, when a Spanish bank, BBVA, was forced to solicit bids from other audit firms and switch from an audit relationship with Deloitte's Spanish firm due to European. - The purpose of this paper is to introduce the problem of the lack of auditor independence in the Egyptian context, how it might affect the audit quality, through assessing reasons behind the voluntary switching of auditors, whether this switch is in the side of improving audit quality or not and the suggestion of the mandatory auditor rotation as a solution to such a problem. Transaction Log files- the log file stores altered data before it is committed to the database. This is an advantage over the per-deployment logging configuration as the redeploy is not required for logging changes to take affect. log file, rebooted, and the services all started. Note that there is a very small time slice between copying the file and truncating it, so some logging data might be lost. The file name must be unique within the database. x you can now expand VMDK's on the fly since the VCSA takes advantage of LVM. So you have an Audit Table to track what is being done on that table. Your can track unauthorized actions or the leakage of information. Start studying Security + Test 3. By default, the audit system logs audit messages to the /var/log/audit/audit. Do NOT ever forget to rotate logs. -Summary: Integer overflow lead to heap overflow +Summary: Heap Overflow due to integer overflows [2016-06-20 08:54 UTC] nguyenvuhoang199321 at gmail dot com Another code block that lead to integer overflow when check encrypt data is not a block cipher. Every successful audit is based on sound planning and an atmosphere of constructive involvement and communication between the client and the auditor. This type of audit log is disabled by default due to the high volume of data gathered. You can collect, store, index, search, correlate, visualize, analyze and report on any machine-generated data to identify and resolve operational and security issues in a. If I try to rotate a bone in pose mode or a mesh part in object mode, I move only a thin black line around the rotation- Stack Exchange Network Stack Exchange network consists of 175 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Q&A for Graphic Design professionals, students, and enthusiasts. Set log_rotation_size to 0 to disable size-based log rotation, as it makes the log file name difficult to predict. In a max heap, each node's children must be less than itself. That was what I needed! I deleted the audit. For security reasons, please log out and exit your web browser when you are done accessing services that require authentication!. The Administration Server writes configuration auditing messages to its local log file. Toyo Kogyo has been practicing job rotation for over 20 years as a direct result of the oil embargo of the 1960s. you should be able to see a box where you can enter the Server Name for Remote Syslog under the Application Audit log Settings section. IDW Publishing is raising funds for Full Bleed Vol. You can configure log audit backend using the following kube-apiserver flags:--audit-log-path specifies the log file path that log backend uses to write audit events. when the cPanel Log Rotation about the suPHP Apache module audit logs. x you can now expand VMDK’s on the fly since the VCSA takes advantage of LVM. One per rotation (core and non-core) submitted with the Supervisor’s Report at the end of each rotation and due by 31 May/31 October. This log template is useful for planning ahead or creating a record of the day's events. By default, Tableau Server log file archives are gathered in a zip file called logs. The first Google search says /var/log/syslog but I don't see any information related to sudo there. Thus I have to enable logon audit events through the Registry. SELinux was restricting the access to logrotate on log files in directories which does not have the required SELinux file context type. And Windows Server 2016 provides basic and advanced security auditing and integrates with third-party auditing solutions. Of course, now that I've rebooted the server the SQL Server (MSSQLSERVER) service won't start, and I'm pretty sure it's because the event trace can't audit. Improve speed of showing cards. During S2LA, travelling with the rotation of the earth, would the flight time be longer than LA2S on account of Los Angeles Stack Exchange Network Stack Exchange network consists of 175 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit PepBoys. However, in the new file I cannot find any logs (that I thought would contain logs after the log rotation). Double click the folder to start the log collection wizard. Developer Student Clubs train thousands of student developers globally and work with their communities to solve real-life problems. Lori-Lynne's Coding Coach Blog This is a blog dedicated to Medical Coding professionals,to find help with coding, billing, payment, revenue, medical records issues and other ancillary concerns for those "worker bees" that perform the difficult job of "coding". 39, which can be obtained in the tools/adb39 under the IDE directory. How fast is Earth moving through the universe? Speed has to be measured in relation to something, so where is the point of relation in the universe? If there is something like the center of the universe far beyond our observable universe, this could be used as a point of measure. For the most part, systems typically can take care of managing log files so they do not eventually eat up available hard drive space. Audit fee discounts generally do not lead to a reduced audit effort, or respectively, audit quality is not impaired when client bargaining power is strong. Added log file reopening file editing with Windows GUI. Bulletproof SSL and TLS is a complete guide to deploying secure servers and web applications. conf is the file that controls the configuration of the audit remote logging subsystem. n limit (if configured), and the p4 logrotate command. Using a logging profile also allows for runtime changes to the configuration. 1) with a new service-level option "nsi". when the cPanel Log Rotation about the suPHP Apache module audit logs. Not all do this successfully, however. It doesn't get compromised until they actually do something nefarious, in which cse the audit log on the remote server will show you what was done and by whom - including who it was that disabled auditd. I'm searching help for my son, who is suffering from Autism. Numbers provided for the configurables do not have to be consecutive. Plus, there will be times when your system’s logging will cause unforeseen issues. When this setting is unchecked, audit log ignores the Maximum No. To add a data or transaction log file, click Add. Faster search and retrieval of file audit data with ADAudit Plus's all new DataEngine. Whenever a write to the log file causes its size to exceed the audit_log_rotate_on_size value, the audit log plugin closes the current log file, renames it, and opens a new log file. Workaround: Follow the steps outlined in Knowledge Base article 57917 to fix the issue. From collecting logs in the. To deal with the terabytes of event log data these devices generate, security administrators can use EventLog Analyzer, a powerful log management tool that covers end-to-end event log management. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Log rotation policies dictate how often log files are rotated, that is to say, how long log files are kept based on various criteria. A version's change log includes entries for both the development and production releases of that version. Get unstuck. [RESOLVED][NETWORK ISSUES] Sporadic Issues with the FAUWPA2 wireless network. This chapter contains these topics: Section 23. As a result, the old log file gets zipped into. Thus, for some rotation schemes the log files will be analyzed and reported in their original order. You can purge audit logs before a specified date and time or audit logs that are older than a specified number of days. A: The Commission's rules relating to listed company audit committees (see Release No. Rotation vector does not function well if magnetic field sensor is not calibrated. The sequence number is appended to the log file name upon rotation. This log maintain metrics of internal queues, internal processor, etc. 33-8220, "Standards Related to Listed Company Audit Committees") require audit committees to approve all audit services provided to the company, whether provided by the principal auditor or other firms. This is all due to the fact that the variability of the data has been written in loadings, as their load. This policy provides guidance on the provision of external audit services by any person engaged to perform external audit services in relation to the Mercury. You can either set a schedule for forwarding the local file to the appropriate machine (the Security Management server or a Log Server), or manually import these files using SmartView Tracker. Read this for more details. Companies experienced with in-store audits report that their greatest successes come when the audit data are integrated with follow-up actions. If I start cycling the logs on a daily basis, it seems I'd need to change my server limit to 365 logs at bare minimum. The integrity of the data should be ensured with the use of digital signatures, message digest tools ,and strong access control. For the ACC Network to succeed, it needs unique content (not overflow) New, 32 comments At a time when cord-cutting is hurting every network, how does the ACC create something compelling on cable?. And they can return that load back to the data points any time - now or after rotation. org/buglist. Note that there is a very small time slice between copying the file and truncating it, so some logging data might be lost. Originally there were. SELinux was restricting the access to logrotate on log files in directories which does not have the required SELinux file context type. Create a LOG_BACKUP Maintenance Plan to take backup logs frequently. Please read the topic above about the log file difference between pre and post 4. In the Database files grid, enter a logical name for the file. Many synchronous callback calls in series can quickly overflow the JS call stack. (kkolinko) 49180: Add an option to disable file rotation in JULI FileHandler. Simply put, the OMS Log Analytics Forwarder is an HTTP forward proxy that supports HTTP tunneling through the HTTP CONNECT command. We have judged all submissions, and are pleased to announce the runners up and winner of the 2015 Underhanded C Contest. Information written by the Data Relationship Management Batch Client. The auditing will happen using the Audit API of theserver. For security reasons, please log out and exit your web browser when you are done accessing services that require authentication!. The flush parameter should be set to sync or data. Using unattended-upgrades on Debian and Ubuntu. Log rotation. Upon rebooting this evening, my original document is placing parentheses as expected. log is normally set to growth to infinite. This is an advantage over the per-deployment logging configuration as the redeploy is not required for logging changes to take affect. # tail /var/log/audit/audit. Figure out what internal customer owns the data inside the database, and walk into their office. During S2LA, travelling with the rotation of the earth, would the flight time be longer than LA2S on account of Los Angeles Stack Exchange Network Stack Exchange network consists of 175 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. In spite of that new 11g feature, the housekeeping or purging of those files got not. log files are stored in the same directory. You can configure criteria that determine when a new log file must be created, based on the size of the log file and the age of the log file. Visit PepBoys. 23 Managing Logging. I don't think it is wrong to ask where to go for help if I suspect security issues due to a repeating entry in the security section of the event log. You will then run df -h to see the disk space usage. Lager (as in the beer) is a logging framework for Erlang. Whenever a write to the log file causes its size to exceed the audit_log_rotate_on_size value, the audit log plugin closes the current log file, renames it, and opens a new log file. Critical Server Log Management with EventLog Analyzer's Predefined Reports. Using unattended-upgrades on Debian and Ubuntu. If you do your due diligence and plan log file monitoring properly, you'll be able to troubleshoot problems more easily and you'll be fully prepared for an audit. Five open mouth views in rotation and lateral flexion may not be safe/advisable for acute trauma, where there is neck pain and significant mechanism of injury, and certainly not if there is any altered sensorium. , the 2003 audit) before the new rules became effective. Use: Provides the record of duty status for each driver. They perform a decent job to collect events on running systems but they need to deploy extra piece of software on the target operating systems. Cause: First, make sure the magnetic field sensor is well calibrated. For this variable to take effect, set the audit_log_handler variable to FILE and the audit_log_rotations variable to a value greater than zero. What is the Registry? A. 10 July 20, 2015 · by Hector · in Computer Stuff · 6 Comments Today I faced some issue when using Fortify. On the Server Maintenance page, under Log Files, click Generate Log File Snapshot. Most commonly this is due to multiple appenders attempting to use the same file path and most likely by having multiple independent instances of log4j read the same configuration file, however having the log file open by another process (an editor, backup utility) can also interfere with rolling. This chapter contains these topics: Section 23. However, as of March 9, either of these methods have a 15 day hold on sales and trades unless both people have Steam Mobile Authentication set up. If your setup generally exceeds 50 GB per hour, it's recommended to split the traffic between multiple log collectors. Logs-based metric can also be used to extract values from logs and create a a distribution of the values. For example: \$ p4 configure set serverlog. Options -e. Contents of audit log files produced by the audit log plugin are not encrypted and may contain sensitive information, such as the text of SQL statements. Fluentd allows you to unify data collection and consumption for a better use and understanding of data. If we conceive of an orthogonal rotation such as varimax that means that we want the components to remain uncorrelated after the rotation done. June 13, 2001 CODE OF FEDERAL REGULATIONS 32 Part 800 to End Revised as of July 1, 2001 National Defense Containing a codification of documents of general applicability and future effect As of July 1, 2001 With Ancillaries. It is easy to miss some important comment hidden in the middle of a lengthy discussion due to this (which is why we move comments to chat in the first place). Shop for Signature Design by Ashley Rotation Reclining Loveseat with Console, 9750194, and other Living Room Two Cushion Loveseats at Sofas Unlimited in Central PA. Such account logon events are generated and stored on the domain controller, when a domain user account is authenticated on that domain controller. This is to ensure that space detection is accurate and that no other process comes along and consumes part of it. And they can return that load back to the data points any time - now or after rotation. All staff will be familiar with the medication system irrespective of where they work in Queensland, reducing the risk of errors due to unfamiliarity with different systems. msc or secpol. Otherwise, log files may become huge. Lastly, another set of logs that could be filling up is the Http Proxy log. If it is a cluster, you could try failing back to the other node and see if that fixes the problem, which in my case was the need for a recompile. Each instance has a name, and they are conceptually arranged in a namespace hierarchy using dots (periods) as separators. We have judged all submissions, and are pleased to announce the runners up and winner of the 2015 Underhanded C Contest. The flush parameter should be set to sync or data. , in contrast, the regulator decided against such rotation. • Inventory rotation practices generally improve where in-store audits occur, but this activity can be expensive. I'd like to create my own logrotate rules to rotate it when it hits a certain size (because the files are too large for my liking), but I don't know where to change it. log seems to be getting rotated, but I'm not sure what's causing it because there doesn't seem to be configuration for it in any logrotate config files. Default Log File Locations Managed Domain. Private keys will even form part of securing most block chain technology. 17 - Auditors get Notice to Scrutinize SAB 74 Disclosures. Not specifying this flag disables log backend. The first Google search says /var/log/syslog but I don't see any information related to sudo there. Whereas proponents argue that rotation would strengthen independence and decrease audit market concentration, opponents stress the importance of auditors' learning effects, which would be eliminated by a change in auditors. Basically there's no need to take care about log rotation in the appliance due to symlinks pointing to the current log files. [On] is specified by default. Gardening & Landscaping Stack Exchange is a question and answer site for gardeners and landscapers. The stream mode is my personal choice, because it offers real-time viewing of audit events due to the audit log file being written to in text mode. Job rotation increases the variety. There is a lot of interesting stuff in the Android system log, that is helpful in many ways find root causes of problems identify misbehaving apps How can I view and examine the Android log?. x along with libKML support, looks like I'll have to. Audit Account Logon Events policy defines the auditing of every event generated on a computer, which is used to validate the user attempts to log on to or log off from another computer. When a program crashes, if there is no logging record, you have little chance to understand what happened. * Fixed log_host_retries not logging the host soft state checks (#599) (Jake Omann) * Fixed stalking_options N option to properly log only when a notification is actually sent (#604) (Jake Omann) * Fixed issue with service status totals being zero when servicegroup=all on servicegroup status page (#579) (Jake Omann). [Audit Log]: Enables you to obtain user operation or job history. Oracle Unified Directory provides the following four log rotation policies: 24 Hours time limit rotation policy. The fix ensures that the Previous_gtids_log_event is correctly skipped and that the correct Format_description_log_event is found in existing relay logs after restarting the slave's SQL thread. Three of Jaeger-LeCoultre's most complex watches have been redesigned with pink gold and opaline magnetite dials ahead of SIHH 2017. log and Netlogon. 4 Company Losses Due to Internal Control An internal audit helps you locate areas where your business can save time and money by examining the day-to-day tasks of workers and pointing out. How to fix VCSA /storage/log filesystem out of disk space. The confidentiality can be protected with encryption and access controls and can be stored on CD-ROM'S to prevent loss or modification of the data. location, verified log, edited log, and violations. When you are ready to start recording a log file, type: script screen. Please read the topic above about the log file difference between pre and post 4. when the cPanel Log Rotation about the suPHP Apache module audit logs. audisp-remote. Originally there were. – user5071535 Nov 16 '15 at 18:23. Log files are not encrypted during use or after log rotation. After even more time the log files are compressed with the gzip utility as in the case of the example daemon. These can generate quite a bit and at 500MB+ a log file you can run out of space rather quickly. rules to meet new Common Criteria. Q&A for information security professionals. 33-8220, "Standards Related to Listed Company Audit Committees") require audit committees to approve all audit services provided to the company, whether provided by the principal auditor or other firms. The files produces by this Valve are rolled over nightly at midnight. Model Audit Rule is a financial reporting regulation applicable to insurance companies, and borrows significantly from the Sarbanes Oxley Act of 2002 (see 'key sections' below). Hi all I am trying to add secure and audit logs to logrotate for a client whom wants the logs for a period of 6 months, compressed/zipped weekly for auditing. A buffer overflow occurs when data written to a buffer also corrupts data values in memory addresses adjacent to the destination buffer due to insufficient bounds checking. The purpose of log rotation is to archive and compress old logs so that they consume less disk space, but are still available for inspection as needed. Each call allocates its own buffer, which exists for the duration of the call. of Files: Enter the maximum number of files that you want to include in the log. ini files in Windows, however the problem with. Both binary search trees and binary heaps are tree-based data structures. Remember that the total disk space that's used by Netlogon logging is the size that's specified in the maximum log file size times two (2). Lager (as in the beer) is a logging framework for Erlang. To help debug issues related to request execution or client access to your API, you can enable Amazon CloudWatch Logs to log API calls. SELinux was restricting the access to logrotate on log files in directories which does not have the required SELinux file context type. Read this for more details. A single complete cycle of such motion. Start studying Security + Test 3. The PET ruled that "the custody of the official, printed and authenticated copies of the decrypted ballot images, election returns and audit logs from the protested clustered precincts of the said pilot provinces should remain with the Tribunal for the conduct of the revision proceedings pursuant to the 2010 PET Rules. Typically that'll include the basic Who, What, When. Oracle alert. log_archives setting, and change the location by specifying a new value for basefilepath. There are additional, non-syslog files maintained you should know about. In order to write logs to a pipe, simply create the FIFO at the desired path (man mkfifo(1)), and use that path in the logging configuration directive. (kkolinko). [Counting Log]: Enables you to obtain information about paper consumption and the reduction rate of paper used for printing. The stream mode is my personal choice, because it offers real-time viewing of audit events due to the audit log file being written to in text mode. For example, changes in cPanel & WHM version 65 appear in the change log for version 66. Setting proper permissions on log files is critical. Log rotation. As always, when all eigenvalues are unity, the rotation is a null rotation. Recently I have reviewed XGBoost algorithm and I have noticed that this algorithm can handle missing data (without requiring imputation) in the training phase. This log is useful only for pre-4. you should be able to see a box where you can enter the Server Name for Remote Syslog under the Application Audit log Settings section. Another important log file is Xorg. When a program crashes, if there is no logging record, you have little chance to understand what happened. I'm trying to figure out in which log this information is logged, for check who tried to run a command with sudo for example, but can't find it. This log includes details about an Import operation. Click [OK] when done. Enable Audit Log Rotation: The system reads this option to determine whether it needs to rotate the database audit log files or it needs to continue to create new files. when the cPanel Log Rotation about the suPHP Apache module audit logs. Rotation vector does not function well if magnetic field sensor is not calibrated. Is there a Stack Exchange site where you can ask questions that are marked as off-topic on Stack Overflow because someone asked to recommend or find a tool, library or favorite off-site resource? support site-recommendation recommendation-questions software-recs-se. What is the Registry? A. It should contain fixes made only after November 10th 2004, when the new documentation project for JK was started. Start studying Security + Test 3. • Audits can reinforce the importance of inventory rotation with store personnel. A single complete cycle of such motion. How to increase memory in HP Fortify Audit Workbench 4. This policy provides guidance on the provision of external audit services by any person engaged to perform external audit services in relation to the Mercury. Never pondered that Console could read logs on a remote device. Is there an easy explanation of this?. Added log file reopening file editing with Windows GUI. Connected to: Oracle Database 11g Enterprise Edition Release 11. log, currently it is access-log. Your can track unauthorized actions or the leakage of information. n+1, and the last log file being deleted) the current log file is truncated without disturbing the writing process. I got around this, by deciding that for my application, the rotation was more important than a few missed log-lines and simply doing a copy, then truncate (though, I'm sure that there was more to it than that. Set log_truncate_on_rotation to on so that old log data isn't mixed with the new in the same file. A simple guide to Tomcat logging configurations | MuleSoft. So the solution was to set this on my application log files and it's parent directory:. Ready to implement job rotation in your workplace? Get started with this step-by-step guide. log file, rebooted, and the services all started. The event log is labeled security. The forensic audit, submitted to IL&FS board last week, also found 107 instances of loan ever-greening, loans without collateral, and of links between the IL&FS management and borrowers’ companies. Automatic log rotation ensures that old logs stored using a naming convention so that they may be retained. Whenever a write to the log file causes its size to exceed the audit_log_rotate_on_size value, the audit log plugin closes the current log file, renames it, and opens a new log file. This book, which provides comprehensive coverage of the ever-changing field of SSL/TLS and Web PKI, is intended for IT security professionals, system administrators, and developers, with the main focus on getting things done. From fairytales to hollywood blockbusters, “the rule of three” (Latin-"omne trium perfectum") principle suggests things that come in threes are inherently more humorous, satisfying and effective th. Manner of rotation of auditor by the companies. Anecdotally, I've seen some "real world" log sources that do encourage users to apply this directive. Question 7 Q : Assume that in the previous situation, the "concurring" partner had only been out one year ( e. In addition to capturing CloudTrail logs in a specified Amazon S3 bucket for long-term analysis, you can perform real-time analysis by configuring CloudTrail to send logs to CloudWatch Logs. The Audit Contracting Process The State Auditor is required by New Mexico law to conduct annual financial audits of all government agencies in New Mexico. This is because 65 is the development version whose changes are released in the production version 66. This log includes details about an Import operation. Everything seems to be perfect. Information about the activity will be stored in the XML log file where each event will have its NAME field, its own unique RECORD_ID field and a TIMESTAMP field. log What if several other events have been recently recorded, finding the specific information would be so difficult, but using ausearch , you can pass the -k flag with the key value you specified in the audit rule to view all log messages concerning events to do with accessing or modifying /etc/passwd file. For other rotation schemes the original log file order will not be honored, which can lead to reporting matched log file records in altered order (the problem does not happen if log files have different last modification times). You can use logrotate to handle most other generic log rotation tasks,. It's time for a reality check. Sometimes this is useful after adding new entries to a logrotate config file, or if old log files have been removed by hand, as the new files will be created, and log- ging will continue correctly. I'd like to create my own logrotate rules to rotate it when it hits a certain size (because the files are too large for my liking), but I don't know where to change it. I also needed to deal with the log disk at 100% capacity. Gardening & Landscaping Stack Exchange is a question and answer site for gardeners and landscapers. 4 users of audit to Ranger 0. Some other types of Log files may be there depending on your installed packages. 33-8220, "Standards Related to Listed Company Audit Committees") require audit committees to approve all audit services provided to the company, whether provided by the principal auditor or other firms. It is also possible that yaw axis fails due to video taking.
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# Thermal Physics and Differentials
A hypothetical substance has a compressibility k = a /V and a volume expansivity
B = 2bT /V , where a and b are constants and V is the molar volume. Show that the
equation of state is:
V = bT2 - aP + constant
To be honest I'm not entirely sure what I'm actually supposed to be doing with this question. Do I treat it as an ideal gas and therefore use that equation or is that completely wrong.
So far I have integrated each of the equations in respect to T, giving me,
(integral of) a/V dT = aT/V + c
(integral of) 2bT/V dT = bT^2/V + c
But now I'm stuck and can't seem to find a relevant relationship between P and what I've got there.
## Answers and Replies
vela
Staff Emeritus
Science Advisor
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Education Advisor
Your first equation is wrong. Compressibility has to do with the change in volume due to a change in pressure, not temperature.
So do I integrate both in terms of dP?
vela
Staff Emeritus
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Homework Helper
Education Advisor
Both what?
What are the (mathematical) definitions of compressibility and expansivity?
Well I think they're k = 1/P and B = 1/T but I've been given equations for both in the question.
vela
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Sorry, I meant the general definitions in terms of derivatives.
To be honest I really don't know. I've become very confused with all the derivatives and general definitions.
vela
Staff Emeritus
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Well, you should start by looking those up in your textbook or notes. If you don't know the basic definitions of the quantities involved in the problem, it's no surprise that the problem is confusing.
http://en.wikipedia.org/wiki/Material_properties_(thermodynamics [Broken])
Last edited by a moderator:
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The Clustering Coefficient of a Scale-Free Random Graph
Eggemann, Nicole
Noble, Steven D.
Description
We consider a random graph process in which, at each time step, a new vertex is added with m out-neighbours, chosen with probabilities proportional to their degree plus a strictly positive constant. We show that the expectation of the clustering coefficient of the graph process is asymptotically proportional to log n/n. Bollob\'as and Riordan have previously shown that when the constant is zero, the same expectation is asymptotically proportional to ((log n)^2)/n.
Comment: 17 pages. Submitted
Keywords
Mathematics - Combinatorics, 05C80 (Primary) 68R10, 60C05 (Secondary)
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#### Archived
This topic is now archived and is closed to further replies.
# Dynamic array of structures causing errors at runtime
This topic is 5218 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
I doubt this is the proper methodology for doing it, so if someone could point a better way to do this, I''d much appreciate it. I have a text file containing geometry data in this form:
2
-2.00 2.00 -5.00 2 0.00 0.00
-2.00 0.00 -5.00 2 0.00 1.00
2.00 0.00 -5.00 2 1.00 1.00
2.00 2.00 -5.00 2 1.00 1.00
2.00 0.00 -5.00 2 1.00 0.00
-2.00 2.00 -5.00 2 0.00 1.00
The first line indicates an integer declaring the number of triangles within the file. During file-loading, I use the following bit to read that first line in:
fscanf(Worldat, "%d", &numtri);
numvert = (numtri * 3)+1;
In this way, I turn numtri, number of triangles, into the number of vertices, numvert. Now, I have a structure called Vector declared in a header file:
typedef struct{
float x;
float y;
float z;
} Vector;
extern Vector* Geometry;
Back in the file-loading code, I try to declare a dynamic array of this structure to hold the information in the file:
Vector *Geometry;
Geometry = new Vector[numvert];
for (int i=0; i<numvert; i++) {
fscanf(Worldat, "%f %f %f %d %f %f", &Geometry[i].x, &Geometry[i].y, &Geometry[i].z);
}
Everything compiles beautifully. However, at run-time, I get one of those nice "EngineTest has caused an error EngineTest.exe" boxes. On Debug, tells me there''s an "Unhandled Exception in EngineTest.exe: 0X0000005: Access Violation" which then takes me to a place in something called fassign in what I can only assume is a .c file belonging to the compiler. Why is it doing this? I''ve never done anything like this before, and everytime I''ve come across an unhandled exception like this, it''s been a really simple mistake on my part. I''d love to learn from it, but everything I''ve read tells me I''m doing this right. Much thanks and appreciation in advance to whomever can help me out!
*bump*
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My suspicion is that you need to read the other fields into temporary variables as well. You have 6 %-thingies in your scanf, so the internal varargs handling is expecting 6 more parameters to be passed, but you only supply 3. Varargs are really low-level, nasty stuff lacking in error checking, so my guess is that it''s grabbing random memory pointers beyond the end of the stack and trying to assign the rest of the file data through them, to random locations in memory - and bang you''re dead.
typedef union { int foo; float bar;} junk;junk j;for (int i=0; i<numvert; i++) {fscanf(Worldat, "%f %f %f %d %f %f", &Geometry[i].x, &Geometry[i].y, &Geometry[i].z, &junk, &junk, &junk); }
(And I assume you have good reason for using C in the first place.)
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A simple solution using etoolbox. Note you'll have to write the same command for every type of matrix environment. I also use the nccmath package to use \mfrac (medium-sized fractions), a \dfrac in a matrix would be too big, in my opinion: \documentclass{article} \usepackage{mathtools, nccmath, etoolbox} ...
4
Here's a possibility: \documentclass{article} \usepackage{amsmath,array} \makeatletter \def\env@dmatrix{\hskip -\arraycolsep \let\@ifnextchar\new@ifnextchar \extrarowheight=2ex \array{*\c@MaxMatrixCols{>{\displaystyle}c}}} \newenvironment{dmatrix} {\env@dmatrix} {\endarray\hskip-\arraycolsep} \newenvironment{bdmatrix} {\left[\env@dmatrix} ...
0
A solution is to proceed as follow : \documentclass{article} \begin{document} $\begin{matrix} {\displaystyle \lim_x} \end{matrix}$ \end{document} Hope it'll help you.
1
One option using blkarray: The code: \documentclass{article} \usepackage{blkarray} \usepackage{amsmath} \begin{document} \[ \begin{blockarray}{ccccc} x_{1} & x_{2} & \cdots & x_{n} \\ \begin{block}{[cccc|c]} a_{11} & a_{12} & \cdots & a_{1n} & b_{1} \\ a_{21} & a_{22} & \cdots & a_{2n} & b_{2} \\ ...
3
I hope the following is what you're looking for. \documentclass{article} \usepackage{amsmath,mleftright} \mleftright % eliminate whitespace inserted by \left and \right \begin{document} \[ \left( \begin{bmatrix}M_{11}&M_{12}\\M_{21}&M_{22}\end{bmatrix} + \begin{bmatrix}A_{11}&A_{12}\\A_{21}&A_{22}\end{bmatrix} \right) \begin{bmatrix} ...
2
Summarizing some of my earlier comments: Use the standalone package with the option preview in order to avoid getting an error message about a missing $symbol. Don't use $$in a LaTeX document to start and end displaymath mode, as it's quite deprecated. See Why is [ ... ] preferable to$$ ...$\$? for more information on this subject. The matrix ...
Top 50 recent answers are included
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• anonymous
I completely understand the prof's explanation about the Lagrangian with one constraint - but I don't understand why the two constraint formulation works: i.e. L = f(x, y, z) + lambda1 g(x, y, z) + lambda2 h(x, y, z) = 0 Here f is the min/max function, and g and h are the two constraints. If there was no h, then the prof's explanation in lecture 39 makes completely sense in "matching up" the gradients of f and g with lambda1 as the scaler (where f = - lambda1 x g). But, I don't understand how the second constraint can simply be "added" to the Lagrangian equation. Thank you!
OCW Scholar - Multivariable Calculus
Looking for something else?
Not the answer you are looking for? Search for more explanations.
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# Written divison
18 views
### FileEX
Dec 29, 2021, 12:02:52 PM12/29/21
to MathJax Users
Hello, how can I achieve this effect? Exactly the same
I use MathJax in HTML
### Murray
Dec 30, 2021, 3:04:46 AM12/30/21
to MathJax Users
This Stackoverflow post has close to what you are looking for. Hopefully you could modify it for your purposes.
Regards
Murray
### FileEX
Dec 30, 2021, 11:09:56 AM12/30/21
to MathJax Users
I don't want a "long division" symbol. I want the same syntax like on screens
### Davide Cervone
Jan 7, 2022, 12:38:07 PMJan 7
Here's one hack that would do it, if you are using LaTeX input.
$$\def\S{\kern .5em} \def\top#1{\rlap{\raise1.5em{\underline{\S\, #1\S\,\mathstrut}}}} \def\row#1#2{\phantom{\S#1 00}\llap{#2}} \def\urow#1#2{\row{#1}{\underline{\S#2\S\mathstrut}}} \def\rows#1#2#3{\row{#1}{#2\S}\\\urow{#1}{#3}\\} \frac{1}{8}= \begin{array}[t]{l} \top{0{,}125} \rows{-}{1\rlap{{}:8}}{-0} \rows{-0}{10}{-8} \rows{-00}{20}{-16} \rows{-000}{40}{-40} \row{-00}{==} \end{array}$$
The first argument to the \rows, \urow, and \row macros is an indentation amount, and the second (and third) are the contents of the various rows. You don't need to provide the definitions more than once, as long as they are done before the first array that uses them. You could add these to your MathJax configuration using the tex.macros object, if you wanted to have them available all the time (though they would have to be converted to the proper format for that).
Davide
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# 5.3: Continuity Equation
In this chapter and the next three chapters, the conservation equations will be applied to the control volume. In this chapter, the mass conservation will be discussed. The system mass change is
$\dfrac{D\,m_{sys}}{Dt} = \dfrac{D}{Dt} \int_{V_{sys}} \rho dV = 0 \label{mass:eq:Dmdt} \tag{1}$
The system mass after some time, according Figure 5.1, is made of
$m_{sys} = m_{c.v.} + m_a - m_{c} \label{mass:eq:mSysAfter} \tag{2}$
The change of the system mass is by definition is zero. The change with time (time derivative of equation (??) results in
$0 = \dfrac{D\,m_{sys}}{Dt} = \dfrac{d\,m_{c.v.}}{dt} + \dfrac{d\, m_a}{dt} - \dfrac{d\,m_{c}}{dt} \label{mass:eq:mSysAfterdt} \tag{3}$
The first term in equation (??) is the derivative of the mass in the control volume and at any given time is
$\label{mass:eq:masst} \dfrac{d\,m_{c.v.} (t)}{dt} = \dfrac{d}{dt} \int_{V_{c.v.}}\rho\,dV \tag{4}$
and is a function of the time.
Fig. 5.3 Schematics of velocities at the interface.
The interface of the control volume can move. The actual velocity of the fluid leaving the control volume is the relative velocity (see Figure 5.3). The relative velocity is
$\overrightarrow{U_r} = \overrightarrow{U_f} - \overrightarrow{U_{b}} \label{mass:eq:relativeU} \tag{5}$
Where $$U_f$$ is the liquid velocity and $$U_b$$ is the boundary velocity (see Figure 5.3). The velocity component that is perpendicular to the surface is
$\label{mass:eq:reltiveA} U_{rn} = -\hat{n}\cdot \overrightarrow{U_r} = U_r \, \cos \theta \tag{6}$
Where $$\hat{n}$$ is an unit vector perpendicular to the surface. The convention of direction is taken positive if flow out the control volume and negative if the flow is into the control volume. The mass flow out of the control volume is the system mass that is not included in the control volume. Thus, the flow out is
$\label{mass:eq:madta} \dfrac{d\, m_a}{dt} = \int_{S_{cv}} \rho_s \,U_{rn} dA \tag{7}$
It has to be emphasized that the density is taken at the surface thus the subscript $$s$$. In the same manner, the flow rate in is
$\label{mass:eq:madtb} \dfrac{d\, m_b}{dt} = \int_{S_{c.v.}} \rho_s\,U_{rn} dA \tag{8}$
It can be noticed that the two equations (??) and (??) are similar and can be combined, taking the positive or negative value of $$U_{rn}$$ with integration of the entire system as
$\label{mass:eq:mdtCombine} \dfrac{d\, m_a}{dt} - \dfrac{d\, m_b}{dt} = \int_{S_{cv}} \rho_s\,U_{rn} \, dA \tag{9}$
applying negative value to keep the convention. Substituting equation (??) into equation (??) results in
Continuity
$\label{mass:eq:intS} \dfrac{d}{dt} \int_{c.v.} \rho_s dV = -\int_{S_{cv}} \rho\,U_{rn} \, dA \tag{10}$
Equation (??) is essentially accounting of the mass. Again notice the negative sign in surface integral. The negative sign is because flow out marked positive which reduces of the mass (negative derivative) in the control volume. The change of mass change inside the control volume is net flow in or out of the control system.
Fig. 5.4 Schematics of flow in in pipe with varying density as a function time for example 5.1.}
The next example is provided to illustrate this concept.
Example 5.1
The density changes in a pipe, due to temperature variation and other reasons, can be approximated as
$\nonumber \dfrac{\rho(x,t)}{\rho_0} = \left( 1 - \dfrac{x}{L\dfrac{}{}}\right)^2 \cos {\dfrac{t}{t_0}}. \tag{11}$
The conduit shown in Figure 5.4 length is $$L$$ and its area is $$A$$. Express the mass flow in and/or out, and the mass in the conduit as function of time. Write the expression for the mass change in the pipe.
Solution 5.1
Here it is very convenient to choose a non-deformable control volume that is inside the conduit $$dV$$ is chosen as $$\pi\,R^2\,dx$$. Using equation (??), the flow out (or in) is
\begin{align*}
\dfrac{d}{dt} \int_{c.v.} \rho dV =
\dfrac{d}{dt} \int_{c.v.}
\overbrace{\rho_0 \left(1 - \dfrac{x}{L} \right)^2 \cos\left( \dfrac{t}{t_0} \right)} ^{\rho(t)}
\overbrace{\pi\,R^2\, dx}^{dV}
\end{align*}
The density is not a function of radius, $$r$$ and angle, $$\theta$$ and they can be taken out the integral as
\begin{align*}
\dfrac{d}{dt} \int_{c.v.} \rho dV =
\pi\,R^2 \dfrac{d}{dt} \int_{c.v.}\rho_0 \left(1-\dfrac{x}{L\dfrac{}{}} \right)^2 \cos \left(
\dfrac{t}{t_0 \dfrac{}{}}\right) dx
\end{align*}
which results in
\begin{align*}
\text{Flow Out} = \overbrace{ \pi\,R^2}^{A}
\dfrac{d}{dt} \int_{0}^{L} \rho_0 \left(1-\dfrac{x}{L \dfrac{}{}}\right)^2 \cos {\dfrac{t}{t_0}} dx =
- \dfrac{\pi\,R^2\,L\, \rho_0}{3\,t_0} \, \sin\left( \dfrac{t}{t_0} \right)
\end{align*}
The flow out is a function of length, $$L$$, and time, $$t$$, and is the change of the mass in the control volume.
### Contributors
• Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.
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# Axiom of Probablity.
The third axiom of Probablity states that,
If $$A_1,A_2,\ldots$$ be a finite or infinite sequence of pairwise mutually exclusive events, then
$$P\left(\bigcup A_i\right)=\sum P(A_i)$$
We know that, $$P\left(\bigcup A_i\right)\le 1$$ but what guarantees that, $$\sum P(A_i)\le 1$$ for infinite case.
I am an beginner in this topic,so pardon me, if this question is a trivial one
First of all the definition of countable additivity of a probability measure says that $$\sum\limits_{i=1}^{\infty} P(A_i)$$ converges and the sum is equal to $$P(A)$$ (if $$A_i$$'s are mutually exlcusive events with union $$A$$). Convergence is part of the definition .
The second point is if $$a_i \geq 0$$ for all $$i$$ and $$\sum\limits_{i=1}^{n} a_i \leq a$$ for al $$n$$ then automatically $$\sum\limits_{i=1}^{\infty} a_i$$ converges and $$\sum\limits_{i=1}^{\infty} a_i \leq a$$.
• Will you elaborate more?I am not familiar with the notion of measure. – who Dec 14 '19 at 6:43
The fact that it (by assumption) equals the LHS does!
The intuition for why we should assume these add up to the LHS in our axioms for probability, is because the events are mutually exclusive. If we think in terms of Venn-diagrams, the regions corresponding to the events do not overlap so the area of the union is just the sum of the areas.
• Can we extend this to any collection (uncountable) of mutually exclusive events? – who Dec 14 '19 at 12:13
• @who No, the third axiom only applies to countable collections. We want situations like distributions on the real line where the probability $X=a$ is zero for any $a\in \mathbb R,$ so we don't require uncountable additivity... it's too strong and doesn't let us do what we want to do. Why we should demand countably infinite instead of just finite additivity is more subtle, but keep in mind the way to look at it is that not requiring countable additivity (and only requiring finite additivity) would be a generalization of the usual probability theory. – spaceisdarkgreen Dec 14 '19 at 17:06
• @who One way to look at it is that additivity requirements accord with our intuition and we want to require as much as we can without going overboard and ruining our ability to work with continuous distributions. Another way is that the additional 'distributions' we get when we weaken to finite additivity are very abstract and it's unclear they do anything useful for us, so why not exclude them? Another reason is more mathematical: there are very nice properties associated with countable additivity. Not everyone agrees: see bof's answer here – spaceisdarkgreen Dec 14 '19 at 17:16
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## 21.47 Weakly contractible objects
An object $U$ of a site is weakly contractible if every surjection $\mathcal{F} \to \mathcal{G}$ of sheaves of sets gives rise to a surjection $\mathcal{F}(U) \to \mathcal{G}(U)$, see Sites, Definition 7.40.2.
Lemma 21.47.1. Let $\mathcal{C}$ be a site. Let $U$ be a weakly contractible object of $\mathcal{C}$. Then
1. the functor $\mathcal{F} \mapsto \mathcal{F}(U)$ is an exact functor $\textit{Ab}(\mathcal{C}) \to \textit{Ab}$,
2. $H^ p(U, \mathcal{F}) = 0$ for every abelian sheaf $\mathcal{F}$ and all $p \geq 1$, and
3. for any sheaf of groups $\mathcal{G}$ any $\mathcal{G}$-torsor has a section over $U$.
Proof. The first statement follows immediately from the definition (see also Homology, Section 12.7). The higher derived functors vanish by Derived Categories, Lemma 13.17.9. Let $\mathcal{F}$ be a $\mathcal{G}$-torsor. Then $\mathcal{F} \to *$ is a surjective map of sheaves. Hence (3) follows from the definition as well. $\square$
It is convenient to list some consequences of having enough weakly contractible objects here.
Proposition 21.47.2. Let $\mathcal{C}$ be a site. Let $\mathcal{B} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ such that every $U \in \mathcal{B}$ is weakly contractible and every object of $\mathcal{C}$ has a covering by elements of $\mathcal{B}$. Let $\mathcal{O}$ be a sheaf of rings on $\mathcal{C}$. Then
1. A complex $\mathcal{F}_1 \to \mathcal{F}_2 \to \mathcal{F}_3$ of $\mathcal{O}$-modules is exact, if and only if $\mathcal{F}_1(U) \to \mathcal{F}_2(U) \to \mathcal{F}_3(U)$ is exact for all $U \in \mathcal{B}$.
2. Every object $K$ of $D(\mathcal{O})$ is a derived limit of its canonical truncations: $K = R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n} K$.
3. Given an inverse system $\ldots \to \mathcal{F}_3 \to \mathcal{F}_2 \to \mathcal{F}_1$ with surjective transition maps, the projection $\mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n \to \mathcal{F}_1$ is surjective.
4. Products are exact on $\textit{Mod}(\mathcal{O})$.
5. Products on $D(\mathcal{O})$ can be computed by taking products of any representative complexes.
6. If $(\mathcal{F}_ n)$ is an inverse system of $\mathcal{O}$-modules, then $R^ p\mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n = 0$ for all $p > 1$ and
$R^1\mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n = \mathop{\mathrm{Coker}}(\prod \mathcal{F}_ n \to \prod \mathcal{F}_ n)$
where the map is $(x_ n) \mapsto (x_ n - f(x_{n + 1}))$.
7. If $(K_ n)$ is an inverse system of objects of $D(\mathcal{O})$, then there are short exact sequences
$0 \to R^1\mathop{\mathrm{lim}}\nolimits H^{p - 1}(K_ n) \to H^ p(R\mathop{\mathrm{lim}}\nolimits K_ n) \to \mathop{\mathrm{lim}}\nolimits H^ p(K_ n) \to 0$
Proof. Proof of (1). If the sequence is exact, then evaluating at any weakly contractible element of $\mathcal{C}$ gives an exact sequence by Lemma 21.47.1. Conversely, assume that $\mathcal{F}_1(U) \to \mathcal{F}_2(U) \to \mathcal{F}_3(U)$ is exact for all $U \in \mathcal{B}$. Let $V$ be an object of $\mathcal{C}$ and let $s \in \mathcal{F}_2(V)$ be an element of the kernel of $\mathcal{F}_2 \to \mathcal{F}_3$. By assumption there exists a covering $\{ U_ i \to V\}$ with $U_ i \in \mathcal{B}$. Then $s|_{U_ i}$ lifts to a section $s_ i \in \mathcal{F}_1(U_ i)$. Thus $s$ is a section of the image sheaf $\mathop{\mathrm{Im}}(\mathcal{F}_1 \to \mathcal{F}_2)$. In other words, the sequence $\mathcal{F}_1 \to \mathcal{F}_2 \to \mathcal{F}_3$ is exact.
Proof of (2). This holds by Lemma 21.23.10 with $d = 0$.
Proof of (3). Let $(\mathcal{F}_ n)$ be a system as in (2) and set $\mathcal{F} = \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n$. If $U \in \mathcal{B}$, then $\mathcal{F}(U) = \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n(U)$ surjects onto $\mathcal{F}_1(U)$ as all the transition maps $\mathcal{F}_{n + 1}(U) \to \mathcal{F}_ n(U)$ are surjective. Thus $\mathcal{F} \to \mathcal{F}_1$ is surjective by Sites, Definition 7.11.1 and the assumption that every object has a covering by elements of $\mathcal{B}$.
Proof of (4). Let $\mathcal{F}_{i, 1} \to \mathcal{F}_{i, 2} \to \mathcal{F}_{i, 3}$ be a family of exact sequences of $\mathcal{O}$-modules. We want to show that $\prod \mathcal{F}_{i, 1} \to \prod \mathcal{F}_{i, 2} \to \prod \mathcal{F}_{i, 3}$ is exact. We use the criterion of (1). Let $U \in \mathcal{B}$. Then
$(\prod \mathcal{F}_{i, 1})(U) \to (\prod \mathcal{F}_{i, 2})(U) \to (\prod \mathcal{F}_{i, 3})(U)$
is the same as
$\prod \mathcal{F}_{i, 1}(U) \to \prod \mathcal{F}_{i, 2}(U) \to \prod \mathcal{F}_{i, 3}(U)$
Each of the sequences $\mathcal{F}_{i, 1}(U) \to \mathcal{F}_{i, 2}(U) \to \mathcal{F}_{i, 3}(U)$ are exact by (1). Thus the displayed sequences are exact by Homology, Lemma 12.29.1. We conclude by (1) again.
Proof of (5). Follows from (4) and (slightly generalized) Derived Categories, Lemma 13.32.2.
Proof of (6) and (7). We refer to Section 21.23 for a discussion of derived and homotopy limits and their relationship. By Derived Categories, Definition 13.32.1 we have a distinguished triangle
$R\mathop{\mathrm{lim}}\nolimits K_ n \to \prod K_ n \to \prod K_ n \to R\mathop{\mathrm{lim}}\nolimits K_ n[1]$
Taking the long exact sequence of cohomology sheaves we obtain
$H^{p - 1}(\prod K_ n) \to H^{p - 1}(\prod K_ n) \to H^ p(R\mathop{\mathrm{lim}}\nolimits K_ n) \to H^ p(\prod K_ n) \to H^ p(\prod K_ n)$
Since products are exact by (4) this becomes
$\prod H^{p - 1}(K_ n) \to \prod H^{p - 1}(K_ n) \to H^ p(R\mathop{\mathrm{lim}}\nolimits K_ n) \to \prod H^ p(K_ n) \to \prod H^ p(K_ n)$
Now we first apply this to the case $K_ n = \mathcal{F}_ n[0]$ where $(\mathcal{F}_ n)$ is as in (6). We conclude that (6) holds. Next we apply it to $(K_ n)$ as in (7) and we conclude (7) holds. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
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The gradient of a line is defined to be the gradient of any interval within the line.
This definition depends on the fact that two intervals on a line have the same gradient.
Suppose AB and PQ are two intervals on the same straight line.
Draw right-angled triangles ABX and PQY with sides AX and PY parallel to the x-axis and sides BX and QY parallel to the y-axis.
Triangle ABX is similar to triangle PQY since the corresponding angles are equal. Therefore:
$$\dfrac{QY}{PY}=\dfrac{BX}{AX}$$.
That is, the intervals have the same gradient.
#### Example 4
Find the gradient of the line passing through the points:
1. A(1, 4) and B(5, 12)
2. C(6, –5) and D(–4, 2)
#### Solution
1. The gradient of the interval $$AB =\dfrac{y_2 − y_1}{x_2 − x_1} =\dfrac{12 − 4}{5 −1} = \dfrac{8}{4} = 2$$
Therefore the gradient of the line AB = 2.
2. The gradient of the interval $$CD =\dfrac{y_2 − y_1}{x_2 − x_1} =\dfrac{2 −(−5)}{−4 − 6} = \dfrac{7}{−10}$$
Therefore the gradient of the line $$CD = −\dfrac{7}{10}$$.
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# Visco-inertial effects
In 1987, Johnson Koplik and Dashen [JKD87] proposed a semi-phenomenological model to describe the complex density of an acoustical porous material with a motionless skeleton having arbitrary pore shapes.
This model is further refined by Pride, Morgan & Gangi [PMG93] (and corrected by D. Lafarge) in 1993 to account for pores with possible constrictions between them. The final expression for $\widetilde{\rho}$ obtained is:
\begin{align} &\widetilde{\rho} =\displaystyle\frac{\rho_0\widetilde{\alpha}\left(\omega\right)} { \phi} \\ &~~ \\ &~~ \\ &\widetilde{\alpha}\left(\omega\right)=\displaystyle\alpha_\infty\left[1+\displaystyle\frac{1}{j \bar{\omega}}\widetilde{F}\left(\omega\right)\right]\\ &~~ \\ &~~ \\\ &\widetilde{F}\left(\omega\right)=1-P+P\sqrt{1+\displaystyle\frac{M}{2{P~}^2}j\bar{\omega}}\\ &~~ \\ &~~ \\ &\bar{\omega}=\displaystyle\frac{\omega \rho_0 k_0 \alpha_\infty}{\eta \phi}\\ &~~ \\ &M=\displaystyle\frac{8 k_0 \alpha_\infty}{\phi\Lambda^2} \\ &~~ \\ & P=\displaystyle\frac{M}{4\left(\displaystyle\frac{\alpha_0}{\alpha_\infty}-1\right)}\\ \end{align}
5 parameters are involved in the calculation of this dynamic density: the static air flow resistivity $\sigma$ (or the static viscous permeability $k_0=\eta/\sigma$), the open porosity $\phi$, the high frequency limit of the tortuosity $\alpha_{\infty}$, the viscous characteristic length $\Lambda$ and the static viscous tortuosity $\alpha_0$.
# Thermal effects
In 1991, Champoux and Allard [CA91] introduced an expression for the dynamic bulk modulus based on the previous work by Johnson et al. This model is further refined by Pride, Morgan & Gangi [PMG93] (and corrected by D. Lafarge) in 1993 to account for pores with possible constrictions between them. Finally, after the work by [LLAT97] (introducing a new parameter to describe the thermal behavior at low frequencies: the static thermal permeability $k'_0$), the final expression obtained for $\widetilde{K}$ is:
\begin{align} & \widetilde{K} =\displaystyle\frac{\gamma P_0}{\phi}\frac{1}{\widetilde{\beta}\left(\omega\right)}\\ &~~ \\ &~~ \\ &\widetilde{\beta}\left(\omega\right)=\displaystyle\gamma - \left(\gamma-1\right)\left[1+\displaystyle\frac{1}{j \bar{\omega}'}\widetilde{F'}\left(\omega\right)\right]^{-1}\\ &~~ \\ &~~ \\ &\widetilde{F'}\left(\omega\right)=1-P~'+P~'\sqrt{1+\displaystyle\frac{M'}{2{P~}'^{2}}j\bar{\omega}'}\\ &~~ \\ &~~ \\ &\bar{\omega}'=\displaystyle\frac{\omega \rho_0 C_P k'_0}{\kappa\phi}\\ &~~ \\ &M'=\displaystyle\frac{8 k'_0}{\phi \Lambda'^2} \\ &~~ \\ &P~'=\displaystyle\frac{M'}{4\left(\alpha'_0-1\right)}\\ \end{align}
4 parameters are invloved in the calculation of this dynamic bulk modulus: the open porosity $\phi$, the thermal characteristic length $\Lambda'$, the static thermal permeability $k'_0$ and the static thermal tortuosity $\alpha'_0$.
# Links with JCA and JCAL models
The Johnson-Champoux-Allard (JCA) and Johnson-Champoux-Allard-Lafarge (JCAL) models are recovered from the JCAPL one by setting, respectively, $M'=P=P~'=1$ and $P=P~'=1$ in the expressions of $\widetilde{\rho}$ and $\widetilde{K}$ reported above.
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# Caught In A Circle
You are trapped in the center of a flat circular area of radius 1000 by an evil maniacal tyrant. You have to walk out by taking steps of size 1. However, for each step, you can only choose the direction to walk in, and the tyrant will dictate if you move in the positive or negative direction.
What is the minimum number of steps you need to take to guarantee that you can reach the circumference?
×
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# Phase Accumulation of Hankel-waves upon propagation
Hankel functions are solutions to the scalar Helmholtz-equation $$\Delta\psi + k_e^2\psi = 0$$ in cylindrical and spherical geometry (with respect to a separated angular dependence). Thus, they are very important describing spherical and cylindrical waves. Here is an example of such a propagation in the spherical case taken from Franz Zotter:
I am searching for a reference that states the phase accumulation of Hankel waves of the form $$F_H^{\mathrm{out/in}}(\mathbf{r}) = H_m^{1/2}(k\rho)\ .$$ Assumed is stationarity with an $e^{-\mathrm{i}\omega t}$ time dependence fixing the meaning of the two different Hankel-waves as outgoing/incoming.
For plane waves one finds that the accumulated phase of a wave in $x$-direction, $$F_p=e^{\mathrm{i}kx}$$ is simply related to its argument, $$\phi_{\mathrm{acc}}(x_1,x_2)=\mathrm{Arg}(F_p(x_2))-\mathrm{Arg}(F_p(x_1)) = k(x_2 - x_1)$$ and it is natural to just use this formula in the Hankel-case, e.g. $$\phi_{\mathrm{acc}}(\rho_1,\rho_2)=\mathrm{Arg}(F_H^{\mathrm{out/in}}(\rho_2))-\mathrm{Arg}(F_H^{\mathrm{out/in}}(\rho_1))$$
However, I was not able to find a suitable reference. Hence my question:
### Is there a reference defining the phase accumulation of Hankel waves?
-
A closed form solution (in terms of more elementary function than Hankel functions) does not exist. However, typically one is only interested in the regime where $k\rho \gg1$, i.e., the asymptotic region far away from the source. There one can use the asymptotic form of the Hankel functions $$H^{1/2}_m (x) \sim \sqrt{\frac2{\pi x}} e^{\pm (ix -i \frac\pi2 m - i\frac\pi4)}.$$ Thus, the accumulated phase is given by $$\phi_\text{acc}(\rho_1,\rho_2) = \pm k(\rho_2 - \rho_1),$$ i.e., the same as for a plane wave.
-
Thank you for your answer. However, in my case a definition of the phase accumulation for all $\rho > 0$ is required. Greets – Robert Filter May 14 '11 at 13:37
@Rober: why would one ever need something like that? – Fabian May 14 '11 at 15:06
For a complete description of the field in the same sense as you need Hankel functions and not only plane waves. – Robert Filter May 14 '11 at 15:22
Ok, there seems no interest in an answer to this question for all $\rho$ so I just accept this one. Greets. – Robert Filter Jun 23 '11 at 8:24
I am not sure how you get outwardly travelling waves from Hanken functions alone without also using Bessel functions. My understanding is that by analogy with plane waves, you have the stading wave solutions eg.
sin(kx)*cos(wt) and
cos(kx)*sin(wt)
and the travelling wave comes from the sum of these two. Similarly in cylindrical geometry you have the standing waves
Bess(kr)*cos(wt) and
Hank(kr)*sin(wt)
and the travelling waves shown in the applets are really the sum of these. I'm not sure if this point is relative to the question being asked, but this is how I understand the pictures.
-
$H^{1/2}_m = J_m \pm N_m$ with respect to an argument. Please have a look at the Wiki sites I was linking. Greets – Robert Filter May 14 '11 at 18:08
OK, right. I guess what I called the Hankel function would actually have to be the Neumann function. – Marty Green May 14 '11 at 20:50
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# Gauss's Law Problem
## Homework Statement
A spherical drop of water has radius R = 5 * 10^-4 cm. The drop has an even surface charge distribution.
Consider an element dA of the surface. This element is exposed to a force from the rest of the surface elements.
The effective field in the surface is equal to the mean value of the effective field just outside and just inside the surface.
Use Gauss's Law to compute the effective field in the surface.
## The Attempt at a Solution
I've done several Gauss's Law exercises before, but in this one I'm not even sure I understand the question.
What bothers me is the line about the "effective field" being the "mean value" of the field just inside and just outside the surface. In one problem I just did, I showed that the electric field inside an even spherical surface distribution is zero, so how can the field on the surface be the "mean" of 0 and a field slightly less than on the surface?
Appreciate any help on this one!
TSny
Homework Helper
Gold Member
What bothers me is the line about the "effective field" being the "mean value" of the field just inside and just outside the surface. In one problem I just did, I showed that the electric field inside an even spherical surface distribution is zero
Good.
so how can the field on the surface be the "mean" of 0 and a field slightly less than on the surface?
I'm not sure I understand the last part your question. You know the field just inside the surface is 0. What is the field just outside the surface? You want to take the mean (average) of the fields just inside and just outside.
Did they give you the amount of charge on the drop?
They only write the charge per area as a sigma, so I have to solve for the electric field in terms of sigma.
So, I basically have to compute the charge just outside the sphere and divide it by 2? Still, I don't understand what they mean with an "effective field"... I figured the field just outside the sphere would be just a tiny amount smaller than the field in the surface, not twice as big!
TSny
Homework Helper
Gold Member
So, I basically have to compute the charge just outside the sphere and divide it by 2?
I think you mean't "field" instead of "charge" here. But, yes, that's right. All you'll need to do is divide the field just outside the drop by 2.
Still, I don't understand what they mean with an "effective field"... I figured the field just outside the sphere would be just a tiny amount smaller than the field in the surface, not twice as big!
Surface charge densities are interesting in that they cause a finite discontinuous jump in the field as you pass from one side of the surface to the other. If you replaced the infinitesimally thin surface layer of charge by a small but finite thickness of charge, then the field would change continuously but rapidly withing the charge layer (from zero on the inner surface of the layer of charge to some finite value at the outer surface of the layer of charge.) The "effective" value of the field within the layer of charge is taken to be the average of the fields at the inner and outer surfaces of the layer. When taking the limit of an infinitesimal thin layer of surface charge, then you get essentially a discontinuous jump in the field. But you still take the effective field to be the average of the fields just inside and outside the surface layer of charge.
So, you'll need to find an expression for the field just outside the drop.
I got the field just outside the sphere to equal sigma / epsilon, which means the field in the surface should equal sigma / 2 epsilon.
Your explanation was interesting, and I tried googling an explanation for why this is true, but without any luck. Is the electric field anywhere always equal to the mean of the EF just outside and just outside the field in question? Say, if this was a solid sphere with a uniform volume distribution instead of an empty shell? When I think about it that last question is obviously true haha, but is it true in general?
Calculating the eletric field on the surface this way was not really hard to do, but I don't think I quite get why it works... Is there a way to find the field on the surface without first calculating the field just outside?
Thanks a lot!!
TSny
Homework Helper
Gold Member
Last edited by a moderator:
collinsmark
Homework Helper
Gold Member
Hi gralla55,
I don't recommend putting much effort into this "mean value of the effective field just outside and just inside the surface." This is the first time I've seen it defined this way. Usually, I see the field defined on the outside and also defined on the inside, but not so much at the very surface itself.
If it helps though, consider the simple, unit Heaviside step function (http://en.wikipedia.org/wiki/Heaviside_step_function). Why is it that H(0) = 1/2? Well, that's a good question. But if you define
$$H(x) = \left\{ \begin{array}{rl} 0 &\mbox{ if x<0} \\ 1 &\mbox{ if x>0} \end{array} \right.$$
and leave H(0) undefined for now, then break it apart through Fourier decomposition, then finally put it back together again (also Fourier decomposition), taking whatever limits need to be taken, I think you'll find that the H(0) = 1/2. Is it a big deal? No, I don't think so. I wouldn't recommend putting too much effort into this.
What *is* important though is that electric field can be, and often is discontinuous. In the case of this problem, it jumps from $\frac{\sigma}{\epsilon_0}$ down to zero all within an infinitesimal length. You can expect electric fields to change a lot over small spaces, in some situations.
On a related note, what is *not* usually discontinuous (and is instead usually continuous) is electric potential. If you find sudden discontinuities in your electric potential functions, it's time to raise an eyebrow.
Last edited:
TSny
Homework Helper
Gold Member
The nice thing about the "effective field", ##E_{eff}##, at the surface is that it provides a direct way to calculate the electrostatic force per unit area, ##f##, on the surface charge: ##f = \sigma E_{eff}##.
Thanks a lot, that really helped!
|
# What's the difference between a negation and a contrapositive?
What’s the difference between a negation and a contrapositive?
I keep mixing them up, but it seems that a contrapositive is a negation where the terms’ order is changed and where there is an imply sign.
#### Solutions Collecting From Web of "What's the difference between a negation and a contrapositive?"
Consider the statement: $$P \implies Q.\tag{1}$$ Then, as you know, the contrapositive of (1) is $$\lnot Q \implies \lnot P.\tag{2}$$
Now, note that statements (1) and (2) are equivalent. I.e., $$\;P \implies Q\equiv \lnot Q \implies \lnot P.$$
One additional observation: Note that the contrapositive of statement $(2)$ is $(1)$, so a contrapositive need not contain any negation symbol $(\lnot)$ at all!
Negating statement (1) (and hence negating (2)) would be a statement to the effect that it is NOT the case that $P \implies Q$. I.e. the negation of statement (1) is given by
$$\lnot (P \implies Q)\tag{3}$$
and since $(P \implies Q) \equiv (\lnot P \lor Q)$, we can write (3) as follows:
$$\lnot(\lnot P \lor Q)\tag{4}$$
By DeMorgan’s, (4) is equivalent to: $$\lnot(\lnot P) \land \lnot Q\tag{5}$$
which yields
$$(P \land \lnot Q)\tag{6}$$
So (3), (4), (5) and (6) are all equivalent expressions, each negating statement (1).
Put another way, the contrapositve of a statement is equivalent to the statement [both a statement and its contrapositive have the same truth-value], while the negation of the statement negates or reverses the truth-value of the original statement.
One last comment:
When one speaks of the contrapositive of a statement, one is usually speaking about the contrapositive of an implication, whereas one can negate any logical statement whatsoever simply by enclosing the original statement in parentheses and negating it, as in statement (3) above.
When you negate a true statement you get a false statement, but the contrapositive of a true statement is still a true statement, formally if $P \Rightarrow Q$ is your statement then $\sim Q \Rightarrow \sim P$ is the contrapositive, where $\sim$ denotes logical NOT and $\Rightarrow$ denotes logical implication. You could use the truth table to convince yourself that they are equivalent, i.e., $(P \Rightarrow Q) \Longleftrightarrow (\sim Q \Rightarrow \sim P)$. Consider the following statement: If today is Monday then we are tired. This is the same as: If we are not tired then today is not Monday.
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posetJoin -- determines the join for two elements of a poset
Synopsis
• Usage:
j = posetJoin(P, a, b)
• Inputs:
• P, an instance of the type Poset,
• a, , an element of the poset
• b, , an element of the poset
• Outputs:
• j, , the least element greater than both $a$ and $b$, if it exists
Description
The join of $a$ and $b$ in $P$, if it exists, is the unique least element greater than both $a$ and $b$.
i1 : B = booleanLattice 3; i2 : posetJoin(B, "001", "100") o2 = {101} o2 : List
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# [NTG-context] counting the words in a TeX document
Mojca Miklavec mojca.miklavec.lists at gmail.com
Sun Aug 6 02:31:31 CEST 2006
```On 8/5/06, Hans Hagen wrote:
> Mojca Miklavec wrote:
> > Hello,
> >
> > I would like to ask how difficult it would be to count the number of
> > words in a TeX/ConTeXt document. If it's too complex, please ignore
> > the rest of the message.
> the way i do such things (and worse trickery) is using pdftotext
>
> you can of course use tex, but then ther ecan be generated words and so and it is insane to use tex (or adapt a tex style) for that; it may help to run with (nondestructive)
>
> \setupalign[nothyphenated]
>
> anyhow, here is a script (i could not locate my normal one)
>
> === wordcount.rb ===
>
> if (file = ARGV[0]) && file && FileTest.file?(file) then
> begin
> system("pdftotext #{ARGV[0]} wc.log")
> data.gsub!(/\d[\.\:]*\w+/o) do ' ' end # remove suffixes
> data.gsub!(/\d/o) do ' ' end # remove numbers
> data.gsub!(/\-\s+/mo) do ' ' end # remove hyphenation
> data.gsub!(/\-/mo) do ' ' end # split compound words
> data.gsub!(/[\.\,\<\>\/\?\\\|\'\"\;\:\]\{\}\{\+\=\-\_\)\(\*\&\^\%\\$\#\@\!\~\`]/mo) do ' ' end
> words = data.split(/\s+/)
> count = Hash.new
> words.each do |w|
> count[w] = (count[w] || 0) + 1
> end
> rescue
> puts("some error #{\$!}")
> else
> puts("words : #{words.size}")
> puts("unique : #{count.size}")
> end
> if ARGV[1] =~ /list/ then
> puts("\n")
> count.sort.each do |k,v|
> puts("#{k} : #{v}")
> end
> end
> end
>
>
> usage: wc filename.pdf [list]
>
> it this kind of stuff is usefull, we can add it to one of the scripts that come with context
Thanks a lot! I guess that's *it*! I always forget about the most
powerful feature of ConTeXt in comparison to LaTeX - scripting can be
added to almost any place (and the user doesn't need to install any
Here's some of my feedback:
- pdftotext is far from being useful for pdf to text conversion
(doesn't handle any accents), but is perfectly suitable for wordcount
- \[ is missing in the last gsub (only the right bracket is deleted)
- something strange (but not critical) happens to en-dashes
But everything else looks like a perfect functionality for ctxtools --wordcount.
> A very crude approach. There is a program called detex
> http://ctan.org/tex-archive/support/detex/ I have not used it, but I
> think that it strips off every command \something from the tex file.
> Then you can filter the file through wc to get a rough estimate of
> the number of words. One approach that will work is
>
> \startstatistics[filename][words|letters|lines]
>
> maps to
>
> \startbuffer[\jobname-statistics-filename]
>
> and
>
> \stopstatistics maps to
>
> \stopbuffer
> \getbuffer[\jobname-statistics-filename]
> \executesystemcommand{detex \jobname-statistics-filename.tmp | wc
> <flags correspondingto words|lines|letters> }
I took a look, but it merely looks like a parser for hardcoded (La)TeX
(someone should correct me if I'm wrong).
However, the fact that abstracts for which one might need wordcount
usually don't have too much trickery involved (they're usually olmost
pure plain text), doing the same, only with a simple ruby script
instead of compiling/installing some external LaTeX-aware C program
> It wasn't too complex for Michael Downes using LaTeX:
>
> \ProvidesFile{wordcount.tex}[2000/09/27 v1.5 Michael Downes]
> % Copyright 2000 Michael John Downes
> % This file has no restrictions on its use, distribution, or sale.
> %
> % If you run LaTeX on wordcount.tex it will prompt you for the name of a
> % document to be counted. For most people, however, it will be more
This solution is more likely to produce better results (just that it
includes slightly more work). It actually runs (La)TeX, just redefines
a few commands before, so that counting the words is then a
straightforward parsing of log files based on the number of some
boxes.
Base on those three answers I got a more clear idea of two (different,
but complementary) methods that might be sensible:
a) ctxtools --wordcount filename[tex|pdf]
to do the wordcount for the whole document using pdftotext + ruby regexp
b)
\usemodule[wordcount]
whatever
\startstatistics[name][words|letters|lines]
some more-or-less plain text
\stopstatistics
whatever
and according to Aditya's idea, run a (ruby) regular expression
(insead of detex) on it which would write the nicely formatted desired
number to the output/log file. (I don't know if it's possible to use
the first approach for the second problem, but it doesn't make sense
to complicate things too much.)
As long as the command names are carefully chosen (and extensible if
the need for more complex behaviour arises in the future), that should
be about everything and it doesn't seem so difficult to implement
after all. (But I would write to the documentation that the resulting
numbers might change slightly in the future if the algorithm for
counting the words is improved.)
Any thoughts?
Thanks a lot,
Mojca
```
|
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# Physicochemical properties, microstructures, nutritional components, and free amino acids of Pleurotus eryngii as affected by different drying methods
## Abstract
In this study, we determined the influences of different drying techniques such as natural air (ND), hot-air (HD), vacuum (VD), infrared (ID), microwave (MD), and freeze drying (FD) methods on the color, shrinkage ratio (SR), rehydration ratio (RR), firmness, crispness, microstructures, nutritional components, and free amino acids of Pleurotus eryngii. The results showed that these parameters were markedly influenced by different drying techniques. Among them, FD was the most effective drying method which retained the main characteristics of the fresh P. eryngii in above mentioned indexes, followed by ND and HD at 40 °C. Finally, despite the least drying time, MD treatment was not suitable to the drying process of P. eryngii slices since it damaged physicochemical properties and caused massive losses of the main nutrients and free amino acids. The results will provide a theoretical basis for industrial processing of P. eryngii.
## Introduction
Pleurotus eryngii, often referred to as the king oyster mushroom, is called Xingbaogu in China. It is one of the important species of edible mushrooms commonly and has been widely cultivated and consumed in China because of its delicious taste, remarkable nutritional values and biological activities including antioxidant, antitumor, immunostimulatory activities, and modulating human gut microbiome1,2,3. The P. eryngii can be widely cultivated in lots of agricultural and industrial wastes conveniently and cheaply4. The production of the cultivated P. eryngii increased substantially at a rate of about 3 × 108 kg/year in Asia5. Fresh P. eryngii slices can be cooked and retained its own firm textures1, and also can be fabricated into canned foods. Nevertheless, fresh P. eryngii is also one of the most perishable food raw materials and readily loses the best edible quality immediately after harvest because of its high in respiration rate, moisture content and nutrients, all of which made it easy to lose water or be attacked by microbes. Its shelf-life only reached about 3 weeks under the appropriate conditions6,7. Therefore, an effective preservation method was required for the industrial production and application of P. eryngii to extend its shelf-life.
Drying is a widely used and comparatively cheaper method for shelf-life extension of highly perishable mushrooms as it limited microbial growth, inhibited enzyme activity, and slowed down many moisture mediated reactions8,9,10. In addition, dried P. eryngii can be also accepted by consumers with a similar high scores like fresh P. eryngii in all acceptability attributes (color, taste, and so on) and a strong purchase decision11. Drying methods are mainly composed of ND, HD, VD, fluidized bed drying, ID, MD, and FD based on different drying mechanisms12,13. For the reason that drying is a complex process involving continuous heat and mass transfer, it may accompany with significant changes in the nutrients, phytochemicals compositions, surface morphology and internal structure as well as physical properties of the products8,9,14,15,16, and even influence in vitro digestibility of substances17. For dried agricultural products, color is one of the most important quality indexes that influenced consumer appetite, acceptance and commodity value of products. The physical characteristics such as firmness, crispness, shrinking, and dehydration capacity are usually used to evaluate the mechanical properties and are defined as the resistance of a material to deformation or penetration, and are influenced by surface and internal structure. Most importantly, these parameters can affect the packaging, transportation and edible quality of dried products. In addition, free amino acids are important taste active compounds in edible fungi, especially Asp and Glu, which are responsible for umami flavour of edible fungi9. Therefore, the choice of the drying methods and processing parameters is of great importance for Pleurotus eryngii. Chen et al. only optimized the technology parameters of Pleurotus eryngii by microwave-vacuum drying based on quality and energy consumption18. Su et al. investigated drying characteristics and water dynamics during microwave hot-air flow rolling drying of Pleurotus eryngii19. Li et al. mainly studied the effects of different drying methods on the tasty compounds but did not investigate the changes in color, physical and texture properties of dried P. eryngii20. Therefore, these studies are still inadequate on the choice of drying methods, drying temperatures as well as research content and parameters for drying processing of P. eryngii, which can not be fully and objectively reflected the effect of different drying methods on sensory, physical and chemical qualities of dried P. eryngii.
The purpose of this study is to assess the influences of different drying methods and temperatures on the physicochemical properties, microstructures, nutritional compositions, and free amino acids of P. eryngii, and recommend a desirable drying method. These will provide valuable information to maximally retain the original nutrients and sensory qualities of P. eryngii during drying process.
## Results and Discussion
### Color determination
Generally speaking, the smaller the total color difference (ΔE), the closer the color to the fresh sample. The color values of fresh and dried P. eryngii slices dried with different methods are shown in Table 1. Except FD, all dried samples exhibited lower lightness (L* value), higher redness (a* value) and yellowness (b* value) compared with fresh P. eryngii, and a significant difference (p < 0.05) was found in L*, a*, and b* values among different drying methods. Similar results were also reported in shiitake mushrooms9,21,22. As expected, the L* values closest to the fresh sample and lowest ΔE values were found in FD samples (Table 1), indicating that the color parameters of FD were close to those of fresh sample. This can be explained by the fact that less oxygen was present in a drying chamber under low temperature and pressure, which led to less intense enzymatic browning reactions and non-enzymatic browning reactions. HD sample at 40 °C (HD40) possessed higher L* and lower ΔE values compared with HD at 60 °C (HD60), VD at 40 °C (VD40), VD at 60 °C (VD60), ID at 40 °C (ID40), ID at 60 °C (ID60) samples, while MD had the lowest L* and the highest ΔE values. In fact, during VD process, less oxygen was present in a drying chamber under low pressure. Hence, the principal cause of color darkening of dried P. eryngii could be due to the heating during drying, causing non-enzymatic browning reactions that depended on heating temperature and time. The change in color of MD P. eryngii slices indicated that browning occurred more intensely because of non-enzymatic browning. Similarly, some studies reported that ID shiitake mushroom, carrot and garlic exhibited better color than MD products14,23. The above results indicated that FD can better retain the color of dried P. eryngii slices.
### Shrinkage ratio
The SR of different dried P. eryngii slices is shown in Fig. 1. VD40, VD60, ID40, ID60 and MD samples had the higher SR ranged from 0.81 to 0.85, but there was no significant difference. The followings were ND, HD40 and HD60 samples with the SR values ranged from 0.72 to 0.73. The lowest SR was found in FD products (0.19), indicating that the volume of P. eryngii slices remained unchanged during FD process. Drying process accompanied with a complicated mass, heat, and momentum transportation within the food products with multiple phases such as solids, liquid water, and gas24. In FD system, an extensive pore network was left by the sublimation of ice, and therefore, little shrinkage took place25. Similar to ND, HD was a convective drying process, and the shrinkage was more obvious compared to FD14. VD process in this study took longer time than HD under the same temperature, which made the product shrink and formed hardened crust. In ID process, the infrared radiation penetrated the product, and then converted to thermal energy through molecular vibration, and these heat was finally absorbed by the materials without heating the surrounding air15,26,27. For MD sample, the microwave energy created a very porous structure of the mushrooms, facilitating the transport of water vapor24, which resulted in cell swelling because water evaporation was accelerated by the preferential absorption of microwave energy. However, negative pressure occurred within the mushrooms, offsetting the puffing process. Hence, the SR of VD, MD and ID samples were the highest. Different results are also reported by other researchers. Baysal et al. reported that SR of MD sample was significantly higher than that of HD and ID samples23, while Tian et al. reported that SR of HD sample was significantly higher than that of VD and MD samples9. In general, VD, ID and MD mushrooms were significantly greater than that of ND and HD mushrooms (p < 0.05) in terms of SR, while the lowest SR was found in FD products.
### Rehydration ratio
The RR of dried P. eryngii slices treated with different drying methods are given in Fig. 1. FD sample showed the highest RR (4.87), followed by ND, HD40 and HD60 mushrooms, which had no significant difference (p < 0.05). The physical and chemical changes during drying process had significant influences on RR, so did the porosity of surface pores and internal structure of the products14. The ND, conventional HD and VD resulted in obvious shrinking of mushrooms, forming a dense structure, and maybe just the collapsed capillaries reduce the water retention ability during rehydration14. Contrary to the results of SR, the RR of MD sample was the lowest (2.04), followed by VD 60, ID 40 and ID 60 samples. Similar results were reported by some studies14,28. In fact, we also found the surface of MD product exhibited serious shrinkage and a hard crust, decreasing the rehydration capacity. There had little effect of drying temperature on rehydration capacity of the P. eryngii slices. However, samples dried by lower temperature had a relatively higher RR than those dried by higher temperature. Similar tendency was also observed in the some studies29,30. This may be associated with more intense moisture change of the samples at higher temperature, which caused more seriously damage to the capillaries and further resulted in modifications of osmotic properties of the cell as well as lower diffusion of water through the surface during rehydration31. In addition, there was moderate negative correlation (p < 0.05) between the RR and SR, and the correlation coefficient was 0.69 (not shown).
### Firmness and crispness
The firmness and crispness of P. eryngii slices dried with different drying methods are illustrated in Fig. 2. Firmness can be related to the force performed by mastication that took part during eating. The higher the firmness value, the harder it was to chew14. The highest firmness was observed for MD sample (6635 N mm−1), followed by VD60, VD40, ID60, ID40, HD60, HD40, ND, and FD samples. According to Fig. 2, MD and VD samples exhibited the highest crispness values, followed by ID samples. FD showed the lowest crispness value, followed by ND and HD. Besides, it can be observed that drying temperature had a positive effect on the firmness while no influence on the crispness. These textural variations could be explained by changes in the plant cell wall, which occurred during drying processing with different temperatures and methods, causing significant decrease in internal pressure8,31. Furthermore, there was a high positive correlation (p < 0.01) between the firmness and the crispness of dried P. eryngii slices, and the correlation coefficient was 0.95 (not shown).
### Microstructure
Figure 3 showed scanning electron microscopy (SEM) micrographs of the surface and lateral section of dried P. eryngii slices with different drying methods. The images showed that drying methods had a obvious effect on the tissue structure of P. eryngii slices. As expected, the FD samples showed the most plenty of pores in the surface and uniform honeycomb networks in the interior, which almost had no collapsed structure. This phenomenon can be explained by the fact thatthe original dimensions of the product were maintained first by freezing during FD process, then the ice was sublimed under a high vacuum, and thus the destruction of FD products cells could be minimized32. It also provided a good way for the infiltration of water during rehydration process, and thus the highest RR, the lowest SR. Compared with FD sample, larger but less pores were seen in the surface of HD samples which indicated more cellular tissue collapse and shrinkage. The morphology in the interior of the HD samples was relatively dense, which partly explained the increasing firmness in HD samples. Compared with HD, VD samples had an irregular surface with a few small cavities, while the organizational structure revealed multiple channels and large irregularities. In VD process, the drying process was conducted under negative vacuum pressure, and thus a reducing boiling point of water made the whole process more rapidly. Finally, a very porous interior structure was formed in the interior of VD samples24. A long time high temperature treatment may cause a collapsed structure, a rapid shrinkage and a hard shell, all of which affected the water absorption capacity of the materials33. ID samples showed a dense surface layer or crust with few holes but a porous and densest internal structure, which may be due to the limited depth of infrared radiation when penetrating from the surface into the interior. The uneven distribution of water diffusion as well as the collapse of the cellular structure on the surface finally resulted in the uneven structures in ID samples14. Different results were reported by Pan et al. who demonstrated that application of ID was better compared to HD for the microstructure of dehydrated banana34. The surface of MD samples stacked together and showed almost invisible pores but had a clear porous internal structure. Microwave energy had the ability of selective heating on the interior portions of the products and created a high steam pressure. The dried surface looks like a barrier to hinder the release of the pressure and the product was thus puffed8, which also led to a hardening texture and a poor rehydration property. Besides, a lot of soluble viscous polysaccharide substances were transported to the surface when ID and MD were applied to P. eryngii. In consequence, a hard crust would be formed in both ID and MD products, just like roasting, and these samples also got the lowest RR and highest firmness and crispness values.
### Nutrient components
The chemical compositions of fresh and dried P. eryngii slices treated with different drying methods are listed in Table 2. The moisture content of fresh P. eryngii was 89.8% based on fresh weight (fw). Largely consistent with the previous results35, other proximate compositions are listed in the following order (g/100 g dry weight (dw)): carbohydrate (63.4 ± 0.5) > protein (22.1 ± 0.2) > ash (6.5 ± 0.2) > reducing sugar (2.9 ± 0.3) > fat (2.5 ± 0.2). As expected, the final moisture contents of dried P. eryngii using different drying methods ranged from 4.2 to 4.9 g/100 g dw and showed no significant difference (p < 0.05).
The results revealed that the ash contents, ranging from 6.3 to 6.9 g/100 g dw, were not affected by drying methods. The protein contents of samples using different drying methods are listed in the following order: FD > fresh, ND and HD40 > HD60 > VD40, VD60 and ID40 > ID60 > MD. There was no significant difference in fat content among the fresh, FD, ND, HD, and VD samples, while ID and MD showed significant lower fat contents. During ID, the rapid heating of mushrooms by infrared radiation may promote fat oxidation15. Unlike ID, heating was generated within the samples in MD, which greatly improved the heating efficiency and also caused fat oxidation. The reducing sugar contents of samples using different drying methods are listed in the following order: fresh, FD, ND and HD40 > HD60, VD40, VD60 and ID40 > ID60 and MD. Contrary to fat content, MD and ID 60 samples got the highest values in carbohydrate contents. The chemical compositions of FD, ND and HD40 samples showed high similarity to fresh P. eryngii.
### Free amino acids
The contents of free amino acids (FAA) compositions of fresh and dried P. eryngii slices treated with different drying methods are given in Table 3. Glutamic acid (Glu), alanine (Ala), phenylalanine (Phe) and lysine (Lys) exhibited higher amounts in all P. eryngii samples, which was similar to the previous reports36,37,38. The content of total FAA (TFAA) in fresh sample was 59.06 mg/g dw, which was much higher than the previous report with 18.07 mg/g dw of TFAA20. The TFAA content decreased significantly (p < 0.05) in the dried samples with a descending order as follows: FD > ND > HD40 > VD40 > HD60 > ID40 > VD60 > ID60 > MD. It was worth mentioning that the FD product contained more Asp, Phe, Tyr and Ser than the fresh mushroom. During FD process, some free amino acids including Glu, His, and Phe, might be released during proteolysis, and after that, the contents of free amino acids decreased due to the strecker degradation of free amino acids and the Maillard reaction39,40,41,42. Eight essential amino acids (EAA) for humans were detected in these samples, and the total contents of EAA (TEAA) ranged from 5.03 to 26.31 mg/g. The lowest concentration was observed in MD sample and the highest in fresh sample. The contents of non-essential amino acids (NEAA) of samples had the same trend with EAA and TFAA. The contribution of total free amino acids to the total crude protein was calculated and showed in Table 3. The percentage of TFAA/protein in different samples ordered similar to the content of TFAA. Heating would promote proteolysis as well as non-enzymatic Maillard reactions between amino acids and reducing sugars in mushrooms during the drying process39. These results agreed with those for protein content, reducing sugar content and color values, suggesting the amount of FAA released from proteolysis was lower than that of FAA lost in Maillard reaction during ND, HD, VD, ID and MD drying process, and the higher temperature can led to mushrooms darken and hardness12.
### Differentiation by cluster analysis
In order to establish differences among different drying methods, a cluster analysis is conducted on FAA contents according to squared Euclidean distance methods (Fig. 4). The results showed that all samples were divided in to five different clusters. Cluster 1 included three drying samples (HD40, VD40 and HD60), which was further divided into two sub-clusters (I and II). The TEAA contents of HD40 and VD40 samples in sub-cluster I showed high similarity due to the same temperature, which was the critical factor of Maillard reaction causing the loss of FAA, while HD60 sample in sub-cluster II was different. The TFAA content of VD40 sample was lower than that of HD40, due to the longer drying time in VD process. ID40, ID60 and VD60 products were brought together in cluster 2. With the Euclidean distance increasing, cluster 1 and 2 got together, which meant that there was a similarity between two clusters. Cluster 3 was grouped with others, which appeared the sample dried by MD with low similarity with others. Fresh sample in cluster 4 exhibited the highest TFAA and high similarity to FD sample, which meant FD process was the most effective drying method to preserve the FAA of P. eryngii. The cluster 5 contained only ND sample. Besides, as the Euclidean distance increased, clusters 4 and 5 got together, which meant that ND sample had more similarities with fresh and FD samples than others.
## Conclusion
In conclusion, the physicochemical properties, microstructures, nutritional compositions, and free amino acids of Pleurotus eryngii were significantly affected by different drying methods. Among them, FD was the most effective drying method and retained the main characteristics of the fresh P. eryngii in physicochemical properties, microstructures, nutritional compositions and free amino acids, followed by ND and HD40. However, it was worth mentioning that FD equipments was rather expensive, and the processing procedure was time and energy consuming. MD treatment damaged the physicochemical properties of P. eryngii slices and resulted in massive losses of the main nutrients and free amino acids. These results will provide a theoretical basis for industrial processing of P. eryngii. It is well known that drying must have an influence of on the bioactivity of the substrate. Therefore, the effects of drying methods on bioactive substances including active polysaccharides and phenols and biological activities of P. eryngii still need to be further studied. In addition, further studies need to be performed on the drying process to improve the nutritional quality and health-caring quality of P. eryngii.
## Materials and Methods
### Materials preparation
Fully mature fresh P. eryngii that is called Xingbaogu in China was harvested from Aokun Biological Agriculture Co. Ltd., Shanxi Province, China. Prior to drying, the fresh P. eryngii samples were cut into slices with thickness of 5.0 ± 0.2 mm, and then were divided into ten portions at random. Among them, one was used for fresh analysis, and other batches were respectively dried with different methods in optimized conditions to get a water content ( ≤ 5% dry base). The moisture content of fresh P. eryngii was 89.8 ± 1.7% (wet basis), which determined with a hot air oven at 105 °C. After cutting, slices were processed immediately.
### Drying process
The P. eryngii slices were dried using the following treatments (1) ND was carried out under natural air flow and ambient temperatures (23–30 °C) for 4 days, (2) HD was conducted in a draught drying cabinet (GZX-9246MBE, Yueming Scientific Instrument Ltd., Shanghai, China) at 40 °C for 13 h or 60 °C for 8 h until constant weight was achieved respectively, (3) VD was realized with a vacuum drier (DZF-6090, Hecheng Instrument Manufacturing Co., Ltd., Shanghai, China), and the samples were dried at 40 °C for 19 h or 60 °C for 12 h with vacuum of 0.08 MPa, (4) ID was performed in a laboratory scale benchtop infrared drier (YHG 600BS, Botai Laboratory Equipment Ltd., Shanghai, China), and the samples were dried at 40 °C for 14 h or 60 °C for 6 h, (5) MD was done in a microwave oven (WD900Y, Galanz Enterprise Group Co. Ltd., Guangdong, China) at 240 W for 10 min, (6) FD was carried out in a freeze drier at 20 °C and 30 Pa for 24 h.
### Color measurements
The CIE color coordinates (L*, a*, b*) of fresh and dried P. eryngii slices were determined with a CM-700D colorimeter (Konica Minolta Optics Inc., Japan). The ΔE was calculated according to the formula21,
$$\Delta E=\sqrt{{({{\rm{L}}}^{\ast }-{{\rm{L}}}_{0})}^{2}+{({{\rm{a}}}^{\ast }-{{\rm{a}}}_{0})}^{2}+{({{\rm{b}}}^{\ast }-{{\rm{b}}}_{0})}^{2}},$$
where L0, a0, and b0 referred to the color reading of fresh samples, which was used as control.
### Shrinkage ratio
The shrinkage ratio (SR) was measured by a displacement method14. The shrinkage ratio of dried P. eryngii slices was calculated according to the formula, SR = (V1 − V2)/V1, where V1 and V2 were the volume (cm3) of the fresh and dried samples, respectively.
### Rehydration ratio
The rehydration ratio (RR) was determined according to the reported method43. Dried slices were weighed (W0) and immersed for 30 min in distilled water at room temperature. Then the rehydrated samples were removed, and the slices were weighed again (W1). The RR was calculated according to the formula, RR = (W1 − W0)/W0.
### Firmness and crispness
Firmness and crispness of dried P. eryngii slices were evaluated by means of a puncture test carried out by a Texture Analyzer. The probe used was cylindrical with a puncture diameter of 10 mm and the parameters were preset as follows: test speed 1.0 mm s−1, pretest speed 3.0 mm s−1, posttest speed 3.0 mm s−1, travel distance of 3.0 mm. Firmness (N mm−1) of samples was defined as the maximum force applied to puncture the mushroom. The crispness was defined as the coordinate of the first peak pressure.
### Microstructure analysis
The surface morphology and internal structure of dried P. eryngii were observed by scanning electron microscopy (Model JSM-5310LV; JEOL Ltd., Tokyo, Japan) at an accelerating voltage of 6.0 kV. The SEM micrographs of the surface and interior of samples were obtained at 100 × magnification and 200 × magnification respectively.
### Nutrient components analysis
Ash, protein and fat were determined following the AOAC procedures44. The nitrogen factor used for crude protein calculation was 4.3813. Fat percentage was determined using a Soxhlet apparatus. The content of reducing sugar was estimated by the 3, 5-dinitrosalicylic acid spectrophotometric method10,16.
### Free amino acids analysis
Free amino acid extraction and analysis were carried out according to the method reported by Li et al.16. The 0.2 g powder samples were added to 20 mL 75% ethanol and shaken with a laboratory rotary shaker at 150 rpm for 30 min at 70 °C. After centrifugation at 10000 g and 4 °C for 15 min, the supernatant was collected, evaporated to dryness. The residual was reconstituted with pH 2.2, 0.2 M sodium citrate loading buffer solution to a final volume of 10 mL. The standard solution and prepared filtrate were analyzed by a automatic amino-acid analyzer (Biochrom Ltd., England). The injection volume and the duration of single run were 20 µL and 60 min, respectively. The amino acids were identified and quantified by comparing peak profiles of the mushroom samples with standard amino acid profiles.
### Statistical analysis
All experiments were carried out in triplicate and the data were expressed as mean ± standard deviation (SD). Correlation coefficient, One-way analysis of variance (ANOVA) and Duncan’s multiple range test were carried out to determine significant differences (p < 0.05) between the means.
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## Acknowledgements
This work was supported the Natural Science Foundation of Shanxi Province, China (Grant No. 201601D011070).
## Author information
Authors
### Contributions
Q.P.H. contributed to the design, planning and coordination of the project; R.L.Y. and Q.L. carried out the experiments and conducted the data analysis; R.L.Y. and Q.P.H. wrote and edited the manuscript. All authors participated in discussion about the results and the manuscript.
### Corresponding author
Correspondence to Qing-Ping Hu.
## Ethics declarations
### Competing interests
The authors declare no competing interests.
Publisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
## Rights and permissions
Reprints and Permissions
Yang, RL., Li, Q. & Hu, QP. Physicochemical properties, microstructures, nutritional components, and free amino acids of Pleurotus eryngii as affected by different drying methods. Sci Rep 10, 121 (2020). https://doi.org/10.1038/s41598-019-56901-1
• Accepted:
• Published:
• ### The drying process of Sarcocornia perennis: impact on nutritional and physico-chemical properties
• M. J. Barroca
• , R. P. F. Guiné
• , S. Ressurreição
• , A. Moreira da Silva
• , M. P. M. Marques
• & L. A. E. Batista de Carvalho
Journal of Food Science and Technology (2020)
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# Reentrant Melting in Laser Field Modulated Colloidal Suspensions
Chakrabarti, J and Krishnamurthy, HR and Sood, AK and Sengupta, S (1995) Reentrant Melting in Laser Field Modulated Colloidal Suspensions. In: Physical Review Letters, 75 (11). pp. 2232-2235.
PDF Reentrant-1.pdf Restricted to Registered users only Download (1023Kb) | Request a copy
## Abstract
We present results from a Monte Carlo simulation study of the phase diagram and the order of the modulated liquid to crystal transition in laser field modulated colloids. We find that for low values of $\beta V_e$, the strength of the modulation potential, the transition is first order, but changes to a continuous transition for high values of $\beta V_e$, in agreement with the conclusions from the density functional theory of laser induced freezing. However, we find in the simulation a novel reentrant laser induced melting transition from the crystal to the modulated liquid phase with increasing $\beta V_e$, unlike the (mean field) density functional phase diagram.
Item Type: Journal Article Copyrigh of this article belongs to the American Physical Society. Division of Physical & Mathematical Sciences > Physics 23 May 2007 19 Sep 2010 04:38 http://eprints.iisc.ernet.in/id/eprint/11062
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Home > Algebra, Uncategorized > IMC 2016 – Day 1 – Problem 2
## IMC 2016 – Day 1 – Problem 2
Problem 2. Let ${k}$ and ${n}$ be positive integers. A sequence ${(A_1,...,A_k)}$ of ${n\times n}$ matrices is preferred by Ivan the Confessor if ${A_i^2 \neq 0}$ for ${1\leq i \leq k}$, but ${A_iA_j = 0}$ for ${1\leq i,j \leq k}$ with ${i \neq j}$. Show that if ${k \leq n}$ in al preferred sequences and give an example of a preferred sequence with ${k=n}$ for each ${n}$.
Sketch of proof: The fact that ${A_iA_j = 0}$ for ${i \neq j}$ should imply somehow that the images of ${A_i}$, i.e. the spaces ${S_i = \{ y = A_ix, x \in \Bbb{R}^n \}}$ do not have much in common. Indeed, let’s suppose that ${y \in S_i \cap S_j}$, then ${y = A_i x_i = A_j x_j}$. If this happens, then ${A_i^2 x_i = A_j^2 x_j = 0}$. Now we know that ${A_i^2 \neq 0}$ for every ${i}$ which implies the existence of ${x_i}$ such that ${A_i^2 x_i \neq 0}$ for every ${i}$.
In this way, the elements ${y_i = A_ix_i}$ are such that ${y_i \in S_i}$ and ${y_i \notin S_j}$ for ${i \neq j}$. Since ${S_i}$ are linear subspaces of ${\Bbb{R}^n}$ this implies, in particular that the family ${\{y_1,...,y_k\}}$ is linearly independent in ${\Bbb{R}^n}$. This implies that ${k \leq n}$.
An example when ${k=n}$ is the family of matrices which have ${1}$ on the ${i}$-th diagonal position and ${0}$ elsewhere.
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Stochastic gradient descent (often abbreviated SGD) is an iterative method for optimizing an objective function with suitable smoothness properties (e.g. differentiable or subdifferentiable). It can be regarded as a stochastic approximation of gradient descent optimization, since it replaces the actual gradient (calculated from the entire data set) by an estimate thereof (calculated from a randomly selected subset of the data). Especially in high-dimensional optimization problems this reduces the very high computational burden, achieving faster iterations in exchange for a lower convergence rate.[1]
While the basic idea behind stochastic approximation can be traced back to the Robbins–Monro algorithm of the 1950s, stochastic gradient descent has become an important optimization method in machine learning.[2]
## Background
Main article: M-estimation
Both statistical estimation and machine learning consider the problem of minimizing an objective function that has the form of a sum:
${\displaystyle Q(w)={\frac {1}{n))\sum _{i=1}^{n}Q_{i}(w),}$
where the parameter ${\displaystyle w}$ that minimizes ${\displaystyle Q(w)}$ is to be estimated. Each summand function ${\displaystyle Q_{i))$ is typically associated with the ${\displaystyle i}$-th observation in the data set (used for training).
In classical statistics, sum-minimization problems arise in least squares and in maximum-likelihood estimation (for independent observations). The general class of estimators that arise as minimizers of sums are called M-estimators. However, in statistics, it has been long recognized that requiring even local minimization is too restrictive for some problems of maximum-likelihood estimation.[3] Therefore, contemporary statistical theorists often consider stationary points of the likelihood function (or zeros of its derivative, the score function, and other estimating equations).
The sum-minimization problem also arises for empirical risk minimization. In this case, ${\displaystyle Q_{i}(w)}$ is the value of the loss function at ${\displaystyle i}$-th example, and ${\displaystyle Q(w)}$ is the empirical risk.
When used to minimize the above function, a standard (or "batch") gradient descent method would perform the following iterations:
${\displaystyle w:=w-\eta \nabla Q(w)=w-{\frac {\eta }{n))\sum _{i=1}^{n}\nabla Q_{i}(w),}$
where ${\displaystyle \eta }$ is a step size (sometimes called the learning rate in machine learning).
In many cases, the summand functions have a simple form that enables inexpensive evaluations of the sum-function and the sum gradient. For example, in statistics, one-parameter exponential families allow economical function-evaluations and gradient-evaluations.
However, in other cases, evaluating the sum-gradient may require expensive evaluations of the gradients from all summand functions. When the training set is enormous and no simple formulas exist, evaluating the sums of gradients becomes very expensive, because evaluating the gradient requires evaluating all the summand functions' gradients. To economize on the computational cost at every iteration, stochastic gradient descent samples a subset of summand functions at every step. This is very effective in the case of large-scale machine learning problems.[4]
## Iterative method
Fluctuations in the total objective function as gradient steps with respect to mini-batches are taken.
In stochastic (or "on-line") gradient descent, the true gradient of ${\displaystyle Q(w)}$ is approximated by a gradient at a single sample:
${\displaystyle w:=w-\eta \nabla Q_{i}(w).}$
As the algorithm sweeps through the training set, it performs the above update for each training sample. Several passes can be made over the training set until the algorithm converges. If this is done, the data can be shuffled for each pass to prevent cycles. Typical implementations may use an adaptive learning rate so that the algorithm converges.[5]
In pseudocode, stochastic gradient descent can be presented as :
• Choose an initial vector of parameters ${\displaystyle w}$ and learning rate ${\displaystyle \eta }$.
• Repeat until an approximate minimum is obtained:
• Randomly shuffle samples in the training set.
• For ${\displaystyle i=1,2,...,n}$, do:
• ${\displaystyle w:=w-\eta \nabla Q_{i}(w).}$
A compromise between computing the true gradient and the gradient at a single sample is to compute the gradient against more than one training sample (called a "mini-batch") at each step. This can perform significantly better than "true" stochastic gradient descent described, because the code can make use of vectorization libraries rather than computing each step separately as was first shown in [6] where it was called "the bunch-mode back-propagation algorithm". It may also result in smoother convergence, as the gradient computed at each step is averaged over more training sample.
The convergence of stochastic gradient descent has been analyzed using the theories of convex minimization and of stochastic approximation. Briefly, when the learning rates ${\displaystyle \eta }$ decrease with an appropriate rate, and subject to relatively mild assumptions, stochastic gradient descent converges almost surely to a global minimum when the objective function is convex or pseudoconvex, and otherwise converges almost surely to a local minimum.[7][8] This is in fact a consequence of the Robbins–Siegmund theorem.[9]
## Example
Let's suppose we want to fit a straight line ${\displaystyle {\hat {y))=\!w_{1}+w_{2}x}$ to a training set with observations ${\displaystyle (x_{1},x_{2},\ldots ,x_{n})}$ and corresponding estimated responses ${\displaystyle ({\hat {y_{1))},{\hat {y_{2))},\ldots ,{\hat {y_{n))})}$ using least squares. The objective function to be minimized is:
${\displaystyle Q(w)=\sum _{i=1}^{n}Q_{i}(w)=\sum _{i=1}^{n}\left({\hat {y_{i))}-y_{i}\right)^{2}=\sum _{i=1}^{n}\left(w_{1}+w_{2}x_{i}-y_{i}\right)^{2}.}$
The last line in the above pseudocode for this specific problem will become:
${\displaystyle {\begin{bmatrix}w_{1}\\w_{2}\end{bmatrix)):={\begin{bmatrix}w_{1}\\w_{2}\end{bmatrix))-\eta {\begin{bmatrix}{\frac {\partial }{\partial w_{1))}(w_{1}+w_{2}x_{i}-y_{i})^{2}\\{\frac {\partial }{\partial w_{2))}(w_{1}+w_{2}x_{i}-y_{i})^{2}\end{bmatrix))={\begin{bmatrix}w_{1}\\w_{2}\end{bmatrix))-\eta {\begin{bmatrix}2(w_{1}+w_{2}x_{i}-y_{i})\\2x_{i}(w_{1}+w_{2}x_{i}-y_{i})\end{bmatrix)).}$
Note that in each iteration (also called update), the gradient is only evaluated at a single point ${\displaystyle x_{i))$ instead of at the set of all samples.
The key difference compared to standard (Batch) Gradient Descent is that only one piece of data from the dataset is used to calculate the step, and the piece of data is picked randomly at each step.
## Notable applications
Stochastic gradient descent is a popular algorithm for training a wide range of models in machine learning, including (linear) support vector machines, logistic regression (see, e.g., Vowpal Wabbit) and graphical models.[10] When combined with the backpropagation algorithm, it is the de facto standard algorithm for training artificial neural networks.[11] Its use has been also reported in the Geophysics community, specifically to applications of Full Waveform Inversion (FWI).[12]
Stochastic gradient descent competes with the L-BFGS algorithm,[citation needed] which is also widely used. Stochastic gradient descent has been used since at least 1960 for training linear regression models, originally under the name ADALINE.[13]
Another stochastic gradient descent algorithm is the least mean squares (LMS) adaptive filter.
## Extensions and variants
Many improvements on the basic stochastic gradient descent algorithm have been proposed and used. In particular, in machine learning, the need to set a learning rate (step size) has been recognized as problematic. Setting this parameter too high can cause the algorithm to diverge; setting it too low makes it slow to converge.[14] A conceptually simple extension of stochastic gradient descent makes the learning rate a decreasing function ηt of the iteration number t, giving a learning rate schedule, so that the first iterations cause large changes in the parameters, while the later ones do only fine-tuning. Such schedules have been known since the work of MacQueen on k-means clustering.[15] Practical guidance on choosing the step size in several variants of SGD is given by Spall.[16]
As mentioned earlier, classical stochastic gradient descent is generally sensitive to learning rate η. Fast convergence requires large learning rates but this may induce numerical instability. The problem can be largely solved[17] by considering implicit updates whereby the stochastic gradient is evaluated at the next iterate rather than the current one:
${\displaystyle w^{\rm {new)):=w^{\rm {old))-\eta \nabla Q_{i}(w^{\rm {new))).}$
This equation is implicit since ${\displaystyle w^{\rm {new))}$ appears on both sides of the equation. It is a stochastic form of the proximal gradient method since the update can also be written as:
${\displaystyle w^{\rm {new)):=\arg \min _{w}\{Q_{i}(w)+{\frac {1}{2\eta ))||w-w^{\rm {old))||^{2}\}.}$
As an example, consider least squares with features ${\displaystyle x_{1},\ldots ,x_{n}\in \mathbb {R} ^{p))$ and observations ${\displaystyle y_{1},\ldots ,y_{n}\in \mathbb {R} }$. We wish to solve:
${\displaystyle \min _{w}\sum _{j=1}^{n}(y_{j}-x_{j}'w)^{2},}$
where ${\displaystyle x_{j}'w=x_{j1}w_{1}+x_{j,2}w_{2}+...+x_{j,p}w_{p))$ indicates the inner product. Note that ${\displaystyle x}$ could have "1" as the first element to include an intercept. Classical stochastic gradient descent proceeds as follows:
${\displaystyle w^{\rm {new))=w^{\rm {old))+\eta (y_{i}-x_{i}'w^{\rm {old)))x_{i))$
where ${\displaystyle i}$ is uniformly sampled between 1 and ${\displaystyle n}$. Although theoretical convergence of this procedure happens under relatively mild assumptions, in practice the procedure can be quite unstable. In particular, when ${\displaystyle \eta }$ is misspecified so that ${\displaystyle I-\eta x_{i}x_{i}'}$ has large absolute eigenvalues with high probability, the procedure may diverge numerically within a few iterations. In contrast, implicit stochastic gradient descent (shortened as ISGD) can be solved in closed-form as:
${\displaystyle w^{\rm {new))=w^{\rm {old))+{\frac {\eta }{1+\eta ||x_{i}||^{2))}(y_{i}-x_{i}'w^{\rm {old)))x_{i}.}$
This procedure will remain numerically stable virtually for all ${\displaystyle \eta }$ as the learning rate is now normalized. Such comparison between classical and implicit stochastic gradient descent in the least squares problem is very similar to the comparison between least mean squares (LMS) and normalized least mean squares filter (NLMS).
Even though a closed-form solution for ISGD is only possible in least squares, the procedure can be efficiently implemented in a wide range of models. Specifically, suppose that ${\displaystyle Q_{i}(w)}$ depends on ${\displaystyle w}$ only through a linear combination with features ${\displaystyle x_{i))$, so that we can write ${\displaystyle \nabla _{w}Q_{i}(w)=-q(x_{i}'w)x_{i))$, where ${\displaystyle q()\in \mathbb {R} }$ may depend on ${\displaystyle x_{i},y_{i))$ as well but not on ${\displaystyle w}$ except through ${\displaystyle x_{i}'w}$. Least squares obeys this rule, and so does logistic regression, and most generalized linear models. For instance, in least squares, ${\displaystyle q(x_{i}'w)=y_{i}-x_{i}'w}$, and in logistic regression ${\displaystyle q(x_{i}'w)=y_{i}-S(x_{i}'w)}$, where ${\displaystyle S(u)=e^{u}/(1+e^{u})}$ is the logistic function. In Poisson regression, ${\displaystyle q(x_{i}'w)=y_{i}-e^{x_{i}'w))$, and so on.
In such settings, ISGD is simply implemented as follows. Let ${\displaystyle f(\xi )=\eta q(x_{i}'w^{old}+\xi ||x_{i}||^{2})}$, where ${\displaystyle \xi }$ is scalar. Then, ISGD is equivalent to:
${\displaystyle w^{\rm {new))=w^{\rm {old))+\xi ^{\ast }x_{i},~{\text{where))~\xi ^{\ast }=f(\xi ^{\ast }).}$
The scaling factor ${\displaystyle \xi ^{\ast }\in \mathbb {R} }$ can be found through the bisection method since in most regular models, such as the aforementioned generalized linear models, function ${\displaystyle q()}$ is decreasing, and thus the search bounds for ${\displaystyle \xi ^{\ast ))$ are ${\displaystyle [\min(0,f(0)),\max(0,f(0))]}$.
### Momentum
Further proposals include the momentum method, which appeared in Rumelhart, Hinton and Williams' paper on backpropagation learning.[18] Stochastic gradient descent with momentum remembers the update Δw at each iteration, and determines the next update as a linear combination of the gradient and the previous update:[19][20]
${\displaystyle \Delta w:=\alpha \Delta w-\eta \nabla Q_{i}(w)}$
${\displaystyle w:=w+\Delta w}$
${\displaystyle w:=w-\eta \nabla Q_{i}(w)+\alpha \Delta w}$
where the parameter ${\displaystyle w}$ which minimizes ${\displaystyle Q(w)}$ is to be estimated, ${\displaystyle \eta }$ is a step size (sometimes called the learning rate in machine learning) and ${\displaystyle \alpha }$ is an exponential decay factor between 0 and 1 that determines the relative contribution of the current gradient and earlier gradients to the weight change.
The name momentum stems from an analogy to momentum in physics: the weight vector ${\displaystyle w}$, thought of as a particle traveling through parameter space,[18] incurs acceleration from the gradient of the loss ("force"). Unlike in classical stochastic gradient descent, it tends to keep traveling in the same direction, preventing oscillations. Momentum has been used successfully by computer scientists in the training of artificial neural networks for several decades.[21] The momentum method is closely related to underdamped Langevin dynamics, and may be combined with Simulated Annealing. [22]
### Averaging
Averaged stochastic gradient descent, invented independently by Ruppert and Polyak in the late 1980s, is ordinary stochastic gradient descent that records an average of its parameter vector over time. That is, the update is the same as for ordinary stochastic gradient descent, but the algorithm also keeps track of[23]
${\displaystyle {\bar {w))={\frac {1}{t))\sum _{i=0}^{t-1}w_{i))$.
When optimization is done, this averaged parameter vector takes the place of w.
AdaGrad (for adaptive gradient algorithm) is a modified stochastic gradient descent algorithm with per-parameter learning rate, first published in 2011.[24] Informally, this increases the learning rate for sparser parameters and decreases the learning rate for ones that are less sparse. This strategy often improves convergence performance over standard stochastic gradient descent in settings where data is sparse and sparse parameters are more informative. Examples of such applications include natural language processing and image recognition.[24] It still has a base learning rate η, but this is multiplied with the elements of a vector {Gj,j} which is the diagonal of the outer product matrix
${\displaystyle G=\sum _{\tau =1}^{t}g_{\tau }g_{\tau }^{\mathsf {T))}$
where ${\displaystyle g_{\tau }=\nabla Q_{i}(w)}$, the gradient, at iteration τ. The diagonal is given by
${\displaystyle G_{j,j}=\sum _{\tau =1}^{t}g_{\tau ,j}^{2))$.
This vector is updated after every iteration. The formula for an update is now
${\displaystyle w:=w-\eta \,\mathrm {diag} (G)^{-{\frac {1}{2))}\odot g}$[a]
${\displaystyle w_{j}:=w_{j}-{\frac {\eta }{\sqrt {G_{j,j))))g_{j}.}$
Each {G(i,i)} gives rise to a scaling factor for the learning rate that applies to a single parameter wi. Since the denominator in this factor, ${\displaystyle {\sqrt {G_{i))}={\sqrt {\sum _{\tau =1}^{t}g_{\tau }^{2))))$ is the 2 norm of previous derivatives, extreme parameter updates get dampened, while parameters that get few or small updates receive higher learning rates.[21]
While designed for convex problems, AdaGrad has been successfully applied to non-convex optimization.[25]
### RMSProp
RMSProp (for Root Mean Square Propagation) is also a method in which the learning rate is adapted for each of the parameters. The idea is to divide the learning rate for a weight by a running average of the magnitudes of recent gradients for that weight.[26] So, first the running average is calculated in terms of means square,
${\displaystyle v(w,t):=\gamma v(w,t-1)+(1-\gamma )(\nabla Q_{i}(w))^{2))$
where, ${\displaystyle \gamma }$ is the forgetting factor.
And the parameters are updated as,
${\displaystyle w:=w-{\frac {\eta }{\sqrt {v(w,t)))}\nabla Q_{i}(w)}$
RMSProp has shown good adaptation of learning rate in different applications. RMSProp can be seen as a generalization of Rprop and is capable to work with mini-batches as well opposed to only full-batches.[27]
Adam[28] (short for Adaptive Moment Estimation) is an update to the RMSProp optimizer. In this optimization algorithm, running averages of both the gradients and the second moments of the gradients are used. Given parameters ${\displaystyle w^{(t)))$ and a loss function ${\displaystyle L^{(t)))$, where ${\displaystyle t}$ indexes the current training iteration (indexed at ${\displaystyle 0}$), Adam's parameter update is given by:
${\displaystyle m_{w}^{(t+1)}\leftarrow \beta _{1}m_{w}^{(t)}+(1-\beta _{1})\nabla _{w}L^{(t)))$
${\displaystyle v_{w}^{(t+1)}\leftarrow \beta _{2}v_{w}^{(t)}+(1-\beta _{2})(\nabla _{w}L^{(t)})^{2))$
${\displaystyle {\hat {m))_{w}={\frac {m_{w}^{(t+1))){1-\beta _{1}^{t))))$
${\displaystyle {\hat {v))_{w}={\frac {v_{w}^{(t+1))){1-\beta _{2}^{t))))$
${\displaystyle w^{(t+1)}\leftarrow w^{(t)}-\eta {\frac ((\hat {m))_{w))((\sqrt ((\hat {v))_{w))}+\epsilon ))}$
where ${\displaystyle \epsilon }$ is a small scalar (e.g. ${\displaystyle 10^{-8))$) used to prevent division by 0, and ${\displaystyle \beta _{1))$ (e.g. 0.9) and ${\displaystyle \beta _{2))$ (e.g. 0.999) are the forgetting factors for gradients and second moments of gradients, respectively. Squaring and square-rooting is done element-wise. The profound influence of this algorithm inspired multiple newer, less well-known momentum-based optimization schemes using Nesterov-enhanced gradients (eg: NAdam[29] and FASFA[30]) and varying interpretations of second-order information (eg: Powerpropagation[31] and AdaSqrt[32]). However, the most commonly used variants are AdaMax,[28] which generalizes Adam using the infinity norm, and AMSGrad,[33] which addresses convergence problems from Adam.
### Backtracking line search
Backtracking line search is another variant of gradient descent. All of the below are sourced from the mentioned link. It is based on a condition known as the Armijo–Goldstein condition. Both methods allow learning rates to change at each iteration; however, the manner of the change is different. Backtracking line search uses function evaluations to check Armijo's condition, and in principle the loop in the algorithm for determining the learning rates can be long and unknown in advance. Adaptive SGD does not need a loop in determining learning rates. On the other hand, adaptive SGD does not guarantee the "descent property" – which Backtracking line search enjoys – which is that ${\displaystyle f(x_{n+1})\leq f(x_{n})}$ for all n. If the gradient of the cost function is globally Lipschitz continuous, with Lipschitz constant L, and learning rate is chosen of the order 1/L, then the standard version of SGD is a special case of backtracking line search.
### Second-order methods
A stochastic analogue of the standard (deterministic) Newton–Raphson algorithm (a "second-order" method) provides an asymptotically optimal or near-optimal form of iterative optimization in the setting of stochastic approximation[citation needed]. A method that uses direct measurements of the Hessian matrices of the summands in the empirical risk function was developed by Byrd, Hansen, Nocedal, and Singer.[34] However, directly determining the required Hessian matrices for optimization may not be possible in practice. Practical and theoretically sound methods for second-order versions of SGD that do not require direct Hessian information are given by Spall and others.[35][36][37] (A less efficient method based on finite differences, instead of simultaneous perturbations, is given by Ruppert.[38]) Another approach to the approximation Hessian matrix is replacing it with the Fisher information matrix, which transforms usual gradient to natural.[39] These methods not requiring direct Hessian information are based on either values of the summands in the above empirical risk function or values of the gradients of the summands (i.e., the SGD inputs). In particular, second-order optimality is asymptotically achievable without direct calculation of the Hessian matrices of the summands in the empirical risk function.
## Notes
1. ^ ${\displaystyle \odot }$ is the element-wise product.
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19. ^ Sutskever, Ilya; Martens, James; Dahl, George; Hinton, Geoffrey E. (June 2013). Sanjoy Dasgupta and David Mcallester (ed.). On the importance of initialization and momentum in deep learning (PDF). In Proceedings of the 30th international conference on machine learning (ICML-13). Vol. 28. Atlanta, GA. pp. 1139–1147. Retrieved 14 January 2016.
20. ^ Sutskever, Ilya (2013). Training recurrent neural networks (PDF) (Ph.D.). University of Toronto. p. 74.
22. ^ Borysenko, Oleksandr; Byshkin, Maksym (2021). "CoolMomentum: A Method for Stochastic Optimization by Langevin Dynamics with Simulated Annealing". Scientific Reports. 11 (1): 10705. arXiv:2005.14605. Bibcode:2021NatSR..1110705B. doi:10.1038/s41598-021-90144-3. PMC 8139967. PMID 34021212.
23. ^ Polyak, Boris T.; Juditsky, Anatoli B. (1992). "Acceleration of stochastic approximation by averaging" (PDF). SIAM J. Control Optim. 30 (4): 838–855. doi:10.1137/0330046.
24. ^ a b Duchi, John; Hazan, Elad; Singer, Yoram (2011). "Adaptive subgradient methods for online learning and stochastic optimization" (PDF). JMLR. 12: 2121–2159.
25. ^ Gupta, Maya R.; Bengio, Samy; Weston, Jason (2014). "Training highly multiclass classifiers" (PDF). JMLR. 15 (1): 1461–1492.
26. ^ Hinton, Geoffrey. "Lecture 6e rmsprop: Divide the gradient by a running average of its recent magnitude" (PDF). p. 26. Retrieved 19 March 2020.
27. ^ Hinton, Geoffrey. "Lecture 6e rmsprop: Divide the gradient by a running average of its recent magnitude" (PDF). p. 29. Retrieved 19 March 2020.
28. ^ a b Kingma, Diederik; Ba, Jimmy (2014). "Adam: A Method for Stochastic Optimization". arXiv:1412.6980 [cs.LG].
29. ^ Dozat, T. (2016). "Incorporating Nesterov Momentum into Adam". S2CID 70293087. ((cite journal)): Cite journal requires |journal= (help)
30. ^ Naveen, Philip (2022-08-09). "FASFA: A Novel Next-Generation Backpropagation Optimizer". dx.doi.org. doi:10.36227/techrxiv.20427852.v1. Retrieved 2022-11-19.
31. ^ Whye, Schwarz, Jonathan Jayakumar, Siddhant M. Pascanu, Razvan Latham, Peter E. Teh, Yee (2021-10-01). Powerpropagation: A sparsity inducing weight reparameterisation. OCLC 1333722169.
32. ^ Hu, Yuzheng; Lin, Licong; Tang, Shange (2019-12-20). "Second-order Information in First-order Optimization Methods". arXiv:1912.09926. ((cite journal)): Cite journal requires |journal= (help)
33. ^ Reddi, Sashank J.; Kale, Satyen; Kumar, Sanjiv (2018). "On the Convergence of Adam and Beyond". arXiv:1904.09237. ((cite journal)): Cite journal requires |journal= (help)
34. ^ Byrd, R. H.; Hansen, S. L.; Nocedal, J.; Singer, Y. (2016). "A Stochastic Quasi-Newton method for Large-Scale Optimization". SIAM Journal on Optimization. 26 (2): 1008–1031. arXiv:1401.7020. doi:10.1137/140954362. S2CID 12396034.
35. ^ Spall, J. C. (2000). "Adaptive Stochastic Approximation by the Simultaneous Perturbation Method". IEEE Transactions on Automatic Control. 45 (10): 1839−1853. doi:10.1109/TAC.2000.880982.
36. ^ Spall, J. C. (2009). "Feedback and Weighting Mechanisms for Improving Jacobian Estimates in the Adaptive Simultaneous Perturbation Algorithm". IEEE Transactions on Automatic Control. 54 (6): 1216–1229. doi:10.1109/TAC.2009.2019793. S2CID 3564529.
37. ^ Bhatnagar, S.; Prasad, H. L.; Prashanth, L. A. (2013). Stochastic Recursive Algorithms for Optimization: Simultaneous Perturbation Methods. London: Springer. ISBN 978-1-4471-4284-3.
38. ^ Ruppert, D. (1985). "A Newton-Raphson Version of the Multivariate Robbins-Monro Procedure". Annals of Statistics. 13 (1): 236–245. doi:10.1214/aos/1176346589.
39. ^ Abdulkadirov, R. I.; Lyakhov, P.A.; Nagornov, N.N. (2022). "Accelerating Extreme Search of Multidimensional Functions Based on Natural Gradient Descent with Dirichlet Distributions". Mathematics. 10 (19): 3556. doi:10.3390/math10193556.
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# Completing the square when the coeffecient of x^2 is not 1
Discussion in 'Math' started by amilton542, Apr 12, 2011.
1. ### amilton542 Thread Starter Active Member
Nov 13, 2010
491
64
I'm having trouble reaching a given answer
2x^2 + 8x - 25 = 0
Answer: -2 + or - sqr/rt 33/2
My workings out:
2(x^2 + 4x - 25/2) = 0
2(x + 4)^2 - 16 - 25/2
-16 - 25/2 = -32/2 - 25/2 = - 57/2
2(x+4)^2 - 57/2 = 0
2x = -4 + or - sqr/rt 57/2
This is where im stuck. I dont no what to do with the coeffeicient, If I divide the 2 to eleminate it I have to divide everything on the right by 2 and that won't give me the answer
2. ### Georacer Moderator
Nov 25, 2009
5,142
1,266
First of all, are you sure your answer is correct? From the 2nd grade root formula I get:
$
a=2,b=8,c=-25\\
\Delta=b^2-4\cdot a \cdot c = 64+ 4 \cdot 2 \cdot 25=264\\
x_{1,2}=\frac{-b \pm \sqrt{\Delta}}{2 \cdot a}=\frac{-8 \pm \sqrt{264}}{4} \simeq -6.025\ and\ 2.0625$
Other than that, you correctly factored out 2 on your first step. Just take the expression in the parenthesis and equate it with 0. You now have another, equivalent equation with 1 as the coefficient of the 2nd grade term.
Last edited: Apr 12, 2011
3. ### Papabravo Expert
Feb 24, 2006
10,025
1,760
The use of the phrase "2nd grade" is kind of amusing. Your meaning is the same as the word "quadratic" as in "quadratic formula" for finding roots or "quadratic term" referring to the term with x^2.
In the US public school system "2nd Grade" refers to the second year of elementary education where children are usually about 7 years of age. I know the trend is to push mathematical concepts into lower grades but I'm pretty sure we are not there yet.
That was a good one.
Last edited: Apr 12, 2011
4. ### amilton542 Thread Starter Active Member
Nov 13, 2010
491
64
http://www.mathtutor.ac.uk/algebra/completingthesquare/text Page 8, excercise 4, question f. The answers for the excercises are below. They've used a different method to the quadratic formula
5. ### Georacer Moderator
Nov 25, 2009
5,142
1,266
@ Papabravo
It was a direct translation from Greek. I 'll remember it next time.
@amilton542
I didn't understand that "+ or -" was supposed to mean $\pm$. Yes, that solution is correct.
I know the way you are trying to solve the problem, and if you can do it for a=1 you can do it for any a. Just factor a out and solve the new quadratic equation that is in the parenthesis and has a=1.
6. ### amilton542 Thread Starter Active Member
Nov 13, 2010
491
64
I do not know how the latex reference works
7. ### jpanhalt AAC Fanatic!
Jan 18, 2008
5,564
839
If you want to do it by completing the square rather than by using the quadratic formula, here is how:
1) Factor out the coefficient of x$^{2}$ to give 2(x$^{2}$ +4x -12.5) = 0
2) Divide both sides by 2 to give: x$^{2}$ +4x -12.5 =0
3) Solve for y in this equation -12.5 +y =4; y=16.5
4) Now add 16.5 to both sides to get: x$^{2}$ +4x +4 = 16.5
5) (x+2)$^{2}$ = 16.5
6) x+2 = (16.5)$^{1/2}$
John
8. ### jpanhalt AAC Fanatic!
Jan 18, 2008
5,564
839
There is a small, but perhaps confusing typo.
$\Delta = b^2 -4\cdot a\cdot c$
John
9. ### Georacer Moderator
Nov 25, 2009
5,142
1,266
Right, typo corrected. Thanks for the correction.
10. ### Papabravo Expert
Feb 24, 2006
10,025
1,760
I know that Greek is your first language and I was not being critical. Translation quite often produces amusing results given the general ability to take account of idiomatic expressions and forms especially in a technical context. One evening while in Italy on assignment the bulb in the reading light in my room went out. The desk clerk was completely nonplussed when I informed him in halting Italian that "the streetlight in my room was not working". My accent was certainly not Piemotese which probably gave him an additional clue.
11. ### Georacer Moderator
Nov 25, 2009
5,142
1,266
Don't worry, no offence was taken whatsoever. Actually, I 'm a very hard person to offend.
I 've had my moments with foreign languages too, and it seems I still do.
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L1 / Variational Distance between distributions [closed]
My statistics knowledge is somewhat poor, so I have to ask one (dumb) question.
Let $\beta$ be a real number in the interval $\big[0, \frac{1}{2}\big)$ and $\mathcal{D}_1, \mathcal{D}_2, \mathcal{D}_3$ be three distributions over a space $\mathcal{X}$, with the property that $\mathcal{D}_1 = \beta \cdot \mathcal{D}_2 + (1-\beta) \cdot \mathcal{D}_3$. What is the statistical variational distance between $\mathcal{D}_1$ and $\mathcal{D}_2$?
Thanks you a lot!
closed as off-topic by Emil Jeřábek, Jeffε, Kaveh, Jan Johannsen, Hsien-Chih Chang 張顯之Aug 18 '17 at 6:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "Your question does not appear to be a research-level question in theoretical computer science. For more information about the scope, please see help center. Your question might be suitable for Computer Science which has a broader scope." – Emil Jeřábek, Jeffε, Kaveh, Jan Johannsen, Hsien-Chih Chang 張顯之
If this question can be reworded to fit the rules in the help center, please edit the question.
Using the relation between total variation and $L_1$/$\ell_1$ distance of the probability/distribution/mass functions, we have \begin{align} d_{\rm TV}(D_1, D_2) &= \frac{1}{2}\lVert D_1-D_2\rVert_1 = \frac{1}{2}\lVert \beta D_2 +(1-\beta)D_3 - D_2\rVert_1\\ &= \frac{1-\beta}{2}\lVert D_3 - D_2\rVert_1 = (1-\beta)d_{\rm TV}(D_2, D_3). \end{align}
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# Compact operator norm estimate
I found the next exercise in Haim Brezis's book Functional Analysis, Sobolev Spaces and Partial Differential Equations. I feel like I solved the problem, but I'm not sure.
The problem is:
Let $E,F$ be two Banach spaces with norms $\|\cdot \|_E,\| \cdot \|_F$. Assume that $E$ is reflexive. Let $T :E \to F$ be a compact operator. Consider on $E$ another norm $| \cdot |$ weaker than $\|\cdot \|_E$, i.e. $|u|\leq C \|u\|_E$. Prove that for every $\varepsilon >0$ ther exists $C_\varepsilon>0$ such that $$\|Tu\|_F \leq \varepsilon \|u\|_E +C_\varepsilon |u|$$
My approach is the following: Assume that the conclusion does not hold. Then there is an $\varepsilon>0$ and a sequence $u_n$ in $E$ such that $$\|Tu_n\|_F > \varepsilon \|u_n\|_E+n|u_n|$$
We can normalize the sequence such that $\|Tu_n\|=1$. Then $u_n$ is bounded in $E$, and because $E$ is reflexive then without loss of generality we can assume that $u_n$ converges weakly to $u \in E$. Since compact operators map weakly convergent sequences onto strongly convergent sequences it follows that $Tu_n \to Tu$ in $F$, so $\|Tu\|_F=1$, which means that $u\neq 0$.
On the other hand $|u_n|<1/n$ so that $u_n$ converges to $0$ in the weaker norm $|\cdot |$. Is this enough to prove that $u=0$ and reach a contradiction?
If my approach does not lead to a good end then what else should I try?
-
Your approach looks good and almost complete. To finish the proof, note that the weak convergence of $u_n$ to $u$ in $(E,\|\cdot\|_E)$ implies weak convergence in $(E,|\cdot|)$ – the assumption $|u|\le C \|u\|_E$ says that $|\cdot|$-continuous functionals are $\|\cdot\|_E$-continuous. Then $u=0$ since the weak topology of $(E,|\cdot|)$ is Hausdorff and $|u_n|\to0$.
I was thinking something on these lines, but I didn't think of the weak topology given by $|\cdot|$. It is a nice idea indeed. Thanks – Beni Bogosel May 31 '12 at 12:46
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# Geometrical Applications of Complex Numbers
The complex numbers $$z_1,z_2,z_3$$ satisfying $$\dfrac{z_1+z_3}{z_2-z_3}=\dfrac{1-i\sqrt{3}}{2}$$ are the vertices of a triangle which is:
a) of area $$0$$
b) equilateral
c) right angled and isosceles
d) obtuse angled
All I got was that from the Rotation Theorem, $$\arg\left(\dfrac{z_1+z_3}{z_2-z_3}\right)=-\pi/3$$ and that $$|z_1+z_3|=|z_2-z_3|$$ Could somebody please show me how to solve this problem? Many thanks!
• If the numerator were $z_1-z_3$ instead of $z_1+z_3$, then the answer would be (b), because the information you're given would imply that two of the sides have equal length and the angle between those two sides is $60^\circ$. – grand_chat Mar 25 '16 at 7:04
• Hint: Shift the origin to any of $z_1, z_2, z_3$ to see the properties of triangle, as angles and sides are invariant under translation of origin. – mea43 Mar 25 '16 at 7:10
• You should edit your post: in fact it's $\frac{\text{z1}+\text{z3}}{\text{z2}-\text{z3}}=e^{-i\frac{\pi }{3}}$ and $\arg \left(\frac{\text{z1}+\text{z3}}{\text{z2}-\text{z3}}\right)=-\frac{\pi }{3}$. As @user21820, I don't think you can know much more about $z1,z2,z3$... – Eddy Khemiri Mar 25 '16 at 10:36
• @mea43 Sir I tried for very long, but couldn't reach anywhere. Could you please help me? – Better World Mar 25 '16 at 11:01
• @EddyKhemiri Edited, thanks for informing me. – Better World Mar 25 '16 at 11:03
There is a problem with the problem. By what you have stated, we know that $(-z_1,z_2,z_3)$ are the vertices of an isosceles triangle with equal sides meeting at $120^\circ$. But then we don't know anything much about $(z_1,z_2,z_3)$, because as you can see translating $(-z_1,z_2,z_3)$ around makes $(z_1,z_2,z_3)$ change shape.
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×
# [Help]Looking for Geometry Package
Hello everyone,
I need a little help. Can anyone give me the link to download a graph sketching software, by which one can draw any and every type of graphs. I have "Geogebra" but that doesn't help me. I need some kindaa perfect graph plotter.
Any kind of help will be heartily appreciated.
Thanks a lot
Note by Sandeep Bhardwaj
1 year, 7 months ago
Sort by:
May I ask you why Geogebra is not helpful to you? · 1 year, 7 months ago
Suppose I want to draw wavy curve graph for the signs of any rational expression, then can it be done using Geogebra ? Honestly, I don't have any idea whether Geogebtra will help with this or not, so I want to know. · 1 year, 7 months ago
I do not know what you mean by "signs" but you can certainly plot rational expressions with Geogebra.
For example, just type y=(x^2+5 x+2)/(x^3 + 7 x + 9) in the input box below. · 1 year, 7 months ago
If I want to draw a graph like this (wavy curve method to get the signs of the rational expressions), can it be done using geogebra, if yes, then how ? · 1 year, 7 months ago
Would you be having the closed form of the function?
For example, here is what you get when you enter the following in the input box: y = (x - 3) (x - 4) (x - 5)^3 / (x - 6)^2 (x - 2)^2 · 1 year, 7 months ago
Desmos or geometry pad both are free and available on mobile and PC. · 1 year, 7 months ago
sorry to write here i am in class 12 please guide me aspiring for 300 + in mains · 1 year, 7 months ago
its for ios i guess · 1 year, 7 months ago
well i dont use much this app but i have an app for graphs called freegracal (searh on app store for graphic calculator) may be it would help. · 1 year, 7 months ago
Graph.tk · 1 year, 7 months ago
You can try this,although it is not available for download it is quite good! · 1 year, 7 months ago
If someone needs to draw freehand graphs i.e. not graph for any particular expression ? · 1 year, 7 months ago
Well, those free-hand graphs you can always draw on a graph paper. Why do you need a software for that?? · 1 year, 7 months ago
I can do that on paper, but I need softcopy for those graphs. and I can't click a pic of that paper work because I need quite clear work. Like If i want to draw an upward facing parabola, then how to do that ? · 1 year, 7 months ago
Free hand graphs,like ogives? · 1 year, 7 months ago
Free hand graphs mean like If i want to draw an upward facing parabola, then how to do that ? · 1 year, 7 months ago
I'm sure you know what the equation of an upward facing parabola is.
Geogebra let's you draw one without the equation though.
For more complicated curves, BeizerCurves looks like a good idea · 1 year, 7 months ago
thank you Agnishom for your help. Also now I'm getting familiar with Geogebra, that also looks good. :) · 1 year, 7 months ago
Ohh well for that one could use paint! · 1 year, 7 months ago
Paint doesn't work. How will you draw an upward parabola on that ? · 1 year, 7 months ago
In the shapes area there is one 'curves' option! · 1 year, 7 months ago
Well. You're quite familiar with paint. Great ! But the images are not sharp in paint. Even drawing a vertical line will not give exactly the line. You can try it out. · 1 year, 7 months ago
Yeah!I won't be online now,sorry! · 1 year, 7 months ago
It's ok. Thanks a lot for the help. · 1 year, 7 months ago
Well, you can use Desmos Graphing calculator for that. Its available on Google play store as an app for android device or you can just visit desmos.com/calculator
Hope that helps!! · 1 year, 7 months ago
Is it available for laptop too ? · 1 year, 7 months ago
Yes.
I just gave you the link to calculator in the above note, which you can access from any browser installed on your laptop.
But I don't know if Desmos is available in download/offline version for the reason that I always open the above link from my laptop whenever I need a graphing calculator. · 1 year, 7 months ago
www.desmos.com · 1 year, 7 months ago
Picture
Sir you might like this app its called maths labs. · 1 year, 7 months ago
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Question
# Solve the fraction: $\dfrac{9}{11}-\dfrac{4}{15}$.
Hint: Find the common denominator of the two fractions. Then subtract the two numerators (top numbers). Put the answer over the denominator and simplify the fraction formed.
Give us two fractions $\dfrac{9}{11}$and $\dfrac{4}{15}$where we need to subtract the second from the first.
The first step towards subtraction is to ensure that the bottom numbers (the denominators) are the same.
Here the denominators are different $\dfrac{9}{11}$and $\dfrac{4}{15}$. Let’s make the denominators the same.
We know the denominators are 11 and 15. They both don’t have any numbers in common. So let’s multiply 11 and 15.
$11\times 15=165$
Now we got the denominator as 165.
First let us take the fraction $\dfrac{9}{11}$. We need to make the denominator of the fraction as 165. So multiply by 15 in the numerator and denominator.
$\dfrac{9\times 15}{11\times 15}=\dfrac{135}{165}-(1)$
Now let us take the fraction $\dfrac{4}{15}$. We need to make the denominator 165.
So let’s multiply by 11 in the numerator and denominator.
$\dfrac{4\times 11}{15\times 11}=\dfrac{44}{165}-(2)$
Now let us subtract (2) from (1).
$\therefore \dfrac{135}{165}-\dfrac{44}{165}=\dfrac{135-44}{165}$
Now let’s subtract the top numbers (the numerators). Put the answer over the same denominator.
$\dfrac{135-44}{165}=\dfrac{91}{165}$
We cannot further simplify the fraction, as they don’t have any common factors.
$\therefore \dfrac{9}{11}-\dfrac{4}{15}=\dfrac{91}{165}$
Note: It can be solved by cross multiplying the numerator and denominator divided by the multiplication of both denominators.
$\dfrac{9}{11}-\dfrac{4}{15}=\dfrac{\left( 9\times 15 \right)-\left( 4\times 11 \right)}{11\times 15}=\dfrac{135-44}{165}=\dfrac{91}{165}$
We get the same answer as we have got in the first method. This is a shortcut for the same.
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# Datasets:circa
Languages: en
Multilinguality: monolingual
Size Categories: 10K<n<100K
Language Creators: crowdsourced
Annotations Creators: expert-generated
Source Datasets: original
Dataset Preview
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Are you employed?
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I'm a veterinary technician.
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I am a fan of Korean food .
I wouldn't say so
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I like to BBQ .
I love to bbq
Yes#Yes#Yes#Yes#Yes
0 (Yes)
0 (Yes)
X wants to know what activities Y likes to do during weekends.
Are you interested in sporting events?
I am interested in sporting events .
Only if my teams are playing.
Yes, subject to some conditions#Yes, subject to some conditions#Yes, subject to some conditions#Yes, subject to some conditions#Yes, subject to some conditions
5 (Yes, subject to some conditions)
3 (Yes, subject to some conditions)
Y has just told X that he/she is considering switching his/her job.
Do you travel any?
I travel .
Once a week.
Yes#Yes#Yes#Yes#Yes
0 (Yes)
0 (Yes)
Y has just told X that he/she is thinking of buying a flat in New York.
Did you check your credit to see if you are able to apply for mortgage?
I did check my credit to see if I am able to apply for mortgage .
I'm paying in cash.
Probably no#Probably no#No#Other#No
-1
1 (No)
X wants to know about Y's food preferences.
Are you still on a Paleo diet?
I am on a Paleo diet .
I have started on the Keto diet.
No#No#No#No#No
1 (No)
1 (No)
X wants to know what sorts of books Y likes to read.
Do you like fantasy books?
I like fantasy books .
I'm not familiar with any.
In the middle, neither yes nor no#In the middle, neither yes nor no#Probably no#Other#Probably no
-1
-1
X wants to know about Y's music preferences.
Do you like jazz music?
I like jazz music .
I think I should have been born in the Jazz Age.
Yes#Probably yes / sometimes yes#Probably yes / sometimes yes#Yes#Yes
0 (Yes)
0 (Yes)
Y has just travelled from a different city to meet X.
Do you still like hiking?
I like hiking .
I still try to hit the trails at least once a month.
Yes#Probably yes / sometimes yes#Yes#Yes#Yes
0 (Yes)
0 (Yes)
Y has just travelled from a different city to meet X.
Do you go bar hoping often?
I go bar hoping often .
I don't drink anymore.
No#No#No#No#No
1 (No)
1 (No)
Y has just told X that he/she is thinking of buying a flat in New York.
Did you consult with your partner/parents?
I did consult with my partner/parents .
I haven't spoken to my partner in ages.
No#Probably no#No#No#No
1 (No)
1 (No)
Y has just moved into a neighbourhood and meets his/her new neighbour X.
How do you like the area?
I do I like the area .
The mountains are awesome.
Other#Probably yes / sometimes yes#Other#Probably yes / sometimes yes#Yes
-1
0 (Yes)
Y has just told X that he/she is thinking of buying a flat in New York.
Do you need a lot of space?
I need a lot of space .
I could do with a small studio.
Probably no#Probably no#No#No#No
1 (No)
1 (No)
Y has just travelled from a different city to meet X.
Want to get a pedicure?
I want to get a pedicure .
My toes are a mess.
In the middle, neither yes nor no#In the middle, neither yes nor no#Probably yes / sometimes yes#Probably yes / sometimes yes#Yes
-1
0 (Yes)
Y has just told X that he/she is considering switching his/her job.
Is your job a manual labor?
My job is a manual labor .
I get a lot of exercise all day at work.
Probably yes / sometimes yes#Yes#In the middle, neither yes nor no#Probably yes / sometimes yes#Yes
-1
0 (Yes)
Y has just moved into a neighbourhood and meets his/her new neighbour X.
Have you met the other neighbors?
I have met the other neighbors .
I have met a few of them
Probably yes / sometimes yes#Yes#Yes#Yes#Yes
0 (Yes)
0 (Yes)
Y has just travelled from a different city to meet X.
Was the travel tiring?
The travel was tiring .
It was exciting
Probably no#No#In the middle, neither yes nor no#No#No
1 (No)
1 (No)
X and Y are colleagues who are leaving work on a Friday at the same time.
Has this work week gone quickly for you?
This work week has gone quickly for I .
It's gone very slowly.
No#No#No#No#No
1 (No)
1 (No)
X and Y are childhood neighbours who unexpectedly run into each other at a cafe.
Have you stayed in touch with anyone from school?
I have stayed in touch with anyone from school .
I really haven't
No#No#No#No#No
1 (No)
1 (No)
Y has just told X that he/she is thinking of buying a flat in New York.
Have you lived in New York before?
I have lived in New York .
I've only been there for vacation.
No#No#No#No#No
1 (No)
1 (No)
X wants to know about Y's food preferences.
Are you in the mood for ramen?
I am in the mood for ramen .
I could go for some Pho.
Probably no#In the middle, neither yes nor no#I am not sure how X will interpret Y’s answer#I am not sure how X will interpret Y’s answer#Probably yes / sometimes yes
-1
2 (In the middle, neither yes nor no)
X and Y are colleagues who are leaving work on a Friday at the same time.
Do you want me to walk you to your car?
I want me to walk I to my car .
That would be really nice, thanks.
Yes#Yes#Yes#Yes#Yes
0 (Yes)
0 (Yes)
X wants to know about Y's food preferences.
Do you like food that are primarily barbecued?
I like food that am primarily barbecued .
it is okay to have sometimes
Yes, subject to some conditions#Probably yes / sometimes yes#In the middle, neither yes nor no#Probably yes / sometimes yes#Probably yes / sometimes yes
3 (Probably yes / sometimes yes)
0 (Yes)
X wants to know about Y's food preferences.
Are you a big meat-eater?
I am a big meat-eater .
I prefer leafy greens.
Probably no#Probably no#No#Probably no#No
4 (Probably no)
1 (No)
Y has just told X that he/she is considering switching his/her job.
Is your job close to home?
My job is close to home .
It's an hour away.
No#No#No#I am not sure how X will interpret Y’s answer#No
1 (No)
1 (No)
Y has just told X that he/she is thinking of buying a flat in New York.
Have you researched the areas you are interested in?
I have researched the areas I am interested in .
I've been looking a lot online.
Probably yes / sometimes yes#Yes#Yes#Yes#Yes
0 (Yes)
0 (Yes)
X wants to know what activities Y likes to do during weekends.
Do you like travelling?
I like travelling .
In first class.
Yes, subject to some conditions#Yes, subject to some conditions#Yes, subject to some conditions#Yes, subject to some conditions#Yes, subject to some conditions
5 (Yes, subject to some conditions)
3 (Yes, subject to some conditions)
Y has just told X that he/she is thinking of buying a flat in New York.
Do you have any pets?
I have pets .
I have a cat.
Yes#Yes#Yes#Yes#Yes
0 (Yes)
0 (Yes)
X wants to know what sorts of books Y likes to read.
Do you enjoy sci-fi books?
I enjoy sci-fi books .
My husband has gotten me into them.
Yes#Yes#Yes#Yes#Yes
0 (Yes)
0 (Yes)
X wants to know what activities Y likes to do during weekends.
Do you like to see live music?
I like to see live music .
if the band is good
Yes, subject to some conditions#Yes, subject to some conditions#Yes, subject to some conditions#Yes, subject to some conditions#Yes, subject to some conditions
5 (Yes, subject to some conditions)
3 (Yes, subject to some conditions)
X wants to know what activities Y likes to do during weekends.
Have you been bowling before?
I have been bowling before .
Years ago, and haven't been back since.
Yes#Yes#Yes#Yes#Yes
0 (Yes)
0 (Yes)
X and Y are childhood neighbours who unexpectedly run into each other at a cafe.
My mum and dad am alive .
I'm an orphan.
I am not sure how X will interpret Y’s answer#No#No#No#No
1 (No)
1 (No)
Y has just moved into a neighbourhood and meets his/her new neighbour X.
Are you new in town?
I am new in town .
I've just got here.
Yes#Yes#Yes#Yes#Yes
0 (Yes)
0 (Yes)
X and Y are colleagues who are leaving work on a Friday at the same time.
I was my day .
Just a typical Friday.
In the middle, neither yes nor no#Other#In the middle, neither yes nor no#I am not sure how X will interpret Y’s answer#Other
-1
2 (In the middle, neither yes nor no)
X wants to know what activities Y likes to do during weekends.
Do you like being outdoors?
I like being outdoors .
I prefer outdoors
Yes#Yes#Yes#Yes#Yes
0 (Yes)
0 (Yes)
X and Y are colleagues who are leaving work on a Friday at the same time.
Do you want to meet up this weekend?
I want to meet up this weekend .
If I have a free moment.
Yes, subject to some conditions#Yes, subject to some conditions#Yes, subject to some conditions#Yes, subject to some conditions#Yes, subject to some conditions
5 (Yes, subject to some conditions)
3 (Yes, subject to some conditions)
X wants to know about Y's food preferences.
Sushi is raw, which is bad for my stomach.
Probably no#In the middle, neither yes nor no#No#Probably no#In the middle, neither yes nor no
-1
1 (No)
X wants to know about Y's music preferences.
Are you interested in trap music?
I am interested in trap music .
I don't know what that is
In the middle, neither yes nor no#No#I am not sure how X will interpret Y’s answer#Probably no#Probably no
-1
1 (No)
X wants to know about Y's music preferences.
Have you heard Taylor Swift's new song?
I have heard Taylor Swift 's new song .
I listened to it yesterday.
Yes#Yes#Yes#Yes#Yes
0 (Yes)
0 (Yes)
X wants to know what sorts of books Y likes to read.
Is reading crime fiction something you like to do?
Something is like to do .
If it's not too creepy.
Yes, subject to some conditions#Yes, subject to some conditions#Yes, subject to some conditions#Yes, subject to some conditions#Yes, subject to some conditions
5 (Yes, subject to some conditions)
3 (Yes, subject to some conditions)
X wants to know what activities Y likes to do during weekends.
Do you enjoy playing any sports?
I enjoy playing sports .
I like to read or watch TV.
Probably no#Probably no#No#Probably no#No
4 (Probably no)
1 (No)
X wants to know about Y's food preferences.
Do you eat seafood?
I eat seafood .
I love catching my own fish and cooking it.
In the middle, neither yes nor no#Yes#Yes#Yes#Yes
0 (Yes)
0 (Yes)
Y has just travelled from a different city to meet X.
Do you enjoy living out of town?
I enjoy living out of town .
I love living in the countryside.
Probably yes / sometimes yes#Yes#Yes#Yes#Yes
0 (Yes)
0 (Yes)
X wants to know what activities Y likes to do during weekends.
Have you been to the art museum?
I have been to the art museum .
I love looking at pictures.
In the middle, neither yes nor no#In the middle, neither yes nor no#Probably yes / sometimes yes#I am not sure how X will interpret Y’s answer#Probably yes / sometimes yes
-1
2 (In the middle, neither yes nor no)
X wants to know about Y's food preferences.
Would you prefer a locally owned restaurant?
I would prefer a locally owned restaurant .
I never really dine at restaurant chains.
Probably yes / sometimes yes#Probably yes / sometimes yes#Yes#Yes#Yes
0 (Yes)
0 (Yes)
Y has just told X that he/she is considering switching his/her job.
Is it an office job?
It is an office job .
It's an assistant position.
In the middle, neither yes nor no#Probably yes / sometimes yes#In the middle, neither yes nor no#In the middle, neither yes nor no#Probably yes / sometimes yes
2 (In the middle, neither yes nor no)
2 (In the middle, neither yes nor no)
X wants to know what sorts of books Y likes to read.
Do you like fantasy books?
I like fantasy books .
Only when their plot is good.
Yes, subject to some conditions#Yes, subject to some conditions#Yes, subject to some conditions#Yes, subject to some conditions#Yes, subject to some conditions
5 (Yes, subject to some conditions)
3 (Yes, subject to some conditions)
Y has just travelled from a different city to meet X.
Looking forward to this weekend?
I am looking forward to this weekend .
No#No#No#No#No
1 (No)
1 (No)
Y has just moved into a neighbourhood and meets his/her new neighbour X.
Are you new to this city?
I am new to this city .
I've been here all my life.
No#No#No#No#No
1 (No)
1 (No)
X wants to know what sorts of books Y likes to read.
Was that biography that you read any good?
That biography was that I read good .
I thought it was forgettable.
Probably no#No#No#No#No
1 (No)
1 (No)
Y has just travelled from a different city to meet X.
Want to go dancing tomorrow night?
Night night .
I'd rather see a movie.
Probably no#No#No#Probably no#Probably no
4 (Probably no)
1 (No)
Y has just told X that he/she is considering switching his/her job.
Have you been looking for awhile?
I have been looking for awhile .
I just started today.
No#No#No#No#No
1 (No)
1 (No)
X wants to know what activities Y likes to do during weekends.
Do you play golf on weekends?
I play golf on weekends .
I vowed I'd never play golf.
No#No#No#No#No
1 (No)
1 (No)
X wants to know what sorts of books Y likes to read.
Do you watch a lot of romance movies?
I watch a lot of romance movies .
I like horror movies.
In the middle, neither yes nor no#In the middle, neither yes nor no#Probably no#Probably no#Probably no
4 (Probably no)
1 (No)
X wants to know what sorts of books Y likes to read.
Have you ever read horror novel books?
I have read horror novel books .
Yes#Yes#Yes#Yes#Yes
0 (Yes)
0 (Yes)
X wants to know about Y's food preferences.
Are you on a budget?
I am on a budget .
I would be at the minute
Probably yes / sometimes yes#Yes#Yes#I am not sure how X will interpret Y’s answer#Yes
0 (Yes)
0 (Yes)
Y has just told X that he/she is thinking of buying a flat in New York.
I would prefer to pay a mortgage more than rent.
In the middle, neither yes nor no#In the middle, neither yes nor no#In the middle, neither yes nor no#Probably yes / sometimes yes#In the middle, neither yes nor no
2 (In the middle, neither yes nor no)
2 (In the middle, neither yes nor no)
X wants to know about Y's food preferences.
Do you like a variety of food?
I like a variety of food .
I like eating good stuff.
In the middle, neither yes nor no#In the middle, neither yes nor no#In the middle, neither yes nor no#Probably yes / sometimes yes#Probably yes / sometimes yes
2 (In the middle, neither yes nor no)
2 (In the middle, neither yes nor no)
X wants to know what sorts of books Y likes to read.
Do you like fantasy novels?
I like fantasy novels .
Disney is cool.
In the middle, neither yes nor no#In the middle, neither yes nor no#I am not sure how X will interpret Y’s answer#I am not sure how X will interpret Y’s answer#Probably yes / sometimes yes
-1
2 (In the middle, neither yes nor no)
Y has just travelled from a different city to meet X.
Do you want some time on your own to explore?
I want time on my own to explore .
I really like poking around small shops off the beaten path.
In the middle, neither yes nor no#In the middle, neither yes nor no#Probably yes / sometimes yes#Probably yes / sometimes yes#Yes
-1
0 (Yes)
X wants to know about Y's food preferences.
Hey since the weather has gotten colder did you want to grab some potato soup?
I did I want to grab potato soup .
that would be good
Yes#Yes#Yes#Yes#Yes
0 (Yes)
0 (Yes)
X wants to know about Y's music preferences.
Have you been to a concert this month?
I have been to a concert this month .
I went to one last night.
Probably yes / sometimes yes#Yes#Yes#Yes#Yes
0 (Yes)
0 (Yes)
X and Y are colleagues who are leaving work on a Friday at the same time.
What are your plans for tonight?
I am my plans for tonight .
My plans fell through.
Other#Other#Other#Other#Other
6 (Other)
4 (Other)
Y has just travelled from a different city to meet X.
Did you bring everything you need?
I did bring everything I need .
I didn't have enough space for everything
No#No#Probably no#No#No
1 (No)
1 (No)
X wants to know about Y's food preferences.
Do you like steak?
I like steak .
Only if it's cooked rare.
Yes, subject to some conditions#Yes, subject to some conditions#Yes, subject to some conditions#Yes, subject to some conditions#Yes, subject to some conditions
5 (Yes, subject to some conditions)
3 (Yes, subject to some conditions)
Y has just told X that he/she is thinking of buying a flat in New York.
Does it have to be in the city itself?
It does have to be in the city .
It can be on the outskirts
No#No#No#In the middle, neither yes nor no#No
1 (No)
1 (No)
Y has just travelled from a different city to meet X.
Want to go to the cinema?
I want to go to the cinema .
I'd prefer to see the sights.
Probably no#Probably no#No#Probably no#No
4 (Probably no)
1 (No)
X and Y are colleagues who are leaving work on a Friday at the same time.
Do you have a date?
I have a date .
I'm meeting my kids.
I am not sure how X will interpret Y’s answer#In the middle, neither yes nor no#No#No#No
1 (No)
1 (No)
Y has just told X that he/she is thinking of buying a flat in New York.
Is is close to the Subway station?
Is is close to the Subway station .
It's just down the street.
Probably yes / sometimes yes#Yes#Yes#Yes#Yes
0 (Yes)
0 (Yes)
Y has just moved into a neighbourhood and meets his/her new neighbour X.
Want to go to the park?
I want to go to the park .
I don't have time at the moment.
No#No#No#No#No
1 (No)
1 (No)
Y has just told X that he/she is thinking of buying a flat in New York.
Will you have to share the flat with anyone?
I will have to share the flat with anyone .
I'm excited to live on my own.
No#No#No#No#No
1 (No)
1 (No)
Y has just told X that he/she is considering switching his/her job.
Do you like to have a flexible schedule or are regular hours better for you?
I am regular hours better for I .
I prefer a rigid routine
Other#Other#Yes#Other#I am not sure how X will interpret Y’s answer
6 (Other)
4 (Other)
Y has just moved into a neighbourhood and meets his/her new neighbour X.
Do you have any kids?
I have kids .
My daughter just turned six.
Yes#Yes#Yes#Yes#Yes
0 (Yes)
0 (Yes)
End of preview (truncated to 100 rows)
# Dataset Card for CIRCA
### Dataset Summary
The Circa (meaning ‘approximately’) dataset aims to help machine learning systems to solve the problem of interpreting indirect answers to polar questions.
The dataset contains pairs of yes/no questions and indirect answers, together with annotations for the interpretation of the answer. The data is collected in 10 different social conversational situations (eg. food preferences of a friend).
The following are the situational contexts for the dialogs in the data.
1. X wants to know about Y’s food preferences
2. X wants to know what activities Y likes to do during weekends.
3. X wants to know what sorts of books Y likes to read.
4. Y has just moved into a neighbourhood and meets his/her new neighbour X.
5. X and Y are colleagues who are leaving work on a Friday at the same time.
6. X wants to know about Y's music preferences.
7. Y has just travelled from a different city to meet X.
8. X and Y are childhood neighbours who unexpectedly run into each other at a cafe.
9. Y has just told X that he/she is thinking of buying a flat in New York.
10. Y has just told X that he/she is considering switching his/her job.
### Languages
The text in the dataset is in English.
## Dataset Structure
### Data Instances
The columns indicate:
1. id : unique id for the question-answer pair
2. context : the social situation for the dialogue. One of 10 situations (see next section). Each
situation is a dialogue between a person who poses the question (X) and the person who
3. question-X : the question posed by X
4. canquestion-X : a (automatically) rewritten version of question into declarative form
Eg. Do you like Italian? --> I like Italian. See the paper for details.
6. judgements : the interpretations for the QA pair from 5 annotators. The value is a list of 5 strings,
separated by the token ‘#’
7. goldstandard1 : a gold standard majority judgement from the annotators. The value is the most common
interpretation and picked by at least 3 (out of 5 annotators). When a majority
judgement was not reached by the above criteria, the value is ‘NA’
8. goldstandard2 : Here the labels ‘Probably yes / sometimes yes’, ‘Probably no', and 'I am not sure how
X will interpret Y’s answer' are mapped respectively to ‘Yes’, ‘No’, and 'In the
middle, neither yes nor no’ before computing the majority. Still the label must be given
at least 3 times to become the majority choice. This method represents a less strict way
of analyzing the interpretations.
### Data Fields
id : 1
context : X wants to know about Y's food preferences.
question-X : Are you vegan?
canquestion-X : I am vegan.
answer-Y : I love burgers too much.
judgements : no#no#no#no#no
goldstandard1 : no (label(s) used for the classification task)
goldstandard2 : no (label(s) used for the classification task)
### Data Splits
There are no explicit train/val/test splits in this dataset.
## Dataset Creation
### Curation Rationale
They revisited a pragmatic inference problem in dialog: Understanding indirect responses to questions. Humans can interpret ‘I’m starving.’ in response to ‘Hungry?’, even without direct cue words such as ‘yes’ and ‘no’. In dialog systems, allowing natural responses rather than closed vocabularies would be similarly beneficial. However, today’s systems are only as sensitive to these pragmatic moves as their language model allows. They create and release the first large-scale English language corpus ‘Circa’ with 34,268 (polar question, indirect answer) pairs to enable progress on this task.
### Source Data
#### Initial Data Collection and Normalization
The QA pairs and judgements were collected using crowd annotations in three phases. They recruited English native speakers. The full descriptions of the data collection and quality control are present in EMNLP 2020 paper. Below is a brief overview only.
Phase 1: In the first phase, they collected questions only. They designed 10 imaginary social situations which give the annotator a context for the conversation. Examples are:
‘asking a friend for food preferences’
Annotators were asked to suggest questions which could be asked in each situation, such that each question only requires a ‘yes’ or ‘no’ answer. 100 annotators produced 5 questions each for the 10 situations, resulting in 5000 questions.
Phase 2: Here they focused on eliciting answers to the questions. They sampled 3500 questions from our previous set. For each question, They collected possible answers from 10 different annotators. The annotators were instructed to provide a natural phrase or a sentence as the answer and to avoid the use of explicit ‘yes’ and ‘no’ words.
Phase 3: Finally the QA pairs (34,268) were given to a third set of annotators who were asked how the question seeker would likely interpret a particular answer. These annotators had the following options to choose from:
* 'Yes'
* 'Probably yes' / 'sometimes yes'
* 'Yes, subject to some conditions'
* 'No'
* 'Probably no'
* 'In the middle, neither yes nor no'
* 'I am not sure how X will interpret Y's answer'
#### Who are the source language producers?
The rest of the data apart from 10 initial questions was collected using crowd workers. They ran pilots for each step of data collection, and perused their results manually to ensure clarity in guidelines, and quality of the data. They also recruited native English speakers, mostly from the USA, and a few from the UK and Canada. They did not collect any further information about the crowd workers.
### Annotations
#### Who are the annotators?
The rest of the data apart from 10 initial questions was collected using crowd workers. They ran pilots for each step of data collection, and perused their results manually to ensure clarity in guidelines, and quality of the data. They also recruited native English speakers, mostly from the USA, and a few from the UK and Canada. They did not collect any further information about the crowd workers.
## Considerations for Using the Data
### Dataset Curators
This dataset is the work of Annie Louis, Dan Roth, and Filip Radlinski from Google LLC.
### Citation Information
@InProceedings{louis_emnlp2020,
author = "Annie Louis and Dan Roth and Filip Radlinski",
title = ""{I}'d rather just go to bed": {U}nderstanding {I}ndirect {A}nswers",
booktitle = "Proceedings of the 2020 Conference on Empirical Methods in Natural Language Processing",
year = "2020",
}
### Contributions
Thanks to @bhavitvyamalik for adding this dataset.
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Rivet is hosted by Hepforge, IPPP Durham
## Rivet analyses reference
### E735_1998_S3905616
Charged particle multiplicity in $p\bar{p}$ collisions at $\sqrt{s} = 1.8\;\text{TeV}$
Experiment: E735 (Tevatron)
Inspire ID: 480349
Status: VALIDATED
Authors:
• Holger Schulz
• Andy Buckley
References:
• Phys.Lett.B435:453-457,1998
Beams: p- p+
Beam energies: (900.0, 900.0) GeV
Run details:
• QCD events, diffractive processes need to be switched on in order to fill the low multiplicity regions. The measurement was done in $|\eta| \lesssim 3.25$ and was extrapolated to full phase space. However, the method of extrapolation remains unclear.
A measurement of the charged multiplicity distribution at $\sqrt{s} = 1.8\;\text{TeV}$. The analysis is reproduced to the best of abilities. There is no theory curve in the paper to compare to.
Source code: E735_1998_S3905616.cc
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 // -*- C++ -*- #include "Rivet/Analysis.hh" #include "Rivet/Projections/ChargedFinalState.hh" #include "Rivet/Projections/TriggerCDFRun0Run1.hh" #include "Rivet/Projections/TriggerUA5.hh" namespace Rivet { /// @brief E735 charged multiplicity in NSD-triggered events class E735_1998_S3905616 : public Analysis { public: /// Constructor E735_1998_S3905616() : Analysis("E735_1998_S3905616") { _sumWTrig = 0; } /// @name Analysis methods //@{ void init() { // Projections declare(TriggerUA5(), "Trigger"); declare(ChargedFinalState(), "FS"); // Histo _hist_multiplicity = bookHisto1D(1, 1, 1); } void analyze(const Event& event) { const bool trigger = apply(event, "Trigger").nsdDecision(); if (!trigger) vetoEvent; const double weight = event.weight(); _sumWTrig += weight; const ChargedFinalState& fs = apply(event, "FS"); const size_t numParticles = fs.particles().size(); _hist_multiplicity->fill(numParticles, weight); } void finalize() { scale(_hist_multiplicity, 1/_sumWTrig); } //@} private: /// @name Weight counter //@{ double _sumWTrig; //@} /// @name Histograms //@{ Histo1DPtr _hist_multiplicity; //@} }; // The hook for the plugin system DECLARE_RIVET_PLUGIN(E735_1998_S3905616); }
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# Circle A has a center at (3 ,2 ) and a radius of 2 . Circle B has a center at (1 ,3 ) and a radius of 4 . Do the circles overlap? If not, what is the smallest distance between them?
Mar 8, 2016
circles overlap
#### Explanation:
First step is to find the distance between the centres using the$\textcolor{b l u e}{\text{ distance formula }}$
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2
where$\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are the coords of 2 points }$
let$\left({x}_{1} , {y}_{1}\right) = \left(3 , 2\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(1 , 3\right)$
d = sqrt((1-3)^2 + (3-2)^2) = sqrt(4+1) = sqrt5 ≈ 2.236
radius of A + radius of B = 2 + 4 = 6
since 6 > 2.236 , circles will overlap
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# What are the main differences between Texmaker and TeXstudio?
Although I've read these two posts:
I would like to ask again specifically about Texmaker and TeXstudio, because they are quite well ranked in the "Big list", and one of them was born as a fork of the other one.
What are the main differences between them? What are the advantages and drawbacks of each of them with respect to the other one?
If you changed to one of them after using the other one, what made you change your mind?
I think these questions are specific and different enough to deserve their own post...
EDIT:
As it can be deduced from my question, I am mainly interested in knowing the experience of people that have used (or at least evaluated) both editors. I know I can read the "features" page in their respective websites, but that's not what I am looking for.
• In my eyes, this question is too broad and mainly asking for opinions and thus off-topic. You can read the feature lists, read the Wiki-comparision or see the two dupes you already link yourself. The main answer to your question is: TeXstudio has more features. Like always this fact is splitting the users into the purists and feature-junkies. If you want information on performance, you will have to download the two gratis tools and test them. – LaRiFaRi Oct 21 '14 at 11:16
• Well, @LaRiFaRi, so there is at least one objective difference: TeXstudio has more features, and that means a different approach. And, yes, I am also asking about performance, but most of all I wanted to understand what was the approach of each one. – Vicent Oct 21 '14 at 11:19
• If you understand German, there is another dupe here: texwelt.de/wissen/fragen/1327/… – LaRiFaRi Oct 21 '14 at 11:41
• In my opinion, TeXstudio has a more powerful auto completion tool. Also, it is easy to change encoding and dictionary, just clicking on the status bar. – Sigur Oct 21 '14 at 11:55
TeXstudio has been forked from Texmaker in 2009, because of the non-open development process of Texmaker and due to different philosophies concerning configurability and features. Originally it was called TeXmakerX because it started off as a small set of extensions to Texmaker with the hope that they would get integrated into Texmaker some day. While at some points you can still see that TeXstudio originates from Texmaker, significant changes in features and the code base have made it to a fully independent program.
This quote explains the main approach of forking Texmaker in the first place. The last sentence indicates that they became two fully independent programs. Only the GUI seems similar but the rest is hard to compare.
On performance: The time consumption will be difficult to measure, but you can do it your self by downloading and testing both. They are both free. The main aspect will be the look and feel while working and depends mainly on the layer 8. I like TeXstudio more, but this is completely opinion based.
You can compare the download size and the space occupied on the system and post your results here. As TeXstudio comes with more features (I hear), I guess it will be bigger.
In order to decide which one sounds best for you, you will have to read the Texmaker feature list, the TeXstudio feature list, the comparison list on Wikipedia, and the answers and comments in our editors big list. Or you come to chat and discuss single topics with other users.
• Imho an absolute dealbraker for texmaker is a missing intelligent paragraph rewrapping and "fit to text" feature for the viewer. Aside the objective arguments texstudio's appearance had me in seconds. I wonder why I had to make this experience at the end of my college... – ManuelSchneid3r Sep 1 '16 at 21:50
Since the questioner asked explicitly for personal experiences, here is mine. I'm using Texmaker and I just learned about TeXstudio. At a first glance, TeXstudio provides all functionality that I'm used to in Texmaker.
My Problems with Texmaker
Texmaker has some performance issues which TeXstudio doesn't have at all.
More precisely, I have a large document with lots of sub documents. Whenever I change a \label, \chapter or \section, Texmaker responds very slowly. I can literally watch every single keystore with about 1s delay. In addition, it also freezes for many seconds if I comment or uncomment a block of \chapter and \input commands (e.g. if I just want to build just certain chapter/sections to speed up the LaTeX build).
What's even more disappointing is that I found posts from 2012 describing exactly the same performance problems, so there's probably little hope they'll fix that anytime soon:
http://latex-community.org/forum/viewtopic.php?f=21&t=21033
My Problems with TeXstudio
The highlighting of invalid/unknown LaTeX command doesn't work well in TeXstudio. There are issues with various commands such as \apptocmd. In addition, it doesn't recognize any of my own macros, probably because I put them into a separate file macros.tex, so it can be reused by other LaTeX documents such as standalone documents for larger TikZ graphics.
It is possible to provide a manual list of known commands, but in my case this is simply too much trouble, so I disabled this highlighting. I also had to disable autocompletion, as that was too annoying with an incomplete list of commands:
• Options / Configure TeXstudio
• Editor / Inline Checking / [ ] Syntax
• Completion / [ ] Automatically start completer when typing LaTeX-Commands
I realize this is an older question, but in case anyone is still interested, I briefly will try to add my own thoughts on your question:
What are the main differences between them? What are the advantages and drawbacks of each of them with respect to the other one?
In my view, the main difference is a substantially higher degree of customizability of TeXstudio in comparison to Texmaker.
This one aspect captures, in my opinion, the most important difference between the two programs. And to me at least, this aspect was also the main factor in choosing one over the other.
I don't know anything about the forking circumstances of TeXstudio, i.e. whether the TeXstudio devs followed forking etiquette or not (as suggested above in a comment). But as far as the current program is concerned, for the customizability reason mentioned above, I much prefer working with TeXstudio ever since I found out it exists -- which actually took a while, since it seems that it's not nearly as famous as Texmaker.
I'll give two examples where customization differs -- two small things that always bothered me in Texmaker, and that just instantly disappeared when switching to TeXstudio:
• Disabling the sidebar/side panel. Both programs show, by default, a sidebar menu containing commonly used commands, e.g. an "italics" button, and so on. However, I prefer to maximize my available screen space, so would rather not see that menu. In Texmaker, this menu can't be removed (at least, last timed I checked). In TeXstudio on the other hand, you can show/hide it. Small detail, but to me it actually mattered.
• Fine-tuning autocomplete commands. In Texmaker, you can add your own commands to the list of autocomplete commands, and there's a list of existing commands that you can edit, i.e. remove commands from the list. But: for some reason, no idea why, there are a few commands that are always suggested, and sometimes in an order that really didn't work well for me (e.g. \citep is always suggested, before \cite). Never found a way to change at least the order in Texmaker, but immediately could change it in TeXstudio.
Other than the issue of customization, the two programs are very similar. TeXstudio has a slightly "brighter" looking standard UI, but I can't say I like it better than the texmaker UI.
In summary, I think your choice between Texmaker and TeXstudio should depend on whether you ever thought that Texmaker offers a too little options/is a bit short on customizability. If not, I'd say go with Texmaker, since it's still the 'standard' as far as I can tell. If, on the other hand, there have been some (minor or major) things that keep nagging you in Texmaker, I highly suggest to give TeXstudio a try.
I am a heavy LaTeX user for science papers and math. I started using TexMaker, 8 years ago I found this processor somewhat "raw" and switched to TexStudio, which use now I acknowledge in my writings. It is my work horse, its stable and friendly, and it has excellent aids for any kind of symbols, handles BibTeX superbly, I now update it daily with Mercurial. Each so often I have dropped by the TeXMaker site and gave it a try. But I find the symbol hints in TeXStudio better. Incidentally, I install TeXLive from CD each time there is a new version, and none of the two processors has any problem to recognize it.
PD: I work under Ubuntu Linux, currently 15.04, I have a MacBook Pro to (run it 99.9% of the time on Ubuntu) and TeXStudio runs great under OS X Yellowstone too.
I've used Texmaker for years until I discovered TeXstudio in my university's lab. At first, both looked and worked pretty much the same, but now I've switched to TeXstudio for these reasons:
1. Easy dictionary language switching: there's a dropdown at the bottom of the screen where you can instantly switch dictionary language, which is very useful for me, since I'm often writing documents in different languages. In Texmaker, this switch required going to Options > Configure and searching for a .dic file.
2. Addition of %TODO tags in file structure: in TeXstudio only, you can add %TODO blablabla to your document and it shows up in your file structure for easy navigation.
My colleagues also like the fact that, in TeXstudio, you can draw math symbols with your mouse and have them translated into LaTeX code.
One controversial point regarding both editors is code completion. I prefer TeXstudio's because it's less intrusive than Texmaker's, which made me break my code way too often. However, if you're a heavy user of the feature, you may find TeXstudio lacking in that department.
I would like to point out one very important difference between Texstudio and TexMaker. I have never been able to get Texmaker to work, all because it doesn't find my Texlive installation. Reconfiguring TexMaker to look for TexLive doesn't work, that is using the Options->Configure Texmaker menu feature and then, even manually, literally changing all the folders to where it can find the Texlive programs, doesn't work. For instance, try using the wrapfig package, it doesn't work, because Texmaker's Latex installation doesn't have it, whereas Texlive does, but you can't get Texmaker to use Texlive. That's very poor. I use Ubuntu 14.04 and have never been able to get Texmaker to work! Whereas Texstudio, found my Texlive right at start up. If you go to its configuration setup, it is correct, it finds my Texlive installation as it should.
I do not know what Texmaker is using as far as Latex installation, but it isn't what should be the standard Latex setup that you install like Texlive. Texmaker has gone off on a tangent into the nether-nether, I do believe.
If you want something that works, right from installation, use Texstudio. The only problem with Texstudio that I can see so far is that it has a very washed out color scheme, and so after many hours of work, the lack of contrast puts too much strain on the eyes, but that's just a personal issue.
• "I have never been able to get Texmaker to work" And your OS is...? Method of installing Texmaker is...? – rbaleksandar Sep 13 '15 at 18:38
• @rbaleksandar He clearly states the OS. dorian: Strange that you have such problems, as far as I remember I've never been able to get Texmaker not to work, either in (K)Ubuntu or Windows. – Torbjørn T. Sep 17 '15 at 11:56
• Ah, sorry, missed the Ubuntu part there. :) – rbaleksandar Sep 17 '15 at 20:10
• @TorbjørnT. You are right, but the problem decription of dorian is still to vague. Also, the first paragraph is not really an answer. It is not clear whether this is an issue with Texmaker (which would be on-topic) or some issue with Ubuntu, or his own installation of Ubuntu (which both would be off-topic here). – vog Sep 18 '15 at 11:02
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# A New Mass Reconstruction Technique for Resonances Decaying to di-tau
Nuclear Instruments and Methods in Physics Research Section A: Accelerators, Spectrometers, Detectors and Associated Equipment, Vol. 654, No. 1. (22 Feb 2011), pp. 481-489, doi:10.1016/j.nima.2011.07.009 Key: citeulike:8468014
## Likes (beta)
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### Abstract
Accurate reconstruction of the mass of a resonance decaying to a pair of $τ$ leptons is challenging because of the presence of multiple neutrinos from $τ$ decays. The existing methods rely on either a partially reconstructed mass, which has a broad spectrum that reduces sensitivity, or the collinear approximation, which is applicable only to the relatively small fraction of events. We describe a new technique, which provides an accurate mass reconstruction of the original resonance and does not suffer from the limitations of the collinear approximation. The major improvement comes from replacing assumptions of the collinear approximation by a requirement that mutual orientations of the neutrinos and other decay products are consistent with the mass and decay kinematics of a $τ$ lepton. This is achieved by minimizing a likelihood function defined in the kinematically allowed phase space region. In this paper we describe the technique and illustrate its performance using $Z/γ^*\toττ$ and $H\toττ$ events simulated with the realistic detector resolution. The method is also tested on a clean sample of data $Z/γ^*\toττ$ events collected by the CDF experiment at the Tevatron. We expect that this new technique will allow for a major improvement in searches for the Higgs boson at both the LHC and the Tevatron.
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The sensitivity of fibroblastoid precursor cells in rat bone marrow to single and fractionated doses of γ rays delivered in vivo was measured. In vitro colonies were classified as being compact or diffuse, and the progenitor cells for both types were slowly cycling in vivo (survival levels after exposure to hydroxyurea were 90 ± 6% and 93 ± 11%, respectively). The progenitor cells forming diffuse colonies were more resistant ($D_{0}=1.39\ {\rm Gy}$) than those forming compact colonies ($D_{0}=0.76\ {\rm Gy}$). The fractionation sensitivities were characterized by an α/β ratio of 12.7 ± 5.5 Gy for diffuse colonies and 4.5 ± 3.0 Gy for compact colonies, respectively. The progenitor cells forming diffuse colonies may contribute more to long-term regeneration after high doses in vivo.
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# Plus One Maths Notes Chapter 11 Conic Sections
Kerala State Board New Syllabus Plus One Maths Notes Chapter 11 Conic Sections.
## Kerala Plus One Maths Notes Chapter 11 Conic Sections
I. Circle
A circle is the set of all points in a plane that are equidistant from a fixed point in the plane. The fixed point is the centre and the fixed distance is the radius.
Equation of a circle with centre origin and radius r is x2 + y2 = r2.
Equation of a circle with centre (h, k) and radius r is (x – h)2 + (y – k)2 = r2.
General form of the equation of a circle is x2 + y2 + 2gx + 2fy + c = 0 with centre (-g, -f) and radius $$\sqrt{g^{2}+f^{2}-c}$$.
II. Conic
A conic is the set of all points in a plane which moves so that the distance from a fixed point is in a constant ratio to its distance from a fixed-line. The fixed point is the focus and fixed line is directrix and the constant ratio is eccentricity, denoted by ‘e’.
III. Parabola (e = 1)
y2 = 4ax
Vertex: (0, 0)
Focus(S): (a, 0)
Length of Latusrectum: (LL’) = 4a
Equation of directrix (DD’) is x = -a
y2 = -4ax
Vertex: (0, 0)
Focus(S): (-a, 0)
Length of Latusrectum (LL’) = 4a
Equation of directrix (DD’) is x = a
x2 = 4ay
Vertex: (0, 0)
Focus(S): (0, a)
Length of Latusrectum (LL’) = 4a
Equation of directrix (DD’) is y = -a
x2 = -4ay
Vertex: (0, 0)
Focus(S): (0, -a)
Length of Latusrectum (LL’) = 4a
Equation of directrix (DD’) is y = a
IV. Ellipse (e < 1)
$$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$, a > b
1. Eccentricity, e = $$\frac{\sqrt{a^{2}-b^{2}}}{a}$$
(ae)2 = a2 – b2 ⇒ c2 = a2 – b2
2. b2 = a2(1 – e2)
3. Length of Latusrectum (LL’) = $$\frac{2 b^{2}}{a}$$
4. Focii, S(ae, 0) and S'(-ae, 0) or S(c, 0), S'(-c, 0)
5. Centre (0, 0)
6. Vertices A(a, 0) and A'(-a, 0)
7. Equation of directrix (DD’) is x = $$\frac{a}{e}$$ and x = $$-\frac{a}{e}$$
8. Length of major axis (AA’) = 2a
9. Length of minor axis'(BB’) = 2b
$$\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$$, a > b
1. Eccentricity, e = $$\frac{\sqrt{a^{2}-b^{2}}}{a}$$
(ae)2 = a2 – b2 ⇒ c2 = a2 – b2
2. b2 = a2(1 – e2)
3. Length of Latus rectum (LL’) = $$\frac{2 b^{2}}{a}$$
4. Focii, S(0, ae) and S'(0, -ae) or S(0, c), S'(0, -c)
5. Centre (0, 0)
6. Vertices A(0, a) and A'(0, -a)
7. Equation of directrix (DD’) is y = $$\frac{a}{e}$$ and y = $$-\frac{a}{e}$$
8. Length of major axis (AA’) = 2a
9. Length of minor axis (BB’) = 2b
V. Hyperbola (e > 1)
$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$
1. Eccentricity, e = $$\frac{\sqrt{a^{2}+b^{2}}}{a}$$
(ae)2 = a2 + b2 ⇒ c2 = a2 + b2
2. b2 = a2(e2 – 1)
3. Length of Latus rectum (LL’) = $$\frac{2 b^{2}}{a}$$
4. Focii, S(ae, 0) and S'(-ae, 0) or S(c, 0), S'(-c, 0)
5. Centre (0, 0)
6. Vertices A(a, 0) and A'(-a, 0)
7. Equation of directrix (DD’) is x = $$\frac{a}{e}$$ and x = $$-\frac{a}{e}$$
$$\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$$
1. Eccentricity, e = $$\frac{\sqrt{a^{2}+b^{2}}}{a}$$
(ae)2 = a2 + b2 ⇒ c2 = a2 + b2
2. b2 = a2(e2 – 1)
3. Length of Latus rectum (LL’) = $$\frac{2 b^{2}}{a}$$
4. Focii, S(0, ae) and S'(0, -ae) or S(0, c), S'(0, -c)
5. Centre (0, 0)
6. Vertices A(0, a) anti A'(0, -a)
7. Equation of directrix (DD’) is y = $$\frac{a}{e}$$ and y = $$-\frac{a}{e}$$
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# Properties
Label 4800.2.f.ba Level $4800$ Weight $2$ Character orbit 4800.f Analytic conductor $38.328$ Analytic rank $1$ Dimension $2$ CM no Inner twists $2$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$4800 = 2^{6} \cdot 3 \cdot 5^{2}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 4800.f (of order $$2$$, degree $$1$$, not minimal)
## Newform invariants
Self dual: no Analytic conductor: $$38.3281929702$$ Analytic rank: $$1$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{-1})$$ Defining polynomial: $$x^{2} + 1$$ Coefficient ring: $$\Z[a_1, a_2, a_3]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 480) Sato-Tate group: $\mathrm{SU}(2)[C_{2}]$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of $$i = \sqrt{-1}$$. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q -i q^{3} - q^{9} +O(q^{10})$$ $$q -i q^{3} - q^{9} + 4 q^{11} + 2 i q^{13} -2 i q^{17} -8 q^{19} + 4 i q^{23} + i q^{27} -6 q^{29} -4 i q^{33} -2 i q^{37} + 2 q^{39} -6 q^{41} -4 i q^{43} + 12 i q^{47} + 7 q^{49} -2 q^{51} -6 i q^{53} + 8 i q^{57} -12 q^{59} -14 q^{61} -12 i q^{67} + 4 q^{69} -2 i q^{73} -8 q^{79} + q^{81} + 4 i q^{83} + 6 i q^{87} -2 q^{89} -14 i q^{97} -4 q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2q - 2q^{9} + O(q^{10})$$ $$2q - 2q^{9} + 8q^{11} - 16q^{19} - 12q^{29} + 4q^{39} - 12q^{41} + 14q^{49} - 4q^{51} - 24q^{59} - 28q^{61} + 8q^{69} - 16q^{79} + 2q^{81} - 4q^{89} - 8q^{99} + O(q^{100})$$
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/4800\mathbb{Z}\right)^\times$$.
$$n$$ $$577$$ $$901$$ $$1601$$ $$4351$$ $$\chi(n)$$ $$-1$$ $$1$$ $$1$$ $$1$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
3649.1
1.00000i − 1.00000i
0 1.00000i 0 0 0 0 0 −1.00000 0
3649.2 0 1.00000i 0 0 0 0 0 −1.00000 0
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
5.b even 2 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 4800.2.f.ba 2
4.b odd 2 1 4800.2.f.j 2
5.b even 2 1 inner 4800.2.f.ba 2
5.c odd 4 1 960.2.a.o 1
5.c odd 4 1 4800.2.a.u 1
8.b even 2 1 2400.2.f.e 2
8.d odd 2 1 2400.2.f.n 2
15.e even 4 1 2880.2.a.i 1
20.d odd 2 1 4800.2.f.j 2
20.e even 4 1 960.2.a.f 1
20.e even 4 1 4800.2.a.ca 1
24.f even 2 1 7200.2.f.b 2
24.h odd 2 1 7200.2.f.bb 2
40.e odd 2 1 2400.2.f.n 2
40.f even 2 1 2400.2.f.e 2
40.i odd 4 1 480.2.a.b 1
40.i odd 4 1 2400.2.a.y 1
40.k even 4 1 480.2.a.e yes 1
40.k even 4 1 2400.2.a.j 1
60.l odd 4 1 2880.2.a.j 1
80.i odd 4 1 3840.2.k.p 2
80.j even 4 1 3840.2.k.k 2
80.s even 4 1 3840.2.k.k 2
80.t odd 4 1 3840.2.k.p 2
120.i odd 2 1 7200.2.f.bb 2
120.m even 2 1 7200.2.f.b 2
120.q odd 4 1 1440.2.a.j 1
120.q odd 4 1 7200.2.a.u 1
120.w even 4 1 1440.2.a.k 1
120.w even 4 1 7200.2.a.bg 1
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
480.2.a.b 1 40.i odd 4 1
480.2.a.e yes 1 40.k even 4 1
960.2.a.f 1 20.e even 4 1
960.2.a.o 1 5.c odd 4 1
1440.2.a.j 1 120.q odd 4 1
1440.2.a.k 1 120.w even 4 1
2400.2.a.j 1 40.k even 4 1
2400.2.a.y 1 40.i odd 4 1
2400.2.f.e 2 8.b even 2 1
2400.2.f.e 2 40.f even 2 1
2400.2.f.n 2 8.d odd 2 1
2400.2.f.n 2 40.e odd 2 1
2880.2.a.i 1 15.e even 4 1
2880.2.a.j 1 60.l odd 4 1
3840.2.k.k 2 80.j even 4 1
3840.2.k.k 2 80.s even 4 1
3840.2.k.p 2 80.i odd 4 1
3840.2.k.p 2 80.t odd 4 1
4800.2.a.u 1 5.c odd 4 1
4800.2.a.ca 1 20.e even 4 1
4800.2.f.j 2 4.b odd 2 1
4800.2.f.j 2 20.d odd 2 1
4800.2.f.ba 2 1.a even 1 1 trivial
4800.2.f.ba 2 5.b even 2 1 inner
7200.2.a.u 1 120.q odd 4 1
7200.2.a.bg 1 120.w even 4 1
7200.2.f.b 2 24.f even 2 1
7200.2.f.b 2 120.m even 2 1
7200.2.f.bb 2 24.h odd 2 1
7200.2.f.bb 2 120.i odd 2 1
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(4800, [\chi])$$:
$$T_{7}$$ $$T_{11} - 4$$ $$T_{13}^{2} + 4$$ $$T_{19} + 8$$ $$T_{23}^{2} + 16$$ $$T_{31}$$
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T^{2}$$
$3$ $$1 + T^{2}$$
$5$ $$T^{2}$$
$7$ $$T^{2}$$
$11$ $$( -4 + T )^{2}$$
$13$ $$4 + T^{2}$$
$17$ $$4 + T^{2}$$
$19$ $$( 8 + T )^{2}$$
$23$ $$16 + T^{2}$$
$29$ $$( 6 + T )^{2}$$
$31$ $$T^{2}$$
$37$ $$4 + T^{2}$$
$41$ $$( 6 + T )^{2}$$
$43$ $$16 + T^{2}$$
$47$ $$144 + T^{2}$$
$53$ $$36 + T^{2}$$
$59$ $$( 12 + T )^{2}$$
$61$ $$( 14 + T )^{2}$$
$67$ $$144 + T^{2}$$
$71$ $$T^{2}$$
$73$ $$4 + T^{2}$$
$79$ $$( 8 + T )^{2}$$
$83$ $$16 + T^{2}$$
$89$ $$( 2 + T )^{2}$$
$97$ $$196 + T^{2}$$
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# Trusted Domain Exchange Environment
In the health system I work for, we have been tasked with taking our old AD domain and making it work within the new domain. Eventually, all computers will be converted to the new domain, but there are steps along the way before that date. One of those stages was to use Exchange from the other domain. We will call this DomainB.com. The domain here at our local hospital is DomainA.com.
We have a trust relationship between DomainA and DomainB. This allows DomainA computers to authenticate against DomainB. However, it isn’t without its issues. When adding Exchange accounts from DomainB into DomainA while still logging into DomainA, Outlook doesn’t play well. The first problem we had was AutoDiscover. AutoDiscover isn’t quite setup correctly on at DomainB as the normal search pattern for the discovery won’t find the autodiscover.xml file with the way the domains are named. And with the trust, SCP doesn’t work correctly. The second problem is passwords. We had to add the password for DomainB into the password vault in Windows 7 (similar to Windows XP). The third problem was the Out-Of-Office (OOF) and Availability Services (AS). Both of these did not work. For the last two months, we have been telling our users that it won’t work until we are in DomainB so they will have to use OWA for those features.
Thursday last week, my CEO pulled me into his office and asked if there was anything that could be done for the Free/Busy (FB) information. I told him the same speel I have been telling everyone else. Gotta wait for DomainB. All night I started crunching ideas. Over the course of Friday and Monday, I finally solved it. All services are now working. So, on to how I did it…
### AutoDiscover
The IT department of DomainB wasn’t particularly helpful and was giving me the same speel I was giving everyone else. Because the OWA URL used a non-standard URL, AutoDiscover would never find the XML file. If you were on DomainB, it would find it via SCP (Service Connection Point). But since we are on DomainA, we don’t have rights to DomainB’s SCPs. So I found a way to force Outlook 2010 to use a local XML file.
HKEY_CURRENT_USER\Software\Microsoft\Office\14.0\Outlook\AutoDiscover\”PreferLocalXML”=dword:1
HKEY_CURRENT_USER\Software\Microsoft\Office\14.0\Outlook\AutoDiscover\”DomainB.com”=”\\path\to\file.xml”
Since AutoDiscover.xml is dynamically generated upon viewing, you can’t really just copy and paste the contents from the real file. You have to redirect.
<?xml version="1.0" encoding="utf-8"?>
<Autodiscover xmlns="http://schemas.microsoft.com/exchange/autodiscover/responseschema/2006">
<Response xmlns="http://schemas.microsoft.com/exchange/autodiscover/outlook/responseschema/2006a">
<Account>
<AccountType>email</AccountType>
<Action>redirectUrl</Action>
<RedirectUrl>https://remote.DomainB.com/autodiscover/autodiscover.xml</RedirectUrl>
</Account>
</Response>
</Autodiscover>
This works by responding back with all the URLs necessary for OOF, FB, AS, etc. This is of course assuming that the Exchange server is correct and has FQDNs. This also pulled other settings such as Personal Archive folders.
Once Outlook reads from this file when setting up an email account, everything is built out without the need for further configuration in Outlook.
|
C. Chocolate Bunny
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
This is an interactive problem.
We hid from you a permutation $p$ of length $n$, consisting of the elements from $1$ to $n$. You want to guess it. To do that, you can give us 2 different indices $i$ and $j$, and we will reply with $p_{i} \bmod p_{j}$ (remainder of division $p_{i}$ by $p_{j}$).
We have enough patience to answer at most $2 \cdot n$ queries, so you should fit in this constraint. Can you do it?
As a reminder, a permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array) and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).
Input
The only line of the input contains a single integer $n$ ($1 \le n \le 10^4$) — length of the permutation.
Interaction
The interaction starts with reading $n$.
Then you are allowed to make at most $2 \cdot n$ queries in the following way:
• "? x y" ($1 \le x, y \le n, x \ne y$).
After each one, you should read an integer $k$, that equals $p_x \bmod p_y$.
When you have guessed the permutation, print a single line "! " (without quotes), followed by array $p$ and quit.
After printing a query do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
• fflush(stdout) or cout.flush() in C++;
• System.out.flush() in Java;
• flush(output) in Pascal;
• stdout.flush() in Python;
• see documentation for other languages.
Exit immediately after receiving "-1" and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
Hack format
In the first line output $n$ ($1 \le n \le 10^4$). In the second line print the permutation of $n$ integers $p_1, p_2, \ldots, p_n$.
Example
Input
3
1
2
1
0
Output
? 1 2
? 3 2
? 1 3
? 2 1
! 1 3 2
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# Two fun facts
Well, it seems that my 15 minutes are up, so let’s get back to some complex analysis. Here are two fun facts about holomorphic functions:
Theorem 1 Suppose ${f\in {\mathcal H}(U)}$, and ${U}$ is connected. Then if ${f}$ constant in a neighborhood of some point ${z_0\in U}$. Then ${f}$ is constant on all of ${U}$.
Proof: Since the set ${\{f(z_0)\}}$ is closed, so is it’s preimage. Let ${V}$ denote this preimage. It suffices to show that ${V}$ has no boundary. Then ${V}$ is open, closed, and non-empty. Since ${U}$ is connected, it follows that ${U=V}$.
Suppose ${V}$ has a boundary, and let ${v}$ be some point on the boundary. Expand a power series about ${v}$:
$\displaystyle f(z)=a_0+a_1(z-v)+a_2(z-v)^2+\cdots.$
Since ${f}$ is holomorphic at ${v}$, this power series must have a non-zero radius of convergence so it is valid on some part of the interior of ${V}$. As we know, the only power series with constant image is the constant series ${f(z)=a_0}$. But now this series is valid in a small ball about ${v}$. This contradicts the assumption that ${v}$ was on the boundary of ${V}$, so ${V}$ must be open, proving the theorem.
$\Box$
Theorem 2 Suppose ${f\in{\mathcal H}(U)}$, and ${U}$ is connected. Suppose we have a sequence of points ${z_1,z_2,\dots}$ converging to ${z_\infty\in U}$ such that ${f(z_1)=f(z_2)=\cdots}$. Then ${f}$ is constant on ${U}$
Proof: By continuity of ${f}$, ${f(z_1)=f(z_2)=\cdots=f(z_\infty)}$. Expanding ${f}$ as a power series about ${z_\infty}$, write
$\displaystyle f(z)=a_0+a_1(z-z_\infty)+a_2(z-z_\infty)^2+\cdots$
Since we have a sequence of points converging to ${z_\infty}$ whose ${f}$-values all agree with ${z_\infty}$, we see that the derivative at ${z_\infty}$ must be
$\displaystyle f'(z_\infty)=\lim_{n\rightarrow\infty}\frac{f(z_n)-f(z_\infty)}{z_n-z_\infty}=\lim_{n\rightarrow\infty}\frac{0}{z_n-z_\infty}=0$
Thus, ${f'(z)=0}$ in a neighborhood of ${z_\infty}$, meaning that the power series: ${f'(z)=a_1+2a_2(z-z_\infty)+3a_3(z-z_\infty)^2+\cdots}$ must in fact be constant. Thus, ${a_n=0}$ for ${n>0}$, meaning that ${f(z)=a_0}$ in a neighborhood of ${z_\infty}$.
Now we can apply the previous theorem to conclude that ${f}$ is in fact constant of all of ${U}$.
$\Box$
|
## Homeostasis is maintained by the _____ which also plays an important role in emotions
Hypothalamus
Limbic System
Thalamus
Reticular Activating System
|
# Is $\| x \| = \sup\{ | \sum_{k=1}^n x_k |\colon n\in \mathbb{N}\}$ a norm in $\ell_1$?
For each $x= (x_n)_{n=1}^\infty \in \ell_1$ set
$$\| x \| = \sup\{ \big| \sum_{k=1}^n x_k \big|\colon n\in \mathbb{N}\}$$
Does this define a norm in $\ell_1$?
-
What have you tried? – Sean Eberhard Apr 11 '13 at 12:58
$\left\| x \right\| = \sup \left| \sum\limits_{k=1}^n x_k \right|$
$\forall \lambda \in \Bbb C, \left\| \lambda x \right\|=\sup \left| \sum\limits_{k=1}^n \lambda x_k \right|=\sup \left| \lambda\sum\limits_{k=1}^n x_k \right|=\sup \left|\lambda\right|\left| \sum\limits_{k=1}^n x_k \right|=\left|\lambda\right|\sup \left| \sum\limits_{k=1}^n x_k \right|=\left|\lambda\right|\left\| x \right\|$
$\forall x, y \in \ell^1,\left\| x+y \right\|=\sup \left| \sum\limits_{k=1}^n x_k+y_k \right|=\sup \left| \sum\limits_{k=1}^n x_k+\sum\limits_{k=1}^ny_k \right|\le \sup\left (\left| \sum\limits_{k=1}^n x_k\right| + \left|\sum\limits_{k=1}^ny_k \right| \right)\le \sup \left| \sum\limits_{k=1}^n x_k\right| + \sup \left|\sum\limits_{k=1}^ny_k \right|=\left\| x \right\|+\left\| y \right\|$
$\forall x \in \ell ^1, \left [\left\| x \right\|=0\right ] \Leftrightarrow \left[\forall n,\sum\limits_{k=1}^n x_k =0 \right] \Leftrightarrow \left[\forall n, \sum\limits_{k=1}^{n+1} x_k- \sum\limits_{k=1}^n x_k = 0\right]\Leftrightarrow \left[\forall n, x_n = 0\right]\Leftrightarrow x = 0$
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# IE 58D Efficient Monte Carlo Simulation
Kredi Bilgisi:
3
Açıklama:
Ability to understand code and use efficient Monte-Carlo simulation algorithms; special emphasis will be laid on the development of variance reduction techniques, on quasi Monte Carlo algorithms and on efficient random variate generation for practical relevant simulation problems. The examples in the course will be taken from queuing applications, disease spread models, risk quantification and option pricing.
Dersi veren öğretim görevlileri:
Syllabus:
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# 3.1 Functions and function notation
Page 1 / 21
In this section, you will:
• Determine whether a relation represents a function.
• Find the value of a function.
• Determine whether a function is one-to-one.
• Use the vertical line test to identify functions.
• Graph the functions listed in the library of functions.
A jetliner changes altitude as its distance from the starting point of a flight increases. The weight of a growing child increases with time. In each case, one quantity depends on another. There is a relationship between the two quantities that we can describe, analyze, and use to make predictions. In this section, we will analyze such relationships.
## Determining whether a relation represents a function
A relation is a set of ordered pairs. The set consisting of the first components of each ordered pair is called the domain and the set consisting of the second components of each ordered pair is called the range . Consider the following set of ordered pairs. The first numbers in each pair are the first five natural numbers. The second number in each pair is twice that of the first.
$\left\{\left(1,\text{\hspace{0.17em}}2\right),\text{\hspace{0.17em}}\left(2,\text{\hspace{0.17em}}4\right),\text{\hspace{0.17em}}\left(3,\text{\hspace{0.17em}}6\right),\text{\hspace{0.17em}}\left(4,\text{\hspace{0.17em}}8\right),\text{\hspace{0.17em}}\left(5,\text{\hspace{0.17em}}10\right)\right\}$
The domain is $\left\{1,\text{\hspace{0.17em}}2,\text{\hspace{0.17em}}3,\text{\hspace{0.17em}}4,\text{\hspace{0.17em}}5\right\}.$ The range is $\left\{2,\text{\hspace{0.17em}}4,\text{\hspace{0.17em}}6,\text{\hspace{0.17em}}8,\text{\hspace{0.17em}}10\right\}.$
Note that each value in the domain is also known as an input value, or independent variable , and is often labeled with the lowercase letter $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ Each value in the range is also known as an output value, or dependent variable , and is often labeled lowercase letter $\text{\hspace{0.17em}}y.$
A function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ is a relation that assigns a single element in the range to each element in the domain . In other words, no x -values are repeated. For our example that relates the first five natural numbers to numbers double their values, this relation is a function because each element in the domain, $\left\{1,\text{\hspace{0.17em}}2,\text{\hspace{0.17em}}3,\text{\hspace{0.17em}}4,\text{\hspace{0.17em}}5\right\},$ is paired with exactly one element in the range, $\left\{2,\text{\hspace{0.17em}}4,\text{\hspace{0.17em}}6,\text{\hspace{0.17em}}8,\text{\hspace{0.17em}}10\right\}.$
Now let’s consider the set of ordered pairs that relates the terms “even” and “odd” to the first five natural numbers. It would appear as
$\left\{\left(\text{odd},\text{\hspace{0.17em}}1\right),\text{\hspace{0.17em}}\left(\text{even},\text{\hspace{0.17em}}2\right),\text{\hspace{0.17em}}\left(\text{odd},\text{\hspace{0.17em}}3\right),\text{\hspace{0.17em}}\left(\text{even},\text{\hspace{0.17em}}4\right),\text{\hspace{0.17em}}\left(\text{odd},\text{\hspace{0.17em}}5\right)\right\}$
Notice that each element in the domain, $\left\{\text{even,}\text{\hspace{0.17em}}\text{odd}\right\}$ is not paired with exactly one element in the range, $\left\{1,\text{\hspace{0.17em}}2,\text{\hspace{0.17em}}3,\text{\hspace{0.17em}}4,\text{\hspace{0.17em}}5\right\}.$ For example, the term “odd” corresponds to three values from the domain, $\left\{1,\text{\hspace{0.17em}}3,\text{\hspace{0.17em}}5\right\}$ and the term “even” corresponds to two values from the range, $\left\{2,\text{\hspace{0.17em}}4\right\}.$ This violates the definition of a function, so this relation is not a function.
[link] compares relations that are functions and not functions.
## Function
A function is a relation in which each possible input value leads to exactly one output value. We say “the output is a function of the input.”
The input values make up the domain , and the output values make up the range .
Given a relationship between two quantities, determine whether the relationship is a function.
1. Identify the input values.
2. Identify the output values.
3. If each input value leads to only one output value, classify the relationship as a function. If any input value leads to two or more outputs, do not classify the relationship as a function.
sebd me some questions about anything ill solve for yall
how to solve x²=2x+8 factorization?
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
hii
Amit
how are you
Dorbor
well
Biswajit
can u tell me concepts
Gaurav
Find the possible value of 8.5 using moivre's theorem
which of these functions is not uniformly cintinuous on (0, 1)? sinx
which of these functions is not uniformly continuous on 0,1
solve this equation by completing the square 3x-4x-7=0
X=7
Muustapha
=7
mantu
x=7
mantu
3x-4x-7=0 -x=7 x=-7
Kr
x=-7
mantu
9x-16x-49=0 -7x=49 -x=7 x=7
mantu
what's the formula
Modress
-x=7
Modress
new member
siame
what is trigonometry
deals with circles, angles, and triangles. Usually in the form of Soh cah toa or sine, cosine, and tangent
Thomas
solve for me this equational y=2-x
what are you solving for
Alex
solve x
Rubben
you would move everything to the other side leaving x by itself. subtract 2 and divide -1.
Nikki
then I got x=-2
Rubben
it will b -y+2=x
Alex
goodness. I'm sorry. I will let Alex take the wheel.
Nikki
ouky thanks braa
Rubben
I think he drive me safe
Rubben
how to get 8 trigonometric function of tanA=0.5, given SinA=5/13? Can you help me?m
More example of algebra and trigo
What is Indices
If one side only of a triangle is given is it possible to solve for the unkown two sides?
cool
Rubben
kya
Khushnama
please I need help in maths
Okey tell me, what's your problem is?
Navin
the least possible degree ?
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# VBT AND Schrodinger(not sure about spelling)
in VBT we consider orbitals
but orbitals were given by Schrodinger and his quantum mechanics.
so we must also consider Heisenberg in it!!
mfb
Mentor
The solutions of the Schrödinger* equation always respect Heisenberg's uncertainty relation.
*that's his name, with dots on the o.
I'm not sure I get the point of this post. Please elaborate.
jfizzix
Gold Member
Schr$\ddot{\text{o}}$dinger and Heisenberg's mechanics are the same mechanics.
Dirac showed this where in the Schr$\ddot{\text{o}}$dinger picture, the states evolve in time, and the observables are stationary, while in the Heisenberg picture, the states are stationary, and the observables evolve in time.
Every calculation gives you the exact same result in both pictures, though one picture might be easier than the other to work with, depending on the problem at hand.
mfb
Mentor
@jfizzix: Ah right, that is another way to interpret the first post.
We'll have to wait for namanjain to make clear what exactly he/she is asking.
in vbt (valance bond theory of covalent bond)
we consider orbitals which are part of Schrodinger, so must not we respect Heisenberg and say it's idiotic to tell electron cloud's position
mfb
Mentor
Heisenberg is a person, not a law. I guess you mean the uncertainty principle?
"Electron cloud" means that the electron has no precise position - and it is uncertain enough to satisfy the uncertainty principle.
jfizzix
Gold Member
The reason Heisenberg's uncertainty relation is obeyed by electrons is because they exhibit wave-particle duality, just as all quantum objects do. The probability distributions that orbitals represent are not those of classical ignorance, but true quantum uncertainty. It's not true that electrons are well defined and we just can't in principle make up an experiment to show us exactly where one is and where it's going (these sorts of interpretations are ruled out by violating Bell inequalities).
The classical perspective of an electron becomes especially hard to maintain with electrons in P-orbitals which have a nodal plane where the probability is precisely zero. The electrons somehow can still go from one side to the other of this nodal plane even though they cannot strictly speaking, pass through.
When dealing with problems like this it's sometimes best to consider these objects (electrons, photons, etc) as a class of their own, because we just don't experience these sort of phenomena in the everyday macroscopic world.
It's not true that electrons are well defined and we just can't in principle make up an experiment to show us exactly where one is and where it's going (these sorts of interpretations are ruled out by violating Bell inequalities).
The classical perspective of an electron becomes especially hard to maintain with electrons in P-orbitals which have a nodal plane where the probability is precisely zero. The electrons somehow can still go from one side to the other of this nodal plane even though they cannot strictly speaking, pass through.
To be clear, electrons are always described by their wavefunctions, which are abstract "probability distributions", but they can be localized, such as the instance of an electron striking a detector. In this case, the wave function is an eigenstate of the position operator and is described the by delta distribution. The "wave/particle duality" thing is kind of a silly idea since 99.9999% of the time when people are actually discussing such things it's always in terms of waves. Please see the following reference for more details:
Quantum mechanics: Myths and facts http://xxx.lanl.gov/abs/quant-ph/0609163
Also, the business with the nodes is often overstated. For one, the one electron wave functions don't describe an electron "moving" anywhere, they're a static entity that describe the probability of an electron's position. In addition, the probability at the nodes isn't identically zero in the most accurate (relativistic) treatment. Please see the very neat article:
Relativistic quantum chemistry: The electrons and the nodes
J. Chem. Educ., 1968, 45 (9), p 558
1 person
jfizzix
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placeholder
## Primary tabs
### Thursday, January 26, 2017
All day
Before 01 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 Conference - CAST - Contact and Symplectic Topology Jan 26, 2017 to Jan 28, 2017 Download the poster in pdf Nantes, from January 26th to January 28th Organization board: Baptiste Chantraine, Vincent Colin, Paolo Ghiggini Scientific board: Jean-François Barraud, Baptiste Chantraine, Kai Cieliebak, Tobias Ekholm Leclercq Jan 26, 2017 By using generating functions, Viterbo constructed spectral invariants for Lagrangians of cotangent bundles in 1992. Ten years later, Oh and Schwarz (independently) adapted the construction to Hamiltonian diffeomorphism groups of quite general manifolds thanks to Floer theory. Since then, spectral invariants were defined in various contexts and yielded a great number of applications of different natures. I will tak about joint work with Frol Zapolsky in which we define spectral invariants for monotone Lagrangians and establish the properties which make them such a useful tool. In particular, given a monotone Lagrangian L, I will show how spectral invariants can be seen as functions defined on the universal cover of the product of the set of Lagrangians Hamiltonian isotopic to L and the Hamiltonian diffeomorphism group. Then, I'll show that these functions are Lipschitz with respect to the natural Hofer distance on this space. Naef Jan 26, 2017 In this talk, I will explain how for a contact manifold the existence of a dynamically convex supporting contact form ensures compactness of Floer moduli spaces and thus allows us to define Rabinowitz Floer homology in a symplectisation. In this setting, the Rabinowitz Floer homology groups give a means to deduce existence results of translated points as introduced by Sandon. This is joint work with Matthias Meiwes. Hutchings Jan 26, 2017 We show that every nondegenerate contact form on a closed connected three-manifold, such that the associated contact structure has torsion first Chern class, has either two or infinitely many simple Reeb orbits. (By previous work, two simple Reeb orbits are possible only when the three-manifold is a sphere or a lens space.) Joint work with Dan Cristofaro-Gardiner and Dan Pomerleano.
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# BuzzFeed
## Response To
#### Kanye Ruins People’s Moments
• Alexander Chavez 3 years ago
Let me try that again - Kanye interrupts Kanye…who interrupts Kanye (you know the deal)
# Facebook Conversations
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I was trying to post this one! (something was wrong with the uploader.. phooey)
• Jaymi M. Let me try that again - Kanye interru... about 3 years ago
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Now Buzzing
|
# Math Help - ellipse word problem
1. ## ellipse word problem
this is the word problem: a stadium is to be built with stands in the form of two concentric ellipses. The outer ellipse is to be 240 yds long and 200 yds wide. The inner ellipse is to be 200 yds long and 100 yds wide. The football field of standard dimensions is to be laid out in the center of the inner ellipse.
I sketched this ellipse with the outer coordinate (240,200) and the inner coordinate (200, 100). I found the equations...outer: (x^2/120^2)+(y^2/100^2)= 1 AND inner: (x^2/100^2)+(y^2/50^2)= 1. I found those numbers because the outer a is 120 and outer b is 100. also the inner a is 100 and inner b is 50.
With all this information I found the inner area and outer area.
Inner: \pi (100)(50)= 15707.96 AND outer: \pi (120)(100)= 37699.11
Next I have to find the clearance will there be between the field and the inner ellipse in the direction of the end line. Im not sure where to begin at this step..
2. By football field I assume you mean American football. The field has dimensions 120yds by 53.3 yards. Since you need the distance from the end lines center a rectangle with these dimensions at the origin. The end lines should be at -60yds and 60yrds. I hope this helps.
3. wouldnt that mean that the endlines would be at -120 and 120?
am I trying to find the distance between the endlines and the inner ellipse?
4. Originally Posted by kassums17
wouldnt that mean that the endlines would be at -120 and 120?
am I trying to find the distance between the endlines and the inner ellipse?
If what you said is true that would make the field
$120 -(-120)=240\text{yrds}$ long twice as long as it needs to be. Draw the picture of the field on you graph to help you visualize it.
5. thank you. the next thing I have to find with this problem is what the area of the stands is? how would I approach solving this?
6. Originally Posted by kassums17
thank you. the next thing I have to find with this problem is what the area of the stands is? how would I approach solving this?
Again look at your diagram. What is the area you want? It is inside the outer ellipse but outside the inner ellipse. You know both of these area's from your first post. How can you find it
7. is there some kind of formula i can use?
maybe my diagram isnt accurate because its hard to depict the coordinates
8. Originally Posted by kassums17
is there some kind of formula i can use?
maybe my diagram isnt accurate because its hard to depict the coordinates
Alright. So in your first post you found the area of the outer ellipse and the inner ellipse correct?
The area of the outer ellipse is too big because we don't want the area inside the inner ellipse. So how much extra area does the outer ellipse have?
Think on this. You have all of the information that you need to answer the question.
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Please use this identifier to cite or link to this item: http://hdl.handle.net/10525/1602
Title: An Algorithmic Solution for Management of Related Text Objects with Application in Phytopharmacy Authors: Dimova, Delyana Keywords: Algorithmic SolutionRelational DatabaseData ProcessingSoftwarePhytopharmacy Issue Date: 2010 Publisher: Institute of Mathematics and Informatics Bulgarian Academy of Sciences Citation: Serdica Journal of Computing, Vol. 4, No 4, (2010), 487p-504p Abstract: This paper presents an algorithmic solution for management of related text objects, in which are integrated algorithms for their extraction from paper or electronic format, for their storage and processing in a relational database. The developed algorithms for data extraction and data analysis enable one to find specific features and relations between the text objects from the database. The algorithmic solution is applied to data from the field of phytopharmacy in Bulgaria. It can be used as a tool and methodology for other subject areas where there are complex relationships between text objects. URI: http://hdl.handle.net/10525/1602 ISSN: 1312-6555 Appears in Collections: Volume 4 Number 4
Files in This Item:
File Description SizeFormat
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# Condition for Denesting of Square Root
## Theorem
Let $a, b \in \Q_{\ge 0}$
Then:
$\exists p, q \in \Q: \sqrt {a + \sqrt b} = \sqrt p + \sqrt q$
iff:
$\exists n \in \N: a^2 - b = n^2$.
## Proof
The proof is split into the necessary condition:
If
$\exists p, q \in \Q: \sqrt {a + \sqrt b} = \sqrt p + \sqrt q$
then
$\exists n \in \N: a^2 - b = n^2$
and the sufficient condition:
If
$\exists n \in \N: a^2 - b = n^2$
then
$\exists p, q \in \Q: \sqrt {a + \sqrt b} = \sqrt p + \sqrt q$
### Sufficient condition
Let $n^2 = a^2 - b$.
As $a^2 = b + n^2$ it follows that:
$a \ge \sqrt b$
Then:
$\displaystyle \sqrt {a + \sqrt b}$ $=$ $\displaystyle \sqrt {\dfrac a 2 + \dfrac {\sqrt {a^2 - b} } 2} + \sqrt {\dfrac a 2 - \dfrac {\sqrt {a^2 - b} } 2}$ Square Root of Sum as Sum of Square Roots $\displaystyle$ $=$ $\displaystyle \sqrt {\dfrac a 2 + \dfrac {\sqrt {n^2} } 2} + \sqrt {\dfrac a 2 - \dfrac {\sqrt {n^2} } 2}$ substituting $n^2 = a^2 - b$ $\displaystyle$ $=$ $\displaystyle \sqrt {\dfrac a 2 + \dfrac n 2} + \sqrt {\dfrac a 2 - \dfrac n 2}$
$\dfrac a 2 + \dfrac n 2$ and $\dfrac a 2 - \dfrac n 2$
are rational.
$\blacksquare$
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## Results (displaying matches 1-50 of 3455) Next
Galois conjugate representations are grouped into single lines.
Label Dimension Conductor Defining polynomial of Artin field $G$ Ind $\chi(c)$
10.69546276192481.30t164.b.a $10$ $11^{4} \cdot 41^{6}$ x6 - x5 - 4x4 + 6x3 - 6x + 5 $S_6$ $1$ $-2$
10.557885504160000.30t164.a.a $10$ $2^{8} \cdot 3^{20} \cdot 5^{4}$ x6 + 3x4 - 4x3 + 9x2 - 6x + 1 $S_6$ $1$ $-2$
10.722204136308736.30t88.a.a $10$ $2^{24} \cdot 3^{16}$ x6 - 3x5 + 3x4 - 6x2 + 6x - 2 $A_6$ $1$ $-2$
10.784294811709696.30t164.b.a $10$ $2^{8} \cdot 3^{12} \cdot 7^{8}$ x6 + 5x4 - 2x3 + 9x2 - 8x + 16 $S_6$ $1$ $2$
10.1719926784000000.30t164.c.a $10$ $2^{24} \cdot 3^{8} \cdot 5^{6}$ x6 - 2x5 - x4 + 6x3 - 2x2 - 4x - 1 $S_6$ $1$ $-2$
10.1719926784000000.30t164.d.a $10$ $2^{24} \cdot 3^{8} \cdot 5^{6}$ x6 - 2x5 - x4 + 6x3 - 2x2 - 4x - 1 $S_6$ $1$ $-2$
10.1757048889305889.30t164.a.a $10$ $3^{4} \cdot 167^{6}$ x6 + 2x2 - x + 1 $S_6$ $1$ $-2$
10.2511765109305169.30t164.a.a $10$ $17^{6} \cdot 101^{4}$ x6 - 2x5 + x4 + x3 - 2x2 + x + 1 $S_6$ $1$ $-2$
10.2845021504820049.30t88.b.a $10$ $3^{14} \cdot 29^{6}$ x6 - 3x3 - 3x + 4 $A_6$ $1$ $-2$
10.2888816545234944.30t164.c.a $10$ $2^{26} \cdot 3^{16}$ x6 - 2x5 + x4 + x2 - 2x - 1 $S_6$ $1$ $-2$
10.2888816545234944.30t164.a.a $10$ $2^{26} \cdot 3^{16}$ x6 - 6x4 - 4x3 + 6x2 - 6 $S_6$ $1$ $-2$
10.3656158440062976.30t164.a.a $10$ $2^{20} \cdot 3^{20}$ x6 + 3x4 - 4x3 - 6x - 2 $S_6$ $1$ $-2$
10.4244902593608889.30t88.a.a $10$ $3^{14} \cdot 31^{6}$ x6 - x3 - 3x2 - 1 $A_6$ $1$ $-2$
10.4324050384430144.30t164.a.a $10$ $2^{6} \cdot 47^{4} \cdot 61^{4}$ x6 - x5 - x3 + x + 1 $S_6$ $1$ $-2$
10.4347986536861696.30t164.a.a $10$ $2^{12} \cdot 101^{6}$ x6 - x5 + x4 - x3 + 3x2 + 1 $S_6$ $1$ $-2$
10.6241849278890625.30t164.a.a $10$ $3^{14} \cdot 5^{6} \cdot 17^{4}$ x6 - 4x3 - 3x2 - 3x - 1 $S_6$ $1$ $-2$
10.6636312096654969.30t164.a.a $10$ $3^{14} \cdot 193^{4}$ x6 - 2x3 - 3x2 - 3x - 1 $S_6$ $1$ $-2$
10.7834102237495296.30t164.b.a $10$ $2^{24} \cdot 3^{4} \cdot 7^{8}$ x6 - 4x4 - 2x3 + x2 - 2x - 5 $S_6$ $1$ $-2$
10.8303765625000000.30t164.a.a $10$ $2^{6} \cdot 3^{12} \cdot 5^{12}$ x6 - 3x5 + 5x3 - 5 $S_6$ $1$ $-2$
10.8303765625000000.30t164.b.a $10$ $2^{6} \cdot 3^{12} \cdot 5^{12}$ x6 - 3x5 + 5x3 - 5 $S_6$ $1$ $-2$
10.8415099419290201.30t164.b.a $10$ $11^{6} \cdot 41^{6}$ x6 - x5 - 4x4 + 6x3 - 6x + 5 $S_6$ $1$ $2$
10.8792367498140625.30t88.a.a $10$ $3^{14} \cdot 5^{6} \cdot 7^{6}$ x6 + 3x4 - 12x2 - 15x + 16 $A_6$ $1$ $-2$
10.9183037782236224.30t164.a.a $10$ $2^{6} \cdot 3461^{4}$ x6 - x5 - 2x4 + 3x3 - x + 1 $S_6$ $1$ $-2$
10.9227446944279201.30t164.a.a $10$ $3^{16} \cdot 11^{8}$ x6 + 3x4 - 2x3 + 6x2 + 9x + 5 $S_6$ $1$ $2$
10.9227446944279201.30t164.b.a $10$ $3^{16} \cdot 11^{8}$ x6 + 3x4 - 2x3 + 6x2 + 9x + 5 $S_6$ $1$ $-2$
10.10000000000000000.30t164.a.a $10$ $2^{16} \cdot 5^{16}$ x6 - 3x5 + 5x4 - 5x3 + 10x2 - 4x - 8 $S_6$ $1$ $-2$
10.10509453369140625.30t164.a.a $10$ $3^{16} \cdot 5^{12}$ x6 - 3x5 + 15x2 - 3x - 1 $S_6$ $1$ $-2$
10.10509453369140625.30t164.b.a $10$ $3^{16} \cdot 5^{12}$ x6 - 3x5 + 15x2 - 3x - 1 $S_6$ $1$ $-2$
10.11555266180939776.30t164.a.a $10$ $2^{28} \cdot 3^{16}$ x6 - 6x4 - 4x3 + 6x2 - 6 $S_6$ $1$ $-2$
10.12548716987355136.30t164.b.a $10$ $2^{12} \cdot 3^{12} \cdot 7^{8}$ x6 + 5x4 - 2x3 + 9x2 - 8x + 16 $S_6$ $1$ $-2$
10.13947137604000000.30t164.a.a $10$ $2^{8} \cdot 3^{20} \cdot 5^{6}$ x6 + 3x4 - 4x3 + 9x2 - 6x + 1 $S_6$ $1$ $-2$
10.14117306610774528.70.a.a 10.14117306610774528.70.a.b $10$ $2^{9} \cdot 3^{14} \cdot 7^{8}$ x7 - 3x6 + 3x5 + 3x4 - 9x3 + 3x2 + x - 3 $A_7$ $0$ $-2$
10.14624633760251904.30t164.a.a $10$ $2^{22} \cdot 3^{20}$ x6 + 3x4 - 4x3 - 6x - 2 $S_6$ $1$ $-2$
10.15813440003753001.30t164.a.a $10$ $3^{6} \cdot 167^{6}$ x6 + 2x2 - x + 1 $S_6$ $1$ $2$
10.16384000000000000.30t164.a.a $10$ $2^{26} \cdot 5^{12}$ x6 - 2x5 + 5x4 - 10x2 + 8x - 6 $S_6$ $1$ $-2$
10.16764094652917824.30t164.a.a $10$ $2^{6} \cdot 3^{12} \cdot 149^{4}$ x6 - x5 + 2x4 + x3 - 2x2 + x + 1 $S_6$ $1$ $-2$
10.16830042327806409.30t88.a.a $10$ $3^{20} \cdot 13^{6}$ x6 - 3x5 + 3x4 - 3x2 + 3x + 2 $A_6$ $1$ $-2$
10.20100618201669201.30t88.b.a $10$ $3^{20} \cdot 7^{8}$ x6 - 3x5 + 3x4 + 9x3 - 18x2 - 9x + 18 $A_6$ $1$ $-2$
10.23713122135310336.30t88.a.a $10$ $2^{18} \cdot 67^{6}$ x6 - 2x4 + x2 - 2x - 1 $A_6$ $1$ $-2$
10.24002626001540481.30t164.a.a $10$ $3^{12} \cdot 461^{4}$ x6 - 3x5 + 3x4 + 2x3 - 6x2 + 3x + 1 $S_6$ $1$ $-2$
10.28158962038780329.30t164.a.a $10$ $3^{14} \cdot 277^{4}$ x6 - 3x4 + 3x2 - 3x + 1 $S_6$ $1$ $-2$
10.30344516420400625.30t164.a.a $10$ $5^{4} \cdot 191^{6}$ x6 - x4 - 3x3 + x + 3 $S_6$ $1$ $-2$
10.31123779291971584.30t164.a.a $10$ $2^{18} \cdot 587^{4}$ x6 - 2x5 + 2x4 - x2 + 2x - 1 $S_6$ $1$ $-2$
10.33364831591822329.30t164.c.a $10$ $3^{14} \cdot 17^{8}$ x6 - x5 - 2x4 + x3 + 14x2 - 17x + 6 $S_6$ $1$ $2$
10.33364831591822329.30t164.d.a $10$ $3^{14} \cdot 17^{8}$ x6 - x5 - 2x4 + x3 + 14x2 - 17x + 6 $S_6$ $1$ $-2$
10.35016268618806336.30t164.a.a $10$ $2^{6} \cdot 3^{4} \cdot 17^{6} \cdot 23^{4}$ x6 - x5 - 2x4 + 3x3 - 2x2 + 3x - 3 $S_6$ $1$ $-2$
10.41489609581920256.30t164.a.a $10$ $2^{24} \cdot 223^{4}$ x6 - 2x5 + 4x3 - 4x2 + 2 $S_6$ $1$ $-2$
10.47090024679574321.30t164.a.a $10$ $14731^{4}$ x6 - x5 + x3 - x2 + 1 $S_6$ $1$ $-2$
10.54419558400000000.30t164.c.a $10$ $2^{18} \cdot 3^{12} \cdot 5^{8}$ x6 + 2x4 - 2x3 - 3x2 - 6x - 3 $S_6$ $1$ $-2$
10.54419558400000000.30t164.d.a $10$ $2^{18} \cdot 3^{12} \cdot 5^{8}$ x6 + 2x4 - 2x3 - 3x2 - 6x - 3 $S_6$ $1$ $-2$
Next
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# Environment Management
Similar to pip, conda is an open source package and environment management system 1. Anaconda is a data science platform that comes with a lot of packages. It uses conda at the core. Unlike Anaconda, Miniconda doesn't come with any installed packages by default. Note that for miniconda, everytime you open up a terminal, conda won’t automatically be available. Run the command below to use conda within miniconda.
# Conda
Let’s first start by checking if conda is installed.
$conda --version conda 4.2.12 To see the full documentation for any command, type the command followed by --help. For example, to learn about the conda update command: $ conda update --help
Once it has been confirmed that conda has been installed, we will now make sure that it is up to date.
$conda update conda Using Anaconda Cloud api site https://api.anaconda.org Fetching package metadata: .... .Solving package specifications: ......... Package plan for installation in environment //anaconda: The following packages will be downloaded: package | build ---------------------------|----------------- conda-env-2.6.0 | 0 601 B ruamel_yaml-0.11.14 | py27_0 184 KB conda-4.2.12 | py27_0 376 KB ------------------------------------------------------------ Total: 560 KB The following NEW packages will be INSTALLED: ruamel_yaml: 0.11.14-py27_0 The following packages will be UPDATED: conda: 4.0.7-py27_0 --> 4.2.12-py27_0 conda-env: 2.4.5-py27_0 --> 2.6.0-0 python: 2.7.11-0 --> 2.7.12-1 sqlite: 3.9.2-0 --> 3.13.0-0 Proceed ([y]/n)? y Fetching packages ... conda-env-2.6. 100% |################################| Time: 0:00:00 360.78 kB/s ruamel_yaml-0. 100% |################################| Time: 0:00:00 5.53 MB/s conda-4.2.12-p 100% |################################| Time: 0:00:00 5.84 MB/s Extracting packages ... [ COMPLETE ]|###################################################| 100% Unlinking packages ... [ COMPLETE ]|###################################################| 100% Linking packages ... [ COMPLETE ]|###################################################| 100% # Environments ## Create In order to manage environments, we need to create at least two so you can move or switch between them. To create a new environment, use the conda create command, followed by any name you wish to call it: # create new environment conda create -n <your_environment> python=2.7.11 ## Clone Make an exact copy of an environment by creating a clone of it. Here we will clone snowflakes to create an exact copy named flowers: conda create --name flowers --clone snowflakes ## List List all environments Now you can use conda to see which environments you have installed so far. Use the conda environment info command to find out $ conda info -e
conda environments:
Verify current environment
Which environment are you using right now – snowflakes or bunnies? To find out, type the command:
conda info --envs
## Remove
If you didn’t really want an environment named flowers, just remove it as follows:
conda remove --name flowers --all
## Share
You may want to share your environment with another person, for example, so they can re-create a test that you have done. To allow them to quickly reproduce your environment, with all of its packages and versions, you can give them a copy of your environment.yml file.
Export the environment file
To enable another person to create an exact copy of your environment, you will export the active environment file.
conda env export > environment.yml
Use environment from file
Create a copy of another developer’s environment from their environment.yml file:
conda env create -f environment.yml
# remove environment
conda remove -n <your_environemnt> --all
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# Finding correlation between two distributions
I have records of census data comprised of complex data types including distributions, such as these two fields from a record, both of which are distributions.
The first is an income distribution:
No income: 1110.0
$1 to$9,999 or loss: 13840.0 ######
$10,000 to$14,999: 9490.0 ####
$15,000 to$24,999: 152145.0 #######################################################################
$25,000 to$34,999: 5465.0 ##
$35,000 to$49,999: 4950.0 ##
$50,000 to$64,999: 6880.0 ###
$65,000 to$74,999: 10420.0 ####
\$75,000 or more: 9100.0 ####
And the second is an age distribution:
< 5: 955.0 #####
5 - 9: 1002.0 #####
10 - 14: 1032.0 #####
15 - 19: 1183.0 ######
20 - 24: 1305.0 #######
25 - 29: 1082.0 ######
30 - 34: 1212.0 ######
35 - 39: 1175.0 ######
40 - 44: 1210.0 ######
45 - 49: 1359.0 #######
50 - 54: 1336.0 #######
55 - 59: 1212.0 ######
60 - 64: 1034.0 #####
65 - 69: 727.0 ####
70 - 74: 618.0 ###
74 - 79: 501.0 ##
80 - 84: 404.0 ##
A full example record can be seen at http://tinypic.com/r/34s2kxj/6
Now what I need to do is create a correlation coefficient matrix for factor analysis. But the correlation matrix needs show correlations between fields in the full records. For example I'd like a correlation coefficient that describes the relationship between age and income, using these two distributions.
Where I'm stumped is how to calculate correlation coefficients for two distributions. For example, if I want to find the correlation coefficient for age distribution and income distribution.
I don't have access to the original data that makes up these distributions, only the summarized data that you see.
Any ideas? Right now I'm investigating using a Q-Q Plot (but I'm having trouble figuring out how to generate an intelligent coefficient from it), a cross-correlation (but I'm not sure a wave function is applicable to this type of data), Ripley's Cross-K function (which seems to be only applicable to temporal data), and fabricating matrices to overlay on the histogram and then calculating correlations between the cells. I've also considered calculating correlation coefficients of the moments of the distribution (mean, skew, kurtosis, std dev), but I've never heard of this and I'm not sure that it's valid.
Any help or a nudge in the right direction would be much appreciated!
-
When all you have are the marginal distributions, as shown here, you have absolutely no information about their correlation. You could simulate a population having these two distributions. Then you could pair ages and incomes by taking the largest with the largest, the second largest with the second largest, and so on, creating a high positive correlation. You could equally well pair the largest income with the smallest age, etc., creating a high negative correlation. Your data say nothing about what pairing actually occurs. – whuber Jul 30 '12 at 16:16
That makes sense to me when looking at two instances of these distributions. However, keep in mind that this is a single record. Doesn't the process of calculating a correlation coefficient only make sense using a large sample population? You wouldn't try to correlate two numbers using only a single data point either. But, over, say 500 of these records, doesn't it make sense that you could find a pattern? – Jason Kolb Jul 30 '12 at 16:21
For example, using just the moments of the distribution such as the mean, couldn't you at the very least find correlations between moments? For example if, instead of the distribution data itself I simply used the mean of the two distributions, and then calculated the Pearson coefficient normally against the means, would that not give me information about how the mean of one distribution correlates to the mean of the other? – Jason Kolb Jul 30 '12 at 16:22
There seems to be some confusion in these comments, Jason. You appear now to be referring to a collection of about 500 joint distributions, not a single distribution. Your original question refers to "correlation" in the sense of a "relationship between age and income": that is, the correlation of a single distribution (for a single record). These relationships could change from one record to another. Now, in the comments, you write of a different correlation, one that measures associations revealed by all 500 records. The two things are not the same! Which one are you really after? – whuber Jul 30 '12 at 16:33
Confounding these two kinds of correlations is known as the ecological fallacy: perhaps that article is a useful nudge? – whuber Jul 30 '12 at 16:38
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# Ω 3
Crossposted from the AI Alignment Forum. May contain more technical jargon than usual.
Summary: There's a "thin" concept of counterfactual that's easy to formalize and a "thick" concept that's harder to formalize.
Suppose you're trying to guess the outcome of a coinflip. You guess heads, and the coin lands tails. Now you can ask how the coin would have landed if you had guessed tails. The obvious answer is that it would still have landed tails. One way to think about this is that we have two variables, your guess and the coin , that are independent in some sense; so we can counterfactually vary while keeping constant.
But consider the variable XOR . If we change to tails and keep the same, we conclude that if we had guessed tails, the coin would have landed heads!
Now this is clearly silly. In real life, we have a causal model of the world that tells us that the first counterfactual is correct. But we don't have anything like that for logical uncertainty; the best we have is logical induction, which just give us a joint distribution. Given a joint distribution over , there's no reason to prefer holding constant rather than holding XOR constant. I want a thin concept of counterfactuals that includes both choices. Here are a few definitions, in increasing generality:
1. Given independent discrete random variables and , such that is uniform, a thin counterfactual is a choice of permutation of for every .
2. Given a joint distribution over and , a thin counterfactual is a random variable independent of and an isomorphism of probability spaces that commutes with the projection to .
3. Given a probability space and a probability kernel , a thin counterfactual is a probability space and a kernel such that .
There are often multiple choices of thin counterfactual. When we say that one of the thin counterfactuals is more natural or better than the others, we are using a thick concept of counterfactuals. Pearl's concept of counterfactuals is a thick one. No one has yet formalized a thick concept of counterfactuals in the setting of logical uncertainty.
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For telecommunications, satellites are better served by rotating around the earth at the same speed as the earth rotates
# For telecommunications, satellites are better served by rotating around the earth at the same speed as the earth rotates
398.7k points
For telecommunications, satellites are better served by rotating around the earth at the same speed as the earth rotates, or 24.0 h. What is the distance from the center of the Earth that the satellite must be placed to achieve this geosynchronous orbit? What is the orbital speed of the satellite?
For telecommunications
389.5k points
Kepler's laws of planetary motion are as follows, 1. The orbit of every planet is an ellipse with the Sun at one of the two foci
2. A line drawn from the Sun to any planet sweeps out equal areas in equal time intervals 3. The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, or
$$T^2=\frac{4 \pi^2}{GM} a^3$$
where T is the period of orbit, M is the mass which the object orbits, and a is the distance from the center of the orbit, G is the gravitation constant,
$G = 6.67 \times 10^{-11} \, m^3 /\, kg\, s^2$
this is especially useful for finding the mass of an object, once its period and average distance are known.
given,
$M_E =5.98 \times 10^24 \, kg$ (mass of the earth)
$G = 6.67 \times 10^{-11} \, m^3 / \,kg\, s^2$
$T = 24.0 \, h = 86400 \, s$
plugging these values into Kepler's third law, and solving for r give,
$r = 4.23 \times 10^7$
this is the distance from the center of the earth to the satellite.
The orbital speed of the satellite can be found by using,
$v=\frac{d}{t}=\frac{2 \pi r}{t} = \frac{2 \pi (4.23 \times 10^7 \,m)}{86400 \,s}=3100 m/s$
Surround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\beta^2-1=0$$
Use LaTeX to type formulas and markdown to format text. See example.
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# How to perform a test using R to see if data follows normal distribution
I have a data set with following structure:
a word | number of occurrence of a word in a document | a document id
How can I perform a test for normal distribution in R? Probably it is an easy question but I am a R newbie.
• @Skarab Maybe I'm totally off, but wouldn't you expect that the frequency of any word will be inversely proportional to its rank in the frequency table of words, according to Zipf's law (j.mp/9er2lv)? In this case, check out the zipfR package. – chl Sep 28 '10 at 9:29
• I agree with @chl - it would be minor miracle if your data was normally distributed. Perhaps another question about what you want to do with the data would be worthwhile. Don't reinvent the wheel! – csgillespie Sep 28 '10 at 9:32
• How could your data be distributed according to a model that gives non zero probability to negative occurrence? – user603 Sep 28 '10 at 15:05
• What is the reason for doing this test? – whuber Sep 28 '10 at 15:12
• I want to estimate if the huge result of the Information Extraction is correct. I want to check if the distribution of the entities found in the text follows my expectations (I know the domain and the text corpus). – Skarab Sep 29 '10 at 9:06
If I understand your question correctly, then to test if word occurrences in a set of documents follows a Normal distribution you can just use a shapiro-Wilk test and some qqplots. For example,
## Generate two data sets
## First Normal, second from a t-distribution
words1 = rnorm(100); words2 = rt(100, df=3)
## Have a look at the densities
plot(density(words1));plot(density(words2))
## Perform the test
shapiro.test(words1); shapiro.test(words2)
## Plot using a qqplot
qqnorm(words1);qqline(words1, col = 2)
qqnorm(words2);qqline(words2, col = 2)
The qqplot commands give:
You can see that the second data set is clearly not Normal by the heavy tails (More Info).
In the Shapiro-Walk normality test, the p-value is large for the first data set (>.9) but very small for the second data set (<.01). This will lead you to reject the null hypothesis for the second.
• Why is it clearly not Normal? – Herman Toothrot Jan 16 '17 at 11:08
• I think the plotted points should lie on the I-III quadrant bisector as closer as they draw a normal distribution. – Campa Mar 1 '18 at 16:19
• More generally (mean != 0), the qqline shall have 1 slope and mu intercept. – Campa Mar 1 '18 at 16:25
• @HermanToothrot it is not Normal when looking at the second plot as there is a very large divergence in the tail values. The QQ plot is a graph of the theoretical quantile (if it was normal) verses the sample quantlie (from the data). If the sample data is normal we expect the observations to be close to line, as they are for the first plot. Also note the very difference scale on the y axis for those plots. – Sheldon Sep 12 '18 at 18:46
Assuming your dataset is called words and has a counts column, you can plot the histogram to have a visualization of the distribution:
hist(words$counts, 100, col="black") where 100 is the number of bins You can also do a normal Q-Q plot using qqnorm(words$counts)
Finally, you can also use the Shapiro-Wilk test for normality
shapiro.test(word\$counts)
Although, look at this discussion: Normality Testing: 'Essentially Useless?'
No test will show you that your data has a normal distribution - it will only be able to show you when the data is sufficiently inconsistent with a normal that you would reject the null.
But counts are not normal in any case, they're positive integers - what's the probability that an observation from a normal distribution will take a value that isn't an integer? (... that's an event of probability 1).
Why would you test for normality in this case? It's obviously untrue.
[In some cases it may not necessarily matter that you can tell your data aren't actually normal. Real data are never (or almost never) going to be actually drawn from a normal distribution.]
If you really need to do a test, the Shapiro-Wilk test (?shapiro.test) is a good general test of normality, one that's widely used.
A more formal way of looking at the normality is by testing whether the kurtosis and skewness are significantly different from zero.
To do this, we need to get:
kurtosis.test <- function (x) {
m4 <- sum((x-mean(x))^4)/length(x)
s4 <- var(x)^2
kurt <- (m4/s4) - 3
sek <- sqrt(24/length(x))
totest <- kurt/sek
pvalue <- pt(totest,(length(x)-1))
pvalue
}
for kurtosis, and:
skew.test <- function (x) {
m3 <- sum((x-mean(x))^3)/length(x)
s3 <- sqrt(var(x))^3
skew <- m3/s3
ses <- sqrt(6/length(x))
totest <- skew/ses
pt(totest,(length(x)-1))
pval <- pt(totest,(length(x)-1))
pval
}
for Skewness.
Both these tests are one-tailed, so you'll need to multiply the p-value by 2 to become two-tailed. If your p-value become larger than one you'll need to use 1-kurtosis.test() instead of kurtosis.test.
If you have any other questions you can email me at [email protected]
• What are the differences, of the above two functions, regarding the kurtosis() and skewness() functions from the moments package? Results using rnorm() samples are different. – Nikos Alexandris Sep 16 '14 at 11:18
In addition to the Shapiro-Wilk test of the stats package, the nortest package (available on CRAN) provides other normality tests.
By using the nortest package of R, these tests can be conducted:
• Perform Anderson-Darling normality test
ad.test(data1)
• Perform Cramér-von Mises test for normality
cvm.test(data1)
• Perform Pearson chi-square test for normality
pearson.test(data1)
• Perform Shapiro-Francia test for normality
sf.test(data1)
Many other tests can be done by using the normtest package. See description at https://cran.r-project.org/web/packages/normtest/normtest.pdf
• @Nick; I've mentioned these test from 'nortest' package found here: (cran.r-project.org/web/packages/nortest/nortest.pdf). Other useful package is 'normtest' as mentioned above. – Dr Nisha Arora Jun 10 '16 at 10:13
• OK, thanks. In my ignorance I assumed that one name was a typo. – Nick Cox Jun 10 '16 at 11:23
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Another proof idea using finite automata
Steve Cook proved three landmark theorems with 1971 dates. The first has been called a “surprising theorem”: that any deterministic pushdown automaton with two-way input tape can be simulated in linear time by a random-access machine. This implies that string matching can be done in linear time, which inspired Donald Knuth and Vaughan Pratt to find a direct algorithm that removes a dependence on the input alphabet size. This was before they learned that James Morris had found it independently, later than but without knowledge of Cook, and their algorithm is now called KMP. Fast string matching has thousands of applications. Second was his characterization of polynomial time by log-space machines with auxiliary pushdowns, a fact which may yet help for proving lower bounds. Then there was his third result, which appeared at STOC 1971, and was given a “slif” (stronger, later, independent form) by Leonid Levin.
Today I want to present a proof of the famous Cook-Levin Theorem that ${\mathsf{SAT}}$ is ${\mathsf{NP}}$-complete, and also mention one used by Ken.
I am currently giving a class on complexity theory, and thought this proof might have some advantage over the usual proofs. There are the original tableau-based proofs, the later circuit-based proofs, and variations of them.
The proof here is based on my favorite kind of objects—well one of them—finite automata. They are remarkably powerful and can be used to give a relatively clean proof. At least I believe the proof is clean for students—I would like to hear any thoughts that you all may have about it. It needs no assumption about oblivious tape access patterns, and does not use lots of complex indexing.
So let’s take a look at the proof. Here is a Wordle from STOC 1971:
## Introduction
My goal is to try and give a proof of Cook’s famous theorem that is easy to follow. There are many proofs already, but I thought I would try and see if I could give a slightly different one, and hope that it is clear. My thought is that sometimes the measure of clarity of a proof is the same as “ownership”: if you wrote it—own it—then it is clear. But here goes a proof that is a bit different.
The overarching idea is to do two things: (i) avoid as much detailed encoding as possible, and (ii) leverage existing concepts that you probably already know.
## Machines
Let ${L}$ be any set in ${\mathsf{NP}}$. Then there is always a one-tape nondeterministic Turing Machine (NTM) ${M}$ that accepts exactly ${L}$ and runs in time ${n^{k}}$ for some fixed ${k}$. For any input ${x}$ this means that ${M}$ accepts ${x}$ if and only if ${x}$ is in ${L}$. An ID ${U}$ of ${M}$ is an encoding of the total state of ${M}$ in the usual way. We have two special ID’s:
• the start ID ${U_{start}}$, which depends on the input, and
• the accepting ID ${U_{accept}}$, which can be made unique and independent of the input.
For ID’s,
$\displaystyle U \vdash V$
means that ${U}$ can reach ${V}$ in one step of ${M}$. This is all standard and is stated here just for review.
Saying that ${A}$ is an FSA means that ${A}$ is a deterministic finite state automaton. The language that ${A}$ accepts is ${L(A)}$. Also for any two strings ${x=x_{1},\dots,x_{n}}$ and ${y=y_{1},\dots,y_{n}}$ of the same length let ${x || y}$ be the shuffle,
$\displaystyle x_{1}\ y_{1}, \dots, x_{n} \ y_{n}.$
If one string is longer, we suppose the other is padded with a special null character to have equal length, and then we shuffle.
You might ask why introduce FSA when our goal is to encode NTM’s? It turns out that our proof will take the following steps:
1. Show that the behavior of an FSA can be encoded by ${\mathsf{GEN}}$${\mathsf{SAT}}$, which is a more powerful version of ${\mathsf{SAT}}$. This version is easier to use and understand.
2. Show that the behavior of a NTM can be reduced to the behavior of FSA’s, and so to ${\mathsf{GEN}}$${\mathsf{SAT}}$.
3. Finally, replace ${\mathsf{GEN}}$${\mathsf{SAT}}$ by regular ${\mathsf{SAT}}$.
## A Slight Generalization of ${\mathsf{SAT}}$
A ${k}$${\mathsf{GEN}}$${\mathsf{SAT}}$ problem ${C}$ is of the following form:
$\displaystyle C = \exists x_{1} \exists x_{2} \cdots \exists x_{m} C_{1} \wedge \cdots \wedge C_{L}.$
Here each ${C_{i}}$ is a general clause, which means it is a Boolean function of the variables ${x_{i}}$‘s. We allow the general-clauses to any Boolean function, but will restrict the number of variables that they can depend to at most ${k}$. The problem ${C}$ is called satisfiable provided there exists Boolean values for the variables ${x_{1},\dots,x_{m}}$ so that each ${C_{i}}$ evaluates to true.
Note that ${k}$${\mathsf{GEN}}$${\mathsf{SAT}}$ is closed under conjunction in the following sense: Suppose that ${A}$ is
$\displaystyle A = \exists x_{1} \exists x_{2} \cdots \exists x_{m} A_{1} \wedge \cdots \wedge A_{K}$
and ${B}$ is
$\displaystyle B = \exists x_{1} \exists x_{2} \cdots \exists x_{m} B_{1} \wedge \cdots \wedge B_{L}.$
Then
$\displaystyle C = \exists x_{1} \exists x_{2} \cdots \exists x_{m} A_{1} \wedge \cdots \wedge A_{K} \wedge B_{1} \cdots \wedge B_{L}$
is the conjunction. This is satisfiable if and only if there are variables ${x_{i}}$ so that both sets of clauses are true. This is a simple, but very useful property of ${k}$${\mathsf{GEN}}$${\mathsf{SAT}}$.
There is nothing mysterious or strange about ${k}$${\mathsf{GEN}}$${\mathsf{SAT}}$. It is just a way of defining a type of search problem. Informally it asks: are there Boolean values (…) that satisfy all of the properties (…)? We have lots of problems like this in mathematics. Consider a system of linear equations over the reals:
$\displaystyle Ax = b,$
where ${A}$ is a matrix, ${x}$ is a vector, and ${b}$ is a vector. The LINEAR problem is: does there exist a vector ${x}$ so that the above is true? We could have written it as
$\displaystyle \exists x Ax = b,$
to make it look like the ${k}$${\mathsf{GEN}}$${\mathsf{SAT}}$ problem. You should, and probably do know, what Gauss knew years ago that LINEAR is “easy”: in modern terms LINEAR is in ${\mathsf{P}}$, that is in polynomial time.
The difference between ${k}$${\mathsf{GEN}}$${\mathsf{SAT}}$ and LINEAR is that while the latter is extremely useful and can be used to model many important problems, ${k}$${\mathsf{GEN}}$${\mathsf{SAT}}$ is “universal.” By universal we mean that any problem from ${\mathsf{NP}}$ can be reduced to solving a ${k}$${\mathsf{GEN}}$${\mathsf{SAT}}$ problem.
## Cook’s Theorem
Our plan is to show that we can encode ${U \vdash V}$ by a ${k}$${\mathsf{GEN}}$${\mathsf{SAT}}$ formula. Then the behavior of ${M}$ for ${n^{c}}$ steps will be encoded by simply using the fact that ${k}$${\mathsf{GEN}}$${\mathsf{SAT}}$ is closed under conjunction.
Theorem 1 The problem ${k}$${\mathsf{GEN}}$${\mathsf{SAT}}$ is ${\mathsf{NP}}$-complete.
Proof: Let ${M}$ be a one-tape NTM that runs in ${N=n^{c}}$ time and let ${x}$ be an input of length ${n}$. Clearly ${M}$ accepts if and only if there are exists a series of ID’s
$\displaystyle S^{1},S^{2},\dots,S^{N}$
of length ${N}$ so that ${S^{1}}$ is the initial state, ${S^{N}}$ is a final accept state, and for each ${i}$,
$\displaystyle S^{i} \vdash S^{i+1}.$.
Let ${S^{i}}$ for all ${i}$ consist of ${N}$ Boolean variables.
Claim: We have ${k}$${\mathsf{GEN}}$${\mathsf{SAT}}$ formulas ${U^{i}}$ over these variables so that (i) ${U^{0}}$ is satisfied precisely when ${S^{1}}$ is the initial state corresponding to the input ${x}$; (ii) ${U^{N}}$ is satisfied precisely when ${S^{N}}$ corresponds to the final state; (iii) and for each ${1 \leq i < N}$, the ${U^{i}}$ is satisfied precisely if
$\displaystyle S^{i} \vdash S^{i+1}.$
Then we claim that the conjunction of the clauses of all the
$\displaystyle U^{1},\dots,U^{N}$
are satisfiable if and only if ${M}$ accepts. This follows directly by the definitions. $\Box$
Thus to complete the proof of the theorem we must show that claim about the existence of ${U^{i}}$‘s is true. But this is easy as we will now show.
## Encoding With FSA
Theorem 2 Let ${M}$ be a one-tape NTM. Then there is an FSA ${A}$ that depends only on ${M}$ so that for all ${U}$ and ${V}$,
$\displaystyle U \vdash V \text{ iff } U||V \in L(A).$
Proof: The FSA ${A}$ just passes over ${U}$ and ${V}$ checking that the tapes are the same, except at the location of the head. There it checks that a legal move of ${M}$ has been used. Because the ID’s are shuffled together this can be done by the automaton. $\Box$
## Encoding With ${k}$–${\mathsf{GEN}}$–${\mathsf{SAT}}$
Theorem 3 Let ${A}$ be a fixed FSA. Then for any input string ${a_{1},\dots,a_{n}}$ there are a set of general-clauses ${C}$ over ${O(n)}$ variables so that
$\displaystyle a_{1},\dots,a_{n} \in L(A) \text{ iff } C \text{ is satisfiable}.$
Proof: The idea is to look at the sequence
$\displaystyle q_{0} a_{1} q_{1} a_{1} \dots q_{n-1} a_{n} q_{n},$
where ${q_{i}}$ are strings of Boolean variables that are long enough to encode the state of the FSA. Then the general-clauses are of three types:
1. One clause that checks that ${q_{0}}$ is the start state of ${A}$.
2. One clause that checks that ${q_{i+1}}$ is a possible state that ${q_{i}}$ can reach after seeing the input bit ${a_{i}}$.
3. One clause that checks that ${q_{n}}$ is an accept state of ${A}$.
Clearly all of these can be done by general-clauses: they are just Boolean tests. $\Box$
## Final Reduction to ${\mathsf{SAT}}$
A ${\mathsf{3SAT}}$ problem ${C}$ is a ${k}$${\mathsf{GEN}}$${\mathsf{SAT}}$ problem except that the clauses are restricted to be disjunctions of at most three variables and their negations. Thus a clause can only be of the form
$\displaystyle (x_{20} \vee x_{30} \vee \bar{x}_{101}),$
for example.
The final step is to note:
Theorem 4 The problem ${k}$${\mathsf{GEN}}$${\mathsf{SAT}}$ can be reduced to ${\mathsf{SAT}}$.
This proves Cook’s Theorem.
## Nand Now For a Shortcut
Ken tells me that especially in Buffalo’s undergraduate theory course, he uses—and gets away with—a big handwave. He states the principle
“Software Can Be Efficiently Burned Into Hardware.”
He uses this to assert that the witness predicate ${R(x,y)}$ for a given ${\mathsf{NP}}$-language ${L}$ can be replaced by a polynomial-sized circuit ${C_n}$ for any input length ${n}$. Since NAND gates are universal, every gate can be a binary NAND. Thus ${x \in L}$ if and only if there exist a ${y}$ and an assignment to every wire in ${C_n}$ that is consistent with every NAND gate, such that the output wire ${w_o}$ has value ${1}$. For the inputs ${u,v}$ and every output ${w}$ of each NAND gate, the assignment must satisfy the clauses
$\displaystyle (u \vee w) \wedge (v \vee w) \wedge (\bar{u} \vee \bar{v} \vee \bar{w}).$
Together with the singleton clause ${(w_0)}$ and clauses fixing the wires for ${x_i}$ to the corresponding bits of the given input string ${x}$, this creates a ${\mathsf{3SAT}}$ formula that is satisfiable if and only if ${x \in L}$.
This proof has some nice features. It shows the completeness of several restricted forms of ${\mathsf{3SAT}}$ where the variables in each clause have the same sign and (with a little more work) assignments must make at least one literal in each clause false as well. But it hides the way computations are represented behind the handwave.
## Open Problems
What do you think of the proof via finite automata?
1. October 10, 2013 9:30 pm
Synchronicity Rules❢
I just started reworking an old exposition of mine on Cook’s Theorem, where I borrowed the Parity Function example from Wilf (1986), Algorithms and Complexity, and translated it into the cactus graph syntax for propositional calculus that I developed as an extension of Peirce’s logical graphs.
By way of providing a simple illustration of Cook’s Theorem, namely, that Propositional Satisfiability is NP-Complete, I will describe one way to translate finite approximations of turing machines into propositional expressions, using the cactus language syntax for propositional calculus that I will describe in more detail as we proceed.
2. October 10, 2013 10:15 pm
I guess I’m a little confused why you are so attached to the Turing machine as a model of computation. If you start with with the fact that Boolean circuits can do polynomial time computation with polynomial-size circuits — something that every CS major understands, since programs can be “compiled” down to machine language and thence to hardware — then we can take Circuit SAT as our root problem. The reduction from Circuit SAT to 3-SAT is extremely simple: just invent additional variables for the internal truth values of the circuit’s wires, and express in CNF form the assertions that each gate in the circuit functions properly, and that the output is true.
Isn’t this 1) a lot simpler, 2) just as rigorous, and 3) a lot easier for modern CS students to understand?
• October 10, 2013 10:22 pm
Chris, That sort of method leads to less clarity on the distinction between P and P/poly.
• October 10, 2013 10:36 pm
I know that’s the usual argument, but I don’t agree. One can always address uniformity in circuit classes after presenting this proof. What a student needs to understand is that for any fixed instance of an NP problem (Hamiltonian Path, Graph Coloring, etc.) it is easy to produce a circuit that checks solutions, and thus reduce the problem to an instance of Circuit SAT.
I agree that you need a uniform program to do this reduction: on the other hand, this reduction is extremely simple in most cases that we discuss. For local problems like Graph Coloring, it’s in DLOGTIME, i.e. local gadget replacement.
Do you really think that starting with single-tape Turing machines is better pedagogically? Doesn’t it just create more formal overhead for the student to wade through, before they get a chance to appreciate what NP-completeness is all about?
October 10, 2013 11:12 pm
“… why you are so attached to the Turing machine as a model of computation.”
Good question, since theoretically any deterministic algorithm would seem to be equally logical for distinguishing between NP and P from a foundational perspective.
• October 11, 2013 12:00 am
Roughly speaking, the importance of the Turing machine model is that we are factoring an infinite machination into an infinite component times a finite component, where it’s the same finite component each time. That amounts to a significant reduction in conceptual complexity.
• October 11, 2013 8:16 am
Cris, exactly what you say is in the last section of the post. When I lecture, I do indeed use the words “…check that every NAND gate works correctly…” and present it as a basic instance of hardware verification. But for lower bound purposes, maybe there is interest in exactly how low one goes for the verification to remain NP-complete.
• October 11, 2013 9:02 am
Hi Ken,
I know, and I like your “software can be burned into hardware” line. I guess my claim is that it’s not that much of a handwave! Yes, we need to teach the idea of uniformity, and at some point we need a uniform model. But we should feel free to use whichever model makes it easier for us to get NP-completeness up and running.
For teaching, my favorite uniform model is the program, written in the student’s favorite programming language. The fact that it doesn’t matter whether we use Java or Lisp for this is a wonderful fact, and the students need to appreciate it (and the hard work that went into proving it in the 1930s, i.e. the equivalence of partial recursive functions, lambda expressions, and Turing machines). But I prefer to start with the “software” proof of NP-completeness, and then to circle back and say “why can we get away with this?” At that point we can talk the equivalence (up to polynomial time) of different models of computation.
• October 11, 2013 12:12 pm
hi CM/all john savages book “models of computation” [free online! very nicely typeset] is partly a massive research project to reformulate/”reframe” most computational/complexity theory in terms of circuits, one that I generally agree with, but which few other researchers seem to have consciously acknowledged although there is much pursuit [allender comes to mind as another authority/expert who has advocated a circuit-centric view over many yrs]. also stasys jukna has contributed strongly to this agenda with his book “boolean function complexity, advances and frontiers”.
the TM model seems to predate circuits, but circuits are arguably a more fundamental or natural way to understand computation in many ways.
its a real ongoing paradigm shift [worthy of a blog somewhere at least?] that is somewhat unrecognized/unremarked on so far. it somewhat reminds me of the major theoretical “teams” in physics eg standard model vs strings etc.
also note that for circuits the P vs NP problem is P/poly vs NP, a stronger assertion, and the concept/distinction of uniform vs nonuniform comes into play.
surely [even after decades of research] one is not really provably superior to the other (circuits/TMs) in all cases, each has its own context, but they exist in a close theoretical coupling/synergy/symbiosis. two sides of the same coin. the yin & yang of complexity theory. and cooks proof is a nice example of this deep interconnection (which is more obvious four decades in retrospect thx in part to massive research in circuit-related complexity classes).
personally I have gone in the direction of circuits esp for complexity class separations & believe the view may eventually/ultimately lead to a P/poly != NP proof. this is because of the deep ties between monotone circuits and extremal set theory….
• October 11, 2013 12:15 pm
oops sorry got that link wrong! here is a NP vs P/poly proof sketch/outline using circuits/hypergraphs etc
• October 12, 2013 12:11 am
vznvzn you are coming out as a crank. If Ken or Lipton, thought your ideas were worthwhile, they would have written about them.
• October 28, 2013 7:46 pm
sincere congratulations J on the feat of combine the converse of argumentum ad populum, genetic fallacy, and argumentum ad hominem all in only two sentences! do they teach this stuff in school any more? but, dont feel too bad, because even elite theorists are known to fall in the same basic traps of reasoning. =)
• October 13, 2013 11:33 pm
No, it’s not a more fundamental or natural way.
Indeed, the very thing that makes computation a natural and interesting notion is exactly that it is independent of the particular choice of implementation. Choose any reasonable implementation, make sure it doesn’t suffer from obvious limitations in power (only finitely many machine states, inability to make an unbounded number of intermediate computations) and you end up with the same notion. So there can’t be a more natural or fundamental way to approach computation since such a way would undermine the very naturalness and interestingness of the notion of computation itself.
I mean it’s like saying that counting pebbles is the fundamental/natural way to understand the natural numbers. It may be true that evolutionarily/historically counting fingers is neurologically easier and occurred first but what makes the numbers the numbers is that they behave the same regardless of what you describing. The best way to understand the numbers is by understanding the numbers!
Indeed, I think all this focus on circuits is holding back the computational complexity theory from looking at more powerful results from number theory and distracts from looking more closely at what makes the difference between the relativizeable complexity results and the non-relativizeable ones.
• November 1, 2013 12:01 pm
ok, will bite, just have to ask
I think all this focus on circuits is holding back the computational complexity theory from looking at more powerful results from number theory…
what results are you referring to, did you have something specific in mind?
ps re J, feel is also guilty of argument from ignorance
3. October 11, 2013 12:53 am
Dear Pip,
Since you are proposing a new proof for the CLT, would you be slightly starting to believe that paper http://www.andrebarbosa.eti.br/The_Cook-Levin_Theorem_is_False.pdf?
4. October 11, 2013 10:24 am
It’s interesting you mention NAND gates and 3SAT.
Recently I tried to come up with a method to encode logic gates (OR and NOT) on Flow Networks with costs, and tried to show how this could maybe be applied to solve 3SAT in poly-time using minimum cost flow algorithm.
Obviously there’s probably some flaw that I don’t see (yet). I will have to implement this and test my theory.
http://polyfree.wordpress.com/2013/10/07/logic-gates-on-flow-networks/
October 11, 2013 10:34 am
I have often heard the sentiment that it should be easy to see, especially for CS students, that programs are equivalent to circuits and hence it suffices to prove Cook’s theorem by proving that Circuit-SAT is reducible to SAT. I don’t think it is any where as simple as that. Typical programs are complicated and have non-local references because of the RAM model and it is easy to get confused by the size of integers, pointer addressing etc etc. The idea of non-determinism, checking solutions vs coming up with them, and a proper understanding of when a language is in NP are, in my experience, not easy to teach.
• October 14, 2013 2:58 pm
If that’s so, then it gives you a good opportunity to teach a little about hardware. Can the students design a circuit that takes an n-bit input x and a log_2 n-bit input i, and returns x with the i’th bit flipped?
If not, then I think we have a larger pedagogical problem ;-)
October 16, 2013 3:31 pm
Yes, they can design such a circuit but this is precisely where the uniformity issues comes up. They can also write code on a regular computer for the same problem with x and the index i given. However the code runs on a fixed size hardware (their laptop) while the circuit designed for the problem depends on n, the size of x. Of course one can say that circuit itself can be generated by a uniform program but then one is really getting back to a uniform model of computation at some level.
6. October 13, 2013 3:30 am
Nice share sir, but i still doubt with this theory
7. October 13, 2013 1:32 pm
Almost surely I missed something in this proof, but since it does not require a lower bound on k, the number of boolean variables in each SAT clause, I think you can just substitute “3-SAT” by “2-SAT” and the theorem would read “2-SAT is NP-Complete”, which is of course not true. Am I missing something here? Maybe the point is that the encoding of the clauses hinted by Theorem 3 requires some clause to have exactly 3 boolean variables?
May 26, 2014 11:40 am
A gate with k inputs and one output requires k+1 variables to describe; thus the CNF clauses need size k+1 (in general). Since one-input gates don’t suffice, we get k+1=3 as the smallest possible clause size in the proof.
October 13, 2013 2:57 pm
I was teaching the last spring my first ever undergraduate course, happened to be “the introduction to TCS”.
Usually in this class the complexity, i.e P vs NP, is barely touched. Most of the time is spent
on regular and context free languages and Turing machines. Anyway, I gave “informal” treatment of Cook-Levin very similar to yours. BTW, in the simple Theorem 3 you need to mention locality in order to get back to Cook-Levin. Actually, the proof from Wikipedia is quite similar. I tried to be very concrete and put the main focus on colorability and maximum clique(because of the connection, Motzkin-Straus, proved in the class, to the continuous world). Colorability is a great motivation for complexity: it is easy to convert an algorithm
for k-colorability to the algorithm for m-colorability, m 2. The students really liked and understood this motivation.
And it can be done without Cook-Levin as for the colorability it is very easy to write down the boolean formula. But, say for maximum clique what is needed is a “short” boolean formula describing all boolean strings with exactly k ones. The case k= 1 is simple and actually is used in Wikipedia proof.
And I gave this formula for “large” k using FSA. Also, similarly to your examples, I talked a lot about linear
equations: gave the algorithm for 2-colorability based on the gauss elimination.
The cool thing about this algorithm: it counts the number of colorings as well; and some of the students were learning
the gauss elimination at the same time in linear algebra class. The connection was a shock!
Thanks a lot for this blog and best regards, Leonid.
October 13, 2013 3:44 pm
Suppose to be “Colorability is a great motivation for complexity: it is easy to convert an algorithm
for k-colorability to the algorithm for m-colorability, m 2.
The students really liked and understood this motivation.”
October 13, 2013 3:49 pm
Somehow some of my sentences in the posts above got lost, with completely kosher words, strange.
October 15, 2013 4:33 pm
How do you encode an instance of SORTING into an instance of 3-SAT? And how do you encode an instance of FACTORING into an instance of 3-SAT? I’d bet the former encoding yields much bigger instances than the latter.
My other guess: this size difference is an explanation of why SORTING is easier than FACTORING. The intuition comes from quantum mechanics. Small objects have unpredictable behavior, but large groups of small objects are deterministic.
October 26, 2013 6:52 am
Is it possible to infer, from the mere probabilistic nature of observing a quantum particle, that a sequence of bits can only be found on a probabilistic basis? If quantum mechanics is the ultimate machine language, then that’s a likely hypothesis.
Thus, whenever a digital data – algorithm or data structure – is observed, this is due to its low-enough Kolmogorov complexity – i.e. the length of the shortest program which outputs this data. Although uncomputable, K-complexity is a function defined on all bit sequences and it plays the role of a probability distribution – just like probability waves in quantum mechanics.
Being undecidable, K-complexity would have been different in another Universe generated by another Big-Bang. In one of those, the Riemann Hypothesis is perhaps an easy lemma for undergraduate students – along with P!=NP – but the fundamental theorem of algebra is undecidable.
October 29, 2013 8:40 am
Well, perhaps not exactly undecidable but a hard conjecture anyway. What is provably true, provably false or provably undecidable in another Universe is the same as in ours, but the hardness of the proofs might be different.
So, whenever you come across a very hard conjecture, it may just be “true for no reason” in the words of Gregory Chaitin. My opinion is that the distribution of easiness and hardness is completely casual, depending on the quantum structure of our Universe. In another one, the same fact might have been true for some reason.
October 29, 2013 8:58 am
“P is true for no reason” means “all proofs of P have zero probability of being found”.
April 5, 2014 2:49 am
I wouldn’t use the automata proof. It is technically nice but it hides the intuition why the theorem is correc, and students intuitively understanding the reason why a statement is correct is very important for me.
I prefer using students programming skills and experience rather than their knowledge about automata theory.
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## itsmichelle29 one year ago The population of a type of local bass can be found using an infinite geometric series where a1 = 94 and the common ratio is one third. Find the sum of this infinite series that will be the upper limit of this population. 121 232 150 141
the infinite series formula is $s_{\infty} = \frac{a_{1}}{1 - r}$ substitute you values and calculate
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# 10.4 Concentrations as Conversion Factors
Learning Objective
• Apply concentration units as conversion factors.
Concentration can be a conversion factor between the amount of solute and the amount of solution or solvent (depending on the definition of the concentration unit). As such, concentrations can be useful in a variety of stoichiometry problems. In many cases, it is best to use the original definition of the concentration unit; it is that definition that provides the conversion factor.
A simple example of using a concentration unit as a conversion factor is one in which we use the definition of the concentration unit and rearrange; we can do the calculation again as a unit conversion, rather than as a definition. For example, suppose we ask how many moles of solute are present in 0.108 L of a 0.887 M NaCl solution. Because 0.887 M means 0.887 mol/L, we can use this second expression for the concentration as a conversion factor:
$0.108\cancel{L\, NaCl}\times \frac{0.887\, mol\, NaCl}{\cancel{L\, NaCl}}=0.0958\, mol\, NaCl$
(There is an understood 1 in the denominator of the conversion factor.) If we used the definition approach, we get the same answer, but now we are using conversion factor skills. Like any other conversion factor that relates two different types of units, the reciprocal of the concentration can be also used as a conversion factor.
Example $$\PageIndex{1}$$
Using concentration as a conversion factor, how many liters of 2.35 M CuSO4 are needed to obtain 4.88 mol of CuSO4?
Solution
This is a one-step conversion, but the concentration must be written as the reciprocal for the units to work out:
$4.88\cancel{mol\, CuSO_{4}}\times \frac{1\, L}{2.35\cancel{mol}}=2.08\, L\, of \, solution$
Exercise $$\PageIndex{1}$$
Using concentration as a conversion factor, how many liters of 0.0444 M CH2O are needed to obtain 0.0773 mol of CH2O?
1.74 L
Of course, once quantities in moles are available, another conversion can give the mass of the substance, using molar mass as a conversion factor.
Example $$\PageIndex{2}$$
What mass of solute is present in 0.765 L of 1.93 M NaOH?
Solution
This is a two-step conversion, first using concentration as a conversion factor to determine the number of moles and then the molar mass of NaOH (40.0 g/mol) to convert to mass:
$0.765\cancel{L}\times \frac{1.93\cancel{mol\, NaOH}}{\cancel{L\, solution}}\times \frac{40.0g\, NaOH}{1\cancel{mol\, NaOH}}=59.1\, g\, NaOH$
Exercise $$\PageIndex{2}$$
What mass of solute is present in 1.08 L of 0.0578 M H2SO4?
6.12 g
More complex stoichiometry problems using balanced chemical reactions can also use concentrations as conversion factors. For example, suppose the following equation represents a chemical reaction:
$\ce{2AgNO3(aq) + CaCl2(aq) → 2AgCl(s) + Ca(NO3)2(aq)}$
If we wanted to know what volume of 0.555 M CaCl2 would react with 1.25 mol of AgNO3, we first use the balanced chemical equation to determine the number of moles of CaCl2 that would react and then use concentration to convert to liters of solution:
$1.25\cancel{mol\, AgNO_{3}}\times \frac{1\cancel{mol\, CaCl_{2}}}{2\cancel{mol\, AgNO_{3}}}\times \frac{1\,L\, solution}{0.555\,\cancel{mol\, CaCl_{2}}}=1.13\, L\, CaCl_{2}$
This can be extended by starting with the mass of one reactant, instead of moles of a reactant.
Example $$\PageIndex{3}$$
What volume of 0.0995 M Al(NO3)3 will react with 3.66 g of Ag according to the following chemical equation?
$\ce{3Ag(s) + Al(NO3)3(aq) → 3AgNO3 + Al(s)}$
Solution
Here, we first must convert the mass of Ag to moles before using the balanced chemical equation and then the definition of molarity as a conversion factor:
$3.66\cancel{g\, Ag}\times \frac{1\cancel{mol\, Ag}}{107.97\,\cancel{g\, Ag}}\times \frac{1\,\cancel{mol\, Al(NO_{3})_{3}}}{3\,\cancel{mol\, Ag}}\times \frac{1\,L\, solution}{0.0995\,\cancel{mol\, Al(NO_{3})_{3}}}=0.114\, L$
The strikeouts show how the units cancel.
Exercise $$\PageIndex{3}$$
What volume of 0.512 M NaOH will react with 17.9 g of H2C2O4(s) according to the following chemical equation?
$\ce{H2C2O4(s) + 2NaOH(aq) → Na2C2O4(aq) + 2H2O(ℓ)}$
0.777 L
We can extend our skills even further by recognizing that we can relate quantities of one solution to quantities of another solution. Knowing the volume and concentration of a solution containing one reactant, we can determine how much of another solution of another reactant will be needed using the balanced chemical equation.
Example $$\PageIndex{4}$$
A student takes a precisely measured sample, called an aliquot, of 10.00 mL of a solution of FeCl3. The student carefully adds 0.1074 M Na2C2O4 until all the Fe3+(aq) has precipitated as Fe2(C2O4)3(s). Using a precisely measured tube called a burette, the student finds that 9.04 mL of the Na2C2O4 solution was added to completely precipitate the Fe3+(aq). What was the concentration of the FeCl3 in the original solution? (A precisely measured experiment like this, which is meant to determine the amount of a substance in a sample, is called a titration.) The balanced chemical equation is as follows:
$\ce{2FeCl3(aq) + 3Na2C2O4(aq) → Fe2(C2O4)3(s) + 6NaCl(aq)}$
Solution
First we need to determine the number of moles of Na2C2O4 that reacted. We will convert the volume to liters and then use the concentration of the solution as a conversion factor:
$9.04\cancel{mL}\times \frac{1\cancel{L}}{1000\cancel{mL}}\times \frac{0.1074mol\, Na_{2}C_{2}O_{4}}{\cancel{L}}=0.000971\, mol\, Na_{2}C_{2}O_{4}$
Now we will use the balanced chemical equation to determine the number of moles of Fe3+(aq) that were present in the initial aliquot:
$0.000971\cancel{mol\, Na_{2}C_{2}O_{4}}\times \frac{2\,mol\, FeCl_{3}}{3\cancel{mol\,Na_{2}C_{2}O_{4}}}=0.000647\,mol\, FeCl_{3}$
Then we determine the concentration of FeCl3 in the original solution. Converting 10.00 mL into liters (0.01000 L), we use the definition of molarity directly:
$M=\frac{mol}{L}=\frac{0.000647\,mol\, FeCl_{3}}{0.01000L}=0.0647\,M\, FeCl_{3}$
Exercise $$\PageIndex{4}$$
A student titrates 25.00 mL of H3PO4 with 0.0987 M KOH. She uses 54.06 mL to complete the chemical reaction. What is the concentration of H3PO4?
$\ce{H3PO4(aq) + 3KOH(aq) → K3PO4(aq) + 3H2O}$
0.0711 M
Fig. 11.5.1 Titration © Thinkstock When a student performs a titration, a measured amount of one solution is added to another reactant.
We have used molarity exclusively as the concentration of interest, but that will not always be the case. The next example shows a different concentration unit being used.
Example $$\PageIndex{5}$$:
H2O2 is used to determine the amount of Mn according to this balanced chemical equation:
$\ce{2MnO4^{-}(aq) + 5H2O2(aq) + 6H+(aq) → 2Mn2+(aq) + 5O2(g) + 8H2O(ℓ)}$
What mass of 3.00% m/m H2O2 solution is needed to react with 0.355 mol of MnO4(aq)?
Solution
Because we are given an initial amount in moles, all we need to do is use the balanced chemical equation to determine the number of moles of H2O2 and then convert to find the mass of H2O2. Knowing that the H2O2 solution is 3.00% by mass, we can determine the mass of solution needed:
$3.55\cancel{mol\, MnO_{4}^{-}}\times \frac{5\cancel{mol\, H_{2}O_{2}}}{2\cancel{mol\,MnO_{4}^{-}}}\times \frac{34.02\cancel{g\, H_{2}O_{2}}}{\cancel{mol\, H_{2}O_{2}}}\times \frac{100\,g\; solution}{3\cancel{g\, H_{2}O_{2}}}=1006\,g\; sol$
The first conversion factor comes from the balanced chemical equation, the second conversion factor is the molar mass of H2O2, and the third conversion factor comes from the definition of percentage concentration by mass.
Exercise $$\PageIndex{5}$$
Use the balanced chemical reaction for MnO4 and H2O2 to determine what mass of O2 is produced if 258 g of 3.00% m/m H2O2 is reacted with MnO4.
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Review question
# For which $x$-values does a function equal its inverse? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R6735
## Solution
A function $f$ is defined by $f:x\rightarrow \dfrac{1}{x+1}$. Write down in similar form expressions for $f^{-1}$
To find the inverse of a function we can write $y=f(x)$ and rearrange to make $x$ the subject. This gives
\begin{align*} y &= \dfrac{1}{x+1}\\ x+1 &= \dfrac{1}{y}\\ x &= \dfrac{1}{y} - 1\\ &= \dfrac{1-y}{y} \end{align*}
Replacing $y$ with $x$ in the right hand side, we obtain the inverse function $f^{-1}:x\rightarrow\frac{1-x}{x}.$
… and $ff$.
The composite function of $f$ with itself can be found by ‘replacing the $x$ in $f(x)$ with $f(x)$’. This gives $f\bigl( f(x) \bigr) = \dfrac{1}{ \left( \dfrac{1}{x+1} \right)+1 }.$
We can now simplify by multiplying top and bottom by $x+1$, yielding
\begin{align*} ff(x) &= \dfrac{1}{ \left( \dfrac{1}{x+1} \right)+1 } \times \dfrac{x+1}{x+1} \\ &= \dfrac{x+1}{1 + (x + 1)} \\ &= \dfrac{x+1}{x+2}. \end{align*}
Our solution then reads $ff:x\rightarrow \frac{x+1}{x+2}.$
It is required to find the values of $x$ for which $(i)$ $f=f^{-1}$, $(ii)$ $f=ff$.
Show that, in each case, the values of $x$ are given by the equation $x^2+x-1=0.$
Using the result from above, we set $f(x) = f^{-1}(x)$, giving
\begin{align*} \dfrac{1}{x+1}&=\dfrac{1-x}{x}\\ 1&=\dfrac{(1-x)(1+x)}{x}\\ x&=1-x^2\\ x^2+x-1&=0 \end{align*}
as required.
Similarly, setting $f(x) = ff(x)$ we have
\begin{align*} \dfrac{1}{x+1} &= \dfrac{x+1}{x+2}\\ 1&=\dfrac{(x+1)^2}{x+2}\\ x+2 &= (x+1)^2\\ x+2 &=x^2+2x+1\\ x^2+x-1&=0. \end{align*}
The question does not require us to solve this equation for $x$, but we could do so using the quadratic formula. We have
\begin{align*} x^2+x-1&=0\\ x &= \dfrac{-1 \pm \sqrt{1^2 - 4 \times 1 \times -1}}{2 \times 1} \\ &= \dfrac{-1 \pm \sqrt{5}}{2} . \end{align*}
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# Violet has received a special order for 110 units of its product. The product normally sells for ...
Violet has received a special order for 110 units of its product. The product normally sells for $2,400 and has the following manufacturing costs: Per unit Direct materials$ 640 Direct labor 340 Variable manufacturing overhead 440 Fixed manufacturing overhead 690 Unit cost $2,110 Assume that Violet has sufficient capacity to fill the order without harming normal production and sales. What minimum price should Violet charge to achieve a$19,800 incremental profit? $1,420$2,110 $1,730$1,600
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# 3.4: Indirect Proofs
Instead of proving $$p \Rightarrow q$$ directly, it is sometimes easier to prove it indirectly. There are two kinds of indirect proofs: proof by contrapositive, and proof by contradiction.
## Proof by Contrapositive
Proof by contrapositive is based on the fact that an implication is equivalent to its contrapositive. Therefore, instead of proving $$p \Rightarrow q$$, we may prove its contrapositive $$\overline{q} \Rightarrow \overline{p}$$. Since it is an implication, we could use a direct proof:
Assume $$\overline{q}$$ is true (hence, assume $$q$$ is false).
Show that $$\overline{p}$$ is true (that is, show that $$p$$ is false).
The proof may proceed as follow:
Prove: $$p \Rightarrow q$$
Proof: We will prove the contrapositive of the stated result.
That is, we will prove $$\overline{q} \Rightarrow \overline{p}$$.
Assume $$q$$ is false, . . .
.
.
.
. . . Therefore $$p$$ is false.
Thus $$\overline{q} \Rightarrow \overline{p}$$.
Therefore, by contraposition, $$p \Rightarrow q.$$
Lemma $$\PageIndex{1}$$
Let $$n$$ be an integer. Show that if $$n^2$$ is even, then $$n$$ is also even.
Proof:
Proof by contrapositive: We want to prove that if $$n$$ is odd, then $$n^2$$ is odd. Let $$n$$ be an odd integer, then $$n=2t+1$$ for some integer $$t$$ by definition of odd. By algebra $n^2 = 4t^2+4t+1 = 2(2t^2+2t)+1.$ Since $$\mathbb{Z}$$ is closed under addition & multiplication, $$2t^2+2t$$ is an integer. Hence $$n^2$$ is odd by definition of odd.
Thus if $$n$$ is odd, then $$n^2$$ is odd.
Therefore, by contraposition, for all integers $$n$$ if $$n^2$$ is even, then $$n$$ is even.
Note: Lemma $$\PageIndex{1}$$ will be used in the proof that $$\sqrt{2}$$ is irrational, later in this section.
Example $$\PageIndex{1}\label{eg:indirectpf-01}$$
Show that if $$n$$ is a positive integer such that the sum of its positive divisors is $$n+1$$, then $$n$$ is prime.
Solution
We shall prove the contrapositive of the given statement. We want to prove that if $$n$$ is composite, then the sum of its positive divisors is not $$n+1$$. Let $$n$$ be a composite number. Then its divisors include 1, $$n$$, and at least one other positive divisor $$x$$ different from 1 and $$n$$. So the sum of its positive divisors is at least $$1+n+x$$. Since $$x$$ is positive, we gather that $1+n+x > 1+n.$ We deduce that the sum of the divisors cannot be $$n+1$$. Therefore, if the sum of the divisors of $$n$$ is precisely $$n+1$$, then $$n$$ must be prime.
## Proof by Contradiction
Another indirect proof is proof by contradiction. To prove that $$p \Rightarrow q$$, we proceed as follows:
Suppose $$p\Rightarrow q$$ is false; that is, assume that $$p$$ is true and $$q$$ is false.
Argue until we obtain a contradiction, which could be any result that we know is false.
How does this prove that $$p\Rightarrow q$$? Assuming that the logic used in every step in the argument is correct, yet we still end up with a contradiction, then the only possible flaw must come from the supposition that $$p\Rightarrow q$$ is false. Consequently, $$p\Rightarrow q$$ must be true.
This is what a typical proof by contradiction may look like:
Proof: Suppose not. That is, suppose $$p$$ is true and $$q$$ is false. Then
. . .
.
.
a contradiction!!
Thus our assumption that $$p$$ is true and $$q$$ is false cannot be true.
Therefore, $$p \Rightarrow q$$ must be true.
There is a more general form for proving a statement $$r$$, which needs not be an implication. To prove the proposition $$r$$ by contradiction, we follow these steps:
Suppose $$r$$ is false.
Argue until we obtain a contradiction.
Proof: Suppose not. That is, suppose $$r$$ is false. Then . . .
.
.
a contradiction!!
Thus our assumption that $$r$$ is false cannot be true.
Therefore, $$r$$ must be true.
Example $$\PageIndex{2}\label{eg:indirectpf-02}$$
Show that if $$P$$ is a point not on a line $$L$$, then there exists exactly one perpendicular line from $$P$$ onto $$L$$.
Solution
Suppose we can find more than one perpendicular line from $$P$$ onto $$L$$. Pick any two of them, and denote their intersections with $$L$$ as $$Q$$ and $$R$$. Then we have a triangle $$PQR$$, where the angles $$PQR$$ and $$PRQ$$ are both $$90^\circ$$. This implies that the sum of the interior angles of the triangle $$PQR$$ exceeds $$180^\circ$$, which is impossible. Hence, there is only one perpendicular line from $$P$$ onto $$L$$.
Example $$\PageIndex{3}\label{eg:indirectpf-03}$$
Show that if $$x^2<5$$, then $$|x|<\sqrt{5}$$.
note: if no set of numbers is specified, the default is the set of real numbers.
Solution
Assume $$x^2<5$$. We want to show that $$|x|<\sqrt{5}$$. Suppose, on the contrary, we have $$|x|\geq\sqrt{5}$$.
By definition, $$|x|=x$$ or $$|x|=-x$$.
So either $$x\geq\sqrt{5}$$, or $$-x\geq\sqrt{5}$$.
The second case, $$-x\geq\sqrt{5}$$ is the same as $$x\leq-\sqrt{5}$$ (by multiplying both sides by negative 1).
If $$x\geq\sqrt{5}$$, then $$x^2\geq5$$, by algebra; note: since x is a positive number the inequality sign does not change.
If $$x\leq-\sqrt{5}$$, we again have $$x^2\geq5$$, by algebra; note: since x is a negative number the inequality sign reverses.
In either case, we have both $$x^2\geq5$$ and $$x^2<5$$ which is a contradiction.
Hence $$|x|<\sqrt{5}$$.
$$\therefore$$ if $$x^2<5$$, then $$|x|<\sqrt{5}$$.
hands-on exercise $$\PageIndex{1}\label{he:indirectpf-01}$$
Prove that if $$x^2\geq49$$, then $$|x|\geq7$$.
Example $$\PageIndex{4}\label{eg:indirectpf-04}$$
Prove $[(p\Rightarrow q) \wedge p] \Rightarrow q$ is a tautology.
Solution
Suppose $$[(p\Rightarrow q) \wedge p] \Rightarrow q$$ is false for some statements $$p$$ and $$q$$. Then we find
• $$(p\Rightarrow q) \wedge p$$ is true, and
• $$q$$ is false.
For the conjunction $$(p\Rightarrow q) \wedge p$$ to be true, we need
• $$p\Rightarrow q$$ to be true, and
• $$p$$ to be true.
Having $$p$$ true and $$p\Rightarrow q$$ true, we must have $$q$$ true. That gives us $$q$$ is true and $$q$$ is false, a contradiction! Thus it cannot be that $$[(p\Rightarrow q) \wedge p] \Rightarrow q$$ is false. Therefore, $$[(p\Rightarrow q) \wedge p] \Rightarrow q$$ is always true, hence it is a tautology.
Example $$\PageIndex{5}\label{eg:indirectpf-05}$$
Prove, by contradiction, that if $$x$$ is rational and $$y$$ is irrational, then $$x+y$$ is irrational.
Solution
Let $$x$$ be a rational number and $$y$$ an irrational number. We want to show that $$x+y$$ is irrational. Suppose, on the contrary, that $$x+y$$ is rational. Then $x+y = \frac{m}{n}$ for some integers $$m$$ and $$n$$, where $$n\neq0$$ by definition of rational. Since $$x$$ is rational, we also have $x = \frac{p}{q}$ for some integers $$p$$ and $$q$$, where $$q\neq0$$ by defintion of rational. It follows by substitution that $\frac{m}{n} = x+y = \frac{p}{q} + y.$ Hence by algebra, $y = \frac{m}{n}-\frac{p}{q} = \frac{mq-np}{nq},$ where $$mq-np$$ and $$nq$$ are both integers because $$\mathbb{Z}$$ is closed under addition and multiplication. Also $$nq\neq0$$ by the Zero Product Property. This makes $$y$$ rational by definition of rational. Now we have $$y$$ is rational and $$y$$ is irrational (by assumption). This is a contradiction! Thus, $$x+y$$ cannot be rational, it must be irrational.
$$\therefore$$ if $$x$$ is rational and $$y$$ is irrational, then $$x+y$$ is irrational.
hands-on exercise $$\PageIndex{2}\label{he:indirectpf-02}$$
Prove that $\sqrt{x+y} \neq \sqrt{x}+\sqrt{y}$ for any positive real numbers $$x$$ and $$y$$.
Hint
The words “for any” suggest this is a universal quantification. Be sure you negate the problem statement properly.
Lemma $$\PageIndex{2}$$
We will use this lemma (along with Lemma 3.4.1) for the proof that $$\sqrt{2}$$ is irrational.
Lemma 3.4.2 Given a rational number, x, x can be written as a fraction $$\frac{m}{n}$$ where $$m,n \in \mathbb {Z}$$ , $$n \neq 0$$ and $$\frac{m}{n}$$ is in lowest terms.
Proof:
Given a rational number, x, x can be written as a fraction $$\frac{a}{b}$$ where $$a,b \in \mathbb {Z}$$ ,$$b \neq 0$$ by definition of rational number.
If $$\frac{a}{b}$$ is not in lowest terms, then $$a$$ and $$b$$ have a common factor. Divide out that common factor to get an equivalent fraction, $$\frac{c}{d}.$$
If $$\frac{c}{d}$$ is not in lowest terms, then $$c$$ and $$d$$ have a common factor. Divide out that common factor to get an equivalent fraction, $$\frac{f}{g}.$$
If $$\frac{f}{g}$$ is not in lowest terms, then $$f$$ and $$g$$ have a common factor. Divide out that common factor to get an equivalent fraction, $$\frac{j}{k}.$$
Continue this process until the numerator and denominator do not have any common factors. Rename the numerator as $$m$$ and the denominator as $$n.$$
Now $$x=\frac{m}{n}$$ and $$\frac{m}{n}$$ is in lowest terms.
$$\therefore$$ a rational number can be written as a fraction in lowest terms.
## The $$\sqrt{2}$$ is irrational.
Prove that $$\sqrt{2}$$ is irrational.
Proof:
Suppose, on the contrary, $$\sqrt{2}$$ is rational. Then we can write $\sqrt{2} = \frac{m}{n}$ for some positive integers $$m$$ and $$n$$ such that $$m$$ and $$n$$ do not share any common divisor except 1 (hence $$\frac{m}{n}$$ is in lowest terms) by Lemma 3.4.2. Squaring both sides and cross-multiplying yields $2n^2 = m^2.$ Since $$\mathbb{Z}$$ are closed under multiplication, $$n^2$$ is an integer and thus $$m^2$$ is even by the definition of even. Consequently, by Lemma 3.4.1, $$m$$ is also even. Then we can write $$m=2s$$ for some integer $$s$$ by the definition of even. By substitution and algebra, the equation above becomes $2n^2 = m^2 = (2s)^2 = 4s^2.$ Hence, $n^2 = 2s^2.$ Since $$\mathbb{Z}$$ are closed under multiplication, $$s^2$$ is an integer and thus $$n^2$$ is even by the definition of even. Consequently, by Lemma 3.4.1, $$n$$ is also even. Even numbers are divisible by 2, by the definition of divides. We have shown that both $$m$$ and $$n$$ are divisible by 2. This contradicts the assumption that $$m$$ and $$n$$ do not share any common divisor. Thus is is not possible for $$\sqrt{2}$$ to be rational.
Therefore, $$\sqrt{2}$$ must be irrational.
hands-on exercise $$\PageIndex{3}\label{he:indirectpf-03}$$
Prove that $$\sqrt{3}$$ is irrational.
Very often, a proof by contradiction can be rephrased into a proof by contrapositive or even a direct proof, both of which are easier to follow. If this is the case, rewrite the proof.
Example $$\PageIndex{6}\label{eg:indirectpf-6}$$
Show that $$x^2+4x+6=0$$ has no real solution. In symbols, show that $$\nexists x\in\mathbb{R},(x^2+4x+6=0)$$.
Solution
Consider the following proof by contradiction:
Suppose there exists a real number $$x$$ such that $$x^2+4x+6=0$$.
Using calculus, it can be shown that the function $$f(x)=x^2+4x+6$$
has an absolute minimum at $$x=-2$$. **Need to show those calculus steps.**
Thus, $$f(x) \geq f(-2) = 2$$ for
any $$x$$. This contradicts the assumption that there exists an $$x$$
such that $$x^2+4x+6=0$$. Thus, $$x^2+4x+6=0$$ has no real solution.
A close inspection reveals that we do not really need a proof by contradiction. The crux of the proof is the fact that $$x^2+4x+6 \geq 2$$ for all $$x$$. This already shows that $$x^2+4x+6$$ could never be zero. It is easier to use a direct proof, as follows.
Using calculus, we see $$f'(x)=2x+4$$. Setting $$f'(x)=0$$ we get $$x=-2$$. Since $$f"(x)=2$$, thus $$f"(-2)=2$$,we find that the function $$f(x)=x^2+4x+6$$ has an
absolute minimum at $$x=-2$$. Therefore, for any $$x$$, we always have
$$f(x) \geq f(-2) = 2$$. Hence, there does not exist any $$x$$ such that
$$x^2+4x+6=0$$.
Do you agree that the second proof (the direct proof) is more elegant?
## Proving a Biconditional Statement
Recall that a biconditional statement $$p\Leftrightarrow q$$ consists of two implications $$p\Rightarrow q$$ and $$q\Rightarrow p$$. Hence, to prove $$p\Leftrightarrow q$$, we need to establish these two “directions” separately.
Example $$\PageIndex{7}\label{eg:indirectpf-7}$$
Let $$n$$ be an integer. Prove that $$n^2$$ is even if and only if $$n$$ is even.
Solution
($$\Rightarrow$$) We first prove that if $$n^2$$ is even, then $$n$$ must be even.
We shall prove its contrapositive: if $$n$$ is odd, then $$n^2$$ is odd. If $$n$$ is odd, then we can write $$n=2t+1$$ for some integer $$t$$ by definition of odd. Then by algebra $n^2 = (2t+1)^2 = 4t^2+4t+1 = 2(2t^2+2t)+1,$ where $$2t^2+2t$$ is an integer since $$\mathbb{Z}$$ is closed under addition and multiplication. Thus, $$n^2$$ is odd. So, if $$n$$ is odd, then $$n^2$$ is odd. By contraposition, if $$n^2$$ is even, then $$n$$ is even.
($$\Leftarrow$$) Next, we prove that if $$n$$ is even, then $$n^2$$ is even.
If $$n$$ is even, we can write $$n=2t$$ for some integer $$t$$ by definition of even. Then $n^2 = (2t)^2 = 4t^2 = 2\cdot 2t^2,$ where $$2t^2$$ is an integer since $$\mathbb{Z}$$ is closed under multiplication. Hence, $$n^2$$ is even. if $$n$$ is even, then $$n^2$$ is even.
$$\therefore \, n^2$$ is even if and only if $$n$$ is even.
hands-on exercise $$\PageIndex{4}\label{he:indirectpf-04}$$
Let $$n$$ be an integer. Prove that $$n$$ is odd if and only if $$n^2$$ is odd.
## Summary and Review
• We can use indirect proofs to prove an implication.
• There are two kinds of indirect proofs: proof by contrapositive and proof by contradiction.
• In a proof by contrapositive, we actually use a direct proof to prove the contrapositive of the original implication.
• In a proof by contradiction, we start with the supposition that the implication is false, and use this assumption to derive a contradiction. This would prove that the implication must be true.
• A proof by contradiction can also be used to prove a statement that is not of the form of an implication. We start with the supposition that the statement is false, and use this assumption to derive a contradiction. This would prove that the statement must be true.
• Sometimes a proof by contradiction can be rewritten as a direct proof. If so, the direct proof is the more direct way to write the proof.
## Exercises
exercise $$\PageIndex{1}\label{ex:indirectpf-01}$$
Let $$n$$ be an integer. Prove that if $$n^2$$ is even, then $$n$$ must be even. Use
(a) A proof by contrapositive (this one is done - see proof of Lemma 3.4.1)
(b) A proof by contradiction.
Remark
The two proofs are very similar, but the wording is slightly different, so be sure you present your proof clearly.
exercise $$\PageIndex{2}\label{ex:indirectpf-02}$$
Let $$n$$ be an integer. Prove that if $$n^2$$ is a multiple of 3, then $$n$$ must also be a multiple of 3. Use
(a) A proof by contrapositive.
(b) A proof by contradiction.
exercise $$\PageIndex{3}\label{ex:indirectpf-03}$$
Let $$n$$ be an integer. Prove that if $$n$$ is even, then $$n^2=4s$$ for some integer $$s$$.
exercise $$\PageIndex{4}\label{ex:indirectpf-04}$$
Let $$m$$ and $$n$$ be integers. Show that $$mn=1$$ implies that $$m=1$$ or $$m=-1$$.
exercise $$\PageIndex{5}\label{ex:indirectpf-05}$$
Let $$x$$ be a real number. Prove by contrapositive: if $$x$$ is irrational, then $$\sqrt{x}$$ is irrational. Apply this result to show that $$\sqrt[4]{2}$$ is irrational, using the assumption that $$\sqrt{2}$$ is irrational.
exercise $$\PageIndex{6}\label{ex:indirectpf-06}$$
Let $$x$$ and $$y$$ be real numbers such that $$x\neq0$$. Prove that if $$x$$ is rational, and $$y$$ is irrational, then $$xy$$ is irrational.
exercise $$\PageIndex{7}\label{ex:indirectpf-07}$$
Prove that $$\sqrt{5}$$ is irrational.
exercise $$\PageIndex{8}\label{ex:indirectpf-08}$$
Prove that $$\sqrt[3]{2}$$ is irrational.
exercise $$\PageIndex{9}\label{ex:indirectpf-09}$$
Let $$a$$ and $$b$$ be real numbers. Prove that if $$a\neq b$$, then $$a^2+b^2 \neq 2ab$$.
exercise $$\PageIndex{10}\label{ex:indirectpf-10}$$
Use contradiction to prove that, for all integers $$k\geq1$$, $2\sqrt{k+1} + \frac{1}{\sqrt{k+1}} \geq 2\sqrt{k+2}.$
exercise $$\PageIndex{11}\label{ex:indirectpf-11}$$
Let $$m$$ and $$n$$ be integers. Prove that $$mn$$ is even if and only if $$m$$ is even or $$n$$ is even.
exercise $$\PageIndex{12}\label{ex:indirectpf-12}$$
Let $$x$$ and $$y$$ be real numbers. Prove that $$x^2+y^2=0$$ if and only if $$x=0$$ and $$y=0$$.
exercise $$\PageIndex{13}\label{ex:indirectpf-13}$$
Prove that, if $$x$$ is a real number such that $$0<x<1$$, then $$x(1-x)\leq\frac{1}{4}$$.
exercise $$\PageIndex{14}\label{ex:indirectpf-14}$$
Let $$m$$ and $$n$$ be positive integers such that 3 divides $$mn$$. Prove that 3 divides $$m$$, or 3 divides $$n$$.
exercise $$\PageIndex{15}\label{ex:indirectpf-15}$$
Prove that the logical formula $(p\Rightarrow q) \vee (p\Rightarrow \overline{q})$ is a tautology.
(See example 3.4.4.)
exercise $$\PageIndex{16}\label{ex:indirectpf-16}$$
Prove that the logical formula $[(p\Rightarrow q) \wedge (p\Rightarrow \overline{q})] \Rightarrow \overline{p}$ is a tautology.
(See example 3.4.4.)
This page titled 3.4: Indirect Proofs is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Harris Kwong (OpenSUNY) .
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# HSPICE -mp -mt options for Ubuntu
I was reading part of the hspice tutorial, I found that, there is a multiprocessing option:
"hspice -mp [process_count] -mt [thread_count] -i input.sp -o output_file" "Note: Running both MP and MT simultaneously is limited only to LINUX."
Is there anyway to use those options on Ubuntu system? I tried it, I got "Syntax error: Bad fd number".
• What was the exact command you have run? This is a generic linux error when tying redirect the IO stream in a shell ubuntuforums.org/showthread.php?t=1460684 – Eugene Sh. Nov 6 '17 at 18:51
• @EugeneSh. Yes, when I was searching it, I noticed many people faced it when they were doing other simulations. I ran this option on linux and Ubunto system. The Ubuntu system gave me this error and linux was fine. Therefore, I thought maybe Ubuntu is the reason. I used: "hspice -mp -i xxx.sp" – Shabnam Nov 6 '17 at 20:04
sudo rm /bin/sh
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# Rindler metric and proper acceleration.
1. Apr 9, 2013
### Scherejg
1. The problem statement, all variables and given/known data
Show that observers whose world-lines are the curves along which only
ξ varies undergo constant proper acceleration with invariant magnitude
ημν(duμ/dτ )(duν/dτ ) = 1/χ2.
2. Relevant equations
ds2 = −a2χ2dξ2 + dχ2 + dy2 + dz2
xμ = {ξ, χ, y, z}
t = χsinh(aξ)
x =χcosh(aξ)
3. The attempt at a solution
The trajectory of this particle in the rest frame would be:
xκ={χsinh(aξ),χcosh(aξ),0,0} as a function of ξ.
dxκ/ dξ = {aχcosh(aξ), aχsinh(aξ),0,0} Velocity of xκ.
Since I want proper time, I would expand dxκ/ dξ into (dxκ/ dξ)*(dξ/dτ)
So I need to define dξ/dτ. For this I would use the metric:
letting ds^2= -dτ^2
dτ/dξ= aχ
γ= dξ/dτ = 1/aχ
The acceleration is then: γ^2{a^2χsinh(aξ),a^2χcosh(aξ),0,0} = d2xk/dτ^2
ημν(duμ/dτ )(duν/dτ ) = γ^4(-1*a^4χ^2sinh^2(aξ) + 1*a^4χ^2cosh^2(aξ))
Which cancels out to: 1/χ^2. I just wanted to make sure that although the math works out, my reasoning is okay.
Where I took ds^2= -dτ^2 my first thought was that dτ^2= (a^2)(χ^2)(dξ)^2 - (V(ξ))^2(dL)^2. I realized that dropping the velocity term would allow me to get the right answer.. so it must = 0. Since the velocity itself is not actually zero I would be inclined to guess that dividing the equation by (dξ)^2 would leave (dL)^2/(dξ)^2 and that must equal zero?
2. Apr 10, 2013
### TSny
Since only ξ changes, that means dχ = dy = dz = 0. I think that's why the "velocity term" is zero. That is,
ds2 = −a2χ22 + dχ2+ dy2 + dz2 = −a2χ22.
So, dτ2 = a2χ22
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# htmltable2latex perl package
This small package provides a mechanism to translate tables in html markup (or better xml format with html table tags) into a corresponding LaTeX code. Not all features of html tables are supported, but basic tables should work fine.
# tbl2ltx($html,$encoding)
The function you will use to convert html to LaTeX. The first parameter is a string containing the xml/html code. This can contain any number of <table> tags. Please note that:
1) The tags are parsed with XML::Parser. This means they have to be well formed xml code.
2) XML::Parser expects a single (global) tag starting. So if you have multiple tables in your $html string wrap them into some arbitrary tag. The corresponding LaTeX code is returned as string. The parser ignores any text not included in <td> cells or outside of the <table> tags. The border'' attribute of the <table> tag is honoured. It is is not 0 a border is drawn in the LaTeX code. The colspan'' attribute of <td> and <th> cells is honoured. Valid values for$encoding are: qw(UTF-8 ISO-8859-1 UTF-16 US-ASCII) or any other you may have configured for the expat xml parser. See man XML::Parser'' for more details.
# Bugs
The parser will only recognize the <table> attribute border'' if it is written in lowercase.
The border of Multicolumn cells is not rendered correct.
The following tags of HTML tables are not supported: <caption> <colgroup> <col>.
# Example
#!/usr/bin/env perl
$html=<<EOF; <html> <table width="337" border="1" style="height: 75px"> <tbody> <tr><td></td><th>In the secönd Column<br /></th></tr> <tr><td colspan="2">More äction in row2 col1</td></tr> </tbody> </table> <table> <tfoot><tr><td>The foot</td></tr></tfoot> <tr><td>Main Part</td><td>second col</td></tr> </table> </html> EOF use htmltable2latex; print tbl2ltx($html,'ISO-8859-1');
will output:
\begin{tabular}{|l|l|}
\hline \\
& \textbf{In the secönd Column} \\
\multicolumn{2}{l}{More äction in row2 col1} \\
\hline
\end{tabular}
\begin{tabular}{ll}
Main Part & second col \\
The foot
\end{tabular}
# Version
Version 1.0 released 2006-01-27
http://llg.cubic.org/tools/htmltable2latex.html
Copyright (c) 2006 Dirk Jagdmann <[email protected]>
This software is provided 'as-is', without any express or implied warranty. In no event will the authors be held liable for any damages arising from the use of this software.
Permission is granted to anyone to use this software for any purpose, including commercial applications, and to alter it and redistribute it freely, subject to the following restrictions:
1. The origin of this software must not be misrepresented; you
must not claim that you wrote the original software. If you use
this software in a product, an acknowledgment in the product
documentation would be appreciated but is not required.
2. Altered source versions must be plainly marked as such, and
must not be misrepresented as being the original software.
3. This notice may not be removed or altered from any source
distribution.
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# Death To Freebees; Or, Freebees Für Alles
Saturday, April 2, 2016
By dreeves
News! We scrapped the overcomplicated concept of Freebees. But don’t panic! By scrapping it we mean that you don’t need to know the term “freebees” or worry about buying them: All goals can now be created with an initial pledge of $0. Can we get a hallelujah? (If you want to cap your pledge at$0, you still need a... »
# What Is Willpower?
Monday, March 21, 2016
By dreeves
Our previous post, “Ego Depletion Depletion,” generated a lot of discussion and I found I was contradicting myself on the question of what willpower is exactly. First a recap, hopefully in plainer English, about what all the fuss is about. A big finding in... »
# Ego Depletion Depletion
Wednesday, March 9, 2016
By dreeves
This is crossposted on Mark Forster’s Get Everything Done blog. The big news in psychology this week is that Baumeister’s Ego Depletion model is bunk. At least it has failed to »
# MongoMapper to Mongoid; Or, Breaking All The Things
Thursday, February 25, 2016
By dreeves
Anyone remember our old blog post from 2012 about accidentally running a query that started deleting our whole database? It’s pretty entertaining and helpfully demarcates the parts that non-nerds should skip. If you’re a non-nerd I’d stick with (the non-nerd parts of) that post. The executive summary of this post... »
# Revealed Preference
Monday, February 15, 2016
By dreeves
The doctrine of revealed preference — that you can infer someone’s utility function based wholly on what they choose to do — has an illustrious history. John Locke said “the actions of men are the best interpreters of their thoughts.” And Ludwig von Mises »
# Post New Year’s Press Roundup
Friday, February 5, 2016
By dreeves
Another month (or so), another swarm of Beeminder buzz. Since new year’s resolution season just ended we got included in a lot of lists (resources for succeeding in college, resources for MBA students, apps to stay motivated in 2016, »
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## Sunday, 10 July 2011
### How much would you pay for the universe?
Well I was going to write a post of the end of an era with the last shuttle fight of the Atlantis. But then I remembered this video I found a while ago and this says everything that needs to be said not that I'm American (as many of his points are based towards Americans and there pride as a nation) but it's still a sad day for space exploration none the less.
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Share
# Explain the Following in Term Reduction of Gain Or Loss of Oxygen with Two Example. - CBSE Class 10 - Science
ConceptChemical Reactions Types of Chemical Reactions - Oxidation and Reduction
#### Question
Explain the following in term Reduction of gain or loss of oxygen with two example.
#### Solution 1
Reduction Reaction: It is a chemical reaction in which loss of oxygen or gain of hydrogen takes place
1)CuO+H2 "heat"/>Cu+H2O
Here, Copper Oxide is reduced to become Copper
2)ZnO+C "heat"/>Zn+CO
Here, Zinc Oxide is reduced to become Zinc
#### Solution 2
In a chemical reaction, when the oxygen is being removed from the compound then it is said to be reduced. Example:
CuO(s)+H2(g)Cu(s)+H2O(l) 2HgO2Hg+O2
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [2]
Solution Explain the Following in Term Reduction of Gain Or Loss of Oxygen with Two Example. Concept: Chemical Reactions - Types of Chemical Reactions - Oxidation and Reduction.
S
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4-velocity vector!
1. Sep 22, 2005
yukcream
I want to know why the 4-velocity vector of a particle is
u'=dx^i/ds where ds=cdt sqrt{1-v^2/c^2}
not dx^i/dt?
also can anyone show me the devide of the 4-acceleration vector,i.e
a' = d^2x'/ds^2?! I dont konw how to write it back in terms of v,c &x^i !!
In addition why the 4-velocity vector is a dimesnsionless quantity? what does it really mean?
thx for help
yukyuk
2. Sep 22, 2005
robphy
One can show that the quantity $$\frac{dx^i}{dt}$$ doesn't transform correctly under a Lorentz Transformation. Hence, it is not a 4-vector.
The quantity $$u^i=\frac{dx^i}{ds}$$ where $$ds=cdt \sqrt{1-v^2/c^2}$$ does transform correctly.
As you have noticed, this quantity is dimensionless. Thus, $$u^i$$ can be regarded as a certain unit 4-vector, namely, the unit timelike 4-vector tangent to the particle's worldline. This normalization is convenient because it absorbs some pesky factors of c. An important role of this unit timelike 4-vector is that it helps decompose ["write as a vector sum"] tensorial quantities into timelike and spacelike parts ["components"] in the frame of the particle.
The 4-acceleration is $$a^i = \frac{d}{ds}u^i=\frac{d}{ds}\frac{d}{ds} x^i$$. This can also be written as $$a^i=u^a \nabla_a u^i$$. To write this in component form, apply the chain rule to $$a^i =\frac{d}{ds}\frac{d}{ds} x^i$$. For simplicity, you'll want to assume that all of the motion is in the x-direction.
Here are some sites to look at
http://farside.ph.utexas.edu/teaching/em/lectures/node115.html
http://vishnu.mth.uct.ac.za/omei/gr/chap2/node4.html
Last edited by a moderator: Apr 21, 2017
3. Sep 22, 2005
Trilairian
Making it unitless is merely someones convention. Not everyone does that. The displacement four-vector $$dx^{\mu }$$Is a rank one tensor, a four-vector and in fact the way dx transforms is usually taken as the definition for what constitutes a tensor. $$ds=cd\tau$$ and ds is obviously invariant and so in that sence so is $$d\tau$$ and tensor multiplied or divided by an invariant remains a tensor and so defining $$U^{\mu }$$ by $$U^{\mu } = dx^{\mu} /d\tau$$ or dx/ds in your case guarantees that you have defined a four-vector. This is then named the velocity four-vector, or 4-velocity vector in your terms. dx/dt is not a four-vector simply because it does not transform like a tensor. The covariant derivative of a tensor is also a tensor and so $$A^{\mu } = DU^{\mu }/d\tau$$ defines a four-vector called the acceleration four-vector, or in your terms the 4-acceleration vector. For intertial frames in special relativity this can be written $$A^{\mu } = d^{2}x^{\mu }/d\tau ^2$$ To write it in terms of v, $$A^{\mu } = (\gamma d/dt)(\gamma dx^{\mu }/dt)$$
$$A^{\mu } = \gamma (\gamma (d^{2}x^{\mu }/dt^{2}) + (dx^{\mu }/dt)(d\gamma /dv)(dv/dt))$$
$$A^{\mu } = \gamma (\gamma (d^2x^{\mu }/dt^{2}) + (dx^{\mu }/dt)((v/c^{2})(\gamma ^{3}))dv/dt)$$
$$A^{\mu } = \gamma ^{2}(a^{\mu } + \gamma ^{2}v^{\mu }((v/c^{2})dv/dt)$$
$$A^{\mu } = \gamma ^{2}(a^{\mu } + \gamma ^{2}v^{\mu }(\mathbf{v}\cdot \mathbf{a}/c^{2}))$$
(Im choosing $$\mathbf{v}\cdot \mathbf{a}$$ to represent the ordinary three component dot product of coordinate velocity and coordinate acceleration and using v and a to represent those even when I choose to give them a fourth element as indicated by the greek index.)
Last edited: Sep 22, 2005
4. Sep 22, 2005
pervect
Staff Emeritus
ds, usually written as $d \tau$, is the proper time of the particle.
The reason that the 4-velocity is the derivative of distance with respect to proper time is to ensure that the 4-velocity is a 4-vector.
4-vectors must have an invariant Lorentz interval. Taking the derivative of (ct,x,y,z) with respect to coordiante time t does not give an invariant lorentz interval, it gives a coordinate dependent quantity. (x,y,z,t) is a 4-vector because it's Lorentz interval is invariant for all observers. To maintain this Lorentz invariance, we have to differentiate with a Lorentz invariant time. The only time available that fits this bill is the proper time of the particle.
I'm usually used to assuming c=1 when dealing with 4-vectors, so I'll let someone else explain the conventions usually used when c is not assumed to be 1.
Note that the "length", i.e. the Lorentz interval, of the 4-velocity is a constant. This implies that the 4-velocity and the 4-acceleration are always perpendicular.
You've already written down the expression for 4-acceleration
$$\left( \frac{d^2 t}{d \tau^2}, \frac{d^2 x}{d \tau^2}, \frac{d^2 y}{d \tau^2}, \frac{d^2 z}{d \tau^2}\right)$$
(modulo some factors of 'c'), so I'm not sure what your question is. Here you write t,x,y,z as functions of $\tau.$. If instead you have x(t), y(t), z(t), you need to find $\tau(t)$, which you can do from the definition of the Lorentz interval, then you need to change varabiles to find x,y, and z in terms of tau, rather than t.
5. Sep 22, 2005
pmb_phy
yukcream - In special relatity should keep in mind that is must be possible to express all of the mathematical quanties used in laws of nature which does not depend on the form of coordinate system chosen. In order for this to work properly each component must have the same dimension.
That the 4-vecocity is dimensionless is a matter of choice and nothing else.
Pete
Last edited: Sep 22, 2005
6. Sep 22, 2005
yukcream
To pervect:
My problem is I don't know how to write a^i as the vector form, like
{gamma, vgamma} for u^i, the 4-velocity vectors!
where gamma= 1/sqrt[1-v^2/c^2]
To Trilairian:
Thank you very much for showing me the devide step by step! I think the dt there is represent the proper time, right?
One more question, does Trilairian's devide give me the spatial component of the 4-acceleration vextor? How about the temporal component?
P.S: Thanks for all who answered my question!
yukyuk
7. Sep 22, 2005
Trilairian
You're welcome. The $$\tau$$ refers to proper time. The t refers to coordinate time(your coordinate's time).
It gives both. For the temporal component plug in $$\mu = 0$$ and use $$a^{0} = 0$$, and $$v^{0} = c$$. (But take note that it seems your instruction uses c = 1, whereas I make no such demand. Such trivial differences in definitions tend to vary from one author to another.)
Last edited: Sep 22, 2005
8. Sep 22, 2005
pervect
Staff Emeritus
Let us suppose you have a particle.
You plot the coordinates of the particle as a function of it's proper time, $\tau$.
You do this by specifying four functions
$\left( t(\tau), x(\tau), y(\tau), z(\tau) \right)$
Then you can easily compute it's 4-velocity, which is the derivative of the above expression with respect to tau, and it's 4-acceleration, which is the second derivative of the above expression with respect to tau, i.e.
4 velocity $$u^i = \frac{d x^i}{d \tau}$$
4 acceleration $$a^i = \frac{d^2 x^i}{d \tau^2}$$
(x(t), y(t), z(t)), rather than the above 4 functions.
You then have to compute $\tau$ by the relationship
$$d\tau^2 = dt^2 - dx^2 - dy^2 - dz^2$$
(This is assuming a flat metric - if you are doing general relativity, you havea to use the more general expression
$$d\tau^2 = g_{ab} dx^a dx^b$$
where you sum over a=0..3 and b=0..3)
It may be worthwile to re-write the above expression to explicitly solve for tau in a Miknowskian space-time, one with a global Lorentz metric:
$$\tau = \int \sqrt{1 - \left( \frac{dx}{dt} \right)^2 - \left( \frac{dy}{dt} \right)^2 - \left( \frac{dz}{dt} \right)^2 } dt$$
This gives you $\tau(t)$. You then have to invert this expression to find $t(\tau)$. Then you can compute the other expressions by substitution, which are $x(t(\tau)), y(t(\tau), z(t(\tau))$
Fortunately, this does not have to be done very often. Usually, one can specify the a^i directly, then find u^i and x^i, and then (if one is interested), convert them into coordinate form via taking $x(t) = x(t(\tau))$, etc.
9. Oct 6, 2005
yukcream
To Trilairian
I know it is too late to ask you, still hope you will answer this!
why a^o is zero? du^0/dt not equal zero ~ where t is the popertime~
why v^o = c? v^0 actually is 1/{1-(v^2/c^2)}
yukyuk
10. Oct 6, 2005
Mortimer
You may want to have a look at http://www.rfjvanlinden171.freeler.nl/4vectors/index.html (of myself). It doesn't encompass acceleration 4-vectors but gives some general considerations regarding the physical interpretation of Minkowski 4-vector components that may answer your other questions.
Last edited by a moderator: Apr 21, 2017
11. Oct 6, 2005
yukcream
To Mortimer:
Thanks for your article but I hope to read an article more related to 4 acceleration as I get in trouble in writing out the component form of 4- acceleration vector~
yukyuk
12. Oct 6, 2005
robphy
If I am understanding the notation correctly, I believe that a^0=0 and v^0=c corresponds to evaluating those 4-vectors in the instantaneous rest-frame of the object being accelerated. In that frame, v is a purely temporal 4-vector (i.e., the spatial velocity is zero, and so gamma is one), and a is a purely spatial 4-vector.
13. Oct 6, 2005
Trilairian
I am using Uppercase A and U for four-vector acceleration and four-vector velocity. I am using lower case $$a^{\mu }$$ and $$v^{\mu }$$ for coordinate acceleration and coordinate velocity and extending their definitions to include a fourth element in the instances where a greek index is used.
$$v^{\mu } = dx^{\mu }/dt$$
$$a^{\mu } = dv^{\mu }/dt$$
For the fourth element:
$$v^{0} = dx^{0}/dt$$
$$v^{0} = dct/dt$$
$$v^{0} = c$$
$$a^{0} = dv^{0}/dt$$
$$a^{0} = dc/dt$$
$$a^{0} = 0$$
14. Oct 6, 2005
Trilairian
As an example this is how you calculate the time element of the acceleration four-vector.
$$A^{\mu } = \gamma ^{2}(a^{\mu } + \gamma ^{2}v^{\mu }(\mathbf{v}\cdot \mathbf{a}/c^{2}))$$
$$A^{0} = \gamma ^{2}(a^{0} + \gamma ^{2}v^{0}(\mathbf{v}\cdot \mathbf{a}/c^{2}))$$
$$A^{0} = \gamma ^{2}(0 + \gamma ^{2}c(\mathbf{v}\cdot \mathbf{a}/c^{2}))$$
$$A^{0} = \gamma ^{4}(\mathbf{v}\cdot \mathbf{a}/c)$$
15. Oct 7, 2005
yukcream
what will be the result of $$\mathbf{v}\cdot\mathbd{a}$$?
will it be zero as velcoity and acceleration are orthognal to each other!
16. Oct 8, 2005
Mortimer
Last edited by a moderator: Apr 21, 2017
17. Oct 9, 2005
yukcream
I have read an article it states that the time-component of 4 acceleration vector is zero, as in 4 velocity vector, $$v^{0}=v$$ so $$a^{0}=0$$. But in fact it is not correct! It should be a non zero quality.
yukyuk
18. Oct 9, 2005
Physics Monkey
yuk,
Because the four velocity is orthogonal to the four acceleration, the time component of the four acceleration will be zero in the rest frame of the particle (where the four velocity has only a time component). This is probably what the article meant, but who knows?
19. Oct 11, 2005
Trilairian
They aren't generally orthogonal. You are mixing them up with Four-vector acceleration $$A^{\mu }$$ and four-vector velocity $$U^{\mu }$$. These are not the same things as coordinate acceleration $$a^{i }$$ and coordinate velocity $$v^{i }$$. You need to look more carefully at the notation that I am using. I clearly defined things and the way I did so caps makes a difference.
Last edited: Oct 11, 2005
20. Oct 11, 2005
Trilairian
No you didn't. It said the time component of the coordinate acceleration is zero, not of the four-vector acceleration.
1st $$v^{\mu }$$ was not the four-vector velocity, $$U^{\mu }$$ was.
2nd that isn't what was said. What was said wat the time component of the coordinate velocity was c. $$v^{0} = c$$
No, that is correct. You're confusing it with $$A^{0}$$. I clearly defined my notation. Feel free to reread it.
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# Brute-force Hash Cracker
I made a hash cracker in Python (for purely educational purposes), but it's really slow (~120 seconds for a 4 character string). How could I speed it up?
Current optimizations and explanations:
• Closures in CharSet.get_advance: These are faster than attribute lookups.
• iter in PasswordCracker.crack: This moves the loop into C.
• CharSet.next as an array.array: Faster than a dict.
Possible future optimizations:
• advance is kind of slow, but I'm not sure how to speed it up.
Code:
import hashlib
from string import printable
from time import time
import itertools
from array import array
ENCODING = "ascii" # utf-8 for unicode support
class CharSet():
def __init__(self, chars):
chars = to_bytes(chars)
self.chars = set(chars)
self.first = chars[0]
self.last = chars[-1]
self.next = array("B", [0] * 256)
for char, next_char in zip(chars, chars[1:]):
self.next[char] = next_char
def update_chars(self, new_chars):
new_chars = to_bytes(new_chars)
new_chars = set(new_chars) - self.chars
if new_chars: # if theres anything new
self.chars |= new_chars
new_chars = list(new_chars)
self.next[self.last] = new_chars[0]
self.last = new_chars[-1]
for char, next_char in zip(new_chars, new_chars[1:]):
self.next[char] = next_char
first = self.first
last = self.last
next_ = self.next
for ind, byte in enumerate(arr):
if byte == last:
arr[ind] = first
else:
arr[ind] = next_[byte]
return hash_(arr)
arr.append(first)
return hash_(arr)
def __init__(self, hash_, chars=None):
self.hash = hash_
if chars is None:
chars = printable
self.char_set = CharSet(chars)
def update_chars(self, string):
self.char_set.update_chars(string)
def crack(self, hashed):
arr = bytearray()
pass
return arr
def to_bytes(string):
if isinstance(string, str):
return bytearray(string, ENCODING)
elif isinstance(string, (bytes, bytearray)):
return string
else:
raise TypeError(f"Cannot convert {string} to bytes")
def get_hasher(hash_):
def hasher(bytes):
return hash_(bytes).digest()
return hasher
md5 = get_hasher(hashlib.md5)
start = time()
end = time()
print(f"Time: {end - start} seconds.")
Profiling results (with password "pww"):
1333313 function calls in 1.500 seconds
Ordered by: standard name
ncalls tottime percall cumtime percall filename:lineno(function)
1 0.000 0.000 1.500 1.500 <string>:1(<module>)
1 0.000 0.000 0.000 0.000 main.py:31(get_advance)
333326 0.394 0.000 1.376 0.000 main.py:35(advance)
1 0.124 0.124 1.500 1.500 main.py:58(crack)
333326 0.311 0.000 0.982 0.000 main.py:74(hasher)
333326 0.265 0.000 0.265 0.000 {built-in method _hashlib.openssl_md5}
1 0.000 0.000 1.500 1.500 {built-in method builtins.exec}
1 0.000 0.000 0.000 0.000 {built-in method builtins.iter}
3 0.000 0.000 0.000 0.000 {method 'append' of 'bytearray' objects}
333326 0.405 0.000 0.405 0.000 {method 'digest' of '_hashlib.HASH' objects}
Profiling results (with password "pwww", extra "w"):
133333314 function calls in 190.800 seconds
Ordered by: standard name
ncalls tottime percall cumtime percall filename:lineno(function)
1 0.000 0.000 190.799 190.799 <string>:1(<module>)
1 0.000 0.000 0.000 0.000 main.py:31(get_advance)
33333326 65.652 0.000 169.782 0.000 main.py:35(advance)
1 21.017 21.017 190.799 190.799 main.py:58(crack)
33333326 40.640 0.000 104.130 0.000 main.py:74(hasher)
33333326 27.957 0.000 27.957 0.000 {built-in method _hashlib.openssl_md5}
1 0.000 0.000 190.800 190.800 {built-in method builtins.exec}
1 0.000 0.000 0.000 0.000 {built-in method builtins.iter}
4 0.000 0.000 0.000 0.000 {method 'append' of 'bytearray' objects}
33333326 35.533 0.000 35.533 0.000 {method 'digest' of '_hashlib.HASH' objects}
1 0.000 0.000 0.000 0.000 {method 'disable' of '_lsprof.Profiler' objects}
• Did you try profiling the code to see where are the bottlenecks? Probably cProfile can do Feb 18, 2018 at 23:55
• @romeu Yep, I profiled it. I'll edit in the results. Feb 18, 2018 at 23:56
• Seems the advance function is the one eating the biggest part of the time, I'd take a look at the for statement, the way you buildit make take some time to enumerate a bytearray millions of times.. Feb 19, 2018 at 1:19
• @romeu I dunno, enumerate seems pretty cheap. Just ran some tests, and it's only about 25% of the time. Feb 19, 2018 at 2:19
• Speculative, but you might be able to use something like itertools.chain( map(lambda byte: next_[byte], itertools.takewhile(lambda byte: byte != last, arr)), (first,) ) to speed up advance? Feb 19, 2018 at 19:34
## Using the Right Tool for the Job
It's easy problem to code but difficult to solve by computer. Better use low level-language like c.
You don't need create passwords manually, better use itertools library.
from hashlib import md5
from time import time
from string import printable
from itertools import product, count
chars = [c.encode(encoding) for c in printable]
for length in count(start=1):
for pwd in product(chars, repeat=length):
yield b''.join(pwd)
def crack(search_hash, encoding):
if md5(pwd).digest() == search_hash:
return pwd.decode(encoding)
if __name__ == "__main__":
encoding = 'ascii' # utf-8 for unicode support
start = time()
end = time()
print(f"Time: {end - start} seconds.")
## Imports
Usually the best option is from x import y, but here you can reduce cache
# import hashlib # usually bad one
# from hashlib import md5 # usually best one
from _md5 import md5 # ugly hack but speed up
• Hey @vaeta, could you back up some times for cracking 3 or 4 characters passwords? in order to compare to original solution? Mar 4, 2018 at 17:50
• @a-romeu I get for 'pwww' about 15s (itertools version) vs 35s (original one) on my terminal. From my point of view, simpler code is more important than speed up. Imho python is wrong tool for this task. Mar 4, 2018 at 18:07
I understand this is for learning purposes, and you are interested in the performance of this specific implementation. Otherwise I would tell you that computing the hashes each time might be a tiny bit slower than storing them.
Would it not be faster to generate the list of possible passwords first? Parallelism and overengineering might make this part slower, I am 99.9% sure, but it would set you up for some nice parallelism for the rest of it.
from itertools import product
passwords = product(printable, repeat = 4)
For me. with range(0,255) instead of printable takes 1.5 seconds.
Then you could use pool.map in multiprocessing.dummy to take you the rest of the way -> generate + check the hashes. (see https://stackoverflow.com/a/28463266/8695782| for reference, I think for the hash generation and checking part parallelism might help). I for one would prefer going towards a lookup-type structure, I want O(1) on retrieval, after generation/storage+reloading.
I can understand why learning purposes might make you not want to store the "rainbow table" and limit memory usage, but remember that when it comes to performance, there will always be speed vs space trade-offs. And speaking of space, why all 255 characters, at least exclude some of the control characters. You can benchmark your code vs https://gizmodo.com/over-560-million-passwords-discovered-by-security-resea-1795254560
• I missed chars = printable :) sorry, I updated my answer to take this into account :) Mar 3, 2018 at 20:37
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• 32
## Goal
Your goal is to recreate a simplified turns prediction for the Monopoly™ game.
You'll be given a list of players with their names and their initial positions, a list of dice rolls, and a list of names of all the game board positions. You'll need to find the latest position for each player after all the given dice rolls.
GAME BOARD
The game is played on a circular game board.
There are 40 positions on the board, indexed from 0 to 39.
BASIC RULES
Each player rolls two dice, and moves forward the sum of the dice rolled positions.
Multiple players can occupy the same position.
If a player reaches position 39 and still needs to move forward, he'll continue from position 0. In other words, positions 38, 39, 0, 1, 2 are contiguous.
Unless something special happens (see below), his turn ends after moving.
DOUBLE RULES
A double refers to a roll where both dice have the same value.
If the player rolls a double, then he rolls the dice and moves again.
If the player rolls a double three times in a row, then he immediately goes to jail (see below) and his turn ends there.
JAIL RULES
The Visit Only / In Jail position has a dual nature: the player is either visiting only, or actually in jail, depending on how he got there.
If at the end of a basic move, the player lands on Go To Jail, then he immediately moves to the position Visit Only / In Jail and is in jail. His turn ends.
If the player rolls a double for the third time in a single turn, then he immediately moves to the position Visit Only / In Jail and is in jail. His turn ends.
If after moving, the player lands on Visit Only / In Jail, he is visiting only and is not in jail.
While the player is in jail, he still rolls the dice on his turn as usual, but doesn't move until either:
(a) he rolls a double, or
(b) he fails to roll a double for three throws (i.e. his previous two turns after moving to jail and his current turn).
If either (a) or (b) happens in the player's turn, then he moves forward the sum of the dice rolled positions and his turn ends. He doesn't roll the dice again even if he has rolled a double.
Note: Visit Only / In Jail is at position 10. Go To Jail is at position 30.
Input
Line 1: An integer P for the number of players
Next P lines: One line for each player, containing the name and the initial position, separated with a space
Next line: An integer D for the number of dice rolls in playing order
Next D lines: One line for each dice roll (space-separated)
Next 40 lines: One line for the name of each board position. Note that these aren't actually needed to solve the puzzle.
Output
P lines: One line for each player, containing the name and the latest position, separated with a space. The output should follow the same order of the players as given in the input.
Constraints
2 ≤ P ≤ 6
Players' names do not contain any spaces.
Players never start on Go To Jail or in jail.
The game board is constant.
Example
Input
2
Horse 0
TopHat 0
6
1 6
2 1
6 4
2 1
4 5
6 5
Go
Mediterranean Avenue
Community Chest
Baltic Avenue
Income Tax
Oriental Avenue
Chance
Vermont Avenue
Connecticut Avenue
Visit Only / In Jail
St. Charles Place
Electric Company
States Avenue
Virginia Avenue
St. James Place
Community Chest
Tennessee Avenue
New York Avenue
Free Parking
Kentucky Avenue
Chance
Indiana Avenue
Illinois Avenue
Atlantic Avenue
Ventnor Avenue
Water Works
Marvin Gardens
Go To Jail
Pacific Avenue
North Carolina Avenue
Community Chest
Pennsylvania Avenue
Short Line
Chance
Park Place
Luxury Tax
Boardwalk
Output
Horse 26
TopHat 17
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# QM 2022
Apr 4 – 10, 2022
Auditorium Maximum UJ
Europe/Warsaw timezone
Proceedings submission deadline extended to August 15, 2022
## Charmed hadron interactions and correlation functions
Apr 7, 2022, 10:00 AM
20m
medium aula A (Auditorium Maximum UJ)
### medium aula A
#### Auditorium Maximum UJ
Oral presentation Correlations and fluctuations
Akira Ohnishi
### Description
Charmed hadron interactions are very important in current exotic hadron physics. For example, charmed pentaquark state ($P_c$) appears around the $\Sigma_c \bar{D}^{(*)}$ thresholds, and then $P_c$'s are suggested to be hadron molecules caused by the attraction between $\Sigma_c$ and $\bar{D}^{(*)}$. While it is difficult to perform charmed hadron scattering, recent femtoscopic studies allows us to get information on charmed hadron interactions. Since the correlation function $C(q)$ is given as the convolution of the source function and the wave function squared, $C(q)$ tells us how much the wave function is enhanced or suppressed from the free case provided that the source function is known.
In this presentation, we discuss the charmed hadron interactions and correlation functions [1]. Specifically, we consider $pD^{\pm}$, $DD^{*}$ and $D\bar{D}^{*}$ pairs, which are related to the (singly) charmed pentaquark state, $T_{cc}$, and $X(3872)$, respectively. As the first step, we use one-range Gaussian potentials whose strength are determined to reproduce the scattering lengths in theoretical models ($DN$ and $\bar{D}N$) or the binding energies ($DD^*$ and $D\bar{D}^*$). The calculated correlation functions show significant dependence on the interactions and the source size. Thus measurement of these correlation functions will judge theoretical models and the molecule picture of exotic hadrons.
[1] Y. Kamiya, T. Hyodo, and A. Ohnishi, in preparation.
### Primary authors
Prof. Tetsuo Hyodo (Tokyo Metropolitan University) Dr Yuki Kamiya (Universit{\"a}t Bonn)
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## Sunday, April 27, 2008
### Has inflation targeting stood the test of time?
Ilian Mihov, Andrew K. Rose (2008). Is Old Money Better than New? Duration and Monetary Regimes. Economics: The Open-Access, Open-Assessment E-Journal, Vol. 2, 2008-13. http://www.economics-ejournal.org/economics/journalarticles/2008-13. [pdf] To excerpt from their conclusion:
Inflation targeting seems like it would be a monetary regime that would compare poorly using the filter of time. After all, IT is a relatively new monetary regime. Nevertheless, we have found that IT seems already to have withstood the test of time; the duration of IT regimes is already as long as or significantly longer than alternatives like fixed exchange rates. Unlike all other monetary frameworks, no country has yet been forced to abandon a regime of inflation targeting in crisis. And this duration matters, since more durable monetary regime are systematically associated with better inflationary outcomes, meaning inflation within a band of (0, 4%). While having an exchange rate fix is better than having no clear quantitative target for monetary policy, inflation targeting is more likely to be associated with good inflationary outcomes. Any extrapolation of the heretofore successful record of inflation targeting remains exactly that: an extrapolation. Still, IT seems to work in both theory and practice, and is spreading quickly. Most importantly, inflation targeting is developing a reputation for durability, something that cannot be said of many alternative monetary regimes. Perhaps the monetary mishaps of the past will soon be seen as the byproduct of antiquated monetary regimes.
LaTeX mathematics works. This means that if you want to say $10 you have to say \$10.
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# Irrational logs and the harmonic series
Consider the series $$S_f = \sum_{x=1}^\infty \frac{f}{x^2+fx}.$$ Goldbach showed that, for integers $f \ge 1$, $$S_f = 1 + \frac12 + \frac13 + \ldots + \frac1f$$ (this follows easily by writing $S_f$ as a telescoping series). Thus $S_f$ is rational for all natural numbers $f \ge 1$. Goldbach claimed that, for all nonintegral (rational) numbers $f$, the sum $S_f$ would be irrational.
Euler showed, by using the substitution $$\frac1k = \int_0^1 x^{k-1} dx,$$ that $$S_f = \int_0^1 \frac{1-x^f}{1-x} dx.$$ He evaluated this integral for $f = \frac12$ and found that $S_{1/2} = 2(1 - \ln 2)$ (this also follows easily from Goldbach's series for $S_f$). Thus Goldbach's claim holds for all $f \equiv \frac12 \bmod 1$ since $S_{f+1} = S_f + \frac1{f+1}$.
Here are my questions:
1. The irrationality of $\ln 2$ was established by Lambert, who proved that $e^r$ is irrational for all rational numbers $r \ne 0$. Are there any (simple) direct proofs?
2. Has Goldbach's claim about the irrationality of $S_f$ for nonintegral rational values of $f$ been settled in other cases?
-
I'm suspicious there are (simple) direct proofs for ln2 being irrational (the standard argument for proving e^r is irrational seems simple enough). I did find this cute proof however: math.sc.edu/~filaseta/gradcourses/Math785/Math785Notes3.pdf – Alex R. Feb 7 '10 at 17:58
It appears that FC answered the second question, but here is another concrete example. If we let f=1/4, then the integral becomes 4-(pi/2)-ln(8). In general, the linear combination mentioned by FC will always have a nonzero algebraic coefficient on pi, except when p/q=1/2 [and that case was dealt with already]. – Pace Nielsen Feb 10 '10 at 20:41
@FC: You should write this as an answer, and claim the bounty. – Theo Johnson-Freyd Feb 10 '10 at 21:07
I agree with Theo's comment. – Franz Lemmermeyer Feb 11 '10 at 6:12
FC's comments led me to the article "Transcendental values of the digamma function", J. Number Theory 125, No. 2, 298-318 (2007) by Ram Murty and N. Saradha, where Thm. 9 states that the values of S_f are transcendental for rational numbers 0 < f < 1. I'd accept his comment as an answer if I could. – Franz Lemmermeyer Feb 11 '10 at 18:47
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### Construction of $n$-variable ($n\equiv 2 \bmod 4$) balanced Boolean functions with maximum absolute value in autocorrelation spectra $< 2^{\frac n2}$
Deng Tang and Subhamoy Maitra
##### Abstract
In this paper we consider the maximum absolute value $\Delta_f$ in the autocorrelation spectrum (not considering the zero point) of a function $f$. In even number of variables $n$, bent functions possess the highest nonlinearity with $\Delta_f = 0$. The long standing open question (for two decades) in this area is to obtain a theoretical construction of balanced functions with $\Delta_f < 2^{\frac n2}$. So far there are only a few examples of such functions for $n = 10, 14$, but no general construction technique is known. In this paper, we mathematically construct an infinite class of balanced Boolean functions on $n$ variables having absolute indicator strictly lesser than $\delta_n = 2^{\frac{n}{2}}-2^{\frac{n+6}{4}}$, nonlinearity strictly greater than $\rho_n = 2^{n-1} - 2^{\frac{n}{2}} + 2^{\frac n2-3} - 5\cdot2^{\frac{n-2}{4}}$ and algebraic degree $n-1$, where $n\equiv 2 \pmod 4$ and $n\geq 46$. While the bound $n \geq 46$ is required for proving the generic result, our construction starts from $n = 18$ and we could obtain balanced functions with $\Delta_f < 2^{\frac{n}{2}}$ and nonlinearity $> 2^{n-1} - 2^\frac{n}{2}$ for $n = 18, 22$ and $26$.
Available format(s)
Category
Secret-key cryptography
Publication info
Preprint. MINOR revision.
Keywords
Absolute IndicatorAutocorrelation SpectrumBalancednessBoolean functionNonlinearity.
Contact author(s)
subho @ isical ac in
History
2016-11-21: revised
See all versions
Short URL
https://ia.cr/2016/1078
CC BY
BibTeX
@misc{cryptoeprint:2016/1078,
author = {Deng Tang and Subhamoy Maitra},
title = {Construction of $n$-variable ($n\equiv 2 \bmod 4$) balanced Boolean functions with maximum absolute value in autocorrelation spectra $< 2^{\frac n2}$},
howpublished = {Cryptology ePrint Archive, Paper 2016/1078},
year = {2016},
note = {\url{https://eprint.iacr.org/2016/1078}},
url = {https://eprint.iacr.org/2016/1078}
}
Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content.
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# In the rectangular coordinate system above, the area of triangular
Author Message
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### Hide Tags
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Joined: 23 Sep 2016
Posts: 231
Re: In the rectangular coordinate system above, the area of triangular [#permalink]
### Show Tags
27 Mar 2018, 01:02
nss123 wrote:
Attachment:
IMAGE PT1.jpg
In the rectangular coordinate system above, the area of triangular region PQR is
(A) 12.5
(B) 14
(C) 10√2
(D) 16
(E) 25
For me here estimation works as following:-
we know base is 7 and by looking at figure we can say maximum height is 4 and minimum is 3 and our triangle height will like somewhere in between so
D and E out as maximum is $$\frac{1}{2}* 7*4= 14$$ and minimum is$$\frac{1}{2}*7*3= 10.5$$
now we can also eliminate C AND B as from the figure itself we know its some where around between so height can be 3.4,3.5 or 3.6 and we can also say$$\sqrt{2}$$ is also not possible then only option A left.
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Joined: 12 Sep 2015
Posts: 2308
Re: In the rectangular coordinate system above, the area of triangular [#permalink]
### Show Tags
19 Apr 2018, 14:30
Expert's post
Top Contributor
nss123 wrote:
In the rectangular coordinate system above, the area of triangular region PQR is
(A) 12.5
(B) 14
(C) 10√2
(D) 16
(E) 25
[Reveal] Spoiler:
Attachment:
IMAGE PT1.jpg
Let's draw a rectangle around the triangle (as shown below) and then subtract from the rectangle's area (28) the areas of the 3 right triangles that surround the triangle in question.
We get the following:
So, the area of PQR = 28 - (3.5 + 6 + 6) = 12.5
Cheers,
Brent
_________________
Brent Hanneson – Founder of gmatprepnow.com
Re: In the rectangular coordinate system above, the area of triangular [#permalink] 19 Apr 2018, 14:30
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# Models of nucleotide substitution
Models of nucleotide substitution are mathematical equations built to predict the probability (or proportion) of nucleotide change expected in a sequence.
## Jukes and Cantor's one-parameter model
JC69 is the simplest of the models of nucleotide substitution.[1] The model assumes that all nucleotides has the same rate (μ) of change to any other nucleotides. The probability that any nucleotide x remains the same from time 0 to time 1 is;
Pxx(1) = 1 − 3μ
Pxx(t) must be read; probability (or proportion, in this case it is equivalent) that x becomes x at time t. For the probability that any nucleotide x changes to any other nucleotide y we write Pxy(t). The probability for time t + 1 is;
Pxx(t + 1) = (1 − 3μ)Pxx(t) + μ(1 − Pxx(t))
The second part of the equation denotes the probability that the nucleotide was changed from time 0 and 1, but then got back to its initial states on time 2. The model can be rewritten in a differential equation with the solution;
$P_{xx(t)} = \frac{1}{4}+\frac{3}{4}e^{-4\mu t}$
Or if we want to know the probability of nuleotide x to change to nucleotide y;
$P_{xy(t)} = \frac{1}{4}-\frac{1}{4}e^{-4\mu t}$
With time, the probability will approach 0.25 (25%).
## Kimura's two-parameters model
Mostly known under the name K80, this model was developed by Kimura in 1980 as it became clear that all nucleotides substitutions weren't occurring at an equal rate. Most often, transitions (changes between A and G or C and T) are more common than transversions.[2]
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# Is there anything written by Newton's roommate Wilkins about him?
I've read that John Wilkins was Newton's room-mate and they lived together for 20 years. Is there anything about Newton written by Wilkins?
By the way, there is nothing easily found on google.
Part of the problem may be that Newton's roommate's name is spelled as Wickins or Wickens, not to be confused with bishop John Wilkins, Newton's older contemporary. Since they did live in the same room for 20 years Wickens's testimony would be precious indeed, but alas, as far as I know, the only things Wickens wrote concerning Newton were not about him but to him, letters. Some of the letters might be in the multi-volume Correspondence of Isaac Newton by Cambridge University Press, but not it seems in the online Newton’s Correspondence. It would not help with the most common speculation concerning Wickens, however. According to Waltz's history of autism:
"Others — e.g. Gribbin, p. 178 [Science: A History 1543-2001, BCA, 2002] have suggested a homosexual relationship with his long-time roommate John Wickens, but there is no hint of this in their surviving correspondence."
Spargo and Pounds in Newton's 'Derangement of the Intellect' quote a remark about Newton by Wickens's son Nicholas, but even that is just a reply to an inquiry forwarded to him by his father rather than some kind of memoir:
"He was turning Grey, I think, at Thirty, and when my Father observed y$$^{\textrm{t}}$$ to him as y$$^{\textrm{e}}$$ Effect of his deep attention of Mind, He would jest w$$^{\textrm{th}}$$ y$$^{\textrm{e}}$$ Experimy$$^{\textrm{ts}}$$ he made so often wth QuickSilver, as if from thence he took so soon that Colour."
The source is a manuscript labeled King's College Library, Cambridge, Keynes MS I37, not something that would be online either.
According to Nova:
A more recent blow was the departure from Cambridge of John Wickens, Newtons room-mate for twenty years. Wickens became the minister of the parish church at Stoke Edith, married and fathered a son called Nicholas. Though they had through much together, the two friends would never meet again, and exchanged no more than a letter or two in the coming years.
Thus there is not likely to be much. Though see this for some anecdotal details.
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# Homotopy Type Theory real geometric algebra > history (Rev #4, changes)
Showing changes from revision #3 to #4: Added | Removed | Changed
# Contents
## Definition
Given an a algebraic sequentially limit Cauchy complete Archimedean ordered field F \mathbb{R}, a algebraic real limit geometric algebra is a F \mathbb{R}-geometric algebra .$A$ such that the limits preserve the ring structure and the grade projection operation:
$\prod_{f:F \to A} \prod_{n:\mathbb{N}} \langle \lim_{x \to c} f(x) \rangle_n = \lim_{x \to c} \langle f(x) \rangle_n$
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# Extending cuspidal representation to more bigger group.
I am thinking of extending an irreducible cuspidal representation to more bigger group. My question is almost same with the earlier one posted by Neal Harris except the only one.
Let me first invoke his original question.
"Let $E/F$ be a quadratic extension of number fields, and let $V$ be an $n$-dimensional Hermitian space over $E$.
Let $\tilde{G} := GU(V)$ and $G := U(V)$. Suppose that $(\pi, V_{\pi})$ is an irreducible cuspidal representation of $G.$
Is there an irreducible cuspidal representation $(\tilde{\pi}, V_{\tilde{\pi}})$ of $\tilde{G}$ such that $V_\pi \subset V_{\tilde{\pi}}|_{G}$? Note that here, the restriction is that of cusp forms, not of the representation itself."
The difference between mine and his is this; While his assumption G=U(2) and $\tilde{G}$=GU(2) hit the condition $\tilde{G}_{der}\subset G\subset\tilde{G}$, a core property after the Mattew's comment on the post, but my situation does not satisfy this.
Because, I am considering the case $G=E^{\times} , \tilde{G}=B^{\times}$ where E/F is quadratic extension of number fields and B is a quaternion algebra over F with a fixed embedding $E \hookrightarrow B$. Is there no hope in this case?
Since this question is very important to me, if you leave some comment or reference treating this, I will be very appreciate to you.
(Harris's original post.
-
## 1 Answer
Looked at a slightly different way: the question of what happens when automorphic forms/repns are restricted to subgroups has complicated answers, in general. It can be treated as a problem in spectral decomposition, say in $L^2$. Another keyword is "period integral" for the integral expressions for the spectral components. Some examples are well-known in other terms: the "Mellin transform" integral that produces the standard L-function for GL(2) cuspforms can be construed (on the critical line!) as computing the spectral components of the restriction of the cuspform to the GL(1) imbedded in the upper left corner. The $GL(n-1)\times GL(n)$ integral formulas amount to computing the $GL(n-1)$ components of the restriction of a cuspform from $GL(n)$.
In other cases, sometimes it is feasible or elementary to compute periods of Eisenstein series, while the corresponding periods for cuspforms are much less elementary. (The Gross-Prasad conjectures were/are an example of this.)
In your specific situation, there are results of Waldspurger that may be approximately what you want.
Given that many of these periods/decomposition coefficients are values of L-functions at the center of the critical strip, except for trivial vanishing due to signs in functional equations, it is non-trivial to prove non-vanishing or vanishing. Thus, to ask that such a period be non-zero for exactly one cuspform would seem to be too much to ask, although perhaps very difficult to prove or disprove.
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Thanks Paul. I will search for Waldspurger's papers. – Jude Dec 25 '12 at 3:18
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Blazars, a class of Active Galactic Nuclei (AGN) characterized by a close orientation of their relativistic outflows (jets) towards the line of sight, are a well established extragalactic TeV $\gamma$-ray emitters. Since 2006, three nearby and TeV bright blazars, Markarian (Mrk) 421, Mrk 501 and 1ES 1959+650, are regularly observed by the MAGIC telescope with single exposures of 30 to 60 minutes. The sensitivity of MAGIC allows to establish a flux level of 30% of the Crab flux for each such observation. In a case of Mrk 421 strong flux variability in different time scales and a high correlation between X-ray/TeV emissions have been observed. In addition, preliminary results on measured light curves from Mrk 501 and 1ES1959+650 in 2007/8 are shown.
C. C., H., K., S., M., T., M., B., E., B., Bonnoli, G., et al. (2009). Monitoring of bright blazars with MAGIC telescope. In ArXiv e-prints (pp.893).
### Monitoring of bright blazars with MAGIC telescope
#### Abstract
Blazars, a class of Active Galactic Nuclei (AGN) characterized by a close orientation of their relativistic outflows (jets) towards the line of sight, are a well established extragalactic TeV $\gamma$-ray emitters. Since 2006, three nearby and TeV bright blazars, Markarian (Mrk) 421, Mrk 501 and 1ES 1959+650, are regularly observed by the MAGIC telescope with single exposures of 30 to 60 minutes. The sensitivity of MAGIC allows to establish a flux level of 30% of the Crab flux for each such observation. In a case of Mrk 421 strong flux variability in different time scales and a high correlation between X-ray/TeV emissions have been observed. In addition, preliminary results on measured light curves from Mrk 501 and 1ES1959+650 in 2007/8 are shown.
##### Scheda breve Scheda completa Scheda completa (DC)
C. C., H., K., S., M., T., M., B., E., B., Bonnoli, G., et al. (2009). Monitoring of bright blazars with MAGIC telescope. In ArXiv e-prints (pp.893).
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Utilizza questo identificativo per citare o creare un link a questo documento: http://hdl.handle.net/11365/43423
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# Why are we still using Continuous Time Fourier Transform when we have Laplace Transform? [duplicate]
I've read that Laplace Transform is more versatile and can cover a broad range of signals compared to Continuous Time Fourier Transform. Then why are we still using Continuous Time Fourier Transform ?
• What you heard is not entirely correct. Read this answer. There are also many related posts over at math.SE. – Matt L. Jan 5 '17 at 8:49
• Note that the Laplace transform of $\sin(\omega_0t)$, $-\infty<t<\infty$ doesn't exist. Also the impulse responses of ideal brickwall filters (ideal low pass, high pass, etc.) have no Laplace transform, while they do have a Fourier transform. – Matt L. Jan 5 '17 at 8:53
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Viraf J. Dalal solutions for Class 9 Simplified ICSE Chemistry chapter 4 - Atomic Structure & Chemical Bonding [Latest edition]
Chapter 4: Atomic Structure & Chemical Bonding
Exercise
Viraf J. Dalal solutions for Class 9 Simplified ICSE Chemistry Chapter 4 Atomic Structure & Chemical BondingExercise
Exercise | Q 1
$\ce{^24_12Mg}$ and $\ce{^26_12Mg}$ are symbols of isotopes of magnesium.
(a) Compare the atoms of these isotopes with respect to :
i. the composition of their nuclei
ii. their electronic configurations
(b) Give reasons why the two isotopes of magnesium have different mass numbers.
Exercise | Q 1
From the symbol $\ce{^4_2He}$ for the element helium, write down the mass number and the atomic number of the element.
Exercise | Q 1
Sulphur has an atomic number of 16 and a mass number of 32. State the number of protons and neutrons present in the nucleus of sulphur.
Exercise | Q 1
Five atoms are labelled V to Z
Atoms Mass Number Atomic Number V 40 20 W 19 9 X 7 3 Y 16 8 Z 14 7
1. Which one of these atoms
(1) contains 7 protons;
(2) has an electronic configuration 2, 7?
2. Write down the formula of the compound formed by atoms X and Y.
Exercise | Q 1
Elements X, Y, Z have atomic numbers 6, 9 and 12 respectively. Which one:
1. forms anion – negative ion;
2. forms cation – positive ion;
3. has 4 electrons in the outermost orbit. [Y, Z, X]
Exercise | Q 1
Write down the electronic configuration of the following
1. 2713X,
2. 3517Y.
Write down the number of electrons in X and neutrons in Y and the formula of the compound formed by X and Y.[XY3]
Exercise | Q 1
According to Dalton’s Atomic Theory, atoms of the same element are identical in all respects. But according to the Modern Atomic Theory, this postulate is proved wrong. Explain.
Exercise | Q 1
What are isotopes?
Exercise | Q 1.1
Define the term of atomic number.
Exercise | Q 1.1
The atom of aluminium is represented by 27Al13. Write down the number of electrons in the different orbits or shells in one atom of aluminium.
Exercise | Q 1.1
Define: Proton
Exercise | Q 1.2
What is the significance of the number of protons found in the atoms of different elements?
Exercise | Q 1.2
Define the term of mass number.
Exercise | Q 1.2
The atom of aluminium is represented by 27Al13. Write down the number of protons in the different orbits or shells in one atom of aluminium.
Exercise | Q 1.3
Define the term of the electron.
Exercise | Q 1.3
The atom of aluminium is represented by 27Al13. Write down the number of neutrons in the different orbits or shells in one atom of aluminium.
Exercise | Q 1.4
The atom of aluminium is represented by 27Al13. Write down the number of the arrangement of electrons in the different orbits or shells in one atom of aluminium.
Exercise | Q 2
Give a simple diagram to show the arrangement of the electrons in an atom of sulphur.
Exercise | Q 2
Chlorine is an element of atomic number 17. It is a mixture of two isotopes having mass number of 35 and 37.
1. What is meant by “atomic number of an element”? What do you understand by an ‘atom’
2. Write down the electronic configuration of the chlorine atom.
3. State the number of protons, electrons, and neutrons in the following isotopes: 35Cl1737Cl17
4. Explain why the two atoms in (iii) above have the same chemical reactions.
5. If molten magnesium chloride is electrolyzed suggest a suitable electrode [anode].
Exercise | Q 2
Ordinary chlorine gas has two isotopes: $\ce{^35_17Cl}$ and $\ce{^37_17Cl}$ in the ratio of 3 : 1. Calculate the relative atomic mass [atomic weight] of chlorine.
Exercise | Q 2
Name the element which does not contain any neutrons in its nucleus.
Exercise | Q 2
What is the relation between the number of protons and the number of electrons in an atom?
Exercise | Q 2.1
Write down the mass number of the atom having 20 neutrons and 15 protons.
Exercise | Q 2.2
Write down the number of neutrons in the nucleus of an atom having atomic number 17 and mass number 37.
Exercise | Q 3
Elements A, B and C have atomic numbers 9, 20 and 10 respectively.
1. State which one is
(1) a non-metal,
(2) a metal,
(3) chemically inert.[A,B,C]
2. Write down the formula of the compound formed by two of the above elements.[BA2]
Exercise | Q 3
What would be the reason for an element to have atoms with different mass numbers?
Exercise | Q 4
Copy and complete the following table relating to the atomic structure of some elements:
Element Atomic Number Mass Number Number of protons Number of Neutrons Number of Electrons Beryllium 4 9 Fluorine 9 10 Sodium 12 11 Aluminium 27 13 Phosphorus 31 15
Exercise | Q 4.1
Define: Proton
Exercise | Q 4.2
Define: Electron
Exercise | Q 4.3
Define: Neutron
Exercise | Q 5.1
The electronic structure [configuration] of fluorine can be written as 2, 7. In a similar way give the electronic configuration of aluminium
Exercise | Q 5.2
The electronic structure [configuration] of fluorine can be written as 2, 7. In a similar way give the electronic configuration of phosphorus
Viraf J. Dalal solutions for Class 9 Simplified ICSE Chemistry Chapter 4 Atomic Structure & Chemical BondingAdditional Questions
State Dalton's atomic theory.
How does the Modern atomic theory contradict and correlate with Dalton's atomic theory?
Explain in brief the experimental proof which led to the discovery of –
Electrons
Explain in brief the experimental proof which led to the discovery of –
Protons
Explain in brief the experimental proof which led to the discovery of –
Atomic nucleus
Explain in brief the experimental proof which led to the discovery of –
Neutrons
State in brief the drawbacks of Rutherford's atomic model correlating them with the postulates of Bohr’s atomic model.
What is meant by the term subatomic particles
What is meant by the term nucleus
What is meant by the term orbits
Define the term of atomic number.
Define the term of mass number.
Represent a proton ‘p’ in term of its symbol showing the subscript and superscript value.
Represent an electron ‘e’ in term of its symbol showing the subscript and superscript value.
Represent a neutron ‘n’ in term of its symbol showing the subscript and superscript value.
What are ‘energy levels’?
Explain the arrangement and distribution of electrons in the various shells with reference to an atom in general and to an atom of potassium ‘3919K’ with special reference to the 2n2 rule.
An element ‘A’ has mass number 23 and atomic number 11. State the –
1. no. of neutrons in its shell,
2. electronic configuration of the element ‘A’.
The following element is given $\ce{_3U}$. State the electronic configuration and state whether it is a metal, non-metal, or inert gas.
The following element is given $\ce{_6V}$. State the electronic configuration and state whether it is a metal, non-metal, or inert gas.
The following element is given $\ce{_9W}$. State the electronic configuration and state whether it is a metal, non-metal, or inert gas.
The following element is given $\ce{_14X}$. State the electronic configuration and state whether it is a metal, non-metal, or inert gas.
The following element is given $\ce{_18Y}$. State the electronic configuration and state whether it is a metal, non-metal, or inert gas.
The following element is given $\ce{_20Z}$. State the electronic configuration and state whether it is a metal, non-metal, or inert gas.
Draw the geometric atomic structure of the following atom showing the number of electrons, protons and neutrons in them:
$\ce{^12_6C}$
Draw the geometric atomic structure of the following atom showing the number of electrons, protons and neutrons in them:
$\ce{^23_11Na}$
Draw the geometric atomic structure of the following atom showing the number of electrons, protons and neutrons in them:
$\ce{^31_15P}$
Draw the geometric atomic structure of the following atom showing the number of electrons, protons and neutrons in them:
$\ce{^39_19K}$
Draw the geometric atomic structure of the following atom showing the number of electrons, protons and neutrons in them:
$\ce{^40_20Ca}$
What are isotopes?
Give the reason why isotopes have the same chemical but different physical properties.
Draw the geometric atomic structure of the three isotopes of hydrogen and the two isotopes of chlorine.
Four elements A, B, C, D are given :
• A shows the presence of 20 neutrons, 17 protons and 17 electrons.
• B shows the presence of 18 neutrons, 17 protons and 17 electrons.
• C shows the presence of 10 neutrons, 9 protons and 10 electrons.
• D shows the presence of 4 neutrons, 3 protons and 2 electrons.
State which of the above is –
1. an anion
2. a cation
3. a pair of isotopes.
Write the formula of the compound formed between D and C.
What are the noble gases?
Give a reason why noble gases have a stable electronic configuration.
Explain the reason for the chemical activity of an atom with reference to its electronic configuration.
Differentiate between the term Stable and unstable electronic configuration.
Differentiate between the term Duplet and octet rule.
Explain the octet rule for the formation of –
Sodium chloride from a sodium atom and a chlorine atom.
Explain the octet rule for the formation of –
Nitrogen molecule from two nitrogen atoms.
Atomic Structure
Viraf J. Dalal solutions for Class 9 Simplified ICSE Chemistry Chapter 4 Atomic Structure & Chemical BondingAtomic Structure
Atomic Structure | Q 1.1
Select the correct answer from the answer in the bracket to complete the sentence.
An element has electronic configuration 2, 8, 1 and 12 neutrons. Its mass no. is ________.
• 11
• 23
• 12
Atomic Structure | Q 1.2
Select the correct answer from the answer in the bracket to complete the sentence.
The maximum number of electrons in M-shell is _______.
• 8
• 32
• 18
Atomic Structure | Q 1.3
Select the correct answer from the answer in the bracket to complete the sentence.
Isotopes have same _________.
• no. of neutrons
• electronic configuration
• atomic masses
Atomic Structure | Q 1.4
Select the correct answer from the answer in the bracket to complete the sentence.
An ________ is capable of independent existence in solution.
• atom
• ion
Atomic Structure | Q 1.5
Select the correct answer from the answer in the bracket to complete the sentence.
An atom with electronic configuration 2, 7 and mass number 19 will have ________ neutrons.
• 8
• 10
• 12
Atomic Structure | Q 2.1
Give a reason
Physical properties of isotopes are different.
Atomic Structure | Q 2.2
Give a reason for the following:
The mass number of an atom is slightly less than the actual atomic mass.
Atomic Structure | Q 2.3
Give a reason for the following:
The shells surrounding the nucleus of an atom are also called ‘energy levels’.
Atomic Structure | Q 2.4
Give a reason for the following:
Helium is chemical extremely unreactive.
Atomic Structure | Q 2.5
Give a reason for the following:
The mass number of an atom is slightly less than the actual atomic mass.
Atomic Structure | Q 3.1
Differentiate between the following term:
Electron and proton
Atomic Structure | Q 3.2
Differentiate between the following term:
Atomic number and mass number
Atomic Structure | Q 3.3
Differentiate between the following term:
Nucleus and nucleons
Atomic Structure | Q 3.4
Differentiate between the following term:
Valence shell and penultimate shell
Atomic Structure | Q 3.5
Differentiate between the term Duplet and octet rule.
Atomic Structure | Q 4.1
Name or state the following.
The three isotopes of hydrogen.
Atomic Structure | Q 4.2
Name or state the following.
Two elements having the same number of protons and electrons but a different number of neutrons.
Atomic Structure | Q 4.3
Name or state the following.
The valency of an element whose electronic configuration is 2, 8, 3.
Atomic Structure | Q 4.4
Name the following:
The shell closest to the nucleus of an atom
Atomic Structure | Q 4.5
Name the following:
An element having valency 'zero'
Atomic Structure | Q 5
State the number of neutrons in each of the atoms A to E. Also state which of the atoms A to E is a metal.
Atomic Structure | Q 6
Match the elements A to E in List 1 with their valencies in List 2 and with their nature in List 3.
List 1 [Elements] List 2 [Valency] List 3 A: At. no. 7, Mass no. 14 1. −3 X: Metal B: Elec. conf. 2, 8 2. +1 Y: Non-metal C: Neutrons 14, electrons 13 3. +3 Z: Noble gas D: Neutrons 22, protons 18 4. +2 E: Elec. config. 2, 8, 1 5. 0
Viraf J. Dalal solutions for Class 9 Simplified ICSE Chemistry chapter 4 - Atomic Structure & Chemical Bonding
Viraf J. Dalal solutions for Class 9 Simplified ICSE Chemistry chapter 4 (Atomic Structure & Chemical Bonding) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CISCE Class 9 Simplified ICSE Chemistry solutions in a manner that help students grasp basic concepts better and faster.
Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. Viraf J. Dalal textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Class 9 Simplified ICSE Chemistry chapter 4 Atomic Structure & Chemical Bonding are Atomic Number (Z), Isotopes, Electrons Distributed in Different Orbits (Shells), Mass Number (A) or Atomic Mass, Nucleus, Neil Bohr’s Model of Atom, Neutrons (n), Bohr-bury Scheme, Valency Electrons, Reason for Chemical Activity of an Atom, Electrovalent (Or Ionic) Bond, Protons (p), Electrons (e), Chemical Bond, Covalent (Molecular) Bond, Atom, Characteristic Properties of Electrovalent Compounds.
Using Viraf J. Dalal Class 9 solutions Atomic Structure & Chemical Bonding exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in Viraf J. Dalal Solutions are important questions that can be asked in the final exam. Maximum students of CISCE Class 9 prefer Viraf J. Dalal Textbook Solutions to score more in exam.
Get the free view of chapter 4 Atomic Structure & Chemical Bonding Class 9 extra questions for Class 9 Simplified ICSE Chemistry and can use Shaalaa.com to keep it handy for your exam preparation
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In rhombus all sides are equal. Possible Answers: Correct answer: Explanation: To find the value of diagonal , we must first recognize some important properties of rhombuses. Find the diagonal of a rhombus if given one-half angle and other diagonal ( , ) : Find the diagonal of a rhombus if given area and other diagonal ( , ) : … Thus, the area of a Rhombus formula can be given as – The Area of a Rhombus Formula is, 4 × area of ∆ AOB The diagonals of a rhombus bisect each other at 90 degrees. The center divides the diagonals into two equal semi-diagonals; Diagonals makes four congruent right triangles, in which the hypotenuse is represented by the rhombus' side, and the cathetus' by the semi-diagonals All the four sides are equal in dimensions same as square. Diagonals are perpendicular; Diagonals meet at a point called rhombus' center. Question: Find the area of a rhombus whose length of the diagonals is 10 cm and 14 cm respectively. 1 Answer: 3 question The endpoints of one diagonal of a rhombus are (0,-8) and (8,-4) - the answers to estudyassistant.com You may need to download version 2.0 now from the Chrome Web Store. A rhombus is a parallelogram in which all sides are congruent. 4 A rhombus is either an equilateral triangle or a slanting square whose sides are equal and the area can be calculated by multiplying both diagonals together and divide the value by two. Rhombus has: All sides of equal length. The Perimeter of a Rhombus. There are several formulas for the rhombus that have to do with its: Sides (click for more detail) All 4 sides are congruent. 10 Therefore, Area of a Rhombus = A = ½ × d 1 × d 2. Diagonals. A rhombus is a type of parallelogram, and what distinguishes its shape is that all four of its sides are congruent. h = height of rhombus. You can find the area in square units of the rhombus by multiplying the lengths of the two diagonals (d 1 and d 2) and dividing by two. Where d 1 and d 2 are the diagonals of the rhombus. Diagonals bisect vertex angles. Using Side Length Set up the formula for perimeter of a rhombus. A quadrilateral is called as a rhombus “if the adjacent sides are equal and also the opposite sides are parallel and no angle is equal to the { 90 }^{ 0 } .. Properties: Diagonals are not equal. Solution: Given, d 1 = 1 cm d 2 = 14 cm Area = (d 1 d 2)/2 = (10 ×14)/2 = 70 cm 2. Given: Area = 60 cm². From the properties we know that diagonals of rhombus bisect each other into equal parts. There are several formulas that can be used to find the area of a rhombus depending on the known parameters. Step-by-step explanation: The formula to compute the area of a rhombus is: Here, p and q = lengths of the two diagonals. All formulas for radius of a circumscribed circle. 2 If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. In a rhombus, each diagonal of a rhombus divides it into two congruent triangles. Here you can read about diagonals, the formula to calculate the number of diagonals, diagonal of square formula, diagonal of rectangle formula, diagonals of rhombus and parallelogram, and some fun properties of diagonals. What is rhombus in math? As we know that, The halves of a diagonal and side of rhombus from a right angle triangle with side as hypotenuse. Now , Putting the values in the above formula. A diagonal is defined as a line segment joining the two opposite vertices of a polygon. Formula of Perimeter of Rhombus. Where, Angles. The length of the other diagonal is 15 cm. The formula for finding the area of a rhombus is A = (d 1 * d 2)/2, where d 1 and d 2 are the lengths of the diagonals. or any sloping line on the rhombus is a diagonal. Using the Base and Height Find the base and the height. Area formula. Explore more mathematical concepts and formulas by visiting BYJU’S. The altitude is a distance at right angles between two parallel sides. Using diagonals. 6 Example 1: On the figure below the diagonals of the rhombus are $\displaystyle AO=2x+1$ and $\displaystyle OC=\frac{1}{2}x+10$. Code to add this calci to your website 6 The length of the side of rhombus can be determined if the diagonals are known. Area of Rhombus Formula. Again, the length of the side also found if area and a diagonal of a rhombus are given. The formula to calculate the area of a rhombus is: A = ½ x d 1 x d 2. where... A = area of rhombus; d 1 = diagonal1 (first diagonal in rhombus, as indicated by red line) d 2 = diagonal2 (second diagonal in rhombus, as indicated by purple line) Calculate the side of a rhombus if given diagonal and one-half angle ( a ) : Calculate the side of a rhombus if given area and angle ( a ) : Calculate the side of a rhombus if given perimeter ( a ) : side of a rhombus : = Digit 1 2 4 6 10 F. =. = Digit x^2+(24/2)^2 =20^2=x^2+144=400. 4 The measure of the another diagonal of a rhombus=2×16=32 units = 4 × (½) × (½) d 1 × (½) d 2 sq. Diagonal of a Rhombus calculator uses Diagonal 2=2*(Area/Diagonal 1) to calculate the Diagonal 2, The Diagonal of a Rhombus formula is defined as twice the area by the other diagonal. The area of the rhombus is the region covered by it in a two-dimensional plane. The two diagonals of a rhombus are bisecting each other. • Diagonal of Square. Area Since the perimeter is of is , and by definition a rhombus has four sides of equal length, each side length of the rhombus is equal to . In fact, if all four sides are equal, it has to be a parallelogram. Opposite sides have to be parallel. Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. First diagonal ( d1 ) ↠ 10 cm Second diagonal ( d2 ) ↠ 8.2 cm. Diagonal of a Rhombus Formula: The area of a rhombus can be calculated with the help of diagonals as given A = ½ × d1 × d2. Try This: Area of Rhombus Calculator Let the length of other diagonal =2x. All formulas for radius of a circle inscribed, All basic formulas of trigonometric identities, Height, Bisector and Median of an isosceles triangle, Height, Bisector and Median of an equilateral triangle, Angles between diagonals of a parallelogram, Height of a parallelogram and the angle of intersection of heights, The sum of the squared diagonals of a parallelogram, The length and the properties of a bisector of a parallelogram, Lateral sides and height of a right trapezoid, Find the diagonal of a rhombus if given side and angle (, Find the diagonal of a rhombus if given side and one-half angle (, Find the diagonal of a rhombus if given side and other diagonal (, Find the diagonal of a rhombus if given one-half angle and other diagonal (, Find the diagonal of a rhombus if given area and other diagonal (. Diagonals are perpendicular. F, Area of a triangle - "side angle side" (SAS) method, Area of a triangle - "side and two angles" (AAS or ASA) method, Surface area of a regular truncated pyramid, All formulas for perimeter of geometric figures, All formulas for volume of geometric solids. The perimeter is the sum of the length of all the 4 sides. The area of the rhombus will be: A = 4 × area of ∆ AOB = 4 × (½) × AO × OB sq. It can be represented as: Area of Rhombus, A = (d1 x d2)/2 square units The sum of the other diagonal is 15 cm rhombuses are … Using side length Set up the for! And just to make things clear, some rhombuses are … Using side length Set up the for. Visiting BYJU ’ S a polygon parallelogram where all four sides are equal we.: 614c98ea4f66f794 • Your IP: 209.236.71.66 • Performance & security by cloudflare, Please complete the security to. Is half the product of its two diagonals of a rhombus whose length of the diagonal. 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# Mole Fraction Calculator
Calculate Mole Fraction
Mole Fraction(Xi) =
$$\begin{array}{l}\frac{Moles\ of\ Solute}{Moles\ of\ Solute + Moles\ of\ Solvent}\end{array}$$
Enter the unknown value as ‘x’
Moles of Solute(i) =
Moles of Solvent =
Mole Fraction(Xi) =
x =
Mole Fraction Calculator is a free online tool that displays the mole fraction for the given chemical component. BYJU’S online mole fraction calculator tool makes the calculation faster, and it calculates the mole fraction of a component in a fraction of seconds.
## How to Use the Mole Fraction Calculator?
The procedure to use the mole fraction calculator is as follows:
Step 1: Enter the moles of solute, solvent and x for the unknown in the respective input field
Step 2: Now click the button “Calculate x” to get the result
Step 3: Finally, the mole fraction of the chemical component will be displayed in the output field
### What is Meant by Mole Fraction?
In chemistry, a mole fraction is a concentration unit which is used to express the concentration of a chemical component. The mole fraction can be calculated by dividing the number of moles of one component of a solution by the total number of moles of all the components of a solution. It is noted that the sum of the mole fraction of all the components in the solution should be equal to 1.
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-2(x-5)-(16-x)=4(2x-1)+7
Simple and best practice solution for -2(x-5)-(16-x)=4(2x-1)+7 equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so dont hesitate to use it as a solution of your homework.
If it's not what You are looking for type in the equation solver your own equation and let us solve it.
Solution for -2(x-5)-(16-x)=4(2x-1)+7 equation:
-2(x-5)-(16-x)=4(2x-1)+7
We move all terms to the left:
-2(x-5)-(16-x)-(4(2x-1)+7)=0
We add all the numbers together, and all the variables
-2(x-5)-(-1x+16)-(4(2x-1)+7)=0
We multiply parentheses
-2x-(-1x+16)-(4(2x-1)+7)+10=0
We get rid of parentheses
-2x+1x-(4(2x-1)+7)-16+10=0
We calculate terms in parentheses: -(4(2x-1)+7), so:
4(2x-1)+7
We multiply parentheses
8x-4+7
We add all the numbers together, and all the variables
8x+3
Back to the equation:
-(8x+3)
We add all the numbers together, and all the variables
-1x-(8x+3)-6=0
We get rid of parentheses
-1x-8x-3-6=0
We add all the numbers together, and all the variables
-9x-9=0
We move all terms containing x to the left, all other terms to the right
-9x=9
x=9/-9
x=-1
`
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LaTeX forum ⇒ Math & Science ⇒ Aligning Math Symbols Topic is solved
Information and discussion about LaTeX's math and science related features (e.g. formulas, graphs).
learningtex
Posts: 2
Joined: Fri Nov 16, 2012 5:49 am
Aligning Math Symbols Topic is solved
Hi there! I'm a beginner to Latex and need some help with aligning subscripts and superscripts in Maths. Here's what I have:
\documentclass{article} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage{geometry} \geometry{a4paper} \usepackage[frenchb]{babel} \title{Remaining Maths Homework} \author{Herp Derp} \date{15 November 2012} \begin{document} \maketitle \Large Exercise 5.1 \normalsize\\\\Q1. (a) The graph of $f(x)$ has a domain $D \in [0,8]$, which is divided into four rectangles of equal base, $x = 2$. The actual area under the graph of $f(x)$ isestimted as: $\sum_{i=1}^{+\infty}{f(x)}$ and $\int_{x_0}^{x_1}$ \end{document}
And here's the output:
Output
example.PNG (17.14 KiB) Viewed 1080 times
As you can see, the $\infty and i=1$ are to the side of the sigma function and not on top of it. How do I solve this?
kaiserkarl13
Posts: 582
Joined: Tue Mar 25, 2008 5:02 pm
Location: Columbia, Missouri, USA
TeX normally typesets summation and product symbols with the indices on the sides in "text" style math fields, but on top/bottom in "display" style math. You can select a math display with $...$, which is equivalent to the displaymath environment. If you want it in text mode, but still want display-sized symbols, use the \displaystyle command (there is also a \textstyl command). I provide an example of both here.
\documentclass{article} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage{geometry} \geometry{a4paper} \usepackage[frenchb]{babel} \title{Remaining Maths Homework} \author{Herp Derp} \date{15 November 2012} \begin{document} \maketitle \Large Exercise 5.1 \normalsize \\ \\Q1. (a) The graph of $f(x)$ has a domain $D \in [0,8]$, which is divided into four rectangles of equal base, $x = 2$. The actual area under the graph of $f(x)$ isestimted [sic] as: $\sum_{i=1}^{+\infty}{f(x)}$ and $\int_{x_0}^{x_1}$ Option 1: \\Q1. (a) The graph of $f(x)$ has a domain $D \in [0,8]$, which isdivided into four rectangles of equal base, $x = 2$. The actual area under thegraph of $f(x)$ is estimated as: $\sum_{i=1}^{+\infty}{f(x)}$ and $\int_{x_0}^{x_1}$ \\Option 2: \\Q1. (a) The graph of $f(x)$ has a domain $D \in [0,8]$, which isdivided into four rectangles of equal base, $x = 2$. The actual area under thegraph of $f(x)$ is estimated as: $\displaystyle\sum_{i=1}^{+\infty}{f(x)}$ and $\displaystyle\int_{x_0}^{x_1}$ \\ \end{document}
learningtex
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Thanks, \displaystyle was just what I wanted.
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• October 14, 2019, 09:00:40 PM
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AuthorTopic: Mission Pro Combo (MissionPro, QMBPro and FMBPro) - Last Release 2011.09.09 (Read 32721 times)
0 Members and 1 Guest are viewing this topic.
gianluca
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Re: Mission Pro Combo (MissionPro, QMBPro and FMBPro) - Last Release 2011.09.09
« Reply #120 on: February 23, 2012, 03:48:08 AM »
Hi pablo, I'm talking about the so called FMB+ which was one of the jsgme options in UP201 and whose readme reads:"This enables the Full Mission Builder Plus". There are inside 53 java classes to teplace the ones contained in "Files".
Cheers
Gianluca
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coenoo1
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• Posts: 227
Re: Mission Pro Combo (MissionPro, QMBPro and FMBPro) - Last Release 2011.09.09
« Reply #121 on: March 07, 2012, 05:24:03 AM »
I have a problem, If I activate the QMBPro in JSGME which standardly comes with DBW 1.71 , none of my maps work, I either get no paratrooper mesh error or: (in this case the vietnam_south map which is also standard issue QMBPro in 1.71 DBW) my whole game crashes at 100%. I do get an error code and will post that later on, since I write this from my iPhone... I like this mod very much cause it essentialy makes FMB easier for me to work with as a softwaren00b
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benitomuso
• SAS Team
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• P.A.L.
Re: Mission Pro Combo (MissionPro, QMBPro and FMBPro) - Last Release 2011.09.09
« Reply #122 on: March 28, 2012, 10:00:09 PM »
People,
finally I had the time to solve the typical problem of EcranWide with QMBPro:
In fact it wasn't a problem of the MissionProCombo, but a result of the use of Ecran Wide with different configuration of the standard one regarding the number of elements of the Comboboxes. The game itself didn't know how to manage the situation and so the problem.
Now I have solved it in its origin, so never again the drop-down or drop-up of the combos will be pulled out of the window.
I'm publishing this here as a test release. I used it in 2 different notebooks in different resolutions and everything worked OK. Please let me know if you see something strange. Now any combination of list lines will be handled correctly, the default one or your one, whatever it was.
Regards,
Pablo
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• o l
Re: Mission Pro Combo (MissionPro, QMBPro and FMBPro) - Last Release 2011.09.09
« Reply #123 on: March 29, 2012, 12:52:16 AM »
Grazie Maestro!
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slejsa
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• Posts: 171
Re: Mission Pro Combo (MissionPro, QMBPro and FMBPro) - Last Release 2011.09.09
« Reply #124 on: June 07, 2012, 02:59:17 AM »
Works it with the current version of UP? I tryed it but new missions are non-functional..
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rotte7teufel
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• Posts: 132
Re: Mission Pro Combo (MissionPro, QMBPro and FMBPro) - Last Release 2011.09.09
« Reply #125 on: June 19, 2014, 03:53:50 AM »
Its a possible to get 16 not 8 plane places in QMBpro using Mission Combo Pro by P.A.L in MA5.3/TFM412 ?
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I know what you're thinking: "Did he fire six shots or only five?" Well, to tell you the truth, in all this excitement, I've kinda lost track myself. But being this is a .44 Magnum, the most powerful handgun in the world, and would blow your head clean off, you've got to ask yourself one question: 'Do I feel lucky?' Well, do ya, punk?
Whiskey_Sierra_972
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• In memory of my beloved hero: Saburo SAKAI!
Re: Mission Pro Combo (MissionPro, QMBPro and FMBPro) - Last Release 2011.09.09
« Reply #126 on: June 19, 2014, 04:21:45 AM »
This is only for 410....
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rotte7teufel
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Re: Mission Pro Combo (MissionPro, QMBPro and FMBPro) - Last Release 2011.09.09
« Reply #127 on: June 19, 2014, 04:33:40 AM »
Whisey_Sierra If You talk its Mission Combo Pro is only for 4.10 youre wrong in here You have version for MA5.3 http://www.mission4today.com/index.php?name=Downloads3&file=details&id=1815 ...... but if You talk about that You can put 16 only in 4.10 version ... thats sad .....for me ;p
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I know what you're thinking: "Did he fire six shots or only five?" Well, to tell you the truth, in all this excitement, I've kinda lost track myself. But being this is a .44 Magnum, the most powerful handgun in the world, and would blow your head clean off, you've got to ask yourself one question: 'Do I feel lucky?' Well, do ya, punk?
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# How good is my prediction?
## 2007-08-13
Dear Professor Mean, I have two time series of data, one actual and one predicted. Since I'm quite new to statistical methods, I would like to know what methods are used to evaluate the different between the two time series. I would like to able to say something like "the predicted values were 70% accurate."
See what others in your area are doing and emulate them, as there is no one measure that is used uniformly. Most formulas are based on the residual.
To compute the residual, subtract the predicted value from the actual value. The residual is used in many statistical models, not just time series. Then there are several statistics that you can compute on the residuals. The simplest is the standard deviation of the residuals. Another possibility is the average absolute residual. The closer that these values are to zero, the better your prediction.
If you are interested summaries that represent a percentage, you might want to consider a relative measures such as the absolute residual divided by the actual time series, as long as the actual time series is never zero or negative. This would give you a percentage error.
Another possibility is to compare the residuals from your prediction to a much simpler prediction (for example, a prediction that uses the mean for every single value). Then the ratio of the variances (the squared standard deviation) of the two predictions is a measure of how well your predictions are doing. Place the variance of the simpler prediction in the denominator. In linear regression, this is known as R-squared or multiple R-squared depending on the context, but it should also work for time series data.
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# A limiting sequence of positive definite matrices
Let $$A\in\mathbb{R}^{n\times n}$$ be a matrix with eigenvalues having (strictly) negative real part. Let $$X\in\mathbb{R}^{n\times n}$$, $$X\succ 0$$, be a positive definite matrix and let $$P\succ 0$$ be the (unique) solution of the following Lyapunov matrix equation $$\tag{1}\label{eq1} AP+PA^\top = - X.$$
My question. Let $$\{X_n\}_{n\ge 0}$$, $$X_n\succ 0$$, be a sequence of positive definite matrices, and let $$\{P_n\}_{n\ge 0}$$, $$P_n\succ 0$$, be the corresponding sequence of solutions of \eqref{eq1}. Suppose that $$\lim_{n\to \infty} X_n=\bar{X}\succeq 0$$ and $$\lim_{n\to \infty} P_n=\bar{P}\succeq 0$$ is singular. I'm wondering whether $$\lim_{n\to \infty} P_n^{-1/2}X_n P_n^{-1/2}$$ always converges to a finite matrix.
A few remarks are in order.
1. If $$\bar{P}$$ is singular, then, $$\bar{X}$$ must be singular as well.
2. If $$A$$ is a scalar matrix, i.e. $$A=\alpha I$$, $$\alpha<0$$, then it is quite easy to see that $$\lim_{n\to \infty} P_n^{-1/2}X_n P_n^{-1/2}$$ converges to a finite matrix.
3. By pre- and post-multiplying \eqref{eq1} by $$P_n^{-1/2}$$, it follows that if $$\lim_{n\to \infty} P_n^{-1/2}A P_n^{1/2}$$ is finite, then $$\lim_{n\to \infty} P_n^{-1/2}X_n P_n^{-1/2}$$ is finite as well.
4. From point 3. it holds $$\mathrm{tr}(P_n^{-1/2}X_n P_n^{-1/2})=-2\,\mathrm{tr}(A)$$ for all $$n$$.
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News
# Abound Solar to Suspend Operations
July 01, 2012 by Jeff Shepard
Abound Solar, a manufacturer of thin-film cadmium telluride (CdTe) photovoltaic modules, announced that the company intends to file a petition for protection under U.S. Bankruptcy Code in Delaware next week. A suspension of operations will impact approximately 125 employees.
Earlier this year, Abound ceased production of its first generation PV module and has been working to scale up manufacturing for its high-efficiency, second generation PV module, which was tested and verified by the National Renewable Energy Laboratory (NREL) to deliver 85 watts per panel and 12.5 percent efficiency. Abound has been in discussions with potential buyers over the last several months, but ended negotiations when the involved parties were unable to come to an agreement on terms.
Abound’s funding has come from $300 million in private investment and$70 million from a DOE loan guarantee program. Abound had used $70 million of the awarded$400 million DOE loan guarantee for construction of solar panel manufacturing lines in Colorado. Abound has not drawn down any further DOE funds since August of 2011 when the DOE determined that challenging market conditions in the solar industry did not merit additional funding risk.
Abound believes that, at scale, its USA-made CdTe panel technology has the ability to achieve lower cost per watt than competing crystalline silicon technology made in China. However, aggressive pricing actions from Chinese solar panel companies have made it very difficult for an early stage startup company like Abound to scale in current market conditions. According to the U.S. Commerce Department, the U.S. solar market has seen the prices for panels drop by more than 50 percent in the past year at a time when the value of imports of Chinese-made solar cells nearly quadrupled from $639 million in 2009 to$3.1 billion in 2011. Abound supports recent initiatives to enforce fair trade with import tariffs, but this action is unfortunately too late for the company.
Abound is appreciative of the significant investment from private investors and the U.S. Department of Energy. Employees should be proud of their continuous innovation and daily efforts to support customers. Abound believes that competitive solar energy remains important to U.S. energy security and job creation; and that longer term, consistent renewable energy policy is critical to encourage further private investment in this sector.
More news and information regarding the latest developments in Smart Grid electronics can be found at Darnell’s SmartGridElectronics.Net.
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# Integration and differential
## Homework Statement
integrate R(t) 2+5sin(4πt/25) to calculate F(t) from 0 to 6
## The Attempt at a Solution
the differential equation is 2+5sin(4πt/25)
so after the integration, i get 2t-5cos(4πt/25)(25/4π)
so i wanna get the value in the interval from 0 to 6)
so i put 2+5sin(4πt/25) on my calculator and made it do the integration,
It gives me 31.8159
but when i F(6)-F(0) on this function, 2t-5cos(4πt/25)(25/4π), i get something entirely different.
My calculator is on radian mode and π is pie
What am i doing wrong?
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# American Institute of Mathematical Sciences
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doi: 10.3934/dcdsb.2022047
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## Synchronization of dynamical systems on Riemannian manifolds by an extended PID-type control theory: Numerical evaluation
1 Dipartimento di Ingegneria dell'Informazione, Università Politecnica delle Marche, Via Brecce Bianche, Ancona, 60131, Italy 2 Graduate School of Information and Automation Engineering, Università Politecnica delle Marche, Via Brecce Bianche, Ancona, 60131, Italy
*Corresponding author: Simone Fiori
Received September 2021 Revised December 2022 Early access March 2022
The present document outlines a non-linear control theory, based on the PID regulation scheme, to synchronize two second-order dynamical systems insisting on a Riemannian manifold. The devised extended PID scheme, referred to as M-PID, includes an unconventional component, termed 'canceling component', whose purpose is to cancel the natural dynamics of a system and to replace it with a desired dynamics. In addition, this document presents numerical recipes to implement such systems, as well as the devised control scheme, on a computing platform and a large number of numerical simulation results focused on the synchronization of Duffing-like non-linear oscillators on the unit sphere. Detailed numerical evaluations show that the canceling contribution of the M-PID control scheme is not critical to the synchronization of two oscillators, however, it possesses the beneficial effect of speeding up their synchronization. Simulation results obtained in non-ideal conditions, namely in the presence of additive disturbances and delays, reveal that the devised synchronization scheme is robust against high-frequency additive disturbances as well as against observation delays.
Citation: Simone Fiori, Italo Cervigni, Mattia Ippoliti, Claudio Menotta. Synchronization of dynamical systems on Riemannian manifolds by an extended PID-type control theory: Numerical evaluation. Discrete and Continuous Dynamical Systems - B, doi: 10.3934/dcdsb.2022047
##### References:
[1] D. P. Atherton, Almost six decades in control engineering, IEEE Control Systems Magazine, 34 (2014), 103-110. [2] A. M. Bloch, An Introduction to Aspects of Geometric Control Theory, in Nonholonomic Mechanics and Control (eds. P. Krishnaprasad and R. Murray), vol. 24 of Interdisciplinary Applied Mathematics, Springer, New York, NY, 2015. [3] F. Bullo and A. D. Lewis, Geometric Control of Mechanical Systems, vol. 49 of Texts in Applied Mathematics, Springer Verlag, New York-Heidelberg-Berlin, 2004. [4] J. C. Butcher, Runge-Kutta Methods, chapter 3, John Wiley & Sons, Ltd, 2016. doi: 10.1002/9781119121534.ch3. [5] G. Chen and X. Yu, Chaos Control – Theory and Applications, Lecture Notes in Control and Information Sciences, Springer, 2003. doi: 10.1007/b79666. [6] L. Cong, J. Mu, Q. Liu, H. Wang, L. Wang, Y. Li and C. Qiao, Thermal noise decoupling of micro-Newton thrust measured in a torsion balance, Symmetry, 13 (2021), 1357. doi: 10.3390/sym13081357. [7] D. N. Das, R. Sewani, J. Wang and M. K. Tiwari, Synchronized truck and drone routing in package delivery logistics, IEEE Transactions on Intelligent Transportation Systems, 1–11. [8] P. Deng, G. Amirjamshidi and M. Roorda, A vehicle routing problem with movement synchronization of drones, sidewalk robots, or foot-walkers, Transportation Research Procedia, 46 (2020), 29-36. doi: 10.1016/j.trpro.2020.03.160. [9] R. Dhelika, A. F. Hadi and P. A. Yusuf, Development of a motorized hospital bed with swerve drive modules for holonomic mobility, Applied Sciences, 11 (2021), 11356. doi: 10.3390/app112311356. [10] S. Fiori, Nonlinear damped oscillators on Riemannian manifolds: Numerical simulation, Communications in Nonlinear Science and Numerical Simulation, 47 (2017), 207–222, URL http://www.sciencedirect.com/science/article/pii/S1007570416304932. doi: 10.1016/j.cnsns.2016.11.025. [11] S. Fiori, Non-delayed synchronization of non-autonomous dynamical systems on Riemannian manifolds and its applications, Nonlinear Dynamics, 94 (2018), 3077-3100. doi: 10.1007/s11071-018-4546-x. [12] S. Fiori, Extension of a PID control theory to Lie groups applied to synchronising satellites and drones, IET Control Theory & Applications, 14 (2020), 2628-2642. doi: 10.1049/iet-cta.2020.0226. [13] S. Fiori, Manifold calculus in system theory and control–Fundamentals and first-order systems, Symmetry, 13 (2021), 2092. doi: 10.3390/sym13112092. [14] R. Fuentes, G. P. Hicks and J. M. Osborne, The spring paradigm in tracking control of simple mechanical systems, Automatica, 47 (2011), 993-1000. doi: 10.1016/j.automatica.2011.01.046. [15] S. Gajbhiye and R. N. Banavar, The Euler-Poincaré equations for a spherical robot actuated by a pendulum, IFAC Proceedings Volumes, 45 (2012), 72–77, URL http://www.sciencedirect.com/science/article/pii/S1474667015337459, 4th IFAC Workshop on Lagrangian and Hamiltonian Methods for Non Linear Control. doi: 10.3182/20120829-3-IT-4022.00011. [16] V. Ghaffari and F. Shabaninia, Synchronization of nonlinear dynamical systems using extended Kalman filter and its application in some well-known chaotic systems, Nonlinear Studies, 25 (2018), 273-286. [17] O. Golevych, O. Pyvovar and P. Dumenko, Synchronization of non-linear dynamic systems under the conditions of noise action in the channel, Latvian Journal of Physics and Technical Sciences, 55 (2018), 70-76. doi: 10.2478/lpts-2018-0023. [18] I. Kovacic and M. J. Brennan, The Duffing Equation: Nonlinear Oscillators and their Behaviour, John Wiley & Sons, Ltd., Chichester, 2011. doi: 10.1002/9780470977859. [19] Y. Li, L. Li and C. Zhang, AMT starting control as a soft starter for belt conveyors using a data-driven method, Symmetry, 13 (2021), 1808. doi: 10.3390/sym13101808. [20] M. A. Magdy and T. S. Ng, Regulation and control effort in self-tuning controllers, IEE Proceedings D – Control Theory and Applications, 133 (1986), 289-292. doi: 10.1049/ip-d.1986.0046. [21] J. Markdahl, Synchronization on Riemannian manifolds: Multiply connected implies multistable, IEEE Transactions on Automatic Control, 66 (2021), 4311-4318. doi: 10.1109/TAC.2020.3030849. [22] A. Návrat and P. Vašík, On geometric control models of a robotic snake, Note di Matematica, 37 (2017), 120-129. doi: 10.1285/i15900932v37suppl1p119. [23] M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, Cambridge, 2000. [24] K. Ojo, S. Ogunjo and A. Olagundoye, Projective synchronization via active control of identical chaotic oscillators with parametric and external excitation, International Journal of Nonlinear Science, 24 (2017), 76-83. [25] J. M. Osborne and G. P. Hicks, The geodesic spring on the Euclidean sphere with parallel-transport-based damping, Notices of the AMS, 60 (2013), 544-556. doi: 10.1090/noti997. [26] Y.-s. Reddy and S.-h. Hur, Comparison of optimal control designs for a 5 MW wind turbine, Applied Sciences, 11 (2021), 8774. doi: 10.3390/app11188774. [27] L. Righetti, Control and Synchronization with Nonlinear Dynamical Systems for an Application to Humanoid Robotics, Ecole Polytechnique Fédérale de Lausanne, 2004, URL https://nyuscholars.nyu.edu/en/publications/control-and-synchronization-with-nonlinear-dynamical-systems-for-. [28] R. W. H. Sargent, Optimal control, Computational and Applied Mathematics, 124 (2000), 361-371. doi: 10.1016/S0377-0427(00)00418-0. [29] M. Shiino and K. Okumura, Control of attractors in nonlinear dynamical systems using external noise: Effects of noise on synchronization phenomena, Discrete and Continuous Dynamical Systems - Series S, 2013 (2013), 685-694. doi: 10.3934/proc.2013.2013.685. [30] K. Sreenath, T. Lee and V. Kumar, Geometric control and differential flatness of a quadrotor UAV with a cable-suspended load, in 52nd IEEE Conference on Decision and Control, 2013, 2269–2274. [31] A. Varga, G. Eigner, I. Rudas and J. K. Tar, Experimental and simulation-based performance analysis of a computed torque control (CTC) method running on a double rotor aeromechanical testbed, Electronics, 10 (2021), 1745. doi: 10.3390/electronics10141745. [32] Y. Wang, Y. Lu and R. Xiao, Application of nonlinear adaptive control in temperature of Chinese solar greenhouses, Electronics, 10 (2021), 1582. doi: 10.1109/CCDC52312.2021.9601368. [33] C. W. Wu, Synchronization in Complex Networks of Nonlinear Dynamical Systems, World Scientific Publishing Co Pte Ltd, Singapore, 2007. doi: 10.1142/6570. [34] M. Zarei, A. Kalhor and M. Masouleh, An experimental oscillation damping impedance control for the Novint Falcon haptic device based on the phase trajectory length function concept, Proceedings of the Institution of Mechanical Engineers, Part C: Journal of Mechanical Engineering Science, 233 (2019), 2663-2672. doi: 10.1177/0954406218799779. [35] Z. Zhang, J. Cheng and Y. Guo, PD-based optimal ADRC with improved linear extended state observer, Entropy, 23 (2021), Paper No. 888, 15 pp. doi: 10.3390/e23070888. [36] Z. Zhong, M. Xu, J. Xiao and H. Lu, Design and control of an omnidirectional mobile wall-climbing robot, Applied Sciences, 11 (2021), 11065. doi: 10.3390/app112211065.
show all references
##### References:
[1] D. P. Atherton, Almost six decades in control engineering, IEEE Control Systems Magazine, 34 (2014), 103-110. [2] A. M. Bloch, An Introduction to Aspects of Geometric Control Theory, in Nonholonomic Mechanics and Control (eds. P. Krishnaprasad and R. Murray), vol. 24 of Interdisciplinary Applied Mathematics, Springer, New York, NY, 2015. [3] F. Bullo and A. D. Lewis, Geometric Control of Mechanical Systems, vol. 49 of Texts in Applied Mathematics, Springer Verlag, New York-Heidelberg-Berlin, 2004. [4] J. C. Butcher, Runge-Kutta Methods, chapter 3, John Wiley & Sons, Ltd, 2016. doi: 10.1002/9781119121534.ch3. [5] G. Chen and X. Yu, Chaos Control – Theory and Applications, Lecture Notes in Control and Information Sciences, Springer, 2003. doi: 10.1007/b79666. [6] L. Cong, J. Mu, Q. Liu, H. Wang, L. Wang, Y. Li and C. Qiao, Thermal noise decoupling of micro-Newton thrust measured in a torsion balance, Symmetry, 13 (2021), 1357. doi: 10.3390/sym13081357. [7] D. N. Das, R. Sewani, J. Wang and M. K. Tiwari, Synchronized truck and drone routing in package delivery logistics, IEEE Transactions on Intelligent Transportation Systems, 1–11. [8] P. Deng, G. Amirjamshidi and M. Roorda, A vehicle routing problem with movement synchronization of drones, sidewalk robots, or foot-walkers, Transportation Research Procedia, 46 (2020), 29-36. doi: 10.1016/j.trpro.2020.03.160. [9] R. Dhelika, A. F. Hadi and P. A. Yusuf, Development of a motorized hospital bed with swerve drive modules for holonomic mobility, Applied Sciences, 11 (2021), 11356. doi: 10.3390/app112311356. [10] S. Fiori, Nonlinear damped oscillators on Riemannian manifolds: Numerical simulation, Communications in Nonlinear Science and Numerical Simulation, 47 (2017), 207–222, URL http://www.sciencedirect.com/science/article/pii/S1007570416304932. doi: 10.1016/j.cnsns.2016.11.025. [11] S. Fiori, Non-delayed synchronization of non-autonomous dynamical systems on Riemannian manifolds and its applications, Nonlinear Dynamics, 94 (2018), 3077-3100. doi: 10.1007/s11071-018-4546-x. [12] S. Fiori, Extension of a PID control theory to Lie groups applied to synchronising satellites and drones, IET Control Theory & Applications, 14 (2020), 2628-2642. doi: 10.1049/iet-cta.2020.0226. [13] S. Fiori, Manifold calculus in system theory and control–Fundamentals and first-order systems, Symmetry, 13 (2021), 2092. doi: 10.3390/sym13112092. [14] R. Fuentes, G. P. Hicks and J. M. Osborne, The spring paradigm in tracking control of simple mechanical systems, Automatica, 47 (2011), 993-1000. doi: 10.1016/j.automatica.2011.01.046. [15] S. Gajbhiye and R. N. Banavar, The Euler-Poincaré equations for a spherical robot actuated by a pendulum, IFAC Proceedings Volumes, 45 (2012), 72–77, URL http://www.sciencedirect.com/science/article/pii/S1474667015337459, 4th IFAC Workshop on Lagrangian and Hamiltonian Methods for Non Linear Control. doi: 10.3182/20120829-3-IT-4022.00011. [16] V. Ghaffari and F. Shabaninia, Synchronization of nonlinear dynamical systems using extended Kalman filter and its application in some well-known chaotic systems, Nonlinear Studies, 25 (2018), 273-286. [17] O. Golevych, O. Pyvovar and P. Dumenko, Synchronization of non-linear dynamic systems under the conditions of noise action in the channel, Latvian Journal of Physics and Technical Sciences, 55 (2018), 70-76. doi: 10.2478/lpts-2018-0023. [18] I. Kovacic and M. J. Brennan, The Duffing Equation: Nonlinear Oscillators and their Behaviour, John Wiley & Sons, Ltd., Chichester, 2011. doi: 10.1002/9780470977859. [19] Y. Li, L. Li and C. Zhang, AMT starting control as a soft starter for belt conveyors using a data-driven method, Symmetry, 13 (2021), 1808. doi: 10.3390/sym13101808. [20] M. A. Magdy and T. S. Ng, Regulation and control effort in self-tuning controllers, IEE Proceedings D – Control Theory and Applications, 133 (1986), 289-292. doi: 10.1049/ip-d.1986.0046. [21] J. Markdahl, Synchronization on Riemannian manifolds: Multiply connected implies multistable, IEEE Transactions on Automatic Control, 66 (2021), 4311-4318. doi: 10.1109/TAC.2020.3030849. [22] A. Návrat and P. Vašík, On geometric control models of a robotic snake, Note di Matematica, 37 (2017), 120-129. doi: 10.1285/i15900932v37suppl1p119. [23] M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, Cambridge, 2000. [24] K. Ojo, S. Ogunjo and A. Olagundoye, Projective synchronization via active control of identical chaotic oscillators with parametric and external excitation, International Journal of Nonlinear Science, 24 (2017), 76-83. [25] J. M. Osborne and G. P. Hicks, The geodesic spring on the Euclidean sphere with parallel-transport-based damping, Notices of the AMS, 60 (2013), 544-556. doi: 10.1090/noti997. [26] Y.-s. Reddy and S.-h. Hur, Comparison of optimal control designs for a 5 MW wind turbine, Applied Sciences, 11 (2021), 8774. doi: 10.3390/app11188774. [27] L. Righetti, Control and Synchronization with Nonlinear Dynamical Systems for an Application to Humanoid Robotics, Ecole Polytechnique Fédérale de Lausanne, 2004, URL https://nyuscholars.nyu.edu/en/publications/control-and-synchronization-with-nonlinear-dynamical-systems-for-. [28] R. W. H. Sargent, Optimal control, Computational and Applied Mathematics, 124 (2000), 361-371. doi: 10.1016/S0377-0427(00)00418-0. [29] M. Shiino and K. 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Wu, Synchronization in Complex Networks of Nonlinear Dynamical Systems, World Scientific Publishing Co Pte Ltd, Singapore, 2007. doi: 10.1142/6570. [34] M. Zarei, A. Kalhor and M. Masouleh, An experimental oscillation damping impedance control for the Novint Falcon haptic device based on the phase trajectory length function concept, Proceedings of the Institution of Mechanical Engineers, Part C: Journal of Mechanical Engineering Science, 233 (2019), 2663-2672. doi: 10.1177/0954406218799779. [35] Z. Zhang, J. Cheng and Y. Guo, PD-based optimal ADRC with improved linear extended state observer, Entropy, 23 (2021), Paper No. 888, 15 pp. doi: 10.3390/e23070888. [36] Z. Zhong, M. Xu, J. Xiao and H. Lu, Design and control of an omnidirectional mobile wall-climbing robot, Applied Sciences, 11 (2021), 11065. doi: 10.3390/app112211065.
Feedback control scheme, where $y_\mathrm{sp}$ denotes the set point, $y_\mathrm{m}$ denotes a measure of the controlled variable $y$, $u$ denotes a control signal and $e$ denotes an error signal, which quantifies the discrepancy between the measured controlled variable and the set point
Synchronization of two Duffing-type oscillators by a M-PID controller. The left-hand panel shows the values of the distance $d(z, x)$ on the top and of the control effort $\sigma$ on the bottom, taken by a hard Duffing oscillator and a soft Duffing oscillator. Instead, the right-hand panel shows the values of the distance $d(z, x)$ on the top and of the control effort $\sigma$ on the bottom, taken by two hard Duffing oscillators. In both cases, the follower is controlled by an M-PID controller with $u_\mathrm{C}\neq 0$ during the first $10$ seconds of the simulation and with $u_\mathrm{C}$ set to zero from $t = 10$ outward. The proportional control coefficient was set to $\kappa_\mathrm{P} = 10$, the integral control coefficient was set to $\kappa_\mathrm{I} = 5$ and the derivative control coefficient was set to $\kappa_\mathrm{D} = 10$
Synchronization of two hard Duffing oscillators where the follower is controlled by a P-controller (a M-PID with only the proportional term). In both panels the black lines are about the follower dynamics while the red lines are referred to the leader dynamics. The left-hand panel shows the trajectories on the sphere. The follower's initial velocity is represented by a green arrow. The right-hand panel shows the values taken by the kinetic energy, the potential energy, the total energy, the control effort and the state-to-state distance over the generated trajectory. The proportional control coefficient value was set to $\kappa_\mathrm{P} = 0.01$
Synchronization of two hard Duffing oscillators where the follower is controlled by a full M-PID controller. Values and graphic elements are as in the Figure 3. The integral control coefficient value was set to $\kappa_\mathrm{I} = 5$, the derivative control coefficient value was set to $\kappa_\mathrm{D} = 10$ and the proportional control coefficient value was set to $\kappa_\mathrm{P} = 10$
Synchronization of two hard Duffing oscillators where the follower is controlled by a M-PI-controller, with only the proportional and integral terms. Values and graphic elements are as in the Figure 3. The integral coefficient value was set to $\kappa_\mathrm{I} = 2.5$ and the proportional term coefficient value was set to $\kappa_\mathrm{P} = 10$
Synchronization of two hard Duffing oscillators where the follower is controlled by a M-PD-controller, with only the proportional and derivative terms. Values and graphic elements are as in the Figure 3. The derivative control coefficient value was set to $\kappa_\mathrm{D} = 10$ and the proportional control coefficient value was set to $\kappa_\mathrm{P} = 10$
Synchronization of two hard Duffing oscillators where the follower is controlled by a M-PID controller: comparison of the effects of two different values of proportional term coefficient. In both panels, the black lines denote the follower's dynamics, while the red lines denote the leader's dynamic. The right-hand panel shows the values taken by the kinetic energy, the potential energy, the total energy, the control effort and the state-to-state distance over the generated trajectory, when the proportional coefficient value was set to $\kappa_\mathrm{P} = 5$. The right-hand panel shows the same quantities when the proportional coefficient value was set to $\kappa_\mathrm{P} = 150$. In both cases, the value of the integral coefficient was set to $\kappa_\mathrm{I} = 5$ and the value of the derivative coefficient was set to $\kappa_\mathrm{D} = 5$
Synchronization of two hard Duffing oscillators where the follower is controlled by M-PID controller: comparison with two different values of derivative term coefficient. Values and graphic elements are as in the Figure 7. The left-hand panel shows results when the derivative coefficient was set to $\kappa_\mathrm{D} = 0.5$, while the right-hand panel shows results when the derivative coefficient was set to $\kappa_\mathrm{D} = 3$. In both cases, the value of the integral coefficient was set to $\kappa_\mathrm{I} = 25$ and of the proportional coefficient was set to $\kappa_\mathrm{P} = 5$
Synchronization of two hard Duffing oscillators where the follower is controlled by M-PID controller: comparison with two different values of integral term coefficient. Values and graphic elements are as in the Figure 7. The left-hand panel shows results when the integral coefficient was set to $\kappa_\mathrm{I} = 3$, while the right-hand panel shows results when the value of the integral coefficient value was set to $\kappa_\mathrm{I} = 25$. In both cases, the derivative coefficient value was set to $\kappa_\mathrm{D} = 5$ and the proportional coefficient was set to $\kappa_\mathrm{P} = 5$
Synchronization of two hard Duffing oscillators where the follower is controlled by a M-PID controller that includes the $u_\mathrm{C}$ term and the initial velocity direction of the follower is not favorable for synchronization (i.e., it is opposed to the leader trajectory). In both panels the black lines indicate the follower's dynamics, while the red lines indicate the leader's dynamics. The left-hand panel shows the trajectories of the two oscillators on the sphere over time. The follower's initial velocity is represented by a green arrow and its initial state by a green open circle, while the leader's initial state is represented by a white open circle. The right-hand panel shows the values taken by the kinetic energy, the potential energy, the total energy, the control effort and the state-to-state distance over the generated trajectory. The proportional control coefficient value is $\kappa_\mathrm{P} = 10$, the integral control coefficient value is $\kappa_\mathrm{I} = 5$ and the derivative control coefficient value is $\kappa_\mathrm{D} = 10$
Synchronization of two hard Duffing oscillators where the follower is controlled by a M-PID controller that does not include the $u_\mathrm{C}$ term and the initial velocity direction of the follower is not favorable for synchronization. Values and graphic elements are as in the Figure 10
Synchronization of two hard Duffing oscillators where the follower is controlled by a M-PID controller endowed with the $u_\mathrm{C}$ term when the initial velocity direction of the follower is favorable for synchronization. In both panels the black lines denote the follower's dynamics, while the red lines denote the leader's dynamics. The left-hand panel shows the trajectories of the two oscillators on the sphere over time. The follower's initial velocity is represented by a green arrow and its initial state by a green open circle, while the leader's initial state is represented by a white open circle. The right-hand panel shows the values taken by the kinetic energy, the potential energy, the total energy, the control effort and the state-to-state distance over the generated trajectory. In particular, the right-bottom panel shows the total control effort $\sigma$ in blue, the PID contribution $\sigma_\mathrm{PID}$ in magenta and the contribution deriving from $u_\mathrm{C}$ term in green color. The proportional control coefficient value was set to $\kappa_\mathrm{P} = 10$, the integral control coefficient value was set to $\kappa_\mathrm{I} = 5$ and the derivative control coefficient value was set to $\kappa_\mathrm{D} = 10$
Synchronization of two hard Duffing oscillators where the follower is controlled by a M-PID controller without the $u_\mathrm{C}$ term when the initial velocity direction of the follower is favorable for synchronization. Values and graphic elements are as in the Figure 12
Synchronization of two hard Duffing oscillators endowed with different reference points, where the follower is controlled by a M-PID controller. In both panels the black lines denote the follower's dynamics, while the red lines denote the leader's dynamics. The left-hand panel shows the trajectories of the two oscillators on the sphere over time. The follower's initial velocity is represented by a green arrow, while the leader's initial velocity is represented by a white arrow. The two different reference points have been represented as two colored open circles on the sphere (green for the leader reference point and white for the follower's one). The right-hand panel shows the values taken by the kinetic energy, the potential energy, the total energy, the control effort and the state-to-state distance over the generated trajectory. In this case, the leader's and follower's starting points, initial velocities and reference points are taken randomly. The proportional control coefficient value was set to $\kappa_\mathrm{P} = 10$, the coefficient of the integral control term value was set to $\kappa_\mathrm{I} = 5$ and the derivative coefficient value was set to $\kappa_\mathrm{D} = 10$
Synchronization of a hard Duffing oscillator and a soft Duffing oscillator, where the follower is controlled by a M-PID controller endowed with the presence of the $u_\mathrm{C}$ term. In both panels the black lines represent the follower's dynamic, while the red ones represent the leader's dynamic. The left-hand panel shows the trajectories of the two oscillators on the sphere over time. The follower's initial velocity is represented by a green arrow. The right-hand panel shows the values taken by the kinetic energy, the potential energy, the total energy, the control effort and the state-to-state distance over the generated trajectory. In particular, in the right-bottom panel, the total control effort $\sigma$ is represented in blue, the PID contribution $\sigma_\mathrm{PID}$ in magenta and the contribution deriving from $u_\mathrm{C}$ term in green color. The value of the proportional control coefficient was set to $\kappa_\mathrm{P} = 10$, the value of the integral control coefficient was set to $\kappa_\mathrm{I} = 5$ and the value of the derivative control coefficient was set to $\kappa_\mathrm{D} = 10$
Synchronization of a hard Duffing oscillator and a soft Duffing oscillator, where the follower is controlled by a M-PID controller without the $u_\mathrm{C}$ term. Values and graphic elements are as in the Figure 15
Synchronization of a hard Duffing oscillator and a soft Duffing oscillator, where the follower is controlled by a M-PID controller and both the follower and the leader are damped. In both panels, the black lines correspond to the follower's dynamics, while the red lines correspond to the leader's dynamics. The left-hand panel shows the trajectories of the two oscillators on the sphere over time. The follower's initial velocity is represented by a green arrow. The right-hand panel shows the values taken by the kinetic energy, the potential energy, the total energy, the control effort and the state-to-state distance over the generated trajectory. The proportional control coefficient value was set to $\kappa_\mathrm{P} = 10$, the integral control coefficient value was set to $\kappa_\mathrm{I} = 5$ and the derivative coefficient value was set to $\kappa_\mathrm{D} = 10$. In this case, the value of the damping coefficient was set to $\mu = 0.1$
Synchronization of a hard Duffing oscillator and a soft Duffing oscillator, where the follower is controlled by a M-PID controller and both the follower and the leader are damped. Values and graphic elements are as in the Figure 17. In this case the value of the damping coefficient was set to $\mu = 0.9$
Synchronization of a hard Duffing oscillator and a soft Duffing oscillator, where the follower is controlled by a M-PID controller, the state/velocity measurements are affected by random disturbances. In both panels the black lines denote the follower's dynamics, while the red lines denote the leader's dynamics. The left-hand panel shows the trajectories of the two oscillators on a sphere over time. The follower's initial velocity is represented by a green arrow. The right-hand panel shows the values taken by the kinetic energy, the potential energy, the total energy, the control effort and the state-to-state distance over the generated trajectory. The proportional control coefficient value was set to $\kappa_\mathrm{P} = 10$, the integral control coefficient value was set to $\kappa_\mathrm{I} = 5$ and the derivative coefficient value was set to $\kappa_\mathrm{D} = 10$. In this case, the value of the coefficient $b_z$ was set to $50$ and the value of the coefficient $b_w$ was set to $0.01$
Synchronization of a hard Duffing oscillator and a soft Duffing oscillator, where the follower is controlled by a M-PID controller and the state/velocity measurements are affected by random disturbances. Values and graphic elements are as in the Figure 19 except that, in this case, the value of the coefficient $b_z$ was set to $300$
Synchronization of an hard Duffing oscillator and a soft Duffing oscillator, where the follower is controlled by a M-PID controller and the state/velocity measurements are affected by random disturbances. Values and graphic elements are as in the Figure 19 except that, in this case, the value of the coefficient $b_z$ was set to $300$ and the value of the coefficient $b_w$ was set to $0.1$
Synchronization of a hard Duffing oscillator and a soft Duffing oscillator, where the follower is controlled by a M-PID controller and the state/velocity measurements are affected by sinusoidal disturbances. In both panels the black lines denote the follower's dynamics, while the red lines denote the leader's dynamics. The left-hand panel shows the trajectories of the two oscillators on the sphere over time. The follower's initial velocity is represented by a green arrow. The right-hand panel shows the values taken by the kinetic energy, the potential energy, the total energy, the control effort and the state-to-state distance over the generated trajectory. The proportional control coefficient value was set to $\kappa_\mathrm{P} = 10$, the integral control coefficient value was set to $\kappa_\mathrm{I} = 5$ and the derivative coefficient value was set to $\kappa_\mathrm{D} = 10$. In this simulation, the noise coefficients were set to $b_z = 300$ and $b_w = 0.01$ and the sinusoidal disturbance's angular frequency was set to $\Omega = 1,000$ rad/s
Synchronization of a hard Duffing oscillator and a soft Duffing oscillator, where the follower is controlled by a M-PID controller and the state/velocity measurements are affected by sinusoidal disturbances. Values and graphic elements are as in the Figure 22 except that the sinusoidal disturbance's angular frequency was set to $\Omega = 5$ rad/s
Synchronization of a hard Duffing oscillator and a soft Duffing oscillator, where the follower is controlled by a M-PID controller. In both panels the black lines denote the follower's dynamics, while the red lines denote the leader's dynamics. The left-hand panel shows the trajectories of the two oscillators on the sphere over time. The right-hand panel shows the values taken by the kinetic energy, the potential energy and the state-to-state distances. The proportional control coefficient value was set to $\kappa_\mathrm{P} = 10$, the integral control coefficient value was set to $\kappa_\mathrm{I} = 5$ and the derivative coefficient value was set to $\kappa_\mathrm{D} = 10$. In this case, the time delay was set to $l = 0.4$ seconds and $h = 0.0002$ was the chosen value of the stepsize
Synchronization of a hard Duffing oscillator and a soft Duffing oscillator, where the follower is controlled by a M-PID controller in the presence of both time-delay and additive noise. Values and graphic elements are as in the Figure 24 and $b_z = 300$ and $b_w = 0.01$. As it can be noticed the total energy of the leader oscillator is constant over time, because it is supposed not to be subjected to external forces. In fact leader's total energy is preserved over time, while follower's total energy is not constant being subjected to the control action of the PID system
Numerical experiments under ideal conditions
Experiment type Control method Results Figures Syncing of two identical oscillators and of two different oscillators. M-PID with $u_\mathrm{C}\neq 0$ only in the first half of time. When the systems are identical, switching off the canceling component does not hinder syncing, while different systems cannot sync without the aid of the canceling component. 2 Syncing of two identical oscillators. Proportional control action only. No sync achieved. 3 Syncing of two identical oscillators. Proportional, integral and derivative control actions. Sync achieved quickly and smoothly. 4 Syncing of two identical oscillators. Proportional and integral control actions. Sync achieved slowly. 5 Syncing of two identical oscillators. Proportional and derivative control actions. Short initial transient compared to the full M-PID case. 6 Syncing of two identical oscillators. Full M-PID with two different values of the proportional term coefficient. Overshoot more apparent for a higher value of the proportional action coefficient. 7 Syncing of two identical oscillators. Full M-PID with two different values of the derivative term coefficient. Quicker convergence for higher value of the derivative action coefficient. 8 Syncing of two identical oscillators. Full M-PID with two different values of the integral term coefficient. Large values of this coefficient entail quicker convergence at the expense of larger oscillations around the set point. 9 Syncing of two identical oscillators. M-PID controller including the canceling term vs. not including the canceling term. Unfavorable initial velocity. Shorter initial transient vs. longer initial transient to convergence. 10 & 11 Syncing of two identical oscillators. M-PID controller including the canceling term vs. not including the canceling term. Favorable initial velocity. Longer initial transient vs. shorter initial transient to convergence. 12 & 13 Syncing of two different oscillators of the same species. Full M-PID controller. Synchronization achieved. 14 Syncing of two different oscillators. M-PID controller with canceling term vs. no canceling term. Syncronization achieved thanks to the canceling term. Absence of the canceling terms makes syncing almost to no avail. 15 & 16
Experiment type Control method Results Figures Syncing of two identical oscillators and of two different oscillators. M-PID with $u_\mathrm{C}\neq 0$ only in the first half of time. When the systems are identical, switching off the canceling component does not hinder syncing, while different systems cannot sync without the aid of the canceling component. 2 Syncing of two identical oscillators. Proportional control action only. No sync achieved. 3 Syncing of two identical oscillators. Proportional, integral and derivative control actions. Sync achieved quickly and smoothly. 4 Syncing of two identical oscillators. Proportional and integral control actions. Sync achieved slowly. 5 Syncing of two identical oscillators. Proportional and derivative control actions. Short initial transient compared to the full M-PID case. 6 Syncing of two identical oscillators. Full M-PID with two different values of the proportional term coefficient. Overshoot more apparent for a higher value of the proportional action coefficient. 7 Syncing of two identical oscillators. Full M-PID with two different values of the derivative term coefficient. Quicker convergence for higher value of the derivative action coefficient. 8 Syncing of two identical oscillators. Full M-PID with two different values of the integral term coefficient. Large values of this coefficient entail quicker convergence at the expense of larger oscillations around the set point. 9 Syncing of two identical oscillators. M-PID controller including the canceling term vs. not including the canceling term. Unfavorable initial velocity. Shorter initial transient vs. longer initial transient to convergence. 10 & 11 Syncing of two identical oscillators. M-PID controller including the canceling term vs. not including the canceling term. Favorable initial velocity. Longer initial transient vs. shorter initial transient to convergence. 12 & 13 Syncing of two different oscillators of the same species. Full M-PID controller. Synchronization achieved. 14 Syncing of two different oscillators. M-PID controller with canceling term vs. no canceling term. Syncronization achieved thanks to the canceling term. Absence of the canceling terms makes syncing almost to no avail. 15 & 16
Numerical experiments under ideal conditions
Experiment type Control method Results Figures Syncing of two different oscillators, either moderately or severely damped. Full M-PID controller. Moderate damping makes syncing possible before collapsing of trajectories, while severe damping prevents synchronization. 17 & 18 Syncing of two different oscillators, where state/velocity measurements are affected by random disturbances. Full M-PID controller. Syncing is achieved, although a larger noise level makes the syncing process slower. Large control efforts required. 19, 20 & 21 Syncing of two different oscillators, where state/velocity measurements are affected by a sinusoidal disturbance. Full M-PID controller. Syncing is achieved in the presence of a very fast-oscillating disturbance, where a slow-oscillating disturbance disrupts the syncing process. 22 & 23 Syncing of two different oscillators in the presence of a (known) observation delay. Full M-PID controller. Synchronization is achieved to the delayed leader's state. 24 Syncing of two different oscillators in the presence of a (known) observation delay and large state/velocity observation disturbance. Full M-PID controller. Syncing is achieved to the delayed leader's state. Large control efforts required. 25
Experiment type Control method Results Figures Syncing of two different oscillators, either moderately or severely damped. Full M-PID controller. Moderate damping makes syncing possible before collapsing of trajectories, while severe damping prevents synchronization. 17 & 18 Syncing of two different oscillators, where state/velocity measurements are affected by random disturbances. Full M-PID controller. Syncing is achieved, although a larger noise level makes the syncing process slower. Large control efforts required. 19, 20 & 21 Syncing of two different oscillators, where state/velocity measurements are affected by a sinusoidal disturbance. Full M-PID controller. Syncing is achieved in the presence of a very fast-oscillating disturbance, where a slow-oscillating disturbance disrupts the syncing process. 22 & 23 Syncing of two different oscillators in the presence of a (known) observation delay. Full M-PID controller. Synchronization is achieved to the delayed leader's state. 24 Syncing of two different oscillators in the presence of a (known) observation delay and large state/velocity observation disturbance. Full M-PID controller. Syncing is achieved to the delayed leader's state. Large control efforts required. 25
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Volume 9 (2013) Article 16 pp. 579-585
Guest Editors' Foreword
The Boolean function $f : \{0,1\}^n\to\{0,1\}$ is one of the most fundamental objects in theoretical computer science. The field of Analysis of Boolean Functions seeks to understand Boolean functions via their Fourier transform and other analytic methods. This field has seen major developments in the 25 years since the watershed paper of Kahn, Kalai, and Linial [KKL88]. It is now an indispensable tool in the theory of computing, with applications in areas as diverse as algorithmic economics [BL85], average-case complexity [O'D04], circuit complexity [LMN93], coding theory [KL95], communication complexity [GKKRW08], computational geometry [MNP07], cryptography [Sie84], derandomization [NN93], genetic algorithms [HW04], inapproximability [Has01], learning theory [Man94], noisy communication [GKS08], property testing [BCHKS96], and quantum computing [BRW08]. Moreover, this area of research has proved to have significant applications in other areas of mathematics that are not as obviously connected with Boolean functions: additive combinatorics [Gre04], Banach spaces [BMW86], extremal combinatorics [ADFS04], Gaussian geometry [Bob97], metric space embeddings [KN06], social choice [Kal02], statistical physics [SS10], and threshold phenomena of random graphs [FriB99].
Over the last decade, a number of powerful themes have emerged in the analysis of Boolean functions: the dichotomy between "juntas" and functions with small influences; noise sensitivity/stability; the "small-set expansion" of the Boolean cube; the role of Gaussian analysis/geometry; probabilistic invariance principles; and regularity lemmas. This deepening understanding of the field has led to quite a few exciting developments in the last five years. To cite just a few examples:
• In the field of inapproximability, two recent STOC "Best Papers": Raghavendra's theory of CSP approximability [Rag10] and Chan's breakthrough on the NP-hardness of Max-kCSP [Cha12].
• In Gaussian geometry, a new proof of Borell's Isoperimetric Inequality [DMN12].
• The developments in percolation theory due to Garban, Pete, and Schramm [GPS10].
• Hatami's characterization of monotone set properties without sharp thresholds [Hat10].
• In concrete complexity, Kane's near resolution of the Gotsman-Linial Conjecture [Kan12].
• In additive combinatorics, Sanders's Quasipolynomial Freiman-Ruzsa Theorem [San12].
In light of these developments, Theory of Computing is devoting a special issue to the field of analysis of Boolean functions. The idea for this special issue was conceived at the Simons Symposium Analysis of Boolean Functions: New Directions and Application held at Caneel Bay, US Virgin Islands, February 5-11, 2012. Submissions were solicited in late 2012; the papers appearing were selected from those submitted, after reviews following the stringent standards of Theory of Computing. The papers will be released individually in upcoming months.
We thank the authors of these papers for their contributions, and the anonymous reviewers for their meticulous and timely work. We would also like to thank Laci Babai, Oded Regev, and Ronald de Wolf for their editorial efforts.
June 10, 2013
Elchanan Mossel
Ryan O'Donnell
Guest Editors
Further resources:
Courses on analysis of Boolean functions:
Other recent and upcoming workshops and semesters on discrete analysis:
Key "classical" papers:
Further lectures, videos, surveys, and other writings:
### List of papers currently accepted for publication in the Special Issue
• Making polynomials robust to noise, by Alexander Sherstov
• Note on low-depth monotone functions, by Daniel Kane
• Hypercontractivity via the entropy method, by Eric Blais and Li-Yang Tan
• Tight Bounds for Monotone Switching Networks via Fourier Analysis, by Siu Man Chan and Aaron Potechin
|
Rating: 5.0
## CIA
### TL;DR
Use Dynamic Programming, with a state of [width, height] and try all reasonable cuts.
### Short Description
Repeatedly cut a bar of choclate, in a way that:
1. resulting pieces match one of the given forms
2. the waste of those *not* matching is as low as possible
Important: Each cut is of the form: Take one piece and cut it with a single straight horizontal or vertical cut into exactly two pieces. Just like breaking the piece of chocolate.
_____________
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|______| |
| |______|
|______| |
| | |
|______|______|
_____________ <- aa
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|________| |
| |____|
|________| |
| | |
|________|____|
|_____________|
### Insights
1. If you look at the final cutting pattern of the chocolate, there is (at least) one cut running entirely from top to bottom or left ro right( line a above). After this cut, there are two pieces, each of which has the same condition on its pattern. (Except for the smallest part)
2. When we have two pieces, we can solve them individually and add their waste.
3. To compute the waste, we only need the current dimensions of the piece. The history of breaking is unimportant.
4. If we do a cut horizontally, one side can have an arbitrary height, but at least one side of the cut (above the red line) has a height that is the sum of some of our given reactangles' heights. (Or to be precise: It is always possible to find a cut like this where no better other option exists. Proof is best done manually on a paper. In short: we can slightly shift the cut position without changing the solution).
### Solution
1. Start by computing all possible heights to cut (according to insight 4):
1. create an array hs of length total_height+1 of type boolean. If the value is true, this height is the sum of some heights.
2. Mark hs[0] = true (height 0 is assumed to be valid)
3. Take the first rectangle with height h, go through the array hs and if hs[i] == true set hs[i + h] = true as well (height i is ok, so i+h is ok as well. Now all multiples of h are marked as valid cuts.
4. Repeat the step (3) with the other rectangles on the same array.
5. We have marked all reasonable cutting heights
6. Run through the array one more time and store all is where hs[i] == true.
7. Complexity of this is O(height * |rectangles|)
8. Do 1-7 again for the widths
2. Now, assume we have a piece of dimensions r x c. If this is a valid rectangle already, the waste is 0. Otherwise, we try all possible heights from step 1, cut the piece and find the best solution for both subpieces. Of all those cuts, we take that one with the smallest sum of results of the subpieces. So, cutting at height h wastes: waste_cut(r, c, cutheight) = waste(height, c) + waste(r - height, c). We try all vertical cuts as well. Of all those options, we take the best.
3. There is one more important thing: The above is too slow because we compute many piece sizes multiple times. (e.g., a bar of size 3 x 8is cut into 3 x 4and 3 x 4, we only need to compute the result for 3 x 4 once (insight 3). We can store the result and use it later if we come along an identical piece again. -> Memoization / Dynamic Programming. To do so, simply create a large 2-dimensional array dpwith r+1 rows and c+1columns. dp[a][b] is the minimal waste possible for a piece of dimensions a x b. Whenever you need to compute a rectangle, check if this is already computed and return it if yes.
4. Complexity of this is O(r x c x (|cut heights| + |cut widths|)). Side note: If there is a rectangle of very small size, this will be too slow, because |cut heights| and |cut widths|will be huge. But the server's rectangles had the same magnitude than the chocolate.
### Implementation
I had a python script to communicate with the server and a C++ code to compute the answer (probably not needed, but , well..)
This code is only the interesting parts of the C++ code. Note, that I did not implement it recursivly as it is indicated above, but I fill the array dp from left upper corner to the lower right corder. In this order, I ensure, that the values I need are already computed beforehand and no check is needed.
// compute all possible cutting heights. chs is the boolean array
// fs[i] is the i'th rectangle. fs[i].first is its height, fs[i].second is its width
for (int x = 0; x < chs.size() - fs[i].first; x++) if (chs[x]) chs[x+fs[i].first] = true;
for (int x = 0; x < cws.size() - fs[i].second; x++) if (cws[x]) cws[x+fs[i].second] = true;
ll solve(ll r, ll c) {
// dp is out array. dp[x][y] holds the best value for piece with size x * y (best = minimal waste)
// currently, the best option we know is wasting everything:
for (ll x = 1; x <= r; x++) for (ll y = 1; y <= c; y++) {
dp[x][y] = x * y;
}
// All pieces matching the rectangles perfectly have zero waste:
for (pll p : fs) dp[p.first][p.second] = 0;
// Now compute the minimal waste for each possible dimension (x * y)
for (ll x = 1; x <= r; x++) for (ll y = 1; y <= c; y++) {
// Try all possible vertical cuts
for (ll w : cutws) {
if (w > y) break; // cuting more than we have
// dp[x][y-w] is the best option for the right side (we already computed this and stored it in dp)
// dp[x][w] is the best option for the left side (we already computed this and stored it in dp)
dp[x][y] = min(dp[x][y], dp[x][y-w] + dp[x][w]);
}
// Try all possible horizontal cuts
for (ll h : cuths) {
if (h > x) break;
dp[x][y] = min(dp[x][y], dp[h][y] + dp[x-h][y]);
}
}
return dp[r][c];
}
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## Subsets and Power Sets
The following is the list of problems for Sections 1.3 and 1.4 of the Book of Proof (pages 14, 16). There is a forum open at the end, so you can ask questions. It is a great way to interact with the instructor and with other students in your class, should you need some assistance with any question. Please, do not post solutions.
### Exercises for Section 1.3
A. List all the subsets of the following sets.
1. $$\{ 1, 2, 3, 4 \}$$
2. $$\{ 1, 2, \emptyset \}$$
3. $$\{ \{ \mathbb{R} \} \}$$
4. $$\emptyset$$
5. $$\{ \emptyset \}$$
6. $$\{ \mathbb{R}, \mathbb{Q}, \mathbb{N} \}$$
7. $$\{ \mathbb{R}, \{ \mathbb{Q}, \mathbb{N} \} \}$$
8. $$\{ \{ 0, 1 \}, \{ 0, 1, \{ 2 \} \}, \{ 0 \} \}$$
B. Write the following sets by listing their elements between braces.
1. $$\{ X : X \subseteq \{ 3, 2, a \} \text{ and } \lvert X \rvert = 2 \}$$
2. $$\{ X \subseteq \mathbb{N} : \lvert X \rvert \leq 1 \}$$
3. $$\{ X : X \subseteq \{ 3, 2, a \} \text{ and } \lvert X \rvert = 4 \}$$
4. $$\{ X : X \subseteq \{ 3, 2, a \} \text{ and } \lvert X \rvert = 1 \}$$
C. Decide if the following statements are true or false. Explain.
1. $$\mathbb{R}^3 \subseteq \mathbb{R}^3$$
2. $$\mathbb{R}^2 \subseteq \mathbb{R}^3$$
3. $$\{ (x,y) : x-1 =0 \} \subseteq \{ (x,y) : x^2-x=0 \}$$
4. $$\{ (x,y) : x^2-x =0 \} \subseteq \{ (x,y) : x-1=0 \}$$
### Exercises for Section 1.4
A. Find the indicated sets.
1. $$\mathscr{P}\big( \{ \{ a, b \}, \{ c \} \} \big)$$
2. $$\mathscr{P} \big( \{ 1, 2, 3, 4 \} \big)$$
3. $$\mathscr{P} \big( \{ \{ \emptyset \}, 5 \} \big)$$
4. $$\mathscr{P} \big( \{ \mathbb{R}, \mathbb{Q} \} \big)$$
5. $$\mathscr{P} \big(\mathscr{P} (\{ 2 \}) \big)$$
6. $$\mathscr{P} \big( \{ 1, 2 \} \big) \times\mathscr{P} (\{ 3 \})$$
7. $$\mathscr{P} \big( \{ a, b \} \big) \times\mathscr{P} \big(\{ 0, 1 \} \big)$$
8. $$\mathscr{P} \big( \{ 1, 2 \} \times \{ 3 \} \big)$$
9. $$\mathscr{P} \big( \{ a, b \} \times \{ 0 \} \big)$$
10. $$\{ X \in\mathscr{P} ( \{ 1, 2, 3 \} ) : \lvert X \rvert \leq 1 \}$$
11. $$\{ X \subseteq\mathscr{P} ( \{ 1, 2, 3 \} ) : \lvert X \rvert \leq 1 \}$$
12. $$\{ X \in\mathscr{P} ( \{ 1, 2, 3 \} ) : 2 \in X \}$$
B. Suppose that $$\lvert A \rvert = m$$ and $$\lvert B \rvert = n$$. Find the following cardinals.
1. $$\big\lvert\mathscr{P} \big(\mathscr{P} \big(\mathscr{P} (A) \big) \big) \big\rvert$$
2. $$\big\lvert\mathscr{P} \big(\mathscr{P} (A) \big) \big\rvert$$
3. $$\big\lvert\mathscr{P} (A \times B) \big\rvert$$
4. $$\big\lvert\mathscr{P}(A) \times\mathscr{P}(B) \big\rvert$$
5. $$\big\lvert \{ X \in\mathscr{P}(A) : \lvert X \rvert \leq 1 \} \big\rvert$$
6. $$\big\lvert\mathscr{P} \big( A \times\mathscr{P}(B) \big) \big\rvert$$
7. $$\big\lvert\mathscr{P} \big(\mathscr{P} \big(\mathscr{P} (A \times \emptyset) \big) \big) \big\rvert$$
8. $$\big\lvert \{ X \subseteq\mathscr{P}(A) : \lvert X \rvert \leq 1 \} \big\rvert$$
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# Solved papers for MGIMS WARDHA MGIMS WARDHA Solved Paper-2014
### done MGIMS WARDHA Solved Paper-2014
• question_answer1) Two particles A and B of masses m and 2m have charges$q$and$2q$respectively. Both particles moving with velocities${{\upsilon }_{1}}$and${{\upsilon }_{2}}$ respectively in the same direction and enter the same magnetic field B acting normally to their direction of motion. If the two forces${{F}_{A}}$and${{F}_{B}}$acting on them are in the ratio of$1:2,$ the ratio of their velocities is
A) $2:1$
B) $3:2$
C) $2:3$
D) $1:1$
• question_answer2) A narrow beam of protons and deuterons, each having the same momentum, enters a region of uniform magnetic field directed perpendicular to their direction of momentum. The ratio of the radii of the circular paths described by them is
A) $2:5$
B) $3:2$
C) $2:3$
D) $1:1$
• question_answer3) A horizontal overhead power line carries a current of, 90 A in east to west direction. The magnitude and direction of the magnetic field due to the current 1.5m below the line is
A) $2.5\times {{10}^{-5}}T$towards north
B) $3\times {{10}^{-5}}T$ towards east
C) $1.2\times {{10}^{-5}}T$towards south
D) $1\times {{10}^{-5}}T$ towards west
• question_answer4) A galvanometer has a resistance of$30\,\Omega$. It gives full scale deflection with a current of 2mA. Calculate, the value of the resistance needed to convert it into an ammeter of range 0-0.3 A. The value of the resistance needed to convert galvanometer into ammeter
A) $0.2\,\,\Omega$
B) $0.3\,\,\Omega$
C) $0.4\,\,\Omega$
D) $0.52\,\,\Omega$
• question_answer5) A tank is filled with water to a height of 12.4 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.3 cm. If water is replaced by a liquid of refractive index 1.6 upto the same height, by what distance would the microscope have to be moved to focus on the needle again
A) 4cm
B) 3cm
C) 2.18 cm
D) 2.05 cm
E) None of these above
• question_answer6) A ray of light falls on a transparent, right-angled isosceles prism made from a glass of refractive index$\sqrt{2}$. The incident ray falls normally on one of the equal sides of this prism. The ray will suffer
A) total internal reflection
B) Interference
C) refraction with angle of refraction$90{}^\circ$
D) None of the above
• question_answer7) The magnifying power of a telescope in its normal adjustment is 20. If the length of the telescope is 105 cm in this adjustment, the focal length of the objective lens is
A) 100cm
B) 5cm
C) 80cm
D) 95cm
• question_answer8) Two convex lenses of same focal lengths but of aperture${{A}_{1}}$and${{A}_{2}}$$({{A}_{2}}<{{A}_{1}}),$are used as the objective lenses in two astronomical telescopes having identical eye-pieces. The ratio of their resolving powers is
A) $\frac{{{A}_{1}}}{{{A}_{2}}}$
B) $1:1$
C) $\frac{{{A}_{2}}}{{{A}_{1}}}$
D) data insufficient
• question_answer9) A beam of light converges at a point P. A concave lens of focal length 16 cm is placed in the path of this beam 12 cm from P. The location of the point, from the lens, at which the beam would now converge is
A) 48 cm
B) 45 cm
C) 80 cm
D) 90 cm
• question_answer10) A nucleus undergoes $\beta -$decay. The mass number and atomic number change as
A) mass number and atomic number both increases
B) mass number remain same and atomic number increases
C) mass number increases and atomic number remain same
D) mass number and atomic number both decreases
• question_answer11) The graph given below showing the variation of stopping potential with the frequency of incident radiation for two different photosensitive materials having work functions ${{W}_{1}}$and ${{W}_{2}}({{W}_{1}}<{{W}_{2}})$ The slope of graph showing the variation of stopping potential with the frequency of incident radiation gives the value of
A) h/e where, h is Plancks constant and e electronic charge
B) h i.e., Plancks constant
C) e i.e., electronic charge
D) None of the above
• question_answer12) Rain drops are spherical in shape, because of
A) surface tension
B) capillarity
C) downward motion
D) acceleration due to gravity
• question_answer13) In Youngs double slit experiment, monochromatic light of wavelength 600 nm illuminates the pair of slits and produces an interference pattern in which two consecutive bright fringes are separated by 10 mm. Another source of monochromatic light produces the interference pattern in which of the two consecutive bright fringes are separated by 8 mm. The wavelength of light from the second source is
A) 448 nm
B) 450 nm
C) 480 nm
D) 580 nm
• question_answer14) Light passes through two polaroids${{P}_{1}}$and ${{P}_{2}}$ with pass axis of ${{P}_{2}}$making an angle 6 with the pass axis of ${{P}_{1}}$. That value of 9 for which the intensity of emergent light is zero.
A) $45{}^\circ$
B) $90{}^\circ$
C) $30{}^\circ$
D) $60{}^\circ$
• question_answer15) If the angle between the axis of polarizer and the analyser is 45 degree, the ratio of the intensities of original light and the transmitted light after passing through the analyser is
A) $2:1$
B) $1:2$
C) $2:3$
D) $1:1$
• question_answer16) The given circuit diagram shows a series L-C-R circuit connected to a variable frequency 230 V, source: $L=5.0\text{ }H,\text{ }C=80\text{ }\mu F,\text{ }R=40\,\Omega$ The frequency and impedance of the circuit at the resonating frequency are
• question_answer17) The instantaneous current and voltage of an AC circuit are given by $I=10sin\text{ }300t\text{ }A\text{ }and\text{ }V=200\text{ }sin\text{ }300t\text{ }V$. The power dissipation in the circuit is
A) 500 W
B) 2000 W
C) 1000W
D) 100W
• question_answer18) In a series L-C-R circuit, the voltages across an inductor, a capacitor and a resistor are 30V, 30V and 60 V respectively. The phase difference between the applied voltage and the current in the circuit is
A) $0{}^\circ$
B) $90{}^\circ$
C) $30{}^\circ$
D) $60{}^\circ$
• question_answer19) An AC source, of voltage$V={{V}_{m}}\sin \omega t,$is applied across a series RC circuit in which the capacitive impedance is a times the resistance in the circuit. The value of the power factor of the circuit is
A) $0$
B) $1/\sqrt{1+{{a}^{2}}}$
C) $\sqrt{1+{{a}^{2}}}$
D) None of these
• question_answer20) Current in a circuit falls steadily from 2.0 A to 0.0 A in 10 ms. If an average emf of 200 V is induced, then the self-inductance of the circuit is
A) 1 H
B) 2H
C) 1/2H
D) 2H
• question_answer21) Volume versus temperature graphs for a given mass of an ideal gas are shown in figure. At two different values of constant pressure. What can be inferred about relation between${{p}_{1}}$and ${{P}_{2}}$?
A) ${{p}_{1}}>{{p}_{2}}$
B) ${{p}_{1}}={{p}_{2}}$
C) ${{p}_{1}}<{{p}_{2}}$
D) data is insufficient
• question_answer22) A rectangular coil of area A, having number of turns Nis rotated at/revolutions per second in a uniform magnetic field B. The maximum emf induced in the coil is
A) $2\pi fNB$
B) $2fNBA$
C) $2\pi fBA$
D) $2\pi fNBA$
• question_answer23) The polarity of the capacitor in the situation described by Fig.
A) Plate A and plate B both positive
B) Plate A negative and plate B positive
C) Plate A positive and plate B negative
D) Plate A and plate B both negative
• question_answer24) A step down transformer operated on a 2.5 kV line. It supplies a load with 20 A. The ratio of the primary winding to the secondary is$10:1$. If the transformer is 90% efficient, the power output is
A) $4.5\times {{10}^{4}}W$
B) $4.5\times {{10}^{2}}W$
C) $6\times {{10}^{4}}W$
D) $6\times {{10}^{4}}W$
• question_answer25) A step down transformer operated on a 2.5 kV line. It supplies a load with 20 A. The ratio of the primary winding to the secondary is$10:1$. If the transformer is 90% efficient, the voltage in the secondary is
A) 225V
B) 550V
C) 250V
D) 200V
• question_answer26) When a given coil is connected to a 200V, 50 Hz AC source, 1 A current flows in the circuit. The inductive reactance of the coil used is
A) $50\sqrt{3}\Omega$
B) $200\sqrt{3}\Omega$
C) $150\sqrt{3}\Omega$
D) $100\sqrt{3}\Omega$
• question_answer27) When a given coil of ohmic resistance$100\,\Omega$and inductive reactance$100\sqrt{3}\,\Omega$is connected to a 200 V DC source, the current flows in the circuit is
A) 1A
B) 1.5 A
C) 2A
D) $100\sqrt{3}\,\Omega$
• question_answer28) If the position of a moving particle, with respect to time be, $x(t)=1+t-{{t}^{2}}$. then, the acceleration of particle is given by (assume all measurements in MKS)
A) $-1\text{ }m{{s}^{-2}}$
B) $-1.5\text{ }m{{s}^{-2}}$
C) $-2\text{ }m{{s}^{-2}}$
D) $-2.5\text{ }m{{s}^{-2}}$
• question_answer29) A ball is dropped from a tower of height h. The duration$({{t}_{0}})$of motion, when it reaches bottom of the tower is given by
A) $\sqrt{\frac{2h}{g}}$
B) $\sqrt{\frac{h}{g}}$
C) $2\sqrt{\frac{h}{g}}$
D) $4\sqrt{\frac{h}{g}}$
• question_answer30) For the given position-time$(x-t)$graph , the interval in which velocity is zero, is
A) $({{t}_{3}}-{{t}_{1}})$
B) $({{t}_{3}}-{{t}_{2}})$
C) $({{t}_{5}}-{{t}_{4}})$
D) $({{t}_{5}}-{{t}_{2}})$
• question_answer31) A person travelling on a straight line moves with an uniform velocity${{\upsilon }_{1}}$for some time and with uniform velocity${{\upsilon }_{2}}$for the next equal interval of time. The average velocity for the entire duration is
A) $({{v}_{1}}+{{v}_{2}})$
B) $({{v}_{1}}+{{v}_{2}})/2$
C) $({{v}_{1}}+{{v}_{2}})/3$
D) $(2{{v}_{1}}+3{{v}_{2}})/3$
• question_answer32) A particle moves along the$x-$axis as $x=u(t-2s)+a{{(t-2s)}^{2}}$. The initial velocity of the particle is
A) $u-2a$
B) $u-4a$
C) $2a-u$
D) $2a-3u$
• question_answer33) The centre of mass of three particles of mass ${{m}_{1}}=1.0\,kg,{{m}_{2}}=2.0\,kg,$and${{m}_{2}}=3.0\,kg$at the comers of an equilateral triangle 1.0 m on a side, as shown in figure
A) $\left( \frac{7}{12}m,\frac{\sqrt{3}}{4}m \right)$
B) $\left( \frac{5}{12}m,\frac{\sqrt{3}}{5}m \right)$
C) $\left( \frac{5}{13}m,\frac{\sqrt{3}}{5}m \right)$
D) $\left( \frac{8}{12}m,\frac{\sqrt{3}}{5}m \right)$
• question_answer34) A system consists of two point masses M and$m(<M)$. The centre of mass of the system is
A) at the middle of m and M
B) nearer to M
C) nearer to m
D) at the position of large mass
• question_answer35) A ball is projected horizontally with a velocity of 5 m/s from the top of a building 19.6 m high. flow long will the ball take to hit the ground?
A) $\sqrt{2}s$
B) 2s
C) $\sqrt{3}s$
D) 3s
• question_answer36) A body of mass 3 kg is acted on by a force which varies as shown in the graph. The momentum acquired is given by
A) zero
B) $5\text{ }N-s$
C) $30N-s$
D) $50\text{ }N-s$
• question_answer37) Boyles law is applicable for an
B) isothermal process
C) isobaric process
D) isochoric process
• question_answer38) Two perfectly elastic particles A and B of equal masses travelling along the line joining, them with velocity 15 m/s and 20 m/s respectively collide. Their velocities after the elastic collision will be (in m/s) respectively.
A) 0 and 25
B) 5 and 20
C) 10 and 15
D) 20 and 15
• question_answer39) A bomb of mass 16 kg at rest explodes into two pieces of masses 4 kg and 12 kg. The velocity of the 12 kg mass is$4\text{ }m{{s}^{-1}}$. The kinetic energy of the other mass is
A) 192J
B) 96J
C) 144 J
D) 288J
• question_answer40) Two particles of masses m^ and 7712 in projectile motion have velocities${{\upsilon }_{1}}$and${{\upsilon }_{2}}({{\upsilon }_{1}}<{{\upsilon }_{2}})$respectively at$t=0$. They collide at time${{t}_{0}}$Their velocities become$\overline{\upsilon }{{}_{1}}$and$\overline{\upsilon }{{}_{2}}$at time$2{{t}_{0}}$while still moving in it. The value of $|({{m}_{1}}\overline{\upsilon }+{{m}_{2}}\overline{\upsilon }{{}_{2}})-({{m}_{1}}{{\overline{\upsilon }}_{1}}+{{m}_{2}}{{\overline{\upsilon }}_{2}})|$is
A) zero
B) $({{m}_{1}}+{{m}_{2}})g{{t}_{0}}$
C) $2({{m}_{1}}+{{m}_{2}})g{{t}_{0}}$
D) $\frac{1}{2}({{m}_{1}}+{{m}_{2}})g{{t}_{0}}$
• question_answer41) If the equation for the displacement of a particle moving on a circular path is given by $\theta =2{{t}^{3}}+0.5,$where$\theta$is in radian and t in second, then the angular velocity of the particle after 2 s from its start is
• question_answer42) A block of mass m is placed on a smooth inclined plane of inclination with the horizontal. The force exerted by the plane on the block has a magnitude
A) $mg$
B) $mg/cos\theta$
C) $mg\text{ }cos\text{ }\theta$
D) $mg\tan \theta$
• question_answer43) A string of negligible mass going over a clamped pulley of mass m supports a block of mass M as shown in the figure. The force on the pulley by the clamp is given by
A) $\sqrt{2}Mg$
B) $\sqrt{2}mg$
C) $\sqrt{{{(M+m)}^{2}}+{{m}^{2}}g}$
D) $\sqrt{{{(M+m)}^{2}}+{{M}^{2}})g}$
• question_answer44) The moment of inertia of a uniform ring of mass$m$and radius r about an axis,$AA$ touching the ring tangentially and lying in the plane of the ring only, as shown in figure , is
A) $\frac{5}{4}m{{r}^{2}}$
B) $\frac{5}{2}m{{r}^{2}}$
C) $\frac{1}{4}m{{r}^{2}}$
D) $\frac{1}{2}m{{r}^{2}}$
• question_answer45) If the disc shown in figure has mass M and it is free to rotate about its symmetrical axis passing through 0, its angular acceleration is
A) $\frac{6F}{MR}$
B) $\frac{3F}{MR}$
C) $\frac{6F}{5MR}$
D) $\frac{4F}{MR}$
• question_answer46) A small ball released from rest from a height h on a smooth surface of varying inclination, as shown in figure. The speed of the ball when it reaches the horizontal part of the surface is
A) ${{v}_{0}}=\sqrt{gh}$
B) ${{v}_{0}}=\sqrt{5gh}$
C) ${{v}_{0}}=\sqrt{2gh}$
D) None of these
• question_answer47) Two masses${{m}_{1}}=5\text{ }kg$and${{m}_{2}}=4.8\text{ }kg$ tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when lift is free to move?
A) $0.2\text{ }m/{{s}^{2}}$
B) $9.8m/{{s}^{2}}$
C) $5m/{{s}^{2}}$
D) $4.8\text{ }m/{{s}^{2}}$
• question_answer48) A particle is acted upon by a conservative force $F=(7\hat{i}-6\hat{j})N$(no other force is acting on the particle). Under the influence of this force, particle moves from (0,0) to$(-3w,4w),$then
A) work done by the force is $3J$
B) work done by the force is $-45\text{ }J$
C) at (0, 0) speed of the particle must be zero
D) at (0, 0) speed of the particle must not be zero
• question_answer49) A full wave rectifier circuit along with the output is shown in the figure. The contribution (s) from the diode 1 is ( are )
A) C
B) AC
C) S.D
D) AB,C.D
• question_answer50) Two identical$p-n$junctions may be connected in series with a battery in three ways as shown in the figure. The potential drops across the two n-p junctions are equal in
A) circuit 1 and circuit 2
B) circuit 2 and circuit 3
C) circuit 3 and circuit 1
D) circuit 1 only
• question_answer51) A transistor is used in common emitter mode as an amplifier, then
A) the base emitter junction is forward unbiased
B) the base emitter junction is reverse biased
C) the input signal is connected in series with the voltage applied to bias of the base emitter junction
D) the input signal is connected in series with the voltage applied to bias the emitter collector junction
• question_answer52) Two point charges$+q$and$+4q$are located at $x=0$and$x=L$respectively. The location of a point on the$x-$axis at which the net electric field due to these two points charges is zero, is
A) $L/3$
B) $2L$
C) $4L$
D) $8L$
• question_answer53) If electric flux entering and leaving an enclosed surface is${{\phi }_{1}}$and${{\phi }_{2}}$respectively, the electric charge inside the enclosed surface will be
A) $({{\phi }_{1}}+{{\phi }_{2}}){{\varepsilon }_{0}}$
B) $({{\phi }_{2}}-{{\phi }_{1}}){{\varepsilon }_{0}}$
C) $({{\phi }_{1}}+{{\phi }_{2}}){{\varepsilon }_{0}}$
D) $({{\phi }_{2}}-{{\phi }_{1}})/{{\varepsilon }_{0}}$
• question_answer54) In a potentiometer experiment, the balancing with a cell is at 240 cm. On shunting the cell with a resistance of$2\,\Omega ,$the balancing length becomes 120 cm. The internal resistance of the cell is
A) $0.5\,\Omega$
B) $1\,\Omega$
C) $2\,\Omega$
D) $4\,\Omega$
• question_answer55) The total current supplied to the circuit by the battery is
A) 1 A
B) 2 A
C) 4 A
D) 6 A
• question_answer56) The displacement y of a particle in a medium can be expressed as $y={{10}^{-6}}\sin (100t+20x+\pi /4)m,$where t is in second and$x$in metre. The speed of the wave is
A) 2000m/s
B) 5 m/s
C) 20 m/s
D) 57cm/s
• question_answer57) If the phase difference between two points is $60{}^\circ$on a wave velocity of 360 m/s and frequency 500 Hz, then path difference between the two points is
A) 1 cm
B) 6 cm
C) 12cm
D) 24cm
• question_answer58) In a wave motion$y=A\sin (kx-\omega t),y$can represent
A) electric field
B) magnetic field
C) displacement
D) pressure
• question_answer59) If${{W}_{1}},{{W}_{2}}$and${{W}_{3}}$represent the work done in moving a particle from A to B along three different paths 1, 2 and 3 respectively in the gravitational field of a point mass m as shown in the figure, find the correct relation between ${{W}_{1}},{{W}_{2}}$and${{W}_{3}}$.
A) ${{W}_{1}}>{{W}_{2}}>{{W}_{3}}$
B) ${{W}_{1}}={{W}_{2}}={{W}_{3}}$
C) ${{W}_{1}}<{{W}_{2}}<{{W}_{3}}$
D) ${{W}_{2}}>{{W}_{2}}>{{W}_{3}}$
• question_answer60) A place where the vertical component of the earths magnetic field is zero has angle of dip is
A) $45{}^\circ$
B) $90{}^\circ$
C) $0{}^\circ$
D) $60{}^\circ$
• question_answer61) 20 mL of 0.5 N$HCl$and 35 mL of 0.1 N $NaOH$are mixed. The resulting solution will
A) be neutral
B) be basic
C) turn phenolphthalein solution pink
D) turn methyl orange red
• question_answer62) The cooling in refrigerator is due to
A) the reaction of the refrigerator gas
B) the expansion of ice
C) the expansion of the gas in the refrigerator
D) the work of the compressor
• question_answer63) Which of the following isomeric heptanes can yield seven different monochlorinated products upon free radical chlorination?
A) 3-methylhexane
B) 2, 2-dimethylpentane
C) 2-methylhexane
D) 2, 3-dimethylpentane
• question_answer64) In an exothermic equilibrium,$A+3BA{{B}_{3}}$all the reactants and products are in gaseous state. The formation of$A{{B}_{3}}$is favoured at
A) low temperature and low pressure
B) low temperature and high pressure
C) high temperature and high pressure
D) high temperature and low pressure
• question_answer65) The compound which is not formed during the dry distillation of a mixture of calcium formate and calcium acetate is
A) methanol
B) propanal
C) propanone
D) ethanol
• question_answer66) A bivalent metal has an equivalent mass of 32. The molecular mass of the metal nitrate is
A) 182
B) 168
C) 192
D) 188
• question_answer67) Arrange the following in the increasing order of their basic strengths $C{{H}_{3}}N{{H}_{2}},{{(C{{H}_{3}})}_{2}}NH,{{(C{{H}_{3}})}_{3}}N,N{{H}_{3}}$
A) $N{{H}_{3}}<{{(C{{H}_{3}})}_{3}}N<{{(C{{H}_{3}})}_{2}}NH<C{{H}_{3}}N{{H}_{2}}$
B) $N{{H}_{3}}<{{(C{{H}_{3}})}_{3}}N<C{{H}_{3}}N{{H}_{2}}<{{(C{{H}_{3}})}_{2}}NH$
C) ${{(C{{H}_{3}})}_{3}}N<N{{H}_{3}}<C{{H}_{3}}N{{H}_{2}}<{{(C{{H}_{3}})}_{2}}NH$
D) $C{{H}_{3}}N{{H}_{2}}<{{(C{{H}_{3}})}_{2}}NH<{{(C{{H}_{3}})}_{3}}N<N{{H}_{3}}$
• question_answer68) The line spectrum of He+ ion will resemble that of
A) hydrogen atom
B) $L{{i}^{+}}ion$
C) helium atom
D) lithium atom
• question_answer69) 0.5 mole of each of${{H}_{2}},S{{O}_{2}}$and$C{{H}_{4}}$are kept in a container. A hole was made in the container. After 3 h, the order of partial pressures in the container will be
A) $pS{{O}_{2}}>p{{H}_{2}}>pC{{H}_{4}}$
B) $pS{{O}_{2}}>pC{{H}_{4}}>p{{H}_{2}}$
C) $p{{H}_{2}}>pS{{O}_{2}}>pC{{H}_{4}}$
D) $p{{H}_{2}}>pC{{H}_{4}}>pS{{O}_{2}}$
• question_answer70) Gold is extracted by hydrometallurgical process, based on its property
A) of being electropositive
B) of being less reactive
C) to form complexes which are water soluble
D) to form salts which are water soluble
• question_answer71) An evacuated glass vessel weighs 50 g when empty, 144.0 g when filled with a liquid of density$0.47\text{ }g\text{ }m{{L}^{-1}}$and 50.5 g when filled with an ideal gas at 760 mm Hg and 300 K. The molar mass of the ideal gas is $[R=0.0821L\text{ }atm\text{ }{{K}^{-1}}mo{{l}^{-1}}]$
A) 61.575
B) 130.98
C) 123.75
D) 47.87
• question_answer72) What is the correct order of acidity from weakest to strongest acid for these compounds
A) I<IV<III<II
B) III<IV<I<II
C) IV<I<III<II
D) II<III<I<IV
• question_answer73) A body of mass$x\text{ }kg$is moving with a velocity of$100\text{ }m{{s}^{-1}}$. Its de-Broglie wavelength is$6.62\times {{10}^{-35}}m$. Hence,$x$is $[h=6.62\times {{10}^{-34}}Js]$
A) 0.1 kg
B) 0.25 kg
C) 0.15 kg
D) 0.2 kg
• question_answer74) In the eclipsed conformation of ethane, the dihedral angle between the hydrogen atoms of adjacent methyl groups is
A) $60{}^\circ$
B) $120{}^\circ$
C) $0{}^\circ$
D) $180{}^\circ$
• question_answer75) A precipitate of$AgCl$is formed when equal volumes of the following are mixed$[{{K}_{sp}}$for$AgCl={{10}^{-10}}$]
A) ${{10}^{-4}}\text{ }M\text{ }AgN{{O}_{3}}and\text{ }{{10}^{-7}}M\text{ }HCl$
B) ${{10}^{-5}}\text{ }M\text{ }AgN{{O}_{3}}and\text{ }{{10}^{-6}}M\text{ }HCl$
C) ${{10}^{-5}}\text{ }M\text{ }AgN{{O}_{3}}and\text{ }{{10}^{-4}}M\text{ }HCl$
D) ${{10}^{-6}}\text{ }M\text{ }AgN{{O}_{3}}and\text{ }{{10}^{-6}}M\text{ }HCl$
• question_answer76) Amongst the following, the compound that can most readily get sulphonated is
A) benzene
B) toluene
C) nitrobenzene
D) chlorobenzene
• question_answer77) The major product of the addition of water molecule to propyne in the presence of mercuric sulphate and dilute sulphuric acid is
A) ethanal
B) 2-propanol
C) propane
D) propanone
• question_answer78) Which of the following is an amphoteric acid?
A) Glycine
B) Salicylic acid
C) Benzoic acid
D) Citric acid
• question_answer79) The crystal with metal deficiency defect is
A) $NaCl$
B) $FeO$
C) $KCl$
D) $ZnO$
• question_answer80) How many moles of Pt may be deposited on the cathode when 0.80 F of electricity is passed through a 1.0 M solution of$P{{t}^{4+}}$?
A) 1.0 mol
B) 0.20 mol
C) 0.40 mol
D) 0.80 mol
• question_answer81) 1 g of a non-volatile, non-electrolyte solute of molar mass 250 g/mol was dissolved in 51.2 g of benzene. If the freezing point depression constant,${{k}_{f}}$of benzene is$5.12\text{ }kg\text{ }K\text{ }mo{{l}^{-1}},$ the freezing point of benzene is lowered by
A) 0.3K
B) 0.5K
C) 0.2K
D) 0.4 K
• question_answer82) A 6% solution of urea is isotonic with
A) 1M solution of glucose
B) 0.05 M solution of glucose
C) 6% solution of glucose
D) 25% solution of glucose
• question_answer83) Pick out the incorrect statement from the following.
A) $sp$ hybrid orbitals are equivalent and are at an angle of$180{}^\circ$with each other
B) $s{{p}^{2}}$hybrid orbitals are equivalent and bond angle between any two of them is$120{}^\circ$
C) $s{{p}^{3}}{{d}^{2}}$hybrid orbitals are equivalent and are oriented towards corners of a regular octahedron
D) $ds{{p}^{2}}$hybrid orbitals are equivalent with a bond angle of$90{}^\circ$between any two of them
• question_answer84) For a chemical reaction$A\to B,$the rate of the reaction is$2\times {{10}^{-3}}$mol$d{{m}^{-3}}{{s}^{-1}},$when the initial concentration is 0.05 mol $d{{m}^{-3}}$. The rate of the same reaction is$1.6\times {{10}^{-2}}mol\text{ }d{{m}^{-3}}{{s}^{-1}}$ when the initial concentration is$0.1\text{ }mol\text{ }d{{m}^{-3}}$. The order of the reaction is
A) 2
B) 0
C) 3
D) 1
• question_answer85) The temperature of the slag zone in metallurgy of iron using blast furnace is
A) $1200-1500{}^\circ C$
B) $1500-1600{}^\circ C$
C) $400-700{}^\circ C$
D) $800-1000{}^\circ C$
• question_answer86) The temperature of the system decreases in an
B) isothermal compression
C) isothermal expansion
• question_answer87) In countries nearer to polar region, the roads are sprinkled with$CaC{{l}_{2}}$. This is
A) to minimise the wear and tear of the roads
B) to minimise the snow fall
C) to minimise pollution
D) to minimise the accumulation of dust on the road
• question_answer88) The correct order of reactivity of the three halides ethyl chloride (I), iso-propyl chloride (II) and benzyl chloride (III) in ${{S}_{N}}1$ reaction is
A) I>II>III
B) III>II>I
C) II>I>III
D) I>III>II
• question_answer89) $C{{H}_{3}}COOH\xrightarrow{LiAl{{H}_{4}}}X\xrightarrow[300{}^\circ C]{Cu}Y\xrightarrow[NaOH]{Dilute}Z$ In the above reaction, Z is
A) butanol
B) aldol
C) ketol
D) acetal
• question_answer90) The reaction in which phenol differs from alcohol is
A) it undergoes esterification with carboxylic acid
B) it reacts with ammonia
C) it forms yellow crystals of iodoform
D) it liberates${{H}_{2}}$with Na metal
• question_answer91) The solubility of a gas in water at 300 K under a pressure of 100 atm is$4\times {{10}^{-3}}kg{{L}^{-1}}$. Therefore, the mass of the gas in kg dissolved 250 mL of water under a pressure of 250 atm at 300 K is
A) $2.5\times {{10}^{-3}}$
B) $2.0\times {{10}^{-3}}$
C) $1.25\times {{10}^{-3}}$
D) $5.0\times {{10}^{-3}}$
• question_answer92) There are 20 naturally occurring ammo acids. The maximum number of tripeptides that can be obtained is
A) 8000
B) 6470
C) 7465
D) 5360
• question_answer93) An example of a sulphur containing amino acid is
A) lysine
B) serine
C) cysteine
D) tyrosine
• question_answer94) The species among the following, which can act as an acid and a base is
A) $HSO_{4}^{-}$
B) $SO_{4}^{2-}$
C) ${{H}_{3}}{{O}^{+}}$
D) $C{{l}^{-}}$
• question_answer95) A compound A has a molecular formula${{C}_{2}}C{{l}_{3}}OH$. It reduces Fehlings solution and on oxidation, gives a monocarboxylic acid B. A can be obtained by the action of chlorine on ethyl alcohol. A is
A) chloroform
B) chloral
C) methyl chloride
D) monochloroacetic acid
• question_answer96) A certain reaction is at equilibrium at$82{}^\circ C$ and the enthalpy change for this reaction is 21.3 kJ. The value of$\Delta S(J{{K}^{-1}}mo{{l}^{-1}})$for the reaction is
A) 55.0
B) 60.0
C) 68.5
D) 120.0
• question_answer97) If in the reaction${{N}_{2}}{{O}_{4}}2NO_{2}^{-};x$ is that part of${{N}_{2}}{{O}_{4}}$which dissociates, then the number of molecules at equilibrium will be
A) 1
B) 3
C) $(1+x)$
D) ${{(1+x)}^{2}}$
• question_answer98) Reaction of diborane with ammonia gives initially
A) ${{B}_{2}}{{H}_{6}}.N{{H}_{3}}$
B) borazole
C) ${{B}_{2}}{{H}_{6}}.3N{{H}_{3}}$
D) ${{[B{{H}_{2}}{{(N{{H}_{3}})}_{2}}]}^{+}}{{[B{{H}_{4}}]}^{-}}$
• question_answer99) An aqueous solution containing 6.5 g of$NaCl$ of 90 purity was subjected to electrolysis. After the complete electrolysis, the solution was evaporated to get solid$NaOH$. The volume of 1M acetic acid required to neutralise$NaOH$ obtained above is
A) $1000c{{m}^{3}}$
B) $2000c{{m}^{3}}$
C) $100c{{m}^{3}}$
D) $200c{{m}^{3}}$
• question_answer100) Corrosion of iron is essentially an electrochemical phenomenon where the cell reactions are
A) Fe is oxidised to$F{{e}^{2+}}$and dissolved oxygen in water is reduced to$O{{H}^{-}}$
B) Fe is oxidised to$F{{e}^{3+}}$and${{H}_{2}}O$is reduced to$O_{2}^{2-}$
C) Fe is oxidised to$F{{e}^{2+}}$and${{H}_{2}}O$is reduced to
D) Fe is oxidised to$F{{e}^{2+}}$and${{H}_{2}}O$is reduced to${{O}_{2}}$
• question_answer101) The hybridisation of the central atom in$Br{{F}_{5}}$molecule is
A) $s{{p}^{3}}$
B) $ds{{p}^{2}}$
C) $s{{p}^{3}}{{d}^{2}}$
D) ${{d}^{3}}s{{p}^{3}}$
• question_answer102) The correct statement with regard to$H_{2}^{+}$and $H_{2}^{-}$is
A) both$H_{2}^{+}$and$H_{2}^{-}$are equally stable
B) both$H_{2}^{+}$and$H_{2}^{-}$do not exist
C) $H_{2}^{-}$is more stable than$H_{2}^{+}$
D) $H_{2}^{+}$is more stable than$H_{2}^{-}$
• question_answer103) A covalent molecule$A{{B}_{3}}$has pyramidal structure. The number of lone pair and bond pair of electrons in the molecule a respectively
A) 2 and 2
B) 0 and 2
C) 3 and 1
D) 1 and 3
• question_answer104) The reaction involved in the oil of winter green test is$Salicylic\text{ }acid\xrightarrow[conc.{{H}_{2}}S{{O}_{4}}]{\Delta }product$. The product is treated with$N{{a}_{2}}C{{O}_{3}}$solution. The missing reagent in the above reaction is
A) phenol
B) $NaOH$
C) ethanol
D) methanol
• question_answer105) The correct order of ionisation energy of C, N, O, F is
A) $F<O<N<C$
B) $F<N<C<O$
C) $C<N<O<F$
D) $C<O<N<F$
• question_answer106) For a reaction$A+2B\xrightarrow[{}]{{}}C,$ the amount of C formed by starting the reaction with 5 moles of A and 8 moles of B is
A) 5 mol
B) 8 mol
C) 16 mol
D) 4 mol
• question_answer107) When plants and animals decay, the organic nitrogen is converted into inorganic nitrogen. The inorganic nitrogen is in the form of
A) ammonia
B) elements of nitrogen
C) nitrates
D) nitrides
• question_answer108) Use of chlorofluoro carbons is not encouraged because,
A) they are harmful to the eyes of people that use it
B) they damage the refrigerators and air conditioners
C) they eat away the ozone in the atmosphere
D) they destroy the oxygen layer
• question_answer109) Which of the following biomolecules contain non-transition metal ion?
A) Vitamin ${{B}_{12}}$
B) Chlorophyll
C) Haemoglobin
D) Insulin
• question_answer110) Among the oxides, $M{{n}_{2}}{{O}_{7}}(I),{{V}_{2}}{{O}_{3}}(II),$ ${{V}_{2}}{{O}_{5}}(III),CrO(IV)$and$C{{r}_{2}}{{O}_{3}}(V),$the basic oxides are
A) I and II
B) II and III
C) III and IV
D) II and IV
• question_answer111) Which of the following complexes exists as pair of enantiomers?
A) $trans-{{[Co{{(en)}_{2}}C{{l}_{2}}]}^{+}}$
B) ${{[Co{{(N{{H}_{3}})}_{2}}C{{l}_{2}}]}^{+}}$
C) $[Co{{\{P{{({{C}_{2}}{{H}_{5}})}_{3}}\}}_{2}}ClBr]$
D) ${{[Cr{{(en)}_{3}}]}^{3+}}$
• question_answer112) Three dimensional molecules with cross links are formed in the case of a
A) thermoplastic
B) thermosetting plastic
C) Both (a) and (b)
D) None of these
• question_answer113) Which of the following is not present in a nucleotide?
A) Cytosine
B) Guanine
D) Tyrosine
A) 4-chloro-3,5-dimethyl phenol
B) 3-chloro-4,5-dimethyl phenol
C) 4-chloro-2,5-dimethyl phenol
D) 5-chloro-3,4-dimethyl phenol
• question_answer115) Alum is a water purifier, because it
A) coagulates the impurities
B) softens hard water
C) gives taste
D) destroys the pathogenic bacteria
• question_answer116) A diabetic person carries a packet or glucose with him always, because
A) glucose reduces the blood sugar level slowly
B) glucose increases the blood sugar level slowly
C) glucose reduces the blood sugar level
D) glucose increases the blood sugar level almost instantaneously
• question_answer117) A emulsifier is a substance which
A) stabilises the emulsion
B) homogenises the emulsion
C) coagulates the emulsion
D) accelerates the dispersion of liquid in liquid
• question_answer118) Which of the following pairs of elements cannot form an alloy?
A) Zn, Cu
B) Fe, Hg
C) Fe, C
D) Hg, Na
• question_answer119) The function of$Fe{{(OH)}_{3}}$in the contact process is
A) to remove arsenic impurity
B) to detect colloidal impurity
C) to remove moisture
D) to remove dust particles
• question_answer120) Consider the following statements
I. Increase in concentration of reactant increases the rate of a zero order reaction II. Rate constant k is equal to collision frequency A, if ${{E}_{a}}=0$ III. Rate constant k is equal to collision frequency A, if ${{E}_{a}}=\infty$ IV. In k us T is a straight line V. In k vs 1 /T is a straight line
Correct statements are
A) I and IV
B) II and V
C) III and IV
D) ll and III
• question_answer121) Function of SER is
A) protein metabolism
B) lipid metabolism
C) carbohydrate metabolism
D) catabolic activity
• question_answer122) The basic principles of embryonic development were first given by
A) Weismann
B) Haeckel
C) Von Baer
D) Malthus
A) fishes and amphibians
B) Annelida and Arthropoda
C) Annelida and Mollusca
D) reptiles and mammals
• question_answer124) Corpus luteum in mammals occurs in
A) brain and connects two cerebral hemispheres
B) ovaries and produces progesterone
C) heart and initiate contraction
D) skin and acts as a pain receptor
• question_answer125) Which of the following helps in the absorption of vitamin$-{{B}_{12}}$?
A) Pepsin
B) Bile juice
C) Castles intrinsic gastric factor
D) Cholecy stokinin
• question_answer126) A fish introduced into a water body to control mosquito population is
A) Gambusia
B) Labeo
C) Notopterus
D) Fistularia
• question_answer127) Patella (knee cap) is an example of
A) replacing bones
B) sesamoid bones
C) visceral bones
D) compact bones
• question_answer128) Which of the following is correctly matched?
A) Enlarged P-wave-Acute myocardial infarction
B) QRS wave flat-Atherosclerotic heart disease
C) Enlarged Q and R-wave-Myocardial infarction
D) Elevated S-T-Enlargement of atria
• question_answer129) Fosciola hepatica (liver fluke) infects its intermediate host and primary host at the following larval stages respectively
A) redia and miracidium
B) cercaria and redia
C) cercaria and metacercaria
D) miracidium and metacercaria
• question_answer130) Match the parts of the human brain with their functions.
Column I Column II A. Cerebrum 1. Controls the pituitary B. Cerebellum 2. Controls vision and hearing C. Hypothalamus 3. Controls the rate of heart beat D. Midbrain 4. Seat of intelligence 5. Maintains body posture
A) A-5 B-4 C-2 D-1
B) A-4 B-5 C-2 D-1
C) A-5 B-4 C-1 D-2
D) A-4 B-5 C-1 D-2
• question_answer131) The endosymbiotic hypothesis of mitochondrial origin states that
A) mitochondria were capable of self-duplication
B) mitochondria were initially aerobic bacteria
C) mitochondria were chloroplast in the beginning
D) None of the above
• question_answer132) Select the correct sequence of stages in the evolution of modern man (Homo sapiens)
A) Australopithecus, Homo habilis, Neanderthal, Cro-magnon man, Modem man
B) Homo habilis, Neanderthal, Australopithecus, Cro-magnon man, Modern man
C) Australopithecus, Neanderthal, Cro-magnon man Homo habilis, Modern man
D) Neanderthal. Australopithecus, Cro-magnon man, Homo habilis, Modern man
• question_answer133) Host cells in response to viral infections produces a low molecular weight protein, that protect other cells against further viral infection. This substance is
A) phytotoxin
B) antibody
C) hormone
D) interferon
• question_answer134) Which of the following branch of biology is correctly matched?
A) Ornithology-Study of mammals
B) Herpetalogy-Study of molluscs
C) Ophiology-Study of snakes
D) Therology-Study of birds
• question_answer135) If a haemophilic carrier female marries a normal man, the chances are that
A) all the daughters will have haemophilia
B) 50% of the daughters will have haemophilia
C) 50% of the sons will have haemophilia
D) all the sons will have haemophilia
• question_answer136) Which one of the given statement is not true for Archaeopteryx?
A) Each wing had three clawed digit
B) Hind feet with five digits
C) Long jointed lizard like tail
D) The codont teeth on both jaws
• question_answer137) A natural form of genetic engineering is exhibited by
A) Haemophilus influenza
B) E. coli
C) Dolly
D) Agrobacterium tumefaciens
• question_answer138) Glands in a vertebrate body originate from
A) ectoderm, mesoderm and endoderm
B) ectoderm and endoderm
C) mesoderm
D) ectoderm
A) substance that activates transcription in cancer cells
B) oncogene promoting cancer
C) substance that produces tumour
D) All of the above
• question_answer140) Identify the incorrect match
A) Mulberry silk-Bombyx mori
B) Tasar silk-Antheraea pernyi
C) Muga silk-Antheraea assama
D) Erisilk-Affacus ricini
• question_answer141) All chordates at one or other stage possess
A) two pairs of pentadactyl limbs
B) pharyngeal gill slits
C) movable jaw
D) vertebral column
• question_answer142) Statement I An injury to the anterior pituitary affects the growth of a person. Statement II Injury to adrenal cortex is not likely to affect secretion of corti Choose the correct option
A) Statement I is correct but II is incorrect
B) Statement II is correct but I is incorrect
C) Both statement are correct
D) Both statements are incorrect
A) biconvex RBCs to increase the surface area
B) biconcave RBCs to increase the surface area
C) spherical RBCs to increase the surface area
D) no RBCs
• question_answer144) Match the Column I and Column II.
Column I Column II A. Squamous 1. Intestinal glands B. Cuboidal 2. Trachea C. Columnar 3. Ovary D. Ciliated 4. Blood vessels 5. Bronchioles
A) A-2, B-5, C-1, D-3
B) A-4, B-3, C-1, D-2
C) A-4, B-3, C-1, D-5
D) A-4, B-5, C-1, D-2
• question_answer145) A person was found to possess reduced haemoglobin content in the blood, caused due to disturbance in formation of RBCs. He is suffering from
A) vitamin$-{{B}_{1}}$defeciency
B) vitamin-A defeciency
C) vitamin$-{{B}_{12}}$defeciency
D) vitamin-E deficiency
A) T H Morgan and W Castle
B) Stern and Hotta
C) Pontecarvo
D) Bridges and Burns
• question_answer147) Which of the following region of root is responsible for increase in length of root?
A) Root cap region
B) Meristematic region
C) Region of elongation
D) Mature region
A) ${{C}_{6}}{{H}_{12}}{{O}_{6}}+6{{O}_{2}}\xrightarrow[{}]{{}}6{{O}_{2}}+6{{H}_{2}}O$
B) ${{C}_{6}}{{H}_{12}}{{O}_{6}}\xrightarrow[{}]{{}}2{{C}_{2}}{{H}_{5}}OH+2C{{O}_{2}}$
C) ${{C}_{6}}{{H}_{12}}{{O}_{6}}\xrightarrow[{}]{{}}2{{C}_{3}}{{H}_{4}}{{O}_{3}}+4H$
D) ${{C}_{3}}{{H}_{4}}{{O}_{3}}+NADH\xrightarrow[{}]{{}}{{C}_{2}}{{H}_{5}}OH+C{{O}_{2}}+NA{{D}^{+}}$
• question_answer149) The protein that reproduce within the living cells are termed as
A) plasmids
B) phages
C) prions
D) prophages
• question_answer150) Water pollution is best assessed by determining
A) DO and BOD
B) BOD and turbidity
C) DO and acidity
D) Hardness and alkalinity
• question_answer151) The correct sequence of cellular growth stages is
A) division$\to$differentiation$\to$elongation
B) differentiation$\to$division$\to$elongation
C) division$\to$elongation$\to$differentiation
D) elongation$\to$differentiation$\to$division
• question_answer152) The organisms that thrives on the remains of dead plants and animals are categorised as
A) carnivore
B) omnivore
C) scavangers
D) predators
• question_answer153) In which of the following, determination of order of bases in a DNA molecule takes place
A) Gene probing
B) Gene splicing
C) Gene mapping
D) Gene sequencing
• question_answer154) A pea with white flower was crossed with another pea with white flower. When they selfed, the${{F}_{2}}-$generation produced purple and white in the ratio of$9:7$. The reason for the result is that
A) It is typical monohybrid Mendelian ratio
B) purple flower colour is dominant over white
C) it is a complementary factor
D) None of the above
• question_answer155) The dark coloured dead wood present in the central region of old tree is
A) spring wood
B) heart wood
C) sap wood
D) cambium
• question_answer156) Diffusion Pressure Deficit (DPD) is equal to the Osmotic pressure substracted by
A) Osmotic Pressure (OP)
B) Turgor Pressure (TP)
C) Suction Pressure (SP)
D) Water Potential$({{\psi }_{w}})$
• question_answer157) Which of the following statement is not true about Griffiths experiment?
A) The experiment was conducted in 1928
B) The experiment occurs with Streptococcus pneumonia
C) R and S-strain of bacteria
D) The experiment proves the process of ransduction
• question_answer158) Oxygen released in photosynthesis comes out by
A) photophosphorylation
B) photolysis of water
C) photorespiration
D) Kelvin cycle
• question_answer159) In protein synthesis the base sequence of mRNA determines the primary structure of polypeptide chain because
A) it determine the sequence of alignment of amino acid charged$tRNA$molecule
B) there is one to one coding relationship between each base on mRNA and amino acid
C) each amino acid is defined by a pair of base on$mRNA$
D) amino acid molecule align directly on the polynucleotide on $mRNA$
• question_answer160) What do you mean to the mutation in which replacement of a nucleotide occur by another similar type?
A) Inversion
B) Transition substitution
C) Transversion substitution
D) Insertion frame-shift
• question_answer161) Which of the following assumption is not belonging to Hardy-Weinberg principle?
A) The population size is very large
B) Random mating is occurring
C) Natural selection takes place
D) No mutation occurs
• question_answer162) Which ion is used in the protoplast fusion with the PEG (polyethylene glycol)?
A) $N{{a}^{+}}$
B) $C{{a}^{++}}$
C) ${{K}^{+}}$
D) $M{{g}^{+}}$
• question_answer163) Which of the following rule states that Avian and mammalian forms are generally of higher size in colder region as compared to warmer region?
A) Bergmanns rule
B) Alien rule
C) Gloger rule
D) Hardy-Weinberg rule
A) catalytic antibodies
B) neutral antibodies
C) non-specific antibodies
D) defermed antibodies
A) all the animals of a place
B) all the animals and plants found in an area
C) human beings in an area
D) animal/plants of a species living in a specific area
• question_answer166) Parthenogenetic embryos in plants are those which are formed by
A) unfertilised eggs
B) fertilised eggs
C) sporophytic cells
D) male gametophyte
• question_answer167) Multiplication of genetically identical copies of a cultiver by asexual reproduction is known as
A) aclonal propagation
B) clonal propagation
C) vegetative propagation
D) polyclonal propagation
• question_answer168) The gene controlling seven traits in pea studied by Mendel were later found to be located on the following number of chromosomes
A) seven
B) four
C) five
D) six
• question_answer169) Glycolate accumulates in chloroplast when there is
A) low$C{{O}_{2}}$
B) high$C{{O}_{2}}$
C) bright light
D) low temperature
• question_answer170) Part of gene which codes for an enzyme is
A) cistron
B) codon
C) exon
D) intron
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1.7
# Cryptography recap
Encryption and hashing was explored extensively in the short course An Introduction to Cryptography. Here, we provide a quick recap.
Generally, we could say that the purpose of cryptography is to keep the system’s data/information secret and secure. There are two main encryption methods: symmetric and asymmetric cryptography.
• Symmetric key encryption makes use of a single key for both the encryption and decryption process. Note that this key needs to be kept secret from third parties and should also be exchanged and shared only between the valid communication parties through a secure communication channel.
• Asymmetric or public key encryption makes use of two keys. The first key is called a public key and can be known to anyone and is used for the encryption. The second key, known as the private key, should be kept secret and is used for the decryption of the exchanged data. Note that in this approach, the public key does not necessarily need to be exchanged over a secure channel.
To have a better understanding of the two methods, we present the encryption and decryption process taking place in each of them in the form of algorithms, involving Alice, Bob and Eve.
### Symmetric cryptography algorithm
The figure below demonstrates a typical communication that uses symmetric cryptography (click image to zoom).
Algorithm steps:
• Alice and Bob agree on the private key
• Alice passes the key to Bob
• Alice creates ciphertext using the private key:
ciphertext = encryptionkey (message)
• Alice sends it through the communication channel
• Bob receives the ciphertext and decrypts it using the private key:
message = decryptionkey (ciphertext)
• Bob receives the original message/plaintext
### Asymmetric cryptography algorithm
The figure below demonstrates a typical communication that uses asymmetric cryptography (click image to zoom):
Algorithm steps
• Alice creates her private key and a public key (similarly for Bob)
• Alice hides her private key and sends out a copy of the public key to Bob (similarly for Bob)
• Bob creates the ciphertext using Alice’s public key (similarly for Alice):
ciphertext = encryptionAlice public key (message)
• Bob sends it through the communication channel (similarly for Alice)
• Alice receives the ciphertext and decrypts it using her private key (similarly for Bob):
message = decryptionAlice private key (ciphertext)
• Alice receives Bob’s original message/plaintext (similarly for Bob)
### Hashing
We could define a hashing function as a special mathematical function, denoted by H, performing one-way encryption. The input is a plain message and the output is an alphanumeric hash value. The mathematical formula of a hash function is as follows:
*H(message) = hash*
It is worth pointing out that a hash function needs to satisfy certain properties in order to be characterised as a strong and well-defined hash function. These properties are:
• Pre-image resistant: This property ensures that given a hash value, the computation of the input message, such that hash= H(message), is hard.
• Avalanche effect: Any small change to the message results in a totally different hash output.
• Second-pre-image resistant: Given a message, it is hard to find message’ (different from message) such that H(message) = H(message’)
• Collision resistant: It is hard to find message and message’ such that H(message) = H(message’).
The satisfaction of these properties could guarantee that the plain message used in the hash function cannot be easily retrieved by unauthorised parties.
### Conclusion
The application of the security principles and methods described here and on the next step could be a good starting point to effectively secure a computer system that is under development.
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Distributions, also known as Schwartz distributions or generalized functions, are objects that generalize the classical notion of functions in mathematical analysis. Distributions make it possible to differentiate functions whose derivatives do not exist in the classical sense. In particular, any locally integrable function has a distributional derivative.
Distributions are widely used in the theory of partial differential equations, where it may be easier to establish the existence of distributional solutions than classical solutions, or where appropriate classical solutions may not exist. Distributions are also important in physics and engineering where many problems naturally lead to differential equations whose solutions or initial conditions are singular, such as the Dirac delta function.
A function ${\displaystyle f}$ is normally thought of as acting on the points in its domain by sending a point x in its domain to the point ${\displaystyle f(x).}$ Instead, distribution theory reinterprets functions as being equivalent to their dual linear functionals: if every nicely behaved test function integrated with a function gives out the same result, this defines a linear functional equivalent to every conventional function.
However, this kind of definition is much laxer, and admits mathematical objects beyond functions. In particular, these kinds of generalised functions can be used to represent singular measures, such as the delta function, and all of its derivatives. Since the distributional framework is localised, linear, and shift-invariant, it can represent almost all of the compositions of the basis waveforms as well. It also admits a full Fourier theory, and since the theory is fully localised, there is no need to have functions fall off in time or shift.
In applications to physics and engineering, test functions are usually infinitely differentiable complex-valued (or real-valued) functions with compact support that are defined on some given non-empty open subset ${\displaystyle U\subseteq \mathbb {R} ^{n))$ (bump functions are examples of test functions). The set of all such test functions forms a vector space that is denoted by ${\displaystyle C_{c}^{\infty }(U)}$ or ${\displaystyle {\mathcal {D))(U).}$
Most commonly encountered functions, including all continuous maps ${\displaystyle f:\mathbb {R} \to \mathbb {R} }$ if using ${\displaystyle U:=\mathbb {R} ,}$ can be canonically reinterpreted as acting via "integration against a test function." Explicitly, this means that ${\displaystyle f}$ "acts on" a test function ${\displaystyle \psi \in {\mathcal {D))(\mathbb {R} )}$ by "sending" it to the number ${\textstyle \int _{\mathbb {R} }f\,\psi \,dx,}$ which is often denoted by ${\displaystyle D_{f}(\psi ).}$ This new action ${\textstyle \psi \mapsto D_{f}(\psi )}$ of ${\displaystyle f}$ is a scalar-valued map, denoted by ${\displaystyle D_{f},}$ whose domain is the space of test functions ${\displaystyle {\mathcal {D))(\mathbb {R} ).}$ This functional ${\displaystyle D_{f))$ turns out to have the two defining properties of what is known as a distribution on ${\displaystyle \mathbb {R} :}$ it is linear and also continuous when ${\displaystyle {\mathcal {D))(\mathbb {R} )}$ is given a certain topology called the canonical LF topology. Distributions like ${\displaystyle D_{f))$ that arise from functions in this way are prototypical examples of distributions, but there are many which cannot be defined by integration against any function. Examples of the latter include the Dirac delta function and distributions defined to act by integration of test functions against certain measures. It is nonetheless still possible to reduce any arbitrary distribution down to a simpler family of related distributions that do arise via such actions of integration.
More generally, a distribution on ${\displaystyle U}$ is by definition a linear functional on ${\displaystyle C_{c}^{\infty }(U)}$ that is continuous when ${\displaystyle C_{c}^{\infty }(U)}$ is given a topology called the canonical LF topology. This leads to the space of (all) distributions on ${\displaystyle U}$, usually denoted by ${\displaystyle {\mathcal {D))'(U)}$ (note the prime), which by definition is the space of all distributions on ${\displaystyle U}$ (that is, it is the continuous dual space of ${\displaystyle C_{c}^{\infty }(U)}$); it is these distributions that are the main focus of this article.
Definitions of the appropriate topologies on spaces of test functions and distributions are given in the article on spaces of test functions and distributions. This article is primarily concerned with the definition of distributions, together with their properties and some important examples.
## History
The practical use of distributions can be traced back to the use of Green functions in the 1830s to solve ordinary differential equations, but was not formalized until much later. According to Kolmogorov & Fomin (1957), generalized functions originated in the work of Sergei Sobolev (1936) on second-order hyperbolic partial differential equations, and the ideas were developed in somewhat extended form by Laurent Schwartz in the late 1940s. According to his autobiography, Schwartz introduced the term "distribution" by analogy with a distribution of electrical charge, possibly including not only point charges but also dipoles and so on. Gårding (1997) comments that although the ideas in the transformative book by Schwartz (1951) were not entirely new, it was Schwartz's broad attack and conviction that distributions would be useful almost everywhere in analysis that made the difference.
## Notation
• ${\displaystyle n}$ is a fixed positive integer and ${\displaystyle U}$ is a fixed non-empty open subset of Euclidean space ${\displaystyle \mathbb {R} ^{n}.}$
• ${\displaystyle \mathbb {N} =\{0,1,2,\ldots \))$ denotes the natural numbers.
• ${\displaystyle k}$ will denote a non-negative integer or ${\displaystyle \infty .}$
• If ${\displaystyle f}$ is a function then ${\displaystyle \operatorname {Dom} (f)}$ will denote its domain and the support of ${\displaystyle f,}$ denoted by ${\displaystyle \operatorname {supp} (f),}$ is defined to be the closure of the set ${\displaystyle \{x\in \operatorname {Dom} (f):f(x)\neq 0\))$ in ${\displaystyle \operatorname {Dom} (f).}$
• For two functions ${\displaystyle f,g:U\to \mathbb {C} ,}$ the following notation defines a canonical pairing:
${\displaystyle \langle f,g\rangle :=\int _{U}f(x)g(x)\,dx.}$
• A multi-index of size ${\displaystyle n}$ is an element in ${\displaystyle \mathbb {N} ^{n))$ (given that ${\displaystyle n}$ is fixed, if the size of multi-indices is omitted then the size should be assumed to be ${\displaystyle n}$). The length of a multi-index ${\displaystyle \alpha =(\alpha _{1},\ldots ,\alpha _{n})\in \mathbb {N} ^{n))$ is defined as ${\displaystyle \alpha _{1}+\cdots +\alpha _{n))$ and denoted by ${\displaystyle |\alpha |.}$ Multi-indices are particularly useful when dealing with functions of several variables, in particular we introduce the following notations for a given multi-index ${\displaystyle \alpha =(\alpha _{1},\ldots ,\alpha _{n})\in \mathbb {N} ^{n))$:
{\displaystyle {\begin{aligned}x^{\alpha }&=x_{1}^{\alpha _{1))\cdots x_{n}^{\alpha _{n))\\\partial ^{\alpha }&={\frac {\partial ^{|\alpha |)){\partial x_{1}^{\alpha _{1))\cdots \partial x_{n}^{\alpha _{n))))\end{aligned))}
We also introduce a partial order of all multi-indices by ${\displaystyle \beta \geq \alpha }$ if and only if ${\displaystyle \beta _{i}\geq \alpha _{i))$ for all ${\displaystyle 1\leq i\leq n.}$ When ${\displaystyle \beta \geq \alpha }$ we define their multi-index binomial coefficient as:
${\displaystyle {\binom {\beta }{\alpha )):={\binom {\beta _{1)){\alpha _{1))}\cdots {\binom {\beta _{n)){\alpha _{n))}.}$
## Definitions of test functions and distributions
In this section, some basic notions and definitions needed to define real-valued distributions on U are introduced. Further discussion of the topologies on the spaces of test functions and distributions are given in the article on spaces of test functions and distributions.
Notation:
1. Let ${\displaystyle k\in \{0,1,2,\ldots ,\infty \}.}$
2. Let ${\displaystyle C^{k}(U)}$ denote the vector space of all k-times continuously differentiable real or complex-valued functions on U.
3. For any compact subset ${\displaystyle K\subseteq U,}$ let ${\displaystyle C^{k}(K)}$ and ${\displaystyle C^{k}(K;U)}$ both denote the vector space of all those functions ${\displaystyle f\in C^{k}(U)}$ such that ${\displaystyle \operatorname {supp} (f)\subseteq K.}$
• Note that ${\displaystyle C^{k}(K)}$ depends on both K and U but we will only indicate K, where in particular, if ${\displaystyle f\in C^{k}(K)}$ then the domain of ${\displaystyle f}$ is U rather than K. We will use the notation ${\displaystyle C^{k}(K;U)}$ only when the notation ${\displaystyle C^{k}(K)}$ risks being ambiguous.
• Every ${\displaystyle C^{k}(K)}$ contains the constant 0 map, even if ${\displaystyle K=\varnothing .}$
4. Let ${\displaystyle C_{c}^{k}(U)}$ denote the set of all ${\displaystyle f\in C^{k}(U)}$ such that ${\displaystyle f\in C^{k}(K)}$ for some compact subset K of U.
• Equivalently, ${\displaystyle C_{c}^{k}(U)}$ is the set of all ${\displaystyle f\in C^{k}(U)}$ such that ${\displaystyle f}$ has compact support.
• ${\displaystyle C_{c}^{k}(U)}$ is equal to the union of all ${\displaystyle C^{k}(K)}$ as ${\displaystyle K\subseteq U}$ ranges over all compact subsets of ${\displaystyle U.}$
• If ${\displaystyle f}$ is a real-valued function on U, then ${\displaystyle f}$ is an element of ${\displaystyle C_{c}^{k}(U)}$ if and only if ${\displaystyle f}$ is a ${\displaystyle C^{k))$ bump function. Every real-valued test function on ${\displaystyle U}$ is always also a complex-valued test function on ${\displaystyle U.}$
The graph of the bump function ${\displaystyle (x,y)\in \mathbb {R} ^{2}\mapsto \Psi (r),}$ where ${\displaystyle r=\left(x^{2}+y^{2}\right)^{\frac {1}{2))}$ and ${\displaystyle \Psi (r)=e^{-{\frac {1}{1-r^{2))))\cdot \mathbf {1} _{\{|r|<1\)).}$ This function is a test function on ${\displaystyle \mathbb {R} ^{2))$ and is an element of ${\displaystyle C_{c}^{\infty }\left(\mathbb {R} ^{2}\right).}$ The support of this function is the closed unit disk in ${\displaystyle \mathbb {R} ^{2}.}$ It is non-zero on the open unit disk and it is equal to 0 everywhere outside of it.
For all ${\displaystyle j,k\in \{0,1,2,\ldots ,\infty \))$ and any compact subsets K and L of U, we have:
{\displaystyle {\begin{aligned}C^{k}(K)&\subseteq C_{c}^{k}(U)\subseteq C^{k}(U)\\C^{k}(K)&\subseteq C^{k}(L)&&{\text{if ))K\subseteq L\\C^{k}(K)&\subseteq C^{j}(K)&&{\text{if ))j\leq k\\C_{c}^{k}(U)&\subseteq C_{c}^{j}(U)&&{\text{if ))j\leq k\\C^{k}(U)&\subseteq C^{j}(U)&&{\text{if ))j\leq k\\\end{aligned))}
Definition: Elements of ${\displaystyle C_{c}^{\infty }(U)}$ are called test functions on U and ${\displaystyle C_{c}^{\infty }(U)}$ is called the space of test functions on U. We will use both ${\displaystyle {\mathcal {D))(U)}$ and ${\displaystyle C_{c}^{\infty }(U)}$ to denote this space.
Distributions on U are continuous linear functionals on ${\displaystyle C_{c}^{\infty }(U)}$ when this vector space is endowed with a particular topology called the canonical LF-topology. The following proposition states two necessary and sufficient conditions for the continuity of a linear functional on ${\displaystyle C_{c}^{\infty }(U)}$ that are often straightforward to verify.
Proposition: A linear functional T on ${\displaystyle C_{c}^{\infty }(U)}$ is continuous, and therefore a distribution, if and only if either of the following equivalent conditions are satisfied:
1. For every compact subset ${\displaystyle K\subseteq U}$ there exist constants ${\displaystyle C>0}$ and ${\displaystyle N\in \mathbb {N} }$ (dependent on ${\displaystyle K}$) such that for all ${\displaystyle f\in C_{c}^{\infty }(U)}$ with support contained in ${\displaystyle K}$,[1][2]
${\displaystyle |T(f)|\leq C\sup\{|\partial ^{\alpha }f(x)|:x\in U,|\alpha |\leq N\}.}$
2. For every compact subset ${\displaystyle K\subseteq U}$ and every sequence ${\displaystyle \{f_{i}\}_{i=1}^{\infty ))$ in ${\displaystyle C_{c}^{\infty }(U)}$ whose supports are contained in ${\displaystyle K}$, if ${\displaystyle \{\partial ^{\alpha }f_{i}\}_{i=1}^{\infty ))$ converges uniformly to zero on ${\displaystyle U}$ for every multi-index ${\displaystyle \alpha }$, then ${\displaystyle T(f_{i})\to 0.}$
### Topology on Ck(U)
We now introduce the seminorms that will define the topology on ${\displaystyle C^{k}(U).}$ Different authors sometimes use different families of seminorms so we list the most common families below. However, the resulting topology is the same no matter which family is used.
Suppose ${\displaystyle k\in \{0,1,2,\ldots ,\infty \))$ and ${\displaystyle K}$ is an arbitrary compact subset of ${\displaystyle U.}$ Suppose ${\displaystyle i}$ an integer such that ${\displaystyle 0\leq i\leq k.}$[note 1] and ${\displaystyle p}$ is a multi-index with length ${\displaystyle |p|\leq k.}$ For ${\displaystyle K\neq \varnothing ,}$ define:
{\displaystyle {\begin{alignedat}{4}{\text{ (1) ))\ &s_{p,K}(f)&&:=\sup _{x_{0}\in K}\left|\partial ^{p}f(x_{0})\right|\\[4pt]{\text{ (2) ))\ &q_{i,K}(f)&&:=\sup _{|p|\leq i}\left(\sup _{x_{0}\in K}\left|\partial ^{p}f(x_{0})\right|\right)=\sup _{|p|\leq i}\left(s_{p,K}(f)\right)\\[4pt]{\text{ (3) ))\ &r_{i,K}(f)&&:=\sup _{\stackrel {|p|\leq i}{x_{0}\in K))\left|\partial ^{p}f(x_{0})\right|\\[4pt]{\text{ (4) ))\ &t_{i,K}(f)&&:=\sup _{x_{0}\in K}\left(\sum _{|p|\leq i}\left|\partial ^{p}f(x_{0})\right|\right)\end{alignedat))}
while for ${\displaystyle K=\varnothing ,}$ define all the functions above to be the constant 0 map.
Each of the functions above are non-negative ${\displaystyle \mathbb {R} }$-valued[note 2] seminorms on ${\displaystyle C^{k}(U).}$
Each of the following families of seminorms generates the same locally convex vector topology on ${\displaystyle C^{k}(U)}$:
{\displaystyle {\begin{alignedat}{4}(1)\quad &\{q_{i,K}&&:\;K{\text{ compact and ))\;&&i\in \mathbb {N} {\text{ satisfies ))\;&&0\leq i\leq k\}\2)\quad &\{r_{i,K}&&:\;K{\text{ compact and ))\;&&i\in \mathbb {N} {\text{ satisfies ))\;&&0\leq i\leq k\}\\(3)\quad &\{t_{i,K}&&:\;K{\text{ compact and ))\;&&i\in \mathbb {N} {\text{ satisfies ))\;&&0\leq i\leq k\}\\(4)\quad &\{s_{p,K}&&:\;K{\text{ compact and ))\;&&p\in \mathbb {N} ^{n}{\text{ satisfies ))\;&&|p|\leq k\}\end{alignedat))} The vector space ${\displaystyle C^{k}(U)}$ is endowed with the locally convex topology induced by any one of the four families of seminorms described above; equivalently, the topology is the vector topology induced by all of the seminorms above (that is, by the union of the four families of seminorms). With this topology, ${\displaystyle C^{k}(U)}$ becomes a locally convex (non-normable) Fréchet space and all of the seminorms defined above are continuous on this space. All of the seminorms defined above are continuous functions on ${\displaystyle C^{k}(U).}$ Under this topology, a net ${\displaystyle (f_{i})_{i\in I))$ in ${\displaystyle C^{k}(U)}$ converges to ${\displaystyle f\in C^{k}(U)}$ if and only if for every multi-index ${\displaystyle p}$ with ${\displaystyle |p| and every compact ${\displaystyle K,}$ the net of partial derivatives ${\displaystyle \left(\partial ^{p}f_{i}\right)_{i\in I))$ converges uniformly to ${\displaystyle \partial ^{p}f}$ on ${\displaystyle K.}$[3] For any ${\displaystyle k\in \{0,1,2,\ldots ,\infty \},}$ any (von Neumann) bounded subset of ${\displaystyle C^{k+1}(U)}$ is a relatively compact subset of ${\displaystyle C^{k}(U).}$[4] In particular, a subset of ${\displaystyle C^{\infty }(U)}$ is bounded if and only if it is bounded in ${\displaystyle C^{i}(U)}$ for all ${\displaystyle i\in \mathbb {N} .}$[4] The space ${\displaystyle C^{k}(U)}$ is a Montel space if and only if ${\displaystyle k=\infty .}$[5] A subset ${\displaystyle W}$ of ${\displaystyle C^{\infty }(U)}$ is open in this topology if and only if there exists ${\displaystyle i\in \mathbb {N} }$ such that ${\displaystyle W}$ is open when ${\displaystyle C^{\infty }(U)}$ is endowed with the subspace topology induced on it by ${\displaystyle C^{i}(U).}$ #### Topology on Ck(K) As before, fix ${\displaystyle k\in \{0,1,2,\ldots ,\infty \}.}$ Recall that if ${\displaystyle K}$ is any compact subset of ${\displaystyle U}$ then ${\displaystyle C^{k}(K)\subseteq C^{k}(U).}$ Assumption: For any compact subset ${\displaystyle K\subseteq U,}$ we will henceforth assume that ${\displaystyle C^{k}(K)}$ is endowed with the subspace topology it inherits from the Fréchet space ${\displaystyle C^{k}(U).}$ If ${\displaystyle k}$ is finite then ${\displaystyle C^{k}(K)}$ is a Banach space[6] with a topology that can be defined by the norm ${\displaystyle r_{K}(f):=\sup _{|p| And when ${\displaystyle k=2,}$ then ${\displaystyle C^{k}(K)}$ is even a Hilbert space.[6] #### Trivial extensions and independence of Ck(K)'s topology from U The definition of ${\displaystyle C^{k}(K)}$ depends on U so we will let ${\displaystyle C^{k}(K;U)}$ denote the topological space ${\displaystyle C^{k}(K),}$ which by definition is a topological subspace of ${\displaystyle C^{k}(U).}$ Suppose ${\displaystyle V}$ is an open subset of ${\displaystyle \mathbb {R} ^{n))$ containing ${\displaystyle U}$ and for any compact subset ${\displaystyle K\subseteq V,}$ let ${\displaystyle C^{k}(K;V)}$ is the vector subspace of ${\displaystyle C^{k}(V)}$ consisting of maps with support contained in ${\displaystyle K.}$ Given ${\displaystyle f\in C_{c}^{k}(U),}$ its trivial extension to V is by definition, the function ${\displaystyle I(f):=F:V\to \mathbb {C} }$ defined by: ${\displaystyle F(x)={\begin{cases}f(x)&x\in U,\\0&{\text{otherwise)),\end{cases))}$ so that ${\displaystyle F\in C^{k}(V).}$ Let ${\displaystyle I:C_{c}^{k}(U)\to C^{k}(V)}$ denote the map that sends a function in ${\displaystyle C_{c}^{k}(U)}$ to its trivial extension on V. This map is a linear injection and for every compact subset ${\displaystyle K\subseteq U}$ (where ${\displaystyle K}$ is also a compact subset of ${\displaystyle V}$ since ${\displaystyle K\subseteq U\subseteq V}$) we have {\displaystyle {\begin{alignedat}{4}I\left(C^{k}(K;U)\right)&~=~C^{k}(K;V)\qquad {\text{ and thus ))\\I\left(C_{c}^{k}(U)\right)&~\subseteq ~C_{c}^{k}(V)\end{alignedat))} If I is restricted to ${\displaystyle C^{k}(K;U)}$ then the following induced linear map is a homeomorphism (and thus a TVS-isomorphism): {\displaystyle {\begin{alignedat}{4}\,&C^{k}(K;U)&&\to \,&&C^{k}(K;V)\\&f&&\mapsto \,&&I(f)\\\end{alignedat))} and thus the next map is a topological embedding: {\displaystyle {\begin{alignedat}{4}\,&C^{k}(K;U)&&\to \,&&C^{k}(V)\\&f&&\mapsto \,&&I(f).\\\end{alignedat))} Using the injection ${\displaystyle I:C_{c}^{k}(U)\to C^{k}(V)}$ the vector space ${\displaystyle C_{c}^{k}(U)}$ is canonically identified with its image in ${\displaystyle C_{c}^{k}(V)\subseteq C^{k}(V).}$ Because ${\displaystyle C^{k}(K;U)\subseteq C_{c}^{k}(U),}$ through this identification, ${\displaystyle C^{k}(K;U)}$ can also be considered as a subset of ${\displaystyle C^{k}(V).}$ Thus the topology on ${\displaystyle C^{k}(K;U)}$ is independent of the open subset U of ${\displaystyle \mathbb {R} ^{n))$ that contains K.[7] This justifies the practice of written ${\displaystyle C^{k}(K)}$ instead of ${\displaystyle C^{k}(K;U).}$ ### Canonical LF topology Main article: Spaces of test functions and distributions See also: LF-space and Topology of uniform convergence Recall that ${\displaystyle C_{c}^{k}(U)}$ denote all those functions in ${\displaystyle C^{k}(U)}$ that have compact support in ${\displaystyle U,}$ where note that ${\displaystyle C_{c}^{k}(U)}$ is the union of all ${\displaystyle C^{k}(K)}$ as ${\displaystyle K}$ ranges over all compact subsets of ${\displaystyle U.}$ Moreover, for every ${\displaystyle k,\,C_{c}^{k}(U)}$ is a dense subset of ${\displaystyle C^{k}(U).}$ The special case when ${\displaystyle k=\infty }$ gives us the space of test functions. ${\displaystyle C_{c}^{\infty }(U)}$ is called the space of test functions on ${\displaystyle U}$ and it may also be denoted by ${\displaystyle {\mathcal {D))(U).}$ Unless indicated otherwise, it is endowed with a topology called the canonical LF topology, whose definition is given in the article: Spaces of test functions and distributions. The canonical LF-topology is not metrizable and importantly, it is strictly finer than the subspace topology that ${\displaystyle C^{\infty }(U)}$ induces on ${\displaystyle C_{c}^{\infty }(U).}$ However, the canonical LF-topology does make ${\displaystyle C_{c}^{\infty }(U)}$ into a complete reflexive nuclear[8] Montel[9] bornological barrelled Mackey space; the same is true of its strong dual space (that is, the space of all distributions with its usual topology). The canonical LF-topology can be defined in various ways. ### Distributions See also: Continuous linear functional As discussed earlier, continuous linear functionals on a ${\displaystyle C_{c}^{\infty }(U)}$ are known as distributions on ${\displaystyle U.}$ Other equivalent definitions are described below. By definition, a distribution on ${\displaystyle U}$ is a continuous linear functional on ${\displaystyle C_{c}^{\infty }(U).}$ Said differently, a distribution on ${\displaystyle U}$ is an element of the continuous dual space of ${\displaystyle C_{c}^{\infty }(U)}$ when ${\displaystyle C_{c}^{\infty }(U)}$ is endowed with its canonical LF topology. There is a canonical duality pairing between a distribution ${\displaystyle T}$ on ${\displaystyle U}$ and a test function ${\displaystyle f\in C_{c}^{\infty }(U),}$ which is denoted using angle brackets by ${\displaystyle {\begin{cases}{\mathcal {D))'(U)\times C_{c}^{\infty }(U)\to \mathbb {R} \\(T,f)\mapsto \langle T,f\rangle :=T(f)\end{cases))}$ One interprets this notation as the distribution ${\displaystyle T}$ acting on the test function ${\displaystyle f}$ to give a scalar, or symmetrically as the test function ${\displaystyle f}$ acting on the distribution ${\displaystyle T.}$ #### Characterizations of distributions Proposition. If ${\displaystyle T}$ is a linear functional on ${\displaystyle C_{c}^{\infty }(U)}$ then the following are equivalent: 1. T is a distribution; 2. T is continuous; 3. T is continuous at the origin; 4. T is uniformly continuous; 5. T is a bounded operator; 6. T is sequentially continuous; • explicitly, for every sequence ${\displaystyle \left(f_{i}\right)_{i=1}^{\infty ))$ in ${\displaystyle C_{c}^{\infty }(U)}$ that converges in ${\displaystyle C_{c}^{\infty }(U)}$ to some ${\displaystyle f\in C_{c}^{\infty }(U),}$ ${\textstyle \lim _{i\to \infty }T\left(f_{i}\right)=T(f);}$[note 3] 7. T is sequentially continuous at the origin; in other words, T maps null sequences[note 4] to null sequences; • explicitly, for every sequence ${\displaystyle \left(f_{i}\right)_{i=1}^{\infty ))$ in ${\displaystyle C_{c}^{\infty }(U)}$ that converges in ${\displaystyle C_{c}^{\infty }(U)}$ to the origin (such a sequence is called a null sequence), ${\textstyle \lim _{i\to \infty }T\left(f_{i}\right)=0;}$ • a null sequence is by definition any sequence that converges to the origin; 8. T maps null sequences to bounded subsets; • explicitly, for every sequence ${\displaystyle \left(f_{i}\right)_{i=1}^{\infty ))$ in ${\displaystyle C_{c}^{\infty }(U)}$ that converges in ${\displaystyle C_{c}^{\infty }(U)}$ to the origin, the sequence ${\displaystyle \left(T\left(f_{i}\right)\right)_{i=1}^{\infty ))$ is bounded; 9. T maps Mackey convergent null sequences to bounded subsets; • explicitly, for every Mackey convergent null sequence ${\displaystyle \left(f_{i}\right)_{i=1}^{\infty ))$ in ${\displaystyle C_{c}^{\infty }(U),}$ the sequence ${\displaystyle \left(T\left(f_{i}\right)\right)_{i=1}^{\infty ))$ is bounded; • a sequence ${\displaystyle f_{\bullet }=\left(f_{i}\right)_{i=1}^{\infty ))$ is said to be Mackey convergent to the origin if there exists a divergent sequence ${\displaystyle r_{\bullet }=\left(r_{i}\right)_{i=1}^{\infty }\to \infty }$ of positive real number such that the sequence ${\displaystyle \left(r_{i}f_{i}\right)_{i=1}^{\infty ))$ is bounded; every sequence that is Mackey convergent to the origin necessarily converges to the origin (in the usual sense); 10. The kernel of T is a closed subspace of ${\displaystyle C_{c}^{\infty }(U);}$ 11. The graph of T is closed; 12. There exists a continuous seminorm ${\displaystyle g}$ on ${\displaystyle C_{c}^{\infty }(U)}$ such that ${\displaystyle |T|\leq g;}$ 13. There exists a constant ${\displaystyle C>0}$ and a finite subset ${\displaystyle \{g_{1},\ldots ,g_{m}\}\subseteq {\mathcal {P))}$ (where ${\displaystyle {\mathcal {P))}$ is any collection of continuous seminorms that defines the canonical LF topology on ${\displaystyle C_{c}^{\infty }(U)}$) such that ${\displaystyle |T|\leq C(g_{1}+\cdots +g_{m});}$[note 5] 14. For every compact subset ${\displaystyle K\subseteq U}$ there exist constants ${\displaystyle C>0}$ and ${\displaystyle N\in \mathbb {N} }$ such that for all ${\displaystyle f\in C^{\infty }(K),}$[1] ${\displaystyle |T(f)|\leq C\sup\{|\partial ^{\alpha }f(x)|:x\in U,|\alpha |\leq N\};}$ 15. For every compact subset ${\displaystyle K\subseteq U}$ there exist constants ${\displaystyle C_{K}>0}$ and ${\displaystyle N_{K}\in \mathbb {N} }$ such that for all ${\displaystyle f\in C_{c}^{\infty }(U)}$ with support contained in ${\displaystyle K,}$[10] ${\displaystyle |T(f)|\leq C_{K}\sup\{|\partial ^{\alpha }f(x)|:x\in K,|\alpha |\leq N_{K}\};}$ 16. For any compact subset ${\displaystyle K\subseteq U}$ and any sequence ${\displaystyle \{f_{i}\}_{i=1}^{\infty ))$ in ${\displaystyle C^{\infty }(K),}$ if ${\displaystyle \{\partial ^{p}f_{i}\}_{i=1}^{\infty ))$ converges uniformly to zero for all multi-indices ${\displaystyle p,}$ then ${\displaystyle T(f_{i})\to 0;}$ #### Topology on the space of distributions and its relation to the weak-* topology The set of all distributions on ${\displaystyle U}$ is the continuous dual space of ${\displaystyle C_{c}^{\infty }(U),}$ which when endowed with the strong dual topology is denoted by ${\displaystyle {\mathcal {D))'(U).}$ Importantly, unless indicated otherwise, the topology on ${\displaystyle {\mathcal {D))'(U)}$ is the strong dual topology; if the topology is instead the weak-* topology then this will be clearly indicated. Neither topology is metrizable although unlike the weak-* topology, the strong dual topology makes ${\displaystyle {\mathcal {D))'(U)}$ into a complete nuclear space, to name just a few of its desirable properties. Neither ${\displaystyle C_{c}^{\infty }(U)}$ nor its strong dual ${\displaystyle {\mathcal {D))'(U)}$ is a sequential space and so neither of their topologies can be fully described by sequences (in other words, defining only what sequences converge in these spaces is not enough to fully/correctly define their topologies). However, a sequence in ${\displaystyle {\mathcal {D))'(U)}$ converges in the strong dual topology if and only if it converges in the weak-* topology (this leads many authors to use pointwise convergence to actually define the convergence of a sequence of distributions; this is fine for sequences but this is not guaranteed to extend to the convergence of nets of distributions because a net may converge pointwise but fail to converge in the strong dual topology). More information about the topology that ${\displaystyle {\mathcal {D))'(U)}$ is endowed with can be found in the article on spaces of test functions and distributions and in the articles on polar topologies and dual systems. A linear map from ${\displaystyle {\mathcal {D))'(U)}$ into another locally convex topological vector space (such as any normed space) is continuous if and only if it is sequentially continuous at the origin. However, this is no longer guaranteed if the map is not linear or for maps valued in more general topological spaces (for example, that are not also locally convex topological vector spaces). The same is true of maps from ${\displaystyle C_{c}^{\infty }(U)}$ (more generally, this is true of maps from any locally convex bornological space). ## Localization of distributions There is no way to define the value of a distribution in ${\displaystyle {\mathcal {D))'(U)}$ at a particular point of U. However, as is the case with functions, distributions on U restrict to give distributions on open subsets of U. Furthermore, distributions are locally determined in the sense that a distribution on all of U can be assembled from a distribution on an open cover of U satisfying some compatibility conditions on the overlaps. Such a structure is known as a sheaf. ### Extensions and restrictions to an open subset Let U and V be open subsets of ${\displaystyle \mathbb {R} ^{n))$ with ${\displaystyle V\subseteq U.}$ Let ${\displaystyle E_{VU}:{\mathcal {D))(V)\to {\mathcal {D))(U)}$ be the operator which extends by zero a given smooth function compactly supported in V to a smooth function compactly supported in the larger set U. The map ${\displaystyle E_{VU}:{\mathcal {D))(V)\to {\mathcal {D))(U)}$ is a continuous injection. A distribution ${\displaystyle S\in {\mathcal {D))'(V)}$ is said to be extendible to U if it belongs to the range of the transpose of ${\displaystyle E_{VU))$ and it is called extendible if it is extendable to ${\displaystyle \mathbb {R} ^{n}.}$[11] For any distribution ${\displaystyle T\in {\mathcal {D))'(U),}$ the restriction ${\displaystyle \rho _{VU}(T)}$ is a distribution in ${\displaystyle {\mathcal {D))'(V)}$ defined as the transpose of ${\displaystyle E_{VU))$ by ${\displaystyle \langle \rho _{VU}T,\phi \rangle =\langle T,E_{VU}\phi \rangle \quad {\text{ for all ))\phi \in {\mathcal {D))(V).}$ Unless ${\displaystyle U=V,}$ the restriction to V is neither injective nor surjective. Lack of surjectivity follows since distributions can blow up towards the boundary of V. For instance, if ${\displaystyle U=\mathbb {R} }$ and ${\displaystyle V=(0,2),}$ then the distribution ${\displaystyle T(x)=\sum _{n=1}^{\infty }n\,\delta \left(x-{\frac {1}{n))\right)}$ is in ${\displaystyle {\mathcal {D))'(V)}$ but admits no extension to ${\displaystyle {\mathcal {D))'(U).}$ ### Gluing and distributions that vanish in a set Theorem[12] — Let ${\displaystyle (U_{i})_{i\in I))$ be a collection of open subsets of ${\displaystyle \mathbb {R} ^{n}.}$ For each ${\displaystyle i\in I,}$ let ${\displaystyle T_{i}\in {\mathcal {D))'(U_{i})}$ and suppose that for all ${\displaystyle i,j\in I,}$ the restriction of ${\displaystyle T_{i))$ to ${\displaystyle U_{i}\cap U_{j))$ is equal to the restriction of ${\displaystyle T_{j))$ to ${\displaystyle U_{i}\cap U_{j))$ (note that both restrictions are elements of ${\displaystyle {\mathcal {D))'(U_{i}\cap U_{j})}$). Then there exists a unique ${\textstyle T\in {\mathcal {D))'(\bigcup _{i\in I}U_{i})}$ such that for all ${\displaystyle i\in I,}$ the restriction of T to ${\displaystyle U_{i))$ is equal to ${\displaystyle T_{i}.}$ Let V be an open subset of U. ${\displaystyle T\in {\mathcal {D))'(U)}$ is said to vanish in V if for all ${\displaystyle f\in {\mathcal {D))(U)}$ such that ${\displaystyle \operatorname {supp} (f)\subseteq V}$ we have ${\displaystyle Tf=0.}$ T vanishes in V if and only if the restriction of T to V is equal to 0, or equivalently, if and only if T lies in the kernel of the restriction map ${\displaystyle \rho _{VU}.}$ Corollary[12] — Let ${\displaystyle (U_{i})_{i\in I))$ be a collection of open subsets of ${\displaystyle \mathbb {R} ^{n))$ and let ${\textstyle T\in {\mathcal {D))'(\bigcup _{i\in I}U_{i}).}$ ${\displaystyle T=0}$ if and only if for each ${\displaystyle i\in I,}$ the restriction of T to ${\displaystyle U_{i))$ is equal to 0. Corollary[12] — The union of all open subsets of U in which a distribution T vanishes is an open subset of U in which T vanishes. ### Support of a distribution This last corollary implies that for every distribution T on U, there exists a unique largest subset V of U such that T vanishes in V (and does not vanish in any open subset of U that is not contained in V); the complement in U of this unique largest open subset is called the support of T.[12] Thus ${\displaystyle \operatorname {supp} (T)=U\setminus \bigcup \{V\mid \rho _{VU}T=0\}.}$ If ${\displaystyle f}$ is a locally integrable function on U and if ${\displaystyle D_{f))$ is its associated distribution, then the support of ${\displaystyle D_{f))$ is the smallest closed subset of U in the complement of which ${\displaystyle f}$ is almost everywhere equal to 0.[12] If ${\displaystyle f}$ is continuous, then the support of ${\displaystyle D_{f))$ is equal to the closure of the set of points in U at which ${\displaystyle f}$ does not vanish.[12] The support of the distribution associated with the Dirac measure at a point ${\displaystyle x_{0))$ is the set ${\displaystyle \{x_{0}\}.}$[12] If the support of a test function ${\displaystyle f}$ does not intersect the support of a distribution T then ${\displaystyle Tf=0.}$ A distribution T is 0 if and only if its support is empty. If ${\displaystyle f\in C^{\infty }(U)}$ is identically 1 on some open set containing the support of a distribution T then ${\displaystyle fT=T.}$ If the support of a distribution T is compact then it has finite order and furthermore, there is a constant ${\displaystyle C}$ and a non-negative integer ${\displaystyle N}$ such that:[7] ${\displaystyle |T\phi |\leq C\|\phi \|_{N}:=C\sup \left\{\left|\partial ^{\alpha }\phi (x)\right|:x\in U,|\alpha |\leq N\right\}\quad {\text{ for all ))\phi \in {\mathcal {D))(U).}$ If T has compact support then it has a unique extension to a continuous linear functional ${\displaystyle {\widehat {T))}$ on ${\displaystyle C^{\infty }(U)}$; this functional can be defined by ${\displaystyle {\widehat {T))(f):=T(\psi f),}$ where ${\displaystyle \psi \in {\mathcal {D))(U)}$ is any function that is identically 1 on an open set containing the support of T.[7] If ${\displaystyle S,T\in {\mathcal {D))'(U)}$ and ${\displaystyle \lambda \neq 0}$ then ${\displaystyle \operatorname {supp} (S+T)\subseteq \operatorname {supp} (S)\cup \operatorname {supp} (T)}$ and ${\displaystyle \operatorname {supp} (\lambda T)=\operatorname {supp} (T).}$ Thus, distributions with support in a given subset ${\displaystyle A\subseteq U}$ form a vector subspace of ${\displaystyle {\mathcal {D))'(U).}$[13] Furthermore, if ${\displaystyle P}$ is a differential operator in U, then for all distributions T on U and all ${\displaystyle f\in C^{\infty }(U)}$ we have ${\displaystyle \operatorname {supp} (P(x,\partial )T)\subseteq \operatorname {supp} (T)}$ and ${\displaystyle \operatorname {supp} (fT)\subseteq \operatorname {supp} (f)\cap \operatorname {supp} (T).}$[13] ### Distributions with compact support #### Support in a point set and Dirac measures For any ${\displaystyle x\in U,}$ let ${\displaystyle \delta _{x}\in {\mathcal {D))'(U)}$ denote the distribution induced by the Dirac measure at ${\displaystyle x.}$ For any ${\displaystyle x_{0}\in U}$ and distribution ${\displaystyle T\in {\mathcal {D))'(U),}$ the support of T is contained in ${\displaystyle \{x_{0})$ if and only if T is a finite linear combination of derivatives of the Dirac measure at ${\displaystyle x_{0}.}$[14] If in addition the order of T is ${\displaystyle \leq k}$ then there exist constants ${\displaystyle \alpha _{p))$ such that:[15]
${\displaystyle T=\sum _{|p|\leq k}\alpha _{p}\partial ^{p}\delta _{x_{0)).}$
Said differently, if T has support at a single point ${\displaystyle \{P\},}$ then T is in fact a finite linear combination of distributional derivatives of the ${\displaystyle \delta }$ function at P. That is, there exists an integer m and complex constants ${\displaystyle a_{\alpha ))$ such that
${\displaystyle T=\sum _{|\alpha |\leq m}a_{\alpha }\partial ^{\alpha }(\tau _{P}\delta )}$
where ${\displaystyle \tau _{P))$ is the translation operator.
#### Distribution with compact support
Theorem[7] — Suppose T is a distribution on U with compact support K. There exists a continuous function ${\displaystyle f}$ defined on U and a multi-index p such that
${\displaystyle T=\partial ^{p}f,}$
where the derivatives are understood in the sense of distributions. That is, for all test functions ${\displaystyle \phi }$ on U,
${\displaystyle T\phi =(-1)^{|p|}\int _{U}f(x)(\partial ^{p}\phi )(x)\,dx.}$
#### Distributions of finite order with support in an open subset
Theorem[7] — Suppose T is a distribution on U with compact support K and let V be an open subset of U containing K. Since every distribution with compact support has finite order, take N to be the order of T and define ${\displaystyle P:=\{0,1,\ldots ,N+2\}^{n}.}$ There exists a family of continuous functions ${\displaystyle (f_{p})_{p\in P))$ defined on U with support in V such that
${\displaystyle T=\sum _{p\in P}\partial ^{p}f_{p},}$
where the derivatives are understood in the sense of distributions. That is, for all test functions ${\displaystyle \phi }$ on U,
${\displaystyle T\phi =\sum _{p\in P}(-1)^{|p|}\int _{U}f_{p}(x)(\partial ^{p}\phi )(x)\,dx.}$
### Global structure of distributions
The formal definition of distributions exhibits them as a subspace of a very large space, namely the topological dual of ${\displaystyle {\mathcal {D))(U)}$ (or the Schwartz space ${\displaystyle {\mathcal {S))(\mathbb {R} ^{n})}$ for tempered distributions). It is not immediately clear from the definition how exotic a distribution might be. To answer this question, it is instructive to see distributions built up from a smaller space, namely the space of continuous functions. Roughly, any distribution is locally a (multiple) derivative of a continuous function. A precise version of this result, given below, holds for distributions of compact support, tempered distributions, and general distributions. Generally speaking, no proper subset of the space of distributions contains all continuous functions and is closed under differentiation. This says that distributions are not particularly exotic objects; they are only as complicated as necessary.
#### Distributions as sheaves
Theorem[16] — Let T be a distribution on U. There exists a sequence ${\displaystyle (T_{i})_{i=1}^{\infty ))$ in ${\displaystyle {\mathcal {D))'(U)}$ such that each Ti has compact support and every compact subset ${\displaystyle K\subseteq U}$ intersects the support of only finitely many ${\displaystyle T_{i},}$ and the sequence of partial sums ${\displaystyle (S_{j})_{j=1}^{\infty },}$ defined by ${\displaystyle S_{j}:=T_{1}+\cdots +T_{j},}$ converges in ${\displaystyle {\mathcal {D))'(U)}$ to T; in other words we have:
${\displaystyle T=\sum _{i=1}^{\infty }T_{i}.}$
Recall that a sequence converges in ${\displaystyle {\mathcal {D))'(U)}$ (with its strong dual topology) if and only if it converges pointwise.
#### Decomposition of distributions as sums of derivatives of continuous functions
By combining the above results, one may express any distribution on U as the sum of a series of distributions with compact support, where each of these distributions can in turn be written as a finite sum of distributional derivatives of continuous functions on U. In other words, for arbitrary ${\displaystyle T\in {\mathcal {D))'(U)}$ we can write:
${\displaystyle T=\sum _{i=1}^{\infty }\sum _{p\in P_{i))\partial ^{p}f_{ip},}$
where ${\displaystyle P_{1},P_{2},\ldots }$ are finite sets of multi-indices and the functions ${\displaystyle f_{ip))$ are continuous.
Theorem[17] — Let T be a distribution on U. For every multi-index p there exists a continuous function ${\displaystyle g_{p))$ on U such that
1. any compact subset K of U intersects the support of only finitely many ${\displaystyle g_{p},}$ and
2. ${\displaystyle T=\sum \nolimits _{p}\partial ^{p}g_{p}.}$
Moreover, if T has finite order, then one can choose ${\displaystyle g_{p))$ in such a way that only finitely many of them are non-zero.
Note that the infinite sum above is well-defined as a distribution. The value of T for a given ${\displaystyle f\in {\mathcal {D))(U)}$ can be computed using the finitely many ${\displaystyle g_{\alpha ))$ that intersect the support of ${\displaystyle f.}$
## Operations on distributions
Many operations which are defined on smooth functions with compact support can also be defined for distributions. In general, if ${\displaystyle A:{\mathcal {D))(U)\to {\mathcal {D))(U)}$ is a linear map which is continuous with respect to the weak topology, then it is possible to extend A to a map ${\displaystyle A:{\mathcal {D))'(U)\to {\mathcal {D))'(U)}$ by passing to the limit.[note 6][citation needed][clarification needed]
### Preliminaries: Transpose of a linear operator
Main article: Transpose of a linear map
Operations on distributions and spaces of distributions are often defined by means of the transpose of a linear operator. This is because the transpose allows for a unified presentation of the many definitions in the theory of distributions and also because its properties are well known in functional analysis.[18] For instance, the well-known Hermitian adjoint of a linear operator between Hilbert spaces is just the operator's transpose (but with the Riesz representation theorem used to identify each Hilbert space with its continuous dual space). In general the transpose of a continuous linear map ${\displaystyle A:X\to Y}$ is the linear map
${\displaystyle {}^{t}A:Y'\to X'\qquad {\text{ defined by ))\qquad {}^{t}A(y'):=y'\circ A,}$
or equivalently, it is the unique map satisfying ${\displaystyle \langle y',A(x)\rangle =\left\langle {}^{t}A(y'),x\right\rangle }$ for all ${\displaystyle x\in X}$ and all ${\displaystyle y'\in Y'}$ (the prime symbol in ${\displaystyle y'}$ does not denote a derivative of any kind; it merely indicates that ${\displaystyle y'}$ is an element of the continuous dual space ${\displaystyle Y'}$). Since A is continuous, the transpose ${\displaystyle {}^{t}A:Y'\to X'}$ is also continuous when both duals are endowed with their respective strong dual topologies; it is also continuous when both duals are endowed with their respective weak* topologies (see the articles polar topology and dual system for more details).
In the context of distributions, the characterization of the transpose can be refined slightly. Let ${\displaystyle A:{\mathcal {D))(U)\to {\mathcal {D))(U)}$ be a continuous linear map. Then by definition, the transpose of A is the unique linear operator ${\displaystyle A^{t}:{\mathcal {D))'(U)\to {\mathcal {D))'(U)}$ that satisfies:
${\displaystyle \langle {}^{t}A(T),\phi \rangle =\langle T,A(\phi )\rangle \quad {\text{ for all ))\phi \in {\mathcal {D))(U){\text{ and all ))T\in {\mathcal {D))'(U).}$
However, since the image of ${\displaystyle {\mathcal {D))(U)}$ is dense in ${\displaystyle {\mathcal {D))'(U),}$ it is sufficient that the above equality hold for all distributions of the form ${\displaystyle T=D_{\psi ))$ where ${\displaystyle \psi \in {\mathcal {D))(U).}$ Explicitly, this means that the above condition holds if and only if the condition below holds:
${\displaystyle \langle {}^{t}A(D_{\psi }),\phi \rangle =\langle D_{\psi },A(\phi )\rangle =\langle \psi ,A(\phi )\rangle =\int _{U}\psi (A\phi )\,dx\quad {\text{ for all ))\phi ,\psi \in {\mathcal {D))(U).}$
### Differential operators
#### Differentiation of distributions
Let ${\displaystyle A:{\mathcal {D))(U)\to {\mathcal {D))(U)}$ be the partial derivative operator ${\displaystyle {\tfrac {\partial }{\partial x_{k))}.}$ In order to extend ${\displaystyle A}$ we compute its transpose:
{\displaystyle {\begin{aligned}\langle {}^{t}A(D_{\psi }),\phi \rangle &=\int _{U}\psi (A\phi )\,dx&&{\text{(See above.)))\\&=\int _{U}\psi {\frac {\partial \phi }{\partial x_{k))}\,dx\\[4pt]&=-\int _{U}\phi {\frac {\partial \psi }{\partial x_{k))}\,dx&&{\text{(integration by parts)))\\[4pt]&=-\left\langle {\frac {\partial \psi }{\partial x_{k))},\phi \right\rangle \\[4pt]&=-\langle A\psi ,\phi \rangle =\langle -A\psi ,\phi \rangle \end{aligned))}
Therefore ${\displaystyle {}^{t}A=-A.}$ Therefore, the partial derivative of ${\displaystyle T}$ with respect to the coordinate ${\displaystyle x_{k))$ is defined by the formula
${\displaystyle \left\langle {\frac {\partial T}{\partial x_{k))},\phi \right\rangle =-\left\langle T,{\frac {\partial \phi }{\partial x_{k))}\right\rangle \qquad {\text{ for all ))\phi \in {\mathcal {D))(U).}$
With this definition, every distribution is infinitely differentiable, and the derivative in the direction ${\displaystyle x_{k))$ is a linear operator on ${\displaystyle {\mathcal {D))'(U).}$
More generally, if ${\displaystyle \alpha }$ is an arbitrary multi-index, then the partial derivative ${\displaystyle \partial ^{\alpha }T}$ of the distribution ${\displaystyle T\in {\mathcal {D))'(U)}$ is defined by
${\displaystyle \langle \partial ^{\alpha }T,\phi \rangle =(-1)^{|\alpha |}\langle T,\partial ^{\alpha }\phi \rangle \qquad {\text{ for all ))\phi \in {\mathcal {D))(U).}$
Differentiation of distributions is a continuous operator on ${\displaystyle {\mathcal {D))'(U);}$ this is an important and desirable property that is not shared by most other notions of differentiation.
If T is a distribution in ${\displaystyle \mathbb {R} }$ then
${\displaystyle \lim _{x\to 0}{\frac {T-\tau _{x}T}{x))=T'\in {\mathcal {D))'(\mathbb {R} ),}$
where ${\displaystyle T'}$ is the derivative of T and ${\displaystyle \tau _{x))$ is translation by x; thus the derivative of T may be viewed as a limit of quotients.[19]
#### Differential operators acting on smooth functions
A linear differential operator in U with smooth coefficients acts on the space of smooth functions on ${\displaystyle U.}$ Given such an operator ${\textstyle P:=\sum _{\alpha }c_{\alpha }\partial ^{\alpha },}$ we would like to define a continuous linear map, ${\displaystyle D_{P))$ that extends the action of ${\displaystyle P}$ on ${\displaystyle C^{\infty }(U)}$ to distributions on ${\displaystyle U.}$ In other words, we would like to define ${\displaystyle D_{P))$ such that the following diagram commutes:
${\displaystyle {\begin{matrix}{\mathcal {D))'(U)&{\stackrel {D_{P)){\longrightarrow ))&{\mathcal {D))'(U)\\[2pt]\uparrow &&\uparrow \\[2pt]C^{\infty }(U)&{\stackrel {P}{\longrightarrow ))&C^{\infty }(U)\end{matrix))}$
where the vertical maps are given by assigning ${\displaystyle f\in C^{\infty }(U)}$ its canonical distribution ${\displaystyle D_{f}\in {\mathcal {D))'(U),}$ which is defined by:
${\displaystyle D_{f}(\phi )=\langle f,\phi \rangle :=\int _{U}f(x)\phi (x)\,dx\quad {\text{ for all ))\phi \in {\mathcal {D))(U).}$
With this notation the diagram commuting is equivalent to:
${\displaystyle D_{P(f)}=D_{P}D_{f}\qquad {\text{ for all ))f\in C^{\infty }(U).}$
In order to find ${\displaystyle D_{P},}$ the transpose ${\displaystyle {}^{t}P:{\mathcal {D))'(U)\to {\mathcal {D))'(U)}$ of the continuous induced map ${\displaystyle P:{\mathcal {D))(U)\to {\mathcal {D))(U)}$ defined by ${\displaystyle \phi \mapsto P(\phi )}$ is considered in the lemma below. This leads to the following definition of the differential operator on ${\displaystyle U}$ called the formal transpose of ${\displaystyle P,}$ which will be denoted by ${\displaystyle P_{*))$ to avoid confusion with the transpose map, that is defined by
${\displaystyle P_{*}:=\sum _{\alpha }b_{\alpha }\partial ^{\alpha }\quad {\text{ where ))\quad b_{\alpha }:=\sum _{\beta \geq \alpha }(-1)^{|\beta |}{\binom {\beta }{\alpha ))\partial ^{\beta -\alpha }c_{\beta }.}$
Lemma — Let ${\displaystyle P}$ be a linear differential operator with smooth coefficients in ${\displaystyle U.}$ Then for all ${\displaystyle \phi \in {\mathcal {D))(U)}$ we have
${\displaystyle \left\langle {}^{t}P(D_{f}),\phi \right\rangle =\left\langle D_{P_{*}(f)},\phi \right\rangle ,}$
which is equivalent to:
${\displaystyle {}^{t}P(D_{f})=D_{P_{*}(f)}.}$
Proof
As discussed above, for any ${\displaystyle \phi \in {\mathcal {D))(U),}$ the transpose may be calculated by:
{\displaystyle {\begin{aligned}\left\langle {}^{t}P(D_{f}),\phi \right\rangle &=\int _{U}f(x)P(\phi )(x)\,dx\\&=\int _{U}f(x)\left[\sum \nolimits _{\alpha }c_{\alpha }(x)(\partial ^{\alpha }\phi )(x)\right]\,dx\\&=\sum \nolimits _{\alpha }\int _{U}f(x)c_{\alpha }(x)(\partial ^{\alpha }\phi )(x)\,dx\\&=\sum \nolimits _{\alpha }(-1)^{|\alpha |}\int _{U}\phi (x)(\partial ^{\alpha }(c_{\alpha }f))(x)\,dx\end{aligned))}
For the last line we used integration by parts combined with the fact that ${\displaystyle \phi }$ and therefore all the functions ${\displaystyle f(x)c_{\alpha }(x)\partial ^{\alpha }\phi (x)}$ have compact support.[note 7] Continuing the calculation above, for all ${\displaystyle \phi \in {\mathcal {D))(U):}$
{\displaystyle {\begin{aligned}\left\langle {}^{t}P(D_{f}),\phi \right\rangle &=\sum \nolimits _{\alpha }(-1)^{|\alpha |}\int _{U}\phi (x)(\partial ^{\alpha }(c_{\alpha }f))(x)\,dx&&{\text{As shown above))\\[4pt]&=\int _{U}\phi (x)\sum \nolimits _{\alpha }(-1)^{|\alpha |}(\partial ^{\alpha }(c_{\alpha }f))(x)\,dx\\[4pt]&=\int _{U}\phi (x)\sum _{\alpha }\left[\sum _{\gamma \leq \alpha }{\binom {\alpha }{\gamma ))(\partial ^{\gamma }c_{\alpha })(x)(\partial ^{\alpha -\gamma }f)(x)\right]\,dx&&{\text{Leibniz rule))\\&=\int _{U}\phi (x)\left[\sum _{\alpha }\sum _{\gamma \leq \alpha }(-1)^{|\alpha |}{\binom {\alpha }{\gamma ))(\partial ^{\gamma }c_{\alpha })(x)(\partial ^{\alpha -\gamma }f)(x)\right]\,dx\\&=\int _{U}\phi (x)\left[\sum _{\alpha }\left[\sum _{\beta \geq \alpha }(-1)^{|\beta |}{\binom {\beta }{\alpha ))\left(\partial ^{\beta -\alpha }c_{\beta }\right)(x)\right](\partial ^{\alpha }f)(x)\right]\,dx&&{\text{Grouping terms by derivatives of ))f\\&=\int _{U}\phi (x)\left[\sum \nolimits _{\alpha }b_{\alpha }(x)(\partial ^{\alpha }f)(x)\right]\,dx&&b_{\alpha }:=\sum _{\beta \geq \alpha }(-1)^{|\beta |}{\binom {\beta }{\alpha ))\partial ^{\beta -\alpha }c_{\beta }\\&=\left\langle \left(\sum \nolimits _{\alpha }b_{\alpha }\partial ^{\alpha }\right)(f),\phi \right\rangle \end{aligned))}
The Lemma combined with the fact that the formal transpose of the formal transpose is the original differential operator, that is, ${\displaystyle P_{**}=P,}$[20] enables us to arrive at the correct definition: the formal transpose induces the (continuous) canonical linear operator ${\displaystyle P_{*}:C_{c}^{\infty }(U)\to C_{c}^{\infty }(U)}$ defined by ${\displaystyle \phi \mapsto P_{*}(\phi ).}$ We claim that the transpose of this map, ${\displaystyle {}^{t}P_{*}:{\mathcal {D))'(U)\to {\mathcal {D))'(U),}$ can be taken as ${\displaystyle D_{P}.}$ To see this, for every ${\displaystyle \phi \in {\mathcal {D))(U),}$ compute its action on a distribution of the form ${\displaystyle D_{f))$ with ${\displaystyle f\in C^{\infty }(U)}$:
{\displaystyle {\begin{aligned}\left\langle {}^{t}P_{*}\left(D_{f}\right),\phi \right\rangle &=\left\langle D_{P_{**}(f)},\phi \right\rangle &&{\text{Using Lemma above with ))P_{*}{\text{ in place of ))P\\&=\left\langle D_{P(f)},\phi \right\rangle &&P_{**}=P\end{aligned))}
We call the continuous linear operator ${\displaystyle D_{P}:={}^{t}P_{*}:{\mathcal {D))'(U)\to {\mathcal {D))'(U)}$ the differential operator on distributions extending P.[20] Its action on an arbitrary distribution ${\displaystyle S}$ is defined via:
${\displaystyle D_{P}(S)(\phi )=S\left(P_{*}(\phi )\right)\quad {\text{ for all ))\phi \in {\mathcal {D))(U).}$
If ${\displaystyle (T_{i})_{i=1}^{\infty ))$ converges to ${\displaystyle T\in {\mathcal {D))'(U)}$ then for every multi-index ${\displaystyle \alpha ,(\partial ^{\alpha }T_{i})_{i=1}^{\infty ))$ converges to ${\displaystyle \partial ^{\alpha }T\in {\mathcal {D))'(U).}$
#### Multiplication of distributions by smooth functions
A differential operator of order 0 is just multiplication by a smooth function. And conversely, if ${\displaystyle f}$ is a smooth function then ${\displaystyle P:=f(x)}$ is a differential operator of order 0, whose formal transpose is itself (that is, ${\displaystyle P_{*}=P}$). The induced differential operator ${\displaystyle D_{P}:{\mathcal {D))'(U)\to {\mathcal {D))'(U)}$ maps a distribution T to a distribution denoted by ${\displaystyle fT:=D_{P}(T).}$ We have thus defined the multiplication of a distribution by a smooth function.
We now give an alternative presentation of multiplication by a smooth function. If ${\displaystyle m:U\to \mathbb {R} }$ is a smooth function and T is a distribution on U, then the product ${\displaystyle mT}$ is defined by
${\displaystyle \langle mT,\phi \rangle =\langle T,m\phi \rangle \qquad {\text{ for all ))\phi \in {\mathcal {D))(U).}$
This definition coincides with the transpose definition since if ${\displaystyle M:{\mathcal {D))(U)\to {\mathcal {D))(U)}$ is the operator of multiplication by the function m (that is, ${\displaystyle (M\phi )(x)=m(x)\phi (x)}$), then
${\displaystyle \int _{U}(M\phi )(x)\psi (x)\,dx=\int _{U}m(x)\phi (x)\psi (x)\,dx=\int _{U}\phi (x)m(x)\psi (x)\,dx=\int _{U}\phi (x)(M\psi )(x)\,dx,}$
so that ${\displaystyle {}^{t}M=M.}$
Under multiplication by smooth functions, ${\displaystyle {\mathcal {D))'(U)}$ is a module over the ring ${\displaystyle C^{\infty }(U).}$ With this definition of multiplication by a smooth function, the ordinary product rule of calculus remains valid. However, a number of unusual identities also arise. For example, if ${\displaystyle \delta }$ is the Dirac delta distribution on ${\displaystyle \mathbb {R} ,}$ then ${\displaystyle m\delta =m(0)\delta ,}$ and if ${\displaystyle \delta ^{'))$ is the derivative of the delta distribution, then
${\displaystyle m\delta '=m(0)\delta '-m'\delta =m(0)\delta '-m'(0)\delta .}$
The bilinear multiplication map ${\displaystyle C^{\infty }(\mathbb {R} ^{n})\times {\mathcal {D))'(\mathbb {R} ^{n})\to {\mathcal {D))'\left(\mathbb {R} ^{n}\right)}$ given by ${\displaystyle (f,T)\mapsto fT}$ is not continuous; it is however, hypocontinuous.[21]
Example. For any distribution T, the product of T with the function that is identically 1 on U is equal to T.
Example. Suppose ${\displaystyle (f_{i})_{i=1}^{\infty ))$ is a sequence of test functions on U that converges to the constant function ${\displaystyle 1\in C^{\infty }(U).}$ For any distribution T on U, the sequence ${\displaystyle (f_{i}T)_{i=1}^{\infty ))$ converges to ${\displaystyle T\in {\mathcal {D))'(U).}$[22]
If ${\displaystyle (T_{i})_{i=1}^{\infty ))$ converges to ${\displaystyle T\in {\mathcal {D))'(U)}$ and ${\displaystyle (f_{i})_{i=1}^{\infty ))$ converges to ${\displaystyle f\in C^{\infty }(U)}$ then ${\displaystyle (f_{i}T_{i})_{i=1}^{\infty ))$ converges to ${\displaystyle fT\in {\mathcal {D))'(U).}$
##### Problem of multiplying distributions
It is easy to define the product of a distribution with a smooth function, or more generally the product of two distributions whose singular supports are disjoint. With more effort it is possible to define a well-behaved product of several distributions provided their wave front sets at each point are compatible. A limitation of the theory of distributions (and hyperfunctions) is that there is no associative product of two distributions extending the product of a distribution by a smooth function, as has been proved by Laurent Schwartz in the 1950s. For example, if ${\displaystyle \operatorname {p.v.} {\frac {1}{x))}$ is the distribution obtained by the Cauchy principal value
${\displaystyle \left(\operatorname {p.v.} {\frac {1}{x))\right)(\phi )=\lim _{\varepsilon \to 0^{+))\int _{|x|\geq \varepsilon }{\frac {\phi (x)}{x))\,dx\quad {\text{ for all ))\phi \in {\mathcal {S))(\mathbb {R} ).}$
If ${\displaystyle \delta }$ is the Dirac delta distribution then
${\displaystyle (\delta \times x)\times \operatorname {p.v.} {\frac {1}{x))=0}$
but,
${\displaystyle \delta \times \left(x\times \operatorname {p.v.} {\frac {1}{x))\right)=\delta }$
so the product of a distribution by a smooth function (which is always well defined) cannot be extended to an associative product on the space of distributions.
Thus, nonlinear problems cannot be posed in general and thus not solved within distribution theory alone. In the context of quantum field theory, however, solutions can be found. In more than two spacetime dimensions the problem is related to the regularization of divergences. Here Henri Epstein and Vladimir Glaser developed the mathematically rigorous (but extremely technical) causal perturbation theory. This does not solve the problem in other situations. Many other interesting theories are non linear, like for example the Navier–Stokes equations of fluid dynamics.
Several not entirely satisfactory[citation needed] theories of algebras of generalized functions have been developed, among which Colombeau's (simplified) algebra is maybe the most popular in use today.
Inspired by Lyons' rough path theory,[23] Martin Hairer proposed a consistent way of multiplying distributions with certain structure (regularity structures[24]), available in many examples from stochastic analysis, notably stochastic partial differential equations. See also Gubinelli–Imkeller–Perkowski (2015) for a related development based on Bony's paraproduct from Fourier analysis.
### Composition with a smooth function
Let T be a distribution on ${\displaystyle U.}$ Let V be an open set in ${\displaystyle \mathbb {R} ^{n},}$ and ${\displaystyle F:V\to U.}$ If ${\displaystyle F}$ is a submersion, it is possible to define
${\displaystyle T\circ F\in {\mathcal {D))'(V).}$
This is the composition of the distribution ${\displaystyle T}$ with ${\displaystyle F}$, and is also called the pullback of ${\displaystyle T}$ along ${\displaystyle F}$, sometimes written
${\displaystyle F^{\sharp }:T\mapsto F^{\sharp }T=T\circ F.}$
The pullback is often denoted ${\displaystyle F^{*},}$ although this notation should not be confused with the use of '*' to denote the adjoint of a linear mapping.
The condition that ${\displaystyle F}$ be a submersion is equivalent to the requirement that the Jacobian derivative ${\displaystyle dF(x)}$ of ${\displaystyle F}$ is a surjective linear map for every ${\displaystyle x\in V.}$ A necessary (but not sufficient) condition for extending ${\displaystyle F^{\#))$ to distributions is that ${\displaystyle F}$ be an open mapping.[25] The Inverse function theorem ensures that a submersion satisfies this condition.
If ${\displaystyle F}$ is a submersion, then ${\displaystyle F^{\#))$ is defined on distributions by finding the transpose map. Uniqueness of this extension is guaranteed since ${\displaystyle F^{\#))$ is a continuous linear operator on ${\displaystyle {\mathcal {D))(U).}$ Existence, however, requires using the change of variables formula, the inverse function theorem (locally) and a partition of unity argument.[26]
In the special case when ${\displaystyle F}$ is a diffeomorphism from an open subset V of ${\displaystyle \mathbb {R} ^{n))$ onto an open subset U of ${\displaystyle \mathbb {R} ^{n))$ change of variables under the integral gives:
${\displaystyle \int _{V}\phi \circ F(x)\psi (x)\,dx=\int _{U}\phi (x)\psi \left(F^{-1}(x)\right)\left|\det dF^{-1}(x)\right|\,dx.}$
In this particular case, then, ${\displaystyle F^{\#))$ is defined by the transpose formula:
${\displaystyle \left\langle F^{\sharp }T,\phi \right\rangle =\left\langle T,\left|\det d(F^{-1})\right|\phi \circ F^{-1}\right\rangle .}$
### Convolution
Under some circumstances, it is possible to define the convolution of a function with a distribution, or even the convolution of two distributions. Recall that if ${\displaystyle f}$ and ${\displaystyle g}$ are functions on ${\displaystyle \mathbb {R} ^{n))$ then we denote by ${\displaystyle f\ast g}$ the convolution of ${\displaystyle f}$ and ${\displaystyle g,}$ defined at ${\displaystyle x\in \mathbb {R} ^{n))$ to be the integral
${\displaystyle (f\ast g)(x):=\int _{\mathbb {R} ^{n))f(x-y)g(y)\,dy=\int _{\mathbb {R} ^{n))f(y)g(x-y)\,dy}$
provided that the integral exists. If ${\displaystyle 1\leq p,q,r\leq \infty }$ are such that ${\textstyle {\frac {1}{r))={\frac {1}{p))+{\frac {1}{q))-1}$ then for any functions ${\displaystyle f\in L^{p}(\mathbb {R} ^{n})}$ and ${\displaystyle g\in L^{q}(\mathbb {R} ^{n})}$ we have ${\displaystyle f\ast g\in L^{r}(\mathbb {R} ^{n})}$ and ${\displaystyle \|f\ast g\|_{L^{r))\leq \|f\|_{L^{p))\|g\|_{L^{q)).}$[27] If ${\displaystyle f}$ and ${\displaystyle g}$ are continuous functions on ${\displaystyle \mathbb {R} ^{n},}$ at least one of which has compact support, then ${\displaystyle \operatorname {supp} (f\ast g)\subseteq \operatorname {supp} (f)+\operatorname {supp} (g)}$ and if ${\displaystyle A\subseteq \mathbb {R} ^{n))$ then the value of ${\displaystyle f\ast g}$ on ${\displaystyle A}$ do not depend on the values of ${\displaystyle f}$ outside of the Minkowski sum ${\displaystyle A-\operatorname {supp} (g)=\{a-s:a\in A,s\in \operatorname {supp} (g)\}.}$[27]
Importantly, if ${\displaystyle g\in L^{1}(\mathbb {R} ^{n})}$ has compact support then for any ${\displaystyle 0\leq k\leq \infty ,}$ the convolution map ${\displaystyle f\mapsto f\ast g}$ is continuous when considered as the map ${\displaystyle C^{k}(\mathbb {R} ^{n})\to C^{k}(\mathbb {R} ^{n})}$ or as the map ${\displaystyle C_{c}^{k}(\mathbb {R} ^{n})\to C_{c}^{k}(\mathbb {R} ^{n}).}$[27]
#### Translation and symmetry
Given ${\displaystyle a\in \mathbb {R} ^{n},}$ the translation operator ${\displaystyle \tau _{a))$ sends ${\displaystyle f:\mathbb {R} ^{n}\to \mathbb {C} }$ to ${\displaystyle \tau _{a}f:\mathbb {R} ^{n}\to \mathbb {C} ,}$ defined by ${\displaystyle \tau _{a}f(y)=f(y-a).}$ This can be extended by the transpose to distributions in the following way: given a distribution ${\displaystyle T,}$ the translation of ${\displaystyle T}$ by ${\displaystyle a}$ is the distribution ${\displaystyle \tau _{a}T:{\mathcal {D))(\mathbb {R} ^{n})\to \mathbb {C} }$ defined by ${\displaystyle \tau _{a}T(\phi ):=\left\langle T,\tau _{-a}\phi \right\rangle .}$[28][29]
Given ${\displaystyle f:\mathbb {R} ^{n}\to \mathbb {C} ,}$ define the function ${\displaystyle {\tilde {f)):\mathbb {R} ^{n}\to \mathbb {C} }$ by ${\displaystyle {\tilde {f))(x):=f(-x).}$ Given a distribution ${\displaystyle T,}$ let ${\displaystyle {\tilde {T)):{\mathcal {D))(\mathbb {R} ^{n})\to \mathbb {C} }$ be the distribution defined by ${\displaystyle {\tilde {T))(\phi ):=T\left({\tilde {\phi ))\right).}$ The operator ${\displaystyle T\mapsto {\tilde {T))}$ is called the symmetry with respect to the origin.[28]
#### Convolution of a test function with a distribution
Convolution with ${\displaystyle f\in {\mathcal {D))(\mathbb {R} ^{n})}$ defines a linear map:
{\displaystyle {\begin{alignedat}{4}C_{f}:\,&{\mathcal {D))(\mathbb {R} ^{n})&&\to \,&&{\mathcal {D))(\mathbb {R} ^{n})\\&g&&\mapsto \,&&f\ast g\\\end{alignedat))}
which is continuous with respect to the canonical LF space topology on ${\displaystyle {\mathcal {D))(\mathbb {R} ^{n}).}$
Convolution of ${\displaystyle f}$ with a distribution ${\displaystyle T\in {\mathcal {D))'(\mathbb {R} ^{n})}$ can be defined by taking the transpose of ${\displaystyle C_{f))$ relative to the duality pairing of ${\displaystyle {\mathcal {D))(\mathbb {R} ^{n})}$ with the space ${\displaystyle {\mathcal {D))'(\mathbb {R} ^{n})}$ of distributions.[30] If ${\displaystyle f,g,\phi \in {\mathcal {D))(\mathbb {R} ^{n}),}$ then by Fubini's theorem
${\displaystyle \langle C_{f}g,\phi \rangle =\int _{\mathbb {R} ^{n))\phi (x)\int _{\mathbb {R} ^{n))f(x-y)g(y)\,dy\,dx=\left\langle g,C_{\tilde {f))\phi \right\rangle .}$
Extending by continuity, the convolution of ${\displaystyle f}$ with a distribution ${\displaystyle T}$ is defined by
${\displaystyle \langle f\ast T,\phi \rangle =\left\langle T,{\tilde {f))\ast \phi \right\rangle ,\quad {\text{ for all ))\phi \in {\mathcal {D))(\mathbb {R} ^{n}).}$
An alternative way to define the convolution of a test function ${\displaystyle f}$ and a distribution ${\displaystyle T}$ is to use the translation operator ${\displaystyle \tau _{a}.}$ The convolution of the compactly supported function ${\displaystyle f}$ and the distribution ${\displaystyle T}$ is then the function defined for each ${\displaystyle x\in \mathbb {R} ^{n))$ by
${\displaystyle (f\ast T)(x)=\left\langle T,\tau _{x}{\tilde {f))\right\rangle .}$
It can be shown that the convolution of a smooth, compactly supported function and a distribution is a smooth function. If the distribution ${\displaystyle T}$ has compact support then if ${\displaystyle f}$ is a polynomial (resp. an exponential function, an analytic function, the restriction of an entire analytic function on ${\displaystyle \mathbb {C} ^{n))$ to ${\displaystyle \mathbb {R} ^{n},}$ the restriction of an entire function of exponential type in ${\displaystyle \mathbb {C} ^{n))$ to ${\displaystyle \mathbb {R} ^{n))$) then the same is true of ${\displaystyle T\ast f.}$[28] If the distribution ${\displaystyle T}$ has compact support as well, then ${\displaystyle f\ast T}$ is a compactly supported function, and the Titchmarsh convolution theorem Hörmander (1983, Theorem 4.3.3) implies that:
${\displaystyle \operatorname {ch} (\operatorname {supp} (f\ast T))=\operatorname {ch} (\operatorname {supp} (f))+\operatorname {ch} (\operatorname {supp} (T))}$
where ${\displaystyle \operatorname {ch} }$ denotes the convex hull and supp denotes the support.
#### Convolution of a smooth function with a distribution
Let ${\displaystyle f\in C^{\infty }(\mathbb {R} ^{n})}$ and ${\displaystyle T\in {\mathcal {D))'(\mathbb {R} ^{n})}$ and assume that at least one of ${\displaystyle f}$ and ${\displaystyle T}$ has compact support. The convolution of ${\displaystyle f}$ and ${\displaystyle T,}$ denoted by ${\displaystyle f\ast T}$ or by ${\displaystyle T\ast f,}$ is the smooth function:[28]
{\displaystyle {\begin{alignedat}{4}f\ast T:\,&\mathbb {R} ^{n}&&\to \,&&\mathbb {C} \\&x&&\mapsto \,&&\left\langle T,\tau _{x}{\tilde {f))\right\rangle \\\end{alignedat))}
satisfying for all ${\displaystyle p\in \mathbb {N} ^{n))$:
{\displaystyle {\begin{aligned}&\operatorname {supp} (f\ast T)\subseteq \operatorname {supp} (f)+\operatorname {supp} (T)\\[6pt]&{\text{ for all ))p\in \mathbb {N} ^{n}:\quad {\begin{cases}\partial ^{p}\left\langle T,\tau _{x}{\tilde {f))\right\rangle =\left\langle T,\partial ^{p}\tau _{x}{\tilde {f))\right\rangle \\\partial ^{p}(T\ast f)=(\partial ^{p}T)\ast f=T\ast (\partial ^{p}f).\end{cases))\end{aligned))}
If ${\displaystyle T}$ is a distribution then the map ${\displaystyle f\mapsto T\ast f}$ is continuous as a map ${\displaystyle {\mathcal {D))(\mathbb {R} ^{n})\to C^{\infty }(\mathbb {R} ^{n})}$ where if in addition ${\displaystyle T}$ has compact support then it is also continuous as the map ${\displaystyle C^{\infty }(\mathbb {R} ^{n})\to C^{\infty }(\mathbb {R} ^{n})}$ and continuous as the map ${\displaystyle {\mathcal {D))(\mathbb {R} ^{n})\to {\mathcal {D))(\mathbb {R} ^{n}).}$[28]
If ${\displaystyle L:{\mathcal {D))(\mathbb {R} ^{n})\to C^{\infty }(\mathbb {R} ^{n})}$ is a continuous linear map such that ${\displaystyle L\partial ^{\alpha }\phi =\partial ^{\alpha }L\phi }$ for all ${\displaystyle \alpha }$ and all ${\displaystyle \phi \in {\mathcal {D))(\mathbb {R} ^{n})}$ then there exists a distribution ${\displaystyle T\in {\mathcal {D))'(\mathbb {R} ^{n})}$ such that ${\displaystyle L\phi =T\circ \phi }$ for all ${\displaystyle \phi \in {\mathcal {D))(\mathbb {R} ^{n}).}$[7]
Example.[7] Let ${\displaystyle H}$ be the Heaviside function on ${\displaystyle \mathbb {R} .}$ For any ${\displaystyle \phi \in {\mathcal {D))(\mathbb {R} ),}$
${\displaystyle (H\ast \phi )(x)=\int _{-\infty }^{x}\phi (t)\,dt.}$
Let ${\displaystyle \delta }$ be the Dirac measure at 0 and ${\displaystyle \delta '}$ its derivative as a distribution. Then ${\displaystyle \delta '\ast H=\delta }$ and ${\displaystyle 1\ast \delta '=0.}$ Importantly, the associative law fails to hold:
${\displaystyle 1=1\ast \delta =1\ast (\delta '\ast H)\neq (1\ast \delta ')\ast H=0\ast H=0.}$
#### Convolution of distributions
It is also possible to define the convolution of two distributions ${\displaystyle S}$ and ${\displaystyle T}$ on ${\displaystyle \mathbb {R} ^{n},}$ provided one of them has compact support. Informally, in order to define ${\displaystyle S\ast T}$ where ${\displaystyle T}$ has compact support, the idea is to extend the definition of the convolution ${\displaystyle \,\ast \,}$ to a linear operation on distributions so that the associativity formula
${\displaystyle S\ast (T\ast \phi )=(S\ast T)\ast \phi }$
continues to hold for all test functions ${\displaystyle \phi .}$[31]
It is also possible to provide a more explicit characterization of the convolution of distributions.[30] Suppose that ${\displaystyle S}$ and ${\displaystyle T}$ are distributions and that ${\displaystyle S}$ has compact support. Then the linear maps
{\displaystyle {\begin{alignedat}{9}\bullet \ast {\tilde {S)):\,&{\mathcal {D))(\mathbb {R} ^{n})&&\to \,&&{\mathcal {D))(\mathbb {R} ^{n})&&\quad {\text{ and ))\quad &&\bullet \ast {\tilde {T)):\,&&{\mathcal {D))(\mathbb {R} ^{n})&&\to \,&&{\mathcal {D))(\mathbb {R} ^{n})\\&f&&\mapsto \,&&f\ast {\tilde {S))&&&&&&f&&\mapsto \,&&f\ast {\tilde {T))\\\end{alignedat))}
are continuous. The transposes of these maps:
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# dependent effect sizes
Power for Meta-Analysis of Dependent Effects
## Power approximations for overall average effects in meta-analysis of dependent effect sizes
Meta-analytic models for dependent effect sizes have grown increasingly sophisticated over the last few decades, which has created challenges for a priori power calculations. We introduce power approximations for tests of average effect sizes based …
## Investigating narrative performance in children with developmental language disorder: A systematic review and meta-analysis
__Purpose__: Speech-language pathologists (SLPs) typically examine narrative performance when completing a comprehensive language assessment. However, there is significant variability in the methodologies used to evaluate narration. The primary aims …
## Meta-Analysis with robust variance estimation: Expanding the range of working models
In prevention science and related fields, large meta-analyses are common, and these analyses often involve dependent effect size estimates. Robust variance estimation (RVE) methods provide a way to include all dependent effect sizes in a single …
## Variance component estimates in meta-analysis with mis-specified sampling correlation
$\def\Pr{{\text{Pr}}} \def\E{{\text{E}}} \def\Var{{\text{Var}}} \def\Cov{{\text{Cov}}}$ In a recent paper with Beth Tipton, we proposed new working models for meta-analyses involving dependent effect sizes. The central idea of our approach is to use a working model that captures the main features of the effect size data, such as by allowing for both between- and within-study heterogeneity in the true effect sizes (rather than only between-study heterogeneity).
## Evaluating meta-analytic methods to detect selective reporting in the presence of dependent effect sizes
Meta-analysis is a set of statistical tools used to synthesize results from multiple studies evaluating a common research question. Two methodological challenges when conducting meta-analysis include selective reporting and correlated dependent …
## What do meta-analysts mean by 'multivariate' meta-analysis?
If you’ve ever had class with me or attended one of my presentations, you’ve probably heard me grouse about how statisticians are mostly awful about naming things.1 A lot of the terminology in our field is pretty bad and ineloquent.
## Sometimes, aggregating effect sizes is fine
In meta-analyses of psychology, education, and other social science research, it is very common that some of the included studies report more than one relevant effect size. For example, in a meta-analysis of intervention effects on reading outcomes, some studies may have used multiple measures of reading outcomes (each of which meets inclusion criteria), or may have measured outcomes at multiple follow-up times; some studies might have also investigated more than one version of an intervention, and it might be of interest to include effect sizes comparing each version to the no-intervention control condition; and it’s even possible that some studies may have all of these features, potentially contributing lots of effect size estimates.
## wildmeta
Cluster-wild bootstrap for meta-regression
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