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How to Value Your Rich Elderly Aunt

Philip Cooper, 26 November 2002, 10 am – 3.30 pm

SECTION 3

Finding the values of definite integrals

When we set up mathematical models the area under a curve has a definite meaning, eg. it represents the sum of money received in a given period of time, the distance travelled by an object in a given time. This means that we are very keen to determine the value of the definite integral (the area under a curve) because it represents something that has a real world interpretation; it is not just a number.

In some models we are lucky and we know the functional form of the curve (ie. we know an equation for it) that we are trying to integrate. This helps a little, but of course it is by no means certain that we will be able to do the integration using one of our standard methods, eg. substitution, integration by parts. If it looks hard, then in today’s world we can try using software, eg. MAPLE, to help find an expression for the integral if it exists. But what can we do if even MAPLE tells us there is no analytic (exact) solution, or if unfortunately we do not have a functional form to integrate, but only a table of values that allows us to plot the function?

There is a whole science devoted to perfecting methods of finding the values of definite integrals. It forms part of the subject of Numerical Analysis.

We have discussed ‘counting the squares underneath the curve’ as a possible method, which you might have met when you first learnt about integration and your teacher wanted to convince you the results you had obtained using the standard methods of integration were indeed correct, or perhaps when you had plotted some results from a physics experiment and needed to find the area under the curve. Of course, you can draw your curve carefully and ‘count the squares’, but this is not likely to give us very accurate answers. However you may already learnt about one method of numerical integration, the Trapezium Rule.

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Remember that the area of a trapezium is $ \frac{(a + b)h}{2} $ where a and b are the lengths of the parallel sides, and h is the height of the trapezium.

You can find out more about the Trapezium Rule if you are not very familiar with it in the Additional Notes, and you can have a go at using it in the next set of tasks if you wish.

Of course with modern computing power if you can find the values of the function for very closely spaced values of the argument (the independent variable), you can get the computer to ‘count the squares’ and get a very accurate answer without the need for an approximate formula. You can do this without a computer but it would take a very long time and you would need a vast army of assistants to help you.

Actuaries need their answers in time to make decisions and the costs of doing the calculations must be less than the benefits accruing from having an accurate answer. However, in many actuarial problems it is not possible to find the value of the function we want to integrate at lots of very closely spaced points because the data to complete these calculations are not available, usually because of the limitations of published life tables. This applies to our problem. So modern day actuaries with the best computers still need to use numerical analysis to find the values of their integrals.

Fortunately the early heroes of actuarial science developed lots of numerical methods that give good answers to these problems.

Many functions of interest in actuarial science are only defined for positive values of the argument, and the functional values are also positive. This is because time is often the argument (and time is positive!) and the function of interest is usually an amount of money (which in most problems remains positive!). The curves are often skewed to the right (have a right hand tail that is fatter than the left hand tail) and have a single ‘hump’ towards the left hand end of the possible range of values of the argument. The function starts at the origin (at time 0 we have no money!) and the time axis is an asymptote to the curve for large values of t (as $ t \rightarrow \infty $ so $ f(t) \rightarrow 0 $ ).

Because the curvature of functions like this often changes very rapidly, it cannot be approximated satisfactorily by a series of straight lines, so the Trapezium Rule is unlikely to give us accurate results.

A hero of both astronomy and actuarial science, G.F. Hardy, working in the 1880’s discovered a numerical integration formula that works well for functions like this. In an early actuarial textbook by George King this formula was numbered ‘39a’ and the name has stuck to this day!

In order to use Hardy’s 39a formula, the function we want to find the area under should have the shape we have described. You then you must find a suitable value for n so that f(7n) is close to zero, where n gives values of the independent variable, t. (Choosing 7n = 7 or 3.5 makes life easy, if they are close to 0!)

So for example we may find that $ f(35) \cong 0 $ so $ n = 5 $ .

We then have to find the values of
$ f(0n=0) $ , $ f(1n=5) $ , $ f(3n=15) $ , $ f(5n=25) $ , $ f(6n=30) $ , $ f(7n=35) $ , and this last value should be very close to zero. (Notice that we don’t actually need the values of f(2n) and f(4n)).

We can then calculate the area we want using Hardy’s 39a formula which states that

\begin{displaymath}\int_{t=0}^{t=\infty} \par f(t) dt = n[0.28f(0) + 1.62f(n) + 2.20f(3n) + 1.62f(5n) \par + 0.56f(6n) + 1.62f(7n)] \end{displaymath}

Now I want you to spend some time trying out the Trapezium Rule, if it is unfamiliar to you, and Hardy’s 39a formula, on functions you have already met, and then have a go at solving our original problem. To do this you will need to use the information in the table and graph towards the end of the second part of this talk, together with which ever method of numerical integration you prefer.

(See Task 3)

 

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© 2002 Millennium Mathematics Project, University of Cambridge