Series Comfort

I ‘blogged a couple days ago about a way to use statistical regressions to develop Maclaurin Series in a way that precalculus or algebra II students could understand. In short, that approach worked because the graph of y=e^x is differentiable–or locally linear as we describe it in my class.  Following is a student solution to a limit problem that proves, I believe, that students can become quite comfortable with series approximations to functions, even while they are still very early in their calculus understanding.

This year, I’m teaching a course we call Honors Calculus.  It is my school’s prerequisite for AP Calculus BC and covers an accelerated precalculus before teaching differential calculus to a group of mostly juniors and some sophomores.  If we taught on trimesters, the precalculus portion would cover the first two trimesters.  My students explored the regression activity above 2-3 months ago and from that discovered the basic three Maclaurin series.

\displaystyle e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\ldots
\displaystyle cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\ldots
\displaystyle sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\ldots

Late last week, we used local linearity to establish L’Hopital’s Rule which I expected my students to invoke when I asked them on a quiz two days ago to evaluate \displaystyle\lim_{x\to 0}\frac{x\cdot sin(x)}{1-cos(x)}.  One student surprised me with his solution.

He didn’t recall L’Hopital’s, but he did remember his series exploration from January.

Had he not made a subtraction error in the denominator, he would have evaluated the limit this way.

\displaystyle\lim_{x\to 0}\frac{x\cdot sin(x)}{1-cos(x)}=\lim_{x\to 0}\frac{x*(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\ldots)}{1-(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\ldots)}

\displaystyle=\lim_{x\to 0}\frac{x^2-\frac{x^4}{3!}+\frac{x^6}{5!}-\ldots}{\frac{x^2}{2!}-\frac{x^4}{4!}+\ldots}

And dividing the numerator and denominator by x^2 leads to the final step.

\displaystyle=\lim_{x\to 0}\frac{1-\frac{x^2}{3!}+\frac{x^4}{5!}-\ldots}{\frac{1}{2!}-\frac{x^2}{4!}+\ldots}

=\displaystyle\frac{1-0+0-\ldots}{\frac{1}{2}-0+0-\ldots}=2

Despite his sign error, he came very close to a great answer without using L’hopital’s Rule at all, and showed an understanding of series utility long before most calculus students ever do.

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