The news is no stranger to third derivatives, although it doesn’t sneak in very often – we’ve mentioned before the October 1996 issue of the Notices of the AMS in which Hugo Rossi wrote, “In the fall of 1972 President Nixon announced that the rate of increase of inflation was decreasing. This was the first time a sitting president used the third derivative to advance his case for reelection.”
Well, just like a blog that doesn’t post for so long that you figure it’s basically dead, and then all of a sudden out of nowhere WHOA there’s a new post (Hi everybody!), the third derivative made a new appearance recently. On the appropriately palindromic- (in the US) date of March 10, 2013, Paul Krugman wrote in the Opinion pages of the New York Times:
People still talk as if the deficit were exploding, as if the United States budget were on an unsustainable path; in fact, the deficit is falling more rapidly than it has for generations, it is already down to sustainable levels, and it is too small given the state of the economy.
Did you catch that? The line about the deficit falling more rapidly than it has been? Let’s take a closer look:
Assume that the National Debt at year t is the original function: D(t). This is positive, since we have debt.
Then the Deficit is the derivative, D'(t). It’s also positive, because the National Debt is increasing.
If the Deficit is falling, that means that the Deficit is a decreasing function, so the derivative of D'(t) – that is, D”(t) – is negative. That would mean the Debt is increasing, but concave down.
But the quote said the Deficit is falling more rapidly than is has been: the derivative of the Deficit is getting more negative, so to speak. In other words, the Deficit itself is decreasing and concave down, which means that D”'(t) is negative.
And so we have a third derivative! Welcome back old friend!
April 2, 2013 at 5:11 am |
Something doesn’t add up here. The way I see it, deficit is not the time derivative of debt.
While deficit is computed on short intervals (yearly) and debt is the accumulation of such deficits, neither of those values has a unit of USD/yr. Deficit may be numerically equal to the time derivative of debt, but not dimensionally.
Analogy : I walk 1 km each day for 10 days. My daily walk (analogous to deficit) is 1 km. My cumulative walk (analogous to debt) is 10 km. My RATE of walking aka mean speed is 1 km/day. The first and the third quantities are numerically same but not dimensionally.
Subtle but crucial difference, I’d say.
Even in the case of the inflation example, I find it hard to see why the final derivative is of third order. Inflation itself is a dimensionless percentage, and to the best of my knowledge, is not the derivative of some quantity with units of time.
Rate of increase of inflation = dN/dt (N = inflation)
Reduction in rate of increase of inflation = d2N/dt2
April 2, 2013 at 6:00 am |
I agree that with deficit (and walking), it’s more natural to do it in discrete quantities, whereas derivatives are continuous, so that makes for some awkwardness and an imperfect analogy. I don’t think of the units as actually being incorrect, but I think the problem might be average vs. instantaneous.
In your daily walk, I *do* think the units can reasonably be km/day [you walk 1 km/day], but that’s more of an average daily speed than an instantaneous one, which I think is the reason it doesn’t quite sit right. The same with deficit and: I think that deficit, while measured in dollars, is an annual measurement so the units are reasonably dollars per year (because it’s how many dollars are added in a year), but because it’s done in 1-year chunks, it doesn’t have the instantaneous aspect that a derivative ought to have.
April 2, 2013 at 6:34 am |
I see what you mean, but the continuum versus discreteness distinction fails to capture the real issue here.
Let us take the conventionally continuous example of water flowing through a pipe at a rate of 1 liter per second.
The analogy would then be :
Deficit = Net volume of water that has flowed in a year = 31536000 liters
Debt = Net volume of water that has flowed till date
Rate of debt = Flow rate of water = 1 liter per second = 31536000 liters per year
Even if fiscal transactions were perfectly continuous like flowing liquids, the distinction between a rate and a short term aggregate would remain. The limits of integration in the deficit and debt case would be different but they would still be dimensionally same. No d/dt.
Sorry if I’m being pedantic here. 😦
April 3, 2013 at 7:58 pm
I think I see what you’re saying, although I feel two ways about it: I can see both “dollars” and “dollars per year” as being appropriate ways to talk about the deficit. [A deficit of $1 Trillion each year for 3 years would be $3 Trillion, and I can see that as 3*$1 Trillion or as (3 years)*($1 Trillion/year).]
But I do agree that wherever the detail of it not matching occurs, it isn’t a perfect analogy. 🙂
April 5, 2013 at 12:06 am |
It’s interesting how high order derivatives appear with lesser frequency than lower order ones … whether in economics or physics. Same with exponents. We don’t see too many y = k. x^12 expressions, for example.
Is it something fundamental to nature or is it because such a relationship would be very hard to come up with based on experiments?
April 5, 2013 at 6:42 am |
the van der waals force has an inverse 6th power.
April 5, 2013 at 7:19 am
Hmmm ….
The folks here have some examples too – http://boards.straightdope.com/sdmb/archive/index.php/t-572473.html
I think the reason could be something like this … at least as far as interpreting experimental data is concerned.
Let’s say there is a 12th power dependence of a physical parameter y on another parameter x =>
y = k.x^12
and the region of interest is x = 1 … 10
y’s rise with increasing x would be too steep to keep track of with adequate dynamic range. We would either see saturation/out-of-range errors with slightly high y’s or if our system is geared for measuring high values, we would not have the ability to measure the lower end values with adequate resolution. Classic range vs sensitivity challenge.
In such a situation, detecting a 12th order dependence would be a very difficult measurement/analysis problem.