Famous limits, 2. de Moivre - Stirling

1. de Moivre

In 1730 Abraham de Moivre published  

to which in 1733 he added a Latin supplement

(Only very few copies are extant, grab one here or here.) Five years later, he included his own English translation of this Latin paper in the 1738 edition of The Doctrine of Chances, and there we can read

Sure enough, the date of 1733-12=1721 is expressly displayed in Miscellanea, where the above-mentioned problem is treated.

What de Moivre gives amounts to:

2. Stirling joins in

 

De Moivre continues his English translation by saying that his worthy and learned friend Mr. James Stirling found that  

a result of which he admits the singular elegancy. Stirling dropped this result casually and without any proof in his 1730 Methodus Differentialis (here) in Exemplum 2 after Propositio 28:

(logarithmo circumferentiae Circuli cujus Radius est Unitas = (to) the logarithm of the circumference of the circle whose radius is unity). De Moivre established it for himself, and delivered the finished result:

As many have remarked: this formula should be called  

de Moivre-Stirling formula

and not Stirling's formula.

3. Two-sided estimates

As with Wallis's formula, we'll give a totally elementary proof of a two-sided estimate, viz.

valid for n=1,2,... For n growing to infinity, the upper estimate decreases to the lower estimate, and we obtain de Moivre-Stirling's formula as a limit.

On the blackboard below we first deduce the estimates à la de Moivre with the constant left undetermined.

And here, at last, is Stirling's constant. We use Wallis's formula in the form given on the last line of this page.

Famous limits, 1. Wallis

In this book from 1656 (consult it here) John Wallis considered, among other things, an interpolation problem which led him to consider numbers of the form

 In Proposition 191, wanting to obtain their limit, he finds what we would write as

This is Wallis's product, rightly famous. More precisely we have for any n=1,2,... the two-sided estimate

of which  Wallis's product is the limiting case. (The lower estimate increases to the upper estimate for n growing to infinity.) Our modern, very elementary, proof of these estimates relies on the integrals

They are obtained by a recursion which starts with integrating by parts. Actually, Wallis's reasoning (though lacking our modern techniques) is not unlike ours; his table reproduced above displays our very distinction in even (pares) and odd (impares). Anyhow, the two-sided estimates can be proved in no time, see blackboard below. For completeness, we even included the integrals required.

Remark. The inequalities last obtained on the blackboard can also be rearranged into

which gives

These estimates are useful in assessing the behaviour of the coefficients in

We learn from them that

Lazy mathematician meets linear 2nd order ODE (2)

Part 2: you're not lucky

As explained here:

1.  Mathematicians are a lazy bunch. 
2.  Facing
   
   y''+a(x)y'+b(x)y=r(x)  (1)
   
   
   you're lucky, if you happen to know a solution φ (not identically zero) of
   y''+a(x)y'+b(x)y=0.  (2)

Today, we suppose you're still lazy but not lucky. First, let's simplify. (Lazy and simple-minded, euh?)
 

As simple as it gets

What if we managed to get rid of y' in the equation (2)? In solving third and fourth order algebraic equations, it helps a lot to eliminate the next-but-highest term. Here, getting rid of y' doesn't help that much, but is does make things a little lighter. OK, let's head for

y''+b(x)y=0.  (3)

Most textbooks provide a simple transformation achieving this reduction, but doing so they require a(x) to have a continuous derivative. As we don't want to go beyond continuity, we'll have to find a different way. On the blackboard below is a simple and explicit way to switch between (2) and (3).

One solution of (3)

To obtain a solution of (3), there is the long established method of successive approximations, to be found in, e.g., E. L. Ince, Ordinary Differential Equations, 1927. This strategy results in a sequence of functions, whose uniform convergence requires to artificially adapt the M-test, which is naturally designed for series. 

The lazy mathematician, of course, prefers a formula over a strategy. Our formula for φ will be a series, whose uniform convergence can be established most naturally by the M-test. Some notation is required. We fix x₀ in the interval under consideration, and for any continuous function f we denote by If the antiderivative of f vanishing in x₀. To save parentheses, we concatenate antiderivatives and products from right to left. Thus, for instance, I²bI²b stands for the function obtained by twice integrating b, multiplying the result by b, and integrating the product twice. 

And here it comes (ta-ta!!!):

 

φ =1-I²b+I²bI²b-I²bI²bI²b+... 

Full proof on this blackboard:

With φ obtained, you made yourself lucky. Proceed as earlier or read on.

Full solution of (3)

While we're at it, why not solve (3) completely? The same proof shows that

ψ =-x₀+x-I²bx+I²bxI²bx-I²bxI²bxI²bx+... 

(we have written x instead of the function mapping x to x) is also a solution of (3). From

φ(x₀)=1,  φ'(x₀)=0, ψ(x₀)=0, ψ'(x₀)=1

it is clear that both solutions are independent, and have a Wronskian determinant 1. Hence, (3) is completely solved, its solutions being

c₁φ+c₂ψ

with 2 arbitrary constants. Examples with x₀=0 (immediately obtained):

Nothing beats our formulas here!

Full solution of the linear 2nd order ODE

If you are lazy but unlucky, you first pass from (1) to

y''+b(x)y=r(x).  (8)

We have kept the notation b(x) and r(x), but both functions have changed in the process of eliminating y'. Instead of b(x) we should consider B(X) as defined on the first blackboard, and change r(x) to R(X) in exactly the same way. With φ and ψ defined as above, the general solution of (8) is (ta-taa!!)

 Extremely lazy proof-by-verification on the blackboard below.

Lazy mathematician meets linear 2nd order ODE (1)

Mathematical laziness

Mathematicians are lazy people, they say so themselves. On the Internet, the string "lazy mathematician" yields some 1600 immediate hits, "mathematics is for lazy people" (coined in 2000 by Peter Hilton) some 300.

On Tanya Khovanova's Math Blog, among other aspects of The Virtue of Laziness, the following is worth citing here.

Mathematicians are driven by laziness. Once ancient mathematicians first solved a quadratic equation, they didn’t want to do it again. So they invented a formula that solves all quadratic equations once and for all.  

 Exactly! The Babylonians found how to solve

ax²+bx+c=0

by writing it as

(x+b/2a)²=(b²-4ac)/(4a²)

which yields first x+b/2a, then 

In any math book, the ancient reasoning is repeated once, and after that the boxed formula is used time and again. Students all over the world know the formula by heart.

Linear 1st order ODE 

Some millennia later, we enter the realm of the ODE (ordinary differential equation), where not an unknown number x but an unknown function y(x) is to be determined. Among the simplest ones is the linear ODE of first order
 

y'+a(x)y=R(x)

where the functions a(x) and R(x) are known and y(x) is to be found. Here again, it suffices to do the reasoning once, to box the formula and apply it every time you need it. (If you're extremely lazy, you could also simply verify the formula, without any reasoning at all.) When I was a student, we had to know this formula (which contains an arbitrary constant c) by heart, just as the one above.

Linear 2nd order ODE, Part 1: you're lucky

The second order version is

 y''+a(x)y'+b(x)y=r(x)

with a(x), b(x) and r(x) known and y(x) unknown. By definition, you are lucky if you happen to know a solution (not identically zero), say φ, of the equation

y''+a(x)y'+b(x)y=0,

with zero as right-hand member. (If you are unlucky, see here.) Any calculus textbook will give you the steps to follow from here:

1. determine a function u(x) such that ψ(x)=u(x)φ(x) is not simply a constant multiple of φ(x) yet satisfies the same equation as φ(x). (Because u(x) is determined from a first order ODE, this step is called reduction of the order.)
2. determine functions v(x) and w(x) such that v(x)φ(x)+w(x)ψ(x) is a solution of the original ODE. (Because v(x) and w(x) replace arbitrary constants—which would yield nothing new—, this step is called variation of the parameters.)

Strangely enough, I could not find any textbook giving the formula by which you could avoid repeating these steps for every new 2nd order ODE. Even the first step is usually given as a strategy, while a simple formula does exist. It is to be found, for instance, in Goursat's 1917 classic Course in mathematical analysis, part Differential Equations:

But the biggest secret, well kept until now, is the general solution which emerges in the end (ta-ta!!!!):

with two arbitrary constants c₁ and c_2.
 
My colleagues used to frown upon my lazy attitude. For reasons I never understood, they preferred their students to eagerly repeat a strategy whose outcome I knew in advance. Like using the Babylonian trick instead of the formula, yes.

 Lest I forget, here is the extremely lazy proof-by-verification.