Tag Archives: surds

Binomial Expansion Variation

Several years ago, I posed on this ‘blog a problem I learned from Natalie Jackucyn:

For some integers A, B, and n, one term of the expansion of (Ax+By)^n is 27869184x^5y^3.  What are the values of A, B, and n?

In this post, I reflect for a moment on what I’ve learned from the problem and outline a solution approach before sharing a clever alternative solution one of my students this year leveraged through her CAS-enabled investigation.

WHAT I LEARNED BEFORE THIS YEAR

Mostly, I’ve loved this problem for its “reversal” of traditional binomial expansion problems that typically give A, B, and n values and ask for either complete expansions or specific terms of the polynomial.  Both of these traditional tasks are easily managed via today’s technology.  In Natalie’s variation, neither the answer nor how you would proceed are immediately obvious.

The first great part of the problem is that it doesn’t seem to give enough information.  Second, it requires solvers to understand deeply the process of polynomial expansion.  Third, unlike traditional formulations, Natalie’s version doesn’t allow students to avoid deep thinking by using technology.

In the comments to my original post, Christopher Olah and a former student, Bryan Spellman, solved the problem via factoring and an Excel document, respectively.  Given my algebraic tendencies, I hadn’t considered Bryan’s Excel “search” approach, but one could relatively easily program Excel to provide an exhaustive search.  I now think of Bryan’s approach as a coding approach to a reasonably efficient search of the sample space of possible solutions.  Most of my students’ solutions over the years essentially approach the problem the same way, but less efficiently, by using one-case-at-a-time expansions via CAS commands until they stumble upon good values for A, B, and n.  Understandably, students taking this approach typically become the most frustrated.

Christopher’s approach paralleled my own.  The x and y exponents from the expanded term show that n=5+3=8.  Expanding a generic (Ax+By)^8 then gives a bit more information.  From my TI-Nspire CAS,

binomial1

so there are 56 ways an x^5y^3 term appears in this expansion before combining like terms (explained here, if needed).  Dividing the original coefficient by 56 gives a^5b^3=497,664, the coefficient of x^5y^3.

binomial2

The values of a and b are integers, so factoring 497,664 shows these coefficients are both co-multiples of 2 and 3, but which ones?  In essence, this defines a system of equations.  The 3 has an exponent of 5, so it can easily be attributed to a, but the 11 is not a multiple of either 5 or 3, so it must be a combination.  Quick experimentation with the exponents leads to 11=5*1+3*2, so 2^1 goes to a and 2^2 goes to b.  This results in a=3*2=6 and b=2^2=4.

WHAT A STUDENT TAUGHT ME THIS YEAR

After my student, NB, arrived at a^5b^3=497,664 , she focused on roots–not factors–for her solution.  The exponents of a and b suggested using either a cubed or a fifth root.

binomial3
binomial3

The fifth root would extract only the value of a if b had only singleton factors–essentially isolating the a and b values–while the cubed root would extract a combination of a and b factors, leaving only excess a factors inside the radical.  Her investigation was simplified by the exact answers from her Nspire CAS software.

binomial4

From the fifth root output, the irrational term had exponent 1/5, not the expected 3/5, so b must have had at least one prime factor with non-singular multiplicity.  But the cubed root played out perfectly.   The exponent–2/3–matched expectation, giving a=6, and the coefficient, 24, was the product of a and b, making b=4.  Clever.

EXTENSIONS & CONCLUSION

Admittedly, NB’s solution would have been complicated if the parameter was composed of something other than singleton prime factors, but it did present a fresh, alternative approach to what was becoming a comfortable problem for me.  I’m curious about exploring other arrangements of the parameters of (Ax+By)^n to see how NB’s root-based reasoning could be extended and how it would compare to the factor solutions I used before.  I wonder which would be “easier” … whatever “easier” means.

As a ‘blog topic for another day, I’ve learned much by sharing this particular problem with several teachers over the years.  In particular, the initial “not enough information” feel of the problem statement actually indicates the presence of some variations that lead to multiple solutions.  If you think about it, NB’s root variation of the solution suggests some direct paths to such possible formulations.  As intriguing as the possibilities here are, I’ve never assigned such a variation of the problem to my students.

As I finish this post, I’m questioning why I haven’t yet taken advantage of these possibilities.  That will change. Until then, perhaps you can find some interesting or alternative approaches to the underlying systems of equations in this problem.  Can you create a variation that has multiple solutions?  Under what conditions would such a variation exist?  How many distinct solutions could a problem like this have?

Softball and the Square Root of 2

Early in my teaching career, I was introduced to the Farey Series by the amazing Henry Pollak at a month-long 1993 Woodrow Wilson Leadership Program in Mathematics on the Mathematics of Change at Princeton University. I’ve inspired many (elementary school to college-level) students over the years with the following simple-to-understand mathematical nugget.  It’s been an especially GREAT HOOK for student athletes and sports fans.

Here’s a simple illustration of how I believe Henry introduced these to us almost 20 years ago.  Let’s say a softball player’s a batting average for the season is 0.387.  How many official at-bats has she had?  A quick-and-dirty response notes 0.387=\frac{387}{1000} from which you could conclude that she has 387 hits in 1000 at-bats.  That’s certainly a valid answer, but isn’t 1000 at-bats an unrealistically large number for one season?  It also ignores the reality the 0.387 is quite likely a rounded or truncated number.  So how can you answer this question when you’re not even sure how much you have of the number you’re approximating?  It’s much simpler than you might think.

Start by finding the simplest two fractions which bound your target number.  In this case, \frac{1}{3}\approx 0.3333<0.387<0.5=\frac{1}{2}.  Create a new fraction closer to the target by adding the two fractional bounds the way many students first learning fraction addition want to do:  add the numerators and add the denominators–\frac{1+1}{3+2}=\frac{2}{5}=0.4.  Notice two things about this fraction:

  1. It’s value is between (but not at the midpoint!) the original two bounding fractions, and
  2. It remains higher than the target 0.387.

Replacing my original upper bound with my new fraction, I now have \frac{1}{3}\approx 0.3333<0.387<0.4=\frac{2}{5}.  Repeat the procedure to get \frac{1+2}{3+5}=\frac{3}{8}=0.375, a new lower bound.  Doing this a few more times leads to

\frac{1}{3}\approx 0.3333<0.387<0.5=\frac{1}{2}
\frac{1}{3}\approx 0.3333<0.387<0.4=\frac{2}{5}=\frac{1+1}{3+2}
\frac{1+2}{3+5}=\frac{3}{8}=0.375<0.387<0.4=\frac{2}{5}
\frac{3+2}{8+5}=\frac{5}{13}\approx 0.3846<0.387<0.4=\frac{2}{5}
\frac{5}{13}\approx 0.3846<0.387<0.3888\approx\frac{7}{18}=\frac{5+2}{13+5}

Notice that each time I added the bounding fractions “the wrong way” to get a new fraction between the original bounds, and this new fraction became either my new upper bound or my new lower bound, and then I repeated the algorithm getting ever closer to my target value of 0.387.  Doing this one more time gives \frac{5+7}{13+18}=\frac{12}{31}\approx 0.387096 !  Forget the earlier guess of 1000 at-bats, I now know that a batting average of 0.387 can be accomplished in just 12 hits in 31 at-bats!

There are two absolutely amazing (to me) facts about this particular fraction.

  • \frac{12}{31} is the smallest-denominator fraction whose decimal approximation to three places rounds to 0.387, meaning 31 is the absolute minimum number of at-bats for the original problem.
  • If you continue the procedure, you will get a long sequence of fractions which all round to 0.387, eventually culminating with the fraction \frac{387}{1000}, bringing you all the way back to the original guess.

This procedure always works.  If you’re looking for an interesting math factoid to hook curious individuals from elementary school to adult, this is enough.

From a mathematical perspective, such fractions form a portion of the Farey Series, are called mediants (NOT medians), and are sometimes called freshman sums in reference to the way young ones often try to add fractions the wrong way.  I was reminded again of Farey Series when I read @benvitale‘s ‘blog post on Approximations of Square Root of 2.  He presents these in a different way that I’m certain relies at its core upon Farey series although I haven’t yet taken the time to fully explore it.

From a theoretical perspective, there’s lots here for older students who really want to know how a mediant works.   Given two fractions where \frac{a}{b}< \frac{c}{d}, it’s relatively simple for a student comfortable with algebra and inequalities to show that \frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}.  That explains how the procedure above quickly narrows the bounds for the target decimal number.  The hard part, as I recall, is proving that this procedure leads to the smallest-denominator fraction that rounds to the target decimal number.

In my brief research for this post, I also discovered some really interesting paradoxes at the end of the Cut-The-Knot article on mediant fractions.  These are absolutely worth reading if you or someone you know adores mathematical surprises.  Cut-The-Knot is a great site for lots of cool math facts and ideas.