Companal - User contributions [en]

MediaWiki:Sidebar

2024-09-06T15:28:40Z

Vipul:

User:Vipul/Sandbox

2024-09-06T15:27:42Z

Vipul:

* <math>\sqrt{7 + 2}!! + 3 = 723</math>
* <math>2^{8 - 1} = 128</math>
* <math>(7 - 4)!! + 4! = 744</math>
* <math>2^7 - 1 = 127</math>
* <math>9^{\sqrt{7 + 2}} = 729</math>
* <math>(7 + 2)^{\sqrt{9}} = 729</math>

User:Vipul/Sandbox

2024-09-06T15:27:23Z

Vipul:

* <math>\sqrt{7 + 2}!! + 3 = 723</math>
* <math>2^{8 - 1} = 128</math>
* <math>\sqrt{7 - 4}!! + 4! = 744</math>
* <math>2^7 - 1 = 127</math>
* <math>9^{\sqrt{7 + 2}} = 729</math>
* <math>(7 + 2)^{\sqrt{9}} = 729</math>

Companal:Enabling site search autocompletion

2024-08-07T21:29:53Z

Vipul: /* How to fix it */

Content copied from [[Ref:Ref:Enabling site search autocompletion]]. Images used are specific to this site (Companal).

Site search autocompletion is currently broken by default on this site. This page includes details on how to get it to work, and what's going on.

==What's wrong with site search autocompletion and how to fix it==

===What's wrong===

When you start typing something in the site search bar, you'll see it stuck at "Loading search suggestions" as shown in the screenshot below:

[[File:Site search autocompletion broken.png]]

Note that the actual search is still working -- you just have to hit Enter after typing the search query and it'll go to the search results page. It's the autocompletion before you hit Enter that is broken.

===How to fix it===

To fix it, you need to follow these steps:

* Write to vipulnaik1@gmail.com asking for a login to the site. Please include the following with your request: preferred username, preferred initial password (you can change it after logging in), real name (if you want it entered), email address to use (if you want an actual email address by which you can be contacted), and whether you want edit access as well. You don't need edit access for enabling site search autocompletion.
* Log in to the site. Then go to [[Special:Preferences]]. Go to the Appearance section and switch the Skin from "Vector (2022)" to "Vector legacy (2010)".
* Make sure to hit "Save" at the bottom.
* Now you can reload the page or load a new page.

Site search autocompletion should now work. Here's an example:

[[File:Site search autocompletion working.png]]

==More background==

We've recently upgraded the MediaWiki version of this wiki from 1.35.13 to 1.41.2 (see [[Special:Version]]). The upgrade allows us to migrate the wiki to a more modern operating system version running PHP 8. With the current setup for MediaWiki 1.41.2, we're in this situation:

* The "Vector legacy (2010)" skin has site search autocompletion working, but it doesn't render well on small screens. Specifically, even on small mobile screens, it still shows the left menu, and doesn't properly use the MobileFrontend extension settings.
* The "Vector (2022)" skin doesn't have site search autocompletion working (see screenshots in preceding section) but it does render fine on mobile devices.

It is possible to set only one default skin (that is applicable to all non-logged-in users and is the default for logged-in users who have not configured a skin for themselves). So, the selection of default skin comes down to whether it's more important for casual users to have the mobile experience working or to have site search autocompletion working. Based on a general understanding of user behavior, we believe that having a usable mobile experience is more important for casual users than having site search autocompletion.

However, for power users who are using the site extensively, site search autocompletion may be important. That's why we've written this page giving guidance on how to set up site search autocompletion.

Companal:429 Too Many Requests error

2024-08-07T20:36:30Z

Vipul:

This content is copied from [[Ref:Ref:429 Too Many Requests error]].

If you get a 429 Too Many Requests error when browsing this site, read on.

You're probably seeing this error because a large number of requests have been made from your IP address over a short period of time. That's probably a lot of requests from you or others who share your IP address (such as your home wi-fi network). Waiting a minute and then retrying should generally work.

If you are an actual human being with a legitimate reason to be browsing the site heavily, first, thank you and sorry about this! We set rate limits to prevent bots, spiders, spammers, and malicious actors from consuming too much of our server's resources so that our server's resources can be devoted to real humans like you. Consider writing to vipulnaik1@gmail.com with your IP address to have the IP address whitelisted. You can get your IP address by [https://www.google.com/search?q=my+ip+address Googling "my IP address"] (scroll down a little bit to where Google includes the IP address in a box). NOTE: If you have both an IPv4 address and an IPv6 address, you should send both; the server supports both IPv4 and IPv6, so either may end up getting used. To check if you have an IPv6 address, try visiting [https://ipv6.google.com/ ipv6.google.com].

If your IP address changes, or you are away from your home network, then you'll get rate-limited again. So if you find yourself getting rate-limited after already having been whitelisted, check if you are on a different IP address than the one for which you requested whitelisting.

Companal:429 Too Many Requests error

2024-08-07T20:34:44Z

Vipul:

This content is copied from [[Ref:Ref:429 Too Many Request error]].

If you get a 429 Too Many Requests error when browsing this site, read on.

You're probably seeing this error because a large number of requests have been made from your IP address over a short period of time. That's probably a lot of requests from you or others who share your IP address (such as your home wi-fi network). Waiting a minute and then retrying should generally work.

If you are an actual human being with a legitimate reason to be browsing the site heavily, first, thank you and sorry about this! We set rate limits to prevent bots, spiders, spammers, and malicious actors from consuming too much of our server's resources so that our server's resources can be devoted to real humans like you. Consider writing to vipulnaik1@gmail.com with your IP address to have the IP address whitelisted. You can get your IP address by [https://www.google.com/search?q=my+ip+address Googling "my IP address"] (scroll down a little bit to where Google includes the IP address in a box). NOTE: If you have both an IPv4 address and an IPv6 address, you should send both; the server supports both IPv4 and IPv6, so either may end up getting used. To check if you have an IPv6 address, try visiting [https://ipv6.google.com/ ipv6.google.com].

If your IP address changes, or you are away from your home network, then you'll get rate-limited again. So if you find yourself getting rate-limited after already having been whitelisted, check if you are on a different IP address than the one for which you requested whitelisting.

Companal:429 Too Many Requests error

2024-08-07T04:24:40Z

Vipul:

This content is copied from [[Ref:Ref:429 Too Many Request error]].

If you get a 429 Too Many Requests error when browsing this site, read on.

You're probably seeing this error because a large number of requests have been made from your IP address over a short period of time. That's probably a lot of requests from you or others who share your IP address (such as your home wi-fi network). Waiting a minute and then retrying should generally work.

If you are an actual human being with a legitimate reason to be browsing the site heavily, first, thank you and sorry about this! We set rate limits to prevent bots, spiders, spammers, and malicious actors from consuming too much of our server's resources so that our server's resources can be devoted to real humans like you. Consider writing to vipulnaik1@gmail.com with your IP address to have the IP address whitelisted. You can get your IP address by [https://www.google.com/search?q=my+ip+address Googling "my IP address"] (scroll down a little bit to where Google includes the IP address in a box). NOTE: If you have both an IPv4 address and an IPv6 address, you may need to send both; the server uses IPv6 if your client has both addresses. To check if you have an IPv6 address, try visiting [https://ipv6.google.com/ ipv6.google.com].

If your IP address changes, or you are away from your home network, then you'll get rate-limited again. So if you find yourself getting rate-limited after already having been whitelisted, check if you are on a different IP address than the one for which you requested whitelisting.

Companal:Enabling site search autocompletion

2024-08-07T04:15:28Z

Vipul:

Content copied from [[Ref:Ref:Enabling site search autocompletion]]. Images used are specific to this site (Companal).

Site search autocompletion is currently broken by default on this site. This page includes details on how to get it to work, and what's going on.

==What's wrong with site search autocompletion and how to fix it==

===What's wrong===

When you start typing something in the site search bar, you'll see it stuck at "Loading search suggestions" as shown in the screenshot below:

[[File:Site search autocompletion broken.png]]

Note that the actual search is still working -- you just have to hit Enter after typing the search query and it'll go to the search results page. It's the autocompletion before you hit Enter that is broken.

===How to fix it===

To fix it, you need to follow these steps:

* Write to vipulnaik1@gmail.com asking for a login to the site. Please include the following with your request: preferred username, preferred initial password (you can change it after logging in), real name (if you want it entered), email address to use (if you want an actual email address by which you can be contacted), and whether you want edit access as well. You don't need edit access for enabling site search autocompletion.
* Log in to the site, and go to [[Special:Preferences]]. Go to the Appearance section and switch the Skin from "Vector (2022)" to "Vector legacy (2010)".
* Make sure to hit "Save" at the bottom.
* Now you can reload the page or load a new page.

Site search autocompletion should now work. Here's an example:

[[File:Site search autocompletion working.png]]

==More background==

We've recently upgraded the MediaWiki version of this wiki from 1.35.13 to 1.41.2 (see [[Special:Version]]). The upgrade allows us to migrate the wiki to a more modern operating system version running PHP 8. With the current setup for MediaWiki 1.41.2, we're in this situation:

* The "Vector legacy (2010)" skin has site search autocompletion working, but it doesn't render well on small screens. Specifically, even on small mobile screens, it still shows the left menu, and doesn't properly use the MobileFrontend extension settings.
* The "Vector (2022)" skin doesn't have site search autocompletion working (see screenshots in preceding section) but it does render fine on mobile devices.

It is possible to set only one default skin (that is applicable to all non-logged-in users and is the default for logged-in users who have not configured a skin for themselves). So, the selection of default skin comes down to whether it's more important for casual users to have the mobile experience working or to have site search autocompletion working. Based on a general understanding of user behavior, we believe that having a usable mobile experience is more important for casual users than having site search autocompletion.

However, for power users who are using the site extensively, site search autocompletion may be important. That's why we've written this page giving guidance on how to set up site search autocompletion.

MediaWiki:Sitenotice

2024-08-07T04:13:22Z

Vipul:

Want site search autocompletion? See [[Project:Enabling site search autocompletion|here]]<br/>
Encountering 429 Too Many Requests errors when browsing the site? See [[Project:429 Too Many Requests error|here]]

Companal:429 Too Many Requests error

2024-08-07T04:11:36Z

Vipul: Created page with "If you get a 429 Too Many Requests error when browsing this site, read on. You're probably seeing this error because a large number of requests have been made from your IP address over a short period of time. That's probably a lot of requests from you or others who share your IP address (such as your home wi-fi network). Waiting a minute and then retrying should generally work. If you are an actual human being with a legitimate reason to be browsing the site heavily, f..."

If you get a 429 Too Many Requests error when browsing this site, read on.

You're probably seeing this error because a large number of requests have been made from your IP address over a short period of time. That's probably a lot of requests from you or others who share your IP address (such as your home wi-fi network). Waiting a minute and then retrying should generally work.

If you are an actual human being with a legitimate reason to be browsing the site heavily, first, thank you and sorry about this! We set rate limits to prevent bots, spiders, spammers, and malicious actors from consuming too much of our server's resources so that our server's resources can be devoted to real humans like you. Consider writing to vipulnaik1@gmail.com with your IP address to have the IP address whitelisted. You can get your IP address by [https://www.google.com/search?q=my+ip+address Googling "my IP address"] (scroll down a little bit to where Google includes the IP address in a box). NOTE: If you have both an IPv4 address and an IPv6 address, you may need to send both; the server uses IPv6 if your client has both addresses. To check if you have an IPv6 address, try visiting [https://ipv6.google.com/ ipv6.google.com].

If your IP address changes, or you are away from your home network, then you'll get rate-limited again. So if you find yourself getting rate-limited after already having been whitelisted, check if you are on a different IP address than the one for which you requested whitelisting.

File:Site search autocompletion working.png

2024-08-07T04:10:16Z

Vipul:

File:Site search autocompletion broken.png

2024-08-07T04:09:58Z

Vipul:

Companal:Enabling site search autocompletion

2024-08-07T04:04:35Z

Vipul: Created page with "Site search autocompletion is currently broken by default on this site. This page includes details on how to get it to work, and what's going on. ==What's wrong with site search autocompletion and how to fix it== ===What's wrong=== When you start typing something in the site search bar, you'll see it stuck at "Loading search suggestions" as shown in the screenshot below: File:Site search autocompletion broken.png Note that the actual search is still working -- y..."

Site search autocompletion is currently broken by default on this site. This page includes details on how to get it to work, and what's going on.

==What's wrong with site search autocompletion and how to fix it==

===What's wrong===

When you start typing something in the site search bar, you'll see it stuck at "Loading search suggestions" as shown in the screenshot below:

[[File:Site search autocompletion broken.png]]

Note that the actual search is still working -- you just have to hit Enter after typing the search query and it'll go to the search results page. It's the autocompletion before you hit Enter that is broken.

===How to fix it===

To fix it, you need to follow these steps:

* Write to vipulnaik1@gmail.com asking for a login to the site. Please include the following with your request: preferred username, preferred initial password (you can change it after logging in), real name (if you want it entered), email address to use (if you want an actual email address by which you can be contacted), and whether you want edit access as well. You don't need edit access for enabling site search autocompletion.
* Log in to the site, and go to [[Special:Preferences]]. Go to the Appearance section and switch the Skin from "Vector (2022)" to "Vector legacy (2010)".
* Make sure to hit "Save" at the bottom.
* Now you can reload the page or load a new page.

Site search autocompletion should now work. Here's an example:

[[File:Site search autocompletion working.png]]

==More background==

We've recently upgraded the MediaWiki version of this wiki from 1.35.13 to 1.41.2 (see [[Special:Version]]). The upgrade allows us to migrate the wiki to a more modern operating system version running PHP 8. With the current setup for MediaWiki 1.41.2, we're in this situation:

* The "Vector legacy (2010)" skin has site search autocompletion working, but it doesn't render well on small screens. Specifically, even on small mobile screens, it still shows the left menu, and doesn't properly use the MobileFrontend extension settings.
* The "Vector (2022)" skin doesn't have site search autocompletion working (see screenshots in preceding section) but it does render fine on mobile devices.

It is possible to set only one default skin (that is applicable to all non-logged-in users and is the default for logged-in users who have not configured a skin for themselves). So, the selection of default skin comes down to whether it's more important for casual users to have the mobile experience working or to have site search autocompletion working. Based on a general understanding of user behavior, we believe that having a usable mobile experience is more important for casual users than having site search autocompletion.

However, for power users who are using the site extensively, site search autocompletion may be important. That's why we've written this page giving guidance on how to set up site search autocompletion.

User:Vipul/Sandbox

2024-08-04T00:41:31Z

Vipul:

* <math>\sqrt{7 + 2}!! + 3 = 723</math>
* <math>2^{8 - 1} = 128</math>
* <math>\sqrt{7 - 4}!! + 4! = 744</math>
* <math>2^7 - 1 = 127</math>
* <math>9^{\sqrt{7 + 2}} = 729</math>

User:Vipul/Sandbox

2024-08-04T00:36:37Z

Vipul:

* <math>\sqrt{7 + 2}!! + 3 = 723</math>
* <math>2^{8 - 1} = 128</math>
* <math>\sqrt{7 - 4}!! + 4! = 744</math>
* <math>2^7 - 1 = 127</math>

User:Vipul/Sandbox

2024-08-04T00:24:09Z

Vipul:

* <math>\sqrt{7 + 2}!! + 3 = 723</math>
* <math>2^{8 - 1} = 128</math>
* <math>\sqrt{7 - 4}!! + 4! = 744</math>

User:Vipul/Sandbox

2024-07-21T21:04:12Z

Vipul:

* <math>\sqrt{7 + 2}!! + 3 = 723</math>
* <math>2^{8 - 1} = 128</math>

User:Vipul/Sandbox

2024-07-21T21:01:19Z

Vipul: Created page with "* <math>\sqrt{7 + 2}!! + 3 = 723</math>"

* <math>\sqrt{7 + 2}!! + 3 = 723</math>

MediaWiki:Sitenotice

2024-07-21T21:00:52Z

Vipul: Blanked the page

Template:Top notice

2009-03-27T11:27:08Z

Vipul:

{{quotation|Welcome to '''{{fullsitetitle}}'''. This is a pre-pre-alpha stage complex analysis wiki primarily managed by [[User:Vipul|Vipul Naik]], a Ph.D. student in mathematics at the University of Chicago. It is part of a broader subject wikis initiative -- see the [[Ref:Main Page|subject wikis reference guide]] for more details.}}

Main Page

2009-03-27T11:26:04Z

Vipul: Replaced content with '{{top notice}}'

MediaWiki:Sidebar

2009-01-16T15:50:13Z

Vipul:

Companal:General disclaimer

2009-01-16T15:40:53Z

Vipul: New page: This is a general disclaimer common to all subject wikis, with the source at Ref:Ref:General disclaimer. For specific hazards of using this particular subject wiki, refer [[{{SITENAME}...

This is a general disclaimer common to all subject wikis, with the source at [[Ref:Ref:General disclaimer]]. For specific hazards of using this particular subject wiki, refer [[{{SITENAME}}:Hazards]].

'''SUBJECT WIKIS MAKE NO GUARANTEE OF VALIDITY'''

Content on individual subject wikis need ''not'', in general, be correct or useful. There are the following general hazards:

* Specific content pages may have wrong or misleading information.
* Content pages may use terminology that is not standard or generally accepted, or differs from terminology or notation in other sources.
* The content is not designed specifically for a particular use, and any use that you put the content to is ''at your own risk''.

Also note that:

* Adding content to subject wikis and reusing content from subject wikis is subject to copyright laws. See [[{{SITENAME}}:Copyrights]] for more details.
* Your activities as a user or editor are tracked and may be used by site administrators and declared third parties. See [[{{SITENAME}}:Privacy policy]] for more details.

Companal:Privacy policy

2009-01-16T01:44:35Z

Vipul: New page: This privacy policy is common to subject wikis. For the original privacy policy, refer Ref:Ref:Privacy policy. ==Privacy for readers== If you are surfing this website, your actions a...

This privacy policy is common to subject wikis. For the original privacy policy, refer [[Ref:Ref:Privacy policy]].

==Privacy for readers==

If you are surfing this website, your actions are logged in our usage logs. These usage logs are accessible to:

* The site's administrators and technical support group. For a full list of administrators, contact [[User:Vipul|Vipul Naik]] by email: vipul@math.uchicago.edu or vipul.wikis@gmail.com. The technical support is [http://www.4am.co.in 4AM].
* The service that hosts the data and servers, which is currently Dreamhost (http://www.dreamhost.com).
* Google Analytics, which has been integrated to collect site statistics. View Google's privacy policy here: http://www.google.com/intl/en_ALL/privacypolicy.html

Your usage logs are not made available to other parties. Aggregated data from logs, such as general usage patterns, may be used by the MedaWiki software as well as by site administrators in decision making. For instance, MediaWiki keeps track of the number of times each page is viewed.

==Privacy for editors==

Editing on subject wikis is generally permitted only for registered users. Registered users must, at the time of registration, provide their real name, and enter basic information about their reason for interest. ''No'' private information such as date of birth, social security or taxation number, or home address is sought.

Regarding personal information:

* The email IDs of registered users are visible to site administrators only. For information about site administrators, contact vipul.wikis@gmail.com or vipul@math.uchicago.edu with the particular subject wiki and the reason for request.
* All editing activity by registered users is recorded on the site and is visible to all site users. However, this information is not indexed by search engines that follows robots.txt.
* For edits made by registered users when logged in, the originating IP addresses for the edits can be accessed only by the site administrators.
* Passwords chosen by registered users are not humanly accessible, even to site administrators.

MediaWiki:Sitenotice

2009-01-15T22:39:58Z

Vipul:

[[Image:Logo.jpg|thumb|75px|right|[http://www.4am.co.in Visit]]]
[[Main Page|{{fullsitetitle}} ({{sitestatus}})]]

'''ALSO CHECK OUT''': <random>[[Groupprops:Main Page|Groupprops]]: The Group Properties Wiki@@@[[Commalg:Main Page|Commalg]]: The Commutative Algebra Wiki@@@[[Diffgeom:Main Page|Diffgeom]]: The Differential Geometry Wiki@@@[[Topospaces:Main Page|Topospaces]]: The Topology Wiki</random>

Template:Sitestatus

2009-01-15T22:35:21Z

Vipul: New page: pre-pre-alpha

pre-pre-alpha

Companal:Copyrights

2009-01-15T22:21:07Z

Vipul: New page: This is a common copyright notice to all subject wikis. Original notice available at Ref:Ref:Copyrights. ==General license information== All content is put up under the [http://creat...

This is a common copyright notice to all subject wikis. Original notice available at [[Ref:Ref:Copyrights]].

==General license information==

All content is put up under the [http://creativecommons.org/licenses/by-sa/3.0/ Creative Commons Attribute Share-Alike 3.0 license (unported)] (shorthand: CC-by-SA). Read the [http://creativehttp://ref.subwiki.org/w/skins/common/images/button_media.pngcommons.org/licenses/by-sa/3.0/legalcode full legal code here].

===Exceptions===

The software engine that runs the wiki is MediaWiki, and this software is licensed under the GNU General Public License, which is distinct from and currently incompatible with the Creative Commons license. Similarly, some of the extensions to the software engine are licensed by their developers using the GPL, which is incompatible with Creative Commons licenses.

More information on the version of MediaWiki and extensions: [[Special:Version]].

In addition, some templates may have been copied or adapted from Wikipedia or other wiki-based websites that use licenses other than Creative Commons licenses. Wikipedia uses the GNU Free Documentation License (GFDL) that, as of now, is incompatible with CC licenses. In such cases, it is clearly indicated on the page that it is licensed differently.

===Short description of the license===

The CC-by-SA license has the following features:

# Content can be used and reused for any purposes, commercial or noncommercial.
# Any public use or reuse of the material requires attribution of the original source (for attribution formats, see later in the document).
# In case of adaptation or creation of derivative works, the newly created works must also be licensed under the same license (CC-by-SA) unless the creator of the derivative work takes ''explicit'' permission from the creator of the original work.

===Fair use and other rights===

The license cannot and does not limit fair use rights. Fair use may include quoting or copying select passages for the purposes of criticism and review.

The license also cannot limit the use of data or information contained in the content on subject wikis. This means that anybody can use or reuse the facts or knowledge they gain from the subject wiki in any way they want.

==Applicability of license to the subject wiki==

===Examples of personal use that do not necessitate use of the license terms===

* Saving, making copies, or taking printouts of pages on the subject wiki, as long as these copies are meant for personal use and are not to be distributed to others.
* Learning facts from the subject wikis and using those facts as part of instruction or research.
* Providing links to pages on the subject wikis.
* Quoting short passages from the subject wiki in a criticism or review (it is preferable, though not legally enforceable, that a link to the original content being reviewed be provided).

===Examples of personal use that necessitate use of the license terms===

* Creating a mirror website by exporting pages or content in bulk: In this case, the mirror website must release the mirrored content under the same license terms and ''also'' attribute the original source with a link. Attribution formats are discussed below.
* Giving handouts of pages or collections of pages to a group of people, such as students in a course: In this case, the handouts should be accompanied by an attribution that makes clear both the source and the original license terms, as well as a link to the Uniform Resource Identifier (URI) of the subject wiki.

Attribution formats:

* For websites: ''This content is from [{{fullurl:Main Page}} {{SITENAME}}]'' or ''This content is from [{{fullurl:Main Page}} {{fullsitetitle}}]''. Preferably, also a link to the specific pages (if any) used. The language of the attribution may be modified somewhat to indicate whether the content has been used as such, or whether changes have been made.
* For print materials: ''This content is from {{SITENAME}}. The site URL is http://{{SITENAME}}.subwiki.org''

Attribution of individuals who have contributed content to the subject wiki entries is ''not'' necessary. (However, this list of individuals can be tracked down, as discussed in the next section).

==Uses not within the ambit of the license or fair use rights==

Those who have contributed all the content to a particular subject wiki entry can republish the same content anywhere else ''without restriction'' and can also release the content elsewhere under a different license, since they own the full copyright to their content.

To use the content on the wiki in a manner that is outside the terms of the license or beyond fair use rights, the individuals who have contributed that particular content need to be contacted. For any particular page, this information can be accessed using the history tab of the page. The users who have made edits to the page are the ones who need to be contacted.

In case of difficulty doing this, please contact vipul.wikis@gmail.com or vipul@math.uchicago.edu for more information.

==Copyright violations==

All users using subject wikis are requested not to violate copyright of other authors or publishers. This includes providing links that may aid and abet copyright violation. Examples of things that may be considered copyright violation:

* Putting up links to personal copies of restricted-access journal articles acquired through a personal or library subscription: This usually goes against the Terms of Service of the subscription. The exception is when the content of the articles is out of copyright.
* Giving links to personal copies, or copies on filesharing systems, of books that are under copyright and where either owning or sharing a copy in that manner is against the terms of copyright.

In case copyright violations are detected, please email vipul.wikis@gmail.com and vipul@math.uchicago.edu to have the matter looked into immediately.

Template:Fullsitetitle

2009-01-15T22:20:05Z

Vipul: New page: Companal, The Complex Analysis Wiki

Companal, The Complex Analysis Wiki

Template:Top notice

2009-01-15T22:01:33Z

Vipul:

{{quotation|Welcome to '''Companal (The Complex Analysis Wiki)'''. This is a pre-pre-alpha stage complex analysis wiki primarily managed by [[User:Vipul|Vipul Naik]], a Ph.D. student in mathematics at the University of Chicago. It is part of a broader subject wikis initiative -- see the [[Ref:Main Page|subject wikis reference guide]] for more details.}}

Template:Top notice

2009-01-15T22:01:24Z

Vipul: New page: {{quotation|Welcome to '''Companal (The Complex Analysis Wiki)'''. This is a pre-pre-alpha stage complex analysis wiki primarily managed by Vipul Naik, a Ph.d. student in ma...

{{quotation|Welcome to '''Companal (The Complex Analysis Wiki)'''. This is a pre-pre-alpha stage complex analysis wiki primarily managed by [[User:Vipul|Vipul Naik]], a Ph.d. student in mathematics at the University of Chicago. It is part of a broader subject wikis initiative -- see the [[Ref:Main Page|subject wikis reference guide]] for more details.}}

Riemann removable singularities theorem

2008-09-12T21:20:25Z

Vipul: Redirecting to Removable singularities theorem

#redirect [[removable singularities theorem]]

Removable singularities theorem

2008-09-12T21:19:54Z

Vipul: New page: {{basic fact}} ==Statement== The '''removable singularities theorem''', sometimes termed the '''Riemann removable singularities theorem''', states that if <math>U</math> is a neighborhoo...

{{basic fact}}

==Statement==

The '''removable singularities theorem''', sometimes termed the '''Riemann removable singularities theorem''', states that if <math>U</math> is a neighborhood of a point <math>z_0</math>, and <math>f:U \setminus \{ z_0 \} \to \mathbb{C}</math> is a [[holomorphic function]] with <math>\lim_{z \to z_0} f(z)</math> existing, then <math>f</math> extends to a holomorphic function on <math>U</math>, by setting <math>f(z_0) = \lim_{z \to z_0} f(z)</math>.

Riemann sphere

2008-09-12T21:14:37Z

Vipul:

{{particular Riemann surface}}

==Definition==

===As a Riemann surface===

The Riemann sphere is the topological space <math>S^2 \subset \R^3</math> (the unit sphere in 3-space) with the following atlas of charts:

* The stereographic projection that maps the complement of the north pole to <math>\mathbb{C}</math>
* The stereographic projection that maps the complement of the south pole to <math>\mathbb{C}</math>, composed with a reflection about the <math>x</math>-axis in <math>\mathbb{C}</math>

The transition map between these two charts is given by:

<math>z \mapsto \frac{1}{z}</math>

===Other descriptions===

The Riemann sphere is viewed in many of the following ways:

* It is the one-point compactification of <math>\mathbb{C}</math>, the one additional point being a ''point at infinity'', denoted <math>\infty</math>. With respect to the stereographic projection, the point at infinity is identified with the north pole.
* It is the [[complex projective line]]: the set of complex lines through the origin in <math>\mathbb{C}^2</math>.

==Automorphism group==

===Description of the automorphism group===

{{further|[[fractional linear transformation]],[[conformal automorphism of Riemann sphere equals fractional linear transformation]]}}

The conformal automorphisms of the Riemann sphere are the [[fractional linear transformation]]s. Viewing the Riemann sphere as <math>\mathbb{C}</math> with a point at infinity, the fractional linear transformations are maps that can be described as:

<math>z \mapsto \frac{az + b}{cz + d}</math>

where <math>ad - bc</math> is nonzero. The fractional linear transformation is often represented by the matrix:

<math>\begin{pmatrix}a & b \\ c & d\end{pmatrix}</math>

and the composition of these transformations corresponds to multiplication of matrices. The fractional linear transformations also have an interpretation in terms of the Riemann sphere as the complex projective line.

The group of fractional linear transformations is the projective general linear group <math>PGL(2,\mathbb{C})</math>.

The proof that this ''is'' the full automorphism group relies on two facts: it acts transitively, and it contains the isotropy subgroup at <math>\infty</math>. {{proofat|[[Conformal automorphism of Riemann sphere equals fractional linear transformation]]}}

===Transitivity of the automorphism group===

{{further|[[Conformal automorphism group is 3-regular on Riemann sphere]]}}

The conformal automorphisms of the Riemann sphere act transitively on it. In fact, more is true. Given any three distinct points <math>z_1, z_2, z_3</math>, and any three distinct points <math>w_1, w_2, w_3</math>, there is a ''unique'' fractional linear transformation mapping <math>z_1</math> to <math>w_1</math>, <math>z_2</math> to <math>w_2</math>, and <math>z_3</math> to <math>w_3</math>. The uniqueness follows from the fact that any fractional linear transformation with three fixed points gives a quadratic equation with three zeros, while the existence follows from a direct argument.

==Isomorphism==

{{further|[[Genus zero Riemann surface is conformally equivalent to Riemann sphere]]}}

The Riemann sphere is the ''only'' Riemann surface, up to conformal equivalence, that is topologically a sphere. In other words, any topological sphere is conformally equivalent to the Riemann sphere. Equivalently, any compact simply connected Riemann surface, or any genus zero Riemann surface, is conformally equivalent to the Riemann sphere.

Homogeneous Riemann surface

2008-09-12T21:00:01Z

Vipul:

{{Riemann surface property}}

==Definition==

A [[Riemann surface]] is termed '''homogeneous''', '''transitive''', '''conformally homogeneous''', or '''conformally transitive''' if it satisfies the following equivalent conditions:

* The [[conformal automorphism group]] acts transitively on the points of the Riemann surface
* Given any two points of the Riemann surface, there is a bijective biholomorphic mapping from the surface to itself that sends the first point to the second.

==Facts==

* A [[compact Riemann surface]] is homogeneous if and only if it has genus zero or one. This follows from the fact that compact Riemann surfaces of higher genus have finite automorphism groups (alternatively, it follows from the fact that higher genus Riemann surfaces have certain special points called [[Weierstrass point]]s). {{further|[[Compact and homogeneous iff genus zero or one]]}}
* Any simply connected Riemann surface is homogeneous. This follows from the classification of simply connected Riemann surfaces by the uniformization theorem: the only simply connected Riemann surfaces are the [[open unit disk]], the [[Riemann sphere]], and the [[complex plane]].

Every compact Riemann surface admits a nonconstant meromorphic function

2008-09-12T20:56:55Z

Vipul: Redirecting to Every compact Riemann surface is a branched cover of the Riemann sphere

#redirect [[Every compact Riemann surface is a branched cover of the Riemann sphere]]

Every compact Riemann surface is a branched cover of the Riemann sphere

2008-09-12T20:56:01Z

Vipul: New page: ==Statement== Let <math>S</math> be a fact about::compact Riemann surface of degree <math>g</math>. Then, <math>S</math> admits a nonconstant meromorphic function of degree at most <m...

==Statement==

Let <math>S</math> be a [[fact about::compact Riemann surface]] of degree <math>g</math>. Then, <math>S</math> admits a nonconstant meromorphic function of degree at most <math>g + 1</math>. In particular, <math>S</math> can be expressed as a branched cover of the Riemann sphere with at most <math>g + 1</math> sheets.

Riemann-Roch theorem

2008-09-12T20:53:56Z

Vipul: New page: ==Statement== Let <math>S</math> be a fact about::compact Riemann surface. Let <math>K</math> be a fact about::canonical divisor for <math>S</math>, and <math>D</math> be any [[fa...

==Statement==

Let <math>S</math> be a [[fact about::compact Riemann surface]]. Let <math>K</math> be a [[fact about::canonical divisor]] for <math>S</math>, and <math>D</math> be any [[fact about::divisor]]. Let <math>L(D)</math> denotes the [[fact about::L-space of a divisor|L-space]] of <math>D</math>, i.e., the vector space of [[fact about::meromorphic function]]s on <math>S</math> that are either zero or have divisor greater than or equal to <math>D</math>. Then:

<math>\operatorname{dim} L(-D) = \operatorname{deg}(D) - g + 1 + \operatorname{dim} L(D - K)</math>.

==Related facts==

===Applications===

* [[Riemann's inequality]]
* [[Genus zero Riemann surface is conformally equivalent to Riemann sphere]]
* [[Every compact Riemann surface is a branched cover of the Riemann sphere]]

Uniformization theorem

2008-09-12T20:44:17Z

Vipul: New page: ==Statement== There are two components to the statement: * Any simply connected fact about::Riemann surface is conformally equivalent to one of these three: the [[fact about::open uni...

==Statement==

There are two components to the statement:
* Any simply connected [[fact about::Riemann surface]] is conformally equivalent to one of these three: the [[fact about::open unit disk]], the [[fact about::complex plane]], the [[fact about::Riemann sphere]].
* Any Riemann surface can be expressed as the quotient of its universal cover (which is one of these) by the action of the fundamental group (the quotient is in the sense of Riemann surfaces).

==Related facts==

===Particular cases/applications===

* [[Genus zero Riemann surface is conformally equivalent to Riemann sphere]]: States that any compact, simply connected Riemann surface is conformally equivalent to the Riemann sphere.
* [[Riemann mapping theorem]]

Classification of simply connected Riemann surfaces

2008-09-12T20:40:58Z

Vipul: Redirecting to Uniformization theorem

#redirect [[Uniformization theorem]]

Simply connected compact Riemann surface is conformally equivalent to Riemann sphere

2008-09-12T20:38:26Z

Vipul: Redirecting to Genus zero Riemann surface is conformally equivalent to Riemann sphere

#redirect [[Genus zero Riemann surface is conformally equivalent to Riemann sphere]]

Simply connected domain that is not the whole plane is conformally equivalent to the open unit disk

2008-09-12T20:36:29Z

Vipul: Redirecting to Riemann mapping theorem

#redirect [[Riemann mapping theorem]]

Genus zero Riemann surface is conformally equivalent to Riemann sphere

2008-09-12T20:35:19Z

Vipul: New page: ==Statement== Any fact about::compact Riemann surface of genus zero (in other words, any compact Riemann surface that is topologically a sphere) is conformally equivalent to the [fact...

==Statement==

Any [[fact about::compact Riemann surface]] of genus zero (in other words, any compact Riemann surface that is topologically a sphere) is conformally equivalent to the [fact about::Riemann sphere]].

Equivalently, any simply connected compact Riemann surface is conformally equivalent to the Riemann sphere.

==Related facts==

* [[Riemann mapping theorem]]: This result has a similar feel to it: it says that any simply connected domain in <math>\mathbb{C}</math> that is not the whole of <math>\mathbb{C}</math> is conformally equivalent to the [[open unit disk]].
* [[Every compact Riemann surface is a branched cover of the Riemann sphere]]

==Facts used==

# [[uses::Riemann-Roch theorem]]
# [[uses::Holomorphic on compact Riemann surface implies constant]]

==Proof==

'''Given''': A compact Riemann surface <math>S</math> of genus zero.

'''To prove''': <math>S</math> is conformally equivalent to the Riemann sphere.

'''Proof''': Pick a point <math>p_0 \in S</math>, and consider the divisor <math>D = p_0</math>. By the Riemann-Roch theorem, we have:

<math>\operatorname{dim} L(-D) = \operatorname{deg}(D) - g + 1 + \operatorname{dim} L(D - K)</math>.

Ignoring <math>\operatorname{dim} L(D-K)</math>, we get:

<math>\operatorname{dim} L(-D) \ge 1 - 0 + 1 = 2</math>.

Thus, the space of meromorphic functions on <math>S</math> with a simple pole at at most one point (namely, <math>p_0 \in S</math>) and no higher order poles, is two-dimensional. Note that the space of ''holomorphic'' functions on <math>S</math> includes only the constant functions, and is thus one-dimensional. Hence, there exists a nonconstant meromorphic function on <math>S</math>, having a simple pole at <math>p_0</math> and no other poles.

In particular, this function gives an analytic mapping from <math>S</math> to the [[Riemann sphere]]. <math>\infty</math> has only one inverse image under this mapping, so the mapping is a degree one analytic mapping, and hence a conformal isomorphism. So, <math>S</math> is conformally equivalent to the Riemann sphere.

Conformal automorphism of complex numbers implies affine map

2008-09-12T19:48:07Z

Vipul: /* Proof using Casorati-Weierstrass */

{{conformal automorphism group computation}}
==Statement==

Suppose <math>f: \mathbb{C} \to \mathbb{C}</math> is a conformal automorphism: in other words, <math>f</math> is an [[entire function]] from <math>\mathbb{C}</math> to <math>\mathbb{C}</math> with entire inverse. Then, <math>f</math> is of the form:

<math>z \mapsto az + b</math>

==Facts used==

* [[Liouville's theorem]]: Any bounded and entire function must be constant
* [[Casorati-Weierstrass theorem]]: This states that near an [[essential singularity]], the image of a holomorphic function is dense in <math>\mathbb{C}</math>
* [[Fundamental theorem of algebra]]
==Proof==

===Proof using Casorati-Weierstrass===

Suppose <math>f: \mathbb{C} \to \mathbb{C}</math> is a conformal map. By composing <math>f</math> with a translation, assume that <math>f(0) = 0</math>. Then, consider the map <math>g: \mathbb{C}^* \to \mathbb{C}</math> given by:

<math>g(z) := f(1/z)</math>

Now consider the singularity of <math>z</math> at <math>0</math>. The following possibilities arise.

* The singularity is a [[removable singularity]]: In that case, it takes a finite complex value, and thus <math>g</math> is bounded in a small neighborhood of <math>0</math>. Thus, <math>f</math> is bounded outside a bounded subset, and since it is continuous, <math>f</math> is globally bounded. [[Liouville's theorem]] (fact (1)) thus forces <math>f</math> to be constant
* The singularity is an [[essential singularity]]: In that case, the Casorati-Weierstrass theorem tells us that the image under <math>g</math> of a small neighborhood of <math>0</math>, is dense in <math>\mathbb{C}</math>. This clearly shows that <math>f</math> isn't a homeomorphism (since it maps a non-dense set to a dense set) so it cannot be a conformal automorphism
* The singularity is a [[pole]]: Thus, we obtain:

<math>g(z) = z^{-n}h(z)</math>

where <math>h</math> is a holomorphic. Plugging back <math>g</math> in terms of <math>f</math> we obtain:

<math>f(z) = z^n h(1/z)</math>

where <math>h</math> is a holomorphic function. But we also have a global power series for <math>f</math>, which must match up with the above, so we see that <math>f</math> is a polynomial of degree at most <math>n</math>. Thus, it remains to check which polynomial maps from <math>\mathbb{C}</math> to <math>\mathbb{C}</math> are conformal.

Suppose <math>f: \mathbb{C} \to \mathbb{C}</math> is conformal and a [[polynomial map]]. Then, <math>f'</math> should vanish nowhere. But <math>f^'</math> is also a polynomial map, and by the [[fundamental theorem of algebra]], <math>f'</math> must be constant. Thus, <math>f</math> must be a linear polynomial.

Conformal automorphism of complex numbers implies affine map

2008-09-12T19:47:31Z

Vipul:

{{conformal automorphism group computation}}
==Statement==

Suppose <math>f: \mathbb{C} \to \mathbb{C}</math> is a conformal automorphism: in other words, <math>f</math> is an [[entire function]] from <math>\mathbb{C}</math> to <math>\mathbb{C}</math> with entire inverse. Then, <math>f</math> is of the form:

<math>z \mapsto az + b</math>

==Facts used==

* [[Liouville's theorem]]: Any bounded and entire function must be constant
* [[Casorati-Weierstrass theorem]]: This states that near an [[essential singularity]], the image of a holomorphic function is dense in <math>\mathbb{C}</math>
* [[Fundamental theorem of algebra]]
==Proof==

===Proof using Casorati-Weierstrass===

Suppose <math>f: \mathbb{C} \to \mathbb{C}</math> is a conformal map. By composing <math>f</math> with a translation, assume that <math>f(0) = 0</math>. Then, consider the map <math>g: \mathbb{C}^* \to \mathbb{C}</math> given by:

<math>g(z) := f(1/z)</math>

Now consider the singularity of <math>z</math> at <math>0</math>. The following possibilities arise.

* The singularity is a [[removable singularity]]: In that case, it takes a finite complex value, and thus <math>g</math> is bounded in a small neighborhood of <math>0</math>. Thus, <math>f</math> is bounded outside a bounded subset, and since it is continuous, <math>f</math> is globally bounded. [[Liouville's theorem]] (fact (1)) thus forces <math>f</math> to be constant
* The singularity is an [[essential singularity]]: In that case, the Casorati-Weierstrass theorem tells us that the image under <math>g</math> of a small neighborhood of <math>0</math>, is dense in <math>\mathbb{C}</math>. This clearly shows that <math>f</math> isn't a homeomorphism (since it maps a non-dense set to a dense set) so it cannot be a conformal automorphism
* The singularity is a [[pole]]: Thus, we obtain:

<math>g(z) = z^{-n}h(z)</math>

where <math>h</math> is a holomorphic. Plugging back <math>g</math> in terms of <math>f</math> we obtain:

<math>f(z) = z^n h(1/z)</math>

where <math>h</math> is a holomorphic function. But we also have a global power series for <math>f</math>, which must match up with the above, so we see that <math>f</math> is a polynomial of degree at most <math>n</math>. Thus, it remains to check which polynomial maps from <math>\mathbb{C}</math> to <math>\mathbb{C}</math> are conformal.

Suppose <math>f: \mathbb{C} \to \mathbb{C}</math> is conformal and a [[polynomial map]]. Then, <math>f^'</math> should vanish nowhere. But <math>f^'</math> is also a polynomial map, and by the [[fundamental theorem of algebra]], <math>f^'</math> must be constant. Thus, <math>f</math> must be a linear polynomial.

Conformal automorphism of Riemann sphere equals fractional linear transformation

2008-09-12T19:38:52Z

Vipul:

{{conformal automorphism group computation}}

==Statement==

Any [[fact about::conformal automorphism]] of the [[fact about::Riemann sphere]] is a [[fact about::fractional linear transformation]], i.e. a map of the form:

<math>z \mapsto \frac{az + b}{cz + d}</math>

Equivalently, the conformal automorphism group of the Riemann sphere is precisely the group of fractional linear transformations: namely, <math>PSL(2,\mathbb{C})</math>.

==Related facts==

* [[Conformal automorphism of disk implies fractional linear transformation]]: The technique used in this proof is the same: first, we show that the fractional linear transformations act transitively; next, we show that they contain the isotropy at <math>0</math>>
* [[Conformal automorphism of complex numbers implies affine map]]

===Related techniques===

* [[Groupprops:Proving product of subgroups]]: A survey article on the group theory wiki about proving that a given group is a product of two subgroups.

==Facts used==

# [[uses::conformal automorphism of complex numbers implies affine map]]
==Proof==

Suppose <math>G</math> denotes the conformal automorphism group of the Riemann sphere, and <math>H</math> the subgroup comprising fractional linear transformations. We want to show that <math>H = G</math>. We show this in two steps:

* <math>H</math> acts transitively on the Riemann sphere.
* <math>H</math> contains the isotropy subgroup at <math>\infty</math>.

With both these facts, observe that for any <math>g \in G</math>, we can find <math>h \in H</math> such that <math>h^{-1}g</math> is in the isotropy subgroup of <math>\infty</math>, which in turn is in <math>H</math>. This forces <math>G = H</math>.

===Transitive action of fractional linear transformations===

We first observe that the group of fractional linear transformations acts transitively on the Riemann sphere. In particular, any point <math>z_0 \in \mathbb{C}</math> can be mapped to <math>\infty</math> can be sent to <math>\infty</math> using the map:

<math>z \mapsto \frac{1}{z - z_0}</math>

===Isotropy at infinity is contained in fractional linear transformations===

The isotropy group at <math>\infty</math> must comprise maps of the form <math>z \mapsto az + b</math>, combining fact (1) with the observation that any conformal automorphism of the Riemann sphere that fixes <math>\infty</math> must be a conformal automorphism of <math>\mathbb{C}</math>.

Conformal automorphism of disk implies fractional linear transformation

2008-09-12T19:29:43Z

Vipul:

{{conformal automorphism group computation}}
==Statement==

Suppose <math>D</math> is the [[fact about::open unit disk]] in <math>\mathbb{C}</math>. Then, any [[fact about::conformal automorphism]] (a bijective [[biholomorphic mapping]] to itself) of <math>D</math> comes as the restriction to <math>D</math> of a [[fact about::fractional linear transformation]].

==Facts used==

# [[uses::Schwarz lemma]]: This states that if <math>f: D \to D</math> is a holomorphic function, then <math>|f'(0)| \le 1</math>, and equality holds if and only if <math>f</math> equals a rotation.

==Proof==

===Proof outline===

Let <math>G = \operatorname{Aut}(D)</math> and <math>H</math> be the subgroup of <math>G</math> comprising those automorphisms that arise by restricting fractional linear transformations to <math>D</math>. We need to show that <math>H = G</math>. We prove this by showing two things:

* <math>H</math> acts transitively on <math>D</math>
* <math>H</math> contains the isotropy subgroup of <math>0</math> in <math>G</math>

Combining these two facts, we see that <math>H = G</math>

===Proof of transitivity===

We'll show that given any element <math>w \in D</math>, there exists an element <math>\sigma \in H</math> such that <math>\sigma(0) = w</math>. This'll show transitivity.

Define:

<math>\sigma := z \mapsto \frac{z - w}{\overline{w}z - 1}</math>

This is a fractional linear transformation. First, we see that this transformation takes the unit circle to itself. Indeed, if <math>|z| = 1</math>, then:

<math>|z - w| = |\overline{z}(z - w)| = |1 - \overline{z}w| = | 1 - \overline{w}z|</math>

Hence the ratio has modulus 1.

Second, observe that since it takes 0 to a point inside the disc, it must by definition send the disc to itself (i.e. it cannot send the disc to the exterior of the circle).

Finally, this fractional linear transformation sends <math>0</math> to <math>w</math>, as we see by putting <math>z = 0</math> in the formula.

===Proof of containing the isotropy===

Suppose <math>f:D \to D</math> is a biholomorphic mapping such that <math>f(0) = 0</math>. Then, by the [[Schwarz lemma]], we have <math>|f'(0)| le 1</math>. Applying the Schwarz lemma to <math>f^{-1}</math>, we see that <math>|f'(0)| \ge 1</math>, hence <math>|f'(0)| = 1</math>, so by the second part of Schwarz lemma, <math>f</math> is a rotation i.e. multiplication by a complex number of unit modulus. Rotations are fractional linear transformations, so we're done.

Maximum modulus principle

2008-09-12T19:19:58Z

Vipul:

{{basic fact}}
{{application of|open mapping theorem}}
==Statement==

Suppose <math>U \subset \mathbb{C}</math> is a [[domain]] (open connected subset). Let <math>f:U \to \mathbb{C}</math> be a [[holomorphic function]]. The '''maximum modulus principle''' (sometimes called the '''maximum principle''') states that if there exists a <math>z_0 \in U</math>, such that for all <math>z \in U</math>, we have:

<math>|f(z)| \le |f(z_0)|</math>

Then, <math>f</math> is a constant function.

==Facts used==

# [[Open mapping theorem]]

==Proof==

'''Given''': <math>f</math> is a nonconstant holomorphic function on a nonempty domain <math>U</math>.

'''To prove''': There does not exist any <math>z_0 \in U</math> with the property that <math>|f(z)| \le |f(z_0)|</math> for every <math>z \in U</math>>

'''Proof''': First, by the [[open mapping theorem]] (fact (1)), <math>f</math> is an open map.

Second, the [[modulus map]] <math>| \cdot |: \mathbb{C} \to [0,\infty)</math> is an open map. Thus, the composite map <math>|f|: U \to [0,\infty)</math> given by <math>z \mapsto |f(z)|</math> is also an open map. Thus, under this map, the image of <math>U</math> must be an open connected subset of <math>[0,\infty)</math>, so it must be of the form <math>[0,a)</math> where <math>a \in \R</math> or <math>a = \infty</math>. Hence, there cannot be a maximum within <math>U</math>.

Holomorphic function on open disk admits globally convergent power series

2008-09-12T19:13:18Z

Vipul:

==Statement==

Suppose <math>U \subseteq \mathbb{C}</math> is an [[open disk]] with center <math>z_0</math> and radius <math>R > 0</math>. Suppose <math>f: U \to \mathbb{C}</math> is a [[holomorphic function]]. Then, there exists a power series:

<math>\sum_{n=0}^\infty a_n(z - z_0)^n</math>

with radius of convergence ''at least'' <math>R</math>, and such that for all <math>z \in U</math>:

<math>f(z) = \sum_{n=0}^\infty a_n(z - z_0)^n</math>.

==Related facts==

* [[Cauchy integral formula]]
* [[Cauchy integral formula for derivatives]]: The formula obtained for the coefficients here is the same as the formula suggested by the Cauchy integral formula for derivatives, although that formula does not directly show that we actually get a holomorphic function.

===Applications===

* [[Holomorphic implies complex-analytic]]

==Facts used==

# [[uses::Cauchy integral formula]]
# A corollary of dominated convergence theorem, allowing for the exchange of an integral and an infinite summation.
==Proof==

'''Given''': <math>U \subseteq \mathbb{C}</math> is an [[open disk]] with center <math>z_0</math> and radius <math>R > 0</math>. Suppose <math>f: U \to \mathbb{C}</math> is a [[holomorphic function]].

'''To prove''': There exists a power series:

<math>\sum_{n=0}^\infty a_n(z - z_0)^n</math>

such radius of convergence ''at least'' <math>R</math>, and such that for all <math>z \in U</math>:

<math>f(z) = \sum_{n=0}^\infty a_n(z - z_0)^n</math>.

'''Proof''': We first show that for any <math>r < R</math>, there exists a power series about <math>z_0</math> convergent at all points within distance <math>r</math> from <math>z_0</math>. Then, we argue that this implies the existence of a ''single'' power series.

By applying the Cauchy integral formula to the disk <math>D</math> of radius <math>r</math>, centered at <math>z_0</math>, we see that for any point <math>z</math> with <math>|z - z_0| < r</math>, we have:

<math>f(z) = \oint_{\partial D} \frac{f(\xi)}{\xi - z} \, d\xi</math>.

To examine the dependence of this integral on <math>z</math>, we use the fact that:

<math>(\xi - z)^{-1} = \xi^{-1} (1 - \frac{z}{\xi})^{-1} = \frac{1}{\xi - z_0} (1 + \frac{z - z_0}{\xi - z_0} + \frac{(z - z_0)^2}{(\xi - z_0)^2} + \dots)</math>

Plugging this in, we see that <math>f(z)</math> can be expressed as the integral of a power series in <math>z</math>, with coefficients depending on <math>\xi</math>:

<math>f(z) = \oint_{\partial D} \left(\sum_{n=0}^\infty \frac{(z-z_0)^nf(\xi)}{(\xi - z_0)^{n+1}} \right) \, d\xi</math>

We now use fact (2) to exchange the summation and the integral, thus obtaining:

<math>f(z) = \sum_{n = 0}^\infty (z - z_0)^n \oint_{\partial D} \frac{f(\xi)}{(\xi - z_0)^{n+1}} \, d\xi</math>.

Thus, we get the required power series for <math>f</math>, with coefficients:

<math>a_n = \frac{f(\xi)}{(\xi - z_0)^{n+1}}</math>

Holomorphic function on open disk admits globally convergent power series

2008-09-12T19:12:34Z

Vipul:

==Statement==

Suppose <math>U \subseteq \mathbb{C}</math> is an [[open disk]] with center <math>z_0</math> and radius <math>R > 0</math>. Suppose <math>f: U \to \mathbb{C}</math> is a [[holomorphic function]]. Then, there exists a power series:

<math>\sum_{n=0}^\infty a_n(z - z_0)^n</math>

such radius of convergence ''at least'' <math>R</math>, and such that for all <math>z \in U</math>:

<math>f(z) = \sum_{n=0}^\infty a_n(z - z_0)^n</math>.

==Related facts==

* [[Cauchy integral formula]]
* [[Cauchy integral formula for derivatives]]: The formula obtained for the coefficients here is the same as the formula suggested by the Cauchy integral formula for derivatives, although that formula does not directly show that we actually get a holomorphic function.

===Applications===

* [[Holomorphic implies complex-analytic]]

==Facts used==

# [[uses::Cauchy integral formula]]
# A corollary of dominated convergence theorem, allowing for the exchange of an integral and an infinite summation.
==Proof==

'''Given''': <math>U \subseteq \mathbb{C}</math> is an [[open disk]] with center <math>z_0</math> and radius <math>R > 0</math>. Suppose <math>f: U \to \mathbb{C}</math> is a [[holomorphic function]].

'''To prove''': There exists a power series:

<math>\sum_{n=0}^\infty a_n(z - z_0)^n</math>

such radius of convergence ''at least'' <math>R</math>, and such that for all <math>z \in U</math>:

<math>f(z) = \sum_{n=0}^\infty a_n(z - z_0)^n</math>.

'''Proof''': We first show that for any <math>r < R</math>, there exists a power series about <math>z_0</math> convergent at all points within distance <math>r</math> from <math>z_0</math>. Then, we argue that this implies the existence of a ''single'' power series.

By applying the Cauchy integral formula to the disk <math>D</math> of radius <math>r</math>, centered at <math>z_0</math>, we see that for any point <math>z</math> with <math>|z - z_0| < r</math>, we have:

<math>f(z) = \oint_{\partial D} \frac{f(\xi)}{\xi - z} \, d\xi</math>.

To examine the dependence of this integral on <math>z</math>, we use the fact that:

<math>(\xi - z)^{-1} = \xi^{-1} (1 - \frac{z}{\xi})^{-1} = \frac{1}{\xi - z_0} (1 + \frac{z - z_0}{\xi - z_0} + \frac{(z - z_0)^2}{(\xi - z_0)^2} + \dots)</math>

Plugging this in, we see that <math>f(z)</math> can be expressed as the integral of a power series in <math>z</math>, with coefficients depending on <math>\xi</math>:

<math>f(z) = \oint_{\partial D} \left(\sum_{n=0}^\infty \frac{(z-z_0)^nf(\xi)}{(\xi - z_0)^{n+1}} \right) \, d\xi</math>

We now use fact (2) to exchange the summation and the integral, thus obtaining:

<math>f(z) = \sum_{n = 0}^\infty (z - z_0)^n \oint_{\partial D} \frac{f(\xi)}{(\xi - z_0)^{n+1}} \, d\xi</math>.

Thus, we get the required power series for <math>f</math>, with coefficients:

<math>a_n = \frac{f(\xi)}{(\xi - z_0)^{n+1}}</math>

Holomorphic function on open disk admits globally convergent power series

2008-09-12T19:12:15Z

Vipul:

==Statement==

Suppose <math>U \subseteq \mathbb{C}</math> is an [[open disk]] with center <math>z_0</math> and radius <math>R > 0</math>. Suppose <math>f: U \to \mathbb{C}</math> is a [[holomorphic function]]. Then, there exists a power series:

<math>\sum_{n=0}^\infty a_n(z - z_0)^n</math>

such radius of convergence ''at least'' <math>R</math>, and such that for all <math>z \in U</math>:

<math>f(z) = \sum_{n=0}^\infty a_n(z - z_0)^n</math>.

==Related facts==

* [[Cauchy integral formula]]
* [{Cauchy integral formula for derivatives]]: The formula obtained for the coefficients here is the same as the formula suggested by the Cauchy integral formula for derivatives, although that formula does not directly show that we actually get a holomorphic function.

===Applications===

* [[Holomorphic implies complex-analytic]]

==Facts used==

# [[uses::Cauchy integral formula]]
# A corollary of dominated convergence theorem, allowing for the exchange of an integral and an infinite summation.
==Proof==

'''Given''': <math>U \subseteq \mathbb{C}</math> is an [[open disk]] with center <math>z_0</math> and radius <math>R > 0</math>. Suppose <math>f: U \to \mathbb{C}</math> is a [[holomorphic function]].

'''To prove''': There exists a power series:

<math>\sum_{n=0}^\infty a_n(z - z_0)^n</math>

such radius of convergence ''at least'' <math>R</math>, and such that for all <math>z \in U</math>:

<math>f(z) = \sum_{n=0}^\infty a_n(z - z_0)^n</math>.

'''Proof''': We first show that for any <math>r < R</math>, there exists a power series about <math>z_0</math> convergent at all points within distance <math>r</math> from <math>z_0</math>. Then, we argue that this implies the existence of a ''single'' power series.

By applying the Cauchy integral formula to the disk <math>D</math> of radius <math>r</math>, centered at <math>z_0</math>, we see that for any point <math>z</math> with <math>|z - z_0| < r</math>, we have:

<math>f(z) = \oint_{\partial D} \frac{f(\xi)}{\xi - z} \, d\xi</math>.

To examine the dependence of this integral on <math>z</math>, we use the fact that:

<math>(\xi - z)^{-1} = \xi^{-1} (1 - \frac{z}{\xi})^{-1} = \frac{1}{\xi - z_0} (1 + \frac{z - z_0}{\xi - z_0} + \frac{(z - z_0)^2}{(\xi - z_0)^2} + \dots)</math>

Plugging this in, we see that <math>f(z)</math> can be expressed as the integral of a power series in <math>z</math>, with coefficients depending on <math>\xi</math>:

<math>f(z) = \oint_{\partial D} \left(\sum_{n=0}^\infty \frac{(z-z_0)^nf(\xi)}{(\xi - z_0)^{n+1}} \right) \, d\xi</math>

We now use fact (2) to exchange the summation and the integral, thus obtaining:

<math>f(z) = \sum_{n = 0}^\infty (z - z_0)^n \oint_{\partial D} \frac{f(\xi)}{(\xi - z_0)^{n+1}} \, d\xi</math>.

Thus, we get the required power series for <math>f</math>, with coefficients:

<math>a_n = \frac{f(\xi)}{(\xi - z_0)^{n+1}}</math>