## Thursday, May 9, 2013

### Taxes as a calculus problem

Around this time of year, no one wants to think any more about taxes, but it makes for a very excellent calculus problem.

## Secants vs. tangents

First, let us say that we have a function t(x), where t is the total tax owed and x is the pre-tax income. What is t '(x), the derivative of t(x)? It is the instantaneous tax rate: at a given income x, t ' is the additional tax paid on earning one additional cent. This is distinct from the effective tax rate, which is t(x) / x. To make this distinction clear, I have my students first work with the current U.S. income tax bracket system:

This is a graph of the current U.S. function for t(x). The x-axis shows pre-tax income; the y-axis shows total tax owed (note: the scales are logarithmic). I ask my students questions about the effective tax rate for someone earning \$93,100 -- a rate represented by the secant line from (93.1, 17.03) to the origin. The effective rate is 17.03/93.1 = .183 = 18.3%. I ask my students to compare that to the instantaneous rate a person earning \$93,099.99 pays on one additional cent -- a rate represented by the tangent line at x = 93.09999. From incomes in the \$44,000 to \$93,100 range, each additional cent is taxed at the same rate. This is the 25% bracket: (17.03 - 4.75)/(93.1 - 44) = .25. If a taxpayer's income falls in this range, she pays a top marginal rate of 25%, and a raise of \$.01 results in .25 cents of additional tax. However, no taxpayer whose income falls in this range pays an effective rate of 25%. The top marginal rate a taxpayer pays never equals her effective rate! In fact, it is the person who earns \$228,100 who pays an effective rate of 25%; she is in the 33% top marginal rate, but a lot of her income is taxed at lower rates.

Here is the graph for t '(x), the derivative of the t(x) above:

Note that the x-axis is still logarithmically scaled pre-tax income; the y-axis is instantaneous tax rate and is normally scaled.

## Modeling, first derivatives, second derivatives, and c

Here is a graph of the effective tax rates, t(x) / x, at the endpoints of each bracket:

The x-axis is logarithmically scaled pre-tax income in thousands of dollars (note: all subsequent formulas assume x is in thousands of dollars). The effective tax rate (total tax paid per \$1,000 pre-tax income) is shown on the y-axis. Intriguingly, the above graph looks linear (but remember, the x-axis is ln(x)). The equation T(x) / x = .076 ln(x) - .17 models the key points fairly well. Thus, the real t(x) can be modeled closely by T(x) = .076 x ln(x) - .17x. And T '(x) would be .076 ln(x) - .094. Here is this T '(x) versus the actual tax brackets:

I also ask them to think about T "(x), which is 0.076 / x. What conclusion can they draw from the fact that, when x > 0, T " > 0? This shows that T '(x) is always increasing, and T (x) is concave up. Of course, they can see from the graph above that T '(x) continues increasing; however, the real tax brackets top out at 35% marginal rate. This function T '(x) eventually equals 1 at about x = 1.8 million, meaning it would hit a 100% instantaneous rate and this high-earner could keep no additional money. At about 5 million, T(x) = x, so this person would have \$0 income after paying taxes! Obviously, this makes no sense; T '(x) should have an asymptote at no greater 1, a conclusion my students reach with little prompting. An asymptote means that T '(x) will always be positive and increasing (progressive) but never reach confiscatory levels.

Furthermore, I ask my students to think about what increasing or decreasing the c in T(x) will do. It will not change T '(x) or T "(x), but it will affect the x- and y-intercepts. Why does it make sense for the y-intercept to be negative? This might correspond to the earned income tax credit: cash payments, in the form of a tax rebate, to people with very low incomes.

## Setting up an ideal function

So, we need to set up an ideal tax function, i(x), such that its derivative will have an asymptote (unlike the linear T '(x)). Generally speaking, the easiest way to accomplish this is by setting up its derivative in the form:

where a and b can be adjusted to change the x- and y-intercepts, and c can alter how quickly the instantaneous tax rate increases (where c > 0). Furthermore, it makes sense to set a = b so that i '(x) > 0 where x > 0, otherwise one would create the unusual situation where additional income pushed the total tax down (i.e., a negative slope on the integral i (x)). The simplest example of i '(x) is

and thus its integral, i(x), would be x - ln(x+1) + c. At this point, I set my students loose to play with the function i '(x). Here is one example that matches up well against the current brackets:

This i '(x) is

I ask my students to compute its antiderivative -- good practice for them -- and compare it to T(x). They look quite similar below \$600,000. After \$600,000, T(x) becomes much bigger, eventually crossing the line y = x. The function i(x), however, looks like it starts to plateau, but appearances are deceptive: i(x) continues to increase, its slope continues to increase, and it is always concave up -- yet it never crosses y = x. Therefore, i(x) could create a progressive tax system without brackets.

Here's the GeoGebra mentioned.

#### 1 comment:

1. Hi, I've actually had similar thoughts about the desirability of a continuous progressive tax rate, and while not going through quite the same process as you have here, had also settled on a logarithmic form. I also got hold of some income distribution data so I could try out the tax and see what kind of revenue it would generate. Turned out a personal and corporate tax rate = R log (Income / Poverty) , where poverty level was about 11K, would have needed an R = 8.83 to raise 3.7 trillion in revenue that year. It would also be a tax cut for every person and corporation making less than 20 million. I have a more complete write-up at https://www.scribd.com/doc/72320709/Log-Tax if you're interested.