# Tag Archives: maclaurin

## 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

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.

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!

## Statistics and Series

I was inspired by the article “Errors in Mathematics Aren’t Always Bad” (Sheldon Gordon, Mathematics Teacher, August 2011, Volume 105, Issue 1) to think about an innovative way to introduce series to my precalculus class without using any of the traditional calculus that’s typically required to derive them.  It’s not a proof, but it’s certainly compelling and introduces my students to an idea that many find challenging in a much less demanding environment.

Following is a paraphrase of an activity I took my students through in January.  They started by computing and graphing a few points on $y=e^x$ near $x=0$.

The global shape is exponential, but this image convinced them to try a linear fit.

Simplifying a bit, this linear regression suggests that $e^x\approx x+1$ for values of x near $x=0$.  Despite the “strength” of the correlation coefficient, we teach our students always to look at the residuals from any attempted fit.  If you have ever relied solely on correlation coefficients to determine “the best fit” for a set of data, the “strength” of $r \approx0.998402$ and the following residual plot should convince you to be more careful.

The values are very small, but these residuals ($res1=e^x-(x+1)$)  look pretty close to quadratic even though the correlation coefficient was nearly 1.  Fitting a quadratic to $(xval,res1)$ gives another great fit.

The linear and constant coefficients are nearly zero making $res1\approx\frac{1}{2}x^2$.  Therefore, a quadratic approximation to the original exponential is $e^x \approx\frac{1}{2}x^2+x+1$.  But even with another great correlation coefficient, hopefully the last step has convinced you to investigate the new residuals, $res2=e^x-(\frac{1}{2}x^2+x+1)$.

And that looks cubic.  Fitting a cubic to $(xval,res2)$ gives yet another great fit.

This time, the quadratic, linear, and constant coefficients are all nearly zero making $res2\approx.167x^3$.  The simplest fraction close to this coefficient is $\frac{1}{6}$ making cubic approximation $e^x \approx\frac{1}{6}x^3+\frac{1}{2}x^2+x+1$.  One more time, check the new residuals, $res3=e^x-(\frac{1}{6}x^3+\frac{1}{2}x^2+x+1)$.

Given this progression and the “flatter” vertex, my students were ready to explore a quartic fit to the res3 data.

As before, only the highest degree term seems non-zero, giving $res3\approx0.04175x^4$.  Some of my students called this coefficient $\frac{1}{25}$ and others went for $\frac{1}{24}$.  At this point, either approximation was acceptable, leading to $e^x \approx\frac{1}{24}x^4+\frac{1}{6}x^3+\frac{1}{2}x^2+x+1$.

My students clearly got the idea that this approach could be continued as far as desired, but since our TI-Nspire had used its highest polynomial regression (quartic) and the decimals were getting harder to approximate, we had enough.  As a final check, they computed a quartic regression on the original data, showing that the progression above could have been simplified to a single step.

If you try this with your classes, I recommend NOT starting with the quartic regression.  Students historically have difficulty understanding what series are and from where they come.  My anecdotal experiences from using this approach for the first time this year suggest that, as a group, my students are far more comfortable with series than ever before.

Ultimately, this activity established for my students the idea that polynomials can be great approximations for other functions at the same time we crudely developed the Maclaurin Series for $e^x\approx\frac{1}{4!}x^4+\frac{1}{3!}x^3+\frac{1}{2!}x^2+x+1$, a topic I’m revisiting soon as we explore derivatives.  We also learned that even very strong correlation coefficients can hide some pretty math.