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I read a very interesting article today from the Notices of the American Mathematical Society (AMS) about the intelligent use of technology in mathematics.  This article, titled “The Misfortunes of a Trio of Mathematicians Using Computer Algebra Systems. Can We Trust in Them?” describes the experiences of these three researchers in using the mathematical software packages Mathematica and Maple.  The details of their research are interesting but the important point for students of mathematics is to be aware of the limitations of the technology you use.  A key quote from the article is:

…even more dramatically, his algorithm yielded different outputs given the same inputs.

A more detailed explanation of what was going wrong:

…given the same matrix, the determinant function can give different values!

The authors do give credit to technology as a groundbreaking aid in modern mathematical research, but as is true in other research disciplines, they recommend using multiple sources.  In this case, checking the results of one mathematical software package against another software package to compare the results:

Having made this criticism, let us stress that software systems have proved very useful to research mathematicians.  Some well-known instances are the proof of the four-color problem by Kenneth Appel and Wolfgang Haken and the Kepler conjecture by Thomas Hales….Software bugs should not prevent us from continuing this mutually beneficial relationship in the future.  However, for the time being, when dealing with a problem whose answer cannot be easily verified without a computer, it is highly advisable to preform the computations with at least two computer algebra systems.

And, for students of mathematics, I would add,

- When dealing with a problem whose answer CAN be easily verified without a computer, do so!

Synthetic Division

We just finished talking about synthetic division in the College Algebra course today and got into a discussion of how to represent the remainder.  For example, given the problem:

\frac{x^4-2x^3-x+10}{x-2}

The answer turns out to be: x^3-1 R: 8 or you can say that the answer is: x^3-1+\frac{8}{x-2}.  This all goes back to the division algorithm which says that given two numbers a and b, then solving the problem \frac{a}{b} means finding q, the quotient and r, the remainder such that a=b*q+r (with r<b) .

If we take the expression a=b*q+r and divide on both sides by b, then we’ll have \frac{a}{b}=\frac{b*q}{b}+\frac{r}{b} or \frac{a}{b}=q+\frac{r}{b}.

Which of these forms we prefer depends on whether we want to say that:

x^4-2x^3-x+10=(x-2)(x^3-1)+8

or

\frac{x^4-2x^3-x+10}{x-2}=x^3-1+\frac{8}{x-2}

Income Inequality

Back in May of 2013, I did a presentation on income inequality that focused on how the Gini index is calculated.

Recently, I came across a graph that shows how the income growth during economic expansions has been distributed over the last 60 years.

Income Growth INEQ

It’s seems clear to me that the change in income tax rates during the Reagan years had a radical effect on income distribution in the US.

In his latest blog post, Robert Reich has a good analysis of some of the other issues that have led to the level of inequality we experience today.

cos(108)

I came across an interesting tidbit of math today in solving a problem from the ever-fruitful, continuously engaging book Mathematics Review Exercises by David P. Smith Jr. and Leslie T. Fagan.

Apparently \sqrt{2-2\cos 108^\circ}=\phi.

In my last post on the Farmer and the Fencing optimization problem, I mentioned that separating the pen into two triangular pieces with a diagonal cross-section makes the problem a lot more complex and requires the use of Lagrangian methods to solve.  After conceiving the problem in 1999-2000 and setting up the Lagrangian system that would solve the problem in 2001-2002 I finally solved it this week by using Wolfram Alpha.  If anyone out there sees a good algebraic solution for the Lagrangian system of three equations with three unknowns, please let me know.

Given 100 feet of fencing, a rectangular pen with length =x, width =y and a diagonal cross-section =\sqrt{x^2+y^2}, the perimeter constraint equation would be:

2x+2y+\sqrt{x^2+y^2}=100

The area equation we want to optimize is:

A=x*y

So, the Lagrangian process says that to maximize a function f(x,y), with a given constraint g(x,y)=C, we need to set up a system based on the equality:

\bigtriangledown f(x,y)=\lambda \bigtriangledown g(x,y)

In this problem, f(x,y)=A=xy and g(x,y)=2x+2y+\sqrt{x^2+y^2}.

So the gradient of f(x,y), or \bigtriangledown f(x,y) is:

\bigtriangledown f(x,y)=(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})

And the gradient of g(x,y), or \bigtriangledown g(x,y) is:

\bigtriangledown g(x,y)=(\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y})

So, going back to the functions we’re working with based on the perimeter and area, the Lagrangian set up would be:

\bigtriangledown f(x,y)=\lambda \bigtriangledown g(x,y)

(y,x)=\lambda (2+\frac{x}{\sqrt{x^2+y^2}}, 2+\frac{y}{\sqrt{x^2+y^2}})

This generates the system of equations:

y=2\lambda+\frac{\lambda x}{\sqrt{x^2+y^2}}

x=2\lambda+\frac{\lambda y}{\sqrt{x^2+y^2}}

2x+2y+\sqrt{x^2+y^2}=100

I must admit once again to resorting to a technological solution for this system.  Using Wolfram Alpha, we get the solution

x=y=\frac{100}{4+\sqrt{2}}\approx 18.47

This makes the cross-piece:

\sqrt{x^2+y^2} =\frac{100}{\sqrt{9+4\sqrt{2}}} \approx 26.12

Which of course means that the optimal area is once again generated by a square!  If we assumed this at the outset of the problem, it becomes much simpler (but less “fun”).  The permimeter in that case is:

4x+x\sqrt{2}=100

and

x=\frac{100}{4+\sqrt{2}}

I’ve been carrying this around with me for nearly 15 years and dug out the work I had originally done a few weeks ago to look at it “with fresh eyes.”  I found a few minor errors and was happy to use Wolfram Alpha to solve the system and finally generate a solution for this problem.

In Defense of Algebra

The American Mathematical Association of Two-Year Colleges (AMATYC) is in the process of putting together a position statement on the Appropriate Use of Intermediate Algebra as a Prerequisite Course that will annul the requirement that students pursuing a four-year degree in disciplines other than the hard sciences take and pass a course in Intermediate Algebra.  While this won’t directly impact students who initially enroll at four-year schools, it will have an enormous impact on math education at the nation’s community and technical colleges.  Some schools may amend their Intermediate Algebra courses without destroying the core topics, but I fear that others will end up going down the same road that many high schools did during the rush to get rid of algebra at the K-12 level during the 1990s.

The “reform math” movement of the 1990s led to a plethora of poorly written NSF-backed “math” textbooks such as Core-Plus and IMP at the high school level and MathLand and Everyday Math at the elementary level.  The repercussions of this misguided effort to reform math education at the K-12 level are still felt today, and in some places the fight continues.  The Seattle School Board was taken to court in 2009-10 over its mathematics textbook selection process and a King County Superior Court judge ruled their selection of the Key Curriculum “Discovery” series to be arbitrary and capricious.  An appeals court in 2011 reversed that decision and the textbook selection was allowed to stand.

The upshot of the use of reform materials at the K-12 level is that developmental math enrollment at the two and four-year colleges and universities has exploded.  The solution – stop teaching algebra there as well.

I respectfully disagree.  Here is the text of an email I’ve sent to AMATYC regarding their position statement:

I am responding to the AMATYC position statement on the Appropriate Use of Intermediate Algebra as a Prerequisite Course.  I have also copied Jerry Kissick, the President of ORMATYC, on this email as I believe that ORMATYC should adopt a position statement on the importance of Beginning and Intermediate Algebra to a Liberal Arts education.

I have been a Mathematics Instructor at Clatsop Community College in Astoria, Oregon for 10 years and taught high school mathematics for 6 years prior to entering the MA program in Mathematics at the University of Maine in the fall of 2000.  It is my professional opinion that Beginning and Intermediate Algebra are appropriate prerequisites for college level mathematics no matter the area of specialization of the student.

Any student expecting to receive a four year degree should have demonstrated an understanding of Beginning and Intermediate Algebra at some point in their academic career.  Students who are enrolled in vocational degree or certificate programs at two-year colleges should have different requirements based on their program of study.

Most four-year colleges and universities require that their students demonstrate an understanding of Beginning and Intermediate Algebra prior to matriculation at the college or university.  This is normally done through the submission of SAT or ACT scores after the completion of high school algebra courses.  At two-year colleges, developmental mathematics courses offer an opportunity for students to demonstrate understanding of Beginning and Intermediate Algebra, after which they may then transfer to a four-year school.

Algebra is one of the foundations of a Liberal Arts education and, together with arithmetic, form the conceptual foundation for all of mathematics.  Algebra is a prerequisite to Statistics.  Statistics is not a separate discipline wholly apart from the rest of mathematics – a true understanding of statistics requires an understanding of Beginning and Intermediate Algebra.

Civil rights activist Bob Moses has spent the past 25 years running the Algebra Project, trying to help minority students achieve a better understanding of algebra because he believes that access to quality algebra instruction is a civil rights issue.  Here is a link to an NPR story on Bob Moses.

The importance of Algebra to a student’s overall education demands that we not abdicate our responsibility to teach the subject, but rather work to ensure that we do a better job teaching the subject to all students.

The reform of California’s K-12 mathematics standards in the 1990s led an explosion of developmental enrollment in the California CC and Cal State systems.  This was not related to the topics in the Beginning and Intermediate Algebra curriculum, but rather to the preparation (or lack thereof) the students received at the K-12 level.  The Complete College America organization has recently been an active proponent of “alternative math pathways.”  I find it telling that the organization is called “Complete College America,” rather than “Educate American Students.”  Their focus appears to be getting students to a diploma without the academic and intellectual underpinnings that a college diploma normally represents.

I do often see students struggling to make sense of the work in their mathematics courses.  However, I also see them mature both emotionally and academically as a result of those struggles.  This is the purpose of education – to help our students to mature intellectually as a result of their engagement with the course material.

Algebra as a discipline has a history that goes several millennia deep in human culture.  The works of Diophantus, Brahmagupta and Al-Khwarizmi are beautiful, ineluctable parts of human heritage, and this is not to mention the more “recent” work of Fibonacci, Cardano, Fermat and Des Cartes.  While their techniques and notation may be archaic, it is the power of their reasoning that shines through across the centuries.

A liberal arts education is not and should not be a form of job training.  Many of my students will not directly use much mathematics in their chosen profession and they know this.  However, they are still interested in being an educated person, which means having the background in a broad range of human knowledge so as to be able to think independently about the world and society in which they live.

Thank you for your consideration of my opinion regarding the AMATYC position statement on the Appropriate Use of Intermediate Algebra as a Prerequisite Course.  I have also attached a series of essays written by individuals who believe as I do that Beginning and Intermediate Algebra are part of the foundation of a Liberal Arts education.

Sincerely,

Richard W. Beveridge

Mathematics Instructor

Clatsop Community College

The links for the attachments I mention in the letter are below:

Daniel Willingham at the Washington Post’s Answer Sheet blog

Jessica Lahey at the New York Times’ Motherlode blog

Nicholas Warner at the Huffington Post

Evelyn Lamb at Scientific American’s Observations blog

Jennifer Ouellette at Scientific American’s Cocktail Party Physics blog

Four years ago I wrote a post about math education that mentioned the problem of “The Farmer and the Fencing.”  The link for the post is here and a sample assignment from an Elementary Algebra (Algebra I) course is here.  I mentioned in that post that I’ve used this problem at a variety of levels of mathematics starting with the 7th grade.

For the 7th grade class, we looked at some possible values for the length and width of the pen that was to be constructed from the fencing and made a table of values showing the length, width and area of the pen for each set of values.

The project from the Elementary Algebra class that is linked above addresses the question from a somewhat more sophisticated level, asking the students to introduce variables for the length and width and then substitute in for one of the variables to arrive at an expression for the area as a function of only one of the dimensions.  The students can then find the maximum area using a graphing calculator.

In a Pre-Calculus class I taught during the 1999-2000 school year, we covered a unit that considered the parabola as a conic section and included applications on projectile motion and maximum and minimum values.  In the Pre-Calculus class, we found the max or min by finding the vertex of the parabola defined at the point x=\frac{-b}{2a} and y=f(\frac{-b}{2a}).  We didn’t need Calculus to determine this, the \frac{-b}{2a} can be defined from the vertex form of the equation of a parabola y=n(x-h)^2+k.  With a little classical algebra, you can show that h=\frac{-b}{2a}.

If you expand y=n(x-h)^2+k

y=n(x^2-2hx+h^2)+k

y=nx^2-2nhx+nh^2+k

Then

\frac{-b}{2a}=\frac{2nh}{2n}=h

I taught Differential Calculus as a grad student at the University of Maine in 2003 and we covered The Farmer and the Fencing from the perspective of Calculus, in which we differentiate the area equation and set the derivative equal to zero to find the maximum area.  The problems can be spruced up a little by dividing the pen into smaller pens using cross-sectional pieces.

In the fundamental problem the maximum area of a simple pen with four sides and no cross-divisions is arrived at by splitting the perimeter equally among the four sides to make a square.  If the pen is cross divided in any way, the maximum area comes from a scenario in which half of the perimeter is allocated to the lengthwise pieces and the other half allocated to the any widths.

If there are m lengths and n widths then the perimeter is represented by ml+nw=P.  The area is A=l*w.  If we isolate the length in the perimeter equation and substitute it into the area equation, we can then differentiate the area equation for an answer for the general problem of maximum area of a pen with mby n crosspieces:

ml+nw=P

ml=P-nw

l=\frac{P-nw}{m}

l=\frac{P}{m}-\frac{nw}{m}

Then substituting into the area equation:

A=l*w

A=(\frac{P}{m}-\frac{nw}{m})*w=\frac{P}{m}*w-\frac{n}{m}*w^2

So, the area equation is then

A=\frac{P}{m}*w-\frac{n}{m}*w^2

Which makes the derivative

A'=\frac{P}{m}-\frac{2n}{m}*w.

Setting this equal to zero and solving for w:

0=\frac{P}{m}-\frac{2n}{m}*w

\frac{P}{m}=\frac{2n}{m}*w

\frac{m}{2n}*\frac{P}{m}=w

\frac{P}{2n}=w

or

\frac{P}{2}=nw

which means that half of the perimeter should be allocated to the n widths and the other half to the m lengths.

For example, if the farmer has 300 feet of fencing, the maximum area is a square that is 75 feet on each side.

FencingSquare

If the pen is split into two smaller pens with a crosswise fence, then the maximum area will allocate 150 feet to the widths (75 feet each) and 150 feet to the three lengths (50 feet each).

FencingCrossPen

When I taught this problem in 1999-2000 I asked myself what might happen if the pen were to be divided on the diagonal creating two triangular pens.  Little did I know how much this would complicate such a simple little problem.

FencingDiagonalThe Briggs and Cochran Calculus textbook includes a similar problem in its section on optimization (problem 10c on pg. 214).  In a key modification that makes the problem solvable for Calculus I students, they eliminate one of the lengths of the figure by placing the pen against a barn.

FencingDiagonalBarnThis eliminates one of the lengths in the problem.  Let’s look at the solution to this problem before we tackle the more general version which requires that both lengths come from the available fencing.  In the problem in the textbook, 200 meters of fencing are to be used, so the constraint on the amount of fencing creates the following equation:

l+2w+\sqrt{l^2+w^2}=200

If we want to maximize the area of the pen then we need to substitute for one of the variables and find the derivative of the area equation:

l*w=A

In order to substitute we need to isolate one of the variables from the constraint equation.  This will involve getting the square root by itself so that we can square both sides of the equation:

l+2w+\sqrt{l^2+w^2}=200

\sqrt{l^2+w^2}=200-2w-l

(\sqrt{l^2+w^2})^2=(200-2w-l)^2

l^2+w^2=(200-2w-l)(200-2w-l)

l^2+w^2=40,000-400w-200l-400w+4w^2+2lw-200l+2lw+l^2

l^2+w^2=40,000-800w-400l+4w^2+4lw+l^2

This is where this problem differs from the general problem, in that the two l^2 terms cancel each other out and allow us to isolate the l

w^2=40,000-800w-400l+4w^2+4lw

We’ll move both terms that contain the l to the same side, so that we can factor out l and then divide through by 400-4w

400l-4lw=40,000-800w+3w^2

l(400-4w)=40,000-800w+3w^2

l=\frac{40,000-800w+3w^2}{(400-4w)}

So now our area equation becomes:

\frac{40,000-800w+3w^2}{(400-4w)}*w=A

OR

\frac{40,000w-800w^2+3w^3}{(400-4w)}=A

Finding the derivative for this is pretty messy and strains the display capabilities of this blog!  We’ll do our best to show the work here:

\frac{(400-4w)(40,000-1600w+9w^2)}{}

\frac{-(40,000w-800w^2+3w^3)(-4)}{16(100-w)^2}=A'

\frac{(16,000,000-640,000w+3600w^2-160,000w+6400w^2-36w^3)}{}

\frac{+160,000w-3200w^2+12w^3)}{16(100-w)^2}=A'

So, combining like terms:

-640,000w-160,000w+160,000w=-640,000w

3600w^2+6400w^2-3200w^2=+6800w^2

-36w^3+12w^3=-24w^3

\frac{(16,000,000-640,000w+6800w^2-24w^3)}{16(100-w)^2}

And then cancel a common factor of 8 in the numerator and denominator:

\frac{(2,000,000-80,000w+850w^2-3w^3)}{2(100-w)^2}

The simplified expression for the derivative is:

\frac{-3w^3+850w^2-80,000w+2,000,000}{2(100-w)^2}=A'

So from this we need to set the numerator equal to zero.  In the textbook, there is no indication that technology will be necessary to complete the problem, but I don’t see any way around using a graphical solution to finding the root of the derivative.  I suppose there is always the cubic formula, but that is truly an anachronism.  Here’s what the graph of the derivative looks like with the root indicated:

GraphicalSolution

From the graph, we can conclude that:

w\approx 38.814

l\approx 55.03

and the crosspiece \sqrt{l^2+w^2}\approx 67.34

We’ll look at the diagonal problem without the barn (which leads to a Lagrangian solution) in another post.

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