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A Tale of Two Subreddits: Change and Continuity in Teaching-Related Online Spaces
Despite social media’s ubiquity in modern life, research on some topics related to educators’ use of social media for professional …
K. Bret Staudt Willet
,
Jeffrey P. Carpenter
Preprint
Project
DOI
Balancing Privacy and Open Science in the Context of COVID-19: A Response to Ifenthaler & Schumacher (2016)
Privacy and confidentiality are core considerations in education. At the same time, using and sharing data—and, more broadly, open …
Joshua M. Rosenberg
,
K. Bret Staudt Willet
Preprint
Project
Teachers on Reddit? Exploring Contributions and Interactions in Four Teaching-Related Subreddits
The website Reddit has received scant attention from educational researchers despite the presence of numerous teaching-related …
K. Bret Staudt Willet
,
Jeffrey P. Carpenter
Project
Source Document
DOI
Identifying Multiple Learning Spaces Within a Single Teacher-Focused Twitter Hashtag
The existing work on teacher-focused Twitter hashtags typically frames each hashtag as a single, unified phenomenon, thereby collapsing …
Spencer P. Greenhalgh
,
Joshua M. Rosenberg
,
K. Bret Staudt Willet
,
Matthew J. Koehler
,
Mete Akcaoglu
Project
DOI
Spam and Educators’ Twitter Use: Methodological Considerations and Challenges
Twitter and other social media have assumed important places in many educators’ professional lives by hosting spaces where new kinds of …
Jeffrey P. Carpenter
,
K. Bret Staudt Willet
,
Matthew J. Koehler
,
Spencer P. Greenhalgh
Project
DOI
Approaches to Mormon Identity and Practice in the #ldsconf Twitter Hashtag
In this paper, we document different expressions of Mormon identity and different approaches to Mormon practice within the #ldsconf …
Spencer P. Greenhalgh
,
K. Bret Staudt Willet
,
Matthew J. Koehler
Project
DOI
Foregrounding Technoethics: Toward Critical Perspectives in Technology and Teacher Education
The Teacher Educator Technology Competencies (TETCs) were created to help all teacher educators support teacher candidates as they …
Daniel G. Krutka
,
Marie K. Heath
,
K. Bret Staudt Willet
Source Document
What Should Be the Role of Social Media in Education?
Social media provide new opportunities for when, how, where, and with whom people learn—venue unimaginable 15 years ago. Today’s …
Christine Greenhow
,
Sarah M. Galvin
,
K. Bret Staudt Willet
DOI
Revisiting How and Why Educators Use Twitter: Tweet Types and Purposes in #Edchat
This study revisits Carpenter and Krutka’s (2014) survey of how and why educators use Twitter, through exploring one of Twitter’s …
K. Bret Staudt Willet
Project
DOI
Educators on the Front Page of the Internet: Education-Related Subreddits as Learning Spaces
Accessing and contributing to online spaces have become regular parts of many educators’ professional activities. However, the popular …
K. Bret Staudt Willet
,
Jeffrey P. Carpenter
PDF
Code
Dataset
Source Document
Communities and Clusters: User Interactions in an Online Discussion Forum for Computer Science Education
Prior research has shown that teacher communities are important for social, cognitive, and affective growth of teachers. Further, …
Sukanya Kannan Moudgalya
,
K. Bret Staudt Willet
PDF
Code
Source Document
Spam, Spam, Spam, Spam: Methodological Considerations and Challenges for Studying Educators’ Twitter Use
Social media have come to play an important role in the professional lives of many educators. Platforms such as Twitter create new …
Jeffrey P. Carpenter
,
Matthew J. Koehler
,
K. Bret Staudt Willet
,
Spencer P. Greenhalgh
PDF
Source Document
Social Scholarship Revisited: Changing Scholarly Practices in the Age of Social Media
This conceptual exploration revisits a key question from earlier work (Greenhow & Gleason, 2014): What is scholarship reconsidered …
Christine M. Greenhow
,
Benjamin Gleason
,
K. Bret Staudt Willet
Project
Source Document
DOI
Tweet, and We Shall Find: Using Digital Methods to Locate Participants in Educational Hashtags
Although researchers have discovered a great deal about
who
uses Twitter for educational purposes,
what
they post about,
when
they post …
Spencer P. Greenhalgh
,
K. Bret Staudt Willet
,
Joshua M. Rosenberg
,
Matthew J. Koehler
Project
Source Document
DOI
Timing is Everything: Comparing Synchronous and Asynchronous Modes of Twitter for Teacher Professional Learning
Spencer P. Greenhalgh
,
K. Bret Staudt Willet
,
Joshua M. Rosenberg
,
Mete Akcaoglu
,
Matthew J. Koehler
PDF
The Career You Save May Be Your Own: Exploring the mathtwitterblogosphere as a Community of Practice
Teachers face numerous challenges: feeling isolated in their classrooms, overwhelmed by the demands of the job, and disappointed in …
K. Bret Staudt Willet
,
Paul N. Reimer
PDF
Project
Source Document
Back to the Gaming Board: Understanding Games and Education through Board Game Reviews
Recent academic research into the use of games for educational purposes has focused almost exclusively on
video
games. In this study, …
K. Bret Staudt Willet
,
Sukanya Kannan Moudgalya
,
Liz Owens Boltz
,
Spencer P. Greenhalgh
,
Matthew J. Koehler
PDF
Source Document
Educating Social Scholars: Examining Novice Researchers' Practices with Social Media
Recent articles in the educational research field have called for a stronger research focus on students’ learning with everyday …
Christine M. Greenhow
,
Benjamin Gleason
,
Holly Marich
,
K. Bret Staudt Willet
PDF
Project
Source Document
A Tweet by Any Other Frame: Three Approaches to Studying Educator Interactions on Twitter
Every day, thousands of educators use Twitter to interact with each other. Yet, the research that examines these educator interactions …
K. Bret Staudt Willet
,
Matthew J. Koehler
,
Spencer P. Greenhalgh
PDF
Project
Source Document
A Tale of Two Twitters: Synchronous and Asynchronous Use of the Same Hashtag
Communication in online settings can occur at the same time (synchronously) or at different times (asynchronously). A new form of …
Joshua M. Rosenberg
,
Mete Akcaoglu
,
K. Bret Staudt Willet
,
Spencer P. Greenhalgh
,
Matthew J. Koehler
PDF
Source Document
Cite
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the number of times to cycle through the training set, useful for small datasetsthe number of images to read in each batchthe base learning rate of the default Adam rate schedulernumber of epochesthe number of initial epochs during which the sizes of the 5 boxes in each cell is forced to match the sizes of the 5 anchors


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These white whales are born dark gray. It can take up to eight years before they turn completely white.
Commonly referred to as the beluga, melonhead, or sea canary due to its high-pitched twitter. They are gregarious and form groups of up to 10 animals on average, although during the summer, they can gather in the hundreds or even thousands in estuaries and shallow coastal areas. They are slow swimmers, but can dive to 700 m (2,300 ft) below the surface.


HALLS CHERRY- menthol lozenge
Mondelez Global LLC
Disclaimer: Most OTC drugs are not reviewed and approved by FDA, however they may be marketed if they comply with applicable regulations and policies. FDA has not evaluated whether this product complies.
Active Ingredient
<(in each drop) Menthol 5.8 mg>
Purposes
<Cough Suppressant, Oral Anesthetic>
Uses
<temporarily relieves:>
- cough due to a cold
- occasional minor irritation or sore throat
Warnings
<Sore throat warning: if sore throat is severe, persists for more than 2 days, is accompanied by fever, headache, rash, swelling, or vomiting, consult a doctor promptly. These may be serious.>
Ask a doctor before use if you have
- persistent or chronic cough such as occurs with smoking, asthma, or emphysema
- cough accompanied by excessive phlegm (mucus)
Stop use and ask a doctor if
- cough persists for more than 1 week, tends to recur, or is accompanied by fever, rash, or persistent headache. These could be signs of a serious condition.
- sore mouth does not improve in 7 days
- irritation, pain, or redness persists or worsens
Keep out of reach of children
Enter section text here
Directions
- adults and children 5 years and over: dissolve 1 drop slowly in the mouth. Repeat every 2 hours as needed.
- children under 5 years: ask a doctor
Other information
- 12 calories per drop
- Contains: SOY
Inactive ingredients
<eucalyptus oil, FDC blue 2, FDC red 40, flavors, glucose syrup, soy lecithin, sucralose, sucrose, water>
Questions
<Call 1-800-524-2854, Monday to Friday, 9AM-6PM Eastern Time or visit our website at www.gethalls.com>
OTC Principal Display Panel
<NDC 12546-109-30 Halls Cherry 30 drops bag


A toddler moment: Last night as I got into bed, I shifted my pillow over. Underneath it was a sippy cup I'd been searching for.
12/29/2009, 8:59:51 AM
Favs: 0
Retweets: 0
link--- layout: tweet ---
A toddler moment: Last night as I got into bed, I shifted my pillow over. Underneath it was a sippy cup I'd been searching for.
12/29/2009, 8:59:51 AM
Favs: 0
Retweets: 0
link


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Windows 10
Home version supports updates, but no upgrades(e.g. 1703).
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Windows 8.1
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Windows 8
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Windows 7
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Windows Server
2012(R2)/2008(R2)/2016
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Welcome to SJTU Windows Server Update Services
-
Automatic Updates overview
With Automatic Updates, Windows routinely checks for updates that can help protect your computer against the latest viruses and other security threats. These high-priority updates are available through the Windows Update Web site and include security updates, critical updates, or service packs.
When you turn on Automatic Updates, you don't have to search for updates online or worry that critical fixes might be missing. Windows automatically downloads and installs them for you, using a schedule that you determine. If you prefer to download and install updates yourself, you can also set up Automatic Updates to notify you whenever any high-priority updates become available.
-
Why install.
Your system may update from campus local server. A lot of internet traffic will be saved. As all patch file signature are examined before installation, this service should not involve additional security problem.
-
Can REMOVE
You can uninstall these service whenever you like.
|


{{sectionTitleLong[0]}}
A Fourier Series is an expansion of any reasonable (see Dirichlet conditions) periodic function
$ f(x) $ in terms of an infinite sums of sines and cosines,
due to the orthogonality relationships of sines and cosines functions.
Any set of functions that form a complete orthogonal system has a Fourier Series counterpart.
By representing functions in sines and cosines, we have a wide range of tools (differentiation and integration) for analysis that might not be possible in the original function domain.
In this visualisation, we aim to provide a visual approach to understanding how some important periodic functions can be composed using sine and cosine functions.
Try changing the number of terms in the Fourier Series. By increasing this, we can see that the function resembles closer and closer to the square function.
By the end of this suite, you would be able to understand the intuition behind Fourier Series, and to do it yourself for any reasonable periodic function.
By representing functions in sines and cosines, we have a wide range of tools (differentiation and integration) for analysis that might not be possible in the original function domain.
In this visualisation, we aim to provide a visual approach to understanding how some important periodic functions can be composed using sine and cosine functions.
Visualisation Guide
On the right, we can see a square function, and a summation of sines and cosine functions with a finite number of terms that represents the square function.Try changing the number of terms in the Fourier Series. By increasing this, we can see that the function resembles closer and closer to the square function.
By the end of this suite, you would be able to understand the intuition behind Fourier Series, and to do it yourself for any reasonable periodic function.
{{sectionTitleLong[1]}}
Given the vectors $ \vec{V}_{n} $ and $ \vec{V}_{m}, $ we define their
inner product as
$$\vec{V}_{n} \cdot \vec{V}_{m} = \sum_{i}V_{n}^{(i)} V_{m}^{(i)}, $$
where $ V_{n}^{(i)} $ is the $ i^{\text{th}} $
component of the vector
$ \vec{V}_{n}. $ If
$$ \vec{V}_{n} \cdot \vec{V}_{m} = 0 \text{ for } m \neq n, $$
then we say the vectors are orthogonal.
If you move in the direction of any vector $ \vec{a}, $ you do not change the position component in the direction of any vector orthogonal to $ \vec{a}. $ In other words, orthogonal vectors are perpendicular.
If $ \vec{V}_{n} \cdot \vec{V}_{m} = \delta_{nm} $ then the vectors are said to be orthonormal. $$ \delta_{nm} = \left\{\begin{matrix}0 \quad \text{if } m\neq n \\ 1 \quad \text{if } m = n \end{matrix}\right. $$ This mean the vectors are orthogonal and conveniently normalised.
Use the visualisation to see that when the basis vectors are not parallel they span $ \mathbb{R}^{2}, $ using these as basis vectors, we can construct any other vector in $ \mathbb{R}^{2}. $
If you can represent any vector in $ \mathbb{R}^{n} $ using $ n $ vectors, then the basis vectors form a complete set. If the set of these vectors are all orthonormal, then we have a complete orthonormal set.
In the same way as before, we will now demonstrate what it means for functions to be complete. We define the functions $ f_{n},f_{m} $ to exist on the domain $ \left[a,b \right]. $
We define their inner product to be $$ \langle f_{n} , f_{m} \rangle = \int_{a}^{b} f_{n}\left(x \right)f_{m}^{*}\left(x \right)dx $$ , where * denotes the complex conjugate of the function. This is commutative, just like the inner product of two vectors. We say the functions are orthogonal if $ \langle f_{n} , f_{m} \rangle = 0 $ for $ m \neq n. $
We then say the functions are orthonormal if $ \langle f_{n} , f_{m} \rangle = \delta_{nm}, $ where $$ \delta_{nm} = \left\{\begin{matrix}0 \quad \text{if } m\neq n \\ 1 \quad \text{if } m = n \end{matrix}\right. $$ We now claim that you can construct any square integrable function from a complete set of basis functions, in the same way vectors could be constructed from a complete set of basis vectors. A square integrable function $ f $ is a function whose inner product with itself, $ \langle f,f \rangle $ is finite.
If you move in the direction of any vector $ \vec{a}, $ you do not change the position component in the direction of any vector orthogonal to $ \vec{a}. $ In other words, orthogonal vectors are perpendicular.
If $ \vec{V}_{n} \cdot \vec{V}_{m} = \delta_{nm} $ then the vectors are said to be orthonormal. $$ \delta_{nm} = \left\{\begin{matrix}0 \quad \text{if } m\neq n \\ 1 \quad \text{if } m = n \end{matrix}\right. $$ This mean the vectors are orthogonal and conveniently normalised.
Use the visualisation to see that when the basis vectors are not parallel they span $ \mathbb{R}^{2}, $ using these as basis vectors, we can construct any other vector in $ \mathbb{R}^{2}. $
If you can represent any vector in $ \mathbb{R}^{n} $ using $ n $ vectors, then the basis vectors form a complete set. If the set of these vectors are all orthonormal, then we have a complete orthonormal set.
In the same way as before, we will now demonstrate what it means for functions to be complete. We define the functions $ f_{n},f_{m} $ to exist on the domain $ \left[a,b \right]. $
We define their inner product to be $$ \langle f_{n} , f_{m} \rangle = \int_{a}^{b} f_{n}\left(x \right)f_{m}^{*}\left(x \right)dx $$ , where * denotes the complex conjugate of the function. This is commutative, just like the inner product of two vectors. We say the functions are orthogonal if $ \langle f_{n} , f_{m} \rangle = 0 $ for $ m \neq n. $
We then say the functions are orthonormal if $ \langle f_{n} , f_{m} \rangle = \delta_{nm}, $ where $$ \delta_{nm} = \left\{\begin{matrix}0 \quad \text{if } m\neq n \\ 1 \quad \text{if } m = n \end{matrix}\right. $$ We now claim that you can construct any square integrable function from a complete set of basis functions, in the same way vectors could be constructed from a complete set of basis vectors. A square integrable function $ f $ is a function whose inner product with itself, $ \langle f,f \rangle $ is finite.
{{sectionTitleLong[2]}}
Consider a function $ f(x) $ which is periodic over
the domain $ -L \leq x < L $
where $ L $ is positive. $ f(x) $
can be represented as a sum
of
sines and cosines with varying
frequencies.
These trigonometric functions form orthogonal basis.
The Fourier series of $ f(x) $ is $$ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \Big[ a_n\cos\Big(\frac{\pi nx}{L}\Big) + b_n\sin\Big(\frac{\pi nx}{L}\Big) \Big] $$ where \begin{align} a_n & = \frac{1}{L}\int_{-L}^{L} f(x) \cos\Big(\frac{\pi nx}{L}\Big) dx, \\ b_n & = \frac{1}{L}\int_{-L}^{L} f(x) \sin\Big(\frac{\pi nx}{L}\Big) dx, \\ \end{align} and $$ n=0,1,2... $$ Note: $ {a_0}/2 $ represents the average value of $ f(x). $
The Fourier series representation of the function $ f(x) $ in exponential (complex) form is $$ f(x) =\sum_{n=-\infty}^{\infty} c_n e^{{in \pi x}/L} \qquad $$ where $$ c_n = \frac{1}{2L}\int_{-L}^{L} f(x) e^{{-in \pi x}/L} dx, $$ Note: The constants in front of the integral(s) of the coefficient for the trigonometric and exponential fourier series are different. Also note the limit of the summations. Follow through the derivation to understand the difference.
These trigonometric functions form orthogonal basis.
The Fourier series of $ f(x) $ is $$ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \Big[ a_n\cos\Big(\frac{\pi nx}{L}\Big) + b_n\sin\Big(\frac{\pi nx}{L}\Big) \Big] $$ where \begin{align} a_n & = \frac{1}{L}\int_{-L}^{L} f(x) \cos\Big(\frac{\pi nx}{L}\Big) dx, \\ b_n & = \frac{1}{L}\int_{-L}^{L} f(x) \sin\Big(\frac{\pi nx}{L}\Big) dx, \\ \end{align} and $$ n=0,1,2... $$ Note: $ {a_0}/2 $ represents the average value of $ f(x). $
The Fourier series representation of the function $ f(x) $ in exponential (complex) form is $$ f(x) =\sum_{n=-\infty}^{\infty} c_n e^{{in \pi x}/L} \qquad $$ where $$ c_n = \frac{1}{2L}\int_{-L}^{L} f(x) e^{{-in \pi x}/L} dx, $$ Note: The constants in front of the integral(s) of the coefficient for the trigonometric and exponential fourier series are different. Also note the limit of the summations. Follow through the derivation to understand the difference.
We begin by assuming that $ f(x) $ can be represented
as a sum of
exponential terms.
$$ f(x) = \sum_{n=-\infty}^{\infty} c_n e^{{in \pi x}/L} $$
Multiplying by $ e^{{-im \pi x}/L}, $ where $ m=0,1,2,..., $
and integrating both sides gives the following:
$$ \int_{-L}^{L} f(x) e^{{-im \pi x}/L} dx = \sum_{n=-\infty}^{\infty} c_n \int_{-L}^{L} e^{{i(n-m) \pi x}/L} dx. $$
Performing the integral on the R.H.S gives
$$ \int_{-L}^L e^{(i\pi x/L)(n-m)}dx =\left\{ \begin{array}{ll} 2L \qquad \text{(if n=m)} \\
0 \qquad \quad \text{(if n≠m)} \end{array} \right \}=2L\cdot\delta_{nm}. $$
This is the orthonormality of complex exponentials.
Now since the only non-zero value of the sum is when $ n=m, $ $$ \int_{-L}^L f(x) e^{-im\pi x/L}dx = 2L \sum_{n=-\infty}^{+\infty}c_n\delta_{mn} = 2L\cdot c_m, $$ hence $$ c_m=\frac{1}{2L}\int_{-L}^{+L}f(x) e^{-im\pi x/L}dx. $$ Note: As $ m $ is an arbitrary integer, it can be assigned as $ n $ for simplicity.
Given the previously derived equation, we can derive the trigonometric form of the Fourier series. By reorganising the sum, we get $$ f(x)= \underbrace{c_0}_\text{n=0} + \sum_{n=1}^{\infty} \big( \underbrace{c_n e^{in\pi x/L}}_\text{Positive n term} + \underbrace{c_{-n} e^{-in\pi x/L}}_\text{Negative n terms} \big). $$ We can regroup the terms in the summation in the following way:
\begin{align}
f(x)= c_0 + \sum_{n=1}^{\infty} \Big[ & \frac{c_n + c_{-n}}{2} \big(e^{in\pi x/L} +e^{-in\pi x/L} \big) \\
& \frac{c_n - c_{-n}}{2} \big(e^{in\pi x/L} -e^{-in\pi x/L} \big) \Big]
\end{align}
Now define $ a_n = c_n + c_{-n} $ and
$ b_n = i(c_n - c_{-n}) $
and use that these exponentials can we written as
sines and cosines.
Note: $ a_0 = 2c_0 $ and $ b_0=0. $
Hence $$ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \Big[ a_n\cos\Big(\frac{\pi nx}{L}\Big) + b_n\sin\Big(\frac{\pi nx}{L}\Big) \Big] $$ $ a_n $ can be derived by using the expression from $ c_n $ \begin{align} a_n = c_n + c_{-n} = & \frac{1}{2L}\int_{-L}^{+L}f(x) e^{-in\pi x/L}dx \\ + & \frac{1}{2L} \int_{-L}^{+L}f(x) e^{in\pi x/L}dx \end{align} Factorise and spot the exponential form of $ \cos. $ \begin{align} & =\frac{1}{L}\int_{-L}^{+L}f(x) \Big[\frac{e^{in\pi x/L} +e^{-in\pi x/L}}{2}\Big]dx \\ & =\frac{1}{L}\int_{-L}^{L} f(x) \cos\Big(\frac{\pi nx}{L}\Big) dx \end{align} Similarly for $ b_n, $ $$ b_n =\frac{1}{L}\int_{-L}^{L} f(x) \sin\Big(\frac{\pi nx}{L}\Big) dx. $$
Now since the only non-zero value of the sum is when $ n=m, $ $$ \int_{-L}^L f(x) e^{-im\pi x/L}dx = 2L \sum_{n=-\infty}^{+\infty}c_n\delta_{mn} = 2L\cdot c_m, $$ hence $$ c_m=\frac{1}{2L}\int_{-L}^{+L}f(x) e^{-im\pi x/L}dx. $$ Note: As $ m $ is an arbitrary integer, it can be assigned as $ n $ for simplicity.
Given the previously derived equation, we can derive the trigonometric form of the Fourier series. By reorganising the sum, we get $$ f(x)= \underbrace{c_0}_\text{n=0} + \sum_{n=1}^{\infty} \big( \underbrace{c_n e^{in\pi x/L}}_\text{Positive n term} + \underbrace{c_{-n} e^{-in\pi x/L}}_\text{Negative n terms} \big). $$ We can regroup the terms in the summation in the following way:
\begin{align}
= \frac{1}{2} \Big[ 2c_n e^{in\pi x/L} + 2c_{-n} e^{-in\pi x/L} & \\
+ \underbrace{c_{-n} e^{in\pi x/L} - c_{-n} e^{in\pi x/L}}_\text{ =0} & \\
+ \underbrace{c_{n} e^{-in\pi x/L} -c_{n} e^{-in\pi x/L}}_\text{ =0} & \Big] \\
\end{align}
\begin{align} = \frac{1}{2} \Big[ c_n e^{in\pi x/L} +c_{-n} e^{in\pi x/L} & \\ + c_n e^{-in\pi x/L} + c_{-n} e^{-in\pi x/L} & \\ + c_n e^{in\pi x/L} - c_{-n} e^{in\pi x/L} & \\ - c_n e^{-in\pi x/L} + c_{-n} e^{-in\pi x/L} & \Big] \\ \end{align}
\begin{align} = \frac{1}{2} \Big[ c_n e^{in\pi x/L} +c_{-n} e^{in\pi x/L} & \\ + c_n e^{-in\pi x/L} + c_{-n} e^{-in\pi x/L} & \\ + c_n e^{in\pi x/L} - c_{-n} e^{in\pi x/L} & \\ - c_n e^{-in\pi x/L} + c_{-n} e^{-in\pi x/L} & \Big] \\ \end{align}
Note: $ a_0 = 2c_0 $ and $ b_0=0. $
Hence $$ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \Big[ a_n\cos\Big(\frac{\pi nx}{L}\Big) + b_n\sin\Big(\frac{\pi nx}{L}\Big) \Big] $$ $ a_n $ can be derived by using the expression from $ c_n $ \begin{align} a_n = c_n + c_{-n} = & \frac{1}{2L}\int_{-L}^{+L}f(x) e^{-in\pi x/L}dx \\ + & \frac{1}{2L} \int_{-L}^{+L}f(x) e^{in\pi x/L}dx \end{align} Factorise and spot the exponential form of $ \cos. $ \begin{align} & =\frac{1}{L}\int_{-L}^{+L}f(x) \Big[\frac{e^{in\pi x/L} +e^{-in\pi x/L}}{2}\Big]dx \\ & =\frac{1}{L}\int_{-L}^{L} f(x) \cos\Big(\frac{\pi nx}{L}\Big) dx \end{align} Similarly for $ b_n, $ $$ b_n =\frac{1}{L}\int_{-L}^{L} f(x) \sin\Big(\frac{\pi nx}{L}\Big) dx. $$
The triangular function is defined by the following:
$$ f(x)=\left\{
\begin{array}{ll}
\frac{-2A}{L} (x+L) \quad (-L \leq x < -L/2) \\
\frac{2A}{L} x \qquad \qquad (-L/2\leq x < L/2) \\
\frac{-2A}{L} (x-L) \quad (L/2\leq x < L)
\end{array}
\right. $$
First notice that $ f(x) $ is odd about $ x=0, $ hence $ a_n=0 $
for all $ n. $ Also, since the $ b_n $ integrand is even about
$ x=0, $
$$ b_n=\frac{2}{L}\int_0^L f(x) \sin \left(\frac{n \pi x}{L}\right)dx. $$
Notice that for even $ n, $ $ b_n $ is odd about
$ x=L/2. $
Hence,
$$ b_n=\frac{2}{L}\int_0^{L} f(x) \sin \left(\frac{n\pi x}{L} \right) dx = 0 \quad \text{for even n}. $$
The integral from $ 0 $ to $ L/2 $ is equal in magnitude but
opposite in sign to the integral from $ L/2 $ to
$ L, $ for
even $ n. $
For odd $ n, $ the function is even around $ L/2, $ so we only have to integrate from $ 0 $ to $ L/2 $ then multiply by $ 2. $ This has made our job much easier. $$ b_n=2\frac{2}{L}\int_0^{L/2} f(x) \sin \left(\frac{n\pi x}{L} \right) dx \quad \text{for odd n} $$ Play around with the visualisation to the right to convince yourself this is true.
For odd $ n, $ the function is even around $ L/2, $ so we only have to integrate from $ 0 $ to $ L/2 $ then multiply by $ 2. $ This has made our job much easier. $$ b_n=2\frac{2}{L}\int_0^{L/2} f(x) \sin \left(\frac{n\pi x}{L} \right) dx \quad \text{for odd n} $$ Play around with the visualisation to the right to convince yourself this is true.
$$ \text{For 0 < x < L/2:} \quad f(x)=\frac{2Ax}{L} $$
Remembering that for even $ n, $ $ b_n=0, $ therefore only
consider odd $ n. $
$$ b_n = \frac{4}{L} \frac{2A}{L}\int_0^{L/2} x \sin \left(\frac{n\pi x}{L} \right) dx $$
This integral can be performed by parts.
$$ b_n=\frac{8A}{L^2} \frac{L^2}{(n\pi)^2} \sin \left(\frac{n\pi x}{L} \right) \biggr |_0^{L/2} $$
Note for odd $ n $:
$$ \sin \left(\frac{n\pi}{2} \right)=(-1)^{\frac{n-1}{2}} $$
Therefore,
$$ b_n=\frac{8A}{n^2\pi^2}(-1)^{\frac{n-1}{2}} $$
and,
$$ f(x)=\sum_{n=1,3,5...}^\infty \frac{8A}{n^2\pi^2} (-1)^{\frac{n-1}{2}} \sin \left(\frac{n\pi x}{L} \right) $$
Finally, using the fact that odd number can be written as $ 2n +1 $ for any integer
$ n, $ we have
$$ f(x)=\sum_{n=1,2,3...}^\infty \frac{8A}{(2n-1)^2\pi^2}(-1)^n \sin \left(\frac{(2n-1)\pi x}{L} \right) $$
$$ u=x \qquad dv= \sin \left(\frac{n\pi x}{L} \right)dx $$
$$ du=dx \qquad v=\frac{-L}{n\pi} \cos \left(\frac{n\pi x}{L} \right) $$
\begin{align}
b_n=\frac{8A}{L^2} \Bigg[ \Big[ \frac{-xL}{n\pi} & \cos \Big(\frac{n\pi x}{L}\Big) \Big] \Bigr|_0^{L/2} \\
-\int_0^{L/2}\frac{-L}{n\pi} & \cos \Big(\frac{n\pi x}{L} \Big)dx \Bigg]
\end{align}
For odd n,
$$ \cos \left(\frac{n\pi}{2} \right)=0 $$
Therefore,
$$ b_n=\frac{8A}{L^2} \Bigg[0+ \int_0^{L/2}\frac{L}{n\pi} \cos \left(\frac{n\pi x}{L}
\right)dx \Bigg] $$
The parabolic function is defined as:
$$ f(x)=Ax^2 \quad \text{for:} -L\leq x< L $$
and is forced to be periodic. $ A $ is a constant.
Therefore, the coefficients $ a_n $ and $ b_n $
are given by
$$ b_n = \frac{1}{L}\int_{-L}^{+L} Ax^2 \sin \left(\frac{n \pi x}{L} \right)dx $$
$$ a_n= \frac{1}{L}\int_{-L}^{+L} Ax^2 \cos \left(\frac{n \pi x}{L} \right)dx $$
The $ b_n $ integral is odd about $ x=0, $ hence
$ b_n=0 $ for all $ n. $ $ f(x) $ is
even about
$ x=0. $ Thus,
$$ a_n=\frac{2}{L}\int_{0}^{+L} Ax^2 \cos \left(\frac{n \pi x}{L} \right)dx. $$
For $ n=0 $:
$$ a_0 = \frac{2}{L} \int _{0}^{+L} Ax^2 dx = \frac{2AL^2}{3} $$
For $ n\neq 0 $:
$$ a_n = \frac{2A}{L}\int_0^L x^2 \cos \left(\frac{n\pi x}{L} \right)dx $$
This equation can now be solved using integration by parts.
$$ a_n = -\frac{4A}{n\pi}\int_{0}^{L} x \sin \left(\frac{n \pi x}{L} \right)dx $$
$$ a_n = \frac{4A}{n\pi} \Bigg[\frac{xL}{n\pi} \cos \left(\frac{n\pi x}{L} \right) \Bigg] \biggr |_0^L $$
$$ = \frac{4AL^2}{n^2\pi^2}(-1)^n $$
Therefore, the fourier series can be written as
$$ f(x) = \frac{AL^2}{3} + \sum_{n=1}^\infty \frac{4AL^2}{n^2\pi^2} (-1)^n \cos \left(\frac{n\pi x}{L} \right) $$
$$ du = 2x\cdot dx \qquad v=\frac{L}{n\pi} \sin \left(\frac{n\pi x}{L} \right) $$
$$ u=x^2 \qquad dv= \cos \left(\frac{n \pi x}{L} \right)dx $$
\begin{align}
a_n = \frac{2A}{L} \Bigg[ \Big[\frac{Lx^2}{n\pi} & \sin \left(\frac{n\pi x}{L} \right) \Big] \biggr |_0^L \\
-\int_0^L\frac{2xL}{n\pi} & \sin \left(\frac{n \pi x}{L} \right)dx \Bigg]
\end{align}
$$ = \frac{2A}{L} \Bigg[0 -\int_0^L\frac{2xL}{n\pi} \sin\left(\frac{n \pi x}{L}\right)dx \Bigg] $$
$$ u = x \qquad dv = \sin \left(\frac{n\pi x}{L} \right) dx $$
$$ du = dx \qquad v = -\frac{L}{n\pi} \cos \left(\frac{n\pi x}{L} \right) $$
\begin{align}
a_n = \frac{-4A}{n\pi} \Bigg[\Big[-\frac{xL}{n\pi} & \cos \left(\frac{n\pi x}{L} \right)\Big] \biggr |_0^L \\
- \int_0^L\frac{-L}{n\pi} & \cos \left(\frac{n\pi x}{L} \right)dx \Bigg]
\end{align}
The Dirac-delta function can be thought of as a infinite spike at $ x=0. $
$$
f(x) = \delta(x)=\left\{
\begin{array}{ll}
+\infty \quad (x=0) \\
0 \qquad (x \neq 0)
\end{array}
\right.
$$
$$ a_n=\frac{1}{L}\int_{-L}^{+L} \delta (x) \cos \left(\frac{n\pi x}{L} \right)dx $$
$$ b_n=\frac{1}{L}\int_{-L}^{+L}\delta(x) \sin \left(\frac{n\pi x}{L} \right)dx $$
$ \delta(x) $ is an even function. Hence $ b_n=0 $ and
$$ a_n=\frac{1}{L} $$
Therefore, the fourier series for positive $ L $ can
be written as
$$ f(x)=\frac{1}{2L}+\sum_{n=1}^\infty \frac{1}{L} \cos \left(\frac{n\pi x}{L} \right). $$
By considering the shifting property of the Dirac-Delta function,
$$ \int_{\alpha-\epsilon}^{\alpha+\epsilon} g(x) \delta(x-\alpha)dx=g(\alpha) $$
For us, $ \alpha=0, $ $ \epsilon=L, $ and $ g(x)= \sin \left(n \pi x/L \right). $
Therefore, $$ a_n=\frac{1}{L}\cos \left(\frac{n\pi \times 0}{L} \right)=\frac{1}{L} $$ $$ b_n=\frac{1}{L}\sin \left(\frac{n\pi \times 0}{L} \right)=0 $$
Therefore, $$ a_n=\frac{1}{L}\cos \left(\frac{n\pi \times 0}{L} \right)=\frac{1}{L} $$ $$ b_n=\frac{1}{L}\sin \left(\frac{n\pi \times 0}{L} \right)=0 $$
The square wave is defined as the following:
$$
f(x)=\left\{
\begin{array}{ll}
-A \quad (-L \leq x < 0) \\
A \quad (0\leq x < L)
\end{array}
\right.
$$
The $ a_{n} $ terms can be found be noting that $ f(x) $ is
odd,
so $ a_n $ is $ 0 $ for all $ n. $
To find the $ b_n $ terms, the integration can be split into two parts: $ -L\leq x <0 $ and $ 0\leq x < L. $
This gives the following integration: \begin{align} b_n & =\frac{1}{L}\int_{-L}^0 -A \sin \left(\frac{n\pi x}{L} \right)dx \\ & + \frac{1}{L}\int_0^L A \sin \left(\frac{n \pi x}{L} \right)dx \end{align} which can be evaluated to: $$ b_n=\frac{-2A}{L}\frac{L}{n\pi} \cos \left(\frac{n\pi x}{L} \right)\biggr |_0^L =\frac{2A}{n\pi}\left[1-\cos\left(n\pi \right) \right] $$ Note that $$ \cos \left(n \pi\right)=(-1)^n $$ Hence $$ b_n=\frac{2A}{n\pi}(1-(-1)^n) $$ Therefore the final fourier series is: $$ f(x)=\sum_{n=1}^{\infty} \frac{2A}{n\pi} (1-(-1)^n)\sin \left(\frac{n\pi x}{L}\right). $$
To find the $ b_n $ terms, the integration can be split into two parts: $ -L\leq x <0 $ and $ 0\leq x < L. $
This gives the following integration: \begin{align} b_n & =\frac{1}{L}\int_{-L}^0 -A \sin \left(\frac{n\pi x}{L} \right)dx \\ & + \frac{1}{L}\int_0^L A \sin \left(\frac{n \pi x}{L} \right)dx \end{align} which can be evaluated to: $$ b_n=\frac{-2A}{L}\frac{L}{n\pi} \cos \left(\frac{n\pi x}{L} \right)\biggr |_0^L =\frac{2A}{n\pi}\left[1-\cos\left(n\pi \right) \right] $$ Note that $$ \cos \left(n \pi\right)=(-1)^n $$ Hence $$ b_n=\frac{2A}{n\pi}(1-(-1)^n) $$ Therefore the final fourier series is: $$ f(x)=\sum_{n=1}^{\infty} \frac{2A}{n\pi} (1-(-1)^n)\sin \left(\frac{n\pi x}{L}\right). $$
The Sawtooth wave is defined as the following:
$$ f(x)=\frac{Ax}{L} \quad (-L \leq x < L). $$
$ f(x) $ is odd, so $ a_n= 0 $
for all $ n. $
The $ b_n $ terms are given by the following integration: \begin{align} b_n & =\frac{1}{L}\int_{-L}^{L} \frac{Ax}{L}\sin \left(\frac{n\pi x}{L} \right)dx \\ & =\frac{2A}{L^{2}}\int_{0}^{L} x \sin \left(\frac{n\pi x}{L} \right)dx. \end{align} Using Integration by parts it can be shown that:
$$ b_n=\frac{2A}{n\pi}(-1)^{n+1}. $$
Therefore the final fourier series is:
$$ f(x)=\sum_{n=1}^\infty \frac{2A}{n\pi}(-1)^{n+1} \sin \left(\frac{n\pi x}{L} \right). $$
The $ b_n $ terms are given by the following integration: \begin{align} b_n & =\frac{1}{L}\int_{-L}^{L} \frac{Ax}{L}\sin \left(\frac{n\pi x}{L} \right)dx \\ & =\frac{2A}{L^{2}}\int_{0}^{L} x \sin \left(\frac{n\pi x}{L} \right)dx. \end{align} Using Integration by parts it can be shown that:
$$ u=x \quad dv=\sin \left(\frac{n\pi x}{L} \right)dx $$
$$ du=dx \quad v=\frac{-L}{n\pi}\cos \left(\frac{n\pi x}{L} \right) $$
\begin{align}
b_n=\frac{2A}{L^{2}} \Bigg[ \frac{-Lx}{n\pi} & \cos \left(\frac{n\pi x}{L} \right)\biggr |_0^L \\
-\int_0^L\frac{-L}{n\pi} & \cos \left(\frac{n\pi x}{L} \right)dx \Bigg]
\end{align}
Note that the second integrand is zero as it is of cosine over a full
period.
$$ b_n=-\frac{2A}{n\pi} \cos \left(n\pi \right)=-\frac{2A}{n\pi}(-1)^n $$
The $ |x| $ function is defined as the following:
$$ f(x)=A|x| \quad (-L \leq x < L). $$
$$ a_n = \frac{1}{L}\int_{-L}^{+L} A|x| \cos \left(\frac{n \pi x}{L} \right)dx $$
$$ b_n = \frac{1}{L}\int_{-L}^{+L} A|x| \sin \left( \frac{n\pi x}{L}\right) dx $$
$ f(x) $ is even about $ x=0. $
Hence $ b_n=0. $
The $ a_n $ integral is even about $ x=0. $ Hence,
$$ a_n = \frac{2}{L}\int_{0}^{+L} Ax \cos \left(\frac{n \pi x}{L}\right)dx. $$
For $ n=0 $
$$ a_0 = \frac{2A}{L}\int_0^L x dx = \frac{2A}{L}\frac{x^2}{2} \biggr |_0^L = AL. $$
For $ n \neq 0 $:
$$ a_n = \frac{2A}{L}\int_{0}^{+L} x \cos \left(\frac{n \pi x}{L} \right)dx. $$
By integrating by parts,
$$ a_n = \frac{2A}{L}\frac{L^2}{(n\pi)^2} \cos \left( \frac{n\pi x}{L} \right)\biggr |_0^L . $$
Hence,
$$ a_n = \frac{2A}{L}\frac{L^2}{(n\pi)^2} \left[(-1)^n-1 \right] = \frac{2AL}{(n\pi)^2}\left[(-1)^n-1 \right]. $$
Therefore the final fourier series is:
$$ f(x)=\frac{AL}{2}+ \sum_{n=1}^\infty \frac{2AL}{(n\pi)^2} \left[(-1)^n-1 \right] \cos \left(\frac{n\pi
x}{L} \right). $$
$$ u=x \qquad dv= \cos\left(\frac{n\pi x}{L}\right)dx $$
$$ du=dx \qquad v=\frac{L}{n\pi} \sin \left(\frac{n\pi x}{L} \right) $$
\begin{align}
a_n = \frac{2A}{L} \Bigg[ \Big[\frac{Lx}{n\pi} & \sin \left( \frac{n \pi x}{L} \right) \Big] \Biggr|_0^L \\
-\int_0^{+L} \frac{L}{n\pi} & \sin \left(\frac{n \pi x}{L} \right)dx \Bigg]
\end{align}
$$ a_n = \frac{2A}{L} \Bigg[ 0 -\int_0^{+L}\frac{L}{n\pi} \sin \left(\frac{n\pi x}{L} \right)dx \Bigg] $$
{{sectionTitleLong[3]}}
The 0th term of the series, $ \frac{a_0}{2}, $ is the
constant term before the sum.
This is labelled as the 0th term so that the series may start indexing its terms from $ n=1. $ $$ f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n \cos\left(\frac{n\pi x}{L}\right)+b_n \sin\left(\frac{n \pi x}{L}\right) $$ If the function is not periodic, then we may cut the function, forcing it to be periodic over some $ -L < x < L. $ This means that we can prescribe a Fourier series that is now valid for the function over the selected range. If we take the limit as $ L \rightarrow \infty, $ then a Fourier transform need be used. This is a different problem entirely!
This is labelled as the 0th term so that the series may start indexing its terms from $ n=1. $ $$ f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n \cos\left(\frac{n\pi x}{L}\right)+b_n \sin\left(\frac{n \pi x}{L}\right) $$ If the function is not periodic, then we may cut the function, forcing it to be periodic over some $ -L < x < L. $ This means that we can prescribe a Fourier series that is now valid for the function over the selected range. If we take the limit as $ L \rightarrow \infty, $ then a Fourier transform need be used. This is a different problem entirely!
{{sectionTitleLong[4]}}
Decomposition of a function into odd and even parts
Representing a function in terms of a trigonometric Fourier series is analogous to representing a vector in terms of basis vectors. Here sines and cosines are our basis functions, we have a method for finding the projection of the function in the direction of each basis function ($ a_{n} $ and $ b_{n} $ are the "projections" of $ f(x) $ onto $ \sin\left(\frac{n\pi x}{L} \right) $ and $ \cos\left(\frac{n \pi x}{L} \right) $ respectively).This visualisation shows how the first ten terms are introduced to the series, notice that usually as the number of terms included in the series increases, new terms become less significant (their amplitude decreases).
Power Spectrum
The power spectrum form is a different representation of a continuous signal.One can think of a signal as a sound wave that is a superposition of different frequencies $ \omega_n, $ each with amplitude $ \alpha_n, $ and relative phases $ \theta_n. $ Any signal of period $ 2L $ can be decomposed in terms of the amplitude and relative phase of its discrete frequency modes. Its two parameters $ \alpha_n $ and $ \theta_n, $ enable us to measure the amount of each frequency $ \omega_n=\frac{n\pi}{L} $contained in the signal, and the relative phase difference between the $ n^{th} $ frequency and a cosine wave centered at $ x=0. $ This is simply a different way to represent a signal rather than decomposing it into its even and odd parts. $$ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty}\alpha_n \cos\bigg(\frac{n\pi x}{L} - \theta_n\bigg) $$ $$ \alpha_n = \sqrt{a^2_n + b^2_n} $$ $$ \theta_n = \tan^{-1} \bigg(\frac{b_n}{a_n}\bigg) $$ For some even functions, when the period transitions from 0 to negative, the value of $ \theta_n $ change from $ 0 $ to $ \pi $ or vice versa. Why do you think this happens?
This happens because the value of $ a_n $ changes sign
($ b_n $ is 0) causing the value of
$ \theta_n $ to change. For $ b_n = 0, $
$ \tan^{-1}(\frac{b_n}{a_n}) $
can be both
0 or $ \pi $ depending on whether $ a_n $ is positive or
negative. Not all even functions exhibit this.
The parabola for example, has its $ a_n $ proportional to
$ L^2, $ so its sign does not change when $ L $ changes
sign.
Orthogonal functions
Two functions $ f $ and $ g $ are said to be orthogonal if their inner product $ \langle f,g\rangle $ is 0. Just like basis vectors in a finite vector space, a set of orthogonal functions can form an infinite basis in a function space and a linear sum of these basis functions can describe any function in that function space.For functions that are periodic over $ -L < x \leq L $ with a period of $ 2L, $ the fourier components $ \rightarrow $ $ \sin(\frac{\pi x}{L}), \cos(\frac{\pi x}{L}), \sin(\frac{2\pi x}{L}), \cos(\frac{2\pi x}{L})... $ and so on form a complete orthogonal set that can be used to describe these functions. This means that increasing the amplitude of one of the components does not affect the other components, so changing the amount of any frequency contained in a signal leaves all the other frequencies unaffected.
{{sectionTitleLong[5]}}
Description
Finally, you can see the components of the series, the power spectrum and the resultant Fourier series displayed all at once. You can select one of the example functions from the previous sections, or input your own custom function, and see how all the elements work together to build a Fourier series.How To Use Custom Functions
To specify a custom function, select custom from the drop down menu and input your desired function into the input for $ f(x). $The period slider forces the function you have specified to be periodic over $ -L < x \leq L. $ For example if $ f(x) = \sin(x) $ with $ L = 1, $ this is the sine function but cut at $ x = \pm 1 $ and forced to be periodic over just this range (not the full $ 2\pi $ period).
The following functions can be used:
- Trigonometric functions (e.g. sin(), cos())
- Trigonometric inverses (e.g. arcsin(), arccos())
- Hyperbolic functions (e.g. sinh(), cosh())
- Hyperbolic inverses (e.g. arsinh(), arcosh())
- Exponentials and logs (e.g. exp(), log())
Syntax
Entering $ 5x $ does not work: instead you should write 5*x.Write exponents as a^(exponent), so $ x^2 $ is written as x^2.
Functions should be real and fully defined over the domain, for example $ log(x) $ is imaginary for values less than zero, so does not work.
$ log(x+1) $ does not work either (with a period of 1), since log is undefined at zero, however $ log(1+x^2) $ does work for all periods, since it is defined over the whole domain and is real.
{{sectionTitleLong[6]}}
In a Fourier Series, if the range over which the function is periodic, $ L $
goes to infinity, this function is no longer periodic.
To treat such a function, we would use Fourier Transform.
At its core, the Fourier Transform converts the function from one domain to another.
For example, by performing a Fourier Transform of a wave packet in the time domain, we can transform the wave packet into the frequency domain. This can make analysis (such as differentiation) much easier than performing it in the original domain.
One major application of Fourier Transform is in sound signal editing. Sound data is picked up by a microphone which records the pressure variation of air against time. When a Fourier Transform is performed, the relative intensities of the constituent frequencies that make up the signal is broken down, making it easier for sound correction and timbre analysis.
The Fourier Transform graph gives the relative "intensity" of the respective frequencies that make up the top-hat graph.
At its core, the Fourier Transform converts the function from one domain to another.
For example, by performing a Fourier Transform of a wave packet in the time domain, we can transform the wave packet into the frequency domain. This can make analysis (such as differentiation) much easier than performing it in the original domain.
One major application of Fourier Transform is in sound signal editing. Sound data is picked up by a microphone which records the pressure variation of air against time. When a Fourier Transform is performed, the relative intensities of the constituent frequencies that make up the signal is broken down, making it easier for sound correction and timbre analysis.
Visualisation Guide
On the left, we see a top-hat function that we intend to perform a Fourier Transform on. The graph on the right shows the Fourier Transform of the top hat function. We have represented the top hat graph in the frequency space.The Fourier Transform graph gives the relative "intensity" of the respective frequencies that make up the top-hat graph.


REVIEW: A Bug's Life By Luke Buckmaster<EMAIL_ADDRESS>Featuring the voices of: Dave Foley, Kevin Spacey, Julia Louis-Dreyfus, Phyllis Diller, Brad Garrett, Jonathan Harris, Bonnie Hun, David Hyde Pierce, Madeline Kahn, Richard Kind, Denis Leary, Hayden Panettiere, Joe Ranft, John Ratzenberger Directors: John Lasseter, Andrew Stanton Screenplay: Andrew Stanton, Donald McEnery, Bob Shaw
Australian theatrical release: December 3, 1998
On the Buckmaster scale of 0 stars (bomb), to 5 stars (a masterpiece): 3 and a half stars
Critics everywhere have compared Disney Pixar's A Bug's Life to Dreamworks' Antz. Don't expect me to be any different; it is virtually impossible not to compare these two pretty special films. As I was wracking my brain trying to figure out why I liked Antz more than this, it suddenly hit me: Antz deals with its themes in more of a theoretical sense, whilst A Bug's Life is very practical. For example, in Antz, the enemy was depicted as the colony itself - their inability to think for themselves. But in this film, the enemy is in the very practical form of ruthless grasshoppers.
It is rare for Disney films to have a good amount of depth and meaning (The Hunchback of Notre Damne is the only in recent memory), but A Bug's Life - a potentially brilliant film - is just a little too thin an experience. But without being hyper critical, this is a very good and solid film, which will no doubt score big with the Christmas box office.
The dangerous Hopper (voice of Kevin Spacey) and his ruthless band of grasshoppers threaten to wipe out a colony of ants, which is led by the ant queen (Phyllis Diller) and Princess Atta (Julia Louis-Dreyfus). Flik (Dave Foley, of TV's NewsRadio) is an ant whose experimental views on technology get him in deep trouble, but he rises to the challenge of saving the colony by enlisting the help of "warrior bugs." Unfortunately, he mistakes circus-performing insects for warriors, and disaster brews as Hopper's threats become closer to reality.
One of the foremost reasons to see A Bug's Life is its clever jokes, some of them parodies of our modern world. When a fallen leaf interrupts a group of ants collecting food, one ant screams "I'm lost! Where's the line?" Rescue workers quickly arrive at the scene. "This is nothing compared to the twig of '93!"
Many other gags, courtesy of script writers Andrew Stanton, Donald McEnery and Bob Shaw, evolve around the crazy personalities of the circus performers. A repressed stick insect, a male ladybug and a chubby caterpillar (who is eager to transform into a "beautiful butterfly") are among the delightful characters delivering the jokes.
Kevin Spacey plays Hopper with a wickedly placid presence. His voice is controlled but his anger is fierce, a little reminiscent of his role as a serial killer in Se7en. Spacey is the best part of A Bug's Life, as this superb actor proves that not being seen doesn't remove any power from his performances. Dave Foley plays a similar role to his part in NewsRadio; acting as the sane center - the instigator that all the other characters can react around. David Hyde Pierce's voice is instantly recognizable from the hit sitcom Frasier, and is a welcomed presence who delivers his trademark personality of intelligence amongst confusion. Almost everyone else seems uniformly average, ranging from the lifeless Julia Louis-Dreyfus to the over the top Madeline Kahn (playing the caterpillar).
A Bug's Life may have had it a little easy from critics. Its plot is lightweight, but it is not hard to be sucked into the films illustrious computer animation. These visuals are often a delight, although I was surprised to find that they are significantly less detailed than those in Antz (especially when we get a peek at the world beneath the ground). A Bug's Life also looks much more spherical, with a smooth and shiny design.
Definitely the best moments of A Bug's Life are near its end, in the final battle between Flick and Hopper, and the delightful roll of film through the credits (when an original gag is staged, which gives us an insight into the film's real actors). Whether you prefer these types of films as theoretical or practical, sophisticated or slim, Antz and A Bug's Life are both terrifically entertaining. Now it's just a case of whether or not Disney can provide depth and meaning in the same package.
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What Makes Bayesian Methods Different and Why Should I Care?
suid
infant-safety
pediatrics
epidemiology
statistics
Author
Daniel P. Hall Riggins, MD
Published
September 8, 2022
In my last post, I showed how to estimate risk of sudden unexpected infant death (SUID) in census tracts using Bayesian methods. The Bayesian approach is a separate branch of statistics from the Frequentist approach that most of us learned in school. Historically, Frequentist methods were so dominantly favored, many people may not have been aware that the statistics they were learning were termed “Frequentist” and that there was an alternative “Bayesian” approach available.
My objective for this post is to outline enough of the Bayesian approach to explain why it is better suited to the use-case of estimating SUID risk in small areas, then to give a set of examples to build intuition around how the method is working under the hood.
For a more in-depth introduction to Bayesian statistics, check out the textbook “Bayes Rules! An Introduction to Applied Bayesian Modeling”, which is available online for free. I adapted the source code from it’s companion package {bayesrules} to develop the visualizations in this post.
The Challenge
Epidemiology works with seemingly idiosyncratic denominators when representing incidence or risk. For example, the overall incidence of SUID in Cook County, IL for 2014 was 88.3 cases per 100,000 live births. While it would be just as mathematically valid to represent it as 0.000883 cases per single live birth, that doesn’t make sense in human terms. When considering a rare event, you want to use a large pool of people as your reference denominator.
Let’s assume the incidence of 88.3 accurately represents the underlying risk of SUID in the county for 2014. If we know that there were 140,000 live births in the next five years, then we would expect there to be about 124 cases of SUID.
\[Expected \space New \space Cases \space in \space the \space County = Incidence \times Live \space Births =\]
\[\frac{88.3 \space cases}{100,000 \space live \space births} \times 140,000 \space live \space births \approx 124 \space cases \]
But let’s zoom in to a single census tract where there were 240 live births over the same time period. Using the formula above, we would expect to observe zero cases. However, given what we know about incidence for the whole county, we wouldn’t observe zero cases and assume the underlying risk for that individual tract was actually zero. Estimating risk is hard when an event is rare and/or the reference pool of people is small.
Why Use Bayesian Methods to Solve this Challenge?
The “Frequentist” approach to statistics uses data in isolation from “prior knowledge”. If we observe 0 cases out of 240 live births in a census tract, our estimate of the underlying risk is 0 cases per 100,000 live births–even if our knowledge about the county as a whole suggests the underlying risk is closer to 88.3. Furthermore, if we want to make inferences about our data, Frequentist methods require a certain threshold sample size (for example n = 30) to be accurate. If the pool of live births in a census tract is something small like 13 total, than we expect up front that our estimate is going to be out of whack.
In contrast, Bayesian methods weigh the observed data against prior knowledge. If we have a lot of observed data, the final estimate will rely mostly on that data and look similar to the Frequentist estimate. But if we don’t observe a lot of data, the prior knowledge weighs more heavily into the estimate and will diverge away from the Frequentist estimate.
Examples
Another important distinguishing characteristic of Bayesian methods, is that they represent the uncertainty of an estimate using a distribution. In my previous post, we used a “Beta distribution” and outlined our process for specifying its parameters to represent our prior knowledge as follows:
The shaded red area represents the full range of plausible SUID risk values represented by our prior knowledge. The point where the curve is highest corresponds to the most plausible value (0) and we can average over the whole curve to get an expected risk of 88.3. These are useful point estimates but we hold onto the whole distribution to say the true risk could vary anywhere from 0 to 2500. Next, we can overlay a range of plausible values based on hypothetical data where we observed 1 case of SUID in 270 live births:
The data alone thinks the most plausible risk values are much higher than our prior understanding suggested. Bayesian methods help us weigh between these two somewhat conflicting sources of information by combining the prior expectations and the observed data to give us a “posterior” estimate:
I’ve taken out the shaded regions for visual clarity, but the area under each curve is still representing the most plausible values. Look how the purple posterior curve distributes itself in between the red and blue ones. Bayesian statistics weighed our prior expectations against the data and estimated the true risk to be somewhere in between.
To get some better intuition about what’s happening, let’s look at some extreme scenarios. First, what if we observed zero cases in a census tract with very few live births?
Code
plot_beta_binom_variation(0.189, 213, 0, 10)
In this example, the data gave us very little information so the prior expectation and posterior estimation curves look almost identical. The blue curve is very spread out, indicating a wide range of plausible risk values based on the data alone.
Code
plot_beta_binom_variation(0.189, 213, 0, 1000)
This example considers a census tract where we still observed zero cases, but there were many live births, meaning there were more opportunities for a case of SUID to happen. Given there were so many observed live births, the data weighed much more heavily into the posterior, which estimated the true risk (16) to be lower than our prior expectations (88.3).
Code
plot_beta_binom_variation(0.189, 213, 5, 1000)
In this example, we still observed 1000 live births, so the data heavily weighed into our posterior, but we increased the case count to 5, so the estimated true risk increased to a whopping 429.
Code
plot_beta_binom_variation(0.189, 213, 5, 50)
And finally, in this example, we still observed 5 cases, but we decreased the number of live births to 50. Using the data alone, we calculated the risk to be very high, but the posterior curve hung back toward the prior curve because the weight of our observed data was less than when we observed 1000 live births.
Wrap-Up
Some people argue that Bayesian statistics introduce too much subjectivity by letting us incorporate “arbitrary” prior expectations. Bayesians counter that statistical analysis always involves subjective interpretation. When we look at risk estimates based on the data alone, we adjust our trust in those numbers based on external knowledge. Bayesian methods simply give us a way to more formally define and weigh those expectations into our estimates. In the case of SUID risk for small populations in small geographic areas, it’s really helpful to moderate extreme estimates created by limited data using the Bayesian prior.


$DE5E/56926: Write Jump from $C8BB, $CAC3, $DD81, $DDF6, $E047, $E3B3, $E3D4, $E4ED: DE5E: A9 90 LDA #$90 ; command code for writing DE60: 8D 4D 02 STA $024D DE63: D0 26 BNE $DE8B Previous Page / Next Page


Airport
Departures
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Scottish Rugby Union pay fine and express regret over typhoon comments
by Odd ANDERSENLONDON — The Scottish Rugby Union has paid a £70,000 ($90,000) fine and "expressed its regret" for comments made in relation to the threatened cancellation of their World Cup pool match with hosts Japan due to Typhoon Hagibis, World Rugby said on Thursday.
Scotland were desperate for the game to be played because if it had been cancelled it would have counted as a 0-0 draw and they would have been eliminated.
SRU chief executive Mark Dodson had said Scotland would not become "collateral damage" of the typhoon and threatened legal action if the game did not go ahead.
The typhoon was one of the most powerful storms Japan has seen in many years and more than 80 people were killed.
The match did go ahead and Japan's 28-21 victory on October 13 saw them progress and Scotland fail to reach the World Cup quarter-finals for only the second time in their history.
World Rugby said at the time the comments were "inappropriate and ill-judged".
"World Rugby can confirm that the Scottish Rugby Union has expressed its regret and has confirmed it will not challenge World Rugby further on this matter," the federation said in a statement on Thursday.
"The Scottish Rugby Union has agreed to pay a donation of £70,000 to World Rugby and the matter is now closed."
© 2019 AFP


Sort candidates by an external array of polIDs.
Array
:
a new array of candidates, sorted. Does not mutate original array.
sortByPolIDs({ candidates, polIDs }) //=> sortedCandidates
Normalizes a party string.
(String)
party string, e.g. 'DEM'
(Any)
String
:
a normalized party string
normalizeParty('DEM') //=> 'dem'
normalizeParty('gop') //=> 'gop'
normalizeParty('green') //=> 'ind'


Consent form and information
Description
This research is part of a Master Thesis Project in the topic of memory-based conversational agents.
The agent is developed to be a question answering component, answering self-disclosure type questions.
This is part of a research study on how people interact with a ‘personal question answering’ chatbot.
After the conversation you will be asked to answer a number of questions.
The questions are about rating how you perceive the agent in terms of its natural-ness (human-like), engagingness, etc., on a 5-point scale.
You will also be asked to fill in your age and your estimated English proficiency.
Your participation will take approximately 4-7 minutes.
Risks and benefits
All messages sent to the chatbot will be saved and may be used by the current or a future version of the system. The stored messages may be used to extend the existing question/answer retrieval database. Data given in the survey may be presented in a Master Thesis report & presentation. There is no discomfort involved in this study, nor any risks. We cannot and do not guarantee or promise that you will receive any benefits from this study.
Requirements to participate
You must be 18 years or older to participate.
Basic conversational proficiency in the English language is desired.
Participant's rights
Please understand your participation is voluntary and you have the right to withdraw your consent or discontinue participation at any time without giving any reasons. You have the right to refuse to answer particular questions. You have the right to refuse afterwards (within 24 hours) to allow your data to be used for the research. If you wish to discontinue your participation and have your data removed, please remember your User ID: {{ session['userID'] }}.
All your data collected in this test:
- Is anonymous.
- will be stored in such a way that unauthorized access will be minimized.
- will not be disclosed to third parties.
Contact information
Further information or Withdrawal: If you have any questions or wish to discontinue your participation and for your data to be removed, please contact me by e-mail<EMAIL_ADDRESS>Remember to include your User ID {{ session['userID'] }} if you wish for your data to be removed.
Independent Contact: If you are not satisfied with how this study is being conducted, or if you have any concerns, complaints, or general questions about the research or your rights as a participant, please contact the Ethics Committee, Faculty of EEMCS, University of Twente, tel:<PHONE_NUMBER>, email<EMAIL_ADDRESS>Cookies
Cookies are only used to store necessary data for the chatbot system and user test to be able to run smoothly, such as your User ID and message history with the chatbot.
Instructions
Please ask the agent at least 5 questions to try to get to know the agent, as if it was a real person. Please send complete sentences one at a time.
When to stop asking questions to the agent is decided by you. Afterwards, please fill in a short 10 question survey to finish the test.


--- title: Contact heading: Contact us description: Please reach out to learn more or volunteer with us map: latitude: 41.777114 longitude: -87.586793 address: "6460 S Stony Island Ave, Chicago, IL 60637" zoom: 14 ---
info@artificechicago.org


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Usually used in conjunction with multi-line text blocks. ## Syntax
"""<#String (Allows Variables)#>"""
""" This is a multi line string Second Line """


---
layout: post
title: Nikon Cheap Thrills
tags:
- Articles
- film
- General Photography
- Kodak Portra 160VC
- lenses
- Nikon
- Nikon FE
status: publish
type: post
published: true
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ghostpool_post_image: http://photo.rwboyer.com/wp-content/uploads/2011/08/P1020222-ico.jpg
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More than a couple of you here have decided to shoot just a little bit of film, either with some of your dusty old 35mm gear or with some new found bargain gem from eBay. One lucky person is shooting one of my old Nikon F3's.
There are plenty of great deals on Nikon 35mm film bodies out there, although as I mentioned last week the trend seems to be somewhat up in terms of prices. I don't see near the number of amazing deals that I saw as little as a few months ago. Instead of me generally promoting that you go out and grab some of these for next to nothing, I thought I would start to give a few very specific recommendations.
First off a plain-Jane Nikon FE is probably the best deal around. For some reason the ultra-cheap FE2 of last year seems to be few and far between now. The FE maybe a better choice anyway - you can shoot Non-AI lenses on it. The only thing you really give up is top shutter speed and a slightly strange AE lock compared to the FE2.
Nikon FE's are still available in fantastic/near mint shape somewhere between $50-$100. Really nice Nikon FM's seem to be a bit more and far fewer - are they drying up? The FE maybe even a better choice than an FM in any case. It has aperture priority AE, a fantastic range of slow shutter speeds, most likely more accurate shutter speeds across the range, and a shutter speed dial that is unbelievabley easy but positive to change with one finger while looking through the viewfinder. The FM shutter speed dial is stiffer and seems somehow smaller. Yes the FE needs batteries but they last almost forever on this camera. Spare button cells take up NO room in your pocket, bag, wallet, whatever. The only thing I really like better about the FM is maybe the viewfinder display because it's easier to see the meter in darker environments.
Moving on to lenses on the cheap. There are a bunch of them I would recommend and will post some here but today I want to focus on a fantastically under-rated, under-priced, work of mechanical art. The Non-AI Nikkor 28mm f3.5.
I accidently aquired two of them. One of each major variety. One of them I bought on an impulse for $30 on eBay. How can a lens not be worth $30?  Yes the case and original caps came with it as well. Considering the price of the package, I consider it a freebie.
Here are the two. The first one is the newer, more familiar Nikkor variety. It's multi-coated and has fantastic ergonomics. A perfect balance of damping and focus lightness that makes it a joy to use. The other one has the more dated orginial look of Nikkons first SLR lenses. It's single coated and has a more damped feel to the focus but still impecable. The older one is actually in mint condition. Not a mark on any of the paint, like it was never used.
[image title="P1020222" size="full" id="3336" align="none" linkto="viewer" ]
Both of them are surpisingly good picture makers. In most circumstances you cannot tell them apart on film. These are nothing like budget lenses of today. They came from an era in Nikon where there was no difference in the mechanics, workmanship, or handling of Nikkors. A Nikkor was a Nikkor. Built to last a lifetime and a work of art in and of itself. I swear both of these has the workmanship, fit and finish, and materials that in some ways are better than the newer Leica lenses I have used. All the markings are engraved. No play. Amazing.
I was never interested in these lenses - or any 28mm - when I was buying new Nikon AI/AIS lenses. I bought a 35mm f2 and a 35mm f 1.4. I swear that these lenses are at least as sharp, contrasty, good color, less CA, and possibly less distortion thatn either of those. You do have to put up with a modest aperture but in practice it really is no different than an f2.8 28mm that for some reason brings a lot more money.
Here is a quick snap shot on one or the other of them - don't remember which, just to see if it worked.
[image title="2011-007-18cn" size="full" id="3334" align="none" linkto="viewer" ]
Here is another - again don't remember now which was which.
[image title="2011-001-13cn2" size="full" id="3337" align="none" linkto="viewer" ]
If you really want the convenience of AI you can try to find one that has been factory converted, or even buy the AI version - still a good deal, just not quite as good. I am sure that you can find one like either of these for less than 50 bucks. Maybe even an AI version but you will have to put a little more time and patience into it. Will it ever be worth less? Doubt it.
For my money I would definitly opt for one of these vs the more expensive but pretty much the same f2.8. Will you notice a difference in use - I doubt it.
The really strange thing about the 28 f3.5 is that it has a very large front element compared to the rest of the elements in the design and a very large front element for the maximum aperture. So as an added bonus it's very cool looking on the front of a nice metal 35mm SLR.
RB


{% extends "layout.html" %} {% block content %}
To create a new chore to store in the database, simple log-in and select the Chores page. From here
you will simple need to click the add button. This will bring you to a new screen where you
will be prompted to enter information such as the name of the chore and it's point value.
Passwords
When choosing a password, make sure think carefully. If any of your children can figure out your password,
They will be able to manipulate a large portion of this application, rendering much of it's capabilities pointless.
As your dealing with those who are familiar to you, it might be good to go above and beyond the usual mix of 8 or
more numbers and letters. While it would be ill advised to use a password you may have employed on multiple other
sites, definitely don't use anything too complicated to remember either.


Check out what I've been working on lately
A webapp that uses IBM Watson sentiment analysis to determine the inspirational measure of quotes. Built with Ruby on Rails.
A webapp for organizing and planning events with your neighbors and friends. Built with Ruby + Sinatra.
Notate your speed chess games by taking photos of each move. OpenCV + Django + React Native.
A tiny money transfer service which uses the MasterCard SendP2P API via SMS. Ruby on Rails with Twilio.
Send a joke to your phone via SMS. Sinatra + Heroku + Twilio.
a repo of skills I have published to the Alexa Skill store. Written with Node.js.


They inform what projects I take on, what actions I perform, and what ideas I explore.
In circumstances of great opportunity and risk, what we value determines whether we succeed or fail.
I solve complex problems by building for-profit companies.
For-profit because I believe that aligning value creation and value capture incentives is essential for maintaining focus.
My process enables me to tackle ambiguous problems in a world filled with uncertainty. I believe that creating processes which enable long-term value creation strategies ensures sustainable success.
Every solution starts by understanding the problem to solve.
I collaboratively develop potential solutions in an agile, lean manner.
Gathering feedback about what customers think is vital to making something that people love!
Based on the feedback, I repeat the process until the problem is solved.
Joint project leader, UX and UI Designer
There is a mismatch between student expectations and the realities of living in rented accomodation. We found that 9 in 10 students had problems in their accomodation that needed fixing, and 60% had problems that it taken more than 4 days to get fixed.
StuRent is an information product for 1st and 2nd year students at Bristol University looking to move into private rented accommodation in Clifton, Redland or Stokes Croft in the following academic year. It contains:
Co-Founder and Director, Technology Lead, Joint UX and UI Designer
Current education system in self-awareness, well-being and life skills is failing young people who need it. With the biggest youth cohort in human history, 1.8 billion people, scalability is essential as current institutions are failing to keep up with demand. We found that 76% of young people feel their school does not prepare them for life.
Merudite is a digital tool for personalised mental wellbeing and life skills education. It offers young people a way to independently learn and engage with learning materials that are uniquely tailored to their personality type. This helps them to enhance or compensate for their unique strengths and weaknesses. It also provides institutions with tools to track the progress of those in their care in a scalable digital way.
Joint Project Leader
As highlighted in the picture to the left, which shows one frame from the game ‘The Division 2’ (released in 2018), the cost of producing 3D assets for a variety of applications is large and growing.
A piece of software that can synthesise and stylise new 3D assets, learning from the data which creative companies have stored in order to create new designs. Using techniques such as Generative Adversarial Networks and Transfer Learning, we believe this problem is solvable with AI methods.
I have facilitated and mentored students designing Android apps in how to solve the needs of their clients, including a non-profit for improving international student experiences.
I provided guidance on developing technical specifications, Spring framework, performing unit and user testing and presenting and pitching their final project.
I regularly mentor a global cohort of students in how to develop a growth mindset, learn from failure, how to network, how to develop lean business models, and how to ideate successfully. This is part of my role as an mentor for the Bristol Futures Online Course, one of the university's flagship online programmes, which I have been with helping with from the start.
As part of this I have presented for and represented the Bristol Futures Online Course at events, such as the Collaborative Communities Showcase where I explored internship opportunities and funding.
In summer 2019 I travelled to Bangalore, India, to work with Infosys and IIITB and develop a solution to the question “how do we ensure that new technologies benefit all citizens of cities across the globe?” I facilitated an international team creating and pitching a solution: an app to utilise local resources to aid non-profits, to a panel of judges.
I achieved accreditation in Inclusive Leadership from Common Purpose for excellence in this programme.
I obtained the Bristol PLUS Award for excelling at my work experience as a research intern, applying myself in my leadership role as Secretary of Manor Hall JCR and developing my teamwork, communication and business skills via workshops run by external organisations such as NatWest Startup Accelerator.
Fewer than 5% of Bristol graduates achieve this award.
I choose to define my own way of operating in this world.
Therefore, I write on Medium to figure out what I actually think about difficult things. I generally write about self-improvement, productivity and technology. I write for publications with a combined active follower base of around 600k people, including Medium's largest: The Startup.


Display: Inline vs. Inline-Block
Positioning elements correctly on a page is one of the biggest challenges in CSS. There are so many selectors, properties and values, it can all be quite overwhelming sometimes.
Let's take a look at one common question - what's the difference between display: inline and display: inline-block? Both specify how elements should fit in relation to the surrounding content - but which one should you use, and when?
In order to answer this question, we should take a step back and make sure we understand the difference between inline and block-level elements. Here's a quick refresher:
An example of this would be when we use the <strong> tag. This fits within the flow of the text and doesn't cause the text within the tag to sit by itself on a new line.
One example of a block-level element is the <p> tag. This causes the text within the tags to begin on a new line. Whatever follows the paragraph will also begin on its own new line.
Generally speaking, block-level elements may contain inline elements and other block-level elements, however inline elements can't contain block-level elements.
Now that we've covered the difference between inline and block, lets look at the difference between inline and inline-block. The names give us our first clue.
Display: inline is essentially what we covered earlier in the article. Content will stay on the same line if possible, only wrapping onto a new line if there's not enough space. We can specify margins and padding for these elements, however only the left and right values will be respected - margins and padding on the top and bottom will overlap with other content. Here's an example:
Display: inline-block, as the name suggests, finds a happy medium between the inline and block-level characteristics. In some way it gives us the best of both worlds. Inline-block allows us to specify the height and width of elements, which we can't do with inline. Margins and padding on the top and bottom of elements are also respected, unlike what we see with the inline value. The key difference between block and inline-block is that the latter doesn't cause the element to appear on a new line. Inline-block elements can sit beside one another as friends and equals. Here's an example of inline-block in action:


abstract val identityKeys: List<PublicKey>
List of current and previous (rotated) identity keys, which member can still use to sign unspent transactions on ledger.


Table 2 Descriptive characteristics of participants in feasibility study and pilot randomised controlled trial
|
|
Pilot CRCT |
|
|
Pilot CRCT |
|
Feasibility study |
Control |
Intervention |
Feasibility study :0 |
Feasibility study :1 |
Intervention :0 |
Intervention :1 |
| N schools |
1 |
1 |
2 |
1 |
1 |
2 |
2 |
| N participants invited* |
234 |
138 |
458 |
234 |
234 |
458 |
458 |
| N parent opt out |
9 |
6 |
23 |
9 |
9 |
23 |
23 |
| N student opt out |
13 |
0 |
8 |
13 |
13 |
8 |
8 |
| N non-attendance |
29 |
17 |
82 |
29 |
29 |
82 |
82 |
| N assented |
183 |
115 |
345 |
183 |
183 |
345 |
345 |
| N 2 waves measured |
160* |
115 |
285 |
160 |
160 |
285 |
285 |
| N 2 waves AG |
57 |
68 |
152 |
57 |
57 |
152 |
152 |
| N 2 waves ≥3d AG |
52 |
43 |
112 |
52 |
52 |
112 |
112 |
| Age |
13.7 (0.4) |
13.1 |
13.2 (0.4) |
13.7 |
0.4 |
13.2 |
0.4 |
|
|
(0.3) |
|
|
|
|
|
| Sex N (% male) |
71 (43.3%) |
50 |
164 (47.7%) |
71 |
43.3 |
164 |
47.7 |
|
|
(43.5) |
|
|
|
|
|
| Height (cm) |
165.8 (8.8) |
161.8 |
162.6 (8.5) |
165.8 |
8.8 |
162.6 |
8.5 |
|
|
(7.0) |
|
|
|
|
|
| Weight (kg) |
58.7 (12.7) |
53.0 |
53.4 (10.6) |
58.7 |
12.7 |
53.4 |
10.6 |
|
|
(10.6) |
|
|
|
|
|
| BMI z-score |
0.63 (1.2) |
0.52 |
0.44 (1.1) |
0.63 |
1.2 |
0.44 |
1.1 |
|
|
(1.1) |
|
|
|
|
|
| Per cent of |
26.9 |
22.7 |
24.1 |
26.9 |
26.9 |
24.1 |
24.1 |
| overweight/obese |
|
|
|
|
|
|
|
| Mean (SD) unless otherwise stated. *Not all participants given accelerometer; 113 participants at baseline, 123 at follow-up and 87 at both baseline and follow-up. AG, actigraph; BMI, body mass index; CRCT, cluster-randomised controlled trial. |
corpus-oa-uclii-01/10.1136.bmjopen-2016-12335/tables/table2/table.svg.html
|
Table 4 Average daily minutes in MVPA by study group at baseline and postintervention, and preliminary intervention effect of GoActive pilot trial
|
Control (SD) |
Intervention (SD) |
Difference adjusted for baseline (95% CI) |
Control (SD) :0 |
Control (SD) :1 |
Intervention (SD) :0 |
Intervention (SD) :1 |
Difference adjusted for baseline (95% CI) :0 |
Difference adjusted for baseline (95% CI) :1 |
Difference adjusted for baseline (95% CI) :2 |
Difference adjusted for baseline (95% CI) :3 |
| Feasibility study |
|
|
|
|
|
|
|
|
|
|
|
| MVPA (baseline) |
|
60.7 (27.5) |
|
|
|
60.7 |
27.5 |
|
|
|
|
| MVPA (postintervention) |
|
61.3 (25.6) |
|
|
|
61.3 |
25.6 |
|
|
|
|
| Pilot trial |
|
|
|
|
|
|
|
|
|
|
|
| MVPA (baseline) |
48.6 (15.4) |
51.9 (15.3) |
|
48.6 |
15.4 |
51.9 |
15.3 |
|
|
|
|
| MVPA (postintervention) |
42.1 (15.0) |
49.4 (18.2) |
|
42.1 |
15.0 |
49.4 |
18.2 |
|
|
|
|
| MVPA (change) |
-6.5 (14.0) |
-2.5 (15.4) |
5.1 (1.1 to 9.2) p=0.014 |
-6.5 |
14.0 |
-2.5 |
15.4 |
5.1 |
1.1 |
9.2 |
0.014 |
| *School-level clustering not taken into account due to insufficient clusters. MVPA, minutes in moderate-to-vigorous physical activity. |
corpus-oa-uclii-01/10.1136.bmjopen-2016-12335/tables/table5/table.svg.html
|
Table 5 Secondary outcomes at baseline and postintervention; results show change adjusted for baseline
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Difference adjusted for baseline (95% CI) |
:0 |
:1 |
:0 |
:1 |
:0 |
:1 |
:0 |
:1 |
Difference adjusted for baseline (95% CI) :0 |
Difference adjusted for baseline (95% CI) :1 |
Difference adjusted for baseline (95% CI) :2 |
Difference adjusted for baseline (95% CI) :3 |
| Types of PA |
19.2 (12.8) |
14.0 (9.4) |
19.8 (15.2) |
16.6 (14.0) |
2.3 (-0.2 to 4.7) p=0.07 |
19.2 |
12.8 |
14.0 |
9.4 |
19.8 |
15.2 |
16.6 |
14.0 |
2.3 |
-0.2 |
4.7 |
0.07 |
| Self-efficacy for PA |
17.7 (0.4) |
17.2 (3.6) |
17.8 (3.0) |
17.6 (3.2) |
0.3 (-0.4 to 1.0) p=0.36 |
17.7 |
0.4 |
17.2 |
3.6 |
17.8 |
3.0 |
17.6 |
3.2 |
0.3 |
-0.4 |
1.0 |
0.36 |
| Peer support |
6.3 (2.6) |
5.3 (1.9) |
5.9 (2.2) |
5.5 (2.2) |
0.5 (0.1 to 0.9) p=0.03 |
6.3 |
2.6 |
5.3 |
1.9 |
5.9 |
2.2 |
5.5 |
2.2 |
0.5 |
0.1 |
0.9 |
0.03 |
| Friendships |
2.8 (1.1) |
2.9 (1.0) |
2.8 (1.1) |
2.9 (1.1) |
-0.1 (-0.3 to 0.1) p=0.37 |
2.8 |
1.1 |
2.9 |
1.0 |
2.8 |
1.1 |
2.9 |
1.1 |
-0.1 |
-0.3 |
0.1 |
0.37 |
| Well-being |
44.5 (0.9) |
43.3 (1.0) |
45.0 (0.5) |
45.5 (0.5) |
1.8 (0.1 to 3.4) p=0.04 |
44.5 |
0.9 |
43.3 |
1.0 |
45.0 |
0.5 |
45.5 |
0.5 |
1.8 |
0.1 |
3.4 |
0.04 |
| Shyness |
13.9 (3.5) |
14.0 (3.7) |
13.7 (3.4) |
13.7 (3.3) |
-0.3 (-0.9 to 0.4) p=0.43 |
13.9 |
3.5 |
14.0 |
3.7 |
13.7 |
3.4 |
13.7 |
3.3 |
-0.3 |
-0.9 |
0.4 |
0.43 |
| Sociability |
13.5 (2.0) |
13.9 (1.9) |
13.7 (2.1) |
14.0 (1.8) |
0.1 (-0.4 to 0.5) p=0.74 |
13.5 |
2.0 |
13.9 |
1.9 |
13.7 |
2.1 |
14.0 |
1.8 |
0.1 |
-0.4 |
0.5 |
0.74 |
| Barriers to PA |
29.7 (5.1) |
28.7 (5.3) |
29.1 (5.2) |
28.4 (5.4) |
0.1 (-1.1 to 1.2) p=0.91 |
29.7 |
5.1 |
28.7 |
5.3 |
29.1 |
5.2 |
28.4 |
5.4 |
0.1 |
-1.1 |
1.2 |
0.91 |
| Analyses not clustered for school as insufficient clusters. PA, physical activity. |
corpus-oa-uclii-01/10.1136.bmjopen-2016-12335/tables/table6/table.svg.html
|


VATA BHOY
Ex-Celtic ace Rudi Vata’s son Rocco scores for Ireland in Poland victory
by Ewan PatonEX-CELTIC ace Rudi Vata's son Rocco has scored on his start for the Republic of Ireland.
The Hoops youngster netted in a 2-1 win over Poland in the match in Galway.
It was youngster Rocco's first start for the under-15s.
SunSport exclusively revealed on Tuesday night that Rocco was set to make his debut.
And he did as he helped his country to two wins in two games at Eamonn Deacy Park.
Jason Donohue's team had previously beaten the same opponent on Tuesday.
Kevin Zefi netted his second in successive games after number ten Rocco Vata had opened the scoring for Ireland.
Vata Snr would have been proud of his son and the rest of the boys as they came from behind to win.
Poland pounced from a corner kick on six minutes and opened the scoring.
However, towards the end of the first half, Ireland missed a penalty but Rocco was on hand to tap into the net on the rebound to level proceedings.
Then on 64 minutes the Irish completed the comeback when Zefi found himself in the right place to fire in from around 22 yards out.
Dad and former Celtic hero Vata won 59 caps for Albania and lifted the Scottish Cup with the Hoops in 1995.
Rocco 14 is in the Hoops academy. He also qualifies for Albania, Scotland and Montenegro.
He can represent the Irish through his Dublin-born gran Ann on his mother’s side.
Vata, 50, now an agent and TV pundit, said: “When I joined Celtic, I met my new team-mates for the first time in Dublin.
“My debut was against Sligo Rovers because we were on a pre-season tour.
“This is part of my history so it is special for my son to be in Ireland now.
“Nobody in the Albanian FA contacted me about Rocco. I know that Scotland have watched him too, but the Irish association was the first to get in contact.
“His granny is the happiest Irish woman in Scotland!"


The F5 load balancer from F5 Networks, Inc. directs traffic away from servers that are overloaded or down to other servers that can handle the load. The F5 load balancer extension collects key performance metrics from an F5 load balancer and presents them in the AppDynamics Metric Browser.
This extension works only with the standalone machine agent.
https://f5-host:f5-port/mgmt/tm/ltm/pool/stats
MachineAgent/monitors
directoryMachineAgent/monitors/F5Monitor/config.yml
and update the following.
Yaml is very strict with the syntax, please make sure that you validate your YML
here
metricPrefix: "Custom Metrics|F5 Monitor|" #or metricPrefix: "Server|Component:$TIER_NAME|Custom Metrics|F5 Monitor|"
servers: - uri: "https://server1:8443" name: "Server1" username: "user" password: "password" - uri: "https://server2:443" name: "server2" username: "user" passwordEncrypted: "y444543gt3="
If passwordEncrypted
is used, make sure to update the encryptionKey
in config.yml
By default, everything is included as shown in the filters with the .*
regex.
This will report a lot of data. Set the correct filters to make sure that you only collect the data you will need
filter: pools: includes: [".*"] poolMembers: includes: [".*"] rules: includes: [".*"] ... ...Please refer to the
config.yml
for the complete configuration.
Token-based authentication can be used in BIG IP v12+.
Please refer to F5 documentation for more details.
To enabled token based auth, please modify the config.yml
as shown
servers: - uri: "https://server1:8443" name: "Server1" username: "user" password: "password" authType: "TOKEN" loginReference: ""
loginReference
from your system administrator.
It looks something like https://localhost/mgmt/cm/system/authn/providers/ldap/-id-/login
loginReference
blankselfLink
of the guest user(created in step #1) from the response JSONcurl -i -k -u <adminuser>:<pwd> https://f5-ip/mgmt/shared/authz/users
link
in the --data
argument json with the value of selfLink
from the response of step #2curl -i -k -u <adminuser>:<pwd> --request PATCH \ --data '{"userReferences":[{"link":"https://localhost/mgmt/shared/authz/users/guestuser"}]}' \ https://f5-ip/mgmt/shared/authz/roles/iControl_REST_API_User
This should give access to the user created in step #1 to invoke the rest api to invoke GET. If the guest role doesn't work, try "manager" role in step#1
To avoid setting the clear text password in the config.yml, please follow the process to encrypt the password and set the encrypted password and the key in the config.yml
java -cp "appd-exts-commons-1.1.2.jar" com.appdynamics.extensions.crypto.Encryptor myKey myPassword
passwordEncrypted
and encryptionKey
fields in config.yml
https://f5-host:f5-port/mgmt/tm/ltm/pool/stats
-Dappdynamics.agent.maxMetrics=5000
if there is a metric limit reached error in the logs.
If you dont see the expected metrics, this could be the cause.
<MachineAgent>/conf/logging/log4j.xml
and update the level of the appender com.appdynamics
to debug
Let it run for 5-10 minutes and attach the logs to a support ticket
Workbench is a feature by which you can preview the metrics before registering it with the controller. This is useful if you want to fine tune the configurations. Workbench is embedded into the extension jar.
To use the workbenchjava -jar /path/to/MachineAgent/monitors/F5Monitor/f5-monitoring-extension.jarThis starts an http server at
http://host:9090/
. This can be accessed from the browser.
#Get the stats curl http://localhost:9090/api/stats #Get the registered metrics curl http://localhost:9090/api/metric-paths
config.yml
and validate it from the browser or the APIThe metrics will be reported under the tree Application Infrastructure Performance|$TIER|Custom Metrics|F5 Monitor|
Please refer to this screenshot to view the complete list of metrics reported by the F5 extension. These metrics can be filtered, please refer to the installation section for details.
Always feel free to fork and contribute any changes directly via GitHub.
For any questions or feature request, please contact AppDynamics Help.
| Version | 2.4.7 |
| Machine Agent Compatibility | 4.0+ |
| Tested With F5 Version | 12.x |
| Last Update | 6/8/17 |


References
1
Local perceptions towards biological conservation in the community of Vila Velha, Pernambuco, Brazil
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