text:
### Problem/Motivation

When using the big_pipe module with JavaScript, under certain circumstances, placeholders generated from #lazy_builder callbacks to history_attach_timestamp in comments cause Drupal.behaviors for the entire document to be detached. This issue became apparent after custom theming the comments field to put the comment form above the list of comments rather than below it. The user-facing issue was that, shortly after being attached to the comment_body field, the CKEditor was subsequently detached, leaving a plain textarea field, and producing the following warning in the browser's developer console:

[CKEDITOR] Error code: editor-incorrect-destroy.

The specific reason for this issue is as follows. After making a theme change to move the comment form above the list of existing comments, the order of content with placeholders changed in the document, putting the placeholder for the comment form above the history_attach_timestamp. Big Pipe uses the order of the placeholders in the markup to set the order in which it renders the

As the above code sample demonstrates, the script tags with the identifier data-big-pipe-placeholder="callback=history_attach_timestamp&args[0]=2&token=15cf9f98" appear after the one with the identifier data-big-pipe-placeholder="callback=comment.lazy_builders%3ArenderForm&args[0]=node&args[1]=2&args[2]=field_question_comments&args[3]=comment&token=6313718a", and as a result the code in the big_pipe.js file evaluates the script tags with callback=history_attach_timestamp after the script tag containing the JSON for the comment form placeholder. This logic appears in the following code excerpt from bigPipeProcessDocument() in big_pipe.js:

    $(context).find('script[data-big-pipe-replacement-for-placeholder-with-id]') .once('big-pipe') .each(bigPipeProcessPlaceholderReplacement);  When a given node has more than one comment (the above example has two, which is why there are two tags with data-big-pipe-placeholder="callback=history_attach_timestamp&args[0]=2&token=15cf9f98"), the comment module generates identical #lazy_builder callbacks to history_attach_timestamp for all of them, which big_pipe in turn uses to generate identical placeholders. The identical placeholders become an issue when the bigPipeProcessPlaceholderReplacement() function simulates an ajax success response using the identical script tags. The success() method in the ajax system invokes the simulated command defined in the JSON content of the script tags, which uses the replaceWith method. This triggers invocation of Drupal.AjaxCommands.prototype.insert() command, which detaches behaviors attached to any DOM elements in the context of the placeholder: Drupal.detachBehaviors(wrapper.get(0), settings);  The first time this executes on a placeholder generated for the #lazy_builder callback for history_attach_timestamp, the value of wrapper.get(0) is

. However, a few lines below that, the jQuery replaceWith method is triggered:  // Add the new content to the page. wrapper[method](new_content);  After this line executes, the DOM node

 is replaced with an empty div (

), because the value of response.data in the simulated ajax response is an empty string. The problem is that the next time this code executes on an identical placeholder from the script tags at the bottom of the document body, all instances of these placeholder divs in the DOM have been replaced with empty divs. The value for the wrapper variable is now an empty object, because the selector in response.selector refers to DOM nodes that have been removed: // The value of response.selector is [data-big-pipe-placeholder-id="callback=history_attach_timestamp&args[0]=2&token=15cf9f98"] var wrapper = response.selector ?$(response.selector) : $(ajax.wrapper);  Then, a few lines down, when Drupal.detachBehaviors() runs, the value of wrapper.get(0) evaluates to undefined: Drupal.detachBehaviors(wrapper.get(0), settings);  Drupal.detachBehaviors() sets the context in which it operates to the entire document if the context is undefined; see the first line of the function:  Drupal.detachBehaviors = function (context, settings, trigger) { context = context || document; settings = settings || drupalSettings; trigger = trigger || 'unload'; var behaviors = Drupal.behaviors; // Execute all of them. for (var i in behaviors) { if (behaviors.hasOwnProperty(i) && typeof behaviors[i].detach === 'function') { // Don't stop the execution of behaviors in case of an error. try { behaviors[i].detach(context, settings, trigger); } catch (e) { Drupal.throwError(e); } } } };  The result is that behaviors are detached from the entire document, causing behaviors that were attached earlier in the execution process to be removed. ### Proposed resolution Proposed solution is to modify \Drupal\comment\CommentViewBuilder::buildComponents() to prevent duplicate copies of the placeholder from being generated. The #lazy_builder callback in the render array that causes Big Pipe to generate the placeholder seems to have one purpose: to add the timestamp for the comments' parent node to the drupalSettings object. The #lazy_builder for the history_attach_timestamp in the comment view builder is:  // Embed the metadata for the comment "new" indicators on this node.$build[$id]['history'] = [ '#lazy_builder' => ['history_attach_timestamp', [$commented_entity->id()]],
'#create_placeholder' => TRUE,
];

The callback function for history_attach_timestamp() is:

function history_attach_timestamp($node_id) {$element = [];
$element['#attached']['drupalSettings']['history']['lastReadTimestamps'][$node_id] = (int) history_read($node_id); return$element;
}

The history_attach_timestamp() callback function always returns a render array that is empty, except for the settings being added to drupalSettings. Since the settings are per node ID and are the same for all comments attached to the same node, subsequent invocations of this callback function merely result in this property of the drupalSettings object being overwritten with the same value.

Proposed code change is to set this #lazy_builder only once for the entire comment list:

diff --git a/core/modules/comment/src/CommentViewBuilder.php b/core/modules/comment/src/CommentViewBuilder.php
index 9b47b7c..435bae5 100644
--- a/core/modules/comment/src/CommentViewBuilder.php
+++ b/core/modules/comment/src/CommentViewBuilder.php
@@ -148,18 +148,18 @@ public function buildComponents(array &$build, array$entities, array $displays,$build[$id]['#attached']['library'][] = 'comment/drupal.comment-by-viewer'; if ($this->moduleHandler->moduleExists('history') && $this->currentUser->isAuthenticated()) {$build[$id]['#attached']['library'][] = 'comment/drupal.comment-new-indicator'; - - // Embed the metadata for the comment "new" indicators on this node. -$build[$id]['history'] = [ - '#lazy_builder' => ['history_attach_timestamp', [$commented_entity->id()]],
-          '#create_placeholder' => TRUE,
-        ];
}
}
if ($build[$id]['#comment_threaded']) {
// The final comment must close up some hanging divs.
$build[$id]['#comment_indent_final'] = $current_indent; } + + // Embed the metadata for the comment "new" indicators on this node. +$build['history'] = [
+      '#lazy_builder' => ['history_attach_timestamp', [$commented_entity->id()]], + '#create_placeholder' => TRUE, + ]; } /**  Patch file attached. ### Remaining tasks Patch needs review. ### User interface changes None. ### API changes None. ### Data model changes None. Files: CommentFileSizeAuthor #321.02 KBdanmuzyka #3012.95 KBdanmuzyka #2413.85 KBWim Leers ## Comments danmuzyka created an issue. See original summary.  Status: Needs review » Needs work FileSize 1.43 KB 945 bytes Corrected patch attached.  Status: Needs work » Needs review  Version: 8.0.x-dev » 8.1.x-dev Drupal 8.0.6 was released on April 6 and is the final bugfix release for the Drupal 8.0.x series. Drupal 8.0.x will not receive any further development aside from security fixes. Drupal 8.1.0-rc1 is now available and sites should prepare to update to 8.1.0. Bug reports should be targeted against the 8.1.x-dev branch from now on, and new development or disruptive changes should be targeted against the 8.2.x-dev branch. For more information see the Drupal 8 minor version schedule and the Allowed changes during the Drupal 8 release cycle. FileSize 1.43 KB 290 bytes Updating patch file for 8.1.x branch.  Issue summary: View changes I've updated the IS with steps to reproduce, and also verified myself that this patch fixes the issue using those steps. +++ b/core/modules/comment/src/CommentViewBuilder.php @@ -142,18 +142,23 @@ public function buildComponents(array &$build, array $entities, array$displays,
if ($this->moduleHandler->moduleExists('history') &&$this->currentUser->isAuthenticated()) {
$build[$id]['#attached']['library'][] = 'comment/drupal.comment-new-indicator';

Can this bit now be moved below as well since it is the same if check?

FileSize
550 bytes

This patch is for manual testing purposes only. It moves the comment form above the list of comments in Bartik.

 Version: 8.1.x-dev » 8.2.x-dev
FileSize
4.57 KB

I started on a javascript test, but did not get far enough to reproduce the error. For some reason, the custom test theme declared here isn't being used, so the comment form is still below the comments.

 Component: comment.module » big_pipe.module

After further investigation, this issue seems to be related more generally to the way that Big Pipe processes placeholders. There are circumstances under which the original patch to the comment module won't fix the issue (e.g., if there are multiple comment fields on the same page), so a better solution seems to be for Big Pipe not to add duplicate copies of the placeholder JSON in the page footer.

I'm attaching a new patch that fixes this in Big Pipe instead of Comment, and includes JavaScript functional tests.

 Status: Needs review » Needs work
 Status: Needs work » Needs review
FileSize
6.79 KB
183 bytes

Rerolled patch.

@danmuzyka could you attach patch that just contains the new test to illustrate the failure without the fix?

Sure, here is the patch file with the test only.

 Status: Needs review » Needs work
 Status: Needs work » Needs review Issue tags: +BigPipe in Core

This is looking good to me, both the fix and the test. Back to NR and tagging in hopes of getting a few more eyes on this for review.

 Status: Needs review » Needs work

Nice work on the patch!

I would like to skip re-rendering placeholders that have already been replaced though - if possible and not just skip the JS replacement part.

 Title: Duplicate placeholders generated from #lazy_builder callbacks to history_attach_timestamp in comments cause JavaScript errors when using Big Pipe » BigPipe unnecessarily renders and sends multiple-occurrence placeholders multiple times when using JS — can cause JS errors Issue tags: +JavaScript, +Needs tests
1. Reproduced, thanks for the clear STR!
2. The solution in the IS is wrong: we should not modify CommentViewBuilder. Having multiple lazy builders map to the same placeholder is very much intentional, and is an architectural requirement: the whole point is that if independent code paths need the same placeholder, that we only replace that placeholder once.
3. The problem is that there really is only a single placeholder for history_attach_timestamp() (verify yourself by putting a breakpoint at \Drupal\big_pipe\Render\Placeholder\BigPipeStrategy::doProcessPlaceholders()), yet BigPipe wants to render each occurrence of a placeholder separately. i.e. the problem is \Drupal\big_pipe\Render\BigPipe::getPlaceholderOrder(). That code relies on parsing to determine which placeholders to replace. This means the solution in #14 is also wrong.

Patch attached.

This should still get a \Drupal\big_pipe\Tests\BigPipeTest::testBigPipeJsMultipleOccurrencePlaceholders() test. That would be a generic test. However, it's awesome that the current patch already contains a functional JS regression test! I'd like to keep that, but rename it to be specifically for regression tests.

1. index 3e67737..6f32c6c 100644
--- a/core/modules/big_pipe/src/Render/BigPipe.php

The changes here should be reverted.

2. +++ b/core/modules/big_pipe/tests/src/FunctionalJavascript/BigPipePlaceholderTest.php
@@ -0,0 +1,116 @@
+ * @group comment

s/comment/big_pipe/

3. +++ b/core/modules/big_pipe/tests/src/FunctionalJavascript/BigPipePlaceholderTest.php
@@ -0,0 +1,116 @@
+class BigPipePlaceholderTest extends JavascriptTestBase {

BigPipeRegressionTest

4. +++ b/core/modules/big_pipe/tests/src/FunctionalJavascript/BigPipePlaceholderTest.php
@@ -0,0 +1,116 @@
+    // Clear the theme registry.
+    $this->container->set('theme.registry', NULL);  This should not be necessary. 5. +++ b/core/modules/big_pipe/tests/src/FunctionalJavascript/BigPipePlaceholderTest.php @@ -0,0 +1,116 @@ + // Ensure an article node type exists. +$this->createContentType(['type' => 'article']);
+    $this->addDefaultCommentField('node', 'article'); + + // Enable CKEditor. + FilterFormat::create([ + 'format' => 'full_html', + 'name' => 'Full HTML', + 'weight' => 1, + 'filters' => [], + ])->save(); +$settings['toolbar']['rows'] = [
+      [
+        [
+          'items' => [
+          ],
+        ],
+      ],
+    ];
+    $editor = Editor::create([ + 'format' => 'full_html', + 'editor' => 'ckeditor', + ]); +$editor->setSettings($settings); +$editor->save();
+
+    $admin_user =$this->drupalCreateUser([
+      'use text format full_html',
+    ]);
+    $this->drupalLogin($admin_user);

Can you move this into the test method?

6. +++ b/core/modules/big_pipe/tests/src/FunctionalJavascript/BigPipePlaceholderTest.php
@@ -0,0 +1,116 @@
+  public function testCommentForm() {

testCommentFormRegression_2698811()

FileSize
1.07 KB

Oops, forgot the patch.

 Status: Needs work » Needs review
FileSize
9.81 KB
8.83 KB

Here's an updated patch file with the new, generic test, and an interdiff of it relative to the patch I posted in #12.

Question:

Will this also work when JS is off?

As there we can only replace one placeholder at a time ...

 Issue tags: -Needs tests
FileSize
10.22 KB
7.34 KB

#20: splendid work! :) Thanks so much!

Here's first a reroll that fixes all my nits. That avoids needless back-and-forth. Zero code changes, just renaming, simplifying, strictifying.

FileSize
11.55 KB
4.05 KB

Will this also work when JS is off?

Yes, this already was only a problem when JS was turned on. My analysis in #18 already applies only to JS placeholders: getPlaceholderOrder() only runs for JS placeholders.

We could look into the no-JS placeholder case too… and I decided to do so here.

In order to test that better, it's important that we can prove that no-JS placeholders are rendered multiple times. An easy way to test that, is by doing something evil: using a static inside a lazy builder, so we can increment a counter. Then the JS version would show "Count=1. Count=1. Count=1." The no-JS version would show "Count=1. Count=2. Count=3." And that's exactly what happens.

Attached patch expands that test coverage, and fails.

FileSize
13.85 KB
2.63 KB

And here are then the changes for \Drupal\big_pipe\Render\BigPipe::sendNoJsPlaceholders().

As far as I'm concerned, this is RTBC. If Fabian agrees, he can RTBC this right away.

 Version: 8.2.x-dev » 8.1.x-dev

Oh, and this should go in 8.1, not 8.2. BigPipe is experimental, so we can commit this to 8.1 :)

Also testing #24 against 8.1.

@danmuzyka & @jhedstrom: I just wanted to say thanks again for your excellent work here. These are much higher quality comments and patches than usual, so please know it is very much appreciated. I've repaid the favor by helping to land this issue swiftly. I'm sorry for not having seen it sooner, if I would have, then I'd have helped land it right then!

The last submitted patch, 23: big_pipe_multi_occurence_js_placeholder-2698811-23.patch, failed testing.

 Status: Needs review » Reviewed & tested by the community

RTBC - Yes, indeed fantastic work on the patches and tests.

It is great to have both a JS regression test (yeah!) and functional tests.

Also good that we now no longer re-replace multiple occurences, which will help for all "token" based placeholders, so a nice perf improvement, too.

It is great to have both a JS regression test (yeah!) and functional tests.

Indeed!

This seems like a perfect model:

1. have thorough, comprehensive tests that are abstract (unit tests and functional tests where necessary)
2. have concrete regression tests for every regression ever encountered, and ensure they have a test in a fully functional environment (and thus functional JS tests)

At least, I personally think that's the perfect model. It allows me to have very strong confidence that it works as intended.

 Status: Reviewed & tested by the community » Needs review
FileSize
12.95 KB
1.52 KB

Thanks @Wim Leers and @Fabianx !

If you feel like fast-tracking any other core bugs, I have two other open ones at the moment: ;-)

I think in the latest patch, the callback function placeholderContent() is no longer needed. I'm attaching an updated patch that removes it.

#30: Could you make interdiffs with git diff OR with diff -u?

They are very hard to read atm. ...

FileSize
1.02 KB

Sorry, this one should be better...

 Status: Needs review » Reviewed & tested by the community

Indeed, back to RTBC

 Status: Reviewed & tested by the community » Needs work

The last submitted patch, 30: big_pipe_multi_occurence_js_placeholder-2698811-30.patch, failed testing.

 Status: Needs work » Reviewed & tested by the community

Those fails do not look related to this patch.

• catch committed 5e86905 on 8.2.x
Issue #2698811 by danmuzyka, Wim Leers, jhedstrom, Fabianx: BigPipe...
• catch committed f9cd621 on 8.1.x
Issue #2698811 by danmuzyka, Wim Leers, jhedstrom, Fabianx: BigPipe...
 Status: Reviewed & tested by the community » Fixed

Agreed the patch and tests look great. Committed/pushed to 8.2.x and cherry-picked to 8.1.x. Thanks!

Also committed & pushed to the contrib module for Drupal 8.0: http://drupalcode.org/project/big_pipe.git/commit/f60bceb.

 Status: Fixed » Closed (fixed)

Automatically closed - issue fixed for 2 weeks with no activity.


text:
# Why Not Work for the Mine? Costs and Benefits of Extractive Economies in Remote Aboriginal Australia

From the Series: The Pilbara Crisis: Resource Frontiers in Western Australia

Western Australia comprises two resource frontiers. Both share a foundation in resource extraction, but while one is driven by global demand, the other is structured by the dynamics of local indigenous economies. This “other” frontier is often referred to as the customary sector of contemporary Aboriginal Australia, which interacts with state and market contexts to form a hybrid economy (Altman 2001). To the confusion of outsiders who do not fully appreciate the complexities of navigating these three sectors, individuals often avoid lucrative market opportunities such as working for local mining operations and instead allocate time to the most traditional activity in the customary economic sector: hunting and gathering. In an attempt to explain this lack of engagement with the global resource frontier, we approach these dynamics with a simple question: why not work for the mine? Drawing on fieldwork with the Martu, an Aboriginal population who have native title over vast estates from the Pilbara to the Percival Lakes and from the Kimberley to Lake Disappointment, we illustrate how engaging in market-oriented activities results in significant costs not apparent in simple economic calculations.

At some basic level, an individual’s choice of work is contingent on the benefits the individual receives from his or her labor. Martu workers in the remote community of Parnngurr can earn approximately A$500 per week (Codding, Bird, and Bird 2015; Codding et al., forthcoming), significantly more than the national median of A$362 per week for an Aboriginal or Torres Strait Islander person over the age of fifteen (Australian Bureau of Statistics 2012). This income derives from a number of sources, including community development projects, the Martu Ranger work program, and producing paintings for sale through the Martumili Artists Cooperative. In addition to these work opportunities, people also supplement their income by foraging for wild resources, the most customary activity of the hybrid economy.

The average adult in Parnngurr community forages 23 of every 100 days and many forage every other day (Bliege Bird et al., forthcoming). On a typical foraging bout, an individual can gain between 2,500 and 3,500 kilocalories for about two and a half hours of work (Codding, Bird, and Bird 2015; Codding et al., forthcoming). Converting these calories to dollars through the equivalent price of food from the community store shows that the average person could add roughly A$45 or A$60 per week to their cash income by foraging three or four days a week,— a roughly 10 percent increase in weekly salary. Importantly, this conversion likely underestimates the dietary importance of foraging, which contributes between 20 and 50 percent of total calories consumed per capita in the community (Bliege Bird et al., forthcoming).

However, if Martu values were defined in simple economic terms, an individual would be better off if he or she left the community for training and work in one of the local mines. Mines in and around Martu country include BHP Billiton’s massive iron ore operation in Newman, Newcrest Mining’s Telfer gold mine, and Cameco Mining’s uranium facility at Kintyre. Working for one of these mines, an individual could increase his or her weekly income to around A\$1000 per week. However, this income comes at a significant opportunity cost.

If the Martu gave up customary components of their economic lives, they would also give up as much as half of their calorie intake from fresh, healthy bush foods. Coupled with decreased time out on Country, this would also diminish individual wellness: recent studies have shown that “caring for country”—activities that involve time on the land—produce significant health benefits (Burgess et al. 2009). Perhaps more importantly, in full engagement with the mining economy, an individual gives up extensive family and ritual obligations, severing ties to the land and people that are central to the identity and personhood of many Martu (Bird 2009). In place of family, such a person would have to negotiate interactions with non-Aboriginal coworkers who are, at best, insensitive to their backgrounds and, at worst, may be outright hostile. Non-Aboriginal supervisors often misunderstand the demands on Aboriginal workers as a result of obligations to family, to the land, and to their ancestors. Neglecting these obligations is no small thing. Doing so essentially means that Martu have to give up who they are.

Moreover, giving up being Martu has consequences far beyond the individual and his or her identity. Maintaining a customary component of Martu economies feeds back to provide important benefits to the public at large (Moritz, Ens, and Altman 2015). In order to hunt, Martu burn their country in a way that increases biodiversity (Bliege Bird et al. 2008), buffers the region from the effects of climate-driven wildfires (Bliege Bird et al. 2012), and maintains habitat for endemic fauna (Codding et al. 2014). Such ecosystem services save the public far more than the government currently invests in remote Aboriginal communities (Bliege Bird et al., forthcoming; Codding, Bird, and Bird 2015).

Taken together, the Martu have plenty of reasons not to work for the mine, and other Australians have plenty of reasons to support Martu decisions to engage in traditional economic pursuits on the “other” resource frontier. However, this support is not typically forthcoming. The Western Australian government prioritizes the globally driven resource frontier at the cost of the locally driven traditional economy, which threatens to have dire consequences (Langton 2012). Beyond questions of preserving cultural heritage, the issue of whether the “other” resource frontier is recognized will partially determine the future sovereignty of the original Australians and the future of Western Australia’s economic and ecological landscape.

## References

Altman, Jon C. 2001. “Sustainable Development Options on Aboriginal Land: The Hybrid Economy in the Twenty-First Century.” Discussion Paper, no. 226. Canberra: Centre for Aboriginal Economic Policy Research.

Australian Bureau of Statistics. 2012. “Census of Population and Housing: Characteristics of Aboriginal and Torres Strait Islander Australians, 2011.”

Bird, Douglas W. “The Inherent Value of Foraging.” Arena Magazine, no. 98: 30–33.

Bliege Bird, Rebecca, Douglas W. Bird, Brian F. Codding, Christopher H. Parker, and James Holland Jones. 2008. “The ‘Fire Stick Farming’ Hypothesis: Australian Aboriginal Foraging Strategies, Biodiversity and Anthropogenic Fire Mosaics.” Proceedings of the National Academy of Sciences 105, no. 39: 14796–801.

_____, Brian F. Codding, Peter G. Kauhanen, and Douglas W. Bird. 2012. “Aboriginal Hunting Buffers Climate-Driven Fire-Size Variability in Australia's Spinifex Grasslands.” Proceedings of the National Academy of Sciences 109, no. 26: 10287–292.

_____, Brian F. Codding, Brooke Scelza and Douglas W. Bird. Forthcoming. “The Dynamics of Indigenous Livelihoods and Ecologies in the Western Desert of Australia: Economic, Social, and Environmental Values of Traditional Subsistence.” Journal of Rural Studies.

Burgess, Christopher P., Fay H. Johnston, Helen L. Berry, Joseph McDonnell, Dean Yibarbuk, Charlie Gunabarra, Albert Mileran, and Ross S. Bailie. 2009. “Healthy Country, Healthy People: The Relationship between Indigenous Health Status and ‘Caring for Country’.” Medical Journal of Australia 190, no. 10: 567–72.

Codding, Brian F., Rebecca Bliege Bird, Peter G. Kauhanen and Douglas W. Bird. 2014. “Conservation or Co-Evolution? Intermediate Levels of Aboriginal Burning and Hunting Have Positive Effects on Kangaroo Populations in Western Australia.” Human Ecology 42, no. 5: 659–69.

_____, Douglas W. Bird, and Rebecca Bliege Bird. 2015. “The Real Cost of Closing Remote Communities.” Arena Magazine, no. 135: 5–6.

_____, Rebecca Bliege Bird, Douglas W. Bird, and David W. Zeanah. Forthcoming. “Alternative Aboriginal Economies: Martu Livelihoods in the Twenty-First Century.” In Why Forage? Hunting and Gathering in the Twenty-First Century, edited by Brian F. Codding and Karen Kramer. Santa Fe, N.M.: SAR Press.

Langton, Marcia. 2012. “2012 Boyer Lectures. The Quiet Revolution: Indigenous People and the Resources Boom.” Radio National, November 18–December 2.

Moritz, Craig, Emilie-Jane Ens, and Jon Altman. 2015. “Remote Indigenous Communities are Vital for our Fragile Ecosystems.” The Conversation, March 12.

format: plain


text:
How do you do this problem?

Which points on the graph of y=4-(x^2) are closest to the points (0,2)?

2. Originally Posted by PrincessIsrael
How do you do this problem?

Which points on the graph of y=4-(x^2) are closest to the points (0,2)?
Employ the distance formula.

3. Welcome to the Math Help Forum!

To find the point on $y=4-x^2$ closest to $(0,2)$, we first do as VonNemo said and use the distance formula to obtain a function of the distance $s$ from a point on the curve to $(0,2)$ as a function of $x$:

$s=\sqrt{(x-0)^2+((4-x^2)-2)^2}=\sqrt{x^2+(2-x^2)^2}.$

We now find the value of $x$ for which $s$ is a minimum. To do this, we differentiate $s$ and find all critical points. As $s$ is differentiable everywhere and has no boundary points, the only critical points will be those at which $s'=0$. As $s$ approaches infinity in both directions of the $x$-axis, at one of these critical points we must find our minimum.

Hope this helps!


text:
Mathematical Sciences Research Institute

Home » Workshop » Schedules » Mixing Times for the Simple Exclusion Process with Open Boundaries

Mixing Times for the Simple Exclusion Process with Open Boundaries

[HYBRID WORKSHOP] Connections and Introductory Workshop: Universality and Integrability in Random Matrix Theory and Interacting Particle Systems, Part 2 September 20, 2021 - September 24, 2021

September 20, 2021 (02:10 PM PDT - 03:00 PM PDT)
Speaker(s): Evita Nestoridi (Princeton University)
Location: MSRI: Simons Auditorium, Online/Virtual
Primary Mathematics Subject Classification No Primary AMS MSC
Secondary Mathematics Subject Classification No Secondary AMS MSC
Video
In the simple exclusion process on a finite segment, a particle is allowed to move to the right at rate $p$ and to the left at rate $q$, provided that the selected site is empty. In joint work with Nina Gantert and Domink Schmid, we study mixing times of the symmetric and asymmetric simple exclusion process on the segment where particles are allowed to enter and exit at the endpoints. We consider different regimes depending on $p,q$ as well as on the entering and exiting rates, and show that the process exhibits pre-cutoff and in some special cases even cutoff.


text:
# Thursday, November 6, 2014¶

Pulled in CPAS de Châtelet to production server and migrated the database.

Fixed a bug in lino.core.actions which caused the toggle_choice action to fail. toggle_choice is our first action with parameters but no window (i.e. lino.core.actions.Action.no_params_window set to True).

lino.modlib.humanlinks : uncle, cousin etc are now addable_link_types.

In docs/tickets/104 there was a last detail:

The welfare.newcomers.AvailableCoaches table no longer displays coaches only for newcomers. So you can now use this table on an active client to assign another coach.


text:
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# Stochastic particle approximation of the Keller-Segel equation and two-dimensional generalization of Bessel processes

3 MATHRISK - Mathematical Risk handling
Inria Paris-Rocquencourt, UPEM - Université Paris-Est Marne-la-Vallée, ENPC - École des Ponts ParisTech
Abstract : The Keller-Segel partial differential equation is a two-dimensional model for chemotaxis. When the total mass of the initial density is one, it is known to exhibit blow-up in finite time as soon as the sensitivity $\chi$ of bacteria to the chemo-attractant is larger than $8\pi$. We investigate its approximation by a system of $N$ two-dimensional Brownian particles interacting through a singular attractive kernel in the drift term. In the very subcritical case $\chi<2\pi$, the diffusion strongly dominates this singular drift: we obtain existence for the particle system and prove that its flow of empirical measures converges, as $N\to\infty$ and up to extraction of a subsequence, to a weak solution of the Keller-Segel equation. We also show that for any $N\ge 2$ and any value of $\chi>0$, pairs of particles do collide with positive probability: the singularity of the drift is indeed visited. Nevertheless, when $\chi<2\pi N$, it is possible to control the drift and obtain existence of the particle system until the first time when at least three particles collide. We check that this time is a.s. infinite, so that global existence holds for the particle system, if and only if $\chi\leq 8\pi(N-2)/(N-1)$. Finally, we remark that in the system with $N=2$ particles, the difference between the two positions provides a natural two-dimensional generalization of Bessel processes, which we study in details.
Type de document :
Article dans une revue
Domaine :
Liste complète des métadonnées

https://hal-enpc.archives-ouvertes.fr/hal-01171481
Contributeur : Benjamin Jourdain Connectez-vous pour contacter le contributeur
Soumis le : vendredi 3 juillet 2015 - 17:44:05
Dernière modification le : jeudi 20 janvier 2022 - 17:29:31
Archivage à long terme le : : mercredi 26 avril 2017 - 00:22:17

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### Citation

Nicolas Fournier, Benjamin Jourdain. Stochastic particle approximation of the Keller-Segel equation and two-dimensional generalization of Bessel processes. Annals of Applied Probability, Institute of Mathematical Statistics (IMS), 2017, 27 (5), pp.2807-2861. ⟨10.1214/16-AAP1267⟩. ⟨hal-01171481⟩

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# Doubt on the equation 8.11.9 ( Green's function's construction ). How do they take limit of the integral?

Reference:

Linear Partial Differential Equations for Scientists and Engineers(Authors:-Tyn Myint-U, Lokenath Debnath)

I understood till the red box. I have doubt on the equation in the red box. How do they take limits of the integral? I don't understand. Please help me. What is the motivation behind the Green's function? Already there is a 'Variation Parameter' Technique is there no for finding the particular integral. Please explain.


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# Tutorial 4 MD Simulation on Two Atoms of Lennard-Jonesium

## 4.1 Introduction

“Lennard-Jonesium” is an idealized model for the interaction of two uncharged atoms. The interaction potential is described by the Lennard-Jones 6-12 potential. This potential combines the weakly attractive London dispersion forces and the strongly repulsive exchange (atomic overlap) forces.

This tutorial carries out an MD simulation on two atoms of Lennard-Jonesium, restricted to move in only one dimension. This will introduce you to the dynamic behavior of non-covalently-linked atoms, and to the Lennard-Jones potential.

For two atoms separated by a distance r,

$$$E_{LJ}(r) = E_0 \left[\left(\frac{r_0}{r}\right)^{12} - 2\left(\frac{r_0}{r}\right)^6\right] \tag{4.1}$$$

In this exercise, the ideal distance is $$r_0 = 3Å$$, and the well depth is $$E_0 = 0.1$$ kJ/mol.

As seen in the graph, the negative r-6 potential is weakly attractive, and it dominates when $$r > r_0$$, the ideal contact distance. The positive r-12 term is strongly repulsive and dominates when $$r < r_0$$.

Create a new folder for your work on this tutorial. Download the following files and put them in that folder:

## 4.3 Prediction

An MD simulation mimics the behavior of real masses, in this case two atoms acting under the influence of the Lennard-Jones force between them. At the start of the simulation ($$t = 0$$), the two atoms are at rest on the x-axis, separated by a distance of $$8Å$$. Their motion is confined to that single dimension: throughout the simulation, $$y = 0$$ and $$z = 0$$. What do you think they will do as time passes?

To predict the behavior, just ask yourself, “What would two real masses do if they were sitting, at rest, on the Lennard-Jones energy surface shown above, and then released?” You are to express your prediction by describing the trajectory of the system, drawing a graph showing $$r(t)$$.

What’s important here is for you to think about the system for a while and make a guess about its behavior. It’s not important that your prediction be right. You will learn more from this simulation if you make a prediction – even if it’s wrong – than you will if you simply run it without thinking about what might happen.

The simplest way to think about this system is to imagine one atom as a ball sitting on the van der Waals energy surface produced by the other atom, as in the figure. The position of the single ball (r) represents the distance between the two atoms. (This is an example of a simple coordinate transformation. In this case the motions of the two atoms, $$x_1(t)$$ and $$x_2(t)$$ are reduced to a single coordinate, $$r(t)$$. You may find it easier to think about the motion of this “ball” in a fixed potential than it is to think about the motions of the two atoms.)

Gravity pulls the ball downward. If you answer the following questions, you should be able to predict how the ball will move, that is, how the distance between the two atoms will change as time passes.

• At the start of the simulation, do you expect the atoms to move, or not? In which direction? (That is, do you expect the two atoms to move toward one another, or away from one another, or to stay fixed?)

• Do you expect the motion to be monotonic (in one direction, forever)? Or do you expect the motion to reverse direction? If so, how far will the atoms go before changing direction?

• And what will happen then? Will the atoms return to their original positions or not?

• Will the motion be periodic, or not? If it is, do you expect it to be damped by friction? Will the atoms eventually come to rest? If so, where?

Once you’ve made your final guess, make your prediction semi-quantitative by sketching a rough graph of $$r(t)$$, that is, with time on the horizontal axis and r on the vertical axis. Again, it doesn’t matter whether your guess is right or not – what’s important is for you to think about the likely behavior of the system. This will help you understand the results of the simulation, once it’s been run.

## 4.4 Running the simulation

Throughout this tutorial, avoid rotating the structure in the Chimera window. This is a one-dimensional system with $$y = 0$$ and $$z = 0$$ for all atoms in the trajectory. If you rotate the structure, you will not be able to see motion in the x direction correctly. You can reset the view by clicking on the Reset View button in the Chimera sidebar.

1. Run an MD simulation in Sophia using the input files listed above. (If you need help setting up and running the simulation, follow the instructions in Tutorial 1). The recipe should produce a simulation with

• a box length of $$10$$ nm
• a cutoff of $$1$$ nm
• a stepsize of $$1$$ fs
• $$10000000$$ steps ($$10$$ ns)
• an output interval of $$100$$ steps ($$100$$ fs)
2. Create a plot of $$r(t)$$, the distance between the two atoms using Chimera’s distance measurement tool and the Sophia Trajectory Viewer window (refer to Tutorial 1 for detailed instructions).

## 4.5 Analysis

1. Compare the plot of $$r(t)$$ with the sketch you made of your predicted behavior of $$r(t)$$. Write a brief description of any differences, explaining the reasoning behind your prediction, and the reason the system behaved differently than you expected.

2. Examine the time-dependence of $$E_{pot}$$, $$E_{kin}$$ and $$E_{tot}$$. Explain the behavior.

3. Look at $$E_{tot}$$ in detail. How well is it conserved?

4. What is the period of the motion?

## 4.6 Challenge

Can you come up with an analytic expression for the period as a function of the Lennard-Jones parameters $$E_0$$ and $$r_0$$, and the initial position $$r(t=0)$$? This is a difficult problem, but it has a solution.

Show that the period predicted by your analytic expression is equal to the period observed in the simulation.


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# Order of Topics in Introductory Proofs Class

Beginning next semester I am teaching a course in proofs and mathematical problem solving at my local university. For some background, the university is primarily a commuter university and the students I will be teaching will be mostly math education, and computer science students.

I am teaching from the book Introduction to Mathematical Thinking by Scott A. Vanstone and Will J. Gilbert. The departmental course outline suggests starting at Chapter 2 - which would mean beginning with topics such as divisibility, diophantine equations, integers in different bases and prime numbers before talking about logic, sets, proofs and methods of proofs (this stuff is given in chapter 1).

While I prefer starting in chapter 1 I see that there may be some wisdom to starting with very concrete topics and moving foward. But if I do not explain proofs and logic from the start I will be unable to prove any of the small intermediate results that are given in chapter 2.

Additionally, I was thinking of opening the class with an excercise using missing entries of a sudoku puzzle and having the students explain what numbers go in the missing entries and writing and explanation of why they believe it. I want to use this exercise to introduce direct proof, proof by cases and proof by contradiction to the students.

My questions are:

1) I was wondering if someone has experienced teaching a course of this nature and has any recommendations on the suggested progression for the beginning of the course.

2) Assuming I start with chapter 2, would it still be a good idea to start with the sudoku exercise - and if so, how would I naturally segue from it into the rest of the class.

I teach such a course to mathematics majors, a somewhat different audience, and I find there are always some who are little unclear about "if-then" vs. "and", quantifiers, and free/bound variables. I mean that in a practical sense -- they confuse themselves, "accidentally" as it were. Theoretical instruction in formal logic helps only a few; they need examples to clarify formal logic, at which point formal logic might be useful to them.

I teach the formal stuff and proof structures and so on, but I have to keep in mind the pitfalls above that my students are facing. I haven't thought about what examples can be shown with sudoku, but it seems a good thing to use, if you can create good examples with them. (It will be fun for many of them, but don't be surprised if not all find it so. It's a hard criterion, to please everybody.) Now you're talking about proof by cases and proof by contradiction, but the pitfalls I mentioned seem more elementary to me. They seem like they should be addressed before getting to cases and contradiction. Probably that can be done with sudoku. (I'll give an example of mine later.)

How I start. All my students have had at least a year of calculus. Many of them are still quite good at calculus. So I use continuous functions and differentiable functions for examples. I ask them to come up with functions that are continuous (C) or not at, say, $x=1$, or differentiable (D) or not. I ask them how many combinations are possible and help them see all four, (C, D), (C, not D), etc. Now not all four are in fact achievable. Why? Someone will say because a differentiable function is continuous. Then we look at "$f$ is D and $f$ is C" and "If $f$ is D, then $f$ is C." Before addressing the implication, I make sure they all agree it is true (because it is a theorem, it has a proof). I put the examples they come up with in a table with the logical values T/F in columns for C and D. Then I go over the meaning of "and" and "if-then". The relation to the rest of the course, which you asked about, is this. I use and reuse the examples when we talk about proving "if-then", whenever confusion arises, and whenever I feel clarification is needed.

I also use other examples. Using a non-theorem is good for illustrating how to show an implication is false. A simple one is "if $x^2 > 4$, then $x > 2$", where $x$ represents an integer or a real number (but be sure to include negative numbers).

Even if those examples are not appropriate for your students, my hope is that it might give you an idea how to use the sudoko puzzles. Elementary number theory is also a good source of simple examples, which may be why it is the recommended starting point by the department.

• As far as sudoku's relation to proofs this can be seen when solving certain puzzle cells. A number of the methods used for more difficult puzzles mentally resemble direct proof, proof by cases, and proof by contradiciton. For more information on how this works: web.maths.unsw.edu.au/~csg/papers/sudoku-logic.pdf Also, I hope that by getting them to explain themselves that it takes them outside their math comfort zone and thinking about how you would explain something that had been an obvious or guess and check exercise (exactly what math has been like for many students before proofs) – Ryan Dec 30 '14 at 5:44


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# Definitive answer to existence of real exponents

Well, I've been searching through this fórum and I know this question has been answered many times. But the answers I see, are kinda circular (I think). Let's start by the natural case.

• Natural case

As the name says, this is natural to think. We can define $a^x \cdot a^y$ as $a^{x+y}$ because it is something we can observe and that is definitively true. When we think about the $0$ case for na expoent, and define it by being $1$ for every number except $0$ itself (not gonna enter in the $0^0$ discussion in this post) we can think in many ways. The best way to define this strongly as being equals to one, should be by taking the limit from both sides, but we don't have a definition yet, for a continuous function $f(x) = a^x$ so we're closed in the natural case. If we observe this behavior:

$$2^3 = 8$$ $$2^2 = 4$$ $$2^1 = 2$$ Then maybe $2^0$ should be $1$ because in the left side we're subtracting $1$ for the exponent, and in the right side we are dividing by the base of the exponent. So let's proceed to other definitions and see if we can still stay with the definition of the zero power.

• Integer case

If we think the same way, we can create a definition that works for both the natural case, and our new case: the integers. If we keep subtracting that table in the left side, we're gonna get to the negative case. So if we accept: $$2^0 = 1$$ And subtract 1 from the exponent on the left side, and divide by the base in the right side, we reach: $$2^{-1} = \frac{1}{2}$$ And if we continue, we'll see that we can define, for every integer exponent:

$$a^y = a\cdot a \cdots a \qquad\text{a times itself y times, for every y>0, y \in \mathbb Z }$$ $$a^y = 1 \qquad\text{for y = 0}$$ $$a^y = \frac{1}{a\cdot a \cdots a} \qquad\text{1 over a times itself y times, for every y<0, y \in \mathbb Z}$$

• Rational case Well, here's what I've been reading:

Define $a^{\frac{1}{n}} = c$ as the n-th root of $c$. So, for example, $a^{\frac{1}{2}} = \sqrt{a}$ This definition only makes sense to me if we ACCEPT that $$a^{b} \cdot a^{c} = a^{b+c}\tag {for b and c \in R}$$ then $$a^{\frac{1}{n}} = b$$ so $$(a^{\frac{1}{n}})^n = c^n$$ then $$a = c^n$$ As we can see, $a$ was the n-th root of $c$, but this only makes sense if we accept that the law for natural numbers hold for the rational case. We can, of course, 'take the limit' (we cannot take the limit yet, because it's not in the reals) of both sides and verify that by our acception, when the exponent $\frac{1}{n}$ gets closer to $1$ (in other words, $n$ gets closer to $1$) we get the number that multiplied by itself $1$ times is equal to itself, wich is just the number itself. In other words: $a^{\frac{1}{n}} = a$ when $n$ gets closer and closer to $1$. So it seems that we have good reasons to believe that the sum property holds for rational exponentes. But since math is highly rigorous we cannot ever accept this only because we have good reasons. Just because something don't makes sense under some sets, we cannot define it. So is there something that really proves that the sum rule rolds for rationals, and even real exponents? And what about imaginary and complex exponents? What can we say about it?

• Well, if you define $$e^x = \sum\limits_{n = 0}^{\infty} \frac{x^n}{n!}$$ and prove the rules of exponents from there, the concerns go away. Alternatively, you can define exponents with real numbers via a limit, based on the rational case. – user61527 Jan 2 '14 at 2:06
• @T.Bongers but I need to prove the rational case first – user108425 Jan 2 '14 at 2:10
• Why does the definition $a^{1/n} = \sqrt[n]{a}$ only make sense if we accept that $a^{b+c} = a^c \cdot b^c$ for $a$, $b$ rational? I didn't quite get that. – Ulrik Jan 2 '14 at 2:13
• @Svinepels I though i wanted to say that $a^{\frac{1}{n}}$ is the number that can be raised to the $n$ exponente. Don't know why I wrote this D; – user108425 Jan 2 '14 at 2:15

## 2 Answers

I'm not completely sure if I got your point, but here's how I would continue defining rational exponents.

Definition: For $r = m/n \in \mathbb{Q}$, define $a^r = (\sqrt[n]{a})^{m}$. This definition is easily seen to be compatible with the special case where $r$ is an integer.

We can now prove the following: $a^{r+s} = a^r a^s$ for $r,s \in \mathbb{Q}$ (1):

Let $r = m/n$ and $s = p/q$. Then $$a^{r+s} = a^{(mq+np)/nq} = \left( \sqrt[nq]{a} \right)^{mq+np} = \left( \sqrt[nq]{a} \right)^{mq} \left( \sqrt[nq]{a} \right)^{np} = a^{mq/nq}a^{nq/np} = a^{m/n}a^{q/p} = a^ra^s$$

As you can see, we didn't need to assume beforehand that (1) was true, but we could prove it right from the definition.

Extending the definition to real exponents is worse though...

• In order for $a^{m/n}$ to be well defined we need to prove that if $m/n=p/q$ then $a^{m/n}=a^{p/q}$. (This is not hard, of course). I would roll it into the definition of $a^{1/r}$, though, and say that when $a\ge 0$, $a^{m/n}$ is the unique nonnegative solution for $x$ to $x^n=a^m$. So roots would simply be an abbreviation for $a^{1/r}$. – Henning Makholm Jan 2 '14 at 3:30
• But isn't this the same as accepting that: ${(a^{\frac{1}{n}})}^n = a$? How can you prove it? – Lucas Zanella Jan 2 '14 at 3:39
• It can be proved that there exists a unique, positive solution to the equation $x^n = a$ for every $a > 0$. Since it is unique, we give it a name: $\sqrt[n]{a}$. And now, since we have defined $a^{1/n} = \sqrt[n]{a}$, the number $a^{1/n}$ is by definition solution to the equation $x^n = a$, i.e. $\left( a^{1/n} \right)^n = a$. – Ulrik Jan 2 '14 at 3:53
• @Svinepels what's the reason to say that $a^{1/n} = \sqrt[n]{a}$? – Lucas Zanella Jan 2 '14 at 16:56
• @LucasZanella What do you mean by "reason"? It certainly is a special case of our definition of $a^{m/n}$. – Ulrik Jan 2 '14 at 23:00

It's not a question of “belief” or “acceptance”, it's a matter of being self-consistent and self-coherent, inasmuch as the latter are characteristics of truth, of which mathematics is intended to be part of, as opposed to it being a useless and incoherent system of falsehoods and inconsistencies.


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# Birthday paradox: How to estimate the probability of two or more people in a group of 30 sharing a birthday?

I might be overthinking this. I generated the output in R and 5 of my 10 samples were successful, so that's 50%. Given that, if I am to estimate the probability of two or more people in a group of 30 sharing a birthday, what is my total sample? Should I be using combinations?

How are you generating your birthdays? To generate 23 birthdays:

dates = sample(1:365, 23, replace = TRUE)

To see if 2 or more share the same birthday:

length(dates) != length(unique(dates)) # TRUE if there are duplicates

How often is the above TRUE?

dupe_count = 0
runs = 1000000
for (i in 1:runs) {
dates = sample(1:365, 23, replace = TRUE)
if (length(dates) != length(unique(dates))) {
dupe_count = dupe_count + 1
}
}
print(dupe_count / runs)

[1] 0.508158

This closely matches the theoretical value of 50.7% in the wikipedia page

• With the expression round(runif(23, 1, 365)) you do not generate dates uniformly: the chances of $1$ or $365$ are only half the chances of any other numbers between them. This error won't noticeably affect the answer in this particular example, but repeating it in other situations with smaller ranges of numbers could introduce serious errors that might be difficult to detect. Interestingly, a two-sided test of your result yields p=3.6%, intimating it might be wrong: it's a tiny bit too large. – whuber Feb 8 '18 at 22:53
• @whuber Good catch. I replaced it with sample(1:365, 23, replace = TRUE) – reisner Feb 8 '18 at 23:12
• p = 1-cumprod(seq(365,266,by=-1) / 365) plot(1:100,p,type='o') abline(h=0.5,col='red') – HEITZ Feb 8 '18 at 23:55


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### Software Requirements

SoftwareRequirements

 Android iOS Windows MacOS with best with Chrome Chrome Chrome Chrome support fullscreen? Yes. Chrome/Opera No. Firefox/ Sumsung Internet Not yet Yes Yes cannot work on some mobile browser that don't understand JavaScript such as..... cannot work on Internet Explorer 9 and below

### Credits

Lee Tat Leong slightly remixed by lookang (This email address is being protected from spambots. You need JavaScript enabled to view it.); This email address is being protected from spambots. You need JavaScript enabled to view it.

### end faq

http://iwant2study.org/lookangejss/02_newtonianmechanics_3dynamics/ejss_model_AtwoodMachine2wee/AtwoodMachine2wee_Simulation.xhtml

### Apps

https://play.google.com/store/apps/details?id=com.ionicframework.atwoodmachineapp294376&hl=en

### Sample Learning Goals

Investigate the relation between the mass and the acceleration of the system

• keep the external force (hanging mass) constant
• add mass to the cart

### Equations

The Atwood machine (or Atwood's machine) was invented in 1784 by the English mathematicianGeorge Atwood as a laboratory experiment to verify the mechanical laws of motion with constant acceleration. Atwood's machine is a common classroom demonstration used to illustrate principles of classical mechanics.

The idea frictionlessl Atwood Machine consists of two objects of mass (Mass of cart M plus additional mass added Madd ) and m (hanging mass), connected by an inextensible massless string over an ideal massless pulley[1]

When M+Madd = m, the machine is in neutral equilibrium or net force is equal to zero, regardless of the position of the weights.

When M+Madd ≠ m the system of masses experience uniform acceleration.

We are able to derive an equation for the acceleration by using force analysis or free body diagram. If we consider a massless, inextensible string and an ideal frictionless pulley.

As a sign convention, we assume that a is positive when downward for m and to the right for M+Madd

By considering free body diagram of M+Madd we get

F = ma

T - f = (M+Madd)a

where

T is the tension pointing towards the right on the cart and added mass system

f is the friction force on the cart and added mass system on the wheels (assume to be a single force pointing left, thus the minus sign)

a is the acceleration of the cart and added mass system

Similarly, by considering the free body diagram of the hanging mass

F = ma by continous motion, downwards as positive,

mg - T = ma

by subsitiution,

T = (M+Madd)a +f =mg -ma

we get

$$a = \frac{mg - f}{M+M_{add}+m}$$

to get an equation of removing f, consider the forces of the cart and added mass system in the vertical direction

F = ma

R - (M+Madd) g = 0

in addtional, the frictional model is related by this equation

f ≤ μR

in motion, assume f = μR

therefore, the equatio of acceleration can be simplified to

$$a = \frac{mg - \mu R}{M+M_{add}+m}$$

$$a = \frac{mg - \mu (M+M_{add})g}{M+M_{add}+m}$$

m is the hanging mass
g is thegravitational acceleration
μ is the coefficient of friction
M is the mass of the cart

### For Teachers

This simulation is to recreate Walter Fendt's Newton's Second Law applet (http://www.walter-fendt.de/ph14e/n2law.htm). The purpose is to align with Modelling Instructions approach to formulate F = ma with experiment conducted with motion sensor rather than light gate.

 example data set collected across varying m from 0.30 to 0.50 kg which result in increasing amount of curve in the parabolic of the model form, x = x0 + vo*t +0.5*a*t^2Modified Atwood Machine JavaScript HTML5 Applet Simulation Model by Tat Leong Lee and Loo Kang Wee

 example data set collected across varying addM from 0 to 0.30 kg which result in decreasing amount of curve in the parabolic of the model form, x = x0 + vo*t +0.5*a*t^2Modified Atwood Machine JavaScript HTML5 Applet Simulation Model by Tat Leong Lee and Loo Kang Wee

### 6.4 Dynamics: Map of key concepts and ideas covered at upper secondary level Motion of objects Key inquiry question:

#### Why do objects on Earth and in the Universe move the way they do?

1. Mass, weight and density

• Mass is a measure of the amount of substance in a body. A body’s mass resists a change in the state of rest or motion of the body (inertia).

• Density of a substance is the mass per unit volume of the substance i.e. density = mass / volume (ρ = m/V). Density can be measured as the mass of 1 cm3 of any substance.

• Mass, a measure of the amount of substance in an object has a magnitude only (scalar), whereas weight, which is the force acting on a body in a gravitational field, has both magnitude and direction (vector).

• Using Newton’s 2nd law, Force = mass x acceleration; Weight = mass x acceleration due to gravity (or gravitational field strength) (W = mg).

2. Newton’s laws of motion

• Newton’s Laws of motion may be applied to: describe the effect of balanced and unbalanced forces on a body; describe the ways in which a force may change the motion of a body; and identify actionreaction pairs acting on two interacting bodies.

• Newton’s 1st law states that an object remains at rest or continues with constant speed in a straight line when no resultant force acts on it (no resultant force means that all the forces acting on the object are balanced). • Newton’s 2nd law states that the resultant force on a body = the mass of the body x acceleration of the body (F = ma). The direction of the acceleration is the same as the direction of the resultant force acting on the body.

• Newton’s 3rd law states that the forces of two bodies on each other (action-reaction pair) are always equal and act along the same line in opposite directions. The two forces (action-reaction pair) are of the same type. Force always appears in pair. The existence of a single force is impossible.

3. Free-body diagrams

• Free-body diagrams and vector graphical diagrams may be used to represent and analyse the forces acting on a body.

• A free-body diagram shows the forces acting on a body only, not the forces the body exerts on other bodies.

• The resultant force acting on the body can be found using graphical methods (parallelogram method or ‘head-to-tail’ method).

### Students’ prior knowledge of Dynamics and Gravitational field

Primary level:

Students learn about mass (a measure of the amount of matter in a body) and volume (the amount of space that a body occupies) and the use of appropriate apparatus to measure these quantities (e.g. use of a lever balance, an electronic balance, a measuring cylinder, a syringe, and a measuring jug). However, the concept of density is not introduced although students do simple experiments to investigate the ability of objects of different materials (plastics, wood, rubber and metals) to float/sink in water.

Students recognize that objects have weight because of the gravitational force between them and the Earth and that weight is different at different places, and can be measured using a spring balance or a weighing scale.

Lower secondary level:

Students learn that:

• the density of a substance is the mass of the substance per unit volume, and can be used to predict whether objects sink or float.

• gravity exists between any two objects (e.g. ball and Earth). The weight of an object depends on the force of gravity pulling on that object. Students’ common misconceptions and learning difficulties in Dynamics and Gravitational field

Newton’s first law:

Students often think that:

• a force is required to maintain an object in its motion;

• if there is no motion, there is no force acting;

• constant speed results from a constant force;

• friction (instead of inertia) causes objects to resist a change in its state of rest or motion.

Newton’s second law:

Students often think that:

• a larger velocity means a larger resultant force;

• acceleration implies increasing force;

• greater mass implies greater force;

• a force cannot move an object unless it is greater than the object’s weight;

• heavier objects fall faster than light objects.

Newton’s third law:

Students often think that force is a single physical quantity associated with a single object rather than as an interaction between two objects which must therefore exist as an action reaction pair. They have difficulty in understanding that:

• two objects of greatly differing masses (e.g. Earth and us) exert forces of equal magnitude on each other;

• the normal force on an object and the weight of the object do not always have equal magnitudes;

• gravity acts on an object all the time (not just when it is falling).

[text]

### Other Resources

1. http://www.walter-fendt.de/ph14e/n2law.htm
2. https://www.geogebra.org/m/hh9CGse3#material/K6afWTwQ
3. https://www.geogebra.org/m/ztUKm6QV Moment of Inertia: Rolling and Sliding Down an Incline by  ukukuku


text:
• # question_answer If $a<b<c<d$, then the roots of the equation $(x-a)(x-c)+2(x-b)(x-d)=0$ are [IIT 1984] A) Real and distinct B) Real and equal C) Imaginary D) None of these

Given equation can be rewritten as $3{{x}^{2}}-(a+c+2b+2d)x+(ac+2bd)=0$ Its discriminant D $={{(a+c+2b+2d)}^{2}}-4.3(ac+2bd)$ $={{\left\{ (a+2d)+(c+2b) \right\}}^{2}}-12(ac+2bd)$      $={{\left\{ (a+2d)-(c+2b) \right\}}^{2}}+4(a+2d)(c+2b)-12(ac+2bd)$     $={{\left\{ (a+2d)-(c+2b) \right\}}^{2}}-8ac+8ab+8dc-8bd$ $={{\left\{ (a+2d)-(c+2b) \right\}}^{2}}+8(c-b)(d-a)$ which is +ve, since $a<b<c<d$. Hence roots are real and distinct.


text:
# Free groups as free product of infinite cyclic groups

Let $$S$$ be an arbitrary set (countable or uncountable). It is clear that the free abelian group generated by $$S$$ is isomorphic to the direct sum

$$\bigoplus_{s\in S}\mathbb{Z}.$$

Is the free group generated by $$S$$ then isomorphic to

$$\ast_{s\in S}\mathbb{Z}$$ where $$\ast$$ denotes the free product of groups?

• Yes, since $\mathbb{Z}$ is also the free group on 1 generator. Dec 28 '20 at 14:15

It's a general pattern based on the free-forgetful adjunction: the free functor preserves colimits, in particular coproducts, so since $$S=\coprod_{s\in S}1$$ where $$1$$ is a set with one element, then $$F(S)=\coprod_{s\in S}F(1)\,,$$ and the coproduct in the category of groups is free product and $$F(1)=\Bbb Z$$.


text:
Meaning of intermediate value theorem.

Definition of intermediate value theorem in the Definitions.net dictionary. The calculator will find all numbers c (with steps shown) that satisfy the conclusions of the Mean Value Theorem for the given function on the given interval. Get help with your Intermediate value theorem homework.
Solution of exercise 4 Using Bolzano's theorem, show that the equation: x³+ x − 5 = 0, has at least one solution for x = a such that 1 < a < 2. Let f be a polynomial function.The Intermediate Value Theorem states that if $f\left(a\right)$ and $f\left(b\right)$ have opposite signs, then there exists at least one value c between a and b for which $f\left(c\right)=0$. A General Note: Intermediate Value Theorem. Here f(a) is a “0-th degree” Taylor polynomial. I understand that the intermediate value theorem lets you confirm that a solution for f(c) exists within a certain interval, but what would you do for a problem that asks you to approximate a certain value (in my case, a cubic function's roots) using the intermediate value theorem? Intermediate Value Theorem. The Intermediate Value theorem is … Practice questions. Intermediate Value Theorem. The Intermediate Value Theorem. Then there exists at least a … I'm having a lot of trouble deciphering the notation in this proof of the mean value theorem in several variables. Remember that the Mean Value Theorem only gives the existence of such a point c, and not a method for how to ﬁnd c. We understand this equation as saying that the diﬀerence between f(b) and f(a) is given by an expression resembling the next term in the Taylor polynomial. All three have to do with continuous functions on closed intervals. The Average Value Theorem is about continuous functions and integrals . equality.
The Intermediate Value Theorem is useful for a number of reasons. The Intermediate Value Theorem basically says that the graph of a continuous function on a … If a function is continuous in [a, b] then it attains all the values between f (a) and f (b) including f (a) and f (b) ... Lagrange’s Mean Value Theorem. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. The intermediate value theorem says the following: Suppose f(x) is continuous in the closed interval [a,b] and N is a number between f(a) and f(b) . Information and translations of intermediate value theorem in the most comprehensive dictionary definitions resource on the web. This is an extremely useful fact which holds in general for di erentiable functions, not only at maximum values but at minimum values as well. This theorem is used to prove statements about a function on an interval starting from local hypotheses about derivatives at points of the interval. What does intermediate value theorem mean? For g(x) = x 3 + x 2 – x, find all the values c in the interval (–2, 1) that satisfy the Mean Value Theorem. That is, at time t 0, when f(t) reaches its maximum value, we have f0(t 0) = 0.

First of all, it helps to develop the mathematical foundations for calculus. This theorem is in fact the general version of Rolle’s theorem. The following practice questions ask you to find values that satisfy the Mean Value Theorem in a given interval. The Mean Value Theorem is about differentiable functions and derivatives. We already know from the definition of continuity at a point that the graph of a function will not have a hole at any point where it is continuous. It follows, by the Intermediate Value Theorem and the fact that v is a continuous function, that we must have v(t 0) = 0. For example, on page 2 of this link we see an example of why the multivariable mean value equality fails & a claim that the best we can do is to find an inequality, yet the pages I've posted provide an equality. Why the Intermediate Value Theorem may be true Statement of the Intermediate Value Theorem Reduction to the Special Case where f(a) 0 In Conclusion.

In mathematics, the mean value theorem states, roughly, that for a given planar arc between two endpoints, there is at least one point at which the tangent to the arc is parallel to the secant through its endpoints.. Since it verifies the intermediate value theorem, the function exists at all values in the interval [1,5].

Mean Value Theorem Calculator. Show Instructions. See the explanation. In fact, the IVT is a major ingredient in the proofs of the Extreme Value Theorem (EVT) and Mean Value Theorem … My reason is that to prove the Mean Value Theorem (MVT) requires the Intermediate Value Theorem and differentiability of the function (an additional assumption) From this perspective, continuity is the key element of what both these theorems say, just so happens the …


text:
Which of these two textbook equations of geodesic deviation is correct?

My previous question Geodesic deviation on a 2-sphere - is this the right track? got shot down as “off topic”, so I'm having a second stab at it.

Misner et al's Gravitation (p34) gives the geodesic deviation equation as$$\frac{D^{2}\xi^{\alpha}}{D\tau^{2}}+R_{\phantom{\mu}\beta\gamma\delta}^{\alpha}\frac{dx^{\beta}}{d\tau}\xi^{\gamma}\frac{dx^{\delta}}{d\tau}=0,$$ with the right-hand side $\xi$ index $\gamma$ equal to the second lower index on the Riemann tensor. Lambourne's Relativity, Gravitation and Cosmology (p185), on the other hand, gives $$\frac{D^{2}\xi^{\mu}}{D\lambda^{2}}+R_{\phantom{\mu}\alpha\beta\gamma}^{\mu}\xi^{\alpha}\frac{dx^{\beta}}{d\lambda}\frac{dx^{\gamma}}{d\lambda}=0,$$ with the right-hand side $\xi$ index $\alpha$ equal to the first lower index on the Riemann tensor.

My question is, which of these two equations is correct?

I tried to answer this question myself by using the two equations to calculate the geodesic deviation on the surface of a unit 2-sphere. With Misner's equation (substituting $\lambda$ for $\tau$) I got $$\frac{D^{2}\xi^{\theta}}{D\lambda^{2}}=\left(\sin^{2}\theta\right)\left(u^{\phi}u^{\theta}\right)\xi^{\phi}-\left(\sin^{2}\theta\right)\left(u^{\phi}u^{\phi}\right)\xi^{\theta}$$ and $$\frac{D^{2}\xi^{\phi}}{D\lambda^{2}}=\xi^{\theta}\left(u^{\theta}u^{\phi}\right)-\xi^{\phi}\left(u^{\theta}u^{\theta}\right).$$

You can see my calculation on my previous question Geodesic deviation on a 2-sphere - is this the right track? With Lambourne's equation I got

$$\frac{D^{2}\xi^{\theta}}{D\lambda^{2}}=0$$ and $$\frac{D^{2}\xi^{\phi}}{D\lambda^{2}}=0.$$ This didn't seem right to me so I concluded that Lambourne's equation is incorrect.

-
The proper method of dealing with a closed question is to edit the closed question, not write it out again (albeit slightly modified). If you merge this question with your other, closed one, I will be willing to nominate for reopening. – Kyle Kanos Aug 28 '14 at 14:38
There are two different conventions for the Riemann tensor, and the tensors can differ by a minus sign between the two conventions. Without telling us which version the two books are using, this question, as written, is unanswerable, since the two expressions you give also differ by a minus sign. – Jerry Schirmer Aug 28 '14 at 14:51
@JerrySchirmer: Lambourne uses +--- and MTW use -+++ – John Rennie Aug 28 '14 at 15:17
@JohnRennie: yes, but there is an additional convention on the Riemann tensor, depending on whether $\nabla_{[a}\nabla_{b]}\omega_{c} = R_{abc}{}^{d}\omega_{d}$ or whether $\nabla_{[a}\nabla_{b]}\omega_{c} = R^{d}{}_{abc}{}\omega_{d}$. But looking at these two expressions again, the second one is wrong, as it is always identically zero. – Jerry Schirmer Aug 28 '14 at 15:19
@JerrySchirmer: oops, yes, sorry I didn't read your comment carefully enough. I have both books but don't know them well enough to find the relevant sections without a prolonged search. I guess I'll admit defeat :-) – John Rennie Aug 28 '14 at 15:23

Despite my comment, on second look your second equation, attributed to Lambourne, is always identically zero. This is because you multiply the symmetric tensor

$$\frac{dx^{\beta}}{d\lambda}\frac{dx^{\gamma}}{d\lambda}$$

against $R^{\mu}{}_{\nu\beta\gamma}$, and the riemann tensor is antisymmetric on those last two indices, and tracing a symmetric tensor against an antisymmetric tensor always gives zero.

-
@Peter4075: all that's going on with that is the fact that the Riemann tensor is defined by $\nabla_{a}\nabla_{b}\omega_{c} - \nabla_{b}\nabla_{a}\omega_{c}$. It's an arbitrary choice whether you put the fourth index that you contract on $\omega$ first or last, so this could be either equal to $R_{abc}{}^{d}\omega_{d}$ or $R^{d}{}_{abc}\omega_{d}$. Using the properties of the Riemann Tensor, you can juggle indices around, and show that these two definitions are equal to each other, except that they differ by a minus sign. – Jerry Schirmer Aug 28 '14 at 17:53
No, they can't! Both expressions are exactly the same 4-vector --- they are the spacetime co-ordinates of a particular point particle. – gj255 Aug 29 '14 at 9:13
@Peter4075 it's just because $\frac{dx^\beta}{d\lambda}\frac{dx^\gamma}{d\lambda}=\frac{dx^\gamma}{d\lambda} \frac{dx^\beta}{d\lambda}$ :) – Danu Aug 29 '14 at 13:46
@Peter4075: they're two copies of the SAME set of four functions. – Jerry Schirmer Aug 29 '14 at 15:18
So if, for eg, we had $x,y,z$ coordinates, the two terms would multiply out to give a symmetric matrix $$\frac{dx^{\beta}}{d\lambda}\frac{dx^{\gamma}}{d\lambda}=\left(\begin{array}{ccc} \frac{dx}{d\lambda}\frac{dx}{d\lambda} & \frac{dx}{d\lambda}\frac{dy}{d\lambda} & \frac{dx}{d\lambda}\frac{dz}{d\lambda}\\ \frac{dy}{d\lambda}\frac{dx}{d\lambda} & \frac{dy}{d\lambda}\frac{dy}{d\lambda} & \frac{dy}{d\lambda}\frac{dz}{d\lambda}\\ \frac{dz}{d\lambda}\frac{dx}{d\lambda} & \frac{dz}{d\lambda}\frac{dy}{d\lambda} & \frac{dz}{d\lambda}\frac{dz}{d\lambda} \end{array}\right)$$ aka a symmetric tensor. – Peter4075 Aug 29 '14 at 18:28

EDIT2: I've just seen Jerry Schirmer's answer, which confirms to me that Lambourne's equation is incorrect. I would mark his as the accepted answer. I'll leave the below for reference, though it is entirely false! Indeed, as pointed out by Peter in the comments, Lambourne uses the same convention as MTW for the Riemann tensor (see equation 3.35). Funnily, this doesn't appear in the August 2012 errata of the book --- perhaps somebody should email him.

In Misner, Thorne and Wheeler, the Riemann tensor is given by $$R^\mu{}_{\alpha \beta \gamma} = \frac{\partial \Gamma^\mu{}_{\alpha \gamma}}{\partial x^\beta} - \frac{\partial \Gamma^\mu{}_{\alpha \beta}}{\partial x^\gamma} + \Gamma^\mu{}_{\sigma \beta} \Gamma^\sigma{}_{\gamma \alpha} - \Gamma^\mu{}_{\sigma \gamma} \Gamma^\sigma{}_{\beta \alpha}$$ I believe Lambourne defines $R^\mu{}_{\alpha \beta \gamma}$ to be precisely the negative of this quantity. Now the Riemann tensor has various symmetries. The ones relevant here are

$$R_{abcd} = - R_{bacd}$$ $$R_{abcd} = - R_{abdc}$$ $$R_{abcd} = + R_{cdab}$$

With these symmetries and the two conventions considered, you see that the two equations of geodesic deviation are identical.

EDIT1: I've just noticed that this isn't right, despite the up-votes. If we think of the four indices on the Riemann tensor in pairs (1 and 2, and 3 and 4), the index on the separation vector sits with the free index in Lambourne, whereas it sits with one of the $\dot{x}$ terms in MTW. This is a significant difference, and in fact I think Lambourne's equation might indeed be incorrect. I just found a PDF of the book and it appears you've reproduced the relevant equation (6.23) faithfully here.

-
I'm struggling to take this in, but Lambourne gives $$R_{\phantom{l}ijk}^{l}=\frac{\partial\Gamma_{ik}^{l}}{\partial x^{j}}-\frac{\partial\Gamma_{ij}^{l}}{\partial x^{k}}+\Gamma_{ik}^{m}\Gamma_{mj}^{l}-\Gamma_{ij}^{m}\Gamma_{mk}^{l}.$$ – Peter4075 Aug 28 '14 at 15:28


text:
# Subscript in molecule over reaction arrow

I am writing a report that requires chemical reactions to be numbered and have used the following code to do so:

\documentclass[12pt,a4paper]{article}
\usepackage{amsmath}
\usepackage{chemformula}

\begin{document}

\begin{subequations}
\begin{eqnarray}
\label{eq:formloss1}
\ch{HCHO + ^.OH -> CHO^. + H2O} \\
\ch{HCHO + \textit{hv_{uv}} ->[O2] CHO^. + HO2} \\
\label{eq:formloss2}
\ch{HCHO + \textit{hv_{uv}} -> CO + H2}
\end{eqnarray}
\end{subequations}
\end{document}

But in the second equation the O2 (which is above the reaction arrow) doesn't subscript the 2, which I need it to do. How can I force it to subscript this? And if it requires changing the packages I'm using, how to I keep the equations numbered?

• According to the manual you should use ->[ O2 ] (note the extra spaces) if you want the 2 subscripted. – John Kormylo Oct 2 '15 at 4:31

From chemformula's manual:

chemformula distinguishes between different types of input. These different parts have to be separated with blanks:

\ch{part1 part2 part3 part4}

The same holds for formulas above or below arrows. This means you have to write

->[ O2 ]

chemformula needs formulas enclosed with spaces to correctly detect wether an input is a formula/a stoichiometric factor/an arrow…

Secondly I'd write $h\nu_{\textup{uv}}$ instead of \textit{hv_{uv}}. By using math mode you'll get the right spacing between the Planck constant and the frequency. Also I'm quite sure you want the frequency \nu and not v (velocity?). The subscript seems to be a label, so it should be upright again.

\documentclass[12pt,a4paper]{article}
\usepackage{amsmath}
\usepackage{chemformula}

\begin{document}

\begin{subequations}
\begin{gather}
\ch{HCHO + ^.OH -> CHO^. + H2O} \label{eq:formloss1} \\
\ch{HCHO + $h\nu_{\textup{uv}}$ ->[ O2 ] CHO^. + HO2} \\
\ch{HCHO + $h\nu_{\textup{uv}}$ -> CO + H2} \label{eq:formloss2}
\end{gather}
\end{subequations}

\end{document}

By the way: do you really need eqnarray? I suggest you read Avoid eqnarray by Lars Madsen and decide for yourself.

I used mhchem package, which solves your problem. Here is the code. You should also install chemgreek as mhchem uses it. (Note that the command for chemical equation is \ce.)

\documentclass[12pt,a4paper]{article}
\usepackage{amsmath}
\usepackage[version=4]{mhchem}

\begin{document}

\begin{subequations}
\begin{eqnarray}
\label{eq:formloss1}
\ce{HCHO + ^.OH -> CHO^. + H2O} \\
\ce{HCHO + $hv_{uv}$ ->[O2] CHO^. + HO2} \\
\label{eq:formloss2}
\ce{HCHO + $hv_{uv}$ -> CO + H2}
\end{eqnarray}
\end{subequations}

\end{document}

• Please don't suggest people to install chemgreek manually! I've had quite a few “bug reports” because people did that instead of updating their distribution and having chemgreek installed and ready to use… – cgnieder Oct 2 '15 at 9:26


text:
# Trends in dwarf early-type kinematics with cluster-centric radius driven by tidal stirring

Benson, A J; Toloba, E; Mayer, L; Simon, J D; Guhathakurta, P (2015). Trends in dwarf early-type kinematics with cluster-centric radius driven by tidal stirring. The Astrophysical Journal, 799(2):171.

## Abstract

We model the dynamics of dwarf early-type galaxies in the Virgo cluster when subject to a variety of environmental processes. We focus on how these processes imprint trends in dynamical state (rotational vs. pressure support as measured by the $\lambda^*_{\rm Re/2}$ statistic) with projected distance from the cluster center, and compare these results to observational estimates. We find a large scatter in the gradient of $\lambda^*_{\rm Re/2}$ with projected radius. A statistical analysis shows that models with no environmental effects produce gradients as steep as those observed in none of the 100 cluster realizations we consider, while in a model incorporating tidal stirring by the cluster potential 34% of realizations produce gradients as steep as that observed. Our results suggest that tidal stirring may be the cause of the observed radial dependence of dwarf early-type dynamics in galaxy clusters.

## Abstract

We model the dynamics of dwarf early-type galaxies in the Virgo cluster when subject to a variety of environmental processes. We focus on how these processes imprint trends in dynamical state (rotational vs. pressure support as measured by the $\lambda^*_{\rm Re/2}$ statistic) with projected distance from the cluster center, and compare these results to observational estimates. We find a large scatter in the gradient of $\lambda^*_{\rm Re/2}$ with projected radius. A statistical analysis shows that models with no environmental effects produce gradients as steep as those observed in none of the 100 cluster realizations we consider, while in a model incorporating tidal stirring by the cluster potential 34% of realizations produce gradients as steep as that observed. Our results suggest that tidal stirring may be the cause of the observed radial dependence of dwarf early-type dynamics in galaxy clusters.

## Statistics

### Citations

Dimensions.ai Metrics
4 citations in Web of Science®
4 citations in Scopus®

### Altmetrics

Detailed statistics

Item Type: Journal Article, refereed, original work 07 Faculty of Science > Institute for Computational Science 530 Physics English February 2015 22 Feb 2016 15:27 14 Feb 2018 11:13 IOP Publishing 1538-4357 Hybrid Publisher DOI. An embargo period may apply. https://doi.org/10.1088/0004-637X/799/2/171 arXiv:1410.5395v1


text:
Skip to Content
Calculus Volume 2

# Key Concepts

Calculus Volume 2Key Concepts

### 6.1Power Series and Functions

• For a power series centered at $x=a,x=a,$ one of the following three properties hold:
1. The power series converges only at $x=a.x=a.$ In this case, we say that the radius of convergence is $R=0.R=0.$
2. The power series converges for all real numbers x. In this case, we say that the radius of convergence is $R=∞.R=∞.$
3. There is a real number R such that the series converges for $|x−a| and diverges for $|x−a|>R.|x−a|>R.$ In this case, the radius of convergence is R.
• If a power series converges on a finite interval, the series may or may not converge at the endpoints.
• The ratio test may often be used to determine the radius of convergence.
• The geometric series $∑n=0∞xn=11−x∑n=0∞xn=11−x$ for $|x|<1|x|<1$ allows us to represent certain functions using geometric series.

### 6.2Properties of Power Series

• Given two power series $∑n=0∞cnxn∑n=0∞cnxn$ and $∑n=0∞dnxn∑n=0∞dnxn$ that converge to functions f and g on a common interval I, the sum and difference of the two series converge to $f±g,f±g,$ respectively, on I. In addition, for any real number b and integer $m≥0,m≥0,$ the series $∑n=0∞bxmcnxn∑n=0∞bxmcnxn$ converges to $bxmf(x)bxmf(x)$ and the series $∑n=0∞cn(bxm)n∑n=0∞cn(bxm)n$ converges to $f(bxm)f(bxm)$ whenever bxm is in the interval I.
• Given two power series that converge on an interval $(−R,R),(−R,R),$ the Cauchy product of the two power series converges on the interval $(−R,R).(−R,R).$
• Given a power series that converges to a function f on an interval $(−R,R),(−R,R),$ the series can be differentiated term-by-term and the resulting series converges to $f′f′$ on $(−R,R).(−R,R).$ The series can also be integrated term-by-term and the resulting series converges to $∫f(x)dx∫f(x)dx$ on $(−R,R).(−R,R).$

### 6.3Taylor and Maclaurin Series

• Taylor polynomials are used to approximate functions near a value $x=a.x=a.$ Maclaurin polynomials are Taylor polynomials at $x=0.x=0.$
• The nth degree Taylor polynomials for a function $ff$ are the partial sums of the Taylor series for $f.f.$
• If a function $ff$ has a power series representation at $x=a,x=a,$ then it is given by its Taylor series at $x=a.x=a.$
• A Taylor series for $ff$ converges to $ff$ if and only if $limn→∞Rn(x)=0limn→∞Rn(x)=0$ where $Rn(x)=f(x)−pn(x).Rn(x)=f(x)−pn(x).$
• The Taylor series for ex, $sinx,sinx,$ and $cosxcosx$ converge to the respective functions for all real x.

### 6.4Working with Taylor Series

• The binomial series is the Maclaurin series for $f(x)=(1+x)r.f(x)=(1+x)r.$ It converges for $|x|<1.|x|<1.$
• Taylor series for functions can often be derived by algebraic operations with a known Taylor series or by differentiating or integrating a known Taylor series.
• Power series can be used to solve differential equations.
• Taylor series can be used to help approximate integrals that cannot be evaluated by other means.
Citation/Attribution

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Citation information

© Mar 30, 2016 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License 4.0 license. The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.


text:
# Can a factor map be a Serre fibration?

Let $D_n$ be an $n$-disc. Is the factor map $p: D_n\to D_n/S^{n-1}\simeq S^n$ a Serre fibration, in other words, can any homotopy $F: [0,1]\times X\to S^n$ be lifted to $\tilde{F}: [0,1]\times X\to D^n$, if a lift $\tilde{f}_0$ of $F(0,\cdot)$ is given?

More generally, if $p(S^{n-1})=\{s_0\}$, can a homotopy $F: [0,1]\times (X,A)\to (S^n, s_0)$ be lifted to $[0,1]\times (X,A)\to (D_n, S^{n-1})$, if $\tilde{f}_0$ is given? Suppose that $X$ is nice, such as a CW or simplicial complex.

• Serre fibrations have homotopy equivalent fibers. The map $p$ has two classes of fibers, the fiber over $[S^n]$ is a sphere, but the fiber over any other point is just a single point. – Dan Rust May 15 '14 at 20:35
• Thanks a lot, Daniel. So, for some homotopy $F$, the $\tilde{F}$ wouldn't exist? Could you give me some reference please? – Peter Franek May 15 '14 at 20:41
• The last sentence of the section on the formal definition of a fibration on wikipedia mentions the homotopy equivalence. I think it may be the case that actually Serre fibrations (which are a larger class of maps than Hurewicz fibrations) only have 'weakly equivalent' fibers but that still leads to a contradiction in this case. – Dan Rust May 15 '14 at 20:46
• Thanks a lot, Daniel, I've found it exactly as you say in Spanier.. However, concerning my second paragraph (the pair version), couldn't that be true? I need it :-)) – Peter Franek May 16 '14 at 14:21


text:
You are viewing a single comment's thread from:

#### RE: Maths Brain Teasers 45:: Can You Eliminate the ODD One Out???

in STEMGeeks2 months ago

The answer is 5246 :-) I won't say why so I don't spoil it for anyone else!

Sort:

I agree with his answer. 5246. I added the digits together and realized they were all a factor of three except 5246. So, that is the odd one out. But, when I realized that all the other numbers could also be divided by 2, I came to the conclusion that all the others could not only be divided by 3 but also by 6. So, while the answer is still 5246, I'm going with the fact that the others can be divided by 6 as the common denominator and thus the reason 5246 is the odd number out.

Spot on my friend!


text:
# An urn contains 3 red and 7 black balls.

An urn contains 3 red and 7 black balls. Players A and B withdraw balls from the urn consecutively until a red ball is selected. Find the probability that A selects the red ball. (A draws the first ball, then B, and so on. There is no replacement of the balls drawn.)
You can still ask an expert for help

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Clelioo
A wins if the first red ball is drawn 1st, 3rd, 5th, or 7th.
We will calculate the number of events for each first apparition of a red ball.(i.e. a red ball is drawn first, therefore there are (9C2) places in which the other 2 red balls can be placed. In other words, there are (9C2) events in which A wins on the first draw)
E(1)=(9C2)
E(3)=(7C2)
E(5)=(5C2)
E(7)=(3C2)
When the sum up the number of favorable events and divide by the number of total events.
S=(10C3)
E(x): The number of favorable events (position of the first red ball)
S: Total number of events (all possible combinations of the balls)
P(A wins)=$\frac{\left(9C2\right)+\left(7C2\right)+\left(5C2\right)+\left(3C2\right)}{10C3}$
P(A wins)=.05833


text:
is below this banner.

Can't find a solution anywhere?

NEED A FAST ANSWER TO ANY QUESTION OR ASSIGNMENT?

You will get a detailed answer to your question or assignment in the shortest time possible.

## Here's the Solution to this Question

Class f cf

0-75 69 69

75-150 137 206

150-225 225 431

225-300 46 477

300-375 88 565

375-400 25 590

400+ 10 600

To find the minimum income of the riches 30% families, we determine the $70^{th}$ percentile given as,

$P_{70}=l+({70\times n\over 100}-cf)\times{c\over f}$ where, $n=600$ and

$l$ is the lower class boundary of the class containing $P_{70}$

$f$ is the frequency of the class containing $P_{70}$

$c$ is the width of the class containing $P_{70}$

$cf$ is the cumulative frequency of the class preceding the class with $P_{70}$.

The class containing $P_{70}$ is,

$({70\times n\over 100})^{th} class=({70\times 600\over 100})=420$. Therefore the class with $P_{70}$ is 150-225

Thus,

$P_{70}=150+(420-206)\times {75\over 225}=150+71.33=221.33\approx 222$

The limits of income of middle 50% of families is same as determining $Q_1$ and $Q_3$ where,

$Q_1$ is the lower limit and $Q_3$ is the upper limit.

Therefore,

$Q_1=l+({n\over4}-cf)\times {c\over f}$ where,

$l$ is the lower class boundary of the class containing $Q_1$

$f$ is the frequency of the class containing $Q_1$

$c$ is the width of the class containing $Q_1$

$cf$ is the cumulative frequency of the class preceding the class with $Q_1$.

The class containing $Q_1$ is,

$({n\over 4})^{th} class={600\over 4}=150$.Therefore, the class with $Q_1$ is 75-150

Thus,

$Q_1=75+(150-69)\times {75\over 137}=119.34\approx 120$

and,

$Q_3=l+({3n\over4}-cf)\times {c\over f}$ where,

$l$ is the lower class boundary of the class containing $Q_3$

$f$ is the frequency of the class containing $Q_3$

$c$ is the width of the class containing $Q_3$

$cf$ is the cumulative frequency of the class preceding the class with $Q_3$.

The class containing $Q_1$ is,

$({3n\over 4})^{th} class={1800\over 4}=450$.Therefore, the class with $Q_3$ is 150-225

Thus,

$Q_3=225+(450-431)\times {75\over 46}=255.98\approx 256$

Therefore, the limits of income of middle 50% of families is 120 and 256.


text:
# Math

Find the reciprocal of (-1/3)(9/2)(6)

Thanks!

1. Chopsticks

Reciprocal means the opposite

Just flip those numbers.

So, do I multiply them once I flip them. or is the answer (-3/1)(2/9)(6)?

3. Bella

first multiply and solve then flip
-6/54 i believe

If I multiply the first two fractions and the answer is 6/54, do I then multiply it by the 6?

5. Bella

you multiply the first two fractions and the answer is -9/6 then you multiply it by the last fraction and you're answer is -54/6
so the reciprocal(flip it around) is -6/54

i don't think you're paying attention Sadie, it's quite simple :)

Thank you for your help this evening.

7. Bella

you're very welcome....goodnight!

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text:
E. Hossam and a Letter
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Hossam bought a new piece of ground with length $n$ and width $m$, he divided it into an $n \cdot m$ grid, each cell being of size $1\times1$.

Since Hossam's name starts with the letter 'H', he decided to draw the capital letter 'H' by building walls of size $1\times1$ on some squares of the ground. Each square $1\times1$ on the ground is assigned a quality degree: perfect, medium, or bad.

The process of building walls to form up letter 'H' has the following constraints:

• The letter must consist of one horizontal and two vertical lines.
• The vertical lines must not be in the same or neighboring columns.
• The vertical lines must start in the same row and end in the same row (and thus have the same length).
• The horizontal line should connect the vertical lines, but must not cross them.
• The horizontal line can be in any row between the vertical lines (not only in the middle), except the top and the bottom one. (With the horizontal line in the top row the letter looks like 'n', and in the bottom row like 'U'.)
• It is forbidden to build walls in cells of bad quality.
• You can use at most one square of medium quality.
• You can use any number of squares of perfect quality.

Find the maximum number of walls that can be used to draw the letter 'H'.

Check the note for more clarification.

Input

The first line of the input contains two integer numbers $n$, $m$ ($1 \le n, m \le 400$).

The next $n$ lines of the input contain $m$ characters each, describing the grid. The character '.' stands for a perfect square, the character 'm' stands for a medium square, and the character '#' stands for a bad square.

Output

Print a single integer — the maximum number of walls that form a capital letter 'H'.

If it is not possible to draw any letter 'H', print $0$.

Examples
Input
2 3
#m.
.#.

Output
0

Input
7 8
...#.m..
..m...m.
.#..#.m#
...m..m.
m.......
..#.m.mm
......m.

Output
16

Note

In the first test case, we can't build the letter 'H'.

For the second test case, the figure below represents the grid and some of the valid letters 'H'. Perfect, medium, and bad squares are represented with white, yellow, and black colors respectively.