Professional Consulting
 Consulting

Because of my diverse skillset, broad experience, and problemsolving nature, I have had the opportunity to work as a consultant with great artists and organizations on a wide variety of projects.
For more information on my consulting services please visit:
davidafeil.com/consulting
Volunteer Labor
 Shrimp Boat Projects

Shrimp Boat Projects is an artistic investigation of the Houston region that explores the connection between a region's identity and its native landscape. As the last form of labor wholly dependent on this landscape, shrimping in Galveston Bay is the project's point of departure. The process of the project melds the daily work aboard a commercial shrimp boat, the F/V Discovery, active participation in the local seafood economy, public programming, and cultural production. Since the summer of 2011, I have had the chance to work with artists Zach Moser and Eric Leshinsky as a tagalong deckhand and repairman on multiple occasions.
 Sweet Acre Farm

Sweet Acre Farm (SAF), started in February 2011, is one acre of land in Mansfield, CT farmed by Charlotte Ross and Jonathan Janeway. They produce diversified vegetables, chicken (for meat), and maple syrup. They have taken the Farmers Pledge (offered through the CT Northeast Organic Farmers Association) and implement farming practices focused on building soilhealth and high nutrient value in their crops. To that aim, they do not use pesticides or herbicides, neither chemical nor organically approved. In the summer of 2011, I spent six weeks on the farm working with and learning from Jonathan and Charlotte as an amateur farmhand and market salesman. I returned in August of 2012 for three weeks to work and learn all over again.
Mathematics
 The Hollow Plate

Starting in 2010, I began to rediscover my love for mathematics. I had not taken a course in Mathematics since 2003, but over the course of the year I began collecting, reading through, and working problems from primers in advanced mathematics, starting with graph theory, moving onto abstract algebra and group theory, formal languages and automata theory, and beyond. In the fall of 2011, I decided to register myself for the GRE Subject Test in Mathematics, which is the test most undergraduate upperclassmen would take as part of the application process for graduate schools. I took the test purely for the reward of seeing how much I had taught myself in the course of the previous year. While preparing for the test, I found numerous blogs that went through previous tests problem by problem, with others commenting on alternate solutions. I decided to start a similar blog, working through the problems found in my books and have found it to be a great way to keep myself focused on the details when writing solutions.
cleanwrinkles.org/math
last updated Mon, 12 Dec 2011 05:48Modern Elementary Differential Equations 2.3  MEDE2.3.16
16. Consider the particular solution of
\(y’ + ay = e^{bx}, y = \frac{e^{bx}}{b + a}\).
Show how a particular solution can be obtained for
\(y’ + ay = x\)
by differentiation with respect to \(b\), followed by \(b = 0\).
Solution:
First we differentiate \(y\) with respect to \(b\) to get:
\(\frac{dy}{db} = \frac{e^{bx}}{b+a}\left(x – \frac{1}{b+a}\right)\)
Next, we find \(y’\) (differentiate with respect to \(x\)) and then differentiate with respect to \(b\):
\(y’ = \frac{be^{bx}}{b+a} = by\)
\(\frac{dy’}{db} = y + b\frac{dy}{db}\)
If we differentiate the righthand side of the original differential equation we simply get \(\frac{d}{db}e^{bx} = xe^{bx}\). We therefore get the new differential equation:
\(\frac{d}{db}\left(y’ + ay\right) = \frac{d}{db}e^{bx}\)
\(y + (b + a)\frac{dy}{db} = xe^{bx}\)
\(\frac{e^{bx}}{b+a} + e^{bx}\left(x – \frac{1}{b+a}\right) = xe^{bx}\)
\(\frac{1}{b+a} + x – \frac{1}{b+a}= x\)
\(x = x\)
Thus the method of differentiating with respect to \(b\) has proven to be a successful method of finding a solution to the new differential equation. We can now proceed to the second part of the question.
If we set \(b = 0\) in our earlier solution for \(\frac{dy}{db}\), we get a new particular solution \(y = \frac{x1}{a}\).
We can check the validity of this solution easily:
\(y’ = \frac{1}{a}\)
\(\frac{1}{a} + \frac{x1}{a} = x\)
Modern Elementary Differential Equations 2.3  MEDE2.3.1
1. Find the solution of
 \(y’ + 2y = x\), \(y(0) = 1\),
 \(y’ – y = \sin x\), \(y(0) = 2\).
Solution:
For part a we use the fact that \(\frac{d}{dx}\left(ye^{2x}\right) = y’e^{2x} + 2ye^{2x}\). Multiplying both sides of the equation by \(e^{2x}\) we get:
\(e^{2x}\left(y’ + 2y\right) = xe^{2x}\)
\(\frac{d}{dx}\left(ye^{2x}\right) = xe^{2x}\)
Integrating between \(0\) and \(x\), we get:
\(ye^{2x} – y(0) = \int_o^x x_1e^{2x_1}dx_1\)
Now we’ll use integration by parts:
\(u = x, u’ = dx, v = e^{2x}, v’ = 2e^{2x}dx, d(uv) = u’v + uv’\)
\(\int_o^x x_1e^{2x_1}dx_1 = \frac{1}{2} \int_0^x uv’ = \frac{1}{2} \left(\int_0^x d(uv) – \int_0^x u’v\right) = \frac{1}{2}\left(xe^{2x} – \frac{1}{2}\left(e^{2x} – 1\right)\right)\)
\(ye^{2x} – 1 = \frac{(2x1)e^{2x}+1}{4}\)
\(y = \frac{2x1 + 5e^{2x}}{4}\)
For part b, we use the same strategy, knowing that \(\frac{d}{dx}\left(ye^{x}\right) = y’e^{x} – ye^{x}\):
\(e^{x}\left(y’ – y\right) = e^{x}\sin x \)
\(\frac{d}{dx}\left(e^{x}y\right) = e^{x}\sin x\)
Again, we integrate between \(0\) and \(x\):
\(e^{x}y – y(0) = \int_0^x e^{x_1}\sin x_1 dx_1 \)
If we use the exponential form of sine,
\(\sin(x) = \frac{1}{2i}\left(e^{ix} – e^{ix}\right)\),
we get:
\(\int_0^x e^{x_1}\sin x_1 dx_1 = \frac{1}{2i}\left(\int_0^x \left(e^{(i1)x_1} – e^{(i+1)x_1}\right)dx_1 \right) = \frac{1}{2i}\left(\frac{1}{i1}e^{(i1)x} + \frac{1}{i + 1}e{(i+1)x} – \frac{1}{i1} – \frac{1}{i + 1}\right)\)
\(e^{x}y – 2 = \frac{(i+1)e^{(i1)x} + (i1)e^{(i+1)x_1} – 2i}{4i}\)
\(y = \frac{(i1)e^{ix} + (1+i)e^{ix} + 2}{4} + 2e^x\)
Using the exponential forms of sine and cosine, we get back to the trigonometric form:
\(y = \frac{2\cos(x) – 2\sin(x) + 2e^x}{4} + 2e^x\)
\(y = \frac{1}{2}\left(\cos(x) + \sin(x) – 5e^x\right)\)
International Mathematics Olympiad 2006  IMO.2006.4
4. Determine all pairs \((x, y)\) of integers such that
\(1 + 2^x + 2^{2x+1} = y^2\).
Solution:
I started off with a bit of circuitous completingthesquare rearrangement:
\(\left(1 + 2^{x1}\right)^2 = 2^{2x2} + 2^x + 1\)
\(\left(1 + 2^{x1}\right)^2 – 2^{2x2} + 2^{2x+1} = y^2\)
\(2^{2x2}\left(2^3 – 1\right) = y^2 – \left(1 + 2^{x1}\right)^2\)
\(2^{2x2}\cdot 7 = \left(y – \left(1 + 2^{x1}\right)\right)\left(y + \left(1 + 2^{x1}\right)\right)\)
Either \((y – 1 – 2^{x1})\) or \((y +1 + 2^{x1})\) must be of the form \(2^a\) for some integer \(a\) and the other must be of the form \(2^b\cdot 7\) for some integer \(b\) where \(a + b = 2x2\).
We can take the difference of both these expression (and thus cancel out the \(y\) terms) in two possible ways.
Here is the first:
\(2^{2x – 2 – b} – 2^b\cdot7 = (y – 1 – 2^{x1}) – (y + 1 + 2^{x1}) = 2 – 2^x\)
\(2^{2x – 2 – b} – 2^b\cdot7 = 2^x – 2\)
\(2^b\cdot7 = 2^x + 2^{2x2b} + 2\)
\(7 = 2^{xb} + 2^{2x22b} + 2^{1b}\)
Since 7 is odd, we know that one of the terms in the sum must be equal to \(2^0 = 1\) and that all the exponential terms must be positive integers. This leaves us with \(b = 1\) or \(b = x1\).
For \(b=1\), we get \(6 = 2^{x1} + 2^{2x – 4}\) which has no integral solutions.
For \(b=x1\), we get \(6 = 2^1 + 2^{2x}\), which yields the solution \(x=0\) and \(y=\pm 2\).
The second difference changes the signs on our first and third terms:
\(7 = – 2^{xb} + 2^{2x22b} – 2^{1b}\)
For this difference, our middle term (the only positive one) must be greater than 7 and therefore it cannot be our \(2^0\) term. Thus our only solution is \(b = 1\), we get:
\(8 = 2^{x1} + 2^{2x4}\)
\(1 = 2^{x4} + 2^{2x7}\)
Using the same logic as before, 1 is an odd number therefore one of the terms must be equal to \(2^0\). The only possibility is for \(x=4\), and yields the solution \(y=\pm 23\).
Thus there are four integer solutions, \((0, \pm 2)\) and \((4,\pm 23)\).