Tag Archives: fractions

Base-x Numbers and Infinite Series

In my previous post, I explored what happened when you converted a polynomial from its variable form into a base-x numerical form.  That is, what are the computational implications when polynomial 3x^3-11x^2+2 is represented by the base-x number 3(-11)02_x, where the parentheses are used to hold the base-x digit, -11, for the second power of x?  

So far, I’ve explored only the Natural number equivalents of base-x numbers.  In this post, I explore what happens when you allow division to extend base-x numbers into their Rational number counterparts.

Level 5–Infinite Series: 

Numbers can have decimals, so what’s the equivalence for base-x numbers?  For starters, I considered trying to get a “decimal” form of \displaystyle \frac{1}{x+2}.  It was “obvious” to me that 12_x won’t divide into 1_x.  There are too few “places”, so some form of decimals are required.  Employing division as described in my previous post somewhat like you would to determine the rational number decimals of \frac{1}{12} gives

Base6

Remember, the places are powers of x, so the decimal portion of \displaystyle \frac{1}{x+2} is 0.1(-2)4(-8)..._x, and it is equivalent to

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

This can be seen as a geometric series with first term \displaystyle \frac{1}{x} and ratio \displaystyle r=\frac{-2}{x}.  It’s infinite sum is therefore \displaystyle \frac{\frac{1}{x}}{1-\frac{-2}{x}} which is equivalent to \displaystyle \frac{1}{x+2}, confirming the division computation.  Of course, as a geometric series, this is true only so long as \displaystyle |r|=\left | \frac{-2}{x} \right |<1, or 2<|x|.

I thought this was pretty cool, and it led to lots of other cool series.  For example, if x=8,you get \frac{1}{10}=\frac{1}{8}-\frac{2}{64}+\frac{4}{512}-....

Likewise, x=3 gives \frac{1}{5}=\frac{1}{3}-\frac{2}{9}+\frac{4}{27}-\frac{8}{81}+....

I found it quite interesting to have a “polynomial” defined with a rational expression.

Boundary Convergence:

As shown above, \displaystyle \frac{1}{x+2}=\frac{1}{x}-\frac{2}{x^2}+\frac{4}{x^3}-\frac{8}{x^4}+... only for |x|>2.  

At x=2, the series is obviously divergent, \displaystyle \frac{1}{4} \ne \frac{1}{2}-\frac{2}{4}+\frac{4}{8}-\frac{8}{16}+....

For x=-2, I got \displaystyle \frac{1}{0} = \frac{1}{-2}-\frac{2}{4}+\frac{4}{-8}-\frac{8}{16}+...=-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}-... which is properly equivalent to -\infty as x \rightarrow -2 as defined by the convergence domain and the graphical behavior of \displaystyle y=\frac{1}{x+2} just to the left of x=-2.  Nice.

Base7

I did find it curious, though, that \displaystyle \frac{1}{x}-\frac{2}{x^2}+\frac{4}{x^3}-\frac{8}{x^4}+... is a solid approximation for \displaystyle \frac{1}{x+2} to the left of its vertical asymptote, but not for its rotationally symmetric right side.  I also thought it philosophically strange (even though I understand mathematically why it must be) that this series could approximate function behavior near a vertical asymptote, but not near the graph’s stable and flat portion near x=0.  What a curious, asymmetrical approximator.  

Maclaurin Series:

Some quick calculus gives the Maclaurin series for \displaystyle \frac{1}{x+2} :  \displaystyle \frac{1}{2}-\frac{x}{4}+\frac{x^2}{8}-\frac{x^3}{16}+..., a geometric series with first term \frac{1}{2} and ratio \frac{-x}{2}.  Interestingly, the ratio emerging from the Maclaurin series is the reciprocal of the ratio from the “rational polynomial” resulting from the base-x division above.  

As a geometric series, the interval of convergence is  \displaystyle |r|=\left | \frac{-x}{2} \right |<1, or |x|<2.  Excluding endpoint results, the Maclaurin interval is the complete Real number complement to the base-x series.  For the endpoints, x=-2 produces the right-side vertical asymptote divergence to + \infty that x=-2 did for the left side of the vertical asymptote in the base-x series.  Again, x=2 is divergent.

It’s lovely how these two series so completely complement each other to create clean approximations of \displaystyle \frac{1}{x+2} for all x \ne 2.

Other base-x “rational numbers”

Because any polynomial divided by another is absolutely equivalent to a base-x rational number and thereby a base-x decimal number, it will always be possible to create a “rational polynomial” using powers of \displaystyle \frac{1}{x} for non-zero denominators.  But, the decimal patterns of rational base-x numbers don’t apply in the same way as for Natural number bases.  Where \displaystyle \frac{1}{12} is guaranteed to have a repeating decimal pattern, the decimal form of \displaystyle \frac{1}{x+2}=\frac{1_x}{12_x}=0.1(-2)4(-8)..._x clearly will not repeat.  I’ve not explored the full potential of this, but it seems like another interesting field.  

CONCLUSIONS and QUESTIONS

Once number bases are understood, I’d argue that using base-x multiplication might be, and base-x division definitely is, a cleaner way to compute products and quotients, respectively, for polynomials.  

The base-x division algorithm clearly is accessible to Algebra II students, and even opens the doors to studying series approximations to functions long before calculus.

Is there a convenient way to use base-x numbers to represent horizontal translations as cleanly as polynomials?  How difficult would it be to work with a base-(x-h) number for a polynomial translated h units horizontally?

As a calculus extension, what would happen if you tried employing division of non-polynomials by replacing them with their Taylor series equivalents?  I’ve played a little with proving some trig identities using base-x polynomials from the Maclaurin series for sine and cosine.

What would happen if you tried to compute repeated fractions in base-x?  

It’s an open question from my perspective when decimal patterns might terminate or repeat when evaluating base-x rational numbers.  

I’d love to see someone out there give some of these questions a run!

Common numerators

As long as I’m leveraging Five Triangles posts, here is another recent one worth discussing.

Too often, I think students believe that the only way to compare fractions is to find common denominators.  In this problem, three of the four given denominators are big enough primes that the common denominator approach would result in some painful enough by-hand computations.

But the pattern in the numerators screams for attention.  Why not find some common numerators and compare the fractions that way?  That approach cracks the problem pretty efficiently.

As a bonus, the common numerator approach also shows that the four given fractions are surprisingly close to each other in size.

Keep thinking …

Developing fractional understanding

Here’s another installment of my infrequent commentaries on my eldest daughter’s math development, this time on an unexpected result when we were talking about equivalent fractions.

I’ve been a firm believer in keeping challenging ideas, games, etc. around my children at all times, some of them intentionally beyond what they’re developmentally ready to handle.  Sometimes I’ll ask leading questions to see if there is any interest; sometimes they pick up an idea or toy again and create their own play rules or ask me to explain how it works.  My ground rule is that they should PLAY.  If my explanations ever bore them, they are welcome to drop it at any time and move on to something more interesting.  Not only does this keep with my mantra that learning should be fun (even if it involves work), but I believe it helps them see that mathematics (and anything else they learn) is about enjoyment and pushing yourself to discover more than what exists within the current boundaries of your understanding.

Several weeks ago

With that philosophy in mind, my 2nd grade daughter had been helping me make some fresh bread one afternoon when we needed a cup of one ingredient, but our cup measurer was dirty.  Whether from school or one of our earlier conversations, she responded something like, ‘No problem.  Just use 2 half-cups.‘  Maybe she actually said  ‘3 third-cups,’ but she clearly had the beginnings of fractions down, so I jumped.

So if I needed a half cup of something and my half-cup measurer was dirty, what else could I use?

It took some conversations, but eventually she drew a circle with a line through the center and shaded one side.  When she drew another line through the center roughly bisecting the original sectors, she declared, “Look, Dad.  Cutting each part in half doesn’t change what you have, it just cuts it into more pieces.  So, \frac{1}{2} must be the same as \frac{2}{4}.”

That was a pretty cool moment of discovery for her, but then she upped the ante.  After some additional thought, she noticed that both parts of the fraction had been doubled, so she applied her rule again and asked if \frac{1}{2} would also be the same as \frac{4}{8}.  Another drawing confirmed her discovery which led to gleeful proclamations that \frac{1}{2}=\frac{2}{4}=\frac{4}{8}=\frac{8}{16}=\frac{16}{32}. She knew she could go on, but what was the point?

I asked if doubling was the only way she could make equivalent fractions, which led to \frac{1}{2}=\frac{3}{6}=\frac{9}{18}. We didn’t go any further that day, but we had already “cooked up” far more than I had anticipated.  Off the top of my head, I don’t recall how elementary school curricula deal with the scope and sequence of teaching equivalent fractions, but it will be difficult for anyone to convince me that my daughter could have had a better experience or more fun learning.

Last weekend

I was putting my daughter to bed and for some reason she asked how much of a year 9 months was.  Being the teacher, I responded to her question with another question:  “Well, how many months are in a year?

So 9 months is \frac{9}{12} of a year?” She asked.

Good job.  Can you think of any other smaller fractions that might be the same as \frac{9}{12}?” It was an innocent question, I thought, trying to get her to take our fraction doubling-trebling idea from earlier in reverse. If she didn’t get it, no big deal, but it was certainly worth asking.  That’s when I got surprised.

Almost immediately, she said, “Four and a half sixths.  Is that right?

Some purists out there might complain that \frac{4.5}{6} isn’t “proper,” but I’ve seen far too many situations where rigid insistence on proper form served instead to stifle creativity far more than to enhance understanding or to encourage deeper exploration or creativity.  I praised the heck out of her solution, letting her know that she had just made a fraction “smaller” for the first time (that I knew of).”  I didn’t mention that her proposed numerator wasn’t whole.  It didn’t matter.

How did you do that?

Easy,” she said, as if her answer would have been obvious to anyone.  “If you can double the parts of a fraction, why can’t you halve them, too?

Why not, indeed?  No matter what they end up looking like.

I asked if there were any other smaller, equivalent fractions.  That took lots more thought and time than I expected, certainly more than her nearly instantaneous \frac{4.5}{6} had required.  Eventually, she asked if I could hold out 2 fingers beside her 10 so that she could look at 12 fingers together.  A little more thought led to her grouping the 12 fingers into 4 equal groups and a claim that \frac{9}{12}=\frac{3}{4}.  I still don’t completely understand her long explanation, but it wasn’t as clear (to me, anyway) that she was saying that 3 of her sub-groups contained the original 9 fingers (or months), and so 9 of an original 12 individual months was the same as 3 of 4 equal sub-groups of months.

When she woke the next morning, she asked if we could play any more fraction games

Conclusion

As students of all ages are learning, we need to reserve space for them to think and be creative.  They should be allowed to give correct mathematical solutions, even if those answers aren’t the arithmetic solutions our “trained” minds expect.  For me, I expected my daughter to say \frac{9}{12}=\frac{3}{4}, if she answered at all.  I was far more delighted to hear \frac{9}{12}=\frac{4.5}{6} than she will ever know, and I’m convinced that she’s becoming more confident and capable because she was allowed to do so.

Nested surds

A few months ago, I noted the following problem on a separate ‘blog for a textbook a colleague and I created to fill a need at our school.

Assuming the pattern continues forever, what is the value of \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...}}}} ?

As you know, I encourage my students to be creative and solve problems in ways that make the most sense to them.  Following are three different approaches my students used when presented with this problem.  All three are essentially identical invocations of the problem’s self-similarity, but I liked the differences in how they approached the problem.

STOP!!!
SOLUTION ALERT!!
DO NOT READ ANY FURTHER IF YOU WANT TO SOLVE THESE PROBLEMS ON YOUR OWN.

Approach 1:
Notice that other than the outermost square root and addition, the entire problem is repeated inside the problem.  Therefore, if X=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...}}}}, then

X=\sqrt{1+X}
X^2-X-1=0
X=\frac{1\pm\sqrt{5}}{2}

The value of X is obviously positive, so X=\frac{1+\sqrt{5}}{2}=\phi, the golden ratio!  How’s that for an unexpected surprise?!

Approach 2:
Two students thought in terms of sequences.  Each assumed a first term of a_1=\sqrt{1} and iterated using a_n=\sqrt{a_{n-1}}.

Recognizing the decimal equivalent of \phi, the first student concluded that the iterated surd appeared to be the golden ratio, but said in a class discussion the following day that while he had compelling evidence for an answer, he knew he hadn’t proven his solution.

Approach 3:
My second student who thought of a sequence approach reached the a_n=\sqrt{a_{n-1}} statement and realized that if she was able to iterate forever and this problem actually had an answer, then there would be a point where L=a_{n-1}=a_n where L is the limit of the iteration.  At this point, she wrote L=\sqrt{1+L} and proceeded as in Approach 1 to get L=\phi.

Some readers might argue that Approaches 1 and 3 really aren’t distinct, but I maintain that the students who used Approach 1 (the clear majority) were thinking purely algebraically and took advantage of the problem’s self-similarity, while the girl who used Approach 3 was thinking numerically and found her answer using limits.

Minor Extension:
For fun, I also assigned 1+\displaystyle\frac{1}{1+\frac{1}{1+\frac{1}{1+...}}}.  At first, this problem looks very different from the one discussed above, but its closed form delighted some (and disturbed others!)  with its connection.  I hope you enjoy!