Sometimes the wrong answer is more interesting than the right answer. If the wrong approach almost works, it may be fun to understand why.
Suppose you want to figure a price so that the final price including tax has a given value. For example, say you want a T-shirt to sell for $10 after adding 8% sales tax. A common mistake would be to subtract 8% from $10 and sell the shirt for $9.20 plus tax. But this makes the price with tax $9.94. Observe two things: (1) the result was wrong, but (2) it wasn’t far off.
The correct solution would be to divide $10 by 1.08. Then when we add 8%, i.e. multiply by 1.08, we get $10. That says we should price the shirt at $9.26 before tax. That explains (1). But what about (2), that is, why was the result nearly correct? Subtracting a percentage p amounts to multiplying by (1-p). But we should have multiplied by 1/(1+p). We’ve stumbled on the fact that for small p, 1/(1+p) approximately equals 1 – p.
With a little experimentation we might discover a couple more things. First, we notice that multiplying by 1-p always gives a smaller result than multiplying by 1/(1+p), i.e. the common mistake always gets the price too low. Also, with a little experimentation we might notice that the difference between 1/(1+p) and (1-p) gets smaller as p gets smaller. Not only that, making p a little smaller makes the difference between 1/(1+p) and (1-p) a lot smaller. In summary, we noticed that 1/(1+p) – (1-p) is
- positive, and
- gets small faster than p gets small.
What accounts for these observations? The Taylor series for 1/(1+p) says that for |p| < 1,
1/(1+p) = 1 – p + p2 + p3 – p4 + …
The error in truncating a Taylor series is roughly equal to the first term that was left out, so the error in approximating 1/(1+p) by (1-p) is roughly p2. That explains why the difference between 1/(1+p) and (1-p) is small, positive, and decreases with p. For example, we should expect that cutting p in half reduces the difference by a factor of four.
The Taylor series argument only necessarily holds for p sufficiently small. If we go back and calculate 1/(1+p) – (1-p) directly we find it’s p2/(1+p). This confirms that 1/(1+p) is greater than 1-p for all p larger than -1.