# College Math Teaching

## June 15, 2016

### Elementary Math in the news: elections

Filed under: calculus, elementary mathematics, news — Tags: — collegemathteaching @ 9:11 pm

Ok, mostly I am trying to avoid writing up the painful details of a proposed mathematics paper.
But I do follow elections relatively closely. In the California Democratic primary, CNN called the election for Hillary Clinton late on June 7; at the time she lead Bernie Sanders 1,940,588-1,502,043, which is a margin of 438,537 votes. Percentage wise, the lead was 55.8-43.2, or 12.6 percentage points.

But due to mail in balloting and provisional ballot counting, there were still many votes to count. As of this morning, the totals were:

2,360,266-1,887,178 for a numerical lead of 473,088 votes. Percentage wise, the lead was 55.1-44.0, or 11.1 percentage points.

So, the lead grew numerically, but shrunk percentage wise.

“Big deal”, you say? Well, from reading social media, it is not obvious (to some) how a lead can grow numerically but shrink as a percentage.

Conceptually, it is pretty easy to explain: suppose one has an election involving 1100 voters who MUST choose between candidates. Say the first 100 votes that are counted happened to come from a strongly pro-Hillary group, and the tally after 100 was 90 Hillary, 10 Bernie. Then suppose the next 1000 was closer, say 550 for Hillary and 450 for Bernie. Then the lead grew by 100 votes (80 to 180) but the percentage lead shrunk from 80 percentage points to a 16.36 percentage point lead (58.18 to 41.82 percent). And it is easy to see that if the rest of the vote was really 55 percent Hillary, her percent of the vote would asymptotically shrink to close to 55 percent as the number of votes counted went up.

So, how might one have students model it? Let $H(t), B(t)$ be increasing functions of $t$ which represent the number of votes for Hillary and Bernie as a function of time. Assume no mistakes, hence $H(t), B(t)$ can be assumed to be increasing functions. So we want a case there $D(t) = H(t)-B(t)$ is an increasing function but $P(t) = \frac{H(t)}{H(t)+ B(t)}$ decreases with time.

Without calculus: rewrite $P(t) = \frac{1}{1+\frac{B(t)}{H(t)}}$ and note that $P(t)$ decreases as $\frac{B(t)}{H(t)}$ increases; that is, as $B(t)$ outgrows $H(t)$. But $H(t)$ must continue to outgrow $B(t)$. That is, the new ballots must still include more Hillary Bernie ballots, but the ratio of Bernie ballots to Hillary ballots must be going down.

If we use some calculus, we see that $H'(t)$ must exceed $B'(t)$ but to make $P(t)$ decrease, use the quotient rule plus a tiny bit of algebra to conclude that $H'(t)B(t)-B'(t)H(t)$ must be negative, or that $\frac{B'(t)}{B(t)} > \frac{H'(t)}{H(t)}$. That is, the Bernie ballots must be growing at a higher percentage rate than the Hillary ballots are.

None of this is surprising, but it might let the students get a feel of what derivatives are and what proportional change means.