Dane DeSutter http://www.danedesutter.com Sun, 20 May 2012 08:47:25 +0000 en hourly 1 http://wordpress.org/?v=3.3.2 Language Immersion While You Browse http://www.danedesutter.com/2012/05/09/language-immersion-while-you-browse/ http://www.danedesutter.com/2012/05/09/language-immersion-while-you-browse/#comments Wed, 09 May 2012 20:23:11 +0000 admin http://www.danedesutter.com/2012/05/09/language-immersion-while-you-browse/ This new extension authored by Google for its Chrome browser is quickly becoming one of my favorites. The concept is simple: load a webpage and the Language Immersion for Chrome extension translates it to the degree of your choosing — beginner to fluent — and you can learn a language immersively while you browse. Just as a foreword, I speak German so my analysis here is primarily based on using this tool between German-English.

Some things I love about this feature include the options to set your level of difficulty as well as the pronunciation of translated text on a mouseover. The TTS feature is of particular merit because not only are the Google voices top quality, they've actually slowed them down a bit within Google Translate ostensibly for exactly the purpose of learning.

That said, Google is the brand of beta, and there are still a few kinks to work out here. Some of these kinks lie specifically with Google Translate as a product, while others are plugin specific.

While I'm slowly starting to find Translate's translations a bit more realistic, they are still far from perfect. I still notice genders and cases getting mixed up, and of course this would be unapparent to any non-native speaker. The translated syntax also can get messy on anything other than "fluent," and that is in part just because English uses a subject, verb, object construction fairly rigidly, whereas other languages have either a different structure or are more flexible in this respect. This can lead to misplaced verbs and can alter subject-verb agreement. In German, for example, this plugin is unlikely to teach you that sentence structures like indirect object, verb, subject are common and perfectly legal, e.g. Er hat mir ein iPhone verschenkt = Mir hat er ein iPhone verschenkt.

Within the plugin itself, the mouseover feature for pronunciation is great, however the TTS voice for some reason rather than relying on the user selected language output tries to determine the language semantically. So if the voice runs across a word that exists in another language or a name that is clearly of a different nationality, it switches to the voice of that language for the rest of the excerpt. It leads to some odd results when a Spanish voice engine is trying to read German.

Aside from my minor grievances, this tool has a lot of potential for anyone looking for a cost free (take that Rosetta Stone!) way to pick up some new vocabulary in a new language.

#education

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]]> http://www.danedesutter.com/2012/05/09/language-immersion-while-you-browse/feed/ 2 Harvard and MIT's Brainchild: edX http://www.danedesutter.com/2012/05/02/harvard-and-mits-brainchild-edx/ http://www.danedesutter.com/2012/05/02/harvard-and-mits-brainchild-edx/#comments Wed, 02 May 2012 17:34:25 +0000 admin http://www.danedesutter.com/2012/05/02/harvard-and-mits-brainchild-edx/ The joint effort between MIT and Harvard called edX was announced today. It is going to be a service dedicated to providing high calibre educational content for students around the world. Students who can show a significant level of mastery will be awarded a certificate from edX, although it will not bear the names of either institution.

This is not the first post-secondary foray into online services like edX. Similar startups with other Ivy Leagues in the early 2000s folded because the projects became too costly to produce and maintain.

Whatever edX's fate may be, I'm chomping at the bit to give it a try.

#education

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MIT news
Joint partnership builds on MITx and Harvard distance learning; aims to benefit campus-based education and beyond.

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]]> http://www.danedesutter.com/2012/05/02/harvard-and-mits-brainchild-edx/feed/ 0 Derivative of e^x http://www.danedesutter.com/2012/04/30/derivative-of-ex/ http://www.danedesutter.com/2012/04/30/derivative-of-ex/#comments Mon, 30 Apr 2012 23:27:51 +0000 admin http://www.danedesutter.com/?p=3519 The derivative of any continuous, differentiable function f\left(x\right) can be found by the limit,

f^{\prime}\left(x\right)=\underset{h\rightarrow0}{\lim}\frac{f\left(x+h\right)-f\left(x\right)}{h}

In this case, the function we are trying to differentiate is,

f\left(x\right)=e^{x}

The limit definition also requires that we find f\left(x+h\right), so we plug x+h
into e^{x}:

f\left(x+h\right)=e^{x+h}

Then replace f\left(x+h\right) and f\left(x\right) in the limit definition with our specific cases.

\frac{d}{dx}\left(e^{x}\right)=\underset{h\rightarrow0}{\lim}\frac{e^{x+h}-e^{x}}{h}

From the algebraic Properties of Exponents, we know that,

x^{a}x^{b}=x^{a+b}

We can make use of an Algebra trick to rewrite this in a more useful way.

=\underset{h\rightarrow0}{\lim}\frac{e^{x}e^{h}-e^{x}}{h}

Factoring out the e^{x},

=\underset{h\rightarrow0}{\lim}\frac{e^{x}\left(e^{h}-1\right)}{h}

The e^{x} can now be pulled out of the limit since it does not depend on h and is effectively constant.

=e^{x}\underset{h\rightarrow0}{\lim}\frac{e^{h}-1}{h}

The rest of this derivation relies on us taking this definition as axiomatic.

e^{h}=\underset{n\rightarrow\infty}{\lim}\left(1+\frac{h}{n}\right)^{n}

(Justification for this when h=1 can be found here.)

This definition of e^{h} allows us to use the Binomial Theorem to expand the right hand side inside the limit.

\left(1+\frac{h}{n}\right)^{n}=\sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array}\right)1^{n-k}\left(\frac{h}{n}\right)^{k}

Take note that the Binomial Coefficient \scriptsize\left(\begin{array}{c}n\\k\end{array}\right) is just a placeholder notation for:

\left(\begin{array}{c}n\\k\end{array}\right)=\frac{n!}{k!\left(n-k\right)!}

The value 1^{n-k} is going to always equal one, since one raised to any power should return the value one. Simplifying our expansion,

=\sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array}\right)\frac{h^{k}}{n^{k}}

Apparently then,

e^{h}=\underset{n\rightarrow\infty}{\lim}\left[\sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array}\right)\frac{h^{k}}{n^{k}}\right]

If we then expand the summation and the Binomial Coefficients,

=\underset{n\rightarrow\infty}{\lim}\left[\left(\begin{array}{c}n\\0\end{array}\right)+\left(\begin{array}{c}n\\1\end{array}\right)\frac{h}{n}+\left(\begin{array}{c}n\\2\end{array}\right)\frac{h^{2}}{n^{2}}+\cdots+\left(\begin{array}{c}n\\n\end{array}\right)\frac{h^{n}}{n^{n}}\right]

=\underset{n\rightarrow\infty}{\lim}\left[1+h+\frac{n!}{2!\left(n-2\right)!}\frac{h^{2}}{n^{2}}+\cdots+\frac{n!}{k!\left(n-k\right)!}\frac{h^{k}}{n^{k}}+\cdots+\frac{h^{n}}{n^{n}}\right]

The next important simplification we can make is to reduce n!/\left(n-k\right)!

=\underset{n\rightarrow\infty}{\lim}\left[1+h+\frac{\left(n\right)\left(n-1\right)\left(n-2\right)!}{2!\left(n-2\right)!}\frac{h^{2}}{n^{2}}+\cdots+\frac{\left(n\right)\left(n-1\right)\cdots\left(n-k+1\right)\left(n-k\right)!}{k!\left(n-k\right)!}\frac{h^{k}}{n^{k}}+\cdots+\frac{h^{n}}{n^{n}}\right]

Which would leave us with:

=\underset{n\rightarrow\infty}{\lim}\left[1+h+\frac{\left(n\right)\left(n-1\right)}{n^{2}}\frac{h^{2}}{2!}+\cdots+\frac{\left(n\right)\left(n-1\right)\cdots\left(n-k+1\right)}{n^{k}}\frac{h^{k}}{k!}+\cdots+\frac{h^{n}}{n^{n}}\right]

Now if we multiplied out the linear products left over from reducing our factorials, we should notice that the degree of the numerator will always match the degree of the denominator in terms of n. When we take the limit then as n\rightarrow\infty, each of these coefficients becomes 1. For example,

\underset{n\rightarrow\infty}{\lim}\frac{\left(n\right)\left(n-1\right)}{n^{2}}=1

Therefore,

e^{h}=1+h+\frac{h^{2}}{2!}+\frac{h^{3}}{3!}+O\left(h^{4}\right)

Where O\left(h^{4}\right) is shorthand notation to indicate all the remaining terms that have a degree \geq4. Plugging this into our original limit,

\frac{d}{dx}\left(e^{x}\right)=e^{x}\underset{h\rightarrow0}{\lim}\frac{1+h+\frac{h^{2}}{2!}+\frac{h^{3}}{3!}+O\left(h^{4}\right)-1}{h}

We can then cancel out the positive and negative 1 on either end of the numerator, and divide by h. This becomes,

=e^{x}\underset{h\rightarrow0}{\lim}\left[1+\frac{h}{2!}+\frac{h^{2}}{3!}+O\left(h^{3}\right)\right]

With direct substitution, the limit reduces:

\frac{d}{dx}\left(e^{x}\right)=e^{x}\left(1+0+0+\cdots\right)

=e^{x}\left(1\right)

And thus,

\frac{d}{dx}\left(e^{x}\right)=e^{x}

]]> http://www.danedesutter.com/2012/04/30/derivative-of-ex/feed/ 0 Derivative of x^n http://www.danedesutter.com/2012/04/30/derivative-of-xn/ http://www.danedesutter.com/2012/04/30/derivative-of-xn/#comments Mon, 30 Apr 2012 17:44:20 +0000 admin http://www.danedesutter.com/?p=3474 The derivative of any continuous, differentiable function f\left(x\right) can be found by the limit,

f^{\prime}\left(x\right)=\underset{h\rightarrow0}{\lim}\frac{f\left(x+h\right)-f\left(x\right)}{h}

In this case, the function we are trying to differentiate is,

f\left(x\right)=x^{n}

The limit definition also requires that we find f\left(x+h\right), so we plug x+h into x^{n}:

f\left(x+h\right)=\left(x+h\right)^{n}

Then replace f\left(x+h\right) and f\left(x\right) in the limit definition with our specific cases.

\frac{d}{dx}\left(x^{n}\right)=\underset{h\rightarrow0}{\lim}\frac{\left(x+h\right)^{n}-x^{n}}{h}

To solve this, we will need to recognize that \left(x+h\right)^{n} can be expanded using the Binomial Expansion Theorem.

\left(x+h\right)^{n}=\sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array}\right)x^{n-k}h^{k}

Take note that the Binomial Coefficient \scriptsize\left(\begin{array}{c}n\\k\end{array}\right) is just a placeholder notation for:

\left(\begin{array}{c}n\\k\end{array}\right)=\frac{n!}{k!\left(n-k\right)!}

Expanding this out, we would get the quotient,

=\underset{h\rightarrow0}{\lim}\frac{\left(\begin{array}{c}n\\0\end{array}\right)x^{n}+\left(\begin{array}{c}n\\1\end{array}\right)x^{n-1}h+\left(\begin{array}{c}n\\2\end{array}\right)x^{n-2}h^{2}+\cdots+\left(\begin{array}{c}n\\n\end{array}\right)h^{n}-x^{n}}{h}

Next, note that \scriptsize\left(\begin{array}{c}n\\0\end{array}\right)=1, and \scriptsize\left(\begin{array}{c}n\\1\end{array}\right)=n. This leads to,

=\underset{h\rightarrow0}{\lim}\frac{x^{n}+nx^{n-1}h+\left(\begin{array}{c}n\\2\end{array}\right)x^{n-2}h^{2}+\cdots+\left(\begin{array}{c}n\\n\end{array}\right)h^{n}-x^{n}}{h}

Then we cancel out the positive and negative x^{n} and divide by h,

=\underset{h\rightarrow0}{\lim}\left[nx^{n-1}+\left(\begin{array}{c}n\\2\end{array}\right)x^{n-2}h+\cdots+\left(\begin{array}{c}n\\n\end{array}\right)h^{n-1}\right]

Next do a direct substitution for h=0,

=nx^{n-1}+0+\cdots+0

And simplifying,

\frac{d}{dx}\left(x^{n}\right)=nx^{n-1}

 

]]> http://www.danedesutter.com/2012/04/30/derivative-of-xn/feed/ 0 TED Ed http://www.danedesutter.com/2012/04/29/ted-ed/ http://www.danedesutter.com/2012/04/29/ted-ed/#comments Sun, 29 Apr 2012 17:13:14 +0000 admin http://www.danedesutter.com/2012/04/29/ted-ed/ Great series of microlectures by the same guys who bring us TED talks. I've always been a fan of the TED lecture: it's short, concise, and it holds the viewer's attention for just the right amount of time.

This begs the question (for me) if this format could work for undergraduate lesson design, treating information as plugins of a bigger framework. The more self-contained the plugin, the more reasonable it is to reintroduce it for review and motivation of new ideas.

#education

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Lessons Worth Sharing
Use engaging videos on TED-Ed to create customized lessons. You can use, tweak, or completely redo any lesson featured on TED-Ed, or create lessons from scratch based on any video from YouTube.

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]]> http://www.danedesutter.com/2012/04/29/ted-ed/feed/ 0 Literature for the Attention Deficit http://www.danedesutter.com/2012/04/24/literature-for-the-attention-deficit/ http://www.danedesutter.com/2012/04/24/literature-for-the-attention-deficit/#comments Tue, 24 Apr 2012 17:37:24 +0000 admin http://www.danedesutter.com/2012/04/24/literature-for-the-attention-deficit/ Discovered "Electric Literature" today. It's a short fictional story periodical. Personally, as someone who always struggled to find the spark in my literature classes, this reinvention of what fiction "should" be is really intriguing. Also, the cover art is wonderful!

#education

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Electric Literature
Electric Literature 1. Michael Cunningham Jim Shepard Lydia Millet T Cooper Diana Wagman. Electric Literature 2. Colson Whitehead Lydia Davis Pasha Malla Marisa Silver Stephen O'Connor. Electric L…

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]]> http://www.danedesutter.com/2012/04/24/literature-for-the-attention-deficit/feed/ 0 Proving Euler’s Identity http://www.danedesutter.com/2012/04/22/proving-eulers-identity/ http://www.danedesutter.com/2012/04/22/proving-eulers-identity/#comments Mon, 23 Apr 2012 02:50:50 +0000 admin http://www.danedesutter.com/?p=1055 Euler’s Identity is arguably one of the most elegant theorems in mathematics. It states that,

e^{i\pi}+1=0

The physicist Gauss famously said that the very nature of this identity should prove to be a litmus test for great mathematical minds; he believed that any spiring mathematician who saw this identity and was not immediately aware of its implications would never be great.

The following will be a demonstration of why Euler’s Identity must be true. A learning project based on this concept will soon follow.

Prerequisite knowledge for this includes:

Calculus

  1. Taylor Series
  2. Differentiation
    1.  Trigonometric Functions
    2. Exponential and Transcendental Functions

Algebra

  1. Complex Numbers

To provide a demonstration of Euler’s Identity, we must first be familiar with the Taylor Series. The Taylor Series is an approximation technique that can be applied to continuous functions, by adding higher ordered derivatives to the traditional linearization of a function at that point.

It can be easily shown that when given the point \left(c,f(c)\right) and the slope of the tangent line at that point, f^{\prime}\left(c\right), that the tangent line at this point has the equation,

L\left(x\right)-f\left(c\right)=f^{\prime}\left(c\right)\left(x-c\right)

L\left(x\right)=f\left(c\right)+f^{\prime}\left(c\right)\left(x-c\right)

To increase the accuracy of the Taylor approximation, we add in the concavity of the function through the second derivative.

T_{2}\left(x\right)=f\left(c\right)+f^{\prime}\left(c\right)\left(x-c\right)+\frac{f^{\prime\prime}\left(c\right)}{2!}\left(x-c\right)^{2}

The addition of the character of higher and higher order derivatives of f\left(x\right) pushes our approximation closer to the original function. The pattern continues and we can find the n-th polynomial approximation,

T_{n}\left(x\right)=f(c)+f^{\prime}\left(c\right)\left(x-c\right)+\frac{f^{\prime\prime}\left(c\right)}{2!}\left(x-c\right)^{2}+\cdots+\frac{f^{\left(n\right)}\left(c\right)}{n!}\left(x-c\right)^{n}

If we let n\rightarrow\infty, our Taylor approximation becomes the original function. As a series,

f\left(x\right)=\sum_{n=0}^{\infty}\frac{f^{\left(n\right)}\left(c\right)}{n!}\left(x-c\right)^{n}

We will use the special case of the Taylor Series known as the McLaurin Series where the approximation is centered around c=0. To verify this identity we will need the Taylor expansions about the origin for e^{x}, \cos\left(x\right), and \sin\left(x\right). This requires us to know the following derivatives:

\frac{d}{dx}e^{x}=e^{x}

\frac{d}{dx}\pm\sin\left(x\right)=\pm\cos\left(x\right)

\frac{d}{dx}\pm\cos\left(x\right)=\mp\sin\left(x\right)

The resulting series:

e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+\cdots

\cos\left(x\right)=\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}x^{2n}}{\left(2n\right)!}=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+\cdots

\sin\left(x\right)=\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}x^{1+2n}}{\left(1+2n\right)!}=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+\cdots

We now introduce e^{i\pi},

e^{i\pi}=1+i\pi+\frac{\left(i\pi\right)^{2}}{2!}+\frac{\left(i\pi\right)^{3}}{3!}+\frac{\left(i\pi\right)^{4}}{4!}+\cdots

=1+i\pi+\left(i\right)^{2}\frac{\pi^{2}}{2!}+\left(i\right)^{3}\frac{\pi^{3}}{3!}+\left(i\right)^{4}\frac{\pi^{4}}{4!}+\cdots

Simplifying the powers of i,

e^{i\pi}=1+i\pi-\frac{\pi^{2}}{2!}-i\frac{\pi^{3}}{3!}+\frac{\pi^{4}}{4!}+i\frac{\pi^{5}}{5!}-\frac{\pi^{6}}{6!}+\cdots

Grouping the terms without a factor of i in the first set of parentheses, and those with i in the second set,

=\left(1-\frac{\pi^{2}}{2!}+\frac{\pi^{4}}{4!}-\frac{\pi^{6}}{6!}+\cdots\right)+i\left(\pi-\frac{\pi^{3}}{3!}+\frac{\pi^{5}}{5!}+\cdots\right)

The first terms in parentheses are nothing more than the Taylor expansion of \cos\left(\pi\right). Similarly, the terms in the second set of parentheses are a Taylor expansion of \sin\left(\pi\right). Thus becoming,

e^{i\pi}=\cos\left(\pi\right)+i\sin\left(\pi\right)

Simplifying with the fact that \cos\left(\pi\right)=-1 and \sin\left(\pi\right)=0,

e^{i\pi}=-1

Or written in its more famous convention,

e^{i\pi}+1=0

]]> http://www.danedesutter.com/2012/04/22/proving-eulers-identity/feed/ 0 Textbook vs. eBook http://www.danedesutter.com/2012/03/30/textbook-vs-ebook/ http://www.danedesutter.com/2012/03/30/textbook-vs-ebook/#comments Fri, 30 Mar 2012 20:17:23 +0000 admin http://www.danedesutter.com/2012/03/30/textbook-vs-ebook/ I wonder if the Apple iBook will sweep education like they're hoping. I would advocate for something more open, but you can't argue with a good product.

#education

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How Higher Education Is Going Digital [INFOGRAPHIC]
Etextbooks and online learning communities are just a few of the ways colleges and universities are dabbling in digital. This graphic breaks it down.

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]]> http://www.danedesutter.com/2012/03/30/textbook-vs-ebook/feed/ 0 Learning to Unlearn http://www.danedesutter.com/2012/02/21/learning-to-unlearn/ http://www.danedesutter.com/2012/02/21/learning-to-unlearn/#comments Tue, 21 Feb 2012 21:32:12 +0000 admin http://www.danedesutter.com/2012/02/21/learning-to-unlearn/ In Duke professor Dr. Cathy Davidson's Book "Now You See It: How the Brain Science of Attention Will Transform the Way We Live, Work, and Learn" she stresses a concept of Learning to Unlearn. The idea is simple, but in practice very difficult: we need to break away from old methods of education and explore the new.

Davidson argues that this is a necessary process to move education into the Information Age. She stresses that old paradigms we've relied on since the turn of the century aren't going to work going forward; we must be aware of how young students, immersed in technology from the very beginning, are building their cognition. In many ways, they are learning in ways very alien to what pre-Internet students experienced.

How has educational conditioning biased your reasoning abilities? See how long this takes you. It may take longer than you'd think.

#education

Reshared post from +Ulrike Friedrich

hm ich habe dafür keine 10 Minuten gebraucht…

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]]> http://www.danedesutter.com/2012/02/21/learning-to-unlearn/feed/ 2 Water Droplets Orbiting Knitting Needle in Space http://www.danedesutter.com/2012/02/08/water-droplets-orbiting-knitting-needle-in-space/ http://www.danedesutter.com/2012/02/08/water-droplets-orbiting-knitting-needle-in-space/#comments Wed, 08 Feb 2012 14:13:46 +0000 admin http://www.danedesutter.com/2012/02/08/water-droplets-orbiting-knitting-needle-in-space/ Video of the electric force pushing charged water droplets into cylindrical spiral paths around a charged teflon/nylon knitting needle in zero gravity. See if the path looks like what you would have predicted. Very cool!

#education

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