8.1: Warming Up

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Let’s start with an everyday example, so that we do not have to handle mathematical or physical complexity along with learning the new tool.

8.1.1 Everyday example of easy cases

One August, upon becoming eligible for one of the tax benefits at work, I chose to put $500 in an account for the calendar year’s health costs (a feature of America’s bureaucratic health system). The payroll office advised me that they would deduct $125 each month for the rest of the year—namely, for August through December. As August is the eighth month and December the twelfth month, the amount looked correct:

\[\frac{$500}{\textrm{12th month -8th month}} = \frac{$500}{\textrm{4 months}} = \frac{$125}{\textrm{month}}.\]

However, in January, the payroll office advised me that they should have deducted only $100 per month.

Which amount was correct?

The simplest way to decide is the method of easy cases. Don’t directly solve the hard problem of computing the correct deduction in general. Instead, imagine a simpler world in which I started deducting for the year in December. In this easy case, with only one month providing the year’s deduction, the answer requires no calculation: Deduct $500 per month.

Then use the easy case and this result to check the proposed recipe, which predicted $125 per month when deductions started in August. In the easy case, when deductions start in December, the starting and ending months are both the twelfth month, so the recipe predicts an infinite deduction:

\[\frac{$500}{\textrm{12th month -12th month}} = \frac{$500}{\textrm{0 months}} = \frac{$ \: infinity}{\textrm{month}}.\]

They do not pay me enough to survive that recipe. It needs an adjustment

\[\frac{$500}{\textrm{12th month -12th month + 1 month}} = \frac{$500}{\textrm{1 month}} = \frac{$500}{\textrm{month}}.\]

Applying this modified recipe to an August instead of a December start, the denominator is 5 months rather than 4 months, and the deduction is $100 per month. The revised advice from the payroll office was correct.

This analysis contains several features that we can abstract away and use to simplify difficult problems. First, the easy cases are specified by values of a dimensionless quantity. Here, it is the difference $m_{2} - m_{1}$ between the first and last month numbers. Second, for particular values of the dimensionless quantity—for the easy cases—the problem has an obvious answer. Here, the easy case is $m_{2} -m_{1} = 0$. Third, understanding the easy cases transfers to the hard cases. Here, our understanding of the case $m_{2} - m_{1} =0$ shows that the denominator should be $m_{2} - m_{1} + 1$ rather than $m_{2}-m_{1}$.

8.1.2 Easy cases of the shared-birthday probability

The same easy-cases reasoning helps us check more abstruse formulas. As an example, cast your mind back to the birthday paradox of Section 4.4.

There, we used proportional reasoning to explain why we need only 23 people, rather than 183 people, in a room before two people are likely to share a birthday. Checking the prediction required the exact probability of a shared birthday:

\[p_{shared} = 1 -(1-\frac{1}{365}) (1 - \frac{2}{365}) ... (1- \frac{(n-1)}{365}).\]

The last part of Problem 4.23 asked why, with n people in the room, the last factor in this probability contains (n − 1)/365 rather than n/365.

The simplest answer comes from the easy case n = 1. With one person in the room, the probability of sharing a birthday is zero. In this easy case, the prediction of the candidate formula with n in the numerator is easy to test. Its last factor would be 1 − 1/365:

\[p_{\textrm{candidate with } n} = 1 - (1 - \frac{1}{365}).\]

This probability is 1/365, which is incorrect. Thus, with n people in the room, the last factor in the probability of a shared birthday cannot contain n/365. In contrast, the candidate formula with (n − 1)/365 in the last factor correctly predicts zero probability when n = 1:

\[p_{\textrm{candidate with } n-1} = 1 - (1 - \frac{0}{365}) = 0.\]

This candidate must be correct.

Exercise $\PageIndex{1}$: Sine or cosine by easy cases

For a mass sliding down an inclined plane, is its acceleration along the plane $g sin \theta$ or $g cos \theta$, where $\theta$ is the angle the plane makes with horizontal? Use an easy case to decide. Assume (the easy case of!) zero friction.

Exercise $\PageIndex{2}$: Friction by easy cases

For the mass of Problem 8.1, relax the assumption of zero friction. If the coefficient of sliding friction is $\mu$, choose the block’s correct acceleration along the plane: (a) $g(sin \theta \mu cos \theta) $, (b) $g (1+ \mu) sin \theta$, (c) $g(sin \theta - \mu cos \theta$), or (d) $g(1-\mu) sin \theta$.