# 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.